Let $E\subset \mathbb R^2$ be compact, convex and connected. For $p_1,\ldots, p_n>0$ with
$$\sum_{i=1}^n p_i=1,$$
and a probability measure $\nu$ supported on $E$ of density $f$, we consider
$$\max_{X\in E^n :~ \mu_X\preceq \nu}~~ \left(\int_E |x|^2 \mu_X(dx)\equiv \sum_{i=1}^n p_i|x_i|^2\right),\quad\quad (\ast)$$
where $\preceq$ denotes the convex order, and for each $X=(x_1,\ldots, x_n)\in E^n$
$$\mu_X(dx):=\sum_{i=1}^n p_i\delta_{x_i}(dx).$$
It is shown that $(\ast)$ admits a maximiser. Can we find a formula, a characterisation or a numerical scheme for the maximiser?
It is straightforward to verify $\Omega:=\{X\in E^n : \mu_X\preceq \nu\}$ is convex and the objective function to maximise is also convex. Therefore, the maximiser must be lying on the set of extremal points of $\Omega$. Intuitively, we need to pick $X\in \Omega$ such that $|x_i|$ is as large as possible. On the other hand, we may believe that
$$\Omega\cap \partial E^n=\emptyset. \quad (\mbox{How to prove it?})$$
For each $\epsilon>0$, define
$$E^n_\epsilon:=\{X\in E^n: d(X,\partial E^n):= \inf_{Y\in \partial E^n}|X-Y| \ge \epsilon\}.$$
Can we find/estimate some $\epsilon^*$ such that $\Omega\subset E^n_{\epsilon^*}$? If so, does this help to guess the $X\in \Omega$ "as large as possible"?
PS : We say that two probability measures $\mu,\nu$ satisfy $\mu\preceq \nu$ if for any convex function $h:\mathbb R^2\to\mathbb R$ it holds
$$\int h d\mu \le \int h d\nu.$$
