Compact set in the weak topology

Question

Let $A \subset E$ be a subset of a Banach space $E$. I wonder if it is true that if $A$ is a compact set with the weak topology $\sigma(E, E')$, then $A$ is bounded.

Answer 1

Since $A$ is compact, it follows that, for each $f \in E'$, $\{f(x) \, \mid \, x \in A\}$ is bounded. (Why?)

Let $J : E \to (E')'$ be the canonical map so that $[J(x)](f) = f(x)$. Recall that $J$ is an isometry, which is a fancy way of saying $\|J(x)\| = \|x\|$.

By the first paragraph, $\{J(x) \, \mid \, x \in A\}$ is pointwise bounded. The Banach-Steinhaus Theorem (or Uniform Boundedness Principle) implies $\{J(x) \, \mid \, x \in A\}$ is actually bounded in $(E')'$. In view of the second paragraph, that means $A$ is bounded.