This course is intended as an introduction to the analysis of elliptic partial differential equations. The objective is to provide a large overview of the different aspects of elliptic partial differential equations and their modern treatment. Besides variational and Schauder methods we study the unique continuation property and the stability for Cauchy problems. The derivation of the unique continuation property and the stability for Cauchy problems relies on a Carleman inequality. This inequality is efficient to establish three-ball type inequalities which are the main tool in the continuation argument.

We know that historically a central role in the analysis of partial differential equations is played by their fundamental solutions. We added an appendix dealing with the construction of a fundamental solution by the so-called Levi parametrix method.

We tried as much as possible to render this course self-contained. Moreover each chapter contains many exercices and problems. We have provided detailed solutions of these exercises and problems.

The most parts of this course consist in an enhanced version of courses given by the author in both undergraduate and graduate levels during several years.

Remarks and comments that can help to improve this course are welcome.

Chapter 1

Sobolev spaces

We define in this chapter the W k,p -spaces when k is non negative integer and 1 ≤ p < ∞ and we collect their main properties. We adopt the same approach as in the books by H. Brézis [2] and M. Willem [13]. That is we get around the theory of distributions. We precisely use the notion of weak derivative which in fact coincide with the derivative in the distributional sense. We refer the interested reader to the books by L. Hörmander [3] and C. Zuily [14] for a self-contained presentation of the theory of distributions.

The most results and proofs of this chapter are borrowed from [2, 13].

L p spaces

In this chapter Ω denotes an open subset of R n . As usual, we identify a measurable function f on Ω by its equivalence class consisting of functions that are equal almost everywhere to f . For simplicity convenience the essential supremum and the essential minimum are simply denoted respectively by the supremum and the minimum.

Let dx be the Lebesgue measure on R n . Define

L 1 (Ω ) = f : Ω → C measurable and Ω | f (x)|dx < ∞ .
Theorem 1.1. (Lebesgue's dominated convergence theorem) Let ( f m ) be a sequence in L 1 (Ω ) satisfying (a) f m (x) → f (x) a.e. x ∈ Ω , (b) there exists g ∈ L 1 (Ω ) so that for any m | f m (x)| ≤ g(x) a.e. x ∈ Ω . Then f ∈ L 1 (Ω ) and

Ω | f m -f |dx → 0. Theorem 1.2. (Fubini's theorem) Let f ∈ L 1 (Ω 1 × Ω 2 ), where Ω i is an open subset of R n i , i = 1, 2. Then f (x, •) ∈ L 1 (Ω 2 ) a.e. x ∈ Ω 1 , Ω 2 f (•, y)dy ∈ L 1 (Ω 1 )
and f (•, y) ∈ L 1 (Ω 1 ) a.e. y ∈ Ω 2 ,

Ω 1 f (x, •)dx ∈ L 1 (Ω 2 ).
Moreover

Ω 1 dx Ω 2 f (x, y)dy = Ω 2
dy

Ω 1 f (x, y)dx = Ω 1 ×Ω 2 f (x, y)dxdy.
Recall that L p (Ω ), p ≥ 1, denotes the Banach space of measurable functions f on Ω satisfying | f | p ∈ L 1 (Ω ). We endow L p (Ω ) with its natural norm

f p;Ω = f L p (Ω ) = Ω | f | p dx 1/p .
Let L ∞ (Ω ) denotes the Banach space of bounded measurable functions f on Ω . This space equipped with the norm

f ∞;Ω = f L ∞ (Ω ) = sup | f |.
When there is no confusion f L p (Ω ) is simply denoted by f p . The conjugate exponent of p, 1 < p < ∞, is given by the formula p = p p -1 .

Note that we have the identity 1 p + 1 p = 1.

If p = 1 (resp. p = ∞) we set p = ∞ (resp. p = 1).

The following inequalities will be very useful in the rest of this text.

Proposition 1.1. (Hölder's inequality) Let f ∈ L p (Ω ) and g ∈ L p (Ω ), where 1 ≤ p ≤ ∞. Then f g ∈ L 1 (Ω ) and

Ω | f g|dx ≤ f p g p .
Proposition 1.2. (Generalized Hölder's inequality) Fix k ≥ 2 an integer. Let 1 < p j < ∞, 1 ≤ j ≤ k so that 1 p 1 + . . .

+ 1 p k = 1.
If u j ∈ L p j (Ω ), 1 ≤ j ≤ k, then ∏ k j=1 u j ∈ L 1 (Ω ) and

Ω k ∏ j=1 |u j |dx ≤ k ∏ j=1 u j p j .
Proposition 1.3. (Interpolation inequality) Let 1 ≤ p < q < r < ∞ and 0 < λ < 1 given by

1 q = 1 -λ p + λ r .
If u ∈ L p (Ω ) ∩ L r (Ω ) then u ∈ L q (Ω ) and u q ≤ u 1-λ p u λ r .

As usual if E ⊂ R n is a measurable set, its measure is denoted by |E|.

Proposition 1.4. Let 1 ≤ p < q < ∞ and assume that |Ω | < ∞. If u ∈ L q (Ω ) then u ∈ L p (Ω ) and

u p ≤ |Ω | 1 p -1 q u q .
We collect in the following theorem the main properties of the L p spaces.

Theorem 1.3. (i) L p (Ω ) is reflexive for any 1 < p < ∞.

(ii) L p (Ω ) is separable for any 1 ≤ p < ∞.

(iii) Let 1 ≤ p < ∞ and u ∈ [L p (Ω )] , the dual of L p (Ω ). Then there exists a unique g u ∈ L p (Ω ) so u, f = u( f ) = Ω f g u dx for any f ∈ L p (Ω ).

Furthermore u [L p (Ω )] = g u L p (Ω ) .

Theorem 1.3 (iii) says that the mapping u → g u is an isometric isomorphism between [L p (Ω )] and L p (Ω ). Therefore, we always identify in the sequel the dual of L p (Ω ) by L p (Ω ).

The following theorem gives sufficient conditions for a subset of L p (Ω ) to be relatively compact.

Theorem 1.4. (M. Riesz-Fréchet-Kolmogorov) Let F ⊂ L p (Ω ), 1 ≤ p < ∞, admitting the following properties. (i) sup f ∈F f L p (Ω ) < ∞. (ii) For any ω Ω , we have

lim h→0 τ h f -f L p (ω) = 0,
where τ h f (x) = f (xh), x ∈ ω, provided that |h| sufficiently small. (iii) For any ε > 0, there exists ω Ω so that, for every f ∈ F , f L p (Ω \ω) < ε. Then F is relatively compact in L p (Ω ).

We end the present section by a density result, where C c (Ω ) denotes the space of continuous functions with compact supported in Ω .

Theorem 1.5. C c (Ω ) is dense in L p (Ω ) for any 1 ≤ p < ∞.

Approximation by smooth functions

Recall that L p loc (Ω ), 1 ≤ p ≤ ∞, is the space of measurable functions f : Ω → R satisfying f χ K ∈ L p (Ω ), for every K Ω . Here χ K is the characteristic function of K. We usually say that L p loc (Ω ) is the space of locally p-integrable functions on Ω Define the space of test functions by

D(Ω ) = {ϕ ∈ C ∞ (Ω ); supp(ϕ) Ω }.
We use the notation ∂ i = ∂ ∂ x i and, if α = (α 1 , . . . , α n ) ∈ N n is a multi-index, we set ∂ α = ∂ α1 1 . . . ∂ α n n . Proposition 1.5. The function f defined by f (x) = e 1/x , x < 0, 0, x ≥ 0, belongs to C ∞ (R).

Proof. We show, by using an induction with respect to the integer m, that f (m) (0) = 0, f (m) (x) = P m (1/x) e 1/x , x < 0, (

where P m is a polynomial. Clearly, the claim is true for m = 0. Assume then that (1.1) holds for some m. Then

lim x<0, x→0 f (m) (x) -f (m) (0) x = lim x<0, x→0 P m (1/x) e 1/x x = 0,
showing that f (m+1) (0) = 0. Finally, we have, for x < 0,

f (m+1) (x) = - 1
x 2 P m (1/x) + P m (1/x) e 1/x = P m+1 (1/x) e 1/x . The proof is then complete.

We call a sequence of mollifiers any sequence of functions (ρ m ) satisfying ϕ(x + εαy)ϕ(xy) ε < ∞.

0 ≤ ρ m ∈ D(R n ), supp(ρ m ) ⊂ B (0, 1/m) ,
We deduce from Lebesgue's dominated convergence theorem that

∂ α (ϕ * f )(x) = ω ∂ α ϕ(x -y) f (y)dy = (∂ α ϕ) * f (x).
The general case follows by using an induction in |α|.

Define, for |y| < 1/m, the translation operator τ y as follows (ii) If f ∈ L p loc (Ω ), 1 ≤ p < ∞ then we have, for sufficiently large m,

τ y : f ∈ L 1 loc (Ω ) → τ y f : τ y f (x) = f (x -y), x ∈ Ω m .
ρ m * f -f L p (ω) ≤ sup |y|<1/m τ y f -f L p (ω) .
Proof. Note first that ω ⊂ Ω m provided that m is sufficiently large. Let f ∈ C(Ω ). As

B(0,1/m)
ρ m (y)dy = 1, we obtain

|ρ m * f (x) -f (x)| = B(0, 1 m ) ρ m (y)( f (x -y) -f (x))dy ≤ sup |y|< 1 m sup x∈ω |τ y f (x) -f (x)|.
That is (i) holds. Next, let f ∈ L p loc (Ω ), 1 ≤ p < ∞. We have by Hölder's inequality, for any x ∈ ω, Proof. Let f ∈ C(Ω ) and ω be so that ω ω Ω . Then (i) follows from the fact that f is uniformly continuous on ω.

|ρ m * f (x) -f (x)| = B(0,1/m) ρ m (y)[ f (x -y) -f (x)]dy
Next, let f ∈ L p loc (Ω ), 1 ≤ p < ∞ and ε > 0. According to Theorem 1.5, we find g ∈ C c ( ω) so that fg L p ( ω) ≤ ε.

By (i), there exists 0 < δ < dist(ω, ω) so that sup x∈ω |τ y g(x)g(x)| ≤ ε for any |y| < δ .

Thus we have, for |y| < δ ,

τ y f -f L p (ω) ≤ τ y f -τ y g L p (ω) + τ y g -g L p (ω) + g -f L p (ω) ≤ 2 g -f L p ( ω) + |ω| 1/p sup x∈ω τ y g(x) -g(x)
≤ (2 + |ω| 1/p )ε.

Hence (ii) follows because ε > 0 is arbitrary.

A combination of Lemmas 1.1 and 1.2 yields the following regularization theorem. Then f = 0 a.e. in Ω .

Proof. We have in particular ρ m * f = 0, for every m. The proof is completed by applying Theorem 1.6 (ii).

Theorem 1.8. D(Ω ) is dense L p (Ω ) for any 1 ≤ p < ∞.

Proof. Let f ∈ L p (Ω ) and ε > 0. Since C c (Ω ) is dense L p (Ω ) by Theorem 1.5, we find g ∈ C c (Ω ) so that gf p ≤ ε/2.

As supp(ρ m * g) ⊂ B (0, 1/m)+supp(g) (see Exercise 1.5), there exists a compact set K ⊂ Ω such that supp(ρ m * g) ⊂ K if m ≥ m 0 , for some integer m 0 . By Proposition 1.6, ρ m * g ∈ D(Ω ), and from Theorem 1.6, ρ m * g converges to g in L p (Ω ) (note that g has compact support). Hence, we find k ≥ m 0 so that ρ k * gg p ≤ ε/2, from which we deduce that (ρ k * g)f p ≤ ε.

Here is another application of the regularization theorem Lemma 1.3. Let K ⊂ Ω be compact. There exists ϕ ∈ D(Ω ) so that 0 ≤ ϕ ≤ 1 and ϕ = 1 in K.

Proof. Set, for r > 0,

K r = {x ∈ R n ; dist(x, K) ≤ r}.
Let ε > 0 so that K 2ε ⊂ Ω . Let g be the function which is equal to 1 in K ε and equal to 0 in R n \ K ε . Fix m > 1/ε. Then the function ϕ = ρ m * g takes its values between 0 and 1, it is equal to 1 in K and equal to 0 in R n \ K 2ε . Finally, ϕ ∈ D(Ω ) by Proposition 1.6.

Theorem 1.9. (Partition of unity) Let K be a compact subset of R n and Ω 1 , . . . , Ω k be open subsets of R n so that K ⊂ ∪ k j=1 Ω j . Then there exist ψ 0 , ψ 1 , . . .

ψ k ∈ D(R n ) so that (a) 0 ≤ ψ j ≤ 1, 0 ≤ j ≤ k, ∑ k j=0 ψ j = 1, (b) supp(ψ 0 ) ⊂ R n \ K, ψ j ∈ D(Ω j ), 1 ≤ j ≤ k.
Proof. By Borel-Lebesgue's theorem K can be covered by finite number of balls B(x m , r m ), 1 ≤ m ≤ , such that, for any m, B(x m , 2r m ) ⊂ Ω j(m) for some 1 ≤ j(m) ≤ k. If 1 ≤ j ≤ k then set

K j = j(m)= j B(x m , 2r m ), K 0 = m=1 B(x m , r m ), Ω = m=1 B(x m , 2r m ).
According to Lemma 1.3, there exists, where 1 ≤ j ≤ k, ϕ j ∈ D(Ω j ) satisfying 0 ≤ ϕ j ≤ 1 and ϕ j = 1 sur K j . There exists also ϕ ∈ D(Ω ) satisfying 0 ≤ ϕ ≤ 1 and ϕ = 1 in K 0 . Take then ϕ 0 = 1ϕ and, for 0 ≤ j ≤ k,

ψ j = ϕ j ∑ k i=0 ϕ i .
Then it is straightforward to check that ψ 0 , ψ 1 , . . . ψ k have the expected properties.

1.3 Weak derivatives

Weak derivatives

The notion of weak derivative consists in generalizing the derivative in classical sense by means of an integration by parts formula. Precisely if f , g ∈ L 1 loc (Ω ) and α ∈ N n , we say that g = ∂ α f in the weak sense provided that Ω f ∂ α ϕdx = (-1) |α| Ω gϕdx for any ϕ ∈ D(Ω ).

Note that by the cancellation theorem ∂ α f is uniquely determined.

In light of to this definition we can state the following result. Proof. We may assume, without loss of generality, that α = (0, . . . , 0, 1). As supp( f ) is compact, it is contained in R n-1 ×]a, b[ for some finite interval ]a, b[. We naturally extend f and g by 0 outside Ω . We still denote by f and g these extensions. Using the notation x = (x ,

x n ) ∈ R n-1 × R, we get R g(x , x n )dx n = b a g(x , x n )dx n = f (x , a) -f (x , b) = 0.
Whence, we obtain by applying Fubini's theorem 1 Sobolev spaces

Ω gdx = R n gdx = R n-1 dx R gdx n = 0.
This completes the proof.

Theorem 1.10. (Du Bois-Reymond lemma) Let f , g ∈ C(Ω ) and |α| = 1. Then g = ∂ α f in the weak sense if and only if g = ∂ α f in the classical sense.

Proof. If g = ∂ α f in the weak sense then, according to Proposition 1.7, we have in the classical sense

∂ α (ρ m * f ) = ρ m * g. Hence ε 0 ρ m * g(x + tα)dt = ε 0 ∂ α (ρ m * f )(x + tα)dt = ρ m * f (x + εα) -ρ m * f (x).
That is we have

ρ m * f (x + εα) = ρ m * f (x) + ε 0 ρ m * g(x + tα)dt. (1.2)
From the regularization theorem ρ m * f (resp. ρ m * g) tends to f (resp. g) uniformly in any compact subset of Ω , as m → ∞. We get by passing to the limit in (1.2) Thus, g = ∂ α f in the weak sense.

f (x + εα) = f (x) +
We close this section by an example. Let, for -n < θ ≤ 1,

f (x) = |x| θ and f ε (x) = (|x| 2 + ε) θ /2 , ε > 0.
The function f ε is continuous in R n and therefore it is measurable in R n . As ( f ε ) converges a.e. (in fact everywhere except at 0), we deduce that f is measurable. On the other hand, for any R > 0, we have by passing to polar coordinates B(0,R)

|x| θ dx = S n-1 dσ R 0 r n+θ -1 dr < ∞,
where S n-1 is the unit sphere of R n . We have, when θ ≤ 0, 0 ≤ f ε ≤ f , and since f ε converges a.e. to f , we may apply Lebesgue's dominated convergence theorem. Therefore, ( f ε ) converges to f in L 1 loc (R n ). We proceed similarly in the case θ ≥ 0. We note that f ε ≤ f 1 , 0 < ε ≤ 1, and we conclude that we still have the convergence of ( f ε ) to f in L 1 loc (R n ).

We get by simple computations that f ε ∈ C ∞ (R n ) and

∂ j f ε = θ x j (|x| 2 + ε) θ -2 2 .
Hence |∂ j f ε | ≤ θ |x| θ -1 and then ∂ j f ε converges in L 1 loc (R) to g = θ x j |x| θ -2 if θ > 1n. In light of the closing lemma, we obtain that g = ∂ j f in the weak sense.

W k,p spaces

If k ≥ 1 is an integer and 1 ≤ p < ∞, we define the Sobolev W k,p (Ω ) by W k,p (Ω ) = {u ∈ L p (Ω ); ∂ α u ∈ L p (Ω ) for any |α| ≤ k}.

Here ∂ α u is understood as the derivative in the weak sense.

We endow this space with the norm

u W k,p (Ω ) = u k,p = ∑ |α|≤k Ω |D α u| p dx 1/p .
The space H k (Ω ) = W k,2 (Ω ) is equipped with the scalar product

(u|v) H k (Ω ) = ∑ |α|≤k (∂ α u|∂ α v) L 2 (Ω ) .
Define also the local Sobolev space W k,p loc (Ω ) as follows W k,p loc (Ω ) = {u ∈ L p loc (Ω ); u| ω ∈ W k,p (ω, for each ω Ω }.

We say that the sequence (u m ) converges to u in W k,p loc (Ω ) if, for every ω Ω , we have u mu W k,p (ω) → 0, as m → ∞.

Consider L p (Ω , R d ) that we equip with its natural norm Then it is not difficult to check that L p (Ω , R d ) is isometrically isomorphic to L p (Γ , dµ). An extension of Theorem 1.3 to a measure space with positive σ -finite measure yields the following result.

Lemma 1.6. Any closed subspace of L p (Ω , R d ) is a separable Banach space.

We get by applying Riesz's representation theorem 1 to L p (Γ , µ) the following lemma.

A : W k,p (Ω ) → L p (Ω , R d ) : u → Au = (∂ α u) |α|≤k
is isometric, i.e. Au p = u k,p .

Theorem 1.11. For 1 ≤ p < ∞ and k ≥ 1 an integer, the Sobolev space W k,p (Ω ) is separable Banach space.

Proof. According to the closing lemma, F = A(W k,p (Ω )) is closed in L p (Ω , R d ).

The theorem follows then from Lemma 1.6.

Theorem 1.12. Let k ≥ 1 be an integer and 1 < p < ∞. For any Φ ∈ W k,p (Ω ) , there exists a unique ( f α ) ∈ L p (Ω , R d ) so that ( f α ) p = Φ and Φ, u = Φ(u) = ∑ |α|≤k f α D α u, for all u ∈ W k,p (Ω ). 1 Riesz's representation theorem Let (X, dµ) be a measure space, where dµ is positive σ -finite measure. For each Φ ∈ [L p (X, µ)] , there exists a unique g Φ ∈ L p (X, µ) so that

Φ( f ) = Φ, f = X f g Φ dµ for any f ∈ L p (X, dµ).
Furthermore Φ = g Φ p .

1.4 W k,p spaces Proof. By Hahn-Banach's extension theorem 2 , there exists a continuous linear form extending Φ to L p (Ω , R d ) without increasing its norm. The theorem follows then from Lemma 1.7.

Let k ≥ 1 be an integer and 1 ≤ p < ∞. Denote the closure of D(Ω ) in W k,p (Ω ) by W k,p 0 (Ω ). The space W k,2 0 (Ω ) is usually denoted by H k 0 (Ω ). Let W -k,p (Ω ) be the space of continuous linear forms given as follows

Φ : D(Ω ) → R : u → ∑ |α|≤k Ω g α ∂ α udx,
when D(Ω ) is seen as a subspace of W k,p (Ω ) and where (g α ) ∈ L p (Ω , R d ).

We endow W -k,p (Ω ) with its natural quotient norm

Φ W -k,p (Ω ) = Φ -k,p = inf (g α ) p ; Φ, u = ∑
|α|≤k Ω g α ∂ α udx for any u ∈ D(Ω ) .

The space W -k,2 (Ω ) is usually denoted by H -k (Ω ).

Theorem 1.13. For any integer k ≥ 1 and 1 ≤ p < ∞, the space W -k,p (Ω ) is isometrically isomorphic to W k,p 0 (Ω ) .

Proof. If Φ ∈ W -k,p (Ω ) then there exists (g α ) ∈ L p (Ω , R d ) so that

Φ, u = ∑
|α|≤k Ω g α ∂ α udx, for any u ∈ D(Ω ).

Hence | Φ, u | ≤ (g α ) p u k,p for any u ∈ D(Ω ).

We can extend uniquely Φ to W k,p 0 (Ω ) by density. We still denote this extension by Φ. Moreover Φ 2 Hahn-Banach's extension theorem. Let V be a real normed vector space with norm • . Let V 0 be a subspace of V and let Φ 0 : V 0 → R be a continuous linear form with norm

Φ 0 V 0 = sup x∈V 0 , x ≤1 Φ 0 (x).
Then there exists Φ ∈ V extending Φ 0 so that

Φ V = Φ 0 V 0 .
Conversely, let Φ ∈ W k,p 0 (Ω ) . Then by Hahn-Banach's extension theorem Φ has an extension, still denoted by Φ, to W k,p (Ω ) that does not increase its norm. But from Theorem 1.12 there exists (g α ) ∈ L p (Ω , R d ) so that (g α ) p = Φ Whence, Φ ∈ W -k,p (Ω ) and

Φ W -k,p (Ω ) ≤ (g α ) p = Φ W k,p 0 (Ω )
. This completes the proof.

Next, we extend some classical rules of differential calculus to weak derivatives. Proposition 1.8. (Derivative of a product) if u ∈ W 1,1 loc (Ω ) and f ∈ C 1 (Ω ) then f u ∈ W 1,1 loc (Ω ) and

∂ j ( f u) = f ∂ j u + ∂ j f u.
Proof. If u m = ρ m * u then we have in the classical sense

∂ j ( f u m ) = f ∂ j u m + ∂ j f u m .
By the regularization theorem, u m → u and

∂ j u m = ρ m * ∂ j u → ∂ j u in L 1 loc (Ω ). Thus f u m → f u, ∂ j ( f u m ) = f ∂ j u m + ∂ j f u m → f ∂ j u + ∂ j f u in L 1 loc (Ω ).
The expected result follows then from the closing lemma. This proposition will be used to prove the following result.

Theorem 1.14. If 1 ≤ p < ∞ then W 1,p 0 (R n ) = W 1,p (R n ). Proof. It is sufficient to prove that D(R n ) is dense in W 1,p (R n ). We use truncation and regularization procedures. We fix θ ∈ C ∞ (R) satisfying 0 ≤ θ ≤ 1 and

θ (t) = 1, t ≤ 1, θ (t) = 0, t ≥ 2.
We define a truncation sequence by setting

θ m (x) = θ (|x|/m) , x ∈ R n .
Let u ∈ W 1,p (R n ). The formula giving the weak derivative of a product shows that u m = θ m u ∈ W 1,p (R n ). With the help of Lebesgue's dominated convergence theorem one can check that u m converges to u in W 1,p (R n ) θ m converges a.e. to u because θ m tends to 1 and we have |θ m u| ≤ |u|. On the other hand, supp(∂ α θ m ) ⊂ {m ≤ |x| ≤ 2m} and ∂ α u m = θ m D α u + D α θ m u . This construction guarantees that the support of u m is contained in B(0, 2m).

We now proceed to regularization. From the previous step we need only to consider u ∈ W 1,p (R n ) with compact support. Let K be a compact subset of de R n so that, for any m, the support of u m = ρ m * u is contained in K. As u m ∈ C ∞ (R n ) by Proposition 1.6, we deduce that u m belongs to D(R n ). We have from the regularization theorem

u m → u, ∂ j u m = ρ m * ∂ j u → ∂ j u in L p (R n ).
The proof is then complete.

Let Ω and ω be two open subsets of R n . Recall that f : ω → Ω is a diffeomorphism if f is bijective, it is continuously differentiable and satisfies J f (x) = det(∂ j f i (x)) = 0, for every x ∈ ω.

The following result follows from the density of C c (Ω ) in L 1 (Ω ) and the classical formula of change of variable for smooth functions.

ω → Ω be a diffeomorphism. If u ∈ W 1,1 loc (Ω ) then u • f ∈ W 1,1 loc (ω) and ∂ j (u • f ) = n ∑ k=1 (∂ k u • f ) ∂ j f k .
Proof. Let u m = ρ m * u and v ∈ D(ω). We have according to the definition of weak derivatives

(u m • f ) (y)∂ j v(y)dy = Ω u m (x) (∂ j v • g) (x)|det(J g (x))|dx = - Ω n ∑ k=1 ∂ k u m (x) (∂ j f k • g) (x) (v • g) (x)|det(J g (x))|dx = ω n ∑ k=1 (∂ k u m • f ) (y)∂ j f k (y)v(y)dy.
(1.3) Now, we have by the regularization theorem

u m → u, ∂ j u m → ∂ j u in L 1 loc (ω).
We pass to the limit, when m → ∞, in the second and the third members of (1.3). We get

Ω u(x) (∂ j v • g) (x)|det(J g (x))|dx = - Ω n ∑ k=1 ∂ k u(x) (∂ j f k • g) (x) (v • g) (x)
(Ω ). If M = sup | f | < ∞ then f • u ∈ W 1,1
loc (Ω ) and

∂ j ( f • u) = f • u ∂ j u.
Proof. If u m = ρ m * u then we have in the classical sense

∂ j ( f • u m ) = f • u m ∂ j u m .
So, by the regularization theorem, we obtain

u m → u, ∂ j u m → ∂ j u in L 1 loc (Ω ).
If ω Ω we get subtracting if necessary a subsequence we may assume that the convergence holds also almost everywhere in ω again from the regularization theorem

ω | ( f • u m ) -( f • u) |dx ≤ M ω |u m -u|dx → 0, ω | f • u m ∂ j u m -f • u ∂ j u|dx ≤ M ω |∂ j u m -∂ j u|dx + ω | f • u m -f • u ||∂ j u|dx → 0. Hence f • u m → f • u, f • u m ∂ j u m → f • u ∂ j u in L 1 loc (
Ω ) from Lebesgue's dominated convergence theorem. The proof is then completed by using the closing lemma.

Corollary 1.1. If u ∈ W 1,1 loc (Ω ) then u + , u -, |u| ∈ W 1,1 loc (Ω ) and

∂ j |u| =    ∂ j u in {u > 0}, -∂ j u in {u < 0}, 0, in {u = 0}. Proof. Let, for ε > 0, f ε (t) = (t 2 + ε 2 ) 1/2 and v =    ∂ j u in {u > 0}, -∂ j u in {u < 0}, 0 in {u = 0}.
We get by using Proposition 1.10

∂ j ( f ε • u) = u (u 2 + ε 2 ) 1/2 ∂ j u. Hence f ε • u → |u|, ∂ j ( f ε • u) → v in L 1 loc
(Ω ) and ∂ j |u| = v according to the closing lemma. Finally, for completing the proof we observe that 2u + = |u| + u and 2u -= |u|u.

Extension and trace operators

We start by extension operators using the simple idea of reflexion. If ω is an open subset of R n-1 and 0 < δ ≤ +∞, we let

Q = ω×] -δ , +δ [, Q + = ω×]0, δ [.
For an arbitrary u : Q + → R, we define σ u and τu on Q by

σ u(x , x n ) = u(x , x n ) if x n > 0, u(x , -x n ) if x n < 0, and τu(x , x n ) = u(x , x n ) if x n > 0, -u(x , -x n ) if x n < 0. Lemma 1.8. (Extension by reflexion) Let 1 ≤ p < ∞. If u ∈ W 1,p (Q + ) then σ u ∈ W 1,p (Q) and σ u L p (Q) ≤ 2 1/p u L p (Q + ) , σ u W 1,p (Q) ≤ 2 1/p u W 1,p (Q + ) .
Proof. We first prove that

∂ j σ u = σ ∂ j u, 1 ≤ j ≤ n -1. If v ∈ D(Q), we have Q σ u∂ j vdx = Q + uD j wdx, with w(x , x n ) = v(x, x n ) + v(x , -x n ). (1.4) Fix η ∈ C ∞ (R) satisfying η(t) = 0 if t < 1 2 , 1 if t > 1,
and set η m = η(mt).

As η m (x n )w(x , x n ) ∈ D(Q + ), we obtain

Q + uη m D j w = Q + u∂ j (η m w) = - Q + ∂ j uη m w.
Il light of Lebesgue's dominated convergence theorem, we can pass to the limit, when m tends to infinity, in the first and third terms. We obtain

Q + u∂ j wdx = - Q + ∂ j uwdx = - Q σ ∂ j uvdx
which, combined with (1.4), entails

Q σ u∂ j vdx = - Q σ ∂ j uvdx.
That is we have

∂ j (σ u) = σ ∂ j u, 1 ≤ j ≤ n -1. (1.5)
Next, we prove that ∂ n (σ u) = τ∂ n u. For this purpose, we note that

Q σ u∂ n vdx = Q + u∂ n wdx, with w(x , x n ) = v(x, x n ) -v(x , -x n ). (1.6)
As w(x , 0) = 0, there exists C 0 > 0 so that | w(x ,

x n )| ≤ C 0 |x n | in Q + .
Using the fact that η m (x n ) w(x , x n ) ∈ D(Q + ), we find

Q + u∂ n (η m w)dx = - Q + ∂ n uη m wdx. But ∂ n (η m w) = η m D n w + mη (mx n ) w.
Let C 1 = η ∞ . We get, observing that w has a compact support, that there exists a compact subset K of ω so that

Q + mη (mx n )u wdx ≤ C 0 C 1 m K×]0,1/m[ |u|x n dx ≤ C 0 C 1 K×]0,1/m[ |u|dx → 0.
We get by applying again Lebesgue's dominated convergence theorem

Q + u∂ n wdx = - Q + ∂ n u wdx = - Q τ∂ n uvdx.
This identity together with (1.6) imply

Q σ u∂ n vdx = Q + ∂ n u wdx = - Q τ∂ n uvdx.
In other words, we demonstrated that ∂ n (σ u) = τ∂ n u.

(1.7)

Finally, identities (1.5) and (1.7) yield the expected result.

We use in the sequel the following notations

D(Ω ) = {u| Ω ; u ∈ D(R n )}, R n + = {(x , x n ) ∈ R n ; x ∈ R n-1 , x n > 0}.
Lemma 1.9. (Trace inequality) Let 1 ≤ p < ∞. We have, for u ∈ D(R n + ),

R n-1 |u(x , 0)| p dx ≤ p u p-1 L p (R n + ) ∂ n u L p (R n + ) .
Proof. Consider first the case 1 < p < ∞. Let u ∈ D(R n + ). As u has compact support, for each x ∈ R n-1 , we find y n = y n (x ) so that u(x , y n ) = 0, y n ≥ y n . Whence

|u(x , 0)| p = - y n 0 p|u(x , x n )| p-1 ∂ n u(x , x n )dx n and hence |u(x , 0)| p ≤ ∞ 0 p|u(x , x n )| p-1 |∂ n u(x , x n )|dx n .
We get by applying Fubini's theorem and then Hölder's inequality

R n-1 |u(x , 0)| p dx ≤ p R n + |u| p-1 |∂ n u|dx ≤ p R n + |u| (p-1)p dx 1 p R n + |∂ n u| p dx 1 p ≤ p R n + |u| p dx 1-1 p R n + |∂ n u| p dx 1 p
, which yields the expected inequality. The proof in the case p = 1 is quite similar to that of the case 1 < p < ∞.

Proposition 1.11. Let 1 ≤ p < ∞. There exists a unique linear bounded operator

γ 0 : W 1,p (R n + ) → L p (R n-1 )
satisfying γ 0 u = u(•, 0), for each u ∈ D(R n + ).

Proof. If u ∈ D(R n + ), we set γ 0 u = u(•, 0). From Lemma 1.9, we have

γ 0 u L p (R n-1 ) ≤ p 1 p u W 1,p (R n + ) .
We deduce, using the theorem of extension by reflexion and the density of D(R n ) in W 1,p (R n ) (a consequence of Theorem 1.14), that D(R n + ) is dense W 1,p (R n + ). We complete the proof by extending γ 0 by density. Proposition 1.12. (Integration by parts) Let 1 ≤ p < ∞. If u ∈ W 1,p (R n + ) and v ∈ D(R n + ) then

R n + v∂ n udx = - R n + ∂ n vudx - R n-1
γ 0 vγ 0 udx and

R n + v∂ j udx = - R n + ∂ j vudx, 1 ≤ j ≤ n -1.
Proof. Assume first that u ∈ D(R n + ). The classical integration by parts formula yields, for each

x ∈ R n-1 , +∞ 0 v(x , x n )∂ n u(x , x n )dx n = - +∞ 0 u(x , x n )∂ n v(x , x n )dx n -u(x , 0)v(x , 0).
We then obtain by applying Fubini's theorem

R n + v∂ n udx = - R n + ∂ n vudx - R n-1
v(x , 0)u(x , 0)dx , that we write in the form

R n + v∂ n udx = - R n + ∂ n vudx - R n-1
γ 0 vγ 0 udx .

(1.8)

We know from the proof of Proposition 1.11 that D(R n + ) is dense in W 1,p (R n + ). So if u ∈ W 1,p (R n + ) then we may find a sequence (u m ) in D(R n + ) that converges to u in W 1,p (R n + ). This and the fact that γ 0 is a bounded operator from W 1,p (R n + ) into L p (R n-1 ) entail that γ 0 u m converges to γ 0 u in L p (R n-1 ). We obtain from (1.8)

R n + v∂ n u m dx = - R n + ∂ n vu m dx - R n-1 γ 0 vγ 0 u m dx .
We then pass to the limit, when m → ∞, to get the first formula. The second formula can be established analogously.

Hereafter, if u is a function defined on R n + then its extension to R n by 0 is denoted by u. Proposition 1.13. Let 1 ≤ p < ∞ and u ∈ W 1,p (R n + ). The following assertions are equivalent.

(i) u ∈ W 1,p 0 (R n + ). (ii) γ 0 u = 0. (iii) u ∈ W 1,p (R n ) and ∂ j u = ∂ j u, 1 ≤ j ≤ n. Proof. If u ∈ W 1,p 0 (R n + ) then there exits a sequence (u m ) in D(R n + ) converging to u in W 1,p (R n + ). Therefore γ 0 u m → γ 0 u dans L p (R n-1 ). But γ 0 u m = 0, for each m. Whence, γ 0 u = 0. That is (i) implies (ii).
If γ 0 u = 0 then by Proposition 1.12 we have, for every

v ∈ D(R n ), R n v∂ j udx = R n ∂ j vudx, 1 ≤ j ≤ n.
In other words, we proved that (ii) implies (iii). Assume finally that (iii) holds. If (θ m ) is the truncation sequence introduced in the proof of Theorem 1.14 then the sequence u m = θ m u converges to u in W 1,p (R n ) and u m has its support contained in B(0, 2m) ∩ R n + . We are then reduced to consider the case where additionally u has a compact support in R n + . Let y m = (0, . . . , 0, 1/m) and v m = τ y m u. Since ∂ j v m = τ y m ∂ j u, the continuity of translation operators guarantees that v m → u in W 1,p (R n + ). That is we are lead to the case where u has a compact support in R n + . Therefore, we may find a compact subset

K of R n + so that, for each m, supp(ρ m * u) ⊂ K. As w m = ρ m * u ∈ C ∞ (R n + ), we have w m ∈ D(R n + ). According to the regularization theorem, w n tends to u in W 1,p (R n + ). In consequence, u ∈ W 1,p 0 (R n + )
and then (iii) entails (i). This completes the proof.

Prior to considering extension and trace theorems for an arbitrary domain of R n , we introduce the definition of an open subset of class C k . We say that an open subset 3 and δ > 0 so that, modulo a rigid transform,

Ω of R n is of class C k if, for each x ∈ Γ = ∂ Ω , we can find a neighborhood U of x in R n , an open subset ω of R n-1 , ψ ∈ C k (ω)
U = {(y , ψ(y ) + t); y ∈ ω, |t| < δ }, Ω ∩U = {(y , ψ(y ) + t); y ∈ ω, 0 < t < δ }, Γ ∩U = {(y , ψ(y )); y ∈ ω}.
In other words, an open subset Ω is of class C k if any point of its boundary admits a neighborhood U so that U ∩ Ω coincide with the epigraph of a function of class C k .

We leave to the reader to check that, with the aid of the implicit function theorem, the above definition of an open subset of R n of class C k is equivalent to the following one: let

Q = {x = (x , x n ) ∈ R n ; |x | < 1 and |x n | < 1}, Q + = Q ∩ R n + , Q 0 = {x = (x , x n ) ∈ R n ; |x | < 1 and x n = 0}. Ω is said of class C k , k ≥ 1 is an integer if, for each x ∈ Γ , there exists a neighbor- hood U of x in R n and a bijective mapping Φ : Q → U satisfying Φ ∈ C k (Q), Φ -1 ∈ C k (U), Φ(Q + ) = U ∩ Ω , Φ(Q 0 ) = U ∩ Γ .
If the open subset Ω is of class C k and has bounded boundary then there exist (think to the compactness of Γ ) a finite number of open subsets of R n , U 1 , . . . ,U , open subsets of R n-1 , ω 1 , . . . , ω , functions ψ 1 , . . . ψ and positive real numbers δ 1 , . . . , δ satisfying all the conditions of the preceding definition and are so that

Γ ⊂ j=1 U j .
By the theorem of partition of unity, there exist φ 0 , . .

. , φ ∈ C ∞ (R n ) satisfying (i) 0 ≤ φ j ≤ 1, 0 ≤ j ≤ , ∑ j=0 φ j = 1, (ii) supp(φ 0 ) ⊂ R n \ Γ , φ j ∈ D(U j ), j = 1, . . . , .
P : W 1,p (Ω ) → W 1,p (R n ) so that Pu| Ω = u.
Proof. We use the notations that we introduced above in the definition of Ω of class C k with bounded boundary. Fix u ∈ W 1,p (Ω ) and 1 ≤ j ≤ . From the change of variable formula, we have

u y , ψ j (y ) + t ∈ W 1,p (ω j ×]0, δ j [).
The reflexion extension lemma then entails

u y , ψ j (y ) + |t| ∈ W 1,p (ω j ×] -δ j , δ j [). Thus v j y , y n = u y , ψ j (y ) + |y n -ψ j (y )| ∈ W 1,p (U j ).
For 1 ≤ j ≤ , we can easily check that

v j W 1,p (U j ) ≤ C 0 u W 1,p (Ω ∩U j ) ,
where the constant C 0 > 0 is independent of u.

Let (φ j ) the partition of unity defined as above. Set U 0 = Ω , v 0 = u and, for 0 ≤ j ≤ , let

u j (x) = φ j (x)v j (x), x ∈ U j , 0, x ∈ R n \U j .
By the formula of the derivative of a product, we obtain that u j ∈ W 1,p (R n ) and

u j W 1,p (R n ) ≤ C 1 u W 1,p (Ω ) ,
where the constant C 1 > 0 is independent of u. Define

Pu = ∑ j=0 u j ∈ W 1,p (R n ) .
Then there exists a constant C, independent of u, so that

Pu W 1,p (R n ) ≤ C u W 1,p (Ω ) .
Furthermore, we have

Pu(x) = ∑ j=0 ψ j (x)u(x) = u(x), x ∈ Ω ,
as expected.

Remark 1.1. In the case where Ω is a cube of R n (which is not of class C 1 ), the extension operator can easily be constructed by using extensions by reflexion and a localization argument. Proof. Let u ∈ W 1,p (Ω ). From Theorem 1.14, there exists a sequence

(v m ) in D(R n ) converging to Pu in W 1,p (R n ). Therefore, u m = v m | Ω tends to u in W 1,p (Ω ).
Let Ω be a bounded open subset of class C 1 with boundary Γ . For u ∈ C(Γ ), the formula

Γ u(γ)dγ = ∑ j=1 ω j (φ j u)(y , ψ j (y )) 1 + |∇ψ j (y )| 2 dy
defines an elementary integral.

Theorem 1.18. (Trace theorem) Let Ω be a domain of class C 1 with bounded boundary Γ . For 1 ≤ p < ∞, there exists a unique bounded operator

γ 0 : W 1,p (Ω ) → L p (Γ ) so that γ 0 u = u |Γ if u ∈ D(Ω ).
Proof. Fix u ∈ D(Ω ) and 1 ≤ j ≤ . There exists ϕ j ∈ D(U j ) so that 0 ≤ ϕ j ≤ 1 and ϕ j = 1 in supp(φ j ). With the help of the change of variable formula and the formula of the derivative of a product, we get v j y ,t = (ϕ j u) y , ψ j (y ) + t ∈ W 1,p (ω j ×]0, δ j [) .

As v j has a compact support in ω j × [0, δ j [, Proposition 1.11 implies

ω j |v j (y , 0)| p dy ≤ C 0 v j p W 1,p (ω j ×]0,δ j [) ≤ C 1 u W 1,p (Ω ) .
That is we proved, where we set γ 0 u = u| Γ ,

γ 0 u L p (Γ ) ≤ C u W 1,p (Ω ) .
We end up the proof by noting that D(Ω ) is dense in W 1,p (Ω ) (Theorem 1.17).

Let Ω be a open subset of class C 1 with bounded boundary Γ . Define the unit exterior normal vector to Γ at γ ∈ Γ ∩U j by ν(γ) = (∇ψ j (y ), -1)

1 + |∇ψ j (y )| 2 . Theorem 1.19. (Divergence theorem) Let Ω an open bounded set of class C 1 with boundary Γ . If V ∈ W 1,1 (Ω , R n ) then Ω divV dx = Γ γ 0 V • νdγ.
Proof. Follows from the classical divergence theorem, which is valid when V ∈ C 1 Ω , R n , and the density of

C 1 Ω , R n in W 1,1 (Ω , R n ).
Proposition 1.14.

Let Ω an open bounded set of class C 1 , 1 ≤ p < ∞ and u ∈ W 1,p (Ω ). The following assertions are equivalent. (i) u ∈ W 1,p 0 (Ω ). (ii) γ 0 u = 0. (iii) There exits a constant C > 0 so that

Ω uD i ϕdx ≤ C ϕ L p (Ω ) , ϕ ∈ D(R n ), 1 ≤ i ≤ n. (iv) u ∈ W 1,p (R n ) and ∂ i u = ∂ i u, 1 ≤ i ≤ n,
where, as before, u denotes the extension of u by 0 outside Ω .

Proof. (i) implies (ii): if u ∈ W 1,p 0 (Ω ) then u is the limit in W 1,p 0 (Ω ) of a sequence (u m ) of elements of D(Ω ). As γ 0 is continuous from W 1,p 0 (Ω ) into L p (Γ ) and γ 0 u m = 0, we deduce immediately that γ 0 u = 0. (ii) implies (iii): if ϕ ∈ D(R n ) and 1 ≤ i ≤ n then the divergence theorem yields Ω u∂ i ϕdx = - Ω ∂ i uϕdx + Γ γ 0 (uϕ)ν i dγ = - Ω ∂ i uϕ. That is we have (iii) with C = ∇u L p (Ω ,R n ) . (iii) implies (iv): for ϕ ∈ D(R n ) and 1 ≤ i ≤ n, we have R n u∂ i ϕdx = Ω u∂ i ϕdx ≤ C ϕ L p (R n ) .
This and Riesz's representation theorem show that there exists

g i ∈ L p (R n ) so that R n u∂ i ϕdx = R n g i ϕdx.
Thus ∂ i u = g i and then u ∈ W 1,p (R n ). Finally, from the identities

R n u∂ i ϕdx = - R n ∂ i uϕdx = - Ω ∂ i uϕdx = - R n ∂ i uϕdx we get ∂ i u = ∂ i u.
(iv) implies (i): by using local cards and partition of unity, we are reduced to the case Ω = R n + . The result follows then from Proposition 1.13.

Imbedding Theorems

Let us first explain briefly how to use an homogeneity argument to get an information on the validity of a certain inequality. Assume then that we can find a constant C > 0 and 1 ≤ q < ∞ so that, for any u ∈ D(R n ),

u L q (R n ) ≤ C ∇u L p (R n ,R n ) .
We get by substituting u by u λ (x) = u(λ x), λ > 0,

u L q (R n ) ≤ Cλ 1+ n q -n p ∇u L p (R n ,R n ) .
This implies that we must have necessarily p < n and

q = p * = np n -p . Lemma 1.10. (Sobolev inequality) For 1 ≤ p < n, there exists a constant c = c(p, n) > 0 so that, for every u ∈ D(R n ), u L p * (R n ) ≤ c ∇u L p (R n ,R n ) .
Proof. We prove by induction in n that, for any u ∈ D(R n ),

u n/(n-1) ≤ n ∏ j=1 ∂ j u 1/n 1 .
(1.9)

If n = 2, we have

u(x) = u(x 1 , x 2 ) = x 1 -∞ ∂ 1 u(t, x 2 )dt = x 2 -∞ ∂ 2 u(x 1 , s)ds. Whence |u(x)| 2 ≤ x 1 -∞ |∂ 1 u(t, x 2 )|dt x 2 -∞ |∂ 2 u(x 1 , s)|ds ≤ R |∂ 1 u(t, x 2 )|dt R |∂ 2 u(x 1 , s)|ds.
Integrating side by side each member of the preceding inequality over R 2 . We obtain

u 2 ≤ ∂ 1 u 1 ∂ 2 u 1 . Assume now that (1.9) is valid until some n ≥ 2. If u ∈ D(R n+1 ) then, for any t ∈ R, R n |u(x,t)| n/(n-1) dx (n-1)/n ≤ n ∏ j=1 R n |∂ j u(x,t)|dx 1/n .
We find by applying generalized Hölder's inequality

R n dt R n |u(x,t)| n/(n-1) dx (n-1)/n ≤ n ∏ j=1 ∂ j u 1/n 1 .
(1.10)

On the other hand, since u(x,t) = t -∞ ∂ n+1 u(x, s)ds, we have

|u(x,t)| (n+1)/n ≤ R ∂ n+1 u(x, s)ds 1/n |u(x,t)|.
Hölder's inequality then yields

R n |u(x,t)| (n+1)/n dx ≤ ∂ n+1 u 1/n 1 R n |u(x,t)| n/(n-1) dx (n-1)/n .
Integrating over R with respect to t and using (1.10) in order to obtain

u (n+1)/n (n+1)/n ≤ n+1 ∏ j=1 ∂ j u 1/n 1 .
That is

u (n+1)/n ≤ n ∏ j=1 ∂ j u 1/(n+1) 1
.

Fix u ∈ D(R n ) and λ > 1. Inequality (1.9) applied to |u| λ and Hölder's inequality give

u λ λ n/(n-1) ≤ λ u λ -1 (λ -1)p n ∏ j=1 D j u 1/n p . Note that ∂ j |u| λ = λ |u| λ -1 ∂ j u . The choice of λ satisfying λ n/(n -1) = (λ -1)p yields u p * ≤ λ n ∏ j=1 ∂ j u 1/n p ≤ c ∇u p ,
which is the expected inequality.

Lemma 1.11. (Morrey inequality) Let n < p < ∞ and λ = 1n/p. There exists a constant c = c(p, n) > 0 so that, for every u ∈ D(R n ) and any x, y ∈ R n , we have

|u(x) -u(y)| ≤ c|x -y| λ ∇u L p (R n ,R n ) , u ∞ ≤ c u W 1,p (R n ) .
Proof. Let Q be a cube containing 0 and having each side parallel to axes and is of length

r. Let u ∈ D(R n ). For x ∈ Q, we have u(x) -u(0) = 1 0 ∇u(tx) • xdt. Hence |u(x) -u(0)| ≤ 1 0 n ∑ j=1 |∂ j u(tx)||x j |dt ≤ r n ∑ j=1 1 0 |∂ j u(tx)|dt. If m(u, Q) = 1 |Q| Q u(x)dx,
we get by integrating the last inequality over Q

|m(u, Q) -u(0)| ≤ r |Q| Q dx n ∑ j=1 1 0 |∂ j u(tx)|dt ≤ 1 r n-1 1 0 dt n ∑ j=1 Q |∂ j u(tx)|dx ≤ 1 r n-1 1 0 dt n ∑ j=1 tQ |∂ j u(y)| dy t n .
Observe that, as Q is convex, we have tQ = tQ + (1 -t){0} ⊂ Q. We obtain then by applying Hölder's inequality

|m(u, Q) -u(0)| ≤ n r n-1 ∇u L p (Q) n 1 0 (tr) n/p t n dt = nr λ λ ∇u L p (Q) n .
By making a translation, we can substitute 0 by an arbitrary x ∈ R n in such a way that

|m(u, Q) -u(x)| ≤ nr λ λ ∇u L p (Q) n . (1.11)
We find by taking r = 1 in this inequality

|u(x)| ≤ |m(u, Q)| + n λ ∇u L p (Q) n ≤ C 0 u W 1,p (Q) ≤ C 0 u W 1,p (R n ) . Let x, y ∈ R n . Then r = 2|x -y| in (1.11) gives |u(x) -u(y)| ≤ |m(u, Q) -u(x)| + |m(u, Q) -u(y)| ≤ n2 1+λ λ |x -y| λ ∇u L p (Q) n ≤ C 1 |x -y| λ Du L p (R n ) .
The proof is then complete. (i) If 1 ≤ p < n and if p ≤ q ≤ p * then W 1,p (Ω ) ⊂ L q (Ω ) and the imbedding is continuous.

Define C 0 (R n ) = {u ∈ C(R n ); u(x) → 0 as |x| → +∞} and C 0 (Ω ) = {u| Ω ; u ∈ C 0 (R n )}.
(ii) If n < p < ∞ and λ = 1n/p then W 1,p (Ω ) ⊂ C 0 (Ω ), the imbedding is continuous and there exists a constant c = c(p, n) > 0 so that, for every u ∈ W 1,p (Ω ) and any x, y ∈ Ω , we have

|u(x) -u(y)| ≤ c|x -y| λ u W 1,p (Ω ) . Proof. Let 1 ≤ p < n and u ∈ W 1,p (R n ). By Theorem 1.14, we find a sequence (u m ) in D(R n ) converging to u in W 1,p (R n ). Sobolev's inequality then gives u m -u L p * (R n ) ≤ c ∇(u m -u ) L p (R n ) n . Hence (u m ) is a Cauchy sequence L p * (R n ). As u m → u in L p (R n ), we deduce that u m → u in L p * (R n ). Therefore u L p * (R n ) ≤ c ∇u L p (R n ,R n ) .
Let P be the extension operator corresponding to Ω and v ∈ W 1,p (Ω ). Then

v L p * (Ω ) ≤ Pv L p * (R n ) ≤ c ∇Pv L p (R n ) n ≤ c 0 v W 1,p (Ω ) .
If p ≤ q ≤ p * , we define 0 ≤ λ ≤ 1 by

1 q = 1 -λ p + λ p *
and we apply the interpolation inequality in Proposition 1.3. We conclude that

v L q (Ω ) ≤ v 1-λ L p (Ω ) v λ L p * (Ω ) ≤ c λ 0 v W 1,p (Ω ) .
We proceed similarly for the case p > n.

If u ∈ W 1,p (R n ), we pick (u m ) a se- quence in D(R n ) converging to u in W 1,p (R n
) and a.e. in R n . We apply the Morrey inequality to u m and pass then to the limit, as m → ∞. We obtain

|u(x) -u(y)| ≤ c|x -y| λ ∇u L p (R n ) , a.e. x, y ∈ R n .
(1.12)

Now, substituting if necessary u by a continuous representative 4 , we may assume that u ∈ C 0 (R n ) and the last inequality holds for any x, y ∈ R n . We end up the proof by using, as in the previous case, the extension operator P corresponding to Ω .

One obtains by applying recursively Theorem 1.20 the following corollary.

Corollary 1.2. Let Ω be an open subset of R n of class C 1 with bounded boundary. (i) If 1 ≤ p < n/m and if p ≤ q ≤ p * = np/(n -mp) then W m,p (Ω ) ⊂ L q (Ω ) and the imbedding is continuous. (ii) If n/m < p < ∞, W m,p (Ω ) ⊂ C k 0 (Ω ), where k = [m -n/p], C k 0 (Ω ) = {u; ∂ α u ∈ C 0 (Ω )
for each α ∈ N so that |α| ≤ k} and the embedding is continuous. In addition, if mn/p in non integer, there exists a constant c = c(p, n, m) > 0 so that, for every u ∈ W m,p (Ω ) and any x, y ∈ Ω ,

|∂ α u(x) -∂ α u(y)| ≤ c|x -y| λ u W m,p (Ω ) for all |α| = k, with λ = m -n/p -[m -n/p].
Prior to stating the Rellich-Kondrachov imbedding theorem, we prove the following lemma. Lemma 1.12. Let Ω be an open subset of R n of class C 1 with bounded boundary Γ , ω Ω and u ∈ W 1,1 (Ω ). For |y| < dist(ω,Γ ), we have

τ y u -u L 1 (ω) ≤ |y| ∇u L 1 (Ω ,R n ) .
Proof. As D(Ω ) is dense in W 1,1 (Ω ), it is enough to prove the lemma when u ∈ D(Ω ). In that case we have

τ y u(x) -u(x) = 1 0 ∇u(x -ty) • ydt ≤ |y| 1 0 |∇u(x -ty)|dt.
Thus, where |y| < dist(ω,Γ ),

τ y u -u L 1 (ω) ≤ |y| ω dx 1 0 |∇u(x -ty)|dt ≤ |y| 1 0 dt ω |∇u(x -ty)|dx ≤ |y| 1 0 dt ω-ty |∇y(z)|dz ≤ |y| ∇u L 1 (Ω ,R n )
and hence the expected inequality holds.

Theorem 1.21. (Rellich-Kondrachov imbedding theorem) Let Ω be a bounded open subset of R n of class C 1 . (a) If 1 ≤ p < n and if 1 ≤ q < p * then W 1,p (Ω ) ⊂ L q (Ω ) and the imbedding is compact. (b) If n < p < ∞ then W 1,p (Ω ) ⊂ C(Ω ) and the imbedding is compact. Proof. (a) Let 1 ≤ p < n.
We are going to show that B, the unit ball of W 1,p (Ω ), satisfies the assumption of Theorem 1.4 in L q (Ω ) provided that 1 ≤ q < p * . This will implies that B will be relatively compact in L q (Ω ). (i) From Theorem 1.20 and Proposition 1.4, we derive that

u L q (Ω ) ≤ u L p * (Ω ) |Ω | 1/q-1/p * ≤ C, for any u ∈ B.
(ii) Let ω Ω and define 0 ≤ λ < 1 so that

1 q = 1 -λ 1 + λ p * .
If |y| < dist(ω,Γ ), the interpolation inequality in Proposition 1.3 and Lemma 1.12 yield, for every u ∈ B,

τ y u -u L q (ω) ≤ τ y u -u 1-λ L 1 (ω) τ y u -u λ L p * (ω) ≤ |y| 1-λ ∇u 1-λ L 1 (Ω ,R n ) 2 u L p * (ω) λ ≤ c|y| 1-λ ,
where we used that

∇u L 1 (Ω ,R n ) ≤ ∇u L p (Ω ,R n ) |Ω | 1-1/p . (iii) Let ε > 0. There exists ω Ω such that 5 u L q (Ω \ω) ≤ u L p * (Ω \ω) |Ω \ ω| 1/q-1/p * ≤ c|Ω \ ω| 1/q-1/p * ≤ ε. (b) Let p > n.
We have from the Sobolev imbedding theorem

u C(Ω ) ≤ c u W 1,p (Ω ) for any u ∈ B, and
|u(x) -u(y)| ≤ c u W 1,p (Ω ) |x -y| λ ≤ c|x -y| λ ,
for any x, y ∈ Ω and u ∈ B. This means that B satisfies the assumptions of Ascoli-Arzela's theorem 6 . Whence B is relatively compact in C(Ω ).

Theorem 1.22. (Poincaré's inequality) Let 1 ≤ p < ∞. If there exists an isometry A : R n → R n so that A(Ω ) ⊂ R n-1 ×]0, a[ then, for any u ∈ W 1,p 0 (Ω ), u L p (Ω ) ≤ a 2 ∇u L p (Ω ,R n ) . Proof. Fix 1 < p < ∞. If u ∈ D(]0, a[) then Hölder's inequality implies |v(x)| ≤ 1 2 a 0 |v (x)|dx ≤ a 1/p 2 a 0 |v (x)| p dx 1/p . Hence a 0 |v(x)| p dx ≤ a p/p 2 p a a 0 |v (x)| p dx = a p 2 p a 0 |v (x)| p dx.
According to our assumption on Ω , we may assume, without loss of generality, that Ω is of the form Ω = R n-1 ×]0, a[. So, for u ∈ D(Ω ), we deduce from the last inequality and Fubini's theorem that

Ω |u| p dx = R n-1 dx a 0 |u(x , x n )| p dx n ≤ a p 2 p R n-1 dx a 0 |∂ n u(x , x n )| p dx n = a p 2 p Ω |∂ n u| p .
The expected inequality follows by using that D(Ω ) is dense in W 1,p 0 (Ω ). 5 One can take ω of the form

ω = Ω m = {x ∈ Ω ; dist(x,Γ ) > 1/m} .
6 Ascoli-Arzela's theorem. Let K = (K, d) be a compact metric space and let F be a bounded subset of C(K). Assume that F is uniformly equicontinuous, i.e. for any ε > 0, there exists δ > 0 so that

d(x, y) < δ =⇒ |u(x) -u(y)| < ε, for any u ∈ F .
Then F is relatively compact in C(K).

The case p = 1 can be established analogously.

We close this chapter by some comments. The Sobolev spaces H s (R n ), s ∈ R, can be defined using the Fourier transform. With the help of local cards and a partition of unity, one can build the H s spaces on a submanifold of R n . The Sobolev spaces H s (Ω ), where Ω is an open subset of R n , can constructed by interpolation from H k (Ω ) spaces. The reader is referred to the book by J.-L. Lions and E. Magenes [8] for more details.

More generally we define the fractional order Sobolev space W s,p (Ω ), 0 < s < 1 and 1 ≤ p < ∞, as follows

W s,p (Ω ) = f ∈ L p (Ω ); | f (x) -f (y)| |x -y| s+n/p ∈ L p (Ω × Ω ) .
Observe that W s,p (Ω ) can be seen as an interpolated space between W 1,p (Ω ) and L p (Ω ).

For arbitrary non integer s > 1, we set

W s,p (Ω ) = f ∈ W k,p (Ω ); ∂ α f ∈ W t,p (Ω ) for any |α| = k , where k = [s] is the integer part of s and t = s -[s].
Again, using a partition of unity and local cards, one can define W s,p spaces on a submanifold of R n . The reader can find in the book by P. Grisvard [2, Chapter 1] a detailed study of the W s,p spaces.

Exercises and problems

1.1. Let 1 ≤ p < q < ∞ and Ω be an open subset of R n satisfying |Ω | < ∞. Show that if u ∈ L q (Ω ) then u ∈ L p (Ω ) and u p ≤ |Ω | 1/p-1/q u q . 1.2. (Generalized Hölder's inequality) Let 1 < p j < ∞, 1 ≤ j ≤ k, so that 1 p 1 + . . . 1 p k = 1 and let u j ∈ L p j (Ω ), 1 ≤ j ≤ k. Prove that ∏ k j=1 u j ∈ L 1 (Ω ) and Ω k ∏ j=1 |u j |dx ≤ k ∏ j=1 u j p j . 1.3. (Interpolation inequality) Let 1 ≤ p < q < r < ∞ and 0 < λ < 1 given by 1 q = 1 -λ p + λ r .
Prove that if u ∈ L p (Ω ) ∩ L r (Ω ) then u ∈ L q (Ω ) and

u q ≤ u 1-λ p u λ r .
1.4. (Support of measurable function) Let Ω be an open subset of R n and f : Ω → R be a measurable function. Assume that there exists a family (ω i ) i∈I of open subsets of Ω so that, for each i ∈ I, f = 0 a.e. in ω i . Set ω = ∪ i∈I ω i . Prove that there exists a countable set J ⊂ I such that ω = ∪ i∈J ω i . Conclude then that f = 0 a.e. in ω.

If ω is the union of all open subset of Ω in which f = 0 a.e., the closed set supp( f ) = Ω \ ω is called the support of the measurable function f .

1.5. Let f ∈ L 1 (R n ) and g ∈ L p (R n ), 1 ≤ p ≤ ∞. a) Prove that y → f (x -y)g(y) is absolutely integrable in R n , a.e. x ∈ R n .
Define the convolution product of f and g by

( f * g)(x) = R n f (x -y)g(y)dy. Show that f * g ∈ L p (R n ) and f * g p ≤ f 1 g p .
Hint: consider first the case p = 1 for which we can apply Tonelli's theorem [START_REF] Ladyzhenskaja | Linear and quasilinear elliptic equations[END_REF] . In a second step, reduce the case 1 < p < ∞ to that of p = 1. b) Demonstrate that supp( f * g) ⊂ supp( f ) + supp(g).

1.6. Prove that the function x α admits a weak derivative belonging to L 2 (]0, 1[) if and only if α > 1/2.

1.7.

Let Ω be an open bounded R n so that there exists a sequence (Ω i ) 1≤i≤k of open subsets that are pairwise disjoint and

Ω = ∪ k i=1 Ω i . Assume moreover that, for each i, Ω i is piecewise of class C 1 . Set C 1 pie (Ω , (Ω i ) 1≤i≤k ) = u : Ω → R; u |Ω i ∈ C 1 (Ω i ), 1 ≤ i ≤ k .
(x) = | ln |x|| α belongs to H 1 B 1/2 for 0 < α < 1/2, but its is unbounded. (b) Assume that n ≥ 3. Show that the function u(x) = |x| -β is in H 1 (B 1 ) provided that 0 < β < (n -2)/2, but its is unbounded. 1.9. Let B be the unit ball of R n . (a) Compute the values of α = 0 for which |x| α ∈ W 1,p (B) (resp. |x| α ∈ W 1,p (R\B)). (b) Show that x/|x| ∈ W 1,p (B) n if and only if p < n. 1.10. Let a, b ∈ R and u ∈ H 1 (]a, b[). a) For x, y ∈]a, b[, show that u(x) 2 + u(y) 2 -2u(x)u(y) ≤ (b -a) b a u (x) 2 dx.
b) Integrate with respect to x and then with respect to y to deduce that there exists a constant c > 0 so that

b a u(x) 2 dx ≤ c b a u (x) 2 dx + b a u(x)dx 2 . 1.11. (a) Let v ∈ C 1 c (R) and G(s) = |s| p-1 s, s ∈ R, with 1 ≤ p < ∞. Let w = G(v) ∈ C 1 c (R) . Use the relation w(x) = x -∞ w (t)dt to show that |v(x)| ≤ p 1/p v 1/p p v 1/p p ≤ e 1/e v 1/p p v 1/p p ≤ e 1/e ( v p + v p ).
(b) Deduce that there exists a constant c ≥ 0 so that, for any 1 ≤ p < ∞ and any u ∈ W 1,p (R), we have

u L ∞ (R) ≤ c u W 1,p (R) .
1.12. Let n ≥ 2 and 1 ≤ p < n. Set p * = np/(np) and q = p(n -1)/(np). Demonstrate that, for every u ∈ D(R n ),

R n-1 |u(x , 0)| q dx ≤ q u q-1 p * ∂ n u p .

1.13.

