Proof. It suffices to show that (iii) implies (i). As before, we can assume that the shift is unweighted.

Let $y\in X$ be a non-zero vector satisfying the condition in (iii). In order to show that B is hypercyclic, fix $p\geq 0$ and ɛ > 0. Since y ≠ 0, there is some $q\in \mathbb{Z}$ such that $y_q\neq 0$, and we can assume that $y_q=2$.

Since $(e_n)_n$ is a basis, the projections $z\mapsto z_ne_n$, $n\in\mathbb{Z}$, are equicontinuous on X by the Banach–Steinhaus theorem. There is, therefore, some η > 0 such that, for any $z\in X$ with $\|z\| \lt \eta$ and any $n\in\mathbb{Z}$,

(3)\begin{equation} \|z_ne_n\| \lt \varepsilon. \end{equation}

By making η smaller, if necessary, we can assume that for any $z\in X,$

(4)\begin{equation} \|z-y\| \lt \eta \ \Longrightarrow \ |z_q|\geq 1. \end{equation}

By hypothesis, there is then some $x\in X$ and a subsequence $(n_k)_k$ such that, for all $k\geq 1$,

\begin{equation*} \|B^{n_k}x-y\| \lt \frac{\eta}{2}, \end{equation*}

where we may assume that $n_1\geq 2p-2q$. By (4), it follows that, for all $k\geq 1$,

(5)\begin{equation} |x_{q+n_k}|=|[B^{n_k}x]_q|\geq 1. \end{equation}

By the basis assumption, there is also some $N\in\mathbb{N}$ such that $w:=\sum_{n=-N}^N y_ne_n$ satisfies $\|y-w\| \lt \frac{\eta}{2}$ and hence that, for all $k\geq 1$,

(6)\begin{equation} \|B^{n_k}x-w\| \lt \eta. \end{equation}

Note that

\begin{equation*} [B^{n_k}x-w]_{q+n_1-n_k}=x_{q+n_1} - w_{q+n_1-n_k}= x_{q+n_1} \end{equation*}

whenever k is so large that $n_k \gt q+n_1+N$. Then (3), (5) and (6) imply that, for such k,

\begin{equation*} \|e_{q+n_1-n_k}\|=\Big\|\frac{1}{x_{q+n_1}}[B^{n_k}x-w]_{q+n_1-n_k}e_{q+n_1-n_k}\Big\| \lt \varepsilon, \end{equation*}

where we have used that $\|az\|\leq \|z\|$ for any scalar a with $|a|\leq 1$ and any $z\in X$, an inequality that follows from the definition of F-norm.

When we set

\begin{equation*} n:= p-q+n_k-n_1, \end{equation*}

then we have for all sufficiently large k that $n\geq 0$ and

(7)\begin{equation} \|e_{p-n}\|=\|e_{q+n_1-n_k}\| \lt \varepsilon. \end{equation}

On the other hand, since $x \in X$, we have that $x_je_j\to 0$ in X as $j\to\infty$; with (5) this implies that

\begin{equation*} e_{q+n_k}= \frac{1}{x_{q+n_k}}x_{q+n_k}e_{q+n_k} \to 0 \end{equation*}

as $k\to\infty$, and therefore by the continuity of B

(8)\begin{equation} \|e_{p+n}\| = \|e_{q+2p-2q+n_k-n_1}\| = \|B^{n_1-2p+2q}e_{q+n_k}\| \lt \varepsilon \end{equation}

whenever k is sufficiently large, where we have noted that $n_1\geq 2p-2q$.

Altogether, we have found some $n\geq 0$ such that (7) and (8) hold. Thus, B is hypercyclic by Theorem 3.1.

Proof. The non-hypercyclicity follows from a well-known characterization, see [Reference Grosse-Erdmann and Peris Manguillot23, Theorem 4.13]. As for the additional assertions, Chan and Sanders [Reference Chan and Sanders13, proof of Corollary 3.3] have shown that B_w is weakly hypercyclic on $\ell^2(\mathbb{Z})$; since this space is continuously and densely included in $c_0(\mathbb{Z})$ in the norm sense and therefore also in the weak sense, B_w is also weakly hypercyclic on $c_0(\mathbb{Z})$. Moreover, by Bès, Chan and Sanders [Reference Bès, Chan and Sanders7], every weakly sequentially hypercyclic backward shift on $c_0(\mathbb{Z})$ is hypercyclic.

