# viscosity solution to complex Hessian equations on compact Hermitian
manifolds

Jingrui Cheng, Yulun Xu

###### Abstract.

We prove the existence of viscosity solutions to complex Hessian equations on
a compact Hermitian manifold that satisfy a determinant domination condition.
This viscosity solution is shown to be unique when the right hand is strictly
monotone increasing in terms of the solution. When the right hand side does
not depend on the solution, we reduces it to the strict monotonicity of the
solvability constant.

##  1\. introduction

The goal of this note is to study the existence and uniqueness of viscosity
solutions to complex Hessian equations on a closed Hermitian manifold.

Let $(M,\omega_{0})$ be a closed Hermitian manifold. In local coordinates, we
can write $\omega_{0}=\sqrt{-1}g_{i\bar{j}}dz_{i}\wedge d\bar{z}_{j}$. Let
$\chi$ be a real $(1,1)$ form on $M$ and in local coordinates we can write it
as: $\chi=\sqrt{-1}\chi_{i\bar{j}}dz_{i}\wedge d\bar{z}_{j}$. For any $C^{2}$
function $\varphi:M\rightarrow\mathbb{R}$, we obtain a new real $(1,1)$ form:
$\chi+dd^{c}\varphi$ and we consider equations for $\varphi$ that may be
written in the form:

(1.1) $F(\chi+dd^{c}\varphi)=h,\,\,\,h=e^{G(x)}\text{ or $e^{G(x,\varphi)}$},$

where $G(x)$ or $G(x,\varphi)$ is some right hand side one prescribes. The
reason we write the right hand side in the form $e^{G}$ is to emphasize that
it is strictly positive. In the above,

(1.2) $F(\chi+dd^{c}\varphi)=f(\lambda[\chi+dd^{c}\varphi]),$

where $\lambda[\chi+dd^{c}\varphi]$ are the eigenvalues of
$A^{i}_{j}=g^{i\bar{k}}\big{(}\chi_{j\bar{k}}+\varphi_{j\bar{k}}\big{)}$ and
$f(\lambda_{1},\cdots,\lambda_{n})$ is a smooth symmetric function. Such
equations have been studied extensively in the literature, going back to the
work of Caffarelli-Nirenberg-Spruck [1] on the Dirichlet problem in the real
case, when $\omega_{0}$ is the Euclidean metric and $M$ is a domain in
$\mathbb{R}^{n}$.

We assume that the function $f$ in (1.2) is defined in a closed convex
symmetric cone $\Gamma\subset\mathbb{R}^{n}$, with
$\Gamma\subset\\{\lambda\in\mathbb{R}^{n}:\sum_{i=1}^{n}\lambda_{i}>0\\}$ and
containing the first octant $\Gamma_{n}$. Therefore, we need to assume that

(1.3) $\lambda[\chi+dd^{c}\varphi]\in\Gamma.$

In addition, we are going to assume that:

######  Assumption 1.1.

  1. (1)

$\frac{\partial f}{\partial\lambda_{i}}\geq 0,\,\,1\leq i\leq n$, $f$ is
concave and $>0$ on $\Gamma$ and $f=0$ on $\partial\Gamma$.

  2. (2)

$\lambda[\chi](x)\in Int(\Gamma)$ for any $x\in M$.

  3. (3)

(determinant domination condition) $f$ is a positive homogeneous function with
degree 1, and there exist constants $c_{0}>0$, such that $f(\lambda)\geq
c_{0}(\Pi_{i=1}^{n}\lambda_{i})^{\frac{1}{n}}$ for
$\lambda\in\Gamma_{n}=\\{\lambda\in\mathbb{R}^{n}:\lambda_{i}>0,\,\,1\leq
i\leq n\\}$.

In the above, the assumption that $\frac{\partial f}{\partial\lambda_{i}}\geq
0$ implies the ellipticity of the equation (1.1), (1.3). The determinant
domination condition is motivated by the important work of Guo-Phong-Tong
[18], which developed a unified PDE approach to $L^{\infty}$ estimate which
satisfies:

(1.4) $\sum_{i}\lambda_{i}\frac{\partial f}{\partial\lambda_{i}}\leq
C_{0}f,\,\,\Pi_{i=1}^{n}\frac{\partial f}{\partial\lambda_{i}}\geq
c_{0},\text{ for some $c_{0},\,C_{0}>0$}.$

[18] also observed that determinant domination condition implies (1.4). There
are many examples which satisfy the Assumption 1.1 above. The most well-known
example is probably $f(\lambda)=\sigma_{k}^{\frac{1}{k}}(\lambda),\,1\leq
k\leq n$, defined on $\Gamma_{k}:=\\{\lambda:\sigma_{i}(\lambda)\geq 0,\,1\leq
i\leq k\\}$, where $\sigma_{k}(\lambda)$ is the $k$-th symmetric polynomial of
$\lambda$. These $\sigma_{k}$ equations have been extensively studied. See
[5], [6], [14], [15] and references therein. However, there are other examples
satisfying Assumption 1.1 which are less studied, and we just name a few here:

  1. (1)

($\sigma_{k}$-equation for $(n-1)-$plurisubharmonic function)
$f(\lambda)=\sigma_{k}^{\frac{1}{k}}(\tilde{\lambda})$, where
$\tilde{\lambda}_{i}=\frac{1}{n-1}\sum_{j\neq i}\lambda_{j}$.

  2. (2)

($p$-fold sum operator)
$f(\lambda)=\big{(}\Pi_{|J|=p}\lambda_{J}\big{)}^{\frac{1}{N}}$, where
$\lambda_{J}=\lambda_{j_{1}}+\lambda_{j_{2}}+\cdots+\lambda_{j_{p}}$ and
$N={n\choose p}$.

The first example with $k=n$ was studied by Tosatti-Weinkove [21] and the
second example was considered by Harvey-Lawson [13] related to
$p$-geometry/$p$-potential theory. Moreover, we have the following general
result due to Leonid Gurvits [11] (see also [12] for a proof) that produces a
large number of examples of $f$ satisfying the determinant domination
condition.

######  Proposition 1.2.

Let $p(x)$ be a homogeneous polynomial of degree $N$ on $\mathbb{R}^{n}$.
Denote $e=(1,\cdots,1)\in\mathbb{R}^{n}$. Assume that:

  1. (1)

All coefficients of $p$ are $\geq 0$,

  2. (2)

$p(e)>0$,

  3. (3)

$\frac{\partial p}{\partial x_{1}}(e)=\frac{\partial p}{\partial
x_{2}}(e)\cdots=\frac{\partial p}{\partial x_{n}}(e)=k$ for some $k>0$.

Then $p(x)^{\frac{1}{N}}\geq c(x_{1}\cdots x_{n})^{\frac{1}{n}}$ for some
$c>0$ on $\\{x_{1}>0,\cdots,x_{n}>0\\}$.

However, the Inverse $\sigma_{k}$-equation does not satisfy the above
determinant domination condition.

For a general Hessian equation without the determinant domination condition,
the apriori estimates will usually require the existence of a subsolution.
Szekelyhidi [20] derived apriori estimates up to $C^{2,\alpha}$ assuming the
existence of a $\mathcal{C}$-subsolution. In order to prove existence of
solutions, the recent work of Guo-Song [10] shows that one needs a more
delicate notion of subsolution. It is no trivial issue to determine whether
such subsolutions exist. Assuming determinant dominationion condition
alleviates this issue since we will always have 0 as a subsolution.

Since there has been many works on the solvability in the smooth category, it
is a natural question to find weak solutions. Most of the previous works have
centered around $\sigma_{k}$-equation using pluripotential theory. The work by
Lu [16] studied the existence and uniqueness of viscosity solutions to
$\sigma_{k}$-equations on bounded domains in $\mathbb{C}^{n}$ as well as
homogeneous Hermitian manifolds. We generalize this result and prove:

######  Theorem 1.1.

Assume that Assumption 1.1 holds:

  1. (1)

Let $G\in C(M)$, then there exists a constant $c\in\mathbb{R}$, and
$\varphi\in C(M)$ that solves the following equation in the viscosity sense:

$F(\chi+dd^{c}\varphi)=e^{G+c},\,\,\,\lambda[\chi+dd^{c}\varphi]\in\Gamma.$

  2. (2)

Let $G(x,u)\in C(M\times\mathbb{R})$. Assume that $G$ is monotone increasing
in $u$, and $f(\lambda[\chi])(x)<e^{G(x,C_{0})}$ for some $C_{0}\in\mathbb{R}$
and any $x\in M$. Then there exists $\varphi\in C(M)$ that solves the
following equation in the viscosity sense:

$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)},\,\,\,\lambda[\chi+dd^{c}\varphi]\in\Gamma.$

For uniqueness, we have:

######  Theorem 1.2.

Let $G\in C(M\times\mathbb{R})$. Assume that $G(x,u_{1})<G(x,u_{2})$ for any
$x\in M$ and $u_{1}<u_{2}$. Then there exists at most one viscosity solution
to:

$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)},\,\,\lambda[\chi+dd^{c}\varphi]\in\Gamma.$

When the right hand side does not depend on $\varphi$, the uniqueness is more
subte. As the first step, we prove:

######  Proposition 1.3.

Assume that Assumption 1.1 holds. Let $G\in C(M)$, then there exists a unique
$c\in\mathbb{R}$ such that the following equation is solvable in the viscosity
sense:

$F(\chi+dd^{c}\varphi)=e^{G+c},\,\,\lambda[\chi+dd^{c}\varphi]\in\Gamma.$

We are not able to prove the uniqueness of the solution $\varphi$ and the main
obstable seems to be a lack of understanding of the constant $c$ that makes
the equation solvable in the viscosity sense. Indeed, for $G\in C(M)$, we may
denote $c(G)$ to be the above said (unique) constant. It is not very hard to
see that $G_{1}\geq G_{2}$ implies $c(G_{1})\leq c(G_{2})$. The question that
is of interest to us is whether this monotonicity is strict. More precisely:

######  Question 1.4.

Assume that $G_{1}\geq G_{2}$, $G_{1}\neq G_{2}$. Do we have
$c(G_{1})<c(G_{2})$?

We show that an affirmative answer to Question 1.4 will lead to the uniqueness
of viscosity solution. More precisely:

######  Proposition 1.5.

Let $G\in C(M)$. Assume that for any $G^{\prime}\leq G,\,G^{\prime}\neq G$ one
has $c(G^{\prime})>c(G)$, then there is at most one solution to the following
equation in the viscosity sense:

$F(\chi+dd^{c}\varphi)=e^{G+c},\,\,\lambda[\chi+dd^{c}\varphi]\in\Gamma,\,\,\sup_{M}\varphi=0.$

As to Question 1.4, we observe that the answer is yes if $Q_{1}$ or $Q_{2}$ is
smooth:

######  Proposition 1.6.

If $Q_{1}$ or $Q_{2}$ is smooth, then the answer to Question 1.4 is
affirmative.

Next we explain our strategy of proof of the above results. For the existence
proof, we first approximates the right hand side $G$ with a sequence of smooth
right hand side $G_{j}$. The apriori estimates developed in Szekelyhidi [20]
allows us to solve:

$F[\chi+dd^{c}\varphi_{j}]=e^{G_{j}(x)+c_{j}}\text{ or
$e^{G_{j}(x,\varphi)}$}.$

In order to get a viscosity solution, all we need is to prove that
$\varphi_{j}$ converges uniformly (at least up to a subsequence). Using that
$\varphi_{j}$ are all subharmonic and that $\varphi_{j}$ are uniformly
bounded, we see that $\varphi_{j}$ is precompact in $L^{1}$. In order to
improve the convergence to uniform convergence, we need the following
stability estimate (roughly stated): there exists $a>0$, such that for any
$v\in C^{2}(M)$, $\lambda[\chi+dd^{c}v]\in\Gamma$,

(1.5) $\sup_{M}(v-\varphi)\leq C||(v-\varphi)_{+}||_{L^{1}}^{a}.$

The proof of (1.5) is really a variant of the $L^{\infty}$ estimate by Guo-
Phong-Tong [18].

For the uniqueness proof, we need to consider the sup/inf convolution, adapted
to manifolds. We use the sup/inf convolution considered in Cheng-Xu[2], and
show that it gives a semi-convex/concave approximation of the viscosity
solution.

Finally we explain the organization of this paper.

In Section 2, we explain some basic notations , definitions and some
preliminary results that we need later on.

In Section 3, we prove the existence of solution with smooth right hand side.

In Section 4, we prove the stability result, that allows us to improve the
$L^{1}$ convergence to $L^{\infty}$ convergence, thereby proving the existence
of a viscosity solution.

In Section 5, we address the uniqueness issues.

##  2\. Notations and preliminaries

In the following, we denote
$d^{c}=\frac{\sqrt{-1}}{2}(\bar{\partial}-\partial)$, so that one has
$\sqrt{-1}\partial\bar{\partial}=dd^{c}$. The advantage of working with $d$
and $d^{c}$ is that they are real operators.

The notion of viscosity solution we work with is consistent with the usual
definition in the literature:

######  Definition 2.1.

Let $\varphi\in C(M)$ and $\Gamma$ be a closed convex symmetric cone in
$\mathbb{R}^{n}$ that contains the first octant. We say that
$\lambda[\chi+dd^{c}\varphi]\in\Gamma$ in the viscosity sense, if for any
$x_{0}\in M$, and $C^{2}$ function $P$ defined in a neighborhood of $x_{0}$
touching $\varphi$ from above at $x_{0}$, one has:

$\lambda[\chi+dd^{c}P](x_{0})\in\Gamma.$

Sometimes we call such a function to be $\Gamma$-subharmonic.

Let $f(\lambda_{1},\cdots,\lambda_{n})$ and $\Gamma$ be as described in
Section 1, we will put $F(\chi+dd^{c}\varphi)=f(\lambda[\chi+dd^{c}\varphi])$.
We may interchangably use both notations in the following.

######  Definition 2.2.

Let $\varphi$ be a $\Gamma$-subharmonic function and $G(x,u)\in
C(M\times\mathbb{R})$. We say that $\varphi$ is a viscosity solution to:

$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)},\,\,\,\,\lambda[\chi+dd^{c}\varphi]\in\Gamma.$

if the following hold:

  1. (1)

For any $x_{0}\in M$ and any $C^{2}$ function $P$ defined in a neighborhood of
$x_{0}$ that touches $\varphi$ from above at $x_{0}$, one has:

$F(\chi+dd^{c}P)(x_{0})\geq e^{G(x_{0},P(x_{0}))}.$

  2. (2)

For any $x_{0}\in M$ and any $C^{2}$ function $P$ defined in a neighborhood of
$x_{0}$ that touches $\varphi$ from below at $x_{0}$, either
$\lambda[\chi+dd^{c}P](x_{0})\notin\Gamma$, or $F[\chi+dd^{c}P](x_{0})\leq
e^{G(x_{0},P(x_{0}))}$.

We will also need the following theorem of Gaudochon [9]:

######  Theorem 2.1.

Let $\omega$ be a Hermitian metric on $M$. Then there exists a unique function
$v\in C^{\infty}(M)$ such that $\inf_{M}v=0$,
$dd^{c}\big{(}e^{(n-1)v}\omega^{n-1}\big{)}=0$.

In general, let $\chi$ be a real $(1,1)$ form on $M$ with
$\lambda[\chi]\in\Gamma$. We choose coordinates on an open subset of $M$.
Define:$F^{i\bar{j}}=\big{(}\frac{\partial F}{\partial
h_{a\bar{b}}}(\chi)\big{)}^{-1}_{ji}.$ That is, $F^{i\bar{j}}\frac{\partial
F}{\partial h_{q\bar{j}}}(\chi)=\delta_{iq}$. We define:

(2.1) $\Omega=\sqrt{-1}F^{i\bar{j}}dz_{i}\wedge d\bar{z}_{j}.$

Then one can verify that the above defined $\Omega$ is actually independent of
the choice of coordinates. Moreover, from ellipticity, one sees that
$\Omega>0$, hence $\Omega$ defines a Hermitian metric. Another thing we note
that the function $\det g_{i\bar{j}}\det(\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi))$ is also independent of the choice of coordinates.
Moreover, for any $u\in C^{2}(M)$, the following formula holds:

(2.2) $\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi)u_{i\bar{j}}\frac{\omega_{0}^{n}}{n!}=\det
g_{i\bar{j}}\det(\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi))\frac{\Omega^{n-1}}{(n-1)!}\wedge dd^{c}u.$

##  3\. existence of solution with smooth right hand side

In this section, our goal is to establish:

######  Theorem 3.1.

  1. (1)

Assume that $G(x,u)$ is smooth, and $G_{u}(x,u)>0$. Then there exists a unique
solution to

$F\big{(}\chi+dd^{c}\varphi\big{)}=e^{G(x,\varphi)}.$

  2. (2)

Assume that $G(x)$ is smooth, then there is a unique $c\in\mathbb{R}$ and
$\varphi\in C^{\infty}(M)$ that solves:

$F(\chi+dd^{c}\varphi)=e^{G(x)+c},\,\,\sup_{M}\varphi=0.$

The above result probably exists somewhere in the literature, but we were
unable to locate the exact reference. The uniqueness part (of both $\varphi$
and $c$) is an easy consequence of maximum principle.

For the existence part, we are going to use a continuity path. The continuity
path when $G_{u}>0$ will be:

(3.1)
$F(\chi+dd^{c}\varphi)=e^{(1-t)(\varphi+B_{0}(x))+tG(x,\varphi)},\,\,t\in[0,1].$

If we denote $G(t,x,u)=(1-t)(u+B_{0}(x))+tG(x,u)$, then we see that
$G_{u}(t,x,u)$ remains positive. In the above, $C_{0}(x)$ is chosen so as to
make sure $\varphi=0$ solves the equation with $t=0$, namely:

$F(\chi)=e^{B_{0}(x)}.$

When the right hand side does not depend on $\varphi$, we are going to use the
following continuity path:

(3.2) $F(\chi+dd^{c}\varphi)=e^{(1-t)B_{0}(x)+tG(x)+c_{t}},\,\,t\in[0,1].$

Note that in the above, there is a unique constant $c_{t}$ that makes the
equation solvable, for each $t\in[0,1]$. Clearly $c_{0}=0$ and $\varphi=0$
when $t=0$. It only remains to establish the openness and closeness.

###  3.1. Openness of the coutinuity path

First we consider the openness of (3.1). We are going to set up the nonlinear
mapping as follows:

$\begin{split}\mathcal{F}:&\mathbb{R}\times C^{k,\alpha}(M)\rightarrow
C^{k-2,\alpha}(M)\\\
&(t,\varphi)\mapsto\log\big{(}F(\chi+dd^{c}\varphi)\big{)}-((1-t)(\varphi+B_{0}(x))+tG(x,\varphi)).\end{split}$

In the above, $k$ is sufficiently large. By implicit function theorem, all we
need is to verify that $D_{\varphi}\mathcal{F}(t_{0},\varphi_{t_{0}})$ is an
invertible map from $C^{k,\alpha}(M)$ to $C^{k-2,\alpha}(M)$, where
$t_{0}\in[0,1]$ and $\varphi_{t_{0}}$ is the solution to (3.1) corresponding
to $t_{0}$. We can compute that:

$\begin{split}D_{\varphi}\mathcal{F}(t_{0},\varphi_{t_{0}}):&C^{k,\alpha}(M)\rightarrow
C^{k-2,\alpha}(M),\\\
&\psi\mapsto\frac{1}{F(\chi+dd^{c}\varphi_{t_{0}})}\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi+dd^{c}\varphi_{t_{0}})\partial_{i\bar{j}}\psi-((1-t)+tG_{u})\psi.\end{split}$

Let us denote $L=D_{\varphi}\mathcal{F}(t_{0},\varphi_{0})$ and we have:

######  Lemma 3.1.

The operator $L$ is invertible from $C^{k,\alpha}(M)$ to $C^{k-2,\alpha}(M)$.

###### Proof.

Injectivity is quite easy to see. Indeed, one can look at the point where
$\psi$ achieves positive maximum or negative minimum. One can take a
coordinate chart near that point, so that locally $Lu=0$ could be written as:

$a_{ij}(x)\partial_{ij}\psi+c\psi=0,\,\,c<0.$

We see by (strong) maximum principle that $u$ must be a constant and clearly
this constant must be zero.

