# Decay of the Fourier transform of the surface measure of the sphere via uncertainty

I'm working through Tao's Recent Progress on the Restriction Conjecture notes (http://arxiv.org/abs/math/0311181). Currently, I'm working on problem 2.4, which will eventually allow us to compute the decay of the Fourier transform of the surface measure of the sphere $S^{n-1}$, but using uncertainty instead of stationary phase. Specifically, the problem I'm working on is this:

"Let $R >>1$ and let $\psi$ be a radial bump function adapted to the annular region $N_{R^{-1}}(S^{n-1}) = \{ \xi: 1 - R^{-1} \le |\xi| \le 1 + R^{-1} \}$; by this we mean that $\psi(x)$ depends only on the magnitude r = |x| of $x$, is supported in the annulus $N_{R^{-1}}(S^{n-1})$, and obeys the estimate

$\sup_{r > 0} |\partial_r^k \psi(r)| \le C_K R^k$

for all $k \ge 0$. Show that $\hat{\psi}(x) = O(R^{-\frac{n-1}{2}})$ for all $|x|$~ $R$"

I'm a little lost beginning this. My first thought was because $\psi$ is radial, it is most natural to try and decompose it on small annular regions of thickness $\frac{1}{kR}$ for some integer k and then try to estimate $\hat{\psi}$ along each of these. But the only thing I can really think of using for these kinds of estimates would be an integration by parts, and that doesn't seem to be getting me anywhere.

A direction on where to start this would be very appreciated!

Edit:

$\hat{\psi}$ is the Fourier transform of $\psi$ (normalized so that it's an isometry on $L^2$; specifically, $\hat{\psi}(x) = \int_{\mathbb{R}^d} \psi(\xi) e^{-2\pi i \xi \cdot x} d\xi$)

• How is $\hat{\psi}$ defined? I'm sure it's in the paper, but you'll get a better response if you include it here. – Semiclassical Aug 3 '14 at 3:46

Following the hint in Tao's notes, decompose the annulus in bump functions $\zeta_{\theta}$ adapted to disks of thickness $1/R$ and radius $\mathcal{O}(1/\sqrt{R})$, because of the geometry of the surface. The Fourier transform of each $\zeta_{\theta}$ is roughly $R^{-(d+1)/2}\chi_{T_{\theta}}$ (indeed, I should multiply this function by $e^{i<\theta,x>}$, but this is irrelevant in this case), where $T_{\theta}$ is a tube of length $R$ and width $\sqrt{R}$ centered at the origin. Using the fact the the directions of the tubes are $1/\sqrt{R}$-separated, you see that if $|x|\sim R$, then there is essentially only one tube contributing to $\hat{\psi}$, thus $|\hat{\psi}(x)|\sim R^{-(d+1)/2}$ (it shouldn't be $R^{-(d-1)/2}$, as you'll see). Actually, $\hat{\zeta_{\theta}}$ is not compactly supported, so use the information on the derivatives of $\zeta_{\theta}$ to get a good estimate on the decay of the Fourier transform.
When you blur out $d\sigma$ using an aproximation to the identity supported in $B(0,1/R)$, you get $R\psi$, so we conclude that $|\hat{d\sigma}(x)|\sim |R\psi(x)|\sim R^{-(d-1)/2}$ whenever $|x|\sim R$.