If a real-valued function $u$ is harmonic on a ball $B_{2r}(x)$ in $\mathbb{R}^n$, how would one show that

$$\sup_{B_r(x)}u^2\leq\frac{2^n}{|B_{2r}(x)|}\int_{B_{2r}(x)}u^2(y) dy$$


$\def\vol{\operatorname{vol}}$For $y \in B_r(x)$ by the mean value property and Hölder \begin{align*} u^2(y) &= \left(\frac 1{\vol B_r(y)}\int_{B_r(y)}u(z)\,dz\right)^2\\ &\le \frac 1{\vol B_r(y)^2} \int_{B_r(y)} u^2(z)\,dz \cdot \int_{B_r(y)} 1^2\,dz\\ &= \frac 1{\vol B_r(y)} \int_{B_r(y)} u^2(z)\,dz \end{align*} Now, as $B_r(y) \subseteq B_{2r}(x)$ and $\vol B_{2r} = 2^n \vol B_r$, we have \begin{align*} {u^2(y)} &\le \frac {2^n}{\vol B_{2r}(x)}\int_{B_{2r}(x)} u^2(z)\, dz \end{align*} And hence, as $y \in B_r(x)$ was arbitrary \[ \sup_{B_r(x)} u^2 \le \frac {2^n}{\vol B_{2r}(x)}\int_{B_{2r}(x)} u^2(z)\, dz \]

  • $\begingroup$ Shouldn't the second $=$ be $\le$? $\endgroup$ – Julián Aguirre Oct 20 '12 at 8:24
  • $\begingroup$ @JuliánAguirre Of course you're right, thanks. $\endgroup$ – martini Oct 20 '12 at 14:28

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