# Polynomial Convexity and Polynomial approximations of certain sets in
$\mathbb{C}^{2n}$ with non-isolated CR-singularities

Golam Mostafa Mondal  Department of Mathematics and Statistics, Indian
Institute of Science Education and Research Kolkata, Mohanpur – 741 246
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we first consider the graph of $(F_{1},F_{2},\cdots,F_{n})$ on
$\overline{\mathbb{D}}^{n},$ where
$F_{j}(z)=\bar{z}^{m_{j}}_{j}+R_{j}(z),j=1,2,\cdots,n,$ which has non-isolated
CR-singularities if $m_{j}>1$ for some $j\in\\{1,2,\cdots,n\\}.$ We show that
under certain condition on $R_{j},$ the graph is polynomially convex and
holomorphic polynomials on the graph approximates all continuous functions. We
also show that there exists an open polydisc $D$ centred at the origin such
that the set
$\\{(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)):z\in\overline{D},m_{j}\in\mathbb{N},j=1,\cdots,2n\\}$
is polynomially convex; and if $\gcd(m_{j},m_{k})=1~{}~{}\forall j\not=k,$ the
algebra generated by the functions
$z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1},\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}$
is dense in $\mathcal{C}(\overline{D}).$ We prove an analogue of Minsker’s
theorem over the closed unit polydisc, i.e, if
$\gcd(m_{j},m_{k})=1~{}~{}\forall j\not=k,$ the algebra
$[z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}},\cdots,\bar{z_{n}}^{m_{2n}};{\overline{\mathbb{D}}^{n}}]=\mathcal{C}(\overline{\mathbb{D}}^{n}).$
In the process of proving the above results, we also studied the polynomial
convexity and approximation of certain graphs.

###### Key words and phrases:

Polynomial convexity, polynomial approximation, totally real, CR-singularity

###### 2010 Mathematics Subject Classification:

Primary: 32E20

##  1\. Introduction and statements of the results

Let $D$ be an open polydisc in $\mathbb{C}^{n}$ with center at the origin, and
by $\mathcal{C}(\overline{D}),$ we denote the set of all continuous complex
valued functions on $\overline{D}.$ For
$f_{1},f_{2},\cdots,f_{N}\in\mathcal{C}(\overline{D}),$ we denote by
$[f_{1},f_{2},\cdots,f_{N};\overline{D}]$ the uniform algebra generated by
$f_{1},f_{2},\cdots,f_{N}$ on $\overline{D}.$ In this article, we report our
investigation to the following question:

######  Question 1.1.

Let $\nu_{1},\cdots,\nu_{n}$ be positive integers. Under what conditions on
$f_{1},\cdots,f_{n},$ can one conclude

$\displaystyle[z^{\nu_{1}}_{1},\cdots,z^{\nu_{n}}_{n},f_{1},f_{2},\cdots,f_{n};\overline{D}]=\mathcal{C}(\overline{D})?$

In this discussion we first focus on the case when $\nu_{j}=1~{}~{}\forall
j\in\\{1,2,\cdots,n\\}.$ This problem seems relatively easier because the
underlined set is the graph of $(f_{1},\cdots,f_{n})$ over the closed polydisc
$\overline{D}.$ We present a brief literature survey on this. The problem is
quite well studied for $n=1.$ The following result by Mergelyan [21] played a
vital role in creating some interest in this question.

######  Result 1.2 (Mergelyan).

Let $D$ be an open disc in $\mathbb{C}$ and let $f$ be a continuous real-
valued function on $\overline{D}$. If for each $a$ in $\overline{D},$
$f^{-1}\left(f(a)\right)$ has no interior and does not separate $\mathbb{C}$,
then $[z,f;\overline{D}]=\mathcal{C}(\overline{D}),$ where
$[z,f;\overline{D}]$ denotes the algebra generated by the functions $z$ and
$f$ with complex coefficients.

We now state a couple of result due to Wermer [34].

######  Result 1.3 (Wermer).

Let $D$ be an open unit disc in $\mathbb{C}$ with center at the origin. If
$f(z)=\bar{z}+R(z),$ where

$\displaystyle\left|R(z)-R(a)\right|<|z-a|$ (1.1)

for all $a,z$ in $\overline{D}$ with $a\neq z,$ then
$[z,f;\overline{D}]=\mathcal{C}(\overline{D}).$

######  Result 1.4 (Wermer).

Fix $\delta_{0}>0.$ Let $g$ be function defined in the disc
$\\{z\in\mathbb{C}:|z|<\delta_{0}\\}$ and have continuous partial derivatives
up to the second order there. Assume

$\displaystyle\frac{\partial g}{\partial\bar{z}}(0)\neq 0.$ (1.2)

Then there exist $\delta,$ $0<\delta<\delta_{0},$ such that for
$D=\\{z:|z|\leq\delta\\},$ $[z,g;D]=C(D).$

There are several results due to Preskenis [28, 29, 26], O’Farrell, Preskenis
and Walsh [25], (see Sanabria [31] for a nice survey) that generalize Wermer’s
results.

Before going further, we discuss the relation between the above mentioned
approximation problem with polynomial convexity, which is a fundamental notion
in several complex variables. Let $K$ be a compact subset of $\mathbb{C}^{n}.$
The polynomial convex hull of $K$ is denoted by $\widehat{K}$ and defined by
$\widehat{K}:=\left\\{\alpha\in\mathbb{C}^{n}:|p(\alpha)|\leq\max_{K}|p|~{}~{}\forall
p\in\mathbb{C}[z_{1},z_{2},\cdots,z_{n}]\right\\}.$ Clearly,
$K\subset\widehat{K}.$ We say $K$ is polynomially convex if $\widehat{K}=K.$
In $\mathbb{C},$ $\hat{K}=K$ if and only if $\mathbb{C}\setminus K$ is
connected. Any convex compact subset of $\mathbb{C}^{n}$ is polynomially
convex. In general, it is difficult to deduce if a given compact subset in
$\mathbb{C}^{n}$ is polynomially convex. A closed subset $E$ is locally
polynomially convex at $p\in E$ if there exists $r>0$ such that
$E\cap\overline{B(p,r)}$ is polynomially convex. Let $K$ be a compact set in
$\mathbb{C}^{n}$ and let $\mathcal{C}(K)$ be the class of of all continuous
complex-valued functions on $K$. By $\mathscr{P}(K)$, we denote the space of
all those functions on $K$ which are uniform limits of polynomials in
$z_{1},z_{2},\cdots,z_{n}$. One of the fundamental question in the theory of
uniform algebras is to characterize compact subset of $\mathbb{C}^{n}$ for
which

$\mathscr{P}(K)=\mathcal{C}(K).$ (1.3)

From the theory of commutative Banach algebras (see [13] for details), we
notice that

$\mathscr{P}(K)=\mathcal{C}(K)\Longrightarrow\widehat{K}=K.$

Therefore, polynomial convexity is a necessary condition for all compacts $K$
of $\mathbb{C}^{n}$ having property (1.3). Lavrentiev [20] showed that for
$K\subset\mathbb{C},$ $\mathscr{P}(K)=\mathcal{C}(K)$ if and only if
$\widehat{K}=K$ and $int(K)=\emptyset.$ In higher dimension, no such
characterization is known. The condition $\mathbb{C}\setminus K$ connected
generalizes to polynomial convexity of $K.$ A generalization of
$int(K)=\emptyset$ condition is that the compact $K$ is totally real except a
small set of points. Recall that a $C^{1}$-smooth submanifold $M$ of
$\mathbb{C}^{n}$ is said to be totally real at $p\in M$ if $T_{p}M\cap
iT_{p}M=\\{0\\},$ where the tangent space $T_{p}M$ is viewed as a real linear
subspace of $\mathbb{C}^{n}.$ Otherwise, we say that $M$ has complex tangent
at $p.$ The manifold $M$ is said to be totally real if it is totally real at
all points of $M$ and a subset $K$ of $\mathbb{C}^{n}$ is said to be totally
real if it is locally contained in a totally real manifold. Result 1.4 due to
Wermer says that a totally real submanifold of $\mathbb{C}^{2}$ is locally
polynomially convex. H$\ddot{o}$rmander and Wermer [18] generalized this
result for smooth totally real submanifold of $\mathbb{C}^{n}.$ Smoothness is
reduced to $C^{1}$ due to Harvey and Wells [15, 16]. For polynomially convex
set $K$, there are several papers, for instance see [1, 2, 27, 32, 34], that
describe situations when (1.3) holds. For $n>1,$ the most general result known
in this direction is the following due to O’Farrell, Preskenis and Walsh [27]:

######  Result 1.5 (O’Farrell, Preskenis and Walsh).

Let K be a compact polynomially convex subset of $\mathbb{C}^{n}$. Assume that
$E$ is a closed subset of $K$ such that $K\setminus E$ is locally contained in
totally-real manifold. Then

$\mathscr{P}(K)=\\{f\in\mathcal{C}(K):f|_{E}\in\mathscr{P}(E)\\}.$

From Result 1.5, we can say that any compact polynomially convex subset of a
totally real submanifold of $\mathbb{C}^{n}$ enjoys property (1.3). Therefore,
to answer the approximation question mentioned in the beginning or (1.3), one
requires to answer certain polynomial convexity question. Gorai [14], Chi [5],
Zajec [35] gave some results regarding the polynomial convexity of a compact
that lies in a totally real submanifold of $\mathbb{C}^{n}.$ When the graph
has certain isolated points with complex tangents (CR-singularity), the
question of local polynomial convexity near the CR-singularity becomes more
complicated. The study has been initiated with Bishop’s [4] work of attaching
analytic discs. Forstnerič-Stout [12] proved local approximation near certain
type of CR-singularity. In this paper, we will focus on sets with certain
particular forms which has certain CR-singularity. First, we mention a result
by Bharali [3], which was a motivating factor for our study.

######  Result 1.6 (Bharali).

Let $\overline{D}$ be a closed disc in $\mathbb{C}$ with center at the origin.
Let $F$ be a function defined by
$F(z)=\bar{z}^{m}+R(z),z\in\overline{D},\text{ where }m\in\mathbb{Z}_{+}$ and
$R$ satisfies

$\displaystyle|R(z)-R(\xi)|<\left|z^{m}-\xi^{m}\right|~{}~{}\forall
z,\xi\in\overline{D}:z^{m}\neq\xi^{m}.$

Then $\Gamma:=Gr_{\overline{D}}(F),\text{the graph of F over }\overline{D}$ is
polynomially convex. Additionally, we can conclude that
$[z,F]_{\overline{D}}=\mathcal{C}(\overline{D})$ in the following cases:

  * •

whenever m=1 (with no further conditions on R)

  * •

if $m\geq 2,R\in\mathcal{C}^{1}(\overline{D})$ and there exist
$\alpha\in(0,1)$ such that $R$ satisfies the stronger estimate:

$|R(z)-R(\xi)|<\alpha\left|z^{m}-\xi^{m}\right|~{}~{}\forall
z,\xi\in\overline{D}:z^{m}\neq\xi^{m}.$

######  Remark 1.7.

In Result 1.6, the graph has a CR-singularity at the origin.

We now present a generalization of Result 1.6 for $n\geq 2.$ This case differs
from Result 1.6 in the sense that the set of points where the tangent space of
the corresponding graph fails to be totally real is non-isolated. It is the
graph over an analytic variety of $D$ of co-dimension one.

######  Theorem 1.8.

Let $D$ be an open polydisc in $\mathbb{C}^{n}$ with centre at the origin and
let $\Omega$ be a neighborhood of $\overline{D}$. Let
$F_{1},F_{2},\cdots,F_{n}$ are functions on $\Omega$ defined by
$F_{j}(z)=\bar{z}^{m_{j}}_{j}+R_{j}(z)\text{ where }m_{j}\in\mathbb{N}\text{
for all }j=1,2,\cdots,n,$ and $R=(R_{1},R_{2},\cdots,R_{n})$ satisfies

$\displaystyle|R(z)-R(\xi)|\leq
c\left(\sum_{j=1}^{n}\left|z_{j}^{m_{j}}-\xi_{j}^{m_{j}}\right|^{2}\right)^{\frac{1}{2}},$
(1.4)

$\text{for all
}z=(z_{1},\cdots,z_{n}),~{}~{}\xi=(\xi_{1},\cdots,\xi_{n})\in\Omega~{}~{}\text{
and for some }c\in(0,1).$ Then $\Gamma:=Gr_{\overline{D}}(F),$ the graph of
$F$ over $\overline{D}$ is polynomially convex. Additionally,  if  R is
$\mathcal{C}^{1}$-smooth on $\Omega$ then we can conclude that

$\displaystyle[z_{1},\cdots,z_{n},\bar{z_{1}}^{m_{1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{n}}+R_{n}(z);\overline{D}]=C(\overline{D}).$

We now discuss the situation where $\nu_{j}\neq 1$ for at least one
$j\in\\{1,\cdots,n\\}.$ For $n=1,$ in [22], Minsker proved the following.

######  Result 1.9 (Minsker).

Let $k,l\in\mathbb{N}$ such that $\gcd(k,l)=1$ and let $\mathbb{D}$ be an open
unit disc in $\mathbb{C}.$ Then
$[z^{k},\bar{z}^{l};\overline{\mathbb{D}}]=C(\overline{\mathbb{D}}).$

In [6], De Paepe gave the following generalization of Result 1.9:

######  Result 1.10.

Let $F(z)=z^{m}(1+f(z)),$ $G(z)=\bar{z}^{n}(1+g(z))$ where $f$ and $g$ are
functions defined in a neighborhood of the origin, of class $C^{1},$ with
$f(0)=0,$ $g(0)=0.$ If $\gcd(m,n)=1$ and if $D$ is a sufficiently small closed
disc around $0$ then $[F,G;D]=C(D).$

Similar type of result can be found in O’Farrell-Preskenis [24], De Paepe [6,
7, 8, 9, 10], De Paepe and Wiegerinck [11], O’Farrell and De Paepe [23], Chi
[5]. But we did not find any result when $\nu_{j}>1,$ for at least one
$j\in\\{1,2,\cdots,n\\},$ and $n>1$ in the literature. In this article, we
present the following result. In this case, the corresponding graph has non-
isolated CR-singularity.

