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Having finally found a reliable Care Agency, I am still so dismayed at the actual care system. They are all (well mostly) lovely ladies and come in smiling and actually turn up on time. They are usually very efficient, speedy and do all they need to do, but they just don't talk. I can understand why my mum feels so isolated and lonely.
The carer arrives, says hello, goes about her business with a flurry, then departs. Absolutely no communication in the meantime, unless they are talking to someone on their phone. So the elderly person just sits there all day on their own virtually isolated from the rest of the world. Yes their needs may be met, if they are lucky, but a few minutes talking about their day, the weather outside or even asking how they are would go such a long, long way.
Age UK provide a befriending service. We did try it a couple of times and it was really good, but the people tend not to stay around.
Again, having mum come to live with me has brought home the harsh reality of actually living on your own when you are housebound – just sitting in front of the tv all day and having no friends still alive or anyone to chat to. The evenings are the worse.
My mum is lucky in that she does have one friend who pops in once a week and her sister visits, plus me now, but what about all those other people who don't have this. There was a 'running group' I heard about once, where runners would pop into an elderly neighbour on their route. I don't know whatever happened to that, and I can imagine there are privacy/security/safeguarding and all sorts of other issues to consider, but honestly if you live next to an elderly neighbour, I cant stress enough the importance of actually popping in if only once a week to just have a quick chat. Just do it.
If only we all had a crystal ball.


UFC welterweight star Leon Edwards was in town for a workshop at Mixed Martial Arts gym TT MMA this week, and DubaiLAD had the chance to talk to him about a potential fight with Jorge Masvidal, his seven-fight win streaks, sticking to his UK roots and on the bad blood between Conor McGregor and Khabib Nurmagomedov.
Welcome to Dubai Leon! What are your impressions of the city?
Yeah, this is my first time in Dubai. It's a beautiful city. I'm looking forward to coming back again in the near future. To see a gym like this in Dubai is perfect for me. I love it here.
You're currently on a seven-fight win streak, with your last two fights coming against two formidable opponents such as Donald Cerrone and Gunnar Nelson. We know you have been campaigning for a fight with Jorge Masvidal after his win versus Darren Till. Considering you're on a seven-fight win streak, coupled with the drama that happened between you and Masvidal backstage, surely, this is the fight UFC needs to make for you. How close is this fight to being made?
I think this fight should be next. From what I'm hearing, they'll probably make Jorge Masvidal vs Ben Askren. I think that's the fight they're trying to make. But I feel the fans and the media want me to fight Masvidal in July. So when I first finish my holiday, I'll go back and speak with my management team and to the UFC, and I'll go from there. But I think that fight should be my next.
You've been vying for the spot as UK's number one martial artist, with Darren Till losing his last two fights and Michael Bisping retired. Where do you think you currently stand?
I feel I'm number one in the UK now. I'm fighting these tough guys and I'm beating these guys and I'm on a seven-fight win streak in the hardest division in MMA. I feel I've pulled off number one.
Would you consider fighting Darren Till in a best British grudge match?
We're not too friendly as we used to. My aim is to be a world champion. I'm not really focusing on Darren Till. It is to be a world champion.
We know you're focused on a road map towards the title. How far do you think you are from a title shot?
I feel I'm one or two fights away. I think I've got a fight in July, and then by the end of the year or early next year, I should be, and I could be the world champion.
A lot of British fighters have moved to America to hone their skills but you've stuck to your roots. What made you decide to do that?
I've travelled before, even trained AKA. But now, I believe in myself. I believe in my team back home. I've been back home after the last several fights and the results are proving it. I mean, I'd love to be a world champion from the UK, not to travel to America and then fight from there and be a world champion. I'd like to win fighting from the UK, to inspire the kids. You can do it where you come from. You know, I made it. That's my aim, to win the world title from the UK.
Have you been following the back and forth between Khabib and McGregor on Twitter? Do you think they've gone too far?
I think it's crossed the line now. For the last two weeks, he's (Conor McGregor) has been talking about Khabib's wife, and I feel he's crossed the line. I feel it doesn't make sense anymore now. Khabib has beaten him already. I don't see why is it going on.


text:
\section{Introduction}
Beurling characterized all invariant subspaces for the shift
operator on the Hardy space $H^2$ in terms of inner functions
\cite{1}. If $\varphi$ and $\phi$ are inner functions such that
$\varphi{H^{2}}\subset{\phi{H^{2}}}$, then we have $\varphi$ is
divisible by $\phi$ \cite{H}. In fact, the converse is also true
\cite{B1}.

In this paper, $\Omega$ denotes a bounded finitely connected
region in the complex plane and $R(\Omega)$ denotes the algebra of
rational functions with poles off $\overline{\Omega}$.

For a Hilbert space $K$ and a shift operator $S_{K}$ on a
vector-valued Hardy space $H^{2}(\Omega,K)$, Y.S. Kim
characterized every $R(\Omega)$-invariant subspace $M$ for the
operator $S_{K}$, that is, invariant under $u(S_{K})$ for every
$u\in{R(\Omega)}$, in terms of quasi-inner functions \cite{K};
$M=\psi{H}^{2}(\Omega,{K}^{\prime})$ for some quasi-inner function
$\psi:\Omega\rightarrow{L(K^{\prime},K)}$ and a Hilbert space
$K^\prime$. Even though a quasi-inner function is defined as an
operator-valued function in \cite{K}, by the Riesz representation
theorem, we also provide a definition of a scalar-valued
quasi-inner function.

For quasi-inner functions
$\varphi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$ and
$u\in{H^{\infty}(\Omega)}$, we provide relationships between
operator-valued and scalar-valued quasi-inner functions (Theorem
\ref{31}). In addition, by using multiplication operator on a
vector-valued Hardy space, we characterize quasi-inner functions
(Corollary \ref{270}).

For quasi-inner functions $\theta\in{H^{\infty}(\Omega)}$ and
$\varphi\in{H^{\infty}(\Omega,K)}$, we provide definitions of the
following two cases ;

(1) $\theta$ is divisible by $\varphi$.

(2) $\varphi$ is divisible by $\theta$.

With these definitions, we characterize those divisibilities by
comparing $R(\Omega)$-invariant subspaces,
$\theta{H^{2}(\Omega,K)}$ and $\varphi{H^{2}(\Omega,K)}$.
Furthermore, the following result of Theorem \ref{34}( also
Theorem \ref{65}) is similar to that of Proposition \ref{33} which
is in case of scalar-valued inner functions on the open unit disk
; for any quasi-inner functions $\theta\in{H^{\infty}(\Omega)}$
and $\varphi\in{H^{\infty}(\Omega,L(K))}$, the following
assertions are equivalent:

(a) $\theta{|}{\varphi}$.

(b)
$\varphi{H}^{\infty}(\Omega,K)\subset\theta{H}^{\infty}(\Omega,K)$.

(c) $\varphi{H}^{2}(\Omega,K)\subset\theta{H}^{2}(\Omega,K)$.

(d) There is a $\lambda>{0}$ such that
$\varphi(z)\varphi(z)^{*}\leq{\lambda^{2}|\theta(z)|^{2}I_{K}}$
for any $z\in{\Omega}$.

The author would like to express her gratitude to her thesis
advisor, Professor Hari Bercovici.

\section{Preliminaries and Notation}
In this paper, $\mathbb{C}$, $\overline{M}$, and $L(H)$ denote the set of
complex numbers, the (norm) closure of a set $M$, and the set of
bounded linear operators from $H$ to $H$ where $H$ is a Hilbert
space, respectively.

\subsection{Inner functions}

Let $\textbf{D}$ be the open unit disc. We denote by $H^{\infty}$
the Banach space of all bounded analytic functions
$\phi:\textbf{D}\rightarrow{\mathbb{C}}$ with the norm
$\norm\phi_{\infty}=$ sup$\{\norm{\phi(z)}:z\in{\textbf{D}}\}.$

Let $\theta$ and $\theta^{\prime}$ be two functions in
$H^{\infty}$. We say that $\theta$ \emph{divides}
$\theta^{\prime}$(or $\theta$$\mid$$\theta^{\prime}$) if
$\theta^{\prime}$ can be written as
$\theta^{\prime}=\theta\cdot\phi$ for some $\phi\in {H^{\infty}}$.
We will use the notation $\theta\equiv{\theta}^{\prime}$ if
$\theta$$\mid$${\theta}^{\prime}$ and
${\theta}^{\prime}$$\mid$${\theta}$.

 Recall that a function
$u\in{H^\infty}$ is \emph{inner} if $\mid$$u(e^{it})$$\mid$$=1$
almost everywhere on $\partial{\textbf{D}}$.
 By Beurling's theorem on
invariant subspaces of the Hardy spaces, for any inner function
$\theta\in{H^{\infty}}$, we have that $\theta{H}^{2}$ is an
invariant subspace for the shift operator
$S:H^{2}\rightarrow{H^{2}}$ defined by $(Sf)(z)=zf(z)$ for
$f\in{H^2}$.
 Generally, the
following Proposition shows how to define a \emph{divisor} of an
inner function in $H^{\infty}$ by using $S$-invariant subspaces.

\begin{prop}(\cite{B1})\label{33}
For any inner functions $\theta$ and $\theta^\prime$ in
$H^{\infty}$, the following assertions are equivalent:

(a) $\theta\mid{\theta^{\prime}}$.

(b) $\theta^{\prime}H^{\infty}\subset\theta{H}^{\infty}$.

(c) $\theta^{\prime}H^{2}\subset\theta{H}^{2}$.

