Professeur Aydin Aytuna

Professeur Dan Coman

Keywords: Problème de Dirichlet, Opérateur de Monge-Ampère, Mesure de Hausdorff-Riesz, Fonction m-sousharmonique, Opérateur hessien, Capacité, Module de continuité, Principe de comparaison, Théorème de stabilité, Domaine strictement hyperconvexe lipschitzien, Domaine strictement m-pseudoconvexe Dirichlet problem, Monge-Ampère operator, Hausdorff-Riesz measure, m-subharmonic function, Hessian operator, Capacity, Modulus of continuity, Comparison principle, Stability theorem, Strongly hyperconvex Lipschitz domain, Strongly m-pseudoconvex domain barriers . . . . . . . . . . . . . . . . . . . . . 4.5.3 Hölder continuity for radially symmetric solution . . . . . . . .

Cette thèse est consacrée à l'étude de la régularité des solutions des équations de Monge-Ampère complexes ainsi que des équations hessiennes complexes dans un domaine borné de C n . Dans le premier chapitre, on donne des rappels sur la théorie du pluripotentiel. Dans le deuxième chapitre, on étudie le module de continuité des solutions du problème de Dirichlet pour les équations de Monge-Ampère lorsque le second membre est une mesure à densité continue par rapport à la mesure de Lebesgue dans un domaine strictement hyperconvexe lipschitzien. Dans le troisième chapitre, on prouve la continuité hölderienne des solutions de ce problème pour certaines mesures générales. Dans le quatrième chapitre, on considère le problème de Dirichlet pour les équations hessiennes complexes plus générales où le second membre dépend de la fonction inconnue. On donne une estimation précise du module de continuité de la solution lorsque la densité est continue. De plus, si la densité est dans L p , on démontre que la solution est Höldercontinue jusqu'au bord.

Chapter 0 Introduction

In this thesis we study the regularity of solutions to the Dirichlet problem for complex Monge-Ampère equations and, more generally, for complex Hessian equations in a bounded domain of C n .

Pluripotential theory became a branch of mathematical research in the last decades and the complex Monge-Ampère equation was studied extensively by many mathematicians.

Two influential works have been the work by Yau [START_REF] Yau | On the Ricci curvature of a compact Kähler manifold and the complex Monge-Ampère equation I[END_REF] on non-degenerate equations on compact Kähler manifolds, and by Bedford-Taylor [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] on generalized weak solutions in the sense of pluripotential theory. They proved [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] that the complex Monge-Ampère operator has a sense for a non-smooth locally bounded plurisubharmonic function and there exists a continuous solution to the Dirichlet problem in a bounded strongly pseudoconvex domain with smooth boundary.

Since then, there has been considerable further progress, it was proved in [START_REF] Caffarelli | The Dirichlet problem for nonlinear second order elliptic equations. II. Complex Monge-Ampère, and uniformly elliptic equations[END_REF] the smoothness of the solution to the Dirichlet problem in the case of non-degenerate smooth density and smooth boundary data.

Ko lodziej demonstrated [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF][START_REF] Ko Lodziej | A sufficient condition for solvability of the Dirichlet problem for the complex Monge-Ampère operator[END_REF] that the Dirichlet problem still admits a unique weak continuous solution when the right hand side of the complex Monge-Ampère equation is a measure satisfying some sufficient condition which is close to be best possible. Furthermore, for the degenerate complex Monge-Ampère equation on compact Kähler manifolds he established [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF] a uniform a priori estimate which generalizes the celebrated a priori estimate of Yau [START_REF] Yau | On the Ricci curvature of a compact Kähler manifold and the complex Monge-Ampère equation I[END_REF].

A viscosity approach to the complex Monge-Ampère equation has been developed by Eyssidieux, Guedj and Zeriahi in [START_REF] Eyssidieux | Viscosity solutions to degenerate complex Monge-Ampère equations[END_REF] on compact Kähler manifolds and they compare viscosity and potential solutions. In the local context, Wang [START_REF] Wang | A viscosity approach to the Dirichlet problem for complex Monge-Ampère equations[END_REF] studied the existence of a viscosity solution to the Dirichlet problem for the complex Monge-Ampère equation and estimated the modulus of continuity of the solution in terms of that of a given subsolution and of the right hand side.

Some results have been known about the Hölder regularity of the solution to this problem for measures absolutely continuous with respect to the Lebesgue measure. Bedford and Taylor [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] studied the Hölder continuity of the solution by means of Hölder continuity of the density and the boundary data. Guedj, Ko lodziej and Zeriahi [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF] established Hölder regularity of solutions for L p -densities bounded near the boundary of strongly pseudoconvex domain. In the compact case, there are many works in this area [START_REF] Ko Lodziej | Hölder continuity of solutions to the complex Monge-Ampère equation with the right-hand side in L p : the case of compact Kähler manifolds[END_REF][START_REF] Pham | Hölder continuity of solutions to the Monge-Ampère equations on compact Kähler manifolds[END_REF][START_REF] Demailly | Hölder continuous solutions to Monge-Ampère equations[END_REF] which exceed the scope of this thesis.

We are also interested in studying the complex Hessian equation in a bounded domain of C n . This equation corresponds to the elementary symmetric function of degree 1 ≤ m ≤ n. When m = 1, this equation corresponds to the Poisson equation which is classical. The case m = n corresponds to the complex Monge-Ampère equation.

The complex Hessian equation is a natural generalisation of the complex Monge-Ampère equation and has some geometrical applications. For examples, this equation appears in problems related to quaternionic geometry [START_REF] Alesker | Quaternionic Monge-Ampère equation and Calabi problem for HKT-manifolds[END_REF] and in the work [START_REF] Székelyhidi | Gauduchon metrics with prescribed volume form[END_REF] for solving Gauduchon's conjecture. Its real counterpart has been developed in the works of Trudinger, Wang and others (see for example [W09]). This all gives us a strong motivation to study the existence and regularity of weak solutions to complex Hessian equations.

The complex Hessian equation is a new subject and is much more difficult to handle than the complex Monge-Ampère equation (e.g. the m-subharmonic functions are not invariant under holomorphic change of variables, for m < n). Despite these difficulties, the pluripotential theory which was developed for the complex Monge-Ampère equation can be adapted to the complex Hessian equation [B l05, DK14, Lu12, SA12]. B locki [B l05] introduced some elements of the potential theory for m-subharmonic functions and proved the existence of continuous solution for the homogeneous Dirichlet problem in the unit ball. Dinew and Ko lodziej [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF] used pluripotential techniques adapted for the complex Hessian equation to settle the question of the existence of weak solutions to the Dirichlet problem. H. C. Lu introduced in [START_REF] Lu | Equations Hessiennes complexes[END_REF][START_REF] Lu | A variational approach to complex Hessian equations in C n[END_REF] finite energy classes of m-subharmonic functions and developed a variational approach to complex Hessian equations. The nondegenerate complex Hessian equation on compact Kähler manifold with smooth density has been studied in [START_REF] Hou | Complex Hessian equation on Kähler manifold[END_REF], [START_REF] Hou | A second order estimate for complex Hessian equations on a compact Kähler manifold[END_REF], [START_REF] Jbilou | Complex Hessian equations on some compact Kähler manifolds[END_REF] and the degenerate case was treated in [START_REF] Lu | Solutions to degenerate complex Hessian equations[END_REF] and [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF]. H.C. Lu persisted in investigating a viscosity approach to complex Hessian equations in his paper [START_REF] Lu | Viscosity solutions to complex Hessian equations[END_REF]. Now we will present an overview of the main results of this thesis. First, for the sake of convenience we recall some notations. We denote by dV 2n the Lebesgue measure in C n and L p (Ω) stands for the usual L p -space with respect to the Lebesgue measure in a bounded domain Ω. We use d = ∂ + ∂ and d c = (i/4)( ∂ -∂), where ∂ and ∂ are the usual differential operators. Here and subsequently, we use the notation : C 0,β ( Ω) = {v ∈ C( Ω); v β < +∞}, for 0 < β ≤ 1, and the β-Hölder norm is given by

v β = sup ¶ |v(z)| : z ∈ Ω© + sup ® |v(z) -v(y)| |z -y| β : z, y ∈ Ω, z = y
´.

We mean by C k,β ( Ω), with k ≥ 1 and 0 < β ≤ 1, the class of functions which have continuous partial derivatives of order less than k, and whose k-th order partial derivatives satisfy a Hölder condition of order β.

The Dirichlet problem for complex Monge-Ampère equations. It asks for a function, u, plurisubharmonic on Ω and continuous on Ω such that (0.0.1) (dd c u) n = f dµ, and u = ϕ on ∂Ω, where ϕ ∈ C(∂Ω), µ is a nonnegative finite Borel measure on Ω and 0 ≤ f ∈ L 1 (Ω, µ).

In Chapter 2, we consider this problem in a bounded strongly hyperconvex Lipschitz domain of C n with continuous densities with respect to the Lebesgue measure. Then we prove in Section 2.5 a sharp estimate for the modulus of continuity of the solution.

Theorem 0.0.1. Let Ω ⊂ C n be a bounded strongly hyperconvex Lipschitz domain, ϕ ∈ C(∂Ω) and 0 ≤ f ∈ C( Ω). Assume that ω ϕ is the modulus of continuity of ϕ and ω f 1/n is the modulus of continuity of f 1/n . Then the modulus of continuity of the unique solution U to (0.0.1) has the following estimate

ω U (t) ≤ η(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), ω f 1/n (t), t 1/2 },
where η is a positive constant depending on Ω.

In [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF], Guedj, Ko lodziej and Zeriahi proved the Hölder continuity of the solution to (0.0.1) when ϕ ∈ C 1,1 (∂Ω) and f ∈ L p (Ω), for p > 1, is bounded near the boundary ∂Ω. Recently N.C. Nguyen [N14] proved that the solution is Hölder continuous when the density f satisfies a growth condition near ∂Ω. Our next result in Chapter 3 concerns the Hölder regularity of the solution when the density is merely in L p (Ω), p > 1. Moreover, we improve the Hölder exponent while p ≥ 2 by using the relation between real and complex Monge-Ampère operators.

Theorem 0.0.2. Let Ω ⊂ C n be a bounded strongly hyperconvex Lipschitz domain. Assume that ϕ ∈ C 1,1 (∂Ω) and f ∈ L p (Ω) for some p > 1. Then the unique solution U to (0.0.1) is γ-Hölder continuous on Ω for any 0 < γ < 1/(nq+1) where 1/p+1/q = 1. Moreover, if p ≥ 2, then the solution U is Hölder continuous on Ω of exponent less than min{1/2, 2/(nq+1)}.

In the same chapter, we study the Hölder regularity of the solution to the Dirichlet problem for a Hausdorff-Riesz measure of order 2n -2 + ǫ, with 0 < ǫ ≤ 2, that is a non-negative Borel measure satisfies the condition µ(B(z, r) ∩ Ω) ≤ Cr 2n-2+ǫ , ∀z ∈ Ω, ∀0 < r < 1, for some positive constant C. These measures are singular with respect to the Lebesgue measure, for 0 < ǫ < 2, and there are many nice examples (see Example 3.5.6).

More precisely, we prove in Section 3.5 the following theorems.

Theorem 0.0.3. Let Ω be a bounded strongly hyperconvex Lipschitz domain in C n and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ, for 0 < ǫ ≤ 2. Suppose that ϕ ∈ C 1,1 (∂Ω) and 0 ≤ f ∈ L p (Ω, µ) for some p > 1, then the unique solution to the Dirichlet problem (0.0.1) is Hölder continuous on Ω of exponent ǫγ/2, for any 0 < γ < 1/(nq + 1) where 1/p + 1/q = 1.

When the boundary data is merely Hölder continuous, we can still prove the Hölder regularity of the solution using the last theorem. Theorem 0.0.4. Let Ω be a bounded strongly hyperconvex Lipschitz domain in C n and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ, for 0 < ǫ ≤ 2. Suppose that ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ), for some p > 1, then the unique solution to the Dirichlet problem (0.0.1) is Hölder continuous on Ω of exponent ǫ ǫ+6 min{α, ǫγ}, for any 0 < γ < 1/(nq + 1) where 1/p + 1/q = 1.

Moreover, when Ω is a smooth strongly pseudoconvex domain the Hölder exponent will be ǫ ǫ+2 min{α, ǫγ}, for any 0 < γ < 1/(nq + 1).

A natural question is that if we have a Hölder continuous subsolution to the Dirichlet problem, can we get a Hölder continuous solution in the whole domain? This question is still open in the local case (see [START_REF] Demailly | Hölder continuous solutions to Monge-Ampère equations[END_REF] for a positive answer in the compact setting). However, we prove some particular case. Theorem 0.0.5. Let µ be a nonnegative finite Borel measure on a bounded strongly hyperconvex Lipschitz domain Ω. Let also ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ), p > 1. Assume that there exists a Hölder continuous plurisubharmonic function w in Ω such that (dd c w) n ≥ µ. If, near the boundary, µ is Hausdorff-Riesz of order 2n -2 + ǫ for some 0 < ǫ ≤ 2, then the solution U to (0.0.1) is Hölder continuous on Ω.

The Dirichlet problem for complex Hessian equations. It consists in finding a function u which is m-subharmonic in Ω and continuous on Ω such that (0.0.2) (dd c u) m ∧ β n-m = f dV 2n and u = ϕ on ∂Ω, where ϕ ∈ C(∂Ω) and 0 ≤ f ∈ L 1 (Ω). We first prove in Chapter 4 a sharp estimate for the modulus of continuity of the solution when the density is continuous and depends on the unknown function.

Theorem 0.0.6. Let Ω be a smoothly bounded strongly m-pseudoconvex domain in C n , ϕ ∈ C(∂Ω) and 0 ≤ F ∈ C( Ω × R) be a nondecreasing function in the second variable. Then the modulus of continuity ω U of the solution U to

       u ∈ SH m (Ω) ∩ C( Ω), (dd c u) m ∧ β n-m = F (z, u)dV 2n in Ω, u = ϕ on ∂Ω,
satisfies the following estimate

ω U (t) ≤ γ(1 + F 1/m L ∞ (K) ) max{ω ϕ (t 1/2 ), ω F 1/m (t), t 1/2 },
where γ is a positive constant depending only on Ω, K = Ω × {a}, a = sup ∂Ω |ϕ| and ω F 1/m (t) is given by

ω F 1/m (t) := sup y∈[-M,M ] sup |z 1 -z 2 |≤t |F 1/m (z 1 , y) -F 1/m (z 2 , y)|,
with M = a + 2 diam(Ω) 2 sup Ω F 1/m (., -a).

For densities in L p (Ω), p > n/m, N. C. Nguyen [N14] proved that the solution to (0.0.2) is Hölder continuous when the density f satisfies some condition near the boundary. Here, we prove the general case. Theorem 0.0.7. Let Ω ⊂ C n be a bounded strongly m-pseudoconvex domain with smooth boundary, ϕ ∈ C 1,1 (∂Ω) and 0 ≤ f ∈ L p (Ω), for some p > n/m. Then the solution to (0.0.2), U ∈ C 0,α ( Ω) for any 0 < α < γ 1 , where γ 1 is a constant depending on m, n, p defined by (4.5.1).

Moreover, if p ≥ 2n/m then the solution to the Dirichlet problem U ∈ C 0,α ( Ω), for any 0 < α < min{ 1 2 , 2γ 1 }.

In the particular case of radially symmetric solution in the unit ball, we are able to find a better Hölder exponent which turns out to be optimal. Theorem 0.0.8. Let B be the unit ball and 0 ≤ f ∈ L p (B) be a radial function, where p > n/m. Then the unique solution U for (0.0.2) with zero boundary values is given by the explicit formula

U(r) = -B 1 r 1 t 2n/m-1 Ç t 0 ρ 2n-1 f (ρ)dρ

Basic facts in pluripotential theory

In this section, some useful facts from pluripotential theory will be stated and then used throughout this thesis. For further information about pluripotential theory, see for example [START_REF] Klimek | Pluripotential Theory[END_REF], [START_REF] Demailly | Potential theory in several complex variables[END_REF], [START_REF] Ko Lodziej | The complex Monge-Ampère equation and pluripotential theory[END_REF] and [START_REF] Guedj | Degenerate complex Monge-Ampère equations[END_REF]. Note that, with a domain we mean a nonempty, open and connected set.

Definition 1.1.1. Let Ω ⊂ R n be a domain. An upper semicontinuous function u : Ω → R ∪ {-∞} is said to be subharmonic if, for every relatively compact open subset U of Ω and every continuous function h : Ū → R that is harmonic on U , we have the implication

u ≤ h on ∂U ⇒ u ≤ h on U.
It is well known in several complex variables that the class of subharmonic functions is very large and the fact that the property of being subharmonic is then not invariant under biholomorphic mappings. This fact motivates the theory of plurisubharmonic functions and pluripotential theory.

In pluripotential theory one therefore studies a smaller class of subharmonic functions whose composition with biholomorphic mappings are subharmonic. This class is precisely the class of plurisubharmonic functions that will be defined below.

Definition 1.1.2. A function u : Ω → R ∪ {-∞} is called plurisubharmonic (briefly psh) if it is upper semicontinuous in Ω and subharmonic on the intersection of Ω with any complex line {a + bξ; ξ ∈ C} where a, b ∈ C n . We denote by P SH(Ω) the set of all plurisubharmonic functions in Ω. We state here some basic properties of psh functions.

Proposition 1.1.3.

1. If u, v ∈ P SH(Ω) then λu + ηv ∈ P SH(Ω), ∀λ, η ≥ 0.

2. If u ∈ P SH(Ω) and χ : R → R is convex increasing function then χ • u ∈ P SH(Ω).

3. Let {u j } j∈N be a decreasing sequence of psh functions in Ω. Then u := lim j→+∞ u j is psh function in Ω.

If u ∈ P SH(Ω)

then the standard regularizations u ǫ = u * ρ ǫ are psh in Ω ǫ := {z ∈ Ω| dist(z, ∂Ω) > ǫ}, for 0 < ǫ ≪ 1.

Let U be a non-empty proper open subset of Ω, if u ∈ P SH(Ω), v ∈ P SH(U ) and lim sup

z→y z∈U v(z) ≤ u(y) for every y ∈ ∂U ∩ Ω, then the function

w = ® max{u, v} in U, u in Ω \ U,
is psh in Ω.

6. Let {u α } ⊂ P SH(Ω) be locally uniformly bounded from above and u = sup u α . Then the upper semi-continuous regularization u * is psh and equal to u almost everywhere.

One of the important reasons to study plurisubharmonic functions is that we can use them to define pseudoconvex domains.

Definition 1.1.4. A domain Ω ⊂ C n is called pseudoconvex if there exists a continuous plurisubharmonic function ϕ in Ω such that {z ∈ Ω; ϕ(z) < c} ⋐ Ω, for all c ∈ R.
An important class of pseudoconvex domains is the class of hyperconvex domains.

Definition 1.1.5. A domain Ω ⊂ C n is called hyperconvex if there exists a negative continuous plurisubharmonic function ψ in Ω such that {z ∈ Ω; ψ(z) < c} ⋐ Ω, for all real c < 0.
It is known that the Hartogs triangle is a pseudoconvex domain but not hyperconvex. However, Demailly [START_REF] Demailly | Mesures de Monge-Ampère et mesures pluriharmoniques[END_REF] proved that any pseudoconvex domain with Lipschitz boundary is a hyperconvex domain.

The complex Monge-Ampère operator

Let ∂, ∂ be the usual differential operators, d = ∂ + ∂ and

d c = (i/4)( ∂ -∂). Then dd c = (i/2)∂ ∂.
If u ∈ C 2 (Ω) is a plurisubharmonic function, then the complex Monge-Ampère operator is defined by

(dd c u) n = (dd c u) ∧ ... ∧ (dd c u) = det Ç ∂ 2 u ∂z j ∂ zk å β n , where β := dd c |z| 2 = (i/2) n j=1 dz j ∧ d zj is the standard Kähler form in C n . Note that β n = n! dV 2n where dV 2n = (i/2) n dz 1 ∧ dz 1 ∧ ... ∧ dz n ∧ dz n is the usual volume form on R 2n or C n .
For n = 1, we have dd c u = (1/4)∆udV 2 and we know that the Laplace operator is well defined on all subharmonic functions. In the case n ≥ 2 the complex Monge-Ampère operator can not be extended in a meaningful way to the whole class of plurisubharmonic functions and still have the range contained in the class of nonnegative Borel measures (see Example 3.1 in [START_REF] Kiselman | Sur la définition de l'opérateur de Monge-Ampère complexe[END_REF]).

In 1976, Bedford and Taylor in their seminal work proved that the complex Monge-Ampère operator is well-defined on locally bounded plurisubharmonic functions. They defined inductively the following closed nonnegative current

dd c u 1 ∧ dd c u 2 ∧ ... ∧ dd c u n := dd c (u 1 dd c u 2 ∧ ... ∧ dd c u n ),
where u 1 , u 2 , ..., u n ∈ P SH(Ω) ∩ L ∞ loc (Ω). Furthermore, Cegrell [START_REF] Cegrell | The general definition of the complex Monge-Ampère operator[END_REF] introduced and investigated the largest class of plurisubharmonic functions on which the operator (dd c .) n is well-defined.

The following inequality, named Chern-Levine-Nirenberg inequality, gives a bound on the local mass of the non-negative measure dd c u 1 ∧ ... ∧ dd c u n in terms of L ∞ -norms of u j 's and hence ensures that these measures dd c u 1 ∧ ... ∧ dd c u n , where

u j ∈ P SH(Ω) ∩ L ∞ loc (Ω), j = 1, ..., n, are Radon measures. Proposition 1.2.1. Let K ⋐ U ⋐ Ω, where K is compact and U is open. Let u j ∈ P SH(Ω) ∩ L ∞ loc (Ω), j = 1, 2, ..., n. Then there exists a constant C depending on K, U, Ω such that dd c u 1 ∧ ... ∧ dd c u n K ≤ C u 1 L ∞ (U ) ... u n L ∞ (U ) .
In [START_REF] Bedford | A new capacity for plurisubharmonic functions[END_REF] Bedford and Taylor showed that the complex Monge-Ampère operator is continuous with respect to monotone sequences of locally bounded plurisubharmonic functions. Later, Xing [Xi96] found out that the convergence in capacity (defined below) entails the convergence of corresponding Monge-Ampère measures and he showed that this condition is quite sharp in some case.

Let Ω be a bounded domain in C n . For a Borel subset K of Ω, we introduce the Bedford-Taylor capacity

Cap(K, Ω) = sup K (dd c u) n ; u ∈ P SH(Ω), -1 ≤ u ≤ 0 .
By proposition 1.2.1, it is clear that the capacity is finite when K is relatively compact in Ω. Definition 1.2.2. A sequence u j of functions defined in Ω is said to converge in capacity to u if for any t > 0 and

K ⋐ Ω lim j→∞ Cap(K ∩ {|u -u j | > t}, Ω) = 0.
The complex Monge-Ampère operator is continuous with respect to sequences of locally uniformly bounded psh functions converging in capacity.

Theorem 1.2.3. Let (u j k ) ∞ j=1 , k = 1, ...

, n be a locally uniformly bounded sequence of psh functions in Ω and u

j k → u k ∈ P SH(Ω) ∩ L ∞ loc (Ω) in capacity as j → +∞ for k = 1, ..., n. Then lim j→∞ dd c u j 1 ∧ ... ∧ dd c u j n = dd c u 1 ∧ ... ∧ dd c u n
in the weak sense of currents in Ω.

We mention some useful theorems about the quasi-continuity of psh functions and the maximum principle. Theorem 1.2.4. Let u be a psh function in Ω. Then for all ǫ > 0, there exists an open set G ⊂ Ω such that Cap(G, Ω) < ǫ and u| (Ω\G) is continuous.

Theorem 1.2.5. Let u, v ∈ P SH(Ω) ∩ L ∞ loc (Ω).
Then we have the following inequality in the sense of Borel measures in Ω

(dd c max{u, v}) n ≥ 1 {u≥v} (dd c u) n + 1 {u<v} (dd c v) n .
One of the most effective tools in pluripotential theory is the following comparison principle Theorem 1.2.6. Assume that u, v

∈ P SH(Ω) ∩ L ∞ loc (Ω) are such that lim inf z→∂Ω (u(z) - v(z)) ≥ 0, then {u<v} (dd c v) n ≤ {u<v} (dd c u) n . Corollary 1.2.7. Assume that u, v ∈ P SH(Ω) ∩ L ∞ loc (Ω) are such that lim inf z→∂Ω (u(z) - v(z)) ≥ 0. If (dd c u) n ≤ (dd c v) n as Radon measures on Ω, then v ≤ u in Ω.
Finally, we introduce Dinew's inequality for mixed Monge-Ampère measures [START_REF] Dinew | An inequality for mixed Monge-Ampère measures[END_REF].

Theorem 1.2.8. Let u, v ∈ P SH(Ω) ∩ L ∞ (Ω). Let also f, g ∈ L 1 (Ω)
be nonnegative functions such that the following inequalities hold,

(dd c u) n ≥ f dV 2n , (dd c v) n ≥ gdV 2n . Then (dd c u) k ∧ (dd c v) n-k ≥ f k n g n-k
n dV 2n , k = 1, ..., n.

Basic facts about m-subharmonic functions

In this section, we briefly recall some facts from linear algebra and basic results from potential theory for m-subharmonic functions. We refer the reader to [B l05, SA12, Lu12, DK12, Lu13a, N13, DK14, Lu15] for more details and recent results.

We set

H m (λ) = 1≤j 1 <...<jm≤n λ j 1 ...λ jm , where λ = (λ 1 , ..., λ n ) ∈ R n . Thus (t + λ 1 )...(t + λ n ) = n m=0 H m (λ)t n-m for t ∈ R, where H 0 (λ) = 1.
We denote by Γ m the closure of the connected component of {H m > 0} containing (1, 1, ..., 1). One can show that

Γ m = {λ ∈ R n : H m (λ 1 + t, ..., λ n + t) ≥ 0, ∀t ≥ 0}.
It follows from the identity

H m (λ 1 + t, ..., λ n + t) = m p=0 Ç n -p m -p å H p (λ)t m-p , that Γ m = {λ ∈ R n : H j (λ) ≥ 0, ∀1 ≤ j ≤ m}. It is clear that Γ n ⊂ Γ n-1 ⊂ ... ⊂ Γ 1 , where Γ n = {λ ∈ R n : λ i ≥ 0, ∀i}.
By the paper of Gårding [G59], the set Γ m is a convex cone in R n and H 1/m m is concave on Γ m . By Maclaurin's inequality, we get

Ç n m å -1/m H 1/m m ≤ Ç n p å -1/p H 1/p p ; 1 ≤ p ≤ m ≤ n.
Let H be the vector space over R of complex Hermitian n × n matrices. For any A ∈ H, let λ(A) = (λ 1 , ..., λ n ) ∈ R n be the eigenvalues of A. We set

Hm (A) = H m (λ(A)).

Now, we define the cone

Γm := {A ∈ H : λ(A) ∈ Γ m } = {A ∈ H : Hj (A) ≥ 0, ∀1 ≤ j ≤ m}.
Let α be a real (1,1)-form determined by

α = i 2 i,j a i j dz i ∧ dz j ,
where A = (a i j ) is a Hermitian matrix. After diagonalizing the matrix A = (a i j ), we see that

α m ∧ β n-m = Sm (α)β n ,
where β is the standard Kähler form in C n and Sm (α) = m!(n-m)! n!

Hm (A). The last equality allows us to define

Γm := {α ∈ C (1,1) : α ∧ β n-1 ≥ 0, α 2 ∧ β n-2 ≥ 0, ..., α m ∧ β n-m ≥ 0},
where C (1,1) is the space of real (1,1)-forms with constant coefficients.

Let M : C m (1,1) → R be the polarized form of Sm , i.e. M is linear in every variable, symmetric and M (α, ..., α) = Sm (α), for any α ∈ C (1,1) . The Gårding inequality (see [G59]) asserts that

(1.3.1) M (α 1 , α 2 , ..., α m ) ≥ Sm (α 1 ) 1/m ... Sm (α m ) 1/m , α 1 , α 2 , ..., α m ∈ Γm . Proposition 1.3.1. ([B l05]). If α 1 , ..., α p ∈ Γm , 1 ≤ p ≤ m, then we have α 1 ∧ α 2 ∧ ... ∧ α p ∧ β n-m ≥ 0.
Let us set Σ m := {α ∈ Γm of constant coefficients such that Sm (α) = 1}.

Recall the following elementary lemma whose proof is included for the convenience of the reader.

Lemma 1.3.2. Let α ∈ Γm . Then the following identity holds

Sm (α) 1/m = inf ® α ∧ α 1 ∧ ... ∧ α m-1 ∧ β n-m β n ; α i ∈ Σ m , ∀i
´.

Proof. Let M be a polarized form of Sm defined by

M (α, α 1 , ..., α m-1 ) = α ∧ α 1 ∧ ... ∧ α m-1 ∧ β n-m β n , for α 1 , ..., α m-1 ∈ Σ m , α ∈ Γm . By Gårding's inequality (1.3.1), we have M (α, α 1 , ..., α m-1 ) ≥ Sm (α) 1/m .
Then we obtain that

Sm (α) 1/m ≤ inf ® α ∧ α 1 ∧ ... ∧ α m-1 ∧ β n-m β n ; α i ∈ Σ m , ∀i
´.

Now, setting α

1 = ... = α m-1 = α Sm(α) 1/m , we can ensure that M (α, α 1 , ..., α m-1 ) = Sm (α) 1/m .
This completes the proof of lemma.

Aspects about m-subharmonic functions. Let Ω ⊂ C n be a bounded domain. Let also β := dd c |z| 2 be the standard Kähler form in C n .

Definition 1.3.3. ([B l05]

). Let u be a subharmonic function in Ω. 1) For smooth case, u ∈ C 2 (Ω) is said to be m-subharmonic (briefly m-sh) if the form dd c u belongs pointwise to Γm . 2) For non-smooth case, u is called m-sh if for any collection α 1 , α 2 , ..., α m-1 ∈ Γm , the inequality

dd c u ∧ α 1 ∧ ... ∧ α m-1 ∧ β n-m ≥ 0
holds in the weak sense of currents in Ω.

We denote by SH m (Ω) the set of all m-sh functions in Ω. B locki observed that up to a point pluripotential theory can by adapted to m-subharmonic functions. We recall some properties of m-sh functions.

Proposition 1.3.4 ([B l05]).

1.

P SH = SH n ⊂ SH n-1 ⊂ ... ⊂ SH 1 = SH. 2. If u, v ∈ SH m (Ω) then λu + ηv ∈ SH m (Ω), ∀λ, η ≥ 0. 3. If u ∈ SH m (Ω) and γ : R → R is convex increasing function then γ • u ∈ SH m (Ω).
4. Let {u j } j∈N be a decreasing sequence of m-subharmonic functions in Ω. Then u := lim j→+∞ u j is m-subharmonic function in Ω.

