C Labourie
email: labourie.camille@ucy.ac.cy

E Milakis
email: emilakis@ucy.ac.cy

The calibration method for the thermal insulation functional

Keywords: AMS Subject Classifications: 49K10, 35R35 Thermal Insulation, Calibration Method, Free Boundary Problems

We provide minimality criteria by construction of calibrations for functionals arising in the theory of Thermal Insulation.

Introduction

Calibrations for free-discontinuity problems

Free-discontinuity problems consists in minimizing the energy E(u, K) of a pair composed of a function u ∈ C 1 (R n \ K) and a set K of dimension n -1. The energy presents a competition between the Dirichlet energy of u in R n \ K and the surface energy of K. The set K is interpreted as an hypersurface (with possibly singularities) where u jumps between different values. This notion of pair composed of a function and of its discontinuity set is alternatively formalized by the space SBV (special functions with bounded variations). The model case of this kind of problem is the Mumford-Shah functional coming from image segmentation.

These problems generally present two kind of Euler-Lagrange equations. Considering small perturbations of u (with the discontinuity set K being fixed), one obtains that u satisfy a PDE with a boundary condition on each connected component of R n \K. As an example, minimizers of the Mumford-Shah functional are harmonic functions satisfying a Neumann boundary condition. Considering small perturbations of K under diffeormophisms, one obtains an equation that deals with the mean curvature of K. However, these equations do not entirely characterise the minimizers. We also point out that the minimizer may not be unique. We summarize these difficulties as a lack of convexity of the functional E.

In [START_REF] Alberti | The calibration method for the Mumford-Shah functional and free-discontinuity problems[END_REF], Alberti, Bouchitté, Dal Maso have introduced a sufficient condition for minimality by adapting the calibration method that was known for minimal hypersurfaces. Here is a simplified summary. Let us consider a competitor (u, K). We define the complete graph Γ u of u as the boundary of the subgraph of u. It is the reunion of the graph of u on R n \ K and of vertical sides above K (the discontinuities of u). We let ν Γu denote the normal vector to Γ u pointing into the subgraph of u. A calibration for (u, K) is a divergence-free vector field ϕ : R n × R → R n × R such that

E(u, K) = ˆΓu ϕ • ν Γu dH n-1 (1)
and for all competitor (v, L),

E(v, L) ≥ ˆΓv ϕ • ν Γv dH n-1 . (2)
One observes that the Gauss-Green theorem and the divergence-free property imply

ˆΓu ϕ • ν Γu dH n-1 = ˆΓv ϕ • ν Γv dH n-1 (3)
so the existence of such a vector field proves that (u, K) is a minimizer.

In [START_REF] Alberti | The calibration method for the Mumford-Shah functional and free-discontinuity problems[END_REF]Lemma 3.7], the authors provide four conditions to ensure the properties (1), [START_REF] Alt | Existence and regularity for a minimum problem with free boundary[END_REF] above and it is convenient to take these conditions as a definition of calibrations. Minimality criterias for the Mumford-Shah functional are proved by construction of calibrations in [START_REF] Alberti | The calibration method for the Mumford-Shah functional and free-discontinuity problems[END_REF], [START_REF] Maso | Local calibrations for minimizers of the Mumford-Shah functional with rectilinear discontinuity sets[END_REF], [START_REF] Mora | Local calibrations for minimizers of the Mumford-Shah functional with a regular discontinuity set[END_REF], [START_REF] Mora | Local calibrations for minimizers of the Mumford-Shah functional with a triple junction[END_REF], [START_REF] Morini | Global calibrations for the non-homogeneous Mumford-Shah functional[END_REF]. The principle of calibrations also inspired a fast primal-dual algorithm to minimize the Mumford-Shah functional ( [START_REF] Chambolle | Convex representation for lower semicontinuous envelopes of functionals in L 1[END_REF], [START_REF] Bischof | An algorithm for minimizing the Mumford-Shah functional[END_REF]).

In general, we don't know if calibrations exist for minimizers of this kind of problem. This question is related to the non-existence of a duality gap (see [START_REF] Alberti | The calibration method for the Mumford-Shah functional and free-discontinuity problems[END_REF]Section 3.13], [START_REF] Chambolle | Convex representation for lower semicontinuous envelopes of functionals in L 1[END_REF] and also [START_REF] Bouchitté | A duality theory for non-convex problems in the calculus of variations[END_REF]). There is no general recipe to follow and the construction can be very difficult. The crack-tip is famous example of minimizer of the Mumford-Shah functional for which a calibration has not been found.

The thermal insulation functional

A free boundary problem related to thermal insulation was recently studied by , [START_REF] Kriventsov | A free boundary problem related to thermal insulation: flat implies smooth[END_REF]) and also Bucur-Luckhaus-Giacomini ( [START_REF] Bucur | Monotonicity formula and regularity for general free discontinuity problems[END_REF], [START_REF] Bucur | Shape optimization problems with Robin conditions on the free boundary[END_REF]). Relaxing the problem in SBV , it consists in minimizing

E(u) = ˆ|∇u| 2 dx + β ˆJu (u -) 2 + (u + ) 2 dH n-1 + γ 2 L n ({ u > 0 }),
where β, γ > 0 and the competitors are functions u ∈ SBV (R n ) such that u = 1 on a given bounded open set Ω ⊂ R n and 0 ≤ u ≤ 1 on R n . Here J u is the set of all jump points of u, that is the points x for which there exist two real numbers u -< u + and a (unique) vector ν u (x) ∈ S n-1 such that lim r→0 Br∩H + u(y)u -dy = 0 (4a)

lim r→0 Br∩H - u(y) -u + dy = 0, (4b)
where

H + = { y ∈ R n | (y -x) • ν u (x) > 0 } (5a) H -= { y ∈ R n | (y -x) • ν u (x) < 0 } . ( 5b
)
The function u can be interpreted as the temperature (which is fixed to 1 on Ω) and J u as an isolating layer which has no width and no thermal conductivity.

Caffarelli-Kriventsov and Bucur-Luckhaus-Giacomini have shown the existence of minimizers in [START_REF] Caffarelli | A free boundary problem related to thermal insulation[END_REF] and [START_REF] Bucur | Monotonicity formula and regularity for general free discontinuity problems[END_REF], [START_REF] Bucur | Shape optimization problems with Robin conditions on the free boundary[END_REF]. They prove a non-degeneracy property: there exists 0 < δ < 1 (depending only on n and Ω) such that such that spt(u) ⊂ B(0, δ -1 ) and

u ∈ { 0 } ∪ [δ, 1] L n -a.e. on R n . ( 6
)
They also prove that the jump set J u is essentially closed, H n-1 (J u \J u ) = 0, and that it satisfies uniform density estimates. In [START_REF] Kriventsov | A free boundary problem related to thermal insulation: flat implies smooth[END_REF], it was proven that the jump set is locally the union of the graphs of two C 1,α functions provided that it is trapped between two planes which are sufficiently close. The approaches in these papers highlighted the similarities and the differences between the thermal insulation problem and the Mumford-Shah functional.

In particular, for the thermal insulation problem, one has to deal with an harmonic function satisfying a Robin boundary condition at the boundary rather than a Neumann boundary condition.

In [START_REF] Labourie | Higher integrability of the gradient for the Thermal Insulation problem[END_REF], we have shown the higher integrability of the gradient for minimizers of the thermal insulation problem, an analogue of De Giorgi's conjecture for the Mumford-Shah functional, and deduced that the singular part of the free boundary has Hausdorff dimension strictly less than n -1. Variants of this problem have been studied in [START_REF] Buttazzo | An optimization problem for thin insulating layers around a conducting medium[END_REF], [START_REF] Bucur | Two optimization problems in thermal insulation[END_REF], [START_REF] Bucur | Symmetry breaking for a problem in optimal insulation[END_REF], [START_REF] Bucur | Shape optimization of a thermal insulation problem[END_REF]. A numerical implementation has been proposed in [START_REF] Bogosel | Numerical implementation in 1D and 2D of a shape optimization problem with Robin boundary conditions[END_REF].

Summary of results

The purpose of this article is to provide minimality criteria for the thermal insulation functional by construction of calibrations. Besides minimality criteria, we are also simply interested in understanding what calibrations look like for this kind of problem. The article is divided into two parts. In the first part, we fix an open set A of R n and we consider the homogeneous functional

E 0 (u) = ˆA|∇u| 2 dx + β ˆJu∩A (u + ) 2 + (u -) 2 dH n-1 , (7)
where u ∈ SBV (A). In Theorem 2.1, we present a sufficient condition so that a non-negative harmonic function u :

A → R is a minimizer of E 0 among all competitors v ∈ SBV (A) with { v ̸ = u } ⊂⊂ A.
The condition is also necessary in dimension one but not in higher dimension. An anologous questions had been studied for the homogeneous Mumford-Shah functional

F 0 (u, A) = ˆA|∇u| 2 dx + βH n-1 (J u ) (8)
by Chambolle without calibrations ([22, Theorem 3.1(i)]) and Alberti, Bouchitté, Dal Maso with calibrations ([1, Section 4.6]).

In the second part, we come back to the full thermal insulation functional,

E(u) = ˆ|∇u| 2 dx + β ˆJu (u + ) 2 + (u -) 2 dH n-1 + γ 2 L n ({ u > 0 }) (9)
where

u ∈ SBV (R n ) is such that u = 1 on a given bounded open set Ω.
The section starts with an informal discussion about the case Ω = B(0, 1).

The relevant competitors should be either the indicator function 1 Ω or an harmonic function supported in a bigger ball. We can make explicit computations with these competitors to find minimality criterias. Then we prove and generalize these criterias to other domains Ω using calibrations. In Theorem 3.3 and 3.7, we prove two sufficient conditions which imply that 1 Ω is a minimizer. In Theorem 3.8, we come back back to the case Ω = B(0, 1) and prove a sufficient condition so that an harmonic function supported in a bigger ball is a minimizer.

Acknowledgement. We would like to thank Antoine Lemenant and Antonin Chambolle for helpful discussions about extensions of normal vector fields.

