Weakly supercyclic operators

Available online at www.sciencedirect.com

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J. Math. Anal. Appl. 292 (2004) 148–159

www.elsevier.com/locate/jma

Weakly supercyclic operators

Rebecca Sanders

Department of Mathematics and Statistics, Bowling Green State University, Bowling Green, OH 43403, US

Received 4 August 2003

Submitted by J.H. Shapiro

Abstract

Inspired by a recent result that a weakly hypercyclic operator may fail to be norm hypercyclshow there exists a weakly supercyclic operator that fails to be norm supercyclic. Moreover, declassical result of Hilden and Wallen that every unilateral weighted backward shift is supercycshow such an operator may have a weakly supercyclic vector that is not a norm supercyclicIn addition to these results, we extend a result of Kitai by showing a hyponormal operator cannoweakly hypercyclic. 2003 Elsevier Inc. All rights reserved.

Keywords:Hypercyclic operator; Supercyclic operator; Hyponormal operator; Bilateral weighted shift;Unilateral weighted shift; Weak topology

1. Introduction

Let X be a separable, infinite dimensional Banach space overC. A bounded linearoperatorT :X → X is norm hypercyclicif there exists a vectorx in X such that its orbitOrb(T , x) = {T nx: n � 0} is norm dense inX. If the set{λT nx: n � 0, λ ∈ C} is normdense inX for some vectorx, then the operatorT is norm supercyclic. In each casesuch a vectorx is called anorm hypercyclic vectorand anorm supercyclic vectorfor T ,respectively. By considering density in the weak topology instead of the norm topologcan discuss weak hypercyclicity and weak supercyclicity. That is, the operatorT :X → X

is weakly hypercyclicif Orb(T , x) is weakly dense inX for some vectorx. Similarly,T is weakly supercyclicif the set{λT nx: n � 0, λ ∈ C} is weakly dense inX for somevectorx. Such a vectorx is called aweakly hypercyclic vectorand aweakly supercyclic

E-mail address:[email protected].

0022-247X/$ – see front matter 2003 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2003.11.049

R. Sanders / J. Math. Anal. Appl. 292 (2004) 148–159 149

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vector for T , respectively. Lastly, the operatorT :X → X is cyclic if the linear span ofOrb(T , x), denoted by spanOrb(T , x), is norm dense inX for some vectorx and sucha vectorx is acyclic vectorfor T .

It is obvious that for either norm or weak topology, hypercyclicity implies supercyity, which in turn implies cyclicity. It is interesting to study the relation of those typecyclicity if we mix the norm and weak topologies. In these directions, we first notecyclicity in the weak topology is equivalent to cyclicity in the norm topology becauseweak closure of the convex set spanOrb(T , x) coincides with the norm closure. Howeveit has been shown in [3, Corollary 3.3] that there exists a weakly hypercyclic operatofails to be norm hypercyclic, and so it naturally follows to ask whether there exists a wsupercyclic operator that fails to be norm supercyclic.

In Section 2, we show similarities and differences between norm supercyclicitweak supercyclicity. Moreover, we show there exists weakly supercyclic operators thto be norm supercyclic. Hilden and Wallen [8, Theorem 3] in 1974 proved that any ueral weighted backward shift on�p with 1 � p < ∞ is norm supercyclic, and thus weaksupercyclic. It follows to ask if every weakly supercyclic vector for a unilateral weigbackward shift is also a norm supercyclic vector. In Section 3, we show that if the wsequence of a unilateral weighted backward shift on�p with 1 < p < ∞ is increasing,then the shift has many weakly supercyclic vectors that are not norm supercyclic; seorem 3.1. In Kitai’s dissertation [9, Corollary 4.5], she showed a hyponormal operatoHilbert space cannot be norm hypercyclic. From this, Chan and Sanders [3, Questiasked if there exists a weakly hypercyclic hyponormal operator on a Hilbert space. In Stion 4, we show a hyponormal operator on a Hilbert space cannot be weakly hypersee Theorem 4.5.

2. Weak supercyclicity

In a separable, infinite dimensional Banach space overC, the weak topology is strictlyweaker than the norm topology. However, norm supercyclic operators and weaklycyclic operators share many of the same properties. For example, for a boundedoperatorT :X → X, it immediately follows from the definitions that every norm supcyclic vector forT is also a cyclic vector forT . Similarly, every weakly supercyclic vectois also a weakly cyclic vector and hence a cyclic vector, as we have mentionedIntroduction.