Let Ω be the open subset of R 2 given by 0 < x < 1 and 0

< y < x β with β > 2. Let v(x) = x α . Prove that v ∈ H 1 (Ω ) if and only if 2α + β > 1 ; while v ∈ L 2 (∂ Ω ) if and only if 2α > -1. Conclude. 1.14. (a) Prove the Caffarelli-Kohn-Nirenberg's theorem: let 1 < p < ∞ and α + n > 0. Then, for any u ∈ D(R n ), R n |u| p |x| α dx ≤ p p (α + n) p R n |x • ∇u| p |x| α dx ≤ p p (α + n) p R n |∇u| p |x| α+p dx.
Hint: Regularize |x| α to show that div(|x| α x) = (α + n)|x| α in the weak sense. (b) Deduce the Hardy's inequality: let 1 < p < n. Then, for every u

∈ W 1,p (R n ), we have u/|x| ∈ L p (R n ) and u |x| p ≤ p n -p ∇u p . 1.15. (a) Let u ∈ L 1 (R n ) and v ∈ W 1,p (R n ) with 1 ≤ p ≤ ∞. Prove that u * v ∈ W 1,p (R n ) and ∂ i (u * v) = u * ∂ i v, 1 ≤ i ≤ n.
Let Ω be an open subset of R n . For a function u defined on Ω , denote by u its extension by 0 outside Ω , i.e.

u(x) = u(x) if x ∈ Ω , 0 if x ∈ R n \ Ω . (b) Let u ∈ W 1,p (Ω ), u m = ρ m * u, ω Ω and ϕ ∈ D(Ω ) satisfying 0 ≤ ϕ ≤ 1 and ϕ = 1 in a neighborhood of ω.
(i) Show the following claims:

ρ m * ϕu = ρ m * u in ω if m sufficiently large. ∂ i (ρ m * ϕu) → ϕ∂ i u + ∂ i ϕu in L p (R n ). Deduce that ∂ i (ρ m * u) → ∂ i u in L p (ω).
(ii) Prove the Friedrichs's theorem: let u ∈ W 1,p (Ω ) with 1 ≤ p < ∞. Then there exists a sequence (u m ) in D(R n ) so that, for any ω Ω ,

u m | Ω → u in L p (Ω ), ∇u m | ω → ∇u| ω in L p (ω) n . (c) Let u, v ∈ W 1,p (Ω ) ∩ L ∞ (Ω ) with 1 ≤ p ≤ ∞. Show that uv ∈ W 1,p (Ω ) ∩ L ∞ (Ω ) and ∂ i (uv) = ∂ i uv + u∂ i v, 1 ≤ i ≤ n.
1.16. In this exercise we only use the definition of

W 1,p 0 (Ω ), that is W 1,p 0 (Ω ) is the closure of D(Ω ) in W 1,p (Ω ). Let Ω be an open subset of R n of class C 1 having bounded boundary Γ and 1 ≤ p < ∞. Let G ∈ C 1 (R) so that |G(t)| ≤ t, t ∈ R, and
G(t) = 0 if |t| ≤ 1, G(t) = t if |t| ≥ 2. (a) Let u ∈ W 1,p (Ω ). If u has a compact support, show then that u ∈ W 1,p 0 (Ω ). (b) Let u ∈ W 1,p (Ω ) ∩C(Ω ) satisfying u = 0 on Γ . (i) Assume that u has a compact support and set u m = G(mu)/m. Check that u m ∈ W 1,p (Ω ), u m → u in W 1,p (Ω ) and supp(u m ) ⊂ {x ∈ Ω ; |u(x)| ≥ 1/m}.
Deduce that u m ∈ W 1,p 0 (Ω ) (and hence u ∈ W 1,p 0 (Ω )). (ii) Show that the result in (i) still holds without the assumption that u has a compact support. (c) Prove that

W 1,p 0 (Ω ) ∩C(Ω ) = {u ∈ W 1,p (Ω ) ∩C(Ω ); γ 0 u = 0} 8 ,
where γ 0 is the trace operator introduced in Theorem 1.18.

1.17. (a) Let ϕ ∈ D(]0, 1[). Prove the following inequalities:

|ϕ(x)| 2 ≤ x 1/2 0 |ϕ (t)| 2 dt, x ∈ [0, 1/2], |ϕ(x)| 2 ≤ (1 -x) 1 1/2 |ϕ (t)| 2 dt, x ∈ [1/2, 1]. Deduce then that 1 0 |u(x)| 2 dx ≤ 1 8 1 0 |u (x)| 2 dx, for any u ∈ H 1 0 (]0, 1[). Set C = sup 1 0 |u(x)| 2 dx 1 0 |u (x)| 2 dx ; u ∈ H 1 0 (]0, 1[), u = 0
and consider the boundary value problem, where

f ∈ L 2 (]0, 1[), -u (x) -ku(x) = f (x), x ∈]0, 1[, u(0) = u(1) = 0. (1.13) A solution of (1.13) is a function u ∈ H 2 (]0, 1[) ∩ H 1 0 (]0, 1[
) satisfying the first identity in (1.13) a.e. x ∈]0, 1[. (b) Prove that if kC < 1 then (1.13) has at most one solution. Hint: as a first step, we can show that if u is a solution of (1.12) then, for every ϕ ∈ D(]0, 1[),

1 0 u (x)ϕ (x)dx -k 1 0 u(x)ϕ(x)dx = 1 0 f (x)ϕ(x)dx.
(c) Compute the non trivial solutions, for k = 0, of the boundary value problem

u (x) + ku(x) = 0, x ∈]0, 1[ and u(0) = u(1) = 0,
and deduce from it that [START_REF] Lions | Problèmes aux limites non homogènes et applications[END_REF] We have in fact (see Proposition 1.14) (c) Show that, if I is bounded and 1 < p ≤ ∞, then the imbedding W 1,p (I) → C(I) is compact. Hint: use Arzela-Ascoli's theorem.

W 1,p 0 (Ω ) = {u ∈ W 1,p (Ω ); γ 0 u = 0}. 1 π 2 ≤ C ≤ 1

(Poincaré-Wirtinger inequality)

Let Ω be a bounded domain of R n . Prove that there exists a constant C > 0, only depending on Ω , so that, for every u ∈ H 1 (Ω ), we have

u -u L 2 (Ω ) ≤ C ∇u L 2 (Ω ,R n ) , where u = 1 |Ω | Ω u(x)dx.
Hint: Show first that it is enough to establish the above inequality when u = 0. Proceed then by contradiction.

1.20. (The space H 1/2 (Γ )) Let Ω be a bounded domain of R n of class C 1 and recall that the trace operator γ 0 :

H 1 (Ω ) → L 2 (Γ ), defined by γ 0 (u) = u |Γ , u ∈ D(Ω ), is bounded. Set H 1/2 (Γ ) = γ 0 H 1 (Ω ) . Define on H 1/2 (Γ ) the quotient norm v H 1/2 (Γ ) = min{ u H 1 (Ω ) ; u ∈ H 1 (Ω ) and γ 0 (u) = v}, v ∈ H 1/2 (Γ ).
Prove that, for any v ∈ H 1/2 (Γ ), there exists a unique

u v ∈ H 1 (Ω ) so that v H 1/2 (Γ ) = u v H 1 (Ω ) .
Chapter 2

Variational solutions

This chapter is mainly devoted to study existence and uniqueness of variational solutions of elliptic partial differential equations. It contains also some classical properties of weak solutions of elliptic equations. Amongst the properties we establish, there are the maximum principle, Harnack inequalities and the unique continuation across a non characteristic hypersurface. This chapter can be completed by the following classical textbooks [1, 2, 5, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14]. We just quote these few references, but of course there are many other excellent textbooks dealing with the analysis of elliptic partial differential equations.

Section 2.1, Section 2.2, Subsections 2.3.2 and 2.3.3 are largely inspired by the book of Brézis [2]. Subsections 2.4.1 and 2.4.2 were based on the book of Gilbarg-Trudinger, while Subsection 2.4.3 was prepared from both the book of Bénilan [1] and Gilbarg-Trudinger [5].

Stampacchia's theorem and Lax-Milgram lemma

We recall the projection theorem on closed convex set of a Hilbert space.

Theorem 2.1. Let H be a real Hilbert space with scalar product (•|•), and let K be a closed nonempty convex subset of H. (i) For any u ∈ H, there exists a unique P K u ∈ K so that

u -P K u = min v∈K u -v ,
where • is the norm associated to the scalar product (•|•). Moreover P K u is characterized by P K u ∈ K and (u -P K u|v -P K u) ≤ 0 for any v ∈ K. P K u is called the projection of u on K. (ii) P K : u ∈ H → P K u ∈ H is a contractive operator, i.e.

P K u -P K v ≤ u -v for all u, v ∈ H.
The projection on a closed subspace is characterized by the following proposition.

Proposition 2.1. Let H be as in Theorem 2.1 and let E be a closed subspace of H. If u ∈ H then P E u is characterized by

P E u ∈ E and (u -P E u, v) = 0 any v ∈ E.
Furthermore, P E is a linear operator.

We recall that a bilinear form a : H × H → R is continuous if and only if there exists a constant C > 0 so that

|a(u, v)| ≤ C u v for any u, v ∈ H.
The bilinear form a is said coercive if we can find α > 0 with the property that a(u, u) ≥ α u 2 for every u ∈ H.

Theorem 2.2. (Stampacchia's theorem) Let H be a real Hilbert space with scalar product (•|•). Let a be a coercive and continuous bilinear form on H × H and K be a closed convex subset of H. For every Φ ∈ H , there exists a unique u ∈ K so that

a(u, v -u) ≥ Φ(v -u) for any v ∈ K. (2.1)
Moreover, if a symmetric then u is characterized by u ∈ K and

1 2 a(u, u) -Φ(u) = min v∈K 1 2 a(v, v) -Φ(v) .
Proof. By Riesz-Fréchet's representation theorem 1 , we find a unique f ∈ H so that

Φ, v = ( f , v) for any v ∈ H.
On the other hand, for arbitrary fixed u ∈ H, the mapping v → a(u, v) is linear continuous form on H. Therefore, again by Riesz-Fréchet's representation theorem, there exists Au ∈ H such that a(u, v) = (Au, v) for every v ∈ H. It is then not hard to check that A is a linear operator from H into H and satisfies 1 Riesz-Fréchet's representation theorem. Let H be a real Hilbert space with scalar product (•|•).

If Φ ∈ H then there exists a unique f ∈ H so that

Φ(v) = ( f , v) for any v ∈ H. Furthermore, Φ = f . Au ≤ C u , for any u ∈ H, (2.2)
and (Au, u) ≥ α u2 , for any u ∈ H.

(2.3)

In consequence, the problem (2.1) is reduced to find u ∈ K so that

(Au, v -u) ≥ ( f , v -u), for any v ∈ K. (2.4)
Let θ > 0 be a constant that we will fix later in the proof and note that inequality (2.4) is equivalent to the following one.

([θ f -θ Au + u] -u, v -u) ≤ 0 for any v ∈ K. (2.5)
That is

u = P K (θ f -θ Au + u). For v ∈ K, set T v = P K (θ f -θ Av + v).
Then we are reduced to show that T has a unique fixed point.

We have, according to Theorem 2.1,

T v -Tw ≤ (v -w) -θ (Av -Aw) for all v, w ∈ K. Hence T v -Tw 2 ≤ v -w 2 -2θ (Av -Aw, v -w) + θ 2 Av -Aw 2 ≤ v -w 2 (1 -2θ α + θ 2 C 2 ),
where we used (2.2) and (2.3). The value of θ that minimize 1 -2θ α + θ 2 C 2 is equal α/C 2 . We have for this choice of θ T v -Tw ≤ k vw for all v, w ∈ K.

Here k = (1α 2 /C 2 ) 1/2 < 1. By Banach's fixed point theorem 2 , T has a unique fixed point u ∈ K, that is u = Tu. Assume next that a symmetric. Then (u, v) → a(u, v) defines a new scalar product on H and the associated norm a(u, u) 1/2 is equivalent to the initial norm on H. Therefore H is a Hilbert space with respect to this new norm. Once again, in light of Riesz-Fréchet's theorem, we find f ∈ H so that d(Mx, My) ≤ kd(x, y) for any x, y ∈ M.

Then T has a unique fixed point u, i.e. u = Tu.

Φ, v = a(g, v) for any v ∈ H.

As a consequence of this, (2.1) takes the form a(gu, vu) ≤ 0 for any v ∈ H.

(2.6)

Then, according to Theorem 2.1, (2.6) is reduced to find u ∈ K satisfying min v∈K a(gv, gv)

a(u, v) = Φ(v) for every v ∈ H.
Furthermore, if a is symmetric then u is characterized by

1 2 a(u, u) -Φ(u) = min v∈H 1 2 a(u, u) -Φ(u) .

Elements of the spectral theory of compact operators

In this section, E, F are two Banach spaces and U denotes the unit ball of E.

An operator A ∈ L (E, F) is said compact whenever A(U) is relatively compact. The subset of L (E, F) consisting in compact operators is denoted by K (E, F). For simplicity convenience we set

K (E) = K (E, E). Theorem 2.3. K (E, F) is a closed subspace of L (E, F). Proof. Note first that is not difficult to see that K (E, F) is a subspace of L (E, F). Next, let A ∈ L (E, F) be the limit in L (E, F) of a sequence (A k ) in K (E, F).
For an arbitrary ε > 0, we are going to show that A(U) can be covered by finite number of ball

B(y i , ε) in F, implying, since F is complete, that A(U) is relatively compact. Fix k so that A k -A < ε/2. As A k (U) is relatively compact, A k (U) ⊂ ∪ i∈I B ( f i , ε/2)
, where I is finite. We deduce from this that A(U) ⊂ ∪ i∈I B( f i , ε).

We say that A ∈ L (E, F) is of finite rank if dim R(A) < ∞, where R(A) is the range of A. Observing that finite rank operators are compact, we get immediately from Theorem 2.3 the following corollary.

Corollary 2.2. If A ∈ L (E, F) is the limit in L (E, F) of finite rank operators, then A ∈ K (E, F).

It is straightforward to check that the composition of bounded operator and compact operator is again a compact operator. We have precisely the following result. Proof. Assume that A ∈ K (E, F). If U is the unit ball of F , we are going to prove that A * (U ) is relatively compact in E . We pick a sequence (v m ) in U and we show that (A * (v m )) admits a convergent subsequence. Consider then the compact metric space M = A(U) and K ⊂ C(M) given by

K = {ϕ m ; ϕ m (x) = v m |x , x ∈ M}.
It is easy to check that K possesses the assumption of Arzela-Ascoli's theorem. Hence, we can subtract from (ϕ m ) a subsequence (ϕ k ) converging to ϕ ∈ C(M). We have in particular

sup u∈U | v k |Au -ϕ(Au)| → 0 when k → +∞, from which we obtain sup u∈U | v k |Au -v |Au | → 0 if k, → +∞. That is A * v k -A * v → 0 as k, → +∞. Whence, since E is complete, (A * v k ) converges in E .
Conversely, assume that A * ∈ K (F , E ). From the previous step, we obtain that A * * ∈ K (E , F ) and consequently A * * (U) is relatively compact in F . But A(U) = A * * (U) and F is identified isomophicly and isometrically to a subspace of F . Thus A(U) is relatively compact in F.

Recall that the kernel of A ∈ L (E, F) is defined by

N(A) = {u ∈ E; Au = 0}.
For X ⊂ E, we use in the sequel the notation X ⊥ = {ϕ ∈ E ; ϕ|u = 0 for any u ∈ X}.

Similarly, if Φ ⊂ E we set Φ ⊥ = {u ∈ E; ϕ|u = 0 for any ϕ ∈ Φ}.

if R(I -A) = E. (d) dim(I -A) = dim(I -A * ).
Fredholm's alternative theorem is useful tool to solve the equation

u -Au = f .
(2.7) Theorem 2.5 says that we have the following alternative.

• For any f ∈ E, (2.7) has unique solution,

• or else u -Au = 0 admits p linearly independent solutions and in that case (2.7) is solvable if and only f satisfies p orthogonality conditions, which means precisely that f ∈ N(I -A * ) ⊥ .

The following theorem will be used in the proof of Fredholm's alternative.

Theorem 2.6. (Riesz's theorem) If U is compact then E is of finite dimension.

We need to introduce the notion of topological supplement. Let G be a closed subspace of E. We say that a subspace L of E is a topological supplement of G if L is closed, G ∩ L = {0} and G + L = E. In that case any z ∈ E has a unique decomposition z = x + y with x ∈ G and y ∈ L. One can check that the projectors z → x et z → y define linear bounded operators. We know that any subspace of finite dimension or finite co-dimension admits a topological supplement. On the other hand one can check that any closed subspace of Hilbert space possesses a topological supplement.

We shall need also the following result in the proof of Theorem 2.5.

Theorem 2.7. Let A ∈ L (E, F). The following assertions are equivalent.

(a) R(A) is closed. (b) R(A * ) is closed. (c) R(A) = N(A * ) ⊥ . (d) R(A * ) = N(A) ⊥ . Proof (of Theorem 2.5). (a) If E 1 = N(I -A) and if U 1 is the unit ball of E 1 , then U 1 ⊂ A(U). Whence E 1 is of finite dimension by Riesz's theorem. (b) Let ( f m ) be a sequence in R(I -A) with f m = u m -Au m ,
for each m, that converges to f ∈ E. We want to check that f ∈ R(I -A). To this end, let d m = dist(u m , N(I -A)). As N(I -A) is of finite dimension, there exists v m ∈ N(I -A) so that d m = u mv m . Note that we have

f m = (u m -v m ) -A(u m -v m ).
(2.8)

We claim that u mv m is bounded. Otherwise, (u mv m ) would admit a subsequence

(u k -v k ) so that u k -v k → ∞ as k → ∞. Let w k = u k -v k u k -v k .
From (2.8), w k -Aw k → 0, as k → ∞. Since A is compact, subtracting again if necessary a subsequence, we may assume that Aw k → z. Hence w m → z and z ∈ N(I -A).

On the other hand, we have

dist(w k , N(I -A)) = dist(u k , N(I -A)) u k -v k = 1.
Passing to the limit, as k → ∞, to deduce that dist(z, N(I -A)) = 1 which is impossible. That is we proved that u mv m is bounded and using once more that A is compact to conclude that there exists a subsequence

(A(u k -v k )) converging to h ∈ E. This and (2.8) entail that u k -v k → f + h. Whence, if g = f + h then g -Ag = f .
In other words, we proved that f ∈ R(I -A) and hence R(I -A) is closed. We deduce from Theorem 2.7 that

R(I -A) = N(I -A * ) ⊥ and R(I -A * ) = N(I -A) ⊥ .
(c) We first prove that N(I -A) = {0} implies that R(I -A) = E. We proceed again by contradiction. To this end, we assume that

E 1 = R(I -A) = E.
We have that E 1 is a Banach space and A(E 1 ) ⊂ E 1 . Hence, A| E 1 ∈ K (E 1 ) and

E 2 = (I -A)(E 1 ) is a closed subspace of E 1 . Moreover E 2 = E 1 because I -A is injective. Let E m = (I -A) m (E).
Then (E m ) is a sequence of strictly decreasing closed subspaces. By Riesz's lemma 3 we find a sequence (u m ) satisfying u m ∈ E m , u m = 1 and dist(u m , E m+1 ) ≥ 1/2. Thus Thus u = 0. We then apply (c) to B to conclude that R(I -B) = E. This not possible since there exists f ∈ L, f ∈ R(Λ ) ; the equation u -Bu = f does not have a solution Indeed, if it has a solution u then, as previously, we should have

Au -Au m = -(u -Au ) + (u m -Au m ) + (u -u m ). If > m, we have E +1 ⊂ E ⊂ E m+1 ⊂ E m and consequently -(u -Au ) + (u m -Au m ) + u ∈ E m+1 . Whence, Au -Au m ≥ 1/2 which is impossible since A is compact. We end up concluding that R(I -A) = E. Conversely, if R(I -A) = E then Theorem 2.7 allows us to get that N(I -A * ) = R(I -A) ⊥ = {0}. But since A * ∈ K (E )
(I -A)u = Λ Pu + f . But (I -A)u ∈ R(I -A) and Λ Pu + f ∈ L. Hence Λ Pu + f = 0 because L is a topological supplement of R(I -A). That is, f = -Λ Pu ∈ R(Λ ) which is absurd . In other words d * ≤ d. This result applied to A * , yields dimN(I -A * * ) ≤ dim(I -A * ) ≤ dimN(I -A).
We get by using N(I -

A * * ) ⊃ N(I -A) that d = d * note that A * * is an extension of A : E ⊂ E → E .
Define the resolvent set of A ∈ L (E) by ρ(A) = {λ ∈ R; (Aλ I) is bijective from E onto E}. 4 5 The spectrum of σ (A) is the complement of the resolvent set, i.e. σ (A) = R \ ρ(A). We say that λ is an eigenvalue of A and write λ ∈ ev(A) if N(Aλ I) = {0}. The subspace N(Aλ I) is called the eigenspace associated to λ .

We make the following remarks.

• If λ ∈ ρ(A) then (A -λ I) -1 ∈ L (E).
• We have ev(A) ⊂ σ (A). Apart the case dim(E) < ∞ for which we have ev(A) = σ (A), the inclusion is in general strict. Indeed, one can find λ so that N(Aλ I) = {0} and R(Aλ I) = E. This is for instance the case when E =2 and Au = (0, u 1 , . . . , u m , . . .) when u = (u 1 , . . . , u m , . . .) (the right shift operator).

Proposition 2.3. σ (A) is compact with σ (A) ⊂ [-A , A ],
where A denotes the norm of A in L (E).

Proof. Let λ ∈ R so that |λ | > A . From Banach's fixed point theorem, for any f ∈ E, there exists a unique u ∈ E so that u

= (1/λ )(Au -f ), that is (A -λ I)u = f . Hence A -λ I is bijective and consequently σ (A) ⊂ [-A , A ].
Next we show that ρ(A) is open. This will imply that σ

(A) = R \ ρ(A) is closed. Let λ ∈ ρ(A). If f ∈ E, solving the problem Au -µu = f is equivalent to find a solution of the equation u = (A -λ I) -1 ((µ -λ )u + f ).
We deduce by applying once again Banach's fixed point theorem that the last equation has a unique solution whenever

|µ -λ | (A -λ I) -1 < 1. That is λ - 1 (A -λ I) -1 , λ + 1 (A -λ I) -1 ⊂ ρ(A).
This completes the proof.

Proposition 2.4. Let A ∈ K (E) with dim(E) = ∞. Then 0 ∈ σ (A) and σ (A) \ {0} = ev(A) \ {0}.
Proof. If 0 does not belong to σ (A), then A would be bijective and hence I = AA -1 would be compact. This would imply that U is compact and then, by Theorem 2.6, E would be of finite dimension. This leads to the expected contradiction.

Let λ ∈ σ (A) \ {0}. If λ is not an eigenvalue of A, we would have N(Aλ I) = {0} and hence R(Aλ I) = E by Fredholm's alternative. That is we would have λ ∈ ρ(A) (by Banach's theorem) contradicting the fact that λ belongs to the spectrum of A.

We now give a more precise description of the spectrum of compact operators. Prior to doing that we prove the following lemma.

Lemma 2.1. Let A ∈ K (E) and let (λ m ) be a sequence of distinct reals numbers so that λ m → λ and λ m ∈ σ (A) \ {0}, for any m. Then λ = 0. In other words, the elements of σ (A) \ {0} are isolated.

Proof. We know from Proposition 1.4 that λ m ∈ ev(A). Let then e m ∈ E, e m = 0 so that (Aλ m I)e m = 0. Define E m = span{e 1 , . . . , e m }. Let us prove E m ⊂ E m+1 strictly for each m. To this end, it is enough to check that e 1 , . . . , e m are linearly independent. We proceed by induction in m. Assume that the result is true for some m and that e m+1 = ∑ m i=1 α i e i . We have then

Ae m+1 = m ∑ i=1 λ i α i e i = m ∑ i=1 λ m+1 α i e i .
Hence α i (λ iλ m+1 ) = 0, 1 ≤ i ≤ m. As λ i 's are distinct, we derive that α i = 0, 1 ≤ i ≤ m, and consequently E m ⊂ E m+1 strictly for each m.

On the other hand, it is clear that (Aλ m I)E m ⊂ E m-1 . By Riesz's lemma we can find a sequence (u m ) satisfying u m ∈ E m , u m = 1 and dist(u m , E m-1 ) ≥ 1/2, for each m ≥ 2. Let 2 ≤ m < in such a way that

E m-1 ⊂ E m ⊂ E -1 ⊂ E . We have Au λ - Au m λ m = Au -λ u λ - Au m -λ m u m λ m + u -u m ≥ dist(u , E -1 ) ≥ 1/2.
If λ m → λ = 0 we get a contradiction since, by compactness, (Au m ) admits a convergent subsequence.

Theorem 2.8.

Let A ∈ K (E) with dim(E) = ∞. Then σ (A) = {0}, or else σ (A) \ {0} is finite, or else σ (A) \ {0} consists in a sequence converging to 0. Proof. For m ≥ 1, σ (A) ∩ {λ ∈ R; |λ | ≥ 1/m}
is empty or else it is finite. Otherwise, it would contain an infinite distinct points entailing, as σ (A) is compact, that this set has an accumulation point in σ (A) which contradicts Lemma 2.1. If the case where σ (A) \ {0} consists in infinite points, we can order these points in a sequence converging to 0.

Our next objective is to provide a spectral decomposition of self-adjoint compact operators. We suppose then that E = H is a Hilbert space, with scalar product (•|•), and A ∈ L (H). Identifying H with its dual H , we may consider that A * is an element of L (H). In this case, we say that A is self-adjoint if A * = A, i.e.

(Au|v) = (u|Av) for any u, v ∈ H. Proof. If λ > M then λ ∈ ρ(A). Indeed, from (Au|u) ≤ M u 2 for any u ∈ H, we get

([λ I -A]u, u) ≥ (λ -M) u 2 = α u 2 for all u ∈ H, with α = λ -M > 0.
Hence λ I -A is bijective according to Lax-Milgram's lemma. Next, we show that M ∈ σ (A). We proceed by contradiction. We assume then that M ∈ ρ(A). In that case the symmetric continuous bilinear form a(u, v) = ([MI -A]u|v) defines a new scalar product on H. Whence we obtain by applying Cauchy-Schwarz's inequality

|([MI -A]u|v)| ≤ ([MI -A]u|u) 1/2 ([MI -A]v|v) 1/2 for every u, v ∈ H.
In particular, for any u ∈ H, we have

Mu -Au = sup v∈H, v =1 |([MI -A]u|v)| ≤ C([MI -A]u, u) 1/2 .
(2.9)

Let (u k ) be a sequence satisfying u k = 1 and (Au k |u k ) → M. We deduce from (2.9) that Mu k -Au k converges to 0 and hence u k = (MI -A) -1 (MI -A)u k → 0 which is impossible because u k = 1. This yields the expected contradiction.

The proof for m is obtained by substituting A by -A.

Corollary 2.3. If A is self-adjoint and σ (A) = {0} then A = 0.

Proof. We have from Proposition 2.5 that (Au|u) = 0, for any u ∈ H, and hence

2(Au|v) = (A(u + v)|u + v) -(Au|u) -(Av, v) = 0 for any u, v ∈ H.
Whence A = 0.

The final result concerning the abstract spectral theory is a fundamental result showing that in a separable Hilbert space we can diagonalize any self-adjoint compact operator.

Theorem 2.9. Let H be a separable Hilbert space and let A ∈ K (H) be self-adjoint. Then H admits a Hilbertian basis consisting of eigenvectors of A.

Proof. Let (λ m ) m≥1 be the sequence of distinct eigenvalues of A except 0. Set

λ 0 = 0, E 0 = N(A) and E m = N(A -λ m I), m ≥ 1. Then 0 ≤ dim(E 0 ) ≤ ∞ and 0 < dim(E m ) < ∞, m ≥ 1.
Let us prove that H is the Hilbertian sum of (E m ) m≥0 . We first note that the subspaces E m are pairwise orthogonal. Indeed, if u ∈ E m and v ∈ E , m = , then, since Au = λ m u and Av = λ v, we have

(Au|v) = λ m (u|v) = (u, Av) = λ (u|v) implying (u, v) = 0.
Next, we show that the subspace spanned by (E m ) m≥0 , denoted by F, is dense in H. Clearly A(F) ⊂ F and therefore A(F ⊥ ) ⊂ F ⊥ . To see this, we observe that if u ∈ F ⊥ and v ∈ F then (Au|v) = (u|Av). The operator B = A| F ⊥ is then selfadjoint and compact. Let claim that σ (B) = {0}. Since otherwise we would find λ ∈ σ (B) \ {0}, that is λ ∈ ev(B). Whence there exists u ∈ F ⊥ , u = 0, satisfying Bu = λ u and consequently λ is equal to one of the eigenvalues λ m and u ∈ F ⊥ ∩ E m . Thus u = 0 which leads to a contradiction. We conclude by applying Corollary 2.3 that B = 0 and then F ⊥ ⊂ N(A) ⊂ F. We deduce that F ⊥ = {0} which means exactly that F is dense in H.

Finally, in each E m we choose a Hilbertian basis consisting of eigenvectors of A. The union of all these eigenvectors form a Hilbertian basis of H consisting in eigenvectors of A.

The rest of this section is inspired by [9, Section 6.2]. To apply the abstract spectral theory to elliptic boundary value problems we need to formulate such spectral problems, via the variational formulation, in an abstract way involving bilinear forms. In such general framework we consider V and H two infinite dimensional Hilbert spaces with V continuously and densely imbedded in H. The norm and the scalar product on H are denoted by (•|•) and | • |, while the norm on V is denoted by • .

As V in continuously imbedded in H, there exists a c > 0 so that |v| ≤ c v for any v ∈ V.

(2.10)

Let a : V × V → R be a continuous symmetric bilinear form and consider the spectral problem : find the values of λ ∈ R so that there exists u ∈ V , u = 0, satisfying the equation

a(u, v) = λ (u|v) for any v ∈ V. (2.11)
We assume in addition that a is V -elliptic, i.e. there exists a constant α > 0 such that a(v, v) ≥ α v 2 for every v ∈ V,

Define A ∈ L (H,V ) by a(Au, v) = (u|v), for any v ∈ V. (2.12)
This definition has a sense since, noting that for any u ∈ H the linear form v → Φ(v) = (u, v) is continuous on V by (2.10), the problem (2.12) admits a unique solution Au ∈ V according to Lax-Milgram's Lemma (here V is endowed with scalar product given by a). It is clear that A defines a linear map from H into V . On the other hand (2.10) yields

Φ = sup v =1 (u, v) v ≤ c|u| in such a way that Au ≤ 1 α Φ ≤ c α |u|,
where α is the V -ellipticity constant of a.

The advantage in introducing the operator A is that (2.11) can be converted into the following spectral problem : find u ∈ V , u = 0, so that u = λ Ãu.

(2.13)

Here à = AI, I being the canonical imbedding of V into H.

Lemma 2.2. If I the canonical imbedding of V into H is compact and the bilinear form a is V -elliptic, then à ∈ K (V ).

This lemma follows readily from Proposition 2.2 because à = AI with I ∈ K (V, H) and A ∈ L (H,V ).

Lemma 2.3. If the bilinear form a is continuous, symmetric and V -elliptic, then à ∈ L (V ) is self-adjoint when V is endowed with the scalar product a(•, •). Moreover, A is positive in the sense that a( Ãv, v) > 0 for any v ∈ V , v = 0.

Proof. Using that a is symmetric, we obtain from (2.12) that, for all u, v ∈ V ,

a( Ãu|v) = (u|v) = (v|u) = a(u, Ãv),
showing that à is self-adjoint whenever V is endowed with the scalar product a(•, •). The positivity of à follows from the fact that a(

Ãv, v) = |v| 2 > 0 for any v ∈ V , v = 0.
Theorem 2.10. Assume that V is compactly imbedded in H and the bilinear form a is continuous, symmetric and V -elliptic. Then the eigenvalues of (2.11) form an non decreasing sequence converging to ∞:

0 < λ 1 ≤ λ 2 ≤ . . . ≤ λ m ≤ . . .
and there exists an orthonormal Hilbertian basis of H consisting of eigenvectors w m so that, for any m ≥ 1,

a(w m , v) = λ m (w m |v) for any v ∈ V.
Moreover, the sequence (λ -1/2 m w m ) form an orthonormal Hilbertian basis of V for the scalar product a(•, •).

Proof. By Lemma 2.3, à is self-adjoint and positive. We apply then Theorem 2.9 to deduce that the spectrum of à consists in a sequence (µ m ) of non increasing of positive reals numbers converging to 0, and there exists an orthonormal Hilbertian basis of V for the scalar product a(•, •) consisting in eigenvectors v m so that Ãv m = µ m v m .

(2.14)

We deduce that the eigenvalues of (2.11) are given by

λ m = 1 µ m .
We get from (2.12) and (2.14)

a(v m , v) = λ m a( Av m , v) = λ m (v m |v) for any v ∈ V. Let w m = λ 1/2 m v m .
We check that (w m ) form an Hilbertian orthonormal basis of H. We first observe that, from (2.12), we have

(w m |w ) = 1 λ m a(w m , w ) = λ λ m a(v m , v ) = δ m .
On the other hand, if u ∈ H satisfies (u|w m ) = 0, for any m ≥ 1, then (u|v) = 0 for any v ∈ V because (v m ) form an Hilbertian orthonormal basis of V . Next, using the fact that V is dense in H to deduce that u = 0 and hence (w m ) form an orthonormal basis of H.

Remark 2.1. We can weaken the V -ellipticity condition in Theorem 2.10. Precisely, we can substitute the V -ellipticity condition by the following one : there exist α > 0 and λ ∈ R so that

a(v, v) + λ |v| 2 ≥ α v 2 for any v ∈ V.
In that case, (u, v) → a(u, v) + λ (u|v) possesses the assumptions of Theorem 2.10.

Under the assumptions and notations of Theorem 2.10, as (w m ) is an orthonormal basis of H, we have, for any u ∈ H,

u = ∑ m≥1 (u, w m )w m (2.15)
and

|u| 2 = ∑ m≥1 (u, w m ) 2 .
(2.16) Also, as

(v m ) = (λ -1/2 m w m
) is an orthonormal basis of V endowed with the scalar product a(•, •), we have for each

v ∈ V a(v, v) = ∑ m≥1 a(v, v m ) 2 = ∑ m≥1 λ -1 m a(v, w m ) 2 .
Hence we get from (2.12)

a(v, v) = ∑ m≥1 λ m (v|w m ) 2 .
(2.17)

We now give a characterization of the eigenvalues. Define the Rayleigh quotient by

R(v) = a(v, v) |v| 2 , v ∈ V, v = 0. (2.18)
We obtain from (2.17)

R(w m ) = λ m , m ≥ 1. (2.19)
For v = ∑ i≥1 α i w i , a non zero element in V , we have by virtue of (2.19)

R(v) = ∑ i≥1 λ i α 2 i ∑ i≥1 α 2 i ≥ λ 1 .
We deduce from this the following characterization of the first eigenvalue:

λ 1 = min v∈V, v =0 R(v).
Let V m be the subspace of V spanned by the eigenvectors w 1 , . . . , w m and let V ⊥ m the orthogonal of V m in V with respect to the scalar product a(•, •), i.e.

V ⊥ m = {v ∈ V ; a(v, w i ) = 0, 1 ≤ i ≤ m}.
Note that we have also

V ⊥ m = {v ∈ V ; (v|w i ) = 0, 1 ≤ i ≤ m}. If v = ∑ i≥1 α i w i ∈ V ⊥ m-1 then α i = 0, 1 ≤ i ≤ m -1. Thus, R(v) = ∑ i≥m λ i α 2 i ∑ i≥m α 2 i ≥ λ m ,
which entails in light of (2.19)

λ m = min v∈V ⊥ m-1 , v =0 R(v).
(2.20)

In fact we have another useful characterization of the eigenvalues λ m as shows the following theorem.

Theorem 2.11. (Min-max principle) Under the assumption of Theorem 2.10, we have

λ m = min E m ∈V m max v∈E m , v =0 R(v),
where V m is the set of all subspaces E m of V of dimension m.

Proof.

If E m = V m then, for v = ∑ m i=1 α i w i = 0, we have R(v) = ∑ m i=1 λ i α 2 i ∑ m i=1 α 2 i ≤ λ m ,
which implies by virtue of (2.20)

max v∈V m , v =0 R(v) = λ m . We claim that λ m ≤ max v∈E m , v =0 R(v)
for any E m ∈ V m . Indeed, we can choose v ∈ E m , v = 0 in such a way that

(v|w i ) = 0, 1 ≤ i ≤ m -1, i.e. v ∈ E m ∩ V ⊥ m-1 .
We obtain from (2.20) that λ m ≤ R(v). This completes the proof.

Variational solutions for model problems

Variational solutions

We show by a simple model how to build a variational formulation associated to a boundary value problem.

Let Ω be a bounded domain of R n of class C 1 with boundary Γ . If f ∈ L 2 (Ω ) we consider the problem of finding a function u defined on Ω and satisfying boundary value problem

-∆ u = f in Ω , (2.21) u = 0 on Γ . (2.22)
Assume that (2.21)-(2.22) admits a solution u ∈ H 2 (Ω ). We multiply each side of (2.21) by ϕ ∈ D(Ω ) and then we integrate over Ω . We obtain

- Ω ∆ uϕdx = Ω f ϕdx.
But the divergence theorem yields

Ω div(ϕ∇u)dx = Ω ∆ uϕdx + Ω ∇u • ∇ϕdx = 0. Hence Ω ∇u • ∇ϕdx = Ω f ϕdx. As D(Ω ) is dense H 1 0 (Ω ), we deduce that Ω ∇u • ∇vdx = Ω f vdx for any v ∈ H 1 0 (Ω ). ( 2
Ω ∇u • ∇ϕdx = Ω f ϕdx for any ϕ ∈ D(Ω ).
We get from the definition of weak derivatives

Ω -∆ uϕdx = Ω f ϕdx, for any ϕ ∈ D(Ω ),
and hence -∆ u = f by the cancellation theorem. Therefore u is the solution of (2.21) and (2.22). Consider now the Neumann boundary value problem

-∆ u + u = f in Ω , (2.24)
∂ ν u = 0 on Γ , (2.25)
where ∂ ν u = ∇u • ν denotes the derivative along the unit exterior normal vector ν.

If u ∈ H 2 (Ω ) is a solution of (2.24) and (2.25), we multiply each side of (2.24) by v ∈ H 1 (Ω ) and then we integrate over Ω . In light of the boundary condition (2.25), the divergence theorem gives

Ω ∇u • ∇vdx + Ω uvdx = Ω f vdx for any v ∈ H 1 (Ω ).
(2.26)

As before we substitute the boundary value problem (2.24) and (2.25) by the following one:

for f ∈ L 2 (Ω ) find u ∈ H 1 (Ω ) satisfying (2.26). Conversely, if u ∈ H 1 (Ω ) is a solution of (2.26) then Ω ∇u • ∇ϕdx + Ω uvdx = Ω f ϕdx for any ϕ ∈ D(Ω ).
If we admit that u, the solution of this variational problem (2.26), belongs to H 2 (Ω ) then as in the Dirichlet case we prove that u satisfies (2.24). On the other hand, we get by choosing ϕ ∈ D(Ω ) in (2.26)

Γ ∂ ν uϕ = 0, for any ϕ ∈ D(Ω ),
and admitting also that

D(Γ ) = {ψ = ϕ| Γ ; ϕ ∈ D(R n )} is dense in L 2 (Γ ), we obtain Γ ∂ ν uw = 0, for any w ∈ L 2 (Γ ),
implying that (2.25) holds. We now give a general framework that generalize the previous two examples. Let Ω be a bounded domain of R n and let V be a closed subspace of H 1 (Ω ) satisfying

H 1 0 (Ω ) ⊆ V ⊆ H 1 (Ω ).
Therefore V is Hilbert space when its is endowed with the norm of H 1 (Ω ).

Pick a i j ∈ L ∞ (Ω ), 1 ≤ i, j ≤ n, a 0 ∈ L ∞ (Ω ) and set a(u, v) = Ω n ∑ i, j=1
a i j ∂ j u∂ i v + a 0 uv dx.

If A = (a i j ) then the last identity takes the form

a(u, v) = Ω (A ∇u • ∇v + a 0 uv) dx. Simple computations show |a(u, v)| ≤ C u H 1 (Ω ) v H 1 (Ω ) ,
where

C = max i j a i j L ∞ (Ω ) + a 0 L ∞ (Ω ) . That is the bilinear form a is continuous on H 1 (Ω ) × H 1 (Ω ).
Assume moreover that the following ellipticity condition holds : there exists α > 0 so that (A ξ , ξ ) ≥ α|ξ | 2 a.e. in Ω , for any ξ ∈ R n .

We also assume that there exists α 0 > 0 so that a 0 ≥ α 0 a.e. in Ω .

Under these assumptions, we have

a(u, u) ≥ α ∇u 2 2 + α 0 u 2 2 ≥ min(α, α 0 ) u 2 V ,
in such a way that a is V -elliptic. Pick f ∈ L 2 (Ω ) and set

Φ(v) = Ω f vdx. The linear form v → Φ(v) is continuous in L 2 (Ω ) and therefore it is also continuous in V .
As a is continuous and V -elliptic, we find by applying Lax-Milgram's lemma a unique u ∈ V satisfying a(u, v) = Φ(v) for any v ∈ V.

(2.27)

Let us interpret the problem we just solved. We make the extra assumption that the solution of (2.27) belongs to H 2 (Ω ). In light of the definition of weak derivatives, we have

Ω (Lu -f )ϕ = 0 for any ϕ ∈ D(Ω ).
Here L is the differential operator with variable coefficients which is given as follows

Lw = - n ∑ i, j=1 ∂ i (a i j ∂ j w) + a 0 w.

Then the cancellation theorem yields

Lu = f a.e. in Ω .

Summing up we get that the solution (2.27) satisfies the following conditions:

u ∈ V, (2.28) Lu = f a.e. in Ω , (2.29) a(u, v) = Ω Luvdx for any v ∈ V. (2.30)
Conversely, we see immediately that if u is a solution of (2.28)-(2.30) then u is also a solution of (2.27). In other words, the solution of (2.27) is characterized by (2.28)-(2.30).

We return back to the two examples we discussed in the beginning of this subsection. We choose V = H 1 0 (Ω ) endowed with the equivalent norm w V = ∇w L 2 (Ω ,R n ) (follows from Poincaré's inequality). Then we have that (2.23) admits a unique solution u ∈ H 1 0 (Ω ). Similarly, taking V = H 1 (Ω ) equipped with the the norm of H 1 (Ω ), we get that (2.26) has a unique solution u ∈ H 1 (Ω ).

We end this subsection by a spectral problem associated to the elliptic operator L:

Lu = λ u.

In addition of the previous assumptions on a, we assume that the matrix A (x) = (a i j (x)) is symmetric for a.e. x ∈ Ω . We obtain by applying Theorem 2.10 a non decreasing sequence (λ m ), λ m → ∞, and an orthonormal basis (w m ) of L 2 (Ω ), w n ∈ V for each m, consisting in eigenvectors so that

a(w m , v) = λ m (w m |v) for any v ∈ V.
(2.31)

Here (•|•) is the usual scalar product of L 2 (Ω ).
As we have done for (2.27), we show that the solution of (2.31) is characterized by

w m ∈ V, Lw m = λ m w m a.e. in Ω , a(w m , v) = Ω Lw m vdx for any v ∈ V.

H 2 -regularity of variational solutions

We limit our study to the Dirichlet problem. The following proposition will be useful in the sequel.

Proposition 2.6. Let Ω be an open subset R n , 1 < p ≤ ∞ and u ∈ L p (Ω ). The fol- lowing properties are equivalent. (i) u ∈ W 1,p (

Ω ).

(ii) There exits a constant C > 0 so that

Ω u∂ i ϕdx ≤ C ϕ L p (Ω ) for any ϕ ∈ D(Ω ), i = 1, . . . n.
(iii) There exists a constant C > 0 so that, for any ω Ω and any h ∈ R n with |h| < dist(ω, Ω c ), we have

τ h u -u L p (ω) ≤ C|h|.
Furthermore, we can take C = ∇u L p (Ω ) n in (ii) and (iii).

Proof. It is straightforward to check that (i) entails (ii).

Let us prove that (ii) implies (i). By assumption the linear form

Φ : D(Ω ) → Ω u∂ i ϕ
is continuous when D(Ω ) is endowed with the norm of L p (Ω ). Therefore, we can extend Φ by density to a continuous linear form, still denoted by Φ, on L p (Ω ). From Riesz's representation theorem there exists v i ∈ L p (Ω ) so that

Φ, ϕ = Ω v i ϕdx for any ϕ ∈ D(Ω ), 1 ≤ i ≤ n. Whence u ∈ W 1,p (Ω ).
Next, we proceed to the proof of (i) entails (iii

). Let us first consider u ∈ D(R n ). If h ∈ R n then u(x + h) -u(x) = 1 0 ∇u(x + th) • hdt. Hence |τ h u(x) -u(x)| p ≤ |h| p 1 0 |∇u(x + th)| p dt and then ω |τ h u(x) -u(x)| p dx ≤ |h| p ω dx 1 0 |∇u(x + th)| p dt ≤ |h| p 1 0 dt ω |∇u(x + th)| p dx ≤ |h| p 1 0 dt ω+th |∇u(y)| p dy. Fix |h| < dist(ω, Ω c ). Then there exists ω Ω such that ω + th ⊂ ω for any t ∈ [0, 1]. Thus τ h u -u p L p (ω) ≤ |h| p ω |∇u| p dx. (2.32) If u ∈ W 1,p (Ω ), p = ∞, by Friedrichs's theorem (see Exercise 1.5), there exists (u m ) a sequence in D(R n ) so that u m → u in L p (Ω )
and ∇u m → ∇u dans L p (ω, R n ) for any ω Ω . Apply (2.32) to u m and pass to the limit, when m → ∞ to get (iii). When p = ∞, we apply the case p < ∞ and then we pass to the limit when p → ∞.

We complete the proof by showing that (iii) implies (ii). Take then ω so that supp(ϕ)

⊂ ω Ω . Let h ∈ R n satisfying |h| < dist(ω, Ω c ). Then (iii) yields Ω (τ h u -u)ϕdx ≤ C|h| ϕ L p (Ω ) . But Ω (τ h u -u)ϕdx = Ω u(τ -h ϕ -ϕ)dx. Thus Ω u τ -h ϕ -ϕ h dx ≤ C ϕ L p (Ω ) .
We derive then (ii) by choosing h = te i , t ∈ R, and passing to the limit as t goes to 0.

We use the following definition of an open set of class C k (see comments in Chapter 1 for an equivalent definition). Define

R n + = {x = (x , x n ) ∈ R n ; x n > 0}, Q = {x = (x , x n ) ∈ R n ; |x | < 1 and |x n | < 1}, Q + = Q ∩ R n + , Q 0 = {x = (x , x n ) ∈ R n ; |x | < 1 and x n = 0}. Recall that Ω is of class C k , k ≥ 1 is an integer, if for any x ∈ Γ = ∂ Ω there exists U a neighborhood of x dans R n and a bijective mapping H : Q → U so that H ∈ C k (Q), H -1 ∈ C k (U), H(Q + ) = U ∩ Ω , H(Q 0 ) = U ∩ Γ . Theorem 2.12. Let Ω be an open subset of R n of class C 2 with bounded boundary Γ or else Ω = R n + . For f ∈ L 2 (Ω ), let u ∈ H 1 0 (Ω ) satisfies Ω ∇u • ∇v + Ω uv = Ω f v for any v ∈ H 1 0 (Ω ).
(2.33)

Then u ∈ H 2 (Ω ) and u H 2 (Ω ) ≤ C f L 2 (Ω ) ,
where the constant C only depends on Ω .

Proof. The proof consists in several steps. We first consider the case Ω = R n and then the case Ω = R n + . For the general case the H 2 interior regularity is obtained from the case Ω = R n while the regularity at the boundary is deduced from that of the case Ω = R n + by using local cards and a partition of unity.

• The case Ω = R n . For h ∈ R n , h = 0, set d h u = τ h u -u |h| .
That is

(d h u)(x) = τ h u(x) -u(x) |h| .
We find by taking in (2.33

) v = d -h d h u R n |∇d h u| 2 dx + R n |d h u| 2 dx = R n f d -h d h u.
Note that we have use the fact that

R n d -h w 1 w 2 dx = R n w 1 d h w 2 dx. Therefore d h u 2 H 1 (R n ) ≤ f L 2 (R n ) d -h d h u L 2 (R n ) . (2.34)
On the other hand,

d -h v L 2 (R n ) ≤ ∇v L 2 (R n ,R n ) for all v ∈ H 1 (R n ). (2.35)
To see this, we recall (see Proposition 2.6) that

d -h v L 2 (ω) ≤ ∇v L 2 (R n ,R n ) for all ω R n and h ∈ R n ,
d h u 2 H 1 (R n ) ≤ f L 2 (R n ) d h u H 1 (R n )
and consequently

d h u H 1 (R n ) ≤ f L 2 (R n ) .
In particular,

d h ∂ j u L 2 (R n ) ≤ f L 2 (R n ) , j = 1, . . . , n.
By Proposition 2.6 we deduce that

∂ j u ∈ H 1 (R n ) and then u ∈ H 2 (R n ). • The case Ω = R n + .
We use again translations but in the present case only in the tangential directions. That is directions of the form h ∈ R n-1 × {0}. In this case, we say that h is parallel to the boundary and we write h Γ . Note that

u ∈ H 1 0 (Ω ) =⇒ τ h u ∈ H 1 0 (Ω ) for any h Γ , which means that H 1 0 (Ω ) is invariant under tangential translations. Let h Γ . We get by taking v = d -h (d h u) in (2.33) Ω |∇d h u| 2 dx + Ω |d h u| 2 dx = Ω f d -h d h u. That is d h u 2 H 1 (Ω ) ≤ f L 2 (Ω ) d -h d h u L 2 (Ω ) .
(2.36)

We shall need the following lemma.

Lemma 2.4. We have

d h v 2 L 2 (Ω ) ≤ ∇v 2 L 2 (Ω ,R n ) for any v ∈ H 1 (Ω ) and h Γ . Proof. If v ∈ D(R n + )
, using that Ω + th = Ω for any t ∈ R and h Γ , we have similarly to the proof of Proposition 2.6

d h u 2 L 2 (Ω ) ≤ ∇u 2 L 2 (Ω ,R n ) for any h Γ .
The expected inequality follows since

D(R n + ) is dense in H 1 (Ω ) (see the proof of Proposition 1.11).
In light of the inequality in Lemma 2.4 and (2.36), it follows

d h u 2 H 1 (Ω ) ≤ f L 2 (Ω ) for any h Γ . (2.37) Let 1 ≤ j ≤ n, 1 ≤ k ≤ n -1, h = |h|e k and ϕ ∈ D(Ω ). We have Ω d h (∂ j u)ϕdx = - Ω ud -h (∂ j ϕ)dx
and by (2.37) we obtain

Ω d h (∂ j u)ϕdx ≤ f L 2 (Ω ) ϕ L 2 (Ω ) .
We then get by passing to the limit, when h → 0,

Ω u∂ 2 jk ϕdx ≤ f L 2 (Ω ) ϕ L 2 (Ω ) , 1 ≤ j ≤ n, 1 ≤ k ≤ n -1. (2.38)
Let us finally prove that

Ω u∂ 2 n ϕdx ≤ f L 2 (Ω ) ϕ L 2 (Ω ) for any ϕ ∈ D(Ω ).
(2.39)

We deduce directly from equation (2.33) and (2.38) that

Ω u∂ 2 n ϕdx ≤ n-1 ∑ i=1 Ω u∂ 2 i ϕdx + Ω ( f -u)ϕdx ≤ C f L 2 (Ω ) ϕ L 2 (Ω ) .
Inequalities (2.38) and (2.39) give

Ω u∂ jk ϕdx ≤ C f L 2 (Ω ) ϕ L 2 (Ω ) , 1 ≤ j, k ≤ n, for all ϕ ∈ D(Ω ).
Whence u ∈ H 2 (Ω ). Note that, by Hahn-Banach's extension theorem and Riesz-Fréchet's representation theorem, there exist

f jk ∈ L 2 (Ω ) so that Ω u∂ jk ϕdx = Ω f jk ϕ for any ϕ ∈ D(Ω ).
• General case. For simplicity convenience, we assume that

Ω is bounded. As Ω is of class C 2 , there exist U i , 1 ≤ i ≤ k,
an open subset of R n and a bijective mapping

H i : Q → U i so that H i ∈ C 2 (Q), H -1 i ∈ C 2 (U i ), H i (Q + ) = U i ∩ Ω , H i (Q 0 ) = U i ∩ Γ (2.40)
and Γ ⊂ k i=1 U i . Let θ 0 , . . . θ k be a partition of unity so that

θ i ∈ D(U i ), 1 ≤ i ≤ k θ 0 ∈ C ∞ (R n ), supp(θ 0 ) ⊂ R n \ Γ 0 ≤ θ i ≤ 1, 0 ≤ i ≤ k and k ∑ i=0 θ i = 1 in R n . Since Ω is bounded, we have θ 0 | Ω ∈ D(Ω ). Write u = ∑ k i=0 θ i u and let us first check that θ 0 u ∈ H 2 (Ω ) (interior regularity). As θ 0 | Ω ∈ D(Ω ), θ 0 u extended by 0 outside Ω belongs to H 1 (R n ). Simple compu- tations show that θ 0 u is the variational solution in R n of the equation -∆ (θ 0 u) + θ 0 u = θ 0 f -2∇θ 0 • ∇u -(∆ θ 0 )u = g, with g ∈ L 2 (R n ). We get by applying the case Ω = R n that θ 0 u ∈ H 2 (R n ) and θ 0 u H 2 (R n ) ≤ C f L 2 (Ω ) + u H 1 (Ω ) . Thus θ 0 u H 2 (R n ) ≤ C f L 2 (Ω ) because u H 1 (Ω ) ≤ f L 2 (Ω ) by (2.33). Next, we prove that θ i u ∈ H 2 (Ω ), 1 ≤ i ≤ k (boundary regularity). Fix i, 1 ≤ i ≤ k,
and, for simplicity convenience, we use the notations

θ = θ i , H = H i et U = U i
, where H i and U i are the same as in (2.40). We write x = H(y) and then y =

H -1 (x) = J(x). Since θ ∈ D(U) and v = θ u ∈ H 1 0 (Ω ∩ U), we easily check that v is the variational solution in Ω ∩U of the equation -∆ v + v = θ f -2∇θ • ∇u -(∆ θ )u = g, where g ∈ L 2 (R n ) and g L 2 (Ω ∩U) ≤ C f L 2 (Ω )
. Precisely, we have

Ω ∩U ∇v • ∇ϕdx = Ω ∩U gϕdx, for any ϕ ∈ H 1 0 (Ω ∩U).
(2.41)

We now make a change of variable in order to transform v| Ω ∩U to a function defined on Q + . For doing that, we set

w(y) = v(H(y)), y ∈ Q + , or equivalently w(J(x)) = v(x), x ∈ Ω ∩U.
We use the following lemma, whose proof is given later, to convert (2.41) to a variational problem in Q + .

Lemma 2.5. Under the notations above, we have w ∈

H 1 0 (Q + ) and n ∑ k, =1 Q + a k ∂ k w∂ ψdy = Q + gψdy for all ψ ∈ H 1 0 (Q + ), (2.42
)

where g = (g • H)|Jac(H)| ∈ L 2 (Q + ) and the functions a k ∈ C 1 (Q + ) satisfy the ellipticity condition n ∑ k, =1 a k (y)ξ k ξ l ≥ α|ξ | 2 , for any y ∈ Q + , ξ ∈ R n ,
for some constant α > 0.

We prove that w ∈ H

2 (Q + ) and w H 2 (Q + ) ≤ C g L 2 (Ω )
. This will imply that

θ u ∈ H 2 (Ω ∩U) and then θ u ∈ H 2 (Ω ). Moreover, θ u H 2 (Ω ) ≤ C f L 2 (Ω ) .
As in the case Ω = R n + we use tangential translations. We choose in (2.42)

ψ = d -h (d h w) with h Q 0 recall that supp(w) ⊂ {(x , x n ); |x | < 1 -δ , 0 < x n < 1 -δ } for some δ > 0 . We get ∑ k, Q + d h (a k ∂ k w)∂ (d h w)dx = Q + gd -h (d h w)dx. But Q + gd -h (d h w)dx ≤ g L 2 (Q + ) ∇(d h w) L 2 (Q + ,R n ) by Lemma 2.4. Hence ∑ k, Q + d h (a k ∂ k w)∂ (d h w)dx ≤ g L 2 (Q + ) ∇(d h w) L 2 (Q + ,R n ) .
(2.43)

On the other hand,

d h (a k ∂ k w)(y) = a k (y + h)∂ k (d h w)(y) + (d h a k )(y)∂ k w(y)
and hence

∑ k, Q + d h (a kl ∂ k w)∂ (d h w)dx ≥ α ∇(d h w) 2 L 2 (Q + ,R n ) (2.44) -C w H 1 (Q + ) ∇(d h w) L 2 (Q + ,R n ) .
A combination of (2.43) and (2.44) yields

∇(d h w) L 2 (Q + ) n ≤ C( w H 1 (Q + ) + g L 2 (Q + ) ),
and since w H 1 (Q + ) ≤ C g L 2 (Q + ) (a consequence of (2.42)), we have

∇(d h w) L 2 (Q + ,R n ) ≤ C g L 2 (Q + ) .
(2.45)

Similarly to the case Ω = R n + , we deduce from (2.45)

Q + ∂ k w∂ ψdx ≤ C g L 2 (Q + ) ψ L 2 (Q + ) (2.46) for any ψ ∈ C 1 c (Q + ), (k, ) = (n, n). To complete the proof of w ∈ H 2 (Q + ) and w H 2 (Q + ) ≤ C g L 2 (Q + ) it is sufficient to check that Q + ∂ n w∂ n ψdx ≤ C g L 2 (Q + ) ψ L 2 (Q + ) for any ψ ∈ C 1 c (Q + ). (2.47)
We apply (2.42) in which we substitute ψ ∈ C 1 c (Q + ) by ψ/a nn , where we note that

a nn > α > 0. Whence Q + a nn ∂ n w∂ n ψ a nn dx = Q + gψ a nn -∑ (k, ) =(n,n) a k ∂ k w∂ ψ a nn dx. Then Q + ∂ n w∂ n ψdx = Q + ∂ n a nn a nn ∂ n wψdx + Q + gψ a nn dx + ∑ (k, ) =(n,n) ∂ k w∂ l a k ψ a nn dx -∑ (k, ) =(n,n) Q + ∂ k w∂ a k ψ a nn dx.
This identity together with (2.46), in which we substituted ψ by a k ψ/a nn , yield

Q + ∂ n w∂ n ψdx ≤ C( w H 1 (Q + ) + g L 2 (Q + ) ψ L 2 (Q + ) .
This gives (2.47).

Proof (of Lemma 2.5). Let ψ ∈ H 1 0 (Q + ) and set ϕ(x) = ψ(J(x)), x ∈ Ω ∩U. Then ϕ ∈ H 1 0 (Ω ∩U) and ∂ j v(x) = ∑ k ∂ k w(J(x))∂ j J k (x), ∂ j ϕ(x) = ∑ ∂ ψ(J(x))∂ j J (x).
Thus,

Ω ∩U ∇v(x) • ∇ϕ(x)dx = Ω ∩U n ∑ j,k, =1 ∂ j J k (x)∂ j J (x)∂ k w(J(x))∂ ψ(J(x))dx = Q + n ∑ j,k, =1 ∂ j J k (H(y))∂ j J (H(y))∂ k w(y)∂ ψ(y)|Jac H(y)|dy
by the classical formula of change of variable. Therefore

Ω ∩U ∇v(x) • ∇ϕ(x)dx = Q + n ∑ k, =1 a k ∂ k w(y)∂ ψ(y)dy, (2.48)
where

a k (y) = n ∑ j=1 ∂ j J k (H(y))∂ j J (H(y))|Jac H(y)|.
Note that a k ∈ C 1 (Q + ) and the ellipticity condition fulfills

min ξ ∈R n , |ξ |=1 n ∑ k, =1 a k (y)ξ k ξ = min ξ ∈R n , |ξ |=1 |JacH(y)| n ∑ j=1 n ∑ k=1 ∂ j J k (H(y))ξ k 2 ≥ α,
with α > 0, because the Jacobian matrices Jac H and Jac J are non singular.

On the other hand, we have

Ω ∩U g(x)ϕ(x)dx = Q + g(H(y))ψ(y)|Jac H(y)|dy.
(2.49)

A combination of (2.48), (2.49) and (2.41) then implies (2.42) and completes the proof.

Maximum principle

In this subsection, Ω is an arbitrary open subset of R n . As we have mentioned in Chapter 1, we use inf Ω f et sup Ω f respectively for inf ess Ω f and sup ess Ω f .

Theorem 2.13. Let f ∈ L 2 (Ω ) and u ∈ H 1 (Ω ) ∩C(Ω ) so that Ω ∇u • ∇ϕ + Ω uϕ = Ω f ϕ for any ϕ ∈ H 1 0 (Ω ).
(2.50)

Then min{inf Γ u, inf Ω f } ≤ u(x) ≤ max{sup Γ u, sup Ω f } for any x ∈ Ω . Remark 2.2. When Ω is of class C 1 the condition u ∈ C(Ω )
is unnecessary because in that case the trace u| Γ is well defined as an element of L 2 (Γ ). Also, when u ∈ H 1 0 (Ω ) we can drop the condition u ∈ C(Ω ) (we refer to the maximum principle in the next section).

Proof (of Theorem 2.13). We use Stampacchia

's truncation method. Fix G ∈ C 1 (R) so that (i) |G (s)| ≤ M for any s ∈ R, (ii) G is increasing in ]0, +∞[, (iii) G(s) = 0 for any s ≤ 0. Let K = max{sup Γ u, sup Ω f }. If K = +∞ there is nothing to prove. Assume then that K < +∞ and let v = G(u -K).
We consider separately two cases :

(a) |Ω | < ∞ and (b) |Ω | = ∞.
In case (a), we apply Proposition 1.10 to

f (t) = G(t -K) -G(-K) to deduce that v ∈ H 1 (Ω ). On the other hand, v ∈ H 1 0 (Ω ) because v ∈ C(Ω ) and v = 0 on Γ . We obtain by taking v as test function in (2.50) Ω G (u -K)|∇u| 2 dx + Ω G(u -K)udx = Ω f G(u -K)dx and hence Ω G (u -K)|∇u| 2 dx + Ω G(u -K)(u -K)dx = Ω ( f -K)G(u -K)dx. But f -K ≤ 0, G(u -K) ≥ 0 and G (u -K) ≥ 0. Whence Ω G(u -K)(u -K)dx ≤ 0.
The assumptions on G yield tG(t) ≥ 0 for any t ∈ R. Then the last inequality implies (u -K)G(u -K) = 0 a.e. in Ω and consequently u ≤ K a.e. in Ω .

We complete the proof of (a) by repeating the precedent analysis with u substituted by -u.

We now proceed to the proof of case (b). Note that in this case we have necessarily

K ≥ 0. Otherwise K < 0 would imply that | f (x)| ≥ K a.e. in Ω because f (x) ≤ K = -|K| a.e. in Ω which is impossible since f ∈ L 2 (Ω ) and |Ω | = ∞. Fix K > K and set v = G(u -K).
Similarly to the precedent case, we check that v ∈ H 1 (Ω ) ∩C(Ω ) and v = 0 on Γ . In particular, v ∈ H 1 0 (Ω ). Taking v as test function in (2.50), we obtain

Ω G (u -K)|∇u| 2 dx + Ω G(u -K)udx = Ω f G(u -K)dx.
(2.51) But, as -K < 0, we have G(-K) = 0 and hence

G(u -K) = G(u -K) -G(-K) ≤ M|u|.
On the other hand,

Ω KG(u -K)dx = [u≥ K] KG(u -K)dx ≤ M [u≥ K] K|u| ≤ K Ω |u| 2 , where u ≥ K = {x ∈ Ω ; u(x) ≥ K}. Therefore G(u -K) ∈ L 1 (Ω ).
We get from (2.51) that

Ω (u -K)G(u -K)dx ≤ Ω ( f -K)G(u -K)dx ≤ 0.
Thus u ≤ K a.e. in Ω and then u ≤ K a.e. in Ω because K > K was chosen arbitrarily.

