2.1. Topological dynamics of linear operators

Let $T:\mathcal {H}\to \mathcal {H}$ be a continuous linear operator acting on a separable Banach space $\mathcal {H}$.

We recall that in the setting of Banach spaces, $T$ is topologically transitive if and only if $T$ admits a hypercyclic vector $\varphi$, i.e., $\varphi \in \mathcal {H}$ such that the orbit $\{\varphi ,\,T\varphi ,\,T^{2}\varphi \ldots \}$ is dense in $\mathcal {H}$.

Definition 2.3 A vector $\varphi \in \mathcal {H}$ is called frequently hypercyclic if for each non-empty open set $U\subseteq \mathcal {H}$, the set of integers $\mathcal {N}(\varphi ,\,U)=\{n\in \mathbb {N}: T^{n}\varphi \in U\}$ has positive lower density, that is,

\[ \liminf_{N\to\infty} \frac{1}{N}\#\left\{1\le n\le N: T^{n}\varphi\in U\right\}>0\cdot \]

The operator $T$ is called frequently hypercyclic if it admits a frequently hypercyclic vector.

The following Frequently Hypercyclic Criterion was provided by Bonilla and Grosse-Erdmann in [Reference Bonilla and Grosse-Erdmann8, theorem 2.1]. It is a strengthened version of the original criterion obtained by Bayart and Grivaux in [Reference Bayart and Grivaux3, theorem 2.1]. Its simplified reformulation is stated below in the context that we use.

2.2. Measurable dynamics

Definition 2.7 Let $p\ge 1$. The composition operator $T_f$ induced by a measurable system $(X,\,\mathcal {B},\,\mu ,\,f)$ is the map $T_f:L^{p}(X)\to L^{p}(X)$ defined by

\[ T_f: \varphi\to\varphi\circ f. \]

It is well known that (*) guarantees that $T_f$ is a continuous linear operator. We refer the reader to [Reference Singh and Manhas19] for a detailed exposition on compositions operators.

We say that a set $A\subseteq X$ is $f$-invariant if $f^{-1}(A)=A$.

In the sequel, we let $\mathcal {B}(W)=\{B\cap W: B\in \mathcal {B}\}$.

Notice that in the definition of dissipative system (see Definition 2.8), we do not require that $W$ has finite measure.

3.1. The summability condition and the frequent hypercyclicity criterion

Let $(X,\,\mathcal {B},\,\mu ,\,f)$ be a measurable system. We say that $f$ satisfies the Summability Condition (SC) if for each $\epsilon >0$ and $B\in \mathcal {B}$ with $\mu (B)<\infty$, there exists a measurable set $B'\subseteq B$ such that

(SC)\begin{equation} \mu\big(B{\setminus }B'\big)<\epsilon\quad\textrm{and}\quad\sum_{n\in\mathbb{Z}} \mu\big(f^{n}(B')\big)<\infty. \end{equation}

The following result shows that the Summability Condition (SC) is the natural translation of the FHC to the composition operator framework.

Now we will provide a list of results that show how Condition (SC) is useful to characterize when a transformation $f$ is dissipative and when the composition operator $T_f$ is frequently hypercyclic or chaotic. The results below are true for all $p\ge 1$.

In the sequel, we will need the following definition. Let $X$ be a metric space and $f:X\to X$ be a map. We say that $x\in X$ is recurrent (for $f$) if for every open set $U$ containing $x$, there exists an integer $n\ge 1$ such that $f^{n}(x)\in U$.

3.2. Bounded distortion and frequent hypercyclicity

Theorem 3.6 (Necessary condition for frequent hypercyclicity) Let $(X,\,\mathcal {B},\,\mu ,\, f)$ be a measurable system with associated composition operator $T_f$ frequently hypercyclic. Then for every wandering set $W$ with positive finite $\mu$-measure, the following inequality holds

\[ \sum_{n\in\mathbb{Z}} \left\Vert\dfrac{\mathrm{d}\mu}{\mathrm{d}\big(\mu\circ f^{n}\big)}\bigg|_W \right\Vert_{\infty}^{{-}1}<\infty\cdot \]

We say that $f$ is ergodic if every $f$-invariant set $A\in \mathcal {B}$ satisfies $\mu (A)=0$ or $\mu (X{\setminus }A)=0$.

