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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 $\text{dim}(X)\geq 2$?
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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.

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