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| ON COMPLETELY SPECTRAL OPERATORS |
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Citation： |
Liu LONGFU.ON COMPLETELY SPECTRAL OPERATORS[J].Chinese Annals of Mathematics B,1981,2(1):61~64 |
| Page view： 878
Net amount： 983 |
Authors： |
Liu LONGFU; |
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| Abstract： |
In this paper we first introduce the concept of completely spectral operators and discuss its various properties; next, we consider spectral synthesis for this kind of operators and partly generalize Wermer’s result [1] on spectral synthesis for completely normal operators; finally, we investigate the problem of unitary equivalence of resolutions of the identity for completely spectral operators.
Let \[\mathcal{X}\] be a separable Hilbert space and \[\mathcal{B}(\mathcal{X})\] the algebra of all bounded linear operators on \[\mathcal{X}\]. Let Lat T denote the lattice of all invariant subspaces of \[T \in \mathcal{B}(\mathcal{X})\]
Definition 1. An operator \[T \in \mathcal{B}(\mathcal{X})\] is completely spectral if T is spectral
and reductive.
Proposition 1. T is a completely spectral operator if and only if T* is a completely spectral operator.
Proposition 2. If T is a completely spectral operator and \[\mathcal{U} \in \] Lat T, then the restriction of T to \[\mathcal{U}\] is a completely spectral operator.
Proposition 3. Let T be a spectral operator. If any pair of complementary subspaces of T is invariant under T, then T is similar to a completely spectral operator.
Proposition 4. If T is a completely spectral operator, then T = N+Q, where N is a normal operator and Q is a quasinilpotent operator commuting with N. Moreover, every \[\mathcal{U} \in \] LatT reduces both N and Q.
Difinition 2. We say that spectral synthesis holds for T, provided that for any \[\mathcal{U} \in \]LatT, which is not {0}, the set of root vectors of T contained in \[\mathcal{U}\] is complete in \[\mathcal{U}\].
Theorem 1, Let T be a completely spectral operator and let the set pf. root vectors of T be complete in \[\mathcal{U}\]. Then the spectral synthesis holds for T.
The following lemma generalizes Theorem 5 of Dunford's [2].
Lemma. Let\[{E_{\text{1}}}(\sigma ),{E_2}(\sigma )\] be resolutions of the identity for bounded spectral operators T1, T2 respectively, on Banach space \[\mathcal{U}\], and let A be a bounded linear operator on \[\mathcal{U}\],If AT1 = T2A, then \[A{E_{\text{1}}}(\sigma ) = {E_2}(\sigma )A\].
Theoerm 2. Let T1, T2 be completely spectral operators. Their resolutions of the identity are \[{E_{\text{1}}}(\sigma ),{E_2}(\sigma )\], respectively. If T1 is a quasi-affine transform of T2, then ,\[{E_{\text{1}}}(\sigma )\] and \[{E_2}(\sigma )\] are unitarily equivalent. |
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