3.2. Conze’s theorem

Before stating Conze’s theorem, let us introduce some terminology.

In what follows, by a measure-preserving dynamical system $(X, \mathcal B, \mu , T)$ , we mean a pair consisting of a probability space $(X, \mathcal B, \mu )$ and a measurable map $T:X\to X$ such that $\mu \circ T^{-1}=\mu $ . Note that here, we are departing from our standing notation: X is an abstract space, not necessarily a topological vector space, and hence, T is not necessarily a linear operator. This ambiguity is in fact intentional and it should cause no confusion.

Given a measure-preserving dynamical system $(X, \mathcal B, \mu , T)$ , we denote by ${U_T:L^2(X, \mathcal B, \mu )\to L^2(X, \mathcal B, \mu )}$ the associated Koopman operator, which is defined by

$$ \begin{align*} U_Tf:= f\circ T.\end{align*} $$

This is an isometry of $L^2(X, \mathcal B, \mu )$ , and a unitary operator if T is bijective and bimeasurable, in which case we say that T is an automorphism of $(X, \mathcal B, \mu )$ or that the measure-preserving dynamical system $(X, \mathcal B, \mu , T)$ is invertible.

We stress a technical point: all probability spaces $(X, \mathcal B, \mu )$ under consideration will be assumed to be standard Borel, that is, the underlying measurable space $(X,\mathcal B)$ is isomorphic to $(Z, \mathcal B(Z))$ for some Polish space Z, where $\mathcal B(Z)$ is the Borel $\sigma $ -algebra of Z. Additionally, when X is already a Polish space, we assume that $\mathcal B$ is the Borel $\sigma $ -algebra of X.

A unitary operator $U:H\to H$ acting on a Hilbert space H is said to have Lebesgue spectrum if there exists a family of vectors $(f_i)_{i\in I}$ in H such that $\{U^n f_i:\ n\in \mathbb {Z},\ i\in I\}$ is an orthonormal basis of H. Observe that if H is separable, the family $(f_i)$ has to be finite or countably infinite. When there exists a countably infinite such family $(f_i)$ , we say that U has countable Lebesgue spectrum. Finally, we say that an invertible measure-preserving dynamical system $(X,\mathcal B,\mu , T)$ has (countable) Lebesgue spectrum if the restriction of the Koopman operator $U_T$ to $ L^2_0(X,\mathcal B,\mu ):=\{ f\in L^2(X,\mathcal B,\mu )\,:\, \int _{X} f\, d\mu =0\}$ has (countable) Lebesgue spectrum. We may also say that T itself has (countable) Lebesgue spectrum.

Conze’s theorem from [Reference Conze20] now reads as follows.

Theorem 3.3. Let $(X,\mathcal B,\mu , T)$ be an invertible measure-preserving dynamical system and assume that T has Lebesgue spectrum. If $(n_k)_{k\geq 0}$ is an increasing sequence of integers with positive lower density then, for any $f\in L^1(\mu ),$

$$ \begin{align*}\frac 1N\sum_{k=0}^{N-1} f(T^{n_k}x)\xrightarrow{N\to\infty}\int_X f\, d\mu\quad\text{ for } \mu\text{-almost every } x\in X.\end{align*} $$

Getting back to the case where X is a separable Banach space, it is easy to check that if $T\in \mathfrak L(X)$ and if one can find a T-invariant measure with full support $\mu $ such that $(X,\mathcal B,\mu , T)$ satisfies the assumptions of Conze’s theorem, then T is hereditarily frequently hypercyclic. However, Conze’s theorem is only stated for automorphisms and, in our context, it would be rather restrictive to confine ourselves to the case of automorphisms. We will get round this problem thanks to the notion of factor. Recall that a measure-preserving dynamical system $(X,\mathcal B,\mu ,T)$ is a factor of a measure-preserving dynamical system $(Y ,\mathcal C,\nu ,S)$ or that $(Y ,\mathcal C,\nu ,S)$ is an extension of $(X,\mathcal B,\mu ,T)$ , if (possibly after deleting two sets of measure $0$ in X and Y) there exists a measurable map $\pi :Y\to X$ such that $\pi \circ S=T\circ \pi $ and $\mu =\nu \circ \pi ^{-1}$ . Using suitable extensions, we will be able to apply Conze’s theorem to general operators T via the following corollary.

Proof. It is enough to prove the result for a single pair $(A,V)$ , where $A\subset \mathbb {N}$ has positive lower density and V is a non-empty open set.

Let $(n_k)_{k\geq 1}$ be the increasing enumeration of A. Let also $W:= \pi ^{-1}(V)$ , where ${\pi :(Y ,\mathcal C,\nu ,S)\to (X,\mathcal B,\mu ,T)}$ is a factor map, so that $\nu (W)=\mu (V)>0$ , and let

$$ \begin{align*}\Omega:=\bigg\{y\in Y:\ \frac 1N\sum_{k=1}^N \mathbf 1_{{W}}(S^{n_{k}}y)\xrightarrow{N\to\infty}\mu(V)\bigg\}.\end{align*} $$

By Conze’s theorem, we know that $\nu (\Omega )=1$ . Since we are working with standard Borel spaces, it follows that the set $\pi (\Omega )$ is $\mu $ -measurable (being an analytic set, it is universally measurable) and that $\mu ( \pi (\Omega ))=1$ . So it is enough to show that every $x\in \pi (\Omega )$ is such that $\underline {\textrm {dens}}( \mathcal N_T(x,V)\cap A)>0$ .

Let $x\in \pi (\Omega )$ and write x as $x=\pi (y)$ for some $y\in \Omega $ . By definition of $\Omega $ , we know that the set $\{k\geq 1:\ S^{n_{k}}y\in {W}\}$ has positive lower density; enumerate it as an increasing sequence $(m_{k})_{k\geq 1}.$ Then, $B:=\{n_{m_{k}}:\ k\geq 1\}$ has positive lower density, it is contained in A and $S^n y\in \pi ^{-1}(V)$ for all $n\in B$ . This means that $T^n x\in V$ for all $n\in B,$ which concludes the proof.

3.3. Natural extensions

To apply Corollary 3.4, we need a simple way of extending a given measure-preserving dynamical system $(X,\mathcal B,\mu ,T)$ to a measure-preserving dynamical system $(\widetilde X,\widetilde {\mathcal B}, \widetilde \mu , \widetilde T)$ , where $\widetilde T$ is an automorphism of $(\widetilde X,\widetilde {\mathcal B},\widetilde \mu )$ . Fortunately, there is a canonical procedure doing precisely that. Consider the set

$$ \begin{align*}\widetilde X:=\{(x_k)_{k\geq 0}\in X^{\mathbb{Z}_+}:\ T(x_{k+1})=x_k\textrm{ for all }k\geq 0\},\end{align*} $$

and for $k\geq 0,$ let $\pi _k:\widetilde X\to X$ denote the projection onto the kth coordinate of $\widetilde X.$ Endow $\widetilde X$ with the smallest $\sigma $ -algebra $\widetilde {\mathcal B}$ which makes every projection $\pi _k$ measurable. The Rokhlin’s natural extension of T is the measurable transformation $\widetilde T:\widetilde X\to \widetilde X$ defined by

$$ \begin{align*}\widetilde T(x_0, x_1, \ldots):=(T(x_0),x_0,x_1,\ldots).\end{align*} $$

One can prove that there exists a unique probability measure $\widetilde \mu $ on $(\widetilde X,\widetilde {\mathcal B})$ such that ${\widetilde \mu \circ \pi _k^{-1}=\mu }$ for all $k\geq 0$ . (Here, the fact $(X,\mathcal B)$ is a standard Borel space is needed.) Then, it is a simple exercise to check that $\widetilde T:\widetilde X\to \widetilde X$ is an automorphism of $(\widetilde X,\widetilde B, \widetilde \mu )$ such that $\pi _k\circ \widetilde T=T\circ \pi _k$ for all $k\geq 0$ . In particular, $(X,\mathcal B,\mu ,T)$ is a factor of $(\widetilde { X},\widetilde { \mathcal B},\widetilde \mu ,\widetilde T)$ as witnessed by $\pi _0: \widetilde X\to X$ . Note also that if X is a Polish space, then $\widetilde X$ is a Borel subset of the Polish space $X^{\mathbb {Z}_+}$ endowed with the product topology, and $\widetilde {\mathcal B}$ is the Borel sigma-algebra of $\widetilde X$ . For more details, see for example [Reference Urbański, Roy and Munday56, §8.4] or [Reference Nicol and Petersen46].

