# Fourier transform of the Fourier transform?

Can someone supply some resources on the Fourier transform of the fourier transform or clarify it for me?

It has been let known to me 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.

• If i recall correctly, Fourier transform of a Fourier transform of $f(x)$ gives us the function $f(-x)$ – Jakobian May 31 '18 at 19:35
• @Adam That's exactly what i'm looking for. Where can I find a more detailed source with explanation? ALso, this doesn't make sense to me intuitively – Goldname May 31 '18 at 19:36
• I'll check my notes. It's like i said. It's a direct conclusion from the formula for Fourier inverse transform – Jakobian May 31 '18 at 19:39
• en.wikipedia.org/wiki/Fourier_inversion_theorem – Hans Lundmark May 31 '18 at 19:46
• You can prove it using $2\pi\delta(x)=\int_\mathbb{R}\exp ikx dk$. – J.G. May 31 '18 at 19:52

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 $-x$
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.
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.