This question is a follow up on my latest MO post which was addressed kindly by Iosif Pinelis. What is new here is that I need to correct the assumption by including a missing hypothesis. The context required me to look into spherical harmonics. That is why.

Let $u(x)$ be a homogeneous harmonic polynomial in the unit ball $B_1(0)\subset\mathbb{R}^n$ with $u(0)=0$.

For $0<r\leq1$, consider the average of its Dirichlet integral $$A(r):=\frac1{\vert B_r(0)\vert}\int_{B_r(0)}\vert\nabla u\vert^2dx,$$ and the average of the square function on the boundary $$B(r):=\frac1{\vert \partial B_r(0)\vert}\int_{\partial B_r(0)}u^2d\sigma.$$

I would like to ask:

QUESTION. Is this true? The ratio $\frac{r^2A(r)}{B(r)}$ is a constant in $r$.

  • 1
    $\begingroup$ Yes, using that $|\nabla u|^2 = \text{div}(u\nabla u)$, integrating by parts, and using that $u_r = k r^{-1}u$ where $k$ is the degree of the polynomial, one sees that $r^2A(r) = nkB(r)$. This is an instance of the Almgren monotonicity formula, which says that $r^2A/B$ is non-decreasing in $r$, and constant if and only if $u$ is homogeneous. $\endgroup$ Jun 21, 2022 at 23:37

1 Answer 1


In fact, this is true for any homogeneous polynomial $u$ (not identically $0$), be $u$ harmonic or not.

Indeed, let $m\ge1$ be the degree of such a polynomial $u$. Then $$u(tx)=t^m u(x)$$ and $$v(tx)=t^{2m-2} v(x)$$ for all real $t$, where $v:=|\nabla u|^2$. So, for $B_r:=B_r(0)$, $$\int_{B_r}v(x)\,dx =\int_{B_1}v(ry)\,r^n\,dy = r^{2m-2+n}\int_{B_1}v(y)\,dy,$$ whence $$A(r)=a_{n,u}r^{2m-2},$$ where $a_{n,u}$ is a nonnegative real constant depending only on $n$ and $u$.

Similarly, $$B(r)=b_{n,u}r^{2m},$$ where $b_{n,u}$ is a positive real constant depending only on $n$ and $u$.

Now the desired result immediately follows.


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