If

u ∈ SH m (Ω) then the standard regularizations u ǫ = u * ρ ǫ are m-subharmonic in Ω ǫ := {z ∈ Ω| dist(z, ∂Ω) > ǫ}, for 0 < ǫ ≪ 1. 6. Let U be a nonempty proper open subset of Ω. If u ∈ SH m (Ω), v ∈ SH m (U )
, and lim z→y z∈U v(z) ≤ u(y) for every y ∈ ∂U ∩ Ω, then the function

w = ® max{u, v} in U, u in Ω \ U, is m-sh in Ω.
7. Let {u α } ⊂ SH m (Ω) be locally uniformly bounded from above and u = sup u α . Then the upper semi-continuous regularization u * is m-sh and equal to u almost everywhere.

The following example was presented by S. Dinew in the international conference in complex analysis and geometry AGC-2013 in Monastir (Tunisia).

Example 1.3.5. Let A be a nonnegative constant and define in C n the function

u(z) = -1 (Im(z 1 ) 2 + Im(z 2 ) 2 + ... + Im(z n ) 2 ) A . We claim that u is m-sh in C n when A ≤ n-2m 2m and m ≤ ⌊ n 2 ⌋. In fact, set v ǫ (z) = Im(z 1 ) 2 + Im(z 2 ) 2 + ... + Im(z n ) 2 + ǫ,
and χ : R + → R -such that χ(t) = -t -A . An easy computation shows that

(dd c (χ • v ǫ )) k ∧ β n-k = k 2 k-1 χ ′′ (v ǫ )(χ ′ (v ǫ )) k-1 dv ǫ ∧ d c v ǫ ∧ β n-1 + (χ ′ (v ǫ )) k 2 k β n .
Hence we get

(dd c (χ • v ǫ )) k ∧ β n-k = A k 2 k (n -1)!v -k(A+1) ǫ (n -2k(A + 1))dV 2n .
Then we can conclude that for any ǫ > 0 the function χ

• v ǫ is m-sh in C n if we have A ≤ (n -2m)/(2m) and m ≤ ⌊n/2⌋.
Since χ is increasing and v ǫ decreases as ǫ tends to zero, we get χ

• v ǫ ց u in C n , thus this yields u ∈ SH m (C n ) when A ≤ (n -2m)/(2m) and m ≤ ⌊n/2⌋.
The following example shows that SH m (Ω) is not invariant under a holomorphic mapping.

Example 1.3.6. We define the function

u(z) = |z 1 | 2 + |z 2 | 2 - 1 2 |z 3 | 2 , z ∈ C 3 .
A simple computation shows that u ∈ SH 2 (C 3 ) and u / ∈ P SH(C 3 ). Let f be a holomorphic mapping from

C 3 to C 3 such that f (z) = (z 1 , z 2 , √ 2z
3 ). Then it is easy to see that u • f is subharmonic but not 2-subharmonic.

For locally bounded m-subharmonic functions, we can inductively define a closed nonnegative current (following Bedford and Taylor for plurisubharmonic functions).

dd c u 1 ∧ ... ∧ dd c u p ∧ β n-m := dd c (u 1 dd c u 2 ∧ ... ∧ dd c u p ∧ β n-m ), where u 1 , u 2 , ..., u p ∈ SH m (Ω) ∩ L ∞ loc (Ω), p ≤ m.
In particular, we define the nonnegative Hessian measure of u ∈ SH m (Ω) ∩ L ∞ loc (Ω) to be

H m (u) := (dd c u) m ∧ β n-m .
We can also use the following identity

du ∧ d c u := (1/2)dd c (u + C) 2 -(u + C)dd c u,
where C is big enough, to define the nonnegative current

du 1 ∧ d c u 1 ∧ dd c u 2 ∧ ... ∧ dd c u p ∧ β n-m , where u 1 , ..., u p ∈ SH m (Ω) ∩ L ∞ loc (Ω), p ≤ m.
One of the most important properties of m-subharmonic functions is the quasicontinuity. Every m-subharmonic function is continuous outside an arbitrarily small open subset. The m-Capacity is used to measure the smallness of these sets.

Definition 1.3.7. Let E ⊂ Ω be a Borel subset. The m-capacity of E with respect to Ω is defined to be Cap m (E, Ω) := sup E (dd c u) m ∧ β n-m ; u ∈ SH m (Ω), -1 ≤ u ≤ 0 .
The m-capacity shares the same elementary properties as the capacity introduced by Bedford and Taylor (see [START_REF] Sadullaev | Potential theory in the class of msubharmonic functions[END_REF][START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF][START_REF] Lu | A variational approach to complex Hessian equations in C n[END_REF]).

Proposition 1.3.8. 1. Cap m (E 1 , Ω) ≤ Cap m (E 2 , Ω), if E 1 ⊂ E 2 . 2. Cap m (E, Ω) = lim j→∞ Cap m (E j , Ω), if E j ↑ E. 3. Cap m (E, Ω) ≤ Cap m (E j , Ω), for E = ∪E j .
Definition 1.3.9. A sequence u j of functions defined in Ω is said to converge with respect to Cap m to a function u if for any t > 0 and K ⋐ Ω,

lim j→+∞ Cap m (K ∩ {|u -u j | > t}, Ω) = 0.
The following results can be proved by repeating the arguments in [START_REF] Ko Lodziej | The complex Monge-Ampère equation and pluripotential theory[END_REF].

Theorem 1.3.10. Let (u j k ) ∞ j=1 , k = 1, ..., m be a locally uniformly bounded sequence of msh functions in Ω and

u j k → u k ∈ SH m (Ω) ∩ L ∞ loc (Ω) in Cap m as j → +∞ for k = 1, ..., m. Then dd c u j 1 ∧ ... ∧ dd c u j m ∧ β n-m ⇀ dd c u 1 ∧ ... ∧ dd c u m ∧ β n-m .
Importantly, the complex Hessian operator is continuous with respect to the decreasing convergence.

Theorem 1.3.11. If u j ∈ SH m (Ω)∩L ∞ (Ω) is a sequence decreasing to a bounded function u in Ω, then (dd c u j ) m ∧ β n-m converges to (dd c u) m ∧ β n-m in the weak sense of currents in Ω.
Theorem 1.3.12. Every m-subharmonic function u defined in Ω is quasi-continuous. This means that for any positive number ǫ one can find an open set U ⊂ Ω with Cap m (U, Ω) < ǫ and such that u| Ω\U is continuous.

Theorem 1.3.13. Let {u j k } ∞ j=1 be a locally uniformly bounded sequence of m-subharmonic functions in Ω for k = 1, 2, ..., m and let

u j k ↑ u k ∈ SH m (Ω) ∩ L ∞ loc almost everywhere as j → ∞ for k = 1, 2, ..., m. Then dd c u j 1 ∧ ... ∧ dd c u j m ∧ β n-m ⇀ dd c u 1 ∧ ... ∧ dd c u m ∧ β n-m .
). Let u ∈ SH m (Ω) ∩ L ∞ loc (Ω). Then H m (u) = 0 in Ω if and only if u is m-maximal. Theorem 1.3.16 (Integration by parts). Let u, v ∈ SH m (Ω)∩L ∞ loc (Ω) such that lim z→∂Ω u = lim z→∂Ω v = 0. Then Ω udd c v ∧ T = Ω vdd c u ∧ T, where T = dd c u 1 ∧ ... ∧ dd c u m-1 ∧ β n-m and u 1 , ..., u m-1 ∈ SH m (Ω) ∩ L ∞ loc (Ω).
Theorem 1.3.17. For u, v ∈ SH m (Ω) ∩ L ∞ loc (Ω), we have

(dd c max{u, v}) m ∧ β n-m ≥ 1 {u>v} (dd c u) m ∧ β n-m + 1 {u≤v} (dd c v) m ∧ β n-m ,
where 1 E is the characteristic function of a set E.

Theorem 1.3.18. Let Ω be a bounded domain in

C n and u, v ∈ SH m (Ω) ∩ L ∞ loc (Ω) be such that lim inf ζ→∂Ω (u -v)(ζ) ≥ 0. Then {u<v} (dd c v) m ∧ β n-m ≤ {u<v} (dd c u) m ∧ β n-m . Corollary 1.3.19. Under the same assumption of Theorem 1.3.18, if (dd c u) m ∧ β n-m ≤ (dd c v) m ∧ β n-m as Radon measures on Ω, then v ≤ u in Ω. Corollary 1.3.20. Let Ω be a bounded domain in C n and u, v ∈ SH m (Ω) ∩ L ∞ loc (Ω) be such that lim z→∂Ω u(z) = lim z→∂Ω v(z) and u ≤ v in Ω. Then Ω (dd c v) m ∧ β n-m ≤ Ω (dd c u) m ∧ β n-m .

Cegrell's inequalities for m-subharmonic functions

Let Ω be a bounded m-hyperconvex domain, that is, there exists a bounded continuous m-sh function ϕ : Ω → R -such that {ϕ < c} ⋐ Ω, for all c < 0.

We recall the definition of the class E 0 m (Ω).

Definition 1.3.21. We let E 0 m (Ω) denote the class of bounded functions v in SH m (Ω) such that lim z→∂Ω v(z) = 0 and Ω (dd

c v) m ∧ β n-m < +∞.
This class was introduced by Cegrell in [START_REF] Cegrell | Pluricomplex energy[END_REF], for m = n, and was considered by Lu in [START_REF] Lu | A variational approach to complex Hessian equations in C n[END_REF].

Lemma 1.3.22. Let u, v, v 1 , ..., v m-1 ∈ E 0 m (Ω) and T = dd c v 1 ∧ ... ∧ dd c v m-1 ∧ β n-m . Then we have Ω (-u)dd c v ∧ T ≤ Å Ω (-u)dd c u ∧ T ã 1/2 Å Ω (-v)dd c v ∧ T ã 1/2 . Proof. It is enough to note that (u, v) := Ω (-u)dd c v ∧ T
is symmetric semi positive bilinear form (using integration by parts). the required inequality follows from the classical Cauchy-Schwarz inequality for the form (u, v).

The following proposition was proved by induction in [START_REF] Cegrell | The general definition of the complex Monge-Ampère operator[END_REF] for plurisubharmonic functions and we can do the same argument for m-sh functions.

Proposition 1.3.23. Suppose that h, u 1 , u 2 ∈ E 0 m (Ω), p, q ≥ 1 such that p + q ≤ m and T = dd c g 1 ∧ ... ∧ dd c g m-p-q ∧ β n-m , where g 1 , ..., g m-p-q ∈ E 0 m (Ω). Then we get

Ω -h(dd c u 1 ) p ∧ (dd c u 2 ) q ∧ T ≤ ï Ω -h(dd c u 1 ) p+q ∧ T ò p p+q ï Ω -h(dd c u 2 ) p+q ∧ T ò q p+q .
Proof. We first prove the statement for p = q = 1. Thanks to the Cauchy-Schwarz inequality, we have

Ω -hdd c u 1 ∧ dd c u 2 ∧ T = Ω -u 1 dd c u 2 ∧ dd c h ∧ T ≤ ï Ω -u 1 dd c u 1 ∧ dd c h ∧ T ò 1/2 ï Ω -u 2 dd c u 2 ∧ dd c h ∧ T ò 1/2 = ï Ω -h(dd c u 1 ) 2 ∧ T ò 1/2 ï Ω -h(dd c u 2 ) 2 ∧ T ò 1/2 .
The general case follows by induction in the same way as in [START_REF] Cegrell | The general definition of the complex Monge-Ampère operator[END_REF].

We will need in this thesis the following particular case.

Corollary 1.3.24. Let u 1 , u 2 ∈ E 0 m (Ω). Then we have

Ω dd c u 1 ∧ (dd c u 2 ) m-1 ∧ β n-m ≤ ï Ω (dd c u 1 ) m ∧ β n-m ò 1 m ï Ω (dd c u 2 ) m ∧ β n-m ò m-1 m .
For m = n, we have the following result proved by Cegrell [START_REF] Cegrell | The general definition of the complex Monge-Ampère operator[END_REF].

Corollary 1.3.25. Let u 1 , u 2 ∈ E 0 (Ω). Then we have

Ω dd c u 1 ∧ (dd c u 2 ) n-1 ≤ ï Ω (dd c u 1 ) n ò 1 n ï Ω (dd c u 2 ) n ò n-1 n .
Chapter 2

Modulus of continuity of the solution to the Dirichlet problem 2.1 Introduction

Let Ω be a bounded domain in C n . Given ϕ ∈ C(∂Ω) and 0 ≤ f ∈ L 1 (Ω), we consider the following Dirichlet problem:

Dir(Ω, ϕ, f ) :      u ∈ P SH(Ω) ∩ C( Ω), (dd c u) n = f β n in Ω, u = ϕ on ∂Ω,
This problem was studied in the last decades by many authors. When Ω is a bounded strongly pseudoconvex domain with smooth boundary and f ∈ C( Ω), Bedford and Taylor showed that Dir(Ω, ϕ, f ) has a unique continuous solution U := U(Ω, ϕ, f ). Furthermore, it was proved in [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] that U ∈ Lip α ( Ω) when ϕ ∈ Lip 2α (∂Ω) and f 1/n ∈ Lip α ( Ω) (0 < α ≤ 1). In the nondegenerate case, i.e. 0 < f ∈ C ∞ ( Ω) and ϕ ∈ C ∞ (∂Ω), Caffarelli, Kohn, Nirenberg and Spruck proved in [START_REF] Caffarelli | The Dirichlet problem for nonlinear second order elliptic equations. II. Complex Monge-Ampère, and uniformly elliptic equations[END_REF] that U ∈ C ∞ ( Ω). However a simple example of Gamelin and Sibony shows that the solution is not, in general, better than C 1,1 -smooth when f ≥ 0 and smooth (see [START_REF] Gamelin | Subharmonicity for uniform algebras[END_REF]). Krylov proved that if ϕ ∈ C 3,1 (∂Ω) and

f 1/n ∈ C 1,1 ( Ω), f ≥ 0 then U ∈ C 1,1 ( Ω) (see [Kr89]).
For B-regular domains, B locki [B l96] proved the existence of a continuous solution to the Dirichlet problem Dir(Ω, ϕ, f ) when 0 ≤ f ∈ C( Ω).

In this chapter which is based on my paper [START_REF] Charabati | Hölder regularity for solutions to complex Monge-Ampère equations[END_REF], we consider the more general case where Ω is a bounded strongly hyperconvex Lipschitz domain for which the boundary does not need to be smooth (see the definition below) and we study the existence and regularity of solutions to Dir(Ω, ϕ, f ) when 0 ≤ f ∈ C( Ω).

The principal result in this chapter gives a sharp estimate for the modulus of continuity of the solution in terms of the modulus of continuity of the data ϕ, f . Theorem 2.1.1. Let Ω ⊂ C n be a bounded strongly hyperconvex Lipschitz domain, ϕ ∈ C(∂Ω) and 0 ≤ f ∈ C( Ω). Assume that ω ϕ is the modulus of continuity of ϕ and ω f 1/n is the modulus of continuity of f 1/n . Then the modulus of continuity of the unique solution U to Dir(Ω, ϕ, f ) has the following estimate

ω U (t) ≤ η(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), ω f 1/n (t), t 1/2 },
where η is a positive constant depending on Ω.

Remark 2.1.2. Here we will use an alternative description of the solution given by Theorem 2.3.2 to get an optimal control for the modulus of continuity of this solution in a strongly hyperconvex Lipschitz domain. This result was suggested by E. Bedford [START_REF] Bedford | Survey of pluri-potential theory, Several Complex Variables[END_REF] and proved in the case of strictly convex domains with f = 0 [START_REF] Bedford | Levi flat hypersurfaces in C 2 with prescribed boundary: stability[END_REF].

We also consider the case when the density in the Dirichlet problem depends on the unknown function:

(2.1.1)

       u ∈ P SH(Ω) ∩ C( Ω), (dd c u) n = F (z, u)β n in Ω, u = ϕ on ∂Ω,
where F : Ω × R → R + is a continuous function and nondecreasing in the second variable. We can prove a sharp estimate for the modulus of continuity of the solution to (2.1.1). Since the proof is similar to the one of Theorem 4.1.1 for complex Hessian equations, we do not mention it in this chapter.

Theorem 2.1.3. Let Ω be a bounded strongly hyperconvex Lipschitz domain in C n , ϕ ∈ C(∂Ω) and 0 ≤ F ∈ C( Ω × R) be a nondecreasing function in the second variable. Then there exists a unique continuous solution U to (2.1.1) and its modulus of continuity satisfies the following estimate

ω U (t) ≤ γ(1 + F 1/n L ∞ (K) ) max{ω ϕ (t 1/2 ), ω F 1/n (t), t 1/2 },
where γ is a positive constant depending only on Ω, K = Ω × {a}, a = sup ∂Ω |ϕ| and ω F 1/n (t) is given by

ω F 1/n (t) := sup y∈[-M,M ] sup |z 1 -z 2 |≤t |F 1/n (z 1 , y) -F 1/n (z 2 , y)|, with M := a + 2 diam(Ω) 2 sup Ω F 1/n (., -a).

Basic facts

Definition 2.2.1. A bounded domain Ω ⊂ C n is called a strongly hyperconvex Lipschitz (briefly SHL) domain if there exist a neighborhood Ω ′ of Ω and a Lipschitz plurisubharmonic defining function ρ : Ω ′ → R such that 1. Ω = {z ∈ Ω ′ ; ρ(z) < 0} and ∂Ω = {ρ = 0}, 2. there exists a constant c > 0 such that dd c ρ ≥ cβ in Ω in the weak sense of currents.

Example 2.2.2.

1.

Let Ω be a strictly convex domain, that is, there exists a Lipschitz defining function ρ such that ρ -c|z| 2 is convex for some c > 0. It is clear that Ω is a strongly hyperconvex Lipschitz domain.

2. A smooth strongly pseudoconvex bounded domain is a SHL domain (see [START_REF] Henkin | Theory of functions on complex manifolds[END_REF]).

3. The nonempty finite intersection of strongly pseudoconvex bounded domains with smooth boundary in C n is a bounded SHL domain. In fact, it is sufficient to set ρ = max{ρ i }. More generally a finite intersection of SHL domains is a SHL domain.

4. The domain

Ω = {z = (z 1 , • • • , z n ) ∈ C n ; |z 1 | + • • • + |z n | < 1} (n ≥
(det Q) 1 n (det H) 1 n = (det(H 1/2 .Q.H 1/2 )) 1 n ≤ 1 n tr(H 1/2 .Q.H 1/2 ). Then (det Q) 1 n (det H) 1 n ≤ 1 n tr(Q.H).
Consequently, we have

(det Q) 1 n ≤ inf{tr(H.Q) : H ∈ H + n and det(H) = n -n }.
Since Q ∈ H + n , we diagonalize it, then we get A = (λ ii ) ∈ H + n such that Q = P.A.P -1 where P is the transformation matrix. One can find a matrix H = (α ii ) ∈ H + n such that det(H) = n -n and (det A) 1 n = tr(A.H). Indeed, it suffices to set

α ii = ( i λ ii ) 1 n nλ ii .
Finally, (det Q)

1 n = (det A) 1 n = tr(H.A) =
h j k.δ k j = tr(H).
Using the inequality of arithmetic and geometric means, we have :

1 = (det I) 1 n ≤ tr(H), hence ∆ H (|z| 2 ) ≥ 1 for every matrix H ∈ H + n and det(H) = n -n .
The following result is well known (see [B l96]), but we will give here an alternative proof using ideas from the theory of viscosity due to Eyssidieux, Guedj and Zeriahi [START_REF] Eyssidieux | Viscosity solutions to degenerate complex Monge-Ampère equations[END_REF].

Proposition 2.2.8. Let u ∈ P SH(Ω) ∩ L ∞ (Ω) and 0 ≤ f ∈ C(Ω). Then the following conditions are equivalent:

(1) ∆ H u ≥ f 1/n in the weak sense of distributions, for any H ∈ H + n and det H = n -n . (2) (dd c u) n ≥ f β n in the weak sense of currents on Ω.

Proof. First, suppose that u ∈ C 2 (Ω). Then by Lemma 2.2.6 the inequality

∆ H u = n j,k=1 h j k ∂ 2 u ∂z j ∂ zk ≥ f 1/n , ∀H ∈ H + n , det(H) = n -n , is equivalent to Ç det( ∂ 2 u ∂z j ∂ zk ) å 1/n ≥ f 1/n .
The latter means that (dd c u) n ≥ f β n .

(1)⇒(2). Let (ρ ǫ ) be the standard family of regularizing kernels with supp ρ ǫ ⊂ B(0, ǫ) and B(0,ǫ) ρ ǫ = 1. Then the sequence u ǫ = u * ρ ǫ decreases to u, and we see that (1) implies ∆ H u ǫ ≥ (f 1/n ) ǫ . Since u ǫ is smooth, we use the first case and get (dd c u ǫ ) n ≥ ((f 1/n ) ǫ ) n β n , hence by applying the convergence theorem of Bedford and Taylor (Theorem 7.4 in [START_REF] Bedford | A new capacity for plurisubharmonic functions[END_REF]) we obtain (dd c u) n ≥ f β n .

(2)⇒(1). Fix x 0 ∈ Ω, and let q be a C 2 -function in a neighborhood B of x 0 such that u ≤ q in this neighborhood and u(x 0 ) = q(x 0 ). First step: We will show that dd c q x 0 ≥ 0. Indeed, for every small enough ball B ′ ⊂ B centered at x 0 , we have

u(x 0 ) -q(x 0 ) ≥ 1 V (B ′ ) B ′ (u -q)dV 2n , therefore 1 V (B ′ ) B ′ qdV 2n -q(x 0 ) ≥ 1 V (B ′ ) B ′ udV 2n -u(x 0 ) ≥ 0.
Since q is C 2 -smooth and the radius of B ′ tends to 0, it follows from Proposition 3.2.10 in [H94] that ∆q x 0 ≥ 0. For every positive definite Hermitian matrix H with det H = n -n , we make a linear change of complex coordinates T such that tr(HQ) = tr( Q) where Q = (∂ 2 q/∂w j ∂ wk ) and q = q • T -1 . Then ∆ H q(x 0 ) = tr(H.Q) = tr( Q) = ∆q(y 0 ). Indeed, we first make a unitary transformation T 1 such that tr(H.Q) = tr(S.Q 1 ) where S is a diagonal matrix with positive eigenvalues λ 1 , ..., λ n and Q 1 := ∂ 2 q 1 /∂x j ∂ xk with

q 1 = q • T -1
1 . Then we do another linear transformation T 2 :

C n → C n such that T 2 (x 1 , ..., x n ) := Ç x 1 √ λ 1 , ..., x n √ λ n å . Let us set q = q 1 • T -1 2 . We get that tr(S.Q 1 ) = λ 1 ∂ 2 q 1 ∂x 1 ∂ x1 + ... + λ n ∂ 2 q 1 ∂x n ∂ xn = ∂ 2 q ∂w 1 ∂ w1 + ... + ∂ 2 q ∂w n ∂ wn = tr( Q).
Hence ∆ H q(x 0 ) ≥ 0 for every H ∈ H + n and det H = n -n , so dd c q x 0 ≥ 0. Second step: We claim that (dd c q) n x 0 ≥ f (x 0 )β n . Suppose that there exists a point x 0 ∈ Ω and a C 2 -function q which satisfies u ≤ q in a neighborhood of x 0 and u(x 0 ) = q(x 0 ) such that (dd c q) n x 0 < f (x 0 )β n . We put

q ǫ (x) = q(x) + ǫ Ä x -x 0 2 -r 2 /2 ä for 0 < ǫ ≪ 1 small enough, we see that 0 < (dd c q ǫ ) n x 0 < f (x 0 )β n .
Since f is lower semi-continuous on Ω, there exists r > 0 such that

(dd c q ǫ ) n x ≤ f (x)β n , x ∈ B(x 0 , r). Then (dd c q ǫ ) n ≤ f β n ≤ (dd c u) n in B(x 0 , r) and q ǫ = q + ǫ r 2 2 ≥ q ≥ u on ∂B(x 0 , r), hence q ǫ ≥ u on B(x 0 , r) by the comparison principle. But q ǫ (x 0 ) = q(x 0 ) -ǫ r 2 2 = u(x 0 ) -ǫ r 2 2 < u(x 0 ), a contradiction.
Hence, from the first part of the proof, we get ∆ H q(x 0 ) ≥ f 1/n (x 0 ) for every point x 0 ∈ Ω and every C 2 -function q in a neighborhood of x 0 such that u ≤ q in this neighborhood and u(x 0 ) = q(x 0 ).

Assume that f > 0 and f ∈ C ∞ (Ω), then there exists

g ∈ C ∞ (Ω) such that ∆ H g = f 1/n . Hence ϕ = u -g is ∆ H -subharmonic (by Proposition 3.2.10', [H94]), from which it follows that ∆ H ϕ ≥ 0 and ∆ H u ≥ f 1/n .
In case f > 0 is merely continuous, we observe that

f = sup{w; w ∈ C ∞ , f ≥ w > 0}, so (dd c u) n ≥ f β n ≥ wβ n . Since w > 0 is smooth, we have ∆ H u ≥ w 1/n . Therefore, we get ∆ H u ≥ f 1/n . In the general case 0 ≤ f ∈ C(Ω), we observe that u ǫ (z) = u(z) + ǫ|z| 2 satisfies (dd c u ǫ ) n ≥ (f + ǫ n )β n ,
and so

∆ H u ǫ ≥ (f + ǫ n ) 1/n . Letting ǫ converge to 0, we get ∆ H u ≥ f 1/n for all H ∈ H + n and det H = n -n .
As a consequence of Proposition 2.2.8, we give an alternative description of the classical Perron-Bremermann family of subsolutions to the Dirichlet problem Dir(Ω, ϕ, f ). Definition 2.2.9. We denote by V(Ω, ϕ, f ) the family of subsolutions of Dir(Ω, ϕ, f ), that is

V(Ω, ϕ, f ) = {v ∈ P SH(Ω) ∩ C( Ω), v| ∂Ω ≤ ϕ and ∆ H v ≥ f 1/n , ∀H ∈ H + n , det H = n -n }.
Remark 2.2.10. We observe that V(Ω, ϕ, f ) = ∅. Indeed, let ρ be as in Definition 2.2.1 and A, B > 0 big enough, then Aρ -B ∈ V(Ω, ϕ, f ).

Furthermore, the family

V(Ω, ϕ, f ) is stable under finite maximum, that is if u, v ∈ V(Ω, ϕ, f ) then max{u, v} ∈ V(Ω, ϕ, f ). It is enough to show that (2.2.2) ∆ H (max{u, v}) ≥ min{∆ H u, ∆ H v} We set µ := min{∆ H u, ∆ H v} and suppose that µ({z; u(z) = v(z)}) = 0. Then in the open set Ω 1 = {u < v}, we have ∆ H (max{u, v}) = ∆ H v ≥ µ, and a similar consequence in the set Ω 2 = {v < u}. Since µ(Ω\(Ω 1 ∪Ω 2 )) = 0 and ∆ H (max{u, v}) ≥ 0, we get ∆ H (max{u, v}) ≥ min{∆ H u, ∆ H v}.
In the general case, we replace v by v + ǫ, where ǫ > 0 is a small constant, then max{u, v + ǫ} → max{u, v}. Thus ∆ H (max{u, v + ǫ}) converges to ∆ H (max{u, v}) in the sense of distributions.

We set µ = min{∆ H u, ∆ H (v + ǫ)}, by the first case the inequality is true for max{u, v + ǫ} for all ǫ > 0 such that µ({z; u(z) = v(z) + ǫ}) = 0. On the other hand, µ({z; u(z) = v(z) + ǫ}) = 0 for all ǫ > 0 except at most countably many ǫ > 0, then we obtain (2.2.2) by passing to the limit when ǫ → 0 (avoiding these countably many values of ǫ > 0).

The Perron-Bremermann envelope

Bedford and Taylor proved in [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] that the unique solution to Dir(Ω, ϕ, f ) in a bounded strongly pseudoconvex domain with smooth boundary, is given as the Perron-Bremermann envelope

u = sup{v; v ∈ B(Ω, ϕ, f )}, where B(Ω, ϕ, f ) = {v ∈ P SH(Ω) ∩ C( Ω) : v| ∂Ω ≤ ϕ and (dd c v) n ≥ f β n }.
Thanks to Proposition 2.2.8, we get the following corollary Corollary 2.3.1. The two families V(Ω, ϕ, f ) and B(Ω, ϕ, f ) coincide, that is

V(Ω, ϕ, f ) = B(Ω, ϕ, f ).
The context of this section is classical and follows the main scheme of Bedford and Taylor's approach. A simplification of their proof was given by Demailly (for the homogeneous case ([De89])) and by B locki for the general case (see [B l96]). Here we will prove the following theorem using an alternative description of the Perron-Bremermann envelope in a bounded SHL domain.

Theorem 2.3.2. Let Ω ⊂ C n be a bounded SHL domain, 0 ≤ f ∈ C( Ω) and ϕ ∈ C(∂Ω).
Then the Dirichlet problem Dir(Ω, ϕ, f ) has a unique solution U. Moreover the solution is given by

U = sup{v; v ∈ V(Ω, ϕ, f )},
where V is defined in Definition 2.2.9 and ∆ H is the Laplacian associated to a positive definite Hermitian matrix H as in (2.2.1).

The uniqueness of the solution to Dir(Ω, ϕ, f ) is a consequence of the comparison principle (Corollary 1.2.7).

The first step to prove this theorem is to ensure that U ∈ V(Ω, ϕ, f ). For this purpose, we use the argument of Walsh (see [START_REF] Walsh | Continuity of envelopes of plurisubharmonic functions[END_REF] and [B l96] ) to prove the continuity of the upper envelope.

Continuity of the upper envelope

Proposition 2.3.3. Let Ω ⊂ C n be a bounded SHL domain, 0 ≤ f ∈ C( Ω) and ϕ ∈ C(∂Ω). Then the upper envelope U = sup{v; v ∈ V(Ω, ϕ, f )} belongs to V(Ω, ϕ, f ) and U = ϕ on ∂Ω.
Proof. Let g ∈ C 2 ( Ω) be an approximation of ϕ such that |g -ϕ| ≤ ǫ on ∂Ω, for fixed ǫ > 0. Let also ρ be the defining function as in Definition 2.2.1 and A > 0 large enough such that

v 0 := Aρ + g -ǫ belongs to V(Ω, ϕ, f ) and ∆ H v 0 ≥ max{sup Ω f 1/n , 1}. A similar construction gives that v 1 := -Bρ + g + ǫ is plurisuperharmonic in Ω when B > 0 is big enough. We claim that U ≤ v 1 in Ω. Suppose that v ∈ V(Ω, ϕ, f ), then v -v 1 ≤ ϕ -g -ǫ ≤ 0 on ∂Ω. Hence, by the maximum principle we get v -v 1 ≤ 0 in Ω. This yields U ≤ v 1 in Ω. Consequently, we get v 0 ≤ U ≤ v 1 . Then on the boundary ∂Ω we have ϕ -2ǫ ≤ g -ǫ ≤ U ≤ g + ǫ ≤ ϕ + 2ǫ.
Letting ǫ tend to 0, we obtain that U = ϕ on ∂Ω and lim z→ξ U(z) = ϕ(ξ) for all ξ ∈ ∂Ω.