To prove surjectivity, we only need to show that $L$ is surjective from
$H^{2}$ to $L^{2}$. With the right hand side in $C^{k-2,\alpha}(M)$, one can
use the standard elliptic regularity theory and work locally to improve the
pre-image to be in $C^{k,\alpha}(M)$. In order to show the surjectivity of $L$
from $H^{2}$ to $L^{2}$, it will not work well to consider $L$, but we will
need to consider $\tilde{L}=e^{(n-1)v_{0}}L$, where $v_{0}$ is the Gaudochon
factor of the following Hermitian metric:

(3.3) $\tilde{\Omega}=F^{-\frac{1}{n-1}}(\chi+dd^{c}\varphi_{t_{0}})(\det
g\det(\frac{\partial F}{\partial
h_{i\bar{j}}})(\chi+dd^{c}\varphi_{t_{0}}))^{\frac{1}{n-1}}\Omega.$

That is, $dd^{c}\big{(}e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\big{)}=0$. In the
above, $\Omega$ is defined in (2.1), evaluated at
$\chi+dd^{c}\varphi_{t_{0}}$. The existence of $v_{0}$ was first proved by
Gaudochon [9] ( Fu-Wang-Wu also give a proof in [8]). We will be able to show
that $\tilde{L}$ is surjective (so $L$ is surjective) if we can show that:

  1. (1)

$Im(\tilde{L})$ is a closed subspace of $L^{2}$. For this we just need to show
$||\psi||_{H^{2}}\leq C||\tilde{L}\psi||_{L^{2}}$.

  2. (2)

$Ker(\tilde{L}^{*})=0$, where $\tilde{L}^{*}:L^{2}\rightarrow H^{-2}$ is the
adjoint map of $\tilde{L}$ and $H^{-2}$ is the dual space of $H^{2}$.

To prove (1), first we observe that $L$ can be re-written as:

$L\psi=\frac{1}{F}\frac{\det g\det(\frac{\partial F}{\partial
h_{i\bar{j}}})\frac{\Omega^{n-1}}{(n-1)!}\wedge
dd^{c}\psi}{\frac{1}{n!}\omega_{0}^{n}}-((1-t)+tG_{u})\psi.$

Therefore

(3.4)
$(\tilde{L}\psi)\frac{1}{n!}\omega_{0}^{n}=e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\wedge
dd^{c}\psi-e^{(n-1)v_{0}}((1-t)+tG_{u})\psi\frac{1}{n!}\omega_{0}^{n}.$

Now we multiply both sides by $\psi$ and integrate, we see that:

$\begin{split}&\int_{M}\psi\tilde{L}\psi\frac{1}{n!}\omega_{0}^{n}=\int_{M}e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\wedge(\psi
dd^{c}\psi)-\int_{M}e^{(n-1)v_{0}}((1-t)+tG_{u})\psi^{2}\frac{\omega_{0}^{n}}{n!}\\\
&=\int_{M}e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\wedge(dd^{c}(\frac{\psi^{2}}{2})-d\psi\wedge
d^{c}\psi)-\int_{M}e^{(n-1)v_{0}}((1-t)+tG_{u})\psi^{2}\frac{\omega_{0}^{n}}{n!}\\\
&=-\int_{M}e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\wedge d\psi\wedge
d^{c}\psi-\int_{M}e^{(n-1)v_{0}}((1-t)+tG_{u})\psi^{2}\frac{\omega_{0}^{n}}{n!}.\end{split}$

At this point, Using the Cauchy-Schwarz inequality, we see that:

(3.5) $||\psi||_{L^{2}}+||\nabla\psi||_{L^{2}}\leq
C||\tilde{L}\psi||_{L^{2}}.$

To get the estimate for the second derivative, we go back to (3.4) and re-
write it as:

$\begin{split}&d\big{(}e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\wedge
d^{c}\psi\big{)}=(\tilde{L}\psi)\frac{1}{n!}\omega_{0}^{n}+d\big{(}e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\big{)}\wedge
d^{c}\psi\\\
&+e^{(n-1)v_{0}}((1-t)+tG_{u})\psi\frac{\omega_{0}^{n}}{n!}.\end{split}$

Note that the right hand side is in $L^{2}$ now. Writing this equation in
local coordinates would be of the form:

$\partial_{i}(a_{ij}\partial_{j}u)=k,\,\,k\in L^{2}.$

The standard estimate would give: $||u||_{H^{2}(B_{\frac{1}{2}})}\leq
C(||u||_{L^{2}(B_{1})}+||k||_{L^{2}(B_{1})})$. This estimate combined with
(3.5) gives what we need.

Next we show that $\tilde{L}^{*}$ is injective. we can compute that:

$(\tilde{L}^{*}v)=\frac{dd^{c}\big{(}ve^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\big{)}}{\frac{1}{n!}\omega_{0}^{n}}-e^{(n-1)v_{0}}((1-t)+tG_{u})v,\,\,v\in
L^{2}.$

If $\tilde{L}^{*}v=0$, then this equation written locally takes the form:

$\partial_{ij}(a_{ij}v)-hv=0,\,\,h\in C^{\infty}(B_{1}),a_{ij}\in
C^{\infty},\,\,v\in L^{2}.$

Using elliptic theory, we can improve the regularity of $v$ to $C^{\infty}$,
but then $v$ satisfies:

$e^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\wedge dd^{c}v+2dv\wedge
d^{c}(e^{(n-1)v_{0}}\tilde{\Omega}^{n-1})-e^{(n-1)v_{0}}((1-t)+tG_{u})v\frac{\omega_{0}^{n}}{n!}=0.$

In local coordinates, the above equation reads:

$a_{ij}\partial_{ij}v+b_{i}\partial_{i}v+cv=0,\,\,c<0.$

If one looks at this equation in a neighborhood of the point where $v$
achieves positive maximum or negative minimum, we see from strong maximum
principle that $v=0$. ∎

Now we consider the openness of (3.2). We set up the nonlinear mapping in a
similar way:

$\begin{split}\mathcal{F}&:\mathbb{R}\times\mathbb{R}\times
C_{0}^{k,\alpha}(M)\rightarrow C^{k-2,\alpha}(M)\\\
&(t,c,\varphi)\mapsto\log\big{(}F(\chi+dd^{c}\varphi)\big{)}-(1-t)B_{0}(x)-tG(x)-c.\end{split}$

In the above:

$C_{0}^{k,\alpha}(M)=\\{h\in C^{k,\alpha}(M):\int_{M}h\omega_{0}^{n}=0\\}.$

Assume that (3.7) is solvable with $t=t_{0}$, that is, there exist
$(c_{t_{0}},\varphi_{t_{0}})$ such that
$\mathcal{F}(t_{0},c_{t_{0}},\varphi_{t_{0}})=0$. By implicit function
theorem, we just need to verify the linearized map
$D_{(c,\varphi)}\mathcal{F}$ defines a bijective map from $\mathbb{R}\times
C_{0}^{k,\alpha}(M)$ to $C^{k-2,\alpha}(M)$. One can compute that:

$\begin{split}D_{(c,\varphi)}\mathcal{F}&:\mathbb{R}\times
C_{0}^{k,\alpha}(M)\rightarrow C^{k-2,\alpha}(M)\\\
&(\lambda,u)\mapsto\frac{1}{F(\chi+dd^{c}\varphi_{t_{0}})}\frac{\partial
F}{\partial
h_{i\bar{j}}}(\chi+\sqrt{-1}\partial\bar{\partial}\varphi_{t_{0}})\partial_{i\bar{j}}u-\lambda.\end{split}$

Denote this linear operator to be $\mathcal{L}$. We then have:

######  Lemma 3.2.

$\mathcal{L}$ is bijective from $\mathbb{R}\times C_{0}^{k,\alpha}(M)$ to
$C^{k-2,\alpha}(M)$.

###### Proof.

First, similar as before, $\mathcal{L}$ may be written as:

$\mathcal{L}(\lambda,u)=\frac{1}{F}\frac{\det g\det(\frac{\partial F}{\partial
h_{i\bar{j}}})\frac{\Omega^{n-1}}{(n-1)!}\wedge
dd^{c}u}{\frac{1}{n!}\omega_{0}^{n}}-\lambda=\frac{\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\wedge
dd^{c}u}{\frac{1}{n!}\omega_{0}^{n}}-\lambda.$

Here $\tilde{\Omega}$ is defined by (3.3). Assume that
$\mathcal{L}(\lambda,u)=0$, we need to show that $\lambda=0,\,u=0$. Let
$v_{0}$ be the Gauduchon factor of $\tilde{\Omega}$, so that
$dd^{c}(e^{(n-1)v_{0}}\tilde{\Omega}^{n-1})=0$. Multiplying by
$e^{(n-1)v_{0}}$ and integrating gives:

$0=\int_{M}e^{(n-1)v_{0}}\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\wedge
dd^{c}u-\int_{M}\lambda
e^{(n-1)v_{0}}\frac{\omega_{0}^{n}}{n!}=-\lambda\int_{M}e^{(n-1)v_{0}}\frac{\omega_{0}^{n}}{n!}.$

This proves $\lambda=0$, so that one has $\tilde{\Omega}^{n-1}\wedge
dd^{c}u=0$. Then one can use the strong maximum principle (similar to the
argument of part (1)) to show that $u$ is a constant. But we are requiring its
integral $=0$, so $u=0$.

Next we show that $\mathcal{L}$ is surjective. Just as before, it will be
sufficient to show that $\mathcal{L}$ is surjective from $\mathbb{R}\times
H_{0}^{2}$ to $L^{2}$. Here $H_{0}^{2}$ denotes the element $u\in H^{2}$ such
that $\int u\omega_{0}^{n}=0$. Also it will be necessary to consider
$\tilde{\mathcal{L}}(\lambda,u)=e^{(n-1)v_{0}}\tilde{\mathcal{L}}$. We will be
able to show $\tilde{\mathcal{L}}$ is surjective (hence $\mathcal{L}$), if we
can show:

  1. (1)

$Im(\tilde{\mathcal{L}})$ is a closed subspace of $L^{2}$. For this we just
need to show that $||u||_{H^{2}}+|\lambda|\leq
C||\tilde{\mathcal{L}}(u,\lambda)||_{L^{2}}$.

  2. (2)

$Ker(\tilde{\mathcal{L}}^{*})=0$, where
$\tilde{\mathcal{L}}^{*}:L^{2}\rightarrow H^{-2}\times\mathbb{R}$ is the
adjoint map of $\mathcal{L}$.

To see (1), first note that one has:

(3.6)
$\tilde{\mathcal{L}}(\lambda,u)\frac{\omega_{0}^{n}}{n!}=e^{(n-1)v_{0}}\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\wedge
dd^{c}u-\lambda e^{(n-1)v_{0}}\frac{\omega_{0}^{n}}{n!}.$

Integrating both sides, we get $|\lambda|\leq
C||\tilde{\mathcal{L}}(\lambda,u)||_{L^{1}}\leq
C^{\prime}||\tilde{\mathcal{L}}(\lambda,u)||_{L^{2}}$. Then we multiply (3.6)
by $u$, and use that $udd^{c}u=dd^{c}(\frac{u^{2}}{2})-du\wedge d^{c}u$, we
integrate and find that:

$\int_{M}u\tilde{L}(\lambda,u)\frac{\omega_{0}^{n}}{n!}=-\int_{M}e^{(n-1)v_{0}}\tilde{\Omega^{n-1}}{(n-1)!}\wedge
du\wedge d^{c}u-\int_{M}\lambda ue^{(n-1)v_{0}}\frac{\omega_{0}^{n}}{n!}.$

This way we get:

$||\nabla
u||_{L^{2}}^{2}\leq\varepsilon||u||_{L^{2}}^{2}+C_{\varepsilon}(|\lambda|^{2}+||\tilde{\mathcal{L}}(\lambda,u)||_{L^{2}}^{2}),\,\,\,\forall\varepsilon>0.$

On the other hand, since $\int_{M}u\omega_{0}^{n}=0$, we may use Poincare
inequality to see that $||\nabla u||_{L^{2}}\geq c_{0}||u||_{L^{2}}$. This way
we obtain that

$||u||_{L^{2}}+||\nabla u||_{L^{2}}\leq
C||\tilde{\mathcal{L}}(\lambda,u)||_{L^{2}}.$

Then we get the second derivative estimates by writing (3.6) under local
coordinates, and argue in the same way as part (1).

Next, we can find the adjoint map is:

$\begin{split}\tilde{\mathcal{L}}^{*}&:L^{2}\rightarrow\mathbb{R}\times
H^{-2}\\\
&g\mapsto(-\int_{M}ge^{(n-1)v_{0}}\frac{\omega_{0}^{n}}{n!},dd^{c}\big{(}ge^{(n-1)v_{0}}\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\big{)}).\end{split}$

If $dd^{c}\big{(}ge^{(n-1)v_{0}}\tilde{\Omega}^{n-1}\big{)}=0$ with $g\in
L^{2}$, then one can improve the regularity of $g$ and see that $g\in
C^{\infty}$. But then

$0=\int_{M}gdd^{c}(ge^{(n-1)v_{0}}\tilde{\Omega}^{n-1})=-\int_{M}e^{(n-1)v_{0}}dg\wedge
d^{c}g\wedge\tilde{\Omega}^{n-1}.$

This implies $g$ is a constant. On the other hand, from
$\int_{M}ge^{(n-1)v_{0}}\omega_{0}^{n}=0$, we see that $g=0$. This proves
$\tilde{\mathcal{L}}^{*}$ is injective. ∎

###  3.2. Closeness of the continuity path

The required apriori estimates we need can be found in Szekelyhidi [20], who
proved that:

######  Theorem 3.2.

Consider the Hessian equation
$f\big{(}\lambda[\chi+\sqrt{-1}\partial\bar{\partial}u]\big{)}=h(x)$, where
$f$ and $h$ satisfy the following assumptions:

  1. (1)

$\frac{\partial f}{\partial\lambda_{i}}>0$ and $f$ is convex.

  2. (2)

$\sup_{\partial\Gamma}f<\inf_{M}h$,

  3. (3)

For any $\sigma<\sup_{\Gamma}f$ and $\lambda\in\Gamma$ we have
$\lim_{t\rightarrow\infty}f(t\lambda)>\sigma$.

Suppose $u$ is a (smooth) solution with $\sup_{M}u=0$ and $\underline{u}$ is a
$\mathcal{C}$-subsolution, then we have an estimate $||u||_{2,\alpha}\leq C$,
where $C$ depends on the given data $M,\,g,\,\chi,\,h$ and the subsolution
$\underline{u}$.

We are going to apply this result to (3.2). Note that by evaluating at the
minimum and maximum of $\varphi$, we see that $c_{t}$ is actually uniformly
bounded in $t$. So we would be able to apply Theorem 3.2 if we can show the
existence of a $\mathcal{C}$-subsolution.

Indeed, a $\mathcal{C}$-subsolution simply means a (smooth) function
$\underline{u}$, such that for each $x\in M$, the following set is bounded:

(3.7) $\\{\lambda^{\prime}\in\Gamma:f(\lambda^{\prime})=h(x),\,\,\text{ and
}\lambda^{\prime}-\lambda(\chi+\sqrt{-1}\partial\bar{\partial}\underline{u})\in\Gamma_{n}\\}.$

In our setting, the situation is simple and we are going to see that
$\underline{u}=0$ will be a $\mathcal{C}$-subsolution. More precisely:

######  Lemma 3.3.

For any $C_{0}>0$ and any $x\in M$, the following set is bounded:
$\\{\lambda^{\prime}\in\Gamma:f(\lambda^{\prime})\leq C_{0},\text{ and
}\lambda^{\prime}-\lambda(\chi)\in\Gamma_{n}\\}$. Moreover, one can estimate
the diameter in terms of $f$, $C_{0}$, $\chi$ and the background metric $g$.

###### Proof.

Denote
$\Gamma_{\infty}=\\{(\lambda_{1},\cdots,\lambda_{n-1}):\text{$(\lambda_{1},\cdots,\lambda_{n})\in\Gamma$
for some $\lambda_{n}$}\\}.$ For any
$\lambda^{\prime}=(\lambda_{1},\cdots,\lambda_{n-1})\in\Gamma_{\infty}$, we
consider the limit

$\lim_{\lambda_{n}\rightarrow+\infty}f(\lambda_{1},\cdots,\lambda_{n}).$

Trudinger [22] proved that either this limit is infinite for all
$\lambda^{\prime}\in\Gamma_{\infty}$ or this limit is finite for all
$\lambda^{\prime}\in\Gamma_{\infty}$. Moreover, this convergence is uniform on
any compact subset of $\Gamma_{\infty}$. The proof essentially follows from
the concavity of $f$. In our setting, we will have the above limit is infinite
for all $\lambda^{\prime}\in\Gamma_{\infty}$. Indeed, since $f$ satisfies the
determinant domination condition, we see that, for
$\lambda_{1}>0,\cdots,\lambda_{n-1}>0$, we have:
$f(\lambda_{1},\cdots,\lambda_{n})\geq
c\big{(}\Pi_{i=1}^{n}\lambda_{i}\big{)}^{\frac{1}{n}}\rightarrow+\infty$ as
$\lambda_{n}\rightarrow+\infty$.

From this, we see that, for each $1\leq i\leq n$, we have:

$\lim_{t\rightarrow+\infty}f\big{(}\lambda[\chi]+t\mathbf{e}_{i}\big{)}=+\infty,$

where $\mathbf{e}_{i}$ is the standard basis in $\mathbb{R}^{n}$. Moreover,
since $\lambda[\chi]$ is strictly contained in $Int(\Gamma)$ as $x$ varies
over $M$, the above convergence is uniform on $M$. Using the concavity of $f$
again, we see that:

$\lim_{\mu\in\Gamma_{n},\,\mu\rightarrow\infty}f\big{(}\lambda[\chi]+\mu\big{)}=+\infty,$

and this convergence is uniform on $M$. ∎

Therefore, we may use Theorem 3.2 to conclude that $||\varphi||_{2,\alpha}$ is
uniformly bounded in (3.2). From the standard elliptic theory, we see that
$\varphi$ is uniformly bounded in any higher order norm.

Now let us consider the continuity path (3.1). First we show that $\varphi$ is
bounded from above, uniform for $t\in[0,1]$. Indeed, we fix any $t\in[0,1]$,
and assume that $\varphi$ achieves maximum at $x_{0}$, then at this point, we
have

(3.8) $F(\chi)(x_{0})\geq
F(\chi+dd^{c}\varphi)(x_{0})=e^{(1-t)(\varphi+B_{0}(x_{0}))+tG(x_{0},\varphi)}.$

Our assumption on $G$ was that there exists $C_{0}>0$ large enough, such that
for all $\varphi\geq C_{0}$, one has

$F(\chi)(x)<e^{G(x,\varphi)},\,\,\varphi\geq C_{0}.$

Therefore, there exists $C_{0}^{\prime}$, such that for any
$t\in[0,1],\,\varphi\geq C_{0}^{\prime}$, one has:

(3.9) $F(\chi)(x)<e^{(1-t)(\varphi+B_{0}(x))+tG(x,\varphi)}.$

(3.8) and (3.9) combined shows that $\varphi(x_{0})\leq C_{0}^{\prime}$, hence
$\sup_{M}\varphi\leq C_{0}^{\prime},\,\,t\in[0,1]$.

Once we have a bound of $\sup_{M}\varphi$, we see that the right hand side of
(3.1) is uniformly bounded. Using Lemma 3.3, we see that the following set is
uniformly bounded in $x\in M$ and $t\in[0,1]$:

$\\{\lambda^{\prime}\in\Gamma:\,\,f(\lambda^{\prime})=e^{(1-t)(\varphi+B_{0}(x))+tG(x,\varphi)}\text{
and $\lambda^{\prime}-\lambda[\chi]\in\Gamma_{n}$}\\}$

Therefore, Theorem 3.2 carries over in this case (indeed, the monotone
increasing dependence on $\varphi$ actually helps with the estimates) and
gives $||\varphi||_{2,\alpha}$ is uniformly bounded. From standard elliptic
estimates, we see that $\varphi$ is uniformly bounded in any higher order
norm.

