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In functional analysis,

Definition A:

for any normed linear space $(X, \| \cdot \| )$, the weak star topology $\sigma (X^*, X)$ on $X^*$ is generated by the collection of seminorms $\{ p_x \, | \, x \in X\}$, defined by $$p_x (f) = |f(x)|.$$

In probability theory (more specifically from the book "Probability measures on Metric Spaces" written by Parthasarathy),

Definition B:

for any metric space $X$, let $\mathcal{M} (X)$ denote the space of measures defined on $\mathcal{B} (X)$ and let $C(X)$ be the space of all bounded real valued continuous functions on $X$, equipped with the sup norm. Then the weak topology on the space $\mathcal{M} (X)$ is generated by the base of open neighbourhoods at a point $\mu$ defined by $$ \bigg\{ \nu \in \mathcal{M} (X) \, \Bigg| \, \, \bigg| \int_X f_i \, d \nu - \int_X f_i \, d \mu \bigg| < \epsilon_i, \, \, i= 1,2,\ldots, k \bigg\},$$ where $f_1, \ldots, f_k \in C(X)$ and $\epsilon_1 , \ldots, \epsilon_k >0$.

Here is what I don't understand:

If $X$ is a compact metric space, then by the representation theorem of bounded linear functionals on $C(X)^*$, then for any $\Lambda \in C(X)^*$, there exists a unique Borel measure $\mu \in \mathcal{M}(X)$ such that $$ \Lambda_\mu (f) := \Lambda (f) = \int_X f \,d \mu, \quad \forall f \in C(X),$$ and that $$\| \Lambda_\mu \| = \mu (X).$$ Thus, if we identify each element $\mu \in \mathcal{M} (X)$ by $\Lambda_\mu \in C(X)^*$, Definitions A and B are the same.

However, for any general metric space $X$ that is NOT necessarily compact, $\mathcal{M} (X)$ and $C(X)^*$ are not necessarily in isometric isomorphism. (Or is there a representation result in greater generality?)

Is Definition B slightly more general than Definition A to cater for the needs in probability theory? If this is the case, then some functional analytic results might not be applicable to weak convergence theory in probability..... Any ideas?

Richard
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1 Answers1

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For any metric space $X$, we can certainly think of any finite measure as a linear functional on $C(X)$, so there is a natural map $\mathcal{M}(X) \to C(X)^*$ which is an isometry. You are right that for non-compact $X$, this map is not surjective. (The "extra" elements of $C(X)^*$ can, I believe, be characterized as regular finitely additive measures. Dunford and Schwartz have a discussion of these objects. They are typically pathological and need the axiom of choice to construct, and probabilists usually don't want to deal with them.)

The probabilist's "weak" topology on $\mathcal{M}(X)$ can be seen as the subspace topology induced by equipping $C(X)^*$ with its weak-* topology, and considering $\mathcal{M}(X)$ as a subset of it. One also commonly studies the even smaller subset $\mathcal{P}(X)$ containing only the probability measures.

You are right that some functional analytic results about the weak-* topology on $C(X)^*$ do not directly apply to $\mathcal{M}(X)$, or need different proofs.

If $X$ is non-compact but is locally compact and separable, we can consider instead $C_0(X)$, the space of continuous functions $f : X \to\mathbb{R}$ which "vanish at infinity", i.e. the closure in $C(X)$ of the compactly supported functions $C_c(X)$. (Some authors use $C_0(X)$ for compactly supported functions, and use $C_\infty(X)$ for functions vanishing at infinity.) There is an isometry $\mathcal{M}(X) \to C_0(X)^*$ and there is a version of the Riesz representation theorem stating that it is surjective. But the weak-* topology on $C_0(X)^*$ doesn't give the probabilist's weak topology on $\mathcal{M}(X)$; in particular, $\mathcal{P}(X)$ is not weak-* closed in $C_0(X)^*$.

Nate Eldredge
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  • The Riesz–Markov–Kakutani theorem says that there is a surjective isometry between $C_0 (X)^$ and the space of all regular* signed Borel measures on $X$. Here $\mathcal M(X)$ can contain non-regular signed Borel measures. – Analyst Oct 21 '22 at 09:31
  • So I'm confused by "There is an isometry $\mathcal{M}(X) \rightarrow C_0(X)^$ and there is a version of the Riesz representation theorem stating that it is surjective*". Could you elaborate more? – Analyst Oct 21 '22 at 09:32
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    @Analyst: You're right, I need the extra hypothesis that $X$ is separable. A locally compact separable metric space is Polish, and in a Polish space, all $\sigma$-finite Borel measures are regular. I've edited to add this. – Nate Eldredge Oct 21 '22 at 13:41
  • Thank you so much for your explanation! – Analyst Oct 21 '22 at 13:42