######  Theorem 1.11.

Let $\mathbb{D}^{n}$ be the open unit polydisc in $\mathbb{C}^{n}$ and
$m_{j}\in\mathbb{N}\text{ for all }j=1,2,\cdots,2n$ and $\gcd(m_{i},m_{j})=1$
for all $i\neq j.$ Then

$\displaystyle[z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}},\cdots,\bar{z_{n}}^{m_{2n}};\overline{\mathbb{D}}^{n}]=C(\overline{\mathbb{D}}^{n}).$

Let $\delta$ be a positive real number. By $D(\delta),$ we denote the open
polydisc $D(\delta):=\\{z\in\mathbb{C}^{n}:|z_{j}|<\delta,j=1,\cdots,n\\}$ in
$\mathbb{C}^{n}$ with polyradius $(\delta,\cdots,\delta).$

######  Theorem 1.12.

Let $m_{1},m_{2},\cdots,m_{2n}$ be positive integers. Fix $\delta_{0}>0.$
Consider a map $R=(R_{1},R_{2},\cdots,R_{n})$ with values in $\mathbb{C}^{n},$
defined and of class $C^{1}$ in $D(\delta_{0})$. Suppose that there is a
constant $0<c<1$ such that

  1. (i)

$|R(z)-R(\xi)|\leq
c\left(\sum_{j=1}^{n}\left|z_{j}^{m_{n+j}}-\xi_{j}^{m_{n+j}}\right|^{2}\right)^{\frac{1}{2}}~{}~{}\forall
z=(z_{1},\cdots,z_{n}),~{}~{}\xi=(\xi_{1},\cdots,\xi_{n})\in D(\delta_{0});$

  2. (ii)

$R_{j}(z)\sim o(|z_{j}|^{m_{n+j}})\text{ as }z_{j}\to 0,\text{ for all
}j=1,2,\cdots,n.$

Then there exist $\delta,$ $0<\delta<\delta_{0},$ such that for
$K:=\\{(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)):z\in\overline{D(\delta)}\\},\mathscr{P}(K)=\mathcal{C}(K).$
Furthermore, if $\gcd(m_{i},m_{j})=1$ $\forall i\neq j,$ then

$\displaystyle[z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z);\overline{D(\delta)}]=C(\overline{D(\delta)}).$

######  Remark 1.13.

Our proof restricts $R_{j}$ to have the property. We do not know what happens
when $R_{j}(z)\sim o(|z|^{m_{n+j}})\text{ for all }j=1,2,\cdots,n.$

######  Remark 1.14.

If $m_{j}=1~{}\forall j=1,\cdots,n,$ then, Theorem 1.8 says that property (ii)
is not required, and in this case, we will have a global result.

Let $X$ be a subset of $\mathbb{C}^{n}.$ $X$ is said to be stratified totally
real set if there exists finitely many closed sets $X_{j},j=1,\cdots,N$ such
that $X_{j}\setminus X_{j-1}$ is a totally real set and
$X_{0}\subset\cdots\subset X_{N}=X$ with $X=\cup_{j}X_{j}.$ We now provide few
remarks about the proof.

  * (i)

For the proof of Theorem 1.8, we first show that graph of $F$ is polynomially
convex by extending a result presented in [3, Proposition 1.7] to higher-
dimension. For uniform approximation, we first locate the set of non-totally
real points (Proposition 2.5), and then we give a totally real stratification
for the graph. Finally, we conclude Theorem 1.8 by applying Result 2.4 due to
Samuelsson and Wold [30].

  * (ii)

To prove Theorem 1.11, we first use a suitable proper holomorphic map $\Psi$
from $\mathbb{C}^{2n}$ to $\mathbb{C}^{2n}$ such that the preimage of
$X:=\\{(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}},\cdots,\bar{z_{n}}^{m_{2n}})\in\mathbb{C}^{2n}:z\in\overline{\mathbb{D}}^{n}\\}$
is the finite union of $X_{I},$ where each $X_{I}$ is the graph of some linear
function. Applying Stone-Weierstrass approximation theorem, we show that
$\mathscr{P}(X_{I})=\mathcal{C}(X_{I})$ for each $I.$ After that repeated use
of Kallin’s lemma (Result 2.2) gives the polynomial convexity of
$\cup_{I}X_{I}.$ Approximation then follows from Result 2.3.

  * (iii)

The method of the proof of Theorem 1.12 is similar to that of Theorem 1.8. In
this case, we also find a proper holomorphic map
$\Phi:\mathbb{C}^{2n}\to\mathbb{C}^{2n}$ such that the preimage of
$X:=\\{(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)):z\in\overline{D}\\}$
is the finite union of $X_{I},$ where $D$ is some small polydisc centred at
the origin. Using Theorem 1.8, we show that
$\mathscr{P}(X_{I})=\mathcal{C}(X_{I})$ for each $I.$ We find a holomorphic
polynomial $p:\mathbb{C}^{2n}\to\mathbb{C}$ such that each $p(X_{I})$ is
contained in an angular sector $\omega_{I}\subset\mathbb{C}$ such that
$\omega_{I}\cap\omega_{J}=\\{0\\}~{}~{}\forall I\neq J$ and
$p^{-1}\\{0\\}\cap(\cup_{I}X_{I})$ is polynomially convex. Then by repeated
application of Kallin’s lemma (Result 2.2), we conclude that
$\mathscr{P}(\cup_{I}X_{I})=\mathcal{C}(\cup_{I}X_{I}).$ Since
$z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)$
separates points $\overline{D}$, we conclude Theorem 1.12, by applying Result
2.3.

The paper is organized as follows. Section 2 collects some earlier results
that we will be using in this paper. We also state and proof of a result that
characterizes complex points of certain graphs. We also discuss a result that
gives a class of continuous functions having polynomially convex graphs. In
Section 3, we give a proof of Theorem 1.8. Section 4 and Section 5 are devoted
to the proof of Theorem 1.11 and Theorem 1.12 respectively.

##  2\. Technical Results

We begin this section by mentioning few known results that will be used in the
proofs of our theorems. The first one is due to Hörmander [17]. Let ${\sf
psh}(\Omega)$ denotes the collection of all plurisubharmonic functions on
$\Omega.$

######  Result 2.1 (Hörmander).

Let $K$ be a compact subset of a pseudoconvex open set
$\Omega\subset\mathbb{C}^{n}$. Then
$\widehat{K}_{\Omega}=\widehat{K}_{\Omega}^{P}$, where
$\widehat{K}_{\Omega}^{P}:=\left\\{z\in\Omega:u(z)\leq\sup\nolimits_{K}u\;\forall
u\in{\sf psh}(\Omega)\right\\}$ and
$\widehat{K}_{\Omega}:=\big{\\{}z\in\Omega:|f(z)|\leq\sup\nolimits_{z\in
K}|f(z)|\;\forall f\in\mathcal{O}(\Omega)\big{\\}}$.

In case $\Omega=\mathbb{C}^{n},$ Result 2.1 says that the polynomially convex
hull of $K$ is the same as the plurisubharmonically convex hull of $K$.

We now state a couple of results from Stouts book [33]. The first one [33,
Theorem 1.6.19] is a lemma due to Eva Kallin [19] and is often referred to as
Kallin’s lemma.

######  Result 2.2.

Let $X_{1}$ and $X_{2}$ be compact, polynomially convex subset of
$\mathbb{C}^{n}.$ Let $p$ be a polynomial such that
$\widehat{p(X_{1})}\cap\widehat{p(X_{2})}\subset\\{0\\}.$ If the set
$p^{-1}(0)\cap(X_{1}\cup X_{2})$ is polynomially convex, then the set
$X_{1}\cup X_{2}$ is polynomially convex. If, in addition,
$\mathscr{P}(X_{j})=\mathcal{C}(X_{j}),j=1,2,$ then $\mathscr{P}(X_{1}\cup
X_{2})=\mathcal{C}(X_{1}\cup X_{2}).$

######  Result 2.3.

If $F:\mathbb{C}^{n}\to\mathbb{C}^{n}$ is a proper holomorphic map, and if
$X\subset\mathbb{C}^{n}$ is a compact set, then the set $X$ is polynomially
convex if and only if the set $F^{-1}(X)$ is polynomially convex, and
$\mathcal{P}(X)=\mathcal{C}(X)$ if and only if
$\mathcal{P}(F^{-1}(X))=\mathcal{C}(F^{-1}(X)).$

We now state an approximation theorem on stratified totally real set due to
Samuelsson and Wold [30]. Let $\mathcal{O}(X_{0})$ be the collection of all of
all holomorphic function on $X_{0}.$

######  Result 2.4.

Let $X$ be a polynomially convex compact set in $\mathbb{C}^{n}$ and assume
that there are closed sets $X_{0}\subset\cdots\subset X_{N}=X$ such that
$X_{j}\setminus X_{j-1},$ $j=1,\cdots,N,$ is a totally real set. Then
$[z_{1},\cdots,z_{n}]_{X}=\\{f\in\mathcal{C}(X):f|_{X_{0}}\in\mathcal{O}(X_{0})\\}.$
In particular, if $C(X_{0})=\mathcal{O}(X_{0})$ then
$C(X)=[z_{1},\cdots,z_{n}]_{X}.$

The next lemma that we state and prove gives a characterization when the graph
$\Gamma$ in the statement of Theorem 1.8 is totally real. This will play a
vital role in our proof of Theorem 1.8.

Let $D$ be an open polydisc in $\mathbb{C}^{n}$ with center at the origin and
let $\Omega$ be a neighborhood of $\overline{D}$. Let
$F_{1},F_{2},\cdots,F_{n}$ are functions defined by
$F_{j}(z)=\bar{z}^{m_{j}}_{j}+R_{j}(z)\text{ on }\Omega\text{ where
}m_{j}\in\mathbb{N}\text{ for all }j=1,2,\cdots,n$ and
$R=(R_{1},R_{2},\cdots,R_{n})$ satisfies (1.4). Also assume that $R$ is
$\mathcal{C}^{1}$-smooth on $\Omega.$ We define a map
$\Phi:\Omega\to\mathbb{C}^{2n}$ by

$\displaystyle\Phi(z_{1},\cdots,z_{n}):=(z_{1},\cdots,z_{n},F_{1}(z),\cdots,F_{n}(z)).$

For $k\leq n,$ let $\widetilde{\Omega}$ be an open set in $\mathbb{C}^{k}$ and
$g:\widetilde{\Omega}\hookrightarrow\Omega$ be an embedding defined by
$g(z_{1},\cdots,z_{k}):=(z_{1},\cdots,z_{k},0,\cdots,0).$ Set $M:=(\Phi\circ
g)(\widetilde{\Omega})=Gr_{g(\widetilde{\Omega})}(F),$ which is a real
submanifold of $\mathbb{C}^{2n}$ of dimension $2k.$

######  Proposition 2.5.

$M$ has complex tangents at $(\Phi\circ g)(z)$ if and only if
$z\in\\{(z_{1},\cdots,z_{k})\in\widetilde{\Omega}:z_{1}\cdots z_{k}=0\\}.$

###### Proof.

First, we assume that $M$ has complex tangent at some point
$z\in\widetilde{\Omega}$. Take $w=\left(\Phi\circ g\right)(z).$ Let $T_{w}M$
be the tangent space to $M$ at $w$ viewed as a real-linear subspace of
$\mathbb{C}^{2n}.$ Since $\Phi\circ g$ is an embedding of
$\widetilde{\Omega},$ we have

$\displaystyle T_{w}M=\left\\{d(\Phi\circ
g)|_{z}(v):v\in\mathbb{C}^{k}\right\\}.$

Since $T_{w}M$ contains non-trivial complex subspace, then there exists a non-
zero vector $\eta\in T_{w}M$ such that $i\eta$ also belongs to $T_{w}M.$ We
take $\eta=d(\Phi\circ g)|_{z}(v^{0}),$ for some
$v^{0}\in\mathbb{C}^{k}\setminus\\{0\\}$ and $i\eta=d(\Phi\circ
g)|_{z}(\omega^{0}),$ for some $\omega^{0}\in\mathbb{C}^{k}\setminus\\{0\\}.$
Therefore, we obtain that

$\displaystyle d(\Phi\circ g)|_{z}(\omega^{0})=id(\Phi\circ g)|_{z}(v^{0}).$
(2.1)

We denote $z^{\prime}:=(z_{1},\cdots,z_{k},0,\cdots,0)\in\mathbb{C}^{n}\text{
for }z=(z_{1},\cdots,z_{k})\in\mathbb{C}^{k}.$

$\displaystyle(\Phi\circ
g)(z_{1},\cdots,z_{k})=(z_{1},\cdots,z_{k},0,\cdots,0,\bar{z_{1}}^{m_{1}}+R_{1}(z^{\prime}),\cdots,\bar{z_{k}}^{m_{k}}+R_{k}(z^{\prime}),R_{k+1}(z^{\prime}),\cdots,R_{n}(z^{\prime})).$

The matrix representation of $d(\Phi\circ g)|_{z}$ is

$\begin{pmatrix}\begin{matrix}I_{k}\\\ \end{matrix}&\vline&0_{k\times k}\\\
\hline\cr\begin{matrix}0_{(n-k)\times k}\end{matrix}&\vline&0_{(n-k)\times
k}\\\ \hline\cr\begin{matrix}R_{z}(z^{\prime})\\\
\end{matrix}&\vline&\begin{matrix}A(z^{\prime})+R_{\bar{z}}(z^{\prime})\\\
\end{matrix}\end{pmatrix}_{2n\times 2k},$