Another property shared by norm and weak supercyclic operators involve the densitysupercyclic vectors. Ifx is a norm supercyclic vector for the operatorT :X → X, then sois every vector in the norm dense set{λT nx: n � 0, λ ∈ C \ {0}}. Thus, the existence oone norm supercyclic vector forT implies that the set of all norm supercyclic vectors foT

is norm dense inX. Using a similar argument, it follows that if the operatorT :X → X isweakly supercyclic, then the set of weakly supercyclic vectors forT is weakly dense inX.In fact, we have something even stronger.

Proposition 2.1. Let T :X → X be a bounded linear operator. IfT is weakly supercyclicthen the set of weakly supercyclic vectors forT is norm dense inX.

150 R. Sanders / J. Math. Anal. Appl. 292 (2004) 148–159

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Proof. Let x be a weakly supercyclic vector forT . Observe that for any nonzero polynmial p, we have

p(T )X ⊆ p(T ){λT nx: n � 0, λ ∈ C

}wk = {

λT np(T )x: n � 0, λ ∈ C}wk.

Thus,p(T )x is a weakly supercyclic vector forT whenever the operatorp(T ) has a weaklydense range. Now, the operatorp(T ) has a weakly dense range if and only ifp(α) �= 0for all eigenvaluesα of the adjointT ∗ of T . A proof of this result can be found in [10Lemma 2]. These two facts combined give us that the set

A = {p(T )x: p(α) �= 0 for all eigenvaluesα of T ∗}

is contained in the set of weakly supercyclic vectors forT , and it suffices to show this seA is norm dense inX.

Recall thatX is a locally convex space under the weak topology. Thus, by a rof Peris [10, Lemma 1], the adjointT ∗ of T has at most one eigenvalue. IfT ∗ has noeigenvalues, then the setA is given by span{T nx: n � 0} \ {0}, and this subspace is weakdense and hence norm dense inX. If T ∗ has one eigenvalueα, then the setA is given by{p(T )x: p(α) �= 0}, and again this set is norm dense inX because it is norm dense in thsubspace span{T nx: n � 0}. �

If X is a Banach space overC, then X ⊕ C is a Banach space under the no‖x ⊕ α‖ = ‖x‖ + |α|. Herrero [7, Lemma 3.2] showed that ifx is a norm hypercyclicvector forT :X → X, thenx ⊕ 1 is a norm supercyclic vector forT ⊕ IC. González et al[6, Theorem 5.2] extended the result by showing the vectorx ⊕ µ is a norm supercyclicvector forT ⊕ αIC if and only if µ �= 0 andx is a norm hypercyclic vector for(1/α)T .Using similar ideas, a weak topology version of these results can be established.

Theorem 2.2. LetT :X → X be a bounded linear operator and letα be a nonzero scalaThe vectorx ⊕µ in X ⊕ C is a weakly supercyclic vector forT ⊕αIC if and only ifµ �= 0andx is a weakly hypercyclic vector forα−1T . Therefore,T ⊕ αIC is weakly supercyclicif and only ifα−1T is weakly hypercyclic.

Proof. To begin, we show the statement holds true forα = µ = 1. Supposex is a weaklyhypercyclic vector forT . To showx ⊕ 1 is a weakly supercyclic vector forT ⊕ IC, choosey ⊕ β in X ⊕ C with β �= 0, ε > 0, andx∗

1, . . . , x∗m in (X ⊕ C)∗. Sincex is a weakly

hypercyclic vector forT , there exists a integern � 0 such that∣∣⟨T nx − β−1y, x∗i

⟩∣∣ < |β|−1ε for 1 � i � m,

wherex∗1, . . . , x∗

m in X∗ are given byx∗i (z) = x∗

i (z ⊕ 0). Thus, for 1� i � m, we have∣∣⟨β(T ⊕ IC)n(x ⊕ 1) − y ⊕ β, x∗i

⟩∣∣ = ∣∣⟨(βT nx − y) ⊕ 0, x∗i

⟩∣∣ = ∣∣⟨βT nx − y, x∗i

⟩∣∣= |β|∣∣⟨T nx − β−1y, x∗

i

⟩∣∣ < ε.

