Bourbaki

Proposition 1

Tag 01GZ, permanent. Cite it as [Bourbaki, Tag 01GZ].

Lie Groups and Lie Algebras, chapter IX, § 8. Fourier transform

Every square integrable complex function ff on G is the sum in the Hilbert space L2(G)L^2(G) of the family (fu)u∈G^(f_u)_{u\in\widehat{G}} where, for all h∈Gh\in G and all u∈G^u\in \widehat{G},

fu(h)=⟨u(h)∣u(f)⟩ f_u(h) =\langle u(h)|u(f)\rangle

=d(u)∫Gf(g=d(u)\int_Gf(g)Tr(u(gh−1)u(gh^{-1}))dg=d(u)∫Gf(ghdg=d(u)\int_Gf(gh)Tr(u(g)u(g))dgdg. (10)

For all u∈G^u\in \widehat{G} choose an orthonormal basis BuBu of EuE_u, and denote by (uij(g))(u_{ij}(g)) the matrix of u(g)u(g) in this basis. Prop. 1 also means that the family of functions √d(u)‾uij\surd\overline{d(u)}u_{ij}, for uu in G^\widehat{G} and i,ji, j in BuB_u, is an orthonormal basis of the space L2(G)L^2(G).

If ff is an integrable function on G such that the family (fu)(f_u) is uniformly summable, then the sum of this family is a continuous function which coincides almost everywhere with ff; in other words, if we assume in addition that ff is continuous, then for all h∈Gh\in G, f(h)=∑u∈G^d(u)∫Gf(ghf(h) =\sum_{u\in\widehat{G}}d(u)\int_Gf(gh)Tr(u(g)u(g))dgdg. (11)

Conversely, let A∈F(G^)A\in F( \widehat{G}); if the family (FuA)u∈G^(\mathscr{F}_uA)_{u\in\widehat{G}} is uniformly summable, the function

g→∑u∈G^(FuA)(g)=∑u∈G^d(ug \rightarrow \sum_{u\in\widehat{G}}(\mathscr{F}_uA)(g) =\sum_{u\in\widehat{G}}d(u)Tr(Auu(g)−1_uu(g)^{-1})

is a continuous function on G whose Fourier cotransform is A.

Let ff be an integrable function on G, and let s∈Gs\in G. Denote by γ(s)f\gamma (s)f and δ(s)f\delta (s)f the functions on G defined by γ(s)f=εs∗f,δ(s)f=f∗εs−1\gamma (s)f=\varepsilon_s*f,\delta (s)f=f*\varepsilon_{s^{-1}}, that is,

(γ(s)f)(g)=f(s−1g),(δ(s)f)(g)=f(gs)(\gamma (s)f)(g) =f(s^{-1}g),(\delta (s)f)(g) =f(gs) for g∈Gg\in G,

(Chap. III, §3, no. 4 and Integration, Chap. VII, §1, no. 1). We have

u(γ(s)f)=∫Gf(s−1g)u(g)dg=∫Gf(g)u(sg)dg u(\gamma (s)f) =\int_Gf(s^{-1}g)u(g)dg=\int_Gf(g)u(sg)dg

so

u(γ(s)f)=u(s)u(f)(12) u(\gamma (s)f) =u(s)u(f) \tag{12}

and similarly

u(δ(s−1)f)=u(f)u(s)(13) u(\delta (s^{-1})f) =u(f)u(s) \tag{13}

When G is commutative, G^\widehat{G} is the underlying set of the dual group of G (Spectral Theories, Chap. II, §1, no. 11),d(u)=1,d(u) = 1 for all u∈G^u\in \widehat{G}, and we recover the definitions of the Fourier transform given in Spectral Theories, Chap. II.

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.

Cited by

Cites

  • Integration, Chap. VII, §1, no. 1 not in this corpus: INT VII
  • Spectral Theories, Chap. II, §1 not in this corpus: TS II
  • Chap. III, §3, no. 4
Transcribed from the printed volume. Not translated.