If $A$ is bounded the $\overline{A}^{weak}$ is bounded

Question

Let $E$ normed sapce and $A$ bounded subset of $E$

$\overline{A}^{weak}$ denotes the weak closure of $A$

I just have a small question , does this result always hold , or does it require $E$ to be a Banach space ?

Answer 1

If this result holds for all Banach spaces, it holds for all normed spaces by the following argument: Let $\hat E$ be the completion of $E$. This is a Banach space and every bounded linear functional on $E$ extends uniquely to a bounded linear functional on $\hat E$. In other words, the dual spaces are "the same". In particular, the weak closure of $A$ inside $E$ is contained in the weak closure of $A$ inside $\hat E$. Since boundedness does not depend on the ambient space, you can deduce the boundedness of the weak closure of $A$ inside $E$ from the boundedness of the weak closure of $A$ inside $\hat E$.

But in this case it is probably easier to go through the proof and see that completeness of $E$ is not needed ...