failing the Hypercyclicity Criterion on classical Banach spaces

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Hypercyclic operators

failing the Hypercyclicity Criterion on classical Banach spaces

F. Bayart

, É. Matheron

Laboratoire Bordelais d’Analyse et de Géométrie, UMR 5467, Université Bordeaux 1, 351 Cours de la Libération, 33405 Talence Cedex, France

Received 14 November 2006; accepted 2 May 2007 Available online 20 June 2007

Communicated by K. Ball

Abstract

By a recent result of M. De La Rosa and C. Read, there exist hypercyclic Banach space operators which do not satisfy the Hypercyclicity Criterion. In the present paper, we prove that such operators can be con- structed on a large class of Banach spaces, includingc₀(N)or^p(N).

©2007 Elsevier Inc. All rights reserved.

Keywords:Hypercyclic operators; Hypercyclicity Criterion

1. Introduction

LetXbe a topological vector space overK=RorC. A continuous linear operatorT ∈L(X) is said to behypercyclic if there exists some x ∈X whose T-orbit {Tⁿ(x); n∈N}is dense inX. For example, the derivation operator and the nontrivial translation operators on the space of entire functions are hypercyclic [3,14], and ifBis the usual backward shift on²(N), then 2B is hypercyclic [15]. We refer to [7,8] for much more on hypercyclicity.

IfXis a separable Fréchet space, it follows from the Baire Category Theorem that an operator T ∈L(X)is hypercyclic if and only if it is topologically transitive, which means that for each

* Corresponding author.

E-mail addresses:frederic.bayart@math.u-bordeaux1.fr (F. Bayart), etienne.matheron@math.u-bordeaux1.fr (É. Matheron).

0022-1236/$ – see front matter ©2007 Elsevier Inc. All rights reserved.

doi:10.1016/j.jfa.2007.05.001

pair (U, V ) of nonempty open subsets of X, one can find n∈N such that Tⁿ(U )∩V = ∅. Using this, one can formulate a very useful criterion for hypercyclicity, which is known asthe Hypercyclicity Criterion. This criterion was first isolated by C. Kitai [12], and then refined by several authors. We state it in the form given in [8], which is the same as in [2].

The Hypercyclicity Criterion.LetXbe a separable Fréchet space, and letT ∈L(X). Assume one can find dense sets D, D⊂X and an increasing sequence of integers(nk)such that the following properties hold:

(1) T^nk(z)→0for allz∈D;

(2) For eachz∈D, one can find a sequence(x_k)⊂Xsuch thatx_k→0andT^nk(x_k)→z. ThenT is topologically transitive, and hence hypercyclic.

Conditions (1) and (2) give in fact a stronger result: it is not hard to check that ifT satisfies the Hypercyclicity Criterion, then the operatorT ⊕T (acting onX⊕X) is topologically transitive, i.e. hypercyclic. It was shown by J. Bès and A. Peris [2] that the converse is also true: ifT ⊕T is hypercyclic, thenT satisfies the Hypercyclicity Criterion. The following problem, originally posed by D. Herrero [10] in theT ⊕T form, has been recognized as one of the most exciting questions in linear dynamics; see e.g. [2,6,8] or [16].

Problem.Does every hypercyclic operator on a separable Fréchet space satisfy the Hypercyclic- ity Criterion?Equivalently, isT ⊕T hypercyclic wheneverT is?

Very recently, this problem was solved in the negative by M. De La Rosa and C. Read [5], who constructed a Banach spaceX and a hypercyclic operator T ∈L(X)such thatT ⊕T is not hypercyclic. Their construction may be very roughly described as follows. One starts with a vector spaceF having an algebraic basis(f_i)_i∈N, and with the linear operatorS:F→F defined by Sf_i =f_i+1. Then one defines a norm · onF in such a way that S is continuous and hypercyclic on(F, · ), and moreover · is in some sense maximal with respect to these properties. The desired Banach spaceX is the completion of (F, · )andT is the extension ofStoX.

