for any $i,j\in\\{1,\ ...\ ,k\\}$.

###### Proof of Lemma A.4.

Since $B{B}^{\dagger}$ is Hermitian and

$\displaystyle(B{B}^{\dagger})_{{i\overline{j}}}=\sum_{p}B_{ip}\overline{B_{pj}},$
(A.31)

we only need to prove the estimate for $A$. Suppose the maximum eigenvalue of
$A$ is $\sigma$. Then we can find a unit vector

$\displaystyle{\mathbb{v}}=(v_{1},\ ...\ v_{k})$ (A.32)

so that

$\displaystyle{\mathbb{v}}A{\mathbb{v}}^{\dagger}=\sigma.$ (A.33)

Here unit vector means ${\mathbb{v}}{\mathbb{v}}^{\dagger}=1$. It follows that

$\displaystyle\sigma\leq\sum_{i,j=1}^{k}|v_{i}||A_{{i\overline{j}}}||v_{j}|\leq\sum_{i,j=1}^{k}|A_{{i\overline{j}}}|\leq
k^{2}\max|A_{{i\overline{j}}}|.$ (A.34)

Therefore (A.30) implies $A<1$. ∎

With the lemma above, we need to show
${\mathcal{U}}_{{\alpha\overline{\beta}}}$ and $\mathcal{V}_{{\alpha\beta}}$
are both small. For ${\mathcal{U}}_{{\alpha\overline{\beta}}}$, we have

$\displaystyle|{\mathcal{U}}_{{\alpha\overline{\beta}}}|\leq\rho
C_{2}(\frac{8}{m}+64\frac{\rho}{m^{2}}),$ (A.35)

where $C_{2}$ depends on the $C^{2}$ norm of $\Phi$. In (A.35), we used the
estimate (A.19). We already required that $\rho\leq m$, so from (A.35), we get

$\displaystyle|{\mathcal{U}}_{{\alpha\overline{\beta}}}|\leq
100C_{2}\frac{\rho}{m}.$ (A.36)

Therefore, to satisfy (A.28), we require that

$\displaystyle\rho\leq\frac{m^{2}}{200C_{2}}.$ (A.37)

For $\mathcal{V}_{{\alpha\beta}}$, similarly, we have

$\displaystyle|\mathcal{V}_{{\alpha\beta}}|\leq
100{\widetilde{C_{2}}}\frac{\rho}{m}.$ (A.38)

Here the constant ${\widetilde{C_{2}}}$ depends on the $C^{2}$ norm of
$\Phi-q$.

To prove (A.38), we need to show the $C^{2}$ derivatives of $q$ can be
controlled by $\rho$ and the thickness of $\mathcal{R}$. Suppose
$B_{r}(z)\subset\mathcal{R}$, where $r$ is the thickness of $\mathcal{R}$.
Then we can estimate $D^{2}q(z)$ since $Dq$ is harmonic in $B_{r}(z)$. The
following estimate is standard:

$\displaystyle|D^{2}q(z)|\leq\frac{C_{n}}{r}\max_{B_{r}}|\nabla q|,$ (A.39)

where $C_{n}$ depends on the dimension $n$. Since $q$ is quadratic, we know
$D^{2}q$ is constant. Then it follows that

$\displaystyle|D^{2}q|\leq\frac{C_{n}\rho}{r}.$ (A.40)

Since $\rho\leq m$ and $m$ depends on the $C^{2}$ norm of $\Phi$, we have
${\widetilde{C_{2}}}$ depends on the $C^{2}$ norm of $\Phi$ and the thickness
of $\mathcal{R}$.

Summing up, (A.36) and (A.38) can both be satisfied by choosing

$\displaystyle\rho=\frac{m^{2}}{C},$ (A.41)

where $C$ is a constant depending on the $C^{2}$ norm of $\Phi$ and the
thickness of $\mathcal{R}$. We choose $C$ large enough so that
$\frac{m}{C}\leq\frac{1}{8}$ because $\rho$ depends on the $C^{2}$ norm of
$\Phi$. Then (A.41) also implies $\rho\leq\frac{m}{8}$. The lemma is proved. ∎

######  Lemma A.5.

Suppose that $\Omega$ is a bounded domain with smooth boundaries and that the
robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi\in
C^{\infty}({\overline{\Omega}})$ is greater than $\rho$ at ${\bf
p}\in{\overline{\Omega}}$. Then the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\frac{\rho}{C}$
at ${\bf p}$, for a constant $C$ depending on the diameter of $\Omega.$

###### Proof of Lemma A.5.

We assume the Hermitian metric on $\Omega$ is
$dz^{i}\otimes\overline{dz^{i}}$.

First, we estimate $|\nabla\Phi({\bf p})|$. The assumption that the robustness
of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\rho$ at
${\bf p}$ implies that for any linear function $l$ with $|\nabla l|\leq\rho$,

$\displaystyle\nabla(\Phi-l)\neq 0\ \ \ \ \ \ \ \ \text{ at }{\bf p}.$ (A.42)

So we have $|\nabla\Phi({\bf p})|>\rho$.

For the convexity estimate, we choose a set of coordinates $(z^{\alpha},\tau)$
so that ${\bf p}=0$, $\Phi_{\alpha}({\bf p})=0$ and the metric is still
$dz^{i}\otimes\overline{dz^{i}}$. Then $\\{\tau=0\\}$ is the complex tangent
plane of $\\{\Phi=\Phi({\bf p})\\}$ at ${\bf p}$. Let

$\displaystyle
q_{\mathbb{v}}=\text{Re}\left[\sum_{\alpha=1}^{n-1}v_{\alpha}(z^{\alpha})^{2}\frac{1}{4\text{
Diameter}(\Omega)}\right],$ (A.43)

for any ${\mathbb{v}}=(v_{\alpha})\in\mathbb{C}^{n-1}$, with
$|v_{\alpha}|\leq\rho$, for $\alpha=1,\ ...\ ,n-1$. Then,

$\displaystyle|\nabla q_{\mathbb{v}}|\leq\rho\ \ \ \ \ \ \ \ \text{ in
}\Omega,$ (A.44)

and, therefore, $q_{\mathbb{v}}\in{\mathscr{Q}}_{\rho}$. So, $\Phi-
q_{\mathbb{v}}$ is strongly $\mathfrak{q}\mathbb{C}$-convex, because of the
condition that the robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$
is greater than $\rho$. It follows that $\nabla(\Phi-q_{\mathbb{v}})\neq 0$ in
$\Omega$, so level sets of $\Phi-q_{\mathbb{v}}$ are smooth surfaces. Since
$\nabla q_{\mathbb{v}}({\bf p})=0$, $\\{\tau=0\\}$ is also the complex tangent
plane of

$\displaystyle\left\\{\Phi-q_{\mathbb{v}}=(\Phi-q_{\mathbb{v}})({\bf
p})\right\\}$ (A.45)

at ${\bf p}$. Therefore, the second order Taylor expansion of
$\Phi|_{\\{\tau=0\\}}-q_{\mathbb{v}}$ at $0$ is strongly convex. This says the
degree of convexity of the second order Taylor expansion of
$\Phi|_{\\{\tau=0\\}}$ at $0$ is greater than $\frac{\rho}{4\text{
Diameter}(\Omega)}$. Then it follows that the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than

$\displaystyle\rho\cdot\min\left\\{1,\frac{1}{4\text{
Diameter}(\Omega)}\right\\}.$ (A.46)

∎

######  Lemma A.6.

Let $g_{{i\overline{j}}}dz^{i}\otimes\overline{dz^{j}}$ be a constant
coefficient Hermitian metric on $\mathbb{C}^{n}$. Let
$G=(g_{{i\overline{j}}})$. Suppose that $A$ is a hermitian matrix and that $B$
is a symmetric matrix, satisfying

$\displaystyle A$ $\displaystyle>\delta G,$ (A.47) $\displaystyle
B{\overline{A^{-1}}}\ {\overline{B}}{A^{-1}}$ $\displaystyle<1-\delta,$ (A.48)

for a positive constant $\delta$, with $\delta<1$. Then the modulus of
convexity of

$\displaystyle
P(z)=A_{{i\overline{j}}}z^{i}\overline{z^{j}}+\text{Re}(B_{ij}z^{i}z^{j})$
(A.49)

is greater than $\frac{\delta^{2}}{2}$.

###### Proof of Lemma A.6.

From (A.48), we easily get

$\displaystyle B{\overline{A^{-1}}}\
{\overline{B}}{A^{-1}}<(1-\frac{\delta}{2})^{2}$ (A.50)

and, consequently,

$\displaystyle B{{\left[\overline{A}(1-\frac{\delta}{2})\right]}^{-1}}\
{\overline{B}}\left[A(1-\frac{\delta}{2})\right]^{-1}<1.$ (A.51)

According to (A.47),

$\displaystyle A(1-\frac{\delta}{2})<A-\frac{\delta^{2}}{2}G;$ (A.52)

then, (A.51) implies

$\displaystyle
B{{\left[\overline{A}-\frac{\delta^{2}}{2}\overline{G}\right]}^{-1}}\
{\overline{B}}\left[A-\frac{\delta^{2}}{2}G\right]^{-1}<1.$ (A.53)

From (A.51) to (A.53), we used the following linear algebra lemma:

######  Lemma A.7.

Suppose $K$ and $H$ are two positive definite Hermitian matrices, with $K>H$.
Then for a symmetric matrix $B$ we have the maximum eigenvalue of
$B\overline{K^{-1}}\ {\overline{B}}K^{-1}$ is smaller or equal to the maximum
eigenvalue of $B\overline{H^{-1}}\ {\overline{B}}H^{-1}$.

The lemma above can be easily proved by simultaneously diagonalize $K$ and
$H$; we omit the proof here.

Then using (A.53) and $A-\frac{\delta^{2}}{2}G>0$, we have

$\displaystyle\left(A-\frac{\delta^{2}}{2}G\right)_{{i\overline{j}}}z^{i}\overline{z^{j}}+\text{Re}(B_{ij}z^{i}z^{j})=P-\frac{\delta^{2}}{2}g_{{i\overline{j}}}z^{i}\overline{z^{j}}$
(A.54)

is a strictly convex function on $\mathbb{C}^{n}$. Therefore, the modulus of
convexity of $P$ is greater than $\frac{\delta^{2}}{2}$ ∎

######  Remark A.8.

The proof of Lemma A.6 can be visualized in a simplified case of $n=1$.
Suppose $a\in\mathbb{R}^{+}$ and $b\in\mathbb{R}$. If $a>\delta$ and
$\frac{b^{2}}{a^{2}}<1-\delta$, then the modulus (or degree) of convexity of
$P(z)=a|z|^{2}+\text{Re}(bz^{2})$ is greater than $\frac{\delta^{2}}{2}$. In
Figure 3 the shadowed area is the set

$\displaystyle
F=\\{a>\delta\\}\cap\left\\{\frac{b^{2}}{a^{2}}<1-\delta\right\\}$ (A.55)

Figure 3: Visualizing the Convexity Estimate

We can easily see for a small enough $c>0$, the cone $C=\\{|b|<a-c\\}$
contains $F$, which says the modulus of convexity of $P$ is greater than $c$.
Through computation, we find we can choose $c=\frac{\delta^{2}}{2}$.

Lemma A.6 proves that for a function

$\displaystyle
P(z)=A_{{i\overline{j}}}z^{i}\overline{z^{j}}+\text{Re}(B_{ij}z^{i}z^{j})$
(A.56)

the positivity of $A$ and $B\overline{A^{-1}}\ \overline{B}A^{-1}<1$ implies a
convexity estimate for $P$; the following lemma proves an estimate in the
opposite direction;

######  Lemma A.9.

Suppose the modulus (or, equivalently, the degree) of convexity of $P(z)$ is
greater than $\delta>0$. Then

$\displaystyle B\overline{A^{-1}}\ \overline{B}A^{-1}<1-\frac{\delta}{C},$
(A.57)

where $C$ depends on the $C^{2}$ norm of $P$.

###### Proof of Lemma A.9.

Using Autonne-Takagi factorization (Corollary 4.4.4(c) of [HJ13], or Lemma A.3
of [H22N]), we diagonalize $A,B$ simultaneously so that

$\displaystyle A=I,\ B=\text{diag}(\lambda_{1},\ ...\ ,\lambda_{n-1}),$ (A.58)

with $\lambda_{\mu}\geq 0.$ Suppose the Hermitian metric on $\mathbb{C}^{n-1}$
is $G_{\alpha\overline{\beta}}dz^{\alpha}\otimes\overline{dz^{\beta}}$. The
modulus of convexity of $P$ is greater than $\delta$ implies $P-\delta
G_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}$ is strongly
convex. Suppose that

$\displaystyle AG^{-1}\leq C_{2}.$ (A.59)

Then $C_{2}$ depends on $|D^{2}P|_{G}$. This implies

$\displaystyle G\geq\frac{1}{C_{2}}A,$ (A.60)

and, with the current coordinate, we have

$\displaystyle G\geq\frac{1}{C_{2}}I.$ (A.61)

So, $P-\frac{\delta}{C_{2}}|z|^{2}$ is strongly convex. Therefore,

$\displaystyle\sigma_{\mu}<1-\frac{\delta}{C_{2}},$ (A.62)

and it follows that

$\displaystyle B\overline{A^{-1}}\
\overline{B}A^{-1}<\left(1-\frac{\delta}{C_{2}}\right)^{2}.$ (A.63)

$P-G_{{\alpha\overline{\beta}}}z^{\alpha}\overline{z^{\beta}}$ is strongly
convex implies $A>\delta G$; therefore, $C_{2}\geq\delta$. So we can choose
$C=2C_{2}.$

∎

We also need the following lemma which shows the modulus or robustness of
$\mathfrak{q}\mathbb{C}$-convexity is weakly preserved when taking limit:

######  Lemma A.10.

Let $\Omega$ be a domain with smooth boundary, and let ${\bf p}$ be a point in
$\overline{\Omega}$. Let $\\{{\Phi^{\varepsilon}}\\}_{{\varepsilon}\in[0,1)}$
be a family of smooth functions defined on $\overline{\Omega}$, with

$\displaystyle{\Phi^{\varepsilon}}\rightarrow\Phi^{0},\
D{\Phi^{\varepsilon}}\rightarrow D\Phi^{0},\
D^{2}{\Phi^{\varepsilon}}\rightarrow D^{2}\Phi^{0},\ \text{ at }{\bf p}.$
(A.64)

Suppose that the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
$\Phi^{\varepsilon}$ at ${\bf p}$ is greater than $\rho$, for any
${\varepsilon}\in(0,1)$. Then for any $0\leq{\widetilde{\rho}}<\rho$, the
robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi^{0}$ is greater than
${\widetilde{\rho}}$ at ${\bf p}$.

