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Suppose $x_0 \in \mathbb{R}^2$ and $u \in C^2(\bar{B}_R(x_0))$ is a non-negative harmonic function. I want to show that $$|\nabla u(x_0)| \leq \frac{2}{R}u(x_0)$$

I know that since $u$ is harmonic it satisfied the mean value property. I'm given two mean value properties (which I earlier proved are equivalent):

$$u(x)=\frac{1}{2\pi r}\int_{B_r(x)}u(y) \, dS(y)$$

and $$u(x)=\frac{1}{\pi r^2}\int_{B_r(x)}u(y) \, dy$$

I think using these is probably the way to get the result, I'm not sure how though.

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  • $\begingroup$ @Dr.MV: but such a $v$ is not guaranteed to be non-negative, so it's not so simple. $\endgroup$ – Anthony Carapetis Mar 29 '17 at 15:27
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    $\begingroup$ @AnthonyCarapetis I missed the non-negative condition on $u$. So, yes. $\endgroup$ – Mark Viola Mar 29 '17 at 15:54
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Rotate your coordinates so that $\nabla u(x_0)$ is in the $e_1$ direction. Note that since $u$ is harmonic, its partial derivative $\partial_1 u$ is too; so the mean value property (ball version) plus the divergence theorem give

$$ \partial_1u(x_0) = \frac{1}{\pi R^2}\int_{B_R} \partial_1u = \frac{1}{\pi R^2}\int_{B_R} {\rm div}(u e_1) = \frac{1}{\pi R^2}\int_{\partial B_R} u e_1 \cdot n.$$

Estimating $|e_1 \cdot n | \le 1$ and applying the mean value property (this time the sphere version) along with the fact $u \ge 0$ we get

$$ |\partial_1 u(x_0)| \le \frac{1}{\pi R^2}\int_{\partial B_R} |u|=\frac{2 \pi R}{\pi R^2}u(x_0).$$

The LHS here is exactly $|\nabla u(x_0)|$ by our choice of coordinates, and the RHS is exactly $\frac 2 R u(x_0)$.

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  • $\begingroup$ Why is $\partial_1u = \operatorname{div}(ue_1)$? $\endgroup$ – MHW Mar 29 '17 at 15:59
  • $\begingroup$ @Si.0788: from the definition of the divergence. The first component of $ue_1$ is $u$ and all the other components are zero. $\endgroup$ – Anthony Carapetis Mar 29 '17 at 16:08
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It is not hard to prove that if $u \geq 0$ is harmonic on $B(x_0,R)$, then for any $x \in B(x_0,R)$ we have $$u(x) \leq \frac{R+||x-x_0||}{R-||x-x_0||}u(x_0).$$ This you can do by using the mean value property you mentioned above and very simple triangle and reverse triangle inequality of the norm. But then $$u(x)-u(x_0) \leq \left(\frac{R+||x-x_0||}{R-||x-x_0||}-1 \right)u(x_0)=\frac{2||x-x_0||}{R-||x-x_0||}u(x_0).$$ Now, take the norms and divide on both sides by $||x-x_0||$ and let $x \to x_0.$

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  • $\begingroup$ Hm, I am suspicious now. I haven't used that the function is of class $C^2$ and I don't think that the limit of $\frac{||u(x)-u(x_0)||}{||x-x_0||}$ always exists. $\endgroup$ – Hua Ying Mar 2 '18 at 16:20

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