Power bounded strictly cyclic operators
| Autor: | Erik J. Rosenthal |
|---|---|
| Rok vydání: | 1978 |
| Předmět: | |
| Zdroj: | Proceedings of the American Mathematical Society. 72:276-280 |
| ISSN: | 1088-6826 0002-9939 |
| Popis: | We show that a power bounded hereditarily strictly cyclic operator on Hilbert space is similar to a contraction. We also show that certain "almost unitary" operators are not strictly cyclic. Recall that the operator T is power bounded if there exists a positive real number M such that II TII < M for n = 1, 2, 3 .... Sz.-Nagy [16] proved that every compact power bounded operator is similar to a contraction and asked whether the hypothesis of compactness can be removed. Foguel [3] (see also Halmos [5]) showed that it cannot be removed-there is a power bounded operator that is not similar to a contraction. (It is still not known if every polynomially bounded operator is similar to a contraction. See Ghatage [4] for some related results.) Given an operator T, let d( T) denote the uniformly closed algebra generated by T and the identity. Lambert [11] defined T to be strictly cyclic if there exists a vector xo such that {Axo: A EE d(T)} is the entire space. T is hereditarily strictly cyclic if its restriction to every invariant subspace is strictly cyclic. There are a number of known results about strictly cyclic operators-see [6], [7], [8], [10], [11], [12], [13], [14]. In this paper we prove that a power bounded hereditarily strictly cyclic operator is similar to a contraction. We require three lemmas. LEMMA 1. If T is a strictly cyclic operator, and if 'T is an invariant subspace of T, then the compression of T to 6Th' is strictly cyclic. PROOF. Let P be the projection onto 6T', and let e be a strictly cyclic vector for T; i.e., {Ae: A E d((T)} = SC. Letf = Pe and m = e f. Then it suffices to show that {PAf: A E d((T)} = 6XL Note that PAm = 0 for every A in d((T). Now, given x E (%6', choose A in d((T) such that x = Ae. Then PAf = PA(f + m) = PAe = Px = x, and the proof is complete. LEMMA 2. If T is hereditarily strictly cyclic and power bounded, and if Received by the editors August 23, 1977. AMS (MOS) subject classifications (1970). Primary 47B99. ? American Mathematical Society 1978 276 This content downloaded from 157.55.39.177 on Fri, 18 Nov 2016 04:17:49 UTC All use subject to http://about.jstor.org/terms POWER BOUNDED STRICTLY CYCLIC OPERATORS 277 a(T) = {X0}, where IR01 = 1, then T acts on a one-dimensional space. PROOF. By replacing T by 90 T, we may assume that Xo = 1. Lambert [11] proved that T strictly cyclic implies that every point in the spectrum of T is compression spectrum. Thus, 1 is an eigenvalue for T*. So choose a unit vector e such that T*e = e, and let T=(l I ) \A BJ be the decomposition of T with respect to the decomposition of SC = V{e} ( {e})'. We claim that {e})' = {O}. If not, since T is hereditarily strictly cyclic and B = Tj{e}', B is strictly cyclic. Also, a(B) = {l}, so there is a unit vector fl e with B*f = f. Then decompose T with respect to the decomposition SC = V{e,f} f ) {e, f } ': T=C D Since T is power bounded, (a 1) is power bounded. Since (I = I A must be 0. Since { e, f }' E Lat T, (O 1) is strictly cyclic by Lemma 1. But the identity is strictly cyclic only on a one-dimensional space, so { e}' must be {0}. We should perhaps note that the above lemma implies the following well known fact. THEOREM I (SZ.-NAGY). A power bounded operator acting on a finite dimensional space is similar to a contraction. PROOF. By the Jordan canonical form theorem, it suffices to prove the theorem for a single Jordan block, so consider an operator |
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