text:
\section{Introduction}
In this paper, we study the cyclicity problem with respect to the forward shift operator $S_b$ acting on the de Branges--Rovnyak space $\HH(b)$, associated to a function $b$ belonging to the closed unit ball of $H^\infty$ and satisfying $\log(1-|b|)\in L^1(\mathbb T)$. This problem of cyclicity has a long and outstanding history and many efforts have been dedicated to solving it in various reproducing kernel Hilbert spaces. It finds its roots in the pioneering work of Beurling who showed that cyclicity of a function $f$ in the Hardy space $H^2$ is equivalent to $f$ being outer. Brown and Shields studied the cyclicity problem in the Dirichlet spaces $\mathcal{D}_\alpha$  for polynomials that do not have zeros inside the disc, but that may have some zeros on its boundary. Such functions are cyclic in $\mathcal{D}_\alpha$ if and only if $\alpha \leq 1$. They also proved that the set of zeros on the unit circle (in the radial sense) of cyclic functions in the Dirichlet space $\mathcal{D}_\alpha$  has zero logarithmic capacity, and this led them to ask whether any outer function with this property is cyclic \cite{BS1}. This problem is still open although there has been relevant contributions to the topic by a number of authors; e.g. see \cite{B2, BC1, EFKR1, EFKR2, HS1, RS1}. We also mention the paper \cite{MR3475457} where the authors prove the Brown--Shields conjecture in the context of some particular Dirichlet type spaces $\mathcal D(\mu)$, which happen to be related to our context of de Branges--Rovnyak spaces \cite{MR3110499}.
\par\smallskip
The de Branges--Rovnyak spaces $\HH(b)$ (see the precise definition in Section~\ref{sec2}) have been introduced by L. de Branges and J. Rovnyak in the context of model theory (see \cite{MR0215065}). A whole class of Hilbert space contractions is unitarily equivalent to $S^*|\HH(b)$, for an appropriate function $b$ belonging to the closed unit ball of $H^{\infty}$. Here $S$ is the forward shift operator on $H^2$ and $S^*$, its adjoint, is the backward shift operator on $H^2$. The space $\HH(b)$ is invariant with respect to $S^*$ for every $b$ in the closed unit ball of $H^\infty$, and $S^*$ defines a bounded operator on $\HH(b)$, endowed with its own Hilbert space topology.
On the contrary, $\HH(b)$ is invariant with respect to $S$ if and only if $\log(1-|b|)\in L^1(\mathbb T)$, i.e. if and only if $b$ is a non-extreme point of the closed unit ball of $H^{\infty}$. See \cite[Corollary 25.2]{FM2}. In \cite{MR3309352}, it is proved that if $\log(1-|b|)\in L^1(\mathbb T)$, the cyclic vectors of $S^*|\HH(b)$ are precisely the cyclic vectors of $S^*$ which live in $\HH(b)$. Note that cyclic vectors of $S^*$ have been characterised by Douglas--Shapiro--Shields \cite{MR203465}. The result of \cite{MR3309352} is based on a nice description, due to Sarason, of closed invariant subspaces of $S^*|\HH(b)$. See also \cite[Corollary 24.32]{FM2}. Unfortunately an analogous description of closed invariant subspaces of $S_b=S|\HH(b)$ remains an unsolved and difficult problem. In \cite{MR4246975}, Gu--Luo--Richter give an answer in the case where $b$ is a rational function which is not inner, generalising a result of Sarason \cite{MR847333}.
\par\smallskip
The purpose of this paper is to study the cyclic vectors of $S_b$ when the function $b$ is such that $\log(1-|b|)\in L^1(\mathbb T)$. In Section~\ref{sec2}, we present a quick overview of some useful properties of de Branges--Rovnyak spaces. Then, in Section~\ref{sec3}, we give some general facts on cyclic vectors and completely characterise holomorphic functions in a neighborhood of the closed unit disc which are cyclic for $S_b$. In Section~\ref{sec4}, we give a characterisation of cyclic vectors for $S_b$ when $b$ is rational (and not inner). Of course, this characterisation can be derived from the description,
given in  \cite{MR4246975},
of invariant subspaces of $S_b$ when $b$ is a non-inner rational function. Nevertheless, we will give  a more direct and easier proof of this characterisation. Finally, Section~\ref{sec5} will be devoted to the situation where $b=(1+I)/2$, where $I$ is a non-constant inner function such that the associated model space $K_I=\HH(I)$ has an orthonormal basis of reproducing kernels.

\par\bigskip
 \section{Preliminaries on $\HH(b)$ spaces}\label{sec2}
\subsection{Definition of de Branges-Rovnyak spaces} Let
 $$\operatorname{ball}(H^{\infty}) := \Big\{b \in H^{\infty}: \|b\|_{\infty} = \sup_{z \in \D} |b(z)| \leq 1\Big\}$$ be the closed unit ball of $H^{\infty}$, the space of bounded analytic functions on the open unit disk $\D$, endowed with the sup norm.
For $b \in \operatorname{ball}(H^{\infty})$, the \emph{de Branges--Rovnyak space} $\HH(b)$ is the reproducing kernel Hilbert space on $\mathbb D$ associated with the positive definite kernel  $k_{\lambda}^b$, $\lambda\in\D$, defined as
\begin{equation}\label{eq:original kernel H(b)}
k^{b}_{\lambda}(z) = \frac{1 -  \overline{b(\lambda)}b(z)}{1 - \overline{\lambda} z}, \quad  z \in \D.
\end{equation}

It is known that $\HH(b)$ is contractively contained in the well-studied Hardy space $H^2$ of analytic functions $f$ on $ \D $ for which
$$\|f\|_{H^2} := \Big(\sup_{0 < r < 1} \int_{\T} |f(r \xi)|^2 dm(\xi)\Big)^{\frac{1}{2}}<\infty,$$
where $m$ is the normalised Lebesgue measure on the unit circle $\T = \{\xi \in \C: |\xi| = 1\}$ \cite{Duren, garnett}. For every $f \in H^2$, the radial limit
$\lim_{r \to 1^{-}} f(r \xi) =: f(\xi)$ (even the non-tangential limit $f(\xi):=\lim_{\substack{z\to\xi\\  \sphericalangle}}f(z)$) exists for $m$-a.e. $\xi \in \T$, and
\begin{equation}\label{knnhHarsysu}
\|f\|_{H^2}  = \Big(\int_{\T} |f(\xi)|^2 dm(\xi)\Big)^{\frac{1}{2}}.
\end{equation}
Though $\HH(b)$ is contractively contained in $H^2$, it is generally not closed in the $H^2$ norm. It is known that $\HH(b)$ is closed in $H^2$ if and only if $b=I$ is an inner function, meaning that $|I(\zeta)|=1$ for a.e. $\zeta\in\mathbb T$. In this case, $\HH(b)=K_I=(I H^2)^\perp$ is the so-called \emph{model space} associated to $I$. Note that $K_I=H^2\cap I\overline{zH^2}$ (see \cite[Proposition 5.4]{MR3526203}), and then $K_I=\ker T_{\bar{I}}$, where $T_{\bar{I}}$ is the Toeplitz operator with symbol $\bar I$ defined on $H^2$ as $T_{\bar I}f=P_+(\bar I f)$, where $P_+$ denotes the orthogonal projection from $L^2$ onto $H^2$.

\par\smallskip
We refer the reader to the book \cite{MR1289670} by Sarason and to the monograph \cite{FM1}, \cite{FM2} by Fricain and Mashreghi for an in-depth study of de Branges-Rovnyak spaces and their connections to numerous other topics in operator theory and complex analysis.
\par\smallskip
In this paper, we will always assume that $b$ is a non-extreme point of $\operatorname{ball}(H^{\infty})$, which is equivalent to requiring that $\log(1-|b|)\in L^1(\mathbb T)$. Under this assumption, there is a unique outer function $a$, called \emph{the pythagorean mate} for $b$, such that $a(0)>0$ and $|a|^2+|b|^2=1$ a.e. on $\mathbb T$. There are two important subspaces of $\HH(b)$ which can be defined via this function $a$. The first one is the space $\MM(a)=aH^2$, equipped with the range norm
\[
\|af\|_{\MM(a)}=\|f\|_2,\qquad f\in H^2.
\]
The second one is $\MM(\bar a)=T_{\bar a}H^2$, equipped also with the range norm
\[
\|T_{\bar a}f\|_{\MM(\bar a)}=\|f\|_2,\qquad f\in H^2.
\]
Note that since $a$ is outer, the Toeplitz operator $T_{\bar a}$ is one-to-one and so the above norm is well defined. It is known that $\MM(a)$ is contractively contained into $\MM(\bar a)$, which itself is contractively contained into $\HH(b)$. See \cite[Theorem 23.2]{FM2}. Note that $\MM(a)$ is not necessarily closed in $\HH(b)$. See \cite[Theorem 28.35]{FM2} for a characterisation of closeness of $\MM(a)$ in $\HH(b)$-norm. There is also an important relation between $\HH(b)$ and $\MM(\bar a)$ which gives a recipe to compute the norm in $\HH(b)$. Indeed, if $f\in H^2$, then $f\in\HH(b)$ if and only if there is a function $f^+\in H^2$ satisfying $T_{\bar b}f=T_{\bar a}f^+$ (and then $f^+$ is necessarily unique). Moreover, in this case, we have
\[
\|f\|_b^2=\|f\|_2^2+\|f^+\|_2^2.
\]
Also,
\begin{equation}\label{Inconnu}
 \langle f,g \rangle_b=\langle f,g \rangle_2+\langle f^+,g^+ \rangle_2\quad\textrm{ for every }f,g\in\HH(b).
\end{equation}
See \cite[Theorem 23.8]{FM2}. Finally, let us recall that $\HH(b)=\MM(\bar a)$ if and only if $(a,b)$ forms a corona pair, that is
\[
\tag{HCR} \qquad \inf_{\mathbb D}(|a|+|b|)>0.
\]
See \cite[Theorem 28.7]{FM2}.

\par\smallskip
A crucial fact on de Branges-Rovnyak space is that the space $\HH(b)$ is invariant with respect to the shift operator $S: f \mapsto zf$ if and only if the function $b$ is non-extreme. Since we will consider in this paper only the case where $b$ is non-extreme, $\HH(b)$ is  indeed invariant by $S$, and $S$ defines a bounded operator on $\HH(b)$, endowed with its own Hilbert space topology, which we will denote by $S_b$. The functions $z^n$ belong to $\HH(b)$
 for every $n\ge 0$. Actually, we have
\begin{equation}\label{eq:density-polynomial}
\mbox{Span}(z^n:n\geq 0)=\HH(b),
\end{equation}
where $\mbox{Span}(A)$ denotes the closed linear span generated by vectors from a certain family $A$.
In other words, the polynomials are dense in $\HH(b)$. See \cite[Theorem 23.13]{FM2}.
Note that \eqref{eq:density-polynomial} exactly means that the constant function $1$ is cyclic for $S_b$.
\par\smallskip
Another tool which will turn out to be useful when studying the cyclicity for the shift operator is the notion of \emph{multiplier}. Recall that the set $\mathfrak M(\HH(b))$ of multipliers of $\HH(b)$ is defined as
\[
\mathfrak M(\HH(b))=\{\varphi\in\mbox{Hol}(\mathbb D): \varphi f\in \HH(b),\forall f\in\HH(b)\}.
\]
Using the closed graph theorem,  it is easy to see that when $\varphi\in \mathfrak M(\HH(b))$, then $M_\varphi$, the multiplication operator by $\varphi$, is bounded on $\HH(b)$. The algebra of multipliers is  a Banach algebra when equipped with the norm $\|\varphi\|_{\mathfrak M(\HH(b)}=\|M_\varphi\|_{\mathcal L(\HH(b))}$.  Using standard arguments, we see that $\mathfrak M(\HH(b))\subset H^\infty\cap\HH(b)$. In general, this inclusion is strict. See \cite[Example 28.24]{FM2}. However, we will encounter below a situation (when $b$ is a rational function which is not a finite Blaschke product) where we have the equality $\mathfrak M(\HH(b))= H^\infty\cap\HH(b)$.

\par\smallskip

\subsection{Some properties of the reproducing kernels of $H^2$ in de Branges-Rovnyak spaces}
Recall that we are supposing that $b$ is non-extreme. If we denote by $k_\lambda(z)=(1-\overline{\lambda}z)^{-1}$ the reproducing kernel of $H^2$ at the point $\lambda\in\D$, then $k_\lambda$ belongs to $\HH(b)$ and
\begin{equation}\label{eq:density-crk}
\mbox{Span}(k_\lambda:\lambda\in\mathbb D)=\HH(b).
\end{equation}
See \cite[Corollary 23.26]{FM2} or \cite[Lemma 7]{MR3390195}. We also know (see \cite[Theorem 23.23]{FM2}) that $bk_\lambda\in \HH(b)$ for every $\lambda\in\mathbb D$, and that  for every $f\in\HH(b)$ we have
\begin{equation}\label{eq1EZD:lem-completeness}
\langle f,k_\lambda \rangle_b=f(\lambda)+\frac{b(\lambda)}{a(\lambda)}f^+(\lambda)\quad\mbox{and}\quad \langle f,bk_\lambda\rangle_b=\frac{f^+(\lambda)}{a(\lambda)}\cdot
\end{equation}
Using these two equations, we can produce an interesting complete family in $\HH(b)$ which will be of use to us.