The homogenous functional

Statement of the problem and calibrations

We fix a parameter β > 0. Given a Borel set S ⊂ R n and a SBV function u in a neighborhood of S, we define

E 0 (u, S) = ˆS|∇u| 2 dx + β ˆJu∩S (u + ) 2 + (u -) 2 dH n-1 . (10)
Let V be an open set and let u : V → R be a a non-negative harmonic function. We are interested in finding a sufficient condition so that for all v ∈ SBV (V ) with { v ̸ = u } ⊂⊂ V , we have

E 0 (u, V ) ≤ E 0 (v, V ). (11)
It is equivalent to require that for all open set A ⊂⊂ V , for all v ∈ SBV (V )

with v = u in V \ A, we have E 0 (u, A) ≤ E 0 (v, A). ( 12
)
Theorem 2.1. Let A be a bounded open set of R n with Lipschitz boundary Let u : A → R be a non-negative harmonic function which has a C 1 extension in open set V containing A. We assume that there exists

0 ≤ m ≤ M such that m ≤ u ≤ M in A and (M -m) 2 -(M -δ) 2 ≤ β 0 (m 2 + δ 2 ), (13)
where

β 0 = β (M -m) sup A |∇u| and δ = M 1 + β 0 . ( 14
)
Then u is a minimizer of

E 0 (v, A) among all v ∈ SBV (V ) such that v = u in V \ A.
Remark 2.2. The condition ( 13) can be written

ˆδ m 2(M -t) dt ≤ β 0 (m 2 + δ 2 ), (15)
to highlight its similarity with the condition ´s r ϕ x (x, t) dt ≤ β(r 2 + s 2 ) in the Definition 2.6 below (the definition of calibrations). The function

s → β 0 (m 2 + s 2 ) - ˆs m 2(M -t) dt (16)
attains its minimum at s = M 1+β 0 so ( 15) is also equivalent to say that for all

s ∈ [0, M ], ˆs m 2(M -t) dt ≤ β 0 (m 2 + s 2 ). (17)
Remark 2.3. Let us consider the case where n = 1, A =]0, h[ (where h > 0) and u is an affine function whose graph joins (0, m) and (h, M ). Then the theorem condition is necessary. In this cases β 0 = βh, δ = M 1+βh and the condition (13) amounts to

h -1 (M -m) 2 ≤ β(m 2 + δ 2 ) + h -1 (M -δ) 2 . ( 18
)
One recognizes that the right-hand side is the Dirichlet energy of u. The left-hand side is the energy of the jump function whose graph joins (0, m), (0, δ) and (h, M ).

Remark 2.4. The condition ( 13) is trivially satisfied if m ≥ δ, which can be simplified to

m ≥ β -1 sup A |∇u|. (19)
Remark 2.5. For the homogeneous Mumford-Shah functional,

F 0 (u) = ˆA|∇u| 2 + βH n-1 (J u ), (20)
Chambolle found the condition

(M -m) sup A |∇u| ≤ β (see [22, Theorem 3.1 (i)]
). This is somewhat analogous to our condition because this can be rewritten

ˆM m 2(M -t) dt ≤ β 0 (21)
where

β 0 = β (M -m) sup A |∇u| .
We are going to state the notion of calibrations associated to our problem. The existence of such a vector field implies that u is a minimizer of E 0 (v, A)

among v ∈ SBV (V ) such that v = u in V \ A.
This is a consequence of [1, Section 2 and 3] but we provide a detailed explanation in Appendix B. We underline that our Dirichlet problem is different from the one of [1, Definition 3.1] because our boundary condition is one-sided (a competitor v might jump on ∂A). This is why we define ϕ up to ∂A × [0, M ]. Definition 2.6. Let A be a bounded open set of R n with Lipschitz boundary, let V be an open set containing A and let u ∈ SBV (V ) be such that

0 ≤ u ≤ M on A (for some M > 0). A calibration for u in A × [0, M ] is a Borel map ϕ = (ϕ x , ϕ t ) : A × [0, M ] → R n × R (22)
which is bounded and approximately regular on A × [0, M ], divergence-free on A×]0, M [ in the sense of distribution and such that (a) ϕ t (x, t) ≥ 1 4 |ϕ x (x, t)| 2 for L n -a.e. x ∈ A and every t ∈ [0, M ];

(b) ´s r ϕ x (x, t) dt ≤ β(r 2 +s 2 ) for H n-1 -a.e. x ∈ A and every r, s ∈ [0, M ];

(a') ϕ x (x, u) = 2∇u and ϕ t (x, u) = |∇u| 2 for L n -a.e. x ∈ A; (b') ´u+ u -ϕ x (x, t) dt = β (u -) 2 + (u + ) 2 ν u for H n-1 -a.e. x ∈ J u ∩ A.
With regard to Theorem 2.1, the candidate function u is smooth so we have J u = ∅ and we don't need to check (b'). We conclude this section with two miscellaneous remarks.

Remark 2.7 (Scaling). We detail the scaling properties of E 0 . We write E 0 (u, β, A) to explicit the parameters β in the definition of E 0 . For all M ∈ R, we have

E 0 (M u, β, A) = M 2 E 0 (u, β, A), (23)
and for all h > 0,

E 0 (u h , βh, h -1 A) = h 2-n E 0 (u, β, A). ( 24
)
where u h (x) = u(h • x). Thus, for all h > 0 and M ≥ 0,

E 0 (M -1 u h , βh, h -1 A) = M -2 h 2-n E 0 (u, β, A). ( 25
)
Remark 2.8 (Slope along a jump). This remark is an example of application of [1, Lemma 2.5] that we will use repeatedly in the next constructions. We consider three C 1 functions σ, u, v : R n → R. We work in R n × R and we introduce the hypersurface

H = { (x, t) ∈ R n × R | t = σ(x) } (26)
and the vector field

ϕ = (2∇u, |∇u| 2 ) for t > σ(x) (2∇v, |∇v| 2 ) for t < σ(x). (27)
A normal vector field to H is (-∇σ, 1) and we have along H,

ϕ(x, σ(x) + ) • -∇σ 1 = ϕ(x, σ(x) -) • -∇σ 1 . ( 28
)
if and only if ∇σ = 1 2 (∇u + ∇v). Therefore, ϕ is divergence-free in the sense of distributions provided that u, v are harmonic and σ = 1 2 (u + v) (modulo an additive constant).

The one dimensional case

A short analysis

We fix constants a < b and 0 ≤ m ≤ M . We consider the functional We are looking for the minimizer(s), depending on m and M . Here is a summary (without details). In the case m = M , the affine (constant) competitor is the only minimizer and from now on, we assume m < M . First, we try to find the jump competitors which have the least energy. It is never convenient to do more than one jump. At each side of the jump, the optimal value of the function corresponds to the Robin condition ∂ ν u = βu (where ν is the inward unit normal vector to the side). The optimal location of the jump may be x 0 = 0 or some x 0 ∈]0, 1[. It cannot be x 0 = 1 though and this comes from the fact that m < M . If m < M 1+β , the best location is x 0 = 0 and the jump is going from m to M 1+β . If m > M 1+β , the best location is a certain x 0 ∈]0, 1[ but then the affine competitor has necessarily a smaller energy. We conclude that the affine competitor is a minimizer if and only if it is better than the jump at x = 0, that is

E 0 (u) = ˆ|∇u| 2 dx + β ˆJu (u + ) 2 + (u -) 2 dH n-1 (29)
(M -m) 2 ≤ βm 2 + M - M 1 + β 2 . ( 30
)
We define δ = M 1+β and we observe that (30) is equivalent to

ˆδ m 2(M -t) dt ≤ β(m 2 + δ 2 ). (31)
Now, we try to guess the calibration in the limit case

ˆδ m 2(M -t) dt = β(m 2 + δ 2 ). (32)
Notice that we have necessarily m < δ. There are two minimizers; the first one is the affine function u whose graph joins the points (0, m), (1, M ) and the second one is the jump function v whose graph joins the points (0, m), (0, δ), (

= (2(M -δ), (M -δ) 2 ) on t = m + (M -δ)x. Finally, we should have ˆδ m ϕ x (0, t) dt = β(m 2 + δ 2 ). (33)
A simple solution is to set

ϕ = 2(M -t) 1 -x , M -t 1 -x 2 (34)
for t ≥ m + (1m)x. Next, we try to determine ϕ x (0, t) for t ∈ [0, m]. We have necessarily

ˆm 0 ϕ x (0, t) dt = ˆδ 0 ϕ x (0, t) dt - ˆδ m ϕ x (0, t) dt (35) ≤ βδ 2 -β(m 2 + δ 2 ) (36) ≤ -βm 2 . ( 37
)
We suggest to set ϕ x (0, t) = -2βt on [0, m]. Then we try to extend ϕ in a simple way while respecting the requirements of calibrations and we arrive at

ϕ =                2(M -t) 1-x , M -t 1-x 2 if m + τ x ≤ t ≤ M 2(M -m), [M -m] 2 if m + σx ≤ t ≤ m + τ x -2βm, β 2 m 2 if m ≤ t ≤ m + σx -2βt, β 2 m 2 if 0 ≤ t ≤ m. ( 38
)
where τ = Mm and σ = 1 2 ((Mm)βm). The slope σ has been chosen as in Remark 2.8. In the next section, we generalize this construction when there is only an inequality in (32).

0 1 0 m δ M φ x = 2(M -t) 1-x φ x = -2βt φ x = 2(M -m) φ x = -2βm x t u m m + σx v

The construction

Context. Let 0 ≤ m ≤ M and let u : [0, 1] → R be the affine function such that u(0) = m and u(1) = M . We assume that

ˆδ m 2(M -t) dt ≤ β(m 2 + δ 2 ), (39)
where δ = M 1+β . We build a calibration for u in [0, 1] × [0, M ]. The construction brings into play an intermediary constant 0

≤ λ ≤ β such that ˆδ m 2(M -t) dt ≤ λm 2 + βδ 2 . ( 40
)
We will also need the conditions λm ≤ Mm and λm ≤ βM 1+β . If m ≤ δ, these two conditions follows from the fact that λ ≤ β. Otherwise, it suffices to take λ = 0 for example. Finally, we define

ϕ =                2(M -t) 1-x , M -t 1-x 2 if m + τ x ≤ t ≤ M 2(M -m), [M -m] 2 if m + σx ≤ t ≤ m + τ x -2λm, λ 2 m 2 if m ≤ t ≤ m + σx -2λt, λ 2 m 2 if 0 ≤ t ≤ m. (41)
where τ = M -m and σ = 1 2 ((M -m)-λm). Notice that we have 0 ≤ σ ≤ τ because 0 ≤ λm ≤ Mm. The slope σ has been chosen as in Remark 2.8.

