# A Maximum Rank Theorem for Solutions to the Homogenous Complex Monge-Ampère
Equation in a $\mathbb{C}$-Convex Ring

Jingchen Hu  Jingchen Hu

###### Abstract

Suppose $\Omega_{0},\Omega_{1}$ are two bounded strongly $\mathbb{C}$-convex
domains in $\mathbb{C}^{n}$, with $n\geq 2$ and
$\Omega_{1}\supset\overline{\Omega_{0}}$. Let
$\mathcal{R}=\Omega_{1}\backslash\overline{\Omega_{0}}$. We call $\mathcal{R}$
a $\mathbb{C}$-convex ring. We will show that for a solution $\Phi$ to the
homogenous complex Monge-Ampère equation in $\mathcal{R}$, with $\Phi=1$ on
$\partial\Omega_{1}$ and $\Phi=0$ on $\partial\Omega_{0}$,
$\sqrt{-1}\partial\overline{\partial}\Phi$ has rank $n-1$ and the level sets
of $\Phi$ are strongly $\mathbb{C}$-convex.

##  1 Introduction

###  1.1 Related Research and Our Main Results

The convexity of level sets of the solution to a partial differential equation
in a ring-shaped domain has been studied for a long time. In general, the
following Dirichlet problem is considered:

######  Problem 1.1 (The Dirichlet Problem in a Ring-Shaped Domain).

Suppose $\Omega_{0},\Omega_{1}$ are bounded domains with smooth boundaries in
$\mathbb{R}^{n}$, with $\overline{\Omega_{0}}\subset\Omega_{1}$. Let
$\mathcal{R}=\Omega_{1}\backslash\overline{\Omega_{0}}$ and let $\mathcal{F}$
be a partial differential operator on $\mathcal{R}$. Find $\Phi$ satisfying

$\displaystyle\mathcal{F}(\Phi)=0\ \ \ \ \ \ $ $\displaystyle\text{ in
}\mathcal{R},$ (1.1) $\displaystyle\Phi=0\ \ \ \ \ \ $ $\displaystyle\text{ on
}\partial\Omega_{0},$ (1.2) $\displaystyle\Phi=1\ \ \ \ \ \ $
$\displaystyle\text{ on }\partial\Omega_{1}.$ (1.3)

Let $\Omega_{t}=\Omega_{0}\cup\\{\Phi<t\\}$, for $t\in(0,1)$. We ask does
$\Omega_{t}$ inherit some properties from $\Omega_{0}$ and $\Omega_{1}$? In
many cases, the answer is affirmative.

One particular case, which attracts many attentions, is when $\Omega_{0}$ and
$\Omega_{1}$ are both convex and $\mathcal{F}$ is the Laplacian (or the
$p$-Laplacian). In this case, we can show $\Omega_{t}$ are strongly convex for
$t\in(0,1)$, and there are several different approaches.

First, there is a “macroscopic” approach, where a “macroscopic” auxiliary
function is introduced to gauge the convexity. For example, extending the idea
of [G55] [G57I] [G57II], in [L77] the following function is considered:

$\displaystyle\tilde{\Phi}(x)=\inf_{x\in\
\overline{yz}}\max\\{\Phi(y),\Phi(z)\\},$ (1.4)

where the infimum is taken over all $y,z\in\mathbb{R}^{n}$ such that $x$ lies
on the closed line segment from $y$ to $z$. Here $\Phi$ is considered to be a
function on $\mathbb{R}^{n}$ with $\Phi=0$ in $\Omega_{0}$ and $\Phi=1$ in
$\Omega_{1}^{c}$. It’s easy to see that the convexity of the level sets is
equivalent to $\tilde{\Phi}=\Phi$. In [RR89], Rosay and Rudin considered the
following function

$\displaystyle
Q(x,y)=\frac{\Phi(x)+\Phi(y)}{2}-\Phi(\frac{x+y}{2})-c|x-y|^{2},$ (1.5)

where $c$ is a constant. In [W18], the following two-point function and its
variants are considered:

$\displaystyle Q(x,y)=(\nabla\Phi(x)-\nabla\Phi(y))\cdot(x-y);$ (1.6)

it’s obvious that when restricted to the set

$\displaystyle\Sigma=\\{(x,y)\in\overline{\mathcal{R}}|\Phi(x)=\Phi(y)\\}$
(1.7)

the strong convexity of the level sets is equivalent to the positivity of $Q$.

The second approach uses the constant rank theorem, and it is often referred
to as a “microscopic” approach. In [K90], it was shown that the second
fundamental form of the level sets of $\Phi$ has constant rank, providing the
level sets are convex; then combing with a deformation argument, the strong
convexity can be proved. The method has been studied and generalized by many
authors, for example [BGMX11] [GX13] [BG09] [SW16].

The third approach is to directly estimate the Gaussian or principle
curvatures of the level sets. It was first proved by [L83] and [OS83] that
when $n=2$, the curvature of level curves of $\Phi$ attains minimum on
$\partial\mathcal{R}$. In general dimension, [MOZ10] proved when $\mathcal{F}$
is the $p$-Laplacian and providing $p,n$ satisfy some conditions,

$\displaystyle|\nabla\Phi|^{n+1-2p}K,\ \ |\nabla\Phi|^{1-p}K\text{\ \ or }K$
(1.8)

attain their minimums on $\partial\mathcal{R}$. Here, $K$ is the Gaussian
curvature of the level sets of $\Phi.$ As a consequence, the convexity of
$\Omega_{0}$ and $\Omega_{1}$ implies the convexity of $\Omega_{t}$. A similar
estimate for the principle curvature was proved in [ZZ12].

Beyond the Laplacian and the real convexity in Euclidean spaces, there are
many possible directions to generalize the forementioned results.

First, we may ask if $\mathbb{R}^{n}$ is replaced by a space form, can we have
similar results. This was first proved by [P92] in the case of $n=2$ in a
Poincare disc and in the case of general dimension by [MZ14].

Another question is when $\mathcal{F}(\Phi)=0$ is the minimal surface
equation, does the convexity of $\Omega_{0}$ and $\Omega_{1}$ imply the
convexity of $\Omega_{t}$. The answer is affirmative. Let $S$ be a minimal
surface in the 3 dimensional Euclidean space bounded by two plane curves
$r_{1},r_{2}$, lying in parallel planes. The following result is proved in
[S56]: If $r_{1},r_{2}$ are convex curves, then the intersection of $S$ by a
plane parallel to the planes of $r_{1},r_{2}$ is again a convex curve. In the
situation of general dimension and that $S$ is a minimal graph, it was proved
by [K90]; it has also been generalized to minimal graphs over space forms by
[WZ17].

Using the constant rank theorems developed by [BGMX11] and [GX13], we know
when $\mathcal{F}$ satisfies some structural conditions, the convexity of
$\Omega_{0}$ and $\Omega_{1}$ implies the convexity of $\Omega_{t}$. This can
be proved via a continuity argument. But these structure conditions exclude
some important equations, for example the real and complex homogenous Monge-
Ampère equations. We may ask, if $\Omega_{0}$ and $\Omega_{1}$ are both
strongly convex and $\mathcal{F}$ is the real Monge-Ampère operator, is it
true that $\Omega_{t}$ are all strongly convex? In this case, a smooth
solution can be constructed explicitly, and we can answer the question
positively. Using methods from complex analysis, in [D98] and [LLS98] it was
proved that when $\mathcal{F}$ is the complex Monge-Ampère operator and
$\Omega_{0}$ and $\Omega_{1}$ are both convex, then $\Omega_{t}$ are all
convex.

We may also ask, instead of the real convexity, when $\Omega_{0}$ and
$\Omega_{1}$ satisfy other convexity conditions, does $\Omega_{t}$ satisfy the
same convexity condition? For example, in complex spaces the following
concepts are natural generalizations of the convexity in real spaces:

######  Definition 1.2 ($\mathbb{C}$-Convexity and Linear Convexity
[E][Z86][APS04][J08]).

In $\mathbb{C}^{n}$, a domain $E$ is said to be $\mathbb{C}$-convex if for any
complex line $l\in\mathbb{C}^{n}$, the intersection $E\cap l$ is both
connected and simply connected; it is called linearly convex if $E^{c}$ is a
union of complex hyperplanes.

######  Remark 1.3.

When $n=1$, any connected and simply connected domain in $\mathbb{C}$ is
$\mathbb{C}$-convex; in this case, a hyperplane is a point, so any domain in
$\mathbb{C}$ is considered linearly convex. When $n>1$ and $E$ is a bounded
domain with a $C^{1}$ boundary, the two notions are equivalent.

We can ask when $\Omega_{0}$ and $\Omega_{1}$ are both $\mathbb{C}$-convex
with smooth boundaries and $\mathcal{F}$ is some linear or non-linear
operator, is $\Omega_{t}$ $\mathbb{C}$-convex? Or more generally, we may
consider the geometric convexity introduced by [HL12]. It seems there has been
no result in this direction.

In this paper, we are concerned with the situation where $\mathcal{F}$ is the
complex Monge-Ampère operator and $\Omega_{0}$ and $\Omega_{1}$ are both
strongly $\mathbb{C}$-convex. The definition of strong $\mathbb{C}$-convexity
is in the following:

######  Definition 1.4.

Suppose $E$ is a bounded connected domain in $\mathbb{C}^{n}$ with a $C^{1,1}$
boundary. Then, at any point ${\bf p}\in\partial E$, $\partial E$ can be
locally represented as the graph of a $C^{1,1}$ function: after a linear
change of coordinates, which makes ${\bf p}=0$, we can find $\delta>0$ and a
$C^{1,1}$ function $\rho_{\bf p}$, so that

$\displaystyle\partial
E\cap\left\\{|\text{Re}(z^{n})|<\delta,|\text{Im}(z^{n})|^{2}+\sum_{\alpha=1}^{n-1}|z^{\alpha}|^{2}<\delta^{2}\right\\}$
(1.9) $\displaystyle=$ $\displaystyle\left\\{(z^{\alpha},\rho_{\bf
p}(z^{\alpha},s)+\sqrt{-1}s)\big{|}\sum_{\alpha=1}^{n-1}|z^{\alpha}|^{2}+s^{2}<\delta^{2}\right\\},$
(1.10)

with $\rho_{\bf p}(0)=0$ and $\nabla\rho_{\bf p}(0)=0$. We say $E$ is strongly
$\mathbb{C}$-convex if for any ${\bf p}$ we can find $\mu>0$, so that

$\displaystyle\rho_{\bf
p}(z^{\alpha},0)\geq\mu\sum_{\alpha=1}^{n-1}|z^{\alpha}|^{2}.$ (1.11)

######  Remark 1.5.

If a bounded connected domain in $\mathbb{C}^{n}$ with a $C^{1,1}$ boundary is
strongly $\mathbb{C}$-convex, then it’s $\mathbb{C}$-convex; this can be
proved by Proposition 2.5.8 of [APS04]. We also note that in (1.11) if we
require

$\displaystyle\rho_{\bf
p}(z^{\alpha},s)\geq\mu(\sum_{\alpha=1}^{n-1}|z^{\alpha}|^{2}+s^{2}),$ (1.12)

then this is just the usual notion of strong convexity.

We study the following Dirichlet problem:

######  Problem 1.6 (The Dirichlet Problem for the Homogenous Complex Monge-
Ampère Equation).

Suppose $\Omega_{0},\Omega_{1}$ are two strongly $\mathbb{C}$-convex domains
with smooth boundaries in $\mathbb{C}^{n}$, $n\geq 2$, with
$\overline{\Omega_{0}}\subset\Omega_{1}$. Let
$\mathcal{R}=\Omega_{1}\backslash\overline{\Omega_{0}}$. Find a continuous
$\Phi:\ \overline{\mathcal{R}}\rightarrow\mathbb{R}$, plurisubharmonic in
$\mathcal{R}$, satisfying

$\displaystyle\left(\sqrt{-1}\partial\overline{\partial}\Phi\right)^{n}=0\ \ \
\ \ \ $ $\displaystyle\text{ in }\mathcal{R},$ (1.13) $\displaystyle\Phi=0\ \
\ \ \ \ $ $\displaystyle\text{ on }\partial\Omega_{0},$ (1.14)
$\displaystyle\Phi=1\ \ \ \ \ \ $ $\displaystyle\text{ on
}\partial\Omega_{1}.$ (1.15)

The homogenous complex Monge-Ampère equation in a ring shaped domain has been
studied by many authors, including [M79] [M82] [G02]. In particular, it was
proved in [G02] that Problem 1.6 has a $C^{1,1}$ solution providing it has a
subsolution, which essentially depends on the fact that $\Omega_{0}$ is
holomorphically convex in $\Omega_{1}$. This condition is satisfied in our
case since $\Omega_{0}$ is strongly $\mathbb{C}$-convex and, therefore,
polynomially convex; we explain this in Appendix B.

######  Remark 1.7.

Since the solution $\Phi$ to the Dirichlet problem above may only be a
$C^{1,1}$ function, its level sets $\Omega_{t}$ may only have $C^{1,1}$
boundaries. Therefore it’s necessary to introduce the strong
$\mathbb{C}$-convexity for domains with $C^{1,1}$ boundaries.

######  Remark 1.8.

The Dirichlet problems in a ring-shaped domain for other nonlinear equations
has also been studied. For example, the real homogenous sigma-$k$ equation was
studied in [MZ22]; the complex homogenous sigma-$k$ equation was studied in
[GMZ22]. Using the estimates in a ring-shaped domain, they derived the
estimates for the corresponding Green’s function.

We have the following result:

######  Theorem 1.9 ($\mathbb{C}$-Convexity of Subevel Sets and the Maximum
Rank Property).

Suppose $\Phi$ is a solution to Problem 1.6. Then $|\nabla\Phi|$ has a
positive lower bound, and $\sqrt{-1}\partial\overline{\partial}\Phi$ has rank
$n-1$ in the weak sense. Let

$\displaystyle\Omega_{t}=\\{\Phi\leq t\\}\cup\Omega_{0},\ \ \ \ \ \
\text{for}\ t\in(0,1).$ (1.16)

Then $\Omega_{t}$ are all strongly $\mathbb{C}$-convex.

######  Remark 1.10.

Since we only know $\Phi$ is $C^{1,1}$,
$\text{rank}(\sqrt{-1}\partial\overline{\partial}\Phi)=n-1$ should be
understood in the weak sense: In Section 6, we can show

$\displaystyle\left(\sqrt{-1}\partial\overline{\partial}\Phi\right)^{n-1}\wedge\omega_{0}>0$
(1.17)

in the sense of distribution, where

$\displaystyle\omega_{0}=\sqrt{-1}\delta_{{i\overline{j}}}dz^{i}\wedge\overline{dz^{j}}.$
(1.18)

(1.17) says the rank of $\sqrt{-1}\partial\overline{\partial}\Phi$ is at least
$n-1$ in the weak sense, and (1.13) says the the rank of
$\sqrt{-1}\partial\overline{\partial}\Phi$ is at most $n-1$, also in the weak
sense. Therefore, we say the rank of
$\sqrt{-1}\partial\overline{\partial}\Phi$ is $n-1$ in the weak sense.

In addition, we also prove that

$\displaystyle\sqrt{-1}\partial\overline{\partial}e^{\Phi}>0,$ (1.19)

i.e. $e^{\Phi}$ is strictly plurisubharmonic.

######  Remark 1.11.

In the case of $n=1$, a $\mathbb{C}$-convex domain is a connected and simply
connected domain, and there is no essential content to the name strongly
$\mathbb{C}$-convex domain. The Monge-Ampère operator in this case is simply
the Laplacian. Actually, we can prove the following result. Suppose
$\Omega_{0}$ and $\Omega_{1}$ are both connected and simply connected domains
in $\mathbb{C}$, with smooth boundary. Then $\nabla\Phi\neq 0$ everywhere, and
$\Omega_{t}$ are all connected and simply connected. The proof is easy; we
only need to consider the function $z\Phi_{z}$, and notice that

$\displaystyle(z\Phi_{z})_{{z\overline{z}}}=0\ \ \ \ \ \ \ \ \text{ on
}\mathbb{C},$ (1.20)

where $z$ is the coordinate on $\mathbb{C}$. When $\Omega_{0}$ and
$\Omega_{1}$ are concentric balls, $z\Phi_{z}$ is non-zero. Also, $z\Phi_{z}$
can not be zero on the $\partial\Omega_{0}$ and $\partial\Omega_{1}$ because
of the boundary gradient estimate. Therefore, we can prove the result using a
deformation argument.

Another closely related topic is the convexity of level sets of Green’s
functions. The earliest result known to the author is Caratheodory’s proof
that level curves of the Green’s function for a bounded convex sets in the
plane are convex; the proof can be found in the book of Ahlfors [A74]. The
$3$-dimensional case was proved by Gabriel [G57I]. In the case of general $n$
dimension, suppose $G$ is the Green’s function in $\Omega$ with $G<0$ in
$\Omega$ and $G=0$ on $\partial\Omega$; Jia-Ma-Shi proved that
$(-G)^{\frac{1}{2-n}}$ is strictly convex in $\Omega\backslash\\{0\\}$ for
$n\geq 3$, where $0$ is the singular point of $G$ [S15] [JMS22]. In the case
of pluricomplex Green’s function, it was proved by Lempert (page 462 of [L81],
Lemma 5.3 of [L85]) that in a strongly convex (or strongly linearly convex)
domain with a smooth boundary the sublevel sets of pluricomplex Green’s
functions are convex (or strongly linearly convex). It’s interesting that in
this case level sets of Green’s functions are geodesic balls with respect to
the Kobayashi metric.

There has been a large amount of research on the convexity of level sets of
solutions of partial differential equations; we refer to [MO10] [M16] [CM18]
for better surveys.

###  1.2 Main Ideas of the Proof

First, we observe that on $\mathbb{C}^{n-1}$, a quadratic function

$\displaystyle
f(z)=A_{\alpha{\overline{\beta}}}z^{\alpha}z^{\overline{\beta}}+\text{Re}\left(B_{\alpha\beta}z^{\alpha}z^{\beta}\right),$
(1.21)

with $\alpha,\beta\in\\{1,\ ...\ ,n-1\\}$, is strongly convex if and only if
$A>0$ and

$\displaystyle\text{ the maximum eigenvalue of }B\overline{A^{-1}}\
\overline{B}A^{-1}<1.$ (1.22)

Here, for strongly convex we mean the real Hessian of $f$ is strictly positive
definite. This is the main idea of [H22N] and [H22A], where we proved that in
a product manifold, providing that the boundary value satisfies a convexity
condition, the solution to the homogenous complex Monge-Ampère equation has
Hessian of the maximum rank. In this paper, we apply this idea to the study of
the convexity of level sets of solutions to the homogenous complex Monge-
Ampère equation in a ring-shaped domain in $\mathbb{C}^{n}$.

Suppose $\nabla\Phi\neq 0$ at a point ${\bf p}$. We choose a set of
coordinates $\left\\{z^{\alpha},\tau\right\\}$, where $\alpha,\beta\in\\{1,\
...\ ,n-1\\}$ and $\tau$ is the $n$-th coordinate, so that $\Phi_{\alpha}=0$
and $\Phi_{\tau}\neq 0$ at ${\bf p}$ . Then, at ${\bf p}$, we let

$\displaystyle{\mathcal{A}}_{\alpha{\overline{\beta}}}=\Phi_{\alpha{\overline{\beta}}},\
\ \ \ \ \ \ {\mathcal{B}}_{\alpha\beta}=\Phi_{\alpha\beta}.$ (1.23)

When the level sets of $\Phi$ are strongly pseudoconvex, we have
${\mathcal{A}}>0$, and we let

$\displaystyle{\mathcal{K}}_{\alpha}^{\theta}={\mathcal{B}}_{\alpha\mu}{\mathcal{A}}^{\mu{\overline{\eta}}}\
\overline{{\mathcal{B}}_{\eta\zeta}}{\mathcal{A}}^{\theta{\overline{\zeta}}}.$
(1.24)

Using matrix notation, denoting
${\mathcal{A}}=({\mathcal{A}}_{{\alpha\overline{\beta}}}),\
{\mathcal{B}}=({\mathcal{B}}_{{\alpha\beta}})$ and
${\mathcal{K}}=({\mathcal{K}}_{\alpha}^{\beta})$, the above is equivalent to

$\displaystyle{\mathcal{K}}={\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}{\mathcal{A}}^{-1}.$ (1.25)

When the maximum eigenvalues of ${\mathcal{K}}$ is smaller than 1, we let

$\displaystyle\sigma_{\Phi}=\text{tr}(I-{\mathcal{K}})^{-1}.$ (1.26)

We note that ${\mathcal{A}}$, ${\mathcal{B}}$ and ${\mathcal{K}}$ all depend
on the choice of coordinate, but the eigenvalues of ${\mathcal{K}}$ are
invariant under a complex linear change of coordinates; therefore,
$\sigma_{\Phi}$ is invariant under a complex linear change of coordinates. So
$\sigma_{\Phi}$ is a well-defined function in $\overline{\mathcal{R}}$. The
upper bound of $\sigma_{\Phi}$ provides an estimate on the
$\mathbb{C}$-convexity of the level sets of $\Phi$. How to quantitatively
gauge the $\mathbb{C}$-convexity will be discussed in Section 2.

If we assume the solution to Problem 1.6 is smooth and do a formal
computation, which is what we did in Section E, we will find

$\displaystyle\sigma_{\Phi}\leq\max_{\partial\mathcal{R}}\sigma_{\Phi}\ \ \ \
\ \ \ \ \text{ in }\mathcal{R}.$ (1.27)

We can consider the above estimate as a weak version of the constant rank
theorem: providing the level sets of $\Phi$ are strongly $\mathbb{C}$-convex,
we have an estimate for the $\mathbb{C}$-convexity.