\begin{Lemma}\label{Lem:completenes}
Let $b$ be a non-extreme point of the closed unit ball of $H^{\infty}$, and let
 $c$ be a complex number with $|c|<1$. Then
\[
\mbox{Span}(k_\mu-cbk_\mu:\mu\in\mathbb D)=\HH(b).
\]
\end{Lemma}

\begin{proof}
Let $h\in\HH(b)$, and assume that for every $\mu\in\mathbb D$, $h$ is orthogonal in $\HH(b)$ to  $k_\mu-cbk_\mu$. According to \eqref{eq1EZD:lem-completeness}, we have
\[
0=h(\mu)+\frac{b(\mu)}{a(\mu)}h^+(\mu)-\overline{c}\frac{h^+(\mu)}{a(\mu)}\cdot
\]
This can be rewritten as $ah=-bh^++\overline{c}h^+$. Multiplying this equality by $\bar b$ and using the fact that $|a|^2+|b|^2=1$ a.e. on $\mathbb T$, we obtain
\[
a(\bar b h-\bar a h^+)=-(1-\overline{c}\bar b)h^+.
\]
Note that $|1-\overline{c}\bar b|\geq 1-|c|>0$, and so the last identity can be written as
\[
\frac{\bar bh-\bar ah^+}{1-\overline{c}\bar b}=-\frac{h^+}{a}\cdot
\]
On the one hand, this equality says that $\frac{h^+}{a}$ belongs to $ L^2$ and since $a$ is outer, we have $\frac{h^+}{a}\in H^2$. See \cite[page 43]{MR1864396}. On the other hand, by definition of $h^+$, the function $\bar bh-\bar ah^+$ belongs to $\overline{H^2_0}$ and since $(1-\overline{c}\bar b)^{-1}$ is in $ \overline{H^\infty}$, we also have $\frac{h^+}{a}\in \overline{H^2_0}$. Then $\frac{h^+}{a}$ belongs to $ H^2\cap \overline{H^2_0}=\{0\}$. Finally we get that $h^+=0$ and thus that $h=0$.
\end{proof}

\par\smallskip

\subsection{Boundary evaluation points on $\HH(b)$}
An important tool in the cyclicity problem will be the boundary evaluation points for $\HH(b)$. It is known that the description of these points depends on the inner-outer factorisation of $b$. Recall that any $b$ in $\operatorname{ball}(H^{\infty})$ can be decomposed as
\begin{equation} \label{E:dec-b-bso}
b(z)= B(z) S_\sigma(z) O(z), \qquad z \in \D,
\end{equation}
where
\[
B(z) = \gamma\,\prod_{n\ge 1} \left(\, \frac{|a_n|}{a_n}\frac
{a_n-z}{1-\overline{a}_nz} \,\right) \quad\textrm{is a Blaschke product,}
\]
with  $|\gamma|=1$, $a_n\in\D$ for every $n\ge1$,  and $\sum_{n\ge 1}(1-|a_n|)<+\infty$,
\[
S_\sigma(z) = \exp\left( -\int_\T \frac{\xi+z}{\xi-z} \, d\sigma(\xi) \right) \quad\textrm{is a singular inner function,}
\]
with $\sigma$ a positive finite Borel measure on $\T$ which is singular with respect to the Lebesgue measure, and
\[
O(z) = \exp\left( \int_\T \frac{\xi+z}{\xi-z} \log|b(\xi)| \,dm(\xi) \right)
\]
is the outer part of $b$.
Now, let $E_0(b)$ be the set of points $\zeta\in\mathbb T$ satisfying the following condition:

 \begin{equation}
  \sum_n\frac{1-|a_n|^2}{|\zeta-a_n|^{2}}+\int_{\T} \frac{d\sigma(\xi)}{|\zeta-\xi|^{2}}+\int_{\T} \frac{\big| \log|b(\xi)| \big|}{|\zeta-\xi|^{2}} \,\, dm(\xi)<\infty.
 \end{equation}

It is proved in \cite{MR2390675} that for every point $\zeta\in\mathbb T$, every function $f\in \HH(b)$ has a non-tangential limit at $\zeta$ if and only if $\zeta\in E_0(b)$.  This is also equivalent to the property that $b$ has an {\emph{angular derivative (in the sense of Carath\'eodory) at $\zeta$}, meaning that $b$ and $b'$ both have a non-tangential limit at $\zeta$ and $|b(\zeta)|=1$.
Moreover, in this case, the linear map
\begin{equation}\label{eq:radial-limit-ADC}
f\longmapsto f(\zeta):=\lim_{\substack{z\to\zeta\\  \sphericalangle}}f(z)
\end{equation}
is bounded on $\HH(b)$.
The function $k_\zeta^b$ defined by
\[
 k_{\zeta}^b(z)=\frac{1 -  \overline{b(\zeta)}b(z)}{1 - \overline{\zeta} z}, \quad  z \in \D,
\]
belongs to $\HH(b)$, and
\[
 \langle f,k_{\zeta}^b\rangle_b=f(\zeta)\quad\textrm{ for every } f\in\HH(b).
\]
We call the function $k_\zeta^b$ \emph{the reproducing kernel of $\HH(b)$ at the point $\zeta$}, and \eqref{eq:radial-limit-ADC} means that the reproducing kernels $k_{z}^b$ tend weakly to $k_{\zeta}^b$ as $z\in\D$ tends non-tangentially to $\zeta$. See \cite[Theorem 25.1]{FM2}. There is also a nice connection between the boundary evaluation points and the point spectrum of $S_b^*$ in the case where $b$ is a non-extreme point in $\operatorname{ball}(H^{\infty})$: for $\zeta\in\mathbb T$, we have that
\begin{equation}\label{point-spectrum-boundary}
\bar\zeta \mbox{ is an eigenvalue for }S_b^* \mbox{ if and only if }  b \mbox{ has an angular derivative at }\zeta.
\end{equation}
\par\smallskip

The boundary evaluation points play a particular role in the description of certain orthogonal basis of  reproducing kernels in model spaces $K_I$, the
so-called {\emph{Clark basis}}. Given an inner function $I$ and $\alpha\in\T$, recall that by Herglotz theorem, there is a unique finite positive Borel measure $\sigma_{\alpha}$ on $\mathbb T$, singular with respect to the Lebesgue measure, such that
\begin{equation}\label{def-clark-measure}
\frac{1-|I(z)|^2}{|\alpha-I(z)|^2}=\int_{\mathbb T}\frac{1-|z|^2}{|\xi-z|^2}\,d\sigma_{\alpha}(\xi),\qquad z\in\mathbb D.
\end{equation}
The collection $(\sigma_{\alpha})_{\alpha\in\T}$ is the family of \emph{Clark measures} of $I$.
\par\smallskip
Let $E_\alpha=\{\zeta\in E_0(I): I(\zeta)=\alpha\}.$ By \cite[Theorem 21.5]{FM2}, the point $\zeta$ belongs to $E_\alpha$ if and only if the measure $\sigma_\alpha$ has an atom at $\zeta$. In this case,
\begin{equation}\label{eq:nice-formula-derivative-norme-kernel-clark-measure}
\sigma_\alpha(\{\zeta\})=\dfrac{1}{|I'(\zeta)|}=\dfrac{1}{\|k_{\zeta}^I\|_2^2}\cdot
 \end{equation}
 See \cite[Theorems 21.1 and 21.5]{FM2}.
When $\sigma_\alpha$ is a discrete measure, its support is exactly the set $E_\alpha$, which is necessarily countable, and we write it as
\begin{equation}
E_\alpha=\{\zeta_n:n\geq 1\}=\{\zeta\in E_0(I): I(\zeta)=\alpha\}.
\end{equation}
Then, in this case, Clark  proved in \cite{MR301534} that the family $\{k_{\zeta_n}^I: n\geq 1\}$ forms an orthogonal basis of $K_I$ (and the family $\{{k_{\zeta_n}^I}/{\|k_{\zeta_n}^I\|_2}: n\geq 1\}$ forms an orthonormal basis of $K_I$). It is called \emph{the Clark basis of $K_I$} associated to point $\alpha\in\mathbb T$.

\subsection{A description of $\HH(b)$ when $b$ is a rational function}
Although the contents of the space $\HH(b)$ may seem mysterious for a general non-extreme $b \in \operatorname{ball}(H^{\infty})$, it turns out that when $b$ is a rational function (and not a finite Blaschke product -- in which case $b$ is an inner function, and thus extreme),  the description of $\HH(b)$ is quite explicit. Since our $b$ is a non-extreme point of $\operatorname{ball}(H^{\infty})$, it admits a pythagorean mate $a$, which is also a rational function. In fact, the function $a$ can be obtained from the Fej\'{e}r--Riesz theorem  (see \cite{MR3503356}). Let $ \zeta_1, \dots, \zeta_n $ denote the {\em distinct} roots of $ a $ on $ \T $, with corresponding  multiplicities $ m_1, \dots, m_n $, and define the polynomial $ a_1 $ by
\begin{equation}\label{eq:definition of a}
a_1(z): =\prod_{k=1}^n (z-\zeta_k)^{m_k}.
\end{equation}
Results from  \cite{MR3110499, MR3503356} show that $\HH(b)$ has an explicit description as
\begin{equation}\label{eq:formula for H(b)}
\HH(b)=a_1H^2 \oplus \P_{N-1}=\mathcal{M}(a_1) \oplus \P_{N-1},
\end{equation}
where $ N=m_1+\dots+m_n $, and $\P_{N-1}$ denotes the set of polynomials of degree at most $N-1$. Since $a/a_1$ is invertible in $H^\infty=\mathfrak M(H^2)$, note that $\mathcal M(a)=\mathcal M(a_1)$. The notation $\oplus$ above denotes a topological direct sum in $\HH(b)$. But this sum may not be an orthogonal one. See \cite{MR3503356}. In particular, $\mathcal{M}(a_1) \cap \P_{N - 1} = \{0\}$. Moreover, if $ f\in\HH(b) $ is decomposed with respect to \eqref{eq:formula for H(b)} as
\begin{equation}\label{uUUiipPPS}
 f=a_1\widetilde{f}+p, \quad  \mbox{where $\widetilde{f}  \in H^2$ and  $p \in \P_{N - 1}$},
 \end{equation}
an equivalent norm on $ \HH(b) $  (to the natural one induced by the positive definite kernel $k_{\lambda}^{b}$, $\lambda\in\D$, above)  is
\begin{equation}\label{eq:norm in h(b)}
\vvvert a_1\widetilde f+p\vvvert^{2}_{b}:=\|\widetilde{f}\|^2_{H^2}+\|p\|^2_{H^2}.
\end{equation}
Note that the functions $\widetilde{f}  \in H^2$ and  $p \in \P_{N - 1}$ appearing in the decomposition (\ref{uUUiipPPS}) are unique.
 It is important to note that $ \vvvert \cdot\vvvert_b $ is only equivalent to the original norm $\|\cdot\|_b$ associated to the kernel in~\eqref{eq:original kernel H(b)}, and its scalar product as well as the reproducing kernels and the adjoints of operators defined on $\HH(b)$ will be different. However, the cyclicity problem for $S_b$ does not depend on the equivalent norm we consider. So, in the rational case, there is no problem to work  with the norm given by \eqref{eq:norm in h(b)}.
 \par\smallskip
 \par\smallskip
 Note also that when the zeros $\zeta_1,\ldots,\zeta_n$ of the polynomial $a_1$ are simple (i.e. when $m_k=1$, $1\le k\le n$), then the space $\HH(b)$ coincides with a Dirichlet type space $\mathcal{D}(\mu)$, where $\mu$ is a finite sum of Dirac masses at the points $\zeta_k$, $1\le k\le n$. See \cite{MR3110499}. So our results are also connected to the works \cite{EFKR1} and \cite{EFKR2} on the cyclicity problem for Dirichlet spaces.

 \par\smallskip
 Using \eqref{uUUiipPPS} and the standard estimate that any $g \in H^2$ satisfies
 \begin{equation}\label{gggbigggooooo}
 |g(z)| \leq \frac{\|g\|_{2}}{\sqrt{1 - |z|^2}} \quad \mbox{for all $z \in \D$},
 \end{equation}
 we see that for fixed $1 \leq k \leq n$ and for each $f \in \HH(b)$ we have
 \begin{equation}\label{q}
 f(\zeta_k) = \lim_{\substack{z\to\zeta_k\\  \sphericalangle}}f(z) = p(\zeta_k),
 \end{equation}
 where $f = a_1 \widetilde{f} + p$ with $\widetilde{f} \in H^2$ and $p \in \P_{N - 1}$. In particular,
 \begin{equation}\label{eq:E0b-rationnel}
 E_0(b)=\{\zeta_k:1\leq k\leq n\}.
 \end{equation}
 Finally, let us mention that when $b \in \operatorname{ball}(H^{\infty})$ is a rational function and not a finite Blaschke product, then $\mathfrak M(\HH(b))=H^\infty\cap \HH(b)$.  See \cite{MR3967886}.

 \par\smallskip
 \subsection{A description of $\HH(b)$ when $b=(1+I)/2$, with $I$ an inner function}
 There is another situation where we have an explicit description of the space \(\HH(b)\): this is when \(b=\frac{1+I}{2}\) and \(I\) is an inner function with \(I\not\equiv 1\). In this case, \(b\) is a non-extreme point of \(\textrm{ball}(H^{\infty})\), and its Pythagorean mate (up to a unimodular constant) is \(a=\frac{1-I}{2}\). Moreover, \((a,b)\) satisfies (HCR), since
\(|a|^{2}+|b|^{2}\geq\frac{1}{2}\) on $\mathbb D$. In particular \(\HH(b)=\mathcal M(\bar a)\), with equivalent norms.
\par\smallskip
Under the assumption that $I(0)\not = 0$, it is proved in \cite{MR3850543} that
\begin{equation}\label{eq:orthogonalite-hb-ma-KI}
\HH(b)=\mathcal M(a)\stackrel{\perp}{\oplus}_b  K_I,
\end{equation}
where the direct sum \(\stackrel{\perp}{\oplus}_b\)  is orthogonal with respect to the $\HH(b)$ norm.
In particular, every $f\in\HH(b)$ can be written in a unique way as
\begin{equation}\label{decomposition2-hb}
f=(1-I)g_1+g_2,\qquad \mbox{with }g_1\in H^2\mbox{ and }g_2\in K_I.
\end{equation}
It turns out that the same proof holds without any assumption on the value of $I(0)$. For completeness's sake, we present it in Lemma \ref{machin} below.
We also give an equivalent norm on $\HH(b)$ analogue to (\ref{eq:norm in h(b)}).

