## Szemerédi’s theorem, frequent hypercyclicity and multiple recurrence

My co-author George Costakis and I have recently uploaded to arxiv our paper “Szemerédi’s theorem, frequent hypercyclicity and multiple recurrence”. As I’m invited to talk about this subject next month, I will try to give here a general overview of the paper, the notions therein and the main ideas involved in the proofs. Our main objective in this paper is to relate some notions in linear dynamics to more classical notions from topological dynamics. In particular we show that frequently Cesàro hypercyclic operators are necessarily topologically multiply recurrent. The main tool we use to prove this result is Szemerédi’s theorem on arithmetic progressions in sets of positive density. In order to motivate this theorem, I will have to define many standard notions from linear dynamics as well as corresponding notions from topological dynamics. Before discussing the main result and (some of) its applications, I will try to give a picture of hypercyclic operators and their properties, as well as examples of natural’ operators which are hypercyclic.

— 1. Introduction: notions of hypercyclicity —

First of all, I will review some basic notions from linear dynamics that will be quite central throughout the exposition. I refer the reader to the excellent book of Bayart and Matheron (Bayart and Matheron, 2009) where most of this material is drawn from anyways. We will state several classical results here omitting the proof. If no other reference is given, this means the proof can be found in (Bayart and Matheron, 2009).

— 1.1. Hypercyclic operators —

We will work on a separable Banach space ${X}$ over ${{\mathbb R}}$ or ${{\mathbb C}}$. We will always use the symbol ${T:X\rightarrow X}$ to denote a bounded linear operator acting on ${X}$. In what follows I will just write ${X}$, ${T}$, without any further comment, assuming always that these symbols have the meaning described above.

The most central notion in linear dynamics is that of hypercyclicity.

Definition 1 The orbit of a vector ${x\in X}$ under ${T}$ (or the ${T}$-orbit) is the set

$\displaystyle \textnormal{Orb}(x,T) := \{ x,Tx, T^2x, T^3x, \ldots\}.$

The operator T is said to be hypercyclic if there is some vector ${x\in X}$ such that the set ${\textnormal{Orb}(x,T)}$ is dense in ${X}$. Such a vector ${x}$ will be called a hypercyclic vector for ${T}$ (or a ${T}$-hypercyclic vector).

Some remarks are in order. First of all let us point out that these definitions only make sense if the space ${X}$ is separable. On the other hand, hypercyclicity is an infinite dimensional phenomenon; there are no hypercyclic operators on a finite-dimensional space ${X\neq \{0\}.}$ To see this quickly think of a square matrix in its Jordan normal form.

An easy consequence of these definitions is that whenever an operator ${T}$ is hypercyclic, we must have ${\|T\|>1}$. Moreover, whenever ${T}$ is an invertible operator, ${T}$ is hypercyclic if and only if ${T^{-1}}$ is hypercyclic. These facts will be used in the discussion below .

The definition of hypercyclicity does not require any linear structure. It makes sense for an arbitrary continuous map ${T:Y\rightarrow Y}$ acting on a topological space ${Y}$.

The most general setup linear dynamics is that of an arbitrary separable topological vector space ${X}$. We will stick however to the case of a Banach space to simplify the exposition, the generalizations being mostly of a technical nature.

The notion of hypercyclicity is strictly stronger (though relevant) than that of cyclicity. Recall from classical operator theory that an operator ${T}$ is called cyclic if there exists a vector ${x\in X}$ (a cyclic vector for ${T}$) such that the linear span of

$\displaystyle \textnormal{Orb}(x,T)=\{P(T)x: P\mbox{ polynomial}\},$

is dense in ${X}$. This notion is related to the invariant subspace problem; the operator ${T}$ lacks (non-trivial) invariant closed subspaces if and only if every non-zero vector ${x\in X}$ is cyclic for ${T}$.

Likewise, the notion of hypercyclicity is closely related to the invariant subset problem. It is an easy observation that an operator ${T}$ lacks non-trivial invariant subsets if and only if every non-zero vector ${x\in X}$ is hypercyclic for ${T}$. P. Enflo first answered the question in the negative for a constructing a rather peculiar Banach space. After that, C.J. Read has proved that there is an operator ${T}$ on ${\ell^1({\mathbb N})}$ for which every non-zero vector ${x}$ is hypercyclic. So the invariant subspace problem has a negative solution on ${\ell^1({\mathbb N})}$. However the problem remains open in the case of Hilbert spaces.

— 1.2. Universal sequences of operators —

We will be interested in the following generalization of hypercyclicity to families of continuous linear operators ${(T_k)_{k\in{\mathbb N}}}$, where each ${T_k:X\rightarrow Y}$ and ${X, Y}$ are two topological spaces.

Definition 2 The family ${(T_k)_{k\in{\mathbb N}}}$ is called universal if there exists a ${x\in X}$ such that the set ${\{T_k x: k\in{\mathbb N}\},}$ is dense in ${X}$.

Of course hypercyclicity is a special case of universality, where the family of operators is defined as the iterates of a fixed operator ${T}$ and ${X=Y}$ is a topological vector space.

— 1.3. Cesàro Hypercyclicity —

In (León-Saavedra, 2002), F. León-Saavedra introduced the notion of Cesàro hypercyclicity.

Definition 3 An operator ${T}$ is called Cesàro hypercyclic if its Cesàro orbit, that is the set

$\displaystyle \bigg\{\frac{1}{n+1} \sum_{k=0} ^{n} T^k x, n\in {\mathbb N}\bigg\},$

is dense in ${X}$. Such a vector ${x}$ will be called Cesàro hypercyclic for ${T}$.

Saavedra showed in (León-Saavedra, 2002) that ${T}$ is Cesàro hypercyclic if and only if there is a vector ${y\in X}$ such that the set

$\displaystyle \bigg\{ \frac{1}{n}T^n y:n\in N.\bigg\},$

is dense in ${X}$. Observe that this means that the family of operators ${\{\frac{1}{n}T^n\}_{n\in{\mathbb N}}}$ is universal. We stress here that, in general, the notions of hypercyclicity and Cesàro hypercyclicity are not ordered’; hypercyclicity does not imply Cesàro hypercyclicity and vice versa.

— 1.4. How to prove that an operator is hypercyclic —

This first characterization of hypercyclicity comes from topological dynamics and is often referred to as Birkhoff’s transitivity theorem’.