Corollary 2.4. Let f ∈ L 2 (Ω ) and u ∈ H 1 (Ω ) ∩C(Ω ) satisfying (2.50). Then (u ≥ 0 on Γ ) and ( f ≥ 0 in Ω ) =⇒ (u ≥ 0 in Ω ) u L ∞ (Ω ) ≤ max u L ∞ (Γ ) , f L ∞ (Ω ) . In particular, u L ∞ (Ω ) ≤ u L ∞ (Γ ) if f = 0 and u L ∞ (Ω ) ≤ f L ∞ (Ω ) if u = 0 on Γ .

Uniqueness of continuation across a non characteristic hypersurface

We first establish a Carleman inequality with a convex weight.

In this subsection the gradient and the Laplace operator with respect to the variable x ∈ R n-1 are denoted respectively by ∇ and ∆ .

Let θ = θ (x ), x ∈ R n-1 , be a C 4 function defined on a neighborhood of the origin satisfying θ (0) = 0 and ∇ θ (0

) = 0. Set ϕ(x , x n ) = (x n -1) 2 + |x | 2 , (x , x n ) ∈ R n-1 × R
and consider the partial differential operator P = P 0 + P 1 with

P 0 = ∆ + 1 + |∇ θ | 2 ∂ 2 n + 2∇ θ • ∇ ∂ n .
Let P 1 be a first order partial differential operator of the variable x ∈ R n , i.e.

P 1 = n ∑ i=1 b i (x)∂ i + c(x),
where b i and c are measurable and bounded functions in a neighborhood of the origin.

Theorem 2.14. There exists a neighborhood U of 0 in R n and two constants τ 0 > 0 and C > 0 so that

U e 2τϕ (Pu) 2 dx ≥ C τ U e 2τϕ |∇u| 2 dx + τ 3 U e 2τϕ u 2 dx (2.52)
for all τ ≥ τ 0 and u ∈ H 2 0 (U ).

Proof. Introduce the operator L = e τϕ P 0 e -τϕ . Then straightforward computations show that

L = ∑ i, j a i j ∂ i j -2τB • ∇ + τc 1 + τ 2 c 2 ,
where

(a i j ) = I n-1 ∇ θ t ∇ θ 1 + |∇ θ | 2 I n-1 denotes the identity matrix of R n-1 , B = ∇ ϕ + ∂ n ϕ∇ θ ∂ n ϕ(1 + |∇ θ | 2 ) + ∇ θ • ∇ ϕ , c 1 = -∆ ϕ -1 + |∇ θ | 2 ∂ 2 n ϕ, c 2 = |∇ ϕ| 2 + 1 + |∇ θ | 2 (∂ n ϕ) 2 + 2∂ n ϕ∇ ϕ • ∇ θ .
It is not hard to check that the formal adjoint of L si given by 7

L * = n ∑ i, j=1 a i j ∂ 2 i j + 2 n ∑ i, j=1 (∂ i a i j )∂ j + 2τB • ∇ + n ∑ i, j=1 ∂ i j a i j + 2τdivB + τc 1 + τ 2 c 2 .
Therefore, the self-adjoint and skew-adjoint parts 8 L + and L -of L are respectively given as follows

L + = n ∑ i, j=1 a i j ∂ 2 i j + n ∑ i, j=1 ∂ i a i j ∂ j + 1 2 n ∑ i, j=1 ∂ 2 i j a i j + τdivB + τc 1 + τ 2 c 2 , L -= - n ∑ i, j=1 ∂ i a i j ∂ j -2τB • ∇ -τdivB - 1 2 n ∑ i, j=1 ∂ 2 i j a i j .
For τ > 0, we introduce the new notations

B + = (B + 1 , . . . , B + n ), B + j = n ∑ i=1 ∂ i a i j , B - τ = (B - τ,1 , . . . , B - τ,n ), B - τ, j = -2B j - 1 τ ∑ i ∂ i a i j , c + τ = c 2 + 1 τ (c 1 + divB) - 1 2τ 2 n ∑ i, j=1 ∂ 2 i j a i j , c - τ = -divB - 1 2τ n ∑ i, j=1 ∂ 2 i j a i j .
With these new notations, L + and L -take the form

L + = n ∑ i, j=1 a i j ∂ 2 i j + B + • ∇ + τ 2 c + τ , L -= τB - τ • ∇ + τc - τ .
Pick v an arbitrary C ∞ function with a compact support in a neighborhood of 0. If [•, •] denotes the usual commutator [START_REF] Raviart | Introduction à l'analyse numérique des équations aux dérivées partielles[END_REF] we have the following decomposition [START_REF] Ladyzhenskaja | Linear and quasilinear elliptic equations[END_REF] Recall that the formal adjoint of L, denoted by L * , is determined according to the relation

Luvdx = uL * vdx for all C ∞ compactly supported functions u and v. 8 Recall that X is self-adjoint if X = X * ; it is skew-adjoint if X = -X * . 9 If X and Y are two operators then [X,Y ] = XY -Y X. v[L + , L -]v = 6 ∑ i=1 I i ,
where

I 1 = τ v n ∑ i, j=1 a i j ∂ 2 i j , B - τ • ∇ vdx, I 2 = τ v n ∑ i, j=1 a i j ∂ 2 i j , c - τ vdx, I 3 = τ v B + • ∇, B - τ • ∇ vdx, I 4 = τ v B + • ∇, c - τ vdx, I 5 = τ 3 v c + τ , B - τ • ∇ vdx, I 6 = τ 3 v c + τ , c - τ vdx.
We have

I 6 = 0. (2.53)
Until the end of this proof C denotes a generic constant independent of τ.

As [B + • ∇, c - τ ] = B + • ∇c - τ , we obtain I 4 ≥ -τ v 2 dx if τ 1.
(2.54)

Here and henceforth the notation τ 1 means that τ is sufficiently large. For I 2 , we get by making integrations by parts that

I 2 ≥ -C τ 2 v 2 + |∇v| 2 dx if τ 1.
(2.55)

Making once again integrations by parts, we find

I 3 = τ v B + • ∇, B - τ • ∇ vdx = τ v B + • ∇B - τ -B - τ • ∇B + • ∇vdx = τ 2 B + • ∇B - τ -B - τ • ∇B + • ∇v 2 dx = - τ 2 div B + • ∇B - τ -B - τ • DB + v 2 dx. Hence I 3 ≥ -Cτ v 2 dx if τ 1.
(2.56)

For I 5 , we first compute -B - τ • ∇c + τ near (x , x n ) = 0. We obtain -B - τ • ∇c + τ |(x ,x n )=0 = 16 + O 1 τ if τ 1.
Since

I 5 = τ 3 v c + τ , B - τ • ∇ vdx = -τ 3 B - τ • ∇c + τ v 2 ,
we deduce

I 5 ≥ Cτ 3 v 2 dx if τ
1 and v has a small support.

(2.57)

Next, we estimate I 1 . In light of the identity

n ∑ i, j=1 a i j ∂ 2 i j , B - τ • ∇ = n ∑ i, j,k=1 a i j ∂ 2 i j B - τ,k ∂ k + 2 n ∑ i, j,k=1 a i j ∂ i B - τ,k ∂ 2 k j - n ∑ i, j,k=1 B - τ,k ∂ k a i j ∂ 2 i j
we can split I 1 into three terms :

I 1 = τ(J 1 + J 2 + J 3 ),
where

J 1 = n ∑ i, j,k=1 a i j ∂ 2 i j B - τ,k ∂ k vvdx, J 2 = 2 n ∑ i, j,k=1 a i j ∂ i B - τ,k ∂ k j vvdx, J 3 = - n ∑ i, j,k=1 B - τ,k ∂ k a i j ∂ 2 i j vvdx.
We have

J 1 = n ∑ i, j,k=1 a i j ∂ 2 i j B - τ,k ∂ k vvdx = - 1 2 n ∑ i, j,k=1 ∂ k a i j ∂ 2 i j B - τ,k v 2 dx. Thus J 1 ≥ -C v 2 dx if τ 1.
(2.58)

For J 2 , we find

J 2 = 2 n ∑ i, j,k=1 a i j ∂ i B - τ,k ∂ 2 k j vvdx = -2 n ∑ i, j,k=1 ∂ j a i j ∂ i B - τ,k ∂ k vvdx -2 ∑ i, j,k a i j ∂ i B - τ,k ∂ k v∂ j v ≥ -C v 2 dx -2 n ∑ i, j,k=1 a i j ∂ i B - τ,k ∂ k v∂ j vdx.
On the other hand, we easily check that

n ∑ i=1 a i j ∂ i B - τ,k = -4δ jk + O 1 τ
and consequently

J 2 ≥ C -v 2 dx + |∇v| 2 dx if τ
1 and v with a small support. (2.59)

Proceeding similarly we obtain

J 3 ≥ -C v 2 dx + n ∑ i, j,k=1 B - τ,k ∂ k a i j ∂ i v∂ j vdx.
But we can show in a straightforward manner that

n ∑ k=1 B - τ,k ∂ k a i j = O 1 τ in a neighborhood of 0. Whence J 3 ≥ -C v 2 dx + O 1 τ |∇v| 2 dx if τ 1 and v with a small support. (2.60)
Inequalities (2.58) to (2.60) imply

I 1 ≥ Cτ -v 2 dx + |∇v| 2 dx . (2.61)
In light of inequalities (2.53) to (2.57) and (2.61) there exists U a neighborhood of 0, two constants τ 0 > 0 and C > 0 so that

U v L + , L -vdx ≥ C τ U |∇v| 2 dx + τ 3 v 2 dx , τ ≥ τ 0 and v ∈ D(U ). Using U (Lv) 2 dx = U L + v 2 dx + U L -v 2 dx + U L + vL -vdx + U L -vL + vdx ≥ U L + vL -vdx + U L -vL + vdx = U L + L -v -L -L + v vdx ≥ U v L + , L -vdx, we find U (Lv) 2 ≥ C τ U |∇v| 2 dx + τ 3 v 2 dx , τ ≥ τ 0 and v ∈ D(U ). (2.62) Let v = e τϕ u, u ∈ D(U ).
We have e τϕ ∇u = ∇v -τv∇ϕ and hence

e 2τϕ |∇u| 2 ≤ 2 |∇v| 2 + τ 2 v 2 |∇ϕ| 2 ≤ C |∇v| 2 + τ 2 v 2 .
As e τϕ P 0 u = Lv, we deduce from (2.62)

U e 2τϕ (P 0 u) 2 ≥C τ U e 2τϕ |∇u| 2 + τ 3 e 2τϕ u 2 dx, (2.63)
τ ≥ τ 0 and u ∈ D(U ).

Finally from

U e 2τϕ (Pu) 2 dx ≥ 1 2 U e 2τϕ (P 0 u) 2 dx - U e 2τϕ (P 1 u) 2 dx ≥ 1 2 U e 2τϕ (P 0 u) 2 -C U e 2τϕ |∇u| 2 dx + U e 2τϕ u 2 dx and (2.63) it follows, changing τ 0 if necessay, that U e 2τϕ (Pu) 2 dx ≥ C τ U e 2τϕ |∇u| 2 dx + τ 3 U e 2τϕ u 2 , τ ≥ τ 0 and u ∈ D(U ).
We get the expected inequality by using that D(U ) is dense in H 2 0 (U ).

Let P be an elliptic operator of the form

P = ∆ + n ∑ i=1 b i (x)∂ i + c(x),
where the functions b i and c are bounded measurable on an open subset D of R n .

Theorem 2.15. Let ψ ∈ C 4 (D) and x 0 ∈ D so that ∇ψ(x 0 ) = 0. There exits

V a neighborhood of x 0 in D so that, if u ∈ H 2 (V ), Pu = 0 and u = 0 in {x ∈ V , ψ(x) < 0}, then u = 0 in V .
Proof. Making a translation if necessary we may assume that x 0 = 0. Also, changing the coordinate system, we are reduced to ∇ ψ(0) = 0 and ∂ n ψ(0) = 0. With the help of the implicit function theorem the equation ψ(x) = 0 in a neighborhood of 0 is equivalent to x n = µ(x ) in a neighborhood of 0, where µ is a C 4 function defined in a neighborhood of 0 and satisfies µ(0

) = 0, ∇ µ(0) = 0.
Observe that these transformations leave invariant the principal part P because the Laplace operator is invariant under orthogonal transformations.

We now make the following change of variable

(x , x n ) → (x , x n -µ(x ) + |x | 2 ) = (x , x n + θ (x )),
which transform P into P, where P has the same form as in the beginning of this subsection.

Note also that under this change of variable supp(u)

⊂ {x; ψ(x) ≥ 0} is trans- formed into supp(u) ⊂ {x; x n ≥ |x | 2 }. Recall that ϕ(x , x n ) = (x n -1) 2 + |x | 2
and define E + by

E + = {(x , x n ); 0 ≤ x n < 1 et x n ≥ |x | 2 }.
It is not hard to see that

E + \{0} ⊂ {(x , x n ); ϕ(x , x n ) < ϕ(0, 0) = 1}.
Let U be as in Theorem 2.14. Reducing U if necessary we may assume that

U ∩ E + ⊂ {(x , x n ); ϕ(x , x n ) ≤ ϕ(0, 0)}. Let u ∈ H 2 (U ) satisfying Pu = 0 and supp(u) ⊂ E + . Let w ∈ D(U ), w = 1 in a neighborhood U 0 ⊂ U of the origin and set v = wu. As v ∈ H 2 0 (U ), Theorem 2.14 implies U e 2τϕ (Pv) 2 dx = U \U 0 e 2τϕ (Pv) 2 dx = E + ∩(U \U 0 ) e 2τϕ (Pv) 2 dx ≥ Cτ 3 E + ∩U e 2τϕ v 2 dx if τ 1.
(2.64)

Let ε > 0 so that

E + ∩ (U \U 0 ) ⊂ {(x , x n ); ϕ(x , x n ) < ϕ(0, 0) -ε},
and V ⊂ U a neighborhood of the origin chosen in such a way that

E + ∩ V ⊂ {(x , x n ); ϕ(x , x n ) ≥ ϕ(0, 0) - ε 2 }.
Then (2.64) entails

V v 2 = E + ∩V v 2 ≤ e -ετ Cτ 3 E + ∩(U \U 0 ) (Pv) 2 if τ 1. Whence v = 0 in V and hence u = 0 in V too.
Theorem 2.15 can be used to obtain a global uniqueness of continuation for the operator P from an interior data.

Theorem 2.16. Assume that Ω is connected. Let u ∈ H 2 (Ω ) satisfying Pu = 0. Let ω be a nonempty open subset of Ω . If u = 0 in ω then u is identically equal to zero.

Proof. Let Ω 0 the greatest open set in which u vanishes (a.e.). We claim that Ω 0 = Ω (modulo a set of zero measure). Otherwise, Ω \Ω 0 would be nonempty. As Ω is connected Ω ∩ ∂ Ω 0 would be also nonempty. Fix then x 1 ∈ Ω ∩ ∂ Ω 0 and assume that B(x 1 , 3r) ⊂ Ω , where B(x 1 , 3r) is the ball with center x 1 and radius 3r. Let

K = ∂ (B(x 1 , 3r) ∩ Ω 0 ), x 2 ∈ B(x 1 , r) ∩ Ω 0 and d = dist (x 2 , K)(≤ r). We easily check that d = dist (x 2 , ∂ Ω 0 ). Therefore B(x 2 , d) is contained in Ω 0 and ∂ Ω 0 ∩ B(x 2 , d) is nonempty. We choose then x 0 in ∂ Ω 0 ∩ ∂ B(x 2 , d). Apply Theorem 2.15 with ψ(x) = |x -x 2 | 2 -d 2 in
order to get that there exists V a neighborhood of x 0 such that u = 0 in V (a.e.). But this contradicts the maximality of Ω 0 .

We apply Theorem 2.16 to obtain the uniqueness of continuation from Cauchy data. [START_REF] Sauvigny | Partial differential equations. 1. Foundations and integral representations[END_REF] . Now as u vanishes on B \ Ω , we get from Theorem 2.16 that u is identically equal to zero.

Corollary 2.5. Assume that Ω is a bounded domain of class C 2 with boundary Γ . Let Σ an nonempty open subset of Γ . Let u ∈ H 2 (Ω ) satisfying Pu = 0 and u = ∂ ν u = 0 on Σ 10 . Then u is identically equal to zero. Proof. Let B be a ball centered at a point of Σ so that B ∩ Γ = B ∩ Σ . The condition u = ∂ ν u = 0 on Σ entails that the extension by 0 of u in B \ Ω , still denoted by u, is in H 2 (B ∪ Ω )

General elliptic operators in divergence form

In this section Ω is an open bounded domain of R n . Consider the differential operator in divergence form with measurable bounded coefficients given by

Lu = - n ∑ i=1 ∂ i n ∑ j=1 a i j ∂ j u + c i u + n ∑ i=1 d i ∂ i u + du. (2.65)
In all of this section L is assumed to be elliptic in the sense that there exists a constant λ > 0 so that

n ∑ i, j=1 a i j (x)ξ i ξ j ≥ λ |ξ | 2 a.e. x ∈ Ω , for any ξ ∈ R n . (2.66)
Assume moreover that there exist two constants Λ > 0 and µ > 0 so that

n ∑ i, j=1 |a i j (x)| 2 ≤ Λ 2 , λ -2 n ∑ i=1 |c i (x)| 2 + |d i (x)| 2 (2.67) + λ -1 |d(x)| ≤ µ 2 a.e. x ∈ Ω .
We associate to L the bilinear form

L (u, v) = n ∑ i, j=1 a i j ∂ j u∂ i v + n ∑ i=1 c i u∂ i v + d i ∂ i uv + duv.
For f ∈ L 2 (Ω ), consider the equation

Lu = f in Ω . (2.68) We say that u ∈ W 1,1 loc (Ω ) is a weak solution of (2.68) if Ω L (u, v)dx = Ω f vdx, v ∈ D(Ω ). u ∈ W 1,1 loc (Ω ) is said a sub-solution (resp. super-solution) of (2.68) whenever Ω L (u, v)dx ≤ (resp. ≥) Ω f vdx, v ∈ D(Ω ), v ≥ 0.

Weak maximum principle

Let u ∈ H 1 (Ω ). We say that u ≤ 0 on

Γ = ∂ Ω if u + = max(u, 0) ∈ H 1 0 (Ω ) 12 . If u, v ∈ H 1 (Ω ) the notation u ≤ v on Γ will mean that u -v ≤ 0 on Γ . Let u ∈ H 1 (Ω ). Define then sup Γ u as follows sup Γ u = inf{k ∈ R; u ≤ k on Γ }.
Theorem 2.17. Under the assumption d

+ ∑ n i=1 ∂ i c i ≥ 0 13 if u ∈ H 1 (Ω ) is a sub- solution of Lu = 0 in Ω then sup Ω u ≤ sup Γ u + .
Proof. We seek a contradiction by assuming that

= sup Γ u + < k * = sup Ω u.
We have as u is a sub-solution of Lu = 0 in Ω [START_REF] Sauvigny | Partial differential equations. 2. Functional analytic methods[END_REF] Observe that u + ∈ H 1 (Ω ) follows from Corollary 1.1. [START_REF] Willem | Analyse fonctionnelle élémentaire[END_REF] This condition is to be understood in the following sense

Ω dϕ - n ∑ i=1 c i ∂ i ϕ dx ≥ 0 for any ϕ ∈ D(Ω ), ϕ ≥ 0. Ω L (u, v)dx = Ω n ∑ i, j=1 a i j ∂ j u∂ i v + n ∑ i=1 d i -c i ∂ i uv dx + Ω n ∑ i=1 c i ∂ i (uv) + duv dx ≤ 0, v ∈ D(Ω ), v ≥ 0.
By density the previous inequality still holds for any

v ∈ H 1 0 (Ω ), v ≥ 0. Let v ∈ H 1 0 (Ω ) so that v ≥ 0 and uv ≥ 0. Since d + ∑ n i=1 ∂ i c i ≥ 0 we deduce from the last inequality Ω n ∑ i, j=1 a i j ∂ j u∂ i v + n ∑ i=1 d i -c i ∂ i uv dx ≤ - Ω n ∑ i=1 c i ∂ i (uv) + duv dx ≤ 0. That is Ω n ∑ i, j=1 a i j ∂ j u∂ i vdx ≤ Ω n ∑ i=1 c i -d i ∂ i uvdx.

This implies

Ω n ∑ i, j=1 a i j ∂ j u∂ i vdx ≤ C Ω v|∇u|dx, (2.69)
where the constant C only depends on the L ∞ -norms of c i and d i .

Let (k m ) be a non decreasing sequence so that ≤ k m < k * for every m and k m converges to k * . Then (2.69

) with v = v m = (u -k m ) + gives Ω n ∑ i, j=1 a i j ∂ j v m ∂ i v m dx ≤ C Ω v m |∇v m |dx, ( 2
Ω |∇v m | 2 dx ≤ C λ v m L 2 (A m ) ∇v m L 2 (Ω ) n
and hence

∇v m L 2 (Ω ) n ≤ C λ v m L 2 (A m ) . (2.71) But (see Lemma 1.10) v m L 2 (A m ) ≤ |A m | 1/n v n L 2n/(n-2) (A m ) ≤ K(n)|A m | 1/n ∇v m L 2 (A m ,R n ) if n > 2,
and

v m L 2 (A m ) ≤ K(2) ∇v m L 1 (A m ) n ≤ K(2)|A m | 1/2 ∇v m L 2 (A m ,R n ) if n = 2,
where the constant K(n) only depends on n. These two inequalities together with

(2.71) yield v m L 2 (Ω ) ≤ K|A m | 1/n v m L 2 (A m ) with K = CK(n)/λ . By assumption, v m L 2 (A m ) = 0 for any m. Therefore |A m | ≥ K -n for any m. (2.72) In particular, if A = ∩ m A m then |A| = lim |A m | ≥ K -n and supp(|∇u|) ⊃ A. (2.73) On the other hand, if B m = [u ≥ k m ] we have |[u = k * ]| = | ∩ m B m | = lim m |B m |.
But A m ⊂ B m for any m. Hence [u = k * ] ⊃ A and then ∇u = 0 in A14 . We get the expected contradiction by comparing this with (2.73).

We have as an immediate consequence of Theorem 2.17 the following uniqueness result.

Corollary 2.6. Assume that the assumptions of Theorem 2.17 hold. If u ∈ H 1 0 (Ω ) is a weak solution of Lu = 0 in Ω then u = 0.

The Dirichlet problem

We aim in this section to prove the following existence result. Theorem 2.18. Assume that d + ∑ n i=1 ∂ i c i ≥ 0. For any ϕ ∈ H 1 (Ω ) and g, f i ∈ L 2 (Ω ), i = 1, . . . , n, the generalized Dirichlet problem

Lu = g + ∑ n i=1 ∂ i f i in Ω , u = ϕ on Γ , admits a unique solution u ∈ H 1 (Ω ).
Proof. We first reduce the initial problem to a problem with zero boundary condition. Set w = uϕ. Then clearly w ∈ H 1 0 (Ω ) and in light of (2.65)

Lw = ĝ + n ∑ i=1 ∂ i fi , where ĝ = g - n ∑ i=1 d i ∂ i ϕ -dϕ ∈ L 2 (Ω ), fi = f i + n ∑ j=1 a i j ∂ j ϕ + c i ϕ ∈ L 2 (Ω ), i = 1, . . . , n.
Introduce the notations H = H 1 0 (Ω ), g = (g, f 1 , . . . , f n ) and

F(v) = Ω gv + n ∑ i=1 f i ∂ i v dx, v ∈ H . Then we have F ∈ H * because |F(v)| ≤ g L 2 (Ω ,R n+1 ) v H 1 0 (Ω ) , v ∈ H .
We now establish that L + σ (•|•) is continuous and coercive for some σ > 0.

Prior to doing that we prove the following lemma.

Lemma 2.6.

L (u, u) ≥ λ 2 Ω |∇u| 2 dx -λ µ 2 Ω u 2 dx.
(2.74)

Proof. We have from (2.65) that

L (u, u) = Ω n ∑ i, j=1 a i j ∂ i u∂ j u + n ∑ i=1 c i -d i u∂ i u -du 2 dx.
From the elementary inequality

αβ ≤ εα 2 + 1 4ε β 2 ,
we deduce that

|c i ||∂ i u||u| ≤ λ 4 |∂ i u| 2 + 1 λ |c i | 2 |u| 2 , |d i ||∂ i u||u| ≤ λ 4 |∂ i u| 2 + 1 λ |d i | 2 |u| 2 .
These two inequalities, (2.66) and (2.67) imply

L (u, u) ≥ Ω λ |∇u| 2 - λ 2 |∇u| 2 -λ ν 2 u 2 dx = λ 2 Ω |∇u| 2 dx -λ ν 2 Ω u 2 dx.
This proves the lemma.

For σ ∈ R, let L σ given by L σ u = Lu + σ u. According to Lemma 2.6, L σ , the bilinear form associated to L σ , is coercive provided that σ is sufficiently large.

Next, we consider the operator I : H → H * defined by

Iu(v) = Ω uvdx, v ∈ H .
Lemma 2.7. The operator I is compact.

Proof. Write I = I 1 I 2 , where I 2 is the canonical imbedding of H into L 2 (Ω ) and

I 1 : L 2 (Ω ) → H * is given by I 1 u(v) = Ω uvdx, v ∈ L 2 (Ω ).
By Sobolev imbedding theorems I 2 is compact and since I 1 is bounded I is also compact.

Fix σ 0 so that L σ 0 is coercive in the Hilbert space H . Note that the equation Lu = F, for u ∈ H , is equivalent to the following one

L σ 0 + σ 0 Iu = F.
But L σ 0 is an isomorphism from H onto H * Lax-Milgram's lemma. Therefore, the last equation is equivalent to the following one

u + σ 0 L -1 σ 0 Iu = L -1 σ 0 F. (2.75)
The operator T = -σ 0 L -1 σ 0 I is compact by Lemma 2.7. In consequence, the existence of u ∈ H satisfying

Lu = g + n ∑ i=1 ∂ i f i in Ω , u = 0 on ∂ Ω
is guaranteed by Fredholm's alternative whenever the equation Lu = 0 in Ω , u = 0 on ∂ Ω has only u = 0 as a solution. This is true by Corollary 2.6.

We now describe the spectrum of the operator L. We can check in a straightforward manner that L * , the formal adjoint of L, is given by

L * u = - n ∑ j=1 ∂ j n ∑ i=1 a i j ∂ i u + d j u + n ∑ i=1 c i ∂ i u + du. As L * (u, v) = L (v, u) for any u, v ∈ H = H 1 0 (Ω )
we derive that L * is also the adjoint of L in the Hilbert space H , and the same is valid if L is substituted by L σ .

The argument we used in the proof of Theorem 2.18 enables us to claim that the solvability of the equation L σ u = F is equivalent to the solvability of the following one

u + (σ 0 -σ )L -1 σ 0 Iu = L -1 σ 0 F. Since T * σ , the adjoint of the operator T σ = (σ 0 - σ )L -1
σ 0 I, is given by T * σ = (σ 0σ )(L * σ 0 ) -1 I, we can apply Fredholm's alternative in order to obtain the following result.

Theorem 2.19. There exists a countable set Σ ⊂ R with no accumulation point so

that if σ ∈ Σ the Dirichlet problem L σ u (resp. L * σ u) = g + ∑ n i=1 ∂ i f i in Ω , u = ϕ on Γ admits a unique solution provided that g, f i ∈ L 2 (Ω ) and ϕ ∈ H 1 (Ω ). Let σ ∈ Σ , then the subspace of solutions of the homogenous equation L σ u (resp. L * σ u) = 0, u = 0 on Γ is of finite dimension (> 0), and the equation L σ u = g + ∑ n i=1 ∂ i f i in Ω , u = ϕ on Γ has a solution if and only if n ∑ i, j=1 Ω g -d i ∂ i ϕ -dϕ + σ ϕ v -f i + a i j ∂ j ϕ + c i ϕ ∂ i v dx = 0 for any v satisfying L * σ v = 0, v = 0 on Γ . Finally, if d + ∑ n i=1 ∑ ∂ i c i ≥ 0 then Σ ⊂ (-∞, 0).

Harnack inequalities

We first prove a Harnack inequality for sub-solutions. Here the constant C only depends on n and the L ∞ -norm of the coefficients of λ -1 L.

Proof. The proof consists in three steps.

In this proof C is a generic constant only depending on n and the L ∞ -norm of the coefficients of λ -1 L.

First step. Pick K a compact subset of Ω , 0 < r < dist(K,Γ ) and q > 1. We are going to prove that if u + ∈ L q (K + B(0, r)) then u + ∈ L n q (K), and

u + L n q (K) ≤ [C (r, q)] 1/q u + L q (K+B(0,r)) .
(2.77)

Here and henceforth, n = n/(n -1) is the conjugate exponent of n. Let

η = dist(•, Ω \(K + B(0, r)) dist(•, Ω \(K + B(0, r)) + dist(•, K) . (2.78) It is not hard to check that η is Lipschitz continuous with Lipschitz constant equal to 1/r, supp(η) ⊂ K + B(0, r), 0 ≤ η ≤ 1 and η = 1 in K. In particular, η ∈ W 1,∞ (R n ). Let θ be a Borelian function defined on R so that θ = 0 in ] -∞, 0], θ > 0 in ]0, +∞[ and set v = η 2 u -∞ θ (s)ds.
We have v ∈ H 1 (Ω ), supp(v) ⊂ K + B(0, r) and v ≥ 0. The function v is then the limit of a sequence of (v k ) belonging to D(Ω ), v k ≥ 0 for any k. As u is a subsolution of Lu = 0 in Ω , we have

Ω L (u, v k )dx ≤ 0 for all k.
We get by passing to the limit, when k → ∞,

Ω L (u, v)dx ≤ 0.
On the other hand, elementary computations give

L (u, v) = n ∑ i, j=1 θ (u)η 2 a i j ∂ j u∂ i u + n ∑ j=1 v 1 2 ∑ i a i j ∂ i η + d j η + c j ηv 2 ηθ 1/2 (u)∂ j u + n ∑ i=1 2c i η∂ i η + dη 2 v 3 , with v 1 = θ -1/2 (u) u -∞ θ (s)ds, v 2 = uθ 1/2 (u), v 3 = u u -∞ θ (s)ds. Hence Ω n ∑ i, j=1 θ (u)η 2 a i j ∂ j u∂ i udx ≤ - Ω n ∑ j=1 v 1 2 n ∑ i=1 a i j ∂ i η + d j η + c j ηv 2 ηθ 1/2 (u)∂ j udx - Ω n ∑ i=1 2c i η∂ i η + dη 2 v 3 dx.

This together with the convexity inequality

|AB| ≤ (a/2)A 2 +(1/2a)B 2 , a > 0, entail Ω n ∑ i, j=1 θ (u)η 2 a i j ∂ j u∂ i udx ≤ λ 2 Ω θ (u)η 2 |∇u| 2 + 1 2λ Ω n ∑ j=1 v 1 2 ∑ i a i j ∂ i η + d j η + c j ηv 2 2 dx - Ω n ∑ i=1 2c i η∂ i η + dη 2 v 3 dx. But (see (2.66)) Ω n ∑ i, j=1 θ (u)η 2 a i j ∂ j u∂ i udx ≥ λ Ω θ (u)η 2 |∇u| 2 dx. Whence λ 2 Ω θ (u)η 2 |∇u| 2 dx ≤ 1 2λ Ω n ∑ j=1 v 1 2 n ∑ i=1 a i j ∂ i η + d j η + c j ηv 2 2 dx - Ω n ∑ i=1 2c i η∂ i η + dη 2 v 3 dx. (2.79)
Introduce the auxiliary function

w = u -∞ θ (s) 1/2 ds. Then |∇w| 2 = θ (u)|∇u| 2 , (2.80)
and from Cauchy-Schwarz's inequality, we have

w 2 = u 0 θ (s) 1/2 ds 2 ≤ u u 0 θ (s)ds = v 3 . (2.81) On the other hand, if v 4 = v 1 + v 2 then v 2 4 ≥ 4v 1 v 2 = 4v 3 ≥ 4w 2 . (2.82)
Using estimates (2.79) to (2.81), the fact that 0 ≤ η ≤ 1 and |∇η| ≤ 1/r in order to deduce

η∇w L 2 (Ω ) ≤ C 1 + 1 r v 4 L 2 (K+B(0,r)) . (2.83) But ∇(ηw) = η∇w + w∇η. Hence |∇(ηw)| L 2 (Ω ) = |∇(ηw)| L 2 (K+B(0,r)) ≤ η|∇w| L 2 (K+B(0,r)) + 1 r w L 2 (K+B(0,r)) .
This and (2.82) imply

|∇(ηw)| L 2 (Ω ) ≤ η|∇w| L 2 (K+B(0,r)) + 1 2r v 4 L 2 (K+B(0,r)) .
In light of (2.83), the last inequality yields

|∇(ηw)| L 2 (Ω ) ≤ C 1 + 1 r v 4 L 2 (K+B(0,r)) .
(2.84)

Taking into account that ηw ∈ W 1,1 0 (R n ), we get from Lemma 1.10

ηw 2 L 2n (Ω ) = (ηw) 2 L n (Ω ) ≤ C ∇(ηw) 2 L 1 (Ω ) ≤ 2C ηw L 2 (Ω ) ∇w L 2 (Ω ) n , for some constant C = C(n).
This inequality together with (2.82) and (2.84) imply

w L 2n (K) ≤ C 1 + 1 r 1/2 v 4 L 2 (K+B(0,r)) . (2.85)
We complete the proof of this first step by choosing θ (s) = χ {s>0} q 2 s q-2 /4. Then w = (u + ) q/2 and v 4 = q 2 2(q -1)

(u + ) q/2 .
We deduce from (2.85) that if u + ∈ L q (K + B(0, r)) then u + ∈ L qn (K + B(0, r)) and

u + L qn (K) ≤ C 1 + 1 r q 4 (q -1) 2 1/q u + L q (K+B(0,r)) .
(2.86)

Second step. We claim that u + ∈ L p loc (Ω ) for any p > 1. Indeed, as u + ∈ L 2 loc (Ω ), we have that u + ∈ L 2n loc (Ω ) by (2.86). We iterate k times (2.86) to obtain is a simple manner that u ∈ L 2(n ) k loc (Ω ) for any positive integer k and hence u ∈ L p loc (Ω ) for every p > 1.

Third step. We employ Moser's iterative method to establish (2.76). Let K be a given compact subset of Ω , 0 < r < dist(K,Γ ), p > 1, and set

ρ k = r 2 k , r k = ρ 1 + . . . + ρ k = r(1 -1/2 k ) and q k = p(n ) k .
Fix a positive integer k and define the sequence of compacts K i by

K k = K, . . . , K i-1 = K i + B(0, ρ i ), . . . , K 0 = K + B(0, r).
We have in light of (2.86)

u + L q i (K i ) ≤ C(ρ i , q i-1 ) 1/q i-1 u + L q i-1 (K i-1 )
and then

u + L q k (K) ≤ k ∏ i=1 [C(ρ i , q i-1 )] 1/q i-1 u + L p (K+B(0,r k )) ≤ k ∏ i=1 [C(ρ i , q i-1 )] 1/q i-1 u + L p (K+B(0,r)) .
On the other hand, elementary computations show

[C(ρ i , q i-1 )] 1/q i-1 ≤ [C(r, p)] (n ) 1-i /p (2n ) (i-1)(n ) 1-i /p . Therefore k ∏ i=1 C(ρ i , q i-1 ) 1/q i-1 ≤ [C(r, p)] α k (2n ) β k , with α k = n p (1 -(n ) -k ) and β k = n(n -1) p (1 + (k -1)(n ) -k -k(n ) 1-k ). Thus u + L q k (K) ≤ C(r, p) α k (2n ) β k u + L p (K+B(0,r)) .
(2.87)

We obtain by passing to the limit in (2.87), when k tends to ∞,

u + L ∞ (K) = lim k→+∞ u + L q k (K) ≤ C(r, p) n/p u + L p (K+B(0,r))
as expected.

We are now going to prove the following Harnak inequality for non negative super-solutions.

Theorem 2.21. Let u ∈ H 1 loc (Ω ) be a super-solution of Lu = 0 in Ω . Let x 0 ∈ Ω , 0 < r ≤ r 0 < dist(x 0 ,Γ )/4, p < n and assume that u is non negative in B(x 0 , 4r). Then u L p (B(x 0 ,2r)) ≤ Cr n/p u a.e. in B(x 0 , r),
where the constant C only depends on the L ∞ -norm of the coefficients of λ -1 L, n, p, r and r 0 .

Proof. The proof consists in three steps. First step. Fix ε > 0 and let 0 < r ≤ r 0 . Let K be a compact subset of Ω with K ⊂ B(x 0 , 3r). In the sequel, η is the function defined in (2.78). Set

u ε = u + ε and v = η 2 (α + 1)u α+1 ε with α > -1.
As we have done in the first step of the proof of Theorem 2.20, we get by using

L (u, v) ≥ 0 λ 2 Ω η 2 |∇u| 2 u α+2 ε dx ≤ Ω 1 2λ n ∑ j=1 1 (α + 1)u α/2 ε 2 n ∑ i=1 a i j ∂ i η + d i η - u u α/2+1 ε c i η 2 dx + Ω u u α+1 ε η α + 1 2 n ∑ i=1 c i ∂ i η + dη dx.
When α = 0 we use the fact that 0 ≤ u/u ε ≤ 1 a.e. in Ω to deduce from the last inequality η|∇u| u ε

L 2 (Ω ) ≤ C η L 2 (Ω ) + ∇η L 2 (Ω ,R n ) . (2.88) If α > 0, noting that u u α+1 ε ≤ 1 u α ε and u u α/2+1 ε ≤ 1 u α/2 ε , we easily check that |∇u| u α/2+1 ε L 2 (K) ≤ η|∇u| u α/2+1 ε L 2 (Ω ) ≤ C 1 + 1 r 1 u α/2 ε L 2 (K+B(0,r))
.

But ∇u u α/2+1 ε = - 2 α ∇ 1 u α/2 ε . Hence ∇ 1 u α/2 ε L 2 (K) ≤ Cα 1 + 1 r 1 u α/2 ε L 2 (K+B(0,r))
.

This and Sobolev imbedding theorems yield similarly to the first step of the proof of Theorem 2.20 15 that 15 Observe that

1 u ε L α (•) = 1 u α/2 ε 2/α L 2 (•) . 1 u ε L αn (K) ≤ C 1 + 1 r α 1/α 1 u ε L α (K+B(0,r)) .
(2.89)

Finally, if -1 < α < 0 we prove in a similar manner to that used to establish (2.89) that

u ε L β n (K) ≤ C 1 + 1 r β 1/β u ε L β (K+B(0,r)) , (2.90)
with β = -α.

We apply Moser's iterative scheme to (2.89) for completing this step . We obtain

1 u ε L ∞ (K) ≤ C 1 + 1 r α n/α 1 u ε L α (K+B(0,r))
, α > 0.

(2.91)

Second step. Let x 0 ∈ Ω and 0 < r ≤ r 0 < dist(x 0 ,Γ )/4.
For simplicity convenience, we use in this step B(r) instead of B(x 0 , r). Let B R be a ball of R n with radius R. If R ≤ r we apply (2.88) with K = B R ∩ B(3r) and r substituted by R (note that ∇u ε = ∇u and η = 1 in K) 16 . We obtain, with

w = ln u ε - 1 |B(3r)| B(3r) ln u ε , ∇w L 2 (B R ∩B(3r)) n ≤ C 1 + 1 R |B R ∩ B(3r) + B(0, R)| 1/2 ≤ C 1 + 1 R |B 2R | 1/2 ≤ C(1 + r 0 ) 1 R |B 2R | 1/2 ≤ MR n/2-1 .
Here and until the end of this step, M is a generic constant only depending on the L ∞ -norm of the coefficients of λ -1 L, n and r 0 . Hence

∇w L 1 (B R ∩B(3r),R n ) ≤ |B R ∩ B(3r)| 1/2 ∇w L 2 (B R ∩B(3r),R n ) ≤ MR n-1 . If R > r then an application of (2.88) with K = B R ∩ B(3r) gives ∇w L 2 (B R ∩B(3r),R n ) ≤ C 1 + 1 r |B R ∩ B(3r) + B(0, r)| 1/2 ≤ C(1 + r 0 ) 1 r |B 4r | 1/2 ≤ Mr n/2-1 and hence ∇w L 1 (B R ∩B(3r),R n ) ≤ Mr n-1 ≤ MR n-1 .
That is we proved

∇w L 1 (B R ∩B(3r),R n ) ≤ MR n-1 for any ball B R of R n .
This inequality together with Theorem 2.22. 17 (F.

John -L. Nirenberg) Let ω be an open convex subset of R n . Let f ∈ W 1,1 (ω) satisfying ω f = 0. Assume that there exists a constant A > 0 so that |∇ f | L 1 (ω∩B R ) ≤ AR n-1 for any ball B R of R n .
Then there exist σ 0 > 0 and D > 0, only depending on n, such that e -q ln u ε +qm dx = e qm B(3r) u -q ε dx,

ω e σ | f |/A dx ≤ Ddiam(ω) n , where σ = σ 0 |ω|diam(ω) -n .
where m = 1 |B(3r)| B(3r) ln u ε dx.
But by Jensen's inequality 18 , ln being concave, 17 A proof can be found in [5, Theorem 7.21 in page 166]. 18 Jensen's inequality. Let (M , A , µ) a probability space and ϕ a convex function from R into R. For any µ-integrable and real-valued function f , we have

ϕ M f dµ ≤ M ϕ • f dµ. e qm = e 1 |B(3r)| B(3r) ln u q ε dx ≥ e ln 1 |B(3r)| B(3r) u q ε dx = 1 |B(3r)| B(3r) u q ε dx. Thus B(3r) e q|w| dx ≥ 1 |B(3r)| B(3r) u q ε dx B(3r) u -q ε dx
and hence

B(3r) e q|w| dx 1/q ≥ 1 |B(3r)| 1/q u ε L q (B(3r)) 1 u ε L q (B(3r)) .
In light of (2.92), this inequality implies

u ε L q (B(3r)) 1 u ε L q (B(3r)) ≤ Cr 2n/q .
(2.93)

Third step. Let 0 < p < n . We write (2.90) with β = p/n , K = B(2r) and r substituted by r/k, k is a positive integer. We get

u ε L p (B(2r)) ≤ C 1 + k r p n n /p u ε L p/n (B(2r+r/k)) .
We have similarly

u ε L p (B(2r+r/k)) ≤ C 1 + k r p (n ) 2 (n ) 2 /p u ε L p/(n ) 2 (B(2r+2r/k)) .
Hence

u ε L p (B(2r)) ≤ C 1 + k r p n n /p × C 1 + k r p (n ) 2 (n ) 2 /p u ε L p/(n ) 2 (B(2r+2r/k)) .
This inequality yields in a straightforward manner

u ε L p (B(2r)) ≤ k ∏ i=1 C 1 + k r p (n ) i (n ) i /p u ε L p/(n ) k ((B(3r)) ≤ Mr -(∑ k i=1 (n ) i )/p u ε L p/(n ) k ((B(3r)) ≤ Mr n(1-(n ) k )/p u ε L p/(n ) k ((B(3r)) .
If we choose k so that p ≤ q(n ) k then, from Hölder's inequality, we obtain

u ε L p/(n ) k ((B(3r)) ≤ |B(3r)| (q(n ) k -p)/(qp) u ε L q ((B(3r)) ,
from which we deduce u ε L p (B(2r)) ≤ Mr n/p-n/q u ε L q ((B(3r)) .

(2.94)

We now combine (2.91) (with α = q and K = B(r)), (2.93) and (2.94) in order to get 1

u ε L ∞ (B(r)) u ε L p (B(2r)) ≤ Mr n/p
and hence u ε L p (B(2r)) ≤ Mr n/p u ε a.e. in B(r).

(2.95)

The expected inequality follows by passing to the limit, when ε goes to 0, in (2.95).

As we have done before, for simplicity convenience, we use respectively sup and inf instead of supess et infess.

The following Harnack's inequality for positive solution follows readily from Theorems 2.20 and 2.24.

Theorem 2.23. Let u ∈ H 1 loc (Ω ) be a positive weak solution of Lu = 0 in Ω . Then for any x 0 ∈ Ω and 0 < r ≤ r 0 < dist(x 0 ,Γ )/4, we have sup

B(x 0 ,r) u ≤ C inf B(x 0 ,r) u.
(2.96)

where the constant C only depends on the L ∞ -norm of the coefficients of λ -1 L, n, r 0 and dist(x 0 ,Γ )r 0 .

We deduce from this theorem the following result.

Corollary 2.7. Assume that Ω is connected. Let u ∈ H 1 loc (Ω ) be a non negative weak solution of Lu = 0 in Ω . Then, for any compact subset K of Ω , we have

sup K u ≤ C inf K u.
where the constant C only depends on the L ∞ -norm of the coefficients of λ -1 L, n and dist(K,Γ )19 .

Proof. Using the connexity of Ω and the compactness of K, we find N balls

B(x i , 4r i ) contained in Ω , (B(x i , r i )) cover K and B(x i , r i ) ∩ B(x i+1 , r i+1 ) = / 0, i = 1, . . . , N -1.
Let k and be two indices so that sup

K u = sup K∩B(x k ,r k ) u and inf K u = inf K∩B(x ,r ) u.
Changing the order of the elements of the sequence (x i ) if necessary, we may assume that k ≤ . If k = the conclusion is straightforward from (2.96). When k < we have, once again from (2.96), sup

K∩B(x k ,r k ) u ≤ sup B(x k ,r k ) u ≤ C inf B(x k ,r k ) u ≤ C sup K∩B(x k+1 ,r k+1 ) u.
We get by iterating these inequalities sup

K∩B(x k ,r k ) u ≤ C -k+1 inf B(x ,r ) u ≤ C -k+1 inf K∩B(x ,r ) u.
This completes the proof.

We end this subsection by showing how one can use Harnak's inequality in Theorem 2.24 for non negative super-solutions in order to obtain a strong maximum principle for weak solutions.

Theorem 2.24. Let u ∈ H 1 loc (Ω ) be a weak super-solution of Lu = 0 in Ω . Assume that Ω is connected and one of the following two assumptions is satisfied.

(i) d = -∑ n i=1 ∂ i c i . (ii) d + ∑ n i=1 ∂ i c i ≥ 0 and sup Ω u ≥ 0. Then u is constant in Ω or else, for any compact subset K of Ω , we have sup K u < sup Ω u. Proof. Let M = sup Ω u. If M = +∞ then the conclusion is straightforward because u + ∈ L ∞ loc (Ω ). Assume now that M < +∞ and let w = M -u. If w vanishes then u is constant. If w does not vanish then clearly w ∈ H 1 loc (Ω ), w ≥ 0 and Ω L (w, v)dx = Ω L (M, v)dx = M Ω d + n ∑ i=1 c i ∂ i vdx ≥ 0, v ∈ D(Ω ), v ≥ 0,
if condition (i) or (ii) holds. In other words, w is a non negative super-solution of Lu = 0 in Ω . We proceed by contradiction by assuming that there exist a compact subset K of Ω so that inf K w = 0. Let V be the greatest open set containing K in which w vanishes (a.e.). As Ω is connected and w is non identically equal to zero, Ω ∩ ∂V is nonempty. Therefore, we find a ball B(y, r) ⊂ V so that B(y, 2r) ⊂ Ω and ∂ B(y, r) ∩ ∂V = / 0. We conclude by applying Theorem 2.20 that w vanishes on V ∪ B(y, 2r) (a.e.). But this contradicts the definition of V and completes the proof.

Exercises and problems

2.1. Show that in infinite dimensional Banach space a linear compact operator is never invertible with bounded inverse.

2.2.

Recall that 2 is the Hilbert space of real sequence x = (x m ) m≥1 so that ∑ m≥1 x 2 m < ∞. This space in endowed with the scalar product x|y = ∑ m≥1 x m y m . Let (a m ) m≥1 be a real bounded sequence satisfying sup m≥1 |a m | ≤ C < ∞. Define the linear operator A : 2 → 2 by Ax = (a m x m ) m≥1 . Prove that A is bounded and it is compact if and only if lim m→+∞ a m = 0.

2.3.

Let H = L 2 (0, 1) and let A be the linear operator from H into H defined by

(A f )(x) = (x 2 + 1) f (x). Check that A is bounded, positive definite i.e. (A f , f ) > 0 for any f ∈ H, f = 0, where (•|•) is the scalar product of H and self-adjoint, but it is non compact. Show that A has no eigenvalues. Prove finally that A -λ I is invertible with bounded inverse if and only if λ ∈ [1, 2].
2.4. Let E, F be two Banach spaces and A ∈ L (E, F). Assume that E is separable and reflexive. Prove that A is compact if and only if any sequence (x n ) in E so that (x n ) converges weakly to x ∈ E then the sequence (Ax n ) converges strongly to Ax.

2.5. Let (X, µ) et (Y, ν) be two measure spaces and k ∈ L 2 (X × Y, µ × ν). For f ∈ L 2 (X, µ), set (A f )(y) = X k(x, y) f (x)dµ(x).
Show that A is linear bounded operator from L 2 (X, µ) into L 2 (Y, ν). We say that A is a kernel operator with kernel k.

Prove that if L 2 (X, µ) is separable this is for instance the case for L 2 (Ω , dx) with Ω an open subset of R n and dx is the Lebesgue measure then A is compact. Hint: use Exercise 2.4.

-∆ u = f in Ω , ∂ ν u = 0 on Γ . (2.97) Let u ∈ C 2 (Ω ).
Show that u is a solution of (2.97) if and only if u satisfies

Ω ∇u • ∇vdx = Ω f vdx for any v ∈ C 1 (Ω ).
Deduce that a necessary condition ensuring the existence of a solution u

∈ C 2 (Ω ) of (2.97) is Ω f dx = 0. 2.7. Let Ω be an open bounded subset of R n with boundary Γ , f ∈ L 2 (Ω ) and V ∈ C 1 (Ω , R n ) satisfying div (V ) = 0.
Demonstrate that there exists a unique variational solution of the following convection-diffusion boundary value problem

-∆ u +V • ∇u = f in Ω , u = 0 on Γ . (2.98) 2.8.
Let Ω be a bounded open subset of R n of class C 1 with boundary Γ . a) Prove that there exist a constant C > 0 so that

v L 2 (Ω ) ≤ C v L 2 (Γ ) + ∇v L 2 (Ω ,R n ) for any v ∈ H 1 (Ω ).
Hint: proceed by contradiction. b) Let f ∈ L 2 (Ω ) and g ∈ L 2 (Γ ). Prove the existence and uniqueness of a variational solution of the boundary value problem Laplace operator with Fourier boundary condition :

-∆ u = f in Ω , ∂ ν u + u = g on Γ .
(2.99)

2.9. Compute the eigenvalues and the eigenfunctions of the Laplace operator with Dirichlet boundary condition in the case where Ω =]0, 1[ Note that if ϕ is an eigenfunction, then by the elliptic regularity ϕ belongs to C ∞ ([0, 1]) . Prove by using the spectral decomposition of this operator, that the series

∑ k≥1 a k sin(kπx) converges in L 2 (0, 1) if and only if ∑ a 2 k < ∞ and in H 1 (0, 1) if and only if ∑ k≥1 k 2 a 2 k < ∞. 2.10. Consider the cube Ω =]0, 1 [× . . . ×]0, n [, where i > 0, 1 ≤ i ≤ n.
Compute the eigenvalues and the eigenfunctions of the Laplace operator on Ω with Dirichlet boundary condition. Hint: use the method of separation of variables.

2.11.

Let Ω a bounded domain of R n . Recall the Poincaré's inequality : there exists a constant C > 0 so that for any u ∈ H 1 0 (Ω ) we have

Ω u 2 dx ≤ C Ω |∇u| 2 dx.
(2.100)

Prove that the best constant in (2.100) is exactly 1/λ 1 , where λ 1 is the first eigenvalue of the Laplace operator on Ω with Dirichlet boundary condition.

2.12. We consider the eigenvalue problem for the Schrödinger operator with a quadratic potential Q(x) = Ax • x, where A is a symmetric positive definite matrix a model of harmonic oscillator

-∆ u + Qu = λ u in R n . (2.101) Let H = L 2 (R n ) and define V = {v ∈ H 1 (R n ) so that |x|v(x) ∈ L 2 (R n )}.
(a) Prove that V is a Hilbert space when it is endowed with the scalar product

u, v V = R n ∇u • ∇vdx + R n |x| 2 uvdx.
Hint: if B i = B(0, i), i = 1, 2, we can first establish that there exists a constant C > 0 so that, for any u ∈ V , we have

B 2 u 2 dx ≤ C B 2 |∇u| 2 dx + B 2 \B 1 u 2 dx .
(b) Show that the imbedding of V into H is compact and deduce from it that there exists a nondecreasing sequence (λ k ) of real numbers converging to ∞ and a Hilbertian basis (ϕ k ) of L 2 (R n ) consisting respectively of eigenvalues and eigenfunctions associated to the boundary value problem (2.101).

2.13. Let 0 < β < 2π and

Ω = {(x, y) ∈ R 2 ; x = r cos θ , y = r sin θ , 0 < r < 1, 0 < θ < β }.
Consider the boundary value problem

∆ u = 0 in Ω u = u 0 on Γ = ∂ Ω , (2.102)
where

u 0 (x, y) = v 0 (r, θ ) = 0 if θ = 0, β , sin(θ π/β ) if r = 1.
Use the method of separation of variables to find the explicit solution of (2.102). Then show that this solution belongs to H 2 (Ω ) if and only if β < π.

2.14.

Let Ω be a bounded domain of R n of class C 1 with ∂ Ω = Γ 1 ∪ Γ 2 , where Γ 1 and Γ 2 are disjoint and closed.

By mimicking the proof of Theorem 2.19, prove that there exists a unique bounded operator t 0 :

H 1 (Ω ) → L 2 (Γ 1 ) so that t 0 w = w| Γ 1 if w ∈ D(Ω ), and a bounded operator (t 1 ,t 2 ) : H 2 (Ω ) → L 2 (Γ 1 ) × L 2 (Γ 2 ) so that (t 1 ,t 2 )w = (∂ ν w| Γ 1 , ∂ ν w| Γ 2 ) if w ∈ D(Ω ).
We use in the sequel the notations w| Γ 1 and ∂ ν w| Γ 2 respectively instead of t 0 w and t 2 w.

Define the vector space

V = {u ∈ H 1 (Ω ); w| Γ 1 = 0}. (a) (i) Show that V is a closed subspace of H 1 (Ω ).
(ii) Demonstrate that there exists a constant C > 0 so that

w L 2 (Ω ) ≤ C ∇w L 2 (Ω ,R n ) for all w ∈ V.
Pick f ∈ L 2 (Ω ) and consider the boundary value problem

   -∆ u = f in Ω , u| Γ 1 = 0, ∂ ν u| Γ 2 = 0 (2.103)
and the variational problem : find u ∈ V satisfying

Ω ∇u • ∇vdx = Ω f vdx, for any v ∈ V. (2.104) (b) Prove that u ∈ V ∩ H 2 (Ω )
is a solution of (2.103) if and only if u is a solution of (2.104) we can admit that D(Γ 2 )20 is dense in L 2 (Γ 2 ) and note that {w ∈ D(Ω ); w| Γ 1 = 0} is dense in V . (c) Show that (2.104) admits a unique solution u ∈ V . Consider next the following spectral problem : find the values µ ∈ R for which there exists a solution u ∈ V , u = 0, of the problem

Ω ∇u • ∇vdx = µ Ω uvdx for any v ∈ V.
(2.105) (d) Show that the eigenvalues of (2.105) consists in a non deceasing sequence converging to +∞ :

0 < µ 1 ≤ µ 2 ≤ . . . ≤ µ m ≤ . . .
and there exists a Hilbetian basis (w n ) of L 2 (Ω ) consisting in eigenfunctions so that

Ω ∇w m • ∇vdx = µ m Ω w m vdx for any v ∈ V.
(e) Denote by (λ m ) m≥1 the sequence of eigenvalues of the spectral problem : find the values of λ ∈ R for which there exists a solution u ∈ H 1 0 (Ω ), u = 0, of the problem

Ω ∇u • ∇vdx = λ Ω uvdx for all v ∈ H 1 0 (Ω ).
Check that we have µ m ≤ λ m , for every m ≥ 1.

(Hölder regularity of weak solutions) Let L be the divergence form differential operator

Lu = - n ∑ i=1 ∂ i n ∑ j=1 a i j ∂ j u + c i u + n ∑ i=1 d i ∂ i u + du.
Assume that L has bounded coefficients and that conditions (2.66) and (2.67) hold. We make additionally the assumption

d = -∑ n i=1 ∂ i c i . Let u ∈ H 1 loc (Ω ) be a weak solution of Lu = 0 in Ω . Let x 0 ∈ Ω and 0 < r 0 < dist(x 0 ,Γ )/4. For 0 < r ≤ r 0 , define m(r) = inf B(r) u, M(r) = sup B(r) u and ω(r) = M(r) -m(r),
where B(r) denotes the ball of center x 0 and radius r. a) Prove that there exists a constant C > 0, only depending on the L ∞ -norm of the coefficients of λ -1 L, n and r 0 , so that

B(2r) (u -m(4r)) ≤ Cr n (m(r) -m(4r)), B(2r) (M(4r) -u) ≤ Cr n (M(4r) -M(r)). Deduce that ω(r) ≤ γω(4r) with γ = C -1 C . b) Establish the inequality ω(r) ≤ γ k ω(4r 0 ), r 0 4 k < r ≤ r 0 4 k-1 ,
and deduce from it that ω(r) ≤ Mr α for some constants M > 0 and α > 0.

(Weak form of Kato's inequality) Consider the divergence form operator

Lu = - n ∑ i=1 ∂ i n ∑ j=1 a i j ∂ j u + c i u + n ∑ i=1 d i ∂ i u + du
with bounded coefficients and a = (a i j ) is positive definite a.e. in Ω .

Let

f ∈ L 1 loc (Ω ) and u ∈ H 1 loc (Ω ) be a weak subsolution of Lu = f . Prove that v = u + ∈ H 1 loc (Ω ) is a weak sub-solution of of Lv = (χ {u>0} + µ χ {u=0} ) f in Ω for any µ ∈ [0, 1]. Hint: one can use first as a test function ϕ ε = φ θ (u/ε), with ε > 0, φ ∈ D(Ω ), φ ≥ 0 and θ ∈ C 1 (R) satisfying θ ≥ 0, θ (0) = µ, θ = 0 on ] -∞, -1], θ = 1 on [1,
+∞[. Pass then to the limit when ε goes to 0. 2.17. Consider the boundary value problem

-∆ u = F(u) + f in Ω , u = 0 on Γ , (2.106)
where F : R → R is continuous and non increasing function satisfying : there exist a > 0, b > 0 so that |F(s)| ≤ a + b|s| for any s ∈ R.

Let f ∈ L 2 (Ω ). We say that u ∈ H 1 0 (Ω ) is a variational solution of (2.106) if Ω ∇u • ∇vdx = Ω F(u)vdx + Ω f vdx for all v ∈ H 1 0 (Ω ). a) Prove that if u ∈ H 1 0 (Ω )
is a variational solution of (2.105) then there exists a constant C > 0, only depending on Ω , F(0) and f 2 , so that

∇u 2 ≤ C.
Hint: use that s(F(s) -F(0)) ≤ 0 for any s, which is a consequence of the fact that F is non increasing. b) Show that (2.106) has at most one variational solution.

Introduce the mapping T :

L 2 (Ω )×[0, 1] → L 2 (Ω ) : (w, λ ) → T (w, λ ) = u
, where u is the variational solution of the boundary value problem

-∆ u = λ (F(w) + f ) in Ω , u = 0 on Γ .
c) Check that T is compact. Then prove with the help of Leray-Schauder's fixed point theorem21 that T (•, 1) possesses a fixed point. Note that a fixed point of T (•, 1) is a solution of the variational problem (2.106).

Chapter 3

Classical solutions

In this chapter we show existence and uniqueness of classical solutions of elliptic partial differential equations under Dirichlet boundary condition. The approach is based only on interior Schauder estimates without any use of boundary estimates.

The original ideas are due to J. H. Michael [2] with an improvement by D. Gilbarg and N. S. Trudinger [5, Section 6.5, p. 112]. The content of this chapter is largely inspired by the lecture notes of a course given by M. V. Safonov at the university of Minesotta during the academic years 2003 and 2004.

Hölder spaces

Let Ω be a domain of R n , n ≥ 1. As usual, for k ∈ N, C k (Ω ) denotes the space of continuous functions u in Ω together with their derivatives ∂ u, | | ≤ k, where

∂ = ∂ 1 1 . . . ∂ n n if = ( 1 , . . . , n ).
We set for convenience ∂ 0 u = u.

Introduce the notations

|u| 0 = |u| 0;Ω = sup Ω |u|, [u] k,0 = [u] k,0;Ω = max | |=k |∂ u| 0;Ω . (3.1)
Define C k,0 (Ω ), where k ∈ N, as the subset of functions u ∈ C k (Ω ) satisfying

|u| k = |u| k,0 = |u| k,0;Ω = k ∑ j=0 [u] j,0;Ω < ∞. (3.2)
It is not hard to check that C k,0 (Ω ) endowed with the norm | • | k is a Banach space. Let 0 < α ≤ 1. We say that u is Hölder continuous, with exponent α, in Ω if the quantity 105

[u] α = [u] α;Ω = sup x,y∈Ω , x =y |u(x) -u(y)| |x -y| α (3.3) is finite. Set then [u] k,α = [u] k,α;Ω = max | |=k [∂ u] α;Ω . (3.4)
Define the Hölder space C k,α (Ω ), k ∈ N and 0 < α ≤ 1, as the Banach space of functions u ∈ C k (Ω ) with finite norm

|u| k,α = |u| k,α;Ω = |u| k,0;Ω + [u] k,α;Ω . (3.5)
We define in a similar manner the Hölder C k,α (Ω ).

We use for simplicity convenience

C α instead of C 0,α , 0 < α < 1. Let u, v ∈ C α (Ω ) with 0 < α ≤ 1.
Using the elementary inequality

|u(x)v(x) -u(y)v(y)| ≤ |u(x)||v(x) -v(y)| + |v(y)||u(x) -u(y)|, (3.1
) and (3.3) we easily obtain

[uv] α ≤ |u| 0 [v] α + |v| 0 [u] α . (3.6) Also, if k ∈ N and u ∈ C k+1,0 (Ω ) ∩C k,1 (Ω ) then it is straightforward to check that [u] k+1,0;Ω ≤ [u] k,1;Ω . (3.7)
We now establish other inequalities when

Ω = B r = {x ∈ R n ; |x -x 0 | < r},
δ h u(x) = δ 1 h,1 δ 2 h,2 . . . δ n h,n with δ h, j u(x) = u(x + he j ) -u(x) h ,
where (e 1 , e 2 , . . . , e n ) is the canonical basis of R n . Using the mean-value theorem, we can easily check that there exists y ∈ B ρ so that δ h u(x) = ∂ u(y). Inequality (3.9) is then a straightforward consequence of the following estimate

|δ h, j u(x)| ≤ 2 h |u| 0 .
The proof is then complete.

We now prove the following interpolation inequality.

Theorem 3.1. Let j, k ∈ N and 0 ≤ α, β ≤ 1 so that j + β < k + α. Let u ∈ C k,α (B r ).
Then, for any ε > 0, we have

r j+β [u] j,β ;B r ≤ εr k+α [u] k,α;B r +C(ε)|u| 0;B r , (3.10
)

with C(ε) = C(ε, n, k, α, β ).
Proof. Making the transformation x → (xx 0 ) /r we may assume that x 0 = 0 and r = 1. We distinguish four cases, where ε > 0 is fixed.

