Supercyclicity and Hypercyclicity of an Isometry Plus a Nilpotent

Abstract

Suppose that 𝑋 is a separable normed space and the operators 𝐴 and 𝑄 are bounded on 𝑋. In this paper, it is shown that if 𝐴𝑄=𝑄𝐴, 𝐴 is an isometry, and 𝑄 is a nilpotent then the operator 𝐴+𝑄 is neither supercyclic nor weakly hypercyclic. Moreover, if the underlying space is a Hilbert space and 𝐴 is a co-isometric operator, then we give sufficient conditions under which the operator 𝐴+𝑄 satisfies the supercyclicity criterion.

1. 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^𝑛𝑥∶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. If 𝑇=𝐴+𝑄, then (i)𝜎(𝑇)=𝜎(𝐴), (ii)𝜎_𝑝(𝑇)=𝜎_𝑝(𝐴), (iii)𝜎_𝑎𝑝(𝑇)=𝜎_𝑎𝑝(𝐴).

Proof. (i) Suppose that 𝜆∉𝜎(𝐴). Then it is easily seen that (𝑇−𝜆)⁻¹=^𝑁_𝑘=1(−1)^𝑘−1(𝐴−𝜆)^−𝑘𝑄^𝑘−1(1.2) which implies that 𝜆∉𝜎(𝑇). Consequently, 𝜎(𝑇)⊆𝜎(𝐴). Since 𝐴=𝑇−𝑄, a similar argument shows that 𝜎(𝐴)⊆𝜎(𝑇).
(ii) If 𝜆∈𝜎_𝑝(𝐴), there exits 𝑥≠0 such that 𝐴𝑥=𝜆𝑥. Therefore, 𝑇𝑄^𝑁−1𝑥=𝐴𝑄^𝑁−1𝑥=𝜆𝑄^𝑁−1𝑥.(1.3) Now, if 𝑄^𝑁−1𝑥≠0, then 𝜆∈𝜎_𝑝(𝑇); otherwise, 𝑇𝑄^𝑁−2𝑥=𝐴𝑄^𝑁−2𝑥=𝜆𝑄^𝑁−2𝑥.(1.4) Also, if 𝑄^𝑁−2𝑥≠0 then 𝜆∈𝜎_𝑝(𝑇); otherwise, consider 𝑄^𝑁−3𝑥 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 𝑇𝑥_𝑛−𝜆𝑥_𝑛⟶0as𝑛⟶+∞.(1.5) Therefore, 𝐴𝑄^𝑁−1𝑥_𝑛−𝜆𝑄^𝑁−1𝑥_𝑛⟶0as𝑛⟶+∞.(1.6) Suppose that there is 𝑐₁>0 so that 𝑄^𝑁−1𝑥_𝑛>𝑐₁(1.7) for all 𝑛≥1; then 𝐴𝑦_𝑛−𝜆𝑦_𝑛→0 as 𝑛→+∞ where 𝑦_𝑛=𝑄^𝑁−1𝑥_𝑛𝑄^𝑁−1𝑥_𝑛(1.8) 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 𝑄^𝑁−1𝑥_𝑛⟶0as𝑛⟶+∞.(1.9) So by (1.5), 𝐴𝑄^𝑁−2𝑥_𝑛−𝜆𝑄^𝑁−2𝑥_𝑛⟶0as𝑛⟶+∞.(1.10) Now, if there is a constant 𝑐₂>0 such that 𝑄^𝑁−2𝑥_𝑛>𝑐₂(1.11) for all 𝑛, then 𝐴𝑧_𝑛−𝜆𝑧_𝑛→0 where 𝑧_𝑛=𝑄^𝑁−2𝑥_𝑛𝑄^𝑁−2𝑥_𝑛(1.12) which implies that 𝜆∈𝜎_𝑎𝑝(𝐴). Otherwise, we can assume, without loss of generality, that 𝑄^𝑁−2𝑥_𝑛→0 as 𝑛→∞ and by (1.5) 𝐴𝑄^𝑁−3𝑥_𝑛−𝜆𝑄^𝑁−3𝑥_𝑛⟶0(1.13) 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.