##  4\. stability estimate and existence of viscosity solution

The key stability result we need is the following:

######  Proposition 4.1.

Let $v$ be a bounded smooth $\Gamma$-subharmonic function, and let $\varphi$
be a smooth solution to (1.1), (1.3) with $h=e^{G_{0}(x)}$ and
$\sup_{M}\varphi=0$. Then for any $p_{0}>1$, and any
$a<\frac{p_{0}-1}{np_{0}+p_{0}-1}$ we have

$||(v-\varphi)_{+}||_{L^{\infty}}\leq C||(v-\varphi)_{+}||_{L^{1}}^{a}.$

Here $C$ depends only on $||v||_{L^{\infty}}$, $||e^{nG_{0}}||_{L^{p_{0}}}$,
the background metric and the choice of $a$.

Let us postpone the proof for the moment, and we first use this proposition to
prove existence.

First we look at the case when $G$ depends only on $x$. We can take a sequence
of smooth $G_{j}(x)$ that converges to $G(x)$ uniformly. By Theorem 3.1, there
exists $\varphi_{j}\in C^{\infty}(M)$ which is strictly $\Gamma$-subharmonic,
that solves:

(4.1)
$F(\chi+dd^{c}\varphi_{j})=e^{c_{j}+G_{j}(x)},\,\,\sup_{M}\varphi_{j}=0.$

Our first goal will be to show that $\varphi_{j}$ is uniformly bounded. For
this we first need to estimate the constants $c_{j}$. Evaluating at the
maximum and minimum point of $\varphi_{j}(x)$ respectively, we find that:

$|c_{j}|\leq\max|G_{j}|+\max\log F(\chi).$

Since $G_{j}$ approximates $G$ uniformly, we see that $\max|G_{j}|$, hence
$c_{j}$ is uniformly bounded. Next, the uniform $C^{0}$ bound of $\varphi_{j}$
follows from Szekelyhidi’s work [20].

Now we look at the situation when $G(x,u)$ is monotone increasing in $u$. We
can approximate $G(x,u)$ uniformly by a sequence of $G_{j}(x,u)$, such that:

  1. (1)

Each $G_{j}(x,u)$ is $C^{\infty}$ smooth in $x$ and $u$,

  2. (2)

$\frac{\partial G_{j}}{\partial u}(x,u)>0$,

  3. (3)

There exists $C_{0}>0$, such that for all $j$, $x\in M$ and $u>C_{0}$, one has
$F(\chi)(x)<e^{G_{j}(x,u)}$.

We can use Theorem 3.1 to conclude that one can solve:

(4.2) $F(\chi+dd^{c}\varphi_{j})=e^{G_{j}(x,\varphi_{j})}.$

Evaluating at the maximum point of $\varphi_{j}$, from point (3) above we see
that

$\sup_{M}\varphi_{j}\leq C_{0}.$

Once we have the uniform upper bound of $\varphi_{j}$, we see that the right
hand side $e^{G_{j}(x,\varphi_{j})}$ is uniformly bounded from above, from
which we can deduce a uniform lower bound for $\varphi_{j}$ following
Szekelyhidi [20] or Guo-Phong-Tong [18].

Next we show that there is uniform $L^{2}$ bound for $\nabla\varphi$:

######  Lemma 4.2.

There exists a constant $C$, depending only on the $C^{0}$ bound of $G$, and
also the Hermitian metric $\omega_{0}$, such that

$\int_{M}d^{c}\varphi_{j}\wedge d\varphi_{j}\wedge\omega_{0}^{n-1}\leq C.$

###### Proof.

We use the fact that each $\varphi_{j}$ is strictly $\Gamma$-subharmonic, and
that
$\Gamma\subset\Gamma_{1}:=\\{\lambda\in\mathbb{R}^{n}:\sum_{i=1}^{n}\lambda_{i}\geq
0\\}$. This would imply that:

$\Delta_{\omega_{0}}\varphi_{j}\geq-n,\,\,\,\text{ where
$\Delta_{\omega_{0}}\varphi_{j}=g^{p\bar{q}}\partial_{p\bar{q}}\varphi_{j}$}.$

This is equivalent to:

$dd^{c}\varphi_{j}\wedge\omega_{0}^{n-1}\geq-\omega_{0}^{n}.$

Let $C_{1}>0$ be large enough such that $\varphi_{j}+C_{1}\geq 0$. Then you
multiply this to both sides above:

$-\int_{M}d\varphi_{j}\wedge
d^{c}\varphi_{j}\wedge\omega_{0}^{n-1}+\int_{M}(\varphi_{j}+C_{1})d^{c}\varphi_{j}\wedge
d\omega_{0}^{n-1}\geq-\int_{M}(\varphi_{j}+C_{1})\omega_{0}^{n}.$

For the middle term, one has

$\begin{split}&\int_{M}(\varphi_{j}+C_{1})d^{c}\varphi_{j}\wedge
d\omega_{0}^{n-1}=\int_{M}d^{c}\big{(}\frac{(\varphi_{j}+C_{1})^{2}}{2}\big{)}\wedge
d\omega_{0}^{n-1}=-\int_{M}d\big{(}\frac{(\varphi_{j}+C_{1})^{2}}{2}\big{)}\wedge
d^{c}\omega_{0}^{n-1}\\\
&=\int_{M}\frac{(\varphi_{j}+C_{1})^{2}}{2}dd^{c}\omega_{0}^{n-1}.\end{split}$

Therefore, we see that the integral $\int_{M}d^{c}\varphi_{j}\wedge
d\varphi_{j}\wedge\omega_{0}^{n-1}$ is uniformly bounded. ∎

Hence we are in a position to apply the Rellich compact embedding theorem to
conclude that there is a subsequence $\varphi_{j_{k}}$ that converges in
$L^{1}$.

Before we proceed with the proof, first we explain how to use this to obtain
existence:

######  Corollary 4.3.

Theorem 1.1 holds.

###### Proof.

First we prove part (1). We have shown that one can find smooth solutions to

$F(\chi+dd^{c}\varphi_{j})=e^{G_{j}+c_{j}},\,\,\lambda[\chi+dd^{c}\varphi_{j}]\in\Gamma,\,\,\sup_{M}\varphi_{j}=0.$

Now we can use Proposition 4.1, and take $v=\varphi_{j_{l}}$,
$\varphi=\varphi_{j_{k}}$ to get:

$||(\varphi_{j_{l}}-\varphi_{j_{k}})_{+}||_{L^{\infty}}\leq
C||(\varphi_{j_{l}}-\varphi_{j_{k}})_{+}||_{L^{1}}^{\mu}.$

Switching the choice of $v$ and $\varphi$ gives:

$||\varphi_{j_{k}}-\varphi_{j_{l}}||_{L^{\infty}}\leq
C||\varphi_{j_{k}}-\varphi_{j_{l}}||_{L^{1}}^{\mu}.$

That is, the subsequence $\varphi_{j_{k}}$ actually converges uniformly.
Therefore, it is easy to see that their uniform limit will solve the limit
equation in the viscosity sense.

Now we look at part (2). We have shown that one can find smooth solutions to

$F(\chi+dd^{c}\varphi_{j})=e^{G_{j}(x,\varphi_{j})},\,\,\lambda[\chi+dd^{c}\varphi_{j}]\in\Gamma.$

Moreover, we have seen that $\varphi_{j}$ has uniform $C^{0}$ bound. Denote
$\tilde{c}_{j}=\sup_{M}\varphi_{j}$ and define
$\tilde{\varphi}_{j}=\varphi_{j}-\tilde{c}_{j}$, so that $\tilde{\varphi}_{j}$
solves:

$F(\chi+dd^{c}\tilde{\varphi}_{j})=e^{G_{j}(x,\tilde{c}_{j}+\tilde{\varphi}_{j})}.$

We have seen that one can take a subsequence such that $\varphi_{j_{k}}$
converges in $L^{1}$ and $c_{j_{k}}$ converges. So that
$\tilde{\varphi}_{j_{k}}$ converges in $L^{1}$. On the other hand, using
Proposition 4.1 will allow us to conclude that $\tilde{\varphi}_{j_{k}}$
converges uniformly, which in turn implies that $\varphi_{j_{k}}$ converges
uniformly and we can conclude that the limit function will be the viscosity
solution. ∎

The rest of the section is devoted to the proof of Proposition 4.1. First we
need a more refined estimate on the constant $c$ that allows one to solve
$f(\lambda[\chi+dd^{c}\varphi])=e^{G+c}$. We have:

######  Lemma 4.4.

Let $G(x)$ be a smooth function on $M$. Let $\varphi$ be the smooth solution
to:

$f(\lambda[\chi+dd^{c}\varphi])=e^{G+c},$

for some constant $c\in\mathbb{R}$. Then

  1. (1)

$c$ can be estimated from above in terms of the structural condition on $f$,
the background manifold and metric $(M,\omega_{0})$, and an upper bound on
$||e^{G}||_{L^{2n}(\omega_{0})}$.

  2. (2)

$c$ can be estimated from below in terms of the function $f$, the background
metric $\omega_{0}$, and a lower bound on $\int_{M}e^{G}\omega_{0}^{n}$.

###### Proof.

First we prove the upper bound. Let $x_{0}$ be the minimum point of $\varphi$
on $x_{0}$. Then we can find $r_{0}>0$ small enough, and a coordinate system
on $B_{2r_{0}}(x_{0})$, such that $x_{0}$ corresponds to $z=0$, and that

$g_{i\bar{j}}\geq c_{0}(|z|^{2})_{ij}=c_{0}\delta_{ij}\text{ on
$B_{2r_{0}}(x_{0})$ and under this coordinate}.$

We wish to apply the Alexandrov maximum principle to $\varphi+c_{0}|z|^{2}$ on
$B_{r_{0}}(x_{0})$ as follows:

$\inf_{B_{r_{0}}(x_{0})}\big{(}c_{0}|z|^{2}+\varphi\big{)}\geq\inf_{\partial
B_{r_{0}}(x_{0})}\big{(}c_{0}|z|^{2}+\varphi\big{)}-C_{n}r_{0}\bigg{(}\int_{B_{r_{0}}(x_{0})\cap\mathcal{C}_{+}}\det
D^{2}\big{(}c_{0}|z|^{2}+\varphi\big{)}\bigg{)}^{\frac{1}{2n}}.$

In the above, $\mathcal{C}_{+}$ is the subset of $B_{2r_{0}}(x_{0})$ such that
$D^{2}(c_{0}|z|^{2}+\varphi)\geq 0$.

We can estimate further from the above. On the set $\mathcal{C}_{+}$:

$\begin{split}&\det D^{2}\big{(}c_{0}|z|^{2}+\varphi)\leq
2^{2n}\big{(}\det(c_{0}|z|^{2}+\varphi)_{i\bar{j}}\big{)}^{2}\leq
2^{2n}\big{(}\det(g_{i\bar{j}}+\varphi_{i\bar{j}})\big{)}^{2}\\\ &\leq
C\big{(}f(\lambda[\chi+dd^{c}\varphi])\big{)}^{2n}=Ce^{2nG+2nc}.\end{split}$

In the second inequality above, we used that
$g_{i\bar{j}}\geq(c_{0}|z|^{2})_{i\bar{j}}$ on $B_{2r_{0}}(x_{0})$. In the
third inequality above, we used the determinant domination condition (item (4)
of Assumption 1.1).

On the other hand, since $x_{0}$ is the global minimum point of $\varphi$, we
get

$\inf_{B_{r_{0}}(x_{0})}(c_{0}|z|^{2}+\varphi)-\inf_{\partial
B_{r_{0}}(x_{0})}(c_{0}|z|^{2}+\varphi)\leq-c_{0}r_{0}^{2}.$

Therefore we get that:

$-c_{0}r_{0}^{2}\geq-
C_{n}r_{0}\bigg{(}\int_{B_{r_{0}(x_{0})}}Ce^{2nG+2nc}\bigg{)}^{\frac{1}{2n}}.$

This gives a lower bound of $c$ with the said dependence.

Next we estimate the $c$ from above. From concavity of $f$, we see that:

(4.3) $e^{G+c}=f(\lambda[\chi+dd^{c}\varphi])\leq
f(\lambda[\chi])+\frac{\partial F}{h_{i\bar{j}}}(\chi)\varphi_{i\bar{j}}.$

In the above, one can use (2.2) to see that:

$\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi)\varphi_{i\bar{j}}\frac{\omega_{0}^{n}}{n!}=\det
g_{i\bar{j}}\det(\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi))\Omega^{n-1}\wedge dd^{c}\varphi.$

Denote $\tilde{\Omega}=\big{(}\det g_{i\bar{j}}\det(\frac{\partial F}{\partial
h_{i\bar{j}}}(\chi))\big{)}^{\frac{1}{n-1}}\Omega$ and let $v_{0}$ be the
Gaudochon factor of $\tilde{\Omega}$, namely
$dd^{c}(e^{(n-1)v_{0}}\tilde{\Omega}^{n-1})=0$. Hence from (4.3), we see that:

$\begin{split}\int_{M}e^{G+(n-1)v_{0}}e^{c}\frac{\omega_{0}^{n}}{n!}&\leq\int_{M}e^{(n-1)v_{0}}f(\lambda[\chi])\frac{\omega_{0}^{n}}{n!}+\int_{M}e^{(n-1)v_{0}}\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\wedge
dd^{c}\varphi\\\
&=\int_{M}e^{(n-1)v_{0}}f(\lambda[\chi])\frac{\omega_{0}^{n}}{n!}.\end{split}$

This gives an upper bound of $c$ with the said dependence. ∎

Now let $0<\delta<1$, let $\chi_{j}:\mathbb{R}\rightarrow\mathbb{R}_{>0}$ be a
sequence of smooth functions such that $\chi_{j}\geq\max(0,x)$,
$\chi_{j}\rightarrow\max(0,x)$ as $x\rightarrow\infty$. Let $s>0$, $\kappa>1$
to be determined, we put:

(4.4)
$A_{\delta,s,\kappa,j}=\bigg{(}\int_{M}\chi_{j}\big{(}(1-\delta)v-\varphi-s\big{)}^{\kappa}e^{n\kappa
G_{0}}\omega_{0}^{n}\bigg{)}^{\frac{1}{\kappa}}.$

Here $G_{0}$ is the right hand side appearing in Proposition 4.1. Let
$\psi_{\delta,s,j,\kappa}$ be the solution to the following problem:

(4.5)
$\big{(}\omega_{0}+\sqrt{-1}\partial\bar{\partial}\psi_{\delta,s,j,\kappa}\big{)}^{n}=\frac{\chi_{j}\big{(}(1-\delta)v-\varphi-s\big{)}}{A_{\delta,s,j,\kappa}}e^{nG_{0}+b_{\delta,s,j,\kappa}},\,\,\sup_{M}\psi_{\delta,s,j,\kappa}=0.$

Note that the function on the right hand side
$\frac{\chi_{j}\big{(}(1-\delta)v-\varphi-s\big{)}}{A_{\delta,s,j,\kappa}}e^{nG_{0}}$
is uniformly bounded in $L^{\kappa}$, we may apply Lemma 5.9 of [15] to get:

$b_{\delta,s,j,\kappa}\geq-C,$

We can also get the same result from Lemma 4.4 if $\kappa\geq 2$. Here $C$
depends only on the background manifold and metric. We have the following
Moser-Trudinger type inequality:

######  Lemma 4.5.

There exists $c_{1}>0$ small enough, $C_{2}>0$ large enough, both depend only
on the structural constant of $f$, the background metric $\omega_{0}$ and the
choice of $\kappa>1$, such that on the set $\\{(1-\delta)v-\varphi-s>0\\}$,
one has

(4.6)
$c_{1}A_{\delta,s,j,\kappa}^{-\frac{1}{n}}\big{(}(1-\delta)v-\varphi-s\big{)}^{\frac{n+1}{n}}\leq-\psi_{\delta,s,j,\kappa}+C_{2}A_{\delta,s,j,\kappa}\delta^{-(n+1)}.$

###### Proof.

We define, with $\varepsilon_{0}>0,\,\Lambda>0$ to be determined:

$\Phi=\varepsilon_{0}\big{(}(1-\delta)v-\varphi-s\big{)}-(-\psi_{\delta,s,j,\kappa}+\Lambda)^{\frac{n}{n+1}}.$

In the following, we will simply denote $\psi_{\delta,s,j,\kappa}$ by $\psi$
for simplicity. The linearized operator is given by $L=\frac{\partial
F}{\partial h_{j\bar{k}}}(\chi+dd^{c}\varphi)\partial_{j\bar{k}}$ and we may
compute:

$\begin{split}&\partial_{j\bar{k}}\Phi=\varepsilon_{0}\big{(}(1-\delta)v_{j\bar{k}}-\varphi_{j\bar{k}}\big{)}+\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}\psi_{j\bar{k}}+\frac{n}{(n+1)^{2}}(-\psi+\Lambda)^{-\frac{n+2}{n+2}}\psi_{j}\psi_{\bar{k}}\\\
&\geq\varepsilon_{0}\big{(}(1-\delta)v_{j\bar{k}}-\varphi_{j\bar{k}}\big{)}+\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}\psi_{j\bar{k}}.\end{split}$

Therefore:

(4.7)
$\begin{split}&L\Phi\geq\varepsilon_{0}(1-\delta)Lv-\varepsilon_{0}L\varphi+\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}L\psi.\\\
&\geq\varepsilon_{0}(1-\delta)\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(\chi_{j\bar{k}}+v_{j\bar{k}})-\varepsilon_{0}\frac{\partial
F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(\chi_{j\bar{k}}+\varphi_{j\bar{k}})\\\
&+\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(g_{j\bar{k}}+\psi_{j\bar{k}})+\varepsilon_{0}\delta\frac{\partial
F}{\partial h_{j\bar{k}}}(\chi+dd^{c}\varphi)\chi_{j\bar{k}}\\\
&-\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)g_{j\bar{k}}.\end{split}$

Now we use that $\chi\in\Gamma_{\omega_{0}}$, so that there exists $c_{*}>0$,
such that one can write $\chi=c_{*}\omega_{0}+\tilde{\chi}$ with
$\lambda(\tilde{\chi})\in\Gamma$. With this observation, we see that:

(4.8) $\begin{split}&\varepsilon_{0}\delta\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)\chi_{j\bar{k}}-\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}\frac{\partial
F}{\partial h_{j\bar{k}}}(\chi+dd^{c}\varphi)g_{j\bar{k}}\\\
&\geq\varepsilon_{0}\delta\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)\tilde{\chi}_{j\bar{k}}+\frac{\partial
F}{\partial h_{j\bar{k}}}(\chi+dd^{c}\varphi)(\varepsilon_{0}\delta
c_{*}-\frac{n}{n+1}\Lambda^{-\frac{1}{n+1}}\big{)}g_{j\bar{k}}.\end{split}$

Using the lemma below, we may conclude that:

(4.9) $\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(\chi_{j\bar{k}}+v_{j\bar{k}})\geq
0,\,\,\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)\tilde{\chi}_{j\bar{k}}\geq 0.$

Combining (4.7)-(4.9), we see that:

(4.10) $\begin{split}&L\Phi\geq-\varepsilon_{0}\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(\chi_{j\bar{k}}+\varphi_{j\bar{k}})+\frac{n}{n+1}(-\psi+\Lambda)^{-\frac{1}{n+1}}\frac{\partial
F}{\partial h_{j\bar{k}}}(\chi+dd^{c}\varphi)(g_{j\bar{k}}+\psi_{j\bar{k}})\\\
&+(\varepsilon_{0}\delta
c_{*}-\frac{n}{n+1}\Lambda^{-\frac{1}{n+1}})\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)g_{j\bar{k}}.\end{split}$

Next we use that $f$ has homogeneity one to see that:

$\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(\chi+dd^{c}\varphi)_{j\bar{k}}=\sum_{i}\frac{\partial
f}{\partial\lambda_{i}}\lambda_{i}=f.$

Also we note that:

$\begin{split}&\frac{\partial F}{\partial
h_{j\bar{k}}}(\chi+dd^{c}\varphi)(g_{j\bar{k}}+\psi_{j\bar{k}})\geq
n\big{(}\det\frac{\partial F}{\partial
h_{i\bar{j}}}\cdot\det(g_{i\bar{j}}+\psi_{i\bar{j}})\big{)}^{\frac{1}{n}}\\\
&\geq
c_{1}A_{\delta,s,j,\kappa}^{-\frac{1}{n}}\big{(}(1-\delta)-\varphi-s\big{)}_{+}^{\frac{1}{n}}F(\chi+dd^{c}\varphi)e^{\frac{b_{\delta,s,j,\kappa}}{n}}.\end{split}$

In the above, we noted that:

$\det(g_{i\bar{j}}+\psi_{i\bar{j}})=\frac{\chi_{j}((1-\delta)v-\varphi-s)}{A_{\delta,s,j,\kappa}}e^{nG_{0}+b_{\delta,s,j,\kappa}}\geq\frac{\big{(}(1-\delta)v-\varphi-s\big{)}_{+}}{A_{\delta,s,j,\kappa}}e^{nG_{0}+b_{\delta,s,j,\kappa}}.$

To proceed further, assume that $\Phi$ achieves positive maximum at $p\in M$.
We obtain the following when evaluated at $p$:

Therefore, when evaluated at $p$, we get:

(4.11) $\begin{split}0&\geq
L\Phi\geq-\varepsilon_{0}f+c_{2}A_{\delta,s,j,\kappa}^{-\frac{1}{n}}\big{(}(1-\delta)-\varphi-s\big{)}_{+}^{\frac{1}{n}}(-\psi+\Lambda)^{-\frac{1}{n+1}}f\\\
&+(\varepsilon_{0}\delta
c_{*}-\frac{n}{n+1}\Lambda^{-\frac{1}{n+1}})\frac{\partial F}{\partial
h_{j\bar{k}}}g_{j\bar{k}}.\end{split}$

In the above, $c_{2}$ depends only on the structural constants of $f$ and also
the background metric. Here we used that $b_{\delta,s,j,\kappa}$ has a
universal lower bound (except depending on the choice of $\kappa>1$). Now we
can choose the parameters $\varepsilon_{0},\,\Lambda$ so that:

$\frac{1}{2}\varepsilon_{0}\delta=\frac{n}{n+1}\Lambda^{-\frac{1}{n+1}},\,\,\,-2\varepsilon_{0}+c_{2}A_{\delta,s,j,\kappa}^{-\frac{1}{n}}\varepsilon_{0}^{-\frac{1}{n}}=0.$

With this choice, we get a contradiction from (4.11). Indeed, since
$\Phi(p)>0$, we see that at $p$, one has

$((1-\delta)v-\varphi-s)^{\frac{1}{n}}(-\psi+\Lambda)^{-\frac{1}{n+1}}>\varepsilon_{0}^{-\frac{1}{n}}.$

With the above made choice of $\varepsilon_{0},\,\Lambda$, one can see that
the right hand side of (4.11) is $>0$, which is a contradiction. ∎

In the above proof, we noted the following lemma.

