A QUANTITATIVE INTERPRETATION OF THE FREQUENT HYPERCYCLICITY CRITERION

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A QUANTITATIVE INTERPRETATION OF THE FREQUENT HYPERCYCLICITY CRITERION

Romuald Ernst, A Mouze

To cite this version:

Romuald Ernst, A Mouze. A QUANTITATIVE INTERPRETATION OF THE FREQUENT HY- PERCYCLICITY CRITERION. 2016. �hal-01395490v2�

HYPERCYCLICITY CRITERION

R. ERNST, A. MOUZE

Abstract. We give a quantitative interpretation of the Frequent Hypercyclicity Criterion.

Actually we show that an operator which satisfies the Frequent Hypercyclicity Criterion is necessarilyA-frequently hypercyclic, whereArefers to some weighted densities sharper than the natural lower density. In that order, we exhibit different scales of weighted densities that are of interest to quantify the “frequency” measured by the Frequent Hypercyclicity Criterion. Moreover we construct an example of unilateral weighted shift which is frequently hypercyclic but notA-frequently hypercyclic on a particular scale.

1. Introduction

The notion of frequent hypercyclicity was introduced in the context of linear dynamics by Bayart and Grivaux in 2006 [1], [2]. This latter is now a central notion in that field and is highly connected to combinatorics, number theory and ergodic theory. Let Xbe a metrizable and complete topological vector space andL(X) be the space of continuous linear operators on X. An operatorT ∈L(X) is said to behypercyclic if there exists x∈X such that for any non-empty open set U ⊂X, the return set {n≥0 :Tⁿx ∈U} is non-empty or equivalently infinite. Such a vector x is called a hypercyclic vector for T. Furthermore an operator T is called frequently hypercyclic if there exists x ∈ X such that for any non-empty open set U ⊂X, the set of integersnsatisfying Tⁿx∈U has positive lower density, i.e.

lim inf

N→+∞

#{k≤N :T^kx∈U}

N >0,

where as usual # denotes the cardinality of the corresponding set. Thus the notion of frequent hypercyclicity extends the classical hypercyclicity and appraises how often the orbit of a hypercyclic vector visits every non-empty open set. In the sequel we denote by Nthe set of positive integers and for any x∈X and any subsetU ⊂Xwe setN(x, U) :={n∈N:Tⁿx∈ U}.Given a subsetE⊂N,we define itslower and upper densities respectively by

d(E) = lim inf

N→+∞

#{k≤N :k∈E}

N and d(E) = lim sup

N→+∞

#{k≤N :k∈E}

N .

In other words, an operator T ∈ L(X) is hypercyclic (resp. frequently hypercyclic) if there exists x ∈ X such that for any non-empty open set U ⊂ X, the set N(x, U) is non-empty (resp. has positive lower density). To prove that an operator is hypercyclic we have at our disposal the so-called Hypercyclicity Criterion (see [4], [10] and the references therein). In the same spirit, Bayart and Grivaux stated the Frequent Hypercyclicity Criterion, which ensures that an operator is frequently hypercyclic [2]. Let us recall it here.

2010Mathematics Subject Classification. 47A16, 37B50.

Key words and phrases. frequently hypercyclic operators, weighted densities.

1

Theorem 1.1. LetT be an operator on a separable Fr´echet spaceX.If there is a dense subset X0 of X and a map S:X0 →X0 such that, for any x∈X0,

(i)

+∞

X

n=0

Tⁿx converges unconditionally,

(ii)

+∞

X

n=0

Sⁿx converges unconditionally, (iii) T Sx=x,

thenT is frequently hypercyclic.

We already know that the above result does not characterize frequently hypercyclic oper- ators. Indeed Bayart and Grivaux have exhibited a frequently hypercyclic weighted shift on c₀ that does not satisfy this criterion [3]. A natural question arises: what does the Frequent Hypercyclicity Criterion really quantify? In order to answer this question, B`es, Menet, Peris and Puig recently generalized the notion of hypercyclic operators by introducing the concept of A-frequent hypercyclicity, where A refers to a family of subsets of N satisfying suitable conditions [8]. In particular,A has to satisfy the following separation condition:

A contains a sequence (A_k) of disjoint sets such that for anyj∈A_k, any j⁰ ∈A_k⁰,j6=j⁰, we have |j⁰−j| ≥max(k, k⁰).

In this abstract framework, they also obtain an A-Frequent Hypercyclicity Criterion and prove that the Frequent Hypercyclicity Criterion has very strong consequences in the sense that ifT satisfies the Frequent Hypercyclicity Criterion, then T also satisfies theA-Frequent Hypercyclicity Criterion for any suitable familyA. Bonilla and Grosse-Erdmann also studied specific notions related to the concept ofA-frequent hypercyclicity in the article [7].

