Supercyclicity and Hypercyclicity of an Isometry Plus a Nilpotent

and Applied Analysis 3 Suppose that there is c1 > 0 so that ∥∥QN−1xn ∥∥ > c1 1.7 for all n ≥ 1; then Ayn − λyn → 0 as n → ∞where yn QN−1xn ∥QN−1xn ∥∥ 1.8 which, in turn, implies that λ ∈ σap A . Now, if 1.7 does not hold, then we can assume, without loss of generality, that xn n satisfies QN−1xn −→ 0 as n −→ ∞. 1.9 So by 1.5 , AQN−2xn − λQN−2xn −→ 0 as n −→ ∞. 1.10 Now, if there is a constant c2 > 0 such that ∥∥QN−2xn ∥∥ > c2 1.11 for all n, then Azn − λzn → 0 where zn QN−2xn ∥QN−2xn ∥∥ 1.12 which implies that λ ∈ σap A . Otherwise, we can assume, without loss of generality, that QN−2xn → 0 as n → ∞ and by 1.5 AQN−3xn − λQN−3xn −→ 0 1.13 as n → ∞. The procedure continues to conclude that λ ∈ σap A . SinceA T −Q, by a similar method σap A ⊆ σap T . In the remaining results of this section, the operatorsA and T are as in Proposition 1.1. Corollary 1.2. Suppose that X is a normed space. Then T − λI is bounded below where |λ|/ 1. Proof. SinceA is an isometry, σap T σap A ⊆ ∂D. In fact, let λ ∈ σap A ; then |λ| ≤ ‖A‖ 1; moreover, there exists a sequence xn n inXwith ‖xn‖ 1 and so A−λI xn → 0 if n → ∞. Therefore, 0 ≤ 1 − |λ| ≤ ‖ A − λI xn ‖ −→ 0 as n −→ ∞, 1.14 4 Abstract and Applied Analysis and so |λ| 1. Now, if |λ|/ 1, then λ / ∈ σap T and so T − λ is bounded below. Corollary 1.3. Suppose that X is an infinite dimensional Banach space. Then the operator T on X is not a compact operator. Proof. If T is a compact operator, then 0 ∈ σ T σ A . Thus D ⊆ σ T which contradicts the fact that the spectrum of a compact operator is at most countable. Proposition 1.4. If the operators T and A are defined on a normed space X, then ker T − λ ⊆ ker A − λ for every scalar λ. Proof. Fix λ ∈ C and suppose that Tx λx for some nonzero vector x. By Proposition 1.1, λ ∈ σp A which implies that |λ| 1. Therefore, if n > N − 1, we have ‖x‖ ‖Tnx‖ ∥∥∥∥A n− N−1 N−1 ∑ k 0 ( n k ) QkAN−1−kx ∥∥∥∥ 2 ∥∥∥∥ N−1 ∑ k 0 ( n k ) QkAN−1−kx ∥∥∥∥ 2 ( n N − 1 )2∥∥∥∥ N−1 ∑ k 0 N − 1 ! n −N 1 ! k! n − k ! Q kAN−1−kx ∥∥∥∥ 2 . 1.15


Introduction
Let x be a vector in a separable normed space X and T an operator on X.The orbit of x under T is defined by orb T, x {T n x : n 0, 1, 2, . ..}.

