# Trace Inequalities, isocapacitary inequalities and regularity of the complex
Hessian equations

Jiaxiang $\text{Wang}^{1}$, Bin $\text{Zhou}^{2}$ Jiaxiang Wang: Beijing
International Center for Mathematical Research, Peking University, Beijing,
100871, P. R. China  wangjx<EMAIL_ADDRESS>Bin Zhou: School of Mathematical
Sciences, Peking University, Beijing 100871, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study the relations between trace inequalities(Sobolev and
Moser-Trudinger type), isocapacitary inequalities and the regularity of the
complex Hessian and Monge-Ampère equations with respect to a general positive
Borel measure. We obtain a quantitative characterization for these relations
through properties of the capacity minimizing functions.

###### Key words and phrases:

Complex Monge-Ampère equations, plurisubharmonic functions, Sobolev
inequality, Moser-Trudinger inequality.

$1$ Partially supported by China post-doctoral grant BX2021015.

$2$ Partially supported by National Key R$\&$D Program of China
SQ2020YFA0712800 and NSFC Grant 11822101.

##  1\. introduction

Sobolev and Moser-Trudinger type inequalities play an important role in both
PDE and geometry. On one hand, these inequalities are widely used in the study
of existence and regularity of solutions to partial differential equations. On
the other hand, people also discovered that they are equivalent to
isoperimetric and isocapacitary inequalities [Ma]. Despite the classical
Sobolev and Moser-Trudinger inequalities, the analogous inequalities for a
series of fully nonlinear equations with variational structure have been
developed, including both real and complex Hessian equations [W1, TrW, TiW,
AC20]. In particular, the Moser-Trudinger type inequality for the complex
Monge-Ampère equations has been established [BB, C19, GKY, AC19, WWZ1]. The
Moser-Trudinger type inequality can also be related to the Skoda integrability
of plurisubharmonic functions [DNS, DN, Ka, DMV]. In this paper, we study the
relations between trace inequalities(Sobolev and Moser-Trudinger type),
isocapacitary inequalities and the regularity of the complex Hessian and
Monge-Ampère equations with respect to a general nonnegative Borel measure
$\mu$. Our results generalize the classical trace inequalities [AH]. Here
‘trace’ refers to that $\mu$ lives on a domain $\Omega$ and is a surface
measure on a smooth submanifold in $\Omega$. The trace and isocapacitary
inequalities for the real Hessian equations were obtained by [XZ].

Let $\Omega\subset{\mathbb{C}}^{n}$ be a pseudoconvex domain with smooth
boundary $\partial\Omega$ and $\omega$ be the Kähler form associated to the
standard Euclidean metric. Let $\mathcal{PSH}_{k}(\Omega)$ be the
$k$-plurisubharmonic functions on $\Omega$ and $\mathcal{PSH}_{k,0}(\Omega)$
be the set of functions in $\mathcal{PSH}_{k,0}(\Omega)$ with vanishing
boundary value. Let $\mathcal{F}_{k}(\Omega)$ be set of $k$-plurisubharmonic
functions which can be decreasingly approximated by functions in
$\mathcal{PSH}_{k,0}(\Omega)\cap C(\overline{\Omega})\cap C^{2}(\Omega)$
[C06]. For $u\in\mathcal{F}_{k}(\Omega)$, denote the $k$-Hessian energy by

$\mathcal{E}_{k}(u)=\int_{\Omega}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}$

and the norm by

$\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}:=\left(\int_{\Omega}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}\right)^{\frac{1}{k+1}}.$

In particular, when $k=n$, we write
$\|u\|_{\mathcal{PSH}_{0}(\Omega)}=\|u\|_{\mathcal{PSH}_{n,0}(\Omega)}$ for
simplicity.

Let $\mu$ be a nonnegative Borel measure with finite mass on
$\Omega\subset{\mathbb{C}}^{n}$. We will consider the trace inequalities and
isocapacitary inequalities with respect to $\mu$ as in the classical case
[AH]. We recall the capacity for plurisubharmonic functions. The relative
capacity for plurisubharmonic functions was introduced by Bedford-Taylor [BT,
B]. For a Borel subset $E\subset\Omega$, the $k$-capacity is defined as

$\text{Cap}_{k}(E,\Omega)=\sup\left\\{\left.\int_{E}(dd^{c}v)^{k}\wedge\omega^{n-k}\,\right|\,v\in\mathcal{PSH}_{k}(\Omega),-1\leq
v\leq 0\right\\}.$

Throughout the context, we will use $|\cdot|$ to denote the Lebesgue measure
of a Borel subset. We follow [Ma] to define the capacity minimizing function
with respect to $\mu$

$\nu_{k}(s,\Omega,\mu):=\inf\left\\{\left.\text{Cap}_{k}(K,\Omega)\,\right|\,\mu(K)\geq
s,K\Subset\Omega\right\\},\ \ \ 0<s<\mu(\Omega).$

Then we denote

(1.1) $\displaystyle
I_{k,p}(\Omega,\mu):=\begin{cases}\int_{0}^{\mu(\Omega)}\left(\frac{s}{\nu_{k}(s,\Omega,\mu)}\right)^{\frac{p}{k+1-p}}\,ds,&0<p<k+1,\\\\[5.0pt]
\displaystyle\sup_{t>0}\left\\{\frac{t}{\nu_{k}(t,\Omega,\mu)^{\frac{p}{k+1}}}\right\\},&p\geq
k+1,\end{cases}$

and

(1.2) $\displaystyle
I_{n}(\beta,\Omega,\mu):=\sup\left\\{s\exp\left(\frac{\beta}{\nu_{n}(s,\Omega,\mu)^{\frac{q}{n+1}}}\right):\
0<s<\mu(\Omega)\right\\}.$

Note that if $\mu(P)>0$ for a $k$-pluripolar subset $P\subset\Omega$, then
$I_{k,p}(\Omega,\mu)$, $I_{n}(\beta,\Omega,\mu)=+\infty$. Therefore, we only
need to consider those measures which charge no mass on pluripolar subsets.
The main result of this paper is as follows.

######  Theorem 1.1.

Suppose $\Omega\subset{\mathbb{C}}^{n}$ is a smooth, $k$-pseudoconvex domain,
where $1\leq k\leq n$.

  1. (i)

The Sobolev type trace inequality

(1.3)
$\displaystyle\sup\left\\{\frac{\|u\|_{L^{p}(\Omega,\mu)}}{\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}}:u\in\mathcal{F}_{k}(\Omega),0<\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}<\infty\right\\}<+\infty$

holds if and only if $I_{k,p}(\Omega,\mu)<+\infty$. Moreover, the Sobolev
trace map

$Id:\,{\mathcal{F}}_{k}(\Omega)\hookrightarrow L^{p}(\Omega,\mu),\ 1<p<\infty$

is a compact embedding if and only if

(1.4) $\displaystyle\lim_{s\to
0}\frac{s}{\nu_{k}(s,\Omega,\mu)^{\frac{p}{k+1}}}\to 0.$

  2. (ii)

When $k=n$, for $q\in[1,\frac{n+1}{n}]$ and $\beta>0$, the Moser-Trudinger
type trace inequality

(1.5)
$\displaystyle\sup\left\\{\int_{\Omega}\exp\left(\beta\left(\frac{-u}{\|u\|_{\mathcal{PSH}_{0}(\Omega)}}\right)^{q}\right)\,d\mu:u\in\mathcal{F}_{n}(\Omega),0<\|u\|_{\mathcal{PSH}_{0}(\Omega)}<\infty\right\\}$

holds if and only if $I_{n}(\beta,\Omega,\mu)<+\infty$.

######  Remark 1.2.

(a) For $p\geq k+1$, $\beta>0$, the conditions
$I_{k,p}(\Omega,\mu),I_{n}(\beta,\Omega,\mu)<+\infty$ are equivalent to the
following isocapacitary type inequalities

(1.6) $\displaystyle\mu(K)\leq
I_{k,p}(\Omega,\mu)\cdot\text{Cap}_{k}(K,\Omega)^{\frac{p}{k+1}},$ (1.7)
$\displaystyle\mu(K)\leq
I_{n}(\beta,\Omega,\mu)\cdot\exp\left(-\frac{\beta}{\text{Cap}_{n}(K,\Omega)^{\frac{q}{n+1}}}\right)$

for $K\Subset\Omega$. By [K96, DK], it is known that when $\mu$ is Lebesgue
measure

(1.8) $\displaystyle|E|$ $\displaystyle\leq$ $\displaystyle
C_{\lambda,\Omega}\cdot\text{Cap}_{k}(E,\Omega)^{\lambda},\
\lambda<\frac{n}{n-k},\ 1\leq k\leq n-1,$ (1.9) $\displaystyle|E|$
$\displaystyle\leq$ $\displaystyle
C_{\beta,n,\Omega}\cdot\exp\left(-\frac{\beta}{\text{Cap}_{n}(E,\Omega)^{\frac{1}{n}}}\right),\
0<\beta<2n.$

It is still open whether $\beta$ can attain $2n$. By a result in [BB], the
conclusion is true for subsets $E\Subset\Omega$ with
${\mathbb{S}}^{1}$-symmetry(invariant under the rotation
$e^{{\sqrt{-1}}\theta}z$ for all $\theta\in{\mathbb{R}}^{1}$), when
$\Omega\subset{\mathbb{C}}^{n}$ is a ball centered at the origin. See Remark
4.4 for more explanations.

(b) By the arguments of [BB, Section 5], (1.5) is equivalent to

$\displaystyle\sup\left\\{\int_{\Omega}\exp\left(k(-u)-\frac{n^{n}k^{1+n}{\mathcal{E}}_{n}(u)}{(n+1)^{1+n}\beta^{n}}\right)^{q}\,d\mu:\forall
k>0,u\in\mathcal{F}_{n}(\Omega)\right\\}<+\infty.$

(c) We can also prove the quasi Moser-Trudinger type trace inequality is
equivalent to the quasi Brezis-Merle type trace inequality(see Theorem 4.5).
Note that the proof for the case $\mu$ is Lebesgue measure in [BB] used
thermodynamical formalism and a dimension induction argument. Our proof here
uses the isocapacitary inequality (1.7).

######  Example 1.3.

Let $\mu$ be the measure with singularities of Poincarè type, i.e.,

$d\mu=\frac{1}{\prod_{j=1}^{d}|z^{j}|^{2}\left(1-\log|z^{j}|\right)^{1+\alpha}}\,dz^{1}\wedge
d\bar{z}^{1}\wedge\cdots\wedge dz^{n}\wedge d\bar{z}^{n},\ \ \ \text{where
}d\leq n,\ \alpha>0.$

nn the unit disk $\mathbb{D}^{n}\subset\mathbb{C}^{n}$, According to [DL,
Lemma 4.1], we have

$\mu(K)\leq C\cdot{\text{Cap}_{n}(K,\mathbb{D}^{n})}^{\alpha}.$

By Theorem 1.1, a Sobolev type inequality with respect to $d\mu$ holds for
plurisubharmonic functions.

Now we turn to the corresponding equations. Consider the Dirichlet problem

(1.10) $\begin{cases}(dd^{c}u)^{k}\wedge\omega^{n-k}=d\mu\ \ &\text{in\
$\Omega$,}\\\ u=\varphi,&\text{on }\partial\Omega\end{cases}$

for a nonnegative Borel measure $\mu$. In a seminal work [K98], Kołodziej
obtained the $L^{\infty}$-estimate and existence of continuous solutions for
the complex Monge-Ampère equation when $d\mu$ is dominated by a suitable
function of capacity, especially for $d\mu\in L^{p}(\Omega)$. Furthermore, the
solution is shown to be Hölder continuous under certain assumptions on
$\Omega$ and $\varphi$ [GKZ]. Generalizations to the complex Hessian equations
were made by [DK, Ngu]. These results were established by pluripotential
theory. In [WWZ2], the authors present a new PDE proof for the complex Monge-
Ampère equation with $d\mu\in L^{p}(\Omega)$ based on the Moser-Trudinger type
inequality. By Theorem 1.1, we have

######  Theorem 1.4.

Let $\mu$ be a non-pluripolar, nonnegative Radon measure with finite mass.
Then the following statements are equivalent:

  1. (i)

There exist $0<\delta<\frac{1}{k}$ and a constant $C>0$ depending on $\mu$ and
$\Omega$ such that for any Borel subset $E\subset\Omega$, the Dirichlet
problem

(1.11)
$\displaystyle\begin{cases}(dd^{c}u)^{k}\wedge\omega^{n-k}=\chi_{E}\,d\mu,\ \
\ &\text{in }\Omega,\\\ u=0,&\text{on }\partial\Omega,\end{cases}$

admits a continuous solution $u_{E}\in\mathcal{PSH}_{k,0}(\Omega)$ such that

(1.12) $\displaystyle\|u_{E}\|_{L^{\infty}(\Omega,\mu)}\leq C\mu(E)^{\delta}.$

Here $\chi_{E}$ is the characteristic function of $E$.