On the other hand, the notion of frequent hypercyclicity measures the frequency and the length of the intervals when iterates of a hypercyclic vector visits every non-empty open set in a very specific way, that is given by the natural density. Actually, there are many types and notions of densities different from the natural one. Our goal is to give quantified consequences to the Frequent Hypercyclicty Criterion in terms of weighted densities. To that purpose, we consider a special kind of lower weighted densities, generalizing the natural one but sharper than this one, by using the formalism of matrix summability methods. For such a matrix A, we use the concept of A-density and A-frequent hypercyclicity (see Definitions 2.2 and 2.11 below). These general densities were already used in the context of linear dynamics to study frequently universal series [11]. In the present paper, we show that the Frequent Hypercyclicity Criterion gives a stronger conclusion than frequent hypercyclicity, that we quantify thanks to explicit weighted densities on different scales. We refer the reader to Proposition 3.5 and Theorem 4.12 below. Therefore an operator which satisfies the Frequent Hypercyclicity Criterion is necessarily A-frequently hypercyclic, where A refers to some weighted densities sharper than the natural lower density.

Let us return to the formalism of A-frequent hypercyclicity. For instance A could be a family of subsets with positive given lower weighted density satisfying the aforementioned separation property. However from [8], there is an underlying question: does there exist a frequently hypercyclic operator not beingA-frequently hypercyclic? We give a positive an- swer by constructing an unilateral weighted shift on c0 which is frequently hypercyclic but notA-frequently hypercyclic with respect to some A-densities covered by the criterion (see

Theorem 5.4 below).

The paper is organized as follows: in Section 2 we introduce some densities that will be of interest in the sequel and some properties on these densities. Section 3 is devoted to an improvement of the Frequent Hypercyclicity Criterion for a certain scale of weighted densities.

In Section 4, we modify this proof in order to obtain a stronger result to the criterion. Finally in Section 5, we exhibit a new example, inspired by [5], of an operator which is frequently hypercyclic although it does not satisfy the Frequent Hypercyclicity Criterion. To ensure this latter property, we will show that this operator is notA-frequently hypercyclic for some suitable matrixA.

2. Densities: preliminary results

In this section, we state some definitions and results we shall need throughout the paper.

Let us first introduce the concept of summability matrix and its connections with some kind of densities on subsets ofN.

Definition 2.1. Asummability matrixis an infinite matrixM = (mn,k) of complex numbers.

Let us recall that, if (x_n) is a sequence and M = (m_n,k) is a summability matrix, then by M xwe denote the sequence ((M x)₁,(M x)₂, . . .) where (M x)_n =P+∞

k=1m_n,kx_k.The matrix M is called regular if the convergence of x to c implies the convergence of M x to c. By a well-known result of Toeplitz (see for instance [12]),M is regular if and only if the following three conditions hold:

(2.1)











(i) lim

n→+∞m_n,k= 0, for allk∈N, (ii) lim

n→+∞

X

k≥1

m_n,k = 1, (iii) sup_nP

k≥1|m_n,k|<∞.

Freedman and Sember showed that every regular summability matrixM with non-negative real coefficients defines a densityd_M on subsets ofN, called lowerM-density [9].

Definition 2.2. For a regular matrix M = (m_n,k) with non-negative coefficients and a set E ⊂N,the lowerM-density ofE,denoted d_M(E),is defined by

d_M(E) = lim inf

n→+∞

+∞

X

k=1

m_n,k1E(k),

and the associated upper M-density, denoted by dM(E),is defined by dM(E) = 1−d_M(N\E).

Remark 2.3. For a non-negative regular matrix M = (m_n,k), Proposition 3.1 of [9] ensures that the upper M-density of any setE⊂N is given by

d_M(E) = lim sup

n→+∞

+∞

X

k=1

m_n,k1E(k).

Let (α_k)k≥1 be a non-negative sequence such thatPn

k=1α_k →+∞ asntends to∞.Then, we deal with the special case of A-density where we write A =

α_k/Pn j=1α_j

, when A = (α_n,k) withα_n,k=α_k/Pn

j=1αj for 1≤k≤nandα_n,k = 0 fork > n.It is easy to check that

A is a non-negative regular summability matrix. In summability theory the transformation given byx= (xn)7→Axis called the Riesz mean (A, αn).Here the associatedA-density can be viewed as a weighted density with respect to the non-negative weight sequence (α_k)k≥1. Definition 2.4. A summability matrix A =

α_k/Pn j=1αj

as above will be called anad- missible matrix. We define its summatory function ϕ_α as follows: ϕ_α : N → R+, ϕ_α(n) = P

k≤nα_k.

Example 2.5. (1) If α_k = 1, k = 1,2, . . . , then the summability matrix A is the well- known Ces`aro matrix and d_A is the natural lower density.

(2) Ifα_k = 1/k, k= 1,2, . . . , d_Ais the so-called lower logarithmic density, which is derived from the well-known logarithmic summability method. We haveϕ_α(k)∼log(k),ask tends to +∞.

(3) The special case α_k = k^r, r ≥ −1, for k = 1,2, . . . , generalizes both the natural density (r= 0) and the logarithmic density (r =−1). Clearly we have ϕ_α(k)∼ ^k_r+1^r+1, ask tends to +∞,when r >−1.

(4) Ifα_k=e^kr,0< r <1,fork= 1,2, . . . ,an easy calculation gives ϕ_α(k)∼ ^k1−r_r e^kr,as ktends to +∞.

(5) Ifα_k=e^k, fork= 1,2, . . . ,then ϕα(k)∼ _e−1^e e^k,asktends to +∞.

(6) If α₁ = 1 and α_k = e^k/logr(k), r > 0, for k = 2,3, . . . , a summation by parts gives ϕα(k)∼log^r(k)e^k/logr(k),ask tends to +∞.

