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Let

  • $X$ be a metric space,
  • $\mathcal M(X)$ the space of all finite signed Borel measures on $X$, and
  • $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions on $X$.

Then $\mathcal C_b(X)$ is a real Banach space with supremum norm $\|\cdot\|_\infty$. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$. Then $(\mathcal M(X), [\cdot])$ is a Banach space. Let $\mathcal M(X)^* := (\mathcal M(X))^*$ be the continuous dual. Let $\mu_n,\mu \in \mathcal M(X)$.

  • We define the first type of weak convergence by $$ \mu_n \overset{1}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X), $$ Let $\sigma(\mathcal M(X), \mathcal C_b(X))$ be the topology induced by $\overset{1}{\rightharpoonup}$.

  • We define the second type of weak convergence by $$ \mu_n \overset{2}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \varphi(\mu_n) \to \varphi (\mu) \quad \forall \varphi \in \mathcal M(X)^*, $$ Let $\sigma(\mathcal M(X), \mathcal M(X)^*)$ be the topology induced by $\overset{2}{\rightharpoonup}$.

Of course, we have $\mu_n \overset{2}{\rightharpoonup} \mu \implies [\mu] \le \liminf_n [\mu_n]$. Also, we can prove that $\mu_n \overset{1}{\rightharpoonup} \mu \implies [\mu] \le \liminf_n [\mu_n]$.

Are there some conditions (locally compact, separable, Polish,...) on $X$ that ensure [$\mu_n \overset{1}{\rightharpoonup} \mu \implies \mu_n \overset{2}{\rightharpoonup} \mu$] or [$\mu_n \overset{2}{\rightharpoonup} \mu \implies \mu_n \overset{1}{\rightharpoonup} \mu$]?

Thank you so much for your elaboration!

I posted this question on MSE, but it seems to receive no answer. So I post it here.

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  • $\begingroup$ For general metric spaces: What are functions vanishing at infinity? $\endgroup$
    – Dieter Kadelka
    Nov 4, 2022 at 11:53
  • $\begingroup$ @DieterKadelka A function $f \in \mathcal C_b (X)$ vanishing at infinity means that for every $\epsilon>0$ the set $\{x:|f(x)| \geq \epsilon\}$ is compact. $\endgroup$
    – Akira
    Nov 4, 2022 at 11:58
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    $\begingroup$ One direction seems to be simple since $C_0(X) \subset C_b(X) \subset M^*(X)$. We always have that $\mu_n \to^2 \mu$ implies $\mu_n \to^1 \mu$. For the other direction you need strong assumptions, I think. Maybe you find something in the old book of Z. Semadeni, Banach spaces of continuous functions I (1971). $\endgroup$
    – Dieter Kadelka
    Nov 4, 2022 at 12:14
  • $\begingroup$ @DieterKadelka Do you mean by $\mathcal C_b(X) \subset \mathcal M(X)^*$ that there is an isometrically isomorphic embedding from $\mathcal C_b(X)$ to $\mathcal M(X)^*$? $\endgroup$
    – Akira
    Nov 4, 2022 at 12:25
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    $\begingroup$ @JochenGlueck I have removed unnecessary parts. $\endgroup$
    – Akira
    Nov 4, 2022 at 12:34

1 Answer 1

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An example. $X = [0,1]$ with the usual metric. $\mathcal C[0,1] = \mathcal C_b[0,1] = \mathcal C_0[0,1]$. $\mathcal C[0,1]^* = \mathcal M[0,1]$. Let $\mu_n$ be the unit point-mass at $1/n$ and $\mu$ the unit point-mass at $0$. Show $\mu_n \overset{1}{\rightharpoonup} \mu $ is true but $\mu_n \overset{2}{\rightharpoonup} \mu$ is false.

Whatever "some conditions" to insure $\big[\mu_n \overset{1}{\rightharpoonup} \mu \implies \mu_n \overset{2}{\rightharpoonup} \mu\big]$ are, they are not satisfied by $[0,1]$.

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  • $\begingroup$ The space $\mathcal M(X)^*$ is quite hard to imagine. Could you provide a map $\varphi \in \mathcal M(X)^*$ such that $\varphi (\mu_n)$ does not converge to $\varphi (\mu)$? $\endgroup$
    – Akira
    Nov 4, 2022 at 12:39
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    $\begingroup$ @Akira: Let $f$ be any bounded measurable function on $X$. Then $\mu \to \int f~d\mu$ is in $M(X)^*$. Choose $f = 1_{\{0\}}$ f.i. $\endgroup$
    – Dieter Kadelka
    Nov 4, 2022 at 13:15
  • $\begingroup$ The entire space $\mathcal M(X)^*$ is, indeed, hard to imagine. But there are simple examples (as Dieter showed) of elements of $\mathcal M(X)^*$ not in $\mathcal C(X)$. A more difficult question would be: exhibit an infinite comapct $X$ where$ \big[\mu_n \overset{1}{\rightharpoonup} \mu \implies \mu_n \overset{2}{\rightharpoonup} \mu\big]$ is true. Surely you should do that before you ask for conditions under which it is true. $\endgroup$
    – Gerald Edgar
    Nov 4, 2022 at 15:18

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