Bessel potential space: Proof of completeness I want to know a proof that the (one-dimensional) Bessel potential space (for $p=2$)
$$H^s(\mathbb{R})=\left\{f:\mathbb{R}\to\mathbb{C}:\int_{\mathbb{R}}(1+\lvert \xi\rvert^2)^{\frac{s}{2}}\lvert \hat{f}\rvert^2\,d\xi<\infty\right\}$$
is complete, even if $s<0$. I tried it myself and the case $s\ge 0$ is easy, but I can't figure out the case $s<0$. It's easy to find an isometry mapping $H^s$ into $L^2$ or $H^{-s}$ using Fourier-multipliers, because the spaces we want to map into are complete and we can use the representation theorem of Riesz. But all this approach yields for me is that the completion of $H^s$ is isomorphic to a Hilbert space.
Does anybody know a proof? I am aware that it's (probably) possible using Triebel-Lizorkin-spaces, but that seems excessive. Is there a trick involving dual spaces or closed images?
 A: Seems to me that the definition given for $H^s$ can't be right for $s<0$; it "must" be that $H^s$ is actually the space of tempered distributions $f$ such that $\hat f\in L^2(\mu)$, where $d\mu(\xi)=(1+|\xi|^2)^{s/2}\,d\xi$.
Assuming so, this is easy: Say $(f_n)$ is Cauchy in $H^s$. Then $(\hat f_n)$ is Cauchy in $L^2(\mu)$. So $\hat f_n\to g$ in $L^2(\mu)$. Now $g\in L^2(\mu)$ implies that $g$ is a tempered distribution, so $g=\hat f$ for some tempered distribution $f$, and now $f_n\to f$ in $H^s$ by definition.

Come to think of it, this proves that $H^s$ is not a space of functions (unless it's actually not complete). Say $\mathcal S$ is the Schwarz space. Then we certainly have $\mathcal S\subset H^s$, and the argument above shows that the completion of $\mathcal S$ is the space of tempered distributions that I describe.
And no, those distributions are not all functions. There exists a tempered distribution $f$ so that $\hat f$ is a function and in fact $\hat f(\xi)=\xi^{10}$. If $s<0$ is small enough (large enough? close enough to $-\infty$) then $f\in H^s$. But $f$ is certainly not a function.
