= Solution
The <distributional derivative> is defined by $\langle u',\phi\rangle=-\langle u,\phi'\rangle$. Since <differentiation under the integral sign> gives $(\widehat\phi)'=\widehat{-i\xi\phi}$, we obtain
$$
\begin{aligned}
\langle\widehat{u'},\phi\rangle
&=\langle u',\widehat\phi\rangle
=-\langle u,(\widehat\phi)'\rangle
=\langle u,\widehat{i\xi\phi}\rangle\\
&=\langle\widehat u,i\xi\phi\rangle
=\langle i\xi\widehat u,\phi\rangle.
\end{aligned}
$$
This proves the <Fourier transform of a derivative> identity $\boxed{\widehat{u'}=i\xi\widehat u}$ for all <tempered distributions>.
If $u,u'\in H^s(\mathbb R)$, their <Fourier transforms> are represented by <locally square-integrable functions>, and the identity identifies $\widehat{u'}$ with $i\xi\widehat u$ <almost everywhere>. Hence the <Sobolev norms> satisfy
$$
\begin{aligned}
\|u\|_{H^{s+1}}^2
&=\int(1+\xi^2)^s(1+\xi^2)|\widehat u|^2\,d\xi\\
&=\int(1+\xi^2)^s|\widehat u|^2\,d\xi
+\int(1+\xi^2)^s|\widehat{u'}|^2\,d\xi,
\end{aligned}
$$
so $\boxed{u\in H^{s+1}(\mathbb R),\quad\|u\|_{H^{s+1}}^2=\|u\|_{H^s}^2+\|u'\|_{H^s}^2}$ for every real $s$.
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