\begin{Lemma}\label{machin}
 Let \(I\) be an inner function with \(I\not\equiv 1\), and let \(b=(1+I)/2\). Then the following assertions hold:
 \begin{enumerate}
  \item [\emph{(i)}] \(\HH(b)=(1-I)H^{2}\stackrel{\perp}{\oplus}_bK_{I}\), where
  \(\stackrel{\perp}{\oplus}_b\) denotes an orthogonal direct sum in \(\HH(b)\);
  \item[\emph{(ii)}] if for \(f=(1-I)g_{1}+g_{2}\in\HH(b)\), \(g_{1}\in H^{2}\), \(g_{2}\in K_{I}\), we define
  \[
|||f|||_{b}^{2}=||g_{1}||_{2}^{2}+||g_{2}||_{2}^{2},
\]
then \(|||\,.\,|||_b\) is a norm on \(\HH(b)\) which is equivalent to \(||\,.\,||_{b}\).
 \end{enumerate}
\end{Lemma}

\begin{proof}
 (i) We have \(\HH(b)=\mathcal{M}(\bar{a})\) with equivalent norms, where \(a=\frac{1-I}{2}\) is the Pythago\-rean mate of \(b\). Also,
 \(
\frac{\bar{a}}{a}=\frac{1-\bar{I}}{1-I}=-\bar{I}
\)
a.e. on \(\T\), and thus \(T_{\bar{a}/a}=-T_{\bar{I}}\). Hence \(\ker T_{\bar{a}/a}=\ker T_{\bar{I}}=K_{I}\). Moreover, \(T_{a/\bar{a}}=-T_{I}\) has closed range, and thus
\begin{align}
 H^{2}=\textrm{Ran}(T_{I})\stackrel{\perp}{\oplus}\ker(T_{I}^{*})=\textrm{Ran}(T_{I})\stackrel{\perp}{\oplus}\ker(T_{\bar{I}})
 &=T_{a/\bar{a}}H^{2}\stackrel{\perp}{\oplus}K_{I}\label{Eq 1}
\end{align}
(the sign $\stackrel{\perp}{\oplus}$ denotes here an orthogonal direct sum in $H^2$).
Using now the fact that \(T_{\bar{a}}\) is an isometry from \(H^{2}\) onto \(\mathcal M(\bar{a})=T_{\bar{a}}H^{2}\) (equipped with the range norm), applying \(T_{\bar{a}}\) to the equation (\ref{Eq 1}), and using the identity \(T_{\bar{a}}\,T_{a/\bar{a}}=T_{a}\), we obtain
\[
\HH(b)=\mathcal M(\bar{a})=\mathcal{M}(a)\stackrel{\perp}{\oplus}_{\,\bar{a}}T_{\bar{a}}\,K_{I},
\]
where the notation \(\stackrel{\perp}{\oplus}_{\,\bar{a}}\) represents an orthogonal direct sum with respect to the range norm on \(\mathcal M(\bar{a})\). Since
\(T_{\bar{I}}\,K_{I}=\{0\}\) and \(\bar{a}=(1-\bar{I})/2\), we have \(T_{\bar{a}}\,K_{I}=(Id-T_{\bar{I}})\,K_{I}=K_{I}\), and so
\begin{equation}\label{Eq 2}
 \HH(b)=\mathcal M(\bar{a})=\mathcal{M}(a)\stackrel{\perp}{\oplus}_{\,\bar{a}}K_{I}=(1-I)H^{2}\stackrel{\perp}{\oplus}_{\,\bar{a}}K_{I}.
\end{equation}
It now remains to prove that the direct sum in this decomposition of \(\HH(b)\) is in fact orthogonal with respect to the \(\HH(b)\) norm.
\par\medskip
Let \(f\in H^{2}\) and \(g\in K_{I}\). Our aim is to show that \(\langle{(1-I)f},{g}\rangle_b=0\). Note that
\[
T_{\bar{b}}\,g=T_{(1+\bar{I})/2}\,g=\dfrac{1}{2}\,g=T_{\bar{a}}\,g
\]
from which it follows that
\begin{equation}\label{Eq 3}
 g^{+}=g.
\end{equation}
Moreover, since \(\bar{b}\,a=-\bar{a}\,b\) a.e. on \(\T\), we have
\[
T_{\bar{b}}\,\bigl ((1-I)f \bigr)=P_{+}\,(2\bar{b}\,af)=-P_{+}(2\bar{a}\,bf)=T_{\bar{a}}\,(-2bf),
\]
whence we get
\begin{equation}\label{Eq 4}
 \bigl ((1-I)f \bigr)^{+}=-2bf=-(1+I)f.
\end{equation}
By (\ref{Inconnu}), (\ref{Eq 3}) and (\ref{Eq 4}), it follows that
\begin{align*}
 \langle{(1-I)f},{g}\rangle_{b}&=\langle{(1-I)f},{g}\rangle_{2}-\langle{2bf},{g}\rangle_{2}
 =\langle{(1-I-2b)f},{g}\rangle_{2}
 =-2\,\langle{If},{g}\rangle_b=0
\end{align*}
because \(g\) belongs to \(K_{I}\).
\par\smallskip
(ii) Since \(\HH(b)=(1-I)H^{2}\stackrel{\perp}{\oplus}_{\,b}K_{I}\), we have
\[
||(1-I)g_{1}+g_{2}||_{b}^{2}=||(1-I)g_{1}||_{b}^{2}+||g_{2}||_{b}^{2},\qquad g_{1}\in H^2,\ g_{2}\in K_{I}.
\]
But observe that by (\ref{Inconnu}) and (\ref{Eq 4}) we have
\[
||(1-I)g_{1}||_{b}^{2}=||(1-I)g_{1}||_{2}^{2}+||(1+I)g_{1}||_{2}^{2}=4||g_{1}||_{2}^{2},
\]
while we get from (\ref{Inconnu}) and (\ref{Eq 3}) that \(||g_{2}||_{b}^{2}=2||g_{2}||_{2}^{2}\). Thus
\[
||(1-I)g_{1}+g_{2}||_{b}^{2}=4||g_{1}||_{2}^{2}+2||g_{2}||_{2}^{2},
\]
and from this the norm \(|||\,.\,|||_{b}\) is easily seen to be equivalent to \(||\,.\,||_{b}\).
\end{proof}

\par\smallskip
The next result is an analogue of \eqref{q} for the case where $b=(1+I)/2$ with respect to decomposition \eqref{decomposition2-hb}.

 \begin{Lemma}\label{lem:existence-limite-radiale}
 Let $I$ be an inner function, $I\not\equiv 1$, and let $b=(1+I)/2$. Let $\zeta\in E_0(I)$ be such that $I(\zeta)=1$. Then $\zeta\in E_0(b)$. Moreover, if $f=(1-I)g_1+g_2$, $g_1\in H^2$ and $g_2\in K_I$, then $f(\zeta)=g_2(\zeta)$.
 \end{Lemma}
 \begin{proof}
 As mentioned above, since $\zeta\in E_0(I)$ the function $g_2$ has a non-tangential limit at the point $\zeta$. Thus it remains to prove that $(1-I)g_1$ has a zero non-tangential limit at $\zeta$. To this purpose,  write for $z\in\mathbb D$
 \[
 \begin{aligned}
 (1-I(z))g_1(z)=&\frac{1-\overline{I(\zeta)}I(z)}{1-\overline{\zeta}z}(1-\overline{\zeta}z)g_1(z)\\
 =&k_{\zeta}^I(z) \,\overline{\zeta}\,(\zeta-z)g_1(z)\\
 =&\langle k_{\zeta}^I,k_{z}^I\rangle_2  \,\overline{\zeta}\,(\zeta-z)g_1(z).
 \end{aligned}
 \]
 Now, since $\zeta\in E_0(I)$, $k_{z}^I$ tends weakly to $k_\zeta^I$ as $z$ tends to $\zeta$ non-tangentially. Hence
 \[
 \lim_{\substack{z\to \zeta\\ \sphericalangle}}\langle k_{\zeta}^I,k_{z}^I\rangle_2=\|k_\zeta^I\|_2^2<\infty.
 \]
 Moreover, using the estimate \eqref{gggbigggooooo}, we obtain that
 \[
 \lim_{\substack{z\to \zeta\\ \sphericalangle}}(\zeta-z)g_1(z)=0,
 \]
from which it follows that
\[
 \lim_{\substack{z\to \zeta\\ \sphericalangle}}(1-I(z))g_1(z)=0.
\]
 \end{proof}
 In the case where $b=(1+I)/2$ and $I$ is an inner function with $I\not\equiv 1$, there is no complete characterisation of multipliers for $\HH(b)$. Nevertheless, we have at our disposal a sufficient condition which will be useful for our study of cyclicity. Before stating this result (Lemma \ref{Lem:multiplier-b-1+I}) on multipliers, we recall a well-known property of model spaces, of which we provide a proof for completeness's sake.

 \begin{Lemma}\label{Lem:model-space-KI.KIcontenu dans KI^2}
 Let $I$ be an inner function and let $f\in K_I$ and $g\in K_I\cap H^\infty$. Then $fg\in K_{I^2}$.
 \end{Lemma}

 \begin{proof}
Using that $K_I=H^2\cap I \overline{zH^2}$, write $f=I\overline{z\widetilde{f}}$ and $g=I\overline{z\widetilde{g}}$, with $\widetilde{f}, \widetilde{g}\in H^2$. Since $g\in H^\infty$, we indeed have $|\widetilde{g}|=|g|\in L^\infty(\mathbb T)$, and thus $\widetilde{g}\in H^\infty$. Moreover, $f g \in H^2$, and
 \[
 f g=I^2 \overline{z^2 \widetilde{f}\widetilde{g}},
 \]
 whence $f g\in H^2\cap I^2\overline{zH^2}=K_{I^2}$.
\end{proof}

 In the case where $b=(1+I)/2$, the de Branges-Rovnyak space $\mathcal H(b)$ contains a sequence of model spaces.
 \begin{Lemma}\label{Lem:sequence-model-spaces-contained-in-DBR}
Let $I$ be an inner function, $I\not\equiv 1$, and let $b=(1+I)/2$.  Then the following assertions hold:
\begin{enumerate}
\item the function $I$ is a multiplier of $\HH(b)$;
\item for every $n\geq 1$, $K_{I^n}\subset \HH(b)$.
\end{enumerate}
 \end{Lemma}

 \begin{proof}
 (a): Let $f\in\HH(b)$. According to \eqref{decomposition2-hb}, we can decompose $f$ as $f=(1-I)g_1+g_2$ with $g_1\in H^2$ and $g_2\in K_I$. Then
 \[
 If=(1-I)(Ig_1)+I g_2=(1-I)(Ig_1-g_2)+g_2
 \]
 and $Ig_1-g_2\in H^2$ and $g_2\in K_I$. Thus, using one more time \eqref{decomposition2-hb}, it follows that  $If\in\HH(b)$.

 (b): We argue by induction. For $n=1$, the property follows from Lemma~\ref{machin}. Assume that for some $n\geq 1$, $K_{I^n}\subset \HH(b)$. It is known that $K_{I^{n+1}}=K_I\oplus I K_{I^n}$. See \cite[Lemma 5.10]{MR3526203}. The conclusion now follows from the induction assumption and (a).
\end{proof}

 Here is now our sufficient condition for $f\in\HH(b)$ to be a multiplier of $\HH(b)$.

 \begin{Lemma}\label{Lem:multiplier-b-1+I}
 Let $I$ be an inner function, $I\not\equiv 1$, and let $b=(1+I)/2$. Assume that $f$ decomposes as $f=(1-I)g_1+g_2$, with $g_1\in H^\infty$ and $g_2\in H^\infty\cap K_I$. Then $f\in\mathfrak M(\HH(b))$.
 \end{Lemma}

 \begin{proof}
 We need to show that for every $\varphi\in\HH(b)$, we have $\varphi f\in\HH(b)$. According to \eqref{decomposition2-hb}, write $\varphi=(1-I)\varphi_1+\varphi_2$, with $\varphi_1\in H^2$ and $\varphi_2\in K_I$. Then
 \[
 \varphi f=(1-I)\varphi_1 f+\varphi_2 f.
 \]
 Since $f\in H^\infty$, $\varphi_1 f\in H^2$, and so the first term $(1-I)\varphi _1 f$ belongs to $(1-I)H^2$ which is contained in $\HH(b)$. Thus it remains to prove that $\varphi_2 f\in\HH(b)$. In order to deal with this term, write
 \[
 \varphi_2 f=(1-I)\varphi_2 g_1+g_2\varphi_2,
 \]
 and as before, since $g_1\in H^\infty$, the term $(1-I)\varphi _2 g_1$ belongs to $(1-I)H^2$, and so to $\HH(b)$. It remains to prove that $g_2\varphi_2\in\HH(b)$. Lemma~\ref{Lem:model-space-KI.KIcontenu dans KI^2} implies that $g_2\varphi_2\in K_{I^2}$, and the conclusion follows now directly from Lemma~\ref{Lem:sequence-model-spaces-contained-in-DBR}.

 \end{proof}

\section{Some basic facts on cyclic vectors for the shift operator}\label{sec3}
Recall that if $T$ is a bounded operator on a Hilbert space $\HH$, then a vector $f\in\HH$ is said to be cyclic for $T$ if the linear span of the orbit of $f$ under the action of $T$ is dense in $\HH$, i.e. if
\[
\mbox{Span}(T^n f:n\geq 0)=\overline{\{p(T)f:p\in\mathbb C[X]\}}=\HH.
\]
When $T=S_b$ is the shift operator on $\HH(b)$, we have $p(S_b)f=pf$ for every $f\in\HH(b)$ and every polynomial $p\in\C[X]$. Thus a function $f\in\HH(b)$ is cyclic for $S_b$ if and only if
\[
\overline{\{pf:p\in\mathbb C[X]\}}=\HH(b).
\]
In fact, it is sufficient to approximate the constant function $1$ by elements of the form $pf$, $p\in\C[X]$, to get that $f$ is cyclic for $S_b$.

\begin{Lemma}\label{Lem:cyclicite-constant1}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$ and $f\in\HH(b)$. Then the following assertions are equivalent:
\begin{enumerate}
\item $f$ is cyclic for $S_b$.
\item There exists a sequence of polynomials $(p_n)_n$ such that
\[
\|p_nf-1\|_b\to 0,\mbox{ as }n\to \infty.
\]
\end{enumerate}
\end{Lemma}

\begin{proof}
Follows immediately from the density of polynomials in $\HH(b)$ and the boundedness of $S_b$.
\end{proof}

The general meaning of our next result is that the set of zeros of a cyclic vector $f\in\HH(b)$ for $S_b$ cannot be too large.
\begin{Lemma}\label{Lem1:CS}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$ and $f\in\HH(b)$. Assume that $f$ is cyclic for $S_b$. Then we have the following properties:
\begin{enumerate}
\item $f$ is outer;
\item for every $\zeta\in E_0(b)$, $f(\zeta)\neq 0$.
\end{enumerate}
\end{Lemma}

\begin{proof}
(a) Since $f$ is cyclic for $S_b$, there exists a sequence of polynomials $(p_n)_n$ such that
\begin{equation}\label{eq-cyclicity-1-approximated}
\|p_nf-1\|_b\to 0,\mbox{ as }n\to \infty.
\end{equation}
Now, using the fact that $\HH(b)$ is contractively contained into $H^2$, we get that
\[
\|p_nf-1\|_2\to 0,\mbox{ as }n\to \infty.
\]
That proves that $f$ is cyclic for $S$ in $H^2$, and so $f$ is outer by Beurling's theorem.

