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Let $X$ be an LCH space and $C_0(X)$ the set of continuous vanishing functions on $X$. If $C_0(X)$ is given the structure of a Banach space with the sup-norm, then its weak topology is given by the set of Radon measures $M(X)$ of finite total variation. One has $f_\alpha \to f$ if and only if $\int f_\alpha ~d\mu \to \int f~d\mu$ for all $\mu \in M(X)$.

My question: Is there a characterization for weak convergence in $C_0(X)$?

Weak convergence is at least as strong as pointwise convergence (because of the Dirac measures). I have an example showing that it is not the same as pointwise convergence in general.

Evan DeCorte
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    When X is compact, a sequence of continuous functions converges weakly iff it converges pointwise and is uniformly bounded. But that's not really what you asked for. – Nate Eldredge Jun 07 '14 at 14:57
  • Nate, this is already interesting for me. Thanks for your comment. – Evan DeCorte Jun 07 '14 at 16:09
  • @NateEldredge Does not your remark answer the question, given that the elements of $C_0(X)$ extend continuously to the one-point compactification of $X$? –  Jul 21 '14 at 06:24
  • @This is much healthier: You're right, it does work in the locally compact case also. I'm not sure why I was thinking it didn't. But it only characterizes weak convergence of sequences, not nets. Nevertheless, I guess I'll post it as an answer later. – Nate Eldredge Jul 21 '14 at 10:41

1 Answers1

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For sequences, there is the following simple characterization:

Proposition. Suppose $f_n, f \in C_0(X)$. The following are equivalent:

  1. $f_n \to f$ weakly;

  2. $f_n \to f$ pointwise and $\sup_n \|f_n\|_\infty < \infty$.

Proof. $1 \implies 2$: The evaluation maps $f \mapsto f(x)$ are continuous linear functionals, so $f_n \to f$ pointwise, and $\sup_n \|f_n\| < \infty$ follows from the uniform boundedness principle.

$2 \implies 1$: If $f_n \to f$ pointwise and $\sup_n \|f_n\| < \infty$, then for any signed or complex measure $\mu$, we have $\int f_n \,d\mu \to \int f\,d\mu$ by dominated convergence, using $\sup_n \|f_n\|$ as the dominating function. (For positive measures $\mu$, this is the classical dominated convergence theorem. For signed measures, use the Jordan decomposition $\mu = \mu^+ - \mu^-$. For complex measures, take real and imaginary parts.) $\quad\square$

This characterization does not work for nets. For example, a weakly convergent net need not be bounded.

You can find this result (for $X$ compact, but the proof is the same) as Example 2 of IV.5 of Reed and Simon, volume 1 (Methods of Mathematical Physics: Functional Analysis, Revised and Enlarged Edition).

Nate Eldredge
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  • In the compact case I agree with this proof, but what would happen on the following example? Take $X=\mathbb{R}$ and let ${f_n}$ be a sequence of bumps of height $1$ moving to $\infty$. I think the application of dominated convergence would fail here. – Evan DeCorte Jul 21 '14 at 15:20
  • @Philip: No, I think we're okay. Remember that $\mu$ is a finite measure. – Nate Eldredge Jul 21 '14 at 20:28
  • @NateEldredge I know this was a long time ago but could you explain the bit about how $\sup_n |f_n| < \infty$ follows from the Uniform Boundedness Principle? That would be much appreciated. – inkievoyd Mar 08 '18 at 04:07
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    @inkievoyd: Since $f_n \to f$ weakly, we have that for each $\ell \in C_0(X)^$, the scalar sequence $\ell(f_n)$ converges (to $\ell(f)$); in particular it is a bounded sequence. So if we view $f_n$ as a continuous linear functional on $C_0(X)^$ (that is, we use the natural embedding of $C_0(X)$ into $C_0(X)^{}$), then ${f_n}$ is a pointwise bounded sequence of continuous linear functionals on $C_0(X)^*$. The uniform boundedness principle gives $\sup_n |f_n|_{C_0(X)^{}} < \infty$. But the natural embedding is an isometry, by Hahn-Banach, so $|f_n|{C_0(X)^{**}} = |f_n|\infty$. – Nate Eldredge Mar 08 '18 at 15:03
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    @inkievoyd: This argument goes through in any Banach space: a weakly convergent sequence is always bounded in norm. – Nate Eldredge Mar 08 '18 at 15:04