However, the solution $\Phi$ to Problem 1.6 is only $C^{1,1}$ in general. So
we need to consider a perturbed Dirichlet problem, which has a smooth
solution. Instead of considering the usually used non-degenerate Monge-Ampère
equation

$\displaystyle\det(\Phi_{{i\overline{j}}})={\varepsilon},$ (1.28)

we consider the following Hessian quotient equation:

######  Problem 1.12 (The Dirichlet Problem for a Hessian Quotient Equation).

Let $\mathcal{G}_{{i\overline{j}}}dz^{i}\otimes\overline{dz^{j}}$ be a
constant coefficient Hermitian metric on $\mathbb{C}^{n}$; then,
${\omega_{\mathcal{G}}}=\sqrt{-1}\mathcal{G}_{ij}dz^{i}\wedge\overline{dz^{j}}$
is the Kähler form. Find $\Phi$ satisfying

$\displaystyle\left(\sqrt{-1}\partial\overline{\partial}\Phi\right)^{n}$
$\displaystyle={\varepsilon}\left(\sqrt{-1}\partial\overline{\partial}\Phi\right)^{n-1}\wedge{\omega_{\mathcal{G}}}\
\ \ \ \ \ $ $\displaystyle\text{in }\mathcal{R},\ \ $ (1.29)
$\displaystyle\sqrt{-1}\partial\overline{\partial}\Phi$ $\displaystyle>0\ \ \
\ \ \ $ $\displaystyle\text{in }\mathcal{R},\ \ $ (1.30) $\displaystyle\Phi$
$\displaystyle=0\ \ \ \ \ \ $ $\displaystyle\text{\ \ on }\partial\Omega_{0},$
(1.31) $\displaystyle\Phi$ $\displaystyle=1\ \ \ \ \ \ $ $\displaystyle\text{\
\ on }\partial\Omega_{1}.$ (1.32)

Here ${\varepsilon}$ is a positive constant.

It was proved by [G02] that the problem above has a subsolution providing
$\overline{\Omega_{0}}$ is holomorphically convex in $\Omega_{1}$; in Appendix
B, we argue that this condition is satisfied in our situation. Then applying
Theorem 1.2 of [GL12], we know this problem has a unique and smooth solution.
Using the method of [G98], we know the solution $\Phi$ has a uniform $C^{2}$
estimate which is independent of the lower bound of ${\varepsilon}$; we
include a proof of this fact in Appendix C since it’s not clearly stated
elsewhere to the knowledge of the author. So if $\Phi^{\epsilon}$ is the
solution to the problem above, we know $\Phi^{\epsilon}$ converges to a
solution to Problem 1.6 as $\epsilon$ goes to $0$. Therefore, an estimate for
the solution to Problem 1.6 can be derived from that of Problem 1.12.

###  1.3 Notation and Conventions of the Paper

In this paper, we will only allow complex linear change of coordinates on
$\mathbb{C}^{n}$. We use Roman letters to denote indices run from $1$ to $n$.
Sometimes, we need to specialize an $n-1$ dimensional complex subspace. For
example, we may specialize the complex tangent plane to the level sets of a
function. In this situation, we use the Greek letters, except $\tau$, to
denote indices go from $1$ to $n-1$; the letter $\tau$ is used to denote the
$n$-th coordinate.

We will introduce a convenient notation to indicate terms in equations. The
term (*.*)k indicate the $k$-th term in equation (*.*), including the sign.
For example,

$\displaystyle(\ref{711121})_{2}=A_{\alpha{\overline{\beta}}}z^{\alpha}z^{\overline{\beta}},\
\ \ \ \ \ \ \ (\ref{711105_RR})_{4}=-c|x-y|^{2}.$ (1.33)

For matrices with real eigenvalues, we denote that

$\displaystyle c_{1}>M(c_{1}\geq M)\ \ \text{ or }M>c_{2}(M\geq c_{2})$ (1.34)

if the maximum eigenvalues of $M<c_{1}(\leq c_{1})$ or the minimum eigenvalues
of $M>c_{2}(\geq c_{2})$.

Suppose $\Phi$ is a smooth function with $\nabla\Phi\neq 0$ around a point
${\bf p}$. Then the level set $\\{\Phi=\Phi({\bf p})\\}$ is a smooth
hypersurface around ${\bf p}$. We denote the complex tangent plane of
$\\{\Phi=\Phi({\bf p})\\}$ at ${\bf p}$ by

$\displaystyle T^{\mathbb{C}}_{\Phi,{\bf p}}.$ (1.35)

For a ring-shaped domain
$\mathcal{R}=\Omega_{1}\backslash\overline{\Omega_{0}}$, we call

$\displaystyle\min\left\\{\min_{x\in\partial\Omega_{0},y\in\partial\Omega_{1}}|x-y|,\
\max\left\\{r|B_{r}(z)\subset\overline{\mathcal{R}}\text{ for some
}z\in\mathcal{R}\right\\}\right\\}$ (1.36)

the thickness of $\mathcal{R}$. Let $\Phi^{\varepsilon}$ be the solution to
Problem 1.12, and let
$\mathcal{G}_{i\overline{j}}dz^{i}\otimes\overline{dz^{j}}$ be the Hermitian
metric on $\mathbb{C}^{n}$. If a constant $C$ depends on

  * •

the modulus of $\mathbb{C}$-convexity of $\mathcal{R}$,

  * •

the diameter of $\mathcal{R}$,

  * •

the thickness of $\mathcal{R}$,

  * •

the upper bound of
$\max_{\overline{\mathcal{R}}}|D^{2}\Phi^{\varepsilon}|_{\mathcal{G}}$,
providing ${\varepsilon}$ small enough, derived in Appendix C,

  * •

the lower bound of
$\min_{\partial\mathcal{R}}|\nabla\Phi^{\varepsilon}|_{\mathcal{G}}$,
providing ${\varepsilon}$ small enough, derived in Appendix B,

we say $C$ depends on the geometry of $\mathcal{R}$.

##  2 The $\mathbb{C}$-Convexity and The Auxiliary Functions

In this section, we will introduce some concepts which quantitatively
characterize the $\mathbb{C}$-convexity of sublevel sets of $\Phi$ and
introduce the auxiliary functions.

###  2.1 Quantitative Characterizations of the $\mathbb{C}$-Convexity

First, we introduce the convexity characterization of quadratic polynomials on
$\mathbb{C}^{n-1}$.

######  Definition 2.1.

Suppose that $g_{\alpha{\overline{\beta}}}dz^{\alpha}\otimes
dz^{\overline{\beta}}$ is a constant coefficient Hermitian metric on
$\mathbb{C}^{n-1}$. Then we say the modulus of convexity of a quadratic
polynomial

$\displaystyle
P(z)=A_{{\alpha{\overline{\beta}}}}z^{\alpha}z^{{\overline{\beta}}}+\text{Re}\left(B_{\alpha\beta}z^{\alpha}z^{\beta}\right)+L$
(2.1)

is greater than $\mu$, for $\mu\geq 0$, if

$\displaystyle
P(z)-L>\mu\left(g_{\alpha{\overline{\beta}}}z^{\alpha}z^{\overline{\beta}}\right)\
\ \ \ \ \ \ \ \text{ on }\mathbb{C}^{n-1}\backslash\left\\{0\right\\}.$ (2.2)

Here $A$ is a Hermitian matrix, $B$ is a complex valued symmetric matrix, and
$L$ is a linear function. We say that $P$ is strongly convex if and only if
the modulus of convexity of $P$ is greater than $0$.

We say the degree of convexity of $P$ is greater than $\delta$, for
$\delta\geq 0$, if for any

$\displaystyle\mathcal{W}(z)=\text{Re}\left(W_{{\alpha\beta}}z^{\alpha}z^{\beta}\right),$
(2.3)

with

$\displaystyle\left(W_{\alpha\beta}g^{\beta{\overline{\gamma}}}\
\overline{W_{\gamma\nu}}g^{\theta{\overline{\nu}}}\right)\leq\delta^{2},$
(2.4)

$P-\mathcal{W}$ is strongly convex.

######  Remark 2.2.

The degree and the modulus of convexity are equivalent; this is proved by
Lemma A.2 of [H22N]. In Appendix A, we will provide an independent proof:
Lemma A.1.

In literature, a function with convex level sets is often called quasi-convex.
Therefore, we introduce the following concept of strongly
quasi-$\mathbb{C}$-convexity:

######  Definition 2.3.

Suppose that $\Omega$ is a bounded domain with smooth boundaries and that
$\Phi$ is a smooth function defined on $\overline{\Omega}$. For a point ${\bf
p}\in\overline{\Omega}$, we say $\Phi$ is strongly quasi-$\mathbb{C}$-convex
at ${\bf p}$ if

$\displaystyle\nabla{\Phi}({\bf p})\neq 0$ (2.5)

and

$\displaystyle\text{the second order Taylor expansion of
}\Phi|_{T^{\mathbb{C}}_{\Phi,{\bf p}}}\text{ at ${\bf p}$ is strongly
convex.}$ (2.6)

Here, ${T^{\mathbb{C}}_{\Phi,{\bf p}}}$ is the complex tangent plane of
$\left\\{\Phi=\Phi({\bf p})\right\\}$ at ${\bf p}$. We say $\Phi$ is strongly
quasi-$\mathbb{C}$-convex on $\Gamma\subset\overline{\Omega}$ if it’s strongly
quasi-$\mathbb{C}$-convex at any ${\bf p}\in\Gamma$.

######  Remark 2.4.

The main goal of this paper is to prove the solution $\Phi$ to Problem 1.12 is
strongly quasi-$\mathbb{C}$-convex. In the following, for convenience, we will
denote quasi-$\mathbb{C}$-convexity by $\mathfrak{q}\mathbb{C}$-convexity.

To gauge the $\mathfrak{q}\mathbb{C}$-convexity, we introduce the following
concepts:

######  Definition 2.5 (Modulus of $\mathfrak{q}\mathbb{C}$-convexity).

Suppose $g_{i\overline{j}}dz^{i}\otimes\overline{dz^{j}}$ is a constant
coefficient metric on $\mathbb{C}^{n}$. We say the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\delta\geq 0$ if

$\displaystyle|\nabla\Phi|_{g}>\delta$ (2.7)

and

$\displaystyle\begin{split}\text{the modulus of convexity of the second order
Taylor expansion}&\\\ \text{ of }\Phi|_{T^{\mathbb{C}}_{\Phi,{\bf p}}}&\text{
at ${\bf p}$ is greater than $\delta$.}\end{split}$ (2.8)

######  Definition 2.6 (Robustness of $\mathfrak{q}\mathbb{C}$-convexity).

Suppose $g_{i\overline{j}}dz^{i}\otimes\overline{dz^{j}}$ is a constant
coefficient Hermitian metric on $\mathbb{C}^{n}$. Let

${\mathscr{Q}}_{\epsilon}=\left\\{q\ |\ \text{pluriharmonic quadratic
polynomials with }|\nabla q|_{g}\leq\epsilon\right\\}.$ (2.9)

We say the robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is
greater than $\epsilon\geq 0$ if for any $q\in{\mathscr{Q}}_{\epsilon}$,
$\Phi-q$ is strongly $\mathfrak{q}\mathbb{C}$-convex. In this paper, when we
say quadratic functions we include linear functions.

######  Remark 2.7.

An easy and useful observation is

$\displaystyle{\mathscr{Q}}_{{\varepsilon}_{1}}+{\mathscr{Q}}_{{\varepsilon}_{2}}={\mathscr{Q}}_{{\varepsilon}_{1}+{\varepsilon}_{2}}.$
(2.10)

So if the robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is
greater than ${\varepsilon}$ and $q\in{\mathscr{Q}}_{\delta}$ with
$\delta<{\varepsilon}$, then the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi-q$ is greater than
${\varepsilon}-\delta$.

######  Remark 2.8.

The name robustness means how robust the $\mathfrak{q}\mathbb{C}$-convexity is
under perturbations. If the modulus of $\mathfrak{q}\mathbb{C}$-convexity of
$\Phi$ is greater than $0$, than the robustness of
$\mathfrak{q}\mathbb{C}$-convexity is also greater than $0$, and vise versa;
the two measurements are actually comparable in the sense of Lemma A.3 and
Lemma A.5.

For the quantitative characterization of the $\mathbb{C}$-convexity of a
domain, we have the following concept of the modulus of
$\mathbb{C}$-convexity:

######  Definition 2.9 (Modulus of $\mathbb{C}$-convexity of a Domain).

Suppose $\Omega$ is a bounded domain in $\mathbb{C}^{n}$ with a $C^{1,1}$
boundary and that $dz^{i}\otimes\overline{dz^{i}}$ is the metric on
$\mathbb{C}^{n}$. Then, at any point ${\bf p}\in\partial E$, $\partial E$ can
be locally represented as the graph of a $C^{1,1}$ function: after a unitary
change of coordinates, which makes ${\bf p}=0$, we can find $\delta>0$ and a
$C^{1,1}$ function $\rho_{\bf p}$, so that

$\displaystyle\partial
E\cap\left\\{|\text{Re}(z^{n})|<\delta,|\text{Im}(z^{n})|^{2}+\sum_{\alpha=1}^{n-1}|z^{\alpha}|^{2}<\delta^{2}\right\\}$
(2.11) $\displaystyle=$ $\displaystyle\left\\{(z^{\alpha},\rho_{\bf
p}(z^{\alpha},s)+\sqrt{-1}s)\big{|}\sum_{\alpha=1}^{n-1}|z^{\alpha}|^{2}+s^{2}<\delta^{2}\right\\},$
(2.12)

with $\rho_{\bf p}(0)=0$ and $\nabla\rho_{\bf p}(0)=0$. We say the modulus of
$\mathbb{C}$-convexity of $E$ is greater than $\mu\geq 0$ if

$\displaystyle\rho_{\bf p}(z^{\alpha},0)>\mu\sum_{\alpha}|z^{\alpha}|^{2},$
(2.13)

for $(z^{\alpha})\neq 0$ when $\delta$ is small enough.

######  Remark 2.10.

A domain is strongly $\mathbb{C}$-convex if and only if its modulus of
$\mathbb{C}$-convexity is greater than $0$. If the inequality (2.13) is
replaced by

$\displaystyle\rho_{\bf
p}(z^{\alpha},s)>\mu(\sum_{\alpha}|z^{\alpha}|^{2}+s^{2}),$ (2.14)

then $\mu$ can be any number smaller than the minimum principle curvature of
$\partial\Omega$. In Section A, we will show if the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $0$, then the
modulus of $\mathbb{C}$-convexity of level sets of $\Phi$ is greater than $0$;
that’s to say, the strong $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ implies
the strong $\mathbb{C}$-convexity of sublevel sets of $\Phi$.

###  2.2 Auxiliary Functions

Let $\Phi$ be a solution to equation (1.29). We denote
$\mathcal{H}=(\Phi_{{i\overline{j}}})$ and
$\mathcal{G}=(\mathcal{G}_{{i\overline{j}}})$. Then (1.29) becomes

$\displaystyle\text{tr}(\mathcal{H}^{-1}\mathcal{G})=\frac{1}{{\varepsilon}}.$
(2.15)

We apply $\partial_{i}$ to the equation above and get

$\displaystyle\text{tr}(\mathcal{H}^{-1}\mathcal{H}_{i}\mathcal{H}^{-1}\mathcal{G})=0.$
(2.16)

Let

$\displaystyle
L^{p{\overline{q}}}={\varepsilon}^{2}\Phi^{p{\overline{v}}}\mathcal{G}_{u{\overline{v}}}\Phi^{u{\overline{q}}};$
(2.17)

then, (2.16) becomes

$\displaystyle L^{p{\overline{q}}}(\Phi_{i})_{p{\overline{q}}}=0,$ (2.18)

and $\mathcal{L}=L^{{i\overline{j}}}\partial_{{i\overline{j}}}$ is the
linearization operator of (1.29). Moreover, $\mathcal{L}$ has bounded
coefficient, independent of ${\varepsilon}$. Let $\mathcal{G}=(\delta_{\ij})$
and $\mathcal{H}=\text{diag}(\lambda_{1},\ ...\ ,\lambda_{n})$. Then equation
(2.15) gives

$\displaystyle\sum_{i=1}^{n}\frac{1}{\lambda_{i}}=\frac{1}{{\varepsilon}},$
(2.19)

so $\lambda_{i}>{\varepsilon}$ for any $i$. With this coordinate,
$L^{{i\overline{j}}}=\delta_{ij}\frac{{\varepsilon}^{2}}{\lambda_{i}^{2}}<1$.
In the following, we will construct some quantities related to the estimates
and apply $\mathcal{L}$ to them.

For the lower bound estimate of $|\nabla\Phi|_{\mathcal{G}}$, we let

$\displaystyle
S={\varepsilon}\Phi_{i}\Phi_{{\overline{j}}}\Phi^{{i\overline{j}}};$ (2.20)

we can show $|\nabla\Phi|_{\mathcal{G}}\geq S$. So a positive lower bound for
$S$ implies a positive lower bound for $|\nabla\Phi|_{\mathcal{G}}$.

For the convexity estimate, we construct a tensor ${\mathcal{K}}$. If $\Phi$
is strongly $\mathfrak{q}\mathbb{C}$-convex, we know $\nabla\Phi\neq 0$
anywhere. For a point ${\bf p}\in\overline{\mathcal{R}}$, we choose a set of
coordinates $\left\\{z^{\alpha},\tau\right\\}$, where $\alpha,\beta\in\\{1,\
...\ ,n-1\\}$ and $\tau$ is the $n$-th coordinate, so that $\Phi_{\alpha}=0$
and $\Phi_{\tau}\neq 0$ at ${\bf p}$ . Then, around ${\bf p}$, we let

$\displaystyle{\mathcal{A}}_{\alpha{\overline{\beta}}}=\Phi_{\alpha{\overline{\beta}}}-\frac{\Phi_{\alpha}}{\Phi_{\tau}}\Phi_{\tau{\overline{\beta}}}-\frac{\Phi_{{\overline{\beta}}}}{\Phi_{\overline{\tau}}}\Phi_{\alpha{\overline{\tau}}}+\frac{\Phi_{\alpha}\Phi_{{\overline{\beta}}}}{|\Phi_{\tau}|^{2}}\Phi_{{\tau\overline{\tau}}},$
(2.21)
$\displaystyle{\mathcal{B}}_{\alpha\beta}=\Phi_{\alpha\beta}-\frac{\Phi_{\alpha}}{\Phi_{\tau}}\Phi_{\tau\beta}-\frac{\Phi_{\beta}}{\Phi_{\tau}}\Phi_{\alpha\tau}+\frac{\Phi_{\alpha}\Phi_{\beta}}{(\Phi_{\tau})^{2}}\Phi_{\tau\tau}.$
(2.22)

Obviously, at ${\bf p}$ we have

$\displaystyle{\mathcal{A}}_{{\alpha\overline{\beta}}}=\Phi_{{\alpha\overline{\beta}}},\
\ \ \ \ \ \ \ {\mathcal{B}}_{{\alpha\beta}}=\Phi_{{\alpha\beta}}.$ (2.23)

Let ${\mathcal{A}},{\mathcal{B}}$ be the matrices

$\displaystyle{\mathcal{A}}=({\mathcal{A}}_{{\alpha\overline{\beta}}}),\ \ \ \
\ \ \ \ {\mathcal{B}}=({\mathcal{B}}_{{\alpha\beta}}).$ (2.24)

Since $\Phi$ is strongly $\mathfrak{q}\mathbb{C}$-convex, the second order
Taylor expansion of $\Phi|_{{T^{\mathbb{C}}_{\Phi,{\bf p}}}}$ at ${\bf p}$ is
strongly convex; then, we have ${\mathcal{A}}>0$ around ${\bf p}$, and we let

$\displaystyle{\mathcal{K}}_{\alpha}^{\theta}={\mathcal{B}}_{\alpha\mu}{\mathcal{A}}^{\mu{\overline{\eta}}}\
\overline{{\mathcal{B}}_{\eta\zeta}}{\mathcal{A}}^{\theta{\overline{\zeta}}}.$
(2.25)

Using matrix notation, the above is equivalent to

$\displaystyle{\mathcal{K}}={\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}{\mathcal{A}}^{-1},$ (2.26)

where ${\mathcal{K}}$ is the matrix $({\mathcal{K}}_{\alpha}^{\theta})$.
${\mathcal{K}}$ can also be considered as a tensor: at ${\bf p}$,
${\mathcal{K}}$ is a complex linear automorphism of
${T^{\mathbb{C}}_{\Phi,{\bf p}}}$. Therefore, eigenvalues of ${\mathcal{K}}$
are independent of the choice of coordinate.

Using Lemma A.9, we know that the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\delta>0$
implies ${\mathcal{K}}<1-\frac{\delta}{C}$, where $C$ is a constant; using
Lemma A.6, we know ${\mathcal{A}}>\delta$ and ${\mathcal{K}}<1-\delta$ implies
that the modulus of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater
than $\frac{\delta^{2}}{C}$ for a constant $C$. So an estimate for the minimum
eigenvalue of $I-{\mathcal{K}}$ implies an estimate for the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$.

In the following, we will provide an upper bound estimate for
$\text{tr}(I-{\mathcal{K}})^{-1}$, which implies a positive lower bound
estimate for the minimum eigenvalue of $I-{\mathcal{K}}$. Actually, we can
show the estimate is still valid if we add a pluriharmonic quadratic
polynomial to $\Phi$; this provides an estimate for the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$.

##  3 The Main Computation Result

The main goal of this section is to prove the following computation results.
These results allow us to use the maximum principle to derive the
corresponding estimates in Section 4 and Section 5.

######  Proposition 3.1.

Suppose that $\Phi$ is a solution to equation (1.29) and that

the robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than
$\delta$. (3.1)

Then if ${\varepsilon}$ is small enough, depending on $\delta$, we have

$\displaystyle
L^{p{\overline{q}}}\left(-\log(S)+\sqrt{{\varepsilon}}\left[W+\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}}\right]\right)_{p{\overline{q}}}\geq
0,$ (3.2) $\displaystyle
L^{p{\overline{q}}}\left(\text{tr}\left[I-{\mathcal{K}}\right]^{-1}+\sqrt{{\varepsilon}}(W+\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}})+{\varepsilon}^{\frac{3}{4}}V\right)_{p{\overline{q}}}\geq
0.$ (3.3)

In above, $L,\ S$ and ${\mathcal{K}}$ are introduced in section 2.2, and
$W=\Phi_{{i\overline{j}}}\mathcal{G}^{{i\overline{j}}}$,
$V=\Phi_{\ij}\mathcal{G}^{j{\overline{k}}}\
\overline{\Phi_{kl}}\mathcal{G}^{i{\overline{l}}}$.