Theorem 4 (Brkhoff’s transitivity theorem) Let ${T}$ be a continuous linear operator on a separable Banach space ${X}$. Then ${T}$ is hypercyclic if and only if it is topologically transitive; that is, for every pair of open sets ${U,V\subset X}$, there exists ${n\in{\mathbb N}}$ such that ${T^n(U)\cap V\neq \emptyset}$.

A byproduct of the proof of Theorem 4 is that the set of ${T}$-hypercyclic vectors, ${HC(T)}$, is a dense ${G_\delta}$ subset of ${X}$.

Actually Birkhoff’s theorem is true in a much more general context but I won’t pursue that here. It is important however that no linearity is necessary in Theorem 4. As a result, when one adds linearity, the following handy criterion becomes available.

Definition 5 (Hypercyclicity criterion) Let ${X}$ be a separable Banach space and ${T:X\rightarrow X}$ a bounded linear operator. We say that ${T}$ satisfies the hypercyclicity criterion if there exists an increasing sequence of positive integers ${(n_k)}$, two dense sets ${D_1,D_2\subset X}$ and a sequence of maps ${S_{n_k}:D_2\rightarrow X}$ such that:

(i) ${T^{n_k}x\rightarrow 0}$ for any ${x\in D_1}$,

(ii) ${S_{n_k}y \rightarrow 0}$ for any ${y\in D_2}$,

(iii) ${T^{n_k} S_{n_k} y\rightarrow y}$ for any ${y\in D_2}$.

Using Theorem 4 one can prove the following:

Theorem 6 Let ${T}$ be a continuous linear operator on a separable Banach space ${X}$. Suppose that ${T}$ satisfies the hypercyclicity criterion 5. Then ${T}$ is hypercyclic.

Definition 5 and Theorem 6 are originally due to Kitai (Kitai, 1982), in the case that ${n_k=k\in{\mathbb N}}$ and ${D_1=D_2}$. The criterion was then evolved by R.Gethner and J. H. Shapiro in (Gethner and Shapiro, 1987) and J. Bès (Bès, 1998).

It was a long-standing question whether every hypercyclic operator satisfies the hypercyclicity criterion. This problem was recently resolved in the negative by M. De La Rosa and C.J. Read. It is not hard to show (and it was known) that the hypercyclicity criterion is equivalent to the operator ${T \oplus T :X^2\rightarrow X^2}$ being hypercyclic. In topological dynamics this property is referred to as ${T}$ being weakly mixing. This problem was recently resolved in the negative in (de la Rosa and Read, 2009) and later in (Bayart and Matheron, 2007) for all classical Banach spaces.

A consequence of the hypercyclicity criterion 5 and Theorem 6 is the following result, which highlights the connection between linear dynamics and spectral theory. Roughly speaking, the following Godefroy-Shapiro criterion states that an operator which has a large supply’ of eigenvectors is hypercyclic. See (Godefroy and Shapiro, 1991).

Theorem 7 (Godefroy-Shapiro criterion) Let ${T}$ be a continuous linear operator on a separable Banach space ${X}$. Suppose that ${\cup_{|\lambda|<1}\textnormal{Ker}(T-\lambda I)}$ and ${\cup_{|\lambda|>1}\textnormal{Ker}(T-\lambda I)}$ both span a dense subspace of ${X}$. Then ${T}$ is hypercyclic.

— 1.5. Examples of hypercyclic operators —

We will now use the previous hypercyclicity criteria to show that some very natural operators are hypercyclic. We will also take the chance to define some classes of operators which I want to discuss later on, in relevance to our main theorem.

Example 1 Let ${\mathbb H({\mathbb C})}$ denote the space of all entire functions on ${C}$ endowed with the topology of uniform convergence on compact sets. Now ${\mathbb H({\mathbb C})}$ is not a Banach space but it is a separable Frèchet space so all the notions and theorems discussed above go through. We consider the derivative operator ${D:f\mapsto f'}$. To see this, apply the hypercyclicity criterion with ${n_k:=k}$ and

$\displaystyle D_1=D_2:=\{P:{\mathbb C}\rightarrow{\mathbb C}, P \mbox{ polynomial}\}.$

Now the operator ${S_k}$ in the hypercyclicity criterion needs to be defined as a sort of (asymptotic) right inverse of the derivative operator so it is natural to define ${S[f](z):=\int_0 ^z f(w)dw}$ and ${S_k:= S^k}$. Then we have that ${D^k z^m \rightarrow 0}$ as ${k\rightarrow +\infty}$ for every monomial ${z^m}$ so that takes care of (i) in the hypercyclicity criterion. Condition (iii) is trivial to verify since ${D^k S_k=I}$ on ${D_2}$. Finally, in order to check the validity of condition (ii) in the hypercyclicity criterion we need to see that ${S_k(z^p)\rightarrow 0}$ as ${k\rightarrow +\infty},$ for every positive integer ${p}$. However, we readily see that

$\displaystyle S_k(z^p)=\frac{1}{p+1}S_{k-1}(z^{p+1})=\ldots=\frac{p!}{(p+k)!}z^{p+k},$

from which we easily conclude that ${S_k(z^p)\rightarrow 0}$ uniformly on compact subsets of ${{\mathbb C}}$.

Example 2 Let us now consider the Hilbert space ${\ell^2({\mathbb N})}$. The backward shift operator is defined by ${B(x_0,x_1,\ldots)=(x_1,x_2,\ldots)}$. Observe that this operator can never be hypercyclic since ${\|B\|=1}$ so the orbit of any vector under ${B}$ stays inside the unit ball. However, the operator ${\lambda B}$ is hypercyclic for every ${\lambda \in {\mathbb C}}$ with ${|\lambda|>1}$. Again it is an easy exercise to check the validity of the hypercyclicity criterion with ${n_k:=k}$ and ${D_1=D_2:c_{00}}$, where ${c_{00}}$ is the space of all finitely supported sequences. Again ${S_k:= S^k}$ where ${S}$ is the natural candidate, the right inverse of ${B}$ which in this case is the forward shift operator defined as ${S(x_0,x_1,\ldots):=(0,x_0,x_1,\ldots)}$.

Our last example one the one hand illustrates the Godefroy-Shapiro criterion and on the other hand gives an introduction to a class of operators I would like to consider later on in the discussion.

Example 3 Here we consider a Hilbert space ${(\mathcal H,\|\cdot\|)}$ of analytic functions ${f:\mathbb D \rightarrow {\mathbb C}}$, where ${\mathbb D}$ is the open unit disk of the complex plane. The space ${\mathcal H}$ is pretty general but we require the following two conditions:

• ${\mathcal H \neq \{0\}}$, and
• for every ${z\in\mathbb D}$, the point evaluation functionals ${f\mapsto f(z), f\in \mathcal H},$ are bounded.

