\section{Introduction} This paper presents a fairly general version of the well-known Gleason-Kahane-$\dot{\text{Z}}$elazko theorem in the spirit of Mashreghi and Ransford \cite[Theorem 3.1]{MR1}. In effect, we characterize—in Theorem \ref{GKZ-p}—a class of linear functionals as point evaluations on the vector space of all polynomials 
$\cl P$. We do not make any topological assumptions. We also wish to state that our techniques are quite elementary. We then apply our Theorem \ref{GKZ-p} to present—in Theorem \ref{GKZ-m}—a version of the GKZ theorem for a class of metric spaces of complex-valued functions on an open disc centered at the origin. This theorem can be deemed to be an extension of Theorem C, enunciated below.

The classical GKZ theorem is a remarkable theorem from the viewpoints of elegance and its import. It assures the multiplicativity of a linear functional on a Banach algebra simply by assuming that the functional is non-zero on invertible elements and maps the identity to the number 1. The theorem was originally proved for commutative Banach algebras by Kahane and $\dot{\text{Z}}$elazko \cite{KZ} and independently by Gleason \cite{Gle}. Later, $\dot{\text{Z}}$elazko \cite{Zel} extended the proof to a general Banach algebra and he even relaxed the assumption of the algebra possessing an identity element. 

\vspace{.2 cm}

\noindent {\bf Theorem A.} {\bf (Gleason-Kahane-$\dot{\text{Z}}$elazko theorem)} {\em Let $\cl A$ be a unital complex Banach algebra, and let $F:\cl A \to \bb C$ be a linear functional such that $F(1)=1$ and $F(x)\ne 0$ for every invertible element $x$ in $\cl A$. Then $F(xy)=F(x)F(y)$ for every $x, \ y\in \cl A.$}  

The original proofs of the theorem had relied on a rather technical result about entire functions till Roitman and Sternfeld 
\cite{RS} provided a simple as well as elementary proof of the GKZ theorem. Since then, the result has attracted a fair amount of interest as evidenced by the appearance of \cite{CHMR, MR, MR1, MRR} and many other articles and books \cite{AD, Fra, Kai}. This series of papers has essentially produced versions or analogues of the GKZ theorem on various normed spaces that have largely been connected to function spaces.

Our interest in the GKZ Theorem stems from the work of Mashreghi and Ransford \cite{MR1} who produced a generalization of the GKZ type theorem to modules over Banach algebras. They then used their generalization to characterize linear functionals on the Hardy spaces that are non-zero on outer functions as point evaluations. This is a true analogue of the GKZ theorem on Hardy spaces since any function $g$ in a Hardy space for which there exists an $h$ in the space such that $gh$ = 1 has to be outer and a point evaluation on a Hardy space is multiplicative whenever the multiplication makes sense.

\vspace{.2 cm}

\noindent{\bf Theorem B.} \cite[Theorem 2.1]{MR1} {\em Let $0<p\le \infty$ and let $F:H^p(\bb D) \to \bb C$ be a linear functional such that $F(1)=1$ and $F(g)\ne 0$ for every outer function $g$ in $H^p(\bb D).$ Then there exists a scalar 
$w\in \bb C$ such that $F(f)=f(w)$ for all $f$ in $H^p(\bb D.)$}

\vspace{.2 cm}

In the same paper \cite{MR1} Mashreghi and Ransford also prove the following GKZ type theorem for a wide variety 
of Banach spaces.