(7) Leths be the real function defined byhs= log.log^(s),with log^(s)= log◦log· · · ◦log, where log appears stimes. If α_k=e^k/hs(k), l∈N, s≥2,fork large enough, again a summation by parts givesϕα(k)∼hs(k)e^k/hs(k),ask tends to +∞.

In the following sections, we shall use the following definitions connected to Example 2.5.

Definition 2.6. We denote by

(1) C_r the admissible matrix C_r =

k^r/P_n

j=1j^r

, r ≥ −1;

(2) Ar the admissible matrixAr=

e^kr/Pn j=1e^jr

, r≥0;

(3) B_r the admissible matrix B_r = (α_k/P_n

j=1α_j) with α₁ = 1 and α_k = e^k/logr(k), for k≥2, r≥0;

(4) Let h_s be the real function defined by h_s = log.log^(s), with log^(s) = log◦log· · · ◦ log, where log appears s times. We denote by Bes the admissible matrix Bes = (α_k/Pn

j=1α_j) withα_k=e^k/hs(k),fork large enough ands≥2.

For any subset E ⊂ N, we can write E as a strictly increasing sequence (n_k) of positive integers. It is well-known thatd(E) = lim infk→+∞ k

nk which allows to deduce the following simple fact: d(E)>0 if and only if the sequence ⁿ_k^k

is bounded [10]. The following lemma extends this remark to suitableA-densities.

Lemma 2.7. Let (α_k) be a non-negative sequence such that P

k∈Nα_k = +∞. Assume that the sequence (αn/Pn

j=1αj) converges to zero as n tends to +∞. Let (nk) be an increasing sequence of integers forming a subsetE ⊂N.Then, we have

d_A(E) = lim inf

k→+∞

P_k

j=1α_nj Pnk

j=1αj

! ,

where d_A is the A-density given by the summability matrix A= (α_k/Pn j=1α_j).

Proof. Let us considern_k≤N < n_k+1,then Pk+1

j=1αnj

Pnk+1

j=1 αj

− α_nk+1 Pnk+1

j=1 αj

= Pk

j=1αnj

Pnk+1

j=1 αj

≤ P

nj≤Nαnj

PN

j=1α_j ≤ Pk

j=1αnj

Pnk

j=1αj

.

Thus, we deduce

d_A(E) = lim inf

N→+∞

P

nj≤Nα_nj PN

j=1αj

!

= lim inf

k→+∞

Pk j=1α_nj Pn_k

j=1α_j

! .

In the present paper, we are mainly interested in sharper A-densities than the classical natural density. From this point of view, the following lemma gives some conditions to ensure that the sequence (α_k) leads to a sharper density.

Lemma 2.8. Let (α_k) and (β_k) be non-negative sequences such that P

k∈Nα_k=P

k∈Nβ_k= +∞.Assume that the sequence(α_k/β_k)is eventually decreasing to zero. LetA= (α_k/Pn

j=1αj) and B = (β_k/Pn

j=1β_j) be the associated admissible matrices. Then, for every subset E⊂N, we have

d_B(E)≤d_A(E)≤dA(E)≤dB(E).

Proof. Let E be a subset of N.For every n≥1,let us define Λ^α_E(n) = Pn

k=1αk1E(k) (resp.

Λ^β_E(n) =Pn

k=1β_k1E(k)). In particular, one may observe that Λ^α

N =ϕ_α. Now, let N ≥1 be an integer such that the sequence (αk/βk)k≥N is decreasing. Then for every n≥N + 1,we have

n

X

k=N+1

α_k1E(k) =

n−1

X

k=N+1

Λ^β_E(k) α_k

β_k −α_k+1 β_k+1

+ Λ^β_E(n)αn

β_n −Λ^β_E(N)α_N+1 β_N+1

=

n−1

X

k=N+1

Λ^β_E(k) ϕ_β(k)ϕβ(k)

αk

β_k −αk+1

β_k+1

+Λ^β_E(n)

ϕ_β(n)ϕβ(n)αn

βn

−Λ^β_E(N)αN+1

β_N+1. Moreover, since (α_k/β_k) is a non-negative decreasing sequence andP

α_k = +∞,we deduce dA(E) = lim sup

n→+∞

"

(ϕα(n))⁻¹

N

X

k=1

αk1E(k) +

n

X

k=N+1

αk1E(k)

!#

= lim sup

n→+∞

"

(ϕ_α(n))⁻¹

n−1

X

k=N+1

Λ^β_E(k) ϕβ(k)ϕ_β(k)

α_k βk

−α_k+1 βk+1

+Λ^β_E(n)

ϕβ(n)ϕ_β(n)α_n βn

!#

≤ sup

k>N

Λ^β_E(k) ϕ_β(k)

!

lim sup

n→+∞

"

(ϕ_α(n))⁻¹

n−1

X

k=N+1

ϕ_β(k) α_k

β_k −α_k+1 β_k+1

+ϕ_β(n)α_n β_n

!#

.

Since Pn−1

k=N+1ϕ_β(k)

αk

βk −^α_β^k+1

k+1

+ϕ_β(n)^α_βⁿ

n =ϕ_β(N + 1)^α_β^N+1

N+1 +Pn

k=N+2α_k,we get dA(E)≤ sup

k>N

Λ^β_E(k) ϕ_β(k)

! .

