Hardy’s uncertainty principle for Gabor transform on compact extensions of $\mathbb {R}^n$

Abstract

We prove in this paper a generalization of Hardy’s theorem for Gabor transform in the setup of the semidirect product $\mathbb {R}^n\rtimes K$, where K is a compact subgroup of automorphisms of $\mathbb {R}^n$. We also solve the sharpness problem and thus obtain a complete analogue of Hardy’s theorem for Gabor transform. The representation theory and Plancherel formula are fundamental tools in the proof of our results.

1 Introduction

It is a well-known fact in classical Fourier analysis that an integrable function f defined on the real line and its Fourier transform $\hat{ f }$ cannot be simultaneously and sharply localized unless $f=0$ almost everywhere. This property of functions is widely known as the uncertainty principle in Fourier analysis. The following result of Hardy makes the rather vague statement above precise (see [11]):

Theorem 1

Let p, q, c be positive real numbers and f a measurable function on $\mathbb {R}^n$ such that:

(i) $\vert f(x)\vert \le c \text {e}^{-p\pi \Vert x\Vert ^2}, \ \ x\in \mathbb {R}^n$,

(ii) $\vert \hat{f}(y)\vert \le c \text {e}^{-q\pi \Vert y\Vert ^2}, \ \ k\in \mathbb {R}^n$.

If $pq > 1$, then $f= 0$ a.e. If $pq = 1$ then $f(t)=C\text {e}^{-p\pi \Vert x\Vert ^2}$, for some constant C. If $pq < 1$, then any finite linear combination of Hermite functions satisfies (i) and (ii).

Here the Fourier transform $\hat{f}$ is defined by

$$\begin{aligned} \hat{f}(y)=\int _{\mathbb {R}^n}f(x)\hbox {e}^{-2i\pi x y}dx,\ y\in \mathbb {R}^n, \end{aligned}$$

where $xy=\sum _{j=1}^n x_jy_j$, and $\Vert x\Vert =\sqrt{ x^2}$ is the Euclidean norm.

Naturally, there has been some effort to prove Hardy-like theorems for various connected Lie groups G. Specifically, analogues and variants of Hardy’s theorem have been shown for motion groups [22, 23], compact extensions of $\mathbb {R}^n$ [1], non-compact connected semisimple Lie groups G with finite center [6, 18, 20, 21] and nilpotent Lie groups [5, 12, 19, 24].

Unlike the classical Fourier transform, the continuous Gabor transform gives a simultaneous representation of the space and the frequency variables. Let $\psi \in L^2(\mathbb {R}^n)$ be a fixed non-zero function usually called a window function. The Gabor transform of a function $f\in L^2(\mathbb {R}^n)$ with respect to the window function $\psi$ is defined on $\mathbb {R}^n\times \widehat{\mathbb {R}}^n$ by

$$\begin{aligned} \mathcal {G}_\psi f(x,w): =\int _{\mathbb {R}^n} f(y)\overline{\psi }(y-x)e^{-2i\pi yw }dy. \end{aligned}$$

According to [9], we have for all $f_1$, $f_2$, $\psi _1$, $\psi _2\in L^2(\mathbb {R}^n)$ the functions $\mathcal {G}_{\psi _1}f_1$ and $\mathcal {G}_{\psi _2}f_2$ belong to $L^{2}(\mathbb {R}^n\times \hat{\mathbb {R}}^n)$ and

$$\begin{aligned} \langle \mathcal {G}_{\psi _1}f_1,\mathcal {G}_{\psi _2}f_2\rangle _{L^2(\mathbb {R}^n\times \hat{\mathbb {R}}^n)}=\langle f_1 , f_2\rangle _{L^2(\mathbb {R}^n)}\overline{\langle \psi _1,\psi _2\rangle }_{L^2(\mathbb {R}^n)}. \end{aligned}$$

(1)

It has been shown in the early 2000 s that many uncertainty principles for the Fourier transform have a counterpart for the continuous Gabor transform (see [2, 10]). We specify that a Hardy-type theorem has been established in [10, Theorem 2.6.2].

Theorem 2

Let $f,\psi \in L^2(\mathbb {R}^n)$. Assume that

$$\begin{aligned} \Big \vert \mathcal {G}_\psi f(x,w)\Big \vert \le c \text {e}^{-{\pi \over 2} (p \Vert x\Vert ^2+q \Vert w\Vert ^2)}, \end{aligned}$$

for some constants $p, q, c>0$. Then three cases can occur.

(i) If $pq > 1$, then either $f\equiv 0$ or $\psi \equiv 0$.

(ii) If $pq = 1$ and $\mathcal {G}_\psi f$ is not zero almost everywhere, then both f and $\psi$ are multiples of some time-frequency shift of the Gaussian $\text {e}^{-p\pi \Vert x\Vert ^2}$.

(iii) If $pq < 1$, then the decay condition is satisfied whenever f and $\psi$ are finite linear combinations of Hermite functions.

In 2012, the continuous Gabor transform for separable locally compact unimodular group of type I has been introduced by Farashahi and Kamyabi-Gol [7]. A brief description is given in Section 2. One should notice that, in the Euclidean setting, the continuous Gabor transform has many symmetries which are lost in the Lie group setting (the dual of G does not identify with G) and this is then a serious obstacle for stating uncertainty principles for the continuous Gabor transform. However, some attempts to extend Theorem 2 on special classes of non-Abelian Lie groups have already been made. Recently, analogues of Hardy’s theorem for Gabor transform have been established for locally compact abelian groups having noncompact identity component and groups of the form $\mathbb {R}^n\times K$, where K is a compact group having irreducible representations of bounded dimension (see [3]). On the other hand, the author and K. Abid [17] proved an analogue of Hardy’s theorem for Gabor transform on connected, simply connected nilpotent Lie groups. However, in the last two references the results obtained concern only the case $pq>1$. In this paper, we prove a generalization of Hardy’s theorem for Gabor transform on a general compact extension $\mathbb {R}^n\rtimes K$, where K is a compact subgroup of automorphisms of $\mathbb {R}^n$, providing evidence to the three cases cited above. The proof of our result, which is given in Section 3, exploits Hardy’s theorem for $\mathbb {R}^n$ and representation theory and the harmonic analysis of $\mathbb {R}^n\rtimes K$.

2 Backgrounds

2.1 Continuous Gabor transform

Let G be a separable locally compact unimodular group of type I, and let dg be its Haar measure. We endow the unitary dual of G with the Mackey Borel structure. We denote by $L^p(G)$ the space of $L^p$-functions on G for $p\ge 1$, and we define

$$\begin{aligned} \pi (f)=\int _Gf(g)\pi (g)dg,\quad \pi \in \hat{G}, \, f\in L^1(G). \end{aligned}$$

Then by the abstract Plancherel theorem, there exists a unique Borel measure $\rho$ on $\hat{G}$ such that for any function $f\in L^1(G)\cap L^2(G)$,

$$\begin{aligned} \int _G \vert f(g)\vert ^2dg=\int _{\hat{G}}\Vert \pi (f) \Vert _{HS}^2d\rho (\pi ), \end{aligned}$$

where $\Vert \pi (f) \Vert _{HS}=\big (\text {tr}\big (\pi (f)^*\pi (f)\big )\big )^{1/2}$ denotes the Hilbert-Schmidt norm of $\pi (f)$.

Let $f \in C_c(G)$, the set of all continuous complex-valued functions on G with compact supports, and $\psi$ a fixed nonzero function in $L^2(G)$, usually called window function. For $(x,\pi )\in G\times \hat{G}$, the continuous Gabor transform of f with respect to the window function $\psi$ is defined as a measurable field of operators on $G\times \hat{G}$ by

$$\begin{aligned} \mathcal {G}_\psi f(x,\pi ) :=\int _Gf(g)\overline{\psi }(x^{-1}g)\pi (g)dg. \end{aligned}$$

Let $f_\psi ^x$ be the function defined on G by

$$\begin{aligned} f_\psi ^x(g)=f(g)\overline{\psi } (x^{-1}g),\qquad \forall g\in G. \end{aligned}$$

Then, $f_\psi ^x\in L^1(G)\cap L^2(G)$ and

$$\begin{aligned} \pi (f_\psi ^x)=\int _Gf_\psi ^x(g)\pi (g)dg=\int _Gf(g)\overline{\psi }(x^{-1}g)\pi (g)dg=\mathcal {G}_\psi f(x,\pi ). \end{aligned}$$

(2)

By the Plancherel theorem, $\mathcal {G}_\psi f(x,\pi )$ is a Hilbert-Schmidt operator for all $x \in G$ and for almost all $\pi \in \hat{G}$. Furthermore,

$$\begin{aligned} \int _G \int _{\hat{G}} \Vert \mathcal {G}_\psi f(x,\pi )\Vert _{HS}^2d\rho (\pi )\, dx=\Vert \psi \Vert _2^2\Vert f \Vert _2^2. \end{aligned}$$

(3)

Thus, the continuous Gabor transform $\mathcal {G}_\psi : f\mapsto \mathcal {G}_\psi f$ ($f\in C_c(G)$) is a multiple of an isometry. So, we can extend $\mathcal {G}_\psi$ uniquely to a bounded linear operator on $L^2(G)$ which we still denote by $\mathcal {G}_\psi$ and this extension satisfies (3) for each $f \in L^2(G)$.