We prove that for all x and all r ≤ s,

ˆs r ϕ x (x, t) dt ≤ β(r 2 + s 2 ). (42)
We have globally

ϕ x ≥ -2λt ≥ -2βt (43)
so it suffices to control ´s r ϕ x dt from above. Let us fix t. We are going to see that the function

r → β(r 2 + s 2 ) - ˆs r ϕ x (x, t) dt (44) is non-decreasing on [0, M ]. If 0 ≤ r ≤ m, β(r 2 + s 2 ) - ˆs r ϕ x (x, t) dt = βr 2 -λr 2 + (. . .) (45)
where (. . .) does not depend on r. The right-hand side is non-decreasing on

[0, m] because λ ≤ β. If m ≤ r ≤ m + σx, β(r 2 + s 2 ) - ˆs r ϕ x (x, t) dt = βr 2 -2λmr + (. . .) (46)
where (. . .) does not depend on r. The right-hand side is non-decreasing on [m, m + σx] because λ ≤ β. Finally, the function in non-decreasing on [m + σx, M ] because ϕ x is non-negative on this interval. We conclude that it suffice to show that for all x and t,

ˆs 0 ϕ x (x, t) dt ≤ βs 2 . ( 47
)
This is trivial if s ≤ m so we assume s > m. There exists x 0 such that s = m + (Mm)x 0 . The function

x → ˆs r ϕ x (x, t) dt (48)
is non-decreasing on [x 0 , 1] because the part where ϕ x ≤ 0 gets bigger at the expense of the part where ϕ x ≥ 0. Now, it suffices to show that for all x and

m + (M -m)x ≤ s ≤ M , ˆs 0 ϕ x (x, t) dt ≤ βs 2 . ( 49
)
We apply the divergence theorem ([1, Lemma 2.4]) in the polygone delimited by the cycle (x, 0), (x, s), (1, M ), (0, 1), (0, 0). We get

ˆs 0 ϕ x (x, t) dt = - (M -s) 2 1 -x + (M -m) 2 -λm 2 + λ 2 m 2 x (50) so βs 2 - ˆs 0 ϕ x (x, t) dt = βs 2 + (M -s) 2 1 -x -λ 2 m 2 x + λm 2 -(M -m) 2 . (51) The function s → βs 2 + (M -s) 2 1-x is minimal for s = M 1+β(1-x)
and its the minimum value is βM 2 1+β(1-x) . We have therefore

βs 2 - ˆs 0 ϕ x (x, t) dt ≥ βM 2 1 + β(1 -x) -λ 2 m 2 x + λm 2 -(M -m) 2 . ( 52
)
The function x

→ βM 1+β(1-x) -λ 2 m 2 x is non-decreasing on [0, 1] because λm ≤ βM 1+β .
We are thus led to

βs 2 - ˆs 0 ϕ x (x, t) dt ≥ βM 2 1 + β + λm 2 -(M -m) 2 (53)
As βM 2 1+β = βδ 2 + (Mδ) 2 , the previous quantity can be rewritten

λm 2 + βδ 2 - ˆδ m 2(M -t) dt ( 54
)
and this is non-negative by assumption.

Proof of Theorem 2.1

Context. Let A be a bounded open set of R n with Lipschitz boundary. Let u : A → R be a non-negative harmonic function which has C 1 extension in a neighborhood of A. We assume that for some 0 ≤ m ≤ M , we have

m ≤ u ≤ M in A and ˆδ m 2(M -t) dt ≤ β 0 (m 2 + δ 2 ), (55)
where

β 0 = β (M -m) sup A |∇u| and δ = M 1 + β 0 . (56)
We build a calibration for

u in A × [0, M ].
The construction brings into play an intermediary constant 0

≤ λ ≤ β 0 such that ˆδ m 2(M -t) dt ≤ λm 2 + β 0 δ 2 . ( 57
)
We also need the conditions λm ≤ Mm and λm ≤ β 0 M 1+β 0 . If m ≤ δ, these two conditions follow from the fact that λ ≤ β 0 . Otherwise, it suffices to take λ = 0 for example. Finally, we define

ϕ = φ x • ∇u(x) M -m , φ t • ∇u(x) M -m 2 , (58)
where

φ =                2(M -m) M -t M -u , (M -m) 2 M -t M -u 2 if u(x) ≤ t ≤ M 2(M -m), [M -m] 2 if σ(x) ≤ t ≤ u(x) -2λm, λ 2 m 2 if m ≤ t ≤ σ(x) -2λt, λ 2 m 2 if 0 ≤ t ≤ m. (59)
and

σ(x) = m + 1 2 (u -m)(1 -λm M -m ) (60) = m + 1 2 u -m M -m (M -m -λm). (61)
As 0 ≤ λm ≤ Mm, we see that m ≤ σ(x) ≤ u(x). We prove that for all x and r ≤ s,

ˆs r φ x (x, t) dt ≤ β 0 (r 2 + s 2 ). (62)
This is a minor variant of the one-dimensional case. We have globally

φ x ≥ -2λt ≥ -2β 0 t (63)
so it suffices to control ´s r φ x dt from above. Let us fix t. We are going to see that the function

r → β 0 (r 2 + s 2 ) - ˆs r φ x (x, t) dt (64) is non-decreasing on [0, M ]. If 0 ≤ r ≤ m, β 0 (r 2 + s 2 ) - ˆs r φ x (x, t) dt = β 0 r 2 -λr 2 + (. . .) (65)
where (. . .) does not depend on r. The right-hand side is non-decreasing on

[0, m] because λ ≤ β 0 . If m ≤ r ≤ σ(x), β 0 (r 2 + s 2 ) - ˆs r φ x (x, t) dt = β 0 r 2 -2λmr + (. . .) (66)
where (. . .) does not depend on r. The right-hand side is non-decreasing on [m, σ(x)] because λ ≤ β 0 . Finally, the function in non-decreasing on [u(x), M ] because φ x is non-negative on this interval. We conclude that it suffice to show that for all x and s,

ˆs 0 φ x (x, t) dt ≤ β 0 s 2 . ( 67
)
This is trivial if s ≤ m so we assume s > m. There exists x 0 such that s = u(x 0 ). Since σ ≤ u, we have in particular σ(x 0 ) ≤ s. For x such that σ(x) ≤ s ≤ u(x), we compute

ˆs 0 φ x (x, t) dt = -2λm 2 -2λm(σ -m) + 2(M -m)(s -σ) (68) = -2λm -2λm(σ -m) + 2(M -m)(s -m) + 2(M -m)(m -σ) (69) = -2λm 2 -2(σ -m)(M -m + λm) + 2(M -m)(s -m). (70)
As u(x 0 ) ≤ u(x) and 0 ≤ λm ≤ Mm, we have σ(x 0 ) ≤ σ(x) and it follows that

ˆs 0 φ x (x, u) ≤ ˆs 0 φ x (x 0 , u) d. (71)
Now, it suffices to show that for all x and u(x

) ≤ s ≤ m, ˆs 0 φ x (x, t) dt ≤ β 0 s 2 . ( 72
)
We compute

ˆs 0 φ x (x, t) dt = -λm 2 -2λm(σ -m) + 2(M -m)(u -σ) + M -m M -u (M -u) 2 -(M -s) 2 . ( 73
)
By definition

σ -m = 1 2 u -m M -m (M -m -λm) (74) u -σ = 1 2 u -m M -m (M -m + λm) (75) so -2λm(σ -m) + 2(M -m)(u -σ) = u -m M -m ((M -m) 2 + λ 2 m 2 ). ( 76
)
We also write

(M -m)(M -u) = (M -m) 2 + (M -m)(m -u) (77) = (M -m) 2 -(M -m) 2 u -m M -m (78)
and we arrive at

ˆs 0 φ x (x, t) dt = -λm 2 + λ 2 m 2 u -m M -m + (M -m) 2 - M -m M -u (M -s) 2 . (79) Therefore β 0 s 2 - ˆs 0 φ x (x, t) dt = β 0 s 2 + M -m M -u (M -t) 2 -λ 2 m 2 u -m M -m + λm 2 -(M -m) 2 . ( 80
)
For a fixed x, the function s

→ β 0 s 2 + M -m M -u (M -s) 2 is minimal for s = M 1 + β 0 M -u M -m (81)
and its minimum value is

β 0 M 2 1 + β 0 M -u M -m . ( 82
)
We have therefore

β 0 s 2 - ˆs 0 φ x (x, t) dt ≥ β 0 M 2 1 + β 0 M -u M -m -λ 2 m 2 u -m M -m + 2λm 2 -(M -m) 2 . ( 83
)
The function

x → β 0 M 2 1 + β 0 M -x M -m -λ 2 m 2 x -m M -m (84)
is non decreasing on [m, M ] because λm ≤ β 0 M 1+β 0 . We are thus led to

β 0 s 2 - ˆs 0 φ x (x, t) dt ≥ β 0 M 2 1 + β 0 + λm 2 -(M -m) 2 (85)
As

β 0 M 2 1+β 0 = β 0 δ 2 + (M -δ) 2
, the right-hand side can be rewritten

λm 2 + β 0 δ 2 - ˆδ m 2(M -t) dt. ( 86
)
3 The full functional

Definition of the problem and calibrations

We fix a bounded open set Ω ⊂ R n with Lipschitz boundary. We fix parameters β > 0 and γ > 0.

Given u ∈ SBV (R n ) such that u = 1 on Ω and 0 ≤ u ≤ 1 in R n , we define E(u) = ˆ|∇u| 2 dx + β ˆJu (u + ) 2 + (u -) 2 dH n-1 + γ 2 L n ({ u > 0 }). ( 87
)
It is shown in [14, Theorem 4.2, Corollary 3.3] that at least one minimizer exists and that there exists δ > 0 (depending only on Ω, β, γ) such that Ω ⊂⊂ B(0, δ -1 ) and all minimizers have a compact support in B(0, δ -1 ). So we can work without loss of generality with the competitors that have a compact support in B(0, δ -1 ). Now, we state the notion of calibration associated to this problem. The existence of such a vector field implies that u is a minimizer of

E(v) among v ∈ SBV (R n ) such that v = 1 on Ω and 0 ≤ v ≤ 1 in R n (see Appendix B). We define ϕ on (R n \ Ω) × [0, 1] but it would be enough for ϕ to be defined on B(0, δ -1 ) \ Ω × [0, 1]. Definition 3.1. Let Ω be a bounded open set of R n with Lipschitz boundary, let u ∈ SBV (R n ) be such that u = 1 on Ω, 0 ≤ u ≤ 1 in R n and u has a compact support. A calibration for u is a Borel map ϕ = (ϕ x , ϕ t ) : (R n \ Ω) × R → R n × R (88)
which is bounded and approximately regular on

(R n \ Ω) × [0, 1], divergence- free on (R n \ Ω)×]0, 1[ in the sense of distribution and such that (a) ϕ t (x, t) ≥ 1 4 |ϕ x (x, t)| 2 -γ 2 1 { t>0 } (t) for L n -a.e. x ∈ R n \ Ω and every t ∈ [0, 1]; (b) ´s r ϕ x (x, t) dt ≤ β(r 2 + s 2 ) for H n-1 -a.e. x ∈ R n \ Ω and every r, s ∈ [0, 1]; (a') ϕ x (x, u) = 2∇u and ϕ t (x, u) = |∇u| 2 -γ 2 1 { t>0 } (u) for L n -a.e. x ∈ R n \ Ω; (b') ´u+ u -ϕ x (x, t) dt = β (u -) 2 + (u + ) 2 ν u for H n-1 -a.e. x ∈ J u .
We are interested in finding minimality criteria in the case Ω = B(0, 1) or Ω convex. Similar questions were studied by De Pauw and Smets for the Mumford-Shah functional ( [START_REF] Pauw | On explicit solutions for the problem of Mumford and Shah[END_REF]), without calibrations.

