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## Weakly Compact means bounded

Let $E$ be a Banach space, and $A \subset E$ be a compact set in the weak topology. Then $A$ is bounded.

Proof: The result becomes trivial after proving the following lemma:

Lemma: For a Banach space $G$ and one of its subsets $B$, if we have $f(B)$ is bounded in $\Bbb{R} (\Bbb{C})$ for any linear functional $f \in G^*$ then $B$ is also bounded.

Since all $f \in E^*$ are continuous with respect to the weak topology, and $A$ is a compact set, it follows that $f(A)$ is compact and therefore bounded for all $f \in E^*$. Therefore $A$ is bounded.

Let’s prove the given lemma. We shall use Banach-Steinhaus uniform boundedness principle. Define

$T_b(f)=\langle f,b \rangle,\ T_b : E^* \to \Bbb{R}$. By the hypothesis of the lemma, we get $\sup_{b \in B}|T_b(f)|<\infty,\ \forall f \in E$. Next, using uniform boundedness principle there exists a constant $c$ such that $\|T_b\| \leq c,\ \forall b \in B$. But this means that $|\langle f,b \rangle | \leq c \|f\|,\ \forall b \in B$. Taking supremum over $f$ with $\|f\| \leq 1$, we get that $\|b\| \leq c$ which means that $B$ is bounded. This ends the proof of the lemma, and the finishes the proof.

The dual of the lemma is also true. Namely

Lemma 2: Let $G$ be a Banach space and $B^* \subset G^*$ one of its subsets. Assume that for every $x \in G$ the set $\langle B^*,x \rangle=\{ \langle f,x \rangle : f \in B^*\}$ is bounded (in $\Bbb{R}$). Then $B^*$ is also bounded.

The two lemmas are both corollaries to the Banach Steinhaus Principle, and I’ve presented them as I found them in Haim Brezis’ book, Functional Analysis, Sobolev spaces and Partial Differential Equations, Springer 2011, an excellent book, by the way.