2.2 Compact extensions of $\mathbb {R}^n$

Let $G=\mathbb {R}^n\rtimes K$ be a semidirect product of $\mathbb {R}^n$ and a compact subgroup K of the automorphisms group Aut$(\mathbb {R}^n)$. In the whole paper, $\mathbb {R}^n$ is equipped with an Euclidean scalar product which embeds the compact group K as a subgroup of orthogonal transformations (for details see [15]). The multiplication law in G is given by

$$\begin{aligned} (a,k)\cdot (b,h)=(a+kb, kh), \end{aligned}$$

for $(a,k),(b,h)\in G$. We fixe once for all a Haar measure dg on G by $dg=dad\mu (k)$, where da denotes the Lebesgue measure on $\mathbb {R}^n$ and $d\mu (k)$ the normalized Haar measure on K. Let us remark that the compactness of K leads to the proof that da is invariant under the action of K on $\mathbb {R}^n$ given by $\mathbb {R}^n\ni a\mapsto k^{-1}ak$, for $k\in K$.

By Mackey’s little group theory [16], the set $\hat{G}$ is given by the following procedure. Let $\gamma$ be a non-zero real linear form on $\mathbb {R}^n$ and let $\chi _\gamma$ be the unitary character of $\mathbb {R}^n$ defined by $\chi _\gamma (a)=\text {e}^{-2i\pi \langle \gamma ,a\rangle }$, $a\in \mathbb {R}^n$. The naturel action $g\cdot \gamma$ on $\mathbb {R}^n$ is given by $\langle g\cdot \gamma ,a\rangle =\langle \gamma ,g^{-1}ag\rangle$ for $g\in G$ and $a\in \mathbb {R}^n$. If G acts on $\widehat{\mathbb {R}^n}$ by $g\cdot \chi _\gamma (a)= \chi _\gamma (g^{-1}ag)$, then $g\cdot \chi _\gamma = \chi _{g\cdot \gamma }$. We identify $\widehat{\mathbb {R}^n}$ with $\mathbb {R}^n$ by the mapping $\mathbb {R}^n\ni \gamma \mapsto \chi _\gamma \in \widehat{\mathbb {R}^n}$.

Let $K_\gamma$ be the set of all $k\in K$ such that $k\cdot \chi _\gamma = \chi _\gamma$, then $G_\gamma =\mathbb {R}^n\rtimes K_\gamma$ is the stabilizer of $\chi _\gamma$ in G. Let us take the normalized Haar measure $d\mu _\gamma$ on $K_\gamma$ and a K-invariant measure $d\dot{\mu }_\gamma$ on $K/K_\gamma$ such that

$$\begin{aligned} \int _K\varphi (k)d\mu (k)=\int _{K/K_\gamma }\int _{K_\gamma }\varphi (kk')d\mu _\gamma (k')d\dot{\mu }_\gamma (kK_\gamma ). \end{aligned}$$

Noting that the measure $d\dot{\mu }_\gamma$ is normalized so that $\int _{K/K_\gamma }d\dot{\mu }_\gamma =1$. Let $d\overline{\gamma }$ be the image of the Lebesgue measure on $\mathbb {R}^n/K$ by the canonical projection $\mathbb {R}^n\ni \gamma \mapsto \overline{\gamma }=K\cdot \gamma \in \mathbb {R}^n/K$ such that

$$\begin{aligned} \int _{\mathbb {R}^n}\varphi (\gamma )d\gamma =\int _{\mathbb {R}^n/K}\int _K\varphi (k\cdot \gamma )d\mu (k)d\overline{\gamma }. \end{aligned}$$

On the other hand, let $(\sigma ,\mathcal {H}_\sigma )$ be a unitary and irreducible representation of $K_\gamma$ and $\mathcal {H}_{\gamma , \sigma }$ be the completion of the vector space of all continuous mapping functions $\varphi :K\rightarrow \mathcal {H}_{\sigma }$ for which

$$\begin{aligned} \varphi (kk')=\sigma (k')^{-1} \varphi (k), \ \ \forall k\in K, \ \forall k'\in K_\gamma , \end{aligned}$$

with respect to the norm

$$\begin{aligned} \Vert \varphi \Vert _2=\left( \int _K\Vert \varphi (k)\Vert ^2_{\mathcal {H}_\sigma }d\mu (k)\right) ^{1\over 2}. \end{aligned}$$

The induced representation $\pi _{\gamma ,\sigma }:=\text {Ind}_{G_\gamma }^G(\chi _\gamma \otimes \sigma )$, realized on $\mathcal {H}_{\gamma , \sigma }$ by

$$\begin{aligned} \pi _{\gamma ,\sigma }(a,k)\varphi (h)=\chi _\gamma (h^{-1}ah)\varphi (k^{-1}h)=\chi _{h\cdot \gamma }(a)\varphi (k^{-1}h), \end{aligned}$$

(4)

for $\varphi \in \mathcal {H}_{\gamma ,\sigma }$, $(a,k)\in G$ and $h\in K$, is an irreducible unitary representation of G. Furthermore, every infinite dimensional irreducible unitary representation of G is equivalent to some representation $\pi _{\gamma ,\sigma }$.

We note that, every irreducible unitary representation $\tau$ of K extends trivially to an irreducible representation (also denoted by $\tau$) of the entire group G, defined by

$$\begin{aligned} \tau (a,k)=\tau (k), \quad a\in \mathbb {R}^n \ \ \text {and} \ \ k\in K. \end{aligned}$$

According to [13, 14], the Plancherel formula for $f\in L^1(G)\cap L^2(G)$ is given by

$$\begin{aligned} \int _G\vert f(a,k)\vert ^2da d\mu (k)=\int _{\mathbb {R}^n/K}\sum _{\sigma \in \hat{K}_\gamma }\Vert \pi _{\gamma , \sigma }(f) \Vert _{HS}^2d\overline{\gamma }, \end{aligned}$$

where

$$\begin{aligned} \pi _{\gamma ,\sigma }(f)=\int _G f(a,k)\pi _{\gamma ,\sigma }(a,k)da d\mu (k) \end{aligned}$$

is a kernel operator. Its kernel is defined for $(s,u)\in K/K_\gamma \times K/K_\gamma$ as

$$\begin{aligned} H(f,\gamma , \sigma )(s,u)= \int _{K_\gamma }f(\cdot , svu^{-1})^{\widehat{}}(s\cdot \gamma )\sigma (v)d\mu _\gamma (v), \end{aligned}$$

(5)

where $f(\cdot , svu^{-1})^{\widehat{}}$ denotes the partial Fourier transform of the function f with respect to the Euclidean variable.

3 The main result

The main motivation of the present study is to generalize Theorem 2, writing down a generalized analogue of Hardy’s uncertainty principle for Gabor transform on $G=\mathbb {R}^n\rtimes K$. Before stating our main result, we need some notations. For every $x, w \in \mathbb {R}^n$, we denote by $\mathcal {M}_w$ and $\mathcal {T}_x$ the modulation and the translation operators defined respectively on $L^2(\mathbb {R}^n)$ by

$$\begin{aligned}{} & {} \forall z\in \mathbb {R}^n,\quad \mathcal {M}_wf(z)=e^{2i\pi zw}f(z),\\{} & {} \forall z\in \mathbb {R}^n,\quad \mathcal {T}_xf(z)=f(z-x). \end{aligned}$$

Then we deduce that,

$$\begin{aligned} \forall z\in \mathbb {R}^n,\quad \mathcal {M}_w(\mathcal {T}_xf)(z)=e^{2i\pi zw}f(z-x), \end{aligned}$$

and

$$\begin{aligned} \forall z\in \mathbb {R}^n,\quad \mathcal {T}_x(\mathcal {M}_wf)(z)=e^{-2i\pi xw}e^{2i\pi zw}f(z-x). \end{aligned}$$

On the other hand, for a measurable function $\varphi$ on G, let

$$\begin{aligned} \varphi (\cdot ,k)(a):=\varphi (a,k),\quad (a,k)\in G. \end{aligned}$$

Our main result is the following:

Theorem 3

Let p and q be positive real numbers. Let $f,\psi \in L^2(G)$ be such that

$$\begin{aligned} \Vert \mathcal {G}_\psi f (g,\pi _{\gamma ,\sigma })\Vert _{HS}\le \phi _\gamma (k,\sigma ) e^{-{\pi \over 2} (p \Vert a\Vert ^2+q\Vert \gamma \Vert ^2)}, \end{aligned}$$

(6)

for all $g=(a,k)\in G$, $\gamma \in \mathbb {R}^n$ and $\sigma \in \hat{K}_\gamma$, with $\Vert \phi _\gamma \Vert _{L^2(K\times \hat{K}_\gamma )}\le C$ for some psitive constant C independent of $\gamma$. Then three cases can occur.

(i) If $pq>1$, then either $f\equiv 0$ or $\psi \equiv 0$.

(ii) If $pq=1$ and $\mathcal {G}_{\psi (\cdot ,k)} f(\cdot ,h)\not \equiv 0$ for each $k,h\in K$ such that the Gabor transform of $f(\cdot ,h)$ with respect to $\psi (\cdot ,k)$ is well defined, then for all $a\in \mathbb {R}^n$ and almost all $h\in K$,

$$\begin{aligned} f(a,h)= C_{1}(h)\mathcal {M}_{\lambda _{1}(h)}\mathcal {T}_{\delta _{1}(h)}\text {e }^{-\pi p\Vert a\Vert ^{2} } \\ \text {and}\quad \psi (a,h)= C_{2}(h)\mathcal {M}_{\lambda _{2}(h)}\mathcal {T}_{\delta _{2}(h)}\text {e }^{-\pi p\Vert a\Vert ^{2} }, \end{aligned}$$

where $C_j\in L^2(K)$ and $\lambda _j, \delta _j$ are functions from K to $\mathbb {R}^n$, $j:=1,2$.

