International Journal of Applied Mathematics————————————————————–Volume 24 No. 2 2011, 245-249

HEREDITARILY HYPERCYCLIC PAIRS

B. Yousefi1, M. Habibi2 §

1Department of MathematicsPayame-Noor University

Shahrake Golestan, P.O. Box: 71955-1368, Shiraz, IRANe-mail: bahmann@spnu.ac.ir2Department of Mathematics

Islamic Azad University, Branch of DehdashtP.O. Box 7571763111, Dehdasht, IRAN

e-mail: habibi.m@iaudehdasht.ac.ir

Abstract: In this paper we will give some necessary and sufficient conditionsfor a pair of operators to be Hereditarily.

AMS Subject Classification: 47A16, 46A32, 47B37Key Words: hereditarily hypercyclic, pair of operators, hypercyclicity crite-rion

1. Introduction

Let T1, T2 be commutative bounded linear operators on a Banach space X . Forthe pair T = (T1, T2) put Γ = {Tm

1 T n2 : m,n ≥ 0}. For x ∈ X the orbit of x

under T is the set Orb(T , x) = {S(x) : S ∈ Γ}, that is

Orb(T , x) = {Tm1 T n

2 (x) : m,n ≥ 0}.

The vector x is called a hypercyclic vector for T if the set Orb(T , x) is densein X , that is

Orb(T , x) = {Tm1 T n

2 (x) : m,n ≥ 0} = X .

Let {m(k,1)}∞k=1, {m(k,2)}∞k=1 be increasing sequences of non-negative inte-gers. The pair T = (T1, T2) is called hereditarily hypercyclic with respect to

Received: January 4, 2011 c© 2011 Academic Publications§Correspondence author

246 B. Yousefi, M. Habibi

{m(k,1)}∞k=1 and {m(k,2)}∞k=1, if for all subsequences {m′(k,1)}∞k=1 and {m′

(k,2)}∞k=1

of {m(k,1)}∞k=1 and {m(k,2)}∞k=1 respectively, the sequence {Tm′

(k,1)

1 Tm′

(k,2)

2 } is hy-percyclic. In the other hand, there exists a vector x in X such that

{Tm′

(k,1)

1 Tm′

(k,2)

2 (x)} = X .

A nice criterion namely the Hypercyclicity Criterion is used in the proofof our main theorem. It was developed independently by Kitai, Gethner andShapiro. This criterion has been used to show that hypercyclic operators arisewithin the class of composition operators, weighted shifts and adjoints of mul-tiplication operators. For some source on this topics, see [1]-[16]. Note that, alloperators in this paper are commutative operator. By (i, j) we mean that aninteger number related to the numbers i and j, and not a pair.

2. Main Results

Note that a pair T = (T1, T2) is said to satisfy the Hypercyclicity Criterion ifit holds in the hypothesis of the following theorem.

Theorem 1. (The Hypercyclicity Criterion) Let X be a separable Ba-nach space and T = (T1, T2) is a pair of continuous linear mappings on X . Ifthere exist two dense subsets Y and Z in X , and strictly increasing sequences{mk}∞k=1, {nj}∞j=1, such that:

1. Tmk1 T

nj

2 → 0 on Y as k → ∞ and j → ∞,2. There exist functions {St : Z → X} such that for every z ∈ Z, Stz → 0,and Tmk

1 Tnj

2 Stz → z,then T is a hypercyclic n-tuple.

Theorem 2. A pair T = (T1, T2) is hereditarily hypercyclic with respectto increasing sequences of non-negative integers {m(k,1)}∞k=1, {m(k,2)}∞k=1 if andonly if for all given any two open sets U , V, there exist some positive integersM , N such that

Tm1 T n

2 (U)⋂

V �= φ

for any m > M and n > N .

Proof. Let T = (T1, T2) be hereditarily hypercyclic pair with respect to in-creasing sequences of non-negative integers {mk,1}∞k=1 and {mk,2}∞k=1. Supposethat there exist some open sets U , V such that

Tm(i,1)

1 Tm(i,2)

2 (U)⋂

V = φ

HEREDITARILY HYPERCYCLIC PAIRS 247

for some subsequence {m′(k,1)}∞k=1, {m′

(k,2)}∞k=1 of {m(k,1)}∞k=1, {m(k,2)}∞k=1 re-spectively. Since the 2-tuple T = (T1, T2) is hereditarily hypercyclic with re-

spect to {m(k,1)}∞k=1, {m(k,2)}∞k=1, thus {Tm′

(k,1)

1 Tm′

(k,2)

2 } is hypercyclic and sowe get a contradiction.