(a) The case j = k and 0 = β < α:

Fix z ∈ B 1 , | | = k and ρ ∈ (0, 1). Let x ∈ B 1 so that z ∈ B ρ = B ρ (x) ⊂ B 1 . By Lemma 3.1, there exists y ∈ B ρ so that |∂ u(z)| ≤ |∂ u(z) -∂ u(y)| + |∂ u(y)| ≤ |z -y| α [u] k,α + 2k ρ k |u| 0 ≤ (2ρ) α [u] k,α + 2k ρ k |u| 0 .
As z ∈ B 1 and | | = k are arbitrary, we conclude that

[u] k,0 ≤ (2ρ) α [u] k,α + 2k ρ k |u| 0 , 0 < ρ < 1. (3.11)
The expected inequality follows by taking ρ = min(1,

ε 1/α )/2. (b) The case j = k and 0 < β < α: from definition (3.4), we find | | = k and x, y ∈ B 1 such that 1 2 [u] k,β ≤ |∂ u(x) -∂ u(y)| |x -y| β ≤ |x -y| α-β [u] k,α . (3.12) If |x -y| ≤ (ε/2) 1/(α-β ) we have [u] k,β ≤ ε[u]
k,α and (3.10) is satisfied. Otherwise, the first inequality in (3.12) yields

[u] k,β ≤ 4|x -y| -β [u] k,0 ≤ C 0 [u] k,0 ,
where

C 0 = C 0 (ε) = 4(ε/2) β /(α-β ) .
The last inequality and (3.11) with ρ = min(1, (ε/C 0 ) 1/α )/2 entails (3.10).

(c) The case j < k and 0 < α: we find by applying (a) with ε = 1, α = 1 and j in place of k

[u] j,0 ≤ [u] j+1,0 +C|u| 0 .

On the other hand, for 0 < β ≤ 1, [u] j,β ≤ C[u] j+1,0 by (3.8). Hence, in any case (i.e. 0 ≤ β ≤ 1) we have the estimate

[u] j,β ≤ C([u] j+1,0 + |u| 0 ),
and by iteration, we get

[u] j,β ≤ C 0 [u] k,0 + |u| 0 ,
where C 0 = C 0 (n, k). As before this inequality yields (3.10).

(d) The case α = 0: since j + β < α + k = k, we have j ≤ k -1 and 0 ≤ β ≤ 1. The three preceding cases with α = 1 give

[u] j,β ≤ ε[u] k-1,1 +C(ε)|u| 0 (3.13)
for all ε > 0. But, from (3.8), we have Furthermore, (u m ) converges to u in C j,β (B r ) if j + β < k + α.

[u] k-1,1 ≤ C[u] k,0 . In light of this estimate, ( 3
Proof. As before we can assume that r = 1. Since B 1 is convex, C k,α (B 1 ) is clearly continuously imbedded in C 0,α (B 1 ). In light of Ascoli-Arzela's threorem, we deduce that (u m ) converges to u in C 0 (B 1 ). Pick j + β < k + α, ε 0 > 0 and set ε = ε 0 /(4A).

Then (3.10) applied to u m 1 -u m 2 entails [u m 1 -u m 2 ] j,β ≤ ε[u m 1 -u m 2 ] k,α + N(ε)|u m 1 -u m 2 | 0 ≤ ε 0 2 +C 0 |u m 1 -u m 2 | 0 , with C 0 = C 0 (ε 0 ) does not depend of m 1 and m 2 . But the sequence (u m ) is convergent in C 0 (B 1 )
. Whence, there exists m 0 = m 0 (ε 0 ) an integer so that

C 0 |u m 1 -u m 2 | 0 < ε 0 2 for m 1 , m 2 > m 0 . Thus [u m 1 -u m 2 ] j,β ≤ ε 0 for m 1 , m 2 > m 0 .
In other words, (u m ) is a Cauchy sequence in C j,β (B 1 ) and consequently it converges to u in C j,β (B 1 ). We finally note that the estimate

|u| k,α ≤ A = sup m |u m | k,α is straightforward.
We need introducing weighted Hölder spaces. For k ∈ N,

0 < α ≤ 1, γ ∈ R and u ∈ C k (Ω ) set [u] (γ) k,α;Ω = [u] (γ) k,α = sup x∈Ω d k+α+γ (x)[u] k,α;B(x) , (3.15) with d(x) = 1 2 dist(x,Γ ), B(x) = B d(x) (x). (3.16
)

Here Γ = ∂ Ω . Denote by C k;γ (Ω ) = C k,0;γ (Ω ) the Banach space of functions u ∈ C k (Ω ) with finite norm u (γ) k,0 = u (γ) k,0;Ω = k ∑ j=0 [u] (γ) j,0;Ω .
(3.17)

Define the weighted Hölder space C k,α;γ (Ω ) as the Banach space of functions u ∈ C k (Ω ) having finite norm

u (γ) k,α = u (γ) k,α;Ω = u (γ) k,0;Ω + [u] (γ) k,α;Ω . (3.18) Lemma 3.2. Let Ω ⊂ B 2r , r ≥ 1 and γ ≥ 0. Then C k,α (Ω ) ⊂ C k,α;γ (Ω ) and u (γ) k,α;Ω ≤ r k+α+γ |u| k,α;Ω . (3.19)
Proof. Follows from (3.15) by observing that d(x) ≤ r.

Lemma 3.3. Let β , γ ∈ R, 0 < α ≤ 1, u ∈ C 0,α;β (Ω ) and v ∈ C 0,α;γ (Ω ). Then [uv] (γ+β ) 0,α ≤ [u] (β ) 0,0 [v] (γ) 0,α + [v] (γ) 0,0 [u] (β ) 0,α (3.20)
and uv

(γ+β ) 0,α ≤ u (β ) 0,α v (γ) 0,α . (3.21)
Proof. We have from (3.6) We show by proceeding as in the preceding proof

d β +γ [uv] α;B ≤ d β |u| 0;B • d γ [v] α;B + d γ |v| 0;B • d β [u] α;B ,
[u] (γ) k,α ≤ C(n)[u] (γ) k+1,0 , 0 < α ≤ 1, (3.22)
for all u ∈ C k+1,0;γ (Ω ).

The following identity follows from the definition of Hölder weighted norms:

max | |= j [∂ u] (γ+ j) k-j,α = [u] (γ) k,α , 0 ≤ j ≤ k, 0 ≤ α ≤ 1, (3.23)
for all u ∈ C k,α;γ (Ω ). Theorem 3.1 can be used to obtain an interpolation inequality for the weighted Hölder spaces C k,α;γ (Ω ). In Theorem 3.1, with r = d = d(x), B r = B(x), we multiply each side of (3.10) by d γ and then we take the sup in x ∈ Ω . We get the following result.

Theorem 3.2. Let j, k ∈ N and 0 ≤ α, β ≤ 1 so that j + β < k + α. Let u ∈ C k,α;γ (Ω ),
where Ω is bounded domain of R n and γ ∈ R. We have, for any ε > 0, Recall that the norm of C 0;γ (Ω ) is given by

[u] (γ) j,β ;Ω ≤ ε[u] (γ) k,α;Ω +C(ε)|u| (γ) 0,0;Ω , (3.24) with C(ε) = C(ε, n, k, α, β ).
[u] (γ) 0,0 = sup Ω d γ (x) sup B(x)
|u|.

(3.26)

Let us compare this norm with the following one

u (γ) = u (γ) Ω = sup Ω d γ x |u(x)|, d x = dist(x,Γ ) = 2d(x). (3.27)
We shall use the following elementary inequality (γ) 0,0 and u (γ) are equivalent on C 0;γ (Ω ). We have precisely

1 2 d y < d x < 2d y for all x ∈ Ω , y ∈ B(x) = B d(x) (x). ( 3
2 -γ u (γ) ≤ [u] (γ) 0,0 ≤ 2 |γ|-γ u (γ) . (3.29)
Proof. The first inequality of (3.29) follows from

2 -γ d γ x |u(x)| = d γ (x)|u(x)| ≤ d γ (x) sup B(x)
|u|.

On the other hand, (3.27) and (3.28) imply

|u(y)| ≤ d -γ y u (γ) ≤ 2 |γ| d -γ x u (γ) = 2 |γ|-γ d -γ (x) u (γ) ,
for all y ∈ B(x), and hence the second inequality in (3.29) holds.

Remark 3.1. In the classical Schauder interior estimates (see [5, Chapter 6]), the notation

[u] (γ)
k,α is used for

A = max | |=k sup x,y∈Ω d k+α+γ x,y |∂ u(x) -∂ u(y)| |x -y| α = sup δ >0 δ k+α+γ [u] k,α;Ω δ , (3.30)
where 0

< α ≤ 1, k + α + γ ≥ 0, d x,y = min(d x , d y )
and

Ω δ = {x ∈ Ω ; d x = dist(x,Γ ) > δ }. (3.31)
One can prove that when Ω is a Lipschitz domain the semi-norm

k,α given by (3.15) and that given by (3.30) are equivalent whenever k + α + γ ≥ 0. We have in particular

C[u] k,α ≤ [u] (-k-α) k,α;Ω ≤ [u] k,α , (3.32)
where the constant C only depends on k, α and Ω .

In the case where k + α + γ < 0, we have A < ∞ only for polynomials of degree at most equal to k ; while [u] (γ) k,α < ∞ for larger class of functions u. If for instance k + α + γ < 0 ≤ k + 1 + γ and u ∈ C k+1 (B 1 ), then by (3.22) we have

[u] (γ) k,α ≤ C[u] (γ) k+1,0 ≤ C[u] k+1 .
(3.33)

Equivalent semi-norms on Hölder spaces

Let Ω be a bounded domain of R n and k ∈ N. Denote by P k the set of all polynomials of degree less or equal to k. The Taylor polynomial, of degree k at y ∈ R, corresponding to the smooth function u is given as follows

T y,k u(x) = ∑ | |≤k ∂ u(y) ! (x -y) ∈ P k . (3.34) Lemma 3.5. Let u ∈ C k,α (Ω ), 0 < α ≤ 1. Then, for any x, y ∈ Ω so that [x, y] 1 ⊂ Ω , we have |u(x) -T y,k u(x)| ≤ C(n)[u] k,α |x -y| k+α . (3.35)
Proof. By Taylor's formula there exists ξ ∈ [x, y] so that

u(x) = T y,k-1 u(x) + ∑ | |=k ∂ u(ξ ) ! (x -y) . Therefore u(x) -T y,k u(x) = ∑ |l|=k ∂ u(ξ ) -∂ u(y) ! (x -y) ≤ C(n) max | |=k ∂ u(ξ ) -∂ u(y) |x -y| k . But from (3.4) max | |=k ∂ u(ξ ) -∂ u(y) ≤ [u] k,α |ξ -y| α ≤ [u] k,α |x -y| α .
Then result follows.

Corollary 3.3. Let k ∈ N, 0 < α ≤ 1 and u ∈ C k,α (B ρ ), where B ρ = B ρ (x 0 ). Then E k [u; B ρ ] = inf p∈P k sup B ρ |u -p| ≤ C(n)[u] k,α ρ k+α . (3.36) Lemma 3.6. Let k ∈ N, 0 < α ≤ 1 and u ∈ C k,α (B ρ )
, where B ρ = B ρ (x 0 ). Then, for any ε > 0, we have

ρ -α max | |=k osc B ρ ∂ u ≤ ε[u] k,α;B ρ +C(ε)ρ -k-α E k [u; B ρ ], (3.37)
where

C(ε) = C(ε, n, k, α) is a constant and osc X f = sup X f -inf X f . Proof. Noting that osc f ≤ 2 sup | f |, (3.10) with r = ρ, j = k and β = 0 gives 1 2 ρ -α max | |=k osc B ρ ∂ u ≤ ρ -α [u] k,0;B ρ ≤ ε[u] k,α;B ρ +C(ε)ρ -k-α sup B ρ |u|.
In this inequality we substitute u by up, p ∈ P k . We obtain

1 2 ρ -α max | |=k osc B ρ ∂ u ≤ ρ -α [u] k,0;B ρ ≤ ε[u] k,α;B ρ +C(ε)ρ -k-α sup B ρ |u -p|.
In the right hand side of the last inequality we take the infimum over p ∈ P k and then we substitute ε by ε/2. The expected inequality then follows.

Theorem 3.3. Let k ∈ N, 0 < α ≤ 1, γ ∈ R so that k + α + γ ≥ 0 and u ∈ C k (Ω ) with finite semi-norm [u] (γ) k,α (see (3.15)). Set M (γ) k,α = M (γ) k,α [u; Ω ] = sup x∈Ω d k+α+γ (x) sup ρ∈(0,d(x)] ρ -k-α E k [u, B ρ (x)], (3.38
C 1 [u] (γ) k,α ≤ M (γ) k,α ≤ C 2 [u] (γ) k,α .
(3.39)

Here C 1 = C 1 (n, k, α, γ) > 0 and C 2 = C 2 (n) > 0 are two constants.
Proof. By Corollary 3.3, for any x ∈ Ω and ρ ∈ (0, d(x)], we have

d k+α+γ (x)ρ -k-α E k [u, B ρ (x)] ≤ Cd k+α+γ (x)[u] k,α;B ρ (x) ≤ C[u] (γ) k,α;Ω ,
implying the second inequality in (3.39).

To prove the first inequality in (3.39), we fix

x 0 ∈ Ω , d = d(x 0 ) = dist(x 0 ,Γ )/2, | | = k and x, y ∈ B d (x 0 ) so that A = [u] (γ) k,α ≤ 2d k+α+γ |∂ u(x) -∂ u(y)| |x -y| α . (3.40)
We distinguish two cases: (a

) ρ = |x -y| < d 2 and (b) ρ = |x -y| ≥ d 2 .
For case (a), it is not hard to see that x and y belong to a ball B ρ (z) ⊂ B(x 0 ) = B d (x 0 ). As d/2 ≤ d(z), we deduce by using (3.2) k,α , in order to obtain, for any ε > 0,

A ≤ C 0 d k+α+γ (z)ρ -α osc B ρ (z) ∂ u, ( 3
A ≤ C 0 εd k+α+γ (z)[u] k,α;B(z) +C(ε)d k+α+γ (z)ρ -k-α E k [u, B ρ (z)] ≤ C 0 εA +C(ε)M (γ) k,α .
Upon taking ε = 1/(2C 0 ), the expected inequality follows, i.e.

C 1 A ≤ M (γ) k,α .
With some simplifications we prove analogously to the preceding theorem the following result.

Theorem 3.4. Let k ∈ N, 0 < α ≤ 1 and u ∈ C k (R n ) having finite semi-norm [u] k,α;R n (see (3.4)). Let M k,α = M k,α [u; R n ] = sup x∈R n , ρ>0 ρ -k-α E k [u, B ρ (x)].
(3.42)

The semi-norms [u] k,α and M k,α are then equivalent. Precisely, we have

C 1 [u] k,α ≤ M k,α ≤ C 2 [u] k,α , (3.43)
where

C 1 = C 1 (n) > 0 and C 2 = C 2 (n, k, α) > 0 are two constant.

Maximum principle

In this section Ω is again an open subset of R n . We aim to derive some properties of the following linear partial differential operator satisfying the assumptions listed below.

Lu = n ∑ i, j=1 a i j ∂ 2 i j u + n ∑ i=1 b i ∂ i u + cu, u ∈ C 2 (Ω ). (3.44)
Here a i j , b i et c are continuous functions. We assume in the sequel that the following assumptions are satisfied on a domain varying eventually.

(a) (Ellipticity condition) The matrix A = (a i j ) is symmetric and there exists a constant ν ∈ (0, 1] so that By classical result from linear algebra, as A is symmetric, there exists an orthogonal matrix P so that P * AP = Λ , where Λ is the diagonal matrix whose diagonal elements consists in the eigenvalues λ 1 , . . . λ n of A (note that P * = P -1 ). As an orthogonal transformation leave invariant the Euclidian scalar product on R n , for

ν|ξ | 2 ≤ n ∑ i, j=1 a i j (x)ξ i ξ j ≤ ν -1 |ξ | 2 , x ∈ Ω , ξ ∈ R n . ( 3
η = P * ξ , we have ξ = Pη, |ξ | = |η| and n ∑ i, j=1 a i j ξ i ξ j = ξ * Aξ = η * P * APη = η * Λ η = n ∑ k=1 λ k η 2 k . Hence (3.45) is valid for any ξ ∈ R n if and only if λ k ∈ [ν, ν -1 ],
for any k = 1, 2, . . . , n. Using these facts, it is not hard to deduce from (3.45) that

a ii ≥ ν, |a i j | ≤ ν -1 , i, j = 1, 2, . . . , n, 2 (3.47) n ∑ i, j=1 a i j ξ i ξ j ≤ n ∑ i=1 a ii |ξ | 2 , ξ ∈ R n . 3 (3.48)
Also, the identity P * AP = Λ implies A = PΛ P * and then

a i j = n ∑ k=1 λ k ξ k i ξ k j , (3.49)
where ξ 1 , . . . ξ n are the column vectors of P * .

Lemma 3.7. Fix r > 0. Then there exists v 0 ∈ C ∞ (B r ), B r = B r (0), so that

Lv 0 ≤ -1 in B r . (3.50) Furthermore, 0 < v 0 ≤ C 0 = C 0 (ν, K, r) in B r , v 0 = 0 on ∂ B r . (3.51) Proof. Consider the function cosh(λ |x|), λ > 0 (of class C ∞ ). Then ∂ i cosh(λ |x|) = λ sinh(λ |x|)ξ i , with ξ = |x| -1 x, ∂ 2 i j cosh(λ |x|) = λ 2 cosh(λ |x|)ξ i ξ j + λ |x| -1 sinh(λ |x|)(δ i j -ξ i ξ j ).
Noting that |ξ | = 1, sinht < cosht and using (3.45) to (3.48), we obtain

(L -c) cosh(λ |x|) = n ∑ i, j=1 a i j ∂ 2 i j + n ∑ i=1 b i ∂ i cosh(λ |x|) ≥ λ 2 cosh(λ |x|) n ∑ i, j=1 a i j ξ i ξ j + λ sinh(λ |x|) n ∑ i=1 b i ξ i ≥ cosh(λ |x|)(λ 2 ν -λ K) ≥ λ (λ ν -K) ≥ 1,
2 To get a ii ≥ ν it is enough to take ξ = (δ k i ) in (3.45). While to prove |a i j | ≤ ν -1 we proceed as follows: if we take in (3.45), where k and are given, ξ i = 0 for i = k and i = , then

a kk ξ 2 k + a ξ 2 l + 2a k ξ k ξ ≤ ν -1 |ξ | 2
and hence (a ii ≥ ν for each i)

2a kl ξ k ξ l ≤ ν -1 |ξ | 2 .
The result follows for

ξ k = 1 √ 2 sgn(a k ) and ξ = 1 √ 2 . 3 For ξ ∈ R n and η = P * ξ , we have n ∑ i, j=1 a i j ξ i ξ j = n ∑ k=1 λ k η 2 k ≤ n ∑ k=1 λ k |η| 2 = n ∑ k=1 λ k |ξ | 2 = Tr(A)|ξ | 2 = n ∑ k=1 a kk |ξ | 2 .
for a well chosen λ = λ (ν, K).

Set v 0 = cosh(λ r)-cosh(λ |x|). Clearly, v 0 satisfies to (3.51) with C 0 = cosh(λ r). On the other hand, as c ≤ 0, we have

Lv 0 ≤ (L -c)v 0 = -(L -c) cosh(λ |x|) ≤ -1 in B r .
That is v 0 satisfies also to (3.50).

Theorem 3.5. (Weak maximum principle) Assume that c = 0. Let u ∈ C 2 (Ω )∩C(Ω ) satisfying Lu ≥ 0 (Lu ≤ 0) in Ω . Then sup Ω u = sup Γ u inf Ω u = inf Γ u . ( 3

.52)

Here Γ = ∂ Ω .

Proof. We first claim that if Lu > 0 in Ω then u can not attain it maximum at a point in Ω . Otherwise, we would find x 0 ∈ Ω (where u attains its maximum) so that

∂ i u(x 0 ) = 0, 1 ≤ i ≤ n, and n ∑ i, j=1 ∂ 2 i j u(x 0 )ξ i ξ j ≤ 0 for any ξ ∈ R n .
This inequality together with (3.49) entail

Lu(x 0 ) = ∑ i j a i j ∂ 2 i j u(x 0 ) = ∑ k λ k ∑ i j ∂ 2 i j u(x 0 )ξ k i ξ k j ≤ 0.
But this contradicts the fact that Lu > 0 in Ω .

In the case Lu ≥ 0, let v 0 be the function in Lemma 3.7 with B r ⊃ Ω . We have then, for all ε > 0, L(u

-εv 0 ) ≥ ε > 0. Whence sup Ω (u -εv 0 ) = sup Γ (u -εv 0 )
from the preceding case. Upon passing to the limit as ε goes to zero, we end up getting sup

Ω u = sup Γ u.
The proof is then complete.

An application of Theorem 3.5 to L 0 u = Lucu ≥ -cu ≥ 0 in Ω + = {u > 0} ⊂ Ω gives, where u ± = max(±u, 0), the following corollary.

Corollary 3.4. Let u ∈ C 2 (Ω ) ∩C(Ω ) satisfies Lu ≥ 0 (Lu ≤ 0) in Ω . Then sup Ω u + = sup Γ u + sup Ω u -= sup Γ u -. (3.53)
This corollary applied to uv yields the following result. A consequence of this theorem is

Theorem 3.7. Let f ∈ C(Ω ) and u ∈ C 2 (Ω ) ∩C(Ω ) so that Lu = f in Ω . Then sup Ω |u| ≤ sup Γ |u| +C 0 sup Ω | f |, (3.54)
where

C 0 = C 0 (ν, K, diam(Ω )) > 0 is a constant. Proof. We have Ω ⊂ B r for some ball B r = B r (x 0 ), r = diam(Ω ). Let v(x) = sup Γ |u| + sup Ω | f |v 0 (x -x 0 ),
where v 0 is as in Lemma 3.7. We find by using (3.50) and (3.51)

Lv(x) ≤ sup Ω | f |Lv 0 (x -x 0 ) ≤ -sup Ω | f | ≤ ± f (x) = ±Lu(x),
v ∈ C 2 (B) ∩C(B) such that Lv < f (resp. Lv > f ) in B, the inequality v ≥ w (resp. v ≤ w) on ∂ B implies v > w (resp. v < w) in B.
Observe that w is a sub-solution of Lu = f if and only if -w is a super-solution of Lu = f . Therefore it is enough to consider sub-solutions. Remark 3.2. If a i j , b i , c, f belong to C(Ω ) then any sub-solution w ∈ C 2 (Ω ) satisfies Lw ≥ f in Ω . Otherwise, we would find a ball B Ω so that Lw < f in B. The choice of v = w contradicts then the definition of a sub-solution. The converse is also true without the continuity condition on a i j , b i , c and f . Indeed, let w ∈ C 2 (Ω ) so that Lw ≥ f in Ω . Let B Ω a ball and v ∈ C 2 (B) ∩ C(B) such that Lv < f and v ≥ w on ∂ B. As L(wv) > 0, the function wv can not attain, by Theorem 3.5, its maximum at a point in Ω . Therefore

w -v < sup ∂ B (w -v) ≤ 0 in B. Lemma 3.8. Let w ∈ C(Ω ) be a sub-solution of Lu = f in Ω . Then, for any v ∈ C 2 (Ω ) ∩C(Ω ) so that Lv < f in Ω , the inequality v ≥ w on Γ implies v > w in Ω . If Lv ≤ f in Ω and v ≥ w on Γ then v ≥ w in Ω .
Proof. We proceed by contradiction. Assume then that there exists v ∈ C 2 (Ω ) ∩ C(Ω ) so that Lv < f in Ω , v ≥ w on Γ and that v > w in Ω does not hold. Hence, we find y ∈ Ω so that 0 ≥ -µ = min Ω (vw) = v(y)w(y).

(3.55)

For a ball B r = B r (y) Ω , we have

L(v + µ) = Lv + cµ ≤ Lv < f in B r , v + µ ≥ w on ∂ B r .
But w is a sub-solution. Then v + µ > w in B r and hence v(y) + µ > w(y). This inequality contradicts (3.55) and consequently v > w in Ω .

In the case Lv ≤ f in Ω , consider the function v 0 given by the Lemma 3.7, which is defined in B r = B r (0) ⊃ Ω . Thus the function

v ε = v + εv 0 , ε > 0, satisfies Lv ε ≤ f -ε < f in Ω , v ε ≥ v ≥ w on Γ .
Therefore v ε > w in Ω by the preceding case. We get v ≥ w in Ω , upon passing to the limit as ε goes to 0. Theorem 3.8. Let w ∈ C(Ω ) and assume that, for any y ∈ Ω , there exists a subsolution w y of Lu = f in a ball B y so that y ∈ B y Ω , w y ≤ w in B y , w y (y) = w(y).

L(v + µ) = Lv + cµ ≤ Lv in B y , v + µ ≥ w ≥ w y on ∂ B y
and v + µ > w y in B r (because w y is a sub-solution of Lu = f in B y B r ). In particular, v(y) + µ > w y (y) = w(y), which contradicts (3.57) and completes the proof.

Corollary 3.5. Let w 1 , w 2 be two sub-solutions of Lu = f in the respective domains

Ω 1 , Ω 2 . Assume that w 1 ≥ w 2 on Ω 1 ∩ ∂ Ω 2 and w 2 ≥ w 1 on Ω 2 ∩ ∂ Ω 1 . (3.58) Then w defined in Ω 1 ∪ Ω 2 by w(x) =    w 1 (x), x ∈ Ω 1 \Ω 2 , w 2 (x), x ∈ Ω 2 \Ω 1 , max(w 1 (x), w 2 (w)), x ∈ Ω 1 ∩ Ω 2 , (3.59)
belongs to C(Ω ) and it is a sub-solution of Lu = f in Ω .

We now use this corollary to construct special sub-solutions of the equation

Lu = d β -2 in Ω , with d = d x = dist(x, ∂ Ω ), 0 < β < 3.
(3.60)

We first consider the particular case

Ω = B R = B R (0), with R > 0 fixed. In the sequel, d = R -|x|, x ∈ B R .
Lemma 3.9. Fix 0 < β < 1 and R > 0. Then, for any Ω = B R = B R (0), there exists a sub-solution w ∈ C(Ω ) of the equation (3.60) so that w = 0 on Γ and

0 ≥ w ≥ -C 1 d β in Ω , (3.61)
where

C 1 = C 1 (ν, K, β , R) is a constant. Proof. Since d = R -|x|, we have ∂ i d β = -β d β -1 ξ i and ∂ 2 i j d β = β (β -1)d β -2 ξ i ξ j + β |x| -1 d β -1 (ξ i ξ j -δ i j ), with ξ = |x| -1 x. As |ξ | = 1, we get from (3.45) to (3.48) Ld β ≤ n ∑ i, j=1 a i j ∂ 2 i j + n ∑ i=1 b i ∂ i d β ≤ β (β -1)d β -2 n ∑ i, j=1 a i j ξ i ξ j -β d β -1 n ∑ i=1 b i ξ i ≤ β d β -2 [(β -1)ν + Kd],
where we used that β -1 < 0. In light of the last inequality, we can choose two constants β 0 > 0 and δ 0 ∈ (0, R/2), depending on ν , K, β and R, in such a way that

Ld β < -β 0 d β -2 , 0 < d < 2δ 0 . (3.62)
On the other hand, there exists, by Lemma 3.7,

v 0 ∈ C ∞ (B R ) so that Lv 0 ≤ -1, 0 ≤ v 0 ≤ C 0 (3.63)
in Ω , for some constant

C 0 = C 0 (ν, K, R). Set w 1 = -C 1 d β , w 2 = -C 1 δ β 0 -δ β -2 0 v 0 , (3.64)
with C 1 = max β -1 0 , (2 β -1) -1 δ -2 0 C 0 , and decompose Ω as follows

Ω = Ω 1 ∪ Ω 2 , with Ω 1 = Ω ∩ {d < 2δ 0 }, Ω 2 = Ω ∩ {d > δ 0 }. (3.65)
We obtain from (3.62) and (3.63)

Lw 1 ≥ C 1 β 0 d β -2 ≥ d β -2 in Ω 1 , Lw 2 ≥ δ β -2 0 ≥ d β -2 in Ω 2 , w 1 -w 2 = δ β -2 0 v 0 ≥ 0 on Ω 1 ∩ ∂ Ω 2 = {d = δ 0 }, w 2 -w 1 ≥ C 1 (2 β -1)δ β 0 -δ β -2 0 v 0 ≥ δ β -2 0 (C 0 -v 0 ) ≥ 0 in Ω \Ω 1 .
Therefore w 1 and w 2 satisfy the assumptions of Corollary 3.5 with

f = d β -2 . Whence, w defined by (3.59) is a sub-solution of Lu = d β -2 in Ω . Note that w 2 ≥ w 1 in Ω \Ω 1 yields 0 ≥ w ≥ w 1 ≥ -C 1 d β in Ω .
In other words, we proved (3.61). We end up the proof by remarking that w = w 1 = 0 on Γ .

We next extend Lemma 3.9 to domains having the exterior sphere property. We say that Ω has the exterior sphere property if for any y ∈ ∂ Ω there exists a ball

B = B R (z) ⊂ R n \ Ω satisfying Ω ∩ B = {y},
where R > 0 does not depend on y. Theorem 3.9. Lemma 3.9 can be extended to a bounded domain of Ω possessing the exterior sphere property, with a constant

C 1 = C 1 (ν, K, β , R, diam(Ω )) in (3.61).
Proof. Assume that Ω has the exterior sphere property. Then

d = d x = dist(x,Γ ) = min z∈Z (|x -z| -R), x ∈ Ω , where Z = {z ∈ R n ; dist(z, Ω ) = R}. If h(x) = |x| -R then d β = d β x = min z∈Z h β (x -z). (3.66)
We prove similarly to (3.62)

Lh β < -β 0 h β -2 , 0 < h < 2δ 0 , (3.67)
with constants β 0 > 0 and δ 0 only depending on ν, K, β and R.

Fix y ∈ Ω 1 = Ω ∩{d < 2δ 0 }, choose z ∈ Z such that h(y-z) = d y and set w y (x) = -h β (x -z). Then (3.67) is equivalent to Lw y (x) > β 0 d β -2
x (3.68) at x = y. By continuity, (3.68) remains true in a ball B y so that y ∈ B y Ω 1 . Moreover

w y ≤ -d β in B y , w y (y) = -d β y .
We conclude from Theorem 3.8 that -d β is a sub-solution of Lu = β 0 d β -2 in Ω 1 .

We can then substitute (3.62) by this inequality in the proof of Lemma 3.9. On the other hand, (3.63) holds in a B r ⊃ Ω , r = diam(Ω ). The remaining part of the proof is identical to that of Lemma 3.9.

Corollary 3.6. Under the assumptions of Theorem 3.

6, if u ∈ C 2 (Ω ) ∩ C(Ω ) is a solution of Lu = f in Ω and u = 0 sur Γ then u (-β ) Ω ≤ C 1 f (2-β ) Ω , (3.69)
where

0 < β < 1, C 1 = C 1 (ν, K, β , R, diam(Ω ))
is the constant in (3.61) and the norms • (γ) are defined in (3.27).

Proof. Set A = f (2-β ) Ω = sup d 2-β x | f (x)|.
Let w be a sub-solution of Lu = d β -2 in Ω given by Theorem 3.9. As Ad β -2 ≥ f , Aw is a sub-solution of Lu = f in Ω . We have u = w = 0 on Γ . Hence, u ≥ Aw ≥ -C 1 Ad β in Ω by Lemma 3.8 and (3.61). Since the inequalities above hold trivially when u is substituted by -u we get that |u| ≤ C 1 Ad β in Ω . That is, we have

u (-β ) = sup Ω d -β |u| ≤ C 1 A as expected.

Some estimates for Harmonic functions

Let Ω be a domain of R n . We say that u ∈ C 2 (Ω ) is harmonic (sub-harmonic, superharmonic) in Ω if it satisfies where ω n = 2π n/2 /Γ (n/2) is the Lebesgue measure of S n-1 .

∆ u = 0 (≥ 0, ≤ 0) in Ω . ( 3
The proof of this theorem will be given in Chapter 4.

We going to show that the mean-value theorem can be used to establish interior regularity of harmonic functions. We fix

ζ ∈ C ∞ (R n ) depending only on |x| such that ζ ≥ 0 in R n , ζ = 0 in R n \ B 1 (0) and R n ζ (x)dx = 1.
(3.73)

Let u ∈ L 1 loc (Ω ), ε > 0 and recall the definition

u (ε) (x) = B 1 (0) u(x -εy)ζ (y)dy = ε -n Ω u(z)ζ (ε -1 (x -z))dz, (3.74) for x ∈ Ω ε = {x ∈ Ω ; d x = dist(x,Γ ) > ε}.
Lemma 3.10. (a) We have, for any ε > 0,

u (ε) ∈ C ∞ (Ω ε ) and [u (ε) ] k,0;Ω ε = max | |=k sup Ω ε |∂ u (ε) | ≤ Cε -k sup Ω |u|, k ∈ N, (3.75
)

where C = C(n, k) is a constant. (b) Assume that |u(x) -u(y)| ≤ ω(ε) for any x, y ∈ Ω such that |x -y| ≤ ε. Then sup Ω ε |u (ε) -u| ≤ ω(ε). (3.76)
In particular, sup

Ω ε |u (ε) -u| ≤ ε α [u] α;Ω , 0 < α ≤ 1. (3.77) (c) If u ∈ C k (Ω ) then ∂ u (ε) = (∂ u) (ε) in Ω ε , for any | | ≤ k. Moreover, |u (ε) | k,α;Ω ε ≤ |u| k,α;Ω , k ∈ N, 0 ≤ α ≤ 1. (3.78)
In (3.77) and (3.78) we used the same notations as in (3.1) to (3.4).

Proof. This lemma is a consequence of (3.74). Indeed, we have, for

| | = k and x ∈ Ω ε , ∂ u (ε) (x) = ε -n-k B ε (x) u(z)∂ ζ (ε -1 (x -z))dz and hence |∂ u (ε) (x)| ≤ ε -n-k mes(B ε (x))[ζ ] k,0 sup Ω |u|. Whence (3.75) follows with C = mes(B 1 )[ζ ] k,0 . While (3.76) is a consequence of u (ε) (x) -u(x) = B 1 [u(x -εy) -u(x)]ζ (y)dy, x ∈ Ω ε .
Finally, (3.78) is immediate from the first inequality in (3.74).

Lemma 3.11. If u is harmonic in Hence, we find by using ζ (y)dy = 1

Ω then u = u (ε) in Ω ε , ε > 0. Furthermore, u ∈ C ∞ (Ω ) and max | |=k sup Ω d k x |∂ u| ≤ C sup Ω |u|, k ∈ N, (3.79
u (ε) (x) = u(x) 1 0 dr |y|=r ζ (y)ds y = u(x) B 1 (0) ζ (y)dy = u(x), x ∈ Ω ε .
We complete the proof by noting that (3.79) is obtained by taking ε = d x = dist(x,Γ ) in (3.75).

The remaining part of this section is devoted to the Dirichlet problem in a ball, for an operator that can be deduced of the Laplace operator by an orthogonal transformation. Consider then an operator of the form

L 0 u = n ∑ i, j=1 a i j ∂ 2 i j u,
where a i j are constants, for each i, j and the matrix (a i j ) is symmetric and positive definite. By diagonalizing the matrix (a i j ) one can easily check that L 0 can be transformed into the Laplace operator by means of a change of variable.

We study first the following Dirichlet problem

L 0 u = n ∑ i, j=1 a i j ∂ 2 i j u = f in B r = B r (x 0 ), u = ϕ on ∂ B r , (3.82)
when f and ϕ are polynomials.

Lemma 3.12. Let f ∈ P k and ϕ ∈ P k+2 be given 4 . Then (3.82) admits a unique solution u in C 2 (B r ) ∩C(B r ). This solution takes the form u = ϕ + (r 2 -|x| 2 )g, with g ∈ P k .

Proof. We may assume, without loss of generality, that r = 1 and x 0 = 0. The uniqueness is contained in 3.6 (comparison principle). To prove the existence we set u = v + ϕ. Then (3.82) is transformed into the equation

L 0 v = f 0 in B 1 , v = 0 on ∂ B 1 , (3.83)
with f 0 = f -L 0 ϕ ∈ P k . Consider then the linear map

T : P k → P k : p → T p = L 0 1 -|x| 2 p .
If T p = 0 then u = 1 -|x| 2 p is a solution (3.82) with f = 0 and ϕ = 0. By uniqueness u = 0 and hence p = 0. Therefore T is injective, and since P k is of finite dimension, T is also surjective, i.e. T P k = P k . Hence, T g = f 0 , for some g ∈ P k , and v = 1 -|x| 2 g is a solution of (3.83).

This lemma will be useful to extend the existence of solutions of (3.82) for f and ϕ in a larger class than polynomials. Lemma 3.13. Let 0 < α ≤ 1, γ ∈ R and let (u m ) m≥1 be a bounded sequence in C 2,α;γ (Ω ), i.e. u m (γ) 2,α ≤ A for any m, for some constant A > 0. Assume that the following limits exist Proof. From Stone-Weierstrass's theorem, we find a sequences of polynomials (ϕ m ) so that |ϕ -ϕ m | ≤ 1/m in B r . We conclude then, by applying Lemma 3.12, that there exists a sequence of polynomials (u m ) satisfying

L 0 u m = 0 in B r , u m = ϕ m on ∂ B r . (3.85)
The comparison principle then yields sup

B r |u i -u j | ≤ sup ∂ B r |ϕ i -ϕ j | ≤ 1 i + 1 j -→ 0 as i, j → ∞.
That is (u m ) is a Cauchy sequence in C(B r ). Therefore it converges to some u ∈ C(B r ).

As we have said above there exists a bijective linear map y = Ax transforming L 0 v(x) = 0 in B r to ∆ v(y) = 0 in Ω = A(B r ). Therefore, estimates (3.79) and (3.80) for v(y) in Ω produce analogous estimates for v(x) in B r , with a constant C = C(n, ν, k). In particular, these estimates are valid for each u m with a constant C independent of m. Passing to the limit, as m goes to infinity, we get u = lim u m ∈ C ∞ (B r )-note that according to (3.79), (∂ u m ) is a Cauchy sequence for each -satisfies the same estimates as u m . We end up the proof by using Lemma 3.13, with γ = 0 and α = 1. We obtain

L 0 u = lim L 0 u m = 0. Corollary 3.7. Let r > 0, x 0 ∈ R n 0 = {x ∈ R n ; x n = 0} and set B + r = B + r (x 0 ) = {x ∈ B r (x 0 ); x n > 0}. ( 3
ϕ(x 1 , . . . , x n-1 , -x n ) = -ϕ(x 1 , . . . , x n-1 , x n ), (x 1 , . . . , x n-1 , x n ) ∈ B + r \ Γ .
The result follows then by using existence, uniqueness and estimates (3.79) and (3.80) for the solution of the Cauchy problem

∆ u = 0 in B r , u = ϕ on ∂ B r .
(b) We proceed similarly to (a) by using in that case the even extension of ϕ, still denoted by ϕ :

ϕ(x 1 , . . . , x n-1 , -x n ) = ϕ(x 1 , . . . , x n-1 , x n ), (x 1 , . . . , x n-1 , x n ) ∈ B + r \ Γ .
The proof is then complete.

Interior Schauder estimates

Let Ω be a domain of R n . As in the preceding section,

L = n ∑ i, j=1 a i j ∂ 2 i j + n ∑ i=1 b i ∂ i + c, (3.90)
We assume additionally to the assumptions in the preceding section that the coefficients of L satisfy the following regularity condition : there exist 0 < α < 1 and

K 1 > 0 so that max i, j a i j (0) 0,α , max i b i (1) 0,α , c (2)
0,α ≤ K 1 , (3.91)
where the norm • (γ) 0,α is defined in (3.18).

Theorem 3.12. For any γ ∈ R and u ∈ C 2,α;γ (Ω ), we have f = Lu ∈ C 0,α;γ+2 (Ω ) and f

(γ+2) 0,α ≤ C 1 u (γ)
2,α .

(3.92)

Here C 1 = C 1 (n, K 1 ) is a constant.
Proof. We have by (3.23)

∂ u (γ+ j) k-j,α ≤ u (γ) k,α , 0 ≤ j ≤ k, 0 < α ≤ 1. (3.93)
This estimate, combined with (3.21) (Lemma 3.3) and (3.93), implies

a i j ∂ 2 i j u (γ+2) 0,α ≤ a i j (0) 0,α ∂ 2 i j u (γ+2) 0,α ≤ K 1 u (γ) 2,α ,
for any i, j. In an analogous way, we have also in light of (3.21)

b i ∂ i u (γ+2) 0,α ≤ b i (1) 0,α ∂ i u (γ+1) 0,α ≤ K 1 u (γ) 1,α ≤ C u (γ) 2,α , cu (γ+2) 0,α ≤ c (2) 0,α u (γ+2) 0,α ≤ C u (γ) 2,α ,
where C = C(n, K 1 ) is a constant. The last three estimates yield (3.92).

Remark 3.3. Note that for the lower order terms we have in light of (3.22)

n ∑ i=1 b i ∂ i u + cu (γ+2) 0,α ≤ N(n, K 1 ) u (γ) 2,0 .
(3.94)

We now give a result saying that for an operator with Hölder continuous coefficients the norms of u and f = Lu in (3.92) are "almost" equivalent. Theorem 3.13. We have, for any γ ∈ R and u ∈ C 2,α;γ (Ω ),

u (γ) 2,α ≤ C [u] (γ) 0,0 + [ f ] (γ+2) 0,α , (3.95)
where f = Lu and C = C(n, ν, K, K 1 , α, γ)5 is a constant.

Proof. In this proof C is generic constant only depending on n, ν, K, K 1 , α and γ.

On the other hand, to simply the notations, we set

U 2,α = [u] (γ) 2,α , U k = [u] (γ) k,0 , F α = [ f ] (γ+2) 0,α , F 0 = [ f ] (γ+2) 0,0 . (3.96) Let γ ∈ R and u ∈ C 2,α;γ (Ω ) be given. Assume first that b i = 0 and c = 0 in Ω . Fix y ∈ Ω , d = d y = 1 2 dist(y,Γ ), ρ ∈ (0, d] and ε ∈ (0, 1 2 ].
Let r = ρ/ε. We distinguish two cases: (a) r ≤ d and (b) r > d. In case (a), let

a i j 0 = a i j (y), L 0 = n ∑ i, j=1 a i j 0 ∂ 2 i j , ϕ = u -T y,2 u,
and denote by v the solution of the problem

L 0 v = n ∑ i, j=1 a i j 0 D i j v = 0 in B r = B r (y), v = ϕ on ∂ B r . (3.97) By Theorem 3.7 v ∈ C ∞ (B r ) ∩C(B r ) and [v] 3,0;B r/2 ≤ Cr -3 sup B r |v| = Cr -3 sup ∂ B r |ϕ|.
As ρ = εr ≤ r/2, we apply first Corollary 3.3, (3.8) to v in B ρ and then apply Lemma 3.5 to u in B r . We then get

ρ -2-α E 2 [v; B ρ ] ≤ Cρ 1-α [v] 3,0;B r/2 ≤ Cρ 1-α r -3 sup ∂ B r |ϕ| ≤ Cρ 1-α r α-1 [u] 2,α;B r = Cε 1-α [u] 2,α;B r . Since r ≤ d, we obtain from the definition of U 2,α = [u] (γ) 2,α (see (3.15)) d 2+α+γ ρ -2-α E 2 [v; B ρ ] ≤ Cε 1-α U 2,α .
(3.98)

Next, we estimate ϕv in B r . We have

L 0 (ϕ -v) = L 0 ϕ = n ∑ i, j=1 a i j 0 [∂ 2 i j u(x) -∂ 2 i j u(y)].
We get by noting that

f = ∑ n i, j=1 a i j ∂ 2 i j u L 0 (ϕ -v) = L 0 (ϕ) = n ∑ i, j=1 a i j 0 -a i j (x) ∂ 2 i j u(x) + f (x) -f (y). (3.99)
In view of notations (3.96), inequality (3.91) yields that we have, for any x ∈ B r = B r (y) ⊂ B d ,

a i j 0 -a i j (x) = a i j (y) -a i j (x) ≤ r α a i j α;B r ≤ d -α r α K 1 , ∂ 2 i j u(x) ≤ [u] 2,0;B r ≤ d -2-γ U 2 , | f (x) -f (y)| ≤ r α [ f ] α;B r ≤ d -2-α-γ r α F α .
These inequalities together with (3.99) entail

|L 0 (ϕ -v)| ≤ Ar α in B r , (3.100)
with

A = d -2-α-γ (n 2 K 1 U 2 + F α ). (3.101) If w(x) = Ar α 2nν r 2 -|x -y| 2 then L 0 w ≤ -Ar α ≤ -|L 0 (ϕ -v)| in B r = B r (y), w = ϕ -v = 0 on ∂ B r .
We deduce from this and the comparison principle sup

B ρ |ϕ -v| ≤ sup B r |ϕ -v| ≤ sup B r |w| = A 2nν r 2+α .
Since r = ρ/ε, (3.101) gives

d 2+α+γ ρ -2-α sup B ρ |ϕ -v| ≤ Cε -2-α (U 2 + F α ). (3.102)
On the other hand,

E 2 [u; B ρ ] ≤ E 2 [v; B ρ ] + E 2 [ϕ -v; B ρ ] ≤ E 2 [v; B ρ ] + sup B ρ |ϕ -v|.
This inequality together with (3.98) and (3.102) imply

d 2+α+γ ρ -2-α E 2 [u; B ρ ] ≤ Cε 1-α U 2,α +Cε -2-α (U 2 + F α ). (3.103)
We now consider case (b):

r = ρ/ε > d. We have d 2+α ρ -2-α < ε -2-α and d γ E 2 [u; B ρ ] ≤ d γ sup B ρ |u| ≤ U 0 .
In consequence, the left hand side of (3.103) is less or equal to Cε -2-α U 0 and hence (3.103) is satisfied for both cases (a) and (b). As y ∈ Ω and 0 < ρ ≤ d = d(y) can be chosen arbitrarily, we conclude

M (γ) 2,α ≤ Cε 1-α U 2,α +Cε -2-α (U 2 +U 1 +U 0 + F α ), (3.104)
for any ε > 0, where the semi-norm

M (γ)
2,α is defined in (3.38). By virtue of Theorem 3.3, this inequality still holds when

M (γ) 2,α is substituted by U 2,α , i.e. U 2,α ≤ Cε 1-α U 2,α +Cε 2-α (U 2 +U 1 +U 0 + F α ).
Thus, there exists ε = ε(n, ν, K, K 1 , α, γ) > 0 (take for instance 2Cε 1-α ≤ 1) for which

U 2,α ≤ C(U 2 +U 1 +U 0 + F α ). (3.105)
In other words, we proved that (3.105) holds when b i = 0 et c = 0. For the general case, we write Lu = f in the form

n ∑ i, j=1 a i j ∂ 2 i j u = f 0 = f - n ∑ i=1 b i ∂ i u -cu.
It follows from the remark following Theorem 3.12

[ f 0 ] (2+γ) 0,α ≤ [ f ] (2+γ) 0,α +C(U 2 +U 1 +U 0 ).
We have, according to (3.105) and the interpolation inequality in Theorem 3.2,

U 2 +U 1 ≤ εU 2,α +C(ε)U 0 . Whence u (γ) 2,α = U 2,α +U 2 +U 1 +U 0 ≤ N(U 0 + F α ),
which completes the proof.

We assume henceforward that Ω possesses the exterior sphere property with some R > 0.

Theorem 3.14. Let β ∈ (0, 1) and u ∈ C 2,α;-β (Ω ). Then f = Lu ∈ C 0,α;2-β (Ω ) and

C 1 f (2-β ) 0,α ≤ u (-β ) 2,α ≤ C 2 f (2-β ) 0,α , (3.106
)

with constants C 1 = C 1 (n, K 1 ) and C 2 = C 2 (n, ν, K, K 1 , α, β , R, diam(Ω )).
Proof. The first inequality is contained in Theorem 3.12. Prior to proving the second inequality, we note that according to Lemma 3.4 the norms [w]

(-β ) 0,0 and w (-β ) are equivalent. On the other hand, we have by Corollary 3.6 the estimate

u (-β ) ≤ C f (-β ) .
We obtain then by applying Theorem 3.13 with γ = -β u

(-β ) 2,α ≤ C [ f ] (2-β ) 0,0 + [ f ] (2-β ) 0,α = C f (2-β ) 0,α ,
This proves the second inequality in (3.106).

Remark 3.4. The two inequalities in (3.106) show that the linear operator

L : u ∈ C 2,α;-β (Ω ) → f = Lu ∈ C 0,α;2-β (Ω )
is bounded ; the mapping f → u = L -1 f defines also a bounded operator on L(C 2,α;-β (Ω )) ⊂ C 0,α;2-β (Ω ). As 0 < β < 1, it is not hard to check that w ∈ C 2,α;-β (Ω ) vanishes on Γ . In other words, solving in C 2,α;-β (Ω ) ⊂ C 2,α loc (Ω ) ∩ C(Ω ) the Dirichlet problem

Lu = f in Ω , u = 0 on Γ , (3.107)
for f ∈ C α (Ω ) ⊂ C 0,α;2-β (Ω ) is reduced to the surjectivity of L, i.e.

L(C 2,α;-β (Ω )) = C 0,α;2-β (Ω ).

(3.108)

We establish in the next section that (3.108) holds when Ω = B r . The general case will be discussed in the last section.

The Dirichlet problem in a ball

In this section, the ball B r = B r (x 0 ) ⊂ R n and the constants α, β ∈ (0, 1) are fixed. Consider then the Dirichlet problem

η ε = 0 in B r \ B r-ε , η ε = 1 in B r-3ε , et |∇η ε | ≤ C/ε, (3.111)
where the constant C is independent on ε. It is not difficult to check that (3.111) holds for the regularization

η ε = h (ε)
ε of the function h ε given by h ε = 1 in B r-2ε and h = 0 elsewhere. We obtain from (3.22) and (3.111)

η ε (0) 0,α ≤ C η ε (0) 1,0 ≤ C(n), ε > 0. (3.112)
Now, f ε = η ε f satisfies the assumptions of the second step and we have from inequalities (3.21) and (3.112)

f ε 2 = η ε f (2-β ) 0,α ≤ η ε (0) 0,α f (2-β ) 0,α ≤ C f 2
for sufficiently small ε. By (3.110) the solutions u ε of ∆ u ε = f ε are bounded in B 1 and hence

|u ε (x)| ≤ (r -|x|) β u ε (-β ) ≤ (r -|x|) β u ε 1 ≤ C(r -|x|) β (3.113)
for any x ∈ B r and small ε > 0.

In order to study the convergence of u ε , as ε → 0, we observe that

∆ (u ε -u ε ) = (η ε -η ε ) f = 0 in B r-3ε if 0 < ε < ε.
In light of (3.113), maximum principle's yields sup

B r |u ε -u ε | = sup B r \B r-3ε |u ε -u ε | ≤ Cε β if 0 < ε < ε.
Hence (u ε ) in bounded is B 1 and converges in C(B r ). We conclude similarly to the second step that u = lim u ε ∈ B 1 and ∆ u = f . Fourth step. We use the continuity method to treat the general case. For 0 ≤ t ≤ 1, set L t = ∆ +t(L -∆ ). We have in particular L 0 = ∆ and L 1 = L. The assumptions on the coefficients of L still valid for the coefficients of L t with the same constants ν, K et K 1 . Consequently, the inequalities in (3.110) hold for f = L t u, 0 ≤ t ≤ 1.

From the third step, L 0 (B 1 ) = B 2 . Assume that L s (B 1 ) = B 2 for some s ∈ [0, 1]. Then again (3.110) tells us that L s possesses a bounded inverse L -1 s : B 2 → B 1 . For t ∈ [0, 1] and f ∈ B 2 , L t u = f is equivalent to the following equation

L s u = f + (t -s)(L -∆ )
and hence it is also equivalent to the following equation

u = Tu = L -1 s f + (t -s)L -1 s (L -∆ )u.
Once again (3.110) implies

Tu -T v 1 = |t -s| L -1 s (L -∆ )(u -v) 1 ≤ C 2 |t -s| (L -∆ )(u -v) 2 ≤ 2C 1 C 2 |t -s| u -v 1 , for any u, v ∈ B 1 . Whence, if |t -s| < δ = (2C 1 C 2 ) -1 then T : B 1 → B 1
is strictly contractive. Therefore, there exists u ∈ B 1 such that u = Tu or equivalently L t u = f . As f is chosen arbitrary, we have L t (B 1 ) = B 2 provided that |t -s| < δ . Dividing [0, 1] into sub-intervals of length less or equal to δ , we deduce that L t (B 1 ) = B 2 for any t ∈ [0, 1]. We have, in particular for t = 1, that L(B 1 ) = B 2 . This completes the proof. Remark 3.5. If ϕ = 0, the existence of a solution of (3.109) in C 2 (B r ) ∩C(B r ) is not guaranteed. To see this, we consider the following one dimensional problem

u -x -2 u = 0 in (0, 1), u(0) = u(1) = 1, (3.114)
where (0, 1) is considered as a ball B r = B r (x 0 ), r

= x 0 = 1/2. If u ∈ C 2 (B r ) ∩C(B r )
then by (3.114) we have

u ∼ x -2 , u ∼ x -1 , u ∼ ln 1 x as x → 0 + , contradicting u(0) = 1.
Theorem 3.16. Assume that a i j , b i , c, f belong to C α (B r ) and ϕ ∈ C(B r ). Then the Dirichlet problem (3.109) admits a unique solution u ∈ C 2,α;0 (B r ) ∩C(B r ).

Proof. Consider first the case ϕ ∈ C 3 (B r ). If u = v + ϕ then (3.109) is equivalent to the following equation

Lv = f 0 in B r , v = 0 on ∂ B r , (3.115)
where

f 0 = f -Lϕ ∈ C α (B r ) ⊂ B 2 .
The last theorem guarantees the solvability of (3.115) and therefore the solvability of (3.109) in B 1 ⊂ C 2,α;0 (B r ) ∩C(B r ).

For the general case we approximate ϕ ∈ C(B r ) by sequence of polynomials (ϕ m ) so that |ϕϕ m | ≤ 1/m in B r with m ≥ 1. As in Theorem 3.13, the solutions of the problems:

Lu m = f in B r , u m = ϕ m on ∂ B r , define a sequence (u m ) converging in C(B r ) to u ∈ C(B r
). Once again Theorem 3.13 with γ = 0 yields u m (0)

2,α ≤ A, for any m, where the constant A is independent on m. Lemma (3.13) then implies that u ∈ C 2,α;0 (B r ) and Lu = f in B r .

Dirichlet problem on a bounded domain

We extend the results of the preceding section to the Dirichlet problem on bounded domain. Consider then the boundary value problem

Lu = f in Ω , u = ϕ on ∂ Ω . (3.116)
The assumptions on L are those of the preceding section. Furthermore, we assume that Ω has the exterior sphere property. Theorem 3.17. There exists, for any f ∈ B 2 = C 0,α;2-β (Ω ), a unique u ∈ B 1 = C 2,α;-β (Ω ) satisfying Lu = f in Ω . In other words, L sends

B 1 onto B 2 , i.e. L(B 1 ) = B 2 . Proof. Fix f ∈ B 2 = C 0,α;2-β (Ω ) and let A = f (2-β )
0,α;Ω . We prove the existence of a solution u ∈ B 1 of Lu = f in Ω by using a variant of Perron's method for sub-solutions.

We split the proof in four steps.

First step. We have, for x ∈ Ω ,

| f (x)| ≤ d x 2 β -2 [ f ] (2-β ) 0,0 ≤ Ad β -2 x , with d x = dist(x,Γ ). (3.117)
There exists by Theorem 3.9 a sub-solution

w ∈ C(Ω ) of Lu = d β -2 x in Ω satisfying 0 ≥ w ≥ -C 1 d β x in Ω .
We have according to estimate (3.117) that U 0 = Aw and -U 0 are respectively sub-solution and super-solution of Lu = f in Ω and

0 ≥ U 0 (x) ≥ -C 1 Ad β x , x ∈ Ω , (3.118)
where

C 1 = C 1 (ν, K, β , R, diam(Ω )) is a constant. Second step.
Starting from U 0 we construct a sequence of sub-solutions (U k ) according to the following scheme : fix (y j ) a dense sequence in Ω and consider a sequence of the form (x 1 , x 2 , x 3 , . . .) = (y 1 , y 1 , y 2 , y 1 , y 2 , y 3 , . . .) in such a way that each y j appears infinitely many times in the sequence (x k ). Denote

d k = dist(x k ,Γ )/2 and B k = B(x k ) = B d k (x k ), k ≥ 1.
By virtue of Theorem 3.16, there exists u ∈ C 2,α;0 (B 1 ) ∩C(B 1 ) a solution of the problem

Lu 1 = f in B 1 , u 1 = U 0 on ∂ B 1 .
As U 0 is a sub-solution of Lu = f in Ω ⊃ B 1 , we have u 1 ≥ U 0 in B 1 . Define then the function U 1 as follows

U 1 = u 1 on B 1 , U 1 = U 0 in Ω \ B 1 .
We have U 1 ∈ C(Ω ) and U 1 ≥ U 0 in Ω . Moreover, Corollary 3.5 guarantees that U 1 is a sub-solution of Lu = f in Ω . Repeating this construction for k = 2, 3, . . .: u k is the solution of the problem

Lu k = f in B k , u k = U k-1 on ∂ B k and define U k by U k = u k in B k , U k = U k-1 in Ω \ B k .
We obtain in that manner a sequence

U 0 ≤ U 1 ≤ U 2 . . . ≤ U k ≤ . . . (3.119) of sub-solution of Lu = f in Ω . As -U 0 is a super-solution of Lu = f in Ω , we have also u k ≤ -U 0 in B k , U k ≤ -U 0 in Ω , k ≥ 1. (3.120)
Hence, there exists

u(x) = lim k→∞ U k (x), x ∈ Ω . (3.121)
Third step. We prove that u ∈ C 2,α loc (Ω ) and Lu = f in Ω . We get by using inequalities (3.118) to (3.121)

|u(x)| ≤ sup k |U k (x)| ≤ C 1 Ad β x , x ∈ Ω . (3.122) Fix j ≥ 1, d = d(y j ) = dist(y j , ∂ Ω )/2, B = B d (y j
) and choose (x k i ) a sub-sequence of (x k ) so that x k i = y j for any i ≥ 1. Then B k i = B and the function

u k i ∈ C 2,α;0 (B) ∩ C(B) satisfies Lu k i = f in B, for any i ≥ 1. We have [ f ] (2) 0,α,B ≤ d 2+α [ f ] α;B ≤ d β [ f ] 0,α;B ≤ Ad β
, and from Theorem 3.13 with γ = 0, and (3.122) we obtain

u k i (0) 2,α;B ≤ C sup B |u k i | + [ f ] (2) 0,α;B ≤ CAd β , i ≥ 1.
We then get, by applying Lemma 3.13, u = lim u k i ∈ C 2,α;0 (B), Lu = f in B and

u (0) 2,α;B ≤ CAd β . (3.123)
But ∪ j B(y j ) = Ω . Hence, u ∈ C 2,α loc (Ω ) and Lu = f in Ω . Fourth step. We show in this last step that u ∈ B 1 = C 2,α;-β (Ω ). We first note that (3.123) entails

2 ∑ k=0 d k [u] k,0;B d/2 + d 2+α [u] 2,α;B d/2 ≤ CAd β , (3.124)
where

B d/2 = B d/2 (y j ) is the ball of radius d/2 = dist(y j , ∂ Ω )/4.
We now evaluate the norm of u in C 2,0;-β (Ω ). By (3.15) and (3.16) we have, for any k ≥ 0, that there exist

x 0 ∈ Ω , | | = k and x ∈ B(x 0 )(= B d x 0 (x 0 )) so that 1 2 [u] (-β ) k,0;Ω ≤ d k-β (x 0 )|∂ u(x)|. (3.125)
As (y j ) is dense in Ω , we find y j sufficiently close to x in such a way that x ∈ B d/2 (y j ), d(x) ≤ 2d, where d = d(y j ). Then inequalities (3.123) and (3.124) imply

[u] (-β ) k,0;Ω ≤ 2 k+1 d k-β [u] k,0;B d 2 ≤ CA, k ≥ 0. (3.126) It remains to estimate [u] (-β )
k,α;Ω . As previously, there exist

x 0 ∈ Ω , | | = k and x, y ∈ B(x 0 ) so that 1 2 [u] (-β ) 2,α;Ω ≤ d 2+α-β (x 0 ) |∂ u(x) -∂ u(y)| |x -y| α . ( 3
(-β ) 2,α;Ω ≤ 2 3+α d 2+α-β [u] 2,α;B d 2 ≤ CA. Finally, if |x -y| ≥ d(x 0 ) 4 then (3.127) implies [u] (-β ) 2,α;Ω ≤ 2 (4 α ) d 2-β (x 0 )|∂ u(x) -∂ u(y)| ≤ 4 1+α d 2-β (x 0 )[u] 2,0;B(x 0 ) ≤ 4 1+α [u] (-β ) 2,0;Ω ≤ CA.
This completes the proof.

Theorem 3.18. Assume that a i j , b i , c, f belong to C α (Ω ) and ϕ ∈ C(Ω ). Then Dirichlet problem (3.116) admits a unique solution u ∈ C 2,α;0 (Ω ) ∩C(Ω ).

Proof. Quite similar to that of Theorem 3.16 with B r substituted by Ω .

Exercises and problems

3.1. Let 0 < α ≤ 1 and f ∈ D(R). Set [ f ] α = sup x∈R, h =0 | f (x + h) -f (x)| |h| α , [ f ] * α = sup x∈R, h =0 | f (x + h) -2 f (x) + f (x + h)| |h| α . Prove that [ f ] * α ≤ 2[ f ] α and, for 0 < α < 1, [ f ] α ≤ C[ f ] * α ,
where the constant C does not depend of α. Hint: consider the operators T h f (x) = f (x + h), I f (x) = f (x) and use the following identities

T h -I = 1 2 (T 2 h -I) -(T 2 h -I) 2 , T 2 h = T 2h . 3.2. (a) Let 0 < α < 1 and u ∈ C 0 c (R n ) so that U α = [u] 0,α := sup x =y |u(x) -u(y)| |x -y| α < ∞
and, for ε > 0, denote by u (ε) the regularization of u :

u (ε) (x) = R n u(x -εy)ϕ(y)dy. (i) Show that |∂ u (ε) | ≤ Cε α-k U α , for any ∈ N n , | | = k ≥ 1, where C = C(ϕ). (ii) If ∂ ε = ∂ ∂ ε , prove that |∂ ε u (ε) | ≤ Cε α-1 U α , where C = C(n, ϕ) is a constant. Hint: use R n ∇ϕ(y) • ydy = -n. (b) Let u, v ∈ C 0 (R) with support in (-1, 1) so that U α < ∞ and V β = [v] 0,β < ∞,
for some constants α, β ∈ (0, 1). Set then w = u * v. (i) Assume that α + β < 1. Show that [w] 0,α+β < ∞. Hint: use the identity w = u (1) * v (1)w 1w 2 , where

w 1 = 1 0 ∂ ε u (ε) * v (ε) dε, w 2 = 1 0 ∂ ε v (ε) * u (ε) dε. (ii) If α + β > 1, demonstrate that w ∈ C 1 c (R) and [w ] 0,α+β -1 < ∞.
Hint: if w 1 is as above then use the approximation

w 1,δ = w 1 = 1 δ ∂ ε u (ε) * v (ε) dε, 0 < δ < 1.

3.3.