###### Proof of Lemma A.10.

We denote the metric on $\mathbb{C}^{n}$ by
$\mathcal{G}_{i\overline{j}}dz^{i}\otimes\overline{dz^{j}}$. In the following,
all the computations are performed at the point ${\bf p}$.

Because of the $\mathfrak{q}\mathbb{C}$-convexity assumption for
${\Phi^{\varepsilon}}$, for any $q\in{\mathscr{Q}}_{{\widetilde{\rho}}}$ and
${\varepsilon}>0$, ${\Phi^{\varepsilon}}-q$ is strongly
$\mathfrak{q}\mathbb{C}$-convex with the robustness greater than
$\rho-{\widetilde{\rho}}$. This implies

$\displaystyle|\nabla({\Phi^{\varepsilon}}-q)|_{\mathcal{G}}>\rho-{\widetilde{\rho}},$
(A.65)

for any ${\varepsilon}\in(0,1)$; then, because of (A.64), we have

$\displaystyle|\nabla({\Phi^{0}}-q)|_{\mathcal{G}}\geq\rho-{\widetilde{\rho}}>0.$
(A.66)

Therefore, we can choose a set of coordinates $(z^{\alpha},\tau)$, so that

$\displaystyle\partial_{\alpha}(\Phi^{0}-q)=0.$ (A.67)

This implies $\partial_{\tau}(\Phi^{0}-q)\neq 0$, and when ${\varepsilon}$ is
small enough, $\partial_{\tau}(\Phi^{\varepsilon}-q)\neq 0$. Let
$\Psi^{\varepsilon}={\Phi^{\varepsilon}}-q$; for each $\Psi^{\varepsilon}$,
with ${\varepsilon}$ small enough, we define

$\displaystyle{\mathcal{A}}^{{\varepsilon}}_{{\alpha\overline{\beta}}}=\Psi^{\varepsilon}_{{\alpha\overline{\beta}}}({\bf
p})-\Psi^{\varepsilon}_{\alpha{\overline{\tau}}}({\bf
p})\overline{\left(\frac{\Psi^{\varepsilon}_{\beta}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf
p})}\right)}-\Psi^{\varepsilon}_{\tau{\overline{\beta}}}({\bf
p})\left(\frac{\Psi^{\varepsilon}_{\alpha}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf
p})}\right)+\Psi^{\varepsilon}_{{\tau\overline{\tau}}}({\bf
p})\left(\frac{\Psi^{\varepsilon}_{\alpha}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf
p})}\right)\overline{\left(\frac{\Psi^{\varepsilon}_{\beta}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf p})}\right)},$ (A.68)
$\displaystyle{\mathcal{B}}^{{\varepsilon}}_{{\alpha\beta}}=\Psi^{\varepsilon}_{{\alpha\beta}}({\bf
p})-\Psi^{\varepsilon}_{\alpha\tau({\bf
p})}{\left(\frac{\Psi^{\varepsilon}_{\beta}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf
p})}\right)}-\Psi^{\varepsilon}_{\tau\beta}({\bf
p})\left(\frac{\Psi^{\varepsilon}_{\alpha}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf
p})}\right)+\Psi^{\varepsilon}_{\tau\tau}({\bf
p})\left(\frac{\Psi^{\varepsilon}_{\alpha}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf
p})}\right){\left(\frac{\Psi^{\varepsilon}_{\beta}({\bf
p})}{\Psi^{\varepsilon}_{\tau}({\bf p})}\right)}.$ (A.69)

Because the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
${\Phi^{\varepsilon}}-q$ is greater than $\rho-{\widetilde{\rho}}$, we can use
Lemma A.5 and get

$\displaystyle({\mathcal{A}}^{{\varepsilon}}_{{\alpha\overline{\beta}}})\geq\frac{\rho-{\widetilde{\rho}}}{C}\left[\delta_{{\alpha\beta}}+\left(\frac{\Psi^{\varepsilon}_{\alpha}}{\Psi^{\varepsilon}_{\tau}}\right)\overline{\left(\frac{\Psi^{\varepsilon}_{\beta}}{\Psi^{\varepsilon}_{\tau}}\right)}\
\right],\ \ \ \ \ \ \ \ \text{ for any small enough ${\varepsilon}>0$,}$
(A.70)

for a constant $C>0$ independent of ${\varepsilon}$; using Lemma A.9, we also
get

$\displaystyle{\mathcal{B}}^{\varepsilon}\overline{({\mathcal{A}}^{{\varepsilon}})^{-1}}\
\overline{{\mathcal{B}}^{\varepsilon}}({\mathcal{A}}^{{\varepsilon}})^{-1}\leq
1-\frac{\rho-{\widetilde{\rho}}}{C_{K}},\ \ \ \ \ \ \ \ \text{ for any small
enough ${\varepsilon}>0$,}$ (A.71)

for a constant $C_{K}$, also independent of ${\varepsilon}$. Since
$\Psi_{\tau}^{\varepsilon}\neq 0$, for ${\varepsilon}$ small enough, condition
(A.64) implies

$\displaystyle{\mathcal{A}}^{{\varepsilon}}\rightarrow{\mathcal{A}}^{0},\ \ \
\ \ {\mathcal{B}}^{{\varepsilon}}\rightarrow{\mathcal{B}}^{0}.$ (A.72)

Therefore, we have

$\displaystyle({\mathcal{A}}^{0}_{{\alpha\overline{\beta}}})\geq\frac{\rho-{\widetilde{\rho}}}{C}\left[\delta_{{\alpha\beta}}\right],$
(A.73) $\displaystyle{\mathcal{B}}^{0}\overline{({\mathcal{A}}^{0})^{-1}}\
\overline{{\mathcal{B}}^{0}}({\mathcal{A}}^{0})^{-1}\leq
1-\frac{\rho-{\widetilde{\rho}}}{C_{K}};$ (A.74)

then, using Lemma A.6, we get $\Psi^{0}$ is strongly
$\mathfrak{q}\mathbb{C}$-convex.

The argument above is valid for any $q\in{\mathscr{Q}}_{\widetilde{\rho}}$, so
the robustness of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than
${\widetilde{\rho}}$. ∎

The following Lemmas show the relation between the
$\mathfrak{q}\mathbb{C}$-convexity of a function and the
$\mathbb{C}$-convexity of its level sets.

######  Lemma A.11.

Let $\Omega$ be a strongly $\mathbb{C}$-convex domain with a smooth boundary
and with the modulus of $\mathbb{C}$-convexity greater than $\mu$. Suppose
that $F\in C^{\infty}({\overline{\Omega}})$ satisfies

$\displaystyle F=0\ \ \ \ \ \ \ \ \text{ on }\Gamma,$ (A.75) $\displaystyle
F_{{\mathbb{n}}}>\sigma\ \ \ \ \ \ \ \ \text{ on }\Gamma,$ (A.76)

where $\Gamma$ is an open set of $\partial\Omega$, and ${\mathbb{n}}$ is the
exterior unit normal vector of $\partial\Omega$. Then the modulus of
$\mathfrak{q}\mathbb{C}$-convexity of $F$ is greater than
$\min\\{1,\sqrt{2}\mu\\}\sigma$ on $\Gamma$.

###### Proof of Lemma A.11.

The proof is straightforward, we only need to use the implicit function
theorem.

Let ${\bf p}$ be a point on $\Gamma$. We choose coordinates
$(z^{\alpha},\tau)$, with $\tau=t+\sqrt{-1}s,$ so that ${\bf p}=0$ and
$\\{t=0\\}$ is the tangential plane of $\partial\Omega$ at ${\bf p}$. Suppose
that around ${\bf p}$

$\displaystyle\Omega=\left\\{t>\rho(z^{\alpha},s)\right\\}.$ (A.77)

Then we have $\rho(0)=0$, $\nabla\rho=0$ and
$\partial_{t}=-\sqrt{2}{{\mathbb{n}}}$, at ${\bf p}$. $F=0$ on $\Gamma$
implies

$\displaystyle F(z^{\alpha},\rho(z^{\alpha},s)+\sqrt{-1}s)=0.$ (A.78)

The implicit function theorem gives

$\displaystyle\begin{split}&\text{the second order Taylor expansion of
}F|_{\\{\tau=0\\}}\text{ at ${\bf p}$}\\\
=&\sqrt{2}|F_{\mathbb{n}}|\cdot\text{the second order Taylor expansion of
}\rho|_{\\{s=0\\}}\text{ at $0$}.\end{split}$ (A.79)

Therefore, the modulus of $\mathbb{C}$-convexity of $\Omega$ is greater than
$\mu$ implies the modulus of convexity of the second order Taylor expansion of
$F|_{\\{\tau=0\\}}\text{ at ${\bf p}$}$ is greater than
$\sqrt{2}|F_{\mathbb{n}}|\cdot\mu.$ Then the result follows. ∎

######  Lemma A.12.

Let $\Phi$ be a smooth function in
$\mathcal{R}=\Omega_{1}\backslash\overline{\Omega_{0}}$, where $\Omega_{0}$
and $\Omega_{1}$ are strongly $\mathbb{C}$-convex domains with smooth
boundaries and $\overline{\Omega_{0}}\subset\Omega_{1}$. Suppose that $\Phi$
satisfies

$\displaystyle\Phi=0\ \ \ \ \ \ \ \ \text{ on }\partial\Omega_{0},$ (A.80)
$\displaystyle\Phi=1\ \ \ \ \ \ \ \ \text{ on }\partial\Omega_{1},$ (A.81)

and the modulus of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater
than $\mu>0$. Then

$\displaystyle\Omega_{t}=\left\\{\Phi<t\right\\}\cup\overline{\Omega_{0}}$
(A.82)

are all strongly $\mathbb{C}$-convex with the modulus of
$\mathbb{C}$-convexity greater than

$\displaystyle\frac{\mu}{\max_{\overline{\mathcal{R}}}|\nabla\Phi|}.$ (A.83)

###### Proof of Lemma A.12.

First, the condition that $\Phi$ is strongly $\mathfrak{q}\mathbb{C}$-convex
implies that $\nabla\Phi\neq 0$ anywhere in $\mathcal{R}$; therefore,

$\displaystyle 0<\Phi<1\ \ \ \ \ \ \ \ \text{ in }\mathcal{R},$ (A.84)

and all the level sets of $\Phi$ are smooth.

Let ${\bf p}$ be a point in $\mathcal{R}$. We choose coordinate
$(z^{\alpha},\tau)$, with $\tau=t+\sqrt{-1}s$, so that the metric is
$dz^{i}\otimes\overline{dz^{i}}$ and that at ${\bf p}$

$\displaystyle\Phi_{\alpha}=0,\ \ \ \Phi_{s}=0,\ \ \ \Phi_{t}=-|\nabla\Phi|.$
(A.85)

We also assume that ${\bf p}=0.$ Then, around ${\bf p}$, the level set
$\\{\Phi=\Phi({\bf p})\\}$ is the graph of a function $\rho$:

$\displaystyle\\{\Phi=\Phi({\bf
p})\\}=\left\\{(z^{\alpha},\rho(z^{\alpha},s)+\sqrt{-1}s)\right\\},$ (A.86)

with $\rho(0)=0$. Similar to the proof of Lemma A.11, we have

$\displaystyle\begin{split}&\text{the second order Taylor expansion of
}\Phi|_{\\{\tau=0\\}}\text{ at ${\bf p}$}\\\ =&\Phi({\bf p})+|\nabla\Phi({\bf
p})|\cdot\text{the second order Taylor expansion of }\rho|_{\\{s=0\\}}\text{
at $0$}.\end{split}$ (A.87)

Therefore, the modulus of $\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is than
$\mu$ implies the modulus of convexity of the second order Taylor expansion of
$\rho|_{\\{s=0\\}}$ at $0$ is greater than $\frac{\mu}{|\nabla\Phi({\bf
p})|}$. Then we let ${\bf p}$ be an arbitrary point in
$\overline{\mathcal{R}}$ and get the result. ∎

##  Appendix B Construction of a Subsolution and Gradient Estimates

In this appendix, we will construct a subsolution to Problem 1.12 (Lemma B.1).
The existence of a subsolution is the key to the solvability of Problem 1.12
and Problem 1.6. In addition, we will show for the solution
${\Phi^{\varepsilon}}$ to Problem 1.12
$|\nabla\Phi^{\varepsilon}|_{\mathcal{G}}$ has a positive lower bound on
$\partial\mathcal{R}$ (Lemma B.3).

######  Lemma B.1.

Suppose $\mathcal{R}=\Omega_{1}\backslash\overline{\Omega_{0}}$, with
$\Omega_{0}$ and $\Omega_{1}$ satisfying the requirements of Problem 1.12.
Then for $\sigma>0$ small enough, we can find $\Psi\in
C^{\infty}(\overline{\mathcal{R}})$ satisfying

$\displaystyle\sqrt{-1}\partial\overline{\partial}\Psi\geq\sigma{\omega_{\mathcal{G}}},\
\ \ \ \ \ \ \ $ $\displaystyle\text{ in }\mathcal{R},$ (B.1)
$\displaystyle\Psi=0,\ \ \ \ \ \ \ \ $ $\displaystyle\text{ on
}\partial\Omega_{0},$ (B.2) $\displaystyle\Psi=1,\ \ \ \ \ \ \ \ $
$\displaystyle\text{ on }\partial\Omega_{1},$ (B.3)

and

$\displaystyle\Psi_{\mathbb{n}}\geq\sigma\ \ \ \ \ \ \ \ \ \text{ on
}\partial\Omega_{0},$ (B.4)

here ${\mathbb{n}}$ is the exterior unit normal vector on
$\partial\Omega_{0}$.

