1

Consider $V$ a normed vector space, and equip its topological dual $V^*$ with the usual operator norm. Consider also the weak* topology on $V^*$ defined by the seminorms $(p_A)_{A\subset V \text{finite}}$ defined as follows : $\forall f \in V \quad p_A(f) = \sup_{x\in A} |f(x)|$

It says in my lecture notes that $\bar B(0,1)$ in $V^*$ is compact for the weak topology. And the proof uses Tychonoff's theorem. What I do not understand, is that the justification for using the theorem is that $\bar B(0,1)$ can be injected into $[0,1]^{[0,1]}$. I really don't see how this can canonically be done.

James Well
  • 1,209

1 Answers1

2

$\bar{B}(0,1)$ is weak-$*$ compact. This result is called Banach-Alaoglu theorem.

First, try to understand that the weak-$*$ topology of $V^*$ is precisely the product topology of $R^V$ induced on the subset $V^*$.

Second, let us consider $K:=\underset{x\in V}\prod [-\Vert x\Vert,\Vert x\Vert]$. Due to the Tychonoff's theorem, since $[-\Vert x\Vert,\Vert x\Vert]$ is compact in $R$, it turns out that $K$ is compact with respect to the product topology of $R^V$.

Finally, try to prove that $\bar{B}(0,1)$ is a closed subset of $K$. Consequently, $\bar{B}(0,1)$ will be compact for the product topology induced on $V^*$, that is, the weak-$*$ topology.

Edited: I have changed $[0,1]^V$ by $K:=\underset{x\in V}\prod [-\Vert x\Vert,\Vert x\Vert]$.

  • Actually, I don't get how $\bar B(0,1)$ is in $[0,1]^V$, since they are linear they reach all reals except the zero one. But if you mean via some kind of identification, the closest I can get to is seing that it is a subset of $[-1,1]^V$ because you could identify any linear form $L$ with $L'(x)=\frac{L(x)}{||x||}$ which goes into $[-1,1]$ when $|||L||| \leq 1$. – James Well Oct 12 '16 at 13:15
  • 1
    You're right. In fact, you can consider $K:=\underset{x\in V}\prod [-\Vert x\Vert,\Vert x\Vert]$. Then $\bar{B}(0,1)\subset K$. By the Tychonoff's theorem $K$ is also compact. I am going to edit my answer, and use $K$ rather than $[0,1]^V$. –  Oct 12 '16 at 14:02
  • Cool yeah I get that so $K$ is compact in $\mathbb{R}^V$ thx to the theorem, as a product of compact sets of $\mathbb{R}$ , and all closed subsets of a compact set are compact, right ? (True for any topology ?) – James Well Oct 13 '16 at 12:25
  • Remember, you also have to prove that $\bar{B}(0,1)$ is closed in $K$. And yes, every closed subset of a compact set, is compact too. It is very easy to prove that it is true for any topology. –  Oct 13 '16 at 12:36