text:
\section{Introduction}

Let $D$ be a domain in $\mathbb C^n$. An automorphism of $D$ is a biholomorphic self-map. The set of all automorphisms of $D$ makes a group under composition. We denote the automorphism group by $\mathrm{Aut}(D)$. The topology on $\mathrm{Aut}(D)$ is that of uniform convergence on compact sets (i.e., the compact-open topology).

It is a standard and classical result of H. Cartan that if $D$ is a bounded domain in $\mathbb C^n$ and the automorphism group of $D$ is noncompact then there exist a point $x \in D$, a point $p \in \partial D$, and automorphisms $\varphi_j \in \mathrm{Aut}(D)$ such that $\varphi_j(x) \to p$. In this circumstance we call $p$ a {\it boundary orbit accumulation point}.

In $1993$, Greene and Krantz \cite{GK} posed a conjecture that for a smoothly
bounded pseudoconvex domain admitting a non-compact automorphism group, the point orbits can
accumulate only at a point of finite type in the sense of Kohn, Catlin, and D'Angelo (see \cite{D, Kr01} for this concept). For this conjecture, we refer the reader to \cite{IK}.

One of the evidence for the correctness of Greene-Krantz's conjecture is provided in \cite{Ka94}. H. Kang \cite{Ka94} proved that the automorphism group $\mathrm{Aut}(E_P)$ is compact, where $E_P$ is a special kind of  Hartogs  domains
$$
E_P=\{(z_1,z_2)\in \mathbb C^2\colon  |z_1|^2+P(z_2)<1\}\Subset \mathbb C^2,
$$
where $P$ is a real-valued, $\mathcal{C}^\infty$-smooth, subharmonic function satisfying:
\begin{itemize}
\item[(i)] $P(z_2)>0$ if $z_2\ne 0$,
\item[(ii)] $P$ vanishes to infinite order only at the origin.
\end{itemize}
Note that $E_P$ is of infinite type along the points $(e^{i\theta},0)\in bE_P$ and $(e^{i\theta},0)$ are the only points of infinite type.

Recently, S. Krantz \cite{Kr12} showed that the domain
$$
\Omega:=\{z\in \mathbb C^n\colon |z_1|^{2m_1}+ |z_2|^{2m_2}+\cdots+ |z_{n-1}|^{2m_{n-1}}+\psi(|z_n|)<1\},
$$
where the $m_j$ are positive integers and where $\psi$ is a real-valued, even, smooth, monotone-and-convex-on-$[0,+\infty)$ function of a real variable with $\psi(0)=0$ that vanishes to infinite order at $0$, has compact automorphism group. In fact, the only automorphisms of $\Omega$ are the rotations in each variable separately (cf. \cite{GK, IK06}).

We would like to emphasize here that the automorphism group of a domain in $\mathbb C^n$ is not easy to describe explicitly; besides, it is unknown in most cases. In this paper, we are going to compute the automorphism group of  an infinite type model
$$
\Omega_P:=\{(z_1,z_2)\in \mathbb C^2\colon \rho(z_1,z_2)= \mathrm{Re}~z_1+P(z_2)<0\},
$$
where  $P:\mathbb C\to \mathbb R$ is a $\mathcal{C}^\infty$-smooth function satisfying:
\begin{itemize}
\item[(i)] $P(z)=q(|z|)$ for all $z\in \mathbb C$, where $q:[0,+\infty)\to \mathbb R$ is a function with $q(0)=0$ such that it is strictly increasing and convex on $[0,\epsilon_0)$ for some $\epsilon_0>0$, and
\item[(ii)] $P$ vanishes to infinite order at $0$.
\end{itemize}
It is easy to see that $(it,0), t\in \mathbb R$, are points of infinite type in $b \Omega_P$, and hence $\Omega_P$ is of infinite type.

In order to state the first main result, we recall the following terminology. A {\it holomorphic vector field} in $\mathbb C^n$ takes the form
$$
H = \sum_{k=1}^n h_k (z) \frac{\partial}{\partial z_k}
$$
for some functions $h_1, \ldots, h_n$ holomorphic in $z=(z_1, \ldots, z_n)$.  A smooth real
hypersurface germ $M$ (of real codimension 1) at $p$ in $\mathbb C^n$ takes a defining
function, say $\rho$, such that $M$ is represented by the equation $\rho(z)=0$.  The holomorphic vector field $H$ is said to be {\it tangent} to $M$ if its real part $\hbox{Re }H$ is tangent to $M$, i.e., $H$ satisfies the equation
\begin{equation}\label{maineq}
(\hbox{Re }H)\rho(z) = 0~\text{for all}~z\in M.
\end{equation}

The first aim of this paper is to prove the following theorem, which is a characterization of tangential holomorphic vector fields.
\begin{theorem}\label{Th1} Let $P:\mathbb C\to \mathbb R$ be a $\mathcal{C}^\infty$-smooth function satisfying
\begin{itemize}
\item[(i)] $P(z)=q(|z|)$ for all $z\in \mathbb C$, where $q:[0,+\infty)\to \mathbb R$ is a function with $q(0)=0$ such that it is strictly increasing and convex on $[0,\epsilon_0)$ for some $\epsilon_0>0$, and
\item[(ii)] $P$ vanishes to infinite order at $0$.
\end{itemize}
If $H=h_1(z_1,z_2)\frac{\partial}{\partial z_1}+h_2(z_1,z_2)\frac{\partial}{\partial z_2}$ with $H(0,0)=0$ is holomorphic in $\Omega_P\cap U$, $\mathcal{C}^\infty$-smooth in $\overline{\Omega_P}\cap U$, and tangent to $b \Omega_P\cap U$, where $U$ is a neighborhood of $(0,0)\in \mathbb C^2$, then $H=i\beta z_2\frac{\partial}{\partial z_2}$ for some $\beta\in \mathbb R$.
\end{theorem}

 In the case that the tangential holomorphic vector field $H$ is holomophic in a neighborhood of the origin, Theorem \ref{Th1} is already proved in \cite{By1, HN}. Here, since the tangential holomorphic vector field $H$ in Theorem \ref{Th1} is only holomorphic inside the domain, it seems to us that some key techniques in \cite{By1} could not use for our situation.  To get around this difficulty, we first employ the Schwarz reflection principle to show that the holomorphic functions $h_1, h_2$ must vanish to finite order at the origin. Then the equation (\ref{maineq}) implies that $h_1\equiv 0$. Therefore, from Chirka's curvilinear Hargtogs' lemma the proof finally follows (see the detailed proof in Section \ref{S2}).

We now note that $\mathrm{Aut}(\Omega_P)$ is noncompact since it contains biholomorphisms
$$
(z_1,z_2)\mapsto (z_1+is, e^{it}z_2 ),~s,t\in \mathbb R.
$$
Let us denote by $\{R_t\}_{t\in \mathbb R}$ the one-parameter subgroup of $\mathrm{Aut}(\Omega_P,0)$ generated by the holomorphic vector field $H_R(z_1,z_2)=iz_2\frac{\partial }{\partial z_2}$, that is,
$$
R_t(z_1,z_2)=\big(z_1,e^{it}z_2\big), ~\forall t\in\mathbb R.
$$
In addition, denote by $T_s(z_1,z_2)=(z_1+is,z_2)$ for $s\in \mathbb R$.

To state the second main result, we need the following definitions.
Recall that the Kobayashi metric $K_D$ of $D$ is defined by
$$K_D(\eta,X):=\inf\{\frac{1}{R}|\ \exists f: \Delta\to D~\text{such that} f(0)=\eta, f'(0)=R X\},$$
where $\eta\in D$ and $X\in T^{1,0}_\eta\mathbb C^n$, where $\Delta_r$ is a disc with center at the origin and radius $r>0$ and $\Delta:=\Delta_1$.

The following definition derives from work of X. Huang (\cite{Hu95}).
\begin{define} Let $D$ be a domain in $\mathbb C^n$ with $\mathcal{C}^2$-smooth boundary $bD$ and $z_0$ be a boundary point.  For a $\mathcal{C}^1$-smooth monotonic increasing function $g: [1,+\infty)\to [1,+\infty)$, we say that $D$ is \emph{$g$-admissible} at $z_0$ if there exists a neighborhood $V$ of $z_0$ such that
$$
K_{D}(z,X)\gtrsim g(\delta_D^{-1}(z))|X|
$$
for any $z\in V\cap D$ and $X\in T^{1,0}_z\mathbb C^n$, where $\delta_D(z)$ is the distance of $z$ to $bD$.
\end{define}
\begin{remark}
\begin{itemize}
\item[(i)] It is proved in \cite[p.93]{Ber} (see also in \cite{Sib81}) that if there exists a plurisubharmonic peak function at $z_0$, then
there exists a neighborhood $V$ of $z_0$ such that
$$
K_D(z,X)\leq K_{D\cap V}(z,X)\leq  2K_D(z, X),
$$
for any $z\in V\cap D$ and $X\in T^{1,0}_z\mathbb C^n$.
\item[(ii)] If $D$ is $\mathcal{C}^\infty$-smooth pseudoconvex of finite type, then $D$ is $t^\epsilon$-admissible at any boundary point for some $\epsilon>0$ (cf. \cite{Cho92}). Recently, T. V. Khanh \cite{Khanh13} proved that a certain pseudoconvex domain of infinite type is also $g$-admissible for some function $g$.
\end{itemize}
\end{remark}

\begin{define}[see \cite{Khanh13}] Let $D\subset \mathbb C^n$ be a $\mathcal{C}^2$-smooth domain. Assume that $D$ is pseudoconvex near $z_0\in bD$. For a $\mathcal{C}^1$-smooth monotonic increasing function $u: [1,+\infty)\to [1,+\infty)$ with $u(t)/t^{1/2}$ decreasing, we say that a domain $D$ has the $u$-property at the boundary $z_0$ if there  exist a neighborhood $U$ of $z_0$ and a family of $\mathcal{C}^2$-functions $\{\phi_\eta\}$ such that
\begin{itemize}
\item[(i)] $|\phi_\eta|<1$, $\mathcal{C}^2$, and plurisubharmonic on $D$;
\item[(ii)] $i\partial\bar \partial \phi_\eta \gtrsim u(\eta^{-1})^2 Id$ and $|D\phi_\eta|\lesssim \eta^{-1}$ on $U\cap \{z\in D\colon -\eta<r(z)<0\}$, where $r$ is a $\mathcal{C}^2$-defining function for $D$.
\end{itemize}
Here and  in what follows, $\lesssim$ and $\gtrsim$ denote inequalities up to a positive constant multiple. In addition, we use $\approx $ for the combination of $\lesssim$ and $\gtrsim$.

\end{define}
\begin{define} [see \cite{Khanh13}]We say that a domain $D$ has the strong $u$-property at the boundary $z_0$ if it has the $u$-property with $u$ satisfying the following:
\begin{itemize}
\item[(i)] $\int\limits_t^{+\infty}\frac{da}{au(a)}$ for some $t>1$ and denote by $(g(t))^{-1}$ this finite integral;
\item[(ii)] The function $\frac{1}{\delta g\Big(1/\delta^\eta\Big)}$ is decreasing and $\int\limits_0^d\frac{1}{\delta g\Big(1/\delta^\eta)\Big)}d\delta<+\infty$ for $d>0$ small enough and for some $0<\eta<1$.
\end{itemize}
\end{define}
\begin{define} We say that $\Omega_P$ satisfies the condition (T) at $\infty$ if one of following conditions holds
\begin{itemize}
\item[(i)]$\lim_{z\to \infty}P(z)=+\infty$;
\item[(ii)] The function $Q$ defined by setting $Q(\zeta):=P(1/\zeta)$ can be extended to be $\mathcal{C}^\infty$-smooth in a neighborhood of $\zeta=0$, $\Omega_Q$ has the strong $\tilde u$-property at $(-r,0)$ for some function $\tilde u$, where $r=\lim_{z\to \infty} P(z)$, and $b\Omega_P$ and $b\Omega_Q$ are not isomorphic as CR maniflod germs at $(0,0)$ and $(-r,0)$ respectively.
\end{itemize}