Where

$R_{z}(z^{\prime}):=\begin{pmatrix}\frac{\partial
R_{1}}{\partial{z_{1}}}(z^{\prime})&\frac{\partial
R_{1}}{\partial{z_{2}}}(z^{\prime})&\cdots&\frac{\partial
R_{1}}{\partial{z_{k}}}(z^{\prime})\\\\[6.45831pt] \frac{\partial
R_{2}}{\partial{z_{1}}}(z^{\prime})&\frac{\partial
R_{2}}{\partial{z_{2}}}(z^{\prime})&\cdots&\frac{\partial
R_{2}}{\partial{z_{k}}}(z^{\prime})\\\\[6.45831pt]
\vdots&\vdots&\cdots&\vdots\\\ \frac{\partial
R_{n}}{\partial{z_{1}}}(z^{\prime})&\frac{\partial
R_{n}}{\partial{z_{2}}}(z^{\prime})&\cdots&\frac{\partial
R_{n}}{\partial{z_{k}}}(z^{\prime})&\\\ \\\ \end{pmatrix}_{n\times
k},~{}~{}R_{\bar{z}}(z^{\prime}):=\begin{pmatrix}\frac{\partial
R_{1}}{\partial\bar{{z_{1}}}}(z^{\prime})&\frac{\partial
R_{1}}{\partial\bar{z_{2}}}(z^{\prime})&\cdots&\frac{\partial
R_{1}}{\partial\bar{{z_{k}}}}(z^{\prime})\\\\[6.45831pt] \frac{\partial
R_{2}}{\partial\bar{{z_{1}}}}(z^{\prime})&\frac{\partial
R_{2}}{\partial\bar{z_{2}}}(z^{\prime})&\cdots&\frac{\partial
R_{2}}{\partial\bar{{z_{k}}}}(z^{\prime})\\\\[6.45831pt]
\vdots&\vdots&\cdots&\vdots\\\ \frac{\partial
R_{n}}{\partial\bar{{z_{1}}}}(z^{\prime})&\frac{\partial
R_{n}}{\partial\bar{z_{2}}}(z^{\prime})&\cdots&\frac{\partial
R_{n}}{\partial\bar{{z_{k}}}}(z^{\prime})\\\ \\\ \end{pmatrix}_{n\times k},$

and

$\displaystyle
A(z^{\prime})=\begin{pmatrix}m_{1}\bar{z_{1}}^{m_{1}-1}&0&0&\cdots&0\\\\[6.45831pt]
0&m_{2}\bar{z_{2}}^{m_{2}-1}&0&\cdots&0\\\\[6.45831pt]
\vdots&\vdots&\vdots&&\vdots\\\
0&0&0&\cdots&m_{k}\bar{z_{k}}^{m_{k}-1}\\\\[6.45831pt]
0&0&0&\cdots&0\\\\[6.45831pt] \vdots&\vdots&\vdots&&\vdots\\\
0&0&0&\cdots&0\\\ \end{pmatrix}_{n\times k},$ (2.2)

For any vector $\beta$ in $\mathbb{C}^{k},$

$d(\Phi\circ g)|_{z}(\beta)=\begin{pmatrix}\begin{matrix}I_{k}\\\
\end{matrix}&\vline&0_{k\times k}\\\ \hline\cr\begin{matrix}0_{(n-k)\times
k}\end{matrix}&\vline&0_{(n-k)\times k}\\\
\hline\cr\begin{matrix}R_{z}(z^{\prime})\\\
\end{matrix}&\vline&\begin{matrix}A(z^{\prime})+R_{\bar{z}}(z^{\prime})\\\
\end{matrix}\end{pmatrix}_{2n\times 2k}\begin{pmatrix}\beta\\\
\overline{\beta}\end{pmatrix}.$

This implies

$d(\Phi\circ
g)|_{z}(\beta)=\left(\beta,0,\cdots,0,R_{z}(z^{\prime})\beta+R_{\bar{z}}({z^{\prime}})\bar{\beta}+A(z^{\prime})\bar{\beta}\right).$

Hence, (2.1) gives us that

$(\omega^{0},0,\cdots,0,R_{z}(z^{\prime})\omega^{0}+R_{\bar{z}}(z^{\prime})\bar{\omega^{0}}+A(z^{\prime})\bar{\omega^{0}})=i(v^{0},0,\cdots,0,R_{z}(z^{\prime})v^{0}+R_{\bar{z}}(z^{\prime})\bar{v^{0}}+A({z^{\prime}})\bar{v^{0}}).$

It follows that $\omega^{0}=iv^{0}$ and

$(R_{z}(z^{\prime})\omega^{0}+R_{\bar{z}}(z^{\prime})\bar{\omega^{0}}+A(z^{\prime})\bar{\omega^{0}})=i(R_{z}(z^{\prime})v^{0}+R_{\bar{z}}(z^{\prime})\bar{v^{0}}+A({z^{\prime}})\bar{v^{0}}).$

Putting $\omega^{0}=iv^{0}$ in the above equation, we obtain that

$\displaystyle R_{\bar{z}}(z^{\prime})\bar{v^{0}}+A(z^{\prime})\bar{v^{0}}=0.$
(2.3)

By Taylor’s formula, for
$z=(z_{1},\cdots,z_{k})\in\widetilde{\Omega},~{}\vartheta=(\vartheta_{1},\cdots,\vartheta_{k})\in\mathbb{C}^{k},\text{
and }\epsilon$ real,

$\displaystyle(R\circ g)(z+\varepsilon\vartheta)-(R\circ
g)(z)=\big{(}(R_{1}\circ g)(z+\varepsilon\vartheta)-(R\circ
g)(z),\cdots,(R\circ g)(z+\varepsilon\vartheta)-(R_{n}\circ g)(z)\big{)}$
$\displaystyle=\bigg{(}\sum_{j=1}^{k}\bigg{[}\frac{\partial(R_{1}\circ
g)(z)}{\partial z_{j}}(\varepsilon\vartheta_{j})+\frac{\partial(R_{1}\circ
g)(z)}{\partial\bar{z_{j}}}(\varepsilon\bar{\vartheta_{j}})\bigg{]}+o(\varepsilon),\cdots,\sum_{j=1}^{k}\bigg{[}\frac{\partial(R_{n}\circ
g)(z)}{\partial z_{j}}(\varepsilon\vartheta_{j})+$
$\displaystyle\frac{\partial(R_{n}\circ
g)(z)}{\partial\bar{z_{j}}}(\varepsilon\bar{\vartheta_{j}})\bigg{]}+o(\varepsilon)\bigg{)}$
$\displaystyle=\left(\sum_{j=1}^{k}\bigg{[}\frac{\partial
R_{1}(z^{\prime})}{\partial z_{j}}(\varepsilon\vartheta_{j})+\frac{\partial
R_{1}(z^{\prime})}{\partial\bar{z_{j}}}(\varepsilon\bar{\vartheta_{j}})\bigg{]}+o(\varepsilon),\cdots,\sum_{j=1}^{k}\bigg{[}\frac{\partial
R_{n}(z^{\prime})}{\partial z_{j}}(\varepsilon\vartheta_{j})+\frac{\partial
R_{n}(z^{\prime})}{\partial\bar{z_{j}}}(\varepsilon\bar{\vartheta_{j}})\bigg{]}+o(\varepsilon)\right)$
$\displaystyle=R_{z}(z^{\prime})\epsilon\vartheta+R_{\bar{z}}(z^{\prime})\epsilon\bar{\vartheta}+o(\epsilon),~{}~{}\text{
where }z^{\prime}=(z,0)\in\Omega.$ (2.4)

Applying (1.4), we have, from (2), that

$\displaystyle|R_{z}(z^{\prime})\epsilon\vartheta+R_{\bar{z}}(z^{\prime})\epsilon\bar{\vartheta}+o(\epsilon)|$
$\displaystyle\leq
c\left(\sum_{j=1}^{k}\left|(z_{j}+\epsilon\vartheta_{j})^{m_{j}}-z_{j}^{m_{j}}\right|^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq
c\left(\sum_{j=1}^{k}|\epsilon\vartheta_{j}|^{2}\left(\sum_{l=0}^{m_{j}-1}\left|z_{j}+\epsilon\vartheta_{j}\right|^{l}\left|z_{j}\right|^{m_{j}-l-1}\right)^{2}\right)^{\frac{1}{2}}.$

Taking $\epsilon\to 0,$ we get

$\displaystyle\left|R_{z}(z^{\prime})\vartheta+R_{\bar{z}}(z^{\prime})\bar{\vartheta}\right|$
$\displaystyle\leq
c\left(\sum_{j=1}^{k}|\vartheta_{j}|^{2}\left(\sum_{l=0}^{m_{j}-1}\left|z_{j}\right|^{l}\left|z_{j}\right|^{m_{j}-l-1}\right)^{2}\right)^{\frac{1}{2}},$

i.e.

$\displaystyle\left|R_{z}(z^{\prime})\vartheta+R_{\bar{z}}(z^{\prime})\bar{\vartheta}\right|$
$\displaystyle\leq
c\left(\sum_{j=1}^{k}\left(m_{j}|\vartheta_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}\text{
for all }\vartheta\in\mathbb{C}^{k}.$ (2.5)

Since (2.5) is true for all $\vartheta\in\mathbb{C}^{n},$ we now replace
$\vartheta$ by $i\vartheta$ to get

$\displaystyle\left|R_{z}(z^{\prime})\vartheta-
R_{\bar{z}}(z^{\prime})\bar{\vartheta}\right|$ $\displaystyle\leq
c\left(\sum_{j=1}^{k}\left(m_{j}|\vartheta_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}\text{
for all }\vartheta\in\mathbb{C}^{k}.$ (2.6)

In view of (2.5) and (2.6), we get that

$\displaystyle\left|R_{\bar{z}}(z^{\prime})\bar{\vartheta}\right|$
$\displaystyle\leq
c\left(\sum_{j=1}^{k}\left(m_{j}|\vartheta_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}\text{
for all }\vartheta\in\mathbb{C}^{k}.$ (2.7)

Therefore, from (2.3) and (2.7), we get that

$\displaystyle\left|A(z^{\prime})\overline{v^{0}}\right|$ $\displaystyle\leq
c\left(\sum_{j=1}^{k}\left(m_{j}|v^{0}_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}},\text{
where
}v^{0}=(v^{0}_{1},v^{0}_{2},\cdots,v^{0}_{k})\in\mathbb{C}^{k}\setminus\\{0\\}.$
(2.8)

This implies

$\displaystyle\left(\sum_{j=1}^{k}\left(m_{j}|v^{0}_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq
c\left(\sum_{j=1}^{k}\left(m_{j}|v^{0}_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}.$
(2.9)

Now, if
$\left(\sum_{j=1}^{k}\left(m_{j}|v^{0}_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}\neq
0,$ then we obtain that $c\geq 1,$ which is a contradiction to our assumption.
Hence,
$\left(\sum_{j=1}^{k}\left(m_{j}|v^{0}_{j}|\left|z_{j}\right|^{m_{j}-1}\right)^{2}\right)^{\frac{1}{2}}=0.$
Again, viewing $A(z^{\prime})$ as a $\mathbb{C}$-linear map from
$\mathbb{C}^{k}$ to $\mathbb{C}^{n}$ with $v^{0}\in\ker A(z^{\prime}),$ from
(2.2) we get that $\operatorname{rank}A(z^{\prime})<k$ and hence
$z_{1}z_{2}\cdots z_{k}=0.$

Conversely, assume that
$p\in\left\\{z=(z_{1},z_{2},\cdots,z_{k})\in\widetilde{\Omega}:z_{1}z_{2}\cdots
z_{k}=0\right\\}.$ We need to show that $M$ has complex tangent at $(\Phi\circ
g)(p).$ Without loss of generality, we take
$p=(p_{1},p_{2},\cdots,p_{k-1},0).$ Suppose $M$ does not have complex tangent
at $(\Phi\circ g)(p)$. Thus, we have
$(A(p^{\prime})+R_{\bar{z}}(p^{\prime}))v\neq 0~{}~{}\forall
v\in\mathbb{C}^{k}\setminus\\{0\\},$ where $p^{\prime}=(p,0)\in\Omega.$ Choose
$v=(0,0,\cdots,0,1)\in\mathbb{C}^{k}$. Since $A(p^{\prime})v=0,$ we have
$R_{\bar{z}}(p^{\prime})v\neq 0.$ Using (2.7), we get that
$|R_{\bar{z}}(p^{\prime})v|=0.$ This is a contradiction. Therefore, $M$ has a
complex tangent at $(\Phi\circ g)(p).$ ∎

Before going to the next section, we state and prove a lemma that will be
useful in the proof of Theorem 1.12.

######  Lemma 2.6.

Assume that $m_{j}\in\mathbb{N}\text{ and
}\alpha_{j}\in\mathbb{N}\cup\\{0\\},j=1,\cdots,2n,$ with
$\gcd\left(m_{i},m_{j}\right)=1$ for ${i\neq j},$ and
$\gcd(m_{i},\alpha_{j})=1~{}~{}\forall\alpha_{j}\neq 0.$ Let
$(t_{1},t_{2},\cdots,t_{2n}),(t^{\prime}_{1},t^{\prime}_{2},\cdots,t^{\prime}_{2n})\in\left\\{0,1,\cdots,m_{1}-1\right\\}\times\cdots\times\left\\{0,1,\cdots,m_{2n}-1\right\\}.$
Assume that $\\{i_{1},\cdots i_{k}\\}\subset\\{1,\cdots,2n\\}$ and if there
exist $j_{0}\in\\{1,\cdots,2n\\}\setminus\\{i_{1},\cdots i_{k}\\}$ such that
$t_{j_{0}}\neq t^{\prime}_{j_{0}},$ and $\alpha_{j_{0}}\neq 0,$ then

$\displaystyle{{\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}\alpha_{j}}{m_{j}}}}\neq{{\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t^{\prime}_{j}\alpha_{j}}{m_{j}}}}.$

###### Proof.

Assume that there exist
$j_{0}\in\\{1,\cdots,2n\\}\setminus\\{i_{1},\cdots,i_{k}\\}$ such that
$t_{j_{0}}\not=t^{\prime}_{j_{0}},$ $\alpha_{j_{0}}\not=0$ and

$\displaystyle{{\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}\alpha_{j}}{m_{j}}}}={{\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t^{\prime}_{j}\alpha_{j}}{m_{j}}}}.$

Therefore,

$\displaystyle\frac{\alpha_{j_{0}}}{m_{j_{0}}}\left(t_{j_{0}}^{\prime}-t_{j_{0}}\right)={{\displaystyle\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k},j_{0}\\}\\\
\end{subarray}}\frac{\alpha_{j}}{m_{j}}\bigg{(}t_{j}-t^{\prime}_{j}}}\bigg{)}$
$\displaystyle=\frac{{{\displaystyle\sum_{\begin{subarray}{c}j\notin\\{i_{1},\cdots,i_{k},j_{0}\\}\\\
\end{subarray}}\bigg{(}\alpha_{j}m_{1}m_{2}\cdots\widehat{m_{j}}\cdots
m_{n}(t_{j}-t^{\prime}_{j})}}\bigg{)}}{\displaystyle\prod_{\begin{subarray}{c}j\notin\\{i_{1},\cdots,i_{k},j_{0}\\}\\\
\end{subarray}}m_{j}},$

where $\widehat{m_{j}}$ denote the absence of $j^{th}$ term. This implies

$\displaystyle
m_{j_{0}}\bigg{(}{{\sum_{\begin{subarray}{c}j\notin\\{i_{1},\cdots,i_{k},j_{0}\\}\\\
\end{subarray}}\alpha_{j}m_{1}m_{2}\cdots\widehat{m_{j}}\cdots
m_{n}(t_{j}-t^{\prime}_{j})}}\bigg{)}=\alpha_{j_{0}}(t^{\prime}_{j_{0}}-t_{j_{0}})\prod_{\begin{subarray}{c}j\notin\\{i_{1},\cdots,i_{k},j_{0}\\}\\\
\end{subarray}}m_{j},$

From above, we can say that $m_{j_{0}}$ divides
$\displaystyle\alpha_{j_{0}}(t^{\prime}_{j_{0}}-t_{j_{0}})\prod_{\begin{subarray}{c}j\notin\\{i_{1},\cdots,i_{k},j_{0}\\}\\\
\end{subarray}}m_{j}.$ Since $\gcd(m_{i},m_{j})=1$ for $i\neq j$ and
$\gcd(m_{i},\alpha_{j})=1,$ we get that $m_{j_{0}}$ divides
$(t^{\prime}_{j_{0}}-t_{j_{0}}).$ This is not possible because $0\leq
t_{j_{0}},t^{\prime}_{j_{0}}<m_{j_{0}}.$

∎

In [3], Bharali introduced a technique to study the polynomial convexity of
graphs of functions in one variable. We observe here that the same technique
can be generalized to higher-dimension. We will now state and prove this
result. For that, we need some notations.