######  Lemma 4.6.

Let $A$, $B$ be two Hermitian matrices such that
$\lambda(g^{i\bar{k}}A_{j\bar{k}}),\,\lambda(g^{i\bar{k}}B_{j\bar{k}})\in\Gamma$.
Define $F(h)=f\big{(}\lambda(g^{i\bar{k}}h_{j\bar{k}})\big{)}$. Then we have:

$\frac{\partial F}{\partial h_{j\bar{k}}}(A)B_{j\bar{k}}\geq 0.$

###### Proof.

We first prove this statement, assuming that $g_{i\bar{j}}=\delta_{ij}$ and
that $A$ is diagnal. With this assumption, we then have:

$\frac{\partial F}{\partial h_{j\bar{k}}}(A)=\frac{\partial
f}{\partial\lambda_{j}}(\lambda(A))\delta_{jk}.$

Without loss of generality, we may assume that
$\lambda_{1}\leq\lambda_{2}\leq\cdots\lambda_{n}$, so that $\frac{\partial
f}{\partial\lambda_{1}}\geq\frac{\partial
f}{\partial\lambda_{2}}\cdots\frac{\partial f}{\partial\lambda_{n}}$.
Therefore, from above we see that

$\frac{\partial F}{\partial h_{j\bar{k}}}(A)B_{j\bar{k}}=\frac{\partial
f}{\partial\lambda_{j}}(\lambda(A))B_{j\bar{j}}\geq\frac{\partial
f}{\partial\lambda_{j}}(\lambda(A))\mu_{j}.$

In the above, $\mu_{j}$ are the eigenvalues of $B$, also listed in the
increasing order. The inequality above used the Horn-Shur lemma, which says
that the vector $(B_{1\bar{1}},\cdots,B_{n\bar{n}})$ is contained in the
convex hull of $(\mu_{\sigma(1)},\cdots,\mu_{\sigma(n)})$, where $\sigma$ is a
permutation of the indices. Moreover, since $\frac{\partial
f}{\partial\lambda_{j}}$ is in the decreasing order, $\frac{\partial
f}{\partial\lambda_{j}}\mu_{j}$ will be minimized if $\mu_{j}$ is in the
increasing order, hence the inequality above. Now it only remains to show
$\frac{\partial f}{\partial\lambda_{j}}(\lambda(A))\mu_{j}\geq 0$. Indeed, we
just need to note that $\frac{\partial
f}{\partial\lambda_{j}}(\lambda(A))\mu_{j}=\frac{d}{dt}|_{t=0}f(\lambda(A)+t\mu)$.
Note that $t\mapsto f(\lambda(A)+t\mu)$ is concave, and is bounded from below
on $[0,\infty)$, hence $\frac{d}{dt}|_{t=0}f(\lambda(A)+t\mu)\geq 0$.

Next we explain how to reduce the general case to the special case considered
above.

First we observe that one can reduce to the case when
$g_{i\bar{j}}=\delta_{ij}$. Indeed, we may assume that there is an invertible
$n\times n$ matrix, such that $g=P\overline{P^{T}}$ (that is,
$g_{i\bar{j}}=P_{ir}\overline{P_{jr}}$). On the other hand,

$(g^{i\bar{k}}A_{j\bar{k}})^{i}_{j}=\big{(}(g^{-1})_{ki}A_{j\bar{k}}\big{)}^{i}_{j}=(A\cdot
g^{-1})_{ij}=(A\cdot\overline{(P^{T})^{-1}}\cdot P^{-1})_{ij}$

Therefore, if we define $\tilde{A}=P^{-1}A\overline{(P^{T})^{-1}}$, then one
has $\lambda((g^{i\bar{k}}A_{j\bar{k}})_{j}^{i})=\lambda(\tilde{A})$.
Likewise, we define $\tilde{B}=P^{-1}B\overline{(P^{T})^{-1}}$, then one gets
$\lambda(\tilde{A}),\,\lambda(\tilde{B})\in\Gamma$. Also we define
$\tilde{F}(h)=f\big{(}\lambda(h)\big{)}$, where $h$ is a Hermitian matrix and
$\lambda(h)$ means the usual eigenvalue of $h$, then we have
$F(h)=\tilde{F}\big{(}P^{-1}h\overline{(P^{T})^{-1}}\big{)}$. From this we may
calculate:

$\begin{split}&\frac{\partial F}{\partial
h_{j\bar{k}}}(A)B_{j\bar{k}}=\frac{\partial\tilde{F}}{\partial
h_{p\bar{q}}}(\tilde{A})\frac{\partial(P^{-1}h\overline{(P^{T})^{-1}})_{p\bar{q}}}{\partial
h_{j\bar{k}}}P_{ja}\tilde{B}_{a\bar{b}}\overline{(P^{T})_{ak}}\\\
&=\frac{\partial\tilde{F}}{\partial
h_{p\bar{q}}}(\tilde{A})(P^{-1})_{pj}\overline{(P^{-1})_{qk}}P_{j\bar{a}}\tilde{B}_{a\bar{b}}\overline{(P^{T})_{ak}}=\frac{\partial\tilde{F}}{\partial
h_{a\bar{b}}}(\tilde{A})\tilde{B}_{a\bar{b}}.\end{split}$

Therefore, we see that, as long as we can verify the lemma with
$\tilde{A},\,\tilde{B},\,\tilde{F}$ (which is equivalent to taking $g=I$), the
general statement would follow.

Next, we explain why we can assume that $A$ is diagonal. Indeed, from the
definition of $F$, one has $F(h)=F(Uh\overline{U^{T}})$, for any unitary
matrix $U$. Therefore one has

$\frac{\partial F}{\partial h_{i\bar{j}}}(h)=\frac{\partial F}{\partial
h_{p\bar{q}}}(Uh\overline{U^{T}})U_{pi}\overline{U_{qj}}.$

Now we choose $h=A$, and choose $U$ to be the unitary matrix such that
$UA\overline{U^{T}}$ is diagonal. Then one gets:

$\frac{\partial F}{\partial h_{i\bar{j}}}(A)B_{i\bar{j}}=\frac{\partial
F}{\partial
h_{p\bar{q}}}(UA\overline{U^{T}})U_{pi}\overline{U_{qj}}B_{i\bar{j}}.$

All we need to do is to consider $\tilde{B}=UB\overline{U^{T}}$ in place of
$B$, and they have the same eigenvalues. ∎

As a consequence of Lemma 4.5, we see that:

######  Corollary 4.7.

Denote

(4.12)
$A_{\delta,s,\kappa}=\bigg{(}\int_{M}\big{(}(1-\delta)v-\varphi-s\big{)}_{+}^{\kappa}e^{\kappa
nG_{0}}\omega_{0}^{n}\bigg{)}^{\frac{1}{\kappa}}.$

There exists $\beta_{0}>0$, $C>0$ which depends only on the background
manifold and metric, as well as the choice of $\kappa>1$, such that

$\int_{M}\exp\big{(}\beta_{0}A_{\delta,s,\kappa}^{-\frac{1}{n}}\big{(}(1-\delta)v-\varphi-s\big{)}_{+}^{\frac{n+1}{n}}\big{)}\omega_{0}^{n}\leq\exp\big{(}CA_{\delta,s,\kappa}\delta^{-(n+1)}\big{)}.$

###### Proof.

Using the following lemma, we see that there exists $\alpha_{0}>0$, $C>0$ such
that for any $\psi\in PHS(M,\omega_{0})$ with $\sup_{M}\psi=0$, one has:

$\int_{M}e^{-\alpha_{0}\psi}\omega_{0}^{n}\leq C.$

Now one multiplies both sides of (4.6) with $\alpha_{0}$, raise to the
exponential, and integrate on $M$, we get that

$\int_{M}\exp\big{(}\beta_{0}A_{\delta,s,j,\kappa}^{-\frac{1}{n}}\big{(}(1-\delta)v-\varphi-s\big{)}_{+}^{\frac{n+1}{n}}\big{)}\omega_{0}^{n}\leq\exp\big{(}CA_{\delta,s,j,\kappa}\delta^{-(n+1)}\big{)}.$

Note that from the definition of $A_{\delta,s,j,\kappa}$, we see that
$A_{\delta,s,j,\kappa}\rightarrow A_{\delta,s,\kappa}$ in (4.4) as
$j\rightarrow\infty$. ∎

In the above, we used the following lemma:

######  Lemma 4.8.

Let $(M,\omega_{0})$ be a compact Hermitian manifold. Then there exists
$\alpha_{0}>0,\,C>0$, such that for any $\psi\in PSH(M,\omega_{0})$ with
$\sup_{M}\psi=0$, one has that

$\int_{M}e^{-\alpha_{0}\psi}\omega_{0}^{n}\leq C.$

Using the above estimate, we can get the pointwise upper bound of
$(1-\delta)v-\varphi-s$. For this we have:

######  Lemma 4.9.

Assume that $\delta>0,\,s_{0}>0$ are chosen so that
$A_{\delta,s_{0},\kappa}\leq\delta^{n+1}$. Assume that $e^{nG_{0}}\in
L^{p_{0}}(M,\omega_{0}^{n})$ for some $p_{0}>1$. Define
$\Omega_{\delta,s}=\\{x\in M:(1-\delta)v-\varphi-s>0\\}$ and put
$u(s)=\int_{\Omega_{\delta,s}}e^{nG_{0}}\omega_{0}^{n}$. Then for any
$0<\delta_{*}<\frac{1}{n}$, we may choose $\kappa>1$ sufficiently close to 1,
such that there exists $C_{*}>0$, depending on $\delta_{*}$,
$||e^{nG_{0}}||_{L^{p_{0}}}$, the background metric and the choice of
$\kappa$, such that

$tu(s+t)\leq C_{*}u(s)^{1+\delta_{*}},$

for any $s\geq s_{0}$, $t>0$. The above choice of $\kappa$ depends only on
$\delta_{*},\,n,\,p_{0}$.

###### Proof.

Let $s\geq s_{0}$, so that $A_{\delta,s,\kappa}\leq
A_{\delta,s_{0},\kappa}\leq\delta^{n+1}$. We then obtain from Corollary 4.7
that for any positive integer $q$, one has that:

(4.13)
$\int_{M}\big{(}(1-\delta)v-\varphi-s\big{)}_{+}^{\frac{n+1}{n}q}\omega_{0}^{n}\leq
C(q)A_{\delta,s}^{\frac{q}{n}}.$

Therefore,

(4.14)
$\begin{split}&A_{\delta,s,\kappa}=||((1-\delta)v-\varphi-s)e^{nG_{0}}||_{L^{\kappa}(\Omega_{\delta,s})}\\\
&\leq||((1-\delta)v-\varphi-s)||_{L^{\frac{n+1}{n}q}(\Omega_{\delta,s})}||e^{nG_{0}}||_{L^{q^{\prime}}(\Omega_{\delta,s})}\\\
&\leq
C(q)A_{\delta,s,\kappa}^{\frac{1}{n+1}}||e^{nG_{0}}||_{L^{q^{\prime}}(\Omega_{\delta,s})}\leq
C(q)A_{\delta,s,\kappa}^{\frac{1}{n+1}}||e^{nG_{0}}||^{\lambda}_{L^{1}(\Omega_{\delta,s})}||e^{nG_{0}}||^{1-\lambda}_{L^{p_{0}}}.\end{split}$

In the second line above, the $q$ will be chosen sufficiently large and
$q^{\prime}$ is such that
$\frac{1}{\kappa}=\frac{n}{n+1}\frac{1}{q}+\frac{1}{q^{\prime}}$. By choose
$\kappa>1$ sufficiently close to 1 and $q$ sufficiently large, we may make
$q^{\prime}>1$ arbitrarily close to 1. This just follows from Hölder’s
inequality.

In the first inequality of the third line, we used (4.13). In the second
inequality of the third line, $0<\lambda<1$ satisfies:
$\frac{1}{q^{\prime}}=\lambda+\frac{1-\lambda}{p_{0}}$. By making
$q^{\prime}>1$ sufficiently close to 1, we may make $\lambda$ as close to 1 as
we want.

Hence we see from (4.14) that, for any $\varepsilon>0$:

$A_{\delta,s,\kappa}\leq
C_{\varepsilon}||e^{nG_{0}}||_{L^{1}(\Omega_{\delta,s})}^{\frac{n+1}{n}-\varepsilon}.$

Here $C_{\varepsilon}$ above depends on $||e^{nG_{0}}||_{L^{p_{0}}}$ as well.

On the other hand, since $(1-\delta)v-\varphi-s>t$ on $\Omega_{s+t}$, one has:

$A_{\delta,s,\kappa}\geq t||e^{nG_{0}}||_{L^{\kappa}(\Omega_{\delta,s+t})}\geq
tc_{1}||e^{nG_{0}}||_{L^{1}(\Omega_{\delta,s+t})}.$

Hence the result follows. ∎

We need to use the following Lemma of De Giorgi:

######  Lemma 4.10.

Let $\phi:[0,\infty)\rightarrow[0,\infty)$ be an decreasing and continuous
function, such that there exists $\mu>0$, $B_{0}>0$, $s_{0}\geq 0$, such that
for any $r>0,\,s\geq s_{0}$, one has:

$r\phi(s+r)\leq B_{0}\phi(s)^{1+\mu}.$

Then $\phi(s)\equiv 0$ for $s\geq
s_{0}+\frac{2B_{0}\phi(s_{0})^{\mu}}{1-2^{-\mu}}.$

###### Proof.

We can choose a sequence $\\{s_{k}\\}_{k\geq 1}$ by induction:

$s_{k+1}-s_{k}=2B_{0}\phi(s_{k})^{\mu}.$

Then we choose $s=s_{k},\,r=s_{k+1}-s_{k}$, we see that

$\phi(s_{k+1})\leq\frac{B_{0}\phi(s_{k})^{1+\mu}}{s_{k+1}-s_{k}}\leq\frac{1}{2}\phi(s_{k}).$

That is, $\phi(s_{k})\leq 2^{-k}\phi(s_{0})$, hence:

$s_{k+1}-s_{k}\leq 2B_{0}\phi(s_{0})^{\mu}2^{-k\mu}.$

Therefore,

$\sum_{k=0}^{\infty}(s_{k+1}-s_{k})\leq\sum_{k=0}^{\infty}2B_{0}\phi(s_{0})^{\mu}2^{-k\mu}=\frac{2B_{0}\phi(s_{0})^{\mu}}{1-2^{-\mu}}.$

This implies that $s_{k}$ is increasing and bounded from above. Hence we see
that $s_{k}\rightarrow s_{\infty}$ and we see that $\phi\equiv 0$ for $s\geq
s_{\infty}$, and $s_{\infty}\leq
s_{0}+\frac{2B_{0}\phi(s_{0})^{\mu}}{1-2^{-\mu}}$. ∎

Combining Lemma 4.9 and 4.10, we see that

######  Lemma 4.11.

Let $0<\delta<1$, $s_{0}>0$ be chosen so that
$A_{\delta,s_{0},\kappa}\leq\delta^{n+1}$. Then for any
$0<\nu<\frac{1}{n}(1-\frac{1}{p_{0}})$, we may choose $\kappa>1$ sufficiently
close to 1, such that there exists $C>0$, depending only on $\nu$,
$||e^{nG_{0}}||_{L^{p_{0}}}$, the background metric and the choice of $\kappa$
such that

$\sup_{M}\big{(}(1-\delta)v-\varphi\big{)}\leq
s_{0}+Cvol\big{(}\Omega_{\delta,s_{0}}\big{)}^{\nu}.$

###### Proof.

Combining Lemma 4.9 and 4.10, and keep in mind that
$u(s)=\int_{\Omega_{\delta,s}}e^{nG_{0}}\omega_{0}^{n}$, we get: for any
$0<\delta_{*}<\frac{1}{n}$:

(4.15) $\sup_{M}((1-\delta)v-\varphi)\leq
s_{0}+C_{*}\big{(}\int_{\Omega_{\delta,s_{0}}}e^{nG_{0}}\omega_{0}^{n}\big{)}^{\delta_{*}}.$

Here $C$ depends only on the choice of $\delta_{*}$,
$||e^{nG_{0}}||_{L^{p_{0}}}$, the background metric and the choice of
$\kappa>1$. We then apply Hölder’s inequality to see that:

$\int_{\Omega_{\delta,s_{0}}}e^{nG_{0}}\omega_{0}^{n}\leq||e^{nG_{0}}||_{L^{p_{0}}}vol(\Omega_{\delta,s_{0}})^{1-\frac{1}{p_{0}}}.$

Plug this to (4.15) gives us the result. ∎

With a little more work, we wish to get rid of the $\delta$ in the above
estimate:

######  Lemma 4.12.

Assume that $0<\delta<1$, and $s_{0}>0$ are chosen so that:

  1. (1)

$s_{0}\geq 2\delta||v||_{L^{\infty}}$,

  2. (2)

$A_{\delta,s_{0},\kappa}\leq\delta^{n+1}$.