Through this whole section, we only need to do the computation at a point
${\bf p}$. To simplify the computation, we choose coordinates so that
$\Phi_{\alpha}({\bf p})=0$, for $\alpha\in\\{1,\ ...\ ,n-1\\}$, and
$(\Phi_{{i\overline{j}}})({\bf p})$ is diagonal. Actually, we can make
$\Phi_{{\alpha\overline{\beta}}}=\delta_{\alpha\beta}$ and $\Phi_{\tau}=1$,
where $\tau$ denotes the $n$-th coordinate. This will be explained in the next
paragraph. We note that after this choice of coordinate, $\mathcal{G}$ cannot
keep diagonal; we can find $C_{\delta}>0$ so that

$\displaystyle C_{\delta}I\geq\mathcal{G}\geq\frac{1}{C_{\delta}}I,$ (3.4)

where $I$ is the identity matrix and $C_{\delta}$ depends on $\delta$ and the
$C^{2}$ norm of $\Phi$. To prove this, we need to use
$(\Phi_{\alpha{\overline{\beta}}})>\delta(\mathcal{G}_{\alpha{\overline{\beta}}})$
and $|\Phi_{\tau}|_{\mathcal{G}}>\delta$, which follow from the apriori
assumption (3.1).

At a point ${\bf p}$, we choose coordinates
$\\{\widetilde{z^{\alpha}},\tilde{\tau}\\}$, so that
$\Phi_{\widetilde{z^{\alpha}}}=0$, $\Phi_{\tilde{\tau}}\neq 0$,
$\Phi_{{\widetilde{z^{\alpha}}}\overline{\widetilde{z^{\beta}}}}=\delta_{\alpha\beta}\lambda_{\alpha}$
and $g_{{i\overline{j}}}=\delta_{ij}$, with $\lambda_{\alpha}>0$. This can be
done because of the apriori assumption (3.1). Then we find
$\\{z^{\alpha},\tau\\}$, with

$\displaystyle{\widetilde{z^{\alpha}}}=A_{\mu}^{\alpha}z^{\mu}+B^{\alpha}\tau,$
(3.5) $\displaystyle\widetilde{\tau}=E_{\gamma}z^{\gamma}+D\tau.$ (3.6)

so that

$\displaystyle\Phi_{\alpha}=0,$ (3.7) $\displaystyle\Phi_{\tau}=1,$ (3.8)
$\displaystyle\Phi_{\alpha{\overline{\beta}}}=\delta_{\alpha\beta}$ (3.9)

and

$\displaystyle\left(\Phi_{{i\overline{j}}}\right)\text{ is diagonal.}$ (3.10)

Condition (3.7) requires

$\displaystyle
0=\Phi_{\alpha}=\Phi_{{\widetilde{z^{\mu}}}}\frac{\partial{\widetilde{z^{\mu}}}}{\partial{{z^{\alpha}}}}+\Phi_{\widetilde{\tau}}\frac{\partial{\widetilde{\tau}}}{\partial{{z^{\alpha}}}}=\Phi_{\widetilde{\tau}}E_{\alpha},$
(3.11)

so we choose $E_{\alpha}=0$. Condition (3.8) requires

$\displaystyle
1=\Phi_{\tau}=\Phi_{{\widetilde{z^{\mu}}}}\frac{\partial{\widetilde{z^{\mu}}}}{\partial{{\tau}}}+\Phi_{\widetilde{\tau}}\frac{\partial{\widetilde{\tau}}}{\partial{{\tau}}}=\Phi_{\widetilde{\tau}}D,$
(3.12)

so we choose $D=\frac{1}{\Phi_{\widetilde{\tau}}}$. Condition (3.9) requires

$\displaystyle\delta_{\alpha\beta}=\Phi_{{\alpha\overline{\beta}}}=\Phi_{\widetilde{z^{\mu}}\overline{\widetilde{z^{\gamma}}}}A^{\mu}_{\alpha}\overline{\left(A^{\gamma}_{\beta}\right)}.$
(3.13)

Such a matrix $A$ can be find since
$\left(\Phi_{\widetilde{z^{\mu}}\overline{\widetilde{z^{\gamma}}}}\right)>\delta
I$, according to (3.1). Condition (3.10) requires

$\displaystyle
0=\Phi_{{{z^{\alpha}}}\overline{{\tau}}}=\left(\Phi_{\widetilde{z^{\mu}}}\frac{\partial{\widetilde{z^{\mu}}}}{\partial
z^{\alpha}}\right)_{\overline{\tau}}=\Phi_{{\widetilde{z^{\mu}}}\overline{\widetilde{z^{\gamma}}}}A^{\mu}_{\alpha}\overline{B^{\gamma}}+\Phi_{{\widetilde{z^{\mu}}}\overline{\widetilde{\tau}}}A^{\mu}_{\alpha}\overline{D}.$
(3.14)

Since
$\Phi_{{\widetilde{z^{\mu}}}\overline{\widetilde{z^{\gamma}}}}=\delta_{\mu\gamma}\lambda_{\mu}$,
we choose

$\displaystyle
B^{\gamma}=-D\frac{\Phi_{\overline{\widetilde{z^{\mu}}}{\widetilde{\tau}}}}{\lambda_{\mu}}.$
(3.15)

Since the constants $A,B,D,E$ only depend on $\delta$ and the second order
derivatives of $\Phi$, the estimate (3.4) is valid.

Denoting $\Phi_{\tau{\overline{\tau}}}=T$, the equation becomes

$\displaystyle\frac{\mathcal{G}_{{\tau\overline{\tau}}}}{T}+\sum_{\alpha}\mathcal{G}_{\alpha{\overline{\alpha}}}=\frac{1}{{\varepsilon}}.$
(3.16)

Therefore $T$ is comparable to ${\varepsilon}$, providing ${\varepsilon}$ is
small enough. More precisely, we have

$\displaystyle\frac{{\varepsilon}}{C_{\delta}}\leq
T\leq{\varepsilon}C_{\delta},$ (3.17)

providing ${\varepsilon}\leq\frac{1}{C_{\delta}}$ for a large enough
$C_{\delta}$ depending on $\delta$ and the $C^{2}$ norm of $\Phi$.

During the computation, most importantly, we need to control third derivatives
of $\Phi$. In Section 3.1 and 3.2, we will find quadratic forms
${\text{I}_{\Phi}},{\text{II}_{\Phi}}$ of third derivatives and use them to
control third derivatives: we will prove

$\displaystyle|\Phi_{\alpha\beta{\overline{\gamma}}}|{\varepsilon}+|\Phi_{\alpha\beta{\overline{\tau}}}|+|\Phi_{\alpha{\overline{\beta}}\tau}|+\frac{1}{{\varepsilon}}|\Phi_{\alpha\tau{\overline{\tau}}}|+\frac{1}{{\varepsilon}}|\Phi_{\tau\tau{\overline{\tau}}}|\leq
C_{\delta}\sqrt{{\text{I}_{\Phi}}}$ (3.18)

and

$\displaystyle|\Phi_{\alpha\beta\tau}|+|\Phi_{\alpha\tau\tau}|+|\Phi_{\tau\tau\tau}|\leq
C_{\delta}(C_{\delta}{\text{I}_{\Phi}}+{\text{II}_{\Phi}})^{\frac{1}{2}}.$
(3.19)

Here, $C_{\delta}$ is a constant depending on the $\delta$ of (3.1) and the
$C^{2}$ norm of $\Phi$. For clarity, we list the estimates for third
derivatives in the following table.

| Pure Holomorphic Derivatives | Mixed Derivatives
---|---|---
Containing No $\tau$ |  $\Phi_{\alpha\beta\gamma}$ Not Used | ${\varepsilon}|\Phi_{\alpha\beta{\overline{\gamma}}}|\leq C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$
Containing One $\tau$ | $|\Phi_{\alpha\beta\tau}|\leq C_{\delta}(C_{\delta}{\text{I}_{\Phi}}+{\text{II}_{\Phi}})^{\frac{1}{2}}$ | $|\Phi_{\alpha\beta{\overline{\tau}}}|+|\Phi_{\alpha{\overline{\beta}}\tau}|\leq C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$
Containing Two $\tau$’s | $|\Phi_{\alpha\tau\tau}|\leq C_{\delta}(C_{\delta}{\text{I}_{\Phi}}+{\text{II}_{\Phi}})^{\frac{1}{2}}$ | $\frac{1}{{\varepsilon}}|\Phi_{\alpha\tau{\overline{\tau}}}|+|\Phi_{{\overline{\alpha}}\tau\tau}|\leq C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$
Containing Three $\tau$’s | $|\Phi_{\tau\tau\tau}|\leq C_{\delta}(C_{\delta}{\text{I}_{\Phi}}+{\text{II}_{\Phi}})^{\frac{1}{2}}$ | $|\Phi_{\tau\tau{\overline{\tau}}}|\leq{\varepsilon}C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$

Table 1: Third Derivatives

###  3.1 Proving (3.2)

In this section, we prove (3.2). First, we compute
$L^{p{\overline{q}}}W_{p{\overline{q}}}$; this will produce a positive
quadratic form, which can be used to control third derivatives of mixed type.
Actually, we will see the only third derivatives involved in section 3.1 are
of mixed type.

We apply $\partial_{{\overline{j}}}$ to (2.16) and get

$\displaystyle\text{tr}\left(\mathcal{H}^{-1}\mathcal{H}_{{i\overline{j}}}\mathcal{H}^{-1}\mathcal{G}\right)=\text{tr}\left(\mathcal{H}^{-1}\mathcal{H}_{i}\mathcal{H}^{-1}\mathcal{H}_{{\overline{j}}}\mathcal{H}^{-1}\mathcal{G}+\mathcal{H}^{-1}\mathcal{H}_{\overline{j}}\mathcal{H}^{-1}\mathcal{H}_{i}\mathcal{H}^{-1}\mathcal{G}\right).$
(3.20)

Using index summation and operator $\mathcal{L}$, this becomes

$\displaystyle
L^{p{\overline{q}}}(\Phi_{{i\overline{j}}})_{p{\overline{q}}}=L^{p{\overline{q}}}\Phi_{p{\overline{v}}i}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u{\overline{j}}}+L^{p{\overline{q}}}\Phi_{p{\overline{v}}{\overline{j}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}ui}.$
(3.21)

Since $\mathcal{G}^{{i\overline{j}}}$ are constants, we get

$\displaystyle
L^{p{\overline{q}}}W_{p{\overline{q}}}=L^{p{\overline{q}}}\Phi_{p{\overline{v}}i}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u{\overline{j}}}\mathcal{G}^{{i\overline{j}}}+L^{p{\overline{q}}}\Phi_{p{\overline{v}}{\overline{j}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}ui}\mathcal{G}^{{i\overline{j}}}.$
(3.22)

We denote

$\displaystyle L^{p{\overline{q}}}W_{p{\overline{q}}}={\text{I}_{\Phi}}.$
(3.23)

Using (3.4) and the expression of $L^{{i\overline{j}}}$, we know

$\displaystyle{\varepsilon}^{2}C_{\delta}\mathcal{H}^{-2}\geq(L^{{i\overline{j}}})\geq{\varepsilon}^{2}\frac{1}{C_{\delta}}\mathcal{H}^{-2}.$
(3.24)

We plug (3.24) and (3.4) into (3.22) and get

$\displaystyle{\text{I}_{\Phi}}\geq\frac{{\varepsilon}^{2}}{C_{\delta}}\delta_{ij}\delta_{uv}\delta_{pq}\Phi^{u{\overline{u}}}\Phi^{p{\overline{p}}}\Phi^{q{\overline{q}}}\left(\Phi_{p{\overline{v}}i}\Phi_{{\overline{q}}u{\overline{j}}}+\Phi_{p{\overline{v}}{\overline{j}}}\Phi_{{\overline{q}}ui}\right).$
(3.25)

Choosing $p=i=u=\tau$, we get

$\displaystyle|\Phi_{\tau\tau{\overline{\tau}}}|^{2}\leq{\varepsilon}C_{\delta}{\text{I}_{\Phi}};$
(3.26)

choosing $p=i=\tau,\ u=\alpha$, we get

$\displaystyle|\Phi_{\alpha\tau{\overline{\tau}}}|^{2}+|\Phi_{{\overline{\alpha}}\tau\tau}|^{2}\leq
C_{\delta}{\text{I}_{\Phi}};$ (3.27)

choosing $p=\tau,\ i=\beta,\ u=\alpha$, we get

$\displaystyle|\Phi_{\alpha\beta{\overline{\tau}}}|^{2}+|\Phi_{\alpha{\overline{\beta}}\tau}|^{2}\leq
C_{\delta}{\text{I}_{\Phi}};$ (3.28)

choosing $p=\gamma,\ i=\beta,\ u=\alpha$, we get

$\displaystyle|\Phi_{\alpha\beta{\overline{\gamma}}}|^{2}\leq\frac{1}{{\varepsilon}^{2}}C_{\delta}{\text{I}_{\Phi}}.$
(3.29)

However, the estimates for $\Phi_{\tau{\tau\overline{\tau}}}$ and
$\Phi_{\alpha{\tau\overline{\tau}}}$ are weaker than we want (comparing to
(3.18)). To get a better estimate, we need to use the equation of $\Phi$. In
(2.16), we let $i=\nu\in\\{1,\ ...\ ,n-1\\}$ and get

$\displaystyle\frac{{\varepsilon}^{2}}{T^{2}}\mathcal{G}_{{\tau\overline{\tau}}}\Phi_{\nu\tau{\overline{\tau}}}+\frac{{\varepsilon}^{2}}{T}\mathcal{G}_{\alpha{\overline{\tau}}}\Phi_{\nu\tau{\overline{\alpha}}}+\frac{{\varepsilon}^{2}}{T}\mathcal{G}_{\tau{\overline{\alpha}}}\Phi_{\nu{\overline{\tau}}\alpha}+{\varepsilon}^{2}\mathcal{G}_{\alpha{\overline{\mu}}}\Phi_{\nu\mu{\overline{\alpha}}}=0.$
(3.30)

According to the estimates (3.28) and (3.29), we have

$\displaystyle|\Phi_{\nu\tau{\overline{\alpha}}}|+|\Phi_{\nu{\overline{\tau}}\alpha}|+{\varepsilon}|\Phi_{\nu\mu{\overline{\alpha}}}|\leq
C_{\delta}\sqrt{{\text{I}_{\Phi}}}.$ (3.31)

Plugging these and the estimate (3.17) into (3.30), when ${\varepsilon}$ is
small enough, we have

$\displaystyle|\Phi_{\nu\tau{\overline{\tau}}}|\leq
C_{\delta}{\varepsilon}\sqrt{{\text{I}_{\Phi}}}.$ (3.32)

Similarly, in (2.16), we let $i=n$ and get

$\displaystyle\frac{{\varepsilon}^{2}}{T^{2}}\mathcal{G}_{{\tau\overline{\tau}}}\Phi_{\tau\tau{\overline{\tau}}}+\frac{{\varepsilon}^{2}}{T}\mathcal{G}_{\alpha{\overline{\tau}}}\Phi_{\tau\tau{\overline{\alpha}}}+\frac{{\varepsilon}^{2}}{T}\mathcal{G}_{\tau{\overline{\alpha}}}\Phi_{\tau{\overline{\tau}}\alpha}+{\varepsilon}^{2}\mathcal{G}_{\alpha{\overline{\mu}}}\Phi_{\tau\mu{\overline{\alpha}}}=0.$
(3.33)

Then using estimates (3.27) (3.28) and (3.32), we get

$\displaystyle|\Phi_{\tau\tau{\overline{\tau}}}|\leq
C_{\delta}{\varepsilon}\sqrt{{\text{I}_{\Phi}}}.$ (3.34)

The term $\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}}$ in (3.2) is used
to produce a positive constant term. We have

$\displaystyle
L^{p{\overline{q}}}(\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}})_{p{\overline{q}}}=L^{p{\overline{q}}}\mathcal{G}_{p{\overline{q}}}={\varepsilon}^{2}\text{tr}\left(\mathcal{G}\mathcal{H}^{-1}\mathcal{G}\mathcal{H}^{-1}\right).$
(3.35)

We apply the Cauchy inequality and get

$\displaystyle\text{tr}\left(\mathcal{G}\mathcal{H}^{-1}\mathcal{G}\mathcal{H}^{-1}\right)\geq\frac{1}{n}\text{tr}(\mathcal{G}\mathcal{H}^{-1})^{2}=\frac{1}{n{\varepsilon}^{2}}.$
(3.36)

In the last equality, we used equation (2.15). Therefore,
$L^{p{\overline{q}}}(\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}})_{p{\overline{q}}}\geq\frac{1}{n}$.

Then we compute $L^{p{\overline{q}}}\left(-\log(S)\right)$; we will show it’s
greater than $-C_{\delta}{\varepsilon}({\text{I}_{\Phi}}+1)$. Directly apply
$\mathcal{L}$ to $-\log(S)$, and we get

$\displaystyle
L^{p{\overline{q}}}(-\log(S))_{p{\overline{q}}}=-\frac{L^{p{\overline{q}}}{S_{p{\overline{q}}}}}{S}+\frac{L^{p{\overline{q}}}S_{p}S_{\overline{q}}}{S^{2}}.$
(3.37)

In the following, we compute the two terms on the right-hand side of (3.37)
separately. First of all, we notice that according to the choice of
coordinate, at the point ${\bf p}$

$\displaystyle
S={\varepsilon}\Phi_{\tau}\Phi^{{\tau\overline{\tau}}}\Phi_{{\overline{\tau}}}=\frac{{\varepsilon}}{T}.$
(3.38)

We apply $\partial_{p}$ to $S$ and get

$\displaystyle
S_{p}={\varepsilon}\Phi_{ip}\Phi^{{i\overline{j}}}\Phi_{{\overline{j}}}-{\varepsilon}\Phi_{i}\Phi^{i{\overline{v}}}\Phi_{u{\overline{v}}p}\Phi^{u{\overline{j}}}\Phi_{{\overline{j}}}+{\varepsilon}\Phi_{p}.$
(3.39)

At the point ${\bf p}$, the expression is simpler:

$\displaystyle S_{p}=\frac{{\varepsilon}}{T}\Phi_{\tau
p}-\frac{{\varepsilon}}{T^{2}}\Phi_{{\tau\overline{\tau}}p}+{\varepsilon}\Phi_{p}.$
(3.40)

So, at ${\bf p}$, we have

$\displaystyle\frac{L^{p{\overline{q}}}S_{p}S_{\overline{q}}}{S^{2}}=L^{p{\overline{q}}}(\Phi_{\tau
p}-\frac{1}{T}\Phi_{{\tau\overline{\tau}}p}+T\Phi_{p})(\Phi_{{\overline{\tau}}{\overline{q}}}-\frac{1}{T}\Phi_{{\tau\overline{\tau}}{\overline{q}}}+T\Phi_{{\overline{q}}}).$
(3.41)

According to the estimate (3.18) and the $C^{1},C^{2}$ estimate of $\Phi$,
$\Phi_{\tau p}-\frac{1}{T}\Phi_{{\tau\overline{\tau}}p}+T\Phi_{p}$ is bounded
for any $p$. We also notice that $|L^{p{\overline{q}}}|$ is smaller than
$C_{\delta}{\varepsilon}$ when $p$ or $q$ is not $n$. So, we have

$\displaystyle\frac{L^{p{\overline{q}}}S_{p}S_{\overline{q}}}{S^{2}}\geq$
$\displaystyle
L^{{\tau\overline{\tau}}}(\Phi_{\tau\tau}-\frac{1}{T}\Phi_{\tau{\tau\overline{\tau}}}+T)(\Phi_{{\overline{\tau}}{\overline{\tau}}}-\frac{1}{T}\Phi_{{\tau\overline{\tau}}{\overline{\tau}}}+T)-C_{\delta}{\varepsilon}$
(3.42) $\displaystyle\geq$ $\displaystyle
L^{{\tau\overline{\tau}}}(\Phi_{\tau\tau}-\frac{1}{T}\Phi_{\tau{\tau\overline{\tau}}})(\Phi_{{\overline{\tau}}{\overline{\tau}}}-\frac{1}{T}\Phi_{{\tau\overline{\tau}}{\overline{\tau}}})-C_{\delta}{\varepsilon}.$
(3.43)

Then we compute $\frac{L^{p{\overline{q}}}S_{p{\overline{q}}}}{S}$. We apply
$\frac{1}{S}L^{p{\overline{q}}}\partial_{{\overline{q}}}$ to the three terms
on the right-hand side of (3.39), then sum over $p$. For the first term we
have

$\displaystyle\frac{1}{S}L^{p{\overline{q}}}({\varepsilon}\Phi_{ip}\Phi^{{i\overline{j}}}\Phi_{{\overline{j}}})_{{\overline{q}}}=L^{p{\overline{q}}}\Phi_{\tau
p{\overline{q}}}-L^{p{\overline{q}}}\Phi_{ip}\Phi^{i{\overline{v}}}\Phi_{\tau{\overline{v}}{\overline{q}}}+TL^{p{\overline{q}}}\Phi_{ip}\Phi^{{i\overline{j}}}\Phi_{{\overline{q}}{\overline{j}}}.$
(3.44)

In above, we used that $\Phi_{{\tau\overline{\tau}}}=T$,
$S=\frac{{\varepsilon}}{T}$, $\Phi_{\tau}=1$ and $\Phi_{\alpha}=0$. We also
observe that the first term on the right-hand side of (3.44) is $0$ because
$L^{{i\overline{j}}}\partial_{{i\overline{j}}}$ is the linearized operator.

