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Let

  • $(E,\mathcal E,\lambda)$ be a measure space;
  • $f:E\to[0,\infty)^3$ be $\mathcal E$-measurable with $\|f\|\in\mathcal L^2(\lambda)$;
  • $\tilde p:=\alpha_1f_1+\alpha_2f_2+\alpha_3f_3$ for some $\alpha_1,\alpha_2,\alpha_3\ge0$ with $\alpha_1+\alpha_2+\alpha_3=1$, $$b:=\int\tilde p\:{\rm d}\lambda\in(0,\infty)$$ and $p:=b^{-1}\tilde p$;
  • $\mu:=p\lambda$ and $$W:=\left\{w:E\to[0,\infty)^I:w\text{ is }\mathcal E\text{-measurable and }\{p>0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}\right\};$$
  • $I$ be a finite nonempty set
  • $q_i,r$ be positive probability densities on $(E,\mathcal E,\lambda)$ for $i\in I$
  • $J$ be a finite nonempty set and $h_j:E\to[0,\infty)$ be $\mathcal E$-measurable for $j\in J$ with $$c:=\sup_{j\in J}\left\|h_j\right\|_\infty<\infty;$$
  • $(E',\mathcal E',\lambda')$ be a measure space;
  • $\varphi_i:E'\to E$ be bijective and $(\mathcal E',\mathcal E)$-measurable with \begin{equation}\lambda'\circ\varphi_i^{-1}=q_i\lambda'\end{equation} and $$a^{(g)}_i:=\int\lambda({\rm d}y)\frac{|g(y)|^2}{q_i(y)r\left(\varphi_i^{-1}(y)\right)}\;\;\;\text{for }g\in\mathcal L^2(\lambda)$$ for $i\in I$.

How can we solve the saddle-point problem $$\inf_{w\in W}\sup_{\substack{j\in J\\k\in\{1,\:2,\:3\}}}\sum_{i\in I}a^{(h_jf_k)}_i\int w_i\:{\rm d}\mu\tag1$$ (at least numerically)?

Note that for fixed $g\in\mathcal L^2(\lambda)$, a minimizer $w\in W$ of $\sum_{i\in I}a^{(g)}_i\int w_i\:{\rm d}\mu$ is given by $$w^{(g)}_i\equiv\delta_{ij^{(g)}}\;\;\;i\in I,$$ where $\delta$ denotes the Kronecker delta and $$j^{(g)}:=\min\underset{i\in I}{\operatorname{arg\:min}}\:a^{(g)}_i$$ ($\operatorname{arg\:min}$ is treated as being set-valued and we break ties by selecting the smallest index).

We may clearly bound $(1)$ above by $$c\inf_{w\in W}\sum_{i\in I}a^{(\left\|f\right\|)}_i\int w_i\:{\rm d}\mu\tag2$$ which is minimized by $w^{(\left\|f\right\|)}$ as described before, but is this the best we can do?

BTW, note that each $w\in W$ satisfies $\sum_{i\in I}\int w_i\:{\rm d}\mu=1$.

EDIT: Don't know if it is helpful, but in my application, $E=M^k$, where $k\in\mathbb N$, $M\subseteq\mathbb R^3$ is a 2-dimensional manifold sufficiently regular to admit a well-defined surface measure $\sigma_M$, $\mathcal E={\mathcal B(M)}^{\otimes k}$ and $\lambda=\sigma_M^{\otimes k}$. Moreover, $E'=[0,1)^d$ for some $d\in\mathbb N$, $\mathcal E'={\mathcal B([0,1))}^{\otimes d}$ and $\lambda'$ is the restriction of the $d$-dimensional Lebesgue measure to $\mathcal E'$.

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  • $\begingroup$ What are the sets $I$, $J$ a subset of? $\endgroup$ – Drew Brady Dec 31 '19 at 7:22
  • $\begingroup$ @DrewBrady They are just some finite nonempty subsets. So, you can assume $I=\{1,\ldots,k\}$, $J=\{1,\ldots,l\}$ for some $k,l\in\mathbb N$. $\endgroup$ – 0xbadf00d Dec 31 '19 at 7:33
  • $\begingroup$ And presumably $l \leq k$. $\endgroup$ – Drew Brady Dec 31 '19 at 7:35
  • $\begingroup$ @DrewBrady No, why do you expect that? There's no relation between $I$ and $J$. The whole point is just that there's a finite number ($|J|\cdot3$) over which the supremum in $(3)$ is taken. $\endgroup$ – 0xbadf00d Dec 31 '19 at 8:40
  • $\begingroup$ Look at your equation (1) and then explain to me why under these identifications you can possibly not have $l \leq k$. $\endgroup$ – Drew Brady Jan 2 '20 at 6:21

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