We will need the following lemma. Recall that if X is a complex Fréchet space, a Borel probability measure $\mu $ on X is said to be Gaussian if every continuous linear functional on X has a complex symmetric Gaussian distribution when considered as a random variable on $(X,\mathcal B,\mu )$ .

Proof. It is clear that $\widetilde X$ is a closed linear subspace of $X^{\mathbb {Z}_+}$ (and hence a Fréchet space) and that $\widetilde T$ is an invertible continuous linear operator on $\widetilde X$ .

Let $\phi $ be a continuous linear functional on $\widetilde X$ . By the Hahn–Banach theorem, one can extend $\phi $ to a continuous linear functional $\Phi $ on $X^{\mathbb {Z}_+}$ . Since $X^{\mathbb {Z}_+}$ is endowed with the product topology, this linear functional $\Phi $ has the form

$$ \begin{align*} \Phi=\sum_{k=0}^N x^*_k\circ \pi_k,\end{align*} $$

where $x_0^*,\ldots , x_N^*\in X^*$ . Moreover, by definition of $\widetilde X$ , we have $\pi _k=T^{N-k}\circ \pi _N$ on $\widetilde X$ for $k=0,\ldots ,N$ ; so we may write

$$ \begin{align*} \phi =x^*\circ \pi_N\quad \text{where } x^*=\sum_{k=0}^N x_k^*\circ T^{N-k}\in X^*.\end{align*} $$

Hence, $\widetilde \mu \circ \phi ^{-1}=( \widetilde \mu \circ \pi _N^{-1})\circ (x^*)^{-1}=\mu \circ (x^*)^{-1}$ . Since $\mu $ is a Gaussian measure, it follows that $\widetilde \mu $ is Gaussian as well.

3.4. Gaussian measures and countable Lebesgue spectrum

In this section, we state a general result about Gaussian linear measure-preserving dynamical systems. This result looks very much like [Reference Cornfeld, Fomin and Sinai21, Theorem 14.3.2], but it is not clear to us that it can be formally deduced from it; so we will give a more or less complete proof. However, this proof is not necessary for understanding the proof of Theorem 3.1, so it is postponed to §3.6 for the sake of better readability. This means that a reader who is ready to take Proposition 3.6 on faith can safely ignore that section.

Recall that if $(X,\mathcal B, \mu , T)$ is a measure-preserving dynamical system then, for any $f\in L^2(\mu )$ , there is a unique positive Borel measure $\sigma _{f}$ on $\mathbb {T}$ with non-negative Fourier coefficients

$$ \begin{align*} \widehat{\,\sigma_{f}}(n)=\langle U_T^n f,f\rangle_{L^2(\mu)},\quad n\geq 0.\end{align*} $$

This follows from Herglotz’s theorem and from the fact that the Koopman operator ${U_T:L^2(\mu )\to L^2(\mu )}$ is a contraction operator (so that the map $\mathbb {Z}_+\ni n\mapsto \langle U_T^n f,f\rangle _{L^2(\mu )}$ can be extended to a positive-definite function on $\mathbb {Z}$ , see for example [Reference Szőkefalvi-Nagy, Foias, Bercovici and Kérchy55, p. 28]). The measure $\sigma _f$ is the spectral measure of f with respect to $U_T$ .

Proposition 3.6 has the following consequence for not necessarily invertible Gaussian linear measure-preserving dynamical systems.

Proof. Let $(\widetilde X,\widetilde {\mathcal B},\widetilde {\mu }, \widetilde {T})$ be the natural extension. By Lemma 3.5, we know that $\widetilde T$ is an invertible linear operator and $\widetilde \mu $ is a Gaussian measure. Hence, by Proposition 3.6, it is enough to show that for any continuous linear functional $\phi $ on $\widetilde X$ , the spectral measure $\sigma _{\phi }$ (with respect to $U_{\widetilde T}$ ) is absolutely continuous with respect to the Lebesgue measure.

By the proof of Lemma 3.5, one can find $N\in \mathbb {Z}_+$ and $x^*\in X^*$ such that $\phi =x^*\circ \pi _N$ on $\widetilde X$ , where $\pi _N:X^{\mathbb {Z}_+}\to X$ is the projection onto the Nth coordinate. Now, we apply the following purely formal fact, which holds true for any measure-preserving dynamical system $(X,\mathcal B,\mu , T)$ .

Fact 3.8. Let $f\in L^2(X,\mathcal B,\mu )$ and $F:=f\circ \pi _N\in L^2(\widetilde X, \widetilde B, \widetilde \mu )$ . Then, the spectral measure $\sigma _{F}$ with respect to $U_{\widetilde T}$ is equal to $\sigma _f$ , the spectral measure of f with respect to $U_T$ .

Proof of Fact 3.8

For all $n\geq 0$ ,

$$ \begin{align*} \langle U_{\widetilde T}^n F , F\rangle_{L^2(\widetilde\mu)}=\langle F \circ\widetilde T^n,F\rangle_{L^2(\widetilde \mu)}&=\langle f\circ\pi_N\circ\widetilde T^n,f\circ\pi_N\rangle_{L^2(\widetilde \mu)}\\ &=\langle f\circ T^n \circ\pi_N,f\circ\pi_N\rangle_{L^2(\widetilde \mu)}\\ &=\langle f\circ T^n ,f\rangle_{L^2(\mu)}\\&= \langle U_T^n f, f\rangle_{L^2(\mu)}.\\[-37pt] \end{align*} $$

By Fact 3.8, $\sigma _\phi =\sigma _{x^*}$ is absolutely continuous with respect to the Lebesgue measure, which concludes the proof of Corollary 3.7.

3.5. Proof of Theorem 3.1

We can now state and prove very easily the following more precise version of Theorem 3.1.

3.6. Proof of Proposition 3.6

In this section, we give the promised proof of Proposition 3.6.

3.6.1. Two facts concerning unitary operators

We will need several results on unitary operators. These results are certainly well known, but since they are rather difficult to locate in the literature, we provide some details.

Let U be a unitary operator acting on a complex separable Hilbert space $H.$ By Herglotz’s theorem, for any $f\in H$ , there exists a unique positive and finite Borel measure $\sigma _f$ on $\mathbb {T}$ , called the spectral measure of f with respect to U, such that

$$ \begin{align*} \text{ for all } n\in\mathbb{Z}\;:\; \langle U^n f,f\rangle=\widehat{\sigma_f}(n).\end{align*} $$

Note in particular that $\sigma _f$ is equal to the Lebesgue measure $\unicode{x3bb} $ if and only if the sequence $(U^nf)_{n\in \mathbb {Z}}$ is orthonormal. This explains the terminology ‘Lebesgue spectrum’.

Denote by $C(f)$ the cyclic subspace generated by $f,$ that is, $C(f)=\overline {\textrm {span}}(U^n f: n\in \mathbb Z)$ . With this notation, one form of the spectral theorem reads as follows (see for example [Reference Conway19, Ch. IX], [Reference Parry47, Appendix, §2] or [Reference Queffélec49]): there exists a finite or infinite sequence of vectors $f_i\in H, 0\leq i<m$ , where $m\in \mathbb N\cup \{\infty \}$ , such that

$$ \begin{align*}H=\bigoplus_{0\leq i< m} C(f_i)\quad\textrm{ and }\quad\sigma_{f_0}\gg\sigma_{f_1}\gg\cdots\gg\sigma_{f_i}\gg\cdots.\end{align*} $$

Moreover, these measures are essentially unique in the following sense: for any other sequence $(g_i)_{0\leq i<m'}$ satisfying $H=\bigoplus _{0\leq i<m'}C(g_i)$ and ${\sigma _{g_0}\gg \sigma _{g_1}\gg \cdots \gg \sigma _{g_i}\gg \cdots }$ , we have $m'=m$ and $\sigma _{f_i}\sim \sigma _{g_i}$ for all $i.$ The maximal spectral type of T is then defined as (the equivalence class of) the measure $\sigma _{f_0}$ .

Observe that for any $f\in H,$ the restriction of U to $C(f)$ is unitary equivalent to the multiplication operator $M_{\sigma _f}:L^2(\mathbb {T},\sigma _f)\to L^2(\mathbb {T},\sigma _f)$ defined by $M_{\sigma _f}u (z):= zu(z)$ , $z\in \mathbb T$ .