Applying $\frac{1}{S}L^{p{\overline{q}}}\partial_{{\overline{q}}}$ to the
second term on the right-hand side of (3.39), we get

$\displaystyle\frac{1}{S}L^{p{\overline{q}}}(-{\varepsilon}\Phi_{i}\Phi^{i{\overline{v}}}\Phi_{u{\overline{v}}p}\Phi^{u{\overline{j}}}\Phi_{{\overline{j}}})_{{\overline{q}}}=$
$\displaystyle-L^{p{\overline{q}}}\Phi_{\tau
p{\overline{q}}}+\frac{1}{T}L^{p{\overline{q}}}\Phi_{{\overline{\tau}}s{\overline{q}}}\Phi^{s{\overline{v}}}\Phi_{\tau{\overline{v}}p}-\frac{1}{T}L^{p{\overline{q}}}\Phi_{{\tau\overline{\tau}}p{\overline{q}}}$
(3.45) $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+\frac{1}{T}L^{p{\overline{q}}}\Phi_{u{\overline{\tau}}p}\Phi^{u{\overline{k}}}\Phi_{\tau{\overline{k}}{\overline{q}}}-L^{p{\overline{q}}}\Phi_{{\overline{\tau}}up}\Phi^{u{\overline{j}}}\Phi_{{\overline{j}}{\overline{q}}}.$
(3.46)

Similar to (3.44), in above we used $\Phi_{\alpha}=0$, $\Phi_{\tau}=1$ and
$\Phi^{{\tau\overline{\tau}}}=\frac{1}{T}$; we also have (3.45)${}_{2}=0$
because $L^{{i\overline{j}}}\partial_{{i\overline{j}}}$ is the linearized
operator. Here we used the convention introduced in Section 1.3. We let the
$i,{\overline{j}}$ of (3.21) be $\tau,{\overline{\tau}}$ and get

$\displaystyle
L^{p{\overline{q}}}\Phi_{{\tau\overline{\tau}}p{\overline{q}}}=L^{p{\overline{q}}}\Phi_{p{\overline{v}}\tau}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u{\overline{\tau}}}+L^{p{\overline{q}}}\Phi_{p{\overline{v}}{\overline{\tau}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u\tau}.$
(3.47)

So

$\displaystyle(\ref{516331new})_{4}=-\frac{1}{T}L^{p{\overline{q}}}\Phi_{p{\overline{v}}\tau}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u{\overline{\tau}}}-\frac{1}{T}L^{p{\overline{q}}}\Phi_{p{\overline{v}}{\overline{\tau}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u\tau},$
(3.48)

and we see (3.48)2 cancels with $(\ref{516331new})_{3}$.

Applying $\frac{1}{S}L^{p{\overline{q}}}\partial_{{\overline{q}}}$ to the last
term of (3.39), we get

$\displaystyle\frac{1}{S}L^{p{\overline{q}}}({\varepsilon}\Phi_{p})_{{\overline{q}}}=\frac{T}{{\varepsilon}}\cdot{\varepsilon}\cdot
L^{p{\overline{q}}}\Phi_{p{\overline{q}}}=T\cdot{\varepsilon}^{2}\text{tr}(\mathcal{G}\mathcal{H}^{-1}).$
(3.49)

In above, we used the definition of $L$ and that $S=\frac{{\varepsilon}}{T}$
at ${\bf p}$. Then, using equation (2.15), we get

$\displaystyle
T\cdot{\varepsilon}^{2}\text{tr}(\mathcal{G}\mathcal{H}^{-1})=T{\varepsilon}.$
(3.50)

We put the results above together and get

$\displaystyle\frac{L^{p{\overline{q}}}S_{p{\overline{q}}}}{S}=TL^{p{\overline{q}}}\left(\Phi_{ip}-\frac{1}{T}\Phi_{ip{\overline{\tau}}}\right)\Phi^{{i\overline{j}}}\left(\Phi_{{\overline{q}}{\overline{j}}}-\frac{1}{T}\Phi_{\tau{\overline{j}}{\overline{q}}}\right)-\frac{1}{T}L^{p{\overline{q}}}\Phi_{p{\overline{v}}{\overline{\tau}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u\tau}+T{\varepsilon}.$
(3.51)

A careful but straightforward computation shows

$\displaystyle(\ref{516339})_{2}\leq
L^{{\tau\overline{\tau}}}\left(\Phi_{\tau\tau}-\frac{1}{T}\Phi_{\tau\tau{\overline{\tau}}}\right)\Phi^{{i\overline{j}}}\left(\Phi_{{\overline{\tau}}{\overline{\tau}}}-\frac{1}{T}\Phi_{\tau{\overline{\tau}}{\overline{\tau}}}\right)+{\varepsilon}C_{\delta}({\text{I}_{\Phi}}+1).$
(3.52)

Then (3.52)2 can cancel with (3.43)1.

Summing up, we find

$\displaystyle
L^{p{\overline{q}}}\left(-\log(S)+\sqrt{{\varepsilon}}\left[W+\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}}\right]\right)_{p{\overline{q}}}\geq-{\varepsilon}C_{\delta}({\text{I}_{\Phi}}+1)+\sqrt{{\varepsilon}}({\text{I}_{\Phi}}+\frac{1}{n}).$
(3.53)

Therefore, when we choose ${\varepsilon}$ small enough, the above is non-
negative.

######  Remark 3.2.

When ${\varepsilon}$ approximates to zero, the $S$ we constructed here
approximates to the $S$ of section E.2, and the operator $\mathcal{L}$ here
approximates to the leaf-wise Laplacian there. The formal computation of
section E.2 shows $S$ is subharmonic along leaves; the computation here
suggests the same result.

###  3.2 Proving (3.3)

Suppose that ${\mathscr{A}}$ and ${\mathscr{B}}$ are matrix-valued functions
defined on a domain in $\mathbb{C}^{n}$, where ${\mathscr{A}}$ is positive
definite and Hermitian and ${\mathscr{B}}$ is symmtric, and suppose

$\displaystyle{\mathscr{L}}={\mathscr{L}}^{p{\overline{q}}}\partial_{p{\overline{q}}}$
(3.54)

is an elliptic operator. If ${\mathscr{A}}$ and ${\mathscr{B}}$ satisfy

$\displaystyle{\mathscr{L}}^{p{\overline{q}}}\partial_{p{\overline{q}}}{\mathscr{A}}={\mathscr{L}}^{p{\overline{q}}}\partial_{p}{\mathscr{A}}{\mathscr{A}}^{-1}\partial_{\overline{q}}{\mathscr{A}}+{\mathscr{L}}^{p{\overline{q}}}\partial_{{\overline{q}}}{\mathscr{B}}\overline{{\mathscr{A}}^{-1}}\partial_{p}\overline{{\mathscr{B}}},$
(3.55)
$\displaystyle{\mathscr{L}}^{p{\overline{q}}}\partial_{p{\overline{q}}}{\mathscr{B}}={\mathscr{L}}^{p{\overline{q}}}\partial_{p}{\mathscr{A}}{\mathscr{A}}^{-1}\partial_{\overline{q}}{\mathscr{B}}+{\mathscr{L}}^{p{\overline{q}}}\partial_{{\overline{q}}}{\mathscr{B}}\overline{{\mathscr{A}}^{-1}}\partial_{p}\overline{{\mathscr{A}}},$
(3.56)

then for

$\displaystyle{\mathscr{K}}={\mathscr{B}}\overline{{\mathscr{A}}^{-1}}\
\overline{{\mathscr{B}}}{\mathscr{A}}^{-1}$ (3.57)

we have

$\displaystyle{\mathscr{L}}^{p{\overline{q}}}\left[\text{tr}({\mathscr{K}}^{r})\right]_{p{\overline{q}}}\geq
0$ (3.58)

providing ${\mathscr{K}}<1-\frac{1}{2r}$. This is the main idea of the paper
[H22N].

Actually, the following is also true:

$\displaystyle{\mathscr{L}}^{p{\overline{q}}}\left[\text{tr}(I-{\mathscr{K}})^{-1}\right]_{p{\overline{q}}}\geq
0,$ (3.59)

providing ${\mathscr{K}}<1$.

In this paper, we generalize this idea to the study of the
$\mathbb{C}$-convexity of level sets of $\Phi$. Instead of (3.55) and (3.56),
we can prove, for ${\mathcal{A}}$ and ${\mathcal{B}}$ defined in Section 2.2,

$\displaystyle|L^{p{\overline{q}}}\partial_{p{\overline{q}}}{\mathcal{A}}-L^{p{\overline{q}}}\partial_{p}{\mathcal{A}}{\mathcal{A}}^{-1}\partial_{\overline{q}}{\mathcal{A}}-L^{p{\overline{q}}}\partial_{{\overline{q}}}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\partial_{p}\overline{{\mathcal{B}}}|\leq{\varepsilon}C_{\delta}(C_{\delta}(1+{\text{I}_{\Phi}})+{\text{II}_{\Phi}}),$
(3.60)
$\displaystyle|L^{p{\overline{q}}}\partial_{p{\overline{q}}}{\mathcal{B}}-L^{p{\overline{q}}}\partial_{p}{\mathcal{A}}{\mathcal{A}}^{-1}\partial_{\overline{q}}{\mathcal{B}}-L^{p{\overline{q}}}\partial_{{\overline{q}}}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\partial_{p}\overline{{\mathcal{A}}}|\leq{\varepsilon}C_{\delta}(C_{\delta}(1+{\text{I}_{\Phi}})+{\text{II}_{\Phi}});$
(3.61)

then, similar to (3.59), we can show

$\displaystyle
L^{p{\overline{q}}}\left[\text{tr}(I-{\mathcal{K}})^{-1}\right]_{p{\overline{q}}}\geq-{\varepsilon}C_{\delta}(C_{\delta}(1+{\text{I}_{\Phi}})+{\text{II}_{\Phi}}),$
(3.62)

providing ${\mathcal{K}}<1$. Here

$\displaystyle{\text{II}_{\Phi}}=L^{p{\overline{q}}}\left(\Phi_{ij}\mathcal{G}^{j{\overline{u}}}\
\overline{\Phi_{uv}}\mathcal{G}^{i{\overline{v}}}\right)_{p{\overline{q}}}.$
(3.63)

In the following, we denote

$\displaystyle V=\Phi_{ij}\mathcal{G}^{j{\overline{u}}}\
\overline{\Phi_{uv}}\mathcal{G}^{i{\overline{v}}}.$ (3.64)

If (3.62) is valid, we have

$\displaystyle\begin{split}L^{p{\overline{q}}}&\left(\text{tr}(I-{\mathcal{K}})^{-1}+{\varepsilon}^{\frac{1}{2}}(W+\mathcal{G}_{i\overline{j}}z^{i}\overline{z^{j}})+{\varepsilon}^{\frac{3}{4}}V\right)_{p{\overline{q}}}\geq\\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
-{\varepsilon}C_{\delta}(C_{\delta}(1+{\text{I}_{\Phi}})+{\text{II}_{\Phi}})+{\varepsilon}^{\frac{1}{2}}({\text{I}_{\Phi}}+\frac{1}{n})+{\varepsilon}^{\frac{3}{4}}{\text{II}_{\Phi}}.\end{split}$
(3.65)

So, when ${\varepsilon}$ is small enough, we have that the right-hand side of
(3.65) is non-negative; then, the maximum principle implies a positive lower
bound estimate for the minimum eigenvalue of $I-{\mathcal{K}}$.

In the following, we first provide computations about ${\text{II}_{\Phi}}$ and
prove (3.19); then, we prove (3.60) and (3.61); finally, we prove (3.62).

Simply using Leibniz rule, we have

$\displaystyle L^{p{\overline{q}}}V_{p{\overline{q}}}=$ $\displaystyle
L^{p{\overline{q}}}\Phi_{ijp}\overline{\Phi_{uvq}}\mathcal{G}^{i{\overline{v}}}\mathcal{G}^{j{\overline{u}}}+L^{p{\overline{q}}}\Phi_{ij{\overline{q}}}\overline{\Phi_{uv{\overline{p}}}}\mathcal{G}^{i{\overline{v}}}\mathcal{G}^{j{\overline{u}}}$
(3.66)
$\displaystyle+L^{p{\overline{q}}}\Phi_{ijp{\overline{q}}}\overline{\Phi_{uv}}\mathcal{G}^{i{\overline{v}}}\mathcal{G}^{j{\overline{u}}}+L^{p{\overline{q}}}\Phi_{ij}\Phi_{{\overline{u}}{\overline{v}}p{\overline{q}}}\mathcal{G}^{i{\overline{v}}}\mathcal{G}^{j{\overline{u}}}.$
(3.67)

Let

$\displaystyle{\text{III}_{\Phi}}=L^{p{\overline{q}}}\Phi_{ijp}\overline{\Phi_{uvq}}\mathcal{G}^{i{\overline{v}}}\mathcal{G}^{j{\overline{u}}}.$
(3.68)

Then

$\displaystyle{\text{III}_{\Phi}}\geq\frac{1}{C_{\delta}}\left(|\Phi_{\tau\alpha\beta}|^{2}+|\Phi_{\tau\tau\alpha}|^{2}+|\Phi_{\tau\tau\tau}|^{2}\right),$
(3.69)

for a constant $C_{\delta}$ large enough depending on $\delta.$ We will show

$\displaystyle|L^{p{\overline{q}}}\Phi_{ijp{\overline{q}}}\overline{\Phi_{uv}}\mathcal{G}^{i{\overline{v}}}\mathcal{G}^{j{\overline{u}}}|\leq
C_{\delta}{\text{I}_{\Phi}}.$ (3.70)

Then it follows that

$\displaystyle{\text{II}_{\Phi}}+C_{\delta}{\text{I}_{\Phi}}\geq{\text{III}_{\Phi}}$
(3.71)

and the estimate (3.19) is valid. For (3.70), we have

$\displaystyle
L^{p{\overline{q}}}\Phi_{ijp{\overline{q}}}=L^{p{\overline{q}}}\Phi_{iu{\overline{q}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{v}}jp}+L^{p{\overline{q}}}\Phi_{ju{\overline{q}}}\Phi^{u{\overline{v}}}\Phi_{i{\overline{v}}p},$
(3.72)

so we only need to show

$\displaystyle|L^{p{\overline{q}}}\Phi_{iu{\overline{q}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{v}}jp}|\leq
C_{\delta}{\text{I}_{\Phi}},$ (3.73)
$\displaystyle|L^{p{\overline{q}}}\Phi_{ju{\overline{q}}}\Phi^{u{\overline{v}}}\Phi_{i{\overline{v}}p}|\leq
C_{\delta}{\text{I}_{\Phi}}.$ (3.74)

We will only prove (3.73); then, by simply switching $i$ with $j$ we can get
(3.74). Since $(\Phi_{{i\overline{j}}})$ is diagonal, we need to show

$\displaystyle|L^{p{\overline{q}}}\Phi_{i\alpha{\overline{q}}}\Phi_{{\overline{\alpha}}jp}|\leq
C_{\delta}{\text{I}_{\Phi}}$ (3.75)

and

$\displaystyle|L^{p{\overline{q}}}\Phi_{i\tau{\overline{q}}}\Phi_{{\overline{\tau}}jp}|\leq{\varepsilon}C_{\delta}{\text{I}_{\Phi}}.$
(3.76)

According to the definition of operator $L$, the left-hand side of (3.75) has
the following decomposition:

$\displaystyle
L^{p{\overline{q}}}\Phi_{i\alpha{\overline{q}}}\Phi_{{\overline{\alpha}}jp}=\left({\varepsilon}\Phi^{p{\overline{v}}}{\Phi_{{\overline{\alpha}}jp}}\right)\mathcal{G}_{u{\overline{v}}}\left({\varepsilon}\Phi^{u{\overline{q}}}{\Phi_{i\alpha{\overline{q}}}}\right).$
(3.77)

We can infer from (3.18) and (3.17) that both
$|{\varepsilon}\Phi^{p{\overline{v}}}{\Phi_{{\overline{\alpha}}jp}}|$ and
$|{\varepsilon}\Phi^{u{\overline{q}}}{\Phi_{i\alpha{\overline{q}}}}|$ are
smaller than $C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$, so (3.75) is valid.
Similarly the left-hand side of (3.76) has the following decomposition

$\displaystyle
L^{p{\overline{q}}}\Phi_{i\tau{\overline{q}}}\Phi_{{\overline{\tau}}jp}=\left({\varepsilon}\Phi^{p{\overline{v}}}{\Phi_{{\overline{\tau}}jp}}\right)\mathcal{G}_{u{\overline{v}}}\left({\varepsilon}\Phi^{u{\overline{q}}}{\Phi_{i\tau{\overline{q}}}}\right).$
(3.78)

We have that both
$|{\varepsilon}\Phi^{p{\overline{v}}}{\Phi_{{\overline{\tau}}jp}}|$ and
$|{\varepsilon}\Phi^{u{\overline{q}}}{\Phi_{i\tau{\overline{q}}}}|$ are
smaller than ${\varepsilon}C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$, so
(3.76) is valid. Therefore, (3.69) is valid and (3.19) is proved.

To prove (3.60) and (3.61), we introduce two intermediate quantities $A$ and
$B$ in a neighborhood of ${\bf p}$: let

$\displaystyle A=\Phi_{\alpha{\overline{\beta}}},\ \ \ \ \ \ \ \
B=\Phi_{\alpha\beta}.$ (3.79)

Because of the assumption (3.1) and that $\Phi\in
C^{\infty}(\overline{\mathcal{R}})$, $A$ is invertible in a small neighborhood
of ${\bf p}$. Here we only need to compute at ${\bf p}$ and don’t need to
worry about the size of the neighborhood. Moreover, we point out that
$A={\mathcal{A}}$ and $B={\mathcal{B}}$ at ${\bf p}$.

We can show $A,B$ satisfy the estimates similar to (3.60) and (3.61)

$\displaystyle|L^{p{\overline{q}}}\partial_{p{\overline{q}}}A-L^{p{\overline{q}}}\partial_{p}AA^{-1}\partial_{\overline{q}}A-L^{p{\overline{q}}}\partial_{{\overline{q}}}B\overline{A^{-1}}\partial_{p}\overline{B}|\leq{\varepsilon}C_{\delta}{\text{I}_{\Phi}},$
(3.80)
$\displaystyle|L^{p{\overline{q}}}\partial_{p{\overline{q}}}B-L^{p{\overline{q}}}\partial_{p}AA^{-1}\partial_{\overline{q}}B-L^{p{\overline{q}}}\partial_{{\overline{q}}}B\overline{A^{-1}}\partial_{p}\overline{A}|\leq{\varepsilon}C_{\delta}{\text{I}_{\Phi}},$
(3.81)

and $A,B$ are close to ${\mathcal{A}},{\mathcal{B}}$ in the sense that the
following relations are valid at ${\bf p}$:

$\displaystyle|L^{i{\overline{j}}}\partial_{i{\overline{j}}}(A-{\mathcal{A}})|$
$\displaystyle\leq{\varepsilon}C_{\delta}(1+{\text{I}_{\Phi}}),$ (3.82)
$\displaystyle|L^{i{\overline{j}}}\partial_{i{\overline{j}}}(B-{\mathcal{B}})|$
$\displaystyle\leq{\varepsilon}C_{\delta}(C_{\delta}(1+{\text{I}_{\Phi}})+{\text{II}_{\Phi}}),$
(3.83) $\displaystyle\partial_{i}A$ $\displaystyle=\partial_{i}{\mathcal{A}},$
(3.84)
$\displaystyle|{\varepsilon}\Phi^{p{\overline{q}}}\partial_{\overline{q}}{\mathcal{B}}-{\varepsilon}\Phi^{p{\overline{q}}}\partial_{\overline{q}}B|$
$\displaystyle\leq{\varepsilon}C_{\delta},$ (3.85)
$\displaystyle|{\varepsilon}\Phi^{p{\overline{v}}}\partial_{p}\overline{B}|+|{\varepsilon}\Phi^{p{\overline{v}}}\partial_{p}A|$
$\displaystyle\leq C_{\delta}\sqrt{{\text{I}_{\Phi}}}.$ (3.86)

With the relations above, in (3.80) and (3.81) we can replace $A,B$ by
${\mathcal{A}},{\mathcal{B}}$ and get (3.60) and (3.61). For this, we only
need to explain how to handle terms containing first derivatives of $A$ and
$B$ in (3.80) and (3.81); indeed, we only explain how to handle
$L^{p{\overline{q}}}\partial_{\overline{q}}B\overline{A^{-1}}\partial_{p}\overline{B}$,
the treatment of other terms is simpler. We have the following decomposition:

$\displaystyle
L^{p{\overline{q}}}\partial_{\overline{q}}B\overline{A^{-1}}\partial_{p}\overline{B}=\left({\varepsilon}\Phi^{u{\overline{q}}}\partial_{{\overline{q}}}B\right)\overline{A^{-1}}\left({\varepsilon}\Phi^{p{\overline{v}}}\partial_{p}\overline{B}\right)\mathcal{G}_{u{\overline{v}}}.$
(3.87)

Then using (3.85) and (3.86), we find

$\displaystyle|\left({\varepsilon}\Phi^{u{\overline{q}}}\partial_{{\overline{q}}}B\right)\overline{A^{-1}}\left({\varepsilon}\Phi^{p{\overline{v}}}\partial_{p}\overline{B}\right)\mathcal{G}_{u{\overline{v}}}-\left({\varepsilon}\Phi^{u{\overline{q}}}\partial_{{\overline{q}}}{\mathcal{B}}\right)\overline{{\mathcal{A}}^{-1}}\left({\varepsilon}\Phi^{p{\overline{v}}}\partial_{p}\overline{{\mathcal{B}}}\right)\mathcal{G}_{u{\overline{v}}}|\leq{\varepsilon}C_{\delta}(1+\sqrt{{\text{I}_{\Phi}}});$
(3.88)

therefore

$\displaystyle|L^{p{\overline{q}}}\partial_{\overline{q}}B\overline{A^{-1}}\partial_{p}\overline{B}-L^{p{\overline{q}}}\partial_{\overline{q}}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\partial_{p}\overline{{\mathcal{B}}}|\leq{\varepsilon}C_{\delta}(1+\sqrt{{\text{I}_{\Phi}}}).$
(3.89)

