# There exists a $\mu\in C_0(\Omega\times \mathbb{R}^n)^*$ such that $\mu_{j_k}(f) → \mu(f)$

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 $$\overline{A}\subset C_0(\Omega \times \mathbb{R}^n)^*$$.

• 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? – edamondo Feb 23 at 10:30
• The weak* topology! Thanks for pointing that out; I've edited the response. – Andrew McMillan Feb 23 at 14:07
• 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"? – edamondo Feb 23 at 15:13
• 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. – Andrew McMillan Feb 23 at 16:10
• via the fact that it is just a topological space or a complete space? Sorry, I am confused. – edamondo Feb 23 at 18:29