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%\fancyhead[CE]{M. A. Ardalani and  S. Haftbaradaran} 
\fancyhead[CO]{On the multiplication operators and multipliers on weighted spaces}



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{\noindent Journal of Mathematical Extension \\
Vol. XX, No. XX, (2020), pp-pp }\\
ISSN: 1735-8299\\
URL: http://www.ijmex.com\\
\vspace*{9mm}

\begin{center}

{\Large \bf 
On the multiplication operators and multipliers on weighted spaces of holomorphic functions on the upper halfplane\\}
%{\bf Do You Have a Subtitle? \\ If so, Write It Here} 


\let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX }

{\bf Mohammad Ali Ardalani $^*$ }\vspace*{-2mm}\\
\vspace{2mm} {\small  University of Kurdistan} \vspace{2mm}

{\bf  Saeed Haftbaradaran\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\
\vspace{2mm} {\small   University of Kurdistan} \vspace{2mm}

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{\footnotesize
\begin{quotation}
{\noindent \bf Abstract.} In this paper, we find necessary and sufficient conditions such that under these conditions a self map multiplication operator between weighted spaces of holomorphic functions is Fredholm or closed range operator. We also obtain a characterization of multipliers for certian type of weights operators between weighted spaces of holomorphic functions on the upper halfplane. Our results will remain valid for any simply connected domain in the complex plane instead of the upper halfplane .
\end{quotation}
\begin{quotation}
\noindent{\bf AMS Subject Classification:} 46E15; 47B38

\noindent{\bf Keywords and Phrases:} Weighted spaces, holomorphic functions, multiplication operators, upper half-plane.
\end{quotation}}

\section{Introduction}
\label{intro} % It is advised to give each section and subsection a unique label.
 Let $\mathbb{D}=\{z\in\mathbb{C}: \mid z\mid<1\}$ and $\mathbb{G}=\{\omega\in\mathbb{C}: Im\ \omega>0\}$ be the open unit disc and upper half plane respectively. Also, let $O$ be an open subset of $\mathbb{C}$. By a weight we mean a continuous function $\upsilon: O\longrightarrow (0,\infty)$. For a holomorphic function $f:O\longrightarrow\mathbb{C}$, we define the weighted sup-norm
\begin{equation*}
\|f\|_{\upsilon}=\sup_{z\in O}\mid f(z)\mid \nu(z)
\end{equation*}
and the weighted spaces
 \begin{equation*}
 H_{\upsilon}(O)=\{f:O\longrightarrow\mathbb{C}: f\ \text{is holomorphic}, \|f\|_{\upsilon}<\infty\}, 
 \end{equation*}
 \begin{equation*}
 H_{\upsilon_{0}}(O)=\{f\in H_{\upsilon}(O): \mid f(z)\mid\upsilon(z)\ \text{vanishes at infinity}\}.
\end{equation*}
Throughout this paper, we deal with the cases $O=\mathbb{D}$ or $O=\mathbb{G}$. In the case $ H_{\upsilon_{0}}(\mathbb{G}),\ \mid f(z)\mid\upsilon(z)$ vanishes at infinity if for any $\epsilon>0$ there is a compact subset $K$ of $\mathbb{G}$ such that $\mid f(z)\mid\upsilon(z)<\epsilon$ for all $z\in \mathbb{G}\setminus K$. In $H_{\upsilon_{0}}(\mathbb{D}), \mid f(z)\mid\upsilon(z)$ vanishes at infinity is equivalent to $\lim_{\mid z\mid\rightarrow1}\mid f(z)\mid\upsilon(z)=0$ (uniform limit). It is easy to see that $ H_{\upsilon}(O)$ and $ H_{\upsilon_{0}}(O)$ are Banach spaces.\\
A weight $\upsilon:\mathbb{G}\longrightarrow(0,\infty)$ is called a standard on $\mathbb{G}$ weight if $ \lim_{r\rightarrow 0}\upsilon(ir)=0$ and $\upsilon(\omega_{1})\leq\upsilon(\omega_{2})$ whenever $Im\ \omega_{1}\leq Im\ \omega_{2}$. We say a standard weight weight $\upsilon$ on $\mathbb{G}$ satisfies condition $(*)$ if 
\begin{equation*}
\sup_{n\in\mathbb{Z}}\frac{\upsilon(2^{k+1} i)}{\upsilon(2^{k} i)}<\infty.
\end{equation*}
A standard weight $\upsilon$ satisfies $(**)$ if
\begin{equation*}
\inf_{n\in\mathbb{N}}\sup_{k\in\mathbb{Z}}\frac{\upsilon(2^{k} i)}{\upsilon(2^{k+n} i)}<1.
\end{equation*}
Condition  $(*)$ is equivalent to $ \frac{\upsilon(ti)}{\upsilon(si)}\leq C(\frac{t}{s})^\beta$ whenever $s\leq t$ for some constants $C>0$ and $\beta>0$ and condition $(**)$ is equivalent to $\frac{\upsilon(ti)}{\upsilon(si)}\geq d(\frac{t}{s})^{\gamma}$ whenever $0<s\leq t$ for some constants $d, \gamma>0$. (see Lemma 1.6 of \cite{1}).\\
A weight $\upsilon:\mathbb{D}\longrightarrow(0,\infty)$ is called a standard weight on $\mathbb{D}$ if $\upsilon$ is radial (i.e $\upsilon(z)=\upsilon(\mid z\mid)$) and $ \lim_{\mid z\mid\rightarrow 1^{-}}\upsilon(z)=0$. We say a standard weight weight $\upsilon$ on $\mathbb{D}$ satisfies condition $(*)^{'}$ if 
\begin{equation*}
\inf_{n\in\mathbb{N}}\frac{\upsilon(1-2^{-n-1})}{\upsilon(1-2^{-n})}>0.
 \end{equation*}
A standard weight $\upsilon$ on $\mathbb{D}$ satisfies condition $(**)^{'}$  if 
\begin{equation*}
\inf_{k\in\mathbb{N}}\limsup_{n\rightarrow\infty}\frac{\upsilon(1-2^{-n-k})}{\upsilon(1-2^{-n})}<1.
\end{equation*}
\begin{remark}
Note that the space $ H_{\upsilon}(\mathbb{G})$ (and therefore $ H_{\upsilon_{0}}(\mathbb{G})$) will make sense whenever $H_{\upsilon}(G)\neq\{0\}$. There is a result of Stanev \cite{9} which states that $H_{\upsilon}(\mathbb{G})\neq\{0\}$ if and only if there exsist $a,b>0$ such that $\upsilon(it)\leq ae^{bt}\ t>0$. Therefore, throughout this paper we always assume our weights satisfies Stanev condition. For example weights which satisfy condition $(*)$, satisfy Stanev condition.  
\end{remark}
 In this paper, we intend to obtain some results concerning to the self-map multiplication operator $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ defiened by $M_{\varphi}(f)=f\varphi$ for each $f\in H_{\upsilon}(\mathbb{G})$  where $\varphi:\mathbb{G}\longrightarrow\mathbb{C}$ is a nonconstant holomorphic function. For this objective, we use wellknown results of \cite{4}, conformal map $\alpha:\mathbb{D}\longrightarrow G$ defined by $\alpha(z)=\frac{1+z}{1-z}i$, Some Lemmas and proper arguments in order to transfer the results to the case of upper halfplane.\\ 
\section{Preliminaries}