Informal computations with Ω = B(0, 1)

We present informal computations in the case Ω = B 1 (the open unit ball centred at the origin). We introduce for r > 0,

Γ(r) =      r -1 if n = 1 ln(r) if n = 2 1 n-2 (1 -r 2-n ) if n ≥ 3. (89)
Given x ∈ R n , we write r for |x|. Thus, x → Γ(r) is an harmonic positive function on R n \ B 1 which is 0 on ∂B 1 .

We assume without proof that the only relevant competitors u are of the following form. Either u is the indicator function of B 1 . Either there exists R > 1 and 0 < δ < 1 such that u is radial continuous on B R , u = 1 on B 1 , u = δ on ∂B R and u = 0 on R n \ B R . For a fixed R > 1, the first Euler-Lagrange equation says that u should be harmonic in B R \ B 1 and that δ = δ(R) should be determined by the Robin boundary condition

-∂ r u = βu on ∂B R . ( 90
)
We find

δ(R) = 1 1 + βR n-1 Γ(R) (91)
and

u(x) =      1 for |x| ≤ 1 1 -βδ(R)R n-1 Γ(x) for 1 ≤ |x| ≤ R 0 for |x| > R. (92)
We plot an example of δ and u on Figure 1.

1 R 0 δ(R) 1 r = |x| t u δ Figure 1: n = 2, β = 3 and R = 2
An integration by parts combined with the Dirichlet condition u = 1 on ∂B 1 and the Robin condition -∂ r u = βu on ∂B R shows that

ˆBR \B 1 |∇u| 2 dx + β ˆ∂B R u 2 dH n-1 = β ˆ∂B R u dH n-1 . ( 93
)
We conclude that the energy of u is

E = nω n βR n-1 δ(R) + ω n γ 2 R n . (94)
Now, we consider E as functions of R ∈ [1, +∞[ that we try to optimize. We observe that nω n R n-1 βδ(R) is the flux of the vector field x → βδ(r)e r through ∂B R . We compute

div(δ(r)e r ) = δ ′ (r) + (n -1) r δ(r) (95) = -β - (n -1) r δ(r) 2 (96) so E(R) = E(1) + ˆBR \B 1 γ 2 -β 2 - (n -1)β r δ(r) 2 dx, (97)
where r means |x|. This shows in particular that

E ′ (R) = nω n R n-1 γ 2 -β 2 - (n -1)β R δ(R) 2 . ( 98
)
The critical radii R > 1 are characterised by the equations

β 2 - β(n -1) R δ(R) 2 = γ 2 . ( 99
)
Remark 3.2. As expected, this coincides with the second Euler-Lagrange equation

|∇u| 2 + γ 2 + βu 2 (H -2β) = 0 (100)
where H = (n -1)R -1 is the mean scalar curvature of ∂B R with respect to -e r (see [START_REF] Ambrosio | Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs[END_REF]Definition 7.32], not to be confused with the arithmetic mean of the principal curvatures which is equal to R -1 ). A proof of this formula for C 1,α surfaces is presented in [START_REF] Kriventsov | A free boundary problem related to thermal insulation: flat implies smooth[END_REF]Theorem 15.1].

Depending on the parameters β, γ, the function R → E(R) may not be convex and the condition E ′ (R) = 0 may not suffice to characterize minimizers. See Figure 2.

Three sufficient conditions of minimality

An sufficient condition for 1 B(0,1) to be a minimizer is that the function r → E(r) is non-decreasing on [1, +∞[. This is equivalent to the fact that for all r ≥ 1,

β 2 - (n -1)β r δ(r) 2 ≤ γ 2 . ( 101
)
In particular, it suffices that β ≤ γ. In the two next theorems, we generalize the sufficient conditions β ≤ γ and (101) to other domains. We apply the divergence theorem ([1, Lemma 2.4]) in the domain X and with the function u = 1 F . For all bounded approximately regular vector field

ϕ on R n \ Ω with div(ϕ) ∈ L ∞ (R n \ Ω), we have ˆE\Ω div(ϕ) dx = ˆ∂ * E\Ω ϕ • ν E dH n-1 - ˆ∂Ω\∂ * E ϕ • ν dH n-1 . ( 102
)
We pick ϕ = ν and we get

ˆ∂ * E\∂Ω (ν • ν E ) dH n-1 = P (Ω) (103)
but since |ν| ≤ 1, we deduce that P (Ω) ≤ P (E).

Reciprocally, if Ω is outward miminizing, its outward unit normal vector field admits at least a weak extension ν ∈ L ∞ (R n \ Ω; R n ) such that |ν| ≤ 1 and which is divergence-free on R n \ Ω in the sense of distribution. Here, weak extension means that the normal component of ν on ∂Ω is defined in a functional sense (this notion is due to Anzellotti, [START_REF] Anzellotti | Pairings between measures and bounded functions and compensated compactness[END_REF]). For the interested reader, we provide a proof of existence of ν communicated by A. Chambolle in Appendix C.

This weak extension ν may not be approximately regular. If we use it to build the calibration ϕ in Section 3.2, ϕ may not be approximately regular either and its normal component may not be well-defined in the usual divergence theorem ([1, Lemma 2.4]). However, for vector fields with L ∞ distributional divergence (and in particular for divergence-free vectorfields), it is always possible to define the trace of the normal component in the sense of Anzellotti [START_REF] Anzellotti | Pairings between measures and bounded functions and compensated compactness[END_REF] and the divergence theorem also holds true for this definition. The notion of calibration may thus be adapted to non approximately regular vector fields. It would be more difficult to check that a vector field implies minimality because we would no longer have pointwise requirements such as (a), (b), (a'), (b') in Definition 3.1. But the calibration that we present in Section 3.2 is quite simple and the competitor is just an indicator function.

This hints that the condition β ≤ γ should be sufficient for the minimality of 1 Ω , provided that Ω is a bounded open set of Lipschitz boundary which is outward minimizing. The proof should be a matter of rewriting things in the formalism of Anzellotti but we do not try in this paper which is quite long already.

Before stating the second theorem, we generalize the previous function Γ to other domains Ω. The proof of the next lemma is postponed in Appendix A because this is not a new result. (ii) Γ = 0 on ∂Ω, (iii) For all x ∈ R n \ Ω, ∇Γ(x) ̸ = 0 (iv) Γ has a C 1 extension up to the boundary and for all x ∈ ∂Ω, there exists t > 0 such that ∇Γ(x) = tν(x), where ν(x) is the outward unit normal vector of Ω at x.

Theorem 3.7.
Let Ω be a non-empty C 2 star-shaped bounded open set of R n and let Γ be a function as in Lemma 3.6 (modulo a positive multiplicative constant). We define for x ∈ R n \ Ω,

ν(x) = ∇Γ |∇Γ| and δ(x) = 1 1 + β|∇Γ| -1 Γ (104)
and we assume that div(ν) is bounded. If for all x ∈ R n \ Ω,

β 2 -βdiv(ν) δ 2 ≤ γ 2 , ( 105
)
then 1 Ω has a calibration.

In the case Ω = B(0, 1), we have ν = e r and the condition (105) just amounts to (101). The proof of Theorem 3.7 is postponed in Section 3.3.

Finally, we come back to Ω = B(0, 1) and the notations of Section 3.1.1. In particular, we refer to (89) and (91) for the definition of Γ and δ. We are going to see that if β ≥ n - 1 2 , the Euler-Lagrange equations characterize minimizers. In view of the formula

E ′ (r) = nω n r n-1 γ 2 -β 2 - (n -1)β r δ(r) 2 , (106)
it suffices to show that the function

r → β 2 - (n -1)β r δ(r) 2 (107)
is decreasing on [1, +∞[. We write

β 2 - (n -1)β r δ(r) = β β - n -1 r (r n-1 δ(r)) 2 r 2n-2 . ( 108
) If β ≥ n -1, the function r → r n-1 δ(r) is decreasing on [1, +∞[ because (r n-1 δ) ′ = -r n-1 β 2 - β(n -1) r δ(r) 2 . ( 109
) If β ≥ n -1 2 , the function r → β -n-1 r 1 r 2n-2 is decreasing on [1, +∞[ because β - n -1 r 1 r 2n-2 ′ = -2 (n -1) r 2n-1 β - n -1 2 r (110)
This proves our claim. In the next theorem, we build a calibration corresponding to this criteria.

Theorem 3.8. Let Ω be the open ball B(0, 1). We assume that β ≥ n -1 2 and that there exists R ≥ 1 such that

β 2 - β(n -1) R δ(R) 2 = γ 2 , (111)
Then the function

u(x) =      1 for |x| ≤ 1 1 -βδ(R)R n-1 Γ(x) for 1 ≤ |x| ≤ R 0 for |x| > R, (112)
has a calibration.

The proof of Theorem 3.8 is postponed in Section 3.4.

Proof of Theorem 3.3

Let Ω be a C 1 bounded open set of R n (we will add more assumptions as we advance in the construction). We want to build a calibration for 1 Ω and we search for a continuous function ϕ

: (R n \ Ω) × [0, 1] → R n × R which is divergence-free in R n \ Ω and such that (a) ϕ t (x, t) ≥ 1 4 |ϕ x (x, t)| 2 -γ 2 1 { t>0 } (t) for all x ∈ R n \ Ω and t ∈ [0, 1]; (b) ´s r ϕ x (x, t) dt ≤ β(r 2 + s 2 ) for all x ∈ R n \ Ω and r, s ∈ [0, 1];
(a') ϕ x (x, 0) = 0 and ϕ t (x, 0) = 0 for all x ∈ R n \ Ω;

(b') ´1 0 ϕ x (x, t) dt = -βν(x) for all x ∈ ∂Ω, where ν is the outward unit normal vector field of Ω.