To prove the other direction, supposex ⊕ 1 is a weakly supercyclic vector forT ⊕ IC.Choosey in X, ε > 0 andx∗, . . . , x∗

m in X∗. Selectδ with 0 < δ < 1 such that
1


r

t

sult.the]

show,i-

δ + δ

1− δ(M + δ) < ε, (2.1)

whereM = max{|〈y, x∗i 〉|: 1 � i � m}. Sincex ⊕ 1 is a weakly supercyclic vector fo

T ⊕ IC, there exists integern � 0 and a nonzero scalarλ such that∣∣⟨λ(T ⊕ IC)n(x ⊕ 1) − y ⊕ 1, x∗i

⟩∣∣ < δ for 1 � i � m + 1,

wherex∗1, . . . , x∗

m, x∗m+1 in (X ⊕ C)∗ are given byx∗

i (z ⊕ β) = x∗i (z) for 1 � i � m and

x∗m+1(z ⊕ β) = β . Thus, we have

|λ − 1| = ∣∣⟨λ(T ⊕ IC)n(x ⊕ 1) − y ⊕ 1, x∗m+1

⟩∣∣ < δ, (2.2)

and for 1� i � m, we have∣∣⟨λT nx − y, x∗i

⟩∣∣ = ∣∣⟨λ(T ⊕ IC)n(x ⊕ 1) − y ⊕ 1, x∗i

⟩∣∣ < δ. (2.3)

Therefore,∣∣⟨T nx − y, x∗i

⟩∣∣ �∣∣⟨λT nx − y, x∗

i

⟩∣∣ + |λ|−1|1− λ|∣∣⟨λT nx, x∗i

⟩∣∣< δ + |λ|−1|1− λ|(∣∣⟨y, x∗

i

⟩∣∣ + δ)

by (2.3)

< δ + δ(1− δ)−1(∣∣⟨y, x∗i

⟩∣∣ + δ)

by (2.2)

< ε by (2.1).

To complete the proof, observe thatx ⊕ µ is a weakly supercyclic vector forT ⊕ αIC

if and only if µ �= 0 andµ−1x ⊕1 is a weakly supercyclic vector forα−1T ⊕ IC. From thework above, the vectorµ−1x ⊕1 is a weakly supercyclic vector forα−1T ⊕ IC if and onlyif µ−1x is a weakly hypercyclic vector forα−1T . The result now follows from the facthatµ−1x is a weakly hypercyclic vector forα−1T if and only if x is a weakly hypercyclicvector forα−1T . �

Combining Theorem 2.2 with the norm topology version, we get the following reThe operatorT ⊕IC is weakly supercyclic but fails to be norm supercyclic if and only ifoperatorT is weakly hypercyclic but fails to be norm hypercyclic.Chan and Sanders [3provided examples of weakly hypercyclic operators on�p(Z) with 2 � p < ∞ that failedto be norm hypercyclic. The operatorT : �p(Z) → �p(Z) given byT eα = eα−1 if α � 0 andT eα = 2eα−1 if α � 1 is one such operator. Since�p(Z) ⊕ C is isometrically isomorphicto �p(Z), it follows that there exists weakly supercyclic operators on�p(Z) for 2 � p < ∞that fail to be norm supercyclic.

The existence of weakly supercyclic but not norm supercyclic operators candifferences between the weak topology and the norm topology on�p(Z). For examplewhen p = 2, the space�2(Z) is a separable, infinite dimensional Hilbert space. For ether type of supercyclicity, to provef is a supercyclic vector forT it suffices to show{g ∈ �2(Z): ‖g‖2 = 1} contained in the closure of{λT nf : n � 0, λ ∈ C}. In the caseof the norm topology, we havef is a norm supercyclic vector forT if and only if{‖λT nf ‖−1

2 λT nf : n � 0, λ ∈ C \ {0}} is norm dense in the unit sphere of�2(Z). Theforward implication follows from the fact that the set{‖h‖−1

2 h: h ∈ A} is norm dense inthe unit sphere of�2(Z) whenever the setA is norm dense in�2(Z). It is interesting to ask


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if we have a similar result for weak supercyclicity. Before we see an answer, we firsto prove the following result about subsets of the unit sphere of�2(Z).

Proposition 2.3. Let B be a subset of the unit sphere of�2(Z). Then the setB is weaklydense in the unit sphere of�2(Z) if and only if it is norm dense in the unit sphere of�2(Z).