We will prove that U is continuous on Ω. Fix ǫ > 0 and z 0 in a compact set K ⋐ Ω. Thanks to the continuity of v 1 and v 0 on Ω, one can find δ > 0 such that for any z 1 , z 2 ∈ Ω we have

|v 1 (z 1 ) -v 1 (z 2 )| ≤ ǫ, |v 0 (z 1 ) -v 0 (z 2 )| ≤ ǫ, if |z 1 -z 2 | ≤ δ.
Let a ∈ C n such that |a| < min{δ, dist(K, ∂Ω)}. Since U is the upper envelope, we can find ṽ ∈ V(Ω, ϕ, f ) such that ṽ(z 0 + a) ≥ U(z 0 + a)ǫ. Let us set v = max{ṽ, v 0 }.

Hence, for all z ∈ Ω and w ∈ ∂Ω such that |z -w| ≤ δ we get

-3ǫ ≤ v 0 (z) -ϕ(w) ≤ v(z) -ϕ(w) ≤ v 1 (z) -ϕ(w) ≤ 3ǫ.
This implies that

(2.3.1) |v(z) -ϕ(w)| ≤ 3ǫ, if |z -w| ≤ δ.
Then for z ∈ Ω and z + a ∈ ∂Ω, we have

v(z + a) ≤ ϕ(z + a) ≤ v(z) + 3ǫ.
We define the following function

v 1 (z) = ® v(z) ; z + a / ∈ Ω, max{v(z), v(z + a) -3ǫ} ; z + a ∈ Ω, which is well defined, plurisubharmonic on Ω, continuous on Ω and v 1 ≤ ϕ on ∂Ω. Indeed, if z ∈ ∂Ω, z+a / ∈ Ω then v 1 (z) = v(z) ≤ ϕ(z). On the other hand, if z ∈ ∂Ω and z+a ∈ Ω then we have, from (2.3.1), that v(z +a)-3ǫ ≤ ϕ(z), so v 1 (z) = max{v(z), v(z +a)-3ǫ} ≤ ϕ(z). Moreover, we note by (2.2.2) that ∆ H v 1 (z) ≥ min(f 1/n (z), f 1/n (z + a)) if z, z + a ∈ Ω.
Let ω be the modulus of continuity of f 1/n . Then we conclude that

(2.3.2) ∆ H v 1 (z) ≥ f 1/n (z) -ω(|a|) in Ω. Now, let us define v 2 = v 1 + ω(|a|)(v 0 -v 0 L ∞ ( Ω) ). It is clear that v 2 ∈ P SH(Ω) ∩ C( Ω) and v 2 ≤ ϕ on ∂Ω. Furthermore, using (2.3.2) we see that ∆ H v 2 = ∆ H v 1 + ω(|a|)∆ H v 0 ≥ f 1/n . This yields that v 2 ∈ V(Ω, ϕ, f ).
For small enough |a| we can assume ω(|a|) ≤ ǫ/ v 0 and infer that

U(z 0 ) ≥ v 1 (z 0 ) + ω(|a|)v 0 (z 0 ) -ω(|a|) v 0 ≥ v(z 0 + a) -5ǫ ≥ U(z 0 + a) -6ǫ.
The last inequality is true for every z 0 ∈ K, then U is continuous on Ω.

It follows from Choquet's lemma that there exists a sequence (u

j ) in V(Ω, ϕ, f ) such that U = (sup j u j ) * .
As the family V(Ω, ϕ, f ) is stable under the operation maximum, we can assume that the sequence (u j ) is increasing almost everywhere to U, then

u j → U in L 1 (Ω). Hence ∆ H U = lim ∆ H u j ≥ f 1/n for all H ∈ H + n , detH = n -n , this implies U ∈ V(Ω, ϕ, f ).
In order to verify that (dd c U) n = f β n in Ω, we first ensure this statement when Ω = B the unit ball in C n . For this end, we introduce the following theorem, which is due to Bedford and Taylor [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF], to prove that the second order derivatives of U are locally bounded under extra assumptions. Here the presentation is derived from Demailly [START_REF] Demailly | Potential theory in several complex variables[END_REF].

Regularity in the case of the unit ball

Theorem 2.3.4. Suppose that Ω = B is the unit ball in C n , f 1/n ∈ C 1,1 ( B) and ϕ ∈ C 1,1 (∂B).
Then the second order partial derivatives of U are locally bounded, in particular

U ∈ C 1,1 loc (B).
Proof. First, we assert that U ∈ C 0,1 ( B). Actually, let φ be a

C 1,1 -extension of ϕ to B2 := B(0, 2) such that φ C 1,1 ( B2 ) ≤ C ϕ C 1,1 (∂B)
for some positive constant C (see [START_REF] Gilbarg | Elliptic partial differential equations of second order[END_REF]).

Let us set

A ≫ 1 such that u 1 = A(|z| 2 -1) + φ is plurisubharmonic on B 2 and u 2 = A(1 - |z| 2 ) + φ is plurisuperharmonic on B 2 . For A big enough, one can note that u 1 ≤ U ≤ u 2 on B by the comparison principle. We set ũ(z) = ® U(z) ; z ∈ B, u 1 (z) ; z ∈ B2 \ B.
Since u 1 = U = ϕ on ∂B, we get a well defined plurisubharmonic function ũ on B 2 and ũ ≤ max{u 1 , u 2 } on B2 . Then for all z ∈ ∂B and |h| small we get

ũ(z + h) ≤ ϕ(z) + C 1 max{ u 1 C 1 ( B2 ) , u 2 C 1 ( B2 ) }|h| ≤ ϕ(z) + C 2 |h|,
where

C 2 = C 1 (A + C ϕ C 1,1 (∂B) ). Since f 1/n ∈ C 1,1 ( B)
, there exists a constant B such that

|f 1/n (z) -f 1/n (y)| ≤ B|z -y|.
Now, let us define the function

û(z) = ũ(z + h) -C 2 |h| + B|h|(|z| 2 -1). It is clear that û ∈ P SH(B) ∩ C( B), û| ∂B ≤ ϕ and ∆ H û ≥ f 1/n for all H ∈ H + n and det H = n -n . Thus we have û ∈ V(B, ϕ, f ) and û ≤ U on B. This implies that ũ(z + h) -U(z) ≤ (C 2 + B)|h| on B.
By changing h into -h, we conclude that

|U(z + h) -U(z)| ≤ (C 2 + B)|h|,
for z ∈ B and |h| small. This yields that U C 0,1 ( B) ≤ (C 2 + B).

Second step, we estimate the following expression

U(z + h) + U(z -h) -2U(z).
But this expression is not defined in the whole ball B, thus we use the automorphism of the unit ball. For a ∈ B, we define a holomorphic automorphism T a of the unit ball as follows;

T a (z) = P a (z) -a + » 1 -|a| 2 (z -P a (z)) 1 -z, a ; P a (z) = z, a a |a| 2 ,
where ., . denote the Hermitian product in C n . Let h = az, a z. Then we get for |a| ≪ 1 that

T a (z) = z -h + O(|a| 2 ),
where O(|a| 2 ) is bounded and converges to 0 when |a| tends to 0, i.e. O(|a| 2 ) ≤ c|a| 2 , for some positive constant c which is uniform for z ∈ B. The determinant of Jacobian matrix of T a is given by

det T ′ a (z) = 1 + (n + 1) z, a + O(|a| 2 ). Then det T ′ a (z) 2/n = 1 + 2(n + 1) n z, a + O(|a| 2 ).
Let g ∈ C 0,1 ( B), so it is easy to see that

(2.3.3) |g • T a (z) -g(z -h)| ≤ g C 0,1 ( B) .|T a (z) -z + h| ≤ c 1 g C 0,1 ( B) .|a| 2 .
Since f 1/n ∈ C 1,1 ( B), we get by Taylor's expansion

f 1/n • T a (z) = f 1/n (z -h + O(|a| 2 )) = f 1/n (z) -Df 1/n (z).h + O(|a| 2 ).
We set ψ(z, a) = -Df 1/n (z).h, then

f 1/n • T a (z) = f 1/n (z) + ψ(z, a) + O(|a| 2 ).
A simple computation yields that the following expression

I := | det T ′ a | 2/n (f 1/n • T a ) + | det T ′ -a | 2/n (f 1/n • T -a ),
can be estimated as follows

I ≥ 2f 1/n - 4(n + 1) n | z, a ψ(z, a)| -c 2 |a| 2 .
There exists

c 3 > 0 depending on f 1/n C 1,1 ( Ω) such that | z, a ψ(z, a)| ≤ c 3 |z|.|a| 2 ≤ c 3 |a| 2 .
Hence we get

| det T ′ a | 2/n (f 1/n • T a ) + | det T ′ -a | 2/n (f 1/n • T -a ) ≥ 2f 1/n -c 4 |a| 2 .
A similar computation yields that the following inequality holds on ∂B

(2.3.4) ϕ • T a + ϕ • T -a ≤ 2ϕ + c 4 |a| 2 ,
where c 4 is large and depending also on

ϕ C 1,1 (∂B) . Let us consider v a (z) := (U • T a + U • T -a )(z). We observe that ∆ H (U • T a ) ≥ | det T ′ a | 2/n (f 1/n • T a ), then we get ∆ H v a ≥ | det T ′ a | 2/n (f 1/n • T a ) + | det T ′ -a | 2/n (f 1/n • T -a ) ≥ 2f 1/n -c 4 |a| 2 . Let us put v(z) := 1 2 v a (z) - c 4 2 |a| 2 (2 -|z| 2 ) ∈ P SH(B) ∩ C( B).
It follows from (2.3.4) that v ≤ ϕ on ∂B. Moreover, we have

∆ H v = 1 2 ∆ H v a + c 4 2 |a| 2 ∆ H (|z| 2 ) ≥ f 1/n - c 4 2 |a| 2 + c 4 2 |a| 2 ≥ f 1/n , for every H ∈ H + n , det H = n -n . Hence v ∈ V(B, ϕ, f ), in particular v ≤ U . Consequently, 1 2 v a (z) -c 4 |a| 2 ≤ 1 2 v a (z) - c 4 2 |a| 2 (2 -|z| 2 ) ≤ U.
Hence, we get

(U • T a + U • T -a )(z) -2U(z) ≤ 2c 4 |a| 2 .
Applying (2.3.3) with g = U, we obtain

(2.3.5)

U(z -h) + U(z + h) -2U(z) ≤ (U • T a + U • T -a )(z) -2U(z) + 2c 1 U C 0,1 ( B) .|a| 2 ≤ (2c 4 + 2c 1 U C 0,1 ( B) ).|a| 2 ≤ c 5 |a| 2 .
Since h = az, a z, the inverse linear map h → a has a norm less than 1/(1 -|z| 2 ). Indeed, using the Cauchy-Schwarz inequality, we have

|h| ≥ ||a| -| z, a |.|z|| ≥ ||a| -|z| 2 |a|| ≥ |a|(1 -|z| 2 ).
Thus we conclude that

U(z + h) + U(z -h) -2U(z) ≤ c 5 (1 -|z| 2 ) 2 |h| 2 .
Let us fix a compact K ⊂ B. For z ∈ K and |h| small enough we obtain by taking a convolution with a regularizing kernel ρ ǫ , for small enough ǫ > 0, that

U ǫ (z + h) + U ǫ (z -h) -2U ǫ (z) ≤ c 5 (1 -(|z| + ǫ) 2 ) 2 |h| 2 . Since U ǫ ∈ P SH ∩ C ∞ (B ǫ )
where B ǫ is the ball of radius 1ǫ and thanks to Taylor's expansion of degree two of u ǫ , we infer

D 2 U ǫ (z).h 2 ≤ c 5 (1 -(|z| + ǫ) 2 ) 2 |h| 2 . Let us set A := 2c 5 dist(K, ∂B) 2 .
Then for all z ∈ K and h ∈ C n with small enough norm we get

D 2 U ǫ (z).h 2 ≤ A|h| 2 .
The plurisubharmonicity of U ǫ yields

D 2 U ǫ (z).h 2 + D 2 U ǫ (z).(ih) 2 = 4 j,k ∂ 2 U ǫ ∂z j ∂ zk .h j hk ≥ 0. Hence D 2 U ǫ (z).h 2 ≥ -D 2 U ǫ (z).(ih) 2 ≥ -A|h| 2 .
Therefore, we have

|D 2 U ǫ (z)| ≤ A; ∀z ∈ K.
We know that the dual space of L 1 (K) is L ∞ (K), hence by applying the Alaoglu-Banach theorem, there exists a bounded function g such that D 2 U ǫ converges weakly to g in L ∞ (K).

On the other hand, D 2 U ǫ → D 2 U in the sense of distributions, then we get D 2 U = g. Finally, the second order derivatives of U exist almost everywhere and are locally bounded in B with

D 2 U L ∞ (K) ≤ A,
where A = 2c 5 / dist(K, ∂B) 2 and c 5 depends on

U C 0,1 ( B) , ϕ C 1,1 (∂B) and f 1/n C 1,1 ( B) . Thus we conclude that U ∈ C 1,1 loc (B).
Remark 2.3.5. Dufresnoy [Du89] proved that the C 1,1 -norm of U does not explode faster than 1/dist(., ∂B) as we approach to the boundary. In general, U can not belong to C 1,1 ( B), the next example shows that there is a necessary loss in the regularity up to the boundary.

Example 2.3.6. Let B ⊂ C 2 and ϕ(z, w) = (1 + Re(w)) 1+ǫ ∈ C 2,2ǫ (∂B) for small ǫ > 0.

We consider the following Dirichlet problem:

     u ∈ P SH(Ω) ∩ C( Ω), (dd c U) 2 = 0 in B, U = ϕ on ∂B.
Then U(z, w) = (1 + Re(w)) 1+ǫ is the solution to this problem. One can observe that U belongs to C 1,ǫ ( B) ∩ C 1,1 loc (B) but it is not C 1,1 -smooth on B. This can be seen by a radial approach to the boundary point (z 0 , w 0 ) = (0, -1).

We will prove in the following proposition that the Perron-Bremermann envelope is the solution to the Dirichlet problem in the unit ball B.

Proposition 2.3.7. Suppose 0 ≤ f 1/n ∈ C 1,1 ( B) and ϕ ∈ C 1,1 (∂B). Then the envelope U is the solution to the Dirichlet problem Dir(B, ϕ, f ).

Proof. We have proved that U ∈ C 1,1 loc (B) and U ∈ V(B, ϕ, f ). It remains to show that (dd c U) n = f β n . Proof by contradiction, suppose that there exists a point z 0 ∈ B at which U has second order partial derivatives and satisfies

(dd c U) n (z 0 ) > (f (z 0 ) + ǫ)β n ,
for some ǫ > 0. Then by Proposition 2.2.8 we have

∆ H U(z 0 ) > (f (z 0 ) + ǫ) 1/n ,
for all H ∈ H + n and det(H) = n -n . Using the Taylor expansion at z 0 , we get

U(z 0 + ξ) = U(z 0 ) + DU(z 0 ).ξ + 1 2 j,k ∂ 2 U ∂z j ∂z k (z 0 )ξ j ξ k + + 1 2 j,k ∂ 2 U ∂ zj ∂ zk (z 0 ) ξj ξk + j,k ∂ 2 U ∂z j ∂ zk (z 0 )ξ j ξk + o(|ξ| 2 ). = U(z 0 ) + ReP (ξ) + L(ξ) + o(|ξ| 2 ),
where P is a complex polynomial of degree 2, then ReP is pluriharmonic and

L(ξ) = j,k ∂ 2 U ∂z j ∂ zk (z 0 )ξ j ξk > 0.
Let us fix

s := Ç f (z 0 ) + ǫ/2 f (z 0 ) + ǫ å 1/n < 1.
One can find δ, r > 0 small enough such that B(z 0 , r) ⋐ B and for |ξ| = r, we have

U(z 0 ) + ReP (ξ) + sL(ξ) + δ ≤ U(z 0 + ξ).
We define the function

v(z) = ® U(z) ; z / ∈ B(z 0 , r), max{U(z), v 1 (z)} ; z ∈ B(z 0 , r), where v 1 (z) := U(z 0 ) + ReP (z -z 0 ) + sL(z -z 0 ) + δ is a psh function in B(z 0 , r). It is clear that v is well defined psh in B and satisfies v = ϕ on ∂B. We claim that ∆ H v ≥ f 1/n for all H ∈ H + n and det(H) = n -n . Indeed, in the ball B(z 0 , r) we note ∆ H v 1 ≥ s∆ H L(z -z 0 ) = s j,k ∂ 2 U ∂z j ∂ zk (z 0 )h k j > s(f (z 0 ) + ǫ) 1/n = (f (z 0 ) + ǫ/2) 1/n .
Since f is uniformly continuous in B, shrinking r if necessary, we can get that f Proof. We choose a sequence of functions (f j ) such that 0 < f j ∈ C ∞ ( B) and f j decreases to f uniformly on B. We also find a sequence C ∞ -smooth functions ϕ j such that ϕ j increases to ϕ uniformly on ∂B. Thanks to the last proposition, there exists a continuous solution U j to the Dirichlet problem Dir(B, ϕ j , f j ). Hence, by the comparison principle, we can conclude that the sequence U j is increasing. Fix ǫ > 0 and since f k converges uniformly to f , we find j 0 > 0 such that

(z 0 )+ǫ/2 ≥ f (z) for z ∈ B(z 0 , r), hence ∆ H v 1 (z) ≥ f 1/n (z) in B(z 0 , r). Consequently, it follows from (2.2.2) that ∆ H v ≥ f 1/n . Thus we infer v ∈ V(B, ϕ, f ) and v ≤ U in B. But we observe that v(z 0 ) = U(z 0 ) + δ > U(z 0 ),
f j ≤ f k + ǫ n in B for all k ≥ j ≥ j 0 . Then we note for all k ≥ j ≥ j 0 that (dd c (U k + ǫ(|z| 2 -1))) n ≥ (f k + ǫ n )β n ≥ f j β n = (dd c U j ) n .
Moreover, we can find j 1 large enough such that ϕ j + ǫ ≥ ϕ k on ∂B for all k ≥ j ≥ j 1 . Then for k ≥ j ≥ max{j 0 , j 1 } we have

(dd c (U k + ǫ(|z| 2 -1))) n ≥ (dd c U j ) n in B,
and

U k + ǫ(|z| 2 -1) ≤ U j + ǫ on ∂B.
Hence by the comparison principle we get that for all k ≥ j ≥ max{j 0 , j 1 }

U k -U j ≤ 2ǫ -ǫ|z| 2 ≤ 2ǫ in B.
On the other hand, U j ≤ U k , so we infer

U k -U j L ∞ ( B) ≤ 2ǫ.
This implies that the sequence (U j ) converges uniformly in B.

Let us put u = lim j→∞ U j which is continuous on B, plurisubharmonic on B and u = ϕ on ∂B. Moreover, (dd c U j ) n converges to (dd c u) n in the weak sense of currents, then

(dd c u) n = f β n . Consequently, u is a candidate in the Perron-Bremermann envelope, i.e. u ∈ V(B, ϕ, f ) and u ≤ U in B. Once again the comparison principle yields u ≥ U in B. Finally, we conclude u = U in B and (dd c U) n = f β n in B.
Proof of Theorem 2.3.2 . We already know as in Proposition 2.3.

3 that U ∈ P SH(Ω)∩ C( Ω), U = ϕ on ∂Ω and (dd c U) n ≥ f β n in Ω. It remains to prove that (dd c U) n = f β n in Ω.
We use the balayage procedure as follows; Fix a ball B 0 ⊂ Ω. Thanks to Corollary 2.3.8, there exists a unique solution ψ to Dir(B 0 , U, f ), that is

(dd c ψ) n = f β n in B 0 and ψ = U on ∂B 0 .
By the comparison principle U ≤ ψ on B 0 . Let us define the function

v(z) = ® ψ(z) ; z ∈ B 0 , U(z) ; z ∈ Ω \ B 0 , which belongs to V(Ω, ϕ, f ) and v = U = ϕ on ∂Ω. In particular v ≤ U, hence ψ ≤ U in B 0 . Consequently, ψ = U in B 0 . Then (dd c U) n = (dd c ψ) n = f β n in B 0 . Since B 0 is an arbitrary ball in Ω, we infer that (dd c U) n = f β n in Ω.

Stability estimates

Proposition 2.3.9. Let ϕ 1 , ϕ 2 ∈ C(∂Ω) and f 1 , f 2 ∈ C( Ω). Then the solutions U 1 = U(Ω, ϕ 1 , f 1 ), U 2 = U(Ω, ϕ 2 , f 2 ) satisfy the following stability estimate (2.3.6) U 1 -U 2 L ∞ ( Ω) ≤ d 2 f 1 -f 2 1/n L ∞ ( Ω) + ϕ 1 -ϕ 2 L ∞ (∂Ω) , where d := diam(Ω).
Proof. Let us fix z 0 ∈ Ω and define

v 1 (z) = f 1 -f 2 1/n L ∞ ( Ω) (|z -z 0 | 2 -d 2 ) + U 2 (z), and v 2 (z) = U 1 (z) + ϕ 1 -ϕ 2 L ∞ (∂Ω) . It is clear that v 1 , v 2 ∈ P SH(Ω) ∩ C( Ω), v 1 ≤ v 2 on ∂Ω and (dd c v 1 ) n ≥ (dd c v 2 ) n in Ω.
Hence, by the comparison principle, we get v 1 ≤ v 2 in Ω. Then we conclude that

U 2 -U 1 ≤ d 2 f 1 -f 2 1/n L ∞ ( Ω) + ϕ 1 -ϕ 2 L ∞ (∂Ω)
. By reversing the roles of U 1 and U 2 , we get the inequality (2.3.6).

Remark 2.3.10. We will need in the sequel an estimate, proved by B locki in [B l93], for the L n -L 1 stability of solutions to the Dirichlet problem Dir(Ω, ϕ, f )

(2.3.7) U 1 -U 2 L n (Ω) ≤ λ(Ω) ϕ 1 -ϕ 2 L ∞ (∂Ω) + r 2 4 f 1 -f 2 1/n L 1 (Ω) ,
where r = min{r ′ > 0 : Ω ⊂ B(z 0 , r ′ ) for some z 0 ∈ C n }.

The modulus of continuity of Perron-Bremermann envelope

Recall that a real function ω on [0, l], 0 < l < ∞, is called a modulus of continuity if ω is continuous, subadditive, nondecreasing and ω(0) = 0.

In general, ω fails to be concave, we denote by ω the minimal concave majorant of ω. We denote by ω ψ the optimal modulus of continuity of the continuous function ψ which is defined by

ω ψ (t) = sup |x-y|≤t |ψ(x) -ψ(y)|.
The following property of the minimal concave majorant ω is well known (see [START_REF] Korneichuk | Precise constant in Jackson's inequality for continuous periodic functions[END_REF] and [START_REF] Charabati | Modulus of continuity of solutions to complex Hessian equations[END_REF]).

Lemma 2.4.1. Let ω be a modulus of continuity on [0, l] and ω be the minimal concave majorant of ω. Then ω(ηt) ≤ ω(ηt) ≤ (1 + η)ω(t) for any t > 0 and η > 0.

Proof. Fix t 0 > 0 such that ω(t 0 ) > 0. We claim that

ω(t) ω(t 0 ) ≤ 1 + t t 0 , ∀t ≥ 0.
For 0 < t ≤ t 0 , since ω is nondecreasing, we have

ω(t) ω(t 0 ) ≤ ω(t 0 ) ω(t 0 ) ≤ 1 + t t 0 .
Otherwise , if t 0 ≤ t ≤ l, by Euclid's Algorithm, we write t = kt 0 + α, 0 ≤ α < t 0 and k is natural number with 1 ≤ k ≤ t/t 0 . Using the subadditivity of ω, we observe that

ω(t) ω(t 0 ) ≤ kω(t 0 ) + ω(α) ω(t 0 ) ≤ k + 1 ≤ 1 + t t 0 .
Let l(t) := ω(t 0 ) + t t 0 ω(t 0 ) be a straight line, then ω(t) ≤ l(t) for all 0 < t ≤ l. Therefore,

ω(t) ≤ l(t) = ω(t 0 ) + t t 0 ω(t 0 ),
for all 0 < t ≤ l. Hence, for any η > 0 we have

ω(ηt) ≤ ω(ηt) ≤ (1 + η)ω(t).

Modulus of continuity of the solution

Now, we will start the first step to establish an estimate for the modulus of continuity of the solution to Dir(Ω, ϕ, f ). For this purpose, it is natural to investigate the relation between the modulus of continuity of U and the modulus of continuity of a subbarrier and a superbarrier. We prove the following:

Proposition 2.4.2. Let Ω ⊂ C n be a bounded SHL domain, ϕ ∈ C(∂Ω) and 0 ≤ f ∈ C( Ω). Suppose that there exist v ∈ V(Ω, ϕ, f ) and w ∈ SH(Ω) ∩ C( Ω) such that v = ϕ = -w on ∂Ω, then the modulus of continuity of U satisfies ω U (t) ≤ (d 2 + 1) max{ω v (t), ω w (t), ω f 1/n (t)},
where d := diam(Ω).

Proof. Let us set g(t) := max{ω v (t), ω w (t), ω f 1/n (t)}. As v = ϕ = -w on ∂Ω, we have for all z ∈ Ω and ξ ∈ ∂Ω that -g(|z -ξ|) ≤ v(z) -ϕ(ξ) ≤ U(z) -ϕ(ξ) ≤ -w(z) -ϕ(ξ) ≤ g(|z -ξ|). Hence (2.4.1) |U(z) -U(ξ)| ≤ g(|z -ξ|), ∀z ∈ Ω, ∀ξ ∈ ∂Ω.
Fix a point z 0 ∈ Ω. For any vector τ ∈ C n with small enough norm, we set Ω -τ := {zτ ; z ∈ Ω} and define in Ω ∩ Ω -τ the function

v 1 (z) = U(z + τ ) + g(|τ |)|z -z 0 | 2 -d 2 g(|τ |) -g(|τ |),
which is a well defined psh function in Ω ∩ Ω -τ and continuous on Ω ∩ Ω-τ . By (2.4.1), if

z ∈ Ω ∩ ∂Ω -τ we can see that (2.4.2) v 1 (z) -U(z) ≤ g(|τ |) + g(|τ |)|z -z 0 | 2 -d 2 g(|τ |) -g(|τ |) ≤ 0. Moreover, we assert that ∆ H v 1 ≥ f 1/n in Ω ∩ Ω -τ for all H ∈ H + n , det H = n -n . Indeed, we have ∆ H v 1 (z) ≥ f 1/n (z + τ ) + g(|τ |)∆ H (|z -z 0 | 2 ) ≥ f 1/n (z + τ ) + g(|τ |) ≥ f 1/n (z + τ ) + |f 1/n (z + τ ) -f 1/n (z)| ≥ f 1/n (z)
for all H ∈ H + n and det H = n -n . Hence, by the above properties of v 1 , we find that

V τ (z) = ® U(z) ; z ∈ Ω \ Ω -τ , max{U(z), v 1 (z)} ; z ∈ Ω ∩ Ω -τ ,
is a well defined function and belongs to

P SH(Ω) ∩ C( Ω). It is clear that ∆ H V τ ≥ f 1/n for all H ∈ H + n , det H = n -n . We claim that V τ = ϕ on ∂Ω. If z ∈ ∂Ω \ Ω -τ then V τ (z) = U(z) = ϕ(z). On the other hand, if z ∈ ∂Ω ∩ Ω -τ , we get by (2.4.2) that V τ (z) = max{U(z), v 1 (z)} = U(z) = ϕ(z). Consequently, V τ ∈ V(Ω, ϕ, f
) and this implies that

V τ (z) ≤ U(z); ∀z ∈ Ω.
Then for all z ∈ Ω ∩ Ω -τ we have

U(z + τ ) + g(|τ |)|z -z 0 | 2 -d 2 g(|τ |) -g(|τ |) ≤ U(z).
Hence,

U(z + τ ) -U(z) ≤ (d 2 + 1)g(|τ |) -g(|τ |)|z -z 0 | 2 ≤ (d 2 + 1)g(|τ |).
Reversing the roles of z + τ and z, we get

|U(z + τ ) -U(z)| ≤ (d 2 + 1)g(|τ |), ∀z, z + τ ∈ Ω.
Thus, finally,

ω U (|τ |) ≤ (d 2 + 1) max{ω v (|τ |), ω w (|τ |), ω f 1/n (|τ |)}.
Remark 2.4.3. Let H ϕ be the harmonic extension of ϕ in a bounded SHL domain Ω. We can replace w in the last proposition by H ϕ . It is known in the classical harmonic analysis (see [START_REF] Aikawa | Modulus of continuity of the Dirichlet solutions[END_REF]) that the harmonic extension H ϕ does not have, in general, the same modulus of continuity of ϕ.

Let us define, for small positive t, the modulus of continuity

ψ α,β (t) = (-log(t)) -α t β
with α ≥ 0 and 0 ≤ β < 1. It is clear that ψ α,0 is weaker than the Hölder continuity and ψ 0,β is the Hölder continuity. It was shown in [START_REF] Aikawa | Hölder continuity of the Dirichlet solution for a general domain[END_REF] that ω Hϕ (t) ≤ cψ 0,β (t) for some c > 0 if ω ϕ (t) ≤ c 1 ψ 0,β (t) for β < β 0 where β 0 < 1 depends only on n and the Lipschitz constant of the defining function ρ. Moreover, a similar result was proved in [START_REF] Aikawa | Modulus of continuity of the Dirichlet solutions[END_REF] for the modulus of continuity ψ α,0 (t). However, the same argument of Aikawa gives that ω Hϕ (t) ≤ cψ α,β (t) for some c > 0 if ω ϕ (t) ≤ c 1 ψ α,β (t) for α ≥ 0 and 0 ≤ β < β 0 < 1. This leads us to the conclusion that if there exists a barrier v to the Dirichlet problem such that v = ϕ on ∂Ω and ω v (t) ≤ λψ α,β (t) with α, β as above, then the last proposition gives

ω U ≤ λ 1 max{ψ α,β (t), ω f 1/n (t)},
where λ 1 > 0 depends on λ and diam(Ω).

