[proofplan] The bound has two regimes. For bounded $|x|$, the integral is uniformly bounded by the total mass of $d\sigma$, which is finite, so $|\widehat{d\sigma}(x)| \le \sigma(S^{n-1})$ directly yields the inequality once the constant $C$ is taken large enough. For $|x|$ large, the decay comes from the curvature of the sphere. Rotational invariance reduces the calculation to the case $x = \lambda e_n$ with $\lambda = |x|$, $e_n$ the north pole of $S^{n-1}$. The integral becomes a one-parameter family of oscillatory integrals on the sphere with phase $\omega \mapsto -e_n \cdot \omega$, whose only critical points are the two poles $\pm e_n$, both non-degenerate (the Hessian on the tangent space is $\pm \mathrm{Id}_{T_{\pm e_n} S^{n-1}}$). The [Stationary Phase Lemma](/theorems/???) on the $(n-1)$-dimensional sphere then produces the decay rate $\lambda^{-(n-1)/2}$. Combining the two regimes gives the uniform bound $|\widehat{d\sigma}(x)| \lesssim (1 + |x|)^{-(n-1)/2}$. [/proofplan] [step:Reduce to $x = \lambda e_n$ via rotational invariance] The surface measure $d\sigma$ is invariant under the orthogonal group $O(n)$: for every $R \in O(n)$ and every Borel set $E \subseteq S^{n-1}$, $\sigma(R E) = \sigma(E)$. Equivalently, for every integrable $g : S^{n-1} \to \mathbb{C}$, \begin{align*} \int_{S^{n-1}} g(R\omega)\, d\sigma(\omega) = \int_{S^{n-1}} g(\omega)\, d\sigma(\omega). \end{align*} Fix $x \in \mathbb{R}^n \setminus \{0\}$ and let $R \in O(n)$ be any rotation with $R^\top x = |x|\, e_n$, where $e_n = (0, \ldots, 0, 1) \in S^{n-1}$. Substituting $\omega' = R^{-1}\omega$ (a measure-preserving change of variables on $S^{n-1}$), \begin{align*} \widehat{d\sigma}(x) = \int_{S^{n-1}} e^{-i x \cdot \omega}\, d\sigma(\omega) = \int_{S^{n-1}} e^{-i x \cdot R\omega'}\, d\sigma(\omega') = \int_{S^{n-1}} e^{-i (R^\top x) \cdot \omega'}\, d\sigma(\omega') = I(\lambda), \end{align*} where we have set $\lambda := |x| \ge 0$ and \begin{align*} I : [0, \infty) &\to \mathbb{C} \\ \lambda &\mapsto \int_{S^{n-1}} e^{-i\lambda \omega_n}\, d\sigma(\omega), \qquad \omega = (\omega_1, \ldots, \omega_n). \end{align*} Hence $\widehat{d\sigma}(x) = I(|x|)$, and it suffices to prove $|I(\lambda)| \le C(1 + \lambda)^{-(n-1)/2}$. [/step] [step:Handle the bounded regime $0 \le \lambda \le 1$ by the elementary bound] For all $\lambda \ge 0$, \begin{align*} |I(\lambda)| \le \int_{S^{n-1}} |e^{-i\lambda\omega_n}|\, d\sigma(\omega) = \sigma(S^{n-1}) = \frac{2\pi^{n/2}}{\Gamma(n/2)} =: \sigma_n. \end{align*} For $0 \le \lambda \le 1$, $(1 + \lambda)^{(n-1)/2} \le 2^{(n-1)/2}$, hence \begin{align*} |I(\lambda)| \le \sigma_n \le \sigma_n\, 2^{(n-1)/2}\, (1 + \lambda)^{-(n-1)/2}. \end{align*} [/step] [step:Localise the large-$\lambda$ integral by a smooth partition of unity around the two poles] Fix $\lambda \ge 1$. The phase \begin{align*} \Phi : S^{n-1} &\to \mathbb{R} \\ \omega &\mapsto -\omega_n \end{align*} is the restriction of the linear function $\omega \mapsto -\omega_n$ on $\mathbb{R}^n$ to $S^{n-1}$. The differential of $\Phi$ on $S^{n-1}$ at a point $\omega \in S^{n-1}$ is the projection of $-e_n$ onto $T_\omega S^{n-1} = \omega^\perp$: \begin{align*} d\Phi_\omega = -e_n + (e_n \cdot \omega)\, \omega = -e_n + \omega_n\, \omega. \end{align*} Hence $d\Phi_\omega = 0$ if and only if $e_n = \omega_n\, \omega$, which (taking norms) forces $\omega_n^2 = 1$, i.e. $\omega = \pm e_n$. So $\Phi$ has exactly two critical points on $S^{n-1}$, namely the poles $\pm e_n$. Choose $\delta \in (0, 1/2)$ small enough