\end{define}

The second aim of this paper is to show the following theorem.
\begin{theorem} \label{Th2}
Let $P:\mathbb C\to \mathbb R$ be a $\mathcal{C}^\infty$-smooth function satisfying
\begin{itemize}
\item[(i)] $P(z)=q(|z|)$ for all $z\in \mathbb C$, where $q:[0,+\infty)\to \mathbb R$ is a function with $q(0)=0$ such that it is strictly increasing and convex on $[0,\epsilon_0)$ for some $\epsilon_0>0$,
\item[(ii)] $P$ vanishes to infinite order at $0$, and
\item[(iii)] $P$ vanishes to finite order at any $z\in \mathbb C^*:=\mathbb C\setminus\{0\}$.
\end{itemize}
Assume that $\Omega_P$ has the strong $u$-property at $(0,0)$ and $\Omega_P$ satisfies the property (T) at $\infty$. Then
$$
\mathrm{Aut}(\Omega_P)=\{(z_1,z_2)\mapsto (z_1+is, e^{it}z_2 )\colon s,t \in \mathbb R\}.
$$
\end{theorem}

\begin{remark}\label{remark}
Let $\Omega_P$ be as in Theorem \ref{Th2} and let $P_\infty(b \Omega_P)$ the set of all points in $b \Omega_P$ of D'Angelo infinite type. It is easy to see that $P_\infty(b \Omega_P)=\{(it,0)\colon t\in \mathbb R\}$. Moreover, since $\Omega_P$ is invariant under any translation $(z_1,z_2)\mapsto (z_1+it,z_2),~t\in \mathbb R$, it satisfies the trong $u$-property at $(it,0)$ for any $t\in \mathbb R$.
\end{remark}
\begin{remark}\label{remark4}
 Let $P$ be a function defined by $P(z_2)=\exp(-1/|z_2|^\alpha)$ if $z_2\ne 0$ and $P(0)=0$, where $0<\alpha<1$. Then by \cite[Corollary 1.3]{Khanh13}, $\Omega_P$ has $\log^{1/\alpha}$-property at $(it,0)$ and thus it is $\log^{1/\alpha-1}$-admissible at $(it,0)$ for any $t\in \mathbb R$. Furthermore, a computation shows that  if $0<\alpha<1/2$, then $\Omega_P$ has the strong $\log^{1/\alpha}$-property at $(it,0)$ for any $t\in \mathbb R$.
\end{remark}
\begin{example}\label{ex1}
Let $E_j$, $j=1,\ldots,3$, be domains defined by
$$
E_j:=\{(z_1,z_2)\in \mathbb C^2\colon \rho(z_1,z_2)= \mathrm{Re}~z_1+P_j(z_2)<0\},
$$
where $P_j$ are defined by
\begin{equation*}
\begin{split}
P_1&=\psi(|z_2|)e^{-1/|z_2|^{\alpha}}+(1-\psi(|z_2|))\frac{1}{|z_2|^{2m}},\\
P_2&=\psi(|z_2|)e^{-1/|z_2|^{\alpha}}+(1-\psi(|z_2|))e^{-1/|z_2|^{\beta}},\\
P_2&=\psi(|z_2|)e^{-1/|z_2|^{\alpha}}+(1-\psi(|z_2|))|z_2|^2
\end{split}
\end{equation*}
if $z_2\ne 0$ and $P(0)=0$, where $0<\alpha,\beta<1/2, m\in \mathbb N^*)$ with $\beta\ne \alpha$ and $\psi(t)$ is a $\mathcal{C}^\infty$-smooth cut-off function such that $\psi(t)=1$ if $|t|<a$ and $\psi(t)=0$ if $|t|>b~(0<a<b)$. It follows from Remark \ref{remark4} and a computation that $E_j,~j=1,\ldots,3,$ have the strong $\log^{1/\alpha}$-property and satisfy the property (T) at $\infty$. Therefore, by Theorem \ref{Th2} we conclude that
$$
\mathrm{Aut}(E_j)=\{(z_1,z_2)\mapsto (z_1+is, e^{it}z_2 )\colon s,t \in \mathbb R\},~j=1,\ldots,3.
$$
\end{example}

We explain now the idea of proof of Theorem \ref{Th2}. Let $f\in \mathrm{Aut}(\Omega_P)$ be an arbitrary. We show that there exist $t_1,t_2\in \mathbb R$ such that $f,f^{-1}$ extend smoothly to $b\Omega_P$ near $(it_1,0)$ and $(it_2,0)$ respectivey and $(it_2,0)=f(it_1,0)$ (cf. Lemma \ref{lem6}). Replacing $f$ by $T_{-t_2}\circ f\circ T_{t_1}$, we may assume that $f,f^{-1}$ extend smoothly to $b\Omega_P$ near the origin and $f(0,0)=(0,0)$. Next, we consider the one-parameter subgroup $\{F_t\}_{t\in \mathbb R}$ of $\mathrm{Aut}(\Omega_P)\cap \mathcal{C}^\infty(\overline{\Omega_P}\cap U)$ defined by $F_t=f\circ R_{-t}\circ f^{-1}$. By employing Theorem \ref{Th1}, there exists a real number $\delta$ such that $F_t=R_{\delta t}$ for all $t\in\mathbb R$. Using the property that $P$ vanishes to infinite order at $0$, it is proved that $f=R_{t_0}$ for some $t_0\in \mathbb R$ (see the detailed proof in Section $4$). This finishes our proof.

This paper is organized as follows. In Section $2$,  we prove Theorem \ref{Th1}. In Section $3$,  we prove several lemmas to be used mainly in the proof of Theorem \ref{Th2}. Section $4$ is devoted to the proof of Theorem \ref{Th2}. Finally, two lemma are given in Appendix.

\section{Holomorphic vector fields tangent to an infinite type model}\label{S2}

This section is devoted to the proof of Theorem \ref{Th1}. Assume that $P:\mathbb C\to \mathbb R$ is a $\mathcal{C}^\infty$-smooth function satisfying (i) and (ii) as in Introduction.

Then we consider a nontrivial holomorphic vector field $H=h_1(z_1,z_2)\frac{\partial}{\partial z_1}+h_2(z_1,z_2)\frac{\partial}{\partial z_2}$ defined on $\Omega_P\cap U$, where $U$ is a neighborhood of the origin. We only consider $H$ is tangent to $b \Omega_P\cap U$. This means that they satisfy the identity
\begin{equation}\label{eq1}
(\mathrm{Re}~H)\rho(z_1,z_2)=0,~\forall~ (z_1,z_2)\in b \Omega_P\cap U.
\end{equation}
By a simple computation, we have
\begin{equation*}
\begin{split}
 \rho_{z_1}(z_1,z_2)&= 1,\\
\rho_{z_2}(z_1,z_2)&= P'(z_2),
\end{split}
\end{equation*}
and the equation (\ref{eq1}) can thus be rewritten as
\begin{equation}\label{eq2}
\begin{split}
&\mathrm{Re} \Big[h_1(z_1,z_2)+P'(z_2)h_2(z_1,z_2)\Big ]=0
\end{split}
\end{equation}
for all $(z_1,z_2)\in b \Omega_P\cap U$.

Since $\Big(it-P(z_2), z_2\Big)\in b \Omega_P$ for any $t \in \mathbb R$ with $t$ small enough, the above equation again admits a new form
\begin{equation}\label{eq3}
\begin{split}
&\mathrm{Re}\Big[ h_1\big(it-P(z_2),z_2\big)+P'(z_2)h_2\big(it-P(z_2),z_2\big)\Big]=0
\end{split}
\end{equation}
for all $z_2\in\mathbb C$ and for all $t\in\mathbb R$ with $|z_2|<\epsilon_0$ and $|t|<\delta_0$, where $\epsilon_0>0$ and $\delta_0>0$ are small enough.

\begin{lemma}\label{lem6} We have that
 $\frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(z_1,0)$ can be extended to be holomorphic in a neighborhood of $z_1=0$ for every $m,n\in\mathbb N$.
\end{lemma}
\begin{proof}
Since $\nu_0(P')=+\infty$, it follows from (\ref{eq3}) with $t=0$ that $\mathrm{Re} h_1(it,0)=0$ for all $t\in (-\delta_0,\delta_0)$. By the Schwarz reflection principle, $h_1(z_1,0)$ can be extended to a holomorphic function on a neighborhood of $z_1=0$. For any $m,n\in\mathbb N$, taking $\frac{\partial^{m+n}}{\partial t^m\partial z_2^n}\mid_{z_2=0}$ of both sides of the equation (\ref{eq3}) one has
$$
 \mathrm{Re}\Big[i^m \frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(it,0)\Big]=0
$$
for all $t\in (-\delta_0,\delta_0)$. Again by the Schwarz reflection principle, $\frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(z_1,0)$ can be extended to be holomorphic in a neighborhood of $z_1=0$, which completes the proof.
\end{proof}
\begin{corollary} \label{coro2}
If $h_1$ vanishes to infinite order at $(0,0)$, then $h_1\equiv 0$.
\end{corollary}
\begin{proof}
Since $h_1$ vanishes to infinite order at $(0,0)$, $\frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(z_1,0)$ also vanishes to infinite at $z_1=0$ for all $m,n\in\mathbb N$. Moreover, by Lemma \ref{lem6} these functions are holomorphic in a neighborhood of $z_1=0$. Therefore, $\frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(z_1,0)\equiv 0$ for every $m,n\in\mathbb N$.

Expand $h_1$ into the Taylor series at $(-\epsilon,0)$ with $\epsilon>0$ small enough so that
$$
h_1(z_1,z_2)=\sum_{m,n=0}^\infty\frac{1}{m!n!} \frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(-\epsilon,0)(z_1+\epsilon)^m z_2^n
$$

Since $\frac{\partial^{m+n}}{\partial z_1^m\partial z_2^n} h_1(-\epsilon,0)=0$ for all $m,n\in\mathbb N$, $h_1\equiv 0$ on a neighborhood of $(-\epsilon,0)$, and thus $h_1\equiv 0$ on $\Omega_P$.

\end{proof}

\begin{proof}[Proof of Theorem \ref{Th1}]

Denote by $D_P(r):=\{z_2\in \mathbb C: |z_2|< q^{-1}(r) \}~(r>0)$. For each $z_1$ with $\mathrm{Re}(z_1)<0$, we have
\begin{equation}\label{eq4}
h_1(z_1,z_2)=\sum_{n=0}^\infty a_n(z_1)z_2^n, ~\forall ~ z_2\in D_P(-\mathrm{Re}(z_1)),
\end{equation}
where $a_n(z_1)=\frac{\partial^n}{\partial z_2^n}h_1(z_1,0)$ for every $n\in \mathbb N$. Since $h_1\in \mathrm{Hol}(\Omega_P\cap U)\cap \mathcal{C^\infty}(\overline{\Omega_P}\cap U)$, $a_n\in Hol(\mathcal{H}\cap U_1)\cap \mathcal{C}^\infty(\overline{\mathcal{H}}\cap U_1)$ for every $n=0,1,\ldots$, where $\mathcal{H}:=\{z_1\in \mathbb C\colon \mathrm{Re}(z_1)<0\}$ and $U_1$ is a neighborhood of $z_1=0$ in $\mathbb C_{z_1}$. Moreover, expanding the function $g_{z_1}(z_2):=h_1(z_1,z_2)$ into the Fourier series we can see that (\ref{eq4}) still holds for all $z_2\in \overline{D_P(-\mathrm{Re}(z_1))}$. Therefore, the function $h_1(it-P(z_2),z_2)$ can be rewritten as follows:
$$
h_1(it-P(z_2),z_2)=\sum_{n=0}^\infty a_n(it-P(z_2))z_2^n,
$$
for all $(t,z_2)\in (-\delta_0,\delta_0)\times \Delta_{\epsilon_0}$, where $\delta_0>0, \epsilon_0>0$ are small enough.

Similarly, we also have
$$
h_2(it-P(z_2),z_2)=\sum_{n=0}^\infty b_n(it-P(z_2))z_2^n
$$
for all $(t,z_2)\in (-\delta_0,\delta_0)\times \Delta_{\epsilon_0}$, where $b_n \in Hol(\mathcal{H}\cap U_1)\cap \mathcal{C}^\infty(\overline{\mathcal{H}}\cap U_1)$ for every $n=0,1,\ldots$.

Now we shall prove that $h_1\equiv 0$. Indeed, aiming for contradiction, we suppose that $h_1\not\equiv 0$. If $h_1$ vanishes to infinite order at $(0,0)$, then by Corollary \ref{coro2} one gets $h_1\equiv 0$. So, $h_1$ vanishes to finite order at $(0,0)$. It follows from (\ref{eq3}) that $h_2$ also vanishes to finite order at $(0,0)$, for otherwise $h_1$ vanishes to infinite order at $(0,0)$.

Denote by
\begin{equation}\label{eq6}
\begin{split}
m_0&:=\min\Big\{m\in \mathbb N \colon \frac{\partial^{m+n}}{\partial^m z_1 \partial^n z_2}h_1(0,0)\ne 0~\text{for some}~ n\in\mathbb N\Big\}, \\
n_0&:=\min\Big\{n\in \mathbb N \colon \frac{\partial^{m_0+n}}{\partial^{m_0} z_1 \partial^n z_2}h_1(0,0)\ne 0\Big\}, \\
k_0&:=\min\Big\{m\in \mathbb N \colon \frac{\partial^{k+l}}{\partial^k z_1 \partial^l z_2}h_2(0,0)\ne 0~\text{for some}~ l\in\mathbb N\Big\}, \\
l_0&:=\min\Big\{l\in \mathbb N \colon \frac{\partial^{k_0+l}}{\partial^{k_0} z_1 \partial^l z_2}h_2(0,0)\ne 0\Big\}.
\end{split}
\end{equation}

Since $\nu_0(P)=+\infty$, one obtains that
\begin{equation}\label{eq6}
\begin{split}
h_1(i\alpha P(z_2)-P(z_2),z_2)&= a_{m_0,n_0} (i\alpha-1)^{m_0}(P(z_2))^{m_0}\big(z_2^{n_0}+o(|z_2|^{n_0}\big),\\
h_2(i\alpha P(z_2)-P(z_2),z_2)&= b_{k_0,l_0} (i\alpha-1)^{k_0}(P(z_2))^{k_0}\big(z_2^{l_0}+o(|z_2|^{l_0}\big),
\end{split}
\end{equation}
where $a_{m_0,n_0}:=\frac{\partial^{m_0+n_0}}{\partial^{m_0} z_1 \partial^{n_0} z_2}h_1(0,0)\ne 0,~b_{k_0,l_0}:=\frac{\partial^{k_0+l_0}}{\partial^{k_0} z_{l_0} \partial^{l_0} z_2}h_2(0,0) \ne 0$, and $\alpha\in\mathbb R$ will be chosen later.

