text:
\section{Introduction}

\subsection{Classical setting} Let $\mathbb{D}$ denote the unit disc in $\mathbb{C}$ and $\mathbb{T}$ the unit circle. Let $\mathcal{O}(\mathbb{D})$ denote the set of holomorphic functions on $\mathbb{D}$. For $1\leq p<\infty$, the ordinary Hardy space is defined as

\begin{displaymath}
\mathcal{H}^p(\mathbb{T}) = \left\{  f \in \mathcal{O}(\mathbb{D}) ~\text{and}~ \norm f_{\mathcal{H}^p} < \infty  \right\},
\end{displaymath}
where
\begin{displaymath}
\norm f_{\mathcal{H}^p}^{p} =  \underset{0 \leq r < 1}{\text{sup}} \int_{0}^{2\pi} \abs{f(re^{i\theta})}^p d\theta.
\end{displaymath}

It is known that (\cite{DurenHp70}) functions in $\mathcal{H}^p(\mathbb{T})$ have boundary limits almost everywhere, i.e., for almost every $\theta \in [0,2\pi]$
\begin{displaymath}
{f}(e^{i\theta}) = \underset{r \rightarrow 1^{-}}{\text{lim~}} f(re^{i\theta})
\end{displaymath}
exists.
Moreover,
\begin{displaymath}
\norm {{f}}_{L^p(\mathbb{T})} =  \norm f_{\mathcal{H}^p(\mathbb{T})}
\end{displaymath}
where $L^p(\mathbb{T})$ is defined using the standard Lebesgue measure (denoted by $d\theta$) on the unit circle.
It is also known that $\mathcal{H}^p(\mathbb{T})$ is a closed subspace of $L^p(\mathbb{T})$. In particular, for $p=2$, the orthogonal projection operator, called the Szeg\"o projection operator exists;
\begin{displaymath}
\mathbf{S}: L^2(\mathbb{T}) \longrightarrow \mathcal{H}^2(\mathbb{T}).
\end{displaymath}
The operator $\mathbf{S}$ is an integral operator with the kernel $S(z, w)$ (called the Szeg\"o kernel) and for $f\in L^2(\mathbb{T})$,
$$\mathbf{S}f(z)=\int_{\mathbb{T}}S(z,w)f(w)d\theta.$$

It follows from the general theory of reproducing kernels that for any orthonormal basis $\{e_n(z)\}_{n = 0}^{\infty}$ for $\mathcal{H}^2(\mathbb{T})$, the Szeg\"o kernel is given by
\begin{displaymath}
S(z,w)=\sum_{n=0}^{\infty}e_n(z)\overline{e_n(w)}.
\end{displaymath}

\subsection{Weighted setting}
Let $g(z)$ be a holomorphic function on $\mathbb{D}$ that is continuous on $\overline{\mathbb{D}}$ and has no zeros inside $\mathbb{D}$. We set $\mu(z)=|g(z)|^2$ and define weighted Hardy spaces and weighted Szeg\"o projections using the function $\mu(z)$ as a weight on $\mathbb{T}$.

For $1 \leq p < \infty$, we define the weighted Lebesgue and Hardy spaces with respect to $\mu$ as;
\begin{displaymath}
L^p(\mathbb{T}, \mu) = \lbrace  f \text{ measurable function on $\mathbb{D}$ and }  \norm{f}_{p,\mu} < \infty \rbrace
\end{displaymath}
where
\begin{eqnarray*}
\norm{f}_{p, \mu}^p = \int_{\mathbb{T}} \abs{f(w)}^p \mu(w) d\theta
= \int_{\mathbb{T}} \abs{f(w) (g(w))^{2/p}}^p d\theta
\end{eqnarray*}
and
\begin{displaymath}
\mathcal{H}^p(\mathbb{T}, \mu) = \lbrace f \in \mathcal{O}(\mathbb{D}) \text{ such that } \norm{f}_{\mathcal{H}^p, \mu} < \infty\rbrace
\end{displaymath}
where
\begin{displaymath}
\norm{f}_{\mathcal{H}^p, \mu}^p = \underset{0 \leq r < 1}{\text{sup}}~~ \int_0^{2\pi} \abs{f(re^{i\theta}) (g(re^{i\theta}))^{2/p}}^p d\theta.
\end{displaymath}

Note that, $\displaystyle{f \in \mathcal{H}^p(\mathbb{T}, \mu)}$ implies $\displaystyle{f(z) (g(z))^{2/p} \in \mathcal{H}^p(\mathbb{T})}$ and which in turn gives that  $\displaystyle{f(z) (g(z))^{2/p}}$ has almost everywhere boundary limits.  Hence so does  $\displaystyle{f(z)}$.  Additionally, $\displaystyle{\norm{f}_{\mathcal{H}^p, \mu} = \norm{f}_{p, \mu}}$. Furthermore, $\displaystyle{L^p(\mathbb{T}, \mu)}$  is a Banach space and $\displaystyle{\mathcal{H}^p(\mathbb{T}, \mu)}$ is a closed subspace of  $\displaystyle{L^p(\mathbb{T}, \mu)}$.

In particular, again when $p = 2$, we obtain the weighted Szeg\"o projection
\begin{displaymath}
\mathbf{S_{\mu}}: L^2(\mathbb{T}, \mu) \longrightarrow \mathcal{H}^2(\mathbb{T}, \mu).
\end{displaymath}
Following the similar theory, we note that $\mathbf{S_{\mu}}$ is an integral operator
\begin{displaymath}
\mathbf{S_{\mu}}f(z) = \int_{\mathbb{T}} S_{\mu}(z, w) f(w) \mu(w) d\theta.
\end{displaymath}
If ~~$\{f_n(z)\}_{n = 0}^{\infty}$ is an orthonormal basis for $\mathcal{H}^2(\mathbb{T}, \mu)$ then
\begin{displaymath}
S_{\mu}(z,w)=\sum_{n=0}^{\infty}f^{}_n(z)\overline{f^{}_n(w)}.
\end{displaymath}

We are interested in the action of $\mathbf{S}_{\mu}$ on $L^p(\mathbb{T},\mu)$. By definition, $\mathbf{S}_{\mu}$ is a bounded operator from $L^2(\mathbb{T},\mu)$ to $L^2(\mathbb{T},\mu)$. The problem we investigate is the boundedness of $\mathbf{S}_{\mu}$ from $L^p(\mathbb{T},\mu)$ to $L^p(\mathbb{T},\mu)$ for other values of $p\in(1,\infty)$. Note that for any given weight $\mu$ as above, we can associate an interval $I_{\mu}\subset(1,\infty)$ such that $\mathbf{S}_{\mu}$ is bounded from $L^p(\mathbb{T},\mu)$ to $L^p(\mathbb{T},\mu)$ if and only if $p\in I_{\mu}$. By definition, $2\in I_{\mu}$ and by duality and interpolation, $I_{\mu}$ is a conjugate symmetric interval around 2. Namely, if some $p_0>2$ is in $I_{\mu}$, so is $q_0$ where $\frac{1}{q_0}+\frac{1}{p_0}=1$.

In the classical setting, i.e., $\mu\equiv 1$, the Szeg\"o projection operator is bounded from $L^p(\mathbb{T})$ to $L^p(\mathbb{T})$ for any $1<p<\infty$, see \cite[page 257]{ZhuBook}.

