Fourier transform on Schwartz space leads to identity.

Question

Let $f$ be a function in the Schwartz space and $F(f) = \left(\frac{1}{2 \pi}\right)^{n/2} \int_{R^n} e^{i \langle x,y \rangle } f(x) dx $ the Fourier transform. Is it true that $F^4(f) = f$? I know that the inverse of $F$ in the Schwartz space is $\left(\frac{1}{2 \pi}\right)^{n/2} \int_{R^n} e^{-i \langle x,y \rangle} f(x) dx $ but can't see how I can use that in my expression.

Answer 1

Hint: show first that $F^2 f(x) = f(-x)$. You can do this from your knowledge of what $F^{-1}$ is without extra work using the almost symmetry of the Fourier transform and its inverse.