Corollary 1.2. Suppose that 𝒳 is a normed space. Then 𝑇−𝜆𝐼 is bounded below where |𝜆|≠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 (𝐴−𝜆𝐼)(𝑥_𝑛)→0 if 𝑛→∞. Therefore, 0≤1−|𝜆|≤(𝐴−𝜆𝐼)𝑥_𝑛⟶0as𝑛⟶∞,(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 ‖𝑥‖²=𝑇^𝑛𝑥²=𝐴^{𝑛−(𝑁−1)𝑁−1}_𝑘=0𝑛𝑘𝑄^𝑘𝐴^{𝑁−1−𝑘}𝑥²=^𝑁−1_𝑘=0𝑛𝑘𝑄^𝑘𝐴^{𝑁−1−𝑘}𝑥²=𝑛𝑁−1^{2𝑁−1𝑘=0}(𝑁−1)!(𝑛−𝑁+1)!()!𝑄𝑘!𝑛−𝑘^𝑘𝐴^{𝑁−1−𝑘}𝑥².(1.15) Consequently, ‖𝑥‖≥𝑛𝑄𝑁−1^𝑁−1𝑥−^𝑁−2_𝑘=0(𝑁−1)!(𝑛−𝑁+1)!()!𝑄𝑘!𝑛−𝑘^𝑘𝐴^{𝑁−1−𝑘}𝑥.(1.16) Since lim_𝑛→∞(𝑁−1)!(𝑛−𝑁+1)!𝑘!(𝑛−𝑘)!=0(1.17) for every 0≤𝑘≤𝑁−2, we conclude that ‖𝑄^𝑁−1𝑥‖=0. 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.

Proof. Let 𝑥 and 𝑦 be eigenvectors of 𝑇 corresponding to distinct eigenvalues 𝜆₁ and 𝜆₂. So, 𝑇𝑥=𝜆₁𝑥 and 𝑇𝑦=𝜆₂𝑦. By Proposition 1.4, 𝐴𝑥=𝜆₁𝑥 and 𝐴𝑦=𝜆₂𝑦 which implies that |𝜆₁|=|𝜆₂|=1. Suppose that ⟨⋅⟩ denotes the inner product of 𝒳. Then 0=𝜆₁𝑥+𝜆₂𝑦²−𝑥+𝑦²=2Re𝜆₁𝜆₂−1𝑥,𝑦.(1.18) Replacing 𝑦 by 𝑖𝑦, we obtain Im(𝜆₁𝜆₂−1)⟨𝑥,𝑦⟩=0; consequently, 𝜆₁𝜆₂−1𝑥,𝑦=𝜆₁𝜆₂−1𝑥,𝑦=0.(1.19) But 𝜆₁≠𝜆₂, and so ⟨𝑥,𝑦⟩=0.

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^𝑛𝑥‖≤𝑐 for all 𝑛≥1, then 𝑄𝑥=0. In particular, if 𝑇 is power bounded, then 𝑄=0.

Proof. Since the sequence (‖𝑇^𝑛𝑥‖)_𝑛 is bounded, an argument similar to the proof of the Proposition 1.4 shows that 𝑄𝑥=0.