(d) $\mid$$\theta^{\prime}(z)$$\mid\leq{c}\mid$$\theta(z)$$\mid$
for some $c>{0}$ and for all $z\in{\textbf{D}}$.
\end{prop}

\subsection{Hardy spaces}
We refer to \cite{R2} for basic facts about Hardy space, and
recall here the basic definitions. Let $\Omega$ be a bounded
finitely connected region in the complex plane.
\begin{defn}
The space ${\: H^{2}(\Omega)}$ is defined to be the space of
analytic functions $f$ on $\Omega$ such that the subharmonic
function $|f|^{2}$ has a harmonic majorant on $\Omega$. For a
fixed $z_{0}$ $\in\Omega$, there is a norm on
$H^{2}$$($$\Omega$$)$ defined by

      $\|f\|$=inf$\{ u(z_{0})^{1/2}$: $u$ is a harmonic majorant of
      $|f|^{2}\}$.
\end{defn}

 Let $m$ be the harmonic measure for the point $z_{0}$, let
 $L^{2}(\partial{\Omega})$ be the $L^{2}$-space of complex valued
 functions on the boundary of $\Omega$ defined with respect to $m$,
 and let $H^{2}(\partial{\Omega})$ be the set of
 functions $f$ in $L^{2}(\partial{\Omega})$ such that
 $\int_{\partial{\Omega}} f(z)g(z) dz$ = 0 for every $g$ that is
 analytic in a neighborhood of the closure of $\Omega$.
 If $f$ is in $H^{2}(\Omega)$, then there is a function $f^{\ast}$ in
 $H^{2}(\partial{\Omega})$ such that $f({z})$ approaches
 $f^{\ast}(\lambda_{0})$ as $z$ approaches $\lambda_{0}$
 nontangentially, for almost every $\lambda_{0}$ relative to $m$. The map
 $f\rightarrow{f^{\ast}}$ is an isometry from $H^{2}(\Omega)$ onto
 $H^{2}(\partial{\Omega})$. In this way, $H^{2}(\Omega)$ can be
 viewed as a closed subspace of $L^{2}(\partial{\Omega})$.

 A function $f$ defined on $\Omega$ is in $H^{\infty}(\Omega)$ if
 it is holomorphic and bounded. $H^{\infty}(\Omega)$ is a closed
 subspace of $L^{\infty}({\Omega})$ and it is a Banach
 algebra if endowed with the supremum norm. Finally, the mapping
 $f\rightarrow{f^{\ast}}$ is an isometry of $H^{\infty}(\Omega)$
 onto a week$^{*}$-closed subalgebra of
 $L^{\infty}(\partial{\Omega})$.

\begin{defn} If $K$ is a Hilbert space, then $H^{2}(\Omega$,$K)$ is
defined to be the space of analytic functions
$f:\Omega\rightarrow{K}$ such that the subharmonic function
${\norm{f}}^{2}$ is majorized by a harmonic function $\nu$. Fix a
point $z_{0}$ in $\Omega$ and define a norm on $H^{2}(\Omega$,$K)$
by

$\norm{f}$=inf $\{{\nu(z_{0})}^{1/2}$ : $\nu$ is a harmonic
majorant of ${\norm{f}}^{2}\}$.
\end{defn}

We will work on this vector-valued Hardy space
$H^{2}(\Omega$,$K)$. Note that $H^{2}(\Omega,K)$ can be identified
with a closed subspace of the space $L^{2}(\partial\Omega$,$K)$ of
square integrable $K$-valued functions on $\partial\Omega$. Define
a shift operator
$S_{K}:H^{2}(\Omega,K)\rightarrow{H^{2}(\Omega,K)}$ by
$(S_{K}f)(z)=zf(z)$.

\section{Quasi-Inner Functions}
Let $R(\Omega)$ denote the algebra of rational functions with
poles off $\overline{\Omega}$, and $T$ be an operator in $L(H)$
such that $\sigma(T)\subset\overline{\Omega}$. Then a closed
subspace $M$ is said to be $R(\Omega)$\emph{-invariant}
(\emph{rationally invariant}) for the operator $T$, if it is
invariant under $u(T)$ for any function $u\in{R(\Omega)}$.

To characterize every $R(\Omega)$-invariant subspace for the shift
operator $S_{K}$, quasi-inner function was defined in \cite{K}.

\begin{defn}\label{26}Let $K$ and $K^{\prime}$ be Hilbert spaces and let
${H}^{\infty}(\Omega,L(K,K^{\prime}))$ be the Banach space of all
analytic functions $\Phi:\Omega\rightarrow{L(K,K^{\prime})}$ with
the supremum norm. For
$\varphi\in{H}^{\infty}(\Omega,L(K,K^{\prime}))$, we will say that
$\varphi$ is \emph{quasi-inner} if there exists a constant $c>0$
such that
\[\norm{\varphi(z)k}\geq{c\norm{k}}\]
for every $k\in{K}$ and almost every $z\in\partial\Omega$.
\end{defn}

Even though a quasi-inner function is defined as an
operator-valued function, by the Riesz representation theorem, we
can identify $L(\mathbb{C})$ with $\mathbb{C}$. Thus we have the following
definition of a scalar-valued quasi-inner function;

\begin{defn}\label{27}
For $\theta\in{H^{\infty}(\Omega)}$, we will say that $\theta$ is
\emph{quasi-inner} if there exists a constant $c>0$ such that
\[|\theta(z)|\geq{c}\]
for almost every $z\in\partial\Omega$.

\end{defn}

\begin{prop}\label{30}
Let $K$ and $K^{\prime}$ be Hilbert spaces with $\dim K= \dim
{K}^{\prime}= n(<\infty)$.

If $\varphi\in{H}^{\infty}(\Omega,L(K,K^{\prime}))$ is a
quasi-inner function, then $\varphi(z)$ is invertible a.e. on
$\partial\Omega$.
\end{prop}
\begin {proof}

Since $\varphi\in{H}^{\infty}(\Omega,L(K,K^{\prime}))$, $\varphi$
has nontangential limits for any $z$ $\in$
$\partial\Omega\setminus$$A$ with $m(A)=0$. For a fixed $z_{0}$
$\in$ $\partial\Omega\setminus$$A$, since for some $c>0$ and $a$
$\in$ $K$,
\begin{center}$c\norm{a}\leq\norm{\varphi(z_{0})a}$,\end{center} the range of
$\varphi(z_{0})$ is closed, and $\varphi(z_{0})$ is one-to-one. By
the first isomorphism theorem, \begin{center}$\varphi(z_{0})K$
$\cong$ the range of $ \varphi(z_{0})$.\end{center} Thus, $K$ and
the range of $ \varphi(z_{0})$ have the same dimension.

Since $\dim$ $K$ = $\dim$ $K^{\prime}$ and the range of $
\varphi(z_{0})$ is a closed subspace of $K^{\prime}$, we conclude
that the range of $ \varphi(z_{0})$ is $K^{\prime}$, that is,
$\varphi(z_{0})$ is one-to-one and onto. Thus $\varphi(z)$ is
invertible for $z$ $\in$ $\partial\Omega\setminus$$A$.\end{proof}

\begin{cor}\label{48}
Let $K$ and $K^{\prime}$ be Hilbert spaces with dim $K=$ dim
$K^{\prime}=$ n$(<\infty)$. If
$\varphi\in{H}^{\infty}(\Omega,L(K,K^{\prime}))$ is quasi-inner,
and $\psi\in{H}^{\infty}(\Omega,L(K^{\prime},K))$ such that
$\varphi(z)\psi(z)=
 u(z)I_{K^{\prime}}$ for $u\in{{H}^{\infty}}(\Omega)$, $u\neq{0}$,
 then $\psi(z)\varphi(z)=
 u(z)I_{K}$.
\end{cor}
\begin{proof}
By Proposition \ref{30} (a), $\varphi(z)$ is invertible a.e. on
$\partial\Omega$. From the equation $\varphi(z)\psi(z)=
 u(z)I_{K^{\prime}}$, we obtain \begin{center}$\psi(z)=
 \varphi(z)^{-1}[u(z)I_{K^{\prime}}]=u(z)\varphi(z)^{-1}I_{K^{\prime}}=u(z)I_{K}\varphi(z)^{-1}$\end{center}
for a.e.  $z\in\partial\Omega$. It follows that for a.e.
$z\in\partial\Omega$, $\psi(z)\varphi(z)=
 u(z)I_{K}$ which proves this Corollary.
\end{proof}
In the next theorem, we have another results of relationships
between operator-valued and scalar-valued quasi-inner functions.

\begin{thm}\label{31}
(a) If $\varphi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$ and
$u\in{H^{\infty}(\Omega)}$ are quasi-inner functions such that
\begin{center}$\varphi(z)\psi(z)=u(z)I_{\mathbb{C}^n}$\end{center} where
$\psi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$, then $\psi$ is also
quasi-inner. \vskip0.1cm (b) Conversely, if
$\varphi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$ and
$\psi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$ are quasi-inner functions
such that \begin{center}$\varphi(z)\psi(z)=u(z)I_{\mathbb{C}^n}$ or
$\psi(z)\varphi(z)=u(z)I_{\mathbb{C}^n}$\end{center} for some
$u\in{H^{\infty}(\Omega)}(u\neq{0})$, then u is quasi-inner.
\end{thm}
\begin{proof}
(a) Since $\varphi$ and $u$ are quasi-inner functions, there is
$m_{1}>{0}$ such that $m_{1}\leq{|}u(z){|}$ a.e.
$z\in{\partial\Omega}$, and there exists $c_{i}(>0)(i=1,2)$ such
that for $h\in{\mathbb{C}^{n}}$,
$c_{1}\norm{h}\leq\norm{\varphi(z)h}\leq{c_{2}}\norm{h}$ a.e.
$z\in\partial\Omega$. Since $\varphi(z)\psi(z)=u(z)I_{C^{n}}$, for
$h\in{\mathbb{C}^{n}}$, $m_{1}\norm{h}\leq$
$|{u}(z)|\norm{h}=\parallel$$\varphi(z)\psi(z)h$$\parallel\leq{c_{2}}\norm{{\psi}(z)h}$.
Thus, for $h\in{\mathbb{C}^{n}}$,
\begin{equation}\label{84} \frac{m_{1}}{c_{2}}\norm{h}\leq\norm {\psi(z)h}\end{equation} a.e. $z\in{\partial\Omega}$.

From (\ref{84}), we conclude that $\psi$ is also quasi-inner.