Hence lettingN →+∞,we obtaind_A(E)≤d_B(E).

The other inequality is obtained using the relations d_A(E) = 1−dA(N\E) and d_B(E) =

1−dB(N\E).

From now on, we are interested in densities given by special admissible matrices given in Definition 2.6. In this case, Lemma 2.8 leads to the following inequalities.

Lemma 2.9. For every subset E ⊂N and for any 0 < r ≤r⁰, 0 < s ≤s⁰ <1, 1 < t ≤t⁰, 2≤l≤l⁰,we have

d_A1(E) =d_B0(E)≤d

Be_l0(E)≤d

Bel(E)≤d_Bt(E)≤d_B

t0(E) and

d_B

t0(E)≤d_A

s0(E)≤d_As(E)≤d_C

r0(E)≤d_Cr(E)≤d(E).

Moreover, observe that a subset E of Npossesses a strictly positive natural lower density if and only if it has a strictly positive lowerC_r-density, for anyr >−1.

Lemma 2.10. Let r > −1. Then, for every subset E ⊂ N the following assertions are equivalent

(i) d_Cr(E)>0, (ii) d(E)>0.

Proof. Let (n_k)⊂Nbe the increasing sequence of elements ofE. We divide the proof in two cases.

Case r ≥ 0: Lemma 2.8 gives d_Cr(E) ≤ d(E), hence (i) ⇒ (ii). For the other implication, assume thatd(E)>0.This means that the sequence_n

j

j

is bounded (see Lemma 2.7). We deduce that there exists an integerM ≥1 such that for everyk∈N,k≤nk≤M k and

n_k

X

j=1

j^r≤

M k

X

j=1

j^r ≤ (M k+ 1)^r+1

r+ 1 ≤M^r+1(k+ 1)^r+1 r+ 1 . Using the fact thatPk

j=1j^r ∼ ^k_r+1^r+1 ∼ ^(k+1)_r+1^r+1,asktends to +∞(cf Example 2.5) we deduce that there exists C >0 such that

nk

X

j=1

j^r≤CM^r+1

k

X

j=1

j^r. Therefore we have

nk

X

j=1

j^r≤CM^r+1

k

X

j=1

(nj)^r and lim infk→+∞

_Pk

j=1(nj)^r P_nk

j=1j^r

>0 which is sufficient to conclude thanks to Lemma 2.7.

Case−1< r <0: As in the previous case, Lemma 2.8 gives the implication (ii)⇒(i) because d_Cr(E) ≥ d(E). For the converse, assume thatd_Cr(E) > 0. According to Lemma 2.7, there existsC >0 such that

n_k

X

j=1

j^r≤C

k

X

j=1

(n_j)^r.

Using the inequalityj≤n_j and since r <0, we get (n_k+ 1)^r+1

r+ 1 − 1 r+ 1 ≤

nk

X

j=1

j^r ≤C

k

X

j=1

(n_j)^r ≤C

k

X

j=1

j^r≤C

k^r+1 r+ 1− 1

r+ 1+ 1

.

The positivity ofr+ 1 now ensures that the sequence ⁿ_k^k

is bounded andd(E)>0.

Finally let us also extend the definition of frequently hypercyclic operators using the general notion of A-densities introduced before.

Definition 2.11. Let A =

α_k/Pn j=1α_j

be an admissible matrix. Using the notations of Section 1, an operatorT ∈L(X) is said to beA-frequently hypercyclic if there existsx ∈X such that for any non-empty open setU ⊂X,the set N(x, U) has positive lowerA-density.

3. Frequent hypercyclicity criterion: classical construction

According to Lemma 2.10, a frequently hypercyclic operator is necessarily C_r-frequently hypercyclic for anyr >−1.So, even if this is obvious, the Frequent Hypercyclicity Criterion allows to obtain theB_r-frequent hypercyclicity too. A careful examination of the well-known proof of this criterion leads to a more precise result. Indeed, in the classical proof of the Frequent Hypercyclicity Criterion the following constructive lemma plays a prominent role [10, Lemma 9.5].

Lemma 3.1. There exist pairwise disjoint subsets A(l, ν), l, ν ≥ 1, of N of positive lower density such that, for any n∈A(l, ν) and m∈A(k, µ), we have that n≥ν and

|n−m| ≥ν+µ if n6=m.

The proof of this result is based on a specific partition of Nusing the dyadic representation n = P+∞

j=0a_j2^j = (a₀, a₁, . . .) of any positive integer. Actually the authors define the sets I(l, ν), l, ν ≥ 1, as the sets of all n ∈ N whose dyadic representation has the form n = (0, . . . ,0,1, . . . ,1,0,∗) withl−1 leading zeros, exactly followed byν ones, then one zero and an arbitrary remainder. Let δ_k =ν, ifk ∈ I(l, ν) for some l ≥ 1. Then they construct the following strictly increasing sequence (n_k) of positive integers by setting

n_k= 2

k−1

X

i=1

δ_i+δ_k, k≥1.

This construction clearly ensures that for any integersi, j,withi6=j,the separation condition stated in Lemma 3.1 holds, that is

|n_i−n_j| ≥δ_i+δ_j.

