Bourbaki

Example

Tag 09L5, permanent. Cite it as [Bourbaki, Tag 09L5].

Integration, VIII, p. 45, § 4. Convolution of measures and functions

We take G=RG = \mathbf{R}. Let us define a function Fn∈K(R)F_n \in \mathscr{K}(\mathbf{R}) by

Fn(x)=(1−x2)nif x∈[−1,1]Fn(x)=0if x∉[−1,1]. F_n(x) = (1 - x^2)^n \quad \text{if } x \in [-1, 1] \\ F_n(x) = 0 \qquad \text{if } x \notin [-1, 1].
Let An=∫−1+1Fn(x)dxA_n = \int_{-1}^{+1} F_n(x) dx, and Gn=An−1FnG_n = A_n^{-1} F_n. It is immediate that the measures Gn(x)dxG_n(x) dx satisfy the conditions of §2, No. 7, Cor. 1 of Lemma 4. Let μ\mu be a measure on R\mathbf{R} whose support is contained in [−1/2,1/2][-1/2, 1/2]. Then
(μ∗Gn)(x)=∫RGn(x−y)dμ(y)=An−1∫−1/21/2Fn(x−y)dμ(y). (\mu * G_n)(x) = \int_{\mathbf{R}} G_n(x - y) d\mu(y) = A_n^{-1} \int_{-1/2}^{1/2} F_n(x - y) d\mu(y).
If −1/2≤x≤1/2-1/2 \leq x \leq 1/2, then
(μ∗Gn)(x)=An−1∫−1/21/2[1−(x−y)2]ndμ(y), (\mu * G_n)(x) = A_n^{-1} \int_{-1/2}^{1/2} [1 - (x - y)^2]^n d\mu(y),
therefore μ∗Gn\mu * G_n coincides on [−1/2,1/2][-1/2, 1/2] with a polynomial. In particular, if ff is a continuous function with support contained in [−1/2,1/2][-1/2, 1/2], then f∗Gnf * G_n coincides in [−1/2,1/2][-1/2, 1/2] with a polynomial; moreover, by Prop. 5 (iv), and §2, No. 7, Cor. 3 of Lemma 4, f∗Gnf * G_n converges uniformly to ff. If ff is of class CrC^r, the derivatives Ds(f∗Gn)D^s(f * G_n) tend uniformly to DsfD^s f for 0≤s≤r0 \leq s \leq r.

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.