weak*closure of {f:||f||=1} in dual.

Question

What is the weak* closure of {f:||f||=1}? I am sure this set is not closed in weak* topology. So what is the weak* closure of this set. Thanks.

Answer 1

The weak$^{*}$ closure of $\{f\in X^{*}:\|f\|=1\}$ is the unit ball $\{f\in X^{*}:\|f\|\leq 1\}$ for infinite dimensional Banach spaces $X$. To prove this, let $f\in X^{*}$ be a functional with $\|f\|\leq 1$. Let $D$ be the set of all finite dimensional subspaces of $X$. Then $D$ is a directed set under inclusion. Let $f_{d}\in X^{*}$ be a linear functional where $f_{d}=f$ on the subspace $D$ and where $\|f_{d}\|=1$. Take note that for $x\in X$ and $d\in D$ with $x\in d$, we have $f_{e}(x)=f_{d}(x)=f(x)$ whenever $d\subseteq e,e\in D$. Therefore $f_{d}(x)\rightarrow f(x)$. We conclude that $f_{d}\rightarrow f$ in the weak$^{*}$-topology. Therefore, since the set $\{f\in X^{*}:\|f\|\leq 1\}$ is closed in the weak$^{*}$-topology, the set $\{f\in X^{*}:\|f\|\leq 1\}$ is the weak$^{*}$-closure of $\{f\in X^{*}:\|f\|=1\}$.