It remains to show that B_w has a norm dense set of weakly sequentially recurrent vectors. Note that this is much stronger than saying that B_w is weakly sequentially recurrent. We start by showing that there are weakly sequentially recurrent vectors arbitrarily close to e ₀. To see this, let $(n_k)_{k\geq 1}$ be a sequence of integers with $n_1\geq 1$ and such that, for every $k\geq 2$,

(9)\begin{equation} n_k\geq\sum_{j=1}^{k-1}n_j+k. \end{equation}

We then let

\begin{equation*} A= \{n_{j_1}+n_{j_2}+\cdots+n_{j_l}: 1\leq j_1 \lt j_2 \lt \cdots \lt j_l,\ l\geq 1\} \end{equation*}

and

\begin{equation*} x=e_0+\sum_{n\in A} \frac{1}{2^n}e_n. \end{equation*}

Clearly, x belongs to $c_0(\mathbb{Z})$, and since n ₁ is the minimal element of A we have that

(10)\begin{equation} \|x-e_0\| \leq \frac{2}{2^{n_1}}. \end{equation}

Since, for any $n,m\geq 0$,

\begin{equation*} B_w^m\Big(\frac{1}{2^n}e_n\Big) = \begin{cases} e_{n-m}, &\text{if } m\geq n,\\ \frac{1}{2^{n-m}}e_{n-m}, &\text{if } m \lt n, \end{cases} \end{equation*}

we find that, for any $k\geq 1$,

(11)\begin{equation} B_w^{n_k} x = e_{-n_k}+\sum_{\substack{n\in A\\n \lt n_k}} e_{n-n_k} + e_0 +\sum_{\substack{n\in A\\n_k \lt n \lt n_{k+1}}} \frac{1}{2^{n-n_k}}e_{n-n_k}+\sum_{\substack{n\in A\\n\geq n_{k+1}}} \frac{1}{2^{n-n_k}}e_{n-n_k}; \end{equation}

note that $n_k\in A$.

First, if $n\in A$, $n \lt n_k$, then we have in view of (9) that there are $j_1 \lt j_2 \lt \cdots \lt j_l \lt k$ with

\begin{equation*} n=n_{j_1}+n_{j_2}+\cdots+n_{j_l}\leq \sum_{j=1}^{k-1}n_j\leq n_k-k, \end{equation*}

so that $n-n_k\leq -k$. Thus the first two terms of the right-hand side in (11) tend weakly to 0, because, for any $y\in \ell^1(\mathbb Z)$,

\begin{equation*} \Big|\Big\langle e_{-n_k}+\sum_{\substack{n\in A\\n \lt n_k}} e_{n-n_k}, y\Big\rangle\Big|\leq \sum_{n\leq -k}|y_n|\to 0. \end{equation*}

Next, (9) also implies that $n\in A$, $n_k \lt n \lt n_{k+1}$ if and only if there are $j_1 \lt j_2 \lt \cdots \lt j_l \lt k$ such that

\begin{equation*} n=n_{j_1}+n_{j_2}+\cdots+n_{j_l}+n_k, \end{equation*}

so that the corresponding numbers $m=n-n_k$ cover exactly the elements in A that are less than n_k. Consequently,

\begin{equation*} e_0 +\sum_{\substack{n\in A\\n_k \lt n \lt n_{k+1}}} \frac{1}{2^{n-n_k}}e_{n-n_k} = e_0 + \sum_{\substack{m\in A\\m \lt n_k}} \frac{1}{2^{m}}e_{m}\to x \end{equation*}

in norm and hence also weakly.

Finally, if $n\in A$, $n\geq n_{k+1}$, then by (9)

\begin{equation*} n-n_k\geq n_{k+1}-n_k\geq k, \end{equation*}

which implies that

\begin{equation*} \sum_{\substack{n\in A\\n\geq n_{k+1}}} \frac{1}{2^{n-n_k}}e_{n-n_k}\to 0, \end{equation*}

again in norm and hence weakly.

Altogether, we have found that

(12)\begin{equation} B_w^{n_k}x\to x \end{equation}

in the weak topology.

To finish the proof, we use the fact that B_w is invertible. Given any scalars a_n, $n=-N,\ldots,N$, $N\geq 1$, the (weak) continuity of B_w implies with (12) that

\begin{equation*} B_w^{n_k}\Big(\sum_{n=-N}^N a_n B_w^n x\Big)=\sum_{n=-N}^N a_n B_w^n (B_w^{n_k} x) \to \sum_{n=-N}^N a_n B_w^n x \end{equation*}

in the weak topology, hence $\sum_{n=-N}^N a_n B_w^n x$ is weakly sequentially recurrent. Moreover, we have by (10) that

\begin{equation*} \Big\|\sum_{n=-N}^N a_n B_w^n x - \sum_{n=-N}^N a_n B_w^n e_0 \Big\|\leq \frac{2}{2^{n_1}}\Big\|\sum_{n=-N}^N a_n B_w^n\Big\|. \end{equation*}

Now note that any finite sequence can be written in the form $\sum_{n=-N}^N a_n B_w^n e_0$ for some $N\geq 1$ and some scalars a_n. Since one may choose n ₁ arbitrarily large, the density of the finite sequences in $c_0(\mathbb{Z})$ then implies that the weakly sequentially recurrent vectors for B_w form a dense set.