that the spherical caps $V_\pm := \{\omega \in S^{n-1} : |\omega \mp e_n| < 2\delta\}$ are disjoint and each lies in a single chart of $S^{n-1}$. Choose $\chi_+, \chi_-, \chi_0 \in C^\infty(S^{n-1})$ forming a partition of unity, $\chi_+ + \chi_- + \chi_0 \equiv 1$ on $S^{n-1}$, with $\operatorname{supp}\chi_\pm \subset V_\pm$, $\chi_\pm \equiv 1$ on $\{|\omega \mp e_n| < \delta\}$, and $\operatorname{supp}\chi_0 \subset S^{n-1} \setminus (\{|\omega - e_n| < \delta/2\} \cup \{|\omega + e_n| < \delta/2\})$. Decompose \begin{align*} I(\lambda) = I_+(\lambda) + I_-(\lambda) + I_0(\lambda), \qquad I_\bullet(\lambda) := \int_{S^{n-1}} e^{-i\lambda\omega_n}\, \chi_\bullet(\omega)\, d\sigma(\omega), \quad \bullet \in \{+, -, 0\}. \end{align*} [/step] [step:Show that the off-pole piece $I_0(\lambda)$ decays faster than any polynomial] On $\operatorname{supp}\chi_0$, $\omega \ne \pm e_n$, so $d\Phi_\omega = -e_n + \omega_n\,\omega \ne 0$ on this closed subset of $S^{n-1}$. Cover $\operatorname{supp}\chi_0$ by finitely many coordinate charts $(U_\alpha, \varphi_\alpha)$ of $S^{n-1}$ and a subordinate smooth partition of unity $(\rho_\alpha)$. In each chart $\varphi_\alpha : U_\alpha \to \mathbb{R}^{n-1}$, the pulled-back integral becomes \begin{align*} \int_{\varphi_\alpha(U_\alpha)} e^{-i\lambda\,\Phi \circ \varphi_\alpha^{-1}(u)}\, (\rho_\alpha \chi_0)\!\left(\varphi_\alpha^{-1}(u)\right) J_\alpha(u)\, d\mathcal{L}^{n-1}(u), \end{align*} where $J_\alpha$ is the smooth surface-measure density of $d\sigma$ in the chart. The amplitude $(\rho_\alpha \chi_0) \circ \varphi_\alpha^{-1} \cdot J_\alpha \in C_c^\infty(\varphi_\alpha(U_\alpha))$ has compact support, and the gradient $\nabla(\Phi \circ \varphi_\alpha^{-1})$ is non-zero on this support (because $d\Phi_\omega \ne 0$ there). Apply the [Rapid Decay Away from Critical Points](/theorems/3201) theorem in $\mathbb{R}^{n-1}$: for every $N \ge 0$ there is a constant $C_{N,\alpha}$ such that \begin{align*} \left| \int_{\varphi_\alpha(U_\alpha)} e^{-i\lambda\,\Phi \circ \varphi_\alpha^{-1}(u)}\, (\rho_\alpha\chi_0)(\varphi_\alpha^{-1}(u))\, J_\alpha(u)\, d\mathcal{L}^{n-1}(u) \right| \le C_{N,\alpha}\, \lambda^{-N}. \end{align*} Summing over the finitely many charts gives, for every $N \ge 0$, a constant $C_N$ with \begin{align*} |I_0(\lambda)| \le C_N\, \lambda^{-N} \qquad \text{for all } \lambda \ge 1. \end{align*} In particular, taking $N = (n-1)/2$ when $n$ is odd, or $N = \lceil (n-1)/2 \rceil$ otherwise, gives $|I_0(\lambda)| \lesssim \lambda^{-(n-1)/2}$. [/step] [step:Compute $I_\pm(\lambda)$ via the Stationary Phase Lemma in the $(n-1)$-dimensional tangent coordinates] Consider $I_+$; the argument for $I_-$ is identical with the roles of $e_n$ and $-e_n$ swapped (same Hessian eigenvalues up to a sign in the phase). On the cap $V_+$, parametrise $S^{n-1}$ near $e_n$ by orthogonal projection onto the tangent hyperplane $T_{e_n}S^{n-1} = e_n^\perp \cong \mathbb{R}^{n-1}$: \begin{align*} \varphi : B_{\mathbb{R}^{n-1}}(0, 2\delta) &\to V_+ \subset S^{n-1} \\ y &\mapsto (y, \sqrt{1 - |y|^2}). \end{align*} Here $|y| < 2\delta < 1$, so $\sqrt{1 - |y|^2}$ is smooth in $y$. The pull-back of $d\sigma$ under $\varphi$ has the explicit smooth density \begin{align*} J(y) = \frac{1}{\sqrt{1 - |y|^2}}, \end{align*} because the Riemannian volume element on $S^{n-1}$ in these coordinates equals $(1 - |y|^2)^{-1/2}\, d\mathcal{L}^{n-1}(y)$. The phase pulled back to $\mathbb{R}^{n-1}$ is \begin{align*} \widetilde\Phi : B_{\mathbb{R}^{n-1}}(0, 2\delta) &\to \mathbb{R} \\ y &\mapsto -\omega_n = -\sqrt{1 - |y|^2}. \end{align*} Compute the Taylor expansion at $y = 0$: \begin{align*} \widetilde\Phi(y) = -1 + \frac{|y|^2}{2} + O(|y|^4), \qquad \nabla\widetilde\Phi(0) = 0, \qquad H_{\widetilde\Phi}(0) = I_{n-1}, \end{align*} the $(n-1) \times (n-1)$ identity matrix. So $y = 0$ is a non-degenerate critical point of $\widetilde\Phi$ with Hessian signature $(n-1, 0)$ and $|\det H_{\widetilde\Phi}(0)| = 1$. Define the smooth, compactly supported amplitude \begin{align*} \psi_+ : \mathbb{R}^{n-1} &\to \mathbb{R} \\ y &\mapsto \chi_+(\varphi(y))\, J(y), \end{align*} extended by zero outside $B_{\mathbb{R}^{n-1}}(0, 2\delta)$, so $\psi_+ \in C_c^\infty(\mathbb{R}^{n-1})$ with $\psi_+(0) = J(0) = 1$. Then \begin{align*} I_+(\lambda) = \int_{\mathbb{R}^{n-1}} e^{-i\lambda\, \widetilde\Phi(y)}\, \psi_+(y)\, d\mathcal{L}^{n-1}(y). \end{align*} The hypotheses of the [Stationary Phase Lemma](/theorems/???) (in the form valid in $\mathbb{R}^{n-1}$) are met: $\widetilde\Phi$ is smooth on a neighbourhood of $\operatorname{supp}\psi_+$, has a unique non-degenerate critical point at $y = 0$ in $\operatorname{supp}\psi_+$ (we shrank $\delta$ to ensure this), and $\psi_+ \in C_c^\infty(\mathbb{R}^{n-1})$. The lemma gives \begin{align*} I_+(\lambda) = e^{-i\lambda\widetilde\Phi(0)}\!\left(\frac{2\pi}{\lambda}\right)^{(n-1)/2} \frac{e^{-i\pi\,\mathrm{sgn}(H_{\widetilde\Phi}(0))/4}}{|\det H_{\widetilde\Phi}(0)|^{1/2}}\, \psi_+(0) + O(\lambda^{-(n+1)/2}). \end{align*} With $\widetilde\Phi(0) = -1$, $\mathrm{sgn}(H_{\widetilde\Phi}(0)) = n-1$, $|\det H_{\widetilde\Phi}(0)| = 1$, $\psi_+(0) = 1$, this reads \begin{align*} I_+(\lambda) = e^{i\lambda}\, (2\pi/\lambda)^{(n-1)/2}\, e^{-i\pi(n-1)/4} + O(\lambda^{-(n+1)/2}). \end{align*} Hence $|I_+(\lambda)| \le A_+\, \lambda^{-(n-1)/2}$ for all $\lambda \ge 1$, with $A_+ = A_+(n)$ a constant depending only on $n$. The argument for $I_-$ is the mirror image: parametrise near $-e_n$ by $y \mapsto (y, -\sqrt{1-|y|^2})$; the phase is $\omega_n \mapsto -\omega_n = +\sqrt{1-|y|^2}$, again with a non-degenerate critical point at $y = 0$ (now a maximum of the phase, Hessian $-I_{n-1}$). The Stationary Phase Lemma yields $|I_-(\lambda)| \le A_-\, \lambda^{-(n-1)/2}$ for $\lambda \ge 1$, with $A_- = A_-(n)$. [/step] [step:Combine the three estimates to conclude] For $\lambda \ge 1$, summing the three pieces and using $|I_0(\lambda)| \lesssim \lambda^{-(n-1)/2}$ from the off-pole step, \begin{align*} |I(\lambda)| \le |I_+(\lambda)| + |I_-(\lambda)| + |I_0(\lambda)| \le (A_+ + A_- + C_{(n-1)/2})\, \lambda^{-(n-1)/2}. \end{align*} Since $\lambda \ge 1$ implies $1 + \lambda \le 2\lambda$, we have $\lambda^{-(n-1)/2} \le 2^{(n-1)/2}\, (1 + \lambda)^{-(n-1)/2}$, giving \begin{align*} |I(\lambda)| \le C^{\mathrm{large}}\, (1 + \lambda)^{-(n-1)/2} \qquad \text{for } \lambda \ge 1, \end{align*} with $C^{\mathrm{large}} = 2^{(n-1)/2}\, (A_+ + A_- + C_{(n-1)/2})$. Combining with the bounded-regime estimate $|I(\lambda)| \le \sigma_n\, 2^{(n-1)/2}\, (1 + \lambda)^{-(n-1)/2}$ for $0 \le \lambda \le 1$, and recalling $\widehat{d\sigma}(x) = I(|x|)$, \begin{align*} |\widehat{d\sigma}(x)| \le C\, (1 + |x|)^{-(n-1)/2} \qquad \text{for all } x \in \mathbb{R}^n, \end{align*} where $C = C(n) := \max\bigl(\sigma_n\, 2^{(n-1)/2},\, C^{\mathrm{large}}\bigr)$ depends only on the ambient dimension $n$. This is the asserted bound, and the asymptotic $\widehat{d\sigma}(x) = O(|x|^{-(n-1)/2})$ as $|x| \to \infty$ is its immediate consequence. [/step]