Conversely, suppose that {m(ki,1)}∞i=1 and {m(ki,2)}∞i=1 are arbitrary subse-quences of {m(k,1)}∞k=1 and {m(k,2)}∞k=1 respectively, and U , V are open sets inX satisfying

Tm(k,1)

1 Tm(k,2)

2 (U)⋂

V �= φ

for any m(k,j) > Mj , j = 1, 2. So there exist (i, j), large enough for j = 1, 2such that m(ki,j) > Mj for j = 1, 2 and

Tm(ki,1)

1 Tm(ki,2)

2 (U)⋂

V �= φ.

This implies that {Tm(ki,1)

1 Tm(ki,2)

2 } is hypercyclic and so the 2-tuple T =(T1, T2) is indeed hereditarily hypercyclic with respect to the sequences {m(k,1)}∞k=1,{m(k,2)}∞k=1.

Theorem 3. If a pair T = (T1, T2 is hereditarily hypercyclic with re-spect to an increasing sequences of non-negative integers {mk}∞k=1, {nj}∞j=1

with supk(mk+1) − mk)) < ∞ and supk(nk+1) − nk) < ∞, then the pair T ishereditarily hypercyclic with respect to the entire sequence.

Proof. Let T = (T1, T2) be hereditarily hypercyclic pair with respect to thesequences {mj}∞j=1, {nj}∞j=1 such that supj(mj+1 − mj) < ∞ and supj(nj+1 −nj) < ∞. We will show that T is hereditary hypercyclic with respect to theentire sequence. For this let U and V be two nonempty open sets in X . We willshow that there exist integers M,N such that

Tm1 T n

2 (U)⋂

V �= φ

for any m > M and n > N , which by Theorem 2 implies that T is in-deed hereditarily hypercyclic with respect to the entire sequence. Put M =sup{(mj+1 − mj : j ≥ 0} and N = sup{(nj+1 − nj : j ≥ 0}. For each integers0 ≤ i ≤ M and 0 ≤ j ≤ N , set Ui,j = U and Vi,j = T−i

1 T−j2 (V). Since T is

hereditarily hypercyclic with respect to {mj}∞j=1 and {nj}∞j=1, by Theorem 2.2,for all integers 0 ≤ i ≤ M and 0 ≤ j ≤ N , there exist Mj , Nj ∈ N such that

Tmk1 T

np

2 (Ui,j)⋂

Vi,j �= φ

for all k > Mi and p > Nj. Let Mo = {Mi : i = 0, 1, 2, ...,M} and No ={Nj : j = 0, 1, 2, ..., N}. Then M0 = mm0 and N0 = nn0 for some integers

248 B. Yousefi, M. Habibi

0 ≤ m0 ≤ M and 0 ≤ n0 ≤ N . Now we can show that

Tm1 T n

2 (Uit)⋂

Vit �= φ

for all m > M0 and n > N0. In fact if m > M0 and n > N0, then there existk1 > m0, k2 > n0, 0 ≤ i ≤ M and 0 ≤ j ≤ N such that m = mk1 + i andn = nk2 + j. Note that

Tmk11 T

nk22 (Ui,j)

⋂Vi,j �= φ

for all k1 > m0 and k2 > n0. Hence

Tmk11 T

nk22 (U)

⋂T−i

1 T−j2 (V) = T

mk1+i

1 Tnk2

+j

2 (Ui,j)⋂

(Vi,j) �= φ

and soTm

1 T n2 (U)

⋂(V) = T

mk1+i

1 Tnk2

+j

2 (Ui,j)⋂

(Vi,j) �= φ

for all m > M0 and n > N0. So T is hereditarily hypercyclic with respect tothe entire sequence and the proof is complete.

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HEREDITARILY HYPERCYCLIC PAIRS 249

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