Although the definition of the norm · is not extraordinarily complicated, it is not clear whether the Banach space constructed in [5] can be identified with a “classical” space. In the present paper, we show that one can construct hypercyclic operators whose direct sum with themselves are not hypercyclic on many classical spaces, includingc₀(N)or^p(N), 1p <∞.

We need the following definition: if(e_i)_i∈Nis a linearly independent sequence in some vec- tor space, then theforward shift associated to(e_i)is the linear operatorS:E→Edefined by S(ei)=ei+1, whereE=span{ei; i∈N}. Our main result reads as follows.

Theorem 1.1.LetXbe a Banach space. AssumeXhas a normalized unconditional basis(e_i)_i∈N whose associated forward shift is continuous. Then there exists a hypercyclic operatorT ∈L(X) such thatT ⊕T is not hypercyclic.

From this, we deduce immediately:

Corollary 1.2.There exist hypercyclic operators onc₀(N)or^p(N),1p <∞which do not satisfy the Hypercyclicity Criterion. In particular, one can find such operators on a Hilbert space.

2. Algebraic preliminary

To prove Theorem 1.1, we will need a kind of “non-hypercyclicity criterion” for a direct sumT ⊕T. There is in fact a very simple algebraic obstruction, which is contained in the fol- lowing easy lemma. This idea appears in [5], in a rather different formulation.

Lemma 2.1.LetAbe a commutative algebra endowed with a topologyτ, and letnbe a semi- norm onAsuch that the map(p, q)→pq is continuous from(A, τ )×(A, τ )into(A,n). Let alsoa, a, b, b∈Aand assume there exist three sequences(p_n),(q_n), and(r_n)inAsuch that p_n→a,q_n→b,r_np_n→aandr_nq_n→b. Thenn(ab−ab)=0.

Proof. Just writep_n(r_nq_n)=(r_np_n)q_nand use the assumptions. 2

Corollary 2.2.Assume the algebraAhas a unit, and that for anya, a, b, b, one can find(pn), (q_n),(r_n)as above. Thenn=0.

Proof. Apply 2.1 witha∈Aarbitrary,b=1anda=b=0. 2 We will apply 2.2 to an algebraAof the form

K[T]e₀:=

P (T )e₀; P polynomial

=span

Tⁱe₀; i∈N ,

whereT:Z→Zis a linear map on some vector spaceZ, ande₀∈Zis such that the vectorsTⁱe₀ are linearly independent. This happens in particular ifT is a linear operator on some infinite- dimensional Fréchet spaceXand ife₀is a cyclic vector forT. The product onK[T]e₀is defined by

P (T )e₀·Q(T )e₀=P Q(T )e₀.

Notice thatK[T]e₀ has a unit, namelye₀. WhenT is a Fréchet space operator with cyclic vectore₀, the algebraK[T]e₀inherits the topology of the underlying Fréchet spaceX.

Corollary 2.3. Let X be an infinite-dimensional Fréchet space, and let T ∈L(X) be cyclic with cyclic vectore₀. Assume there exists a nonzero linear functionalφ:K[T]e₀→Ksuch that the map(x, y)→φ(x·y)is continuous onK[T]e₀×K[T]e₀. ThenT ⊕T is not hypercyclic onX×X.

Proof. Suppose on the contrary thatT ⊕T is hypercyclic. Then the set of hypercyclic vectors forT ⊕T is dense inX⊕X. Leta, b, a, b∈K[T]e₀be arbitrary. One can find a sequence of (T⊕T )-hypercyclic vector(x_n⊕y_n)⊂X⊕Xsuch thatx_n⊕y_n→a⊕b. Then one can choose a sequence of integer(k_n)such thatT^knx_n⊕T^kny_n→a⊕b. Finally, sincee₀is cyclic forT, we may find polynomialsP_n,Q_nsuch thatT^knP_n(T )e₀⊕T^knQ_n(T )e₀−T^knx_n⊕T^kny_n→0.

