
> 
> 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?



How to prove the reverse Hölder inequality for Laplace equations?
The inequality is scale invariant and holds for a ball of any radius. It follows by a standard argument that is the inductive step in what's known as Moser iteration.


The constant $C$ below can change from line to line but always depends only on the dimension. Let $B = B(0,r)$ and $2B = B(0,2r)$. Let $\chi$ be a smooth compactly supported function on $2B$ that is identically $1$ on $B$ and satisfies
$$ \|\nabla\chi\|\_\infty \le \frac{C}{r} $$


The Sobolev inequality states that there exists a constant $C$, for any smooth compactly supported function $f$ on $2B$,
$$
\left(\int\_{2B} |f|^{\frac{2d}{d-2}}\right)^{\frac{d-2}{d}} \le C\int\_{2B} |\nabla f|^2
$$
Suppose $\Delta u = 0$. First, observe that, if you integrate by parts, then for any constant $a > 0$,
\begin{align\*}
\int\_{2B} \chi^2|\nabla u|^2 &= \int\_{2B} - 2(a^{-1} u\nabla\chi)\cdot(a\chi\nabla u) \\
&\le a^{-2}\int\_{2B} \chi^2|\nabla u|^2 + a^{2}\int\_{2B} u^2|\nabla\chi|^2.
\end{align\*}
In particular, if we set, say, $a = 2$, then
$$
\int\_{2B} \chi^2|\nabla u|^2
\le C\int\_{2B} u^2|\nabla\chi|^2.
$$
It now follows that
\begin{align\*}
\left(\int\_{B} |u|^{\frac{2d}{d-2}}\right)^{\frac{d-2}{d}} &\le \left(\int\_{2B} |\chi u|^{\frac{2d}{d-2}}\right)^{\frac{d-2}{d}}\\
& \le C\int\_{2B} |\nabla (\chi u)|^2\\
&\le C\int\_{2B} \chi^2|\nabla u|^2 + |\nabla\chi|^2u^2\\
&\le C\int\_{2B} |\nabla\chi|^2u^2\\
& \le C\|\nabla\chi\|\_\infty^2\int\_{2B} u^2\\
&= \frac{C}{r^2}\int\_{2B} u^2.
\end{align\*}
Since $\Delta(\partial\_ku) = 0$, the estimate above holds for $\partial\_ku$. The desired estimate can be derived from this.