2. 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 𝑃_𝑘(𝑛)=𝑥₀+𝑥₁𝑛+𝑥₂𝑛²+⋯+𝑥_𝑘𝑛^𝑘(2.1) 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 𝑃₁(𝑛)=𝑥₀+𝑥₁𝑛. It is easily seen that for every 𝑛≥1𝑃₁(≤1𝑛+1)2𝑃₁(𝑛)+𝑃₁(𝑛+2).(2.2) Since lim_𝑛→∞‖𝑃₁(𝑛)‖=+∞, there is a positive integer 𝑖 such that 𝑃₁(𝑖)<𝑃₁(𝑖+1).(2.3) This fact coupled with (2.2) implies that 0<𝑃₁(𝑖+1)−𝑃₁(𝑖)≤𝑃₁(𝑛+1)−𝑃₁(𝑛)(2.4) for every 𝑛≥𝑖. Therefore, the sequence (‖𝑃₁(𝑛)‖)_𝑛≥𝑖 is increasing. Suppose that (‖𝑃_𝑘(𝑛)‖)_𝑛 is eventually increasing and let 𝑃_𝑘+1(𝑛)=𝑥₀+𝑥₁𝑛+⋯+𝑥_𝑘+1𝑛^𝑘+1,(2.5) where 𝑥_𝑘+1≠0. Since lim_𝑛→∞𝑥₁+𝑥₂(𝑛+1)+⋯+𝑥_𝑘+1(𝑛+1)^𝑘=+∞,(2.6) using the induction hypothesis there is a positive integer 𝑗 such that for every 𝑛≥𝑗𝑥₁+𝑥₂(𝑛+1)+⋯+𝑥_𝑘+1(𝑛+1)^𝑘≥max2𝑥₀,𝑥₁+𝑥₂𝑛+⋯+𝑥_𝑘+1𝑛^𝑘.(2.7) Therefore, 𝑃_𝑘+1(𝑛+1)𝑃_𝑘+1(𝑛≥(𝑥)𝑛+1)₁+𝑥₂(𝑛+1)+⋯+𝑥_𝑘+1(𝑛+1)^𝑘−𝑥₀𝑛𝑥₁+𝑥₂𝑛+⋯+𝑥_𝑘+1𝑛^𝑘+𝑥₀≥1(2.8) 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, 𝐴𝑄=𝑄𝐴. 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 𝑇^𝑛𝑥=^𝑁−1_𝑘=0𝑛𝑘𝑄^𝑘𝐴^{𝑁−1−𝑘}𝑥,(2.9) and so by Lemma 2.1, the sequence (‖𝑇^𝑛𝑥‖)_𝑛 is eventually increasing. Suppose that 𝑥₀ 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_𝑖𝑇^𝑛𝑖𝑥₀→𝑥. Moreover, since the sequence (‖𝑇^𝑛𝑥₀‖)_𝑛 is eventually increasing, we have ‖𝛼_𝑖𝑇^𝑛𝑖𝑥₀‖≤‖𝛼_𝑖𝑇^𝑛𝑖+1𝑥₀‖ 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 𝑇⁻¹=𝐴⁻¹+𝑃,(2.10) where 𝑃=^𝑁−1_𝑘=1(−1)^𝑘𝐴^−(𝑘+1)𝑄^𝑘.(2.11) Since 𝑃^𝑁=0, by a similar argument the sequence (‖𝑇^−𝑛𝑥‖)_𝑛 is eventually increasing for every 𝑥∈𝒳. But 𝑇⁻¹ is also supercyclic (see [1, Theorem 1.12]); therefore, ‖𝑥‖≤𝑇⁻¹𝑥(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 𝑇^∗𝑛𝑥^∗=^𝑁−1_𝑘=0𝑛𝑘𝑄^∗𝑘𝐴^{∗𝑛−𝑘}𝑥^∗(2.13) for every 𝑥^∗∈𝒳^∗ and every positive integer 𝑛. If ker𝑄^∗≠{0}, then there is a nonzero 𝑥^∗∈𝒳^∗ such that ‖𝑇^∗𝑛𝑥^∗‖=‖𝐴^∗𝑛𝑥^∗‖≤‖𝑥^∗‖ because ‖𝐴^∗‖=‖𝐴‖=1. Now, suppose that 𝑥₀ is a weakly hypercyclic vector for 𝑇. Since orb(𝑇,𝑥₀) is weakly dense in 𝒳 and 𝑥^∗ is nonzero, the set {𝑥^∗(𝑇^𝑛𝑥₀)∶𝑛≥0} is dense in ℂ. But 𝑥^∗𝑇^𝑛𝑥₀=𝑇^∗𝑛𝑥^∗𝑥₀≤𝑇^∗𝑛𝑥^∗𝑥₀≤𝑥^∗𝑥₀(2.14) for all 𝑛≥0, which is a contradiction. If ker𝑄^∗={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 𝑙²(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_𝑛𝑒_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 𝑄²=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 lim^𝑛_{𝑛→∞𝑗=1}1+𝑤_𝑗^𝑛_𝑗=111+𝑤_−𝑗=0.(2.15) But ∏^∞_𝑗=1(1+𝑤_𝑗) is finite, because ∑^∞_𝑗=1𝑤_𝑗<∞. Furthermore, ∏^∞_𝑗=11/(1+𝑤_−𝑗)=0, because ^∞_𝑗=111−1+𝑤_−𝑗=^∞_𝑗=1𝑤_−𝑗1+𝑤_−𝑗=^∞_𝑗=1𝑤_{−(2𝑗−1)}1+𝑤_{−(2𝑗−1)}=^∞_𝑗=11/2𝑗−1=1+1/2𝑗−1^∞_𝑗=112𝑗=∞(2.16) (see [7, pages 299 and 300]). Therefore, (2.15) holds.

Example 2.4. Consider the isometric operator 𝐴 on 𝑙²(ℤ) defined by 𝐴𝑒_𝑛=𝑒_𝑛−1 and the weighted shift operator 𝑄 defined by 𝑄𝑒_𝑛=𝑤_𝑛𝑒_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 𝑄²=0 and 𝐴𝑄≠𝑄𝐴. Also, since 1+𝑤₁1+𝑤₂⋯1+𝑤_𝑛≥𝑤₁+𝑤₂+⋯+𝑤_𝑛(2.17) for all 𝑛≥1, and ∑^∞_𝑛=1𝑤_𝑛=∞, we conclude that lim_𝑛→∞1+𝑤₁1+𝑤₂⋯1+𝑤_𝑛=∞.(2.18) Furthermore, lim_𝑛→∞1+𝑤₋₁1+𝑤₋₂⋯1+𝑤_−𝑛<∞,(2.19) because ^∞_𝑛=1𝑤_−𝑛<∞.(2.20) Hence, using Corollary 10.27 of [1], we observe that the operator 𝐴+𝑄 is weakly hypercyclic.

