Home > Functional Analysis > Compact operator maps weakly convergent sequences into strong convergent sequences

Compact operator maps weakly convergent sequences into strong convergent sequences


Suppose T \in \mathcal{L}(E,F) is a compact operator and (u_n) is a sequence in E such that u \rightharpoonup u (i.e. converges weakly). Prove that Tu_n \to Tu strongly in F.

Proof: Pick f a continuous linear functional on F. Then f \circ T is a continuous linear functional on E, and \langle f , Tu_n \rangle \to \langle f, Tu \rangle. Therefore, since f was an arbitrary linear functional, it follows that Tu_n \rightharpoonup Tu weakly in F.

Since (u_n) is weakly convergent, it follows that it is bounded, and therefore, by compactness of T, for any subsequence (u_{n_k}) the sequence (Tu_{n_k} has a convergent subsequence (strongly) in F. Since Tu_{n_k} \rightharpoonup Tu, that convergent subsequence must converge to Tu. Therefore, we have the following property for (Tu_n):

Every subsequence of (Tu_n) contains a convergent subsequence with limit equal to Tu.

This is enough to conclude that Tu_n \to Tu strongly in F.

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