  2. (ii)

There exists $p\geq k+1$ such that $I_{k,p}(\Omega,\mu)<+\infty$.

More precisely, $\delta$ and $p$ can be determined mutually by
$p=\frac{k+1}{1-k\delta}$.

######  Remark 1.5.

(1) The conclusion from (ii) to (i) in the above theorem also holds with
general continuous boundary value.

(2) When $d\mu$ is an integrable function, once we have the
$L^{\infty}$-estimate for complex Monge-Ampère equation, the
$L^{\infty}$-estimate of many other equations, including the complex Hessian
equations and $p$-Monge-Ampère operations [HL09], can be derived by a simple
comparison. In real case, this is indicated by [W2]. For example, we consider
the complex $k$-Hessian equation

(1.13) $\begin{cases}(dd^{c}u)^{k}\wedge\omega^{n-k}=f\,\omega^{n},\ \ \
&\text{in }\Omega,\\\ u=\varphi,&\text{on }\partial\Omega.\end{cases}$

Suppose $d\mu=f\,\omega^{n}$ with $f\in L^{\frac{n}{k}}(\log
L^{1})^{n+\varepsilon}$. Let $v$ be the solution to the complex Monge-Amperè
equation

$\begin{cases}(dd^{c}v)^{n}=f^{\frac{n}{k}}\,\omega^{n},\ \ \ &\text{in
}\Omega,\\\ v=\varphi,&\text{on }\partial\Omega.\end{cases}$

By elementary inequalities, $v$ is a subsolution to (1.13). Then
$\|u\|_{L^{\infty}(\Omega)}\leq\|v\|_{L^{\infty}(\Omega)}\leq C$. However,
when $\mu$ is a general measure, this comparison does not work.

It is also interesting to ask when the solution is Hölder continuous. In
[DKN], Dinh, Kołodziej and Nguyen introduced a new condition on $d\mu$ and
proved that it is equivalent to the Hölder continuity for the complex Monge-
Ampère equation. We give a pure PDE proof as well as for the complex Hessian
equations, based on the Sobolev type inequality for complex Hessian operators
and the arguments in [WWZ2].

As in [DKN], we denote by $W^{*}(\Omega)$ the set of functions $f\in
W^{1,2}(\Omega)$ such that

$df\wedge d^{c}f\leq T$

for some closed positive $(1,1)$-current $T$ of finite mass on $\Omega$.
Define a Banach norm by

$\|f\|_{*}:=\|f\|_{L^{1}(\Omega)}+\min\left\\{\|T\|_{\Omega}^{\frac{1}{2}}\,\big{|}\,T\text{
as above}\right\\},$

where the mass of $T$ is defined by
$\|T\|_{\Omega}:=\int_{\Omega}T\wedge\omega^{n-1}$.

######  Theorem 1.6.

Suppose $1\leq k\leq n$, $\Omega$ is a $k$-pseudoconvex domain with smooth
boundary. Let $\mu$ be a Radon measure with finite mass. Let
$\gamma\in\left(\frac{(n-k)(k+1)}{2nk+n+k},k+1\right]$. The following
statements are equivalent:

  1. (i)

The Dirichlet problem (1.10) admits a solution $u\in
C^{0,\gamma^{\prime}}(\Omega)$ with

$0<\gamma^{\prime}<\frac{(2nk+n+k)\gamma-(n-k)(k+1)}{(n+1)k\gamma+(n+1)k^{2}+nk+k}.$

  2. (ii)

There exists $C>0$ such that for every smooth function $f\in W^{*}(\Omega)$
with $\|f\|_{*}\leq 1$,

(1.14) $\displaystyle\mu(f):=\int_{\Omega}f\,d\mu\leq
C\|f\|_{L^{1}(\Omega)}^{\gamma}.$

The structure of the paper is as follows: Section 2 is devoted to a review on
the relative capacity for $k$-plurisubharmonic functions. In particular, we
obtain several equivalent definitions for the capacity. Section 3, we
establish the capacitary estimates for level sets of $k$-plurisubharmonic
functions, which is the main tool in the proof of Theorem 1.1. Theorem 1.1
will be proved case by case in Section 4. In the last section, we apply
Theorem 1.1 to the Dirichlet problem for the complex Hessian equations to
prove Theorem 1.4 and 1.6.

##  2\. On relatively capacities and relatively extremal functions

In this section, we recall the relative capacity for plurisubharmonic
functions [BT, B].

For a Borel subset $E\subset\Omega$, the $k$-capacity is defined as

$\text{Cap}_{k}(E,\Omega)=\sup\left\\{\left.\int_{E}(dd^{c}v)^{k}\wedge\omega^{n-k}\,\right|\,v\in\mathcal{PSH}_{k}(\Omega),-1\leq
v\leq 0\right\\}.$

It is well known that the $k$-capacity can be characterized by the relatively
$k$-extremal function $u_{k,E,\Omega}^{*}$, which is the upper regularization
of

$u_{k,E,\Omega}:=\sup\left\\{v\,\left|v\in\mathcal{PSH}_{k}(\Omega),v\leq-1\text{
on }E,v<0\text{ on }\Omega\right.\right\\}.$

We will usually write $u_{k,E}^{*}$ for simplicity if there is no confusion.
When $E=K$ is a compact subset, we have $-1\leq u_{k,K}^{*}\leq 0$, the
complex Hessian measure $(dd^{c}u_{K}^{*})^{k}\wedge\omega^{n-k}=0$ on
$\Omega\setminus K$. Moreover, $\text{Cap}_{k}(K,\Omega)=0$ if
$u_{k,K}^{*}>-1$ on $K$. The following well-known fact shows
$u_{k,K}^{*}\in\mathcal{PSH}_{k,0}(\Omega)$.

######  Lemma 2.1.

If $K\subset\Omega$ is a compact subset, we have
$u_{k,K}^{*}\big{|}_{\partial\Omega}=0$.

###### Proof.

By [KR, Proposition 1.2], there exists an exhaustion function $\psi\in
C^{\infty}(\Omega)\cap\mathcal{PSH}_{k,0}(\Omega)$. By the maximum principle
we have $-a:=\sup_{K}\psi<0$. Let $\hat{\psi}:=\frac{\psi}{a}\leq-\chi_{K}$,
then we get $\hat{\psi}\leq u_{k,K}^{*}$ on $\Omega$. By the fact
$\hat{\psi}(\xi)\to 0$ as $\xi\to z\in\partial\Omega$, we get the result. ∎

Suppose $\Omega$ is $k$-hyperconvex so that $\mathcal{PSH}_{k,0}(\Omega)$ is
non-empty. Inspired by [XZ] for the real Hessian equations, we consider
several capacities defined as follows.

######  Definition 2.2.

(i) Let $K$ be a compact subset of $\Omega$. Define

(2.1)
$\displaystyle\widetilde{\text{Cap}}_{k,1}(K,\Omega)=\sup\left\\{\left.\int_{K}(-v)(dd^{c}v)^{k}\wedge\omega^{n-k}\,\right|\,v\in\mathcal{PSH}_{k,0}(\Omega),-1\leq
v\leq 0\right\\},$ (2.2)
$\displaystyle\widetilde{\text{Cap}}_{k,2}(K,\Omega):=\inf\left\\{\left.\int_{\Omega}(dd^{c}v)^{k}\wedge\omega^{n-k}\,\right|\,v\in\mathcal{PSH}_{k,0}(\Omega),v\big{|}_{K}\leq-1\right\\},$
(2.3)
$\displaystyle\widetilde{\text{Cap}}_{k,3}(K,\Omega):=\inf\left\\{\left.\int_{\Omega}(-v)(dd^{c}v)^{k}\wedge\omega^{n-k}\,\right|\,v\in\mathcal{PSH}_{k,0}(\Omega),v\big{|}_{K}\leq-1\right\\}.$

(ii) For an open subset $O\subset\Omega$ and $j=1$, $2$, $3$, let

$\widetilde{\text{Cap}}_{k,j}(O,\Omega):=\sup\left\\{\left.\widetilde{\text{Cap}}_{k,j}(K,\Omega)\,\right|\,\text{compact
}K\subset O\right\\}.$

(iii) For a Borel subset $E\subset\Omega$ and $j=1$, $2$, $3$, let

$\widetilde{\text{Cap}}_{k,j}(E,\Omega):=\inf\left\\{\left.\widetilde{\text{Cap}}_{k,j}(O,\Omega)\,\right|\,\text{open
$O$ with }E\subset O\subset\Omega\right\\}.$

In order to show the equivalence of capacities defined above, we need the
following well-known comparison principle.

######  Lemma 2.3.

Suppose $\Omega$ is a $k$-hyperconvex domain with $C^{1}$-boundary. Let $u$,
$v\in{\mathcal{F}}_{k}(\Omega)$. If $\lim_{z\to\xi\in\partial\Omega}(u-v)\geq
0$, $u\leq v$ in $\Omega$, then there hold

$\int_{\Omega}(dd^{c}u)^{k}\wedge\omega^{n-k}\geq\int_{\Omega}(dd^{c}v)^{k}\wedge\omega^{n-k},\
\ \
\int_{\Omega}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}\geq\int_{\Omega}(-v)(dd^{c}v)^{k}\wedge\omega^{n-k}.$

###### Proof.

It is a direct consequence of the integration by parts and the smooth
approximation for functions in ${\mathcal{F}}_{k}(\Omega)$. ∎

######  Lemma 2.4.

Suppose $1\leq k\leq n$, and $K\subset\Omega$ is a compact subset. Then

(2.4)
$\displaystyle\widetilde{\text{Cap}}_{k,1}(K,\Omega)=\int_{K}(-u_{k,K}^{*})(dd^{c}u_{k,K}^{*})^{k}\wedge\omega^{n-k}.$

###### Proof.

First, by definition we have

$\widetilde{\text{Cap}}_{k,1}(K,\Omega)\geq\int_{K}(-u_{k,K}^{*})(dd^{c}u_{k,K}^{*})^{k}\wedge\omega^{n-k}.$

To reach the reversed inequality, we choose $\\{K_{j}\\}$ to be a sequence of
compact subsets of $\Omega$ with smooth boundaries $\partial K_{j}$ such that

$K_{j+1}\subset K_{j},\ \ \ \bigcap_{j=1}^{\infty}K_{j}=K.$

Using the smoothness of $\partial K_{j}$, the relatively extremal function
$u_{j}:=u_{k,K_{j}}^{*}=u_{k,K_{j}}\in C(\overline{\Omega})$. Note that
$u_{j}\uparrow v$ and $u_{k,K}^{*}=v^{*}$. Then for
$u\in\mathcal{PSH}_{k,0}(\Omega)$ such that $-1\leq u\leq 0$, we have $u\geq
u_{j}$ on $K_{j}$. Therefore, by Lemma 2.3,

$\displaystyle\int_{K}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}$ $\displaystyle\leq$
$\displaystyle\int_{\\{u_{j}\leq u\\}}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}$
$\displaystyle\leq$ $\displaystyle\int_{\\{u_{j}\leq
u\\}}(-u_{j})(dd^{c}u_{j})^{k}\wedge\omega^{n-k}$ $\displaystyle\leq$
$\displaystyle\int_{\Omega}(-u_{j})(dd^{c}u_{j})^{k}\wedge\omega^{n-k}=\int_{K_{j}}(-u_{j})(dd^{c}u_{j})^{k}\wedge\omega^{n-k}.$

Taking $j\to\infty$, we obtain

$\int_{K}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}\leq\int_{\Omega}(-u_{k,K}^{*})(dd^{c}u_{k,K}^{*})^{k}\wedge\omega^{n-k}=\int_{K}(-u_{k,K}^{*})(dd^{c}u_{k,K}^{*})^{k}\wedge\omega^{n-k}.$

