Lemma 8

Let $f_1\in \mathscr{K}_{\chi}(G)$ and $f_2\in \mathscr{L}_{\chi}^2(G)$.

a) The function $f$ on G defined by $f(g) =\langle f_1|\boldsymbol{\gamma }_{G,\chi}(g)f_2\rangle$ for every $g\in G$ belong to $\mathscr{L}_{\overline{\chi}}^2(G)$ and satisfies $N_2(f)\leqslant N_1(f_1)N_2(f_2)$;

b) The function $f$ on G defined by $f(g) =\langle f_1|\boldsymbol{\delta }_{G,\chi}(g)f_2\rangle$ for every $g\in G$ belong to $\mathscr{L}_{\chi}^2(G)$ and satisfies $N_2(f)\leqslant N_1(f_1)N_2(f_2)$.

Let us prove a), the proof of assertion b) being similar. The function $f$ is continuous, hence $\mu$-measurable. For every $z\in Z$ and every $g\in G$, we have

(formula (7), p. 417), hence $f\in \mathscr{F}_{\overline{\chi}}(G)$. Prop. 5 of V, p. 413 implies that it now suffices to prove that $N_2(f)\leqslant N_1(f_1)N_2(f_2)$.

Suppose first that $f_2$ belongs to $\mathscr{K}_{\chi}(G)$. For every $g\in G$, by definition we have

where the function $\overline{f_1}\boldsymbol{\gamma }_{G,\chi}(g)f_2$ is identified with a function on $G/Z$.

Define a function $f_3$ on G by putting $f_3(g) = 0$ if $f_1(g) = 0$ and $f_3(g) =f_1(g)|f_1(g)|^{-1/2}$ otherwise. The function $f_3$ belongs to $\mathscr{F}_{\chi}(G)$ and satisfies $f_1=|f_1|^{1/2}f_3$; it is $\mu$-measurable and zero outside a compact modulo Z, since $f_1$ is. Since $|f_1|^{1/2}\in \mathscr{K}_1$(G), it follows that

for every $g\in G$.

Let $g\in G$. By the Cauchy-Schwarz inequality, we have

Since we have $|f_3|^2=|f_1|$ in $\mathscr{K}(G/Z)$, it follows, on integrating over $G/Z$, that

The function on $G/Z\times G/Z$ deduced from the function

by passing to quotients is $(\nu \otimes \nu$)-measurable and has compact support, hence it is $(\nu \otimes \nu$)-moderate (INT, V, p. 4, § 5, No.$^o2$, Def. 2). By Prop. 7 of INT, V, p. 93, § 8, No.$^o3$, it follows that

Consequently, we have $N_2(f)^2\leqslant N_1(f_1)^2N_2(f_2)^2$, which establishes the required property in this case.

Let us consider the general case. Let $u$ denote the linear mapping of $\mathscr{K}_{\chi}(G)$ into $\mathscr{L}_{\overline{\chi}}^2(G)$ which associates $f$ to $f_2$. Let $f_2\in \mathscr{L}_{\chi}^2(G)$ and let $(f_{2,n})_{n\in\mathbf{N}}$ be a sequence in $\mathscr{K}_{\chi}(G)$ converging to $f_2$ in $\mathscr{L}_{\chi}^2(G)$. Let $f_n=u(f_{2,n})$; the sequence $(f_n)_{n\in\mathbf{N}}$ is a Cauchy sequence in $\mathscr{L}_{\overline{\chi}}^2(G)$ since the preceding case implies that $N_2(f_n-f_m)\leqslant N_1(f_1)N_2(f_{2,n}-f_{2,m})$ for all $n$ and $m$ in $\mathbf{N}$. Let $f\in \mathscr{L}_{\overline{\chi}}^2(G)$ be such that $(f_n)$ converges to $f$ (Prop. 3, c) of V, p. 409). Since $N_2(f_n)\leqslant N_1(f_1)N_2(f_{2,n})$ for every $n\in \mathbf{N}$, it follows that $N_2(f)\leqslant N_1(f_1)N_2(f_2)$.

There exists a subsequence $(f_{n_k})_{k\in\mathbf{N}}$ such that $f_{n_k}(g)$ converges to $f(g)$ for every $g\in G$ outside a subset of G negligible modulo Z (loc. cit., b)). But on the other hand, for every $g\in G$, one has

Consequently, one has $f(g) =\langle f_1|\boldsymbol{\gamma }_{G,\chi}(g)f_2\rangle$ for every $g$ outside a subset of G negligible modulo Z. Since $f\in \mathscr{L}_{\overline{\chi}}^2(G)$ and $N_2(f)\leqslant N_1(f_1)N_2(f_2)$, the lemma is proved.