We look for ϕ x first and then we will derive ϕ t . The starting point is to define ϕ x on the jump part of the complete graph of 1 Ω in such way that t → ϕ x (x, t) is linear; we define for x ∈ ∂Ω and 0 ≤ t ≤ 1,

ϕ x (x, t) = -2βtν(x). (113)
Let us assume that ν has a continuous extension on R n \ Ω. We can extend the previous formula and define for all x ∈ R n \ Ω and 0 ≤ t ≤ 1,

ϕ x (x, t) = -2βtν(x). (114)
We have then for x ∈ R n \ Ω and for r, s

∈ [0, 1], ˆs r ϕ x dt ≤ β(r 2 + s 2 )|ν(x)| (115) because |ϕ x (x, t)| ≤ 2βt|ν(x)|.
The requirement (b) leads us to assume that |ν| ≤ 1 in R n \Ω. Next, we assume that ν has a L ∞ distributional divergence in R n \Ω. We have then div(ϕ) = 2βtdiv(ν)+∂ t ϕ t in the sense of distribution so the conditions div(ϕ) = 0 and ϕ t (x, 0) = 0 impose

ϕ t (x, t) = βt 2 div(ν). (116)
The last requirement

ϕ t ≥ 1 4 |ϕ x | 2 -γ 2 1 { t>0 } (t) amounts to γ 2 ≥ β 2 |ν| 2 -βdiv(ν). (117)
In view of |ν| ≤ 1, the simplest choice is to assume div(ν) = 0 and γ ≥ β.

Proof of Theorem 3.7

Let Ω be a C 1 bounded open set of R n (we will add more assumptions as we advance in the construction). We are more careful than in the previous section and we define ϕ in two pieces. However, we still arrange ϕ to be be globally continuous.

The principle is the same as before. We let ν denote the outward unit normal vector field of Ω. The starting point is to define ϕ x (x, t) for x ∈ ∂Ω and 0 ≤ t ≤ 1 by ϕ x (x, t) = -2βtν(x). We assume that ν has a continuous extension on R n \ Ω such that |ν| ≤ 1 and ν is C 1 on R n \ Ω. We also consider a continuous function δ :

R n \ Ω → [0, 1] which is C 1 on R n \ Ω and such that δ = 1 on ∂Ω and 0 < δ < 1 on R n \ Ω. We define for x ∈ R n \ Ω 0 1 φ x = -2βtν(x) φ x = -2(1 -t) βδ(x)ν(x) 1-δ(x) x t δ and 0 ≤ t ≤ δ(x), ϕ x = -2βtν (118)
ϕ t = 2 div(ν) (119)
and δ(x) < t ≤ 1,

ϕ x = -2(1 -t) βδν 1 -δ (120)
ϕ t = -(1 -t) 2 div βδν 1 -δ -C(x) (121)
where C(x) will be chosen so that ϕ is continuous on (R n \Ω)×[0, 1]. Observe that this is already the case for ϕ x . We find

C(x) = -(1 -δ) 2 div βδν 1 -δ -βδ 2 div(ν). (122)
However we compute

div δν 1 -δ = div(δν) -δ 2 div(ν) (1 -δ) 2 (123)
so this simplifies to C(x) = -βdiv(δν). With regard to approximate regularity, the definition of ϕ t on ∂Ω × [0, 1] does not matter. Indeed, let M be an hypersurface of R n × R. Then for H n -a.e. (x 0 , t 0 ) ∈ M ∩ (∂Ω × [0, 1]), the vector n 0 = (ν(x 0 ), 0) is a normal vector to M at x and ϕ • n 0 = ϕ x . In order for ϕ to be bounded, it suffices that div(ν) and div(δν) are bounded. Finally, the requirement

ϕ t ≥ 1 4 |ϕ x | 2 -γ 2 1 { t>0 } (t) amounts to β 2 |ν| 2 -βdiv(ν) δ 2 ≤ γ 2 (124a) -βdiv(δν) ≤ γ 2 . ( 124b
)
It is tempting to choose δ and ν in such a way that

β 2 |ν| 2 -βdiv(ν) δ 2 = -βdiv(δν). (125)
In that case, div(δν) is bounded provided that div(ν) is bounded. According to (123), the equality (125) is equivalent to

β|δν| 1 -δ 2 = -div βδν 1 -δ (126)
A natural solution is to assume that Ω is C 2 star-shaped, to consider the function Γ of Lemma 3.6 and to choose δ, ν in such a way that

βδν 1 -δ = ∇Γ Γ . ( 127
)
We suggest to define

ν(x) = ∇Γ |∇Γ| (128) and δ(x) = 1 1 + β|∇Γ| -1 Γ . ( 129
)
In conclusion, the conditions (124) simplify to

β 2 -βdiv(ν) δ 2 ≤ γ 2 . ( 130
)
3.4 Proof of Theorem 3.8

Let Ω be the open ball B(0, 1). We assume that β ≥ n -1 2 and that there exists R ≥ 1 such that

β 2 - β(n -1) r δ(r) 2 = γ 2 , (131)
We have seen just before the statement of 3.8 that when β ≥ n -1 2 , the function

r → β 2 - (n -1)β r δ(r) 2 (132)
is decreasing on [1, +∞[. Therefore, (131) implies that for all r ≥ R,

β 2 - β(n -1) r δ(r) 2 ≤ γ 2 , (133)
The case R = 1 has already been dealt with in Theorem 3.7 so we can assume R > 1.

We recall the notations. Given x ∈ R n , we write r for |x|. We define for r ≥ 1,

Γ(r) =      r -1 if n = 1 ln(r) if n = 2 1 n-2 (1 -r 2-n ) if n ≥ 3. ( 134
)
and

δ(r) = 1 1 + βr n-1 Γ(r) . ( 135
)
We define for x ∈ R n ,

u(x) =      1 for |x| ≤ 1 1 -βδ(R)R n-1 Γ(r) for 1 ≤ |x| ≤ R 0 for |x| > R. ( 136
)
For r ≥ 0, we write u(r) for the value of u on ∂B r . Thus, we consider u as a function of the real variable r ∈ [0, +∞[. Now, we list a few useful formulas. For 1 < |x| < R, we have

∇u = -βδ(R) R r n-1 e r (137)
and

∇u 1 -u = - βδ 1 -δ e r = - 1 r n-1 Γ(r) e r . ( 138
)
With a slight abuse of notations, we consider that ∇u is defined on 1 ≤ |x| ≤ R by (137). We observe that

div ∇u 1 -u = |∇u| 1 -u 2 . ( 139
)
or equivalently

-div βδe r 1 -δ = βδ 1 -δ 2 . ( 140
)
According to (123), the line ( 140) is also equivalent to

-βdiv(δe r ) = β 2 - β(n -1) r δ(r) 2 . ( 141
)
We are going to define the calibration. Although we define ϕ x and ϕ t in parallel, the relevant part is really ϕ x . The function ϕ t is derived as usual. We consider a continuous function ρ :

[1, R] → R which is C 1 on [1, R] and such that δ(R) ≤ ρ ≤ u. 1 R 0 δ(R) 1 φ x = 2(1 -t) ∇u 1-u φ x = -2βte r φ x = 2∇u φ x = -2βδ(R)e r φ x = -2(1 -t) βδe r 1-δ φ x = -2βte r r = |x| t u δ(R) ρ δ
We fix x such that 1 ≤ |x| ≤ R. We define for 0 ≤ t ≤ δ(R),

ϕ x = -2βte r ϕ t = (n -1)βt 2 r , (143)
for δ(R) ≤ t < ρ(r)

ϕ x = -2βδ(R)e r ( 144
)
ϕ t = 2(n -1)βδ(R)t r - (n -1)βδ(R) 2 r (145) for ρ(r) < t ≤ u(r) ϕ x (x, t) = 2∇u (146) = -2βδ(R) R r n-1 e r ( 147
)
ϕ t (x, t) = |∇u| 2 -γ 2 (148) = β 2 δ(R) 2 R r 2n-2 -γ 2 (149)
and for u(r) ≤ t ≤ 1,

ϕ x (x, t) = 2(1 -t) ∇u 1 -u (150) = -2(1 -t) βδe r 1 -δ (151) ϕ t (x, t) = (1 -t) 2 |∇u| 1 -u 2 -γ 2 (152) = (1 -t) 2 βδ 1 -δ 2 -γ 2 (153)
Next, we fix x such that |x| ≥ R and we use the same formula as in Section 3.3. We define for t ≤ δ(r),

ϕ x = -2βte r ( 154
)
ϕ t = (n -1)βt 2 r ( 155
)
and for t ≥ δ(r),

ϕ x (x, t) = -2(1 -t) βδe r 1 -δ (156) ϕ t (x, t) = (1 -t) 2 βδ 1 -δ 2 -β 2 - β(n -1) r δ 2 (157)
When n = 1, the function ρ is given by the constant δ(R). When n ≥ 2, it is given by the following complicated formula that the reader can ignore for the moment,

ρ(r) = δ(R) 2 + βδ(R)r 2 R r 2n-2 Γ R r R r n-1 -1 - δ(R)r 2n β - n -1 R R r n -1 R r n-1 -1 . (158)
Regardless of ρ, many properties can be checked. The vector field ϕ is bounded. It is continuous outside the graph

{ (x, t) | t = ρ(r), 1 ≤ r < R } . ( 159
)
We point out that it is continuous through |x| = R because

γ 2 = β 2 - β(n -1) R δ(R) 2 (160)
and because for |x| = R, ∇u(x) = -βδ(R)e r . The function ϕ is divergence free in the interior of each part. In order for ϕ to be divergence-free in R n \Ω (in the sense of distributions), we have to choose ρ in such a way that for all

1 < |x| < R ϕ(x, ρ(r) -) • -ρ ′ (r)e r 1 = ϕ(x, ρ(r) + ) • -ρ ′ (r)e r 1 . (161)
This will imply that ϕ is approximately regular on (R n \ Ω) × [0, 1] by [1, Remark 2.6]. Next, we are going to deal with the the approximately regularity of ϕ on on ∂Ω × [0, 1] as in the previous section. Let M be an hypersurface of R n × R. For H n -a.e. (x 0 , t 0 ) ∈ M ∩ (∂Ω × [0, 1]), we have t 0 ̸ = ρ(x 0 ) and the vector n 0 = (ν(x 0 ), 0) is a normal vector to M at x. Thus ϕ • n 0 = ϕ x and we conclude by continuity of ϕ x in an neighborhood of (x 0 , t 0 ).

We check the values of ϕ(x, u(x)). It is clear from the construction that for 1 < |x| < R,

ϕ x (x, u) = 2∇u
(162)

ϕ t (x, u) = |∇u| 2 -γ 2 , ( 163
) that for |x| > R, ϕ x (x, u) = 0 = 2∇u (164) ϕ t (x, u) = 0 = |∇u| 2 (165)
and that for |x| = R,

ˆδ(R) 0 ϕ x (x, t) dt = βδ(R) 2 . ( 166
)
The requirement

ϕ t ≥ 1 4 |ϕ x | 2 -γ 2 1 { t>0 } (t) holds true because for every r ≥ R, γ 2 ≥ β 2 - β(n -1) r δ(r) 2 . ( 167
)
It is left to compute ρ and to check that ´t s ϕ x ≤ β(s 2 + t 2 ). From now on, the proof is quite computational and the reader is free to skip.