Now it follows from (\ref{eq3}) with $t=\alpha P(z_2)$ that
\begin{equation}\label{eq7}
\begin{split}
\mathrm{Re} &\Big[ a_{m_0 n_0}(i\alpha -1)^{m_0}(P(z_2))^{m_0}\big(z_2^{n_0}+o(|z_2|^{n_0})\big)+
 b_{k_0 l_0}(i\alpha -1)^{k_0}(z_2^{l_0}+o(|z_2|^{l_0}) \\
&\times (P(z_2))^{k_0} P'(z_2)   \Big ]=0
\end{split}
 \end{equation}
for all $z_2\in \Delta_{\epsilon_0}$ and for all $\alpha \in \mathbb R$ small enough. We note that in the case $n_0=0$ and $\mathrm{Re}(a_{m_0 0})=0$, $\alpha$ can be chosen in such a way that $\mathrm{Re}\big( (i\alpha-1)^{m_0}a_{m_0 0}\big)\ne 0$. Then the above equation yields that $k_0>m_0$. Furthermore, since $P$ is rotational, it follows that $\mathrm{Re}(iz_2P'(z_2))\equiv 0$ (see \cite[Lemma $4$]{Kim-Ninh}), and hence we can assume that $\mathrm{Re}(b_{10})\ne 0$ for the case that $k_0=1,l_0=0$. However, (\ref{eq7}) contradicts Lemma $3$ in \cite{Kim-Ninh}. Therefore, $h_1\equiv 0$.

Granted $h_1\equiv 0$, (\ref{eq3}) is equivalent to
\begin{equation}\label{eq8}
\begin{split}
\mathrm{Re} \Big[ P'(z_2)h_2(it-P(z_2),z_2)   \Big ]=0
\end{split}
 \end{equation}
for all $(t,z_2)\in (-\delta_0,\delta_0)\times \Delta_{\epsilon_0}$. Thus, for each $z_2\in \Delta^*_{\epsilon_0}$ the function $g_{z_2}$ defined by setting $g_{z_2}(z_1):=h_2(z_1,z_2)$ is holomorphic in $\{z_1\in \mathbb C:\mathrm{Re}(z_1)<-P(z_2) \}$ and $\mathcal{C}^\infty$-smooth up to the real line $\{z_1\in \mathbb C: \mathrm{Re}(z_1)=-P(z_2)\}$. Moreover, $g_{z_2}$ maps this line onto the real line $\mathrm{Re}(P'(z_2)w)=0$ in the complex plane $\mathbb C_w$. Thus, by the Schwarz reflection principle, $g_w$ can be extended to be holomorphic in a neighborhood $U$ of $z_1=0$ in the plane $\mathbb C_{z_1}$. (The neighborhood $U$ is independent of $z_2$.)

Now our function $h_2$ is holomorphic in $z_1\in U$ for each $z_2\in\Delta^*_{\epsilon_0}$ and holomorphic in $(z_1,z_2)$ in the domain $\{(z_1,z_2)\in \mathbb C^2: \mathrm{Re}(z_1)<0,~ |z_2|<q^{-1}(-\mathrm{Re}(z_1))\}$. Therefore, it follows from Chirka's curvilinear Hartogs' lemma (see \cite{Ch}) that $h_2$ can be extended to be holomorphic in a neighborhood of $(0,0)$ in $\mathbb C^2$. Moreover, by (\ref{eq8}) and by \cite[Theorem $3$]{HN} we conclude that $h_2(z_1,z_2)\equiv i\beta z_2$ for some $\beta\in\mathbb R^*$. So, the proof is complete.

\end{proof}
\section{Extension of automorphisms }

If $f: D\to \mathbb C^N$ is a continuous map on a domain $D\subset \mathbb C^n$  and $z_0\in \partial D$, we denote by $\mathcal{C}(f,z_0)$ the cluster set of $f$ at $z_0$:
$$
\mathcal{C}(f,z_0)=\{w\in \mathbb C^N \colon w=\lim f(z_j), z_j\in D, \text{and}~\lim z_j=z_0 \}.
$$
\begin{define}[see \cite{Bel81}]
When $\Gamma$ be an open subset of the boundary of a smooth domain $D$, we say that $\Gamma$ satisfies \emph{local condition} $\mathrm{R}$ if for each $z\in \Gamma$, there is an open set $V$ in $\mathbb{C}^n$ with $z\in V$ such that for each $s$, there is an $M$ such that
$$
\mathrm{P}\big(W^{s+M}(D\cap V)\big)\subset W^{s}(D\cap V).
$$
We say that $D$ satisfies \emph{local condition} $\mathrm{R}$ at $z_0\in bD$ if there exists an open subset of the boundary $bD$ containing $z_0$ and satisfying local condition $\mathrm{R}$.
\end{define}

\begin{define}
Let $D, G$ be domains in $\mathbb C^n$ and let $F: [0,+\infty)\to [0,+\infty)$ be an inceasing function with $F(0)=0$. Let $z_0\in bD$ and $w_0\in bG$. We say that $D,G$ satisfies the property $(D,G)_{(z_0,w_0)}^F$ if for each proper holomorphic mapping $f: D\to G$, there exist neighborhoods $U$ and $V$ of $z_0$ and $w_0$ respectively such that
$$
d_G(f(z))\leq  F(d_D(z))
$$
for any $z\in U\cap D$ such that $f(z)\in V\cap G$.
\end{define}
For the case $D$ and $D$ are bounded pseudoconvex domains with generic corners, D. Chakrabarti and K. Verma \cite[Propsition 5.1]{CV14} proved there exists a $\delta\in (0,1)$ such that
$$
(d_D(z))^{1/\delta}\lesssim d_G(f(z))\lesssim   (d_D(z))^\delta
$$
for all $z\in D$, which is a generalization of  \cite{DF79,Ber91}. Consequently, $D,G$ satisfies the property $(D,G)_{(z_0,w_0)}^F$, where $F(t)=t^\delta$, for any $z_0\in bD$ and $w_0\in bG$.

We now recall the general H\"{o}lder continuity (see \cite{Khanh13}). Let $f$ be an increasing function such that $\lim_{t\to +\infty}f(t)=+\infty$. For $\Omega\subset \mathbb C^n$, define $f$-H\"{o}lder space on $\bar \Omega$ by
$$
\Lambda^f(\overline{\Omega})=\{u\colon \|u\|_\infty+\sup_{z,z+h\in \bar\Omega}f(|h|^{-1})|u(z+h)-u(z)|<+\infty\}.
$$
Note that the $f$-H\"{o}lder space includes the standard H\"{o}lder space $\Lambda_\alpha(\overline{\Omega})$ by taking $f(t)=t^\alpha$ with $0<\alpha<1$.

The following lemma is a slight generalization of \cite[Theorem 1.4]{Khanh13}.
\begin{lemma} \label{lem2}
 Let $D$ and $G$ be domains in $\mathbb C^n$ with $\mathcal{C}^2$-smooth boundaries. Let $g:[1,+\infty)\to [1,+\infty)$ and $F: [0,+\infty)\to [0,+\infty)$ be nonnegative increasing functions with $F(0)=0$ such that the function $\frac{1}{\delta g\Big(1/F\big(\delta\big)\Big)}$ is decreasing and $\int\limits_0^d\frac{1}{\delta g\Big(1/F\big(\delta\big)\Big)}d\delta<+\infty$ for $d>0$ small enough. Assume that $D$ and $G$ satisfies the property $(D,G)_{(z_0,w_0)}^F$ and $G$ is $g$-admissible at $w_0$. Let $f:D\to G$ be a proper map such that $w_0\in \mathcal{C}(f,z_0)$. Then there exist neighborhoods $U$ and $V$ of $z_0$ and $w_0$, respectively, such that $f$ can be extended as a general H\"{o}der continuous map $\hat f: U\cap \overline{D}\to V\cap \overline{G}$ with a rate $h(t)$ defined by
$$
(h(t))^{-1}:=\int^{t^{-1}}_0\frac{1}{\delta g\Big(1/F\big(\delta\big)\Big)}d\delta.
$$
\end{lemma}
\begin{proof}
Since $G$ is $g$-admissible at $w_0$, using the Schwarz-Pick lemma for the Kobayashi metric and the upper bound of Kobayashi metric, we obtain the following estimate
$$
g(\delta^{-1}_G(f(z)))|f'(z)X|\lesssim K_{G}(f(z),f'(z)X)\leq K_D(z,X)\lesssim \delta^{-1}_D(z) |X|
$$
for any $z\in D\cap  U$ such that $f(z)\in V\cap G$ and $X\in T^{1,0}\mathbb C^n$. Moreover, since the property $(D,G)_{(z_0,w_0)}^F$ holds, we may assume that
$$
\delta_{G}(f(z))\leq F(\delta_D(z))
$$
for any $z\in D\cap  U$ such that $f(z)\in V\cap G$. Therefore,
\begin{equation}\label{22-10}
|f'(z)X|\lesssim \frac{1}{\delta_D(z)g\Big(1/F\big(\delta_D(z)\big)\Big)} |X|
\end{equation}
for any $z\in D\cap U$ such that $f(z)\in V\cap G$ and $X\in T^{1,0}\mathbb C^n$.

By using the Henkin's technique (see \cite{Ber92, Suk}), we are going to prove that $f$ extends continuously to $z_0$. Indeed, suppose that $f$ does not extend continuously to $z_0$: there are an open ball $B\subset V$ (with center at $w_0$) and a neighborhoods basis $U_j$ of $z_0$ such that $f(D\cap U_j)$ is connected and never contained in $B$. Then, since $w_0\in \mathcal{C}_\Omega(f,z_0)$, there exists a sequence $\{z_j'\}$, $z_j'\in U_j$ such that $f(z_j')\in \partial B$ and $\lim f(z_j')=w_0'\in \partial B\cap\partial G$.

Let $\{z_j\}\subset \Omega\cap U$ such that $\lim z_j=z_0$ and $\lim f(z_j)=w_0$. Let $l_j:=|z_j-z_j'|$ and $\gamma_j:[0,3l_j]\to \Omega\cap U$ be a $\mathcal{C}^1$-path such that:
\begin{itemize}
\item[(a)] $\gamma_j(0)=z_j$ and $\gamma_j(3l_j)=z_j'$.
\item[(b)] $\delta_\Omega(\gamma(t))\geq t$ on $[0,l_j]$; $\delta_\Omega(\gamma(t))\geq l_j$ on $[l_j,2l_j]$; $\delta_\Omega(\gamma(t))\geq 3l_j-t$ on $[2l_j,3l_j]$.
\item[(c)] $\|\frac{d\gamma_j(t)}{dt}\|\lesssim 1$, $t\in [0,3l_j]$.
\end{itemize}
(See \cite[Prop. 2, p. 203]{HL84}).

Choose $t_j\in [0,3l_j]$ such that $f\circ \gamma_j([0,t_j])\subset \bar B$ and $f\circ \gamma_j(t_j)\in \partial B$. It follows from (\ref{22-10}), (b) and (c) that $|f(z_j)-f\circ \gamma_j(t_j)|\lesssim 1/h(1/l_j)+1/g\Big(1/F\big(l_j\big)\Big)\to 0$ as $j\to \infty$: a contradiction. Hence, $f$ extends continuously to $z_0$.

We may now assume that $f(D\cap U)\subset G\cap V$ and apply \cite[Lemma 1.4]{Khanh13} for proving that $f$ can be extended to a $h$-H\"{o}der continuous map $\hat f: \overline{D}\cap  U\to \overline{G}\cap V$ with the rate $h(t)$ defined by
$$
((h(t))^{-1}:=\int^{t^{-1}}_0 \frac{1}{\delta g\Big(1/F\big(\delta\big)\Big)} d\delta.
$$
\end{proof}