Proof. Let S be the unit sphere of�2(Z). Since the weak topology on�2(Z) is strictlyweaker than the norm topology, it suffices to show that if the setB is weakly dense inS,then it is norm dense inS. Observe that for any vectorsf,g in S and for anyε > 0, wehave

‖f − g‖22 < ε2

if and only if

2− ε2 < 2 Re〈f,g〉.Therefore,

‖f − g‖2 < ε if and only if 1− ε2/2 < Re〈f,g〉. (2.4)

Let g be any vector inS and letε > 0. Consider the weakly open setU = {f ∈ �2(Z):|〈f − g,g〉| < ε2/2}. Since the setB is weakly dense inS, there exists a vectorh in B

such that|〈h − g,g〉| < ε2/2. Thus, we have∣∣Re〈h,g〉 − 1∣∣ = ∣∣Re

(〈h,g〉 − 1)∣∣ �

∣∣〈h,g〉 − 1∣∣ = ∣∣〈h − g,g〉∣∣ < ε2/2.

This implies that−ε2/2 < Re〈h,g〉 − 1, or equivalently, 1− ε2/2 < Re〈h,g〉. Hence,from (2.4), we get‖h − g‖2 < ε. �

Since the weak closure of the unit sphere of�2(Z) is the closed unit ball [4, p. 128from Proposition 2.3, we get that a subset of the unit sphere in�2(Z) is weakly dense inthe closed unit ball if and only if it is norm dense in the unit sphere. On the other hanprovide the following result in contrast with the norm topology.

Corollary 2.4. There exists a weakly dense setA of nonzero vectors in�2(Z) such that theset{‖h‖−1

2 h: h ∈ A} fails to be weakly dense in the unit sphere of�2(Z).

Proof. Let S be the unit sphere of�2(Z), and letf be a weakly supercyclic vector foa weakly supercyclic operatorT that fails to be norm supercyclic. LetA be the weaklydense set given byA = {λT nf : n � 0, λ ∈ C \ {0}}. To see the set{‖h‖−1

2 h: h ∈ A} is notweakly dense inS, observe that if the set is weakly dense inS, then, by Proposition 2.3it is also norm dense inS. However, this implies that the set{β‖h‖−1

2 h: h ∈ A,β ∈ C} ={λT nf : n � 0, λ ∈ C} is norm dense in�2(Z) which is a contradiction. �

From the proof of Corollary 2.4, it follows that iff is a weakly supercyclic vector foT : �2(Z) → �2(Z) such that{‖λT nf ‖−1

2 λT nf : n � 0, λ ∈ C\ {0}} is weakly dense in theunit sphere of�2(Z), thenf is a norm supercyclic vector forT .


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With Theorem 2.2, we can show that some properties for norm supercyclic operatnot necessarily hold for weakly supercyclic operators. For example, an invertible opT :X → X is norm supercyclic if and only ifT −1 is norm supercyclic. For a proof of thresult, see [1] or [11, Corollary 2.4]. Theorem 2.2 provides a counterexample to thisin the weak topology.

Corollary 2.5. For 2� p < ∞, there exists an invertible operator on�p(Z) that is weaklysupercyclic but its inverse is not weakly supercyclic.

Proof. Let T : �p(Z) → �p(Z) be the weakly hypercyclic operator given byT eα = eα−1 ifα � 0 andT eα = 2eα−1 if α � 1; see [3]. By Theorem 2.2, the operatorT ⊕ IC is weaklysupercyclic. However, since‖T −1‖ = 1, the operatorT −1 is not weakly hypercyclic, anso(T ⊕ IC)−1 = T −1 ⊕ IC is not weakly supercyclic. �

The proof Salas provided in [11, Corollary 2.4] to showT is norm supercyclic if andonly if T −1 is norm supercyclic used the Baire category theorem which is not avain the weak topology. Thus, it is not surprising that the inverse of a weakly supercoperator is not necessarily weakly supercyclic.

