Bourbaki

Exercise 15, § 3

Tag 02NT, permanent. Cite it as [Bourbaki, Tag 02NT].

Spectral Theories, II, p. 314

Bourbaki marks this one as harder than the rest.

Let μ\mu denote a Haar measure on G and B\mathfrak{B} the filter of neighbourhoods of ee in G^\widehat{G}.

a) Let U be an integrable open neighbourhood of ee in G. There exists a function a∈L1(G),a⩾0a\in L^1(G),a\geqslant 0, with integral 1, vanishing outside U, such that F(a)∈L1(G^)\mathscr{F}(a)\in L^1(\widehat{G}), and

∫Ga(x)2dx⩽2μ(U) \int_Ga(x)^2dx\leqslant \frac{2}{\mu(U)}
√ \surd

(Let V be a compact neighbourhood of ee such that V⊂U,μ(U)<2‾μ(V)V\subset U,\mu(U)<\overline{2}\mu(V) ; let W be a symmetric compact neighbourhood of ee such that VW2⊂U^2\subset U; take a=λf∗ga=\lambda f*g, where λ\lambda is a constant >0>0, and where f,gf, g are the characteristic functions of the sets VW and W.) b) If f∈L∞(G)f\in L^{\infty}(G), let A(f)A(f) denote the set of elements of G^\widehat{G} which belong to the weakly closed translation-invariant vector subspace of L∞(G)L^{\infty}(G) generated by ff. If f,g∈L∞(G)f, g\in L^{\infty}(G), one has

A(fg)⊂A(f)A(g)‾ A(f g)\subset \overline{A(f)A(g)}

(Let U be a neighbourhood of ee. Now prove that fgf g is the weak limit of linear combinations of elements of G^\widehat{G} belonging to A(fA(f)UA(gg)UU.)

c) Let g∈L∞(G)g\in L^{\infty}(G), K be a compact subset of G^\widehat{G}, ff a function of L1(G)L^1(G) such that F(f)\mathscr{F}(f) vanishes on A=A(g)A = A(g) and outside K. For every V∈BV\in \mathfrak{B}, let ω(V)\omega (V) be the least upper bound of ∣F(f)∣|\mathscr{F}(f)| on AV. If

lim infBω(V)2μ_{\mathfrak{B}}\omega (V)^2\mu((AVμ(V)\mu(V)- AA)∩K\cap K)=0= 0,

then ⟨f,g⟩=0\langle f, g\rangle = 0. (Let ε>0\varepsilon >0. There exists a compact subset H of G such that ∫G−H∣f∣dx⩽ε\int_{G-H}|f|dx\leqslant \varepsilon. Next there exists an open neighbourhood U of ee in G^\widehat{G} such that ∣⟨x,x^⟩−1∣⩽ε|\langle x,\widehat{x}\rangle -1|\leqslant \varepsilon for x∈H,x^∈Ux\in H,\widehat{x}\in U. Applying a) to G^\widehat{G} and U, one obtains a function a∈L1(G^)a\in L^1(\widehat{G}). Let b=F(a)∈L1(G)b=\mathscr{F}(a)\in L^1(G). Using b), now prove that F(bg)\mathscr{F}(bg) vanishes outside AU. On the other hand,

∣∫Gfgdx−∫Gfgbdx∣⩽ε(∥f∥1+2)∥g∥∞ |\int_Gf g dx-\int_Gf gb dx|\leqslant \varepsilon (\|f\|_1+ 2)\|g\|_{\infty}

and, since f,gb∈L2(G)f, gb\in L^2(G),

∫fgbdx=∫F(f)F(gb)dx^=∫F(f)F(gb)dx^ \int f gb dx=\int\mathscr{F}(f)\mathscr{F}(gb)d\widehat{x}=\int\mathscr{F}(f)\mathscr{F}(gb)d\widehat{x}

G G^\widehat{G} (AU-A)∩K\cap K

Majorize this last expression by :

2∥g∥∞ω(U)√μ2\|g\|_{\infty}\omega (U)\surd\mu((AU√μ(U)‾−A\surd\overline{\mu(U)}-A)∩K\cap K)

to conclude.)

d) Let I be a closed ideal of L1(G)L^1(G), A = V(I),f,f a function of L1(G)L^1(G) such that F(f)\mathscr{F}(f) vanishes on A, S the set of points where F(f)\mathscr{F}(f) vanishes, S′S' the boundary of S, B the largest perfect set contained in S∩AS\cap A. For V∈BV\in \mathfrak{B}, let ω(V)\omega (V) be the supremum of ∣F(f)∣|\mathscr{F}(f)| on AV. It is supposed that every point of B possesses a neighbourhood K in G^\widehat{G} such that

lim infBω(V)2μ_{\mathfrak{B}}\omega (V)^2\mu((AVμ(V)\mu(V)- AA)∩K\cap K)=0= 0.

Then f∈If\in I. (Use the fact that L1(G)L^1(G) satisfies Ditkin's condition, and reason as in Prop. 5 of I, p. 94; let G′=G^∪{∞}G'=\widehat{G}\cup \{\infty \} be the Alexandroff compactification of G^\widehat{G}, and let N be the set of χ∈G′\chi \in G' such that F(f)\mathscr{F}(f) does not belong to F(I)\mathscr{F}(I) in a neighbourhood of χ\chi. First now prove that N⊂B∪{∞}N\subset B\cup \{\infty \}. Then, using c), now prove that N⊂{∞}N\subset \{\infty \}. Finally, now prove that N=∅.)N =\emptyset .) e) Let I be a closed ideal of L1(Rn)L^1(\mathbf{R}^n), A = V(I),f,f a function of L1(Rn)L^1(\mathbf{R}^n) such that F(f)\mathscr{F}(f) vanishes on A. For h>0h >0, let AhA_h be the set of points of Rn\mathbf{R}^n outside A and at a distance from A less than hh, and let ω(h)=\omega (h) = supx∈Ah∣F(f)(x)∣_{x\in A_h}|\mathscr{F}(f)(x)|. If

lim infh→0ω(h)2μ(Ah)hn=0_{h\rightarrow 0}\frac{\omega(h)^2\mu(A_h)}{h^n}= 0

then f∈If\in I. (Use d).)

f) Let f∈L1(R),α∈]0,1[f\in L^1(\mathbf{R}),\alpha \in ]0,1[, and suppose ∫R∣y∣α∣f(y)∣dy<+∞\int_{\mathbf{R}}|y|^{\alpha}|f(y)|dy <+\infty. Then there exists a constant kk such that ∣F(f)(x+h)−F(f)(x)∣⩽khα|\mathscr{F}(f)(x+h)-\mathscr{F}(f)(x)|\leqslant kh^{\alpha} for every x∈Rx\in \mathbf{R} and every h∈Rh\in \mathbf{R}.

g) Let I be a closed ideal of L1(R),fL^1(\mathbf{R}),f a function of L1(R)L^1(\mathbf{R}) such that F(f)\mathscr{F}(f) vanishes on V(I). If

∫R∣y1/2f(y)∣dy<+∞ \int_{\mathbf{R}}|y^{1/2}f(y)|dy <+\infty

then one has f∈If\in I. (Use e) and f).)

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Machine translation of the English, by gpt-5.4. Not checked by a person.