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!