Suppose that each function χ(s,⋅) is bounded. Then γχ leaves LCp(X,β) stable, and the linear representation γχ of G in LCp(X,β) is continuous.
Let f∈LCp(X,β). Then
∫∗∣χ(s−1,x)f(s−1x)∣pdβ(x) ≤x∈Xsupχ(s−1,x)p−1∫∗∣f(s−1x)∣pχ(s−1,x)dβ(x) =x∈Xsupχ(s−1,x)p−1∫∣f(x)∣pdβ(x), therefore
γχ(s)f∈LCp(X,β), and
∥γχ(s)∥≤x∈Xsupχ(s−1,x)1/q, (5) where
q denotes the exponent conjugate to
p. If
f∈K(X), then
γχ(s)f tends to
γχ(s0)f in
K(X), hence in
LCp(X,β), as
s tends to
s0. Therefore the representation
γχ of
G in
LCp(X,β) is continuous (No. 1, Prop. 2).
Properties analogous to those of Nos. 3, 4, 5 hold if G operates on the right in X.
In particular, if one regards G as operating on itself by left or right translations, and if one takes χ=1, one obtains the left and right regular representations of G in C(G), K(G), K(G), C′(G), M(G), M1(G). If one takes β to be a left (resp. right) Haar measure on G, and if one takes χ=1, one obtains the left (resp. right) regular representation of G in LCp(G,β).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.