Let
- $(E,\mathcal E,\lambda)$ be a measure space;
- $p$ be a probability density on $(E,\mathcal E,\lambda)$ and $\mu:=p\lambda$
- $\kappa$ be a Markov kernel on $(E,\mathcal E)$ symmetric with respect to $\mu$ and $$(\kappa f)(x):=\int\kappa(x,{\rm d}y)f(x,y)$$ for $\mathcal E^{\otimes2}$-measurable $f:E^2\to\mathbb R$ with $(\kappa|f|)(x)<\infty$ for all $x\in E.$
Now, let $$\iota g:=\left(E^2\ni(x,y)\mapsto g(x)-g(y)\right)\;\;\;\text{for }g:E\to\mathbb R.$$
Are we able to show that $$L(g):=\int\mu({\rm d}x)\int\kappa(x,{\rm d}y)|(\iota g)(x,y)|^2\;\;\;\text{for }g\in L^2(\mu)$$ is continuous?
In particular, I'd like to show that $$\sup_{g\in L^2(\mu)}L(g)=\sup_{\substack{g\in L^2(\mu)\\g\text{ is bounded}}}L(g)\tag1.$$ The latter should follow from the continuity, since any $g$ may be approximated by $1_{\{\:|g|\:\le\:n\:\}}g$.
EDIT: Note that, by a Fubini-like theorem, If $f:E^2\to\mathbb R$ is $\mathcal E$-measurable and either $f\ge0$ or $f\in\mathcal L^1(\mu\otimes\kappa)$, then $$\int\kappa f\:{\rm d}\mu=\int f\:{\rm d}(\mu\otimes\kappa)\tag2.$$ Moreover, if $p\ge1$ and $g\in\mathcal L^p(\mu)$, then $$\int|\iota g|^p\:{\rm d}(\mu\otimes\kappa)\le 2^p\int|g|^p\:{\rm d}\mu<\infty\tag3$$ by Jensen's inequality and hence $\iota g\in\mathcal L^p(\mu\otimes\kappa)$.
For any finite measure $\nu$, $\nu f:=\int f\:{\rm d}\nu$ is a bounded linear functional on $L^p(\nu)$ for all $p\ge1$. In particular, $\mu\otimes\kappa$ is a bounded linear functional on $L^p(\mu\otimes\kappa)$ for all $p\ge1$. By $(3)$, $\iota$ is a bounded linear operator from $L^p(\mu)$ to $L^p(\mu\otimes\kappa)$ for all $p\ge1$. Thus, by $(2)$, $$L^p(\mu)\ni g\mapsto\int\mu({\rm d}x)\int\kappa(x,{\rm d}y)|(\iota g)(x,y)|^p=\left\|\iota g\right\|_{L^p(\mu\otimes\kappa)}^p$$ is bounded (and hence continuous).