Kernel of bounded operator $C_0(\mathbb{R})\to C_0(\mathbb{R})$

Question

Let $T:C_0(\mathbb{R})\to C_0(\mathbb{R})$ be a bounded linear operator, where $C_0(\mathbb{R})$ is the space of continuous functions on the real line vanishing at the infinity equipped with the supremum norm.

In general it is not true that $T$ admits a kernel $K$ in the sense $$ (Tf)(x) = \int K(x,y) f(y) d y. $$ (for example if $T$ is the identity then $K$ should coincide with the Dirac delta). Is it true that for any operator $T$ there is some "family" of measures $K(x,dy)$ such that $$ (Tf)(x) = \int K(x,dy) f(y). $$ What conditions should a "family" of measures $K(x,dy)$ satisfy so that the above formula defines a bounded operator $T:C_0(\mathbb{R})\to C_0(\mathbb{R})$?

Answer 1

Existence of measures $K(x,\cdot)$ such that $$(Tf)(x)=\int_{\mathbb R} K(x,dy) f(y)$$ for all $f\in C_0(\mathbb R)$ follows from the Riesz-Markov representation theorem just because $f\mapsto (Tf)(x)$ are continuous linear functionals on $C_0(\mathbb R)$. For open sets $A\subseteq\mathbb R$ an inner approximation of $1_A=\sup\{\varphi_n:n\in\mathbb N\}$ with $\varphi_n\in C_c(\mathbb R)$ together with , e.g., a monotone class argument should imply that $K$ is indded a kernel, i.e., $x\mapsto K(x,A)$ is measurable for every $A$.

Vice versa, if you have a kernel $K$ and you define $Tf$ by the displayed formula, you will need continuity of $x\mapsto K(x,A)$ for all $A$ and probaly some kind of tightness to have $Tf\in C_0(\mathbb R)$ of each $f\in C_0(\mathbb R)$. Continuity should then be for free by the closed graph theorem.