We take G=R. Let us define a function Fn∈K(R) by
Fn(x)=(1−x2)nif x∈[−1,1]Fn(x)=0if x∈/[−1,1]. Let
An=∫−1+1Fn(x)dx, and
Gn=An−1Fn. It is immediate that the measures
Gn(x)dx satisfy the conditions of §2, No. 7, Cor. 1 of Lemma 4. Let
μ be a measure on
R whose support is contained in
[−1/2,1/2]. Then
(μ∗Gn)(x)=∫RGn(x−y)dμ(y)=An−1∫−1/21/2Fn(x−y)dμ(y). If
−1/2≤x≤1/2, then
(μ∗Gn)(x)=An−1∫−1/21/2[1−(x−y)2]ndμ(y), therefore
μ∗Gn coincides on
[−1/2,1/2] with a polynomial. In particular, if
f is a continuous function with support contained in
[−1/2,1/2], then
f∗Gn coincides in
[−1/2,1/2] with a polynomial; moreover, by Prop. 5 (iv), and §2, No. 7, Cor. 3 of Lemma 4,
f∗Gn converges uniformly to
f.
If f is of class Cr, the derivatives Ds(f∗Gn) tend uniformly to Dsf for 0≤s≤r.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.