Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|cite|improve this question

1 Answer 1

$\def\vol{\operatorname{vol}}$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 \[ \sup_{B_r(x)} u^2 \le \frac {2^n}{\vol B_{2r}(x)}\int_{B_{2r}(x)} u^2(z)\, dz \]

share|cite|improve this answer
Shouldn't the second $=$ be $\le$? – Julián Aguirre Oct 20 '12 at 8:24
@JuliánAguirre Of course you're right, thanks. – martini Oct 20 '12 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.