Iran J Sci Technol Trans Sci https://doi.org/10.1007/s40995-017-0325-7

RESEARCH PAPER

A Note on Topological and Strict Transitivity Mohammad Ansari1,2 • Karim Hedayatian1 • Bahram Khani Robati1 • Abbas Moradi1 Received: 20 October 2015 / Accepted: 18 June 2016 Ó Shiraz University 2017

Abstract Let X be a normed linear space and LðXÞ be the algebra of continuous linear operators on X. We give a necessary condition for topological transitivity of subsets of LðXÞ which gives a necessary condition for hypercyclicity and supercyclicity of a single operator on X. Also, we prove the strict transitivity of some particular families of operators on locally convex spaces and Hilbert spaces. Keywords Hypercyclic  Supercyclic  Topologically transitive  Strictly transitive

1 Introduction Let X be a topological vector space over the field of complex numbers C. Throughout this paper, we assume that dim X [ 1. Denote by LðXÞ, the algebra of all continuous linear operators on X. By an operator, we always mean a continuous linear operator. We say that a set C  LðXÞ is hypercyclic if there exists some x 2 X, called a hypercyclic vector, for which orbðC; xÞ ¼ fTx : T 2 Cg is a dense subset of X. The set C is said to be supercyclic if C:orbðC; xÞ ¼ fkTx : T 2 C; k 2 Cg is a dense subset of X for some x 2 X which is called a supercyclic vector. C is called cyclic if spanðorbðC; xÞÞ is dense in X for some x 2 X that is called a cyclic vector for C. An operator T 2 LðXÞ is called hypercyclic (supercyclic, cyclic) if C ¼ fT n : n 2 N0 g is hypercyclic (supercyclic, cyclic), where N0 ¼ f0; 1; 2; 3; . . .g and T 0 ¼ I, the & Bahram Khani Robati [email protected] Mohammad Ansari [email protected] Karim Hedayatian [email protected] Abbas Moradi [email protected] 1

Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71467-13565, Iran

2

Present Address: Islamic Azad University of Gachsaran, Dogonbadan, Iran

identity operator on X. In this case, we write orbðT; xÞ instead of orbðC; xÞ. By HCðCÞ, SCðCÞ, and CðCÞ, we mean, respectively, the set of all hypercyclic vectors, supercyclic vectors, and cyclic vectors for C. These are denoted by HCðTÞ, SCðTÞ, and CðTÞ when C ¼ fT n : n 2 N0 g. Two historic examples of hypercyclic operators are the derivative operator D on HðCÞ which is due to Maclane (Maclane 1952), and T ¼ 2B, twice the backward shift on ‘2 ðNÞ, due to Rolewics (1969). The reader can also see (Bayart and Matheron 2009; Grosse-Erdmann and Peris 2011) for more information. Recall that HðCÞ is the space of analytic functions on C and ‘2 ðNÞ is the space of all sequences ðan Þn in C for which jjðan Þn jj22 ¼ P1 2 n¼0 jan j \1: In Sect. 2, we investigate topological transitivity and hypercyclicity of sets of operators. In Sect. 3, we give a necessary condition for topological transitivity of subsets of LðXÞ where X is a normed linear space. We apply it for some classes of operators on Banach and Hilbert spaces. In the last section, some particular families of operators are proved to be strictly transitive; a small subset of unitary operators on a Hilbert space and a small subclass of involutions together with scalar operators on a locally convex space. By a small subset, we mean those operators whose distance to the identity operator is at most 1 in the sense we will define in the relevant section.

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2 Topological Transitivity and Hypercyclicity An operator T 2 LðXÞ is called topologically transitive if for each pair of non-empty open sets U; V in X there exits some n 2 N0 such that T n ðUÞ \ V 6¼ ;. The author in (Grosse-Erdmann 1999) has used this notion for an arbitrary family of operators which in the special case of dealing with a single operator T, the relevant family is in fact the set fT n : n 2 N0 g. We use the name topological transitivity for this generalization as well. Definition 1 Let X be a topological vector space and C  LðXÞ. The set C is said to be topologically transitive if for each pair of non-empty open sets U, V in X there exists some T 2 C such that TðUÞ \ V 6¼ ;. The following proposition is a direct consequence of Theorem 1 of (Grosse-Erdmann 1999).

