# Supercyclicity and Hypercyclicity of an Isometry Plus a Nilpotent

Authors: S. Yarmahmoodi; K. Hedayatian; B. Yousefi
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Suppose thatX 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.


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## Body

## 1. Introduction

Letx be a vector in a separable normed space 𝒳 and T an operator on 𝒳.  The orbit of x under T is defined by (1.1)orb(T,x)={Tnx:n=0,1,2,…}.We recall that a vectorx 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, σap(T), σp(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 QN=0 and QN-1≠0. From now on, we assume that Q is a nilpotent operator of order N≥1 unless stated otherwise.Proposition 1.1.
Suppose that𝒳 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)
σp(T)=σp(A),(iii)
σap(T)=σap(A).Proof.
(i) Suppose thatλ∉σ(A). Then it is easily seen that
(1.2)(T-λ)-1=∑k=1N(-1)k-1(A-λ)-kQk-1
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) Ifλ∈σp(A), there exits x≠0 such that Ax=λx. Therefore,
(1.3)TQN-1x=AQN-1x=λQN-1x.
Now, if QN-1x≠0, then λ∈σp(T); otherwise,
(1.4)TQN-2x=AQN-2x=λQN-2x.
Also, if QN-2x≠0 then λ∈σp(T); otherwise, consider QN-3x and continue this process to conclude that Tx=Ax=λx which implies that λ∈σp(T). Hence, σp(A)⊆σp(T). Moreover, since A=T-Q, using a similar method, we get σp(T)⊆σp(A).
(iii) Letλ∈σap(T); then there exists a sequence (xn)n in 𝒳 such that ∥xn∥=1 and
(1.5)Txn-λxn⟶0asn⟶+∞.
Therefore,
(1.6)AQN-1xn-λQN-1xn⟶0asn⟶+∞.
Suppose that there is c1>0 so that
(1.7)‖QN-1xn‖>c1
for all n≥1; then Ayn-λyn→0 as n→+∞ where
(1.8)yn=QN-1xn‖QN-1xn‖
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
(1.9)QN-1xn⟶0asn⟶+∞.
So by (1.5),
(1.10)AQN-2xn-λQN-2xn⟶0asn⟶+∞.
Now, if there is a constant c2>0 such that
(1.11)‖QN-2xn‖>c2
for all n, then Azn-λzn→0 where
(1.12)zn=QN-2xn‖QN-2xn‖
which implies that λ∈σap(A). Otherwise, we can assume, without loss of generality, that QN-2xn→0 as n→∞ and by (1.5)
(1.13)AQN-3xn-λQN-3xn⟶0
as n→∞. The procedure continues to conclude that λ∈σap(A). Since A=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 that𝒳 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 in 𝒳 with ∥xn∥=1 and so (A-λI)(xn)→0 if n→∞. Therefore,
(1.14)0≤1-|λ|≤‖(A-λI)(xn)‖⟶0asn⟶∞,
and so |λ|=1. Now, if |λ|≠1, then λ∉σap(T) and so T-λ is bounded below.Corollary 1.3.
Suppose that𝒳 is an infinite dimensional Banach space. Then the operator T on 𝒳 is not a compact operator.Proof.
IfT 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 operatorsT and A are defined on a normed space 𝒳, then ker(T-λ)⊆ker(A-λ) for every scalar λ.Proof.
Fixλ∈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
(1.15)‖x‖2=‖Tnx‖2=‖An-(N-1)∑k=0N-1(nk)QkAN-1-kx‖2=‖∑k=0N-1(nk)QkAN-1-kx‖2=(nN-1)2‖∑k=0N-1(N-1)!(n-N+1)!k!(n-k)!QkAN-1-kx‖2.
Consequently,
(1.16)‖x‖≥(nN-1)[‖QN-1x‖-∑k=0N-2(N-1)!(n-N+1)!k!(n-k)!‖QkAN-1-kx‖].
Since
(1.17)lim⁡n→∞⁡(N-1)!(n-N+1)!k!(n-k)!=0
for every 0≤k≤N-2, we conclude that ∥QN-1x∥=0. Continue the above process to get Qx=0, and so Ax=λx.Corollary 1.5.
If𝒳 is a Hilbert space, then the eigenvectors of T corresponding to distinct eigenvalues are orthogonal.Proof.
Letx 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
(1.18)0=‖λ1x+λ2y‖2-‖x+y‖2=2Re(λ1λ2¯-1)〈x,y〉.
