Erratum to “The Hölder continuous subsolution theorem for complex Hessian equations”

Erratum to “The Hölder continuous subsolution theorem for complex Hessian equations”

Tome 8 (2021), p. 779-789.

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ERRATUM TO

“THE HÖLDER CONTINUOUS SUBSOLUTION THEOREM FOR COMPLEX HESSIAN EQUATIONS”

by Amel Benali & Ahmed Zeriahi

Abstract. — The statement and the proof of Theorem B in [BZ20] are not complete, except when the boundary datumgvanishes identically. We give here the correct version of the state- ment as well as a complete proof.

Résumé(Erratum à « Le théorème des sous-solutions Hölder continues pour les équations hes- siennes complexes »)

L’énoncé et la preuve du théorème B dans [BZ20] ne sont pas complets, sauf dans le cas où la donnée au bordgs’annule identiquement. Nous donnons ici la version correcte de l’énoncé ainsi qu’une démonstration complète.

Contents

1. Introduction. . . 779

2. A new version of [BZ20, Lem. 4.2]. . . 781

3. Proof of Theorem B’. . . 786

References. . . 789

1. Introduction

LetΩbCⁿbe a bounded domain andman integer such that16m6n. We deno- te bySH_m(Ω) the set of m-subharmonic functions onΩ. The Dirichlet problem we are concerned with is the following: findU ∈ SHm(Ω)∩C⁰(Ω)such that

(1.1)

((dd^cU)^m∧β^n−m=µ onΩ,

U_|∂Ω=g on∂Ω,

where the boundary datumg∈C⁰(∂Ω)andµis a positive Borel measure onΩwith finite massµ(Ω)<+∞.

Mathematical subject classification (2020). — 31C45, 32U15, 32U40, 32W20, 35J96.

Keywords. — Complex Monge-Ampère equations, complex Hessian equations, Dirichlet problem, obstacle problems, maximal subextension, capacity.

The Hölder continuous subsolution problem asks basically if the Dirichlet prob- lem (1.1) admits a Hölder continuous solution when it admits a Hölder continuous subsolution, provided that the boundary datumg is Hölder continuous.

This qualitative statement is true and was essentially proved in [BZ20, Th. B] and is essentially correct when the boundary datum g belongs toC^1,1(∂Ω). However its proof in the general case relies on [BZ20, Lem. 4.2] whose proof is not complete (see Lemma 2.2 below).

Here is the correct version of Theorem B of [BZ20].

Theorem B’. — Let Ω b Cⁿ be a bounded strongly m-pseudoconvex domain and µ a positive Borel measure on Ω with finite mass. Assume that there exists ϕ∈ E_m⁰(Ω)∩ C^α(Ω)with 0< α <1 such that

(1.2) µ6(dd^cϕ)^m∧β^n−m, weakly onΩ, andϕ|∂Ω≡0.

Then for any boundary datum g Hölder continuous on ∂Ω, the Dirichlet problem (1.1) admits a unique solution U = U_g,µ which is Hölder continuous on Ω. More precisely,

(1) ifg∈ C^1,1(∂Ω), thenU ∈ C^α0(Ω)for any α⁰ such that 0< α⁰<2γ⁰(m, n, α)α^m/2^m,

where

(1.3) γ⁰(m, n, α) := mα

m(m+ 1)α+ (n−m)[(2−α)m+α],

(2) ifg∈ C^2α(∂Ω), then U ∈ C^α00(Ω) for anyα⁰⁰ such that 0< α⁰⁰< γ⁰⁰(m, n, α)α^m/2^m,

where

(1.4) γ⁰⁰(m, n, α) := α

m(m+ 1)α+ (n−m)[(2−α)m+α]·

The first statement of Theorem B’ is the same as that of [BZ20, Th. B] when the boundary datumg ∈ C^1,1(∂Ω), except for the exponent. In fact the one given in the statement of [BZ20, Th. B] is not correct due to an unfortunate misprint at the end of its proof in page 1005, where [BZ20, Prop. 2.20] was applied with a wrong exponent.

On the other hand, when g is only assumed to be Hölder continuous on ∂Ω, we had to find a new argument because in this case we do not know if the bounds given by (2.2) in Lemma 2.1 below still hold. This will be done in Lemma 2.2 below, but unfortunately this leads to a worse exponent.

