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\mbox{}\vspace*{.1cm}

\begin{center}
{\bf {\large A Condition of Reflexivity on some Sequence Spaces}} \vspace*{%
.3cm} \\[0pt]
{\bf Sh. Khoshdel \qquad B. Yousefi}\\[0pt]
{\small Department of Mathematics, Payame Noor University\\[0pt]
P.O. Box: 19395-3697, Tehran, Iran \\[0pt]
Email: khoshdel@pnu.ac.ir, b$_{-}$yousefi@pnu.ac.ir \\[0pt]
}
\end{center}

\vspace*{.3cm}\newline
{\bf Abstract:} In this paper we will give sufficient conditions for the
powers of the multipliction operator $M_{z}$ to be reflexive on formal
Laurent sequence spaces. \vspace*{.3cm}\newline
{\bf AMS Subject Classification:} 47B37; 47L10.\newline
{\bf Keywords:} Banach space of Laurent series associated with a sequence $%
\beta$, bounded point evaluation, reflexive operator \vspace*{.3cm}

\begin{center}
{\large {\bf 1. Introduction}}
\end{center}

\vspace*{.3cm} Let $\{\beta(n)\}^{\infty}_{n=-\infty}$ be a sequence of
positive numbers satisfying $\beta(0)=1$. If $1 < p < \infty$, the space $%
L^{p}(\beta)$ consists of all Laurent power series 
\[
f(z)=\sum\limits^{\infty}_{n=-\infty}\hat{f}(n)z^{n}
\]
such that the norm 
\[
\|f\|^{p}=\|f\|^{p}_{\beta}=\sum^{\infty}_{n=-\infty}|\hat{f}(n)|^{p}
\beta(n)^{p} 
\]
is finite. These are reflexive Banach spaces with the norm $\|\cdot\|_{\beta}
$ and $L^{p}(\beta)^{*}=L^{q}(\beta^{-1})$, where ${\frac{1}{p}}+{\frac{1}{q}%
}=1$ ([18]). Let $\hat{f}_{k}(n)=\delta_{k}(n)$. So $f_{k}(z)=z^{k}$ and
then $\{f_{k}\}_{k}$ is a basis for $L^{p}(\beta)$ such that $%
\|f_{k}\|=\beta(k)$. We denote the set of multipliers 
\[
\{\varphi \in L^{p}(\beta):\; \varphi L^{p}(\beta)\subseteq L^{p}(\beta)\}
\]
by $L^{p}_{\infty}(\beta)$ and the linear operator of multiplication by $%
\varphi$ on $L^{p}(\beta)$ by $M_{\varphi}$.

We say that a complex number $\lambda$ is a bounded point evaluation on $%
L^{p}(\beta)$ if the functional $e(\lambda):L^{p}(\beta)\longrightarrow %
\mbox{$\rm\bf\kern0.2em\rule{0.06em}{1.45ex}\kern-0.3em C$}$ defined by $%
e(\lambda)(f)=f(\lambda)$ is bounded.

Recall that if $E$ is a separable Banach space and $A\in B(E)$, then $Lat(A)$
is by definition the lattice of all invariant subspaces of $A$, and $Alg
Lat(A)$ is the algebra of all operators $B$ in $B(E)$ such that $Lat(A)
\subset Lat(B)$. For the algebra $B(E)$, the weak operator topology is the
one induced by the family of seminorms 
\[
p_{x^{*},x}(A)=|<Ax,x^{*}>|
\]
where $x\in E$, $x^{*} \in E^{*}$ and $A \in B(E)$. Hence $%
A_{\alpha}\longrightarrow A$ in the weak operator topology if and only if $%
A_{\alpha}x \longrightarrow Ax$ weakly. Also similarly $A_{\alpha}%
\longrightarrow A$ in the strong operator topology if and only if $%
A_{\alpha}x \longrightarrow Ax$ in the norm topology. An operator $A$ in $%
B(E)$ is said to be reflexive if 
\[
Alg Lat(A)=W(A),
\]
where $W(A)$ is the smallest subalgebra of $B(E)$ that contains $A$ and the
identity $I$ and is closed in the weak operator topology. For some source on
weighted sequence spaces, we refer the reader to [1 -- 20]. \vspace*{.3cm}

