Proof Note that, according to condition $(2)$ , T has dense range in X. Now, let $U,V$ be nonempty open subsets of X. There exist nonempty open subsets $U',V'$ of X and a $0$ -neighborhood W such that

$$ \begin{align*}U'+W\subset U\quad \text{and}\quad V'+W\subset V.\end{align*} $$

Since $\widetilde {\mathcal {F}}$ is a filter and

$$ \begin{align*}N(U,V) \supset N(U'+W,V'+W)\supset N(U',W)\cap N(W,V'),\end{align*} $$

it is enough to show that $N(U',W)\in \widetilde {\mathcal {F}}$ and $N(W,V')\in \widetilde {\mathcal {F}}$ . Let $N\in \mathbb {N}_0$ . Since T has dense range, the sets $T^{-N}U'$ and $T^{-N}V'$ are nonempty open sets. Choose $x\in X_0\cap T^{-N}U'$ and $y\in Y_0\cap T^{-N}V'$ . We start by showing that $N(T^{-N}U',W)\in \widetilde {\mathcal {F}}_{+}$ . For any $M\in \mathbb {N}_0$ , by $(1)$ , there exists $A_M\in \mathcal {F}$ such that, for every $n\in A_M$ ,

$$ \begin{align*}I_nx\in T^{-N}U' \quad \text{and} \quad T^nI_nx\in \bigcap_{k=0}^{M}T^{-k}W.\end{align*} $$

Note that, for any $n\in A_M+[\![0,M]\!]$ , there exists $k\in [\![0,M]\!]$ such that $n-k\in A_M$ , hence

$$ \begin{align*}I_{n-k}x \in T^{-N}U' \quad\text{and} \quad T^{n-k}I_{n-k}x\in T^{-k}W,\end{align*} $$

therefore, $I_{n-k}x \in T^{-N}U'$ and $T^{n}I_{n-k}x\in W$ , so $n\in N(T^{-N}U',W)$ . Thus, $A_M+[\![0,M]\!]\subset N(T^{-N}U',W)$ holds for any arbitrary $M\in \mathbb {N}_0$ . Consequently, $N(T^{-N}U',W) \in \widetilde {\mathcal {F}}_{+}$ . Thus, there is $B\in \mathcal {F}$ such that $B+[\![0,2N]\!]\subset N(T^{-N}U',W)$ . Therefore,

$$ \begin{align*}(B +[\![-N,N]\!])\cap\mathbb{N}_0 \subset (N(T^{-N}U',W)-N)\cap \mathbb{N}_0 \subset N(U',W),\end{align*} $$

and, since N was arbitrary, we deduce that $N(U',W)\in \widetilde {\mathcal {F}}$ .

On the other hand, by $(2)$ , there is $B\in \mathcal {F}$ such that, for every $n\in B$ ,

$$ \begin{align*}S_n y\in \bigcap_{k=0}^{2N}T^{-k}W \quad\text{and}\quad T^n S_n(y)\in T^{-N}V'.\end{align*} $$

In particular, for any $n\in (B+[\![-N,N]\!])\cap \mathbb {N}_0$ , there exists $k\in [\![-N,N]\!]$ such that $n-k\in B$ and $N-k\in [\![0,2N]\!]$ , thus

$$ \begin{align*}S_{n-k}y\in T^{k-N}W \quad \text{and} \quad T^{n-k}S_{n-k}y\in T^{-N}V',\end{align*} $$

hence

$$ \begin{align*}T^{N-k}S_{n-k}y\in W \quad \text{and} \quad T^{n}(T^{N-k}S_{n-k}y)\in T^{-N}V',\end{align*} $$

from which we deduce that $n\in N(W,V')$ . Therefore,

$$ \begin{align*}(B+[\![-N,N]\!])\cap \mathbb{N}_0 \subset N(W,V').\end{align*} $$

Since N was arbitrary, we obtain that $N(W,V')\in \widetilde {\mathcal {F}}$ . This finishes the proof.

Proof We will only prove the equivalences in the case where $X = \ell ^p(V,\mu )$ with $1 < p < +\infty $ . A similar argument can be made to deduce the cases where $X = \ell ^1(V,\mu )$ and $X = c_0(V,\mu )$ . It is clear that $(2)\Rightarrow (3)$ and $(2)\Rightarrow (5)$ . Moreover, $(1)\Rightarrow (2)$ since $\widetilde {\mathcal {F}} \subset \mathcal {F}$ . Suppose that B is $\mathcal {F}$ -transitive; thus, it is transitive (equivalently, hypercyclic). Using [Reference Grosse-Erdmann and Papathanasiou22, Theorem 4.3], we can conclude that B is weakly mixing. Thus, B is both $\mathcal {F}$ -transitive and weakly mixing, which is equivalent to saying that B is $\widetilde {\mathcal {F}}$ -transitive, according to [Reference Bès, Menet, Peris and Puig9, Lemma 2.3]. Hence, $(2)\Rightarrow (1)$ . Moreover, it clear that $(4)\Leftrightarrow (6)$ when $\mathcal {F}$ is a filter.

