Let
- $(E,\mathcal E,\lambda)$ and $(E',\mathcal E',\lambda')$ be measure spaces
- $f\in L^2(\lambda)$
- $I$ be a finite nonempty set
- $\varphi_i:E'\to E$ be bijective $(\mathcal E',\mathcal E)$-measurable with $$\lambda'\circ\varphi_i^{-1}=q_i\lambda\tag1$$ for $i\in I$
- $p,q_i:E\to[0,\infty)$ be $\mathcal E$-measurable with $$\int p\:{\rm d}\lambda=\int q_i\:{\rm d}\lambda=1$$ for $i\in I$
- $\mu:=p\lambda$
- $w_i:E\to[0,1]$ be $\mathcal E$-measurable
- $\tau':(I\times E')^2\to[0,\infty)$ be $(2^I\otimes\mathcal E')^{\otimes2}$-measurable with $$\sum_{j\in I}\int\lambda'({\rm d}y')\tau'((i,x'),(j,y'))=1\tag2$$ for all $(i,x')\in I\times E'$
- $$\tau_{i,\:j}(x,y):=w_i(x)q_j(y)\tau'\left(\left(i,\varphi_i^{-1}(x)\right),\left(j,\varphi_j^{-1}(y)\right)\right)$$ for $x,y\in E$ and $i,j\in I$
- $$\alpha_{i,\:j}(x,y):=\begin{cases}\displaystyle\min\left(1,\frac{p(y)w_j(y)q_i(x)}{p(x)w_i(x)q_j(y)}\right)&\text{, if }p(x)w_i(x)q_j(y)>0\\1&\text{, otherwise}\end{cases}$$ for $x,y\in E$ and $i,j\in I$
- $$k(x,y):=\sum_{i,\:j\:\in\:I}\alpha_{i,\:j}(x,y)\tau_{i,\:j}(x,y)$$ for $x,y\in E$
Assume $$\{q_i=0\}\subseteq\{w_ip=0\}\;\;\;\text{for all }i\in I,\tag3$$ $$\{p=0\}\subseteq\{f=0\}\tag4$$ and $$\{pf\ne0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}.\tag5$$
I want to determine which choice of $(w_i)_{i\in I}$ is maximizing the quantity $$\int\mu({\rm d}x)\int\lambda({\rm d}y)k(x,y)|f(x)-f(y)|^2\tag6$$ subject to the constraints $(3)$ and $(5)$. How can we do that?
Note that, by definition, $$p(x)k(x,y)=\sum_{i,\:j\:\in\:I}\min(p(x)w_i(x)q_j(y),p(y)w_j(y)q_i(x))\tau'\left(\left(i,\varphi_i^{-1}(x)\right),\left(j,\varphi_j^{-1}(y)\right)\right)\tag7$$ for all $x,y\in E$.
EDIT: Maybe we can rewrite the quantity using that $2\min(a,b)=a+b-|a-b|$ for all $a,b\in\mathbb R$ as I've done it for this question: https://stats.stackexchange.com/q/424032/222528.
