Does $\mu_n \overset{1}{\rightharpoonup} \mu$ necessarily imply $\mu_n \overset{2}{\rightharpoonup} \mu$?

Question

Let

• $X$ be a metric space,
• $\mathcal M(X)$ the space of all finite signed Borel measures on $X$,
• $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions on $X$, and
• $\mathcal C_0(X)$ be the space of real-valued continuous functions on $X$ that vanish at infinity.

Then $\mathcal C_b(X)$ and $\mathcal C_0(X)$ are 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 $\mathcal M(X)^* := (\mathcal M(X))^*$ and $\mathcal C_b(X)^* := (\mathcal C_b(X))^*$ be the continuous duals. 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 $\overset{2}{\rightharpoonup}$.

Then we have

Theorem: If $\mu_n \overset{1}{\rightharpoonup} \mu$ then $\{[\mu_n] \mid n \in \mathbb N\}$ is bounded.

Proof: It's well-known that $$ \mathcal M(X) \to \mathcal C_b(X)^*, \nu \mapsto \left (L_\nu :f \mapsto \int_X f \mathrm d \nu \right). $$ is an isometrically isomorphic embedding. This implies $[\nu] = \|L_\nu\|$. By uniform boundedness principle, it suffices to show that $(L_{\mu_n} (f))_n$ is bounded for each $f \in \mathcal C_b(X)$. This is indeed true because $L_{\mu_n} (f) \to L_{\mu} (f)$ for each $f \in \mathcal C_b(X)$. This completes the proof.

My question:

Does $\mu_n \overset{1}{\rightharpoonup} \mu$ necessarily imply $\mu_n \overset{2}{\rightharpoonup} \mu$? If not, what if $X$ is locally compact (and possibly separable)?

Thank you so much for your elaboration!

Answer 1

While they might not be the exact same thing, in spirit you are comparing weak and weak$^*$ convergence.

Let $X=[0,1]$ (the example works exactly the same if we take $X=\{0\}\cup\{\frac1n:\ n\}$), $$\mu_n=\delta_{1/n},\qquad\qquad \mu=\delta_{\{0\}}.$$ For any $f\in C_b(X)$, $$ \int_Xf\,d\mu_n=f\big(\frac1n\big)\xrightarrow[n\to\infty]{}f(0)=\int_Xf\,d\mu. $$ Now let $\varphi\in M(X)^*$ be given by $\varphi(\eta)=\eta(\{0\})$. This is linear, and bounded since $|\eta(\{0\})|\leq|\eta(X)|\leq\|\eta\|$. And $$ \varphi(\mu_n)=0,\qquad\qquad n\in\mathbb N, $$ while $$ \varphi(\mu)=\delta_{\{0\}}(\{0\})=1. $$ What happens is that in the #$1$ convergence you are testing against continuous functions, while in the #$2$ convergence you are testing against all bounded Borel functions and possibly more; this is a stronger notion of convergence.