Bourbaki

Lemma 7

Tag 0369, permanent. Cite it as [Bourbaki, Tag 0369].

Théories spectrales, chapter IV, § 5. Opérateurs partiels normaux et théorème spectral

Let f∈D(R)f\in \mathscr{D}(\mathbf{R}) and let f~∈D(C)\widetilde{f}\in \mathscr{D}(\mathbf{C}) be an almost analytic extension of ff. For ε>0\varepsilon >0, define fεf_{\varepsilon} on R\mathbf{R} by

fε(x)=−12iπ∫R(f~(y+iε)y−x+iε−f~(y−iε)y−x−iε)dy f_{\varepsilon}(x) =-\frac{1}{2i\pi}\int_{\mathbf{R}}(\frac{\widetilde{f}(y + i\varepsilon)}{y-x + i\varepsilon}-\frac{\widetilde{f}(y-i\varepsilon)}{y-x-i\varepsilon})dy

Then fεf_{\varepsilon} is continuous and bounded, and fεf_{\varepsilon} converges to ff in Cb(R)\mathscr{C}_b(\mathbf{R}) when ε\varepsilon tends to 0.

The continuity of fεf_{\varepsilon} is a consequence of INT, IV, p. 144, § 4, no3^o3, cor. 1, since f~\widetilde{f} has compact support. Moreover, if rr is such that the support of f~\widetilde{f} is contained in [−r,r]×R[-r, r]\times \mathbf{R}, one has

∣fε(x)∣⩽2∥f~∥∞∫−rr1√(x−y)2+ε2dy⩽4rε∥f~∥∞ |f_{\varepsilon}(x)|\leqslant 2\|\widetilde{f}\|_{\infty}\int_{-r}^r\frac{1}{\surd(x-y)^2 + \varepsilon^2}dy\leqslant \frac{4r}{\varepsilon}\|\widetilde{f}\|_{\infty}

therefore fεf_{\varepsilon} is bounded.

The Taylor expansion to order 1 (FVR, I, p. 30, prop. 3) and formula (13) prove that there exist M⩾0M\geqslant 0 and a function ϱ1\varrho_1 on R2\mathbf{R}^2 such that

f~(y+iγ)=f(y)+iγf′(y)+γ2ϱ1(y;γ)\widetilde{f}(y+i\gamma ) =f(y) +i\gamma f'(y) +\gamma^2\varrho_1(y;\gamma ), and ∣ϱ1(y;γ)∣⩽M|\varrho_1(y;\gamma )|\leqslant M,

for every y∈Ry\in \mathbf{R} and γ∈R\gamma \in \mathbf{R}. Since the mapping y↦f~(y+iγ)y\mapsto \widetilde{f}(y+i\gamma ) has support contained in [−r,r][-r, r] for every γ∈R\gamma \in \mathbf{R}, the mapping y↦ϱ1(y;γ)y\mapsto \varrho_1(y;\gamma ) has support contained in [−r,r][-r, r], for every γ∈R\gamma \in \mathbf{R}.

Let gεg_{\varepsilon} be the function defined on R2\mathbf{R}^2 by

gε(x,y)=f~(y+iε)y−x+iε−f~(y−iε)y−x−iε g_{\varepsilon}(x, y) =\frac{\widetilde{f}(y + i\varepsilon)}{y-x + i\varepsilon}-\frac{\widetilde{f}(y-i\varepsilon)}{y-x-i\varepsilon}

For every (x,y)∈R2(x, y)\in \mathbf{R}^2 and ε>0\varepsilon >0, one obtains

gε(x,y)=−2iε(x−y)2+ε2f(y)+2iε(y−x)(x−y)2+ε2f′(y) g_{\varepsilon}(x, y) =-\frac{2i\varepsilon}{(x-y)^2 + \varepsilon^2}f(y) +\frac{2i\varepsilon(y-x)}{(x-y)^2 + \varepsilon^2}f'(y)
+ε2(ϱ1(y;ε)y−x+iε−ϱ1(y;−ε)y−x−iε) +\varepsilon^2(\frac{\varrho_1(y; \varepsilon)}{y-x + i\varepsilon}-\frac{\varrho_1(y;-\varepsilon)}{y-x-i\varepsilon})