The following lemma is a local version of Fefferman's theorem (see \cite{Bel81}).
\begin{lemma}\label{lem3}
Suppose that $D$ and $G$ are $\mathcal{C}^\infty$-smooth domains in $\mathbb  C^n$ satisfying local condition $\mathrm{R}$ at $z_0\in bD$ and $w_0\in bG$ respectively. Assume that $D$ and $G$ are pseudoconvex near $z_0$ and $w_0$ respectively. Let $g:[1,+\infty)\to [1,+\infty)$ and $F: [0,+\infty)\to [0,+\infty)$ be nonnegative increasing functions with $F(0)=0$ such that the function $\frac{1}{\delta g\Big(1/F\big(\delta\big)\Big)}$ is decreasing and $\int\limits_0^d\frac{1}{\delta g\Big(1/F\big(\delta\big)\Big)}d\delta<+\infty$ for $d>0$ small enough. Suppose that $D$ and $G$ satisfy the property $(D,G)^F_{(z_0,w_0)}$. Let $f$ be a biholomorphic mapping of $D$ onto $G$ such that $w_0\in \mathcal{C}(f,z_0)$. Then $f$ extends smoothly to $bD$ in some neighborhood of the point $z_0$.
\end{lemma}
\begin{proof}
By Lemma \ref{lem2}, we may assume that there exist neighborhoods $U$ and $V$ of $z_0$ and $w_0$ respectively such that $f$  extends continuously to $U\cap \bar D$. Moreover, we may assume that $f(U\cap D)=V\cap G$ and $U\cap D$ and $V\cap G$ are  bounded $\mathcal{C}^\infty$-smooth pseudoconvex domains. Therefore, the proof follows from the theorem in \cite[Section 7]{Bel81}.
\end{proof}
\begin{lemma}\label{lem4}
Let $D \subset \mathbb C^n$ be a $\mathcal{C}^2$-smooth domain and let $0<\eta<1$. Assume that $D$ is pseudoconvex near $z_0\in bD$ and $D$ has $u$-property at $z_0$, where  $u: [1,+\infty)\to [1,+\infty)$ is a smooth monotonic increasing function with $u(t)/t^{1/2}$ decreasing and $\int\limits_{t_0}^{+\infty} \frac{da}{au(a)}<+\infty$ for some $t_0>1$. Then $D$ is  $g$-admissible at $z_0$, where $g$ is a function defined by
$$
(g(t))^{-1}=\int_t^{+\infty} \frac{da}{au(a)},~ t_0\leq t<+\infty.
$$
Moreover, the property $(D,G)_{(z_0,w_0)}^{F_2}$ holds for any $\mathcal{C}^2$-smooth domain $G\subset \mathbb C^n$ and $w_0\in bG$, where $F_2(t):=c_2 t^\eta,~t>0$, for some $c_2>0$.
\end{lemma}
\begin{proof} Let $D \subset \mathbb C^n$ be a $\mathcal{C}^2$-smooth domain. Assume that $D$ is pseudoconvex near $z_0\in bD$ and $D$ has the $u$-property at $z_0$, where  $u: [1,+\infty)\to [1,+\infty)$ is a smooth monotonic increasing function with $u(t)/t^{1/2}$ decreasing and $\int_{t_0}^{+\infty} \frac{da}{au(a)}<+\infty$ for some $t_0>1$. It follows from \cite[Theorem 1.2]{Khanh13} that $D$ is  $g$-admissible at $z_0$, where $g$ is a function defined by
$$
(g(t))^{-1}=\int_t^{+\infty} \frac{da}{au(a)},~ t_0\leq t<+\infty.
$$
Denote by $\tilde g$ the functions defined by
\begin{equation*}
\begin{split}
\tilde g(\delta)=\frac{1}{g^{-1}(1/(\gamma \delta))},
\end{split}
\end{equation*}
for any $0<\delta<\delta_0$, where $\gamma, \delta_0$ sufficiently small. By \cite[Theorem 3.1]{Khanh13} and the proof of \cite[Theorem 2.1]{Khanh13}, there exists a family $\psi_w(z)$ as in Lemma \ref{lemma12} in Appendix, where $F_1:=c_1 \tilde{g}^\eta$ and $F_2(t):=c_2 t^\eta,~t>0$, for some $0<\eta<1$ and $c_1,c_2>0$. Therefore, it follows from Lemma \ref{lemma12} in Appendix that the property $(D,G)_{(z_0,w_0)}^{F_2}$ holds for any $\mathcal{C}^2$-smooth domain $G\subset \mathbb C^n$ and $w_0\in bG$. This finishes the proof.
\end{proof}
By the definition of strong $u$-property, lemmas \ref{lem3} and \ref{lem4}, we obtain the following corollary.
\begin{corollary}\label{Co2}
Suppose that $D$ and $G$ are $\mathcal{C}^\infty$-smooth domains in $\mathbb  C^n$ satisfying the local condition R at $z_0\in bD$ and $w_0\in bG$ respectively. Suppose that $D$ and $G$ are pseudoconex near $z_0$ and $w_0$ respectively. Assume that $D$ (resp. $G$) has the strong $u$-property at $z_0$ (resp. strong $\tilde u$-property at $w_0$).
Let $f$ be a biholomorphic mapping of $D$ onto $G$ such that $w_0\in \mathcal{C}(f,z_0)$. Then $f$ and $f^{-1}$ extend smoothly to $bD$ in some neighborhoods of the points $z_0$ and $w_0$, respectively.
\end{corollary}
\begin{remark}\label{Re5} Suppose that $D$ is $\mathcal{C}^\infty$-smooth pseudoconvex of finite type near $z_0\in bD$. Then  $D$ has the $t^\epsilon$-property at $z_0$ for some $\epsilon>0$ (cf. \cite{Cho92, Khanh13}). Moreover, a computation shows that the strong $t^\epsilon$-property at $z_0$. In addition, $D$ satisfies the local condition R at $z_0$ (cf. \cite{BL80}).
\end{remark}

By Corollary \ref{Co2} and Remark \ref{Re5}, we obtain the following corollary which is proved by A. Sukhov.
\begin{corollary}[See Corollary 1.4 in \cite{Suk}]\label{Cor2}
Suppose that $D$ and $G$ are $\mathcal{C}^\infty$-smooth domains in $\mathbb  C^n$. Suppose that $D$ and $G$ are pseudoconex of finite type near $z_0\in bD$ and $w_0\in bG$ respectively. Let $f$ be a biholomorphic mapping of $D$ onto $G$ such that $w_0\in \mathcal{C}(f,z_0)$. Then $f$ and $f^{-1}$ extend smoothly to $bD$ in some neighborhoods of the points $z_0$ and $w_0$, respectively.
\end{corollary}

It is well-known that any accumulation orbit boundary point is pseudoconvex (cf. \cite{GK91}). The following lemma says that the pseudoconvexity is invariant under any biholomorphism.
\begin{lemma}\label{le3}
 Let $D,G$ be $\mathcal{C}^2$-smooth domains in $\mathbb C^n$ and let $z_0\in bD$ and $w_0\in bG$. Let $f:D\to G$ be a biholomorphism such that $w_0\in \mathcal{C}(f,z_0)$. If $D$ is pseudoconvex at $z_0$, then $G$ is also pseudoconvex at $w_0$.
\end{lemma}
\begin{proof}
Since $w_0\in \mathcal{C}(f,z_0)$, we may assume that there exists a sequence $\{z_j\}\subset D$ such that $z_j\to z_0$ and $f(z_j)\to w_0$ as $j\to \infty$. Assume the contrary, that $G$ is not pseudoconvex at $w_0$. Then there is a compact set $K\Subset G$ such that the holomorpphic hull $\hat{K}$ of $K$ contains $V\cap G$, where $V$ is a small neighborhood of $w_0$. (Recall that $\hat{K}:=\{z\in G\colon |g(z)|\leq \max_{K}|g|,~\forall~g:G\to \mathbb C~\text{holomorphic} \}$.)  Consequently, $f(z_j)\in\hat{K}$ for every $j\geq j_0$, where $j_0$ is big enough.

Denote by $L:=f^{-1}(K)$. Then $L$ is a compact subset in $D$. We shall prove that $z_j\in \hat{L}$ for every $j\geq j_0$ and hence the proof follows. Indeed,
let $g:D\to \mathbb C$ be any holomorphic function. Then since $f(z_j)\in \hat{K}$ for every $j\geq j_0$, we have
$$
|g\circ f^{-1}(f(z_j))|\leq \max_{K} |g\circ f^{-1}|,~\forall~j\geq j_0.
$$
This implies that
$$
|g(z_j)|\leq \max_{K} |g\circ f^{-1}|=\max_{f^{-1}(K)}|g|=\max_{L}|g|,~\forall~j\geq j_0.
$$
Therefore, $z_j\in \hat{L}$ for every $j\geq j_0$, and thus the proof is complete.
\end{proof}
\begin{lemma}\label{lem6} Let $\Omega_P$ be as in Theorem \ref{Th2} and let $f\in \mathrm{Aut}(\Omega_P)$ be arbitrary. Then there exist $t_1,t_2\in\mathbb R$ such that $f$ and $f^{-1}$ extend to be locally $\mathcal{C}^\infty$-smooth up to the boundaries near $(it_1,0)$ and $(it_2,0)$, respectively, and $f(it_1,0)=(it_2,0)$.
\end{lemma}
\begin{proof}
We shall follow the proof of \cite[Lemma $3.2$]{Ber94}. Let $\phi: \Omega_P \to \Delta $ be the function defined by
$$
\phi(z_1,z_2)=\frac{z_1+1}{z_1-1}.
$$
Then we see that $\phi$ is continuous on $\overline{\Omega_P}$ such that $|\phi(z)|<1$ for $z=(z_1,z_2)\in \Omega_P$ and tends to $1$ when $z_1 \to \infty$. Let $f: \Omega_P\to \Omega_P$ be an automorphism. We claim that there exists $t_1\in \mathbb R$ such that $\lim_{x\to 0^-} \inf |\pi_1 \circ f (x+it_1,0)|<+\infty$. Here, $\pi_1,\pi_2$ are the projections of $\mathbb C^2$ onto $\mathbb C_{z_1}$ and $\mathbb C_{z_2}$, respectively, i.e. $\pi_1(z)=z_1$ and $\pi_2(z)=z_2$. Indeed, if this would not be the case, the function $\phi\circ f$ would be equal to $1$ on the half plane $\{(z_1,z_2)\in \mathbb C^2\colon\mathrm{Re}~z_1<0, z_2=0\}$ and this is impossible since $|\phi(z)|<1$ for every $z\in \Omega_P$. Therefore, we may assume that there exists a sequence $x_k<0$ such that $\lim_{k\to \infty} x_k=0$ and $\lim_{k\to \infty } \pi_1\circ f (x_k+it_1,0)=w_1^0\in \overline{\mathcal{H}}$.

We shall prove that, after taking some subsequence if necessary, $\lim_{k\to \infty}\pi_2\circ f (x_k+it_1,0)=w_2^0$ for some $w_2^0\in \mathbb C$. Indeed, arguing by contradiction we assume that  $\pi_2\circ f (x_k+it_1,0)\to \infty$ as $k\to \infty$. Because of the convergence of $\{\pi_1\circ f (x_k+it_1,0)\}$, the sequence $\{P(\pi_2\circ f (x_k+it_1,0))\}$ is bounded, which is a contradition if $\lim_{z_2\to \infty}P(z_2)=+\infty$. Therefore, after taking some subsequence if necessary, we may assume that
$$
 \lim_{k\to \infty} P(\pi_2\circ f (x_k+it_1,0))=r\geq 0.
$$

Define $\psi(w_1,w_2)=(w_1,1/w_2)$. Then the map $\psi\circ f$ is well-defined near $(it_1,0)$ and
$$
 \lim_{k\to \infty } \psi\circ f (x_k+it_1,0)=(w_1^0,0).
$$
Moreover, the defining function for $\psi\circ f(\Omega_P\cap U)$ near $(w_1^0,0)$, where $U$ is a small neighborhood of $(it_1,0)$, is
$$
\mathrm{Re}~w_1+Q(w_2)<0,
$$
where
\[
Q(w_2)=
\begin{cases}
P(1/w_2) &~\text{if}~ w_2\ne 0\\
r& \text{if}~ w_2=0.
\end{cases}
\]

Notice that $\psi\circ f$ is a local biholomorphism on $\Omega_P\cap U$. Since $\Omega_P\cap U$ is pseudoconvex near $(0,0)$, $\psi\circ f(\Omega_P\cap U)$ is pseudoconvex near $(-r,0)$. Moreover, the domain
$$
\Omega_{Q}=\{(w_1,w_2)\in \mathbb C^2\colon \mathrm{Re}~w_1+Q(w_2)<0\}
$$
has the strong $\tilde u$-property at $(w_1^0,0)$. Therefore, it follows from Corollary \ref{Co2} that the local biholomorphisms $\psi\circ f$ and $(\psi\circ f)^{-1}$ can be extended to be $\mathcal{C}^\infty$-smooth up to the boundaries in neighborhoods of $(it_1,0)$ and $(w_1^0,0)$, respectively. However, $b\Omega_P$ and $b\Omega_Q$ are not isomorphic as CR maniflod germs at $(0,0)$ and $(-r,0)$ respectively. This is a contradiction.