The purpose of this note is to construct weights $\mu$ on $\mathbb{T}$ for which the corresponding interval $I_{\mu}$ can be any open interval larger than $\{2 \}$ but smaller than $(1, \infty)$.

\begin{theorem}\label{S_0bounded}
For any given $p_0>2$, there exists a weight $\mu$ on $\mathbb{T}$ such that $I_{\mu}=(q_0,p_0)$ where $\frac{1}{q_0}+\frac{1}{p_0}=1$, i.e., the weighted Szeg\"o projection $\mathbf{S}_{\mu}$ is bounded on $L^p(\mathbb{T},\mu)$ if and only if $q_0<p<p_0$.
\end{theorem}

We prove this theorem similar to the proof of the analogous statement for weighted Bergman projections in \cite{Zeytuncu13} with modifications from Bergman kernels to Szeg\"o kernels. The main ingredient is the theory of $A_p$ weights on $\mathbb{T}$.

When the weighted Szeg\"o projection $\mathbf{S}_{\mu}$ is bounded on $\displaystyle{}$ $\displaystyle{L^p(\mathbb{T}, \mu)}$ for some $p$,
one can identify the dual space of the weighted Hardy space $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \mu \right)}$.  However, when $\displaystyle{\mathbf{S}_{\mu}}$ fails to be bounded, a different approach is needed to identify the dual spaces.  In the third section, we address this issue and describe the dual spaces of weighted Hardy spaces.

The following notations are used in the rest of the note. We denote by $\displaystyle{f(z) \simeq g(z)}$ when  $\displaystyle{c \cdot g(z) \leq f(z) \leq C \cdot g(z)}$ for some positive constants $c$ and $C$ which are independent of $z$.  Similarly we denote by  $\displaystyle{f(z) \lesssim g(z)}$ when  $\displaystyle{f(z) \leq C\cdot g(z)}$ for some positive constant $C$. We use $d\theta$ for the Lebegue measure on the unit circle $\mathbb{T}$. When we integrate functions (that are also defined on the unit disc) on $\mathbb{T}$, instead of writing $e^{i\theta}$, we keep $z$ and $w$ as the variables.

\section{Proof of Theorem \ref{S_0bounded}}

\subsection{Relation between weighted kernels} The particular choice of $\mu_{}(z)$ indicates the following relation between the weighted Szeg\"o kernels ${S_{\mu}}(z,w)$ and the ordinary Szeg\"o kernel ${S}(z,w)$.

\begin{proposition}\label{WeightedNonweighted}
For $\mu(z)=|g(z)|^2$ as above, the following relation holds
\begin{equation} \label{3.5}
S\left(z, w \right) = g_{}(z)  S_{\mu}\left( z, w \right) \overline{g_{}(w)}.
\end{equation}
\end{proposition}

\begin{proof}
Let $\displaystyle{\{e_n(z)\}_{n=0}^{\infty}}$ be an orthonormal basis for $\mathcal{H}^2\left( \mathbb{T}\right)$. Since $g_{}(z)$ does not vanish inside $\mathbb{D}$, each $\frac{e_n(z)}{g_{}(z)}$ is a holomorphic function on $\mathbb{D}$ and is in $\mathcal{H}^2(\mathbb{T},\abs{g_{}}^2)$ by construction.  Following the orthonormal properties of $\displaystyle{e_n(z)}$'s we have
\begin{displaymath}
\left<\frac{e_n(z)}{g(z)_{}},\frac{e_m(z)}{g(z)_{}}\right>_{\mu}=\left<e_n(z),e_m(z)\right> = \delta_{n,m},
\end{displaymath}
where $\displaystyle{}$ $\displaystyle{\delta_{n, m}}$ is the Kronecker delta.

Also for any $f$ in $\mathcal{H}^2(\mathbb{T},\abs{g_{}}^2)$, $(f\cdot g)_{}$ is in $\mathcal{H}^2(\mathbb{T})$ and hence can be written as a linear combination of the $e_n(z)$'s.  Consequently so can $f$ using $\frac{e_n(z)}{g(z)}$'s.  Hence,$\displaystyle{\left\{{e_n}(z)/{g(z)_{}}\right\}}_{n=0}^{\infty}$ is an orthonormal basis for $\mathcal{H}^2(\mathbb{T},\abs{g_{}}^2)$.

Therefore, using the basis representation of the Szeg\"o kernels we obtain
\begin{align*}
S(z,w)&=\sum_{n=0}^{\infty}e_n(z)\overline{e_n(w)}=g_{}(z)\left(\sum_{n=0}^{\infty}\frac{e_n(z)}{g_{}(z)}\overline{\frac{e_n(w)}{g_{}(w)}}\right)\overline{g_{}(w)}\\
&=g_{}(z)S_{\mu}(z,w)\overline{g_{}(w)}.
\end{align*}
\end{proof}

\subsection{$A_p$ weights on $\mathbb{T}$}

For $p \in (1, \infty)$, a weight $\mu$ on $\mathbb{T}$ is said to be in $A_p(\mathbb{T})$ if

\begin{displaymath}
\underset{I \subset \mathbb{T}}{\underset{I}{\text{sup}}} \left(\frac{1}{\abs{I}}\int_{I} \mu(\theta) d\theta\right) \left( \frac{1}{\abs{I}}  \int_{I} \mu(\theta)^{\frac{-1}{p - 1}} d\theta\right)^{p-1} < \infty,
\end{displaymath}
where $I$ denotes intervals in $\mathbb{T}$.

These weights are used to characterize the $L^p$ regularity of the ordinary Szeg\"o projection on weighted spaces.  The following result appears in
\cite{Garnett}  and is used in \cite[2.3]{LanzaniStein04} in connection with a conformal map based approach to the investigation of the unweighted Szeg\"o projection for a general domain.

\begin{theorem}\label{Ap}
The ordinary Szeg\"o projection $\mathbf{S}$ is bounded from $\displaystyle{L^p(\mathbb{T}, \mu)}$ to $\displaystyle{L^p(\mathbb{T}, \mu)}$ if and only if $\mu \in A_p(\mathbb{T})$.
\end{theorem}

\begin{proof}

This result is an immediate consequence of the fact that the Szeg\"o kernel of the unit disc agrees with the Cauchy kernel (see \cite{KerzmanStein}) together with the classical weighted theory for the latter, see also \cite{Garnett}.

\end{proof}

The following theorem follows from Proposition \ref{3.5} and Theorem \ref{Ap}.

\begin{proposition}\label{Main1}
For $1 < p < \infty$ and $\mu(z)=|g(z)|^2$ as above the following are equivalent.
\begin{enumerate}
\item $\mathbf{S_{\mu}}$ is bounded from $\displaystyle{L^p(\mathbb{T}, \abs{g_{}}^2)}$ to    $\displaystyle{}$     $\displaystyle{L^p(\mathbb{T}, \abs{g_{}}^2)}$.
\item $\mathbf{S}$ is bounded from $\displaystyle{L^p(\mathbb{T}, \abs{g_{}}^{2 - p})}$ to    $\displaystyle{}$     $\displaystyle{L^p(\mathbb{T}, \abs{g_{}}^{2 - p})}$.
\item $\abs{g_{}}^{2 - p} \in A_p(\mathbb{T})$.
\end{enumerate}
\end{proposition}

\begin{proof}
Theorem \ref{Ap} gives the equivalence of (2) and (3).  We show the equivalence of (1) and (2).  Using the relation between the kernels from the previous proposition, we obtain the following relation between the corresponding operators:
\begin{displaymath}
g_{}(z) \left({\bf{S}}_{\mu} f\right)(z) = \left({\bf S}(f\cdot g_{})\right)(z) \qquad \text{for} ~~ f \in L^2(\mathbb{T}, \abs{g_{}}^2).
\end{displaymath}

Indeed, suppose (2) is true.  Then
\begin{eqnarray*}
\norm{{\bf S}_{\mu} f}_{p, \abs{g_{}}^2}^p &=& \int_{\mathbb{T}} \abs{({\bf S}_{\mu} f)(w)}^p \abs{g_{}(w)}^2d\theta \\
&=& \int_{\mathbb{T}}\abs{({\bf S}(f\cdot g_{}))(w)}^p \abs{g_{}(w)}^{2-p}d\theta = \norm{{\bf S}(f\cdot g_{})}^p_{p, \abs{g_{}}^{2 - p}}\\
&\lesssim& \norm{f\cdot g}^p_{p, \abs{g_{}}^{2 - p}} = \norm{f}^p_{p, \abs{g_{}}^2}
\end{eqnarray*}
which proves (1).