(iii) If $pq<1$, then there are infinitely many linearly independent pairs $(f,\psi )$ satisfying (6).

3.1 Some lemmas

The results in the following lemma are quite standard.

Lemma 1

Let $f,\psi \in L^2(\mathbb {R}^n)$ and $\xi , \lambda , y, z \in \mathbb {R}^n$. Then,

(i) $\mathcal {G}_{(\mathcal {M}_\xi \mathcal {T}_z\psi )}(\mathcal {M}_\lambda \mathcal {T}_yf)(x,w) = e^{2i\pi x\xi } e^{-2i\pi y(w-\lambda +\xi )} \mathcal {G}_\psi f(x-y+z,w-\lambda +\xi ).$

In particular, $\mathcal {G}_\psi (\mathcal {M}_\lambda \mathcal {T}_yf)(x,w)=e^{-2i\pi yw } e^{2i\pi y\lambda } \mathcal {G}_\psi f(x-y,w-\lambda )$.

(ii) $\mathcal {G}_\psi f(-x,-w)=e^{-2i\pi x w}\overline{\mathcal {G}_f\psi (x,w)}.$

(iii) Let $F(x,w)= \mathcal {G}_\psi f(x,w) \mathcal {G}_\psi f(-x,-w)e^{2i\pi xw}.$ Then,

$$\begin{aligned} \hat{F}(\nu ,\theta )=F(-\theta ,\nu ),\quad \nu , \theta \in \mathbb {R}^n. \end{aligned}$$

Now, we shall give two lemmas which are required to prove Theorem 3. Let $g=(a,k)$ be an element of G and $f,\psi \in L^2(G)$. For $h\in K$, let $(f^g_\psi )_h$ be the complex valued function defined on $\mathbb {R}^n$ by

$$\begin{aligned} (f^g_\psi )_h(c):=f^g_\psi (\cdot , h)(c)=f^g_\psi (c,h)=f(c,h)\overline{\psi }\big ((a,k)^{-1}(c,h)\big ). \end{aligned}$$

It is easy to see that $f^g_\psi \in L^1(G)$ for all $g\in G$, it sufficient to use Cauchy–Schwarz inequality. Moreover by [4, Lemma 3.1], we have

$$\begin{aligned} \mathcal {G}_\psi f(g,\pi _l)=\pi _l(f^g_\psi ), \end{aligned}$$

(7)

for all $g\in G$. We should also mention that $f^g_\psi \in L^2(G)$, for almost all $g\in G$. In fact,

$$\begin{aligned} \int _G\int _G\vert f^g_\psi (x)\vert ^2dx\, dg =\int _G\int _G\vert f(x)\vert ^2\vert \psi (g^{-1}x)\vert ^2dx\, dg =\Vert f \Vert _2^2\Vert \psi \Vert _2^2<\infty . \end{aligned}$$

Then obviously $\int _G\vert f^g_\psi (x)\vert ^2dx<\infty$, for almost all $g\in G$. By setting $f^{a,k}_\psi =f^g_\psi$, we have the following lemma.

Lemma 2

Let $f,\psi \in L^2(G)$ meet the condition (6) of Theorem 3. Then

$$\begin{aligned} I(f,\psi , a):= \int _K \int _K \Bigg ( \int _{\mathbb {R}^n} \big \vert (f^{a,k}_\psi )_h(c)\big \vert dc\Bigg )^2 d\mu (h)d\mu (k) <\infty , \end{aligned}$$

for all $a\in \mathbb {R}^n$.

Proof

By using (6), we have

$$\begin{aligned}{} & {} \int _K\int _{\mathbb {R}^n}\int _{\mathbb {R}^n/K} \sum _{\sigma \in \hat{K}_\gamma }(1+\Vert a \Vert ^2) \Vert \mathcal {G}_\psi f\big ((a,k),\pi _{\gamma ,\sigma }\big )\Vert _{HS}^2d\bar{\gamma }\, da \, d\mu (k) \\{} & {} \quad \le \int _K\int _{\mathbb {R}^n}\int _{\mathbb {R}^n/K} \sum _{\sigma \in \hat{K}_\gamma }(1+\Vert a \Vert ^2) \phi _\gamma ^2(k,\sigma ) e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \gamma \Vert ^2)} d\bar{\gamma }\, da \, d\mu (k) \\{} & {} \quad \le C\int _{\mathbb {R}^n} (1+\Vert a \Vert ^2) e^{-{\pi } p \Vert a\Vert ^2} da \int _{\mathbb {R}^n/K}e^{-{\pi } q\Vert \gamma \Vert ^2}d\bar{\gamma } \\{} & {} \quad = C\int _{\mathbb {R}^n} (1+\Vert a \Vert ^2) e^{-{\pi } p \Vert a\Vert ^2} da \int _{\mathbb {R}^n/K}\int _K e^{-{\pi } q\Vert k\cdot \gamma \Vert ^2}d\mu (k)\, d\bar{\gamma } \\{} & {} \quad = C\int _{\mathbb {R}^n} (1+\Vert a \Vert ^2) e^{-{\pi } p \Vert a\Vert ^2} da\int _{\mathbb {R}^n} e^{-{\pi } q\Vert \gamma \Vert ^2}d{\gamma }<\infty . \end{aligned}$$

By (7) and the Plancherel formula, we obtain

$$\begin{aligned}{} & {} \infty >\int _K\int _{\mathbb {R}^n}\int _{\mathbb {R}^n/K} \sum _{\sigma \in \hat{K}_\gamma }(1+\Vert a \Vert ^2) \big \Vert \pi _{\gamma ,\sigma }\big (f^ {a,k}_\psi \big )\big \Vert _{HS}^2d\bar{\gamma }\, da \, d\mu (k) \\{} & {} \quad =\int _K\int _{\mathbb {R}^n} \int _K\int _{\mathbb {R}^n}(1+\Vert a \Vert ^2)\left| f(c,h){\psi }\big ((a,k)^{-1}(c,h)\big )\right| ^2 dc\,d\mu (h)\, da \, d\mu (k)\\{} & {} \quad =\int _K\int _{\mathbb {R}^n} \int _K\int _{\mathbb {R}^n}(1+\Vert a \Vert ^2)\left| f(c,h){\psi }\big (-k^{-1}(a-c),k^{-1}h\big )\right| ^2dc\,d\mu (h)\, da \, d\mu (k)\\{} & {} \quad =\int _K\int _{\mathbb {R}^n} \int _K\int _{\mathbb {R}^n}(1+\Vert a \Vert ^2)\left| f(c,h){\psi }\big (-kh^{-1}a+kh^{-1}c,k\big )\right| ^2dc\,d\mu (h)\, da \, d\mu (k)\\{} & {} \quad =\int _K\int _{\mathbb {R}^n} \int _K\int _{\mathbb {R}^n}(1+\Vert a-kh^{-1}c \Vert ^2)\left| f(c,h){\psi }(a,k)\right| ^2dc\,d\mu (h)\, da \, d\mu (k).\end{aligned}$$

Therefore,

$$\begin{aligned} \left| {\psi }(a,k)\right| ^2\int _K\int _{\mathbb {R}^n}(1+\Vert a-kh^{-1}c \Vert ^2)\left| f(c,h)\right| ^2dc\,d\mu (h) <\infty , \end{aligned}$$

for almost all $a\in \mathbb {R}^n$ and $k\in K$. As $\psi$ is non identically zero, there exists $a_0,k_0$ such that $\psi ( a_0,k_0) \ne 0,$ and

$$\begin{aligned} \int _K\int _{\mathbb {R}^n}(1+\Vert a_0-k_0h^{-1}c \Vert ^2)\left| f(c,h)\right| ^2dc\,d\mu (h) <\infty . \end{aligned}$$

(8)

On the other hand, we have

$$\begin{aligned} I(f,\psi , a)= & {} \int _K\int _K\left( \int _{\mathbb {R}^n}\left| f(c,h){\psi }\big ((a,k)^{-1}(c,h)\big )\right| dc \right) ^2d\mu (h)\, d\mu (k)\\= & {} \int _K\int _K\left( \int _{\mathbb {R}^n}\left| f(c,h){\psi }\big (-k^{-1}a+k^{-1}c,k^{-1}h\big )\right| dc \right) ^2d\mu (h)\, d\mu (k)\\\le & {} \int _K\int _K\left( \int _{\mathbb {R}^n}{\left| {\psi }\big (-k^{-1}a+k^{-1}c,k^{-1}h\big )\right| ^2\over 1+\Vert a_0-k_0h^{-1}c \Vert ^2}dc\right) \\{} & {} \times \left( \int _{\mathbb {R}^n} \Big ( 1+\Vert a_0-k_0h^{-1}c \Vert ^2\Big ) \Big \vert f(c,h)\Big \vert ^2 dc\right) d\mu (h)\, d\mu (k) \end{aligned}$$