Let Ω 1 and Ω 2 be two bounded open subsets of R n so that

Ω 1 Ω 2 . For k = 1, 2, let u k ∈ C 2 (Ω k ) ∩C(Ω k ) satisfying u k > 0, Lu k = n ∑ i, j=1 a i j ∂ 2 i j u k = λ k u k in Ω k , u k = 0 on ∂ Ω k ,
where λ k is a constant and the coefficients a i j = a i j (x) belong to C(Ω 2 ) and satisfy

a i j = a ji , ν|ξ | 2 ≤ ∑ i, j a i j ξ i ξ j ≤ ν -1 |ξ | 2 for any ξ ∈ R n ,
for some constant ν ∈ (0, 1]. Prove that λ 1 < λ 2 < 0. Hint: we can apply the maximum principle to the function

v = u 1 /u 2 in Ω 1 . 3.4. Let u ∈ C 2 (B 1 ), B 1 = {x ∈ R n ; |x| < 1}, so that u = 0 on ∂ B 1 . Prove that B 1 u 2 dx ≤ C B 1 (∆ u) 2 dx,
the constant C only depends on the dimension n. Hint: use Poincaré's inequality

B 1 u 2 dx ≤ C 0 (n) B 1 |∇u| 2 dx.
3.5. Let u, v ∈ C 2 (R n ) be two harmonic functions so that u(tx) = t a u(x), v(tx) = t b v(x) for any x ∈ R n and t > 0, with constants a = b. Establish the orthogonality relation

∂ B 1 (0) uvds = 0.
3.6. Let f be a continuous and bounded function on R so that

[ f ] α = sup | f (t) -f (s)| |t -s| α ; t, s ∈ R t = s < ∞,
where α ∈ (0, 1) is a constant. Let u(x) = u(x 1 , x 2 ) be the solution of the Laplace equation

∆ u = 0 in R 2 + = {x = (x 1 , x 2 ); x 2 > 0}, with boundary condition u(x 1 , 0) = f (x 1 , 0), x 1 ∈ R.
Recall that u is explicitly given by the formula

u(x 1 , x 2 ) = 1 π R x 2 f (t) (x 1 -t) 2 + x 2 2 dt. a) Check that, for (x 1 , x 2 ), (y 1 , x 2 ) ∈ R 2 + , we have |u(x 1 , x 2 ) -u(y 1 , x 2 )| ≤ 1 π R x 2 t 2 + x 2 2 dt [ f ] α |x 1 -y 1 | α = [ f ] α |x 1 -y 1 | α . b) If (y 1 , x 2 ), (y 1 , y 2 ) ∈ R 2 + then prove that |u(y 1 , x 2 ) -u(y 1 , y 2 )| ≤ 1 π R |s| α s 2 + 1 ds [ f ] α |x 2 -y 2 | α .
Hint: note that

u(x 1 , x 2 ) = 1 π R f (x 1 + sx 2 ) s 2 + 1 ds. c) Deduce that [u] α ≤ C[ f ] α ,
the constant C only depends on α.

3.7.

Fix p ∈ (0, 1). a) Let q such that q -2 = qp. Compute the constant c = c(n, p) for which v(x) = c|x| q , x ∈ R n , is a solution of the equation

∆ v = v p in R n . Let u ∈ C 2 (R n ) satisfying u > 0, ∆ u = u p in R n . b) Check that max |x|≤r u(x) = max |x|=1 u(x) for any r > 0. c) Assume that there exists r > 0 so that u < v in ∂ B r (0). Prove that u ≤ v in Ω = {v < u} ∩ B r (
.8. Let u ∈ C 2 (B 2 (0)) so that u > 0, ∆ u = 0 in B 2 (0).
B 1/2 (x)
u, for any x ∈ B 1 (0),

with a constant C 1 = C 1 (n). (b) Prove the estimate sup B 1/2 (x) u ≤ C 2 u(x) for any x ∈ B 1 (0), with C 2 = C 2 (n) is a constant.
(c) Conclude that we have the following estimate sup

B 1 (0) |∇(ln u)| ≤ C,
for some constant C = C(n).

3.9.

Let Ω be a bounded domain of R n with Lipschitz boundary. Prove the following interpolation inequality, where ε > 0 is arbitrary,

Ω |∇u| 2 dx ≤ ε Ω (∆ u) 2 dx + 1 4ε Ω u 2 dx, u ∈ H 1 0 (Ω ) ∩ H 2 (Ω ).
3.10. Let u be a harmonic function in B = B(x 0 , 1). Prove the gradient estimate

|∇u(x 0 )| ≤ n sup B u -u(x 0 ) .
Hint: as ∆ u = 0 is invariant under rotation we may assume that |∇u(x 0 )| = -∂ n u(x 0 ).

Apply the mean value theorem to the harmonic function Mu in B r = B(x 0 , r), 0 < r < 1, with M = sup B u. Get then the estimate

|∂ n u(x 0 )| ≤ 1 ω n r n ∂ B (M -u)dσ (x). 3.11. Let u ∈ C ∞ (R n ) so that ∆ u(x) = 0, |u(x)| ≤ C|x| α in R n ,
where C and α are two positive constants. Prove that u is a polynomial of degree less or equal to [α]. i) Prove that there exists ρ > 0 sufficiently large in such a way that ∆ v > 0 in D. Fix this ρ. Verify then that, for a given ε > 0,

∆ (u -u(x 0 ) + εv) > 0 in D.
ii) Show that, for sufficiently small ε, uu(x 0 ) + εv ≤ 0 in ∂ D. Deduce that

∂ ν u(x 0 ) > 0. b) (Strong maximum principle) Let Ω be a domain of R n and u ∈ C(Ω ) ∩ C 2 (Ω )
satisfying ∆ u ≥ 0. Prove the following claim : if u is non constant, then u can not attain its maximum at a point of Ω . Hint : Proceed by contradiction by using Hopf's lemma and 3.13. Let K ≥ 0 be a constant.

(a) Prove that the nonlinear equation

U + K|U | + 1 = 0 in (-1, 1), U(±1) = 0 admits a unique even solution U ∈ C 2 ([-1, 1]). (b) Let p ∈ C([-1, 1]) and |p(t)| ≤ K. Let u ∈ C 2 ([-1, 1]
) be a solution of the boundary value problem

u + pu + 1 = 0 in (-1, 1), u(±1) = 0.
Check that 0 ≤ u ≤ U in (-1, 1). Deduce that the nonlinear equation in (a) admits a unique solution.

Let

Ω = {x = (x 1 , x 2 ) ∈ R 2 ; x 1 > 0, x 2 > 0} and let u ∈ C 2 (Ω ) be a solution of ∆ u = 0 in Ω , u = 0 on ∂ Ω satisfying |u(x)| ≤ c 1 + c 2 |x|
in Ω , where c 1 et c 2 are two positive constants. Prove that u is identically equal to zero. Hint: we can first extend u to R 2 and then use the interior estimate of derivatives of harmonic functions in balls centered at the origin (see Lemma 3.11).

3.15.

Let Ḃ = {x ∈ R n ; 0 < |x| < 1}, n ≥ 2, and let u ∈ C 2 Ḃ ∩C B be a harmonic function in Ḃ so that u(x) = o(Γ (x)) when x → 0, where Γ is the fundamental

I = {λ ≥ 0; B(x 0 , λ ) ⊂ {u < M}}.
(i) As u is continuous, I nonempty.

(ii) I is an interval because if λ ∈ I and λ ≤ λ then B(x 0 , λ ) ⊂ B(x 0 , λ ). On the other hand, as z ∈ B(x 0 , δ ) and u(z) = M, we have I ⊂ [0, δ ].

(iii) For λ ∈ I and x ∈ B(x 0 , λ ), we have

|x -z| ≤ |x -x 0 | + |z -x 0 | < λ + δ ≤ 2δ and hence B(x 0 , λ ) ⊂ Ω for any λ ∈ I. (iv) I is closed: let (λ n ) be a sequence in I converging to λ ∈ [0, δ ].
If there exists n 0 so λ n 0 ≥ λ then B(x 0 , λ ) ⊂ B(x 0 , λ n 0 ) and hence λ ∈ I. Otherwise, we would have λ n < λ , for any n. Let then x ∈ B(x 0 , λ ). Since λ n converges to λ , there exists an n 0 so that |x -

x 0 | < λ n 0 < λ . Whence, x ∈ {u < M}. We deduce then that B(x 0 , λ ) ⊂ {u < M}. That is λ ∈ I.
In conclusion, there exists R > 0 such that I = [0, R]. Then clearly B(x 0 , R) ⊂ {u < M}. We have also that ∂ B(x 0 , R) ∩ F is non empty. Otherwise, a simple argument based on the continuity of u and the compactness of ∂ B(x 0 , R) would imply that B(x 0 , R + ε) ⊂ {u < M}, for some ε > 0, and consequently R + ε ∈ I which is impossible. solution of the operator -∆ :

Γ (x) =      Γ (n/2) 2(n-2)π n/2 |x| 2-n if n ≥ 3, -1 2π ln |x| if n = 2.
Fix r ∈ (0, 1) and denote by B r the ball with center 0 and radius r. Consider then v ∈ C ∞ (B r ) ∩C(B r ) the solution of the boundary value problem

∆ v = 0 in B r , v = u on ∂ B r .
(a) Prove that, for any ε > 0, there exists δ ∈ (0, r) so that

|u -v| ≤ εΓ in B δ \ {0}.
(b) Deduce that |u -v| ≤ εΓ in B r \ B δ . Hint: apply the comparison principle to both u and v ± = v ± εΓ . (c) Conclude that u admits a harmonic extension in the whole unit ball B.

3.16. For r > 0, let Ω r = {x = (x 1 , x 2 ) ∈ R 2 ; |x 2 | < x 1 < r} and, for 1 ≤ i, j ≤ 2, let a i j be continuous functions in R 2 so that the matrix a = (a i j ) is symmetric and satisfies to the ellipticity condition

ν|ξ | 2 ≤ a(x)ξ • ξ ≤ ν -1 |ξ | 2 , for all x, ξ ∈ R 2 ,
where ν ∈ (0, 1] is a constant. Set

L = n ∑ i, j=1 a i j ∂ 2 i j .
(a) Prove that there exists µ = µ(ν

) et λ = λ (ν) for which v(x 1 , x 2 ) = (1 - x 2 1 ) µ cosh(λ x 2 ) satisfies Lv ≥ 0, in (-1, 1) × R.
Until the end of this exercise, we fix µ and λ is such a way that the last inequality holds.

Let

u ∈ C 2 (Ω R ), R > 0 is given, so that Lu = 0 in Ω R , u = 0 on {|x 2 | = x 1 } ∩ ∂ Ω R .
We want to prove that

M r = sup Ω r |u| ≤ 2r R 1+α M R for 0 < 2r ≤ R,
where the constant α only depends on ν.

(b) Check that is enough to prove that

M r ≤ 2 -(1+α) M 2r for 0 < 2r ≤ R.
Without loss of generality we may assume that r = 2 in the last inequality. Set then

U(x) = U(x 1 , x 2 ) = 1 4 M 4 x 1 ± u(x 1 , x 2 ), in Ω 4 = {|x 2 | < x 1 < 4}. (c) If c = c(ν) = 1/(4 cosh(3λ )), show that U(x 1 , x 1 ) ≥ cM 4 v(x 1 -2, x 2 ), in Ω = {1 < x 1 < 3, |x 2 | < x 1 }.
Then deduce that there exists α = α(ν) so that M 2 ≤ 2 -(1+α) M 4 .

3.17. Consider the operator

L = n ∑ i, j=1 a i j (x)∂ 2 i j + n ∑ i=1 b i (x)∂ i ,
where the matrix a(x) = (a i j (x)) is symmetric and fulfills the ellipticity condition

ν|ξ | 2 ≤ a(x)ξ • ξ ≤ ν -1 |ξ | 2 for all x, ξ ∈ R n ,
with ν ∈ (0, 1], and |b i (x)| ≤ K, 1 ≤ i ≤ n, for some constant K. Assume moreover that a i j and b i are 1-periodic. That is, a i j (x + z) = a i j (x) and b

i (x + z) = b i (x), for any x ∈ R n and z ∈ Z n . (a) Let x 0 ∈ R n and set B r = B(x 0 , r). Prove that if u ∈ C 2 (B 4r ) satisfies Lu = 0 in B 4r then sup B r (M -u) ≤ C(M -u(x 0 )) with M ≥ sup B 4r u,
where

C = C(n, ν, K, r) is a constant. Let u ∈ C 2 (R n ) be a bounded solution of Lu = 0 in R n . Fix z ∈ Z n and let v(x) = u(x + z) -u(x), x ∈ R n . (b) If M 1 = sup R n v and m 1 = inf R n v,
show that we cannot have neither M 1 > 0, nor m 1 < 0 (hence v is identically equal to zero and consequently u is 1-periodic). (c) Conclude that u is constant. Hint: as u is 1-periodic, it is enough to check that u is constant in the unit cube Q = [0, 1) n .

Let u be a harmonic function in the unit ball

B 1 = {x ∈ R n ; |x| < 1}. (a) Use the analyticity of u in B 1 to show that u(x) = ∑ k≥0 P k (x), (3.128)
in a neighborhood of 0 ∈ R n where, for each k, P k is a homogenous polynomial of degree k ; that is, P(λ x) = λ k P k (x), for any λ ∈ R, x ∈ R n and ∆ P k = 0. (b) Prove that the polynomials P k in (3.128) are two by two orthogonal in L 2 (B 1 ), i.e.

B 1

P j P k = 0 if j = k.
(c) Let r ∈ (0, 1), B r = {x ∈ R n ; |x| < r} and let (p j ) be a family of polynomials, p j of degree j, so that sup B r |up j | → 0 as j → +∞.

(Note that such family exists by Stone-Weierstrass's theorem).

(i) Show that there exists h j , a harmonic polynomial of degree j, such that h j = p j on ∂ B r . Deduce that sup

B r

|uh j | → 0 as j → +∞.

Hint: use the maximum principle. Therefore, uh j L 2 (B r ) → 0 as j → +∞.

(ii) Let S j = P 1 + . . . + P j be the orthogonal projection of u on E j , the subspace of harmonic polynomials of degree ≤ j. Check that

u -h j 2 L 2 (B r ) = u -S j 2 L 2 (B r ) + S j -h j 2 L 2 (B r )
and deduce that u -S j L 2 (B r ) → 0 as j → +∞.

(d) Demonstrate that the series in (3.128) converges uniformly in any ball B r , with r ∈ (0, 1). Hint: use that any harmonic function h in Ω coincide with its regularization h (ε) in

Ω ε = {x ∈ Ω ; dist(x, ∂ Ω ) > ε}.
Chapter 4

Classical inequalities, Cauchy problems and unique continuation

The first part of this chapter is essentially dedicated to some classical inequalities for harmonic functions. We precisely establish three-ball, three-sphere and doubling inequalities for harmonic functions. We limited, for sake of clarity, ourselves to harmonic functions but these kind of inequalities are in fact true for general elliptic operators but the proofs are more involved in that case. We refer the reader to the paper by R. Brummelhuis [1] for the three-sphere inequality and to N. Garofalo and F.-H. Lin [3, 4] for the doubling inequality. The rest of this chapter is devoted to three-ball inequalities that we apply in various situation for establishing stability inequalities for Cauchy problems. Our results rely on a Carleman estimate for a family of elliptic operators depending on a parameter and a generalized Poincaré-Wirtinger inequality. The results in this part improve substantially those in [2, Chapter 2].

Classical inequalities for harmonic functions

In this section, Ω is a bounded domain of R n (n ≥ 2) with boundary Γ and

H (Ω ) = {u ∈ C 2 (Ω ); ∆ u = 0}.
Here ∆ is the usual Laplace operator acting as follows

∆ u = n ∑ i=1 ∂ 2 i u, u ∈ C 2 (Ω ).
The ball and the sphere of centrer ξ and radius r are respectively denoted by B(ξ , r) and S(ξ , r).

The first result we prove is the following three-ball inequality Theorem 4.1. We have, for any u ∈ H (Ω ), ξ ∈ Ω and 0

< r 1 < r 2 < r 3 < r ξ = dist(ξ ,Γ ), u L 2 (B(ξ ,r 2 )) ≤ u α L 2 (B(ξ ,r 3 )) u 1-α L 2 (B(ξ ,r 1 )
) . Here α = (r 2r 1 )/(r 3r 1 ).

Proof. For simplicity convenience, we use in this proof the following notations

S(r) = S(ξ , r), B(r) = B(ξ , r)
and, where 0 < r < r ξ ,

H(r) = S(r) u 2 (x)dS(x), K(r) = B(r) u 2 (x)dx. As H(r) = S n-1 u 2 (ξ + ry)r n-1 dS(y),
we get

H (r) = n -1 r H(r) + 2 S n-1 u(ξ + ry)∇u(ξ + ry) • yr n-1 ds(y) (4.1) = n -1 r H(r) + 2 S(r) u(x)∂ ν u(x)ds(x).
But according to Green's formula and taking into account that ∆ u = 0 in Ω we have

S(r) u(x)∂ ν u(x)ds(x) = B(r) ∆ u(x)u(x)dx + B(r) |∇u(x)| 2 dx (4.2) = B(r) |∇u(x)| 2 dx.
This in (4.1) yields

H (r) = n -1 r H(r) + 2 B(r) |∇u(x)| 2 dx ≥ 0. (4.3) Whence K(r) = r 0 H(ρ)dρ ≤ r 0 H(r)dr = rH(r). (4.4) Define F(r) = ln K(r), 0 < r < r ξ .
We have

F (r) = H(r) K(r) and F (r) = H (r)K(r) -H(r) 2 H(r) 2 .
We obtain in light of (4.4)

F (r) ≥ rH (r)H(r) -H(r) 2 H(r) 2 .
This and (4.3) imply

F (r) ≥ n -2 + 2r H(r) 2 B(r) |∇u(x)| 2 dx ≥ 0.
Therefore F is convex.

As F is convex and

r 2 = αr 3 + (1 -α)r 1 , we deduce that ln K(r 2 ) = F(r 2 ) ≤ αF(r 3 ) + (1 -α)F(r 1 ) = ln K(r 3 ) α + ln K(r 1 ) 1-α = ln K(r 3 ) α K(r 1 ) 1-α . Thus K(r 2 ) ≤ K(r 3 ) α K(r 1 ) 1-α ,
which leads immediately to the expected inequality.

We next establish the so-called doubling inequality.

Theorem 4.2. Let u ∈ H (Ω ), ξ ∈ Ω and 0 < r < r ξ = dist(ξ ,Γ ). There exits a constant C > 0, depending on u, ξ and r, so that, for any 0 < r ≤ r/2, we have

u L 2 (B(ξ ,2r)) ≤ C u L 2 (B(ξ ,r)) .
Proof. Let S(r), B(r), H and K be as in the preceding proof and set

D(r) = B(r) |∇u(x)| 2 dx, 0 < r < r ξ .
Taking into account that

D(r) = r 0 S n-1 |∇u(ξ + ty)| 2 t n-1 dS(y)dt, we obtain D (r) = S(r) |∇u(y)| 2 dS(y) implying D (r) = 1 r S(r) |∇u(y)| 2 (x -ξ ) • ν(y)dS(y).
We get by applying the divergence theorem

D (r) = 1 r B(r) div |∇u(x)| 2 (x -ξ ) dx (4.5) = n r B(r) |∇u(x)| 2 dx + 1 r B(r) ∇(|∇u(x)| 2 ) • (x -ξ )dx.
On the other hand, we obtain by making an integration by parts

B(r) ∂ j (∂ i u(x)) 2 (x j -ξ j )dx = 2 B(r) ∂ i u(x)∂ 2 i j u(x)(x j -ξ j )dx = -2 B(r) ∂ 2 ii u(x)(x j -ξ j ) -2 B(r) ∂ i u(x)∂ j u(x)δ i j dx + 2 S(r) ∂ i u(x)∂ j u(x)(x j -ξ j )ν i dS(x).
Hence

B(r) ∇(|∇u(x)| 2 ) • (x -ξ )dx = -2 B(r) |∇u(x)| 2 dx + 2r S(r) (∂ ν u(x)) 2 dS(x).
This in (4.5) yields

D (r) = n -2 r D(r) + 2L(r) (4.6) with L(r) = S(r) (∂ ν u(x)) 2 dS(x).
Introduce now the so-called frequency function

N(r) = rD(r) H(r) .
Elementary computations show that

N (r) N(r) = 1 r + D (r) D(r) - H (r) H(

r) .

As ∆ u = 0 we have

∆ (u 2 ) = 2|∇u| 2 .
Applying Green's formula we find out 2

B(r) |∇u(x)| 2 dx = B(r) ∆ (u 2 )dx = 2 S(r) u(x)∂ ν u(x)dS(x).
That is

D(r) = S(r) u(x)∂ ν u(x)dS(x). (4.7)
(This is can be derived directly from (4.2)).

We have from (4.3)

H (r) H(r) = n -1 r + 2 D(r) H(r) (4.8)
and (4.6) entails

D (r) D(r) = n -2 r + 2 L(r) D(r) . (4.9) Therefore N (r) N(r) = 2 L(r) D(r) -2 D(r) H(r) = 2 L(r)H(r) -D(r) 2 D(r)H(r) . (4.10)
According to Cauchy-Schwarz's inequality, (4.7) yields

D(r) 2 ≤ L(r)H(r).
This in (4.10) entails N (r) ≥ 0.

In other words we proved that N is non-decreasing.

Next, from ln

H(r) r n-1 = H (r) H(r) - n -1 r
and (4.8) we deduce that ln

H(r) r n-1 = 2 D(r) H(r) = 2 N(r) r .
Fix r ≤ r ξ . Then, bearing in mind that N is non-decreasing, we get ln

H(r) r n-1 ≤ 2 N(r) r , 0 < r ≤ r.
Thus, with 0 < r 1 < r 2 ≤ r,

r 2 r 1 ln H(r) r n-1 dr = ln H(r 2 )r n-1 1 H(r 1 )r n-1 2 ≤ ln r κ 2 r κ 1 ,
where κ = 2N(r). In consequence

H(r 2 ) ≤ r 2 r 1 κ+n-1 H(r 1 ).
From this we get

K(r 2 ) = r 2 1 0 H(sr 2 )ds ≤ r 2 r 2 r 1 κ+n-1 1 0 H(sr 1 )ds = r 2 r 1 κ+n K(r 1 ).
The doubling inequality holds by taking r 2 = 2r 1 in the preceding inequality. That is we have, for any 0 < r ≤ 2 -1 r, K(2r) ≤ CK(r). We say that u vanishes of infinite order at ξ if

K(r) = O(r N ), for any N ∈ N. (4.12)
We have as a consequence of the doubling inequality in Theorem 4.2 the following strong unique continuation property for harmonic functions. Proof. In this proof K is as in the proof of Theorem 4.1.

We get, for sufficiently small r, by applying recursively (4.11)

K(r) ≤ CK(2 -1 r) ≤ . . .C K(2 -r) = C (2 -r) N [(2 -r) -N K(2 -r)].
Fix first and N 0 so that C (2 -r) N remains bounded for any N ≥ N 0 . Whence, as (2 -r) -N K(2 -r) tends to zero as N converges to ∞, we obtain that K(r) = 0. That is u = 0 in B(r). The proof is completed by using Theorem 2.16 in Chapter 2.

The calculations we carried out in the proof of the preceding theorems can used to obtain a three-sphere inequality. Indeed, using ln But from the preceding proof, we observed that L(r)H(r) -D(r) 2 ≥ 0. In consequence ln

H(r) r n-1 = 2 N(r) r , N (r) N(r) = 1 r + D (r) D(r) - H (r
H(r) r n+1 ≥ 0,
showing in particular that r → ln H(r) r n+1 is convex. We can then state the following result. Theorem 4.3. We have, for any u ∈ H (Ω ), ξ ∈ Ω and 0

< r 1 < r 2 < r 3 < r ξ = dist(ξ ,Γ ), u L 2 (S(ξ ,r 2 )) ≤ r 2 r 3 (n+1)α r 2 r 1 (n+1)(1-α) u α L 2 (S(ξ ,r 3 )) u 1-α L 2 (S(ξ ,r 1 )) ,
with α = (r 2r 1 )/(r 3r 1 ).

We establish in the remaining part of this section the mean-value identities and their consequences.

u(ξ ) = n |B(r)| B(r) u(x)dx. (4.14)
Here S(r) = S(ξ , r) and B(r) = B(ξ , r).

Proof. Define

I(r) = 1 |S(r)| S(r) u(x)dx, 0 < r < r ξ . Since I(r) = 1 |S n-1 | S n-1 u(ξ + ry)dS(y), (4.15)
we have

I (r) = 1 |S n-1 | S n-1 ∇u(ξ + ry) • ydS(y) = 1 |S n-1 | S n-1 ∂ ν u(ξ + ry)dS(y) = 1 |S n-1 | S n-1 ∂ ν u(x)dS(x).
We then get by applying then the divergence theorem

I (r) = 1 |S(r)| B(r) ∆ u(x)dx = 0.
That is I is constant and from (4.15) we have that I can be extended by continuity at r = 0 by posing I(0) = u(ξ ). Whence I(r) = u(ξ ) or equivalently

u(ξ ) = 1 |S(r)| S(r) u(x)dx.
This is exactly (4.13).

One gets using again (4.15)

r 0 S n-1 u(ξ + ty)t n-1 dtdS(y) = r n |S n-1 | n u(ξ ),
from which we deduce

u(ξ ) = n |B(r)| B(r) u(x)dx.
In other words, we proved (4.14).

We now apply the mean-value inequality (4.14) to obtains the strong maximum principle for harmonic functions. In the sequel

H c (Ω ) = H (Ω ) ∩C 0 (Ω ).
Theorem 4.5. Let u ∈ H c (Ω ) be non constant. Then u can not achieve its maximum or its minimum at an interior point.

Proof. Set M = max Ω u and

Ω M = {x ∈ Ω ; u(x) = M}.
Note that Ω M is closed by the continuity of u. We claim that, as u is non constant, Ω M is empty. If Ω M was non empty then for ξ ∈ Ω M we have u(ξ ) = M. Using one more time (4.14) in which we substitute u by u -M. We obtain, where 0 < r < r ξ is fixed,

0 = u(ξ ) -M = n |B(r)| B(r) (u(x) -M)dx ≤ 0.
Thus u -M = 0 in B(r) implying that Ω M would be also open. Therefore, we would have Ω M = Ω and consequently u is constant, which contradicts our assumption. The case of an interior minimum can be treated similarly.

We now apply again (4.14) to establish a Harnak type inequality for harmonic functions. We obtain by applying twice (4.14)

u(x 1 ) = n r n |S n-1 | B(x 1 ,r) u(x)dx ≤ n r n |S n-1 | B(2r) u(x)dx and u(x 2 ) = n 3 n r n |S n-1 | B(x 2 ,3r) u(x)dx ≥ n 3 n r n |S n-1 | B(2r) u(x)dx.
Whence u(x 1 ) ≤ 3 n u(x 2 ) which means that and (PP * u|u) = (P * u|P * u) + P * u|Qu = P * u 2 + QP * u|u .

Hence

Pu 2 = ℜ(P * Pu|u) + ℜ QPu|u , (4.18)

P * u 2 = ℜ(PP * u|u) -ℜ QP * u|u . (4.19)
We obtain by taking the difference side by side of (4.18) and (4.19)

Pu 2 -P * u 2 = ℜ([P * , P]u|u) + ℜ Ru|u , with R = QP + QP * . Since P * = -P -Pϕ, R = 2iℑ(QP) -QPϕ = 2iℑ(a 1 a 2 )∂ τ -QPϕ. Consequently Pu 2 -P * u 2 = ℜ([P * , P]u|u) + ℜ (2iℑ(a 1 a 2 )∂ τ -QPϕ)u|u as expected. As usual, (x 1 , x 2 ) ∈ R 2 is identified with z = x 1 + ix 2 ∈ C. When P = ∂ 1 + i∂ 2 = 2∂ z , P = ∂ 1 -i∂ 2 = 2∂ z , P * = -2(∂ z + ∂ z ϕ), [P * , P] = -4[∂ z + ∂ z ϕ, ∂ z ] = 4∂ z ∂ z ϕ = ∆ ϕ, QPϕ = (ν 1 + iν 2 )(∂ 1 ϕ -i∂ 2 ϕ) = ∂ ν ϕ -i∂ τ ϕ.
In light of these identities, we have as a consequence of Lemma 4.1

Corollary 4.2.

(1) We have, for any real-valued u ∈ C 1 (Ω ),

4 ∂ z u 2 = 4 (∂ z + ∂ z ϕ)u 2 + (∆ ϕu|u) -∂ ν ϕu|u . (4.20)
In particular,

Ω (∆ ϕ)u 2 e ϕ dx ≤ Ω |∇u| 2 e ϕ dx + Γ (∂ ν ϕ)u 2 e ϕ dσ .
+ 2 Γ (∆ u(∂ ν u) -(∂ 2 12 u)J∇u • ν)e ϕ dσ .
Here J = 0 1 1 0 .

Proof. (1) is immediate from Lemma 4.1. To prove (2) we substitute in (4.19) u by ∂ z u. We then get (4.22) since

ℜ(2i∂ τ ∂ z u∂ z u) = ℜ(2i∂ τ ∂ z u∂ z u) = 1 2 ∆ u(∂ ν u) - 1 2 (∂ 2 12 )uJ∇u • ν.
This completes the proof.

≤ Ω |∆ u| 2 e ϕ dx + Γ ∂ ν ϕ(u 2 + |∇u| 2 )e ϕ dσ . + 2 Γ (∆ u(∂ ν u) -(∂ 2 12 u)J∇u • ν)e ϕ dσ .
Fix 0 < α < 1. We assume in the remaining part of this section that Ω is of class C 2,α . Let γ be a closed subset of Γ with non empty interior so that Γ 0 = Γ \ γ = / 0.

Lemma 4.2.

There exists ϕ 0 ∈ C 2 (Ω ) possessing the properties:

∆ ϕ 0 = 0 in Ω , ϕ 0 = 0 on Γ 0 , ∂ ν ϕ 0 < 0 on Γ 0 , ϕ 0 ≥ 0 on γ.
Proof. Fix Γ0 , an open subset of Γ so that Γ 0 ⊂ Γ0 and Γ \ Γ 0 = / 0. Pick χ ∈ C 2,α (Γ ) non identically equal to zero, satisfying χ = 0 on Γ0 and χ ≥ 0 on γ. Since Ω is of class C 2,α by [5, Theorem 6.8, page 100] there exits a unique ϕ 0 ∈ C 2,α (Ω ) solving the BVP ∆ ϕ 0 = 0 in Ω , ϕ 0 = χ on Γ .

From the strong maximum principle ϕ 0 > 0 in Ω (see Theorem 4.5) and, bearing in mind that ϕ 0 = χ = 0 on Γ0 , we have ∂ ν ϕ 0 < 0 on Γ0 ⊃ Γ 0 according to Hopf's lemma (see Exercise 3.12).

Let

Ψ (ρ) = |ln ρ| -1/2 + ρ, ρ > 0,
and

Ψ (0) = 0. Proposition 4.2. Let M > 0. There exists a constant C = C(M, Ω , γ) so that, for any real-valued function u ∈ C 2 (Ω ) satisfying ∆ u = 0 in Ω and u C 2 (Ω ) ≤ M, we have Γ (u 2 + |∇u| 2 )dσ 1/2 ≤ CΨ γ (u 2 + |∇u| 2 )dσ 1/2 . ( 4

.24)

Proof. For s > 0 let ϕ = ϕ 1 + sϕ 0 , where ϕ 0 is as in Lemma 4.2, and ϕ 1 is the solution of the BVP ∆ ϕ 1 = 2 in Ω , ϕ 1 = 0 on Γ . Let θ = min

Γ 0 |∂ ν ϕ 0 |, c = max Ω |ϕ 0 |, c 0 = max Γ |∂ ν ϕ 0 |, c 1 = max Γ |∂ ν ϕ 1 | and I = γ (u 2 + |∇u| 2 )dσ , J = Γ 0 (u 2 + |∇u| 2 )dσ . Then (4.25) implies 0 ≤ (-sθ + c 1 )J + (sc 0 + c 1 )e cs I + 2M |γ|e cs √ I + 2 |Γ 0 |M 2 . (4.26)
In the rest of this proof, C = C(M, Ω , γ) and C j = C j (M, Ω , γ) are generic constants.

When s ≥ s 0 = 2c 1 /θ and I ≤ 1, we get from (4.26) .

C 1 J ≤ e C 0 s √ I + 1 s . ( 4
(4.28)

In the case where I ≥ κ 2 , we have Proof. We have from Green's formula

J ≤ CM 2 ≤ CM 2 κ 2 I . ( 4
0 = Ω ∆ uudx = - Ω |∇u| 2 dx + Γ ∂ ν uudσ .
Whence

Ω |∇u| 2 dx = Γ ∂ ν uudσ ≤ 2 Γ (u 2 + (∂ ν u) 2 )dσ ≤ 2 Γ (u 2 + |∇u| 2 )dσ .
Therefore (4.30) follows from (4.24).

Lemma 4.3.

There exists a constant C = C(Ω ) > 0 so that, for any real-valued function u ∈ H 1 (Ω ), we have

Ω u 2 dx ≤ C Ω |∇u| 2 dx + Γ u 2 dσ . (4.31)
Proof. We proceed by contradiction. We assume then that, for each integer k ≥ 1, there exists

u k ∈ H 1 (Ω ) so that Ω u 2 k dx > k Ω |∇u k | 2 dx + Γ u 2 k dσ . (4.32) If v k = u k / u k L 2 (Ω ) then (4.32) gives Ω |∇v k | 2 dx + Γ v 2 k dσ < 1 k .
Then the sequence (v k ) is bounded in H 1 (Ω ) and L 2 (Γ ). Subtracting a subsequence if necessary, we may assume that v k converges to v, weakly in H 1 (Ω ), strongly in L 2 (Ω ) and weakly in L 2 (Γ ). Using the lower semi-continuity of a norm with respect to the weak topology, we obtain

Ω |∇v| 2 dx + Γ v 2 dσ ≤ lim inf Ω |∇v k | 2 dx + Γ v 2 k dσ = 0. Hence v = 0. But 1 = v k L 2 (Ω ) → v L 2 (Ω ) = 1.
This leads to the expected contradiction.

The following corollary is a consequence of Corollary 4.3, Lemma 4.3 and the identity

|∇u| 2 = (∂ τ u) 2 + (∂ ν u) 2 .
Corollary 4.4. Let M > 0. There exists a constant C = C(M, Ω , γ) so that, for any real-valued function u ∈ C 2 (Ω ) satisfying ∆ u = 0 in Ω and u C 2 (Ω ) ≤ M, we have

u H 1 (Ω ) ≤ CΨ u H 1 (γ) + ∂ ν u L 2 (γ) .
This result is nothing but a logarithmic stability of the Cauchy problem for harmonic functions with data on γ.

This corollary contains obviously the uniqueness of continuation from the Cauchy data on γ.

A Carleman inequality for a family of operators

Let Ω be a bounded domain of R n with Lipschitz boundary Γ . Let I be an arbitrary set and consider the family of operators

L t = div(A t ∇ •), t ∈ I ,
where for each t ∈ I the matrix A t = (a i j t ) is a symmetric with coefficients in W 1,∞ (Ω ).

We assume that there exist κ > 0 and κ ≥ 1 so that

κ -1 |ξ | 2 ≤ A t (x)ξ • ξ ≤ κ|ξ | 2 , x ∈ Ω , ξ ∈ R n , t ∈ I , (4.33) and n ∑ k=1 n ∑ i, j=1 ∂ k a i j t (x)ξ i ξ j ≤ κ|ξ | 2 , x ∈ Ω , ξ ∈ R n , t ∈ I . (4.34) Pick 0 ≤ ψ ∈ C 2 (Ω ) without critical points in Ω and let ϕ = e λ ψ .
Theorem 4.7. (Carleman inequality) There exist three positive constants C, λ 0 and τ 0 , only depending on ψ, Ω , κ and κ, so that

C Ω λ 4 τ 3 ϕ 3 v 2 + λ 2 τϕ|∇v| 2 e 2τϕ dx ≤ Ω (L t v) 2 e 2τϕ dx + Γ λ 3 τ 3 ϕ 3 v 2 + λ τϕ|∇v| 2 e 2τϕ dσ , (4.35)
for all v ∈ H 2 (Ω ), t ∈ I , λ ≥ λ 0 and τ ≥ τ 0 .

Proof. Since the dependance of the constants will be uniform with respect to t ∈ I , we drop for simplicity the subscript t in L t and its coefficients. Let Φ = e -τϕ and w ∈ H 2 (Ω ). Then straightforward computations give

Pw = [Φ -1 LΦ]w = P 1 w + P 2 w + cw,
where

P 1 w = aw + div (A∇w), P 2 w = B • ∇w + bw, with a = a(x, λ , τ) = λ 2 τ 2 ϕ 2 |∇ψ| 2 A , B = B(x, λ , τ) = -2λ τϕA∇ψ, b = b(x, λ , τ) = -2λ 2 τϕ|∇ψ| 2 A , c = c(x, λ , τ) = -λ τϕdiv (A∇ψ) + λ 2 τϕ|∇ψ| 2 A .
Here

|∇ψ| A = A∇ψ • ∇ψ.
We obtain by making integrations by parts

Ω aw(B • ∇w)dx = 1 2 Ω a(B • ∇w 2 )dx = - 1 2 Ω div(aB)w 2 dx + 1 2 Γ a(B • ν)w 2 dσ (4.36)
and

Ω div (A∇w)(B • ∇w)dx = - Ω A∇w • ∇(B • ∇w)dx + Γ (B • ∇w)(A∇w • ν)dσ = - Ω B ∇w • A∇wdx - Ω ∇ 2 wB • A∇wdx + Γ (B • ∇w)(A∇w • ν)dσ . (4.37)
Here B = (∂ j B i ) is the Jacobian matrix of B and ∇ 2 w = (∂ 2 i j w) is the Hessian matrix of w.

But

Ω B i ∂ 2 i j wa ik ∂ k wdx = - Ω B i a ik ∂ 2 ik w∂ j wdx - Ω ∂ i B i a ik ∂ k w∂ j wdx + Γ B i ν i a jk ∂ k w∂ j wdσ . Therefore Ω ∇ 2 wB • A∇wdx = - 1 2 Ω div(B)A + à ∇w • ∇wdx + 1 2 Γ |∇w| 2 A (B • ν)dσ , (4.38) with à = ( ãi j ), ãi j = B • ∇a i j .
It follows from (4.37) and (4.38)

Ω div (A∇w)B • ∇wdx = 1 2 Ω -2AB + div(B)A + Ã ∇w • ∇wdx + Γ (B • ∇w) (A∇w • ν) dσ - 1 2 Γ |∇w| 2 A (B • ν)dσ . ( 4

.39)

A new integration by parts yields

Ω div (A∇w)bwdx = - Ω b|∇w| 2 A dx - Ω w∇b • A∇wdx + Γ bw(A∇w • ν)dσ .
This and the following inequality where

- Ω w∇b • A∇wdx ≥ - Ω (λ 2 ϕ) -1 |∇b| 2 A w 2 dx - Ω λ 2 ϕ|∇w| 2 A dx imply Ω div (A∇w)bwdx ≥ - Ω (b + λ 2 ϕ)|∇w| 2 A dx - Ω (λ 2 ϕ) -1 |∇b| 2 A w 2 dx + Γ bw(A∇w • ν)dσ . ( 4
f = - 1 2 div(aB) + ab -(λ 2 ϕ) -1 |∇b| 2 A -c 2 , F = -AB + 1 2 div(B)A + Ã -(b + λ 2 ϕ)A, g(w) = 1 2 aw 2 (B • ν) - 1 2 |∇w| 2 A (B • ν) + (B • ∇w)(A∇w • ν) + bw(A∇w • ν).
We deduce, by using the elementary inequality (pq) 2 ≥ p 2 /2q 2 , p > 0, q > 0, that

Pw 2 2 ≥ ( P 1 w + P 2 w 2 -cw 2 ) 2 ≥ 1 2 P 1 w + P 2 w 2 2 -cw 2 2 ≥ Ω P 1 wP 2 wdx - Ω c 2 w 2 dx.
In light of (4.41), we obtain

Pw 2 2 ≥ Ω f w 2 dx + Ω F∇w • ∇wdx + Γ g(w)dσ . (4.42)
By straightforward computations, there exist four positive constants C 0 , C 1 , λ 0 and τ 0 , only depending on ψ, Ω , κ and κ, such that, for all λ ≥ λ 0 and τ ≥ τ 0 ,

f ≥ C 0 λ 4 τ 3 ϕ 3 , Fξ • ξ ≥ C 0 λ 2 τϕ|ξ | 2 , for any ξ ∈ R n , |g(w)| ≤ C 1 λ 3 τ 3 ϕ 3 w 2 + λ τϕ|∇w| 2 . Hence C Ω (λ 4 τ 3 ϕ 3 w 2 + λ 2 τϕ|∇w| 2 )dx ≤ Ω (Pw) 2 dx + Γ (λ 3 τ 3 ϕ 3 w 2 + λ τϕ|∇w| 2 )dσ . Finally, w = Φ -1 v, v ∈ H 2 (Ω ),

Three-ball inequalities

The following Caccioppoli's type inequality will be useful in the sequel. The notations and the assumptions are those of the preceding section. Lemma 4.4. Let 0 < k < . There exists a constant C > 0, only depending on Ω , k, , κ and κ, so that, for any x ∈ Ω , 0 < ρ < dist(x,Γ )/ and u ∈ H 1 (Ω ) satisfying L t u ∈ L 2 (Ω ) in Ω , for some t ∈ I , we have

C B(x,kρ) |∇u| 2 dy ≤ 1 ρ 2 B(x, ρ) u 2 dy + B(x, ρ) (L t u) 2 dy. ( 4

.43)

Proof. We give the proof for k = 1 and = 2. That for arbitrary k and is similar.

Let x ∈ Ω , 0 < ρ < dist(x,Γ )/2, t ∈ I and u ∈ H 1 (Ω ) satisfying L t u ∈ L 2 (Ω ). Then Ω n ∑ i, j=1 a i j t ∂ i u∂ j vdy = Ω L t uvdy for any v ∈ C 1 0 (Ω ). (4.44) Pick χ ∈ C ∞ 0 (B(x, 2ρ
)) so that 0 ≤ χ ≤ 1, χ = 1 in a neighborhood of B(x, ρ) and |∂ γ χ| ≤ cr -|γ| for |γ| ≤ 2, where c is a constant not depending on ρ. Therefore, identity (4.44) with v = χu gives

Ω χ n ∑ i; j=1 a i j t ∂ i u∂ j udy = - Ω u n ∑ i, j=1 a i j t ∂ i u∂ j χdy + Ω χuL t udy = - 1 2 Ω n ∑ i, j=1 a i j t ∂ i u 2 ∂ j χdy + Ω χuL t udy = 1 2 Ω u 2 n ∑ i j=1 ∂ i a i j t ∂ j χ dy + Ω χuL t udy. But Ω χ n ∑ i, j=1 a i j t ∂ i u∂ j udy ≥ κ Ω χ|∇u| 2 dy. Whence C B(x,ρ) |∇u| 2 dy ≤ 1 ρ 2 B(x,2ρ) u 2 dy + B(x,2ρ) (L t u) 2 dy.
This is the expected inequality.

Consider L = div(A∇ •),
where A = (a i j ) is a symmetric matrix with W 1,∞ (Ω ) entries satisfying: there exist κ > 0 and κ ≥ 1 so that

κ -1 |ξ | 2 ≤ A(x)ξ • ξ ≤ κ|ξ | 2 , x ∈ Ω , ξ ∈ R n , (4.45) and d ∑ k=1 d ∑ i, j=1 ∂ k a i j (x)ξ i ξ j ≤ κ|ξ | 2 , x ∈ Ω , ξ ∈ R n . (4.46)
Theorem 4.8. Let 0 < k < < m. There exist C > 0 and 0 < γ < 1, only depending on Ω , k, , m, κ and κ, so that, for any v

∈ H 1 (Ω ) satisfying Lv ∈ L 2 (Ω ) in Ω , y ∈ Ω and 0 < r < dist(y,Γ )/m, we have C v L 2 (B(y, r)) ≤ v L 2 (B(y,kr)) + Lv L 2 (B(y,mr)) γ v 1-γ L 2 (B(y,mr)) .
Proof. As in the preceding lemma we give the proof when k = 3/2, = 2 and m = 7/2. The proof of arbitrary k, and m is similar. Let v ∈ H 1 (Ω ) satisfying Lv ∈ L 2 (Ω ) and set B(s) = B(0, s), s > 0. Fix y ∈ Ω and 0 < r < r y = 2dist(y,Γ )/7 (≤ 2diam (Ω )/7) .

Let w(x) = v(rx + y), x ∈ B (7/2) .

Straightforward computations show

L r w = div(A r ∇w) = r 2 Lv(rx + y) in B (7/2) , (4.47) where A r (x) = (a i j r (x)), a i j r (x) = a i j (rx + y). It is not hard to see that the family (A r ) satisfies (4.45) and (4.46) uniformly with respect to r ∈ (0, r y ).

Set

U = {x ∈ R n ; 1/2 < |x| < 3} , K = {x ∈ R n ; 1 ≤ |x| ≤ 5/2} .
and pick χ ∈ C ∞ 0 (U) satisfying 0 ≤ χ ≤ 1 and χ = 1 in a neighborhood of K . We get by applying Theorem 4.7 to χw, with Ω is substituted by U, that for any λ ≥ λ 0 and any τ ≥ τ 0 , C B( 2)\B( 1)

λ 4 τ 3 ϕ 3 w 2 + λ 2 τϕ|∇w| 2 e 2τϕ dx ≤ B(3) (L r (χw)) 2 e 2τϕ dx. (4.48) We have L r (χw) = χL r w + Q r (w) with Q r (w) = ∂ j χa i j r ∂ i w + ∂ j (a i j r w)∂ i w + a i j r ∂ 2 i j χw, supp (Q r w)) ⊂ {1/2 ≤ |x| ≤ 1} ∪ {5/2 ≤ |x| ≤ 3} and (Q r w) 2 ≤ Λ (w 2 + |∇w| 2 ),
where the constant Λ is independent of r. We introduce the temporary notations

P = B(3/2) w 2 dx + B(7/2) (L r w) 2 dx, Q = C B(2) w 2 dx, R = B(7/2)
w 2 dx.

Then (4.53) becomes

Q ≤ e ατ P + e -β τ R, τ ≥ τ 0 . (4.54)

Let

τ 1 = ln(R/P) α + β . If τ 1 ≥ τ 0 then τ = τ 1 in (4.54) yields Q ≤ 2P α α+β R β α+β . (4.55)
If τ 1 < τ 0 , we have R < e (α+β )τ 0 P and then

Q ≤ R = R α α+β R β α+β ≤ e ατ 0 P α α+β R β α+β . (4.56)
Summing up, we find that in any case one of inequalities (4.55) and (4.56) holds. That is in terms the original notations

C w L 2 (B(2)) ≤ w L 2 (B(3/2)) + L r w L 2 (B(7/2)) γ w 1-γ L 2 (B(7/2)) , (4.57)
with γ = α α + β .
We derive in a straightforward manner from (4.47) and (4.57) that

C v L 2 (B(y,2r)) ≤ v L 2 (B(y,3r/2)) + Lv L 2 (B(y,7r/2)) γ v 1-γ L 2 (B(y,7r/2)) .
This is the expected inequality.

Prior to establishing the three-sphere inequality for the gradient we prove a generalized Poincaré-Wirtinger type inequality. For this purpose, let O be an arbitrary open bounded subset of R n . Define, for f ∈ L 2 (O) and E ⊂ O Lebesgue-measurable with non zero Lebesgue measure |E|,

M E ( f ) = 1 |E| E f (x)dx.
Proposition 4.3. There exits a constant C > 0, only depending on O so that, for any f ∈ H 1 (O) and any Lebesgue-measurable set E with non zero Lebesgue measure, we have

f -M E ( f ) L 2 (O) ≤ C |O| 1/2 |E| 1/2 ∇ f L 2 (O,R n ) . (4.58)
Proof. A simple application of Cauchy-Schwarz inequality gives

M E ( f ) L 2 (O) ≤ |O| 1/2 |E| 1/2 f L 2 (O) . (4.59)
Inequality (4.59) with E = O and f substituted by f -M E ( f ) yields

M O ( f -M E ( f )) L 2 (O) ≤ f -M E ( f ) L 2 (O) . (4.60)
On the other hand, by the classical Poincaré-Wirtinger's inequality (see Exercise 1.19) there exists a constant C, only depending on O, so that

f -M O ( f ) L 2 (O) ≤ C ∇ f L 2 (O,R n ) . (4.61)
Now, as

M E (M O ( f )) = M O ( f ), we have f -M E ( f ) = f -M O ( f ) -M E ( f -M O ( f )).
We then obtain in light of (4.59)

f -M E ( f ) L 2 (O) ≤ 1 + |O| 1/2 |E| 1/2 f -M O ( f ) L 2 (O) implying f -M E ( f ) L 2 (O) ≤ 2 |O| 1/2 |E| 1/2 f -M O ( f ) L 2 (O) . Whence f -M E ( f ) L 2 (O) ≤ 2C |O| 1/2 |E| 1/2 ∇ f L 2 (O,R n ) .
The proof is then complete.

Theorem 4.9. Let 0 < k < < m. There exist C > 0 and 0 < γ < 1, only depending on Ω , k, , m, κ and κ, so that, for any v ∈ H 1 (Ω ) satisfying Lv ∈ L 2 (Ω ), y ∈ D and 0 < r < dist(y,Γ )/m, we have

C ∇v L 2 (B(y, r),R n ) ≤ ∇v L 2 (B(y,kr),R n ) + Lv L 2 (B(y,mr)) γ ∇v 1-γ L 2 (B(y,mr),R n ) .
Proof. We keep the same notations as in the preceding proof. We take k = 1, = 2 and m = 3. The proof for arbitrary 0 < k < < m is the same. An application of the generalized Poincaré-Wirtinger's inequality in Proposition 4.3 then yields

ρ = 1 |B(1)| B(1) w(x)dx, B (1)
(w -ρ) 2 dx ≤ C B (1)
|∇w| 2 dx, (4.62)
B(3) (w -ρ) 2 dx ≤ C B(3) |∇w| 2 dx. (4.63)
On the other hand (4.50), in which w is substituted by wρ, gives

C B(2) |∇w| 2 dx ≤ e ατ B( 1
) (w -ρ) 2 + |∇w| 2 dx + B (3)
(L r w) 2 dx (4.64)

+ e -β τ B(3) (w -ρ) 2 + |∇w| 2 dx.
In view of (4.62) and (4.63) in (4.64), we get

C B(2) |∇w| 2 dx ≤ e ατ B( 1
)
|∇w| 2 dx + B(3) (L r w) 2 dx + e -β τ B(3) |∇w| 2 dx.
The rest of the proof is similar to that of Theorem 4.8.

Stability of the Cauchy problem

The following lemma will useful in the sequel.

Lemma 4.5. Let (η k ) be a sequence of real numbers satisfying 0 < η k ≤ 1, k ∈ N, and

η k+1 ≤ c(η k + b) γ , k ∈ N,
for some constants 0 < γ < 1, b > 0 and c ≥ 1. Then

η k ≤ C(η 0 + b) γ k , (4.65)
where C = (2c) 1/(1-γ) .

Proof. Note first that (4.65) is trivially satisfied when

η 0 + b ≥ 1. Assume then that η 0 + b < 1. As b < cb γ < c(η k + b) γ , k ∈ N, we obtain η k+1 + b ≤ 2c(η k + b) γ . (4.66)
If τ k = η k + b then (4.66) can rewritten as follows

τ k ≤ 2cτ γ k , k ∈ N.
An induction in k yields

η k ≤ (2c) 1+γ+...+γ k-1 τ γ k 0 ≤ (2c) 1/(1-γ) (η 0 + b) γ k .
The proof is then complete.

Note that, as Ω is Lipschitz, it has the uniform cone property. Whence, there exist R > 0 and θ ∈]0, π/2[ so that to any x ∈ Γ corresponds ξ = ξ ( x) ∈ S n-1 with the property that

C ( x) = {x ∈ R n ; 0 < |x -x| < R, (x -x) • ξ > |x -x| cos θ } ⊂ Ω .
Proposition 4.4. Let 0 < α ≤ 1. There exist ω Ω , depending only on Ω and three constants C > 0, c > 0 and β > 0, depending only on Ω , κ, κ and α, so that for any 0 < ε < 1:

(1) for any u ∈ H 1 (Ω ) ∩C 0,α (Ω ) with Lu ∈ L 2 (Ω ), we have C u L ∞ (Γ ) ≤ e c/ε u L 2 (ω) + Lu L 2 (Ω ) + ε β [u] α + u L 2 (Ω ) , (2) for any u ∈ C 1,α (Ω ) with Lu ∈ L 2 (Ω ), we have C ∇u L ∞ (Γ ,R n ) ≤ e c/ε ∇u L 2 (ω,R n ) + Lu L 2 (Ω ) + ε β [∇u] α + ∇u L 2 (Ω ,R n ) .
Here

[∇u] α = max 1≤i≤n [∂ i u] α .
Proof. Fix x ∈ Γ and let ξ = ξ ( x) be as in the definition of the uniform cone property. Let

x 0 = x + (R/2)ξ , d 0 = |x 0 -x| and ρ 0 = d 0 sin θ /3. Note that B(x 0 , 3ρ 0 ) ⊂ C ( x).
Define the sequence of balls (B(x k , 3ρ k )) as follows

   x k+1 = x k -α k ξ , ρ k+1 = µρ k , δ k+1 = µδ k , where δ k = |x k -x|, ρ k = κδ k , α k = (1 -µ)δ k , with κ = sin θ /3, µ = 1 -κ.
This definition guarantees that, for each k, B(x k , 3ρ k ) ⊂ C ( x) and

B(x k+1 , ρ k+1 ) ⊂ B(x k , 2ρ k ). (4.67)
Let u ∈ H 1 (Ω ) ∩ C 0,α (Ω ) with u = 0 and Lu ∈ L 2 (Ω ). From Theorem 4.8, we have

u L 2 (B(x k ,2ρ k )) ≤ C u 1-γ L 2 (B(x k ,3ρ k )) u L 2 (B(x k ,ρ k )) + Lu L 2 (B(x k ,3ρ k )) γ and then u L 2 (B(x k ,2ρ k )) ≤ C u 1-γ L 2 (Ω )) u L 2 (B(x k ,ρ k )) + Lu L 2 (Ω ) γ But B(x k+1 , ρ k+1 ) ⊂ B(x k , 2ρ k ). Hence, u L 2 (B(x k+1 ,ρ k+1 )) ≤ C u 1-γ L 2 (Ω ) u L 2 (B(x k ,ρ k )) + Lu L 2 (Ω ) γ , or equivalently u L 2 (B(x k+1 ,ρ k+1 )) u L 2 (Ω ) ≤ C u L 2 (B(x k ,ρ k )) u L 2 (Ω ) + Lu L 2 (Ω ) u L 2 (Ω ) γ .
Substituting if necessary C by max(C, 1), we may assume that C ≥ 1. Lemma 4.5 then yields

u L 2 (B(x k ,ρ k )) u L 2 (Ω ) ≤ C u L 2 (B(x 0 ,ρ 0 )) u L 2 (Ω ) + Lu L 2 (Ω ) u L 2 (Ω ) γ k .
This inequality can be rewritten in the following form

u L 2 (B(x k ,ρ k )) ≤ C u L 2 (B(x 0 ,ρ 0 )) + Lu L 2 (Ω ) γ k u 1-γ k L 2 (Ω ) . (4.68)
An application of Young's inequality, for ε > 0, gives

C u L 2 (B(x k ,ρ k )) ≤ ε -1/γ k u L 2 (B(x 0 ,ρ 0 )) + Lu L 2 (Ω ) + ε 1/(1-γ k ) u L 2 (Ω ) .
(4.69) We have, by using the Hölder continuity of u,

|u( x)| ≤ [u] α | x -x| α + |u(x)|, x ∈ B(x k , ρ k ). Whence |S n-1 |ρ n k |u( x)| 2 ≤ 2[u] 2 α B(x k ,ρ k ) | x -x| 2α dx + 2 B(x k ,ρ k ) |u(x)| 2 dx, or equivalently |u( x)| 2 ≤ 2|S n-1 | -1 ρ -n k [u] 2 α B(x k ,ρ k ) | x -x| 2α dx + B(x k ,ρ k ) |u(x)| 2 dx .
As δ k = µ k δ 0 , we have

| x -x| ≤ | x -x k | + |x k -x| ≤ δ k + ρ k = (1 + κ)δ k = (1 + κ)µ k δ 0 . Therefore |u( x)| 2 (4.70) ≤ 2 [u] 2 α (1 + κ) α δ α 0 µ 2αk + |S n-1 | -1 (κd 0 ) -n µ -nk u 2 L 2 (B(x k ,ρ k )) . Let ω = x∈Γ B(x 0 ( x), ρ 0 )
and introduce the following temporary notations

M = [u] α + u L 2 (Ω ) , N = u L 2 (ω) + Lu L 2 (Ω ) .
Then (4.70) yields

C|u( x)| ≤ Mµ αk + µ -nk/2 u L 2 (B(x k ,ρ k )) . (4.71)
A combination of (4.69) and (4.71) entails

C u L ∞ (Γ ) ≤ µ -nk/2 ε -1/γ k N + µ αk + µ -nk/2 ε 1/(1-γ k ) M, ε > 0.
In this inequality we take ε > 0 in such a way that

µ αk = µ -nk/2 ε 1/(1-γ k ) . That is ε = µ (n/2+α)k(1-γ k ) . We obtain C u L ∞ (Γ ) ≤ µ αk-k γ k ( n 2 +α) N + µ αk M.
For t > 0, let k be the integer so that k ≤ t < k + 1. Bearing in mind that 0 < µ, γ < 1, we deduce by straightforward computations from the preceding inequality

C u L ∞ (Γ ) ≤ µ -e ct N + µ αt M.
Take e ct = 1/ε, we end up getting

C u L ∞ (Γ ) ≤ e c/ε N + ε β M, 0 < ε < 1,
which is the expected inequality in (1).

We omit the proof of (2) which is quite similar of that of (1). We have only to apply Theorem 4.9 instead of Theorem 4.8. Proposition 4.5. Let ω Ω and ω Ω be non empty. There exist C > 0 and β > 0, only depending on Ω , κ, κ, ω and ω, so that, for any u ∈ H 1 (Ω ) satisfying Lu ∈ L 2 (Ω ) and ε > 0, we have

C u L 2 ( ω) ≤ ε β u L 2 (Ω ) + ε -1 u L 2 (ω) + Lu L 2 (Ω ) , (4.72)
C ∇u L 2 ( ω,R n ) ≤ ε β ∇u L 2 (Ω ,R n ) + ε -1 ∇u L 2 (ω,R n ) + Lu L 2 (Ω ) . (4.73)
Proof. We limit ourselves to the proof of (4.72). That of (4.73) is similar. Fix x 0 ∈ ω and x ∈ ω. There exists a sequence of balls B(x j , r), r > 0, j = 0, . . . , N, so that

       B(x 0 , r) ⊂ ω, B(x j+1 , r) ⊂ B(x j , 2r), j = 0, . . . , N -1, x ∈ B(x N , r), B(x j , 3r) ⊂ Ω , j = 0, . . . , N.
We give in the end of this proof the construction of such sequence of balls.

We get from Theorem 4.8

u L 2 (B(x j ,2r)) ≤ C u 1-γ L 2 (B(x j ,3r)) u L 2 (B(x j ,r)) + Lu L 2 (Ω ) γ , 1 ≤ j ≤ N,
for some constants C > 0 and 0 < γ < 1, only depending on Ω , κ and κ. We obtain by proceeding as in the proof of Proposition 4.4

u L 2 (B(x N ,2r)) ≤ C u 1-γ N L 2 (B(x j ,3r)) u L 2 (B(x 0 ,r)) + Lu L 2 (Ω ) γ N .
Combined with Young's inequality this estimate yields

C u L 2 (B(x N ,2r)) ≤ ε β u L 2 (Ω ) + ε -1 u L 2 (ω) + Lu L 2 (Ω ) ,
where

β = γ N 1 -γ N .
As ω is compact, it can be covered by a finite number of balls B(x N , r) that we denote by B(x 1 N , r), . . . , B(x N , r). Hence

u L 2 ( ω) ≤ ∑ i=1 u L 2 (B(x N ,r)) . Whence C u L 2 ( ω) ≤ ε β u L 2 (Ω ) + ε -1 u L 2 (ω) + Lu L 2 (Ω ) ,
We complete the proof by showing how we construct the sequence of balls B(x j , r). Let γ : [0, 1] → Ω be a continuous path joining x 0 to x. That is γ is a continuous function so that γ(0) = x 0 and γ(1) = x. Fix r > 0 so that B(x 0 , r) ⊂ ω and 3r

< dist(γ([0, 1]), R n \ Ω ). Let t 0 = 0 and t k+1 = inf{t ∈ [t k , 1]; γ(t) ∈ B(γ(t k ), r)}, k ≥ 0.
We claim that there exists an integer N ≥ 1 so that γ(1) ∈ B(x N , r). If this claim does not hold, we would have γ(1) ∈ B(γ(t k ), r), for any k ≥ 0. Now, as the sequence (t k ) is non decreasing and bounded from above by 1 it converges to t ≤ 1. In particular, there exists an integer

k 0 ≥ 1 so that γ(t k ) ∈ B(γ(t), r/2), k ≥ k 0 . But this contradicts the fact that |γ(t k+1 ) -γ(t k )| = r, k ≥ 0. Let x k = γ(t k ), 0 ≤ k ≤ N. Then x = γ(1) ∈ B(x N , r), and since 3r < dist(γ([0, 1]), R n \ Ω ), we have B(x k , 3r) ⊂ Ω . Finally, if |y -x k+1 | < r then |y -x k | ≤ |y -x k+1 | + |x k+1 -x k | < 2r.
In other words, B(x k+1 , r) ⊂ B(x k , 2r).

Proposition 4.6. Let Γ 0 be a non empty open subset of Γ . There exist ω 0 Ω , depending only on Ω and Γ 0 , and two constants C > 0 and γ > 0, depending only on Ω , κ and κ, so that, for any u ∈ H 1 (Ω ) satisfying Lu ∈ L 2 (Ω ) and ε > 0, we have

C u H 1 (ω 0 ) ≤ ε γ u H 1 (Ω ) + ε -1 u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) .
Proof. Let x ∈ Γ 0 be arbitrarily fixed and let R > 0 so that B( x, R) ∩ Γ ⊂ Γ 0 . Pick x 0 in the interior of R n \ Ω sufficiently close to x is such a way that ρ = dist(x 0 , K) < R, where K = B( x, R) ∩ Γ 0 (think to the fact that Ω is on one side of its boundary). Fix then r > 0 in order to satisfy B(x 0 , ρ + r) ∩ Γ ⊂ Γ 0 and B(x 0 , ρ + θ r) ∩ Ω = / 0, for some 0 < θ < 1.

