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...
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.