Then fε is continuous and bounded, and fε converges to f in Cb(R) when ε tends to 0.
The continuity of fε is a consequence of INT, IV, p. 144, § 4, no3, cor. 1, since f has compact support. Moreover, if r is such that the support of f is contained in [−r,r]×R, one has
∣fε(x)∣⩽2∥f∥∞∫−rr√(x−y)2+ε21dy⩽ε4r∥f∥∞
therefore fε is bounded.
The Taylor expansion to order 1 (FVR, I, p. 30, prop. 3) and formula (13) prove that there exist M⩾0 and a function ϱ1 on R2 such that
f(y+iγ)=f(y)+iγf′(y)+γ2ϱ1(y;γ), and ∣ϱ1(y;γ)∣⩽M,
for every y∈R and γ∈R. Since the mapping y↦f(y+iγ) has support contained in [−r,r] for every γ∈R, the mapping y↦ϱ1(y;γ) has support contained in [−r,r], for every γ∈R.
therefore F(f′)F(σε) converges to 0 in L1(R) when ε tends to 0. Since f′ and σε belong to L2(R), Prop. 14 of II, p. 223 implies that f′∗σε=F(F(f′)F(σε)) converges to 0 in Cb(R).
For every ε>0, the positive measure δε⋅dx on R has total mass 1 (cf. FVR, III, p. 7). The set of measures δε⋅dx for ε>0 and the filter induced on this set by the filter of neighbourhoods of 0 in R+∗ satisfy the hypotheses of Lemma 4 of INT, VIII, p. 137, § 2, no7. The measures δε⋅dx therefore converge in M1(R) to the point measure ε0 when ε tends to 0. There follows f∗δε→f in Cb(R) (INT, VIII, p. 163, § 4, no4). The lemma is proved.
Let μ denote the Lebesgue measure on C.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.