# About the Fourier transform of the surface measure of the unit sphere

Let $d\sigma$ denote the surface measure on $\mathbb{S}^{n-1}$. To compute its Fourier transform $$\hat{d\sigma}(\xi)=\int e^{-i x\cdot \xi}\, d\sigma(x),$$ a standard technique (cfr. Folland's "Real Analysis" first edition, exercise 19, chapter 8: "Topics in Fourier Analysis") is based on the observation that $$\tag{1} \lvert x\rvert^2\,d\sigma = d\sigma,$$ because $d\sigma$ is supported on the unit sphere. Fourier transforming identity (1) one gets $$\tag{2} -\Delta_\xi \hat{d\sigma}=\hat{d\sigma},$$ so the sought Fourier transform satisfies the Helmholtz equation. Moreover, $d\sigma$ is spherically symmetric, so equation (2) can be explicitly integrated with initial conditions $$\hat{d\sigma}(0)=\lvert \mathbb{S}^{n-1}\rvert\quad \text{and}\quad \frac{\partial\hat{d\sigma}}{\partial \rho}(0)=0,\quad \text{where}\ \rho=\lvert\xi\rvert.$$

Question. The measure $d\mu=g\,d\sigma$, where $g\ge 0$ is a smooth function on the sphere, is supported on $\mathbb{S}^{n-1}$. Applying the same computation as above we get for its Fourier transform that $$\tag{!!}\hat{d\mu}= \frac{\|g\|_{L^1(\mathbb{S}^{n-1})}}{\lvert \mathbb{S}^{n-1}\rvert} \hat{d\sigma},$$ which, by uniqueness of the Fourier transform, implies that $g$ must be constant. This is absurd. Where is the fallacy in this reasoning?

• You don't have spherical symmetry anymore. How did you get the transform without it?
– user147263
Oct 27, 2014 at 22:16
• @WeaponofChoice: Right! That was the missing piece. The computation works if $d\mu$ is a spherically symmetric measure concentrated on the unit sphere, and clearly any such measure is equal to $d\sigma$ up to a constant. Publish this as an answer so that I can upvote & accept it. Oct 27, 2014 at 22:42

For general $d\mu = g\,d\sigma$, you don't have spherical symmetry, so the same computation as for $d\sigma$ does not apply.