  * •

For a compact set $K\subset\mathbb{C}^{n},$ by $\mathcal{O}(K;\mathbb{C}^{N})$
we define the set of holomorphic functions from some neighborhood of $K$ to
$\mathbb{C}^{N}.$ If $N=1,$ we simply denote it $\mathcal{O}(K);$ And for fix
$\xi\in K,$ we define a sub-class of $\mathcal{O}(K;\mathbb{C}^{N})$ by
$\mathcal{O}_{\xi}(K;\mathbb{C}^{N}):=\left\\{F\in\mathcal{O}(K;\mathbb{C}^{N}):F(\xi)=0\right\\}.$

  * •

For $A\in\mathcal{O}(K)\text{ and
}F=(f_{1},\cdots,f_{N})\in\mathcal{O}(K;\mathbb{C}^{N})$ by $A(z)F(z)$, we
mean $\left(A(z)f_{1}(z),\cdots,\right.$ $\left.A(z)f_{N}(z)\right)$ and by
$P(z)Q(z),$ we mean $\sum_{j=1}^{N}p_{j}(z)q_{j}(z)$ for
$P=(p_{1},\cdots,p_{N})$ and $Q=(q_{1},\cdots,q_{N});$

  * •

Let $K$ be a subset of $\mathbb{C}^{n}$ and
$F:=(f_{1},f_{2},\cdots,f_{N}):K\to\mathbb{C}^{N}$ be a function. We will
denote the graph of $F$ over $K$ by $Gr_{K}F$ or by
$Gr_{K}(f_{1},f_{2},\cdots,f_{N}).$

######  Proposition 2.7.

Let $N\geq n\geq 1$ and $F\in\mathcal{C}(\overline{\Omega};\mathbb{C}^{N}),$
where $\Omega$ is a bounded domain in $\mathbb{C}^{n}$ with
$\overline{\Omega}$ is polynomially convex, and let $\xi\in\overline{\Omega}.$
Suppose there exist a constant $p\geq 2,$ a nowhere vanishing function
$A\in\mathcal{O}(\overline{\Omega}),$ and mappings
$G,H\in\mathcal{O}_{\xi}(\overline{\Omega};\mathbb{C}^{N})$ such that

$\displaystyle\left|A(z)F(z)-A(\xi)F(\xi)+G(z)\right|^{p}\leq{\sf
Re}\left(H(z)\left(A(z)F(z)-A(\xi)F(\xi)+G(z)\right)\right)~{}~{}\forall
z\in\overline{\Omega}.$ (2.10)

Then $Gr_{\overline{\Omega}}(F)$ is polynomially convex.

###### Proof.

Since $A\in\mathcal{O}(\overline{\Omega})$ and
$G,H\in\mathcal{O}_{\xi}(\overline{\Omega};\mathbb{C}^{N}),$ there exists a
neighborhood $U$ of $\overline{\Omega}$ such that the functions $A,G,H$ are
holomorphic on $U.$ Since $\overline{\Omega}$ is polynomially convex, there
exist an open polynomial polyhedron $\triangle$ such that
$\overline{\Omega}\subset\triangle\subset U.$ Define a map
$\Psi:\triangle\times\mathbb{C}^{N}\to\mathbb{R}$ by

$\displaystyle\Psi(z,w):=\left|A(z)w-A(\xi)F(\xi)+G(z)\right|^{p}-{\sf
Re}\left(H(z)\left(A(z)w-A(\xi)F(\xi)+G(z)\right)\right).$

Since ${\sf Re}\left(H(z)\left(A(z)w-A(\xi)F(\xi)+G(z)\right)\right)$ is
pluriharmonic in $z$ and $w$ on $\triangle\times\mathbb{C}^{N},$ $\Psi$ is
plurisubharmonic on $\triangle\times\mathbb{C}^{N}.$ We set
$\Gamma:=Gr_{\overline{\Omega}}(F)$. Since $\overline{\Omega}$ is polynomially
convex, we can say that
$\widehat{\Gamma}\Subset{\overline{\Omega}}\times\mathbb{C}^{N}.$ We consider
the set

$\displaystyle\mathcal{S}:=\big{\\{}(z,w)\in{\overline{\Omega}}\times\mathbb{C}^{N}$
$\displaystyle:\Psi(z,w)\leq 0\big{\\}}.$

We now claim that $\widehat{\Gamma}\subset\mathcal{S}.$ To prove this claim
assume $(z_{0},w_{0})\in\widehat{\Gamma}$ but
$(z_{0},w_{0})\notin\mathcal{S}.$ Therefore, $\Psi(z_{0},w_{0})>0$ and by
assumption (2.10), we get that $\sup_{\Gamma}\Psi(z,w)\leq 0.$ Since
$\triangle$ is an open Runge and Stein neighborhood of $\overline{\Omega}$
with $\Psi\in{\sf psh}(\triangle\times\mathbb{C}^{N})$ such that
$\Psi(z_{0},w_{0})>\sup_{\Gamma}\Psi(z,w),$ we get that
$(z_{0},w_{0})\notin\widehat{\Gamma}_{\mathcal{O}(\triangle\times\mathbb{C}^{N})}$
and so $(z_{0},w_{0})\notin\widehat{\Gamma}.$ Therefore, we obtain that
$\widehat{\Gamma}\subset\mathcal{S}.$ This also implies that

$\displaystyle\left|A(z)w-A(\xi)F(\xi)+G(z)\right|^{p}$
$\displaystyle\leq|H(z)|\left|A(z)w-A(\xi)F(\xi)+G(z)\right|~{}~{}\forall(z,w)\in\widehat{\Gamma}.$

Since $p\geq 2$, we obtain that

$\displaystyle\widehat{\Gamma}\subset\left\\{(z,w)\in\overline{\Omega}\times\mathbb{C}^{N}:\left|A(z)w-A(\xi)F(\xi)+G(z)\right|\leq|H(z)|^{\frac{1}{p-1}}\right\\}.$
(2.11)

Our next claim is:
$\widehat{\Gamma}\cap\left(\\{\xi\\}\times\mathbb{C}^{N}\right)=\\{(\xi,F(\xi))\\}.$
To prove this claim, assume
$(\alpha,\beta)\in\widehat{\Gamma}\cap\left(\\{\xi\\}\times\mathbb{C}^{N}\right).$
This implies $\alpha=\xi.$ Since $(\xi,\beta)\in\widehat{\Gamma},$ we get from
(2.11) that

$\displaystyle\left|A(\xi)\beta-A(\xi)F(\xi)+G(\xi)\right|\leq|H(\xi)|^{\frac{1}{p-1}}.$

In view of the assumption $H(\xi)=0=G(\xi),$ we obtain that
$|A(\xi)||\beta-F(\xi)|=0.$ Since $A$ is nowhere vanishing, we obtain
$\beta=F(\xi).$ This proves the claim. Since, by assumption, for each
$\xi\in\overline{\Omega},$ such functions $A,G,H$ exists, therefore, we obtain
that
$\widehat{\Gamma}=\widehat{\Gamma}\cap\left(\cup_{\xi\in{\overline{\Omega}}}\left(\\{\xi\\}\times\mathbb{C}^{N}\right)\right)=\cup_{\xi\in{\overline{\Omega}}}\\{(\xi,F(\xi))\\}=\Gamma.$

∎

##  3\. The proof of Theorem 1.8

First, we state and prove a lemma that is crucial to our proof.

######  Lemma 3.1.

Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$ such that
$\widehat{\overline{\Omega}}=\overline{\Omega}$ and let
$F_{1},F_{2},\cdots,F_{n}$ be functions on $\overline{\Omega}$ defined by
$F_{j}(z)=\bar{z}^{m_{j}}_{j}+R_{j}(z),~{}~{}j=1,\cdots,n,$ and
$m_{j}\in\mathbb{N}.$ Assume that there exists a real number $c\in(0,1)$ such
that the map $R=(R_{1},R_{2},\cdots,R_{n})$ satisfies

$\displaystyle|R(z)-R(\xi)|\leq
c\left(\sum_{j=1}^{n}\left|z_{j}^{m_{j}}-\xi_{j}^{m_{j}}\right|^{2}\right)^{\frac{1}{2}}~{}~{}\forall
z,\xi\in\overline{\Omega}.$

Then $Gr_{\overline{\Omega}}(F_{1},F_{2},\cdots,F_{n})$ is polynomially
convex.

###### Proof.

Fixing $\xi\in\overline{\Omega}$ we define
$\Psi_{\xi}:\overline{\Omega}\to\mathbb{C}$ by

$\displaystyle\Psi_{\xi}(z)$
$\displaystyle:=\sum_{j=1}^{n}\left(z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right)\left(F_{j}(z)-F_{j}(\xi)\right).$

Putting $F_{j}(z)=\bar{z}^{m_{j}}_{j}+R_{j}(z),$ we get that

$\displaystyle\Psi_{\xi}(z)=\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi_{j}^{m_{j}}\right|^{2}+\sum_{j=1}^{n}\left(z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right)\left(R_{j}(z)-R_{j}(\xi)\right)~{}~{}\forall
z\in\overline{\Omega}.$ (3.1)

Therefore, the real part of $\Psi_{\xi}(z)$ is

$\displaystyle{\sf
Re}(\Psi_{\xi}(z))=\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi_{j}^{m_{j}}\right|^{2}+{\sf
Re}\left(\sum_{j=1}^{n}\left(z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right)\left(R_{j}(z)-R_{j}(\xi)\right)\right).$

We now compute:

$\displaystyle{\sf
Re}\left(\sum_{j=1}^{n}\left(z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right)\left(R_{j}(z)-R_{j}(\xi)\right)\right)$
$\displaystyle\leq\sum_{j=1}^{n}\left|\left(z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right)\left(R_{j}(z)-R_{j}(\xi)\right)\right|$
$\displaystyle\leq\left(\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right|^{2}\right)^{\frac{1}{2}}|R(z)-R(\xi)|$
$\displaystyle\leq
c\left(\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right|^{2}\right).$

Here we use Cauchy-Schwarz for the second inequality and the last inequality
follows from (1.4). From above calculations, we have

$\displaystyle-c\left(\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right|^{2}\right)\leq{\sf
Re}\left(\sum_{j=1}^{n}\left(z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right)\left(R_{j}(z)-R_{j}(\xi)\right)\right)$
$\displaystyle\leq
c\left(\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right|^{2}\right).$

Therefore, we obtain that

$\displaystyle
0\leq(1-c)\left(\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right|^{2}\right)\leq{\sf
Re}(\Psi_{\xi}(z))\leq(1+c)\left(\sum_{j=1}^{n}\left|z^{m_{j}}_{j}-\xi^{m_{j}}_{j}\right|^{2}\right).$
(3.2)

Let us compute:

$\displaystyle\left|F(z)-F(\xi)\right|$
$\displaystyle=\left(\sum_{j=1}^{n}\left|\bar{z_{j}}^{m_{j}}+R_{j}(z)-\bar{\xi_{j}}^{m_{j}}-R_{j}(\xi)\right|^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq\left(\sum_{j=1}^{n}\left(\left|\bar{z_{j}}^{m_{j}}-\bar{\xi_{j}}^{m_{j}}\right|+\left|R_{j}(z)-R_{j}(\xi)\right|\right)^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq\left(\sum_{j=1}^{n}\left|\bar{z_{j}}^{m_{j}}-\bar{\xi_{j}}^{m_{j}}\right|^{2}\right)^{\frac{1}{2}}+\left(\sum_{j=1}^{n}\left|R_{j}(z)-R_{j}(\xi)\right|^{2}\right)^{\frac{1}{2}}.$

Here we use Minkowski inequality for the last inequality. Using (1.4), we
obtain from above that

$\displaystyle\left|F(z)-F(\xi)\right|^{2}\leq(1+c)^{2}\sum_{j=1}^{n}\left|{z_{j}}^{m_{j}}-{\xi_{j}}^{m_{j}}\right|^{2}~{}~{}\forall(z,\xi)\in\overline{\Omega}\times\overline{\Omega}.$
(3.3)

In view of (3.2) and (3.3), we get that

$\displaystyle\left|F(z)-F(\xi)\right|^{2}$
$\displaystyle\leq\frac{(1+c)^{2}}{(1-c)}{\sf
Re}(\Psi_{\xi}(z))~{}~{}\forall(z,\xi)\in\overline{\Omega}\times\overline{\Omega}.$
(3.4)

Therefore, we obtain that $|F(z)-F(\xi)|^{2}\leq\text{C}{\sf
Re}\left(\Psi_{\xi}(z)\right),$ where C is a constant on the right side of
(3.4) which is independent of $z.$ Hence, by Proposition 2.7, we conclude that
$Gr_{\overline{\Omega}}(F)$ is polynomially convex. ∎

###### Proof of Theorem 1.8.

In view of Lemma 3.1, we know that $\Gamma:=Gr_{\overline{D}}(F),$ the graph
of $F$ over $\overline{D}$ is polynomially convex.