Define

ψ(x) = ln (ρ + r) 2 |x -x 0 | 2 . Then |∇ψ(x)| = 2 |x -x 0 | ≥ 2 ρ + r , x ∈ B(x 0 , ρ + r) ∩ Ω . Pick χ ∈ C ∞ 0 (B(x 0 , ρ + r)), χ = 1 in B(x 0 , (1 + θ )r/2) and |∂ α χ| ≤ κ, |α| ≤ 2, for some constant κ. Let u ∈ H 1 (Ω ) satisfying Lu ∈ L 2 (Ω ).
As in proof of Theorem 4.8, we have

L(χu) = χLu + Q(u) with Q(u) 2 ≤ C u 2 + |∇u| 2 and supp(Q) ⊂ B(x 0 , ρ + r) \ B(x 0 , (1 + θ )r/2) := D.
It follows, from Theorem 4.7 applied to v = χu in the domain Ω ∩ B(x 0 , ρ + r), where λ ≥ λ 0 is fixed and τ ≥ τ 0 , that

C B(x 0 ,ρ+θ r)∩Ω e 2τϕ u 2 dx ≤ D∩Ω e 2τϕ (u 2 + |∇u| 2 )dx + B(x 0 ,ρ+r)∩Ω e 2τϕ (Lu) 2 dx + B(x 0 ,ρ+r)∩Γ e 2τϕ (u 2 + |∇u| 2 )dσ . But ϕ(x) = e λ ln (ρ+r) 2 |x-x 0 | 2 = (ρ + r) 2λ |x -x 0 | 2λ . Whence Ce τϕ 0 B(x 0 ,ρ+θ r)∩Ω u 2 dx ≤ e τϕ 1 D∩Ω (u 2 + |∇u| 2 )dx (4.74)
+ e τϕ 2 B(x 0 ,ρ+r)

(Lu) 2 dx + e τϕ 2 B(x 0 ,ρ+r)∩Γ (u 2 + |∇u| 2 )dσ ,
where

ϕ 0 = 2(ρ + r) 2λ (ρ + θ r) 2λ , ϕ 1 = 2(ρ + r) 2λ (ρ + (1 + θ )r/2) 2λ , ϕ 2 = 2(ρ + r) 2λ ρ 2λ . Let α = 2rλ (1 -θ )(ρ + r) 2λ (ρ + (1 + θ )r/2) 2λ +1 and β = 4λ θ r(ρ + d) 2λ ρ 2λ +1
.

Elementary computations show that

ϕ 0 -ϕ 1 ≥ α and ϕ 2 -ϕ 0 ≤ β .
These inequalities in (4.74) yield

C B(x 0 ,ρ+θ r)∩Ω u 2 dx ≤ e -ατ D∩Ω (u 2 + |∇u| 2 )dx (4.75) + e β τ B(x 0 ,ρ+r)∩Ω (Lu) 2 dx + e β τ B(x 0 ,ρ+r)∩Γ (u 2 + |∇u| 2 )dσ . Let ω 0 ω 1 B(x 0 + ρ + θ r) ∩ Ω . Then Caccioppoli's inequality gives C ω 0 |∇u| 2 dx ≤ ω 1 u 2 dx + ω 1 (Lu) 2 dx. (4.76)
Using (4.76) in (4.75) we obtain

C ω 0 (u 2 + |∇u| 2 )dx ≤ e -ατ Ω (u 2 + |∇u| 2 )dx + e β τ Ω (Lu) 2 dx + e β τ Γ 0 (u 2 + |∇u| 2 )dσ .
We complete the proof similarly to that of Theorem 4.8.

We shall need the following lemma.

Lemma 4.6. There exists a constant C > 0 so that, for any u

∈ H 1 (Ω ) with Lu ∈ L 2 (Ω ), we have C u H 1 (Ω ) ≤ Lu L 2 (Ω ) + u H 1/2 (Γ ) . (4.77) Proof. Fix u ∈ H 1 (Ω ) and let F ∈ H 1 (Ω ) so that F |Γ = u |Γ and F H 1 (Ω ) = u H 1/2 (Γ ) (
Ω A∇v • ∇wdx = - Ω A∇F • wdx - Ω Luwdx, w ∈ H 1 0 (Ω ) (4.78)
has a unique solution v ∈ H 1 0 (Ω ). We obtain by taking w = v in (4.78)

κ Ω |∇v| 2 dx ≤ A∇v∇v = Ω A∇F • ∇v - Ω Luwdx.
Combined with Poincaré's inequality this estimate yields in a straightforward man-

ner v H 1 (Ω ) ≤ C F H 1 (Ω ) + Lu L 2 (Ω ) , (4.79)
for some constant C = C(κ, Ω ).

On the other hand, we get by using one more time (4.78), where ũ

= v + F, L(u -ũ) = 0 in Ω and u -ũ ∈ H 1 0 (Ω )
which leads immediately to ũ = u. Therefore in light of F H 1 (Ω ) = u H 1/2 (Γ ) and (4.79) inequality (4.77) follows.

Theorem 4.10. Let Γ 0 be an open subset of Γ and 0 < α ≤ 1. There exist C > 0, c > 0 and β > 0, only depending on Ω , κ, κ, α and Γ 0 , so that, for any u

∈ C 1,α (Ω ) satisfying Lu ∈ L 2 (Ω ), we have C u H 1 (Ω ) ≤ ε β u C 1,α (Ω ) + e c/ε u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) (4.80) for every 0 < ε < 1. Proof. Let u ∈ C 1,α (Ω ) satisfying Lu ∈ L 2 (Ω ).
From Lemma 4.6, we have

C u H 1 (Ω ) ≤ Lu L 2 (Ω ) + u H 1/2 (Γ ) .
In this proof, C > 0 and c > 0 are generic constants only depending on Ω , κ, κ, α and Γ 0 .

By Proposition 4.4, and noting that W 1,∞ (Γ ) is continuously embedded in H 1/2 (Γ ), we find β > 0 and ω Ω so that

C u H 1 (Ω ) ≤ ε β u C 1,α (Ω ) + e c/ε u H 1 (ω) + Lu L 2 (Ω ) , 0 < ε < 1. (4.81)
On the other hand by Proposition 4.6 there exist ω 0 Ω and γ > 0 so that, for any ε 1 > 0, we have

C u H 1 (ω 0 ) ≤ ε γ 1 u C 1,α (Ω ) (4.82) + ε -1 1 u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) .
But by Proposition 4.5 there is δ > 0 such that

C u H 1 (ω) ≤ ε δ 2 u C 1,α (Ω ) + ε -1 2 u H 1 (ω 0 ) + Lu L 2 (Ω ) , ε 2 > 0. (4.83) Estimate (4.82) in (4.83) gives C u H 1 (ω) ≤ (ε δ 2 + ε -1 2 ε γ 1 ) u C 1,α (Ω ) + ε -1 1 ε -1 2 u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) + ε -1 2 Lu L 2 (Ω )
.

ε 1 = ε (γ+1)/δ 2
in this estimate yields, where ρ = (γ + δ + 1)/δ ,

C u H 1 (ω) ≤ ε δ 2 u C 1,α (Ω ) + ε -ρ 2 u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (D) + ε -1 2 Lu L 2 (Ω ) ,
which in combination with (4.81) entails

C u H 1 (Ω ) ≤ (ε β + ε δ 2 e c/ε ) u C 1,α (Ω ) + ε -ρ 2 e c/ε u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) + (ε -1 2 + 1)e c/ε Lu L 2 (Ω ) . Therefore C u H 1 (Ω ) ≤ (ε β + ε δ 2 e c/ε ) u C 1,α (Ω ) + ε -ρ 2 + ε -1 2 + 1 e c/ε u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) .
We end up getting the expected inequality by taking ε 2 = e -2c/(εδ ) .

Theorem 4.10 can be used to get logarithmic stability estimate for the Cauchy problem.

Corollary 4.6. Let Γ 0 be an open subset of Γ and 0 < α ≤ 1. There exist C > 0, β > 0, only depending on Ω , κ, κ, α and Γ 0 , so that, for any u ∈ C 1,α (Ω ) satisfying u = 0 and Lu ∈ L 2 (Ω ), we have

u H 1 (Ω ) u C 1,α (Ω ) ≤ CΦ β u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) u C 1,α (Ω ) .
Here Φ β (s) = | ln s| -β + s, s > 0, and

Φ β (0) = 0.
This corollary is a direct consequence of Theorem 4.10 and the following lemma.

Lemma 4.7. Let α > 0, β > 0, c > 0, a > 0 and s > 0 be given constants. There exists a constant C > 0, only depending on α, β , c > 0, a and s, so that, for any a ∈ (0, a] and b > 0, the relation We can prove similarly to Corollary 4.6 the following consequence of Theorem 4.10.

a ≤ s -α + e cs b β , s ≥ s, (4.84)
Corollary 4.7. Let Γ 0 be an open subset of Γ , 0 < α ≤ 1 and fix M > 0. There exist C > 0, β > 0, only depending on Ω , κ, κ, α and Γ 0 , C depending also on M, so that, for any u

∈ C 1,α (Ω ) satisfying u C 1,α (Ω ) ≤ M and Lu ∈ L 2 (Ω ), we have u H 1 (Ω ) ≤ CΦ β u L 2 (Γ 0 ) + ∇u L 2 (Γ 0 ,R n ) + Lu L 2 (Ω ) .
Here Φ β is as in Corollary 4.6.

Exercises and problems

4.1. Let u ∈ H (B(0, 3)) satisfying u(0) = 0. (a) Let v = u 2 .
Check that ∆ v ≥ 0 and deduce from it that there exists a constant C > 0, only depending on n, so that

v L ∞ (B(0,r)) ≤ C v L 2 (B(0,2r)) , 0 < r ≤ 1. (b) Let 0 < ε < 1.
(i) Show that there exists C u,ε , depending on u and ε, so that

u L ∞ (B(0,r)) ≤ C u,ε u L 2 (B(0,2r)) , ε ≤ r ≤ 1.
(ii) Deduce then that there exists Cu,ε , depending on u and ε, so that

u L ∞ (B(0,r)) ≤ Cu,ε u L 2 (B(0,r)) , ε ≤ r ≤ 1.
(c) Prove that there exists a constant C u > 0, depending on u, so that

u L ∞ (B(0,r)) ≤ C u u L 2 (B(0,r)) , 0 < r ≤ 1.
Hint: use the second mean-value identity.

4.2.

Let Ω , D 1 and D 2 be three bounded domains of

R n of class C ∞ with D i Ω , i = 1, 2. Denote the boundary of Ω by Γ . Let ϕ ∈ C ∞ (Γ ) non identically equal to zero. For i = 1, 2 let u i ∈ H 2 (Ω ) ∩ C ∞ (D i ) ∩ C ∞ (Ω \ D i ) be the solution of the boundary value problem -∆ u + χ D i u = 0 in Ω , u = ϕ in Γ ,
where

χ D i is the characteristic function of D i , i = 1, 2.
We make the following assumptions:

(A) Ω 0 = Ω \D 1 ∪ D 2 , D 0 = D 1 ∩ D 2 are connected and S = ∂ Ω 0 ∩ ∂ D 0 has nonempty interior. (B) There exists γ a non empty open subset of Γ so that ∂ ν u 1 = ∂ ν u 2 on γ. (a) Show that u 1 = u 2 in Ω 0 and then u 1 = u 2 in D 0 . (b) (i) Assume that ω = D 2 \D 1 = / 0. Check that u = u 2 -u 1 ∈ H 2 0 (ω) and ∆ u = u 2 in ω. Then show that u satisfies -∆ 2 u + ∆ u = 0 in ω.
(ii) Deduce that u = 0 in ω and conclude that ω = / 0. (c) Prove that

D 1 = D 2 . 4.3. Let B denotes the unit ball of R n . In this exercise, σ ∈ C 2 (B) and β ∈ C(B) satisfy σ 0 ≤ σ ≤ σ 1 , |β | ≤ β 0 ,
where 0 < σ 0 ≤ σ 1 and

β 0 > 0 are fixed constants. Let u ∈ C 2 (B) so that -div(σ ∇u) + β u = 0 in B.
Define for 0 < r < 1

H(r) = S(r) σ (x)u 2 (x)dS(x), D(r) = B(r) σ (x)|∇u(x)| 2 + β (x)u 2 (x) dx.
Here B(r) (resp. S(r)) is the ball (resp. sphere) of center 0 and radius r.

(a) Prove that

H (r) = n -1 r H(r) + H(r) + 2D(r), (4.88)
D (r) = n -2 r D(r) + D(r) + 2H(r) + D(r) + Ĥ(r), (4.89)
where

H(r) = S(r) u 2 (x)∇σ (x) • ν(x)dS(x), H(r) = S(r) σ (x)(∂ ν u(x)) 2 dS(x), Ĥ(r) = S(r) β (x)u 2 (x)dx, D(r) = B(r) |∇u(x)| 2 ∇σ (x) • xdx, D(r) = -2 B(r) β (x)u(x)x • ∇u(x)dx - n -2 r B(r) β (x)u 2 (x)dx. (b) Let, for 0 < r < 1, K(r) = B(r) σ (x)u 2 (x)dx. (i) Show that if β ≥ 0 then K(r) ≤ e σ 1 /σ 0 n H(r).
(ii) Assume that σ = 1 (in that case we can take

σ 0 = σ 1 = 1). Demonstrate then that K(r) ≤ rH(r), 0 < r < r 0 = min 1, (n -1)β -1 0 1/2 .
Recall that the frequency function N is defined by

N(r) = rD(r) H(r)
and the following identity holds

N (r) N(r) = 1 r + D (r) D(r) - H (r) H(r) . (4.90)
(c) (i) Assume that β = 0. Prove that, for 0 < r < 1 and 0 < r ≤ r, N(r) ≤ e 2σ 1 /σ 0 N(r).

ii) Show that, under the assumption that β ≥ 0 or σ = 1, we have

N(r) ≤ C max(N(r 0 ), 1),
where the constant C > 0 only depends on Ω , σ 1 /σ 0 and β 0 . Hint: we can establish a preliminary result. Set I = {r ∈ (0, δ ); N(r) > max(N(r 0 ), 1)} and observe that I is a countable union of open intervals:

I = ∞ i=1 (r i , s i ).
Show then that Ĥ(r)

D( r) and D(r) D(r)
are bounded on each (r i , s i ) by a constant independent on i.

4.4.

Let Ω be a C 1,1 bounded domain of R n with boundary Γ . We admit the following theorem which is contained in [5, Theorem 9.15 and Lemma 9.17].

Theorem 4.11. Let 1 < p < ∞.

(1) For any f ∈ L p (Ω ), there exists a unique u

∈ W 2,p (Ω ) ∩ W 1,p 0 (Ω ) satisfying -∆ u = f in Ω .
(2) There exists C > 0, depending on Ω and p, so that, for any u ∈ W 2,p (Ω ) ∩ W 1,p 0 (Ω ), we have

u W 2,p (Ω ) ≤ C ∆ u L p (Ω ) . (4.91)
(a) Let A be the unbounded operator defined on

L 2 (Ω ) by Au = -∆ u, u ∈ D(A) = H 1 0 (Ω ) ∩ H 2 (Ω ). Fix 0 < α < 1, λ ∈ σ (A)
and φ ∈ D(A) an eigenfunction associated to λ . (i) Assume that n < 4. Prove that φ ∈ W 2,p (Ω ), for any 1 < p < ∞, and

φ W 2,p (Ω ) ≤ Cλ 2 φ L 2 (Ω ) , (4.92)
where the constant C only depends on Ω and n. Show that φ ∈ C 1,α (Ω ) and deduce from (4.92) that

φ C 1,α (Ω ) ≤ Cλ 2 φ L 2 (Ω ) ,
where the constant C depends only on Ω , n and α.

(ii) Consider the case 4 ≤ n < 8. Prove that φ ∈ W 2,q 0 (Ω ), with q 0 = 2n n-4 if 4 < n < 8 and q 0 = 2p/(2p) for an arbitrary fixed 1 < p < 2 if n = 4, and

φ W 2,q 0 (Ω ) ≤ Cλ 2 φ L 2 (Ω ) ,
where the constant C only depends on Ω and n. Proceed then as in (i) in order to obtain

φ C 1,α (Ω ) ≤ Cλ 3 φ L 2 (Ω ) ,
the constant C depends only on Ω , n and α.

(iii) Prove that, for any n ≥ 1, we have φ ∈ C 1,α (Ω ) and there exists an non negative integer m = m(n) so that

φ C 1,α (Ω ) ≤ Cλ m φ L 2 (Ω ) ,
the constant C only depends on Ω , n and α. Hint:

write n = 4k + , k ∈ N 0 , ∈ {0, 1, 2, 3}. (b) (i) Let D be a Lipschitz bounded domain of R m , m ≥ 1, E D and 0 < α ≤ 1.
Demonstrate that there exist three constants β > 0, c > 0 and C > 0, only depending on D, E and α, so that for any v ∈ C 1,α (D) ∩ H 2 (D) and 0 < ε < 1 we have

C v H 1 (D) ≤ ε β u C 1,α (D) + e c/ε v L 2 (E) + ∆ v L 2 (D) . (4.93)
(ii) Let ω Ω and 0 < α ≤ 1. Prove that there exist β > 0, c > 0 and C > 0, only depending on Ω , ω and α, so that, for any

λ > 0, u ∈ C 1,α (Ω ) ∩ H 2 (Ω ) and 0 < ε < 1, we have that Ce -2 √ λ u L 2 (Ω ) ≤ ε β u C 1,α (Ω ) + e c/ε u L 2 (ω) + (∆ + λ )u L 2 (Ω ) . (4.94)
Hint: apply (i) to v(x,t) = u(x)e λt , (x,t) ∈ D = Ω × (0, 1). c) (i) We use the same notations as in (a). If m is the integer in (a) (iii) then show that

Ce -2 √ λ φ L 2 (Ω ) ≤ λ m ε β φ L 2 (Ω ) + e c ε φ L 2 (ω) , 0 < ε < 1. (4.95)
The constants C, c and β are the same as in (ii).

(ii) Conclude that there exists a constant κ > 0, only depending on n, Ω and ω, so that

e -e κ √ λ ≤ φ L 2 (ω)
φ L 2 (Ω ) .

4.5.

Let Ω be a C 1,1 bounded domain of R n with boundary Γ . We recall the following simple version of an interpolation inequality due to G. Lebeau and L. Robbiano [6] Theorem 4.12. Let ω Ω and D = Ω × (0, 1). There exist two constants C > 0 and β ∈ (0, 1), only depending on n, Ω and ω so that,

for v = v(x,t) ∈ H 2 (D) ∩ H 1 0 (D), we have that v H 1 (Ω ×(1/4,3/4)) ≤ C v 1-β H 1 (D) ∆ v L 2 (D) + ∂ t v(0, •) L 2 (ω) β . (4.96)
Henceforth, ω Ω , C is a generic constant only depending on n, Ω and ω, and β is as in Theorem 4.12. a) Let u ∈ H 1 0 (Ω ) and λ > 0. Prove that

u H 1 (Ω ) ≤ Ce 3 √ λ 4 u 1-β H 1 (Ω ) (∆ + λ )u L 2 (Ω ) + u L 2 (ω) β . Hint: apply Theorem 4.12 to v(x,t) = u(x)e √ λt , (x,t) ∈ D = Ω × (0, 1). b) Let A be the unbounded operator, defined on L 2 (Ω ), by Au = -∆ u, u ∈ D(A) = H 1 0 (Ω ) ∩ H 2 (Ω ). Fix λ ∈ σ (A) and φ ∈ D(A) an eigenfunction associated to λ . Show that e -κ √ λ ≤ φ L 2 (ω) u L 2 (Ω ) ,
with κ = 11/(4β ) -2.

4.6.

Let Ω is a bounded domain of R n . a) (i) In this exercise we use the same assumptions and notations as in Section 4.4.

Prove the following variant of Theorem 4.7.

Theorem 4.13. Let 0 < k < < m and Λ > 0. There exist C > 0 and 0 < γ < 1, that can depend only on Ω , Λ , k, , m, κ and κ, so that, for any v

∈ H 1 (Ω ) satisfying Lv ∈ L 2 (Ω ) in Ω together with (Lv) 2 ≤ Λ v 2 + |∇v| 2 in Ω , y ∈ Ω and 0 < r < dist(y,Γ )/m, we have C v L 2 (B(y, r)) ≤ v γ L 2 (B(y,kr)) v 1-γ L 2 (B(y,mr)) .
(ii) Establish the following result Proposition 4.7. Let ω Ω and ω ⊂ Ω be non empty. There exist C > 0 and β > 0, that can depend on ω and ω, so that, for any u ∈ H 2 (Ω ) satisfying

(Lu) 2 ≤ Λ u 2 + |∇u| 2 in Ω , we have C u L 2 ( ω) ≤ ε β u L 2 (Ω ) + ε -1 u L 2 (ω) , C ∇u L 2 ( ω,R n ) ≤ ε β ∇u L 2 (Ω ,R n ) + ε -1 ∇u L 2 (ω,R n )
for any ε > 0.

b) (Calderón's theorem) Let g : R → R be continuous and satisfies |g(s)| ≤ c|s|, s ∈ R, for some constant c > 0. Let u ∈ H 2 (Ω ) satisfying Lu = g(u) and there exist a nonempty ω Ω so that u = 0 in ω. Prove that u is identically equal to zero.

Solutions of Exercises and Problems

Exercices and Problems of Chapter 1

1.1 If u ∈ L q (Ω ), we get by applying Hölder's inequality

Ω |u| p dx ≤ Ω |u| p(q/p) dx p/q Ω dx 1-p/q = Ω |u| q dx p/q |Ω | 1-p/q .
Hence u ∈ L p (Ω ) and u p ≤ |Ω | 1/p-1/q u q .

1.2 We proceed by induction in k. The case k = 2 corresponds to classical Hölder's inequality. Assume then that the inequality holds for some integer k ≥ 2 and let 1 ≤ p 1 , . . . ,1 ≤ p k , 1 ≤ p k+1 so that

1 p 1 + . . . 1 p k + 1 p k+1 = 1.
The assumption that the inequality holds for k entails

Ω k+1 ∏ j=1 |u j |dx ≤ k-1 ∏ j=1 u j p j u k u k+1 q , (5.1)
where q is defined by

1 q = 1 p k + 1 p k+1 .
On the other hand, since p k+1 /q is the conjugate exponent of p k /q, the classical Hölder's inequality yields

u k u k+1 q q = |u k u k+1 | q 1 ≤ |u k | q p k /q |u k+1 | q p k+1 /q .
That is 185 u k u k+1 q ≤ u k p k u k+1 p k+1 .

This in (5.1) give the expected inequality.

1.3

We get from Hölder's inequality

u q q = |u| q(1-λ ) |u| qλ 1 ≤ Ω |u| q(1-λ )[p/q(1-λ )] dx q(1-λ )/p Ω |u| qλ (r/qλ ) dx q/r
, where we used that r/(qλ ) is the conjugate exponent of p/[q(1λ )].

Whence u ∈ L q (Ω ) and

u q ≤ u 1-λ p u λ r .
1.4 For each integer j ≥ 1, set

K j = {x ∈ ω; dist(x, R n \ ω) ≥ 1/ j and |x| ≤ j} .
We have ω = ∪ j K j , and as K j is compact, we can covert it by finite number of ω i , say K j ⊂ ∪ i∈I j ω i , with I j ⊂ I finite. Hence J = ∪ j I j is countable and we have ω = ∪ i∈J ω i . We know that f = 0 in ω i \ A i with A i of zero measure for each i.

Therefore, f = 0 in ω \ (∪ i∈J A i ). That is f = 0 a.e. in ω because ∪ i∈J A i is of zero measure.
1.5 (a) The case p = ∞ is obvious. Next, we consider the case p = 1. Set h(x, y) = f (xy)g(y).

Then we have

R n |h(x, y)|dx = |g(y)| R n | f (x -y)|dx = f 1 |g(y)| < ∞ a.e. y ∈ R n and R n dy R n |h(x, y)|dx = f 1 g 1 < ∞.
We conclude by applying Tonelli's theorem that h ∈ L 1 (R n × R n ). Then, in light of Fubini's theorem, we obtain

R n |h(x, y)|dy < ∞ a.e. x ∈ R n and R n dx R n |h(x, y)|dy ≤ f 1 g 1 < ∞.
This completes the proof for the case p = 1.

We proceed now to the proof of the case 1 < p < ∞.

From the case p = 1, y → | f (x -y)||g(y)| p is integrable in R n , a.e. x ∈ R n . That is | f (x -y)| 1/p |g(y)| ∈ L p y (R n ) a.e. x ∈ R n . Since | f (x -y)| 1/p ∈ L p (R n
), we deduce by using Hölder's inequality that

| f (x -y)||g(y)| = | f (x -y)| 1/p |g(y)|| f (x -y)| 1/p ∈ L 1 y (R n ) a.e. x ∈ R n and R n | f (x-y)||g(y)|dy ≤ R n | f (x -y)||g(y)| p dy 1/p f 1/p 1 ∈ L 1 y (R n ) a.e. x ∈ R n .
In other words,

|( f * g)(x)| p ≤ (| f | * |g| p )(x) f p/p 1 a.e. x ∈ R n .
The result for the case p = 1 enables us to deduce that f * g ∈ L p (R n ) and

f * g p p ≤ f 1 g p p f p/p 1 . Thus f * g p ≤ f 1 g p . (b) Fix x ∈ R n so that y → f (x -y)g(y) is integrable. Then we have ( f * g)(x) = R n f (x -y)g(y)dy = (x-supp( f ))∩supp(g) f (x -y)g(y)dy. If x ∈ supp( f ) + supp(g) then (x -supp( f )) ∩ supp(g) = / 0 and ( f * g)(x) = 0. Whence ( f * g)(x) = 0 a.e. in R n \ (supp( f ) + supp(g)).
In particular, ( f * g)(x) = 0 a.e. in R n \ supp( f ) + supp(g) and hence supp( f * g) ⊂ supp( f ) + supp(g).

1.6 First, note that f (x) = x α belongs to L 2 (]0, 1[) if and only if α > -1/2. According to the definition of weak derivatives, g = f means that

1 0 g(x)ϕ(x)dx = - 1 0 x α ϕ (x) for any ϕ ∈ D(]0, 1[). Let ϕ ∈ D(]0, 1[) with supp(ϕ) ⊂]a, b[⊂]0, 1[ where 0 < a < b < 1. Then 1 0 x α ϕ (x) = b a x α ϕ (x)dx = - b a αx α-1 ϕ(x)dx = 1 0 αx α-1 ϕ(x)dx. Whence g(x) = αx α-1 and then g ∈ L 2 (]0, 1[) if and only if α > 1/2.
1.7 (a) Introduce the notations I i = { j; 1 ≤ j ≤ k, j = i and Ω j ∩ Ω i = / 0} and, for j ∈ I i , set Γ j i = Ω j ∩ Ω i . We denote by ν j i the unit normal vector field on

Γ j i directed from Ω i to Ω j . Let f ∈ C 1 pie (Ω , (Ω i ) 1≤i≤k
) and ϕ ∈ D(Ω ). Green's formula on each Ω i gives, with

f i = f | Ω i , Ω f ∂ ϕdx = k ∑ i=1 Ω i f i ∂ ϕdx = - k ∑ i=1 Ω i ∂ f i ϕdx + k ∑ i=1 Γ i f i ϕν dσ = - k ∑ i=1 Ω i ∂ f i ϕdx + k ∑ i=1 ∑ j∈I i Γ j i f i ϕ(ν j i ) dσ . If J = {(i, j); 1 ≤ i, j ≤ k, i = j and j ∈ I i }, we deduce Ω f ∂ ϕdx = - k ∑ i=1 Ω i ∂ f i ϕdx + ∑ (i, j)∈J Γ j i ( f i (ν j i ) + f j (ν i j ) )ϕdσ . But ν j i = -ν i j .
Hence

Ω f ∂ ϕdx = - k ∑ i=1 Ω i ∂ f i ϕdx + ∑ (i, j)∈J Γ j i ( f i -f j )ϕ(ν j i ) dσ . (5.2) If in addition f ∈ C Ω then Ω f ∂ ϕdx = - k ∑ i=1 Ω i ∂ f i ϕdx.
Hence g defined by g

| Ω i = ∂ f i in Ω i , 1 ≤ i ≤ k, is in L 2 (Ω ) because ∂ f i ∈
C Ω i and we have in the weak sense

∂ f = g , 1 ≤ ≤ k. (b) Let f ∈ C 1 pie (Ω , (Ω i ) 1≤i≤k
). Assume that there exists, for fixed 1 ≤ ≤ k, g ∈ L 2 (Ω ) so that ∂ f = g in the weak sense, i.e.

Ω f ∂ ϕdx = - Ω g ϕdx = - k ∑ i=1 Ω i g ϕdx for any ϕ ∈ D(Ω ).
This and (5.2) entail

k ∑ i=1 Ω i (g -∂ f i )ϕdx + ∑ (i, j)∈J Γ j i ( f i -f j )ϕ(ν j i ) dσ = 0. (5.3)
We deduce, by choosing ϕ to be the extension by 0 of a function in D(Ω i ),

Ω i (g -∂ f i )ϕdx = 0 for any ϕ ∈ D(Ω i ).
Thus, g | Ω i = ∂ f i by the cancellation theorem. We come back to (5.3) to conclude that ∑ (i, j)∈J Γ j i ( f if j )ϕ(ν j i ) dσ = 0 for any ϕ ∈ D(Ω ) and 1 ≤ ≤ n. (5.4) This implies that f i = f j for any (i, j) ∈ J. Otherwise, we would find (i, j) de J, 1 ≤ ≤ n and B r ⊂ Ω a ball centred at a point in Γ j i so that B r does not intersect any other Γ k and ( f if j )(ν j i ) does not vanish and is of constant sign in B r ∩Γ j i . If we choose ϕ ∈ D(Ω ) satisfying 0 ≤ ϕ ≤ 1, ϕ = 1 in B r/2 and suppϕ ⊂ B r in (5.4) we obtain

Γ j i ∩B r/2 ( f i -f j )(ν j i ) dσ = 0.
This yields the expected contradiction.

1.8 (a) For α > 0, we have

u(x) = |ln |x|| α = ln |x| 2 2 α . Therefore ∇u(x) = -α |ln |x|| α-1 x |x| 2 , x = 0 and hence B 1/2 |∇u| 2 dx = B 1/2 α| ln |x|| α-1 |x| 2 dx.
We obtain by passing to polar coordinates where we make the change of variable s =ln r in order to get the last integral. We conclude that u ∈ H 1 (B 1/2 ) whenever the last integral is convergent. This is the case if 2(α -1) < -1 or equivalently α < 1/2. Finally, if α > 0 then u is unbounded. (b) Let 0 < β < (n -2)/2 and u(x) = |x| -β . Note that u is unbounded because β > 0. On the other hand, we have

∇u(x) = -β |x| -(β +2) x, x = 0.
If S n-1 denotes the unit sphere of R n , we obtain by passing to spherical coordinates

B 1 |∇u| 2 dx = β 2 B 1 |x| -2(β +1) dx = β 2 |S n-1 | 1 0 r n-1-2(β +1) dr. Thus u ∈ H 1 (B 1 ) if n -2β -3 > -1 or equivalently β < (n -2)/2.
1.9 Let 1 ≤ i, j ≤ n.

(a) We have ∂ i (|x| α ) = α|x| α-2 x i for any x = 0 as a consequence of the formula ∂ i |x| = x i /|x|. We obtain by passing to spherical coordinates Therefore |x| α ∈ W 1,p (B) if and only if n + p(α -1) > 0.

We have similarly

R n \B |x| α p dx = S n-1 dσ (ω) +∞ 1 r n-1+pα dr and R n \B |∂ i (|x| α )| p dx = S n-1 |ω i | p dσ (ω) +∞ 1
r n-1+p(α-1) dr.

We deduce that |x| α ∈ W 1,p (R n \ B) if and only if n + pα < 0. (b) We have Whence (u k ) is a Cauchy sequence in L ∞ (R). Therefore it converges to u in L ∞ (R).

∂ j x i |x| = δ i j |x| - x i x j |x| 3 . Whence B x i |x| p dx = S n-1 |ω i | p dσ (ω)
The expected result is obtained by passing to the limit, as k → +∞, in u k L ∞ (R) ≤ c u k W 1,p (R) .

1.12 As in the proof of Lemma 1.9, we have

|u(x , 0)| q ≤ q +∞ 0 |u(x , x n )| q-1 |∂ n u(x , x n )|dx n .
We obtain then by applying Fubini's theorem

R n-1 |u(x , 0)| q dx ≤ q R n |u(x)| q-1 |∂ n u(x)|dx.
We end up getting by using Hölder's inequality

R n-1 |u(x , 0)| q dx ≤ q ∂ n u p R n |u(x)| (q-1)p dx 1/p .
The result then follows by noting that a simple computation yields (q -1)p = p * .

1.13

We have

Ω |v| 2 dxdy = 1 0 dx x β 0 x 2α dy = 1 0 x 2α+β dx. Whence v ∈ L 2 (Ω ) if and only if 2α + β > -1. Observing that ∂ y v = 0 and ∂ x v = αx α-1 , we prove similarly that v ∈ H 1 (Ω ) if and only if 2(α -1) + β > -1 or equivalently 2α + β > 1.
On the other hand,

∂ Ω |v| 2 dσ = 1 + 1 0 x 2α (1 + β 2 x 2β -1 ) 1/2 dx. Since β > 2, (1 + β 2 x 2β -1 ) 1/2 is bounded in [0, 1]. We then deduce that v ∈ L 2 (∂ Ω )
if and only if 2α > -1. Now, as β > 2, there exists α so that 1β < 2α < -1. In that case, v ∈ H 1 (Ω ) and v ∈ L 2 (∂ Ω ). This result shows that for the domain Ω the trace operator

C(Ω ) → L 2 (∂ Ω ) : u → u| ∂ Ω
does not admit a bounded extension to H 1 (Ω ).

1.14 (a) Let, for ε > 0,

∂ i (|x| 2 + ε) α/2 x i = α(|x| 2 + ε) α/2-1 x 2 i + (|x| 2 + ε) α/2 . Hence div (|x| 2 + ε) α/2 x = α(|x| 2 + ε) α/2-1 |x| 2 + n(|x| 2 + ε) α/2 .
It is not hard to check that, in

L 1 loc (R n ), (|x| 2 + ε) α/2 → |x| α and div[(|x| 2 + ε) α/2 x] → (α + n)|x| α .
By the closing lemma, we have div(|x| α x) = (α + n)|x| α in the weak sense. For u ∈ D(R n ), we get from the divergence theorem

R n div [|u| p |x| α x] dx = 0. Thus (α + n) R n |u| p |x| α dx = -p R n x • ∇u|u| p-1 |x| α dx.
This and Hölder's inequality entail

R n |u| p |x| α dx ≤ p α + n R n |x • Du||u| p-1 |x| α dx ≤ p α + n R n |u| (p-1)p |x| α dx 1-1/p R n |x • ∇u| p |x| α dx 1/p ≤ p α + n R n |u| p |x| α dx 1-1/p R n |x • ∇u| p |x| α dx 1/p
, and hence the result follows.

(b) Let u ∈ W 1,p (R n ). In light of the fact that D(R n ) is dense in W 1,p (R n ), there exists a sequence (u m ) in D(R n ) converging to u in W 1,p (R n ).
We find, by applying (a) with α = -p,

u m 1 |x| - u m 2 |x| p ≤ p n -p ∇u m 1 -∇u m 2 p . (u m /|x|) is then a Cauchy sequence in L p (R n ). As u m → u in L p (R n ), u m /|x| → u/|x| in L p (R n ).
We have also

u m |x| p ≤ p n -p ∇u m p .
Passing to the limit, as m goes to ∞, we end up getting

u |x| p ≤ p n -p ∇u p .
1.15 (a) As in the preceding proof, using the density of D(R n ) in W 1,p (R n ), we obtain that there exists a sequence

(v m ) in D(R n ) converging in W 1,p (R n ) to v.
A slight modification of the proof of Proposition 1.6 enables us to get

∂ i (u * v m ) = u * ∂ i v m , 1 ≤ i ≤ n.
On the other hand,

u * v m -u * v p ≤ u 1 v m -v p and u * ∂ i v m -u * ∂ i v p ≤ u 1 ∂ i v m -∂ i v p . Whence, u * v m converges to u * v in L p (R n ) and ∂ i (u * v m ) converges to u * ∂ i v in L p (R n ). The closing lemma then yields u * v ∈ W 1,p (R n ) and ∂ i (u * v) = u * ∂ i v, 1 ≤ i ≤ n. (b) (i) We have supp(ρ m * ϕu -ρ m * u) = supp(ρ m * (1 -ϕ)u). But supp(ρ m * (1 -ϕ)u) ⊂ B 0, 1 m + supp((1 -ϕ)u) supp((1 -ϕ)u) ⊂ supp(1 -ϕ).
In consequence,

supp(ρ m * ϕu -ρ m * u) ⊂ B 0, 1 m + supp(1 -ϕ) ⊂ R n \ ω
provided that m is sufficiently large. We obtain from (a)

∂ i (ρ m * ϕu) → ∂ i (ϕu), in L p (R n ).
As ϕ has compact support in Ω , we can check that

∂ i ϕu = ∂ i (ϕu) = ∂ i ϕu + ϕ∂ i u and then ∂ i (ρ m * ϕu) → D i ϕu + ϕD i u in L p (R n ).
In particular,

∂ i (ρ m * ϕu) → ∂ i u in L p (ω),
and since ρ m * ϕu = ρ m * u in ω, we deduce

∂ i (ρ m * u) → ∂ i u in L p (ω).
(ii) If (θ m ) is a truncation sequence then u m = θ m (ρ m * u) satisfies the required conditions. (c) From Friedrichs's theorem and its proof, there exist two sequences (u m ) and

(v m ) in D(R n ) so that u m → u, v m → v in L p (Ω ) and a.e. in Ω , ∇u m → ∇u, ∇v m → ∇v in L p (ω), for any ω Ω , u m ∞ ≤ u ∞ , v m ∞ ≤ v ∞ .
We have, for any ϕ ∈ D(Ω ),

Ω u m v m ∂ i ϕdx = - Ω (∂ i u m v m + u m ∂ i v m )ϕdx, 1 ≤ i ≤ n.
In light of the dominated convergence theorem, we can pass to the limit, when m goes to ∞. We get

Ω uv∂ i ϕdx = - Ω (∂ i uv + u∂ i v)ϕdx, 1 ≤ i ≤ n,
for any ϕ ∈ D(Ω ), and the expected result follows.

1.16 (a) Fix ω an open set so that supp(u) ⊂ ω Ω and pick ϕ ∈ D(Ω ) so that ϕ = 1 in supp(u). According to Friedrichs's theorem (see Exercise 1.15), there exists a sequence

(φ m ) in D(R n ) such that φ m → u in L p (Ω ) and ∇φ m → ∇u in L p (ω, R n ).
Whence, ϕφ m → ϕu in W 1,p (Ω ), ϕu ∈ W 1,p 0 (Ω ) and then u ∈ W 1,p 0 (Ω ). (b) (i) We have u m ∈ W 1,p (Ω ) by Proposition 1.10. On the other hand, we easily check, with the aid of the dominated convergence theorem, that u m → u in W 1,p (Ω ). As G(mu) = 0 if |u| ≤ m, we then get supp(u m ) ⊂ {x ∈ Ω ; |u(x)| > 1/m} . Now, as u = 0 on Γ , u m has compact support in Ω . Therefore, u m ∈ W 1,p 0 (Ω ) by (a). (ii) Let (θ m ) be the truncation sequence in the proof of Theorem 1.10. By (i), θ m u ∈ W 1,p 0 (Ω ), and as θ m u → u in W 1,p (Ω ), we conclude that u ∈ W 1,p 0 (Ω ). (c) We saw in (b) that if u ∈ W 1,p (Ω ) ∩ C Ω is such that u = 0 on Γ then u ∈ W 1,p 0 (Ω ). Conversely, if u ∈ W 1,p 0 (Ω ) ∩C Ω then, according to the definition of W 1,p 0 (Ω ), u is the limit in W 1,p (Ω ) of a sequence of elements in D(Ω ). Using that the trace operator γ 0 is bounded from W 1,p (Ω ) into L p (Γ ) and that γ 0 u m = 0, we obtain γ 0 u = 0.

1.17 (a) We get, by using ϕ(x) = x 0 ϕ (t)dt and applying then Cauchy-Schwarz's inequality,

|ϕ(x)| 2 ≤ x x 0 |ϕ (t)| 2 ≤ x 1/2 0 |ϕ (t)| 2 , x ∈ 0, 1 2 .
Similarly, Cauchy-Schwarz's inequality applied to the identity ϕ (b) Let u be a solution of (1.13). Multiply each side of the first equation of (1.13) by ϕ ∈ D(]0, 1[) to derive that

(x) = 1 x ϕ (t)dt yields |ϕ(x)| 2 ≤ (1 -x) 1 x |ϕ (t)| 2 ≤ (1 -x) 1 1 2 |ϕ (t)| 2 , x ∈ 1 2 , 1 .
- 1 0 u (x)ϕ(x)dx -k 1 0 u(x)ϕ(x)dx = 1 0 f (x)ϕ(x)dx.
But, from the definition of derivatives in the weak sense, we have

- 1 0 u (x)ϕ(x)dx = 1 0 u (x)ϕ (x)dx. Hence 1 0 u (x)ϕ (x)dx -k 1 0 u(x)ϕ(x)dx = 1 0 f (x)ϕ(x)dx.
(5.5)

Let now u 1 and u 2 be two solutions of (1.13). As (5.5) is satisfied for both u 1 and u 2 , we deduce, for any ϕ ∈ D(]0, 1[), that

1 0 u (x)ϕ (x)dx -k 1 0 u(x)ϕ(x)dx = 0, with u = u 1 -u 2 .
Once again, by the density of D(]0, 1[) in H 1 0 (]0, 1[), we can choose ϕ = u k , where (u k ) is a sequence in D(]0, 1[) converging in H 1 0 (]0, 1[) to u. After passing to the limit, as k goes to ∞, we get In particular, if k = π2 and if u is a solution (1.13) then u + sin(πx) is also a solution of (1.13) according to (b). But this holds only if kC ≥ 1 or equivalently π 2 C ≥ 1. We already know from (a) that C ≤ 1/8. In conclusion, we proved that 1 In other words, we proved that B is equi-continuous and therefore it is relatively compact in C I by Arzela-Ascoli's theorem.

π 2 ≤ C ≤

1.19

Observing that ∇(uu) = ∇u, it is not hard to see that we are reduced to prove the following result: there exists a constant C > 0 so that, for any u ∈ V = {v ∈ H 1 (Ω ); Ω vdx = 0}, we have

u L 2 (Ω ) ≤ C ∇u L 2 (Ω ,R n ) .
V is a closed subspace of H 1 (Ω ) that we endow with norm of H 1 (Ω ).

If the above inequality does not hold, we would find a sequence (u k ) in V so that

u k L 2 (Ω ) > k ∇u k L 2 (Ω ,R n ) .
Note that we can always assume that u k H 1 (Ω ) = 1 for each k. Therefore, Subtracting a subsequence if necessary, we can also assume that u k converges strongly in L 2 (Ω ) and weakly in H 1 (Ω ). But, as ∇u k converge strongly to 0 in L 2 (Ω ), we deduce that ∇u = 0 by the closing lemma. Whence u is a.e. equal to a constant c. This and the fact that 0

= Ω u k dx → Ω udx
entail that c = 0 and then u = 0. In particular, u k H 1 (Ω ) → 0 which contradicts u k H 1 (Ω ) = 1, for each k.

1.20 Pick v ∈ H 1/2 (Γ ) and let K v be the closed convex set of H 1 (Ω ) given by

K v = {v ∈ H 1 (Ω ); γ 0 (u) = v}.
According to the projection theorem, there exists a unique P K v (0) ∈ H 1 (Ω ) so that

0 -P K v (0) H 1 (Ω ) = min u∈K v 0 -u H 1 (Ω ) .
That is

u v H 1 (Ω ) = min u∈K v u H 1 (Ω ) = v H 1/2 (Γ ) ,
where

u v = P K v (0).
Exercises and Problems of Chapter 2

2.1 Let E and F be two infinite dimensional Banach spaces. If A ∈ K (E, F) admitted an inverse A -1 ∈ L (F, E) then AA -1 = I would be compact. But this is impossible by Riesz's theorem.

2.2

The operator A is bounded since, for any x ∈ 2 , we have

Ax 2 2 = ∑ m≥1 (a m x m ) 2 ≤ sup m≥1 a 2 m ∑ m≥1 x 2 m ≤ C 2 x 2 2 .
Assume that lim m→+∞ a m = 0. Let (x ) be a sequence in the closed unit ball of 2 and set, for each , y = Ax . To prove that A is compact we construct a subsequence of (y ) converging in 2 . Let y ,0 = y . By induction in k, if y ,k is constructed then, as the sequence y ,k m is bounded, there exists y . We extract a diagonal subsequence by setting z = y , . We claim that (z ) is a Cauchy sequence in 2 . Indeed, for ε > 0, as a m tends to 0, there exists an integer p so that for any m > p, we have that |a m | < ε. Therefore, for each , we have

∑ m≥p z m 2 ≤ ε 2 ∑ m x m 2 ≤ ε 2 .
On the other hand, since z m is convergent, there exits 0 such that

∑ m≤p |z m -z m | 2 ≤ ε 2 for any , ≥ 0 .
A combination of these two inequalities then yields

∑ m |z m -z m | 2 ≤ 3ε 2 for any , ≥ 0 .
Thus (z ) is a Cauchy sequence in 2 . Using that 2 is complete we end up getting that (z ) is a convergent subsequence of (y k ).

Conversely, assume that the sequence (a m ) does not converge to 0. Hence there exits a constant C > 0 so that, for any integer m, we find an integer k > m with the property that |a k | > C. Define the sequences (x ) in 2 and (k ) in N so that x m = δ m k , |a k | > C, and the sequence (k ) is increasing. Set y = Ax . Then the sequence (x ) in bounded in 2 , while (y ) does not admit any convergent sub-sequence. Indeed, for any = , we have yy 2 > 2C. In other words, the operator A is non compact.

We have

A f 2 H = 1 0 (x 2 + 1) 2 f (x) 2 dx ≤ x 2 + 1 2 L ∞ (0,1) f 2 H = 4 f 2 H
and then A is bounded. It is clear that A is self-adjoint and

(A f , f ) = 1 0 (x 2 + 1) f (x) 2 dx > 0 for any f ∈ H, f = 0.
That is A is positive.

If f ∈ H is an eigenvector of A corresponding to the eigenvalue λ then, for any g ∈ H, we have 1 0

(x 2 + 1) f (x)g(x)dx = (A f , g) = λ ( f , g) = λ 1 0 f (x)g(x)dx
(x 2 + 1 -λ ) f (x)g(x)dx = 0.
Choosing g = (x 2 + 1λ ) f , we get that (x 2 + 1λ ) f H = 0. Consequently, (x 2 + 1λ ) f = 0 a.e. in (0, 1) implying that f = 0 a.e. in (0, 1). This contradicts the fact that f is an eigenvector. Whence A does not admit any eigenvalue.

To prove that Aλ I is invertible, if g ∈ H we seek f ∈ H satisfying (Aλ I) f = g. That is, we want to find f ∈ H such that

(x 2 + 1 -λ ) f (x) = g(x)
a.e. in (0, 1). Therefore, if Aλ I is invertible then necessarily f = (Aλ I) -1 g is given by

f (x) = (x 2 + 1 -λ ) -1 g(x)
a.e. in (0, 1).

The inverse of (x 2 + 1λ ) is well defined except at the endpoints. Hence, f (x) is well defined a.e. x ∈ (0, 1).

If λ ∈ [1, 2], then m(λ ) = min [0,1] |x 2 + 1 -λ | > 0. Hence A -λ I is invertible with bounded inverse: (A -λ I) -1 g H ≤ m(λ ) -1 g H .
For λ ∈ [1, 2], if (Aλ I) was invertible then (x 2 + 1λ ) -1 would be an element of H. This can not be true because otherwise, since

1 x 2 + 1 -λ = 1 (x - √ λ -1)(x + √ λ -1)
, we would have

1 x 2 + 1 -λ ∼ 1 2 √ λ -1(x - √ λ -1) as x → λ -1. Hence 1 0 1 (x 2 + 1 -λ ) 2 dx = +∞.
This leads to the expected contradiction.

2.4

Assume that A is compact and let (x m ) be a sequence in E converging weakly to x. We show that the only limit point for the strong topology of the sequence

(Ax m ) is Ax. Indeed, if Ax ψ(m) → y then ϕ, Ax ψ(m) = ϕ • A, x ψ(m) → ϕ • A, x = ϕ, Ax ,
for any ϕ ∈ F , where we used that (x m ) converges weakly to x. Therefore, ϕ, Axy = 0 for any ϕ ∈ F , and hence y = Ax. On the other hand, we know that any weakly convergent sequence are bounded. Then (x m ) is bounded, and since A is compact, (Ax m ) admits Ax as limit point. Whence (Ax m ) converges to Ax. 3 Conversely, as E is reflexive, the closed unit ball is compact for the weak topology. To prove that the image by A of the closed unit ball of E is relatively compact, it is sufficient to show that for any arbitrary sequence (x m ) of E so that x m ≤ 1, the sequence (Ax m ) has a convergent subsequence. As (x m ) belongs to the closed unit ball, it has a limit point for the weak topology and therefore a convergent subsequence because the weak topology is metrizable. That is we have a subsequence

x ψ(m) converging weakly to x ∈ E. Using the assumption on A, we deduce that Ax ψ(m) converges strongly to Ax. In other words, (Ax m ) admits a limit point and hence A is compact.

2.5

We have by applying Cauchy-Schwarz's inequality

|A f (y)| ≤ X |k(x, y)| 2 dµ(x) 1/2 X | f (x)| 2 dµ(x) 1/2 . Hence Y |A f (y)| 2 dν(y) ≤ Y X |k(x, y)| 2 dµ(x) dν(y) X | f (x)| 2 dµ(x). As k ∈ L 2 (X × Y, µ × ν),
Thus A f L 2 (Y ) ≤ k L 2 (X×Y ) f L 2 (X) .
Since A is clearly a linear map, we conclude that A is bounded linear operator from L 2 (X) into L 2 (Y ) and its norm is less or equal to k L 2 (X×Y ) . When L 2 (X) (reflexive) is separable, it is sufficient, by the preceding exercise, to show that if ( f n ) converges weakly to f in L 2 (X) then (A f n ) converges strongly to A f in L 2 (Y ). Substituting f n by f nf , we may assume that ( f n ) converges weakly to 0. For such a sequence (A f n (y)) converges to 0 = A f (y), for any y so that x → k(x, y) ∈ L 2 (X), i.e. we have

|A f n (y)| ≤ f n L 2 (X) K(y) with K(y) = X |k(x, y)| 2 dµ(x) ∈ L 2 (Y ).
But, the sequence ( f n ) being weakly convergent, it is therefore bounded. We apply then the dominated convergence theorem to get that Y |A f (y)| 2 dν(y) tends to 0. In other words, (A f n ) converges strongly to 0 in L 2 (Y ) and hence A is compact.

2.6

Assume that u is a solution of (2.97). Multiply each side of the first equation of (2.97) by v ∈ C 1 (Ω ) and then integrate over Ω . Apply then the divergence theorem to the resulting identity to obtain

Ω ∇u • ∇vdx - Γ ∂ ν uvdσ = Ω f vdx.
As ∂ ν u = 0 on Γ , we deduce

Ω ∇u • ∇vdx = Ω f vdx for any v ∈ C 1 (Ω ).
(5.6)

Conversely, if u is a solution of (5.6) then, again by the divergence theorem,

Ω (-∆ u -f )vdx + Γ ∂ ν uvdσ = 0 for any v ∈ C 1 (Ω ),
from which we deduce first that -∆ u = f in Ω and then ∂ ν u = 0 on Γ . Choosing v = 1 in (5.6), we obtain that if there exists a solution of class C 2 then necessarily Ω f dx = 0.

2.7

We multiply each side of the first equation of (2.98) by v ∈ H 1 0 (Ω ) and then we integrate over Ω . The divergence theorem then enables us to obtain the following variational problem: find u ∈ H 1 0 (Ω ) satisfying

a(u, v) = Φ(v), for any v ∈ H 1 0 (Ω ), with a(u, v) = Ω (∇u • ∇v +V • ∇uv)dx and
Φ(v) = Ω f vdx.
It is not hard to check that a and Φ are continuous. In order to apply Lax-Milgram's lemma, we need to prove that a is coercive. We have

Ω V • ∇uudx = 1 2 Ω div (u 2 V )dx = 1 2 Γ u 2 V • ν = 0. Whence a(u, u) = Ω ∇u • ∇udx
and hence a is coercive in H 1 0 (Ω ).

(a)

We proceed by contradiction. Assume then that, for each m, there exists

v m ∈ H 1 (Ω ) so that v m L 2 (Ω ) > m( v m L 2 (Γ ) + ∇v m L 2 (Ω ,R n ) ).
Substituting v m L 2 (Ω ) by v m / v m L 2 (Ω ) , we may assume that v m L 2 (Ω ) = 1. Then (v m ) is bounded in H 1 (Ω ) and, hence by Rellich's theorem, there exists (v p ) a subsequence of (v m ) converging strongly in L 2 (Ω ) to v ∈ L 2 (Ω ). Additionally, ∇v p converges to 0 in L 2 (Ω , R n ). In light of the closing lemma, we get that v ∈ H 1 (Ω ) and ∇v = 0. Thus v is a.e. equal to a constant in Ω . On the other hand, as the trace operator w ∈ H 1 (Ω ) → w| Γ ∈ L 2 (Γ ) is bounded we deduce that v = 0 on Γ and consequently v = 0 in Ω . But this contradicts the fact that v L 2 (Ω ) = 1.

(b) We easily obtain, for v ∈ H 1 (Ω ),

Ω ∇u • ∇vdx - Γ ∂ ν uvdσ = Ω f vdx. But -∂ ν u = u -g on Γ . That is we have a variational problem in the form a(u, v) = Φ(v), where a(u, v) = Ω ∇u • ∇vdx + Γ uvdσ and
Φ(v) = Ω f vdx + Γ gvdσ .
The existence and uniqueness of a solution of the variational problem is obtained by applying Lax-Milgram's Lemma. It is straightforward to check that a and Φ are continuous. While the coercivity of a follows from (a) because

a(u, u) = ∇u L 2 (Ω ,R n ) + u L 2 (Γ ) .
2.9 If ϕ (of class C ∞ ) is an eigenfunction then ϕ is a solution of the ordinary differential equation, with λ > 0,

ϕ + λ ϕ = 0.
The solutions of this equation are of the form

ϕ = A sin( √ λ x) + B cos( √ λ x).
The boundary conditions ϕ(0) = ϕ(1) = 0 imply B = 0 and √ λ = kπ for some k ∈ Z. Therefore, the eigenfunctions of the Laplace operator with Dirichlet boundary condition are given by ϕ k = sin(kπx), k ≥ 1, each ϕ k corresponding to the eigenvalue

λ k = k 2 π 2 .
Apply Theorem 1.9 to H = L 2 (0, 1), V = H 1 0 (0, 1) and a(u, v) = 1 0 u v dx to deduce that the sequence ϕ k forms an orthonormal basis of L 2 (0, 1) and the sequence (ϕ k /(kπ)) forms an orthonormal basis of H 1 0 (0, 1) for the scalar product ( f , g) = 1 0 f g dx. Whence, ∑ a k sin(kπx) converges in L 2 (0, 1) if and only if ∑ a 2 k < ∞ and in H 1 0 (0, 1) if and only if ∑ k 2 a 2 k < ∞. 2.10 Let (ϕ k ), ϕ k = sin(kπx) for each k, be the sequence of eigenfunctions of the Laplace operator in ]0, 1[ under Dirichlet boundary condition. For 1 ≤ p ≤ n and k ≥ 1, set ϕ p,k (x) = ϕ k (x/ p ). Finally, define, for k 1 ≥ 1, . . . , k n ≥ 1,

ψ k 1 ,...,k n (x 1 , . . . , x n ) = ϕ 1,k 1 (x 1 ) . . . ϕ n,k n (x n ).
We can easily check that ψ k 1 ,...,k n is an eigenfunction for the Laplace operator under Dirichlet boundary condition corresponding to the eigenvalue

λ k 1 ,...,k n = n ∑ i=1 k i π i 2 .
To complete the proof we have to show that ψ k 1 ,...,k n forms a basis in L 2 (Ω ), i.e. if w ∈ L 2 (Ω ) is so that

(w, ψ k 1 ,...,k n ) = 0 for all k 1 ≥ 1, . . . , k n ≥ 1, (5.7)
then w = 0. We proceed by induction in the dimension n. The result is true for n = 1 by the preceding exercise. Assume then that the result holds in dimension n -1.

Introduce the function y ∈ L 2 (]0, n [) defined by

y(x n ) = Ω w(x , x n ) ∏ i<n ϕ k i (x i )dx ,
with Ω =]0, 1 [× . . . ×]0, n-1 [ and x = (x 1 , . . . , x n-1 ). By (5.7), for each k ≥ 1,

n 0 y(x n )ϕ n,k (x n )dx n = 0.
As (ϕ n,k ) k forms a basis in L 2 (]0, n [), y(x n ) = 0 a.e. in ]0, n [. Hence, for a.e.

x n ∈ ]0, n [, w x n (x ) = w(x , x n ) ∈ L 2 (Ω ) is so that Ω w x n (x ) ∏ i<n ϕ k i (x i )dx = 0
and by induction's assumption w x n = 0. The proof is then complete.

2.11

Follows readily from the min-max principle:

λ 1 = min Ω |∇u| 2 dx Ω u 2 dx ; u ∈ H 1 0 (Ω ), u = 0 .
2.12 We first prove that V is a Hilbert space. It is clear that •, • defines a scalar product on V . It remains to show that V is complete for the norm associated to this scalar product. This norm is denoted by • V . Proceeding by contradiction, we can easily show that there exists a constant C > 0 so that, for any u ∈ V ,

B 2 u 2 dx ≤ C B 2 |∇u| 2 dx + B 2 \B 1 u 2 dx ,
where B i is the ball centered at 0 with radius i = 1, 2. Then

u H 1 (R n ) ≤ C u V .
Therefore, if (u m ) is a Cauchy sequence in V then it is also a Cauchy sequence in

H 1 (R n ). Thus, there exists u ∈ H 1 (R n ) so that u m converges to u in H 1 (R n ). Also as (|x|u m ) is a Cauchy sequence in L 2 (R n ) it converges in L 2 (R n ) to some v ∈ L 2 (R n ).
On the other hand, |x|u m converges weakly to |x|u. Indeed, for ϕ ∈ D(R n ),

lim m→+∞ R n |x|u m ϕdx = R n |x|uϕdx = R n
vϕdx from which we deduce that v = |x|u and u m converges to u in V . We now prove that V is compactly imbedded in L 2 (R n ). Let (u m ) be a bounded sequence in V , u m V ≤ M. Bearing in mind that H 1 (B) is compactly imbedded in L 2 (B), B an arbitrary open ball of R n , we conclude that (u m ) admits a convergent subsequence denoted again by

(u m ) converging to u in L 2 (B). Moreover, |x|u ∈ L 2 (R n ). Whence R n (u -u m ) 2 dx < |x|<R (u -u m ) 2 dx + 1 R 2 |x|>R |x| 2 (u -u m ) 2 dx < |x|<R (u -u m ) 2 dx + 2M R 2 . Then lim sup m→+∞ R n (u -u m ) 2 dx ≤ 2M R 2 for any R > 0.
Therefore (u m ) converges to u in L 2 (R n ). That is, we proved that V in compactly embedded in L 2 (R n ).

Next, we observe that the bilinear form

a(u, v) = R n (∇u • ∇v + Q(x)uv)dx
is clearly continuous and coercive on V . Apply then Theorem 3.9 to conclude that there exists a Hilbertian basis of L 2 (R n ) consisting in eigenfunctions, corresponding to a sequence of positive eigenvalues converging to infinity.

2.13

Writing u(x, y) = v(r, θ ), we get

∆ u(x, y) = 1 r ∂ r (r∂ r v(r, θ )) + 1 r 2 ∂ 2 θ 2 v(r, θ ).
If we seek v in the form v(r, θ ) = f (r)g(θ ), we find, after making straightforward computations, that f and g are the respective solutions of the equations

-g (θ ) = λ g(θ ), 0 < θ < β , g(0) = g(β ) = 0, (5.8)
and r 2 f (r) + r f (r)λ f (r) = 0, 0 < r < 1.

(5.9) By Exercise 2.9, (5.8) admits as solutions

g k (θ ) = sin kπ β θ , 0 ≤ θ ≤ β , k ≥ 1,
and

λ = kπ β 2 , k ≥ 1.
For the equation (5.9), we look for solutions of the form f (r) = r γ4 . After some computations, we get two systems of solutions

f + (r) = r kπ/β , f -(r) = r -kπ/β .
Using the boundary conditions at θ = 0, β , we obtain two families of solutions

u +,k = r kπ/β sin kπ β θ , u -,k = r -kπ/β sin kπ β θ .
In light of the boundary condition at r = 1, we end up getting that there are only two possible solutions

u + = r π/β sin π β θ , u -= r -π/β sin π β θ .
Next we use the relations

∂ x u = cos θ ∂ r - sin θ r ∂ θ v, ∂ y u = sin θ ∂ r + cos θ r ∂ θ v to deduce that ∂ x u ± et ∂ y u ± are of the form ψ ± (θ )r ±π/β -1 with ψ ± a function of class C ∞ .
In consequence, we get that u + ∈ H 1 (Ω ) and u -∈ H 1 (Ω ) because r π/β -1 ∈ L 2 ((0, 1), rdr) and r -π/β -1 ∈ L 2 ((0, 1), rdr). Thus the only solution belonging to

H 1 (Ω ) is u = u + . We can similarly check that ∂ 2 xx u, ∂ 2 xy u, ∂ 2 yy u are of the form φ (θ )r π/β -2 where φ is a function of class C ∞ . Therefore, u ∈ H 2 (Ω ) if and only if r π/β -2 ∈ L 2 ((0, 1), rdr), or equivalently u ∈ H 2 (Ω ) if and only if β < π. 2.14 (a) (i) V is closed because V = t -1
1 {0} and t 1 is bounded. (ii) If the inequality does not hold, we would find a sequence (w n ) in V so that

w n L 2 (Ω ) = 1 and ∇w n L 2 (Ω ,R n ) ≤ 1 n .
In particular, (w k ) would be a bounded sequence in H 1 (Ω ). As H 1 (Ω ) is compactly imbedded in L 2 (Ω ), (w k ) would admit a subsequence, still denoted by (w k ), converging in L 2 (Ω ) to w ∈ L 2 (Ω ). On the other hand, since

∇w k L 2 (Ω ,R n ) ≤ 1/k, ∇w k tends to 0 in L 2 (Ω , R n ).
In light of the closing lemma, we deduce that w ∈ H 1 (Ω ) and ∇w = 0 a.e. in Ω . Hence w is equal a.e. to a constant. Now, t 1 being bounded, we get that 0

= t 1 w n → 0 = t 1 w. Thus w = 0, contradicting 1 = w k L 2 (Ω ) → 1 = w L 2 (Ω ) . (b) If u ∈ V ∩ H 2 (Ω ) is a solution (2.103) and if v ∈ V then by the divergence theorem - Ω f vdx = Ω ∆ uvdx = - Ω ∇u • ∇vdx + Γ 1 ∂ ν uvdσ + Γ 2 ∂ ν uvdσ . As ∂ ν u| Γ 2 = 0 and v| Γ 1 = 0, we conclude that Ω f vdx = Ω ∇u • ∇vdx.
That is u is a solution of (2.104).

Conversely, assume u ∈ V ∩ H 2 (Ω ) is a solution of (2.104). Then choosing v ∈ D(Ω ) in (2.103), and applying again the divergence theorem, we easily obtain

Ω (-∆ u -f )v = 0 for any v ∈ D(Ω ).
Hence -∆ u = f a.e. in Ω . Next, we take in (2.104) v ∈ {w ∈ D(Ω ); w| Γ 1 = 0}. We get, by using once again the divergence theorem,

Γ 2 ∂ ν uvdσ = 0 for any v ∈ D(Γ 2 ). Whence Γ 2 D ν uvdσ = 0 for any v ∈ L 2 (Γ 2 )
by the density of D(Γ 2 ) in L 2 (Γ 2 ). In consequence, we obtain that ∂ ν u| Γ 2 = 0 and u is a solution of (2.103).

(c) Set

a(u, v) = Ω ∇u • ∇vdx, u, v ∈ V.
By (a) (i), a defines an equivalent scalar product on V . As v ∈ V → Ω f vdx belongs to V , Riesz-Fréchet's representation theorem enables us to deduce that (2.104) admits a unique solution u ∈ V . (d) Is immediate from Theorem 2.2 applied to a, V and H = L 2 (Ω ).

