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.



weak*closure of {f:||f||=1} in dual.
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 $\{f\in X^{\*}:\|f\|=1\}$.


