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## 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 $F$.