Harnack's inequality for gradient

Question

How to prove the following:

If u is a function harmonic and positive in the disk |z| < R, then $|∇u(0)| ≤ 2u(0)/R$?

Any hint is welcome. Thanks in advance.

Answer 1

Let $v(z)=u(Rz)$ so we are in the disc of radius $1$ and $|∇v(0)|=R|∇u(0)|, v(0)=u(0)$, so we need to prove $|∇v(0)| \le 2v(0)$.

Now we can do a proof using Schwarz Lemma and one using Herglotz.

Using Herglotz there is a positive finite measure $\mu(t)$ on the circle st:

$v(re^{i\theta})=\frac{1}{2\pi}\int_0^{2\pi} P(r, \theta-t)d\mu(t)$ where $P(r,\theta)=\frac{1-r^2}{1-2r\cos \theta+r^2}$ is the Poisson kernel which we also can write as $1+2\sum_{n \ge 1}r^n\cos n\theta$

But then for $z \ne 0$, $|∇v(z)|^2=v_r^2+\frac{1}{r^2}v_{\theta}^2$ and since by differentiating the Poisson kernel (using second form is easier to see as we have absolute convergence for $r<1$ so we can differentiate term by term) $\frac{1}{r}v_{\theta} \to 0$ uniformly in $\theta$ as $r \to 0$, while $v_r(re^{i\theta}) \to \frac{1}{2\pi}\int_0^{2\pi}2\cos(\theta-t)d\mu(t)$, and $|\frac{1}{2\pi}\int_0^{2\pi}2\cos(\theta-t)d\mu(t)| \le \frac{1}{2\pi}\int_0^{2\pi}2d\mu(t)=2v(0)$ it follows that for any $\epsilon >0$ there is $\delta$ s.t $|z|<\delta$ implies $|∇v(z)| \le 2v(0)+\epsilon$ and we are done by continuity.

To use Schwarz, wlog we make one more assumption, namely that $v(0)=1$ (by positivity $v(0)=c>0$ and dividing by it doesn't change anything). Then we let $f=v+iw, f(0)=1$ the analytic completion of $v$ and by C-R we have $f'(0)=v_x(0)-iv_y(0)$ so $|f'(0)|=|∇v(0)|$. But now $B=\frac{f-1}{f+1}$ is a Schwarz (in the analytic not distribution sense) function on the unit disc ($B(0)=0, |B|<1$), so $|B'(0)| \le 1$ Since $f=\frac{1+B}{1-B}, f'=\frac{2B'}{(1-B)^2}$, $f'(0)=2B'(0)$ so $|∇v(0)|=|f'(0)|\le 2=2v(0)$ and we are done again!

Answer 2

Yet another proof is via mean value property for harmonic functions. Note that $\nabla u$ is harmonic as well. Therefore, by mean value property \begin{align*} |\nabla u(0)| &= \frac{1}{\pi R^2}\left|\int_{B_R(0)} \nabla u(x)\,dx \right| \tag{1} \\&= \frac{1}{\pi R^2}\left|\int_{\partial B_R(0)} u \cdot \hat{\mathbb{n}}\,d\sigma \right| \tag{2} \\ & \le \frac{1}{\pi R^2}\int_{\partial B_R(0)} u \,d\sigma \tag{3} \\&= \frac{2\pi R}{\pi R^2}u(0) = \frac{2}{R}u(0) \tag{4}\end{align*}

where, in lines $(1)$ and $(4)$ we used mean value property of harmonic functions, in line $(2)$ we used integration by parts and in line $(3)$ triangle inequality with the fact that $u$ is positive in ball.