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 of this kind 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 $Bbb{F}_qbig( sqrt{delta} big)$.
We view elements of $Bbb{F}_qbig( sqrt{delta} big)$
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 $Bbb{F}_qbig( sqrt{delta} big)$-Fourier transform
$widehat{f}$ of a complex-valued function $f: Bbb{F}_qbig( sqrt{delta} big) longrightarrow Bbb{C}$ by
begin{equation}
widehat{f}(z) :=
{1 over q} , sum_{w in Bbb{F}_qbig( sqrt{delta} big)} ,
f(w) , psi Big(-mathrm{Tr}(zw) Big)
end{equation}
If we endow the function space $Bbb{C}big( Bbb{F}_qbig( sqrt{delta} big) big)$ with the hermitian inner product
begin{equation}
langle f , g rangle := {1 over q} , sum_{w in Bbb{F}_qbig( 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
$Bbb{F}_qbig( sqrt{delta} big)$-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 $Bbb{F}_qbig( sqrt{delta} big)$-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 $w in Bbb{F}_qbig( sqrt{delta} big)$, and some overall scalar factor $c in Bbb{C}$?
thanks, ines.