$\textbf{Problem}$ Let $\Omega$ be open and connected in $\mathbb{R}^n$ for $n\geq 2$, and suppose that $u \in C^2(\Omega)$. Prove that the following statements are all equivalent. \begin{align*} (\textrm{i}) \quad u \textrm{ is harmonic in } \Omega\\ (\textrm{ii}) \quad \textrm{If } \overline{B_r(x)} \subset \Omega, \textrm{ then}\\ &u(x)=\frac{1}{\textrm{Vol} \partial B_r(x)}\int _{\partial B_r(x)} u(y)dS(y).\\ (\textrm{iii}) \quad \textrm{If } \overline{B_r(x)} \subset \Omega, \textrm{ then}\\ &u(x)=\frac{1}{\textrm{vol}B_r(x)}\int_{B_r(x)} u(y) dy. \end{align*}

I proved (i) and (ii) are equivalent. I want to know how to prove (i) and (iii) are equivalent.

I knew that this problem was already uproaded. However, I don't know the Poisson's Integral fromula on $\mathbb{R}^n$....

Any help is appreciated..


Let's see that $1)$ implies $3)$: \begin{align}\frac{1}{\textrm{vol}B_r(x)}\int_{B_r(x)} u(y) dy&=\frac{1}{\textrm{vol}B_r(x)} \int_0^r\int_{\partial B_s(x)}u(y)dS(y)ds\\ &=\frac{1}{\textrm{vol}B_r(x)} \int_0^r\textrm{vol}\partial B_s(x)u(x)ds\\ &=u(x)\frac{1}{\textrm{vol}B_r(x)}\textrm{vol}B_r(x)=u(x) .\end{align} The second equality is given by the already proved equivalence between $1)$ and $2)$.
Let's see that $3)$ implies $1)$:
Define $$\phi(r)=\frac{1}{\textrm{vol}\partial B_r(x)}\int_{\partial B_r(x)}u(y)dS(y).$$ You can check (and probably you already did proving the first equivalence) that $$\phi'(r)=\frac{r}{n\textrm{vol}B_r(x)}\int_{B_r(x)}\Delta u(y)dy.$$ Assume that $\Delta u$ is not identically $0$, we can find a point $x$ and an open ball $B_\rho(x)$ such that $\Delta u(y)>0$ in $B_\rho(x)$ (you can use the same argument if $\Delta u(y)<0$). This implies that $\phi'(s)>0$ for $0<s\leq\rho$, thus $\phi$ is strictly increasing in this interval. Finally, you can conclude \begin{align}\frac{1}{\textrm{vol}B_\rho(x)}\int_{B_\rho(x)} u(y) dy&=\frac{1}{\textrm{vol}B_\rho(x)} \int_0^\rho\int_{\partial B_s(x)}u(y)dS(y)ds\\ &=\frac{1}{\textrm{vol}B_\rho(x)} \int_0^\rho\textrm{vol}\partial B_s(x)\phi(s)ds\\ &>\phi(0)=u(x),\end{align} which is a contradiction.

  • $\begingroup$ Nice answer, is it possible to prove separately that $(ii) \implies (iii)$? $\endgroup$
    – Math nerd
    Aug 2 '20 at 5:12
  • $\begingroup$ Hello, in 1=>3, could someone explain the first equality by any chance? $\endgroup$
    – Raphaël
    Mar 24 at 16:36

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