# Does $\mu_n \overset{*}{\rightharpoonup} \mu$ (or $\mu_n \rightharpoonup \mu$) imply $\{[\mu_n] \mid n \in \mathbb N\}$ is bounded?

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 $$\mu_n,\mu \in \mathcal M(X)$$. We define weak convergence by $$\mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X),$$ and weak$$^*$$ convergence by $$\mu_n \overset{*}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_0 (X).$$

Now let $$\mu_n,\mu \in \mathcal M(X)$$. If $$\mu_n \rightharpoonup \mu$$, then $$\mu_n (X) \to \mu (X)$$ and thus $$\{\mu_n(X) \mid n \in \mathbb N\}$$ is bounded. On the other hand, $$\mu_n \overset{*}{\rightharpoonup} \mu$$ does not necessarily imply $$\mu_n (X) \to \mu (X)$$.

My questions:

1. Does $$\mu_n \rightharpoonup \mu$$ imply $$\{[\mu_n] \mid n \in \mathbb N\}$$ is bounded?
2. Does $$\mu_n \overset{*}{\rightharpoonup} \mu$$ imply $$\{[\mu_n] \mid n \in \mathbb N\}$$ is bounded?

Thank you so much!

• Did you tried the uniform boundedness principle?
– Feng
Commented Nov 3, 2022 at 10:52
• @Feng Thank you so much! So the answer of 1. is yess. How about 2.? Commented Nov 3, 2022 at 10:58
• It seems to me that they are both correct and the proofs are the same. Although it sounds strange. Do we have $$[\mu]=\sup_{f\neq 0\\ f\in C_b}\frac{\int f\,d\mu}{\|f\|_\infty}=\sup_{f\neq 0\\ f\in C_0}\frac{\int f\,d\mu}{\|f\|_\infty}?$$
– Feng
Commented Nov 3, 2022 at 11:15

This was initially intended to be a reply to geetha290krm's supposed counterexample to (2) (showing it is $$\textbf{not}$$ in fact a valid counterexample, and that in the context of $$X$$ being locally compact Hausdorff, (2) will always in fact be true), but was too long for a comment, so I wrote it up as an answer.

Lets work in $$\mathbb{R}_{> 1}$$, for simplicity (with the Lebesgue measure).

Let $$0 \leq \chi_{n} \leq \frac{1}{n}$$ be a bump function equal to $$1/n$$ on $$[n + \epsilon_{n}, n+1 - \epsilon_{n}]$$, with support contained in $$[n + \epsilon_n^2 ,n + 1 - \epsilon_n^2] \subset (n,n+1)$$ for each $$n$$ , where $$\frac{1}{4}> \epsilon_{n} > 0$$ are chosen sufficiently small so that $$\epsilon_{n} \downarrow 0$$ as $$n \uparrow \infty$$, finally notice that because $$\frac{1}{4} > \epsilon_{n} > 0$$, $$(1 - 2 \epsilon_{n}) \geq \frac{1}{2}$$ for all $$n = 1,2,3,...$$

Clearly then, $$\chi_n \in C_c(\mathbb{R}_{>1}) \subset C_0(\mathbb{R}_{>1})$$ for each $$n$$, and we have $$|| \sum_{k=m}^n \chi_{k} ||_{u} \leq \frac{1}{m}$$, for all $$1 \leq m < n$$, so that $$\sum_{k \geq 1} \chi_{k} \rightarrow_{||.||_{u}} f$$, where $$f \in C_0(\mathbb{R}_{>1})$$ and is non-negative everywhere.

Writing $$\mathbb{R}_{>1} = (1,2) \cup (2,3) \cup (3,4) \cup ...$$, we see that $$\sum_{k \geq 1} \chi_k \restriction (n,n+1) = \chi_n$$, as $$\text{supp}(\chi_{k}) \subset (k,k+1)$$ for all $$k \geq 1$$, so that only $$\chi_n$$ may be non-zero when $$f$$ is restricted to $$(n,n+1) \subset \mathbb{R}_{>1}$$.

geetha290krm claims that the measures $$\mu_n = n d \lambda \restriction (n,n+1)$$, which are indeed in $$\mathcal{M}(\mathbb{R}_{>1})$$ converge to $$0$$ in the weak$$^{\ast}$$-sense.