Proof. $(a)\implies (c)$ Assume that $T_f$ is frequently hypercyclic. Let $x\in X$ be an atom of $\mu$. Since $f$ is dissipative and ergodic, we have that $W=\{x\}$ is a wandering set that generates $X$, thus $\mu (X)=\sum \nolimits _{n\in \mathbb {Z}}\mu (f^{n}(W))$. Applying Theorem 3.6 and using the fact that $\textrm {d}\mu /\textrm {d}(\mu \circ f^{n})$ on $W$ is equal to the constant $\mu (W)/\mu (f^{n}(W))$, we obtain that

\[ \frac{\mu(X)}{\mu(W)}=\sum_{n\in\mathbb{Z}}\frac{ \mu\big(f^{n}(W)\big)}{\mu(W)}=\sum_{n\in\mathbb{Z}} \left\Vert\dfrac{\mathrm{d}\mu}{\mathrm{d}\big(\mu\circ f^{n}\big)}\bigg|_W \right\Vert_{\infty}^{{-}1}<\infty, \]

implying that $\mu$ is finite.

$(c)\implies (b)$ It follows from Theorem 3.3 that $f$ satisfies Condition (SC) and hence it is chaotic.

$(b)\implies (a)$ As $f$ is dissipative and $T_f$ chaotic, by Corollary 3.5, we have that $f$ satisfies Condition (SC) and hence it is frequently hypercyclic.

5.1. Proof of Theorem 3.1

Proof of $(SC)\implies (FHC)$. Assume that $f$ satisfies (SC). We will show that $T_f$ satisfies (FHC). By Lemma 5.2, $f$ is dissipative. Let $W$ be a wandering set such that $X=\bigcup_{n\in \mathbb {Z}} f^{n}(W)$. Let $\mathcal {H}=L^{p}(X)$ and $\mathcal {H}_0\subseteq \mathcal {H}$ be defined as follows: $\varphi \in \mathcal {H}_0$ if and only if there exist $m\ge 1,\, a_1,\,\ldots ,\,a_m\in \mathbb {R}$, $k_1,\,\ldots ,\,k_m\in \mathbb {Z}$, and pairwise disjoint measurable sets $B_1,\,\ldots ,\,B_m$ such that $\varphi =\sum \nolimits _{i=1}^{m} a_i \chi _{B_i}$ and, for each $1\le i\le m$, $B_i\subseteq f^{k_i}(W)$ and $\sum \nolimits _{n\in \mathbb {Z}} \mu (f^{n}(B_i) )<\infty$.

We will now show that $\mathcal {H}_0$ is dense in $\mathcal {H}$. Given $\epsilon >0$ and $\psi \in \mathcal {H}$, let $\varphi =\sum \nolimits _{j=1}^{r} a_j\chi _{D_j}\in \mathcal {H}\backslash \{0\}$ be such that $D_1,\,\ldots ,\,D_r$ are pairwise disjoint measurable sets with finite positive $\mu$-measures and $\Vert \varphi -\psi \Vert _p<\epsilon /2$. Set $M=\max \{\vert a_j\vert : 1\le j\le r\}$, then $M>0$. As $f$ satisfies Condition (SC) and $X=\bigcup_{n\in \mathbb {Z}}f^{n}(W)$, there exist an integer $N>0$ and measurable sets $C_1,\,\ldots ,\,C_r$ such that, for each $1\le j\le r$, $C_j\subseteq D_j\cap \bigcup_{\vert n\vert \le N} f^{n}(W)$,

\[ \mu(D_j{\setminus}C_j)<\left(\dfrac{\epsilon}{2 r M}\right)^{p} \text{ and }\sum_{n\in\mathbb{Z}} \mu\big(f^{n}(C_j)\big)<\infty. \]

Let $\varphi ' =\sum \nolimits _{j=1}^{r} a_j \chi _{C_j}$. Since $C_j\subseteq \bigcup_{\vert n\vert \le N} f^{n}(W)$, we have that each $C_j$ is the union of finitely many pairwise disjoint sets $B_i$ satisfying $B_i\subseteq f^{k_i}(W)$ for some $-N\le k_i\le N$. In this way, $\varphi '\in \mathcal {H}_0$. Moreover,

\[ \Vert \varphi -\varphi '\Vert_p=\left\Vert \sum_{j=1}^{r} a_j \big(\chi_{D_j}-\chi_{C_j}\big)\right\Vert_p= \left\Vert \sum_{j=1}^{r} a_j \chi_{D_j{\setminus C_j}}\right\Vert_p\le M\sum_{j=1}^{r} \left[\mu(D_j{\setminus} C_j)\right]^{1/p}<\dfrac{\epsilon}{2}. \]

In this way, $\Vert \varphi ' -\psi \Vert _p<\epsilon$, completing the proof of the denseness of $\mathcal {H}_0$.