(e) Let V m (resp. W m ) be the set consisting in all subspaces of V (resp. H 1 0 (Ω )) of dimension m. Since W m ⊂ V m we have according to the min-max's formula that max

v∈F m , v =0 Ω |∇v| 2 dx Ω v 2 dx ≥ min E m ∈V m max v∈E m , v =0 Ω |∇v| 2 dx Ω v 2 dx = µ m for any F m ∈ W m .
Whence

λ m = min E m ∈W m max v∈E m , v =0 Ω |∇v| 2 dx Ω u 2 dx ≥ µ m .
2.15 (a) As um(4r) and M(4r)u are two non negative solutions of Lu = 0 in B(4r), we can apply Theorem 2.24 for both. We obtain, for p = 1,

B(2r) (u -m(4r)) ≤ Cr n (m(r) -m(4r)), (5.10)
B(2r) (M(4r) -u) ≤ Cr n (M(4r) -M(r)), (5.11)
where the constant C only depends on the L ∞ -norm of the coefficients of λ -1 L, n and r 0 . We add side by side inequalities (5.10) and (5. (5.12) Fix now 0 < r ≤ r 0 and let k be the integer so that

r 0 4 k < r ≤ r 0 4 k-1 .
We obtain from (5.12)

ω(r) ≤ ω r 0 4 k-1 ≤ γ k ω(4r 0 ), Ω L (u + , φ )dx ≤ Ω f (χ {u>0} + µ χ {u=0} )φ dx, as expected.
2.17 (a) We obtain, by taking v = u in (2.106),

Ω |∇u| 2 dx = Ω F(u)udx + Ω f udx = Ω (F(u) -F(0))udx + Ω F(0)udx + Ω f udx. But (F(u) -F(0))u ≤ 0. Hence Ω |∇u| 2 dx ≤ Ω F(0)udx + Ω f udx.
Cauchy-Schwarz's inequality then yields

∇u 2 L 2 (Ω ,R n ) ≤ |F(0)||Ω | 1/2 + f L 2 (Ω ) u L 2 (Ω ) .
In light of Poincaré's inequality, there exists µ = µ(Ω ) so that

u L 2 (Ω ) ≤ µ ∇u L 2 (Ω ,R n ) and hence ∇u L 2 (Ω ,R n ) ≤ µ(|F(0)||Ω | 1 2 + f L 2 (Ω ) ) = C.
(b) Let u 1 and u 2 be two variational solutions. Then

Ω ∇u 1 • ∇(u 1 -u 2 )dx = Ω F(u 1 )(u 1 -u 2 )dx + Ω f (u 1 -u 2 ), Ω ∇u 2 • ∇(u 1 -u 2 )dx = Ω F(u 2 )(u 1 -u 2 )dx + Ω f (u 1 -u 2 ).
Subtracting side by side these two identities, we get

Ω |∇u 1 -∇u 2 | 2 dx = Ω (F(u 1 ) -F(u 2 ))(u 1 -u 2 )dx. But (F(u 1 ) -F(u 2 ))(u 1 -u 2 ) ≤ 0, because F is non decreasing. Whence ∇(u 1 - u 2 ) = 0 in Ω . Therefore, since u 1 -u 2 ∈ H 1 0 (Ω ), we deduce that u 1 = u 2 in Ω . (c) Let B be the unit ball of L 2 (Ω ). For (λ , w) ∈ [0, 1] × B, we easily check, similarly to (a), that u = T (λ , w) satisfies ∇u L 2 (Ω ,R n ) ≤ µλ F(w) L 2 (Ω ) + f L 2 (Ω ) ≤ µ F(w) L 2 (Ω ) + f L 2 (Ω ) . But F(w) L 2 (Ω ) ≤ a|Ω | 1/2 + b w L 2 (Ω ) ≤ a|Ω | 1/2 + b. Then ∇u L 2 (Ω ,R n ) ≤ µ a|Ω | 1/2 + b + f L 2 (Ω ) .
In other words, T (

[0, 1] × B) is bounded in H 1 0 (Ω ). Since H 1 0 (Ω ) is compactly imbedded in L 2 (Ω ), it follows that T is compact.
In order to show that T satisfies the assumptions Leray-Schauder's theorem, it is sufficient to check that the set

{u ∈ L 2 (Ω ); u = T (λ , u) for some λ ∈ [0, 1]} is bounded in L 2 (Ω ).
In view of Poincaré's inequality, it is enough to prove that this set is bounded in H 1 0 (Ω ). Repeating the estimates in (a), in which we substitute F and f respectively by λ F and λ f , we deduce, for u = T (λ , u),

∇u 2 L 2 (Ω ,R n ) ≤ µλ (|F(0)||Ω | 1/2 + f L 2 (Ω ) ) ≤ µ |F(0)||Ω | 1/2 + f L 2 (Ω ) = C.
Exercises and Problems of Chapter 3

3.1 Let 0 < α ≤ 1. We first note that, as a consequence of the following identities

(T -h -I) f ∞ = T -h (I -T h ) f ∞ = (I -T h ) f ∞ .
we obtain

[ f ] α = sup h>0 (I -T h ) f ∞ h α .
On the other hand, we have

T h -2I + T -h = T -h (T 2 h -2T h + I) = T -h (T h -I) 2 . Whence [ f ] * α = sup h>0 (T h -I) 2 f ∞ h α . As (T h -I)g ∞ ≤ 2 g ∞ , we conclude that [ f ] * α = sup h>0 (T h -I) 2 f ∞ h α = sup h>0 (T h -I)(T h -I) f ∞ h α ≤ 2 sup h>0 (T h -I) f ∞ h α = 2[ f ] α .
Assume now that 0 < α < 1. Using the following two identities

T h -I = 1 2 (T 2 h -I) -(T h -I) 2 , T 2 h = T 2h ,
we get

(T h -I) f ∞ = 1 2 (T 2h -I) f -(T h -I) 2 f ∞ ≤ 1 2 (T 2h -I) f ∞ + 1 2 (T h -I) 2 f ∞ .
Hence, for h > 0,

(T h -I) f ∞ h α ≤ 2 α-1 (T 2h -I) f ∞ (2h) -α + (T h -I) 2 f ∞ 2h -α ≤ 2 α-1 [ f ] α + 2 -1 [ f ] * α . Whence [ f ] α ≤ 2 α-1 [ f ] α + 2 -1 [ f ] * α . As 0 < α < 1, we have 2 α-1 < 1. Therefore [ f ] α ≤ C[ f ] * α with C = 2 -1 1 -2 α-1 = 2 2 -2 α . 3.2 (a) (i) We first prove that u (ε) α ≤ U α . We have u (ε) (x + h) -u (ε) (x) ≤ R n |u(x + h -εy) -u(x -εy)|ϕ(y)dy ≤ h α U α R n ϕ(y)dy = h α U α .
Hence the result follows.

Let k ≥ 1 be an integer and

∈ N n such that | | = k. Then ∂ u (ε) (x) = ε -n ∂ R n u(y)ϕ( x -y ε )dy = ε -n-| | R n u(y)∂ ϕ( x -y ε )dy = ε -| | R n u(x -εy)∂ ϕ(y)dy -ε -| | R n u(x)∂ ϕ(y)dy,
where we used that ∂ ϕ(y)dy = 0 (see Lemma 1.5). We proceed as above to deduce from the last term that |∂ u

(ε) | ≤ Cε α-k U α with C = C(ϕ, k) = max | |=k R n |∂ ϕ(y)|dy.
(ii) We get by taking the derivative under the integral

∂ ε u (ε) (x) = -nε -n-1 R n u(y)ϕ( x -y ε )dy -ε -n-2 R n u(y)∇ϕ( x -y ε ) • (x -y)dy = -nε -1 R n u(x -εy)ϕ(y)dy -ε -1 R n u(x -εy)∇ϕ(y) • ydy = -nε -1 R n u(x -εy)ϕ(y)dy + nε -1 u(x) -nε -1 u(x) -ε -1 R n u(x -εy)∇ϕ(y) • ydy = -nε -1 R n [u(x -εy) -u(x)]ϕ(y)dy -ε -1 R n [u(x -εy) -u(x)]∇ϕ(y) • ydy,
where we used

R n ∇ϕ(y) • ydy = -n. (5.14)
The last term in these identities gives ∇ ε u (ε) ≤ Cε α-1 , where

C = C(n, ϕ) = n + R n |∇ϕ(y) • y|dy.
Note that (5.14) can be simply established by using the divergence theorem: ϕ(y)dy

= -n. (b) As suppϕ(ε -1 •) ⊂ (-ε, ε), u (ε) and v (ε) are in C ∞ c (-2, 2), 0 ≤ ε ≤ 1. We decompose w as follows w(x) = u * v(x) = u (0) * v (0) = u (1) * v (1) - 1 0 ∂ ε [u (ε) * v (ε) (x)]dε = u (1) * v (1) -w 1 (x) -w 2 (x), where w 1 (x) = 1 0 ∂ ε u (ε) * v (ε) (x)dε, w 2 (x) = 1 0 u (ε) * ∂ ε v (ε) dε. (i) Assume 0 < α + β < 1. As u (1) * v (1) belongs to C ∞ (-2, 2), it is sufficient to prove that [w i ] 0,α+β < ∞, for i = 1, 2. We prove [w 1 ] 0,α+β < ∞.
By interchanging the roles of u and v we get also that [w 2 ] 0,α+β < ∞. We are then reduced to prove that there exits a constant C > 0 so that

|w 1 (x + h) -w 1 (x)| ≤ Ch α+β for any h > 0.
Since this result is obvious if h ≥ 1, it is enough to consider the case 0 < h ≤ 1. Recall that we defined the operator T h by T h f (x) = f (x + h). We have

w 1 (x + h) -w 1 (x) = (T h -I)w 1 (x) = 1 0 ∂ ε u (ε) * (T h -I)v (ε) (x)dε. Clearly, (T h -I)v (ε) (x) ≤ h β [v (ε) ] 0,β ≤ h β V β .
On the other hand, according to the mean-value theorem, we have

(T h -I)v (ε) (x) ≤ h ∂ x v (ε) ∞ ≤ Chε β -1 V β ,
and as

|∂ ε u (ε) | ≤ Cε α-1 U α , we get |w 1 (x + h) -w 1 (x)| ≤ h 0 + 1 h |D ε u (ε) | * |(T h -I)v (ε) |(x)dε ≤ CU α V β h β h 0 ε α-1 dε + h ∞ h ε α+β -2 dε ≤ CU α V β 1 α + 1 1 -(α + β ) h α+β . (ii) Consider now the case 1 < α + β < 2. We show that w = u * v ∈ C 1 c (R) and [w ] 0,α+β -1 < ∞.
As before, it sufficient to give the proof with w 1 instead of w. To this end, we introduce the following approximation of w 1 :

w 1,δ (x) = 1 δ ∂ ε u (ε) * v (ε) (x)dε, 0 < δ < 1.
We have from the estimates in (a)

|w 1 (x) -w 1,δ (x)| = δ 0 ∂ ε u (ε) * v (ε) (x)dε ≤ Cδ α ,
where the constant C is independent on δ . Hence w 1,δ converges uniformly in R to w 1 , when δ tends to 0. But

∂ x w 1,δ = 1 δ ∂ ε u (ε) * ∂ x v (ε) (x)dε
and hence, for some constant C independent on δ ,

∂ x w 1,δ 1 -∂ x w 1,δ 2 = δ 2 δ 1 ∂ ε u (ε) * ∂ x v (ε) (x)dε| ≤ C| δ 2 δ 1 ε α+β -2 dε| ≤ C|δ α+β -1 1 -δ α+β -1 2 |. It follows that (∂ x w 1,δ ) is a uniform Cauchy sequence in R (note that α +β -1 > 0). In consequence, w 1 ∈ C 1 c (R) and ∂ x w 1 = lim δ →0 ∂ x w 1,δ . For w 1,δ , we have w 1,δ (x + h) -w 1,δ (x) = (T h -I)w 1,δ (x) = 1 δ ∂ ε u (ε) * (T h -I)∂ x v (ε) (x)dε.
Similarly to (i), we give two estimates for the term (T h -I)v (ε) (x). In the second estimate we use the mean-value theorem. These estimates are the following ones

(T h -I)∂ x v (ε) (x) ≤ 2 ∂ x v (ε) ∞ ≤ CV β ε β -1 , (T h -I)∂ x v (ε) (x) ≤ h ∂ 2 x v (ε) ∞ ≤ ChV β ε β -2 .
Fix h ∈ (0, 1) and set = max(h, δ ). Then

∂ x w 1,δ (x + h) -∂ x w 1,δ (x) ≤ δ + 1 ∂ ε u (ε) * (T h -I)∂ x v (ε) (x)dε ≤ CU α V β δ ε α+β -2 dε + h 1 ε α+β -3 dε ≤ CU α V β δ ε α+β -2 dε + h +∞ h ε α+β -3 dε ≤ CU α V β 1 α + β -1 + 1 2 -α -β h α+β -1 .
Passing to the limit in the first term of these inequalities, when δ goes to 0, we get

|∂ x w 1 (x + h) -∂ x w 1 (x)| ≤ CU α V β 1 α + β -1 + 1 2 -α -β h α+β -1 .
This completes the proof of (ii).

Let us first prove by contradiction that λ

k < 0. Indeed, if λ k ≥ 0 then Lu k = λ k u k ≥ 0. Hence sup Ω k u k = sup ∂ Ω k u k = 0,
by the maximum principle. But this contradicts the fact that u k > 0 in Ω k . We now prove, again by using a contradiction, that λ 1 < λ 2 . Assume then that λ 1 ≥ λ 2 . As u 2 > 0 in Ω , we have that v = u 1 /u 2 belongs to C 2 (Ω ) ∩C(Ω ) and it is a solution of the equation

Lv = ∑ i, j ãi j ∂ 2 i j v + ∑ i bi ∂ i v + cv = 0 in Ω 1 , with ãi j = a i j u 2 , bi = ∑ j a i j ∂ j u 2 , c = (λ 2 -λ 1 )u 2 .
Since c ≤ 0 and v = 0 on ∂ Ω 1 , we obtain v = 0 in Ω 1 , which contradicts the fact that v > 0 in Ω 1 and completes the proof.

3.4

We obtain from Poincaré's inequality

B 1 u 2 dx ≤ C 0 (n) B 1 |∇u| 2 dx
and we get from the divergence theorem

B 1 |∇u| 2 dx = - B 1 ∆ uudx.
These two inequalities together with Cauchy-Schwarz's inequality imply

B 1 u 2 dx ≤ -C 0 (n) B 1 ∆ uudx ≤ C 0 (n) B 1 (∆ u) 2 dx 1/2 B 1 u 2 dx 1/2
and hence the result follows.

For

x ∈ ∂ B 1 (0), we have ν = ν(x) = x. In consequence, u(x + tν) = u((1 + t)x) = (1 + t) a u(x), x ∈ ∂ B 1 (0). Hence ∂ ν u(x) = lim t→0 u(x + tν) -u(x) t = au(x), x ∈ ∂ B 1 (0).
In a similar manner, we have also

∂ ν v(x) = bv(x) if x ∈ ∂ B 1 (
0). We get by applying the divergence theorem

0 = B 1 (0) [v∆ u -u∆ v]dx = ∂ B 1 (0) [uD ν v -D ν uv] ds = ∂ B 1 (0) (b -a)uvds.
We deduce, as a = b,

∂ B 1 (0) uvds(x) = 0. 3.6 (a) A simple change of variable implies u(x 1 , x 2 ) = 1 π R x 2 f (x 1 -t) t 2 + x 2 2 dt. Then |u(x 1 , x 2 ) -u(y 1 , x 2 )| ≤ 1 π R x 2 | f (x 1 -t) -f (y 1 -t)| t 2 + x 2 2 dt and hence |u(x 1 , x 2 ) -u(y 1 , x 2 )| ≤ 1 π R x 2 t 2 + x 2 2 dt [ f ] α |x 1 -y 1 | α . But 1 π R x 2 t 2 + x 2 2 dt = 1 π R 1 s 2 + 1 ds = 1. Therefore |u(x 1 , x 2 ) -u(y 1 , x 2 )| ≤ [ f ] α |x 1 -y 1 | α . (5.15) (b) The change of variable t = x 1 + sx 2 yields u(x 1 , x 2 ) = 1 π R f (x 1 + sx 2 ) s 2 + 1 ds and consequently |u(y 1 , x 2 ) -u(y 1 , y 2 )| ≤ 1 π R | f (y 1 + sx 2 ) -f (y 1 + sy 2 )| s 2 + 1 ds. As | f (y 1 + sx 2 ) -f (y 1 + sy 2 )| ≤ [ f ] α |s| α |x 2 -y 2 | α , we find |u(y 1 , x 2 ) -u(y 1 , y 2 )| ≤ 1 π R |s| α s 2 + 1 ds [ f ] α |x 2 -y 2 | α . (5.16) (c) Let x = (x 1 , x 2 ), y = (y 1 , y 2 ) ∈ R 2 + . Then |u(x) -u(y)| = |u(x 1 , x 2 ) -u(y 1 , y 2 )| ≤ |u(x 1 , x 2 ) -u(y 1 , x 2 )| + |u(y 1 , x 2 ) -u(y 1 , y 2 )|.
This (5.15) and (5.16) entail

|u(x) -u(y)| ≤ C[ f ] α |x -y| α , with C = 1 + 1 π R |s| α s 2 + 1 ds. That is [u] α ≤ C[ f ] α . 3.7 Fix p ∈ (0, 1). (a) We have ∂ i v = cq|x| q-2 x i and ∂ 2 i v = cq|x| q-2 + cq(q -2)|x| q-4 x 2 i and ∆ v = cq(n + q -2)|x| q-2 = cq(n + q -2)|x| qp .
It is then sufficient to take c = [q(n + q -2)] 1/(p-1) in order to get (c) Assume that there exists r > 0 such that u < v in ∂ B r (0). As sup

∆ v = v p . ( b
∆ u = u p > v p = ∆ v in Ω , we conclude that u ≤ v in Ω , which is impossible. Therefore u ≥ v in ∂ B r (
B 1/2 (x) u ≤ C 2 inf B 1/2 (x) u ≤ C 2 u(x) for any x ∈ B 1 (0), (5.18)
where

C 2 = C 2 (n) is a constant. ( c
) A combination of (5.17 In other words, we proved

|∇(ln u(x))| = |∇u(x) u(x) ≤ C for any x ∈ B 1 (0), with C = C 1 C 2 .
3.9 Let u ∈ H 1 0 (Ω ) ∩ H 2 (Ω ). We have by the divergence theorem

Ω |∇u| 2 = - Ω ∆ uu.
Therefore, we get by using the convexity inequality |ab| ≤ a

2 /2 + b 2 /2 Ω ∆ uu ≤ Ω √ 2ε∆ u 1 √ 2ε u ≤ ε Ω (∆ u) 2 + 1 4ε Ω u 2 .
Hence

Ω |∇u| 2 = - Ω ∆ uu ≤ ε Ω (∆ u) 2 + 1 4ε Ω u 2 .

3.10

We have by applying the mean-value theorem

-∂ n u(x 0 ) = ∂ n (M -u)(x 0 ) = 1 ω n r n B r ∂ n (M -u)dx.
Whence, we obtain from the divergence theorem, applied to to the last term,

-∂ n u(x 0 ) = 1 ω n r n ∂ B r (M -u)ν n dσ (x) ≤ 1 ω n r n ∂ B r (M -u)dσ (x).
Then an application of the mean value theorem to the last term gives

-∂ n u(x 0 ) ≤ 1 ω n r n ∂ B r (M -u)dσ (x) = nω n r n-1 ω n r n (M -u(x 0 )) = n r (M -u(x 0 )).
Hence the result follows by letting r → 1.

Let

x ∈ R n and r > 0. As u is harmonic we can apply Lemma 3.11. We obtain in particular, for any k ∈ N, When -k + α < 0, we obtain, by letting in this inequality r → +∞, that ∂ u(x) = 0 for all | | = k. In other words, u is a polynomial of degree less or equal to [α], the integer part of α.

max | |=k r k |∂ u(x)| ≤ C sup B(x,
3.12 (a) (i) We have ∂ i e -ρr 2 = -2ρx i e -ρr 2 . Therefore

∆ v = ∑ i ∂ 2 i v = (4r 2 ρ 2 -2nα)e -ρr 2 . But r = |x -y| ≥ |y -x 0 | -|x -x 0 | ≥ R -R = δ in D. Whence ∆ v = ∑ i ∂ 2 i ≥ (4δ 2 ρ 2 -2nρ)e -αr 2 in D.
Thus, ∆ v > 0 in D provided that ρ is chosen sufficiently large. Then, for ε > 0, we have

∆ (u -u(x 0 ) + εv) = ∆ u + ε∆ v > 0 in D.
(ii) It is not hard to check that uu(x 0 ) + εv ≤ 0 on ∂ D if ε is chosen sufficiently large. We get then, by applying the maximum principle, that w = uu(x 0 ) + εv ≤ 0 in D and, as w(x 0 ) = 0, we deduce that w attains its maximum at x 0 . Therefore ∂ ν w(x 0 ) ≥ 0. In consequence,

∂ ν u = ∂ ν w -ε∂ ν v = ∂ ν w + 2Rρεe -ρR 2 > 0.
(b) Set M = max u and make the assumption that the (closed) set F = {x ∈ Ω ; u(x) = M} is nonempty. As u is non constant Ω \ F is also non empty. By Lemma 3.14 there exists a ball B so that B ⊂ Ω , B ∩ F = / 0 and ∂ B ∩ F = / 0. We get then, by applying Hopf's lemma with B and y ∈ ∂ B ∩ F, that ∂ ν u(y) > 0. But this contradicts ∂ i u(y) = 0, 1 ≤ i ≤ n, and consequently y ∈ Ω and u attains its maximum at y.

(a)

Consider an even solution U. Then U (0) = 0, and as U < 0, we get U < 0 in (0, 1). Whence U satisfies U -KU + 1 = 0 in (0, 1). The solution of this equation is of the form

U(t) = αe Kt + β + t K .
Using that U (0) = 0, U(1) = 0 and the fact that U is even, we deduce

U(t) = 1 K 2 (e K -K -e K|t| + K|t|), t ∈ [-1, 1]. This solution is clearly of class C 2 because e |x| -|x| = 1 + x 2 2! + ∑ j≥3 |x| j j! .
(b) Since Lu = u + pu = -1 < 0, the maximum principle entails that u attains its minimum in [-1, 1] at x = ±1. Hence u ≥ 0. On the other hand,

L(U -u) ≤ (-K|U | + pU ) ≤ 0.
Then the maximum principle enables us to assert that U -u attains its null minimum at x = ±1 and hence u ≤ U.

If U 1 and U 2 are two solutions of the non linear equation, we apply the preceding results with p = Ksgn(U 1 ) and U = U 2 . We conclude that U 1 ≤ U 2 . Interchanging the roles of U 1 and U 2 , we obtain U 2 ≤ U 1 and therefore U 1 = U 2 .

3.14

We extend by reflexion u to an odd function that we still denote by u. This extension belongs to

C 2 (R 2 ), satisfies ∆ u = 0 in R 2 and |u(x)| ≤ c 1 + c 2 |x|. If B r (x)
denotes the ball of radius r and center x then by Lemma 3.11, we have max i, j

|∂ 2 i j u(x)| ≤ max i, j sup B r (x) |∂ 2 i j u| ≤ Cr -2 sup B 2r (x) |u| ≤ Cr -2 (c 1 + 2c 2 r
), for all r > 0.

Letting r → +∞, we deduce ∂ 2 i j u = 0, for each i and j. In consequence, u is of the form

u(x 1 , x 2 ) = a + bx 1 + cx 2 .
Using that u = 0 on ∂ Ω , we end up getting u = 0.

3.15 (a) As Γ (x) → +∞ when x → 0, we get, for any ε > 0, that there exists δ > 0 so that |u -v| ≤ εΓ sur B δ \ {0}.

(b) In light of the fact that v = u sur ∂ B r , we obtain from (a)

v -= v -εΓ ≤ u ≤ v + = v + εΓ , sur ∂ (B r \ B δ ).
We then deduce, by applying the maximum principle, that v

-≤ u ≤ v + on B r \ B δ . That is |u -v| ≤ εΓ , on B r \ B δ .
(c) We make successively in the last inequality δ → 0 and then ε → 0. We get u = v in B r \ {0}. In other words, v is the extension of u in the whole B.

(a) We easily check that

∂ 2 11 v = [4µ(µ -1)x 2 1 -2µ(1 -x 2 1 )]w, ∂ 2 12 v = -2µλ x 1 (1 -x 2 1 ) tanh(λ x 2 )w, ∂ 2 22 v = λ 2 (1 -x 2 1 ) 2 w
In light of the inequality R/2 < 2 k r, we deduce that 2 -k < 2r/R and hence 2 -k(1+α) ≤ (2r/R) 1+α . We obtain the expected result by combining the last inequality and the right hand side of the third inequality in (5.21). (c) Let w = U(x 1 , x 2 ) -cM 4 v(x 1x 2 ), (x 1 , x 2 ) ∈ Ω . We have Lw ≤ 0 in Ω 4 , and since w ≥ 0 on ∂ Ω , we get w ≥ 0 in Ω by applying the maximum principle. In particular, we have

2 -1 M 4 ± u(2, x 2 ) = U(2, x 2 ) ≥ cM 4 for |x 2 | ≤ 2. For α = α(ν) given by 2 -1 -c = 2 -1-α , we obtain then that 2 -1-α M 4 ± u ≥ 0 on ∂ Ω 2 (Note that u = 0 in {|x 2 | = x 1 }). The maximum principle yields ±u ≤ 2 -1-α M 4 in Ω 2 and hence M 2 = sup Ω 2 |u| ≤ 2 -1-α M 4 .
3.17 (a) Since L(Mu) = 0, there exists by Harnack's inequality (Theorem 2.8)

C = C(n, ν, K, r) so that sup B r (M -u) ≤ C inf B r (M -u) ≤ C(M -u(x 0 )).
b) Assume that M 1 > 0 and fix k ≥ 1 an integer. Pick then r > k|z|. We apply (a) to v and M = M 1 in an arbitrary B 2r = B(x 0 , 2r). We obtain sup

B r (M 1 -v) ≤ C 1 (M 1 -v(x 0 )), where C 1 = C 1 (n, ν, K) is a constant. Choose now x 0 in such a way that C 1 (M 1 - v(x 0 )) ≤ M 1 /2. This choice entails v ≥ 1 2 M 1 in B r . Using x 0 + jz ∈ B r , 0 ≤ j ≤ k, we obtain u(x 0 + (k + 1)z) -u(x 0 ) = k ∑ j=0 [u(x 0 + ( j + 1)z) -u(x 0 + jz)] = k ∑ j=0 v(x 0 + jz) ≥ k + 1 2 M 1 .
The right hand side of the last inequality tends to +∞, when k → +∞, contradicting that u is bounded. Therefore M 1 ≤ 0. In the preceding results substituting z by -z we can see that we have also sup

R n [u(x -z) -u(x)] ≤ 0. But sup R n [u(x -z) -u(x)] = sup R n [u(x) -u(x + v)] = sup R n (-v) = -inf R n v = -m 1 .
Thus m 1 ≥ 0 entailing that M 1 = m 1 = 0. (c) As u is periodic, we have

M = sup R n u = u(x 0 ),
for some x 0 ∈ Q = [0, 1) n . Let B r the ball of center 0 and radius r. Since Q ⊂ B r for r ≥ √ n, we can apply again (a) to deduce

0 ≤ M -u ≤ C(M -u(x 0 )) = 0 in B r .
Hence u = M in Q ⊂ B r and, using again that u is periodic, we conclude that u = M in R n .

3.18 (a) In light of the analyticity of u in B 1 , we find a ball B δ = B(0, δ ) in such a way that u coincide in B δ with its Taylor series at the origin, i.e.

u(x) = ∑ k≥0 P k (x), x ∈ B δ ,
where, for each k, P k is homogenous polynomial of degree k. Whence

0 = ∆ u(x) = ∑ k≥0 ∆ P k (x), x ∈ B δ .
This entails that ∆ P k = 0 for each k. (b) (We provide a direct proof. However we can also adapt the proof of Exercice 3.5) For r ∈ (0, 1), the exterior normal vector at x ∈ ∂ B r is nothing but ν = r -1 x. From P k (λ x) = λ k P(x), k ≥ 0, we get

P k (x + λ ν) = P((1 + λ r -1 )x) = (1 + λ r -1 ) k P k (x)
and consequently

∂ ν P k (x) = d dλ P(x + λ ν) λ =0 = k r P k (x).
We obtain then, by applying the divergence's theorem with k, ≥ 0,

0 = B r (P k ∆ P -P ∆ P k )dx = ∂ B r (P k ∂ ν P -P ∂ ν P k )dσ = -k r ∂ B r P k P dσ .
Then ∂ B r P k P dσ = 0 if k = and hence (c) (i) The existence of harmonic polynomial of degree k so that h k = p k in ∂ B r is guaranteed by Lemma 3.12. By the maximum principle we have (u is also harmonic)

sup B r |u -h k | = sup ∂ B r |u -h k | = sup ∂ B r |u -p k | ≤ sup B r |u -p k | → 0 when k → +∞, and then u -h k L 2 (B r ) → 0 when k → +∞. (5.22) (ii) As u -S k is orthogonal in L 2 (B r ) to S k -h k ∈ E k , we find u -h k 2 L 2 (B r ) = u -S k 2 L 2 (B r ) + S k -h k 2 L 2 (B r ) .
This and (5.22) imply

u -S k L 2 (B r ) ≤ u -h k L 2 (B r ) → 0 when k → +∞. (d) Fix 0 < r < R < 1 and let ϕ ∈ D(R n ) so that ϕ ≥ 0 in R n , ϕ = 0 for |x| ≥ 1, ϕ(x)dx = 1. Note that we can choose ϕ = ϕ 0 (|x|). For 0 < ε < R-r, define ϕ ε (x) = ε -n ϕ(ε -1 x).
We have, according to the properties of harmonic functions in Lemma 3.11,

u(x) = u (ε) (x) = u * ϕ ε (x) = B R u(y)ϕ ε (x -y)dy, x ∈ B r ,
and a similar formula holds for S k . We deduce by applying Cauchy-Schwarz's inequality sup

B r |u -S k | = sup B r B R (u(y) -S k (y))ϕ ε (x -y)dy ≤ ϕ ε L 2 (R n ) u -S k L 2 (B R ) .
We end up getting that u -

S k L 2 (B R ) → 0, as k → +∞, implies sup B R |u -S k | → 0 as k → +∞.
Exercises and Problems of Chapter 4

4.1 (a) We have

∆ v = 2u∆ u + 2|∇u| 2 = 2|∇u| 2 ≥ 0.
We obtain, by applying Harnak's inequality for sub-solutions (Theorem 2.20), that

v L ∞ (B(0,r)) ≤ C v L 2 (B(0,2r)) .
(b) Form (a), we get

u 2 L ∞ (B(0,r)) = v L ∞ (B(0,r)) ≤ C u 2 L 2 (B(0,2r)) ≤ C u L ∞ (B(0,2)) u L 2 (B(0,2r)) . Hence u L ∞ (B(0,r)) ≤ C u,ε u L 2 (B(0,2r)) , 1/2 ≤ r ≤ 1, (5.23)
where C u is given by

C 2 u = C u L ∞ (B(0,2)) u L ∞ (B(0,ε)) .
Note that according the uniqueness of continuation u L ∞ (B(0,ε)) = 0. (c) In light of (5.23), the doubling inequality entails

u L ∞ (B(0,r)) ≤ C u u L 2 (B(0,r)) , ε ≤ r ≤ 1, (5.24) (d) Fix ε > 0 so that u L ∞ (B(0,ε)) ≤ 2|u(0)|. ( 5

.25)

By the second mean-value identity

u(0) = n |B(0, ε)| B(0,ε) udx and hence |u(0)| ≤ n |B(0, ε)| 1/2 u L 2 (B(0,ε)) . This in (5.25) yields u L ∞ (B(0,ε)) ≤ 2n |B(0, ε)| 1/2 u L 2 (B(0,ε)) .
(5.26)

A combination of (5.24) and (5.26) gives the expected inequality.

4.2 (a) For u = u 1u 2 , we have

∆ u = 0 in Ω 0 , u = 0 on Γ , ∂ ν u = 0 on γ.
We get, by applying Corollary 2.5, that u = 0 in Ω 0 .

As -∆ u + u = 0 in D 0 and u = ∂ ν u = 0 in S, we obtain, by using again Corollary 2.5, that u = 0 in D 0 .

(

) (i) By (a), u = 0 in Ω 0 ∪D 0 . In particular, u = ∂ ν u on ∂ ω. Whence u ∈ H 2 0 (ω). We have -∆ u 2 + u 2 = 0 and ∆ u 1 = 0 in ω. In consequence, ∆ u = u 2 in ω. b
Now, as u, u 2 ∈ C ∞ (ω), we can apply ∆ to each member of the last identity. We obtain

∆ 2 u = ∆ u 2 = ∆ u 2 -∆ u 1 = ∆ u in ω.
(ii) From the divergence theorem, we get

ω ∆ uudx = - ω ∆ 2 uudx = ω (∆ u) 2 dx - ∂ ω ∂ ν ∆ uudσ (x) + ∂ ω ∆ u∂ ν udσ (x) = ω (∆ u) 2 dx, ω ∆ uudx = - ω |∇u| 2 dx + ∂ ω ∂ ν uudσ (x) = - ω |∇u| 2 dx. Hence ω (∆ u) 2 dx + ω |∇u| 2 dx = 0.
Therefore ∇u = 0 in ω. Thus u = 0 in ω because u ∈ H 2 0 (ω). If ω = / 0 then u = 0 in ω would entail that u 2 = 0 in ω and hence u 2 = 0 in Ω by Theorem 2.16. In consequence, we would have ϕ = 0 which is impossible.

(c) By (b), we can not have neither

D 2 \ D 1 = / 0 nor D 1 \ D 2 = / 0. We end up getting that D 1 = D 2 .

(a) A simple change of variable yields

H(r) = S(1)
σ (ry)u 2 (ry)r n-1 dS(y).

Hence

H (r) = n -1 r H(r) + S (1)
∇(σ u 2 )(ry) • yr n-1 dS(y) = n -1 r H(r) + S (1)
u 2 ∇σ (ry) • yr n-1 dS(y) + S

σ ∇(u 2 )(ry) • yr n-1 dS(y)

= n -1 r H(r) + S(r) u 2 ∇σ (x) • ν(x)dS(x) + S(r) σ (x)∇(u 2 )(x) • ν(x)dS(x) = n -1 r H(r) + H(r) + S(r) σ ∇(u 2 )(x) • ν(x)dS(x).
Identity (4.88) will follow if we prove that

2D(r) = S(r) σ ∇(u 2 )(x) • ν(x)dS(x). Since div(σ ∇u) = β u, we get div(σ ∇(u 2 )) = 2udiv(σ ∇u) + 2σ |∇u| 2 = 2σ |∇u| 2 + 2β u 2 .
We obtain by applying the divergence theorem

2D(r) = B(r) div(σ (x)∇(u 2 )(x))dx = S(r) σ (x)∇(u 2 )(x) • ν(x)dS(x). ( 5
σ (x)|∇u(x)| 2 dS(x) + S(r) β (x)u 2 (x)dS(x) = 1 r S(r) σ (x)|∇u(x)| 2 x • ν(x)dS(x) + S(r) V (x)u 2 (x)dS(x).
Then an application of the divergence theorem gives

D (r) = 1 r B(r) div(σ (x)|∇u(x)| 2 x)dx + S(r) β (x)u 2 (x)dS(x). Therefore D (r) = 1 r B(r) |∇u(x)| 2 div(σ (x)x)dx + 1 r B(r) σ (x)x • ∇(|∇u(x)| 2 )dx + S(r) β (x)u 2 (x)dS(x) implying D (r) = n r D(r) + 1 r D(r) + 1 r B(r) σ (x)x • ∇(|∇u(x)| 2 )dx + Ĥ(r).
(5.28)

On the other hand,

B(r) σ (x)x j ∂ j (∂ i u(x)) 2 dx = 2 B(r) σ (x)x j ∂ 2 i j u∂ i u(x)dx = -2 B(r) ∂ i [∂ i u(x)σ (x)x j ] ∂ j u(x)dx + 2 S(r) σ (x)∂ i u(x)x j ∂ j u(x)ν i (x)dS(x) = -2 B(r) ∂ 2 ii u(x)σ (x)x j ∂ j u(x)dx -2 B(r) ∂ i u(x)∂ j u(x)∂ i [σ (x)x j ] dx + 2 S(r) σ (x)∂ i u(x)x j ∂ j u(x)ν i (x)dS(x).
Thus, taking into account that σ

∆ u = -∇σ • ∇u + β u, B(r) σ (x)x • ∇(|∇u(x)| 2 )dx = -2 B(r) σ (x)|∇u(x)| 2 dx -2 B(r) β (x)u(x)x • ∇u(x)dx + 2r S(r) σ (x)(∂ ν u(x)) 2 dS(x).
This identity in (5.28) yields

D (r) = n -2 r D(r) + 1 r D(r) -2 B(r) β (x)u(x)x • ∇u(x)dx - n -2 r B(r) β (x)u 2 (x)dx + 2H(r) + Ĥ(r).
That is we proved (4.89). (b) (i) Assume that β ≥ 0. Since

H(r) = 1 r S(r) σ (x)u 2 (x)x • ν(x)dS(x),
we get by applying the divergence theorem

H(r) = 1 r B(r) div σ (x)u 2 (x)x dx. (5.29)
Hence

H (r) = - 1 r H(r) + 1 r S(r) div σ (x)u 2 (x)x dS(x) = n -1 r H(r) + S(r) ∂ ν σ (x)u 2 (x)dS(x) + 2 S(r) σ (x)∂ ν u(x)u(x)dS(x). But S(r) σ (x)∂ ν u(x)u(x)dS(x) = B(r) div(σ (x)∇u(x))u + B(r) σ (x)|∇u| 2 dx = B(r) σ (x)|∇u(x)| 2 + β (x)u 2 (x) dx = D(r).
Therefore

H (r) = n -1 r H(r) + 2D(r) + S(r) ∂ ν σ (x)u 2 (x)dS(x) ≥ S(r) ∂ ν σ (x)u 2 (x)dS(x) ≥ - σ 1 σ 0 H(r),
where we used that H(r) ≥ 0 and D(r) ≥ 0. Then N(r) ≤ C max(N(r 0 ), 1), r ∈ I , for some constant C > 0, only depending on Ω , σ 1 /σ 0 and β 0 . The proof is completed by noting that N ≤ max(N(r 0 ), 1) on (0, r 0 ) \ I . Here we used the following H 2 a priori estimate: w H 2 (Ω ) ≤ C Ω ∆ u L 2 (Ω ) if w ∈ H 2 (Ω ).

Choose p > 1 so that 2n/p = 1 + α, i.e. p = n/(1α). In that case, W 2,p (Ω ) is continuously imbedded in C 1,α (Ω ). Therefore φ C 1,α (Ω ) ≤ Cλ 2 φ L 2 (Ω ) .

(ii) Assume that 4 ≤ n < 8. As H 2 (Ω ) is continuously imbedded in L q 0 (Ω ), q 0 = 2n/(n -4), 4 < n < 8. Also, as H 2 (Ω ) is continuously embedded in W 2,p (Ω ), 1 < p < 2, we deduce that H 2 (Ω ) is continuously imbedded in L q 0 (Ω ), q 0 = 2p/(2-p), for some fixed 1 < p < 2, when n = 4.

Then it follows from Theorem 4.11 that φ ∈ W 2,q 0 (Ω ) and φ W 2,q 0 (Ω ) ≤ Cλ φ L q 0 (Ω )

≤ Cλ 2 φ L 2 (Ω ) .

But 2n/q 0 = 4n/2 > 0 if 4 < n < 8 and 2n/q 0 = 4(p -1)/p if n = 4. Hence, W 2,q 0 (Ω ) is continuously imbedded in L p (Ω ) for 1 < p < ∞. We deduce by repeating the argument in (i) that φ ∈ C 1,α (Ω ) and φ C 1,α (Ω ) ≤ Cλ 3 φ L 2 (Ω ) .

(iii) In light of (ii), we can make an induction argument to deduce that if n = 4 j + , with m ≥ 1 and ∈ {0, 1, 2, 3}, then

φ C 1,α (Ω ) ≤ Cλ 2+ j φ L 2 (Ω ) .
That is we have

φ C 1,α (Ω ) ≤ Cλ m(n) φ L 2 (Ω ) .
where m(n) -2 is the unique non negative integer j so that n/4j ∈ [0, 1). (b) (i) Similarly to the proof of (4.81), there exists E 0 ⊂ D so that for 0 < ε < 1 and v ∈ H 2 (D)

C v H 1 (D) ≤ ε β v C 1,α (D) + e c/ε v H 1 (E 0 ) + ∆ v L 2 (D) .
Combined with Caccioppoli's inequality this inequality yields, where F 0 E 0 , C v H 1 (D) ≤ ε β u C 1,α (D) + e c/ε v L 2 (F 0 ) + ∆ v L 2 (D) .

(5.40)

On the other hand, we have from Proposition 4.5

C v L 2 (F 0 ) ≤ ε γ 1 v L 2 (D) + ε -1 1 u L 2 (E) + ∆ v L 2 (Ω ) , ε 1 > 0.
Taking in this inequality ε 1 = e -c/(γε) ε β /γ , we find C v L 2 (F 0 ) ≤ e -c/ε ε β v L 2 (D) + e c/(γε) ε -β /γ u L 2 (E) + ∆ v L 2 (Ω ) .

This and (5.40) yields the expected inequality.

(ii) We apply (i) with v(x,t) = u(x)e √ λt , (x,t) ∈ D = Ω × (0, 1) and E = ω × (0, 1). Noting that 1 ≤ e λt L 2 (0,1) ≤ e where k = 2(1 + m) and

ℵ = φ L 2 (ω)
φ L 2 (Ω ) .

If ℵ < e -c , we find 0 < ε < 1 so that ε β e -c/ε = ℵ. This particular choice of ε in (5.41) yields

Ce -k √ λ ≤ 1 (-ln ℵ) β .
The expected inequality then follows. When ℵ ≥ e -c the expected inequality is obviously satisfied. Suppose first that β 0 + β 1 < n. We split I 0 into two terms I 0 = J 0 + J 1 , where

1 n -α 0 -α 1 d n-α 0 -α 1 |x -y| n-α 0 -α 1 -2 n-α 0 -α 1 if α 0 + α 1 = n,
J 0 = Ω ∩B(x 0 ,η)
[ f 0 (x, z)f 0 (x 0 , z)] f 1 (z, y)dz,

J 1 = Ω \B(x 0 ,η)
[ f 0 (x, z)f 0 (x 0 , z)] f 1 (z, y)dz. |x 0 -z| -n+β 0 dz , from which we deduce that there exists η 0 so that, for any 0 < η ≤ η 0 , we have

|J 0 | ≤ ω n c 2 (2 β 0 + 1) β 0 δ 2 -n+β 1 η β 0 ≤ ε 4 . (A.5)
Let b > 0 so that Ω ⊂ B(x 0 , b). As f 0 is uniformly continuous in Ω ∩ B(x 0 , η/2) × Ω \ B(x 0 , η) , there exits η 1 ≤ η/2 so that ,for any |xx 0 | ≤ η 1 and z ∈ Ω \ B(x 0 , η), we have For η = min(η 0 , η 1 ), we get by combining (A.5) and (A.6)

|I 0 | ≤ ε/2, |x -x 0 | ≤ η |y -y 0 | < δ 4 .
Proceeding similarly to I 1 , we obtain that there exists η * so that While for the third term we find B(0,t)\B(0,2)

|u -w| -n+β 0 |w| -n+β 1 dw ≤ 2 n-β 1 2 n-β 0 B(0,t)\B(0,2)

|w| -n+β 0 +β 1 dw = 2 n+β 0 t -n+β 0 +β 1 ω n β 0 + β 1 -n = 2 n+β 0 (2η) -n+β 0 +β 1 ω n (β 0 + β 1 -n)|x -y| -n+β 0 +β 1 .
We find by collecting all these inequalities

I ≤ Cη -n+β 0 +β 1 , |x -x 0 | ≤ η/2, |y -x 0 | ≤ η/2,
where the constant C only depends on n, β 0 and β 1 . This shows that in particular I 0 → 0 when (x, y) → (x 0 , x 0 ). On the other hand, since (A.6) still holds when δ /4 is substituted by |yx 0 |, we prove analogously that I 1 → 0 when (x, y) → (x 0 , x 0 ). We proceed now to the proof of (ii). Fix x 0 ∈ Ω \ {y}. Let δ = |x 0 -y| and 0 < η ≤ δ /4. Denote the canonical basis of R n by (e 1 , . . . , e n ). For |t| ≤ η we have f (x 0 + te i , y)f (x 0 , y)t Ω ∂ x i f 0 (x 0 , z) f 1 (z, y)dz = I 0 + I 1 , with

I 0 = Ω ∩B(x 0 ,η)
[ f 0 (x 0 + te i , z)f 0 (x 0 , z) -t∂ x i f 0 (x 0 , z)] f 1 (z, y)dz,

I 1 = Ω \B(x 0 ,η)
[ f 0 (x 0 + te i , z)f 0 (x 0 , z) -t∂ x i f 0 (x 0 , z)] f 1 (z, y)dz.

If z ∈ B(x 0 , η) then f (x 0 + te i , z)f (x 0 , z) -t∂ x i f 0 (x 0 , z) = t Noting that B(x 0 , η) ⊂ B((x 0 + tse i , 2η) and |z -y| ≥ η/2, we get similarly to I 1 that there exits η 0 ≤ δ /4 so that, for any |t| ≤ η 0 , we have

|I 0 | ≤ |t|ε/2.
On the other hand, using the continuity of ∂ x i f , we can mimic the proof used for estimating I 1 . We find η 1 > 0 so that, for |t| ≤ η 1 , we have

|I 1 | ≤ |t|ε/2.
In light of the last two inequalities, we can assert that ∂ x i f (x 0 , y) exists and

∂ x i f (x 0 , y) = Ω ∂ x i f 0 (x 0 , z) f 1 (z, y)dz.
The proof of (ii) is then complete.

Next, we proceed to the proof of (iii). Let Let R = diam(Ω ). Then

Λ 0 |x 1 -z| -n | f 1 (z, y)| ≤ c Λ 0 |x 1 -z| -n |z -y| -n+β 1 ≤ c2 n-β 1 ω n |x 1 -y| -n+β 1 R 2d dr r ≤ c2 n-β 1 ω n |x 1 -y| -n+β 1 ln R 2d .
On the other hand, since 2|zx 1 | > |x 1 -y| for z ∈ Λ 1 , we get

Λ 1 |x 1 -z| -n | f 1 (z, y)|dz ≤ c Λ 1 |x 1 -z| -n |z -y| -n+β 1 ≤ c2 n |x -y 1 | -n B(y,|x 1 -y|/2) |z -y| -n+β 1 dz ≤ c 2 n+β 1 ω n β 1 |x 1 -y| -n+β 1
and hence

Ω \B(x 1 ,2d) |x 1 -z| -n | f 1 (z, y)|dz ≤ C|x 1 -y| -n+β 1 .
We have similarly We have

Σ 0 |x 1 -z| -n+β 0 | f 1 (z, y)| ≤ c Σ 0 |x 1 -z| -n+β 0 |z -y| -n+β 1 dz ≤ c2 n-β 1 |x 1 -y| -n+β 1 B(x 1 ,2d) |x 1 -z| -n+β 0 dz ≤ C|x 1 -y| -n+β 1 d β 0 = C|x 1 -x 2 | β 0 |x 1 -y| -n+β 1 .
As before, using 2|z - The result then follows by applying the cancellation theorem.

If the kernel K of the operator A given by (A.10) is of the form K(x, y) = B(x, y) |x -y| α , for some complex-valued function B ∈ L ∞ (Ω × Ω ) and 0 < α < n, we say that A is a weakly singular integral operator.

The following lemma will be useful in sequel. The proof is then complete.

Theorem A.2. Any weakly singular integral operator on L 2 (Ω ) is compact.

Proof. Let A a be weakly singular integral operator. Then there exists 0 < α < n and B ∈ L ∞ (Ω × Ω ) so that, for any f ∈ L 2 (Ω ), we have In a first step we prove that A is bounded. Pick f ∈ L 2 (Ω ). Then according to Lemma A.3 we obtain

Ω Ω |B(x, y)| |x -y| α | f (y)|dxdy ≤ C B ∞ Ω | f (y)|dy ≤ C|Ω | 1/2 B ∞ f 2 ,
where C is the constant in Lemma A.3. Therefore, with reference to Fubini's theorem, we get that the integrals Here and until the end of this proof, C denotes a generic constant only depending on n, Ω , α and B ∞ .

Inequality (A.13) being valid for a.e. x ∈ Ω , we can integrate over Ω with respect to x. We find We now prove that A is compact. We split A into two integral operators A = A ε + R ε , ε > 0 is given, where the operators A ε and R ε have as respective kernels K ε (x, y) = K(x, y)χ(|x -y|), L ε (x, y) = B(x, y)(1χ ε (|x -y|).

Here χ ε is the characteristic function of the interval [ε, +∞).

Since K ε ∈ L 2 (Ω × Ω ), A ε is compact (see Exercise 2.5). On the other hand, similarly to the proof of Lemma A. In light of Theorem 2.3, the compactness of A follows then readily.

As a straightforward consequence of Theorem A.10 and Theorem 2.8 we have the following result.

Theorem A.3. Let A be a weakly singular operator of the form (A.10). Then σ (A) = {0}, or else σ (A)\{0} is finite, or else σ (A)\{0} consists in a sequence converging to 0.

Let A be a weakly singular operator with kernel K, λ = 0 and g ∈ L 2 (Ω ). Consider then the Fredholm integral equation of the second kind Ω K(x, y) f (y)dyλ f (x) = g(x), a.e. x ∈ Ω .

(A.14)

The result we state now is a direct consequence of Fredholm's alternative. Proof. Pick χ ∈ C ∞ ([0, ∞)) satisfying 0 ≤ χ ≤ 1, χ(t) = 0 for t ≤ 1/2 and χ(t) = 1 for t ≥ 1. Define then, for j ≥ 1, K j (x, y) = χ( j|x -y|)K(x, y) if x = y, 0 if x = y and denote by A j the integral operator with kernel K j . It is clear that K j is continuous and there exists a constant C > 0 so that, for each j, we have

A f (x) -A j f (x) ≤ C f C(Ω ) Ω ∩B(x,1/ j) |x -y| -α ≤ C f C(Ω ) j -α
for each x ∈ Ω . Whence A f ∈ C(Ω ) as the uniform limit of the sequence continuous functions (A j f ). Moreover A -A j B(C(Ω )) ≤ C j -α . The proof will be completed by showing that an integral operator with continuous kernel is compact. We then consider A as an integral operator with kernel K ∈ C(Ω × Ω ). Let ε > 0. Since K is uniformly continuous, there exists η > 0 so that, for any x, y, z ∈ Ω satisfying |x -z| ≤ η, we have Considering C Ω ,C Ω as dual system with respect to the usual scalar product of L 2 (Ω ), we get that Theorem A.4 is also valid for weakly singular operators acting on C Ω . We refer to [2, Chapter 4] for a general Fredholm's alternative for dual systems.

Proposition A.1. For i = 0, 1, let A i be a weakly singular integral operator with kernel K i satisfying L0

x H(x, y) ≤ C 1 |x -y| -n+α , (x, y) ∈ Σ . (A.26)

Proof. (i) is immediate from (A.16), (A.18) and (A.22).

To prove (ii), we note that from (A.24) we have L0

x H(x, y) = L0

x H(x, y) -L 0 x H(x, y)

= n ∑ i, j=1
a i j (x)a i j (y) ∂ 2 x i x j H(x, y).

Hence L0

x H(x, y) ≤ Λ |x -y| α n ∑ i, j=1

∂ 2

x i x j H(x, y) .

This together with (A.23) entail the expected inequality.

We have as an immediate consequence of this lemma:

Corollary A.1. Assume in addition that a i j ∈ C 0,α (Ω ), 1 ≤ i, j ≤ n and ρ n (y, x) -1

ρ n (x, y) .

Noting that a k , 1 ≤ k, ≤ n, belong also to C 0,α (Ω ), the first term in the right hand side of the last identity is clearly estimated by C|x -y| -n+α+1 . To complete the proof of (A.28), we establish the estimate 1 ρ n (y, x) -1

ρ n (x, y) ≤ C|x -y| -n+α , x = y.

Invoking the mean-value theorem, we find θ ∈ (0, 1) so that ρ n (x, y)ρ n (y, x) = n 2 A -1 (y) -A -1 (x) (xy)|xy × θ A -1 (y)(xy)|xy + (1θ ) A -1 (x)(xy)|xy n/2-1 .

Inequality (A.28) then follows.

We have from the preceding calculations ρ n (y, x) -1

ρ n (x, y) .

We have

∂ x j 1 ρ n (y, x) - 1 ρ n (x, y) = n n ∑ =1 a j (y) ρ n+2 (x, y) - a j (x) ρ n+2 (y, x) (x -y ) - n 2ρ n+2 (y, x) ∂ x j A -1 (x)(x -y)|x -y .
Splitting the first term on the right hand side into two ones, we can mimic the proof of (A.28) in order to estimate this term by C|x -y| -n-1+α . While the second term in the last inequality is clearly estimated by C|x -y| -n . Returning back to I 3 , we find that it is estimated by C|x -y| -n+α . The proof is then complete.

Lemma A.6. Let 0 < α ≤ 1 and Λ > 0. Assume that a i j ∈ C 0,1 (Ω ), b i , c ∈ C 0,α (Ω ) and max 1≤i, j≤n a i j C 0,1 (Ω ) + max 1≤i≤n b i C 0,α (Ω ) + c C 0,α (Ω ) ≤ Λ .

Then there exists a constant C > 0, depending on n, diam(Ω ), µ α and λ , so that a i j (x)a i j (y) ∂ 2 x i x j H(x, z)

+ n ∑ i, j=1
a i j (y)a i j (z) ∂ 2 x i x j H(x, z) -∂ 2 x i x j H(y, z) .

In light of (A.23), we can proceed as in the preceding lemma. With the aid of (A.31) and the identity Lemma A.9. We have g ∈ C(Ω × Ω ).

(x k -z k )(x -z ) -(y k -z k )(y -z ) = (x k -y k )(x -z ) -(y k -z k )(x -y ),
Proof. We claim that there exists C > 0 so that, for any y, z ∈ Ω , we have We proceed by contradiction. So if our claim does not hold we would find two sequences (y j ) and (z j ) in Ω so that τ j = max In particular, the sequence of functions ( f j ) given by f j (x) = f (x, y j )f (x, z j ) τ j , x ∈ Ω , converges uniformly in Ω to 0. Define then two sequences of functions (u j ) and (v j ) by u j (x) = g(x, y j )g(x, z j )

τ j , x ∈ Ω , v j (x) = Ω K(x, z)u j (z)dz = (Au j )(x), x ∈ Ω .
As u j C(Ω ) = 1 and A is compact, subtracting a subsequence if necessary, we may assume that v j = Au j converges to v in C(Ω ). But We need to take partial derivatives of both sides of this identity. For doing that, we first rewrite this identity in a different form. Fix y ∈ Ω and set Ω η = Ω \ B(y, η). Note that if x = y and η is sufficiently small x ∈ B(y, η). We write Proceeding as for the first order derivative, we get by substituting H(x, y) by ∂ x j H(x, z) + ∂ z j H(z, x),

∂ x j Γ k (x, y) =
∂ x i Ω [∂ x j H(x, z) + ∂ z j H(z, x)]G k (z, y)dz = Ω [∂ 2
x i x j H(x, z) + ∂ x i z j H(z, x)]G k (z, y)dz. Since G k (•, y) ∈ C 0,β (Λ k η ) we can apply Theorem A.6 with Ω substituted by Λ k η . We get

∂ x i Λ k η ∂ z j H(z, x)G k (z, y)dz = lim ε→0 Λ k η \B(x,ε)
∂ 2

x i z j H(z, x)g(z, y)dz

+ G k (x, y) S n-1
ξ i Ψ j (x, ξ )dσ (ξ ).

We have a i j (x)∂ 2 x i z j H(z, x)G k (z, y)dz -G k (x, y).

Finally, we have

∂ x i B(x k ,η) ∂ z j H(z, x)G k (z, y)dz = B(x k ,η) ∂ 2 x i z j H(z, x)G k (z, y)dz.
Assembling all these calculations, we end up getting

1 /Proposition 1 . 6 .

 116 (|x| 2 -1) dx.It is straightforward to check that (ρ m ) defined byρ m = m n ρ(mx), x ∈ R n , is a sequence of mollifiers. Let f ∈ L 1 loc (Ω ) and ϕ ∈ C c (R n ) satisfying supp(ϕ) ⊂ B (0, 1/m) with m ≥ 1. Define the convolution ϕ * f on Ω m = {x ∈ Ω ; dist(x, ∂ Ω) > 1/m} by ϕ * f (x) = Ω ϕ(xy) f (y)dy = B(0,1/m) ϕ(y) f (xy)dy. If f ∈ L 1 loc (Ω ) and ϕ ∈ D(R n ) is so that supp(ϕ) ⊂ B (0, 1/m), for m ≥ 1, then ϕ * f ∈ C ∞ (Ω m ) and, for every α ∈ N n , ∂ α (ϕ * f ) = (∂ α ϕ) * f . Proof. Assume first that |α| = ∑ α i = 1. Let x ∈ Ω m and r > 0 such that B(x, r) ⊂ Ω m . Whence, ω = B (x, r + 1/m) Ω and, for 0 < |ε| < r, ϕ * f (x + εα)ϕ * f (x) ε = ω ϕ(x + εαy)ϕ(xy) ε f(y)dy. But lim ε→0 ϕ(x + εαy)ϕ(xy) ε = ∂ α ϕ(xy) and sup y∈Ω , 0<|ε|<r

Lemma 1 . 1 .

 11 Let ω Ω . (i) If f ∈ C(Ω ) then we have, for sufficiently large m, sup x∈ω |ρ m * f (x)f (x)| ≤ sup |y|<1/m sup x∈ω τ y f (x)f (x) .

1 p ρ m (y) 1 p

 11 [ f (xy)f (x)]dy ≤ B(0,1/m) ρ m (y) | f (xy)f (x)| p dy 1/p .We get by applying Fubini's theoremω ρ m * f (x)f (x) p dx ≤ ω dx B(0,1/m) ρ m (y) f (xy)f (x) y) f (xy)f (x) p dx ≤ B(0,1/m) ρ m (y)dy sup |z|<1/m ω | f (xz)f (x)| p dx ≤ sup |z|<1/m ω | f (xz)f (x)| p dx.Assertion (ii) then follows. Lemma 1.2. (Continuity of translation operator) Let ω Ω . (i) If f ∈ C(Ω ) then lim y→0 sup x∈ω τ y f (x)f (x) = 0. (ii) If f ∈ L p loc (Ω ), 1 ≤ p < ∞, then lim y→0 τ y ff L p (ω) = 0.

Theorem 1 . 6 .

 16 (i) If f ∈ C(Ω ) then ρ m * f converges to f uniformly in any compact subset of Ω . (ii) If u ∈ L p loc (Ω ), 1 ≤ p < ∞, then ρ m * f converges to f in L p loc (Ω ). Theorem 1.7. (Cancellation theorem) Let f ∈ L 1 loc (Ω ) satisfying Ω f ϕdx = 0 for all ϕ ∈ D(Ω ).

Lemma 1 . 4 .Lemma 1 . 5 .

 1415 (Closing lemma) If ( f m ) converges to f in L 1 loc (Ω ) and (g m ) converge vers g in L 1 loc (Ω ) and if g m = ∂ α f m in the weak sense, for any m, then g = ∂ α f in the weak sense.Proposition 1.7. Let f , g ∈ L 1 loc (Ω ) and α ∈ N n . If g = ∂ α f in the weak sense then in Ω m = {x ∈ Ω ; dist(x, ∂ Ω ) > 1/m} we have ∂ α (ρ m * f ) = ρ m * g. Proof. As ρ m * f ∈ C ∞ (Ω m ) by Proposition 1.6, we have, for x ∈ Ω m , ∂ α (ρ m * f )(x) = Ω ∂ α x ρ m (xy) f(y)dy = (-1) |α| Ω ∂ α y ρ m (xy) f (y)dy. Whence ∂ α (ρ m * f )(x) = (-1) 2|α| Ω ρ m (xy)g(y)dy = ρ m * g(x) as expected. Let f , g ∈ C(Ω ) and |α| = 1. If f has a compact support and g = ∂ α f in the classical sense then Ω gdx = 0.

ε 0 g

 0 (x + tα)dt. Whence, g = ∂ α f in the classical sense. Conversely, if g = ∂ α f in the classical sense then by virtue of Lemma 1.5 we have 0 = Ω ∂ α ( f ϕ)dx = Ω ( f ∂ α ϕ + gϕ)dx for every ϕ ∈ D(Ω ).

.

  Let Γ = {1, . . . , d} × Ω and define onL = {u : Γ → R; u(s, •) ∈ C c (Ω ), 1 ≤ s ≤ d}

Lemma 1 . 7 .

 17 Let 1 < p < ∞ and Φ ∈ [L p (Ω , R d )] . Then there exists a unique ( f s ) ∈ L p (Ω , R d ) so that Φ, (v s ) = d ∑ s=1 Ω f s v s dx. Moreover ( f s ) p = Φ . Let 1 ≤ p < ∞,k ≥ 1 an integer and d = ∑ |α|≤k 1. Then the mapping

≤

  (g α ) p in such a way that Φ

  = ∑ |α|≤k Ω g α D α udx for all u ∈ D(Ω ).

Theorem 1 .

 1 15. If u ∈ L 1 (Ω ) then (u • f )|J f | ∈ L 1 (ω) and ω u( f (x))|J f (x)|dx = Ω u(y)dy.Proposition 1.9. (Change of variable formula) Let ω and Ω be two open subsets of R n and let f :

ω

  (u m • f ) (y)∂ j v(y)dy =ω n ∑ k=1 (∂ k u m • f ) (y)∂ j f k (y)v(y)dy.If g = f -1 then Theorem 1.15 yields ω

  |det(J g (x))|dx and henceω (u • f ) (y)∂ j v(y)dy =u • f ) (y)∂ j f k (y)v(y)dy by Theorem 1.15. The proposition is then proved. Proposition 1.10. (Derivative of a composition) Let f ∈ C 1 (R) and u ∈ W 1,1 loc

Theorem 1 .

 1 16. (Extension theorem) Let 1 ≤ p < ∞ and let Ω be an open subset of R n of class C 1 with bounded boundary. There exists a bounded operator

Theorem 1 .

 1 17. (Density theorem) Let 1 ≤ p < ∞ and Ω be an open subset of class C 1 with bounded boundary. Then D(Ω ) is dense W 1,p (Ω ).

Theorem 1 .

 1 20. (Sobolev imbedding theorem) Let Ω be an open subset of R n of class C 1 with bounded boundary.

( a ) 1 . 8 .

 a18 Prove that any function from C(Ω ) ∩C 1 pie (Ω , (Ω i ) 1≤i≤k ) admits a weak derivative belonging to L 2 (Ω ). (b) Does an arbitrary function from C 1 pie (Ω , (Ω i ) 1≤i≤k ) admits a weak derivative belonging to L 2 (Ω )? Denote by B r the ball of R n with center 0 and radius r. (a) Let n = 2. Prove that the function u

8 . 1 . 18 .

 8118 In this exercise, I =]a, b[ is an interval and 1 ≤ p ≤ ∞. (a) Let g ∈ L 1 loc (I). Fix c ∈ I and set f (x) = x c g(t)dt, x ∈ I. (i) Prove that f ∈ C(I). Hint: use Lebesgue's dominated convergence theorem. (ii) Let ϕ ∈ D(I). With the aid of the identity I f ϕ dx = -)ϕ (x)dt, and Fubini's theorem, demonstrate that I f ϕ dx = -I gϕdx. (b) Let u ∈ W 1,p (I). Use (a) to show that there exists ũ ∈ C(I) so that u = ũ a.e. in I and ũ(y)ũ(x) = y x u (t)dt for all x, y ∈ I.

