Analysis Problem of the Day 16

Today’s problem appeared as Problem 3 on the UCLA Fall 2016 Analysis Qual:

Problem 3: Let $X$ be a compact metric space, and let $\mathcal{P}(X)$ be the space of Borel probability measures on $X.$

a) Let $\phi: X \to [0,\infty]$ be a lower semi-continuous function. Show that if $\mu_n$ converges weak-* to $\mu,$ then

$\int_X \phi d\mu \leq \liminf_{n \to \infty} \int_X \phi d\mu_n.$

b) For $K \subset \mathbb{R}^d$ compact, show that the function $E: \mathcal{P}(K) \to [0,\infty]$ defined by

$E(\mu) = \int_K \int_K \frac{1}{|x-y|} d\mu(x) d\mu(y)$

attains its minimum on $\mathcal{P}(K).$

Solution: Part a) is a particular case of a famous result from probability known as the Portmanteau lemma. For brevity, we will not reproduce the proof the theorem here. However, a) follows as an immediate consequence of the theorem.

For b), we first note that $\frac{1}{|x-y|}$ is a continuous (thus in particular lower semi-continuous) function from $K$ to $[0,\infty].$ Since $E$ is non-negative, it achieves a (possibly infinite) infimum on $\mathcal{P}(K),$ and in particular there exists a sequence of measures $\mu_n \in \mathcal{P}(K)$ s.t. $E(\mu_n) \to \inf_{\mathcal{P}(K)} E.$ Now, $C(K)$ is separable since $K$ is compact, so the unit ball in $C(X)^*$ is weak-* sequentially compact, so by Banach-Alaouglu there exists a weak-* convergent subsequence $\mu_n \to \mu.$ It is easy to see that $\mu(K)=1,$ so $\mu \in \mathcal{P}(K).$ Then, by a), $E(\mu) \leq \inf_{\nu \in \mathcal{P}(K)} E(\nu) \implies E(\mu) = \inf_{\nu \in \mathcal{P}(K)} E(\nu),$ which implies that the infimum is achieved on $\mathcal{P}(K).$

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