Settingp_n:=P_n(T )e₀,q_n:=Q_n(T )e₀andr_n:=T^kne₀, one can apply Corollary 2.2 with the semi-normndefined byn(z)= |φ(z)|, and this gives a contradiction sinceφ=0. 2

Remark.The above proof gives in fact the formally stronger conclusion thatT ⊕T cannot have a dense set of cyclic vectors: just note that the proof still works if one replacesT^kn byR_n(T ), for some polynomial R_n. However, this is not a real strengthening, for ifT is a hypercyclic operator such thatT ⊕T is cyclic, thenT ⊕T is actually hypercyclic by a result of S. Grivaux [6, Proposition 4.1].

From now on,X is a Banach space with a normalized unconditional basis (e_i)_i∈N whose associated forward shift is continuous. We putc₀₀:=span{e_i; i∈N}. In view of Corollary 2.3, our main result will be proved if we are able to construct a linear operatorT:c₀₀→c₀₀ and a nonzero linear functionalφ:c₀₀→Ksuch that the following properties hold.

(a) span{Tⁱe₀; i∈N} =span{e_i;i∈N}; in other wordsK[T]e₀=c₀₀. (b) The set{Tⁱe0; i∈N}is dense inc00.

(c) T is continuous.

(d) The map(x, y)→φ(x·y)is continuous onc₀₀×c₀₀.

Indeed, (c) allows to extend T to a continuous linear operator onX, which is hypercyclic with hypercyclic vectore₀by (b), and whose direct sum with itself is not hypercyclic by (d) and Corollary 2.3.

The operatorT and the linear functionalφwill be constructed in the next two sections. They will both depend on a sequence of positive numbers(a_n)_n0 tending to infinity and on an in- creasing sequence of integers(bn)_n0. For convenience, we assume thata0=1 andb0=0. The conditions needed on(an)and(bn)will be specified later.

It will turn out that the sequence(an)can be first chosen arbitrarily, but that(bn)will then have to grow much faster than(a_n). Thus, we could put e.g.a_n=2ⁿfrom the very beginning, but since this would not simplify the proof, we do not specify an explicit value fora_n.

Notations.IfP is a polynomial, we denote by deg(P )the degree ofP and by|P|1the sum of the moduli of the coefficients ofP. We choose a countable dense setQ⊂K, and we fix once and for all an enumeration(P_n)_n∈Nof the set of all polynomials with coefficients inQ, withP₀=0.

We will assume from the beginning thatb_n>deg(Pn)for alln∈N. 3. The operatorT

One can associate to(an)and(bn)a unique linear mapT:c00→c00satisfying the following two properties:

T e_i=2ei+1 ifi∈ [b_n−1, b_n−2]; (1) T^bn(e₀)=P_n(T )e₀+ 1

an

e_bn for alln. (2)

Indeed, writing

T^bne0=T^bn−bn−1T^bn−1e0

=T^bn−bn−1

Pn−1(T )e0+ 1 a_n−1eb_n−1

=2^{bn−bn−1−1}

a_n−1 T (eb_n−1)+T^bn−bn−1Pn−1(T )e0,

we have to set

T eb_n−1= a_n−1 2^{bn−bn−1−1}

1

a_neb_n+Pn(T )e0−T^bn−bn−1Pn−1(T )e0

; (3)

and since deg(Pn) < b_n and deg(Pn−1) < b_n−1, the operatorT is well defined by formulae (1) and (3).

By definition ofT, we have{P (T )e₀; degPN} =span{e₀;. . .;e_N}for allN∈N, so that K[T]e0=c00. It follows that the set{Pn(T )e0; n∈N}is dense inc00, and hence (by (2)) that the set{Tⁱe0; i∈N}is dense as well. In other words, the first two conditions needed to prove Theorem 1.1 are satisfied, whatever the choice of(a_n)and(b_n)may be.

In the remainder of this section, we intend to show that if the sequences(a_n)and(b_n)are suitably chosen, then the operator T is continuous. We will make use of the ¹-norm on c₀₀ associated to the basis(e_i); this norm will be denoted by · 1. Thus, if x =

ix_ie_i, then x1=

i|x_i|. IfEis a finite-dimensional subspace ofc₀₀, we denote byT_|E1,1the norm of the operatorT_|E:(E, · 1)→(c00, · 1).

