For $f = u +iv$ analytic in $\mathbb{D}$, with $|u| <1$, prove that $|f'(0)| \le 2$. Let $f = u +iv$ be analytic in the unit disc $\mathbb{D}$, with $|u| <1$ on $\mathbb{D}$. Prove that $|f'(0)| \le 2$.
This problem appears on an old complex analysis qualifying exam that I am working on. I've tried several things so far but I have not proved it just yet. 
First, I tried applying the Cauchy estimates to $g(z) = e^{f(z)},$ getting 
$$|f'(0)||e^{f(0)}| =|f'(0)|e^{u(0)}  \le \frac{\operatorname{max}_{z \in \mathbb{D}}{|e^{f(z)}|}}{r} = \frac{\operatorname{max}_{z \in \mathbb{D}}{e^{u(z)}}}{r} \le \frac{e}{r}$$
for any $0 < r < 1$. Letting $r \to 1$ and dividing by $e^{u(0)}$ ($\ge \frac{1}{e}$) yields $|f'(0)| \le e^2$. Darn! Still a ways off from the bound I want.
Another fact I know is that $|f'(0)| = |u_x(0) - iu_y(0)| \le |u_x(0)| + |u_y(0)|$. Is there some tractable way to get an estimate on $|u_x(0)|$ and $|u_y(0)|$ just knowing that $|u|< 1$ on $\mathbb{D}$? 
Any hints or solutions are greatly appreciated.
 A: Let's state something more general and broadly useful.
Principle of subordination. Let $\phi:\mathbb D\to\Omega$ be a bijective holomorphic map. Suppose that $f$ is holomorphic in $\mathbb D$,   $f(\mathbb D)\subseteq \phi(\mathbb D)$ and $f(0)=\phi(0)$. Then $|f'(0)|\le |\phi'(0)|$.
Proof: Apply Schwarz's lemma to $\phi^{-1}\circ f$. $\quad \Box$
To finish off the problem, you could play with maps onto the vertical strip $|\operatorname{Re}w|<1$. This is a bit tiresome, but you'll get a better bound than $2$, namely $4/\pi$. (Cf. Maximize absolute value of complex logarithm where the special case $f(0)=0$ is solved).  
But to get $|f'(0)|\le 2$, it suffices to use maps onto half-planes $ \operatorname{Re}w <1$ and $ \operatorname{Re}w >-1$, which are simpler to write. For definiteness,   assume that $ \operatorname{Re}f(0) \ge 0$ (otherwise consider $-f$). Then $\phi$ should be a Möbius map from $\mathbb D$ onto $ \operatorname{Re}w <1$ such that $\phi(0)=f(0)$. Write down this map and check that $|\phi'(0)|\le 2$.
A: Let $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ be the Taylor expansion of $f$ at $0$, whose radius of convergence is at least $1$. Then 
$$f'(0)=a_1\tag{1}$$ and 
$$u(z)=\frac{1}{2}\sum_{n=0}^\infty(a_nz^ n+\bar{a}_n\bar{z}^n)\tag{2}.$$
From $(1)$ and $(2)$ we know that for every $0<r<1$,
$$f'(0)=\frac{1}{\pi i}\int_{|z|=r}\frac{u(z)}{z^2}dz\tag{3}.$$
From $|u|<1$ it follows that
$$|f'(0)|\le \frac{1}{\pi}\int_{|z|=r}\frac{|u(z)|}{r^2}|dz|\le\frac{2}{r}\tag{4}.$$
Letting $r\to 1$ in $(4)$, we obtain $|f'(0)|\le 2$.

Edit: Starting from $(3)$, if we replace the rough estimate in $(4)$ with the idea in Sam's answer, we may obtain the optimal upper bound $\frac{4}{\pi}$ as follows.
Let $\theta\in\Bbb R$ be such that $f'(0)=|f'(0)|e^{i\theta}$, then by $(3)$,
$$|f'(0)|=\frac{e^{-i\theta}}{\pi i}\int_{|z|=r}\frac{u(z)}{z^2}dz=\frac{1}{\pi r}\int_0^{2\pi}u(re^{it})e^{-i(t+\theta)}dt\tag{5}.$$
The left hand side of $(5)$ is real, so the imaginary part of the right hand side of $(5)$ vanishes, i.e. 
 $$|f'(0)|=\frac{1}{\pi r}\int_0^{2\pi}u(re^{it})\cos(t+\theta)dt\le\frac{1}{\pi r}\int_0^{2\pi}|\cos(t+\theta)|dt=\frac{4}{\pi r}\tag{6}.$$
where the intermediate inequality is due to $|u|<1$. Letting $r\to 1$ in $(6)$, we obtain 
$$|f'(0)|\le \frac{4}{\pi}.$$
A: Note that $f' = u_x - iu_y$, so that $|f'|^2 = (u_x)^2 + (u_y)^2 = |\nabla u|^2$ in terms of the gradient of $u$. We have $\Delta u = u_{xx} + u_{yy} = 0$ by the Cauchy Riemann equations. In particular, this implies that for any fixed unit length vector $\nu\in \mathbb R^2$, the function $\nu \cdot \nabla u$ is again harmonic. 
By the mean value principle this implies that $\nu\cdot \nabla u(0) = \frac{1}{\pi r^2} \int_{\mathbb D_r} \nu\cdot \nabla u$ for any $r>0$. An application of the divergence theorem shows that 
$$\nu\cdot \nabla u(0) = \frac{1}{\pi r^2} \int_{S_r} u\,  \nu \cdot d\vec\sigma = \frac{1}{\pi r^2} \int_0^{2\pi} u(r\cos\theta, r\sin\theta)  \; \nu \cdot (r\cos\theta, r\sin\theta)^t\, d\theta.$$
Choosing $\nu$ such that $|\nabla u(0)| = \nu\cdot \nabla u(0)$, we therefore obtain the estimate (using $|u|< 1$)
$$|f'(0)| = |\nabla u(0)| \le \frac{1}{\pi r}\int_0^{2\pi} \left|\nu \cdot (\cos(\theta), \sin(\theta))^t\right| \, d\theta  = \frac{4}{\pi r}.$$
Now let $r\to 1$.