\label{sec:2}
In this section, we recall some definitions, notations and theorems which are necessary in the rest of this paper.
For more details, we will refer the reader to the suitable references. We denote the space of all bounded holomorphic functions on $\mathbb{G}$ by $H^{\infty}(\mathbb{G})$. A sequence $(\omega_{n})$ in $\mathbb{G}$ is called an interpolating sequence if for any bounded sequence $(\beta_{n})$ there is an $f\in H^{\infty}(\mathbb{G})$ for which $f(\omega_{n})=\beta_{n}$ for all $n\in\mathbb{N}$. Equivalently, $(\omega_{n})$  is an interpolating sequence if the bounded linear operator $T:H^{\infty}(\mathbb{G})\longrightarrow\ell^{\infty}$ defined by $(Tf)(z_{j})=f(z_{j})$ is an onto operator.\\

The maximal ideal space of $H^{\infty}(\mathbb{G})$ which is denoted by $M(H^{\infty}(\mathbb{G}))$ is the collection of all nonzero hemomorphisms of $ H^{\infty}(\mathbb{G})\longrightarrow\mathbb{C}$, whenever $H^{\infty}(\mathbb{G})$ is endowed with weak* topology as a subset of $H^{\infty}(\mathbb{G})^*$.\\ 
The pseudohyperboilc distance between two points $m$ and $n$ in $M(H^{\infty}(\mathbb{G}))$ is defined by 
\begin{equation*}
\rho(m,n)=\sup\{\mid \hat{f}(n)\mid: f\in H^{\infty}(\mathbb{G}), \hat{f}(m)=0, \|f\|_{\infty}\leq 1\}
\end{equation*}
where $ \|f\|_{\infty}=\sup\{\mid f(\omega)\mid :\omega\in \mathbb{G}\}$ and $\hat{f}$ is the Gelfand transform of $f$ . Also  For $\alpha,\omega\in\mathbb{G},\rho(\omega,\alpha)=\mid\varphi_{\alpha}(\omega)\mid$ where $\varphi_{\alpha}(\omega)=\frac{\alpha-\omega}{\alpha-\overline{\omega}}$.\\
 For the sake of simplicity, we denote the Gelfand transform of $f$ by $f$ itself futher on.
Let $F$ be a family of complex valued functions on a set $X$, a subset $\Gamma$ of $X$ is called a boundary for $F$ if  for each $f\in F$, there is an $x\in\Gamma$ such that $\mid f(x)\mid=\sup\{\mid f(y)\mid: y\in X\}$.
Shilov proved that if $A$ is a function algebra on a locally compact space $X$, then there exsists a unique minimal( intersection of all boundaries of $A$) closed boundary for $A$. This minimal boundary is called the\textbf{ Shilov boundary}.\\
The Gleason part of $ m\in M(H^{\infty}(\mathbb{G})) $ is defined by $ P(m)=\{n\in M(H^{\infty}(\mathbb{G})): \rho(m,n)<1 \}$. The set of trivial Gleason parts $ \{m\in M(H^{\infty}(\mathbb{G})): P(m)=\{m\}\} $ is a closed subset of $ M(H^{\infty}(\mathbb{G}))$
that contains properly the shilov boundary $ \Gamma(H^{\infty}(\mathbb{G})) $ of $(H^{\infty}(\mathbb{G}) $.See \cite{6} \\
A function $F$ of the following form 
\begin{equation*}
F(\omega)=e^{i\gamma}\exp(\frac{1}{\pi i}\int_{-\infty}^{\infty}\frac{(1+t\omega)\log g(t)}{(t-\omega)(1+t^2)}dt)
\end{equation*}
is an\textbf{ outer function} for the upper halfplane $\mathbb{G}$. Here $\gamma$ is a real number, $g(t)\geq0$ is a measurable function and
\begin{equation*}
\int_{-\infty}^{\infty}\frac{\log g(t)}{1+t^2}dt>-\infty,\ \ \ \     \int_{-\infty}^{\infty}\frac{[g(t)]^p}{1+t^2}dt<\infty.
\end{equation*}
A \textbf{compactification} of a space $X$ is an ordered pair $(K,h)$ where $K$ is a compact Hausdorff space and $h$ is an embedding of $X$ as a dense subset of $K$.\\ 
For a weight $\upsilon$ the function
\begin{align*}
\tilde{\upsilon}(z)&=\frac{1}{\sup\{\mid h(z)\mid: h\in H_{\upsilon}(O),  \|h\|_{\upsilon}<\infty\}}\\
&=\inf\{\frac{1}{\mid h(z)\mid}: h\in H_{\upsilon}(O),  \|h\|_{\upsilon}<\infty\}
\end{align*}
is called the \textbf{associated weight}. It is wellknown that (see \cite{2}):\\ 
- $\|f\|_{\upsilon}=\|f\|_{\tilde{\upsilon}}$ for each $f\in  H_{\upsilon}(O)$.\\
- For any $z\in O$, there is an $ h\in H_{\upsilon}(O)$ with $\|h\|_{\upsilon}\leq 1$ such that $\tilde{\upsilon}(z)=\frac{1}{\mid h(z)\mid}$.\\
- $\tilde{\upsilon}(z)\leq\upsilon(z)$ for all $z\in O$.\\
  A weight $\upsilon$ is called  an \textbf{essential weight} if there exists a constant $C>0$ such that $\tilde{\upsilon}(z)\leq C\upsilon(z)$ for all $z\in O$.\\
 Let $\upsilon$ be a standard weight on $\mathbb{G}$. In \cite{1}, it has been shown that $\tilde{\upsilon}(\omega)=\tilde{\upsilon}(i Im\ \omega)$ and $\tilde{\upsilon}(it)\geq\tilde{\upsilon}(is)$ whenever $t>s>0$. \\
 We conclude this section by recalling following definitions.\\ 
Bounded linear operator $T:X\longrightarrow Y(X$ and $Y$ are normed spaces) is called Fredholm operator if it has the closed range and dim $\ker(T)$ and $\dim\frac{Y}{Im\ T}<\infty$. Spectrum of bounded linear operator $T:X\longrightarrow X$ which is denoted by $\sigma(T)$ is defined by as follow.
\begin{equation*}
\sigma(T)=\{\lambda\in\mathbb{C}: \lambda I-T\ \text{is not invertible}\}.
\end{equation*} 
The essential spectrum of $T$ which is denoted by $\sigma_{e}(T)$ is defined by
\begin{equation*}
\sigma_{e}(T)=\{\lambda\in\mathbb{C}: \lambda I-T\ \text{is not a Fredhom operator}\}.
\end{equation*} 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%You can also see Sect.~\ref{intro} for more details about JME. about JME format see Sub Sect.~\ref{subsec:1}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%----------------------------------------------------------------------------------------------------------------------
