resolvent set


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resolvent set

[ri′zäl·vənt ′set]
(mathematics)
Those scalars λ for which the operator T- λ I has a bounded inverse, where T is some linear operator on a Banach space, and I is the identity operator.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
References in periodicals archive ?
The spectrum of an operator H will be denoted by [SIGMA](H) and its resolvent set by [??](H).
We assume that 0 [member of] [rho](A), where [rho](A) is the resolvent set of A and the analytic semigroup {S(t), t [greater than or equal to] 0} in X is uniformly bounded; that is, [[parallel]S(t)[parallel].sub.X] [less than or equal to] M for some positive constant M [greater than or equal to] 1.
The results obtained showed that the cocycle has an exponential dichotomy if and only if the associated evolution semigroup is hyperbolic and if and only if the imaginary axis is contained in the resolvent set of the generator of the evolution semigroup.
For z in the resolvent set of T, re(T), and y in X we consider the Fredholm integral problem of the second kind
Then we get [lambda] = N + r [member of] [rho]([L.sub.0]) and [lambda] = n - r [member of] [rho]([L.sub.0]), where r > 0, [rho]([L.sub.0]) is resolvent set of the operator [L.sub.0], i.e., [rho]([L.sub.0]) = C\[sigma]([L.sub.0]) and [sigma]([L.sub.0]) is the spectrum of the operator [L.sub.0].
The analytic model represents a subnormal operator as the multiplication by the independent variable of a space of vector-valued functions that are analytic on the resolvent set of its normal extension, says Xia.
[3] if A is a closed linear operator with domain D(A) defined on a Banach space E and [alpha] > 0, the we say that A is the generator of an [alpha]-resolvent family if there exists [omega] [greater than or equal to] 0 and a strongly continuous function [S.sub.[alpha]] : [R.sub.+] [right arrow]L(E) such that {[[lambda].sub.[alpha]] : Re([lambda]) > [omega]} [subset] [rho](A)) ([rho](A) being the resolvent set of A) and
We denote by [rho](A) the resolvent set of A and by [sigma](A) the spectrum of A.
If [[OMEGA].sub.a] (A) is connected, then (it has no bounded component, and hence) it has just one component, namely itself, and hence the resolvent set [rho](A) intersects [[OMEGA].sub.a](A).
The last theorem that we state in this section shows that the point spectrum, continuous spectrum, and resolvent set of a self-adjoint operator A and each of its associated left-definite operators [A.sub.r] (r > 0) are identical.
The spaces [X.sub.1] and [X.sub.-1] are defined as follows: [X.sub.1] := (D(A), [[paralle]*[parallel].sub.1]), where [[parallel]x[parallel].sub.1] := ([lambda]I-A)x[parallel], x [member of] D(A) (for some [lambda] fixed in the resolvent set [rho](A) of A), and [X.sub.-1][parallel] ([lambda]I-A)x[parallel], x [member of] D(A) (for some A fixed in the resolvent set [rho](A) of A), and [X.sub.-1] is the completion of X with respect to the norm [parallel]x[[parallel].sub.-1].
In principle, the comments from the previous section imply that, because of (5.11), the scheme Solve with the scaling D would indeed work for each fixed 7 in the resolvent set with optimal complexity.