1995
DOI: 10.2307/2154774
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Power Regular Operators

Abstract: Abstract.We show that for a wide class of operators T on a Banach space, including the class of decomposable operators, the sequence {||T''x||1/'1}^°¡1 converges for every x in the space to the spectral radius of the restriction of T to the subspace \Z™=0{T"x} .

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Cited by 4 publications
(5 citation statements)
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“…But x is a cyclic vector for T , and, since T is power regular [4], this contradicts our assumption that T has spectral radius 1.…”
Section: Theorem 2 Suppose That T ∈ L(x) Has Bishop's Property (β) mentioning
confidence: 80%
See 1 more Smart Citation
“…But x is a cyclic vector for T , and, since T is power regular [4], this contradicts our assumption that T has spectral radius 1.…”
Section: Theorem 2 Suppose That T ∈ L(x) Has Bishop's Property (β) mentioning
confidence: 80%
“…Therefore, since T | XT (K) also has property (β), we conclude that σ(T ) = σ su (T | XT (K) ) ⊂ K. Because operators with Bishop's property (β) are power regular [4], that is, lim n→∞ T n x 1/n exists for every x ∈ X, it follows that for a cyclic vector x, the sequence ( T n x 1/n ) n≥1 converges to the spectral radius of T .…”
Section: Corollary 1 Suppose That X Is Separable and That T ∈ L(x)mentioning
confidence: 85%
“…Recall that an operator S in B(H) is power regular if lim n→∞ S n h 1 n exists for every h ∈ H. It is known that compact operators, normal operators and decomposable operators are power regular (see [1] and references therein). It is important in operator theory that which operators are power regular.…”
Section: Resultsmentioning
confidence: 99%
“…On the other hand, it follows from [5] that S is power-regular , that is, the sequence ( S n x 1/n ) n≥0 converges for all x ∈ H. In particular, ( S n e 0 1/n ) n≥0 converges; so,…”
Section: Which Unilateral Weighted Shift Operators Have Bishop's Propmentioning
confidence: 99%