Construction of barriers

In this subsection, we will construct a subsolution to the Dirichlet problem with the boundary value ϕ and estimate its modulus of continuity.

Proposition 2.4.4. Let Ω ⊂ C n be a bounded SHL domain. Assume that ϕ ∈ C(∂Ω) and 0 ≤ f ∈ C( Ω), then there exists a subsolution v ∈ V(Ω, ϕ, f ) such that v = ϕ on ∂Ω and the modulus of continuity of v satisfies the following inequality

ω v (t) ≤ λ(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), t 1/2 }, where λ > 0 depends on Ω.
Observe that we do not assume any smoothness on ∂Ω.

Proof. First of all, fix ξ ∈ ∂Ω. We claim that there exists v ξ ∈ V(Ω, ϕ, f ) such that v ξ (ξ) = ϕ(ξ). It is sufficient to prove that there exists a constant C > 0 depending on Ω such that for every point ξ ∈ ∂Ω and ϕ ∈ C(∂Ω), there is a function

h ξ ∈ P SH(Ω) ∩ C( Ω) satisfying (1) h ξ (z) ≤ ϕ(z), ∀z ∈ ∂Ω, (2) h ξ (ξ) = ϕ(ξ), (3) ω h ξ (t) ≤ Cω ϕ (t 1/2 ).
Assume this is true. We fix z 0 ∈ Ω and write

K 1 := sup Ω f 1/n ≥ 0. Hence ∆ H (K 1 |z -z 0 | 2 ) = K 1 ∆ H |z -z 0 | 2 ≥ f 1/n , ∀H ∈ H + n , det H = n -n .
We also set

K 2 := K 1 |ξ -z 0 | 2 . Then for the continuous function φ(z) := ϕ(z) -K 1 |z -z 0 | 2 + K 2 , we have h ξ such that (1)-(3) hold. Then the desired function v ξ ∈ V(Ω, ϕ, f ) is given by v ξ (z) = h ξ (z) + K 1 |z -z 0 | 2 -K 2 . Thus h ξ (z) ≤ φ(z) = ϕ(z) -K 1 |z -z 0 | 2 + K 2 on ∂Ω, so v ξ (z) ≤ ϕ on ∂Ω and v ξ (ξ) = ϕ(ξ). Moreover, it is clear that ∆ H v ξ = ∆ H h ξ + K 1 ∆ H (|z -z 0 | 2 ) ≥ f 1/n , ∀H ∈ H + n , det H = n -n .
Furthermore, using the hypothesis of h ξ , we can control the modulus of continuity of v ξ

ω v ξ (t) = sup |z-y|≤t |v ξ (z) -v ξ (y)| ≤ ω h ξ (t) + K 1 ω |z-z 0 | 2 (t) ≤ Cω φ(t 1/2 ) + 4d 3/2 K 1 t 1/2 ≤ Cω ϕ (t 1/2 ) + 2dK 1 (C + 2d 1/2 )t 1/2 ≤ (C + 2d 1/2 )(1 + 2dK 1 ) max{ω ϕ (t 1/2 ), t 1/2 },
where d := diam(Ω). Hence, we conclude that

ω v ξ (t) ≤ λ(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 },
where λ := (C + 2d 1/2 )(1 + 2d) is a positive constant depending on Ω. Now we will construct h ξ ∈ P SH(Ω) ∩ C( Ω) which satisfies the three conditions above. Let B > 0 be large enough such that the function

g(z) = Bρ(z) -|z -ξ| 2
is psh in Ω. Let ωϕ be the minimal concave majorant of ω ϕ and define

χ(x) = -ω ϕ ((-x) 1/2 ), which is a convex nondecreasing function on [-d 2 , 0]. Now fix r > 0 so small that |g(z)| ≤ d 2 in B(ξ, r) ∩ Ω and define for z ∈ B(ξ, r) ∩ Ω the function h(z) = χ • g(z) + ϕ(ξ).
It is clear that h is a continuous psh function on B(ξ, r) ∩ Ω and we see that h(z) ≤ ϕ(z) if z ∈ B(ξ, r) ∩ ∂Ω and h(ξ) = ϕ(ξ). Moreover by the subadditivity of ωϕ and Lemma 2.4.1 we have

ω h (t) = sup |z-y|≤t |h(z) -h(y)| ≤ sup |z-y|≤t ωϕ ï |z -ξ| 2 -|y -ξ| 2 -B(ρ(z) -ρ(y)) 1/2 ò ≤ sup |z-y|≤t ωϕ î (|z -y|(2d + B 1 )) 1/2 ó ≤ C.ω ϕ (t 1/2 ),
where C := 1 + (2d + B 1 ) 1/2 depends on Ω.

Recall that ξ ∈ ∂Ω and fix 0 < r 1 < r and

γ 1 ≥ 1 + d/r 1 such that -γ 1 ωϕ î (|z -ξ| 2 -Bρ(z)) 1/2 ó ≤ inf ∂Ω ϕ -sup ∂Ω ϕ, for z ∈ ∂Ω ∩ ∂B(ξ, r 1 ). Set γ 2 = inf ∂Ω ϕ. Then γ 1 (h(z) -ϕ(ξ)) + ϕ(ξ) ≤ γ 2 for z ∈ ∂B(ξ, r 1 ) ∩ Ω. Now set h ξ (z) = ® max[γ 1 (h(z) -ϕ(ξ)) + ϕ(ξ), γ 2 ] ; z ∈ Ω ∩ B(ξ, r 1 ), γ 2 ; z ∈ Ω \ B(ξ, r 1 ),
which is a well defined psh function on Ω, continuous on Ω and such that h ξ (z) ≤ ϕ(z) for all z ∈ ∂Ω. Indeed, on ∂Ω ∩ B(ξ, r 1 ) we have

γ 1 (h(z) -ϕ(ξ)) + ϕ(ξ) = -γ 1 ωϕ (|z -ξ|) + ϕ(ξ) ≤ -ω ϕ (|z -ξ|) + ϕ(ξ) ≤ ϕ(z).
Hence it is clear that h ξ satisfies the three conditions above.

We have just proved that for each ξ ∈ ∂Ω, there is a function

v ξ ∈ V(Ω, ϕ, f ) with v ξ (ξ) = ϕ(ξ) and ω v ξ (t) ≤ λ(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 }. Set v(z) = sup {v ξ (z); ξ ∈ ∂Ω} . Since 0 ≤ ω v (t) ≤ λ(1 + K 1 ) max{ω ϕ (t 1/2
), t 1/2 }, we see that ω v (t) converges to zero when t converges to zero. Consequently, v ∈ C( Ω) and v = v * ∈ P SH(Ω). Thanks to Choquet's lemma, we can choose a nondecreasing sequence (v j ), where v j ∈ V(Ω, ϕ, f ), converging to v almost everywhere. This implies that

∆ H v = lim j→∞ ∆ H v j ≥ f 1/n , ∀H ∈ H + n , det H = n -n . It is clear that v(ξ) = ϕ(ξ) for any ξ ∈ ∂Ω. Finally, v ∈ V(Ω, ϕ, f ), v = ϕ on ∂Ω and ω v (t) ≤ λ(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 }.
Remark 2.4.5. If we assume that Ω has a smooth boundary and ϕ is C 1,α -smooth for 0 < α ≤ 1, then it is possible to construct a (1 + α)/2-Hölder continuous barrier v to the Dirichlet problem Dir(Ω, ϕ, f ) (see [BT76, Theorem 6.2]). Here, for a bounded SHL domain, if ϕ ∈ C 1,1 (∂Ω) we can find a Lipschitz barrier to Dir(Ω, ϕ, f ). It is enough to take v := Aρ + φ where φ is an extension of ϕ to Ω and A ≫ 1.

Corollary 2.4.6. Under the same assumption of Proposition 2.4.4, there exists a plurisuperharmonic function ṽ ∈ C( Ω) such that ṽ = ϕ on ∂Ω and

ω ṽ(t) ≤ λ(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), t 1/2 },
where λ > 0 depends on Ω.

Proof. We can perform the same construction as in the proof of Proposition 2.4.4 for the function

ϕ 1 = -ϕ ∈ C(∂Ω); then we get v 1 ∈ V(Ω, ϕ 1 , f ) such that v 1 = ϕ 1 on ∂Ω and ω v 1 (t) ≤ λ(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), t 1/2 }.
Hence, we set ṽ = -v 1 which is a plurisuperharmonic function on Ω, continuous on Ω and satisfying ṽ = ϕ on ∂Ω and

ω ṽ(t) ≤ λ(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), t 1/2 },
where λ > 0 is a constant depending on Ω.

Proof of main results

Proof of Theorem 2.1.1

Thanks to Proposition 2.4.4, we have a subsolution v ∈ V(Ω, ϕ, f ) with v = ϕ on ∂Ω and

ω v (t) ≤ λ(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), t 1/2 }.
From Corollary 2.4.6, we get w ∈ P SH(Ω) ∩ C( Ω) such that w = -ϕ on ∂Ω and

ω w (t) ≤ λ(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), t 1/2 },
where λ > 0 is a constant. Applying the Proposition 2.4.2 we obtain the required result, that is

ω U (t) ≤ η(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (t 1/2 ), ω f 1/n (t), t 1/2 },
where η > 0 depends on Ω.

Corollary 2.5.1. Let Ω be a bounded SHL domain in C n . Let ϕ ∈ C 0,α (∂Ω) and 0 ≤ f 1/n ∈ C 0,β ( Ω), 0 < α, β ≤ 1. Then the solution U to the Dirichlet problem Dir(Ω, ϕ, f ) belongs to C 0,γ ( Ω) for γ = min{β, α/2}.

The following example illustrates that the estimate of ω U in Theorem 2.1.1 is optimal.

Example 2.5.2. Let ψ be a concave modulus of continuity on [0, 1] and

ϕ(z) = -ψ[ » (1 + Rez 1 )/2], for z = (z 1 , z 2 , ..., z n ) ∈ ∂B ⊂ C n .
It is easy to show that ϕ ∈ C(∂B) with modulus of continuity

ω ϕ (t) ≤ Cψ(t), for some C > 0. Let v(z) = -(1 + Rez 1 )/2 ∈ P SH(B) ∩ C( B) and χ(λ) = -ψ( √ -λ) be a convex increasing function on [-1, 0]. Hence we see that u(z) = χ • v(z) ∈ P SH(B) ∩ C( B),
and satisfies (dd c u) n = 0 in B and u = ϕ on ∂B. The modulus of continuity of U, ω U (t), has the estimate

C 1 ψ(t 1/2 ) ≤ ω U (t) ≤ C 2 ψ(t 1/2 ),
for C 1 , C 2 > 0. Indeed, let z 0 = (-1, 0, ..., 0) and z = (z 1 , 0, ..., 0) ∈ B where z 1 = -1 + 2t and 0 ≤ t ≤ 1. Hence, by Lemma 2.4.1, we see that

ψ(t 1/2 ) = ψ[ » |z -z 0 |/2] = ψ[ » (1 + Rez 1 )/2] = |U(z) -U(z 0 )| ≤ 3ω U (t).
Finally, it is natural to try to relate the modulus of continuity of U := U(Ω, ϕ, f ) to the modulus of continuity of U 0 := U(Ω, ϕ, 0) the solution to Bremermann problem in a bounded SHL domain.

ω U (t) ≤ C(1 + f 1/n L ∞ ( Ω) ) max{ω U 0 (t), ω f 1/n (t)}.
Proof. First, we search for a subsolution v ∈ V(Ω, ϕ, f ) such that v| ∂Ω = ϕ and estimate its modulus of continuity. Since Ω is a bounded SHL domain, there exists a Lipschitz defining function ρ on Ω. Define the function

v(z) = U 0 (z) + Aρ(z),
where

A := f 1/n L ∞ /c and c > 0 is as in Definition 2.2.1. It is clear that v ∈ V(Ω, ϕ, f ), v = ϕ on ∂Ω and ω v (t) ≤ Cω U 0 (t),
where

C := γ(1 + f 1/n L ∞ ( Ω)
) and γ ≥ 1 depends on Ω. On the other hand, by the comparison principle we get that U ≤ U 0 . So,

v ≤ U ≤ U 0 in Ω and v = U = U 0 = ϕ on ∂Ω.
Thanks to Proposition 2.4.2, there exists λ > 0 depending on Ω such that

ω U (t) ≤ λ max{ω v (t), ω U 0 (t), ω f 1/n (t)}.
Hence, for some C > 0 depending on Ω,

ω U (t) ≤ C(1 + f 1/n L ∞ ( Ω) ) max{ω U 0 (t), ω f 1/n (t)}.

Estimate of the ψ-norm of the solution

Definition 2.5.4. Let ψ be a modulus of continuity, E ⊂ C n be a bounded set and g ∈ C ∩ L ∞ (E). We define the norm of g with respect to ψ ( briefly, ψ-norm) as follows:

g ψ := sup z∈E |g(z)| + sup z =y∈E |g(z) -g(y)| ψ(|z -y|) .
Proposition 2.5.5. Let Ω ⊂ C n be a bounded SHL domain, ϕ ∈ C(∂Ω) with modulus of continuity ψ 1 and f 1/n ∈ C( Ω) with modulus of continuity ψ 2 . Then there exists a constant C > 0 depending on Ω such that

U ψ ≤ C(1 + f 1/n L ∞ ( Ω) ) max{ ϕ ψ 1 , f 1/n ψ 2 },
where

ψ(t) = max{ψ 1 (t 1/2 ), ψ 2 (t)}.
Proof. By hypothesis, we see that

ϕ ψ 1 < ∞ and f 1/n ψ 2 < ∞. Let z = y ∈ Ω. By Theorem 2.1.1, we get |U(z) -U(y)| ≤ η(1 + f 1/n L ∞ ( Ω) ) max{ω ϕ (|z -y| 1/2 ), ω f 1/n (|z -y|)} ≤ η(1 + f 1/n L ∞ ( Ω) ) max{ ϕ ψ 1 , f 1/n ψ 2 }ψ(|z -y|),
where

ψ(|z -y|) = max{ψ 1 (|z -y| 1/2 ), ψ 2 (|z -y|)}. Hence sup z =y∈ Ω |U(z) -U(y)| ψ(|z -y|) ≤ η(1 + f 1/n L ∞ ( Ω) ) max{ ϕ ψ 1 , f 1/n ψ 2 },
where η ≥ d 2 + 1 and d = diam(Ω) (see Proposition 2.4.2). From Proposition 2.3.9, we note that

U L ∞ ( Ω) ≤ d 2 f 1/n L ∞ ( Ω) + ϕ L ∞ (∂Ω) ≤ η max{ ϕ ψ 1 , f 1/n ψ 2 }.
Then we can conclude that

U ψ ≤ 2η(1 + f 1/n L ∞ ( Ω) ) max{ ϕ ψ 1 , f 1/n ψ 2 }.
Chapter 3

Hölder continuity of solutions for general measures

Introduction

In this chapter, we are interested in studying the regularity of solutions to the following Dirichlet problem:

Dir(Ω, ϕ, f dµ) :      u ∈ P SH(Ω) ∩ C( Ω), (dd c u) n = f dµ in Ω, u = ϕ on ∂Ω,
where µ is a nonnegative finite Borel measure on a bounded SHL domain Ω, 0 ≤ f ∈ L p (Ω, µ) for p > 1, and ϕ ∈ C(∂Ω). Ko lodziej demonstrated [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF][START_REF] Ko Lodziej | A sufficient condition for solvability of the Dirichlet problem for the complex Monge-Ampère operator[END_REF] the existence of a weak continuous solution to this problem as soon as µ is dominated by a suitable function of capacity on a bounded strongly pseudoconvex domain with smooth boundary.

We consider in this thesis the class of measures satisfying (3.3.1) and ensure Ko lodziej's existence theorem in a bounded SHL domain. More precisely, we prove the following.

Theorem 3.1.1. Let µ be a measure satisfying Condition H(τ ) for some τ > 0 on a bounded SHL domain Ω ⊂ C n and ϕ ∈ C(∂Ω). Then there exists a unique continuous solution to Dir(Ω, ϕ, dµ).

Then we investigate the Hölder continuity of the solution in several cases. In the case of the Lebesgue measure, we have estimated in Chapter 2 the modulus of continuity of the solution in terms of the modulus of continuity of the boundary data ϕ and the density f in a bounded SHL domain.

Guedj, Ko lodziej and Zeriahi proved [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF] that the solution to Dir(Ω, ϕ, f dV 2n ) is Hölder continuous on Ω when f ∈ L p (Ω), p > 1, is bounded near the boundary of strongly pseudoconvex domain and ϕ ∈ C 1,1 (∂Ω). Recently, N. C. Nguyen [N14] proved the Hölder continuity when the density satisfies a growth condition near the boundary.

Here, we deal the case of L p -density without assuming any condition near the boundary.

Theorem 3.1.2. Let Ω ⊂ C n be a bounded SHL domain. Assume that ϕ ∈ C 1,1 (∂Ω) and f ∈ L p (Ω) for some p > 1. Then the unique solution U to Dir(Ω, ϕ, f dV 2n ) is γ-Hölder continuous on Ω for any 0 < γ < 1/(nq + 1) where 1/p + 1/q = 1. Moreover, if p ≥ 2, then the solution U is Hölder continuous on Ω of exponent less than min{1/2, 2/(nq + 1)}.

In the case of singular measures with respect to the Lebesgue measure, there is no study about the regularity of solution in a bounded domain in C n (see [START_REF] Pham | Hölder continuity of solutions to the Monge-Ampère equations on compact Kähler manifolds[END_REF] for regularity of solutions in the compact case). We will consider the case of measures having densities in L p , for p > 1, with respect to Hausdorff-Riesz measures which are defined in (3.5.5).

We prove the Hölder continuity of the solution while the boundary data belongs to C 1,1 (∂Ω).

Theorem 3.1.3. Let Ω be a bounded SHL domain in C n and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ for 0 < ǫ ≤ 2. Suppose that ϕ ∈ C 1,1 (∂Ω) and 0 ≤ f ∈ L p (Ω, µ) for some p > 1, then the unique solution to Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω of exponent ǫγ/2 for any 0 < γ < 1/(nq + 1) and 1/p + 1/q = 1.

This result generalizes the one proved in [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF][START_REF] Charabati | Hölder regularity for solutions to complex Monge-Ampère equations[END_REF] from which the main idea of our proof originates.

When the boundary data is merely Hölder continuous we state the regularity of the solution using the previous theorem. Theorem 3.1.4. Let Ω be a bounded SHL domain in C n and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ for 0 < ǫ ≤ 2. Suppose that ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ) for some p > 1, then the unique solution to Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω of exponent ǫ ǫ+6 min{α, ǫγ} for any 0 < γ < 1/(nq +1) and 1/p+1/q = 1. Moreover, when Ω is a smooth strongly pseudoconvex domain the Hölder exponent will be ǫ ǫ+2 min{α, ǫγ}, for any 0 < γ < 1/(nq + 1).

In the case of the Lebesgue measure, i.e. ǫ = 2, in a smooth strongly pseudoconvex domain we get the Hölder exponent min{α/2, γ} which is better than the one obtained in [START_REF] Baracco | Hölder regularity of the solution to the complex Monge-Ampère equation with L p density[END_REF].

Our final purpose concerns how to get the Hölder continuity of the solution to the Dirichlet problem Dir(Ω, ϕ, f dµ) by means of the Hölder continuity of a subsolution to Dir(Ω, ϕ, dµ) for some special measure on Ω. Most of the content of this chapter will be found in my papers [START_REF] Charabati | Hölder regularity for solutions to complex Monge-Ampère equations[END_REF] and [START_REF] Charabati | Regularity of solutions to the Dirichlet problem for Monge-Ampère equations[END_REF].

Stability theorem

Definition 3.2.1. A nonnegative finite Borel measure µ on Ω is said to satisfy Condition H(∞) if for any τ > 0 there exists a positive constant A depending on τ such that

µ(K) ≤ ACap(K, Ω) 1+τ ,
for any Borel subset K of Ω.

Before announcing the stability theorem, let us prove some useful lemmas.

Lemma 3.2.2. Let v 1 , v 2 ∈ P SH(Ω) ∩ L ∞ (Ω) be such that lim inf z→∂Ω (v 1 -v 2 )(z) ≥ 0.
Then for all t, s > 0, we have

t n Cap({v 1 -v 2 < -s -t}, Ω) ≤ {v 1 -v 2 <-s} (dd c v 1 ) n . Proof. Fix v ∈ P SH(Ω) such that -1 ≤ v ≤ 0.
Then for any t, s > 0, we have

{v 1 -v 2 < -s -t} ⊂ {v 1 -v 2 < -s + tv} ⊂ {v 1 -v 2 < -s} ⋐ Ω.
The comparison principle yields that

t n {v 1 -v 2 <-s-t} (dd c v) n ≤ {v 1 -v 2 <-s-t} (dd c (v 2 + tv)) n ≤ {v 1 -v 2 <-s+tv} (dd c (v 2 + tv)) n = {v 1 <-s+v 2 +tv} (dd c (-s + v 2 + tv)) n ≤ {v 1 <-s+v 2 +tv} (dd c v 1 ) n ≤ {v 1 -v 2 <-s} (dd c v 1 ) n .
Taking the supremum over all such functions v gives the required result.

Lemma 3.2.3. Let g : R + → R + be a decreasing right continuous function. Assume that there exist τ, B > 0 such that

(3.2.1) tg(s + t) ≤ B[g(s)] 1+τ , for all s, t > 0. Then g(s) = 0 for all s ≥ s ∞ , where s ∞ := 2B[g(0)] τ 1-2 -τ .
Proof. We define by induction an increasing sequence (s j ) ∈ R N + as follows.

s 0 := 0,

s j := sup{s > s j-1 : g(s) > g(s j-1 )/2}, ∀j ≥ 1.
It is clear that for any s > s j we have g(s) ≤ g(s j-1 )/2. As g is right continuous, we conclude that g(s j ) ≤ g(s j-1 )/2. Hence, we infer

(3.2.2) g(s j ) ≤ g(0) 2 j .
Let us set M := 2B[g(0)] τ > 0 and M j := 2 -jτ M for j ≥ 1. We apply (3.2.1) for s j and M j , then it follows from (3.2.2) that

g(s j + M j ) ≤ B M j g(s j ) 1+τ ≤ g(s j )/2.
Consequently, we get s j+1 ≤ s j + M j since g(s) > g(s j )/2 for any s ∈ (s j , s j+1 ). In the same way we can see that s 1 ≤ M . Thus the sequence (s j ) is bounded from above with limit

j≥0 (s j+1 -s j ) ≤ M + j≥1 M j = M 1 -2 -τ =: s ∞ .
Then the lemma follows.

The following weak stability estimate, proved in [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF] for the Lebesgue measure, plays an important role in our work. A similar, but weaker, estimate was established by Ko lodziej [START_REF] Ko Lodziej | Equicontinuity of families of plurisubharmonic functions with bounds on their Monge-Ampère masses[END_REF] and in the compact setting it was proved by Eyssidieux, Guedj and Zeriahi [START_REF] Eyssidieux | Singular Kähler-Einstein metrics[END_REF]. Here we show that this estimate is still true for any measure µ satisfying Condition H(∞). Theorem 3.2.4. Let µ satisfy Condition H(∞) on a bounded domain Ω ⊂ C n and 0 ≤ f ∈ L p (Ω, µ), p > 1. Suppose that v 1 , v 2 are two bounded psh functions in Ω such that lim inf z→∂Ω (v 1v 2 )(z) ≥ 0 and (dd c v 1 ) n = f dµ. Fix r ≥ 1 and 0 < γ < r/(nq + r), 1/p + 1/q = 1. Then there exists a constant C = C(r, γ, n, q) > 0 such that

(3.2.3) sup Ω (v 2 -v 1 ) ≤ C(1 + f η L p (Ω,µ) )||(v 2 -v 1 ) + || γ L r (Ω,µ) ,
where (v 2v 1 ) + = max{v 2v 1 , 0} and η = 1 n + γq r-γ(r+nq) . In order to prove this theorem we need the following proposition. Proposition 3.2.5. Under the same assumption of Theorem 3.2.4 and for any α > 0, there exists a positive constant

C 1 = C 1 (n, q, α) such that for all ǫ > 0, sup Ω (v 2 -v 1 ) ≤ ǫ + C 1 f 1/n L p (Ω,µ) [Cap({v 1 -v 2 < -ǫ}, Ω)] α .
Proof. Let us set g(s) := Cap({v 1v 2 < -s -ǫ}, Ω) 1/n . By applying Lemma 3.2.2 we conclude that

t n Cap({v 1 -v 2 < -ǫ -s -t}, Ω) ≤ {v 1 -v 2 <-ǫ-s} (dd c v 1 ) n ≤ {v 1 -v 2 <-ǫ-s} f dµ ≤ f L p (Ω,µ) µ({v 1 -v 2 < -ǫ -s}) 1/q
Since µ satisfies Condition H(∞), we find a positive constant C depending on n, q and α such that

t n Cap({v 1 -v 2 < -ǫ -s -t}, Ω) ≤ C f L p (Ω,µ) [Cap({v 1 -v 2 < -ǫ -s}, Ω)] 1+αn .
Therefore, this yields that tg(s

+ t) ≤ B[g(s)] 1+αn ,
where

B := C1/n f 1/n L p (Ω,µ) . Now, it follows from Lemma 3.2.3 that Cap({v 1 -v 2 < -ǫ -s ∞ }, Ω) = 0. Hence v 2 -v 1 ≤ ǫ + s ∞ almost
everywhere and then the inequality holds everywhere in Ω. Consequently, we have sup

Ω (v 2 -v 1 ) ≤ ǫ + C 1 f 1/n L p (Ω,µ) [Cap({v 1 -v 2 < -ǫ}, Ω)] α ,
where C 1 depends only on n, q and α.

Proof of Theorem 3.2.4. Applying Lemma 3.2.2 with s = t = ǫ and using Hölder inequality, we infer

Cap({v 1 -v 2 < -2ǫ}, Ω) ≤ ǫ -n {v 1 -v 2 <-ǫ} f dµ ≤ ǫ -n-r/q Ω (v 2 -v 1 ) r/q + f dµ ≤ ǫ -n-r/q f L p (Ω,µ) (v 2 -v 1 ) + r/q L r (Ω,µ) .
Fix α > 0 to be chosen later and apply Proposition 3.2.5 to get sup

Ω (v 2 -v 1 ) ≤ 2ǫ + C 1 ǫ -α(n+r/q) f α+1/n L p (Ω,µ) (v 2 -v 1 ) + αr/q L r (Ω,µ) .
We set ǫ := (v 2v 1 ) + γ , where 0 < γ < r/(nq + r) is fixed and

α := γq r -γ(r + nq) .
Then we get sup

Ω (v 2 -v 1 ) ≤ C(1 + f α+1/n L p (Ω,µ) )||(v 2 -v 1 ) + || γ L r (Ω,µ) ,
where C > 0 depends on n, q, γ and r.

Remark 3.2.6. When µ satisfies only the condition in Definition 3.3.1 below, we can get some stability estimate. Suppose that v 1 , v 2 are two bounded psh functions in Ω such that lim inf z→∂Ω (v 1 -v 2 )(z) ≥ 0 and (dd c v 1 ) n = dµ. Fix r ≥ 1, then there exists a constant C = C(r, τ, n) > 0 such that (3.2.4) sup

Ω (v 2 -v 1 ) ≤ C||(v 2 -v 1 ) + || γ L r (Ω,µ) ,
where (v 2v 1 ) + = max{v 2v 1 , 0} and γ := τ r n+τ (n+r) .

Existence of solutions

This section is devoted to explain the existence of continuous solutions to the Dirichlet problem Dir(Ω, ϕ, µ) for measures µ dominated by Bedford-Taylor's capacity, as in (3.3.1) below, on a bounded SHL domain.

Definition 3.3.1. A finite Borel measure µ on Ω is said to satisfy Condition H(τ ) for some fixed τ > 0 if there exists a positive constant A such that

(3.3.1) µ(K) ≤ ACap(K, Ω) 1+τ ,
for any Borel subset K of Ω.

Ko lodziej [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF] demonstrated the existence of a continuous solution to Dir(Ω, ϕ, µ) when µ verifies (3.3.1) and some local extra condition in a bounded strongly pseudoconvex domain with smooth boundary. Furthermore, he disposed of the extra condition in [Ko99] using Cegrell's result [START_REF] Cegrell | Pluricomplex energy[END_REF] about the existence of a solution in the energy class F 1 .

Here, the existence of continuous solutions to Dir(Ω, ϕ, µ) in a bounded SHL domain follows from the lines of Ko lodziej and Cegrell's arguments in [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF][START_REF] Cegrell | Pluricomplex energy[END_REF].

First of all, we prove the existence of continuous solutions to the Dirichlet problem for measures having densities in L p (Ω) with respect to the Lebesgue measure.

Theorem 3.3.2. Let Ω ⊂ C n be a bounded SHL domain, ϕ ∈ C(∂Ω) and 0 ≤ f ∈ L p (Ω), for some p > 1. Then there exists a unique solution U to the Dirichlet problem Dir(Ω, ϕ, f dV 2n ).

Proof. Let (f j ) be a sequence of smooth functions on Ω which converges to f in L p (Ω). Thanks to Theorem 2.3.2, there exists a function U j ∈ P SH(Ω) ∩ C( Ω) such that U j = ϕ on ∂Ω and (dd c U j ) n = f j dV 2n in Ω. We claim that

(3.3.2) U k -U j L ∞ ( Ω) ≤ A(1 + f k η L p (Ω) )(1 + f j η L p (Ω) ) f k -f j γ/n L 1 (Ω) ,
where 0 ≤ γ < 1/(q + 1) is fixed, η := 1 n + γq n-γn(1+q) , 1/p + 1/q = 1 and A = A(γ, n, q, diam(Ω)).

Indeed, by the stability theorem 3.2.4 and for r = n, we get that sup

Ω (U k -U j ) ≤ C(1 + f j η L p (Ω) ) (U k -U j ) + γ L n (Ω) ≤ C(1 + f j η L p (Ω) ) U k -U j γ L n (Ω) ,
where 0 ≤ γ < 1/(q + 1) is fixed and C = C(γ, n, q) > 0.

Hence by the L n -L 1 stability theorem in [B l93] (see our Remark 2.3.10),

U k -U j L n (Ω) ≤ C f k -f j 1/n L 1 (Ω) ,
where C depends on diam(Ω). Then, from the last two inequalities and reversing the role of U j and U k , we deduce

U k -U j L ∞ (Ω) ≤ C Cγ (1 + f k η L p (Ω) )(1 + f j η L p (Ω) ) f k -f j γ/n L 1 (Ω) .
Since U k = U j = ϕ on ∂Ω, the inequality (3.3.2) holds. As f j converges to f in L p (Ω), there is a uniform constant B > 0 such that

U k -U j L ∞ ( Ω) ≤ B f k -f j γ/n L 1 (Ω) .
This implies that the sequence U j converges uniformly in Ω. Set

U = lim j→+∞ U j .
It is clear that U ∈ P SH(Ω) ∩ C( Ω), U = ϕ on ∂Ω. Moreover, (dd c U j ) n converges to (dd c U) n in the sense of currents, thus (dd c U) n = f dV 2n in Ω. The uniqueness of the solution follows from the comparison principle.