Proof. We consider the same weights as in the previous example,

\begin{equation*} w_n= \begin{cases} 2, &\text{if } n \gt 0,\\ 1, &\text{if } n\leq 0. \end{cases} \end{equation*}

We demand here slightly more of the sequence $(n_k)_{k\geq 1}$: let $n_1\geq 2$ and, for every $k\geq 2$,

\begin{equation*} n_k\geq\sum_{j=1}^{k-1}n_j+n_{k-1}+k. \end{equation*}

Let again

\begin{equation*} A= \{n_{j_1}+n_{j_2}+\cdots+n_{j_l}: 1\leq j_1 \lt j_2 \lt \cdots \lt j_l,\ l\geq 1\}. \end{equation*}

It then follows that, for any $n\in A$ with $n \lt n_k$, $k\geq 1$,

(13)\begin{equation} -n_k \lt n-n_k \lt -n_{k-1}. \end{equation}

We consider the sets of negative integers

\begin{equation*} I_k = [-n_k,-n_{k-1}),\ \ k\geq 1 \end{equation*}

with $n_0=0$, and the numbers $p_k:=\text{card}(I_k)=n_k-n_{k-1}\geq 2$, $k\geq 1$.

The idea of our construction is to replace the space $c_0(\mathbb{Z})$ of Example 3.6 by a smaller reflexive Banach sequence space X which is big enough so that the argument of the example continues to hold. With this aim, we regard the space X of all sequences $x=(x_n)_{n\in\mathbb{Z}}$ such that

\begin{equation*} \|(x_n)_{n\geq 0}\|^2_2 + \sum_{k=1}^\infty \|(x_n)_{n\in I_k}\|_{p_k}^2 \lt \infty, \end{equation*}

with the canonical norm. Then X is reflexive as the direct $\ell^2$-sum of the reflexive spaces $\ell^2(\mathbb{N}_0)$ and $\ell^{p_k}(I_k)$, $k\geq 1$, and $(e_n)_n$ is an unconditional basis.

By considering separately the sequences $x\in X$ with support inside and outside $\{-n_k:k\geq 0\}$, one sees that the weighted backward shift B_w maps X into X and is, therefore, continuous; similarly, B_w is seen to be invertible. And since $\|e_n\|=1$ for all $n\in \mathbb{Z}$, B_w cannot be hypercyclic in view of [Reference Grosse-Erdmann and Peris Manguillot23, Theorem 4.13].

To see that it is not even weakly sequentially hypercyclic, note that X is continuously and densely included in $c_0(\mathbb{Z})$ in the norm sense and, therefore, also in the weak sense. Thus, if B_w were weakly sequentially hypercyclic on X, it would also be so on $c_0(\mathbb{Z})$, which is not the case by the previous example. By the same argument, since B_w is weakly hypercyclic on $\ell^2(\mathbb{Z})$ by [Reference Bayart and Matheron5, Corollary 10.28], it is weakly hypercyclic on the larger space X that contains $\ell^2(\mathbb{Z})$ as a dense subspace.

Now, let again

\begin{equation*} x=e_0+\sum_{n\in A} \frac{1}{2^n}e_n. \end{equation*}

Then x belongs to X, and we have that

(14)\begin{equation} \|x-e_0\| \leq \frac{1}{2^{n_1}}\frac{2}{\sqrt{3}}. \end{equation}

The identity (11) for $B_w^{n_k}x$, $k\geq 1$, continues to hold, and the sum of the last three terms on the right-hand side converges to x in norm, and hence weakly: replacing the norm of $c_0(\mathbb{N})$ by that of $\ell^2(\mathbb{N})$ preserves the validity of our previous argument.

Let us now show that

\begin{equation*} e_{-n_k}+\sum_{\substack{n\in A\\n \lt n_k}} e_{n-n_k}\to 0 \end{equation*}

weakly in X as $k\to\infty$. To see this, let $y=(y_n)_{n\in\mathbb{Z}}\in X^*$. Then

\begin{equation*} \|(y_n)_{n\geq 0}\|^2_2 + \sum_{k=1}^\infty \|(y_n)_{n\in I_k}\|_{p^*_k}^2 \lt \infty, \end{equation*}

where $\frac{1}{p_k^*}+ \frac{1}{p_k}=1$, $k\geq 1$. We have by (13) that for any $n\in A$, $n \lt n_k$, $k\geq 1$,

\begin{equation*} n-n_k\in I_k, \end{equation*}

and also $-n_k\in I_k$. Thus

\begin{equation*} \Big|\Big\langle e_{-n_k}+\sum_{\substack{n\in A\\n \lt n_k}} e_{n-n_k}, y\Big\rangle\Big|\leq \sum_{n\in I_k}|y_n|\leq \text{card}(I_k)^{1/p_k}\|(y_n)_{n\in I_k}\|_{p^*_k}\to 0; \end{equation*}

note that $p_k=\text{card}(I_k)$. This establishes the previous claim.

Altogether we have again that

\begin{equation*} B_w^{n_k}x\to x \end{equation*}

weakly in X.

From there, we conclude the proof with (14) as for Example 3.6.

Proof. Let $(x^{(k)})_k$ and $(y^{(k)})_k$ be sequences of vectors in X such that $(x^{(k)})_k$ converges weakly to 0 and that $(y^{(k)})_k$ is coordinatewise controlled by $(x^{(k)})_k$. Note that then $(y^{(k)})_k$ is coordinatewise convergent to 0.