 \vskip0.1cm (b) Since
$\varphi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$ and
$\psi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$ are quasi-inner functions,
there exist $m_{1}(>0)$, and $m_{2}(>0)$ such that for
$h\in{\mathbb{C}^{n}}$, $m_{1}\norm{h}\leq\norm{\varphi(z)h}$ and
$m_{2}\norm{h}\leq\norm{\varphi(z)h}$ a.e. $z\in\partial\Omega$.
Then
\begin{center}$\norm{\varphi(z)\psi(z)h}\geq{m_1}\norm{\psi(z)h}\geq{m_{1}m_{2}\norm{h}}$\end{center}
and so $\norm{\varphi(z)\psi(z)}\geq{m_{1}m_{2}}$. Let
$m^{\prime}=m_{1}m_{2}$. Since
$\mid$$u(z)$$\mid=\norm{\varphi(z)\psi(z)}$, it is proven.
\end{proof}

\begin{prop}\label{49}
Let $\varphi\in$ $H^{\infty}(\Omega,L(K,K^{\prime}))$. If
$\varphi$ is a quasi-inner function, then
$\varphi{H}^{2}(\Omega,K)$ is closed subspace of
$H^{2}(\Omega,K^{\prime})$.
\end{prop}
\begin{proof}
Clearly $\varphi{H}^{2}(\Omega,K)\subset{H}^{2}(\Omega,
K^{\prime})$. Let $\{\varphi{f}_{n}\}$ be a Cauchy sequence in
$\varphi{H}^{2}(\Omega,K)$. If $f_{n}^{\ast}$ is the nontangential
limits of $f_{n}$, then since $\varphi$ is a quasi-inner function,
there exists a constant $c>0$ such that
\begin{equation}\label{50} \norm{\varphi(z)f_{n}^{\ast}(z) -
\varphi(z)f_{m}^{\ast}(z)}\geq c \norm{f_{n}^{\ast}(z) -
f_{m}^{\ast}(z)} \end{equation} for a.e. $z$ $\in$
$\partial\Omega$.

Since $\norm{\varphi{f}_{n}^{\ast}}_{H^{2}(\partial\Omega,
K)}=\norm{\varphi{f}_{n}}_{H^{2}(\Omega, K)}$, $\{
\varphi{f}_{n}^{\ast}\}$ is a Cauchy sequence in
$\varphi{H}^{2}(\partial\Omega,K)$, from the equation (\ref{50})
we conclude that $\{ f_{n}^{\ast}\}$ is also a Cauchy sequence in
$H^{2}(\partial\Omega, K)$. We know that $H^{2}(\partial\Omega,
K)$ is a Banach space. Thus
$\lim_{n\rightarrow\infty}f_{n}^{\ast}$ exists.

If $\lim_{n\rightarrow\infty}
f_{n}^{\ast}={f}^{\ast}\in{H}^{2}(\partial\Omega, K)$, then there
is unique $f\in{H}^{2}(\Omega,{K})$ such that
$\lim_{n\rightarrow\infty} f_{n}=f$ and so
$\lim_{n\rightarrow\infty}\varphi{f}_{n}=\varphi{f}\in
\varphi{H}^{2}(\Omega,K^{\prime})$ which proves this Proposition.
\end{proof}

Furthermore, by using these quasi-inner functions, we have a
generalization of Beurling's theorem as following:

\noindent\textbf{Theorem A}. (Theorem 1.5 in \cite{K}) Let $K$ be
a Hilbert space. Then a closed subspace $M$ of ${H}^{2}(\Omega,
K)$ is $R(\Omega)$-invariant for $S_{K}$ if and only if there is a
Hilbert space $K^{\prime}$ and a quasi-inner function
$\varphi:\Omega\rightarrow{L(K^{\prime},K)}$ such that
$M=\varphi{H}^{2}(\Omega,K^{\prime})$.

Also, we compare two $R(\Omega)$-invariant subspaces for $S_{K}$.
\begin{prop} \cite{K}\label{23} Let $\varphi_{1}
:\Omega\rightarrow{L(K_{1},K)}$ and
$\varphi_{2}:\Omega\rightarrow{L(K_{2},K)}$ be quasi-inner
functions.

Then the subspaces $\varphi_{1}H^{2}(\Omega,K_{1})$ and
$\varphi_{2}H^{2}(\Omega,K_{2})$ of $H^{2}(\Omega,K)$ are equal if
and only if there exist functions
 $\varphi\in{H}^{\infty}(\Omega,L(K_{1},
K_{2}))$ and $\psi\in{H}^{\infty}(\Omega,L(K_{2},K_{1}))$ such
that $\varphi\psi=I_{K_{2}}$, $\psi\varphi=I_{K_{1}}$ and
$\varphi_{1}(z)=\varphi_{2}(z)\varphi(z)$ for any $z\in\Omega$.
\end{prop}

Since we have two kinds of quasi-inner functions, we have two
kinds of $R(\Omega)$-invariant subspaces for $S_{K}$. One of them
is generated by a scalar-valued quasi-inner function, and the
other one is generated by an operator-valued quasi-inner function.
We will also compare these two $R(\Omega)$-invariant subspaces for
$S_{K}$ in Theorem \ref{34}.

Let $K_1$ and $K_2$ be separable Hilbert spaces. To characterize
quasi-inner functions, we define a \emph{multiplication operator}
for a given function $\psi\in{H^{\infty}}(\Omega,L(K_{1},K{2}))$.
A \emph{multiplication operator}
$M_{\psi}:H^{2}(\Omega,K_{1})\rightarrow{H^{2}(\Omega,K_{2})}$ is
defined by
\[M_{\psi}(g)(z)=\psi(z)g(z)\]
for all $g$ in $H^{2}(\Omega,K_{1})$. We can easily check that
$\norm{M_{\psi}}=\norm{\psi}_{\infty}$.

Recall an important property of this multiplication operator :
\begin{prop}\label{38} \cite{K}
Let $K_{1}$ and $K_{2}$ be separable Hilbert spaces. If
$T:{H}^{2}(\Omega,K_{1})\rightarrow{H}^{2}(\Omega,K_{2})$ is a
bounded linear operator such that $TS_{K_{1}}=S_{K_{2}}T$, then
there is a function $\psi\in$$H^{\infty}(\Omega,L(K_{1},K_{2}))$
such that $T=M_{\psi}$, where $M_{\psi}(g)(z)=\psi(z)g(z)$ for
$g\in$${H}^{2}(\Omega,K_{1})$, we have
$\norm{T}=\norm{\psi}_{\infty}$.
\end{prop}

\begin{cor}\label{270}
Let $\varphi\in{H^{\infty}(\Omega,L(K_{1},K_{2}))}$.

(a) If $\varphi$ is quasi-inner, then $M_{\varphi}$ is one-to-one
and has closed range.

(b) If
$M_{\varphi}:H^{2}(\Omega,K_{1})\rightarrow{H^{2}(\Omega,K_{2})}$
is invertible, then $\varphi$ is quasi-inner.
\end{cor}
\begin{proof}
$(a)$ By Proposition \ref{49},
$M_{\varphi}H^{2}(\Omega,K_{1})=\varphi{H^{2}(\Omega,K_{1})}$ is
closed. Since $\varphi$ is quasi-inner, for some $c>0$,
$\norm{\varphi(z)a}\geq{c\norm{a}}$ for a.e. $z\in\partial\Omega$
and for any $a\in{K_{1}}$. Thus condition $(a)$ implies
$\norm{M_{\varphi}f}\geq{c\norm{f}}$ for
$f\in{H^{2}(\Omega,K_{1})}$. It follows that $M_{\varphi}$ is
one-to-one. \vskip0.2cm

 $(b)$ Since
$\varphi{H^{2}(\Omega,K_{1})}$ is is $R(\Omega)$-invariant for
$S_{K_2}$, by Theorem \textbf{A},
\begin{center}$\varphi{H^{2}(\Omega,K_{1})} =\varphi_{1}{H}^{2}(\Omega,K_{0})$\end{center}
for a Hilbert space $K_{0}$ and a quasi-inner function
$\varphi_{1}:\Omega\rightarrow{L(K_{0},K_{2})}$.

Define an operator
$T:{H^{2}(\Omega,K_{1})}\rightarrow{H^{2}(\Omega,K_{0})}$ as
follows. For $f\in{H^{2}(\Omega,K_{1})}$, $Tf=g$ such that
$\varphi{f}=\varphi_{1}g$. Since $\varphi_{1}$ is a quasi-inner
function, by $(a)$, $T$ is well-defined. Since
$S_{K_{0}}T=TS_{K_{1}}$, by Proposition \ref{38},
$T=M_{\varphi_2}$ for a function
$\varphi_{2}\in$$H^{\infty}(\Omega,L(K_{1},K_{2}))$. It follows
that
\begin{equation}\label{37}\varphi(z)=\varphi_{1}(z)\varphi_{2}(z)\end{equation} for any $z\in\Omega$.

Since
$M_{\varphi}:H^{2}(\Omega,K_{1})\rightarrow{H^{2}(\Omega,K_{2})}$
is invertible, so is $T=M_{\varphi_2}$. Note that the
invertibility of $T$ is equivalent to the invertibility of
$\varphi_{2}(z)$ for any $z$ in $\Omega$.  It follows that
$\varphi_{2}(z)$ is bounded below for any $z$ in $\Omega$.

Since $\varphi_1$ is quasi-inner, $\varphi_1(z)$ is also bounded
below almost every $z\in{\partial\Omega}$.

