Bourbaki

Exercise 28, § 3

Integration, VIII

Bourbaki marks this one as harder than the rest.

Let G1,G2G_1, G_2 be two locally compact groups, β1,β2\beta_1, \beta_2 left Haar measures on G1,G2G_1, G_2 and A1A_1 (resp. A2A_2) the topological algebra (over R\mathbf{R}) L1(G1,β1)L^1(G_1, \beta_1) (resp. L1(G2,β2)L^1(G_2, \beta_2)). Let TT be an algebra isomorphism of A1A_1 onto A2A_2, such that the relation f≥0f \geq 0 almost everywhere is equivalent to T(f)≥0T(f) \geq 0 almost everywhere.

a) Now prove that TT is a topological algebra isomorphism. (First note that if a decreasing sequence (fn)(f_n) of elements of A1A_1 tends to 00 almost everywhere, the same is true of the sequence of the T(fn)T(f_n); deduce that for every sequence (fn)(f_n) bounded above in A1A_1, T(sup⁡(fn))=sup⁡(T(fn))T(\sup(f_n)) = \sup(T(f_n)) and conclude with the aid of Fatou's lemma).

b) Now prove that TT extends in one and only one way to an isomorphism of the topological algebra M1(G1)\mathscr{M}^1(G_1) onto the topological algebra M1(G2)\mathscr{M}^1(G_2), and that there exist a topological isomorphism uu of G1G_1 onto G2G_2 and a continuous homomorphism χ\chi of G1G_1 into R+∗\mathbf{R}_+^* such that T(εs)=χ(s)εu(s)T(\varepsilon_s) = \chi(s)\varepsilon_{u(s)} (use Exercise 15 a) and Exercise 19 b); to prove that uu and χ\chi are continuous, note that TT defines an isomorphism of the algebra L(A1)\mathscr{L}(A_1) of continuous endomorphisms of the topological vector space A1A_1 onto the analogous algebra L(A2)\mathscr{L}(A_2) and that this isomorphism is bicontinuous for the topology of simple convergence; on the other hand, observe that s→δ(s−1)s \to \delta(s^{-1}) is an isomorphism of GG onto a multiplicative subgroup of L(A1)\mathscr{L}(A_1).

c) Deduce from b) that one has, for every f∈A1f \in A_1,

(T(f))(t)=χ(u−1(t))f(u−1(t)) (T(f))(t) = \chi(u^{-1}(t))f(u^{-1}(t))

for every t∈G2t \in G_2.

Recall (Alg., Chapter III, 3rd ed., § 1) that there are finite groups G1,G2G_1, G_2 such that the algebras L1(G1,β1)L^1(G_1, \beta_1) and L1(G2,β2)L^1(G_2, \beta_2) are isomorphic without G1G_1 and G2G_2 being isomorphic.

Solution

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