When is an operator $T$ on $L^1(\mu)$ of the form $(Tf)(x)=\int\mu({\rm d}y)p(x,y)f(y)$?

Question

Let $(E,\mathcal E,\mu)$ be a $\sigma$-finite measure space and $T\in\mathfrak L(\mathcal L^1(\mu))$.

I would like to know which conditions on $T$ would be sufficient to conclude that there is a (possibly signed) transition kernel $\kappa$ on $(E,\mathcal E)$ s.t. $$(Tf)(x)=\int\kappa(x,{\rm d}y)f(y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal L^1(\mu).\tag1$$

On page 159 of Analysis of Heat Equations on Domains it is claimed that $$\left\|T\right\|_{\mathfrak L(L^1(\mu),\:L^\infty(\mu))}<\infty\tag2$$ is equivalent to $(1)$ with the additional claim that $\kappa$ has a density with respect to $\mu$, i.e. $$\kappa(x,B)=\int_B\mu({\rm d}y)p(x,y)\;\;\;\text{for all }(x,B)\in E\times\mathcal E\tag3$$ for some $\mathcal E^{\otimes2}$-measurable $p:E^2\to\mathbb R$.

How do we see this? And is there a weaker condition which at least implies the existence of a kernel $\kappa$ (which possibly has no density wrt $\mu$)?

Answer 1

We are looking at different instances of Dunford–Pettis Kernel Representation Theorem (not to be confused with other theorems/properties named after them). The numbering of the statements refers to [1] (see below). In citing the results I am specializing them to the case that all measure spaces are $\sigma$-finite, countably generated, and countably separated. Among other things, these assumptions imply that $L^1$-spaces are separable, that $L^\infty=(L^1)^*$, and that we may regard naturally $L^1$ as a subset of $(L^\infty)^*$. For more general spaces, see [1].

$(S,\mathcal E,\alpha)$ and $(T,\mathcal F,\beta)$ are measure spaces.

Theorem 2.2.4: $U\in \mathfrak L(L^1(S), L^\infty(T))$ if and only if there exists $x_\cdot\in \mathfrak L^\infty_0(S)[L^\infty(T), L^1(T)]$ such that $$U(f)=\int_S x_s f(s) d\alpha.$$ (Bochner $L^1(T)$-valued integral.) The function $x_\cdot$ is necessarily essentially bounded and essentially unique, and $\|U\|=\mathrm{esssup}_S \|x_\cdot\|$.

Here, for a Banach space $X$ and for $\Gamma\subset X^*$ any closed linear manifold, $\mathfrak L^p_0(S)[X, \Gamma]$ is the space of functions $x_\cdot$ with the following properties (p. 327, B):

1. $x_\cdot\colon S\to X$
2. $s\mapsto f(x_s)$ is measurable for every $f\in \Gamma$
3. $s\mapsto f(x_s)$ is an element of $L^p(S)$ for every $f\in \Gamma$
4. for each $E\in\mathcal E$ with $\alpha E<\infty$ there exists $x_E\in X$ such that $$f(x_E)=\int_E f(x_s)d\alpha, \qquad f\in\Gamma$$

The OP is interested in the case $S=T$. The representation of $x_\cdot$ by a kernel is a consequence of the separability of (the range of) $U$, discussed in Thm. 2.2.6. If $U$ is separable, then the kernel is automatically a measurable function on $\mathcal E\otimes \mathcal F$ (i.e. of the form $p$, rather than $\kappa$, in the op).

Operators in $\mathfrak L(L^1(S), L^1(T))$ may not be represented in this way, but a representation theorem for them is also available (Thm. 2.3.9) only in the case $(T,\beta)=(\mathbb R,\lambda)$. The crucial point is that in this case the representation of $U$ involves differentiating the kernel.

Depending on the level of generality for $S$ and $T$, the proofs of the above statements can be quite involved. The idea is essentially to approximate $U$ by operators with finite-dimensional range, for which the kernel representation is not hard to show. In general, this requires some notion of separability (for either $S$, $T$, $U$, in [1] on a case by case basis).

[1] N. Dunford, B.J. Pettis, Linear Operations on Summable Functions, Trans. AMS, 1940

Answer 2

This is a partial answer as it pertains to positive kernels

First some generalities: Suppose $(S,\mathscr{S})$ to $(T,\mathscr{T})$ are measurable spaces.

• A kernel $\nu$ from $S$ to $K$ is a function $\nu:S\times\mathscr{T}\rightarrow[0,\infty]$ such that for any $B\in \mathscr{T}$, $\nu(\cdot,B):S\rightarrow[0,\infty]$ is $\mathscr{S}$-measurable.
• A kernel from $S$ to $T$ is $\sigma$--finite if there is a partition of $T$, $\{B_n:n\in\mathbb{N}\}\subset\mathscr{T}$ such that $\nu(\cdot,B_n)$ is finite for all $s\in S$.

If $\mu$ and $\nu$ are $\sigma$-finite (positive) kernels from spaces $(S,\mathscr{S})$ to $(T,\mathscr{T})$, and $\mathscr{T}$ is countable generated, the de Possel-Doob's theorem (Kallenberg, O., Random Measures and Applications, Springer 2017, Section 1.5) implies that a function $X:(S\times T,\mathscr{S}\otimes\mathscr{T})\rightarrow[0,\infty]$ such that for all $B\in\mathscr{T}$, $$\nu(s,B)=\int_BX(x,t)\mathbb{1}_{\{X<\infty\}}\mu(s,dt)+\nu(s,B\cap\{X=\infty\})$$ The singular part $\mathbb{1}_{\{X=\infty\}}\cdot\nu$ would be zero iff $\nu(s,B)=0$ whenever $\mu(s,B)$.

1. Suppose $\mu$ is in fact a measure on $(T,\mathscr{T})$
2. $\eta$ is a measure on $(S,\mathscr{S})$ such that $\eta\big(\nu(S)\big)=\int_S\nu(s,S)\,\eta(ds)<\infty$, and $\eta(\nu(B))=0$ whenever $\mu(B)=0$.

Then $\eta(\{X=\infty\})=0$ and so $\nu$ has the desired property.