# Can we find a minimizer of $\sum_{i=1}^k\int a_iw_i$ satisfying $\sum_{i=1}^kw_i=1$?

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 $$\mathcal E'$$.

• What are the sets $I$, $J$ a subset of? – Drew Brady Dec 31 '19 at 7:22
• @DrewBrady They are just some finite nonempty subsets. So, you can assume $I=\{1,\ldots,k\}$, $J=\{1,\ldots,l\}$ for some $k,l\in\mathbb N$. – 0xbadf00d Dec 31 '19 at 7:33
• And presumably $l \leq k$. – Drew Brady Dec 31 '19 at 7:35
• @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. – 0xbadf00d Dec 31 '19 at 8:40
• Look at your equation (1) and then explain to me why under these identifications you can possibly not have $l \leq k$. – Drew Brady Jan 2 '20 at 6:21