Bilinear Harmonic Function

$u(x)$ is harmonic in $\Omega\subset\mathbb{R}^n$, and suppose $B_c(x_0)\subset\subset\Omega$. If $a\le b\le c, ac=b^2$, then

$\displaystyle \int_{|\omega|=1}u(x_0+a\omega)u(x_0+c\omega)\mathrm{d}\omega=\int_{|\omega|=1}u^2(x_0+b\omega)\mathrm{d}\omega$

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I like this theorem, it is beautiful  and concise. The proof was kind of trivial if apply Poisson Kernel bravely. Let’s make it more general.

Set $S=\{|\omega|=1\}$, thus define $\xi(\rho_1,\rho_2)=\int_{S}\phi_1(\rho_1 x)\phi_2(\rho_2 x)$, if we are given that $\alpha_1\alpha_2=\beta_1\beta_2$, then we are going to prove:

$\xi(\alpha_1,\alpha_2)=\xi(\beta_1,\beta_2)$

consider unit vector $a,b$, by Poisson Integral,

$\phi(\alpha a)=\int_{S_b}K(\alpha a,\beta b)\phi(\beta b)$,

where

$\displaystyle K(\alpha a,\beta b)=\frac{\beta^{n-2}(\beta^2-\alpha^2)}{C_n(\alpha^2+\beta^2-2\alpha\beta\cos(a,b))^{n/2}}$,

since $\alpha_1\alpha_2=\beta_1\beta_2$, thus we assume $\kappa=\frac{\alpha_1}{\beta_2}=\frac{\beta_1}{\alpha_2}$, we know that $K$ is homogeneous, hence

$K(\alpha_1 a,\beta_1 b)=K(\kappa\beta_2 a,\kappa\alpha_2 b)=K(\beta_2 a,\alpha_2 b)$.

$\xi(\alpha_1,\alpha_2)=\int_{S_a}\int_{S_b}K(\alpha_1 a,\beta_1 b)\phi_1(\beta_1 b)\phi_2(\alpha_2 a)$

which is

$\xi(\alpha_1,\alpha_2)=\int_{S_b}\int_{S_a}K(\beta_2 a,\alpha_2 b)\phi_2(\alpha_2 a)\phi_1(\beta_1 b)$

The inner integral is $\phi_2(\beta_2 a)$.

Done.

$\Box$