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?
Best Answer
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.