Fall 2016 - Problem 3

Solution.

1. Since $\phi$ is lower-semicontinuous, there exists a sequence $\set{f_n}_n$ of continuous functions on $X$ which increase pointwise to $\phi$, and so we have

   $$\int f_k \,\diff\mu_n \leq \int \phi \,\diff\mu_n \implies \liminf_{n\to\infty} \int f_k \,\diff\mu_n \leq \liminf_{n\to\infty} \int \phi \,\diff\mu_n.$$

   for all $n, k \in \N$. By Riesz representation, $\p{C\p{X}}^*$ is the space of positive Borel measures on $X$. Hence, weak-* convergence of $\set{\mu_n}_n$ means

   $$\liminf_{n\to\infty} \int f_k \,\diff\mu_n = \int f_k \,\diff\mu,$$

   since each $f_k$ is continuous. Since the $f_k$ are increasing, this means that $f_1 \leq f_k$, and because $X$ is compact, $f_1$ is bounded. Hence, the $f_k$ are bounded uniformly from below, so we may apply monotone convergence to get

   $$\lim_{k\to\infty} \int f_k \,\diff\mu = \int \phi \,\diff\mu.$$

   Hence,

   $$\int \phi \,\diff\mu \leq \liminf_{n\to\infty} \int \phi \,\diff\mu_n.$$

2. Let $I = \inf_{\mathcal{P}\p{K}} E$. If $I = \infty$, then any measure on $\mathcal{P}\p{K}$ is a minimizer, e.g., a normalized Lebesgue measure will work. Otherwise, let $\set{\mu_n}_n \subseteq \mathcal{P}\p{K}$ be a sequence such that $E\p{\mu_n} \to I$ as $n \to \infty$. By Banach-Alaoglu (the $\mu_n$ all hae norm $1$), there exists a measure $\mu \in \mathcal{P}\p{K}$ such that, passing to a subsequence if necessary, $\mu_n \to \mu$ weakly-*.

   We claim that $\mu_n \otimes \mu_n \to \mu \otimes \mu$ weakly-* in $\p{C\p{K \times K}}^*$ as well. Indeed, by Stone-Weierstrass, the set

   $$A = \set{\sum_{k=1}^n c_kf_k\p{x}g_k\p{x} \st n \in \N,\ c_k \in \R,\ f_k, g_k \in C\p{K}}$$

   is uniformly dense in $C\p{K \times K}$. Moreover,

   $$\begin{aligned} \lim_{n\to\infty} \int_K \int_K \sum_{k=1}^N c_k f_k\p{x} g_k\p{y} \,\diff\mu_n\p{x} \,\diff\mu_n\p{y} &= \lim_{n\to\infty} \sum_{k=1}^N c_k \p{\int_K f_k\p{x} \,\diff\mu_n\p{x}} \p{\int_K g_k\p{y} \,\diff\mu_n\p{y}} \\ &= \sum_{k=1}^N \sum_{k=1}^N c_k \p{\int_K f_k\p{x} \,\diff\mu\p{x}} \p{\int_K g_k\p{y} \,\diff\mu\p{y}} \\ &= \int_K \int_K \sum_{k=1}^N c_k f_k\p{x} g_k\p{y} \,\diff\mu\p{x} \,\diff\mu\p{y}, \end{aligned}$$

   so $\mu_n \otimes \mu_n \to \mu \otimes \mu$ on a dense set, hence on all of $C\p{K \times K}$. Finally, observe that $\frac{1}{\abs{x - y}}$ is a lower-semicontinuous function on $K \times K$: if $x \neq y$, then it is continuous there, and otherwise, if $x = y$, then the function tends to $\infty$ when approaching these points. Thus, by (1),

   $$\begin{aligned} E\p{\mu} &= \int_K \int_K \frac{1}{\abs{x - y}} \,\diff\mu\p{x} \,\diff\mu\p{y} \\ &\leq \liminf_{n\to\infty} \int_K \int_K \frac{1}{\abs{x - y}} \,\diff\mu_n\p{x} \,\diff\mu_n\p{y} \\ &= \liminf_{n\to\infty} E\p{\mu_n} \\ &= I, \end{aligned}$$

   so $\mu$ is a minimizer.