Bourbaki

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 μ\mu on X, the intersection X∩u−1(B)X \cap u^{-1}(B) is μ\mu-measurable. Then u is continuous.

First assume that F is the base field. For every compact subset X of E and every measure μ\mu on X, the restriction of uu to X is μ\mu-measurable (INT, IV). Suppose that uu is not continuous. Then we can find a sequence of points (xn)(x_n) in E such that ∑n∥xn∥<∞\sum_n \| x_n \| < \infty and ∣u(xn)∣≥n|u(x_n)| \geq n for every integer nn. Consider the mapping g:(tn)↦∑ntnxng : (t_n) \mapsto \sum_n t_n x_n from the cube C=[0,1]NC = [0, 1]^{\mathbf{N}} into E; it is clear that gg is continuous. Hence f=u∘gf = u \circ g is measurable for every measure on C (INT, V); in particular for the measure μ\mu which is the product of Lebesgue measures on the factors of C. Hence there exists a compact subset D of C such that μ(D)>12\mu(D) > \frac{1}{2} and such that the restriction of ff to D is continuous, hence also bounded. Let M be the upper bound of ∣f∣|f| on D and let p∈Np \in \mathbf{N} be such that p≥4Mp \geq 4M. Let s=(sn)s = (s_n) and t=(tn)t = (t_n) be two points of D such that sn=tns_n = t_n for all n≠pn \neq p. Then

f(s)−f(t)=u(∑nsnxn−∑ntnxn)=(sp−tp)u(xp). f(s) - f(t) = u(\sum_n s_n x_n - \sum_n t_n x_n) = (s_p - t_p) u(x_p) .

Since ∣f(s)−f(t)∣≤2M|f(s) - f(t)| \leq 2M and ∣u(xp)∣≥p≤4M|u(x_p)| \geq p \leq 4M, we get

∣sp−tp∣≤12. |s_p - t_p| \leq \frac{1}{2} .

The Lebesgue-Fubini theorem (INT, V, 2nd ed., § 8, No. 3, cor. 2 of prop. 7) implies that μ(D)≤12\mu(D) \leq \frac{1}{2}; this gives a contradiction. Hence uu is continuous.

In the general case, for every v∈F′v \in F', the linear form v∘uv \circ u is continuous, by the preceding argument. Let (xn)n∈N(x_n)_{n \in \mathbf{N}} be a sequence of points of E tending to 0; then the sequence (u(xn))n∈N(u(x_n))_{n \in \mathbf{N}} tends to 0 in F, if F is assigned the topology σ(F,F′)\sigma(F, F'); hence this sequence is bounded for σ(F,F′)\sigma(F, F') and so it is bounded in F (III, p. 27, cor. 3). Since E is bornological (III, p. 12, prop. 2); the linear mapping u:E→Fu : E \to F is continuous. \*

1^1 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.

Cites

Transcribed from the printed volume. Not translated.