\section{Main results}
In this section we recall theorems on the  boundedness and invertiblity of multiplication operators $ M_{\varphi} $ between  weighted spaces of holomophic functions on the upper halfplane. Then we characterize closed range and Fredholm self  map multiplication operators between weighted spaces of holomophic functions on the upper halfplane. 
Before that we prove some lemmas which have a major role in the proof of main results of this paper. Firstly, we begin by recalling the following elementary fact.
%----------------------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{1}
Let $(a_{n})$ be a sequence of real positive numbers such that $\lim_{n\rightarrow\infty}a_{n}=0$. then $(a_{n})$ has a subsequence $(a_{n_{k}})$ such that $\sum a_{n_{k}}$ is convergent.
\end{lemma}
%------------------------------------------------------------------------------------------------------------------------------------------ 
%-----------------------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{12}
Every sequence in $\mathbb{G}$ has a subsequence which is an interpolating sequence in $\mathbb{G}$. 
\end{lemma}
\begin{proof}
By Theorem 1.1 of chapter VII. of \cite{6} a sequence $(\omega_{j})$ in $\mathbb{G}$ is an interpolating sequence if and only if there is a $\delta>0$ such that for each $k=1,2, ...$
\begin{equation*}
\prod_{j=1_{j\neq k}}^\infty\mid\frac{\omega_{k}-\omega_{j}}{\omega_{k}-\overline{\omega}_{j}}\mid\geq\delta.
\end{equation*}
Now if the sequence $(\omega_{j})$ has iterative elements we eliminate them. Call the new sequence $({\omega_{j}})$ again. If this sequence is an interpolating sequence we are done. Otherwise, we claim this sequence has a subsequence which is an interpolating sequence. By Theorem 5.15 of \cite{7}
\begin{equation}\label{2}
\prod_{j=1_{j\neq k}}^\infty\mid\frac{\omega_{k}-\omega_{j}}{\omega_{k}-\overline{\omega}_{j}}\mid>0\ \Leftrightarrow  \sum 1-\mid\frac{\omega_{k}-\omega_{j}}{\omega_{k}-\overline{\omega}_{j}}\mid<\infty.
\end{equation}
Since $(\omega_{j})$ is a sequence in $\mathbb{G}$, so $\mid\frac{\omega_{k}-\omega_{j}}{\omega_{k}-\overline{\omega}_{j}}\mid<1$. Also, it easy to see that $\lim_{j\rightarrow\infty}1-\mid\frac{\omega_{k}-\omega_{j}}{\omega_{k}-\overline{\omega}_{j}}\mid=0$. Hence $\sum 1-\mid\frac{\omega_{k}-\omega_{j}}{\omega_{k}-\overline{\omega}_{j}}\mid$ satisfies the assumptions of Lemma \ref{1}. This means there is a subseuence of  $(\omega_{j})$ which satifies the left hand side of \ref{2} and therefore this subsequence is an interpolating sequence. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  
\end{proof}
%----------------------------------------------------------------------------------------------------------------------------------
%--------------------------------------------------------------------------------------------------------------------
\begin{lemma}\label{3}
i) $\mathbb{\overline{D}}$ is a compactification of $\mathbb{G}$.\\ 
ii) $ H^{\infty}(\mathbb{G}) $ is isometric isomorphic to $ H^{\infty}(\mathbb{D}) $.
\end{lemma}
\begin{proof}
i) Evidently, $\alpha^{-1}:\mathbb{G}\longrightarrow\mathbb{D}$ defined by $\alpha^{-1}(\omega)=\frac{\omega-i}{\omega+i}=z$ is a homeomorphism from $\mathbb{G}$ onto $\mathbb{D}$. So $\overline{\alpha^{-1}(\mathbb{G})}=\overline{D}$ which is a compact set. Hence, the pair $(\alpha^{-1},\overline{D})$ is a compactification of $\mathbb{G}$.\\
ii) It is enough to define $T:H^{\infty}(\mathbb{D})\longrightarrow H^{\infty}(\mathbb{G})$ by $T(f)=f\circ\alpha^{-1}$. Since $\alpha^{-1}$ is an onto map so
\begin{equation*}
\|T(f)\|_{\infty}=\sup\{\mid(f\circ\alpha^{-1}(\omega)\mid:\ \omega\in G\}=\sup\{\mid f(z)\mid: z\in\mathbb{D}\}=\|f\|_{\infty}.
\end{equation*}
\end{proof}\\
%-----------------------------------------------------------------------------------------------------------------------------------
Lemma \ref{3} implies immidately following two corolllaries. 
%--------------------------------------------------------------------------------------------------------
\begin{corollary}{\label{6}}
$\overline{\mathbb{G}}=\mathbb{G}\cup\partial^\infty \mathbb{G}$ is homeomorphic with $\overline{\mathbb{D}}$.
\end{corollary}
%---------------------------------------------------------------------------------------------------------
%------------------------------------------------------------------------------
\begin{corollary}\label{7}
$M(H^{\infty}(\mathbb{G}))$ is isomorphic to $M(H^{\infty}(\mathbb{D})).\ L^{\infty}( \partial \mathbb{D}) $ is isomorphic to $ L^{\infty}(\partial^{\infty}\mathbb{G}) $. This is true for Shilov boundary, trivial(nontrivial) Gleason parts 
of $ H^{\infty}(\mathbb{G}) $ and $ H^{\infty}(\mathbb{D}) $
\end{corollary}
%---------------------------------------------------------------------------------------------
Also, the disc algebra $A(\mathbb{D})$ is isomorphic to the $A(\mathbb{G})$, the algebra of all holomorphic functions on $\mathbb{G}$ and continuous on $\overline{\mathbb{G}}$.
\begin{lemma}\label{9}