Suppose now that U has countable Lebesgue spectrum and let $(f_i)_{i\in \mathbb {N}}$ be a sequence in H such that $\{U^n f_i:\ n\in \mathbb Z,\ i\in \mathbb {N}\}$ is an orthonormal basis of $H.$ Then, $H=\bigoplus _{i\in \mathbb {N}}C(f_i)$ and $\sigma _{f_i}=\unicode{x3bb} $ for all $i\in I$ . Hence, U is unitarily equivalent to $M_{\unicode{x3bb} }^{(\infty )}:=\bigoplus _{i=1}^\infty M_\unicode{x3bb} $ acting on $\bigoplus _{i=1}^\infty L^2(\mathbb {T},\unicode{x3bb} )$ . Conversely, it is clear that if $U\cong M_{\unicode{x3bb} }^{(\infty )}$ , then U has countable Lebesgue spectrum.

By the uniqueness part of the spectral theorem, it follows in particular that the maximal spectral type of a unitary operator with countable Lebesgue spectrum is the Lebesgue measure. The next lemma is a kind of converse.

Lemma 3.10. Let U be a unitary operator on a complex separable Hilbert space H satisfying the following two conditions.

1. (a) There exists a closed subspace $K\subset H$ such that $U(K)=K$ and $U_{| K}:K\to K$ has countable Lebesgue spectrum.

2. (b) The maximal spectral type of U is the Lebesgue measure.

Then, $U:H\to H$ has countable Lebesgue spectrum.

Proof. We will use the following notation: if $\sigma $ is a positive finite Borel measure on $\mathbb {T}$ and $N\in \mathbb {N}\cup \{\infty \}$ , we denote by $M_{\sigma }^{(N)}$ the operator $\bigoplus _{0\leq j<N} M_\sigma $ acting on $\bigoplus _{0\leq j<N} L^2(\mathbb {T},\sigma )$ . Recall also that $\unicode{x3bb} $ is the Lebesgue measure on $\mathbb {T}$ .

By condition (a), we know that

$$ \begin{align*} U_{| K}\cong M_{\unicode{x3bb}}^{(\infty)}.\end{align*} $$

Consider now the operator $U_{| K^\perp }:K^\perp \to K^\perp $ . By condition (b), its maximal spectral type is absolutely continuous with respect to the Lebesgue measure $\unicode{x3bb} $ . By the ‘second formulation’ of the spectral theorem (see for example [Reference Queffélec49] or [Reference Conway19, Ch. IX]), there exist pairwise disjoint Borel sets $\Delta _\infty ,\Delta _1,\Delta _2,\ldots $ and positive finite measures $\mu _\infty ,\mu _1,\mu _2,\ldots $ on $\mathbb {T}$ supported on $\Delta _\infty ,\Delta _1,\Delta _2,\ldots $ and absolutely continuous with respect to $\unicode{x3bb} $ , such that $U_{|K^\perp }$ is unitarity equivalent to $M_{\mu _\infty }^{(\infty )}\oplus M_{\mu _1}^{(1)}\oplus M_{\mu _2}^{(2)}\oplus \cdots .$ Note that some of the measures $\mu _n$ may be $0$ . For $n\in \{ \infty \}\cup \mathbb {N}$ , we may write $\mu _n=f_n\, \unicode{x3bb} $ for some non-negative function $f_n\in L^1(\mathbb {T})$ and we may assume that $\Delta _n=\{ f_n>0\}$ . Hence, if we set ${\nu _n:=\mathbf {1}_{\Delta _n}\, \unicode{x3bb} ,}$ we get $M_{\mu _n}\cong M_{\nu _n}$ since the measures $\mu _n$ and $\nu _n$ are equivalent. Therefore,

$$ \begin{align*} U_{|K^\perp}\cong M_{\nu_\infty}^{(\infty)}\oplus M_{\nu_1}^{(1)}\oplus M_{\nu_2}^{(2)}\oplus\cdots.\end{align*} $$

Now, let $E:=\mathbb {T}\backslash (\Delta _\infty \cup \bigcup _{n\geq 1}\Delta _n)$ and $\nu :=\mathbf {1}_{E}\, \unicode{x3bb} .$ Then, $\unicode{x3bb} =\nu +\nu _\infty +\sum \limits _{n\in \mathbb {N}}\nu _n$ so that

$$ \begin{align*} M_\unicode{x3bb}\cong M_\nu\oplus M_{\nu_\infty}\oplus M_{\nu_1}\oplus \cdots.\end{align*} $$

Consequently, we obtain

$$ \begin{align*} U&\cong U_{|K}\oplus U_{| K^\perp}\\ &\cong M_\nu^{(\infty)}\oplus M_{\nu_\infty}^{(\infty)}\oplus M_{\nu_1}^{(\infty)}\oplus M_{\nu_2}^{(\infty)}\oplus \cdots\\ &\qquad\qquad \oplus M_{\nu_\infty}^{(\infty)}\oplus M_{\nu_1}^{(1)}\oplus M_{\nu_2}^{(2)}\oplus\cdots\\ &\cong M_\nu^{(\infty)} \oplus M_{\nu_\infty}^{(\infty)}\oplus M_{\nu_1}^{(\infty)}\oplus M_{\nu_2}^{(\infty)} \cdots\\ &\cong M_\unicode{x3bb}^{(\infty)}, \end{align*} $$

which means that U has countable Lebesgue spectrum.

The following observation is useful to check assumption (a) in Lemma 3.10.

Lemma 3.11. Let $(f_i)_{i\in I}$ be a countably infinite family of vectors in H. Assume that the cyclic subspaces $C(f_i)$ are pairwise orthogonal and let $K:=\oplus _{i\in I} C(f_i)$ . If $\sigma _{f_i}$ is equivalent to the Lebesgue measure for all $i\in I$ , then $U_{| K}$ has countable Lebesgue spectrum.

Proof. For each $i\in I$ , there exists $g_i\in C(f_i)$ such that $\sigma _{g_i}$ is exactly equal to the Lebesgue measure and for any such $g_i$ , we have $C(g_i)=C(f_i)$ ; see for example [Reference Parry47, pp. 93–94]. Hence, $\{U^ng_i\,:\, i\in I, n\in \mathbb {Z}\}$ is an orthonormal basis of K.

We will also need the following fact.

Lemma 3.12. Let U be a unitary operator acting on a separable Hilbert space H and let $\sigma $ be a finite, positive Borel measure on $\mathbb {T}$ . If there exists a dense set $\mathcal D\subset H$ such that $\sigma _f\ll \sigma $ for all $f\in \mathcal D$ , then $\sigma _f\ll \sigma $ for all $f\in H$ .

Proof. Denote by $M(\mathbb {T})$ the space of all complex Borel measures on $\mathbb {T}$ . By [Reference Queffélec49, Corollary 2.2], the map $f\mapsto \sigma _f$ is continuous from H into $M(\mathbb {T})$ endowed with the norm topology. The result follows immediately.

3.6.2. Proof of Proposition 3.6

Before really starting the proof, we need to recall some basic facts concerning the $L^2$ -space of a Gaussian measure. In what follows, $\mu $ is a Gaussian measure on the (separable) complex Fréchet space X.

Let $\mathcal G:=\overline {\textrm {span}}\,(\langle x^*,\cdot \rangle :\ x^*\in X^*)$ , where the closure is taken in $L^2(X,\mathcal B,\mu )$ . This is a Gaussian subspace of $L^2(X,\mathcal B,\mu )$ , in the sense that any function in $\mathcal G$ has symmetric complex Gaussian distribution. Moreover, $\mathcal G$ is clearly $U_T$ -invariant.

For $k\geq 0,$ let us denote by $\mathcal G^k$ the space of homogeneous polynomials of degree k in the elements of $\mathcal G$ , with $\mathcal G^0=\mathbb C.$ The subspaces $\mathcal G^k$ , $k\geq 0$ , are linearly independent (which is not obvious) and one can orthonormalize them thanks to the so-called Wick transform $f\mapsto \; \pmb :{f}\pmb :$ which is defined on $ \bigcup _{k\geq 0} \mathcal G^k$ as follows:

$$ \begin{align*} \pmb:{f}\pmb:\;=\begin{cases} f&\text{if } f \text{ is constant},\\ f- P_k f\quad&\text{if } f\in\mathcal G^k, k\geq 1, \end{cases} \end{align*} $$

where we denote by $ P_k$ the orthogonal projection onto ${\overline {\mathrm {span}}}\,({\mathcal {G}}^i:0\leq i\leq k-1).$

With this notation, we have the orthogonal decomposition

$$ \begin{align*} L^2(X,{\mathcal{B}},\mu)=\bigoplus_{k=0}^{\infty}\overline{\pmb:{{\mathcal{G}}^k}\pmb:}.\end{align*} $$

Moreover, for any $f_1,\ldots ,f_k, g_1,\ldots ,g_k\in \mathcal G$ , the scalar product $ \langle \pmb :{f_1\cdots f_k}\pmb :\; ,\, \pmb :{g_1\cdots g_k}\pmb :\rangle $ is given by

(3.1) $$ \begin{align} \langle \pmb:{f_1\cdots f_k}\pmb:\; ,\, \pmb:{g_1\cdots g_k}\pmb:\rangle=\sum_{\mathfrak s\in\mathfrak S_k}\langle f_{\mathfrak s(1)},g_1\rangle\cdots \langle f_{\mathfrak s(k)},g_k\rangle,\end{align} $$

where $\mathfrak S_k$ is the permutation group of $\{ 1,\ldots ,k\}$ .