proposition. Thus, there is a polynomial p such that pðCÞT 6¼ TpðCÞ. Hence, CT 6¼ CT and we are done. Example 1 Let X ¼ C00 ðNÞ, the space of all finitely supported sequences in the complex plane, and T ¼ 2B, twice the backward shift on X. Then T is a locally nilpotent operator which is not nilpotent. If C ¼ fTg0 then C is not hypercyclic by Proposition 2. We show that C is topologically transitive by proving that T is topologically transitive (fT n : n 2 N0 g  fTg0 ). Let U; V be two non-empty open sets in X. Let x ¼ ða1 ; a2 ; . . .; ak ; 0; 0; . . .Þ 2 U, y ¼ ðb1 ; b2 ; . . .; bm ; 0; 0; . . .Þ 2 V, and e [ 0 satisfy Dðx; eÞ ¼ fz : jjz  xjj\eg  U. Choose n [ k large enough such that 2n jjyjj\e. If z ¼ ða1 ; a2 ; . . .; ak ; 0; 0; . . .; 2n b1 ; 2n b2 ; . . .; 2n bm ; 0; 0; . . .Þ;

where 2n b1 is in the ðn þ 1Þth position, it can be easily seen that z 2 U and T n z ¼ y. Thus, T n ðUÞ \ V 6¼ ;.

Proposition 2 Let X be a topological vector space and T 2 LðXÞ be a locally nilpotent operator which is not nilpotent. Then fTg0 , the commutant of T, is not hypercyclic.

Suppose T 2 LðXÞ is a hypercyclic operator with hypercyclic vector x. Then for every positive integer k, orbðT; T k xÞ ¼ orbðT; xÞnfx; Tx; . . .; T k xg. So, orbðT; T k xÞ is dense in X for all k 2 N. This shows that orbðT; xÞ  HCðTÞ and hence HCðTÞ is dense in X. Hence, the hypercyclicity of T implies that T is topologically transitive by the above remark and whenever X is second countable and Baire, T is hypercyclic if and only if T is topologically transitive. Assume that T 2 LðXÞ is supercyclic and x 2 SCðTÞ. Then, fkT n x : k 2 C; n 2 N0 g ¼ fkT n x : k 2 C; n 2 Ng [ Cx and since Cx is nowhere dense, the set fkT n x : k 2 C; n 2 Ng  RanðTÞ is dense in X and so T is a denserange operator. Then, an easy use of the mathematical induction shows that T k is also dense-range for all k 2 N. The result given in the following remark is a particular case of Proposition 1 of Grosse-Erdmann (1999) where the authors mention that it is due to Peris.

Proof To get a contradiction, suppose that x is a hypercyclic vector for fTg0 : Since T is a locally nilpotent operator, there is some n 2 N such that T n x ¼ 0. Since T is not nilpotent, there is some y 2 X such that T n y 6¼ 0. On the other hand, there is a net ðTj Þj in fTg0 such that Tj x ! y. Then 0 ¼ Tj T n x ¼ T n Tj x ! T n y which is a contradiction. Thus, fTg0 is not hypercyclic.

Remark 2 If X is a topological vector space and T 2 LðXÞ, then SCðTÞ is either empty or dense in X. To see this, let x 2 SCðTÞ and k 2 N. Then, we have C:orbðT; T k xÞ ¼ T k ðC:orbðT; xÞÞ  T k ðC:orbðT; xÞÞ ¼ T k ðXÞ: Hence, C:orbðT; T k xÞ ¼ X which shows that T k x 2 SCðTÞ for all for all pairs k2N and so kT k x 2 SCðTÞ ðk; kÞ 2 ðCnf0gÞ  N. Thus, SCðTÞ is dense in X.

Proposition 1 Let X be a second countable Baire topological vector space and C  LðXÞ. Then, C is topologically transitive if and only if HCðCÞ ¼ X. Remark 1 The ‘‘if’’ part of Proposition 1 holds for every topological vector space X. To see this, let U and V be two non-empty open sets in X. Since HCðCÞ ¼ X, there exists x 2 U \ HCðCÞ. This implies that orbðC; xÞ ¼ X and therefore, there exists T 2 C such that Tx 2 orbðC; xÞ \ V; hence Tx 2 TðUÞ \ V. Example 1 presents a topologically transitive set of operator which is not hypercyclic. Recall that an operator T 2 LðXÞ is called locally nilpotent if for every x 2 X there is some n 2 N such that T n x ¼ 0.