We will summarize the steps of the proof of Theorem 3.1.1.

• We approximate µ by non-negative measures µ s having bounded denstities with respect to the Lebesgue measure and preserving the total mass on Ω.

• We find solutions U s to Dir(Ω, ϕ, µ s ) in a bounded SHL domain Ω using Theorem 3.3.2.

• We prove that the measures µ s are uniformly dominated by capacity. Then, we can ensure that the solutions U s are uniformly bounded on Ω.

• We set U := (lim sup U s ) * which is a candidate to be the solution of Dir(Ω, ϕ, µ).

• The delicate point is then to show that (dd c U s ) n converges to (dd c U) n in the weak sense of measures. For this purpose, we invoke Cegrell's techniques [START_REF] Cegrell | Pluricomplex energy[END_REF] to ensure that

Ω U s dµ → Ω U dµ,
and

Ω |U s -U| dµ s → 0,
when s → +∞.

• Finally, we assert the continuity of this solution in Ω. We will control the L ∞ -norm of U s . For this end, we first prove that µ s are uniformly dominated by Bedford-Taylor's capacity.

Suppose first that

The following lemma is due to S. Ko lodziej [START_REF] Ko Lodziej | Some sufficient conditions for solvability of the Dirichlet problem for the complex Monge-Ampère operator[END_REF].

Lemma 3.3.3. Let E ⋐ Ω be a Borel set. Then for any D > 0 there exists t 0 > 0 such that Cap(K y , Ω) ≤ DCap(K, Ω), |y| < t 0 , where K ⊂ E and K y := {x; xy ∈ K}.

Proof. Without loss of generality we can assume that K is compact and K ⋐ E. We define w y := u * Ky (x + y), where u Ky is the extremal function of K y defined by

u Ky := sup{v ∈ P SH(Ω) : v ≤ 0 on Ω, v ≤ -1 on K y }.
For any 0 < c < 1/2, we set Ω c := {u * E < -c}. Let A ≫ 1 be such that Aρ ≤ u E in Ω. Since ρ ≤ -c/(2A) for any x ∈ Ω c/2 , we can find t 0 := t 0 (E, Ω) such that x + y ∈ Ω for any |y| < t 0 . Therefore,

g(x) := ® max{w y (x) -c, (1 + 2c)u * E (x)} ; x ∈ Ω c/2 , (1 + 2c)u * E (x) ; x ∈ Ω \ Ω c/2 ,
is a well defined bounded psh function in Ω.

Since K ⋐ E and u * E = -1 on a neighborhood of K, we infer that w yc ≥ (1 + 2c)u * E there. Hence, we have

Cap(K, Ω) ≥ (1 + 2c) -n K (dd c g) n = (1 + 2c) -n K (dd c w y ) n = (1 + 2c) -n Ky (dd c u * Ky ) n = (1 + 2c) -n Cap(K y , Ω).
Consequently, we obtain Cap(K y , Ω) ≤ (1 + 2c) n Cap(K, Ω), for any |y| < t 0 .

Lemma 3.3.4. Let Ω be a bounded SHL domain and µ be a compactly supported measure satisfying Condition H(τ ) for some τ > 0. Then there exist s 0 > 0 and B = B(n, τ ) > 0 such that for all s > s 0 the measures µ s , defined above, satisfy

µ s (K) ≤ BCap(K, Ω) 1+τ ,
for all Borel subsets K of Ω.

Proof. Let us set δ s := diam I s j . We define for large s ≫ 1 a regularizing sequence of measures μs = µ * ρ s , where ρ s ∈ C ∞ 0 (B(0, 2δ s )) is a radially symmetric non-negative function such that

ρ s = 1 2Vol(B(0, δ s ))
on B(0, δ s ), and B(0,2δs)

ρ s dV 2n = 1.
For all Borel subsets K ⊂ Ω, we get

μs (K) = j K∩I s j Ç B(x,2δs) ρ s (x -y)dµ(y) å dV 2n ≥ j K∩I s j Ç B(x,δs) ρ s (x -y)dµ(y) å dV 2n ≥ j K∩I s j Ç µ(B(x, δ s )) 2Vol(B(x, δ s )) å dV 2n ≥ 1 2(2n) n τ 2n j K Ç µ(I s j ) d 2n s χ I s j å dV 2n = µ s (K) 2(2n) n τ 2n ,
where τ 2n is the volume of the unit ball in C n . We set K y := {x; xy ∈ K}, for y ∈ C n . Then, by Lemma 3.3.3, we find t 0 > 0 and

s 0 > 1/t 0 such that Cap(K y , Ω) ≤ 2Cap(K, Ω), |y| < t 0 ,
for any Borel set K ⊂ ∪ s>s 0 suppµ s ⋐ Ω.

We infer for all s > s 0 and K ⊂ Ω, that

μs (K) ≤ sup |y|<1/s µ(K y ) ≤ A sup |y|<1/s Cap(K y , Ω) 1+τ ≤ 2 1+τ ACap(K, Ω) 1+τ .
This completes the proof.

Proposition 3.3.5. There exists a uniform constant C > 0 such that

U s L ∞ ( Ω) ≤ C,
for all s > s 0 , where s 0 is as in Lemma 3.3.4.

Proof. We owe the idea of the proof to Benelkourchi, Guedj and Zeriahi [START_REF] Benelkourchi | A priori estimates for weak solutions of complex Monge-Ampère equations[END_REF] in a slightly different context. Without loss of generality we can assume ϕ = 0 in Dir(Ω, ϕ, µ) and µ(Ω) ≤ 1. Let us fix s > s 0 . It follows from Lemma 3.3.4 that there exists a uniform constant B = B(n, τ ) > 0 so that the following inequality holds for all Borel sets K ⊂ Ω,

µ s (K) ≤ BCap(K, Ω) 1+τ .
We define for k > 0,

g(k) := - 1 n ln(Cap{U s < -k}).
This function is increasing on [0, +∞] and g(+∞) = +∞. We claim that

(3.3.3) ln t + (1 + τ )g(k) -ln B/n ≤ g(k + t),
for all t, k > 0. Indeed, Lemma 3.2.2 yields that (3.3.4)

t n Cap({U s < -k -t}) ≤ µ s ({U s < -k}) ≤ BCap({U s < -k}) 1+τ .
Now we define an increasing sequence (k j ) as follows

k j+1 := k j + B 1/n e 1-τ g(k j ) , for all j ∈ N,
where k 0 = 2. We claim that g(k 0 ) ≥ 0. To get this end, we apply the inequality (3.3.4) for t = k = 1, then we get

Cap({U s < -2}) ≤ µ s ({U s < -1}) ≤ µ(Ω) ≤ 1.
We apply (3.3.3) with t = t j = k j+1k j and k = k j to get that

g(k j ) ≥ j + g(k 0 ) ≥ j.
Thus g(k j ) goes to +∞ as j goes to +∞.

Let us set k

∞ := lim N →+∞ k N . Then g(k ∞ ) = +∞. We claim that k ∞ is bounded by an absolute constant independent of U s . k ∞ = lim N →+∞ N -1 0 (k j+1 -k j ) + 2 = lim N →+∞ N -1 0 (B 1/n e 1-τ g(k j ) ) + 2 ≤ lim N →+∞ eB 1/n N -1 0 e -τ j + 2 ≤ eB 1/n /(1 -e -τ ) + 2 =: M (n, τ ).
For any k ≥ k ∞ , we conclude that g(k) = +∞, hence

Cap({U s < -k}) = 0 for all k ≥ k ∞ .
This means that for any s > s 0 the function U s is bounded from below by an absolute constant -k ∞ ≥ -M (n, τ ).

Thanks to Proposition 3.3.5, the sequence (U s ) is uniformly bounded. Passing to a subsequence we can assume that U s converges in L 1 loc (Ω) (see Theorem 4.1.9 in [H83]). Let us set U := (lim sup U s ) * ∈ P SH ∩ L ∞ (Ω). Hence U s converges to U almost everywhere in Ω with respect to the Lebesgue measure dV 2n . Lemma 3.3.6. Let µ be a finite Borel measure on Ω. Suppose that U s ∈ P SH(Ω) ∩ C( Ω) converges to U ∈ P SH(Ω)∩L ∞ (Ω) almost everywhere with respect to the Lebesgue measure and U s L ∞ ( Ω) ≤ C, for some uniform constant C > 0. Then, we have

(3.3.5) lim s→+∞ Ω U s dµ = Ω Udµ, and
(3.3.6) lim s→+∞ Ω |U s -U|(dd c U s ) n = 0.
Proof. Since U s is uniformly bounded in L 2 (Ω, dµ), there exists a subsequence, for which we keep the same notation, (U s ) converges weakly to v 1 in L 2 (Ω, dµ). In particular, U s converges to v 1 almost everywhere with respect to dµ and

Ω U s dµ → Ω v 1 dµ.
By Banach-Saks' Theorem there exists a subsequence U s such that (1/M ) M s=1 U s converges to v 2 in L 2 (Ω, dµ) and hence there exists a subsequence such that f M = (1/M ) M s=1 U s converges to v 2 almost everywhere with respect to dµ, when M → +∞. Hence v 1 = v 2 almost everywhere with respect to dµ and we have

Ω ( sup N ≥M f M ) * dµ = Ω sup N ≥M f M dµ → Ω v 2 dµ = Ω v 1 dµ.
On the other hand, f M → U in L 2 (Ω, dV 2n ) and so (sup N ≥M f M ) * ց U everywhere in Ω and thus Ω ( sup

N ≥M f M ) * dµ → Ω Udµ.

Then we get lim

s→+∞ Ω U s dµ = Ω v 1 dµ = Ω v 2 dµ = lim M →+∞ Ω ( sup N ≥M f M ) * dµ = Ω Udµ.
So as to prove (3.3.6), we define

v s (x) = 1 τ 2n (2nd s ) 2n |ξ|≤2nds |U(x + ξ) -U s (x + ξ)|dV 2n ,
where τ 2n is the volume of the unit ball in C n and d s = diam(Ω)/3 s . Then we see that

Ω |U s -U|(dd c U s ) n = j µ(I s j ) d 2n s I s j |U -U s |dV 2n ≤ j τ 2n (2n) 2n I s j v s (x)dµ(x) ≤ τ 2n (2n) 2n Ω v s (x)dµ(x).
We claim that Ω v s (x)dµ(x) → 0 as s → +∞. Indeed, we note that

v s (x) = 1 τ 2n (2nd s ) 2n |ξ|≤2nds |U(x + ξ) -sup j≥s U j (x + ξ) + sup j≥s U j (x + ξ) -U s (x + ξ)|dV 2n ≤ 1 τ 2n (2nd s ) 2n |ξ|≤2nds (sup j≥s U j (x + ξ) -U(x + ξ))dV 2n + 1 τ 2n (2nd s ) 2n |ξ|≤2nds sup j≥s U j (x + ξ)dV 2n - 1 τ 2n (2nd s ) 2n |ξ|≤2nds U s (x + ξ)dV 2n ≤ 2 τ 2n (2nd s ) 2n |ξ|≤2nds (sup j≥s U j (x + ξ)) * dV 2n -U(x) -U s (x).
It stems from the monotone convergence theorem and (3.3.5) that

Ω v s (x)dµ(x) → 0, s → +∞.
Proof of Theorem 3.1.1. We can assume, by passing to a subsequence in (3.3.6), that

Ω |U s -U|(dd c U s ) n ≤ 1/s 2 . Consider Ũs := max{U s , U -1/s} ∈ P SH(Ω) ∩ L ∞ ( Ω).
It follows from Hartogs' lemma that Ũs → U in Bedford-Taylor's capacity. In fact, we prove that for any Borel set K ⊂ Ω such that U| K is continuous we have Ũs converges uniformly to U on K. Since Ũs → U in L 1 loc (Ω) and by Theorem 4.1.9 in [H83] we get

lim s→+∞ sup K ( Ũs -U) = 0.
Thereby, we conclude that Ũs -U L ∞ (K) → 0, as s → +∞.

Thus the convergence in capacity of Ũs to U comes immediately from the quasicontinuity of U. Now, since Ũs is uniformly bounded for all s > s 0 as in Proposition 3.3.5, we get by Theorem 1.2.3 that (dd c Ũs ) n converges to (dd c U) n in the weak sense of currents. We need now to compare (dd c Ũs ) n and (dd c U s ) n following [START_REF] Guedj | The weighted Monge-Ampère energy of quasiplurisubharmonic functions[END_REF]. It is known that

(dd c Ũs ) n ≥ 1 {Us≥U-1/s} (dd c U s ) n .
Our assumption implies that 1 {Us<U-1/s} (dd c U s ) n → 0. Indeed, 1), hence we get by letting s → +∞ that

0 ≤ {Us<U-1/s} (dd c U s ) n ≤ s Ω |U s -U|(dd c U s ) n ≤ 1/s. Therefore, 0 ≤ (dd c U s ) n ≤ (dd c Ũs ) n + o(
(dd c U) n ≥ dµ.
Now, we prove that

Ω (dd c U) n = Ω dµ.
Actually, let v be the continuous solution to the Dirichlet problem for the homogeneous Monge-Ampère equation with the boundary data ϕ. From the comparison principle we get U s ≤ v for all s > 0 and so U ≤ v in Ω. Since the continuous function v -U s equals to zero on ∂Ω, we find a neighborhood of ∂Ω such that v -U s < 1/s there. Hence, U -1/s ≤ v -1/s < U s in this neighborhood and so that Ũs = U s there. Now, we get by Stokes' theorem

Ω (dd c Ũs ) n = Ω (dd c U s ) n = Ω dµ.
By the weak convergence of measures, we obtain

Ω (dd c U) n ≤ Ω dµ.
This complete the proof of Theorem 3.1.1 when µ has compact support in Ω.

For the general case, when µ is only satisfying Condition H(τ ). Let χ j is a nondecreasing sequence of smooth cut-off function, χ j ր 1 in Ω, we can do the same argument and get solutions U j to the Dirichlet problem for the measures χ j µ. By Lemma 3.3.5, the solutions U j are uniformly bounded. We set U := (lim sup U j ) * ∈ P SH(Ω) ∩ L ∞ ( Ω) and the last argument yields that U is the required bounded solution to Dir(Ω, ϕ, µ).

It remains to prove the continuity of the solution U in Ω. It is clear that

(3.3.7) lim z→ξ U(z) = ϕ(ξ), ∀ξ ∈ ∂Ω.
Let us fix K ⊂ Ω and let u j be the standard regularization of U. We extend ϕ to a continuous function on Ω. For all small d > 0 we can find by (3.3.7) an open set K d ⊃ K and j 0 > 0 such that

ϕ < U + d/2 and u j < ϕ + d/2 in a neighborhood of ∂K d , ∀j ≥ j 0 .
Hence u j < U + d in a neighborhood of ∂K d for all j ≥ j 0 and then lim inf

z→ζ (U(z) + d -u j (z)) ≥ 0, for all ζ ∈ ∂K d .
We claim that the set {u j -U > 2d} is empty for any j ≥ j 0 . Otherwise, we will get a contradiction following similar techniques to those in Lemma 3.2.3 and Lemma 3.3.5 as follows. Let us set v 1 := U + d and v 2 := u j . We define for s ≥ 0 the function g(s) := Cap({v 1v 2 < -s}) and an increasing sequence (k m ) such that k 0 := 0 and

k m := sup{k > k m-1 ; g(k) > g(k m-1 )/e}.
Hence we get g(k m ) ≤ g(k m-1 )/e. Let N be an integer so that k N ≤ d and

g(d) ≥ g(k N )/e. By Lemma 3.2.2 we obtain (d -k N ) n g(d) ≤ µ({v 1 -v 2 < -k N }) ≤ Ae 1+τ g(d) 1+τ .
Then we get

(3.3.8) d -k N ≤ A 1/n e (1+τ )/n g(d) τ /n . Now, let t := k -k m-1 where 0 < k m-1 < k ≤ d such that g(k) > g(k m-1
)/e. We infer again by Lemma 3.2.2 that

t n g(k) ≤ µ({v 1 -v 2 < -k m-1 } ≤ Aeg(k)g(k m-1 ) τ . Hence, t ≤ (Ae) 1/n g(k m-1 ) τ /n . Letting k → k - m , we get t m := k m -k m-1 ≤ (Ae) 1/n g(k m-1 ) τ /n .
Then we have

k N = m=N m=1 t m ≤ (Ae) 1/n m=N m=1 g(k m-1 ) τ /n ≤ (Ae) 1/n N g(0) τ /n .
By the definition of convergence by capacity, we get for j ≥ j 0 that g(0) is very small so that k N ≤ d/2. Then (3.3.8) yields that

d/2 ≤ A 1/n e (1+τ )/n g(d) τ /n .
Since d > 0 is fixed and g(d) = Cap({u j -U > 2d}) goes to zero when j goes to +∞, we obtain a contradiction in the last inequality.

Hölder continuity of solutions

We introduce in this section the basic ingredients of proofs of main theorems. Let µ be a measure satisfying Condition H(∞), 0 ≤ f ∈ L p (Ω, µ), p > 1 and ϕ ∈ C(∂Ω). Thanks to Theorem 3.1.1, we denote by U the continuous solution to Dir(Ω, ϕ, f dµ) and consider

U δ (z) := sup |ζ|≤δ U(z + ζ), z ∈ Ω δ ,
where Ω δ := {z ∈ Ω; dist(z, ∂Ω) > δ}.

To ensure the Hölder continuity of the solution in Ω, we need to control the L ∞ -norm of U δ -U in Ω δ .

It will be shown in Lemma 3.4.3 that the Hölder norm of the solution U can be estimated by using either sup

Ω δ (U δ -U) or sup Ω δ ( Ûδ -U), where Ûδ (z) := 1 τ 2n δ 2n |ζ-z|≤δ U(ζ)dV 2n (ζ), z ∈ Ω δ ,
and τ 2n is the volume of the unit ball in C n . It is clear that Ûδ is not globally defined in Ω, so we extend it with a good control near the boundary ∂Ω. To this end, we assume the existence of ν-Hölder continuous function v such that v ≤ U in Ω and v = U on ∂Ω. Then, we present later the construction of such a function. Lemma 3.4.1. Let Ω be a bounded SHL domain and ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1. Assume that there is a function v ∈ C 0,ν ( Ω) for 0 < ν ≤ 1, such that v ≤ U in Ω and v = ϕ on ∂Ω. Then there exist δ 0 > 0 small enough and c 0 > 0, depending on Ω, ϕ C 0,α (∂Ω) and v C 0,ν ( Ω) , such that for any 0 < δ 1 ≤ δ < δ 0 the function

Ũδ 1 = max{ Ûδ 1 , U + c 0 δ ν 1 } in Ω δ , U + c 0 δ ν 1 in Ω \ Ω δ ,
is plurisubharmonic in Ω and continuous on Ω, where ν 1 = min{ν, α/2}.

Proof. If we prove that Ûδ 1 ≤ U + c 0 δ ν 1 on ∂Ω δ , then the required result can be obtained by the standard gluing procedure. Thanks to Corollary 2.4.6, we find a plurisuperharmonic function ṽ ∈ C 0,α/2 ( Ω) such that ṽ = ϕ on ∂Ω and ṽ

C 0,α/2 ( Ω) ≤ C ϕ C 0,α (∂Ω) ,
where C depends on Ω. From the maximum principle we see that U ≤ ṽ in Ω and ṽ = ϕ on ∂Ω.

Fix z ∈ ∂Ω δ , there exists ζ ∈ C n with ζ = δ 1 such that Ûδ 1 (z) ≤ U(z + ζ). Hence, we obtain Ûδ 1 (z) -U(z) ≤ U(z + ζ) -U(z) ≤ ṽ(z + ζ) -v(z).
We choose

ζ 0 ∈ C n , with ζ 0 = δ, so that z + ζ 0 ∈ ∂Ω. Since ṽ(z + ζ 0 ) = v(z + ζ 0 ), we infer ṽ(z + ζ) -v(z) ≤ [ṽ(z + ζ) -ṽ(z + ζ 0 )] + [v(z + ζ 0 ) -v(z)] ≤ 2 ṽ C 0,α/2 ( Ω) δ α/2 + v C 0,ν ( Ω) δ ν ≤ c 0 δ ν 1 , where c 0 := 2C ϕ C 0,α (∂Ω) + v C 0,ν ( Ω) .
Moreover, we can conclude from the last argument that

(3.4.1) |U(z 1 ) -U(z 2 )| ≤ 2c 0 δ ν 1 , for all z 1 , z 2 ∈ Ω \ Ω δ such that |z 1 -z 2 | ≤ δ.
Remark 3.4.2. When ϕ ∈ C 1,1 (∂Ω), the last lemma holds for ν 1 = ν. Indeed, let φ be a C 1,1 -extension of ϕ to Ω. We define the plurisuperharmonic Lipschitz function ṽ := -Aρ + φ, where A ≫ 1 and ρ is the defining function of Ω. Hence, the constant c 0 in Lemma 3.4.1 will depend only on Ω, ϕ C 1,1 (∂Ω) and v C 0,ν ( Ω) .

Lemma 3.4.3. Given 0 < α < 1, the following conditions are equivalent.

1. There exist δ ′ , A > 0 such that for any

0 < δ ≤ δ ′ , U δ -U ≤ Aδ α on Ω δ . 2. There exist δ ′′ , B > 0 such that for any 0 < δ ≤ δ ′′ , Ûδ -U ≤ Bδ α on Ω δ .
Proof. Since Ûδ ≤ U δ , we get immediately the implication (1) ⇒ (2). In order to prove (2) ⇒ (1) we need to show that there exist δ ′ , A > 0 such that

ω(δ) := sup z∈Ω δ [(U δ -U)(z)] ≤ Aδ α .
Fix δ Ω > 0 small enough so that Ω δ = ∅ for δ ≤ δΩ := (C + 2)δ Ω where C > 0 is a constant to be chosen later. Since U is uniformly continuous on Ω, we have for any fixed 0 < δ < δΩ ,

ν(δ) := sup δ<t≤ δΩ ω(t)t -α < +∞.
We claim that there exists δ ′ > 0 small enough such that for any 0 < δ ≤ δ ′ , we have

ω(δ) ≤ Aδ α with A = 4c 0 (C + 3) α + e 4 (C + 1) α B + ν(δ Ω ),
where c 0 is as in Lemma 3.4.1. Assume that this is not the case. Then there exists δ ≤ δ Ω such that ω(δ) > Aδ α .

Let us set

δ := sup{t < δ Ω ; ω(t) > At α }. Then (3.4.2) ω(δ) δ α ≥ A ≥ ω(t) t α for all t ∈ [δ, δΩ ].
Since U is continuous on Ω, we find

z 0 ∈ Ωδ , ζ 0 ∈ Ω such that |z 0 -ζ 0 | ≤ δ and ω(δ) = U(ζ 0 ) -U(z 0 ).
We assert that dist(z 0 , ∂Ω) > (C + 2)δ. In fact, if dist(z 0 , ∂Ω) ≤ (C + 2)δ and z 1 ∈ ∂Ω such that dist(z 0 , z 1 ) = dist(z 0 , ∂Ω), then we have by (3.4.1) that

ω(δ) = U(ζ 0 ) -U(z 1 ) + U(z 1 ) -U(z 0 ) ≤ 4c 0 (C + 3) α δ α < Aδ α .
This is a contradiction. Now we apply (3.4.2) for t = (C + 2)δ and hence we get

U(ζ 0 ) -U(z) ≤ (C + 2) α ω(δ) for all z ∈ B(z 0 , (C + 1)δ).
As

B 1 := B(ζ 0 , Cδ) ⊂ B 2 := B(z 0 , (C + 1)δ), we can write (3.4.3) Û(C+1)δ (z 0 ) = 1 τ 2n (C + 1) 2n δ 2n B 2 U(z)dV 2n (z) = Å C C + 1 ã 2n 1 τ 2n C 2n δ 2n B 1 U(z)dV 2n (z) + 1 τ 2n (C + 1) 2n δ 2n B 2 \B 1 U(z)dV 2n (z) ≥ Å C C + 1 ã 2n U(ζ 0 ) + [U(ζ 0 ) -(C + 2) α ω(δ)] Ç 1 - Å C C + 1 ã 2n å = U(ζ 0 ) -(C + 2) α Ç 1 - Å C C + 1 ã 2n å ω(δ) = U(z 0 ) + Dω(δ),
where

D := 1 -(C + 2) α 1 - Ä C C+1 ä 2n . We have D ≥ e -4 if α ≤ 1 log(C + 2) log Ñ 1 -e -4 1 - Ä C C+1 ä 2n é =: α.
Hence, we infer Û(C+1)δ (z 0 ) ≥ U(z 0 ) + e -4 ω(δ).

By (2), the last inequality is equivalent to

ω(δ) ≤ e 4 B(C + 1) α δ α < Aδ α .
This is a contradiction and hence our claim is true. It remains to show that for any fixed 0 < α < 1 we can find C > 0 such that α > α. For this end, we choose C := n/x with 0 < x < 1 and note that

Ç n/x n/x + 1 å 2n ≥ e -2x for all n ∈ N. Hence this yields that α ≥ 1 log(n/x + 2) log Ç 1 -e -4 1 -e -2x
å .

Since the function

g(x) := log(1 -e -4 ) -log(1 -e -2x ) log(n/x + 2)
is continuous on ]0, 1[ and lim x→0 g(x) = 1, we can find x > 0 small enough such that g(x) ≥ α. This completes the proof.

Theorem 3.4.4. Let Ω be a bounded SHL domain and let µ be a finite Borel measure on Ω satisfying Condition H(∞). Suppose that ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1, and 0 ≤ f ∈ L p (Ω, µ) for p > 1. Then the solution U to Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω of exponent 1 λ min{ν, α/2, τ γ}, for any γ < 1/(nq+1) and 1/p+1/q = 1, if the two following conditions hold: (i) there exists v ∈ C 0,ν ( Ω), for 0 < ν ≤ 1, such that v ≤ U in Ω and v = ϕ on ∂Ω, (ii) and Ûδ 1 -U L 1 (Ω δ ,µ) ≤ cδ τ , where c, τ > 0 and 0 < δ 1 = δ λ , for some λ ≥ 1. Moreover, if ϕ ∈ C 1,1 (∂Ω) then the Hölder exponent of U will be 1 λ min{ν, τ γ}. Proof. It follows from Lemma 3.4.1 that there exist c 0 > 0 and δ 0 > 0 so that

Ũδ 1 = max{ Ûδ 1 , U + c 0 δ ν 1 } in Ω δ , U + c 0 δ ν 1 in Ω \ Ω δ ,
belongs to P SH(Ω) ∩ C( Ω), for 0 < δ 1 ≤ δ < δ 0 and ν 1 = min{ν, α/2}. By applying Theorem 3.2.4 with v 1 := U + c 0 δ ν 1 and v 2 := Ũδ 1 , we infer that sup

Ω δ ( Ûδ 1 -U -c 0 δ ν 1 ) ≤ sup Ω ( Ũδ 1 -U -c 0 δ ν 1 ) ≤ c 1 (1 + f η L p (Ω,µ) ) ( Ũδ 1 -U -c 0 δ ν 1 ) + γ L 1 (Ω,µ)
,

where η := 1/n + γq/[1 -γ(1 + nq)], c 1 = c 1 (n, q, γ) and 0 < γ < 1/(nq + 1) is fixed. Since Ũδ 1 = U + c 0 δ ν 1 in Ω \ Ω δ and ( Ũδ 1 -U -c 0 δ ν 1 ) + L 1 (Ω,µ) ≤ Ûδ 1 -U L 1 (Ω δ ,µ) .
We conclude that sup

Ω δ ( Ûδ 1 -U) ≤ c 0 δ ν 1 + c 1 (1 + f η L p (Ω,µ) ) Ûδ 1 -U γ L 1 (Ω δ ,µ)
.

By hypotheses we have

sup

Ω δ ( Ûδ 1 -U) ≤ c 0 δ ν 1 + c 1 c γ (1 + f η L p (Ω,µ) )δ τ γ . Let us set c 2 := (c 0 + c 1 c γ )(1 + f η L p (Ω,µ)
). We derive from the last inequality that sup

Ω δ ( Ûδ 1 -U) ≤ c 2 δ min{ν 1 ,τ γ} .
This means that Ûδ -U ≤ c 2 δ 1 λ min{ν 1 ,τ γ} in Ω δ 1/λ . Hence, by Lemma 3.4.3, there exists c 3 , δ0 > 0 such that for all 0 < δ < δ0 we have

(3.4.4) U δ -U ≤ c 3 δ 1 λ min{ν 1 ,τ γ} in Ω δ 1/λ .
Thus, (3.4.4) and (3.4.1) yield the Hölder continuity of U on Ω of exponent 1 λ min{ν, α/2, τ γ}, for any γ < 1/(nq + 1) and 1/p + 1/q = 1.

Finally, if ϕ ∈ C 1,1 (∂Ω), we get that the Hölder exponent is 1 λ min{ν, τ γ}, since ν 1 = ν (see Remark 3.4.2).

We prove in the following proposition that the total mass of Laplacian of the solution is finite when the boundary data is C 1,1 -smooth. Proposition 3.4.5. Let µ be a finite Borel measure satisfying Condition H(τ ) on Ω and ϕ ∈ C 1,1 (∂Ω). Then the solution U to Dir(Ω, ϕ, dµ) has the property that

Ω ∆U ≤ C,
where C > 0 depends on n, Ω, ϕ C 1,1 (∂Ω) and µ(Ω).

Proof. Let U 0 be the solution to the Dirichlet problem Dir(Ω, 0, dµ). We first claim that the total mass of ∆U 0 is finite in Ω. Indeed, let ρ be the defining function of Ω. Then by Corollary 1.3.25 we get (3.4.5)

Ω dd c U 0 ∧ (dd c ρ) n-1 ≤ Å Ω (dd c U 0 ) n ã 1/n Å Ω (dd c ρ) n ã (n-1)/n ≤ µ(Ω) 1/n Å Ω (dd c ρ) n ã (n-1)/n .
Since Ω is a bounded SHL domain, there exists a constant c > 0 such that dd c ρ ≥ cβ in Ω. Hence, (3.4.5) yields

Ω dd c U 0 ∧ β n-1 ≤ 1 c n-1 Ω dd c U 0 ∧ (dd c ρ) n-1 ≤ µ(Ω) 1/n c n-1 Å Ω (dd c ρ) n ã (n-1)/n
. Now we note that the total mass of complex Monge-Ampère measure of ρ is finite in Ω by the Chern-Levine-Nirenberg inequality and since ρ is psh and bounded in a neighborhood of Ω. Therefore, the total mass of ∆U 0 is finite in Ω.