Acknowledgements. — We thank the referee for his careful reading of this note and his useful suggestions that helped to improve the presentation of the mains results.

2. A new version of[BZ20, Lem. 4.2]

Here we will give a new version of [BZ20, Lem. 4.2] and its complete proof following essentially the same scheme. Before stating this new version, we need to prove the following result which provides the argument missing in the proof of [BZ20, Lem. 4.2].

Lemma2.1. — Let ΩbCⁿ be a bounded strongly m-pseudoconvex domain and k an integer such that06k6m−1.

(1) Assume that ψ ∈ SHm(Ω)∩L^∞(Ω), u, v ∈ SHm(Ω)∩L^∞(Ω) satisfy u6 v onΩ and for anyζ∈∂Ω,limz→ζ(u(z)−v(z)) = 0. Then

(2.1)

Z

Ω

dd^cv∧(dd^cψ)^k∧β^n−k−16 Z

Ω

dd^cu∧(dd^cψ)^k∧β^n−k−1.

(2) Assume thatg∈C^1,1(∂Ω). Then there exists a constantM⁰ =M⁰(m, n, g)>0 such that for anyψ∈ E_m⁰(Ω) andv∈ SHm(Ω, R)withv|∂Ω=g, we have

(2.2)

Z

Ω

dd^cv∧(dd^cψ)^k∧β^n−k−16(R+M⁰)H_m(ψ)^k/m.

The first statement is well known and was stated in [BZ20, Cor. 2.9]. Since its proof there was not complete, we will give a different proof here.

Proof

(1) Fixε >0. From the hypothesis, the exists a compact subsetKbΩsuch that u>v−εonΩrK. Thenvε := max{u, v−ε} ∈ SHm(Ω)∩L^∞(Ω)and vε=uon ΩrK. We claim that

(2.3)

Z

Ω

dd^cv_ε∧(dd^cψ)^k∧β^n−k−1= Z

Ω

dd^cu∧(dd^cψ)^k∧β^n−k−1.

Indeed we have in the sense of currents onΩ,

dd^cvε∧(dd^cψ)^k∧β^n−k−1−dd^cu∧(dd^cψ)^k∧β^n−k−1=dd^cT, whereT := (v_ε−u)(dd^cψ)^k∧β^n−k−1.

Since T is a current of order 0 with compact support in Ω, it follows that R

Ωdd^cT = 0, which proves (2.3). Now observe that(vε)increases tov as εdecreases to0. Therefore, taking the limit in (2.3) asε→0 and using the monotone continuity of the Hessian operators, we obtain the inequality (2.1).

(2) Let us first assume thatg|∂Ω≡0. Thenv|∂Ω≡0and we can apply Cegrell type inequalities [BZ20, Lem. 2.8] and use the normalization mass condition, to conclude that

Ik(v, ψ)6Hm(v)^1/mHm(ψ)^k/m6R Hm(ψ)^k/m. This proves the inequality (2.2).

Now assume that g ∈ C^1,1(∂Ω) and fix k such that 1 6 k 6 m. There exists G∈C^1,1(Ω)such thatG=gon∂Ω. SinceΩis stronglym-pseudoconvex, it admits a defining functionρwhich is stronglym-subharmonic onΩ. Then we can find a large constant L >0 such thatw:=Lρ+Gism-subharmonic on Ω and for 16k6m,

(dd^cw)^k∧β^n−k 6L⁰_mβⁿ pointwise almost everywhere inΩfor some uniform constant L⁰_m>0.

By the bounded subsolution theorem ([Ngu12]) there exists v₀ ∈ E_m⁰(Ω) solution to the equation(dd^cv₀)^m∧β^n−m= (dd^cv)^m∧β^n−m with boundary values v₀ ≡0.

The functionve:=v0+w ism-subharmonic and bounded on Ω, (dd^cev)^m∧β^n−m>

(dd^cv)^m∧β^n−monΩandev=g=von∂Ω. By the comparison principle we getve6v onΩ. Therefore, by the formula (2.1) we conclude that

Ik(v, ψ)6Ik(ev, ψ).

It suffices to estimateIk(ev, ψ)by a uniform constant. We have Ik(ev, ψ) =Ik(v0, ψ) +Ik(w, ψ).

Sincev0|∂Ω≡0, from the previous case it follows that Ik(v0, ψ)6

Z

Ω

(dd^cv0)^m∧β^n−m

1/mZ

Ω

(dd^cψ)^m∧β^n−m k/m

6Hm(ψ)^k/mR.

