## Weakly Compact means bounded

Let be a Banach space, and be a compact set in the weak topology. Then is bounded.

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

Lemma:For a Banach space and one of its subsets , if we have is bounded in for any linear functional then is also bounded.

Since all are continuous with respect to the weak topology, and is a compact set, it follows that is compact and therefore bounded for all . Therefore is bounded.

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

. By the hypothesis of the lemma, we get . Next, using uniform boundedness principle there exists a constant such that . But this means that . Taking supremum over with , we get that which means that 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 be a Banach space and one of its subsets. Assume that for every the set is bounded (in ). Then 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.