Analysis Problem of the Day 95

Today’s problem appeared as Problem 7 on the Texas A&M August 2021 Analysis Qual:

Problem 7. Let $A \subset C([0,1])$ be a weakly compact subset. Show that $A$ is a norm-compact subset of $L^2([0,1]).$

Solution: By the Eberlein-Smulian theorem, a weakly compact subset of a Banach space is weakly sequentially compact. Thus, any sequence $\{f_n\} \subset A$ has a weakly convergent subsequence $f_{n_k} \to f,$ i.e. $\int f_{n_k}d\nu \to \int f d\nu$ for all Borel probability measures $\nu$ on $[0,1].$ In particular, letting $\nu_t = \delta(x-t)$ for $t \in [0,1],$ we conclude that $f_{n_k} \to f$ pointwise. Moreover, note that since $f_{n_k}$ is a weakly convergent sequence in a Banach space, by the uniform boundedness principle it is uniformly bounded in $C([0,1]).$ Thus, by the dominated convergence theorem, $f_n \to f$ in $L^p([0,1])$ for all $1 \leq p < \infty,$ implying that any sequence in $A$ has a convergent subsequence in $L^2,$ and since compactness equals sequential compactness in metric spaces, we conclude that $A$ is compact in $L^p([0,1])$ for all $1 \leq p < \infty$ (and thus also for $p=2$ ).

Remark: This is equivalent to the statement that the inclusion $i: C([0,1]) \to L^2([0,1])$ being a completely continuous operator, i.e. one sending weakly convergent to norm convergent sequences. In fact, in a reflexive Banach space, an operator is completely continuous if and only if it is compact.

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