Now when (1) is true,
\begin{eqnarray*}
\norm{{\bf S}_{} f}_{p, \abs{g_{}}^{2 - p}}^p &=& \int_{\mathbb{T}} \abs{({\bf S}_{} f)(w)}^p \abs{g_{}(w)}^{2 - p} d\theta\\
&=& \int_{\mathbb{T}}\abs{({\bf S}_{\mu}(f/g_{}))(w)}^p \abs{g_{}(w)}^2d\theta = \norm{{\bf S}_{\mu}(f/g_{})}^p_{p, \abs{g_{}}^2}\\
&\lesssim& \norm{f/g_{}}^p_{p, \abs{g_{}}^2} = \norm{f}^p_{p, \abs{g_{}}^{2 - p}}
\end{eqnarray*}
and hence (2) is true.

\end{proof}

We can now present a family of weights that behave as claimed in Theorem \ref{S_0bounded}.

\begin{theorem}\label{Main2}
For $\alpha \geq 0 $, let
$g_{\alpha}(z) = (z - 1)^{\alpha}$ and $\mu_{\alpha}(z) =\abs{g_{\alpha}(z)}^2$. Then the weighted Szeg\"o projection operator $\mathbf{S}_{\mu_{\alpha}}$ is bounded on $L^p(\mathbb{T},\mu_{\alpha})$ if and only if $p \in \left( \frac{2\alpha + 1}{\alpha + 1}, \frac{2\alpha + 1}{\alpha} \right)$.
\end{theorem}

\begin{remark}
Theorem \ref{Main2} is a quantitative version of Theorem \ref{S_0bounded} and therefore we  obtain a proof of Theorem \ref{S_0bounded} when we prove Theorem \ref{Main2}.
\end{remark}

\begin{remark}
Note that as $\alpha\to 0^+$ the interval $\left( \frac{2\alpha + 1}{\alpha + 1}, \frac{2\alpha + 1}{\alpha} \right)$ approaches to $(1,\infty)$ and as $\alpha\to\infty$ the interval $\left( \frac{2\alpha + 1}{\alpha + 1}, \frac{2\alpha + 1}{\alpha} \right)$ approaches to $\{2\}$. Hence, any conjugate symmetric interval around $2$ can be achieved as the boundedness range of a weighted Szeg\"o projection.
\end{remark}

\begin{proof}
First note that on intervals $I$ with  $\theta = 0 \notin I$, the weight $\abs{g_{\alpha}(z)}^{2 - p}= \abs{z - 1}^{\alpha(2 - p)} \simeq C$.  Therefore, both integrals in  the $A_p(\mathbb{T})$ condition are finite and hence so is the supremum over all such intervals when $p$ is in the given range.  On intervals that contain $z = 0$ we have the following.

\subsubsection*{Step 1}
We show that for the weights $\omega(z) = \abs{g_{\alpha}(z)}^{2 - p} = \abs{z - 1}^{\alpha(2 - p)}$, the second integral in the $A_p(\mathbb{T})$ condition diverges for arcs $I = (-\epsilon, \epsilon)$ if and only if  $p$ is outside the given region.

For intervals $I = (-\epsilon, \epsilon)$ with small  $\epsilon$ and $p \leq  \frac{2\alpha + 1}{\alpha + 1}$,
\begin{eqnarray*}
 \int_I \omega(z)^{\frac{1}{1 - p}} d\theta &=& \int_{-\epsilon}^{\epsilon} \abs{e^{i\theta} - 1}^{\frac{\alpha (2 - p)}{(1 - p)}} d\theta \\
 &=& \int_{-\epsilon}^{\epsilon} \left( \sqrt{2(1 - \cos(\theta))} \right)^{\frac{\alpha (2 - p)}{(1 - p)}} d\theta \\
 &\simeq&   \int_{-\epsilon}^{\epsilon}  \left( \theta \right)^{\frac{\alpha (2 - p)}{(1 - p)}} d\theta ~=~ \infty \\
\end{eqnarray*}
because \qquad $\displaystyle{\frac{\alpha(2 - p)}{(1 - p)} \leq -1}$.
Hence $\omega \not \in A_p(\mathbb{T})$ for such $p$.

Also  when $p \geq \frac{2\alpha + 1}{\alpha}$,
\begin{eqnarray*}
 \int_I \omega(z) d\theta
 &\simeq&   \int_{-\epsilon}^{\epsilon}  \left( \theta \right)^{\alpha (2 - p)} d\theta ~=~ \infty\\
\end{eqnarray*}
because $\displaystyle{\alpha(2 - p) \leq -1}$.
Hence $\omega \not \in A_p(\mathbb{T})$ for  $p \geq \frac{2\alpha + 1}{\alpha}$ either.

The same calculations show convergence of all integrals for  $p$ in the desired range.

\subsubsection*{Step 2}

We show that for $p\in (\frac{2\alpha + 1}{\alpha + 1},\frac{2\alpha + 1}{\alpha})$ and any (general) interval $I = (\theta_0 - R, \theta_0 + R)$ with $\theta_0 \neq 0$ the integrals in the $A_p$ condition is finite.  We consider two cases.

\subsubsection*{Case 1}

$\displaystyle{I \cap \text{Arc}(0, 2R) = \emptyset}$.

On such intervals $I$,
$\displaystyle{3R < \theta_0}$ ~and so~ $\displaystyle{2\theta_0/3  \leq \theta_0 - R \leq\theta \leq \theta_0 + R \leq 4\theta_0/3}$ ~giving ~ $\theta \simeq \theta_0$.  So, $\displaystyle{\omega = \abs{z - 1}^{\alpha(2 - p)} \simeq \theta_{0}^{\alpha(2 - p)}}$

Therefore,

\begin{eqnarray*}
\frac{1}{\abs{I}} \int_I \omega(z) d\theta
 &\lesssim&  \frac{1}{2R} \int_I   \theta_0^{\alpha(2 - p)} ~d\theta ~=~  \theta_0^{\alpha(2 - p)}. \\
\end{eqnarray*}
and
\begin{eqnarray*}
\left( \frac{1}{\abs{I}} \int_I \left( \omega(z) \right)^{\frac{1}{1 - p}} d\theta \right)^{p - 1}
 &\lesssim& \left(  \frac{1}{2R} \int_I \theta_0^{\frac{\alpha(2 - p)}{1 - p}} ~d\theta \right)^{p - 1}\\
 &=& \theta_0^{-\alpha(2 - p)}
\end{eqnarray*}

Hence the supremum over all such intervals is finite

\subsubsection*{Case 2}

$\displaystyle{I \cap \text{Arc}(0, 2R) \neq \emptyset}$.