We next show that $\sum \nolimits _{n\ge 1} T_f^{n}\varphi$ is unconditionally convergent for all $\varphi =\sum \nolimits _{i=1}^{m} a_i \chi _{B_i} \in \mathcal {H}_0$. Since $B_i\subseteq f^{k_i}(W)$, we have that the sets $f^{n}(B_i)$, $n\in \mathbb {Z}$, are pairwise disjoint. Hence, for each sequence of integers $1\le n_1< n_2<\ldots$, we have that

(1)\begin{equation} \sum_{j\ge 1} T_f^{n_j}\varphi=\sum_{i=1}^{m} a_i \sum_{j\ge 1} \chi_{f^{{-}n_j}(B_i)}=\sum_{i=1}^{m} a_i \chi_{\bigcup_{j\ge 1}f^{{-}n_j}(B_i)}. \end{equation}

Since $\sum \nolimits _{n\in \mathbb {Z}}\mu (f^{n}(B_i))<\infty$ for all $1\le i\le m$, we have that $\chi _{\bigcup_{j\ge 1}f^{-n_j}(B_i)}\in L^{p}(X)$. Thus, $\sum \nolimits _{j\ge 1}T_f^{n_j}\varphi$ converges for each sequence of integers $1\le n_1< n_2<\ldots\ .$ Therefore, $\sum \nolimits _{n\ge 1} T_f^{n}\varphi$ is unconditionally convergent and Condition (a) in Theorem 2.4 is true.

Given $\varphi =\sum \nolimits _{i=1}^{m} a_i \chi _{B_i}\in \mathcal {H}_0$, let $S\varphi =\sum \nolimits _{i=1}^{m} a_i \chi _{f(B_i)}.$ Since $f$ is bijective, bimeasurable and non-singular, we have that $S\varphi \in \mathcal {H}_0$ and $S:\varphi \to S\varphi$ is a self-map on $\mathcal {H}_0$. Moreover, $TS\varphi =\sum \nolimits _{i=1}^{m} a_i \chi _{f^{-1}(f(B_i))}=\varphi$, showing that Condition (c) in Theorem 2.4 is true. By proceeding as in (5.1), we can show that $\sum \nolimits _{n\ge 1}S^{n}\varphi$ is unconditionally convergent for all $\varphi \in \mathcal {H}_0$. We have proved that $T_f$ satisfies (FHC).

Proof of $(FHC)\implies (SC)$ when $p\ge 2$. Assume the hypothesis, i.e., $T_f$ satisfies the FHC. Let $\mathcal {H}_0 \subseteq L^{p}(X)$ be as in the statement of (FHC). Let $0<\epsilon <\tfrac {1}{2}$ and $B\in \mathcal {B}$ with $\mu (B)<\infty$. By the denseness of $\mathcal {H}_0$, there is $\varphi \in \mathcal {H}_0$ such that

\[ \left\Vert \varphi-\chi_B \right\Vert_p< {\epsilon^{\left(1+(1/p)\right)}}. \]

Let

\[ B'=\left\{x\in B: \vert\varphi(x)-1\vert \le{\epsilon}\right\}. \]

Then,

\begin{align*} {\epsilon}\left[\mu(B{\setminus}B')\right]^{1/p}&\le \bigg(\int_{B{\setminus B'}} \vert \varphi-1\vert^{p}\,\textrm{d}\mu\bigg)^{1/p}\nonumber\\&=\bigg(\int_{B{\setminus B'}} \vert \varphi-\chi_B\vert^{p}\,\textrm{d}\mu\bigg)^{1/p}\le \Vert \varphi-\chi_B\Vert_p< {\epsilon^{\left(1+(1/p)\right)}},\end{align*}

showing that $\mu (B{\setminus }B')<\epsilon$.