$(3)\Rightarrow (4)$ . Let C be a bounded subset of $\ell ^p(V,\mu )\setminus \{0\}$ such that $N(W,C)\in \mathcal {F}$ , for every nonempty weakly open subset W of $\ell ^p(V,\mu )$ . Set $M:=\sup \{\|f\|_{p,\mu }:\, f\in C\}<+\infty $ . Let $N\in \mathbb {N}$ and let $F\subset V$ be a finite subset. Then

$$ \begin{align*}W=\{f\in\ell^p(V,\mu):\, |\langle f-(MN+1)\sum_{u\in F}e_u,e_v\rangle|<1,\,\forall v\in F\},\end{align*} $$

is a weakly open neighborhood of $(MN+1)\sum _{u\in F}e_u$ . By the hypothesis, $N(C,W)\in \mathcal {F}$ . Since $\mathcal {F}$ is a Furstenberg family, we need only to prove that

(4.1) $$ \begin{align} N(C,W)\subset \bigcap_{v\in F}\Big\{n\in\mathbb{N}_0:\, \Big(\underset{u\in\mathrm{Chi}^n(v)}{\sum}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}>N\Big\}. \end{align} $$

Let $n\in N(C,W)$ . There exists then $f\in C$ such that $B^{n}f\in W$ , thus for every $v\in F$

$$ \begin{align*}|(B^{n}f)(v)-(MN+1)|<1,\end{align*} $$

hence, by using Hölder’s inequality, we obtain

$$ \begin{align*} MN&<\sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)|\\ &\leq \Big( \sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)\mu_u|^p\Big)^{1/p}\Big( \sum_{u\in\mathrm{Chi}^{n}(v)}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}\\ &\leq \|f\|_{p,\mu}\Big( \sum_{u\in\mathrm{Chi}^{n}(v)}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}\\ &\leq M\,\Big( \sum_{u\in\mathrm{Chi}^{n}(v)}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast},\\ \end{align*} $$

therefore

$$ \begin{align*}\Big( \sum_{u\in\mathrm{Chi}^{n}(v)}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}>N,\quad \forall v\in F,\end{align*} $$

and so (4.1) holds.

Let us show that $(4)\Rightarrow (2)$ . Assume that $(4)$ holds, that is, for every $N>0$ and every finite subset $F\subset V$ , we have

$$ \begin{align*}I(F,N):= \bigcap_{v\in F}\Big\{n\in\mathbb{N}_0:\, \Big(\underset{u\in\mathrm{Chi}^n(v)}{\sum}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}>N\Big\}\in\mathcal{F}.\end{align*} $$

Let $\mathcal {B}$ be the filter base consisting of all subsets $I(F,N)$ of $\mathbb {N}_0$ , where $N>0$ and $F\subset V$ is finite. We define

$$ \begin{align*}\mathcal{F}_{\mathcal{B}}=\{A\subset \mathbb{N}_0: B\subset A \text{ for some } B\in \mathcal{B}\}.\end{align*} $$

In other words, $\mathcal {F}_{\mathcal {B}}$ is the filter generated by $\mathcal {B}$ . We will show that B satisfies the $\mathcal {F}_{\mathcal {B}}$ -transitivity criterion, which implies that B is $\mathcal {F}$ -transitive since $\mathcal {F}_{\mathcal {B}}\subset \mathcal {F}$ . Set $\mathcal {D}=\mathrm {span}\{e_v:\, v \in V\}$ . Note that $\mathcal {D}$ is dense in $\ell ^p(V,\mu )$ . Let $f=\sum _{u\in F}f(u) e_u\in \mathcal {D}$ , with $F\subset V$ finite. Let $\varepsilon>0$ and $\mathcal {U}:=B(0,\varepsilon )$ be the open ball of center $0$ and radius $\varepsilon $ in $\ell ^p(V,\mu )$ . Let us start by proving that

$$ \begin{align*}N(f,\mathcal{U}):=\{n\in\mathbb{N}_0:\, B^{n}f\in \mathcal{U}\}\in \mathcal{F}_{\mathcal{B}}.\end{align*} $$

To achieve this, it is enough to find a number $N\geq 1$ such that $I(F,N) \subset N(f,\mathcal {U})$ . First, select an integer $n_0\in \mathbb {N}$ large enough such that $\mathrm {Chi}^{n}(V)\cap F=\emptyset $ for every integer $n\geq n_0$ . Then, we set

$$ \begin{align*}N:=1+\underset{v\in F}{\max}\,\underset{0\leq k<n_0}{\max}\Big(\underset{u\in\mathrm{Chi}^k(v)}{\sum}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}.\end{align*} $$

Note that, by using Proposition 2.2 and the boundedness of $B^{k}$ for every $0\leq k<n_0$ , N is finite. Now, suppose that $n\in I(F,N)$ . This implies that $n\geq n_0$ . Thus, for every $v\in V$ , we have $\mathrm {Chi}^n(v)\cap F=\emptyset $ , hence, $(B^{n} f)(v)=0$ . This implies that $n\in N(f,\mathcal {U})$ . Therefore, we have shown that $I(F,N) \subset N(f,\mathcal {U})$ , and consequently, $N(f,\mathcal {U})\in \mathcal {F}_{\mathcal {B}}$ since $\mathcal {F}_{\mathcal {B}}$ is a Furstenberg family. Hence, the first condition of the $\mathcal {F}_{\mathcal {B}}$ -transitivity criterion holds, where the maps $I_n$ correspond to the identity map.