The second condition assures that convergence in ${\mathcal H}$ implies pointwise convergence on ${\mathbb D}$. By the boundedness of holomorphic functions on compact sets and the uniform boundedness principle the second condition amounts to requiring that convergence in ${\mathcal H}$ implies uniform convergence on compact subsets of ${\mathbb D}$. The reader is thus encouraged to think of the Hardy space ${\mathbb H^2(\mathbb D)}$ or the Bergman space ${A^2(\mathbb D)}$ in the place of ${\mathcal H}$, keeping in mind however that interesting phenomena occur outside these two particular cases.

A feature of ${\mathcal H}$ that we will use is the existence of a reproducing kernel. In particular, For each ${z\in \mathbb D}$, the boundedness of the point evaluation functionals and the Riesz representation theorem provide a unique function ${k_z\in\mathcal H}$, the reproducing kernel of ${\mathcal H}$ at ${z}$, such that

$\displaystyle f(z)=\langle f,k_z\rangle, \quad f\in\mathcal H.$

Recall that a function ${\phi:\mathbb D \rightarrow {\mathbb C}}$ is called a multiplier of ${\mathcal H}$ if ${\phi f\in\mathcal H}$ for every ${f\in \mathcal H}$. Such a ${\phi}$ defines a multiplication operator ${M_\phi:\mathcal H\rightarrow \mathcal H}$ in terms of the formula

$\displaystyle M_\phi(f)=\phi f, \quad f\in\mathcal H.$

By the boundedness of point evaluation functionals and the closed graph theorem it follows that ${M_\phi}$ is a bounded linear operator on ${\mathcal H}$. Moreover, every multiplier ${\phi}$ is a bounded holomorphic function, this is,

$\displaystyle \|\phi\|_\infty:=\sup_{z\in\mathbb D}|\phi(z)|<+\infty.$

Observe that for every ${z\in \mathbb D}$ and every ${n\in{\mathbb N}}$ we have that

$\displaystyle |\phi(z)|^n|f(z)| \leq \| M_\phi ^n f \| =\| M_\phi ^n (f)\| \leq \| M_\phi\|^n \|f\|.$

Remembering that there is at least one ${f\in\mathcal H}$ which is not identically ${0}$ we conclude that ${\|\phi\|_\infty \leq \|M_\phi\|}$. Thus every multiplier ${\phi}$ is a bounded holomorphic function with ${\|\phi\|_\infty \leq \|M_\phi\|}$. The opposite is not always true under our assumptions as can be seen by considering for example the Dirichlet space of holomorphic functions on ${\mathbb D}$, that is the space of all functions ${f:\mathbb D\rightarrow {\mathbb C}}$ such that

$\displaystyle \|f\|^2 _{\textnormal Dir}:=|f(0)|^2+ \int_{\mathbb D}|f'(z)|^2 dA(z)<+\infty.$

Here ${dA(z)}$ denotes area measure. In the Dirichlet space not every bounded holomorphic function is a multiplier.

In general it is not difficult to see that a multiplication operator is never hypercyclic. The situation is quite different for the adjoints of multiplication operators. In order to make the statement of the following theorem more clear we require the extra assumption that every holomorphic function is a multiplier of ${\mathcal H}$ such that ${\|M_\phi\|=\|\phi\|_\infty}$. This extra assumption is automatically satisfied in the case of the Hardy space ${\mathbb H^2(\mathbb D)}$ or the Bergman space ${A^2(\mathbb D)}$ but not in the Dirichlet space. The following theorem is from (Godefroy and Shapiro, 1991).

Theorem 8 (Godefroy, Shapiro) Assume that ${\mathcal H}$ is a Hilbert space of holomorphic functions as above. Furthermore assume that every bounded holomorphic function is a multiplier of ${\mathcal H}$ such that ${\|M_\phi\|=\|\phi\|_\infty}$. Then the adjoint multiplication operator ${M^* _\phi}$ is hypercyclic if and only if ${\phi}$ is non-constant and ${\phi(\mathbb D)\cap \mathbb T\neq \emptyset}$.

Proof: We first prove that if ${\phi(\mathbb D)\cap \mathbb T\neq \emptyset}$ then ${M_\phi ^*}$ is hypercyclic. For ${z\in\mathbb D}$ we consider the reproducing kernel ${k_z\in\mathcal H}$. Since

$\displaystyle \begin{array}{rcl} \langle f, M_\phi ^* (k_z)\rangle & =& \langle M_\phi (f), k_z \rangle = \phi(z)f(z)\\ \\ & =& \phi(z)\langle f,k_z\rangle=\langle f, \overline{\phi(z)}k_z\rangle, \end{array}$

for every ${f\in\mathcal H}$, we conclude that ${M_\phi ^* (k_z)=\overline{\phi(z)}k_z}$ for every ${z\in\mathbb D}$. That is, for every ${z\in\mathbb D}$, ${k_z}$ is an eigenvector of ${M_\phi ^*}$ with corresponding eigenvalue ${\overline{\phi(z)}}$. Now let ${U:=\{z\in\mathbb D : |\phi(z)|<1\}}$ and ${V:=\{ z \in\mathbb D : |\phi(z)|>1\}}$. Since ${\phi}$ is non-constant and ${\phi(\mathbb D) \cap \mathbb T \neq \emptyset}$ we have that both ${U,V}$ are non-empty open sets (by the open-mapping theorem for analytic functions ${\phi(\mathbb D)}$ is an open set). By the Godefroy Shapiro criterion, in order to show that ${M^* _\phi}$ is hypercyclic it suffices to show that ${\{k_z:z\in U\}}$ and ${\{k_z:z\in V\}}$ both span a dense subset of ${\mathcal H}$. Indeed, assume that there exists a function ${f\in \mathcal H}$ which is orthogonal to all ${k_z}$ either for all ${z\in U}$ or for all ${z\in V}$. In either case ${f}$ vanishes on a non-empty open set and thus is identically zero.

