# Interior gradient estimate harmonic function: decay estimate

I found this problem Proposition but I am completely stuck. Let $$u$$ be an harmonic function satisfying $$\int_{B_1(0)}|\nabla u|^2 \mathrm{d}x \leq 1,$$ where $$B_1(0)$$ is the unitary ball in $$\mathbb{R}^n$$. Then $$\frac{1}{r^n}\int_{B_r(0)}|\nabla u - \nabla u(0)|^2\mathrm{d}x \leq C(n)r^2 \leq \frac{1}{2}$$ for $$r\leq \theta(n)$$. Obviously I understand how to prove the final bound $$1/2$$, but I don't get how to obtain the quadratic decay bound.

• What are $n, \theta(n), C(n)$? Commented Nov 4, 2018 at 17:55
• $n$ is the dimension of the space and $\theta(n), C(n)$ are dimensional constant. Commented Nov 5, 2018 at 11:13
• Thanks for the clarification. Commented Nov 5, 2018 at 16:07

Theorem: Let $$0. There are constant $$C(n)$$ such that $$\int_{B_r(0)} [f-f(0)]^2 \leq C(n)\left(\frac{r}{s}\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2$$ for all harmonic function $$f$$ on $$B_s(0)$$. The constant $$C(n)$$ can be chosen to be $$4^{n+2}$$.
noting that, since $$u$$ is harmonic, $$D_iu$$ is harmonic too.
Lemma: If $$f$$ is harmonic on $$B_{2r}(0)$$, then $$\sup_{B_r(0)} f^2\leq2^n-\!\!\!\!\!\!\!\int_{B_{2r}(0)} f^2$$ Proof of Lemma: For $$y\in B_r(0)$$, Cauchy-Schwarz with the mean-value property of harmonic function gives $$[f(y)]^2=\left[-\!\!\!\!\!\!\int_{B_r(y)} f\right]^2\leq-\!\!\!\!\!\!\int_{B_r(y)} f^2-\!\!\!\!\!\!\int_{B_r(y)} 1=-\!\!\!\!\!\!\int_{B_r(y)} f^2$$. Now expand the volume of integration to $$B_{2r}(0)$$.
Proof of Theorem: If $$4r\geq s$$, then we have simply $$\int_{B_r(0)}[f-f(0)]^2\leq\left(\frac{4r}s\right)^{n+2}\int_{B_r(0)}[f-f(0)]^2\leq 4^{n+2}\left(\frac rs\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2.$$ Otherwise, we have $$r\leq s\leq 4r$$ and we need to do a little more work. Separately bounding each component of $$\nabla f$$ using Lemma, we have $$\sup_{B_r(0)}\lvert\nabla f\rvert^2\leq2^n-\!\!\!\!\!\!\!\int_{B_{2r}(0)} \lvert\nabla f\rvert^2$$ and so \begin{align*} -\!\!\!\!\!\!\int_{B_r(0)}[f-f(0)]^2 &\leq -\!\!\!\!\!\!\int_{B_r(0)}\left(r\sup_{B_r(0)}\lvert\nabla f\rvert\right)^2\\ &=r^2\sup_{B_r(0)}\lvert\nabla f\rvert^2\\ &\leq r^2\sup_{B_{s/4}(0)}\lvert\nabla f\rvert^2\\ &\leq r^2 2^n-\!\!\!\!\!\!\!\int_{B_{s/2}(0)}\lvert\nabla f\rvert^2\\ &\leq r^2 2^n \frac{2^n}{s^2}-\!\!\!\!\!\!\!\int_{B_s(0)}[f-f(0)]^2 \end{align*} where the last inequality is by Caccioppolli. Rearranging, we obtain $$\int_{B_r(0)}[f-f(0)]^2\leq 4^n\left(\frac rs\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2$$