# 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?

• Yes it is, See Read & Simon Mathematical Physics Volume II
– Paul
May 6, 2021 at 17:15
• This is the "Schwartz" with a "t"... :) May 6, 2021 at 18:41

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.
• No, I don't think so. OP is already assuming/knows that the Fourier transform of a Schwarz function is Schwarz. Hence, they right $F:\mathcal{S}(\mathbb{R}^n)\rightarrow \mathcal{S}(\mathbb{R}^n)$. The question is whether this operator is invertible. May 6, 2021 at 17:34
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.