Finally they define the sets A(l, ν) = {n_k;k ∈ I(l, ν)} and they prove that these sets have positive lower density since (n_k) does and the sets I(l, ν) are arithmetic sequences. Actually we are going to prove that the sequence (n_k) has positive lower B2-density. To do that, we start by giving an exact formula for this sequence that will allow to obtain easily its asymptotic behavior. We obtain the following result, whose proof will be given later (see Lemma 4.8 below).

Lemma 3.2. If k= 2ⁿ+Pn−1

i=0 αi2ⁱ with αi ∈ {0; 1} for every0≤i≤n−1, then n_k= 4k−2



 X

i∈I_k

L_i(i+ 1)



−δ_k,

where Ik stands for the set of integers i such that αi is the first non-zero integer of a block (of consecutive non-zero coefficients) having length L_i in the dyadic decomposition of k.

From this lemma, we deduce the following estimate using the same notations.

Proposition 3.3. The sequence (n_k) satisfies the following estimate n_k−4k =O(log²(k)).

Moreover this estimate is optimal in the following sense: there exists an increasing sub- sequence(λn) of positive integers such that the sequence ⁿ_log^λj2^−4λ(λj)^j converges to a non-zero real number.

Proof. According to Lemma 3.2, we have, for k = 2^l+Pl−1

i=0α_i2ⁱ with α_i ∈ {0; 1} for every 0≤i≤l−1,

n_k= 4k−2



 X

i∈I_k

L_i(i+ 1)



−δ_k,

where Ik = {i ∈ N; αi 6= 0 andαi−1 = 0}, with the conventions α−1 = 0, αl = 1 and for i∈I_k, L_i = min{j;α_i+j = 0}.Obviously we deduce

n_k≤4k−2 log₂(k)−1.

Notice that we have the equalityn_k = 4k−2(log₂(k) + 1)−1 for k= 2^l. On the other hand, we can write

X

i∈I_k

L_i(i+ 1) = X

i1<i2<···<i_mk

L_ij(i_j+ 1), whereIk={i₁ < i2 <· · ·< imk}.Observe that we have

in+Lin+ 1≤in+1 forn= 1, . . . , m_k−1 and Li_mk =l−im_k+ 1.

Sinceimk ≤l,we get X

i1<i2<···<i_mk

L_ij(i_j+ 1)≤





mk−1

X

j=1

(i_j+1−(i_j+ 1))(i_j+ 1)



+ (l+ 1−i_mk)(i_mk+ 1)≤(l+ 1)². By construction we have

δ_k≤log₂(k) + 1.

Since log₂(k)≤l≤log₂(k) + 1,we conclude

4k−2(log₂(k) + 2)²−log₂(k)−1≤n_k≤4k−2 log₂(k)−1 and the estimaten_k−4k=O(log²k) holds. Finally let us consider λ_j =Pj

l=02^2l.An easy calculation gives

n_λj = 4λ_j −2

j

X

l=0

(2l+ 1)−1 = 4λ_j −2j²−4j−3.

Sinceλ_j = (4^j+1−1)/3,the sequence_n

λj−4λj

log²(λj)

converges to a non-zero real number.

We now prove that the sequence (n_k) constructed above not only has positive lower density but has also positive lowerB2-density.

Lemma 3.4. We haved_B2((nk))>0.

Proof. Using (6) from Example 2.5, we have d_B2(n_k) = lim inf

k→+∞

Pk

j=1e^nj/log2(nj) log²(n_k)e^nk/log2(nk)

! .

According to Proposition 3.3, there exists a constantC >0 such that, forN large enough, Pk

j=Ne^nj/log2(nj) log²(n_k)e^nk/log2(nk) ≥

Pk

j=Ne^{(4j−Clog2(j))/log2(4j−Clog2(j))} log²(4k)e^4k/log2(4k) . A summation by parts gives

k

X

j=N

e^{(4j−Clog2(j))/log2(4j−Clog2(j))}∼ log²(k)

4 e^{(4k−Clog2(k))/log2(4k−Clog2(k))}, ask→+∞.

Finally, a similar computation as those needed for Example 2.5, yields Pk

j=Ne^{(4j−Clog2(j))/log2(4j−Clog2(j))}

log²(4k)e^4k/log2(4k) ∼ e^−C

4 , ask→+∞,

which finishes the proof.

Lemma 3.4 allows us to show that the Frequent Hypercyclicity Criterion gives a strength- ened result.

Proposition 3.5. Let T be an operator on a separable Fr´echet space X. If there is a dense subsetX0 of X and a map S :X0→X0 such that, for any x∈X0,

(i)

+∞

X

n=0

Tⁿx converges unconditionally,

(ii)

+∞

X

n=0

Sⁿx converges unconditionally, (iii) T Sx=x,

thenT isB₂-frequently hypercyclic.

The proof of this result is the same as the classical proof of the Frequent Hypercyclicity Criterion. Indeed, from Lemma 3.4, we can deduce that the setsA(l, ν) not only have positive lower density but even have positive lowerB2-density. We won’t detail the proof here because we will prove a stronger result in Section 4.

Thanks to Lemma 2.9, one may actually deduce the following corollary proving that the scale defined by matricesAr is not fine enough to exhibit the limit in term of densities of the Frequent Hypercyclicity Criterion.