We next show that $\sum \nolimits _{n \ge 1} \mu (f^{-n}(B')) < \infty$. Proceeding as earlier we have that for any $n\ge 1$,

\[ (1-{\epsilon})\left[\mu\big(f^{{-}n}(B')\big)\right]^{1/p} \le \Bigg(\int_{f^{{-}n}(B')} \left\vert \varphi\circ f^{n}\right\vert^{p} \textrm{d}\mu\Bigg)^{1/p} \]

implying that

\[ \sum_{n\ge 1} \mu\big(f^{{-}n}(B')\big)< 2^{p}\sum_{n\ge 1}\Bigg(\int_{f^{{-}n}(B')} \left\vert \varphi\circ f^{n}\right\vert^{p} \textrm{d}\mu\Bigg) \le 2^{p}\sum_{n\ge 1} \Vert T_f^{n}\varphi\Vert_p^{p}. \]

Since $\sum \nolimits _{n\ge 1} T_f^{n}\varphi$ converges unconditionally and $p\ge 2$, by Theorem 5.1 we have that

\[ \sum_{n\ge 1} \mu\big(f^{{-}n}(B')\big)<\infty. \]

We next show that $\sum \nolimits _{n \ge 1} \mu (f^{n}(B')) < \infty$. As $\{\varphi ,\, S\varphi ,\, \ldots ,\, S^{n}\varphi \}\subset \mathcal {H}_0$ and $T_fS$ is the identity map on $\mathcal {H}_0$, we have that $T_f^{n}S^{n}(\varphi ) = \varphi$ for all $n \ge 1$. Since $f$ is bijective, bimeasurable and non-singular, we have that

\[ \left [T_f^{n}S^{n}(\varphi)|_B = \varphi|_B \right ] \Longrightarrow \left [S^{n}(\varphi)\circ f^{n}| _B = \varphi|_B \right ] \Longrightarrow \left [ S^{n} (\varphi)|_{f^{n}(B)} = \varphi \circ f^{{-}n}|_{f^{n}(B)} \right ]. \]

Using this we have that

\begin{align*} (1-{\epsilon})\left[\mu\big(f^{n}(B')\big)\right]^{1/p} &\le\Bigg(\int_{f^{n}(B')} \left\vert \varphi\circ f^{{-}n}\right\vert^{p} \textrm{d}\mu\Bigg)^{1/p}\nonumber\\&=\Bigg(\int_{f^{n}(B')}\left\vert S^{n} (\varphi)\right\vert^{p} \textrm{d}\mu\Bigg)^{1/p} \le \left\Vert S^{n}\varphi\right\Vert_p.\end{align*}

Since $\sum \nolimits _{n\ge 1} S_f^{n}\varphi$ converges unconditionally and $p\ge 2$, by Lemma 5.1 we have that

\[ \sum_{n\ge 1} \mu\big(f^{n}(B')\big)\le 2^{p}\sum_{n\ge 1}\left\Vert S^{n}(\varphi)\right\Vert_p^{p}<\infty, \]

which completes the proof.

5.2. Proof of Theorem 3.2

Proof of $(a) \implies (b)$. Suppose that $f$ satisfies Condition (SC). Then, by Lemma 5.2, $f$ is dissipative. By Theorem 3.1, $T_f$ satisfies (FHC). Hence, by Theorem 2.4, $T_f$ is frequently hypercyclic, topologically mixing and chaotic. In particular, $T_f$ has a dense set of periodic points.

Proof of $(b) \Longrightarrow (a)$. Since $f$ is dissipative, there exist a wandering set $W\in \mathcal {B}$ of positive $\mu$-measure such that $X=\bigcup_{n\in \mathbb {Z}} f^{n}(W)$. Let $\epsilon >0$ and $B\in \mathcal {B}$ with $\mu (B)<\infty$. Let $n_0\in \mathbb {N}$ be such that if

\[ B_0=B\cap\bigcup_{\vert k\vert\le n_0}f^{k}(W),\text{ then } \mu(B{\setminus} B_0)<\dfrac{\epsilon}{2}. \]

By hypothesis, there exists a periodic point $\varphi \in L^{p}(X)$ of $T_f$ such that

\[ \Vert \varphi-\chi_{B_0}\Vert_p^{p}\le \dfrac{1}{4^{p}}\cdot \dfrac{\epsilon}{2}. \]

For $B'=\{x\in B_0: |\varphi (x)|>{3}/{4}\}$, we have that $\mu (B_0{\setminus } B')\le {\epsilon }/{2}$ because

\[ \dfrac{1}{4^{p}} \mu(B_0{\setminus}B')\le\int_{B_0{\setminus}B'} \left| \varphi-1\right|^{p} \textrm{d}\mu\le \Vert \varphi-\chi_{B_0}\Vert_p^{p}\le \dfrac{1}{4^{p}}\cdot \dfrac{\epsilon}{2}, \text{ implying } \mu\big(B_0{\setminus}B'\big)\le \frac{\epsilon}{2}. \]

