Fourier transform of the Fourier transform? Can someone refer me on the Fourier transform of the fourier transform or clarify it for me? 
It is known that the F.T. of the F.T. is some small modification of the original function itself, but I can't find anything about this online. 
 A: The Fourier transform of $f$ is:
$$F(\omega)=\mathscr{F}(f(t))\{\omega\}=\int_{-\infty}^{\infty} \mathrm{d}t f(t) e^{-i \omega t}$$
Transforming it again:
$$g(\tau)=\mathscr{F}(F(\omega))\{\tau\}=\int_{-\infty}^{\infty} \mathrm{d} \omega e^{-i \tau \omega}\int_{-\infty}^{\infty} \mathrm{d}t f(t) e^{-i \omega t}$$
Changing the order of integrations:
$$=\int_{-\infty}^{\infty} \mathrm{d}t f(t) \int_{-\infty}^{\infty} \mathrm{d} \omega e^{-i \omega t} e^{-i \omega \tau}$$
And $\mathscr{F}(e^{i a t})\{\omega\}=2 \pi \delta(\omega -a)$:
$$=\int_{-\infty}^{\infty} \mathrm{d}t f(t) 2 \pi \delta(t + \tau)$$
$$=2 \pi\int_{-\infty}^{\infty} \mathrm{d}t f(t) \delta(t + \tau)$$
$$=2 \pi f(-\tau)$$
But you might get a different result with a different definition of the FT.
A: We have that $$ f(x) = \int\limits_{\mathbb{R}^n} \mathfrak{F}(f)(s)e^{2\pi\cdot isx} ds $$
And from this
$$ f(x) = \mathfrak{ F}\circ\mathfrak{ F}(f)(-x) $$
Because all we do, is take the Fourier transform of $\mathfrak{F}(f)(s)$ with respect to $-x$
A: It may be easier to see in the finite case. The
Discrete Fourier Transform
is given by multiplication by the $\, n \!\times\! n\,$ DFT (Vandermonde) matrix. Define
$\, e_n(x) := \exp(2\pi\sqrt{-1}x/n),\,$ and matrix $\, T(n) := \{\frac1{\sqrt{n}}e_n(i j)\}_{i,j}.\,$ The
matrix $\,A(n) := T(n)^2\,$ is 
 $\, A(n)_{i,j} = \frac1n\!\sum_{k=0}^{n-1} e_n((i\!+\!j)k)\,$
but $\, d_n(j) := \sum_{k=0}^{n-1} e_n(jk) \,$ evaluates to $\, d_n(j) = n\,$ if
$\, j \equiv 0 \pmod n\,$ and $\, d_n(j) = 0\,$ otherwise. This implies that
$\, A(n)_{i,j} = 1\,$ if  $\, i \equiv -j \pmod n\,$ and  $\,A(n)_{j,j} = 0\,$ otherwise.
Thus, $\,A(n)\,$ takes any vector $\,(x_0,x_1,\dots,x_{n-1})\,$ to
$\,(x_0,x_{n-1},\dots,x_1).\,$ 
 This corresponds to $\,\mathscr{F}^2\!: f(x) \mapsto f(-x)\,$ in the continuous Fourier transform case.
