Weakly compact set is bounded in norm

Question

I'm doing Ex 3.1 in Brezis's book of Functional Analysis.

Let $(E, | \cdot |)$ be a normed space and let $A \subset E$ be a subset that is compact in the weak topology $\sigma\left(E, E'\right)$. Prove that $A$ is bounded.

I have found a proof by Principle of Uniform Boundedness here. I'm lucky to find my own proof :v

Could you have a check on my attempt?

I posted my proof separately so that I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

Answer 1

One can prove a stronger result without resorting to the UBP (but an appeal to Baire's category theorem is needed): let $X$ denote a lc tvs and let $K$ be a $\sigma(X,X')$-compact in $X$. Then $K$ is bounded in the original topology (let us call it $\tau$)

To prove the result, we proceed as follows: Given $V$ a $\tau-$neighbourhood of $0$, let $I$ be a absolutely convex $\tau$-neighbourhood of $0$ such that $\overline{I}\subset V$. By the bipolar theorem, $(I^\circ)_{\circ}=\overline{I}$. Since $K$ is $\sigma(X,X')$ compact every linear functional is bounded on $K$. Moreover $I^\circ$ is $\sigma(X',X)$ compact by Banach-Alaoglu so an application of the Baire category theorem implies that $$M:=\sup_{f\in I^\circ} |f(K)|<\infty$$ so now we must have $$\frac 1M K\subset {I^\circ}_\circ=\overline I\subset V$$ proving the claim.

Answer 2

We have $| \cdot|$ is l.s.c. in norm topology and convex, so $| \cdot|$ is l.s.c. in weak topology. Then $|\cdot|$ is bounded on the weakly compact set $A$.