(using Cauchy–Shwartz inequality)

$$\begin{aligned}{} & {} \le \int _K\int _K\left( \int _{\mathbb {R}^n}\left| {\psi }\big (-k^{-1}a+k^{-1}c,k^{-1}h\big )\right| ^2dc\right) \\{} & {} \quad \times \left( \int _{\mathbb {R}^n} \Big ( 1+\Vert a_0-k_0h^{-1}c \Vert ^2\Big ) \Big \vert f(c,h)\Big \vert ^2 dc\right) d\mu (h)\, d\mu (k)\\{} & {} =\int _K\int _K\left( \int _{\mathbb {R}^n}\left| {\psi }\big (c,k\big )\right| ^2dc\right) \\{} & {} \quad \times \left( \int _{\mathbb {R}^n} \Big ( 1+\Vert a_0-k_0h^{-1}c \Vert ^2\Big ) \Big \vert f(c,h)\Big \vert ^2 dc\right) d\mu (h)\, d\mu (k)\\{} & {} =\Vert \psi \Vert ^2_2\int _K\int _{\mathbb {R}^n} \Big ( 1+\Vert a_0-k_0h^{-1}c \Vert ^2\Big ) \Big \vert f(c,h)\Big \vert ^2 dc\, d\mu (h), \end{aligned}$$

which is finite by (8). $\square$

Lemma 3

For all $a,\gamma \in \mathbb {R}^n$,

$$\begin{aligned} \int _K \int _K\left| \widehat{\big (f_\psi ^{a,k} \big )_h}(\gamma )\right| ^2d\mu (h)\, d\mu (k)\le C e^{-{\pi } ( p\Vert a\Vert ^2+q\Vert \gamma \Vert ^2)}. \end{aligned}$$

Proof

Let E(a) be the function defined on $\mathbb {R}^n$ by

$$\begin{aligned} E(a)(c)=\int _K\int _K ((f_\psi ^{a,k})_h*(f_\psi ^{a,k})_h^\star )(c)d\mu (h)\, d\mu (k), \end{aligned}$$

where $c\in \mathbb {R}^n$ and $(f_\psi ^{a,k})_h^\star (c)=\overline{(f_\psi ^{a,k})_h(-c)}$. Then $E(a)\in L^1(\mathbb {R}^n)$, for all $a\in \mathbb {R}^n$. In fact for all $a\in \mathbb {R}^{n}$,

$$\begin{aligned} \int _{\mathbb {R}^n}\vert E(a)(c)\vert dc\le & {} \int _{\mathbb {R}^{n}}\int _K\int _K\int _{\mathbb {R}^{n}}\big \vert (f_\psi ^{a,k})_h(t)\big \vert \big \vert (f_\psi ^{a,k})_h(t-c)\big \vert dt\, d\mu (h)\, d\mu (k)\, dc \\= & {} \int _K\int _K \Bigg (\int _{\mathbb {R}^n}\big \vert (f_\psi ^{a,k})_h(c)\big \vert dc\Bigg )^2d\mu (h)\, d\mu (k)<\infty \end{aligned}$$

(using Lemma 2). Thus,

$$\begin{aligned} \widehat{E(a)}(\gamma )=\int _K\int _K\big \vert \widehat{(f_\psi ^{a,k})_h}(\gamma ) \big \vert ^2d\mu (h)\, d\mu (k),\quad \gamma \in \mathbb {R}^n. \end{aligned}$$

(9)

For $U\in L^1(\mathbb {R}^{n})$, define $U*f_\psi ^{a,k}$ on G by

$$\begin{aligned} U*f_\psi ^{a,k}(c,h)=\int _{\mathbb {R}^{n}}U(t)f_\psi ^{a,k}(c-t,h)\,dt \end{aligned}$$

and then $E(a)_U: \mathbb {R}^n\rightarrow \mathbb {C}$ by

$$\begin{aligned} E(a)_U(c)=\int _K\int _K\Big ((U*f_\psi ^{a,k})_h* \big ((U*f_\psi ^{a,k})_h\big )^\star \Big )(c)\, d\mu (h)\, d\mu (k). \end{aligned}$$

It is not hard to see that

$$\begin{aligned} E(a)_U(c)=\int _K\int _K \big ((U*(f_\psi ^{a,k})_h)*(U*(f_\psi ^{a,k})_h)^\star \big )(c)\, d\mu (h)\, d\mu (k). \end{aligned}$$

Therefore, for every $\eta \in \mathbb {R}^n$

$$\begin{aligned} \widehat{E(a)_U}(\eta )&=\int _K\int _K \Big \vert \big (U*(f_\psi ^{a,k})_h\big )^{\widehat{}}(\eta )\Big \vert ^2\, d\mu (h)\, d\mu (k) \nonumber \\&=\vert \hat{U}(\eta )\vert ^2\int _K\int _K \Big \vert \widehat{(f_\psi ^{a,k})_h}(\eta )\Big \vert ^2\, d\mu (h)\, d\mu (k) =\vert \hat{U}(\eta )\vert ^2\widehat{E(a)}(\eta ). \end{aligned}$$

(10)

By the inversion formula for $\mathbb {R}^n$, we have

$$\begin{aligned} \int _{\mathbb {R}^n}\widehat{E(a)_U}(\eta )\,d\eta= & {} E(a)_U(0) \nonumber \\= & {} \int _K\int _K\int _{\mathbb {R}^n} \big \vert (U*f_\psi ^{a,k})_h (c)\big \vert ^2\,dc\, d\mu (h)\, d\mu (k)\nonumber \\= & {} \int _K\Vert U*f_\psi ^{a,k}\Vert _{L^2(G)}^2d\mu (k). \end{aligned}$$

(11)

On the other hand for $d\in \mathbb {N}$ and $\gamma \in \mathbb {R}^n$, let

$$\begin{aligned} \mathcal {L}_d(\gamma )=\left\{ \eta \in \mathbb {R}^n \, \ \Vert \gamma \Vert -{1\over 2d}\le \Vert \eta \Vert \le \Vert \gamma \Vert +{1\over 2d} \right\} \end{aligned}$$

denote the annulus in $\mathbb {R}^n$ and $v_d$ its volume. For every $d\in \mathbb {N}$, there exists a sequence $(U_{d,m})_m$ of $L^1$-functions on $\mathbb {R}^n$ satisfying following properties:

(i) $0\le \widehat{U_{d,m}}\le 1$.

(ii) $\widehat{(U_{d,m})}_m$ converges pointwise to the characteristic function $\chi _{\mathcal {L}_d(\gamma )}$ of $\mathcal {L}_d(\gamma )$.

As $\widehat{E(a)}$ is continuous and $\mathcal {L}_m(\gamma )$ has volume $v_d$, we have

$$\begin{aligned} \widehat{E(a)}(\gamma )&=\lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathcal {L}_m(\gamma )}\widehat{E(a)}(\eta )\,d\eta =\lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathbb {R}^n}\lim _{m\rightarrow \infty }\widehat{E(a)}(\eta ) \big (\widehat{U_{d,m}}(\eta )\big )^2\,d\eta \\&= \lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathbb {R}^n}\lim _{m\rightarrow \infty }\widehat{E(a)_{U_{d,m}}}(\eta ) \,d\eta \qquad \text {(using (10))} \\&=\lim _{d\rightarrow \infty }v_d^{-1}\lim _{m\rightarrow \infty }\int _K \Vert U_{d,m}*f_\psi ^{a,k} \Vert _{L^2(G)}^2\, d\mu (k) \qquad \text {(using (11))} \\&=\lim _{d\rightarrow \infty }v_d^{-1}\lim _{m\rightarrow \infty } \int _{K} \int _{\mathbb {R}^n/K}\sum _{\sigma \in \hat{K}_\eta }\Vert \pi _{\eta , \sigma }\big ( U_{d,m}*f_\psi ^{a,k}\big ) \Vert _{HS}^2d\bar{\eta }\, d\mu (k). \end{aligned}$$

From [1, p. 732], one has that

$$\begin{aligned} \Vert \pi _{\eta , \sigma }\big ( U_{d,m}*f_\psi ^{a,k}\big ) \Vert _{HS}^2\le \widehat{U_{d,m}}(s_\eta \cdot \eta )^2\Vert \pi _{\eta , \sigma }\big ( f_\psi ^{a,k}\big ) \Vert _{HS}^2, \end{aligned}$$

for some $s_\eta \in K$. It follows, using (6), that

$$\begin{aligned}{} & {} \widehat{E(a)}(\gamma )\le \lim _{d\rightarrow \infty }v_d^{-1}\lim _{m\rightarrow \infty } \int _{K} \int _{\mathbb {R}^n/K}\\{} & {} \quad \sum _{\sigma \in \hat{K}_\eta }\phi _\eta (k,\sigma )^2\widehat{U_{d,m}}(s_\eta \cdot \eta )^2 e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \eta \Vert ^2)}d\overline{\eta }\, d\mu (k)\\{} & {} \quad =C\lim _{d\rightarrow \infty }v_d^{-1}\lim _{m\rightarrow \infty } \int _{\mathbb {R}^n/K}\widehat{U_{d,m}}(s_\eta \cdot \eta )^2 e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \eta \Vert ^2)}d\overline{\eta }\\{} & {} \quad =C\lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathbb {R}^n/K}\chi _{\mathcal {L}_d(\gamma )}(\eta )^2 e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \eta \Vert ^2)}d\overline{\eta }\\{} & {} \quad =C\lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathbb {R}^n/K}\int _K\chi _{\mathcal {L}_d(\gamma )}(h\cdot \eta ) e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \eta \Vert ^2)}d\mu (h)\, d\overline{\eta }\\{} & {} \quad =C\lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathbb {R}^n}\chi _{\mathcal {L}_d(\gamma )}(\eta )e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \eta \Vert ^2)}d{\eta }\\{} & {} \quad =C\lim _{d\rightarrow \infty }v_d^{-1} \int _{\mathcal {L}_d(\gamma )}e^{-{\pi } (p \Vert a\Vert ^2+q\Vert \eta \Vert ^2)}d{\eta }=Ce^{-{\pi } (p \Vert a\Vert ^2+q\Vert \gamma \Vert ^2)}. \end{aligned}$$

Finally, Eq. (9) allows us to conclude. $\square$

3.2 Proof of Theorem 3

For $k,h\in K$, let $f_{k,h}$ and $\psi _{k,h}$ be the complex-valued functions defined on $\mathbb {R}^n$ by

$$\begin{aligned} f_{k,h}(a)=f(a,kh)\quad \text {and} \quad \psi _{k,h}(a)=\psi (k^{-1}a,h). \end{aligned}$$

Then obviously $f_{k,h},\psi _{k,h}\in L^2(\mathbb {R}^n)$, for almost all $h\in K$ and all $k\in K$.