\par\smallskip

(b) Since the functional $f\longmapsto f(\zeta)$ is bounded on $\HH(b)$ for every $\zeta\in E_0(b)$, we deduce  from \eqref{eq-cyclicity-1-approximated} that
\[
|p_n(\zeta)f(\zeta)-1|\to 0,\mbox{ as }n\to \infty
\]
for every $\zeta\in E_0(b)$.
This property implies directly that $f(\zeta)\neq 0$.
\end{proof}

We will encounter in the sequel of the paper some situations where the converse of Lemma~\ref{Lem1:CS} is also true, i.e. where conditions (a) and (b) of Lemma \ref{Lem1:CS} give a necessary and sufficient condition for a function $f\in\HH(b)$ to be cyclic.
\par\smallskip
We now provide some elementary results concerning cyclic functions for $S_b$.

\begin{Lemma}\label{lem:multi-inver}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$. Suppose that $f\in\mathfrak M(\HH(b))$ and that $1/f\in \HH(b)$. Then $f$ is cyclic for $S_b$.
\end{Lemma}
\begin{proof}
Using \eqref{eq:density-polynomial}, we see that there exists a sequence of polynomials $(p_n)_n$ such that
\[
\|p_n-f^{-1}\|_b\to 0,\mbox{ as }n\to\infty.
\]
Now, since $f\in\mathfrak M(\HH(b))$, the multiplication operator by $f$ is bounded on $\HH(b)$, and thus we get that
\[
\|p_nf-1\|_b\to 0,\mbox{ as }n\to \infty,
\]
which by Lemma~\ref{Lem:cyclicite-constant1} implies that $f$ is cyclic for $S_b$.
\end{proof}

In the following result, the set $\mbox{Hol}(\overline{\mathbb D})$ denotes the space of analytic functions in a neighborhood of the closed unit disc $\overline{\mathbb D}$.
\begin{Corollary}\label{cor:cyclicity-holvoisinagedeDbar}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$. Let $f\in \mbox{Hol}(\overline{\mathbb D})$ and assume that $\inf_{\overline{\mathbb D}}|f|>0$. Then $f$ is cyclic for $S_b$.
\end{Corollary}

\begin{proof}
When $b$ is a non-extreme point in $\operatorname{ball}(H^{\infty})$, we have $\mbox{Hol}(\overline{\mathbb D})\subset \mathfrak M(\HH(b))$. See \cite[Theorem 24.6]{FM2}.
Hence $f\in  \mathfrak M(\HH(b))$.  Moreover, the conditions on $f$ also imply that $1/f \in \mbox{Hol}(\overline{\mathbb D})$. In particular, $1/f\in\HH(b)$. It remains to apply Lemma~\ref{lem:multi-inver} in order to get that $f$ is cyclic.
\end{proof}

\begin{Lemma}\label{Lem:product-multiplicateur}
Let $f_1,f_2\in\mathfrak M(\HH(b))$. Then the following assertions are equivalent:
\begin{enumerate}
\item the product function $f_1f_2$ is cyclic for $S_b$;
\item each of the functions $f_1$ and $f_2$ is cyclic for $S_b$.
\end{enumerate}
\end{Lemma}

\begin{proof}
$(a)\implies (b)$: Assume that $f_1f_2$ is cyclic. By symmetry, it suffices to prove that $f_1$ is cyclic. Let $\varepsilon>0$. There exists a polynomial $q$ such that $\|qf_1f_2-1\|_b\leq \epsilon$. Now since the polynomials are dense in $\HH(b)$, we can also find a polynomial $p$ such that $$\|f_2q-p\|_b\leq \frac{\varepsilon}{\|f_1\|_{\mathfrak M(\HH(b))}}\cdot$$ Thus
\[
\begin{aligned}
\|pf_1-1\|_b\leq & \|pf_1-f_1f_2q\|_b+\|f_1f_2q-1\|_b
\leq \|f_1\|_{\mathfrak M(\HH(b))} \|p-f_2q\|_b+\varepsilon
\leq 2\varepsilon,
\end{aligned}
\]
which proves that $f_1$ is cyclic.

$(b)\implies (a)$: Assume that $f_1$ and $f_2$ are cyclic for $\HH(b)$. Let $\varepsilon>0$. There exists a polynomial $p$ such that $\|pf_1-1\|_b\leq \epsilon$. On the other hand, there is also a polynomial $q$ such that $$\|qf_2-1\|_b\leq \frac{\varepsilon}{\|pf_1\|_{\mathfrak M(\HH(b))}}\cdot$$ Now we have
\[
\begin{aligned}
\|pqf_1f_2-1\|_b\leq & \|pqf_1f_2-pf_1\|_b+\|pf_1-1\|_b
&\leq \|pf_1\|_{\mathfrak M(\HH(b))} \|qf_2-1\|_b+\varepsilon
&\leq 2\varepsilon.
\end{aligned}
\]
Hence the function $f_1f_2$ is cyclic.
\end{proof}

Our next result is motivated by the Brown--Shields conjecture and the work \cite{MR3475457} for Dirichlet type spaces $\mathcal D(\mu)$. Indeed,
let $\mu$ be a positive finite measure on $\T$, and let $\mathcal D(\mu)$ be the associated Dirichlet space (i.e. the space of holomorphic functions on $\D$ whose derivatives are square-integrable when weighted against the Poisson integral of the measure $\mu$). It is shown in \cite{MR3110499}, \cite{MR3390195} that in some cases, Dirichlet spaces and de Branges-Rovnyak spaces are connected. More precisely, let $b\in \textrm{ball}(H^{\infty})$ be a rational function (which is not a finite Blaschke product), and let $a$ be its pythagorean mate. Let also $\mu$ be a positive finite measure on $\T$. Then $\mathcal D(\mu)=\HH(b)$ with equivalent norms if and only if the zeros of $a$ on $\T$ are all simple, and the support of $\mu$ is exactly the set of these zeros \cite{MR3110499}.
In the context of Dirichlet spaces, the authors of \cite{MR3475457}
prove the Brown--Shields conjecture when the measure $\mu$ has countable support, using two notions of capacity (which they denote $c_{\mu}(F)$ and $c_{\mu}^a(F)$ respectively) and showing that they are comparable: $c_{\mu}(F)\le c_{\mu}^a(F)\le 4 \,c_{\mu}(F)$ for every $F\subset \mathbb T$ (\cite[Lemma 3.1]{MR3475457}).
In the same spirit, we introduce the following notions of {capacity} in $\HH(b)$-spaces. For a set $F\subset \mathbb T$, we define  $c_1(F)$ and $c_2(F)$ as
\[
c_1(F)=\inf\{\|f\|_b: f\in\HH(b),\,|f|\geq 1\mbox{ a.e. on a neighborhood of }F\},
\]
and
\[
c_2(F)=\inf\{\|f\|_b: f\in\HH(b),\,|f|=1\mbox{ a.e. on a neighborhood of }F\}.
\]
Observe  that $c_1(F)\leq c_2(F)$. We do not know if $c_1(F)$ and $c_2(F)$ are comparable in general in our context of de Branges-Rovnyak spaces.
\par\smallskip

Our next result should be compared to \cite[Lemma 3.2]{MR3475457}.

\begin{Theorem}\label{Thm:capacity}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$ and $\zeta\in\mathbb T$.
Consider the following assertions:
\begin{enumerate}
\item $z-\zeta$ is not cyclic for $S_b$;
\item $\zeta\in E_0(b)$;
\item $c_1(\{\zeta\})>0$;
\item $c_2(\{\zeta\})>0$.
\end{enumerate}
Then $(a)\iff(b)$, $(b)\implies (d)$ and $(c)\implies (a)$.
\end{Theorem}

\begin{proof}  $(b)\implies (a)$: follows immediately from Lemma~\ref{Lem1:CS}.
\par\smallskip
$(a)\implies (b)$: our assumption (a) exactly means that
\[
[z-\zeta]:=\mbox{Span}((z-\zeta)z^n:n\geq 0)\subsetneq \HH(b).
\]
Denote by $\pi$ the orthogonal projection from $\HH(b)$ onto $[z-\zeta]^\perp$. First note that $\pi(1)\neq 0$, otherwise we would have $1\in [z-\zeta]$ and then the function $z-\zeta$ would be cyclic for $S_b$,  which is a contradiction.

Let us now prove that $[z-\zeta]^\perp=\mathbb C \pi(1)$. For every $g\in [z-\zeta]^\perp$ and every $n\geq 0$, we have
\[
0=\langle g,(z-\zeta)z^n\rangle_b = \langle g,z^{n+1}\rangle_b-\overline{\zeta} \langle g,z^n\rangle_b.
\]
From this, we immediately get that
\begin{equation}\label{point}
 \langle g,z^n\rangle_b={\overline{\zeta}}^n \langle g,1\rangle_b, \quad n\ge 0.
\end{equation}
This implies that $\langle \pi(1),1\rangle_b\neq 0$ (otherwise, by (\ref{point}) we would have that $\pi(1)$ is orthogonal to $z^n$ for every $n\ge0$, which implies that $ \pi(1)=0$). Secondly, if we define $c:=\frac{\langle g,1\rangle_b}{\langle \pi(1),1\rangle_b}$, then we have $\langle g-c\pi(1),z^n\rangle_b=0$ for every $n\geq 0$. By the density of polynomials in $\HH(b)$, we deduce that $g=c\pi(1)$, which proves that $[z-\zeta]^\perp$ is of dimension $1$, generated by $\pi(1)$.

Now consider the continuous linear functional $\rho:\mathbb C\pi(1)\longrightarrow \mathbb C$ defined by $\rho(\alpha\pi(1))=\alpha$ for every $\alpha\in\C$. Let us check that for every $n\geq 0$,
\begin{equation}\label{eq13EZ:Thm:capacity}
(\rho\circ\pi)(z^n)=\zeta^n.
\end{equation}
For $n=0$, this is true by definition. Assume that \eqref{eq13EZ:Thm:capacity} is satisfied for some integer $n\geq 0$. Then,
\[
(\rho\circ\pi)(z^{n+1})=(\rho\circ\pi)(z^n(z-\zeta))+\zeta (\rho\circ\pi)(z^n)=\zeta^{n+1}.
\]
By induction, we deduce \eqref{eq13EZ:Thm:capacity} and by linearity, for any polynomial $p$, we have $(\rho\circ \pi)(p)=p(\zeta)$. Now, using the continuity of $\rho$ and $\pi$, we obtain that there exists a constant $C>0$ such that
\[
|p(\zeta)|\leq C \|p\|_b,\qquad \mbox{for any polynomial }p\in\C[X].
\]
Denote by $L_\zeta$ the linear functional defined on $\mathbb C[X]$ by $L_\zeta(p)=p(\zeta)$, $p\in\mathbb C[X]$. Then $L_\zeta$ is continuous on
$\C[X]$ endowed with the topology of
$\HH(b)$. Hence it extends to a continuous linear map on $\HH(b)$. By the Riesz representation theorem, there exists a unique vector $h_\zeta\in\HH(b)$, $h_\zeta\neq 0$,  such that
\[
p(\zeta)=L_\zeta(p)=\langle p,h_\zeta\rangle_b, \qquad \mbox{for any polynomial }p\in\C[X].
\]
Now, note that for any polynomial $p$, we have
\[
\langle p,S_b^* h_\zeta\rangle_b=\langle zp,h_\zeta \rangle_b=\zeta p(\zeta)=\langle p,\overline{\zeta}h_\zeta\rangle_b,
\]
whence, using \eqref{eq:density-polynomial}, $S_b^*h_\zeta=\overline{\zeta}h_\zeta$. In particular, $\overline{\zeta}$ belongs to the point spectrum of $S_b^*$. But by \eqref{point-spectrum-boundary}, this implies that $b$ has an angular derivative at $\zeta$, which is equivalent to the property that $\zeta\in E_0(b)$. Note that the function $h_\zeta$ is in fact the reproducing kernel $k_\zeta^b$ of $\HH(b)$ at the point $\zeta$.
\par\smallskip
$(b)\implies (d)$: assume now that $\zeta\in E_0(b)$. Let $f\in\HH(b)$ be such that $|f|=1$ a.e. on a neighborhood $\mathcal O$ of $\zeta$. Let us consider the inner-outer factorisation of $f=f_i f_o$, where $f_i$ is the inner part and $f_o$ the outer part of $f$. Since by definition $|f_i|=1$ a.e. on $\mathbb T$, we have $|f_o|=1$ a.e. on $\mathcal O$. Moreover, $f_o\in\HH(b)$ and $\|f_o\|_b\leq \|f\|_b$. Indeed, $f_o=T_{\bar f_i}f$, where $T_{\bar f_i}$ is the Toeplitz operator with symbol $\bar f_i$ and $\HH(b)$ is invariant with respect to co-analytic Toeplitz operators. Furthermore,
\[
\|f_o\|_b=\|T_{\bar f_i}f\|_b\leq \|f_i\|_\infty \|f\|_b=\|f\|_b.
\]
See \cite[Theorem 18.13]{FM2}. Since $f_o$ is outer and $\log|f_o|=0$ a.e. on $\mathcal O$, we have
\[
f_o(z)=\lambda\exp\left(\int_{\mathbb T\setminus\mathcal O}\frac{\xi+z}{\xi-z}\log|f_o(\xi)|\,dm(\xi)\right),
\]
for some constant $\lambda\in\mathbb T$. Hence $f_o$ is analytic in a neighborhood of $\zeta$ and in particular, we deduce that $|f_o(\zeta)|=1$. Using now the fact that $\zeta\in E_0(b)$,  we know that there exists a constant $C>0$ such that  $|g(\zeta)|\leq C \|g\|_b$ for every $g\in\HH(b)$. Hence
\[
1=|f_o(\zeta)|\leq C \|f_o\|_b\leq C \|f\|_b
\]
for every function $f\in\HH(b)$ such that $|f|=1$ a.e. on a neighborhood $\mathcal O$ of $\zeta$. We deduce that $c_2(\{\zeta\})\geq C^{-1}>0$.
\par\smallskip
$(c)\implies (a)$: by contradiction, assume that $z-\zeta$ is cyclic for $S_b$. Then, for every $\varepsilon >0$, we can find a polynomial $q$ such that $\|q(z-\zeta)-1\|_b\leq \varepsilon$. Note that the value of the polynomial $q(z-\zeta)-1$ at $\zeta$ is $-1$. So by continuity, we can find a neighborhood $\mathcal O$ of $\zeta$ on $\mathbb T$ such that $|q(z-\zeta)-1|\geq 1/2$ on $\mathcal O$. Hence $|2(q(z-\zeta)-1)|\geq 1$ on $\mathcal O$ and by definition of $c_1(\{\zeta\})$, we obtain that
\[
c_1(\{\zeta\})\leq 2\|q(z-\zeta)-1\|_b\leq 2 \varepsilon.
\]
Since this is true for every $\varepsilon>0$, we deduce that $c_1(\{\zeta\})=0$, which contradicts $(c)$.
\end{proof}

If we knew that $c_1(F)$ and $c_2(F)$ were comparable, assertions (a) to (d) in Theorem \ref{Thm:capacity} would be equivalent. This motivates the following question:

\begin{question}
(i) Does there exist $\kappa>0$ such that  $c_2(F)\le \kappa\, c_1(F)$ for every $F\subseteq\T$?