\vspace{.2 cm}

\noindent {\bf Theorem C.} \cite[Theorem 3.1]{MR1} {\em Let 
$\cl X$ be a Banach space consisting of analytic functions on the open unit disc $\bb D$ satisfying the following properties:
\begin{enumerate}
\item for each $w\in \bb D,$ the evaluation map at $w$ is continuous on $\cl X;$
\item $\cl X$ contains all polynomials and polynomials are dense in $\cl X;$
\item multiplication with the function $z$ is well-defined on $\cl X.$ 
\end{enumerate}
Furthermore, let $\cl Y$ be a subset of $\cl X$ with the following properties:
\begin{enumerate}
\item if $g\in \cl X$ with $0<\inf_{z\in \bb D}|g(z)|\le \sup_{z\in \bb D}|g(z)|<\infty,$ then $g\in \cl Y;$
\item $\cl Y$ contains all linear polynomials $z-\alpha$ with $|\alpha|=1.$
\end{enumerate}
If $F$ is a continuous linear functional on $\cl X$ such that $F(1)=1$ and $F(g)\ne 0$ for each $g\in \cl Y,$ then there exists a scalar $w\in \bb D$ such that $F(f)=f(w)$ for all $f\in \cl X.$}

\vspace{.4 cm}

If we take $\cl X=H^p(\bb D), \ 0<p\le \infty,$ and $\cl Y$ to be the set of outer functions in $\cl X$, then they satisfy the properties stated in Theorem C. Therefore, Theorem B is a special case of Theorem C. But there is an extra assumption of continuity of the linear functional in Theorem C, and the reason the authors could deduce the desired conclusion in 
Theorem B without the continuity assumption  
is the special properties of Hardy spaces which one cannot expect general function spaces to possess. 

Both these theorems motivated us to investigate the possibility of obtaining similar results by knowing the behavior of the linear functional on a simpler class of functions. Our investigation leads us to the following 
theorems. The first one is a characterization of multiplicative linear functionals on 
$\cl P$—the linear space of all polynomials—in terms of its action on a very select class of powers of linear polynomials whose zeros 
lie outside an open disc centered at the origin. For this reason, we shall refer to it as a GKZ theorem for polynomials.

\begin{thm}\label{GKZ-p} Let $\cl P$ denote the set of all complex polynomials and let $F: \cl P\to \bb{C}$ be a linear functional. Suppose $F(1)=1$ and there exists a positive real number $r$ such that 
$F((z-\alpha)^n)\ne 0$ for all $n\ge 1$ and for all complex numbers $\alpha $ with $|\alpha|\ge r.$ Then 
there exists a complex number $w$ with $|w|<r$ such that $F(p)=p(w).$
\end{thm}

As an application, we obtain the following GKZ theorem for metric spaces. Although different, it is in the same spirit 
as Theorem C (\cite[Theorem 3.1]{MR1}). Let $\cl F(r,\bb C)$ denote the set of all complex-valued functions on the open disc with center at $0$ and radius $r.$ Clearly it a vector space over $\bb C$ with respect to pointwise addition and scalar multiplication. 

\begin{thm}\label{GKZ-m}
Let $\cl X$ be a vector subspace of $\mathcal{F}(r,\mathbb{C})$. Suppose $\cl X$ is also a metric space such that: 
\begin{enumerate}[(i)]
\item for each $w$ with $|w|<r$, the point evaluation map $f \mapsto f(w)$ is continuous on $\cl X$;
\item $\cl X$ contains all the polynomials and they are dense in $\cl X.$  
\end{enumerate}
Let $F: \cl X \to \mathbb{C}$ be a continuous linear functional such that $F(1)=1$ and $F((z-\alpha)^n)\ne 0$ whenever $n\ge 1$ and $\alpha$ is a scalar with $|\alpha|\ge r$. Then there exists a scalar $w$ with $|w|<r$ such that 
\begin{equation*}
F(f)=f(w) 
\end{equation*}
for every $f\in \cl X.$
\end{thm} 

If we take $r=1$ in Theorem \ref{GKZ-m}, then the polynomials $(z-\alpha)^n$ for $n\ge 1$ and scalars $\alpha$ with $|\alpha|\ge 1$ are actually outer functions in the Hardy spaces $H^p(\bb D)$. So, similar to Theorem C, our Theorem \ref{GKZ-m} is also an extension of 
Theorem B.  