Since $(x^{(k)})_k$ is weakly bounded, hence bounded, then so is $(y^{(k)})_k$ by unconditionality of the basis. Now the fact that $(e_n)_n$ is shrinking implies that $(y^{(k)})_k$ converges weakly to 0, see [Reference Singer31, Chapter II, Theorem 4.2].

Proof. Let $x\in X$ be the weak limit of $(x^{(k)})_k$. Since we also have coordinatewise convergence, there is a sequence $(N_k)_k$ of integers with $N_k\to\infty$ such that $\sum_{|n|\leq N_k} \|(x_n-x_n^{(k)})e_n\|\to 0$. By decreasing the N_k, if necessary, we can assume by hypothesis that $y^{(k)}$ has support in $[-N_k,N_k]^c$, $k\geq 1$.

We consider $z^{(k)}:=\sum_{|n| \gt N_k} x_n^{(k)}e_n$. Since

\begin{equation*} z^{(k)} = (x^{(k)}-x) + \sum_{|n| \gt N_k} x_ne_n + \sum_{|n|\leq N_k} (x_n-x_n^{(k)})e_n, \end{equation*}

the basis assumption and the definition of $(N_k)_k$ imply that $(z^{(k)})_k$ converges weakly to 0 and that it dominates $(y^{(k)})_k$ coordinatewise. By property (BW), $(y^{(k)})_k$ converges weakly to 0.

Proof. By conjugacy, which also preserves property (BW), it suffices to consider the unweighted shift B.

Let $y\in X$ be of the form $y=\sum_{j=-p}^p y_ne_n$ with $p\geq 0$, and let ɛ > 0. The continuity of the operator

\begin{equation*} T:=\sum_{n=0}^{2p} y_{p-n} B^n \end{equation*}

with $Te_p=y$ implies that there is some η > 0 such that, for all $z\in X$,

(15)\begin{equation} \|z-e_p\| \lt \eta \ \Longrightarrow \ \|Tz-y\| \lt \varepsilon. \end{equation}

For the sequel, we fix these p and η.

Now, by assumption, there is a weakly sequentially recurrent vector x ≠ 0 for B. By applying B sufficiently often and noting that B is weakly continuous, we can assume that

(16)\begin{equation} \delta:=|x_p| \gt 0. \end{equation}

There is then an infinite set $D\subset\mathbb{N}_0$ such that

\begin{equation*} (B^nx)_{n\in D} \end{equation*}

converges weakly to x.

Let $q\geq 1$. By weak recurrence, which implies coordinatewise recurrence, there is some $m_q\in D$, $m_q\geq q$, such that

\begin{equation*} |x_{p+m_q}-x_p|=|[B^{m_q}x-x]_p| \lt \frac{\delta}{2}, \end{equation*}

and hence $|x_{p+m_q}| \gt \frac{\delta}{2}$.

Again by coordinatewise recurrence, there is an infinite set $D_q\subset D$ such that, for any $n\in D_q$,

(17)\begin{equation} |[B^nx-x]_{p+j}| \lt \frac{\delta}{2^{q+1}},\ \ 0\leq j\leq m_q. \end{equation}

In particular, we obtain that

\begin{equation*} |x_{p+n+m_q}-x_{p+m_q}|= |[B^nx-x]_{p+m_q}| \lt \frac{\delta}{4} \end{equation*}

and thus $|x_{p+n+m_q}| \gt \frac{\delta}{4}$. Since $x=\sum_{j\in\mathbb{Z}} x_je_j$ converges in X, we have that $x_je_j\to 0$ as $j\to\infty$ and hence

\begin{equation*} e_{p+n+m_q}= \frac{1}{x_{p+n+m_q}}x_{p+n+m_q}e_{p+n+m_q}\to 0, \ n\to\infty, n\in D_q. \end{equation*}

Applying B up to $2m_q$ times and noting that $q\leq m_q$, we find that for all $n\in D_q$ sufficiently large,

(18)\begin{equation} \|e_{p+n+j}\| \lt \frac{1}{q2^{q+1}}\eta, \ \ |j|\leq q. \end{equation}

We may assume without loss of generality that this is the case for all $n\in D_q$.

We now define inductively a sequence of integers $(n_k)_{k\geq 0}$ by taking $n_0=0$ and then, for every $k\geq 1$,

(19)\begin{align} N_{k-1}&:= \sum_{j=0}^{k-1}n_j, \end{align}

(20)\begin{align} q_k&:= 2N_{k-1}+1, \end{align}

(21)\begin{align} n_{k}&\in D_{q_k}\ \text{and}\ n_{k}\geq N_{k-1}+k. \end{align}

We let

\begin{equation*} A= \{n_{j_1}+n_{j_2}+\cdots+n_{j_m}: 1\leq j_1 \lt j_2 \lt \cdots \lt j_l,\ l\geq 1\} \end{equation*}

and define

(22)\begin{equation} z=e_p+\sum_{n\in A} e_{p+n}. \end{equation}

Our first task is to show that this series converges. We have that

\begin{equation*} \sum_{n\in A} \|e_{p+n}\| = \sum_{k=1}^\infty \sum_{\substack{n\in A\\n_k\leq n \lt n_{k+1}}} \|e_{p+n}\|. \end{equation*}

