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Definition (Schwartz Space) We say $f \in S(\mathbb{R}) \Leftrightarrow \sup_{x \in \mathbb{R}} |x^i D^j f(x)|< \infty \Leftrightarrow \sup_{x \in \mathbb{R}}|(1+|x|)^n D^k f(x)| < \infty $

for all $i,j,k,n \in \mathbb{N_0}$.

Definition (Fourier Trnsformation): Let $f \in S(\mathbb{R})$ be a Schwartz function then, $\widehat{f}(\xi):=\int_{\mathbb{R}}f(x) e^{-2 \pi i x \xi}dx$ is called the Fourier transform of $f$.

I am struggling with the following: Let $f \in S(\mathbb{R})$ then the Fourier transform exists.

$\int_{\mathbb{R}}f(x)dx\leq \int_{\mathbb{R}}|f(x)|dx=\int_{\mathbb{R}}(1+x^2)|f(x)|\frac{1}{1+x^2}dx=\int_{\mathbb{R}}(1+x^2) |f(x)|dx \int_{\mathbb{R}}\frac{1}{1+x^2}dx=(\int_{\mathbb{R}}(1+x^2) |f(x)|dx) \pi$

To be able to conclude that the Fourier Trnsformation does exist on $S(\mathbb{R})$ As far as I know I somehow have to show that $\int_{\mathbb{R}}(1+x^2) |f(x)|dx$ is bounded, then the prove would be finished.

I don't know how to continue from here on. Could someone please show me (in detail) how to continue from here on. (If my attempt is complety wrong, then an attempt that is working.)

Sha Vuklia
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Sigi
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  • Hang on, multiplying the two integrals is blatantly wrong. What you need is to show the function $(1+x^2)|f(x)|$ is bounded. This follows immediately from the definition of Schwartz function. – peek-a-boo Jun 10 '23 at 14:09
  • @peek-a-boo Yeah that with the Integrals was a mistake. I don't really see why $(1+x^2) |f(x)|$ has to be bounded. – Sigi Jun 10 '23 at 14:12
  • First question is $|f|$ bounded? Next question: is $x\mapsto |x^2 f(x)|$ bounded? – peek-a-boo Jun 10 '23 at 14:28
  • @peek-a-boo $|f|$ is bounded since, let $C_{ij}:=sup_{x \in R}|x^iD^jf(x)| $, then from the definiton the result follows. If I use the second equivalent definiton with $C_{20}$ I get that $(1+x^2)|f|$ is bounded too. – Sigi Jun 10 '23 at 14:38
  • Yes, so this is one method of proof. Another method of proof is that you can split up the integral to a region near the origin and away. Near the origin, by continuity, you get a finite contribution. Away from the origin, use the quadratic bound. – peek-a-boo Jun 10 '23 at 14:40
  • @peek-a-boo Thinking about it, to show that $(1+x^2) |f|$ is bounded doesn't do the job to show that the integral exists. Since $\int_{\mathbb{R}}C dx$ does not converge – Sigi Jun 10 '23 at 14:40
  • I refer you to my first comment. You said you realized your mistake. But apparently you didn’t… think about it some more. – peek-a-boo Jun 10 '23 at 14:41
  • @peek-a-boo Oh yeah, since $(1+x^2)|f(x)|$ is bounded and $\frac{1}{1+x^2}$ is bounded too. $\int_{\mathbb{R}}(1+x^2)|f(x)|\frac{1}{1+x^2} dx$ is finite because $(1+x^2)|f(x)| \leq C$ and since $\int_{\mathbb{R}}\frac{1}{1+x^2}dx=\pi$ we get $\int_{\mathbb{R}}(1+x^2)|f(x)|\frac{1}{1+x^2} dx \leq \int_{\mathbb{R}}C \frac{1}{1+x^2} dx=C \pi$ Is that correct? – Sigi Jun 10 '23 at 14:49
  • Yes. Just that you made one true but irrelevant remark “.. and $\frac{1}{1+x^2}$ is bounded too…” Boundedness of this function is irrelevant, what we care about is that this is integrable (btw the fact that its integral is exactly $\pi$ is also irrelevant). – peek-a-boo Jun 10 '23 at 14:55
  • @peek-a-boo Ok, thank you !! – Sigi Jun 10 '23 at 15:04

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