Now, for every $v\in V$ , let $(\delta _{v,n})_n$ be a decreasing sequence of positive real numbers tending to zero. For all $v\in V$ and $n\in \mathbb {N}_0$ , by Lemma 2.4, we have

$$ \begin{align*}\underset{\|x\|_1=1}{\inf}\Big(\underset{u \in \mathrm{Chi}^{n}(v)}{\sum}|x_u\mu_u|^p\Big)^{1/p}=\Big(\underset{u \in \mathrm{Chi}^{n}(v)}{\sum}\,\frac{1}{|\mu_u|^{p^\ast}}\Big)^{-1/p^{\ast}}.\end{align*} $$

There exists then $g_{v,n}\in \mathbb {K}^V$ nonnegative, of support in $\mathrm {Chi}^{n}(v)$ , such that

(4.2) $$ \begin{align} & \sum_{u\in \mathrm{Chi}^{n}(v)}g_{v,n}(u)=1 \quad \text{and} \quad \nonumber\\&\quad\Big(\sum_{u\in \mathrm{Chi}^{n}(v)}|g_{v,n}(u)\,\mu_u|^p\Big)^{1/p}<\Big(\underset{u \in \mathrm{Chi}^{n}(v)}{\sum}\,\frac{1}{|\mu_u|^{p^\ast}}\Big)^{-1/p^{\ast}}+\delta_{v,n}. \end{align} $$

Note that $g_{v,n}\in \ell ^p(V,\mu )$ . Let us define the maps $S_k:\mathcal {D}\to \ell ^p(V,\mu )$ linearly, for $k\geq 0$ , by

$$ \begin{align*}S_ke_v=g_{v,k},\quad \forall v\in V.\end{align*} $$

Let $\varepsilon>0$ and $g=\sum _{u\in F} g(a) e_a\in \mathcal {D}$ , where F is the support of g. Set again $\mathcal {U}:=B(0,\varepsilon )$ and $\mathcal {V}:=B(g,\varepsilon )$ . We want to show that

$$ \begin{align*}A:=\{n\in\mathbb{N}_0:\, (S_ng,B^nS_ng)\in \mathcal{U}\times \mathcal{V}\}\in\mathcal{F}_{\mathcal{B}}.\end{align*} $$

Let $n_0\in \mathbb {N}$ be large enough so that $\frac {\varepsilon }{|g(a)| |F|}-\delta _{a,n_0}>0$ , for every $a\in F$ (where $|F|$ stands for the cardinality of the finite set F). Set

$$ \begin{align*}N=\max\Bigg\{1+\underset{v\in F}{\max}\,\underset{0\leq k<n_0}{\max}\Big(\underset{u\in\mathrm{Chi}^k(v)}{\sum}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}, \dfrac{1}{ \underset{a\in F}{\min}\big\{\frac{\varepsilon}{|g(a)| |F|}-\delta_{a,n_0}\big\}} \Bigg\}.\end{align*} $$

We will show that $I(F,N)\subset A$ . Note that $B^nS_ne_v=B^ng_{v,n}=e_v$ , for every integer $n\geq 0$ . Indeed, by (4.2), we have

$$ \begin{align*} (B^ng_{v,n})(a)&=\sum_{b\in \mathrm{Chi}^n(a)}g_{v,n}(b)\\ &=\begin{cases} 1 & \text{ if } a=v \\ 0 & \text{ if } a\neq v \end{cases}\\ &=e_v(a). \end{align*} $$

Thus, $B^nS_ng=g\in \mathcal {V}$ , for every integer $n\geq 0$ . Let $n\in I(F,N)$ , so as above and by the definition of N, we have $n\geq n_0$ and

$$ \begin{align*}\Big(\underset{u \in \mathrm{Chi}^{n}(a)}{\sum}\,\frac{1}{|\mu_u|^{p^\ast}}\Big)^{-1/p^{\ast}}<\dfrac{1}{N}, \, \forall a\in F,\end{align*} $$

thus, by using (4.2), we obtain:

$$ \begin{align*}\Big(\sum_{u\in \mathrm{Chi}^{n}(a)}|g_{a,n}(u)\,\mu_u|^p\Big)^{1/p}<\dfrac{1}{N}+\delta_{a,n}, \, \forall a\in F,\end{align*} $$

by using again the definition of N, we deduce

$$ \begin{align*}\Big(\sum_{u\in \mathrm{Chi}^{n}(a)}|g_{a,n}(u)\,\mu_u|^p\Big)^{1/p}<\dfrac{\varepsilon}{|g(a)| |F|}, \, \forall a\in F,\end{align*} $$

therefore

$$ \begin{align*} \|S_ng\|_{p,\mu} \leq \sum_{a\in F} |g(a)| \|g_{a,n}\|_{p,\mu} =\sum_{a\in F} |g(a)| \Big( \sum_{b\in\mathrm{Chi}^n(a)}|g_{a,n}(b)\mu_b|^p\Big)^{1/p} <\varepsilon, \end{align*} $$

hence $S_ng\in \mathcal {U}$ . We have shown that $n\in A$ and so $I(F,N)\subset A$ . Consequently, $A\in \mathcal {F}_{\mathcal {B}}$ . This provides the second condition of the $\mathcal {F}_{\mathcal {B}}$ -transitivity criterion. Hence, B is $\mathcal {F}$ -transitive and statement $(2)$ holds.

$(5)\Rightarrow (4)$ . Let us assume that $(5)$ holds. Let F be a finite subset of V and let $N\in \mathbb {N}$ . We want to show that

$$ \begin{align*}I(F,N):= \bigcap_{v\in F}\Big\{n\in\mathbb{N}_0:\, \Big(\underset{u\in\mathrm{Chi}^n(v)}{\sum}\dfrac{1}{|\mu_u|^{p^\ast}}\Big)^{1/p^\ast}>N\Big\}\in\mathcal{F}.\end{align*} $$

Set $g=\sum _{v\in F} e_v\in \ell ^p(V,\mu )$ . Let $U=\{f\in \ell ^p(V,\mu ):\, \|f-g\|_{p,\mu }<1/(2N)\}$ and $W=\{f\in \ell ^p(V,\mu ):\, |\langle f- g, e_{v}\rangle |<1/(2N),\, \forall v\in F\}$ . By the hypothesis, we have $N(U,W)\in \mathcal {F}$ an so $N(U,W)\cap [n,+\infty [\in \mathcal {F}$ for all $n\in \mathbb {N}$ . Let $n_0\in \mathbb {N}$ big enough so that $\mathrm {Chi}^{n}(F)\cap F=\emptyset $ , for all $n\geq n_0$ . We will show that

$$ \begin{align*}N(U,W)\cap [n_0,+\infty[\subset I(F,N).\end{align*} $$

Let $n\in N(U,W)\cap [n_0,+\infty [$ . Thus, $\mathrm {Chi}^{n}(F)\cap F=\emptyset $ and there exists $f\in \ell ^p(V,\mu )$ such that