We now assume that $R$ is $\mathcal{C}^{1}$-smooth on $\Omega.$ We wish to use
Result 2.4 for approximation. For that we require a suitable totally real
stratification of $\Gamma.$ For each subset
$\\{i_{1},\cdots,i_{k}\\}\subset\\{1,\cdots,n\\},$ we define

$\displaystyle\sigma_{(i_{1},\cdots,i_{k})}(z_{1},\cdots,z_{n}):=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}z_{j}.$

We now give a stratification of $\Omega$ as follows:

$\displaystyle Z_{n}$ $\displaystyle:=\Omega;$ $\displaystyle Z_{n-1}$
$\displaystyle:=\left\\{z\in\Omega:\prod^{n}_{\begin{subarray}{c}j=1\\\
\end{subarray}}z_{j}=0\right\\};$ $\displaystyle Z_{n-2}$
$\displaystyle:=\left\\{z\in\Omega:\sigma_{(i_{1})}=0,z_{i_{1}}=0,1\leq
i_{1}\leq n\right\\};$ $\displaystyle Z_{n-(k+1)}$
$\displaystyle:=\left\\{z\in\Omega:\sigma_{(i_{1},\cdots,i_{k})}=0,z_{i_{l}}=0,l=1,\cdots,k,~{}~{}1\leq
i_{1}\neq\cdots\neq i_{k}\leq n\right\\},1\leq k\leq n-1.$

Clearly,

$\displaystyle Z_{0}$
$\displaystyle:=\left\\{z\in\Omega:\sigma_{(i_{1},i_{2},\cdots,i_{n-1})}=0,1\leq
i_{1}\neq i_{2}\neq\cdots\neq i_{n-1}\leq n\right\\}=\\{(0,\cdots,0)\\}.$

We define our stratification of $Gr_{\overline{D}}(F)$ as follows:

$\displaystyle X_{0}$ $\displaystyle:=Gr_{Z_{0}\cap\overline{D}}(F);$
$\displaystyle X_{1}$ $\displaystyle:=Gr_{Z_{1}\cap\overline{D}}(F);$
$\displaystyle\vdots$ $\displaystyle X_{n}$
$\displaystyle:=Gr_{Z_{n}\cap\overline{D}}(F)=Gr_{\overline{D}}(F).$

We claim that $X_{k}\setminus X_{k-1}$ is totally real for each
$k=1,\cdots,n.$ First we take $k=n$ and $(p,F(p))\in X_{n}\setminus X_{n-1}.$
This implies $p=(p_{1},\cdots,p_{n})\in Z_{n}=\Omega.$ By taking
$\widetilde{\Omega}=\Omega$ and $g=Id$ in Proposition 2.5, we get that
$(\Phi\circ g)(\widetilde{\Omega})=Gr_{\Omega}(F)$ and $Gr_{\Omega}(F)$ is
totally real at $(p,F(p))$ if and only if $p_{1}\cdots p_{n}\neq 0$ i.e
$p\notin Z_{n-1}.$ Therefore, $X_{n}\setminus X_{n-1}$ is totally real.
Similarly, fix $k\in\\{1,\cdots,n\\}$ and we take $(p,F(p))\in X_{k}\setminus
X_{k-1}.$ This implies $p\in Z_{k}.$ Then there exists a set
$\\{i_{1},\cdots,i_{k}\\}\subset\\{1,\cdots,n\\}$ such that
$p_{j}=0~{}~{}\forall{j}\in\\{i_{1},\cdots,i_{k}\\}.$ Without loss of
generality, we can assume that that $i_{j}=n-k+j.$ Therefore,
$p=(p_{1},\cdots,p_{n-k},0,\cdots,0).$ Since Proposition 2.5 is true for any
$k\leq n,$ $Gr_{Z_{k}}(F)$ is totally real if and only if $p_{1}\cdots
p_{n-k}\not=0$ i.e $p\notin Z_{k-1}.$ Therefore, $X_{k}\setminus X_{k-1}$ is
totally real.

So far, we have the following:

  * •

$X_{n}$ is polynomially convex;

  * •

$X_{0}\subset\cdots\subset X_{n}=Gr_{\overline{D}}(F)$ with $X=\cup_{j}X_{j};$

  * •

$X_{j}\setminus X_{j-1}$ is totally real $\forall{j}\in\\{1,\cdots,n\\};$ and

  * •

$X_{0}=\\{0\\}.$

We now apply Result 2.4 to conclude that

$[z_{1},\cdots,z_{n},\bar{z_{1}}^{m_{1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{n}}+R_{n}(z);\overline{D}]=C(\overline{D}).$

∎

##  4\. Proof of Theorem 1.11

Before going into the proof of Theorem 1.11, we need to do some preparation
and we need to prove some lemmas.

Let

$\displaystyle
X:=\\{(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}},\cdots,\bar{z_{n}}^{m_{2n}})\in\mathbb{C}^{2n}:z\in\overline{\mathbb{D}}^{n}\\},$

and we define $\Phi:\mathbb{C}^{2n}\to\mathbb{C}^{2n}$ by

$\displaystyle\Phi\left(z_{1},\cdots,z_{n},w_{1},\cdots,w_{n}\right)=\left(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},w^{m_{n+1}}_{1},\cdots,w^{m_{2n}}_{n}\right).$

Therefore, $\Phi^{-1}(X)=\cup_{I}X_{I},\text{ where }$
$I\in\\{0,1,\cdots,m_{1}-1\\}\times\cdots\times\\{0,1,\cdots,m_{2n}-1\\}$ and
for $I=(t_{1},t_{2},\cdots,t_{2n}),$

$\displaystyle X_{I}:=\left\\{\left(e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},\cdots,e^{\frac{2i\pi t_{n}}{m_{n}}}z_{n},e^{\frac{2i\pi
t_{n+1}}{m_{n+1}}}\bar{z_{1}},\cdots,e^{\frac{2i\pi
t_{2n}}{m_{2n}}}\bar{z_{n}}\right):z\in\overline{\mathbb{D}}^{n}\right\\}.$

We will denote $X_{(0,\cdots,0)}$ by $X_{0}.$ For each subset
$\\{i_{1},\cdots,i_{k}\\}\subseteq\\{1,\cdots,2n\\}$ and $X_{I}$ as above, we
denote

$\displaystyle X^{(i_{1},\cdots,i_{k})}_{I}:=\bigg{\\{}\bigg{(}e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},\cdots,e^{\frac{2i\pi t_{n}}{m_{n}}}z_{n},e^{\frac{2i\pi
t_{n+1}}{m_{n+1}}}\bar{z_{1}},$ $\displaystyle,\cdots,e^{\frac{2i\pi
t_{2n}}{m_{2n}}}\bar{z_{n}}\bigg{)}:$ $\displaystyle
z\in\overline{\mathbb{D}}^{n},z_{i_{l}}=0,~{}l=1,\cdots,k\bigg{\\}}.$

We consider the polynomial

$\displaystyle p_{k}(z,w))=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}z_{j}w_{j}.$

We now prove three lemmas that are crucial in the proof of Theorem 1.11.

######  Lemma 4.1.

Let $p_{k}$ and $X^{(i_{1},\cdots,i_{k})}_{I}$ be as above. Then

  * (i)

$p_{k}(X^{(i_{1},\cdots,i_{k})}_{I})\subset L^{(i_{1},\cdots,i_{k})}_{I}$,
where $L^{(i_{1},\cdots,i_{k})}_{I}$ is the half line through the origin with
argument $\left({{\displaystyle 2\pi\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}}{m_{j}}}}\right);$

  * (ii)

$L^{(i_{1},\cdots,i_{k})}_{I}\cap
L^{(i_{1},\cdots,i_{k})}_{J}=\\{0\\}~{}~{}\text{ for distinct
}X^{(i_{1},\cdots,i_{k})}_{I}\text{ and }X^{(i_{1},\cdots,i_{k})}_{J};\text{
we also have }$

  * (iii)

$p_{k}^{-1}\\{0\\}\cap
X^{(i_{1},\cdots,i_{k})}_{I}=\bigg{\\{}\bigg{(}e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},\cdots,e^{\frac{2i\pi t_{n}}{m_{n}}}z_{n},e^{\frac{2i\pi
t_{n+1}}{m_{n+1}}}\bar{z_{1}},\cdots,e^{\frac{2i\pi
t_{2n}}{m_{2n}}}\bar{z_{n}}\bigg{)}:\\\
z\in\overline{\mathbb{D}}^{n},z_{i_{1}}=0,\cdots,z_{i_{k}}=0,\displaystyle\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}z_{j}=0\bigg{\\}}.$

###### Proof.

  * (i)

Let $(z,w)\in X^{(i_{1},\cdots,i_{k})}_{0}.$ Then

$\displaystyle p_{k}(z,w)$ $\displaystyle=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2}$ (4.1)

Writing $p_{k}(z,w)=u+iv,$ we get from (4.1) that $v=0\text{ and }u\geq 0.$
Therefore,

$\displaystyle
p_{k}\left(X^{(i_{1},\cdots,i_{k})}_{0}\right)\subset\\{u+iv\in\mathbb{C}:v=0,u\geq
0\\}.$ (4.2)

Let $I=(t_{1},t_{2},\cdots,t_{2n})$ be an arbitrary element of
$\\{0,1,\cdots,m_{1}-1\\}\times\\{0,1,\cdots,m_{2}-1\\}\times\cdots\times\\{0,1,\cdots,m_{2n}-1\\}.$
For $(z,w)\in X_{I},$ we get that

$\displaystyle p_{k}(z,w)=e^{i\alpha}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2},\text{ where
}\alpha={\displaystyle\left({2\pi\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}}{m_{j}}}\right)}$ (4.3)

Equality (4.3) says that $p_{k}\left(X^{(i_{1},\cdots,i_{k})}_{I}\right)$ lies
on the half real line passing through the origin with argument
$\left({2\displaystyle\pi\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}}{m_{j}}}\right).$

  * (ii)

For each $I,$ we denote each of the above line by
$L^{\\{i_{1},\cdots,i_{k}\\}}_{I}.$ For $\\{i_{1},\cdots,i_{k}\\}=\emptyset,$
we denote these half line $L^{\\{i_{1},\cdots,i_{k}\\}}_{I}$ simply by
$L_{I}.$ By putting
$\alpha_{j}=1~{}~{}~{}\forall{j}\in\\{1,\cdots,2n\\}\setminus\\{i_{1},\cdots,i_{k}\\}$
in Lemma 2.6, we get that $L^{\\{i_{1},\cdots,i_{k}\\}}_{I}\cap
L^{\\{i_{1},\cdots,i_{k}\\}}_{J}=\\{0\\}.$

  * (iii)

Proof follows from (4.3).

∎

######  Lemma 4.2.

Assume that $\\{l_{1},\cdots,l_{k}\\}\subset\\{1,\cdots,n\\}$ and
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}$ is polynomially convex for each
$j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\},$. Then
$\cup_{j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}}\left(\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\right)$
is polynomially convex.

###### Proof.

Without loss of generality, we assume that $1\leq<l_{2}<\cdots<l_{k}\leq n.$
Clearly,

$\displaystyle\cup_{j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}=\cup_{j=1}^{k+1}A_{l_{j}},$

where

$\displaystyle A_{l_{1}}:$
$\displaystyle=\cup_{j=1}^{l_{1}-1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)};$
$\displaystyle A_{l_{r+1}}:$
$\displaystyle=\cup_{j=l_{r}+1}^{l_{r+1}-1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}~{}\text{
for }r=1,\cdots,k-1;\text{ and }$ $\displaystyle A_{l_{k+1}}:$
$\displaystyle=\cup_{j=l_{k+1}}^{n}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}.$

First, we prove that each $A_{l_{j}}$ is polynomially convex. Without loss of
generality, it is enough to show that $A_{l_{1}}$ is polynomially convex. We
show this by induction principle. Given that each
$\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$ is polynomially
convex. Assume that

$\displaystyle
K_{1}:=\cup_{j=1}^{t}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)},~{}~{}t<l_{1}-1$

is polynomially convex. We need to show that

$\displaystyle\cup_{j=1}^{t+1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}=K_{1}\cup
X_{I}^{(l_{1},\cdots,l_{k},t+1)}=:K_{1}\cup K_{2}$

is polynomially convex. For this, we consider the polynomial

$\displaystyle p_{t}(z,w)=\prod^{t}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}w_{j}.$

Then we have the following:

  * •

$p_{t}(K_{1})=\\{0\\}.$

  * •

$p_{t}(K_{2}\setminus K_{1})\not=\\{0\\}:$ using Lemma 4.1, assertion (iii),
we get that for $(z,w)\in K_{2}\setminus K_{1},$ $p_{t}(z,w)=0$ implies
$\displaystyle\prod^{t}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}=0.$ Since $(z,w)\notin
K_{1},$ this is not possible. Therefore, $p_{t}(K_{2}\setminus
E_{1})\not=\\{0\\}$ and $p_{t}(K_{1})\cap p_{t}(K_{2})=\\{0\\}.$

  * •

by Lemma 4.1, assertion (iii), we get that $p_{t}^{-1}\\{0\\}\cap K_{1}=K_{1}$
and $p_{t}^{-1}\\{0\\}\cap K_{2}\subset K_{1}.$ Therefore,
$p_{t}^{-1}\\{0\\}\cap\bigg{(}K_{1}\cup K_{2}\bigg{)}=K_{1}$ is polynomially
convex.

Therefore, using Kallin’s lemma we get that $K_{1}\cup
K_{2}=\cup_{j=1}^{t+1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$
is polynomially convex, and hence, by induction principle, $A_{l_{1}}$ is
polynomially convex.

Now we show that

$\displaystyle\cup_{j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}=\cup_{j=1}^{k+1}A_{l_{j}},$

is polynomially convex. Again, we will apply induction principle. Assume that
$E_{1}:=\cup_{j=1}^{s}A_{l_{j}}$ is polynomially convex for $s<k+1$. We need
to show that

$\displaystyle\cup_{j=1}^{s+1}A_{l_{j}}=E_{1}\cup A_{l_{s+1}}=:E_{1}\cup
E_{2}$

is polynomially convex. We consider the polynomial

$\displaystyle p_{s}(z,w)=\prod^{s}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}w_{j}.$

Then we have the following:

  * •

$p_{s}(E_{1})=\\{0\\}.$

  * •

$p_{s}(E_{2}\setminus E_{1})\not=\\{0\\}:$ using Lemma 4.1, assertion (iii),
we get that for $(z,w)\in E_{2}\setminus E_{1},$ $p_{s}(z,w)=0$ implies
$\displaystyle\prod^{s}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}=0.$ Since $(z,w)\notin
E_{1},$ this is not possible. Therefore, $p_{s}(E_{2}\setminus
E_{1})\not=\\{0\\}$ and $p_{s}(E_{1})\cap p_{s}(E_{2})=\\{0\\}.$

  * •

applying Lemma 4.1, assertion (iii), we get that
$p_{s}^{-1}\\{0\\}\cap(E_{1}\cup E_{2})=E_{1},$ which is polynomially convex.

