How to prove that If $F$ is closed with respect to the weak topology then $F$ is closed with respect to the norm topology…
Weak topology means let $V$ Banach space the weak topology on $V*$ is smallest topology in which each function is continuous
norm topology is the topology generated by $\mathscr{B}=\{B(x,\epsilon):x\in X, \epsilon>0 \}$
and also tell me what are balls that generated weak topology….
thank you somuch
Best Answer
It is called the weak topology because it is weaker than the strong topology. Let $T_w$ be the set of weakly open sets and let $T_s$ be the set of strongly open sets. Let $U$ be the set of all topologies on $V^*$ such that each $f\in V^*$ is continuous. Then $T_w=\cap U,$ and $T_s\in U,$ so $ T_w\subset T_s.$
So: $F$ is weakly closed $\implies V^*$ \ $F \in T_w \implies V^*$ \ $F \in T_s \implies F $ is strongly closed.