Milman–Pettis theorem

Remarks

Let $J \colon X \to X^{**}$ be the canonical embedding, and let $\xi \in X^{**}$ with $\norm{\xi} = 1$. By the Goldstine theorem, $\xi$ is a weak* limit of a net $J(x_\alpha)$ with $\norm{x_\alpha} \le 1$. Given $\varepsilon > 0$, let $\delta$ be as in the definition of uniform convexity, and choose $f \in X^*$ of norm $1$ with $\operatorname{Re} \xi(f) > 1 - \delta$. For late indices $\alpha, \beta$ one has $\operatorname{Re} f(x_\alpha + x_\beta) > 2 - 2\delta$, so $\norm{(x_\alpha + x_\beta)/2} > 1 - \delta$, and uniform convexity gives $\norm{x_\alpha - x_\beta} < \varepsilon$. This shows that $\xi$ is within any given $\varepsilon$ of the closed set $J(X)$, so $\xi \in J(X)$.

Example: $L^p$ spaces for $1 < p < \infty$ are uniformly convex by Clarkson's inequalities, so they are reflexive. Completeness is needed. The space of finitely supported sequences with the norm of $\ell^2$ is an inner product space, hence uniformly convex, but it is not complete and therefore not reflexive. The converse fails: there are reflexive spaces with no equivalent uniformly convex norm.

Going deeper

For a student

The theorem is the usual quick route to the reflexivity of $L^p$ for $1 < p < \infty$, without computing the dual twice. In a uniformly convex space, weak convergence plus convergence of norms implies norm convergence, and closed convex sets have unique nearest points. These facts and reflexivity make such spaces convenient in variational problems. Being reflexive is invariant under isomorphism, so every Banach space isomorphic to a uniformly convex one is reflexive.