In this case,  since $\displaystyle{I \subset \text{Arc}(0, 4R)}$ and $\displaystyle{\alpha(2 - p) + 1 > 0}$ ~when~ $\displaystyle{\frac{2\alpha + 1}{\alpha} > p}$ we have,
 \begin{eqnarray*}
\frac{1}{\abs{I}} \int_I \omega(z) d\theta
 &\simeq&  \frac{1}{8R}~ 2 \int_{0}^{4R}   \theta^{\alpha(2 - p)} ~d\theta ~=~  \frac{1}{4R}~ \frac{\theta^{\alpha(2 - p) + 1}}{[\alpha(2 - p) + 1]}\Big|_{0}^{4R}  = \frac{4R^{\alpha(2 - p)}}{\alpha(2 - p) + 1}.\\
\end{eqnarray*}

Also since   $\displaystyle{\frac{\alpha(2 - p)}{1 - p} + 1 > 0}$ ~~ when ~~ $\displaystyle{\frac{2\alpha + 1}{\alpha + 1} < p}$,
 \begin{eqnarray*}
\left( \frac{1}{\abs{I}} \int_I \omega(z)^{\frac{1}{1 - p}} d\theta \right)^{p - 1}
 &\simeq& \left( \frac{1}{8R}~ 2 \int_{0}^{4R}   \theta^{\frac{\alpha(2 - p)}{1 - p}} ~d\theta \right)^{p - 1} ~=~  \left( \frac{1}{4R}~ \frac{\theta^{\frac{\alpha(2 - p)}{1 - p} + 1} }{\left[\frac{\alpha(2 - p)}{1 - p} + 1\right]} \right)^{p - 1}    \\
 &\simeq& \left(  \frac{R^{\frac{\alpha(2 - p)}{1 - p} } }{\left[\frac{\alpha(2 - p)}{1 - p} + 1\right]} \right)^{p - 1}
 = \frac{2R^{- \alpha(2 - p)}}{\left[\frac{\alpha(2 - p)}{1 - p} + 1\right]^{p - 1}}.\\
\end{eqnarray*}

Therefore, the supremum over all intervals of the type in case two are also finite and  $\omega=|g|^{2 - p}\in A^p(\mathbb{T})$ if and only if $p\in (\frac{2\alpha + 1}{\alpha + 1},\frac{2\alpha + 1}{\alpha})$.

\end{proof}

\begin{remark}
The analog of Theorem \ref{S_0bounded} for domains in $\mathbb{C}^n$ ($n\geq 2$) is an open problem. See \cite{BekolleSzego} for a partial result. Also see \cite{LanzaniStein13} for the regularity on strongly pseudoconvex domains.
\end{remark}

\section{Duality}

In this section, we investigate the duals of Hardy spaces corresponding to weights from the previous section.
For $\alpha \geq 0$ and $\displaystyle{\mu_{\alpha}(z) = \abs{z - 1}^{2\alpha}}$, a consequence of Theorem \ref{Main2} is the following.

\begin{theorem}\label{FixedWeightDualHardy}
Let $\alpha \geq 0$ and $\displaystyle{\mu_{\alpha}(z) = \abs{z - 1}^{2\alpha}}$.  For any $\displaystyle{p \in \left( \frac{2\alpha + 1}{\alpha + 1}, \frac{2\alpha + 1}{\alpha}  \right)}$, the dual space of the weighted Hardy space $\displaystyle{}$ $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha}   \right)}$ can be identified by $\displaystyle{\mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}^{2\alpha}   \right)}$ where $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$ and under the pairing
\begin{displaymath}
\left< f, h  \right> = \int_{\mathbb{T}} f(z) \overline{h(z)} \abs{z - 1}^{2\alpha} d\theta.
\end{displaymath}
\end{theorem}

\begin{proof}
This is a standard argument; however, we present a proof here for completeness.
For a given function $\displaystyle{}$ $\displaystyle{h \in \mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}^{2\alpha}  \right)}$, we define a linear functional on $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha}  \right)}$ by
\begin{displaymath}
{\bf{G}}(f) = \int_{\mathbb{T}} f(z) \overline{h(z)} \abs{z - 1}^{2\alpha} d\theta.
\end{displaymath}
It is clear that by the H\"older's inequality, ${\bf{G}}$ is a bounded functional with the operator norm less than $\displaystyle{\norm{h}_{\mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right)}}$.

For the opposite direction, let ${\bf{G}}$ be a bounded linear functional on $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right)}$.  By the Hahn-Banach theorem, ${\bf{G}}$ extends to a bounded linear functional on $\displaystyle{L^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right)}$.
Now using the duality of $L^p$ spaces, we find a function $\displaystyle{h \in L^q\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right)}$ such that
\begin{displaymath}
{\bf{G}}(f) = \int_{\mathbb{T}} f(z) \overline{h(z)} \abs{z - 1}^{2\alpha} dz \qquad \text{for} \qquad f \in L^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right).
\end{displaymath}
When we restrict ${\bf{G}}$ onto $\displaystyle{L^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha}  \right) \cap \mathcal{H}^2\left( \mathbb{T}, \abs{z - 1}^{2\alpha}  \right)}$
and use self-adjointness of $\displaystyle{{\bf{S}}_{\mu_{\alpha}}}$ we get the following.
\begin{eqnarray*}
{\bf{G}}(f) &=& \int_{\mathbb{T}} f(z)~\overline{h(z)} \abs{z - 1}^{2\alpha} d\theta\\
&=& \int_{\mathbb{T}} ({\bf{S}}_{\mu_{\alpha}}f)(z) ~\overline{h(z)} \abs{z - 1}^{2\alpha}d\theta\\
&=& \int_{\mathbb{T}} f(z) ~\overline{({\bf{S}}_{\mu_{\alpha}}h)(z)} \abs{z - 1}^{2\alpha} d\theta
\end{eqnarray*}
 for $\displaystyle{f \in L^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha}  \right) \cap \mathcal{H}^2\left( \mathbb{T}, \abs{z - 1}^{2\alpha}  \right)}$.

Since the intersection of these two spaces is dense in  $\displaystyle{}$   $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right)}$, we note that ${\bf{G}}$ is represented by the function  $\displaystyle{({\bf{S}}_{\mu_{\alpha}}h)(z)}$ and  $\displaystyle{{\bf{S}}_{\mu_{\alpha}}h \in \mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right) }$ by Theorem \ref{Main2}.
\end{proof}

A natural question arises after this statement.  How can we identify the dual space of the weighted Hardy space, $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}^{2\alpha} \right)}$, when $\displaystyle{p \notin \left( \frac{2\alpha + 1}{\alpha + 1} , \frac{2\alpha + 1}{\alpha} \right)}$?
The answer to this question follows from the following result on the boundedness of the weighted Szeg\"o projection, $\displaystyle{{\bf{S}}_{\mu_{\alpha}}}$.
Similar results for weighted Bergman projections have been presented recently in \cite{ArroussiPau13} and \cite{ConstantinPelaez13}.

\begin{proposition}\label{ChangeScaleAndMeasure}
Let $\displaystyle{}$ $\displaystyle{\alpha \geq 0}$ and $\displaystyle{\mu_{\alpha} = \abs{z - 1}^{2\alpha}}$.  For any $\displaystyle{1 < p < \infty}$, the weighted Szeg\"o  projection $\displaystyle{{\bf{S}}_{\mu_{\alpha}}}$ is bounded on $\displaystyle{L^p\left( \mathbb{T}, \abs{z - 1}^{\alpha p} \right)}$.
\end{proposition}

\begin{remark}
Note that as $p$ varies, changes occur not only the integrability scale but also in the measure.
\end{remark}

\begin{proof}
The proof follows from the relation between the kernels in Proposition \ref{WeightedNonweighted} and the fact that the unweighted Szeg\"o projection $\textbf{S}$ is bounded on $\displaystyle{L^p(\mathbb{T})}$ for $1<p<\infty$.

