**Proof plan.** The argument proceeds by dividing the $H^s$ membership condition $(1 + |\xi|^2)^{s/2}\hat{u} = T_g$ by the weight $(1 + |\xi|^2)^{s/2}$. Since this weight is a smooth, everywhere-positive [function](/page/Function) in $\mathcal{O}_M(\mathbb{R}^n)$, multiplying a [regular distribution](/page/Regular%20Distribution) by its reciprocal produces another regular distribution, yielding the representative $(\hat{u})_{\mathrm{rep}}$. **Step 1: Dividing by the weight.** By the definition of $H^s(\mathbb{R}^n)$, there exists $g \in L^2(\mathbb{R}^n)$ such that $(1 + |\xi|^2)^{s/2}\hat{u} = T_g$ in $\mathcal{S}'(\mathbb{R}^n)$. The reciprocal $(1 + |\xi|^2)^{-s/2}$ belongs to $\mathcal{O}_M(\mathbb{R}^n)$ (it is smooth on $\mathbb{R}^n$ and all its [derivatives](/page/Derivative) have at most polynomial growth). Multiplying both sides by $(1 + |\xi|^2)^{-s/2}$: \begin{align*} \hat{u} = (1 + |\xi|^2)^{-s/2} \cdot T_g. \end{align*} **Step 2: Identifying the representative.** For any $h \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ and any $\psi \in \mathcal{O}_M(\mathbb{R}^n)$, the product $\psi \cdot T_h = T_{\psi h}$ (the regular distribution of the pointwise product $\psi h$). Applying this with $\psi = (1 + |\xi|^2)^{-s/2}$ and $h = g$: \begin{align*} \hat{u} = T_{(\hat{u})_{\mathrm{rep}}}, \qquad (\hat{u})_{\mathrm{rep}}(\xi) := g(\xi)(1 + |\xi|^2)^{-s/2}. \end{align*} Since $g \in L^2(\mathbb{R}^n)$ and $(1 + |\xi|^2)^{-s/2}$ is locally bounded, the product $(\hat{u})_{\mathrm{rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. Hence $\hat{u}$ is a regular distribution. **Step 3: The [integral](/page/Integral) formula.** Since $g(\xi) = (1 + |\xi|^2)^{s/2}(\hat{u})_{\mathrm{rep}}(\xi)$ $\mathcal{L}^n$-a.e., the norm becomes \begin{align*} \|u\|_{H^s}^2 = \|g\|_{L^2}^2 = \int_{\mathbb{R}^n} |g(\xi)|^2\, d\mathcal{L}^n(\xi) = \int_{\mathbb{R}^n} (1 + |\xi|^2)^{s}\, |(\hat{u})_{\mathrm{rep}}(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*} **Step 4: Uniqueness.** If $\hat{u} = T_{h_1} = T_{h_2}$ for $h_1, h_2 \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$, then $T_{h_1 - h_2} = 0$ in $\mathcal{S}'(\mathbb{R}^n)$. Since $\mathcal{S}(\mathbb{R}^n)$ contains $C_c^\infty(\mathbb{R}^n)$, this gives $\int (h_1 - h_2)\phi\, d\mathcal{L}^n = 0$ for all $\phi \in C_c^\infty(\mathbb{R}^n)$, and the fundamental lemma of the [calculus of variations](/page/Calculus%20of%20Variations) gives $h_1 = h_2$ $\mathcal{L}^n$-a.e.