3. Unilateral weighted shifts

Let {eα: α � 0} be the canonical basis for�p = �p(Z+). That is, the vectoreα is thesequence(0, . . . ,0,1,0, . . .) where the 1 is inαth position. For a bounded subset of positreal numbers{wα : α � 1}, the operatorT : �p → �p defined byT eα = wαeα−1 for α � 1and T e0 = 0 is called aunilateral weighted backward shiftwith respect to the weighsequence{wα : α � 1}. In 1974, Hilden and Wallen [8, Theorem 3] showed that evunilateral weighted backward shift on�p with 1 � p < ∞ is norm supercyclic, and hencweakly supercyclic. It naturally follows to ask whether every weakly supercyclic vefor a unilateral weighted backward shift is also a norm supercyclic vector. With a scondition on the weight sequence, we can show that the set of weakly supercyclic vthat fail to be norm supercyclic is norm dense.

Theorem 3.1. Let 1 < p < ∞. If T : �p → �p is a unilateral weighted backward shiwhose weight sequence{wα : α � 1} satisfies the conditionwα � wα+1 for eachα � 1,then the set of weakly supercyclic vectors forT that fail to be norm supercyclic vectors iin fact, norm dense in�p .

Proof. To begin, consider the norm dense setA = {f + ∑mα=0 βαeα : m � 0, β0, . . . , βm

∈ C}. SinceT m+n+1(f +∑mα=0 βαeα) = T m+n+1f , it follows that wheneverf is a weakly

supercyclic vector forT that fails to be norm supercyclic, then so is every vector innorm dense setA. Therefore, the existence of one vector of this type implies that the sall such vectors is norm dense in�p .

To show there exists a weakly supercyclic vector that fail to be norm supercyclicconsider the case when 1< wα � wα+1 for eachα � 1. For eachh ∈ �p , defineh(α) =


ers

e-

〈h, eα〉. Let {hi : i � 1} be a countable set in�p satisfying{hi : i � 1}norm= {g: ‖g‖p � 1},and also that there is a strictly increasing sequence of positive integers(mi)

∞i=1 such that

for eachi � 1,

hi (α) = 0 wheneverα > mi.

Let β = w1 > 1 and setn0 = m0 = 0. Inductively choose a sequence of positive integ(ni)

∞i=1 that satisfy

ni > mi +i−1∑j=0

(nj + mj + 2) and βni > ‖T ‖i+ni−1(1+ 4p)1/p (3.1)

for each i � 1. Let ki = ni+1 − ni − 2 and define the mapS : span{eα : α � 0} →span{eα: α � 0} by lettingSeα = w−1

α+1eα+1 and extending linearly.

Claim 1. For eachi � 1, we have‖Sni (hi + 4eki )‖p � ‖T ‖−i−ni−1 .

Proof. Observe that for anyh ∈ span{eα: α � 0},‖Sh‖p � β−1‖h‖p,

and so,‖Sni (hi + 4eki )‖pp � β−pni ‖hi + 4eki‖p

p. Next, notice that from (3.1) and the dfinition of ki , we getki > mi , and hencehi(ki) = 0. Therefore,‖hi + 4eki‖p

p = ‖hi‖pp +

‖4eki ‖pp � 1+ 4p, where the last inequality holds because‖hi‖p � 1. The claim now fol-

lows from the choice ofni in (3.1). �Since‖T ‖ = sup{wα: α � 1} > 1, by this claim, the sumh = ∑∞

i=1 Sni (hi + 4eki ) isan absolutely convergent series in�p. To prove thath is a weakly supercyclic vector forT ,we show the set{λT nh: n � 0, λ ∈ C} is weakly sequentially dense in�p. First, observethat by the definition ofT and the definition ofki , for anyj � 2, we have

T nj Sni (hi + 4eki ) = T nj −ni (hi + 4eki ) = 0 (3.2)

for i = 1, . . . , j − 1. Hence, for a vectorg ∈ �p , a vectorx ∈ �q , wherep−1 + q−1 = 1,and a nonzero scalarλ, we have∣∣〈λT nj h − g,x〉∣∣ = |λ|∣∣〈T nj h − λ−1g,x〉∣∣

= |λ|∣∣∣∣∣⟨hj + 4ekj − λ−1g +

∞∑i=j+1

T nj Sni (hi + 4eki ), x

⟩∣∣∣∣∣� |λ|(‖hj − λ−1g‖p‖x|q

) + 4∣∣λ〈ekj , x〉∣∣

+ |λ|‖x‖q

∞∑i=j+1

‖T ‖nj∥∥Sni (hi + 4eki )

