Bourbaki

Exercise 17, § 4

Integration, VIII, p. 69

Notations and conventions are those of Exer. 15.

a) Let μ\mu be a bounded measure on G; for the endomorphism γ(μ):f↦μ∗f\gamma(\mu): f \mapsto \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 a cp>0c_p > 0 such that ∥ν∗g∥q≥cp∥g∥q\| \nu * 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. TVS, IV, §4, No. 2, remarks following Cor. 3 of Th. 1).

b) If μ\mu is a measure with base β\beta, and G is not discrete, show that γ(μ)\gamma(\mu) is never surjective for 1≤p≤+∞1 \leq p \leq +\infty (for p<+∞p < +\infty, use a), reducing to the case that the density of μ\mu with respect to β\beta belongs to K(G)\mathscr{K}(G); for p=+∞p = +\infty, argue 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 G is abelian, 1≤p≤21 \leq p \leq 2 and the endomorphism γ(μ)\gamma(\mu) of Lp(G,β)L^p(G, \beta) is surjective, show that it is bijective (using regularization, show that if γ(μ)\gamma(\mu) is not injective then the endomorphism γ(ν)\gamma(\nu) of Lq(G,β)L^q(G, \beta) is not injective: one makes use of Ch. IV, §6, No. 5, Cor. of Prop. 4).

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

Solution

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