###### Proof of Lemma B.1.

Let $V_{\Omega_{0}}$ be the pluricomplex Green’s function of $\Omega_{0}$ with
pole at infinity. Then

$\displaystyle V_{\Omega_{0}}(z)=\sup\left\\{u(z)\ |\ u\in{\mathscr{L}},u\leq
0\text{ in }\Omega_{0}\right\\},$ (B.5)

where ${\mathscr{L}}$ is the family of all plurisubharmonic functions on
$\mathbb{C}^{n}$ with

$\displaystyle u(z)-\log|z|\leq O(1)\text{ as }|z|\rightarrow\infty.$ (B.6)

Using the method of [BT76], we know $V_{\Omega_{0}}$ is continuous in
$\mathbb{C}^{n}$. Actually, using the theory of Lempert, we can even show
$V_{\Omega_{0}}$ is smooth in $\Omega_{0}^{c}$ (see Theorem 5.1 of [L85] and
the first remark following the proof of Lemma 5.3); however, we only need to
use the continuity. We also need to show

$\displaystyle V_{\Omega_{0}}>0\text{ \ \ \ \ \ \ \ \ in
}(\overline{\Omega_{0}})^{c}.$ (B.7)

This depends on the fact that $\Omega_{0}$ is polynomially convex, which is
proved in the following lemma:

######  Lemma B.2.

Suppose $E$ is a bounded $\mathbb{C}$-convex domain with a smooth boundary.
Then $E$ is polynomially convex.

###### Proof of Lemma B.2.

We only need to show that any complex hyperplane in $E^{c}$ can be
continuously moved to infinity; then, we can apply Prop 2.1.9 of [APS04] (also
see Remark 2.1.10).

Assume $0\in E$, then let

$\displaystyle r_{0}=\sup\left\\{r\ |\ B_{r}^{c}\bigcap
E\neq\emptyset\right\\},$ (B.8)

where $B_{r}=\\{|z|\leq r\\}$. Since $E$ is bounded,
$r_{0}\leq\text{diameter}(E)$; since $\partial E$ is smooth, $\partial
B_{r_{0}}$ should contact with $\partial E$. Let ${\bf p}\in B_{r_{0}}\cap
E^{c}$.

With the preparation above, we can take three steps to move a complex
hyperplane ${\mathfrak{l}}\subset E^{c}$ to infinity.

Step 1. We continuously move ${\mathfrak{l}}$ towards $0$ until it contacts
with $\partial E$ at a point ${\bf a}$. Here ${\bf a}$ may not be unique.

Step 2. Let $\gamma=\left\\{\gamma(t)\ |\ t\in[0,1]\right\\}$ be a smooth
curve on $\partial E$ connecting ${\bf p}$ with ${\bf a}$. Then

$\displaystyle\left\\{T^{\mathbb{C}}_{\partial
E,\gamma(t)}\right\\}_{t\in[0,1]}$ (B.9)

is a family of complex hyperplanes connecting $T^{\mathbb{C}}_{\partial E,{\bf
a}}$ and $T^{\mathbb{C}}_{\partial E,{\bf p}}$. Here $T^{\mathbb{C}}_{\partial
E,{\bf z}}$ denotes the complex tangent plane of $\partial E$ at a point ${\bf
z}\in\partial E$.

Step 3. The complex tangent plane of $\partial E$ at ${\bf p}$ can be moved to
infinity since it’s also the complex tangent plane of $\partial B_{r_{0}}$ (we
have $B_{r_{0}}\supset\overline{E}$). ∎

Denote

$\displaystyle\Phi_{\Omega_{0}}(z)=\sup\left\\{|p(z)|^{\frac{1}{\deg(p)}}\
\big{|}\ p\in\mathcal{P}_{\Omega_{0}}\right\\}$ (B.10)

where $\mathcal{P}_{\Omega_{0}}$ is the space of complex polynomials $p$ on
$\mathbb{C}^{n}$ with $\deg(p)\geq 1$ and $|p|\leq 1$ on $\Omega_{0}$. This
function is called Siciak’s extremal function. Since $\Omega_{0}$ is
polynomially convex, for any $z\in\left(\overline{\Omega_{0}}\right)^{c}$ we
can find $p\in\mathcal{P}_{\Omega_{0}}$ so that $|p(z)|>1$; then, it follows
that $\Phi_{\Omega_{0}}(z)>1$.

According to the theory of Zajarjuta-Siciak ([S81], Theorem 5.1.7 of [K91]),

$\displaystyle V_{\Omega_{0}}=\log\Phi_{\Omega_{0}};$ (B.11)

we have $V_{\Omega_{0}}(z)>0$, for $z\notin\overline{\Omega_{0}}$.

In the following, applying the method of [G02], we use $V_{\Omega_{0}}$ and
the distance functions to $\partial\Omega_{0}$ and $\partial\Omega_{1}$ to
construct a strictly plurisubharmonic function in $\mathcal{R}$.

Let

$\displaystyle F(z)$
$\displaystyle=c\left(V_{\Omega_{0}}(z)-c+c^{2}|z|^{2}\right),$ (B.12)
$\displaystyle f_{1}(z)$
$\displaystyle=\frac{1}{c}(e^{-\frac{1}{c}{\text{dist}}(z,\partial\Omega_{1})}-1)+1,$
(B.13) $\displaystyle f_{0}(z)$
$\displaystyle=e^{-\frac{1}{c^{2}}}\left(e^{\frac{1}{c}{\text{dist}}(z,\partial\Omega_{0})}-1\right),$
(B.14)

where $c$ is a small constant to be determined. Since $V_{\Omega_{0}}$ is
plurisubharmonic, $F$ is strongly plurisubharmonic. Let $\Omega_{1}^{r}$ and
$\Omega_{0}^{r}$ be small neighborhoods of $\partial\Omega_{0}$ and
$\partial\Omega_{1}$:

$\displaystyle\Omega_{0}^{r}=\left\\{z\in\mathcal{R}\ |\
{\text{dist}}(z,\Omega_{0})<r,\ z\in\mathcal{R}\right\\},$ (B.15)
$\displaystyle\Omega_{1}^{r}=\left\\{z\in\mathcal{R}\ |\
{\text{dist}}(z,\Omega_{1}^{c})<r,\ z\in\mathcal{R}\right\\}.$ (B.16)

Using Theorem 3.18 of [L11], we know when $c$ and $r$ are small enough $f_{1}$
and $f_{0}$ are strongly plurisubharmonic in $\Omega_{1}^{r}$ and
$\Omega_{0}^{r}$, respectively.

As illustrated by Figure 4, we choose $c$ small enough so that

$\displaystyle F<1=f_{1}\ \ \ \ \ \ \ \ $ $\displaystyle\text{ on
}\partial\Omega_{1},$ (B.17) $\displaystyle F>0>f_{1}\ \ \ \ \ \ \ \ $
$\displaystyle\text{ on }\partial\Omega_{1}^{r}\backslash\partial\Omega_{1},$
(B.18) $\displaystyle F>f_{0}\ \ \ \ \ \ \ \ $ $\displaystyle\text{ on
}\partial\Omega_{0}^{r}\backslash\partial\Omega_{0},$ (B.19) $\displaystyle
F<0=f_{0}\ \ \ \ \ \ \ \ $ $\displaystyle\text{ on }\partial\Omega_{0}.$
(B.20)

∎

Figure 4: Construction of Subsolution

(B.17) can be satisfied since $V_{\Omega_{0}}$ is bounded on a compact set and
$f_{1}=1$ on $\partial\Omega_{1}$; (B.18) can be satisfied since
$V_{\Omega_{0}}>0$ on $\partial\Omega_{1}^{r}\backslash\partial\Omega_{1}$ and
$e^{-\frac{1}{c}{\text{dist}}(z,\partial\Omega_{1})}-1$ becomes negative on
$\partial\Omega_{1}^{r}\cap\mathcal{R}$ when $c$ is small enough; (B.19) can
be satisfied because $V_{\Omega_{0}}>0$ on
$\partial\Omega_{0}^{r}\backslash\partial\Omega_{0}$ and $f_{0}\leq
e^{-\frac{1}{c}}$ providing $c$ small enough; (B.20) can be satisfied since
$V_{\Omega_{0}}$ and $f_{0}$ both equal to $0$ on $\partial\Omega_{0}$.

As we mentioned before, according to the theory of Lempert, $V_{\Omega_{0}}$
is smooth in $\Omega_{0}^{c}$ and, therefore, $F$ is smooth in
$\Omega_{0}^{c}$; however, if we don’t want to depend on the theory of Lempert
and pretend to be unaware of the smoothness of $F$, then we need to mollify
$F$. Note that $V_{\Omega_{0}}$ is a plurisubharmonic function on
$\mathbb{C}^{n}$ with $V_{\Omega_{0}}=0$ in $\Omega_{0}$, so $F$ is strictly
plurisubharmonic on $\mathbb{C}^{n}$ in weak sense. Let $\eta=\eta(|y|)$ be a
cut-off function on $\mathbb{C}^{n}$, with support in $B_{1}$ and
$\int_{\mathbb{C}^{n}}\eta=1$; let

$\displaystyle
F_{\rho}(x)=\int_{\mathbb{C}^{n}}F(x-y)\eta(\frac{|y|}{\rho})\frac{1}{\rho^{2n}}dy.$
(B.21)

Then $F_{\rho}\rightarrow F$ in $C^{0}$ norm as $\rho\rightarrow 0$. When
$\rho$ is small enough, condition (B.17)-(B.20) are still valid after
replacing $F$ by $F_{\rho}$.

Let $h$ be the function constructed in Lemma 3.1 of [G02], with $\delta$ small
enough; then, we let

$\displaystyle\Psi=\left\\{\begin{array}[]{ccc}\frac{F_{\rho}+f_{1}}{2}+h(\frac{F_{\rho}-f_{1}}{2})&\text{
in }\Omega_{1}^{r},\\\
\frac{F_{\rho}+f_{0}}{2}+h(\frac{F_{\rho}-f_{0}}{2})&\text{ in
}\Omega_{0}^{r},\\\ F_{\rho}&\text{ in
}\mathcal{R}\backslash{(\Omega_{1}^{r}\cup\Omega_{0}^{r})}.\end{array}\right.$
(B.25)

According to the computation of Lemma 3.2 of [G02] $\Psi$ is smooth and
strictly plurisubharmonic. Therefore for $\sigma$ small enough, (B.1) is
valid. We also note that $\Psi=f_{0}$ in a small neighborhood of
$\partial\Omega_{0}$, proving $\delta$ is small enough so $\partial_{n}\Psi>0$
on $\partial\Omega_{0}$; then we can choose $\sigma$ small so that
$(\ref{628B4})$ is satisfied.

Because of (B.1), we can choose ${\varepsilon}_{0}$ small enough so that

$\displaystyle\sigma^{n}{\omega_{\mathcal{G}}}^{n}\geq{\varepsilon}_{0}\left(\sqrt{-1}\partial\overline{\partial}\Psi\right)^{n-1}\wedge{\omega_{\mathcal{G}}}$
(B.26)

Then $\Psi$ is a subsolution to Problem 1.12 when
${\varepsilon}\leq{\varepsilon}_{0}$.

Suppose $\Phi^{\varepsilon}$ is a solution to Problem 1.12. Then
$|\nabla\Phi^{\varepsilon}|_{\mathcal{G}}$ has a positive lower bound on
$\partial\Omega_{0}$ since $\Phi^{\varepsilon}\geq\Psi$ in $\mathcal{R}$ and
$\Phi^{\varepsilon}=\Psi$ on $\partial\Omega_{0}$. The estimate of
$|\nabla\Phi^{\varepsilon}|_{\mathcal{G}}$ on $\partial\Omega_{1}$ is very
simple. Let $H$ be the harmonic function in $\mathcal{R}$ with $H=0$ on
$\partial\Omega_{0}$ and $H=1$ on $\partial\Omega_{1}$. Then the maximum
principle implies

$\displaystyle H\geq\Phi^{\varepsilon}\ \ \ \ \ \ \ \ \text{ in
}{\mathcal{R}}$ (B.27)

since $\Phi^{\varepsilon}$ is subharmonic in $\mathcal{R}$; the Hopf maximum
principle implies $H_{{\mathbb{n}}}>0$ on $\partial\Omega_{1}$, where
${\mathbb{n}}$ is the exterior unit normal vector on $\partial\Omega_{1}$.
Therefore, $\Phi_{\mathbb{n}}^{\varepsilon}>0$ on $\partial\Omega_{1}$.

Summing up, we have the following positive lower bound estimate for
$|\nabla\Phi^{\varepsilon}|_{\mathcal{G}}$:

######  Lemma B.3.

Suppose $\Omega_{0}$ and $\Omega_{1}$ are domains satisfying conditions of
Problem 1.12 . Then there is a positive constant $C_{\partial\mathcal{R}}^{g}$
so that when ${\varepsilon}$ is small enough, the solution
$\Phi^{\varepsilon}$ to Problem 1.12 satisfies

$\displaystyle|\nabla{\Phi^{\varepsilon}}|_{\mathcal{G}}\geq
C_{\partial\mathcal{R}}^{g}\ \ \ \ \ \ \ \ \text{ on }\partial\mathcal{R}.$
(B.28)

Now, for a fixed $\mathbb{C}$-convex ring, we have already constructed the
subsolution and derived the desired gradient estimate; however, when
considering a family of domains $\\{\mathcal{R}^{t}\\}$, we need to show the
constants $C_{\partial\mathcal{R}^{t}}^{g}$ can be chosen so that they have a
uniform positive lower bound and that the $C^{3}$ norms of $\Psi^{t}$, the
subsolutions in $\mathcal{R}^{t}$, have a uniform upper bound. So, we need the
following lemma:

######  Lemma B.4.

Let $\\{\Omega_{0}^{t}\\}_{t=0}^{1}$ and $\\{\Omega_{1}^{t}\\}_{t=0}^{1}$ be
two families of strongly $\mathbb{C}$-convex domains with smooth boundaries
and satisfying $\overline{\Omega_{0}^{t}}\subset\Omega_{1}^{t}$ and condition
3 of Lemma D.1. Then we can construct a family of smooth subsolutions $F^{t}$
to Problem 1.12 with $\Omega_{0}=\Omega_{0}^{t}$ and
$\Omega_{1}=\Omega_{1}^{t}$; more precisely, the following conditions are
satisfied:

$\displaystyle\sqrt{-1}\partial\overline{\partial}F^{t}\geq\sigma{\omega_{\mathcal{G}}}\
\ \ \ \ \ \ \ $ $\displaystyle\text{ in
}\Omega_{1}^{t}\backslash\overline{\Omega_{0}^{t}},$ (B.29)
$\displaystyle\partial_{{\mathbb{n}}}F^{t}\geq\sigma\ \ \ \ \ \ \ \ $
$\displaystyle\text{ on }\partial\Omega_{0}^{t},$ (B.30)
$\displaystyle|F^{t}|_{C^{3}}\leq\frac{1}{\sigma},$ (B.31)

where $\sigma$ is a positive constant. Let ${\Phi_{\varepsilon}^{t}}$ be the
solution to Problem 1.12 with $\Omega_{0}=\Omega_{0}^{t}$ and
$\Omega_{1}=\Omega_{1}^{t}$; we have

$\displaystyle|\nabla{\Phi^{t}_{\varepsilon}}|\geq C^{g}\ \ \ \ \ \ \ \ \text{
on }\partial\Omega_{0}^{t}\cup\partial\Omega^{t}_{1},$ (B.32)

when ${\varepsilon}$ is small enough, where $C^{g}$ is a positive constant.

