Integral operator with a measurable kernel

An integral operator with kernel $K$ has the form

(Tf)(x)=\int_Y K(x,y)f(y)\,d\nu(y)

on a specified class of functions for which the integral exists. This kernel is a function of two variables; it is distinct from the nullspace of a linear map.

A direct bound

If $K$ is jointly measurable and $\sup_x\int|K(x,y)|\,d\nu(y)\le M$ , then $\|Tf\|_{\sup}\le M\|f\|_{\sup}$ on bounded measurable functions. For essential norms, use the corresponding almost-everywhere row bound on sigma-finite product spaces. A formula alone does not establish boundedness on a chosen function space; its kernel estimates do.

Examples

On $\mathbb R^n$ , taking $K(x,y)=k(x-y)$ gives convolution. An integral $\int_a^tK(t,s)f(s)\,ds$ is a Volterra operator and respects the time ordering $s\le t$ .