Decay of the Fourier transform of surface measure

Remarks

By a partition of unity, one may assume that the support of $\psi$ meets $S$ in a small piece. This piece is the graph of a smooth function $h$ over a ball in $\R^{n-1}$. Then $\hat\mu(\xi)$ is an oscillatory integral in $n - 1$ variables with phase $-2\pi (x' \cdot \xi' + h(x')\xi_n)$. If $\xi$ is far from the normal directions to the piece, the phase has no critical point and the integral decays faster than any power. If $\xi$ is close to a normal direction, the critical points are nondegenerate, because the curvature hypothesis makes the Hessian of $h$ invertible. Then stationary phase gives the decay $\abs{\xi}^{-(n-1)/2}$.

Example: for the unit sphere $S^2 \subseteq \R^3$ and $\psi = 1$ near $S^2$, a computation in polar coordinates gives $$\hat\sigma(\xi) = 2\pi \int_{-1}^{1} e^{-2\pi i \abs{\xi} t} \, dt = \frac{2 \sin(2\pi\abs{\xi})}{\abs{\xi}},$$ which decays like $\abs{\xi}^{-1} = \abs{\xi}^{-(3-1)/2}$ and no faster.

The curvature hypothesis is needed. If $S$ is the hyperplane $\set{x_n = 0}$, which has curvature zero, then $e^{-2\pi i x \cdot \xi} = 1$ on $S$ when $\xi = (0, \ldots, 0, \xi_n)$. So $\hat\mu(0, \ldots, 0, \xi_n) = \int_S \psi \, d\sigma$ for every $\xi_n$, and there is no decay when this integral is not $0$.

Going deeper

For a student

This decay is the starting point of the Stein–Tomas restriction theorem, whose proof only uses the bound $\abs{\hat\mu(\xi)} \lesssim \abs{\xi}^{-(n-1)/2}$. It also gives the dispersive decay of waves and the asymptotics of Bessel functions, since $\hat\sigma$ for the sphere is a Bessel function. When some principal curvatures vanish, the decay is weaker and is measured by the number of nonvanishing curvatures.