###### Proof.

Using Lemma B.1, for each $s\in[0,1]$ we can find $\Psi^{s}$ and
${{\epsilon}_{s}}$ so that

$\displaystyle\sqrt{-1}\partial\overline{\partial}\Psi^{s}\geq{{\epsilon}_{s}}{\omega_{\mathcal{G}}}\
\ \ \ \ \ \ \ \text{ in }\Omega_{1}^{s}\backslash\overline{\Omega_{0}^{s}},$
(B.33)

and

$\displaystyle 0<\Psi^{s}<1,\ \ \ \ \ \ \ \ \text{ in
}\Omega_{1}^{s}\backslash\overline{\Omega_{0}^{s}}.$ (B.34)

This implies $\Psi^{s}$ are subsolutions to the following Problem B.5 when
$t=s$; therefore, using the method of [G02], the following problem is solvable
when $t=s$:

######  Problem B.5.

Find $M^{t}$ satisfying

$\displaystyle\left[\sqrt{-1}\partial\overline{\partial}M^{t}\right]^{n}=\frac{({{\epsilon}_{s}})^{n}}{2}{\omega_{\mathcal{G}}}^{n}\
\ \ \ \ \ \ \ $ $\displaystyle\text{ in
}\Omega_{1}^{t}\backslash\overline{\Omega_{0}^{t}},$ (B.35) $\displaystyle
M^{t}=1\ \ \ \ \ \ \ \ $ $\displaystyle\text{ on }\partial\Omega_{1}^{t},$
(B.36) $\displaystyle M^{t}=0\ \ \ \ \ \ \ \ $ $\displaystyle\text{ on
}\partial\Omega_{0}^{t}.$ (B.37)

Then we can apply the implicit function theorem and solve the Dirichlet
problem above, for $t,$ with $|t-s|<\delta_{s}$, where $\delta_{s}$ is a small
number. We note that $\delta_{s}$ may depend on ${{\epsilon}_{s}}$. We can
choose $\delta_{s}$ and ${{\epsilon}_{s}}$ small enough so that

$\displaystyle\partial_{\mathbb{n}}M^{t}\geq\frac{{\epsilon}_{s}}{2}\ \ \ \ \
\ \ \ \text{ on }\partial\Omega_{0}^{s},$ (B.38)

and $|M^{t}|_{C^{3}}\leq\frac{2}{{{\epsilon}_{s}}}$.

Then we use the compactness of $[0,1]$: We have

$\displaystyle\bigcup_{s\in[0,1]}\left\\{|t-s|<\delta_{s}\right\\}\supset[0,1],$
(B.39)

so we can find finite $s_{k},$ for $k\in\\{1,\ ...\ ,N\\}$, such that

$\displaystyle\bigcup_{k=1}^{n}\left\\{|t-s_{k}|<\delta_{s_{k}}\right\\}\supset[0,1].$
(B.40)

Therefore, the solution $M^{t}$ to Problem B.5, with
${\epsilon}_{s}={{\epsilon}_{s_{k}}}$ and $|t-s_{k}|<\delta_{s_{k}}$, are
subsolutions of Problem 1.12 satisfying the conditions of Lemma B.4, with

$\displaystyle\sigma=\min\left\\{\frac{{{\epsilon}_{s_{k}}}}{2}\right\\}_{k=1}^{N}.$
(B.41)

The estimate for $|\nabla{\Phi^{t}_{\varepsilon}}|$ follows easily. ∎

##  Appendix C $C^{1,1}$ Estimate Independent of ${\varepsilon}$

In this section we provide a $C^{1,1}$ estimate for solutions of Problem 1.12
which is independent of ${\varepsilon}$, providing ${\varepsilon}$ small
enough. All the ideas are taken from [G98].

######  Proposition C.1.

Suppose that $\Phi$ is a smooth solution to Problem 1.12 with ${\varepsilon}$
small enough. Then $|\nabla\Phi|_{\mathcal{G}}$ and
$|D^{2}\Phi|_{\mathcal{G}}$ have upper bounds which depend on the positive
lower bound of $|\Phi_{n}|$ on $\partial\mathcal{R}$ and the
plurisubharmonicity and the $C^{3}$ norm of $\Psi$ where $\Psi$ is the
subsolution constructed in Section B; in particular, the estimates are
independent of ${\varepsilon}$, providing ${\varepsilon}$ small enough.

###### Proof.

In the following, we only need to do unitary coordinate transformations, so we
let $\mathcal{G}_{i\overline{j}}=\delta_{ij}$. Let $U$ be the solution to the
Laplace equation

$\displaystyle U_{i{\overline{i}}}=0,$ (C.1)

with the same boundary values as those of $\Phi$. Then, obviously, we have

$\displaystyle\Phi\leq U\ \ \ \ \ \ \text{ in }\mathcal{R}$ (C.2)

since $\Phi$ is plurisubharmonic and, therefore, subharmonic. Let $\Psi$ be
the strictly plurisubharmonic function constructed in Section B; suppose that

$\displaystyle\sqrt{-1}\partial\overline{\partial}\Psi\geq{\varepsilon}_{0}{\omega_{\mathcal{G}}}.$
(C.3)

Then $\Psi\leq\Phi$, providing $\epsilon$ small enough. So, we can control the
gradient of $\Phi$ on the boundary since $\Psi=\Phi=U$ on
$\partial\mathcal{R}$ and $\Psi\leq\Phi\leq U$ in $\Omega$.

The interior estimate for gradients can be reduced to the boundary estimate.
Let $\mathcal{G}=(\mathcal{G}_{{i\overline{j}}})$ and
$\mathcal{H}=(\Phi_{{i\overline{j}}})$. Then the equation (1.29) becomes

$\displaystyle\mathcal{H}^{-1}=\frac{1}{{\varepsilon}}.$ (C.4)

Let $X$ be a constant vector field on $\mathbb{C}^{n}$. We apply
$\partial_{X}$ to the equation above and get

$\displaystyle\mathcal{H}^{-1}\mathcal{H}_{X}\mathcal{H}^{-1}=0;$ (C.5)

using operator $L$ (introduced in Section 2.2), this is equivalent to

$\displaystyle L^{{i\overline{j}}}(\Phi_{X})_{{i\overline{j}}}=0.$ (C.6)

Since $L$ is an elliptic operator, an estimate on the $C^{1}$ norm of $\Phi$
in $\overline{\mathcal{R}}$ follows from the usual maximum principle.

We apply $\partial_{X}$ to (C.5) and get

$\displaystyle\mathcal{H}^{-1}\mathcal{H}_{XX}\mathcal{H}^{-1}=2\mathcal{H}^{-1}\mathcal{H}_{X}\mathcal{H}^{-1}\mathcal{H}_{X}\mathcal{H}^{-1},$
(C.7)

which says

$\displaystyle
L^{{i\overline{j}}}(\Phi_{XX})_{{i\overline{j}}}=2(\partial_{X}\Phi)_{i{\overline{q}}}\Phi^{p{\overline{q}}}(\partial_{X}\Phi)_{p{\overline{j}}}L^{{i\overline{j}}}\geq
0.$ (C.8)

Since $\Phi$ is a plurisubharmonic function, we have

$\displaystyle\Phi_{XX}+\Phi_{JXJX}\geq 0.$ (C.9)

So the estimate of second order derivatives of $\Phi$ can be reduced to the
boundary.

Suppose that $0$ is a boundary point and that around $0$ the boundary of
$\mathcal{R}$ is locally the graph of a function: for a small $r>0$

$\displaystyle\partial\mathcal{R}\cap
Q_{r}=\left\\{(z_{\alpha},z_{n}=t+\sqrt{-1}s)|t=\rho(z_{\alpha},s)\right\\}.$
(C.10)

Here $Q_{r}$ is a cube

$\displaystyle Q_{r}=\left\\{|z_{\alpha}|^{2}+|s|^{2}\leq r^{2},|t|\leq
r\right\\},$ (C.11)

and $\rho(0)=0,\nabla\rho(0)=0.$ We recall that Greek indices run from $1$ to
$n-1$. In the following, we will estimate second order derivatives of $\Phi$
at $0$.

Let $Y$ be a constant vector field in $\mathbb{C}^{n}$ parallel to
$\\{t=0\\}$; let

$\displaystyle\mathcal{Y}(z_{\alpha},z_{n})=Y+\partial_{Y}\rho(z_{\alpha},s)\partial_{t}.$
(C.12)

Then the boundary condition $\Phi|_{\partial\mathcal{R}}=\text{constant}$
implies

$\displaystyle\partial_{\mathcal{Y}}\Phi=\partial_{\mathcal{Y}\mathcal{Y}}\Phi=0\
\ \ \ \ \ \text{ on }\partial\mathcal{R}.$ (C.13)

Therefore,

$\displaystyle\Phi_{YY}+\rho_{YY}\Phi_{t}=0\ \ \ \ \ \ \text{ at }0;$ (C.14)

thus, $\Phi_{\alpha{\overline{\beta}}},\ \Phi_{\alpha\beta},\ \Phi_{\alpha
s},\ \Phi_{ss}$ can all be controlled by $|\Phi|_{C^{1}}$ and
$|\rho|_{C^{2}}$, and it remains to estimate $\Phi_{Yt}$ and $\Phi_{tt}$.

The equation (1.29) implies the following relation

$\displaystyle\Phi_{{n\overline{n}}}\cdot\det(\Phi_{\alpha{\overline{\beta}}})={\varepsilon}\Phi_{{n\overline{n}}}\cdot\sigma_{n-2}(\Phi_{\alpha{\overline{\beta}}})+\text{Terms
Consisting of }\Phi_{\alpha{\overline{\beta}}}\text{ and
}\Phi_{\alpha{\overline{n}}}.$ (C.15)

Because of (C.14), $\det(\Phi_{\alpha{\overline{\beta}}})$ has a lower bound,
depending on the $\mathbb{C}$-convexity of the boundary and the lower bound of
$|\Phi_{t}|$, and $\sigma_{n-2}(\Phi_{\alpha{\overline{\beta}}})$ has an upper
bound, depending on the $C^{2}$ regularity of the boundary and the gradient
estimate of $\Phi$. Therefore, when ${\varepsilon}$ is small enough (C.14)
reduces the estimate for $\Phi_{{n\overline{n}}}$ to the estimates for
$\Phi_{\alpha{\overline{n}}}$ and $\Phi_{\alpha{\overline{\beta}}}$. Since
$4\Phi_{{n\overline{n}}}=\Phi_{tt}+\Phi_{ss}$ and $\Phi_{ss}$ has been
estimated, it only remains to estimate $\Phi_{tY}$.

The estimate of $\Phi_{tY}$ depends on the construction of a barrier function
in $Q_{r}\cap\mathcal{G}$. Let

$\displaystyle h=-\Phi_{s}^{2}+A(s^{2}+\sum_{\alpha}|z_{\alpha}|^{2})-B\Psi.$
(C.16)

We will show, for properly chosen $A$ and $B$,

$\displaystyle L^{{i\overline{j}}}h_{{i\overline{j}}}\leq
L^{{i\overline{j}}}(\Phi_{\mathcal{Y}})_{{i\overline{j}}}\ \ \ \ \ \ \text{ in
}Q_{r},$ (C.17)

and

$\displaystyle h\geq\Phi_{\mathcal{Y}}\ \ \ \ \ \ \text{ on }\partial Q_{r}.$
(C.18)

First, we compute $L^{{i\overline{j}}}(\Phi_{\mathcal{Y}})_{{i\overline{j}}}$
and bound it from below. Directly using the definition of $\mathcal{Y}$, we
have

$\displaystyle L^{{i\overline{j}}}(\Phi_{\mathcal{Y}})_{{i\overline{j}}}$
$\displaystyle=L^{{i\overline{j}}}(\Phi_{Y}+\rho_{Y}\Phi_{t})_{{i\overline{j}}}$
(C.19)
$\displaystyle=L^{{i\overline{j}}}\Phi_{Y{i\overline{j}}}+L^{{i\overline{j}}}\Phi_{t{i\overline{j}}}\rho_{Y}+L^{{i\overline{j}}}\rho_{iY}\Phi_{t{\overline{j}}}+L^{{i\overline{j}}}\rho_{{\overline{j}}Y}\Phi_{ti}+L^{{i\overline{j}}}\rho_{Y{i\overline{j}}}\Phi_{t}.$
(C.20)

In above, using (C.6), we know
$L^{{i\overline{j}}}\Phi_{Y{i\overline{j}}}=L^{{i\overline{j}}}\Phi_{t{i\overline{j}}}=0$.
Since $L$ is an operator with bounded coefficients, we have

$\displaystyle L^{{i\overline{j}}}\rho_{Y{i\overline{j}}}\Phi_{t}\leq
C(|\rho|_{C^{3}},|\Phi|_{C^{1}}).$ (C.21)

$\Phi_{t{\overline{j}}}$ is relatively harder to control; here, we couple it
with $\Phi_{s{\overline{j}}}$ and get

$\displaystyle
L^{{i\overline{j}}}\Phi_{t{\overline{j}}}\rho_{iY}=2L^{{i\overline{j}}}\Phi_{n{\overline{j}}}\rho_{iY}+L^{{i\overline{j}}}\sqrt{-1}\Phi_{s{\overline{j}}}\rho_{iY}.$
(C.22)

In above, $L^{{i\overline{j}}}\Phi_{n{\overline{j}}}$ can be well controlled.
Recall that
$L^{{i\overline{j}}}={\varepsilon}^{2}\Phi^{i{\overline{k}}}\mathcal{G}_{l{\overline{k}}}\Phi^{l{\overline{j}}}$;
so,

$\displaystyle
L^{{i\overline{j}}}\Phi_{n{\overline{j}}}={\varepsilon}^{2}\Phi^{i{\overline{q}}}\mathcal{G}_{n{\overline{q}}}.$
(C.23)