In order to prove the other direction first observe that whenever ${M_\phi ^* }$ is hypercyclic, ${\phi}$ is non-constant. Moreover we have that ${\phi(\mathbb D)}$ is connected so it either lies entirely inside, or entirely outside the unit disk. In the first case we have that ${\|M^*_\phi\|=\|M_\phi\|=\|\phi\|_\infty<1}$, thus ${M^* _\phi}$ cannot be hypercyclic. In the complementary case, the function ${1/\phi}$ is a bounded holomorphic function and ${\|1/\phi\|_\infty <1 }$. By the first case, ${M^*_{1/\phi}}$ is not hypercyclic, and since ${M^* _\phi=(M^* _{1/\phi})^{-1}}$, neither is ${M_\phi ^*}$. $\Box$

Example 4 We finish this short list of examples by giving another typical class of hypercyclic operators, namely unilateral and bilateral weighted shifts. Let ${l^2(\mathbb{N})}$ be the Hilbert space of square summable sequences ${x=(x_n)_{n\in \mathbb{N}}}$. Consider the canonical basis ${(e_n)_{n\in \mathbb{N}}}$ of ${l^2(\mathbb{N})}$ and let ${(w_n)_{n\in \mathbb{N}}}$ be a (bounded) sequence of positive numbers. The operator ${B^{\textnormal{uni}} _w:l^2(\mathbb{N})\rightarrow l^2(\mathbb{N})}$ is a unilateral (backward) weighted shift with weight sequence ${(w_n)_{n\in\mathbb N}}$ if ${Te_{n}=w_ne_{n-1}}$ for every ${n\geq 1}$ and ${Te_1=0}$.

Let ${l^2(\mathbb{Z})}$ be the Hilbert space of square summable sequences ${x=(x_n)_{n\in \mathbb{Z}}}$ endowed with the usual ${l^2}$ norm. That is, ${x=(x_n)_{n\in \mathbb{Z}}\in l^2(\mathbb{Z})}$ if ${\sum_{n=-\infty}^{+\infty}|x_n|^2<+\infty }$. Let ${(w_n)_{n\in \mathbb{Z}}}$ be a (bounded) sequence of positive numbers. The operator ${B^\textnormal{bil} _w :l^2(\mathbb{Z})\rightarrow l^2(\mathbb{Z})}$ is a bilateral (backward) weighted shift with weight sequence ${(w_n)_{n\in\mathbb Z}}$ if ${Te_{n}=w_ne_{n-1}}$ for every ${n\in \mathbb{Z}}$. Here ${(e_n)_{n\in \mathbb{Z}}}$ is the canonical basis of ${l^2(\mathbb{Z})}$.

Theorem 9 Let ${B^{\textnormal{uni}} _v ,B^\textnormal{bil} _w}$ be defined as above, with weight sequences ${v=(v_n)_{n\in {\mathbb N}}, w=(w_n)_{n\in {\mathbb Z}}}$ respectively.

(i) ${B^{\textnormal{uni}} _v}$ is hypercyclic if and only if

$\displaystyle \limsup_{n\rightarrow +\infty} (v_1\cdots v_n)=+\infty$

(ii) ${B^\textnormal{bil} _w}$ is hypercyclic if and only if, for any ${q\in{\mathbb N}}$

$\displaystyle \limsup_{n\rightarrow +\infty}(w_1\cdots w_{n+q})=+\infty$

and

$\displaystyle \quad \liminf_{n\rightarrow +\infty}(w_0\cdots w_{-n+q+1})=+\infty.$

— 2. Recurrence, multiple recurrence and hypercyclicity —

Let us consider a bounded linear operator ${T:X\rightarrow X }$ on a separable Banach space ${X}$. We have already seen that saying that an operator is hypercyclic is equivalent to saying that an operator is topologically transitive, that is that for every pair of open sets ${U,V\subset X}$, there is some positive integer ${n}$ such that ${T^nU\cap V\neq \emptyset}$. In what follows I will introduce some notions that come from topological dynamical systems.

— 2.1. Recurrence and Multiple recurrence —

A somewhat weaker notion in topological dynamics is that of recurrence.

Definition 10 The operator ${T:X\rightarrow X}$ is called recurrent if for every open set ${U\subset X}$ there is a ${k\in\mathbb N}$ such that ${U\cap T^{-k}U\neq \emptyset}$.

Clearly every hypercyclic operator ${T}$ is recurrent. Unlike hypercyclicity which is a purely infinite dimensional phenomenon, there are recurrent operators in finite dimensions (consider for example a rotation on the plane).

A recurrent operator has many points whose orbit under ${T}$ asymptotically returns’ to the point. To make this more precise, let us call a vector ${x\in X}$ recurrent vector for ${T}$ if there exists an increasing sequence of positive integers ${n_k}$ such that ${T^{n_k}x \rightarrow x}$ as ${k\rightarrow +\infty}$. It turns out that a recurrent operator has a ${G_\delta}$ dense set of recurrent vectors.

Proposition 11 An operator ${T:X\rightarrow X}$ is recurrent if and only if the set of recurrent vectors for ${T}$ is dense in ${X}$. In this case the set of recurrent vectors for ${T}$ is a ${G_\delta}$ subset of ${X}$.

Proof: Let us first prove the easy implication. That is we assume that ${T}$ has a dense set of recurrent points and let ${U}$ be an open set in ${X}$. Since the recurrent points of ${T}$ are dense, there is a ${y\in U}$ which is recurrent for ${T}$. Take ${\epsilon>0}$ such that ${B:=B(y,\epsilon)\subset U}$. Since ${y}$ is recurrent, there is a ${k\in{\mathbb N}}$ such that ${\| T^ky-y\|<\epsilon}$. Thus ${T^ky\in B\subset U}$. That is we have that ${y\in U\cap T^{-k} (U)}$. Let us now assume that ${T}$ is recurrent. We fix an open ball ${B:=B(x,\epsilon)}$ for some ${x\in X}$ and ${\epsilon_1<1}$. We need to show that there is a recurrent vector in ${B}$. Since ${T}$ is recurrent there exists a positive integer ${k_1}$ such that ${x_1\in T^{-k_1}(B)\cap B}$, for some ${x_1\in X}$.That is we have that ${\|x_1-x\|<\epsilon}$ and ${\|T^kx_1-x\|<\epsilon}$. Since ${T}$ is continuous, there exists ${\epsilon_1<\frac{1}{2}}$ such that ${B_2:=B(x_1,\epsilon_1)\subset B}$ and ${B_2 \subset T^{-k_1}(B)\iff T^{k_1}(B_2)\subset B}$. Now since ${T}$ is recurrent, there is a ${k_2>k_1}$ such that ${x_2\in T^{-k_2}(B_2)\cap B_2}$ for some $x_2\in X$. By continuity again there is an ${\epsilon_2<\frac{1}{2^2}}$ such that ${B_3:=B(x_2,\epsilon_2)\subset B_2}$ and ${B_3\subset T^{-k_2}(B_2)\iff T^{k_2}(B_3) \subset B_2}$. Continuing inductively we construct a sequence ${x_n\in X}$, a strictly increasing sequence of positive integers ${k_n}$ and a sequence of positive real numbers ${\epsilon_n<\frac{1}{2^n}}$, such that

