We have a result that says that the space formed by the linear continuous functionals in $E$ with the weak topology coincides with the topological dual of $E$, that is, $E' = (E,\, \sigma(E,\, E'))'$ for every normed space $E$, but the dual of a normed space is always Banach, then of equality $(E,\, \sigma(E,\, E'))'$ is Banach. If I understood this incorrectly (correct me please), otherwise, how could I show that $(E,\, \sigma(E,\, E'))'$ is Banach without knowing equality $E' = (E,\, \sigma(E,\, E'))'$? This is very confusing to me, it is possible to induce a norm in $(E,\, \sigma(E,\, E'))'$?
2026-04-03 14:05:42.1775225142
A doubt about the weak topology, $(E,\, \sigma(E,\, E'))'$ is Banach for all $E$ normed space?
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Because $(E,\sigma(E,E'))'=E'$, you can obviously put a norm on it. But such norm has nothing to do with the topological structure of $(E,\sigma(E,E'))$, which is not a normed space.