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.
 A: This follows from the scalar version of the estimate:

Theorem: Let $0<r\leq s$.  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.
The proof relies on Caccioppolli inequality and the estimate

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
$$
\int_{B_r(0)}[f-f(0)]^2\leq 4^n\left(\frac rs\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2
$$