(4.3) $$ \begin{align} \|f- g\|_{p,\mu}<\dfrac{1}{2N} \quad\text{and}\quad|\langle B^{n}f- g, e_{v}\rangle|<\dfrac{1}{2N},\quad \forall v\in F. \end{align} $$

Fix now $v\in F$ . Then the rightmost inequality in (4.3) gives

$$ \begin{align*} 0<\dfrac{1}{2}\leq 1- \dfrac{1}{2N} < \sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)|. \end{align*} $$

Since $\mathrm {Chi}^{n}(v)\cap F=\emptyset $ , by using the leftmost inequality in (4.3), we obtain

$$ \begin{align*} \sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)\mu_u|^p<\dfrac{1}{(2N)^p}. \end{align*} $$

Using Hölder’s inequality (for $p>1$ ) and the last two inequalities, we can conclude that:

$$ \begin{align*} \dfrac{1}{2N}&> \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)\mu_u|^p\Big)^{1/p}\\ &\geq \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)|\Big) \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|\mu_u|^{-p^{\ast}}\Big)^{-1/p^{\ast}}\\ &>\dfrac{1}{2} \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|\mu_u|^{-p^{\ast}}\Big)^{-1/p^{\ast}}, \end{align*} $$

thus

$$ \begin{align*}\Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|\mu_u|^{-p^{\ast}}\Big)^{1/p^{\ast}}>N.\end{align*} $$

This holds for all $v\in F$ . Therefore, $n\in I(F,N)$ , hence $N(U,W)\cap [n_0,+\infty [\subset I(F,N)$ . Consequently, $I(F,N)\in \mathcal {F}$ .

Proof It is clear that $(1)\Rightarrow (2)$ . Assume that $(2)$ holds. Let $U,V$ be nonempty open subsets of X. There exist a nonempty open subset $V'$ of X and a $0$ -neighborhood W such that $W+V' \subset V$ . Assume that $U\neq V'$ , otherwise there is nothing to prove. Let $y\in (U-V')\cap X_0$ . Then there exists a nonnegative integer N such that $T^ny\in W$ , for all $n\geq N$ . By the hypothesis, we have $N(V',V')\in \mathcal {F}$ and so $N(V',V')\cap [N,+\infty [\in \mathcal {F}$ . Now, we will show that

$$ \begin{align*}N(V',V')\cap [N,+\infty[ \subset N(U,V).\end{align*} $$

Let $n\in N(V',V')\cap [N,+\infty [$ . Thus, $T^nV'\cap V'\neq \emptyset $ and $T^ny\in W$ . Then there exists $z\in V'$ such that $T^nz\in V'$ . Hence $x=y+z\in U$ and

$$ \begin{align*}T^nx=T^ny+T^nz\in W+V'\subset V,\end{align*} $$

therefore, $n\in N(U,V)$ . Consequently, $N(V',V')\cap [N,+\infty [\subset N(U,V)$ and so $N(U,V)\in \mathcal {F}$ . This finishes the proof.

Proof We will only prove the equivalences in the case where $X = \ell ^p(V,\mu )$ with $1 < p < +\infty $ . A similar argument can be made to deduce the cases where $X = \ell ^1(V,\mu )$ and $X = c_0(V,\mu )$ . As in the proof of Theorem 4.1 and by using [Reference Grosse-Erdmann and Papathanasiou22, Theorem 5.2], we obtain that $(1)\Longleftrightarrow (2)$ . Note that it is clear that $(2)\Rightarrow (4)$ .

Let us show that $(4)\Rightarrow (3)$ . Let F be a finite subset of V and let $N\in \mathbb {N}$ . Set $g=\sum _{v\in F} e_v\in \ell ^p(V,\mu )$ . Let $U=\{f\in \ell ^p(V,\mu ):\, \|f-g\|_{p,\mu }<\frac {1}{2N}\}$ . Let $n_0\in \mathbb {N}$ be big enough so that $\mathrm {Chi}^{n}(F)\cap F=\emptyset $ and $\mathrm {par}^{n}(F)\cap F=\emptyset $ , for all $n\geq n_0$ . By the hypothesis, $N(U,U)\in \mathcal {F}$ and so $N(U,U)\cap [n_0,+\infty [\in \mathcal {F}$ . Since $\mathcal {F}$ is a Furstenberg family, it is enough to prove that

$$ \begin{align*}N(U,U)\cap [n_0,+\infty[\subset I(F,N)\cap J(F,N).\end{align*} $$

We can assume that N is sufficiently large such that $1<|\mu _v|N$ , $\forall v \in F$ . Let $n\in N(U,U)\cap [n_0,+\infty [$ . Then there then a function f in $\ell ^p(V,\mu )$ such that

(5.1) $$ \begin{align} \|f- g\|_{p,\mu}<\dfrac{1}{2N} \quad\text{and}\quad\|B^{n}f- g\|_{p,\mu}<\dfrac{1}{2N}. \end{align} $$

Fix now $v\in F$ . By the rightmost inequality above, we have

(5.2) $$ \begin{align} 0<\dfrac{1}{2}<1-\dfrac{1}{2N|\mu_v|}< \sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)|. \end{align} $$

Since $\mathrm {Chi}^{n}(v)\cap F=\emptyset $ and $F=\operatorname {\mathrm {supp}}(g)$ , by the leftmost inequality in (5.1), we obtain

