I'm struggling with the following question:
Maximize the value of $\frac{\partial}{\partial x}u(0, 0)$ among all harmonic functions $u$ from the complex unit disk $\{z: |z| < 1\}$ to the interval $[-1, 1]$. Prove that your answer is correct.
I suspect that the harmonic function which does the maximizing is just $z \mapsto f(z) = \Re (z)$. Whatever the maximizing function is, we may assume it is odd by replacing it with $\frac{1}{2}(u(z) - u(-z))$. We may also assume that $u_y(0) = 0$, lest we be able to rotate the function to increase the $x$ derivative at 0. But now I'm stuck. Schwarz's lemma I'm sure could be useful.
A bounded harmonic function $u$ in the disc has the Poisson integral representation
$$u(re^{i\theta}) = \frac{1}{2\pi}\int_{-\pi}^\pi \frac{1-r^2}{ 1-2r\cos (\theta -t)+r^2}u(e^{it})\,dt,$$
where $u(e^{it})$ is the boundary function of $u.$ Thus
$$u(x,0) = \frac{1}{2\pi}\int_{-\pi}^\pi (1-x^2)(1-2x\cos t+x^2)^{-1}u(e^{it})\,dt.$$
Now let's differentiate through the integral sign with respect to $x,$ and then set $x=0.$. This looks like it might get complicated, but stay patient and then you will be rewarded when you set $x=0.$ We arrive at
$$u_x(0,0)=\frac{1}{2\pi}\int_{-\pi}^\pi (2\cos t) u(e^{it})\,dt.$$
Therefore, since $|u|\le 1,$ we get
$$u_x(0,0)\le \frac{1}{2\pi}\int_{-\pi}^{\pi }|2\cos t ||u(e^{it})|\,dt \le \frac{1}{2\pi}\int_{-\pi}^\pi 2|\cos t |\,dt =\frac{4}{\pi}.$$
But note this upper bound is acheived if $u=1, t\in (-\pi/2,\pi/2),$ $u=-1, \pi/2 < |t|< \pi.$ Thus the maximum value of $u_x(0,0)$ over all such $u$ is $4/\pi.$