Property of the Fourier transform in Schwartz space Let $F: S(\mathbb{R}^n) \rightarrow S(\mathbb{R}^n)$ the function that associates to the function $f$ its Fourier transform $\hat f$.
Is it a bijection in the Schwartz space?
 A: Yes, one can note that the Fourier Inversion Theorem still holds for continuous functions that are integrable. Since $\mathcal{S}(\mathbb{R}^n)\subset L^1(\mathbb{R}^n)$, the Schwarz functions are indeed integrable.
A: Yes, Fourier transform is a bijection on Schwartz functions. In fact, it is a homeomorphism for the Frechet space topology on Schwartz functions.
The keyword(s) to google for related stuff is "Fourier inversion (transform)"... which gives a two-sided inverse to Fourier transform.
The proof can be found many places. Letting $F$ be Fourier transform and $I$ the inverse transform, the heuristic for the proof of Fourier inversion is
$$
I(Ff)(x) \;=\; \int e^{2\pi ix\xi} Ff(\xi)\;d\xi
\;=\; \int e^{2\pi ix\xi}\int e^{-2\pi i \xi t}f(t)\;dt\;d\xi
$$
$$
=_{???}\; \int f(t) \Big(\int e^{2\pi i\xi(x-t)}\;d\xi\Big)dt
\;=_{???} \int f(t)\,\delta_{x-t}\;dt 
\;=\; f(x)
$$
An immediate problem is that the interchange of integrals is not justified. However, if we are able to believe that the inner integral is $\delta_{x-t}$ (Dirac delta), then this does produce the correct outcome.
To make this into a proof, insertion of a dummy Schwartz function is the usual ploy...
