I want to maximize $$\Phi_g(w):=\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)\left(w_i(x)p(x)q_j(y)\wedge w_j(y)p(y)q_i(x)\right)\sigma_{ij}(x,y)|g(x)-g(y)|^2$$ over all choices of $w=\left(w_1,\ldots,w_{|I|}\right)$ subject to $$\sum_{i\in I}w_i=1\tag1.$$ I guess this is usually solved by the method of Lagrange multipliers, but the shape of the integrand seems to be problematic. What can we do?

If this problem is too hard, are we at least able to find a choice of $w$ yielding a sharp lower bound?

As usual, $a\wedge b:=\min(a,b)$. And it might be useful to rewrite $\Phi_g(w)$ using $2(a\wedge b)=a+b-|a-b|$.

The objects are defined as follows:

• $(E,\mathcal E,\lambda)$ is a measure space
• $I$ is a finite nonempty set
• $p,q_i:E\to[0,\infty)$ are $\mathcal E$-measurable with $$\int p\:{\rm d}\lambda=\int q_i\:{\rm d}\lambda=1\tag2$$ for $i\in I$
• $g\in L^2(p\lambda)$
• $w_i:E\to[0,1]$ is $\mathcal E$-measurable with $$\{q_i=0\}\subseteq\{w_ip=0\}\tag3$$ for $i\in I$ with $$\{p\ne0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}\tag4$$
• $\sigma_{ij}:E^2\to[0,\infty)$ 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 $$\sum_{j\in I}\int\lambda({\rm d}y)q_j(y)\sigma_{ij}(x,y)=1\tag6$$

I'm primarily interested in finding the choice of $w=\left(w_1,\ldots,w_{|I|}\right)$ maximizing $\Phi_g(w)$ and satisfying $(3)$ and $(4)$, but if it's easier to deal with feel free to assume $(1)$ instead of $(4)$.

EDIT: Let's elaborate on the hint given by dchatter. Let $$f:E^2\times{L^2(\lambda)}^I\to\mathbb R\;,\;\;\;((x,y),w)\sum_{i\in I}\sum_{j\in I}\left(w_i(x)p(x)q_j(y)\wedge w_j(y)p(y)q_i(x)\right)\sigma_{ij}(x,y)|g(x)-g(y)|^2.$$ To make everything as simple as possible, assume $I=\{1\}$ (we ignore that $(1)$ immediately implies that necessarily $w_1=1$). Then, as discussed here, $$\partial_wf((x,y),w_1)=\left.\begin{cases}\{\delta_x\}&\text{, if }w_1(x)<w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\\\left\{c\delta_x+(1-c)\frac{p(y)q_1(x)}{p(x)q_1(y)}\delta_y:c\in[0,1]\right\}&\text{, if }w_1(x)=w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\\\left\{\frac{p(y)q_1(x)}{p(x)q_1(y)}\delta_y\right\}&\text{, if }w_1(x)>w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\end{cases}\right\}p(x)q_1(y)\sigma_{11}(x,y)|g(x)-g(y)|^2$$ for all $x,y\in E$ and $w_1\in L^2(\lambda)$, where $\delta_x$ denotes the evaluation functional on $\mathcal L^2(\lambda)$. In the paper of Clarke (Theorem 1 of Section 3), the author shows that $$\partial\Phi_g(w_1)\subseteq\int\lambda^{\otimes2}({\rm d}(x,y))\partial_wf((x,y),w_1)$$ (all derivatives have to be understood in the sense of Clarke's generalized gradient). That means, that for all $\varphi\in\partial F(w_1)$, there is a mapping $\Phi:E^2\to\partial_wf((x,y),w_1)\subseteq{L^2(\lambda)}'$ such that $(x,y)\mapsto\langle\Phi(x,y),v\rangle$ belongs to $L^1(\lambda^{\otimes2})$ and $$\langle\varphi,v\rangle=\int\lambda^{\otimes2}({\rm d}(x,y))\langle\Phi(x,y),v\rangle$$ for all $v\in L^2(\lambda)$. But I don't know how to proceed ...