Let φ be a C 1,1 -extension of ϕ to Ω such that φ C 1,1 ( Ω) ≤ C ϕ C 1,1 (∂Ω) for some C > 0. Now, let v = Aρ + φ + U 0 where A > 0 is big enough such that Aρ + φ ∈ P SH(Ω). By the comparison principle we see that v ≤ U in Ω and v = U = ϕ on ∂Ω. Since ρ is psh in a neighborhood of Ω and ∆U 0 Ω < +∞, we deduce that ∆v Ω < +∞. Then the following lemma completes the proof.

Lemma 3.4.6. Let Ω be a bounded domain in C n . Suppose that v 1 , v 2 are continuous subharmonic function in Ω such that v 1 ≤ v 2 in Ω and v 1 = v 2 on ∂Ω, then we have

Ω dd c v 2 ∧ β n-1 ≤ Ω dd c v 1 ∧ β n-1 .
Proof. First assume that v 1 = v 2 in a neighborhood of ∂Ω. Then Stokes' theorem yields that

Ω dd c v 2 ∧ β n-1 = Ω dd c v 1 ∧ β n-1 .
For the general case, we define the function v ǫ := max{v 2ǫ, v 1 }. Hence we see that v 1 ≤ v ǫ in Ω and v ǫ = v 1 near the boundary ∂Ω. Therefore, we get

Ω dd c v ǫ ∧ β n-1 = Ω dd c v 1 ∧ β n-1 . Since v 1 ≤ v 2 in Ω, we get that v ǫ ր v 2 in Ω. Hence dd c v ǫ ∧ β n-1 converges to dd c v 1 ∧ β n-1
in the weak sense of measures and we conclude that

Ω dd c v 2 ∧ β n-1 ≤ lim inf ǫ→0 Ω dd c v ǫ ∧ β n-1 = Ω dd c v 1 ∧ β n-1 .

Proof of main results

Our first aim is to prove Theorem 3.1.2 by applying Theorem 3.4.4. It is well known that the Lebesgue measure dV 2n satisfies Condition H(∞) (see [Z01]). We first estimate the L 1 -norm of Ûδ -U with respect to the Lebesgue measure as in [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF].

Lemma 3.5.1. ([GKZ08]

). Let ϕ ∈ C 1,1 (∂Ω) and f ∈ L p (Ω), p > 1. Then the solution U to the Dirichlet problem satisfies

Ω δ [ Ûδ (z) -U(z)]dV 2n (z) ≤ Cδ 2 ,
where C is a positive constant depending on n, Ω and f L p (Ω) .

Proof. Let us denote by σ 2n-1 the surface measure of the unit sphere. It follows from the Poisson-Jensen formula, for z ∈ Ω δ and 0 < r < δ, that

1 σ 2n-1 r 2n-1 ∂B(z,r) U(ξ)dσ(ξ) -U(z) = c n r 0 t 1-2n Ç B(z,t) ∆U(ξ) å dt.
Using polar coordinates we obtain for z ∈ Ω δ ,

Ûδ (z) -U(z) = c n δ 2n δ 0 r 2n-1 dr r 0 t 1-2n dt Ç B(z,t) ∆U(ξ) å .
Now we integrate on Ω δ with respect to dV 2n and use Fubini's theorem

Ω δ [ Ûδ (z) -U(z)]dV 2n (z) = c n δ 2n Ω δ δ 0 r 2n-1 dr r 0 t 1-2n dt Ç |ξ-z|≤t ∆U(ξ) å dV 2n (z) = c n δ 2n δ 0 r 2n-1 dr r 0 t 1-2n dt Ω δ Ç B(z,t) ∆U(ξ) å dV 2n (z) ≤ c n δ 2n δ 0 r 2n-1 dr r 0 t 1-2n dt Ω Ç B(ξ,t) dV 2n (z) å ∆U(ξ) ≤ c n Ω ∆Uδ 2
Proposition 3.4.5 yields that the total mass of ∆U is finite in Ω and this completes the proof.

We will introduce here the interplay between the real and complex Monge-Ampère measures which really goes back to Cheng-Yau and was first explained in Bedford's survey [START_REF] Bedford | Survey of pluri-potential theory, Several Complex Variables[END_REF] (see also [START_REF] Cegrell | The Dirichlet problem for the complex Monge-Ampère operator: Stability in L 2[END_REF]). This relation will be useful in the proof of Theorem 3.1.2.

We recall that if u is a locally convex smooth function in Ω, its real Monge-Ampère measure is defined by

M u := det Ç ∂ 2 u ∂x j ∂x k å dV 2n .
When u is only convex, then M u can be defined following Alexandrov [A55] by means of the gradient image as a nonnegative Borel measure on Ω (see [START_REF] Gutiérrez | The Monge-Ampère equation[END_REF], [START_REF] Rauch | The Dirichlet problem for the multidimensional Monge-Ampère equation[END_REF], [START_REF] Gaveau | Méthodes de contrôle optimal en analyse complexe I. Résolution d'équation de Monge-Ampère[END_REF]).

We recall the theorem of existence of convex solution to the Dirichlet problem for the real Monge-Ampère equation, this theorem is due to Rauch and Taylor.

Theorem 3.5.2. ([RT77]

). Let Ω be a strictly convex domain. Assume that ϕ ∈ C(∂Ω) and µ is a nonnegative Borel measure on Ω with µ(Ω) < ∞. Then there is a unique convex u ∈ C( Ω) such that M u = µ in Ω and u = ϕ on ∂Ω. Proposition 3.5.3. Let 0 ≤ f ∈ L p (Ω), p ≥ 2 and u be a locally convex function in Ω and continuous on Ω. If the real Monge-Ampère measure M u ≥ f 2 dV 2n then the complex Monge-Ampère measure satisfies the inequality (dd c u) n ≥ f dV 2n in the weak sense of measures in Ω.

Proof. For a smooth function u, we have

(3.5.1) | det(∂ 2 u/∂z j ∂ zk )| 2 ≥ det(∂ 2 u/∂x j ∂x k ).
Hence, we immediately get that (dd c u) n ≥ f dV 2n (see [START_REF] Cegrell | The Dirichlet problem for the complex Monge-Ampère operator: Stability in L 2[END_REF]). Moreover, it is well known for smooth convex function that

(3.5.2) (M u) 1/n = inf ∆ H u, where ∆ H u := j,k h jk ∂ 2 u ∂x j ∂x k ,
for any symmetric positive definite matrix H = (h jk ) with det H = n -n (see [START_REF] Gaveau | Méthodes de contrôle optimal en analyse complexe I. Résolution d'équation de Monge-Ampère[END_REF], [B l97]). In general case, we will prove that (dd c u) n ≥ f β n weakly in Ω. Indeed, the problem being local, we can assume that u is defined and convex in a neighborhood of a ball B ⊂ Ω. For δ > 0, we set µ δ := M u * ρ δ then µ δ ≥ g δ , where g δ := f 2 * ρ δ (without loss of generality we assume g δ > 0). We may assume that u and µ δ are defined in this neighborhood of B. Let ϕ δ be a sequence of smooth functions on ∂B converging uniformly to u there. Let u δ be a smooth convex function such that M u δ = µ δ in B and u δ = ϕ δ on ∂B. Let ũ ∈ C( B) be a convex function such that M ũ = 0 and ũ = ϕ δ on ∂B. Moreover, let v δ ∈ C( B) be a convex function such that M v δ = µ δ and v δ = 0 on ∂B.

From the comparison principle for the real Monge-Ampère operator (see [START_REF] Rauch | The Dirichlet problem for the multidimensional Monge-Ampère equation[END_REF]), we can infer that

(3.5.3) ũ + v δ ≤ u δ ≤ ũ -v δ .
It follows from Lemma 3.5 in [RT77] that

(3.5.4) (-v δ (x)) 2n ≤ c n (diam(B)) 2n-1 dist(x, ∂B)M v δ (B), x ∈ B.
Then we conclude that {u δ } is uniformly bounded sequence of convex functions, hence there exists a subsequence {u δ j } converging locally uniformly on B. Moreover, (3.5.3) and (3.5.4) imply that {u δ j } is uniformly convergent on B. From the comparison principle it follows that u δ j converges uniformly to u. Since u δ j ∈ C ∞ ( B) and

M u δ j ≥ f 2 * ρ δ j dV 2n , we get that (dd c u δ j ) n ≥ (f 2 * ρ δ j ) 1/2 dV 2n .
Finally, as u δ j converges uniformly to u, we conclude by Bedford and Taylor's convergence theorem that (dd c u) n ≥ f dV 2n .

We prove now Hölder continuity of the solution to the Dirichlet problem Dir(Ω, ϕ, f dV 2n ) with 0 ≤ f ∈ L p (Ω).

Proof of Theorem 3.1.2. We first suppose that f = 0 near the boundary of Ω, that is, there exists a compact K ⋐ Ω such that f = 0 in Ω \ K. To apply Theorem 3.4.4, we establish a Hölder continuous function v such that v ≤ U in Ω and v = ϕ on ∂Ω. Let ρ be the defining function of Ω given by Definition 2.2.1 and φ be a

C 1,1 -extension of ϕ to Ω such that φ C 1,1 ( Ω) ≤ C ϕ C 1,1 (∂Ω)
, for some C > 0. Now, we take A > 0 large enough such that v := Aρ + φ ∈ P SH(Ω) ∩ C 0,1 ( Ω) and v ≤ U in a neighborhood of K. By the comparison principle, we can find that v ≤ U in Ω \ K and hence v ≤ U in Ω and v| ∂Ω = U| ∂Ω = ϕ. Hence, by this construction and Lemma 3.5.1, the two conditions in Theorem 3.4.4 are satisfied. This implies that the solution U is Hölder continuous in Ω of exponent 2γ for any γ < 1/(nq + 1) and 1/p + 1/q = 1.

For the general case, when f ∈ L p (Ω), p > 1. Let us fix a large ball B ⊂ C n so that Ω ⋐ B ⊂ C n . Let f be the trivial extension of f to B. Since f ∈ L p (Ω) is equal to zero near ∂B, the first case yields that the solution h 1 to the following Dirichlet problem (dd c h 1 ) n = f dV 2n in B, and h 1 = 0 on ∂B, is Hölder continuous on B of exponent 2γ. Now, let h 2 denote the solution to the Dirichlet problem in Ω with boundary values ϕh 1 and the zero density. Thanks to Theorem 2.1.1, we infer that h 2 ∈ C 0,γ ( Ω). Therefore, the required barrier will be v := h 1 + h 2 . It is clear that v ∈ P SH(Ω) ∩ C( Ω), v| ∂Ω = ϕ and (dd c v) n ≥ f dV 2n in the weak sense in Ω. Hence, by the comparison principle we get that v ≤ U in Ω and v = U = ϕ on ∂Ω. Moreover, we have v ∈ C 0,γ ( Ω), for any γ < 1/(nq + 1). By applying Theorem 3.4.4, we conclude that the solution U belongs to C 0,γ ( Ω).

In the special case when f ∈ L p (Ω), p ≥ 2. We can improve the Hölder exponent of U by using the relation between the real and complex Monge-Ampère measures. Let us set µ := f 2 dV 2n which is a nonnegative Borel measure on B with µ(B) < ∞. Thanks to Theorem 3.5.2 there exists a unique convex function u ∈ C( B) such that M u = µ in B and u = 0 on ∂B. Hence u is Lipschitz continuous in Ω. By Proposition 3.5.3, we have (dd c u) n ≥ f dV 2n in Ω. We will construct the required barrier as follows. Let h ϕ-u be the solution to the Dirichlet problem with zero density and ϕu boundary data. Then h ϕ-u is Hölder continuous of exponent 1/2 in Ω by Theorem 2.1.1. Now, it is easy to check that v := u + h ϕ-u is psh in Ω and satisfies v = ϕ in ∂Ω and (dd c v) n ≥ f dV 2n in Ω. So, by the comparison principle, we have v ≤ U in Ω. By Theorem 3.4.4 and Lemma 3.5.1, our solution U will be Hölder continuous of exponent min{1/2, 2γ}, for any γ < 1/(nq + 1). Remark 3.5.4. It is shown in [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF] that we cannot expect a better Hölder exponent than 2/(nq) (see also [START_REF] Plis | A counterexample to the regularity of degenerate complex Monge-Ampère equation[END_REF]).

We introduce an important class of Borel measures on Ω containing Riesz measures and closely related to Hausdorff measures which play an important role in geometric measure theory [START_REF] Mattila | Geometry of sets and measures in euclidean spaces[END_REF]. We call such measures Hausdorff-Riesz measures. Definition 3.5.5. A finite Borel measure on Ω is called a Hausdorff-Riesz measure of order 2n -2 + ǫ, for 0 < ǫ ≤ 2 if it satisfies the following condition :

(3.5.5) µ(B(z, r) ∩ Ω) ≤ Cr 2n-2+ǫ , ∀z ∈ Ω, ∀0 < r < 1,
for some positive constant C.

We give some interesting examples of Hausdorff-Riesz measures.

Example 3.5.6.

1. The Lebesgue measure dV 2n on Ω, for ǫ = 2.

2. The surface measure of a compact real hypersurface, for ǫ = 1.

Measures of the type dd

c v ∧ β n-1
, where v is a α-Hölder continuous subharmonic function in a neighborhood of Ω, for ǫ = α.

4. The measure 1 E H 2n-2+ǫ , where H 2n-2+ǫ is the Hausdorff measure and E is a Borel set such that H 2n-2+ǫ (E) < +∞.

5. If µ is a Hausdorff-Riesz measure of order 2n -2 + ǫ, then f dµ is Hausdorff-Riesz of order 2n-2+ǫ ′ , with ǫ ′ := ǫ-(2n-2+ǫ)/p, for any f ∈ L p (Ω, µ), p > (2n-2+ǫ)/ǫ.

The existence of continuous solutions to Dir(Ω, ϕ, f dµ) for such measures follows immediately from Theorem 3.1.1 and the following lemma. Lemma 3.5.7. Let Ω be a bounded SHL domain and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ, for 0 < ǫ ≤ 2. Assume that 0 ≤ f ∈ L p (Ω, µ) for p > 1, then for all τ > 0 there exists D > 0 depending on τ, ǫ, q and diam(Ω) such that for any Borel set K ⊂ Ω, (3.5.6)

K f dµ ≤ D f L p (Ω,µ) [Cap(K, Ω)] 1+τ .
Proof. By the Hölder inequality we have

K f dµ ≤ f L p (Ω,µ) µ(K) 1/q .
Let z 0 ∈ Ω be a fixed point and R := 2 diam(Ω). Hence, Ω ⋐ B := B(z 0 , R). For any Borel set K ⊂ Ω we get, by Corollary 5.2 in [Z04] and Alexander-Taylor's inequality, that

µ(K) ≤ C(T R (K)) ǫ/2 ≤ C exp(-ǫ/2 Cap(K, B) -1/n ) ≤ C exp(-ǫ/2 Cap(K, Ω) -1/n ),
where C > 0 depends on ǫ and diam(Ω). Now, for any τ > 0, we can find D > 0 depending on τ, ǫ, q and diam(Ω) such that

K f dµ ≤ D f L p (Ω,µ) [Cap(K, Ω)] 1+τ .
The first step in the proof of Theorem 3.1.3 is to estimate Ûδ -U L 1 (Ω δ ,µ) , so we present the following lemma. Lemma 3.5.8. Let Ω ⊂ C n be a SHL domain and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ on Ω, for 0 < ǫ ≤ 2. Suppose that 0 ≤ f ∈ L p (Ω, µ), p > 1 and ϕ ∈ C 1,1 (∂Ω). Then the solution U to Dir(Ω, ϕ, f dµ) satisfies

Ω δ [ Ûδ (z) -U(z)]dµ(z) ≤ Cδ ǫ ,
where C is a positive constant depending on n, ǫ, Ω, f L p (Ω,µ) and µ(Ω).

Proof. Following a slight modification in the proof of Lemma 3.5.1, we can get the required inequality.

When ϕ is not C 1,1 -smooth, the measure ∆U may have infinite mass on Ω. Fortunately, we can estimate Ûδ 1 -U L 1 (Ω δ ,µ) for some δ 1 < δ ≤ 1.

We need the following property of a bounded SHL domain.

Lemma 3.5.9. Let Ω be a bounded SHL domain. Then there exist a function ρ ∈ P SH(Ω)∩ C 0,1 ( Ω) such that near ∂Ω we have

(3.5.7) c 1 dist(z, ∂Ω) ≥ -ρ(z) ≥ c 2 dist(z, ∂Ω) 2 ,
for some c 1 , c 2 > 0 depending on Ω. Moreover, dd c ρ ≥ c 2 β in the weak sense of currents on Ω.

Proof. Since Ω is a strongly hyperconvex Lipschitz domain, there exist a constant c > 0 and a defining function ρ such that dd c ρ ≥ cβ in the weak sense of currents on Ω. Let us fix ξ ∈ ∂Ω, then the function defined by ρξ (z) := ρ(z)c/2|z -ξ| 2 is Lipschitz continuous in Ω and satisfies dd c ρξ ≥ (c/2)β in the weak sense of currents on Ω. Hence, ρξ ∈ P SH(Ω) ∩ C 0,1 ( Ω). Set ρ := sup{ρ ξ : ξ ∈ ∂Ω}.

It is clear that ρ ∈ C 0,1 ( Ω) ∩ P SH(Ω) and thus the first inequality in (3.5.7) holds. For any ξ ∈ ∂Ω we have -ρ ξ (z) ≥ (c/2)|z -ξ| 2 , so we infer that

-ρ(z) ≥ (c/2) dist(z, ∂Ω) 2 ,
for any z near ∂Ω.

The last statement follows from the fact that for any ξ ∈ ∂Ω, dd c ρξ ≥ (c/2)β in the weak sense of currents on Ω.

Remark 3.5.10. When Ω is a smooth strongly pseudoconvex domain, we know that the defining function ρ satisfies near the boundary, -ρ ≈ dist(., ∂Ω). Lemma 3.5.11. Let Ω ⊂ C n be a bounded SHL domain and µ be a Hausdorff-Riesz measure of order 2n -2 + ǫ on Ω, for 0 < ǫ ≤ 2. Suppose that 0 ≤ f ∈ L p (Ω, µ), p > 1 and ϕ ∈ C 0,α (∂Ω), α ≤ 1. Then for any small ǫ 1 > 0, we have the following inequality

Ω δ [ Ûδ 1 (z) -U(z)]dµ(z) ≤ Cδ ǫ/2-ǫ 1 ,
where δ 1 = (1/2)δ 1/2+3/ǫ and C is a positive constant depending on n, Ω, ǫ, ǫ 1 and u L ∞ ( Ω) .

Proof. One sees as in the proof of Lemma 3.5.1 that

Ûδ 1 (z) -U(z) = c n δ 2n 1 δ 1 0 r 2n-1 dr r 0 t 1-2n dt Ç B(z,t) ∆U(ξ) å .
Then, we integrate on Ω δ with respect to µ and use Fubini's Theorem

Ω δ [ Ûδ 1 (z) -U(z)]dµ(z) ≤ c n δ 2n 1 δ 1 0 r 2n-1 dr r 0 t 1-2n dt Ω δ-t Ç B(ξ,t) dµ(z) å ∆U(ξ) ≤ c n δ 2n 1 δ 1 0 r 2n-1 dr r 0 t -1+ǫ dt Ω δ-t ∆U(ξ) ≤ c n δ 2n 1 sup Ω δ-δ 1 (-ρ) -(3+ǫ 1 )/2 δ 1 0 r 2n-1 dr r 0 t -1+ǫ dt Ω δ-t (-ρ) (3+ǫ 1 )/2 ∆U(ξ) ≤ c n δ 2n 1 sup Ω δ/2 (-ρ) -(3+ǫ 1 )/2 (-ρ) (3+ǫ 1 )/2 ∆U Ω δ 1 0 r 2n-1 dr r 0 t -1+ǫ dt ≤ c n δ -3-ǫ 1 ǫ(2n + ǫ) δ ǫ 1 (-ρ) (3+ǫ 1 )/2 ∆U Ω ≤ C 1 δ ǫ/2-ǫ 1 (-ρ) (3+ǫ 1 )/2 ∆U Ω ,
where ρ is as in Lemma 3.5.9 and C 1 > 0 is a positive constant depending on ǫ and n.

To complete the proof we demonstrate that the mass (-ρ) (3+ǫ 1 )/2 ∆U Ω is finite. The following idea is due to [START_REF] Baracco | Hölder regularity of the solution to the complex Monge-Ampère equation with L p density[END_REF] with some appropriate modifications. We set for simplification θ := (3 + ǫ 1 )/2. Let ρ η be the standard regularizing kernels with supp ρ η ⊂ B(0, η) and B(0,η) ρ η dV 2n = 1. Hence,

u η = U * ρ η ∈ C ∞ ∩ P SH(Ω η ) decreases to U in Ω. It is clear that u η L ∞ (Ωη) ≤ U L ∞ (Ω)
and the first derivatives of u η have L ∞ -norms less than U L ∞ (Ω) /η. We denote by χ Ωη the characteristic function of Ω η . Since u η ց U in Ω, we have χ Ωη (-ρ) θ ∆u η converges to (-ρ) θ ∆U in the weak sense of measures.

It is sufficient to show that

I := Ωη (-ρ) θ dd c u η ∧ β n-1 ,
is bounded by an absolute constant independent of η. We have by Stokes' theorem

I = ∂Ωη (-ρ) θ d c u η ∧ β n-1 + θ Ωη (-ρ) θ-1 dρ ∧ d c u η ∧ β n-1 . Note that ∂Ωη (-ρ) θ-1 u η d c ρ ∧ β n-1 = Ωη (-ρ) θ-1 du η ∧ d c ρ ∧ β n-1 + + Ωη (-ρ) θ-1 u η dd c ρ ∧ β n-1 -(θ -1) Ωη (-ρ) θ-2 u η dρ ∧ d c ρ ∧ β n-1 .
Hence, we get

I = ∂Ωη (-ρ) θ d c u η ∧ β n-1 + θ ∂Ωη (-ρ) θ-1 u η d c ρ ∧ β n-1 -θ Ωη (-ρ) θ-1 u η dd c ρ ∧ β n-1 + θ(θ -1) Ωη (-ρ) θ-2 u η dρ ∧ d c ρ ∧ β n-1 ≤ C u L ∞ ( Ω) Ç ∂Ωη dσ + Ωη dd c ρ ∧ β n-1 + Ωη (-ρ) θ-2 β n å , ≤ C u L ∞ ( Ω) Ç ∂Ωη dσ + Ω dd c ρ ∧ β n-1 + Ω (-ρ) (-1+ǫ 1 )/2 β n å ,
where dσ = d c ρ ∧ (dd c ρ) n-1 and ρ is the defining function of Ω. Since ρ is psh in a neighborhood of Ω, the second integral in the last inequality is finite. Thanks to Lemma 3.5.9, we have -ρ ≥ c 2 dist(., ∂Ω) 2 near ∂Ω and so the third integral will be finite since ǫ 1 > 0 small enough. Consequently, we infer that I is bounded by a constant independent of η and then this proves our claim.

Corollary 3.5.12. When Ω is a smooth strongly pseudoconvex domain, then Lemma 3.5.11 holds also for δ 1 = (1/2)δ 1/2+1/ǫ .

Proof. Let ρ be the defining function of Ω. In view of Remark 3.5.10 and the last argument, we can estimate (-ρ) 1+ǫ 1 ∆U Ω , for ǫ 1 > 0 small enough, and ensure that this mass is finite. So the proof of Lemma 3.5.11 is still true for more better δ 1 := (1/2)δ 1/2+1/ǫ .

We are in a position to prove the Hölder continuity of the solution to Dir(Ω, ϕ, f dµ) where µ is a Hausdorff-Riesz measure of order 2n -2 + ǫ and ϕ ∈ C 1,1 (∂Ω).

Proof of Theorem 3.1.3. We first assume that f equals to zero near the boundary ∂Ω, then there exists a compact K ⋐ Ω such that f = 0 on Ω \ K. Since ϕ ∈ C 1,1 (∂Ω), we extend it to φ ∈ C 1,1 ( Ω) such that φ C 1,1 ( Ω) ≤ C ϕ C 1,1 (∂Ω) for some constant C. Let ρ be the defining function of Ω and let A ≫ 1 be so that v := Aρ + φ ∈ P SH(Ω) and v ≤ U in a neighborhood of K. Moreover, by the comparison principle, we see that v ≤ U in Ω \ K. Consequently, v ∈ P SH(Ω) ∩ C 0,1 ( Ω) and satisfies v ≤ U on Ω and v = U = ϕ on ∂Ω. It follows from Theorem 3.4.4 and Lemma 3.5.8 that U ∈ C 0,ǫγ ( Ω), for any 0 < γ < 1/(nq+1).

In the general case, fix a large ball B ⊂ C n containing Ω and define a function f ∈ L p (B, µ) so that f := f in Ω and f := 0 in B \ Ω. Hence, the solution to the following Dirichlet problem

     v 1 ∈ P SH(B) ∩ C( B), (dd c v 1 ) n = f dµ in B, v 1 = 0 on ∂B,
belongs to C 0,γ ′ ( B), with γ ′ = ǫγ for any γ < 1/(nq + 1). Let h ϕ-v 1 be the continuous solution to Dir(Ω, ϕ-v 1 , 0). Then, Theorem 2.1.1 implies that h ϕ-v 1 belongs to C 0,γ ′ /2 ( Ω). This enables us to construct a Hölder barrier for our problem. We take

v 2 = v 1 + h ϕ-v 1 . It is clear that v 2 ∈ P SH(Ω) ∩ C 0,γ ′ /2 ( Ω)
and v 2 ≤ U on Ω by the comparison principle. Hence, Theorem 3.4.4 and Lemma 3.5.8 imply that the solution U to Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω of exponent ǫγ/2 for any 0 < γ < 1/(nq +1).

In the case when ϕ is only Hölder continuous, we prove the Hölder regularity of the solution.

Proof of Theorem 3.1.4. Let also v 1 be as in the proof of Theorem 3.1.3 and h ϕ-v 1 be the solution to Dir(Ω, ϕv 1 , 0). In order to apply Theorem 3.4.4, we set v = v 1 + h ϕ-v 1 . Hence, v ∈ P SH(Ω)∩ ∈ C( Ω), v = ϕ on ∂Ω and (dd c v) n ≥ f dµ in Ω. The comparison principle yields v ≤ U in Ω. Moreover, by Theorem 2.1.1, we have h ϕ-v 1 ∈ C 0,γ ′′ ( Ω) with γ ′′ = 1/2 min{α, ǫγ}. Hence, it stems from Theorem 3.4.4 and Lemma 3.5.11 that the solution U is Hölder continuous on Ω of exponent ǫ ǫ+6 min{α, ǫγ}, for any 0 < γ < 1/(nq + 1).

Moreover, when Ω is a smooth strongly pseudoconvex domain and by Corollary 3.5.12 we get more better Hölder exponent ǫ ǫ+2 min{α, ǫγ}, for any 0 < γ < 1/(nq + 1).

Corollary 3.5.13. Let Ω be a finite intersection of strongly pseudoconvex domains in C n . Assume that ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1, and 0 ≤ f ∈ L p (Ω) for some p > 1.

Then the solution U to the Dirichlet problem Dir(Ω, ϕ, f dV 2n ) belongs to C 0,α ′ ( Ω) with α ′ = min{α/2, γ} for any 0 < γ < 1/(nq + 1). Moreover, if ϕ ∈ C 1,1 (∂Ω) the solution U is γ-Hölder continuous on Ω.

The first part of this corollary was proved in Theorem 1.2 in [START_REF] Baracco | Hölder regularity of the solution to the complex Monge-Ampère equation with L p density[END_REF] with the Hölder exponent min{2γ, α}γ and the second part was proved in [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF] and [START_REF] Charabati | Hölder regularity for solutions to complex Monge-Ampère equations[END_REF] (see also [START_REF] Nguyen | Hölder continuous solutions to complex Hessian equations[END_REF][START_REF] Charabati | Modulus of continuity of solutions to complex Hessian equations[END_REF] for the complex Hessian equation).

Our final purpose concerns how to get the Hölder continuity of the solution to the Dirichlet problem Dir(Ω, ϕ, f dµ), by means of the Hölder continuity of a subsolution to Dir(Ω, ϕ, dµ) for some special measure µ on Ω. We suppose here that µ is less than the Monge-Ampère measure of a Hölder continuous psh function and has the behavior of some Hausdorff-Riesz measure near the boundary.