Notice that formula (3) above can be written as

T (eb_n−1)=εneb_n+fn, (4) where

ε_n= an−1

2^{bn−bn−1−1}a_n, and f_n= an−1

2^{bn−bn−1−1}

P_n(T )e₀−T^bn−bn−1P_n−1(T )e₀ .

Since deg(Pn) < b_nand deg(Pn−1) < b_n−1, the vectorf_nis supported on[0, b_n).

Given a positive integern, we shall say that T is convenient up to stagen if the following properties hold:

• ε_kmin(1,_2a ^2−k−3

kPk(T )e01)for allk∈ {1;. . .;n};

• f_k12^−k for allk∈ {1;. . .;n}.

The continuity ofT will be an easy consequence of the next lemma.

Lemma 3.1.Leta₁, . . . , a_n,b₁, . . . , b_nbe given, and assumeT is convenient up to stagen. Then, for any choice ofa_n+1, one can find a positive numberB such thatT is convenient up to stage n+1wheneverbn+1> B.

The following simple remark will be useful for the proof of Lemma 3.1.

Remark.IfT is convenient up to stagen, thenT_{|span{e0;...;ebn+1−2}}1,12.

Proof of Remark. We have T (e_bk−1)1ε_k + f_k12 for all k∈ {1;. . .;n}, so that T (e_i)12 for alli < b_n+1−1. 2

Proof of Lemma 3.1. Recall thatT e_bn+1−1=ε_n+1e_bn+1+f_n+1, withε_n+1=_{2bn+1−bn−}^an 1a_n+1 and

fn+1= a_n

2^{bn+1−bn−1}Pn+1(T )e0−2an

T 2

b_n+1−bn

Pn(T )e0.

If bn+1 is large enough, then εn+1 is small and the first term in fn+1 has norm less than 2⁻ⁿ⁻². Therefore, we just have to check that2an(T /2)^bn+1−bnPn(T )e012⁻ⁿ⁻² ifbn+1 is large enough.

Claim 1.Assumeb_n+1>2bn. Letp∈ [0, b_n), so thatb_kp < b_k+1for somekn−1. Ifr satisfiesb_n−prb_n, then

T^r(e_p)

2^r =ε_k+1. . . ε_n 2^n−k er+p+

n

j=k+1

k+1s<jε_s 2^j−k

T^r+p−bj(f_j) 2^r+p−bj .

Proof of Claim 1. This is proved by reverse induction, starting withk=n−1. Fork+1=n, one has

T^r(e_p)

2^r =2^−rT^r−(bn−p)T^bn−p(e_p)

=2^−rT^r+p−bn2^bn−p−1(ε_ne_bn+f_n)

=1 2

T^r+p−bn

2^r+p−bn(ε_ne_bn+f_n)

=ε_n

2e_r+p+1 2

T^r+p−bn 2^r+p−bn(f_n),

where we have used in the last line the inequalitiesr+p <2bn< b_n+1. Thus, the formula holds whenk=n−1.

Assume the result is known for all pairs(p, r)withp∈ [b_k+1, b_k+2)andb_k+1−pr b_k+1. Ifp∈ [b_k, b_k+1)andb_k−prb_k, then, writing

T^r(e_p) 2^r =1

2

T^r−bk+1+p

2^r−bk+1+p(εk+1eb_k+1+fk+1),

we conclude by using the induction hypothesis withp=b_k+1andr=r+p−b_k+1. 2 Claim 2.Ifu∈c₀₀is supported on[0, b_n)and ifiis a positive integer such that(i+1)bnb_n+1, one can write

T^ibn(u) 2^ibn =v_i+

p

T^rp(up) 2^rp , where the sum is finite,

pup1 ^u₂i¹, supp(up)⊂ [0, bn), rp < ibn, supp(vi)⊂ [bn, (i+1)bn)andv_i1^ε₂ⁿ(1+¹₂+ · · · +₂i¹−1)u1.

