I would like to know whether we are able to reduce the following optimization problem to the pointwise optimization of the integrand (or how we can solve it otherwise): Maximize $$\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)\left(w_i(x)p(x)q_j(y)\wedge w_j(y)p(y)q_i(x)\right)\sigma_{ij}(x,y)|g(x)-g(y)|^2\tag1$$ (as usual, $a\wedge b:=\min(a,b)=\frac{a+b-|a-b|}2$ for $a,b\in\mathbb R$) with respect to the family $(w_i)_{i\in I}$ and subject to the constraints $$\{q_i=0\}\subseteq\{w_ip=0\}\;\;\;\text{for all }i\in I\tag2$$ and $$\{p\ne0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\},\tag3$$ where
- $(E,\mathcal E,\lambda)$ is a measure space;
- $I$ is a finite nonempty set;
- $p,q_i:E\to[0,\infty)$ are $\mathcal E$-measurable with $$\int p\:{\rm d}\lambda=\int q_i\:{\rm d}\lambda=1\tag4$$ for $i\in I$;
- $g\in L^2(p\lambda)$;
- $w_i:E\to[0,1]$ is $\mathcal E$-measurable for $i\in I$;
- $\sigma_{ij}:E^2\to[0,\infty)$ is $\mathcal E^{\otimes2}$-measurable for $i,j\in I$ with $$\sigma_{ij}(x,y)=\sigma_{ji}(y,x)\;\;\;\text{for all }x,y\in E\text{ and }i,j\in I\tag5$$ and $$\sum_{j\in I}\int\lambda({\rm d}y)q_j(y)\sigma_{ij}(x,y)=1\;\;\;\text{for all }x\in E\text{ and }i\in I.\tag6$$
It's clear that increasing the integrand will increase the integral, but my problem is the product form of the integral. I guess maximizing the integrand pointwise yields a maximizing function $f:E\times E\to[0,\infty)$ which is not necessarily the product of two functions $E\to[0,\infty)$ ...
Using $(5)$, we may rewrite $(1)$ as \begin{equation}\begin{split}&\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)w_i(x)p(x)q_j(y)\sigma_{ij}(x,y)|g(x)-g(y)|^2\\&\;\;\;\;-\frac12\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)\left|w_i(x)p(x)q_j(y)-w_j(y)p(y)q_i(x)\right|\sigma_{ij}(x,y)|g(x)-g(y)|^2.\end{split}\tag7\end{equation}
Remark: If the derivation of a closed form of an analytical solution is not possible, I want to solve the problem at least numerically. Moreover, if this is still too hard, can we find a (sharp) lower bound which is easier to maximize?