This yields

$\widetilde{\text{Cap}}_{k,1}(K,\Omega)\leq\int_{K}(-u_{K}^{*})(dd^{c}u_{K}^{*})^{k}\wedge\omega^{n-k},$

thereby completing the proof. ∎

######  Lemma 2.5.

For any Borel set $E\subset\Omega$, we have

$\widetilde{\text{Cap}}_{k,j}(E,\Omega)=\text{Cap}_{k}(E,\Omega),\ \ \
j=1,2,3.$

###### Proof.

By definition, it suffices to prove the equalities when $E=K$ is a compact
subset of $\Omega$. Note that for $u$, $v\in\mathcal{PSH}_{k,0}(\Omega)$ such
that $-1\leq u\leq 0$, $v\big{|}_{K}\leq-1$, we have

$\int_{K}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}\leq\int_{K}(dd^{c}u)^{k}\wedge\omega^{n-k}\leq\int_{K}(dd^{c}u_{k,K}^{*})^{k}\wedge\omega^{n-k}\leq\int_{\Omega}(dd^{c}v)^{k}\wedge\omega^{n-k}.$

Hence we obtain

$\widetilde{\text{Cap}}_{k,1}(K,\Omega)\leq\text{Cap}_{k}(K,\Omega)\leq\widetilde{\text{Cap}}_{k,2}(K,\Omega).$

It suffices to prove

(2.5)
$\widetilde{\text{Cap}}_{k,2}(K,\Omega)\leq\widetilde{\text{Cap}}_{k,3}(K,\Omega)\leq\widetilde{\text{Cap}}_{k,1}(K,\Omega).$

We still choose $\\{K_{j}\\}$ to be a sequence of compact subsets in $\Omega$
with smooth boundaries $\partial K_{j}$ such that

$K_{j+1}\subset K_{j},\ \ \ \bigcap_{j=1}^{\infty}K_{j}=K.$

Then $u_{j}:=u_{k,K_{j}}^{*}=u_{k,K_{j}}\in C(\overline{\Omega})$,
$u_{j}\uparrow v$ with $v^{*}=u_{k,K}^{*}$, and
$u_{j}\big{|}_{K_{j}}\equiv-1$. For any $j$, we have

$\displaystyle\widetilde{\text{Cap}}_{k,2}(K,\Omega)$ $\displaystyle\leq$
$\displaystyle\int_{\Omega}(dd^{c}u_{j})^{k}\wedge\omega^{n-k}$
$\displaystyle=$
$\displaystyle\int_{\Omega}(-u_{j})(dd^{c}u_{j})^{k}\wedge\omega^{n-k}\leq\int_{\Omega}(-v_{j})(dd^{c}v_{j})^{k}\wedge\omega^{n-k},$

where $v_{j}$ is an arbitrary function in $\mathcal{PSH}_{k,0}(\Omega)$ such
that $v_{j}\big{|}_{K_{j}}\leq-1$. This implies
$\widetilde{\text{Cap}}_{k,2}(K,\Omega)\leq\widetilde{\text{Cap}}_{k,3}(K_{j},\Omega)$.
Then by Lemma 2.3, we have

$\widetilde{\text{Cap}}_{k,3}(K_{j+1},\Omega)\leq\int_{\Omega}(-u_{j})(dd^{c}u_{j})^{k}\wedge\omega^{n-k}=\widetilde{\text{Cap}}_{k,1}(K_{j},\Omega)=\text{Cap}_{k}(K_{j},\Omega).$

Letting $j\to\infty$, (2.5) holds by the convergence of
$\text{Cap}_{k}(K_{j},\Omega)$ and the weak continuity of
$(-u_{j})(dd^{c}u_{j})^{k}\wedge\omega^{n-k}$. ∎

##  3\. Capacitary estimates for level sets of plurisubharmonic functions

In this section, we establish a capacitary estimate for level sets of
$k$-plurisubharmonic functions, which will play an important role in the proof
of Theorem 1.1. The analogous estimate for the real Hessian equation was
obtained by [XZ]. This estimate is a generalization of the capacitary
estimates for the Wiener capacity [AH, Chaper 7].

First, we recall the capacitary weak type inequality

(3.1) $\displaystyle
t^{k+1}\text{Cap}_{k}(\left\\{z\in\Omega\,\left|\,u(z)\leq-t\right.\right\\},\Omega)\leq\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1},\
\ \ \forall t>0.$

This inequality was proved by [ACKZ] for $k=n$, and by [Lu] for general $k$.
We prove the capacitary strong type inequality as follows.

######  Theorem 3.1.

Suppose $u\in\mathcal{PSH}_{k,0}(\Omega)\cap C^{2}(\Omega)\cap
C(\overline{\Omega})$. For any $A>1$, we have

(3.2)
$\displaystyle\int_{0}^{\infty}t^{k}\text{Cap}_{k}(\left\\{z\in\Omega\,\left|\,u<-t\right.\right\\},\Omega)\,dt\leq\left(\frac{A}{A-1}\right)^{k+1}\log
A\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1}.$

###### Proof.

We use similar arguments as [Ma] for the Laplacian and [XZ] for the real
Hessian case. For $t>0$, denote

$K_{t}:=\left\\{z\in\Omega\,\left|\,u(z)\leq-t\right.\right\\},\
\Omega_{t}:=\left\\{z\in\Omega\,\left|\,u(z)<-t\right.\right\\}$

and $v_{t}:=u_{k,K_{t}}^{*}$. For a Borel subset $E\subset\Omega$, we denote

$\phi(E):=\frac{\int_{E}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}}{\int_{\Omega}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}}.$

For $A>1$,

$\displaystyle\int_{0}^{\infty}\phi(\Omega_{t}\setminus
K_{At})\,\frac{dt}{t}\leq$ $\displaystyle\int_{0}^{\infty}\phi(K_{t}\setminus
K_{At})\,\frac{dt}{t}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\left(\int_{t}^{At}\frac{d\phi(K_{s})}{ds}\,ds\right)\,\frac{dt}{t}$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\left(\int_{s}^{\frac{s}{A}}\frac{dt}{t}\right)\frac{d\phi(K_{s})}{ds}\,ds$
$\displaystyle=$ $\displaystyle-\log
A\int_{0}^{\infty}\frac{d\phi(K_{s})}{ds}\,ds$ $\displaystyle=$
$\displaystyle\lim_{t\to 0^{+}}\phi(K_{t})\cdot\log A$ $\displaystyle\leq$
$\displaystyle\log A.$

This implies

(3.3) $\displaystyle\int_{0}^{\infty}\left\|u\cdot\chi_{\Omega_{t}\setminus
K_{At}}\right\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1}\,\frac{dt}{t}$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\left(\int_{\Omega_{t}\setminus
K_{At}}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}\right)\,\frac{dt}{t}$
$\displaystyle\leq$
$\displaystyle\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1}\log A.$

Then for $\forall t>0$, we consider

$u^{t}:=\frac{u+t}{(A-1)t},\ \ \ \tilde{u}^{t}:=\max\\{u^{t},-1\\}.$

It is clear that $u^{t},\tilde{u}^{t}\in\mathcal{PSH}_{k,0}(\Omega_{t})\cap
C^{0,1}(\overline{\Omega}_{t})$ and $\tilde{u}^{t}=-1$ on $K_{At}$. We have

$\displaystyle\|\tilde{u}^{t}\|_{\mathcal{PSH}_{k,0}(\Omega_{t})}^{k+1}=$
$\displaystyle\int_{\Omega_{t}}(-\tilde{u}^{t})(dd^{c}\tilde{u}^{t})^{k}\wedge\omega^{n-k}$
$\displaystyle=$ $\displaystyle\int_{\Omega_{t}}d\tilde{u}^{t}\wedge
d^{c}\tilde{u}^{t}\wedge(dd^{c}\tilde{u}^{t})^{k-1}\wedge\omega^{n-k}$
$\displaystyle=$ $\displaystyle\int_{\Omega_{t}\setminus
K_{At}}d\tilde{u}^{t}\wedge
d^{c}\tilde{u}^{t}\wedge(dd^{c}\tilde{u}^{t})^{k-1}\wedge\omega^{n-k}$
$\displaystyle\leq$ $\displaystyle\int_{\Omega_{t}\setminus
K_{At}}\left(-\frac{u}{(A-1)t}\right)(dd^{c}\tilde{u}^{t})^{k}\wedge\omega^{n-k}$
$\displaystyle=$ $\displaystyle(A-1)^{-k-1}t^{-k-1}\int_{\Omega_{t}\setminus
K_{At}}\left(-u\right)(dd^{c}u)^{k}\wedge\omega^{n-k},$

where we have used $\frac{\partial\tilde{u}^{t}}{\partial\nu}\geq 0$ almost
everywhere on $\partial K_{At}$ for $t>0$ at the fourth line. That is,

$\displaystyle\int_{\Omega_{t}}(-\tilde{u}^{t})(dd^{c}\tilde{u}^{t})^{k}\wedge\omega^{n-k}\leq(A-1)^{-k-1}t^{-k-1}\phi(\Omega_{t}\setminus
K_{At})\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1}.$

Now we denote by $\hat{v}_{t}:=u_{k,K_{At},\Omega_{t}}^{*}$ the relatively
extremal function of $K_{At}$ with respect to $\Omega_{t}$. Note that
$\hat{v}_{t}\geq\tilde{u}^{t}$ in $\Omega_{t}$, and
$\hat{v}_{t}=\tilde{u}^{t}=0$ on $\partial\Omega_{t}$. By comparison
principle, we have

$\displaystyle\text{Cap}_{k}(K_{At},\Omega_{t})=$
$\displaystyle\int_{K_{At}}(-\hat{v}_{t})(dd^{c}\hat{v}_{t})^{k}\wedge\omega^{n-k}\leq\int_{\Omega_{t}}(-\tilde{u}^{t})(dd^{c}\tilde{u}^{t})^{k}\wedge\omega^{n-k}$
(3.4) $\displaystyle\leq$
$\displaystyle(A-1)^{-k-1}t^{-k-1}\phi(\Omega_{t}\setminus
K_{At})\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1}.$

Finally, by (3.3), (3) with $\lambda=At$, we obtain

$\displaystyle\int_{0}^{\infty}\lambda^{k}\text{Cap}_{k}(K_{\lambda},\Omega)\,d\lambda$
$\displaystyle\leq$ $\displaystyle
A^{k+1}\int_{0}^{\infty}t^{k}\text{Cap}_{k}(K_{At},\Omega_{t})\,dt$
$\displaystyle\leq$ $\displaystyle A^{k+1}(A-1)^{-k-1}\log
A\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{k+1}.$

∎

##  4\. The Trace Inequalities

In this section, we are going to prove the trace inequalities in Theorem 1.1.

###  4.1. The Sobolev type trace inequality(the case $0<p<k+1$).

First, we show $I_{k,p}(\Omega,\mu)<+\infty$ implies the Sobolev type
inequality. For $u\in\mathcal{F}_{k}(\Omega)$, denote

$S_{k,p}(\mu,u):=\sum_{j=-\infty}^{\infty}\frac{[\mu(K_{2^{j}}^{u})-\mu(K_{2^{j+1}}^{u})]^{\frac{k+1}{k+1-p}}}{\text{Cap}_{k}(K_{2^{j}}^{u},\Omega)^{\frac{p}{k+1-p}}},\
\ \ K_{s}^{u}:=\\{u\leq-s\\}.$

By the elementary inequality $a^{c}+b^{c}\leq(a+b)^{c}$ for $a$, $b\geq 0$ and
$c\geq 1$, we get

$\displaystyle S_{k,p}(\mu,u)\leq$
$\displaystyle\sum_{j=-\infty}^{\infty}\frac{[\mu(K_{2^{j}}^{u})-\mu(K_{2^{j+1}}^{u})]^{\frac{k+1}{k+1-p}}}{[\nu_{k}(2^{j},\Omega,\mu)]^{\frac{p}{k+1-p}}}$
$\displaystyle\leq$
$\displaystyle\sum_{j=-\infty}^{\infty}\frac{[\mu(K_{2^{j}}^{u})]^{\frac{k+1}{k+1-p}}-[\mu(K_{2^{j+1}}^{u})]^{\frac{k+1}{k+1-p}}}{[\nu_{k}(2^{j},\Omega,\mu)]^{\frac{p}{k+1-p}}}$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\frac{1}{[\nu_{k}(s,\Omega,\mu)]^{\frac{p}{k+1-p}}}\,d\left(s^{\frac{k+1}{k+1-p}}\right)=\frac{k+1}{k+1-p}I_{k,p}(\Omega,\mu).$

Then by the strong capacitary inequality (3.2) with $A=n$ and by integration
by parts,