Proof of Claim 2. We prove by induction onithat the result holds for allu. Ifi=1, we apply Claim 1 withr=b_n. Writingu=

px_pe_pand definingk_pbyb_kpp < b_kp+1, we get T^bn(u)

2^bn =v₁+

p<bn

T^p(u_p) 2^p , where

v₁=

p<bn

x_pε_kp+1. . . ε_n

2^n−kp e_bn+p and u_p=x_p

n

j=kp+1

kp+1s<jε_s 2^j−kp

T^bn−bj(f_j) 2^bn−bj . Thenv₁is supported on[b_n,2bn)and

v₁1ε_n 2 p

|x_p| =ε_n 2u1.

Moreover, eachupis supported on[0, bn)because supp(fj)⊂ [0, bj)for allj. Finally, we have up1^|x₂^p|for allp, becausefj1¹₂ for allj andT^bn−bj has norm not greater than 2^bn−bj when restricted to span{e0;. . .;eb_j}. Thus, we get

p

u_p1u1

2 .

Assume the result has been proved foriand that(i+2)bnb_n+1. Applying the casei=1 and then the induction hypothesis to eachu_p, we get

T^(i+1)bn(u)

2^(i+1)bn =T^ibn(v₁) 2^ibn +

p<bn

T^p 2^p

T^ibn(up)

=T^ibn(v1) 2^ibn +

p<b_n

T^p 2^p

v_p,i+

q

T^rp,q(up,q) 2^rp,q

=T^ibn(v₁) 2^ibn +

p<bn

T^p(v_p,i)

2^p +

p<bn q

T^p+rp,q(u_p,q) 2^p+rp,q · SinceT_|span{e0;...;e_bn+1−2}1,12, we have

T^ibn(v₁) 2^ibn +

p<b_n

T^p(vp,i) 2^p

1

ε_nu1

2 +

p<b_n

εnup1

2

1+1

2+ · · · + 1 2ⁱ⁻¹

εnu1

2

1+1

2+ · · · + 1 2ⁱ

.

Moreover,

p,q

up,q1

p<bn

u_p1

2ⁱ u1

2ⁱ⁺¹.

Finally, we havep+rp,q< bn+(i+1)bn=(i+2)bnfor all pairs(p, q). This concludes the proof. 2

Now, we apply Claim 2 with u=2anP_n(T )e₀. Since T is convenient up to stage n, we have v_i1 ε_nu12⁻ⁿ⁻³ for all i such that (i +1)bn b_n+1. Moreover, since T_{|span{e0;...;e}

bn+1−2}1,12 andr_p< ib_nfor allp, we also have

p

T^rp(u_p) 2^rp

1

p

u_p1u1

2ⁱ . Thus, fixingi0with^u1

2ⁱ⁰ 2⁻ⁿ⁻³, any choice ofb_n+1withb_n+1−b_ni0b_nyields the desired result. 2

Corollary 3.2.There exist functionsF_n:Nⁿ×Nⁿ⁺¹→Nsuch that the following holds:ifb_n+1 Fn(b1, . . . , bn, a1, . . . , an+1)for alln, thenT is continuous onc00with respect to the topology ofX.

Proof. By the lemma, it is enough to show that ifT is convenient up to every stagen, thenT is continuous.

We can decomposeT asT =R+K, whereR is a forward weighted shift with a bounded weight sequence, andKis defined by:K(e_bn−1)=f_nfor allnandK(e_i)=0 otherwise. Since the forward shift associated to(e_i)is continuous and since the basis(e_i)is unconditional, the operatorRis continuous.

Sincef_nis supported on[0, bn), we have K(e_b

n−1)

Xmax

e_jX; jb_n−1

· f_n1

for alln1. Since the sequence(e_i)is bounded, it follows that_∞

0 K(e_i)X<∞; and since the sequence of coordinate functionals(e^∗_i)is also bounded (because infie_iX>0), we con- clude that the operatorKis continuous. 2

Remark.It is not difficult to show that the operatorK is compact, as a uniform limit of finite rank operators. Hence,T is a compact perturbation of a weighted shift operator.

4. The linear functionalφ

In this part, we viewc₀₀as span{Tⁱe₀; i∈N}rather than span{e_i; i∈N}. In particular, we will say that a vectorz∈c₀₀is supported on some setI⊂Nifz∈span{Tⁱe₀; i∈I}.