In effect, we claim novelty on four counts:

\begin{enumerate}
\item Our hypothesis in Theorem \ref{GKZ-m} demands checking the linear functional to be non-zero on a much simpler and smaller class of functions, and yet shows it to be a point evaluation.
\item Our proof of Theorem \ref{GKZ-m}, unlike the other versions given in \cite{CHMR, MR, MR1, MRR}, does not depend on the classical GKZ theorem.
\item Our proof of Theorem \ref{GKZ-m} is elementary and simple.
\item We do not require the space $X$ to be a Banach space, and this allows us to include a few more well-known spaces for which the theorem holds true. 
\end{enumerate}

\vspace{.2 cm}

We also prove the following GKZ theorem for the weighted Hardy spaces on which the operator of multiplication with $z$ is strictly cyclic. 

\begin{thm}\label{strictly} Let $M_z,$ the operator of multiplication with $z,$ be strictly cyclic on the weighted Hardy space $H^2(\beta)$. Let $F:H^2(\beta)\to \bb C$ be a linear functional such that $F(1)=1$ and $F(f)\ne 0$ for every invertible $f\in H^2(\beta).$ Then there exists a scalar $w$ with $|w|\le r(M_z)$ such that 
$$
F(f)=f(w)
$$
for all $f\in H^2(\beta).$ 
\end{thm}
 
 \section{A Brief Preview}
The organization of the paper is as follows. We prove Theorem \ref{GKZ-p} in Section \ref{poly} and use it to prove Theorem \ref{GKZ-m} in Section \ref{metric} and Theorem \ref{strictly} in Section \ref{cyclic}. In Example \ref{cont}, 
we give an example that validates our assumption of continuity of the linear functional in Theorem \ref{GKZ-m}. Indeed, we construct an unbounded linear functional on the Hardy space $H^2(\bb D)$ such that it is 1 at the constant function 1 and it is non-zero on polynomials $(z-\alpha)^n$ for every $n\ge 1$ and each $\alpha$ with $|\alpha|\ge 1.$ This functional being unbounded can never be a point evaluation. In Corollary \ref{wcomp0}, we obtain a characterization of weighted composition operators for our setting.

\section{The GKZ theorem for polynomials}\label{poly}
In this Section, we shall prove Theorem \ref{GKZ-p}. Our proof is a beautiful extension of the idea used by Roitman and Sternfeld \cite[Lemma 2]{RS} to give an 
elementary proof of the classical GKZ theorem.  

\vspace{.2 cm}

\noindent {\bf Proof of Theorem \ref{GKZ-p}}.
Fix any $k\ge 0$, and take $n>k$. Set  
\begin{eqnarray}
p(t)&=& F((t-z)^n)\nonumber\\
&=& {n\choose0}t^n-{n\choose1}F(z)t^{n-1}+\dots+(-1)^{n} F(z^n)\label{poly1}
\end{eqnarray}
Then $p(t)$ is a polynomial in $t$ of degree $n$. Let $\lambda_1, \lambda_2,\dots, \lambda_n$ denote the roots (counting the multiplicity) of $p.$ Then the hypothesis $p(t)=F((t-z)^n)\neq 0$ whenever 
$|t|\ge r$ implies that $|\lambda_i|<r$ for $1\le i\le n.$ Further, we can write  
\begin{equation}\label{poly2}
p(t)=\prod_{i=1}^n (t-\lambda_i).
\end{equation} 
Then by comparing the coefficients of $t^{n-1}$ from Equations (\ref{poly1}) and (\ref{poly2}), we obtain  
\begin{equation}\label{poly3}
nF(z)= \sum_{i=1}^{n}\lambda_{i},
\end{equation} 
and comparing the coefficients of $t^{n-k}$ from Equations (\ref{poly1}) and (\ref{poly2}), we obtain  
\begin{equation}\label{poly4}
\sum_{1\le i_{1}<i_{2}<...<i_{k}\le n}\lambda_{i_{1}} \lambda_{i_{2}}\dots\lambda_{i_{k}}={n\choose k}F(z^{k}).
\end{equation}
 
