# Can we reduce the maximization of this integral to the maximization of the integrand?

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 $$\Phi_g(w):=\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)\left(w_i(x)p(x)q_j(y)\wedge w_j(y)p(y)q_i(x)\right)\sigma_{ij}(x,y)|g(x)-g(y)|^2\tag1$$ (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 $$$$\begin{split}&\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)w_i(x)p(x)q_j(y)\sigma_{ij}(x,y)|g(x)-g(y)|^2\\&\;\;\;\;-\frac12\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)\left|w_i(x)p(x)q_j(y)-w_j(y)p(y)q_i(x)\right|\sigma_{ij}(x,y)|g(x)-g(y)|^2.\end{split}\tag7$$$$

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?

EDIT: Let's elaborate on the nonsmooth Lagrange multiplier rule given by Clarke in Theorem 10.47. 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, $$\partial_wf((x,y),w_1)=\left.\begin{cases}\{\delta_x\}&\text{, if }w_1(x)w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\end{cases}\right\}p(x)q_1(y)\sigma_{11}(x,y)|g(x)-g(y)|^2$$ 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 ...