For details on these general facts, we refer to [Reference Peller48, Ch. 8] or [Reference Janson34, Chs. 2 and 3]. We will also need the following lemma (see [Reference Bayart and Grivaux5, Lemma 3.28]).

Lemma 3.13. Let $T\in \mathfrak L(X)$ and let $\mu $ be a T-invariant Gaussian measure on X. For any $f_1,\ldots ,f_k\in \mathcal G$ , we have

$$ \begin{align*} U_T(\pmb:{f_1\cdots f_k}\pmb:)=\; \pmb:{(U_Tf_1)\cdots (U_Tf_k)}\pmb:.\end{align*} $$

In particular, each subspace $\pmb :{\mathcal G^k}\pmb :$ is $U_T$ -invariant.

We can now finally give the proof of Proposition 3.6.

Proof of Proposition 3.6

Let $(X,\mathcal B,\mu , T)$ be a measure-preserving dynamical system, where $T\in \mathcal (X)$ is invertible and the measure $\mu $ is Gaussian, such that for any $x^*\in X^*\subset L^2(\mu )$ , the spectral measure $\sigma _{x^*}$ with respect to $U_T$ is absolutely continuous with respect to the Lebesgue measure. We want to show that $(X,\mathcal B,\mu , T)$ has countable Lebesgue spectrum.

Let us first recall that for any $f,g\in L^2(\mu )$ , there is a unique complex Borel measure $\sigma _{f,g}$ on $\mathbb {T}$ with Fourier coefficients

$$ \begin{align*} \widehat{\sigma_{f,g}}(n)=\langle U_T^n f,g\rangle,\quad n\in\mathbb Z.\end{align*} $$

If $f=g$ , then $\sigma _{f,f}$ is the spectral measure $\sigma _f$ ; and the existence of $\sigma _{f,g}$ for arbitrary functions f and g follows from a polarization argument. Indeed, we have

$$ \begin{align*} \sigma_{f,g}= \frac14\sum_{k=0}^3 \mathbf i^k\sigma_{f+\mathbf i^k g}.\end{align*} $$

This formula and the assumption of the proposition show that $\sigma _{x^*, y^*}$ is absolutely continuous with respect to the Lebesgue measure for any $x^*, y^*\in X^*$ . Note also that the map $(f,g)\mapsto \sigma _{f,g}$ is obviously $\mathbb {R}$ -bilinear.

In what follows, we denote by $\star $ the convolution product for measures on $\mathbb {T}$ and we write the elements of $X^*$ as $f,g, \ldots $ rather than $x^*, y^*, \ldots $ to avoid the proliferation of stars.

Fact 3.14. If $f_1, \ldots ,f_k\in X^*$ , then $ \sigma _{\pmb :{f_1\cdots f_k}\pmb :}= \sum _{\mathfrak s\in \mathfrak S_k} \sigma _{f_{\mathfrak s(1)}, f_1}\star \cdots \star \sigma _{f_{\mathfrak s(k)},f_k}$ .

Proof of Fact 3.14

By Lemma 3.13 and equation (3.1), we have for all $n\in \mathbb {Z}$ :

$$ \begin{align*} \widehat{\sigma}_{\pmb:{f_1\cdots f_k}\pmb:}(n)&=\langle \pmb:{(U_T^nf_1)\cdots (U_T^nf_k)}\pmb:\, ,\, \pmb:{f_1\cdots f_k}\pmb:\rangle\\ &=\sum_{\mathfrak s\in\mathfrak S_k} \langle U_T^nf_{\mathfrak s(1)},f_1\rangle\cdots \langle U_T^nf_{\mathfrak s(k)},f_k\rangle\\ &= \sum_{\mathfrak s\in\mathfrak S_k} \widehat\sigma_{f_{\mathfrak s(1)}, f_1}(n)\cdots \widehat\sigma_{f_{\mathfrak s(k)}, f_k}(n).\\[-3.9pc] \end{align*} $$

Now, let us denote by $\mathcal D$ the set of all functions $f\in L^2(\mu )$ of the form

$$ \begin{align*} f=\sum_{k=1}^N \pmb:{\sum_{j=1}^{m_k}f_{j,1}\cdots f_{j,k}}\pmb:\quad\text{with } m_k\geq 1 \text{ and } f_{j,i}\in X^*.\end{align*} $$

This is a dense linear subspace of $L^2_0(\mu )$ .

Fact 3.15. If $f\in \mathcal D$ , then $\sigma _f$ is absolutely continuous with respect to the Lebesgue measure.

Proof of Fact 3.15

Let $f\in \mathcal D$ , so that $ f=\sum _{k=1}^N f_k$ , where

$$ \begin{align*} f_k= \sum_{j=1}^{m_k}\pmb:{f_{j,1}\cdots f_{j,k}}\pmb:\quad\text{with } f_{j,i}\in X^*, \quad\text{so that } f_k\in {\pmb:{\mathcal G^k}\pmb:}.\end{align*} $$

By orthogonality and $U_T$ -invariance of the subspaces $\pmb :{\mathcal G^k}\pmb :$ , we have for all $n\in \mathbb {Z}$ ,

$$ \begin{align*}\widehat{\sigma_f}(n)=\langle U_T^n f,f\rangle=\sum_{k=1}^N \langle U_T^n f_k,f_k\rangle=\sum_{k=1}^N \widehat{\,\sigma_{f_k}}(n),\end{align*} $$

so that $\sigma _f= \sigma _{f_1}+\cdots +\sigma _{f_N}$ . Hence, it is enough to check that each measure $\sigma _{f_k}$ is absolutely continuous with respect to the Lebesgue measure. However, this is clear by Fact 3.14 and bilinearity of the map $(f,g)\mapsto \sigma _{f,g}$ : indeed, we have

$$ \begin{align*} \sigma_{f_k}= \sum_{j,j'=1}^N\sigma_{ \pmb:{f_{j,1}\cdots f_{j,k}}\pmb:\,,\, \pmb:{f_{j',1}\cdots f_{j',k}}\pmb:}=\sum_{j,j'=1}^N \sum_{\mathfrak s\in\mathfrak S_k} \sigma_{f_{j,\mathfrak s(1)},f_{j',1}}\star\cdots\star \sigma_{f_{j,\mathfrak s(k)},f_{j',k}} \end{align*} $$

and the result follows since all measures $\sigma _{f_{j,\mathfrak s(i)},f_{j',i}}$ are absolutely continuous with respect to the Lebesgue measure.

We can now finish the proof of Proposition 3.6. By Fact 3.15 and Lemma 3.12, the maximal spectral type of $U_{T}$ is absolutely continuous with respect to the Lebesgue measure. Now, take any $f\in X^*\setminus \{ 0\}$ . Then, $\sigma _f$ is a non-zero measure which is absolutely continuous with respect to the Lebesgue measure. It follows that there exists $k_0\geq 1$ such that, for all $k\geq k_0,$ the measure $\sigma _{:f^k:}=k!\, \sigma _f\star \cdots \star \sigma _f$ (k times) is equivalent to the Lebesgue measure (see for example [Reference Newton and Parry45, Lemma 3.6], where the symmetry assumption on the measure is in fact not necessary). Let us set

$$ \begin{align*}K:=\overline{\textrm{span}}\,\{U_{T}^n (\pmb:{f^k}\pmb:) : \ n\in\mathbb Z,\ k\geq k_0\}.\end{align*} $$

By orthogonality and $U_T$ -invariance of the spaces $\pmb :{\mathcal G^k}\pmb :$ , and applying Lemma 3.11, we see that $(U_T)_{| K}$ has countable Lebesgue spectrum. Hence, by Lemma 3.10, $U_{T}$ has countable Lebesgue spectrum.