Now, for $k\geq 1$ and $n\in A$ with $n_k\leq n \lt n_{k+1}$, we have by (19) and (21) that $0\leq j:=n-n_k\leq N_{k-1}$, and hence by (18), (20) and (21) that

\begin{align*} \sum_{\substack{n\in A\\n_k\leq n \lt n_{k+1}}}\| e_{p+n}\|&=\sum_{\substack{n\in A\\n_k\leq n \lt n_{k+1}}}\| e_{p+n_k+(n-n_k)}\| \leq \sum_{0\leq j\leq N_{k-1}}\frac{1}{q_k2^{q_k+1}}\eta\\ &\leq \sum_{0\leq j\leq N_{k-1}}\frac{1}{(2N_{k-1}+1)2^{k}}\eta = \frac{1}{2^{k}}\eta, \end{align*}

which implies the convergence of (22), so that z is a well-defined vector in X. Moreover,

(23)\begin{equation} \|z-e_p\|\leq \eta. \end{equation}

Next, for any $k\geq 1$, we have that

(24)\begin{equation} B^{n_k} z = e_{p-n_k}+\sum_{\substack{n\in A\\n \lt n_k}} e_{p+n-n_k} + e_p +\sum_{\substack{n\in A\\n_k \lt n \lt n_{k+1}}} e_{p+n-n_k}+\sum_{j=k+1}^\infty \sum_{\substack{n\in A\\n_j\leq n \lt n_{j+1}}} e_{p+n-n_k}. \end{equation}

Let’s look at the right-hand side of this identity.

First, we have seen in the proof of Example 3.6 that, for any $k\geq 1$, $\{n-n_k: n \in A, n_k \lt n \lt n_{k+1}\} = \{n \in A, n \lt n_{k}\}$, so that we have

(25)\begin{equation} e_p +\sum_{\substack{n\in A\\n_k \lt n \lt n_{k+1}}}e_{p+n-n_k} = e_p +\sum_{\substack{n\in A\\n \lt n_{k}}} e_{p+n}\to z \end{equation}

in norm and hence weakly.

Secondly, for any $j\geq k+1$,

\begin{equation*} \sum_{\substack{n\in A\\n_j\leq n \lt n_{j+1}}} \|e_{p+n-n_k}\| = \sum_{\substack{n\in A\\n_j\leq n \lt n_{j+1}}} \|e_{p+n_j + (n-n_j -n_k)}\|. \end{equation*}

Since, for any $n\in A$ with $n_j\leq n \lt n_{j+1}$ we have by (19) that

\begin{equation*} -N_{j-1}\leq -n_k\leq m:=n-n_j-n_k\leq n-n_j\leq N_{j-1}, \end{equation*}

we obtain with (18), (20) and (21) that

\begin{equation*} \sum_{\substack{n\in A\\n_j\leq n \lt n_{j+1}}} \|e_{p+n-n_k}\| \leq \sum_{\substack{n\in A\\n_j\leq n \lt n_{j+1}}} \frac{1}{q_{j}2^{q_{j}+1}}\eta\leq \sum_{|m|\leq N_{j-1}}\frac{1}{(2N_{j-1}+1)2^{j}}\eta = \frac{1}{2^{j}}\eta. \end{equation*}

Thus

(26)\begin{equation} \Big\|\sum_{j=k+1}^\infty \sum_{\substack{n\in A\\n_j\leq n \lt n_{j+1}}} e_{p+n-n_k}\Big\|\leq \frac{1}{2^{k}}\eta \to 0. \end{equation}

Thirdly, let x be the initial weakly sequentially recurrent vector. If $n\in A$, $n \lt n_{k}$, then there are $l\geq 1$ and $1\leq j_1 \lt j_2 \lt \cdots \lt j_l\leq k-1$ with

\begin{equation*} n=n_{j_1}+n_{j_2}+\cdots+n_{j_{l-1}}+n_{j_l}. \end{equation*}

Let $m=1,\ldots,l$. Since $n_{j_m} \in D_{q_{j_m}}$ and, for $m\geq 2$,

\begin{equation*} 0\leq n_{j_1}+\cdots + n_{j_{m-1}} \leq N_{j_{m-1}}\leq N_{j_{m}-1}\leq q_{j_{m}}\leq m_{q_{j_m}}, \end{equation*}

we have by (17) that, for any $m\geq 1$,

\begin{align*} |x_{p+ n_{j_1}+\cdots+n_{j_{m-1}}+n_{j_m}} - x_{p+ n_{j_1}+\cdots+n_{j_{m-1}}}| &= |[B^{n_{j_m}}x-x]_{p+n_{j_1}+\cdots+n_{j_{m-1}}}|\\ &\leq \frac{\delta}{2^{q_{j_m}+1}}\leq \frac{\delta}{2^{m+1}}, \end{align*}

where an empty sum is 0. Telescoping we find that

\begin{equation*} |x_{p+n}-x_p|\leq \sum_{m=1}^l \frac{\delta}{2^{m+1}} \leq \frac{\delta}{2} \end{equation*}

and therefore by (16)