Replacing y by iy, we obtain Im(λ1λ2¯-1)〈x,y〉=0; consequently,
(1.19)(λ1λ2-1)〈x,y〉=(λ1λ2¯-1)〈x,y〉=0.
But λ1≠λ2, and so 〈x,y〉=0.Recall that an operatorT is power bounded if there exists some constant c>0 such that ∥Tn∥≤c for all n=1,2,3,….Proposition 1.6.
Let𝒳 be a normed space and x∈𝒳. If there is a constant c>0 such that ∥Tnx∥≤c for all n≥1, then Qx=0. In particular, if T is power bounded, then Q=0.Proof.
Since the sequence(∥Tnx∥)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.
Let𝒳 be a normed space. For nonnegative integers k,n, if
(2.1)Pk(n)=x0+x1n+x2n2+⋯+xknk
is a polynomial in n with coefficients in 𝒳 of degree k, then the sequence (∥Pk(n)∥)n is eventually increasing.Proof.
We prove the lemma by induction onk, the degree of the polynomial Pk(n). For k=1, let P1(n)=x0+x1n. It is easily seen that for every n≥1(2.2)‖P1(n+1)‖≤12(‖P1(n)‖+‖P1(n+2)‖).
Since limn→∞∥P1(n)∥=+∞, there is a positive integer i such that
(2.3)‖P1(i)‖<‖P1(i+1)‖.
This fact coupled with (2.2) implies that
(2.4)0<‖P1(i+1)‖-‖P1(i)‖≤‖P1(n+1)‖-‖P1(n)‖
for every n≥i. Therefore, the sequence (∥P1(n)∥)n≥i is increasing. Suppose that (∥Pk(n)∥)n is eventually increasing and let
(2.5)Pk+1(n)=x0+x1n+⋯+xk+1nk+1,
where xk+1≠0. Since
(2.6)lim⁡n→∞⁡‖x1+x2(n+1)+⋯+xk+1(n+1)k‖=+∞,
using the induction hypothesis there is a positive integer j such that for every n≥j(2.7)‖x1+x2(n+1)+⋯+xk+1(n+1)k‖≥max⁡{2‖x0‖,‖x1+x2n+⋯+xk+1nk‖}.
Therefore,
(2.8)‖Pk+1(n+1)‖‖Pk+1(n)‖≥(n+1)‖x1+x2(n+1)+⋯+xk+1(n+1)k‖-‖x0‖n‖x1+x2n+⋯+xk+1nk‖+‖x0‖≥1
for every n≥j. Hence, the sequence (∥pk+1(n)∥)n≥j is increasing.Theorem 2.2.
Suppose that𝒳 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.
Let𝒳~ 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 Proposition1.4, if x∈𝒳 then
(2.9)‖Tnx‖=‖∑k=0N-1(nk)QkAN-1-kx‖,
and so by Lemma 2.1, the sequence (∥Tnx∥)n is eventually increasing. Suppose that x0 is a supercyclic vector for T. Thus, for any x∈𝒳 there is a sequence (ni)i of positive integers and a sequence (αi)i of scalars such that αiTnix0→x. Moreover, since the sequence (∥Tnx0∥)n is eventually increasing, we have ∥αiTnix0∥≤∥αiTni+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
(2.10)T-1=A-1+P,
where
(2.11)P=∑k=1N-1(-1)kA-(k+1)Qk.
Since PN=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,
(2.12)‖x‖≤‖T-1x‖
for every x∈𝒳. Thus, T is an isometry which implies that it is not a supercyclic operator.
To show that the operatorT is not weakly hypercyclic, note that
(2.13)‖T*nx*‖=‖∑k=0N-1(nk)Q*kA*n-kx*‖
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*(Tnx0):n≥0} is dense in ℂ. But
(2.14)‖x*(Tnx0)‖=‖(T*nx*)(x0)‖≤‖T*nx*‖‖x0‖≤‖x*‖‖x0‖
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 spacelp(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 ofA and Q is essential in the preceding theorem.Example 2.3.
Let(en)n=-∞+∞be the standard orthonormal basis for l2(ℤ). Define the isometric operator A by Aen=en+1 for all n∈ℤ and the weighted shift operator Q by Qen=wnen+1, where w2n=0 for all integers n, w2n-1=1/(2n-1)2 for all n≥1, and w2n-1=1/(1-2n) for all n≤0. Note that Q2=0 and AQ≠QA. Moreover, since 1≤infn(1+wn)≤supn(1+wn)≤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
(2.15)lim⁡n→∞∏j=1n(1+wj)∏j=1n11+w-j=0.
But ∏j=1∞(1+wj) is finite, because ∑j=1∞wj<∞. Furthermore, ∏j=1∞1/(1+w-j)=0, because