Let us take $\displaystyle{f(z) \in L^p\left( \mathbb{T}, \abs{z - 1}^{\alpha p}  \right)}$ and set $$\widetilde{f}(z)=f(z)  \frac{\abs{z - 1}^{2\alpha}}{(z-1)^{\alpha}},$$ then we have $\widetilde{f} \in L^p(\mathbb{T})$. Using this notation, we notice
\begin{eqnarray*}
{\bf{S}}_{\mu_{\alpha}}f (z)&=& \int_{\mathbb{T}} S_{\mu_{\alpha}}(z, w) f(w)\abs{w - 1}^{2\alpha}~ d\theta \\
&=& \frac{(z - 1)^{\alpha}}{(z - 1)^{\alpha}} ~\int_{\mathbb{T}} S_{\mu_{\alpha}}(z, w)  (w-1)^{\alpha} f(w)\frac{\abs{w - 1}^{2\alpha}}{(w-1)^{\alpha}}  ~ d\theta\\
&=& \frac{1}{(z - 1)^{\alpha}} ~\int_{\mathbb{T}} S(z, w) \widetilde{f}(w) ~ d\theta\\
&=& \frac{1}{(z - 1)^{\alpha}} \textbf{S}\left(\widetilde{f}(w) \right)(z),
\end{eqnarray*}
where we invoke Proposition  \ref{WeightedNonweighted} when we pass from the second to the third line. Next by using the fact that the unweighted Szeg\"o projection operator $\textbf{S}$ is bounded on $\displaystyle{L^p(\mathbb{T})}$, we obtain the following.
\begin{eqnarray*}
\norm{{\bf{S}}_{\mu_{\alpha}} f}_{L^p\left( \mathbb{T}, \abs{z - 1}^{\alpha p}  \right)} &=& \int_{\mathbb{T}} \abs{z - 1}^{\alpha p} \frac{1}{\abs{z - 1}^{\alpha p}} ~~ \left\vert
\textbf{S}\left(\widetilde{f}(w) \right)(z)
 \right\vert^p d\theta\\
&=& \norm{
\textbf{S}\left(\widetilde{f}(w) \right)
}^p_{L^p(\mathbb{T})}\lesssim\norm{
\widetilde{f}(w)
}^p_{L^p(\mathbb{T})}\\ &=& \norm{ f(w) \frac{\abs{w - 1}^{2\alpha}}{(w-1)^{\alpha}}}^p_{L^p(\mathbb{T})}\\
&=& \norm{f}^p_{L^p\left( \mathbb{T}, \abs{z - 1}^{\alpha p}  \right)}.
\end{eqnarray*}
This finishes the proof of the proposition.
\end{proof}

Now we can answer the duality question by using Proposition \ref{ChangeScaleAndMeasure}.  Following the same argument in the proof of Theorem \ref{FixedWeightDualHardy} we obtain the following statement.

\begin{theorem}\label{Main3}
Let $\displaystyle{}$  $\displaystyle{\alpha \geq 0}$ and $\displaystyle{\mu_{\alpha} = \abs{z - 1}^{2\alpha}}$.  Then for any $\displaystyle{p \in (1, \infty)}$, the dual space of the weighted Hardy space $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}^{\alpha p}  \right)}$ can be identified by $\displaystyle{\mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}^{\alpha q} \right)}$ where $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$, under the pairing
\begin{displaymath}
\left<  f, h \right> = \int_{\mathbb{T}} f(z) \overline{h(z)} \abs{z - 1}^{2\alpha} ~ d\theta.
\end{displaymath}
\end{theorem}

At first, the two duality results in Theorem \ref{FixedWeightDualHardy} and Theorem \ref{Main3} may seem confusing for
$\displaystyle{p \in \left(\frac{2\alpha + 1}{\alpha + 1}, \frac{2\alpha + 1}{\alpha}   \right)}$.  However, the main point is to note the difference in the exponents of the weights and the way the pairing is defined.  We illustrate these two results in the following example.

\begin{example}
Let us take $\alpha = 1/2$.  Then $\displaystyle{{\bf{S}}_{\abs{z - 1}}}$ is bounded on $\displaystyle{L^p\left( \mathbb{T}, \abs{z - 1}  \right)}$ for $\displaystyle{p \in (4/3, 4)}$.  In particular, for any $\displaystyle{p \in (4/3, 4)}$, the dual space of $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}  \right)}$ can be identified by $\displaystyle{\mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}  \right)}$ where $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$, under the pairing
\begin{displaymath}
\left< f, h \right>_{\abs{z - 1}} = \int_{\mathbb{T}} f(z)~ \overline{h(z)}~ \abs{z - 1} ~ d\theta.
\end{displaymath}

On the other hand, using the second duality result for any $p > 1$, the dual space of $\displaystyle{\mathcal{H}^p\left( \mathbb{T}, \abs{z - 1}  \right)}$ can be identified by $\displaystyle{\mathcal{H}^q\left( \mathbb{T}, \abs{z - 1}^{q/p} \right)}$ when $\displaystyle{\frac{1}{p} + \frac{1}{q} = 1}$, under the pairing
\begin{displaymath}
\left< f, h \right>_{\abs{z - 1}^{2/p}} = \int_{\mathbb{T}} f(z)~ \overline{h(z)}~ \abs{z - 1}^{2/p} ~ d\theta.
\end{displaymath}
\end{example}

\section{Acknowledgment}

We thank the anonymous referee for constructive comments on the proofs of Proposition 3 and Theorem 5 and also for the useful editorial remarks on the exposition of the article. This study started when the first author visited the second author at Texas A\&M University, we thank the Department of Mathematics for the support.

redpajama_set_name: RedPajamaArXiv


Oxbow's Cavy Cuisine - Adult Guinea Pig Food is made from timothy grass meal, which includes the high fiber adult guinea pigs need. However, it contains a lower level of protein, calories and calcium than alfalfa-based pellets. Adult animals' digestive systems do not require the same level as that of younger animals. Cavy Cuisine - Adult Guinea Pig contains stabilized Vitamin C and an optimal calcium-to-phosphorus ratio that helps maintain a healthy body. When fed with Oxbow grass hays and treats, Cavy Cuisine - Adult Guinea Pig provides complete stabilized nutrition. Sizes: 5 lb, 10 lb, 50 lb.
The Toughest on the Market and a Wise Investment for Serious Professionals and Pet Enthusiasts. Durable construction. Includes eight 2 ft-wide panels totaling 16 feet in length. All pens have a zinc electroplate finish, added after welding, to reinforce protection and increase the pen's value.
Four Paws old "drop-pin" standard crates have been replaced by these new, single door deluxe crates. This lightweight, fold and carry series contains 6 size of crates.
Constructed of top quality single-ply nylon in four stylish colors.
Pet Lodge Exercise Pen 30"
Double door wire crate. Secure latches. Rounded corners. Crate sets up in seconds and folds down for storage. Perfect for home, training and travel. Includes removable, easy to clean. Miller Manufacturing.
Fresh 'n Clean Scented Shampoo cleans away dirt and leaves a long-lasting original fresh and clean floral fragrance that clings to the hair for up to two weeksAloe vera soothes the skin and extra conditioners make the coat shiny, manageable and free of mats and tangles 18 oz.


On the corner of High and Federal Streets stands a reproduction of the Grellet house, built by area lawyer Louis Colaguori. The original house stood on the corner of High and Library Streets, and was the home of Etienne de Grellet du Mabillier, better known as Stephen Grellet.
The house later served as a boarding house for some years, its most notable tenant being Miss Kitty Balester, a student at nearby Van Rensselaer Seminary, who went on to marry author Rudyard Kipling. In the mid-twentieth century, the house was moved a short distance down High Street. Only one wall of the original house remains, serving as the north wall of a building which once housed the local offices of Public Service Electric and Gas. Those offices are now housed in the new Grellet house, along with the Colaguori law offices.