It remains to estimate Ik(w, ψ). Since w ∈C^1,1(Ω), it follows that dd^cw 6M3β pointwise almost everywhere onΩ, hence by [BZ20, Lem. 2.8] , we have

Z

Ω

dd^cw∧(dd^cψ)^k∧β^n−k−16M⁰ Z

Ω

(dd^cψ)^k∧β^n−k

6M⁰Hm(ψ)^k/m6M⁰Hm(ψ)^k/m,

whereM⁰ =M⁰(g)>0 depends on the uniform bound ofdd^cG.

To state the main lemma which will replace [BZ20, Lem. 4.2], let us fix some nota- tions.

FixR>1and denote bySHm(Ω, R)) the set of bounded negativem-subharmonic functionswonΩsuch thatR

Ω(dd^cw)^m∧β^n−m6R^m. On the other hand, we denote byE_m⁰(Ω, R) =SHm(Ω, R)∩ E_m⁰(Ω).

Lemma2.2. — Let ΩbCⁿ be a bounded strongly m-pseudoconvex domain and ϕ∈ E_m⁰(Ω)∩ C^α(Ω), with 0< α61. Then for everyk such that 16k6m, there exists a constant Cek =C(k, m, ϕ,e Ω)>0 such that for every u, v ∈ SHm(Ω, R) such that u=v on∂Ω, we have

(2.4) Z

Ω

|u−v|(dd^cϕ)^k∧β^n−k6Cek(1 +ku−vk^m−1_∞ )R[ku−vk1]^αek, provided that ku−vk161, whereαek:=α^k/m2^k.

Furthermore ifg∈C^1,1(∂Ω), for anyksuch that16k6m, there exists a constant C_k =C(k, m, ϕ, g,Ω)>0 such that for every u, v ∈ SH_m(Ω, R) with u=g =v on

∂Ω, we have

(2.5)

Z

Ω

|u−v|(dd^cϕ)^k∧β^n−k6C_kR[ku−vk₁]^αk, provided that ku−vk1:=R

Ω|u−v|βⁿ 61, whereαk:= 2(α/2)^k.

The proof follows the same scheme as that of [BZ20, Lem. 4.2]. More precisely, we will first start by assumingg|∂Ω≡0, since the proof is exactly the same in this case.

Then we will consider the caseg ∈C^1,1(∂Ω), which can be reduced to the previous case using Lemma 2.1. Finally we will treat the general case by emphasizing the new ingredient to be used to fill the gap in the proof of [BZ20, Lem. 4.2].

Proof. — Let us first recall the main steps of the proof of [BZ20, Lem. 4.2]. We can reduce to the case whenu=vnear the boundary ∂Ωandu>vonΩ.

We approximate ϕ by smooth functions. As in [BZ20], we extend ϕas a Hölder continuous function of orderαonCⁿ and denote byϕ_δ(0< δ < δ₀) the usual smooth approximants ofϕonCⁿ. We know thatϕδ ∈ SHm(Ωδ)∩ C^∞(Cⁿ).

To prove the required estimates, we will argue by induction onksuch that06k6m.

Fix06k6m−1andδ₀>0small enough and write forδsuch that0< δ < δ₀, Z

Ω

(u−v)(dd^cϕ)^k+1∧β^n−k−1=A(δ) +B(δ),

where

A(δ) :=

Z

Ω

(u−v)dd^cϕδ∧(dd^cϕ)^k∧β^n−k−1,

and

B(δ) :=

Z

Ω

(u−v)dd^c(ϕ−ϕ_δ−κδ^α)∧(dd^cϕ)^k∧β^n−k−1,

where we recall thatϕ−κδ^α6ϕδ6ϕ+κδ^αonΩby hypothesis.

The first termA(δ)is estimated as follows. Observe that we have dd^cϕδ 6M1κδ^α−2β

pointwise on Ω, where M1 > 0 is a uniform bound on the second derivatives of χ.

Then sinceu>v we deduce that (2.6) |A(δ)|6M₁κδ^α−2

Z

Ω

(u−v)(dd^cϕ)^k∧β^n−k.

We now estimate the second termB(δ)following the arguments of [BZ20, Lem. 4.2]

with a slight modification.