Shilov boundary of $H^{\infty}(\mathbb{G})$ is maximal ideal space of $L^\infty(\partial^\infty \mathbb{G})$.
\end{lemma}
\begin{proof}
We know that $\mathbb{G}$ is homeomorphic to $\mathbb{D}$. The map  $\alpha:\mathbb{D}\longrightarrow \mathbb{G}$ maps $\partial^{\infty}\mathbb{D}$ onto $\partial^\infty \mathbb{G}$ homemorphically. Also, $H^{\infty}(\mathbb{G})$ and $H^{\infty}(\mathbb{D})$ are isomorphic spaces and for a function $f\in H^{\infty}(\mathbb{G}),\ \mid f\circ\alpha(z)\mid$ attins its supremum through the points $z\in\mathbb{D}$, tending to some point $z_{0}\in\partial\mathbb{D}$ if and only if $\mid f(\omega)\mid$ attains its supremum thourgh the points $\omega=\alpha(z)\in G$ tending to some points $\omega_{0}=\alpha(z_{0})\in \partial^\infty G$ and vice versa. So the shilov boundaries of $H^{\infty}(G)$ and $ H^{\infty}(\mathbb{D})$ are homemorphic. By Corollary \ref{7} $L^\infty(\partial\mathbb{D})$ and $L^\infty(\partial^\infty \mathbb{G})$ are isomorphic. From these facts and the result which states that the Shilov boundary of $H^\infty(\mathbb{D})$ is the maximal ideal space of $L^\infty(\partial\mathbb{D})$ (see V.1.7 of \cite{6} and page 169 of \cite{7}) lemma is proved.
\end{proof}\\
\begin{lemma}\label{10}
Let $X$ be the maximal ideal space of $L^\infty(\partial^\infty \mathbb{G})$. As $E$ varies over the measurable subsets of the $\partial^\infty \mathbb{G}$ the open closed sets $\{\phi\in X: \hat{\chi}_{E}(\phi)=0\}$ give a basis for the topology of $X$, where $\hat{\chi}$ is the Gelfand transform of characteristic function. In particular $X$ is totally disconnected.
\end{lemma}
\begin{proof}
By a similar lemma on $L^\infty(\partial\mathbb{D})$ (see page 169 of \cite{7}) and arguments similar to what has been done in Lemma \ref{9} we have done.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\end{proof}
\begin{lemma}
A function $\varphi\in H^{\infty}(\mathbb{G})$ does not vanish at any point of the Shilov boundary of $H^{\infty}(\mathbb{G})$ if and only if $\mid\varphi\mid$ is essentially bounded away from zero on $\partial^\infty \mathbb{G}$. 
\end{lemma}
\begin{proof}
$\mid\varphi\mid$ is essentially bounded away from zero means that for some $\epsilon>0,\ \mid\varphi(\omega)\mid\geq\epsilon$ a.e on $\partial^\infty \mathbb{G}$. Now if $\varphi$ is zero on any point of the shilov boundary of $H^{\infty}(\mathbb{G})$ by Lemma \ref{10} it vanishes on a totally disconnected open-closed set of positive Lebesgue measure.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\end{proof}
\begin{remark}
Indeed the maximal ideal space of $L^\infty(\partial^\infty \mathbb{G}),\ X$ is a closed subset of the maximal ideal space of $H^\infty(\mathbb{G}),\ M(H^\infty(\mathbb{G}))$ and because of the injection $H^\infty(\mathbb{G})\hookrightarrow L^\infty(\partial^\infty \mathbb{G}),\ X$ is a Shilov boundary of $H^\infty(\mathbb{G})$. For more details see page 184 of \cite{6} .
\end{remark}\\
Now, we continue by  recalling  some results from \cite{10}. For the sake of completeness, here, we state a modified proof for Theorem \ref{4}.  
%---------------------------------------------------------------------------------------------------------------------------
\begin{theorem}{\label 4}
Let $\upsilon$ be a weight on $\mathbb{G}$. The following statements are equivalent.\\
(a) $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ is bounded.\\
(b) $\varphi\in H^\infty(\mathbb{G})$.\\
If $M_{\varphi}$ is bounded then $\|M_{\varphi}\|=\|\varphi\|_{\infty}$. Besides, theorem is also true for $H_{\upsilon_{0}}(\mathbb{G})$ instead of $H_{\upsilon}(\mathbb{G})$. 
\end{theorem}
\begin{proof}
(b)$\Rightarrow(a)$: It is obvious, since for arbitrary $f\in H_{\upsilon}(\mathbb{G})$ we have 
\begin{align*}
\|M_{\varphi}(f)\|_{\upsilon}&=\sup\{\mid f(\omega)\mid\mid\varphi(\omega)\mid\upsilon(\omega): \omega\in \mathbb{G}\}\\
&\leq\sup\{\mid f(\omega)\mid\upsilon(\omega): \omega\in \mathbb{G}\}\sup\{\mid\varphi(\omega)\mid: \omega\in \mathbb{G}\}\\
&\leq\|f\|_{\upsilon}\|\varphi\|_{\infty}.
\end{align*}
Hence, $\|M_{\varphi}\|\leq \|\varphi\|_{\infty}$.\\ 
(a)$\Rightarrow(b)$: $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ is bounded. So the adjoint map $M_{\varphi}^*:H_{\upsilon}(\mathbb{G})^*\longrightarrow H_{\upsilon}(\mathbb{G})^*$ is bounded too and $\|M_{\varphi}\|_{\upsilon}=\|M_{\varphi}^*\|_{\upsilon}$. Now let $\omega\in \mathbb{G}$ be given. Evidently, the evaluational function $\delta_{\omega}:H_{\upsilon}(\mathbb{G})\longrightarrow \mathbb{C}$ defined by $\delta_{\omega}(f)=f(\omega)$ belongs to the dual space $H_{\upsilon}(\mathbb{G})^*$. Note that 
\begin{equation*}
\mid\varphi(\omega)\mid=\frac{\|M_{\varphi}^*(\delta_{\omega})\|}{\|\delta_{\omega}\|}\leq\|M_{\varphi}^*\|=\|M_{\varphi}\|<\infty.
\end{equation*}
Therefore, 
\begin{equation*}
\sup\{\mid\varphi(\omega)\mid: \omega\in G\}=\|\varphi\|_{\infty}\leq\|M_{\varphi}\|<\infty.
\end{equation*}
Where by $\|M_{\varphi}\|$ and $\|M_{\varphi}^*\|$ we mean the operator norm of linear operators $M_{\varphi}$ and $M_{\varphi}^*$.