(27)\begin{equation} |x_{p+n}| \geq \frac{\delta}{2}. \end{equation}

When we now define

\begin{equation*} y^{(k)} := e_p + \sum_{\substack{n\in A\\n \lt n_{k}}} e_{p+n}, \ \ k\geq 1, \end{equation*}

then it follows from (27) that the sequence $(y^{(k)})_{k\geq 1}$ is coordinatewise controlled by $(x)_k$, a constant sequence of vectors in X. Since the operator B is a simple shift, we deduce that the sequence $(B^{n_k}y^{(k)})_{k\geq 1}$ is coordinatewise controlled by the sequence $(B^{n_k}x)_{k\geq 1}$. Now, since each n_k belongs to D, $(B^{n_k}x)_{k\geq 1}$ is weakly convergent. And finally, since

\begin{equation*} B^{n_k}y^{(k)}=e_{p-n_k} + \sum_{\substack{n\in A\\n \lt n_{k}}} e_{p+n-n_k} \end{equation*}

and, for any $n\in A$, $n \lt n_{k}$, by (21),

\begin{equation*} p+n-n_k \leq p+N_{k-1}-n_k \leq p-k, \end{equation*}

we see that, for any $j\in\mathbb{Z}$, $([B^{n_k}y^{(k)}]_j)_{k\geq 1}$ is eventually 0. By property (BW), via Lemma 3.10, this implies that

(28)\begin{equation} e_{p-n_k} + \sum_{\substack{n\in A\\n \lt n_{k}}} e_{p+n-n_k}\to 0 \end{equation}

in the weak topology.

Altogether, we see from (24), (25), (26) and (28) that

\begin{equation*} B^{n_k}z\to z \end{equation*}

in the weak topology.

To finish the proof, we apply the (weakly) continuous operator $T=\sum_{n=0}^{2p} y_{p-n} B^n$. Since it commutes with B, we have that

\begin{equation*} B^{n_k}Tz\to Tz \end{equation*}

in the weak topology, and by (15) and (23) that

\begin{equation*} \|Tz-y\| \lt \varepsilon. \end{equation*}

We have, therefore, shown that in every neighbourhood of any finite sequence there exists a weakly sequentially recurrent vector. Since the finite sequences are dense in X, the theorem is proved.

Question 3.12. Let B_w be a weighted backward shift on a reflexive Banach sequence space over $\mathbb{Z}$ in which $(e_n)_n$ is a basis. Does the existence of a non-zero weak sequential limit point imply weak sequential recurrence or even the existence of a norm dense set of weakly sequentially recurrent vectors?

Proof. Note first that since $B_w^*(e_n^*)=w_{n+1}e^*_{n+1}\in Z$ it follows that $B_w^*|_Z:Z\to Z$ is a well-defined operator. This implies that $B_w=(B_w^*|_{Z})^*$ and hence B_w is weak*-continuous. Also, weak*-convergence implies coordinatewise convergence.

Let us now show that the analogue of Lemma 3.10 holds. Thus let $(x^{(k)})_k$ and $(y^{(k)})_k$ be sequences of vectors in X so that $(x^{(k)})_k$ converges weak*, $(y^{(k)})_k$ is coordinatewise controlled by $(x^{(k)})_k$ and, for every $n\in \mathbb{Z}$, $(y^{(k)}_n)_k$ is eventually 0. It follows from the Banach–Steinhaus theorem that $(x^{(k)})_k$ is bounded in X, and hence also $(y^{(k)})_k$ by unconditionality of the basis in X.

Now let $z\in Z$ and ɛ > 0. Since $(e_n^*)_n$ is a basis of Z, there is some $N\in\mathbb{N}$ such that

\begin{equation*} \Big\|\sum_{|n|\geq N} z_ne_n^*\Big\| \lt \frac{\varepsilon}{1+ \sup_{k\geq 1}\| y^{(k)}\|}. \end{equation*}

Since $y^{(k)}\to 0$ coordinatewise, there is then some $K\geq 1$ such that, for any $k\geq K$,

\begin{equation*} |\langle y^{(k)},z\rangle|\leq \Big|\sum_{|n| \lt N} y^{(k)}_nz_n\Big| + \| y^{(k)}\| \Big\|\sum_{|n|\geq N} z_ne_n^*\Big\| \lt 2\varepsilon. \end{equation*}

Thus, $(y^{(k)})_k$ converges weak* to 0.

Now the proof can be done exactly as that of Theorem 3.11: it suffices to replace throughout ‘weak’ and ‘weakly’ by ‘weak*-’.

Proof. Suppose that B_w is not hypercyclic. Then there are some $q\geq 0$ and ɛ > 0 such that, for all $n\geq 0$,

(29)\begin{equation} \max\big(|w_{q+1} \cdots w_{q+n}|^{-1} , |w_{q-n+1}\cdots w_{q}|\big)\geq \varepsilon \end{equation}

by Theorem 3.1 and using the conjugacy explained in § 1.