Thus,  
\begin{eqnarray}
&&\left(nF(z)\right)^{k}-k!{n\choose k}F(z^{k})\nonumber\\ 
&=&\left(\sum_{i=1}^{n} \lambda_i\right)^{k} - k! \sum_{1\le i_{1}<i_{2}<...<i_{k}\le n}\lambda_{i_{1}} \lambda_{i_{2}}\dots\lambda_{i_{k}}\nonumber\\
&=&\sum_{\substack{j_{1}+j_{2}+\dots+j_{n}=k \\ j_i\ge0} }\frac{k! \, (\lambda_{1}^{j_{1}}\lambda_{2}^{j_2}\dots\lambda_{n}^{j_n})}{j_{1}!\, j_{2}!\,\dots\,j_{n}!}-k!\sum_{1\le i_{1}<i_{2}<\dots<i_{k}\le n}\lambda_{i_{1}} \lambda_{i_{2}}\dots\lambda_{i_{k}}\label{poly5}.
\end{eqnarray}
Clearly, each term in the later summation of Equation (\ref{poly5}) also appears in the first summation of Equation (\ref{poly5}). Also, $|\lambda_i|<r$ for each $i.$ Therefore, if we let  
 $A_{\#}$ denote the number of terms in the expression $\displaystyle\sum_{\substack{j_{1}+j_{2}+\dots+j_{n}=k \\ j_i\ge0} }\frac{k!}{j_{1}!\, j_{2}!\,\dots\,j_{n}!} (\lambda_{1}^{j_{1}}\lambda_{2}^{j_2}\dots\lambda_{n}^{j_n})$
and $B_{\#}$ denote the number of terms in the expression $\displaystyle\sum_{1\le i_{1}<i_{2}<\dots<i_{k}\le n}\lambda_{i_{1}} \lambda_{i_{2}}\dots\lambda_{i_{k}}$, then 
\begin{equation*}
\left\vert(nF(z))^{k}-k!{n\choose k}F(z^{k})\right\vert\le k!(A_\#-B_\#)r^k.
\end{equation*}
Now using standard facts from combinatorics, 
\begin{eqnarray}
A_\#-B_\# &=&{n+k-1\choose n-1}- {n\choose k} \nonumber\\
&=&\frac{(n+k-1)!}{k!(n-1)!}-\frac{n!}{k!(n-k)!}
\nonumber \\
&=&\frac{(n+k-1)(n+k-2)\dots(n+k-(k+1))!}{k!(n-1)!}-\frac{n(n-1)\dots(n-(k-1))(n-k)!}{k!(n-k)!}
\nonumber\\
&=& \frac{n^k}{k!}\left[\Big (1+\frac{k-1}{n}\Big)\Big(1+\frac{k-2}{n}\Big)\dots1-\Big(1-\frac{1}{n}\Big)\dots\Big (1-\frac{k-1}{n}\Big)\right].
\nonumber
\end{eqnarray}
Therefore, 
\begin{eqnarray}
\left\vert(nF(z))^{k}-k!{n\choose k}F(z^{k})\right\vert &\le& k!(A_\#-B_\#)r^k
\nonumber\\
&=&n^k \left[\Big (1+\frac{k-1}{n}\Big)\Big(1+\frac{k-2}{n}\Big)\dots 1
-\Big(1-\frac{1}{n}\Big)\dots\Big (1-\frac{k-1}{n}\Big)\right]r^k.
\nonumber
\end{eqnarray}
Lastly, dividing both sides by $n^k$ and letting $n \to \infty$, we conclude 

\begin{equation*}
	F(z^k)=F(z)^k.
\end{equation*}
Hence, we have established that $F(z^k)=(F(z))^k$ for all $k\ge 0.$  

Let $w=F(z).$ Then $F(p)=p(w)$ for every polynomial $p\in \cl P.$ Finally, $F(z-w)=F(z)-wF(1)=0,$ which implies that $|w|<r.$ This completes the proof.  
  
