I'm interested in proving the following:
Let $u$ harmonic in $\Omega $ and let $ B_{\rho} (x) \subset {\Omega} $ an open ball, then
$$ \left |{ u_{x_{i}}(x) }\right | \leq{ \displaystyle\frac{\gamma_{n} M}{\rho} }, $$ where $ u_ {x_{i}} = \frac {\partial u} {\partial x_{i}}$, $M=sup \left\{{ \left |{u(x)}\right | : x \in B_{\rho}(x) }\right\}$ , $\gamma _{n} = \displaystyle\frac{2n \omega _{n-1}}{(n-1) \omega _{n}}$, and, $\omega_{k}$ is the area of the unit sphere in $\mathbb{R} ^{k}$.
My attempt:
Since $u$ is harmonic, we know that, for each $B_{r}(x) \subset{\Omega}$ we have that $$ u(x)=\displaystyle\frac{1}{Vol(B_{r}(x))} \displaystyle\int_{B_{r}(x)}^{} u(y) dy. $$ However, if $u$ is harmonic then $u_{x_{i}}$ is also harmonic, thus $$ u_{x_{i}}(x)=\displaystyle\frac{1}{Vol(B_{r}(x))} \displaystyle\int_{B_{r}(x)}^{} u_{x_{i}}(y) dy . $$ If we use Divergence theorem, the above equation becomes $$ u_{x_{i}}(x)=\displaystyle\frac{1}{Vol(B_{r}(x))} \displaystyle\int_{\partial B_{r}(x)}^{} e_{i}u(y) dS, $$ where $e_{i}$ is the vector which has a $1$ in the $i$-th coordinate and zeros in all the others, it follows that $$ \left |{ u_{x_{i}} (x) }\right | \leq{ \displaystyle \frac{M_{r} Vol(\partial B_{r}(x))}{Vol(B_{r}(x))} } = \displaystyle\frac{n M_{r}}{r}, $$ where $M_{r}=sup \left\{{ \left |{u(x)}\right | : x \in B_{r}(x) }\right\}$.
Of course, I definitely want to take $ r = \frac {\rho (n-1) \omega_ {n}} {2 \omega_ {n-1}} $, but, as you can see, the problem with this idea is that I need to ensure that , for this choice of $r$, we have that $B_{r} (x) \subset B _ {\rho} (x) $. However, this is generally false :( , if we take for instance $n=3$, we have that $\omega_{3}=4 \pi$, $\omega_{2}= 2 \pi$ and so $r=2 \rho > \rho$ . On the other hand, the hint of the book is consider the Poisson integral formula:
$$ u(y)=\displaystyle\frac{\rho ^2 - \left\|{y}\right\| ^2 }{\rho \omega _{n}} \displaystyle\int_{\partial B_{\rho} (0) }^{} \displaystyle\frac{u(x)}{\left\|{x-y}\right\| ^{n} } dS_{x}, $$ But, I definitely don't see why this formula is more useful :( .
Any help?
Thanks in advance.