1.1
We recall that a vector x in X is cyclic for an operator T on X if the closed linear span of orb T, x is X; it is supercyclic, if the set of all scalar multiples of the elements of orb T, x is dense in X; also it is said to be weakly hypercyclic if orb T, x is weakly dense in X.An operator T is called cyclic, supercyclic, or weakly hypercyclic operator, respectively, if it has a cyclic, supercyclic, or weakly hypercyclic vector.Recently, the cyclicity of operators has attracted much attention from operator theorists.For a good source on this topic, see 1 .
Hilden and Wallen in 2 proved that isometries on Hilbert spaces with dimension more than one are not supercyclic.Ansari and Bourdon in 3 and Miller in 4 independently proved this fact on Banach spaces.Moreover, recently it is shown in 5 that m-isometric operators on Hilbert spaces, which forms a larger class than isometries, are neither supercyclic nor weakly hypercyclic.In this paper, it is shown that an isometry plus a nilpotent on normed spaces are neither supercyclic nor weakly hypercyclic if they commute.We also discuss this fact when the underlying space is a Hilbert space and the isometry is replaced by a co-isometry.We begin with some elementary properties of such operators.In what follows, as usual, for an operator T , σ ap T , σ p T , and σ T are denoted, respectively, the approximate point spectrum, point spectrum, and spectrum of T .Also, D denotes the open unit disc.Recall that an operator Q on a normed space X is a nilpotent operator of order N ≥ 1 if Q N 0 and Q N−1 / 0. From now on, we assume that Q is a nilpotent operator of order N ≥ 1 unless stated otherwise.Proposition 1.1.Suppose that X is a normed space, and A ∈ B X is an isometry such that AQ QA.
Proof.i Suppose that λ / ∈ σ A .Then it is easily seen that ii If λ ∈ σ p A , there exits x / 0 such that Ax λx.Therefore, x and continue this process to conclude that Tx Ax λx which implies that λ ∈ σ p T .Hence, σ p A ⊆ σ p T .Moreover, since A T − Q, using a similar method, we get σ p T ⊆ σ p A .
iii Let λ ∈ σ ap T ; then there exists a sequence x n n in X such that x n 1 and Therefore, Suppose that there is c 1 > 0 so that for all n ≥ 1; then Ay n − λy n → 0 as n → ∞ where which, in turn, implies that λ ∈ σ ap A .Now, if 1.7 does not hold, then we can assume, without loss of generality, that x n n satisfies 1.9 So by 1.5 , Now, if there is a constant c 2 > 0 such that for all n, then Az n − λz n → 0 where 12 which implies that λ ∈ σ ap A .Otherwise, we can assume, without loss of generality, that Q N−2 x n → 0 as n → ∞ and by 1.5 as n → ∞.The procedure continues to conclude that λ ∈ σ ap A .Since A T −Q, by a similar method σ ap A ⊆ σ ap T .
In the remaining results of this section, the operators A and T are as in Proposition 1.1.
Proof.Since A is an isometry, σ ap T σ ap A ⊆ ∂D.In fact, let λ ∈ σ ap A ; then |λ| ≤ A 1; moreover, there exists a sequence x n n in X with x n 1 and so 1.14 and so |λ| 1.Now, if |λ| / 1, then λ / ∈ σ ap T and so T − λ is bounded below.
Corollary 1.3.Suppose that X is an infinite dimensional Banach space.Then the operator T on X is not a compact operator.
Proof.If T is a compact operator, then 0 ∈ σ T σ A .Thus D ⊆ σ T which contradicts the fact that the spectrum of a compact operator is at most countable.Proposition 1.4.If the operators T and A are defined on a normed space X, then ker T − λ ⊆ ker A − λ for every scalar λ.
Proof.Fix λ ∈ C and suppose that Tx λx for some nonzero vector x.By Proposition 1.1, λ ∈ σ p A which implies that |λ| 1.Therefore, if n > N − 1, we have

1.15
Consequently, Continue the above process to get Qx 0, and so Ax λx.Corollary 1.5.If X is a Hilbert space, then the eigenvectors of T corresponding to distinct eigenvalues are orthogonal.
Recall that an operator T is power bounded if there exists some constant c > 0 such that T n ≤ c for all n 1, 2, 3, . ... Proposition 1.6.Let X be a normed space and x ∈ X.If there is a constant c > 0 such that T n x ≤ c for all n ≥ 1, then Qx 0. In particular, if T is power bounded, then Q 0.
Proof.Since the sequence T n x n is bounded, an argument similar to the proof of the Proposition 1.4 shows that Qx 0.

