Newton's Theorem

Question

For my Measure & Integration course, I've been asked to prove the following:

Let $g$ be a function taking values on $\Bbb{R_+}$, $f: \Bbb{R}^3 \to \Bbb{R}$ such that $f(x) = g(|x|)$. Suppose that $f \in L^1(\Bbb{R}^3, dx)$, and let

$$ \Phi(x) = \int_{\Bbb{R}^3} \frac{f(y)}{|x-y|}dy. $$

For all $r=|x| > 0$,

a) Show that $$\Phi(x) = \frac{4π}{r}\int_0^r g(s)s^2\ \mathrm ds+ 4\pi\int_r^{\infty}g(s)s\ \mathrm ds.$$

b) Deduce that $$\Phi(x) = \int_{\Bbb{R}^3} \frac{f(y)}{max\{|x|,|y|\}}\mathrm dy.$$

c) Show that $$\mu\{x: \Phi(x) > t\} < \infty\qquad\text{for all }t>0,$$ where $\mu$ denotes the Lebesgue measure

d) Conclude that $\Phi(x) = \Phi^*(x)$, where $\Phi^*(x)$ is the symmetric decreasing rearrangement of the function $\Phi(x)$.

After a lot of torture I managed to prove part a), but I'm completely stuck on the following parts. Any advice would be very very welcome!

for part b, you can break up the integral according to whether |y|>|x| then write both pieces in polar coordinates — Simon Segert, Jan 05 '21 at 18:51
Is this related? https://math.stackexchange.com/q/1429782/27978 — copper.hat, Jan 06 '21 at 21:59
seems related but I don't really understand how; the tools used to solve that question are too advanced for what I've seen in my class unfortunately — Lupetto1927, Jan 07 '21 at 07:08

score 2 · Accepted Answer · answered Jan 11 '21 at 23:06

In part a, you showed that $\Phi$ itself is radial, it depends only on $r = |x|$. Thus, $\Phi(x)$ is equal to its average value of the sphere of radius $|x|$.

I.e.

$$ \Phi(x) = \frac{1}{4\pi |x|^2}\int_{|u|=|x|} \Phi(u) du $$ which by Fubini is $$ = \frac{1}{4\pi |x|^2} \int_{\mathbb{R}^3}f(y) \int_{|u|=|x|} \frac{1}{|u-y|}du\, dy. $$

This inner integral is a standard computation of a surface integral. You can see it's calculation e.g. here Surface integral of $f(x) = \frac{1}{ \Vert x -x_0 \Vert } $ over sphere. Thus the last line is $$ \frac{1}{4\pi |x|^2} \int_{\mathbb{R}^3}f(y) \frac{{4\pi |x|^2}}{\max(|x|, |y|)} dy = \int_{\mathbb{R}^3}f(y) \frac{1}{\max(|x|, |y|)} dy $$ as desired for part b.

By part b, we know that as $x \to \infty$ the integrand tends to $0$. Since the integrands with $|x|>1$ are dominated by the integrand with for a fixed $x_0$ on the unit sphere, we can apply dominated convergence to see that $\Phi \to 0$ as $|x| \to \infty$ (important that $f \in L^1$ here). This obviously implies part c as the measure in question is dominated by a that of a big ball.

For part d, just note that $\Phi$ is symmetric by b, and clearly decreasing in $|x|$ by c, so it is its own decreasing symmetric rearrangement.

Newton's Theorem

1 Answers1