My question is a follow-up to Abdelmalek Abdesselam's recent post

What makes Gaussian distributions special? Local field version?

asking about various characterizations of (real-valued) Gaussian distributions which remain valid for other analogues of Gaussian distributions/functions (e.g. in the p-adic context). One interesting characterization arises with Babenko-Beckner's fine-tuning of the Hausdorff inequality (see https://en.wikipedia.org/wiki/Babenko–Beckner_inequality). For real numbers $s, t$ with ${1 \over s} + {1 \over t} = 1$ and $1 < s \leq 2$ it is known that the Fourier transform $f \mapsto \hat{f}$ maps $L^s(\Bbb{R}^n)$ to $L^t(\Bbb{R}^n)$ and satisfies the inequality

\begin{equation} \| \hat{f} \, \|_t \ \leq \ \Big( s^{1 \over s} \, t^{-{1 \over t}} \Big)^{n \over 2} \, \| f \|_s \quad \left( {\scriptstyle \begin{array}{l} \text{Babenko} \\ \text{Beckner} \\ \text{inequality} \end{array}} \right) \end{equation}

When $s = t = 2$ this inequality becomes an equality which is valid for all $f \in L^2(\Bbb{R}^n)$. For $s < 2$ equality is achieved if and only if $f$ is a Gaussian function.

My question concerns an analogue of this inequality for finite fields: Let $q$ be a power of a prime $p$ and let $\Bbb{F}_q$ be the finite field with $q$ elements. Choose a non-square $\delta \in \Bbb{F}_q$ and form the quadratic extension $\Bbb{F}_q\big( \sqrt{\delta} \big)$. We view elements of $\Bbb{F}_q\big( \sqrt{\delta} \big)$ as linear combinations of the form $z = x + \sqrt{\delta} y$ with $x, y \in \Bbb{F}_q$ subject to the usual rules of addition and multiplication. Conjugation and norm are expressed, respectively, as $\bar{z} = x - \sqrt{\delta} y$ and $\mathrm{N}(z)= x^2 - \delta y^2$. Furthermore define $\mathrm{Tr}(z) := z + \bar{z}$. Choose any non-trivial additive character $\psi: \Bbb{F}_q \longrightarrow \Bbb{C}^*$ and define the $\Bbb{F}_q\big( \sqrt{\delta} \big)$-Fourier transform $\widehat{f}$ of a complex-valued function $f: \Bbb{F}_q\big( \sqrt{\delta} \big) \longrightarrow \Bbb{C}$ by

\begin{equation} \widehat{f}(z) \ := \ {1 \over q} \, \sum_{w \in \Bbb{F}_q\big( \sqrt{\delta} \big)} \, f(w) \, \psi \Big(-\mathrm{Tr}(zw) \Big) \end{equation}

If we endow the function space $\Bbb{C}\big[ \Bbb{F}_q\big( \sqrt{\delta} \big) \big]$ with the hermitian inner product

\begin{equation} \langle f , g \rangle \ := \ {1 \over q} \, \sum_{w \in \Bbb{F}_q\big( \sqrt{\delta} \big)} \, f(w) \, \overline{g(w)} \end{equation}

then Plancherel holds, i.e. $\| \widehat{f} \, \|_2 = \| f \|_2$ and the Babenko-Beckner inequality should take the form

\begin{equation} (\dagger) \quad \| \widehat{f} \, \|_t \ \leq \ \|f \, \|_s \end{equation}

for any pair of real numbers $s,t$ with ${1 \over s} + {1 \over t} = 1$ and $1 < s \leq 2$. This is a finite field rendering of a more general version of the Babenko-Beckner inequality that holds for finite abelian groups (see for example https://www.e-periodica.ch/cntmng?pid=ens-001:2000:46::190). As a side note, I would very keen to learn what shape this equality takes in the non-abelian setting.

For $s<2$ inequality is not strict. Indeed $(\dagger)$ becomes an equality for what I'll call the $\Bbb{F}_q\big( \sqrt{\delta} \big)$-Gaussian functions defined by $G_x(z) := \psi \big( x\, \mathrm{N}(z) \big)$ with $x \in \Bbb{F}_q$. This is because $G_x$is an eigenfunction of the $\Bbb{F}_q\big( \sqrt{\delta} \big)$-Fourier transform whose eigenvalue will be a unit complex number; indeed the Fourier transform is unitary!

Question: Within the range $1 < s < 2$ does inequality $(\dagger)$ become an equality if and only if $f(z) = c \,G_x(z-w)$ for some parameter $x \in \Bbb{F}_q$, some shift $w \in \Bbb{F}_q\big( \sqrt{\delta} \big)$, and some overall scalar factor $c \in \Bbb{C}$?

thanks, ines.