$\displaystyle B(x_n,\epsilon_n)\subset B(x_{n-1,\epsilon_{n-1}}), \quad T^{k_n}(B(x_n,\epsilon_n))\subset B(x_{n-1},\epsilon_{n-1}).$

Since ${X}$ is complete we conclude by Cantor’s theorem that

$\displaystyle \bigcap_n B(x_n,\epsilon_n)=\{y\},$

for some ${y\in X}$. We also have that ${\|T^{k_n} y -x_{n-1}\|<\epsilon_{n-1}}$, for all ${n}$. Thus we have that for every ${n\in{\mathbb N}}$, ${\|T^{k_n}y-y\|\leq 2\epsilon_{n-1}}$ which means that ${T^{n_k}y\rightarrow y}$ in ${X}$. That is, ${y}$ is a recurrent point in the original ball ${B}$.

Finally, let us write ${\textnormal{Rec(T)}}$ for the set of ${T}$-recurrent vectors. Observe that

$\displaystyle \textnormal{Rec}(T)=\bigcap_{s=1} ^{\infty} \bigcup_{n=0} ^{\infty}\bigg \{ x\in X:\|T^n x -x \| <\frac{1}{s} \bigg\},$

which shows that the set of ${T}$-recurrent vectors is a ${G_\delta}$-set. $\Box$

After (simple) recurrence, let’s now consider multiple recurrence. An operator ${T}$ is called topologically multiply recurrent if for every non-empty open set ${U\subset X}$ and every ${m\in{\mathbb N}}$ there is a ${k\in {\mathbb N}}$ such that

$\displaystyle U\cap T^{-k} U \cap \cdots\cap T^{-mk}U \neq \emptyset.$

Of course a hypercyclic operator is always recurrent. However, there is no reason why a hypercyclic operator should be topologically multiply recurrent in general. This is illustrated in the following proposition.

Proposition 12 (Costakis and Parissis, 2010) There exists a hypercyclic bilateral weighted shift on ${\ell^2({\mathbb Z})}$ which is not topologically multiply recurrent.

— 2.2. Frequent hypercyclicity and Szemerédi’s theorem —

Recently, Bayart and Grivaux introduced in (Bayart and Grivaux, 2005) and (Bayart and Grivaux, 2006) a notion that examines how frequently the orbit of a hypercyclic operator visits a non-empty open set.

Definition 13 An operator ${T:X\rightarrow X }$ is called frequently hypercyclic if there exists a vector ${x\in X}$ such that, for every non-empty open set ${U}$, the set

$\displaystyle \{n\in{\mathbb N}: T^nx\in U\},$

has positive lower density.

This is the strongest form of this definition, using the weakest’ density. There are variations where the lower density is replaced for example by the upper density. Recall that the lower density of a set ${B\subset {\mathbb N}}$ is defined as

$\displaystyle \underline{\textnormal{d}}(B):=\liminf_{N\rightarrow +\infty}\frac{|\{n\in B: n\leq N\}|}{N},$

while the upper density of ${B}$ is

$\displaystyle \overline{\textnormal{d}}(B):=\limsup_{N\rightarrow +\infty}\frac{|\{n\in B: n\leq N\}|}{N}.$

In (Bayart and Grivaux, 2006) a frequent hypercyclicity criterion’ was established. We won’t describe this here but point out one of its applications. Going back to adjoints of multiplication operators, an application of the Bayart-Grivaux frequent hypercyclicity criterion yields the following result:

Example 5 Recall that ${\mathcal H}$ is a non-trivial Hilbert space of holomorphic functions with bounded point evaluation functionals. We consider multiplier operators ${M_\phi}$ with symbol ${\phi\in H^\infty}$. We have the following result which is a corollary of the Bayart-Grivaux criterion

Proposition 14 (Bayart, Grivaux) Assume that ${\mathcal H}$ is a Hilbert space of holomorphic functions as above. Furthermore assume that every bounded holomorphic function is a multiplier of ${\mathcal H}$ such that ${\|M_\phi\|=\|\phi\|_\infty}$. The following are equivalent:

(i) The adjoint multiplication operator ${M^* _\phi}$ is hypercyclic.

(ii) The adjoint multiplication operator ${M^* _\phi}$ is frequently hypercyclic.

(iii) The function ${\phi}$ is non-constant and ${\phi(\mathbb D)\cap \mathbb T\neq \emptyset}$.

The notion of frequent hypercyclicity seems to be the right one in relevance to topological multiple recurrence. In order to illustrate this connection we need Szemerédi’s theorem on arithmetic progressions.

Theorem 15 (Szemerédi) Let ${A}$ be a subset of ${{\mathbb N}}$ with positive upper density. Then ${A}$ contains arbitrarily long arithmetic progressions.

The following proposition is just an easy application of Szemerédi’s theorem:

Proposition 16 Let ${T:X\rightarrow X}$ be a frequently hypercyclic operator. Then ${T}$ is topologically multiple recurrent.

Proof: Let ${U\subset X}$ be an open set and let ${m\in{\mathbb N}}$. Since ${T}$ is frequently hypercyclic, there exists a ${x\in X}$ such that the set

$\displaystyle A:=\{ n\in {\mathbb N}: T^n x\in U\}$

has positive lower density. By Szemerédi’s theorem, ${A}$ contains an arithmetic progression of length ${m}$, that is we have that

$\displaystyle a, a+k, a+2k,\ldots,a+mk\in A.$

This means that

$\displaystyle T^a x\in U\cap T^{-k} U \cap T^{-2k}U\cap\cdots \cap T^{-mk}U,$

that is, ${T}$ is topologically multiply recurrent. $\Box$

— 2.3. Frequently Cesàro hypercyclic operators —

As we have seen earlier, an operator ${T:X\rightarrow X}$ is Cesàro hypercyclic if and only if there exists a ${x\in X}$ such that the set

$\displaystyle \bigg\{ \frac{1}{n}T^nx:n\in {\mathbb N}\bigg\},$

is dense in ${X}$. In accordance to frequently hypercyclicity, Costakis and Ruzsa introduced in (Costakis and Ruzsa, 2010) the notion of a frequently Cesàro hypercyclic operator in the obvious way.

