Let $(E,\mathcal E,\lambda)$, $w:E\to[0,1]$ be $\mathcal E$-measurable, $p,q:E\to[0,\infty)$ be probability densities on $(E,\mathcal E,\lambda)$, $\sigma:E^2\to[0,\infty)$ be symmetric and $\mathcal E^{\otimes2}$-measurable, $f\in L^2(\lambda)$, $c:=\int f\:{\rm d}\lambda$ and $\Lambda\in L^2(p\lambda)$. Can we solve for $$\frac{w(x)p(x)}{q(x)}\left|\frac{f(x)}{p(x)}-c\right|^2\int\lambda({\rm d}y)\underbrace{\frac{w(y)p(y)}{\sigma(x,y)}}_{=:\:g(x,\:y)}=-\frac12\left(\int\lambda({\rm d}y)\frac{|w(y)p(y)|^2}{q(y)\sigma(x,y)}\left|\frac{f(y)}{p(y)}-c\right|^2+\Lambda(x)\right),\tag1$$ for all $x\in E$, for $w$?
Maybe it's useful to write $$\frac{w(x)p(x)}{q(x)}\left|\frac{f(x)}{p(x)}-c\right|^2\left\|g(x,\;\cdot\;)\right\|_{L^2(\lambda)}^2=-\frac12\left(\left\langle\frac{wp}q\left|\frac fp-c\right|^2,g(x,\;\cdot\;)\right\rangle_{L^2(\lambda)}+\Lambda(x)\right),\tag2$$ for all $x\in E$.