In the following, we prove (3.80)-(3.86) one by one. First, we prove (3.80).
We let $i,j$ be $\alpha,\beta$ in (3.21) and get

$\displaystyle
L^{p{\overline{q}}}(\Phi_{\alpha{\overline{\beta}}})_{p{\overline{q}}}=L^{p{\overline{q}}}\Phi_{p{\overline{v}}\alpha}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u{\overline{\beta}}}+L^{p{\overline{q}}}\Phi_{p{\overline{v}}{\overline{\beta}}}\Phi^{u{\overline{v}}}\Phi_{{\overline{q}}u\alpha}.$
(3.90)

Since $\Phi_{{i\overline{j}}}$ is diagonal, we have

$\displaystyle\begin{split}|L^{p{\overline{q}}}(\Phi_{\alpha{\overline{\beta}}})_{p{\overline{q}}}-L^{p{\overline{q}}}\Phi_{p{\overline{\nu}}\alpha}\Phi^{\mu{\overline{\nu}}}\Phi_{{\overline{q}}\mu{\overline{\beta}}}&-L^{p{\overline{q}}}\Phi_{p{\overline{\nu}}{\overline{\beta}}}\Phi^{\mu{\overline{\nu}}}\Phi_{{\overline{q}}\mu\alpha}|\\\
\leq\frac{1}{T}&\left(L^{p{\overline{q}}}\Phi_{p{\overline{\tau}}\alpha}\Phi_{{\overline{q}}\tau{\overline{\beta}}}+L^{p{\overline{q}}}\Phi_{p{\overline{\tau}}{\overline{\beta}}}\Phi_{{\overline{q}}\tau\alpha}\right).\end{split}$
(3.91)

So we need to show the right-hand side of (3.91) can be controlled by
${\varepsilon}C_{\delta}{\text{I}_{\Phi}}$. Similar to the proof of (3.76), we
have

$\displaystyle
L^{p{\overline{q}}}\Phi_{{\overline{\beta}}\tau{\overline{q}}}\Phi_{{\overline{\tau}}\alpha
p}=\left({\varepsilon}\Phi^{p{\overline{v}}}{\Phi_{{\overline{\tau}}\alpha
p}}\right)\mathcal{G}_{u{\overline{v}}}\left({\varepsilon}\Phi^{u{\overline{q}}}{\Phi_{{\overline{\beta}}\tau{\overline{q}}}}\right).$
(3.92)

Using (3.18), we know
$|{\varepsilon}\Phi^{p{\overline{v}}}{\Phi_{{\overline{\tau}}\alpha p}}|$ and
$|{\varepsilon}\Phi^{u{\overline{q}}}{\Phi_{{\overline{\beta}}\tau{\overline{q}}}}|$
are both smaller than
${\varepsilon}C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$, so (3.92) can be
controlled by ${\varepsilon}^{2}C_{\delta}{\text{I}_{\Phi}}$.
$L^{p{\overline{q}}}\Phi_{p{\overline{\tau}}{\overline{\beta}}}\Phi_{{\overline{q}}\tau\alpha}$
can be estimated similarly, and we know (3.80) is valid. The proof of (3.81)
is almost parallel; we only need to replace ${\overline{\beta}}$ by $\beta$ in
the argument above.

For the proof of (3.82), we have

$\displaystyle
L^{{i\overline{j}}}\partial_{{i\overline{j}}}(A-{\mathcal{A}})=$
$\displaystyle
L^{{i\overline{j}}}\Phi_{\alpha{\overline{\tau}}i}\overline{\Phi_{\beta
j}}+L^{{i\overline{j}}}\Phi_{\alpha{\overline{\tau}}{\overline{j}}}{\Phi_{{\overline{\beta}}i}}+L^{{i\overline{j}}}\Phi_{\tau{\overline{\beta}}i}\Phi_{\alpha{\overline{j}}}+L^{{i\overline{j}}}\Phi_{\tau{\overline{\beta}}{\overline{j}}}\Phi_{\alpha
i}$ (3.93) $\displaystyle\ \ \ \ \ \ \ \
+L^{{i\overline{j}}}\Phi_{{\tau\overline{\tau}}}\Phi_{\alpha
i}\overline{\Phi_{\beta
j}}+L^{{i\overline{j}}}\Phi_{{\tau\overline{\tau}}}\Phi_{\alpha{\overline{j}}}\Phi_{{\overline{\beta}}i}.$
(3.94)

Terms in (3.94) can be controlled by ${\varepsilon}C_{\delta}$ since second
derivatives of $\Phi$ are all bounded by $C_{\delta}$ and
$\Phi_{{\tau\overline{\tau}}}$ is comparable to ${\varepsilon}$. For the terms
in the right-hand side of (3.93), we only estimate the first one; we have

$\displaystyle
L^{{i\overline{j}}}\Phi_{\alpha{\overline{\tau}}i}\overline{\Phi_{\beta
j}}=({\varepsilon}\Phi^{i{\overline{q}}}\Phi_{i\alpha{\overline{\tau}}})\mathcal{G}_{p{\overline{q}}}({\varepsilon}\Phi^{p{\overline{j}}}\
\overline{\Phi_{\beta j}}).$ (3.95)

In above, we notice that

$\displaystyle|\Phi^{i{\overline{q}}}\Phi_{\alpha{\overline{\tau}}i}|\leq
C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}},$ (3.96)

which follows from (3.18), and

$\displaystyle|{\varepsilon}\Phi^{p{\overline{j}}}\overline{\Phi_{\beta
j}}|\leq C_{\delta},$ (3.97)

which follows from (3.17). So we have that $|(\ref{611394})|$ is smaller than
${\varepsilon}C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}$. Estimates for
$L^{{i\overline{j}}}\Phi_{\alpha{\overline{\tau}}{\overline{j}}}{\Phi_{{\overline{\beta}}i}}$
$L^{{i\overline{j}}}\Phi_{\tau{\overline{\beta}}i}\Phi_{\alpha{\overline{j}}}$
and
$L^{{i\overline{j}}}\Phi_{\tau{\overline{\beta}}{\overline{j}}}\Phi_{\alpha
i}$ are similar; thus, we proved (3.82).

For the proof of (3.83), we have

$\displaystyle
L^{p{\overline{q}}}\partial_{p{\overline{q}}}\left(B_{\alpha\beta}-{\mathcal{B}}_{\alpha\beta}\right)$
(3.98) $\displaystyle=$ $\displaystyle L^{p{\overline{q}}}\Phi_{\alpha\tau
p}\Phi_{\beta{\overline{q}}}+L^{p{\overline{q}}}\Phi_{\alpha\tau{\overline{q}}}\Phi_{\beta
p}-L^{p{\overline{q}}}\Phi_{\alpha\tau}\Phi_{\beta{\overline{q}}}\Phi_{\tau
p}-L^{p{\overline{q}}}\Phi_{\alpha\tau}\Phi_{\beta
p}\Phi_{\tau{\overline{q}}}$ (3.99)
$\displaystyle+L^{p{\overline{q}}}\Phi_{\beta\tau
p}\Phi_{\alpha{\overline{q}}}+L^{p{\overline{q}}}\Phi_{\beta\tau{\overline{q}}}\Phi_{\alpha
p}-L^{p{\overline{q}}}\Phi_{\beta\tau}\Phi_{\alpha{\overline{q}}}\Phi_{\tau
p}-L^{p{\overline{q}}}\Phi_{\beta\tau}\Phi_{\alpha
p}\Phi_{\tau{\overline{q}}}$ (3.100)
$\displaystyle-L^{p{\overline{q}}}\Phi_{\tau\tau}\Phi_{\alpha
p}\Phi_{\beta{\overline{q}}}-L^{p{\overline{q}}}\Phi_{\tau\tau}\Phi_{\beta
p}\Phi_{\alpha{\overline{q}}}.$ (3.101)

We notice that $(\ref{611A})_{1}$, $(\ref{611A})_{3}$, $(\ref{611A})_{4}$,
$(\ref{611B})_{1}$, $(\ref{611B})_{3}$, $(\ref{611B})_{4}$ and terms of
(3.101) all contain $L^{p{\overline{q}}}\Phi_{k{\overline{q}}}$; using

$\displaystyle|L^{p{\overline{q}}}\Phi_{k{\overline{q}}}|=|({\varepsilon}\Phi^{p{\overline{v}}}\Phi_{k{\overline{q}}})(\mathcal{G}_{u{\overline{v}}}\Phi^{u{\overline{q}}}{\varepsilon})|\leq{\varepsilon},$
(3.102)

and (3.19) we know they can all be controlled by
${\varepsilon}C_{\delta}(C_{\delta}({\text{I}_{\Phi}}+1)+{\text{II}_{\Phi}})^{\frac{1}{2}}$.
For the estimates of $(\ref{611A})_{2}$, we have

$\displaystyle|L^{p{\overline{q}}}\Phi_{\alpha\tau{\overline{q}}}\Phi_{\beta
p}|=|({\varepsilon}\Phi^{k{\overline{q}}}\Phi_{\alpha\tau{\overline{q}}})\mathcal{G}_{k{\overline{s}}}({\varepsilon}\Phi^{p{\overline{s}}}\Phi_{\beta
p})|\leq{\varepsilon}C_{\delta}{\text{I}_{\Phi}}^{\frac{1}{2}}.$ (3.103)

The estimate for $(\ref{611B})_{2}$ follows by switching $\alpha$ with
$\beta$. Therefore, we proved (3.83).

The proof of (3.84) is simple: we only need to notice that ${\mathcal{A}}-A$
is a second order small quantity around ${\bf p}$. This is because
$\Phi_{\alpha{\overline{\tau}}}=0$ and $\Phi_{\beta}=0$ at ${\bf p}$.

For (3.85), we have

$\displaystyle{\varepsilon}\Phi^{p{\overline{q}}}\partial_{\overline{q}}(B_{\alpha\beta}-{\mathcal{B}}_{\alpha\beta})={\varepsilon}\Phi^{p{\overline{q}}}(\Phi_{\alpha\tau}\Phi_{\beta{\overline{q}}}-\Phi_{\beta\tau}\Phi_{\alpha{\overline{q}}}).$
(3.104)

When $q=n$, $\Phi_{\beta{\overline{q}}}=\Phi_{\alpha{\overline{q}}}=0$; when
$q\neq n$, $\Phi^{p{\overline{q}}}$ is bounded. So (3.104) can be controlled
by ${\varepsilon}C_{\delta}$.

For (3.86), we only need to notice that ${\varepsilon}\Phi^{{i\overline{j}}}$
is bound and use (3.18).

With all the estimates above, we know (3.60) and (3.61) are valid; with them,
we will prove (3.62) in the following. Before the proof, we do some
preparations.

We note that

$\displaystyle{\mathcal{K}}^{\dagger}=({\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}{\mathcal{A}}^{-1})^{\dagger}={\mathcal{A}}^{-1}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}={\mathcal{A}}^{-1}{\mathcal{K}}{\mathcal{A}}$ (3.105)

and that ${\mathcal{K}}^{\dagger}={\mathcal{K}}$ at ${\bf p}$ since
${\mathcal{A}}=I$ at ${\bf p}$.

To simplify the computation, we let

$\displaystyle{\mathfrak{B}}_{k}={\mathcal{B}}_{k}-{\mathcal{A}}_{k}{\mathcal{A}}^{-1}{\mathcal{B}}-{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{\mathcal{A}}_{k};$ (3.106)

we note that ${\mathfrak{B}}_{k}$ is symmetric. Then

$\displaystyle{\mathcal{A}}^{-1}{\mathfrak{B}}_{k}\overline{{\mathcal{A}}^{-1}}=\partial_{k}({\mathcal{A}}^{-1}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}).$
(3.107)

We need to compute
$L^{p{\overline{q}}}\partial_{\overline{q}}{{\mathfrak{B}}_{p}}$: directly
applying $L^{p{\overline{q}}}\partial_{\overline{q}}$ to (3.106) we have

$\displaystyle L^{p{\overline{q}}}\partial_{\overline{q}}{{\mathfrak{B}}_{p}}$
$\displaystyle=L^{p{\overline{q}}}\partial_{p{\overline{q}}}{\mathcal{B}}-L^{p{\overline{q}}}{\mathcal{A}}_{p}{\mathcal{A}}^{-1}{\mathcal{B}}_{{\overline{q}}}-L^{p{\overline{q}}}{\mathcal{B}}_{{\overline{q}}}\overline{{\mathcal{A}}^{-1}}\
\overline{\mathcal{A}}_{p}$ (3.108) $\displaystyle-$ $\displaystyle
L^{p{\overline{q}}}{\mathcal{A}}_{p{\overline{q}}}{\mathcal{A}}^{-1}{\mathcal{B}}-L^{p{\overline{q}}}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{A}}}_{p{\overline{q}}}+L^{p{\overline{q}}}{\mathcal{A}}_{p}{\mathcal{A}}^{-1}{\mathcal{A}}_{{\overline{q}}}{\mathcal{A}}^{-1}{\mathcal{B}}+L^{p{\overline{q}}}{\mathcal{B}}\
\overline{{\mathcal{A}}^{-1}}\ \overline{{\mathcal{A}}_{q}}\
\overline{{\mathcal{A}}^{-1}}\ \overline{{\mathcal{A}}}_{p}.$ (3.109)

Using (3.60) and (3.61), we get

$\displaystyle|L^{p{\overline{q}}}\partial_{\overline{q}}{{\mathfrak{B}}_{p}}-L^{p{\overline{q}}}{\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}_{p}{\mathcal{A}}^{-1}{\mathcal{B}}-L^{p{\overline{q}}}{\mathcal{B}}\
\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}_{p}{{\mathcal{A}}^{-1}}\
{{\mathcal{B}}}_{\overline{q}}|\leq
C_{\delta}(C_{\delta}({\text{I}_{\Phi}}+1)+{\text{II}_{\Phi}}).$ (3.110)

For ${\mathscr{A}}$ and ${\mathscr{B}}$ satisfying (3.56), we have

$\displaystyle
L^{p{\overline{q}}}\partial_{p}\left({\mathscr{A}}^{-1}{\mathscr{B}}_{{\overline{q}}}\overline{{\mathscr{A}}^{-1}}\right)=0;$
(3.111)

similarly, for ${\mathcal{A}},{\mathcal{B}}$ satisfying (3.61), we have

$\displaystyle|L^{p{\overline{q}}}\partial_{p}\left({\mathcal{A}}^{-1}{\mathcal{B}}_{{\overline{q}}}\overline{{\mathcal{A}}^{-1}}\right)|\leq
C_{\delta}(C_{\delta}({\text{I}_{\Phi}}+1)+{\text{II}_{\Phi}}).$ (3.112)

When applying $\partial_{\overline{q}}$ and $\partial_{p}$ to ${\mathcal{K}}$,
we get

$\displaystyle{\mathcal{K}}_{\overline{q}}={\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}}\
{\mathcal{B}}{\mathcal{A}}^{-1}+{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathfrak{B}}_{q}}{\mathcal{A}}^{-1};$ (3.113)

its complex conjugation is

$\displaystyle({\mathcal{K}}^{\dagger})_{p}={\mathcal{A}}^{-1}{\mathfrak{B}}_{p}\overline{{\mathcal{A}}^{-1}}\
\overline{\mathcal{B}}+{\mathcal{A}}^{-1}{\mathcal{B}}\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathcal{B}}}_{p}.$ (3.114)

With the preparation above, we can compute
$L^{p{\overline{q}}}\partial_{p{\overline{q}}}\left[\text{tr}(I-{\mathcal{K}})^{-1}\right]$.
Using (3.113), we have

$\displaystyle
L^{p{\overline{q}}}\partial_{p}\partial_{{\overline{q}}}\left[\text{tr}(I-{\mathcal{K}})^{-1}\right]$
(3.115) $\displaystyle=$ $\displaystyle
L^{p{\overline{q}}}\partial_{p}\left[\text{tr}{(I-{\mathcal{K}})}^{-1}\left({\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}}\
\overline{\mathcal{B}}{\mathcal{A}}^{-1}+{\mathcal{B}}\
\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathfrak{B}}_{q}}{\mathcal{A}}^{-1}\right){(I-{\mathcal{K}})}^{-1}\right].$
(3.116)

We insert ${\mathcal{A}}$ and ${\mathcal{A}}^{-1}$ into the equation above to
make ${\mathcal{K}}$ into ${\mathcal{K}}^{\dagger}$ so that the computation is
simpler when we apply $\partial_{p}$; we have

$\displaystyle\text{tr}{(I-{\mathcal{K}})}^{-1}\left({\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}}\
\overline{\mathcal{B}}{\mathcal{A}}^{-1}+{\mathcal{B}}\
\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathfrak{B}}_{q}}{\mathcal{A}}^{-1}\right){(I-{\mathcal{K}})}^{-1}$
(3.117) $\displaystyle=$
$\displaystyle\text{tr}\left({\mathcal{A}}^{-1}{(I-{\mathcal{K}})}^{-1}{\mathcal{A}}\right)\left({\mathcal{A}}^{-1}{\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}}\
\overline{\mathcal{B}}+{\mathcal{A}}^{-1}{\mathcal{B}}\
\overline{{\mathcal{A}}^{-1}}\
\overline{{\mathfrak{B}}_{q}}\right)\left({\mathcal{A}}^{-1}{(I-{\mathcal{K}})}^{-1}{\mathcal{A}}\right)$
(3.118) $\displaystyle=$
$\displaystyle\text{tr}{(I-{\mathcal{K}}^{\dagger})}^{-1}\left(({\mathcal{A}}^{-1}{\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}})\
\overline{\mathcal{B}}+({\mathcal{A}}^{-1}{\mathcal{B}}\
\overline{{\mathcal{A}}^{-1}})\
\overline{{\mathfrak{B}}_{q}}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}.$
(3.119)

Then we apply $L^{p{\overline{q}}}\partial_{p}$ to (3.119) and get that
$(\ref{6113115})$ equals to

$\displaystyle
L^{p{\overline{q}}}\left[\text{tr}{(I-{\mathcal{K}}^{\dagger})}^{-1}\left({{\mathfrak{B}}_{p}}\overline{\mathcal{B}}+{\mathcal{B}}\overline{{\mathcal{B}}}_{p}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}\left({\mathcal{B}}_{\overline{q}}\overline{\mathcal{B}}+{\mathcal{B}}\overline{{\mathfrak{B}}_{q}}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}\right]$
(3.120) $\displaystyle+$ $\displaystyle
L^{p{\overline{q}}}\left[\text{tr}{(I-{\mathcal{K}}^{\dagger})}^{-1}\left({\mathcal{B}}_{\overline{q}}\overline{\mathcal{B}}+{\mathcal{B}}\overline{{\mathfrak{B}}_{q}}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}\left({\mathfrak{B}}_{p}\overline{\mathcal{B}}+{\mathcal{B}}\overline{{\mathcal{B}}}_{p}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}\right]$
(3.121) $\displaystyle+$
$\displaystyle\text{tr}\left(L^{p{\overline{q}}}\partial_{p}({\mathcal{A}}^{-1}{\mathcal{B}}_{\overline{q}}\
\overline{{\mathcal{A}}^{-1}})\overline{\mathcal{B}}(I-{\mathcal{K}})^{-2}\right)+\text{tr}\left({\mathcal{B}}(L^{p{\overline{q}}}\partial_{p}\overline{{\mathfrak{B}}_{q}})(I-{\mathcal{K}})^{-2}\right)$
(3.122) $\displaystyle+$ $\displaystyle
L^{p{\overline{q}}}\text{tr}\left({\mathcal{B}}_{\overline{q}}\overline{{\mathcal{B}}}_{p}(I-{\mathcal{K}})^{-2}\right)+L^{p{\overline{q}}}\text{tr}\left({\mathfrak{B}}_{p}\overline{{\mathfrak{B}}_{q}}(I-{\mathcal{K}})^{-2}\right).$
(3.123)

To get (3.120) and (3.121), we apply $L^{p{\overline{q}}}\partial_{p}$ to the
$(I-{\mathcal{K}})^{-1}$’s of (3.119) and use equation (3.114); to get
(3.122), we apply $L^{p{\overline{q}}}\partial_{p}$ to
${\mathcal{A}}^{-1}{\mathcal{B}}_{\overline{q}}\overline{{\mathcal{A}}^{-1}}$
and $\overline{{\mathfrak{B}}_{q}}$ of (3.119); to get (3.123), we apply
$L^{p{\overline{q}}}\partial_{p}$ to $\overline{\mathcal{B}}$ and
${\mathcal{A}}^{-1}{\mathcal{B}}\ \overline{{\mathcal{A}}^{-1}}$. In above, we
also used ${\mathcal{A}}=I$ and ${\mathcal{K}}={\mathcal{K}}^{\dagger}$ at
${\bf p}$. Because of the apriori assumption (3.1), we have
$(I-{\mathcal{K}})^{-1}\leq C_{\delta}$. So we can use (3.110) and (3.112) to
simplify (3.122) and get

$\displaystyle|(\ref{draft138})-L^{p{\overline{q}}}\text{tr}\left({\mathcal{B}}(\overline{{\mathcal{B}}}_{p}{\mathcal{B}}_{\overline{q}}\overline{{\mathcal{B}}}+\overline{\mathcal{B}}{\mathcal{B}}_{\overline{q}}\overline{\mathcal{B}}_{p})(I-{\mathcal{K}})^{-2}\right)|\leq
C_{\delta}(C_{\delta}({\text{I}_{\Phi}}+1)+{\text{II}_{\Phi}}).$ (3.124)

In the following, we use index summation to simplify (3.120)-(3.123). First,
with a unitary change of coordinate in the direction of $<\partial_{z^{1}},\
...\ ,\partial_{z^{n-1}}>_{\mathbb{C}}$, we make

