I understand that the Sobolev space $H^{s}:=\left\{u\in L^2:=L^2(\mathbb{R}^n): \int_{\mathbb{R} ^n} |(1+|\xi|^2)^{s/2}\widehat{u}(\xi)|^2\,d\xi<\infty\right\}$ is isomorphic to $L^2$ for any $s\in\mathbb{R}$.
Question 1. This is right? The isomorphism is $T:L^2\to H^{s}$ given by $Tu(\xi)=(1+|\xi|^2)^{s/2}\widehat{u}(\xi)$.
The argument to verify the surjectivity of the map $T$ consists in using the fact that $\frac{1}{(1+|\xi|^2)^{s/2}}$ is bounded. Indeed, if $v\in H^{s}$, then $Tu=v $ with $\widehat{u}=\frac{1}{(1 +|\xi|^2)^{s/2}}v$. As $v\in L^2$ by definition of the space $H^{s}$ and $\frac{1}{(1+|\xi|^2)^{s/2}}$ is bounded, then $\widehat{u}$ is in $L^2$, and by Plancherel's theorem, $u$ is in $L^2$.Therefore, $T$ is surjective.
Question 2. If a weighted space is defined, denoted and defined by $H^{s}(f):=\left\{u\in L^2: \int_{\mathbb{R}^n} |(1+f(\xi))^{s/2}\widehat{u}(\xi)|^2\,d\xi<\infty\right\}$.
Would this space also be isomorphic to $L^2$ if $f$ is assumed to be a function such that $\frac{1}{1+f(\xi )^{s/2}}$ is bounded?
Thanks.