Exercise 31, § 1

It is said that a normed space $E$ (real or complex) is uniformly convex if, for every $\varepsilon$ such that $0 < \varepsilon < 2$, there exists $\delta > 0$ such that the relations $\|x\| \leqslant 1, \|y\| \leqslant 1, \|x - y\| \geqslant \varepsilon$ in $E$ imply $\|\frac{1}{2}(x + y)\| \leqslant 1 - \delta$. A uniformly convex space is strictly convex (V, p. 66, exerc. 27). It is said that $E$ is uniformly smooth if, for every $\varepsilon > 0$, there exists $\eta > 0$ such that the relations $\|x\| \geqslant 1, \|y\| \geqslant 1, \|x - y\| \leqslant \eta$ entail the inequality $\|x + y\| \geqslant \|x\| + \|y\| - \varepsilon \|x - y\|$. It is equivalent to saying that, for every $\varepsilon > 0$, there exists $\rho > 0$ such that the relations $\|x\| = 1, \|y\| \leqslant \rho$ entail the inequality

A uniformly smooth space is smooth (V, p. 65, exerc. 26). a) Now prove that if $E$ is uniformly convex, its strong dual $E'$ is uniformly smooth, and that if $E$ is uniformly smooth, $E'$ is uniformly convex; the restriction to the unit sphere $S$ of E of the unique mapping support (V, p. 66, exerc. 29) is then a mapping from S into the unit sphere S’ of E’, continuous for the topologies of E and E’ deduced from the norm. b) Now prove that if E is uniformly convex, and if a filter $\mathfrak{F}$ on E converges to $x_0$ for the topology $\sigma(E, E')$ and is such that $\lim_{\mathfrak{F}} \|x\| = \|x_0\|$, $\mathfrak{F}$ converges to $x_0$ for the initial topology of E. c) Now prove that a Banach space which is uniformly convex or uniformly smooth is reflexive (use b) and c), as well as IV, p. 61, exerc. 12). (Cf. V, p. 71, exerc. 14.) d) Generalize to uniformly convex Banach spaces the first part of Theorem 1 of V, p. 10, as well as Cor. 1 and 2 of V, p. 11.