$\displaystyle{\mathcal{B}}=\text{diag}(\lambda_{1},\ ...\ ,\lambda_{n-1});$
(3.125)

this can be done with Autonne-Takagi factorization (Corollary 4.4.4(c) of
[HJ13]). Then at ${\bf p}$

$\displaystyle
I-{\mathcal{K}}=I-{\mathcal{B}}\overline{{\mathcal{B}}}=\text{diag}(1-\lambda_{1}^{2},\
...\ ,1-\lambda_{n-1}^{2}).$ (3.126)

We denote

$\displaystyle\overline{\mathcal{B}}_{p}=(\overline{\mathcal{B}}_{p;\alpha\beta}),\
\ \ \ \ \ \ \ {\mathfrak{B}}_{p}=({\mathfrak{B}}_{p;\alpha\beta});$ (3.127)

since ${\mathcal{B}}_{\overline{q}}$ and ${\mathfrak{B}}_{p}$ are both
symmetric, we have
$\overline{\mathcal{B}}_{p;\alpha\beta}=\overline{\mathcal{B}}_{p;\beta\alpha}$
and ${\mathfrak{B}}_{p;\alpha\beta}={\mathfrak{B}}_{p;\beta\alpha}$. Then

$\displaystyle(\ref{draft136})=\sum_{p,q,\alpha,\beta}L^{p{\overline{q}}}\left({\mathfrak{B}}_{p;\alpha\beta},\overline{{\mathcal{B}}}_{p;\alpha\beta}\right)M^{1}_{\alpha\beta}\left({\mathfrak{B}}_{q;\alpha\beta},\overline{{\mathcal{B}}}_{q;\alpha\beta}\right)^{\dagger},$
(3.128)

where

$\displaystyle
M^{1}_{\alpha\beta}=\frac{1}{(1-\lambda_{\alpha}^{2})^{2}(1-\lambda_{\beta}^{2})}\left(\begin{array}[]{cc}\lambda_{\beta}^{2}&\lambda_{\alpha}\lambda_{\beta}\\\
\lambda_{\alpha}\lambda_{\beta}&\lambda_{\alpha}^{2}\end{array}\right).$
(3.131)

In (3.121), we transpose
${(I-{\mathcal{K}}^{\dagger})}^{-1}\left({\mathcal{B}}_{\overline{q}}\overline{\mathcal{B}}+{\mathcal{B}}\overline{{\mathfrak{B}}_{q}}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}\left({\mathfrak{B}}_{p}\overline{\mathcal{B}}+{\mathcal{B}}\overline{{\mathcal{B}}}_{p}\right){(I-{\mathcal{K}}^{\dagger})}^{-1}$
and get

$\displaystyle(\ref{draft137})=\sum_{p,q,\alpha,\beta}L^{p{\overline{q}}}\left({\mathfrak{B}}_{p;\alpha\beta},\overline{{\mathcal{B}}}_{p;\alpha\beta}\right)M^{2}_{\alpha\beta}\left({\mathfrak{B}}_{q;\alpha\beta},\overline{{\mathcal{B}}}_{q;\alpha\beta}\right)^{\dagger},$
(3.132)

where

$\displaystyle
M^{2}_{\alpha\beta}=\frac{1}{(1-\lambda_{\alpha}^{2})^{2}(1-\lambda_{\beta}^{2})}\left(\begin{array}[]{cc}\lambda_{\alpha}^{2}&\lambda_{\alpha}\lambda_{\beta}\\\
\lambda_{\alpha}\lambda_{\beta}&\lambda_{\beta}^{2}\end{array}\right).$
(3.135)

Finally, (3.124) gives us

$\displaystyle(\ref{draft138})+(\ref{draft139})=\sum_{p,q,\alpha,\beta}L^{p{\overline{q}}}\left({\mathfrak{B}}_{p;\alpha\beta},\overline{{\mathcal{B}}}_{p;\alpha\beta}\right)M^{3}_{\alpha\beta}\left({\mathfrak{B}}_{q;\alpha\beta},\overline{{\mathcal{B}}}_{q;\alpha\beta}\right)^{\dagger}+F,$
(3.136)

where

$\displaystyle
M^{3}_{\alpha\beta}=\frac{1}{(1-\lambda_{\alpha}^{2})^{2}}\left(\begin{array}[]{cc}1&\\\
&1-2\lambda_{\alpha}^{2}\end{array}\right)$ (3.139)

and $F$ is a term which can be controlled by
$C_{\delta}(C_{\delta}({\text{I}_{\Phi}}+1)+{\text{II}_{\Phi}})$.

Summing up, we have

$\displaystyle
L^{p{\overline{q}}}\partial_{p{\overline{q}}}(\text{tr}(I-{\mathcal{K}})^{-1})=\sum_{p,q,\alpha,\beta}L^{p{\overline{q}}}\left({\mathfrak{B}}_{p;\alpha\beta},\overline{{\mathcal{B}}}_{p;\alpha\beta}\right)M_{\alpha\beta}\left({\mathfrak{B}}_{q;\alpha\beta},\overline{{\mathcal{B}}}_{q;\alpha\beta}\right)^{\dagger}+F,$
(3.140)

where

$\displaystyle
M_{\alpha\beta}=M_{\alpha\beta}^{1}+M_{\alpha\beta}^{2}+M_{\alpha\beta}^{3}=\frac{1}{(1-\lambda_{\alpha}^{2})^{2}(1-\lambda_{\beta}^{2})}\left(\begin{array}[]{cc}1+\lambda_{\alpha}^{2}&2\lambda_{\alpha}\lambda_{\beta}\\\
2\lambda_{\alpha}\lambda_{\beta}&1-\lambda_{\alpha}^{2}+2\lambda_{\alpha}^{2}\lambda_{\beta}^{2}\end{array}\right).$
(3.143)

We can easily check that when $\lambda_{\alpha}$ and $\lambda_{\beta}$ are
both positive and smaller than $1$, $M_{\alpha\beta}$ is positive definite. So
$(\ref{7143114})_{2}$ is positive, and (3.62) follows. With the estimates for
$W$ and $V$ we know (3.3) follows from (3.62).

##  4 Lower Bound Estimates for Norms of Gradients

In this section, using inequality (3.2) of Proposition 3.1. We derive an
apriori estiamte for the lower bound of the norm of the gradient of $\Phi$.
More precisely, we have the following apriori estimate

######  Proposition 4.1.

[Apriori Lower Bound Estimate for the Norm of the Gradient] Suppose $\Phi$ is
a solution to Problem 1.12 with the robustness of
$\mathfrak{q}\mathbb{C}$-convexity greater than $\delta$, and suppose that
${\varepsilon}$ is small enough. Then, there is a positive constant
$C_{\mathcal{R}}^{g}$ dependent of the geometry of $\mathcal{R}$ and
independent of ${\varepsilon}$ and $\delta$ so that

$\displaystyle|\nabla\Phi|_{\mathcal{G}}\geq{C_{\mathcal{R}}^{g}}.$ (4.1)

There are three steps in the proof. First, we show

$\displaystyle|\nabla\Phi|_{\mathcal{G}}^{2}\geq S\ \ \ \ \ \ \ \ \text{ in
}\mathcal{R};$ (4.2)

then, using (3.2), we show

$\displaystyle S\geq\frac{1}{C}\min_{\partial\mathcal{R}}S\ \ \ \ \ \ \ \
\text{ in }\mathcal{R}$ (4.3)

for a positive constant $C$; finally, using the estimate for
$|\nabla\Phi|_{\mathcal{G}}$ on $\partial\mathcal{R}$, which we derive in
Section B, we estimate the lower bound for $S$ on $\partial\mathcal{R}$.

###### Proof of Proposition 4.1.

Step 1. We choose a coordinate $(z^{\alpha},\tau)$, $\alpha=1,\ ...\ ,n-1$, so
that $\mathcal{G}_{{i\overline{j}}}=\delta_{ij}$ and
$\Phi_{{i\overline{j}}}=\delta_{ij}\lambda_{i}$ for a point ${\bf
p}\in\mathcal{R}$. Then

$\displaystyle|\nabla\Phi|_{\mathcal{G}}^{2}=\sum_{i}|\Phi_{i}|^{2},$ (4.4)
$\displaystyle S={\varepsilon}\sum_{i}\frac{|\Phi_{i}|^{2}}{\lambda_{i}}.$
(4.5)

At the point ${\bf p}$, equation (2.15) is

$\displaystyle\sum_{i}\frac{{\varepsilon}}{\lambda_{i}}=1,$ (4.6)

so $\frac{{\varepsilon}}{\lambda_{i}}<1$.Therefore,
$|\nabla\Phi|_{\mathcal{G}}^{2}\geq S$ at ${\bf p}$. Since ${\bf p}$ can be
any point in $\mathcal{R}$, and $|\nabla\Phi|_{\mathcal{G}},\ S$ are both
independent of coordinate, we know (4.2) is valid.

Step 2. When ${\varepsilon}$ is small enough, we can use (3.2) and the maximum
principle and get

$\displaystyle\min_{\partial\mathcal{R}}\log(S)-\max_{\partial\mathcal{R}}\sqrt{{\varepsilon}}\left[W+\mathcal{G}_{{i\overline{j}}}z^{i}\overline{z^{j}}\right]\leq\log(S)\
\ \ \ \ \ \ \ \text{ in }\mathcal{R}.$ (4.7)

The exponential of the inequality above is

$\displaystyle S\geq
e^{-\sqrt{{\varepsilon}}\max_{\partial\mathcal{R}}\left(W+\mathcal{G}_{{i\overline{j}}}z^{i}z^{\overline{j}}\right)}\min_{\partial\mathcal{R}}S,\
\ \ \ \ \ \ \ \text{ in }\mathcal{R}.$ (4.8)

Step 3. To estimate $\min_{\partial\mathcal{R}}S$, we choose a coordinate so
that $\mathcal{G}_{i\overline{j}}=\delta_{ij}$ and $\Phi_{\alpha}=0$ at a
boundary point ${\bf p}$. Then

$\displaystyle
S={\varepsilon}\Phi_{\tau}\Phi^{{\tau\overline{\tau}}}\Phi_{{\overline{\tau}}},\
\ \ |\nabla\Phi|_{\mathcal{G}}^{2}=|\Phi_{\tau}|^{2}\ \ \ \ \ \ \ \ \text{ at
}{\bf p}.$ (4.9)

To find the relation between $|\nabla\Phi|_{\mathcal{G}}^{2}$ and $S$, we need
to estimate ${\varepsilon}\Phi^{{\tau\overline{\tau}}}$. We further require
that $\Phi_{\alpha{\overline{\beta}}}=\delta_{\alpha\beta}\lambda_{\alpha}$ at
${\bf p}$. We have an estimate on the lower bound of $\lambda_{\alpha}$,
depending on the the $\mathbb{C}$-convexity of the boundary and the lower
bound estimate of $|\nabla\Phi|_{\mathcal{G}}$ at ${\bf p}$. A basic linear
algebra computation gives

$\displaystyle\Phi^{{\tau\overline{\tau}}}=(\mathcal{H}^{-1})_{{\tau\overline{\tau}}}=\frac{1}{\Phi_{{\tau\overline{\tau}}}-\sum_{\mu}\frac{|\Phi_{\tau{\overline{\mu}}}|^{2}}{\lambda_{\mu}}}$
(4.10)

and

$\displaystyle\Phi^{\alpha{\overline{\alpha}}}=(\mathcal{H}^{-1})_{\alpha{\overline{\alpha}}}=\frac{1}{\lambda_{\alpha}}+\frac{\frac{|\Phi_{\tau{\overline{\alpha}}}|^{2}}{\lambda_{\alpha}^{2}}}{\Phi_{{\tau\overline{\tau}}}-\sum_{\mu}\frac{|\Phi_{\tau{\overline{\mu}}}|^{2}}{\lambda_{\mu}}}.$
(4.11)

Therefore, at ${\bf p}$ the equation (2.15) is

$\displaystyle\left(1+\sum_{\alpha}\frac{|\Phi_{\tau{\overline{\alpha}}}|^{2}}{\lambda_{\alpha}^{2}}\right)\Phi^{{\tau\overline{\tau}}}+\sum_{\mu}\frac{1}{\lambda_{\mu}}=\frac{1}{{\varepsilon}}.$
(4.12)

From this, we get

$\displaystyle{\varepsilon}\Phi^{{\tau\overline{\tau}}}=\frac{1-{\varepsilon}\sum_{\mu}\frac{1}{\lambda_{\mu}}}{1+\sum_{\alpha}\frac{|\Phi_{\tau{\overline{\alpha}}}|^{2}}{\lambda_{\alpha}^{2}}}.$
(4.13)

When ${\varepsilon}$ is small enough, it’s greater than a positive constant.
Then using (4.9), we know $S$ has a positive lower bound on
$\partial\mathcal{R}$. ∎

##  5 Convexity Estimate

The main goal of this section is to derive the estimate for the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of the solution ${\Phi^{\varepsilon}}$ to
Problem 1.12. In section 5.1, by deforming $\Omega_{0}$ and $\Omega_{1}$ to
concentric balls, we derive an estimate which depends on this deformation
procedure; in section 5.2, we improve the estimate so that it only depends on
the geometry of $\mathcal{R}$; in section 5.3, we estimate
$(\sqrt{-1}\partial\overline{\partial}{\Phi^{\varepsilon}})^{n-1}\wedge{\omega_{\mathcal{G}}}$
and $\sqrt{-1}\partial\overline{\partial}\left(e^{\Phi^{\varepsilon}}\right)$.

###  5.1 Convexity Estimate Using Deformation

Using Lemma D.1, we can construct two families of domains
$\left\\{\Omega_{0}^{\lambda}\right\\}_{\lambda\in[0,1]}$ and
$\left\\{\Omega_{1}^{\lambda}\right\\}_{\lambda\in[0,1]}$ connecting
$\Omega_{0}$ and $\Omega_{1}$ with two concentric balls $B_{r}$ and $B_{R}$
both contained in $\Omega_{0}$. In addition, by properly choosing the
parameter $\lambda$,

$\displaystyle\overline{\Omega^{\lambda}_{0}}\subset\Omega_{1}^{\lambda}\ \ \
\ \ \ \ \ \text{ for any }\lambda\in[0,1],$ (5.1)

can be satisfied. Then we let
$\mathcal{R}^{\lambda}=\Omega_{1}^{\lambda}\backslash\overline{\Omega_{0}^{\lambda}}$
and let $\Phi^{\lambda}_{\varepsilon}$ be the solution to the following
problem:

######  Problem 5.1.

Suppose that
${\omega_{\mathcal{G}}}=\sqrt{-1}\mathcal{G}_{ij}dz^{i}\wedge\overline{dz^{j}}$
be a constant coefficient non-degenerate Kähler form on $\mathbb{C}^{n}$ and
that ${\varepsilon}$ is a positive constant. Find
${\Phi_{\varepsilon}^{\lambda}}$ satisfying

$\displaystyle\left(\sqrt{-1}\partial\overline{\partial}{\Phi_{\varepsilon}^{\lambda}}\right)^{n}$
$\displaystyle={\varepsilon}\left(\sqrt{-1}\partial\overline{\partial}{\Phi_{\varepsilon}^{\lambda}}\right)^{n-1}\wedge{\omega_{\mathcal{G}}}\
\ \ \ \ \ $ $\displaystyle\text{in }\mathcal{R}^{\lambda},\ \ $ (5.2)
$\displaystyle\sqrt{-1}\partial\overline{\partial}{\Phi_{\varepsilon}^{\lambda}}$
$\displaystyle>0\ \ \ \ \ \ $ $\displaystyle\text{in }\mathcal{R}^{\lambda},\
\ $ (5.3) $\displaystyle{\Phi_{\varepsilon}^{\lambda}}$ $\displaystyle=0\ \ \
\ \ \ $ $\displaystyle\text{\ \ on }\partial\Omega_{0}^{\lambda},$ (5.4)
$\displaystyle{\Phi_{\varepsilon}^{\lambda}}$ $\displaystyle=1\ \ \ \ \ \ $
$\displaystyle\text{\ \ on }\partial\Omega_{1}^{\lambda}.$ (5.5)

According to Lemma B.4, there is an ${\epsilon}_{0}$ and
${\underline{\sigma}}$ so that for any ${\varepsilon}\in(0,{\epsilon}_{0}]$
and any $\lambda\in[0,1]$
$|\nabla{\Phi_{\varepsilon}^{\lambda}}|_{\mathcal{G}}$ is greater than
${\underline{\sigma}}$ on $\partial\mathcal{R}^{\lambda}$. Since
$\Omega_{0}^{\lambda}$ and $\Omega_{1}^{\lambda}$ are all strongly
$\mathbb{C}$-convex, we can apply Lemma A.11 and get the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{\lambda}}$ is
greater than ${C}{{\underline{\sigma}}}$ on $\partial\mathcal{R}^{\lambda}$
for a small constant $C$. We then apply Lemma A.3 and get that the robustness
of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi^{\lambda}_{\varepsilon}$ is
greater than $C^{2}{\underline{\sigma}}^{2}$. In above, $C$ depends on the
geometry of $\Omega_{1}^{\lambda}\backslash\overline{\Omega_{0}^{\lambda}}$
for all $\lambda\in[0,1]$.

When $\lambda=0$ and ${\varepsilon}=0$, the solution can be written down
explicitly:

$\displaystyle\Phi^{0}_{0}=\frac{\log|z|-\log r}{\log R-\log r}.$ (5.6)

It’s easy to see that $\Phi^{0}_{0}$ is strongly
$\mathfrak{q}\mathbb{C}$-convex since the level sets of $\Phi^{0}_{0}$ are all
Euclidean balls, and its gradient does not vanish anywhere. This two
properties hold for $\Phi^{0}_{\varepsilon}$ with ${\varepsilon}$ small
enough. Therefore, we can find small constants ${\epsilon}_{1}$ and
$\delta_{0}$ so that the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
$\Phi^{0}_{\varepsilon}$ is greater than $\delta_{0}$, providing
${\varepsilon}\in(0,{\epsilon}_{1}]$.

Let

$\displaystyle\sigma=\frac{1}{2}\min\left\\{C^{2}{\underline{\sigma}}^{2},\
\delta_{0},\ C_{\mathcal{R}}^{g}\right\\}.$ (5.7)

Here $C_{\mathcal{R}}^{g}$ is from Proposition 4.1; we can choose it small
enough, so that it does not depend on $\lambda$. Then for any
${\varepsilon}\in(0,\frac{\min\\{{\epsilon}_{0},{\epsilon}_{1}\\}}{2})$ we let

$\displaystyle\begin{split}L^{\varepsilon}=&\left\\{\lambda\in[0,1]\ |\text{
the robustness of $\mathfrak{q}\mathbb{C}$-convexity\ of
}\Phi_{\varepsilon}^{\lambda}\text{ is greater than
$\sigma$}\right\\},\end{split}$ (5.8)

and we will show $L^{\varepsilon}=[0,1]$, providing ${\varepsilon}$ is small
enough. It’s easy to see that $0\in L^{\varepsilon}$ because
$\sigma<\delta_{0}$; it’s also easy to see that $L^{\varepsilon}$ is open
since $\Omega_{0}$ and $\Omega_{1}$ change smoothly as $\lambda$ changes. It’s
crucial to show that $L^{\varepsilon}$ is right-closed. Suppose that
$[0,l)\subset L^{\varepsilon}$. We will show $l\in L^{\varepsilon}$; we need
to show the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi_{\varepsilon}^{l}}$ is greater than $\sigma$. To do this, we choose
$\delta$ to be a very small constant, which is independent of ${\varepsilon}$,
and apply Proposition 3.1. As illustrated in Figure 1, there are several
steps:

Figure 1: Convexity Estimate via a Continuity Argument

In step 1, using Lemma A.10 we show the robustness
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}$ is greater
than $\sigma-\delta$; in step 2, we show for any
$q\in{\mathscr{Q}}_{\sigma-2\delta}$ the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}-q$ is greater
than $\delta$; in step 3, we show the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}-q$ is actually
larger; in step 4, we show the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}$ is greater
than $\sigma$.

Step 1. According to the assumption, for any $\lambda<l$, the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{\lambda}}$ is
greater than $\sigma$. So, using Lemma A.10, we know the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}$ is greater
than $\sigma-\delta$, by letting $\lambda\rightarrow l$. Here we choose
$0<\delta<\sigma$.

Step 2. Since the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi_{\varepsilon}^{l}}$ is greater than $\sigma-\delta$,
${\Phi_{\varepsilon}^{l}}-q-q_{1}$ is strongly
$\mathfrak{q}\mathbb{C}$-convex, for any $q\in{\mathscr{Q}}_{\sigma-2\delta}$
and $q_{1}\in{\mathscr{Q}}_{\delta}$. Therefore, the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}-q$ is greater
than $\delta$. This allows us to apply Proposition 3.1 to
${\Phi_{\varepsilon}^{l}}-q$ and its variants.

Step 3. In this step, we fix $q$ and denote ${\Phi_{\varepsilon}^{l}}-q$ by
$\Psi$. We will show the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
$\Psi$ is greater than $\frac{\sigma^{2}}{C}$, for a constant $C$, which is
independent of $\sigma,\ \delta$ and ${\varepsilon}$. This is the most crucial
step, and we split it into step 3.1 and step 3.2.

Step 3.1. In this step, we fix a point ${\bf p}\in\mathcal{R}$ and choose
coordinate so that $\Psi_{\alpha}({\bf p})=0$, for $\alpha\in\\{1,\ ...\
,n-1\\}$, and estimate $\Psi_{\alpha{\overline{\beta}}}({\bf p})$. Let
$h\in{\mathscr{Q}}_{\frac{\sigma}{2}}$. Here we require that $4\delta<\sigma$,
so

$\displaystyle\frac{\sigma}{2}<\sigma-2\delta.$ (5.9)

Therefore, $h\in{\mathscr{Q}}_{\sigma-2\delta}$. Let

$\displaystyle\Psi_{m}=\Psi+m(q-h)={\Phi_{\varepsilon}^{l}}-(1-m)q-mh,$ (5.10)
$\displaystyle\Psi^{m}=\Psi+m(q+h)={\Phi_{\varepsilon}^{l}}-(1-m)q+mh.$ (5.11)

Since $q$ and $h$ are both elements of ${\mathscr{Q}}_{\sigma-2\delta}$, the
robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Psi_{m}$ and $\Psi^{m}$
are both greater than $\delta$. Therefore, we can apply Proposition 3.1 and
Proposition 4.1 to $\Psi^{m}$ and $\Psi_{m}$.

