# How to prove the reverse Hölder inequality for Laplace equations?

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?

• You have already proved the (weak) reverse Holder inequality, so you "just" have to apply Gehring's lemma, see p. 130 in the book of Giaquinta--Martinazzi. I personally don't think Gehring's lemma is simple enough that I would rediscover it by myself, but I am also not an expert on these covering arguments. :) Sep 12, 2021 at 14:42

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.