BitMEX has implemented several different order types to assist with users' unique trading and hedging strategies. This section outlines the various order types available with some examples.... How to place an Order on BitMEX.com After your deposit is confirmed by the Bitcoin network, you arrange the trading platform widgets according to your personal preferences and make chart analysis with the tools available; finally you will be able to actually place orders on the BitMEX.com trading platform.
Sierra Chart Engineering - Posts: 67606 : These problems are going to be on the BitMEX side. There really is not much we can do. We are working on direct integration to BitMEX and when that is done, we expect these problems to be resolved or substantially less.... BitMEX like most crypto-exchanges provides a vertically arranged order book, where traders can quickly switch between the different products, and place their order. Apart from that, the platform has an excellent visual appeal.
You have found the BitMEX advanced trading features tutorial, so please have a look around. Here you will find all the info you need to take your skills to the next level. Click here for all the details.... BitMEX like most crypto-exchanges provides a vertically arranged order book, where traders can quickly switch between the different products, and place their order. Apart from that, the platform has an excellent visual appeal.
Sierra Chart Engineering - Posts: 67606 : These problems are going to be on the BitMEX side. There really is not much we can do. We are working on direct integration to BitMEX and when that is done, we expect these problems to be resolved or substantially less.


Get email updates about Penny Zamora delivered directly to your inbox.
So saddened by the news of Penny's passing....Penny always had a smile to brighten a day. She truly was a genuine soul filled with gratitude for other's accomplishments.
Penny always made me feel welcome . What a kind soul!!! I will be surrounding the family with comforting love light prayers.
Until we meet again, Penny!!!
To Rick, the Zamora and Davies family . My heart goes out to you, we all know she was a special one.
Rick, I wish I could just bear hug you right now. I cant image what you are going through, but please know she would want you to lean on others and let people help.
I was sad to hear of your passing. It was my pleasure to get to know you when you and Rick got married. I have missed the friendship we have made since you moved away. But I'm sure your in a much happier place. May God be with you and your family and we will see you soon.
Never miss an update about Penny. Sign up today.


To satisfy various metalworking requirement, King Kiang Li Machinery Industrial Co., Ltd. builds new CNC Machine Four Hard Boxway SV-1250 currently. The President of King Kiang Li Machinery Industrial Co., Ltd., Mr. Chen Chung HUANG said "The major structure of SV-1250 is made of Meehanite cast iron with high rigidity, and the machine housing is designed to Euroscale. "The column is constructed in a 米 shape to achieve strong rib reinforcement, the saddle structure traverses 4 axes, and four hard boxways is applied in SV-1250.", he continuously said. Besides, an A-shape box Base is provided to stabilize the machine, and all inspections are made based on German Standard VDI 3441.


Manufacturing can involve complex processes that require specialist insurance cover.
To protect your business and employees, we provide an insurance solution that reflects the individuality of your business.
Our experienced risk management team acts for some of the UK's most successful manufacturing businesses. From SME package covers to large corporate portfolios, we tailor insurance protection to reduce risk exposure and provide our clients with peace of mind. Speak to one of our professional and friendly experts to see how we can help you.
Speak to our experienced insurance broker who specialises in the manufacturing sector, or get in touch via the enquiry form below.


text:
The Story Behind the Song Part 2: Wouldn't Change a Single Thing
In 1975 I left school. I'd been to a place where if you didn't work they beat the shit out of you. That, combined with a bit of ability, meant that, having spent hours smoking in a local cafe, playing pinball and listening to Slade, I still left school with some average to good qualifications. Just about enough to enable me to apply for Cambridge University, one of the best colleges in the world where it was generally considered that graduates were made for life.
So I left the council house in Canterbury I'd grown up in with my mum, to move to London to live with my Dad in the embarrassingly named and therefore to be lied about on all occasions: Ponsonby Terrace. The idea was that I would go to a college which specialised in getting oddballs into Cambridge. But I was sick to the back teeth with school and did no work whatsoever. They didn't beat the crap out of you there and I duly failed the entrance exam.
Needing some money, I joined old school mate, Jack Black, working in a T shirt printing factory where the boss was a Rod Stewart look a like northerner called John Splain. The day generally started with a cup of tea and a joint and carried on with more joints while listening to the rare decent music on offer in 1975 (mainly the Faces who our leader had modelled himself on).
Our trustworthy boss spent most of the day bunking off, hiding upstairs on a mezzanine floor where he could throw things down onto mine and Jack's heads, thinking we didn't know what he was up to even though he was giggling uncontrollably at the hilarity of his actions.
One day, on hearing that Jack played drums and I played bass, the to be renamed Honest (on account of his oft used phrase "I did, -Honest!" while telling porkies) John Plain invited us over to "jam" with his mates Casino Steel and Matt Dangerfield. They were forming a band to be called "The Boys" who were going to be the next Beatles, if not bigger. An antidote of short, catchy songs to blow away all the rubbish infecting the airwaves and concert halls at the time.
And so I arrived, 18 years old, at 47a Warrington Crescent. There should be a blue plaque there. Inside this damp basement, mould growing on the kitchen wall, was a tiny, four track studio. The electricity was hooked up to a lamp post outside, bypassing the meter. Various intimidating (for this youngster) "adults" (none older than about 26) were taking turns to play the intro bars to Slow Death by the Flamin' Groovies for about four hours at a time. One of them was a curly haired, confident local called Mick Jones. A good looking fella called Billy Idol had made the long way over from Bromley, a quiet Brian James lurked along with various others who would coalesce over the coming months into various bands.
1976 was a hot summer. I remember it being the USA's bicentennial and a bunch of Americans held a large party in the communal garden to the back of 47a. All the nascent punks gatecrashed this feast of free booze and burgers and I ended up passed out on the grass dressed from head to toe in white which by 4 am, when I woke, was various shades of green.
But the summer ended and a deadline I had been pushing to the back of my mind approached. In September I had a place to study Chemical Engineering at University College London. Not Cambridge, but a very good university all the same.
So I left the job printing T shirts and went to register for my course. At the end of the first day I went back home and told my father I had made a decision. With all of my 18 years of worldly experience I had given University a look and didn't like it. I wasn't going back because The Boys were going to be a big band and being a punk was what I was going to do.
My Dad was delighted (Ummmm – no he wasn't).
So, off I trotted to the dole office to see if I could get any money to live on. They decided that, as I had left my job voluntarily and was living at home, they would give me the princely sum of £5.50 per week. Not unreasonably, and as an incentive to get a job or go back to college, my father decided to charge me £5 per week rent. So I had 50p per week to myself and had to get to Warrington Crescent every day to rehearse with The Boys in that little studio, as well as play cards, drink tea, and see if anyone was generous enough to buy me the odd pint in the pub.
So, no choice, I had to walk there and back every day, a round trip of almost three hours as the following Google Maps screen shot shows (it appears that 47 Warrington Crescent is called Venetian House now that all the old Italians, squatters and church property renters have been moved out to make way for the rich of London).
Through sun, wind and rain, in the morning and in the dead of night, a young man, head full of dreams of fame, fortune and women, looking like a star but still on the dole (to paraphrase Ian Hunter) would make his trek to eventual, hoped for success, past some of London's most famous landmarks and richest neighbourhoods, with not a penny in his pocket.
Sometimes I tried hitching but had to ask to be set down immediately as most of the rides were given by old men who wanted something in return I wasn't inclined to give. Boy I was innocent.
One day on my trek up the Edgware Road Mick Jones came running up behind me, ruffled my hair and shouted "Hello Dunc, Alright?" as he ran on and jumped on the number 6 bus which would pass near Warrington Crescent, the place I was headed to. The Clash had just signed to CBS and so he was on a weekly wage. "You rich bastard", I thought, "Being able to afford a bus" as I trudged on.
The young author far left: Dream realised?
But good comes from everything, and forty years later, thinking back to those days of penury a song emerged. A song of hope and dreams.
You can hear it here:
If you don't have Spotify listen here on Bandcamp: Click here
I hope you like it. I'm particularly proud of the last chorus with its answer vocals. And with the guitar riff, if Led Zeppelin played power pop this is how I imagine it would sound!
Posted on August 30, 2017 February 21, 2018 by Duncan Reid and the Big HeadsPosted in SongsTagged Bombs Away, Duncan Reid and the Big Heads, Pop Punk, Power Pop, Punk.
Previous Previous post: That Puts it in Context
Next Next post: The Story Behind the Song Part 3: That's Just the Way it is