\qed

\section{The GKZ Theorem for metric spaces} \label{metric}

We are now ready to prove Theorem \ref{GKZ-m}, our GKZ theorem for metric spaces which also have a linear structure on them. But we do not assume our underlying space to be a topological vector space. 

\vspace{.2 cm}  
 
\noindent {\bf Proof of Theorem \ref{GKZ-m}}. The given hypotheses yields that $F:\cl P\to \bb C$ is linear, $F(1)=1,$ and 
$F((z-\alpha)^n)\ne 0$ for all $n\ge 1$ and scalars $\alpha$ with $|\alpha|\ge r.$ Therefore, using Theorem 
\ref{GKZ-p}, we get a scalar $w$ with $|w|<r$ such that $F(p)=p(w)$ for all polynomials $p\in \cl P.$ 
Now, let $f \in \cl X$. Then there exists a sequence of polynomials $p_n$ that converge to $f$ as $n$ tends to 
infinity. Since $f\mapsto f(w)$ is continuous, we have $p_n(w) \to f(w)$. But, $p_n(w)=F(p_n)$ and $F$ is assumed to be continuous; so, $p_n(w)=F(p_n)$ converges to $F(f)$ as $n$ tends to infinity. Hence, 
$F(f)=f(w).$ This completes the proof.

\qed 

Our space $\cl X$ is much more general than those considered by Mashreghi and Ransford in Theorem C, and thus all the examples mentioned in \cite{MR1} are covered by our theorem as well. In addition, since we do not require $\cl X$ to be a Banach space, the following spaces are also covered in our theorem.

\begin{enumerate}
\item  The Hardy spaces $H^p(\bb D)$ for $0<p<1$;
\item The Bergman spaces $A_p$ for  $0<p<1$.
\end{enumerate}

Furthermore, there is another class of function spaces that is convered by our theorem, namely the weighted Hardy spaces $H^2(\beta)$ with $\liminf(\beta_n)^\frac{1}{n}>0$. We shall discuss these spaces in more details in Section \ref{cyclic}.

We now give the following example to validate the assumption of continuity of the linear functional in Theorem \ref{GKZ-m}.

\begin{ex}\label{cont} Let $\mathcal{B}$ be a Hamel basis for the Hardy space $H^2(\mathbb{D})$ obtained by extending the orthonormal basis $\{z^n:n \ge 0\}$. Since $\mathcal{B}$ is an uncountable set, we can choose a countable subset $\{f_n: n\in \mathbb{N} \}$ of $\mathcal{B}$ such that $f_n\ne z^m$ for any $n$ and $m$. For a fixed $w \in \mathbb{D}$, define $F: \mathcal{B} \to \mathbb{C}$ by
$$  
F(b)=
\begin{cases}
w^n &\quad \text{if}  \quad b =z^n,\\
m\norm{b} &\quad \text{if}  \quad b=f_m,\\
0 &\quad  \quad \text{otherwise},\\
\end{cases}
$$
and extend it linearly to $H^2(\bb D).$ Then $F(1)=1$ and $F(p)=p(w)$ for every polynomial $p.$ This implies    
$F((z-\alpha)^n)=(w-\alpha)^n\ne 0$ whenever $\alpha$ is a scalar with $|\alpha|\ge 1.$ Therefore, $F$ is a linear functional on $H^2(\bb D)$ which satisfies the hypotheses of Theorem \ref{GKZ-m}. But $F(f_n)=n||f_n||$ for each $n$ which implies that $F$ is unbounded.  

