Theorem 2
Tag 07Q8, permanent. Cite it as [Bourbaki, Tag 07Q8].
Topological Vector Spaces, chapter III, § 6. Borel's graph theorem
— Let E be a locally convex space, which is the inductive limit of an increasing sequence of subspaces of E, endowed with the topologies of Fréchet spaces satisfying the first axiom of countability. Suppose that every compact subset of E is contained in one of the and is compact in this space. Let F be a Fréchet space satisfying the first axiom of countability. Then the space is a Lusin space.
The space E is bornological (III, p. 12), hence the space is complete (III, p. 23, prop. 12). The linear mapping is an injection from into the product space ; by virtue of the hypothesis on the compact subsets of E and the definition of the -topologies, is an isomorphism from onto its image (endowed with the topology induced by the product topology); moreover, since is complete, this image is a closed subspace of (GT, II, § 3, No. 4, prop. 8). By GT, IX, § 6, No. 4, it is therefore enough to prove that each of the spaces is a Lusin space. For the rest of the proof, we shall assume that E is a Fréchet space satisfying the first axiom of countability.
Since F is a Fréchet space satisfying the first axiom of countability, it is isomorphic to a closed subspace of a countable product of Banach spaces , each of which is a quotient of F (II, p. 5), hence satisfies the first axiom of countability. The linear mapping is an injection from into the product space , and by using the definition of the -topologies and of the open sets in a product, is an isomorphism from onto its image; moreover, since is complete, this image is a closed subspace of . Therefore it is enough to prove that each of the spaces is a Lusin space (GT, IX, § 6, No. 4), and consequently, we can assume that F is a Banach space satisfying the first axiom of countability.
The space is the union of a countable family of equicontinuous and closed subsets (III, p. 19, cor. 1 and GT, X, § 2, No. 3, prop. 6). But every equicontinuous subset H of is metrizable and satisfies the first axiom of countability (III, p. 18, prop. 6 and GT, X, § 2, No. 4, th. 1); if H is closed, then it is a complete space for the uniform structure induced by that of , since the latter is complete In other words, H is a polish space, and a fortiori a Lusin space; consequently the regular space is a Lusin space (GT, IX, § 6, No. 7, cor. of th. 3).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Cited by
Nothing.
Cites
- IX, § 6, No. 4 not yet on this site
- IX, § 6, No. 7 not yet on this site
- X, § 2, No. 3 not yet on this site
- X, § 2, No. 4 not yet on this site
- II, § 3, No. 4
- III, p. 12
- III, p. 18
- III, p. 19