Proposition 2
Tag 07QB, permanent. Cite it as [Bourbaki, Tag 07QB].
Topological Vector Spaces, chapter III, § 6. Borel's graph theorem
— Let E be a Banach space, F a locally convex space and u a linear mapping from E into F. Assume that for every closed subset B of F, every compact subset X of E and every measure on X, the intersection is -measurable. Then u is continuous.
First assume that F is the base field. For every compact subset X of E and every measure on X, the restriction of to X is -measurable (INT, IV). Suppose that is not continuous. Then we can find a sequence of points in E such that and for every integer . Consider the mapping from the cube into E; it is clear that is continuous. Hence is measurable for every measure on C (INT, V); in particular for the measure which is the product of Lebesgue measures on the factors of C. Hence there exists a compact subset D of C such that and such that the restriction of to D is continuous, hence also bounded. Let M be the upper bound of on D and let be such that . Let and be two points of D such that for all . Then
Since and , we get
The Lebesgue-Fubini theorem (INT, V, 2nd ed., § 8, No. 3, cor. 2 of prop. 7) implies that ; this gives a contradiction. Hence is continuous.
In the general case, for every , the linear form is continuous, by the preceding argument. Let be a sequence of points of E tending to 0; then the sequence tends to 0 in F, if F is assigned the topology ; hence this sequence is bounded for and so it is bounded in F (III, p. 27, cor. 3). Since E is bornological (III, p. 12, prop. 2); the linear mapping is continuous. \*
The results of this section depend on the book of Integration.
Exercises
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Cited by
Nothing.