Let p be a real number ⩾1. The left (resp. right) regular representation of G in Lp(G,μ)(resp. in Lp(G,μ′)) is faithful.
More precisely, let q be the conjugate exponent of p and let g be an element of G such that g=e.
a) There exists a function φ∈K(G), positive and nonzero in Lq(G,μ), such that ⟨φ,γG(p)(g)φ⟩=0;
b) There exists a function φ∈K(G), positive and nonzero in Lq(G,μ′), such that ⟨φ,δG(p)(g)φ⟩=0.
Let us consider the case of the left regular representation γG(p), that of the right regular representation being similar. Assertion a) implies that γG(p) is faithful, for if g=e is an element of G, and if φ is as in a), one has φ=γG(p)(g)φ since ⟨φ,φ⟩=∫Gφ2>0.
Let us therefore prove a). Let g=e in G. Let C be a symmetric compact neighbourhood of e such that g∈/C2 and let φ∈K(G) be a positive continuous function of integral 1 with support contained in C; the function φ is nonzero in Lp(G,μ). Since C∩g−1C=∅, one has
⟨φ,γG(p)(g)φ⟩=∫Gφ(x)φ(g−1x)dμ(x)=0
Let ϱ be a continuous bounded representation of G in a Banach space E. For every f∈L1(G) (resp. f′∈L1(G,μ′)) and every g∈G, one has
(1) ϱ(g)ϱ(f⋅μ)=ϱ(εg∗(f⋅μ))=ϱ(γG(1)(g)f⋅μ),
(2) ϱ(f′⋅μ′)ϱ(g)=ϱ((f′⋅μ′)∗εg)=ϱ(δG(1)(g−1)f′⋅μ′)
(INT, VIII, p. 144, § 3, no2, formula (5)).
Suppose G unimodular, and put μ′=μ. The biregular representation of G in Lp(G) (resp. in Lp(G)) is the representation ϱG(p) of G×G in Lp(G) (resp. in Lp(G)) such that
ϱG(p)(g1,g2)=γG(p)(g1)δG(p)(g2)=δG(p)(g2)γG(p)(g1)
It is a continuous linear representation (Lemma 1 of V, p. 377). It satisfies
(ϱG(p)(g1,g2)f)(x)=f(g1−1xg2)
for every f∈Lp(G), every (g1,g2)∈G×G and every x∈G.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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