redpajama_set_name: RedPajamaCommonCrawl


Apple enthusiasts can expect a 2019 iPhone triple camera to come their way as we shared some days ago. This will be for the highest model, while the iPhone XR successor will have a dual camera, early insider reports say.
The Wall Street Journal reports that Apple's 2019 plans include a triple rear camera on the highest model and a dual camera on the iPhone XR successor. According to the WSJ, this bit of news is from "people familiar with the matter".
This simply means that we can expect that the triple camera will be on the "iPhone 11 Plus" model. In any case, we do not know the official names Apple will use for their smartphone models this year.
Also, this is unconfirmed news, but then 2019 is shaping up to be a serious year of competition in the camera phone space. Already, it appears that 48-megapixel cameras will be a thing this year. We have seen a handful of announcements already – Honor V20, HiSense U30, Huawei Nova 4, UMIDIGI S3 Pro, and Redmi Note 7.
And 2019 has only just started. A number of the announced 48 MP camera phones have a triple camera setup. Will we see a 48 megapixel lens on the 2019 iPhone triple camera?
Is 2019 Going To Bring iPhone Triple Camera With A 48MP Lens?
Should we be ready to see a 48MP camera from Apple? If past trends are anything to go by, that is not likely. Apple tends to avoid the megapixels numbers game. A triple camera on an iPhone will likely not hit the 48MP spot.
A 2019 iPhone triple camera is more likely to include a wide-angle lens though, as this has been rumoured to be a feature that Apple is working on.
It is a long wait till the official announcements, but we are sure to be treated to plenty of links between now and then.


text:
June 26, 2018 February 25, 2019 Muhammad liaqat1 Comment on Top 10 Famous Chinese Actresses In Hollywood
Top 10 Famous Chinese Actresses In Hollywood.
Hello friends how are you all? Today we are going to talk about the Top 10 Famous Chinese Actresses In Hollywood. There are many Chinese actresses in Hollywood but we choose some of them. Everybody does not agree with this list please comments and give your precious feedback to us. So friends here is the list of the Top 10 Famous Chinese Actresses In Hollywood.
10.Zhang Ziyi.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Zhang Ziyi is at no 10. She is the 10th Famous Chinese Actresses In Hollywood and also on our list. She is a Chinese actress and model with international fame. Zhang Ziyi was born on 9 February 1979 in Beijing, China and now she is 39 years old. She considered one of the best-known Asian actresses in the West. She became famous in the Western world for her leading role in Ann Lee's martial arts film Crouching Tiger and Hidden Dragon. Zhang works in many movies like The Grandmaster, Memoirs of a Geisha, House of Flying Daggers, and more. She is one of the famous Chinese actresses in Hollywood.
9.Lucy Alexis Liu.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Lucy Alexis Liu is at no 9. She is the 9th Famous Chinese Actresses In Hollywood and also on our list. She is an American Chinese actress, voice actress, and director. Lucy Alexis was born on 2 December 1968 in Jackson Heights, New York City, New York, United States and now she is 49 years old. She is a producer, singer, and artist. She has American nationality. Her parents are Tom Liu and Cecilia Liu. She works in the movies names Elementary, Charlie's Angels, Kill Bill Volume 1, and more. She is one of the famous Chinese actresses in Hollywood.
8.Michelle Yeoh.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Michelle Yeoh is at no 8. She is the 8th Famous Chinese Actresses In Hollywood and also on our list. She is a Hong Kong-based Malaysian Chinese actress. Tan Sri Michelle Yeoh Choo-Kheng is the full name of Michelle Yeoh. She was born on 6 August 1962 in Ipoh, Malaysia and now she is 55 years old. She appeared in the movies like Tomorrow Never Dies, Reign of Assassins, The Lady, and more. Her upcoming movie is Crazy Rich Asians. Her TV shows are Star Trek Discovery and Marco Polo. She is one of the famous Chinese actresses in Hollywood.
7.Vivian Wu.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Vivian Wu is at no 7. She is the 7th Famous Chinese Actresses In Hollywood and also on our list. She is a Chinese actress with international fame. Wu Junmei is the other name of Vivian Wu. She was born on 5 February 1966 in Shanghai, China and now she is 52 years old. She got her education from the Hawaii Pacific University. Her parents are Zhu Manfang and Wu Cheng Ye. She works in the movies like The Pillow Book, The Soong Sisters, The Last Emperor, and more. Her TV shows are A Bright Shining Lie, Vanishing Son, Strange World, and more. She is one of the famous Chinese actresses in Hollywood.
6.Ling Bai.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Ling Bai is at no 6. She is the 6th Famous Chinese Actress In Hollywood and also on our list. She is a Chinese-American actress. Ling Bai was born on 10 October 1966 in Chengdu, China and now she is 51 years old. She works in many movies like Crank High Voltage, Red Corne, The Crow, and more. She won the Hong Kong Film Award for Best Supporting Actress and more. Her TV shows are The Monkey King, Celebrity Rehab with Dr. Drew, and But Can They Sing?. She is one of the famous Chinese actresses in Hollywood.
5.Ming-Na Wen.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Ming-Na Wen is at no 5. She is the 5th Famous Chinese Actresses In Hollywood and also on our list. She is a Chinese-American actress and voice actress. Ming-Na was born on 20 November 1963 in Coloane, Coloane Village, Macau and now she is 54 years old. She has the American, Chinese, and Macanese nationality. She works in many movies and TV series like Agents of S.H.I.E.L.D, Stargate Universe, Mulan, and more. Ming-Na is one of the famous Chinese actresses in Hollywood.
4.Chen Chong.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Chen Chong is at no 4. She is the 4th Famous Chinese Actress In Hollywood and also on our list. She is a Chinese American actress and film director. Chen Chong was born on 26 April 1961 in Shanghai, China and now she is 57 years old. She is also a screenwriter and film producer. She got her education at the Shanghai Film Academy. Her parents are Zhang Anzhong and Chen Xingrong. She works in many movies and TV shows names The Last Emperor, Marco Polo, Lust Caution, and more. She is one of the famous Chinese actresses in Hollywood.
3.Tsai Chin.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Tsai Chin is at no 3. She is the 3rd Famous Chinese Actresses In Hollywood and also on our list. She is a Chinese-born British actress and director. Tsai Chin was born on 30 November 1936 in Tianjin, China and now she is 81 years old. She is also a teacher and author. She wrote the book Daughter of Shanghai. Her parents are Zhou Xinfang and Ju Lilian. She works in the movies like Now You See Me 2, The Joy Luck Club, You Only Live Twice, and more. She is one of the famous Chinese actresses in Hollywood.
2.Nancy Kwan.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Nancy Kwan is at no 2. She is the 2nd Famous Chinese Actress In Hollywood and also on our list. She is a Hong Kong-born American actress. Nancy Ka Shen Kwan is the full name of Nancy Kwan. She was born on 19 May 1939 in Hong Kong and now she is 78 years old. She is the daughter of Marquita Scott and Kwan Wing Hong. She works in many movies names The World of Suzie Wong, Dragon The Bruce Lee Story, Flower Drum Song, and more. Nancy Kwan is one of the famous Chinese actresses in Hollywood.
1.Anna May Wong.
In the list of the Top 10 Famous Chinese Actresses In Hollywood, Anna May Wong is at no 1. She is the Famous Chinese Actress In Hollywood and also on our list. She was a Chinese American actress. Anna May was born on 3 January 1905, Los Angeles, California, United States and she has died on 3 February 1961, Santa Monica, California, United States. She won many awards like Hollywood Walk of Fame – Motion Picture 1700 Vine Street and more. She works in many movies like Shanghai Express, The Thief of Bagdad, The Toll of the Sea, and more. Anna May was one of the famous Chinese actresses in Hollywood.
Top 10 Most Tallest Models In The World
October 28, 2017 June 1, 2019 Arina
Top 10 Most Beautiful Hottest Chinese Actresses In The World
December 21, 2017 February 25, 2019 Arina
Top 10 Richest Bollywood Actresses Of 2018
January 19, 2018 February 25, 2019 Arina
1 thought on "Top 10 Famous Chinese Actresses In Hollywood"
Pingback: Top 10 Hottest Male Korean Singers In The World 2018


