# Tools for “infinite-dimensional linear programming”

I was wondering, whether you could point me to some tools with which I could tackle the following "infinite-dimensional linear programming" problem:

Notation:

$a=1,2,\ldots, A$, $x\in\Omega:=\left\{x\in\mathbb R^n|\sum_{i=1}^nx_i=1, x_i\geq 0, \forall i\right\}$

$r_a:\Omega\rightarrow [r_-,r_+]\subset\mathbb R$, $p_a:\Omega\times\Omega\rightarrow \mathbb R^+, \int_\Omega dy\ p_a(x,y)=1,\forall x,a$

$\pi_a:\Omega\rightarrow \mathbb R^+$

The Problem:

Given $r=(r_1,\ldots,r_A)$ and $p=(p_1,\ldots,p_A)$.

Let $\pi=(\pi_1,\ldots,\pi_A)$ and define

$\Pi(r,p)=\left\{\pi|\int_\Omega dx\sum_a\pi_a(x)=1 \land\int_\Omega dx\sum_a\left(\pi_a(y)- p_a(x,y)\pi_a(x)\right)=0,\forall y \right\}.$

Find $\pi^*$ such that

$\pi^*=\arg\max_{\pi\in\Pi(r,p)}\int_\Omega dx\sum_a r_a(x)\pi_a(x)$

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Something is fishy: the volume of the simplex in question is well below $1$, so the integral operators are very strongly contracting, reducing the domain to a bunch of identically $0$ functions. Are you sure you wrote what you wanted to write? In any case, if you replace the measure, the operators are compact, so $\sum_a\pi_a$ is an element of a finite-dimensional space of functions. So the first step is to figure out what that subspace is. –  fedja Jun 1 '13 at 0:03
Thanks for your comment. There was indeed something wrong: the range of $p_a$ and $\pi_a$ (now corrected). Furthermore I forgot to mention a constraint on $p_a$ (added). This may be trivial, but I don't see why the compactness implies that $\sum_a\pi_a$ is an element of a finite-dimensional space of functions. –  bfrank Jun 2 '13 at 12:50