Let $f(x,y)$ be a harmonic function on the unit disk, such that $|f| \leqslant M$. Prove that $\lvert\nabla f(0)\rvert \leq 2M$ using the formula $$f(z)=\frac 1 {2\pi}\int _0^{2 \pi} f(e^{i \phi})P_r(\theta-\phi)d\phi$$
Here $P_r(\theta) =\frac{1-|z|^2}{|1-z|^2} = \frac{1-r^2}{1+r^2-2r \cos \theta}$.
Differentiating the formula yields $$\frac{\partial f(z)}{\partial x}=\frac 1 {2\pi}\int _0^{2 \pi} f(e^{i \phi}) \frac{\partial P_r(\theta-\phi)}{dx}d\phi$$
$$|\frac{\partial f(z)}{\partial x}| \leqslant \frac M {2\pi}\int _0^{2 \pi} | \frac{\partial P_r(\theta-\phi)}{dx}|d\phi$$
and similarly for $y$, so
$$|\nabla f(z)|=|\frac 1 {2\pi}\int _0^{2 \pi} f(e^{i \phi}) \nabla P_r(\theta-\phi)d\phi| \leqslant \frac M {2\pi} |\int _0^{2 \pi} \nabla P_r(\theta-\phi)d\phi|$$$$ \leqslant\frac M {2\pi} \int _0^{2 \pi} |\nabla P_r(\theta-\phi)|d\phi$$ but the two explicit derivatives inside the gradient in the integral are complicated and do not seem to be bounded by $\frac 12$, which would have allowed the estimation $|\nabla f(0)| \leq |f_x(0)|+|f_y(0)| \leq 2M$.
To compute the gradient I think we may differentiate as a complex function and then substitute $ze^{-i \phi}$: $$\nabla P = \nabla \text{Re} (\frac{1+z}{1-z})=\text{Re}(\frac d {dz}\frac{1+z}{1-z})=\text{Re}(\frac {2}{(1-z)^2})$$
So for $\phi=0, z=0$ we are not bounded by 1, and so pointwise estimates are not enough. How can I proceed, then?
If we use normalized arc length measure on the circle, we can write the Poisson kernel as
$$P(z,\zeta) = \frac{1-|z|^2}{|z-\zeta|^2} = (1-|z|^2)(|z-\zeta|^2)^{-1}$$
for $|z|<1, |\zeta|=1.$ Then, with $\zeta = (\zeta_1, \zeta_2),$ we have
$$ \frac{\partial P(z,\zeta)}{\partial x} = -2x(|z-\zeta| ^2)^{-1}+ (1-|z|^2)(-1)(|z-\zeta|^2)^{-2}\cdot 2(x- \zeta_1).$$
At $z=0$ this equals $2\zeta_1.$ Thus
$$\frac{\partial u}{\partial x}(0) = \frac{1}{2\pi}\int_0^{2\pi} 2\cos t \cdot u(t)\, dt.$$
In absolute value this is no more than $\dfrac{4M}{\pi}.$ The estimate for $|\partial u/\partial y(0)|$ is the same. Therefore $|\nabla u (0)| \le \sqrt 2\dfrac{4M}{\pi}.$ Since $\sqrt 2\dfrac{4}{\pi} <2$ by a bit, we're done.