\noindent
(ii) Is it true that $c_1(F)>0$ if and only if $c_2(F)>0$?
\end{question}

\begin{Remark}
It can be easily seen from Theorem \ref{Thm:capacity} that the condition $\inf_{\overline{\mathbb D}}|f|>0$ in Corollary~\ref{cor:cyclicity-holvoisinagedeDbar} is not necessary for $f$ to be cyclic in $\HH(b)$. Indeed, let $b(z)=\frac{1+z}{2}S_{\delta_1}(z)$, where $S_{\delta_1}$ is the singular inner function associated to $\delta_1$, the Dirac measure at point $1$. See \eqref{E:dec-b-bso}. It is clear that
\[
\int_{\mathbb T}\frac{d\delta_1(\xi)}{|\zeta_0-\xi|^2}=\frac{1}{|\zeta_0-1|^2}=\infty
\]
when $\zeta_0=1$. Hence $1\notin E_0(b)$. Therefore, by Theorem~\ref{Thm:capacity}, the function $z-1$ is cyclic for $S_b$ while $\inf_{\overline{\mathbb D}}|z-1|=0$.
\end{Remark}

\begin{Corollary}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$. Let $p$ be a polynomial.
The following assertions are equivalent:
\begin{enumerate}
\item $p$ is cyclic for $S_b$.
\item $p(z)\neq 0$ for every $z\in\mathbb D\cup E_0(b)$.
\end{enumerate}
\end{Corollary}

\begin{proof}
$(a)\implies (b)$: follows immediately from Lemma~\ref{Lem1:CS}.

$(b)\implies (a)$: factorise the polynomial $p$ as $p(z)=c\prod_{j=1}^n (z-\zeta_j)$, where by definition the roots $\zeta_j$ belong to $\mathbb C\setminus (\mathbb D\cup E_0(b))$. On the one hand, if $|\zeta_j|>1$, then, according to Corollary~\ref{cor:cyclicity-holvoisinagedeDbar}, the function $z-\zeta_j$ is cyclic for $S_b$. On the other hand, if $|\zeta_j|=1$, then $\zeta_j\not\in E_0(b)$ and Theorem~\ref{Thm:capacity} implies that  the function $z-\zeta_j$ is also cyclic for $S_b$. Thus, for every $1\leq j\leq n$, the function $z-\zeta_j$ is cyclic and it follows from Lemma~\ref{Lem:product-multiplicateur} that $p$ itself is cyclic for $S_b$.  \end{proof}

This result can be slightly generalised:

\begin{Corollary}\label{Cor:cyclicite-fonction-holomorphe-dans-Dbar}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$. Let $f\in\mbox{Hol}(\overline{\mathbb D})$.
The following assertions are equivalent:
\begin{enumerate}
\item $f$ is cyclic for $S_b$.
\item $f$ is outer and $f(\zeta)\neq 0$ for every $\zeta\in E_0(b)$.
\end{enumerate}

\end{Corollary}

\begin{proof}
$(a)\implies (b)$: follows from Lemma~\ref{Lem1:CS}.

$(b)\implies (a)$: since $f$ is outer and $f\in\mbox{Hol}(\overline{\mathbb D})$, $f$ does not vanish on the unit disc and has at most a finite number of zeros on $\mathbb T$ (otherwise by compactness and the uniqueness principle for holomorphic functions, $f$ would vanish identically). Let $\zeta_1,\zeta_2,\dots,\zeta_n$ be the (possible) zeros of $f$ on $\mathbb T$. Then there exists a function $g\in\mbox{Hol}(\overline{\mathbb D})$ with $\inf_{\overline{\mathbb D}}|g|>0$ such that
\[
f(z)=\prod_{j=1}^n (z-\zeta_j) g(z),\qquad z\in\overline{\mathbb D}.
\]
Our assumption implies that for every $1\leq j\leq n$, $\zeta_j\notin E_0(b)$, and thus by Theorem~\ref{Thm:capacity}, the function $z-\zeta_j$ is cyclic for $S_b$. Moreover, by Corollary~\ref{cor:cyclicity-holvoisinagedeDbar}, the function $g$ is also cyclic. Now it follows from Lemma~\ref{Lem:product-multiplicateur} that $f$ itself is cyclic for $S_b$.
\end{proof}

\begin{Example}\label{example-rkhardycyclic}
Let $b$ a non-extreme point in $\operatorname{ball}(H^{\infty})$. For every $\lambda\in\mathbb D$, $k_\lambda$ is a cyclic vector for $S_b$. Indeed, it is clear that $k_\lambda(z)=(1-\overline{\lambda}z)^{-1}$ satisfies the conditions of Corollary~\ref{Cor:cyclicite-fonction-holomorphe-dans-Dbar}. Hence $k_\lambda$ is cyclic. \
\par\smallskip
In particular, by \eqref{eq:density-crk}, the set of cyclic vectors for $S_b$ spans a dense subspace in $\HH(b)$.
\end{Example}

\par\medskip
\section{The rational case}\label{sec4}

The main result of this section is a characterisation of cyclic functions for $S_b$ when $b$ is a rational function which is not a finite Blaschke product. As mentioned already in the Introduction, this result can be derived from the work \cite{MR4246975} by Luo -- Gu -- Richter, but we provide here an elementary proof, the ideas of which will turn out to be also relevant to the case where $b=(1+I)/2$ (see Section \ref{sec5} below). Note that Theorem \ref{Thm:rational-case} extends a result proved in \cite{MR3309352}
in the particular case where $b(z)=(1+z)/2$.

\begin{Theorem}\label{Thm:rational-case}
Let $b\in \operatorname{ball}(H^{\infty})$ and assume that $b$ is rational (but not a finite Blaschke product). Let $a_1$ be the associated polynomial given by \eqref{eq:definition of a}, and let $f\in \HH(b)$. Then the following assertions are equivalent:
\begin{enumerate}
\item $f$ is cyclic for $S_b$.
\item $f$ is an outer function and for every $1\leq k\leq n$, $f(\zeta_k)\neq 0$.
\end{enumerate}
\end{Theorem}

\begin{proof}
$(a)\implies (b)$: according to \eqref{eq:E0b-rationnel}, we know that $E_0(b)=\{\zeta_k:1\leq k\leq n\}$. Hence this implication follows from Lemma~\ref{Lem1:CS}.

$(b)\implies (a)$: according to \eqref{uUUiipPPS}, write $f=a_1\widetilde{f}+p$, where $\widetilde f\in H^2$ and $p\in\P_{N-1}$. By \eqref{q}, $p(\zeta_k)\neq 0$, $1\leq k\leq n$.
Let now $r\in\P_{N-1}$ be the unique polynomial satisfying the following interpolation properties:  for every $1\leq k\leq n$,
\[
r^{(j)}(\zeta_k)=\begin{cases}
\frac{1}{p(\zeta_k)}& \mbox{if }j=0 \\
-\frac{1}{p(\zeta_k)}\,{\sum_{\ell=0}^{j-1}\binom{j}{\ell}r^{(\ell)}(\zeta_k)p^{(j-\ell)}(\zeta_k)}& \mbox{if }1\leq j\leq m_{k}-1.
\end{cases}
\]
This polynomial $r$ can be constructed using Hermite polynomial interpolation, see for instance \cite[Chapter 1, E. 7]{MR1367960}.
By Leibniz's rule, we easily see that for every $1\leq k\leq n$ and $0\leq j\leq m_k-1$, we have $(rp-1)^{(j)}(\zeta_k)=0$. Hence $a_1$ divides the polynomial $rp-1$. In other words, there exists a polynomial $q$ such that $rp-1=a_1 q$. Using that $f$ is outer, we can find a sequence of polynomials $(q_n)_n$ such that $\|q_nf+r\widetilde{f}+q\|_2\to 0$ as $n\to \infty$. Define now a sequence of polynomials $(p_n)$ by $p_n=a_1q_n+r$, $n\geq 1$. Observe that
\[
\begin{aligned}
p_nf-1=&(a_1q_n+r)f-1=a_1q_nf+r(a_1\widetilde{f}+p)-1\\
=&a_1(q_nf+r\widetilde{f})+rp-1=a_1(q_nf+r \widetilde{f}+q).
\end{aligned}
\]
Then it follows from \eqref{eq:norm in h(b)} that
\[
\vvvert p_nf-1\vvvert_b=\vvvert a_1(q_nf+r \widetilde{f}+q) \vvvert_b=\|q_nf+r \widetilde{f}+q\|_2\to 0\,\textrm{ as } n \to \infty.
\]
Therefore $f$ is cyclic for $S_b$.
 \end{proof}

\begin{Example}
Let $b(z)=\frac{1}{2}(1-z^2)$. Then it is proved in \cite{MR3503356} that $a(z)=c(z-i)(z+i)$, for some constant $c$. Thus, according to Theorem~\ref{Thm:rational-case}, a function $f\in\HH(b)$ is cyclic for $S_b$ if and only if $f$ is outer, $f(i)\neq 0$ and $f(-i)\neq 0$.
\end{Example}

\section{The case where $b=(1+I)/2$}\label{sec5}
Our main result in this section is the following:

\begin{Theorem}\label{Thm:main-b-1+I}\label{sec5-th}
Let $I$ be an inner function, $I\not\equiv 1$, and assume that its Clark measure $\sigma_1$ associated to point $1$ (defined in \eqref{def-clark-measure}) is a discrete measure. Let $\{\zeta_n:n\geq 1\}=\{\zeta\in E_0(I): I(\zeta)=1\}$. Let $b=(1+I)/2$, and $f\in\HH(b)$ which we decompose according to \eqref{decomposition2-hb} as $f=(1-I)g_1+g_2$, with $g_1\in H^2$, $g_2\in K_I$. Assume that:
\begin{enumerate}
\item $g_1,g_2\in H^\infty$;
\item $f$ is outer;
\item we have
\[
\sum_{n\geq 1}\frac{1}{|f(\zeta_n)|^2 |I'(\zeta_n)|}<\infty.
\]
\end{enumerate}
Then $f$ is cyclic for $S_b$.
\end{Theorem}