text:
Conservative conference 2009
This article is more than 10 years old
Cameron defends alliance with controversial Polish party
• Tory leader says Law and Justice party is not homophobic
Deborah Summers and Allegra Stratton
Wed 7 Oct 2009 10.26 EDT First published on Wed 7 Oct 2009 10.26 EDT
Michał Kamiński, centre, speaks at the European Conservatives and Reformists Group fringe meeting at the Conservative party conference. Photograph: Martin Argles
David Cameron has today defended the Conservative alliance with a controversial Polish party in the European parliament, insisting it was not homophobic.
In an interview with Sky News, the Tory leader denied that the Law and Justice party – which opposes gay marriage – was prejudiced towards gay people.
Asked how he would respond to a letter from celebrities including Stephen Fry, which called on him to oppose "vile rhetoric" about gay rights from the Law and Justice party, Cameron said: "I'll respond to the letter very clearly.
"I don't believe the Law and Justice party are homophobic. Poland is a Catholic country. Most of the parties in Poland do take a stance like that on issues like gay marriage.
"I don't happen to agree with them, but they're not a homophobic party."
Last nigh, Ben Summerskill, the chief executive of the gay rights group Stonewall, pulled out of a gay pride event at the Tory conference in protest at appearances by Michał Kamiński, the head of the Law and Justice party and the leader of the Tories' new bloc in the European parliament, and Roberts Zile, the leader of the Latvian For Fatherland and Freedom party, another member of the new group.
"There is no doubt the progress that has been made in the last couple of years has genuinely been historic," Summerskill told Channel 4 News, referring to Tory attitudes to gay rights. "But the event tonight has been overshadowed by the presence, not just at conference but on the same platform as some senior members of the party, of people of such extreme and offensive views."
Cameron's alliance came under renewed pressure as the representative body of British Jews wrote to the Tory leader raising questions about their partnership.
A letter was sent from Vivian Wineman, the president of the Board of Deputies of British Jews, in a sign of the Jewish community's concerns over Cameron's pact with the Polish and Latvian parties.
In response, the Tories appeared dismissive about the discontent, saying in a statement that the board had been swayed by "politically motivated allegations made by the Labour party".
Kamiński and Zile both appeared and defended their political careers at a fringe meeting at the Tory conference in Manchester yesterday.
The pair received only gentle questioning, and Kamiński had been expected to speak at a late afternoon event held by the Conservative Friends of Israel, but did not appear. He did, however, earlier attend lunch with the group, where the Israeli ambassador to the UK, Ron Prosser, was present.
Last week, the board said it had looked into Kamiński and found nothing objectionable in his background, but its decision to write to Cameron suggests misgivings had become substantial.
The foreign secretary, David Miliband, said at the Labour party conference that Kamiński had an "antisemitic past" and highlighted the fact that members of Zile's party take part in annual celebrations of a Latvian unit of the Waffen-SS.
Zile said: "What has surprised me on this issue was I never thought this would come from a western democratic party. I would expect it to come from Moscow or the Kremlin, as it does from time to time.
"What happened was during the second world war [Latvian] people were often conscripted against their will to fight for both sides, Germans and Russians. Once a year these people commemorate their war dead, the people they fought alongside. They don't in any way commemorate Hitler or the Nazi regime. We would never do business with anyone who we thought glorified the SS."
The shadow minster for Europe, Mark Francois, who appeared on the same platform as Kamiński at a fringe meeting yesterday, said: "This is a slur which comes from the Soviet era, that was thrown by the Soviet authorities at the Latvian people. They in no way whatsoever commemorate anything to do with Hitler or the Nazi regime. The Latvian ambassador in London has chided the foreign secretary for using this tack, which basically dates from the days of Soviet propaganda."
Francois added: "All the parties in the group are signatories to the Prague protocol and we are perfectly happy with all of our partners that they believe in a liberal society and full respect for human rights."
Although the fringe event was carefully stage-managed – terse political lines trotted out and limited time for questions – there was one unfortunate mistake. The basement room in which delegates gathered to hear the controversial Tory allies was in Manchester's Midland Hotel, a building Hitler is said to have liked so much that he would have made it his northern residence if he had invaded.
Earlier in the day, comedians Eddie Izzard and Stephen Fry, both Labour supporters, published a letter condemning the Tories' alliance. The views of the Poles and Latvians were, they said, at odds with the Conservatives' new social liberalism.
A Tory spokesman said: "We're concerned that the politically motivated allegations made by the Labour party and their allies have been noted by the Board of Deputies.
"All these allegations have repeatedly been shown to be false. We will talk to the Board of Deputies about the lack of evidence behind them and their partisan nature. People should take account of the fact that the Polish chief rabbi has said that his remarks have been misrepresented, that the Latvian foreign minister has called David Miliband about his remarks, that the Latvian ambassador in London has spoken to the Foreign Office about the matter."
Conservative conference
Labour accused of misrepresenting views of Tories' new European allies
Tory frontbencher Nick Herbert says his party should not be judged by the partners it chooses in Europe, and accuses Labour of using gay rights as a 'political pawn'
Do the Tories know the truth about Michal Kaminski?
David Cameron's EU alliance questioned by Jewish group
Freedom party hits back at David Miliband claims of neo-Nazism
Once no self-respecting politician would have gone near people such as Kaminski


Bandwan College Admission 2018, Bandwan Mahavidyalaya Admission 2018 Merit List Download. Bandwan Mahavidyalaya (Purulia) online application for admission to BA (Honours & General) Pass Course for the academic session 2018-2019 through online www.bandwanmahavidyalaya.in. Admission in BA Honours course in Bengali, History, Sanskrit, Santali. Bandwan Mahavidylaya Admission 2018 Online Application & Merit List available on www.bandwanmahavidyalaya.in. Bandwan Mahavidyalaya Admission 2018 Eligibility Criteria, How to apply process, Selection process, Merit list, Waiting list, Provisional merit list is given bellow.
Bandwan Mahavidyalaya is affiliated to university of Sidho-Kanho-Birsa. Bandwan Mahavidyalaya situated in the place of Jungle Mahal area in the District of Purulia. Bandwan Mahavidyalaya has been established in the year 2010 with introduction of three year degree courses in Arts people of the locality donated more than 5 acrs.
BA Honours in - Bengali, History, Sanskrit, Santali.
Reservation of seats SC, ST, OBC A, OBC B and Physically Challenged candidates will be accordance to the existing rules of the Government of West Bengal. A certificate from the competent authority must be produced during counselling for claiming such reservation. No application will be allowed to take advantage of more than one quota of reservation. Candidates having all cast & category certificate issued by the States other than West Bengal will be treated as General Category.
The Merit list of valid application will be published online website of college on date as per the published admission scheduled, the Merit list lists will be prepared in the descending order of merit index. In case of a tie,precedence will be given to higher subject marks, wherever applicable.
Bandwan Mahavidyalaya Admission 2018 Online Application Form are available after announced HS Result 2018.
Regular visit for more updates of - Bandawn College Merit List 2018, Bandawan Mahavidyalaya 1st, 2nd, 3rd Merit List 2018.