Bourbaki

Exercise 5, § 4

Integration, VIII, p. 65

Bourbaki marks this one as harder than the rest.

Let G be a locally compact group countable at infinity, β\beta a left Haar measure on G, a∈R+∗a \in \mathbf{R}_+^*, and (νu)0≤u≤a(\nu_u)_{0 \leq u \leq a} a family of positive measures on G satisfying the following conditions: (i) ν0=εe\nu_0 = \varepsilon_e; νu+v=νu∗νv\nu_{u+v} = \nu_u * \nu_v if u+v≤au + v \leq a; νu=νu∨\nu_u = \nu_u^\vee; (ii) for 0<u≤a0 < u \leq a, νu=fu⋅β\nu_u = f_u \cdot \beta with fu≥0f_u \geq 0 lower semi-continuous; (iii) νu\nu_u is a vaguely continuous function of uu. a) Let f∈K+(G)f \in \mathscr{K}_+(G). Show that, for 0<u≤a0 < u \leq a,

∫(fu/2∗f)(z)2dβ(z)=∫f(x)(fu∗f)(x)dβ(x). \int (f_{u/2} * f)(z)^2 d\beta(z) = \int f(x)(f_u * f)(x) d\beta(x).

b) Let f∈K(G)f \in \mathscr{K}(G). Show that fu/2∗ff_{u/2} * f has β\beta-integrable square for 0<u≤a0 < u \leq a, and that (1) again holds.

c) Let f∈K(G)f \in \mathscr{K}(G) be such that νa∗f=0\nu_a * f = 0. Show that f=0f = 0. (One has νa/2n∗f=0\nu_{a/2^n} * f = 0 for every integer n>0n > 0 by b), therefore f=0f = 0 by §3, No. 3, Remark 1).)

d) Let ν∈C′(G)\nu \in \mathscr{C}'(G) be such that νa∗ν=0\nu_a * \nu = 0. Show that ν=0\nu = 0. (Regularize ν\nu by functions in K(G)\mathscr{K}(G) and apply c).)

Solution

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