Definition 17 An operator ${T:X\rightarrow X}$ is called frequently Cesàro hypercyclic if there is a vector ${x\in X}$ such that, for every open set ${U\subset X}$, the set

$\displaystyle \bigg\{n\in{\mathbb N}: \frac{1}{n}T^nx\in U\bigg\},$

has positive lower density.

In contrast with Cesàro hypercyclic operators, frequently Cesàro hypercyclic operators are always hypercyclic:

Theorem 18 (Costakis and Ruzsa, 2010) Let ${T:X\rightarrow X}$ be a frequently Cesàro hypercyclic operator. Then ${T}$ is hypercyclic.

As in the case of frequently hypercyclic operators, frequently Cesàro hypercyclic operators are always topologically multiply recurrent. However, this is not so obvious any more.

Theorem 19 (Costakis and Parissis, 2010) Let ${T:X\rightarrow X}$ be a frequently Cesàro hypercyclic operator. Then ${T}$ is topologically multiply recurrent.

The hypothesis of the previous theorem is optimal in the sense that a Cesàro hypercyclic is not in general topologically multiply recurrent.

Proposition 20 (Costakis and Parissis, 2010) There exists a Cesàro hypercyclic bilateral weighted shift on ${\ell^2(\mathbb Z)}$ which is not recurrent, and hence not topologically multiply recurrent.

Before giving the actual proof of Theorem 19, let us try to repeat the simple argument used in the proof of Proposition 16. We begin by fixing a positive integer ${m}$ and an open set ${U}$. We will assume that ${U}$ is a ball, say ${U=B(x_o,\epsilon)}$. We need to show that there exists some vector ${y\in X}$ with

$\displaystyle y\in U\cap T^{-1}U\cap\cdots\cap T^{-mk}U\neq \emptyset,$

or, in other words, that there is a ${y\in U}$ such that

$\displaystyle T^ky,T^{2k}y,\ldots,T^{mk}y\in U.$

By the hypothesis and Szemerédi’s theorem there is a vector ${x\in X}$ and an arithmetic progression of length ${m}$

$\displaystyle a,a+k,\ldots,a+mk,$

such that

$\displaystyle \frac{1}{a}T^ax,\frac{1}{a+k}T^{a+k}x,\ldots,\frac{1}{a+mk}T^{a+mk}x\in U.$

In this case it is not obvious which is the natural candidate for the vector ${y}$ but let’s take ${y:=\frac{1}{a}T^ax\in U}$. We then have for ${1\leq j\leq m}$

$\displaystyle T^{jk}y=\frac{1}{a} T^{a+jk}x = \frac{a+jk}{a} \frac{1}{a+jk}T^{a+jk}x:=\frac{a+jk}{a} w_j,$

where we know that all the ${w_j}$‘s are in ${U}$. We can then naively estimate

$\displaystyle \begin{array}{rcl} \|T^{jk}y- x_o\|& = & \bigg\| \frac{a+jk}{a} w_j-x_o\bigg\|\leq \bigg\| \frac{a+jk}{a} w_j-w_j\bigg\|+\|w_j -x_o\| \\ \\ &\leq& \bigg|\frac{a}{a+jk}-1\bigg| \|w_j\|+\epsilon \leq \frac{mk}{a}(\|x_o\|+\epsilon)+\epsilon. \end{array}$

There are two problems here. The first is that we cannot control the factor ${\frac{mk}{a}(\|x_o\|+\epsilon)}$. The second is that even if we could, say we had ${\frac{mk}{a}(\|x_o\|+\epsilon)<\epsilon}$, this estimate would give us that ${\|T^{jk}y-x_o\|<2\epsilon}$ which is one ${\epsilon}$ too large. The second problem is easy to deal with. We just start with a smaller ball inside our original set and carry out this reasoning for the smaller ball. In the proof given below we will consider two cases. In the first we will just assume that ${\frac{mk}{a}(\|x_o\|+\epsilon)}$ is small. In the complementary case, we will appropriately use the information that ${\frac{mk}{a}(\|x_o\|+\epsilon)}$ is large!

Proof of Theorem 19: Let ${U}$ be any non-empty open set in ${X}$. We fix a non-zero vector ${y\in U}$ and take a positive number ${\epsilon <1}$ such that ${B(y,\epsilon )\subset U}$. Without loss of generality we may assume that ${\| y\| > \epsilon}$. Consider the ball ${B(y,\delta)}$ with

$\displaystyle \delta =\frac{\epsilon^2}{100\| y\|} .$

Observe that ${B(y,\delta )\subset B(y,\epsilon)}$. Since ${T}$ is a frequently Ces\{a}ro hypercyclic operator there exists ${x\in X}$ such that the set

$\displaystyle A=\bigg\{ n\in\mathbb{N}: \frac{T^n}{n}x \in B(y,\delta ) \bigg\},$

has positive lower density. By Szemerédi’s theorem the set ${A}$ contains an arithmetic progression of length ${2m+1}$, i.e. there exist positive integers ${a,k}$ such that

$\displaystyle a, a+k, \ldots , a+2mk\in A.$

Therefore the vectors

$\displaystyle \begin{array}{rcl} v&:=&\frac{T^a}{a}x, \quad v_j:=\frac{T^{a+jk}}{a+jk}x=\frac{a}{a+jk}T^{jk}v, \\ \\ w&:=&\frac{T^{a+mk}}{a+mk}x, \quad w_j:=\frac{T^{a+mk+jk}}{a+mk+jk}x=\frac{a+mk}{a+mk+jk}T^{jk}w, \end{array}$

${j=1,\ldots ,m,}$ belong to ${B(y,\delta )}$.

As promised, we will consider two cases depending on the values of the ratio ${k/a}$ of the step over the first term of the arithmetic progression provided by Szemerédi’s theorem:

Case 1. ${\frac{k}{a}\leq \frac{\epsilon}{2(\epsilon +\| y\|)m}}$.

We define the vector ${u}$ as

$\displaystyle u:=v\in B(y,\epsilon ).$

Then we have

$\displaystyle \begin{array}{rcl} \| T^{kj}u-v_j \| &=&\bigg\| \frac{a+jk}{a}v_j-v_j\bigg\|=\bigg| 1-\frac{a+jk}{a} \bigg| \| v_j\| \\ \\ &\leq& \frac{mk}{a}\| v_j\| < \frac{mk(\epsilon+\| y\|) }{a} \leq \frac{\epsilon}{2}, \end{array}$

for every ${j=1,\ldots ,m}$. Since

$\displaystyle \| v_j-y\| <\delta=\frac{\epsilon^2}{100\| y\|}<\frac{\epsilon}{100},$

we conclude that

$\displaystyle \| T^{kj}u-y\|<\epsilon$

and therefore

$\displaystyle T^{jk} u \in U \quad \textrm{for every}\,\, j=1,2,\ldots ,m,$

as we wanted to show.