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{proof}
%-------------------------------------------------------------------------------------------------------------------
%------------------------------------------------------------------------------------------------------------------------------
\begin{remark}
From now on we always assume $\varphi\in H^{\infty}(G)$. Because this assumption is equivalent to that $M_{\varphi}$ is welldefined.
Also note that the proof of (a)$\Rightarrow(b)$ in Theorem \ref{4} can not be done by classical arguments on the supremums as the following example shows us. 
\end{remark}
%----------------------------------------------------------------------------------------------
%-----------------------------------------------------------------------------------------------------------------
\begin{example}
Let $\upsilon(\omega)=e^{-Im\ \omega},\ f(\omega)=e^{i\omega}$ and $\varphi(\omega)=e^{-i\omega}$ be given. Then
\begin{equation*}
\|f\|_{\upsilon}=\sup\{\mid e^{i\omega}\mid e^{-Im\ \omega}: \omega\in \mathbb{G}\}=\sup\{e^{-2Im\ \omega}: \omega\in \mathbb{G}\}\leq1
\end{equation*}
and
\begin{equation*}
\sup\{\mid \varphi(\omega)\mid\ \mid f(\omega)\mid \varphi(\omega)}: \omega\in \mathbb{G}\}=\sup\{e^{-Im\ \omega}: \omega\in \mathbb{G}\}\leq1
\end{equation*}
but $\varphi\notin H^{\infty}(\mathbb{G})$. Since
\begin{equation*}
 \sup\{\mid e^{-i\omega}\mid : \omega\in \mathbb{G}\}=\sup\{e^{Im\ \omega}: \omega\in \mathbb{G}\}=\infty.
\end{equation*}
\end{example}
\begin{theorem}\label{5}
Let $\upsilon$ be a weight on $\mathbb{G}$. The following statements are equivalent.\\
(a) $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ is invertible.\\
(b) $\frac{1}{\varphi}\in H^{\infty}(\mathbb{G})$ (or equivalently there exsists a $K>0$ such that $\forall\ \omega\in \mathbb{G} \mid\varphi(\omega)\mid\geq K$).
\end{theorem}
\begin{proof} See \cite{10}.
\end{proof}
\begin{remark}
Since $\varphi\in  H^{\infty}(\mathbb{G})$ then $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ is a bounded linear operator. So $\sigma(M_{\varphi})$ is defined. 
\begin{theorem}
\sigma(M_{\varphi})=\overline{\varphi(\mathbb{G})}=\varphi(M( H^{\infty}(\mathbb{G}))).
\end{theorem}\\
\begin{proof}
Proof of the first equality can be found in \cite{10}. For proving the last equivality note that  $\overline{\varphi(\mathbb{D})}=\varphi(M( H^{\infty}(\mathbb{D})))$ (see pages 159-162 of \cite{7}). Now using Corollary \ref{6} and Corollary \ref{7} we conculude $\overline{\varphi(\mathbb{G})}=\varphi(M( H^{\infty}(\mathbb{G})))$.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  
\end{proof}\\
\begin{corollary}
$M_{\varphi}$ is not a compact opertator.
\end{corollary}
\begin{proof}
See \cite{10}.
\end{proof}\\
In the next Theorem\label{8} we state a necessry and sufficient condition such that $M_{\varphi}$ be a Fredholm operator. Since proof is very similar to the proof in \cite{4}, we do not repeat it again. we only explain necessary changes in order to transfer the proof to the upper half-plane case. Note that condition on $\varphi$ in theorem \ref{8} is completely different from the case of unit disc since compact subsets of $\mathbb{G}$ have wide variety in comparison with compact subsets of $\mathbb{D}$.
\begin{theorem}\label{8}
Let $\upsilon$ be a weight on $\mathbb{G}$ and $\varphi\in  H^{\infty}(\mathbb{G})$. The operator $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ is Fredhom if and only if there exsist $\epsilon>0$ and a compact subset $K\subset \mathbb{G}$ such that $\mid\varphi(\omega)\mid>\epsilon$ for all $\omega\in \mathbb{G}\setminus K$. Consequently $ \sigma_{e}(M_{\varphi})=\varphi(M(H^{\infty}(\mathbb{G})))\setminus \mathbb{G}$. The same holds for $H_{\upsilon_{0}}(\mathbb{G})$ instead of $H_{\upsilon}(\mathbb{G})$ whenever $H_{\upsilon_{0}}(\mathbb{G})\neq\{0\}$. 
\end{theorem}
\begin{proof}
Firstly, suppose that $M_{\varphi}$ is Fredholm but there is a sequence $(\omega_{n})\subset \mathbb{G}$ such that $\mid\varphi(\omega_{n})\mid\rightarrow0$ whenever $\omega_{n}\rightarrow\partial^{\infty}\mathbb{G}$. By Lemma \ref{12} we can assume that  $(\omega_{n})$ is an interpolating sequence in  $H^{\infty}(\mathbb{G})$. Rest of the proof is quite similar. Last assertion of the theorem follows from Corollary \ref{7} and equivality $ \sigma_{e}(M_{\varphi})=\varphi(M(H^{\infty}(\mathbb{D}))\setminus\mathbb{D})$ in \cite{3}.\ \ \ \ \ \ \  
\end{proof}
With a similar proof as in \cite{4} we have:
\begin{theorem}
Let $\varphi\in H^{\infty}(\mathbb{G})$. If $\upsilon$ and $w$ are two weights on $\mathbb{G}$ and $u:=\frac{\upsilon}{w}$ is equivalent to an essential weight. Then every closed range map $M_{\varphi}:H_{\upsilon}(\mathbb{G})\longrightarrow H_{\upsilon}(\mathbb{G})$ has also closed range as map $M_{\varphi}:H_{w}(\mathbb{G})\longrightarrow H_{w}(\mathbb{G})$. An analogus result holds for $H_{\upsilon_{0}}(\mathbb{G})$ and $H_{w_{0}}(\mathbb{G})\neq\{0\}$. 
\end{theorem}
\begin{theorem}
The map $M_{\varphi}:H^\infty(\mathbb{G})\longrightarrow H^\infty(\mathbb{G})$ has a closed range if and only if $\varphi$ does not vaish at any point of Shilov boundary $\Gamma(H^\infty(\mathbb{G}))$ of $H^\infty(\mathbb{G})$.
