How can we show this inequality for a product integral? Let


*

*$(E,\mathcal E,\lambda)$ be a measure space;

*$p,q$ be positive probability densities on $(E,\mathcal E,\lambda)$;

*$\mu:=p\lambda$;

*$\sigma:E^2\to[0,\infty)$ be symmetric and $\mathcal E^{\otimes2}$-measurable with $$\int\lambda({\rm d}y)q(y)\sigma(x,y)=1\;\;\;\text{for all }x\in E\tag1.$$
Can we show that there is a $c\ge0$ such that \begin{equation}\begin{split}&\int\mu({\rm d}x)\left|\int\lambda({\rm d}y)\left(\min\left(q(y),\frac{p(y)}{p(x)}q(x)\right)\sigma(x,y)-p(y)\right)(g(y)-g(x))\right|^2\\&\;\;\;\;\;\;\;\;\;\;\;\;\le c\int\mu({\rm d}x)|g(x)|^2\end{split}\tag2\end{equation} for all nonnegative $g\in\mathcal L^2(\mu)$.
Since I don't see a clever way to apply an inequality other than Jensen's inequality, I've tried to first apply Jensen's inequality and then apply the binomial formula and Fubini's theorem: \begin{equation}\begin{split}&\int\mu({\rm d}x)\left|\int\lambda({\rm d}y)\left(\min\left(q(y),\frac{p(y)}{p(x)}q(x)\right)\sigma(x,y)-p(y)\right)(g(y)-g(x))\right|^2\\&\;\;\;\;\le\int\mu({\rm d}x)\int\mu({\rm d}y)|g(y)-g(x)|^2\left|\min\left(\frac{q(y)}{p(y)},\frac{q(x)}{p(x)}\right)\sigma(x,y)-1\right|^2\\&\;\;\;\;=2\int|g|^2\:{\rm d}\mu-2\left|\int g\:{\rm d}\mu\right|^2\\&\;\;\;\;\;\;\;\;+2\int\mu({\rm d}x)|g(x)|^2\int\mu({\rm d}y)\left|\min\left(\frac{q(y)}{p(y)},\frac{q(x)}{p(x)}\right)\right|^2|\sigma(x,y)|^2\\&\;\;\;\;\;\;\;\;-4\int\mu({\rm d}x)|g(x)|^2\int\mu({\rm d}y)\min\left(\frac{q(y)}{p(y)},\frac{q(x)}{p(x)}\right)\sigma(x,y)\\&\;\;\;\;\;\;\;\;-2\int\mu({\rm d}x)g(x)\int\mu({\rm d}y)g(y)\left|\min\left(\frac{q(y)}{p(y)},\frac{q(x)}{p(x)}\right)\right|^2|\sigma(x,y)|^2\\&\;\;\;\;\;\;\;\;+4\int\mu({\rm d}x)g(x)\int\mu({\rm d}y)g(y)\min\left(\frac{q(y)}{p(y)},\frac{q(x)}{p(x)}\right)\sigma(x,y)\end{split}\tag3\end{equation} We may drop some of the subtracted terms (remembering that $g$ is nonnegative), but I don't see which one we can drop at the moment. Moreover, we could use that $2\min(a,b)=a+b-|a-b|$ and write (using Fubini's theorem once again) \begin{equation}\begin{split}&\int\mu({\rm d}x)g(x)\int\mu({\rm d}y)g(y)\min\left(\frac{q(y)}{p(y)},\frac{q(x)}{p(x)}\right)\sigma(x,y)\\&\;\;\;\;=\int\mu({\rm d}x)g(x)\int\lambda({\rm d}y)q(y)\sigma(x,y)g(y)\\&\;\;\;\;\;\;\;\;-\frac12\int\mu({\rm d}x)g(x)\int\mu({\rm d}y)g(y)\left|\frac{q(y)}{p(y)}-\frac{q(x)}{p(x)}\right|\sigma(x,y).\end{split}\tag4\end{equation}
 A: (1) implies $$\int |\sigma(x,y)|q(y)\,d\lambda(y)<\infty.$$
Let (2) now denote the left hand side of inequality (2) above. Then apply Cauchy-Schwarz :
$$(2)\leq \int d\mu(x)\left\lbrace\int |g(y)-g(x)|^2\,d\mu(y)\cdot \int \left(1+\frac{q(y)}{p(y)}|\sigma(x,y)|\right)^2d\mu(y)\right\rbrace.$$ If $q(\cdot)|\sigma(x,\cdot)|\in L^2(\lambda)$ with norm bounded above by a constant $c>0$ for $\mu$-a.e $x\in X$, then 
$$(2)\leq c\int d\mu(x)\left(|g(x)|^2+\int|g(y)|^2\,d\mu(y)\right)\leq 2c\int |g(x)|^2\,d\mu(x).$$ A different application of Cauchy Schwarz gives : 
$$(2)\leq \int d\mu(x)\left\lbrace\int |g(y)-g(x)|^2\left(1+\frac{q(y)}{p(y)}|\sigma(x,y)|\right)\,d\mu(y)\cdot \int \left(1+\frac{q(y)}{p(y)}|\sigma(x,y)|\right)d\mu(y)\right\rbrace.$$ If $q(\cdot)|\sigma(x,\cdot)|\in L^\infty(\lambda)$ for $\mu$-a.e $x\in X$  again such an inequality holds. Under these hypotheses the non-negativity of $g$ or symmetry of $\sigma$ are not needed.
In general, such an inequality does not hold. For simplicity suppose $p=q$. For suitable $g\in L^2(\mu)$, one can get a $\sigma$ such that for a positive $\mu$-measure set of points $x\in X$, $$\int (\sigma(x,y)-1)(g(y)-g(x))\,d\mu(y)=\infty.$$ For example, if $X=\mathbb{R}^2$, $p$ is a radially symmetric distribution, then, for point $x\in B(0,1)$, the unit ball around the origin, one can construct $g$ and $\sigma$ such that both $\sigma(x,\cdot)$ and $g(\cdot)-g(x)$ are in $L^1(\mu)$ but the product is not.
Edit: Easier (counter) examples can be found with $X=\mathbb{Z}$ and $\lambda$ the counting measure. Take $p$ and $q$ to be geometric sequences (suitably normalized). Choose an increasing (in absolute value) sequence $g$ in $L^2(\mu)$. Then $$\int (\sigma(0,y)-1)(g(y)-g(0))\,d\mu(y)=\infty,$$ should be easier to arrange (and this contradicts (2) because the last integral should be finite everywhere). 