Now suppose, $F$ is a point evaluation at some point $\zeta.$ Then $\zeta=F(z)=w\in \bb D.$ This implies that 
$|F(f)|=|f(w)|\le ||f||||k_w||$ for all $f$ in $H^2(\bb D),$ where $k_w$ is the kernel function at $w.$  This means $F$ is bounded, which is a contradiction. Hence, $F$ cannot be a point evaluation.
\end{ex}

As in \cite{MR1}, we also obtain the following characterization of weighted composition operators for our setting. The proof is similar to the proof of Theorem 3.2 in \cite{MR1}, we include it here for completeness. Let $Hol(\bb D)$ denote the set of all holomorphic functions on the open unit disc $\bb D$. We assume that $Hol(\bb D)$ is given its usual $Fr\grave{e}chet$-space topology.

\begin{cor}\label{wcomp0} Let $\cl X$ be a vector subspace of $\cl F(r, \bb C)$ containing all polynomials, and let it has a metric such that point evaluations on it are continuous and polynomials are dense in it.  Let $T: \cl X \to Hol(\mathbb{D})$ be a continuous linear map such that $\left(T(z-\alpha)^n\right)(\zeta)\ne0$ for all $n\ge 0,$ for all scalars $\alpha$ with 
$|\alpha|\ge r,$ and for all $\zeta\in \bb D$. Then there exist holomorphic functions $\phi: \mathbb{D} \to \mathbb{D}$ and $ \psi: \mathbb{D} \to \mathbb{C}\setminus\{0\}$ such that 
\begin{equation}\label{wcomp}
Tf=\psi.(f\circ{\phi}),
\end{equation}
 for all $f\in \cl X.$
\end{cor}

\begin{proof}
Set $\psi=T(1)$. Then, using the hypotheses, $\psi$ is holomorphic on $\bb D$ and does not assume zero at any point. Define $\phi:=(Tz)/\psi$. Then $\phi$ is clearly a holomorphic function on $\mathbb{D}$. Now we claim that $\phi$ and $\psi$ satisfy Equation (\ref{wcomp}).
	
Fix a $ \zeta \in \mathbb{D}$ and define $T_\zeta: \cl X \to \bb C$ by $T_\zeta(f) =(Tf)(\zeta)/\psi(\zeta)$. 
We can easily verify, using the assumptions on $T,$ that $T_\zeta$ is continuous, linear, and is non-zero on polynomials of the form $(z-\alpha)^n$ whenever $|\alpha|\ge r.$ Also, $T_\zeta(1)=1.$ Then, using Theorem \ref{GKZ-m}, we get a scalar $w_{\zeta}$ with $|w_{\zeta}|<r$ such that 
$T_\zeta(f)=f(w_{\zeta})$ for all $f\in \cl X$. But, $T_\zeta(f)=(Tf)(\zeta)/\psi(\zeta)$ for all $f\in \cl X$, therefore 
$Tf(\zeta)=\psi(\zeta)f(w_{\zeta}).$ In particular, $(Tz)(\zeta)= \psi(\zeta)w_{\zeta}$ which implies that 
$\phi(\zeta) = w_{\zeta}.$ Hence,  $Tf=\psi.(f\circ \phi)$ for all $f\in \cl X$.
\end{proof}
 
 \begin{remark} We note that Corollary \ref{wcomp0} will hold even when we replace $Hol(\mathbb{D})$with the space $C(\mathbb{D})$ of all complex-valued continuous functions on $\mathbb{D}$ considered with 
 the supremum norm. Of course, in such a case, we can guarantee only the continuity of the functions $\phi$ and $\psi $ in Equation (\ref{wcomp}).
 \end{remark}

\section {The GKZ theorem and Strictly cyclic shifts} \label{cyclic}
In this section, we shall use Theorem \ref{GKZ-p} to prove Theorem \ref{strictly}. Recall that Theorem \ref{strictly} 
is our GKZ theorem for weighted Hardy spaces on 
which the operator of multiplication by the function $z$ is strictly cyclic. Before we proceed further, we want to remind that all weighted Hardy spaces with $\liminf(\beta_n)^\frac{1}{n}>0$ are already covered by Theorem \ref{GKZ-m}. However, there are weighted Hardy spaces  for which $\liminf(\beta_n)^\frac{1}{n}=0$ and on which the operator of multiplication with $z$ is strictly cyclic. All such spaces clearly lie outside the purview of Theorem 
\ref{GKZ-m}; so we do have something new to offer here. We shall give one such example at the end.  