Corollary 1 Cyclic operators do not commute with nonnilpotent locally nilpotent operators. Proof Assume that C is a cyclic operator. The cyclicity of C (by definition) is equivalent to the hypercyclicity of C ¼ fpðCÞ : p apolynomialg: Suppose that T is a locally nilpotent operator which is not nilpotent. Since C is hypercyclic, it can not be a subset of fTg0 by the above

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3 The Infimum Lemma and its Consequences The following lemma gives a necessary condition for topological transitivity of a subset C of LðXÞ when X is a normed linear space.

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Lemma 1 (The infimum lemma) Let X be a normed linear space and C  LðXÞ. If C is topologically transitive, then inf T2C jjTxyjj 1þjjTjj ¼ 0 for all x; y in X. Proof

Assume that there exist x; y in X and d [ 0 such

that inf T2C jjTxyjj 1þjjTjj [ d. Thus, jjTx  yjj [ dð1 þ jjTjjÞ for all T 2 C. Choose a positive number e\d and put U ¼ Dðx; eÞ and V ¼ Dðy; eÞ. Since TðUÞ \ V 6¼ ; for some T 2 C, there exists z 2 U such that Tz 2 V; then jjx  zjj\e and jjTz  yjj\e. Therefore, jjTx  Tzjj\ejjTjj and by triangle inequality we must have jjTx  yjj\eð1 þ jjTjjÞ, which is a contradiction. Corollary 2

T kh ¼

0 X

cn enk þ

1

N X

2n cn enk ;

1

and this gives jjT k hjj\

0 X

jcn j2 þ 2N

N X

1

jcn j2 \2N jjhjj2 :

1

Let X be a normed linear space and n

xyjj T 2 LðXÞ. If T is hypercyclic then inf n2N0 jjT 1þjjT n jj ¼ 0 for all

x; y 2 X. Proof This is an especial case of Lemma 1 with C ¼ fT n : n 2 N0 g. Since SCðTÞ ¼ HCfkT n : k 2 C; n 2 N0 g, by Remarks 1 and 2, the supercyclicity of T implies that fkT n : k 2 C; n 2 N0 g is topologically transitive. Thus, Lemma 1 gives the following corollary. Corollary 3

0 for arbitrary pair x; y 2 ‘2 ðZÞ. Let e [ 0 and P1 P 2 2 x¼ 1 1 cn en . Since jjxjj ¼ 1 jcn j , there is an inteP1 2 e ger N such that n [ N implies m¼n jcm j \ 3. Put PN h ¼ 1 cn en . For each k  1, we have

Let X be a normed linear space and

T 2 LðXÞ. If T is supercyclic, then

jjkT n xyjj inf ðk;nÞ2CN0 1þjkjjjT n jj

¼

0 for all x; y 2 X.

On the other hand, jjT n jj  jjT n en jj ¼ jj2n e0 jj ¼ 2n which shows that jjT n jj ! 1. Hence, for sufficiently large n, we have

jjT n hjj e 1þjjT n jj \ 3

n

and

n

jjyjj e 1þjjT n jj \ 3.

Finally, we have:

n

jjT x  yjj jjT hjj jjT ðx  hÞjj jjyjj  þ þ \e; n n n 1 þ jjT jj 1 þ jjT jj 1 þ jjT jj 1 þ jjT n jj for large enough integer n and hence, the desired result is obtained. n

xyjj The condition inf n jjT 1þjjT n jj ¼ 0 for all x; y 2 X, implies

that supn jjT n jj ¼ 1 and this, in turn, gives jjT n jj  1 for all n 2 N0 . Then, we have 12 ð1 þ jjT n jjÞ  jjT n jj for all n 2 N0 and so the hypercyclicity of an operator T implies that n