We recall that we are looking for a continuous function ρ :

[1, R] → [0, 1] which is C 1 on [1, R[ and such that δ(R) ≤ ρ ≤ u and such that for 1 < |x| < R, ϕ(x, ρ(r) -) • -ρ ′ (r)e r 1 = ϕ(x, ρ(r) + ) • -ρ ′ (r)e r 1 . (168)
When n = 1, we can take ρ equals to the constant δ(R). In this case, one can check that ϕ is continuous on (R n \ Ω) × [0, 1] and |ϕ x | ≤ 2βt so (168) and ´s r ϕ x ≤ β(r 2 + s 2 ) hold true. We pass to the case n ≥ 2. It will be convenient to express (168) in divergence form. We compute

ϕ(x, ρ -) • -ρ ′ e r 1 = 2βδ(R)ρ ′ + 2(n -1)βδ(R)ρ r - (n -1)βδ(R) 2 r (169) = 2βδ(R)div(ρe r ) -βδ(R) 2 div(e r ). (170)
Next, we compute

ϕ(x, ρ + ) • -ρ ′ e r 1 = 2βδ(R) R r n-1 ρ ′ + β 2 δ(R) 2 R r 2n-2 -γ 2 (171) = 2βδ(R)R n-1 div r 1-n ρe r + β 2 δ(R) 2 R 2n-2 div r 1-n Γ(r)e r -γ 2 (172)
and we observe that we can write

γ 2 = div γ 2 r n e r + c r n-1 e r ( 173
)
where c is any real constant. A natural solution is to choose ρ such that

2βδ(R)ρ -βδ(R) 2 = 2βδ(R) R r n-1 ρ + β 2 δ(R) 2 R n-1 R r n-1 Γ(r) - γ 2 r n + c r n-1 . (174)
We rewrite this

2βδ(R) R r n-1 -1 ρ = -βδ(R) 2 -β 2 δ(R) 2 R n-1 R r n-1 Γ(r) + γ 2 r n - c r n-1 . (175)
We choose c in such a way that the left-hand side cancels at r = R (otherwise ρ would have a singularity at r = R). This yields

2βδ(R) R r n-1 -1 ρ = βδ(R) 2 R r n-1 -1 + β 2 δ(R) 2 R n-1 R r n-1 (Γ(R) -Γ(r)) - γ 2 r n R r n -1 . (176) Remember that Γ(R) -Γ(r) = r 2-n Γ R r so 2βδ(R) R r n-1 -1 ρ = βδ(R) 2 R r n-1 -1 + β 2 δ(R) 2 r R r 2n-2 Γ R r - γ 2 r n R r n -1 . (177)
We conclude that

ρ(r) = δ(R) 2 + βδ(R)r 2 R r 2n-2 Γ R r R r n-1 -1 - γ 2 r 2nβδ(R) R r n -1 R r n-1 -1 . (178) As γ 2 = β 2 -β(n-1) R δ(R) 2 , an alternative expression is ρ(r) = δ(R) 2 + βδ(R)r 2 R r 2n-2 Γ R r R r n-1 -1 - δ(R)r 2n β - n -1 R R r n -1 R r n-1 -1 . (179)
It is clear that ρ is a continuous function of r ∈]1, R[ and we also have

lim r→R ρ(r) = δ(R) because lim t→1 + Γ(t) t n-1 -1 = 1 n -1 (180)
and lim

t→1 + t n -1 t n-1 -1 = n n -1 . (181)
We show that ρ ≥ δ(R). It suffices to show that ρ is decreasing on [1, R[. For 0 ≤ r < R, we write t = R r so that

ρ(r) = δ(R) 2 + βδ(R)Rt n-2 2 t n-1 Γ(t) t n-1 -1 - δ(R) 2n β - n -1 R t -1 t n -1 t n-1 -1 . (182)
According to Lemma A.2, the function t → t n-1 Γ(t)

t n-1 -1 is increasing on ]1, +∞[. It is also easy to see that the function t -1 t n -1 t n-1 -1 is decreas- ing on ]1, +∞[ because t -1 t n -1 t n-1 -1 = 1 + t -1 t -1 t n-1 -1 (183)
and t → t n-1 is convex. We deduce that r → ρ(r) is decreasing on [1, R[. Next, we show that for all 1 ≤ r < R, we have ρ(r) ≤ u(r). Remember that

u = δ(R) 1 + βr R r n-1 Γ R r (184)
so we have to show that for all

1 ≤ r < R, βr R r 2n-2 Γ R r R r n-1 -1 - r n β - n -1 R R r n -1 R r n-1 -1 ≤ 1 + 2βr R r n-1 Γ R r (185)
We rewrite this,

βr R r n-1 Γ R r R r n-1 -1 2 -R r n-1 - βr n R r n -1 R r n-1 -1 ≤ 1 - n -1 n r R R r n -1 R r n-1 -1 . (186)
It suffices to check that for all t > 1,

β t n-1 Γ(t) t n-1 -1 2 -t n-1 - β n t n -1 t n-1 -1 ≤ 0 (187) 1 - n -1 n t -1 t n -1 t n-1 -1 ≥ 0. (188)
The first point can be simplified as

t n-1 Γ(t) t n -1 2 -t n-1 ≤ 1 n (189)
and this follows from Lemma A.2. To prove the second point, we observe that t → t -1 t n -1 t n-1 -1 is decreasing and that its limit when t → 1 is n n-1 . We finally prove that for all |x| ≥ 1 and for all r, s ∈ [0, 1], we have ´s r ϕ x (x, t) dt ≤ β(r 2 + s 2 ). Since ϕ x = -|ϕ x |e r , it amounts to show that for all |x| ≥ 1 and for all s ∈ [0, 1], we have

ˆs 0 |ϕ x (x, t)| dt ≤ βs 2 . ( 190
) If |x| ≥ R, inequality (
|ϕ x (x, t)| ≤ 2|∇u(x)| = 2βδ(R) R r n-1 (191)
whence for s ∈ [ρ(r), 1],

βs 2 - ˆs 0 |ϕ x (x, t)| dt ≥ βs 2 -βδ(R) 2 -2βδ(R)(ρ(r) -δ(R)) -2βδ(R) R r n-1 (s -ρ(r)). (192)
The right hand side function attains its minimum over s ∈ R at s = δ(R) R r n-1 and its corresponding value is

2βδ(R)ρ R r n-1 -1 -βδ(R) 2 R r 2n-2 -1 . (193)
It is non-negative if and only if

ρ(r) ≥ δ(R) 2 R r n-1 + 1 . (194)
and given the formula (179) of ρ, this means

βr R r 2n-2 Γ R r R r n-1 -1 - r n β - n -1 R R r n -1 R r n-1 -1 ≥ R r n-1 . (195)
We rewrite this,

βr R r 2n-2 Γ R r R r n -1 - 1 n ≥ R r n-1 R r n-1 -1 R r n -1 - n -1 n r R . (196)
We see first that the left-hand side is non-negative. Indeed, for all t ≥ 1,

Γ(t) ≥ 1 n -1 t n-1 -1 t n-1 ≥ 1 n t n -1 t n , (197)
where the first inequality comes from Lemma A.2 and the second one comes from the fact that whenever α > β > 0 and t ≥ 1,

1 β t α-β (t β -1) ≥ 1 α (t α -1). (198)
Then, (196) is equivalent to the fact that for all t > 1,

β t 2n-2 Γ(t) t n -1 - 1 n ≥ t n-1 t n-1 -1 t n -1 - n -1 n t -1 (199)
The inequality holds true if β ≥ n -1 2 by the last point of Lemma A.2. In fact, it is necessary for β to be greater than or equal to n- 1 2 because dividing the left-hand side by t -1 and taking the limit t → 1 + yields the value n-1 n β whereas the same operation at the right-hand side yields n-1 n n -1 2 . We leave the details to the interested reader.

Appendices

A The function Γ

We recall and prove Lemma 3.6.

Lemma A.1. Let Ω be a non-empty C 2 star-shaped bounded open set of R n . There exists a continuous function Γ :

R n \ Ω → R such that (i) Γ is harmonic positive on R n \ Ω; (ii) Γ = 0 on ∂Ω, (iii) For all x ∈ R n \ Ω, ∇Γ(x) ̸ = 0
(iv) Γ has a C 1 extension up to the boundary and for all x ∈ ∂Ω, there exists t > 0 such that ∇Γ(x) = tν(x), where ν(x) is the outward unit normal vector of Ω at x.

Proof. Without loss of generality, we assume that 0 ∈ Ω and that Ω is starshaped with respect to 0. We detail the case n ≥ 3. According to [21, Section 3A, Theorem 3.40], there exists a unique function

p ∈ C(R n \ Ω) such that (i) p is harmonic on R n \ Ω, (ii)
p is harmonic at infinity, (iii) p = 1 on ∂Ω.

We refer to [21, Proposition 2.74] for the characterisations of functions which are harmonic at infinity. Now we define Γ(x) = 1p(x) and we review the properties of the lemma. It is clear that Γ is harmonic on R n \ Ω and Γ = 0 on ∂Ω. As p is harmonic at infinity, we have lim x→∞ p(x) = 0 and thus lim x→+∞ Γ(x) = 1. We can then apply the maximum principle to see that Γ > 0 on R n \ Ω. The function Γ has a C 1 extension up to the boundary thanks to the usual regularity results for Dirichlet problems. As Γ is constant on ∂Ω, its tangential derivative is 0 along ∂Ω. And according to the Hopf Lemma, the normal derivative (with respect to the outward normal vector) is > 0 along ∂Ω. This proves that for all x ∈ ∂Ω, there exists t > 0 such that ∇Γ(x) = tν(x). The fact that ∇Γ never vanishes comes from the fact that Ω is star-shaped. Indeed, the function w : x → x • ∇Γ(x) is harmonic on R n \ Ω and ≥ 0 on ∂Ω. In addition, we see that lim x→+∞ w(x) = 0 by applying [START_REF] Folland | Introduction to partial differential equations[END_REF]Proposition 2.75] to the function p. We can use the maximum principle to conclude that w > 0 on R n \ Ω In the case n = 2, we define p as the unique function p ∈ C(R n \ Ω) such that (i) p is harmonic on R n \ Ω, (ii) p is harmonic at infinity, (iii) p(x) = ln(|x|) on ∂Ω.

