If $\Omega$ is a countable set, every linear operator mapping $\mathbb R^\Omega$ to itself can be represented as a matrix $M\colon \Omega\times\Omega\to\mathbb{R}$.
In the case $\Omega$ is uncountable, and if we settle on a particular $\sigma$-algebra $\Sigma$ on $\Omega$, I am wondering if some nice linear operators on $\Omega$ can be represented as a "matrix" $M\colon\Omega\times\Sigma\to\mathbb{R}$.
In particular, consider a measurable space $(\Omega,\Sigma)$, where $\Sigma\subseteq \{0,1\}^\Omega$ is a $\sigma$-algebra for $\Omega$. A (sub-)Markov kernel is a map $L\colon \Omega\times\Sigma\to[0,1]$ such that for each $x\in \Omega$, we have
$$ \int L(x,\mathrm{d}y) \le 1. $$
Now consider a measure $\mu$ on $(\Omega,\Sigma)$. The function space $\mathcal H= \mathcal L_2(\Omega,\mu)$ endowed with the natural inner product \begin{align*} \langle f,g\rangle=\int_\Omega fg\,\mathrm{d}\mu \end{align*} is a Hilbert space, therefore we may speak of linear operators $A$ mapping a subspace $D_A$ of $\mathcal H$ to $\mathcal H$. I am wondering if for every linear operator $A\colon D_A\to\mathcal H$ of which $\mathcal L_1\to \mathcal L_1$ norm is bounded, there are reals $c_1,c_2\in\mathbb{R}$ and Markov kernels $L_1,L_2$ such that $A=c_1L_1 + c_2L_2$. This latter equality is understood as the following: for all $f\in D_A$, we have $fA = c_1 fL_1 + c_2fL_2$
Here we need to allow two Markov kernels because $A$ may not be positivity preserving. Thanks
A sub-Markov kernel, or any linear combination thereof, has the property that it maps bounded functions to bounded functions. So any bounded operator without this property will be a counterexample.
For a specific example, take $\Omega = [0,1]$, $\mu$ Lebesgue measure, and let $h : [0,1] \to \mathbb{R}$ be an unbounded function in $\mathcal{L}^1 \cap \mathcal{L}^2$. For instance, $h(x) = x^{-1/3}$ would do. Then choose a functional which is continuous in both $\mathcal{L}^1$ and $\mathcal{L}^2$ norms, such as for instance the integral, and multiply by $h$. So we could consider $$Af(x) = \int_0^1 f(t)\,dt \cdot h(x).$$ This is a bounded operator with respect to either the $\mathcal{L}^1$ or $\mathcal{L}^2$ norms. If it were of the form $A = c_1 L_1 + c_2 L_2$ then we would have, for instance, $|A1(x)| \le |c_1| + |c_2|$ for (almost?) every $x$. But $A1=h$ which does not have this property.