For fixed $\lambda$, $y\in \mathbb {R}^n$, let $F_{\lambda ,y}(k,h)$ and $K_{\lambda ,y}^{\varphi _1,\varphi _2 }$ be the functions defined on $\mathbb {R}^n\times \mathbb {R}^n$ by

$$\begin{aligned} F_{\lambda ,y}(k,h)(a,\gamma )=\mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(a,\gamma ) \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(-a,-\gamma )e^{2i\pi a \gamma }. \end{aligned}$$

and

$$\begin{aligned} K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,\gamma )=\int _{K}\int _{K}F_{\lambda ,y}(k,h)(a,\gamma ) \varphi _1(k)\varphi _2(h) d\mu (k)\, d\mu (h), \end{aligned}$$

where $\varphi _1,\varphi _2$ are bounded functions on K. Noting that, $F_{\lambda ,y}(k,h)$ is well defined for almost all $h\in K$ and all $k\in K$ and

(12)

Lemma 4

There exists a positive constant $C_1$ such that

$$\begin{aligned} \Big \vert K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,\gamma )\Big \vert \le C_1 \, e^{-\pi \big ( p\Vert a\Vert ^2+q\Vert \gamma \Vert ^2\big )}, \end{aligned}$$

for all $a,\gamma \in \mathbb {R}^n$. Moreover, the constant $C_1$ does not depend on $\lambda$ and y.

Proof

By using Cauchy–Schwartz inequality, we obtain

$$\begin{aligned} \Big \vert K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,\gamma )\Big \vert\le & {} \displaystyle \int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(a,\gamma )\mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(-a,-\gamma )\Big \vert \\{} & {} \times \Big \vert \varphi _1(k)\varphi _2(h)\Big \vert d\mu (h)\, d\mu (k)\\\le & {} \Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty }\displaystyle \int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(a,\gamma )\mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(-a,-\gamma )\Big \vert \\ {}{} & {} \times d\mu (h)\, d\mu (k)\\\le & {} \Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty }\left( \displaystyle \int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(a,\gamma )\Big \vert ^2d\mu (h)\, d\mu (k)\right) ^{\frac{1}{2}}\\{} & {} \quad \times \left( \displaystyle \int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(-a,-\gamma )\Big \vert ^2d\mu (h)\,d\mu (k)\right) ^{\frac{1}{2}}. \end{aligned}$$

Remark that,

$$\begin{aligned}{} & {} \int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(a,\gamma )\Big \vert ^2d\mu (h)\,d\mu (k)\\{} & {} \quad =\int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}f_{k,h}(a-y,\gamma -\lambda )\Big \vert ^2d\mu (h)\,d\mu (k) \\{} & {} \quad =\int _K\int _K\Big \vert \int _{\mathbb {R}^n}f_{k,h}(c)\overline{\psi }_{k,h}(c-a+y)\text {e}^{-2i\pi c(\gamma -\lambda )}dc\Big \vert ^2d\mu (h)\, d\mu (k) \\{} & {} \quad =\int _K\int _K\Big \vert \int _{\mathbb {R}^n}f(c,kh)\overline{\psi }(-k^{-1}a+k^{-1}(c+y),h)\text {e}^{-2i\pi c(\gamma -\lambda )}dc\Big \vert ^2d\mu (h)\, d\mu (k) \\{} & {} \quad =\int _K\int _K\Big \vert \int _{\mathbb {R}^n}f(c,h)\overline{\psi }(-k^{-1}a+k^{-1}(c+y),k^{-1}h)\text {e}^{-2i\pi c(\gamma -\lambda )}dc\Big \vert ^2d\mu (h)\, d\mu (k) \\{} & {} \quad =\int _K\int _K\Big \vert \int _{\mathbb {R}^n}f(c,h)\overline{\psi }\big ((a-y,k)^{-1}(c,h)\big ) \text {e}^{-2i\pi c(\gamma -\lambda )}dc\Big \vert ^2d\mu (h)d\mu (k) \\{} & {} \quad =\int _K\int _K\Big \vert \widehat{(f_\psi ^{a-y,k})_h}(\gamma - \lambda )\Big \vert ^2d\mu (h)\,d\mu (k). \end{aligned}$$

It results that,

$$\begin{aligned} \big \vert K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,\gamma ) \big \vert\le & {} \Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty }\left( \int _K\int _K\big \vert \widehat{(f_\psi ^{a-y,k})_h}(\gamma - \lambda )\big \vert ^2d\mu (h)\,d\mu (k)\right) ^{\frac{1}{2}}\\{} & {} \times \left( \int _K\int _K\big \vert \widehat{(f_\psi ^{-a-y,k})_h}(-\gamma - \lambda )\big \vert ^2d\mu (h)\,d\mu (k)\right) ^{\frac{1}{2}}\\\le & {} C\Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty } \left( e^{-\pi (p\Vert a-y\Vert ^2+q \Vert \gamma -\lambda \Vert ^2)} \right) ^{\frac{1}{2}}\\{} & {} \left( e^{-\pi (p\Vert a+y\Vert ^2+q \Vert \gamma +\lambda \Vert ^2)} \right) ^{\frac{1}{2}} \quad \text {(using Lemma 3)}\\\le & {} C\Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty } e^{-\pi (p\Vert a\Vert ^2+q\Vert \gamma \Vert ^2)}, \end{aligned}$$

which is the desired result. $\square$

Lemma 5

For all $w,\theta \in \mathbb {R}^n$,

$$\begin{aligned} \big \vert \hat{K}_{\lambda ,y}^{\varphi _1,\varphi _2 }(w,\theta )\big \vert \le C_1\, e^{-\pi \big ( p\Vert \theta \Vert ^2+q\Vert w\Vert ^2\big )}. \end{aligned}$$

Proof

By using (iii) in Lemma 1, we have

$$\begin{aligned} \hat{K}_{\lambda ,y}^{\varphi _1,\varphi _2 }(w,\theta )= & {} \displaystyle \int _{K}\int _{K}\hat{F}_{\lambda ,y}(w,\theta )\varphi _1(k)\varphi _2(h) d\mu (k)\, d\mu (h)\nonumber \\= & {} \int _{K}\displaystyle \int _{K}F_{\lambda ,y}(-\theta ,w)\varphi _1(k)\varphi _2(h) d\mu (k)\, d\mu (h)=K_{\lambda ,y}^{\varphi _1,\varphi _2 }(-\theta ,w).\nonumber \\ \end{aligned}$$

(13)

Therefore,

$$\begin{aligned} \big \vert \hat{K}_{\lambda ,y}^{\varphi _1,\varphi _2 }(w,\theta )\big \vert\le & {} \Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty }\displaystyle \int _K\int _K\Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(-\theta ,w)\Big \vert \\{} & {} \times \Big \vert \mathcal {G}_{\psi _{k,h}}(\mathcal {M}_\lambda \mathcal {T}_yf_{k,h})(\theta ,-w)\Big \vert d\mu (h)\, d\mu (k). \end{aligned}$$

As in the proof of the Lemma 4 we can show that,

$$\begin{aligned} \big \vert \hat{K}_{\lambda ,y}^{\varphi _1,\varphi _2 }(w,\theta )\big \vert \le C\Vert \varphi _1 \Vert _{\infty }\Vert \varphi _2 \Vert _{\infty } e^{-\pi \big ( p\Vert \theta \Vert ^2+q\Vert w\Vert ^2\big )}, \end{aligned}$$

which allows us to conclude. $\square$

(i) For fixed $\lambda , y$ and $\theta$ in $\mathbb {R}^n$, let $R_{\lambda ,y,\theta }$ be the function defined on $\mathbb {R}^n$ by

$$\begin{aligned} R_{\lambda ,y,\theta }(a)=K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,.)^{\widehat{}}(\theta ), \end{aligned}$$

where $K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,.)^{\widehat{}}$ is the partial Fourier transform of $K_{\lambda ,y}^{\varphi _1,\varphi _2 }$ with respect to the second variable $\gamma$. It follows, using (13), that

$$\begin{aligned} \widehat{R}_{\lambda ,y,\theta }(w)= \hat{K}_{\lambda ,y}^{\varphi _1,\varphi _2 }(w,\theta ) =K_{\lambda ,y}^{\varphi _1,\varphi _2 }(-\theta ,w). \end{aligned}$$

(14)

There exists a positive constant $C_2$ such that

$$\begin{aligned} \Big \vert R_{\lambda ,y,\theta }(a)\Big \vert \le C_2 \, e^{-p\pi \Vert a \Vert ^2}. \end{aligned}$$

