The topological dual for the weak topology is the topological dual.

Question

Let $E$ be Banach space and $E'$ its topological dual. The weak topology on $E$ is define to be the coarsest topology (with the least amount of open sets) on $E$ such that all element of $E'$ are continuous. Now if we denote by $\widetilde{E'}$ the topological dual of $E$ for the weak topology, we clearly have $E' \subset \widetilde{E'}$. In my functional analysis course, we said that these sets has to be algebrically the same but I don't really understand from where comes the reverse inclusion. Any help ?

Answer 1

The comment of Ruy is correct. Since norm convergence is stronger for elements in $E$ than weak convergence, every weakly continuous functional is norm continuous.

For the reverse, if $\psi:E \to \mathbb{C}$ is a norm continuous functional and $a_\alpha \to a$ converge weakly, we have to see that $\psi(a_\alpha) \to \psi(a)$. But that follows from the fact that $a_\alpha$ converge weakly.