Case 2. ${\frac{k}{a}>\frac{\epsilon}{2(\epsilon +\| y\|)m}}$.

Here we first need to specify a number ${\eta \in \mathbb{R}}$ such that

$\displaystyle 1=\eta \frac{a+jk}{a}+(1-\eta)\frac{a+mk+jk}{a+mk}$

for every ${j=1,\ldots ,m}$. Indeed, solving the above equation for ${\eta}$ we get

$\displaystyle \eta=-\frac{a}{mk}.$

We now define the vector ${u}$ as

$\displaystyle u:=\eta v+(1-\eta)w.$

Then we have

$\displaystyle \begin{array}{rcl} \| u-y\| &\leq& |\eta |\| v-y\| +|1-\eta| \| w-y\| \leq (1+2|\eta |)\delta \\ \\ &=&\bigg(1+2\frac{a}{mk}\bigg)\frac{\epsilon^2}{100\| y\|} < \bigg(1+\frac{4(\epsilon +\| y\| )}{\epsilon} \bigg)\frac{\epsilon^2}{100\| y\| }\\ \\ & =&\frac{\epsilon^2}{100\| y\| }+\frac{4\epsilon^2}{100\| y\| }+\frac{4\epsilon}{100}<\frac{\epsilon}{100}+\frac{4\epsilon}{100}+\frac{4\epsilon}{100}<\epsilon , \end{array}$

that is ${u\in U}$. On the other hand,

$\displaystyle \begin{array}{rcl} T^{kj}u-v_j&=&\eta T^{kj}v+(1-\eta )T^{kj}w-v_j\\ \\ &=&\eta \frac{a+jk}{a}v_j+(1-\eta)\frac{a+mk+jk}{a+mk}w_j\\ \\&&\quad - \eta \frac{a+jk}{a}v_j-(1-\eta )\frac{a+mk+jk}{a+mk}v_j \\ \\ &=& (1-\eta )\frac{a+mk+jk}{a+mk}(w_j-v_j), \end{array}$

for every ${j=1,\ldots ,m}$. The last equality and the above estimates imply

$\displaystyle \begin{array}{rcl} \|T^{kj}u-v_j\| &\leq &(1+|\eta |)\frac{a+2mk}{a+mk}\| w_j-v_j\| \leq (1+\frac{a}{mk})2(2\delta )\\ \\ &\leq&\bigg(1+\frac{2(\epsilon +\| y\| )}{\epsilon } \bigg)\frac{4\epsilon^2}{100\| y\| } \\ \\ &=&\frac{4\epsilon^2}{100\| y\|} +\frac{8\epsilon^2}{100\| y\| }+\frac{8\epsilon }{100}<\frac{\epsilon }{2}, \end{array}$

for every ${j=1,\ldots ,m}$. Let ${j\in \{ 1,\ldots ,m\}}$. Since

$\displaystyle \| v_j-y\| <\frac{\epsilon}{100}$

we conclude that

$\displaystyle T^{kj}u\in B(y,\epsilon ).$

Therefore

$\displaystyle T^{jk} u \in U \quad \textrm{for every} \,\, j=1,2,\ldots ,m.$

This completes the proof of the theorem. $\Box$

— 3. Back to adjoints of multiplication operators. —

We can now give a full characterization of frequent hypercyclicity and multiple recurrence in the case of adjoints of multiplication operators on a non-trivial Hilbert space of holomorphic functions. It turns out that the weaker property of ${M_\phi ^*}$ being recurrent is equivalent to frequent hypercyclicity and thus to every other property we have discussed here.

Proposition 21 (Costakis and Parissis, 2010) Assume that ${\mathcal H}$ is a Hilbert space of holomorphic functions as above. Furthermore assume that every bounded holomorphic function is a multiplier of ${\mathcal H}$ such that ${\|M_\phi\|=\|\phi\|_\infty}$. The following are equivalent:

(i) ${M_\phi ^*}$ is recurrent.

(ii) The adjoint multiplication operator ${M^* _\phi}$ is hypercyclic.

(iii) The adjoint multiplication operator ${M^* _\phi}$ is frequently hypercyclic.

(iv) The adjoint multiplication operator ${M^* _\phi}$ is topologically multiply recurrent.

(v) The function ${\phi}$ is non-constant and ${\phi(\mathbb D)\cap \mathbb T\neq \emptyset}$.

Proof: We have already seen in Theorem 8 and Proposition 14 that conditions (ii), (iii) and (v) are equivalent. Also, by Proposition 16, (iii) implies (iv) and obviously (iv) implies (i). So the proof will be complete if we show for example that (i) implies (v).

Indeed, assume that ${M_{\phi }^*}$ is recurrent. Suppose, for the sake of contradiction, that ${\phi (\mathbb D )\cap \mathbb{T}= \emptyset }$. Since ${\mathbb D}$ is connected, so is ${\phi (\mathbb D )}$; thus, we either have that ${\phi (\mathbb D )\subset \{ z\in \mathbb{C}: |z|<1 \}}$ or ${\phi (\mathbb D )\subset \{ z\in \mathbb{C}: |z|>1 \}}$.

Case 1. ${\phi (\mathbb D )\subset \{ z\in \mathbb{C}: |z|<1 \}}$.

Then we have ${\| M_{\phi }^* \|=\| M_{\phi } \| =\| \phi \|_{\infty} \leq 1}$. We will consider two complementary cases. Assume that there exist ${0<\epsilon<1}$ and a recurrent vector ${g}$ for ${M_{\phi }^*}$ such that

$\displaystyle \| M_{\phi }^*g\| \leq (1-\epsilon)\| g\| .$

The above inequality and the fact that ${\| M_{\phi }^* \|=1}$ imply that for every positive integer ${n}$

$\displaystyle \| (M_{\phi }^*)^ng\| \leq (1-\epsilon)\| g\| .$

On the other hand for some strictly increasing sequence of positive integers ${\{ n_k\}}$ we have ${(M_{\phi }^*)^{n_k}g\rightarrow g}$. Using the last inequality we arrive at ${\| g\| \leq (1-\epsilon )\| g\|}$, a contradiction. In the complementary case we must have ${\| M_{\phi }^*g \| \geq \| g\| }$ for every vector ${g}$ which is recurrent for ${M_{\phi }^*}$. Since the set of recurrent vectors for ${M_{\phi }^*}$ is dense in ${H}$ we get that ${\| M_{\phi }^*h \| \geq \| h\| }$ for every ${h\in H}$. Hence ${\| M_{\phi }^*h \| = \| h\| }$ for every ${h\in H}$. Take now ${z\in \Omega }$ and consider the reproducing kernel ${k_z}$ of ${H}$. We have already seen in the proof of Theorem 8 that ${M_\phi ^*(k_z)=\overline{\phi(z)}k_z}$ where ${k_z}$ is the reproducing kernel at ${z}$. We conclude that

$\displaystyle \| M_{\phi }^*k_z \| =|\phi (z)|\| k_z\|< \| k_z\| .$

However, this is clearly impossible since ${M_{\phi }^*}$ is an isometry.