In this section $\mathcal{G}_{{i\overline{j}}}=\delta_{ij}$, so
$\Phi^{i{\overline{q}}}\mathcal{G}_{k{\overline{q}}}=\Phi^{i{\overline{k}}}$
is a non-negative matrix, with
$\Phi^{i{\overline{i}}}=\frac{1}{{\varepsilon}}$. Therefore,

$\displaystyle|L^{{i\overline{j}}}\Phi_{n{\overline{j}}}\rho_{iY}|\leq{\varepsilon}|\rho|_{C^{2}}.$
(C.24)

For $\Phi_{s{\overline{j}}}\rho_{iY}L^{{i\overline{j}}}$, we use Cauchy
inequality and get

$\displaystyle
L^{{i\overline{j}}}\Phi_{s{\overline{j}}}\rho_{iY}\geq-\frac{1}{2}L^{{i\overline{j}}}\Phi_{s{\overline{j}}}\Phi_{si}-\frac{1}{2}L^{{i\overline{j}}}\rho_{Y{\overline{j}}}\rho_{Yi}.$
(C.25)

So

$\displaystyle
L^{{i\overline{j}}}\Phi_{\mathcal{Y}{i\overline{j}}}\geq-C(|\rho|_{C^{3}},|\Phi|_{C^{1}})-L^{{i\overline{j}}}(\Phi_{s}^{2})_{{i\overline{j}}}.$
(C.26)

The computation for $h$ is straightforward:

$\displaystyle
L^{{i\overline{j}}}h_{{i\overline{j}}}\leq-L^{{i\overline{j}}}(\Phi_{s}^{2})_{{i\overline{j}}}+AC(\mathcal{G})-BL^{{i\overline{j}}}\Psi_{{i\overline{j}}}.$
(C.27)

Using (C.3), we know $L^{{i\overline{j}}}\Psi_{{i\overline{j}}}\geq
C(\mathcal{G}){\varepsilon}_{0}$. So, combining (C.26) and (C.27) we know when
$\frac{B}{A}$ is large enough (C.17) is valid. To make (C.18) valid, we only
need to make $A$ and $B$ both large enough, since $(\Phi_{s})^{2}$ is a second
order small quantity around $0$.

Summing up, we find a uniform estimate on $|\Phi|_{C^{2}}$ as ${\varepsilon}$
goes to zero. ∎

##  Appendix D Deformation

In this section, we explain how to deform a strongly $\mathbb{C}$-convex
domain with a smooth boundary to a Euclidean ball. We need to prove the
following Lemma:

######  Lemma D.1 (Shrinking to a Standard Ball).

Suppose that $E_{1}$ is a strongly $\mathbb{C}$-convex domain with a smooth
boundary. Then we can find a family of strongly $\mathbb{C}$-convex domains
$E_{t}$ with smooth boundaries and with $t\in[0,1]$, so that they satisfy the
following conditions:

  * 1.

$E_{0}$ is a Euclidean ball of small radius contained in $E_{1}$.

  * 2.

$\overline{E_{s}}\subset E_{t}$, for any $s<t$.

  * 3.

$E_{t}$ smoothly deforms from $E_{1}$ to $E_{0}$.

######  Remark D.2.

For the third requirement, we mean the following: Let ${\bf p}$ be a point on
$\partial\Omega_{s}$, for $s\in[0,1]$. Then around ${\bf p}$,
$\partial\Omega_{t}$ is the graph of a function $\rho_{t}$, for $t$ close
enough to $s$, and all the derivatives of $\rho_{t}$ are continuous functions
of $t$.

###### Proof of Lemma D.1.

The following idea is provided to the author by Lászó Lempert through a
private communication.

Given $E_{1}$, we can construct the pluricomplex Green’s function $G$ of
$E_{1}$ with a logarithmic singularity at $0\in E_{1}$ [L84]. Then $G$ is
smooth with $\nabla G\neq 0$ everywhere in $E_{1}\backslash\\{0\\}$, and all
the level sets of $G$ are smooth and strongly $\mathbb{C}$-convex, according
to Lemma 5.3 of [L85].

Let $\mu:{E_{1}^{\ast}}\rightarrow E_{1}$ be the blowing up of $E_{1}$ at $0$,
where

$\displaystyle{E_{1}^{\ast}}=\left\\{(z_{1},\ ...\ ,z_{n},\zeta)\ |\ (z_{1},\
...\ ,z_{n})\in E_{1},\ \zeta=[\zeta_{1},\ ...\
,\zeta_{n}]\in\mathbb{C}P^{n-1},\ \text{with
}z_{i}\zeta_{j}=\zeta_{i}z_{j}\right\\}$ (D.1)

and

$\displaystyle\mu(z_{1},\ ...\ ,z_{n},\zeta)=(z_{1},\ ...\ ,z_{n}).$ (D.2)

Denote $G^{\ast}=\mu^{\ast}G$; then it was proved in [L83] that
$e^{2G^{\ast}}$ is a smooth function on ${E_{1}^{\ast}}$. Here we assume that

$\displaystyle G(z)-\log(|z|)=O(1)\text{\ \ as }z\rightarrow 0.$ (D.3)

Moreover, we have that

$\displaystyle H(z,\zeta)\triangleq\frac{1}{|z|^{2}}e^{2G^{\ast}}$ (D.4)

is smooth on ${E_{1}^{\ast}}$ and $H>0$. Denote $\log(H)=2h$; then $h$ is
smooth on ${E_{1}^{\ast}}$, and

$\displaystyle G^{\ast}=h+\log(|z|).$ (D.5)

We will show

$\displaystyle\frac{1}{c}\left\\{G<\log c\right\\}$ (D.6)

converges to the indicatrix introduced in [L88]. Since

$\displaystyle\left\\{G<\log c\right\\}=\mu\left\\{G^{\ast}<\log c\right\\},$
(D.7)

we only need to study the level sets of $G^{\ast}$. We have

$\displaystyle\frac{1}{c}\\{G^{\ast}<\log(c)\\}=\frac{1}{c}\left\\{h(z,\zeta)+\log(|z|)<\log(c)\right\\}=\left\\{h(cz,\zeta)+\log(|z|)<0\right\\}.$
(D.8)

Here $\frac{1}{c}$ means rescaling in $z$ direction. So
$\frac{1}{c}\left\\{G^{\ast}<\log c\right\\}$ converges to

$\displaystyle\left\\{h(0,\zeta)+\log(|z|)<0\right\\}=\left\\{|z|^{2}\leq\frac{1}{H(0,\zeta)}\right\\}.$
(D.9)

Let $f_{\zeta}$ be an extremal disc so that

$\displaystyle f_{\zeta}(0)=0\text{ \ \ \ \ and \ \ \ \ }\pi
f_{\zeta}^{\prime}(0)=\zeta,$ (D.10)

where $\pi$ is the projection from $\mathbb{C}^{n}\backslash\\{0\\}$ to
$\mathbb{C}P^{n-1}$; then

$\displaystyle\frac{1}{H(0,\zeta)}=\lim_{z\rightarrow
0}\frac{|z|^{2}}{e^{2G^{\ast}}(z,\zeta)}=\lim_{\tau\rightarrow
0}\frac{|f_{\zeta}(\tau)|^{2}}{|\tau|^{2}}=|f_{\zeta}^{\prime}(0)|^{2}.$
(D.11)

So $\mu\left\\{|z|^{2}\leq\frac{1}{H(0,\zeta)}\right\\}$ is the indicatrix.

By slightly generalizing the argument on Page 47 of [L88], we know the
indicatrix for a strongly $\mathbb{C}$-convex domain is strongly convex. So
when $\theta$ is large enough, $\\{G\leq-\theta\\}$ is a strongly convex
domain. We choose $B_{r_{0}}\subset\\{G\leq-\theta\\}$, then
$\\{G\leq-\theta\\}$ is connected with $B_{r_{0}}$ by the following family of
strongly convex domains

$\displaystyle s\\{G\leq-\theta\\}+(1-s)B_{r_{0}}.$ (D.12)

Then the family

$\displaystyle
E^{t}=\left\\{\begin{array}[]{cc}\\{G\leq-2\theta(1-t)\\}&\text{ for
}t\in[\frac{1}{2},1],\\\ 2t\\{G\leq-\theta\\}+(1-2t)B_{r_{0}}&\text{ for
}t\in[0,\frac{1}{2}]\end{array}\right.$ (D.15)

satisfies the condition of Lemma D.1. ∎

##  Appendix E Simplified Proof in Dimension $2$

In this appendix, we provide an estimate for the robustness of
$\mathfrak{q}\mathbb{C}$-convexity in the case of $n=2$, under an unrealistic
assumption that solutions to Problem 1.6 are smooth. But the computations and
arguments in this case are much simpler and straightforward, and they
demonstrates the main idea of this paper. That is the purpose of this
appendix. We will prove the following apriori estimate:

######  Proposition E.1 (Apriori Estimate in $2$-dimension with the Smoothness
Assumption).

Suppose $\Phi$ is a smooth solution to Problem 1.6 in the case of $n=2$, and
$\Phi$ is strongly $\mathfrak{q}\mathbb{C}$-convex. Then the robustness of the
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\epsilon$, where
$\epsilon$ is a constant depending on the geometry of $\mathcal{R}$.

In Section E.1, we explain the foliation structure associated to a strongly
$\mathfrak{q}\mathbb{C}$-convex solution of Problem 1.6, which is not
available when we consider the perturbed problem, Problem 1.12. In the proof
of the above Proposition, we need to first estimate the gradient; this is the
content of Section E.2. We then introduce the auxiliary function
${\mathcal{Q}}_{\Phi}$ in Section E.3, which is a simplification of the
quantity ${\mathcal{K}}$ we constructed in general dimensions. In Section E.4
and E.5, we show $\frac{1}{1-{\mathcal{Q}}_{\Phi}}$ is subharmonic along each
leaf, which is essential for the convexity estimate. In Section E.6 we proved
Proposition E.1 using a continuity estimate.

In this appendix, we let
$\mathcal{G}=\mathcal{G}_{{i\overline{j}}}dz^{i}\otimes dz^{\overline{j}}$ be
a constant coefficient Hermitian metric on $\mathbb{C}^{2}$ and

$\displaystyle{\mathscr{Q}}_{\epsilon}=\left\\{q|\text{quadratic pluriharmonic
polynomials, }|\nabla q|_{\mathcal{G}}\leq\epsilon\right\\}.$ (E.1)

###  E.1 Foliation Structure

Given a solution $\Phi$ to the homogenous complex Monge-Ampère equation in
$\mathbb{C}^{n}$ with the rank of $(\Phi_{{i\overline{j}}})$ being $n-1$,
there is a foliation structure, formed by integrating kernels of
$(\Phi_{{i\overline{j}}})$. This has been discussed in many works including
[L81] [D02] [CFH20]. In our case (n=2), according to the apriori assumption,
given any interior point ${\bf p}$, we can choose a coordinate $(\tau,z)$ on
$\mathbb{C}^{2}$ so that $\Phi_{{z\overline{z}}}({\bf p})\neq 0$. Then we let

$\displaystyle
Z=\partial_{\tau}-\frac{\Phi_{\tau\overline{z}}}{\Phi_{z\overline{z}}}\partial_{z},$
(E.2)

and it satisfies

$\displaystyle[Z,\overline{Z}]=0.$ (E.3)

According to the complex Frobienius Theorem, we can find a holomorphic map
$\varphi(\zeta)=(\zeta,h(\zeta)),$ from a disc in $\mathbb{C}$ to
$\mathbb{C}^{2}$ such that

$\displaystyle\varphi(0)={\bf p},\text{ and
}\varphi_{\ast}(\partial_{\zeta})=\partial_{\tau}-\frac{\Phi_{\tau\overline{z}}}{\Phi_{z\overline{z}}}\partial_{z}.$
(E.4)

At a boundary point the map is from a half disc to $\mathbb{C}^{2}.$ The image
of $\varphi$ is called a local leaf (or, for simplicity, a leaf), and since
through any point we can find a leaf, these leaves form a foliation in
$\mathcal{R}$. In the following, we use $\zeta$ to denote the complex
coordinate on a leaf.

With $\Phi_{z\overline{z}}\neq 0$, the homogenous complex Monge-Ampère
equation for $\Phi$ is equivalent to

$\displaystyle\Phi_{\tau\overline{\tau}}-\frac{\Phi_{\tau\overline{z}}\Phi_{z\overline{\tau}}}{\Phi_{z\overline{z}}}=0$
(E.5)

and also

$\displaystyle\Phi_{\zeta{\overline{\zeta}}}=0.$ (E.6)

This says $\Phi$ is a harmonic function when restricted to a leaf.

We apply $\partial_{z}$ to (E.5) and get

$\displaystyle\Phi_{z{\tau\overline{\tau}}}-\frac{\Phi_{z\tau{\overline{z}}}\Phi_{z\overline{\tau}}}{\Phi_{z\overline{z}}}-\frac{\Phi_{\tau\overline{z}}\Phi_{zz{\overline{\tau}}}}{\Phi_{z\overline{z}}}+\frac{\Phi_{\tau\overline{z}}\Phi_{z\overline{\tau}}\Phi_{z{z\overline{z}}}}{(\Phi_{z\overline{z}})^{2}}=0.$
(E.7)

This is equivalent to

$\displaystyle\Phi_{z\zeta{\overline{\zeta}}}=0,$ (E.8)

which says $\Phi_{z}$ is harmonic when restricted to a leaf.

######  Remark E.2.

Let $G$ be the pluricomplex Green’s function of a strongly $\mathbb{C}$-convex
domain $\Omega$ with a smooth boundary; then, the rank of
$\sqrt{-1}\partial\overline{\partial}G$ is $n-1$ and the kernels of
$\sqrt{-1}\partial\overline{\partial}G$ form an extremal disc (see [L81] and
[L84]). For any point ${\bf z}$ in $\Omega\backslash\\{0\\}$, where $0$ is the
singular point of $G$, we can find a holomorphic map $\varphi$ from the unit
disc $D$ in $\mathbb{C}$ to $\Omega$, with

$\displaystyle\varphi_{\ast}\partial_{\zeta}\in\text{ kernel of
}\sqrt{-1}\partial\overline{\partial}G,\ \ \ \ \ \text{ in
}\Omega\backslash\\{0\\},$ (E.9)

and

$\displaystyle\varphi(D)\ni 0,{\bf z}\ \ \ \ \ \text{and}\ \ \ \ \
\varphi(\partial D)\subset\partial\Omega.$ (E.10)

However, in our situation, we would not have an “extremal cylinder”. An entire
leaf may be an infinite strap. So we only consider local leaves. This is the
main reason why Problem 1.6 does not have a smooth solution in general. For
geodesic problem in the space of Kähler potentials, we discussed a similar
singular phenomenon in [H21].

