Bourbaki

Exercise 8, § 3

Topological Vector Spaces, II, p. 73

It is set in small type as supplementary.

a) Let K be a non-discrete valued division ring of which the absolute value is an ultrametric, non linearly compact (cf. CA, III, § 2, exerc. 15); then there exists a well ordered set I of numbers >0> 0 and a family (B(ρ))ρ∈I(B(\rho))_{\rho \in I} of closed balls in K such that the relation ρ<ρ′\rho < \rho' implies B(ρ)⊂B(ρ′)B(\rho) \subset B(\rho'), that B(ρ)B(\rho) has radius ρ\rho and that the intersection of the B(ρ)B(\rho) is empty (CA, VI, § 5, exerc. 5). For every x∈Kx \in K, there exists ρ∈I\rho \in I such that x∉B(ρ)x \notin B(\rho); show that the number ϕ(x)=\phi(x) = ∣x−y∣|x - y| for a y∈B(ρ)a\ y \in B(\rho) depends neither on y∈B(ρ)y \in B(\rho), nor on ρ∈I\rho \in I such that x∉B(ρ)x \notin B(\rho). If ρ∈I\rho \in I is such that x∈B(ρ)x \in B(\rho) then ϕ(x)≤ρ\phi(x) \leq \rho. This being so, for (x1,x2)∈K2(x_1, x_2) \in K^2, put ∥(x1,x2)∥=∣x1∣\|(x_1, x_2)\| = |x_1| if x2=0x_2 = 0, and ∥(x1,x2)∥=∣x2∣ ϕ(x2−1x1)\|(x_1, x_2)\| = |x_2| \ \phi(x_2^{-1} x_1) if x2≠0x_2 \neq 0. Show that ∥(x1,x2)∥\|(x_1, x_2)\| is an ultranorm on K2K^2 (I, p. 26, exerc. 12) and show that there does not exist any projection of norm 1 of K2K^2 on K×{0}K \times \{0\}.

b) Let KK be a complete non-discrete valued division ring of which the absolute value is an ultrametric and which is linearly compact. Let EE be a vector space of dimension 2 on KK with an ultranorm and let DD be a line in EE; show that for all points x∈Ex \in E, there exists y∈Dy \in D such that d(x,D)=d(x,y)=∥x−y∥d(x, D) = d(x, y) = \|x - y\| (note that the intersection of DD and of a ball of centre xx is a ball in DD).

c) Deduce from a) and b) that for a complete non-discrete valued division ring KK, of which the absolute value is an ultrametric, the following properties are equivalent :

α)\alpha) KK is linearly compact.

β)\beta) For every ultranormed vector space EE on KK, for every vector subspace FF of EE and every continuous linear form ff on FF there exists a continuous linear form gg on EE that extends ff and is such that ∥g∥=∥f∥\|g\| = \|f\|. (Reduce to the case where EE is of dimension 2 and use b).) \*

Solution

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Transcribed from the printed volume. Not translated.