Supercyclicity and Hypercyclicity
We begin this section with a useful lemma.Lemma 2.1.Let X be a normed space.For nonnegative integers k, n, if is a polynomial in n with coefficients in X of degree k, then the sequence P k n n is eventually increasing.
Proof.We prove the lemma by induction on k, the degree of the polynomial P k n .For k 1, let P 1 n x 0 x 1 n.It is easily seen that for every n ≥ 1 Since lim n → ∞ P 1 n ∞, there is a positive integer i such that This fact coupled with 2.2 implies that for every n ≥ i.Therefore, the sequence P 1 n n≥i is increasing.Suppose that P k n n is eventually increasing and let where using the induction hypothesis there is a positive integer j such that for every n ≥ j Therefore, for every n ≥ j.Hence, the sequence p k 1 n n≥j is increasing.
Theorem 2.2.Suppose that X is a normed space, and A ∈ B X is an isometry such that AQ QA.
If T A Q, then the operator T is neither supercyclic nor weakly hypercyclic.
Proof.Let X be the completion of X and T , A, and Q the extensions of T , A, and Q on X, respectively.Thus, T A Q where A is an isometry and Q is a nilpotent operator; moreover, A Q Q A. Also, note that the supercyclicity of the operator T implies the supercyclicity of T .So we can assume, without loss of generality, that X is a Banach space.
As we have seen in the proof of Proposition 1.4, if x ∈ X then and so by Lemma 2.1, the sequence T n x n is eventually increasing.Suppose that x 0 is a supercyclic vector for T .Thus, for any x ∈ X there is a sequence n i i of positive integers and a sequence α i i of scalars such that α i T n i x 0 → x.Moreover, since the sequence T n x 0 n is eventually increasing, we have α i T n i x 0 ≤ α i T n i 1 x 0 for large i.So letting i → ∞, we conclude that x ≤ Tx , for all x in X.On the other hand, the supercyclicity of T implies that it has a dense range and so is invertible.Thus, in light of Proposition 1.1 we see that A is invertible.It is easy to see that where

2.11
Since P N 0, by a similar argument the sequence T −n x n is eventually increasing for every x ∈ X.But T −1 is also supercyclic see 1, Theorem 1.12 ; therefore, x ≤ T −1 x 2.12 for every x ∈ X.Thus, T is an isometry which implies that it is not a supercyclic operator.
To show that the operator T is not weakly hypercyclic, note that for every x * ∈ X * and every positive integer n.If ker Q * / {0}, then there is a nonzero x * ∈ X * such that T * n x * A * n x * ≤ x * because A * A 1. Now, suppose that x 0 is a weakly hypercyclic vector for T .Since orb T, x 0 is weakly dense in X and x * is nonzero, the set {x * T n x 0 : for all n ≥ 0, which is a contradiction.If ker Q * {0}, then Q * 0 and so T A is not a weakly hypercyclic operator.
We remark that there are Banach space isometries which are also weakly supercyclic.Indeed, the unweighted bilateral weighted shift on the space l p Z where p > 2 is weakly supercyclic see 1, Corollary 10.32 .However, the question that whether an isometry plus a nonzero nilpotent which commute with each other, are weakly supercyclic or not is still an open question.
The following examples show that the commutativity of A and Q is essential in the preceding theorem.

Example 2.3. Let e n
∞ n −∞ be the standard orthonormal basis for l 2 Z .Define the isometric operator A by Ae n e n 1 for all n ∈ Z and the weighted shift operator Q by Qe n w n e n 1 , where w 2n 0 for all integers n, w 2n−1 1/ 2n − 1 2 for all n ≥ 1, and w 2n−1 1/ 1 − 2n for all n ≤ 0. Note that Q 2 0 and AQ / QA.Moreover, since 1 ≤ inf n 1 w n ≤ sup n 1 w n ≤ 2, the weighted shift operator T A Q is invertible.To see that T is supercyclic by Theorem 3.4 of [6], it is sufficient to show that (see [7, pages 299 and 300]).Therefore, 2.15 holds.
Example 2.4.Consider the isometric operator A on l 2 Z defined by Ae n e n−1 and the weighted shift operator Q defined by Qe n w n e n−1 , where w 2n 0 for all n ∈ Z, w 2n−1 1/ 2n − 1 , for n ≥ 1, and w 2n−1 1/ 2n − 1 2 for n ≤ 0. Note that Q 2 0 and AQ / QA.Also, since for all n ≥ 1, and ∞ n 1 w n ∞, we conclude that

2.20
Hence, using Corollary 10.27 of [1], we observe that the operator A Q is weakly hypercyclic.

A Co-isometry Plus a Nilpotent
From now on, we assume that H is a separable Hilbert space with orthonormal basis {e n } ∞ n 0 .Recall that the unilateral shift operator S : H → H is given by Se n e n 1 for all n and the backward shift operator B : H → H is defined by Be 0 0 and Be n e n−1 for all n ≥ 1.It is known that the operator B is supercyclic see 1, page 9 .It follows that a co-isometry can be supercyclic.In this section, we give sufficient conditions such that the sum of a co-isometry and a nilpotent is supercyclic on H.