Example 1 shows that the converse of Corollary 2 is not true. In that example, although supjjT n jj ¼ 1, for each vector x there is a positive integer n such that T n x ¼ 0. This condition does not hold in Banach spaces because of the principle of uniform boundedness. The following example shows that the converse of Corollary 2 is not true even for Banach spaces. Before it, we need the following lemma. Note that the bilateral weighted backward shift BW on ‘2 ðZÞ is defined by BW ðen Þ ¼ wn en1 (n 2 Z), where ðen Þn2Z is the canonical basis of ‘2 ðZÞ and W ¼ ðwn Þn2Z is a bounded sequence of positive numbers. Lemma 2 (Theorem 2.1 of Salas 1995) Let BW be a bilateral weighted backward shift on ‘2 ðZÞ, with weight sequence W ¼ ðwn Þn2Z . Then BW is hypercyclic if and only if, for any q 2 N, 1

liminf maxfðw1    wnþq Þ ; ðw0    wnþqþ1 Þg ¼ 0:

inf n2N0 jjTjjTxyjj ¼ 0 for all x; y 2 X: Finally, by choosing n jj y ¼ 0 in this equation, we have Let X be a normed linear space and

Corollary 4

n

xjj T 2 LðXÞ. If T is hypercyclic then inf n2N0 jjT jjT n jj ¼ 0 for all

x 2 X. Proposition 3 Suppose that H is a Hilbert space and T 2 BðHÞ is supercyclic. Then, every vector x 2 H satisn

xjj fying inf n2N0 jjT jjT n jj [ 0 is a cyclic vector for T.

Proof

n

xjj Assume that inf n2N0 jjT jjT n jj ¼ d [ 0. Suppose, to get

a contradiction, that x is not a cyclic vector. Then M ? 6¼ ð0Þ where M ¼ fpðTÞx : p a polynomialg. Choose a nonzero vector y 2 M ? with jjyjj ¼ d. Then we have: inf

jjkT n x  yjj jjkT n xjj þ jjyjj dðjkjjjT n jj þ 1Þ  inf pffiffiffi  inf pffiffiffi [ 0; n n 1 þ jjkT jj 2ð1 þ jjkT jjÞ 2ð1 þ jkjjjT n jj

n!þ1

Example 2 Let T ¼ Bw be the bilateral weighted backward shift on ‘2 ðZÞ where wn ¼ 1 if n  0 and wn ¼ 2 if n  1. For q ¼ 1 and every n  3, we have w0    wnþqþ1 ¼ 1. Thus, maxfðw1    wnþq Þ1 ; ðw0    wnþqþ1 Þg  1 for n  3. Hence, Bw is not hyper-

where the infimum is taken over all pairs ðk; nÞ 2 C  N0 . Hence, the set C ¼ fkT n : k 2 C; n 2 N0 g is not topologically transitive by the infimum lemma. Now, Remark 1 shows that HCðCÞ 6¼ X but, SCðTÞ ¼ HCðCÞ and so Remark 2 shows that T is not supercyclic.

n

xyjj cyclic by the above lemma. We show that inf n2N0 jjT 1þjjT n jj ¼

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An operator T is said to be power bounded if, for some positive number k, we have jjT n jj\k for all n 2 N0 . By a result of Ansari and Bourdon (1997), if T is a powerbounded supercyclic operator on a Banach space X then for every x 2 X, jjT n xjj ! 0 as n ! 1. As a consequence, T may not have a unimodular eigenvalue. The infimum lemma provides a direct proof for this result: if x is a nonzero vector and Tx ¼ ax where jaj ¼ 1 then fpðTÞx : p a polynomialg ¼ Cx. Let M be the complement of Cx. If we choose a non-zero vector y 2 M, then we have: inf

jjkT n x  yjj rðjjkT n xjj þ jjyjjÞ rðjkjjjxjj þ jjyjjÞ [ 0;  inf  inf n n 1 þ jjkT jj 1 þ jjkT jj 1 þ jkjk

Z sup 0\r\1

2p

jf ðreih Þjp dh\1;

0 1

and H ¼ ff 2 HðDÞ : supz2D jf ðzÞj\1g. Also, for 1  p\1 and a [  1, the weighted Bergman space Apa is the set of all f 2 HðDÞ satisfying Z jf ðzÞjp ð1  jzj2 Þa dAðzÞ\1: D

