This question is a follow up on my latest [MO post](https://mathoverflow.net/questions/425179/a-constant-ratio-of-integrals-part-i) 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 $r$.
> 
> 
> 



A constant ratio of integrals? Part II
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.