\begin{proof}
The proof proceeds along the same lines as in the  rational case.
According to Lemma~\ref{Lem:multiplier-b-1+I}, $f\in\mathfrak M(\HH(b))$ and by Lemma~\ref{lem:existence-limite-radiale} we have $f(\zeta_n)=g_2(\zeta_n)$, $n\geq 1$.
\par\smallskip
\noindent
{\emph{First step}}: We claim that there exists a sequence of functions $(\psi_n)_n$ in $\HH(b)$ such that $\|\psi_n f-1\|_b\to 0$ as $n\to\infty$.
\par\smallskip
In order to construct the sequence $(\psi_n)_n$,  let us first consider the function $r$ given by
\[
r=\sum_{n=1}^\infty \frac{1}{f(\zeta_n)}\frac{k_{\zeta_n}^I}{\|k_{\zeta_n}^I\|_2^2}\cdot
\]
Recall that by \eqref{eq:nice-formula-derivative-norme-kernel-clark-measure}, $\|k_{\zeta_n}^I\|_2^2=|I'(\zeta_n)|$.
Combining this with condition (c) and the fact that the family $(k_{\zeta_n}^I/\|k_{\zeta_n}^I\|_2)_n$ forms an orthonormal basis of $K_I$ (since $\sigma_1$ is discrete, Clark's theorem holds true), we see that the series defining the function $r$ is convergent in $K_I$. In others words, $r\in K_I$ and $r(\zeta_n)=1/f(\zeta_n)=1/g_2(\zeta_n)$ for every $n\geq 1$.
\par\smallskip
Let us now prove that $rg_2-1\in (1-I)H^2$. Observe that Lemma~\ref{Lem:model-space-KI.KIcontenu dans KI^2} implies that $rg_2\in K_{I^2}=K_I\oplus IK_I$. Hence there exist $\varphi_1,\varphi_2\in K_I$ such that $rg_2-1=\varphi_1+I\varphi_2-1$. Since $r(\zeta_n)g_2(\zeta_n)-1=0$, we have $\varphi_1(\zeta_n)+\varphi_2(\zeta_n)-1=0$ for every $n\geq 1$.
Note that \((1-\overline{I(0)})^{-1}(1-\overline{I(0)}I)=((1-\overline{I(0)})^{-1} k_0^I \in K_{I}\) and
\[
(1-\overline{I(0)})^{-1}(1-\overline{I(0)}I(\zeta _{n}))=1\qquad\textrm{for every}\ n\ge 1.
\]
Since the family \(\bigl (k_{\zeta _{n}}^{I} \bigr)_{n} \) is complete in \(K_{I}\), and since \(\varphi _{1}(\zeta _{n})+\varphi _{2}(\zeta _{n})=1\) for every \(n\ge 1\), we deduce that
\[
\varphi _{1}+\varphi_{2}=(1-\overline{I(0)})^{-1}(1-\overline{I(0)}I).
\]
Hence
\begin{align*}
 rg_{2}-1&=\varphi _{1}+I\varphi _{2}-1
 =-\varphi _{2}+(1-\overline{I(0)})^{-1}(1-\overline{I(0)}I)+I\varphi _{2}-1\\
 &=(1-I)(-\varphi _{2})+(1-\overline{I(0)})^{-1}(1-\overline{I(0)}I)-1.
\end{align*}
Observe that
\[
(1-\overline{I(0)})^{-1}(1-\overline{I(0)}I)-1=(1-\overline{I(0)})^{-1}\overline{I(0)}(1-I),
\]
from which it follows that
\[
rg_{2}-1=(1-I)(-\varphi _{2}+\overline{I(0)}(1-\overline{I(0)})^{-1}).
\]
This proves that \(rg_{2}-1\) belongs to \((1-I)H^{2}\). Write \(rg_{2}-1\) as
\(rg_{2}-1=(1-I)g_{3}\), with \(g_{3}\in H^{2}\).
\par\smallskip
Using now that $f$ is outer, and that $rg_1\in H^2$ (as $g_1\in H^{\infty}$) we can find a sequence of polynomials $(q_n)_n$ such that $\|q_nf+rg_1+g_3\|_2\to 0$, as $n\to\infty$. We then define for each $n\geq 1$ a function $\psi_n$ as
\[
\psi_n:=(1-I)q_n+r,
\]
where $q_n$ and $r$ are defined above. Note that $\psi_n\in (1-I)H^2+K_I=\HH(b)$ and
\[
\begin{aligned}
\psi_n f-1=&(1-I)q_nf+rf-1\\
=&(1-I)q_nf+(1-I)rg_1+rg_2-1\\
=&(1-I)(q_nf+rg_1+g_3).
\end{aligned}
\]
It follows from Lemma~\ref{machin} that there exists a positive constant $C$ such that
\[
\begin{aligned}
\|\psi_nf-1\|_b =&\|(1-I)(q_nf+rg_1+g_3)\|_b\\
\le& C\, \vvvert (1-I)(q_nf+rg_1+g_3) \vvvert_b\\
=& C\, \|q_nf+rg_1+g_3\|_2,
\end{aligned}
\]
from which it follows that $\|\psi_nf-1\|_b\to 0$ as $n\to\infty$.
\par\smallskip
{\emph{Second step}}: Let us now prove that there exists a sequence of polynomials $(p_{n})_n$ such that $\|p_ {_n}f-1\|_b\to 0$ as $n\to\infty$.
\par\smallskip
By the density of polynomials in $\HH(b)$, we can find a sequence of polynomials $(p_{n})_n$ such that $\|p_{n}-\psi_n\|_b\to 0$ as $n\to\infty$. Now write
\[
\begin{aligned}
\|p_{n}f-1\|_b\leq& \|p_{n}f-\psi_n f\|_b+\|\psi_n f-1\|_b\\
\leq & \|f\|_{\mathfrak M(\HH(b)} \|p_{n}-\psi_n\|_b+\|\psi_n f-1\|_b,
\end{aligned}
\]
and by the choice of the sequence $(p_{n})_n$ and the first step, we get the conclusion of the second step.
\par\smallskip
We finally conclude that $f$ is cyclic for $S_b$.
\end{proof}

\begin{Remark}
If $I$ is an inner function such that, for some $\alpha\in\mathbb T$, its Clark measure $\sigma_\alpha$ is a discrete measure and $I\not\equiv \alpha$, then we may apply Theorem~\ref{sec5-th} replacing $I$ by $\bar\alpha I$ and $b=(1+I)/2$ by $b=(1+\bar\alpha I)/2$.
\end{Remark}

\begin{Example}
Let \(I=S_{\delta _{1}}\) be the inner function associated to the measure \(\delta _{1}\):
\[
I(z)=\exp\Bigl (-\dfrac{1+z}{1-z} \Bigr) .
\]
In this case we can compute explicitly the Clark basis of \(K_{I}\) associated to point $1$. We have \(E_1=\{\zeta\in E_0(I):I(\zeta)=1\}=\{\zeta _{n}\;;\;n\in\mathbb{Z}\}\) with
\[\zeta _{n}=\dfrac{2i\pi n-1}{2i\pi n+1}\quad\textrm{and}\quad I'(\zeta _{n})=-\dfrac{1}{2}(2i\pi n+1)^{2},\ n\in\mathbb{Z}.\]
Therefore, if \(f\in\mathcal{H}(b)\) is outer, with \(f=(1-I)g_{1}+g_{2}\), \(g_{1}\in H^{\infty}\), \(g_{2}\in K_{I}\cap H^{\infty}\), and if
\[
\sum_{n\in\mathbb{Z}}\dfrac{1}{|f(\zeta _{n})|^{2}}\cdot\dfrac{1}{4n^{2}\pi ^{2}+1}<+\infty,
\]
then \(f\) is cyclic for \(S_{b}\).
\end{Example}

\begin{Remark}
There exists a recipe to construct an inner function satisfying the hypothesis of Theorem~\ref{sec5-th}. Let $\sigma$ be a positive discrete measure on $\mathbb T$ and let $H\sigma$ be its Herglotz transform,
\[
H\sigma(z)=\int_{\mathbb T}\frac{\xi+z}{\xi-z}\,d\sigma(\xi),\qquad z\in\mathbb D.
\]
We easily see that $H\sigma$ defines an analytic function on $\mathbb D$ and satisfies $\Re e (H\sigma(z))\geq 0$ for every $z\in\D$. Now define a function $I$ on $\mathbb D$ as
\[
I=\frac{H\sigma-1}{H\sigma+1}\cdot
\]
Since $\Re e H\sigma\geq 0$, it is easy to check that $I\in H^\infty$ and $|I|\leq 1$. Moreover, for every $0<r<1$ and $\zeta\in\mathbb T$, we have
\begin{equation}\label{trick-construction-inner-function}
|I(r\zeta)|=\frac{(\Re e (H\sigma(r\zeta))-1)^2+(\Im m (H\sigma(r\zeta))^2}{(\Re e (H\sigma(r\zeta))+1)^2+(\Im{m}(H\sigma(r\zeta))^2}\cdot
\end{equation}
Since $\sigma$ is a singular measure, it is well-known that for almost all $\zeta\in\mathbb T$, we have
\[
\Re e(H\sigma(r\zeta))=\int_{\mathbb T}\frac{1-r^2}{|\xi-r\zeta|^2}\,d\sigma(\xi)\to 0\quad\mbox{as }r\to 1^-.
\]
See \cite[Corollary 3.4]{FM1}. Moreover, the radial limit of $\Im{m}(H\sigma)$ also exists and is finite for almost all $\zeta\in\mathbb T$. See \cite[page 113]{FM1}. Thus, it follows from \eqref{trick-construction-inner-function} that $|I(\zeta)|=1$ for almost all $\zeta\in\mathbb T$,  meaning that $I$ is an inner function. Of course, we have $I\not\equiv 1$. Now, we easily check that
\[
\frac{1-|I(z)|^2}{|1-I(z)|^2}=\Re e(H\sigma(z))=\int_{\mathbb T}\frac{1-|z|^2}{|\xi-z|^2}\,d\sigma(\xi),
\]
which implies by unicity of the Clark measure that $\sigma_1=\sigma$. Therefore $I$ satisfies the assumptions of Theorem~\ref{sec5-th}.
\end{Remark}

\begin{Corollary}\label{Cor-rk-cyclique}
 Let $I$ be an inner function, $I\not\equiv 1$, and assume that $\sigma_1$ is a discrete measure. Let $b=(1+I)/2$.
 Then  $k_\lambda^b$ is a cyclic vector for $S_b$ for every $\lambda\in\mathbb D$.
\end{Corollary}

\begin{proof}
Let us prove that $k_\lambda^b$ satisfies the assumptions $(a),(b)$ and $(c)$ of Theorem~\ref{sec5-th}. First, using that $b=(1+I)/2$, straightforward computations show that
\[
\frac{1-\overline{b(\lambda)}b(z)}{1-\overline{\lambda}z}=(1-I(z))\frac{1}{4}\cdot\frac{(1-\overline{I(\lambda)})}{1-\overline{\lambda}z}+\frac{1}{2}\cdot\frac{1-\overline{I(\lambda)}I(z)}{1-\overline{\lambda}z}\cdot
\]
In other words, $k_\lambda^b$ can be written as $k_\lambda^b=(1-I)g_1+g_2$, with $g_1=\frac{1}{4}(1-\overline{I(\lambda)})k_\lambda$ and $g_2=\frac{1}{2}k_\lambda^I$. In particular, $g_1,g_2\in H^\infty$ and $k_\lambda^b$ satisfies the assumption $(a)$.
\par\smallskip
Observe now that $\Re e(1-\overline{b(\lambda)}b(z))\geq 0$ and $\Re e(1-\overline{\lambda}z)\geq 0$ for every $z\in\D$, which implies that the functions $1-\overline{b(\lambda)}b(z)$ and $1-\overline{\lambda}z$ are outer. See \cite[page 67]{MR1864396}. So $k_\lambda^b$ is outer as the quotient of two outer functions. It remains to check that $k_{\lambda}^b$ satisfies assumption $(c)$. But
\[
|k_\lambda^b(\zeta_n)|=\left|\frac{1-b(\lambda)}{1-\overline{\lambda}\zeta_n}\right|\geq \frac{|1-b(\lambda)|}{1+|\lambda|},
\]
and the property $(c)$ follows from the fact that
 \[
\sum_{n\geq 1}\frac{1}{|I'(\zeta_n)|}=\sum_{n\geq 1}\sigma_1(\{\zeta_n\})\le\sigma_1(\T)<+\infty.
\]

Thus $k_\lambda^b$ satisfies the assumptions $(a),(b),(c)$ of Theorem~\ref{sec5-th}, and $k_\lambda^b$ is cyclic for $S_b$.

\end{proof}
It should be noted that in Corollary~\ref{Cor-rk-cyclique}, the reproducing kernels $f=k_\lambda^b$, $\lambda\in \mathbb D$, which are cyclic for $S_b$, are such that $1/f\in H^\infty$.
As we already observed in Lemma~\ref{lem:multi-inver},
certain invertibility conditions for $f$ make cyclicity easier. Using Theorem \ref{Thm:main-b-1+I}, we now construct a family of functions $f$ which are cyclic for $S_b$ but are such that $1/f\notin H^2$.

\begin{Example}
 Let $I$ be a non-constant inner function, and assume that $\sigma_1$ is a discrete measure.  Let $b=(1+I)/2$ and $f=(1+I)k_{\lambda}^I$ for some $\lambda\in\D$. Then $f$ is cyclic for $S_b$ and $1/f\notin L^2$.
\end{Example}

\begin{proof}
  First observe that
\[
f=(1-I)(-k_\lambda^I)+2k_\lambda^I,
\]
so that $f=(I-I)g_1+g_2$, with $g_2=-2g_1=2k_\lambda^I \in H^\infty\cap K_I$. In particular, $f$ satisfies condition $(a)$ of Theorem~\ref{Thm:main-b-1+I}. Moreover, the function $f$ is outer as  the product of two outer functions (use the same arguments as in the proof of Corollary~\ref{Cor-rk-cyclique}). Finally, since $|f(\zeta_n)|=2|k_\lambda^I(\zeta_n)|\geq |1-I(\lambda)|$,  $f$ satisfies condition $(c)$. Hence by Theorem~\ref{Thm:main-b-1+I}, $f$ is cyclic for $S_b$.
\par\smallskip
Let us now check that $1/f\not \in L^2$. First observe that there exist two positive constants $C_1$ and $C_2$ such that
\[
C_1 \frac{1}{|1+I(\zeta)|}\le
\frac{1}{|f(\zeta)|}\le C_2 \frac{1}{|1+I(\zeta)|}\quad\mbox{ for a.e. }\zeta\in\T,
\]
because
\[
\frac{1-|I(\lambda)|}{2}\leq |k_\lambda^I(\zeta)|\leq \frac{2}{1-|\lambda|}\cdot
\]
Now assume that $1/f$ belongs to $L^2$. Then  $1/(1+I)\in L^2$. But since $1+I$ is outer, we get that $1/(1+I)\in H^2$. See \cite[page 43]{MR1864396}. As
\[
\frac{1}{1+I}=\overline{\frac{I}{1+I}}\quad\mbox{ for a.e. }\zeta\in\T,
\]
we deduce that $1/(1+I)$ also belongs to $\overline{H^2}$. Thus $1/(1+I)$ is constant, which is a contradiction.
\end{proof}

In the context of Corollary~\ref{Cor-rk-cyclique}, it is easy to see that $(a,b)$ forms a corona pair, and we have seen that $k_\lambda^b$ is cyclic for $S_b$. In fact, this cyclicity result holds true under the (HCR) condition only.

\begin{Proposition}\label{chose}
Let $b$ be a non-extreme point in $\operatorname{ball}(H^{\infty})$, and let $a$ be its pythagorean mate. Assume that $(a,b)$ satisfies (HCR). Then the following. assertions hold:
\begin{enumerate}
\item $k_\lambda^b$ is cyclic for $S_b$ for every $\lambda\in\mathbb D$;
\item If $b$ is furthermore assumed to be outer, then $bk_\lambda$ is also cyclic for $S_b$ for every $\lambda\in\mathbb D$. In particular, $b$ is a cyclic vector for $S_b$.
\end{enumerate}
\end{Proposition}

\begin{proof}
(a) We have $p(S_b)k_\lambda^b=(1-\overline{b(\lambda)}b)p(S_b)k_\lambda$ for every polynomial $p\in\C[X]$. Since $(a,b)$ satisfies (HCR), we have $\HH(b)=\MM(\bar a)$, and $b$ is a multiplier of $\HH(b)$. See \cite[Theorems 28.7 and 28.3]{FM2}. In particular, the multiplication operator $T=M_{1-\overline{b(\lambda)}b}$ is bounded on $\HH(b)$ and we have
\[
p(S_b)k_\lambda^b=Tp(S_b)k_\lambda\qquad \textrm{for every polynomial }p\in\C[X].
\]
Since $k_\lambda$ is cyclic for $S_b$ (see Example~\ref{example-rkhardycyclic}), in order to check that $k_\lambda^b$ is also cyclic for $S_b$ it is sufficient to check that $T$ has dense range. Let $h\in\HH(b)$ be such that $h\perp \mbox{Range}(T)$. Then $h\perp Tk_\mu=k_\mu-\overline{b(\lambda)}bk_\mu$ for every $\mu\in\mathbb D$. Lemma~\ref{Lem:completenes} now implies that $h=0$, proving that $T$ has dense range. It follows that $k_\lambda^b$ is cyclic for $S_b$.