∥∥p

� |λ|(‖hj − λ−1g‖p‖x‖q

) + 4∣∣λ〈ekj , x〉∣∣


+ |λ|‖x‖q

∞∑i=j+1

‖T ‖−i by Claim 1

= |λ|(‖hj − λ−1g‖p‖x‖q

) + 4|λ〈ekj , x〉| + |λ|‖x‖q

‖T ‖j (‖T ‖ − 1)

for any j � 2. Therefore, since{hi : i � 1}norm is the closed unit ball of�p(Z), for anynonzero vectorg in �p , choose a strictly increasing sequence(jm)∞m=1 such that

‖hjm − λ−1g‖p → 0 asm → ∞,

whereλ = ‖g‖p , and it follows thatλT njm h → g weakly asm → ∞ from the estimationabove.

To showh is not a norm supercyclic vector forT , we prove the vectore0 + e1 is not inthe norm closure of the set{λT nh: n � 0, λ ∈ C}.

Claim 2. Let 0 < ε < 1. If for somen � n1 andλ ∈ C, we have|λ〈T nh, eα〉 − 1| < ε forα = 0,1, then‖λT nh‖p � 4(1− ε).

Proof. For eachi � 1, letfi = hi + 4eki . Then

fi (α) =

hi (α) if 0 � α � mi ,

4 if α = ki = ni+1 − ni − 2,

0, otherwise.

Sincen � n1, there existsj � 1 and an integerm satisfying 0� m � nj+1 − nj − 1 suchthatn = nj + m. This gives

T nh = T nSnj fj +∞∑

i=j+1

T nSni fi by (3.2)

= T mfj +∞∑

i=j+1

T nSni fi .

By the definition of the linear mapS and the definition of the vectorfi , for anyi � 1, weget

Sni fi(α) �= 0 only if ni � α � ki + ni = ni+1 − 2. (3.3)

Hence, for anyi � j + 1,

T nSni fi(α) �= 0 only if ni − n � α � ni+1 − n − 2. (3.4)

In particular, sincen < nj+1, we have for anyi � j + 1,

T nSni fi(0) = 0.

Therefore,


on

〈T nh, e0〉 = 〈T mfj , e0〉 +∞∑

i=j+1

〈T nSni fi , e0〉

= fj (m)〈T mem, e0〉 = fj (m)

m−1∏t=0

wm−t . (3.5)

Moreover, since|λ〈T nh, e0〉 − 1| < ε < 1, we getfj (m) �= 0. From the definition of thevectorfj this implies that the integerm must satisfy 0� m � mj or m = kj = nj+1 −nj − 2. If m = kj = nj+1 − nj − 2, then for anyi � j + 1, by (3.4), we get

T nSni fi(1) = 0,

becausen = nj + m = nj+1 − 2 andni − n � 2. Hence,

〈T nh, e1〉 = 〈T mfj , e1〉 +∞∑

i=j+1

〈T nSni fi , e1〉 = fj (m + 1)〈T mem+1, e1〉

= fj (nj+1 − nj − 1)

m−1∏t=0

wm+1−t = 0.

However, this implies that|λ〈T nh, e1〉 − 1| = 1 which contradicts the original assumptithat|λ〈T nh, e1〉 − 1| < ε < 1. Therefore, the integerm must satisfy 0� m � mj . Further-more, from the definition of the vectorfj and (3.5), we get∣∣∣∣∣λhj (m)

m−1∏t=0

wm−t − 1

∣∣∣∣∣ = ∣∣λ〈T nh, e0〉 − 1∣∣ < ε,

and so,

1− ε < |λ|∣∣hj (m)∣∣m−1∏

t=0

wm−t < |λ|m−1∏t=0

wm−t , (3.6)

where the last inequality holds since‖hj‖p � 1. To conclude the claim, observe that

‖λT nh‖pp =

∞∑i=1

‖λT nSni fi‖pp by (3.3)

� ‖λT nSnj ‖pp = |λ|p∥∥T m(hj + 4ekj )

∥∥p

p

� 4p|λ|p‖T mekj ‖pp becausehj (kj ) = 0

= 4p|λ|pm−1∏t=1

wpkj −t becausem � mj < kj

� 4p|λ|pm−1∏t=1

wpm−t becausewα � wα+1

� 4p(1− ε)p by (3.6).