This is false, as we will now show. Indeed, suppose it were true. Then by the definition of convergence in the weak$$^{\ast}$$-sense as defined by the OP, we should have $$\int f d \mu_n \rightarrow_{n \rightarrow \infty} 0 = \int f d\mu$$ , here $$\mu = 0$$ is the zero measure on $$\mathbb{R}_{>1}$$.

However, we have that $$\int f d\mu_n \geq \int_{[n + \epsilon_n, n+1 - \epsilon_n]} f d\mu_n = (1 - 2 \epsilon_n)\frac{n}{n} \geq (1 - 2 \epsilon_n) \geq \frac{1}{2}$$ for all $$n = 1, 2 , 3 , ...$$ with the inequalities all holding because $$f$$ is non-negative, its restriction to each interval $$[n+\epsilon_n, n+1 - \epsilon_n] \subset (n,n+1)$$ is just $$\chi_n$$ restricted to this interval, where it is identically equal to $$\frac{1}{n}$$, and because we chose $$\frac{1}{4} > \epsilon_n > 0$$ for all $$n = 1,2,3,...$$, we indeed have that $$(1 - 2 \epsilon_n) \geq \frac{1}{2}$$ for all $$n = 1,2,3,...$$ (as already established in the first paragraph).

This shows that we do not have $$\int f d \mu_n \rightarrow_{n \rightarrow \infty} \int f d \mu$$, so that geetha290krm's counterexample is invalid, because $$\mu_n$$ does not converge to $$0$$ in the weak$$^{\ast}$$ - sense, as I have demonstrated.

Indeed, there is no counterexample to (2), by the Riesz-Markov theorem , whenever $$X$$ is a locally compact Hausdorff space, since then your $$\mathcal{M}(X)$$ is precisely isometric to $$C_0(X)^{\ast}$$, i.e. the dual space to $$C_0(X)$$, consisting of all bounded $$\mathbb{R}$$-linear functionals $$C_0(X) \rightarrow \mathbb{R}$$, which is a Banach space with the operator norm. By equipping $$C_0(X)^{\ast}$$ with the weak$$^{\ast}$$-topology, and then transporting this topology onto $$\mathcal{M}(X)$$ via the isometric isomorphism $$\mathcal{M}(X) \rightarrow C_0(X)$$ given by $$\mu \rightarrow I_{\mu}$$, where $$I_{\mu}(f) = (f \rightarrow \int_{X} f d \mu)$$, $$I_{\mu} : C_0(X) \rightarrow \mathbb{R}$$, your question for (2) is answered in the affirmative, because it is a general fact that for a Banach space $$X$$, a weak$$^{\ast}$$-convergent sequence $$f_n \in X^{\ast}$$ is uniformly bounded in operator norm.

Therefore to seek a counterexample for (2), we need to find a (metrizable) Hausdorff topological space $$X$$, which is NOT locally compact.

One candidate is $$X = (1,2) \times (2,3) \times (3,4) \times (4,5) \times ...$$, with the usual product metric (inducing the product topology). This is NOT locally compact, so perhaps we could find a counterexample here.

Now perhaps it is possible to adapt geetha290krm's counterexample to this situation. But I have not figured out all the details yet and am not entirely convinced $$X$$ is sufficient to provide a counterexample to (2) for the following reason:

While this space $$X$$ is not locally compact, it is Borel isomorphic to $$[0,1]$$, which is indeed (locally) compact Hausdorff, so perhaps this $$X$$ may not work to produce a counterexample.

So for (2) a counterexample may need to come from a metrizable topological space $$Y$$, that is NOT locally compact, and is either

1: NOT second countable, or

2: NOT $$\textbf{completely}$$ metrizable if it $$\textbf{is}$$ second countable (i.e. if condition 1. does hold)

Either of these conditions will prohibit $$Y$$ from being Borel isomorphic to a Borel subset of $$[0,1]$$, where (in the latter space $$[0,1]$$) the usual Riesz-Markov theorem applies, and this may cause complications.