(b) The proof of (b) proceeds along the same lines of (a). We have
\[
p(S_b)bk_\lambda=Vp(S_b)k_\lambda\qquad \textrm{for every polynomial }p\in\C[X],
\]
where $V=M_b$ is the multiplication operator by $b$. As previously, in order to show that $bk_\lambda$ is cyclic for $S_b$, it is sufficient to check that $V$ has a dense range. Let $h\in\HH(b)$ be such that $h\perp \mbox{Range}(V)$. Then $h\perp Vk_\mu=bk_\mu$ for every $\mu\in\mathbb D$. By \eqref{eq1EZD:lem-completeness}, it then follows that $h^+(\mu)=0$ for every $\mu\in\mathbb D$. Then $h^+=0$ and $T_{\bar b}h=T_{\bar a}h^+=0$. But, since $b$ is outer, $T_{\bar b}$ is one-to-one, which implies that $h=0$. It then follows that $bk_\lambda$ is cyclic for $S_b$.
\end{proof}

We finish the paper with the following question:

\begin{question}
Does Proposition \ref{chose} hold true without the assumption that $(a,b)$ satisfies (HCR)?
\end{question}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


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On September 22, 2018, BAPS Charities hosted the 5th annual community Health Fair in Memphis, TN. Over 124 individuals benefitted from the volunteer services provided by 24 physicians, nurses, and other medical professionals who donated their time to advise participants on preventative care and provide tips to improve general health as well as mental wellbeing.


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U.S. Navy sailor Peter Mims punished for hiding on USS Shiloh
July 18, 2017 / 3:02 PM / CBS News
A U.S. Navy sailor whose disappearance prompted a 50-hour search spanning a 5,500-square-mile area has admitted to hiding for a week, CBS News national security correspondent David Martin reports.
Petty Officer 3rd Class Peter Mims was reported missing from the USS Shiloh, a guided-missile cruiser, on June 8 while the ship was conducting routine operations 180 miles east of Okinawa. He was presumed overboard.
The Japanese navy and coast guard joined the U.S. in searching the Philippine Sea for Mims before suspending search efforts in the water at midnight on June 11.
But the search on the Shiloh continued while the crew began planning a memorial service.
Mims came out of hiding on June 15. He is small in stature and apparently had been able to squeeze into hiding places in the ship's engineering spaces where his shipmates didn't think he could be.
Mims has been punished for abandoning watch and dereliction of duty. The Navy didn't release details of his punishment, but the maximum would be loss of pay and confinement in the brig for six months.
The Navy said it is pursuing additional administrative procedures against Mims, which could mean being kicked out of the service with a bad conduct discharge.
Flights viewing damage to Tonga from huge undersea volcano eruption
These are the top U.S. athletes to watch in the Beijing Olympics
FBI identifies suspect who took hostages at Texas synagogue
First published on July 18, 2017 / 3:02 PM

redpajama_set_name: RedPajamaCommonCrawl


text:
/*
 * 2012-3 Red Hat Inc. and/or its affiliates and other contributors.
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 * http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */
package org.overlord.rtgov.epn.embedded;

import org.overlord.rtgov.ep.EventProcessor;

public class TestEventProcessor3 extends EventProcessor {

    public java.io.Serializable process(String source, java.io.Serializable event, int retriesLeft) throws Exception {
        // TODO Auto-generated method stub
        return null;
    }

}

redpajama_set_name: RedPajamaGithub


text:
Texas Uniform Trade Secrets Act
View cases from Texas
On May 2, 2013, Texas officially signed the Texas Uniform Trade Secret Act ("TUTSA") into law. The act will replace Texas common law, effective September 1, 2013, and apply to all trade secret misappropriation claims filed thereafter.
The Texas legislature amended Section 1, Title 6 of Texas' Civil Practice and Remedies code by adding Chapter 134A - Trade Secrets. The relevant portions of the act are included below:
CHAPTER 134A. TRADE SECRETS
Sec. 134A.001. SHORT TITLE. This chapter may be cited as
the Texas Uniform Trade Secrets Act.
Sec. 134A.002. DEFINITIONS. In this chapter:
(1) "Claimant" means a party seeking to recover damages under this chapter, including a plaintiff, counterclaimant, cross-claimant, or third-party plaintiff. In an action in which a party seeks recovery of damages under this chapter on behalf of another person, "claimant" includes both that other person and the party seeking recovery of damages.
(2) "Improper means" includes theft, bribery, misrepresentation, breach or inducement of a breach of a duty to maintain secrecy, to limit use, or to prohibit discovery of a trade
secret, or espionage through electronic or other means.
(3) "Misappropriation" means:
(A) acquisition of a trade secret of another by a person who knows or has reason to know that the trade secret was acquired by improper means; or
(B) disclosure or use of a trade secret of another without express or implied consent by a person who:
(i) used improper means to acquire knowledge of the trade secret;
(ii) at the time of disclosure or use, knew or had reason to know that the person's knowledge of the trade secret was:
(a) derived from or through a person who had utilized improper means to acquire it;
(b) acquired under circumstances giving rise to a duty to maintain its secrecy or limit its use; or
(c) derived from or through a person who owed a duty to the person seeking relief to maintain its secrecy or limit its use; or
(iii) before a material change of the person's position, knew or had reason to know that it was a trade secret and that knowledge of it had been acquired by accident or
mistake.
(4) "Proper means" means discovery by independent development, reverse engineering unless prohibited, or any other means that is not improper.
(5) "Reverse engineering" means the process of studying, analyzing, or disassembling a product or device to discover its design, structure, construction, or source code
provided that the product or device was acquired lawfully or from a person having the legal right to convey it.
(6) "Trade secret" means information, including a
formula, pattern, compilation, program, device, method, technique,
process, financial data, or list of actual or potential customers
or suppliers, that:
(A) derives independent economic value, actual or potential, from not being generally known to, and not being readily ascertainable by proper means by, other persons who can
obtain economic value from its disclosure or use; and
(B) is the subject of efforts that are reasonable under the circumstances to maintain its secrecy.
Sec. 134A.003. INJUNCTIVE RELIEF.
(a) Actual or threatened misappropriation may be enjoined. On application to the
court, an injunction shall be terminated when the trade secret has ceased to exist, but the injunction may be continued for an additional reasonable period of time in order to eliminate commercial advantage that otherwise would be derived from the misappropriation.
(b) In exceptional circumstances, an injunction may condition future use upon payment of a reasonable royalty for no longer than the period of time for which use could have been
prohibited. Exceptional circumstances include a material and prejudicial change of position before acquiring knowledge or reason to know of misappropriation that renders a prohibitive injunction inequitable.
(c) In appropriate circumstances, affirmative acts to protect a trade secret may be compelled by court order.
Sec. 134A.004. DAMAGES.
(a) In addition to or in lieu of injunctive relief, a claimant is entitled to recover damages for misappropriation. Damages can include both the actual loss caused by misappropriation and the unjust enrichment caused by misappropriation that is not taken into account in computing actual loss. In lieu of damages measured by any other methods, the damages caused by misappropriation may be measured by imposition of liability for a reasonable royalty for a misappropriator's unauthorized disclosure or use of a trade secret.
(b) If wilful and malicious misappropriation is proven by clear and convincing evidence, the fact finder may award exemplary damages in an amount not exceeding twice any award made under Subsection (a).
Sec. 134A.005. ATTORNEY'S FEES. The court may award reasonable attorney's fees to the prevailing party if:
(1) a claim of misappropriation is made in bad faith;
(2) a motion to terminate an injunction is made or resisted in bad faith; or
(3) wilful and malicious misappropriation exists.
Sec. 134A.006. PRESERVATION OF SECRECY. In an action under this chapter, a court shall preserve the secrecy of an alleged trade secret by reasonable means. There is a presumption in favor of granting protective orders to preserve the secrecy of trade secrets. Protective orders may include provisions limiting access to confidential information to only the attorneys and their experts, holding in camera hearings, sealing the records of the action, and ordering any person involved in the litigation not to disclose an alleged trade secret without prior court approval.
Sec. 134A.007. EFFECT ON OTHER LAW.
(a) Except as provided by Subsection (b), this chapter displaces conflicting
tort, restitutionary, and other law of this state providing civil remedies for misappropriation of a trade secret.
(b) This chapter does not affect:
(1) contractual remedies, whether or not based upon misappropriation of a trade secret;
(2) other civil remedies that are not based upon misappropriation of a trade secret; or
(3) criminal remedies, whether or not based upon misappropriation of a trade secret.
(c) To the extent that this chapter conflicts with the Texas Rules of Civil Procedure, this chapter controls. Notwithstanding Section 22.004, Government Code, the supreme court may not amend or adopt rules in conflict with this chapter.
(d) This chapter does not affect the disclosure of public information by a governmental body under Chapter 552, Government Code.
The law also includes a section to ensure its uniform application and construction, which defines the term "theft" (as used in Sec. 134A.002 as an example of "improper means") to mean "unlawfully appropriating property or unlawfully obtaining services as described by Section 31.03, 31.04, 31.06, 31.07, 31.11, 31.12, 31.13, or 31.14, Penal Code."
Texas became the 48th state to adopt some form of the Uniform Trade Secrets Act, leaving New York and Massachusetts as the only remaining holdouts. Prior to Texas' adoption of the TUTSA, the state applied First Restatement of Torts §757 to trade secret misappropriation claims. Texas courts will continue to apply the Restatement to misappropriation claims made either before September 1, 2013, or afterwards if the claim alleges a "continuing misappropriation" that began before the TUTSA's effective date. §757 of the First Restatement of Torts reads as follows:
§ 757. Liability For Disclosure Or Use Of Another's Trade Secret—General Principle
One who discloses or uses another's trade secret, without a privilege to do so, is liable to the other if
(a) he discovered the secret by improper means, or
(b) his disclosure or use constitutes a breach of confidence reposed in him by the other in disclosing the secret to him, or
(c) he learned the secret from a third person with notice of the facts that it was a secret and that
the third person discovered it by improper means or that the third person's disclosure of it was otherwise a breach of his duty to the other, or
(d) he learned the secret with notice of the facts that it was a secret and that its disclosure was made to him by mistake.
See Tewari De-Ox Sys., Inc. v. Mountain States/Rosen, LLC, 637 F.3d 604 (5th Cir. 2011), footnote 2.
Additionally, Texas had previously passed legislature imposing a statute of limitations for trade secrets claims. The TUTSA does not appear to explicitly override, nor conflict with, the statute (included below). Barring additional guidance from either the Texas legislature or courts, the statue should still be good law:
Tex. Civ. Prac. & Rem. Code § 16.010
§ 16.010. Misappropriation of Trade Secrets
(a) A person must bring suit for misappropriation of trade secrets not later than three years after the misappropriation is discovered or by the exercise of reasonable diligence should have been discovered.
(b) A misappropriation of trade secrets that continues over time is a single cause of action and the limitations period described by Subsection (a) begins running without regard to whether the misappropriation is a single or continuing act.


text:
Escape from Death Block 13 (2021) Review
The first trailer for Escape from Death Block 13 showed up nearly three years ago in early 2019. Given all the hype around its star, Charles Bronson lookalike Robert Bronzi (Death Kiss, Once Upon a Time in Deadwood) it seemed to be taking a long time getting released. And now that it's here I still can't see why.
Mick Kovack (Robert Bronzi) has come to America from Hungary to settle the score with his dead brother's boss Renda (Nicholas Turturro, The Penthouse, Angels Fallen). But rather than pay him the money he owes he sets his goons on him. Which doesn't go very well for the goons until he intervenes with a large gun.
They all swear Kovak started it and he's hauled off to jail. And the judge that will hear the case just happens to be related to Renda and will see that he's sent to the badly misnamed Pleasant Hill Penitentiary. Detective Borelli (Lyindaa Russell, 6 Feet Below Hell, Black Mamba) and FBI Agent Langley (Lawrence Hilton-Jacobs, Welcome Back Kotter, Compton's Finest) offer him a deal. If he works as an informant they'll get him out when the time is right. Of course, we all know that's easier said than done.
Escape from Death Block 13 is writer/director Gary Jones (Mosquito, Axe Giant: The Wrath of Paul Bunyan) return to directing after concentrating on his career as an effects man for the past few years. That need for a change of pace may explain why he stepped away from the horror and sci-fi films he was known for in favour of an old-school action film.
And at its heart Escape from Death Block 13 is an old-fashioned prison film, its title even evokes the old Lee Van Cleef film Escape from Death Row. It's all here, the violent inmates and sadistic guards all overseen by the corrupt Warden Jack (Debbie Scaletta, The Curse of Lilith Ratchet, The Kingsbury Run). This means several fights, mostly involving the warden's stooge Bunyan (Chris Hahn, Inoperable, Wrong Turn).
Unfortunately, some of those fight scenes are not exactly brilliantly choreographed. They're far from the worst I've seen, but they're also fairly clunky and shot with very static camera work. I get that Carver didn't have the budget to stage elaborate brawls, but I've seen lower-budget films like Nightshooters do it a lot better too. Thankfully the throwdown between Kovack and Bunyan that sets up the film's climactic riot is given much more care and attention than the other similar scenes.
I wish as much attention had been paid to the riot's CGI blood and gunfire effects. The last half hour of Escape from Death Block 13 is full of action, but the effects take a lot away from them. The best scene, which involves a guard ending up on the wrong end of his cattle prod, is all practical, however.
Like The Shawshank Redemption, Escape from Deathblock 13 was filmed in Ohio's Mansfield Reformatory which gives it an authentically oppressive atmosphere. Unlike that film though they didn't have the budget to fill it with extras which leaves some scenes looking comically underpopulated. Better choices of camera angles could have helped with that, this is one time emphasizing the vastness of the location wasn't a good idea.
As for Bronzi himself, this was the first time I saw him in a film that wasn't directed by Rene Perez. I've read that Perez dubbed someone else's voice in to make him sound more like Bronson. That would explain why he sounds a bit different here and some of his dialogue scenes are cringingly bad. Thankfully once he gets to Pleasant Hill he mostly lets his fists do the talking. His next two films were shot in England by the uber prolific Scott Jeffrey (The Mutation, Bats), which should be an interesting team-up.
Weighing it all out, Escape from Deathblock 13 is an average low-budget action film. Yes, it has some problems, but it also has a lot more going for it than plenty of films that found a distributor a lot quicker. Certainly worth a weekend watch with pizza and plenty of beer.
Uncork'd Entertainment is releasing Escape from Death Block 13 On Demand on November 2nd and DVD on November 23rd. You can check the film's website and Facebook page for details.
Where to watch Escape from Death Block 13
Cowgirls vs. Pterodactyls (2021) Review
By Jim Morazzini / February 7, 2021 October 13, 2021
Review: HOLLAND ROAD MASSACRE: THE LEGEND OF PIGMAN (2020)
By Jim Morazzini / April 8, 2020 September 23, 2021
Review: EXPO (2019)
By Jim Morazzini / July 31, 2019 August 11, 2021


Straight hole with green elevated. Take an extra club. Green slopes from back to front.
The Kitchen People are a well-established company on the Hibiscus Coast and have been designing, manufacturing, and installing high quality kitchens, bathrooms, laundries, wardrobes and custom made joinery for all budgets to the greater Auckland area (central city – Mangawhai) for over 30 years.
Let us help turn your ideas into reality with our designers or yours. Call us for a free no obligation measure and quote in your home, or bring your plans into our showroom where you can see the latest products and materials on offer.
You can decide how much or how little you want to do as we can organise our local tradies to assist or work with your own to create a smooth transition from old to new.
So give us a call we look forward to hearing from you soon. 09 424 3468.