\end{theorem}
\begin{proof}
Let $\varphi^*$ be the a.e limit of $\varphi$ on $\partial^\infty \mathbb{G}$. If $\mid\varphi^*\mid<\epsilon$ on a subset $A\subset\partial^\infty \mathbb{G}$ of positive measure and $\mid\varphi^*\mid\leq 1$ elsewhere, then we take the outer function 
\begin{equation*}
F(\omega)=\exp{\frac{1}{\pi i}\int_{-\infty}^{\infty}\frac{1+tz}{(t-z)(t^2+1)}\log\mid F^*(t)\mid dt}
\end{equation*}
where $\mid F^*\mid=1$ on $A$ and $\mid F^*\mid=\epsilon$ on $\partial^\infty \mathbb{G}\setminus A$ to get $\|M_{\varphi}F\|_{\infty}\leq\epsilon$. Thus if $\mid\varphi\mid$ is not essentially bounded from below then $M_{\varphi}$ is not bounded from below and this is equivalent to $M_{\varphi}$ does not have closed range. Conversely, suppose that $\varphi$ does not vanish at any point of Shilov boundary. Hence, for some $\epsilon>0,\ \mid\varphi(\omega)\mid>\epsilon$ almost everywhere . Now if $M_{\varphi}$ is not of closed range then we can consider an  $f\in H^\infty(\mathbb{G})$ with $\|f\|_{\infty}=1$ and $\|M_{\varphi}f\|_{\infty}\rightarrow0$. Since $\mid\varphi(\omega)\mid>\epsilon$ almost everywhere so we must have $\mid f(\omega)\mid=0$ a.e on $\partial^\infty \mathbb{G}$. But this is a contardiction since $f\in H^\infty(\mathbb{G})\subset N$ (Nevalinna class) and a function in the Nevalinna class can not be zero on a subset of $\partial^{\infty}\mathbb{G}$ of positive Lebesuge measure (see Theorem 2.2 of \cite{5}).\ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \
\end{proof}\\
Here we recall Theorem 3.5 of \cite{4}. 
%----------------------------------------------------------------- Theorem 4.6 ----------------------------------------------------------
\begin{theorem}\label{Bonet}
For any weight $ \upsilon $ on $\mathbb{D}$ (not necessarily radial) there is a closed set $A_{\upsilon},\ \Gamma(H^{\infty}(\mathbb{D}))\subset A_{\upsilon}\subset M(H^{\infty}(\mathbb{D}))\setminus\mathbb{D}$, such that $ M_{\varphi}:H_{ \upsilon}(\mathbb{D})\longrightarrow H_{ \upsilon}(\mathbb{D})$ has closed range if and only if $\varphi$ does not vanish on $A_{\upsilon}$.
\end{theorem}
Now we Prove Theorem \ref{Bonet} for the upper halfplane. 
%------------------------------------------------------------------- Theorem 4.8 ---------------------------------------------------------
\begin{theorem}\label{28}
For any weight $ \upsilon $ on $\mathbb{G}$ there is a closed set $A_{\upsilon},\ \Gamma(H^{\infty}(\mathbb{G}))\subset A_{\upsilon}\subset M(H^{\infty}(\mathbb{G}))\setminus\mathbb{G}$, such that $ M_{\varphi}:H_{ \upsilon}(\mathbb{G})\longrightarrow H_{ \upsilon}(\mathbb{G})$ has closed range if and only if $\varphi$ does not vanish on $A_{\upsilon}$.
\end{theorem}
\begin{proof}
Let $\upsilon$ be a weight on $\mathbb{G}$. then $u=\upsilon\circ\alpha$ is a weight on $\mathbb{D}$ and $T:H_{ \upsilon}(\mathbb{G})\longrightarrow H_{ u}(\mathbb{G})$ defined by $Tg=g\circ\alpha$ is an isometric isomorphism since
\begin{equation*}
\|Tg\|_{u}=\sup_{z\in\mathbb{D}}\mid(g\circ\alpha)(z)\mid(\upsilon\circ\alpha)(z)=\sup_{\omega\in\mathbb{G}}\mid g(\omega)\mid\upsilon(\omega)=\|g\|_{\upsilon}
\end{equation*}
Now consider the following diagram.
\begin{displaymath}
\begin{array}{cccc}
M_{\varphi}: & H_{\upsilon}(\mathbb{G})& \longrightarrow & H_{\upsilon}(\mathbb{G}) \\
& \downarrow T & & \downarrow T \\
M_{\tilde{\varphi}}: & H_{u}(\mathbb{D}) &\longrightarrow & H_{u}(\mathbb{D})\\
\end{array}
\end{displaymath}
where $\tilde{\varphi}=\varphi\circ\alpha$. For any $g\in H_{\upsilon}(\mathbb{G})$\\
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (T^{-1}\circ M_{\tilde{\varphi}}\circ T)(g)&=(T^{1}\circ M_{\tilde{\varphi}})(g\circ\alpha)
=T^{-1}((\varphi\circ\alpha)(g\circ\alpha))=
[(\varphi\circ\alpha)\circ\alpha^{-1}][(g\circ\alpha)\circ\alpha^{-1}]
=\varphi g
=M_{\varphi}(g).$\\
Thus the diagram is commutataive. Therefore $ M_{\varphi}$ has closed range on $H_{\upsilon}(\mathbb{G})$ if and only if $M_{\tilde{\varphi}}$ has closed range on $ H_{u}(\mathbb{D})$. Now let $M_{\varphi}$ be of closed range. Since $M_{\tilde{\varphi}}$ has closed range, Theorem \ref{Bonet} implies that $\tilde{\varphi}$ does not vanish on a closed subset $A_{u=\upsilon\circ\alpha}$ of $\overline{\mathbb{D}}$ and this implies that $\varphi$ does not vanish on a closed subset $A_{\upsilon}$ of $\overline{\mathbb{G}}$ ( an isomorhic map takes a closed set to a closed set). $A_{u}$ satisfies $\Gamma(H^{\infty}(\mathbb{D}))\subset A_{u}\subset M(H^{\infty}(\mathbb{D}))\setminus\mathbb{D}$. Now Lemma \ref{3} and Corollary \ref{7} imply that $A_{\upsilon}$ satisfies $ \Gamma(H^{\infty}(\mathbb{G}))\subset A_{\upsilon}\subset M(H^{\infty}(\mathbb{G}))\setminus\mathbb{G}$.\\
Conversely if $\varphi$ does not vanish on a set $A_{\upsilon}$ in $\overline{\mathbb{G}}$, $\Gamma(H^{\infty}(\mathbb{G}))\subset A_{\upsilon}\subset M(H^{\infty}(\mathbb{G}))\setminus\mathbb{G}$ then $\tilde{\varphi}$ does not vanish on a homeomomorphic set $A_{u}=A_{\upsilon\circ\alpha}$ in $\overline{\mathbb{D}}$ and $\Gamma(H^{\infty}(\mathbb{D}))\subset A_{u}\subset M(H^{\infty}(\mathbb{D}))\setminus\mathbb{D}$. Again by Theorem \ref{Bonet} $M_{\tilde{\varphi}}$ has closed range and equivalently $M_{\varphi}$ has closed range.\ \ \ \ \ \ 