We denote by| · |1the¹-norm associated to the basis(Tⁱe0). Thus, ifz=P (T )e0∈c00, then

|z|1= |P|1.

From now on, we fix some positive numberε∈(0,1), and we assume that deg(Pn)+εb_n< b_n andb_n+1(2+ε)b_nfor alln∈N.

Using the same idea as in [5], we define a linear functionalφ:c₀₀→Kas follows. We put φ(e₀)=1 andφ(Tⁱe₀)=0 ifi∈(0, b1). Ifi∈ [b_k, b_k+1)for somek1, we set:

φ(Tⁱe₀)=

φ(Pk(T )T^i−bke0) ifi∈ [bk, (1+ε)bk)∪ [2bk, (2+ε)bk),

0 otherwise.

Notice thatφ(Tⁱe0)is indeed well defined ifφ(T^je0)is known for allj < i, because deg(Pk)+ i−bk< iand hencePk(T )T^i−bke0is supported on[0, i).

The next lemma collects the properties ofφwhich will be needed below.

Lemma 4.1.The following properties hold for allk1.

(1) φ((T^bk−Pk(T ))z)=0wheneverzis supported on[0, εbk)∪ [bk, (1+ε)bk).

(2) maxi∈[0,b_k)|φ(Tⁱe0)|Nk:=

0<j <kmax(1,|Pk|1)².

Proof. Part (1) is obvious from the definition ofφ. The proof of part (2) is the same as in [5], but we give the details for the sake of completeness. The result is true fork=1 if we give the value 1 to an empty product. Assume the inequality holds fork. Settingφi:= |φ(Tⁱe0)|we have

i∈[maxbk,2bk)φi= max

i∈[bk,(1+ε)bk)

φ

Pk(T )T^i−bke0 |P_k|1· max

j <εbk+deg(Pk)

φ_j |P_k|1·N_k,

becauseεb_k+deg(Pk) < b_k. Similarly, we have

i∈[2bmax_k,b_k+1)φ_i= max

i∈[2b_k,(2+ε)b_k)φ_i|P_k|1 max

j <2bk

φ_j|P_k|²₁N_k.

We conclude that maxi∈[bk,bk+1)φ_iN_k+1, and the result follows by induction. 2

We now prove that the map(x, y)→φ(x·y)is continuous if (a_n)and(b_n)are suitably chosen.

Lemma 4.2.There exist functionsG_n:N×N→Nsuch that the following holds:

if b_nG_n(b_n−1, a_n)for alln1, then

p,q

φ(e_p·e_q)<+∞.

Proof. Ifp∈ [bk, bk+1)andq∈ [bl, bl+1), then the definition ofT gives e_p=a_k

2^u

T^bk−P_k(T ) T^u(e₀), eq=a_l

2^v

T^bl−Pl(T ) T^v(e0),

whereu=p−b_k∈ [0, bk+1−b_k)andv=q−b_l∈ [0, bl+1−b_l). Thus, settingΛ:= {(m, w)∈ N×N; w < b_m+1−b_m}and

y(k,u)(l,v)=

T^bk−Pk(T )

T^bl−Pl(T )

T^u+v(e0), we have to prove that

Σ:=

((k,u),(l,v))∈Λ×Λ

a_ka_l

2^u+vφ(y_(k,u)(l,v))<+∞. We will need the following facts.

Claim 1.Ifkl, then we always have|φ(y_(k,u)(l,v))|M_l:=N_l+2·maxjl(1+ |P_j|1)². Proof of Claim 1. Observe that supp(y(k,u)(l,v))⊂ [0, bk +u+b_l+v] ⊂ [0, b_k+1+b_l+1), hencey_(k,u)(l,v) is supported on[0, bl+2)becausekl andb_l+2>2bl+1. Moreover, one has

|y_(k,u)(l,v)|1(1+ |P_k|1)(1+ |P_l|1). By Lemma 4.1, the result follows. 2 Claim 2.In each of the following two cases, we haveφ(y_(k,u)(l,v))=0:

• k=landu+v+deg(Pk) < εb_k;

• k < landu+v+b_k< εb_l.