(5.3) $$ \begin{align} \sum_{u\in \mathrm{Chi}^{n}(v) } |f(u)\,\mu_u|^p<\dfrac{1}{(2N)^p}. \end{align} $$

By Hölder’s inequality ( $p>1$ ) and (5.2), we get

$$ \begin{align*} \sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)\mu_u|^p &\geqslant \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|f(u)|\Big)^{p} \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|\mu_u|^{\frac{p}{1-p}}\Big)^{1-p}\\ &\geqslant \dfrac{1}{2^p} \Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|\mu_u|^{\frac{p}{1-p}}\Big)^{1-p}, \end{align*} $$

combining this with (5.3), we get

$$ \begin{align*}\Big(\sum_{u\in\mathrm{Chi}^{n}(v)}|\mu_u|^{-p^{\ast}}\Big)^{1/p^{\ast}}>N;\end{align*} $$

this holds for any fixed $v\in F$ , hence $n\in I(F,N)$ . Let us also show that $n\in J(F,N)$ . By contradiction, suppose that there is a $v\in F$ for which it holds

(5.4) $$ \begin{align} \dfrac{1}{|\mu_{\mathrm{par}^{n}(v)}|^{p^\ast}}+\underset{u \in \mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}{\sum}\dfrac{1}{|\mu_u|^{p^{\ast}}}\leq N^{p^\ast}. \end{align} $$

Thus, $\frac {1}{N}\leqslant |\mu _{\mathrm {par}^{n}(v)}|$ . Set

$$ \begin{align*}h=(f-g)\chi_{\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}.\end{align*} $$

One has

(5.5) $$ \begin{align} \|h\|_{p,\mu}\leqslant \|f-g\|_{p,\mu}<\dfrac{1}{2N}. \end{align} $$

Note now that we have

$$ \begin{align*}(B^{n}f)(\mathrm{par}^{n}(v))=d_n+\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}h(u),\end{align*} $$

where $d_n:=\big |\mathrm {Chi}^{n}(\mathrm {par}^{n}(v))\cap F\big |\geq 1$ , where $|\cdot |$ stands for the cardinality. Since $\mathrm {par}^{n}(v)\in V\setminus F$ , by using the rightmost inequality in (5.1), we get

$$ \begin{align*} \dfrac{1}{2N} &>|(B^{n}f)(\mathrm{par}^{n}(v))\,\mu_{\mathrm{par}^{n}(v)}|\\ &\geq \Big(d_n-\big|\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}h(u)\big|\Big)|\mu_{\mathrm{par}^{n}(v)}|\\ & \geq \dfrac{1}{N}\Big(1-|\sum_{u\in\mathrm{Chi}^{n_k}(\mathrm{par}^{n_k}(v))}h_k(u)|\Big), \end{align*} $$

hence

$$ \begin{align*}\dfrac{1}{2}<\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}|h(u)|.\end{align*} $$

By Hölder’s inequality and (5.5), we have

$$ \begin{align*} \dfrac{1}{2N}&> \Big(\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}|h(u)|\Big) \Big(\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}|\mu_u|^{-p^\ast}\Big)^{-1/p^{\ast}}\\ &> \dfrac{1}{2} \Big(\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}|\mu_u|^{-p^\ast}\Big)^{-1/p^{\ast}}, \end{align*} $$

then

$$ \begin{align*}\underset{u \in \mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}{\sum}\dfrac{1}{|\mu_u|^{p^{\ast}}}>N^{p^\ast}.\end{align*} $$

This contradicts (5.4). Therefore, $n\in J(F,N)$ . Consequently, $N(U,U)\cap [n_0,+\infty [\subset I(F,N)\cap J(F,N)$ , and so $I(F,N)\cap J(F,N)\in \mathcal {F}$ .

Note that a minor adjustment to the proof of the implication $(4)\Rightarrow (3)$ allows us to deduce $(2)\Rightarrow (3)$ . Let us show now that $(3)\Rightarrow (2)$ . Let $\mathcal {B}$ be the filter base consisting of all subsets $I(F,N)\cap J(F,N)$ of $\mathbb {N}_0$ , where $N>0$ and $F\subset V$ is finite. Let $\mathcal {F}_{\mathcal {B}}$ be the filter generated by $\mathcal {B}$ , that is,

$$ \begin{align*}\mathcal{F}_{\mathcal{B}}=\{A\subset \mathbb{N}_0: B\subset A \text{ for some } B\in \mathcal{B}\}.\end{align*} $$

We will show that B satisfies the $\mathcal {F}_{\mathcal {B}}$ -transitivity criterion, which implies that B is $\mathcal {F}$ -transitive since $\mathcal {F}_{\mathcal {B}}\subset \mathcal {F}$ . Set $\mathcal {D}=\mathrm {span}\{e_v:\, v\in V\}$ , which is dense in $\ell ^p(V,\mu )$ . For every $v\in V$ and $n\in \mathbb {N}$ , by the continuity of $B^{n}$ , we have

$$ \begin{align*}M_{v,n}:=\dfrac{1}{|\mu_{\mathrm{par}^{n}(v)}|^{p^\ast}}+\underset{u \in \mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}{\sum}\dfrac{1}{|\mu_u|^{p^{\ast}}}<+\infty,\end{align*} $$

then either

(5.6) $$ \begin{align} \dfrac{1}{|\mu_{\mathrm{par}^{n}(v)}|^{p^\ast}}\geq \dfrac{M_{v,n}}{2}, \end{align} $$

or

(5.7) $$ \begin{align} \underset{u \in \mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}{\sum}\dfrac{1}{|\mu_u|^{p^{\ast}}}>\dfrac{M_{v,n}}{2}. \end{align} $$

