 Solution

Let $w(\xi )=(1+|\xi {|}^2{)}^{s/2}$. The map $U:f\mapsto w\hat{f}$ is linear, and $\Vert U(f+g){\Vert }_2\le \Vert Uf{\Vert }_2+\Vert Ug{\Vert }_2$ proves the vector-subspace claim. The stated inner product is $\int w^2\hat{f}\overline{\hat{g}}$. By the Plancherel theorem it is positive definite.

More strongly, $U:H^s\to L^2$ is onto: for $h\in L^2$, $w^{-1}h\in L^2$, since $w\ge 1$, and $f={\mathcal{F}}^{-1}(w^{-1}h)$ satisfies $Uf=h$. Thus $U$ is an isometric linear bijection, and completeness of $L^2$ givesThis argument includes $s=0$ if zero is allowed in the printed ${\mathbb{R}}_{+}$ convention.