I want to prove: If $X$ is "infinite" dimensional normed space, then weak closure of unit ball $S = \{x \in X : \|x\| = 1\}$ is $\{x \in X : \|x\| \leq 1\}$.
I can see the proof here, but I still don't understand why we need "infinite" dimension. My proof is following.
Let $B = \{x \in X : \|x\| \leq 1\}$. Since $B$ is convex so weakly closed, implying that $wk \bar{S} \subseteq B$. Conversely, let $\|x\| \leq 1$ and suppose $U \neq X$ be any weak nbd of $0$ such that $x + U$ is an weak nbd of $x$. To complete the proof, it is enough to show that $x+U$ intersects $S$.
Now, by the definition of weak topology, there exists $x_1^*, \cdots, x_n^* \in X^*$ and $\epsilon_1, \cdots, \epsilon_n$ such that $V: = \bigcap_{i=1}^n \{x \in X : |<x, x_i^*>| \leq \epsilon_j\} \subseteq U$. Define $x^* := \sum_{i=1}^n \frac{x_i^*}{\epsilon_i}$. Then clearly $\{x \in X : |<x, x^*>| < 1\} \subseteq V \subseteq U.$ In particular, $ker(x^*) \subseteq U.$ Since $U$ is proper, so $x^*$ is not a zero functional, so surjective, and so $X / ker(x^*) \cong \mathbb{F}$ holds. Now since $X$ is infinite dimensional, $\text{dim}_{\mathbb{F}}ker(x^*) \geq 1$, so it contains a line (1-dimensional subspace). Since $\|x\| \leq 1$, so $x + ker(x^*)$ must intersects $S$. Since $ker(x^*) \subseteq U$, so $x + U$ intersects $S$, completing the proof.
As I wrote, the fact that $X$ is infinite dimensional is only used to show $ker(x^*)$ contains a line. So it seems that $X$ is "infinite" dimensional is redundant, and I think it can be relaxed to $dim X \geq 2$.
Am I right, or there are any errors in my proof ?
Best Answer
It is not true that $$ \{x\in X:|\langle x^\ast,x\rangle|<1\}\subset V $$ because you can have "cancellation effects". Just because the sum has absolute value less than one does not mean that each summand has.
In fact, the weak topology on a finite-dimensional normed space coincides with the norm topology, in which the unit sphere is closed.