Abstract

We present two examples. One of an operator T such that {T(T − I)}n=1 is precompact in the operator norm and the spectrum of T on the unit circle consists of an infinite number of points accumulating at 1, and the other of an operator T such that {T(T − I)}n=1 is convergent to zero but T is not power bounded. Let A be a Banach algebra and x ∈ A be power bounded. Denote by Γ the unit circle in C. The main aim of this note which may be regarded as an addendum to [2] is to answer a question stated there if the precompactness of {x(x− 1)}n=1 in A implies that 1 6∈ σ(x) ∩ (Γ {1}). The structure of σ(x) in this case has been investigated in [1] and [2] (see the references quoted therein). It was proved in [1] that {x(x − 1)}n=1 is precompact if and only if σ(x)∩ (Γ \{1}) consists of simple poles of x. The example that we present below shows that 1 can belong to the closure of σ(x)∩ (Γ {1}) and therefore the result quoted above is sharp. Example 1. Let λn = e. Define T : l → l by Ten = λnen, n = 1, 2, . . . , where en = (0, . . . , 0, 1, 0, . . .) is the nth standard basis vector of l . Then ‖T‖ = 1 and σ(T ) = {λ1, λ2, . . .}. We claim that {T(T − I)}n=1 is precompact in the algebra of bounded operators on l equipped with the operator norm. This fact can be deduced from the above quoted result of [1], however we will give a simple and direct proof. We will show that for every increasing sequence {nk}k=1 of positive integers we can choose a subsequence {mk}k=1 such that Tk(T − I) is convergent as k → ∞. We construct {mk}k=1 as follows. Since λ nk 2 attains only a finite number of values we choose a subsequence {n2k}k=1 such that λ n2k 2 = λ2 for some λ2. Then, since λ n2k 3 takes on only a finite number of values, we choose from {n2k}k=1 a subsequence {n3k}k=1 such that λ n3k 3 = λ3 for some λ3. We continue this process and define m1 = n1 and mk = nkk for 1991 Mathematics Subject Classification: Primary 46H05, 46H30, 46H35. The paper is in final form and no version of it will be published elsewhere. [381]

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