In fact, from Lemma 4 we have,

$$\begin{aligned} \vert R_{\lambda ,y,\theta }(a)\vert= & {} \left| K_{\lambda ,y}^{\varphi _1,\varphi _2 } (a,.)^{\widehat{}}(\theta )\right| \le \int _{\mathbb {R}^{n}}\left| K_{\lambda ,y}^{\varphi _1,\varphi _2 } (a,\gamma )\right| d\gamma \\\le & {} C_1\int _{\mathbb {R}^{n}} e^{-\pi \big ( p\Vert a\Vert ^2+q\Vert \gamma \Vert ^2\big )} d\gamma =C_2e^{-p\pi \Vert a\Vert ^2}, \end{aligned}$$

where $C_2=C_1\int _{\mathbb {R}^{n}}e^{-\pi q\Vert \gamma \Vert ^2}d\gamma$. On the other hand, by (14) and Lemma 4, we have

$$\begin{aligned} \big \vert \hat{R}_{\lambda ,y,\theta }(w)\big \vert =\left| K_{\lambda ,y}^{\varphi _1,\varphi _2 }(-\theta ,w)\right| \le C_1e^{-q\pi \Vert w\Vert ^2}. \end{aligned}$$

By Hardy’s theorem, this implies $R_{\lambda ,y,\theta }\equiv 0$ and $\hat{R}_{\lambda ,y,\theta }=0$ for all $\lambda , y, \theta \in \mathbb {R}^n$. We then obtain

$$\begin{aligned} K_{\lambda ,y}^{\varphi _1,\varphi _2 }(-\theta ,w)=\int _{K}\displaystyle \int _{K}F_{\lambda ,y}(k,h)(-\theta ,w)\varphi _1(k)\varphi _2(h) d\mu (k)\, d\mu (h)=0, \end{aligned}$$

for any bounded function $\varphi _1$ and $\varphi _2$ on K. Therefore, $F_{\lambda ,y}(k,h)(-\theta ,w)=0$ for all $\lambda , y, \theta$ in $\mathbb {R}^n$ and almost all $w \in \mathbb {R}^n$. As $F_{-\lambda ,-y}(k,h)$ is continuous on $\mathbb {R}^n\times \mathbb {R}^n$,

$$\begin{aligned} \vert F_{-\lambda ,-y}(k,h)(0,0)\vert =\vert G_{\psi _{k,h}}f_{k,h}(y,\lambda )\vert ^2=0 \quad \text {(using (12))}. \end{aligned}$$

Hence, $\mathcal {G}_{\psi _{k,h}}f_{k,h}\equiv 0$. By using (1), we have

$$\begin{aligned} \Vert \psi _{k,h}\Vert _2^2 \Vert f_{k,h}\Vert _2^2 =0, \end{aligned}$$

which implies either $\psi _{k,h}\equiv 0$ or $f_{k,h}\equiv 0$. Observe that,

$$\begin{aligned}{} & {} \int _K\int _K\Vert \psi _{k,h}\Vert _2^2 \Vert f_{k,h}\Vert _2^2 d\mu (k)\, d\mu (h)\\{} & {} \quad =\int _K\int _K\left( \int _{\mathbb {R}^n}\vert f(c,kh)\vert ^2dc \right) \left( \int _{\mathbb {R}^n}\vert \psi (k^{-1}t,h)\vert ^2dt \right) d\mu (k)\, d\mu (h)\\{} & {} \quad =\int _K\int _K\left( \int _{\mathbb {R}^n}\vert f(c,h)\vert ^2dc \right) \left( \int _{\mathbb {R}^n}\vert \psi (t,k^{-1}h)\vert ^2dt \right) d\mu (k)\, d\mu (h)\\{} & {} \quad = \Vert f\Vert _2^2\int _K\int _{\mathbb {R}^n}\vert \psi (t,k)\vert ^2dt d\mu (k)= \Vert f\Vert _2^2 \Vert \psi \Vert _2^2. \end{aligned}$$

This allow us to achieve this case.

(ii) We start by treat the case $p=q=1$. By using Lemmas 4 and 5, the function $K_{\lambda ,y}^{\varphi _1,\varphi _2 }$ verifies the decay conditions of Hardy’s theorem on $\mathbb {R}^n\times \mathbb {R}^n$. Then,

$$\begin{aligned} K_{\lambda ,y}^{\varphi _1,\varphi _2 }(a,\gamma )=C_{\lambda ,y}^{\varphi _1,\varphi _2 } e^{-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )}, \end{aligned}$$

where $C_{\lambda ,y}^{\varphi _1,\varphi _2 }$ is a positive constant.

For $\tau \in \hat{K}$, let $u_{ij}^\tau$ be the matrix coefficients of $\tau$ in an orthonormal basis $\{e_j^\tau \, \ 1\le j\le d_\tau \}$ of its associated Hilbert space $\mathcal {H}_\tau$ of dimension $d_\tau$. In other words, $u_{ij}^\tau (k)=\langle \tau (k)e_i^\tau ,e_j^\tau \rangle$, for each $k\in K$. Peter-Weyl Theorem asserts that the set of functions $\big \{\sqrt{d_\tau } u_{ij}^\tau \, \ \tau \in \hat{K}, \ 1\le i,j\le d_\tau \big \}$ is an orthonormal basis of $L^2(K)$. Allowing now $\varphi _1$ and $\varphi _2$ to vary over this base, we obtain

$$\begin{aligned} \int _K\int _KF_{\lambda ,y}(k,h)(a,\gamma )\overline{ u_{ij}^\tau (k)u_{i'j'}^{\tau '} (h)}d\mu (k)\,d\mu (h) =C_{\lambda ,y}^{\tau ,\tau ',i,j,i',j' } e^{-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )}. \end{aligned}$$

This entails that

$$\begin{aligned} F_{\lambda ,y}(k,h)(a,\gamma )=C_{\lambda ,y}(k,h)e^{-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )}, \end{aligned}$$

(15)

where $C_{\lambda ,y}(k,h)=\sum _{\tau \in \hat{K}}\, \sum _{1\le i,j\le d_\tau }\, \sum _{\tau '\in \hat{K}}\, \sum _{1\le i',j'\le d_{\tau '}} C_{\lambda ,y}^{\tau ,\tau ',i,j,i',j' }u_{ij}^\tau (k)u_{i'j'}^{\tau '} (h)$.

Moreover by using (12),

$$\begin{aligned} C_{\lambda ,y}(k,h)= F_{\lambda ,y}(k,h)(0,0)= e^{4\pi i \lambda y}\big ( \mathcal {G}_{\psi _{k,h}}f_{k,h} \big )^2(-y,-\lambda ). \end{aligned}$$

(16)

$\square$

Lemma 6

There exist $\lambda _0,y_0\in \mathbb {R}^n$ such that $C_{\lambda _0,y_0}(k,h)$ is different to zero whenever it exists.

Proof

There exist $\lambda _0,y_0\in \mathbb {R}^n$ such that $C_{\lambda _0,y_0}\ne 0$, otherwise using (16), we have $\mathcal {G}_{\psi _{k,h}}f_{k,h}=0$, for almost all $h\in K$ and all $k\in K$. Hence, $\psi _{k,h}\equiv 0$ or $f_{k,h}\equiv 0$, for almost all $h\in K$ and all $k\in K$. By taken $k=\text {Id}$, we obtain $\psi (\cdot ,s)\equiv 0$ or $f(\cdot ,t)\equiv 0$, for almost all $s,t\in K$. It results that, $\mathcal {G}_{\psi (\cdot ,s)} f(\cdot ,t)\equiv 0$, contradicting the assumption of the theorem. Now, if there exist $k_0,h_0\in K$ such that $C_{\lambda _0,y_0}(k_0,h_0)=0$, then $\vert F_{\lambda _0,y_0}(k_0,h_0)(a,\gamma )\vert =0$, for all $a,\gamma \in \mathbb {R}^n$. By using (12), we have

$$\begin{aligned} \left| \mathcal {G}_{\psi _{k_0,h_0}} f_{k_0,h_0}(a-y_0,\gamma -\lambda _0) \right| \left| \mathcal {G}_{\psi _{k_0,h_0}} f_{k_0,h_0}(-a-y_0,-\gamma -\lambda _0) \right| =0, \end{aligned}$$

for all $a,\gamma \in \mathbb {R}^n$. Thus, $\mathcal {G}_{\psi _{k_0,h_0}} f_{k_0,h_0}=0$ and $\psi _{k_0,h_0}\equiv 0$ or $f_{k_0,h_0}\equiv 0$. As the Lebesgue measure da is invariant under the action of K on $\mathbb {R}^n$, we obtain $\psi (\cdot ,h_0)\equiv 0$. Therefore for almost all $k,h\in K$, $\mathcal {G}_{\psi {(\cdot ,h_0)}} f{(\cdot ,k)}\equiv 0$ or $\mathcal {G}_{\psi {(\cdot ,h)}} f{(\cdot ,k_0h_0)}\equiv 0$. This contradicts again the hypothesis of the theorem. $\square$

