[proofplan] Part (1) is a Cauchy--Schwarz estimate in polar coordinates: the hypothesis $s < n/2$ guarantees that the weight $|\xi|^{-s}$ is square-integrable near the origin, upgrading $\hat{f}_{T\text{-rep}}$ from $L^1_{\mathrm{loc}}(\mathbb{R}^n \setminus \{0\})$ to $L^1_{\mathrm{loc}}(\mathbb{R}^n)$. Part (2) constructs the canonical representative $\Phi([f]) = \mathcal{F}^{-1}(T_{\hat{f}_{T\text{-rep}}})$, proves it belongs to $[f]$ by showing the difference from any representative is a polynomial (via the [Distributions Supported at a Point](/theorems/451) theorem), and establishes uniqueness among representatives with regular Fourier transform by showing that a regular distribution supported at $\{0\}$ must be zero. Part (3) follows from the injectivity of the regular-distribution identification. [/proofplan] [step:Establish local integrability away from the origin] Let $[f] \in \dot{H}^s(\mathbb{R}^n)$. By definition, $\hat{f}_{T\text{-rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n \setminus \{0\})$ with $|\xi|^s \hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$. Define $g(\xi) := |\xi|^s \hat{f}_{T\text{-rep}}(\xi)$, so $g \in L^2(\mathbb{R}^n)$ and $\hat{f}_{T\text{-rep}}(\xi) = |\xi|^{-s} g(\xi)$ for $\mathcal{L}^n$-a.e. $\xi \neq 0$. On any compact set $K \subset \mathbb{R}^n \setminus \{0\}$, the weight $|\xi|^{-s}$ is bounded, so $\hat{f}_{T\text{-rep}} \in L^2(K) \subset L^1(K)$. Local integrability on all of $\mathbb{R}^n$ therefore reduces to integrability near the origin. [/step] [step:Upgrade to $L^1_{\mathrm{loc}}(\mathbb{R}^n)$ via Cauchy--Schwarz near the origin] Apply the Cauchy--Schwarz inequality over $B(0, 1)$: \begin{align*} \int_{B(0,1)} |\hat{f}_{T\text{-rep}}(\xi)|\, d\mathcal{L}^n(\xi) &= \int_{B(0,1)} |\xi|^{-s}\,|g(\xi)|\, d\mathcal{L}^n(\xi) \\ &\leq \left(\int_{B(0,1)} |\xi|^{-2s}\, d\mathcal{L}^n(\xi)\right)^{1/2} \|g\|_{L^2(\mathbb{R}^n)}. \end{align*} The second factor equals $\|[f]\|_{\dot{H}^s} < \infty$. Converting the first factor to polar coordinates with $r = |\xi|$ and $\omega_{n-1} := \mathcal{H}^{n-1}(\mathbb{S}^{n-1})$: \begin{align*} \int_{B(0,1)} |\xi|^{-2s}\, d\mathcal{L}^n(\xi) = \omega_{n-1}\int_0^1 r^{n-1-2s}\, d\mathcal{L}^1(r). \end{align*} This integral converges if and only if $n - 1 - 2s > -1$, i.e., $s < n/2$. Under this hypothesis, the integral equals $\omega_{n-1}/(n - 2s) < \infty$. Combined with the previous step, $\hat{f}_{T\text{-rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. This proves Part (1). [guided] The hypothesis $s < n/2$ enters precisely here. We need $|\xi|^{-s}$ to be square-integrable near the origin so that Cauchy--Schwarz pairs it with $g = |\xi|^s \hat{f}_{T\text{-rep}} \in L^2$. In polar coordinates, $|\xi|^{-2s}$ becomes $r^{-2s}$ and the volume element contributes $r^{n-1}$, giving an integrand $r^{n-1-2s}$. This is integrable on $(0,1)$ precisely when $n - 1 - 2s > -1$, i.e., $s < n/2$. What fails when $s \ge n/2$? The weight $|\xi|^{-s}$ becomes too singular at the origin: $r^{n-1-2s} \le r^{-1}$, and $\int_0^1 r^{-1}\, d\mathcal{L}^1(r) = +\infty$. In that regime, $\hat{f}_{T\text{-rep}}$ may genuinely fail to be locally integrable at the origin, and the polynomial ambiguity in $\mathcal{S}'/\mathcal{P}$ cannot be removed. The Cauchy--Schwarz step gives the explicit bound: \begin{align*} \int_{B(0,1)} |\hat{f}_{T\text{-rep}}(\xi)|\, d\mathcal{L}^n(\xi) \leq \left(\frac{\omega_{n-1}}{n - 2s}\right)^{1/2} \|[f]\|_{\dot{H}^s}. \end{align*} [/guided] [/step] [step:Prove well-definedness of the canonical representative $\Phi$] The map $\Phi([f]) := \mathcal{F}^{-1}(T_{\hat{f}_{T\text{-rep}}})$ involves the choice of representative $f_0 \in [f]$ used to compute $\hat{f}_{T\text{-rep}}$. [claim:Independence of representative] The $T$-representative $\hat{f}_{T\text{-rep}} \in L^1_{\mathrm{loc}}(\mathbb{R}^n \setminus \{0\})$ depends only on the equivalence class $[f] \in \mathcal{S}'/\mathcal{P}$, not on the choice of representative $f_0 \in [f]$. Furthermore, the regular distribution $T_{\hat{f}_{T\text{-rep}}} \in \mathcal{S}'(\mathbb{R}^n)$ is uniquely determined by $[f]$. [/claim] [proof] If $f_1, f_2 \in \mathcal{S}'(\mathbb{R}^n)$ satisfy $[f_1] = [f_2]$ in $\mathcal{S}'/\mathcal{P}$, then $f_1 - f_2 = p$ for some polynomial $p \in \mathcal{P}$. On the Fourier side, $\hat{f}_1 - \hat{f}_2 = \hat{p}$. By the [Distributions Supported at a Point](/theorems/451) theorem, the Fourier transform of a polynomial is a finite linear combination of derivatives of $\delta_0$: $\hat{p} = \sum_{|\alpha| \leq N} c_\alpha \partial^\alpha \delta_0$ for some $N$ and constants $c_\alpha$. On $\mathbb{R}^n \setminus \{0\}$, every $\partial^\alpha \delta_0$ vanishes (since $\operatorname{supp}(\partial^\alpha \delta_0) = \{0\}$), so $\hat{f}_1|_{\mathbb{R}^n \setminus \{0\}} = \hat{f}_2|_{\mathbb{R}^n \setminus \{0\}}$. The $T$-representative is the $L^1_{\mathrm{loc}}$ function representing this common restriction, so $(\hat{f}_1)_{T\text{-rep}} = (\hat{f}_2)_{T\text{-rep}}$ $\mathcal{L}^n$-a.e. on $\mathbb{R}^n \setminus \{0\}$. By Part (1), this common $T$-representative lies in $L^1_{\mathrm{loc}}(\mathbb{R}^n)$, so the regular distribution $T_{\hat{f}_{T\text{-rep}}} \in \mathcal{S}'(\mathbb{R}^n)$ is uniquely determined. Hence $\Phi([f]) = \mathcal{F}^{-1}(T_{\hat{f}_{T\text{-rep}}})$ is independent of the choice of representative. [/proof] [/step] [step:Show the canonical representative belongs to $[f]$] Let $f_0 \in [f]$ be any representative. Then $\hat{f}_0|_{\mathbb{R}^n \setminus \{0\}} = T_{\hat{f}_{T\text{-rep}}}|_{\mathbb{R}^n \setminus \{0\}}$ by definition of the $T$-representative. Hence $\hat{f}_0 - T_{\hat{f}_{T\text{-rep}}}$ is a tempered distribution supported at $\{0\}$. By the [Distributions Supported at a Point](/theorems/451) theorem: \begin{align*} \hat{f}_0 - T_{\hat{f}_{T\text{-rep}}} = \sum_{|\alpha| \leq N} c_\alpha\,\partial^\alpha \delta_0. \end{align*} Applying $\mathcal{F}^{-1}$ (which maps $\mathcal{S}'$ to $\mathcal{S}'$ by the [Fourier Transform as Automorphism of Schwartz Space](/theorems/228) extended to tempered distributions): \begin{align*} f_0 - \Phi([f]) = \mathcal{F}^{-1}\!