Since u−v = 0near the boundary ∂Ω, i.e., onΩrΩ⁰, whereΩ⁰ bΩis an open set, we can integrate by parts to get the following formula

B(δ) = Z

Ω⁰

(ϕ_δ−ϕ+κδ^α)dd^c(v−u)∧(dd^cϕ)^k∧β^n−k−1,

and then, since06ϕδ−ϕ+κδ^α62κδ^αonΩ, it follows that

|B(δ)|62κδ^α Z

Ω⁰

dd^cv∧(dd^cϕ)^k∧β^n−k−1.

Therefore, we get

(2.7) |B(δ)|62κδ^αI_k⁰(v, ϕ),

where I_k⁰(v, ϕ) := R

Ω⁰dd^cv ∧(dd^cϕ)^k ∧β^n−k−1 < +∞ by Chern-Levine-Nirenberg inequalities (see [Lu12]).

The problem is to estimate the massesI_k⁰(v, ϕ)by a uniform constant which does not depend onΩ⁰bΩ. We could use the obvious inequality

Z

Ω⁰

dd^cv∧(dd^cϕ)^k∧β^n−k−16 Z

Ω

dd^cv∧(dd^cϕ)^k∧β^n−k−1,

to conclude under the finiteness full mass condition Ik(v, ϕ) :=

Z

Ω

dd^cv∧(dd^cϕ)^k∧β^n−k−1<+∞.

This is precisely the statement of Lemma 2.1 wheng∈C^1,1(∂Ω), whose justification was missing in the proof of [BZ20, Lem. 4.2].

So let us proceed to the proof the second statement of the lemma assuming that g∈C^1,1(∂Ω). By (2.7) and the formula (2.2) of Lemma 2.1 applied tov andψ=ϕ, we get

(2.8) |B(δ)|6κ d(m, n)R δ^α,

whered(m, n) =d(m, n, ϕ, g)>0 is a uniform constant.

For simplicity setσk(ϕ) := (dd^cϕ)^k∧β^n−k.Combining the inequalities (2.6), (2.7) and (2.8), we obtain forδ such that0< δ < δ0,

(2.9)

Z

Ω

(u−v)σk+1(ϕ)6M1

κδ^α δ²

Z

Ω

(u−v)σk(ϕ) +κ d(m, n)R δ^α.

To finish the proof of the last statement of the lemma, we argue by induction onk for k such that 0 6 k 6m. When k = 0, the inequality is obviously satisfied with C₀= 1andα₀= 1.

Assume that the inequality holds for some integer06k6m−1, i.e., (2.10)

Z

Ω

(u−v)σk(ϕ)6CkR[ku−vk1]^αk. We will show that there existsC_k+1>0such that

Z

Ω

(u−v)σk+1(ϕ)6Ck+1R[ku−vk1]^αk+1. Indeed (2.9) and (2.10) yields

Z

Ω

(u−v)σk+1(ϕ)6M1CkRκ δ^α−2[ku−vk1]^αk+d(m, n)κRδ^α.

We want to optimize the last estimate by a suitable choice ofδ. Sinceku−vk161, we can takeδ=δ0[ku−vk1]^αk/2< δ0in the last inequality to obtain

Z

Ω

(u−v)σ_k+1(ϕ)6(M₁C_k+d(m, n))κ R ku−vk₁]^αk/2α

6C_k+1R[ku−vk₁]^αk+1,

whereαk+1:=αk(α/2). This proves the last statement of the lemma.

We now proceed to the proof of the first statement. As we saw before the main issue is to estimate uniformly the integrals like I_k⁰(v, ϕ). We do not know whether

the integralsIk(v, ϕ)are finite. We will rather estimate the following integrals which behave much better:

J_k(u, v, ϕ) :=

Z

Ω

(u−v)^m(dd^cϕ)^k∧β^n−k.

By Hölder inequality, we have (2.11)

Z

Ω

(u−v)(dd^cϕ)^k∧β^n−k 6|Ω|^(m−1)/m(Jk(u, v, ϕ))^1/m, where|Ω|is the volume of Ω.

It is then enough to estimate J_k(u, v, ϕ). We will proceed by induction on k (06k6m) to prove the following estimate

(2.12) J_k(u, v)6C_k⁰R(1 +ku−vk^m−1_∞ )ku−vk^α₁^k, whereC_k⁰ =C_k⁰(k, m, nϕ, g)>0is a constant.