###  E.2 Positive Lower Bound Estimate for Norms of Gradients

In this section, we show the norms of the gradients of $\Phi$ have a positive
lower bound. We consider the following quantity:

$\displaystyle
S=\frac{\Phi_{\zeta}\Phi_{\overline{\zeta}}}{\mathcal{G}(\partial_{\zeta},\partial_{\overline{\zeta}})},$
(E.11)

where $\zeta$ is the complex coordinate on a leaf and $\mathcal{G}$ is the
Hermitian metric on $\mathbb{C}$. We know that $S$ is independent of the
choice of coordinate on the leaf. Actually, it’s the square of the norm of the
gradient of the restriction of $\Phi$ to a leaf; as a consequence,

$\displaystyle|\nabla\Phi|_{\mathcal{G}}^{2}>S.$ (E.12)

We will show $S$ has a positive lower bound in $\mathcal{R}$. First, we
estimate $S$ on the boundary; then, we show $\log(S)$ is a superharmonic
function on a leaf. The desired estimate then follows from the maximum
principle.

When estimating $S$ at a boundary point ${\bf p}$, we choose a set of
coordinates $(\tau,z)$ on $\mathbb{C}^{2}$, so that
$\mathcal{G}_{{i\overline{j}}}=\delta_{ij}$ and $\Phi_{z}({\bf p})=0$. Then,
at ${\bf p}$,

$\displaystyle
S=\frac{\left|(\partial_{\tau}-\frac{\Phi_{\tau\overline{z}}}{\Phi_{z\overline{z}}}\partial_{z})\Phi\right|^{2}}{\left|\partial_{\tau}-\frac{\Phi_{\tau\overline{z}}}{\Phi_{z\overline{z}}}\partial_{z}\right|^{2}_{\mathcal{G}}}=\frac{|\Phi_{\tau}|^{2}}{1+\left|\frac{\Phi_{\tau\overline{z}}}{\Phi_{z\overline{z}}}\right|^{2}}.$
(E.13)

Therefore, for a constant $C$ depending on the $C^{2}$ norm of $\Phi$ and the
lower bound of $\Phi_{z\overline{z}}$,

$\displaystyle C|\Phi_{\mathbb{n}}|^{2}\leq S\leq|\Phi_{\mathbb{n}}|^{2},$
(E.14)

where ${\mathbb{n}}$ is the unit normal vector to $\partial_{\mathcal{R}}$ at
${\bf p}$. According to Lemma A.11 the lower bound of $\Phi_{z\overline{z}}$
depends on the $\mathbb{C}$-convexity of the boundary and the gradient of
$\Phi$ on the boundary. The lower bound estimate for $|\Phi_{\mathbb{n}}|$ can
be find in Appendix B. Thus the lower bound for $S$ on the boundary is known.

To get the interior estimate for $S$, we need to show $\log(S)$ is
superharmonic on each leaf. We have

$\displaystyle(\log(S))_{\zeta{\overline{\zeta}}}=\left[\log(\Phi_{\zeta}\Phi_{\overline{\zeta}})-\log\mathcal{G}(\partial_{\zeta},\partial_{{\overline{\zeta}}})\right]_{{\zeta{\overline{\zeta}}}}.$
(E.15)

In the following, we compute
$\left[\log(\Phi_{\zeta}\Phi_{\overline{\zeta}})\right]_{{\zeta{\overline{\zeta}}}}$
and
$-\left[\log\mathcal{G}(\partial_{\zeta},\partial_{{\overline{\zeta}}})\right]_{{\zeta{\overline{\zeta}}}}$
seperately and show that they are both non-positive.

Using $\Phi_{\zeta{\overline{\zeta}}}=0$, which is (E.6), and
$\Phi_{\zeta{\zeta{\overline{\zeta}}}}=0$, which follows from the previous
one, we have

$\displaystyle\left[\log(\Phi_{\zeta}\Phi_{\overline{\zeta}})\right]_{{\zeta{\overline{\zeta}}}}=\left[\frac{\Phi_{\zeta\zeta}\Phi_{\overline{\zeta}}+\Phi_{\zeta}\Phi_{{\zeta{\overline{\zeta}}}}}{\Phi_{\zeta}\Phi_{\overline{\zeta}}}\right]_{{\overline{\zeta}}}=\frac{|\Phi_{\zeta\zeta}|^{2}}{|\Phi_{\zeta}|^{2}}-\frac{|\Phi_{\zeta\zeta}\Phi_{\overline{\zeta}}|^{2}}{|\Phi_{\zeta}|^{4}}=0.$
(E.16)

For the computation of
$\left[\log\mathcal{G}(\partial_{\zeta},\partial_{{\overline{\zeta}}})\right]_{{\zeta{\overline{\zeta}}}}$,
we choose a coordinate $(\tau,z)$, so that $g_{{i\overline{j}}}=\delta_{ij}$.
Then

$\displaystyle\mathcal{G}(\partial_{\zeta},\partial_{\overline{\zeta}})=1+\left|\frac{\Phi_{\tau{\overline{z}}}}{\Phi_{{z\overline{z}}}}\right|^{2}.$
(E.17)

As a locally defined function
$\frac{\Phi_{\tau{\overline{z}}}}{\Phi_{{z\overline{z}}}}$ is holomorphic on
each leaf; actually it is $-h^{\prime}$, where $h$ is the $\tau$-component of
$\varphi$ in (E.4). This can be verified by applying
$\partial_{\overline{\zeta}}$ to
$\frac{\Phi_{\tau{\overline{z}}}}{\Phi_{{z\overline{z}}}}$. In the following,
we denote $\frac{\Phi_{\tau{\overline{z}}}}{\Phi_{{z\overline{z}}}}$ by $f$,
and then

$\displaystyle\mathcal{G}(\partial_{\zeta},\partial_{\overline{\zeta}})=1+|f|^{2};$
(E.18)

we have

$\displaystyle[\log(1+|f|^{2})]_{\zeta{\overline{\zeta}}}=\left(\frac{f_{\zeta}\overline{f}}{1+|f|^{2}}\right)_{\overline{\zeta}}=\frac{\overline{f}_{\overline{\zeta}}f_{\zeta}}{(1+|f|^{2})^{2}}\geq
0.$ (E.19)

To sum up, we have

$\displaystyle(\log S)_{\zeta{\overline{\zeta}}}\leq 0,$ (E.20)

and combining with the boundary estimate, we have a positive lower bound for
$S$. Therefore, $|\nabla\Phi|_{\mathcal{G}}^{2}$ is also bounded from below by
a positive constant since it’s greater or equal to $S$. We denote

$\displaystyle|\nabla\Phi|_{\mathcal{G}}\geq C_{\mathcal{R}}^{g}.$ (E.21)

######  Remark E.3.

A positive lower bound estimate for $|\nabla\Phi|_{\mathcal{G}}$ is crucial in
the study of level sets of $\Phi$; without it, we don’t even know the level
sets of $\Phi$ are submanifolds. It’s interesting to note the following
results. In [MOZ10], Proposition 4.1, they proved that for the solution $u$ of
the $p$-Laplace equation $|\nabla u|$ strictly increases in the direction of
$\nabla u$. In [L81], for an extremal map $f:U\rightarrow D$, where $U$ is the
unit disc in $\mathbb{C}$ and $D$ is a strongly convex domain with a smooth
boundary, it was proved that $f^{\prime}\neq 0$ anywhere; as a consequence,
for the pluricomplex Green’s function $\Phi$, $\nabla\Phi\neq 0$.

###  E.3 The Construction of ${\mathcal{Q}}$ and Its Relation to
$\mathbb{C}$-Convexity

Under the apriori assumption that $\Phi$ is strongly
$\mathfrak{q}\mathbb{C}$-convex, the gradients of $\Phi$ never vanish; then,
for any point ${\bf p}$, we can choose a coordinate $(\tau,z)$ so that
$\Phi_{\tau}\neq 0$, $\Phi_{z}=0$ and ${\bf p}=(0,0)$. That’s to say, the
plane $\\{\tau=0\\}$ is the complex tangent plane of the level set of $\Phi$
at the point ${\bf p}$. In such a coordinate chart, we define

$\displaystyle{\mathfrak{b}_{\Phi}}=\Phi_{zz}-2\frac{\Phi_{z}}{\Phi_{\tau}}\Phi_{\tau
z}+\left(\frac{\Phi_{z}}{\Phi_{\tau}}\right)^{2}\Phi_{\tau\tau},$ (E.22)
$\displaystyle{\mathfrak{a}_{\Phi}}=\Phi_{z\overline{z}}-\frac{\Phi_{z}}{\Phi_{\tau}}\Phi_{\tau\overline{z}}-\frac{\Phi_{\overline{z}}}{\Phi_{\overline{\tau}}}\Phi_{z\overline{\tau}}+\left|\frac{\Phi_{z}}{\Phi_{\tau}}\right|^{2}\Phi_{\tau\overline{\tau}}.$
(E.23)

Because of the apriori assumption that $\Phi$ is strongly
$\mathfrak{q}\mathbb{C}$-convex, when this neighborhood is small enough,
${\mathfrak{a}_{\Phi}}\neq 0$ and we can define

$\displaystyle{\mathcal{Q}_{\Phi}}=\frac{|{\mathfrak{b}_{\Phi}}|^{2}}{{\mathfrak{a}_{\Phi}}^{2}}.$
(E.24)

Actually, ${\mathcal{Q}_{\Phi}}$ is invariant under a complex linear change of
coordinate, even ${\mathfrak{a}_{\Phi}}$ and ${\mathfrak{b}_{\Phi}}$ depend on
the choice of coordinates. $\Phi$ is strongly $\mathfrak{q}\mathbb{C}$-convex
is equivalent to that ${\mathfrak{a}}_{\Phi}>0$ and ${\mathcal{Q}_{\Phi}}<1$.
In Section E.5, we will show $\frac{1}{1-{\mathcal{Q}_{\Phi}}}$ is subharmonic
along each leaf.

We also need to perturb $\Phi$ by pluriharmonic quadratic polynomials, which
does not change the equation but will alter ${\mathcal{Q}_{\Phi}}$. This gives
us the estimate of the robustness of $\mathfrak{q}\mathbb{C}$-convexity of
$\Phi$. For a quadratic polynomial $q$, assuming that $\Phi-q$ is strongly
$\mathfrak{q}\mathbb{C}$-convex and $\partial_{\tau}(\Phi-q)\neq 0$, we
replace $\Phi$ by $\Phi-q$ in (E.22) and (E.23) and let

$\displaystyle{\mathfrak{b}}_{\Phi-q}=(\Phi-q)_{zz}-2\frac{(\Phi-q)_{z}}{(\Phi-q)_{\tau}}(\Phi-q)_{\tau
z}+\left(\frac{(\Phi-q)_{z}}{(\Phi-q)_{\tau}}\right)^{2}(\Phi-q)_{\tau\tau},$
(E.25)
$\displaystyle{\mathfrak{a}}_{\Phi-q}=(\Phi-q)_{{z\overline{z}}}-\frac{(\Phi-q)_{z}}{(\Phi-q)_{\tau}}(\Phi-q)_{\tau{\overline{z}}}-\frac{(\Phi-q)_{\overline{z}}}{(\Phi-q)_{\overline{\tau}}}(\Phi-q)_{z{\overline{\tau}}}+\left|\frac{(\Phi-q)_{z}}{(\Phi-q)_{\tau}}\right|^{2}(\Phi-q)_{{\tau\overline{\tau}}}.$
(E.26)

Then we let

$\displaystyle{\mathcal{Q}}_{\Phi-q}=\frac{|{\mathfrak{b}}_{\Phi-q}|^{2}}{{\mathfrak{a}}_{\Phi-q}^{2}}.$
(E.27)

If $q$ is a pluriharmonic function, $\Phi-q$ is still a solution to the
homogenous complex Monge-Ampère equation, and the subharmonicity of
${\mathcal{Q}}_{\Phi}$ implies the subharmonicity of ${\mathcal{Q}}_{\Phi-q}$.

###  E.4 Equations for ${\mathfrak{a}}$ and ${\mathfrak{b}}$ with a Good
Coordinate Choice

At any point ${\bf p}$, because of the assumption that $\Phi$ is strongly
$\mathfrak{q}\mathbb{C}$-convex, we have

$\displaystyle\ker\sqrt{-1}\partial\overline{\partial}\Phi\oplus\ker\partial\Phi=\mathbb{C}^{2},$
(E.28)

so we can choose a coordinate $(\tau,z)$ so that $\Phi_{z}=0$ and
$\Phi_{{\tau\overline{\tau}}}=0$. Then the homogenous complex Monge-Ampère
equation implies $\Phi_{z\overline{\tau}}=0$. We note that with this
coordinate the coefficients of $\mathcal{G}$ may be very large.

Then, the computations at ${\bf p}$ become simple: (E.5) becomes
$\Phi_{\tau{\overline{\tau}}}=0$, and (E.7) becomes
$\Phi_{z{\tau\overline{\tau}}}=0.$ For any function $f$ defined around ${\bf
p}$, we have $f_{\zeta}=f_{\tau}$, and

$\displaystyle
f_{\zeta{\overline{\zeta}}}=\left(f_{\tau}-\frac{\Phi_{\tau\overline{z}}}{\Phi_{z\overline{z}}}f_{z}\right)_{\overline{\tau}}=f_{{\tau\overline{\tau}}}-\frac{\Phi_{\tau{\overline{z}}{\overline{\tau}}}}{\Phi_{z\overline{z}}}f_{z}=f_{{\tau\overline{\tau}}}.$
(E.29)

In the last equality, we used
$\Phi_{z{\tau\overline{\tau}}}=\Phi_{{\overline{z}}{\tau\overline{\tau}}}=0$.

