*Assume χ a continuous multiplier and each function χ(s,⋅) bounded. Let ρ(s)=supx∈Xχ(s−1,x). Let p and q be two conjugate exponents (1≤p<+∞). Let μ∈Mρ1/q(G) and f∈Lp(X,β)2 Then: (i) μ and f are convolvable; (ii) the convolution product μ∗βf is given locally β-almost everywhere by the formula (3), and is equal locally β-almost everywhere to a function g∈Lp(X,β) such that ∥g∥p≤∥μ∥ρ1/q∥f∥p; (iii) g is equal to the element γχ(μ)f defined by the continuous representation γχ of G in Lp(X,β).
We have
∫∗∥γχ(s)f∥pd∣μ∣(s)≤(∫∗ρ(s)1/qd∣μ∣(s))∥f∥p<+∞ by §2, No. 5, formula (5). On the other hand, the mapping
s↦γχ(s)f of
G into
Lp(X,β) is continuous (§2, No. 5, Prop. 9). Therefore this mapping is
μ-integrable. Let
g=∫G(γχ(s)f)dμ(s)∈Lp(X,β). We have
∥g∥p≤(∫ρ1/q(s)d∣μ∣(s))∥f∥p. Applying the preceding remarks to
∣f∣, one sees that the mapping
s↦εs∗∣f∣ of
G into
Lp(X,β) is
μ-integrable, therefore that, for every
h∈H(X), the mapping
s↦⟨h,εs∗(∣f∣⋅β)⟩ is
μ-integrable. Prop. 7 of §1, No. 5 then proves that
μ and
f⋅β are convolvable. Moreover,
∫Xg(x)h(x)dβ(x)=∫Gdμ(s)∫X(γχ(s)f)(x)h(x)dβ(x)=∫G⟨h,εs∗(f⋅β)⟩dμ(s), and this last integral is equal to
⟨h,μ∗(f⋅β)⟩ by Prop. 7 of §1, No. 5. One therefore sees that
g is a convolution product of
μ and
f. This convolution product is given locally
β-almost everywhere by (3), by Prop. 2 and the
Remark that follows it.
2For a function f, the expression " f∈Lp(X,β) " is an abuse of notation signifying that, depending on the context, the symbol f is to be interpreted either as a function defined β-almost everywhere, or as the equivalence class of such a function for the relation of equality β-almost everywhere. Thus f∈Lp can symbolize either f∈Lp or f˙∈Lp.
By an abuse of notation, it is often one of the functions g of the statement that is denoted μ∗βf, which permits writing
∥μ∗βf∥p≤∥μ∥ρ1/q∥f∥p.
If X is countable at infinity, this style of notation is, moreover, entirely justified.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.