Let $\Omega$ be an open subset of $\mathbb{R}^n$. I have a sequence $\mu_j\in C_0(\Omega\times \mathbb{R}^n)^*$ (topological dual). Now I am told that $(\mu_j)_j$ is weakly* sequentially precompact. I am not even sure what this means but I think it means $\mu_j$ has a subsequence $\mu_{j_k}$ that is weakly* Cauchy. I am not sure either on the meaning of this but I guess it means $\|\mu_{j_k}(f)-\mu_{j_l}(f)\|<\epsilon$ when $k,l$ big enough. Now from here I have to show that there exists some $\mu\in C_0(\Omega\times \mathbb{R}^n)^*$ such that $\mu_{j_k}(f) → \mu(f)$. But I am not sure on why this is the case.

What I know is that $\mathbb{R}$ is complete so we $\mu_{j_k}(f)$ converges to some real number that I can call $\mu(f)$ but then it is not so clear that $\mu\in C_0(\Omega\times \mathbb{R}^n)^*$ for all $f ∈ C_0(Ω \times \mathbb{R}^n)$.


Definition: We say that a sequence $\mu_j$ is weakly* sequentially pre-compact if the weak* closure of that sequence is weakly* sequentially compact.

Definition: We say that $\mu_k\rightarrow \mu$ if and only if $\mu_{jk}(f)\rightarrow \mu(f)$, $\forall f\in C_0(\Omega \times \mathbb{R}^n)$. This is a consequence of the Riesz Representation Theorem and hence what it means to be weakly* convergent.

The result follows once you write down the definition of everything. Let $A:=\{\mu_j\}$ just the set containing every element of the sequence. Since A is weakly* sequentially pre-compact, this implies that $\overline{A}$, the weak* closure of A, is weakly* sequentially compact. Therefore, there exists a weakly* convergent subsequence $\mu_{nk}$ such that $\mu_{nk}\rightarrow \mu\in \overline{A}$.

The only thing that remains to show is that $\mu \in A$ is actually in $C_0(\Omega \times \mathbb{R}^n)^*$. However, the space $C_0(\Omega \times \mathbb{R}^n)^*$ is weak* closed as a topological space and hence $\overline{A}\subset C_0(\Omega \times \mathbb{R}^n)^*$.

  • $\begingroup$ Thank you. In the first definition, is it the closure with respect to the weak* topology? Or the closure with respect to the strong topology? $\endgroup$ – edamondo Feb 23 at 10:30
  • $\begingroup$ The weak* topology! Thanks for pointing that out; I've edited the response. $\endgroup$ – Andrew McMillan Feb 23 at 14:07
  • $\begingroup$ I see now, thank you. But then I am not sure I get the last sentence. Should it be "$C_0(\Omega\times \mathbb{R}^n)*$ is Banach and hence complete" instead of "$C_0(\Omega\times \mathbb{R}^n)*$ is Banach and hence closed"? $\endgroup$ – edamondo Feb 23 at 15:13
  • $\begingroup$ You could also prove the result using the analogous definitions for weak* Cauchy sequences. I was really just getting at the fact that the entire space $C_0(\Omega\times \mathbb{R}^n)^*$ is closed in the weak* topology just via the fact that it's a topological space. $\endgroup$ – Andrew McMillan Feb 23 at 16:10
  • $\begingroup$ via the fact that it is just a topological space or a complete space? Sorry, I am confused. $\endgroup$ – edamondo Feb 23 at 18:29

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