Proof of Theorem 3.1.5. Let Ω 1 ⋐ Ω be an open set such that µ is a Hausdorff-Riesz measure on Ω \ Ω 1 . Let also Ω 2 ⋐ Ω be a a neighborhood of Ω1 . We claim that (3.5.8)

Ω 1 ( Ûδ -U)dµ ≤ Ω 1 ( Ûδ -U)(dd c w) n ≤ C ∆U Ω 2 δ 2λ λ+2n ,
where C depends on Ω 1 and Ω 2 . This estimate was proved in [START_REF] Demailly | Hölder continuous solutions to Monge-Ampère equations[END_REF]. We can assume without loss of generality that Ω 1 := B 1 , Ω 2 := B 2 and -2 ≤ w ≤ -1 in Ω. This implies that h(z) := |z| 2 -4 < w on B 1 , while w < h on B 2 \ B r 0 for some 1 < r 0 < 2. Replacing w by max{w, h}, we can assume that w = h on

B 2 \ B r 0 . Fix χ ∈ C ∞ 0 (C n ) such that χ ≥ 0, χ(z) := χ(|z|), supp χ ⊂ B 1 and B 1 χdV 2n = 1. Let us set w δ (z) := 1 δ 2n B(z,δ) w(y)χ( z -y δ )dV 2n (y).
Since w ∈ P SH(Ω) ∩ C 0,λ (Ω), we obtain that

w δ (z) -w(z) ≤ C 1 δ λ .
Observe that (3.5.9)

∂ 2 w δ ∂z j ∂ zk ≤ C 2 w L ∞ (Ω) δ 2 .
We choose φ ∈ C ∞ 0 (C n ) such that 0 ≤ φ ≤ 1, φ = 1 on B r 1 and supp φ ⊂ B r 2 , where r 0 < r 1 < r 2 < 2. We define

wδ (z) = B 1 w Ä z -δφ(z)y ä χ(y) dV 2n (y). Note that wδ (z) -w(z) = B 1 [w Ä z -δφ(z)y ä -w(z)]χ(y) dV 2n (y) ≤ C 1 δ λ , and wδ (z) = w δ (z) on B r 1 , wδ (z) = w(z) on B 2 \B r 2 .
Fix now any z ∈ B 2 \B r 0 . Since w = h there, we have for any δ < δ 0 ,

∂ 2 wδ ∂z j ∂z k (z) = B 1 ∂ 2 ∂z j ∂z k h Ä z -δφ(z)y ä χ(y) dV 2n (y) = B 1 [δ jk + δO(1)]χ(y) dV 2n (y) = δ jk + δO(1).
If δ is small enough, we conclude that wδ ∈ P SH(B 2 \B r 0 ), ∀δ < δ 0 . Hence wδ is actually plurisubharmonic in all of B 2 . Set

T := n-1 j=0 (dd c w) j ∧ (dd c wδ ε ) n-1-j ,
where ǫ > 0 to be chosen later. From (3.5.9), Lemma 3.5.1 and Stokes' formula we get

Ω 1 ( Ûδ -U)(dd c w) n ≤ B 2 ( Ûδ -U)(dd c w) n = B 2 ( Ûδ -U)[(dd c w) n -(dd c wδ ε ) n ] + B 2 ( Ûδ -U)(dd c wδ ε ) n ≤ B 2 ( Ûδ -U)dd c (w -wδ ε ) ∧ T + C 3 δ 2nε B 2 ( Ûδ -U) dV 2n ≤ B 2 ( wδ ε -w)dd c (U -Ûδ ) ∧ T + C 3 δ 2nε B 2 ∆Uδ 2 ≤ B 2 ( wδ ε -w)dd c U ∧ T + C 3 B 2 ∆Uδ 2(1-nε) ≤ C 1 δ ελ Br 2 dd c U ∧ T + C 3 δ 2(1-nε) B 2 ∆U ≤ C 4 B 2 ∆U[δ ελ w n-1 L ∞ (Ω) + δ 2(1-nε) ] ≤ C 4 B 2 ∆U δ τ ,
where ε = 2 λ+2n and τ = 2λ λ+2n . Now, let μ be a Hausdorff-Riesz measure on Ω of order 2n -2 + ǫ so that μ equals µ in Ω \ Ω 1 . As ϕ is not C 1,1 -smooth, we estimate Ûδ 1 -U L 1 (Ω δ ,µ) with δ 1 := (1/2)δ 1/2+3/ǫ . Then, we have

Ω δ ( Ûδ 1 -U)dµ ≤ Ω 1 ( Ûδ 1 -U)dµ + Ω δ ( Ûδ 1 -U)dμ.
Fix ǫ 1 > 0 small enough. Then, it follows from (3.5.8) and Lemma 3.5.11 that

Ω δ ( Ûδ 1 -U)dµ ≤ Ω 1 ( Ûδ 1 -U)(dd c w) n + Ω δ ( Ûδ 1 -U)dμ ≤ C ∆U Ω 2 δ 2λ λ+2n 1 + C ′ δ ǫ/2-ǫ 1 , where C = C(Ω 1 , Ω 2 , w L ∞ )
is a positive constant and C ′ depends on n, Ω, ǫ, ǫ 1 and U L ∞ ( Ω) . Since the mass of ∆U is locally bounded, there exists a constant C ′′ > 0 such that

Ω δ ( Ûδ 1 -U)dµ ≤ C ′′ δ τ ,
where τ = min{ ǫ 2ǫ 1 , λ(ǫ+6) ǫ(λ+2n) }. The last requirement to apply Theorem 3.4.4 is to construct a function v ∈ C 0,ν ( Ω) for 0 < ν ≤ 1 such that v ≤ U in Ω and v = ϕ on ∂Ω. Let us denote by w 1 the solution to Dir(Ω, 0, f dμ) and h ϕ the solution to Dir(Ω, ϕ, 0). Now, set v = w 1 + h ϕ + Aρ with A ≫ 1 so that v ≤ U in a neighborhood of Ω1 . It is clear that v ∈ P SH(Ω) ∩ C( Ω), v = ϕ on ∂Ω and v ≤ U in Ω by the comparison principle. Moreover, by Theorem 2.1.1, we infer that v ∈ C 0,ν ( Ω), for ν = 1/2 min{ǫγ, α} and any γ < 1/(nq + 1). Finally, we get from Theorem 3.4.4 that U is Hölder continuous on Ω of exponent ǫ ǫ+6 min{α, ǫγ, 2λγ(ǫ+6) ǫ(λ+2n) }.

The following are nice applications of Theorem 3.1.5.

Corollary 3.5.14. Let Ω ⊂ C n be a bounded SHL domain and µ be a finite Borel measure with compact support on Ω. Let also ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ), p > 1.

Assume that there exists a λ-Hölder continuous psh function w in Ω such that (dd c w) n ≥ µ.

Then the solution to the Dirichlet problem Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω of exponent min{α/2, 2λγ λ+2n }, for any γ < 1/(nq + 1) and 1/p + 1/q = 1.

Example 3.5.15. Let µ be a finite Borel measure with compact support on a bounded SHL domain Ω. Let also ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ), p > 1. Suppose that µ ≤ dV n , where dV n is the Lebesgue measure of the totally real part R n of C n , then the solution to the Dirichlet problem Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω of exponent min{α/2, 2γ 1+2n }, for any γ < 1/(nq + 1) and 1/p + 1/q = 1.

Proof. Since R n = {Imz j = 0, j = 1, ..., n}, one can present the Lebesgue measure of the totaly real part R n of C n in the form

Ñ dd c n j=1 (Imz j ) + é n .
Let us set w = n j=1 (Imz j ) + . It is clear that w ∈ P SH(Ω)∩C 0,1 ( Ω) and µ ≤ (dd c w) n on Ω. Corollary 3.5.14 yields that the solution U belongs to C 0,α ′ ( Ω) with α ′ = min{α/2, 2γ 1+2n }, for any γ < 1/(nq + 1).

At the end, we note that a slight modification in the proof of Theorem 3.1.5 enables us to estimate the modulus of continuity of the solution in terms of the modulus of continuity of a subsolution. Remark 3.5.16. Let µ be a measure satisfying the Condition H(∞) on a bounded SHL domain Ω. Let also ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ), p > 1. Assume that there exists a continuous plurisubharmonic function w in Ω such that (dd c w) n ≥ µ. If the measure µ is Hausdorff-Riesz of order 2n -2 + ǫ in Ω \ Ω 1 for some 0 < ǫ ≤ 2, where Ω 1 ⋐ Ω, then the solution U to Dir(Ω, ϕ, f dµ) has the following modulus of continuity

ω U (δ) ≤ C max{δ ν , ω γ w (δ 2-τ 2n )},
where ν = min{ αǫ ǫ+6 , ǫ 2 γ ǫ+6 , τ γ}, 0 < τ < 2 is an arbitrary constant and C is a positive constant depends on Ω, Ω 1 , n, ǫ, U L ∞ (Ω) , w L ∞ (Ω) .

Open questions

• Let ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and let µ be a finite Borel measure on Ω satisfying Condition H(∞). Suppose that the Dirichlet problem Dir(Ω, ϕ, dµ) has a Hölder continuous subsolution in Ω. Is the solution to this problem Hölder continuous in Ω?

We have shown in Theorem 3.1.5 an affirmative answer when µ satisfies some nice condition near ∂Ω.

• Suppose that µ is a finite Borel measure on Ω and it is strongly dominated by capacity, that is, there exist A, B > 0 so that for any Borel set K ⊂ Ω,

µ(K) ≤ Ae -BCap(K,Ω) -1/n .
Suppose that ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1. Does the solution to Dir(Ω, ϕ, dµ) belong to C 0,α ′ ( Ω) for some 0 < α ′ < 1?

A potential theory for the complex Hessian equation was independently developed by Sadullaev and Abdullaev in [START_REF] Sadullaev | Potential theory in the class of msubharmonic functions[END_REF] and H.C. Lu in [START_REF] Lu | Equations Hessiennes complexes[END_REF]. H.C. Lu developed [START_REF] Lu | Viscosity solutions to complex Hessian equations[END_REF] a viscosity approach to the following Dirichlet problem for the complex Hessian equation.

(4.1.2)

       u ∈ SH m (Ω) ∩ C( Ω), (dd c u) m ∧ β n-m = F (z, u)β n in Ω, u = ϕ on ∂Ω,
where F : Ω × R → R + is a continuous function and nondecreasing in the second variable.

Our first main result in this chapter gives a sharp estimate for the modulus of continuity of the solution to the Dirichlet problem (4.1.2). More precisely, we will prove the following theorem.

Theorem 4.1.1. Let Ω be a smoothly bounded strongly m-pseudoconvex domain in C n , ϕ ∈ C(∂Ω) and 0 ≤ F ∈ C( Ω × R) be a nondecreasing function in the second variable. Then the modulus of continuity ω U of the solution U satisfies the following estimate

ω U (t) ≤ γ(1 + F 1/m L ∞ (K) ) max{ω ϕ (t 1/2 ), ω F 1/m (t), t 1/2 },
where γ is a positive constant depending only on Ω, K = Ω × {a}, a = sup ∂Ω |ϕ| and ω F 1/m (t) is given by

ω F 1/m (t) := sup y∈[-M,M ] sup |z 1 -z 2 |≤t |F 1/m (z 1 , y) -F 1/m (z 2 , y)|,
with M := a + 2 diam(Ω) 2 sup Ω F 1/m (., -a). [START_REF] Lu | Viscosity solutions to complex Hessian equations[END_REF] that the solution to (4.1.2) is Hölder continuous on a smooth bounded strongly pseudoconvex domain Ω under conditions of Hölder continuity of ϕ and F .

H.C. Lu proved in

In the case of the complex Monge-Ampère equation, Y. Wang gave a control on the modulus of continuity of the solution assuming the existence of a subsolution and a supersolution with the given boundary data ( [START_REF] Wang | A viscosity approach to the Dirichlet problem for complex Monge-Ampère equations[END_REF]).

Here we do not assume the existence of a subsolution and a supersolution. Actually the main argument in our proof consists in constructing adequate barriers for the Dirichlet problem for the complex Hessian equation (4.1.2) in a strongly m-pseudoconvex domain.

In the case when the density f ∈ L p (Ω) with p > n/m, N.C. Nguyen [N14] proved the Hölder continuity of the solution to (4.1.1) when the boundary data is in C 1,1 (∂Ω) and the density f satisfies a growth condition near the boundary of Ω.

In the case m = n, the author recently proved [START_REF] Charabati | Hölder regularity for solutions to complex Monge-Ampère equations[END_REF] that the solution to the Dirichlet problem (4.1.1) is Hölder continuous on Ω without assuming any condition near the boundary. Using the same idea we can prove a similar result for the complex Hessian equation. Accurately, we have the following theorem.

Theorem 4.1.2. Let Ω ⊂ C n be a bounded strongly m-pseudoconvex domain with smooth boundary, ϕ ∈ C 1,1 (∂Ω) and 0 ≤ f ∈ L p (Ω), for some p > n/m. Then the solution to (4.1.1) U ∈ C 0,α ( Ω) for any 0 < α < γ 1 , where γ 1 is a constant depending on m, n, p defined by (4.5.1).

Moreover, if p ≥ 2n/m then the solution U ∈ C 0,α ( Ω), for any 0 < α < min{ 1 2 , 2γ 1 }.

In the particular case when f ∈ L p (Ω), for p > n/m, and satisfies some condition near the boundary ∂Ω we can get a better exponent.

Theorem 4.1.3. Let Ω ⊂ C n be a strongly m-pseudoconvex bounded domain with smooth boundary. Suppose ϕ ∈ C 1,1 (∂Ω) and 0 ≤ f ∈ L p (Ω) for some p > n/m, such that

f (z) ≤ (h • ρ(z)) m near ∂Ω,
where ρ is the defining function on Ω and 0 ≤ h ∈ L 2 ([-A, 0[), with A ≥ sup Ω |ρ|, is an increasing function. Then the solution U to (4.1.1) is Hölder continuous of exponent α < min{1/2, 2γ 1 }, where γ 1 is a constant defined by (4.5.1).

Finally, we prove Hölder continuity of the radially symmetric solution with a better exponent which turns out to be optimal.

Preliminaries

We define the differential operator

L α : SH m (Ω) ∩ L ∞ loc (Ω) → D ′ (Ω) such that dd c u ∧ α 1 ∧ ... ∧ α m-1 ∧ β n-m = L α uβ n ,
where α 1 , ..., α m-1 ∈ Σ m . In appropriate complex coordinates this operator is the Laplace operator.

Example 4.2.1. Using the Gårding inequality (1.3.1), one can note that L α (|z| 2 ) ≥ 1 for any

α i ∈ Σ m , 1 ≤ i ≤ m -1.
We will prove the following essential proposition by applying ideas from the viscosity theory developed in [START_REF] Eyssidieux | Viscosity solutions to degenerate complex Monge-Ampère equations[END_REF] for the complex Monge-Ampère equation and extended to the complex Hessian equation by H.C.Lu [START_REF] Lu | Viscosity solutions to complex Hessian equations[END_REF]. A similar result to the following proposition, but for m = n, was proved in [B l96] (see also [START_REF] Charabati | Hölder regularity for solutions to complex Monge-Ampère equations[END_REF]). which is a C 2 -function in B and satisfies u ǫ (y ǫ ) = q(y ǫ ) and q ≥ u ǫ in B, then the following inequality holds in y ǫ , (

dd c q) k ∧ β n-k ≥ 0 for 1 ≤ k ≤ m.
This means that (dd c q + δβ) k yǫ ∧ β n-k ≥ 0 for 1 ≤ k ≤ m. Letting ǫ tend to 0, we get

(dd c q + δβ) k x 0 ∧ β n-k ≥ 0 for 1 ≤ k ≤ m.
Since the last inequality holds for any δ > 0, we can get that dd c q x 0 ∈ Γm . Second step: assume that there exist a point x 0 ∈ Ω and a C 2 -function q satisfying u ≤ q in a neighborhood of x 0 and u(x 0 ) = q(x 0 ) such that

(dd c q) m x 0 ∧ β n-m < F (x 0 , u(x 0 ))β n .
Let us set

q ǫ (x) = q(x) + ǫ(|x -x 0 | 2 - r 2 2 ),
which is a C 2 -function and for 0 < ǫ ≪ 1 small enough we have 0

< (dd c q ǫ ) m x 0 ∧ β n-m < F (x 0 , u(x 0 ))β n .
Since F is continuous on Ω × R, there exists r > 0 such that

(dd c q ǫ ) m ∧ β n-m ≤ F (x, u(x))β n in B(x 0 , r).
Hence, we get (dd c q ǫ ) m ∧ β n-m ≤ (dd c u) m ∧ β n-m in B(x 0 , r), and q ǫ = q + ǫr 2 /2 > q ≥ u on ∂B(x 0 , r). It follows from the comparison principle (see [B l05,[START_REF] Lu | Equations Hessiennes complexes[END_REF]) that q ǫ ≥ u in B(x 0 , r). But this contradicts that q ǫ (x 0 ) = u(x 0 )ǫr 2 /2 < u(x 0 ). We have shown that for every point x 0 ∈ Ω, and every C 2 -function q in a neighborhood of x 0 such that u ≤ q in this neighborhood and u(x 0 ) = q(x 0 ), we have (dd c q) m x 0 ∧ β n-m ≥ F (x 0 , u(x 0 ))β n , hence we have L α q x 0 ≥ F 1/m (x 0 , u(x 0 )).

Final step: assume that F > 0 is a smooth function. Then there exists a C ∞ -function, say g such that L α g = F 1/m (x, u) . Hence Theorem 3.2.10' in [H94]

implies that ϕ = u -g is L α -subharmonic, consequently L α u ≥ F 1/m (x, u). In case F > 0 is only continuous, we note that F (z, u) = sup{w ∈ C ∞ , 0 < w ≤ F }. Since (dd c u) m ∧ β n-m ≥ F (x, u)β n , we get (dd c u) m ∧ β n-m ≥ wβ n . As w > 0 is smooth, we see that L α u ≥ w 1/m . Therefore, we conclude L α u ≥ F 1/m (x, u). In the general case 0 ≤ F ∈ C(Ω × R), we observe that u ǫ (z) = u(z) + ǫ|z| 2 satisfies (dd c u ǫ ) m ∧ β n-m ≥ (F (x, u) + ǫ m )β n .
By the last step, we get L α u ǫ ≥ (F (x, u) + ǫ m ) 1/m , then the required result follows by letting ǫ tend to 0. Definition 4.2.3. Let Ω ⊂ C n be a smoothly bounded domain, we say that Ω is strongly m-pseudoconvex if there exist a defining function ρ of Ω (i.e. a smooth function in a neighborhood U of Ω such that ρ < 0 on Ω, ρ = 0 and dρ = 0 on ∂Ω ) and c > 0 such that

(dd c ρ) k ∧ β n-k ≥ cβ n in U, for 1 ≤ k ≤ m.
The existence of a solution U to the Dirichlet problem (4.1.1) was proved in [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF]. This solution can be given by the upper envelope of subsolutions to the Dirichlet problem as in [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] for the complex Monge-Ampère equation.

(4.2.1)

U = sup{v ∈ SH m (Ω) ∩ C( Ω); v| ∂Ω ≤ ϕ and (dd c v) m ∧ β n-m ≥ F (z, v)β n }.
However, thanks to Lemma 4.2.2, we can describe the solution as the following (4.2.2)

U = sup{v ∈ V m (Ω, ϕ, F )},
where the family V m (Ω, ϕ, F ) is defined as

V m = {v ∈ SH m (Ω) ∩ C( Ω); v| ∂Ω ≤ ϕ and L α v ≥ F (z, v) 1/m , ∀α i ∈ Σ m , 1 ≤ i ≤ m -1}.
This family is nonempty and stable under the operation of taking finite maximum.

Observe that the description of the solution in formula (4.2.2) is more convenient, since it deals with subsolutions with respect to a family of linear elliptic operators.

Existence of solutions

At first, Li proved [START_REF] Li | On the Dirichlet problems for symmetric function equations of the eigenvalues of the complex Hessian[END_REF] that there exist smooth solutions to (4.1.1) for smooth positive densities and smooth boundary values. Moreover, it is well known that there exist continuous solutions to (4.1.1) for L p -densities (see [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF]). We can give an alternative proof to the existence of these solutions using an analogue method to the proof of Proposition 3.3.2.

In this section we study the existence of a continuous solution to (4.1.2) following Cegrell [START_REF] Cegrell | On the Dirichlet problem for the complex Monge-Ampère operator[END_REF] and using the Schauder-Tychonoff fixed point theorem.

Let u 1 be the continuous solution to (4.1.1) for the boundary values ϕ and the density f = 0 and let also u 2 be the continuous solution to (4.1.1) for the boundary values ϕ and the density f = max K F (z, t) where K := Ω × {max ∂Ω |ϕ|}.

Let us set

A := {v ∈ SH m (Ω) ∩ L ∞ (Ω); u 2 ≤ v ≤ u 1 }.
This set is convex and compact in the weak topology. We define the operator G : A → A by taking G(v) to be the continuous solution to the Dirichlet problem:

(dd c w) m ∧ β n-m = F (z, v)β n and lim z→∂Ω w(z) = ϕ,
which exists and is unique by [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF]. We claim that this operator is continuous in the L 1 (Ω)-topology. Let v j ∈ A converges to v in the L 1 (Ω)-topology. By passing to a subsequence, we can assume that v j converges pointwise almost everywhere to v. We set

m i (z) := inf j≥i F (z, v j ) and M i (z) := sup j≥i F (z, v j ). It is clear m i (z) ≤ F (z, v i ) ≤ M i (z).
We take ṽi and vi to be the solutions to (4.1.1) with densities m i and M i respectively. Thus, we conclude vi ≤ G(v i ) ≤ ṽi . Hence, (ṽ i ) is decreasing sequence and (v i ) is increasing sequence. So, we put lim ṽ := lim ṽi ∈ SH m (Ω) and v := (lim vi ) * ∈ SH m (Ω). Hence, we infer

(dd c ṽ) m ∧ β n-m = (dd c v) m ∧ β n-m = F (z, v)β n .
The comparison principle implies that ṽ = v. Finally, we get lim

G(v i ) = v = ṽ = G(v)
almost everywhere. Hence G is continuous in the weak topology.

It follows from the Schauder-Tychonoff fixed point theorem that there exists

v ∈ A such that G(v) = v. So that we have a function u ∈ SH m (Ω) ∩ L ∞ (Ω) such that (dd c u) m ∧ β n-m = F (z, u)β n and lim z→ξ u(z) = ϕ(ξ), ∀ξ ∈ ∂Ω.
Since our solution is the unique solution to (4.1.1) for the bounded density f = F (z, u), this implies that u is continuous on Ω.

The uniqueness of the solution to (4.1.2) is a consequence of the comparison principle. Indeed, suppose that there exist two continuous solutions u 1 , u 2 such that the open set V := {u 1 < u 2 } is not empty. Since F is nondecreasing in the second variable, we get that (dd c u 1 ) m ∧ β n-m ≤ (dd c u 2 ) m ∧ β n-m in V and u 1 = u 2 on ∂V . By the comparison principle, we infer that u 1 ≥ u 2 in V . This is a contradiction.

Modulus of continuity of the solution

(1) h ξ (z) ≤ ϕ(z), ∀z ∈ ∂Ω, (2) h ξ (ξ) = ϕ(ξ), (3) ω h ξ (t) ≤ Cω ϕ (t 1/2 ).
Proof. Since Ω is strongly m-pseudoconvex and its defining function ρ is smooth, we can choose B > 0 large enough such that the function

g(z) = Bρ(z) -|z -ξ| 2 ,
is m-subharmonic in Ω. Let ωϕ be the minimal concave majorant of ω ϕ and define

χ(x) = -ω ϕ ((-x) 1/2 ), which is a convex nondecreasing function on [-d 2 , 0]. Now, fix r > 0 so small that |g(z)| ≤ d 2 in B(ξ, r) ∩ Ω and define for z ∈ B(ξ, r) ∩ Ω the function h(z) = χ • g(z) + ϕ(ξ).
It is clear that h is a continuous m-subharmonic function on B(ξ, r)∩Ω and one can observe that h(z) ≤ ϕ(z) if z ∈ B(ξ, r) ∩ ∂Ω and h(ξ) = ϕ(ξ). Moreover, by the subadditivity of ωϕ and Lemma 2.4.1 we have

ω h (t) = sup |z-y|≤t |h(z) -h(y)| ≤ sup |z-y|≤t ωϕ ï |z -ξ| 2 -|y -ξ| 2 -B(ρ(z) -ρ(y)) 1/2 ò ≤ sup |z-y|≤t ωϕ î ((2d + B 1 )|z -y|) 1/2 ó ≤ C.ω ϕ (t 1/2 ),
where C := 1 + (2d + B 1 ) 1/2 is a constant depending on Ω.

Recall that ξ ∈ ∂Ω and fix 0 < r 1 < r and

γ 1 ≥ 1 + d/r 1 such that -γ 1 ωϕ î (|z -ξ| 2 -Bρ(z)) 1/2 ó ≤ inf ∂Ω ϕ -sup ∂Ω ϕ, for z ∈ ∂Ω ∩ ∂B(ξ, r 1 ). Let us set γ 2 = inf ∂Ω ϕ, it follows that γ 1 (h(z) -ϕ(ξ)) + ϕ(ξ) ≤ γ 2 for z ∈ ∂B(ξ, r 1 ) ∩ Ω. Now set h ξ (z) = ® max{γ 1 (h(z) -ϕ(ξ)) + ϕ(ξ), γ 2 } ; z ∈ Ω ∩ (B(ξ, r 1 ), γ 2 ; z ∈ Ω \ B(ξ, r 1 ),
which is a well defined m-subharmonic function on Ω and continuous on Ω. Moreover, it satisfies h ξ (z) ≤ ϕ(z) for all z ∈ ∂Ω. Indeed, on ∂Ω ∩ B(ξ, r 1 ) we have

γ 1 (h(z) -ϕ(ξ)) + ϕ(ξ) = -γ 1 ωϕ (|z -ξ|) + ϕ(ξ) ≤ -ω ϕ (|z -ξ|) + ϕ(ξ) ≤ ϕ(z).
Furthermore, the modulus of continuity of h ξ satisfies

ω h ξ (t) ≤ Cω ϕ (t 1/2 ),
where C := γ 1 C depends on Ω. Hence, h ξ satisfies the conditions (1)-(3), and this completes the proof.

In the following proposition, we establish a barrier to the problem (4.1.2) and estimate its modulus of continuity. Proposition 4.4.2. Let Ω ⊂ C n be a bounded strongly m-pseudoconvex domain with smooth boundary. Assume that ω ϕ is the modulus of continuity of ϕ ∈ C(∂Ω) and 0 ≤ F ∈ C( Ω × R) is nondecreasing in the second variable. Then there exists a subsolution v ∈ V m (Ω, ϕ, F ) such that v = ϕ on ∂Ω and the modulus of continuity of v satisfies the following inequality

ω v (t) ≤ λ max{ω ϕ (t 1/2 ), t 1/2 }, where λ = η(1 + F 1/m L ∞ (K) ), K = Ω × {sup ∂Ω |ϕ|} and η is a positive constant depending on Ω.
Proof. First of all, fix ξ ∈ ∂Ω. We will prove that there exists v ξ ∈ V m (Ω, ϕ, F ) such that v ξ (ξ) = ϕ(ξ). We fix z 0 ∈ Ω and set K 1 := sup K F 1/m . Hence, we have

L α (K 1 |z -z 0 | 2 ) = K 1 L α |z -z 0 | 2 ≥ F 1/m (z, sup ∂Ω |ϕ|), for all α i ∈ Σ m , 1 ≤ i ≤ m -1 and z ∈ Ω. We also set K 2 := K 1 |ξ -z 0 | 2 and define the continuous function φ(z) := ϕ(z) -K 1 |z -z 0 | 2 + K 2 .
we find, by Lemma 4.4.1, a constant C > 0 depending on Ω and a function h ξ satisfying the following conditions:

1) h ξ (z) ≤ φ(z), ∀z ∈ ∂Ω, 2) h ξ (ξ) = φ(ξ), 3) ω h ξ (t) ≤ Cω φ(t 1/2 ). Then the required function v ξ ∈ V m (Ω, ϕ, F ) is given by v ξ (z) := h ξ (z) + K 1 |z -z 0 | 2 -K 2 . It is obvious that v ξ ∈ SH m (Ω) ∩ C( Ω). Since h ξ (z) ≤ φ(z) = ϕ(z) -K 1 |z -z 0 | 2 + K 2 on ∂Ω,
we conclude v ξ (z) ≤ ϕ(z) on ∂Ω and v ξ (ξ) = ϕ(ξ). Moreover, we have

L α v ξ = L α h + K 1 L α |z -z 0 | 2 ≥ F 1/m (z, v ξ ) in Ω.
Furthermore, by the hypothesis on h ξ , we can estimate the modulus of continuity of v ξ :

ω v ξ (t) = sup |z-y|≤t |v(z) -v(y)| ≤ ω h (t) + K 1 ω |z-z 0 | 2 (t) ≤ Cω φ(t 1/2 ) + 4d 3/2 K 1 t 1/2 ≤ Cω ϕ (t 1/2 ) + 2dK 1 (C + 2d 1/2 )t 1/2 ≤ (C + 2d 1/2 )(1 + 2dK 1 ) max{ω ϕ (t 1/2 ), t 1/2 },
where d := diam(Ω). Hence, we have

ω v ξ (t) ≤ η(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 },
where η := (C + 2d 1/2 )(1 + 2d) is a constant depending on Ω.

We have just proved that for each ξ ∈ ∂Ω, there is a function v ξ ∈ V m (Ω, ϕ, F ) such that v ξ (ξ) = ϕ(ξ), and

ω v ξ (t) ≤ η(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 }. Let us set v(z) = sup {v ξ (z); ξ ∈ ∂Ω} .
We have 0 ≤ ω v (t) ≤ η(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 }, thus ω v (t) converges to zero when t converges to zero. Consequently, we get v ∈ C( Ω) and v = v * ∈ SH m (Ω). Thanks to Choquet's lemma, we can choose a nondecreasing sequence (v j ), where v j ∈ V m (Ω, ϕ, F ), converging to v almost everywhere. So that

L α v = lim j→∞ L α v j ≥ F 1/m (z, v), ∀α i ∈ Σ m . It is clear that v(ξ) = ϕ(ξ) for any ξ ∈ ∂Ω. Finally, we get v ∈ V m (Ω, ϕ, F ) such that v = ϕ on ∂Ω and ω v (t) ≤ η(1 + K 1 ) max{ω ϕ (t 1/2 ), t 1/2 }.
Corollary 4.4.3. Under the same assumption of Proposition 4.4.2. There exists a msuperharmonic function ṽ ∈ C( Ω) such that ṽ = ϕ on ∂Ω and

ω ṽ(t) ≤ λ max{ω ϕ (t 1/2 ), t 1/2 },
where λ > 0 is as in Proposition 4.4.2.

Proof. We can do the same construction as in the proof of Proposition 4.4.2 for the function

ϕ 1 = -ϕ ∈ C(∂Ω), then we get v 1 ∈ V m (Ω, ϕ 1 , F ) such that v 1 = ϕ 1 on ∂Ω and ω v 1 (t) ≤ λ max{ω ϕ (t 1/2 ), t 1/2 }.
Hence, we set ṽ = -v 1 which is a m-superharmonic function on Ω, continuous on Ω and satisfying ṽ = ϕ on ∂Ω and ω ṽ(t) ≤ λ max{ω ϕ (t 1/2 ), t 1/2 }.

Proof of Theorem 4.1.1. Thanks to Proposition 4.4.2, we obtain a subsolution v ∈ V m (Ω, ϕ, F ) with v = ϕ on ∂Ω and ω v (t) ≤ λ max{ω ϕ (t 1/2 ), t 1/2 }. From Corollary 4.4.3, we construct a m-superharmonic function ṽ ∈ C( Ω) such that ṽ = ϕ on ∂Ω and ω ṽ(t) ≤ λ max{ω ϕ (t 1/2 ), t 1/2 }, where λ is as in Proposition 4.4.2.

Applying the maximum principle, we get that v(z) ≤ U(z) ≤ ṽ(z) for all z ∈ Ω.

We set g(t) = max{λ max{ω ϕ (t 1/2 ), t 1/2 }, ω Consequently, ũ ≤ U in Ω and then we get the required statement. Now, we assert that L α V ≥ F 1/m (z, V ), for all α i ∈ Σ m . Indeed, We will construct the required barrier as follows. Let h ϕ-u be the upper envelope of V m (Ω, ϕu, 0). Then, thanks to Theorem 4.1.1, h ϕ-u is Hölder continuous of exponent 1/2 in Ω. Now it is easy to check that v := u + h ϕ-u is m-sh in Ω and satisfies v = ϕ in ∂Ω and (dd c v) m ∧ β n-m ≥ f β n on Ω. Hence v ≤ U in Ω by the comparison principle.