In this way,

\[ \mu(B{\setminus}B')= \mu(B{\setminus}B_0)+\mu(B_0{\setminus}B')< \epsilon. \]

Let $N\ge 2n_0+1$ be such that $T_f^{N} \varphi =\varphi \circ f^{N}=\varphi$ $\mu$-a.e. Hence, since $f$ is bijective, we have that $\varphi \circ f^{kN}=\varphi$ $\mu$-a.e. for all $k\in \mathbb {Z}$. Because $B'\subseteq B_0$ and $N\ge 2n_0+1$, we have that the sets in the family $\{ f^{kN}(B'):k\in \mathbb {Z}\}$ are pairwise disjoint. Moreover,

\begin{align*} \left(\dfrac{3}{4}\right)^{p}\sum_{k\in\mathbb{Z}} \mu\big(f^{kN}(B')\big) &\le \sum_{k\in\mathbb{Z}}\int_{f^{kN}(B')} \vert \varphi\circ f^{{-}kN}\vert^{p} \textrm{d}\mu= \sum_{k\in\mathbb{Z}}\int_{f^{kN}(B')} \vert \varphi\vert^{p} \textrm{d}\mu \\ &=\int_{\bigcup_{k\in\mathbb{Z}} f^{kN}(B')} \vert \varphi\vert^{p} \textrm{d}\mu\le \int |\varphi|^{p} \textrm{d}\mu, \end{align*}

showing that $\sum \nolimits _{k\in \mathbb {Z}} \mu (f^{kN}(B'))<\infty$. By (*), we have that

\[ \sum_{n\in\mathbb{Z}}\mu(f^{n}(B'))=\sum_{i=0}^{N-1}\sum_{k\in\mathbb{Z}}\mu(f^{kN-i}(B'))\le N \max\, \{1,c,\ldots,c^{(N-1)}\}\cdot \sum_{k\in\mathbb{Z}}\mu\big(f^{kN}(B')\big)<\infty. \]

This proves that Condition (SC) is true, which concludes the proof of $(b)\implies (a)$.

5.3. Proof of Theorem 3.3

Proof of $(a)\implies (b)$. This is Lemma 5.2.

Proof of $(b)\implies (a)$. Assume $f$ is dissipative and $\mu (X)<\infty$. Then, there exists a wandering set $W$ such that $X=\bigcup_{n\in \mathbb {Z}}f^{n}(W)$. Let $\epsilon >0$ and $B\in \mathcal {B}$ with $\mu (B) < \infty$ be given. Let us verify Condition (SC). Let $N\in \mathbb {N}$ be such that $\mu (B{\setminus } \bigcup_{\vert n\vert \le N} f^{n}(W))<\epsilon$. Then $B'=B\cap \bigcup_{\vert n\vert \le N} f^{n}(W)$ verifies Condition (SC) because $\mu (B{\setminus }B')<\epsilon$ and by the dissipativity of $f$, we have that

\[ \sum_{n\in\mathbb{Z}}\mu\big(f^{n}(B')\big)\le(2N+1)\mu(X)<\infty, \]

which concludes the proof.

5.4. Proof of corollary 3.4

The proof consists in verifying Condition (SC). Hence, to begin with, let $\epsilon >0$ and $B\in \mathcal {B}$ with $0<\mu (B)<\infty$ be given. We may assume that $B$ has no recurrent points. For each $n\ge 1$, set

\[ A_n=\left\{x \in B: \{f(x), f^{2}(x),\ldots\}\cap B\left(x,\frac 1n\right)=\emptyset\right\}, \]

where $B(x,\,1/n)=\{y\in X: d(y,\,x)<1/n\}$, where $d$ is the metric on $X$. Notice that $A_1\subseteq A_2\subseteq \ldots$ and $B=\bigcup_{n=1}^{\infty } A_n$. In this way, there exist $N$ large enough so that $\mu (B{\setminus } A_N)<{\epsilon }/{4}$. Let $K$ be a compact subset of $A_N$ such that $\mu (A_N{\setminus }K)< {\epsilon }/{4}$. Let $U_1,\, U_2,\,\ldots ,\, U_r$ be a finite collection of balls of radius ${1}/{2N}$ such that $K\subseteq \bigcup_{i=1}^{r} U_i$. Since $U_i\cap K\subseteq A_N$, we have that if $x\in U_i\cap K$, then $\{f(x),\,f^{2}(x),\,\ldots \}\cap U_i=\emptyset$. In particular, if $K_i=U_i\cap K$, then the following statements are true:

1. (a) $K=\bigcup_{i=1}^{r} K_i$;

2. (b) $f^{n}(K_i)\cap K_i=\emptyset$ for all $n\ge 1$.

