# Foruier transform of spherical measure is radial

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?

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.
• 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 $\sigma(E)=3m(\bar E)$ Dec 10, 2020 at 9:13