L α v 1 (z) ≥
The last theorem provides us with a Hölder continuous barrier for the Dirichlet problem (4.1.1) with better exponent.

However, when f ∈ L p (Ω) for p > n/m, we can find a Hölder continuous barrier with exponent less than γ 1 .

Proof of Theorem 4.1.2. We first prove that the total mass of ∆U is finite in Ω. Let U 0 be the solution to the Dirichlet problem (4.1.1) with zero boundary values and the density f . We first claim that the total mass of ∆U 0 is finite in Ω. Indeed, let ρ be the defining function of Ω. By Corollary 1.3.24 we obtain Since Ω is a bounded strongly m-pseudoconvex domain, there exists a constant c > 0 such that (dd c ρ) j ∧ β n-j ≥ cβ n in Ω for 1 ≤ j ≤ m. We find A ≫ 1 such that Aρ -|z| 2 is m-sh function. Now, it is easy to see that

Ω dd c U 0 ∧ β n-1 ≤ Ω dd c U 0 ∧ (Add c ρ) m-1 ∧ β n-m .
Hence, the inequality (4.5.2) yields

Ω dd c U 0 ∧ β n-1 ≤ A m-1 ï Ω (dd c U 0 ) m ∧ β n-m ò 1 m ï Ω (dd c ρ) m ∧ β n-m ò m-1 m .
Now, we note that the total mass of complex Hessian measures of ρ and U 0 are finite in Ω. Therefore, the total mass of ∆U 0 is finite in Ω. Let φ be a C 1,1 -extension of ϕ to Ω such that φ C 1,1 ( Ω) ≤ C ϕ C 1,1 (∂Ω) , for some C > 0. Let v = Bρ + φ + U 0 , where B ≫ 1 is so that Bρ + φ ∈ SH m (Ω) ∩ C( Ω). By the comparison principle, we see that v ≤ U in Ω and v = U = ϕ on ∂Ω. Since ρ is smooth in a neighborhood of Ω and ∆U 0 Ω < +∞, we derive that ∆v Ω < +∞. Then, by Lemma 3.4.6, we have ∆U Ω < +∞.

To apply Theorem 4.5.2 we construct a Hölder continuous function v such that v ≤ U in Ω and v = ϕ on ∂Ω. We first assume that f = 0 near the boundary of Ω, that is there exists a compact K ⋐ Ω such that f = 0 in Ω \ K. We set A > 0 large enough so that v := Aρ + φ ∈ SH m (Ω) ∩ C 0,1 ( Ω) and v ≤ U in a neighborhood of K. By the comparison principle, we can find that v ≤ U in Ω \ K and hence v ≤ U in Ω and v| ∂Ω = U| ∂Ω = ϕ. Thus, Theorem 4.5.2 implies that the solution U is Hölder continuous in Ω of exponent α 1 < 2γ 1 , where γ 1 is as in (4.5.1).

For the general case, when 0 ≤ f ∈ L p (Ω), p > n/m. Let us fix a large ball B ⊂ C n such that Ω ⋐ B ⊂ C n . We define f = f in Ω and f = 0 in B \ Ω. Let h 1 to the Dirichlet problem in B with the density f and zero boundary values. Since f ∈ L p (Ω) is bounded near ∂B, h 1 is Hölder continuous on B of exponent α 1 < 2γ 1 by the previous case. Now let h 2 denote the solution to the Dirichlet problem in Ω with boundary values ϕh 1 and zero density. Thanks to Theorem 4.1.1, we infer that h 2 ∈ C 0,α 2 ( Ω), where α 2 = α 1 /2. Therefore, the required barrier will be v = h 1 + h 2 . It is clear that v ∈ SH m (Ω) ∩ C( Ω), v| ∂Ω = ϕ and (dd c v) m ∧ β n-m ≥ f β n in the weak sense in Ω. Hence, by the comparison principle, we get that v ≤ U in Ω and v = U = ϕ on ∂Ω. Moreover, we have v ∈ C 0,α ( Ω) for any α < γ 1 . Hence, when p > n/m, we get by Theorem 4.5.2 that U ∈ C 0,α ( Ω) for any α < γ 1 . Moreover, if p ≥ 2n/m, Theorem 4.5.4 gives the existence of a 1/2-Hölder continuous barrier to the Dirichlet problem. Then, we obtain by Theorem 4.5.2 that U ∈ C 0,α ( Ω) for any α < min{1/2, 2γ 1 }.

We are able to find a better Hölder-exponent of the solution, when the density f ∈ L p (Ω), p > n/m, satisfies the following condition near the boundary ∂Ω, Now consider a C 1,1 -extension φ of ϕ to Ω and choose B ≫ 1 large enough so that Bρ + φ is m-subharmonic in Ω and ṽ := B(v + ρ) + φ ≤ U in a neighborhood of K.

Then ṽ is m-subharmonic in Ω and if B ≥ (1/c) 1/m , then it follows from (4.5.4) that (dd c ṽ) m ∧ β n-m ≥ f β n in Ω \ K.

By the comparison principle, we have ṽ ≤ U on Ω \ K. Consequently, ṽ ≤ U on Ω, ṽ = ϕ on ∂Ω and ṽ ∈ C 0,1/2 ( Ω). As in the proof of Theorem 4.1.2 we have that the total mass of ∆U is finite in Ω. Hence, Theorem 4.5.2 yields that the solution U belongs to C 0,α ( Ω) for any α < min{1/2, 2γ 1 }.

As an example of application of the last result, fix p > n/m, take h(t) := (-t) -α with 0 < α < 1/(pm), t < 0 and define f := (h • ρ) m .

Hölder continuity for radially symmetric solution

Here we consider the case when the right hand side and the boundary data are radial. In this case, Huang and Xu [START_REF] Huang | Regularity of radial solutions to the complex Hessian equations[END_REF] gave an explicit formula for the radial solution of the Dirichlet problem (4.1.1) with f ∈ C( B) (see also [START_REF] Monn | Regularity of the complex Monge-Ampère equation for radially symmetric functions of the unit ball[END_REF] for complex Monge-Ampère equations). Moreover, they studied higher regularity for radial solutions (see also [START_REF] Dieu | Radial symmetric solution of complex Hessian equation in the unit ball[END_REF]).

Here, we will extend this explicit formula to the case when f ∈ L p (B), for p > n/m, is a radial nonnegative function and ϕ ≡ 0 on ∂B. Then, we prove Hölder continuity of the radially symmetric solution. We claim that the sequence {U k } is uniformly bounded and equicontinuous in B. Indeed, let 0 < r < r 1 ≤ 1, we have

|U k (r 1 ) -U k (r)| = B r 1 r 1 t 2n/m-1 Ç t 0 ρ 2n-1 f k (ρ)dρ å 1/m dt ≤ B r 1 r 1 t 2n/m-1
Ç t 0 ρ (2n-1)/q ρ (2n-1)/p f k (ρ)dρ We give an example which illustrates that the Hölder exponent 2 -2n mp given by Theorem 4.1.4 is optimal. mp + δ)-Hölder with δ = (2n/pα)/m. Since α can be chosen arbitrary close to 2n/p, this implies that the optimal Hölder exponent is 2 -2n mp . 2. Observe that when 1 ≤ p < n/m and 2m < α < 2n, then the solution U α is unbounded.

å 1/m dt ≤ C f k 1/m L p (B)
The next example shows that in Theorem 4.1.4, n/m is the critical exponent in order to have a continuous solution. It is clear that f ∈ L n/m (B) \ L n/m+δ (B) for any δ > 0. An elementary computation shows that the corresponding solution U given by the explicit formula (4.1.3) can be estimated by U(z) ≤ C(1 -(1 -log|z|) 1-γ/m ), where C > 0 depends only on n, m and γ. Hence we see that if m/n < γ < m then U goes to -∞ when z goes to 0. In this case the solution U is unbounded.

Open questions

• Let Ω be a smooth bounded strongly m-pseudoconvex domain in C n , ϕ ∈ C(∂Ω). Let also µ be a Hausdorff-Riesz measure on Ω and 0 ≤ f ∈ L p (Ω, µ) for some p > n/m. Does there exist a continuous solution to (4.1.1)? Moreover, if ϕ is Hölder continuous, can we say that U is Hölder continuous in Ω?

Résumé

Cette thèse est consacrée à l'étude de la régularité des solutions des équations de Monge-Ampère complexes ainsi que des équations hessiennes complexes dans un domaine borné de C n . Dans le premier chapitre, on donne des rappels sur la théorie du pluripotentiel. Dans le deuxième chapitre, on étudie le module de continuité des solutions du problème de Dirichlet pour les équations de Monge-Ampère lorsque le second membre est une mesure à densité continue par rapport à la mesure de Lebesgue dans un domaine strictement hyperconvexe lipschitzien. Dans le troisième chapitre, on prouve la continuité hölderienne des solutions de ce problème pour certaines mesures générales. Dans le quatrième chapitre, on considère le problème de Dirichlet pour les équations hessiennes complexes plus générales où le second membre dépend de la fonction inconnue. On donne une estimation précise du module de continuité de la solution lorsque la densité est continue. De plus, si la densité est dans L p , on démontre que la solution est Höldercontinue jusqu'au bord.

Mots-clés

Problème de Dirichlet, Opérateur de Monge-Ampère, Mesure de Hausdorff-Riesz, Fonction m-sousharmonique, Opérateur hessien, Capacité, Module de continuité, Principe de comparaison, Théorème de stabilité, Domaine strictement hyperconvexe lipschitzien, Domaine strictement m-pseudoconvexe.

  this is a contradiction. Corollary 2.3.8. Let B be the unit ball in C n , 0 ≤ f ∈ C( B) and ϕ ∈ C(∂B). Then the upper envelope U is the solution to Dirichlet problem Dir(B, ϕ, f ).

  Proposition 2.5.3. Let Ω be a bounded SHL domain in C n , 0 ≤ f ∈ C( Ω) and ϕ ∈ C(∂Ω). Then there exists a positive constant C = C(Ω) such that

Theorem 3.1. 5 .

 5 Let µ be a finite Borel measure on a bounded SHL domain Ω satisfying Condition H(∞) mentioned below. Let also ϕ ∈ C 0,α (∂Ω), 0 < α ≤ 1 and 0 ≤ f ∈ L p (Ω, µ), p > 1. Assume that there exists a λ-Hölder continuous plurisubharmonic function w in Ω such that (dd c w) n ≥ µ. If, near the boundary, µ is Hausdorff-Riesz of order 2n -2 + ǫ for some 0 < ǫ ≤ 2, then the solution U to Dir(Ω, ϕ, f dµ) is Hölder continuous on Ω. Such a problem is still open for measures without any condition near the boundary of a bounded domain in C n .

  µ has compact support in Ω. Let us consider a subdivision I s of suppµ consisting of 3 2ns congruent semi-open cubes I s j with side d s = d/3 s , where d := diam(Ω) and 1 ≤ j ≤ 3 2ns . Thanks to Theorem 3.3.2, one can find U s ∈ P SH(Ω) ∩ C( Ω) such that U s = ϕ on ∂Ω, and (dd c U s ) n = µ s :=

Theorem 4.1. 4 .

 4 Let f ∈ L p (B) be a radial function, where p > n/m. Then the unique solution U to (4.1.1) with zero boundary values is given by the explicit formula ä -1/m . Moreover, U ∈ C 0,2-2n mp ( B) for n/m < p < 2n/m and U ∈ Lip( B) for p ≥ 2n/m.

  Lemma 4.4.1. Let Ω ⊂ C n be a bounded strongly m-pseudoconvex domain with smooth boundary. Then for every point ξ ∈ ∂Ω and ϕ ∈ C(∂Ω), there exist a constant C > 0 depending only on Ω and a function h ξ ∈ SH m (Ω)∩C( Ω) such that the following conditions hold:

  F 1/m (t)} and d := diam(Ω). Then |U(z) -U(ξ)| ≤ g(|z -ξ|); ∀z ∈ Ω, ∀ξ ∈ ∂Ω.Let us fix a point z 0 ∈ Ω, for any vector τ ∈ C n with small enough norm, we defineV (z, τ ) = ® U(z) ; z + τ / ∈ Ω, z ∈ Ω, max{U(z), v 1 (z)} ; z, z + τ ∈ Ω, where v 1 (z) = U(z + τ ) + g(|τ |)|zz 0 | 2d 2 g(|τ |)g(|τ |) . Observe that if z ∈ Ω, z + τ ∈ ∂Ω, we have (4.4.1) v 1 (z) -U(z) ≤ g(|τ |) + g(|τ |)|zz 0 | 2d 2 g(|τ |)g(|τ |) ≤ 0.Then v 1 (z) ≤ U(z) for z ∈ Ω, z + τ ∈ ∂Ω. In particular, V (z, τ ) is well defined and belongs to SH m (Ω) ∩ C( Ω). We claim thatF 1/m (z 1 , U(x)) -F 1/m (z 2 , U(x)) ≤ ω F 1/m (|z 1z 2 |),for all x, z 1 , z 2 ∈ Ω. Indeed, it is enough to show thatU L ∞ ( Ω) ≤ M := a + 2d 2 sup Ω F 1/m (., -a),with a := sup ∂Ω |ϕ|. By the maximum principle, we have U ≤ a. We set b = sup Ω F 1/m (., -a) and ũ = b(|zz 0 | 2d 2 )a ∈ SH m (Ω) ∩ C( Ω) where z 0 ∈ Ω is a fixed point. Hence, ũ ≤ ϕ on ∂Ω. Since F is nondecreasing in the second variable, we get (dd c ũ) m ∧ β n-m ≥ F (z, -a)β n ≥ F (z, ũ)β n .

  F 1/m (z + τ, U(z + τ )) + g(|τ |)L α (|zz 0 | 2 ) ≥ F 1/m (z + τ, U(z + τ )) + g(|τ |) ≥ F 1/m (z + τ, U(z + τ )) + |F 1/m (z + τ, U(z + τ )) -F 1/m (z, U(z + τ ))| ≥ F 1/m (z, U(z + τ )), ≥ F 1/m (z, v 1 (z)), for all α i ∈ Σ m , 1 ≤ i ≤ m -1. If z ∈ ∂Ω , z + τ / ∈ Ω, then V (z, τ ) = U(z) = ϕ(z). On the other hand, z ∈ ∂Ω, z + τ ∈ Ω, we get by (4.4.1) that V (z, τ ) = max{U(z), v 1 (z)} = U(z) = ϕ(z). Then V (z, τ ) = ϕ(z) on ∂Ω, hence V ∈ V m (Ω, ϕ, F ). Consequently, V (z, τ ) ≤ U(z); ∀z ∈ Ω. This implies that if z ∈ Ω , z + τ ∈ Ω, we have U(z + τ ) + g(|τ |)|zz 0 | 2d 2 g(|τ |)g(|τ |) ≤ U(z). Hence, U(z + τ ) -U(z) ≤ (d 2 + 1)g(|τ |)g(|τ |).|zz 0 | 2 ≤ (d 2 + 1)g(|τ |).Reversing the roles of z + τ and z, we get|U(z + τ ) -U(z)| ≤ (d 2 + 1)g(|τ |); ∀z ∈ Ω. Thus, ω U (t) ≤ (d 2 + 1)g(t).Finally, we haveω U (t) ≤ γ(1 + F 1/m L ∞ (K) ) max{ω ϕ (t 1/2 ), ω F 1/m (t), t 1/2 },where γ := η(d 2 +1), η is a positive constant depending on Ω and K = Ω×{sup ∂Ω |ϕ|}. Remark 4.4.4. When m = n we can get by a slight modification that the proof is still true for a bounded strongly hyperconvex Lipschitz domain in C n and this yields Theorem 2.1.3 in Chapter 2. Theorem 4.1.1 has the following consequence.

Theorem 4.5. 4 .

 4 Let ϕ ∈ C 0,1 (∂Ω) and f ∈ L p (Ω), p ≥ 2n/m. Then there exists v ∈ SH m (Ω) ∩ C 0,1/2 ( Ω) such that v = ϕ on ∂Ω and (dd c v) m ∧ β n-m ≥ f β n in the weak sense of currents. In particular, v ≤ U in Ω.Proof. Let B be a large ball containing Ω and let f be the function defined by f = f on Ω and f = 0 on B \ Ω. Then f ∈ L p (B), p ≥ 2n/m . Let us set µ := f 2n/m (n!) 2n/m dV 2n that is a nonnegative Borel measure on B with µ(B) < ∞. Thanks to Theorem 3.5.2 there exists a unique convex function u ∈ C( B) such that M u = µ in B and u = 0 on ∂B. Hence u is Lipschitz continuous on Ω. By Proposition 4.5.3, we have (dd c u) m ∧ β n-m ≥ f β n in Ω.

Ω

  dd c U 0 ∧(dd c ρ) m-1 ∧β n-m ≤ ï Ω (dd c U 0 ) m ∧ β n-m ò 1 m ï Ω (dd c ρ) m ∧ β n-m ò m-1 m .

f

  (z) ≤ (h • ρ(z)) m near ∂Ω, where 0 ≤ h ∈ L 2 ([-A, 0[) is an increasing function and A ≥ sup Ω |ρ|.Proof of Theorem 4.1.3. Let χ : [-A, 0] → R -be the primitive of h such that χ(0) = 0. It is clear that χ is a convex increasing function. By the Hölder inequality, we see that|χ(t 1 )χ(t 2 )| ≤ h L 2 |t 1t 2 | 1/2 , for all t 1 , t 2 ∈ [-A, 0]. From the hypothesis, there exists a compact K ⋐ Ω such that (4.5.3) f (z) ≤ (h • ρ(z)) m for z ∈ Ω \ K.Then the function v = χ • ρ is m-subharmonic in Ω, continuous on Ω and satisfiesdd c χ • ρ = χ ′′ (ρ)dρ ∧ d c ρ + χ ′ (ρ)dd c ρ ≥ χ ′ (ρ)dd c ρ,in the weak sense of currents in Ω. By Definition 4.2.3, there is a constant c > 0 such that (dd c ρ) m ∧ β n-m ≥ cβ n . Hence the inequality (4.5.3) yields (4.5.4)(dd c v) m ∧ β n-m ≥ c(h • ρ) m β n ≥ c.f β n in Ω \ K.

  Proof of Theorem 4.1.4. Let f k ∈ C( B) be a positive radial symmetric function such that {f k } converges to f in L p (B). Then there exists, by[START_REF] Huang | Regularity of radial solutions to the complex Hessian equations[END_REF], a unique solution U k ∈ C( B) to (4.1.1) with zero boundary values and the density f k , given by the following formula:It is clear that U k converges in L 1 (B) to the function ũ given by the same formula i.e. ũ(r) = -B

  mp ).Since f k converges to f in L p (B), we get f k L p (B) ≤ C 1 , where C 1 > 0 does not depend on k. Hence U k is equicontinuous on B. By Arzelà-Ascoli theorem, there exists a subsequence U k j converges uniformly to ũ. Consequently, ũ ∈ SH m (B) ∩ C( B) and thanks to the convergence theorem for the complex Hessian operator (see[START_REF] Sadullaev | Potential theory in the class of msubharmonic functions[END_REF]) we can see that(dd c ũ) m ∧ β n-m = f β n in B.Moreover, we have |ũ(r 1 )ũ(r)| ≤ C f p ≥ 2n/m we get ũ ∈ Lip( B), and for n/m < p < 2n/m we have ũ ∈ C 0,2-2n mp ( B).

  Example 4.5.5. Let p ≥ 1 be a fixed exponent. Take f α (z) = |z| -α , with 0 < α < 2n/p. Then it is clear that f α ∈ L p (B). The unique radial solution to the Dirichlet problem (4.1.1) with right hand side f α and zero boundary values is given byU α (z) = c(r 2-α/m -1); r := |z| ≤ 1, where c = Ä C m n 2 m+1 n ä -1/m ( 1 2n-α ) 1/m m 2m-α .Then we have 1. If p > n/m then 0 < α < 2m and the solution U α is (2 -2n

Example 4.5. 6 .

 6 Consider the density f given by the formulaf (z) := 1 |z| 2m (1 -log|z|) γ ,where γ > m/n is fixed.

Definition 1.3.14. Let

  Ω be a bounded domain in C n and u ∈ SH m (Ω). We say that u is m-maximal if for every open set G ⋐ Ω and for each upper semicontinuous function v on Ḡ such that v ∈ SH m (G) and v ≤ u on ∂G, we have v ≤ u in G.

	Theorem 1.3.15 ([B l05]

  A bounded domain Ω in C n is called B-regular if for any boundary point z 0 ∈ ∂Ω there exists v ∈ P SH(Ω) such that lim z→z 0 v(z) = 0 and v * | Ω\{z 0 } < 0.

	Proof. For every matrix H ∈ H + n , there is C ∈ H + n such that C 2 = H. We set H 1/2 := n . After diagonalizing the matrix H 1/2 .Q.H 1/2 and by the C, hence H 1/2 .Q.H 1/2 ∈ H + inequality of arithmetic and geometric means, we get
						2)
	is a bounded strongly hyperconvex Lipschitz domain in C n with non-smooth bound-
	ary.
	5. The unit polydisc in C n (n ≥ 2) is hyperconvex with Lipschitz boundary but it is not strongly hyperconvex Lipschitz.
	We recall the definition of B-regular domain in the sense of Sibony ([Sib87], [B l96]).
	Definition 2.2.3. Remark 2.2.4. Any bounded SHL domain is B-regular in the sense of Sibony. Indeed,
	for any boundary point z 0 ∈ ∂Ω it is enough to take v(z) = Aρ -|z -z 0 | 2 where A > 1/c and c > 0 is as in Definition 2.2.1.
	Remark 2.2.5. Kerzman and Rosay [KR81] proved that in a hyperconvex domain there
	exists an exhaustion function which is smooth and strictly plurisubharmonic. Furthermore,
	they proved that any bounded pseudoconvex domain with C 1 -boundary is hyperconvex domain. Later, Demailly [De87] proved that any bounded pseudoconvex domain with
	Lipschitz boundary is hyperconvex. It is obvious that such a domain can contain a germ
	of analytic subvariety in the boundary, hence it can not be a bounded SHL domain (for
	example, we smooth out the boundary of a polydisc) since the condition (2) in Definition
	2.2.1 fails.
	Let Ω ⊂ C n be a bounded domain. If u ∈ P SH(Ω) then dd c u ≥ 0 in the sense of currents. We define
	(2.2.1)	∆ H u :=	n j,k=1	h j	k ∂ 2 u ∂z k ∂ zj	= tr (H.Hess C u)
	for every positive definite Hermitian matrix H = (h j k). We can view ∆ H u as a nonnegative
	Radon measure in Ω.
	The following lemma is elementary and important for what follows (see [Gav77]).
	Lemma 2.2.6. Let Q be a n × n nonnegative Hermitian matrix. Then
	(det Q)	1 n = inf{tr(H.Q) : H ∈ H + n and det(H) = n -n },
	where H + n denotes the set of all positive Hermitian n × n matrices.

  tr(H.P.A.P -1 ) = tr(H.Q).

	Example 2.2.7. We calculate ∆ H (|z| 2 ) for every matrix H ∈ H + n and det H = n -n .
	n
	∆ H (|z| 2 ) =
	j,k=1

Preliminaries

Remerciements

Chapter 4

The Dirichlet problem for complex Hessian equations

Introduction

Let Ω be a bounded domain in C n with smooth boundary and let m be an integer such that 1 ≤ m ≤ n. Given ϕ ∈ C(∂Ω) and 0 ≤ f ∈ C( Ω), we consider the Dirichlet problem for the complex Hessian equation: (4.1.1)

where SH m (Ω) denotes the set of all m-subharmonic functions in Ω and β := dd c |z| 2 is the standard Kähler form in C n .

In the case m = 1, this equation corresponds to the Poisson equation which is classical (see [START_REF] Gilbarg | Elliptic partial differential equations of second order[END_REF]). The case m = n corresponds to the complex Monge-Ampère equation which was intensively studied these last decades by several authors (see [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF], [START_REF] Cegrell | The Dirichlet problem for the complex Monge-Ampère operator: Stability in L 2[END_REF], [START_REF] Cegrell | The Dirichlet problem for the complex Monge-Ampère operator: Perron classes and rotation invariant measures[END_REF], [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF]).

The complex Hessian equation is a new subject and is much more difficult to handle than the complex Monge-Ampère equation (e.g. the m-subharmonic functions are not invariant under holomorphic change of variables, for m < n). Despite these difficulties, the pluripotential theory developed in ([BT82] , [START_REF] Demailly | Potential theory in several complex variables[END_REF], [START_REF] Ko Lodziej | The complex Monge-Ampère equation[END_REF]) for the complex Monge-Ampère equation, can be adapted to the complex Hessian equation.

The Dirichlet problem (4.1.1) was considered by S.Y. Li in [START_REF] Li | On the Dirichlet problems for symmetric function equations of the eigenvalues of the complex Hessian[END_REF]. He proved that if Ω is a bounded strongly m-pseudoconvex domain with smooth boundary (see the definition below), ϕ ∈ C ∞ (∂Ω) and 0 < f ∈ C ∞ ( Ω) then there exists a unique smooth solution to (4.1.1). The existence of continuous solution for the homogeneous Dirichlet problem in the unit ball was proved by Z. B locki [B l05]. Recently, S. Dinew and S. Ko lodziej proved in [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF] that there exists a unique continuous solution to (4.1.1) when 0

Proof. First observe that if u ∈ C 2 (Ω), then by Lemma 1.3.2 we can see that (1) is equivalent to

where α = dd c u is a real (1,1)-form belongs to Γm . The last inequality corresponds to

(1 ⇒ 2) We consider the standard regularization u ǫ of u by convolution with smoothing kernel. We then get

Since u ǫ is smooth, we infer by the observation above that

Letting ǫ → 0, by the convergence theorem for the Hessian operator under decreasing sequence, we get

(2 ⇒ 1) Fix x 0 ∈ Ω and q is a C 2 -function in a neighborhood V ⋐ Ω of x 0 such that u ≤ q in this neighborhood and u(x 0 ) = q(x 0 ). We will prove that

First step: we claim that dd c q x 0 ∈ Γm . If u is smooth, we note that x 0 is a local minimum point of qu, then dd c (qu) x 0 ≥ 0. Hence, we see that (dd c q) k ∧ β n-k ≥ 0 in x 0 , for 1 ≤ k ≤ m. This gives that dd c q x 0 ∈ Γm . If u is non-smooth, let u ǫ be the standard smooth regularization of u. Then u ǫ is m-sh, smooth and u ǫ ց u. Now let us fix δ > 0 and ǫ 0 > 0 such that the neighborhood of x 0 , V ⊂ Ω ǫ 0 . For each ǫ < ǫ 0 , let y ǫ be the maximum point of

where B is a small ball centered at x 0 . Then we have

Assume that y ǫ → y ∈ B and set x = x 0 . By passing to the limit in the last inequality, we derive that 0 ≤ u(y)q(y) -δ|yx 0 | 2 , but q ≥ u in V , then we can conclude that y = x 0 . Let us then define This result was proved by Nguyen in [N14] for the homogeneous case (f ≡ 0). H.C. Lu proved in [START_REF] Lu | Equations Hessiennes complexes[END_REF][START_REF] Lu | Viscosity solutions to complex Hessian equations[END_REF] the Hölder continuity of the solution U under the same assumption of Corollary 4.4.5 in a bounded strongly pseudoconvex domain. A similar result for m = n goes back to Bedford and Taylor in [START_REF] Bedford | The Dirichlet problem for a complex Monge-Ampère equation[END_REF] and the main idea of the proof depends on Walsh's method [START_REF] Walsh | Continuity of envelopes of plurisubharmonic functions[END_REF]. We now give an example to point out that there is a loss in the regularity up to the boundary and show that our result is optimal.

Example 4.4.6. Let ψ be a concave modulus of continuity on [0, 1] and

We can show that ϕ ∈ C(∂B) with modulus of continuity ω ϕ (t) ≤ Cψ(t), for some C > 0. We consider the following Dirichlet problem:

where 2 ≤ m ≤ n is an integer. Then by the comparison principle, U(z) := -ψ[ » (1 + Rez 1 )/2] is the unique solution to this problem. One can observe by a radial approach to the boundary point (-1, 0, ..., 0) that

Hölder continuous solutions for L p -densities

Preliminaries and known results

The existence of a weak solution to the complex Hessian equation in some bounded domain in C n was established in the work of Dinew and Ko lodziej [START_REF] Dinew | A priori estimates for complex Hessian equations[END_REF]. More precisely, let Ω ⋐ C n be a smoothly (m -1)-pseudoconvex domain, ϕ ∈ C(∂Ω) and 0 ≤ f ∈ L p (Ω) for some p > n/m. Then there exists

Recently, N.C. Nguyen in [N14] proved that the Hölder continuity of this solution under some technical conditions: the density f ∈ L p (Ω), p > n/m is bounded near the boundary ∂Ω or f ≤ C|ρ| -mν there and the boundary data ϕ belongs to C 1,1 (∂Ω).

Here we follow the approach proposed in [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF] for the complex Monge-Ampère equation. A crucial role in this approach is played by an a priori weak stability estimate of the solution. This approach has been adapted to the complex Hessian equation in [N14] and [START_REF] Lu | Equations Hessiennes complexes[END_REF]. Here is the precise statement.

In order to simplify the notation, we set from now on for r ≥ 1, (4.5.1)

where p > n/m, 1 ≤ m ≤ n and 1/p + 1/q = 1.

where γ r is as in (4.5.1). Then there exists a uniform positive constant

where (vu) + := max{vu, 0}.

The proof of this stability theorem is similar to the one for the complex Monge-Ampère equation (see Theorem 3.2.4).

The second result gives the Hölder continuity under some additional hypothesis.

Theorem 4.5.2. ([N14]). Let 0 ≤ f ∈ L p (Ω) for p > n/m, and ϕ ∈ C(∂Ω). Let U be the continuous solution to (4.1.1). Suppose that there exists v ∈ C 0,ν ( Ω) for 0

2) If the total mass of ∆U is finite in Ω then U ∈ C 0,α ( Ω) for any α < min{ν, 2γ 1 }, where γ r is defined by (4.5.1) for r ≥ 1. This result is analogue to that proved by Guedj, Ko lodziej and Zeriahi [START_REF] Guedj | Hölder continuous solutions to the complex Monge-Ampère equations[END_REF].

Construction of Hölder barriers

The remaining problem is to construct a Hölder continuous barrier with the right exponent which guarantees one of the conditions in Theorem 4.5.2.

Using the interplay between the real and complex Monge-Ampère measures, we will construct Hölder continuous m-subharmonic barriers for the Dirichlet problem (4.1.1) when f ∈ L p (Ω), p ≥ 2n/m. The following result gives the existence of a 1/2-Hölder continuous m-subharmonic barrier for the problem (4.1.1) when f ∈ L p (Ω), p ≥ 2n/m.