The bijectivity of $f$ and (b) imply that $W$ is a wandering set. Hence,

\[ \sum_{n\in\mathbb{Z}} \mu\big( f^{n}(K_i)\big)\le \mu(X). \]

In this way,

\[ \mu(B{\setminus}K) <\epsilon\quad\textrm{and}\quad \sum_{n\in\mathbb{Z}} \mu\big(f^{n}(K)\big)\le r\mu(X)<\infty. \]

Thus, Condition (SC) is true, which concludes the proof.

5.5. Proof of Theorem 3.6

Throughout this section, $(X,\,\mathcal {B},\,\mu ,\,f)$ will denote a measurable system. Given a measurable set $W\subseteq X$ with finite positive $\mu$-measure, let $(d_n(W))_{n\in \mathbb {Z}}$ be the sequence defined by

(2)\begin{equation} d_n(W)=\left\Vert\dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{n}\big)}\bigg|_W \right\Vert_{\infty}^{{-}1}, \end{equation}

where we set $1/\infty =0$ so that $d_n(W)$ is well defined even in the case in which $\left \Vert {\mathrm {d}\mu }/{\mathrm {d} (\mu \circ f^{n})}\bigg |_W\right \Vert _{\infty }=\infty$.

Proof. By ($\star$), $\mu (f^{n}(B))\le c\mu ( f^{n+1}(B))$ for all $B\in \mathcal {B}$ and $n\in \mathbb {Z}$. Hence, the Radon–Nikodym derivatives $\textrm {d}(\mu \circ f^{n})/\textrm {d}(\mu \circ f^{n+1})$, $n\in \mathbb {Z}$, are bounded by $c$ at $\mu \circ f^{n+1}$-almost every point, and therefore, $\mu$-a.e. In this way, for all $n\in \mathbb {Z}$,

(3)\begin{equation} \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{n+1}\big)}=\dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{n}\big)} \cdot\dfrac{\textrm{d}\big(\mu\circ f^{n}\big)}{\textrm{d}\big(\mu\circ f^{n+1}\big)}\le c \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{n}\big)} \quad \mu\textrm{-a.e.} \end{equation}

Therefore, for all $n\in \mathbb {Z}$, we have that

(4)\begin{equation} \dfrac{1}{d_{n+1}(W)}=\left\Vert \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{n+1}\big)}\bigg|_W\right\Vert_{\infty}\le c \left\Vert \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{n}\big)}\bigg|_W\right\Vert_{\infty}=c \dfrac{1}{d_n(W)}, \end{equation}

showing that $d_{n+1}(W)\ge c^{-1} d_n(W)$ for all $n\in \mathbb {Z}$.

Lemma 5.4 Suppose that $T_f$ is frequently hypercyclic. If $W$ is a wandering set with finite positive $\mu$-measure, then there exists a set $A\subset \mathbb {N}$ with positive lower density such that for each $n\in A,$

\[ \sum_{m\in A} d_{m-n}(W)<2. \]

Proof. Let $\varphi \in L^{p}(X)$ be a frequently hypercyclic vector for $T_f$ and let $W$ be a wandering set with finite positive $\mu$-measure. Let

(5)\begin{equation} A=\Big\{ n\in\mathbb{N}: \left\Vert T_f^{n}\varphi-\chi_W\right\Vert_p^{p}<\frac{1}{2^{p}} \mu(W)\Big\}, \end{equation}

then $A$ has positive lower density.

