# A constant ratio of integrals? Part II

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 $$0, consider the average of its Dirichlet integral $$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.$$

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

• 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. Jun 21, 2022 at 23:37

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 $$u$$.