Corollary 3.6. Under the assumptions of the previous proposition, the operator T is Ar- frequently hypercyclic for every0≤r <1.

The previous result proves that for any 0≤r <1, theA_r-frequent hypercyclicity phenom- enon exists and is even common. On the other hand, one may also notice that the geometric rate of growth (i.e. r = 1) is unreachable in terms of dynamics. More precisely, we have the following result.

Proposition 3.7. There is no A1-frequently hypercyclic operator.

Proof. We argue by contradiction. Assume thatT is aA₁-frequently hypercyclic operator on a Banach space X and x is a A1-frequently hypercyclic vector. Let also U be a non-empty open subset in X. Then, by definition and with Example 2.5, we get

0<lim inf

N→+∞

X

k≤N

e^k PN

j=1e^j1N(x,U)(k) = lim inf

N→+∞(1−e⁻¹)X

k≤N

e^k−N1N(x,U)(k).

Moreover one may remark that asserting that this limit is non-zero implies that the set N(x, U) has bounded gaps. Indeed, if one suppose that N(x, U) has unbounded gaps then there exists a sequence (Ni) and a sequence (pi) tending to +∞ such that for every i ∈ N, {N_i−p_i+ 1;N_i−p_i+ 2;. . .;N_i} ∩N(x, U) =∅. This gives

0<lim inf

i→+∞(1−e⁻¹) X

k≤Ni

e^k−Ni1N(x,U)(k)≤lim inf

i→+∞(1−e⁻¹) X

k≤Ni−pi

e^k−Ni = lim

i→+∞e^−pi−1 = 0 and this contradiction shows that the setN(x, U) has bounded gaps. Let us denote byM an upper bound of the length of these gaps. It suffices to choose V so far from the origin such that the norm ofT forbidsT^k(U) from intersectingV fork≤M. This means that the orbit ofx will never reach the open setV contradicting theA₁-frequent hypercyclicity of x.

On the other hand, observe that the following result holds.

Lemma 3.8. For every 0< r <1, d_Br(n_k) = 0.

Proof. We have to estimate the following limit d_Br(n_k) = lim inf

k→+∞

Pk

j=1e^nj/logr(nj) Pn_k+1−1

j=1 e^j/logr(j)

! .

Remark that by definition of δ_k, there exists an increasing sequence of integers (λ_k)_k∈N such that n_λk+1−n_λk −1 =δ_λk =k+ 1∼log₂(λ_k), ask tends to∞ (consider for example λ_k= 2^k+1−1). Then,

d_Br(nk)≤lim inf

k→+∞



 Pλ_k

j=1e^nj/logr(nj) Pⁿ_λk+1−1

j=1 e^j/logr(j)



≤lim inf

k→+∞





Pⁿλk

j=1e^j/logr(j) Pⁿ_λk+1−1

j=1 e^j/logr(j)



.

Using the estimate (6) from Example 2.5 and the one from Proposition 3.3, we get:

d_Br(n_k)≤lim inf

k→+∞

log^r(n_λk)e^{nλk/logr(nλk)} log^r(nλ_k+1−1)e^{nλk/logr(nλk)}

!

≤lim inf

k→+∞e^{nλk/logr(nλk)−(nλk+δλk)/logr(nλk+δλk)}

We begin by studying the term in the exponent n_λk

log^r(nλ_k) − n_λk +δ_λk

log^r(nλ_k +δλ_k) = n_λk log^r(nλ_k+δλ_k)







1 + log

1 +_n^δλk

λk

log(nλ_k)





r

−

1 + δ_λk nλ_k





which reduces to the following thanks to a Taylor expansion:

nλk

log^r(n_λk+δ_λk)

1

log(n_λk) −1

+o

δλk

log^1+r(n_λk)

.

Now combining the estimateδλ_k ∼log₂(λk) asktends to∞with the one given by Proposition 3.3, we deduce thatδ_λk/log^1+r(n_λk)→0,ask tends to∞.Hence we get

e^{nλk/logr(nλk)−(nλk+δλk)/logr(nλk+δλk)}∼e

nλk logr(nλk+δλk)

1 log(nλk)−1

k→+∞−→ 0.

This proves that d_Br(n_k) = 0.

Notice that Proposition 3.3 combined with Lemma 3.8 do not allow us to conclude to the Br-frequent hypercyclicity or not in the Frequent Hypercyclicity Criterion for 1≤r <2.

4. Further results

In this section, we are going to improve the conclusion of the Frequent Hypercyclicity Cri- terion given by Proposition 3.5. To do this, we will modify the sequence (nk) used in the proof of Lemma 3.1 to obtain a new sequence possessing a positiveA-density for an admissible matrixAdefining a sharper density than the natural density.

Throughout this section, (a_n) will be an increasing sequence of positive integers witha₁ = 1.

Using this sequence we define the functionf :N→N,byf(j) =mfor allj∈ {a_m, . . . , am+1− 1}.In the spirit of the sequence studied in the previous section, we also define the sequence (nk(f)) by induction:

n1(f) =f(1) = 1 andn_k(f) =nk−1(f) +f(δk−1) +f(δ_k) for k≥2.

Clearly we obtain the following equality, for allk≥2,

(4.1) n_k(f) = 2

k−1

X

i=1

f(δ_i) +f(δ_k).