By combining the definition of $A$ with the fact that $W$ is a wandering set, we obtain for each $n\in A$,

(6)\begin{equation} \int_W \left\vert \varphi\circ f^{n}-1\right\vert^{p} \textrm{d}\mu + \sum_{m\neq 0}\int_{f^{m}(W)} \left\vert \varphi\circ f^{n}\right\vert^{p} \textrm{d}\mu \le \left\Vert T_f^{n} \varphi-\chi_W\right\Vert_p^{p} <\dfrac{1}{2^{p}}\mu(W). \end{equation}

In particular, the second term of (5.6) satisfies the following inequality for each $n\in A$,

(7)\begin{equation} \sum_{m\neq 0}\int_{f^{m}(W)} \left\vert \varphi\circ f^{n}\right\vert^{p} \textrm{d}\mu <\dfrac{1}{2^{p}}\mu(W). \end{equation}

Applying the triangle inequality to the first term of (5.6) yields for each $n\in A$,

(8)\begin{equation} \left[\mu(W)\right]^{1/p}- \bigg(\int_W \left\vert\varphi\circ f^{n}\right\vert^{p}\textrm{d}\mu\bigg)^{1/p}\le \bigg(\int _W \left\vert\varphi\circ f^{n}-1\right\vert^{p}\textrm{d}\mu\bigg)^{1/p}<\frac 12\left[\mu(W)\right]^{{1}/{p}}. \end{equation}

Therefore, renaming $n$ by $k$ in (5.8) yields

(9)\begin{equation} \int_W \left\vert\varphi\circ f^{k}\right\vert^{p}\textrm{d}\mu> \frac{1}{2^{p}}\mu(W),\quad \forall k\in A. \end{equation}

In what follows, keep $n\in A$ fixed and let $m\in A-n$ be arbitrary. Setting $k=n+m$ in (5.9) yields $\int _W \left \vert \varphi \circ f^{n+m}\right \vert ^{p}\textrm {d}\mu > {1}/{2^{p}}\mu (W)$. By combining this with (5.7), we arrive at

(10)\begin{equation} \sum_{\substack{ m\in A-n\\ m\neq 0}} \dfrac{\displaystyle\int_{f^{m}(W)} \left\vert \varphi\circ f^{n}\right\vert^{p} \textrm{d}\mu}{\displaystyle\int_{W}\left\vert\varphi\circ f^{n+m}\right\vert^{p}\textrm{d}\mu}<1. \end{equation}

By the Change of Variables Formula, for all $m\in A-n$,

(11)\begin{equation} \int_{f^{m}(W)}\left\vert\varphi\circ f^{n}\right\vert^{p}\textrm{d}\mu=\int_{f^{m}(W)}\left\vert\varphi\circ f^{n+m}\circ f^{{-}m}\right\vert^{p}\textrm{d}\mu=\int_W \left\vert\varphi\circ f^{n+m}\right\vert^{p} d\big(\mu\circ f^{m}\big). \end{equation}

Replacing (5.11) in (5.10) yields

(12)\begin{equation} \sum_{\substack{ m\in A-n\\ m\neq 0}} \dfrac{{\displaystyle\int_W} \left\vert\varphi\circ f^{n+m}\right\vert^{p} d\big(\mu\circ f^{m}\big)}{\displaystyle\int_{W}\left\vert\varphi\circ f^{n+m}\right\vert^{p} \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{m}\big)}\textrm{d}\big(\mu\circ f^{m}\big)}<1. \end{equation}

For all $m\in A-n$,

(13)\begin{equation} \int_{W}\left\vert\varphi\circ f^{n+m}\right\vert^{p} \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{m}\big)}\textrm{d}\big(\mu\circ f^{m}\big)\le\left\Vert \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{m}\big)}\bigg|_W \right\Vert_\infty\cdot \int_W \left\vert\varphi\circ f^{n+m}\right\vert^{p}\textrm{d}\big(\mu\circ f^{m}\big). \end{equation}

Replacing (5.13) in (5.12) and cancelling out the nonzero term $\int _W \left \vert \varphi \circ f^{n+m}\right \vert ^{p}\textrm {d}(\mu \circ f^{m})$ yields

\[ \sum_{\substack{ m\in A-n\\ m\neq 0}} \left\Vert \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{m}\big)}\bigg|_W\right\Vert_\infty^{{-}1}<1. \]

Therefore, after renaming variables, we obtain

\[ \sum_{m\in A} d_{m-n}(W)=\sum_{m\in A} \left\Vert \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{m-n}\big)}\bigg|_W\right\Vert_\infty^{{-}1}=1+\sum_{\substack{ m\in A-n\\ m\neq 0}} \left\Vert \dfrac{\textrm{d}\mu}{\textrm{d}\big(\mu\circ f^{m}\big)}\bigg|_W\right\Vert_\infty^{{-}1}<2. \]

