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## Compact operator maps weakly convergent sequences into strong convergent sequences

Suppose is a compact operator and is a sequence in such that (i.e. converges weakly). Prove that strongly in .

**Proof: **Pick a continuous linear functional on . Then is a continuous linear functional on , and . Therefore, since was an arbitrary linear functional, it follows that weakly in .

Since is weakly convergent, it follows that it is bounded, and therefore, by compactness of , for any subsequence the sequence has a convergent subsequence (strongly) in . Since , that convergent subsequence must converge to . Therefore, we have the following property for :

Every subsequence of contains a convergent subsequence with limit equal to .

This is enough to conclude that strongly in .

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Categories: Functional Analysis
compact, functional, operators

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