Home > Functional Analysis, Topology > Weakly Compact means bounded

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.

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