This ends the proof of Claim 2.�


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If the vectore0 + e1 is in the norm closure of the set{λT nh: n � 0, λ ∈ C}, then foranyε with 0 < ε < 4−1, there existsn � n1 andλ ∈ C such that∥∥λT nh − (e0 + e1)

∥∥p

< ε. (3.7)

Thus, we have|λ〈T nh, eα〉 − 1| < ε for α = 0,1. However, this implies

3 < 4(1− ε)

� ‖λT nh‖p by Claim 2

�∥∥λT nh − (e0 + e1)

∥∥p

+ ‖e0 + e1‖p � ε + 21/p < 3,

which is a contradiction.For the case whenwα � wα+1 for eachα � 1, let β = w1 and chooseγ > 0 such that

γβ > 1. Then the operatorγ T is a unilateral weighted backward shift with the weigsequence{γwα: α � 1}. Moreover, by the choice ofγ , we have 1< γwα � γwα+1 foreachα � 1. Therefore, by the case above, there exists a weakly supercyclic vectorf forγ T that fails to be a norm supercyclic vector. Lastly, since

〈λγ nT nf − h,x〉 = ⟨λ(γ T )nf − h,x

⟩and

‖λT nf − h‖p = ∥∥λγ −n(γ T )nf − h∥∥

p

for anyn � 1, λ ∈ C, h ∈ �p, andx ∈ �q , wherep−1 + q−1 = 1, it follows thatf is also aweakly supercyclic vector forT that fails to be a supercyclic vector.�

Proposition 2.1 gives us that a unilateral weighted backward shift on�p has a normdense set of weakly supercyclic vectors. However, some of these vectors may also bsupercyclic vector. On the other hand, Theorem 3.1 states that the set of weakly supevectors that fail to be norm supercyclic for a unilateral weighted backward shift isnorm dense.

4. Hyponormal operators

Let H be a separable, infinite dimensional Hilbert space. A bounded linear operatT :H → H is hyponormalif T ∗T � T T ∗. For example, the operator in the statemenTheorem 3.1 withp = 2 is the adjoint of a hyponormal shift.

Under the norm topology onH , Kitai [9, Corollary 4.5] showed that a hyponormoperator cannot be norm hypercyclic. Thus, it was natural for Chan and Sanders [3, Qution 5.3] to pose the following question:Does there exist a weakly hypercyclic hyponormoperator on a Hilbert space? In this section, by combining results from Kitai [9] and froDilworth and Troitsky [5], we show that a hyponormal operator cannot be weakly hcyclic, giving a negative answer.

To prove the main result of this section, weneed three lemmas. The first lemma westablished by Dilworth and Troitsky [5] which was used in their proof that the specof a weakly hypercyclic operator must intersect the unit circle.


a set

who

ector

Lemma 4.1. LetX be a Banach space and letc > 1. If {xn: n � 1} is a set inX such that‖xn‖ � cn for eachn, then0 /∈ {xn: n � 1}wk .

A proof of Lemma 4.1 can be found in [5]. Using Lemma 4.1, it can be shown thatsatisfying the conditions of Lemma 4.1 is, in fact, weakly closed.

Lemma 4.2. LetX be a Banach space and letc > 1. If {xn: n � 1} is a set inX such that‖xn‖ � cn for eachn, then{xn: n � 1} is weakly closed inX.

Proof. Let x ∈ {xn: n � 1}wk . Choose a positive integerN such that

‖x‖cn

<1

2

for eachn � N . Let yj = 2c−N(xj+N − x) for eachj � 1. Observe that

‖yj‖ � 2c−N(‖xj+N‖ − ‖x‖) � 2c−N

(cj+N − ‖x‖) = 2cj

(1− c−j−N‖x‖) > cj .

From Lemma 4.2, it follows that 0/∈ {yj : j � 1}wk , and so 0/∈ {xn − x: n > N}wk . How-ever, the zero vector is in the set

{xn − x: n � 1}wk = {xi − x: 1� i � N} ∪ {xn − x: n > N}wk.

Hence,x = xi for somei with 1 � i � N . �The third lemma gives a relationship between a fixed nonzero vectorg in H and all the

other vectors in Orb(T , g) when the operatorT is hyponormal.

Lemma 4.3. Let T :H → H be a hyponormal operator on a Hilbert spaceH . Then forany nonzero vectorg in H, we have

‖T ng‖ � ‖g‖(‖Tg‖

‖g‖)n

for eachn � 0.

