Let $ u\in H^1(2B) $ be a weak solution of $ \Delta u=0 $ in $ 2B $, where $ B=B(0,1) $ is a ball with center $ 0 $ and radius $ 1 $. Then there exists some $ p>2 $ such that

\begin{eqnarray} \left(\frac{1}{|B|}\int_{B}|\triangledown u|^p dx\right)^{1/p}\leq C\left(\frac{1}{|2B|}\int_{2B}|\triangledown u|^2 dx\right)^{1/2}. \end{eqnarray} where $ C $ is an absolute constant.

I recently saw this problem and I want to get the solution of this problem. However, I meet with some troubles in it. Here is my try. First as $ u-\frac{1}{|2B|}\int_{2B}u $ is also a weak solution for the Laplace equation, then by using integration by parts on the function, I can obtain that

\begin{eqnarray}
\frac{1}{|B|}\int_{B}|\triangledown u|^2 dx\leq C\left\{\int_{2B}\left|u-\frac{1}{|2B|}\int_{2B}u\right|^2dx\right\}.
\end{eqnarray}
where $ C $ is an absolute constant. Then, by using the Sobolev-Poincaré inequality I have

\begin{eqnarray}
\left(\frac{1}{|B|}\int_{B}|\triangledown u|^2 dx\right)^{1/2}\leq C\left(\frac{1}{|2B|}\int_{2B}|\triangledown u|^q dx\right)^{1/q}
\end{eqnarray}
where $ q=\frac{2d}{d+2} $. I think it is quite similar to the final result. But I cannot go further. Can you give me some hints or references?

Iwould rediscover it by myself, but I am also not an expert on these covering arguments. :) $\endgroup$