Let $\Omega$ be an open subset of $\mathbb{R}^N$ and $u:\Omega\to \mathbb{R}$ be a harmonic function. I need to prove that, for every $x_0\in\Omega$, the function $$ \rho\mapsto \frac{1}{\rho^N}\int_{|x-x_0|<\rho} |\nabla u|^2 \; dx $$ is non-decreasing on $(0,+\infty)$.

I proved that for any $u\in C^2(\Omega,\mathbb{R})$ (not necessarily harmonic) we have that $$ \frac{d}{d\rho} \left( \frac{1}{\rho^{N-1}}\int_{|x-x_0|=\rho} u \; d\sigma \right) = \frac{1}{\rho^{N-1}}\int_{|x-x_0|<\rho} \Delta u\; dx, $$ and from this, if $u$ is harmonic, that the function $$ \rho \mapsto \frac{1}{\rho^N}\int_{|x-x_0|<\rho} u \; dx $$ is constant on $(0,+\infty)$.

In the book Partial Differential Equations by Evans, Section 8.6, he proves that the function $$ \rho \mapsto \frac{1}{\rho^{N-2}}\int_{|x-x_0|<\rho} |\nabla u|^2 \; dx $$ is non-decreasing, but I can't see (if possible) how to apply the ideas given there to this case.

  • $\begingroup$ $\frac{1}{\rho} \int_{|x-x_0|<\rho} u\textrm{d}x=\rho^{N-1}\frac{1}{\rho^N} \int_{|x-x_0|<\rho} u\textrm{d}x,$ so it's a product of a constant function and a non-decreasing function. $\endgroup$ Oct 30, 2019 at 18:21
  • $\begingroup$ @WoolierThanThou I made a mistake. I edited the question. Thank you! $\endgroup$
    – Albert
    Oct 30, 2019 at 18:23
  • $\begingroup$ Still, if you've proven that it is constant, it is, in particular, non-decreasing. $\endgroup$ Oct 31, 2019 at 8:14
  • $\begingroup$ @WoolierThanThou But The function I need to prove that is non-decreasing has $|\nabla u|^2$ as integrand, but the function I proved to be constant has $u$ as integrand. And it is not true in general that $|\nabla u|^2$ is harmonic when $u$ is. $\endgroup$
    – Albert
    Oct 31, 2019 at 16:00
  • $\begingroup$ Right, whoops... $\endgroup$ Oct 31, 2019 at 16:03

1 Answer 1


Forgive me for writing down part of this in Einstein summation notation, i'm just not used enough to vector calculus to be able to write this out in reasonable length otherwise.

If I'm not mistaken, $\left|\nabla u \right|^2$ is a subharmonic function, i.e. $\Delta\left|\nabla u \right|^2>0$ on $\Omega$. Indeed, \begin{align*} \Delta\left|\nabla u \right|^2 = \nabla \cdot \nabla (\nabla u \cdot \nabla u) = \partial^k\partial_k \left(\partial^ju \partial_j u\right) = 2\partial^k\left(\left(\partial_k\partial^j u \right)\partial_j u\right) = \\ 2\left(\left(\partial^j \partial^k\partial_ku \right)\partial_j u + \left(\partial_k\partial^j u \right)\left(\partial^k\partial_j u \right)\right) = 2\left(\left(\partial^j \Delta u \right)\partial_j u + tr(H_uH_u^T\right)) = 2 tr(H_uH_u^T) > 0, \end{align*} where $H_u$ denotes the Hessian of $u$.

Then $F(\rho) := \frac{1}{\rho^{N-1}} \int_{|x-x_0| = \rho} |\nabla u|^2dx$ is a non-decreasing function by the OP's remark. A change of variables yields

$$\frac{1}{\rho^{N}}\int_{|x-x_0| < \rho} |\nabla u|^2dx = \frac{1}{\rho^{N}} \int_0^\rho r^{n-1}\frac{1}{r^{n-1}}\int_{|y-x_0| = r}|\nabla u|^2dy dr = \\ \frac{1}{\rho^{N}} \int_0^\rho r^{n-1} F(r) dr = \int_0^1 s^n F(s\rho) ds.$$

We can now differentiate inside the integral to obtain

$$\frac{d}{d\rho} \int_0^1 s^n F(s\rho) ds = \int_0^1 s^{n+1} F'(s\rho) ds > 0,$$

finishing the proof.


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