Then for any $0<\nu<\frac{1}{n}(1-\frac{1}{p_{0}})$, there exists $C>0$,
depending only on $\nu$, $||e^{nG_{0}}||_{L^{p_{0}}}$, and the background
metric, such that

$\sup_{M}(v-\varphi)\leq\frac{3s_{0}}{2}+Cs_{0}^{-\nu}||(v-\varphi)_{+}||_{L^{1}(\omega_{0}^{n})}^{\nu}.$

###### Proof.

Let $s_{0}>0,\,0<\delta<1$ be as stated in the lemma. Then we have

$vol(\Omega_{\delta,s_{0}})\leq\frac{1}{s_{0}}\int_{\Omega_{\delta,s_{0}}}\big{(}(1-\delta)v-\varphi\big{)}_{+}\omega_{0}^{n}\leq\frac{1}{s_{0}}\big{(}||(v-\varphi)_{+}||_{L^{1}}+\delta||v||_{L^{\infty}}vol(\Omega_{\delta,s_{0}})\big{)}.$

Since $s_{0}\geq 2\delta||v||_{L^{\infty}}$, one gets that:

$vol(\Omega_{\delta,s_{0}})\leq\frac{2}{s_{0}}||(v-\varphi)_{+}||_{L^{1}}.$

Therefore, using Lemma 4.11, we get:

$\sup_{M}(v-\varphi)\leq\sup_{M}((1-\delta)v-\varphi)+\delta||v||_{L^{\infty}}\leq\frac{3s_{0}}{2}+Cvol(\Omega_{\delta,s_{0}})^{\nu}\leq\frac{3s_{0}}{2}+C2^{\nu}s_{0}^{-\nu}||(v-\varphi)_{+}||_{L^{1}}^{\nu}.$

So the result follows. ∎

For $0<\delta<1$, we define $s_{0}(\delta)$ to be the minimum of $s_{0}$ that
satisfies $s_{0}\geq 2\delta||v||_{L^{\infty}}$ and
$A_{\delta,s_{0},\kappa}\geq\delta^{n+1}$.

At this point, it only remains to estimate $s_{0}(\delta)$, and we have:

######  Lemma 4.13.

For any $0<\delta<1$, we define $s_{0}(\delta)$ to be the smallest $s_{0}$
such that $s_{0}\geq 2\delta||v||_{L^{\infty}}$ and
$A_{\delta,s_{0},\kappa}\leq\delta^{n+1}$. Then for any
$\mu>\frac{np_{0}}{p_{0}-1}$, there exists $C_{\mu}$, depending only on $\mu$,
$||e^{nG_{0}}||_{L^{p_{0}}}$, the background metric and the choice of
$\kappa$, such that

$s_{0}(\delta)\leq\max\big{(}2\delta||v||_{L^{\infty}},C_{\mu}\delta^{-\mu}||(v-\varphi)_{+}||_{L^{1}}\big{)}.$

###### Proof.

Note that from (4.12), $A_{\delta,s,\kappa}$ depends continuously on $s$.
Therefore, with $s_{0}=s_{0}(\delta)$, one must have either
$s_{0}=2\delta||v||_{L^{\infty}}$ or $A_{\delta,s_{0},\kappa}=\delta^{n+1}$.
If the first possibility happens, then we are done. Now we look at what
happens when $A_{\delta,s_{0}}=\delta^{n+1}$. For this, let $\beta>\kappa$, we
can then calculate:

$\begin{split}&A_{\delta,s_{0},\kappa}=\big{(}\int_{\Omega_{\delta,s_{0}}}\big{(}(1-\delta)v-\varphi-s\big{)}_{+}^{\kappa}e^{\kappa
nG_{0}}\omega_{0}^{n}\big{)}^{\frac{1}{\kappa}}\leq||((1-\delta)-\varphi-s)_{+}||_{L^{\frac{\kappa\beta}{\beta-\kappa}}(\omega_{0}^{n})}||e^{nG_{0}}||_{L^{\beta}(\Omega_{\delta,s_{0}})}\\\
&\leq
C(\beta)A_{\delta,s_{0},\kappa}^{\frac{1}{n+1}}\cdot||e^{nG_{0}}||_{L^{p_{0}}}vol(\Omega_{\delta,s_{0}})^{\frac{1}{\beta}-\frac{1}{p_{0}}}\leq
C(\beta)A_{\delta,s_{0},\kappa}^{\frac{1}{n+1}}\cdot||e^{nG_{0}}||_{L^{p_{0}}}(2s_{0}^{-1})^{\frac{1}{\beta}-\frac{1}{p_{0}}}||(v-\varphi)_{+}||_{L^{1}}^{\frac{1}{\beta}-\frac{1}{p_{0}}}.\end{split}$

Using that $A_{\delta,s_{0}}=\delta^{n+1}$, we get that:

$s_{0}\leq
C^{\prime}(\beta)\delta^{-\frac{np_{0}\beta}{p_{0}-\beta}}||(v-\varphi)_{+}||_{L^{1}}.$

By choosing $\beta>1$ as close to 1 as we want, we may make
$\frac{np_{0}\beta}{p_{0}-\beta}$ sufficiently close to
$\frac{np_{0}}{p_{0}-1}$. ∎

From this, the Proposition 4.1 immediately follows:

###### Proof.

(Of Proposition 4.1) Assume that $||(v-\varphi)_{+}||_{L^{1}}<1$ for the
moment, then for any $\mu>\frac{np_{0}}{p_{0}-1}$, we wish to take:

$\delta=||(v-\varphi)_{+}||_{L^{1}}^{\frac{1}{\mu+1}}.$

With this choice, we see from Lamma 4.13 that:

$s_{0}(\delta)\leq C||(v-\varphi)_{+}||_{L^{1}}^{\frac{1}{\mu+1}}.$

Here $C$ depends on $\mu,\,||v||_{L^{\infty}},\,||e^{nG_{0}}||_{L^{p_{0}}}$,
the background metric and the choice of $\kappa$. That is, if we take
$s_{0}=C||(v-\varphi)_{+}||_{L^{1}}^{\frac{1}{\mu+1}}$, it will satisfy
$s_{0}\geq 2\delta||v||_{L^{\infty}}$ and $A_{s_{0},\delta}\leq\delta^{n+1}$.
Hence, we may use Lemma 4.12 to conclude that

$\sup_{M}(v-\varphi)\leq
C\big{(}||(v-\varphi)_{+}||_{L^{1}}^{\frac{1}{\mu+1}}+||(v-\varphi)_{+}||_{L^{1}}^{\frac{\nu\mu}{\mu+1}}\big{)}\leq
C^{\prime}||(v-\varphi)_{+}||_{L^{1}}^{a}.$

If you calculate the exponents more carefully, you will find that one can take
any $a<\frac{p_{0}-1}{np_{0}+p_{0}-1}$. If $||(v-\varphi)_{+}||_{L^{1}}\geq
1$, then the situation is trivial. Indeed, we have estimate for
$||\varphi||_{L^{\infty}}$ from Guo-Phong’s $L^{\infty}$ estimate for
Hermitian case [17]. ∎

##  5\. uniqueness issues

###  5.1. sup/inf convolution of the viscosity solution

The basic strategy to prove uniqueness of viscosity solution will be to
perform sup/inf convolution of the solution, so that we get a sub/sup solution
that is twice differentiable a.e. Then we get the pointwise differential
inequality wherever the solution is twice differentiable.

First we explain what we mean by second order differentiable:

######  Definition 5.1.

Let $U\subset\mathbb{R}^{d}$ be an open set, and $x_{0}\in U$. Let $\varphi$
be a function defined on $U$. We say that $\varphi$ is twice differentiable at
$x_{0}$, if there exists a quadratic polynomial $P_{\varphi,x_{0}}(x)$, such
that

$\lim_{r\rightarrow 0}r^{-2}\sup_{x\in
B_{r}(x_{0})}|\varphi(x)-P_{\varphi,x_{0}}(x)|=0.$

In this case, we call $P_{\varphi,x_{0}}(x)$ to be the touching polynomial of
$\varphi$ at $x_{0}$.

If $\varphi$ is defined on a manifold, then we can define twice
differentiability using coordinate charts, and this notion is clearly
independent of the choice of the coordinate charts.

One thing we observe is that, if a viscosity subsolution/supersolution is
twice differentiable at a point, then at that point, the differential
inequality holds in the classical sense.

######  Lemma 5.2.

Let $G:M\times\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function.

  1. (1)

Let $\varphi\in C(M)$ be a viscosity subsolution to
$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)}$. Assume that $\varphi$ is second
order differentiable at $x_{0}$, then the following holds in the classical
sense:

$F(\chi+dd^{c}P_{\varphi,x_{0}})(x_{0})\geq e^{G(x_{0},\varphi(x_{0}))},$

  2. (2)

Let $\psi\in C(M)$ be a viscosity supersolution to
$F(\chi+dd^{c}\psi)=e^{G(x,\psi)}$. Assume that $\psi$ is second order
differentiable at $x_{0}$, then one has

$F(\chi+dd^{c}P_{\psi,x_{0}})(x_{0})\leq e^{G(x_{0},\psi(x_{0}))},$

###### Proof.

First we prove (1). Since $P_{\varphi,x_{0}}$ is the quadratic Taylor
polynomial of $\varphi$ at $x_{0}$, we see that for any $\varepsilon>0$,
$P_{\varphi,x_{0}}(x)+\varepsilon|x-x_{0}|^{2}$ would touch $\varphi$ from
above at $x_{0}$. By writing $|x-x_{0}|^{2}$, we have chosen some holomorphic
coordinate chart near $x_{0}$. Also we may assume that $\delta_{ij}\leq
Cg_{i\bar{j}}$ on this chart, for some $c>0$. Since $\varphi$ is a viscosity
subsolution, we see that:

$f\big{(}\lambda[\chi+dd^{c}P_{\varphi,x_{0}}(x_{0})+C\varepsilon\omega_{0}]\big{)}\geq
f\big{(}\lambda[\chi+dd^{c}(P_{\varphi,x_{0}}+\varepsilon|x-x_{0}|^{2})]\big{)}\geq
e^{G(x_{0},\varphi(x_{0}))}.$

Let $\varepsilon\rightarrow 0$, we see that
$\lambda[\chi+dd^{c}P_{\varphi,x_{0}}+C\varepsilon\omega_{0}]\rightarrow\lambda[\chi+dd^{c}P_{\varphi,x_{0}}](x_{0})$.
Since $f$ is a coutinuous function, we see that

$f\big{(}\lambda[\chi+dd^{c}P_{\varphi,x_{0}}]\big{)}(x_{0})\geq
e^{G(x_{0},\varphi(x_{0}))}.$

The proof of (2) is similar. Indeed, we know that for any $\varepsilon>0$,
$P_{\psi,x_{0}}-\varepsilon|x-x_{0}|^{2}$ would touch $\psi$ from below. If it
happens that $\lambda[\chi+dd^{c}P_{\psi,x_{0}}]\notin Int(\Gamma)$, then
$f(\lambda[\chi+dd^{c}P_{\psi,x_{0}}])=0$ by default, and the inequality is
trivial. If $\lambda[\chi+dd^{c}P_{\psi,x_{0}}]\in Int(\Gamma)$, then
$\lambda[\chi+dd^{c}P_{\psi,x_{0}}-C\varepsilon\omega_{0}]\in Int(\Gamma)$ for
small enough $\varepsilon$. Then the argument is the same as (1), by sending
$\varepsilon\rightarrow 0$ and using the continuity of $f$. ∎

In general, a viscosity solution is only continuous, and there is no guarantee
that it is twice differentiable anywhere. So our first step would be to find
suitable regularization, that makes the original solution a
subsolution/supsolution, and that it is twice differentiable a.e.

Henceforth we assume that $\varphi\in C(M)$ is a viscosity solution to
$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)}$. We define the sup convolution of
$\varphi$ as follows:

(5.1) $\varphi^{\varepsilon}(z)=\sup_{\xi\in
T_{z}M}\big{(}\varphi(\exp_{z}(\xi))+\varepsilon-\frac{1}{\varepsilon}|\xi|_{z}^{2}\big{)},$

where $\exp_{z}(\xi)$ is the exponential map at $z$, defined using the metric
$\omega_{0}$, and $|\xi|_{z}$ denotes the length of the tangent vector $\xi$,
again using the metric $\omega_{0}$. Similarly we define the inf convolution
of $\varphi$:

(5.2) $\varphi_{\varepsilon}(z)=\inf_{\xi\in
T_{z}M}\big{(}\varphi(\exp_{z}(\xi))-\varepsilon+\frac{1}{\varepsilon}|\xi|_{z}^{2}\big{)}.$

A key step is to show that $\varphi^{\varepsilon}$ and $\varphi_{\varepsilon}$
defined above produces viscosity subsolution and supsolutions, up to a small
error. Also from the definition of $\varphi^{\varepsilon}$ and
$\varphi_{\varepsilon}$, they will be semi-convex and semi-concave, hence
twice differentiable a.e. Next we make this precise. We first verify that
$\varphi^{\varepsilon}$ and $\varphi_{\varepsilon}$ defined above are
subsolution/supsolutions.

Denote $\rho_{\varphi}(r),\,\,0<r<1$ to be the modulus of continuity of
$\varphi$. That is,
$\rho_{\varphi}(r)=\max\\{|\varphi(x)-\varphi(y)|:\,\,d_{g}(x,y)\leq
r,\,x,y\in M\\}$.

######  Proposition 5.3.

Let $\varphi$ be a viscosity solution to
$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)}$ with $G(x,\varphi)$ continuous. Then
there exist continuous functions
$\rho(\varepsilon):(0,1)\rightarrow\mathbb{R}_{>0}$ with $\rho(0+)=0$, and
$\rho_{1}(\varepsilon,a_{1}),\,\rho_{2}(\varepsilon,a_{2}):(0,1)^{2}\rightarrow\mathbb{R}_{>0}$
with $\rho_{i}(0+,0+)=0,\,\,i=1,2$, such that for any $0<\varepsilon<1$,
$0<a_{i}<1,\,i=1,\,2$:

  1. (1)

$\frac{\varphi^{\varepsilon}}{1+a_{1}}$ is $\Gamma$-subharmonic with respect
to $\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}$ in the viscosity sense,
and
$F(\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{\varphi^{\varepsilon}}{1+a_{1}})\geq
e^{G(x,\frac{\varphi^{\varepsilon}}{1+a_{1}})}-\rho_{1}(\varepsilon,a_{1})$ in
the viscosity sense.

  2. (2)

$\frac{\varphi_{\varepsilon}}{1-a_{2}}$ satisfies
$F(\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{\varphi_{\varepsilon}}{1-a_{2}})\leq
e^{G(x,\frac{\varphi_{\varepsilon}}{1-a_{2}})}+\rho_{2}(\varepsilon,a_{2})$ in
the viscosity sense.

Here the functions $\rho,\,\rho_{i},\,i=1,\,2$ are determined by
$\rho_{\varphi}$, $||\varphi||_{L^{\infty}}$, the form $\chi$ and the
background metric.

For the proof of Proposition 5.3, we need to understand how the touching of
$\varphi^{\varepsilon}$ or $\varphi_{\varepsilon}$ from above or below
translates to the touching of $\varphi$.

######  Lemma 5.4.

  1. (1)

Let $\varphi\in C(M)$ and we define $\varphi^{\varepsilon}$ according to
(5.1). Let $x_{0}\in M$ and $P$ is a $C^{2}$ function defined in a
neighborhood $x_{0}$ that touches $\varphi^{\varepsilon}$ from above at
$x_{0}$. Assume that

$\varphi^{\varepsilon}(x_{0})=\varphi\big{(}\exp_{x_{0}}(\xi_{0})\big{)}+\varepsilon-\frac{1}{\varepsilon}|\xi|_{x_{0}}^{2},\text{
for some $\xi_{0}\in T_{x_{0}}M$}.$

Let $\xi(z)\in T_{z}M$ be a smooth vector field defined in a neighborhood of
$x_{0}$ with $\xi(x_{0})=\xi_{0}$. Put $\phi(z)=\exp_{z}(\xi(z))$, then
$z\mapsto P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|_{z}^{2}$ touches
$\varphi\circ\phi$ from above at $x_{0}$.

  2. (2)

Let $\varphi\in C(M)$ and we define $\varphi_{\varepsilon}$ according to
(5.2). Let $x_{0}\in M$ and $P$ is a $C^{2}$ function defined in a
neighgorhood $x_{0}$ that touches $\varphi_{\varepsilon}$ from below at
$x_{0}$. Assume that:

$\varphi_{\varepsilon}(x_{0})=\varphi\big{(}\exp_{x_{0}}(\tilde{\xi}_{0})\big{)}-\varepsilon+\frac{1}{\varepsilon}|\tilde{\xi}|_{x_{0}}^{2},\text{
for some $\tilde{\xi}_{0}\in T_{x_{0}}M$}.$

Let $\tilde{\xi}(z)\in T_{z}M$ be a smooth vector field defined in a
neighborhood of $x_{0}$ with $\tilde{\xi}(x_{0})=\tilde{\xi}_{0}$. Put
$\tilde{\phi}(z)=\exp_{z}(\tilde{\xi}(z))$, then $z\mapsto
P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2}_{z}$ touches
$\varphi\circ\phi$ from below at $x_{0}$.

###### Proof.

We will just prove (1), and the proof of (2) repeats that of (1) almost word
for word. By assumption, we know that $P(z)\geq\varphi^{\varepsilon}(z)$ in a
neighorhood of $x_{0}$. Therefore

$\begin{split}P(z)&\geq\sup_{\xi\in
T_{z}M}\big{(}\varphi(\exp_{z}(\xi))+\varepsilon-\frac{1}{\varepsilon}|\xi|_{z}^{2}\big{)}\geq\varphi(\exp_{z}(\xi(z))+\varepsilon-\frac{1}{\varepsilon}|\xi(z)|_{z}^{2}\\\
&=\varphi\circ\phi(z)+\varepsilon-\frac{1}{\varepsilon}|\xi(z)|_{z}^{2}.\end{split}$

Also we know that equality is achieved when $z=x_{0}$. This follows from that
$\xi(x_{0})=\xi_{0}$, and $\xi_{0}$ achieves the sup by assumption. ∎

Next we wish to choose an appropriate $\phi(z)$, and obtain $\xi(z)$ by
inverting $\exp_{z}$. Denote $w_{0}=\exp_{z}(\xi_{0})$, we hope to define
$\phi(z)$ in a neighhorhood of $x_{0}$ such that:

$\sum_{i,j}g_{i\bar{j}}(w_{0})\frac{\partial\phi_{i}}{\partial
z_{a}}\frac{\partial\bar{\phi}_{j}}{\partial\bar{z}_{b}}=g_{a\bar{b}}(x_{0}),\,\,\text{$\phi$
is holomorphic},\,\,\det D_{z}\phi(x_{0})\neq 0.$

Next we explain why this choice is possible. We first wish to estimate the
smallness of $\xi$. For this we have

######  Lemma 5.5.

  1. (1)

Let $\varphi\in C(M)$ and let $\varphi^{\varepsilon}$ be as defined by (5.1).
Let $x_{0}\in M$ and $\xi_{0}\in T_{x_{0}}M$. Assume that $\xi_{0}$ achieves
the sup in the definition of $\varphi^{\varepsilon}$ at $x_{0}$, then

$|\xi_{0}|_{x_{0}}\leq\varepsilon^{\frac{1}{2}}\big{(}\rho_{\varphi}(C\varepsilon^{\frac{1}{2}})\big{)}^{\frac{1}{2}}.$

  2. (2)

Let $\varphi_{\varepsilon}$ be as defined by (5.2) and $\tilde{\xi}_{0}\in
T_{x_{0}}M$ achieves the inf in the definition of $\varphi_{\varepsilon}$ at
$x_{0}$, then

$|\tilde{\xi}_{0}|_{x_{0}}\leq\varepsilon^{\frac{1}{2}}\big{(}\rho_{\varphi}(C\varepsilon^{\frac{1}{2}})\big{)}^{\frac{1}{2}}.$

Here the constant $C$ appearing above depends only on
$||\varphi||_{L^{\infty}}$ and the background metric.

###### Proof.

Again we will just prove (1), and the proof of (2) repeats that of (1) word
for word. Indeed, we have:

(5.3)
$\varphi^{\varepsilon}(x_{0})=\varphi(\exp_{x_{0}}(\xi_{0}))+\varepsilon-\frac{1}{\varepsilon}|\xi_{0}|_{x_{0}}^{2}\geq\varphi(x_{0})+\varepsilon.$

In the $\geq$ above, we have take $\xi=0$. Therefore, we get:

$|\xi_{0}|_{x_{0}}\leq\varepsilon^{\frac{1}{2}}(2||\varphi||_{L^{\infty}})^{\frac{1}{2}}.$

Denote $C=(2||\varphi||_{L^{\infty}})^{\frac{1}{2}}$, we go back to (5.3) and
obtain that

$|\xi_{0}|_{x_{0}}^{2}\leq\varepsilon\big{(}\varphi(\exp_{x_{0}}(\xi_{0}))-\varphi(x_{0})\big{)}\leq\varepsilon\rho_{\varphi}\big{(}d_{g}(\exp_{x_{0}}(\xi_{0}),x_{0})\big{)}\leq\varepsilon\rho_{\varphi}(|\xi_{0}|_{x_{0}})\leq\varepsilon\rho_{\varphi}(C\varepsilon^{\frac{1}{2}}).$

Hence our result follows. ∎

From Lemma 5.5, we see that, as $\varepsilon\rightarrow 0$, we would have
$\xi_{0},\,\,\tilde{\xi}_{0}\rightarrow x_{0}$.

