Integrating $u^2(x+y)$ over $\partial B_1 (0)$, where $u$ is harmonic

Question

Let $u : \mathbb{R}^n \longrightarrow \mathbb{R}$ be a harmonic function. I must prove that $$\int_{\partial B_1 (0)} u(x+ty) u \left( x+\frac{1}{t}y \right) dS_y = \int_{\partial B_1 (0)} u^2(x+y) dS_y \quad \forall x \in \mathbb{R}^n \ \forall t$$ I've tried using polar coordinates on the left side, tried some change of variables (namely, $z = x + ty$ on the left side and $z = x+y$ on the right one), but they don't seem to help. I also tried using the mean value property on the right side, writing $u^2(x+y) = u(x+y) u(x+y)$. Should I keep trying with these tools or is there something I'm missing?

Answer 1

We can take $x=0.$ Now every harmonic function $u$ on $\mathbb R^n$ has a unique expansion into harmonic homogeneous polynomials. That is,

$$u(y) = \sum_{k=0}^{\infty}P_k(y),$$

where each $P_k$ is a homogeneous harmonic polynomial of degree $k.$ The convergence is uniform on compact subsets. These polynomials have the further property that if $j\ne k,$ then

$$\int_S P_j(\zeta)P_k(\zeta) \,d\sigma (\zeta) = 0.$$

Here $S$ is the unit sphere, and $\sigma$ is surface area measure on $S.$ Your result falls out of this nicely. (Such expansions when $n=2$ follow easily from the power series expansions of entire functions of a complex variable. For $n>2,$ this is not as well known, and there may well be a more elementary approach.)