Example 3 Let X be the Hardy space H p (1  p  1Þ or the weighted Bergman space Apa (1  p\1; a [  1). For 0  r\1, let Ar be the set of all analytic maps / : D ! D satisfying j/ð0Þj  r and Cr ¼ fC/ : / 2 Ar g. 1

where the infimum is taken over all pairs ðk; nÞ 2 C  N0 and r is the positive number arising from the equivalence of the norm on X and the norm defined on the direct sum Cx M. So, if we put C ¼ fkT n : k 2 C; n 2 N0 g then C is not topologically transitive by the infimum Lemma. Then by Remarks 1 and 2, T may not be supercyclic, a contradiction. For a positive number d, we denote by Dc ð0; dÞ the set fy 2 X : jjyjj  dg. Proposition 4 Let X be a Banach space and C be a bounded subset of BðXÞ. If there is a vector x 2 X, a nontrivial complemented subspace M and a positive number d such that orbðC; xÞ  M \ Dc ð0; dÞ then CC is not topologically transitive. Proof By our assumption on C, there is a positive number k such that jjTjj\k for all T 2 C. Choose a non-zero vector y in the complement of M. Then: jjkTx  yjj bðjkjjjTxjj þ jjyjjÞ bðjkjd þ jjyjjÞ  inf  inf [ 0: 1 þ jjkTjj 1 þ jkjk 1 þ jkjk kT2C kT2C k2C inf

Here, b is the positive number arising from the equivalence of the primary norm on X and the norm defined on the direct sum of M and its complement. As a direct consequence of the above proposition, we have Corollary 5 Let X be a Banach space and C be a bounded subset of BðXÞ. If there is a vector x 2 X and a non-zero vector y 2 X such that Tx ¼ y for all T 2 C then CC is not topologically transitive. Denote by IsoðXÞ the set of all linear isometries on X. Corollary 6 Let X be a Banach space. For every pair of non-zero vectors x; y 2 X, let Cxy ¼ fT 2 IsoðXÞ : Tx ¼ yg. Then CCxy is not topologically transitive. Let HðDÞ be the set of all analytic functions on D. Recall that, for 1  p\1 the Hardy space H p is the set of all f 2 HðDÞ for which

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p p Since jjC/ jj  ð1þr 1rÞ , when X ¼ H (MacCluer and Mercer 2þa

p for some con1995, Corollary 3.7), and jjC/ jj  Cð1þr 1rÞ p stant C (independent of /) when X ¼ Aa (Cima and Mercer 1995; MacCluer and Mercer 1995), Cr is bounded. Now, Corollary 5 (by choosing x ¼ y ¼ 1, the constant function 1 on H p or Apa ) shows that CCr is not topologically transitive. Also let x ¼ 1 and y ¼ 0 by the infimum lemma CCr is not topologically transitive.

4 Some Strictly Transitive Sets of Operators Definition 2 A subset C of LðXÞ is said to be strictly transitive if for each pair of non-zero elements x, y in X, there exists some T 2 C such that Tx ¼ y. If C is strictly transitive then Xnf0g  orbðC; xÞ for each non-zero x 2 X. So, orbðC; xÞ ¼ X whenever x 6¼ 0 and hence HCðCÞ ¼ Xnf0g. On the other hand, Xnf0g  orbðC; xÞ shows that C should be uncountable and so, for every T 2 LðXÞ, the semigroup fT n : n 2 N0 g may not be strictly transitive. Example 4 The set of all rank-one operators on any locally convex space X is strictly transitive. Indeed, Let x; y be two non-zero vectors of X and f be a functional in X

such that f ðxÞ ¼ 1. Define T : X ! X by Tz ¼ f ðzÞy, then T is a rank-one operator and Tx ¼ y. For a unitary operator u on a Hilbert space H, define Mu ¼ Ranðu  IÞ. It is easy to see that Mu is a reducing subspace for u. Moreover, for any given cardinal number a  dimH, it is easy to construct a unitary operator u for which dimMu ¼ a (If B is the orthonormal basis of H and B1  B with cardðB1 Þ ¼ a, define uea ¼ ea for ea 2 BnB1 and uea ¼ ea where ea 2 B1 ). Note that dimMu may be regarded as kind of a distance between u and I. With this meaning, the following theorem says that if R is the set of all unitary operators whose distances from I is less than or equal to 1 then Rþ R is