We apply Proposition 4.1 to ${\Phi_{\varepsilon}^{l}}$ and get

$\displaystyle|\nabla{\Phi_{\varepsilon}^{l}}|\geq{C_{\mathcal{R}}^{g}}.$
(5.12)

Then $|\nabla\Psi|>\frac{{C_{\mathcal{R}}^{g}}}{2}$ since
$\sigma<\frac{{C_{\mathcal{R}}^{g}}}{2}$. Since $\Psi_{\alpha}({\bf p})=0$ for
$\alpha\in\\{1,\ ...\ ,n-1\\}$, we have $|\Psi_{\tau}({\bf
p})|>\frac{{C_{\mathcal{R}}^{g}}}{2}$. In the following, all the computation
for $\Psi^{m}$ and $\Psi_{m}$ will be at ${\bf p}$. For
$\partial_{\tau}\Psi_{m}$ we have

$\displaystyle|\partial_{\tau}\Psi_{m}|=|\partial_{\tau}(\Psi+m(q-h))|.$
(5.13)

We choose $h$, so that $\nabla h=0$ at ${\bf p}$. Then

$\displaystyle|\partial_{\tau}\Psi_{m}|\geq\frac{{C_{\mathcal{R}}^{g}}}{2}-m|\nabla
q({\bf p})|\geq\frac{{C_{\mathcal{R}}^{g}}}{2}-m\sigma.$ (5.14)

The above is positive providing $m<1$ since we already assumed that
$\sigma\leq\frac{{C_{\mathcal{R}}^{g}}}{2}$. For $\partial_{\tau}\Psi^{m}$, we
have the same result.

Let

$\displaystyle
v_{\alpha}^{m}=\frac{\partial_{\alpha}(\Psi_{m})}{\partial_{\tau}(\Psi_{m})}=\frac{\partial_{\alpha}(\Psi^{m})}{\partial_{\tau}(\Psi^{m})}=\frac{\Psi_{\alpha}+mq_{\alpha}}{\Psi_{\tau}+mq_{\tau}}.$
(5.15)

Since $\Phi_{\alpha}=0$, we have the estimate that

$\displaystyle|v_{\alpha}^{m}|\leq\frac{m\sigma}{\frac{{C_{\mathcal{R}}^{g}}}{2}-m\sigma}.$
(5.16)

Also, we know $\partial_{\alpha}-v_{\alpha}^{m}\partial_{\tau}$ is a
tangential direction for the level sets of $\Psi_{m}$ and $\Psi^{m}$ at ${\bf
p}$. Therefore, we have

$\displaystyle\frac{|[\Psi+m(q-h)]_{\alpha\alpha}-2v_{\alpha}^{m}[\Psi+m(q-h)]_{\alpha\tau}+(v_{\alpha}^{m})^{2}[\Psi+m(q-h)]_{\tau\tau}|}{\Psi_{\alpha{\overline{\alpha}}}-v_{\alpha}^{m}\Psi_{\tau{\overline{\alpha}}}-\overline{v_{\alpha}^{m}}\Psi_{\alpha{\overline{\tau}}}+|v_{\alpha}^{m}|^{2}\Psi_{\tau{\overline{\tau}}}}<1,$
(5.17)
$\displaystyle\frac{|[\Psi+m(q+h)]_{\alpha\alpha}-2v_{\alpha}^{m}[\Psi+m(q+h)]_{\alpha\tau}+(v_{\alpha}^{m})^{2}[\Psi+m(q+h)]_{\tau\tau}|}{\Psi_{\alpha{\overline{\alpha}}}-v_{\alpha}^{m}\Psi_{\tau{\overline{\alpha}}}-\overline{v_{\alpha}^{m}}\Psi_{\alpha{\overline{\tau}}}+|v_{\alpha}^{m}|^{2}\Psi_{\tau{\overline{\tau}}}}<1.$
(5.18)

We apply the triangle inequality to (5.17) and (5.18) and get

$\displaystyle\frac{m|h_{\alpha\alpha}-2v_{\alpha}^{m}h_{\alpha\tau}+(v_{\alpha}^{m})^{2}h_{\tau\tau}|}{\Psi_{\alpha{\overline{\alpha}}}-v_{\alpha}^{m}\Psi_{\tau{\overline{\alpha}}}-\overline{v_{\alpha}^{m}}\Psi_{\alpha{\overline{\tau}}}+|v_{\alpha}^{m}|^{2}\Psi_{\tau{\overline{\tau}}}}<1.$
(5.19)

Because of the assumption (5.3) and that $q$ is pluriharmonic, we have

$\displaystyle\left(\begin{array}[]{cc}\Psi_{\alpha{\overline{\alpha}}}&\Psi_{\alpha{\overline{\tau}}}\\\
\Psi_{\tau{\overline{\alpha}}}&\Psi_{{\tau\overline{\tau}}}\end{array}\right)\geq
0.$ (5.22)

Therefore,

$\displaystyle|\Psi_{\alpha{\overline{\tau}}}|\leq\left(\Psi_{\alpha{\overline{\alpha}}}\Psi_{{\tau\overline{\tau}}}\right)^{\frac{1}{2}}.$
(5.23)

Since $q$ is pluriharmonic,
$\Psi_{{\tau\overline{\tau}}}=\partial_{{\tau\overline{\tau}}}{\Phi_{\varepsilon}^{l}}$.
Then, using the $C^{2}$ estimate of Appendix C, we have

$\displaystyle 0\leq\Psi_{\tau\overline{\tau}}\leq C_{2},$ (5.24)

for a constant $C_{2}$. So we have

$\displaystyle|\Psi_{\alpha{\overline{\tau}}}|\leq
C_{2}^{\frac{1}{2}}(\Psi_{\alpha{\overline{\alpha}}})^{\frac{1}{2}}.$ (5.25)

We choose $h$, so that, at ${\bf p}$,

$\displaystyle h_{\alpha\alpha}=\frac{\sigma}{2\ \text{
Diameter}(\mathcal{R})}$ (5.26)

and

$\displaystyle h_{\alpha\tau}=0,\ \ h_{\tau\tau}=0.$ (5.27)

We remind that we have already required that $\nabla h({\bf p})=0$. These
conditions guarantee $|\nabla h|\leq\frac{\sigma}{2}$ and, therefore,
$h\in{\mathscr{Q}}_{\sigma-2\delta}$.

In the following, we denote $x=\Psi_{\alpha{\overline{\alpha}}}$ and derive a
positive lower bound estimate for $x$. From (5.19), we get

$\displaystyle\frac{m\sigma}{2\text{
Diameter}(\mathcal{R})}\leq\left(x^{\frac{1}{2}}+|v_{\alpha}^{m}|C_{2}^{\frac{1}{2}}\right)^{2}.$
(5.28)

Then a straightforward computation gives

$\displaystyle\sqrt{m}\left(\left(\frac{\sigma}{2\text{
Diameter}(\mathcal{R})}\right)^{\frac{1}{2}}-\frac{\sqrt{m}\sigma
C^{\frac{1}{2}}_{2}}{{C_{\mathcal{R}}^{g}}-m\sigma}\right)\leq
x^{\frac{1}{2}}.$ (5.29)

We choose

$\displaystyle
m=\frac{1}{2}\max\left\\{\frac{{C_{\mathcal{R}}^{g}}}{2\sigma},\frac{({C_{\mathcal{R}}^{g}})^{2}}{32\sigma
C_{2}\text{ Diameter}(\mathcal{R})}\right\\};$ (5.30)

then,

$\displaystyle x\geq\frac{m\sigma}{8\text{ Diameter}(\mathcal{R})}.$ (5.31)

Summing up, we can find a constant $C$ depending on $\text{
Diameter}(\mathcal{R})$ and ${C_{\mathcal{R}}^{g}}$ so that

$\displaystyle\Psi_{\alpha{\overline{\alpha}}}\geq\frac{\sigma}{C},$ (5.32)

and, therefore,

$\displaystyle\left(\Phi_{{\alpha\overline{\beta}}}\right)\geq\frac{\sigma}{C}.$
(5.33)

Step 3.2. According to the definition of $\sigma$, the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi_{\varepsilon}^{l}}$ is greater
than $2\sigma$ on $\partial\mathcal{R}$. So the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of $\Psi$ is greater than $\sigma$ on
$\partial\mathcal{R}$ since $q\in{\mathscr{Q}}_{\frac{\sigma}{2}}$.

Then, at any point ${\bf z}\in\mathcal{R}$, we can choose coordinate
$(z^{\alpha},\tau)$, so that $\Psi_{\tau}\neq 0$, and, similar to
${\mathcal{A}},{\mathcal{B}}$ we define

$\displaystyle{\mathcal{A}}^{\Psi}_{\alpha{\overline{\beta}}}=\Psi_{\alpha{\overline{\beta}}}-\frac{\Psi_{\alpha}}{\Psi_{\tau}}\Psi_{\tau{\overline{\beta}}}-\frac{\Psi_{{\overline{\beta}}}}{\Psi_{\overline{\tau}}}\Psi_{\alpha{\overline{\tau}}}+\frac{\Psi_{\alpha}\Psi_{{\overline{\beta}}}}{|\Psi_{\tau}|^{2}}\Psi_{{\tau\overline{\tau}}},$
(5.34)
$\displaystyle{\mathcal{B}}^{\Psi}_{\alpha\beta}=\Psi_{\alpha\beta}-\frac{\Psi_{\alpha}}{\Psi_{\tau}}\Psi_{\tau\beta}-\frac{\Psi_{\beta}}{\Psi_{\tau}}\Psi_{\alpha\tau}+\frac{\Psi_{\alpha}\Psi_{\beta}}{(\Psi_{\tau})^{2}}\Psi_{\tau\tau},$
(5.35)

and

$\displaystyle{\mathcal{K}}^{\Psi}={\mathcal{B}}^{\Psi}\overline{({\mathcal{A}}^{\Psi})^{-1}}\
\overline{{\mathcal{B}}^{\Psi}}({\mathcal{A}}^{\Psi})^{-1}.$ (5.36)

Then Lemma A.9 implies

$\displaystyle{\mathcal{K}}^{\Psi}\leq 1-\frac{\sigma}{C}\ \ \ \ \ \ \ \
\text{ on }\partial\mathcal{R},$ (5.37)

for a constant $C>0$. Therefore

$\displaystyle\text{tr}\left(I-{\mathcal{K}}^{\Psi}\right)^{-1}\leq\frac{nC}{\sigma}\
\ \ \ \ \ \ \ \text{ on }\partial\mathcal{R}.$ (5.38)

Then, providing ${\varepsilon}$ is small enough, we can use Proposition 3.1
and get

$\displaystyle\text{tr}\left(I-{\mathcal{K}}^{\Psi}\right)^{-1}\leq\frac{nC{\widetilde{C}}}{\sigma}\
\ \ \ \ \ \ \ \text{ in }\mathcal{R}.$ (5.39)

Here ${\widetilde{C}}$ is a constant depending on the $C^{2}$ norm of $\Psi$.
So we get

$\displaystyle{\mathcal{K}}^{\Psi}\leq 1-\frac{\sigma}{nC{\widetilde{C}}}.$
(5.40)

With estimates (5.33) and (5.40), we can apply Lemma A.6 and get the
robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Psi$ is greater than
$\frac{\sigma^{2}}{C_{0}}$.

Step 4. Now, we know for any $q\in{\mathscr{Q}}_{\sigma-2\delta}$, the
robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi_{\varepsilon}^{l}}-q$ is greater than $\frac{\sigma^{2}}{C_{0}}$.
Therefore, the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi_{\varepsilon}^{l}}$ is greater than
$\sigma-2\delta+\frac{\sigma^{2}}{C_{0}}$. So we choose

$\displaystyle\delta=\frac{\sigma^{2}}{4C_{0}}.$ (5.41)

Then the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi_{\varepsilon}^{l}}$ is greater than $\sigma$. Therefore $L$ is right-
closed, and, as a result $L=[0,1]$, providing ${\varepsilon}$ is small enough
so that the conditions of Proposition 3.1 and 4.1 are satisfied.

We let $\lambda=1$ and get the robustness of
$\mathfrak{q}\mathbb{C}$-convexityof solution $\Phi^{\varepsilon}$ to Problem
1.12 is greater than $\sigma$.

###  5.2 Improving the Convexity Estimate

The estimate of section 5.1 depends on the geometry of a family of
$\mathbb{C}$-convex rings. In this section, we show that the estimate can be
improved so that it only depends on the geometry of $\mathcal{R}$.

According to Lemma B.3, for a constant ${C_{\partial\mathcal{R}}^{g}}$,

$\displaystyle|\nabla{\Phi^{\varepsilon}}|>{C_{\partial\mathcal{R}}^{g}}\ \ \
\ \ \ \ \ \text{ on }\partial\mathcal{R};$ (5.42)

then, using Lemma A.11, the modulus of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi^{\varepsilon}}$ is greater than a constant $\mu_{g}$ on
$\partial\mathcal{R}$; then, we apply Lemma A.3 and get the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of ${\Phi^{\varepsilon}}$ is greater than
$\rho_{\partial\mathcal{R}}$ on ${\partial\mathcal{R}}$, for a constant
$\rho_{\partial\mathcal{R}}$ depending on the geometry of $\mathcal{R}$.

Let

$\displaystyle\tilde{\sigma}=\frac{1}{2}\min\left\\{{\rho_{\partial\mathcal{R}}},\
{C_{\mathcal{R}}^{g}}\right\\},$ (5.43)

and we will show the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi^{\varepsilon}}$ is greater than $\tilde{\sigma}$. Let $\delta$ be a
small constant in
$(0,\min\left\\{\frac{\sigma}{4},\frac{\tilde{\sigma}}{4}\right\\}]$ to be
determined, and let

$\displaystyle\begin{split}I^{\varepsilon}=\left\\{\theta\in[0,\tilde{\sigma}]\
\big{|}\right.&\text{ the robustness of $\mathfrak{q}\mathbb{C}$-convexity\ of
${\Phi^{\varepsilon}}-q$ is greater than $\delta$,}\\\ &\ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{with $q\in{\mathscr{Q}}_{\theta}$
}\right\\};\end{split}$ (5.44)

we will show $I^{\varepsilon}=[0,\tilde{\sigma}]$.

First, we notice that $I^{\varepsilon}\supset[0,\frac{\sigma}{4}]$. It’s also
easy to see that $I^{\varepsilon}$ is open. Most importantly, we can show it’s
right-closed; then $I^{\varepsilon}=[0,\tilde{\sigma}]$ follows.

For the right-closeness, we assume $I^{\varepsilon}\supset[0,l)$; then we show
it contains $l$. This part is completely parallel to the proof of section 5.1.

###  5.3 Estimates Related to the Rank Estimate

In this section, we show for a positive constant $c$, which depends on the
geometry of $\mathcal{R}$,

$\displaystyle(\sqrt{-1}\partial\overline{\partial}\Phi^{\varepsilon})^{n-1}\wedge{\omega_{\mathcal{G}}}\geq
c\ {\omega_{\mathcal{G}}}^{n},$ (5.45)

and

$\displaystyle\sqrt{-1}\partial\overline{\partial}\left(e^{\Phi^{\varepsilon}}\right)\geq
c\ {\omega_{\mathcal{G}}}.$ (5.46)

When ${\varepsilon}$ converges to $0$, these estimates converge to the
estimates for the solution $\Phi$ of the homogenous equation; the estimates
then indicate the rank of $\sqrt{-1}\partial\overline{\partial}\Phi$ is $n-1$
in the weak sense.

We first prove (5.45). We choose a set of coordinates $(z^{\alpha},\tau)$ so
that at a point ${\bf p}$, $\Phi^{\varepsilon}_{\alpha}=0$,
$\Phi^{\varepsilon}_{{\alpha\overline{\beta}}}=\delta_{\alpha\beta}\lambda_{\alpha}$
and $\mathcal{G}_{{i\overline{j}}}=\delta_{\ij}$. Let
$\mathcal{H}=(\Phi^{\varepsilon}_{{i\overline{j}}})$; then,

$\displaystyle\mathcal{H}=\left(\begin{array}[]{cccc}\lambda_{1}&&&\Phi^{\varepsilon}_{1{\overline{\tau}}}\\\
&\lambda_{2}&&\Phi^{\varepsilon}_{2{\overline{\tau}}}\\\ &&\ddots&\vdots\\\
\Phi^{\varepsilon}_{\tau\overline{1}}&\Phi^{\varepsilon}_{\tau\overline{2}}&\cdots&\Phi^{\varepsilon}_{{\tau\overline{\tau}}}\end{array}\right).$
(5.51)

Direct computation gives

$\displaystyle\text{tr}(\mathcal{H}^{-1})=\frac{1}{\mathfrak{t}}\left(1+\sum_{\mu=1}^{n-1}\frac{|\Phi^{\varepsilon}_{\mu{\overline{\tau}}}|^{2}}{\lambda_{\mu}^{2}}\right)+\sum_{\mu=1}^{n-1}\frac{1}{\lambda_{\mu}}$
(5.52)

and

$\displaystyle\det(\mathcal{H})=\mathfrak{t}\prod_{\mu=1}^{n-1}\lambda_{\mu},$
(5.53)

where

$\displaystyle\mathfrak{t}=\Phi^{\varepsilon}_{{\tau\overline{\tau}}}-\sum_{\mu=1}^{n-1}\frac{|\Phi^{\varepsilon}_{\mu{\overline{\tau}}}|^{2}}{\lambda_{k}}.$
(5.54)

Then, using equation (2.15), we get

$\displaystyle\frac{1}{\mathfrak{t}}\left(1+\sum_{\alpha=1}^{n-1}\frac{|\Phi^{\varepsilon}_{\alpha{\overline{\tau}}}|^{2}}{\lambda_{k}^{2}}\right)\leq\frac{1}{{\varepsilon}}.$
(5.55)

Therefore, ${\varepsilon}\leq\mathfrak{t}$, and

$\displaystyle\det(\mathcal{H})\geq{\varepsilon}\prod_{\mu=1}^{n-1}\lambda_{\mu}.$
(5.56)

We have shown the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi^{\varepsilon}}$ has a uniform positive lower bound, so
$\lambda_{\alpha}\geq{\widetilde{\sigma}}$, for $\alpha\in\\{1,\ ...\
,n-1\\}$, for a constant ${\widetilde{\sigma}}$, which depends on the geometry
of $\mathcal{R}$. Then, it follows

$\displaystyle\det(\mathcal{H})\geq{\varepsilon}({\widetilde{\sigma}})^{n-1}.$
(5.57)

Therefore,

$\displaystyle(\sqrt{-1}\partial\overline{\partial}\Phi^{\varepsilon})^{n}\geq{\varepsilon}({\widetilde{\sigma}})^{n-1}\
{\omega_{\mathcal{G}}}^{n}.$ (5.58)

Using equation 2.15, we know at the point ${\bf p}$

$\displaystyle(\sqrt{-1}\partial\overline{\partial}\Phi^{\varepsilon})^{n-1}\wedge{\omega_{\mathcal{G}}}\geq({\widetilde{\sigma}})^{n-1}\
{\omega_{\mathcal{G}}}^{n}.$ (5.59)

Since the above inequality is independent of the choice of coordinate, we know
it’s valid anywhere in $\mathcal{R}$.

We can also show $e^{{\Phi^{\varepsilon}}}$ is a strictly pluri-subharmonic
function and derive an estimate for the minimum eigenvalue of the matrix
$[(e^{\Phi^{\varepsilon}})_{{i\overline{j}}}]$.

Direct computation gives

$\displaystyle(e^{{\Phi^{\varepsilon}}})_{{i\overline{j}}}=e^{{\Phi^{\varepsilon}}}\left(\Phi^{\varepsilon}_{{i\overline{j}}}+\Phi^{\varepsilon}_{i}\Phi^{\varepsilon}_{\overline{j}}\right).$
(5.60)

We adopt the coordinates and the notations for the previous estimate. Then

$\displaystyle\det[(e^{{\Phi^{\varepsilon}}})_{{i\overline{j}}}]=e^{n{\Phi^{\varepsilon}}}\det\left({\widetilde{\mathcal{H}}}\right),$
(5.61)

where

$\displaystyle{\widetilde{\mathcal{H}}}=\mathcal{H}+\text{diag}(0,\ ...\
,|{\Phi^{\varepsilon}_{\tau}}|^{2}).$ (5.62)

Again, by direction computation, we have

$\displaystyle\det({\widetilde{\mathcal{H}}})\geq|{\Phi^{\varepsilon}_{\tau}}|\prod_{\mu=1}^{n-1}\lambda_{\mu}.$
(5.63)

According to the construction of the lower solution in Appendix B, we have
${\Phi^{\varepsilon}}\geq 0$, so the right-hand side of
$(\ref{623549})\geq\det({\widetilde{\mathcal{H}}})$. Therefore, we have

$\displaystyle\det[(e^{{\Phi^{\varepsilon}}})_{{i\overline{j}}}]\geq|{\Phi^{\varepsilon}_{\tau}}|^{2}({\widetilde{\sigma}})^{n-1}.$
(5.64)

Combining with the $C^{0}$ and $C^{2}$ estimates for ${\Phi^{\varepsilon}}$,
we have the lower bound estimate for the minimum eigenvalue of
$[(e^{{\Phi^{\varepsilon}}})_{{i\overline{j}}}]$.

##  6 Approximating to the Homogenous Equation

In this section, we let ${\varepsilon}$ go to zero and get an estimate for the
modulus of $\mathbb{C}$-convexity of level sets of solutions to Problem 1.6.

