I am currently trying to understand the Fouriertransform of Schwartz functions. There are two proofs I'm trying to figure out.
- $\mathcal{F}$ is a continous operator $\mathcal{S}(\mathbb R) \rightarrow \mathcal{S}(\mathbb R)$.
- $\mathcal{F}$ is a automorphism on $\mathcal{S}( \mathbb R)$.
At first we prove, that $\forall f \in \mathcal{S}( \mathbb R) :\widehat f \in \mathcal{S}( \mathbb R)$.
This proof is a use of the Riemann Lebesgue Lemma and the fact, that the derivatives of the FT can be expressed as multiplication with polynomials.
That $\mathcal F$ is continuous follows:
from the Riemann Lebesgue Lemma and the Fact, that $ \mathcal{S}( \mathbb R) \subset L^1( \mathbb R) $ and $\mathcal{S}( \mathbb R) \subset C_0( \mathbb R)$
Now my question here is, if any arguments can be simplifyed. Or if the continuity already follows from the continuity of $\mathcal{F}: L^1 \rightarrow C_0$.
For the proof of the automorphism property:
The Plancherel theorem gives us an isomorphism $\mathcal F :L^2 \rightarrow L^2$. So we deduce that $\mathcal F$ is injective on $\mathcal S \subset L^2$.
For the surjevtivity we use the inversion theorem which gives us a $\mathcal F^{-1}$ for all $f \in L^1$ where $\widehat f \in L^1$ (hence for all $\mathcal S$ functions).
At this point im not sure if my proof is 100% correct or if it could be simplyfied as well.