 Solution

The distributional derivative is defined by $⟨u^{\prime },\varphi ⟩=-⟨u,{\varphi }^{\prime }⟩$. Since differentiation under the integral sign gives $(\hat{\varphi }{)}^{\prime }=\widehat{-i\xi \varphi }$, we obtainThis proves the Fourier transform of a derivative identity $\boxed{\widehat{u^{\prime }}=i\xi \hat{u}}$ for all tempered distributions.

If $u,u^{\prime }\in H^s(\mathbb{R})$, their Fourier transforms are represented by locally square-integrable functions, and the identity identifies $\widehat{u^{\prime }}$ with $i\xi \hat{u}$ almost everywhere. Hence the Sobolev norms satisfyso $\boxed{u\in H^{s+1}(\mathbb{R}),\Vert u{\Vert }_{H^{s+1}}^2=\Vert u{\Vert }_{H^s}^2+\Vert u^{\prime }{\Vert }_{H^s}^2}$ for every real $s$.