3.7. The frequent hypercyclicity criterion again

The tools introduced in the previous sections allow us to give another proof of Theorem 2.1. Arguably, this proof is much less elementary. Let us say that an operator $T\in \mathfrak L(X)$ has countable Lebesgue spectrum after extension if there exists a T-invariant Borel probability measure $\mu $ on X with full support such that $(X,\mathcal B,\mu ,T)$ is a factor of a measure-preserving dynamical system which has countable Lebesgue spectrum. By Corollary 3.4, any such operator T is hereditarily frequently hypercyclic.

One can also prove the following ‘probabilistic’ version of Proposition 3.16. Let us say that a sequence $(x_n)_{n\in \mathbb {Z}}$ is a bilateral backward orbit for an operator T if $Tx_n=x_{n-1}$ for all $n\in \mathbb {Z}$ .

The (almost sure) convergence of the bilateral series $\sum g_n x_n$ means that both series

$$ \begin{align*}\sum_{n\geq 0} g_n x_n \quad \textrm{ and }\quad\sum_{n<0} g_n x_n\end{align*} $$

are (almost surely) convergent.

Proof. Let $\mu $ be the distribution of the random variable $\xi :=\sum _{n\in \mathbb {Z}} g_n x_n$ . This is a Gaussian measure, which has full support since $\overline {\mathrm {span}}( x_n\,:\, n\in \mathbb {Z})=X$ and which is T-invariant because $(x_n)$ is a bilateral backward orbit for T. By Corollary 3.7, it is enough to show that for any $x^*\in X^*\subset L^2(\mu )$ , the spectral measure $\sigma _{x^*}$ of $x^*$ with respect to $U_T$ is absolutely continuous with respect to the Lebesgue measure.

By orthogonality of the Gaussian variables $g_k$ , we have for all $n\geq 0$ ,

$$ \begin{align*} \widehat{\sigma_{x^*}}(n)=\langle U_T^n x^*, x^*\rangle&=\sum_{k\in\mathbb{Z}} \langle x^*, T^n x_k\rangle\,\overline{\langle x^*, x_k\rangle}\\ &=\sum_{k\in\mathbb{Z}} \langle x^*, x_{k-n}\rangle\,\overline{\langle x^*, x_k\rangle}. \end{align*} $$

The series is absolutely convergent since the almost sure convergence of the scalar Gaussian series $\sum \langle x^*, x_k\rangle \, g_k$ implies that $\sum _{k\in \mathbb {Z}} \vert \langle x^*, x_k\rangle \vert ^2<\infty $ .

Now, let $\varphi \in L^2(\mathbb {T})$ be the function with Fourier coefficients $\widehat \varphi (k):= \langle x^*, x_{-k}\rangle $ , $k\in \mathbb Z$ , that is,

$$ \begin{align*} \varphi(z)\sim \sum_{k\in\mathbb{Z}} \langle x^*, x_{-k}\rangle \,z^k;\end{align*} $$

and let $g:=\vert \varphi \vert ^2\in L^1(\mathbb {T})$ . By definition, we have for all $n\geq 0$ ,

$$ \begin{align*} \widehat g(n)=\sum_{k\in\mathbb{Z}} \widehat{\overline\varphi}(k)\,\widehat{\varphi}(n-k)=\sum_{k\in\mathbb{Z}} \overline{\langle x^*, x_k\rangle}\, \langle x^*, x_{k-n}\rangle.\end{align*} $$

Since two positive measures on $\mathbb {T}$ with the same non-negative Fourier coefficients must be equal, it follows that $\sigma _{x^*}= g(\unicode{x3bb} )\, d\unicode{x3bb} $ , which concludes the proof.

We mentioned above that Proposition 3.17 is a probabilistic version of Proposition 3.16; let us be a little bit more explicit. The following fact (which was observed independently by A. López-Martínez) can be extracted from the proof of [Reference Agneessens1, Theorem 4.9].

In view of that, the next result is an improvement of Proposition 3.16 when X is a Banach space with non-trivial cotype.

We now show that, at least for operators on complex Hilbert spaces, Theorem 3.1 is ‘strictly stronger’ than Theorem 2.1. This answers a very natural question asked by the referee.

Finally, the ergodic-theoretic point of view allows us to give an ‘extreme’ example of a hereditarily frequently hypercyclic operator T which does not satisfy the FHCC.

3.8. Perfect spanning and hereditary UFHC

The links between properties of unimodular eigenvectors of an operator T and frequent hypercyclicity of T have been very much studied since [Reference Bayart and Grivaux5]. The strongest available result may be the following (see [Reference Bayart and Matheron8, Reference Grivaux29]).

Let X be a separable complex Fréchet space and let $T\in \mathfrak L(X)$ . If the $\mathbb {T}$ -eigenvectors of T are perfectly spanning, then T is frequently hypercyclic and, in fact, there exists a Gaussian T-invariant measure $\mu $ with full support such that T is weakly mixing with respect to $\mu $ .

The perfect spanning assumption means that for any countable set $N\subset \mathbb {T}$ , the eigenvectors of T with eigenvalues in $\mathbb {T}\setminus N$ span a dense linear subspace of X; equivalently (see [Reference Grivaux and Matheron30, Proposition 6.1]), the $\mathbb {T}$ -eigenvectors of T are $\sigma $ -spanning for some continuous probability measure $\sigma $ on $\mathbb {T}$ . It is plausible that under this assumption, the operator T is, in fact, hereditarily frequently hypercyclic; but we are very far from being able to prove that. We would be already happy enough if we could weaken the assumptions of Theorem 3.1 and prove that for any complex Banach space X, an operator $T\in \mathfrak L(X)$ is hereditarily frequently hypercyclic as soon as the $\mathbb {T}$ -eigenvectors of T are spanning with respect to the Lebesgue measure—but again, this seems out of reach for the moment. However, we do have the following result.

For the proof, we will need the following variant of [Reference Conze20, Lemme 5].

Lemma 3.25. Let $(X, \mathcal B,\mu , T)$ be a measure-preserving dynamical system and assume that T is weakly mixing with respect to $\mu $ . Let also $(n_k)_{k\geq 1}$ and $(k_i)_{i\geq 0}$ be two increasing sequence of integers. Assume that $n_{k_i}=O(k_i)$ as $i\rightarrow \infty $ . Then, for any measurable set $V\subset X$ ,

$$ \begin{align*} \frac1{k_i}\sum_{k=1}^{k_i} \mathbf 1_V\circ T^{n_k}\xrightarrow{L^2} \mu(V)\quad\text{as } i\to\infty.\end{align*} $$

Proof. The proof is similar to that of the classical Blum–Hanson theorem [Reference Blum and Hanson15]. We have

$$ \begin{align*} \bigg\Vert \frac1{k_i}\sum_{k=1}^{k_i} \mathbf 1_V\circ T^{n_k}-\mu(V)\bigg\Vert_2^2&= \frac1{k_i^2} \sum_{r,s=1}^{k_i} ( \mu( T^{-n_r}(V)\cap T^{-n_s}(V)) -\mu(V)^2)\\ &=\frac2{k_i^2} \sum_{1\leq r<s\leq k_i} ( \mu( V\cap T^{-(n_s-n_r)}(V)) -\mu(V)^2) +O\bigg(\frac1{k_i}\bigg). \end{align*} $$

So it is enough to check that

(3.2) $$ \begin{align} \sum_{1\leq r<s\leq k_i} \vert \mu( V\cap T^{-(n_s-n_r)}(V)) -\mu(V)^2\vert=o(k_i^2). \end{align} $$

In what follows, we put

$$ \begin{align*} \gamma_{s,r} :=\vert\mu( V\cap T^{-(n_s-n_r)}(V)) -\mu(V)^2\vert.\end{align*} $$

Since T is weakly mixing with respect to $\mu $ , there is a set $D\subseteq \mathbb {N}$ with $\mathrm {dens} (D)=1$ such that

(3.3) $$ \begin{align} \mu( V\cap T^{-d}(V))-\mu(V)^2\to 0\quad\text{as } d\to\infty, d\in D.\end{align} $$

Write

$$ \begin{align*} \sum_{1\leq r<s\leq k_i} \gamma_{s,r}=\sum_{r=1}^{k_i}\sum_{\substack{r<s\leq k_i \\ n_s-n_r\in D}} \gamma_{s,r}+ \sum_{r=1}^{k_i}\sum_{\substack{r<s\leq k_i \\ n_s-n_r\not\in D}} \gamma_{s,r}=:\alpha_i+\beta_i. \end{align*} $$

Using equation (3.3) and since $\gamma _{r,s}\leq 1$ for all $r,s$ , it is not hard to check that

$$ \begin{align*}\sum_{\substack{r<s\leq k_i \\ n_s-n_r\in D}} \gamma_{s,r}=o(k_i)\quad\text{ as } i\to\infty, \text{ uniformly in } r;\end{align*} $$

and it follows that $\alpha _i=o(k_i^2)$ . Moreover, since $\mathrm {dens}(D)=1$ , we see that

$$ \begin{align*} \beta_i\leq \sum_{r=1}^{k_i} \# \{ s\in (r,k_i] : n_s-n_r\not\in D\} \leq k_i\times \# ( (0,n_{k_i}]\setminus D) =k_i\times o(n_{k_i});\end{align*} $$

so $\beta _i=o(k_i^2)$ since we are assuming that $n_{k_i}=O(k_i)$ . This proves equation (3.2).

