English

Compactness of integral operators and uniform integrability on measure spaces

Functional Analysis 2022-01-25 v1

Abstract

Let (E,E,μ)(E,\mathcal E,\mu) be a measure space and G ⁣:E×E[0,]G\colon E\times E\to [0,\infty] be measurable. Moreover, let F ⁣ui\mathcal F\!_{ui} denote the set of all qE+q\in\mathcal E^+ (measurable numerical functions q0q\ge 0 on EE) such that {G(x,)q ⁣:xE}\{G(x,\cdot)q\colon x\in E\} is uniformly integrable, and let F ⁣co\mathcal F\!_{co} denote the set of all qE+q\in\mathcal E^+ such that the mapping fG(fq):=G(,y)f(y)q(y)dμ(y)f\mapsto G(fq) :=\int G(\cdot,y) f(y) q(y)\,d\mu(y) is a compact operator on the space Eb\mathcal E_b of bounded measurable functions on EE (equipped with the sup-norm). It is shown that F ⁣ui=F ⁣co\mathcal F\!_{ui}=\mathcal F\!_{co} provided both F ⁣ui\mathcal F\!_{ui} and F ⁣co\mathcal F\!_{co} contain strictly positive functions.

Keywords

Cite

@article{arxiv.2201.09080,
  title  = {Compactness of integral operators and uniform integrability on measure spaces},
  author = {Wolfhard Hansen},
  journal= {arXiv preprint arXiv:2201.09080},
  year   = {2022}
}
R2 v1 2026-06-24T08:58:39.515Z