If k= 0, we have J₀(u, v)6ku−vk^m−1_∞ ku−vk1. Assume the estimate (2.12) is proved for some integer06k6m−1. To prove it for the integerk+ 1, we write as before forδsuch that0< δ < δ0,

J_k+1(u, v) =A⁰(δ) +B⁰(δ), where

A⁰(δ) :=

Z

Ω

(u−v)^mdd^cϕδ∧(dd^cϕ)^k∧β^n−k,

and

B⁰(δ) :=

Z

Ω

(u−v)^mdd^c(ϕ−ϕδ−κδ^α)∧(dd^cϕ)^k∧β^n−k.

The first termA⁰(δ)is estimated as before using (2.12), i.e.,

(2.13) |A⁰(δ)|6M1

κδ^α

δ² Jk(u, v).

We need to estimate the second termB⁰(δ). Sinceu−v= 0near the boundary, we can integrate by parts to get the following formula

(2.14) B⁰(δ) = Z

Ω

(ϕδ−ϕ+κδ^α) (−dd^c[(u−v)^m])∧(dd^cϕ)^k∧β^n−k−1.

Ifm= 1, thenk= 0. Since dd^cu∧βⁿ⁻¹>0weakly on Ω, it follows that

−dd^c[(u−v)∧βⁿ⁻¹6dd^cv∧βⁿ⁻¹,

weakly onΩ. Since06ϕ_δ−ϕ+κδ^α62κδ^α, it follows from (2.14) that

|B⁰(δ)|62κδ^α Z

Ω

dd^cv∧βⁿ⁻¹62R κ δ^α. Ifm>2, a simple computation shows that

−dd^c[(u−v)^m] = −m(u−v)^m−1dd^c(u−v)

−m(m−1)(u−v)^m−2d(u−v)∧d^c(u−v) 6 −m(u−v)^m−1dd^c(u−v),

Sincedd^cu∧(dd^cϕ)^k∧β^n−k−1>0 weakly onΩ, it follows that

−dd^c[(u−v)^m]∧(dd^cϕ)^k∧β^n−k−16m(u−v)^m−1dd^cv∧(dd^cϕ)^k∧β^n−k−1, weakly onΩ. Since06ϕ_δ−ϕ+κδ^α62κδ^α, it follows from (2.14) that

(2.15) |B⁰(δ)|62m κ δ^α Z

Ω

(u−v)^m−1dd^cv∧(dd^cϕ)^k∧β^n−k−1.

Recall that β =dd^cψ0, where ψ0(z) :=|z|²−r²₀, where r0 >0 is chosen so that ψ060 onΩ. Then the inequality (2.15) implies that

(2.16) |B⁰(δ)|62mκδ^α Z

Ω

(u−v)^m−1dd^cv∧(dd^cϕ)^k∧(dd^cψ)^m−k−1∧β^n−m.

Sincev, ϕ, ψ₀60 onΩ, repeating the integration by parts(m−1)times, we deduce from (2.16) that

(2.17) |B⁰(δ)|62m!κ δ^αkϕk^k_∞kψ0k^m−k−1_∞ Z

Ω

(dd^cv)^m∧β^n−m.

Combining the inequalities (2.13) and (2.17), we obtain forδsuch that0< δ < δ0, Z

Ω

(u−v)^mσ_k+1(ϕ)6M₁κδ^α−2J_k(u, v) +d⁰(m, n)Rδ^α,

where d⁰(m, n) = d⁰(m, n, ϕ, g) > 0 is a uniform constant. Applying the induction hypothesis (2.12), we get

Z

Ω

(u−v)^mσk+1(ϕ)6κM1Ckδ^α−2(1 +ku−vk^m−1_∞ )ku−vk^α₁^k+d⁰(m, n)Rδ^α. We want to optimize the last estimate. Sinceku−vk161, we can take

δ=δ0[ku−vk1]^αk/2< δ0

in the last inequality to obtain Z

Ω

(u−v)^mσk+1(ϕ)6 κM1Ck(1 +ku−vk^m−1_∞ ) +Rd⁰(m, n)

ku−vk^α₁^k/2α

6Ck+1(1 +ku−vk^m−1_∞ )Rku−vk^α₁^k+1,

where αk+1 := αk(α/2) and and Ck+1 := κM1δ^α−2₀ Ck+d⁰(m, n). This proves the estimate (2.12) for k+ 1. Taking into account the inequality (2.11) we obtain the estimate of the lemma with appropriate constants. This finishes the proof of the

second part of the lemma.