Proposition 2 . 2 .

 22 Let G be another Banach space. If A ∈ L (E, F) and B ∈ K (F, G) or A ∈ K (E, F) and B ∈ L (F, G), then BA ∈ K (E, G). Define the adjoint of an operator A ∈ L (E, F) as the unique operator A * ∈ L (F , E ) given by the relation v|Au = A * v|u for any u ∈ E and v ∈ F . Here •|• denotes the duality pairing both between E and E and between F and F . Theorem 2.4. A ∈ K (E, F) if and only if A * ∈ K (F , E ).

Theorem 2 . 5 .

 25 (Fredholm's alternative) Let A ∈ K (E). Then the following assertions hold. (a) N(I -A) is of finite dimension. (b) R(I -A) = N(I -A * ) ⊥ and therefore R(I -A) is closed. (c) N(I -A) = {0} if and only

  the preceding result is applicable when A is substituted by A * . That is we have R(I -A * ) = E . We get by applying one more time Theorem 2.7 that N(I -A) = R(I -A * ) ⊥ = {0}. (d) Set d = dimN(I -A) and d * = dimN(I -A * ). Let us first show that d * ≤ d. We proceed once again by contradiction. Assume then that d < d * . As N(I -A) is of finite dimension, it admits a topological supplement in E. This yields that there exists a bounded projector P from E into N(I -A). On the other hand, R(I -A) = N(I -A * ) ⊥ has finite co-dimension d * and hence R(I -A) admits in E a topological supplement, denoted by L, of dimension d * . As d < d * there exists an injective linear map Λ : N(I -A) → L which is non surjective. If B = A +Λ P then B ∈ K (E) because Λ P is of finite rank. Next, we prove that N(I -B) = {0}. If 0 = u -Bu = (u -Au) -Λ Pu then (I -A)u = 0 and Λ Pu = 0 (because (I -A)u ∈ R(I -A) and Λ Pu ∈ L).

Proposition 2 . 5 .

 25 Let A ∈ L (H) be a self-adjoint operator and set m = inf u∈H, u =1 (Au|u) and M = sup u∈H, u =1 (Au|u). Then σ (A) ⊂ [m, M] and m, M ∈ σ (A).

  .70) where we used that ∇v m = χ [u>k m ] ∇u. If A m = supp(|∇v m |) then (2.66) and (2.70) entail

Theorem 2 .

 2 20. Let u ∈ H 1 loc (Ω ) be a sub-solution of Lu = 0 in Ω . Then u + ∈ L ∞ loc (Ω ) and, for any compact subset K of Ω , 0 < r < dist (K,Γ ) and p > 1, we have u ≤ [C(r, p)] n/p u + L p (K+B(0,r)) a.e. in K, (2.76) where C(r, p) = C 1 + 1 r p 4 (p -1) 2 .

2. 6 .

 6 Let Ω be an open subset of R n of class C 1 with boundary Γ and consider the boundary value problem for the Laplace operator with Neumann boundary condition, where f ∈ C(Ω ),

. 8 ) 3 . 1 .

 831 the ball of center x 0 ∈ R n and radius r > 0. Let u ∈ C k+1,0 (B r ), | | = k and x, y ∈ B r . Then an application of the mean-value theorem yields |∂ u(x) -∂ u(y)| ≤ C|x -y|[u] k+1,0;B r , where the constant C only depends on n. This inequality combined with (3.3) and (3.4) implies, for any u ∈ C k+1,0 (B r ), k ∈ N and 0 < α ≤ 1, [u] k,α;B r ≤ C(n)r 1-α [u] k+1,0;B r . (3Lemma Let u ∈ C k,0 (B r ). Then, for any ball B ρ = B ρ (x) ⊂ B r , ρ > 0 and | | = k, there exists y ∈ B ρ so that |∂ u(y)| ≤ 2k ρ k |u| 0,B r . (3.9) Proof. Let h = ρ/k, | | = k and consider the operator

Corollary 3 . 1 .

 31 .10) is deduced from (3.13) with ε is substituted by ε/C. Let k ∈ N, 0 < α ≤ 1 and (u m ) a bounded sequence in C k,α (B r ). Assume that (u m (x)) converges for any x ∈ B r . Then u = lim m→+∞ u m ∈ C k,α (B r ) and |u| k,α ≤ A = sup m |u m | k,α . (3.14)

|uv|.

  where d = d(x) and B = B(x) are given by (3.16). We get then (3.20) by multiplying each side of the last inequality by d α and then taking the sup in x ∈ Ω . We get similarly from |uv| 0 ≤ |u| 0 |v| 0 This inequality and (3.20) entail (3.21).

Corollary 3 . 2 .

 32 Let k ∈ N, 0 < α ≤ 1 and let (u m ) be a sequence of C k,α;γ (Ω ). Assume that (u m (x)) converges for any x ∈ Ω . Then u = lim m→+∞ u m ∈ C k,α;γ (Ω ) and u(γ) k,α ≤ A = sup m u m (γ) k,α .(3.25)This corollary is immediate from Corollary 3.1 applied to the balls B r = B(x) ⊂ Ω .

.28) Lemma 3 . 4 .

 34 The norms[u]

)

  with d(x) = dist(x,Γ )/2 and E k is defined by (3.36). Then the semi-norms [u] are equivalent. Precisely, we have

  .41) with C 0 = C 0 (k, α, γ). This inequality is also trivially satisfied in case (b) for z = x 0 , ρ = d and C 0 = 2 1+α . Hence, in any case (3.41) holds for 0 < ρ ≤ d(z). Using Lemma 3.6, the definition of A and M (γ)

  .45) (b) c ≤ 0 and there exists a constant K ≥ 0 so that n ∑ i=1 |b i (x)| ≤ K, x ∈ Ω . (3.46)

Theorem 3 . 6 .

 36 (Comparison principle) Let u, v ∈ C 2 (Ω ) ∩C(Ω ) so that Lu ≥ Lv in Ω and u ≤ v on Γ . Then u ≤ v in Ω . In particular, Lu = Lv in Ω and u = v on Γ imply that u = v in Ω .

  and v ≥ |u| on Γ . Hence |u| ≤ v in Ω by the comparison principle. Therefore (3.54) is satisfied with the same constant C 0 as in (3.51). Let us now introduce the notion of sub-solution and super-solution. A function w ∈ C(Ω ) is said a sub-solution (resp. super-solution) of Lu = f in Ω if for any ball B Ω and any function

(3. 56 )

 56 Then w is a sub-solution of Lu = f in Ω . Proof. If the result is not true we would find a ball B Ω and v ∈ C 2 (B) ∩ C(B) so that Lv < f in B and v ≥ w on ∂ B ; but the inequality v > w does not hold in B. Thus, there exists y ∈ B so that 0 ≥ -µ = min B (vw) = v(y)w(y). (3.57) Fix a ball B r = B r (y) B ∩ B y . Then

1 ω

 1 .70) In other words, (sub, super) harmonic functions are (sub, super) solutions of the Laplace equation ∆ u = 0. Since div(∇u) = 0 in Ω , we have according to the diverdomain ω Ω , where ν is the exterior unit normal vector field on ∂ ω. A consequence (3.71) is the following mean-value theorem Theorem 3.10. Let u ∈ C 2 (Ω ) satisfies ∆ u = 0 in Ω . Then, for any ball B = B r (x 0 ) Ω , we have u(x 0 ) =

  ) where C = C(n, k) is a constant. Using the semi-norms defined in (3.15) and (3.16) we can rewrite (3.79) in the form εy)ζ (y)ds y , (3.81) for x ∈ Ω ε . As ζ is constant on {|y| = r} = ∂ B r (0), we get from (3.72) |y|=r u(xεy)ds y = u(x)ω n r n-1 = u(x) |y|=r ds y .

  u(x) = lim m→∞ u m (x), f (x) = lim m→∞ Lu m (x) for any x ∈ Ω . Then u ∈ C 2,α;γ (Ω ), u (γ) 2,α ≤ A and Lu = f in Ω . Proof. Fix x ∈ Ω . Let d = dist(x,Γ )/2 and B = B d (x). As (u m ) m≥1 is bounded in C 2,α;γ (Ω ), it is also bounded in C 2,α (B) and converges to u. By Corollary 3.1, we have u ∈ C 2,α (B) and u m → u in C 2,α (B). Whence Lu = lim m→∞ Lu m = f in Ω . The rest of the proof is contained in Corollary 3.2. Theorem 3.11. Let B r = B r (x 0 ) and ϕ ∈ C(B r ). Then the Dirichlet problem L 0 u = n ∑ i, j=1 a i j ∂ 2 i j u = 0 in B r , u = ϕ on ∂ B r (3.84) admits a unique solution u ∈ C ∞ (B r ) ∩ C(B r ). Furthermore, estimates (3.79) and (3.80) hold with Ω = B r and C = C(n, ν, k).

2 ≤

 2 .86) (a) For any ϕ ∈ C B + r satisfying ϕ = 0 on Γ = R n 0 ∩ B r (x 0 ), the Dirichlet problem ∆ u = 0 in B + r , u = ϕ on ∂ B + r (3.87) admits a unique solution u ∈ C ∞ (B + r ∪ Γ ) ∩C B + r . Moreover, [u] k,0;B + r/C(n, k)r -k sup B + r |u|, k ∈ N. (3.88) (b) For any ϕ ∈ C B + r , the equation ∆ u = 0 in B + r with boundary condition∂ n u = 0 on Γ , u = ϕ on ∂ B + r \ Γ (3.89) admits a unique solution u ∈ C ∞ (B + r ∪ Γ ) ∩C B +r and estimate (3.88) is satisfied. Proof. (a) We extend ϕ, on ∂ B r , to the odd function, still denoted by ϕ, defined by

  0) (an open subset containing 0). Obtain a contradiction by noting that max |x|≤r u(x) ≥ cr 2 1-p for any r > 0.

3

 3

  for any B 2r (x) ⊂ B 2 (0), where C 0 = C 0 (n) is a constant. Hint: deduce from Lemma 3.11 that |∇u(x)| ≤ C 1 sup

3. 12 .

 12 We say that the bounded open set ω of R n has the interior ball property at x 0 ∈ ∂ ω if there exists B ⊂ ω an open ball so that ∂ B ∩ ∂ ω = {x 0 }. One can prove that any C 2 bounded open subset of R n has the interior ball property at each point of its boundary. a) (Hopf's lemma) Let Ω be a bounded subset of R n admitting the interior ball property at x 0 ∈ ∂ Ω , u ∈ C 1 (Ω ) ∩C 2 (Ω ) satisfying ∆ u ≥ 0 in Ω and u(x 0 ) > u(x), for any x ∈ Ω . Let B = B(y, R) ⊂ Ω so that ∂ B ∩ ∂ Ω = {x 0 } and B = B (x 0 , R ), R < R. Set r = |x -y|, D = B ∩ B v(x) = e -ρr 2e -ρR 2 .

Lemma 3 . 14 . 6

 3146 Let u ∈ C(Ω ) and set M = max Ω u. Assume that F = {x ∈ Ω ; u(x) = M} and Ω \F are nonempty. Then there exists an open ball B so that B ⊂ Ω , B∩F = / 0 and ∂ B ∩ F = / 0.

(4. 11 )

 11 Here C = 2 κ+n .

Corollary 4 . 1 .

 41 If u ∈ H vanishes of infinite order at some ξ ∈ Ω then u = 0.

  r)H(r) -D(r)2 ) rD(r)H(r) .

Theorem 4 . 4 .

 44 Let u ∈ H (Ω ) and ξ ∈ Ω . Then we have for any 0 < r < r ξ = dist(ξ ,Γ )u(ξ ) = 1 |S(r)| S(r) u(x)dS(x),(4.13)

Theorem 4 . 6 .

 46 Let 0 ≤ u ∈ H (Ω ), ξ ∈ Ω and 0 < 4r < r ξ = dist(ξ ,Γ ). Then max B(r) u ≤ 3 n min B(r) u Proof. Pick x 1 , x 2 ∈ B(r) so that u(x 1 ) = max B(r) u and u(x 2 ) = min B(r) u.

( 2 )

 2 We have, for any real-valued u ∈ C 2 (Ω ), Ω ∆ ϕ|∇u| 2 e ϕ dx ≤ Ω |∆ u| 2 e ϕ dx + Γ ∂ ν ϕ|∇u| 2 e ϕ dσ (4.22)

  The sum, side by side, of inequalities (4.21) and (4.22) yields Proposition 4.1. (Carleman inequality) We have, for any real-valued u ∈ C 2 (Ω ), Ω (∆ ϕ)u 2 e ϕ dx + Ω (∆ ϕ -1)|∇u| 2 e ϕ dx (4.23)

Proposition 4 . 2 Ω u 2 e 2 +

 4222 1 with that ϕ and an arbitrary u ∈ C 2 (Ω ) satisfying ∆ u = 0 in Ω and u C 2 (Ω ) ≤ M yields ϕ dx + Ω |∇u| 2 e ϕ dx ≤ Γ ∂ ν ϕ(u 2 + |∇u| 2 )e ϕ dσ + 2M Γ |∇u|e ϕ dσ . |∇u| 2 )e ϕ dσ + 2M Γ |∇u|e ϕ dσ . (4.25)

  .27) Let κ 0 = min 1, e -C 0 s 0 /s 0 . If √ I ≤ κ = min(κ 0 , 1) then there exists s * ≥ s 0 so that s * e c 0 s * = 1/ √ I . Therefore s = s * in (4.27) gives J ≤ C ln √ I -1

Corollary 4 . 3 .

 43 Let M > 0. There exists C = C(M, Ω , γ) so that, for any real-valued function u ∈ C 2 (Ω ) satisfying ∆ u = 0 in Ω and u C 2 (Ω ) ≤ M, we have Ω

Corollary 4 . 5 .

 45 Let u ∈ C 2 (Ω ) be a real-valued function satisfying ∆ u = 0 in Ω . If u = ∂ ν u = 0 on γ then u is identically equal to zero.

P 1

 1 .40) Now a combination of (4.36), (4.39) and (4.40) leads Ω wP 2 wdx -Ω c 2 w 2 dx ≥ Ω f w 2 dx + Ω F∇w • ∇wdx + Γ g(w)dσ , (4.41)

  implies a ≤ C | ln b| -α + b . (4.85) Proof. Let b = s -α e -cs . Assume that b ≤ b. Then, since the mapping s → s -α e -cs is decreasing, there exists s 0 ≥ s so that s -α 0 e -cs 0 = b. In particular, b -1 = s α 0 e cs 0 ≤ e Cs 0 , or equivalently s -1 0 ≤ C| ln b| -1 . (4.86) On the other hand, if b ≥ b then 84) with s = s 0 , (4.86) and (4.87) yield (4.85).

B 1/ 2 |∇u| 2 dx = 2πα 2

 222

B 1 0

 1 |∂ i (|x| α )| p dx = S n-1 |ω i | p dσ (ω)r n-1+p(α-1) dr.

1 |δ 1 0≤ e 1 /

 111 i jω i ω j | p dσ (ω) r n-1-p dr.In consequence, x/|x| ∈ W 1,p (B, R n ) if and only if p < n.1.10 (a) We assume, without loss of generality, that x ≤ y. Then we haveu(x) 2 + u(y) 2 -2u(x)u(y) = (u(x)u(y)) 2 = ) 2 + u(y) 2 -2u(x)u(y) ≤ (ba) b a u (t) 2 dt.(b) We obtain by integrating with respect to x b a u(x) 2 dx + (ba)u(y) 2 -2u(y) Hence the result follows. 1.11 (a) Clearly, G is of C 1 and G (t) = p|t| p-1 for any t ∈ R. Whence, w is of class C 1 and w = p|v| p-1 v . As v has a compact support and G(0) = 0, we obtain that w has also a compact support. Assume that 1 < p. Then |w(x)| ≤ R p|v| p-1 |v |dt. Then Hölder's inequality yields |v(x)| p ≤ p Since p ∈ [0, +∞[→ p 1/p attains its maximum at p = e, the last inequality gives |v(x)| ≤ e 1/e v e v W 1,p (R) .The expected inequality is obvious when p = 1. In fact we have simply in that case |v(x)| ≤ v 1 .(b) Set c = e 1/e . Let u ∈ W 1,p (R) and (u k ) ∈ C 1 c (R) converging to u in W 1,p (R). By (a) we have, for every k and ,uu k L ∞ (R) ≤ c u u k W 1,p (R) .1 Young's inequality. Let a and b be two positive reals numbers, p > 1 and p the conjugate exponent of p.

2 ( 1 -x)dx = 1 8 ,

 218 the last two inequalities and the density of D(]0, 1[) in H 1 )| 2 dx for any u ∈ H 1 0 (]0, 1[).

1 0 1 0

 11 |u (x)| 2 dxk |u(x)| 2 dx = 0. On the other hand, the definition of C yields C )| 2 dx and hence u = 0 if Ck < 1. (c) The non trivial solutions, for k = 0, of the boundary value problem u (x) + ku(x) = 0, x ∈]0, 1[ and u(0) = u(1) = 0 are of the form u(x) = sin( √ kx), with k = n 2 π 2 .

1 8 . 1 .u

 181 18 (a) (i) Let x ∈ I and (x m ) be a sequence in I converging to x. Let J ⊂ I a compact interval containing x and x m , for each m. We check that χ ]x,x m [ g 2 converges a.e. to 0 (indeed if t = x is such that |g(t)| < ∞, then t ∈]x, x m [ for sufficiently large m) and |χ ]x,x m [ g| ≤ |g| a.e.. An application of the dominated convergence theorem givesf (x m )f (x) = χ ]x,x m [ gdt → 0. (ii) As I f ϕ dx = -Set u = x c u (t)dt,where c ∈ I is arbitrarily fixed. By (a) (i), u ∈ C I , and by (a) (ii), we have ϕdx for any ϕ ∈ D(I). That is I (uu) ϕdx = 0 for any ϕ ∈ D(I) and hence uu = k a.e. in I by the closing lemma, where k is a constant. The function ũ = u + k satisfies then the required properties. (c) Let u be an element of B, the unit ball of W 1,p (I). Using that u has a continuous representative by (b), we can write u(y)u(x) = y x u (x)dx. Apply Hölder's inequality to the right hand side of this identity in order to deduce that |u(y)u(x)| ≤ u L p (I) |x -y| 1/p ≤ |x -y| 1/p , for any x, y ∈ I.

.

  We set then y ,k+1 = yϕ m,k ( ),k m m

  we get from Fubini's theorem that X |k(x, y)| 2 dµ(x) is finite a.e. y ∈ Y and Y X |k(x, y)| 2 dµ(x) dν(y) = X×Y |k(x, y)| 2 dµ(x) ⊗ dν(y).

  11) to get M(4r)m(4r) ≤ C [(M(4r)m(4r)) -(M(r)m(r))] . That is ω(4r) ≤ C(ω(4r)ω(r)). Hence ω(r) ≤ γω(4r), with γ = (C -1)/C. (b) We obtain by iterating the last inequality ω r 4 k-1 ≤ γ k ω(4r), k ≥ 0 is an integer.

  for any r > 0 follows readily from the maximum principle because ∆ u ≥ 0.

  0) for all r > 0. That is we proved the following estimate max |x|≤r u(x) ≥ cr 2/(1-p) for all r > 0.3.8 (a) Let x be such that B 2r (x) ⊂ B 2 (0). By Lemma 3.11, we have that ify ∈ B r (x) then d y |∂ i u(y)| ≤ C sup B 2r (x)u, with a constant C = C(n), where d y = dist(y, ∂ B 2r (x)). As d y ≥ r, we conclude r|∂ i u(y)| ≤ C sup B 2r (x) for any B 2r (x) ⊂ B 2 (0), with a constant C 0 = C 0 (n). We easily deduce from the last estimate take r = 1/4 |∇u(x)| ≤ C 1 sup B 1/2 (x) u for any x ∈ B 1 (0) (5.17) with a constant C 1 = C 1 (n). (b) We have according to Harnack's inequality (Theorem 2.8)

  ) and (5.18) entail|∇u(x)| ≤ C 1 C 2 u(x), for any x ∈ B 1 (0).

  r) |u|, with a constant C = C(n, k). This and the assumption on u imply max | |=k r k |∂ u(x)| ≤ Cr α and hence max | |=k |∂ u(x)| ≤ Cr -k+α .

  dσ = 0, if k = .

σ

  (ty)|∇u(ty)| 2 + β (ty)u 2 (ty) t n-1 dS(y)dt. Hence D (r) = S(1) σ (ry)|∇u(ty)| 2 + β (ry)u 2 (ry) r n-1 dS(y) = S(r)

From

  (b), we get D(r) ≤ β 0 rD(r) + β 0 + c n -2 r H(r) .Combined with H(r) < rD(r), this estimate yieldsD(r) D(r) ≤ β 0 (1 + β 0 + c(n -2)) .(5.39)In light of the previous comments, (5.38) and (5.39) we haveN (r) N(r) ≥ -c ,the constant c only depends on Ω , σ 1 /σ 0 and β 0 . Hence N(r) ≤ e c s j N(s j ) ≤ e c r 0 max(N(r 0 ), 1), r ∈ (r j , s j ).

4. 4 ≤

 4 (a) (i) Since n < 4, H 2 (Ω ) is continuously imbedded in L p (Ω ) for 1 ≤ p < ∞.By Theorem 4.11, we have φ ∈ W 2,p (Ω ) andφ W 2,p (Ω ) ≤ Cλ φ L p (Ω ) Cλ u H 2 (Ω ) ≤ Cλ 2 φ L 2 (Ω ) .

√ λ and v C 1

 1 ,α (D) ≤ κe √ λ u C 1,α (Ω )for some universal constant κ, we deduce from (4.93) that (4.94) holds for v.(c) (i) Follows immediately from (4.94) with u = φ and the estimate in (a) (iii).(ii) It is straightforward to check that (4.95) implies the following inequalityCe -k √ λ ≤ ε β + e c/ε ℵ, 0 < ε < 1,(5.41)

4. 5 L 2 1 ≤ 2 α 1 Be| α 1 ≤ ω n 2 α 0 n 3 n-α 1 .

 5211131 (a) Using the following inequalities e √ λ /4 ≤ e λt L 2 (1/4,3/4) , e λt L 2 (0,1) ≤ e √ λ , we easily get, by applying (4.96) to v(x,t) = u(x)e λ )u L 2 (Ω ) + u L 2 (ω) obtain by taking u = φ in (5.42)φ L 2 (Ω ) ≤ Cλ 1-β e (ω) ,This inequality implies the expected one in a straightforward manner.4.6 (a) (i)The proof is obtained by slight modifications of that of Theorem 4.8. In the sequel the notations are those of the proof of Theorem 4.8. We have L r (χw) = χL r w + Q r (w) and(χL r w) 2 ≤ Λ w 2 + |∇w| 2 ,the constant Λ only depends on Ω and Λ . Before obtaining an inequality similar to (4.48) when using Carleman inequality in Theorem 4.7, we absorb Λ w 2 + |∇w| 2 by the left hand side by modifying λ 0 and τ 0 if necessary. In that case, we have an Proof. Let x, y ∈ Ω with x = y. Write z = x + |x -y|η. Hence |x -z| = |x -y||η| and |z -y| = |zx + x -y| = ||x -y|η + x -y| = |x -y| η + xy |x -y| . If e = x-y |x-y| then the last identities yield |z -y| = |x -y||η + e|.We have, where d = 2diam(Ω ),Ω dz |x -z| α 0 |z -y| α 1 ≤ 1 |x -y| -n+α 0 +α 1 B 0, d |x-y| dη |η| α 0 |η -e| α 1 . If η ∈ B(e, 1/2) then |η| ≥ |e| -|η -e| ≥ 1/2. Whence B(e,1/2) dη |η| α 0 |η -e| α For η ∈ B 0, d |x-y| \ B(0, 2) we have |e| = 1 ≤ |η| -|e| ≤ |η -e| and hence |η| ≤ |η -e| + |e| ≤ 2|η -e|. Thus B 0, d |x-y| \B(0,2) dη |η| α 0 |η -e| α 1 ≤ B 0, d |x-y| \B(0,2) dη |η| α 0 +α 1 ≤ ω n d |x-y| 2 r n-1-α 0 -α 1 dr. dln 2ln |x -y| if α 0 + α 1 = n,

we obtain | f (x 1 ,

 1 y)f (x 2 , y)| ≤ C|x 1x 2 | µ 1 |x 1 -y| -n+µ 0 + |x 2 -y| -n+µ 0 , for any x 1 , x 2 , y ∈ Ω , y = x j , j = 1, 2. Proof. (i) Fix (x 0 , y 0 ) ∈ Σ . If δ = |x 0y 0 | > 0, pick then 0 < η ≤ δ /4 and ε > 0. For (x, y) ∈ Σ , we write f (x, y)f (x 0 , y 0 ) = I 0 + I 1 , with I 0 = Ω [ f 0 (x, z)f 0 (x 0 , z)] f 1 (z, y)dz, I 1 = Ω f 0 (x 0 , z)[ f 1 (z, y)f 1 (z, y 0 )]dz.

1 ≤ c 2 δ 2 -n+β 1 |x 2 -n+β 1 B| ≤ c 2 δ 2 -n+β 1 B

 1212121 Assume that |xx 0 | < η and |y -y 0 | < δ /4. For z ∈ B(x 0 , η), we have |x -z| < 2η and |z -y| ≥ |x 0y 0 | -|y 0 -y| -|zx 0 | ≥ δ /2.Whence, where z ∈ B(x 0 , η),|[ f 0 (x, z)f 0 (x 0 , z)] f 1 (z, y)| ≤ c 2 |x -z| -n+β 0 + |x 0 -z| -n+β 0 |z -y| -n+β -z| -n+β 0 + |x 0 -z| -n+β 0and consequently|J 0 | ≤ c 2 δ (x 0 ,η) |x -z| -n+β 0 + |x 0 -z| -n+β 0 dz. (A.4) But B(x 0 , η) ⊂ B(x, 2η). Hence (A.4) yields |J 0 (x,2η)|x -z| -n+β 0 dz + B(x 0 ,η)

| f 0 1 .

 01 (x, z)f 0 (x 0 , z)| ≤ ε 4ℵ , If |xx 0 | ≤ η 1 then |J 1 | ≤ cε 4ℵ B(x 0 ,b) |z -y| -n+β 1 dz (A.6) ≤ cε 4ℵ B(y,b+5δ /4) |z -y| -n+β 1 dz ≤ ε/4.

|B 2 -n+β 0 +β 1 ω n β 1 .

 21 f (x, y)f (x 0 , y 0 )| ≤ ε, |xx 0 | ≤ η * |yy 0 | ≤ η * .We now consider the caseβ 0 + β 1 > n. Fix η > 0. Let (x, y) ∈ Ω × Ω , x = y so that |xx 0 | ≤ η and |x -y| ≤ η. Then, for z ∈ B(x 0 , η), we have |z -y| ≤ |zx 0 | + |x 0 -x| + |x -y| ≤ 3η and hence |z -y| |x -y| ≤ 3η |x -y| = t. Note that |x -y| ≤ η entails 3 ≤ t. With u = xy |x -y| , the substitution z = y + |x -y|w yields I = B(x 0 ,η) |x -z| -n+β 0 |y -z| -n+β 1 dz ≤ |x -y| -n+β 0 +β 1 B(0,t)|u -w| -n+β 0 |w| -n+β 1 dw.We decompose the last integral into three termsThe first term in right hand side of this identity is estimated as followsB(0,1/2) |u -w| -n+β 0 |w| -n+β 1 dw ≤ 2 n-β 0We have similarly for the second termB(0,2)\B(0,1/2) |u -w| -n+β 0 |w| -n+β 1 dw ≤ 2 n-β 1 B(0,2)\B(0,1/2)|u -w| -n+β 0 dw = 2 -n+β 1 3 β 0 ω n β 0 .

1 0[ 1 0( 1 0|

 111 ∂ x i f (x 0 + ste i , z) -∂ x i f 0 (x 0 , z)] ds. Whence | f 0 (x 0 + te i , z)f 0 (x 0 , z) -t∂ x i f 0 (x 0 , z)| ≤ |t| |∂ x i f 0 (x 0 + ste i , z)| + |∂ x i f 0 (x 0 , z)|) ds ≤ |t|κ (x 0 + ste i ) -z| -n+β + |x 0 -z| -n+β ds.

x 1 ,Ω 1 =ΩΛ 1 =

 111 x 2 ∈ Ω , x 1 = x 2 , d = |x 1x 2 | and y ∈ Ω \ {x 1 , x 2 }. Then f (x 1 , y)f (x 2 , y) = Ω \B(x 1 ,2d) + Ω ∩B(x 1 ,2d) [ f 0 (x 1 , z)f 0 (x 2 , z)] f 1 (z, y)dz.We deduce from this identity| f (x 1 , y)f (x 2 , y)| ≤ C(|x 1x 2 | δ J 0 + J 1 , \B(x 1 ,2d) |x 1 -z| -n + |x 2 -z| -n | f 1 (z, y)|dz, J ∩B(x 1 ,2d) |x 1 -z| -n+β 0 + |x 2 -z| -n+β 0 | f 1 (z, y)|dz. Define Λ 0 = {z ∈ Ω \ B(x 1 , 2d); 2|y -z| ≥ |x 1 -y|}, {z ∈ Ω \ B(x 1 , 2d); 2|y -z| < |x 1 -y|}.

Ω

  \B(x 1 ,2d) |x 2 -z| -n | f 1 (z, y)|dz ≤ C|x 1 -y| -n+β 1 . Whence J 0 ≤ C |x 1 -y| -n+β 1 + |x 2 -y| -n+β 1 . (A.8)We now estimate J 1 . Define for this purposeΣ 0 = {z ∈ Ω ∩ B(x 1 , 2d); 2|y -z| ≥ |x 1 -y|}, Σ 1 = {z ∈ Ω ∩ B(x 1 ,2d); 2|y -z| < |x 1 -y|}.

x 1 |

 1 > |x 1 -y| and |y -z| ≤ 2d for z ∈ Σ 1 , we obtain Ω φ (y)(A * g)(y)dy = Ω Ω K(x, y)g(x)dx φ (y)dy.

Lemma A. 3 .

 3 Let 0 < α < n. Then there exists a constant C > 0, depending on n, Ω and α, so thatsup x∈Ω Ω dx |x -y| α ≤ C.Proof. Choose R > 0 in such a way that Ω ⊂ B(y, R) for any y ∈ Ω . Then we get by passing to spherical coordinatesΩ dx |x -y| α ≤ B(y,R) dx |x -y| α = ω n R 0 r n-α-1 dr = ω n R n-α nα and hence sup x∈Ω Ω dx |x -y| α ≤ ω n R n-α nα .

(

  A f )(x) = Ω B(x, y) |x -y| α f (y)dy a.e. x ∈ Ω .

ΩB 3 Ω 2 ≤ C Ω 1

 321 (x, y) |x -y| α f (y)dy, Ω |B(x, y)| |x -y| α | f (y)|dy and Ω |B(y, x)| |x -y| α | f (y)|dy (A.11) exist for a.e. x ∈ Ω .Also, asΩ Ω | f (y)| 2 |x -y| α dxdy = Ω | f (y)| 2 Ω 1 |x -y| α dx dy ≤ C Ω | f (y)| 2 dy < ∞, the integral Ω | f (y)| 2 |x -y| α dy (A.12)exists for a.e. x ∈ Ω . This follows again from Fubini's theorem. Whence, for x ∈ Ω so that the integrals in (A.11) and (A.12) exist, we get by applying Cauchy-Schwarz's inequality and Lemma A.B(x, y) |x -y| α f (y)dy |x -y| α/2 | f (y)| |x -y| α/2 dy (A.13) ≤ C Ω | f (y)| 2|x -y| α dy.

  A ∈ B(L 2 (Ω )).

  3, we have sup x∈Ω B(x,ε)1 |x -y| α dy ≤ Cε n-α .In light of this estimate, we can carry out the same calculation as for A in order to get, for a.e. x ∈ Ω ,|(R ε f ) (x)| ≤ Ω ∩B(x,ε) B(x, y) |x -y| α | f (y)|dy ≤ C B(x,ε) 1 |x -y| α dy Ω | f (y)| 2 |x -y| α dy ≤ Cε n-α Ω | f (y)| 2 |x -y| α dy,from which we deduce, as we have done for A,R ε f 2 ≤ Cε n-α f 2 and hence R ε B(L 2 (Ω )) ≤ Cε n-α .Thus A -A ε B(L 2 (Ω )) → 0 as ε → 0.

Theorem A. 4 .

 4 Let A be a weakly singular operator with kernel K, λ = 0 and g ∈ L 2 (Ω ). Then the integral equation (A.14) has a unique solution f ∈ L 2 (Ω ), or else the homogenous equation Ω K(y, x)h(y)dyλ h(x) = 0 a.e. x ∈ Ω has exactly p linearly independent solutions h 1 , . . . , h p . In that case, (A.14) is solvable if and only if g satisfies the following orthogonality relations Ω g(x)h j (x)dx = 0, j = 1, . . . , p.Next, when in the kernel K of the weakly singular integral operator A is so thatB ∈ C(Ω × Ω \ D) ∩ L ∞ (Ω × Ω )then we are going to show that A acts as a compact operator on C(Ω ). Note that, for each f ∈ C(Ω ), we have|K(x, y) f (y)| ≤ B L ∞ (Ω ×Ω ) f C(Ω ) |x -y| -α , x, y ∈ Ω , x = y.Therefore, as an improper integral,(A f )(x) = Ω K(x,y) f (y)dy exits for any x ∈ Ω . Theorem A.5. The weakly singular operator A : C(Ω ) → C(Ω ) is compact.

  |K(x, y) -K(z, y)| ≤ ε/|Ω |.Thus, for an arbitraryf ∈ C(Ω ) with f C(Ω ) ≤ 1, |A f (x) -A f (z)| ≤ ε, provided that |x -z| ≤ η. In other words, F = {A f ; f ∈ C(Ω ), f C(Ω ) ≤ 1}is relatively compact by Arzela-Ascoli's theorem and the result follows.

- 1 ω

 1 max 1≤i, j≤n [a i j ] α + max 1≤i≤n |b i (x)| + |c(x)| ≤ Λ , x ∈ Ω ,for some constant 0 < α ≤ 1 and Λ > 0. Then there exists a constant C, only depending on n, diam(Ω ), µ and Λ , so that|L x H(x, y)| ≤ C|x -y| -n+α , (x, y) ∈ Σ . (A.27) Lemma A.5. Let 0 < α ≤ 1, Λ > 0 and assume that a i j ∈ C 1 (Ω )(⊂ C 0,α (Ω )), 1 ≤ i, j ≤ n and max 1≤i, j≤n a i j C 1 (Ω ) ≤ Λ .Then there exists a constant C > 0, only depending on n, diam(Ω ), α, Λ and Ω , so that∂ x k H(x, y) + ∂ y k H(y, x) ≤ C|x -y| -n+α+1 , (x, y) ∈ Σ , (A.28) ∂ 2 x j x k H(x, y) + ∂ 2 x j y k H(y, x) ≤ C|x -y| -n+α , (x, y) ∈ Σ . (A.29)Proof. We have by (A.22)∂ x k H(x, y) + ∂ y k H(y, x) = n d(y)ρ n (x, y) n ∑ =1 a k (y)(xy ) +1ω n d(x)ρ n (y, x) x)(yx ) 1

1 = 1 ω 1 ρ

 111 ∂ 2 x j x k H(x, y) + ∂ 2 x j y k H(y, x) = I 1 + I 2 + I 3 ,withI n ρ n (x, y) a k j (x) x)(yx )∂ x j 1 ρ n (y, x) n (x, y)It is straightforward to check that |I 1 + I 2 | ≤ C|x -y| -n+α . To estimate I 3 , we first compute the term∂ x j 1

1 ω

 1 |L x H(x, z) -L x H(y, z)| ≤ |x -y| α |x -z| -n + |y -z| -n , (A.30) for all x, y, z ∈ Ω satisfying |x -z| ≥ 2|x -y|.Proof. Recall thatρ(x, z) = A -1 (z)(xz)|xz 1/2and∂ x i ρ(x, z) = 1 ρ(x, z) n ∑ j=1 a i j (z)(x jz j ).Fix x, y ∈ Ω and s > 0. Define thenθ (t) = ρ(x + t(yx), z) -s , t ∈ [0, 1].Since there exits τ ∈ (0, 1) so that θ (1)θ (0) = θ (τ), we get ,wherew = x + τ(yx), ρ(y, z) -sρ(x, z) -s = -s ρ s+2 (w, z) A -1 (z)(wz)|yx . Assume that |x -z| ≥ 2|x -y|. Then |w -z| ≤ |x -y| < |x -z|/2 and hence |w -z| ≥ |z -x|/2. Therefore ρ(y, z) -sρ(x, z) -s ≤ C|x -y||x -z| -(s+1) . (A.31)Note that (A.31) with s = n -2, n ≥ 3, and (A.20) yield|c(y)H(y, z)c(x)H(x, z)| ≤ C |x -y| α |x -z| -n+2 +C|x -y||x -z| -n+1and hence|c(y)H(y, z)c(x)H(x, z)| ≤ C|x -y| α |x -z| -n . (A.32)We have the same inequality for n = 2. From (A.22), we have∂ x i H(x, z) = -1 ω n d(z)ρ n (x, z) n ∑ j=1 a i j (z)(x jz j ).Whence∂ x i H(y, z) -∂ x i H(x, z) = 1 ω n d(z) ρ -n (x, z)ρ -n (y, z) n ∑ j=1 a i j (z)(y jz j ) + n d(z)ρ n (x, z) n ∑ j=1 a i j (z)(y jx j ).As |y -z| ≤ 3|x -z|/2, we obtain, by applying (A.25) and (A.31) with s = n,|∂ x i H(y, z) -∂ x i H(x, z)| ≤ C|x -y||x -z| -n . (A.33)It follows from (A.25) and (A.33)b i (y)∂ x i H(y, z)b i (x)∂ x i H(x, z) ≤ C |x -y||x -z| -n + |x -y| α |x -z| -n+1and consequently b i (y)∂ x i H(y, z)b i (x)∂ x i H(x, z) ≤ C|x -y| α |x -z| -n .

  (y)∂ 2 x i x j H(y, z) ≤ C|x -y||x -z| -n . (A.35)The expected inequality is obtained by putting together (A.32), (A.34) and (A.35).Ω f (x, y)φ (x)dx = Ω (A * ) j (z) [K(z, y) + (z, y)] dz = Ω φ (z) [K(z, y) + (z, y)] dz. = φ (y) + Ω φ (z) (z, y)dz.This and (A.44) entailΩ f (x, y)φ (x)dx = 0, φ ∈ N(I -A * ), y ∈ Ω .This orthogonality relation at hand, we can apply Fredholm's alternative to deduce that the integral equationg(x, y) -Ω K(x, z)g(z, y)dz = f (x, y) (A.46)has a unique solution g(•, y) ∈ C(Ω ) orthogonal to N(I -A).

  max x∈Ω |g(x, y)g(x, z)| ≤ C max x∈Ω | f (x, y)f (x, z)|from which the result follows.

  x∈Ω |g(x, y j )g(x, z j )| > j max x∈Ω | f (x, y j )f (x, z j )| = jt j , j ≥ 1.

  u jv j = f j by (A.46). Whence, u j converges also to u = v in C(Ω ) and hence u = Au or equivalently u ∈ N(I -A). On the other hand, we know that u j ∈ N(I -A) ⊥ . Therefore0 = Ω u j (x)u(x)dx, j ≥ 1, that yields 0 = lim j→∞ Ω u j (x)u(x)dx = u L 2 (Ω ) .We end up getting the expected contradiction by noting that 1= u j C(Ω ) → u C(Ω ) = 1. Define G(x, y) = g(x, y) + K(x, y) + (x, y) + j-1 ∑ s=1 Ω K s (x, z) [K(z, y) + (z, y)] dz.We note that the term belongs toC Ω \ {x 1 , . . . , x m } × Ω while the other terms are in C Ω × Ω \ D .Moreover, we can check that G is a solution of integral equationG(x, y) -Ω K(x, z)G(z, y)dz = K(x, y) + (x, y). (A.47)Fix y ∈ Ω and let x 0 = y. Let η sufficiently small in such a way thatB(x i , η) ∩ B(x k , η) = / 0, for 0 ≤ i, j ≤ m, i = j and B(x i , η) ⊂ Ω \ Ω 0 , j = 1, . . . m.We decompose G as follows y) = g(x, y) + K(x, y) + 0 (x, x, z) [K(z, y) + 0 (z, y)] dz,G 0 i (x, y) = -K(x, x i ) f i (y), j = 1, . . . , m, G 1 i (x, y) = -x, z)K(z, x i ) f i (y)dz, j = 1, . . . , m. Define Λ j η = Ω \ B(x i , η), j = 0, . . . , m.Noting that the coefficients of L are in C 0,α (Ω ), it not hard to check that G 0 i (•, y) belongs to C 0,α (Λ i η ), i = 0, . . . m. On the other hand, in view of Lemma A.6 and the estimate |K(z,x i )| ≤ C|xx i | -n+α , i = 0, . . . , m,we deduce by applying (iii) of Theorem A.1 that G 1 i (•, y) belongs to C 0,β (Λ η ), for any β ∈ (0, α), and consequentlyG i (•, y) = G 0 i (•, y) + G 1 i (•, y) ∈ C 0,β (Λ η ) for i = 0, . . . m.Consider the functionΓ k (x, y) = Ω H(x, z)G k (z,y)dz, k = 0, . . . m. According to Theorem A.1 (i), Γ k belongs to C(Ω 0 × Ω 0 \ D) and, by Lemma A.4 and Theorem A.1 (ii), Γ i (•, y) ∈ C 1 (Ω 0 \ {y}) with ∇ x Γ 0 (x, y) = Ω ∇ x H(x, z)G k (z, y)dz.

  Ω η [∂ x j H(x, z) + ∂ z j H(z, x)]G k (z, y)dz -Ω η ∂ z j H(z, x)G k (z, y)dz + B(x k ,η) ∂ x j H(x, z)G k (z, y)dz.We have in mind to apply the following Michlin's theorem Theorem A.6. Let Ω be a bounded open subset of R n and w ∈ C(Σ ) of the form w(x, y) = |x -y| -n+1 Ψ x, xy |x -y| , (x, y) ∈ Σ , with Ψ ∈ C 1 (Ω × S n-1 ). Let β ∈ (0, 1) and u ∈ C 0,β (Ω ). Then v(x) = Ω w(x, y)u(y)dy belongs to C 1 (Ω ) and we have, for x ∈ Ω , ∇v(x) = lim ε→0 Ω \B(x,ε)∇ x w(x, y)u(y)dy + u(x)S n-1 ξΨ (x, ξ )dσ (ξ ).

  (x)(z kx k ) and ρ(z, x) = A -1 (zx)|zx 1/2 ,we have∂ z j H(z, x) = |x -y| -n+1 Ψ j x, zx |z -x| , with Ψ j (x, ξ ) = -n ∑ k=1 a jk (x) ω n d(x) ξ k A -1 (x)ξ |ξ -n/2 .

2 = - 1 ω/ 2 .can be proved 1 that 1 ω

 2121 (x)ξ i Ψ j (x, ξ ) = -∑ i, j a i j (x) n ∑ k=1 a jk (x) ω n d(x) ξ i ξ k A -1 (x)ξ |ξ -n/n d(x) A -1 (x)ξ |ξ -nIt n d(x) S n-1 A -1 (x)ξ |ξ -n/2 dσ (ξ ) = 1.Thusn ∑ i, j=1 a i j (x)∂ x i Ω η ∂ z j H(z, x)G k (z, y)dz =

  L x Γ k (x, y) = Ω L x H(x, z)G k (z, y)dz -G k (x, y), x ∈ Ω 0 \ {y}.(A.48)

  .127) If |x -y| < d(x 0 )/4, we choose y j such that x, y ∈ B d/2 (y j ), d(x 0 ) ≤ 2d.

	Then in-
	equalities (3.124) and (3.127) enable us obtaining the following estimate
	[u]

  in the previous inequality gives (4.35) in straightforward manner. Observe that Theorem 4.7 holds for complex-valued v ∈ H 2 (Ω ). In that case we have to substitute v 2 and L t v respectively by |v| 2 and |L t v| 2 .

	Remark 4.1.

  9λ e 5λ , β = e 5λe 11λ /4 .

	C	B(2)	w 2 + |∇w| 2 dx ≤ e ατ	B(1)	w 2 + |∇w| 2 dx +	B(3)	(L r w) 2 dx	(4.50)
										+ e -β τ	w 2 + |∇w| 2 dx,
										B(3)
	where
	α = e On the other hand, we have by Caccioppoli's inequality (4.43)
				C	B(1)	|∇w| 2 dx ≤	B(3/2)	w 2 dx +	B(3/2)	(L r w) 2 dx,	(4.51)
				C	B(3)	|∇w| 2 dx ≤	B(7/2)	w 2 dx +	B(7/2)	(L r w) 2 dx	(4.52)
	Inequalities (4.51) and (4.52) in (4.50) yield
	C	B(2)	w 2 dx ≤ e ατ		B(3/2)	w 2 dx +	B(7/2)	(L r w) 2 dx + e -β τ	B(7/2)	w 2 dx. (4.53)
										Therefore, fixing λ and changing τ 0 if
	necessary, (4.47) implies for τ ≥ τ 0
		C	w 2 + |∇w| 2 e 2τϕ dx ≤	w 2 + |∇w| 2 e 2τϕ dx
			B(2)							B(1)
			+	B(3)	(L r w) 2 e 2τϕ dx +	{5/2≤|x|≤3}	w 2 + |∇w| 2 e 2τϕ dx.	(4.49)
	We get by taking ψ(x) = 9 -|x| 2 in (4.49), which is without critical points in U,
	that for τ ≥ τ 0

  by Exercise 1.20, F exists and it is unique). According to Lax-Milgram's Lemma the variational problem

Sobolev spaces

Recall that C k (ω) = {u| ω ; u ∈ C k (R n )}.

If A is a negligible set of R n so (1.12) holds for any x, y ∈ R n \ A then, as R n \ A is dense in R n , u| R n \A admits a unique continuous extension to R n .

Tonelli's theorem For i = 1, 2, let Ω i be an open subset of R n i . Let h : Ω 1 × Ω 2 → R be a measurable function so that Ω 2 |h(x, y)|dy < ∞, a.e. x ∈ Ω 1 , and Ω 1 dx Ω 2 |h(x, y)|dy < ∞. Then h ∈ L 1 (Ω 1 × Ω 2 ).

Banach's fixed point theorem. On a complete metric space M with metric d, let T : M → M be so that there exists a constant 0 < k < 1 with the property that

Riesz's lemma Let M be a closed subspace of E so that M = E. Then, for any ε > 0, there exists u ∈ E satisfying u = 1 and dist(u, M) ≥ 1ε.

Variational solutions

For simplicity convenience we considered the resolvent set as a subset of R. But the more appropriate framework should be to consider the resolvent set as a subset of C.

 5 Note that if Aλ I is bijective, then (Aλ I) -1 ∈ L (E) by de Banach's theorem. Banach's theorem. Any bijective bounded operator between two Banach spaces admits a bounded inverse.

u| Γ stands for the trace of u on Γ in the sense of Theorem 1.18.

u and ∂ ν u exist, in the trace sense, as elements of L 2 (Γ ).

This fact can be proved by using divergence theorem and the definition of weak derivatives.

Note that ∇u = 0 in any set where u is constant. This an immediate consequence of Corollary 1.1.

Remark that η in (2.88) is built from K (see formula (2.78)).

Recall that dist(K,Γ ) = inf{d(x, y); (x, y) ∈ K × Γ }.

Recall that D(Γ 2 ) = {v = u| Γ 2 ; u ∈ D(R n )}.

Leray-Schauder's fixed point theorem. Let X be a Banach space and let T : X × [0, 1] → X be a compact mapping (i.e. T is continuous and sends the bounded sets of X × [0, 1] into relatively compact sets of X). If T (•, 0) = 0 and if {x ∈ X; x = T (x, λ ) for some λ ∈ [0, 1]} is bounded then T (•, 1) admits a fixed point.

Here [x, y] = {z = tx + (1t)y; 0 ≤ t ≤ 1}.

Recall that P k is the vector space of polynomials of degree less or equal to k.

Here ν, α, K et K 1 are the constants appearing in the assumptions on the coefficients of L.

Proof. As Ω is connected, the boundary of F in Ω contains a least a point in Ω . Since F is closed in Ω there exits z ∈ F so that any neighborhood of z contains at least a point where u < M. Let δ > 0 such that B(z, 3δ ) ⊂ Ω and let x 0 ∈ B(z, δ ) so that u(x 0 ) < M. Consider then the set

Classical inequalities, Cauchy problems and unique continuation

Here ]x, x m [ is the interval with endpoints x and x m .

Recall that in a compact space a sequence admitting a unique limit point is convergent.

We can also observe that (5.9) is an Euler's equation and solve it by setting s = ln r and h(s) = f (r).

See [1, Appendix

2, page 289].

Lu = f in B r , u = ϕ on ∂ B r .

(3.109)

We assume that L fulfills the assumptions of the preceding section. We first consider the case ϕ = 0 and then the case ϕ ∈ C(B r ).

Theorem 3.15. For any f ∈ B 2 = C 0,α;2-β (B r ), there exists a unique u ∈ B 1 = C 2,α;-β (B r ) satisfying Lu = f in B r . In other words, L maps B 1 onto B 2 , i.e. L(B 1 ) = B 2 .

As we have mentioned in Remark 3.4, any function u ∈ B 1 vanishes on ∂ B r . Such u belongs to C(B r ) and u = 0 sur ∂ B r . Theorem 3.15 ensures the existence of a solution of (3.109) when f ∈ B 2 ⊃ C α (B r ) and ϕ = 0.

Proof. Note first that the uniqueness is a straightforward consequence of the comparison principle. Before proceeding to the proof of the existence, we introduce some notations. The natural norms of B 1 and B 2 are denoted respectively by • 1 and • 2 . Under these new notations, (3.106) takes the form

(3.110)

The proof of existence consists in four steps. First step. L = ∆ and f ∈ C 1 (B r ). By Stone-Weierstrass's theorem, we find a sequence of polynomials ( f m ) converging to f in C 1 (B r ). For each m, Lemma 3.12 guarantees the existence of u m ∈ B 1 so that ∆ u m = f m . As 2β > 0, f m converges to f in C 0,α;2-β (B r ). But, we have from (3.110)

Hence, (u m ) is a Cauchy sequence in B 1 . Therefore, it converges to u ∈ B 1 satisfying ∆ u = f . Second step. L = ∆ and f ∈ C α (B r ) vanishing in a neighborhood of ∂ B r . If ε > 0 is sufficiently small the regularization f (ε) of f given by (3.74) is well defined. By the first step, there exists u ε ∈ B 1 satisfying ∆ u ε = f (ε) . We get in light of (3.78) and (3.110) that (u ε ) is uniformly bounded in B 1 :

On the other hand, f (ε) converges to f in C(B r ) by (3.77). The comparison principle (Theorem 3.7) then entails sup

Whence, there exists lim ε→0 u ε = u ∈ C(B r ). An application of Lemma 3.13 yields u ∈ B 1 and ∆ u = lim ∆ u ε = lim f (ε) = f in B r . Third step. L = ∆ and f ∈ B 2 . Consider, for small ε > 0, the auxiliary function

This is the expected inequality.

Two-dimensional Cauchy problems

Let Ω be a bounded domain of R 2 with Lipschitz boundary Γ . Fix ϕ ∈ C 2 (Ω ; R) and equip L 2 (Ω , e ϕ dx) and L 2 (Γ , e ϕ dσ ) respectively by the following scalar products

The norm associated to the scalar product (•|•) is simply denoted by • .

Pick a 1 , a 2 ∈ C and define the differential operator

The formal adjoint of P is given by

The unit normal vector to Γ pointing outward Ω is denoted by

Since Γ is Lipschitz ν(σ ) is well defined for a.e. σ ∈ Γ . The unit tangent vector to Γ is given by τ(σ ) = (-ν 2 (σ ), ν 1 (σ )).

Let Q be the operator Q = a 1 ν 1 + a 2 ν 2 . We obtain from Green's formula

We recall that [A, B] = AB -BA denotes the usual commutator of the operators A and B. Lemma 4.1. We have, for any u ∈ C 1 (Ω ),

(4.17)

Proof. We get by applying twice (4.16)

where we used that ω is a non decreasing function. But ln(r 0 /r) ln 4 < k.

Thus, γ k ≤ e ln(r 0 /r) ln γ/ ln 4 = r 0 r ln γ/ ln 4 because γ ≤ 1. Therefore, ω(r) ≤ Mr α with α =ln γ/ ln 4 and M = ω(4r 0 )r -α 0 .

2.16

We have

where

When ε → 0, we have

On the other hand,

According to the dominated convergence theorem, we can pass to the limit when ε → 0 in (5.13). We obtain with w = (1x 2 1 ) µ-2 cosh(λ x 2 ). Under the notations

we have

Hence

)]a 11 w.

We obtain from the ellipticity condition

Then there exists µ * = µ * (ν) sufficiently large in such a way that R(µ * , λ , r) ≥ R(µ * , 0, r) ≥ 0, for any 0 ≤ r ≤ 1 2 and λ .

Furthermore, there exists λ * = λ * (ν) so that R(µ * , λ * , r) ≥ 0, for any 1 2 ≤ r ≤ 1.

We then choose µ = µ * and λ = λ * . We obtain Lv ≥ R(µ, λ , r)w ≥ 0 in (-1, 1)×R. (b) We prove that

(5.20) For all 0 < r ≤ R, there exists a non negative integer k so that R/2 < 2 k r ≤ R. If k = 0, we have R < 2r and (5.20) is obvious. If k ≥ 1, we obtain by iterating (5.19)

(5.21)

Consequently, r → e r κ H(r) is non decreasing, where κ = σ 1 /σ 0 . We obtain from this r 0

we end up getting

(ii) Assume that σ = 1. Using Green's formula, we get in a straightforward manner

(5.30)

(5.31) in (5.30) yields

(c) (i) Assume that β = 0. By formulas (4.88) and (4.89) and identity (4.90), we have

But, in light of (5.27), we have

We find by applying Cauchy-Schwarz's inequality

(5.34) This and (5.33) lead

and similarly D(r) ≤ σ 1 σ 0 dD(r).

(5.37)

We get from (5.35), (5.36) and (5.37)

that is to say (e κr N(r)) ≥ 0.

Consequently N(r) ≤ e κ(r-r) N(r) ≤ e κr N(r).

(ii) We first note that (5.34) remains true without any condition. On the other hand, D = 0 when σ = 1 and (5.37) is clearly satisfied if β ≥ 0. Therefore, we need only to estimate the terms Ĥ(r)/D(r) and D(r)/D(r).

Set

Observe that H(r) = 0 for any r ∈ (0, 1). Otherwise, we would have H(r) = 0 for some r ∈ (0, 1) and therefore u would be identically equal to zero by the unique continuation property. Hence N is continuous and I is an open subset of R. Consequently, I is a countable union of open intervals:

Appendix A

Building a fundamental solution by the parametrix method

We aim in this appendix to construct a fundamental solution of a general elliptic operators. We followed the paper by Kalf [1] where he constructed a fundamental solution using a method introduced by E. E. Levi [3]. This method consists in building a fundamental solution as a perturbation of the canonical parametrix, which is roughly speaking a fundamental solution corresponding to constant coefficients elliptic operator. The problem is then reduced to solve an integral equation with a weakly singular kernel. Proceeding in this way, the main difficulty is to guarantee an orthogonality relation appearing if one wants to use Fredholm's alternative. We overcome this difficulty by deforming the right hand side of the integral equation in order to comply with the orthogonality condition appearing is Fredholm's alternative.

The paper by Kalf [1] contains many historical comments and remarks starting from the founding paper by E. E. Levi. Some technical results we used in this appendix are borrowed from the books of R. Kress [2] and H. Triebel [4].

A.1 Functions defined by singular integrals

We start with the following technical lemma.

Lemma A.1. Let 0 ≤ α i < n, i = 0, 1. There exists a constant C > 0, depending on n, Ω , α 0 and α 1 , so that, for any x, y ∈ Ω with x = y, we have

Putting together all these inequalities, we get the expected result.

We use hereafter the notations

for some constants c > 0 and

belongs to C 0 (Σ ) and

and f can be extended by continuity in Ω × Ω when β 0 + β 1 > n.

(ii) Suppose in addition of the assumptions in (i) that f 0 (•, y) ∈ C 1 (Ω \ {y}) for all y ∈ Ω and

for some constants κ > 0 and β > 0. Then, for any y ∈ Ω , we have f (•, y) ∈ C 1 (Ω \ {y}) and

(iii) Assume additionally to the assumptions in (i) that f 0 satisfies the following estimate: there exists c > 0 and δ ∈]0, 1] so that

and µ 1 = min(δ , µ 0 ), then there exits C > 0 and so that

Doing the same with x 1 substituted by x 2 and noting that B(x 1 , 2d) ⊂ B(x 2 , 3d), we end up getting

(A.9)

The expected inequality follows by combining (A.7), (A.8) and (A.9).

A.2 Weakly singular integral operators

Let K : Ω × Ω → C be a measurable function. Consider the integral operator acting on L 2 (Ω ) as follows

The function K is usually called the kernel of the operator A.

Lemma A.2. Assume that K, the kernel of the operator A given by (A.10), have the property that for any f ∈ L 2 (Ω ) it holds Suppose furthermore that A ∈ B(L 2 (Ω )). Then the adjoint of A is given by

Proof. Let φ ∈ D(Ω ). In light of the assumptions on K, an application of Fubini's theorem allows us to get

Furthermore, there exists a constant C > 0 only depending on n, Ω , α 0 , α 1 , B 0 ∞ and B 1 ∞ so that, for any x, y ∈ Ω , we have

According to (A.1), for a.e. x ∈ Ω , the integral

is integrable in Ω × Ω with respect to (y, z). For such a point x ∈ Ω , it follows from Fubini's theorem

We complete the proof by using Lemma A.1 and noting that, when

Let K be as in the preceding proof and assume that

. This and estimate (A.15) show that K is the kernel of a weakly integral operator acting on C(Ω ).

A.3 Canonical parametrix

Let Ω be a bounded domain of R n of class C 2 . In that case, for any j + β < k + α ≤ 2 with 0 ≤ α, β ≤ 1 and positive integers j and k, we know that C j,β (Ω ) is continuously imbedded in C k,α (Ω ) (see [5, Lemma 6.35, page 135]).

Consider A = (a i j ) ∈ C(Ω , R n 2 ) satisfying the ellipticity condition

where µ ≥ 1 is a constant and (•|•) denotes the Euclidian scalar product of R n . Let b i , 1 ≤ i ≤ n, and c belong to C(Ω ), and consider the non-divergence form operator defined by

Here B = (b 1 , . . . , b n ).

Define d(x) = det(A). Then (A.16) yields in straightforward manner that

which in light of (A.16) satisfies

If A -1 = (a i j ) then clearly

Consider the function defined, for t > 0, by

Here

Define the function H, for x ∈ R n and y ∈ Ω with x = y, by

In light of (A.17) and (A.18), we have

if n = 2 and 0 < µ|x -y| < 1. Using (A.19), we get

from which we deduce

A straightforward computation yields

It follows from this identity that H(•, y) is the solution of the equation

Henceforth, Σ is as in Section A.1.

Lemma A.4. (i) There exists a constant C 0 , only depending on µ and n, so that

for some constants 0 < α ≤ 1 and Λ > 0. Then there exists a constant C 1 , only depending on n, µ and Λ , so that

Let us now give the precise definition of a parametrix and a fundamental solution. To this end, we assume from now on that, for some fixed 0 < α ≤ 1 and Λ > 0, the following assumptions fulfill.

.

Assume moreover that the ellipticity condition (A.16) holds. Under all these assumptions, we can compute the adjoint of L. We find

Here B = (B 1 , . . . , B n ).

and, for any ϕ ∈ D(Ω ), we have

A parametrix P satisfying, for any y ∈ Ω ,

is called a fundamental solution for L relative to Ω .

Proposition A.2. H(x, y) is a parametrix for L relative to Ω .

Proof. From the definition of H we easily see that H(•, y) ∈ C 2 (Ω \ {y}), for any y ∈ Ω . The fact that H(•, y) ∈ L 1 (Ω ) (resp. L x P(•, y) ∈ L 1 (Ω )), for any y ∈ Ω , is immediate from (A.20) (resp. Corollary A.1) and Lemma A.3. Fix y ∈ Ω and ε > 0 sufficiently small so that B(y, ε) Ω . Let ϕ ∈ D(Ω ). Starting from the identity, where u, v are arbitrary in

we get by applying Gauss's theorem

We will show in the next section that

which yields in a straightforward manner the expected identity.

The parametrix H(x, y) constructed in this section is usually called the canonical parametrix.

A.4 Fundamental solution

In this section, Ω is a bounded domain of R n of class C 2 and assume that assumptions (A.36) to (A.39) together with (A.16) hold.

Denote by H(x, y) the canonical parametrix constructed in the previous section and let K = L x H(x, y). Consider then A the weakly singular integral operator with the kernel K(x, y) acting on C(Ω ).

Introduce N = {φ ∈ N(I -A * ); supp(φ ) ⊂ Ω }.

Let P be the orthogonal projection on N and L = L |D(Ω ) .

Lemma A.7. We have that N = R(PL ).

For all ϕ ∈ D(Ω ) and ψ ∈ R(PL ) ⊥ , we have

In other words, ψ is a weak solution of L * ψ = 0. Since ψ has a compact support, according to Theorem 4.5, ψ is identically equal to zero. Therefore R(PL ) ⊥ = {0} and hence the result follows.

By Lemma A.7, if (φ 1 , . . . φ p ) is a fixed basis of N then φ j = PL ϕ j = Lϕ j , ϕ j ∈ D(Ω ), 1 ≤ j ≤ p. Set Then

where we used that Lϕ j = φ j .

In other words, the following orthogonality relation holds

Whence there exists d ∈ R m+1 , d = 0, so that ψ = ∑ m+1 =1 d ψ vanishes in Ω \ C . In particular, ψ ∈ N which is impossible since ψ is non identically equal to zero.

We fix in the sequel x 1 , . . . x m ∈ Ω \C satisfying (A.41). Therefore, for any y ∈ Ω , there exists a unique ( f 1 (y), . . .

Note that the second equality follows from 0 (x, y) = -p ∑ j=1 φ j (x)φ j (y).

By Cramer's method for solving linear systems, we can easily check that f k ∈ C(Ω ).

In light of (A.40), (A.42), using that N ⊕ N ⊥ = N(I -A * ) and φ (x ) = 0 for φ ∈ N , = 1, . . . , m, we get

Clearly L x R(x, y) = (x, y). (A.45)

By induction in j ≥ 1, define K 1 (x, y) = K(x, y) and

Then it is straightforward to check that K j is the kernel of A j . Moreover, we get by applying Theorem A.1 that K j ∈ C(Ω × Ω ) provided that j is sufficiently large. We fix then j so that K j ∈ C(Ω × Ω ) and we set

In light of the properties of the canonical parametrix, we deduce, once again from Theorem A.1, that f ∈ C(Ω × Ω ). Furthermore, for φ ∈ N(I -A * ), and bearing in mind that φ = (A * ) j φ , we have Using (A.23), (A.49) and the fact that L x R(x, y) = (x, y), we find, for any y ∈ Ω 0 ,

In light of Proposition A.2, the definition of R and the properties of H and G collected above, we can state the following ultimate result Theorem A.7. F is a fundamental solution of L relative to Ω 0 satisfying: for any β ∈ (0, α) if n = 2 and β = α if n ≥ 3, we find a constant C > 0, only depending on n, Ω , α and Λ , so that, for any x, y ∈ Ω 0 with x = y, we have

Index

Ascoli