# Weak closure of unit sphere is unit ball - a question about the hypotheses.

A homework problem I recall from functional analysis was to prove that the weak closure of the unit sphere, $$S$$, in an infinite-dimensional real normed vector space is the unit ball, $$B$$.

Looking back at what I turned in, I argued as follows:

Note that $$S$$ would be weakly dense in $$B$$ if, for any nonempty (relatively) weakly open subset $$U\subset B$$, one has $$S\cap U\neq\emptyset$$. Let $$U$$ be such a subset and let $$x_{0}\in U\subset B$$. Fixing $$\epsilon>0$$ and $$x^{*}\in X^{*}$$, one has by continuity, that the inverse image $$V_{*}^{\epsilon}:=(x^{*})^{-1}[(\langle x^{*},x_{0}\rangle-\epsilon,\langle x^{*},x_{0}\rangle+\epsilon)]$$ is weakly open, and hence, $$U\cap V_{*}^{\epsilon}$$ is (relatively) weakly open in $$B$$, and contains $$x_{0}$$. As long as $$x^{*}$$ does not vanish identically, it's kernel has codimension $$1$$, so since $$\text{dim}(X)=\infty$$, one must have that $$\text{ker}(x^{*})$$ is nontrivial. Then, finding a nonzero $$\xi\in\text{ker}(x^{*})$$, one has $$x_{0}+t\xi\in S$$ for some $$t\in\mathbb{R}$$. Finally, this yields $$|\langle x^{*},x_{0}\rangle-\langle x^{*},x_{0}+t\xi\rangle|=|t|\cdot|\langle x^{*},\xi\rangle|=0<\epsilon$$ which means $$x_{0}+t\xi\in V_{*}^{\epsilon}$$.

Now, I have two questions:

1. If we knew that $$V_{*}^{\epsilon}\subset U$$, we'd be done. Why can we assume this? (It seems in some of the proofs I've seen elsewhere, this is assumed WLOG)
2. Why do we need $$\text{dim}(X)=\infty$$? We are using the fact that $$X/\text{ker}(x^{*})\cong\mathbb{R}$$ so if the kernel were trivial, wouldn't this still be a contradiction as long as $$\text{dim}(X)\geq 2$$?

## 1 Answer

Nevermind - I have answered my own questions.

1. $$V_{w}=\Big{\{} \bigcap_{j=1}^{n} (x_{j}^{*})^{-1}[(a_{j},b_{j})] \text{ }\Big{|}\text{ } x_{1}^{*},\ldots, x_{n}^{*}\in X^{*}\Big{\}}$$ is a base for the weak topology on $$X$$. Thus, we may find some $$V\in V_{w}$$ so that $$x_{0}\in V\subset U$$, and in particular, this means that for some $$\epsilon>0$$, we have that $$V_{\epsilon}=\Big{\{}x\in X \text{ } \Big{|}\text{ } |\langle x_{j}^{*},x_{0}-x\rangle|<\epsilon \text{ for all } j=1,\ldots,n\Big{\}}\subset U$$

2. The fact that $$\text{dim}(X)=\infty$$ is then required to find a nonzero $$\xi\in\bigcap_{j=1}^{n}\text{ker}(x^{*}_{j})$$, and then we may proceed as above.