Proposition 16

The measure $\widehat{\alpha}$ on $\widehat{H} =\widehat{G}/H^{\bot}$ is equal to $\widehat{\beta /}\widehat{\gamma}$.

Let us fix a nonzero $f\in \mathscr{K}(G)$. For $x\in G$ and $\chi \in \widehat{G}$, put

The function $\varphi$ is continuous on $G\times \widehat{G}$ (INT, IV, §4, n$^o3$, cor. 1 of Theorem 2). For fixed $x$, $\varphi (x, \chi )$ depends only on the class of $\chi$ in $\widehat{G}/H^{\bot}=\widehat{H}$. For fixed $\chi$, $\langle \chi , x\rangle \varphi (x, \chi )$ depends only on the class of $x$ in $G/H$, and the function $\dot{x}\mapsto \langle \chi , x\rangle \varphi (x, \chi )$ on $G/H$ has compact support.

Let $x\in G$. The function $\dot{\chi}\mapsto \varphi (x, \chi )$ on $\widehat{H}$ is the Fourier cotransform of the function $y\mapsto f(xy)$ on H. The latter is square-integrable, hence by the Plancherel formula (23) of II, p. 217, one has

Let $\chi \in \widehat{G}$. The function $\dot{x}\mapsto \langle \chi , x\rangle \varphi (x, \chi )$ belongs to $\mathscr{K}(G/H)$, hence to $L^1(G/H)$. Its Fourier cotransform is the function on $H^{\bot}$ whose value at $\xi \in H^{\bot}$ is

by formula (33). Therefore

by the Plancherel formula again.

One then finally computes

$\int_{\widehat{G}}|\overline{\mathscr{F}}_G(f)|^2d\widehat{\beta}=\int_G|f|^2d\beta$ (by (23))

$=\int_{G/H}d\gamma ( \dot{x})\int_H|f(xy)|^2d\alpha (y)$ (by (33))

$=\int_{G/H}d\gamma ( \dot{x})\int_{\widehat{G}/H^{\bot}}|\varphi (x, \chi )|^2d\widehat{\alpha}( \dot{\chi})$ (by (35))

$=\int_{\widehat{G}/H^{\bot}}d\widehat{\alpha}( \dot{\chi})\int_{H^{\bot}}|\overline{\mathscr{F}}_G(f)(\chi \xi )|^2d\widehat{\gamma}(\xi )$ (by (36)),

where INT, V, §8, No.$^o3$, prop. 5 has been applied to the positive continuous function $( \dot{x},\dot{\chi})\mapsto |\varphi (x, \chi )|^2$ on $G/H\times \widehat{G}/H^{\bot}$.

Comparing this equality with the integration formula (33) for the group $\widehat{G}$, one then concludes that the Haar measures $\widehat{\alpha}$ and $\widehat{\beta /}\widehat{\gamma}$ coincide.