2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ You don't have spherical symmetry anymore. How did you get the transform without it? $\endgroup$
    – user147263
    Oct 27 '14 at 22:16
  • $\begingroup$ @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. $\endgroup$ Oct 27 '14 at 22:42
2
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ For completeness: MathOverflow question Fourier transform of the unit sphere gives references for the spherically symmetric case. $\endgroup$
    – user357151
    Mar 14 '17 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.