TITLE: fourier analytic proofs
QUESTION [6 upvotes]: While searching through Mathoverflow, I found out a fourier analytic proof of the Isoperimetric Inequality.Also, by google search I found a fourier analytic proof of Quadratic Reciprocity theorem.I know of the fourier analytic approach used by combinatorialists like Ben Green. But what are the other fourier analytic proofs of some of the well known classical theorems other than what I have mentioned above specially those which admit a starkly different proofs.

REPLY [9 votes]: The sign of the quadratic Gauss sum $\tau$ can be obtained from
the spectrum of the discrete Fourier transform $\Phi$: the trace of $\Phi$
gives $\tau$, and $\det\Phi$ distinguishes $\tau$ from $-\tau$.
Recall that for an odd prime $p$ the quadratic Gauss sum
can be defined by
$$
\tau = \sum_{n=0}^{p-1} \zeta^{n^2}
$$
where $\zeta = e^{2\pi i / p}$.  It is elementary that $|\tau|^2 = p$
and that $\tau$ is real or pure imaginary according as $p \equiv 1 \bmod 4$
or $p \equiv -1 \bmod 4$.  In fact $\tau$ is always $+\sqrt p$
in the former case, and $+i\sqrt p$ in the latter, but this is
notoriously tricky to prove.
One trick is to recognize $\tau$
as the trace of the discrete Fourier transform on ${\bf C}^p$,
which has matrix
$$
\Phi = (\zeta^{mn})_{m,n=0}^{p-1}.
$$
Now $\Phi^2$ is the matrix whose $(m,n)$ entry is $p$ if $m+n \equiv 0 \bmod p$
and $0$ otherwise (this is tantamount to discrete Fourier inversion).
This matrix has eiganvalues $+1$ and $-1$ with multiplicity $(p+1)/2$
and $(p-1)/2$ respectively.  Hence $\Phi$ has eigenvalues $i^k \sqrt p$
($k=0,1,2,3$) with multiplicities $m_k$ satisfying
$m_0 + m_2 = (p+1)/2$ and $m_1 + m_3 = (p-1)/2$, and then
$\tau = \sum_{k=0}^3 m_k i^k \sqrt p$.
Since we already know $\tau$ up to sign there are only two possibilities:
if $p \equiv 1 \bmod 4$ then $m_0$ or $m_2$ is $(p+3)/4$
and the other three $m_k$ are $(p-1)/4$, while
if $p \equiv -1 \bmod 4$ then $m_1$ or $m_3$ is $(p-3)/4$
and the other three $m_k$ are $(p+1)/4$.
We are to show that the odd man out is always
$m_0$ in the former case and $m_3$ in the latter.
In each case we can decide the correct choice by computing the
sign (a.k.a. argument) of $\det \Phi = p^{p/2} \prod_{k=0}^3 i^{k m_k}$.
We can do this because $\Phi$ is a Vandermonde matrix, whence $\det\Phi$
has the product expansion $\prod_{0 \leq m < n < p} (\zeta^n - \zeta^m)$.
Each factor $\zeta^n - \zeta^m$ is a positive real multiple of
$\exp((m+n+\frac12)\pi i)$.  It soon follows that
$\det\Phi = i^{(1-p)/2} p^{p/2}$ (we already knew $\left|\det\Phi\right|$
because each eigenvalue has absolute value $\sqrt{p}$),
and conclude as desired that $\tau = \sqrt{p}$ when $p \equiv 1 \bmod 4$
while $\tau = i\sqrt{p}$ when $p \equiv -1 \bmod 4$.
[This looks like a known but not very well-known argument
that is easier to rediscover than to find in the literature.
What is the original source?]