Therefore, by equation (\ref{37}), for any $a\in{K_{1}}$, there is
a constant $c>{0}$ such that
\begin{center}$\norm{\varphi(z)a}\geq{c\norm{a}}$\end{center} for
a.e.$z\in\partial\Omega$.
\end{proof}

\section{Quasi-inner Divisors}

Let $K$ be a Hilbert space. The time has come to consider
divisibilities between a function in $H^{\infty}(\Omega)$ and a
function in $H^{\infty}(\Omega,L(K))$ or between functions in
$H^{\infty}(\Omega,L(K))$.
\begin{defn}\label{55}If $\theta\in{H^{\infty}(\Omega)}$ and
$\varphi\in{H^{\infty}(\Omega,L(K))}$, then we say that $\theta$
\emph{divides} $\varphi$ (denoted $\theta|\varphi)$ if $\varphi$
can be written as
\begin{center}$\varphi=\theta\cdot\phi^{\prime}$\end{center} for
some $\phi^{\prime}\in{H^{\infty}(\Omega,L(K))}$.\end{defn}

\begin{defn}\label{86}If $\theta\in{H^{\infty}(\Omega)}$ and
$\varphi\in{H^{\infty}(\Omega,L(K))}$, then we say that $\varphi$
\emph{divides} $\theta$ (denoted $\varphi|\theta)$ if there exists
$\psi\in{H}^{\infty}(\Omega,L(K))$ satisfying the following
relations $;$
\begin{center}$\varphi(z)\psi(z)=
 \theta(z)I_{K}$\end{center}  and \begin{center}$\psi(z)\varphi(z)=
 \theta(z)I_{K}$\end{center} for $z\in\Omega$.\end{defn}

\begin{defn}\label{56}
If $\varphi$ and $\varphi^{\prime}$ are functions in
$H^{\infty}(\Omega,L(K))$, then we say that $\varphi$ is a
\emph{left divisor} of $\varphi^\prime$ if
\begin{center}$\varphi^{\prime}(z)=\varphi(z)\varphi^{\prime\prime}(z)$ (denoted
$\varphi{|}_{l}\varphi^\prime$)\end{center} for some
$\varphi^{\prime\prime}\in{H^{\infty}(\Omega,L(K))}$, and
 we say that $\varphi$
is a \emph{right divisor} of $\varphi^\prime$ if
\begin{center}$\varphi^{\prime}(z)=\varphi^{\prime\prime}(z)\varphi(z)$ (denoted
$\varphi{|}_{r}\varphi^\prime$)\end{center} for some
$\varphi^{\prime\prime}\in{H^{\infty}(\Omega,L(K))}$. We will say
$\varphi$ \emph{divides} $\varphi^\prime$ if $\varphi$ is not only
a left divisor but also a right divisor of $\varphi^\prime$.
\end{defn}

\begin{defn}\label{58}
Let $\varphi$ be a functions in $H^{\infty}(\Omega,L(K))$. A
quasi-inner function $\theta\in{H^{\infty}(\Omega)}$ is called the
\emph{greatest quasi-inner divisor} of $\varphi$ if $\theta$
divides $\varphi$ and every (complex-valued) quasi-inner divisors
of $\varphi$ is a divisor of $\theta$. \emph{The greatest
quasi-inner divisor} of $\varphi$ is denoted by $D(\varphi)$.
\end{defn}

With definitions of divisibility, we provide the following results
similar to Proposition \ref{33}.

\begin{thm}\label{34}
For any quasi-inner functions $\theta\in{H^{\infty}(\Omega)}$ and
$\varphi\in{H^{\infty}(\Omega,L(K))}$, the following assertions
are equivalent:\vskip0.2cm

(a) $\theta{|}{\varphi}$.

(b)
$\varphi{H}^{\infty}(\Omega,K)\subset\theta{H}^{\infty}(\Omega,K)$.

(c) $\varphi{H}^{2}(\Omega,K)\subset\theta{H}^{2}(\Omega,K)$.

(d) There is a $\lambda>{0}$ such that
$\varphi(z)\varphi(z)^{*}\leq{\lambda^{2}|\theta(z)|^{2}I_{K}}$
for any $z\in{\Omega}$, where $I_K$ is the identity function on
$K$.

\end{thm}
\begin{proof}
If $\theta{|}{\varphi}$, $\varphi=\theta\varphi_{1}$ for some
$\varphi_{1}\in{H^{\infty}(\Omega,L(K))}$. Then
\begin{center}$\varphi{H}^{\infty}(\Omega,K)=\theta\varphi_{1}{H}^{\infty}(\Omega,K)
\subset\theta{H}^{\infty}(\Omega,K)$.\end{center} Thus (a) implies
(b).

Conversely, suppose that
$\varphi{H}^{\infty}(\Omega,K)\subset\theta{H}^{\infty}(\Omega,K)$.
Then
\[\varphi^{\ast}{H}^{\infty}(\partial\Omega,K)\subset\theta^{\ast}{H}^{\infty}(\partial\Omega,K).\]

Let $\{b_{i}:i\in{I}\}$ be an orthonormal basis of $K$ and
$g_{i}\in{H}^{\infty}(\partial\Omega,K)$ defined by
$g_{i}(z)=b_{i}(i\in{I})$. Since
$\varphi^{\ast}{H}^{\infty}(\partial\Omega,K)\subset\theta^{\ast}{H}^{\infty}(\partial\Omega,K)$,
there is $f_{i}\in{{H}^{\infty}(\partial\Omega,K)}$ such that
$\varphi^{\ast}{g_{i}}=\theta^{\ast}{f_{i}}$, i.e. for $i\in{I}$,
\begin{equation}\label{66}\varphi^{\ast}(z)b_{i}=\theta^{\ast}(z)f_{i}(z).\end{equation}
Define $\varphi_{1}:\partial\Omega\rightarrow{L(K)}$ by for
$i\in{I}$,
\begin{equation}\label{67}\varphi_{1}(z)b_{i}=f_{i}(z).\end{equation}
For $i\in{I}$, define
$\varphi_{i}\in{H^{\infty}(\partial\Omega,L(K))}$ by
$\varphi_{i}(z)b_{j}=\delta_{ij}f_{i}(z) (j\in{I})$, where
$\delta_{ij}=\left\{\begin{array}{ll}
                                         1 & \mbox{if $i=j$}\\0 &
                                         \mbox{otherwise.}\end{array}
                                  \right. $ Then
                                  \begin{equation}\label{b2}\varphi_{1}=\sum_{i\in{I}}\varphi_{i}.\end{equation}

\noindent By (\ref{66}) and (\ref{67}), for each $i\in{I}$,
$\varphi^{\ast}(z)b_{i}=\theta^{\ast}(z)\varphi_{1}(z)b_{i}$, and
so
\begin{center}$\varphi^{\ast}=\theta^{\ast}\varphi_{1}$.\end{center}

 To prove that $(b)$
implies $(a)$, we have to show that
$\varphi_{1}\in{H^{\infty}(\partial\Omega,L(K))}$. Since
$\theta\in{H^{\infty}}$ is a quasi-inner function, there is $c>0$
such that $|\theta(z)|\geq{c}$ for every
$z\in{A}\subset\partial\Omega$ with $m(A)=0$. For any $x\in{K}$
with $\norm{x}=1$ and $z\in{A}$,
\begin{equation}\label{b3}\norm{\varphi_{1}(z)x}=\frac{\norm{\varphi^{\ast}(z)x}}{|\theta^{\ast}(z)|}\leq
\frac{\norm{\varphi}_{\infty}}{c}.\end{equation} From (\ref{b2})
and (\ref{b3}), we conclude that
\begin{center}$\varphi_{1}\in{H^{\infty}(\partial\Omega,L(K))}$.\end{center}

  By the same way as above,
$(a)\Leftrightarrow(c)$ is proven. We begin to prove
$(a)\Leftrightarrow(d)$. If $\theta{|}{\varphi}$,
$\varphi=\theta\varphi_{1}$ for some
 $\varphi_{1}\in{H^{\infty}(\Omega,L(K))}$. Then
\begin{center}$\varphi(z)\varphi(z)^{*}=\theta(z)\varphi_{1}(z)$
$\varphi_{1}(z)^{*}\overline{\theta(z)}
\leq{\norm{\varphi_{1}}_{\infty}^{2}|\theta(z)|^{2}I_{K}}$.\end{center}
Let $\lambda=\norm{\varphi_{1}}_{\infty}$. Since $\varphi$ is
quasi-inner, $\varphi\neq{0}$ and so $\lambda>0$. Thus (a) implies
(d).\vskip0.2cm

Conversely, suppose that for any $z\in{\Omega}$,
\begin{equation}\label{69}\varphi(z)\varphi(z)^{*}\leq{\lambda^{2}|\theta(z)|^{2}I_{K}}\end{equation}
for some $\lambda>{0}$. For each $z\in{\Omega}$, we will define a
linear mapping $F_{z}\in{L(K)}$. Let
\begin{center}$A=\{z\in{\Omega}:\theta(z)=0\}$\end{center} and
\begin{center}$B=\{z\in{\Omega}:\theta(z)\neq0\}$.\end{center} If $z\in{A}$, then let
$F_{z}=0$. If $z\in{B}$, then range of $\overline{\theta(z)}I_{K}$
is $K$ and so we can define a linear mapping $F_{z}$ from $K$ to
range
 of $\varphi(z)^{*}$ by \begin{center}$F_{z}(\overline{\theta(z)}f)=\varphi(z)^{*}f$\end{center}
for $f\in{K}$.

Since
$\norm{F_{z}(\overline{\theta(z)}f)}^{2}=\norm{\varphi(z)^{*}f}^{2}
=(\varphi(z)\varphi(z)^{*}f,f)\leq\lambda^{2}(|\theta(z)|^{2}f,f)=\lambda^{2}\norm{\theta(z)f}^{2}$,
that is,
\begin{equation}\label{70}\norm{F_{z}(\overline{\theta(z)}f)}\leq{\lambda\norm{\theta(z)f}},\end{equation}
$F_{z}$ is well-defined for $z\in{B}$. By definition of $F_{z}$,
if $z\in{B}$,
\begin{equation}\label{q}\theta(z)F_{z}^{\ast}=\varphi(z).\end{equation} If
$z\in{A}$, by (\ref{69}) $\norm{\varphi(z)}=0$ and so
$\varphi(z)=0(z\in{A})$. Thus $\theta(z)F_{z}^{*}=\varphi(z)$ for
any $z\in{\Omega}$.

Define a function $F:\Omega\rightarrow{K}$ by
\begin{center}$F(z)=F_{z}^{*}$.\end{center} Then by equation (\ref{q}), \begin{center}$\varphi(z)=\theta(z)F(z)$\end{center}
for $z\in{\Omega}$. To finish this proof, we have to prove that
$F\in{H^{\infty}(\Omega,L(K))}$. From inequality (\ref{70}), we
have
\begin{equation}\label{c3}\norm{F}_{\infty}\leq{\lambda}\end{equation} and so
$F=\frac{\varphi}{\theta}$ has only removable singularities in
$\Omega$. Thus $F$ can be defined on
$\{z\in{\Omega:\theta(z)=0}\}$ so that $F$ is analytic and
\begin{center}$\varphi=\theta{F}$.\end{center} From (\ref{c3}),
$F\in{H^{\infty}(\Omega,L(K))}$ which proves $(d)\Rightarrow(a)$.
\end{proof}

Note that for $\varphi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$, we can
think $\varphi(z)(z\in{\Omega})$ as an $n\times{n}$ matrix, say
$(\varphi_{ij}(z))_{n\times{n}}$.