Another observation that follows from Lemma 5.5 is that both
$\varphi^{\varepsilon}$ and $\varphi_{\varepsilon}$ approximate $\varphi$
uniformly.

######  Lemma 5.6.

$\varphi^{\varepsilon}$ and $\varphi_{\varepsilon}$ approximate $\varphi$
uniformly as $\varepsilon\rightarrow 0$. More precisely, for any $x_{0}\in M$:

$\begin{split}&\varphi(x_{0})+\varepsilon\leq\varphi^{\varepsilon}(x_{0})\leq\varphi(x_{0})+\varepsilon+\rho_{\varphi}(C\varepsilon^{\frac{1}{2}}),\\\
&\varphi(x_{0})-\varepsilon-\rho_{\varphi}(C\varepsilon^{\frac{1}{2}})\leq\varphi_{\varepsilon}(x_{0})\leq\varphi(x_{0})-\varepsilon.\end{split}$

Here $C$ is the same constant as in Lemma 5.5.

###### Proof.

By taking $\xi=0$, we see that
$\varphi^{\varepsilon}(x_{0})\geq\varphi(x_{0})+\varepsilon$,
$\varphi_{\varepsilon}(x_{0})\leq\varphi(x_{0})-\varepsilon$.

For the other inequality, we note that:

$\begin{split}&\varphi^{\varepsilon}(x_{0})=\varphi(\exp_{x_{0}}(\xi_{0}))+\varepsilon-\frac{1}{\varepsilon}|\xi_{0}|_{x_{0}}^{2}\leq\varphi(\exp_{x_{0}}(\xi_{0}))+\varepsilon\\\
&\leq\varphi(x_{0})+\rho_{\varphi}\big{(}d_{g}(x_{0},\exp_{x_{0}}(\xi_{0}))\big{)}+\varepsilon\leq\varphi(x_{0})+\rho_{\varphi}(C\varepsilon^{\frac{1}{2}})+\varepsilon.\end{split}$

∎

Now we choose a coordinate chart near $x_{0}$ such that $x_{0}$ is represented
by the origin, and that:

$g_{a\bar{b}}(x_{0})=\delta_{ab}.$

Let $N$ be an invertible $n\times n$ matrix, such that

(5.4)
$\sum_{i,j}N_{ia}\bar{N}_{jb}g_{i\bar{j}}(w_{0})=\delta_{ab}=g_{a\bar{b}}(x_{0}),\,\,\,w_{0}=\exp_{x_{0}}(\xi_{0}).$

Moreover, we may assume that:

(5.5) $|N-I|\leq C_{n}|g_{i\bar{j}}(w_{0})-\delta_{ij}|\leq
C_{n}Cd_{g}(w_{0},x_{0})\leq C_{n}C|\xi_{0}|_{x_{0}}.$

In the above, the $C$ depends only on the background metric. Now we wish to
take:

(5.6) $\phi(z)=\exp_{x_{0}}(\xi_{0})+N\cdot z.$

Next, we need to show the existence of a local vector field such that
$\exp_{z}(\xi(z))=\phi(z)$.

######  Lemma 5.7.

  1. (1)

There exists a smooth vector field $\xi(z)\in T_{z}M$, defined in a
neighborhood of $x_{0}$, such that

$\exp_{z}(\xi(z))=\phi(z).$

  2. (2)

There exist a constant $C>0$, depending only on the background metric, such
that:

$|D_{z}\xi|(x_{0})\leq C|\xi_{0}|_{x_{0}},\,\,|D_{z}^{2}\xi|(x_{0})\leq
C|\xi_{0}|_{x_{0}},\,\,|D_{z}^{2}(|\xi(z)|_{z}^{2})|\leq
C|\xi_{0}|_{x_{0}}^{2}.$

We postpone the proof of this lemma for the moment and explain first how to
use this lemma to finish the proof of Proposition 5.3. We still need one more
lemma, which justifies our choice of the map $\phi$.

######  Lemma 5.8.

Let $\phi$ be define by (5.5) in a neighborhood of $x_{0}$ with $N$ given by
(5.4). Denote $w_{0}=\exp_{x_{0}}(\xi_{0})$, then for any function $Q$ defined
in a neighborhood of $x_{0}$, one has:

$\sum_{k}g^{i\bar{k}}(Q\circ\phi^{-1}\big{)}_{j\bar{k}}(w_{0})=\sum_{a,b,p,q}N_{iq}g^{q\bar{b}}Q_{a\bar{b}}(x_{0})(N^{-1})_{aj}.$

###### Proof.

This is really a straightforward calculation. Indeed,

(5.7)
$\begin{split}&\sum_{k}g^{i\bar{k}}(w_{0})\big{(}Q\circ\phi^{-1}\big{)}_{j\bar{k}}(w_{0})=\sum_{a,b}\sum_{k}g^{i\bar{k}}(w_{0})Q_{a\bar{b}}(x_{0})\frac{\partial(\phi^{-1})_{a}}{\partial
z_{j}}\frac{\partial(\phi^{-1})_{\bar{b}}}{\partial\bar{z}_{k}}\\\
&=\sum_{a,b}\sum_{k}g^{i\bar{k}}(w_{0})(N^{-1})_{aj}\overline{(N^{-1})_{bk}}Q_{a\bar{b}}(x_{0})\\\
&=\sum_{a,b}\sum_{k}\sum_{p,q}\bar{N}_{kp}g^{q\bar{p}}(x_{0})N_{iq}(N^{-1})_{aj}\overline{(N^{-1})_{bk}}Q_{a\bar{b}})(x_{0})\\\
&=\sum_{a,b,p,q}N_{iq}g^{q\bar{b}}(x_{0})Q_{a\bar{b}})(x_{0})(N^{-1})_{aj}.\end{split}$

Some explanations are in order. In the first equality, we noted that $\phi(z)$
as given by (5.5) is holomorphic. In the third equality, we used (5.4). In the
last equality, we noted that
$\sum_{k}\bar{N}_{kp}\overline{(N^{-1})_{bk}}=\delta_{pb}$. ∎

Another thing we observe is that:

######  Lemma 5.9.

Denote $w_{0}=\exp_{x_{0}}(\xi_{0})$, then for any $i,\,j$, we have

$|g^{i\bar{k}}\chi_{j\bar{k}}(w_{0})-N_{iq}g^{q\bar{b}}\chi_{a\bar{b}}(x_{0})(N^{-1})_{aj}|\leq
C|\xi_{0}|_{x_{0}}.$

In the above, $C$ depends only on the background metric and the form $\chi$.

###### Proof.

To see this, we simply write this as a telescoping sum:

$\big{(}g^{i\bar{k}}\chi_{j\bar{k}}(w_{0})-g^{i\bar{k}}\chi_{j\bar{k}}(x_{0})\big{)}+\big{(}\delta_{iq}g^{q\bar{b}}\chi_{a\bar{b}}(x_{0})\delta_{aj}-N_{iq}g^{q\bar{b}}\chi_{a\bar{b}}(x_{0})(N^{-1})_{aj}\big{)}.$

The first bracket above is clearly bounded by $Cd_{g}(w_{0},x_{0})\leq
C^{\prime}|\xi_{0}|_{x_{0}}$. For the second bracket, we need to use (5.5) to
see that it is also bounded by $C|\xi_{0}|_{x_{0}}$. ∎

###### Proof.

(Of Proposition 5.3) First we prove (1). Let $P$ be a $C^{2}$ function that
touches $\frac{\varphi^{\varepsilon}}{1+a_{1}}$ from above at $x_{0}$, which
is the same as saying $(1+a_{1})P$ touches $\varphi^{\varepsilon}$ from above
at $x_{0}$. We need to show that:

(5.8)
$\begin{split}&\lambda\big{(}\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}P\big{)}\in\Gamma,\\\
&f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}P](x_{0})\big{)}\geq
e^{G(x_{0},P(x_{0}))}-\rho_{1}(\varepsilon,a_{1}).\end{split}$

The choice of $\rho(\varepsilon),\,\rho_{1}(\varepsilon,a_{1})$ will be made
clear later on. First we can see from Lemma 5.4, part (1) that the function
$(1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|_{z}^{2}$ touches
$\varphi\circ\phi$ from above. Note that $\phi(x_{0})=\exp_{x_{0}}(\xi_{0})$
and that $\det D_{z}\phi(x_{0})\neq 0$, we see that $\phi$ defines an
invertible map between a neighborhood of $x_{0}$ and a neighborhood of
$\exp_{x_{0}}(\xi_{0})$.

Therefore, we see that
$\big{(}(1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}\big{)}\circ\phi^{-1}(w)$
touches $\varphi$ from above at $\phi(x_{0})=\exp_{x_{0}}(\xi_{0})$.
Therefore, using that $\varphi$ is a viscosity solution, we see that: with
$w_{0}=\exp_{x_{0}}(\xi_{0})$,

(5.9)
$\begin{split}&\lambda[\chi+dd^{c}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\circ\phi^{-1}](w_{0})\in\Gamma,\\\
&f\big{(}\lambda[\chi+dd^{c}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\circ\phi^{-1}](w_{0})\big{)}\geq
e^{G(w_{0},((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\circ\phi^{-1}(w_{0}))}.\end{split}$

We wish to show that (5.9) implies (5.8). For this, we may calculate, for
fixed $i,\,j$:

(5.10)
$\begin{split}&g^{i\bar{k}}\big{(}\chi_{j\bar{k}}+\big{(}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi|^{2}_{z})\circ\phi^{-1}\big{)}_{j\bar{k}}\big{)}(w_{0})\\\
&=g^{i\bar{k}}\chi_{j\bar{k}}(w_{0})+N_{iq}g^{q\bar{b}}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})_{a\bar{b}}(x_{0})(N^{-1})_{aj}\\\
&\leq
N_{iq}g^{q\bar{b}}(\chi_{a\bar{b}}(x_{0})+C|\xi_{0}|_{x_{0}}g_{a\bar{b}}(x_{0}))(N^{-1})_{aj}\\\
&+N_{iq}g^{q\bar{b}}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})_{a\bar{b}}(x_{0})(N^{-1})_{aj}\end{split}$

In the equality above, we used Lemma 5.8 with
$Q=(1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|_{z}^{2}$. The
inequality above follows from Lemma 5.9. Then we see from (5.10) that:

(5.11)
$\lambda[\chi+dd^{c}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\circ\phi^{-1}]\leq\lambda\big{(}\chi+C|\xi_{0}|_{x_{0}}\omega_{0}+dd^{c}((1+a_{1})P+\frac{1}{\varepsilon}|\xi(z)|_{z}^{2})\big{)}.$

The meaning of the above inequality is that the difference belongs to
$\Gamma_{n}$. Moreover, we may use Lemma 5.5 and 5.7 to see that:

(5.12)
$\begin{split}&\chi+C|\xi_{0}|_{x_{0}}\omega_{0}+dd^{c}((1+a_{1})P+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\\\
&\leq\chi+C^{\prime}(\rho_{\varphi}^{\frac{1}{2}}(C^{\prime\prime}\varepsilon^{\frac{1}{2}})+\rho_{\varphi}(C^{\prime\prime}\varepsilon^{\frac{1}{2}}))\omega_{0}+(1+a_{1})dd^{c}P.\end{split}$

Therefore, if we now put
$\rho(\varepsilon)=C^{\prime}(\rho_{\varphi}^{\frac{1}{2}}(C^{\prime\prime}\varepsilon^{\frac{1}{2}})+\rho_{\varphi}(C^{\prime\prime}\varepsilon^{\frac{1}{2}}))$,
we see that:

$\lambda\big{(}\chi+\rho(\varepsilon)\omega_{0}+(1+a_{1})dd^{c}P\big{)}\in\Gamma.$

This is exactly the first statement of (5.8). For the second statement, the
above calculation already implies that:

(5.13)
$\begin{split}&f\big{(}\lambda[\chi+dd^{c}((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\circ\phi^{-1}](w_{0})\big{)}\leq
f\big{(}\lambda\big{(}\chi+\rho(\varepsilon)\omega_{0}+(1+a_{1})dd^{c}P\big{)}\big{)}\\\
&=(1+a_{1})f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}P]\big{)}.\end{split}$

The equality here uses that $f$ is of homogeneous one. On the other hand,

(5.14)
$\begin{split}&\frac{1}{1+a_{1}}e^{G(w_{0},((1+a_{1})P-\varepsilon+\frac{1}{\varepsilon}|\xi(z)|^{2}_{z})\circ\phi^{-1}(w_{0}))}=\frac{1}{1+a_{1}}e^{G(w_{0},((1+a_{1})P(x_{0})-\varepsilon+\frac{1}{\varepsilon}|\xi_{0}|^{2}_{x_{0}}))}\\\
&\geq e^{G(x_{0},P(x_{0}))}-\rho_{1}(\varepsilon,a_{1}).\end{split}$

Here we again used Lemma 5.5 on the estimate of $|\xi_{0}|_{x_{0}}$. Combining
(5.13) and (5.14) gives the second statement of (5.8). So far we have proved
the first part of Proposition 5.3. The second part of Proposition 5.3 is
proved similarly, which we sketch briefly.

Let $P$ be a $C^{2}$ function defined in a neighborhood of $x_{0}$, which
touches $\frac{\varphi_{\varepsilon}}{1-a_{2}}$ from below at $x_{0}$. We need
to show that:

(5.15) $\begin{split}&\text{Either
$\lambda\big{(}\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}P\big{)}(x_{0})\notin\Gamma$,
or}\\\
&f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}P](x_{0})\big{)}\leq
e^{G(x_{0},P(x_{0})}+\rho_{2}(\varepsilon,a_{2}).\end{split}$

Since $(1-a_{2})P$ touches $\varphi_{\varepsilon}$ from below at $x_{0}$, we
see from Lemma 5.4, part (2) that
$\big{(}(1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2}_{z}\big{)}\circ\tilde{\phi}^{-1}(w)$
touches $\varphi$ from below at
$\tilde{\phi}(x_{0})=\exp_{x_{0}}(\tilde{\xi}_{0})$ (which we denote as
$\tilde{w}_{0}$ from now on). Here $\tilde{\xi}_{0}$, $\tilde{\xi}(z)$,
$\tilde{\phi}(z)$ is defined in the same way as $\varphi^{\varepsilon}$, hence
satisfy the same estimates as $\xi_{0},\,\xi(z),\,\phi(z)$. Therefore, we may
conclude, as before:

(5.16) $\begin{split}&\text{Either
$\lambda[\chi+dd^{c}((1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2}_{z})\circ\tilde{\phi}^{-1}](\tilde{w}_{0})\notin\Gamma$,
or}\\\
&f\big{(}\lambda[\chi+dd^{c}((1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|_{z}^{2})\circ\tilde{\phi}^{-1}](\tilde{w}_{0})\leq
e^{G(\tilde{w}_{0},((1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2})\circ\tilde{\phi}^{-1}(\tilde{w}_{0}))}.\end{split}$

So we just need to deduce (5.15) from (5.16). Similar calculations as in the
proof of part (1) will show that:

(5.17)
$\lambda\big{(}\chi-\rho(\varepsilon)\omega_{0}+dd^{c}(1-a_{2})P\big{)}(x_{0})\leq\lambda[\chi+dd^{c}((1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2}_{z})\circ\tilde{\phi}^{-1}](\tilde{w}_{0}).$

Here you can actually make $\rho(\varepsilon)$ to be the same as part (1).
Therefore, if
$\lambda[\chi-\rho(\varepsilon)\omega_{0}+dd^{c}(1-a_{2})P]\in\Gamma$, it will
imply
$\lambda[\chi+dd^{c}((1-a_{2})P+\varepsilon+\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2}_{z})\circ\tilde{\phi}^{-1}](\tilde{w}_{0})\in\Gamma$.
Moreover, in this case, we also have:

$\begin{split}&f\big{(}\lambda\big{(}\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}P\big{)}\big{)}\leq\frac{1}{1-a_{2}}f\big{(}\lambda[\chi+((1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2}_{z})\circ\tilde{\phi}^{-1}]\big{)}(\tilde{w}_{0})\\\
&\leq\frac{1}{1-a_{2}}e^{G(\tilde{w}_{0},((1-a_{2})P+\varepsilon-\frac{1}{\varepsilon}|\tilde{\xi}(z)|^{2})\circ\tilde{\phi}^{-1}(\tilde{w}_{0}))}\end{split}$

The same calculation as in the proof of part (1) shows that one can estimate
the right hand side from above by
$e^{G(x_{0},P(x_{0}))}+\rho_{2}(\varepsilon,a_{2})$ (again we note that
$\tilde{\phi}^{-1}(\tilde{w}_{0})=x_{0}$). ∎

Now let us prove Lemma 5.7

###### Proof.

(Of Lemma 5.7) The existence part of the vector field $\xi(z)$ is a result of
implicit function theorem. Let $U$ be an open subset of $TM$, such that
$(x_{0},\xi_{0})\in U$ and that $\exp_{z}(\xi),\,\phi(z)$ is inside the
coordinate chart near $x_{0}$. We consider the following map:

$\mathcal{F}:U\rightarrow\mathbb{C}^{n},\,\,\,(z,\xi)\mapsto\exp_{z}(\xi)-\phi(z).$

On the right hand side above, we have identified $\exp_{z}(\xi)$ and $\phi(z)$
with points in $\mathbb{C}^{n}$ using the coordinate chart, so that the
subtraction makes sense. Moreover,

$D_{\xi}\mathcal{F}|_{(z,\xi)(q)=(x_{0},\xi_{0})}=D_{\xi}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}(q),\,\,\,q\in
T_{x_{0}}M.$

We know that $D_{\xi}(\exp_{z}\xi)|_{z=x_{0},\,\xi=0}(q)=q$, hence
$D_{\xi}(\exp_{z}\xi)_{z=x_{0},\,\xi=\xi_{0}}$ would be non-singular, since
$\xi_{0}$ is very close to 0 due to Lemma 5.5. Moreover, we also know that
$\mathcal{F}(x_{0},\xi_{0})=0$. Therefore, we may conclude from implicit
function theorem that there is a neighborhood $V$ of $x_{0}$, and a vector
field $\xi(z),\,z\in V$, such that $\mathcal{F}(z,\xi(z))=0$. Namely
$\phi(z)=\exp_{z}(\xi(z))$.