$\displaystyle\int_{\Omega}(-u)^{p}\,d\mu=$
$\displaystyle\int_{\Omega}\int_{0}^{\infty}\chi_{[0,-u]}(s^{p})\,d\left(s^{p}\right)\,d\mu$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mu(K_{s}^{u})d\left(s^{p}\right)$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}s^{p}\,d\mu(K_{s}^{u})$
$\displaystyle\leq$
$\displaystyle\sum_{j=-\infty}^{\infty}2^{(j+1)p}\cdot[\mu(K_{2^{j}}^{u})-\mu(K_{2^{j+1}}^{u})]$
$\displaystyle\leq$
$\displaystyle{S_{k,p}(\mu,u)}^{\frac{k+1-p}{k+1}}\cdot\left(\sum_{j=-\infty}^{\infty}2^{j(k+1)}\text{Cap}_{k}(K_{2^{j(k+1)}}^{u},\Omega)\right)^{\frac{p}{k+1}}$
$\displaystyle\leq$
$\displaystyle(k+1){S_{k,p}(\mu,u)}^{\frac{k+1-p}{k+1}}\cdot\left(\int_{0}^{\infty}s^{k}\text{Cap}(K_{s}^{u},\Omega)\,ds\right)^{\frac{p}{k+1}}$
$\displaystyle\leq$ $\displaystyle
C(n,k,\mu,p)\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{p},$

thereby completing the proof.∎

On the other hand, suppose there exists $p\in(0,k+1)$ such that the inequality
(1.3) holds. That is, for any $u\in{\mathcal{F}}_{k}(\Omega)$

$\sup_{t>0}\\{t\mu(K_{t}^{u})^{\frac{1}{p}}\\}\leq\|u\|_{L^{p}(\Omega,\mu)}\leq
C\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}.$

By the definition of $\nu_{k}$, for any integer $j<\frac{\log\mu(\Omega)}{\log
2}$, we can choose a compact subset $K_{j}\subset\Omega$ such that

$\displaystyle\mu(K_{j})\geq 2^{j},\ \text{Cap}_{k}(K_{j},\Omega)\leq
2\nu_{k}(2^{j},\Omega,\mu).$

Then by Lemma 2.5, we can choose $u_{j}\in{\mathcal{F}}_{k}(\Omega)$ such that

$\displaystyle u_{j}\big{|}_{K_{j}}\leq-1,\ {\mathcal{E}}_{k}(u_{j})\leq
2\text{Cap}_{k}(K_{j},\Omega).$

Then for integers $i$, $m$ with $-\infty<i<m$ and $\frac{\log\mu(\Omega)}{\log
2}\in(m,m+1]$, let

$\gamma_{j}:=\left(\frac{2^{j}}{\nu_{k}(2^{j},\Omega,\mu)}\right)^{\frac{1}{k+1-p}},\
\ \forall j\in[i,m]\text{\ and \ }u_{i,m}:=\sup_{i\leq j\leq
m}\\{\gamma_{j}u_{j}\\}.$

Since $u_{i,m}\in{\mathcal{F}}_{k}(\Omega)\cap C(\overline{\Omega})$, we have

${\mathcal{E}}_{k}(u_{i,m})\leq
C_{n,k}\sum_{i}^{m}\gamma_{j}^{k+1}{\mathcal{E}}_{k}(u_{j})\leq
C_{n,k}\sum_{i}^{m}\gamma_{j}^{k+1}\nu_{k}(2^{j},\Omega,\mu).$

Note that for $i\leq j\leq m$ we have

$2^{j}<\mu(K_{j})\leq\mu(K_{\gamma_{j}}^{u_{i,m}}).$

This implies

$\displaystyle\|u_{i,m}\|^{p}_{\mathcal{PSH}_{k,0}(\Omega)}$
$\displaystyle\geq$ $\displaystyle
C^{-p}C_{n,k,p}\int_{\Omega}|u_{i,m}|^{p}\,d\mu$ $\displaystyle\geq$
$\displaystyle
C\int_{0}^{\infty}\left(\inf\left\\{t\,\left|\,\mu(K_{t}^{u_{i,m}})\leq
s\right.\right\\}\right)^{p}\,ds$ $\displaystyle\geq$ $\displaystyle
C\sum_{j=i}^{m}\left(\inf\left\\{t\,\left|\,\mu(K_{t}^{u_{i,m}})\leq
2^{j}\right.\right\\}\right)^{p}\cdot 2^{j}$ $\displaystyle\geq$
$\displaystyle C\sum_{j=i}^{m}\gamma_{j}^{p}2^{j}$ $\displaystyle\geq$
$\displaystyle
C\frac{\sum_{j=i}^{m}\gamma_{j}^{p}2^{j}}{\sum_{j=i}^{m}\gamma_{j}^{k+1}\nu_{k}(2^{j},\Omega,\mu)^{\frac{p}{k+1}}}\|u_{i,m}\|^{p}_{\mathcal{PSH}_{k,0}(\Omega)}$
$\displaystyle\geq$ $\displaystyle
C\frac{\sum_{j=i}^{m}2^{\frac{j(k+1)}{k+1-p}}\nu_{k}(2^{j},\Omega,\mu)^{-\frac{p}{k+1-p}}}{\left(\sum_{j=i}^{m}2^{\frac{j(k+1)}{k+1-p}}\nu_{k}(2^{j},\Omega,\mu)^{-\frac{p}{k+1-p}}\right)^{\frac{p}{k+1}}}\|u_{i,m}\|^{p}_{\mathcal{PSH}_{k,0}(\Omega)}$
$\displaystyle=$ $\displaystyle
C\left(\sum_{j=i}^{m}2^{\frac{j(k+1)}{k+1-p}}\nu_{k}(2^{j},\Omega,\mu)^{-\frac{p}{k+1-p}}\right)^{\frac{k+1-p}{k+1}}\|u_{i,m}\|^{p}_{\mathcal{PSH}_{k,0}(\Omega)}.$

Consequently,

$\displaystyle I_{k,p}(\Omega,\mu)$ $\displaystyle\leq$
$\displaystyle\lim_{i\to-\infty}\sum_{j=i}^{m}2^{\frac{(j+1)(k+1)}{k+1-p}}\nu_{k}(2^{j},\Omega,\mu)^{-\frac{p}{k+1-p}}+\int_{2^{m}}^{\mu(\Omega)}\left(\frac{s}{\nu_{k}(s,\Omega,\mu)}\right)^{\frac{p}{k+1-p}}\,ds$
$\displaystyle\leq$
$\displaystyle(1+\mu(\Omega))\lim_{i\to-\infty}\sum_{j=i}^{m}2^{\frac{(j+1)(k+1)}{k+1-p}}\nu_{k}(2^{j},\Omega,\mu)^{-\frac{p}{k+1-p}}<+\infty.\qed$

###  4.2. The Sobolev type trace inequality(the case $p\geq k+1$).

First, we assume $I_{k,p}(\Omega,\mu)<+\infty$ and prove the Sobolev type
inequality. For $u\in\mathcal{F}_{k}(\Omega)$, we have

$\displaystyle\int_{\Omega}(-u)^{p}\,d\mu=$ $\displaystyle
p\int_{0}^{\infty}\mu(K_{t})t^{p-1}\,dt$ $\displaystyle\leq$ $\displaystyle
pI_{k,p}(\Omega,\mu)\int_{0}^{\infty}{\text{Cap}_{k}(K_{t},\Omega)}^{\frac{p}{k+1}}t^{p-1}\,dt$
$\displaystyle=$ $\displaystyle
pI_{k,p}(\Omega,\mu)\int_{0}^{\infty}t^{p-1-k}t^{k}{\text{Cap}_{k}(K_{t},\Omega)}^{\frac{p-1-k}{k+1}+1}\,dt$
$\displaystyle\leq$ $\displaystyle
pI_{k,p}(\Omega,\mu)\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{p-1-k}\int_{0}^{\infty}t^{k}\text{Cap}_{k}(K_{t},\Omega)\,dt$
$\displaystyle\leq$
$\displaystyle\left[pI_{k,p}(\Omega,\mu)\left(\frac{A}{A-1}\right)^{k+1}\log
A\right]\cdot\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}^{p}.$

The sufficient part is proved.

On the contrary, let $u=u_{k,K}^{*}$ the relatively capacitary potential with
respect to a compact set $K\subset\Omega$, then

$\mu(K)^{\frac{1}{p}}\leq\|u_{k,K}^{*}\|_{L^{p}(\Omega)}\leq
C\cdot\text{Cap}_{k}(K,\Omega)^{\frac{1}{k+1}}.$

By Remark 1.2 (c), $I_{k,p}(\Omega,\mu)<+\infty$.

###  4.3. Compactness

In this section, we are going to consider the compactness of the embedding
induced by inequalities (1.3). First, we recall the Poincaré type inequality
for complex Hessian operators.

######  Theorem 4.1.

[Hou, AC20] Suppose $\Omega$ is a pseudoconvex domain with smooth boundary,
and $1\leq l<k\leq n$. Then there exists a uniform constant $C>0$ depending on
$k$, $l$ and $\Omega$ such that

(4.1) $\displaystyle\|u\|_{\mathcal{PSH}_{l,0}(\Omega)}\leq
C\|u\|_{\mathcal{PSH}_{k,0}(\Omega)},\ \ \ \forall
u\in\mathcal{PSH}_{k,0}(\Omega).$

######  Corollary 4.2.

Suppose $K$ is a compact subset of $\Omega$ with smooth boundary $\partial K$,
and $1\leq l<k\leq n$. Then we have

(4.2) $\displaystyle\text{Cap}_{l}(K,\Omega)\leq
C\cdot{\text{Cap}_{k}(K,\Omega)}^{\frac{l+1}{k+1}}.$

###### Proof.

Let $u\in\mathcal{PSH}_{k,0}(\Omega)$ such that $u\leq-\chi_{K}$. By
$\mathcal{PSH}_{k,0}(\Omega)\subset\mathcal{PSH}_{l,0}(\Omega)$, we have

$\displaystyle\text{Cap}_{l}(K,\Omega)=\widetilde{\text{Cap}}_{l,3}(K,\Omega)$
$\displaystyle\leq$
$\displaystyle\int_{\Omega}(-u)(dd^{c}u)^{l}\wedge\omega^{n-l}$
$\displaystyle\leq$ $\displaystyle
C\left[\int_{\Omega}(-u)(dd^{c}u)^{k}\wedge\omega^{n-k}\right]^{\frac{l+1}{k+1}}.$

Then (4.2) follows by taking the infimum. ∎

Now, we are in position to show the compactness of the Sobolev trace
inequality. First we need the following compactness theorem for classical
Sobolev embedding.

######  Theorem 4.3.

[Ma] Suppose $p\geq 2$, the embedding

$Id:\,W^{1,2}_{0}(\Omega)\to L^{p}(\Omega,\mu)$

is compact if and only if

(4.3) $\displaystyle\lim_{s\to 0}\frac{s}{\nu_{1}(s,\mu)^{\frac{p}{2}}}\to 0.$

By Corollary 4.2, condition (1.4) implies (4.3).

Proof of compactness in Theorem 1.1(i). First, we consider the ’if’ part. For
a sequence $u_{j}\in\mathcal{F}_{k}(\Omega)$ with bounded
$\|u_{j}\|_{\mathcal{PSH}_{k,0}(\Omega)}$, we can obtain the boundedness of
$\|u_{j}\|_{W^{1,2}_{0}(\Omega)}$ by Theorem 4.1. Then by Theorem 4.3, we can
find a subsequence $\\{u_{j_{l}}\\}$ such that $u_{j_{l}}$ converges to $u$ in
$L^{\frac{p+2}{2}}(\Omega,\mu)$. Hence, we have

$\displaystyle\int_{\Omega}|u_{j_{l}}-u|^{p}\,d\mu\leq$
$\displaystyle\left(\int_{\Omega}|u_{j_{l}}-u|^{\frac{p+2}{2}}\,d\mu\right)^{\frac{1}{2}}\cdot\left(\int_{\Omega}|u_{j_{l}}-u|^{\frac{3p-2}{2}}\,d\mu\right)^{\frac{1}{2}}$
(4.4) $\displaystyle\leq$ $\displaystyle
C\|u_{j_{l}}-u\|_{L^{\frac{p+2}{2}}(\Omega,\mu)}\cdot(\|u_{j_{l}}\|_{\mathcal{PSH}_{k,0}(\Omega)}^{\frac{3p-2}{4}}+\|u\|^{\frac{3p-2}{4}}_{\mathcal{PSH}_{k,0}(\Omega)})\to
0,$

thereby completing the proof.