Let $F$ be a non-empty family of functions in $H^(\Omega)$. Recall
the definition of the greatest common quasi-inner divisor $\theta$
of $F$ which means that $\theta$ divides every element in $F$, and
$\theta$ is a multiple of any other common quasi-inner divisors of
$F$.
 By the same way as the proof of
Proposition 2.3.4 in \cite{20}, we have the following result.

\begin{prop}\label{24}
Let $F$ be a non-empty family of functions in $H^(\Omega)$. Then
the greatest common quasi-inner divisor $\theta$ of $F$, denoted
by $\bigwedge{F}$, exists.

\end{prop}

\begin{lem}\label{68}
 If
$\varphi=[\varphi_{ij}]_{i,j=1}^{n}$ is a function in
${H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$, then
$D(\varphi)=\bigwedge{\varphi_{ij}}$.
\end{lem}
\begin{proof}
Let $\bigwedge{\varphi_{ij}}=\theta_{1}$. If
$\theta\in{H^{\infty}(\Omega)}$ is a quasi-inner function such
that $\theta|\varphi$, then $\varphi=\theta\phi$ for some
$\phi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$. Thus
\begin{center}$(\varphi_{ij})_{n\times{n}}=(\theta\phi_{ij})_{n\times{n}}$\end{center} for
each $i, j=1,2,\cdot\cdot\cdot,n$. Thus
$\varphi_{ij}=\theta\phi_{ij}$ for each $i,
j=1,2,\cdot\cdot\cdot,n$. Since
$\phi\in{H^{\infty}(\Omega,L(\mathbb{C}^{n}))}$,
$\phi_{ij}\in{H^{\infty}(\Omega)}$ and so $\theta|{\varphi_{ij}}$
 for each $i,
j=1,2,\cdot\cdot\cdot,n$. By definition of greatest common
divisor, we can conclude that $\theta|\theta_{1}$. Clearly
$\theta_{1}|\varphi$. Thus $D(\varphi)=\theta_{1}.$
\end{proof}

We have another result similar to Theorem \ref{34}.

\begin{thm}\label{65}
For any quasi-inner functions $\theta\in{H^{\infty}(\Omega)}$ and
$\varphi\in{H^{\infty}(\Omega,L(K))}$, the following assertions
are equivalent:

(a) $\varphi{|}{\theta}$.

(b)
$\theta{H}^{\infty}(\Omega,K)\subset\varphi{H}^{\infty}(\Omega,K)$.

(c) $\theta{H}^{2}(\Omega,K)\subset\varphi{H}^{2}(\Omega,K)$.

(d) There is a $\lambda>{0}$ such that
$|\theta(z)|^{2}I_{K}\leq{\lambda^{2}\varphi(z)\varphi(z)^{\ast}}$
for any $z\in{\Omega}$.

\end{thm}
\begin{proof}
This theorem is proven by the same way as Theorem \ref{34}.
\end{proof}

\begin{prop}\label{35}
For any quasi-inner functions $\varphi$ and $\varphi^\prime$ in
$H^{\infty}(\Omega,L({\mathbb{C}^n}))$ and $H^{\infty}(\Omega,L(C^{n}))$
respectively, the following assertions are equivalent:

(a) $\varphi{|}_{l}{\varphi^\prime}$.

(b)
$\varphi^{\prime}{H}^{\infty}(\Omega,{\mathbb{C}}^{n})\subset\varphi{H}^{\infty}(\Omega,{\mathbb{C}^n})$.

(c)
$\varphi^{\prime}{H}^{2}(\Omega,{\mathbb{C}}^{n})\subset\varphi{H}^{2}(\Omega,{\mathbb{C}^n})$.
\end{prop}
\begin{proof}
In the same way as the proofs of Theorem \ref{34}, it is proven
that (a) and (b) are equivalent and so are (a) and (c).
\end{proof}
\begin{cor}\label{36}
Let $\varphi$ and $\varphi^\prime$ be quasi-inner functions in
$H^{\infty}(\Omega,L({\mathbb{C}^n}))$. Then
$\varphi{|}_{l}{\varphi^\prime}$ and
$\varphi^{\prime}{|}_{l}{\varphi}$ if and only if there exist a
function $\psi$ in

\noindent${H^{\infty}(\Omega,L(K_{1},K_{2}))}$ such that $\psi(z)$
is invertible for any $z\in{\Omega}$ with $\sup
\norm{\psi(z)^{-1}}<\infty$ and
$\varphi^{\prime}(z)=\varphi(z)\psi(z)$ for any $z\in\Omega$.
\end{cor}
\begin{proof}
By Proposition \ref{35}, we can conclude that
$\varphi{|}_{l}{\varphi^\prime}$ and
$\varphi^{\prime}{|}_{l}{\varphi}$ if and only if
$\varphi^{\prime}{H}^{2}(\Omega,{\mathbb{C}}^{n})=\varphi{H}^{2}(\Omega,{\mathbb{C}^n})$.
Then by Proposition \ref{23}, this Corollary is proven.
\end{proof}

For quasi-inner functions $\varphi$ and $\varphi^\prime$ in
${H^{\infty}(\Omega,L({\mathbb{C}^n}))}$, $\varphi(z)$ is a bounded
operator on the Hilbert space $\mathbb{C}^n$. When we consider some
relationships between $\varphi|_{l}{\varphi^\prime}$
 ($\varphi|_{r}{\varphi^\prime}$) and
$\varphi(z)|_{l}{\varphi^\prime}(z)
(\varphi(z)|_{r}{\varphi^\prime}(z)$, respectively$)$ for each
$z\in{\Omega}$, first of all, we obtain easily if
$\varphi|_{l}\varphi^{\prime}(\varphi|_{r}{\varphi^\prime}$) then
\[\varphi(z)|_{l}{\varphi^\prime}(z)(\varphi(z)|_{r}{\varphi^\prime}(z),
\texttt{ respectively})\] for each $z\in{\Omega}$.

Suppose that $z\in{\Omega}$ if and only if
$\overline{z}\in{\Omega}$. Then for
$\varphi\in{H^{\infty}(\Omega,L({\mathbb{C}^n}))}$, we define a new
function $\varphi^{\sim}\in{H^{\infty}(\Omega,L({\mathbb{C}^n}))}$ by
\begin{center}$\varphi^{\sim}(z)=\varphi(\overline{z})^{*}$\end{center}
for $z\in{\Omega}$.
\begin{lem}\label{61}
Let $\varphi_{1}$ and $\varphi_{2}$ be functions in
${H^{\infty}(\Omega,L({\mathbb{C}^n}))}$. Suppose that $z\in{\Omega}$ if
and only if $\overline{z}\in{\Omega}$. Then following statements
are equivalent:

(a) $\varphi_{1}|_{l}\varphi_{2}$.

(b) $\varphi_{1}^{\sim}|_{r}\varphi_{2}^{\sim}$.
\end{lem}
\begin{proof}
($\Rightarrow$) Since $\varphi_{1}|_{l}\varphi_{2}$, there is a
function $\varphi_{3}\in{H^{\infty}(\Omega,L({\mathbb{C}^n}))}$ such that
\begin{center}$\varphi_{2}(z)=\varphi_{1}(z)\varphi_{3}(z)$\end{center} for $z\in{\Omega}$.
By our assumption,
$\varphi_{2}(\overline{z})=\varphi_{1}(\overline{z})\varphi_{3}(\overline{z})$
for $z\in{\Omega}$. Thus
\begin{center}$\varphi_{2}(\overline{z})^{*}=\varphi_{3}(\overline{z})^{*}\varphi_{1}(\overline{z})^{*}$\end{center}
for $z\in{\Omega}$ and so
$\varphi_{1}^{\sim}|_{r}\varphi_{2}^{\sim}$.

($\Leftarrow$) By the same way as the proof of ($\Rightarrow$),
this implication can be proven.
\end{proof}
From Lemma \ref{61}, we have the following result.
\begin{cor}\label{62}
Let $\varphi_{1}$ and $\varphi_{2}$ be functions in
${H^{\infty}(\Omega,L({\mathbb{C}^n}))}$. Suppose that $z\in{\Omega}$ if
and only if $\overline{z}\in{\Omega}$. Then following statements
are equivalent:\vskip0.2cm

(a) $\varphi_{1}|\varphi_{2}$.

(b) $\varphi_{1}|_{r}\varphi_{2}$ and
$\varphi_{1}^{\sim}|_{r}\varphi_{2}^{\sim}$.

\end{cor}

redpajama_set_name: RedPajamaArXiv


Although the Corporation was founded in 2004, the actual business started in February 2007. We are a Small scale Cosmetic Manufacturing Company.So far we have made only samples to test two of our products.They both have received enthusiastic consumer responses. We are planning to have a bigger Lab prepare and package one of the products. We will launch that one first, since the revenues will be more sizeable. We need financial help in the forms of LOC, or a $30,000 loan or Angel Investor Aid. My credit is fair.. We have not had any Sales therefore no Revenues yet. Please help us stop the bleeding..We need to finally start making some money.
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So you have an Accountant?? A Spread sheet showing EXPENSES??
Who is on payroll?? A Cash Flow Projection??
How long do you need the funds for?? and How and When will I get paid back??
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Hello Luckiest!! Have you seen my answer?. Please let me know.