Phoenix is the capital and most populous city of the U.S. You'll find plenty of places to visit near Phoenix: Boyce Thompson Arboretum (in Superior), Scottsdale (Taliesin West & McDowell Sonoran Preserve) and St. Anne's Roman Catholic Church (in Gilbert). The adventure continues: examine the collection at Musical Instrument Museum, see the world from above with a helicopter tour, enjoy unforgettable views with a balloon tour, and explore the galleries of Desert Botanical Garden.
For where to stay, traveler tips, ratings, and more tourist information, read our Phoenix trip planner .
In March, daily temperatures in Phoenix can reach 82°F, while at night they dip to 57°F. Finish your sightseeing early on the 28th (Thu) to allow enough time to travel back home.


text:
Santa Isabel princesa de Hungría (Castillo de Saróspatak, Borsod-Abaúj-Zemplén, 7 de julio de 1207-Marburgo, Alemania, 17 de noviembre de 1231) fue una princesa húngara y terciaria franciscana. Canonizada por la Iglesia Católica y nombrada patrona de la orden tercera de san Francisco de Asís.

Isabel se quedó viuda siendo aún joven, dedicó su riqueza a los pobres, construyó hospitales y allí atendió personalmente a los necesitados. A partir de su canonización en 1236 se convirtió en un símbolo de caridad cristiana para toda Europa, extendiéndose su culto muy rápidamente y profundamente desde los territorios germánicos, polacos, húngaros, checos, hasta los italianos, ibéricos y franceses.

Biografía
Isabel nació en 1207 como hija del rey Andrés II de Hungría y su esposa Gertrudis de Andechs-Merania. Su madre era hermana de la religiosa que posteriormente será conocida como Santa Eduviges de Silesia. Santa Isabel creció en la corte húngara junto a sus hermanos los príncipes Béla, Colomán y Andrés. En 1215 su padre tomó una nueva esposa dos años después del asesinato de su madre, y nació una única hija, Violante de Hungría (la posterior esposa del rey Jaime I de Aragón). En 1221 Isabel se casó con el landgrave Luis de Turingia-Hesse y según los registros y leyendas, el matrimonio estuvo caracterizado por un amor correspondido y felicidad. A Luis no le preocupaba demasiado el reparto de su riqueza entre los pobres que Isabel solía llevar a cabo, ya que creía que la labor caritativa de su esposa le traería una recompensa eterna; se la venera en Turingia como santa. Luis fue un aliado cercano y defensor acérrimo de los Hohenstaufen, y en particular del emperador germánico Federico II, por lo cual en la primavera de 1226, cuando Turingia se vio asolada por inundaciones, hambre y la peste, Luis representó a Federico II en la Dieta de Cremona. En esta ocasión, Isabel asumió el control de sus asuntos y repartió limosnas por todo su territorio, incluso dando vestidos y adornos de la corte a los pobres. Debajo del castillo de Wartburgo, hizo construir un hospital con 28 camas, y visitaba todos los días a los enfermos para atenderlos. En esa misma época, el inquisidor Konrad von Marburg se convirtió en su director espiritual.

La vida de Isabel cambió radicalmente cuando Luis murió a causa de la plaga el 11 de septiembre de 1227 en Otranto (Italia) cuando se dirigía a unirse a la Sexta Cruzada conducida por Federico II. Pocos días después, el 29 de septiembre, Santa Isabel dio a luz a su hija, la beata Gertrudis de Altenberg, la cual fue enviada a un claustro de las monjas Premonstratenses junto a Wetzlar, donde fue criada como religiosa y murió a una edad avanzada como abadesa en 1300.

Isabel murió en Marburgo, bien debido a agotamiento físico o a una enfermedad, cuando contaba solo 24 años de edad. Fue canonizada por el Papa Gregorio IX en 1235, hallándose presente en la ceremonia el propio emperador Federico II Hohenstaufen. Su carta papal puede verse en el Schatzkammer de la Deutschordenskirche en Viena, Austria. El día de Pentecostés (28 de mayo) del año 1235, durante la ceremonia de canonización, se la llamó "la mujer más grande de la Edad Media alemana". Su cuerpo se colocó sobre un magnífico altar dorado —que hoy todavía puede visitarse— en la Elisabethkirche (Iglesia de Santa Isabel, en alemán) en Marburgo. Ahora es una iglesia protestante, pero cuenta con espacios reservados a la fe católica. Marburgo se convirtió en el centro de la Orden Teutónica, que adoptó a Santa Isabel como su segunda patrona. La Orden permaneció en Marburgo hasta que Napoleón I de Francia la disolvió en 1803.

En la cultura

Expansión de su culto
El culto a Santa Isabel se extendió muy rápidamente justo después de su canonización, adoptándose en territorios germánicos, húngaros, polacos y checos. El nombre de Isabel cobró enorme popularidad tras esto y en los territorios ibéricos. Tras el matrimonio de Violante de Hungría y Jaime I de Aragón, se convirtió en uno de los nombres españoles más tradicionales entre la monarquía y la gente común por igual.

Luego del matrimonio de la princesa María (la hija del rey Esteban V de Hungría) con Carlos II de Nápoles y Sicilia en 1270, el culto de Santa Isabel se extendió igualmente con gran fuerza a los territorios italianos, quedando numerosos frescos sobre su ciclo de leyendas como el del domo de Santa Maria Donna Regina Nuova.

Casi inmediatamente después de su muerte, Santa Isabel pasó a ser una santa patrona de la Orden de los Caballeros Teutónicos junto a la Virgen María y San Jorge.

También es patrona de la Orden Franciscana Seglar (antiguamente conocida como Tercera Orden Franciscana o Venerable Orden Tercera de San Francisco) junto con San Luis, Rey de Francia. Santa Isabel mantuvo durante años frecuente trato con los hermanos franciscanos de la zona y, una vez abandonada la vida de la Corte, llegó a profesar la Regla franciscana e instituir una fraternidad junto con cuatro compañeras, anteriormente sus sirvientas.

La capilla más pequeña del mundo (de 1,96 metros cuadrados) está dedicada a Santa Isabel de Hungría y se encuentra en el monumento Castillo de Colomares de Benalmádena, provincia de Málaga, España.

Santa Isabel de Hungría es la Santa Patrona de la ciudad de Bogotá, capital de la República de Colombia.

Es la patrona de las enfermeras en España. Se celebra su fiesta y Día de las Enfermeras en España el 17 de noviembre.

Reliquia de su cráneo

La reliquia del cráneo de Santa Isabel de Hungría llegó a Bogotá de manos de Mons. Luis Zapata de Cárdenas, quien fuese hacia el año 1573 el Arzobispo Metropolitano de Santafé de Bogotá.

Antes de su destino final en Colombia, la reliquia ya había recorrido un largo camino. La reforma protestante en los estados alemanes había causado, entre otras cosas, la profanación de los restos de la santa, antes venerados en la Catedral de Marburgo. Pero un descendiente de Isabel de Hungría, Felipe I de Hesse, las retuvo consigo hasta 1548, cuando el Rey Carlos V lo obliga a restituirlos. Parte de las reliquias fueron entregadas a Ana de Austria, la cuarta esposa de Felipe II, quien se las confía al Arzobispo Zapata de Cárdenas.

La reliquia, resguardada en una caja de plata en forma del busto de la Santa, permanece en la bóveda del tesoro de la Catedral primada de Colombia

Representaciones

Santa Isabel se suele representar comúnmente como una dama vestida con ropajes reales y corona cargando un cesto lleno de rosas. Esto es una alegoría al milagro de las rosas que ocurrió cuando estaba obsequiando pan a los pobres y fue sorprendida por su cuñado Conrado de Turingia. Para que no tuviese que mentir, o fuese castigada por el noble germánico, según la leyenda Dios hizo convertir el contenido del cesto en rosas.

Igualmente es común hallar representaciones de Santa Isabel, donde está rodeada de gente pobre o enferma, aunque casi como motivo fijo se halla el cesto lleno de rosas o de hogazas de pan.

Literatura

La obra de teatro áureo "El ejemplo de virtudes y Santa Isabel Reina de Hungría", firmada por "Una dama sevillana", se basa en esta historia y exalta la devoción religiosa de la santa, especialmente, después de la muerte de su marido, cuando decide consagrarse a Dios.

Véase también
 Santa Isabel de Hungría curando a los tiñosos, cuadro realizado en 1672 por el pintor español Bartolomé Esteban Murillo. El pintor era terciario franciscano del .
 Isabel de Aragón (reina de Portugal)
 Casilda de Toledo
 Diego de Alcalá

Referencias

Enlaces externos

 Biografía de Santa Isabel de Hungría en Santoral Católico de www.divinavoluntad.net
 Catholic Encyclopedia: Santa Isabel de Hungría (en inglés)
 Elisabeth of Thuringia gravesite (en inglés)
 Saint Elisabeth of Hungary - París (en francés)
 El Papa relanza al mundo el testimonio de caridad de Santa Isabel de Hungría
 Carta del Papa en el VII centenario de santa Isabel de Hungría
 Podcast audio IVOOX sobre el ideario de santa Isabel de Hungría
 Podcast audio en speaker sobre las reliquias de santa Isabel de Hungría
 Enlace al twiter de la orden tercera de san Francisco de Asís
 Directorio franciscano sobre la Orden franciscana secular

Nobles de Hungría del siglo XIII
Landgravinas de Alemania
Consortes de la casa de Árpád
Santas
Santos nobles
Santos franciscanos
Santos católicos de Hungría del siglo XIII
Personas canonizadas por Gregorio IX
Santos de la Comunión anglicana
Santos luteranos
Húngaras del siglo XIII
Nacidos en Bratislava
Princesas reales de la casa de Árpád
Nobles alemanas del siglo XIII

redpajama_set_name: RedPajamaWikipedia


The Kenyan president's official chair had a temporary new occupant last week – and Kenyans aren't happy. The offending rear belongs to gospel star Bahati, whose unprecedented breach of protocol won't be repeated any time soon. By NJERI KIMANI.
It was business as usual for Kenyan gospel superstar Bahati as he performed in front of a large audience at the launch of the Jubilee Party at a Nairobi stadium. The crowd was on its feet as he launched into Mama, his best-known hit, and even President Uhuru Kenyatta – seated in the front row seat of honour – looked as though he was enjoying himself.
A true showman, Bahati decided to get the president more involved. He shimmied over and coaxed the smiling Kenyatta from his seat. Emboldened, Bahati thought he'd go back for the president's wife, first lady Margaret Kenyatta. The song is about Kenyan mothers, after all, so it was only fitting to include her.
She took a little more convincing than her husband, so Bahati sat in the empty chair next to her to try to talk her into getting on her feet.
Except that it wasn't an empty chair, exactly. It was the president's chair, an ornate wooden and red leather number with the presidential seal branded into the back rest.
In Kenya, this is an extraordinary breach of protocol. No one sits in the president's chair except the president himself, and Bahati's transgression did not go unnoticed by Kenyatta, who stood awkwardly on the stage as the musician sang from his throne. At one point, Bahati even had his feet up on the stool reserved for the president's refreshments.
Afterwards, Bahati joked about the incident, describing himself as Kenya's "shortest-serving president" on social media. But not everyone thought it was funny.
On both social and traditional media, Bahati came in for plenty of criticism.
"Society operates on strict social rules, some of which are unwritten. Sociologists talk of social distance defining relationships. There are invisible lines that must never be crossed. That line was crossed at Saturday's launch of the Jubilee Party," said an editorial the Standard, which went on to describe Bahati as "highly disrespectful".
"No one is even allowed to be near the dais without the security personnel knowledge. The president's actions are well choreographed. [Security] know when he will sit and when he will stand and how he will move. They must have been gotten completely unaware," said Samson Cherambos, who was head of security for former president Daniel Arap Moi. Cherambos said the security breach exposed Kenyatta to an increased risk of attack.
The intense backlash eventually forced a sheepish Bahati to apologise on national TV.
Bahati said he had completely forgotten protocol as he got caught up in the moment. He was relieved, however, that State House seems to have brushed off the incident.
"I was in constant fear that I would be jailed any time. I was just waiting for the security officers to come and arrest me for breach of protocol. I knew I would be jailed any time and I'm grateful it hasn't happened," he said.
Joseph Omondi, Nakuru Centre of Accountability Forum Chair, said Kenyans should follow the lead of the president and relax a little.
"Our president is a cool guy and doesn't take some of these things seriously. However, some people still harbour the school of thought from traditionalists who take the issue of presidency seriously. They feel that everything around him should be given some modicum of respect. This includes the seat and even the water he takes," said Omondi.
If anything, Omondi argues, the incident was good for the president's image, showing him as a "friendly man" who can adapt with grace to any situation.