For the ’only if’ part, we assume $\mathcal{PSH}_{k,0}(\Omega)$ is compactly
embedded into $L^{p}(\Omega,\mu)$. Similar to the linear case [Ma], we will
show that for any $s>0$, there exists $\varepsilon(s)$ satisfying
$\varepsilon(s)\to 0$, as $s\to 0$, such that for any compact subset
$K\subset\Omega$ with $\mu(K)<s$, it holds

(4.5)
$\displaystyle\left(\int_{K}(-u)^{p}\,d\mu\right)^{\frac{1}{p}}\leq\varepsilon(s)\|u\|_{\mathcal{PSH}_{k,0}(\Omega)}$

holds for all $u\in\mathcal{PSH}_{k,0}(\Omega)$.

We show this by contradiction. Assume there exists a sequence of compact
subsets $K_{i}$ such that $\mu(K_{i})\to 0$ and
$\\{u_{j}\\}_{j=1}^{\infty}\subset\mathcal{PSH}_{k,0}(\Omega)$, and a uniform
constant $c>0$ such that

$\int_{K_{i}}(-u_{j})^{p}\,d\mu\geq
c\|u_{j}\|_{\mathcal{PSH}_{k,0}(\Omega)}^{p}.$

By scaling, we may assume $\|u_{j}\|_{\mathcal{PSH}_{k,0}(\Omega)}=1$. Then by
the compactness of the embedding, there is a subsequence, still denoted by
$\\{u_{j}\\}$, which converges to $u\in L^{p}(\Omega,\mu)$ in $L^{p}$-sense.
Now we consider a sequence of cut-off functions $\eta_{j}\in
C^{\infty}_{0}(\Omega)$ such that $\text{supp}(\eta_{j})\subset K_{j}$. Then
by $\mu(K_{i})\to 0$, $\eta_{j}^{\frac{1}{p}}u_{j}$ converges to $0$ in
$L^{p}(\Omega,\mu)$ sense. Hence, $\|u_{j}\|_{L^{p}(K_{j},\mu)}\to 0$ as
$j\to\infty$, which makes a contradiction.

Finally, for any $K\Subset\Omega$ with $\mu(K)<s$, by letting $u=u_{k,K}^{*}$
in the (4.5), we have

$\frac{\mu(K)}{\text{Cap}_{k}(K,\Omega)^{\frac{p}{k+1}}}\leq\varepsilon(s)\to
0.\qed$

###  4.4. The Moser-Trudinger type trace inequality(Proof of Theorem 1.1(ii))

Let $\mu$ be a positive Randon measure,

$\displaystyle\int_{\Omega}\exp\left(\beta\left(\frac{-u}{\|u\|_{\mathcal{PSH}_{0}(\Omega)}}\right)^{q}\right)\,d\mu$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{\infty}\frac{\beta^{i}}{i!}\int_{\Omega}\left(\frac{-u}{\|u\|_{\mathcal{PSH}_{0}(\Omega)}}\right)^{qi}\,d\mu$
$\displaystyle=$
$\displaystyle\sum_{i<\frac{n+1}{q}}\frac{\beta^{i}}{i!}\int_{\Omega}\left(\frac{-u}{\|u\|_{\mathcal{PSH}_{0}(\Omega)}}\right)^{qi}\,d\mu+\sum_{i\geq\frac{n+1}{q}}\frac{\beta^{i}}{i!}\int_{\Omega}\left(\frac{-u}{\|u\|_{\mathcal{PSH}_{0}(\Omega)}}\right)^{qi}\,d\mu$
$\displaystyle=:$ $\displaystyle I+II.$

Since $I_{n}(\beta,\Omega,\mu)<+\infty$ implies $I_{k,p}(\Omega,\mu)$, we have
$I\leq C$. It suffices to estimate $II$. We have

$\displaystyle II=$
$\displaystyle\sum_{i\geq\frac{n+1}{q}}\frac{\beta^{i}}{i!}\|u\|_{\mathcal{PSH}_{0}(\Omega)}^{-qi}\int_{\Omega}(-u)^{qi}\,d\mu$
$\displaystyle=$
$\displaystyle\sum_{i\geq\frac{n+1}{q}}\frac{\beta^{i}}{i!}\|u\|_{\mathcal{PSH}_{0}(\Omega)}^{-qi}\int_{0}^{\infty}t^{qi}\frac{d\mu(K_{t})}{dt}\,dt$
$\displaystyle=$
$\displaystyle\sum_{i\geq\frac{n+1}{q}}\frac{\beta^{i}}{i!}\|u\|_{\mathcal{PSH}_{0}(\Omega)}^{-qi}\int_{0}^{\infty}\mu(K_{t})\,d(t^{qi})$
$\displaystyle\leq$
$\displaystyle\sum_{i\geq\frac{n+1}{q}}\frac{\beta^{i}}{i!}\int_{0}^{\infty}\frac{\text{Cap}(K_{t},\Omega)}{t^{-n}\|u\|^{n+1}}\left(\frac{\mu(K_{t})}{\text{Cap}(K_{t},\Omega)^{\frac{qi}{n+1}}}\right)\,dt$
$\displaystyle\leq$ $\displaystyle\beta
q\|u\|^{-n-1}\int_{0}^{\infty}\sum_{i=0}^{\infty}\frac{\beta^{i}}{i!}\left(\frac{\mu(K_{t})}{\text{Cap}(K_{t},\Omega)^{\frac{qi}{n+1}}}\right)\,t^{n}\text{Cap}(K_{t},\Omega)\,dt$
$\displaystyle=$ $\displaystyle\beta
q\|u\|^{-n-1}\int_{0}^{\infty}\mu(K_{t})\exp\left(\frac{\beta}{\text{Cap}(K_{t},\Omega)^{\frac{q}{n+1}}}\right)t^{n}\text{Cap}(K_{t},\Omega)\,dt,$
$\displaystyle\leq$ $\displaystyle\beta
qI_{n}(\beta,\Omega,\mu)\left(\frac{A}{A-1}\right)^{k+1}\log A,$

where we have used (3.1) at the fourth line and (3.2) at the last line. ∎

######  Remark 4.4.

When the measure $d\mu$ is Lebesgue measure and $\Omega=B_{1}$ is the unit
ball centered at the origin, as proved in [K96], there holds

(4.6) $|K|\leq
C_{\lambda,\Omega,n}\exp\left\\{-\frac{\lambda}{\text{Cap}_{n}(K,\Omega)^{\frac{1}{n}}}\right\\}$

for any $0<\lambda<2n$. In particular, when $K=B_{r}$ with $r<1$, by standard
computations,

$|B_{r}|e^{\frac{\lambda}{\text{Cap}_{n}(B_{r},\Omega)^{\frac{1}{n}}}}=C_{n}r^{2n-\lambda}.$

Hence (4.6) does not hold when $\lambda>2n$. It is natural to ask if $\lambda$
can attain the optimal constant $2n$.

As shown in [BB], for those $u\in\mathcal{PSH}_{0}(\Omega)$ with
${\mathbb{S}}^{1}$-symmetry, i.e. $u(z)=u(e^{{\sqrt{-1}}\theta}z)$ for all
$\theta\in{\mathbb{R}}^{1}$, (1.5) holds with $\beta=2n$. Then by the proof of
Theorem 1.1 (ii) with ${\mathbb{S}}^{1}$-symmetry, there holds

$|E|\leq
C\exp\left\\{-\frac{2n}{\text{Cap}_{n}(E,\Omega)^{\frac{1}{n}}}\right\\}$

for any ${\mathbb{S}}^{1}$-invariant subset $E$. Moreover, by [BB14], the
Schwartz symmetrization $\hat{u}$ of $u_{E}^{*}$ is plurisubharmonic and has
smaller $\|\cdot\|_{\mathcal{PSH}_{0}(\Omega)}$-norm. This leads to

$|E|e^{\frac{2n}{\text{Cap}_{n}(E,\Omega)^{\frac{1}{n}}}}\leq|B_{r}|e^{\frac{2n}{\text{Cap}_{n}(B_{r},\Omega)^{\frac{1}{n}}}}\equiv
C_{n}|B_{1}|.$

###  4.5. The Brezis-Merle type trace inequality

Similar to [BB], we can obtain a relationship between the Brezis-Merle type
trace inequality and the Moser-Trudinger type inequality.

######  Theorem 4.5.

The Moser-Trudinger type trace inequality (1.5) holds for any
$0<\lambda<\beta$ for some $\beta>0$ and $q\in[1,\frac{n+1}{n}]$ if and only
if for any $0<\lambda<\beta$, the Brezis-Merle type trace inequality

(4.7)
$\displaystyle\sup\left\\{\int_{\Omega}\exp\left(\lambda\left(\frac{-u}{{\mathcal{M}}[u]^{\frac{1}{n}}}\right)^{\frac{nq}{n+1}}\right)\,d\mu:u\in\mathcal{F}_{n}(\Omega),0<\|u\|_{\mathcal{PSH}_{0}(\Omega)}<\infty\right\\}<+\infty$

holds. Here ${\mathcal{M}}[u]:=\int_{\Omega}(dd^{c}u)^{n}$.

###### Proof.

First, we prove the if part. By Theorem 1.1(iv), there is $C>0$ such that for
any compact subset $K\subset\Omega$

$\mu(K)e^{\lambda\frac{1}{\text{Cap}_{n}(K,\Omega)^{\frac{q}{n+1}}}}\leq C.$

For $u\in{\mathcal{F}}_{k}(\Omega)$, we denote $K_{t}:=\\{u\leq-t\\}$, $t>0$.
By comparison principle, we have

${\mathcal{M}}[u]=\int_{\Omega}(dd^{c}u)^{n}\geq
t^{n}\int_{\Omega}(dd^{c}u_{K_{t}})^{n}=t^{n}\text{Cap}_{n}(K_{t},\Omega),\
\forall t>0.$

Therefore,

$\mu(K_{t})\leq
Ce^{-\lambda\frac{1}{\text{Cap}_{n}(K,\Omega)^{\frac{q}{n+1}}}}\leq
Ce^{-\lambda\frac{t^{\frac{nq}{n+1}}}{{\mathcal{M}}[u]^{\frac{q}{n+1}}}}.$

Then for every $0<\varepsilon<\lambda$,

$\displaystyle\int_{\Omega}e^{(\lambda-\varepsilon)\frac{(-u)^{\frac{nq}{n+1}}}{{\mathcal{M}}[u]^{\frac{q}{n+1}}}}\,d\mu$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\frac{(\lambda-\varepsilon)^{j}}{j!}\int_{\Omega}\frac{(-u)^{\frac{nqj}{n+1}}}{{\mathcal{M}}[u]^{\frac{qj}{n+1}}}\,d\mu$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\frac{(\lambda-\varepsilon)^{j}}{j!}\int_{0}^{\infty}\frac{\mu(K_{t})}{{\mathcal{M}}[u]^{\frac{qj}{n+1}}}\,d\left(t^{\frac{nqj}{n+1}}\right)$
$\displaystyle\leq$ $\displaystyle
C\sum_{0}^{\infty}\frac{(\lambda-\varepsilon)^{j}}{j!}\int_{0}^{\infty}e^{-\lambda\frac{t^{\frac{nq}{n+1}}}{{\mathcal{M}}[u]^{\frac{q}{n+1}}}}d\left(\frac{t^{\frac{nqj}{n+1}}}{{\mathcal{M}}[u]^{\frac{qj}{n+1}}}\right)$
$\displaystyle\leq$
$\displaystyle\sum_{j=0}^{\infty}\frac{(\lambda-\varepsilon)^{j}}{\lambda^{j}}=\frac{\lambda}{\varepsilon}-1.$

Next, we show the only if part. For a Borel subset $E\subset\Omega$, we can
apply the Brezis-Merle trace inequality to $u_{E}^{*}$ to obtain the condition
(1.7), which implies the Moser-Trudinger type inequality. ∎

##  5\. The Dirichlet problem

###  5.1. The continuous solution(Proof of Theorem 1.4)

First, we prove (ii) $\Rightarrow$ (i). Suppose $E\subset\Omega$ is a Borel
subset, and $u$ solves

(5.1)
$\displaystyle\begin{cases}(dd^{c}u)^{k}\wedge\omega^{n-k}=\chi_{E}\,d\mu,\ \
\ &\text{in }\Omega,\\\ u=0,&\text{on }\partial\Omega,\end{cases}$

where $\mu$ is a positive Radon measure.