Assume that y with $\sup\{n\geq 0: y_n\neq 0\}=\infty$ is a weak limit point of the orbit of some vector x. By weak continuity of B_w, we may assume that $y_q\neq 0$, and then that $y_q=2$.

Moreover, suppose that there is some $n\geq 0$ such that $B_w^nx=y$. If we also had $B_w^mx=y$ for some m > n, $B_w^nx$ would be a non-zero periodic point, which implies that B_w is hypercyclic by Corollary 3.3. Thus, replacing x by $B_w^{n+1}x$, we see that we can assume that y does not belong to the orbit of x.

Define $\Lambda =\{n\geq 0: |[B_w^{n}x]_q| \gt 1\}$. Since y is a weak limit point of $(B_w^{n}x)_{n}$, Λ is an infinite set. Moreover, $\{B_w^{n}x:n\in\Lambda\}$ is not weakly closed because its weak closure contains y, which is outside this set.

Now we can follow the proof of [Reference Bayart and Matheron5, Proposition 10.20] to show that $\{B_w^{n}x:n\in\Lambda\}$ is weakly closed, thus arriving at a contradiction.

Indeed, for any $n\in\Lambda$, we have that

\begin{align*} \|B_w^{n}x\|^p & \ge \sum_{m\in\Lambda} |w_{q+m-n+1} \cdots w_{q+m}|^{p}|x_{q+m}|^p \\ & = \sum_{m\in\Lambda, m\le n} |w_{q+m-n+1} \cdots w_{q}|^{p}|[B_w^{m}x]_q|^p + \sum_{m\in\Lambda, m \gt n} |w_{q+1} \cdots w_{q+m-n}|^{-p}|[B_w^{m}x]_q|^p \\ & \ge \sum_{m\in\Lambda, m\le n} |w_{q+m-n+1} \cdots w_{q}|^{p} + \sum_{m\in\Lambda, m \gt n} |w_{q+1} \cdots w_{q+m-n}|^{-p}. \end{align*}

Define the infinite matrix $(c_{n,m})_{n,m\in \Lambda}$ by $c_{n,m}=|w_{q+m-n+1} \cdots w_{q}|^{p}$ if $ m\le n$ and $c_{n,m}=|w_{q+1} \cdots w_{q+m-n}|^{-p}$ if m > n. Then by (29), we can apply [Reference Bayart and Matheron5, Lemma 10.22] to conclude that $\sum_{n\in\Lambda}(\sum_{m\in\Lambda}c_{n,m})^{-r} \lt \infty$ for any r > 1, and thus $\sum_{n\in\Lambda}\|B_w^{n}x\|^{-rp} \lt \infty$. Taking r > 1 such that rp < 2, [Reference Bayart and Matheron5, Corollary 10.7] implies that $\{B_w^{n}x:n\in \Lambda\}$ is weakly closed, which is what we need to show.

Question 3.16. Is every weak*-hypercyclic weighted backward shift on $\ell^1(\mathbb{Z})$ hypercyclic?

Proof. By conjugacy, which also preserves property (BC), it suffices to consider the unweighted shift B.

Suppose that there are $x,y\in X$ with y ≠ 0 such that the hypothesis holds. Then there is some $q\in\mathbb{Z}$ such that $y_q\neq 0$, where we can assume that $y_q=2$. Also, by the second part of the hypothesis, we can assume that, for all $k\geq 1$,

(30)\begin{equation} |x_{q+n_k}|=|[B^{n_k}x]_q|\geq 1. \end{equation}

The remainder of the proof is in two stages. We first show the following claim: for any $r\geq 1$, ɛ > 0 and $m\geq 0$ there are $l\geq 1$ and $k\geq l$ such that

(31)\begin{equation} \|e_{q+n_l-n_k}\|_r \lt \varepsilon \quad \text{and}\quad \|e_{q+n_k-n_l+m}\|_r \lt \varepsilon. \end{equation}

To see this, note that by unconditionality of the basis there is some $s\geq 1$ and a constant C > 0 such that, whenever $z\in X$ and $|a_n|\leq 1$ for all $n\in \mathbb{Z}$, then

(32)\begin{equation} \Big\|\sum_{n\in\mathbb{Z}} a_nz_ne_n\Big\|_r\leq C\|z\|_s. \end{equation}

Now let $K=\sup_{k\geq 1}\|B^{n_k}x\|_s$, and let N and M be given by property (BC), applied to r and $\frac{\varepsilon}{CK}$. We first have, for any $k\geq 1$,

\begin{align*} 1 &\geq \frac{1}{K}\| B^{n_k} x\|_s = \frac{1}{K}\Big\| \sum_{n\in\mathbb{Z}} x_n e_{n-n_k}\Big\|_s= \frac{1}{K}\Big\| \sum_{n\in\mathbb{Z}} x_{q+n} e_{q+n-n_k}\Big\|_s\geq \Big\| \frac{1}{CK} \sum_{l=m+1}^{m+N} e_{q+n_l-n_k}\Big\|_r, \end{align*}

where the last inequality follows from (30) and (32). Then (BC) tells us that for any $k\geq 1$ with $n_k\geq q+n_{m+N}-M$ there is some $l_k\in[m+1, m+N]$ such that

\begin{equation*} \| e_{q+n_{l_k}-n_k}\|_r \lt \varepsilon. \end{equation*}

On the other hand, since $x\in X$, we have that $x_{q+n_k}e_{q+n_k}\to 0$ as $k\to \infty$, so that $e_{q+n_k}\to 0$ by (30) and thus $B^{n_l-m}e_{q+n_k}\to 0$ for any $l\in[m+1,m+N]$; note that $n_l-m\geq 0$. Then there is some $k\geq m+N$ such that

\begin{equation*} \|e_{q+n_k-n_{l_k}+m}\|_r=\| B^{n_{l_k}-m}e_{q+n_k}\|_r \lt \varepsilon. \end{equation*}