3. Proof of TheoremB’

We are now ready to prove Theorem B’ using Lemma 2.2.

Proof. — The strategy of the proof is the same as that of [BZ20, Th. B], but for convenience of the reader we will recall the main steps.

We know that the Dirichlet problem (1.1) admits a unique bounded solutionu∈ SHm(Ω)∩L^∞(Ω) by [Ngu12].

To prove thatuis Hölder continuous with some exponent 0< θ <1, it is enough to consider the standard regularizationuδ :=u ? χδ, for δsuch that 0< δ < δ0 and to prove that there exists a constantL₁>0and0< δ₁< δ₀ such that for anyδsuch that0< δ < δ₁, and any z∈Ω_δ, we have

(3.1) sup

Ωδ

(uδ−u)6Cδ^θ,

provided we prove that u is Hölder continuous near the boundary with the same exponentθ. We refer the reader to [Zer20] for a complete proof of this fact.

Recall that uδ can be defined forδ such that0 < δ 6δ0 (whereδ0 > 0 is small enough) by the following formula : forz∈Cⁿ,

uδ(z) = Z

Ω

u(ζ)χδ(z−ζ)dλ2n(ζ),

where λ_2n is the Lebesgue measure on Ω. Observe that u_δ is smooth on Cⁿ and m-subharmonic onΩδ.

The first step is to use the Hölder continuity of the boundary datumgto construct global barriers to show thatuis Hölder continuous near the boundary and to deduce boundedm-subharmonic global approximants(ue_δ)0<δ<δ0 ofuonΩwhich satisfy the following inequalities:

(3.2) 06euδ(z)−u(z)6uδ(z)−u(z)6ueδ(z)−u(z) +κδ^α, z∈Ωδ,

andeuδ >uonΩandeuδ =uonΩrΩδ forδsuch that0< δ < δ0. We refer to [BZ20]

for this construction.

Sinceeuδ =unear the boundary∂Ω, it follows from Stokes formula that forδsuch that0< δ < δ0, we have

Z

Ω

(dd^cueδ)^m∧β^n−m= Z

Ω

(dd^cu)^m∧β^n−m6µ(Ω)<∞.

The second step is to use stability estimates. Since euδ = u on ΩrΩδ and µ 6 (dd^cϕ)^m∧β^n−m onΩ, we can apply [BZ20, Prop. 2.20] with the exponentτ given in [BZ20, Th. A], to deduce that for anyγsuch that

0< γ < γ⁰(m, n, α) := mα

m(m+ 1)α+ (n−m)[(2−α)m+α], there exists a constantD_γ>0 such that any0< δ < δ₀,

(3.3) sup

Ω

(ueδ−u)6Dγ

Z

Ω

(ueδ−u)dµ γ

.

To prove (3.1), we need to estimate R

Ω(ueδ − u)dµ in terms of the integral R

Ω(ue_δ−u)dλ_2n, whereλ_2n is the Lebesgue measure.

Sinceµ6(dd^cϕ)^m∧β^n−m, this is possible by applying Lemma 2.2 if we can ensure that k(ueδ−uk1:=R

Ω(euδ−u)dλ2n 61 forδ >0 small enough. This is clearly true sinceeuδ decreases touasδdecreases to0, but we will need a quantitative estimate.

Indeed by [GKZ08] (see also [BZ20, Lem. 2.3]), we know that (3.4)

Z

Ω

(uδ−u)dλ2n6Bk∆ukΩ_δδ².

To complete the proof of our theorem, it suffices to estimate the massk∆ukΩ_δ :=

R

Ωδdd^cu∧βⁿ⁻¹in terms ofδ. We will consider the two cases separately.

(1) Assume first thatg∈C^1,1(∂Ω). By the inequality (2.2) of Lemma 2.1, we have forδsuch that0< δ < δ0,

k∆ukΩ_δ 6 Z

Ω

dd^cu∧βⁿ⁻¹6d(m, n) M⁰+µ(Ω)^1/m

<+∞.

