Bourbaki

Exercise 19, § 1

Topological Vector Spaces, V, p. 65

Bourbaki marks this one as harder than the rest.

Let EE be an infinite dimensional complex normed space. Suppose that there exists a bijective mapping ω\omega from the set M\mathfrak{M} of closed vector subspaces of EE onto itself, having the properties listed in exerc. 18. a) Show that there exists a semi-linear mapping uu from EE onto its dual E′E' (for the automorphism ξ↦ξ‾\xi \mapsto \overline{\xi} of C\mathbf{C}) defined up to a scalar factor and such that u(M)=(ω(M))∘u(M) = (\omega(M))^{\circ} for every M∈MM \in \mathfrak{M}. (Proceed as in exerc. 18; using IV, p. 65, exerc. 16, show that uu is a semi-linear mapping relative to the identity automorphism of C\mathbf{C} or to the automorphism ξ↦ξ‾\xi \mapsto \overline{\xi}; finally prove that the first case cannot occur since ⟨x,u(x)⟩≠0\langle x, u(x) \rangle \neq 0 for x≠0x \neq 0.) b) If we put ⟨y∣x⟩=⟨x,u(y)⟩\langle y|x \rangle = \langle x, u(y) \rangle, show that ⟨x∣y⟩=⟨y∣x⟩\langle x|y \rangle = \langle y|x \rangle and that ⟨x∣x⟩\langle x|x \rangle has the same sign on the set of all x≠0x \neq 0 (same method as in exerc. 18). c) Finally show that the topology defined by the norm ⟨x∣x⟩1/2\langle x|x \rangle^{1/2} is finer than the initial topology T0\mathscr{T}_0 on EE, and that these two topologies are identical when EE is complete for T0\mathscr{T}_0; in the latter case, ω(M)\omega(M) is the orthogonal complement of MM in the hilbertian space EE.

Solution

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Transcribed from the printed volume. Not translated.