The previous lemma and (15) imply that $\big \vert F_{\lambda _0,y_0}(k,h)(a,\gamma )\big \vert \ne 0$, for all $a,\gamma \in \mathbb {R}^n$. This in turn will imply that

$$\begin{aligned} \left| \mathcal {G}_{\psi _{k,h}} f_{k,h}(a-y_0,\gamma -\lambda _0) \right| \left| \mathcal {G}_{\psi _{k,h}} f_{k,h}(-a-y_0,-\gamma -\lambda _0) \right| \ne 0, \end{aligned}$$

for all $a,\gamma \in \mathbb {R}^n$. In particular, $\mathcal {G}_{\psi _{k,h}}f_{k,h}(a,\gamma )\ne 0$, for all $a,\gamma \in \mathbb {R}^n$. Thus we may define $H(k,h)(a,\gamma )=\log \left( \mathcal {G}_{\psi _{k,h}}f_{k,h}(-a,-\gamma )\right)$. Combining (12), (15) and (16), we get

$$\begin{aligned}{} & {} e^{4\pi i \lambda y}\big ( \mathcal {G}_{\psi _{k,h}}f_{k,h} \big )^2(-y,-\lambda )e^{-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )}\\{} & {} \quad =e^{2i\pi a \gamma }e^{-2i\pi (\gamma -\lambda ) y}\mathcal {G}_{\psi _{k,h}} f_{k,h}(a-y,\gamma -\lambda )\\{} & {} \quad \quad e^{-2i\pi (-\gamma -\lambda ) y} \mathcal {G}_{\psi _{k,h}} f_{k,h}(-a-y,-\gamma -\lambda ). \end{aligned}$$

By taking $\log$ on both sides, we obtain

$$\begin{aligned}{} & {} 4\pi i \lambda y+2H(k,h)(y,\lambda )-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )\\{} & {} \quad =2i\pi a \gamma -2i\pi (\gamma -\lambda ) y+H(k,h)( y-a,\lambda -\gamma )-2i\pi (-\gamma -\lambda ) y\\{} & {} \qquad +H(k,h)(a+y,\gamma +\lambda )+2i\pi m, \end{aligned}$$

for some $m\in \mathbb {Z}$. Letting $a=\gamma =0$ shows that $m=0$, hence we get the following difference equation,

$$\begin{aligned}{} & {} H(k,h)\big ((y,\lambda )+(a,\gamma )\big )-2H(k,h)(y,\lambda )+H(k,h)\big (( y,\lambda ) -(a,\gamma )\big )\\{} & {} \quad =-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )-2i\pi a \gamma . \end{aligned}$$

The solution of the above equation can be written as

$$\begin{aligned} H(k,h)(a,\gamma )=-\pi \big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )/2-i\pi a\gamma +\alpha (k,h)a+\beta (k,h)\gamma +\varsigma (k,h), \end{aligned}$$

where $\alpha (k,h),\beta (k,h)\in \mathbb {C}^n$ and $\varsigma (k,h)\in \mathbb {C}$. This shows that

$$\begin{aligned} \mathcal {G}_{\psi _{k,h}}f_{k,h}(a,\gamma )=\tilde{C}(k,h) e^{-{\pi \over 2}\big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )-i\pi a\gamma -\alpha (k,h)a-\beta (k,h)\gamma }, \end{aligned}$$

(17)

where $\tilde{C}$ is complex valued function on $K\times K$. Hence,

$$\begin{aligned} \left| \mathcal {G}_{\psi _{k,h}}f_{k,h}(a,\gamma )\right| =\big \vert \tilde{C}(k,h)\big \vert e^{-{\pi \over 2}\big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )-\text {Re}(\alpha (k,h))a-\text {Re}(\beta (k,h))\gamma }. \end{aligned}$$

(18)

The computation in the proof of Lemma 4 shows that,

$$\begin{aligned}{} & {} \int _K \int _K\left| \mathcal {G}_{\psi _{k,h}}f_{k,h}(a,\gamma )\right| ^2d\mu (h)\, d\mu (k)\\{} & {} \quad = \int _K \int _K\left| \widehat{\big (f_\psi ^{a,k} \big )_h}(\gamma )\right| ^2d\mu (h)\, d\mu (k)\le C e^{-{\pi } ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2)}, \end{aligned}$$

for all $a,\gamma \in \mathbb {R}^n$. It follows, using (18), that

$$\begin{aligned} \int _K \int _K\left| \tilde{C}(k,h)\right| ^2 e^{-2\text {Re}(\alpha (k,h))a-2\text {Re}(\beta (k,h))\gamma }d\mu (h)\, d\mu (k)\le C, \end{aligned}$$

for all $a,\gamma \in \mathbb {R}^n$. Hence for all a and all $\gamma$ in the countable set $\mathbb {Z}^n$,

$$\begin{aligned} \left| \tilde{C}(k,h)\right| ^2 e^{-2\text {Re}(\alpha (k,h))a-2\text {Re}(\beta (k,h))\gamma } <\infty , \end{aligned}$$

for almost all $h,k\in K$. This implies that $\left| \tilde{C}(k,h)\right|$ is finite for almost all $h,k\in K$, $\text {Re}(\alpha (k,h))\equiv 0$ and $\text {Re}(\beta (k,h))\equiv 0$. By choosing $\lambda (k,h)=-\alpha (k,h)/2\pi i$ and $\delta (k,h)=\beta (k,h)/2\pi i$ we have for all $a\in \mathbb {R}^n$ and almost all $h,k\in K$,

$$\begin{aligned} \mathcal {G}_{\psi _{k,h}}f_{k,h}(a,\gamma )=\tilde{C}(k,h)e^{2\pi i(\lambda (k,h)a-\delta (k,h)\gamma ) } e^{-{\pi \over 2}\big ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2\big )}e^{-i\pi a\gamma }\quad \text {(using (17))}. \end{aligned}$$

Then it follows from Theorem 1.2 in [8], that for all $a\in \mathbb {R}^n$ and almost all $h,k\in K$,

$$\begin{aligned}{} & {} f_{k,h}(a)=f(a,kh)= \tilde{C}_1(k,h)e^{2\pi i\lambda (k,h)a } e^{-{\pi }\Vert a-\delta (k,h)\Vert ^2},\\{} & {} \text {and} \quad \psi _{k,h}(a)=\psi (k^{-1}a,h)= \tilde{C}_2(k,h)e^{2\pi i\lambda (k,h)a } e^{-{\pi }\Vert a-\delta (k,h)\Vert ^2}, \end{aligned}$$

where $\tilde{C}_2(k,h),\tilde{C}_2(k,h)$ are multiplicative constants depending on k and h. Fix $k_0$ in K such that for almost all $h\in K$, $\text {Re}(\alpha (k_0,h))=\text {Re}(\beta (k_0,h))= 0$. We then obtain

$$\begin{aligned} \psi (a,h)= & {} \tilde{C}_2(k_0,h)e^{2\pi i\lambda (k_0,h)k_0a } e^{-{\pi }\Vert k_0a-\delta (k_0,h)\Vert ^2}\\= & {} \tilde{C}_2(k_0,h)e^{2\pi i k_0^{-1}\lambda (k_0,h)a } e^{-{\pi }\Vert a-k_0^{-1}\delta (k_0,h)\Vert ^2}, \end{aligned}$$

for all $a\in \mathbb {R}^n$ and almost all $h\in K$. Therefore, we may define ${C}_2(h)=\tilde{C}_2(k_0,h)$, $\lambda _2(h)=k_0^{-1}\lambda (k_0,h)$ and $\delta _2(h)=k_0^{-1}\delta (k_0,h)$ and obtain for $\psi$ the form claimed in the theorem. It is obvious that ${C}_2\in L^2(K)$, simce $\psi \in L^2(G)$. On the other hand, we have

$$\begin{aligned} f(a,k_0h)= \tilde{C}_1(k_0,h)e^{2\pi i\lambda (k_0,h)a } e^{-{\pi }\Vert a-\delta (k_0,h)\Vert ^2}, \end{aligned}$$

for all $a\in \mathbb {R}^n$ and almost all $h\in K$. As $d\mu$ is a Haar measure on K, we get

$$\begin{aligned} f(a,h)=\tilde{C}_1(k_0,k_0^{-1}h)e^{2\pi i\lambda (k_0,k_0^{-1}h)a } e^{-{\pi }\Vert a-\delta (k_0,k_0^{-1}h)\Vert ^2}. \end{aligned}$$

By setting ${C}_1(h)=\tilde{C}_1(k_0,k_0^{-1}h)$, $\lambda _1(h)=\lambda (k_0,k_0^{-1}h)$ and $\delta _1(h)=\delta (k_0,k_0^{-1}h)$, we have

$$\begin{aligned} f(a,h)= C_1(h)\mathcal {M}_{\lambda _1(h)}\mathcal {T}_{\delta _1(h)}\text {e }^{-\pi \Vert a\Vert ^2 }, \end{aligned}$$

for all $a\in \mathbb {R}^n$ and almost all $h\in K$.