For the transportation process, we have transportation from source and destination. For each type of transportation, do we need to create two condition type at source and at destination. There is already an exisiting condition type (ZLT0).


text:
Simon Phillips, 14 July 2021 2:18pm
SENSE AND SENSITIVITY
Audio transcript: 14 July at 2:18pm. Black and white image, from the bottom of the picture at the centre a long jetty reaches up and out like an upside down letter "L". It's surrounded by smooth still water with a mirror like quality that meets faint misty land in the distance as a brilliant white misty sky halos above. This image is dark at the bottoms edge, and the jetty begins from the bottom centre giving the impression that some is about to walk out on the jetty, the wood plank of the jetty look grey and weathered with a clear tone difference of many foot steps over time. This image has a reflective feel to it, with the jetty only going so far, do people come here to fish or find some quiet time to reflect?
Audio descriptions of artworks by Simon Phillips in the exhibition, Sense and Sensitivity, with Dawn-joy Leong, 2021. Recordings by DADAA.
Sense and Sensitivity celebrates the unique sensorial place of autistic artistry in the world via the lens of landscape photographer, Simon Phillips, and the deep listening of performance artist Dawn-joy Leong.
In a kaleidoscope of beautiful imagery and subtle sound, this remarkable collaboration privileges the feelings of both artists for the world; how they engage with it through what they see, hear and touch; revealing their exquisite appreciation for themes of diversity, balance, place and timelessness.
Exhibition open 17 November – 15 December 2021.
More Listening
Simon Phillips, 7 December 2020 11:34am
Simon Phillips, 18 March 2021 9:52am
Simon Phillips, 31 August 2021 11:15am
IOTA21 ARTISTS' TALK

redpajama_set_name: RedPajamaCommonCrawl


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text:
Joe Cornish Offered to Direct "King Kong" Prequel "Skull Island"
WeLiveFilm July 31, 2014 0 likes
by Justin Cook
Among one of the more shocking announcements made at San Diego Comic-Con last week was the news that Legendary Pictures would be developing a King Kong prequel entitled Skull Island. Skull Island is of course the fictitious location in which King Kong inhabits in the King Kong movies.
Today it sounds like we have a potential director for the film with Deadline reporting that Legendary has reached out to Joe Cornish to helm the movie, which is due out November 4, 2016. Cornish is primarily known for directing the critically-acclaimed sci-fi action film Attack the Block, as well as co-writing The Adventures of Tintin: The Secret of the Unicorn with Steven Moffat and Edgar Wright. Not to mention the star of Cornish's directorial debut Block, John Boyega, was recently cast in Star Wars: Episode VII.
Cornish is mulling over many potential future projects at the time so it is unclear whether or not he will accept the offer, but considering that he brought such a unique directorial style to Block, which was an extremely entertaining genre mixing movie, he seems like a great fit to direct Skull Island.
Spoiler Sundays (07/20/2014): "The Purge: Anarchy," "Planes: Fire and Rescue," and "Sex Tape"
The Grade Book: "The Purge: Anarchy"
New "Star Wars: The Force Awakens" Trailer Released
"Star Wars: Episode VII" Title Revealed
Weekend Box Office: "Jurassic World" Takes a Bite Out of $500M Record
Opinion – "Should 'Transformers 4' Be Like 'War for Cybertron?'" by Romero Movie News


text:
World's smallest MRI performed on single atoms
(Nanowerk News) Researchers at the Center for Quantum Nanoscience (QNS) within the Institute for Basic Science (IBS) at Ewha Womans University have made a major scientific breakthrough by performing the world's smallest magnetic resonance imaging (MRI). In an international collaboration with colleagues from the US, QNS scientists used their new technique to visualize the magnetic field of single atoms.
An MRI is routinely done in hospitals nowadays as a part of imaging for diagnostics. MRI's detect the density of spins - the fundamental magnets in electrons and protons - in the human body. Traditionally, billions and billions of spins are required for an MRI scan.
The new findings, published today in the journal Nature Physics ("Magnetic resonance imaging of single atoms on a surface"), show that this process is now also possible for an individual atom on a surface. To do this, the team used a Scanning Tunneling Microscope, which consists of an atomically sharp metal tip that allows researchers to image and probe single atoms by scanning the tip across the surface.
Single magnetic atoms are deposited on a surface of magnesium oxide. They are imaged by the magnetic tip of a scanning tunneling microscope which also allows researchers to perform an MRI scan of the atom's magnetic field. (Image: Philip Willke et al)
The two elements that were investigated in this work, iron and titanium, are both magnetic. Through precise preparation of the sample, the atoms were readily visible in the microscope. The researchers then used the microscope's tip like an MRI machine to map the three-dimensional magnetic field created by the atoms with unprecedented resolution.
In order to do so, they attached another spin cluster to the sharp metal tip of their microscope. Similar to everyday magnets, the two spins would attract or repel each other depending on their relative position.
By sweeping the tip spin cluster over the atom on the surface, the researchers were able to map out the magnetic interaction.
Lead author, Dr. Philip Willke of QNS says: "It turns out that the magnetic interaction we measured depends on the properties of both spins, the one on the tip and the one on the sample. For example, the signal that we see for iron atoms is vastly different from that for titanium atoms. This allows us to distinguish different kinds of atoms by their magnetic field signature and makes our technique very powerful."
The researchers plan to use their single-atom MRI to map the spin distribution in more complex structures such as molecules and magnetic materials.
"Many magnetic phenomena take place on the nanoscale, including the recent generation of magnetic storage devices." says Dr. Yujeong Bae also of QNS, a co-author in this study. "We now plan to study a variety of systems using our microscopic MRI."
The ability to analyze the magnetic structure on the nanoscale can help to develop new materials and drugs. Moreover, the research team wants to use this kind of MRI to characterize and control quantum systems. These are of great interest for future computation schemes, also known as quantum computing.
The bright areas mark positions where the atom's magnetic field is the same. (Image: Philip Willke et al)
"I am very excited about these results. It is certainly a milestone in our field and has very promising implications for future research." says Prof. Andreas Heinrich, Director of QNS. "The ability to map spins and their magnetic field with previously unimaginable precision, allows us to gain deeper knowledge about the structure of matter and opens new fields of basic research."
Source: Institute for Basic Science
How to discover new materials quickly
Quantum sensor breakthrough using naturally occurring vibrations in artificial atoms


text:
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15 Mindfulness and Relaxation Apps for Kids with Anxiety
This list of 15 Apps for Kids with Anxiety has done wonders for helping us navigate the negative thinking, difficult social situations, and anxiety.
The A Project
The A project is a community building and conflict resolution charity. We are committed to providing a platform that encompass a fully integrated & multicultural process for togetherness and resolution, regardless of age, ethnicity, religion or culture.We develop innovative and client lead educational themes and…
Lansdowne House57 Berkeley SquareLondonW1J 6ER
The Abbey Community Centre
Provides social and health care activities for a range of BME groups and others ranging from the age of 0 to 90+ years old. Hall/Room Hire for public and private use.
34 GREAT SMITH STREETLONDONSW1P 3BU
Action for Change
Seeks to break the cycle of repeat removal from birth parents. Works directly with parents who have had, or are at risk of having their children removed.
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Email Email WebsiteAdd to My Shortlist
Advance - Domestic Violence Charity
We support women aged 16+ who live in the following 4 boroughs:- Hammersmith and Fulham - Brent - Kensington and Chelsea - Westminster Referrals routes into each service may be different, so please check our "how to refer" page for further details. Cases assessed as…
020 8741 7008 0808 2000 247 - National Domestic Violence Helpline
ADVANCE Advocacy Service
24 hour support to women experiencing Domestic Violence
PO Box 47151LondonW6 6AB
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ADVANCE delivers nationally accredited, quality marked and award-winning services across parts of West London. Their focus is on preventing domestic violence against women and children and supporting women to reduce their offending, whilst challenging and changing how women are treated in the justice system, whether…
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Website has a directory of advice services on subjects including benefits, legal, money and debt, housing and property, age and family, disability and health, immigration and nationality, alcohol and substance misuse, and education.
The Advocacy Project - Our Choice - Finding Out Group and Pop Ups
Facilitates Self Advocacy and consultation groups so that people with learning disabilities can develop the skills that they need to self advocate and promotes their inclusion in the development of services through their participation in the Learning Disability Partnership Board. The Finding Out Group is…
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Advocate for Mental Health
We provide mental health advocacy support to people using or wishing to use mental health services in Westminster and Kensington & Chelsea. To promote equality of access to our services, two special responsibilities posts were created. One is for black afro-Caribbean service users and the…


Everywhere Al Janssen and Brother Andrew have travelled to meet with men and women who are persecuted for the name of Christ, the most urgent request is simply, "Please pray for us." But how do we pray for them? And furthermore, how can you learn to pray so that your prayers impact your life and influence your community, your nation, and the world?
In this powerful booklet by Brother Andrew and Al Janssen, you'll learn practical, biblical lessons about the true effectiveness of prayer from those who suffer persecution for following Jesus. You will come to understand that prayer is the real battle!


text:
PA DEP Fines Range Resources $4.15M for Wastewater Impoundments
September 19, 2014 Energy Companies, Industrywide Issues, Pennsylvania, Range Resources Corp, Regulation, Washington County
Range Resources has just had their knuckles rapped, hard, by the Pennsylvania Dept. of Environmental Protection (DEP) with respect to wastewater/recycled water impoundments (i.e. ponds) they operate in Washington County, PA. Range has been fined the most any company has been fined by the DEP in the modern shale era–$4.15 million. They will also be required to close five of the seven impoundments they've operated in the county (Range was closing them anyway), and make major upgrades to the two remaining impoundments. There's no way to sugarcoat this–Range was taken to the proverbial woodshed by the DEP and got a lot more than a switch to the rear-end…
NJ's Largest Utility Becomes 5th Partner in PennEast Pipeline
Marcellus & Utica Shale Story Links: Mon, Sep 22, 2014


text:
The sudden passing of Tanya Roberts
With a heavy heart I can confirm the death of Tanya Roberts (age 65) last night on January 4, 2021, around 9:30pm PT at Cedars-Sinai Medical Center in Los Angeles CA.
Her 18-year domestic partner, Lance O'Brien, received the phone call from Cedars-Sinai Medical Center last night at their home confirming her passing.
Roberts's cause of death was from a urinary tract infection which spread to her kidney, gallbladder, liver and then blood stream.
Today, January 5, 2021 at 10:00am PT, O'Brien picked up Roberts personal possessions from Cedars-Sinai Medical Center.
Roberts was an animal rights activist, so in lieu of flowers, they are asking for donations to be given in Roberts's name to the ASPCA: https://secure.aspca.org
An online memorial for Tanya Roberts will be announced shortly.
The family ask for privacy as they mourn her death.
Any additional questions, please feel free to call or email me.
Mike Pingel
Rainbow PR
mpingel@msn.com
COPYRIGHT © 2023 TANYA ROBERTS - ALL RIGHTS RESERVED


text:
O Waldfriedhof Stuttgart foi instalado em 1913, pouco antes da eclosão da Primeira Guerra Mundial, de acordo com planos do diretor de obras de Stuttgart Albert Pantle. Está instalado no distrito de Degerloch.

Com área de 30,7 ha, é o maior cemitério de Stuttgart. Contém cerca de 15.000 sepulturas.