Granted the fact that $\lim_{k\to \infty}f(x_k+it_1,0)=w^0:=(w_1^0,w_2^0)\in b\Omega_P$, it follows from Lemma \ref{le3} that $\Omega_P$ is pseudoconvex near $w^0$. Moreover, again Corollary \ref{Co2} ensures that $f$ and $f^{-1}$ extend to be locally $\mathcal{C}^\infty$-smooth up to the boundaries. Hence, $\tau_{w^0}(b\Omega_P)=\tau_{(it_1,0)}(b \Omega_P)=+\infty$, which means that $w^0=(it_2,0)$ for some $t_2\in \mathbb R$ by virtue of Remark \ref{remark}. The lemma is proved.
\end{proof}

\section{Automorphism group of $\Omega_P$}
In this section, we are going to prove Theorem \ref{Th2}. To do this, let $P$ be as in Theorem \ref{Th2}. Let $p(r)$ be a $\mathcal{C}^\infty$-smooth function on $(0,\epsilon_0)~(\epsilon_0>0)$ such that the function
\[
P(z)=
\begin{cases}
e^{p(|z|)}&~\text{if}~z\in \Delta^*_{\epsilon_0} \\
0     &~\text{if}~z=0.
\end{cases}
\]
\begin{remark} \label{remark1} Since $\nu_0(P)=+\infty$, $\lim_{r\to 0^+} p(r)=-\infty$. Moreover, we observe that $\limsup_{r\to 0^+} |r p'(r)|=+\infty$, for otherwise one gets $|p(r)|\lesssim |\log(r)|$ for every $0<r<\epsilon_0$, and thus $P$ does not vanish to infinite order at $0$. In addtion, it follows from \cite[Corollary $1$]{Ninh1} that the function $P(r)p'(r)$ also vanishes to infinite order at $r=0$.
\end{remark}

 In proving Theorem \ref{Th2}, we need the following lemmas.
 \begin{lemma}[See Lemma 2 in \cite{Ninh1}]\label{lemma2} Suppose that there are $0<\alpha\leq 1$ and $\beta>0$ such that
$$
 \lim_{z\to 0}\frac{P(\alpha z)}{P(z)}=\beta.
$$
Then $\alpha=\beta=1$.
\end{lemma}
\begin{lemma}[See Lemma 3 in \cite{Ninh1}]\label{lemma3} Let $\beta\in \mathcal{C}^\infty(\Delta_{\epsilon_0})$ with $\beta(0)=0$. Then
$$
P(z+z\beta(z))-P(z)=P(z)\Big(|z|p'(|z|)\big(\mathrm{Re}(\beta(z)+o(\beta(z))\big)\Big)+o((\beta(z))^2)
$$
for any $z\in  \Delta^*_{\epsilon_0}$ satisfying $z+z\beta(z)\in \Delta_{\epsilon_0}$.
\end{lemma}
In what follows, denote by $\mathcal{H}:=\{z_1\in \mathbb C\colon \mathrm{Re}(z_1)<0\}$ the left half-plane.
\begin{lemma}\label{lemma7}
If $f\in \mathrm{Aut}(\Omega_P\cap U)\cap \mathcal{C}^\infty(\overline{\Omega_P}\cap U)$ satisfying $f_1(z_1,z_2)= a_{01} z_1+\tilde a_0(z_1)$ and $f_2(z_1,z_2)=b_{10}z_2+z_2\tilde b_1(z_1)$, where $a_{01},b_{10}\in \mathbb C^*$ with $b_{10}>0$ and $\tilde a_0,\tilde b_1\in \mathrm{Hol}(\mathcal{H}\cap U_1)\cap \mathcal{C}^\infty(\overline{\mathcal{H}}\cap U_1)$ with $\nu_0(\tilde a_0)\geq 2$ and $\nu_0(\tilde b_1)\geq 1$, where $U$ and $U_1$ are neighborhoods of the origins in $\mathbb C^2$ and $\mathbb C_{z_1}$, respectively, then $a_{01}=b_{10}=1$.
\end{lemma}
\begin{proof}
Since $f(b \Omega_P\cap U)\subset b\Omega_P$, we have
\begin{equation}\label{qtt1}
\begin{split}
\mathrm{Re}\Big(a_{01}\big(it-P(z_2)\big)+\tilde a_0\big(it-P(z_2)\big)\Big)+P\Big(b_{10}z_2+z_2 \tilde b_{1}\big(it-P(z_2)\big)\Big)\equiv 0
\end{split}
\end{equation}
on $\Delta_{\epsilon_0}\times (-\delta_0,\delta_0)$ for some $\epsilon_0,\delta_0>0$. It follows from (\ref{qtt1}) with $z_2=0$ that
$$
\mathrm{Re}(a_{01} it)+o(t)= 0
$$
for every $t\in\mathbb R$ small enough. This yields that $\mathrm{Im}(a_{01})=0$.

On the other hand, letting $t=0$ in (\ref{qtt1}) one has
\begin{equation}\label{qtt2}
\begin{split}
P\Big(b_{10}z_2+z_2O(P(z_2)) \Big)-\mathrm{Re}(a_{01})P(z_2)+o(P(z_2))\equiv 0
\end{split}
\end{equation}
on $\Delta_{\epsilon_0}$. This implies that $\lim_{z_2\to 0} P\big(b_{10} z_2+z_2O(P(z_2))\big)/P(z_2)=\mathrm{Re}(a_{01})=a_{01}> 0$.

By assumption, we can write $P(z_2)=e^{p(|z_2|)}$ for all $z_2\in \Delta^*_{\epsilon_0}$ for some function $p\in\mathcal{C}^\infty(0,\epsilon_0) $ with $\lim_{t\to 0^+}p(t)=-\infty$ such that $P$ vanishes to infinite order at $z_2=0$. Therefore, by Lemma \ref{lemma3} and the fact that $P(z_2)p'(|z_2|)$ vanishes to infinite order at $z_2=0$ (cf. Remark \ref{remark1}), one gets that
 $$
  \lim_{z_2\to 0}\frac{P(b_{10} z_2)}{P(z_2)}=\lim_{z_2\to 0} \frac{P\big(b_{10} z_2+z_2O(P(z_2))\big)}{P(z_2)}=a_{01}> 0.
 $$
Hence, Lemma \ref{lemma2} ensures that $a_{01}=b_{10}=1$, which ends the proof.
\end{proof}

\begin{lemma}\label{lemma6}
If $f\in \mathrm{Aut}\big(\Omega_P\cap U\big)\cap \mathcal{C}^\infty(\overline{\Omega_P}\cap U)$ satisfying $f_1(z_1,z_2) =z_1+\tilde a_0(z_1)$ and $f_2(z_1,z_2)=z_2+z_2 \tilde b_1(z_1)$, where $\tilde a_0\in \mathrm{Hol}(U_1) $ and $\tilde b_1\in \mathrm{Hol}(\mathcal{H}\cap U_1)\cap \mathcal{C}^\infty(\overline{\mathcal{H}}\cap U_1)$ with $\nu_0(\tilde a_0)\geq 2$ and $\nu_0(\tilde b_1)\geq 1$, where $U$ and $U_1$ are neighborhoods of the origins in $\mathbb C^2$ and $\mathbb C_{z_1}$, respectively, then $f=id$.
\end{lemma}
\begin{proof}
Expand $\tilde a_0$ into the Taylor at $0$ we have
$$
\tilde a_0(z_1)=\sum_{k=2}^\infty a_{0k}z_1^k,
$$
where $a_{0k}\in \mathbb C$ for every $k\geq 2$.

Since $f$ preserves $b \Omega_P\cap U$, it follows that
\begin{equation}\label{qqt2}
\begin{split}
\mathrm{Re}\Big(\big(it-P(z_2)\big)+\sum_{k=2}^\infty a_{0k}\big(it-P(z_2)\big)^k \Big)+P\Big(z_2+\tilde b_1\big(it-P(z_2)\big)\Big)\equiv 0,
\end{split}
\end{equation}
or equivalently,
\begin{equation}\label{qqt3}
\begin{split}
P\Big(z_2+z_2\tilde b_1\big(it-P(z_2)\big)\Big)-P(z_2)+ \mathrm{Re}\Big(\sum_{k=2}^\infty a_{0k}\big(it-P(z_2)\big)^k \Big)\equiv 0
\end{split}
\end{equation}
on $\Delta_{\epsilon_0}\times (-\delta_0,\delta_0)$ for some $\epsilon_0,\delta_0>0$.

If $f_1(z_1,z_2)\equiv z_1$, then let $k_1=+\infty$. In the contrary case, let $k_1\geq 2$ be the smallest integer $k$ such that $a_{0k}\ne 0$. Similarly, if $\tilde b_1(z_1)$ vanishes to infinite order at $z_1=0$, then denote by $k_2=+\infty$. Otherwise, let $k_2\geq 1$ be the smallest integer $k$ such that $b_{1k}:=\frac{\partial^k}{\partial z_1^k}\tilde b_1(0)\ne 0$.

Notice that we may choose $t=\alpha P(z_2)$ in (\ref{qqt3})
(with $\alpha \in \mathbb R$ to be chosen later). Then one gets
\begin{equation}\label{qqt4}
\begin{split}
&P\Big(z_2+z_2 b_{1k_2}P^{k_2}(z_2)(\alpha i-1)^{k_2}+z_2o(P^{k_2}(z_2))\Big)-P(z_2)\\
&\quad +\mathrm{Re} \Big(a_{0k_1}P^{k_1}(z_2)(\alpha i-1)^{k_1}+o(P^{k_1}(z_2))\Big)\equiv 0
\end{split}
\end{equation}
on $\Delta_{\epsilon_0}\times (-\delta_0,\delta_0)$. Moreover, by Lemma \ref{lemma3} one obtains that
\begin{equation}\label{qqt4,5}
\begin{split}
& P^{k_2+1}(z_2)|z_2|p'(|z_2|)\Big(\mathrm{Re}\big(b_{1k_2}(\alpha i-1)^{k_2}+g_2(z_2)\big)\Big)\\
&\quad +P^{k_1}(z_2)\mathrm{Re} \Big(a_{0k_1}(\alpha i-1)^{k_1}+g_1(z_2)\Big)\equiv 0
\end{split}
\end{equation}
on $\Delta_{\epsilon_0}$, where $g_1,g_2\in \mathcal{C}^\infty(\Delta_{\epsilon_0})$ with $g_1(0)=g_2(0)=0$.

We remark that $\alpha$ can be chosen so that $\mathrm{Re}\big(b_{1k_2}(\alpha i-1)^{k_2}\ne 0$ and $\mathrm{Re} \big(a_{0k_1}(\alpha i-1)^{k_1}\big)\ne 0$. Furthermore, since $\limsup_{r\to 0^+} |r p'(r)|=+\infty$ (cf. Remark \ref{remark1}), (\ref{qqt4,5}) yields that $k_2+1>k_1$. However, by the fact that $P(z_2)p'(|z_2|)$ vanishes to infinite order at $z_2=0$ (see Remark \ref{remark1}) and by (\ref{qqt4,5}) one has $k_1>k_2$. Hence, we conclude that $k_1=k_2=+\infty$.

Since $k_1=k_2=+\infty$, it follows that $f_1(z_1,z_2)\equiv z_1$ and (\ref{qqt3}) is equivalent to
\begin{equation}\label{qqt5,5}
\begin{split}
P\Big(z_2+\tilde b_1\big(it-P(z_2)\big)\Big)\equiv P(z_2),
\end{split}
\end{equation}
on $\Delta_{\epsilon_0}\times (-\delta_0,\delta_0)$. Since the level sets of $P$ are circles, (\ref{qqt5,5}) implies that $\tilde b_1(z_1)\equiv 0$. Thus, the proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Th2}]
Let $f=(f_1,f_2)\in \mathrm{Aut}(\Omega_P)$. By Lemma \ref{lem6}, there exist $t_1,t_2\in \mathbb R$ such that $f$ and $f^{-1}$ extend smoothly to the boundaries near $(it_1,0)$ and $(it_2,0)$, respectively, and $f(it_1,0)=(it_2,0)$. Replacing $f$ be $T_{-t_2}\circ f \circ T_{t_1}$ we may assume that $f(0,0)=(0,0)$ and there are neighborhoods $U$ and $V$ of $(0,0)$ such that $f$ is a local CR diffeomorphism between $V\cap b \Omega_P$ and $V\cap b \Omega_P$.

For each $t\in \mathbb R$, let us define $F_t$ by setting $F_t:=f\circ R_{-t}\circ f^{-1} $. Then $\{F_t\}_{t\in \mathbb R}$ is a one-parameter subgroup of $\mathrm{Aut}(\Omega_P)\cap \mathcal{C}^\infty(\overline{\Omega_P}\cap U)$.

By Theorem \ref{Th1}, there exists a real number $\delta$ such that $F_t=R_{\delta t}$ for all $t\in \mathbb R$. This implies that
\begin{equation}\label{tte1}
f= R_{\delta t}\circ f\circ R_{t},~\forall t\in \mathbb R .
\end{equation}
We note that if $\delta=0$, then $f=f\circ R_t$ and thus $R_t=id$ for any $t\in \mathbb R$, which is a contradiction. Hence, we can assume that $\delta\ne 0$.

We shall prove that $\delta=-1$. Indeed, by (\ref{tte1}) we have
\begin{equation}\label{tte2}
\begin{split}
 f_2(z_1,z_2)\equiv e^{i\delta t} f_2\big(z_1, z_2 e^{it}\big)
 \end{split}
\end{equation}
on a neighborhood $U$ of $(0,0)\in \mathbb C^2$ and for all $t\in \mathbb R$.

Expand $f_2$ into Taylor series, one obtains that
$$
f_2(z_1,z_2)=\sum_{n=0}^\infty b_{n} (z_1)z_2^n,
$$
where $b_{n},~n=0,2,\ldots$, are in $Hol(\mathcal{H})\cap\mathcal{C}^\infty(\overline{\mathcal{H}})$ and $b_{0}(0)=f_2(0,0)=0$. Hence, Eq. (\ref{tte2}) is equivalent to
\begin{equation}\label{tte3}
\begin{split}
\sum_{n=0}^\infty b_{n}( z_1) z_2^n\equiv \sum_{n=0}^\infty b_{n}(z_1) z_2^n e^{i\delta t+int}
\end{split}
\end{equation}
on $U$ for all $t\in \mathbb R$. This implies immediately that $b_0(z_1)\equiv 0$. Since $f$ is biholomorphism, $b_1(z_1)\not \equiv 0$. Therefore, (\ref{tte3}) yields that $\delta=-1$ and $b_{n}=0$ for every $n\in \mathbb N\setminus \{1\} $. It means that $f_2(z_1,z_2)\equiv z_2b_1(z_1)$.

We conclude that $F_t=R_{-t}$ for all $t\in \mathbb R$. This implies that
\begin{equation}\label{e1}
f= R_{-t}\circ f\circ R_{t},~\forall t\in \mathbb R,
\end{equation}
which implies that
\begin{equation*}
f_1(z_1,z_2)\equiv f_1(z_1,z_2 e^{it})
\end{equation*}
on a neighborhood $U$ of $(0,0)$ in $\mathbb C^2$ for all $t\in \mathbb R$. This yields that $f_1(z_1,z_2)=a_0(z_1)$.