Previously, we have shown that when ${\varepsilon}$ is small enough, the
robustness of $\mathfrak{q}\mathbb{C}$-convexity of ${\Phi^{\varepsilon}}$ is
greater than ${\widetilde{\sigma}}$, for a constant ${\widetilde{\sigma}}$,
depending on the geometry of $\mathcal{R}$. Then using Lemma A.12, we know the
modulus of $\mathbb{C}$-convexity of level sets of $\Phi^{\varepsilon}$ is
greater than a positive constant ${\sigma}$, which depends on the geometry of
$\mathcal{R}$.

According to the equation (1.29) and the $C^{2}$ estimate for
${\Phi^{\varepsilon}}$ in Appendix C, we have

$\displaystyle\det({\Phi^{\varepsilon}}_{{i\overline{j}}})\rightarrow 0\ \ \ \
\ \ \ \ \text{ as }{\varepsilon}\rightarrow 0.$ (6.1)

So $\Phi^{\varepsilon}\rightarrow\Phi^{0}$ at least in $C^{0}$ norm. Then
using interpolation, we know ${\Phi^{\varepsilon}}\rightarrow\Phi^{0}$ in
$C^{1}$ norm. Therefore the uniform positive lower bound estimate for
$|\nabla\Phi^{\varepsilon}|_{\mathcal{G}}$ implies a positive lower bound for
$|\nabla\Phi^{0}|_{\mathcal{G}}$.

We also know $\Phi^{0}$ is a $C^{1,1}$ function, because of the uniform
$C^{1,1}$ estimate of Appendix C. This together with the lower bound estimate
for $|\nabla\Phi^{0}|_{\mathcal{G}}$ implies level sets of $\Phi^{0}$ are
$C^{1,1}$ hypersurfaces.

At a point ${\bf p}\in\mathcal{R}$, we choose coordinates $(z^{\alpha},\tau)$,
with $\tau=t+\sqrt{-1}s,$ so that ${\bf p}=0$, the metric is
$\sum_{i}dz^{i}\otimes\overline{dz^{i}}$ and

$\displaystyle\partial_{\alpha}\Phi^{0}=0,\ \partial_{s}\Phi^{0}=0,\
\partial_{t}\Phi^{0}>0,$ (6.2)

at ${\bf p}$; therefore, when ${\varepsilon}$ is small enough,

$\displaystyle\partial_{t}\Phi^{\varepsilon}>0,$ (6.3)

around ${\bf p}$. Let
${\mathcal{T}}_{\delta}=\left\\{\sum_{\alpha}|z^{\alpha}|^{2}+s^{2}<\delta^{2},\
|t|<\delta\right\\}$. For ${\varepsilon}$ and $\delta$ small enough, the level
set $\left\\{{\Phi^{\varepsilon}}={\Phi^{\varepsilon}}({\bf
p})\right\\}\cap{{\mathcal{T}}_{\delta}}$ is the graph of a $C^{1,1}$
function:

$\displaystyle\begin{split}&\left\\{{\Phi^{\varepsilon}}={\Phi^{\varepsilon}}({\bf
p})\right\\}\cap{{\mathcal{T}}_{\delta}}=\left\\{(z^{\alpha},\rho^{\varepsilon}(z^{\alpha},s)+\sqrt{-1}s)\
\big{|}\ \sum_{\alpha}|z^{\alpha}|^{2}+s^{2}<\delta^{2}\right\\}.\end{split}$
(6.4)

In addition, we know

$\displaystyle|\nabla{\Phi^{\varepsilon}}|\leq C_{2}({\varepsilon}+\delta),\ \
\ \ \ \ \ \ \text{ in }T_{\delta}.$ (6.5)

In the following, we will show when ${\varepsilon}$ and $\delta$ are small
enough,

$\displaystyle\rho^{\varepsilon}(z^{\alpha},0)\geq\frac{\sigma}{C}\sum_{\alpha}|z^{\alpha}|^{2}$
(6.6)

for a constant $C$ when $\sum_{\alpha}|z^{\alpha}|^{2}\leq\delta$. Most
importantly, we need to show that $\delta$ is independent of ${\varepsilon}$,
so when ${\varepsilon}$ converges to zero, we get the convexity estimate for
$\rho^{0}$.

We choose $q=2\text{Re}(z^{\alpha}\partial_{\alpha}{\Phi^{\varepsilon}})$;
then,

$\displaystyle\partial_{\alpha}\left({\Phi^{\varepsilon}}-q\right)({\bf z})=0\
\ \ \ \ \ \ \ \text{ for }\alpha\in\\{1,\ ...\ ,n-1\\}.$ (6.7)

It follows that

$\displaystyle{T^{\mathbb{C}}_{{\Phi^{\varepsilon}}-q,{\bf
z}}}=\\{\zeta\in\mathbb{C}^{n}|\zeta^{n}-z^{n}=0\\}.$ (6.8)

Because the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi^{\varepsilon}}$ is greater than $\sigma$, ${\Phi^{\varepsilon}}-q$ is
strongly $\mathfrak{q}\mathbb{C}$-convex of robustness greater than
$\frac{\sigma}{2}$ providing $\delta$ is small enough. Therefore, the second
order Taylor expansion of ${\Phi^{\varepsilon}}|_{\\{z^{n}=a\\}}$ at any point
in $\\{|z^{\alpha}|^{2}\leq\delta^{2}\\}$, for $|a|<\delta,a\in\mathbb{C}$, is
convex of modulus greater than $\frac{\sigma}{C_{1}}$, for a constant $C_{1}$
depending on the diameter of $\mathcal{R}$, when $\delta$ is small enough. The
convexity estimate for $\rho^{\varepsilon}(z^{\alpha},0)$ then follows since

$\displaystyle\Phi^{\varepsilon}(z^{\alpha},\rho^{\varepsilon}(z^{\alpha},0))=0.$
(6.9)

For the estimate (5.59), we let ${\varepsilon}$ converge to zero. Since
$\Phi^{\varepsilon}\rightarrow\Phi^{0}$ in $C^{1}$ norm and
$\Phi^{\varepsilon}$ are all plurisubharmonic functions, we have

$\displaystyle(\sqrt{-1}\partial\overline{\partial}\Phi^{0})^{n-1}\wedge{\omega_{\mathcal{G}}}\geq\frac{1}{C}\
{\omega_{\mathcal{G}}}^{n},$ (6.10)

in the weak sense for a constant $C$ depending on the geometry of
$\mathcal{R}$.

For the estimate (5.46), since $e^{{\Phi^{\varepsilon}}}$ is a
plurisubharmonic function and $\Phi^{\varepsilon}\rightarrow\Phi^{0}$ in
$C^{1}$ norm, we have

$\displaystyle\sqrt{-1}\partial\overline{\partial}(e^{{\Phi^{0}}})\geq\frac{1}{C}{\omega_{\mathcal{G}}}$
(6.11)

in weak sense for a constant $C$ depending on the geometry of $\mathcal{R}$.

##  Appendix A Linear Algebra Lemmas

In this appendix, we prove some linear algebra lemmas. The goal is to
establish relations between different descriptions of the convexity and
$\mathfrak{q}\mathbb{C}$-convexity.

Lemma A.1 shows that for quadratic polynomials defined on $\mathbb{C}^{n-1}$
the degree and the modulus of convexity are equivalent.

######  Lemma A.1.

The modulus of convexity of a quadratic polynomial on $\mathbb{C}^{n-1}$ is
greater than $\mu\geq 0$ if and only if its degree of convexity is greater
than $\mu$.

###### Proof.

We note that any quadratic polynomial $P$ on $\mathbb{C}^{n-1}$ has the
following form

$\displaystyle
P(z)=A_{{\alpha\overline{\beta}}}z^{\alpha}z^{\overline{\beta}}+\text{Re}(B_{\alpha\beta}z^{\alpha}z^{\beta})+L,$
(A.1)

where $A$ is hermitian, $B$ is symmetric and $L$ is a linear function. Suppose
$G_{\alpha{\overline{\beta}}}dz^{\alpha}\otimes\overline{dz^{\beta}}$ is the
metric on $\mathbb{C}^{n-1}$. We only need to show

$\displaystyle\sup_{B\overline{G^{-1}}\ \overline{B}G^{-1}\leq
1}\text{Re}(B_{\alpha\beta}{z^{\alpha}z^{\beta}})=G_{\alpha{\overline{\beta}}}z^{\alpha}z^{\overline{\beta}}.$
(A.2)

To simplify the computation, we change coordinates, so that
$G_{{\alpha\overline{\beta}}}=\delta_{{\alpha\beta}}$. Then we need to show

$\displaystyle\sup_{B\overline{B}\leq
1}\text{Re}(B_{\alpha\beta}{z^{\alpha}z^{\beta}})=|z|^{2}.$ (A.3)

First, we fix $(z^{\alpha})$ and choose
$B_{\alpha\beta}=\frac{\overline{z^{\alpha}z^{\beta}}}{|z|^{2}}$, then

$\displaystyle\text{Re}(B_{\alpha\beta}z^{\alpha}z^{\beta})=|z|^{2}.$ (A.4)

So

$\displaystyle\sup_{B\overline{B}\leq
1}\text{Re}(B_{\alpha\beta}{z^{\alpha}z^{\beta}})\geq|z|^{2}.$ (A.5)

Then for any symmetric $B$, with $B\overline{B}\leq 1$, after a unitary change
of coordinates, we can make $B=\text{diag}(\sigma_{1},\ ...\ ,\sigma_{n-1})$,
with $\sigma_{\gamma}\in\mathbb{R}^{\geq 0}$ and $\sigma_{\gamma}\leq 1$. This
is the Autonne-Takagi factorization, which is Corollary 4.4.4(c) of [HJ13] (or
see Lemma A.3 of [H22N]). Therefore

$\displaystyle\text{Re}(B_{\alpha\beta}z^{\alpha}z^{\beta})=\sum\sigma_{\gamma}\text{Re}({z^{\gamma}})^{2}\leq|z|^{2};$
(A.6)

as a result,

$\displaystyle\sup_{B\overline{B}\leq
1}\text{Re}(B_{\alpha\beta}{z^{\alpha}z^{\beta}})\leq|z|^{2}.$ (A.7)

(A.5) and (A.7) together gives the equality. ∎

We would also need the following lemma to estimate the robustness of
$\mathfrak{q}\mathbb{C}$-convexity.

######  Lemma A.2.

Suppose that
$G_{\alpha{\overline{\beta}}}dz^{\alpha}\otimes\overline{dz^{\beta}}$ is a
constant coefficient metric on $\mathbb{C}^{n-1}$ and that the modulus of
convexity of

$\displaystyle\mathcal{P}(z^{\alpha})=A_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}+\text{Re}\left(B_{\alpha\beta}z^{\alpha}z^{\beta}\right)$
(A.8)

is greater than $\delta$. Let $G=(G_{\alpha{\overline{\beta}}})$. Then, for
any Hermitian matrix $V$ with

$\displaystyle V\leq\frac{\delta}{2}G$ (A.9)

and any symmetric matrix $W$ with

$\displaystyle W\overline{G^{-1}}\
\overline{W}G^{-1}\leq\frac{\delta^{2}}{4},$ (A.10)

we have that

$\displaystyle\mathcal{P}-V_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}-\text{Re}\left(W_{\alpha\beta}z^{\alpha}z^{\beta}\right)$
(A.11)

is strongly convex.

The idea of this lemma is very intuitive in the case of $n=2$. In this case
$A\in\mathbb{R}^{+}\text{ and }B\in\mathbb{C}.$ $\mathcal{P}$ is convex of
modulus $>\delta$ if and only if $(B,A)$ stays in the cone
$\left\\{A-\delta>|B|\right\\}$, and $\mathcal{P}$ is convex of degree
$>\delta$ if and only if for any $\beta\in[-\delta,\delta]$, $(B,A)$ stays in
the cone $\left\\{A>|B-\beta|\right\\}$. Lemma A.1 basically says

$\displaystyle\left\\{|B|<A-\delta\right\\}=\bigcap_{|\beta|\leq\delta}\left\\{|B-\beta|<A\right\\}.$
(A.12)

For Lemma A.2, we need to show when $(V,W)\in\\{A\geq 0,|B|\leq\delta-A\\}$

$\displaystyle\left\\{|B|<A-\delta\right\\}\subset\left\\{|B-W|<A-V\right\\}.$
(A.13)

As illustrated by Figure 2, the blue cone covers the cone
$\left\\{|B|<A-\delta\right\\}$ providing the corner of the blue cone stays in
the cone $\\{A\geq 0,|B|\leq\delta-A\\}$, the yellow area.

Figure 2: Visualizing the Convexity Estimate: Lemma A.2

###### Proof of Lemma A.2.

Actually, the proof is quite simple. We denote

$\displaystyle\tilde{\mathcal{P}}=\mathcal{P}-V_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}-\text{Re}\left(W_{\alpha\beta}z^{\alpha}z^{\beta}\right).$
(A.14)

Since $\tilde{\mathcal{P}}$ is a quadratic polynomial, it’s strongly convex if
and only if $\tilde{\mathcal{P}}(z)>0$ for $z\neq 0$. So, we need to show

$\displaystyle
V_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}+\text{Re}\left(W_{\alpha\beta}z^{\alpha}z^{\beta}\right)\leq\delta
g_{\alpha{\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}.$ (A.15)

According to the proof of Lemma A.1, we have

$\displaystyle\text{Re}\left(W_{\alpha\beta}z^{\alpha}z^{\beta}\right)\leq\frac{\delta}{2}g_{\alpha{\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}.$
(A.16)

The following inequality follows from the assumption (A.9):

$\displaystyle
V_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}\leq\frac{\delta}{2}g_{\alpha{\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}.$
(A.17)

The two inequalities above implies (A.15) is valid. ∎

With the preparation above, we can establish the relation between the
robustness and modulus of $\mathfrak{q}\mathbb{C}$-convexity. We have the
following Lemma A.3, which proves the robustness of
$\mathfrak{q}\mathbb{C}$-convexity is positive providing the modulus of
$\mathfrak{q}\mathbb{C}$-convexity is positive, and Lemma A.5, which proves an
estimate in the opposite direction.

######  Lemma A.3.

Let $\mathcal{R}$ be the ring-shaped domain of Problem 1.12. Suppose the
modulus of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi\in
C^{\infty}(\overline{\mathcal{R}})$ is greater than $m$ at a point ${\bf
p}\in\overline{\mathcal{R}}$, for $m>0$. Then the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\frac{m^{2}}{C}$
for a constant $C$ depending on the $C^{2}$ norm of $\Phi$ and the thickness
of $\mathcal{R}$.

###### Proof of Lemma A.3.

Let the metric on $\mathbb{C}^{n}$ be
$\sum_{i}dz^{i}\otimes\overline{dz^{i}}$.

We need to show for a small positive constant $\rho$, $\Phi-q$ is strongly
$\mathfrak{q}\mathbb{C}$-convex providing $q\in{\mathscr{Q}}_{\rho}$.

First, according to the definition of modulus of
$\mathfrak{q}\mathbb{C}$-convexity, we have $|\nabla\Phi({\bf p})|>m$, so

$\displaystyle|\nabla(\Phi-q)({\bf p})|>\frac{m}{2}$ (A.18)

for any $q\in{\mathscr{Q}}_{\rho}$, providing $\rho\leq\frac{m}{8}$. In the
following, we show when $\rho$ is small enough, the second order Taylor
expansions of $\Phi|_{T^{\mathbb{C}}_{\Phi,{\bf p}}}$ and
$(\Phi-q)|_{\Phi_{T^{\mathbb{C}}_{\Phi-q,{\bf p}}}}$ at ${\bf p}$ are close to
each other. Then the $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ implies the
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi-q$.

We choose a coordinate $(z^{\alpha},\tau)$, so that
$\partial_{\alpha}\Phi({\bf p})=0$ and the metric is
$\sum_{i}dz^{i}\otimes\overline{dz^{i}}$. Let $\Psi=\Phi-q$; then,

$\displaystyle|\Psi_{\tau}({\bf p})|\geq|\Phi_{\tau}({\bf p})|-|q_{\tau}({\bf
p})|\geq|\nabla\Phi({\bf p})|-|\nabla q({\bf p})|\geq
m-\frac{m}{8}\geq\frac{m}{2}.$ (A.19)

Let

$\displaystyle{\mathcal{A}}_{{\alpha\overline{\beta}}}=\Psi_{{\alpha\overline{\beta}}}({\bf
p})-\Psi_{\alpha{\overline{\tau}}}({\bf
p})\overline{\left(\frac{\Psi_{\beta}({\bf p})}{\Psi_{\tau}({\bf
p})}\right)}-\Psi_{\tau{\overline{\beta}}}({\bf
p})\left(\frac{\Psi_{\alpha}({\bf p})}{\Psi_{\tau}({\bf
p})}\right)+\Psi_{{\tau\overline{\tau}}}({\bf
p})\left(\frac{\Psi_{\alpha}({\bf p})}{\Psi_{\tau}({\bf
p})}\right)\overline{\left(\frac{\Psi_{\beta}({\bf p})}{\Psi_{\tau}({\bf
p})}\right)},$ (A.20)
$\displaystyle{\mathcal{B}}_{{\alpha\beta}}=\Psi_{{\alpha\beta}}({\bf
p})-\Psi_{\alpha\tau}({\bf p}){\left(\frac{\Psi_{\beta}({\bf
p})}{\Psi_{\tau}({\bf p})}\right)}-\Psi_{\tau\beta}({\bf
p})\left(\frac{\Psi_{\alpha}({\bf p})}{\Psi_{\tau}({\bf
p})}\right)+\Psi_{\tau\tau}({\bf p})\left(\frac{\Psi_{\alpha}({\bf
p})}{\Psi_{\tau}({\bf p})}\right){\left(\frac{\Psi_{\beta}({\bf
p})}{\Psi_{\tau}({\bf p})}\right)}.$ (A.21)

Then the second order Taylor expansion of $\Psi|_{{T^{\mathbb{C}}_{\Psi,{\bf
p}}}}$ at ${\bf p}$ is

$\displaystyle{\mathcal{A}}_{{\alpha{\overline{\beta}}}}z^{\alpha}\overline{z^{\beta}}+\text{Re}({\mathcal{B}}_{{\alpha\beta}}z^{\alpha}z^{\beta}).$
(A.22)

Because $q$ is pluriharmonic, we have
$\Psi_{{i\overline{j}}}=\Phi_{{i\overline{j}}}$. Then using
$\Phi_{\alpha}({\bf p})=0$, we have

$\displaystyle{\mathcal{A}}_{{\alpha\overline{\beta}}}=\Phi_{\alpha{\overline{\beta}}}+\Phi_{\alpha{\overline{\tau}}}\overline{\left(\frac{q_{\beta}}{\Psi_{\tau}}\right)}+\Phi_{\tau{\overline{\beta}}}{\left(\frac{q_{\alpha}}{\Psi_{\tau}}\right)}+\Phi_{\tau{\overline{\tau}}}\frac{q_{\alpha}q_{\overline{\beta}}}{|\Psi_{\tau}|^{2}}.$
(A.23)

We denote that
${\mathcal{A}}_{{\alpha\overline{\beta}}}=\Phi_{{\alpha\overline{\beta}}}+{\mathcal{U}}_{{\alpha{\overline{\beta}}}}$.
Similarly, for ${\mathcal{B}}$, we have

$\displaystyle{\mathcal{B}}_{{\alpha\beta}}=\Psi_{\alpha\beta}+\Psi_{\alpha\tau}{\left(\frac{q_{\beta}}{\Psi_{\tau}}\right)}+\Psi_{\tau\beta}{\left(\frac{q_{\alpha}}{\Psi_{\tau}}\right)}+\Psi_{\tau\tau}\frac{q_{\alpha}q_{\beta}}{(\Psi_{\tau})^{2}};$
(A.24)

we denote

$\displaystyle
B_{{\alpha\beta}}=\Phi_{{\alpha\beta}}+\mathcal{V}_{{\alpha\beta}}.$ (A.25)

So,

$\displaystyle\text{the second order Taylor expansion of
}\Psi|_{T^{\mathbb{C}}_{\Psi,{\bf p}}}\text{ at ${\bf p}$}$ (A.26)
$\displaystyle=$ $\displaystyle\text{the second order Taylor expansion of
}\Phi|_{T^{\mathbb{C}}_{\Phi,{\bf p}}}\text{ at ${\bf
p}$}+{\mathcal{U}}_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}+\text{Re}\left((\mathcal{V}_{{\alpha\beta}}-q_{{\alpha\beta}})z^{\alpha}z^{\alpha}\right).$
(A.27)

Because $\Phi$ is strongly $\mathfrak{q}\mathbb{C}$-convex of modulus greater
than $m$, we know the second order Taylor expansion of
$\Psi|_{T^{\mathbb{C}}_{\Psi,{\bf p}}}$ at ${\bf p}$ is convex of modulus
greater than $m$. Then using Lemma A.2, we know when

$\displaystyle({\mathcal{U}}_{{\alpha\overline{\beta}}})\leq\frac{m}{2}$
(A.28)

and

$\displaystyle\left(\sum_{\gamma}\mathcal{V}_{\alpha\gamma}\overline{\mathcal{V}_{\gamma\beta}}\right)\leq\frac{m^{2}}{4},$
(A.29)

(A.26) is strongly convex. In the following, we show (A.28) and (A.29) are
satisfied when $\rho$ is small enough; we need to use the following linear
algebra Lemma:

######  Lemma A.4.

Suppose $A$ is a $k\times k$ Hermitian matrix. Then $A<1$, providing

$\displaystyle|A_{{i\overline{j}}}|<\frac{1}{k^{2}}$ (A.30)

for any $i,j\in\\{1,\ ...\ ,k\\}$. Suppose $B$ is a $k\times k$ complex-valued
matrix. Then $B{B}^{\dagger}<1$ providing $|B_{ij}|<\frac{1}{k^{\frac{3}{2}}}$