To prove the general case where $pq=1$, we apply the following dilation. Let $\varepsilon =(q/p)^{1/4}$, $f_\varepsilon (a,h)=\varepsilon ^{n/2}f(\varepsilon a,h)$ and $\psi _\varepsilon (a,h)=\varepsilon ^{n/2}\psi (\varepsilon a,h)$. Noting that,

$$\begin{aligned} \mathcal {G}_{\psi _\varepsilon } f_\varepsilon \big ((a,k),\pi _{\gamma ,\sigma }\big )= & {} \int _K\int _{\mathbb {R}^n}f_\varepsilon (c,h)\overline{\psi }_\varepsilon (-k^{-1}(a-c),k^{-1}h)\pi _{\gamma ,\sigma }(c,h)dc\, d\mu (h)\\= & {} \varepsilon ^{n}\int _K\int _{\mathbb {R}^n}f(\varepsilon c,h)\overline{\psi }(- k^{-1}(\varepsilon a-\varepsilon c),k^{-1}h)\pi _{\gamma ,\sigma }(c,h)dc\, d\mu (h)\\= & {} \int _K\int _{\mathbb {R}^n}f( c,h)\overline{\psi }(- k^{-1}(\varepsilon a- c),k^{-1}h)\pi _{\gamma ,\sigma }(c/\varepsilon ,h)dc\, d\mu (h)\\= & {} \mathcal {G}_{\psi } f\big ((\varepsilon a,k),\pi _{\gamma /\varepsilon ,\sigma }\big ) \quad \text {(using(4)).} \end{aligned}$$

Therefore for all $(a,k)\in G$, $\gamma \in \mathbb {R}^n$ and $\sigma \in \hat{K}_l$,

$$\begin{aligned} \big \Vert \mathcal {G}_{\psi _\varepsilon } f_\varepsilon \big ((a,k),\pi _{\gamma ,\sigma }\big )\big \Vert _{HS}= & {} \big \Vert \mathcal {G}_{\psi } f\big ((\varepsilon a,k),\pi _{\gamma /\varepsilon ,\sigma }\big )\big \Vert _{HS}\\\le & {} \phi _{\gamma /\varepsilon }(k,\sigma ) e^{-{\pi \over 2} (p \Vert \varepsilon a\Vert ^2+q\Vert \gamma /\varepsilon \Vert ^2)}\\{} & {} =\phi _{\gamma /\varepsilon }(k,\sigma ) e^{-{\pi \over 2}\sqrt{pq} ( \Vert a\Vert ^2+\Vert \gamma \Vert ^2)}. \end{aligned}$$

This implies that $f_\varepsilon$ and $\psi _\varepsilon$ have the required form, as well as f and $\psi$.

(iii) We show in this case that the functions $f_{\zeta _1,r}$ and $\psi _{\zeta _2,r}$ defined on G by

$$\begin{aligned} f_{\zeta _1,r}(c,h)=\zeta _1(h)e^{-{\pi } r \Vert c\Vert ^2} \quad \text {and }\quad \psi _{\zeta _2,r}(c,h)=\zeta _2(h)e^{-{\pi } r \Vert c\Vert ^2} \end{aligned}$$

satisfy condition (6) of Theorem 3 for any $r\in [p,1/q]$ and any $\zeta _1,\zeta _2\in L^2(K)$. Indeed, for $g=(a,k)\in G$, we have

$$\begin{aligned} \left( f_{\zeta _1,r}\right) _{\psi _{\zeta _2,r}}^g(c,h)= & {} f_{\zeta _1,r}(c,h)\overline{\psi _{\zeta _2,r}}\big ((a,k)^{-1}(c,h)\big )\\= & {} f_{\zeta _1,r}(c,h)\overline{\psi _{\zeta _2,r}}\big (-k^{-1}(a-c),k^{-1}h\big )\\= & {} \zeta _1(h)\overline{\zeta _2}(k^{-1}h)e^{-{\pi } r \Vert c\Vert ^2}e^{-{\pi } r \Vert c-a\Vert ^2}. \end{aligned}$$

Thus,

$$\begin{aligned} \left( \left( f_{\zeta _1,r}\right) _{\psi _{\zeta _2,r}}^g(\cdot ,h)\right) ^{\widehat{}}(\gamma )= & {} \zeta _1(h)\overline{\zeta _2}(k^{-1}h) \int _{\mathbb {R}^n}e^{-{\pi } r \Vert c\Vert ^2}e^{-{\pi } r \Vert c-a\Vert ^2}e^{-2i{\pi } \gamma c }dc\\= & {} \zeta _1(h)\overline{\zeta _2}(k^{-1}h)e^{-i{\pi } \gamma a } \int _{\mathbb {R}^n}e^{-{\pi } r \Vert c+a/2\Vert ^2}e^{-{\pi } r \Vert c-a/2\Vert ^2}e^{-2i{\pi } \gamma c }dc\\= & {} \zeta _1(h)\overline{\zeta _2}(k^{-1}h)e^{-i{\pi } \gamma a }e^{-{\pi \over 2} r \Vert a\Vert ^2} \int _{\mathbb {R}^n}e^{-2{\pi } r \Vert c\Vert ^2}e^{-2i{\pi } \gamma c }dc\\= & {} (2r)^{-n/2}\zeta _1(h)\overline{\zeta _2}(k^{-1}h)e^{-i{\pi } \gamma a }e^{-{\pi \over 2} r \Vert a\Vert ^2} e^{-{ {\pi }\Vert \gamma \Vert ^2\over 2r}}. \end{aligned}$$

By using (5), the kernel of the operator $\pi _{\gamma ,\sigma }\left( \left( f_{\zeta _1,r}\right) _{\psi _{\zeta _2,r}}^g\right)$ is defined on $K/K_\gamma \times K/K_\gamma$ by

$$\begin{aligned} H\left( {\left( f_{\zeta _1,r}\right) _{\psi _{\zeta _2,r}}^g},\gamma , \sigma \right) (s,u)= & {} (2r)^{-n/2}e^{-i{\pi } s\cdot \gamma a }e^{-{\pi \over 2} (r \Vert a\Vert ^2+{1\over r}\Vert \gamma \Vert ^2)}\\{} & {} \times \int _{K_\gamma }\zeta _1(svu^{-1})\overline{\zeta _2}(k^{-1}svu^{-1})\sigma (v)d\mu _\gamma (v)\\= & {} (2r)^{-n/2}e^{-i{\pi } s\cdot \gamma a }e^{-{\pi \over 2} (r \Vert a\Vert ^2+{1\over r}\Vert \gamma \Vert ^2)}\sigma \left( \zeta (s,u,k) \right) , \end{aligned}$$

where $\zeta (s,u,k)(v)=\zeta _1(svu^{-1})\overline{\zeta _2}(k^{-1}svu^{-1}),$ for all $v\in K_\gamma$. It follows, using (7), that

$$\begin{aligned}{} & {} \left\| \mathcal {G}_{\psi _{\zeta _2,r}} f_{\zeta _1,r}(g,\pi _{\gamma , \sigma }) \right\| _{HS}=\left\| \pi _{\gamma , \sigma }\Big (\left( f_{\zeta _1,r}\right) _{\psi _{\zeta _2,r}}^g\Big ) \right\| _{HS} \\{} & {} \quad =(2r)^{-n/2}e^{-{\pi \over 2} (r \Vert a\Vert ^2+{1\over r}\Vert \gamma \Vert ^2)}\left( \int _{K/K_\gamma } \int _{K/K_\gamma } \left\| \sigma \left( \zeta (s,u,k) \right) \right\| ^2_{HS}d\dot{\mu }_\gamma (sK_\gamma )d\dot{\mu }_\gamma (uK_\gamma )\right) ^{1\over 2}\\{} & {} \quad \le (2r)^{-n/2}\tilde{\phi }_\gamma (k,\sigma )e^{-{\pi \over 2} (p \Vert a\Vert ^2+q\Vert \gamma \Vert ^2)}, \end{aligned}$$

where $\tilde{\phi }_\gamma (k,\sigma )=\left( \int _{K/K_\gamma } \int _{K/K_\gamma } \left\| \sigma \left( \zeta (s,u,k) \right) \right\| ^2_{HS}d\dot{\mu }_\gamma (sK_\gamma )d\dot{\mu }_\gamma (uK_\gamma )\right) ^{1\over 2}$. Finally, notice that

$$\begin{aligned}{} & {} \int _K\sum _{\sigma \in \hat{K}_\gamma }\tilde{\phi }_\gamma (k,\sigma )^2\\ {}{} & {} \quad =\int _K \int _{K/K_\gamma } \int _{K/K_\gamma } \sum _{\sigma \in \hat{K}_\gamma }\left\| \sigma \left( \zeta (s,u,k) \right) \right\| ^2_{HS}d\dot{\mu }_\gamma (sK_\gamma )d\dot{\mu }_\gamma (uK_\gamma )\, d\mu (k)\\{} & {} \quad =\int _K \int _{K/K_\gamma } \int _{K/K_\gamma }\int _{K_\gamma }\left| \zeta _1(svu^{-1})\overline{\zeta _2}(k^{-1}svu^{-1}) \right| ^2d\mu _\gamma (v)\,\\{} & {} \quad \quad d\dot{\mu }_\gamma (sK_\gamma )d\dot{\mu }_\gamma (uK_\gamma )\, d\mu (k) \end{aligned}$$

(using the Plancherel formula for$K_\gamma$)

$$\begin{aligned}= & {} \Vert \zeta _2\Vert _2^2\int _{K/K_\gamma } \int _{K/K_\gamma }\int _{K_\gamma }\left| \zeta _1(svu^{-1}) \right| ^2d\mu _\gamma (v)\, d\dot{\mu }_\gamma (sK_\gamma )d\dot{\mu }_\gamma (uK_\gamma )\\= & {} \Vert \zeta _2\Vert _2^2\int _{K/K_\gamma } \int _{K}\left| \zeta _1(ku^{-1}) \right| ^2 d\mu (k)\, d\dot{\mu }_\gamma (uK_\gamma )\le \Vert \zeta _2\Vert _2^2\Vert \zeta _1\Vert _2^2, \end{aligned}$$

which is independent of $\gamma$. This completes the proof.