Suppose I have a finite measure space $(X,\Sigma,\mu)$. I have doubts about the following statement:
In a reflexive Banach space weakly compact set implies the norm compact. Is the following statement true?
I have read that the above is true now my doubt is if $X$ is finite-dimensional and if will the above result holds for the $L^p(X)$ case?
I don't think the statement is true.
Take a infinite dimensional reflexive Banach-Space $X$ (for example $L^p(X), 1 < p < \infty$). Now look at the closed unit ball $\overline{B_1(0)} \subset X$.
1.) $\overline{B_1(0)}$ is not norm compact.
Proof:
This is characteristic for an infinte dimensional Banach-space, indeed the following statement is true (Riesz Lemma):
$$\text{dim} \ X = \infty \iff \overline{B_1(0)} \text{ is not norm compact.}$$
2.) $\overline{B_1(0)}$ is weakly compact.
Proof:
Take a sequence $\{x_n\}_{n=0}^{\infty} \subset \overline{B_1(0)}$. Clearly $\{x_n\}_{n=0}^{\infty} $ is bounded, and thus, because $X$ is reflexive, has a weakly converging subsequence $\{x_{n'}\}_{n'=0}^{\infty} \rightharpoonup x \in X$. Because any convex and closed set in $X$ is also weakly sequential closed it follows that $x \in \overline{B_1(0)}$. So $\overline{B_1(0)}$ is weakly sequential compact.
(Eberlein-Šmulian theorem tells you that weakly sequential compact and weakly compact are equivalent in reflexive Banach-spaces).