I'm trying to solve a saddle point problem of the following form: Let

 - $(E,\mathcal E,\lambda)$ be a measure space;
 - $p$ be a probability density on $(E,\mathcal E,\lambda)$, $\mu:=p\lambda$ and $$\iota:\mathcal L^1(\lambda)\to\mathcal L^1(\mu)\;,\;\;\;g\mapsto\left(E\ni x\mapsto\begin{cases}\displaystyle\frac{g(x)}{p(x)}&\text{, if }p(x)>0\\0&\text{, otherwise}\end{cases}\right);$$
 - $W$ be a closed, convex subspace of a $\mathbb R$-Hilbert space $H$ with empty interior and $\left\|w\right\|_H\le1$ for all $w\in W$;
 - $\kappa_w$ be a [Markov kernel](https://en.wikipedia.org/wiki/Markov_kernel#Formal_definition) on $(E,\mathcal E)$ with density$^1$ $k_w$ with respect to $\lambda$ for $w\in W$.

Moreover, let $$\mathcal L^2:=\left\{g:E\to[0,\infty)\mid g\text{ is }\mathcal E\text{-measurable with }\{p=0\}\subseteq\{g=0\}\text{ and }\int_{\{\:p\:>\:0\:\}}\frac{g^2}p\:{\rm d}\lambda<\infty\right\}$$ and $$K:=\left\{\iota g:g\in\mathcal L^2\text{ and }\int g\:{\rm d}\lambda=0\right\}.$$

Assume $\mu$ is reversible$^2$ with respect to $\kappa_w$ for all $w\in W$.

> I want to choose a $w\in W$ minimizing (or at least reducing as much as possible) the quantity $$\sup_{g\in K}L(w,g)\tag1,$$ where $$L(w,g):=\int\mu({\rm d}x)\int\lambda({\rm d}y)k_w(x,y)|g(x)-g(y)|^2\;\;\;\text{for }w\in W\text{ and }g\in L^2(\mu).$$

> **Question**: How can we deal with this problem? Which assumption on the dependence of $k_w$ on $w\in W$ (e.g. Fréchet differentiability) do we need to impose? And if it's too hard to search for a true minimizer, can we find a "nearly optimal" solution (maybe in terms of a minimizer of an upper bound)?

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$^1$ i.e. $k_w:E^2\to[0,\infty)$ is $\mathcal E^{\otimes2}$-measurable and $$\kappa_w(x,B)=\int\lambda({\rm d}y)k_w(x,y)\;\;\;\text{for all }(x,B)\in E\times\mathcal E.$$

$^2$ i.e. $$\int\mu({\rm d}x)\int\kappa_w(x,{\rm d}y)f(x,y)=\int\mu({\rm d}y)\kappa_w(y,{\rm d}x)f(x,y)$$ for all bounded $\mathcal E^{\otimes2}$-measurable $f:E\times E\to\mathbb R$.