Proof. Let $(n_k)_{k\geq 1}$ be the increasing enumeration of A. Since $\overline {\mathrm {dens}}(A)>0$ , one can find an increasing sequence of integers $(k_i)_{i\geq 0}$ such that $n_{k_i}=O(k_i)$ . By Lemma 3.25, one can find a subsequence $(k^{\prime }_i)$ of $(k_i)$ such that $(1/{k^{\prime }_i})\sum _{k=1}^{k^{\prime }_i} \mathbf 1_V\circ T^{n_k}\to \mu (V) \mu $ -almost everywhere. In other words: for $\mu $ -almost every $x\in X$ ,

$$ \begin{align*} \frac1{k^{\prime}_i}\, \#\{ k\in [1, k^{\prime}_i] : n_k\in \mathcal N_T(x,V)\}\to \mu(V).\end{align*} $$

Since $\#\{ k\in [1, k^{\prime }_i] : n_k\in \mathcal N_T(x,V)\}=\#( [1, n_{k^{\prime }_i}]\cap A\cap \mathcal N_T(x,V))$ and $n_{k^{\prime }_i}=O(k^{\prime }_i)$ , it follows that $\overline {\mathrm {dens}}(A\cap \mathcal N_T(x,V))>0$ for $\mu $ -almost every $x\in X$ .

Let us point out one consequence of Theorem 3.24.

5.1. The result

In this section, we use the machinery developed in [Reference Grivaux, Matheron and Menet31], following the construction in [Reference Menet41] of chaotic operators which are not frequently hypercyclic, to produce an operator on $\ell _p(\mathbb {Z}_+)$ , $1\leq p<\infty $ , which is frequently hypercyclic but not hereditarily frequently hypercyclic. We will, in fact, obtain a formally stronger result.

Obviously, if an operator is hereditarily frequently hypercyclic, then it is frequently hypercyclic along any set $A\subset \mathbb {N}$ with positive lower density. Our aim is to prove the following theorem.

With the terminology of [Reference Grivaux, Matheron and Menet31], the operator we are looking for will be a C $_{+,1}$ -type operator. So we will need to recall the definition of C $_{+,1}$ -type operators, and more generally of C-type and C $_{+}$ -type operators. However, before that, we will prove a general result allowing to check in a simple way that an operator is not frequently hypercyclic along some set with positive lower density.

5.2. How not to be hereditarily FHC

The next theorem gives simple conditions ensuring that an operator is not hereditarily frequently hypercyclic.

Proof. Extracting subsequences of $(K_n)$ and $(J_n)$ if necessary, we may assume without loss of generality that $J_{n+1}\ge 2(n+1)^{2^{J_n}}J_n$ for all n.

Let us denote by $\mathcal {F}_n\subset 2^{\mathbb {N}}$ the family of all finite sets $F\subset [0,J_n)$ such that $\#F\ge J_n/2$ . Let $C_n:=\#\mathcal {F}_n$ and let $(F_{n,j})_{0\le j< C_n}$ be an enumeration of $\mathcal {F}_n$ . We set $M_n:=(n+1)^{C_{n}}J_n$ and we remark that $M_n\le (n+1)^{2^{J_n}}J_n\le J_{n+1}/2$ .

We now construct the set A by induction. To start, let

$$ \begin{align*}A_1:=[0,J_1)\cup \bigcup_{0\le j< C_1}\bigcup_{0\le l<2^{j+1}-2^j}((2^j+l)J_1+F_{1,j}).\end{align*} $$

Given $k\ge 2$ , if $A_1,\ldots , A_{k-1}$ have been defined already, we define $A_k$ by setting

$$ \begin{align*} A_{k}:=[M_{k-1},J_{k})\; \cup \; \bigcup_{0\le j<C_{k}}\bigcup_{0\le l<(k+1)^{j+1}-(k+1)^j}((k+1)^j+l)J_{k}+F_{k,j}). \end{align*} $$

Observe that the sets $(k+1)^j+l)J_{k}+F_{k,j}$ involved in this definition are pairwise disjoint, since they are contained in successive intervals. More precisely,

$$ \begin{align*} A_k\cap [sJ_{k}, (s+1)J_{k})= sJ_{k}+F_{k,j}\end{align*} $$

for every $0\le j<C_k$ and every $(k+1)^{j}\le s<(k+1)^{j+1}$ , and $\max (A_k)< (k+1)^{C_k} J_k= M_k$ . Hence, $A_k\subseteq [M_{k-1}, M_k)$ . Finally, we let

$$ \begin{align*} A:=\bigcup_{k\ge 1} A_k. \end{align*} $$

Claim 5.4. We have ${\#(A\cap [0,N])}/({N+1})\ge 1/4$ for all $N\geq 0$ . In particular, $\underline {\textrm {dens}}(A)>0$ .

Proof of Claim 5.4

We will check by induction on $k\ge 1$ that

(5.1) $$ \begin{align} \frac{\#( A\cap [0,N])}{(N+1)}\ge \frac{1}{4} \quad\textrm{ for all } N<M_k. \end{align} $$

If $N<M_1$ , there exists $0\le s < 2^{C_1}$ such that $s J_1\le N<(s+1)J_1$ . Then, ${\#(A\cap [0,N])}/({N+1})\ge 1$ if $s=0$ and

$$ \begin{align*} \frac{\#(A\cap [0,N])}{N+1}\ge \frac{\#(A_1\cap [0,sJ_1])}{(s+1)J_1}\ge \frac{s}{2(s+1)}\ge \frac{1}{4}\quad \text{if } s\ge 1, \end{align*} $$

because the sets involved in the definition of $A_1$ are pairwise disjoint and $\#F_{1,j}\ge J_1/2$ for every $j<C_1$ .

Assume that the inequality (5.1) has been proved for $k-1$ . To get the result for k, it is enough to check that

$$ \begin{align*}\frac{\#(A\cap [0,N])}{N+1}\ge \frac{1}{4} \quad\textrm{ for every } M_{k-1}\le N<M_{k}.\end{align*} $$

If $M_{k-1}\le N< J_{k}$ then, since $[ M_{k-1}, N]\subset [M_{k-1},J_{k})\subset A_{k}$ , we get by the induction assumption that

$$ \begin{align*}\frac{\#(A\cap [0,N])}{N+1}\ge \frac{\#(A\cap [0,M_{k-1}))}{M_{k-1}}\geq \frac{1}{4}\cdot\end{align*} $$

Moreover, we even have ${\#(A\cap [0,J_{k}))}/{J_{k}}\ge \tfrac 12$ since $[M_{k-1},J_{k})\subset A_{k}$ and $M_{k-1}\le J_{k}/2$ . However, if $J_{k}\le N<M_{k}$ , there exists $1\le s < (k+1)^{C_{k}}$ such that $s J_{k}\le N<(s+1)J_{k}$ and we obtain in this case that

$$ \begin{align*}\frac{\#(A\cap [0,N])}{N+1}\ge \frac{\#(A\cap [0,J_{k}))}{(s+1)J_{k}}+\frac{\#(A_{k}\cap [J_{k},sJ_{k}])}{(s+1)J_{k}}\ge \frac{1}{2(s+1)}+\frac{s-1}{2(s+1)}\ge \frac{1}{4}.\end{align*} $$

This proves Claim 5.4.

