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

https://mathoverflow.net/questions/382952/what-makes-gaussian-distributions-special-local-field-version/385411?noredirect=1#comment982188_385411

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 refinement 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 (quadratic extensions of) 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 $\bar{z} := z^q = x - \sqrt{\delta} y$
and $\mathrm{N}(z) := z\bar{z} = x^2 - \delta y^2$. 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 \ := \  \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 $s<2$ the inequality is not strict. Indeed $(\dagger)$ becomes 
<s>an equality</s> for what I'll call the
$\Bbb{F}_q\big( \sqrt{\delta} \big)$-Gaussian functions
defined by $\mathrm{G}_x(z) := \psi \big( x\, \mathrm{N}(z) \big)$ 
where $x \in \Bbb{F}_q$ is a parameter. This is because (1) the values of $\mathrm{G}_x$ are all unit complex numbers and (2) it is **almost** an eigenfunction of the $\Bbb{F}_q\big( \sqrt{\delta} \big)$-Fourier transform. Indeed, let's check to be sure:

\begin{equation}
\begin{array}{ll}
\displaystyle \widehat{\mathrm{G}}_x(z) 
&\displaystyle = \ {1 \over q} 
\sum_{w \in \Bbb{F}_q\big( \sqrt{\delta} \big)}
\psi \big(x \mathrm{N}(w) \big) \, \psi \big(-\mathrm{Tr}(zw) \big) \\
&\displaystyle = \  {1 \over q} 
\sum_{w \in \Bbb{F}_q\big( \sqrt{\delta} \big)}
\psi \big(x w \bar{w} - zw - \bar{z} \bar{w} \big) \\
&\displaystyle = \  {1 \over q} 
\sum_{w \in \Bbb{F}_q\big( \sqrt{\delta} \big)}
\psi \Big( x^{-1} \mathrm{N}\big( xw - \bar{z} \big) - x^{-1} \mathrm{N}(z) \Big) \\
&\displaystyle = \  {1 \over q} \ 
\psi \big(  -x^{-1} \mathrm{N}(z) \big) 
\sum_{w \in \Bbb{F}_q\big( \sqrt{\delta} \big)}
\psi \Big( x^{-1} \mathrm{N}\big( xw - \bar{z} \big) \Big) \\
&\displaystyle = \  {1 \over q} \ 
\mathrm{G}_{-x^{-1}}(z) 
\sum_{u \in \Bbb{F}_q\big( \sqrt{\delta} \big)}
\psi \big( x^{-1} \mathrm{N}(u) \big) \\
&\displaystyle = \  {1 \over q} \ 
\mathrm{G}_{-x^{-1}}(z)\,
\Bigg( 1 \, + \,  
(q+1) \, \sum_{y \in \Bbb{F}_q^*}
\psi \big( x^{-1} y \big) \Bigg) \\
&\displaystyle = \  {1 \over q} \ 
\mathrm{G}_{-x^{-1}}(z) \, \big( -q \big)
\quad \text{($\psi$ is non-trivial!)} \\
&\displaystyle = \ -\mathrm{G}_{-x^{-1}}(z) 
\end{array}
\end{equation}


Note that $\mathrm{G}_x$ will be an eigenfunction of the $\Bbb{F}_q\big( \sqrt{\delta} \big)$-Fourier transform if and only if $x^2 = -1$. 

 

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

thanks, ines.