> 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](https://mathoverflow.net/a/341653/91890). 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](https://mathoverflow.net/a/341920/91890), $$\partial_wf((x,y),w_1)=\left.\begin{cases}\{\delta_x\}&\text{, if }w_1(x)<w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\\\left\{c\delta_x+(1-c)\frac{p(y)q_1(x)}{p(x)q_1(y)}\delta_y:c\in[0,1]\right\}&\text{, if }w_1(x)=w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\\\left\{\frac{p(y)q_1(x)}{p(x)q_1(y)}\delta_y\right\}&\text{, if }w_1(x)>w_1(y)\frac{p(y)q_1(x)}{p(x)q_1(y)}\end{cases}\right\}p(x)q_1(y)\sigma_{11}(x,y)|g(x)-g(y)|^2$$ 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](https://pdf.sciencedirectassets.com/272585/1-s2.0-S0001870800X02502/1-s2.0-0001870881900323/main.pdf?X-Amz-Security-Token=AgoJb3JpZ2luX2VjEMD%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJIMEYCIQDncC3kz2PE7TC8KlgxOrKGXZboogmvjd%2FWr3oL3dR8TgIhAO8c6uJpTlXb9BVyFg091nHZAVaJXBNMXIiwuBN1mXXyKuMDCIn%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEQAhoMMDU5MDAzNTQ2ODY1Igx0g%2BgnzEdovUHHPU8qtwPLOK0uSYKkcP7Epkqzua2wNo0VVHHF379AxjWXRGpFY2cM081ElPcXHr1WZQtCbSIGpdbavxbfPcTqlJuYSevJkksJT%2FlTmwcNxrSw9bpWhjw2DhX%2B3fqk1GRnZ38Lsvfcqhrlt2rNspAWIQe%2F93eD4Sii8P%2BZFbMjgkDuvISlsym0wzORJ%2B%2FTxeH5w4mdIq4gMQ1Vbgw6LV%2FBcxx5sydeS9Vyd3PFh5ZMl5H5uH71ZBRO2gLXBK2RHLGHk5oipEf%2B5EPIrNfnu2eYRQnFFmxrrD5gqAO1ZVhUh%2BrDgpHjNV6XqvSRHFOH7F6LqVJv%2FOlW5UhnrXfewMQVprtXQ8054GHH3WEgx%2FqWQML%2F957vcYRFPrVCrm8VTb7%2B3AtsY8rOcN%2F00%2F11Zg6wd1LYJR8iG714rurZzbEWqOlLEzoJaPNTLP1vwRBkb%2B91%2FLVaT9CjSqULYvMaTldbq%2FMqney8QmpTo6mUQeqYwD2b%2BNk8GXYpIPeKBLy2Mrtnub8BuznaLqv%2FVSBUtdUuqFAPjIc90dsf15KvkFpuriZN4IZb3OJh%2B5Fi%2Bgw29Oa9WF%2BA8K3OfD%2FTEhj8MOG6h%2BwFOrMB1oE6S0%2FJS%2BJ%2BPfsR2be%2F8gKCc4EDj8IVnr%2BZuSOMvbIhoxhywZqpXDvGS8MNgJv4axqZ2OL2CSBcjs7SfOazxHzmcRBqL9veYQi5e0sUL0F6X%2BoD%2FRukrdisUZMuNXF%2FDDE%2BMIcuefsmOHVTd0DR1hUuZTSZi%2B5Oqj%2BSTqhE6YrwX6Ewtx%2BgV1dJMJf91cQlT7wEZU%2FKvVYrOrDE6d4IkUniYnNnIW11n5%2B4metofUYDjhY%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20190918T084923Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYZ5D4I5AJ%2F20190918%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=3794a3cd286dac45255e2a59e02bd53c8def7f8f6c193d95480e0f6d7be8fb67&hash=33ba1fda1dfb5fb50522e5f436a6cd1f37852dec06e1854e80f94e5019494ea5&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=0001870881900323&tid=spdf-d0e3327f-85c2-421d-a45c-8e6f4ab0cf5d&sid=b376c27b7e56944aa038dc543f341a6f5a61gxrqa&type=client) (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 ...