\end{proof}
\begin{remark}
It is worth to be mentioned that all of our results hold in every simply connected domain in the complex plane and can be obtained by methods which we have used here, since all of simply connected domains are homeomorphic (see Chap. 13 of \cite{7}). For example the strip $\{z\in\mathbb{C}: z=x+iy,\ -\frac{\pi}{2}<x<\frac{\pi}{2}\}$ and $\mathbb{C}\setminus\{(x,0):\ x\geq0\}$. Nevertheless the simply connected domain $\mathbb{G}$ is of special interest, since it is frequently easier to handle .
\end{remark}
\section{Multipliers}
In this section we characterize multipliers of $H_{\upsilon}(\mathbb{G})$ for certain type of weights on the upper half-plane. We recall that a sequence $\{\lambda_{n}\}$ is a multiplier of a sequence space $A$ if $\{\lambda_{n}a_{n}\}\in A$ for each $\{a_{n}\}\in A$. Since ant analytic function has a uniqe Talyor coefficients in its expantion , so any space of holomorphic functions can be regarded as a function space particularly, $H_{\upsilon}(\mathbb{D})$. In \cite{12} Shields and Williams proved the following characterization for multipliers of  $H_{\upsilon}(\mathbb{D})$ whenever $\upsilon$ is a normal weight i.e  $\upsilon$ is a standard weight on $\mathbb{D}$ such that there are $\epsilon , k>0$ such that $ 0<\epsilon<k\ \ \frac{\upsilon(r)}{(1-r)^{k}}\nearrow \infty$ as $r\rightarrow 1^{-}$ and  $\frac{\upsilon(r)}{(1-r)^{\epsilon}}\searrow0$ as $r\rightarrow 1^{-}$.
\begin{theorem}\label{29}
Let $\upsilon$ be a normal weight on $\mathbb{D}$. The sequence $\{\lambda_{n}\}$ is a multiplier of $H_{\upsilon}(\mathbb{D})$ if and only if\\
(i) The power series $h(z)=\Sigma_{n=1}^{\infty}\lambda_{n}z^{n}$ converges for $\mid z\mid<1$ and\\
(ii) $M_{1}(h',r)=O(\frac{1}{1-r})$.\\
Where $(ii)$ means $M_{1}(h',r)(1-r)$ is bounded.
\end {theorem}
We intend to obtain a result similar to Theorem \ref{29} for $H_{\upsilon}(\mathbb{G})$ for a special kind of weights on $\mathbb{G}$. We recall the  Definition 1.2  of \cite{11}.  We say a weight $\upsilon$ on $\mathbb{G}$ is of type(II) weight if $\upsilon(\omega)=\upsilon_{1}(\omega)$ ($\mid\omega\mid\leq1$ and $\upsilon_{1}$ is a standard weight on $\mathbb{G}$) and there is a constant $C>0$ such that $\frac{\upsilon(\omega)}{\upsilon(\frac{1}{\omega})}\leq C$. For examples of type(II) weights we refer the reader to Example 1.3 of \cite{11}.
\begin{lemma}\label{30}
Let $\upsilon$ be a type(II) weight which satisfy $(*)$. Put $\tilde{\upsilon}(z)=\upsilon(\alpha(-\mid z\mid))$. Then $\tilde{\upsilon}(z)$ is a radial with on $\mathbb{D}$. Moreover, the map $T$ defined by $(Tf)(z)=f(\alpha(z))$ is an isomorphism from $H_{\upsilon}(\mathbb{G})$ on to $H_{\tilde{\upsilon}}(\mathbb{D})$.
\end{lemma}
\begin{proof}
See Lemma 3.1 of \cite{11}. 
\end{proof}\\
Note that proof of \ref{30} revales that weights $\upsilon$ and $\tilde{\upsilon}$ are equivalent. Before stating main result of this section we need the following lemmas
\begin{lemma}\label{31}
If $f:\mathbb{G}\longrightarrow\mathbb{C}$ is a holomorphic function, then there are $\alpha_{k}\in\mathbb{C}$ such that $f(\omega)=\Sigma_{k=0}^{\infty}\alpha_{k}(\frac{\omega-i}{\omega+i})^{k}$, where the series converges uniformly on compact subsets of  $\mathbb{G}$.
\end{lemma}
\begin{proof}
Clearly $f\circ\alpha:\mathbb{D}\longrightarrow\mathbb{C}$ is analytic. So $(f\circ\alpha)(z)=\Sigma_{k=0}^{\infty}\alpha_{k}z^{k}$ for some $\alpha_{k}$ and series convergence uniformly on the compact subsets of $\mathbb{D}$.  Put $\alpha(z)=\omega$ then $z=\alpha^{-1}(\omega)=\frac{\omega-i}{\omega+i}$. This completes the proof.
\end{proof}
\begin{lemma}
Let $\upsilon$ be a type(II) weight on $\mathbb{G}$ which satisfy $(*)$ and $(**)$. Then $\tilde{\upsilon}(\tilde{\upsilon}$ is as in Lemma \ref{30}) satisfies $(*)'$ and $(**)'$ respectively.
\end{lemma}
\begin{proof}
See Theorem 2.2.3 of \cite{13}.
\end{proof}
\begin{remark}
Lemma \ref{31} enable us to regard $H_{\upsilon}(\mathbb{G})$ as a function space. Also it is easy to see that a weight $\upsilon$ is normal if and only if it satisfies $(*)'$ and $(**)'$. See also section three of  \cite{4}. 
\end{remark}
what we have obtained in this section can be summarized in the follwing theorem
\begin{theorem}
Let $\upsilon$ be a type two weight on $\mathbb{G}$ satisfing $(*)$ and $(**)$. Then $(\lambda_{n})$ is a multiplier on $H_{\upsilon}(\mathbb{G})$ if and only if\\ 
(i) The power series $h(\omega)=\Sigma_{n=1}^{\infty}\lambda_{n}(\frac{\omega-i}{\omega+i})^{n}$ converges for on $\mathbb{G}$ and\\
(ii) $M_{1}(h',r)=O(\frac{1}{1-r})$.\\
\end{theorem}
\begin{proof}
Only note that $h(\omega)=h\circ\alpha(z)$ is again can be considered as an element of $H_{\tilde{\upsilon}}(\mathbb{D})$.
\end{proof} 
%\vspace{4mm}\noindent{\bf Acknowledgements}\\
%\noindent If you'd like to thank anyone, place your comments here.


