Abstract. We present two examples. One of an operator T such that {T

LINEAR OPERATORS

BANACH CENTER PUBLICATIONS, VOLUME 38 INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES

WARSZAWA 1997

A NOTE ON THE DIFFERENCES OF THE CONSECUTIVE POWERS OF OPERATORS

A N D R Z E J ´ S W I E ¸ C H

School of Mathematics, Georgia Institute of Technology Atlanta, Georgia 30332, U.S.A.

E-mail: swiech@math.gatech.edu

Abstract. We present two examples. One of an operator T such that {T

(T − I)}

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)}

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)}

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)}

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 λ

= e

. Define T : l

→ l

by T e

= λ

e

, n = 1, 2, . . . , where e

= (0, . . . , 0, 1, 0, . . .) is the nth standard basis vector of l

. Then kT k = 1 and σ(T ) = {λ

, λ

, . . .}. We claim that {T

(T − I)}

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 {n

}

of positive integers we can choose a subsequence {m

}

such that T

(T − I) is convergent as k → ∞. We construct {m

}

as follows. Since λ

attains only a finite number of values we choose a subsequence {n

}

such that λ

= λ

for some λ

. Then, since λ

takes on only a finite number of values, we choose from {n

}

a subsequence {n

}

such that λ

= λ

for some λ

. We continue this process and define m

= n

and m

= n

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]

382

k > 1. For m

so defined we have

(1) λ

= λ

for k ≥ j, k, j = 1, 2, . . .

Notice that m

≥ k for k = 1, 2, . . . Let ε > 0. Choose n

such that |λ

− 1| < ε/2 for n ≥ n

. Denote by P the orthogonal projection onto span{e

, e

, . . .}. Take any k, l ≥ n

. By the definition of T and (1) it follows that

k(T

− T

)(T − I)xk = k(T

− T

)(T − I)P xk ≤ 2k(T − I)P kkxk ≤ εkxk, where the last inequality follows from the choice of n

. Therefore {T

(T − I)}

is Cauchy. It converges to T

defined by T

e

= λ

e

, n = 1, 2, . . . The claim is proved.

An interesting feature of Example 1 is that the set {T

}

is discrete. In fact, if n − m = p > 0, then

k(T

− T

)e

k = k2e

e

k = 2.

However, if we set λ

= e

2πi

2n2

, n = 1, 2, . . . , in the definition of T , then {T

}

has an accumulation point. To show this observe that, if x = P

n=1

x

e

, kxk ≤ 1, p

= 2

, then

k(T

− I)xk =

∞

X

j=n+1

(λ

− 1)x

e

=  X

j=n+1

|λ

− 1|

|x

|



≤  X

j=n+1

(arg(λ

))



≤



X

j=n+1

16π

2



= 4π

√ 15 1

2

→ 0 as n → ∞.

Therefore the existence of accumulation points depends also on the geometry of the spectrum and not only on its cardinality. Moreover, it trivially follows that if {x

}

has an accumulation point then σ(x) = σ

∪ σ

, where σ

, σ

are two closed sets such that σ

⊂ {z : |z| < 1}, σ

⊂ Γ . To see this, suppose that x

converges as k → ∞ for some sequence {n

}

, i.e. for every ε > 0 there is n

such that kx

− x

k < ε if n

, n

> n

. Let n

> n

. We have σ(x

− x

) = (z

− z

)(σ(x)). However, (2) |z

− z

| ≥ |z|

− |z|

and if we fix n

, the right hand side of (2) can be made arbitrarily close to 1 by choosing

|z| sufficiently close to 1 and then n

sufficiently large. Hence σ(x) cannot approach Γ . We would like to finish with an elementary example of an operator T such that T

(T − I) → 0 as n → ∞ but T is not power bounded. It is similar in the spirit to Example 3.7 in [2], ours is however very explicit.

Example 2. Let T : l

→ l

be defined by T e

= n − 1

n e

+ 1

ln (n + 1) e

, T e

= e

DIFFERENCES OF CONSECUTIVE POWERS

383

for n = 1, 2, . . . Let ε > 0, x = P

n=1

x

e

. Then k(T

− T

)xk

=

∞

X

n=1

 1

ln (n + 1)

 n − 1 n



x

e

− 1 n

 n − 1 n



x

e



2

≤

∞

X

n=1

1 ln

(n + 1)

 n − 1 n



|x

|

+

∞

X

n=1

1 n

 n − 1 n



|x

|

≤

n₀

X

n=1

1 ln

(n + 1)

 n − 1 n



|x

|

+ 1 ln

(n

+ 1)

∞

X

n=n0+1

|x

|

+

n₀

X

n=1

1 n

 n − 1 n



|x

|

+ 1 (n

+ 1)

∞

X

n=n₀+1

|x

|

≤ ε

kxk

by first choosing n

sufficiently large and then taking k large enough. To see that T is not power bounded, observe that

kT

e

k ≥ 1 ln(n + 1)

n

X

j=1

 n − 1 n



. Since (

)

>

for large n (actually it is close to 1/e), we obtain

kT

e

k ≥ n 3 ln(n + 1) .

Hence kT

k ≥ O(n

) for every 0 < α < 1. On the other hand, we have T

= o(n).

References

[1] S. H u a n g, Stability properties characterizing the spectra of operators on Banach spaces, J.

Funct. Anal. 132 (1995), 361–382.

[2] H. C. R ¨ o n n e f a r t h, On the differences of the consecutive powers of Banach algebra ele-

ments, this volume, 297–314.