\begin{center}
{\large {\bf 2. Main Results}}
\end{center}

\vspace*{.3cm} In this section we will investigate the reflexivity of the
powers of the operator $M_{z}$ acting on $L^{p}(\beta)$. First, we note that
the multiplication operator $M_{z}$ on $L^{p}(\beta)$ $(H^{p}(\beta))$ is
unitarily equivalent to an injective bilateral (unilateral) weighted shift
and conversely, every injective bilateral (unilateral) weighted shift is
unitarily equivalent to $M_{z}$ acting on $L^{p}(\beta)$ $(H^{p}(\beta))$
for a suitable choice of $\beta$ (the proof is similar to the case p=2 that
was proved in [3]).\vspace*{.1cm}\vspace*{.1cm}

We use the following notations:\newline
\begin{eqnarray*}
r_0 &=& \overline{\lim} \beta(-n)^{-1/n}, \\
r_1 &=& \underline{\lim} \beta(n)^{1/n}, \\
\Omega_0 &=& \{z \in {\bf C : |z| > r_0\},} \\
\Omega_1 &=& \{z \in {\bf C : |z| < r_1\},} \\
\Omega &=& \Omega_0 \cap \Omega_1 .
\end{eqnarray*}
\vspace*{.1cm}\newline
From now on we consider that $M_z$ is bounded on $L^{p}(\beta)$. \vspace*{%
.3cm}\newline
{\bf Theorem. } Let $0 < r_0 < r_1=1 $ and $\frac{1}{p} + \frac{1}{q} = 1$.
If 
\[
\sum\limits_{n<0} \frac{{r_0}^{nq}} {\beta(n)^q} < \infty \qquad ; \qquad
\sum\limits_{n\geq 0} \frac{1}{\beta(n)^q} < \infty ,
\]
then $M_{z^{k}}$ is reflexive on $L^{p}(\beta)$ for all positive integers $k$%
. \vspace*{.3cm}\newline
{\bf Proof.} Let $X \in AlgLat(M_{z^{k}})$. Since $Lat(M_{z}) \subset
Lat(M_{z^{k}})$, thus $Lat(M_{z}) \subset Lat(X)$. This implies that $X \in
AlgLat(M_z)$. It is well known that $X=M_{\psi}$ for some $\psi \in
L^{p}_{\infty}(\beta)$. Now set 
\[
{\cal N}=H^{\infty}(\Omega_{1}) \bigcap L^{p}_{\infty}(\beta).
\]
Then ${\cal N}\neq \emptyset$, since $1 \in {\cal N}$. It is a closed
subspace of $L^p(\beta)$, since if $\{h_{n}\}_{n}\subset {\cal N}$ and $%
h_{n}\longrightarrow f$ in $L^p(\beta)$, then for all $n$, $\|h_{n}\|_{p}
\leq c_{2}$ for some $c_{2}>0$. Note that $\lambda \in \Omega$ is a bounded
point evaluation on $L^{p}(\beta)$ if and only if $\{\lambda ^{n}/
\beta(n)\}\in \ell ^{q}$ where $\frac{1}{p} + \frac{1}{q} = 1$. Now, since 
\[
c_{3}=\sum\limits_{n<0} \frac{{r_0}^{nq}} {\beta(n)^q} < \infty 
\]
and 
\[
c_{4}=\sum\limits_{n\geq 0} \frac{1}{\beta(n)^q} < \infty ,
\]
each point of $\Omega$ is a bounded point evaluation on $L^p(\beta)$. By
boundedness of point evaluations, for all $\lambda$ in $\Omega$ we have 
\[