######  Theorem 5.1.

Assume there exists $p>k+1$ such that $I_{k,p}(\Omega,\mu)<+\infty$. Let $u$
be a solution to (5.1). Then there exists $C_{1}>0$ depending on $k$, $p$ and
$I_{k,p}(\Omega,\mu)$ such that

(5.2) $\|u\|_{L^{\infty}(\Omega,\mu)}\leq C_{1}\mu(E)^{\delta}$

where $\delta=\frac{p-k-1}{kp}$.

###### Proof.

The proof is similar to [WWZ2]. We denote $\hat{\mu}:=\chi_{E}\cdot\mu$. For
any $s>0$, let $K_{s}:=\\{u\leq-s\\}$ and $u_{s}=u+s$. Denote
$p=\frac{k+1}{1-k\delta}$. By $I_{k,p}(\Omega,\mu)<+\infty$, we can apply
Theorem 1.1 to get

$\left(\int_{K_{s}}\left(\frac{-u_{s}}{\|u_{s}\|_{\mathcal{PSH}_{k,0}(K_{s})}}\right)^{p}d\hat{\mu}\right)^{\frac{1}{p}}\leq
C.$

Then by the equation,

$\displaystyle t\int_{K_{s+t}}(dd^{c}u_{s})^{k}\wedge\omega^{n-k}$
$\displaystyle\leq$ $\displaystyle\int_{K_{s}}(-u_{s})\,d\hat{\mu}$
$\displaystyle\leq$
$\displaystyle\left(\int_{K_{s}}(-u_{s})^{p}d\hat{\mu}\right)^{\frac{1}{p}}\left(\int_{K_{s}}d\hat{\mu}\right)^{1-\frac{1}{p}}$
$\displaystyle=$
$\displaystyle\left(\int_{K_{s}}\left(\frac{-u_{s}}{\|u_{s}\|_{\mathcal{PSH}_{k,0}(K_{s})}}\right)^{p}d\hat{\mu}\right)^{\frac{1}{p}}\left(\int_{K_{s}}d\hat{\mu}\right)^{1-\frac{1}{p}}\cdot\|u_{s}\|_{\mathcal{PSH}_{k,0}(K_{s})},$

which implies

(5.3) $\displaystyle t\hat{\mu}(K_{s+t})\leq
C\hat{\mu}(K_{s})^{(1-\frac{1}{p})\frac{k+1}{k}}=C\hat{\mu}(K_{s})^{1+\delta}.$

In particular,

$\hat{\mu}(K_{s})\leq\frac{C}{s}\hat{\mu}(\Omega)^{1+\delta}$

for some $C>1$. Then choose
$s_{0}=2^{1+\frac{1}{\delta}}C^{1+\frac{1}{\delta}}\hat{\mu}(\Omega)^{\delta}$,
we get $\hat{\mu}(K_{s_{0}})\leq\frac{1}{2}\hat{\mu}(\Omega)$. For any
$l\in\mathbb{Z}_{+}$, define

(5.4) $\displaystyle s_{l}=s_{0}+\sum_{j=1}^{l}2^{-\delta
j}\hat{\mu}(\Omega)^{\delta},\ \ u^{l}=u+s^{l},\ \ K_{l}=K_{s_{l}}.$

Then

$2^{-\delta(l+1)}\hat{\mu}(\Omega)^{\delta}\hat{\mu}(K_{l+1})=(s_{l+1}-s_{l})\hat{\mu}(K_{l+1})\leq
C\hat{\mu}(K_{l})^{1+\delta}.$

We claim that $|K_{l+1}|\leq\frac{1}{2}|K_{l}|$ for any $l$. By induction, we
assume the inequality holds for $l\leq m$. Then

$\displaystyle\hat{\mu}(K_{m+1})\leq$ $\displaystyle
C\hat{\mu}(K_{m})^{1+\delta}\frac{2^{\delta(m+1)}}{\hat{\mu}(\Omega)^{\delta}}$
$\displaystyle\leq$ $\displaystyle
C\left[\left(\frac{\hat{\mu}(K_{0})}{2^{m}}\right)^{\delta}\frac{2^{\delta(m+1)}}{\hat{\mu}(\Omega)^{\delta}}\right]\cdot|K_{m}|$
$\displaystyle\leq$
$\displaystyle\left[C^{1+\delta}\frac{2^{\delta}}{s_{0}^{\delta}}\hat{\mu}(\Omega)^{\delta^{2}}\right]\cdot\hat{\mu}(K_{m})\leq\frac{1}{2}\hat{\mu}(K_{m}).$

This implies that the set

$\hat{\mu}\left(\left\\{x\in\Omega\big{|}u<-s_{0}-\sum_{j=1}^{\infty}\left(\frac{1}{2^{\delta}}\right)^{j}\hat{\mu}(\Omega)^{\delta}\right\\}\right)=0.$

Hence,

$\displaystyle\|u\|_{L^{\infty}(\Omega,\mu)}$ $\displaystyle\leq$
$\displaystyle
s_{0}+\sum_{j=1}^{\infty}\left(\frac{1}{2^{\delta}}\right)^{j}\hat{\mu}(\Omega)^{\delta}$
$\displaystyle=$ $\displaystyle
2^{1+\frac{1}{\delta}}C^{1+\frac{1}{\delta}}\hat{\mu}(\Omega)^{\delta}+\frac{1}{2^{\delta}-1}\hat{\mu}(\Omega)^{\delta}$
$\displaystyle\leq$ $\displaystyle
C\hat{\mu}(\Omega)^{\delta}=C\mu(E)^{\delta}.$

∎

Next, by similar arguments as in [WWZ2], we can get a stability result as well
as the existence of the unique continuous solution $u_{E}$ for (1.11). We need
the stability lemma.

######  Lemma 5.2.

Let $u$, $v$ be bounded $k$-plurisubharmonic functions in $\Omega$ satisfying
$u\geq v$ on $\partial\Omega$. Assume $(dd^{c}u)^{k}\wedge\omega^{n-k}=d\mu$
and $\mu$ satisfies the condition in Theorem 1.4 (ii). Then $\forall\
\varepsilon>0$, there exists $C>0$ depending on $k$, $p$ and
$I_{k,p}(\Omega,\mu)$,

(5.5)
$\sup\limits_{\Omega}(v-u)\leq\varepsilon+C\mu(\\{v-u>\varepsilon\\})^{\delta}.$

###### Proof.

We may suppose $d\mu\in L^{\infty}(\Omega)$ for the estimate since we will
consider the approximation of $\mu$ later in the proof of existence and
continuity. Denote $u_{\varepsilon}:=u+\varepsilon$ and
$\Omega_{\varepsilon}:=\\{v-u_{\varepsilon}>0\\}$. It suffices to estimate
$\sup_{\Omega_{\varepsilon}}|u_{\varepsilon}-v|$.

Note that $\Omega_{\varepsilon}\Subset\Omega$ and $u_{\varepsilon}$ solves

$\begin{cases}(dd^{c}u_{\varepsilon})^{k}\wedge\omega^{n-k}=d\mu\ \ &\text{\
in $\Omega_{\varepsilon}$,}\\\\[-3.0pt] u_{\varepsilon}=v&\text{\ on
${\partial\Omega}_{\varepsilon}$.}\end{cases}$

Let $u_{0}$ be the solution to the Dirichlet problem

$\begin{cases}(dd^{c}u_{0})^{k}\wedge\omega^{n-k}=\chi_{\Omega_{\varepsilon}}d\mu\
\ &\text{\ in $\Omega$,}\\\\[-5.0pt] \ u_{0}=0\hskip 55.0pt\ \ &\text{\ on
${\partial\Omega}$.}\end{cases}$

By the comparison principle we have

$u_{0}\leq u_{\varepsilon}-v\leq 0\ \ \text{in\ $\Omega_{\varepsilon}$}.$

Hence we obtain

$\sup_{\Omega_{\varepsilon}}|u_{\varepsilon}-v|\leq\sup_{\Omega_{\varepsilon}}|u_{0}|\leq
C\mu(\Omega_{\varepsilon})^{\delta}.$

∎

######  Proposition 5.3.

Let $u$, $v$ be bounded $k$-plurisubharmonic functions in $\Omega$ satisfying
$u\geq v$ on $\partial\Omega$. Assume that
$(dd^{c}u)^{k}\wedge\omega^{n-k}=d\mu$ and $\mu$ satisfies the condition in
Theorem 1.4 (ii). Then for $r\geq 1$ and $0\leq\gamma^{\prime}<\frac{\delta
r}{1+\delta r}$, it holds

(5.6) $\sup\limits_{\Omega}(v-u)\leq
C\|\max(v-u,0)\|_{L^{r}(\Omega,d\mu)}^{\gamma}$

for a uniform constant
$C=C(\gamma^{\prime},\|v\|_{L^{\infty}(\Omega,d\mu)})>0$.

###### Proof.

Note that for any $\varepsilon>0$,

$\mu(\\{v-u>\varepsilon\\})\leq\varepsilon^{-r}\int_{\\{v-u>\varepsilon\\}}|v-u|^{r}\,d\mu\leq\varepsilon^{-r}\int_{\Omega}[\max(v-u,0)]^{r}\,d\mu.$

Let $\varepsilon:=\|\max(v-u,0)\|_{L^{r}(\Omega,d\mu)}^{\gamma^{\prime}}$,
where $\gamma^{\prime}$ is to be determined. By Lemma 5.2, we have

$\displaystyle\sup\limits_{\Omega}(v-u)$ $\displaystyle\leq$
$\displaystyle\varepsilon+C\mu(\\{v-u>\varepsilon\\})^{\delta}$
$\displaystyle\leq$
$\displaystyle\|\max(v-u,0)\|_{L^{r}(\Omega,d\mu)}^{\gamma^{\prime}}+C\|\max(v-u,0)\|_{L^{r}(\Omega,d\mu)}^{\delta(r-\gamma^{\prime}r)}.$

Choose $\gamma^{\prime}\leq\frac{\delta r}{1+\delta r}$, where
$0<\delta<\frac{1}{k}$, (5.6) follows from (5.1). ∎

Now we can show the existence of the unique continuous solution for (1.11)
when $\mu$ satisfies the condition in Theorem 1.4 (ii). We consider
$\tilde{\mu}_{\varepsilon}:=\rho_{\varepsilon}*d\mu$ defined on
$\Omega_{\varepsilon}:=\\{x\in\Omega\,|\,\text{dist}(x,\partial\Omega)\leq\varepsilon\\}$,
where $\varepsilon>0$ and $\rho_{\varepsilon}$ is the cut-off function such
that

$\rho_{\varepsilon}\in C^{\infty}_{0}({\mathbb{R}}^{n}),\
\rho_{\varepsilon}=1\ \ \text{in }B_{\varepsilon}(O),\ \text{and
}\rho_{\varepsilon}=0\text{ on
}{\mathbb{R}}^{n}\setminus\overline{B_{2\varepsilon}(O)}.$

Then for every $E\subset\Omega$, we define

$\mu_{\varepsilon}(E):=\tilde{\mu}_{\varepsilon}(E\cap\Omega_{\varepsilon}).$

By the classic measure theory, $\mu_{\varepsilon}$ is absolutely continuous
with respect to Lebesgue measure $\omega^{n}$ with bounded density functions
$f_{\varepsilon}$. We are going to check $\mu_{\varepsilon}$ satisfies the
condition in Theorem 1.4 (ii) uniformly. Denote $K_{y}:=\\{x\in\Omega:x-y\in
K\cap\Omega_{\varepsilon}\\}$. By definition, for every compact subset
$K\subset\Omega$, we have

(5.8)
$\displaystyle\mu_{\varepsilon}(K)=\int_{{\mathbb{C}}^{n}}\int_{K_{y}}\rho_{\varepsilon}(x-y)\,d\mu(x)\,\omega^{n}(y)\leq
C\sup_{|y|\leq 2\varepsilon}\mu(K_{y})\leq C\sup_{|y|\leq
2\varepsilon}\left(\text{Cap}_{k}(K_{y},\Omega)\right)^{p}.$

Let $u_{y}(x):=u_{k,K_{y}}^{*}(x+y)$,
$u_{\varepsilon}(x):=u_{k,\Omega_{\varepsilon}}^{*}(x)$ be the relative
$k$-extreme functions of $K_{y}$, $\Omega_{\varepsilon}$ with respect to
$\Omega$. For any $0<c<\frac{1}{2}$, denote
$\Omega_{c}=\\{u_{\varepsilon}\leq-c\\}$. Let

$g(x):=\begin{cases}\max\left\\{u_{y}(x)-c,(1+2c)u_{\varepsilon}\right\\},\ \
\ &x\in\Omega_{\frac{c}{2}};\\\
(1+2c)u_{\varepsilon},&x\in\Omega\setminus\Omega_{\frac{c}{2}}.\end{cases}$

Note that $g$ is a well-defined $k$-plurissubharmonic function in $\Omega$.
Since $K_{y}\subset\Omega_{\varepsilon}$ for every $|y|\leq\varepsilon$, we
have $u_{y}-c\geq(1+2c)u_{\varepsilon}=-1-2c$ on
$\overline{\Omega_{\varepsilon}}$. Hence for any compact subset
$K\subset\Omega$, we can get

(5.9) $\displaystyle\text{Cap}_{k}(K,\Omega)$ $\displaystyle\geq$
$\displaystyle(1+2c)^{-k}\int_{K\bigcap\Omega_{\varepsilon}}(dd^{c}g)^{k}\wedge\omega^{n-k}=(1+2c)^{-k}\int_{K\bigcap\Omega_{\varepsilon}}(dd^{c}u_{y})^{k}\wedge\omega^{n-k}$
$\displaystyle=$
$\displaystyle(1+2c)^{-k}\int_{K_{y}}(dd^{c}u_{k,K_{y}}^{*})^{k}\wedge\omega^{n-k}=(1+2c)^{-k}\text{Cap}_{k}(K_{y},\Omega).$

By (5.8) and (5.9), we have shown that $\mu_{\varepsilon}$ satisfies the
condition in Theorem 1.4 (ii) uniformly.

