= Solution
Let $w(\xi)=(1+|\xi|^2)^{s/2}$. The map $U:f\mapsto w\widehat f$ is linear, and $\|U(f+g)\|_2\leq\|Uf\|_2+\|Ug\|_2$ proves the vector-subspace claim. The stated <inner product> is $\int w^2\widehat f\,\overline{\widehat g}$. 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 $w\geq1$, and $f=\mathcal F^{-1}(w^{-1}h)$ satisfies $Uf=h$. Thus $U$ is an isometric linear bijection, and completeness of $L^2$ gives
$$
\boxed{H^s(\mathbb R^n)\text{ is a Hilbert space with }\|f\|_{H^s}=\|w\widehat f\|_2.}
$$
This argument includes $s=0$ if zero is allowed in the printed $\mathbb R_+$ convention.
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