\left(\sum_{|\alpha| \leq N} c_\alpha\,\partial^\alpha \delta_0\right) = \sum_{|\alpha| \leq N} c_\alpha\,\mathcal{F}^{-1}(\partial^\alpha \delta_0). \end{align*} Each $\mathcal{F}^{-1}(\partial^\alpha \delta_0)$ is a monomial (up to normalisation constants), as computed in the proof of the [Canonical Isomorphism $\dot{H}^0 \cong L^2$](/theorems/471). Hence $f_0 - \Phi([f])$ is a polynomial, which means $[\Phi([f])] = [f_0] = [f]$. [/step] [step:Establish uniqueness of the canonical representative] Suppose $f_0 \in [f]$ also has the property that $\hat{f}_0 = T_h$ for some $h \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$. Then $\hat{f}_0 - T_{\hat{f}_{T\text{-rep}}}$ is a tempered distribution supported at $\{0\}$ (by the same argument as the previous step). But both $\hat{f}_0 = T_h$ and $T_{\hat{f}_{T\text{-rep}}}$ are regular distributions on $\mathbb{R}^n$, so their difference $T_h - T_{\hat{f}_{T\text{-rep}}} = T_{h - \hat{f}_{T\text{-rep}}}$ is also a regular distribution. A regular distribution supported at $\{0\}$ must be zero: if $\psi \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ satisfies $T_\psi(\varphi) = \int_{\mathbb{R}^n} \psi\,\varphi\, d\mathcal{L}^n = 0$ for all $\varphi \in C_c^\infty(\mathbb{R}^n \setminus \{0\})$, then $\psi = 0$ $\mathcal{L}^n$-a.e. (since $\{0\}$ has $\mathcal{L}^n$-measure zero and $C_c^\infty(\mathbb{R}^n \setminus \{0\})$ is dense in $L^{p'}(K)$ for any compact $K$). Hence $T_h = T_{\hat{f}_{T\text{-rep}}}$, giving $f_0 = \mathcal{F}^{-1}(T_h) = \mathcal{F}^{-1}(T_{\hat{f}_{T\text{-rep}}}) = \Phi([f])$. [/step] [step:Prove injectivity of $\Phi$] Suppose $\Phi([f_1]) = \Phi([f_2])$ for $[f_1], [f_2] \in \dot{H}^s(\mathbb{R}^n)$. Then $T_{(\hat{f}_1)_{T\text{-rep}}} = T_{(\hat{f}_2)_{T\text{-rep}}}$ in $\mathcal{S}'(\mathbb{R}^n)$ (since $\mathcal{F}$ is an isomorphism on $\mathcal{S}'$). Since both sides are regular distributions: \begin{align*} \int_{\mathbb{R}^n} \left[(\hat{f}_1)_{T\text{-rep}}(\xi) - (\hat{f}_2)_{T\text{-rep}}(\xi)\right]\varphi(\xi)\, d\mathcal{L}^n(\xi) = 0 \quad \text{for all } \varphi \in \mathcal{S}(\mathbb{R}^n). \end{align*} Since $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^{p'}$ for any $1 < p' < \infty$, the fundamental lemma of the calculus of variations gives $(\hat{f}_1)_{T\text{-rep}} = (\hat{f}_2)_{T\text{-rep}}$ $\mathcal{L}^n$-a.e. In particular, $|\xi|^s(\hat{f}_1)_{T\text{-rep}} = |\xi|^s(\hat{f}_2)_{T\text{-rep}}$ $\mathcal{L}^n$-a.e., so $\|[f_1] - [f_2]\|_{\dot{H}^s} = 0$ and $[f_1] = [f_2]$. [/step]