Hence the solutions $\\{u_{\varepsilon}\\}$ to (1.11) with
$\\{f_{\varepsilon}\\}$ are uniformly bounded and continuous. Furthermore,
there exists a sequence $\varepsilon_{j}\to 0$ such that
$\lim_{j\to\infty}\|\mu_{\varepsilon_{j}}-\mu\|=0$, where $\|\cdot\|$ denotes
the total variation of a signed measure. Then the proof is finished by the
following well-known result.

######  Proposition 5.4.

Let $\mu_{j}=(dd^{c}\varphi_{j})^{k}\wedge\omega^{n-k}$,
$\mu=(dd^{c}\varphi)^{k}\wedge\omega^{n-k}$ be non-pluripolar nonnegative
Radon measures with finite mass, where $\varphi_{j}$,
$\varphi\in\mathcal{PSH}_{k,0}(\Omega)$ and
$\|\varphi_{j}\|_{L^{\infty}(\Omega,d\mu)}$,
$\|\varphi\|_{L^{\infty}(\Omega,d\mu)}\leq C$. If $\|\mu_{j}-\mu\|\to 0$, then

$\varphi_{j}\to\varphi\text{ in }L^{1}(\Omega,d\mu).$

###### Proof.

When $k=n$, this is Proposition 12.17 in [GZ]. The proof also applies to
$k<n$. We write a sketch of its proof here. Note that by $\|\mu_{j}-\mu\|\to
0$, the measure

$\nu:=2^{-1}\mu+\sum_{j\geq 2}2^{-j}\mu_{j}$

is a well-defined non-pluripolar nonnegative Radon measure. $\mu$, $\mu_{j}$
are absolutely continuous with respect to $\nu$. We may suppose that
$\mu_{j}=f_{j}\nu$, $\mu=f\nu$ and $f_{j}\to f$ in $L^{1}(\Omega,\nu)$. Then
by the weak compactness, there exists a subsequence $\varphi_{j}$ and
$\psi\in\mathcal{PSH}_{k,0}(\Omega)$ such that $\varphi_{j}\to\psi$ in
$L^{1}(\Omega,d\mu)$. Denote $\psi_{j}=\left(\sup_{l\geq
j}\varphi_{l}\right)^{*}$, which converges to $\psi$ decreasingly almost
everywhere with respect to $\mu$. Then by comparison principle, we have

$(dd^{c}\psi_{j})^{k}\wedge\omega^{n-k}\leq(dd^{c}\varphi_{j})^{k}\wedge\omega^{n-k}=d\mu_{j},$

which implies $(dd^{c}\psi)^{k}\wedge\omega^{n-k}\leq d\mu$. To get the
equality, we use the absolutely continuity to obtain

$(dd^{c}\psi_{j})^{k}\wedge\omega^{n-k}\geq\inf_{l\geq j}f_{l}\,d\nu,$

thereby completing the proof. ∎

The proof of (i) $\Rightarrow$ (ii) is simple. Let $\varphi_{E}$ be the
solution to (1.11). Then
$\hat{\varphi}:=\frac{\varphi_{E}}{C\mu(E)^{\delta}}\in\mathcal{PSH}_{k}(\Omega)$,
and $-1\leq\hat{\varphi}\leq 0$. By definition we have

$C^{-k}\mu(E)^{1-k\delta}=\frac{1}{C^{k}\mu(E)^{k\delta}}\int_{E}(dd^{c}\varphi_{E})^{k}\wedge\omega^{n-k}=\int_{E}(dd^{c}\hat{\varphi})^{k}\wedge\omega^{n-k}\leq\text{Cap}_{k}(E,\Omega).$

In view of (1.6), we have completed the proof.

###  5.2. Hölder continuity(Proof of Theorem 1.6)

In this section, we consider the Hölder continuity of the solution. As we
mentioned in the introduction, we will give a pure PDE proof for Theorem 1.6
as in [WWZ2]. For the proof from (i) to (ii), we just follow Proposition 2.4
in [DKN], which is a PDE approach. It suffices to consider the other
direction. Suppose (1.14) holds. In order to obtain the Hölder estimate, we
need an $L^{\infty}$-estimate like (5.2) with the measure replaced by Lebesgue
measure. We suppose the solution $u$ and the measure $d\mu$ in (1.10) are
smooth, and the general result follows by approximation.

First, we show there is an upper bound on $\|u\|_{*}$ under condition (1.14).
Note that by the classical Sobolev inequality and Theorem 4.1, we have

(5.10) $\displaystyle\|u\|_{*}\leq
C\|u\|_{L^{1}(\Omega,\omega^{n})}+C\left(\int_{\Omega}du\wedge
d^{c}u\wedge\omega^{n-1}\right)^{\frac{1}{2}}\leq
C{\mathcal{E}}_{k}(u)^{\frac{1}{k+1}}<+\infty.$

By condition (1.14), we get

(5.11)
$\displaystyle\frac{{\mathcal{E}}_{k}(u)}{\|u\|_{*}}=\int_{\Omega}\left(-\frac{u}{\|u\|_{*}}\right)\,d\mu\leq
C\frac{\|u\|_{L^{1}(\Omega,\omega^{n})}^{\gamma}}{\|u\|_{*}^{\gamma}}.$

This implies

(5.12) $\displaystyle{\mathcal{E}}_{k}(u)^{1-\frac{\gamma}{k+1}}\leq
C\|u\|_{*}^{1-\gamma}.$

Then by (5.10) we get $\|u\|_{*}\leq c_{0}$.

Choose $f=\hat{u}:=\frac{u}{c_{0}}$ in condition (1.14). Then we get

$\int_{\Omega}(-\hat{u})\,d\mu\leq
C\left(\int_{\Omega}(-\hat{u})\,\omega^{n}\right)^{\gamma}.$

Then in the proof of the $L^{\infty}$-estimate(Theorem 5.1), we use a
different iteration. We denote $\Omega_{s}:=\\{\hat{u}\leq-s\\}$,
$p<\frac{n(k+1)}{n-k}$(when $k=n$, we can choose any $p>0$) and
$\hat{u}_{s}:=\hat{u}+s$. Note that

$\displaystyle{\mathcal{E}}_{k,\Omega_{s}}(\hat{u}_{s}):=$
$\displaystyle\int_{\Omega_{s}}(-\hat{u}_{s})\,d\mu\leq
C\left(\int_{\Omega_{s}}(-\hat{u}_{s})\,\omega^{n}\right)^{\gamma}$ (5.13)
$\displaystyle\leq$ $\displaystyle
C\left(\int_{\Omega_{s}}\left(\frac{-\hat{u}_{s}}{{\mathcal{E}}_{k,\Omega_{s}}(\hat{u}_{s})^{\frac{1}{k+1}}}\right)^{p}\,\omega^{n}\right)^{\frac{\gamma}{p}}{\mathcal{E}}_{k,\Omega_{s}}(\hat{u}_{s})^{\frac{\gamma}{k+1}}|\Omega_{s}|^{(1-\frac{1}{p})\gamma}.$

By the Sobolev inequality for the complex Hessian equation with respective to
the Lebesgue measure [AC20], we get

$\displaystyle t|\Omega_{s+t}|\leq\int_{\Omega_{s}}(-\hat{u}_{s})\,\omega^{n}$
$\displaystyle\leq$ $\displaystyle
C\left(\int_{\Omega_{s}}(-\hat{u}_{s})^{p}\,\omega^{n}\right)^{\frac{1}{p}}|\Omega_{s}|^{1-\frac{1}{p}}$
$\displaystyle\leq$ $\displaystyle
C{\mathcal{E}}_{\Omega_{s}}(\hat{u}_{s})^{\frac{1}{k+1}}|\Omega_{s}|^{1-\frac{1}{p}}\leq
C|\Omega_{s}|^{1+\frac{\gamma p-k-1}{(k+1-\gamma)p}}.$

Hence,

$\|\hat{u}\|_{L^{\infty}(\Omega,\omega^{n})}\leq C|\Omega|^{\frac{\gamma
p-k-1}{(k+1-\gamma)p}}.$

By (5.10) and (5.2), we have $\|u\|_{*}\leq
C|\Omega|^{\frac{(p-1)\gamma}{(k+1-\gamma)p}}$. Therefore, for
$1<p\leq\frac{n(k+1)}{n-k}$ when $1\leq k<n$ and $p>1$ when $k=n$,

(5.14) $\|u\|_{L^{\infty}(\Omega,\omega^{n})}\leq C|\Omega|^{\frac{\gamma
p-k-1}{(k+1-\gamma)p}}\|u\|_{*}\leq C|\Omega|^{\frac{2\gamma
p-\gamma-k-1}{(k+1-\gamma)p}},$

where

(5.15) $\delta:=\frac{2\gamma p-\gamma-k-1}{(k+1-\gamma)p}.$

Then by the same arguments as in [WWZ2], we can obtain the Hölder continuity.
For readers’ convenience and the self-containness of this paper, we include a
sketch here.

First, since we have $L^{\infty}$-estimate (5.14) now, we can get the similar
stability results as in Lemma 5.2 and Proposition 5.3, in which we change the
measure $d\mu$ to be the standard Lebesgue measure $\omega^{n}$.

######  Proposition 5.5.

Let $u$, $v$ be bounded $k$-plurisubharmonic functions in $\Omega$ satisfying
$u\geq v$ on $\partial\Omega$. Assume that
$(dd^{c}u)^{k}\wedge\omega^{n-k}=d\mu$ and $d\mu$ satisfies the (1.14). Then
for $r\geq 1$ and $0\leq\gamma^{\prime}<\frac{\delta r}{1+\delta r}$, it holds

(5.16) $\sup\limits_{\Omega}(v-u)\leq
C\|\max(v-u,0)\|_{L^{r}(\Omega,\omega^{n})}^{\gamma}$

for a uniform constant
$C=C(\gamma^{\prime},\|v\|_{L^{\infty}(\Omega,\omega^{n})})>0$. Here $\delta$
is given by (5.15).

For any $\varepsilon>0$, we denote
$\Omega_{\varepsilon}:=\\{x\in\Omega|\,dist(x,{\partial\Omega})>\varepsilon\\}$.
Let

$\displaystyle u_{\varepsilon}(x)$ $\displaystyle:=$
$\displaystyle\sup\limits_{|\zeta|\leq\varepsilon}u(x+\zeta),\
x\in\Omega_{\varepsilon},$ $\displaystyle\hat{u}_{\varepsilon}(x)$
$\displaystyle:=$
$\displaystyle{\mathchoice{{-\mkern-19.0mu\int}}{{-\mkern-16.0mu\int}}{{-\mkern-16.0mu\int}}{{-\mkern-16.0mu\int}}}_{|\zeta-x|\leq\varepsilon}u(\zeta)\,\omega^{n},\
x\in\Omega_{\varepsilon}.$

Since $u$ is plurisubharmonic in $\Omega$, $u_{\varepsilon}$ is a
plurisubharmonic function. For the Hölder estimate, it suffices to show there
is a uniform constant $C>0$ such that $u_{\varepsilon}-u\leq
C\varepsilon^{\alpha^{\prime}}$ for some $\alpha^{\prime}>0$. The link between
$u_{\varepsilon}$ and $\hat{u}_{\varepsilon}$ is made by the following lemma.