Let us notice that, if we setam =mfor everym≥1,then the corresponding sequence (n_k(f)) is the sequence (n_k) of Section 3. From now on, we will omit the notation f in (n_k(f)) for sake of readability. Our purpose is to compute an exact formula for the new sequence (n_k) to understand its asymptotic behavior. First of all, we obtain an expression for the subsequence (n2^am).

Lemma 4.1. For all m∈N, we have

n2^am = 2^am+1

m

X

i=1

1

2^ai−1 −2m+ 1.

Proof. Set ∆^(m)_j ={0≤l≤2^am−1 : δ_l=j}.First let us observe that we have, by definition, for every 1≤j≤a_m,

n2^am = 2

2^am−1

X

k=1

f(δk) +f(δ2^am) = 2

am

X

j=1

f(j)#∆^(m)_j +f(δ2^am).

Thus it suffices to compute the cardinal of the set ∆^(m)_j . It easily follows that #∆^(m)_j = 1 +Pam−j−1

i=0 2^am−j−i. Indeed, we separate the case when the first block of ones in the dyadic decomposition ofl ends on 2^am−1 and the case when the first block of ones ends before. In the first case, we have no choice, there is only one possibility but in the second case we have a certain numberiof zeros at the beginning, then the first block of ones, which is of lengthj, then one zero (because the first block of ones has to be of lengthj) and then we have 2^am−j−i possible choices as shown below.

(

lengtham+1

z }| {

0,0, . . . ,0

| {z }

lengthi

,1,1, . . . ,1,1,0

| {z }

lengthj+1

, ?, ?, . . . , ?, ?,0,0, . . .).

A quick calculation leads to #∆^(m)_j = 1 +Pam−j−1

i=0 2^am−j−i = 2^am−j.Therefore we get n₂am = 2

am

X

j=1

f(j)2^am−j +f(δ₂am) = 2

am

X

j=1

f(j)2^am−j+ 1.

Now we use the link between the values off(j) and the position ofjcompared to the sequence (am) to compute the sum:

n₂^am = 2

am

X

j=1

f(j)2^am−j+ 1

= 2^am+1

am−1

X

j=1

f(j)2^−j+ 2m+ 1.

Let us now split the sum according to the values of f(j):

n2^am = 2^am+1

m−1

X

i=1





ai+1−1

X

j=ai

f(j)2^−j



+ 2m+ 1

= 2^am+1

m−1

X

i=1

i





ai+1−1

X

j=ai

2^−j



+ 2m+ 1

= 2^am+1

m−1

X

i=1

i 1

2^ai−1 − 1 2^ai+1−1

+ 2m+ 1

= 2^am+1

m

X

i=1

1

2^ai−1 −2m+ 1.

We strengthen the previous lemma as follows.

Lemma 4.2. Let m ∈ N. For every q ∈ N such that q < a_m −am−1, the following equality holds

n₂am−q = 2^am−q+1

m−1

X

j=1

1

2^aj−1 −2(m−1) + 1.

Proof. This proof works along the same lines as the proof of Lemma 4.1. Thus, we adapt the preceding proof. It yields

n₂am−q = 2

2^am−q−1

X

j=1

f(δj) +f(δ₂am−q)

= 2

am−q

X

j=1

f(j)2^am−q−j + 1

= 2^am−q+1









m−2

X

i=1 ai+1−1

X

j=ai

i2^−j



+

am−q

X

j=am−1

(m−1)2^−j



+ 1

= 2^am−q+1

m−1

X

j=1

1

2^aj−1 −2(m−1) + 1.

Lemma 4.3. For k= 2ⁿ+Pn−2

i=0 α_i2ⁱ with α_i ∈ {0; 1}, 0≤i≤n−2, and (α₀, . . . , αn−2)6=

(0, . . . ,0),we have

nk−2ⁿ = 2

k−1

X

i=2ⁿ+1

f(δi) +f(δk).

Fork= 2ⁿ+ 2ⁿ⁻¹+Pn−2

i=0 α_i2ⁱ withα_i∈ {0; 1},0≤i≤n−2, we have nk−2ⁿ = 2

k−1

X

i=2ⁿ+1

f(δi)−f(δk−2ⁿ)−2(f(L)−1) + 2f(δk),

where L is the length of the block of one’s containing the coefficient one of2ⁿ in the dyadic decomposition of k.

Proof. We begin by proving the first assertion. We have nk−2ⁿ = 2

k−2ⁿ−1

X

i=1

f(δi) +f(δk−2ⁿ).

Sincek= 2ⁿ+Pn−2

i=0 α_i2ⁱ,observe that for all 2ⁿ+ 1≤i≤k−1 the dyadic decomposition of icontains a one for some 2^lwith 0≤l≤n−2 and the coefficient of 2ⁿ⁻¹ is zero. Therefore, the first block of ones in the dyadic decomposition ofidoes not contain the coefficient of 2ⁿ. Thus, for every such 2ⁿ+ 1≤i≤k−1, we have δi=δi−2ⁿ. This proves the first part of the lemma sinceδ_k=δk−2ⁿ.

To prove the second assertion, we begin by observing that 2

k−1

X

i=2ⁿ+1

f(δ_i) = 2

k

X

i=2ⁿ+1

f(δ_i)−2f(δ_k).

