**Proof plan.** By the [Integral Formula (Theorem 929)](/theorems/929), $\hat{u}$ is a regular distribution with representative $(\hat{u})_{\mathrm{rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. The hypothesis $s \ge 0$ upgrades this to $(\hat{u})_{\mathrm{rep}} \in L^2(\mathbb{R}^n)$, and [Plancherel's theorem](/page/Fourier%20Transform) then gives $u = T_{f_{\mathrm{rep}}}$ for some $f_{\mathrm{rep}} \in L^2$. **Step 1: The Fourier representative is in $L^2$.** By the definition of $H^s(\mathbb{R}^n)$, there exists $g \in L^2(\mathbb{R}^n)$ such that $(1+|\xi|^2)^{s/2}\hat{u} = T_g$. By [Theorem 929](/theorems/929), $\hat{u} = T_{(\hat{u})_{\mathrm{rep}}}$ with $(\hat{u})_{\mathrm{rep}}(\xi) = (1+|\xi|^2)^{-s/2}g(\xi)$ $\mathcal{L}^n$-a.e. Since $s \ge 0$, we have $(1+|\xi|^2)^{-s/2} \le 1$ for all $\xi \in \mathbb{R}^n$. Therefore \begin{align*} |(\hat{u})_{\mathrm{rep}}(\xi)| = (1+|\xi|^2)^{-s/2}|g(\xi)| \le |g(\xi)| \qquad \mathcal{L}^n\text{-a.e.} \end{align*} Since $g \in L^2(\mathbb{R}^n)$, it follows that $(\hat{u})_{\mathrm{rep}} \in L^2(\mathbb{R}^n)$ with $\|(\hat{u})_{\mathrm{rep}}\|_{L^2} \le \|g\|_{L^2} = \|u\|_{H^s}$. **Step 2: $u$ is a regular distribution with $L^2$ representative.** By the [Plancherel theorem](/page/Fourier%20Transform), the $L^2$ Fourier transform $\mathcal{F}_{L^2}: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ is a unitary bijection. Since $(\hat{u})_{\mathrm{rep}} \in L^2(\mathbb{R}^n)$, there exists a unique $f_{\mathrm{rep}} \in L^2(\mathbb{R}^n)$ such that $\hat{f}_{\mathrm{rep}} = (\hat{u})_{\mathrm{rep}}$ $\mathcal{L}^n$-a.e. Now $\hat{u} = T_{(\hat{u})_{\mathrm{rep}}} = T_{\hat{f}_{\mathrm{rep}}} = \widehat{T_{f_{\mathrm{rep}}}}$ (the last equality is the compatibility of the distributional and $L^2$ Fourier transforms on $L^2$ [functions](/page/Function)). Since $\mathcal{F}: \mathcal{S}' \to \mathcal{S}'$ is injective, $u = T_{f_{\mathrm{rep}}}$. **Step 3: The norm formula.** By Theorem 929: \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) = \int_{\mathbb{R}^n}(1+|\xi|^2)^s |\hat{f}_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*} **Step 4: The $L^2$ bound.** For $s \ge 0$, $(1+|\xi|^2)^s \ge 1$, so \begin{align*} \|f_{\mathrm{rep}}\|_{L^2}^2 = \|\hat{f}_{\mathrm{rep}}\|_{L^2}^2 = \int_{\mathbb{R}^n} |\hat{f}_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi) \le \int_{\mathbb{R}^n}(1+|\xi|^2)^s|\hat{f}_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi) = \|u\|_{H^s}^2. \end{align*} Hence $\|f_{\mathrm{rep}}\|_{L^2} \le \|T_{f_{\mathrm{rep}}}\|_{H^s}$. $\square$