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Is the Fourier transform of the Fourier transform of $f(t)$: $$\hat{\hat{f(t)}} = f(-t)$$ or $$\hat{\hat{f(t)}} = 2\pi f(-t)$$ ?

I have read the two versions here and here (respectively) for example.

And also, can I show this from the definition ? I tried this :

$$ \hat{\hat{f(t)}} = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\right)e^{-i\omega t}dt $$ $$= \hat{f(\omega)}\int_{-\infty}^{\infty}e^{-i\omega t}dt$$

but I don't really know where I'm going with this...

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user1234161
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1 Answers1

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UPDATE:

The proof below might be wrong, due to a faulty substituion, as pointed out by Dan in the comments.

I marked the equation in question with a question mark.

Original Proof Attempt:

You reuse the $t$ and $\omega$ variables, also you factor out $\hat{f}(\omega)$ while there is an integration running over it and it is not constant. So your derivation might better start like this:

\begin{align} \hat{\hat{f}}(\omega) &= \int\limits_{-\infty}^{+\infty}\hat{f}(t)e^{-i\omega t}dt \\ &= \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} f(\tau)e^{-it \tau}d\tau \, e^{-i\omega t}dt \\ &= \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} f(\tau)e^{-it (\tau+\omega)}d\tau \, dt \\ \end{align}

Switching integration order gives $$ \begin{align} \hat{\hat{f}}(\omega) &= \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} f(\tau)e^{-it (\tau+\omega)}dt \, d\tau \\ &= \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} e^{-it (\tau+\omega)}dt \, f(\tau) \, d\tau \\ &{? \atop =} \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} e^{-it u}du \, f(\tau) \, d\tau \\ &= \int\limits_{-\infty}^{+\infty} \!\! 2\pi\delta(u) \, f(\tau) d\tau \\ &= 2\pi \int\limits_{-\infty}^{+\infty} \!\! \delta(\tau + \omega) \, f(\tau) d\tau \\ &= 2\pi \int\limits_{-\infty}^{+\infty} \!\! f(\tau) \, \delta(\tau-(-\omega)) \, d\tau \\ &= 2\pi \, f(-\omega) \end{align} $$ where in between a substitution $u = t + \omega$ with $du = dt$ was used.

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