A proof of Lemma 4.3 can be found in the dissertation of Kitai [9, Lemma 4.3],used the definition of a hyponormal operator to show‖T 2g‖ � ‖g‖(‖Tg‖/‖g‖)2 for allnonzero vectorsg in H , and then used induction to establish the result for eachn � 2.Using the lemmas above, we can establish a sufficient condition for the orbit of a vunder a hyponormal operator to be weakly closed.

Proposition 4.4. Let T :H → H be a hyponormal operator on a Hilbert spaceH . If g isa vector inH such that‖Tg‖ > ‖g‖, thenOrb(T , g) is weakly closed inH .

Proof. Let c = ‖Tg‖/‖g‖ > 1. From Lemma 4.3, we have∥∥T n(‖g‖−1g

)∥∥ = ‖g‖−1‖T ng‖ �(‖Tg‖

‖g‖)n

= cn

for eachn � 0. Thus, by Lemma 4.2, it follows that Orb(T ,‖g‖−1g) is weakly closedin H . Therefore, Orb(T , g) is weakly closed inH . �


e

ot be

kly,

ly-.5lic.rm su-

for the

n

.

23

Using Proposition 4.4, we now show the main result of this section.

Theorem 4.5. LetT :H → H be a bounded linear operator on a Hilbert spaceH . If T ishyponormal, thenT is not weakly hypercyclic.

Proof. By way of contradiction, suppose thatg is a weakly hypercyclic vector for thoperatorT . Observe that if‖T n+1g‖ � ‖T ng‖ for eachn � 1, then Orb(T , g) is a normbounded subset ofH . However, this is impossible because a weakly dense set cannnorm bounded. Thus,‖T N+1g‖ > ‖T Ng‖ for someN � 1. Even further, since theg is aweakly hypercyclic vector forT , so is the vectorT Ng. Thus, we may assume that a weahypercyclic vectorg for T has the property that‖Tg‖ > ‖g‖. However, by Proposition 4.4we get that Orb(T , g) is weakly closed which is a contradiction becauseg is a weaklyhypercyclic vector forT . �

A bounded linear operatorT :H → H is normal if T T ∗ = T ∗T . Since every normaoperator is hyponormal, Theorem 4.5 tells us that a normal operator cannot be weakly hpercyclic. Also, since every norm hypercyclicoperator is weakly hypercyclic, Theorem 4also gives us Kitai’s original result that ahyponormal operator cannot be norm hypercycBourdon [2, Theorem 3.1] showed that a hyponormal operator, in fact, cannot be nopercyclic. Hence, we pose the following question.

Question 4.6. Does there exist a weakly supercyclic hyponormal operator?

Acknowledgment

The author would like to thank the referee for providing many valuable comments and simpler proofsmain results.

References

[1] S.I. Ansari, P.S. Bourdon, Some properties of cyclic operators, Acta Sci. Math. 63 (1997) 195–207.[2] P.S. Bourdon, Orbits of hyponormal operators, Michigan Math. J. 44 (1997) 345–353.[3] K.C. Chan, R. Sanders, A weaklyhypercyclic operator that is not normhypercyclic, J. Operator Theory, i

press.[4] J.B. Conway, A Course in Functional Analysis, second ed., Springer-Verlag, New York, 1990.[5] S.J. Dilworth, V.G. Troitsky, Spectrum of a weaklyhypercyclic operator meets theunit circle, Preprint, 2002[6] M. González, F. León-Saavedra, A. Montes-Rodríguez, Semi-Fredholm theory:hypercyclic and supercyclic

subspaces, Proc. London Math. Soc., in press.[7] D.A. Herrero, Limits of hypercyclic and supercyclic operators, J. Funct. Anal. 99 (1991) 179–190.[8] H.M. Hilden, L.J. Wallen, Some cyclic and non-cyclicvectors of certain operators, Indiana Univ. Math. J.

(1974) 557–565.[9] C. Kitai, Invariant closed sets for linear operators, Ph.D. thesis, University of Toronto, 1982.

[10] A. Peris, Multi-hypercyclic operators are hypercyclic, Math. Z. 236 (2001) 779–786.[11] H. Salas, Supercyclicity and weighted shifts, Studia Math. 135 (1999) 55–74.

Weakly supercyclic operators

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