Theorem 3.1. Suppose that A is a co-isometric operator on a Hilbert space H. Then A is supercyclic if and only if
Proof.First assume that ∩ n≥0 A * n H 0 .Then by the von Neumann-Wold decomposition, A * S m for some positive integer m see 8 .Therefore, A B m which is a positive power of a supercyclic operator and so is supercyclic 9 .For the converse, assume that M ∩ n≥0 A * n H / 0 , and let P M denote the orthogonal projection on M. By the von Neumann-Wold decomposition, M is a reducing subspace for A and A * | M is unitary.Since A| M * P M A * | M is also unitary, the operator A| M is not supercyclic.Assume that A is supercyclic, and h g ⊕ k is a supercyclic vector for A, where g ∈ M and k ∈ M ⊥ .If g 0, then H M ⊥ which is impossible; so g / 0. Take f ∈ M, and let ε > 0 be arbitrary.Then there is a nonnegative integer n and a scalar α ∈ C such that Hence, g is a supercyclic vector for A| M which is impossible.
To prove the next theorem, we need the supercyclicity criterion due to H. N. Salas see 10 , or more generally 11 .

Supercyclicity Criterion
Suppose that X is a separable Banach space and T is a bounded operator on X.If there is an increasing sequence of positive integers {n k } k∈N , and two dense sets Y and Z of X such that 1 there exists a function S : Z → Z satisfying TSx x for all x ∈ Z, 2 T n k x • S n k y → 0 for every x ∈ Y and y ∈ Z, then T is supercyclic.Theorem 3.2.Suppose that A is a co-isometry on a Hilbert space H such that ∩ n≥0 A * n H 0 .If AQ QA, then the operator T A Q satisfies the supercyclicity criterion.
Proof.By Corollary 1.2, the operator T * is bounded below and so is left invertible.Consequently, T is a right invertible operator.Let x ∈ ∩ n≥0 T * n H.For every i ≥ 0, there is a vector x N i in H such that T * N i x N i x.Since Q N 0, we have which implies that x ∈ A * i 1 H.Hence, x ∈ ∩ n≥0 T * n H 0 and so the operator T admits a complete set of eigenvectors that is, H ∨ μ∈D r ker T − μ for every positive real number r, where D r {z ∈ C : |z| < r} see 12 , Part A of the lemma .Since T * is bounded below, TT * is invertible.Take S T * TT * −1 .Choose r > 0 so that r < 1/ S , and let Y Z span ker T − μ : μ ∈ D r .

10
Abstract and Applied Analysis Now, if h ∈ Y Z, then as n → ∞.Finally, T n S n h h for every h ∈ H and every n ≥ 0. Thus, the operator T satisfies the supercyclicity criterion.
The Hilbert-Schmidt class, C 2 H , is the class of all bounded operators S defined on a Hilbert space H, satisfying where • is the norm on H induced by its inner product.We recall that C 2 H is a Hilbert space equipped with the inner product S, T tr ST * in which tr S * T denotes the trace of S * T .Furthermore, C 2 H is an ideal of the algebra of all bounded operators on H, see 8 .For any bounded operator B on a Hilbert space H, the left multiplication operator L B and the right multiplication operator R B on C 2 H are defined by L B S BS and R B S SB for all S ∈ C 2 H .It is known that an operator B satisfies the supercyclicity criterion if and only if L B is supercyclic on the space C 2 H see 13, page 37 .In the following proposition, we see that an operator T may satisfy the supercyclicity criterion although R T is not a supercyclic operator on C 2 H . which implies that R A is an isometry.Also, if S ∈ C 2 H , then R N Q S 0. Since R T S R A S R Q S , Theorem 2.2 implies that R T is not supercyclic.
The proof of the following proposition is similar to the proof of the second part of Theorem 2.2, and we omit it.Proposition 3.4.Suppose that X is a normed space and A ∈ B X is a co-isometry such that AQ QA.Then the operator T A Q is not weakly hypercyclic.

Proposition 3 . 3 .
Suppose that H is a Hilbert space and A ∈ B H is a co-isometry such that ∩ n≥0 A * n H 0 and AQ QA.Then the operator T A Q satisfies the supercyclicity criterion but the operator R T is not supercyclic on C 2 H . Proof.By Theorem 3.2, the operator T satisfies the supercyclicity criterion.Moreover, for every S ∈ C 2 H we have