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strictly transitive. Furthermore, in the proof of the theorem we characterize the form of elements in R.

easy to see that uy;a ðxÞ ¼ z where a ¼ 2 and 1ffiffi y¼p x þ p1ffiffi2 z. If b ¼ 6 0, by the Cauchy–Bunyakowsky– 2

Theorem 1 Let H be a Hilbert space and U be the set of all unitary operators on H. If R ¼ fu 2 U : dimMu  1g, then Rþ R is strictly transitive.

b Schwarz inequality we have jbj\1. Let k ¼ jbj , r¼

Proof For a unit vector y 2 H and a complex number a satisfying j1 þ aj ¼ 1, define uy;a 2 BðHÞ by uy;a ðxÞ ¼ x þ a\x; y [ y ðx 2 HÞ. We show that each uy;a is a unitary operator on H. The operator uy;a is surjective; given h 2 H we can write h ¼ by þ z for some b 2 C and z 2 H such that \z; y [ ¼ 0. Then, uy;a ðkÞ ¼ h for k ¼ bð1 þ aÞy þ z. On the other hand, jjuy;a ðhÞjj2 ¼ jjuy;a ðby þ zÞjj2 ¼ jbð1 þ aÞj2 þ jjzjj2 ¼ jjhjj2 . Also, it is clear that dimMuy;a  1 and hence uy;a 2 R. We claim that R ¼ fuy;a : jjyjj ¼ j1 þ aj ¼ 1g. If u 2 R, then there is a unit vector y 2 H such that for every x 2 H, uðxÞ  x ¼ by for some scalar b which depends on x. In particular, uðyÞ  y ¼ ay and so we must have j1 þ aj ¼ 1. If \z; y [ ¼ 0 and uðzÞ ¼ z þ dy for some d 2 C, then jjzjj2 ¼ jjuðzÞjj2 implies that d ¼ 0 and so uðzÞ ¼ z. Now, let x be any vector in H. Write x ¼ cy þ z where c 2 C and \z; y [ ¼ 0. Then, \x; y [ ¼ c and so uðxÞ ¼ cð1 þ aÞy þ z ¼ x þ acy ¼ x þ a\x; y [ y. Hence, u ¼ uy;a . To prove that Rþ R is strictly transitive suppose that x; z are non-zero vectors in H. Without loss of generality, we can assume that jjxjj ¼ jjzjj ¼ 1. Find b 2 C and w 2 H satisfying \x; w [ ¼ 0 such that z ¼ bx þ w. Then, \z; x [ ¼ b. If b ¼ 1, then w ¼ 0 and so we have 2

jb1j ux;0 ðxÞ ¼ z. Otherwise, letting t ¼ 2ð1ReðbÞÞ , we choose a ¼ pffi b1 1ffi p tx þ w. Then it can be readily verified t and y ¼ a t

that

j1 þ aj ¼ 1

(putting

b ¼ r þ si,

we

have

1 pffiffiffiffiffiffiffiffiffiffiffiffi ffi and y ¼ rkx þ rz. Then, 2ð1jbjÞ

jjyjj2 ¼ r 2 ðjkj2 þ jjzjj2  2ReðkbÞÞ ¼ r 2 ð2  2jbjÞ ¼ 1; and for a ¼ 2, we have kuy;a ðxÞ ¼ kx  2r 2 ðkx  z  kjbjx þ jbjzÞ ¼ z:

Hence, CR0 is strictly transitive. Suppose, to get a contradiction, that RR0 is hypercyclic. Then, there exists some vector x 2 H such that B ¼ fruy;a ðxÞ : uy;a 2 R0 ; r 2 Rg is dense in H. Consider the linear functional f : H ! C defined by f ðhÞ ¼ \x; h [ . Then, C ¼ f ðHÞ ¼ f ðBÞ  f ðBÞ and so f ðBÞ is dense in C, but, this is impossible since \x; uy;a ðxÞ [ 2 R for all uy;a 2 R0 . Now, we give an example of an uncountable family of operators which is topologically transitive but not strictly transitive. Example 5 Let U be the group of unitary operators on a Hilbert space H and s0 ¼ Qþ U ¼ fru : u 2 U; r 2 Qþ g , where Qþ is the set of positive rational numbers. Choose pffiffiffi x; y 2 H such that jjxjj ¼ 1; jjyjj ¼ 2. Then, there is no t 2 s0 such that y ¼ tðxÞ and so s0 is not strictly transitive. On the other hand, let U; V be non-empty open subsets of H and x; y be two linearly independent vectors such that x 2 U, y 2 V. Find e [ 0 such that Dðy; eÞ  V. In the proof of Theorem 1, we have seen that there exists some jjyjj u 2 U such that y ¼ jjxjj ux. If r ¼ jjyjj jjxjj 2 Q then ru 2 s0 and