Again, by Kallin’s lemma, we conclude that $\cup_{j=1}^{s+1}A_{l_{j}}$ is
polynomially convex. Therefore, by induction principle, we obtain that
$\cup_{j=1}^{k+1}A_{l_{j}}$ is polynomially convex. ∎

######  Lemma 4.3.

Assume that $\\{l_{1},\cdots,l_{k}\\}\subset\\{1,\cdots,n\\}$ and for each
$j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\},$
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}$ is polynomially convex. Then
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}$ is polynomially convex.

###### Proof.

Since $\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}$ is polynomially convex, by
Lemma 4.2,
$\cup_{j\notin\\{l_{1},\cdots,l_{k}\\}}^{n}\left(\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\right)$
is polynomially convex. We consider the polynomial

$\displaystyle p_{1}(z,w)=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}w_{j}.$

Therefore

  * •

By Lemma 4.1, each $p_{1}\bigg{(}X_{I}^{(l_{1},\cdots,l_{k})}\bigg{)}\subset
L_{I}^{(l_{1},\cdots,l_{k})}$ with $L_{I}^{(l_{1},\cdots,l_{k})}\cap
L_{J}^{(l_{1},\cdots,l_{k})}=\\{0\\},$ where each
$L_{I}^{(l_{1},\cdots,l_{k})}$ is the half real line starting from the origin
with argument $\left({{\displaystyle 2\pi\sum^{2n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}}{m_{j}}}}\right).$

  * •

By Lemma 4.1, assertion (iii), we get that
$p_{1}^{-1}\\{0\\}\cap\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}\bigg{)}=\cup_{j\notin\\{l_{1},\cdots,l_{k}\\}}^{n}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}.$

From above we can say that
$p_{1}^{-1}\\{0\\}\cap\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}\bigg{)}$ is
polynomially convex. Therefore, by Kallin’s lemma, we infer that
$\left(\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}\right)$ is polynomially convex. ∎

We now begin the proof of Theorem 1.11.

###### Proof of Theorem 1.11.

Since $X_{I}$ is the image of a compact subset of
$\mathbb{R}^{2n}\subset\mathbb{C}^{2n}$ under an invertible
$\mathbb{C}$-linear map, $\mathscr{P}(X_{I})=\mathcal{C}(X_{I})$ for all
$I\in\\{0,1,\cdots,m_{1}-1\\}\times\cdots\times\\{0,1,\cdots,m_{2n}-1\\}.$

Next, we wish to show that
$\mathscr{P}(\Phi^{-1}(X))=\mathcal{C}(\Phi^{-1}(X)).$

We consider the polynomial

$\displaystyle
p(z,w)=\prod^{n}_{\begin{subarray}{c}j=1\end{subarray}}z_{j}w_{j}.$

For $\\{i_{1},\cdots,i_{k}\\}=\emptyset,$ in view of Lemma 4.1, we can say
that:

  * •

For each
$I=(t_{1},\cdots,t_{2n})\in\\{0,1,\cdots,m_{1}-1\\}\times\cdots\times\\{0,1,\cdots,m_{n}-1\\},$
$p(X_{I})\subset L_{I},$ where $L_{I}$ is a half-line starting from the origin
with argument $\left({{\displaystyle
2\pi\sum^{2n}_{\begin{subarray}{c}j=1\end{subarray}}\frac{t_{j}}{m_{j}}}}\right).$

  * •

$L_{I}\cap L_{J}=\\{0\\}~{}\text{ for all }{I\neq J}.$

In view of Lemma 4.3, and by application of induction principle on $n,$ we can
say that

$\displaystyle
p^{-1}\\{0\\}\cap\bigg{(}\cup_{I}X_{I}\bigg{)}=\cup^{n}_{j=1}\left(\cup_{I}X_{I}^{(j)}\right)$

is polynomially convex. Since for each
$I\in\left\\{0,1,\cdots,m_{1}-1\right\\}\times\cdots\times\left\\{0,1,\cdots,m_{n}-1\right\\},$
$\mathscr{P}(X_{I})=\mathcal{C}(X_{I}),$ applying Result 2.2 we conclude that

$\mathscr{P}\left(\cup_{I}X_{I}\right)=\mathcal{C}\left(\cup_{I}X_{I}\right),~{}\text{
that is
}\mathscr{P}\left(\Phi^{-1}(X)\right)=\mathcal{C}\left(\Phi^{-1}(X)\right).$

So, by Result 2.3, we get that

$\mathscr{P}(X)=\mathcal{C}(X).$

Clearly,
$z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}},\cdots,\bar{z_{n}}^{m_{2n}}$
separates points on $\overline{\mathbb{D}}^{n}$ because
$\gcd(m_{i},m_{j})=1\text{ for all }i\neq j.$ Therefore,

$\displaystyle[z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}},\cdots,\bar{z_{n}}^{m_{2n}};\overline{\mathbb{D}}^{n}]=C(\overline{\mathbb{D}}^{n}).$

∎

##  5\. Proof of Theorem 1.12

The structure of the proof of Theorem 1.12 is similar to that of Theorem 1.8.
Before going into the proof of Theorem 1.12, we need some preparations,
including some lemmas as in the proof of Theorem 1.8.

We set

$\displaystyle
X:=\\{(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)):z\in
D(\delta_{0})\\},$

and we define $\Phi:\mathbb{C}^{2n}\to\mathbb{C}^{2n}$ by

$\displaystyle\Phi\left(z_{1},\cdots,z_{n},w_{1},\cdots,w_{n}\right)=\left(z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},w_{1},\cdots,w_{n}\right).~{}~{}~{}\Phi\text{
is a proper holomorphic map.}$

We have, $\Phi^{-1}(X)=\cup_{I}X_{I},\text{ where }$
$I\in\\{0,1,\cdots,m_{1}-1\\}\times\cdots\times\\{0,1,\cdots,m_{n}-1\\}$ and
for each $I=(t_{1},t_{2},\cdots,t_{n}),$

$\displaystyle X_{I}:=\left\\{\left(e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},\cdots,e^{\frac{2i\pi
t_{n}}{m_{n}}}z_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)\right):z\in
D(\delta_{0})\right\\}.$

We will denote $X_{(0,\cdots,0)}$ by $X_{0}.$ For
$\\{i_{1},\cdots,i_{k}\\}\subseteq\\{1,\cdots,n\\}$ and $X_{I}$ as above, we
denote

$\displaystyle X^{(i_{1},\cdots,i_{k})}_{I}:=\bigg{\\{}\bigg{(}e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},\cdots,e^{\frac{2i\pi
t_{n}}{m_{n}}}z_{n},\bar{z_{1}}^{m_{n+1}}$
$\displaystyle+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)\bigg{)}:$
$\displaystyle z\in D(\delta_{0}),z_{i_{l}}=0,l=1,\cdots,k\bigg{\\}}.$

We consider the polynomial

$\displaystyle p_{k}(z,w))=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}z_{j}^{m_{n+j}}w_{j},$

where $z=(z_{1},\cdots,z_{n}),w=(w_{1},\cdots,w_{n})\in\mathbb{C}^{n}.$

######  Lemma 5.1.

$p_{k}(X^{(i_{1},\cdots,i_{k})}_{I})\subset\omega^{(i_{1},\cdots,i_{k})}_{I}$,
where $\omega^{(i_{1},\cdots,i_{k})}_{I}$ is a closed sector in the complex
plane with vertex at the origin and $L^{(i_{1},\cdots,i_{k})}_{I}$ as the
angular bisector with argument $\left({\displaystyle
2\pi\sum^{n}_{\begin{subarray}{c}j=1\\\ j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}m_{n+j}}{m_{j}}}\right).$ Furthermore,
$\omega^{(i_{1},\cdots,i_{k})}_{I}\cap\omega^{(i_{1},\cdots,i_{k})}_{J}=\\{0\\}~{}~{}\text{
for distinct }X^{(i_{1},\cdots,i_{k})}_{I}$ and
$X^{(i_{1},\cdots,i_{k})}_{J}.$

###### Proof.

Let $(z,w)\in X^{(i_{1},\cdots,i_{k})}_{0}.$ We now do some computations here:

$\displaystyle p_{k}(z,w)$ $\displaystyle=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\bigg{(}|z_{j}|^{2m_{n+j}}+z_{j}^{m_{n+j}}R_{j}\bigg{)}$
$\displaystyle=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}|z_{j}|^{2m_{n+j}}+\bigg{[}\sum_{1\leq l_{1}\leq
n}\alpha_{(l_{1})}(z)+\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)+\cdots$ $\displaystyle\quad+\sum_{1\leq l_{1}\neq
l_{2}\neq\cdots\neq l_{n-k-1}\leq
n}\alpha_{(l_{1},\cdots,l_{n-k-1})}(z)+\alpha_{(l_{1},\cdots,l_{n-k})}(z)\bigg{]},$
(5.1)

where
$\\{l_{1},\cdots,l_{r}\\}\subset\\{1,\cdots,n\\}\setminus\\{i_{1},\cdots,i_{k}\\}$
and

$\displaystyle\alpha_{(l_{1},\cdots,l_{r})}(z):=\bigg{(}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
j\notin\\{l_{1},\cdots,l_{r}\\}\end{subarray}}|z_{j}|^{2m_{n+j}}\bigg{)}\prod_{\begin{subarray}{c}j\in\\{l_{1},\cdots,l_{r}\\}\\\
\end{subarray}}\bigg{(}{z_{j}}^{m_{n+j}}R_{j}\bigg{)}.$

Since $R_{j}(z)\sim o(|z_{j}|^{m_{n+j}})\text{ as }z_{j}\to 0$ for
$j=1,\cdots,n;$ choose $\varepsilon^{\prime}>0,$ then there exists
$\delta_{j}>0$ such that

$\displaystyle|R_{j}|\leq\varepsilon^{\prime}|z_{j}|^{m_{n+j}}\text{ whenever
}|z_{j}|\leq\delta_{j}~{}~{}\forall{j}=1,2,\cdots,n.$ (5.2)

Taking $\delta^{\prime}=\min_{1\leq j\leq n}\\{\delta_{j}\\},$ we obtain from
(5.2) that

$\displaystyle\left|\sum_{1\leq l_{1}\neq\cdots\neq l_{r}\leq
n}\alpha_{(l_{1},\cdots,l_{r})}(z)\right|$ $\displaystyle\leq\left|\sum_{1\leq
l_{1}\neq\cdots\neq l_{r}\leq n}\bigg{(}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
j\notin\\{l_{1},\cdots,l_{r}\\}\end{subarray}}|z_{j}|^{2m_{n+j}}\bigg{)}\prod_{\begin{subarray}{c}j\in\\{l_{1},\cdots,l_{r}\\}\\\
\end{subarray}}\bigg{(}{z_{j}}^{m_{n+j}}R_{j}\bigg{)}\right|$
$\displaystyle\leq\sum_{1\leq l_{1}\neq\cdots\neq l_{r}\leq
n}\bigg{(}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
j\notin\\{l_{1},\cdots,l_{r}\\}\end{subarray}}|z_{j}|^{2m_{n+j}}\bigg{)}\prod_{\begin{subarray}{c}j\in\\{l_{1},\cdots,l_{r}\\}\\\
\end{subarray}}\left|\bigg{(}{z_{j}}^{m_{n+j}}R_{j}\bigg{)}\right|$
$\displaystyle\leq{n-k\choose
r}{\varepsilon^{\prime}}^{r}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}|z_{j}|^{2m_{n+j}}~{}~{}\forall
z\in\overline{D(\delta^{\prime})}.$ (5.3)

Writing $p_{k}(z)=u+iv,$ we get from (5) that for $(z,w)\in
X^{(i_{1},\cdots,i_{k})}_{0},$

$\displaystyle|v|$ $\displaystyle=|{\sf Im}(p_{k}(z,w))|$
$\displaystyle=\bigg{|}{\sf Im}\bigg{[}\sum_{1\leq l_{1}\leq
n}\alpha_{(l_{1})}(z)+\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)+\cdots$ $\displaystyle\quad+\sum_{1\leq l_{1}\neq
l_{2}\neq\cdots\neq l_{n-k-1}\leq
n}\alpha_{(l_{1},\cdots,l_{n-k-1})}(z)+\alpha_{(l_{1},\cdots,l_{n-k})}(z)\bigg{]}\bigg{|}$
$\displaystyle\leq\bigg{|}\sum_{1\leq l_{1}\leq
n}\alpha_{(l_{1})}(z)\bigg{|}+\bigg{|}\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)\bigg{|}+\cdots+\bigg{|}\sum_{\begin{subarray}{c}1\leq
l_{1}\neq l_{2}\neq\\\ \cdots\neq l_{n-k-1}\leq
n\end{subarray}}\alpha_{(l_{1},\cdots,l_{n-k-1})}(z)\bigg{|}+\bigg{|}\alpha_{(l_{1},\cdots,l_{n})}(z)\bigg{|}$
$\displaystyle\leq\sum_{r=1}^{n-k}{n-k\choose
r}{\varepsilon^{\prime}}^{r}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}.$

We denote $\varepsilon:=\displaystyle\sum_{r=1}^{n-k}{n-k\choose
r}{\varepsilon^{\prime}}^{r}.$ Therefore,

$\displaystyle|v|\leq\varepsilon\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}|z_{j}|^{2m_{n+j}}~{}~{}\forall
z\in\overline{D(\delta^{\prime})}.$ (5.4)

Similarly, for any $(z,w)\in X^{(i_{1},\cdots,i_{k})}_{0},$ we get from (5)
that the real part of $p_{k}$ is

