Proposition 4

Let $f\in \mathscr{L}_E^1(G)$.

a) The set $S_f$ is negligible modulo H and $f^{\pi}\in \mathscr{F}_{\pi}(G)$;

b) Let C be a compact subset of G. If $f$ is continuous and has support contained in C, then $S_f$ is empty, the function $f^{\pi}$ belongs to $\mathscr{K}_{\pi}(G)$ and its support is contained in $C\cdot H$.

The first part of a) follows from INT, VII, p. 57, § 2, n$^o5$, c). Let $w\in H$. If $x\in G$ - $S_f$, then $xw\in G$ - $S_f$ and

whereas if $x\in S_f$, then $f^{\pi}(xw) = 0 =\pi (w)^*f^{\pi}(x)$. The function $f^{\pi}$ therefore belongs to $\mathscr{F}_{\pi}(G)$.

Suppose now that $f$ is continuous and has support contained in C. For every $x\in G$, the mapping $h\mapsto f(xh)$ then belongs to $\mathscr{K}$ (H), hence is integrable, which proves that $S_f=\emptyset$.

Let us prove that $f^{\pi}$ is continuous. Let $x\in G$. Let U be a relatively compact open neighbourhood of $x$ in G.

For every $y\in U$, one has

$H\cap (y^{-1}C\cup x^{-1}C)$

$\leqslant \beta (H\cap U^{-1}C)$ sup$_{h\in U^{-1}C}\|f(yh)-f(xh)\|$,

and the continuity of $f^{\pi}$ then follows from the uniform continuity of $f$ on G.

Finally, if $x\in G$ satisfies $f^{\pi}(x)\not = 0$, there exists $h\in H$ such that $f(xh)\not = 0$; thus $xh$ belong to C and $x$ belong to $C\cdot H$. Since $C\cdot H$ is closed in G, one concludes that the support of $f^{\pi}$ is contained in $C\cdot H$.

Let C be a compact subset of G. Let $u\in \mathscr{K}_+(G)$ be a function such that $u(x)>0$ for every $x\in C$. Let $v$ be the function on G defined by

for every $x\in G$; with the preceding notation, one has $v=u^1$, corresponding to the trivial representation of dimension 1 of H. The function $v$ is continuous and positive; it belong to $\mathscr{F}_1$(G), its support is contained in Supp($u$)$\cdot H$ and one has

(5) $x$inf$^{\in}_{C\cdot H}v(x)>0$

(INT, VII, p. 39–40, § 2, n$^o1$, prop. 1 and lemma 1, a)).