0

It is from Stein's Functional Analysis Chapter 1, Problem 4.

Suppose $X$ is a $\sigma$-compact measurable metric space, and $C_b(X)$ is separable, where $C_b(X)$ denotes the Banach space of bounded continuous functions on $X$ with sup-norm.

If $\{\mu_n\}_{n=1}^\infty$ is a bounded sequence in $M(X)$, then there exists a $\mu\in M(X)$ and a subsequence $\{\mu_{n_j}\}_{j=1}^\infty$, so that $\mu_{n_j}$ convergens to $\mu$ in the following (weak*) sense:

$$\int_X g(x)d\mu_{n_j}(x)\to \int_X g(x)d\mu(x),\quad \text{for all }g\in C_b(X).$$

==========

Here $M(X)$ is the space of finite signed Borel measures on $X$.

stlinex
  • 782

1 Answers1

1

(1) The $\sigma$-compactness of $X$ is redudant.

Proposition. Let $X$ be a metric space. Then $X$ is compact if and only if $C_b(X)$ is separable.

($\implies$) By considering the covering $B(x,\frac{1}{n})$, we see that $X$ contains a countable dense subset $\{x_n\}$. With $f_n(x)=d(x,x_n)$, by Stone-Weierstrass theorem, finite linear combinatoriations of $f_n(x)$ over $\mathbb Q$ is dense in $C_b(X)$. (See this)

($\Longleftarrow$) Assume $X$ is not compact (sequential compact), we have a sequence $P=\{x_n\}$ with no limit points. For each $A\subseteq P$ with $B=P-A$, note that $A,B$ closed and $A\cap B=\emptyset$, we have a continuous function $f_A$ with $0\leq f\leq 1$, $f(A)=1$ and $f(B)=0$ (Urysohn). We have $\Vert f_{A_1}-f_{A_2}\Vert=1$ if $A_1\neq A_2$. (See this)

(2) With $X$ compact, we only need to construct a bounded linear functional $\ell$ to have the finite signed Borel measure $\mu$.

Let $\{g_k\}$ be dense in $C_b(X)$.

Then with a diagnoal process, we can choose a $\mu_{n_j}$ such that $\int_X g_k(x)d \mu_{n_j}(x)$ convergent for all $g_k$. With $$ \begin{aligned} \ell(g_k)&=\lim_{j\to \infty} \int_X g_k(x)d \mu_{n_j}(x),\\ \ell(g)&=\lim_{g_{t}\to g} \ell (g_{t}).\end{aligned}$$ Then $\ell$ is well-defined, bounded and linear.

stlinex
  • 782