$\displaystyle u=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}+$
$\displaystyle{\sf Re}\bigg{[}\sum_{1\leq l_{1}\leq
n}\alpha_{l_{1}}(z)+\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)+\cdots$ $\displaystyle\quad+\sum_{1\leq l_{1}\neq
l_{2}\neq\cdots\neq l_{n-1}\leq n}\alpha_{(l_{1},l_{2},\cdots
l_{n-k-1})}(z)+\alpha_{(l_{1},l_{2},\cdots,l_{n-k})}(z)\bigg{]}.$ (5.5)

We compute:

$\displaystyle{\sf Re}\bigg{[}\sum_{1\leq l_{1}\leq
n}\alpha_{(l_{1})}(z)+\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)+\cdots$ $\displaystyle\quad+\sum_{1\leq l_{1}\neq
l_{2}\neq\cdots\neq l_{n-k-1}\leq
n}\alpha_{(l_{1},l_{2},\cdots,l_{n-k-1})}(z)+\alpha_{(l_{1},l_{2},\cdots,l_{n-k})}(z)\bigg{]}$
$\displaystyle\leq\bigg{|}\sum_{1\leq l_{1}\leq
n}\alpha_{(l_{1})}(z)\bigg{|}+\bigg{|}\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)\bigg{|}+\cdots+\bigg{|}\sum_{\begin{subarray}{c}1\leq
l_{1}\neq l_{2}\neq\\\ \cdots\neq l_{n-k-1}\end{subarray}\leq
n}\alpha_{(l_{1},\cdots,l_{n-k-1})}(z)\bigg{|}+\bigg{|}\alpha_{(l_{1},\cdots,l_{n-k})}(z)\bigg{|}$
$\displaystyle\leq\sum_{r=1}^{n-k}{n-k\choose
r}{\varepsilon^{\prime}}^{r}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}|z_{j}|^{2m_{n+j}}=\varepsilon\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}.$

Therefore

$\displaystyle-\varepsilon\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}\leq$
$\displaystyle{\sf Re}\bigg{[}\sum_{1\leq l_{1}\leq
n}\alpha_{(l_{1})}(z)+\sum_{1\leq l_{1}\neq l_{2}\leq
n}\alpha_{(l_{1},l_{2})}(z)+\cdots$ $\displaystyle\quad+\sum_{1\leq l_{1}\neq
l_{2}\neq\cdots\neq l_{n-k-1}\leq
n}\alpha_{(l_{1},l_{2},\cdots,l_{n-k-1})}(z)+\alpha_{(l_{1},l_{2},\cdots,l_{n-k})}(z)\bigg{]}$
$\displaystyle\leq\varepsilon\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}.$

Hence

$\displaystyle(1-\varepsilon)\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}\leq
u\leq(1+\varepsilon)\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}|z_{j}|^{2m_{n+j}}.$ (5.6)

We choose $\varepsilon^{\prime}$ in such a way such that $\varepsilon<1$. In
view of (5.4) and (5.6), we obtain that

$\displaystyle|v|\leq\varepsilon\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}|z_{j}|^{2m_{n+j}}\leq\left(\frac{\varepsilon}{1-\varepsilon}\right)u,u\geq
0,\forall z\in\overline{D(\delta^{\prime})},\text{ by shrinking
}\delta^{\prime}\text{ if required}.$ (5.7)

Above inequalities says that

$\displaystyle
p_{k}(X^{(i_{1},\cdots,i_{k})}_{0})\subset\left\\{u+iv\in\mathbb{C}:|v|\leq\left(\frac{\varepsilon}{1-\varepsilon}\right)u,~{}~{}u\geq
0\right\\}.$ (5.8)

We set $\theta^{\prime}:=\min_{I\neq J}\left\\{\text{ angle between the lines
}L^{(i_{1},\cdots,i_{k})}_{I}\text{ and
}L^{(i_{1},\cdots,i_{k})}_{J}\right\\},$ and we take
$\theta<\min\\{\theta^{\prime},\frac{\pi}{2}\\}$ where,
$L^{(i_{1},\cdots,i_{k})}_{I}$ is the half line through the origin with
argument ${\left({\displaystyle 2\pi\sum^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}m_{n+j}}{m_{j}}}\right)}$ for
$I=(t_{1},\cdots,t_{n})$ and $L^{(i_{1},\cdots,i_{k})}_{J}$ is the half line
through the origin with argument ${\left({\displaystyle
2\pi\sum^{n}_{\begin{subarray}{c}j=1\\\ j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{s_{j}m_{n+j}}{m_{j}}}\right)}$ for
$J=(s_{1},\cdots,s_{n}).$

Again, we shrink $\varepsilon^{\prime}$ further so that

$\varepsilon<\frac{\tan\left(\frac{\theta}{2}\right)}{1+\tan\left(\frac{\theta}{2}\right)}.$

This implies

$\left(\frac{\varepsilon}{1-\varepsilon}\right)<\tan\left(\frac{\theta}{2}\right).$

The expression (5.8) says that $p_{k}(X^{(i_{1},\cdots,i_{k})}_{0})$ lies in
the closed angular sector $\omega^{(i_{1},\cdots,i_{k})}_{0}$ with vertex at
$0,$ positive real axis as the angular bisector and has an vertex-angle
$2\phi,$ where

$\displaystyle\tan(\phi)=\left(\frac{\varepsilon}{1-\varepsilon}\right)<\tan\left(\frac{\theta}{2}\right)$

i.e $\omega^{(i_{1},\cdots,i_{k})}_{0}$ is an angular sector with vertex at
the origin, positive real axis as the angular bisector and has vertex-angle
$2\phi<\theta.$

Let $I=(t_{1},t_{2},\cdots,t_{n})$ be an arbitrary element of
$\\{0,1,\cdots,m_{1}-1\\}\times\\{0,1,\cdots,m_{2}-1\\}\times\cdots\times\\{0,1,\cdots,m_{n}-1\\}.$
Since $(z,w)\in X_{I},$ we get that

$\displaystyle p_{k}(z,w)$
$\displaystyle=e^{i\alpha}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\bigg{(}|z_{j}|^{2m_{n+j}}+z_{j}^{m_{n+j}}R_{j}\bigg{)},\text{
where }\alpha={\bigg{(}{\displaystyle 2\pi\sum^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}m_{n+j}}{m_{j}}}\bigg{)}}.$

There exist a closed sector $\omega^{(i_{1},\cdots,i_{k})}_{I}$ with vertex at
the origin, $L^{(i_{1},\cdots,i_{k})}_{I}$ as the angular bisector with
argument ${\bigg{(}{\displaystyle 2\pi\sum^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}m_{n+j}}{m_{j}}}\bigg{)}}$ and has vertex-angle
$<\theta.$ Clearly, $L^{(i_{1},\cdots,i_{k})}_{I}$ is obtained by rotation of
the positive real axis at an angle ${\bigg{(}{\displaystyle
2\pi\sum^{n}_{\begin{subarray}{c}j=1\\\ j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\frac{t_{j}m_{n+j}}{m_{j}}}\bigg{)}}.$ By putting
$\alpha_{j}=m_{n+j}~{}~{}~{}\forall{j}\in\\{1,\cdots,n\\}\text{ and
}\alpha_{j}=0~{}~{}\forall{j}\in\\{n+1,\cdots,2n\\}$ in Lemma 2.6, we get that

$\displaystyle\omega^{(i_{1},\cdots,i_{k})}_{I}\cap\omega^{(i_{1},\cdots,i_{k})}_{J}=\\{0\\}~{}~{}\text{
for distinct }X^{(i_{1},\cdots,i_{k})}_{I}\text{ and
}X^{(i_{1},\cdots,i_{k})}_{J}.$ (5.9)

For $\\{i_{1},\cdots,i_{k}\\}=\emptyset,$ then we denote
$\omega^{(i_{1},\cdots,i_{k})}_{I}$ by $\omega_{I}.$

∎

######  Lemma 5.2.

Let $p_{k}$ and $X^{(i_{1},\cdots,i_{k})}_{I}$ be as above. Then

$\displaystyle p_{k}^{-1}\\{0\\}\cap
X^{(i_{1},\cdots,i_{k})}_{I}=\bigg{\\{}\bigg{(}e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},\cdots,$ $\displaystyle e^{\frac{2i\pi
t_{n}}{m_{n}}}z_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)\bigg{)}:$
$\displaystyle
z\in\overline{D(\delta^{\prime})},z_{i_{l}}=0,l=1,\cdots,k,\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}z_{j}=0\bigg{\\}}.$

###### Proof.

Take $w=(w_{1},\cdots,w_{n},w_{n+1},\cdots,w_{2n})\in p_{k}^{-1}\\{0\\}\cap
X^{(i_{1},\cdots,i_{k})}_{I}.$ Then

$\displaystyle w_{j}$ $\displaystyle=e^{\frac{2i\pi
t_{j}}{m_{j}}}z_{j}~{}~{}\text{ and }$ $\displaystyle w_{n+j}$
$\displaystyle=\bar{z_{j}}^{m_{n+j}}+R_{j}(z)~{}~{}\text{ for
}j=1,2,\cdots,n;$

Since $w\in p_{k}^{-1}\\{0\\},$

$\displaystyle\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\ \end{subarray}}e^{\frac{2\pi
it_{j}m_{n+j}}{m_{j}}}z_{j}^{m_{n+j}}\bigg{(}\bar{z_{j}}^{m_{n+j}}+R_{j}\bigg{)}=0.$

This implies

$\displaystyle e^{i\alpha}\bigg{(}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}z_{j}^{m_{n+j}}\bigg{)}\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\bigg{(}\bar{z_{j}}^{m_{n+j}}+R_{j}\bigg{)}=0.$ (5.10)

On $\overline{D(\delta^{\prime})},$
$|R_{j}|\leq\varepsilon^{\prime}|z_{j}|^{m_{n+j}},$ for some
$\varepsilon^{\prime}<1,$ Therefore, we can say that

$\displaystyle\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}\bigg{(}\bar{z_{j}}^{m_{n+j}}+R_{j}\bigg{)}\neq 0.$

Hence, form (5.10), we have

$\displaystyle\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}z_{j}^{m_{n+j}}=0~{}~{}\text{ i.e
}\prod^{n}_{\begin{subarray}{c}j=1\\\ j\notin\\{i_{1},\cdots,i_{k}\\}\\\
\end{subarray}}z_{j}=0.$

∎

######  Lemma 5.3.

Assume that $\\{l_{1},\cdots,l_{k}\\}\subset\\{1,2,\cdots,n\\}$ and
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}$ is polynomially convex for each
$j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}.$ Then

$\displaystyle\cup_{j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$

is polynomially convex.

###### Proof.

Without loss of generality, we assume that $1\leq l_{1}<l_{2}\cdots<l_{k}\leq
n.$ Clearly,

$\displaystyle\cup_{j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}=\cup_{j=1}^{k+1}A_{l_{j}},$

where

$\displaystyle A_{l_{1}}:$
$\displaystyle=\cup_{j=1}^{l_{1}-1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)};$
$\displaystyle A_{l_{r+1}}:$
$\displaystyle=\cup_{j=l_{r}+1}^{l_{r+1}-1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}~{}\text{
for }r=1,\cdots,k-1;\text{ and }$ $\displaystyle A_{l_{k+1}}:$
$\displaystyle=\cup_{j=l_{k+1}}^{n}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}.$

First, we prove that each $A_{l_{j}}$ is polynomially convex. Without loss of
generality, it is enough to show that

$\displaystyle A_{l_{1}}:$
$\displaystyle=\cup_{j=1}^{l_{1}-1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$

is polynomially convex. We will now apply induction principle to show
$A_{l_{1}}$ is polynomially convex. Assume that
$K_{1}:=\cup_{j=1}^{t}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$
is polynomially convex for $t<l_{1}-1$. We need to show that

$\displaystyle\cup_{j=1}^{t+1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}=K_{1}\cup
X_{I}^{(l_{1},\cdots,l_{k},t+1)}=:K_{1}\cup K_{2}$

is polynomially convex. For this, we consider the polynomial

$\displaystyle p_{t}(z,w)=\prod^{t}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}^{m_{n+j}}w_{j}.$

Then we have the following:

  * •

$p_{t}(K_{1})=\\{0\\}.$

  * •

$p_{t}(K_{2}\setminus K_{1})\not=\\{0\\}:$ using Lemma 5.2, we get that for
$(z,w)\in K_{2}\setminus K_{1},$ $p_{t}(z,w)=0$ implies
$\displaystyle\prod^{t}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}=0.$ Since $(z,w)\notin
K_{1},$ this is not possible. Therefore, $p_{t}(K_{2}\setminus
K_{1})\not=\\{0\\}$ and $p_{t}(K_{1})\cap p_{t}(K_{2})=\\{0\\}.$

  * •

Applying Lemma 5.2, we get that $p_{t}^{-1}\\{0\\}\cap K_{1}=K_{1}$ and
$p_{t}^{-1}\\{0\\}\cap K_{2}\subset K_{1}.$ Therefore,
$p_{t}^{-1}\\{0\\}\cap\bigg{(}K_{1}\cup K_{2}\bigg{)}=K_{1}$ is polynomially
convex.

Therefore, using Kallin’s lemma we get that $K_{1}\cup
K_{2}=\cup_{j=1}^{t+1}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$
is polynomially convex, and hence, by induction principle $A_{l_{1}}$ is
polynomially convex.

Again, we will apply induction principle to show $\cup_{j=1}^{k+1}A_{l_{j}}$
is polynomially convex. Assume that $E_{1}:=\cup_{j=1}^{s}A_{l_{j}}$ is
polynomially convex for $s<k+1$. We write $\cup_{j=1}^{s+1}A_{l_{j}}=E_{1}\cup
A_{l_{s+1}}=:E_{1}\cup E_{2}.$

We consider the polynomial

$\displaystyle p_{s}(z,w)=\prod^{s}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}^{m_{n+j}}w_{j}.$

Then we have the following:

  * •

$p_{s}(E_{1})=\\{0\\}.$

  * •

$p_{s}(E_{2}\setminus E_{1})\not=\\{0\\}:$ using Lemma 5.2, we get that for
$(z,w)\in E_{2}\setminus E_{1},$ $p_{s}(z,w)=0$ implies
$\displaystyle\prod^{s}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}=0.$ Since $(z,w)\notin
E_{1},$ this is not possible. Therefore, $p_{s}(E_{2}\setminus
E_{1})\not=\\{0\\}$ and $p_{s}(E_{1})\cap p_{s}(E_{2})=\\{0\\}.$

  * •

Applying Lemma 5.2, we get that $p_{s}^{-1}\\{0\\}\cap E_{1}=E_{1}$ and
$p_{s}^{-1}\\{0\\}\cap E_{2}\subset E_{1}.$ Therefore,
$p_{s}^{-1}\\{0\\}\cap(E_{1}\cup E_{2})=E_{1}$ is polynomially convex.

