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 denote a Haar measure on G and B the filter of neighbourhoods of e in G.
a) Let U be an integrable open neighbourhood of e in G. There exists a function a∈L1(G),a⩾0, with integral 1, vanishing outside U, such that F(a)∈L1(G), and
∫Ga(x)2dx⩽μ(U)2
√
(Let V be a compact neighbourhood of e such that V⊂U,μ(U)<2μ(V) ; let W be a symmetric compact neighbourhood of e such that VW2⊂U; take a=λf∗g, where λ is a constant >0, and where f,g are the characteristic functions of the sets VW and W.) b) If f∈L∞(G), let A(f) denote the set of elements of G which belong to the weakly closed translation-invariant vector subspace of L∞(G) generated by f. If f,g∈L∞(G), one has
A(fg)⊂A(f)A(g)
(Let U be a neighbourhood of e. Now prove that fg is the weak limit of linear combinations of elements of G belonging to A(f)UA(g)U.)
c) Let g∈L∞(G), K be a compact subset of G, f a function of L1(G) such that F(f) vanishes on A=A(g) and outside K. For every V∈B, let ω(V) be the least upper bound of ∣F(f)∣ on AV. If
lim infBω(V)2μ((AVμ(V)- A)∩K)=0,
then ⟨f,g⟩=0. (Let ε>0. There exists a compact subset H of G such that ∫G−H∣f∣dx⩽ε. Next there exists an open neighbourhood U of e in G such that ∣⟨x,x⟩−1∣⩽ε for x∈H,x∈U. Applying a) to G and U, one obtains a function a∈L1(G). Let b=F(a)∈L1(G). Using b), now prove that F(bg) vanishes outside AU. On the other hand,
∣∫Gfgdx−∫Gfgbdx∣⩽ε(∥f∥1+2)∥g∥∞
and, since f,gb∈L2(G),
∫fgbdx=∫F(f)F(gb)dx=∫F(f)F(gb)dx
G G (AU-A)∩K
Majorize this last expression by :
2∥g∥∞ω(U)√μ((AU√μ(U)−A)∩K)
to conclude.)
d) Let I be a closed ideal of L1(G), A = V(I),f a function of L1(G) such that F(f) vanishes on A, S the set of points where F(f) vanishes, S′ the boundary of S, B the largest perfect set contained in S∩A. For V∈B, let ω(V) be the supremum of ∣F(f)∣ on AV. It is supposed that every point of B possesses a neighbourhood K in G such that
lim infBω(V)2μ((AVμ(V)- A)∩K)=0.
Then f∈I. (Use the fact that L1(G) satisfies Ditkin's condition, and reason as in Prop. 5 of I, p. 94; let G′=G∪{∞} be the Alexandroff compactification of G, and let N be the set of χ∈G′ such that F(f) does not belong to F(I) in a neighbourhood of χ. First now prove that N⊂B∪{∞}. Then, using c), now prove that N⊂{∞}. Finally, now prove that N=∅.) e) Let I be a closed ideal of L1(Rn), A = V(I),f a function of L1(Rn) such that F(f) vanishes on A. For h>0, let Ah be the set of points of Rn outside A and at a distance from A less than h, and let ω(h)= supx∈Ah∣F(f)(x)∣. If
lim infh→0hnω(h)2μ(Ah)=0
then f∈I. (Use d).)
f) Let f∈L1(R),α∈]0,1[, and suppose ∫R∣y∣α∣f(y)∣dy<+∞. Then there exists a constant k such that ∣F(f)(x+h)−F(f)(x)∣⩽khα for every x∈R and every h∈R.
g) Let I be a closed ideal of L1(R),f a function of L1(R) such that F(f) vanishes on V(I). If