In the case where (5.6) holds, we set

$$ \begin{align*}I_{n}e_v=e_v,\end{align*} $$

and then

(5.8) $$ \begin{align} \|I_{n}e_v-e_v\|_{p,\mu}=0 \quad \text{and} \quad\|B^{n}I_{n} e_v\|_{p,\mu}=\|e_{\mathrm{par}^{n}(v)}\|_{p,\mu}=| \mu_{\mathrm{par}^{n}(v)}|\leq \Big(\dfrac{2}{M_{v,n}}\Big)^{1/p^\ast}. \end{align} $$

In the case where (5.7) holds, by Lemma 2.4, there exists $h_{v,n}\in \mathbb {K}^V$ nonnegative, of support in $\mathrm {Chi}^{n}(\mathrm {par}^{n}(v))$ such that

(5.9) $$ \begin{align} \sum_{u\in \mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}h_{v,n}(u)=1 \quad \text{and} \quad \Big(\sum_{u\in\mathrm{Chi}^{n}(\mathrm{par}^{n}(v))}|h_{v,n}(u)\,\mu_u|^p\Big)^{1/p}\leq \Big(\dfrac{2}{M_{v,n}}\Big)^{1/p^\ast}. \end{align} $$

Hence, $h_{v,n}\in \ell ^p(V,\mu )$ , and by setting

$$ \begin{align*}I_{n}e_v=e_v-h_{v,n},\end{align*} $$

we obtain

(5.10) $$ \begin{align} \|I_{n}e_v-e_v\|_{p,\mu}\leq \Big(\dfrac{2}{M_{v,n}}\Big)^{1/p^\ast} \quad \text{and} \quad B^{n}I_{n} e_v=0. \end{align} $$

In both cases, we extend linearly on $\mathcal {D}$ the maps $I_{n}$ . Let $g=\sum _{a\in F}g(a) e_a\in \mathcal {D}$ , where $F\subset V$ is its support. We will show that $\mathcal {F}_{\mathcal {B}}\text {-}\underset {n}{\lim }(I_ng,B^nI_ng)=(g,0)$ . Let $\varepsilon>0$ , $\mathcal {U}:=\{f\in \ell ^p(V,\mu ):\, \|f-g\|_{p,\mu }<\varepsilon \}$ and $\mathcal {V}:=\{f\in \ell ^p(V,\mu ):\, \|f\|_{p,\mu }<\varepsilon \}$ . Let us check that

(5.11) $$ \begin{align} J(F,N)\subset\{n\in\mathbb{N}_0:\, (I_ng,B^nI_ng)\in \mathcal{U}\times \mathcal{V}\}, \end{align} $$

where

$$ \begin{align*}N=\max\Big\{\frac{|g(a)|\,|F|2^{1/p^\ast}}{\varepsilon}:\, a\in F\Big\}.\end{align*} $$

Let $n\in J(F,N)$ . We have then $M_{a,n}>N^{p^\ast }$ , for all $a\in F$ . By the definition of N, (5.8) and (5.10), we obtain

$$ \begin{align*} \max \Big\{\|I_{n}g-g\|,\|B^{n}I_{n}g\|_{p,\mu} \Big\} &\leq \sum_{a\in F}|g(a)| \Big(\dfrac{2}{M_{a,n}}\Big)^{1/p^\ast}\\ &< \sum_{a\in F}|g(a)| \dfrac{2^{1/p^\ast}}{N}\\ &< \varepsilon, \end{align*} $$

hence $I_ng\in \mathcal {U}$ and $B^nI_ng\in \mathcal {V}$ . Therefore, (5.11) holds and so

$$ \begin{align*}\{n\in\mathbb{N}_0:\, (I_ng,B^nI_ng)\in \mathcal{U}\times \mathcal{V}\}\in\mathcal{F}_{\mathcal{B}}.\end{align*} $$

Consequently, the first statement of the $\mathcal {F}_{\mathcal {B}}$ -transitivity criterion holds. As for the second statement, we can define the maps $S_n$ as in the proof of Theorem 4.1. By employing the same arguments, we can deduce that

$$ \begin{align*}\{n\in\mathbb{N}_0:\, (B^nS_ng,S_ng)\in \mathcal{U}\times \mathcal{V}\}\in\mathcal{F}_{\mathcal{B}}.\end{align*} $$

This shows that the second statement of the $\mathcal {F}_{\mathcal {B}}$ -transitivity criterion holds. Hence, B is $\mathcal {F}$ -transitive and statement $(2)$ holds.

Proof It is clear that $(1)\Rightarrow (2)\Rightarrow (3)$ . Let us show that $(3)\Rightarrow (1)$ . Assume that T has dense range. Let $\Gamma $ be an unbounded subset of $\mathbb {C}$ . We will show that $T\oplus T$ is $\Gamma $ -supercyclic. Let $U_1$ , $U_2$ , $V_1$ , $V_2$ be nonempty open subsets of X. Since $\ker ^\ast (T)$ is dense in X, there exists $(x_1,x_2)\in U_1\times U_2$ such that $T^nx_1=T^nx_2=0$ for some $n\in \mathbb {N}$ . Since T has dense range, $T^n$ also does; therefore, there exist $y_1,y_2\in X$ such that $(T^ny_1,T^ny_2)\in V_1\times V_2$ . By the unboundedness of $\Gamma $ , there exists $\lambda \in \Gamma $ such that

$$ \begin{align*}\dfrac{1}{\lambda}(y_1,y_2)+(x_1,x_2)\in U_1\times U_2,\end{align*} $$

and

$$ \begin{align*}\lambda(T\oplus T)^n(\dfrac{1}{\lambda}(y_1,y_2)+(x_1,x_2))=(T^ny_1,T^ny_2)+\lambda(T^nx_1,T^nx_2)\in V_1\times V_2.\end{align*} $$

Hence, $T\oplus T$ is $\Gamma $ -supercyclic.