Then we define Γ = ln(|x|)p. As p is harmonic at infinity, it is bounded at infinity and thus lim x→+∞ Γ(x) = +∞. The rest of the proof is the same except that lim x→+∞ x • ∇Γ(x) = 1. The case n = 1 is trivial.

In the case of the unit ball Ω = B(0, 1), the function Γ is given by

Γ(r) =      r -1 if n = 1 ln(r) if n = 2 1 n-2 (1 -r 2-n ) if n ≥ 3. (200)
We isolate a few useful estimates about this function in the following lemma. The reader is free to skip the proof.

Lemma A.2. Let n be an integer ≥ 2 and let Γ be defined as in (200).

(i) For all s, t ≥ 1, Γ(t) -Γ(s) = s 2-n Γ t s .

(ii) The function

t → t n-1 Γ(t) t n-1 -1 (201) is increasing on ]1, +∞[. (iii) For all t ≥ 1, 1 n -1 t n-1 -1 t n-1 ≤ Γ(t) ≤ 1 n t n -1 t n-1 . ( 202
) (iv) For all t > 1, n -1 2 t 2n-2 Γ(t) t n -1 - 1 n ≥ t n-1 t n-1 -1 t n -1 - n -1 n t -1 (203)
Proof. The first point is easy and we pass directly to the second one. If n = 2, the function

t → t n-1 Γ(t) t n-1 -1 = t ln(t) t -1 (204)
The condition (215) holds true when we replace Γ(t) by the lower bound

Γ(t) ≥ 1 n -1 t n-1 -1 t n-1 . (216)
Indeed, the new inequality simplifies to the fact that for all t ≥ 1,

1 n -1 t n-1 (t n-1 -1) ≥ 1 n (t n -1) (217)
and this comes from the fact that whenever α > β > 0 and t ≥ 1,

1 β t α-β (t β -1) ≥ 1 α (t α -1). (218)

B Calibrations

The goal of this section is to recall the main results and definitions of [START_REF] Alberti | The calibration method for the Mumford-Shah functional and free-discontinuity problems[END_REF] which explain why the existence of a vectorfield as in Definitions 2.6 and 3.1 implies the minimality of u.

Let X be an open set of R n and let u ∈ SBV (X). We let E u denote the

subgraph of u E u = { (x, t) ∈ X × R | t ≤ u(x) } . ( 219
)
It has a locally finite perimeter in X × R. We define the complete graph of u, written Γ u , as the measure-theoretic boundary of E u . According to the usual structure theorem, D1 Eu = ν Γu H n-1 Γ u where ν Γu ∈ R n × R is the measure-theoretic inward normal to E u . This notation should not be confused with ν u ∈ R n which is defined on J u . We will rely on [1, Lemma 2.10, Lemma 2.12], summarized in Lemma B.1 below. We refer to [START_REF] Alberti | The calibration method for the Mumford-Shah functional and free-discontinuity problems[END_REF]Definition 2.1] for the definition of an approximately regular vector field. The reader can think of it as a piecewise continuous vector field ϕ whose normal component does not jump through the discontinuity, except on a H n-1 -negligible set. The point of this property is to give a pointwise meaning (except on a H n-1 -negligible set) to the "normal component of ϕ" in the divergence theorem. We are more specifically interested in bounded approximately regular vector fields which are divergence-free in the sense of distribution. The main example is a piecewise smooth vector field which is divergence-free on each separate piece and whose normal component does not jump through the discontinuity. See [1, Lemma 2.6 and Remark 2.8] for a more formal statement.

Lemma B.1. Let X be an open set of R n , let u ∈ SBV (X).

For any Borel map

ϕ : X × R → R n × R, we have ˆX×R ϕ • Du = ˆX ϕ x (x, u) • ∇u(x) -ϕ t (x, u) dx + ˆJu ˆu+ u - ϕ x (x, t) • ν u (x) dt dH n-1 (x), (220)
provided that the right-hand side is well-defined (it suffices that ϕ is bounded and x → ϕ t (x, u(x)) has a bounded support).

2. We assume that X is a bounded open set of Lipschitz boundary and we let ν X denote its inner normal vector field (defined H n-1 -a.e. on ∂X).

Let m ≤ M be two real numbers, let ϕ :

X × [m, M ] → R n × R
, v ∈ SBV (X) such that m ≤ u, v ≤ M , ˆΓu ϕ • ν Γu dH n-1 - ˆΓv ϕ • ν Ev dH n-1 = ˆ∂X ˆv * u * ϕ x (x, t) • ν X (x) dt dH n-1 (x), (221)
where u * , v * are the traces of u and v on ∂X.

The general principle of calibrations for free-discontinuity problems is as follow. Let us say that we minimize an energy E(u) over functions u ∈ SBV (X). Given a candidate minimizer u ∈ SBV (X), we expect a calibration for u to satisfy the following properties: for all competitor v of u, we have

ˆΓv ϕ • ν Γv dH n-1 = ˆΓu ϕ • ν Γu dH n-1 (222a) and ˆΓv ϕ • ν Γv dH n-1 ≤ E(v) (222b)
with equality when v = u. This clearly implies the minimality of u.

We consider A, V , u, M , ϕ as in Definition 2.6. We extend ϕ by ϕ(x, t) = 0 outside A × [m, M ]. The extension may not be approximately regular and divergence-free outside A × [m, M ] but it does not matter. We check the two conditions (222) which in our case means X = V and for all competitor

v ∈ SBV (V ) such that 0 ≤ v ≤ M and v = u in V \ A, we have ˆΓv∩A ϕ • ν Γv dH n-1 = ˆΓu∩A ϕ • ν Γu dH n-1 (223a) and ˆΓv∩A ϕ • ν Γv dH n-1 ≤ E 0 (v, A), (223b)
with equality when v = u. Let us start with (223b). We apply the first item of Lemma B.1 with X = V . We have thus for all v ∈ SBV (V ) such that 0

≤ v ≤ M , ˆΓv∩A ϕ • ν Γv dH n-1 = ˆAϕ x (x, v) • ∇v(x) -ϕ t (x, v) dx + ˆJv∩A ˆv+ v - ϕ x (x, t) • ν v (x) dt dH n-1 (x). ( 224
)
The requirements (a) and (b) allow to bound the right-hand side of (224) and to obtain

ˆΓv∩A ϕ • ν Γv dH n-1 ≤ ˆA|∇v| 2 dx + β ˆJv∩A (v + ) 2 + (v -) 2 dH n-1 , (225)
with equality for v = u. We have proved (223b).

Remark B.2. One can also see, using the requirements (a) and (b), that there is equality in (225) if and only if

(a') ϕ x (x, v) = 2∇v and ϕ t (x, v) = |∇v| 2 for L n -a.e. x ∈ A, (b') ´v+ v -ϕ x (x, t) dt = β (v -) 2 + (v + ) 2 ν v for H n-1 -a.e. x ∈ J v ∩ A.
We deduce that (a'), (b') holds true not only for v = u but for all minimizers, that is for all competitors v such that E 0 (v, A) = E 0 (u, A).

We pass to (223a). We apply the second item of Lemma B.1 with

X = A. For all v ∈ SBV (V ) such that 0 ≤ v ≤ M , we have ˆΓu∩A ϕ • ν Γu dH n-1 = ˆΓv∩A ϕ • ν Γv dH n-1 + ˆ∂A ˆv * u * ϕ x (x, t) • ν A (x) dt dH n-1 (x), (226)
where u * and v * are the traces of u| A and v| A on ∂A. Using the fact that A has a Lipschitz boundary and assuming v = u in V \ A, we are going to show that ˆv *

u * ϕ x (x, t) • ν A (x) dt = ˆv+ v - ϕ x (x, t) • ν v (x) dt - ˆu- u + ϕ x (x, t) • ν u dt. ( 227
)
For H n-1 -a.e. x ∈ A, the function v has a trace at each side of ∂A; the inner trace v * that we have already defined and an outer trace, that we shall write v * . Moreover, for H n-1 -a.e. x ∈ ∂A, we have the following alternative. Either x is a Lebesgue point of v and v

+ (x) = v -(x) = v * (x) = v * (x), or x is a jump point of A with ν v (x) = ±ν A (x), v + (x) = max(v * (x), v * (x)) and v -= min(v * (x), v * (x)
). The vector ν v points toward the higher trace of v and ν A is an inner normal to A so the case ν

v (x) = ν A (x) corresponds to the matching v + (x) = v * (x), v -(x) = v * (x) and the case ν v (x) = -ν A (x) corresponds to the matching v + (x) = v * (x), v -(x) = v * (x). This proves that ˆv * v * ϕ x (x, t) • ν A (x) dt = ˆv+ v - ϕ x (x, t) • ν v (x) dt. (228)
We have similarly, ˆu *

u * ϕ x (x, t) • ν A (x) dt = ˆu+ u - ϕ x (x, t) • ν u (x) dt. (229)
And since u = v in V \ A, we have u * (x) = v * (x) for H n-1 -a.e. x ∈ ∂A. Now, it suffices to substract (229) from ( 228) to obtain (227). The equality (226) can be rewritten

ˆΓu∩A ϕ • ν Γu dH n-1 + ˆJu∩∂A ˆu+ u - ϕ x (x, t) • ν u (x) dt dH n-1 (x) = ˆΓv∩A ϕ • ν Γv dH n-1 + ˆJv∩∂A ˆv+ v - ϕ x (x, t) • ν v (x) dt dH n-1 (x) (230)
Apply ( 220) with X = A to develop ´Γu∩A ϕ•ν Γu dH n-1 . In view of (224), we recognize that the left-hand side of ( 230) is ´Γu∩A ϕ • ν Γu dH n-1 . We reason similarly with the right-hand side and we have proved (223a). We would proceed in the same way with Definition 3.1. We recall that there exists δ > 0 (depending only on Ω, β, γ) such that Ω ⊂⊂ B(0, δ -1 ) and all relevant competitors of (87) have a compact support included in B(0, δ -1 ). We would apply the items of Lemma B.1 with X = B(0, δ -1 ) and X = B(0, δ -1 ) \ Ω.

C Extension of a unit normal vector field

Let X be an open set of R n , let Ω ⊂⊂ X be an open subset of X with Lipschitz boundary. We let ν denote the outward unit normal vector field of Ω (defined H n-1 -a.e. on ∂Ω), and we define A = X \ Ω.

