[proofplan] We establish the isometric isomorphism $\Phi_s^{\mathrm{hom}}: \dot{H}^s \to L^2$ in three stages. Uniqueness of the $T$-representative follows from the [Injectivity of the Canonical Embedding into Distributions](/theorems/450) on $\mathbb{R}^n_0$, and independence across representatives follows from the fact that Fourier transforms of polynomials are supported at $\{0\}$. Surjectivity constructs a preimage for each $g \in L^2$ by defining $h(\xi) = |\xi|^{-s}g(\xi)$ on $\mathbb{R}^n_0$, extending $T_h$ to a tempered distribution via a cutoff, and using the [Distributions Supported at a Point](/theorems/451) theorem to show different extensions give the same equivalence class. Completeness follows from the isometric identification with $L^2$. [/proofplan] [step:Prove uniqueness of the $T$-representative for a fixed representative] If $\hat{f}\big|_{\mathcal{D}'(\mathbb{R}^n_0)} = T_{h_1} = T_{h_2}$ for $h_1, h_2 \in L^1_{\mathrm{loc}}(\mathbb{R}^n_0)$, then $\int (h_1 - h_2)\phi \, d\mathcal{L}^n = 0$ for every $\phi \in C_c^\infty(\mathbb{R}^n_0)$. By the [Injectivity of the Canonical Embedding into Distributions](/theorems/450) applied on the open set $\mathbb{R}^n_0$, this implies $h_1 = h_2$ $\mathcal{L}^n$-a.e. on $\mathbb{R}^n_0$. Since $\{0\}$ has $\mathcal{L}^n$-measure zero, $h_1 = h_2$ a.e. on $\mathbb{R}^n$. [/step] [step:Show independence of the equivalence class representative] Let $f_1 - f_2 \in \mathcal{P}$, so $\hat{f}_1 - \hat{f}_2 = \hat{p}$ for some polynomial $p$. The Fourier transform of a polynomial is supported at the origin: $\operatorname{supp}(\hat{p}) \subseteq \{0\}$. For any $\phi \in C_c^\infty(\mathbb{R}^n_0)$ (supported away from the origin), $\hat{p}(\phi) = 0$ since $\hat{p}$ acts by evaluating derivatives of the test function at $\xi = 0$, weighted by the coefficients of $p$. Therefore $\hat{f}_1\big|_{\mathcal{D}'(\mathbb{R}^n_0)} = \hat{f}_2\big|_{\mathcal{D}'(\mathbb{R}^n_0)}$, and the $T$-representatives coincide $\mathcal{L}^n$-a.e. by the previous step. Steps 1--2 together establish Part 1 of the theorem. [/step] [step:Verify well-definedness and isometry of $\Phi_s^{\mathrm{hom}}$] Part 1 guarantees that $\Phi_s^{\mathrm{hom}}([f]) = |\xi|^s\hat{f}_{T\text{-rep}}$ depends only on $[f]$. The product $|\xi|^s\hat{f}_{T\text{-rep}}(\xi)$ is a pointwise product of the measurable function $|\xi|^s$ (smooth and positive on $\mathbb{R}^n_0$) with $\hat{f}_{T\text{-rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n_0)$; no distributional multiplication is involved since $\{0\}$ has $\mathcal{L}^n$-measure zero. By the definition of membership in $\dot{H}^s$, $|\xi|^s\hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$, and \begin{align*} \|\Phi_s^{\mathrm{hom}}([f])\|_{L^2} &= \||\xi|^s\hat{f}_{T\text{-rep}}\|_{L^2} = \|[f]\|_{\dot{H}^s}, \end{align*} which is the isometry. [/step] [step:Construct a preimage for each $g \in L^2$ to prove surjectivity] Let $g \in L^2(\mathbb{R}^n)$. Define $h: \mathbb{R}^n_0 \to \mathbb{C}$ by $h(\xi) := |\xi|^{-s}g(\xi)$. For any compact $K \subset \mathbb{R}^n_0$, the function $|\xi|^{-s}$ is continuous (hence bounded) on $K$, so $\sup_K |\xi|^{-s} =: C_K < \infty$. By Cauchy--Schwarz: \begin{align*} \int_K |h(\xi)| \, d\mathcal{L}^n(\xi) &\le C_K\,\|g\|_{L^2(K)}\,\mathcal{L}^n(K)^{1/2} < \infty. \end{align*} Hence $h \in L^1_{\mathrm{loc}}(\mathbb{R}^n_0)$, so $T_h \in \mathcal{D}'(\mathbb{R}^n_0)$. Fix a cutoff $\chi \in C^\infty(\mathbb{R}^n)$ with $\chi(\xi) = 0$ for $|\xi| \le 1/2$ and $\chi(\xi) = 1$ for $|\xi| \ge 1$. Define \begin{align*} \hat{f}_0(\varphi) &:= \int_{\mathbb{R}^n} \chi(\xi)\,h(\xi)\,\varphi(\xi) \, d\mathcal{L}^n(\xi) + \int_{\mathbb{R}^n} (1-\chi(\xi))\,h(\xi)\,\varphi(\xi) \, d\mathcal{L}^n(\xi) \end{align*} for $\varphi \in \mathcal{S}(\mathbb{R}^n)$. The first integral converges because $\chi h = \chi|\xi|^{-s}g$ with $\chi|\xi|^{-s}$ bounded, so $\chi h \in L^2(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$. The second integral converges because $(1-\chi)h$ is supported in $\overline{B}(0,1) \setminus \{0\}$ and $h \in L^1_{\mathrm{loc}}(\mathbb{R}^n_0)$. Hence $\hat{f}_0 \in \mathcal{S}'(\mathbb{R}^n)$, and $\hat{f}_0\big|_{\mathcal{D}'(\mathbb{R}^n_0)} = T_h$. If $\hat{f}_1$ is another extension, then $\operatorname{supp}(\hat{f}_0 - \hat{f}_1) \subseteq \{0\}$. By the [Distributions Supported at a Point](/theorems/451) theorem, $\hat{f}_0 - \hat{f}_1 = \sum_{|\alpha| \le N} c_\alpha\,\partial^\alpha\delta_0$. Set $f := \mathcal{F}^{-1}(\hat{f}_0)$. Then $\hat{f}_{T\text{-rep}} = h$ on $\mathbb{R}^n_0$, and \begin{align*} |\xi|^s\hat{f}_{T\text{-rep}}(\xi) &= |\xi|^s \cdot |\xi|^{-s}g(\xi) = g(\xi) \quad \text{for } \mathcal{L}^n\text{-a.e. } \xi. \end{align*} So $[f] \in \dot{H}^s(\mathbb{R}^n)$ and $\Phi_s^{\mathrm{hom}}([f]) = g$. The class $[f]$ is independent of the extension: different extensions yield representatives differing by $\sum c_\alpha\,\mathcal{F}^{-1}(\partial^\alpha\delta_0)$, and each $\mathcal{F}^{-1}(\partial^\alpha\delta_0)$ is a polynomial, so the difference lies in $\mathcal{P}$. [guided] The surjectivity construction has two subtleties. First, $h(\xi) = |\xi|^{-s}g(\xi)$ is only locally integrable on $\mathbb{R}^n_0$, not necessarily on all of $\mathbb{R}^n$ (the singularity at $\xi = 0$ may not be integrable). The cutoff $\chi$ isolates the singularity: $\chi h$ avoids the origin and belongs to $L^2$, while $(1-\chi)h$ is compactly supported away from the origin and locally integrable. The sum defines a tempered distribution $\hat{f}_0$ that agrees with $T_h$ away from the origin. Second, the extension from $\mathcal{D}'(\mathbb{R}^n_0)$ to $\mathcal{S}'(\mathbb{R}^n)$ is not unique. Any two extensions differ by a distribution supported at $\{0\}$, which by the [Distributions Supported at a Point](/theorems/451) theorem is a finite linear combination of $\partial^\alpha\delta_0$. Taking inverse Fourier transforms, $\mathcal{F}^{-1}(\partial^\alpha\delta_0)$ acts as integration against the polynomial $(-i)^{|\alpha|}x^\alpha/(2\pi)^n$. So different extensions yield representatives $f$ differing by a polynomial, which is absorbed by the quotient $\mathcal{S}'/\mathcal{P}$. This ensures $[f] \in \dot{H}^s$ is well-defined regardless of the extension chosen. \begin{align*} |\xi|^s\hat{f}_{T\text{-rep}}(\xi) &= |\xi|^s \cdot |\xi|^{-s}g(\xi) = g(\xi) \quad \mathcal{L}^n\text{-a.e.} \end{align*} [/guided] [/step] [step:Transfer completeness from $L^2$ via the isometric isomorphism] Since $\Phi_s^{\mathrm{hom}}: \dot{H}^s(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ is an isometric isomorphism and $L^2(\mathbb{R}^n)$ is a complete [Hilbert space](/page/Hilbert%20Space), the pullback of the $L^2$ inner product \begin{align*} \langle [f], [g] \rangle_{\dot{H}^s} &:= \langle \Phi_s^{\mathrm{hom}}([f]),\,\Phi_s^{\mathrm{hom}}([g]) \rangle_{L^2} = \int_{\mathbb{R}^n} |\xi|^{2s}\,\hat{f}_{T\text{-rep}}(\xi)\,\overline{\hat{g}_{T\text{-rep}}(\xi)} \, d\mathcal{L}^n(\xi) \end{align*} makes $\dot{H}^s(\mathbb{R}^n)$ a Hilbert space. [/step]