Now we derive the estimates of $\xi(z)$. By differentiation, we see that:

$N=D_{z}\phi(z)|_{z=x_{0}}=D_{z}(\exp_{z}\xi)|_{z=x_{0},\xi=\xi_{0}}+D_{\xi}(\exp_{z}\xi)|_{z=x_{0},\xi=\xi_{0}}D_{z}\xi|_{z=x_{0}}.$

From the lemma below, we know that
$|D_{z}(\exp_{z}\xi)-I|_{z=x_{0},\,\xi=\xi_{0}}\leq C|\xi_{0}|_{x_{0}}$. Hence
we may use (5.5) to see that:

$|N-D_{z}(\exp_{z}(\xi))|_{z=x_{0},\,\xi=\xi_{0}}|\leq C|\xi_{0}|_{x_{0}}.$

Here the $C$ depends only on the background metric. Also we noted that
$D_{\xi}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}$ is invertible, we see that
$|D_{z}\xi|_{z=x_{0}}\leq C|\xi_{0}|_{x_{0}}$. Differentiating once more, we
get:

$\begin{split}&0=D_{zz}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}+2D_{z\xi}(\exp_{z}\xi)|_{z=x_{0},\xi=\xi_{0}}D_{z}\xi|_{z=x_{0}}\\\
&+D_{\xi\xi}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}D_{z}\xi*D_{z}\xi|_{z=x_{0}}+D_{\xi}(\exp_{z}\xi)|_{z=x_{0},\xi=\xi_{0}}D_{z}^{2}\xi|_{z=x_{0}}.\end{split}$

In the above, $D_{z}\xi*D_{z}\xi$ just denotes some quadratic expression of
$D_{z}\xi$. Using the lemma below, we know that:

$|D_{zz}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}\leq
C|\xi_{0}|_{x_{0}},\,\,|D_{z\xi}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}\leq
C,\,|D_{\xi\xi}(\exp_{z}\xi)|_{z=x_{0},\,\xi=\xi_{0}}\leq C.$

From this, we see that

$|D_{z}^{2}\xi|_{z=x_{0}}\leq C|\xi_{0}|_{x_{0}}.$

Finally,

$\begin{split}&D_{z}^{2}\big{(}|\xi(z)|_{z}^{2}\big{)}=D_{z}^{2}\big{(}g_{i\bar{j}}(z)\xi_{i}(z)\bar{\xi}_{j}(z)\big{)}=D_{z}^{2}g_{i\bar{j}}(z)\xi_{i}(z)\bar{\xi}_{j}(z)+D_{z}g_{i\bar{j}}D_{z}\xi_{i}\bar{\xi}_{j}\\\
&+D_{z}g_{i\bar{j}}\xi_{i}D_{z}\bar{\xi}_{j}+g_{i\bar{j}}\big{(}D_{z}^{2}\xi_{i}\bar{\xi}_{j}+\xi_{i}D_{z}^{2}\bar{\xi}_{j}+D_{z}\xi_{i}*D_{z}\bar{\xi}_{j}\big{)}.\end{split}$

Using the above estimates for $D_{z}\xi$ and $D_{z}^{2}\xi$, we see that
$|D_{z}^{2}\xi(z)|_{z=x_{0}}\leq C|\xi_{0}|_{x_{0}}^{2}$. ∎

In the above, we used the following lemma about the Taylor expansion of the
exponential map. This lemma can be found in [4].

######  Lemma 5.10.

The exponential map on a Hermitian manifold has the Taylor expansion in the
following form under local coordinates:

$\exp_{z}(\xi)_{m}=g_{m}(z,\xi)+\sum_{j,k,l}c_{jklm}(\frac{1}{2}\bar{z}_{k}+\frac{1}{6}\bar{\xi}_{k})\xi_{j}\xi_{l}+O\big{(}|\xi|^{2}(|z|+|\xi|)^{2}),$

where

$g_{m}(z,\xi)=z_{m}+\xi_{m}-\sum_{j,l}a_{jlm}z_{j}\xi_{l}+\sum_{j,k,l,p}a_{jlp}a_{kpm}z_{j}z_{k}\xi_{l}-\sum_{j,k,l}b_{jklm}(z_{j}z_{k}\xi_{l}+z_{k}\xi_{j}\xi_{l}+\frac{1}{3}\xi_{j}\xi_{k}\xi_{l}).$

In the above, $(\exp_{z}\xi)_{m}$ denotes the $m$-th component of the
exponential map under local coordinates.

Finally we observe that $\varphi^{\varepsilon}$ and $\varphi_{\varepsilon}$
are twice differentiable a.e. Indeed, one has:

######  Lemma 5.11.

Let $\varphi\in C(M)$, and we define
$\varphi^{\varepsilon},\,\varphi_{\varepsilon}$ according to (5.1) and (5.2).
Let $x_{0}\in M$ and we choose local coordinates in a neighborhood of $x_{0}$.
Then there exists a neighborhood $U$ of $x_{0}$, and $C_{\varepsilon}>0$, such
that: $z\mapsto\varphi^{\varepsilon}(z)+C_{\varepsilon}|z|^{2}$ is convex on
$U$ under the coordinates, and
$z\mapsto\varphi_{\varepsilon}(z)-C_{\varepsilon}|z|^{2}$ is concave on $U$
under the coordinates. In particular, $\varphi^{\varepsilon}$,
$\varphi_{\varepsilon}$ are twice differentiable a.e.

###### Proof.

We just prove that $\varphi^{\varepsilon}$ is semi-convex. The proof that
$\varphi_{\varepsilon}$ is semi-concave follows similar lines. We can choose
$U_{0}$ small enough, such that for any $z,\,w\in U_{0}$, there is a unique
$\xi\in T_{z}M$ such that $\exp_{z}(\xi)=w$. Moreover, we can assume that
$\xi$ depends smoothly on $z$ and $w$, and that $|\xi|^{2}_{z}$ is also smooth
in $z$ and $w$. Therefore, one has, for some neighborhood $U$ of $x_{0}$
(possibly smaller than $U_{0}$):

(5.18) $\begin{split}&\varphi^{\varepsilon}(z)=\sup_{\xi\in
T_{z}M}\big{(}\varphi(\exp_{z}(\xi)+\varepsilon-\frac{1}{\varepsilon}|\xi|^{2}_{z})\big{)}=\sup_{\xi\in
T_{z}M,\,|\xi|<r_{0}}\big{(}\varphi(\exp_{z}(\xi))+\varepsilon-\frac{1}{\varepsilon}|\xi|^{2}\big{)}\\\
&=\sup_{w\in
U_{0}}\big{(}\varphi(w)+\varepsilon-\frac{1}{\varepsilon}|\xi(z,w)|^{2}_{z}\big{)},\,\,z\in
U.\end{split}$

The second equality used Lemma 5.5 on the estimate of $\xi$ that achieves the
sup. In the last inequality, we noted that for some neighborhood $U$ of
$x_{0}$, the image of $U_{0}\ni w\mapsto\xi(z,w)$ will cover $\\{\xi\in
T_{z}M:|\xi|_{z}<r_{0}\\}$ for any $z\in U$. Note that in (5.18), the function
$z\mapsto\varphi(w)+\varepsilon-\frac{1}{\varepsilon}|\xi(z,w)|^{2}_{z}$ is
smooth, and has uniform $C^{2}$ bound (in $z$, uniform with respect to $w$).
Therefore, taking sup will imply that $\varphi^{\varepsilon}$ is semi-convex.
∎

###  5.2. when the right hand side has strict monotonicity

In this subsection, we assume that $G(x,\varphi)$ is continuous and strictly
monotone increasing in $\varphi$. We wish to show that:

######  Theorem 5.1.

Let $G(x,\varphi)$ be continuous, and strictly monotone increasing in
$\varphi$. Then there exists at most one viscosity solution to
$F(\chi+dd^{c}\varphi)=e^{G(x,\varphi)}$.

If not, then there exist two viscosity solutions $\varphi_{1},\,\varphi_{2}$,
and $\varphi_{1}\neq\varphi_{2}$. Without loss of generality, we may assume
that:

(5.19) $\kappa_{0}:=\max_{M}(\varphi_{1}-\varphi_{2})>0.$

Now we consider the sup-convolution, applied to $\varphi_{1}$, and the inf-
convolution, applied to $\varphi_{2}$. We define:

$\kappa_{\varepsilon,a_{1},a_{2}}=\max_{M}\big{(}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}\big{)}.$

Then it is easy to see that, as $\varepsilon,\,a_{1},\,a_{2}\rightarrow 0+$,
$\kappa_{\varepsilon,a_{1},a_{2}}\rightarrow\kappa_{0}$. Assume that the above
max is achieved at $x_{*}$. Then we have:

######  Proposition 5.12.

Assume that $\varepsilon,\,a_{1},\,a_{2}$ are chosen so that
$\frac{c_{*}+\rho(\varepsilon)}{1+a_{1}}<\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}}$.
Assume also that both $(\varphi_{1})^{\varepsilon}$ and
$(\varphi_{2})_{\varepsilon}$ are twice differentiable at $x_{*}$. Then one
has:

  1. (1)

$\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}](x_{*})\in\Gamma$,

  2. (2)

$F(\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}})(x_{*})\leq
F(\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}})(x_{*})$.

In the above, $c_{*}>0$ is the constant that allows one to write
$\chi=\tilde{\chi}+c_{*}\omega_{0}$ with $\lambda(\tilde{\chi})\in\Gamma$.

This proposition allows us to exclude the non-unqueness of viscosity solution,
as long as $(\varphi_{1})^{\varepsilon}$, $(\varphi_{2})_{\varepsilon}$ are
both twice differentiable at $x_{*}$.

###### Proof.

(of Theorem 5.1, assuming $(\varphi_{1})^{\varepsilon}$ and
$(\varphi_{2})_{\varepsilon}$ are twice differentiable at $x_{*}$) We choose
$\varepsilon,\,a_{1},\,a_{2}$ so that
$\frac{c_{*}+\rho(\varepsilon)}{1+a_{1}}<\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}}$.
Combining Proposition 5.3 and Proposition 5.12, we see that at $x_{*}$:

(5.20)
$\begin{split}&f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\big{)}(x_{*})\leq
f(\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}])(x_{*}),\\\
&f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\big{)}(x_{*})\geq
e^{G(x_{*},\frac{(\varphi_{1})^{\varepsilon}(x_{*})}{1+a_{1}})}-\rho_{1}(\varepsilon,a_{1}),\\\
&f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}])\big{)}(x_{*})\leq
e^{G(x_{*},\frac{(\varphi_{2})_{\varepsilon}(x_{*})}{1-a_{2}})}+\rho_{2}(\varepsilon,a_{2}).\end{split}$

Combining the three inequalities, we see that:

(5.21)
$e^{G(x_{*},\frac{(\varphi_{1})^{\varepsilon}(x_{*})}{1+a_{1}})}-\rho_{1}(\varepsilon,a_{1})\leq
e^{G(x_{*},\frac{(\varphi_{2})_{\varepsilon}(x_{*})}{1-a_{2}})}+\rho_{2}(\varepsilon,a_{2}).$

On the other hand,
$\frac{(\varphi_{1})^{\varepsilon}(x_{*})}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}(x_{*})}{1-a_{2}}=\kappa_{\varepsilon,a_{1},a_{2}}\rightarrow\kappa_{0}>0$
as $\varepsilon,\,a_{1},\,a_{2}\rightarrow 0$. This is clearly inconsistent
with (5.21) when $\varepsilon,\,a_{1},\,a_{2}$ are all small enough. ∎

Now we prove Proposition 5.12.

###### Proof.

(Of Proposition 5.12) Since both $(\varphi_{1})^{\varepsilon}$ and
$(\varphi_{2})_{\varepsilon}$ are differentiable at $x_{*}$, and that
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$
achieves maximum at $x_{*}$, we see that:

$dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}\geq
0.$

Moreover,

$\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}-\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}=\tilde{\chi}(\frac{1}{1-a_{2}}-\frac{1}{1+a_{1}})+(\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}}-\frac{c_{*}+\rho(\varepsilon)}{1+a_{1}})\omega_{0}.$

Therefore,

(5.22)
$\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}\geq\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}+c_{1}\tilde{\chi}+\eta.$

In the above, $c_{1}=\frac{1}{1-a_{2}}-\frac{1}{1+a_{1}}>0$,
$\lambda(\tilde{\chi})\in\Gamma$, $\eta\geq 0$. Also by Proposition 5.3, part
(1), we also have
$\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\in\Gamma$.
Therefore, we see that
$\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}]\in\Gamma$.

To prove the second part, we wish to use concavity. Indeed, one has:

$\begin{split}&f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\big{)}(x_{*})\leq
f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}]\big{)}(x_{*})\\\
&+\frac{\partial F}{\partial
h_{i\bar{j}}}(\frac{\chi-\rho(\varepsilon)}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}})\big{(}\frac{\chi_{i\bar{j}}+\rho(\varepsilon)g_{i\bar{j}}}{1+a_{1}}+\frac{(\varphi_{1})_{i\bar{j}}^{\varepsilon}}{1+a_{1}}\\\
&-\frac{\chi_{i\bar{j}}-\rho(\varepsilon)g_{i\bar{j}}}{1-a_{2}}-dd^{c}\frac{((\varphi_{2})_{\varepsilon})_{i\bar{j}}}{1-a_{2}}\big{)}.\end{split}$

So we just need to show that:

$\begin{split}&\frac{\partial F}{\partial
h_{i\bar{j}}}(\frac{\chi-\rho(\varepsilon)}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}})\big{(}\frac{\chi_{i\bar{j}}+\rho(\varepsilon)g_{i\bar{j}}}{1+a_{1}}+\frac{(\varphi_{1})_{i\bar{j}}^{\varepsilon}}{1+a_{1}}\\\
&-\frac{\chi_{i\bar{j}}-\rho(\varepsilon)g_{i\bar{j}}}{1-a_{2}}-\frac{((\varphi_{2})_{\varepsilon})_{i\bar{j}}}{1-a_{2}}\big{)}\leq
0.\end{split}$

By Lemma 4.6, we just need to show that
$\lambda\big{(}\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}-dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}\big{)}\in\Gamma$.
However, one can already see this from (5.22). ∎

Next we look at the general case, without assuming
$(\varphi_{1})^{\varepsilon}$, $(\varphi_{2})_{\varepsilon}$ twice
differentiable at $x_{*}$. For this, we need a perturbation argument from [7]
and [16].

First we choose normal coordinate near $x_{*}$, such that $x_{*}$ is given by
$z=0$. We wish to show that:

######  Lemma 5.13.

There exists a neighborhood $U_{0}$ of $x_{*}$, and there exists a sequence
$p_{k}\in\mathbb{C}^{n}$, $p_{k}\rightarrow 0$, and a sequence
$\delta_{k}>0,\,\delta_{k}\rightarrow 0$, such that one can find a sequence of
points $x_{k}\in U_{0}$, such that

  1. (1)

$x_{k}\rightarrow x_{*}$ as $k\rightarrow\infty$,

  2. (2)

$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-<p_{k},z>-\delta_{k}|z|^{2}$
has local maximum at $x_{k}$,

  3. (3)

Both $(\varphi_{1})^{\varepsilon}$ and $(\varphi_{2})_{\varepsilon}$ are twice
differentiable at $x_{k}$.

Another thing we note is that:

######  Lemma 5.14.

Define
$\psi_{\varepsilon,a_{1},k}(z)=\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-<p_{k},z>-\delta_{k}|z|^{2}$
on $U_{0}$. Then for large enough $k$ ($\varepsilon,\,a_{1}$ is fixed now)
$\psi_{\varepsilon,a_{1},k}$ solves the following equation on $U_{0}$ in the
viscosity sense:

$\begin{split}&\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\psi_{\varepsilon,a_{1},k}]\in\Gamma.\\\
&f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\psi_{\varepsilon,a_{1},k}]\big{)}\geq
e^{G(x,\psi_{\varepsilon,a_{1},k})}-2\rho_{1}(a_{1},\varepsilon).\end{split}$

We first explain how this implies the uniqueness of viscosity solution in the
general case.

###### Proof.

(Of Theorem 5.1, in the general case) We just need to suitably choose
$\varepsilon,\,a_{1},\,a_{2}$, and choose $k$ large enough, then we evaluate
at $x_{k}$. Now we know that
$\psi_{\varepsilon,a_{1},k}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$
achieves maximum at $x_{k}$. Moreover, both $\psi_{\varepsilon,a_{1},k}$ and
$(\varphi_{2})_{\varepsilon}$ are twice differentiable at $x_{k}$. Therefore,
if we follow the argument of Proposition 5.12, we see that if we choose
$\varepsilon,\,a_{1},\,a_{2}$ so that
$\frac{c_{*}+2\rho(\varepsilon)}{1+a_{1}}<\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}}$,
we would be able to conclude that:

$f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\psi_{\varepsilon,a_{1},k}]\big{)}(x_{k})\leq
f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}]\big{)}(x_{k}).$

Therefore, we see that:

(5.23)
$e^{G(x_{k},\psi_{\varepsilon,a_{1},k}(x_{k}))}-2\rho_{1}(a_{1},\varepsilon)\leq
e^{G(x_{k},\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}(x_{k}))}+\rho_{2}(a_{2},\varepsilon).$

Now one passes to limit as $k\rightarrow\infty$. Note that
$\psi_{\varepsilon,a_{1},k}(x_{k})\rightarrow\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}(x_{*})$,
we see that:

$e^{G(x_{*},\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}(x_{*}))}-2\rho_{1}(a_{1},\varepsilon)\leq
e^{G(x_{*},\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}(x_{*}))}+\rho_{2}(a_{2},\varepsilon).$

On the other hand,
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}(x_{*})-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}(x_{*})$
is strictly positive and bounded away from zero as
$\varepsilon,\,a_{1},\,a_{2}\rightarrow 0$. From the strict monotonicity of
$G$, we see a contradiction. ∎

Now it only remains to establish the technicalities Lemma 5.13 and 5.14. We
start with Lemma 5.14.

###### Proof.

(Of Lemma 5.14) Let $x_{0}\in U_{0}$ and let $P$ be a $C^{2}$ function on
$U_{0}$ that touches $\psi_{\varepsilon,a_{1},k}$ from above at $x_{0}$. This
would imply that $P+<p_{k},z>+\delta_{k}|z|^{2}$ touches
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}$ at $x_{0}$. Therefore, we may
use Proposition 5.3, part (1) to conclude that:

$\begin{split}&\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}(P+<p_{k},z>+\delta_{k}|z|^{2})]\in\Gamma,\\\
&f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}(P+<p_{k},z>+\delta_{k}|z|^{2}]\big{)}\\\
&\geq
e^{G(x_{0},(P+<p_{k},z>+\delta_{k}|z|^{2})(x_{0}))}-\rho_{1}(\varepsilon,a_{1}).\end{split}$

On the other hand, it is easy to see that when $k$ is large enough (so that
$\delta_{k}$ is small enough), one has:

$\begin{split}&\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}(P+<p_{k},z>+\delta_{k}|z|^{2})\leq\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}P,\\\
&e^{G(x_{0},(P+<p_{k},z>+\delta_{k}|z|^{2})(x_{0}))}-\rho_{1}(\varepsilon,a_{1})\geq
e^{G(x_{0},P(x_{0}))}-2\rho_{1}(\varepsilon,a_{1}).\end{split}$

Combining, we get:

$\begin{split}&\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}P]\in\Gamma,\\\
&f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}P]\big{)}\geq
e^{G(x_{0},P(x_{0}))}-2\rho_{1}(\varepsilon,a_{1}).\end{split}$

∎

Now we prove Lemma 5.13.

###### Proof.

(Of Lemma 5.13) Without loss of generality, let us assume that
$U_{0}=B_{1}(0)$, $x_{*}=0$ under the local coordinates. Then this lemma
really follows from a lemma of Crandall, Ishii and Lions ([3] lemma A.3),
which states that:

Let $\varphi:\mathbb{R}^{N}\rightarrow\mathbb{R}$ be semi-convex and $\hat{x}$
be a strict local maximum of $\varphi$. For $p\in\mathbb{R}^{N}$, put
$\varphi_{p}(x)=\varphi(x)+<p,x>$. Then for any $r>0,\,\delta>0$, the
following set $K$ has positive measure:

$K=\\{x\in B_{r}(\hat{x}):\text{there exists $p\in B_{\delta}(0)$ such that
$\varphi_{p}(x)$ has a local maximum at $x$}\\}.$

We are going to apply this lemma with
$\varphi=\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-\frac{1}{k}|z|^{2}$.
This function will be semi-convex due to Lemma 5.11. Moreover, we know that
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-\frac{1}{k}|z|^{2}$
has strict maximum at $x_{*}$ (given by $z=0$). So that the above lemma
applies.

