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1988
DOI: 10.2307/2046992
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Inverses of Generators

Abstract: ABSTRACT. Let A be a (possibly unbounded) linear operator on a Banach space X that generates a bounded holomorphic semigroup of angle 0 (0 < 6 < t/2).We show that, if the range of A is dense, then A is one-to-one, and A-1 (defined on the range of A) generates a bounded holomorphic semigroup of angle 9, given by e**"' = fe-^(wA + z)-1-, J 2tt¿over an appropriate curve. When X is reflexive, it is sufficient that A be one-to-one. Introduction.A special class of semigroups of operators that frequently arises in … Show more

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Cited by 16 publications
(23 citation statements)
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“…From this it is easy to see that A generates an analytic, sectorially bounded semigroup if and only if A −1 generates an analytic, sectorially bounded semigroup, provided A −1 exists. Note that without using the above resolvent characterization for sectorially bounded analytic semigroups this result has been proved by deLaubenfels [14]. Thus using Theorem 4.4 we derive the following.…”
Section: Guo and Zwart Ieotmentioning
confidence: 76%
“…From this it is easy to see that A generates an analytic, sectorially bounded semigroup if and only if A −1 generates an analytic, sectorially bounded semigroup, provided A −1 exists. Note that without using the above resolvent characterization for sectorially bounded analytic semigroups this result has been proved by deLaubenfels [14]. Thus using Theorem 4.4 we derive the following.…”
Section: Guo and Zwart Ieotmentioning
confidence: 76%
“…Достаточно ли приведенных условий для того, чтобы A −1 был генератором C 0 -полугруппы? Этот вопрос был поставлен в работе [3], где было доказано, что если оператор A порождает ограниченную голоморфную C 0 -полугруппу и существует оператор A −1 ∈ E , то A −1 также является гене-ратором ограниченной голоморфной C 0 -полугруппы и, в частности, оператор A −1 ∈ G (элементарное доказательство этого утверждения см. в [4]).…”
unclassified
“…Основной результат настоящей статьи дает отрицательный ответ на сформу-лированный выше вопрос из [3] в случае рефлексивного банахова пространства. А именно, показано, что для банахова пространства…”
unclassified
“…The set H 0 consists of generators of semigroups holomorphic in some sector Σ θ and, for β > 0, quasibounded in the terminology in [1, Chap. 9, §1.4].If A ∈ H (θ) has an inverse A −1 ∈ E (i.e., ker A = {0} and the range of A is dense in B), then A −1 also belongs to H (θ) [3,4]; in particular, A −1 ∈ G . Indeed [1, 2], A ∈ H (θ) if and only if the resolvent R(A, λ) exists for λ ∈ Σ π/2+θ and satisfies the estimate…”
mentioning
confidence: 99%
“…If A ∈ H (θ) has an inverse A −1 ∈ E (i.e., ker A = {0} and the range of A is dense in B), then A −1 also belongs to H (θ) [3,4]; in particular, A −1 ∈ G . Indeed [1,2], A ∈ H (θ) if and only if the resolvent R(A, λ) exists for λ ∈ Σ π/2+θ and satisfies the estimate…”
mentioning
confidence: 99%