Let
- $(E,\mathcal E,\lambda)$ be a measure space
- $I$ be a finite set
- $p,q_i,w_i:E\to[0,\infty)$ be $\mathcal E$-measurable for $i\in I$
- $f_{i,\:j}:E^2:\to[0,\infty)$ be $\mathcal E$-measurable for $i,j\in I$
Assume $$\Phi\left(w_1,\ldots,w_{|I|}\right):=\int\lambda^{\otimes2}\left({\rm d}(x,y)\right)\sum_{i,\:j\:\in\:I}\left(p(x)w_i(x)q_j(y)\wedge p(y)w_j(y)q_i(x)\right)f_{i,\:j}(x,y)$$ is well-defined ($a\wedge b:=\min(a,b)$ as usual).
How can we find the maximum of $\Phi$ over all choices of $w_1,\ldots,w_{|I|}$?
Denoting the integrand of $\Phi\left(w_1,\ldots,w_{|I|}\right)$ at fixed $x,y\in E$ by $\varphi_{x,\:y}\left(w_1,\ldots,w_{|I|}\right)$, we see that \begin{equation} \begin{split} &\lim_{t\to0}\frac{\varphi_{x,\:y}\left(w_1,\ldots,w_{i_0}+th,\ldots,w_{|I|}\right)-\varphi_{x,\:y}\left(w_1,\ldots,w_{|I|}\right)}t=\\ &\;\;\;\;\sum_{i\in I\setminus\{i_0\}}\left.\begin{cases}0&\text{, if }p(x)w_i(x)q_{i_0}(y)<p(y)q_i(x)w_{i_0}(y)\\p(y)q_i(x)h(y)&\text{, if }p(x)w_i(x)q_{i_0}(y)>p(y)q_i(x)w_{i_0}(y)\end{cases}\right\}f_{i,\:i_0}(x,y)\\ &\;\;\;\;+\sum_{j\in I\setminus\{i_0\}}\left.\begin{cases}p(x)q_j(y)h(x)&\text{, if }p(x)q_j(y)w_{i_0}(x)<p(y)w_j(y)q_{i_0}(x)\\0&\text{, if }p(x)q_j(y)w_{i_0}(x)>p(y)w_j(y)q_{i_0}(x)\end{cases}\right\}f_{i_0,\:j}(x,y)\\ &\;\;\;\;+\left.\begin{cases}h(x)&\text{, if }w_{i_0}(x)<w_{i_0}(y)\\h(y)&\text{, if }w_{i_0}(x)>w_{i_0}(y)\end{cases}\right\}f_{i_0,\:i_0}(x,y) \end{split}\tag1 \end{equation} for all $\mathcal E$-measurable $h:E\to[0,\infty)$ and $i_0\in I$. However, this expression is quite complicated. We would need to use differentiation under the integral sign first and then set the gradient of all these derivatives to $0$ in order to find the maximum.
Remark: To give some context, this is a first step consideration in my attempt to solve this problem: https://mathoverflow.net/q/338619/91890.