Since $f$ preserves the boundary $b\Omega_P\cap U$, we have

\begin{equation}\label{eq2014-1}
\mathrm{Re}\Big(a_0(is-P(z_2))\Big)+P\Big(z_2b_1(is-P(z_2))\Big)=0
\end{equation}
for all $(z_2,s)\in \Delta_{\epsilon_0}\times (-\delta_0,+\delta_0)$. Letting $z_2=0$ in (\ref{eq2014-1}) one gets
\begin{equation}\label{eq2014-2}
\mathrm{Re}(a_0(is))=0
\end{equation}
for all $s\in (-\delta_0,+\delta_0)$. Hence, by the Schwarz reflection principle $a_0$ extends to be holomorphic in a neighborhood of the origin $z_1=0$; we shall denote the extension by $a_0$ too and the Taylor expansion of $a_0$ at $z_1=0$ is given by
$$
a_0(z_1)=\sum_{m=1}^\infty a_{0m} z_1^m.
$$
Moreover, because $f\in \mathrm{Aut}(\Omega_P)$, it follows that $a_{01}\ne 0$. From (\ref{eq2014-2}), we have
$$
\mathrm{Im}(a_{01})=0.
$$

Next, we are going to show that $b_1(0)\ne 0$. Indeed, suppose otherwise that $\nu_0(b_1)\geq 1$. Then it follows from (\ref{eq2014-1}) with $s=0$ that
$$
\lim_{z_2\to 0}\frac{P(z_2b_1(P(z_2)))}{P(z_2)}=\mathrm{Re}(a_{01})=a_{01}>0,
$$
which is impossible since
$$
\lim_{z_2\to 0}\frac{P(z_2b_1(P(z_2)))}{P(z_2)}=\lim_{z_2\to 0}\frac{P(z_2b_1(P(z_2)))}{z_2b_1(P(z_2)}\lim_{z_2\to 0}\frac{z_2b_1(P(z_2)}{P(z_2)}=0.
$$
Hence, we conclude that
$$
f_2(z_1,z_2)=b_{10} z_2 + z_2 \tilde b_1(z_1),
$$
where $b_{10}\in \mathbb C^*$ and $\tilde b_1\in Hol(\mathcal{H})\cap\mathcal{C}^\infty(\overline{\mathcal{H}})$ with $\tilde b_1(0)=0$. In addition, replacing $f$ by $f\circ R_\theta$ for some $\theta\in \mathbb R$, we can assume that $b_{10}$ is a positive real number.

We now apply Lemma \ref{lemma7} to obtain that $a_{01}=b_{10}=1$. Furthermore, by  Lemma \ref{lemma6} we conclude that $f=id$. Hence, the proof is complete.
\end{proof}
\section{Appendix}
We recall the following lemma, which is a version of the Hopf lemma.
\begin{lemma}[See Lemma 2.3 in \cite{Suk}] \label{hopf}
Let $\Omega\subset \mathbb C^n$ be a bounded domain with $\mathcal{C}^2$ boundary. Let $K\Subset \Omega$ be a compact set nonempty interior, and choose $L>0$. Then there exists $C=C(K,L)>0$ such that for any negative plurisubharmonic function $u$ in $\Omega$ satisfying the condition $u(z)<-L$ on $K$, the following bound holds:
$$
|u(z)|\geq C \delta_\Omega(z)~\text{for}~z\in \Omega.
$$
\end{lemma}
The following lemma is a slight generalization of \cite[Lemma 2.4]{Suk}.
\begin{lemma}\label{lemma12} Let $D$ and $G$ be domains in $\mathbb C^n$ with $\mathcal{C}^2$-smooth boundaries, $z_0\in bD$, $w_0\in bG$, $F_1, F_2:[0,+\infty)\to [0,+\infty)$ are nonnegative functions with $F_1(0)=F_2(0)=0$ such that $F_1$ is increasing. Assume that there is a neighborhood $U$ of $z_0$ such that for each $w\in U\cap bD$, there is a plurisubharmonic function $\psi_w$ such that
\begin{itemize}
\item[(i)] $\lim\limits_{D\times bD\ni(z,w)\to (z_0,z_0)}\psi_w(z)=0$,
\item[(ii)] $\psi_w(z)\leq -F_1(|z-w|)$,
\item[(iii)] $\psi_{\pi(z)}(z)\geq -F_2(\delta_D(z))$
\end{itemize}
for $z\in U\cap D$. Let $f:D\to G$ be a proper map such that $w_0\in \mathcal{C}(f,z_0)$. Then there exist neighborhoods $\tilde U\subset U$ and $V$ of $z_0$ and $w_0$, respectively, such that  $\delta_G(f(z))\lesssim F_2(\delta_D(z))$ for any $z\in \tilde U\cap D$ such that $f(z)\in V\cap G$.
\end{lemma}
\begin{proof}
The proof proceeds along the same lines as in \cite[Section $2$]{Suk}, but for the reader's convenience, we shall give the detailed proof.

For $\epsilon>0$ we consider the open set
$$
D^\epsilon=\{z\in U\cap D\colon \psi_{z_0}>-\epsilon\}.
$$
By virtue of (ii) there exists $\epsilon_0>0$ such that for any $\epsilon\in (0,\epsilon_0]$ one has
$\bar D^\epsilon\subset \bar U$. Hence, the boundary $bD^\epsilon\subset  (\bar U\cap bD) \cup S^\epsilon$, where $S^\epsilon=\{z\in \bar U\cap D\colon \psi_{z_0}(z)=-\epsilon\}$.

We fix $\epsilon\in (0,\epsilon_0/2]$ and choose $\epsilon_1>0$ so that $D\cap (z_0+\epsilon_1\mathbb B)\subset D^\epsilon$, where $\mathbb B$ is the open unit ball in $\mathbb C^n$. Without loss of generality we may assume that the neighborhood $U$ is small enough such that $\delta_D(z)=|z-\pi(z)|$ for $z\in U\cap D$.  We fix a positive number $\delta$ with the properties $\epsilon_1/50<\delta<2\delta<\epsilon_1/10$ and consider the compact set $K=\bar D\cap (z_0+2\delta\bar{\mathbb B})\setminus (z_0+\delta\bar{\mathbb B})$. For $\epsilon_2<\epsilon_1/100$ we have by (ii) that
\begin{equation*}
\begin{split}
\max\{\psi_\zeta(z)\colon z\in K,\zeta \in bD\cap (z_0+\epsilon_2 \bar{\mathbb B})\}&\leq -F_1\Big(d\big(K,bD\cap (z_0+\epsilon_2 \bar{\mathbb B}) \big)\Big)\\
  &\leq -F_1(\delta-\epsilon_2).
\end{split}
\end{equation*}
On the other hand, by (i) one can choose $\epsilon_2$ such that
\begin{equation*}
\begin{split}
-F_1(\delta-\epsilon_2)<\gamma:=\min\{\psi_\zeta(z)\colon z\in D\cap (z_0+\epsilon_2 \bar{\mathbb B}),\zeta \in bD\cap (z_0+\epsilon_2 \bar{\mathbb B})\}.
\end{split}
\end{equation*}
We fix $\epsilon_2>0$. Let $\tau>0$ be such that
$$
-F_1(\delta-\epsilon_2)<-\tau<-\tau/2<\gamma<0.
$$

We consider a smooth nondecreasing convex function $\phi(t)$ with the properties $\phi(t)=-\tau$ for $t\leq -\tau$ and $\phi(t)=t$ for $t>-\tau/2$. We set $\rho_\zeta(z)=\tau^{-1}\phi\circ \psi_\zeta(z)$. Then $\rho_\zeta(z)\mid_K=-1$ for $\zeta \in bD\cap (z_0+\epsilon_2 \bar{\mathbb B}) $, and we can extend $\rho_\zeta(z)$ to $D$ by setting $\rho_\zeta(z)=-1$ for $z\in D\setminus (z_0+2\delta \bar{\mathbb B})$. We obtain a function $\rho_\zeta(z)$, which is a negative continuous plurisubharmonic function on $D$ satisfying $\rho_\zeta(z)=-1$ on $D\setminus (z_0+\delta \mathbb B)$ and $\rho_\zeta(z)=\tau^{-1}\psi_\zeta(z)$ on $D\cap (z_0+\epsilon_2\mathbb B)$ for $\zeta \in bD\cap (z_0+\epsilon_2\mathbb B)$.

There is $\epsilon_3\in (0,\epsilon_2/2)$ such that $\pi(z)\in bD\cap (z_0+\epsilon_2\mathbb B)$ for any $z\in D\cap (z_0+\epsilon_3\bar{\mathbb B})$. We also fix a point $p\in D\cap (z_0+\epsilon_3\bar{\mathbb B})$ and define the function
\begin{equation*}
\varphi_p(w)=
\begin{cases}
\sup\{\rho_{\pi(p)}(z)\colon z\in f^{-1}(w)\}&~\text{for}~ w\in f(D^\epsilon),\\
-1 &~\text{for}~ w\in G\setminus f(D^\epsilon).
\end{cases}
\end{equation*}

Since $f$ is proper, the function $\varphi_p(w)$ is a continuous negative plurisubharmonic function on $G$ ( see \cite[Lemma 2.2]{Suk}).

Let $V$ be a neighborhood of the point $w_0$ such that the surface $V\cap b G$ is smooth. We fix a compact set $K\Subset f(D^{\epsilon_2})\cap V$ with nonempty interior (this is possible since $w_0\in \mathcal{C}(f,z_0)$ and $f(D^{\epsilon_2})$ is an open set). Assume that $2\max_{w\in K} \varphi_p(w)\leq -L=-L(p)$. The by Lemma \ref{hopf}, we have $|\varphi_p(w)|\geq C(L) \delta_G(w)$ for $w\in G\cap V$, where $C=C(L)>0$ depends ony on $L=L(p)$. We now show that $L$ (hence also $C$) can be chosen independent of $p$.

We have
\begin{equation*}
\begin{split}
\max_{w\in K}\varphi_p(w)&=\max\{\rho_{\pi(p)}(z)\colon z\in f^{-1}(w)\cap D^{\epsilon_2}, w\in K\}\\
                      &=\max\{\rho_{\pi(p)}(z)\colon z\in f^{-1}(K)\cap D^{\epsilon_2}\}.
\end{split}
\end{equation*}

Since $f$ is proper, $\mathcal{C}(f,z)\subset b G$ for $z\in U\cap b D$, and thus the set $f^{-1}(K)$ has no limit points on $U\cap b D$. Therefore, the set $K'=\overline{f^{-1}(K)\cap D^{\epsilon_2}}$ is relatively compact in $U\cap D$. If $z\in K'$, then by (iii) we have
\begin{equation*}
\begin{split}
\max\{\psi_{\pi(p)}(z) \colon z\in K'\}&\leq -F_1\Big(\min\{|z-\zeta|\colon z\in  K',\zeta \in b D \cap (z_0+\epsilon_2\mathbb{B})\}\Big)\\
                      &=-F_1\Big(d(K',b D \cap (z_0+\epsilon_2\mathbb{B}))\Big).
\end{split}
\end{equation*}
Since the last quantity does not exceed some constant $-N<0$, we may set
$$
 2\max_{z\in K'} \rho_{\pi(p)}(z)\leq 2\tau^{-1}N:=L,
$$
and $L$ is independent of $p$.

If now $f(p)\in V\cap G$, we have by (iii)
\begin{equation*}
\begin{split}
\delta_G(f(p))&\leq C|\varphi_p(f(p))|\leq C |\rho_{\pi(p)}(p)|\\
                      &\leq CF_2(\delta_D(p)).
\end{split}
\end{equation*}
Since $C>0$ here does not depend on $p$, we have arrived at
$$
\delta_G(f(z))\lesssim F_2(\delta_D(z))
$$
for any $z\in D\cap (z_0+\epsilon_3\mathbb B)$ such that $f(z)\in G\cap V$. This ends the proof.
\end{proof}
\section*{Acknowlegement}
 This work was completed when the author was visiting the Vietnam Institute for Advanced Study in Mathematics (VIASM). He would like to thank the VIASM for financial support and hospitality.  It is a pleasure to thank Tran Vu Khanh and Dang Anh Tuan for stimulating discussions.