Again, by Kallin’s lemma, we conclude that $\cup_{j=1}^{s+1}A_{l_{j}}$ is
polynomially convex. Therefore, by induction principle, we obtained that
$\cup_{j=1}^{k+1}A_{l_{j}}$ is polynomially convex, that is,
$\cup_{j\notin\\{l_{1},\cdots,l_{k}\\}}\left(\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\right)$
is polynomially convex. ∎

######  Proposition 5.4.

Assume that $\\{l_{1},\cdots,l_{k}\\}\subset\\{1,\cdots,n\\}$ and
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}$ is polynomially convex for each
$j\in\\{1,\cdots,n\\}\setminus\\{l_{1},\cdots,l_{k}\\}.$ Then
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}$ is polynomially convex.

###### Proof.

Since $\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}$ is polynomially convex, by
Lemma 5.3,
$\cup_{j\notin\\{l_{1},\cdots,l_{k}\\}}^{n}\bigg{(}\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\bigg{)}$
is polynomially convex. We consider the polynomial

$\displaystyle p(z,w)=\prod^{n}_{\begin{subarray}{c}j=1\\\
j\notin\\{l_{1},\cdots,l_{k}\\}\\\ \end{subarray}}z_{j}^{m_{n+j}}w_{j}.$

Therefore

  * •

by Lemma 5.1,
$p\left(X_{I}^{(l_{1},\cdots,l_{k})}\right)\subset\omega_{I}^{(l_{1},\cdots,l_{k})}~{}~{}~{}\forall
I\in\\{0,1,\cdots,m_{1}-1\\}\times\\{0,1,\cdots,m_{2}-1\\}\times\cdots\times\\{0,1,\cdots,m_{n}-1\\};$

  * •

Since
$\omega_{I}^{(l_{1},\cdots,l_{k})}\cap\omega_{J}^{(l_{1},\cdots,l_{k})}=\\{0\\},$
we have $p\left(X_{I}^{(l_{1},\cdots,l_{k})}\right)\cap
p\left(X_{J}^{(l_{1},\cdots,l_{k})}\right)\subset\\{0\\};$

  * •

by Lemma 5.2,
$p^{-1}\\{0\\}\cap\left(\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}\right)=\bigcup_{j\notin\\{l_{1},\cdots,l_{k}\\}}^{n}\left(\cup_{I}X_{I}^{(l_{1},\cdots,l_{k},j)}\right),$
which is polynomially convex by Lemma 5.3.

Therefore, by Kallin’s lemma, we infer that
$\cup_{I}X_{I}^{(l_{1},\cdots,l_{k})}$ is polynomially convex. ∎

###### Proof of Theorem 1.12.

Take $\delta:=\min\\{\delta^{\prime},\delta_{0}\\}.$ We divide the proof of
this theorem in three steps.

Step I: Showing
$\mathscr{P}(X_{I})=\mathcal{C}(X_{I})~{}~{}~{}\forall_{I}\in\\{0,1,\cdots,m_{1}-1\\}\times\\{0,1,\cdots,m_{2}-1\\}\times\cdots\times\\{0,1,\cdots,m_{n}-1\\}.$

We have

$\displaystyle X_{0}$
$\displaystyle=\\{(z_{1},\cdots,z_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)):z\in\overline{D(\delta)}\\}$
$\displaystyle=Gr_{\overline{D(\delta)}}(\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)).$

For each $I=(t_{1},\cdots,t_{n}),$ we define a map
$\Psi_{I}:\mathbb{C}^{2n}\to\mathbb{C}^{2n}$ by

$\displaystyle\Psi_{I}(z_{1},\cdots,z_{n},w_{1},w_{2},\cdots,w_{n})=\left(e^{\frac{2i\pi
t_{1}}{m_{1}}}z_{1},e^{\frac{2i\pi t_{2}}{m_{2}}}z_{2}\cdots,e^{\frac{2i\pi
t_{n}}{m_{n}}}z_{n},w_{1},w_{2}\cdots,w_{n}\right).$

Therefore, $\Psi_{I}(X_{0})=X_{I}.$ Theorem 1.8 gives us that
$\mathscr{P}(X_{0})=\mathcal{C}(X_{0}).$ Since $\Psi_{I}$ is a biholomorphism,
using Result 2.3, we conclude that $\mathscr{P}(X_{I})=\mathcal{C}(X_{I}).$

Step II: Showing $\mathscr{P}(\Phi^{-1}(X))=\mathcal{C}(\Phi^{-1}(X)).$

We consider the polynomial

$\displaystyle
p(z,w)=\prod^{n}_{\begin{subarray}{c}j=1\end{subarray}}z_{j}^{m_{n+j}}w_{j}.$

For $\\{i_{1},\cdots,i_{k}\\}=\emptyset,$ in view of Lemma 5.1, we get that

  * •

For each
$I=(t_{1},\cdots,t_{n})\in\\{0,1,\cdots,m_{1}-1\\}\times\cdots\times\\{0,1,\cdots,m_{n}-1\\},$
$p(X_{I})\subset\omega_{I},$ where $\omega_{I}$ is a closed sector in the
complex plane with vertex at the origin and $L_{I}$ as the angular bisector
with argument argument ${\bigg{(}{\displaystyle
2\pi\sum^{n}_{\begin{subarray}{c}j=1\end{subarray}}\frac{t_{j}m_{n+j}}{m_{j}}}\bigg{)}};$

  * •

$\omega_{I}\cap\omega_{J}=\\{0\\}~{}\text{ for all }{I\neq J}.$

To apply Kallin’s lemma, we need to show that
$p^{-1}\\{0\\}\cap\bigg{(}\cup_{I}X_{I}\bigg{)}=\cup^{n}_{j=1}\left(\cup_{I}X_{I}^{(j)}\right)$
is polynomially convex. First we focus on the the polynomial convexity of
$\cup_{I}X_{I}^{(j)}\text{ for }j\in{1,\cdots,n}.$ Proposition 5.4 says that
for each $i_{1}\in\\{1,\cdots,n\\},$ $\cup_{I}X_{I}^{(i_{1})}$ is polynomially
convex if for each $i_{2}\in\\{1,\cdots,n\\}\setminus\\{i_{1}\\},$
$\cup_{I}X_{I}^{(i_{1},i_{2})}$ is polynomially convex. Again by Proposition
5.4, $\cup_{I}X_{I}^{(i_{1},i_{2})}$ is polynomially convex if for each
$i_{3}\in\\{1,\cdots,n\\}\setminus\\{i_{1},i_{2}\\},$
$\cup_{I}X_{I}^{(i_{1},i_{2},i_{3})}$ is polynomially convex. Proceeding in
this way, we arrive at this situation that
$\cup_{I}X_{I}^{(i_{1},i_{2},\cdots,i_{n-1})}$ is polynomially convex if for
each $i_{n}\in\\{1,\cdots,n\\}\setminus\\{i_{1},i_{2},\cdots,i_{n-1}\\},$
$\cup_{I}X_{I}^{(i_{1},i_{2},\cdots,i_{n-1},i_{n})}$ is polynomially convex.
But $\cup_{I}X_{I}^{(i_{1},i_{2},\cdots,i_{n-1},i_{n})}=\\{0\\}$ for all $I,$
which is obviously polynomially convex. Hence, for each
$j\in\\{1,\cdots,n\\},$ $\cup_{I}X_{I}^{(j)}$ is polynomially convex. We will
now apply induction on $n,$ to conclude that
$p^{-1}\\{0\\}\cap\left(\cup_{I}X_{I}\right)$ is polynomially convex. Assume
that $K_{1}:=\cup^{k}_{j=1}\left(\cup_{I}X_{I}^{(j)}\right)$ is polynomially
convex. We need to show that
$\cup^{k+1}_{j=1}\left(\cup_{I}X_{I}^{(j)}\right)=K_{1}\cup\left(\cup_{I}X_{I}^{(k+1)}\right)=:K_{1}\cup
K_{2}$ is polynomially convex. For this we again apply Kallin’s lemma. We
consider the polynomial

$\displaystyle
p_{1}(z,w)=\prod^{k}_{\begin{subarray}{c}j=1\end{subarray}}z_{j}^{m_{n+j}}w_{j}.$

Then we have the following:

  * •

$p_{1}(K_{1})=\\{0\\}.$

  * •

$p_{1}(K_{2}\setminus K_{1})\not=\\{0\\}:$ using Lemma 5.2, we get that for
$(z,w)\in K_{2}\setminus K_{1},$ $p_{1}(z,w)=0$ implies
$\displaystyle\prod^{k}_{\begin{subarray}{c}j=1\\\ \end{subarray}}z_{j}=0.$
Since $(z,w)\notin K_{1},$ this is not possible. Therefore,
$p_{1}(K_{2}\setminus K_{1})\not=\\{0\\}$ and $p_{1}(K_{1})\cap
p_{1}(K_{2})=\\{0\\}.$

  * •

applying Lemma 5.2, we get that $p_{1}^{-1}\\{0\\}\cap K_{1}=K_{1}$ and
$p_{1}^{-1}\\{0\\}\cap K_{2}\subset K_{1}.$ Hence
$p_{1}^{-1}\\{0\\}\cap\left(K_{1}\cup K_{2}\right)=K_{1}$ is polynomially
convex.

Therefore, by Kallin’s lemma, $K_{1}\cup K_{2}$ is polynomially convex. Again
from Step I, we get that for each
$I\in\left\\{0,1,\cdots,m_{1}-1\right\\}\times\cdots\times\left\\{0,1,\cdots,m_{n}-1\right\\},$
$\mathscr{P}(X_{I})=\mathcal{C}(X_{I}).$

The above informations allow us to apply Result 2.2 to conclude that

$\mathscr{P}\left(\cup_{I}X_{I}\right)=\mathcal{C}\left(\cup_{I}X_{I}\right),~{}\text{
that is
}\mathscr{P}\left(\Phi^{-1}(X)\right)=\mathcal{C}\left(\Phi^{-1}(X)\right).$

Therefore, by Result 2.3, we get that

$\mathscr{P}(X)=\mathcal{C}(X).$

Step III: Showing that
${z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z)}$
separates points on $\mathbf{\overline{D(\delta)}}.$

Let
$a=(a_{1},a_{2},\cdots,a_{n}),b=(b_{1},b_{2},\cdots,b_{n})\in\mathbf{\overline{D(\delta)}}$
with $a\neq b$. Then there exists a set
$\\{i_{1},\cdots,i_{k}\\}\subseteq\\{1,\cdots,n\\}$ such that $a_{j}\neq
b_{j}~{}~{}\forall j\in\\{i_{1},\cdots,i_{k}\\}$ and $a_{j}=b_{j}~{}~{}\forall
j\notin\\{i_{1},\cdots,i_{k}\\}.$ If $a^{m_{l}}_{l}\neq b^{m_{l}}_{l}\text{
for some }l\in\\{i_{1},\cdots,i_{k}\\}$ then $z^{m_{l}}_{l}$ separates $a$ and
$b.$ Next, we assume that $a^{m_{j}}_{j}=b^{m_{j}}_{j}\text{ for all
}j\in\\{i_{1},\cdots,i_{k}\\}.$ We now show that, for some
$j\in\\{i_{1},\cdots,i_{k}\\},~{}~{}\bar{z_{j}}^{m_{n+j}}+R_{j}(z)$ separates
$a$ and $b.$ If possible, assume that

$\displaystyle\left(\bar{z_{j}}^{m_{n+j}}+R_{j}\right)(a)=(\bar{z_{j}}^{m_{n+j}}+R_{j})(b)~{}~{}~{}\forall{j}\in\\{i_{1},\cdots,i_{k}\\}.$

This implies

$\displaystyle\bar{a_{j}}^{m_{n+j}}-\bar{b_{j}}^{m_{n+j}}=R_{j}(b)-R_{j}(a)~{}~{}\forall{j}\in\\{i_{1},\cdots,i_{k}\\}.$
(5.11)

Now we compute:

$\displaystyle\sqrt{\sum_{j\in\\{i_{1},\cdots,i_{k}\\}}|R_{j}(b)-R_{j}(a)|^{2}}\leq|R(b)-R(a)|$
$\displaystyle\leq
c\left(\sum_{j=1}^{n}\left|b_{j}^{m_{n+j}}-a_{j}^{m_{n+j}}\right|^{2}\right)^{\frac{1}{2}}$
$\displaystyle=c\left(\sum_{j\in\\{i_{1},\cdots,i_{k}\\}}\left|b_{j}^{m_{n+j}}-a_{j}^{m_{n+j}}\right|^{2}\right)^{\frac{1}{2}}.$

Therefore, using (5.11), we obtain from above that

$\displaystyle\sqrt{\sum_{j\in\\{i_{1},\cdots,i_{k}\\}}\left|b_{j}^{m_{n+j}}-a_{j}^{m_{n+j}}\right|^{2}}$
$\displaystyle\leq
c\left(\sum_{j\in\\{i_{1},\cdots,i_{k}\\}}\left|b_{j}^{m_{n+j}}-a_{j}^{m_{n+j}}\right|^{2}\right)^{\frac{1}{2}}.$
(5.12)

Since $a_{j}\neq b_{j},~{}~{}a^{m_{j}}_{j}=b^{m_{j}}_{j}\text{ for all
}j\in\\{i_{1},\cdots,i_{k}\\}$ and $\gcd(m_{j},m_{n+j})=1,$ we can say that

$\displaystyle{\sum_{j\in\\{i_{1},\cdots,i_{k}\\}}\left|b_{j}^{m_{n+j}}-a_{j}^{m_{n+j}}\right|^{2}}\neq
0.$

Therefore, from (5.12) we get that $c\geq 1.$ This is a contradiction to our
hypothesis. Therefore,

$\displaystyle[z^{m_{1}}_{1},\cdots,z^{m_{n}}_{n},\bar{z_{1}}^{m_{n+1}}+R_{1}(z),\cdots,\bar{z_{n}}^{m_{2n}}+R_{n}(z);\overline{D(\delta)}]=C(\overline{D(\delta)}).$

∎

Acknowledgements. I would like to thank Sushil Gorai for the discussion during
the course of this work. This work is supported by an INSPIRE Fellowship (IF
160487) funded by DST.

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