In the following, we discuss the equations satisfied by
${\mathfrak{a}}_{\Phi}$ and ${\mathfrak{b}}_{\Phi}$ at ${\bf p}$ with the
coordinate chosen above. We will drop the subindex $\Phi$ and just denote them
by ${\mathfrak{a}}$ and ${\mathfrak{b}}$. We compute
${\mathfrak{a}}_{\zeta{\overline{\zeta}}}$,
${\mathfrak{b}}_{\zeta{\overline{\zeta}}}$, ${\mathfrak{a}}_{\zeta}$ and
${\mathfrak{b}}_{{\overline{\zeta}}}$ and show that they satisfy the following
equations:

$\displaystyle{\mathfrak{b}}_{\zeta{\overline{\zeta}}}$
$\displaystyle=2\frac{{\mathfrak{b}}_{\overline{\zeta}}{\mathfrak{a}}_{\zeta}}{{\mathfrak{a}}},$
(E.30) $\displaystyle{\mathfrak{a}}_{\zeta{\overline{\zeta}}}$
$\displaystyle=\frac{{\mathfrak{b}}_{\overline{\zeta}}{\overline{\mathfrak{b}}}_{\zeta}}{{\mathfrak{a}}}+\frac{{\mathfrak{a}}_{\zeta}{\mathfrak{a}}_{\overline{\zeta}}}{{\mathfrak{a}}}.$
(E.31)

Using these equations, we show, in Section E.5,
$\frac{1}{1-{\mathcal{Q}}_{\Phi}}$ is subharmonic along any leaves.

To simplify the computation, let $b=\Phi_{zz}$ and $a=\Phi_{{z\overline{z}}}$.
By applying $\partial_{z\overline{z}}$ and $\partial_{zz}$ to (E.5), we have

$\displaystyle b_{\zeta{\overline{\zeta}}}$
$\displaystyle=2\frac{b_{\overline{\zeta}}a_{\zeta}}{a},$ (E.32)
$\displaystyle a_{\zeta{\overline{\zeta}}}$
$\displaystyle=\frac{b_{\overline{\zeta}}{\overline{b}}_{\zeta}}{a}+\frac{a_{\overline{\zeta}}a_{\zeta}}{a}.$
(E.33)

The computation is the same as that of Section 2 of [H22N] and Appendix A of
[H22A]. In the following, we will show, at the point ${\bf p}$,
${\mathfrak{b}}_{\overline{\zeta}}=b_{\overline{\zeta}}$,
${\mathfrak{a}}_{\zeta}=a_{\zeta}$,
${\mathfrak{b}}_{\zeta{\overline{\zeta}}}=b_{\zeta{\overline{\zeta}}}$,
${\mathfrak{a}}_{\zeta{\overline{\zeta}}}=a_{\zeta{\overline{\zeta}}}$. Thus,
(E.30) and (E.31) follow from (E.32) and (E.33).

Subtracting $a=\Phi_{{z\overline{z}}}$ from (E.23), we get

$\displaystyle{\mathfrak{a}}_{\zeta}-a_{\zeta}=\left(-\frac{\Phi_{z}}{\Phi_{\tau}}\Phi_{\tau\overline{z}}-\frac{\Phi_{\overline{z}}}{\Phi_{\overline{\tau}}}\Phi_{z\overline{\tau}}+\left|\frac{\Phi_{z}}{\Phi_{\tau}}\right|^{2}\Phi_{\tau\overline{\tau}}\right)_{\zeta}=0,\
\ \ \ \ \ \text{ at }p.$ (E.34)

It’s zero because $\Phi_{z{\overline{\tau}}}$, $\Phi_{\tau{\overline{z}}}$ and
$\Phi_{z}$ all vanish at the point ${\bf p}$. Similarly, we subtract
$b=\Phi_{zz}$ from (E.22) and get

$\displaystyle{\mathfrak{b}}_{\overline{\zeta}}-b_{\overline{\zeta}}=\left(-2\frac{\Phi_{z}}{\Phi_{\tau}}\Phi_{\tau
z}+\left(\frac{\Phi_{z}}{\Phi_{\tau}}\right)^{2}\Phi_{\tau\tau}\right)_{\overline{\zeta}}.$
(E.35)

On the right-hand side of (E.35), the only non-zero terms are produced by
$\partial_{\overline{\zeta}}$ acting on $\Phi_{z}$ because $\Phi_{z}$ vanishes
at ${\bf p}$. However,
$\partial_{{\overline{\zeta}}}\Phi_{z}=\Phi_{z{\overline{\tau}}}=0$, at ${\bf
p}$, so ${\mathfrak{b}}_{\overline{\zeta}}=b_{\overline{\zeta}}$. Applying
$\partial_{\overline{\zeta}}$ to (E.34), we get

$\displaystyle{\mathfrak{a}}_{\zeta{\overline{\zeta}}}-a_{\zeta{\overline{\zeta}}}=\left(-\frac{\Phi_{z}}{\Phi_{\tau}}\Phi_{\tau\overline{z}}-\frac{\Phi_{\overline{z}}}{\Phi_{\overline{\tau}}}\Phi_{z\overline{\tau}}+\left|\frac{\Phi_{z}}{\Phi_{\tau}}\right|^{2}\Phi_{\tau\overline{\tau}}\right)_{\zeta{\overline{\zeta}}}.$
(E.36)

On the right-hand side of the equation above, $\Phi_{z}$,
$\Phi_{z\overline{\tau}}$ and $\Phi_{\tau\overline{\tau}}$ all vanish at ${\bf
p}$, so

$\displaystyle\left(\left|\frac{\Phi_{z}}{\Phi_{\tau}}\right|^{2}\Phi_{\tau\overline{\tau}}\right)_{\zeta{\overline{\zeta}}}=0,$
(E.37)

since the item in the bracket is a third order small quantity around ${\bf
p}$, and

$\displaystyle{\mathfrak{a}}_{\zeta{\overline{\zeta}}}-a_{\zeta{\overline{\zeta}}}=-\frac{\Phi_{z\tau}}{\Phi_{\tau}}\Phi_{\tau{\overline{z}}{\overline{\tau}}}-\frac{\Phi_{z{\overline{\tau}}}}{\Phi_{\tau}}\Phi_{\tau{\overline{z}}\tau}-\frac{\Phi_{{\overline{z}}\tau}}{\Phi_{\overline{\tau}}}\Phi_{z{\overline{\tau}}{\overline{\tau}}}-\frac{\Phi_{{\overline{z}}{\overline{\tau}}}}{\Phi_{\overline{\tau}}}\Phi_{z{\overline{\tau}}\tau}.$
(E.38)

In (E.38), $\Phi_{z{\overline{\tau}}}$ and $\Phi_{z\tau{\overline{\tau}}}$ all
vanish at ${\bf p}$, so
${\mathfrak{a}}_{\zeta{\overline{\zeta}}}=a_{\zeta{\overline{\zeta}}}$.
Applying $\partial_{\zeta}$ to (E.35) gives

$\displaystyle{\mathfrak{b}}_{\zeta{\overline{\zeta}}}-b_{\zeta{\overline{\zeta}}}=\left(-2\frac{\Phi_{z}}{\Phi_{\tau}}\Phi_{\tau
z}+\left(\frac{\Phi_{z}}{\Phi_{\tau}}\right)^{2}\Phi_{\tau\tau}\right)_{\zeta{\overline{\zeta}}}.$
(E.39)

It vanishes at ${\bf p}$ since $\Phi_{z}$, $\Phi_{z\overline{\tau}}$ and
$\Phi_{z{\tau\overline{\tau}}}$ all vanish at ${\bf p}$, similar to the
previous argument.

Therefore, (E.30) and (E.31) are valid at the point ${\bf p}$. However, we
point out that they depend on the choice of coordinates, but in the next
section, using (E.30) and (E.31), we will show
$\frac{1}{1-{\mathcal{Q}}_{\Phi}}$ is subharmonic along a leaf, which is a
fact independent of the choice of coordinates.

###  E.5 Subharmonicity of $\frac{1}{1-{\mathcal{Q}}_{\Phi}}$ Along Leaves

In this section, we will show $\frac{1}{1-{\mathcal{Q}}_{\Phi}}$ is
subharmonic along a leaf when ${\mathcal{Q}}_{\Phi}<1$. We will drop the
subindex $\Phi$ and denote ${\mathcal{Q}}_{\Phi}$ by ${\mathcal{Q}}$ since we
don’t consider perturbations of $\Phi$ in this section. We will continue to
use the coordinate introduced in Section E.4, where
$\partial_{\zeta}=\partial_{\tau}$ and
$\partial_{{\zeta{\overline{\zeta}}}}=\partial_{{\tau\overline{\tau}}}$ at
${\bf p}$.

We directly apply $\partial_{{\tau\overline{\tau}}}$ to
$\frac{1}{1-{\mathcal{Q}}}$ and get

$\displaystyle\left(\frac{1}{1-{\mathcal{Q}}}\right)_{\tau{\overline{\tau}}}=\left[\frac{1}{(1-{\mathcal{Q}})^{2}}{\mathcal{Q}}_{\tau}\right]_{{\overline{\tau}}}=\frac{1}{(1-{\mathcal{Q}})^{3}}\left[2{\mathcal{Q}}_{\tau}{\mathcal{Q}}_{\overline{\tau}}+{\mathcal{Q}}_{{\tau\overline{\tau}}}(1-{\mathcal{Q}})\right].$
(E.40)

To simplify the computation, we introduce a quantity:

$\displaystyle\beta={\mathfrak{b}}_{\tau}-2\frac{{\mathfrak{b}}{\mathfrak{a}}_{\tau}}{{\mathfrak{a}}}.$
(E.41)

With $\beta$, we have

$\displaystyle\left(\frac{{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\right)_{\tau}=\frac{\beta}{{\mathfrak{a}}^{2}}$
(E.42)

and

$\displaystyle\beta_{\overline{\tau}}=\underbrace{{\mathfrak{b}}_{\tau\overline{\tau}}-2\frac{{\mathfrak{b}}_{\overline{\tau}}{\mathfrak{a}}_{\tau}}{{\mathfrak{a}}}}_{=0,\
\text{ by
(\ref{equation:A_cb_Phi})}}\underbrace{-2\frac{{\mathfrak{a}}_{\tau\overline{\tau}}{\mathfrak{b}}}{{\mathfrak{a}}}+2\frac{{\mathfrak{a}}_{\tau}{\mathfrak{a}}_{\overline{\tau}}{\mathfrak{b}}}{{\mathfrak{a}}^{2}}=-2\frac{\overline{{\mathfrak{b}}}_{\tau}{\mathfrak{b}}_{\overline{\tau}}{\mathfrak{b}}}{{\mathfrak{a}}^{2}}}_{\text{by
(\ref{equation:A_ca_Phi})}}.$ (E.43)

In addition, we observe that (E.30) is equivalent to

$\displaystyle\left(\frac{{\mathfrak{b}}_{\overline{\tau}}}{{\mathfrak{a}}^{2}}\right)_{\tau}=0.$
(E.44)

Then

$\displaystyle{\mathcal{Q}}_{\tau}=\left(\frac{{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\cdot\overline{\mathfrak{b}}\right)_{\tau}=\left(\frac{{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\right)_{\tau}\overline{\mathfrak{b}}+\frac{{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\cdot\overline{\mathfrak{b}}_{\tau}=\frac{\beta}{{\mathfrak{a}}^{2}}\overline{\mathfrak{b}}+\frac{{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\cdot\overline{\mathfrak{b}}_{\tau},$
(E.45)

and

$\displaystyle{\mathcal{Q}}_{\tau}{\mathcal{Q}}_{\overline{\tau}}=\left(\frac{\beta}{{\mathfrak{a}}},\frac{\overline{{\mathfrak{b}}}_{\tau}}{{\mathfrak{a}}}\right)\left(\begin{array}[]{cc}{\mathcal{Q}}&\frac{\overline{{\mathfrak{b}}}^{2}}{{\mathfrak{a}}^{2}}\\\
&\\\
\frac{{\mathfrak{b}}^{2}}{{\mathfrak{a}}^{2}}&{\mathcal{Q}}\end{array}\right)\left(\begin{array}[]{c}\frac{\overline{\beta}}{{\mathfrak{a}}}\\\
\\\
\frac{{\mathfrak{b}}_{\overline{\tau}}}{{\mathfrak{a}}}\end{array}\right).$
(E.52)

For ${\mathcal{Q}}_{{\tau\overline{\tau}}}$, we also make it into a quadratic
form of $\frac{\beta}{{\mathfrak{a}}}$ and
$\frac{\overline{\mathfrak{b}}_{\tau}}{{\mathfrak{a}}}$. Applying
$\partial_{\overline{\tau}}$ to (E.45), we have

$\displaystyle{\mathcal{Q}}_{{\tau\overline{\tau}}}=\left(\beta\cdot\frac{\overline{\mathfrak{b}}}{{\mathfrak{a}}^{2}}+{\mathfrak{b}}\cdot\frac{\overline{\mathfrak{b}}_{\tau}}{{\mathfrak{a}}^{2}}\right)_{\overline{\tau}}=\beta_{\overline{\tau}}\cdot\frac{\overline{\mathfrak{b}}}{{\mathfrak{a}}^{2}}+\beta\cdot\left(\frac{\overline{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\right)_{\overline{\tau}}+{\mathfrak{b}}_{\overline{\tau}}\cdot\frac{\overline{\mathfrak{b}}_{\tau}}{{\mathfrak{a}}^{2}}+{\mathfrak{b}}\cdot\left(\frac{\overline{\mathfrak{b}}_{\tau}}{{\mathfrak{a}}^{2}}\right)_{\overline{\tau}}.$
(E.53)

Using the conjugate of (E.44), the last term of (E.53) is zero; for
$\left(\frac{\overline{\mathfrak{b}}}{{\mathfrak{a}}^{2}}\right)_{\overline{\tau}}$
, we use the conjugate of (E.42); for $\beta_{\overline{\tau}}$, we use the
equation (E.43). Then we have

$\displaystyle{\mathcal{Q}}_{{\tau\overline{\tau}}}=\left(\frac{\beta}{{\mathfrak{a}}},\frac{\overline{{\mathfrak{b}}}_{\tau}}{{\mathfrak{a}}}\right)\left(\begin{array}[]{cc}1&\\\
&1-2{\mathcal{Q}}\end{array}\right)\left(\begin{array}[]{c}\frac{\overline{\beta}}{{\mathfrak{a}}}\\\
\\\
\frac{{\mathfrak{b}}_{\overline{\tau}}}{{\mathfrak{a}}}\end{array}\right).$
(E.59)

