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

  • $\begingroup$ In general the weak topology may fail to be metrizable. $\endgroup$ – DanielWainfleet Feb 21 '19 at 23:24

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.


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