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Spring 2013 - Problem 4

Banach-Alaoglu

Let $K$ be a non-empty compact subset of $\R^3$ . For any Borel probability measure $\mu$ on $K$ , define the Newtonian energy $I\p{\mu} \in \poc{0, \infty}$ by

I\p{\mu} = \int_K \int_K \frac{1}{\abs{x - y}} \,\diff\mu\p{x} \,\diff\mu\p{y}

and let $R_K$ be the infimum of $I\p{\mu}$ over all Borel probability measures $\mu$ on $K$ . Show that there exists a Borel probability measure $\mu$ such that $I\p{\mu} = R_K$ .

Solution.

By definition, there exists a sequence of Borel probability measures $\set{\mu_n}_n$ on $K$ such that $I\p{\mu_n} \to R_K$ . Viewing $\set{\mu_n}_n \subseteq \p{C\p{K}}^*$ , we find via Banach-Alaoglu a Borel probability measure $\mu$ and a subsequence $\set{\mu_{n_k}}_k$ such that for any continuous function $f \in C\p{K}$ ,

\int_K f\p{x} \,\diff\mu_n\p{x} \xrightarrow{n\to\infty} \int_K f\p{x} \,\diff\mu\p{x}.

Next, we will show that $\mu_n \otimes \mu_n \to \mu \otimes \mu$ weakly on $C\p{K \times K}$ . Notice that $A = \set{\sum_{k=1}^n f_k\p{x}g_k\p{y} \mid f_k, g_k \in C\p{K}}$ is closed under addition, scalar multiplication, and multiplication. It also separates points: if $\p{x, y} \neq \p{x', y'}$ , say, $x \neq x'$ , then $x \cdot 1$ separates them. Since $1 \in A$ , it vanishes nowhere, so by Stone-Weierstrass, $A$ is dense in $C\p{K \times K}$ . Hence, given $\epsilon > 0$ and $f \in C\p{K \times K}$ , there exist $g, h \in C\p{K}$ such that $\norm{f\p{x, y} - g\p{x}h\p{y}}_{L^\infty} < \epsilon$ . By weak convergence, if $n$ is large enough, then

\begin{aligned} \abs{\int_{K \times K} g\p{x}h\p{y} \,\diff\mu\p{x} \,\diff\mu\p{x} - \int_{K \times K} g\p{x}h\p{y} \,\diff\mu_n\p{x} \,\diff\mu_n\p{x}} &= \abs{\p{\int_K g \,\diff\mu} \p{\int_K h \,\diff\mu} - \p{\int_K g \,\diff\mu_n} \p{\int_K h \,\diff\mu_n}} \\ &\leq \abs{\p{\int_K g \,\diff\mu} \p{\int_K h \,\diff\mu} - \p{\int_K g \,\diff\mu} \p{\int_K h \,\diff\mu_n}} + \abs{\p{\int_K g \,\diff\mu} \p{\int_K h \,\diff\mu_n} - \p{\int_K g \,\diff\mu_n} \p{\int_K h \,\diff\mu_n}} \\ &\leq \abs{\int_K g \,\diff\mu}\abs{\int_K h \,\diff\mu - \int_K h \,\diff\mu_n} + \abs{\int_K h \,\diff\mu_n}\abs{\int_K g \,\diff\mu - \int_K g \,\diff\mu_n} \\ &\leq \epsilon, \end{aligned}

since $\set{\int_K g \,\diff\mu_n}_n, \set{\int_K h \,\diff\mu_n}_n$ are convergent, hence bounded sequences. Thus,

\begin{aligned} \abs{\int_{K \times K} f \,\diff\p{\mu \otimes \mu} - \int_{K \times K} f \,\diff\p{\mu_n \otimes \mu_n}} &\leq \int_{K \times K} \abs{f - gh} \,\diff\p{\mu \otimes \mu} + \int_{K \times K} \abs{f - gh} + \,\diff\p{\mu_n \otimes \mu_n} + \abs{\int_{K \times K} gh \,\diff\p{\mu \otimes \mu} - \int_{K \times K} gh \,\diff\p{\mu_n \otimes \mu_n}} \\ &\leq \epsilon\mu\p{K}^2 + \epsilon\mu_n\p{K}^2 + \abs{\int_{K \times K} gh \,\diff\p{\mu \otimes \mu} - \int_{K \times K} gh \,\diff\p{\mu_n \otimes \mu_n}} \\ &\leq 3\epsilon \end{aligned}

which proves the claim. To complete the proof, observe that $\p{x, y} \mapsto \frac{1}{\abs{x - y}}$ is lower semicontinuous: it is continuous, except when $x = y$ , where it is infinity. It is also bounded below by $0$ , so there exists a sequence of non-negative continuous functions $\set{f_n}_n$ which increase to it. Thus,

\begin{aligned} \int_{K \times K} f_N\p{x, y} \,\diff\mu\p{x} \,\diff\mu\p{y} &= \lim_{n\to\infty} \int_{K \times K} f_N\p{x, y} \,\diff\mu_n\p{x} \,\diff\mu_n\p{y} \\ &\leq \lim_{n\to\infty} I\p{\mu_n} \\ &= R_K. \end{aligned}

By monotone convergence,

I\p{\mu} \leq \lim_{N\to\infty} \int_{K \times K} f_N\p{x, y} \,\diff\mu\p{x} \,\diff\mu\p{y} \leq R_K.

By definition of $R_K$ , $R_K \leq I\p{\mu}$ , so these are equalities.