What happenes if you apply the Fourier transform for $n$ times to a function?

Question

I know that $\mathcal{F} \{\mathcal{F} \{ \mathcal{F} \{ \mathcal{F} \{ f(t) \}\}\}\} = f(t)$. How can I prove this statement?

Also, whith whom are equal $\mathcal{F} \{ \mathcal{F}\{ f(t) \}\}$ and $\mathcal{F} \{ \mathcal{F} \{\mathcal{F} \{ f(t) \}\}\}$?

Answer 1

Hint: \begin{align} [\mathcal{F}^2 f](x) = f(-x), \ \ [\mathcal{F}^3 f](x) = \mathcal{F}^{-1}f(x)= [\mathcal{F} f](-x) \ \ \text{ and } \ \ [\mathcal{F}^4 f](x) = f(x). \end{align}

Edit: Let me do the first computation. Observe \begin{align} [\mathcal{F}^2f](x)=&\ \int^{\infty}_{-\infty} \left[\int^\infty_{-\infty} f(t)e^{-2\pi i \xi t}\ dt \right]e^{-2\pi i \xi x}\ d\xi\\ =&\ \int^\infty_{-\infty}\int^\infty_{-\infty} f(t) e^{-2\pi i\xi(x+t)}\ dtd\xi\\ =&\ \int^\infty_{-\infty}f(t)\left[\int^\infty_{-\infty} e^{-2\pi i\xi(x+t)}\ d\xi\right] dt\\ =&\ \int^\infty_{-\infty} f(t)\delta(x+t)\ dt = f(-x). \end{align}