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How can I see that $S(\mathbb{R}) \subset H^s(\mathbb{R})$, where the former is Schwartz and the latter is Sobolev space ? This should be obvious according to my notes but unfortunately I can't make an argument why this should be true ...

What I know is that for $f \in S(\mathbb{R})$ I have $\hat{f} \in S(\mathbb{R})$ where $\hat{f}$ stands for the Fourier Transform of $f$. Hence for any $k,m = 0,1,2, \dots$ I can find constants $c_{k,m}$ such that \begin{equation} \sup_x |\xi^k\hat{f}^{(m)}(\xi)| \leq c_{k,m} \end{equation} Now I need to make the step to argue that this implies \begin{equation} \int |(1+\xi^2)^{s/2}\hat{f}(\xi)|^2 \,d\xi < \infty \end{equation} but there I am struggeling. I mean it doesn't help that I can write this as \begin{equation} \int |(1+\xi^2)^{s/2}\hat{f}(\xi)|^2 \,d\xi \quad \leq \quad c^2_{0,0} \int |(1+\xi^2)^{s/2}|^2 \,d\xi \end{equation} since the LHS is still to big to be finite. It's also not enough to say \begin{equation} \int |(1+\xi^2)^{s/2}\hat{f}(\xi)|^2 \,d\xi \quad \leq \quad \tilde{c}_{k,0} \int \,d\xi \end{equation} (where $\tilde{c}_{k,0}$ stands for a constand that I obtain by expanding $(1 + \xi^2)^{s/2}$ and then using the property that $\hat{f}$ decays faster than any polynomial.)

There must be more that I can deduce from the fact that $\hat{f} \in S(\mathbb{R})$ but I seem to be blind.

Any hint would be helpful, many thanks!

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up vote 4 down vote accepted

It's a standard result that the Schwartz space is stable by Fourier transform (in other word, if $f\in\mathcal S(\mathbb R)$ then $\mathcal Ff\in\mathcal S(\mathbb R))$. We can find a constant $C$ such that $\sup_{x\in\mathbb R}|(1+\xi^2)^{\lfloor s/2\rfloor +3}\mathcal F(f)(x)|\leq C$. So we have $$|(1+\xi^2)^{s/2}\mathcal F(f)(x)|\leq (1+\xi^2)^{\lfloor s/2\rfloor +1}\mathcal F(f)(x)\leq C\frac 1{1+x^2},$$ which is square-integrable.

In fact, it can we shown that the Schwartz space is dense in $H^s(\mathbb R)$ for all $s$, and so is $\mathcal D(\mathbb R)$.

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aah, of course that makes perfect sense, thanks a lot ! – harlekin Feb 4 '12 at 19:54
Isn't $S(\mathbb{R}) \subset W^{k,p}(\mathbb{R})$ for all $k, p$, by the basic definition of Schwartz space, no Fourier transforms needed? – Stefan Smith May 26 '12 at 1:33
When $k$ is an integer yes, but here we work with $H^s=W^{s,2}$ with $s$ a real number (and we fix the exponent). – Davide Giraudo May 26 '12 at 7:44
@Davide: Can you take $k$ an integer with $k \geq s$ and conclude $S(\mathbb{R}) \subset H^k(\mathbb{R}) \subset H^s(\mathbb{R})$? – Stefan Smith May 27 '12 at 1:10
Rigorously speaking you have a ``copy'' of $S(\mathbb R)$ inside $H^s(\mathbb R^n)$ for $H^s(\mathbb R^n)$ consists of functionals and the elements of $S(\mathbb R)$ are functions which induces unique functionals which live in $H^s(\mathbb R)$.. – PtF Jan 4 '15 at 0:05

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