Compactness criteria for integral operators in L^infinity and L^1 spaces

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JournalProc Amer Math Soc
DatePublished - Dec 1995
Issue number12
Volume123
Number of pages8
Pages (from-to)3709-3716
Original languageEnglish

Abstract

Let $(S,\Sigma,\mu)$ be a positive measure space, $k:S\times S\to\mathbb{R}$ a
measurable function such that the kernel $|k|$ induces a bounded integral operator
on $L^\infty(S,\Sigma,\mu)$ (equivalently, that $\operatorname{ess.sup}_{t\in S}\int_Sk(s,t)dt<\infty$), 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.

    Research areas

  • Analysis

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