Hence by (3.4) it follows that kuδ−uk1=

Z

Ω

(u_δ−u)dλ_2n 6C_m⁰ δ²61,

for δsuch that 0< δ 6δ₁ with0 < δ₁ < δ₀ and small enough. Therefore, applying the inequality (2.5) of Lemma 2.2, we get forδ such that0< δ < δ1,

Z

Ω

(euδ−u)dµ6Cm

Z

Ω

(euδ−u)(z)dλ2n(z) ^αm

6Cm

Z

Ω_δ

(uδ(z)−u(z)dλ2n(z) αm

,

where the last inequality follows from (3.2). Using (3.4) we deduce that for δ such that0< δ < δ1,

(3.5)

Z

Ω

(ueδ−u)dµ6C_m⁰ δ^2αm.

It follows from (3.2), (3.3) and (3.5) that forδ such that0< δ < δ₁, sup

Ω_δ

(u_δ−u)6sup

Ω

(ue_δ−u) +κδ^α

6C_m⁰⁰δ^2γαm+κδ^α. This proves the required estimate (3.1) withθ= 2γαm< α.

(2) In the general case we need to estimate k∆uk_Ωδ in a different way. Since the defining functionρofΩis smooth and|∇ρ|>0on∂Ω, it follows from Hopf’s lemma that there exists a uniform constant ec1 > 0 such that −ρ(z) > ec1dist(z, ∂Ω) (see [Zer20] for more details). Then

Z

Ω_δ

dd^cu∧βⁿ⁻¹6ec₂δ⁻¹ Z

Ω

(−ρ)dd^cu∧βⁿ⁻¹.

We can assume that u6 0 on Ω. Then by the integration by parts inequality (see [BZ20, (2.5)]), it follows that

(3.6)

Z

Ω_δ

dd^cu∧βⁿ⁻¹6ec₂δ⁻¹ Z

Ω

(−u)dd^cρ∧βⁿ⁻¹6ec₃δ⁻¹osc_Ωu,

whereec2,ec3>0 are uniform constants. Hence by (3.4) and (3.6) we get Z

Ω

(uδ−u)dλ2n6Ce_m⁰ δ61, forδsuch that0< δ6δ2 with0< δ2< δ1 and small enough.

Therefore, applying the inequality (2.4) of Lemma 2.2, we get for δ such that 0< δ < δ2,

Z

Ω

(eu_δ−u)dµ6Ce_m⁰ Z

Ω

(eu_δ−u)dλ_2n(z) αe_m

6Ce_m⁰ Z

Ωδ

(uδ−u)dλ2n

αe_m

,

where the last inequality follows from (3.2). This implies that forδsuch that0< δ < δ₂, (3.7)

Z

Ω

(ueδ−u)dµ6Ce_m⁰⁰ δ^αem.

Therefore, taking into account (3.3) and (3.7), we get forδsuch that0< δ < δ2, sup

Ω_δ

(uδ−u)6Cemδ^γαm/m+κδ^α,

sincekue_δ−uk_∞6osc_Ωu. This proves the required estimate (3.1) withθ=γα_m/m < α.

References

[BZ20] A. Benali&A. Zeriahi– “The Hölder continuous subsolution theorem for complex Hessian equations”,J. Éc. polytech. Math.7(2020), p. 981–1007.

[GKZ08] V. Guedj, S. Kolodziej&A. Zeriahi– “Hölder continuous solutions to Monge-Ampére equa- tions”,Bull. London Math. Soc.40(2008), no. 6, p. 1070–1080.

[Lu12] C. H. Lu– “Équations hessiennes complexes”, PhD Thesis, Université de Toulouse 3, 2012, theses.fr:2012TOU30154.

[Ngu12] N. C. Nguyen– “Subsolution theorem for the complex Hessian equation”,Univ. Iagel. Acta Math.50(2012), p. 69–88.

[Zer20] A. Zeriahi – “Remarks on the modulus of continuity of subharmonic functions”, 2020, arXiv:2007.08399.

Manuscript received 16th October 2020 accepted 8th March 2021

Amel Benali, University of Gabes, Faculty of Sciences of Gabes, LR17ES11, Mathematics and Applications

6072 Gabes, Tunisia

E-mail :amelmath.kabs@gmail.com

Ahmed Zeriahi, Institut de Mathématiques de Toulouse, Université de Toulouse, CNRS, UPS 118 route de Narbonne, 31062 Toulouse cedex 09, France

E-mail :ahmed.zeriahi@math.univ-toulouse.fr Url :https://www.math.univ-toulouse.fr/~zeriahi/