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\begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format.
\bibitem{13} 
M. A. Ardalani, Weighted spaces of holomorphic functions on the upper halfplane, Dissertation, Paderborn (2010).
\bibitem{11}
M. A. Ardalani, Boundedness of self map composition operators for two type of weights on the upper half-plane, {\it Cogent Mathematics}, Vol. 3( 2016).
\bibitem{10} 
M. A. Ardalani, Boundedness and invertibilty of a pointwise multiplication operator on weighted spaces of holomorphic functions on the upper half-plane, {\it Complex and Nonlinear Systems}, Vol. 1, No. 2. Winter 2018,  107-111.
\bibitem{1}
M. A. Ardalani, W. Lusky, Bounded operators on weighted spaces of holomorphic functions on the upper half plane, {\it Studia. Math}, Vol. 209 (3) (2012), 225-234.
\bibitem{2}
M. A. Ardalani, W. Lusky, Weighted spaces of holomorphic functions on the upper half plane, {\it Math. Scandinavica}, Vol. 111(2012), 244-260.
\bibitem{3}
K. D. Bierstedt, J. Bonet and J. Taskinen, Associated weights and spaces of holomorphic functions,{\it Studia Mathematica}, Vol. 127(1998), 137-168.
\bibitem{4}
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J. Bonet, P. Domanski and M. Lindstrom, Pointwise multiplication operators on weighted Banach spaces of analytic functions, {\it Studia Mathematica}, Volume 137 (1999), 177-194. 
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Walter Rudin, {\it Real and complex analysis}, McGraw-Hill, 1987.
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A. L. Shields, D. L. Williams, Bounded projections, duality and multipliers in spaces od analytic functions, {\it Trans. Amer. Math. Soc.}, Vol. 162( 1971), 287-302. 
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M. A. Stanev, Weighted Banach spaces of holomorphic functions in the upper half plane, arXiv:math.FA/9911082 v1 (1999).
 
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{\small

\noindent{\bf Mohammad Ali Ardalani}

\noindent Department of Mathematics

\noindent Associate Professor of Mathematics

\noindent University of Kurdistan

\noindent Sanandaj, Iran

\noindent E-mail:m.ardalani@uok.ac.ir}\\

{\small
\noindent{\bf  Saeed Haftbradaran  }

\noindent  Department of Mathematics

\noindent M.Sc of Mathematics

\noindent University of Kurditan


\noindent Sanandaj, Iran

\noindent E-mail: s.haftbradaran@sci.uok.ac.ir}\\



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