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Let $u:\mathbb{R}^3 \to \mathbb{R}$ a harmionic function, and let $S_r = \{x^2+y^2+z^2 = r^2 \}$

I want to show that if $0 < r_1 <r_2$, then:

$$ \frac{1}{\mathrm{area}(S_{r_1})} \int_{S_{r_1}} u^2 dS \le \frac{1}{\mathrm{area}(S_{r_2})} \int_{S_{r_2}} u^2 dS$$

I know that: $u^2(0) \le \frac{1}{\mathrm{area}(S_r)} \int_{S_r} u^2 dS$

So I can instead try to show $$\frac{1}{\mathrm{area}(S_{r_1})} \int_{S_{r_1}} u^2 dS \le u^2(0)$$

But I don't see how to do it.

Help would be appreciated.

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Theorem(Green's formula). Assume $U$ is a bounded, open subset of $\mathbb R^n$, and $\partial U$ is $C^1$. Let $u,v\in C^2(\bar{U})$. Then $$\int_UDu\cdot Dv\,dx=-\int_U u\Delta v\,dx+\int_{\partial U}\frac{\partial v}{\partial \nu}u\,dS.$$

We can use this to show the $n-$dimensional case.

Set $$\phi(r):=\frac1{\text{area}(S_r)}\int_{S_r}u^2(y)\,dS(y)=\frac1{\text{area}(S_1)}\int_{S_1}u^2(rz)\,dS(z).$$ Then $$\phi'(r)=\frac1{\text{area}(S_1)}\int_{S_1}2u(rz)Du(rz)\cdot z\,dS(z),$$ and consequently, using Green's formula, we compute \begin{align*} \phi'(r)&=2\frac1{\text{area}(S_r)}\int_{S_r}u(y)Du(y)\cdot\frac yr\,dS(y)\\&=2\frac1{\text{area}(S_r)}\int_{S_r}u\frac{\partial u}{\partial \nu}\,dS\\&=2\frac1{\text{area}(S_r)}\left(\int_{B_r}Du\cdot Du\,dx+\int_{B_r}u\Delta u\,dx\right)\\&=2\frac1{\text{area}(S_r)}\int_{B_r}|Du|^2\,dx\geq 0. \end{align*} Hence $\phi(r)$ is increasing.

Addendum: From above we can get $$\frac1{\text{area}(S_r)}\int_{S_r}u^2\,dS=\phi(r)\geq \lim_{r\to 0}\phi(r)=u^2(0),$$ since $\phi$ is continuous at $r=0$, which can be proven using the method in a previous post.

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    $\begingroup$ @ThomasShelby $D(u^2)=2u Du$? $\endgroup$ – Feng Shao Sep 1 at 14:00
  • $\begingroup$ I am not sure exactly how did you get from the second line to the third line, could you elaborate? In other note, if I declare $\phi (r)$ as you did, then I will have $u^2(0) \le \phi (r)$ and then $0 \le \phi '(r)$ and I think it is enough $\endgroup$ – Gabi G Sep 1 at 14:28
  • $\begingroup$ @GabiG You can just differentiate the integrand: $$\frac{d}{dr}(u^2(rz))=2u(rz)\frac{d}{dr}(u(rz))=2u(rz)Du(rz)\cdot z$$ $\endgroup$ – Feng Shao Sep 1 at 15:05
  • $\begingroup$ @ThomasShelby Yes, it was almost a copy from Evans. Thanks for the link. I will add it in the text tomorrow. $\endgroup$ – Feng Shao Sep 1 at 15:07
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For a smooth $f$, $$ \begin{align} \frac{\partial}{\partial r}\frac1{|S_r|}\int_{S_r}f(x)\,\mathrm{d}\sigma(x) &=\frac1{|S_r|}\int_{S_r}n(x)\cdot\nabla f(x)\,\mathrm{d}\sigma(x)\tag1\\ &=\frac1{|S_r|}\int_{B_r}\Delta f(x)\,\mathrm{d}x\tag2\\ \end{align} $$ Explanation:
$(1)$: the derivative of the spherical average is the spherical average of the outward derivative
$(2)$: Divergence Theorem

Furthermore, $$ \Delta f^2=2|\nabla f|^2+2f\Delta f\tag3 $$ Thus, for a harmonic $u$ $$ \begin{align} \frac{\partial}{\partial r}\frac1{|S_r|}\int_{S_r}u(x)^2\,\mathrm{d}\sigma(x) &=\frac1{|S_r|}\int_{B_r}2|\nabla u|^2\,\mathrm{d}x\tag4\\ &\ge0\tag5 \end{align} $$ That is, $\frac1{|S_r|}\int_{S_r}u(x)^2\,\mathrm{d}\sigma(x)$ is a non-decreasing function of $r$.

Thus $u(0)^2\le\frac1{|S_r|}\int_{S_r}u(x)^2\,\mathrm{d}\sigma(x)$ for all $r$, so the suggested attempt probably won't work.

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