h_{n}(\lambda)=<h_{n},e(\lambda)> \; \longrightarrow \;
<f,e(\lambda)>=f(\lambda).
\]
Also, for all $\lambda$ in $\Omega$, 
\begin{eqnarray*}
|h_{n}(\lambda)|&=&|<h_{n},e(\lambda)>| \\
&\leq& \|h_{n}\|_{p} \|e(\lambda)\| \\
&\leq& (c_{3}+c_{4}) \|h_{n}\|_{p},
\end{eqnarray*}
because 
\[
\sup\limits_{\lambda \in \Omega}\|e(\lambda)\| \leq c_{3}+c_{4}.
\]
Thus 
\[
\|h_{n}\|_{\Omega_{1}} = \|h_{n}\|_{\Omega}\leq c_{3} \|h_{n}\|_{p} \leq
c_{2}(c_{3}+c_{4})
\]
for all $n$. This implies that $\{h_{n}\}_{n}$ is a normal family in $%
H^{\infty}(\Omega_{1})$ and by passing to a subsequence if necessary, we may
suppose that $h_{n} \longrightarrow f$ uniformly on compact subsets of $%
\Omega_{1}$. Thus $f \in H^{\infty}(\Omega_{1})$. Note that 
\[
\|M_{h_n}\| \leq c_{1}\|h_n\|_{\Omega_{1}}\leq c_{1}c_{2}(c_{3}+c_{4})
\]
for all $n$, and ball $B({\cal H})$ is compact in the weak operator
topology. Hence $M_{h_{n}}\longrightarrow A$ in the weak operator topology
for some $A \in B({\cal H})$. Since $h_{n}(\lambda)\longrightarrow f(\lambda)
$, we see that $A=M_{f$ and so ${\cal N$is indeed a closed subspace of $%
L^{p}(\beta)$. Now clearly $N\in Lat(M_{z})$, thus $XN\subset N$. Since $1
\in N$, we get 
$$
X1=\psi \in N=H^{\infty}(\Omega_{1}) \bigcap L^{p}_{\infty}(\beta).%
$$
This implies that $M_{P_{n}(\psi)} \rightarrow M_{\psi}$in the weak operator
topology, where 
$$
P_{n}(\psi) = \sum^{n}_{k=0} (1 - \frac{k}{n+1})\hat{\psi}(k)z^{k}, \; n
\geq 0%
$$
(see [3]). For simplicity put $r_{n}=P_{n}(\psi)$and let ${M}_{k} $be the
closed linear span of the set $\{f_{nk}: n\geq 0\} $. We have 
$$
M_{z^{k}} f_{nk}=f_{(n+1)k} \in {M}_{k}%
$$
for all $n\geq 0$. Thus ${M}_{k} \in Lat(M_{z^{k}})$and so ${M}_{k} \in
Lat(M_{\psi})$. Let 
$$
\psi(z)=\sum\limits^{\infty}_{n=0}\hat{\psi}(n)z^{n}.%
$$
Since $1 \in {M}_{k}$, thus $M_{\psi} 1 = \psi \in {M}_{k}$. Hence $\hat{\psi%
}(i)=0$for all $i\neq nk$, $n\geq 0$. Now, by the particular construction of 
$r_{n}$, each $r_{n}$should be a polynomial in $z^{k}$, i.e., $%
r_{n}(z)=q_{n}(z^{k})$for some polynomial $q_{n}$. Thus 
$$
M_{r_{n}}=r_{n}(M_{z})=q_{n}(M_{z^{k}})\rightarrow X%
$$
in the weak operator topology. Hence $X \in W(M_{z^{k}})$. Thus $M_{z^{k}}$%
is reflexive and so the proof is complete. $\Box$\vspace*{.3cm}\newline
}, {\it Fixed Point Theory and Applications}, (2011), doi:
10.1186/1687-1812-2011-16. }

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