Let $s \in \mathbb{R}$, $n \ge 1$, and $u \in H^s(\mathbb{R}^n)$. Then the Fourier transform $\hat{u} \in \mathcal{S}'(\mathbb{R}^n)$ is a regular distribution: there exists a unique $(\hat{u})_{\mathrm{rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ such that
\begin{align*} \hat{u} = T_{(\hat{u})_{\mathrm{rep}}}, \end{align*}
where $T_{(\hat{u})_{\mathrm{rep}}}: \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}$ is the regular distribution $T_{(\hat{u})_{\mathrm{rep}}}(\phi) = \int_{\mathbb{R}^n} (\hat{u})_{\mathrm{rep}}(\xi)\,\phi(\xi)\, d\mathcal{L}^n(\xi)$.

Moreover, the $H^s$ norm admits the integral representation
\begin{align*} \|u\|_{H^s}^2 = \int_{\mathbb{R}^n} (1 + |\xi|^2)^{s}\, |(\hat{u})_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*}

The notation $(\hat{u})_{\mathrm{rep}}$ denotes the representative of the distribution $\hat{u}$. This is not a priori the same as $\widehat{u_{\mathrm{rep}}}$ (the Fourier transform of some function representing $u$), since no such function has been constructed at this stage.