jjyjj2 ¼

ruðUÞ \ V 6¼ ;. Otherwise, choose a positive rational e . Then jjqux  yjj ¼ number q such that jq  rj\ jjxjj

t þ jaj12 t jjwjj2 ¼ tð1 þ 1jbj Þ ¼ 1 and it is easy to see that jb1j2

jjqux  ruxjj ¼ jq  rjjjxjj\e which shows that quðUÞ \ V 6¼ ;. Therefore, s0 is topologically transitive.

2ð1rÞ j1 þ aj ¼ j1 þ rsi1 j ¼ j1bj ¼ 1). jb1j

Also,

2

uy;a ðxÞ ¼ z. Note that the only real solutions of the equation j1 þ aj ¼ 1 are a¼0 and a ¼ 2. Let R0 ¼ fuy;a 2 R : a ¼ 0 or a ¼ 2g. If x; z 2 H, \z; x [ ¼ b, and ruy;a ðxÞ ¼ z for some r 2 R and uy;a 2 R, then rðjjxjj2 þ aj\x; y [ j2 Þ ¼ b. Thus, if b 62 R, then a 62 R and so RR0 may not be strictly transitive. In fact, we have Theorem 2 The set CR0 is strictly transitive but RR0 may not be even hypercyclic. Proof Let x; z be non-zero unit vectors in H. If z ¼ kx for some scalar k 2 C, then kux;0 ðxÞ ¼ z. If x; z are linearly independent, let \z; x [ ¼ b. For the case b ¼ 0, it is

As we defined Mu for a unitary operator u on a Hilbert space H, let X be any topological vector space, T 2 LðXÞ and MT ¼ RanðT  IÞ. Clearly, MT is an invariant subspace for T. Recall that an operator T 2 LðXÞ is said to be an involution if T 2 ¼ I. Denote by InvoðXÞ, the set of all involutions on X and put C ¼ Cnf0g. Theorem 3 If X is a locally convex space, C0 ¼ fT 2 InvoðXÞ : dimMT ¼ 1g and C ¼ C0 [ fkI : k 2 C g; then C is strictly transitive; C0 is topologically transitive but not strictly transitive; If X is assumed to be second countable and Baire then C0 is hypercyclic.

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Proof (i) Let x; y be two non-zero elements of X. If y ¼ kx for some k 2 C then y ¼ kIx and kI 2 C. On the other hand, if x; y are linearly independent let f1 ; f2 be continuous linear functionals on X such that f1 ðxÞ ¼ 1; f1 ðyÞ ¼ 0 and f2 ðyÞ ¼ 1; f2 ðxÞ ¼ 0. Put f ¼ f1 þ f2 and define T : X ! X by Tz ¼ f ðzÞðx þ yÞ  z, so T 2 LðXÞ and Tx ¼ y; Ty ¼ x. Meanwhile, T 2 z ¼ f ðzÞðy þ xÞ  Tz ¼ z, hence T 2 ¼ I and since Tz  z ¼ f ðzÞðx þ yÞ, we have dimMT ¼ 1, so T 2 C. (ii) Let U; V be two non-empty open sets in X. Since dimX [ 1, there exist two linearly independent vectors x; y such that x 2 U and y 2 V. Now, if T is the involution constructed in ðiÞ, then Tx ¼ y and so TðUÞ \ V 6¼ ;. To prove that C0 is not strictly transitive, let y ¼ kx for some k 2 Cnf1; 1g. If Tx ¼ y for some T 2 C0 then x ¼ T 2 x ¼ ky ¼ k2 x which gives k ¼ 1, a contradiction. (iii) This is a consequence of (ii) and Proposition 1.

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