Is the weighted shift strong frequently hypercyclic?

Question

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 $y\in L^{-K}(z)\cap B(x,\epsilon)$.

Answer 1

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.