Let x∈Rx\in \mathbf{R} and ε>0\varepsilon >0. We have

∣ε2∫R(ϱ1(y;ε)y−x+iε−ϱ1(y;−ε)y−x−iε)dy∣⩽2Mε2∫−rrdy√(x−y)2+ε2 |\varepsilon^2\int_{\mathbf{R}}(\frac{\varrho_1(y;\varepsilon)}{y-x + i\varepsilon}-\frac{\varrho_1(y;-\varepsilon)}{y-x-i\varepsilon})dy|\leqslant 2M\varepsilon^2\int_{-r}^r\frac{dy}{\surd(x-y)^2 + \varepsilon^2}
⩽4Mrε \leqslant 4Mr\varepsilon

Consequently, for every x∈Rx\in \mathbf{R}, there follows

fε(x)=−12iπ∫Rgε(x,y)dy=(f∗δε)(x)−12iπ(f′∗σε)(x)+kε(x) f_{\varepsilon}(x) =-\frac{1}{2i\pi}\int_{\mathbf{R}}g_{\varepsilon}(x, y)dy= (f*\delta_{\varepsilon})(x)-\frac{1}{2i\pi}(f'*\sigma_{\varepsilon})(x) +k_{\varepsilon}(x)

where δε\delta_{\varepsilon} and σε\sigma_{\varepsilon} are the functions on R\mathbf{R} defined by

δε(x)=1πεx2+ε2,σε(x)=2iεxx2+ε2 \delta_{\varepsilon}(x) =\frac{1}{\pi}\frac{\varepsilon}{x^2 + \varepsilon^2},\sigma_{\varepsilon}(x) =\frac{2i\varepsilon x}{x^2 + \varepsilon^2}

and ∥kε∥∞⩽2Mrε\|k_{\varepsilon}\|_{\infty}\leqslant 2Mr\varepsilon.

The function F(f′)∈S(R)\mathscr{F}(f')\in \mathscr{S}(\mathbf{R}) is integrable (Prop. 18 of IV, p. 219 and Prop. 13 of IV, p. 213). By Lemma 6, there follows

∫R∣F(f′)(y)F(σε)(y)∣dy=2πε∫R∣F(f′)(y)∣e−2πε∣y∣dy \int_{\mathbf{R}}|\mathscr{F}(f')(y)\mathscr{F}(\sigma_{\varepsilon})(y)|dy= 2\pi \varepsilon \int_{\mathbf{R}}|\mathscr{F}(f')(y)|e^{-2\pi \varepsilon|y|}dy
⩽2πε∫R∣F(f′)(y)∣dy \leqslant 2\pi \varepsilon \int_{\mathbf{R}}|\mathscr{F}(f')(y)|dy

therefore F(f′)F(σε)\mathscr{F}(f')\mathscr{F}(\sigma_{\varepsilon}) converges to 0 in L1(R)L^1(\mathbf{R}) when ε\varepsilon tends to 0. Since f′f' and σε\sigma_{\varepsilon} belong to L2(R)L^2(\mathbf{R}), Prop. 14 of II, p. 223 implies that f′∗σε=F‾(F(f′)F(σε))f'*\sigma_{\varepsilon}=\overline{\mathscr{F}}(\mathscr{F}(f')\mathscr{F}(\sigma_{\varepsilon})) converges to 0 in Cb(R)\mathscr{C}_b(\mathbf{R}).

For every ε>0\varepsilon >0, the positive measure δε⋅dx\delta_{\varepsilon}\cdot dx on R\mathbf{R} has total mass 1 (cf. FVR, III, p. 7). The set of measures δε⋅dx\delta_{\varepsilon}\cdot dx for ε>0\varepsilon >0 and the filter induced on this set by the filter of neighbourhoods of 0 in R+∗\mathbf{R}_+^* satisfy the hypotheses of Lemma 4 of INT, VIII, p. 137, § 2, no7^o7. The measures δε⋅dx\delta_{\varepsilon}\cdot dx therefore converge in M1(R)\mathscr{M}^1(\mathbf{R}) to the point measure ε0\varepsilon_0 when ε\varepsilon tends to 0. There follows f∗δε→ff*\delta_{\varepsilon}\rightarrow f in Cb(R)\mathscr{C}_b(\mathbf{R}) (INT, VIII, p. 163, § 4, no4^o4). The lemma is proved.

Let μ\mu denote the Lebesgue measure on C\mathbf{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.

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.