Maximizing $|f'(0)|$ of an analytic function with $f(1/2)=0$. 
Let f be an analytic function from the unit disk D to the unit disk D. Assume that $f(1/2)=0$, prove $|f'(0)| \leq 25/32$.

I am currently stucked with this problem. I have tried adding some conformal self-map g of the unit disk so that $fg(0)=0$ to apply Schwarz lemma,but this only gives upper bound for $|f'(1/2)|$. Is there any other approach for this problem? 
 A: Where did you find this problem? My solution is a bit tricky...
Write $f(z) = g(z) \cdot \frac{z-\frac12}{\frac12 z - 1}$. Notice $g$ is an analytic function from $D$ (the unit disk) to $D$.
You can easily check that $f^\prime(0) = -\frac34 g(0) + \frac12 g^\prime(0)$. By Cauchy's integral formula, we have
$$ f^\prime(0) = \frac{1}{2 \pi i} \int_{|z|=r} g(z) \cdot \left( -\frac34 \cdot \frac1z + \frac12 \cdot \frac{1}{z^2} \right) \, \rm{d} z =: \frac{1}{2 \pi i} \int_{|z|=r} g(z) \cdot K(z) \, \rm{d} z, $$
for $0 < r < 1$. Taking absolute value, we get
$$ |f^\prime(0)| \le \frac{\| g \|_\infty}{2 \pi} \cdot \int_{|z|=r} |K(z)| \cdot |\rm{d} z|, $$
where $\| g \|_\infty = \sup_{|z|\le 1} |g(z)| \le 1$. If we take the limit $r \to 1$, and perform the integration, we get $|f^\prime(0)| \le 0.836$.
Since $\frac{25}{32} = 0.78125$, we have to improve the method. Notice the we can replace $K(z)$ by any function that has the same principal part (that is, add an analytic function to $K$). The optimal choice turns out to be
$$ \kappa(z) = \frac{(4-3z)^2}{32 z^2} = \frac12 \cdot \frac{1}{z^2} - \frac34 \cdot \frac1z + \frac{9}{32}, $$
but this requires more explanations. Repeating the same argument, we get
$$ |f^\prime(0)| \le \frac{1}{2 \pi} \cdot \int_{|z|=1} |\kappa(z)| \cdot |\rm{d} z| = \frac{1}{2 \pi} \int_{-\pi}^\pi \frac{1}{32}|4-3 e^{it}|^2 \, \rm{d} t = \frac{1}{2 \pi} \int_{-\pi}^\pi \frac{16 - 24 \cos(t) + 9}{32}\, \rm{d} t = \frac{25}{32}.$$
