If $s\in\mathbb R$ define the sobolev space $H^s(\mathbb R^n)$ as $$H^s(\mathbb R^n):=\{u\in \mathcal{S}^{'}(\mathbb R^n): (1+|\xi|^2)^{s/2}\hat{u}\in L^2(\mathbb R^n)\}.$$ I have a doubt concearning this definition. The Fourier transform of $u\in \mathcal{S}^{'}(\mathbb R^n)$ is again a tempered distribution so what would be to show $(1+|\xi|^2)^{s/2} \hat{u}$ is a function? Does that mean I should exhibit $g\in L^2(\mathbb R^n)$ such that $$\langle (1+|\xi|^2)^{s/2}\hat{u}, \phi\rangle=\langle g, \phi\rangle=\int_{\mathbb R^n}g(x)\phi(x)\ dx, $$ for all $\phi\in \mathcal{S}(\mathbb R^n)$? This would be strange because of the term $(1+|\xi|^2)^{s/2}$..
Can anyone give me an example of the calculation involved for showing $u\in \mathcal{S}^{'}(\mathbb R^n)$ is in $H^s(\mathbb R^n)$?
Above $\mathcal{S}(\mathbb R^n)$ is the Schwartz space and $\mathcal{S}^{'}(\mathbb R^n)$ is the space of tempered distributions, the topological dual of $\mathcal{S}(\mathbb R^n)$.
Yes, you must show that the tempered distribution $(1+|\xi|^2)^{s/2}\,\hat{u}$ is given by integration against an $L^2$ function. The multiplication by $(1+|\xi|^2)^{s/2}$ does not change whether or not it is an a.e. pointwise function, but it does affect square integrability. For example, the Fourier transform of Dirac $\delta$ is (up to a constant) the function $1$. This is not in $L^2(\mathbb R^n)$, but $(1+|\xi|^2)^{-r/2}\cdot 1$ is $L^2$ for $r>n$.