Bourbaki

Remark

Tag 07HU, permanent. Cite it as [Bourbaki, Tag 07HU].

Topological Vector Spaces, chapter II, § 6. Weak topologies

— 1) Let (F1,G1),(F2,G2),(F3,G3)(F_1, G_1), (F_2, G_2), (F_3, G_3) be three pairs of spaces in separating duality and consider a sequence of two linear mappings

F1→uF2→vF3 F_1 \xrightarrow{u} F_2 \xrightarrow{v} F_3

that are continuous for the weak topologies corresponding respectively with G1,G2,G3G_1, G_2, G_3; we consider the sequence of transposed mappings

G3→tvG2→tuG1. G_3 \xrightarrow{^tv} G_2 \xrightarrow{^tu} G_1 .

It is clear that t(v∘u)=tu∘tv^t(v \circ u) = ^t u \circ ^t v, therefore the relation v∘u=0v \circ u = 0 is equivalent to tu∘tv=0^t u \circ ^t v = 0. The sequence (5) is exact if, and only if, the three following conditions are satisfied

a) tu∘tv=0^t u \circ ^t v = 0; b) Im⁡(tv)\operatorname{Im}(^t v) is dense in Ker⁡(tu)\operatorname{Ker}(^t u); c) tu^t u is a strict morphism of G2G_2 in G1G_1.

This follows in effect from cor. 3 of II, p. 49 and formulae (3) and (4) of II, p. 47.

2) It must not be thought that when uu is a strict morphism of FF in F1F_1, then tu^t u is necessarily a strict morphism of G1G_1 in GG; in other words uu can be a strict morphism without u(F)u(F) being closed in F1F_1 for σ(F1,G1)\sigma(F_1, G_1). This is shown by the example where FF is a non-closed subspace of F1F_1 and G=G1/F∘G = G_1 / F^\circ, uu being the canonical injection. Similarly, the fact that the sequence (5) is exact does not necessarily imply that (6) is exact, however, if the sequence (5) is exact and if vv is a strict morphism, then the sequence (6) is exact, by the remark 1 and by II, p. 49, cor. 3.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed volume. Not translated.