# Is the weighted shift strong frequently hypercyclic?

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

• This is not a variant of "aperiodicity", it is a variant of topological/metric transitivity. (I don't know a standard term for it, I would perhaps call it "uniform strong transitivity") Good question though. May 1, 2023 at 5:04
• I chose the term "aperiodic" motivated by the theory of discrete Markov processes (finite state space). In this context if the adjacency matrix A is aperiodic (i.e. there exists n such that A^n>0) then the subshift the mentioned property in the post above. Also the associated markov chain is called aperiodic. Maybe is not the most appropriate, but like you mention I also don't know a standard term for it. May 1, 2023 at 17:28
• @VilleSalo I have made some minor edits on the question. Consulting the literature of linear operators I found that the property that I am looking for is closer to the property called "frequently hypercyclic". May 1, 2023 at 17:56
• You are right about aperiodic matrices, indeed the term is not as crazy as I thought. (For matrices a lot of usually different dynamical notions coincide.) May 1, 2023 at 18:22

The answer to the question is no. If the sequence $$\alpha = (\alpha_{n})_{\geqslant 1}$$ satisfies the condition $$c<\alpha_{n} there is no other condition on a sequence $$\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. However, this is a contradiction.
• I'am very glad for you interest in the question I had ask. But I figure out a answer embarrassingly simple. For every $N$, any continuous linear operator dilates(or contracts) a ball of finite radius for a factor of at most $\|T\|^n$. May 12, 2023 at 19:54