######  Lemma 5.6.

(Lemma 4.2 in [Ngu]) Given $\alpha\in(0,1)$, the following two conditions are
equivalent.

(1) There exists $\varepsilon_{0}$, $A>0$ such that for any
$0<\varepsilon\leq\varepsilon_{0}$,

$u_{\varepsilon}-u\leq A\varepsilon^{\alpha}\ \ \text{on\
$\Omega_{\varepsilon}$}.$

(2) There exists $\varepsilon_{1}$, $B>0$ such that for any
$0<\varepsilon\leq\varepsilon_{1}$,

$\hat{u}_{\varepsilon}-u\leq B\varepsilon^{\alpha}\ \ \text{on\
$\Omega_{\varepsilon}$}.$

The following estimate is a generalization of Lemma 4.3 in [Ngu].

######  Lemma 5.7.

Assume $u\in W^{2,r}(\Omega)$ with $r\geq 1$. Then for $\varepsilon>0$ small
enough, we have

(5.17)
$\left[\int_{\Omega_{\varepsilon}}|\hat{u}_{\varepsilon}-u|^{r}\,\omega^{n}\right]^{\frac{1}{r}}\leq
C(n,r)\|\triangle u\|_{L^{r}(\Omega,\omega^{n})}\varepsilon^{2}$

where $C(n,r)>0$ is a uniform constant.

Note that the function $u_{\varepsilon}$ is not globally defined on $\Omega$.
However, by $\varphi\in C^{2\alpha}(\partial\Omega)$, there exist
$k$-plurisubharmonic functions $\\{\tilde{u}_{\varepsilon}\\}$ which decreases
to $u$ as $\varepsilon\to 0$ and satisfies [GKZ]

(5.18)
$\begin{cases}\tilde{u}_{\varepsilon}=u+C\varepsilon^{\alpha}&\text{in}\
\Omega\setminus\Omega_{\varepsilon},\\\\[4.0pt]
\hat{u}_{\varepsilon}\leq\tilde{u}_{\varepsilon}\leq\hat{u}_{\varepsilon}+C\varepsilon^{\alpha}&\text{in}\
\Omega_{\varepsilon},\end{cases}$

where the constant $C$ is independent of $\varepsilon$. Then if $u\in
W^{2,r}(\Omega)$, by choosing $v=\tilde{u}_{\varepsilon}$,
$\gamma^{\prime}<\frac{\delta r}{1+\delta r}$ in Proposition 5.5, and using
Lemma 5.7, we have

$\displaystyle\sup_{\Omega_{\varepsilon}}(\hat{u}_{\varepsilon}-u)$
$\displaystyle\leq$
$\displaystyle\sup_{\Omega}(\tilde{u}_{\varepsilon}-u)+C\varepsilon^{\alpha}$
$\displaystyle\leq$ $\displaystyle
C\|\tilde{u}_{\varepsilon}-u\|_{L^{r}}^{\gamma^{\prime}}+C\varepsilon^{\alpha}$
$\displaystyle\leq$ $\displaystyle C\|\triangle
u\|_{L^{r}(\Omega,\omega^{n})}^{\gamma^{\prime}}\varepsilon^{2\gamma^{\prime}}+C\varepsilon^{\alpha}.$

Hence, once we have $u\in W^{2,r}$ for $r\geq 1$, it holds $u\in
C^{\gamma^{\prime}}$ for $\gamma^{\prime}<\frac{\delta r}{1+\delta r}$, where
$\delta=\frac{2\gamma p-\gamma-k-1}{(k+1-\gamma)p}$ and
$p\leq\frac{n(k+1)}{n-k}$($p\geq 1$ when $k=n$).

Finally, we show that under the assumption of Theorem 1.6, it holds $u\in
W^{2,1}(\Omega)$, i.e., $\triangle u$ has finite mass, and hence $u\in
C^{\gamma^{\prime}}$ for $\gamma^{\prime}<\frac{\delta r}{1+\delta r}$. Let
$\rho$ be the smooth solution to

$\displaystyle\begin{cases}(dd^{c}\rho)^{k}\wedge\omega^{n-k}=\omega^{n},\ \
&\text{in}\ \Omega,\\\\[4.0pt] \rho=0,\ \ &\text{on}\
\partial\Omega.\end{cases}$

We can choose $A>>1$ such that $dd^{c}(A\rho)\geq\omega$. Then by the
generalized Cauchy-Schwartz inequality,

$\displaystyle\int_{\Omega}dd^{c}u\wedge\omega^{n-1}$ $\displaystyle\leq$
$\displaystyle\int_{\Omega}dd^{c}u\wedge[dd^{c}(A\rho)]^{n-1}$
$\displaystyle\leq$ $\displaystyle
A^{k-1}\left(\int_{\Omega}(dd^{c}\rho)^{k}\wedge\omega^{n-k}\right)\cdot\left(\int_{\Omega}(dd^{c}u)^{k}\wedge\omega^{n-k}\right)\leq
C.$

Hence we finish the proof.

######  Example 5.8.

As an example, we consider the Dirichlet problem (1.11) with the measure
$\mu_{S}$ defined by $\mu_{S}(E)={\mathcal{H}}^{2n-1}(E\cap S)$ for a real
hypersurface $S\subset{\mathbb{C}}^{n}$ and $E\subset\Omega$, where
${\mathcal{H}}^{2n-1}(\cdot)$ is the $2n-1$-Hausdorff measure. By Stein-Tomas
Restriction Theorem, for all $f\in C^{\infty}_{0}({\mathbb{C}}^{n})$, there
holds [BS]

$\|f|_{S}\|_{L^{2}(S,\mu_{S})}\leq C\|f\|_{L^{p}(\Omega)},\ \ \ 1\leq
p\leq\frac{4n+2}{2n+3}.$

Therefore, we can apply Theorem 1.6 with $\gamma=1$. This leads to $u\in
C^{\gamma^{\prime}}(\Omega)$ for

$\gamma^{\prime}<\frac{n+k+2}{(n+1)(k+2)}.$

## References

  * [AH] D.R. Adams and L. I. Hedberg, Function spaces and potential theory, Springer Science and Business Media, 2012.
  * [AC19] P. Ahag and R. Czyz, On the Moser-Trudinger inequality in complex space, J. Math. Anal. Appl. 479(2019), 1456-1474.
  * [AC20] P. Ahag and R. Czyz, Poincaré-and Sobolev-type inequalities for complex m-Hessian equations, Results in Math. 75(2020), 1-21.
  * [ACKZ] P. Ahag, U. Cegrell, S. Kołodziej, H. H. Phạm and A. Zeriahi, Partial pluricomplex energy and integrability exponents of plurisubharmonic functions, Adv. Math. 222(2009), 2036-2058.
  * [BT] E. Bedford and B. A. Taylor, A new capacity for plurisubharmonic functions, Acta Math. 149(1982), 1-40.
  * [B] Z. Blocki, The complex Monge-Ampère operator in pluripotential theory, Lecture notes, 1999.
  * [BB] R. J. Berman and B. Berndtsson, Moser-Trudinger type inequalities for complex Monge-Ampère operators and Aubin’s” hypothèse fondamentale”, arXiv:1109.1263, 2011.
  * [BB14] R. J. Berman and B. Berndtsson, Symmetrization of plurisubharmonic and convex functions, Indiana Univ. Math. J. 63(2014), 345-365.
  * [BS] J. G. Bak and A. Seeger, Extensions of the Stein-Tomas theorem, Math. Res. Let. 18(2011), 767-781.
  * [C06] U. Cegrell, Approximation of plurisubharmonic functions in hyperconvex domains. Acta Universitatis Upsaliensis, 2006, 86: 125-129.
  * [C19] U. Cegrell, Measures of finite pluricomplex energy, Ann. Polon. Math. 123(2019), no. 1, 203-213.
  * [DK] S. Dinew and S. Kołodziej, A priori estimates for complex Hessian equations, Anal. PDE 7(2014), 227-244.
  * [DKN] T. C. Dinh, S. Kołodziej and N.C. Nguyen, The complex Sobolev space and Hölder continuous solutions to Monge-Ampère equations, arXiv:2008.00260, 2020.
  * [DL] D. Di Nezza and C. H. Lu, Complex Monge-Ampère equations on quasi-projective varieties, Creille’s J. 727(2017), 145-167.
  * [DMV] T. C. Dinh, G. Marinescu and D.V. Vu, Moser-Trudinger inequalities and complex Monge-Ampère equation, arXiv:2006.07979, 2020.
  * [DN] T. C. Dinh and V. A. Nguyen, Characterization of Monge-Ampère measures with Hölder continuous potentials, J. Func. Anal. 266(2014), 67-84.
  * [DNS] T. C. Dinh, V. A. Nguyen and N. Sibony, Exponential estimates for plurisubharmonic functions, J. Diff. Geom. 84(2010), 465-488.
  * [GKY] V. Guedj, B. Kolev and N. Yeganefar, Kähler-Einstein fillings, J. Lond. Math. Soc. (2) 88(2013), no. 3, 737-760.
  * [GKZ] V. Guedj, S. Kołodziej and A. Zeriahi, Hölder continuous solutions to Monge-Ampère equations, Bull. London Math. Soc. 40(2008), 1070-1080.
  * [GZ] V. Guedj and A. Zeriahi, Degenerate complex Monge-Ampère equations, EMS Tracts in Math. 26(2017), Zurich.
  * [Hou] Z. L. Hou, Convexity of Hessian integrals and Poincaré type inequalities, Proc. Amer. Math. Soc. 138(2010), 2099-2105.
  * [HMT] J. A. Hempel, G. R. Morris and N. S. Trudinger, On the sharpness of a limiting case of the Sobolev imbedding theorem, Bull. Aust. Math. Soc. 3(1970), 369-373.
  * [HL09] F. R. Harvey and Jr. H. B. Lawson, $p$-convexity, $p$-plurisubharmonicity and the Levi problem. Indiana Univ. Math. J. 62(2013), 149-169.
  * [K96] S. Kołodziej, Some sufficient conditions for solvability of the Dirichlet problem for the complex Monge-Ampr̀e operator, Ann. Polon. Math. 65(1996), 11-21.
  * [K98] S. Kołodziej, The complex Monge-Ampère equation, Acta Math. 180(1998), 69-117.
  * [Ka] L. Kaufmann, A Skoda-type integrability theorem for singular Monge-Ampère measures, Michigan Math. J. 66(2017), 581-594.
  * [KR] N. Kerzman and J. P. Rosay, Fonctions plurisouscharmoniques d’exhaustion bornées et domaines taut, Math. Ann. 257(1981), 171-184.
  * [Lu] C. H. Lu, A variational approach to complex Hessian equations in ${\mathbb{C}}^{n}$, J. Math. Anal. Appl. 431(2015), 228-259.
  * [Ma] V. Maz’ya, Lectures on isoperimetric and isocapacitary inequalities in the theory of Sobolev spaces, Contemp. Math. 338(2003), 307-340.
  * [MP] V. Maz’ya and S. Poborchi, Differentiable functions on bad domains, World Sci. Publ., River Edge, NJ xx, 1997.
  * [Ngu] N. C. Nguyen, Hölder continuous solutions to complex Hessian equations, Potential Anal. 41(2014), 887-902.
  * [TiW] G.J. Tian and X.-J. Wang, Moser-Trudinger type inequalities for the Hessian equation, J. Func. Anal. 259(2010), 1974-2002.
  * [TrW] N. S. Trudinger and X.-J. Wang, A Poincaré type inequality for Hessian integrals, Cal. Var. PDEs 6(1998), 315-328.
  * [W1] X.-J. Wang, A class of fully nonlinear elliptic equations and related functionals, Indiana Univ. Math. J. 43(1994), 25-54.
  * [W2] X.-J. Wang, The k-Hessian Equation, Lecture Notes in Math., Vol. 1977, Springer, 2009.
  * [WWZ1] J. X. Wang, X.-J. Wang and B. Zhou, Moser-Trudinger inequality for the complex Monge-Ampère equation, J. Func. Anal. 279(2020), 108765.
  * [WWZ2] J. X. Wang, X.-J. Wang and B. Zhou, A priori estimates for the complex Monge-Ampère equation, Peking Math. J. 4(2021),143-157.
  * [XZ] J. Xiao and N. Zhang, Isocapacity Estimates for Hessian Operators, J. Func. Anal. 267(2014), 579-604.