Let $p,q$ be probability densities on a measure space $(E,\mathcal E,\lambda)$, $\mu:=p\lambda$, $\sigma:E^2\to[0,\infty)$ be $\mathcal E^{\otimes2}$-measurable with $$\int\lambda({\rm d}y)q(y)\sigma(x,y)=1\;\;\;\text{for all }x\in E\tag1$$ and $$(\kappa g)(x):=g(x)+\int_{\left\{\:p\:>\:0\:\right\}}\mu({\rm d}y)(g(y)-g(x))\min\left(\frac{q(y)}{p(y)}\sigma(x,y),\frac{q(x)}{p(x)}\sigma(y,x)\right)$$ for $x\in E$ and $g\in\mathcal L^1(\mu)$.
Let $g,h\in\mathcal L^2(\mu)$ be nonnegative. I want to show that $$\langle\kappa g,h\rangle_{L^2(\mu)}\le\langle g,h\rangle_{L^2(\mu)}.\tag2$$
I actually know that $\kappa$ is a contraction (operator norm $\le1$) on $L^p(\mu)$, for all $p\in[1,\infty]$, and self-adjoint on $L^2(\mu)$. So, $$\sup_{\substack{g,\:h\:\in\:L^2(\mu)\\\left\|g\right\|_{L^2(\mu)},\:\left\|h\right\|_{L^2(\mu)}\:\le\:1}}\left|\langle\kappa g,h\rangle_{L^2(\mu)}\right|=\sup_{g\:\in\:L^2(\mu)\setminus\left\{0\right\}}\frac{\left|\langle\kappa g,g\rangle_{L^2(\mu)}\right|}{\left\|g\right\|_{L^2(\mu)}^2}=\left\|\kappa\right\|_{\mathfrak L^2(\mu)}\le1\tag3.$$
However, any inequality (for example, $\min(a,b)\le\sqrt{ab}$ for all $a,b\ge0$) which comes to my mind seems not sharp enough for $(2)$. So, what can we do?
EDIT: Note that \begin{equation}\begin{split}&\langle\kappa g,h\rangle_{L^2(\mu)}=\langle g,h\rangle_{L^2(\mu)}+\\&\;\;\;\;\int\lambda({\rm d}x)g(x)\left[\int\lambda({\rm d}y)h(y)\min(p(x)q(y)\sigma(x,y),p(y)q(x)\sigma(y,x))-h(x)\int\lambda({\rm d}y)\min(p(x)q(y)\sigma(x,y),p(y)q(x)\sigma(y,x)).\right]\end{split}\tag4\end{equation} So, the second term must be $\le0$ and hence (since $g\ge0$) we need to have \begin{equation}\begin{split}&\int\lambda({\rm d}y)h(y)\min(p(x)q(y)\sigma(x,y),p(y)q(x)\sigma(y,x))\\&\;\;\;\;\;\;\;\;\;\;\;\;\le h(x)\int\lambda({\rm d}y)\min(p(x)q(y)\sigma(x,y),p(y)q(x)\sigma(y,x))\end{split}\tag5\end{equation} for $\lambda$-almost all $x\in E$.
EDIT2: By self-adjointness and the Cauchy-Schwarz inequality, \begin{equation}\begin{split}\langle\kappa g,h\rangle_{L^2(\mu)}&=\frac{\langle\kappa(g+h),g+h\rangle_{L^2(\mu)}-\langle\kappa(g-h),g-h\rangle_{L^2(\mu)}}4\\&\le\frac{\left\|g+h\right\|_{L^2(\mu)}^2-\langle\kappa(g-h),g-h\rangle_{L^2(\mu)}}4.\end{split}\tag65\end{equation} On the other hand, $$\frac{\left\|g+h\right\|_{L^2(\mu)}^2-\left\|g-h\right\|_{L^2(\mu)}^2}4=\langle g,h\rangle_{L^2(\mu)}.\tag7$$ So, the desired inequality holds iff $$\langle\kappa(g-h),g-h\rangle_{L^2(\mu)}\ge\left\|g-h\right\|_{L^2(\mu)}^2,\tag8$$ which seems to be implausible (unless $g=h$).