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Let $f$ be a real function with $\Delta f=0$ on an open ball $B_{2n}(y)\subset\mathbb{R}^N$.

How would I show

$$\int\limits_{B_n(y)}|Df|^2(z)dz\leq Cn\int\limits_{\partial B_n(y)}|Df|^2(z)d\sigma(z)$$

for some constant C, where $\sigma$ is surface measure?

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1 Answer 1

Suppose $z\in B_{n}(y)$. Write $z=y+r\sigma$, where $r\in[0,1]$ and $\sigma\in\partial B_{n}(0)$. Hence $dz=nr^{N-1}$.

Now, we have for $D_{i}f=\frac{\partial f}{\partial x_{i}}$

\begin{eqnarray} \int_{B_{n}(y)}(D_if(z))^{2}dz &=& \int_{\partial B_{n}(0)}\int_{0}^{1}(D_if(y+r\sigma))^2nr^{N-1}drd\sigma \\ &\leq& \int_{\partial B_{n}(0)}\int_0^1(D_if(y+\sigma))^2nr^{N-1}drd\sigma \\ &=&\frac{n}{N}\int_{\partial B_{n}(0)}(D_if(y+\sigma))^2d\sigma\\ &=&\frac{n}{N}\int_{\partial B_{n}(y)}(D_if(z))^2d\sigma(z)\ \end{eqnarray}

In the inequality, i used the fact that $f$ is harmonic which implies that $D_if$ is harmonic and hence must attains its maximum on $\partial B_{n}(y)$. We conclude that $$\sum_{i=1}^N \int_{B_{n}(y)}(D_if(z))^{2}dz\leq\sum_{i=1}^n\frac{n}{N}\int_{\partial B_{n}(y)}(D_if(z))^2d\sigma(z)$$

Therefore, $$\int_{B_{n}(y)}|\nabla f(z)|^{2}dz\leq\frac{n}{N}\int_{\partial B_{n}(y)}|\nabla f(z)|^2d\sigma(z)$$

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