**One sided Shift**


Let be $M$ separable metric space. Consider $X=M^{\mathbb{N}}$ the sequence space equipped with the product metric $d(x,y)=\sum\_{i=1}^\infty |x\_i-y\_i|/2^i$ . Define the shift map $\sigma:X\to X$ by putting in each sequence $x=(x\_i)$, $\sigma(x)\_i=x\_{i+1}$. Then we have that the shift has the following propery: given $x\in X$ and $\epsilon>0$ there exists a uniform nonnegative integer $K$ such that for each $z\in X$ there exist $y\in \sigma^{-K}(z)\cap B(x,\epsilon)$.


**Weighted Shift**


Now $X=\ell^p({\mathbb{N}})$ with $1 \leq p < \infty$
endowed with it standard norm.
We fix values $0 < c < c'$ and a sequence $(\alpha\_n)\_{n \geqslant 1}$ satisfying $\alpha\_n \in (c, c')$ for each $n \in \mathbb{N}$. The weighted shift associated to the sequence $(\alpha\_n)\_{n \geqslant 1}$  

is defined as the linear map $L : X \to X$ given by
$$
L((x\_n)\_{n \geqslant 1}) = (\alpha\_n x\_{n+1})\_{n \geqslant 1} \,.
$$


For each $k,n\in\mathbb{N}$ we define $\beta\_k^n \equiv \alpha\_k \ldots \alpha\_{k+n-1}$, if


\begin{eqnarray}
\label{dnn}
\sum\_n (\beta\_1^n)^{-p}<\infty
\end{eqnarray}
then is a known result that $L$ is **frequently hypercyclic** in $\ell^p(\mathbb{N})$, which means that,there exists $x$ such that for any open set $V$ the set $N(x, V)=\{n: T^nx\in V\}$ has positive lower density, i.e
$$
\liminf\_{n\to \infty} \dfrac{1}{n}\# N(x,V)\cap\{1,\ldots,n \}>0.
$$


**My Question:**


I'am looking for to know if there exist any condition over the weight sequence $(\alpha\_n)$ in order to have a property a a bit more stronger that frequently hypercyclicity for the the weighted shift.


More precisely I would like to know any condition over the weight sequence $(\alpha\_n)$ in order to have the following property(silimar to the ordinary one-sided shift): given $x\in X$ and $\epsilon>0$ a uniform nonnegative integer $K$ such that for each $z\in X$ there exists some $y\in L^{-K}(z)\cap B(x,\epsilon)$.



Is the weighted shift strong frequently hypercyclic?
The answer to the question is no. If the sequence $\alpha = (\alpha\_{n})\_{\geqslant 1}$ satisfies the condition $c<\alpha\_{n}<c^\prime$ there is no other condition on a sequence $\alpha$ for the answer to be affirmative. Let's assume that the answer to the question is affirmative. That is, there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for every open ball $B(x,\epsilon)\subset \ell^p({\mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z\in \ell^p({\mathbb{N}})$ there exists $y=(y\_{n})\_{n\in \mathbb{N}} = (y\_0,y\_1,y\_3,\ldots,y\_n,\ldots)$ satisfying the condition
$$
{\rm (i)}\; y\in\ell^{p}(\mathbb{N}),
\qquad 
{\rm (ii)}\;y\in L^{-K}\_{\alpha}(z) 
\qquad \mbox{ and } \qquad 
{\rm (iii)}\; y\in B(x,\epsilon)
$$
We observe that for every integer $K\geqslant 1$ and every $({\bf u}\_{n})\_{n\in\mathbb{N}}\in \ell^p(\mathbb{N})$ we have, in a susceptible way,
$$
L^{K}\_{\alpha}\Big( ({\bf u}\_{n})\_{n\in\mathbb{N}} \Big) = \left( \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}\Big)\cdot {\bf u}\_{K+n} \right)\_{n\in\mathbb{N}}.
$$
From the observation we made above about the $L\_{\alpha}$ operator the first condition below is an immediate consequence of the assumption we made. Furthermore, the second condition is a consequence of the first condition and the third condition is a consequence of the second condition.


**Condition 1.** there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for any open ball $B(x, \epsilon)\subset \ell^p({ \mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z=(z\_{n})\_{n\in \mathbb{N}}$ there exists $ y=(y\_{n})\_{n\in \mathbb{N}}$ satisfying
$$
{\rm (i)}\; \|y\|\_{p}<\infty,
\qquad 
{\rm (ii)}\;\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}\Big)\cdot y\_{K+n}=z\_{n} \quad n\in\mathbb{N}
\quad \mbox { and } \quad 
{\rm (iii)}\;\| y-x\|\_p<\epsilon 
$$
**Condition 2.** there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for any open ball $B(x, \epsilon)\subset \ell^p( {\mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z=(z\_{n})\_{n\in \mathbb{N}}$ there exists $ y=(y\_{n})\_{n\in \mathbb{N}}$ satisfying
$$
{\rm (i)}\; \sum\_{n=0}^{\infty}|z\_n|^{p}<\infty,
\qquad 
{\rm (ii)}\; y\_{K+n}= \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad 
{\rm (iii)}\;\sum\_{n=0}^{K-1}\left|y\_{n} - x\_n\right|^{p}
+
\sum\_{n=K}^{\infty}\left|\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n} - x\_n\right|^{p} <\epsilon 
$$
**Condition 3.** there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-\{0\}}$
such that for any open ball $B(x, \epsilon)\subset \ell^p( \mathbb{N})$
there exists a $K\in \mathbb{N}$ such that for every pair $z=(z\_{n})\_{n\in \mathbb{N}}$
and $\tilde{z}=(\tilde{z}\_{n})\_{n\in \mathbb{N}}$ in $\ell^p(\mathbb{N})$
there exists $ y=(y\_{n})\_{n\in \mathbb{N}}$ which fits both $z$ and $\tilde{z}$ satisfying


$$
{\rm (i)}\; \sum\_{n=0}^{\infty}|z\_n|^{p}<\infty,
\qquad 
{\rm (ii)}\; y\_{K+n}= \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad 
{\rm (iii)}\sum\_{n=0}^{K-1}\left|y\_{n} - x\_n\right|^{p}
+
\sum\_{n=K}^{\infty}\left|\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n} - x\_n\right|^{p} <\epsilon 
$$
$$
{\rm (iv)}\; \sum\_{n=0}^{\infty}|\tilde{z}\_n|^{p}<\infty,
\qquad 
{\rm (v)}\; y\_{K+n}= \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot \tilde{z}\_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad 
{\rm (vi)}\; \sum\_{n=0}^{K-1}\left|y\_{n} - x\_n\right|^{p}
+
\sum\_{n=K}^{\infty}\left|\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot \tilde{z}\_{n} - x\_n\right|^{p} <\epsilon 
$$


Since $z=(z\_{n})\_{n\in \mathbb{N}}$ and $\tilde{z}=(\tilde{z}\_{n})\_{n\in \mathbb{N}}$ are arbitrary we can choose their $K$ first coordinates equal and the others different such that $\|\tilde{z}-z \|\_p\geqslant 3\epsilon $. Adding member by member the equalities {\rm (ii)} and {\rm (v)} will result in $z=(z\_{n})\_{n\in \mathbb{N}}=\tilde{z}=(\tilde{z}\_{n})\_{n\in \mathbb{N}}$ since $0<c<\alpha\_{n}$. However, this is a contradiction.


