We define the surface measure of the sphere $\sigma$ as the unique measure on $\mathbb{S}^2$ satisfying $$\int_\mathbb{S^2} f \operatorname{d}\hspace{-0.5mm}\sigma:=\int_0^\pi\int_0^{2\pi} f(\sin\phi \cos\theta,\sin\phi \sin\theta,\cos\phi) \sin\phi \operatorname{d}\hspace{-0.5mm}\theta \operatorname{d}\hspace{-0.5mm}\phi$$ for every $f \in L^\infty(\mathbb{S^2})$

Then, we define its Fourier transform as $$\hat\sigma(\xi)=\int_\mathbb{S^2} e^{-2\pi i \langle x, \xi\rangle} \operatorname{d}\hspace{-0.5mm}\sigma(x)$$

So, I'm trying to see that $\hat\sigma$ is a radial function, and I know it must be by a change of variable with a rotation (similar to what is done when you have a radial function and you show its Fourier transform is radial) but I'm not able to do it rigurously since in the definition we go from working on $\mathbb{R}^3$ to $\mathbb{R}^2$.

Could someone prove this fact in detail?


1 Answer 1


It's easier to prove this result using a different definition of $\sigma$, so let's first prove the two definitions are equivalent. I claim that the definition of $\sigma$ given above satisfies $$ \sigma(E) = 3 m(\overline{E}) $$ for all measurable $E \subseteq S^2$, where $m$ is $3$-dimensional Lebesgue measure and $\overline{E} := \{ tx : 0 \leq t \leq 1, x \in E \}$. To prove this, evaluate the right hand side using spherical coordinates: $$ m(\overline{E}) = \int_{\mathbb{R}^3} 1_E = \int_{0}^{\infty} \int_0^\pi \int_0^{2\pi} 1_\overline{E}(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi) r^2 \sin\phi \,d\theta \,d\phi \,dr. $$ By definition, $(r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi) \in \overline{E}$ if and only if $(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) \in E$ and $0 \leq r \leq 1$. Therefore the above is equal to $$ \int_0^1 r^2 \int_0^\pi \int_0^{2\pi} 1_E(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) \sin\phi \,d\theta \,d\phi \,dr = \sigma(E) \cdot \int_0^1 r^2\,dr = \frac13 \sigma(E) $$ as claimed.

Now, in order to prove that $\widehat{\sigma}$ is radial, we need to prove that $\widehat{\sigma}(A\xi) = \widehat{\sigma}(\xi)$ for any $3\times3$ orthogonal matrix $A$. First we have $$ \widehat{\sigma}(A\xi) = \int_{S^2} e^{-2\pi i \langle x, A\xi \rangle} \,d\sigma(x) = \int_{S^2} e^{-2\pi i \langle Ax, \xi \rangle} \,d\sigma(x) $$ because orthogonal matrices move across inner products. Another property of orthogonal matrices is that they preserve distances between points, and therefore they also preserve the $3$-dimensional Lebesgue measure (in the sense that $m(A^{-1} E) = m(E)$ for any measurable $E$). Therefore since we've shown that $\sigma(E)$ is just a constant multiple of $m(\overline{E})$, $\sigma$ must also be preserved by the action of orthogonal matrices, which allows us to conclude $$ \int_{S^2} e^{-2\pi i \langle Ax, \xi \rangle} \,d\sigma(x) = \int_{S^2} e^{-2\pi i \langle x, \xi \rangle} \,d\sigma(x) = \widehat{\sigma}(\xi) $$ as desired.

  • $\begingroup$ Thanks for the answer, it was really useful but I think there is a little mistake because in the spherical change of variables when you calculate the jacobian you get $r^2$ instead of $r$ as you put, so it would be $\sigma(E)=3m(\bar E)$ $\endgroup$
    – Eparoh
    Dec 10, 2020 at 9:13
  • $\begingroup$ You're right, thanks. Edited $\endgroup$
    – Adam
    Dec 10, 2020 at 15:55

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