Research output: Contribution to journal › Article

Journal | Proc Amer Math Soc |
---|---|

Date | Published - Dec 1995 |

Issue number | 12 |

Volume | 123 |

Number of pages | 8 |

Pages (from-to) | 3709-3716 |

Original language | English |

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.

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.

- Analysis

Find related publications, people, projects, datasets and more using interactive charts.