Case 2. ${\phi (\mathbb D )\subset \{ z\in \mathbb{C}: |z|>1 \}}$.

Here ${1/\phi}$ is a bounded holomorphic function satisfying ${\| 1/\phi \|_{\infty }\leq 1}$; therefore, ${M_{\phi}^*}$ is invertible. It is easy to see that if an operator ${T}$ is invertible, then ${T}$ is recurrent if and only if ${T^{-1}}$ is recurrent. Thus the operator ${M_{1/\phi}^{*}=(M_{\phi}^*)^{-1}}$ is recurrent and the proof follows by Case 1. $\Box$

Remark 22 It is easy to see that under the hypotheses of Proposition 21, ${M_{\phi }}$ is never recurrent. On the other hand, suppose that ${\phi}$ is a constant function with ${\phi(z)=a}$ for some ${a\in\mathbb C}$ and every ${z\in \Omega}$. Then we have that ${M_{\phi }}$ (or equivalently ${M_{\phi }^*}$) is recurrent if and only if ${M_{\phi }}$ is topologically multiply recurrent if and only if ${|a|=1}$. In order to prove this it is enough to notice that for every non-zero complex number ${a}$, with ${|a|=1}$, and every positive integer ${m}$, there exists an increasing sequence of positive integers ${\{n_k\}}$ such that ${(a^{n_k},a^{2n_k},\ldots,a^{mn_k})\rightarrow (1,1,\ldots,1).}$

— 4. Some open questions —

I will close this post by suggesting a couple of open problems. For more information you can check the actual paper.

— 4.1. Multipliers on the Dirichlet space. —

First of all, let me come back to the adjoints of multiplication operators. Recall that the Dirichlet space ${\textnormal{Dir}(\mathbb D)}$ is defined as the space  of holomorphic functions ${f:\mathbb D\rightarrow {\mathbb C}}$ such that

$\displaystyle \|f\|^2 _{\textnormal Dir}:=|f(0)|^2+ \int_{\mathbb D}|f'(z)|^2 dA(z)<+\infty.$

The reader might have noticed that throughout the discussion here, I have assumed that the multipliers of the Hilbert space ${\mathcal{H}}$ are exactly the bounded holomorphic functions and that ${\| M_\phi \|=\|\phi \| _\infty}$. Although this is actually the case on the Hardy space ${\mathbb H^2(\mathbb D)}$ or the Bergman space ${A^2(\mathbb D)}$, things are quite different on the Dirichlet space defined before. On the Dirichlet space, not all bounded holomorphic functions are multipliers. In fact the characterization of multipliers on the Dirichlet space is a bit more technical and is due to Stegenga (Stegenga 1980):

Theorem 23 (Stegenga) The function ${\phi}$ is a multiplier for the Dirichlet space ${\textnormal{Dir}(\mathbb D)}$ if and only if ${\phi \in H^\infty(\mathbb D)}$ and the measure ${d\mu_\phi(z)=|\phi'(z)|^2dA(z)}$ is a Carleson measure for the Dirichlet space ${\textnormal{Dir}(\mathbb D)}$.

Of course this theorem doesn’t tell us much if we can’t understand which are the Carleson measures for the Dirichlet space. Here I will just give the definition as the characterization of these measures is completely beyond the scope of this post.

Definition 24 A positive Borel measure ${\mu}$ on ${{ \overline{\mathbb D}}}$ is a Carleson measure for the Dirichlet space if for some positive constant ${c>0}$

$\displaystyle \int_ {\overline { \mathbb D}} |f|^2 d\mu \leq c \| f\|^2 _{\textnormal {Dir}},$

for every ${f\in \textnormal{Dir}(\mathbb D)}$.

Due to the more involved characterization of the multipliers on the Dirichlet space, characterizing when adjoints of multiplication operators on ${\textnormal {Dir}(\mathbb D)}$ are hypercyclic is an open question. It is however known that the condition ${\phi(\mathbb D)\cap \mathbb T\neq \emptyset}$ is no longer necessary, though it is sufficient. An example is provided by the function ${\phi(z)=z}$ on ${\mathbb D}$. On the other hand it is known that $\overline{\phi(\mathbb D)}\cap \mathbb T\neq \emptyset$ is necessary. For this, see for example the PhD thesis of Irina Seceleanu.

— 4.2. Frequently universal sequences of operators. —

Remember that a family of operators ${(T_k)_{k\in{\mathbb N}}}$ on ${X}$ is called universal if there exists a ${x\in X}$ such that the set

$\displaystyle \{T_k x:k\in{\mathbb N}\},$

is dense in ${X}$. The following definition is the natural extension of frequent hypercyclicity to universal families

Definition 25 The family of operators ${(T_k)_{k\in{\mathbb N}}}$ is called frequently universal if there exists a ${x\in X}$ such that for every open set ${U\subset X}$ the set

$\displaystyle \{ k\in N: T_k x\in U\},$

has positive lower density.

Thus saying that an operator ${T}$ is frequently Cesàro hypercyclic amounts to saying that the family ${(\frac{1}{n}T^n)_{n\in N}}$ is frequently universal. Theorem 19 says that if the family ${(\frac{1}{n}T^n)_{n\in N}}$ is frequently universal then ${T}$ is topologically multiply recurrent. However, there is nothing too special about the sequence ${\frac{1}{n}}$. One can consider the family of operators ${\lambda_n T^n}$ where ${\lambda_n}$ is an appropriate sequence of complex numbers.

Under what condition on the sequence of complex numbers ${(\lambda_n)_{n\in{\mathbb N}}}$ one may conclude that ${T}$ is topologically multiply recurrent from the hypothesis that the family ${(\lambda_nT^n)_{n\in N}}$ is frequently universal?

— 5. Bibliography —

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