Since $k\geq m+N\geq l_k$, the claim has been shown.

To finish the proof, let $p\geq 0$ with $p\geq q$, $r\geq 1$ and ɛ > 0. Set $m=2p-2q\geq 0$. By the claim, there are $l\geq 1$ and $k\geq l$ such that (31) holds. We set

\begin{equation*} n=p-q+n_k-n_l\geq 0. \end{equation*}

Then

(33)\begin{equation} \|e_{p-n}\|_r=\|e_{q+n_l-n_k}\|_r \lt \varepsilon \end{equation}

and

(34)\begin{equation} \|e_{p+n}\|_r = \|e_{q+n_k-n_l+2p-2q}\|_r= \|e_{q+n_k-n_l+m}\|_r \lt \varepsilon. \end{equation}

Altogether, for any $p\geq 0$ with $p\geq q$, any $r\geq 1$ and ɛ > 0, we have found some $n\geq 0$ such that (33) and (34) hold. By Theorem 3.1, B is hypercyclic; note that it is clearly enough by the continuity of B that the condition holds for all large p.

Proof. We can assume that the shift is unweighted.

For this proof, it is again convenient to work with seminorms. Thus let $(\|\cdot\|_r)_r$ be an increasing sequence of seminorms that defines the topology of X.

We will show that the characterizing condition of Theorem 3.1 is satisfied, that is, that for any $p\geq 0$, $r\geq 1$ and ɛ > 0 there is some $n\geq 0$ such that

(35)\begin{equation} \|e_{p+n}\|_r \lt \varepsilon\ \ \text{and}\text\ \ \|e_{p-n}\|_r \lt \varepsilon. \end{equation}

First, since the mappings $z\mapsto z_ne_n$, $n\in \mathbb{Z}$, are equicontinuous on X by the basis assumption, there is a constant C > 0 and some $s\geq 1$ such that, for any $z\in X$,

(36)\begin{equation} \|z_ne_n\|_r\leq C\|z\|_s, \ \ n\in \mathbb{Z}. \end{equation}

Now let $x\in X$ satisfy the hypothesis of the theorem. Taking $y=e_p$, we see that there exists a sequence $(n_k)_k$ and a constant K > 0 such that

(37)\begin{equation} \|B^{n_k}x\|_s\leq K, k\geq 1, \ \text{and }\ x_{p+n_k}= [B^{n_k}x]_p \to 1. \end{equation}

Let $\rho \gt \frac{CK}{\varepsilon}$. Applying the hypothesis again, this time with $y=2\rho e_p$, we find some $m\geq 0$ such that

\begin{equation*} |x_{p+m}-2\rho| = |[B^{m}x-2\rho e_p]_p| \lt \rho, \end{equation*}

so that $|x_{p+m}| \gt \rho$. Using (36) and (37), we have that, whenever $n_k\geq m$,

\begin{align*} \|e_{p-(n_k-m)}\|_r & \lt \frac{1}{\rho} \|x_{p+m}e_{p-n_k+m}\|_r\leq \frac{C}{\rho}\Big\|\sum_{n\in\mathbb{Z}} x_{n+m}e_{n-n_k+m}\Big\|_s\\ &= \frac{C}{\rho}\Big\|\sum_{n\in\mathbb{Z}} x_{n}e_{n-n_k}\Big\|_s =\frac{C}{\rho}\|B^{n_k}x\|_s \lt \varepsilon. \end{align*}

On the other hand, since $x\in X$ and $(e_n)_n$ is a basis,

\begin{equation*} x_{p+n_k}e_{p+n_k}\to 0. \end{equation*}

By continuity of B, and by (37), there is some $n_k\geq m$ such that

\begin{equation*} \|x_{p+n_k}e_{p+n_k-m}\|_r = \|B^m(x_{p+n_k}e_{p+n_k})\|_r \lt \frac{\varepsilon}{2} \end{equation*}

and

\begin{equation*} |x_{p+n_k}-1| \lt \frac{1}{2}, \end{equation*}

so that $|x_{p+n_k}| \gt \frac{1}{2}$, and hence

\begin{equation*} \|e_{p+(n_k-m)}\|_r = \frac{1}{|x_{p+n_k}|}\|x_{p+n_k}e_{p+n_k-m}\|_r \lt \varepsilon. \end{equation*}

Altogether, (35) holds with $n=n_k-m \geq 0$.