Let us now get back to the proof of Theorem 5.3. Our aim is to show that under assumptions (a) and (b), T is not hereditarily frequently hypercyclic along the set A that we just constructed. To this end, we consider for any $c>0$ , the open sets

$$ \begin{align*} U_c:=\{y\in X\,:\, |\langle e^*_0, y\rangle|<c\}\quad\mathrm{and}\quad V_c=\{y\in X\,:\, |\langle e^*_0, y\rangle|>c\},\end{align*} $$

and we show that for every $x\in X$ , either $\mathcal N_T(x,U_{1/2})\cap A$ or $\mathcal N_T(x,V_{3/2})\cap A$ has a lower density equal to $0$ . Let $x\in X$ . Since $\vert \langle e_0^*, u\rangle \vert \leq C\, \Vert \pi _{K_n} u\Vert $ for some absolute constant $C>0$ , it follows from assumption (b) that for n sufficiently large, we have

$$ \begin{align*} \mathcal N_T(x,U_{1/2})\cap [0,(n+1)^{2^{J_n}}J_n)&\subset \mathcal N_T(\pi_{K_n}x,U_{3/4})\cap [0,(n+1)^{2^{J_n}}J_n)\\ &\subset \mathcal N_T(x,U_{1})\cap [0,(n+1)^{2^{J_n}}J_n) \end{align*} $$

and

$$ \begin{align*} \mathcal N_T(x,V_{3/2})\cap [0,(n+1)^{2^{J_n}}J_n)&\subset \mathcal N_T(\pi_{K_n}x,V_{4/3})\cap [0,(n+1)^{2^{J_n}}J_n)\\ &\subset \mathcal N_T(x,V_{1})\cap [0,(n+1)^{2^{J_n}}J_n). \end{align*} $$

Moreover, since $\mathcal N_T(x,U_1)\cap \, \mathcal N_T(x,V_1)=\emptyset $ , we have, for any $n\geq 1$ , that

$$ \begin{align*}\mathrm{either}\ \#(\mathcal N_T(x,U_{1})\cap [0,J_n))\le J_n/2 \quad\textrm{ or }\quad \#(\mathcal N_T(x,V_{1})\cap [0,J_n))\le J_n/2.\end{align*} $$

Hence, either $\#(\mathcal N_T(x,U_{1}){\kern-1pt}\cap{\kern-1pt} [0,J_n)){\kern-1pt}\le{\kern-1pt} J_n/2$ for infinitely many n values or $\#( \mathcal N_T(x,{\kern-1pt} V_{1}){\kern-1pt}\cap [0,J_n))\le J_n/2$ for infinitely many n values. Without loss of generality, we assume that $\#( \mathcal N_T(x,U_{1})\cap [0,J_n))\le J_n/2$ for infinitely many n values (the other case being similar). Hence, there exists an increasing sequence $(n_k)_{k\ge 1}$ of integers and a sequence $(j_k)_{k\ge 1}$ of integers such that

$$ \begin{align*}\mathcal N_T(x,U_{1})\cap [0,J_{n_k})\cap F_{n_k,j_k}=\emptyset \quad\textrm{ for every } k\ge 1.\end{align*} $$

Now, since $T^{J_{n_k}} \pi _{K_{n_k}}=\pi _{K_{n_k}}$ by assumption (a), we have for every $k\ge 1$ ,

$$ \begin{align*} \mathcal N_T(x,U_{1/2})\cap [&0,(n_k+1)^{2^{J_{n_k}}}J_{n_k})\\ &\quad\subset \;\mathcal N_T(\pi_{K_{n_k}}x,U_{3/4})\cap [0,(n_k+1)^{2^{J_{n_k}}}J_{n_k})\\ &\quad=\bigcup_{0\le l<(n_k+1)^{2^{J_{n_k}}}} (lJ_{n_k}+ \mathcal N_T(\pi_{K_{n_k}}x,U_{3/4})\cap[0,J_{n_k}) )\\ &\quad \subset \bigcup_{0\le l<(n_k+1)^{2^{J_{n_k}}}} (lJ_{n_k}+ \mathcal N_T(x,U_{1})\cap[0,J_{n_k}) ). \end{align*} $$

Intersecting with A and observing that $(n_k+1)^{j_k+1}\le (n_k+1)^{2^{J_{n_k}}}$ , we get that

$$ \begin{align*} (\mathcal N_T(x,U_{1/2})\cap A)\cap [0,(&n_k+1)^{j_k+1}J_{n_{k}})\\ &\quad\subset A\cap \bigcup_{0\le s<(n_k+1)^{j_k+1}} (sJ_{n_k}+\mathcal N_T(x,U_{1})\cap[0,J_{n_k})). \end{align*} $$

Now, by definition of A, we have

$$ \begin{align*}A\cap \bigcup_{(n_k+1)^{j_k}\le s<(n_k+1)^{j_k+1}}[sJ_{n_k}, (s+1)J_{n_k})= \bigcup_{(n_k+1)^{j_k}\le s<(n_k+1)^{j_k+1}}(sJ_{n_k}+F_{n_k,j_k}).\end{align*} $$

Since $\mathcal N_T(x,U_{1})\cap [0,J_{n_k})\cap F_{n_k,j_k}=\emptyset $ , it follows that

$$ \begin{align*} (\mathcal N_T(x,U_{1/2})\cap A)\cap [(n_k+1)^{j_k}J_{n_k},(n_k+1)^{j_k+1}J_{n_{k}}) =\emptyset, \end{align*} $$

so that

$$ \begin{align*} (\mathcal N_T(x,U_{1/2})\cap A)\cap [0,(n_k+1)^{j_k+1}J_{n_{k}}) \subset [0,(n_k+1)^{j_k}J_{n_k}). \end{align*} $$

So we see that

$$ \begin{align*} \frac{\# ((\mathcal N_T(x,U_{1/2})\cap A)\cap [0,(n_k+1)^{j_k+1}J_{n_{k}}))}{(n_k+1)^{j_k+1}J_{n_{k}}}\le \frac{(n_k+1)^{j_k}J_{n_k}}{(n_k+1)^{j_k+1}J_{n_{k}}}\cdot\end{align*} $$

The right-hand side of this inequality tends to $0$ as n tends to infinity, and this shows that $\underline {\textrm {dens}}(\mathcal N_T(x,U_{1/2})\cap A)=0$ .

5.3. C-type operators

We recall here very succinctly some basic facts concerning C-type operators, and we refer the reader to [Reference Grivaux, Matheron and Menet31, §§6 and 7] for more on this class of operators. In what follows, we denote by $(e_k)_{k\geq 0}$ the canonical basis of $\ell _p(\mathbb {Z}_+)$ , $1\le p<\infty $ .

Let us consider four ‘parameters’ v, w, $\varphi $ and b, where:

• • $v=(v_{n})_{n\ge 1}$ is a sequence of non-zero complex numbers such that $\sum _{n\ge 1}|v_{n}|<\infty $ ;

• • $w=(w_{j})_{j\geq 1}$ is a sequence of complex numbers such that $0<\inf _{k\ge 1} \vert w_k\vert \leq \sup _{k\ge 1}\vert w_k\vert <\infty $ ;

• • $\varphi $ is a map from $\mathbb {Z}_+$ into itself, such that $\varphi (0)=0$ , $\varphi (n)<n$ for every $n\ge 1$ , and the set $\varphi ^{-1}(l)=\{n\ge 0\,;\,\varphi (n)=l\}$ is infinite for every $l\ge 0$ ;

• • $b=(b_{n})_{n\ge 0}$ is a strictly increasing sequence of positive integers such that $b_{0}=0$ and $b_{n+1}-b_{n}$ is a multiple of $2(b_{\varphi (n)+1}-b_{\varphi (n)})$ for every $n\ge 1$ .

If w and b are such that

$$ \begin{align*} \inf_{n\geq 0} \prod_{b_n<j<b_{n+1}}\, \vert w_j\vert>0,\end{align*} $$

then, by [Reference Grivaux, Matheron and Menet31, Lemma 6.2], there is a unique bounded operator $T_{v,\,w,\, \varphi , \,b\,}$ on $\ell _p(\mathbb {Z}_+)$ such that

$$ \begin{align*} T_{v,\,w,\, \varphi, \,b\,}\ e_k= \begin{cases} w_{k+1}\,e_{k+1} & \textrm{if}\ k\in [b_{n},b_{n+1}-1),\; n\geq 0,\\ v_{n}\,e_{b_{\varphi(n)}}-\bigg(\displaystyle\prod_{j=b_{n}+1}^{b_{n+1}-1} w_{j}\bigg)^{ -1 } e_ { b_{n}} & \textrm{if}\ k=b_{n+1}-1,\ n\ge 1,\\ -\bigg(\displaystyle\prod_{j=b_0+1}^{b_{1}-1}w_j\bigg)^{-1}e_0& \textrm{if}\ k=b_1-1. \end{cases} \end{align*} $$

Any such operator $T_{v,\,w,\, \varphi , \,b\,}$ is called a C-type operator. A notable fact to be pointed out immediately is that C-type operators have lots of periodic points; indeed, we have the following fact, which is [Reference Grivaux, Matheron and Menet31, Lemma 6.4].

