Fourier transform of the Fourier transform?

Question

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.

Answer 1

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 $-x$

Answer 2

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.

Answer 3

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.