We recall the functional definition of the "normal component on ∂Ω" introduced by Anzellotti ( [START_REF] Anzellotti | Pairings between measures and bounded functions and compensated compactness[END_REF]). Let z ∈ L ∞ (A; R n ) be a vector field such that div(z) is a finite Radon measure in A. According to [4, Theorem 1.2], there exists a unique function

[z • ν] ∈ L ∞ (∂Ω) such that for all φ ∈ C 1 c (X), ˆAφ d[div(z)] -ˆAz • ∇φ dx = - ˆ∂Ω [z • ν]φ dH n-1 . (231) Moreover, we have ∥[z, •ν]∥ L ∞ (∂Ω) ≤ ∥z∥ L ∞ .
If z was a smooth vectorfield, the function [z • ν] would coincide with the scalar product z • ν on ∂Ω.

In the case div(z) = 0 in A, we can give another interpretation to (231). We extend z by 0 in Ω so (231) says that for all φ ∈

C 1 c (X), ˆX z • ∇φ dx = - ˆ∂Ω [z • ν]φ dH n-1 . (232)
Thus, div(z) is a finite Radon measure in X and is equal to

[z • ν]H n-1 ∂Ω.
As Ω has a Lipschitz boundary, z = 0 in Ω and div(z) is supported on ∂Ω, one can see the pairing (z, D1 + Ω ) (see [29, Definition 3.1]) coincides with -div(z), i.e.

(z,

D1 + Ω ) = -[z • ν]∥D1 Ω ∥. (233)
This means that z

= -[z • ν] D1 Ω ∥D1 Ω ∥ (234)
in a weak sense. Now, we assume that Ω is outward minimizing in X; for all set of finite perimeter E ⊂⊂ X containing Ω, we have P (Ω) ≤ P (E). This property holds true for convex sets but not only. If the curvature of ∂Ω is positive, there exists an open neighborhood X of Ω in which Ω is outward minimizing. According to the co-area formula, we deduce that that for all φ ∈ C 1 c (X) such that φ ≥ 1 on Ω,

P (Ω) ≤ ˆX |∇φ| dx. (235)
Inequality (235) also holds true all u ∈ W 1,1 0 (X) such that u ≥ 1 on Ω. Indeed, one can first replace u by its post-composition with the orthogonal projection onto [0, 1]. This makes sure that u = 1 and Ω and do not increase the total variation. Since Ω has a Lipschitz boundary, the function u has trace 1 on ∂Ω and can be approximated in W 1,1 (X) convergence by functions φ ∈ C 1 c (X) such that such that Ω ⊂⊂ { φ = 1 }. The following lemma and its elementary proof were communicated by A. Chambolle. There is an alternative proof, in a more general setting, in [18, Appendix A].

Proposition C.1. Let X be a bounded open set of R n , let Ω ⊂⊂ X be an open subset of X with Lipschitz boundary. We let ν denote the outward unit normal vector field of Ω (defined H n-1 -a.e. on ∂Ω), and A = X \ Ω. Then there exists a vectorfield z ∈ L ∞ (A; R n ) such that ∥z∥ L ∞ ≤ 1, div(z) = 0 in A and [z • ν] = 1 H n-1 -a.e. on Ω, that is for all φ ∈ C 1 c (X), ˆAz • ∇φ dx = -ˆ∂Ω φ dH n-1 .

(236)

Proof. Without loss of generality, we assume Ω ̸ = ∅. Let p be a real number such that 1 < p < 2. We write p ′ the positive number such that 1 p + 1 p ′ = 1, i.e. p ′ = p/(p -1). We let S p denote the affine subsace

{ u ∈ W 1,p 0 (X) | u = 1 on Ω } . ( 237
)
As we have seen before, the elements of S p can be approximated in W 1,p convergence by functions φ ∈ C 1 c (X) such that Ω ⊂⊂ { φ = 1 }. We can also approximate 1 Ω in strict BV convergence ([3, Definition 3.14]) by such functions. Indeed, Ω has a Lipschitz boundary so for all ε, there exists δ > 0 such that Ω δ ⊂⊂ X and P (Ω δ ) ≤ P (Ω)+ε, where Ω δ = { x ∈ R n | dist(x, Ω) < δ }. We can also assume δ small enough so that |Ω δ \ Ω| ≤ ε. Then, it suffices to mollify 1 Ω δ .

We consider the energy

E p (u) := ˆRn |∇u| p dx (238)
defined for u ∈ S p and we let m p denote its infimum. We can bound m p from above independently from p, by considering any test function φ ∈ C 1 c (X) such that φ = 1 on Ω and |∇φ| ≤ C on R n , where C depends on Ω and X.

Next, we show that lim p→1 m p = P (Ω). We start with lim sup p→1 m p ≤ P (Ω). For all φ ∈ C 1 c (X) such that φ = 1 on Ω, we have

m p ≤ ˆX |∇φ| p dx (239)
We set z p = -|∇u| p-2 ∇u ∈ L p ′ (R n ) and we are going to extract a subsequence of (z p ) that converges to a vector field z when p → 1; it will be the solution to our lemma. We recall that p ′ = p/(p -1) so p ′ → +∞ as p → 1. Fix any 1 < q < ∞ and consider p close enough to 1 so that q ≤ p ′ . Then

We use a diagonalisation argument to extract a subsequence of (z p ) that converges weakly in L q to a vector field z, for all 1 < q < ∞. The norm ∥•∥ L q is lower-semicontinuous with respect to weak convergence in L q (since it can be computed by duality) so for all 1 < q < ∞,

X |z| q dx 1 q ≤ 1 (

  defined over the function u ∈ SBV (R) such that u = m on ] -∞, a[ and u = M on ]b, ∞]. By scaling, it suffices to study the case [a, b] = [0, 1].

Theorem 3 . 3 .

 33 Let Ω be a C 1 bounded open set of R n and assume that its outward unit normal vector field has a continuous extension ν : R n \ Ω → R n such that |ν| ≤ 1 and which is divergence-free on R n \ Ω in the sense of distribution. If β ≤ γ, then 1 Ω has a calibration.

1 EFigure 2 :Remark 3 . 4 .Remark 3 . 5 .

 123435 Figure 2: n = 2, β = 1 and γ = 0.34

Lemma 3 . 6 .

 36 Let Ω be a non-empty C 2 star-shaped bounded open set of R n . There exists a continuous function Γ : R n \ Ω → R such that (i) Γ is harmonic positive on R n \ Ω;

  190) holds true for all s ∈ [0, 1] because we have |ϕ x (x, t)| ≤ 2βt for all t ∈ [0, 1]. Now, we fix 1 ≤ |x| < R and as usual, we write r for |x|. Inequality (190) holds true for all s ∈ [0, ρ(r)] because we have |ϕ x (x, t)| ≤ 2βt for all t ∈ [0, ρ(r)]. Next, we estimate for t ∈ [ρ(r), 1],

whence lim sup p→1 m 1 p 1 p

 p→111 p ≤ ˆX |∇φ| dx.(240)We deduce that lim sup p→1 m p ≤ P (Ω) by approximating 1 Ω in strict BV convergence with functions φ ∈ C 1 c (X) such that φ = 1 on Ω. Reciprocally, for all u ∈ S p , the outward minimality of Ω yieldsP (Ω) ≤ ˆX |∇u| dx (241) ˆX |∇u| p dx(242)whence P (Ω) ≤ |X| 1m p and then P (Ω) ≤ lim inf p→1 m p . The energy E p has a unique minimizer u p ∈ S p . The short and usual proof starts with the fact that for any miniming sequence (v k ), the gradient sequence (∇v k ) is a Cauchy sequence in L p (X) thanks to the uniform convexity of the L p norm. One can use Poincaré inequality to deduce that for all open ball B ⊂ X, the sequence(v km B v k ) is also a Cauchy sequence in L p (B), where m B v k is the average value of v k in B. Similarly, for all open balls B 1 , B 2 ⊂ X, the sequence of real numbers (m B 1 v km B 2 ) is also a Cauchy sequence. So we can fix a ball B 0 ⊂ Ω and see that (v km B 0 v k ) converge in L 1 loc (X), but since v k = 1 on Ω, this just means that v k converge in L 1 loc .The limit is a minimizer and is even unique, again by uniform convexity of the L p norm. Now, we state the Euler-Lagrange associated to this problem. For all function v ∈ S p , we can testv ε = u p ± ε(vu) for all ε > 0 to deduce ˆX |∇u p | p-2 ∇u p • ∇(vu) dx = 0,(243)that isˆX |∇u p | p-2 ∇u p • ∇v = ˆX |∇u p | p dx.(244)An equivalent way to state (244) is that for all v ∈ W 1,p 0 (X) such that v = 0 on Ω, we have ˆX |∇u p | p-2 ∇u p • ∇v dx = 0.

  249)which means that |z| ≤ 1 on X. Passing to the limit in (244) and (245), we see that for all φ ∈ C 1 c (X) with φ = 1 on Ω, we haveˆX z • ∇φ dx = -P (Ω)(250)and for all φ ∈ C 1 c (X) with φ = 0 on Ω, we haveˆX z • ∇φ dx = 0.(251)Observe that |∇u p | p-2 ∇u p = 0 on Ω so z = 0 as well on Ω. The condition (251) implies that div(z) is zero on A. According to the introduction of this section, there exists a function[z • ν] ∈ L ∞ (∂Ω) such that ∥[z • ν]∥ L ∞ (∂Ω) ≤ 1 and for all φ ∈ C 1 c (X), ˆX z • ∇φ dx = -ˆ∂Ω [z • ν]φ dH n-1 .(252)Takingany φ ∈ C 1 c (X) such that φ = 1 on Ω yields ˆ∂Ω [z • ν] dH n-1 = P (Ω) (253) but since, [z • ν] ≤ 1 H n-1 -a.e. on ∂Ω, we actually have [z • ν] = 1 H n-1 -a.e on ∂Ω.

  be a Borel map which is bounded, approximately regular in X × [m, M ] and divergence-free on X×]m, M [ in the sense of distribution. Then for all u

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is increasing because t → t ln(t) is convex. If n ≥ 3, the function

is increasing because

, we also deduce that for all t ≥ 1,

Next, we prove that for all t ≥ 1,

We rewrite this condition

and since both sides equals 0 at t = 1, it suffices to show that the derivative of the left-hand side is greater than or equal to the derivative of the left-hand side on [1, +∞[. We are thus led to show that for all t ≥ 1,

that is

When n = 2, this comes from the concavity of t → ln(t). When n ≥ 3, this comes from the fact that whenever α > β > 0 and t ≥ 1,

We finally show that for all t > 1,

We isolate Γ in (213) and we obtain the equivalent condition

Since both sides equals 0 at t = 1, it suffices to show that the derivative of the left-hand side is greater than or equal to the derivative of the left-hand side on [1, +∞[. This condition simplifies to the fact that for all t ≥ 1, n n -1 2 t 2n-2 Γ(t) ≥ nt n-1 (t n-1 -1) + 1 2 (t n -1).

(215)