Proof It is enough to prove that $(2) \Rightarrow (5) \Rightarrow (6) \Rightarrow (3)$ and $(5)\Rightarrow (7)$ . We will prove these implications when $X=\ell ^p(V,\mu )$ , with $1<p<+\infty $ , the same reasoning works for the other cases. Let us show that $(2)$ implies $(5)$ . Let $f,g\in \ell ^p(V,\mu )$ be nonzero vectors. Let $v_0\in V$ such that $g(v_0)\neq 0$ . Let $(\delta _k)_{k}$ be a decreasing sequence of positive numbers such that $2\delta _k<|g(v_0)|$ . Let $(n_k)_{k}$ be an increasing sequence of nonnegative integers such that

$$ \begin{align*}|\langle B^{n_k}f- g, e_{v_0}\rangle|<\delta_k,\end{align*} $$

that is

$$ \begin{align*}|(B^{n_k}f)(v_0)-g(v_0)|<\delta_k,\end{align*} $$

thus

(7.1) $$ \begin{align} 0<\dfrac{|g(v_0)|}{2}<|g(v_0)|-\delta_k< \sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|f(u)|. \end{align} $$

Since $\|f\|_{p,\mu }<+\infty $ , we obtain

$$ \begin{align*}\sum_{k\geq0}\sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|f(u)\mu_u|^p<+\infty,\end{align*} $$

hence

(7.2) $$ \begin{align} \sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|f(u)\mu_u|^p\underset{k\to+\infty}{\longrightarrow}0. \end{align} $$

By Hölder’s inequality ( $p>1$ ), we get

$$ \begin{align*} \sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|f(u)\mu_u|^p &\geqslant \Big(\sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|f(u)|\Big)^{p} \Big(\sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|\mu_u|^{\frac{p}{1-p}}\Big)^{1-p}\\ &\overset{(7.1)}{\geqslant} \dfrac{|g(v_0)|^p}{2^p} \Big(\sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|\mu_u|^{\frac{p}{1-p}}\Big)^{1-p} \end{align*} $$

hence

$$ \begin{align*}\Big(\sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|\mu_u|^{-p^{\ast}}\Big)^{-1/p^{\ast}}\leqslant \dfrac{2}{|g(v_0)|} \Big(\sum_{u\in\mathrm{Chi}^{n_k}(v_0)}|f(u)\mu_u|^p\Big)^{1/p},\end{align*} $$

combining this with (7.2), we get

$$ \begin{align*}\sum_{u \in \mathrm{Chi}^{n_k}(v_0)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\underset{k\to+\infty}{\longrightarrow}+\infty.\end{align*} $$

Let us show now that $(5)$ implies $(6)$ . Suppose that $(5)$ holds for some sequence $(n_k)_k$ and $v\in V$ . Let $m\in \mathbb {N}$ be such that $v\in \mathrm {Chi}^{m}(\texttt {r})$ . Set $m_k=n_k+m$ . Thus, $\mathrm {Chi}^{n_k}(v)\subset \mathrm {Chi}^{m_k}(\texttt {r})$ and hence

$$ \begin{align*}\sum_{u \in \mathrm{Chi}^{n_k}(v)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\leqslant \sum_{u \in \mathrm{Chi}^{m_k}(\texttt{r})}\dfrac{1}{|\mu_u|^{p^{\ast}}},\end{align*} $$

since the term at the left goes to infinity, as $k \to +\infty $ , we get

$$ \begin{align*}\sum_{u \in \mathrm{Chi}^{m_k}(\texttt{r})}\dfrac{1}{|\mu_u|^{p^{\ast}}} \underset{k\to+\infty}{\longrightarrow}+\infty.\end{align*} $$

Let us show that $(6)$ implies $(3)$ . Let $(n_k)_{k\geq 0}$ be an increasing sequence of positive integers such that

$$ \begin{align*}\sum_{u \in \mathrm{Chi}^{n_k}(\texttt{r})}\dfrac{1}{|\mu_u|^{p^{\ast}}}\underset{k\to+\infty}{\longrightarrow}+\infty.\end{align*} $$

Set $J_k=\mathrm {Chi}^{n_k}(\texttt {r})$ , for $k\in \mathbb {N}$ . By Lemma 2.4, we obtain

$$ \begin{align*}\underset{\|x\|_1=1}{\inf}\Big(\sum_{u\in J_k}|x_u\mu_u|^p\Big)^{1/p}\underset{k\to+\infty}{\longrightarrow}0\end{align*} $$

There exists then an increasing sequence $(k_j)_{j\geq 0}$ of positive integers such that, for each $j\in \mathbb {N}_0$ , we have

$$ \begin{align*}\underset{\|x\|_1=1}{\inf}\Big(\sum_{u\in J_{k_j}}|x_u\mu_u|^p\Big)^{1/p}\leqslant\dfrac{1}{2^{j+1}c_j}, \quad\text{ where } \quad c_j=\max\lbrace 1; \|B^{n_{k_l}}\|, 0\leqslant l\leqslant j-1\rbrace,\end{align*} $$

hence, there exists $f_{k_j}\in \mathbb {K}^{V}$ , of support in $J_{k_j}=\mathrm {Chi}^{n_{k_j}}(\texttt {r})$ such that