redpajama_set_name: RedPajamaArXiv


text:
Opening of new pubs and bars in Birmingham increased 4% in 2019
By Olivia Lawlor on	 January 22, 2020 Data, food and drink, News
25 new pubs and bars opened in Birmingham in 2019 according to new data published by the government.
Birmingham was 8th for the highest number of pubs and bars opening in 2019 according to data published by the Office for National Statistics shows that.
William Stenson, assistant manager at The Colmore in Birmingham, said: "Birmingham, in general, is one of the up and coming cities within the bar trade and beer scene. 2019 was a great year for opening venues and it seems to have exploded the Birmingham beer scene"
West Midlands saw the addition of 750 new pubs and bars being opened in 2019.
The Colmore opened December 2019 on Colmore Row in Birmingham and reached capacity on their opening night.
Originally set to open early 2019, legal issues delayed the opening until the end of the year:
"The building before The Colmore and Lloyds Bank was a Quaker meeting house that had a ban on selling alcohol that was still valid, so we needed to get that changed legally before we could do anything."
The bar is in collaboration with Pivovar, a specialist beer importer, and Thornbridge Brewery, both located north of England.
Mr Stenson said: "Thornbridge beers have become quite common now. You see them in supermarkets but you don't really see their beer in Birmingham in casks, kegs or bottles in venues. They have a good reputation in the North but that's it at the minute."
In the UK, Wiltshire had the highest number of new pubs and bars with 45 openings.
Manchester suffered a loss of 50 fewer establishments from 2018 to 2019.
Despite 25 new pubs and bars adding to competition in Birmingham, there is a good relationship between local pubs, bars and businesses:
"We've had an amazing reception from everyone around here, we didn't want to tread on people's toes. Other bars are supporting us, and we are supporting local businesses. We're building a good relationship."
William Stenson
The opening of 25 new pubs and bars in Birmingham has brought 500 jobs into the city.
The Colmore said they have a lot of staff to allow flexible hours for their staff as well as employing local university students.
The UK has seen an increase in employees in pubs and bars from 2018-2019 including occupations such as bar staff and chefs/cooks increases.
However, the number of people in the UK eating in restaurants, pubs, bars and cafés has decreased by 0.2% from 2018-19.
Mr Stenson said: "Some days there is an awful lot of food orders and then somedays we won't get much at all. I think it's mainly that people come in for drinks and then they notice that we have a food menu."
After being open for a month, he said: "This year is our building year so we can try and get out there, get known and next year we can really shine.
"We're improving our menu and improving staff training on wines to ensure people know that there is something in here for everyone."
Olivia Lawlor
Freelance journalist and MA Data Journalism student. Recent Media and Communications graduate with a particular interest in music, environment and international conflicts/affairs.

redpajama_set_name: RedPajamaCommonCrawl


text:
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Best Induction Loop Devices For TV Watching
Hearing Loss Education and Information
Sereonic Wireless TV Listening Speaker (formerly TV Soundbox) Review
June 7, 2021 April 3, 2015 by Frances Martin
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The Sereonic Wireless TV Listening Speaker is a great device to help people with hearing loss hear and understand dialogue on their television better. It is a lightweight, wireless, portable speaker that you put beside you when you are watching TV.
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The purpose of the Soundbox is to enhance the sound that is coming from the TV. It uses voice enhancing technology, so it is particularly useful for people with hearing loss who have difficulty hearing dialogue or voices. It also effectively reduces background noise that can make it difficult to listen to the TV.
You can customize the sound by adjusting the Hi-Med-Low tone control switch, as well as the volume control. You cannot change the base and treble, but this system isn't designed to be a surround sound system, and that would be more applicable to someone who is looking to create a surround sound type of experience.
Depending on how your TV system is set up, the Sereonic Wireless TV Listening Speaker should be easy to install. You connect the speaker transmitter to your TV, cable box, or music system by connecting it to the headphone jack or the audio-out port using the audio cord that comes with the system. If your TV has a digital optical-out port, then you will have to buy a separate TV connector.
You leave the receiver on the base to charge it, and then when you are ready to use it, you remove it from the bottom and place it near where you are sitting so the sound will be coming to you directly from the receiver. When you aren't using the receiver, you put it back on the base to charge.
To use the device you just turn the volume knob up and down to find your personal sound preference, so it is pretty easy to use especially for people who might also have vision problems or don't like dealing with complicated buttons and remotes
If you have set up a Sereonic Wireless TV Listening Speaker you can use it in a number of ways: you can have only TV speaker working, only sound box working or TV and sound box on at the same time. If you would prefer to use headphones, there is a 3.5mm headphone jack on the side of the speaker. The non-hearing impaired members of the family will be able to watch TV at their average level while the user of the Sereonic Wireless TV Listening Speaker can watch TV with their preferred method.
This TV speaker has been designed to be light (1.7 lb) and portable. You can quickly pick it up and move it around the house with you. It is an excellent design for people with hearing aids who don't want to wear headphones or just for anyone in general who doesn't want to wear headphones.
All of the input and controls have clear labels, and the main volume knob is very easy to use. It is not a complicated piece of technology that will take a long time to understand how to use it. You plug it in and turn the volume knob and you are set.
If you need to use more than one receiver for each base, that is no problem. Each base can be used with up to 50 additional receivers, or if you have multiple TVs and would like a different TV-SB for each TV so that people can watch different channels, this is also possible since the paired receiver will automatically tune to and receive sounds from its dedicated Base.
The TV SB has a battery life of approximately 8 hours. When the battery light indicates that it is running low, you just place the receiver back onto the base to charge. If you are worried that you will forget to charge the receiver or that the battery life isn't long enough for you can purchase an AC adapter and plug it into the receiver.
It will work as a power source and for recharging the battery. That way, you don't need to put the receiver on the base for recharging, and you don't have to rely on the charger system entirely. The battery type that is used is a lithium rechargeable battery, and it is built-in to the receiver, it is not replaceable by the user, and it will last for years. If the battery needs to replaced or serviced, please call the Manufacturers Customer Care Department for instructions.
The Sereonic Wireless TV Listening Speaker has a one-year manufacturer's warranty. If you want more information about it contact Serene Innovations directly at [email protected].
Serene Innovations have an excellent customer service reputation; you shouldn't have any trouble getting help with setup or warranty information by contacting the company directly either through email at [email protected] or by phone or fax at :
You can also contact the company by mail at:
Serene Innovations, Inc.
14731 Carmenita Road
Norwalk, Ca. 90650
Frances Martin
Frances Martin has been an editor at Hearing Hub since 2014, where she researches all of the best products for people with hearing loss. Frances has extensive experience testing and reviewing products with a focus on health.
Latest posts by Frances Martin (see all)
The Best Hearing Amplifiers of 2021 - October 23, 2020
Best TV Soundbars for Hearing Impaired - October 23, 2020
Hearing Devices to Help People in Nursing Homes and Hospitals - October 23, 2020
Categories Amplifier Reviews, Speakers Tags Watching TV, wireless speaker
TV Ears 11290 5.0 Wireless Speaker Review
Audio Fox Wireless TV Speakers Review
5 of the Top Telephone Ring and Voice Amplifiers
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BLJ Hearing Amplifier
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The Best Hearing Amplifiers of 2021
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This Sunday is the first day of winter, and what better way to celebrate and stock up the pantry than with a visit to the Bream Creek Farmers Market?
This Sunday, we are very excited to have our annual Preserve Swap – beginning at 10:30am with demonstrations on the best ways to preserve Tasmania's incredible seasonal produce. We'll have pickling, jam-making, preserving and dehydrating demonstrations, by Tom Westcott, of Inner Pickle, Kate North of Kellevie Kitchen and Eloise Emmett of Eloise Emmett Real Food. This is followed by the Preserve Swap at 11am – simply bring along a jar or bottle of homemade jam, chutney, pickle, relish or sauce and swap it for another!
Lean To Kitchen – new!


Supermassive Games brings its PlayStation Move puzzler to virtual reality.
When Sony announced the PlayStation VR, the repurposing of the six-year-old Move controllers raised more than a few eyebrows. While a clear commercial failure, the bright-orbed, ugly-ish stepsister of Nintendo's Wiimote was an intuitive, technologically superior take on the motion control craze of the late noughties.
With a permanent moratorium for the Move now on hold thanks to the PlayStation VR, notable PS3 titles that supported Sony's glowing wands are now ripe for a virtual reality makeover. Considering the platform's swift initial drift into obscurity however, the pickings are fairly slim.
Of the games that washed ashore among the sea of mini-game-filled bilge, fans of tricky block-building were quick to praise Supermassive Games' Tumble for its motion-controlled, Jenga-meets-Tetris puzzles. Now Tumble VR – a repackaged, enhanced Tumble experience – is one of the gems of the PSVR's launch line-up.
Tumble VR begins with a simple task: stack a bundle of disparate shapes together and make your (very likely leaning) tower of bits and bobs stand as tall as possible. As with most great puzzle games, this idea twists and turns as much as the shapes themselves as you try and piece them together with increasingly mixed results.
Eventually a multitude of factors influence the stability of your precarious creations with slanted bases, tremors, projectiles, wind machines and limbo bars all threatening to turn your efforts into rubble. On a basic level, each block, triangle or in some cases bizarre Kinder Egg-like object has its own physical properties. Metal pieces work well as the base for lighter plastic 'bricks', with a few frictional rubber rectangles in between.
Tumble VR supports both the Dualshock or Move controller, and while both have their own merits and drawbacks, the Move feels the most natural within the context of VR as it allows for one-to-one handling, rotation and angling of each block.
Stepping around a block tower in virtual reality feels like a natural fit for the original mechanics of Tumble on PS3. Players will need to do this to seek out gaps for that last pesky diamond shape that they have to meticulously place without knocking everything to the ground.
In its VR guise, Supermassive – of Until Dawn fame (and its new VR spin-off Rush Of Blood) – has added new challenges and a reworked Versus mode where a second player disrupts the first's handiwork with an assortment of lasers, fans and blasters.
The local multiplayer is a blast in short bursts, although playing through the main challenges with a puzzle-solver nearby can be equally entertaining – if you can tolerate some backseat puzzling and cries of "no, not there... there!".
Arguably the best puzzle variant completely subverts the game's initial goal: the demolition challenges. Armed with just three bombs, Tumble VR pits you against increasingly sturdy block amalgamations with points dished out based on how far each one flies away from its initial position.
While the bridge building and light reflecting challenges veer toward broken controller territory as harder 'Zones' unlock, time targets, hidden blocks and medal rankings offer other avenues of puzzling fun if you get completely stuck.
For all the freedom Tumble VR offers in its new virtual reality guise, there are drawbacks that stem from the tech itself. Stepping around the central pedestal to find an optimum viewpoint can send the PlayStation Camera's sensors into meltdown if you happen to drift out of the small play area, while losing track of the Move controllers as you contort your limbs is a fairly common occurrence.
On a more fundamental level, Tumble VR is crying out for a 'rewind' mechanic. In the latter zones, structures can reach absurd heights and one accidental knock or misjudged block placement can trigger a catastrophic, yes, tumble. A limited-use do-over of the last move would reduce the challenge a tad, but also alleviate frustration.
Tumble VR is easily one of the most comprehensive games among the PSVR's launch line-up, both in terms of longevity and mechanical depth. While not the kind of quintessential puzzle game that can stack up to this year's best (The Witness), Supermassive Games' virtual reality stack-a-thon stands tall on Sony's virtual reality machine.


Applications for federal funds require the submission of standard forms, which have been developed or approved by the Office of Management and Budget (OMB).
The following standard forms are the most common federal forms required by OMB's grants management circulars that will be encountered when preparing applications directly to federal agencies, although some federal agencies have specialized forms that have been approved by OMB. These standard forms can be downloaded in PDF format by clicking on the links below.
Clarifies and updates cost reporting guidelines for award recipients.
The Uniform Guidance and related documents can be found by clicking on the links below.
There are a variety of grant writing resources on the Internet that can provide guidance to the novice grant writer. Be mindful that the quality of advice varies greatly from one source to another. Use internet grant resources with some degree of caution. The links below are just a few of the many resources.


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Marlyne Nayokah Barrett (née Afflack) est une actrice et productrice américaine née le  à Brooklyn, New York.

Après avoir joué dans une quinzaine de séries télévisées, elle se fait connaître par son rôle de Maggie Lockwood, la chef des infirmières dans la série médicale Chicago Med. Il s'agit d'une série dérivée de Chicago Fire et de Chicago PD dans lesquelles l'actrice intervient aussi de manière récurrente.

Biographie
Marlyne Barrett est née le  à Brooklyn, New York.

D'origine haïtienne, ses deux parents sont scientifiques et elle a une sœur qui est médecin. Elle grandit à Montréal entourée de sa famille.

Vie privée
Elle est mariée depuis 2009 à Gavin Barrett. Elle parle couramment français.

Carrière

Animation, théâtre et débuts discrets
Elle fait ses débuts à la télévision à l'automne 1996 sur la chaîne MusiquePlus, à l'animation des émissions musicales Groove et Hip Hop, ainsi que l'émissions sportive S*P*A*M en 1997, et elle a réalisé de nombreuses interviews.

En 2001, elle intègre une troupe de comédiens, la Hudson Shakespeare Company, ils se produisent sur scène, dans des festivals et interprètent des grands classiques comme Hamlet.

Après une succession de petits rôles dans diverses productions cinématographiques (la plupart inédites sur le territoire français), elle décroche, en 2003, le premier rôle féminin de la comédie romantique indépendante, Love, Sex and Eating the Bones. Le film séduit les critiques et permet à Marlyne de se faire remarquer. La production est notamment récompensé lors du Festival du film de Toronto.

La même année, pour le téléfilm dramatique Au-delà des barrières, elle rejoint Whoopi Goldberg et Danny Glover. Le téléfilm est rendu populaire grâce au duo d'acteurs principaux et permet à Marlyne d'enchaîner, par la suite, les apparitions sur le petit écran.

En 2005, elle obtient un rôle mineur dans la comédie romantique Hitch, aux côtés de Will Smith et Eva Mendes. Le film est un énorme succès commercial, il engrange plus de 366 millions de dollars au box office pour un budget de 70 millions de dollars et la critique est également séduite.

Rôles réguliers à la télévision
Parallèlement à ses petites incursions au cinéma, elle développe sa carrière à la télévision : En effet, Marlyne apparaît notamment dans les séries : New York, police judiciaire, New York, cour de justice et New York, unité spéciale, sous la houlette du célèbre Dick Wolf, qu'elle retrouvera quelques années plus tard.

En 2007, elle rejoint Glenn Close, pour un arc de plusieurs épisodes, dans la série télévisée américaine, Damages. Elle sélectionne judicieusement ses interventions télévisuelles puisque celle-ci est également acclamée par les critiques et se retrouve nommée dans de prestigieuses cérémonies.

