A doubt about the weak topology, $(E,\, \sigma(E,\, E'))'$ is Banach for all $E$ normed space?

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 $$(E,\, \sigma(E,\, E'))'$$?

• Another way to approach. If $E$ is a normed space, and $f$ is a functional on $E$, then $f$ is continuous with respect to the norm topology if and only if it is continuous with respect to the weak [i.e. $\sigma(E,E')$] topology. Apr 15, 2021 at 19:43

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.
• I still don't understand it well, look, supposing that you only have this set $(E, \, \sigma(E, \, E'))'$, the question is, would you know how to justify that he is Banach without knowing $E' = (E, \, \sigma(E, \, E'))'$? My question arises exactly from what you said, the weak topology is generally not metrizable. Apr 15, 2021 at 19:14
• Not sure what the problem is. $(E,\sigma(E,E'))$ is not a normed space. So there is no reason to expect that the dual will be normed. What are you aiming for? Apr 15, 2021 at 20:00
• I think I understand what I'm missing, the point is that $(E, \sigma(E, E ′)) '$ is just a notation to indicate the functionals that are continuous when $E$ is equipped with a weak topology, this does not mean that is taking the topological dual of $E$ with the weak topology, because as you said, with the weak topology the space is not normed. Apr 15, 2021 at 21:55
• No, it's the opposite. $F=(E,\sigma(E,E'))$ is a topological vector space (more specifically it's a locally convex space). It's dual $F'$ is the set of continuous linear functionals on $F$. The theorem you mentioned says that the elements of $F'$ are exactly the elements of $E'$. Apr 15, 2021 at 22:35