Bourbaki

Exercise 17, § 3

Integration, VIII

The notation and conventions are those of Exercise 15.

a) Let μ\mu be a bounded measure on GG; for the endomorphism γ(μ):f→μ∗f\gamma(\mu): f \to \mu * f of Lp(G,β)L^p(G, \beta) to be surjective (1≤p<+∞1 \leq p < +\infty), it is necessary and sufficient that there exist cp>0c_p > 0 such that ∥μ~∗g∥q≥cp∥g∥q\|\tilde{\mu} * g\|_q \geq c_p \|g\|_q for all g∈Lq(G,β)g \in L^q(G, \beta) (where 1p+1q=1\frac{1}{p} + \frac{1}{q} = 1) (cf. Esp. vect. top., chap. IV, § 5, exerc. 11).

b) If μ\mu is a base measure β\beta, if GG is not discrete, now prove that γ(μ)\gamma(\mu) is never surjective for 1≤p≤+∞1 \leq p \leq +\infty (for p<+∞p < +\infty, use a)), by reducing to the case where the density of μ\mu with respect to β\beta belongs to K(G)\mathscr{K}(G); for p=+∞p = +\infty, reason directly by observing that for f∈L1(G,β)f \in L^1(G, \beta) and g∈L∞(G,β)g \in L^\infty(G, \beta), f∗gf * g is uniformly continuous for the right uniform structure).

c) If GG is commutative, 1≤p≤21 \leq p \leq 2 and if the endomorphism γ(μ)\gamma(\mu) of Lp(G,β)L^p(G, \beta) is surjective, now prove that it is bijective (using regularization, prove that if γ(μ)\gamma(\mu) is not injective, the endomorphism γ(μ~)\gamma(\tilde{\mu}) of Lq(G,β)L^q(G, \beta) is not injective: one will use chap. IV, § 6, No. 5, cor. of prop. 4).

d) Let G=ZG = \mathbf{Z} and μ=ε1−ε0\mu = \varepsilon_1 - \varepsilon_0; now prove that γ(μ)\gamma(\mu) is injective in the Lp(Z,β)L^p(\mathbf{Z}, \beta) such that p≠+∞p \neq +\infty, but not in L∞(Z,β)L^\infty(\mathbf{Z}, \beta), and is not surjective for any value of pp.

Solution

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