(2.16)∑j=1∞(1-11+w-j)=∑j=1∞w-j1+w-j=∑j=1∞w-(2j-1)1+w-(2j-1)=∑j=1∞1/(2j-1)1+1/(2j-1)=∑j=1∞12j=∞
(see [7, pages 299 and 300]). Therefore, (2.15) holds.Example 2.4.
Consider the isometric operatorA on l2(ℤ) defined by Aen=en-1 and the weighted shift operator Q defined by Qen=wnen-1, where w2n=0 for all n∈ℤ, w2n-1=1/(2n-1), for n≥1, and w2n-1=1/(2n-1)2 for n≤0. Note that Q2=0 and AQ≠QA. Also, since
(2.17)(1+w1)(1+w2)⋯(1+wn)≥w1+w2+⋯+wn
for all n≥1, and ∑n=1∞wn=∞, we conclude that
(2.18)lim⁡n→∞⁡(1+w1)(1+w2)⋯(1+wn)=∞.
Furthermore,
(2.19)lim⁡n→∞⁡(1+w-1)(1+w-2)⋯(1+w-n)<∞,
because
(2.20)∑n=1∞w-n<∞.
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 thatℋ is a separable Hilbert space with orthonormal basis {en}n=0∞. Recall that the unilateral shift operator S:ℋ→ℋ is given by Sen=en+1 for all n and the backward shift operator B:ℋ→ℋ is defined by Be0=0 and Ben=en-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 thatA 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*=Sm for some positive integer m (see [8]). Therefore, A=Bm 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 PM 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)*=PMA*|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
(3.1)‖αAng-f‖≤‖αAn(g⊕k)-(f⊕0)‖<ε.
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 thatX is a separable Banach space and T is a bounded operator on X. If there is an increasing sequence of positive integers {nk}k∈ℕ, and two dense sets Y and Z of X such that(1)
there exists a functionS:Z→Z satisfying TSx=x for all x∈Z,(2)
∥Tnkx∥·∥Snky∥→0 for every x∈Y and y∈Z,then T is supercyclic.Theorem 3.2.
Suppose thatA 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 Corollary1.2, 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 xN+i in ℋ such that T*N+ixN+i=x. Since QN=0, we have
(3.2)x=T*N+ixN+i=∑k=0N+i(N+ik)(A*kQ*N+i-k)(xN+i)=∑k=i+1N+i(N+ik)(A*kQ*N+i-k)(xN+i)
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∈𝔻rker(T-μ) for every positive real number r, where 𝔻r={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
(3.3)Y=Z=span⁡{ker⁡(T-μ):μ∈Dr}.
Now, if h∈Y=Z, then
(3.4)‖Tnh‖‖Snh‖≤|μ|n‖S‖n‖h‖≤(r‖S‖)n‖h‖⟶0
as n→∞. Finally, TnSnh=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(3.5)‖S‖22=∑n=1∞‖Sen‖2<+∞,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 LB and the right multiplication operator RB on 𝒞2(ℋ) are defined by LB(S)=BS and RB(S)=SB for all S∈𝒞2(ℋ). It is known that an operator B satisfies the supercyclicity criterion if and only if LB 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 RT is not a supercyclic operator on 𝒞2(ℋ).Proposition 3.3.
Suppose thatℋ 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 RT is not supercyclic on 𝒞2(ℋ).Proof.
By Theorem3.2, the operator T satisfies the supercyclicity criterion. Moreover, for every S∈𝒞2(ℋ) we have
(3.6)‖RA(S)‖22=‖SA‖22=‖(SA)*‖22=‖A*S*‖22=∑n=1∞‖A*S*en‖2=∑n=1∞‖S*en‖2=‖S‖22,
which implies that RA is an isometry. Also, if S∈𝒞2(ℋ), then RQN(S)=0. Since RT(S)=RA(S)+RQ(S), Theorem 2.2 implies that RT is not supercyclic.The proof of the following proposition is similar to the proof of the second part of Theorem2.2, and we omit it.Proposition 3.4.
Suppose that𝒳 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.



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