We shall first give some definitions and preliminary 
results. Let $\{\beta_n\}_{n=0}^\infty$ be a sequence of positive numbers with $\beta_0=1.$ Let $H^2(\beta)$ 
denote the weighted Hardy space 
consisting of formal power series $f(z)=\sum_{n=0}^\infty a_nz^n$ such that 
$\sum_{n=0}^\infty |a_n|^2\beta_n^2$ is finite. The norm on $H^2(\beta)$ is given by $||f||^2_\beta=\sum_{n=0}^2|a_n|^2\beta_n^2.$ The equality 
$f(z)=\sum_{n=0}^\infty a_nz^n$ simply means that $f$ denotes this formal power series; it does not mean convergence of the series at any complex number $z.$ The sequence $\{z^n\}_{n=0}^\infty$ is an orthogonal basis for $H^2(\beta)$ with $||z^n||_\beta=\beta_n$ for each $n.$ Of course, some functions in $H^2(\beta)$ are entire functions, but in general, the disc $\{w\in \bb C: |w|< \liminf\beta_n^{1/n}\}$ is the largest disc in which every power series in $H^2(\beta)$ converges; hence converges absolutely. If $sup_{n}{\beta_{n+1}/{\beta_n}}$ is finite, then $M_z(z^n)=z^{n+1}$ gives rise to a well-defined bounded operator on $H^2(\beta).$ Clearly, the converse is also true. Henceforth, we shall assume $M_z$ to be a bounded operator on $H^2(\beta),$ and use $r(M_z)$ to denote its spectral radius and $r_0(M_z)$ for $\liminf{\beta_n}^{1/n}.$

Furthermore, let $H^\infty(\beta)$ be the set of all formal power series that multiplies $H^2(\beta)$ into itself. Since the constant function $1$ is in $H^2(\beta), \ H^\infty(\beta)$ is a subset of $H^2(\beta).$ In fact, every $\phi\in H^\infty(\beta)$ defines a bounded operator on $H^2(\beta),$ which maps $f\in H^2(\beta)$ to $\phi f;$ we denote it by $M_\phi.$ In addition, if $\phi, \ \psi\in H^\infty(\beta),$ then $\phi \psi$ (the formal Cauchy 
product) is again in $H^\infty(\beta)$ and $M_{\phi \psi}=M_{\phi} M_\psi.$ This allows us to identify 
$H^\infty(\beta)$ as a commutative subalgebra of $\cl B(H^2(\beta)),$ the Banach algebra consisting of all bounded operators on $H^2(\beta).$ Therefore, for $\phi\in H^\infty(\beta),$ if we define $||\phi||_{\infty}:=||M_\phi||,$ then it becomes a commutative unital Banach algebra. Furthermore, the commutant of $M_z$ equals $H^\infty(\beta).$ 

A bounded operator $T$ on a Hilbert space $\cl H$ is said to be strictly cyclic if there exists $x_0\in \cl H$ such that $\{Rx_0: R\in \cl W[T]\}=\cl H,$ where $\cl W[T]$ is the closure of the algebra of polynomials in $T$ with respect to the weak operator topology. It is proved in \cite[Proposition 31]{Shie} that the operator $M_z$ on $H^2(\beta)$ is strictly cyclic if and only if $H^2(\beta)=H^\infty(\beta)$ (as sets), and in such a case the two norms, namely $||\cdot||_\beta$ and $||\cdot||_\infty,$ are equivalent. Actually, this proposition reveals more about these weighted Hardy spaces. In general, as noted above, we can guarantee the absolute convergence of power in $H^2(\beta)$ only on the disc $\{w: |w|< r_0(M_z)\}$ and $r_0(M_z)\le r(M_z).$ But, if $M_z$ is strictly cyclic, then Proposition 31 from \cite{Shie} establishes that $r(M_z)$ equals $r_0(M_z),$ and each formal power series in $H^2(\beta)$ converges absolutely at every $w$ with $|w|\le r(M_z).$ This in turn implies that the point evaluation at every $w$ with $|w|\le r(M_z)$ defines a bounded linear functional on $H^2(\beta).$  