5.6. Proof of Theorem 3.6

Set $\alpha _n= d_n(W)$ for all $n\in \mathbb {Z}$. By Lemma 5.3, we have that $\alpha _{n+1}\ge c^{-1} \alpha _n$ for all $n\in \mathbb {Z}$. Let $A$ be as in Lemma 5.4 and let $(\beta _n)_{n\in {A}}$ be defined by $\beta _n=\sum \nolimits _{m\in A} \alpha _{m-n}$. Then, by Lemma 5.4, for each $n\in A$,

\[ \beta_n=\sum_{m\in A} \alpha_{m-n}= \sum_{m\in A} d_{m-n}(W)<2. \]

Hence, $(\beta _n)_{n\in {A}}$ is bounded. By Lemma 5.5,

\[ \sum_{n\in\mathbb{Z}} d_n(W)=\sum_{n\in\mathbb{Z}} \alpha_n<\infty. \]

5.7. Proof of Theorem 3.7

Proof of $(a)\implies (b)$. This follows immediately from Theorem 3.2.

Proof of $(b)\implies (a)$. Let $B\in \mathcal {B}$ with $\mu (B)<\infty$ and $\epsilon >0$ be given. We will verify Condition (SC). Let $W$ be a wandering set of finite positive $\mu$-measure satisfying the bounded distortion Conditions (i) and (ii) in Definition 2.11. For each $i\in \mathbb {Z}$, set $W_i=f^{i}(W)$. By (i), there exists $N\in \mathbb {N}$ such that if $B'=B\cap \bigcup_{\vert i\vert \le N}f^{i}(W)$, then $\mu (B{\setminus B'})<\epsilon$. We claim that there exists $K>0$ such that for each integer $-N\le i\le N$, each $D\in \mathcal {B}(W_i)$ with positive $\mu$-measure and each $n\in \mathbb {Z}$,

\[ \frac{\mu(D)}{\mu\big(f^{n}(D)\big)}\le K^{2}\frac{\mu(W_i)}{\mu\big(f^{n}(W_i)\big)}. \]

In fact, since $f$ is bijective, bimeasurable and non-singular, any set $D\in \mathcal {B}(W_i)$ has positive $\mu$-measure if and only if $D=f^{i}(C)$, where $C\in \mathcal {B}(W)$ has positive $\mu$-measure. By Condition (ii) in Definition 2.11, we have that

\begin{align*} \frac{\mu\big(W_i\big)}{\mu\big(f^{n}(W_i)\big)}&=\frac{\mu\big(f^{i}(W)\big)}{\mu\big(f^{n+i}(W)\big)}\\ &=\frac{\mu\big(f^{i}(W)\big)}{\mu(W)}\cdot\frac{\mu(W)} {\mu\big(f^{n+i}(W)\big)}\\ &\ge \frac{1}{K^{2}}\frac{\mu\big(f^{i}(C)\big)} {\mu(C)}\frac{\mu(C)}{\mu\big(f^{n+i}(C)\big)}\\ &=\frac{1}{K^{2}} \frac{\mu(D)}{\mu\big(f^{n}(D)\big)}. \end{align*}

which proves the claim.

Hence, since $D\in \mathcal {B}(W_i)$ is an arbitrary measurable set of positive $\mu$-measure, we obtain

\[ \frac{\mu(W_i)}{\mu\big(f^{n}(W_i)\big)}\ge \frac{1}{K^{2}}\frac{\mathrm{d}\mu}{\mathrm{d}\big(\mu\circ f^{n}\big)}(x),\quad \text{for }\mu-\text{almost every }x\in W_i. \]

Therefore, since $W_i$ is a wandering set and $T_f$ is frequently hypercyclic, by Theorem 3.6 we arrive at

\[ \sum_{n\in\mathbb{Z}}\frac{\mu\big(f^{n}(W_i)\big)}{\mu(W_i)}\le K^{2} \sum_{n\in\mathbb{Z}}\left\Vert\frac{\mathrm{d}\mu}{\mathrm{d}\big(\mu\circ f^{n}\big)}\bigg|_{W_i}\right\Vert_{\infty}^{{-}1}<\infty. \]

In this way,

\[ \sum_{n\in\mathbb{Z}} \mu\big(f^{n}(B')\big)\le \sum_{n\in\mathbb{Z}}\sum_{\vert i\vert\le N} \mu\big(f^{n}(W_i)\big)=\sum_{\vert i\vert\le N}\mu(W_i)\sum_{n\in\mathbb{Z}} \frac{\mu\big(f^{n}(W_i)\big)}{\mu(W_i)}<\infty. \]

Proof of $(a)\iff (c)$. This follows immediately from Corollary 3.5.