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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?

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

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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.

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  • $\begingroup$ 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. $\endgroup$ May 6, 2021 at 17:34
  • $\begingroup$ Oh yep I misinterpreted the question with the way it was worded! $\endgroup$ May 6, 2021 at 17:38
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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...

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