Entre 2006 et 2008, elle rejoint la série criminelle Sur écoute (), diffusée sur la chaîne HBO aux États-Unis. Bien qu'elle n'ait pas connu de succès commercial important, cette série est acclamée par les critiques, reconnue pour son réalisme proche du documentaire,.

En 2009, elle fait partie de la distribution de la série dramatique ambitieuse Kings. Diffusée sur la chaîne NBC, le premier épisode est doté d'un budget de 10 millions de dollars, ce qui représente un coût de production colossal pour le milieu du petit écran : le pitch est vaguement basé sur l'histoire biblique du Roi David transposée à notre époque. Le concept est rejeté par le public et faute d'audiences, la série est annulée à la fin de la première saison.

Par la suite, elle continue d'apparaître en tant qu'invitée pour les shows Bored to Death et The Good Wife et vise un public plus jeune pour Gossip Girl, en 2010.

Retour télévisuel remarqué
L'actrice marque une pause dans sa carrière et revient en 2015, dans American Crime, la création de John Ridley pour le réseau ABC, le temps des deux premiers épisodes.

Puis, elle est choisie pour incarner la chef des infirmières, Maggie Lockwwood, dans la nouvelle série médicale de la chaîne américaine NBC, Chicago Med. Il s'agit de la troisième série créée par Dick Wolf (aussi créateur de New York, unité spéciale et ses dérivés) dont l'intrigue se situe à Chicago et suit le quotidien des médecins et autres personnels de l'hôpital.

Elle est liée à deux autres séries du même créateur, Chicago Fire et Chicago Police Department qui s'intéressent respectivement aux pompiers et aux policiers de la ville de Chicago. La série est un succès critique et commercial avec une moyenne de 8 millions de téléspectateurs aux États-Unis, elle est reconduite pour une seconde saison.

L'actrice signe son retour au cinéma, en 2017, dans le drame indépendant Quest qui raconte l'histoire d'amour passionnelle entre un humble professeur et une toxicomane abusée sexuellement. Elle tient également le rôle-titre du court métrage Night Call, dont elle est la productrice, ou elle interprète un policier qui doit prendre une décision qui va radicalement changer sa vie à la suite d'une patrouille difficile. Dans le même temps, le show médical, Chicago Med est renouvelé grâce à la stabilité de ses audiences.

En 2018, elle rejoint, dans un rôle mineur, la comédie dramatique After Everything, portée par Marisa Tomei et Maika Monroe. Cette production est saluée par la presse.

Filmographie

Cinéma

Longs métrages

2001 : Hidden Agenda de Marc S. Grenier : L'hôtesse de l'air
2001 : Braquages (Heist) de David Mamet : Une femme
2003 : Love, Sex and Eating The Bones de Sudz Sutherland : Jasmine LeJeune
2005 : Hitch, expert en séduction (Hitch) d'Andy Tennant : Stéphanie
2006 : Off the Black de James Ponsoldt : Nancy
2017 : Quest de Santiago Rizzo : Sheila
2018 : After Everything d'Hannah Marks et Joey Power : Dr. Beatty

Courts métrages

1999 : Angelique de Michael Jarvis : Angelique
2017 : Night Call de Amanda Renee Knox : Kadedra Domek (également productrice)

Télévision

Séries télévisées
 2000 : The Wonderful World of Disney : Jody
 2004 : The Jury : Faith Ward
 2005 : New York, cour de justice (Law and Order : Trial by Jury) : Diane Harris
 2005 : Rescue Me : Les Héros du 11 septembre (Rescue Me) : La mère du garçon dans le coma
 2005 / 2006 : New York, unité spéciale (Law and Order : Special Victims Unit) : Sarah Miller / Alma Cordoza
 2005 / 2008 : New York, police judiciaire (Law and Order) : Bocanegra, l'avocate / Alana Sinclair
 2006 : Conviction :  Donner
 2006 - 2008 : Sur écoute (The Wire) : Nerese Campbell
 2007 : Damages : Felicia Marquand
 2009 : Kings : Thomasina
 2009 : Bored to Death : Ava
 2009 : The Good Wife : Thalia Ramsey
 2010 : Gossip Girl : Martha Chamberlai
 2014 : Satisfaction : Lauretta
 2015 : American Crime : Détective Chris Thompson
2015 - présent : Chicago Med : Maggie Lockwood
 2016 : Chicago PD : Maggie Lockwood
 2016 - 2019 : Chicago Fire : Maggie Lockwood

Téléfilm

2003 : Good Fences de Ernest R. Dickerson : Tina

Distinctions
Sauf indication contraire ou complémentaire, les informations mentionnées dans cette section proviennent de la base de données IMDb.

Récompenses
 Jim Thorpe Independent Film Festival 2018 : meilleure actrice dans un court métrage pour Night Call
 SENE Film, Music and Art Festival 2018 : meilleure actrice dans un court métrage pour Night Call
Twin Falls Sandwiches Film Festival 2018 : meilleure actrice pour Night Call

Nominations
 Rahway International Film Festival 2018 : meilleure actrice dans un court métrage pour Night Call
Westfield International Film Festival 2018 :
meilleure actrice dans un court métrage pour Night Call
meilleure court métrage dramatique pour Night Call
Blackbird Film Festival 2019 : meilleure actrice dans un court métrage pour Night Call

Notes et références

Liens externes

Naissance en septembre 1978
Naissance à Brooklyn
Actrice américaine
Productrice américaine
Actrice afro-américaine
Acteur de Chicago Med

redpajama_set_name: RedPajamaWikipedia


Leatherman Multi Tools are single piece with various tools hooked up along with safety locks in order that one could use a particular tool while different instruments will remain unfolded and locked. The precise locks present security against damage as a consequence of sharp blades of the instruments that are not in use. When stored open and free, the multi software folds itself. Some Leatherman tools uses Button locks too, whereas Leatherman Wave contains peculiar liner locks on its blades, which makes it easily accessible whereas different elements of the tool could stay unfolded.
All of those bags work nicely and are usually accompanied by a chain or metal rope strap. This could create a extra casual look if the links within the rope are large or a really elegant look if the strap is extra tightly woven in the steel and smaller. Any mixture of steel, crystals and sequins can work nicely as a whole also. You'll be able to?t go fallacious with these adornments on the arm of a lady wearing slightly black costume. As long as the bag glitters like gold, shimmers with silver and sequins or cascades with crystals you may be good to go!
The footwear is produced using materials of highest high quality that may withstand regular wear and tear. This makes these products very durable as they don't tear or fade simply. Those who purchase this footwear therefore get real worth for his or her cash as they will use it for many years. Cleaning and sustaining these shoes is straightforward as they're straightforward to wash they usually also dry shortly.
More and more people are forming the habit of buying stuff from the ecommerce site and the net outlets. There are various advantages that one should consider that has made online purchasing a profitable and comfy affair. Since that is an internet course of there are not nagging salesman who normally forces you to purchase stuff that you just had no desire to obtain.


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Missionaries of Charity welcomes transgender people in Bangladesh
By LiCAS.news
The Missionaries of Charity in Bangladesh welcomed 50 transgender people in its Home of Compassion to mark the 25th death anniversary of Saint Mother Teresa of Calcutta.
"We called on the Hjiras (transgender people) because they are a neglected group in society in Bangladesh," Sister Bondita, the Home's superior, told AsiaNews.
The missionary sister said Saint Mother Teresa "showed her love and compassion to all those who are neglected and unloved. It is the same sign of love that we also want to show."
The Missionaries of Charity, founded by Saint Mother Teresa in 1950, have over 4,500 nuns in more than 130 countries around the world.
There are an estimated 1.6 million to 4.8 million members of the LGBTQI community in Bangladesh, which has a total population of more than 160 million people.
"We have been overwhelmed by the love, respect, and support of the Sisters of Mother Teresa," said transgender Hor Ram.
Hor Ram said the LGBTQI community is "not well received in society, but the Sisters called on us; they were kind and showed love to us, winning our hearts."
The Catholic Church celebrated the Feast Day of Saint Mother Teresa of Calcutta on Sept. 5.
Mother Teresa died on Sept. 5, 1997, at the age of 87, and was beatified by Pope John Paul II just 6 years later. She was canonized by Pope Francis in 2016.
The United Nations has declared Sept. 5 of every year as the International Day of Charity in memory of Mother Teresa.


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Mayumi Morinaga
A singer songwriter from Saga Prefecture. Specializes in dance music. After appearing in Nippon Television's audition program "Utasuta!" in 2006, debuted in 2007 with the single "ONE MILLION MILES". In 2008, traveled around the globe with just a boombox and held many street performances. Released "NIGHT OF FIRE / Mayumi Morinaga feat. Britney Hamada" which she made in collaboration with comic artist Britney Hamada for the cover compilation album "Pagyaru! Trance". Provided a song for the arcade game "beatmania IIDX" under the name "moimoi", and has been working under other aliases such as MAKI and PRIM. In June 2012, released the 1st album "Glitter / Kamiuta" which featured the title track "Glitter", the ending theme song for the TV Tokyo's animation "FAIRY TAIL".
Official Website: http://mayumimorinaga.com/
Facebook: http://www.facebook.com/pages/ExpMayumi-Morinaga/176339552414315
Twitter: http://twitter.com/MAYUMIMORINAGA
Blog: http://ameblo.jp/mayumi-morinaga/


After six decades and over 200 albums, 85-year-old Country icon Willie Nelson is showing no signs of slowing down anytime soon with the announcement that he's hitting the road again for a nationwide tour that includes a stop here in Evansville!
The Willie Nelson & Family tour will be at the Old National Events Plaza on April 3rd, 2019, with tickets set to go on sale this Friday at 10:00 a.m. at the Old National Events Plaza box office and Ticketmaster.com.
Be sure to listen to Leslie Morgan while you're at work all this week for your chance to win tickets before they go on sale!


Running cross country is all about resolve and courage , yet above all of that ,it is character building. As someone once said , 'when you put yourself on the line in the race and expose yourself to the unknown , you learn things about yourself that are very exciting ' . This is why Cross Country is such an important part of our curriculum . We are glad to report that on April 11, 2018 our school conducted Inter House Cross Country competition for primary group i.e. grade 3-5 .It was a mass participation and all the scholars of all four Houses participated enthusiastically and showed a true spirit of sportsmanship.
The runners were in good condition with 4 to 5 injured cases. Winners were awarded with medals and certificate of merit. Fruit and energy drinks were given to all runners as a refreshment from management side.
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Roquan Smith Requests Trade: Bears 'Refused to Negotiate' Contract 'in Good Faith'
Mike Chiari@@mikechiariTwitter LogoFeatured Columnist IVAugust 9, 2022
Robin Alam/Icon Sportswire via Getty Images
Star linebacker Roquan Smith announced Tuesday that he has requested a trade from the Chicago Bears.
NFL Network's Ian Rapoport tweeted a statement written by Smith:
#Bears All-Pro LB Roquan Smith has requested a trade. pic.twitter.com/x4vmOMhROt
Smith called playing for the Bears a dream come true and wrote that he had wanted to be with the team for his entire career, but his perspective changed because of contract negotiations.
Smith said the Bears have "refused to negotiate in good faith" and that the organization's "focus has been on trying to take advantage of me."
The 25-year-old linebacker added that while he hasn't yet spoken to ownership about his situation and is wondering if they could "salvage" things, he doesn't "see a path back" to the Bears as of now.
Smith, who was the No. 8 pick in the 2018 NFL draft out of Georgia, is set to enter the final year of his rookie contract in 2022 after the Bears exercised the fifth-year option in his deal.
Although he hasn't received a Pro Bowl nod or first-team All-Pro selection during his NFL career, Smith has been a second-team All-Pro each of the past two seasons and has undoubtedly played at an elite level.
He has finished with 100 or more tackles in each of his four NFL campaigns, giving him 524 in 61 career games, as well as 14 sacks and five interceptions.
In 2020, Smith recorded 139 tackles, four sacks and a career-high two interceptions, and he followed that up last season with a career-high 163 tackles to go along with three sacks and one pick, which he returned for a touchdown.
Even at a young age, Smith has been the leader of the Chicago defense, and that was set to be even truer in 2022 given the significant personnel changes that have taken place.
The Bears traded star pass-rusher Khalil Mack to the Los Angeles Chargers, while defensive Akiem Hicks, Eddie Goldman, Alec Ogletree and Danny Trevathan are also gone from last year's team.
If the Bears decide to honor Smith's trade request, many teams are likely to line up for the chance to acquire one of the NFL's top linebackers.
ESPN's Mike Clay listed the Baltimore Ravens, Miami Dolphins, New England Patriots and Denver Broncos as teams that could make the most sense in a trade scenario.
Smith has played well enough to earn a contract in line with the best linebackers in the league. While it seemed likely that the Bears would be the ones to give it to him, they'll have to do damage control in order to convince him to stick around.


Please note: there is no limit to the number of concept notes that one organisation, consortium or family of organisations can submit during one funding round.
However, the maximum combined grant model allowance is 5 live UK Aid Direct grants of different types.
A live grant is a project with more than 9 months left to run from the start date of the UK Aid Direct call for proposals.
Information relating to submitting a full application for an Impact grant (the 2nd stage of the application process), can be found here.
What is expected of an organisation applying for this type of grant?
UK Aid Direct grants portfolio - Thinking of applying?
What types of grants are available through UK Aid Direct?