Bourbaki

Exercise 30, § 1

Topological Vector Spaces, V, p. 67

Let E be a Banach space, E' its strong dual, E'' the strong dual of E', E''' the strong dual of E'', E^{IV} the strong dual of E'''. a) Suppose that E is non-reflexive ; then there exists x′∈E′x' \in E' such that ∥x′∥=1\|x'\| = 1, but that we do not have ⟨x,x′⟩=1\langle x, x' \rangle = 1 for any x∈Ex \in E with ∥x∥=1\|x\| = 1 (IV, p. 57, exerc. 25). On the other hand, there exists a sequence (xn′)(x'_n) of points of E' such that ∥xn′∥=1\|x'_n\| = 1, tending strongly to x′x', and a sequence (xn)(x_n) of points of E such that ∥xn∥=1\|x_n\| = 1 and ⟨xn,xn′⟩=1\langle x_n, x'_n \rangle = 1 for all nn (II, p. 77, exerc. 4). Show that in E'', the sequence (xn)(x_n) does not converge to any point for the topology σ(E′′,E′′′)\sigma(E'', E''') (observe that otherwise it would converge to a point x∈Ex \in E for which ⟨x,x′⟩=1\langle x, x' \rangle = 1). b) Show that it is not possible that x′x' and xn′x'_n are points of smoothness of the unit sphere in E''' (observe that xnx_n, considered as an element of EIVE^{IV} will be the unique element xnIV∈EIVx_n^{IV} \in E^{IV} such that ∥xnIV∥=1\|x_n^{IV}\| = 1 and ⟨xn′,xnIV⟩=1\langle x'_n, x_n^{IV} \rangle = 1 and use exerc. 29). c) Conclude that if E''' is smooth, or if E^{IV} is strictly convex, then E is necessarily reflexive.

Solution

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