Proof of Claim 2. Whenk=l1, we write y(k,u)(k,v)=

T^bk−Pk(T )

T^bk+u+ve0−

T^bk−Pk(T )

Pk(T )T^u+ve0

=

T^bk−P_k(T ) (z₁)−

T^bk−P_k(T ) (z₂).

Thenz1is supported on[bk, (1+ε)bk)ifu+v < εbk, andz2is supported on[0, εbk)ifu+v+ deg(Pk) < εbk. By Lemma 4.1, this gives the first part of the claim. Whenl > k, we just write

y(k,u)(l,v)=

T^bl−Pl(T ) (z),

wherez=(T^bk−P_k(T ))T^u+ve₀is supported on[0, u+v+b_k)⊂ [0, εbl). 2 Now, we write

Σ=Σ1+2Σ2,

whereΣ₁ is the sum over all pairs((k, u), (l, v))withk=l, andΣ₂is the sum over the pairs withl > k.

By the two above claims, we have Σ₁ ^∞

k=0

a_k²M_k

(u,v)∈N×N u+vεb_k−deg(P_k)

1 2^{u+v ∞}

k=0

a_k²M_k

iεb_k−deg(P_k)

i+1 2ⁱ ,

so thatΣ₁<∞providedb_nis always large enough with respect toa_n.

To estimateΣ₂, we use the claims to get Σ₂

∞ k=0l>k

a²_lM_l

u+vεbl−bk

1 2^u+v

^∞

k=0

2^bk

l>k

a_l²M_l

iεbl

i+1 2ⁱ .

Thus, we haveΣ2<∞provided(b_n)is rapidly increasing andb_n is large enough with respect toan. This concludes the proof of the lemma. 2

Corollary 4.3.If(b_n)is rapidly increasing andb_nis large enough with respect toa_n, then the map(x, y)→φ(x·y)is continuous onc00×c00.

Proof. Writingx=

px_pe_pandy=

qy_qe_q, we get φ(x·y)

p,q

|x_p| |y_q|φ(e_p·e_q)C²

p,q

φ(e_p·e_q)xy

for all(x, y)∈c₀₀×c₀₀, whereC=sup_ie_i^∗. 2

Putting together Corollaries 3.2 and 4.3, the proof of our main theorem is now complete.

5. Variations on the main result

It should be clear from the proof that Theorem 1.1 can be formulated in a Fréchet space setting.

More precisely, the result remains valid ifXis a separable Fréchet space with an unconditional basis(e_i)such that the following properties hold true, where(e^∗_i)is the sequence of coordinate functionals: the forward shift associated to(e_i)is continuous, the sequence(e_i)is bounded, and the sequence(e^∗_i)is equicontinuous. However, this does not seem to apply to the most interesting non-Banach examples.

Nevertheless, we do have the following result. Let us denote byH(Ω)the space of all holo- morphic functions on an open setΩ⊂C.

Proposition 5.1.IfΩ⊂Cis a simply connected domain, then there exists a hypercyclic operator onH(Ω)which does not satisfy the Hypercyclicity Criterion.

Proof. We may assume thatΩis a diskD(0, R), where 1< R∞. We will mimic the proof of Theorem 1.1, but the operator has to be slightly modified. Let us denote by(ei)_i∈Nthe “canonical basis” ofH(Ω),ei(z)=zⁱ. If one wants to imitate the proof of Theorem 1.1, one difficulty comes to mind: the operatorT defined above is hypercyclic because _a¹

neb_n→0. But this is no longer true for anarbitrary sequence(an) tending to infinity in the present setting:an must grow faster thanr^bn for any r < R. Thus, an grows in fact much faster than bn, so that one cannot simply reproduce the proof of Theorem 1.1. On the other hand, the highly non-Banach structure ofH(Ω)allows continuous shifts with unbounded weights, and ensures a fast decay of the coordinate functionals associated to(e_i), so one can hope to overcome this difficulty.