Sobolev spaces are Hilbert spaces

Question

Define $H^s(\mathbb{R}^n)=\lbrace u\in \mathcal{S}'(\mathbb{R}^n): (1+\vert y\vert^2)^\frac{s}{2}\hat{u}\in L^2(\mathbb{R}^n)\rbrace$ where $\mathcal{S}'(\mathbb{R}^n)$ is the space of tempered distributions together with the norm $\Vert u\Vert_{H^s}= \Vert (1+\vert y\vert^2)^\frac{s}{2}\hat{u}\Vert_{L^2}$. I've read that $H^s$ are Hilbert spaces, but I am stuck showing that they are complete. If $u_n$ is a Cauchy sequence in $H^s$, then $\hat{u}_n$ is a Cauchy sequence in a weighted $L^2$ space. Therefore we can find $v^*$ with $(1+\vert y\vert^2)^\frac{s}{2}v^*\in L^2$ and $(1+\vert y\vert^2)^\frac{s}{2}\hat{u}_n\rightarrow (1+\vert y\vert^2)^\frac{s}{2}v^*$ in $L^2$. But I don't know how to conclude that $v^*$ is the Fourier transform of a tempered distribution. For $s\geq 0$ the case is clear, but not for $s<0$.

Answer 1

As shown e.g. in Rudins Functional Analysis: Let $1\leq p <\infty$ and $N>0$ with $$\int_{\mathbb{R}^n}\vert (1+\vert x\vert^2)^{-N}u(x)\vert^p dx<\infty.$$ Then $u $ is a tempered distribution.

Therefore your function $v^\ast$ is a tempered distribution for all $s$ and thus we can find a tempered distribution $v$ with $\hat{v}=v^\ast$.