Sepultamentos

Bibliografia

Geral
Brunnen und Staffeln in Stuttgart: Jungfrau mit den Tränenschalen. Stuttgart 2014, online:.
Dieter Buck; Harald Schukraft: Stuttgarter Grenz-Wanderungen. Stadtgeschichtliche Entdeckungstouren. Tübingen 2005, Seite 57–58.
Genres in visual representations. Proceedings of a conference held in 1986 by invitation of the Werner-Reimers-Stiftung in Bad Homburg (Federal Republic of Germany). Leiden 1990, Seite 153, 162, 168 (Ehrenmal für die gefallenen Teilnehmer des Ersten Weltkriegs von Paul Bonatz) online:.
Alexandra Kaiser: Von Helden und Opfern. Eine Geschichte des Volkstrauertags. Campus-Verlag, Frankfurt am Main u. a. 2010, ISBN 978-3-593-39288-2 Seite 93–97, 105-111 (Gefallenengedenkfeiern).
[David Koch]: Der Stuttgarter Waldfriedhof von Oberbaurat Pantle. In: Christliches Kunstblatt für Kirche, Schule und Haus Band 59, 1917, Seite 210–215, 217-219, 221.
Werner Koch; Christopher Koch: Stuttgarter Friedhofsführer. Ein Wegweiser zu Gräbern bekannter Persönlichkeiten. Tübingen 2012, Seite 86–115.
Mammut-Verlag (Herausgeber und Redaktion): Stuttgart, Der Friedhofwegweiser, DIESSEITS und JENSEITS, Stuttgart 2011, Seite 62–65. – Die Broschüre ist kostenlos bei der Infothek im Rathaus erhältlich.
Johannes Merz: Der evangelische Kirchenbau in Württemberg, Band 61, 1919, Seite 322–338, hier: 338, 357, 358 (Kruzifix von Max Natter).
Stuttgarter Straßenbahnen (Hrsg.): Lebenslinien. Mit der Stuttgarter Seilbahn in die 1920er Jahre – auf dem Waldfriedhof durch die Jahrzehnte. Stuttgart 2009, online:. – Die Broschüre liegt bei der Seilbahn aus.
175 Jahre Walcker-Orgelbau. Hausmitteilung Nummer 16, Juni 1956, Seite 71 (Orgeldisposition), online:.
Walcker. Hausmitteilung Nummer 18, August 1957, Seite 20 (Orgelprospekt), online:.
Der Waldfriedhof in Stuttgart. In: Wasmuths Monatshefte für Baukunst und Städtebau Band 1, 1914/1915, Seite 442–443, 454-459, online:.

Otto Herbert Hajek
Hajek-Plastik in Stuttgart aufgestellt. In: art Das Kunstmagazin vom 20. Oktober 2008, .
Monika Bugs: Otto Herbert Hajek im Gespräch mit Monika Bugs. Saarlouis 1998.
Ausst.kat. Bundeskunsthalle Bonn (Hrsg.): O. H. Hajek. Eine Welt der Zeichen. mit einem Werkverzeichnis von Anuschka Koos. Bonn 2000.
Otto Herbert Hajek. Plastiken, Reliefs, Schreibschriften, Zeichnungen, Lithographien. Haus am Waldsee, Berlin-Zehlendorf, [Mai – Juni 1963]. Berlin 1963.
Claus Pese (Redaktion): O. H. Hajek. Die Durchdringung des Lebens mit Kunst. Stuttgart 1987.
Kunst im Öffentlichen Raum. Hajek, Otto Herbert. Stuttgart 2014, .

Bibliografia complementar
Hans Fegers: Rudolf Yelin der Jüngere. In: Reutlinger Geschichtsblätter Neue Folge, Band 22, 1983, Seite 7–10, 23-24.
Wolfgang Hesse: Fritz von Graevenitz' "Mutter der Heimat". Gesinnung bildhauerisch. In: Karlheinz Fuchs (Redaktion): Ausstellungsreihe Stuttgart im Dritten Reich. [3]. Die Machtergreifung. Von der republikanischen zur braunen Stadt. Stuttgart 1983, Seite 47–49.
Erich Heyfelder: Ludwig Habich und seine Bildhauerschule an der Stuttgarter Kunstakademie. In: Schwäbisches Heimatbuch, 1935, Seite 53-66.
Rudolf Pantle: Grabmalplastik Zur Ausstellung für Friedhofkunst im Hoppenlaufriedhof zu Stuttgart. In: Die Plastik Band 4, 1914, Seite 60–62, hier: 61, Tafel 80.
Travertinpark [Hinweistafeln und Karte]. Fenster in die Urzeit. Stuttgart 2014, besonders Seite 3, online:.
Gustav Wais: Stuttgarts Kunst- und Kulturdenkmale. 25 Bilder mit stadtgeschichtlichen, baugeschichtlichen und kunstgeschichtlichen Erläuterungen. Stuttgart [1954].

Ligações externas

Webseite der Stadt Stuttgart, Waldfriedhof.
Gräberfelder der Luftkriegopfer beider Weltkriege auf dem Waldfriedhof.
Gräberfelder der gefallenen Soldaten beider Weltkriege auf dem Waldfriedhof.
Abbildungen der Gedenktafeln, Gedenkstelen und Gedenkobelisken des Ehrenmals für die gefallenen Teilnehmer des Ersten Weltkriegs.
Topografische Karte, Hybridkarte und Luftbilder im Maßstab 1:500, Stadtplan Stuttgart, Suchbegriff: Waldfriedhof.

Cemitérios de Stuttgart
Cemitérios de Baden-Württemberg

redpajama_set_name: RedPajamaWikipedia


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text:
Neil Pharaoh: 'I am sick of the sexism in politics'
The biggest double standard in politics is sexism. And thanks to the current state of Australian parliament, it's now reached epic proportions.
news.com.auAugust 3, 201912:12pm
November 28, 2018. Labor MP Claire O'Neil discusses how the Labor Party deals with bullying and sexism by other parliamentarians. Ms O'Neil says that Julia Bank's treatment by...
November 28, 2018. Labor MP Claire O'Neil discusses how the Labor Party deals with bullying and sexism by other parliamentarians. Ms O'Neil says that Julia Bank's treatment by colleagues would be dealt with differently in the Labor Party. "There are political parties that are actively confronting this problem", Ms O'Neil says. Greens senator Larissa Waters added that she would love to see more women in parliament that represented the broader community which would lead to better decision making. (AAP Video/Marc Tewksbury)
There is a definite culture of sexism in Australian politics, and it's now reached epic proportions, writes Neil Pharaoh. Picture: Kelly Barnes/The Australian.Source:News Corp Australia
When someone you work with restrains you and removes your underwear, when reported to "management" in any normal job, your actions would preclude you from that industry for life. Yet a few months down the track the man who allegedly sexually assaulted a woman became "eligible" for a potential promotion and was being considered for preselection for the Liberal Party. When this is backed up by senior Liberals saying "for every woman that comes forward with a complaint is probably representing at least five more", you know something is broken.
RELATED: The real issue with political sex secrets
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The biggest double standard in politics is sexism.
On all sides of politics, the way we treat women differently to men astounds me. And this is coming from a man who has been involved in politics.
If you got through that far in this opinion piece, you will either be quietly agreeing or telling me I don't believe in "merit"; if the latter, stop reading now.
Why is it when a male is stiff as a board, monotone and boring that we call him "statesmanlike", yet when a female is she is "detached, cold and ruthless"?
Why is it that men don't get asked about who will look after the children, yet women do?
The sexism in politics has reached epic proportions.
Emma Husar stood up after Question Time in the House of Representatives and defended her reputation after she was defamed by a media outlet. Picture: Gary RamageSource:News Corp Australia
On the Labor side, this week we saw the settlement between Emma Husar MP and Buzzfeed. Let's revisit the situation: unproven allegations, six-week media cycle against Emma, no proof, Emma forced to not recontest her seat, political career over, nothing ever proven — female. Take her circumstance versus Greg Barber MLC, where a bullying claim led to a settlement of $56,000 (that is your taxes paying for a bullying settlement for a MP), and he was still able to continue as a MP, even after the "hairy-legged feminist, power pussies" comments (coupled with his "men's room" to boot).
So why can an allegation kill a female MP's career, but not a male? Even when the male has settlement payments for bullying on the taxpayer funded books? Not to mention Barnaby and his affair — all proven, yet none lost their career. Emma? She is gone.
Emma Husar tells 7.30 young, single women are often the subject of rumours in politics. CREDIT: 7:30 Report...
Emma Husar tells 7.30 young, single women are often the subject of rumours in politics. CREDIT: 7:30 Report
Let's look a bit deeper at sexist comments directed at women; Fiona Scott being called "sex appeal" during the Lindsay 2013 campaign, which then clouded her time in office — meaning everywhere she went, the "sex appeal" comment remained. Now, what man has had such an equivalent comment levelled at him? And has it stuck? Exactly. Silence.
Sarah Hanson-Young, in court commentary: "Mr. Leyonhjelm called me a hypocrite because I have sex with men," said the Greens senator during cross-examination over Leyonhjelm's comments about her in the media following a debate in the Senate last year.
"What's sexist about that?" Leyonhjelm's barrister, Tony Morris, QC, replied.
"He wouldn't say it to a man," she replied. Again, double standards of behaviour.
Globally women's participation in parliament is a tad above 24 per cent, yet accounted for only 8 per cent of national leaders and 2 per cent of presidents' posts. In Australia, Labor has 47 per cent and Liberals 23 per cent after the last Federal election.
Still, Fiona Scott insists Tony Abbott's description of her as "sex appeal" damaged her credibility and by default, her career. Picture: Supplied.Source:Supplied
We all know the story of how Julia Gillard was taunted with tag lines like "ditch the witch" and described as "barren" as well as many other names. Name for me a male who has got equivalent levels of vitriol in public debate and discussion.
Julie Bishop — an amazingly capable, talented woman — looked over for Scomo and Dutton. Jane Prentice: lost preselection to a former male staffer, Julian Simmonds.
Ann Sudamalis: again another male, Grant Schultz (whose bullying complaint and review has still not announced its findings).
On Labors ledger, Lauren Palmer lost preselection to James Martin in Hasluck, and Lyndal Howlison in NSW for Brian Owler. Time and time again we walk past more capable and qualified women for men — it can't continue.
Even Bronwyn Bishop and the helicopter affairs stinks of double standard, when a number of male politicians have undertaken similar activities without consequence — Bronwyn had to go after a helicopter flight to a Liberal Party Fundraiser. Yet Tony Abbott charged taxpayers over $3000 to attend the birthday party of Santo Santoro without consequences.
That's right, a birthday party. Again, one standard for women another for men.
"Looking back on my time in politics, I do wish I'd called out the sexism earlier," former Prime Minister Julia Gillard told reporters last year. Picture: Kelly Barnes/The AustralianSource:News Corp Australia
Susan Ley had to resign from a role over a taxpayer funded trip to the Gold Coast to purchase an apartment, something she admitted was within guidelines but failed the pub test. Yet Darren Chester did EXACTLY the same thing for an apartment purchase in Melbourne and yet no consequences for him, no resignation or role reduction. I mean, can the hypocrisy be any more obvious?
I can't tell you the number of times in Labor preselection that I have seen amazingly qualified women looked over for men — let alone discussed it with friends who are Members of the Liberal Party and say they are continually disadvantaged during preselection.
Margaret Fitzherbert (Liberal) undertook professional polling on this issue and found 38 per cent of Liberal preselections think it is OK to ask a women who will look after her children if she is elected into parliament. A question which she rightly says has no right answers (not focused enough on family and too focused on career etc). And in the private sector, that question is illegal.
Former MP Margaret Fitzherbert named her report the Double Hurdle. "The first hurdle is a woman's own self-assessment of her readiness for preselection; the second is a preselection process that often treats men and women very differently, with motherhood identified as a key issue," she said. Picture: Penny StephensSource:News Corp Australia
And don't let the Greens Party claim moral high ground; the majority of their Federal leadership is also white men.
And while Labor is better on Parliamentary benches, peel back the curtain to the backrooms of power and you will find the "powerbrokers" behind the politicians is usually a room full of men.
Look to the number of single mums we have around Cabinet tables in Australia? None that I know of, for I have seen when a single mum wasn't supported to sit in Cabinet because a Leader won't assist with a reduction in portfolios, or offer other support to keep her in the Cabinet. I am sorry, but I want my taxpayer dollars to fund a full-time nanny just so we can have a single mum (or dad) in Cabinet. That is the Australia I want to live in.
Ironically, in the 2019 election there is one seat which was won by Labor. Labor won't learn the lesson in it though. That is: that Fiona Phillips MP won Gilmore, and she had never worked as a staffer or advisor, was a local through and through, had deep strong roots to the community, and was a female running against a male. Coincidence? I think not.
Fiona Phillips delivering her maiden speech in the House of Representatives Chamber, at Parliament House in Canberra. Picture: Kym Smith.Source:News Corp Australia
I have personally alleged sexism within the Labor Party against a certain Parliamentarian — an accusation which after an internal investigation was found to be legitimate and accurate — only to be personally disadvantaged and brought before the disputes committee of the Labor Party for "disloyalty" and "bringing the party into disrepute" (for calling out sexism against a Parliamentarian). How is that for double standards?
If a male bystander receives a harsher penalty for calling out sexism than the person who was independently found to have been sexist, there is an issue. (Oh, and he got away with a simple written apology to the victim, not made public of course).
What I have learned from the many amazing women I have seen survive in politics, who try and try again and again, is that even those who never get preselected, those who don't sit on parliamentary benches, have overcome so much more and are often so much more connected to the community than the men who succeeded them.
We need to change the discussion. We as men need to start calling this for what it is, whether Liberal, Labor or other. It is sexism and it needs to change.
Neil Pharaoh was the Former National co-convener of Rainbow Labor and twice Victorian Labor Candidate for Prahran. Continue the conversation @neilpharaoh
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text:
Q: Phalcon Volt turns html code into string I have html code from my db and I'm trying to print it as {{ blog.content }}. But it renders as string and shows all the html tags etc. How can I make it to render as html?
it looks like:

I am writing my blog post here to show everyone that I can do such things as this: adsdsfadsf and also like this one

htmltags shouldn't be visible. The line above should be rendered as html. Which means bold parts example; should be bold.

A: 1) Create Elements.php (Library or Plugin all are same)

namespace YourAppNameSpace;

use Phalcon\Config;

class Elements extends \Phalcon\Mvc\User\Plugin
{
     public function decodeString($string)
    {
        return html_entity_decode($string);
    }
}

2) Add Elements.php to the service
$di->set('element', function () {
    return new YourAppNameSpace\Elements();
});

3) In Volt File Try this
{{ element.decodeString(blog.content) }}

Hope It Works.. :)

redpajama_set_name: RedPajamaStackExchange


text:
De kabelbaan Stresa-Alpino-Mottarone is een kabelbaan, gelegen in de gemeente Stresa in Piemonte, die het Lago Maggiore verbindt met de top van de Mottarone.

Geschiedenis
Het idee om een kabelbaan aan te leggen om Stresa met de top van Mottarone te verbinden, dateert uit de jaren zestig, toen het bedrijf dat een bestaande bergspoorlijn naar Mottarone beheerde moest overwegen om de bestaande spoorwegverbinding uit 1911 te vervangen. De keuze viel op een kabelbaan, daar dit vervoerstype sneller en efficiënter geacht werd te zijn. Deze keuze resulteerde er in dat in 1963 begonnen werd met de ontmanteling van het lijntje, de Ferrovia del Mottarone. Nadat deze lijn buiten gebruik was gesteld werd begonnen met de bouw van de kabelbaan. Deze werd uitgevoerd door de firma Piemonte Funivie. Na zes jaar werk werd de baan op 1 augustus 1970 in gebruik genomen.

Constructie

Het systeem bestaat uit twee etappes; een eerste gondelbaan van het meer naar het gehucht Alpino, en een tweede etappe van Alpino naar het bergstation op 106 meter beneden de top van de Mottarone.

Vanuit de nabije omgeving van het bergstation gaat een stoeltjeslift tot bovenop de top.

Eerste etappe: Stresa - Alpino
 Wagentype: twee gesloten gondels
 Bouwjaar: 1970
 Onderste station: haven van Stresa, hoogte 205 m. boven zeeniveau
 Bovenste station: bij de botanische tuin in Alpino, hoogte 803 m.
 Reistijd: 20 minuten
 Overbrugde afstand: 2351 meter
 Hoogteverschil tussen onderste en bovenste station: 598 meter

Tweede etappe: Alpino - Mottarone
 Wagentype: twee gesloten gondels
 Bouwjaar: 1970
 Onderste station: bij de botanische tuin in Alpino, hoogte 803 m.
 Bovenste station:  hoogte 1385 m. (106 meter beneden de top van de Mottarone)
 Reistijd: 20 minuten
 Overbrugde afstand: 3020 meter
 Hoogteverschil tussen onderste en bovenste station: 582 meter

Stoeltjeslift

Wagentype: pen stoeltjeslift (twee personen per wagen)
Bouwjaar: 2009
 Onderste station: ca 1387 m.
Bovenste station: 1491 m.
Overbrugde afstand: ca 270 m.
Hoogteverschil: ca 104 m.

Onderhoud
In 2002 onderging de kabelbaan groot onderhoud; in 2009 is er door de firma Leitner een systeem met twee gondels gebouwd dat gaat vanaf het aankomststation van de kabelbaan naar de bergtop van Mottarone op 1491 m boven de zeespiegel. Op de bergtop staat het wegkruis en zijn skipistes aangelegd. Ook is er een Alpyland, een amusementspark dat daar in 2010 is aangelegd en dat onder meer een rodelachtbaan (een bobslee op rails) omvat.

Ongeluk
Op 23 mei 2021 is een gondel van de kabelbaan neergestort; dit resulteerde in  veertien doden en een zwaargewond kind. Een kabel zou op zo'n 5 meter voor het eindpunt zijn geknapt, waarna de cabine naar beneden stortte in een bergachtig, bebost gebied. Een onderzoek is gaande naar de toedracht van het ongeval.. Op 26 mei werden de ingenieur van de kabelbaan, de directeur en het operatieve hoofd van het bedrijf dat de dienst levert gearresteerd. Het noodremsysteem van de kabelbaan zou bewust zijn uitgeschakeld om vertragingen van de kabelbaan te voorkomen. Toen de sleepkabel doorbrak ging het systeem niet in werking en ging de gondel vervolgens met volle snelheid achteruit waardoor deze van de draagkabels ging en neerstortte.

Verbano-Cusio-Ossola
Bouwwerk in Piëmont
Stresa


GWP Project Services used IES TaP to achieve the first Outstanding rating for BREEAM 2011.
GWP Project Services was tasked with helping to transform an old industrial site into an eco-friendly student village for the University of Cumbria. Phase one of the development, consisted of 30 three-storey townhouses, and had to be completed within just eight months.
Meanwhile BREEAM, had just updated its 2008 criteria, which meant that GWP had to quickly educate the entire design team on what was required to meet the new 2011 standard, and get them to submit the required evidence to gain enough credits in time. To do this it decided to deploy IES TaP from the outset of the project.
It quickly became apparent that there was an issue with GWP's energy strategy. Whereas it only needed to achieve an Energy Performance Rating of A under the old criteria, the team now needed to deliver a 40% reduction in CO2 emissions against 2010 building regulations. Comparing previous Outstanding builds against the new criteria showed they were only delivering half the CO2 reduction required.
The main energy consumption within the village was that required to heat cold water to a tepid temperature when it first entered the boilers. By using IES TaP to co-ordinate feedback from a wide range of boiler and energy specialists, to electrical and mechanical engineers, GWP were able to quickly identify that Alpha boilers, which use waste heat from the flu to raise the temperature of incoming cold water, reduced emissions to the extent required within the existing budget and design.
To get a BREEAM Outstanding rating, a score of over 85% was required. IES TaP helped GWP to quickly educate the design team about the new criteria, solve problems and keep to deadlines, which meant the project remained on target to achieve a score of 86.12% and was the first building to reach an Outstanding rating under BREEAM 2011.
IES TaP has made the entire BREEAM assessment process much more efficient and profitable. GWP's clients are delighted with the high ratings they're achieving on their behalf and now that IES TaP has been awarded the BREEAM badge of recognition GWP can export content from the tool directly to BRE without having to produce a separate report, saving even more time. The result is that their assessors can take on more projects than they could before, enabling them to increase profits.