Before we prove Theorem \ref{strictly}, we want to note that one can derive this result using the classical GKZ theorem. Indeed, the assumption $M_z$ is strictly cyclic yields the equality of  
$H^2(\beta)$ and $H^\infty(\beta);$ and then the classical GKZ theorem implies that $F$ must be multiplicative. Further, as recorded in Corollary 1 on Page 94 in {Shie}, when $M_z$ is strictly cyclic, then the multiplicative linear functionals on $H^\infty(\beta)$ are point evaluations with points coming from the closed disc 
$\{w: |w|\le r(M_z)\}$. This establishes the desired conclusion.  

On the other hand, our proof does not rely on the classical GKZ. Instead, it is a beautiful application of Theorem \ref{GKZ-p}. 

\vspace{.2 cm}

\noindent {\bf Proof of Theorem \ref{strictly}}.
 First of all, since $F$ is non-zero on invertible elements and $F(f-F(f))=0,$ therefore $f-F(f)$ cannot be invertible in $H^2(\beta) = H^\infty(\beta)$. This implies that 
$F(f)$ lies in the spectrum of $f$ in $H^\infty(\beta)$ which further implies that $|F(f)|\le ||f||_\infty.$ But $||\cdot||_\infty$ is equivalent to $||\cdot||_\beta,$ therefore $F$ is bounded.   

Now choose a real number $\eta$ such that $\eta>r(M_z)$ and take a scalar $\alpha$ such that 
$|\alpha|\ge \eta.$ This imples that $M_z-\alpha I$ is invertible in $\cl B(H^2(\beta)).$ But, the commutant of $M_z$ equals $H^\infty(\beta),$ therefore $z-\alpha$ is invertible in $H^\infty(\beta)$ which means it is invertible in $H^2(\beta).$ Therefore, $(z-\alpha)^n$ is invertible in $H^2(\beta)$ which implies $F(z-\alpha)^n\ne 0$ for every $n\ge 1.$ Then using Theorem \ref{GKZ-p}, there exists a scalar $w$ with $|w|\le \eta$ such that $F(p)=p(w)$ for every polynomial $p.$ In particular, $w=F(z).$ So the point $w$ is independent of the choice of 
$\eta$ and $|w|\le \eta$ for all $\eta>r(M_z).$ Thus, $|w|\le r(M_z).$ Now, $\{z^n\}_{n=0}^\infty$ is an orthogonal sequence in $H^2(\beta)$ which yields polynomials as a dense set in $H^2(\beta).$ Moreover, point evaluation at $w$ is continuous which establishes that $F(f)=f(w).$    
\qed

\vspace{.2 cm}
 
As promised, we shall now give an example of a weighted Hardy space that is not covered by Theorem \ref{GKZ-m} but covered by Theorem \ref{strictly}. This means we are looking for a weighted Hardy space $H^2(\beta)$ for which $r_0(M_z)=\liminf{\beta_n}^{1/n}=0$ and on which the operator $M_z$ is strictly cyclic.    

Let $\{w_n\}_{n=0}^\infty$ be a sequence of positive real numbers that is monotonically decreasing and square-summable. Set $\beta_0=1$ and $\beta_n=w_0\cdots w_{n-1}$ for all $n\ge 1.$ Then by \cite[Corollary 4, Page 98]{Shie}, the operator $M_z$ on $H^2(\beta)$ is strictly cyclic. Also, since $w_n$ converges to zero, $r_0(M_z)=r(M_z)=0.$ Actually, the operator $M_z$ is unitarily equivalent to the weighted shift operator $T$ on 
$H^2(\bb D)$ which is given by $T(z^n)=w_nz^{n+1}$ for all $n.$ The operator $T^*$ is famously known as a Donoghue operator. 

\subsection*{Acknowledgements} The first and third authors
thank the Mathematical Sciences Foundation,
Delhi for support and facilities needed to complete
the present work.

