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Compactness criteria for integral operators in $ L\sp \infty$ and $ L\sp 1$ spaces


Author: S. P. Eveson
Journal: Proc. Amer. Math. Soc. 123 (1995), 3709-3716
MSC: Primary 47B38; Secondary 47B07, 47G10
DOI: https://doi.org/10.1090/S0002-9939-1995-1291766-8
MathSciNet review: 1291766
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Abstract: Let $ (S,\Sigma ,\mu )$ be a positive measure space, $ k:S \times S \to \mathbb{R}$ be a measurable function such that the kernel $ \vert k\vert$ induces a bounded integral operator on $ {L^\infty }(S,\Sigma ,\mu )$ (equivalently, that $ {\text{ess}}.{\sup _{s \in S}}\vert k(s,t)\vert d\mu (t) < \infty $), and for $ s \in S$ let $ {k_s}(t) = k(s,t)$. We show that it is sufficient for the integral operator T induced by k on $ {L^\infty }(S,\Sigma ,\mu )$ to be compact, that there exists a locally $ \mu $-null set $ N \in \Sigma $ such that the set $ \{ {k_s}:s \in S\} $ is relatively compact in $ {L^1}(S,\Sigma ,\mu )$, and that this condition is also necessary if $ (S,\Sigma ,\mu )$ is separable. In the case of Lebesgue measure on a subset of $ {\mathbb{R}^n}$, we use Riesz's characterisation of compact sets in $ {L^1}({\mathbb{R}^n})$ to provide a more tractable form of this criterion.


References [Enhancements On Off] (What's this?)

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DOI: https://doi.org/10.1090/S0002-9939-1995-1291766-8
Article copyright: © Copyright 1995 American Mathematical Society

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