Combining (E.52) (E.59) with (E.40), we know

$\displaystyle\left(\frac{1}{1-{\mathcal{Q}}}\right)_{\tau\overline{\tau}}=\frac{1}{(1-{\mathcal{Q}})^{3}}\left(\frac{\beta}{{\mathfrak{a}}},\frac{\overline{{\mathfrak{b}}}_{\tau}}{{\mathfrak{a}}}\right)M\left(\frac{\overline{\beta}}{{\mathfrak{a}}},\frac{{\mathfrak{b}}_{\overline{\tau}}}{{\mathfrak{a}}}\right)^{\prime},$
(E.60)

where $M$ is a $2\times 2$ matrix:

$\displaystyle
M=\left(\begin{array}[]{cc}1+{\mathcal{Q}}&2\frac{\overline{\mathfrak{b}}^{2}}{{\mathfrak{a}}^{2}}\\\
\\\
2\frac{{\mathfrak{b}}^{2}}{{\mathfrak{a}}^{2}}&1-{\mathcal{Q}}+2{\mathcal{Q}}^{2}\end{array}\right)$
(E.64)

Direct computation gives

$\displaystyle\det(M)$
$\displaystyle=(1-{\mathcal{Q}})^{2}(1+2{\mathcal{Q}}),$ (E.65)
$\displaystyle\text{tr}(M)$ $\displaystyle=2+2{\mathcal{Q}}^{2}.$ (E.66)

By the definition of ${\mathcal{Q}}$, it’s always nonnegative, so
$1+2{\mathcal{Q}}>0$. Therefore,
$\left(\frac{1}{1-{\mathcal{Q}}}\right)_{\tau\overline{\tau}}\geq 0$, at ${\bf
p}$, providing ${\mathcal{Q}}<1$. We can choose ${\bf p}$ to be any point in
$\mathcal{R}$, so ${\mathcal{Q}}$, as a globally defined function in
$\mathcal{R}$, satisfies

$\displaystyle\left(\frac{1}{1-{\mathcal{Q}}}\right)_{{\zeta{\overline{\zeta}}}}\geq
0,$ (E.67)

along a leaf, providing ${\mathcal{Q}}<1$. Then,

$\displaystyle\frac{1}{1-{\mathcal{Q}}}\leq\max_{\partial_{\mathcal{R}}}\frac{1}{1-{\mathcal{Q}}}\
\ \ \ \ \ \text{ in }\mathcal{R}\text{, \ providing }{\mathcal{Q}}<1\text{ in
}\overline{\mathcal{R}}.$ (E.68)

This is because the maximum of $\frac{1}{1-{\mathcal{Q}}}$ cannot be achieved
at any interior point in $\mathcal{R}$. Suppose $\frac{1}{1-{\mathcal{Q}}}$
achieves its maximum at ${\bf p}\in\mathcal{R}$. Let $\mathcal{L}$ be a local
leaf passing ${\bf p}$. Then the restriction of $\frac{1}{1-{\mathcal{Q}}}$ on
$\mathcal{L}$ also achieves a local maximum at ${\bf p}$, but this contradicts
with the strong maximum principle. It follows that

$\displaystyle{\mathcal{Q}}\leq\max_{\partial_{\mathcal{R}}}{\mathcal{Q}}\ \ \
\ \ \ \text{ in }\mathcal{R}\text{, \ providing }{\mathcal{Q}}<1\text{ in
}\overline{\mathcal{R}}.$ (E.69)

###  E.6 Convexity Estimate

In this section we will show for a small constant $\epsilon_{0}$, $\Phi-q$ is
strongly $\mathfrak{q}\mathbb{C}$-convex for any
$q\in{\mathscr{Q}}_{\epsilon_{0}}$. The constant $\epsilon_{0}$ will depend on
the geometry of $\mathcal{R}$. In this section, we only do unitary change of
coordinate, so we assume the metric is
$\sum_{i}dz^{i}\otimes\overline{dz^{i}}$.

Firstly, we need to choose $\epsilon_{0}$ small enough, so that it’s smaller
than the lower bound estimate for $|\nabla\Phi|_{\mathcal{G}}$ in Section E.2:
we choose

$\displaystyle\epsilon_{0}<C_{\mathcal{R}}^{g}.$ (E.70)

This implies $\nabla(\Phi-q)$ is non-zero for
$q\in{\mathscr{Q}}_{\epsilon_{0}}$ and the level sets of $\Phi$ are smooth.
Then we make $\epsilon_{0}$ smaller so that

$\displaystyle\Phi-q\text{ is strongly $\mathfrak{q}\mathbb{C}$-convex},\text{
on }\partial\mathcal{R}\text{ for any }q\in{\mathscr{Q}}_{\epsilon_{0}}.$
(E.71)

Then using a continuity argument, we will prove $\Phi-q$ is actually strongly
$\mathfrak{q}\mathbb{C}$-convex in $\mathcal{R}$ for any
$q\in{\mathscr{Q}}_{\epsilon_{0}}$. Let

$\displaystyle I_{\epsilon_{0}}=\left\\{\epsilon\in[0,\epsilon_{0}]|\text{ for
any }q\in{\mathscr{Q}}_{\epsilon},\Phi-q\text{ is strongly
$\mathfrak{q}\mathbb{C}$-convex}\right\\}.$ (E.72)

We will show $I_{\epsilon_{0}}$ is non-empty, open and right-closed; thus,
$I_{\epsilon_{0}}=[0,\epsilon_{0}]$.

It’s obvious that $I_{\epsilon_{0}}$ is non-empty since it contains $0$; this
is because of the apriori assumption. The openness can also be proved easily:
if $\Phi-q_{1}$ is strongly $\mathfrak{q}\mathbb{C}$-convex, then we have
${\mathfrak{a}}_{\Phi-q_{1}}>0$; so, ${\mathcal{Q}}_{\Phi-q_{1}-q_{2}}$ is
close to ${\mathcal{Q}}_{\Phi-q_{1}}$ for
$q_{2}\in{\mathscr{Q}}_{\epsilon_{2}}$ with small enough $\epsilon_{2}$.

The right-closeness is more complicated, and we need to use the subharmonicity
of $\frac{1}{1-{\mathcal{Q}}_{\Phi-q}}$. Here, we notice that to derive (E.69)
we only need that $\Phi$ is a solution to the homogenous complex Monge-Ampère
equation and that $\Phi$ is strongly $\mathfrak{q}\mathbb{C}$-convex. For a
pluriharmonic function $q$, $\Phi-q$ is still the solution to the homogenous
complex Monge-Ampère equation; so, if $\Phi-q$ is also strongly
$\mathfrak{q}\mathbb{C}$-convex, we have

$\displaystyle{\mathcal{Q}}_{\Phi-q}\leq\max_{\partial_{\mathcal{R}}}{\mathcal{Q}}_{\Phi-q}\
\ \ \ \ \ \text{ in }\mathcal{R}.$ (E.73)

Suppose $[0,\epsilon_{1})\subset I_{\epsilon_{0}}$ and $\epsilon_{1}\notin
I_{\epsilon_{0}}$. Then we will derive a contradiction. According to the
definition of $I_{\epsilon_{0}}$ (E.72), there is a
$k\in{\mathscr{Q}}_{\epsilon_{1}}$ so that $\Phi-k$ is not strongly
$\mathfrak{q}\mathbb{C}$-convex. This means we can find a point ${\bf
p}\in\mathcal{R}$ so that if we choose a coordinate $(\tau,z)$ around ${\bf
p}$ such that $(\Phi-k)_{z}({\bf p})=0$, then either

$\displaystyle(\Phi-k)_{{z\overline{z}}}({\bf p})=\Phi_{{z\overline{z}}}({\bf
p})=0$ (E.74)

or

$\displaystyle{\mathcal{Q}}_{\Phi-k}({\bf p})=1.$ (E.75)

In the following we will argue none of these can happen.

Let $h$ be an element of ${\mathscr{Q}}_{\frac{\epsilon_{1}}{2}}$ . For
$\theta\in(0,1]$ we consider the following polynomials

$\displaystyle k^{\theta}=(1-\theta)k-\theta h\ \ \ \ \ \text{and}\ \ \ \ \
k_{\theta}=(1-\theta)k+\theta h.$ (E.76)

They all belong to the interior of ${\mathscr{Q}}_{\epsilon_{1}}$. So

$\displaystyle\Phi^{\theta}=\Phi-k^{\theta}\ \ \ \ \ \text{and}\ \ \ \ \
\Phi_{\theta}=\Phi-k_{\theta}$ (E.77)

are both strongly $\mathfrak{q}\mathbb{C}$-convex. Then using a continuity
argument, we have

$\displaystyle{\mathcal{Q}}_{\Phi^{\theta}},{\mathcal{Q}}_{\Phi_{\theta}}\leq\max_{q\in{\mathscr{Q}}_{\epsilon_{0}}}\max_{\partial\mathcal{R}}{\mathcal{Q}}_{\Phi-q}<1\
\ \ \ \ \ \text{for }\theta\in(0,1].$ (E.78)

Because of (E.70), we know the gradient of $\Phi-k$ does not vanish anywhere.
We denote $(\Phi-k)_{\tau}({\bf p})=\nu$; then $\nu>0$. The gradients of
$\Phi^{\theta}$ and $\Phi_{\theta}$ are both close to the gradient of
$\Phi-k$, providing $\theta$ very small. We choose $h$ so that $\nabla h({\bf
p})=0$; then, we have

$\displaystyle\partial_{\tau}\Phi^{\theta}=\partial_{\tau}\Phi_{\theta}=(\Phi-k)_{\tau}+\theta(k+h)_{\tau}=\nu+\theta
k_{\tau},$ (E.79)
$\displaystyle\partial_{z}\Phi^{\theta}=\partial_{z}\Phi_{\theta}=(\Phi-k)_{z}+\theta(k+h)_{z}=\theta
k_{z},$ (E.80)

at the point ${\bf p}$. Let

$\displaystyle\mathfrak{v}_{\theta}=\frac{\theta k_{z}({\bf p})}{\nu+\theta
k_{\tau}({\bf p})};$ (E.81)

then $\partial_{z}-\mathfrak{v}_{\theta}\partial_{\tau}$ is a complex
tangential vector to the level sets of $\Phi-k$, with
$\mathfrak{v}_{\theta}=O(\theta)$. We compute ${\mathcal{Q}}$ in the complex
tangential direction and get

$\displaystyle{\mathcal{Q}}_{\Phi^{\theta}}({\bf
p})=\left|\frac{K_{\theta}+\theta(h_{zz}+2\mathfrak{v}_{\theta}h_{z\tau}+(\mathfrak{v}_{\theta})^{2}h_{\tau\tau})}{\Phi_{{z\overline{z}}}-\mathfrak{v}_{\theta}\Phi_{\tau{\overline{z}}}-\overline{\mathfrak{v}_{\theta}}\Phi_{z{\overline{\tau}}}+|\mathfrak{v}_{\theta}|^{2}\Phi_{{\tau\overline{\tau}}}}\right|^{2},$
(E.82) $\displaystyle{\mathcal{Q}}_{\Phi^{\theta}}({\bf
p})=\left|\frac{K_{\theta}-\theta(h_{zz}+2\mathfrak{v}_{\theta}h_{z\tau}+(\mathfrak{v}_{\theta})^{2}h_{\tau\tau})}{\Phi_{{z\overline{z}}}-\mathfrak{v}_{\theta}\Phi_{\tau{\overline{z}}}-\overline{\mathfrak{v}_{\theta}}\Phi_{z{\overline{\tau}}}+|\mathfrak{v}_{\theta}|^{2}\Phi_{{\tau\overline{\tau}}}}\right|^{2},$
(E.83)

where we denote

$\displaystyle(\Phi-(1-\theta)k)_{zz}-2(\Phi-(1-\theta)k)_{z\tau}\mathfrak{v}_{\theta}+(\Phi-(1-\theta)k)_{\tau\tau}(\mathfrak{v}_{\theta})^{2}=K_{\theta}.$
(E.84)

Here $k$ and $h$ don’t appear in the denominator because they are both
pluriharmonic. Equations above imply

$\displaystyle\lim_{\theta\rightarrow
0}{\mathcal{Q}}_{\Phi^{\theta}}=\lim_{\theta\rightarrow
0}{\mathcal{Q}}_{\Phi^{\theta}}={\mathcal{Q}}_{\Phi-k};$ (E.85)

therefore, (E.75) is impossible.

(E.78) implies

$\displaystyle\sqrt{{\mathcal{Q}}_{\Phi^{\theta}}({\bf
p})}+\sqrt{{\mathcal{Q}}_{\Phi_{\theta}}({\bf p})}\leq 2,\ \ \ \ \ \ \text{
for }\theta\in(0,1].$ (E.86)

The triangle inequality implies

$\displaystyle\left|\frac{\theta(h_{zz}+2\mathfrak{v}_{\theta}h_{z\tau}+(\mathfrak{v}_{\theta})^{2}h_{\tau\tau})}{\Phi_{{z\overline{z}}}-\mathfrak{v}_{\theta}\Phi_{\tau{\overline{z}}}-\overline{\mathfrak{v}_{\theta}}\Phi_{z{\overline{\tau}}}+|\mathfrak{v}_{\theta}|^{2}\Phi_{{\tau\overline{\tau}}}}\right|\leq
1\text{\ \ \ \ \ at ${\bf p}$}.$ (E.87)

If (E.74) is true, then the subharmonicity of $\Phi$ implies

$\displaystyle\Phi_{z{\overline{\tau}}}=\Phi_{\tau{\overline{z}}}=0,\ \ \ \ \
\ \ \ \text{ at }{\bf p}.$ (E.88)

So we get

$\displaystyle\frac{\theta|h_{zz}({\bf p})|+O(\theta^{2})}{O(\theta^{2})}\leq
2$ (E.89)

when $\theta$ is small enough. This is a contradiction since we can choose $h$
so that $h_{zz}({\bf p})\neq 0$ Here we used
$\Phi_{{z\overline{z}}}=\Phi_{z{\overline{\tau}}}=\Phi_{\tau{\overline{z}}}=0$
at ${\bf p}$ and $\mathfrak{v}_{\theta}=O(\theta)$.

So we have proved that $\Phi-k$ is strongly $\mathfrak{q}\mathbb{C}$-convex,
and as a consequence, $I_{\epsilon_{0}}$ is right-closed. Therefore,
$I_{\epsilon_{0}}=[0,\epsilon_{0}]$. This says the robustness of
$\mathfrak{q}\mathbb{C}$-convexity of $\Phi$ is greater than $\epsilon_{0}$.

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Jingchen Hu
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