We will also choose $\delta=\frac{1}{k},\,r=\frac{1}{k}$. So that we can
conclude the set of $x\in B_{\frac{1}{k}}(x_{*})$ such that
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-\frac{1}{k}|z|^{2}+<p,z>$
has local maximum at $x$ for some $|p|\leq\frac{1}{k}$ has positive measure.
Since $(\varphi_{1})^{\varepsilon}$ and $(\varphi_{2})_{\varepsilon}$ are both
twice differentiable a.e., we can find $x_{k}\in B_{\frac{1}{k}}(x_{*})$
belonging to the above said set, such that both $(\varphi_{1})^{\varepsilon}$
and $(\varphi_{2})_{\varepsilon}$ are twice differentiable at $x_{k}$. If we
denote the corresponding $p$ to be $p_{k}$, we see that we are done. ∎

###  5.3. when the right hand side does not depend on $\varphi$

In this subsection, we will assume that the right hand side $G$ depends only
on $x$. In this case, one can only hope to solve
$F(\chi+dd^{c}\varphi)=e^{G+c}$, for some constant $c$. Even though we can
show that there is a unique $c\in\mathbb{R}$ that makes this equation solvable
in the viscosity sense, we still don’t have a good enough understanding of
this constant, which is the main hurdle to a proof of uniqueness of viscosity
solution in this case. Let us start with observing the monotonicity of the
constant, in terms of the right hand side. More specifically, we have:

######  Proposition 5.15.

Let $G_{1},\,G_{2}\in C(M)$ with $G_{1}\geq G_{2}$. Assume that there exist
$c_{i}\in\mathbb{R},\,\varphi_{i}\in C(M),\,i=1,\,2$, that solve
$F(\chi+dd^{c}\varphi_{i})=e^{G_{i}+c_{i}}$ in the viscosity sense. Then one
has $c_{1}\leq c_{2}$.

###### Proof.

Assume that this is false, namely $c_{1}>c_{2}$. This would imply that
$e^{G_{1}+c_{1}}>e^{G_{2}+c_{2}}+\delta_{0}$ on $M$, for some $\delta_{0}>0$.
Heuristically this would lead to a contradiction if one evaluates at $x_{0}$,
where $\varphi_{1}-\varphi_{2}$ achieves maximum. At this point, one would
have $\chi+dd^{c}\varphi_{1}\leq\chi+dd^{c}\varphi_{2}$, so that
$F(\chi+dd^{c}\varphi_{1})\leq F(\chi+dd^{c}\varphi_{2})$ at $x_{0}$.

To proceed rigorously, one needs to perform the sup/inf convolutions
considered in Subsection 4.1. One consideres
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$,
where $(\varphi_{1})^{\varepsilon}$ and $(\varphi_{2})_{\varepsilon}$ are
defined according to (5.1) and (5.2).

Assume that the maximum is achieved at $x_{*}$. If both
$(\varphi_{1})^{\varepsilon}$ and $(\varphi_{2})_{\varepsilon}$ are both twice
differentiable at $x_{*}$, then one has:

$\begin{split}&f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\big{)}(x_{*})\geq
e^{G_{1}(x_{*})+c_{1}}-\rho_{1}(\varepsilon,a_{1}),\\\
&f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}]\big{)}(x_{*})\leq
e^{G_{2}(x_{*})+c_{2}}+\rho_{2}(\varepsilon,a_{2}).\end{split}$

Since
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$
has maximum at $x_{*}$, we see that, if $\varepsilon,\,a_{1},\,a_{2}$ are
chosen so that
$\frac{c_{*}+\rho(\varepsilon)}{1+a_{1}}<\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}}$,
we can follow the argument of Proposition 5.12 to see that:

$f\big{(}\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\big{)}(x_{*})\leq
f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}]\big{)}(x_{*}).$

Then one gets: $e^{G_{1}(x_{*})+c_{1}}-\rho_{1}(\varepsilon,a_{1})\leq
e^{G_{2}(x_{*})+c_{2}}+\rho_{2}(\varepsilon,a_{2})$. This is inconsistent with
$e^{G_{1}+c_{1}}\geq e^{G_{2}+c_{2}}+\delta_{0}$, if one chooses
$\varepsilon,\,a_{1},\,a_{2}$ all small enough, and
$\frac{c_{*}+\rho(\varepsilon)}{1+a_{1}}<\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}}$.

In the general case, we can take a coordinate chart in a neighborhood of
$x_{*}$, and consider
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}-<p_{k},z>-\delta_{k}|z|^{2}$
with $p_{k},\,\delta_{k}\rightarrow 0$. Using the same argument as in the
proof of Theorem 5.1 in the general case, we can find a sequence
$x_{k}\rightarrow x_{*}$ such that both $(\varphi_{1})^{\varepsilon}$ and
$(\varphi_{2})_{\varepsilon}$ are twice differentiable at $x_{k}$, and the
above function has local minimum at $x_{k}$. We still get a contradiction
after evaluating at $x_{k}$ and passing to the limit as $k\rightarrow\infty$.
∎

A direct consequence of the above proposition is that there is a unique
constant $c$ that allows for a viscosity solution. That is,

######  Corollary 5.16.

Let $G\in C(M)$, there is at most one constant $c\in\mathbb{R}$, such that
$F(\chi+dd^{c}\varphi)=e^{G+c}$ is solvable in the viscosity sense.

Because of this result, we may simply denote this constant to be $c(G)$. And
Proposition 5.15 implies that $c(G_{1})\leq c(G_{2})$ whenever $G_{1}\geq
G_{2}$.

The question that is of crucial importance to us is the following:

######  Question 5.17.

Assume that $G_{1}\geq G_{2}$, and $G_{1}\neq G_{2}$. Do we actually have
$c(G_{1})<c(G_{2})$?

In order to justify its importance, we are going to show the uniqueness of
viscosity solution, assuming we have an affirmative answer to Question 5.17.
More precisely:

######  Theorem 5.2.

Let $G\in C(M)$. Assume that for any $G^{\prime}\in C(M),\,G^{\prime}\leq G$
and $G^{\prime}\neq G$, one has $c(G^{\prime})>c(G)$. Then there is at most
one viscosity solution to $F(\chi+dd^{c}\varphi)=e^{G+c}$.

Before presenting the proof in full rigor, let us first explain heuristically
how the above said strict monotonicity helps.

Let $\varphi_{1}$ and $\varphi_{2}$ be two viscosity solutions to the
equation. Denote $E=\\{x\in
M:\varphi_{2}(x)-\varphi_{1}(x)=\min_{M}(\varphi_{2}-\varphi_{1})\\}$. Clearly
$E$ is a compact subset of $M$ and we will be done if we can show $E=M$.
Assume otherwise, we can take $\delta>0$ small enough, such that
$E_{\delta}\neq M$, where $E_{\delta}$ denotes the $\delta$-neighborhood of
$E$.

Now we can define $\tilde{G}\in C(M)$ as follows: $e^{\tilde{G}}=e^{G}$ on
$E_{\frac{\delta}{2}}$, $e^{\tilde{G}}=\frac{1}{2}e^{G}$ outside $E_{\delta}$,
and $\tilde{G}\leq G$ on $M$. Let $\eta$ be a viscosity solution to:

$F(\chi+dd^{c}\eta)=e^{\tilde{G}+\tilde{c}},\,\,\sup_{M}\eta=0.$

From the previous section, we know such $\eta$ exists. From the assumption, we
know that $\tilde{c}>c$. Let $0<r<1$, we consider the minimum of
$\varphi_{2}-((1-r)\varphi_{1}+r\eta)$ and assume that it is achieved at
$x_{\delta,r}$. Evaluating at $x_{\delta,r}$, we would have:

$f\big{(}\lambda[\chi+dd^{c}\varphi_{2}]\big{)}\geq
f\big{(}\lambda[\chi+dd^{c}(1-r)\varphi_{1}+r\eta]\big{)}\geq(1-r)f(\lambda[\chi+dd^{c}\varphi_{1}])+rf(\lambda[\chi+dd^{c}\eta]).$

This would mean that:

(5.24)
$e^{G+c}(x_{\delta,r})\geq(1-r)e^{G+c}(x_{\delta,r})+re^{\tilde{G}+\tilde{c}}(x_{\delta,r}).$

On the other hand, for fixed $\delta$, the minimum of
$\varphi_{2}-((1-r)\varphi_{1}+r\eta)$ can only be achieved in
$E_{\frac{\delta}{2}}$, as long as $r$ is small enough, but then one would
have $e^{\tilde{G}}(x_{\delta,r})=e^{G}(x_{\delta,r})$, and this is
inconsistent with (5.24).

Next we are going to make the above argument rigorous.

As before, in place of $\varphi_{1}$, we wish to consider
$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}$,
$\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}},\,\frac{\eta^{\varepsilon}}{1-a_{2}}$,
with parameters $\varepsilon,\,a_{i},\,i=1,\,2$ small enough, then similar
arguments as before shows that

######  Lemma 5.18.

There exist continuous functions
$\rho(\varepsilon),\,\rho_{i}(\varepsilon,a_{i}),\,i=1,2$ with
$\rho(0+)=0,\,\rho_{i}(0+,0+)=0,\,i=1,2$, such that:

  1. (1)

$\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}$ solves the following in the
viscosity sense:

$\begin{split}&\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}]\in\Gamma,\\\
&F(\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}})\geq
e^{G+c}-\rho_{1}(\varepsilon,a_{1}).\end{split}$

  2. (2)

$\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$ solves
$F(\frac{\chi-\rho(\varepsilon)}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}})\leq
e^{G+c}+\rho_{2}(\varepsilon,a_{2})$ in the viscosity sense.

  3. (3)

$\frac{\eta^{\varepsilon}}{1+a_{1}}$ solves
$\lambda[\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{\eta^{\varepsilon}}{1+a_{1}}]\in\Gamma$
and
$F(\frac{\chi+\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{\eta^{\varepsilon}}{1+a_{1}})\geq
e^{\tilde{G}+\tilde{c}}-\rho_{2}(\varepsilon,a_{1})$ in the viscosity sense.

Next we consider the maximum point of
$((1-r)\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}+r\frac{\eta^{\varepsilon}}{1+a_{1}})-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$.
Assume that the maximum is achieved at $x_{*}$, then we have:

######  Lemma 5.19.

For any fixed $\delta>0$, $x_{*}\in E_{\frac{\delta}{2}}$ as long as
$\varepsilon,\,a_{1},\,a_{2},\,r$ are all small enough.

###### Proof.

We observe that for $x\notin E_{\frac{\delta}{2}}$, there exists
$\delta^{\prime}>0$, such that

$\varphi_{1}-\varphi_{2}\leq\max_{M}(\varphi_{1}-\varphi_{2})-\delta^{\prime},$

On the other hand, we know that as $\varepsilon,\,a_{1},\,a_{2},\,r\rightarrow
0$, we have that
$((1-r)\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}+r\frac{\eta^{\varepsilon}}{1+a_{1}})-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$
will converge to $\varphi_{1}-\varphi_{2}$ uniformly, so if the parameters are
all small enough and $x\notin E_{\frac{\delta}{2}}$, one would have:

$\big{(}(1-r)\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}+r\frac{\eta^{\varepsilon}}{1+a_{1}}\big{)}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}\leq\max_{M}\big{(}\big{(}(1-r)\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}+r\frac{\eta^{\varepsilon}}{1+a_{1}}\big{)}-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}\big{)}-\frac{\delta^{\prime}}{2}.$

This proves $x_{*}\in E_{\frac{\delta}{2}}$. ∎

In order to fix the issue that the functions
$(\varphi_{1})^{\varepsilon},\,\eta^{\varepsilon},\,(\varphi_{2})_{\varepsilon}$
are not twice differentiable at $x_{*}$, we need to consider slight
perturbations of the original function. As before, we take a coordinate chart
near $x_{*}$, and we have the following analogue of Lemma 5.13:

######  Lemma 5.20.

There exists a neighborhood $U_{0}$ of $x_{*}$, and there exists a sequence
$p_{k}\in\mathbb{C}^{n}$, $p_{k}\rightarrow 0$, and a sequence
$\delta_{k}>0,\,\delta_{k}\rightarrow 0$, such that one can find a sequence of
$x_{k}\in U_{0}$, such that if we define
$\psi_{\varepsilon,a_{1},k}:=\frac{(\varphi_{1})^{\varepsilon}}{1+a_{1}}-<p_{k},z>-\delta_{k}|z|^{2}$,
we have:

  1. (1)

$x_{k}\rightarrow x_{*}$ as $k\rightarrow\infty$,

  2. (2)

$((1-r)\psi_{\varepsilon,a_{1},k}+r\frac{\eta^{\varepsilon}}{1+a_{1}})-\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}$
has local maximum at $x_{k}$,

  3. (3)

$(\varphi_{1})^{\varepsilon},\,\eta^{\varepsilon},\,(\varphi_{2})_{\varepsilon}$
are all twice differentiable at $x_{k}$.

With this preparation, we are ready to present the rigorous proof of Theorem
5.2:

###### Proof.

(Of Theorem 5.2) From Lemma 5.18 and the definition of
$\psi_{\varepsilon,a_{1},k}$, we see that $\psi_{\varepsilon,a_{1},k}$ solves
the following equation in the viscosity sense:

(5.25)
$\begin{split}&\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\psi_{\varepsilon,a_{1},k}]\in\Gamma,\\\
&f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\psi_{\varepsilon,a_{1},k}]\big{)}\geq
e^{G+c}-\rho_{1}(\varepsilon,a_{1}).\end{split}$

From Lemma 5.18, part (3), we also have:

(5.26)
$\begin{split}&\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{\eta^{\varepsilon}}{1+a_{1}}]\in\Gamma,\\\
&f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\frac{\eta^{\varepsilon}}{1+a_{1}}]\big{)}\geq
e^{\tilde{G}+\tilde{c}}-\rho_{2}(\varepsilon,a_{1}).\end{split}$

Since $x_{k}\rightarrow x_{*}$ and $x_{*}\in E_{\frac{\delta}{2}}$, we see
that $x_{k}\in E_{\frac{\delta}{2}}$ for large enough $k$. Moreover, since
$\psi_{\varepsilon,a_{1},k}$ and $\eta^{\varepsilon}$ are twice differentiable
at $x_{k}$, we can evaluate (5.25), (5.26) at $x_{k}$ and see that:

(5.27)
$\begin{split}&f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}((1-r)\psi_{\varepsilon,a_{1},k}+r\eta^{\varepsilon})]\big{)}(x_{k})\\\
&\geq(1-r)f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\psi_{\varepsilon,a_{1},k})])(x_{k})+rf\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}\eta^{\varepsilon}]\big{)}(x_{k})-2\rho_{1}(a_{1},\varepsilon)\\\
&\geq(1-r)e^{G+c}(x_{k})+re^{\tilde{G}+\tilde{c}}(x_{k})-2\rho_{1}(a_{1},\varepsilon)=(1-r)e^{G+c}(x_{k})+re^{G+\tilde{c}}(x_{k})-2\rho_{1}(a_{1},\varepsilon).\end{split}$

In the first inequality above, we used the concavity of $f$. In the second
inequality above, we used that $\psi_{\varepsilon,a_{1},k}$ and
$\eta^{\varepsilon}$ are subsolutions. Also since $x_{k}\in
E_{\frac{\delta}{2}}$, we have $\tilde{G}(x_{k})=G(x_{k})$. On the other hand,
if one chooses the parameters such that:

$\frac{c_{*}+2\rho(\varepsilon)}{1+a_{1}}\leq\frac{c_{*}-\rho(\varepsilon)}{1-a_{2}},$

we then have (similar to Proposition 5.12):

(5.28)
$\begin{split}&f\big{(}\lambda[\frac{\chi+2\rho(\varepsilon)\omega_{0}}{1+a_{1}}+dd^{c}((1-r)\psi_{\varepsilon,a_{1},k}+r\eta^{\varepsilon})]\big{)}(x_{k})\leq
f\big{(}\lambda[\frac{\chi-\rho(\varepsilon)\omega_{0}}{1-a_{2}}+dd^{c}\frac{(\varphi_{2})_{\varepsilon}}{1-a_{2}}]\big{)}(x_{k})\\\
&\leq e^{G+c}(x_{k})+\rho_{2}(\varepsilon,a_{2}).\end{split}$

Therefore, we finally obtain that:

(5.29)
$(1-r)e^{G+c}(x_{k})+re^{G+\tilde{c}}(x_{k})-2\rho_{1}(a_{1},\varepsilon)\leq
e^{G+c}(x_{k})+\rho_{2}(\varepsilon,a_{2}).$

First we let $k\rightarrow\infty$ and use that $x_{k}\rightarrow x_{*}$, we
see that

$(1-r)e^{G+c}(x_{*})+re^{G+\tilde{c}}(x_{*})-2\rho_{1}(a_{1},\varepsilon)\leq
e^{G+c}(x_{*})+\rho_{2}(\varepsilon,a_{2}).$

Note that if $r>0$ and fixed, $e^{G+c}(x_{*})$ is strictly less than
$(1-r)e^{G+c}(x_{*})+re^{G+\tilde{c}}(x_{*})$. Thereby we get a contradiction
with (5.29) after $\varepsilon,\,a_{1},\,a_{2}$ are chosen sufficiently small.
∎

Now let us go back to Question 5.17. The only positive examples we know are
the classical ones: if $\omega_{0}$ is a Kähler metric and if $\chi$ is
closed, so the equation could be written as:

$\big{(}\frac{(\chi+dd^{c}\varphi)^{k}\wedge\omega_{0}^{n-k}}{\omega_{0}^{n}}\big{)}^{\frac{1}{k}}=e^{G+c},\,\,\,1\leq
k\leq n.$

Then we can see that:

$e^{kc}=\frac{\int_{M}\chi^{k}\wedge\omega_{0}^{n-k}}{\int_{M}e^{kG}\omega_{0}^{n}}.$

Therefore, we see that Question 5.17 indeed holds in this case. Another
observation we make is that, if either one of them is smooth, then Question
5.17 also has an affirmative answer:

######  Lemma 5.21.

If either $G_{1}$ or $G_{2}$ is smooth, then the answer to Question 5.17 is
yes.

###### Proof.

First we assume that both $G_{1}$ and $G_{2}$ are smooth. Assume otherwise,
that is, $G_{1}$ and $G_{2}$ are both smooth, $G_{1}\geq G_{2}$, $G_{1}\neq
G_{2}$, but still $c(G_{1})=c(G_{2})$. Then we can linearly interpolate
between $G_{1}$ and $G_{2}$, namely we define:

$G_{t}=(t-1)G_{2}+(2-t)G_{1},\,\,t\in[1,2].$

Then $G_{t}$ is a smooth decreasing family of smooth functions. Moreover,
since $G_{1}\geq G_{t}\geq G_{2}$, we can use Proposition 5.15 to see that
$c(G_{t})=c(G_{1})=c(G_{2})$. Now we consider the solution corresponding to
$G_{t}$:

$f\big{(}\lambda[\chi+dd^{c}\varphi_{t}]\big{)}=e^{G_{t}+c(G_{1})}.$

We differentiate the above equation in $t$:

$\frac{\partial F}{\partial
h_{i\bar{j}}}(\lambda[h_{\varphi_{t}}])(\partial_{t}\varphi)_{i\bar{j}}=e^{G_{t}}(G_{2}-G_{1}).$

Therefore, if we define

$\tilde{\Omega}=\big{(}\det g\det(\frac{\partial F}{\partial
h_{i\bar{j}}})(\chi+dd^{c}\varphi_{t})\big{)}^{\frac{1}{n-1}}\Omega,$

then the above equation can be written as:

$\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\wedge
dd^{c}(\partial_{t}\varphi)=e^{G_{t}}(G_{2}-G_{1})\frac{\omega_{0}^{n}}{n!}.$

Now let $v_{t}$ be the Gaudochon factor corresponding to $\tilde{\Omega}$,
that is $dd^{c}\big{(}e^{(n-1)v_{t_{0}}}\tilde{\Omega}^{n-1}\big{)}=0$, then
we have:

$0=\int_{M}e^{(n-1)v_{t}}\frac{\tilde{\Omega}^{n-1}}{(n-1)!}\wedge
dd^{c}(\partial_{t}\varphi)=\int_{M}e^{(n-1)v_{t}+G_{t}}(G_{2}-G_{1})\frac{\omega_{0}^{n}}{n!}.$

This implies $G_{1}=G_{2}$, a contradiction.

Next we consider when only one of $G_{1}$ or $G_{2}$ is smooth. Assume, say,
$G_{1}$ is smooth, $G_{1}\geq G_{2}$, and $G_{1}>G_{2}$ on some open set $U$.
By adding a bump function supported on $U$, it is easy to find a smooth
function $G_{2}^{\prime}$ such that $G_{1}\geq G_{2}^{\prime}\geq G_{2}$, and
$G_{1}\neq G_{2}^{\prime}$. Then from the strict monotonicity in the smooth
case, we can conclude that:

$c(G_{1})<c(G_{2}^{\prime})\leq c(G_{2}).$

The other possibility that $G_{2}$ is smooth can be dealt with similarly. ∎

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