Fact 5.6. If $T=T_{v,\,w,\, \varphi , \,b\,}$ is a C-type operator, then

$$ \begin{align*}T^{2(b_{n+1}-b_n)}e_k=e_k\quad\text{if } k\in [b_n,b_{n+1}), \; n\geq 0.\end{align*} $$

It follows that every finitely supported vector is periodic for $T_{v,\,w,\, \varphi , \,b\,}$ ; in particular, a C-type operator is chaotic as soon as it is hypercyclic.

A C $_{+}$ -type operator is a C-type operator for which the parameters satisfy the following additional conditions: for every $k\geq 1$ :

• • $\varphi $ is increasing on each interval $[2^{k-1},2^{k})$ with $\varphi ([2^{k-1},2^{k}))=[0,2^{k-1})$ , that is,

  $$ \begin{align*}\varphi (n)=n-2^{k-1}\quad\text{for every } n\in[2^{k-1},2^{k});\end{align*} $$

• • the blocks $[b_{n},b_{n+1})$ , $n\in [2^{k-1},2^{k})$ all have the same size, which we denote by $\Delta ^{(k)}$ :

  $$ \begin{align*}b_{n+1}-b_{n}= \Delta ^{(k)}\quad\text{for every } n\in[2^{k-1},2^{k}); \end{align*} $$

• • the sequence v is constant on the interval $[2^{k-1}, 2^{k})$ : there exists $v^{(k)}$ such that

  $$ \begin{align*} v_{n}=v^{(k)}\quad\text{for every } n\in[2^{k-1},2^{k}); \end{align*} $$

• • the sequences of weights $(w_{b_{n}+i})_{1\le i <\Delta ^{(k)}}$ are independent of $n\in [2^{k-1},2^{k})$ : there exists a sequence $(w_{i}^{(k)})_{1\le i<\Delta ^{(k)}}$ such that

  $$ \begin{align*} w_{b_{n}+i}=w_{i}^{(k)} \quad\text{for every } n\in[2^{k-1},2^{k}) \text{ and } 1\le i<\Delta ^{(k)}. \end{align*} $$

Finally, a C $_{+,1}$ -type operator is a C $_{+}$ -type operator whose parameters are such that for all $k\geq 1$ ,

$$ \begin{align*} v^{(k)}=2^{-\tau ^{(k)}}\quad\textrm{and}\quad w_{i}^{(k)}= \begin{cases} 2&\textrm{if}\ \ 1\le i\le \delta ^{(k)},\\ 1&\textrm{if}\ \ \delta ^{(k)}<i<\Delta ^{(k)}, \end{cases} \end{align*} $$

where $(\tau ^{(k)})_{k\ge 1}$ and $(\delta ^{(k)})_{k\ge 1}$ are two increasing sequences of integers with $\delta ^{(k)}<\Delta ^{(k)}$ for each $k\ge 1$ .

These operators have been studied in detail in [Reference Grivaux, Matheron and Menet31, §7]. In particular, we have the following crucial fact [Reference Grivaux, Matheron and Menet31, Theorem 7.1].

Fact 5.7. A C $_{+,1}$ -type operator $T_{v,\,w,\, \varphi , \,b\,}$ is frequently hypercyclic as soon as

(5.2) $$ \begin{align} \limsup\limits_{k\to\infty }\, \dfrac{\delta ^{(k)} -\tau ^{(k)}}{\Delta ^{(k)}}>0. \end{align} $$

5.4. Proof of Theorem 5.2

Let $T=T_{v,\,w,\, \varphi , \,b\,}$ be an operator of C $_{+,1}$ -type on $\ell _{p}(\mathbb {Z}_+)$ , so that v and w are given by

$$ \begin{align*} v^{(k)}=2^{-\tau ^{(k)}}\quad\textrm{and}\quad w_{i}^{(k)}= \begin{cases} 2&\textrm{if}\ \ 1\le i\le \delta ^{(k)},\\ 1&\textrm{if}\ \ \delta ^{(k)}<i<\Delta ^{(k)}. \end{cases} \end{align*} $$

We assume that $\Delta ^{(k)}\in 8\mathbb {N}$ for all $k\ge 1$ , and we choose

$$ \begin{align*}\delta^{(k)}:=\tfrac{1}{4}\Delta^{(k)} \quad \text{and} \quad \tau^{(k)}:=\tfrac{1}{8}\Delta^{(k)}.\end{align*} $$

So the only ‘free’ parameter is now the sequence $(\Delta ^{(k)})_{k\geq 1}$ .

By Fact 5.7, the operator T is frequently hypercyclic (and hence chaotic since it is a C-type operator). So we just have to show that if the sequence $(\Delta ^{(k)})$ is suitably chosen, then T satisfies the assumptions of Theorem 5.3. We will in fact show that this holds as soon as the sequence $(\Delta ^{(k)})$ grows sufficiently rapidly. Let us set

$$ \begin{align*} K_n:=b_{2^{n}}\quad\mathrm{and}\quad J_n:=2\Delta^{(n)}\quad\textrm{ for every }n\ge 1.\end{align*} $$

With this choice of the sequences $(K_n)$ and $(J_n)$ , condition (a) in Theorem 5.3 is satisfied by Fact 5.6. So the only thing to check is condition (b).

Let $\gamma _k:=2^{\,\delta ^{(k-1)}-\tau ^{(k)}}(\Delta ^{(k)})^{1-1/{p}}$ . If $(\Delta ^{(k)})$ grows sufficiently rapidly, then the sequence $(\gamma _k)$ is decreasing and

$$ \begin{align*}2^n \sum_{k\ge n+1}2^{k-1}\gamma_k\le 1\quad\text{for all } n\geq 0.\end{align*} $$

Let us also define a sequence $(\beta _l)_{l\geq 1}$ as follows:

$$ \begin{align*}\beta _{l}:=4\,\gamma_k\quad\text{if } l\in [2^{k-1},2^{k}).\end{align*} $$

Finally, for any $l\geq 1$ , let $P_l$ be the projection of $\ell _p({\mathbb Z}_+)$ defined by

$$ \begin{align*} P_lx=\sum_{k=b_l}^{b_{l+1}-1}x_ke_k \quad\textrm{ for every } x\in\ell_p({\mathbb Z}_+). \end{align*} $$

As in the proof of [Reference Grivaux, Matheron and Menet31, Theorem 7.2], one can show that the following estimate holds for every $k\ge 0$ , every $l\in [2^{k-1},2^k[$ , every $0\le m<l$ and every $0\le j\le \Delta ^{(k)}-\delta ^{(k)}=\tfrac 34\Delta ^{(k)}$ :

$$ \begin{align*}\|P_mT^jP_lx\|\le \frac{\beta_l}{4}\bigg(\prod_{i=\Delta^{(k)}-j+1}^{\Delta^{(k)}-1}|w_i^{(k)}|\bigg) \|P_lx\|\le \frac{\beta_l}{4}\|P_lx\|.\end{align*} $$

Hence, we have for all n and $j\le \tfrac 34\Delta ^{(n+1)}$ ,

$$ \begin{align*} \|\pi_{K_n}T^j (I-\pi_{K_n})x\| &\le \sum_{m< 2^n}\sum_{l\ge 2^n}\|P_mT^j P_lx\|\\ &\le \sum_{m< 2^n}\sum_{l\ge 2^n} \frac{\beta_l}{4}\,\|P_lx\|\\ &\le 2^n \bigg(\sum_{l\ge 2^n}\frac{\beta_l}{4}\bigg)\,\|(I-\pi_{K_n})x\|\\ &\le 2^n \bigg(\sum_{k\ge n+1}2^{k-1}\gamma_k\bigg)\,\|(I-\pi_{K_n})x\|\le \|(I-\pi_{K_n})x\|. \end{align*} $$

So, if we take care to ensure that $\tfrac 34\Delta ^{(n+1)}\ge (n+1)^{2^{J_n}}J_n=(n+1)^{2^{2\Delta ^{(n)}}}2\Delta ^{(n)}$ for all $n\ge 1$ , then condition (b) in Theorem 5.3 is satisfied. This concludes the proof of Theorem 5.2.