Then, as above if the index i is such that the coefficient of 2ⁿ does not belong to the first block of ones (in the dyadic decomposition ofi) then δi = δi−2ⁿ. On the other hand, if the coefficient of 2ⁿ belongs to the first block of ones and since we have an−1 = 1 then i has to be of the form i= Pp

l=02^n−l for p ∈ {1, . . . , L−1} and δi = δi−2ⁿ+ 1. Now let us pick a particular index iof the form i=Pp

l=02^n−l withp∈ {1, . . . , L−1}.We consider two cases:

Case 1: for every j ∈ {2, . . . , f(L)}, we have p+ 1 6=a_j. Then we have f(δ_i) =f(p+ 1) = f(δ_i−1) =f(δi−2ⁿ).

Case 2: there exists an integerj∈ {2, . . . , f(L)}withp+ 1 =aj.Then we getf(δi) =f(aj) = j=f(δ_i−1) + 1 =f(δi−2ⁿ) + 1.

Finally we deduce 2

k−1

X

i=2ⁿ+1

f(δ_i) = 2

k

X

i=2ⁿ+1

f(δ_i)−2f(δ_k) = 2

k−2ⁿ

X

i=1

f(δ_i) + (f(L)−1)

!

−2f(δ_k)

=nk−2ⁿ+f(δk−2ⁿ) + 2(f(L)−1)−2f(δk).

From Lemma 4.3 we deduce the following result.

Lemma 4.4. Let L be any non-zero integer andq be an integer. If k=PL−1

j=0 2^q+j+k⁰ with 0≤k⁰<2^q−1 then either k⁰6= 0 and

nk=nk⁰ +

L−1

X

j=0

n₂^q+j+ 2

L

X

j=2

(f(j)−1) +L,

or k⁰= 0 and

nk=

L−1

X

j=0

n₂^q+j+ 2

L

X

j=2

(f(j)−1) +L−f(L).

Proof. We proceed by induction onL.ForL= 1, setk= 2^q+k⁰. First, observe that ifk⁰ = 0, the result is clear by Lemma 4.3. So assume that 0< k⁰ <2^q−1.We divide n_k into two sums

n_k= 2

2^q−1

X

i=1

f(δi) +f(δ2^q)

!

+ f(δ2^q) + 2

k−1

X

i=2^q+1

f(δi) +f(δ_k)

! .

It suffices to apply Lemma 4.3 to obtain

n_k=n₂^q+f(δ₂^q) +n_k⁰ =n₂^q + 1 +n_k⁰

and we have the desired conclusion. Now chooseL≥2 and suppose that the result holds for every integerl,with 1≤l≤L−1.By Lemma 4.3, we get

nk =

k−1

X

i=1

f(δi) +f(δk)

=

2^q+L−1−1

X

i=1

f(δi) + 2f(δ₂^q+L−1) + 2

k−1

X

i=2^q+L−1+1

f(δi) +f(δk)

=n₂q+L−1 +f(δ₂q+L−1) +n_k−2q+L−1 +f(δ_k−2q+L−1) + 2(f(L)−1)−f(δ_k).

We have f(δ₂^q+L−1) = 1. Moreover suppose that 0 < k⁰ < 2^q−1, then the block of ones containing the coefficient one of 2^q+L−1 is not the first one, thusδ_k=δ_k⁰ =δ_k−2q+L−1.Using the induction hypothesis, we obtain

n_k =n₂^q+L−1 +n_k−2^q+L−1 + 2(f(L)−1) + 1

=n_k⁰+

L−2

X

j=0

n₂^q+j+ 2

L−1

X

j=2

(f(j)−1) +L−1 +n₂^q+L−1 + 2(f(L)−1) + 1

=n_k⁰+

L−1

X

j=0

n₂q+j+ 2

L

X

j=2

(f(j)−1) +L.

On the other hand, in the case k⁰ = 0, the induction hypothesis gives nk=n₂^q+L−1 + 1 +n_k−2^q+L−1 +f(L−1) + 2(f(L)−1)−f(L)

=

L−2

X

j=0

n₂^q+j + 2

L−1

X

j=2

(f(j)−1) + (L−1)−f(L−1) +n₂^q+L−1+ 1 +f(L−1) + 2(f(L)−1)−f(L)

=

L−1

X

j=0

n₂q+j + 2

L

X

j=2

(f(j)−1) +L−f(L).

This completes the proof.

From Lemma 4.4, we immediately get the following result since f(1) = 1.

Lemma 4.5. LetL₁, . . . , L_r be non-zero integers andq₁, . . . , q_r be integers such thatq_i+L_i<

qi+1 for every1≤i≤r−1. For k=Pr i=1

PLi−1

j=0 2^qi+j we have n_k=

r

X

i=1 Li−1

X

j=0

n_2qi+j + 2

r

X

i=1 Li

X

j=1

f(j)−

r

X

i=1

L_i−f(L₁).

We are ready to obtain a general formula for the sequence (n_k). Let us introduce some notations.

Notations 4.6. Let L₁, . . . , L_r be non-zero integers and q₁, . . . , q_r be integers such that q_i+L_i < q_i+1 for every 1≤i≤r−1. We define the integersm_i, t_i, s_i and p_i as follows:

(1) mi is the greatest integer such thatami ≤qi+Li−1, (2) t_i =q_i+L_i−1−a_mi,