(7.3) $$ \begin{align} \|f_{k_j}\|_1=\sum_{u\in \mathrm{Chi}^{n_{k_j}}(\texttt{r})}|f_{k_j}(u)|=1, \end{align} $$

and

(7.4) $$ \begin{align} \Big(\sum_{u\in J_{k_j}}|f_{k_j}(u)\mu_u|^p\Big)^{1/p}\leqslant\dfrac{1}{2^{j}c_j}. \end{align} $$

For each $u\in V$ , set

$$ \begin{align*}f(u)=\begin{cases} g_j(u) &\text{ if } u\in \mathrm{Chi}^{n_{k_j}}(\texttt{r}) \text{ for some } j\\ 0 &\text{ otherwise } \end{cases},\quad \text{where } g_j=|f_{k_j}|. \end{align*} $$

By (7.4), we obtain

$$ \begin{align*}\|f\|_{p,\mu}^{p}=\sum_{u\in V}|f(u)\mu_u|^p=\sum_{j\in\mathbb{N}_0}\sum_{u\in \mathrm{Chi}^{n_{k_j}}(\texttt{r})}|f_{k_j}(u)\mu_u|^p \leqslant \sum_{j\in\mathbb{N}_0}\dfrac{1}{2^{pj}}<+\infty, \end{align*} $$

thus $f\in \ell ^p(V,\mu )$ . Moreover, we have

$$ \begin{align*} \| B^{n_{k_j}}f-e_{\texttt{r}}\|_{p,\mu}&\leq \| B^{n_{k_j}}g_j-e_{\texttt{r}}\|_{p,\mu}+\sum_{ l=j+1} \| B^{n_{k_j}}g_l\|_{p,\mu}\\ &=\sum_{ l=j+1} \| B^{n_{k_j}}g_l\|_{p,\mu}\qquad (\text{by } (7.3))\\ &\leqslant \sum_{ l=j+1} \| B^{n_{k_j}}\| \|f_{k_l}\|_{p,\mu}\\ &< \sum_{ l=j+1}\dfrac{\| B^{n_{k_j}}\|}{2^{l}c_{l}} \qquad (\text{by } (7.4))\\ &\leqslant \sum_{ l=j+1}\dfrac{1}{2^{l}}\underset{j\to+\infty}{\longrightarrow}0, \end{align*} $$

hence condition $(3)$ holds.

Let us show that $(5)$ implies $(7)$ . Set $C=\max \lbrace 2, \|B\|\rbrace $ . By assuming that $(5)$ holds, there exist an increasing sequence $(n_{k})_{k\in \mathbb {N}}$ of positive integers and a vertex $v\in V$ such that

(7.5) $$ \begin{align} \Big(\sum_{u \in \mathrm{Chi}^{n_{k}+k}(v)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\Big)^{-1/p^{\ast}}<C^{-2k}. \end{align} $$

For every $u\in V\setminus \lbrace \texttt {r}\rbrace $ , by Proposition 2.2, we obtain

$$ \begin{align*}C\geqslant \Big(\sum_{v\in \mathrm{Chi}(\mathrm{par}(u))} \left|\frac{\mu_{\mathrm{par}(u)}}{\mu_v }\right|{}^{p^\ast}\Big)^{1/p^{\ast}}\geqslant \left|\frac{\mu_{\mathrm{par}(u)}}{\mu_u }\right|,\end{align*} $$

that is, for every $u\in V\setminus \lbrace \texttt {r}\rbrace $ ,

(7.6) $$ \begin{align} \left|\frac{\mu_u }{\mu_{\mathrm{par}(u)}}\right|\geqslant C^{-1}. \end{align} $$

Fix now $l\in \mathbb {N}_0$ . Let $k\in \mathbb {N}$ be such that $k>l$ . Note that

(7.7) $$ \begin{align} u\in \mathrm{Chi}^{n_{k}+k}(v) \Longleftrightarrow \mathrm{par}^{k-l}(u)\in \mathrm{Chi}^{n_{k}+l}(v), \end{align} $$

and for every $u\in \mathrm {Chi}^{n_{k}+k}(v)$ , we have

$$ \begin{align*}\mu_{u}=\dfrac{\mu_u}{\mu_{\mathrm{par}(u)}}\times \dfrac{\mu_{\mathrm{par}(u)}}{\mu_{\mathrm{par}^2(u)}} \times \cdots\times \dfrac{\mu_{\mathrm{par}^{k-l-1}(u)}}{\mu_{\mathrm{par}^{k-l}(u)}}\times \mu_{\mathrm{par}^{k-l}(u)}.\end{align*} $$

By (7.6), we obtain

$$ \begin{align*}\dfrac{1}{|\mu_{\mathrm{par}^{k-l}(u)}|^{p^{\ast}}}\geqslant C^{(l-k)p^{\ast}}\dfrac{1}{|\mu_{u}|^{p^{\ast}}}.\end{align*} $$

Combining this with (7.7), we obtain

$$ \begin{align*}\Big(\sum_{u \in \mathrm{Chi}^{n_{k}+l}(v)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\Big)^{1/p^{\ast}}\geqslant C^{l-k} \Big(\sum_{u \in \mathrm{Chi}^{n_{k}+k}(v)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\Big)^{1/p^{\ast}}.\end{align*} $$

Thus,

$$ \begin{align*} \Big(\sum_{u \in \mathrm{Chi}^{n_{k}+l}(v)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\Big)^{-1/p^{\ast}}&\leqslant C^{k-l} \Big(\sum_{u \in \mathrm{Chi}^{n_{k}+k}(v)}\dfrac{1}{|\mu_u|^{p^{\ast}}}\Big)^{-1/p^{\ast}}\\ &\overset{(7.5)}{\leqslant} C^{k-l} C^{-2k}\leqslant C^{-k} \underset{k\to+\infty}{\longrightarrow}0. \end{align*} $$