3. 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 𝐵𝑒₀=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 ∩_𝑛≥0𝐴^∗𝑛ℋ=(0).

Proof. First assume that ∩_𝑛≥0𝐴^∗𝑛ℋ=(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≥0𝐴^∗𝑛ℋ≠(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_𝑀𝐴^∗|_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 𝛼𝐴^𝑛𝑔−𝑓≤𝛼𝐴^𝑛𝑔⊕𝑘−𝑓⊕0<𝜀.(3.1) 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^𝑛𝑘𝑥‖⋅‖𝑆^𝑛𝑘𝑦‖→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≥0𝐴^∗𝑛ℋ=(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. Let𝑥∈∩_𝑛≥0𝑇^∗𝑛ℋ. For every 𝑖≥0, there is a vector 𝑥_𝑁+𝑖 in ℋ such that 𝑇^∗𝑁+𝑖𝑥_𝑁+𝑖=𝑥. Since 𝑄^𝑁=0, we have 𝑥=𝑇^∗𝑁+𝑖𝑥_𝑁+𝑖=^𝑁+𝑖_𝑘=0𝑘𝐴𝑁+𝑖_∗𝑘𝑄^{∗𝑁+𝑖−𝑘}𝑥_𝑁+𝑖=^𝑁+𝑖_𝑘=𝑖+1𝑘𝐴𝑁+𝑖_∗𝑘𝑄^{∗𝑁+𝑖−𝑘}𝑥_𝑁+𝑖(3.2) which implies that 𝑥∈𝐴^∗𝑖+1ℋ. Hence, 𝑥∈∩_𝑛≥0𝑇^∗𝑛ℋ=(0) and so the operator 𝑇 admits a complete set of eigenvectors that is, ℋ=∨_{𝜇∈𝔻𝑟}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=spanker𝑇−𝜇∶𝜇∈𝔻_𝑟.(3.3) Now, if ℎ∈𝑌=𝑍, then 𝑇^𝑛ℎ𝑆^𝑛ℎ≤𝜇^𝑛‖𝑆‖^𝑛‖ℎ‖≤(𝑟‖𝑆‖)^𝑛‖ℎ‖⟶0(3.4) as 𝑛→∞. Finally, 𝑇^𝑛𝑆^𝑛ℎ=ℎ for every ℎ∈ℋ and every 𝑛≥0. Thus, the operator 𝑇 satisfies the supercyclicity criterion.

The Hilbert-Schmidt class, 𝒞₂(ℋ), is the class of all bounded operators 𝑆 defined on a Hilbert space ℋ, satisfying‖𝑆‖²₂=^∞_𝑛=1𝑆𝑒_𝑛²<+∞,(3.5)

where ‖⋅‖ is the norm on ℋ induced by its inner product. We recall that 𝒞₂(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, 𝒞₂(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 𝒞₂(ℋ) are defined by 𝐿_𝐵(𝑆)=𝐵𝑆 and 𝑅_𝐵(𝑆)=𝑆𝐵 for all 𝑆∈𝒞₂(ℋ). It is known that an operator B satisfies the supercyclicity criterion if and only if 𝐿_𝐵 is supercyclic on the space 𝒞₂(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 𝒞₂(ℋ).

Proposition 3.3. Suppose that ℋ is a Hilbert space and 𝐴∈ℬ(ℋ) is a co-isometry such that ∩_𝑛≥0𝐴^∗𝑛ℋ=(0) and 𝐴𝑄=𝑄𝐴. Then the operator 𝑇=𝐴+𝑄 satisfies the supercyclicity criterion but the operator 𝑅_𝑇 is not supercyclic on 𝒞₂(ℋ).

Proof. By Theorem 3.2, the operator 𝑇 satisfies the supercyclicity criterion. Moreover, for every 𝑆∈𝒞₂(ℋ) we have 𝑅_𝐴(𝑆)²₂=‖𝑆𝐴‖²₂=(𝑆𝐴)^∗2₂=𝐴^∗𝑆^∗2₂=^∞_𝑛=1𝐴^∗𝑆^∗𝑒_𝑛²=^∞_𝑛=1𝑆^∗𝑒_𝑛²=‖𝑆‖²₂,(3.6) which implies that 𝑅_𝐴 is an isometry. Also, if 𝑆∈𝒞₂(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.

Acknowledgments

The second author is in part supported by a grant from Shiraz University Research Council.