Let $f : \Bbb{D} \to \Bbb{D}$ be a holomorphic function. If $f(0)=f(1/2) = 0$, then prove that $|f'(0)| \le 1/2$
Here is the previous problem, which I was easily able to solve using the Schwarz lemma and Riemann's theorem on removable singularities:
Let $f : \Bbb{D} \to \Bbb{D}$ be holomorphic such that $f(0) = 0$. Prove that $g(z) = \frac{f(z)}{z}$ extends to a holomorphic function on $\Bbb{D}$ satisfying $g : \Bbb{D} \to \Bbb{D}$.
I was told to try to use the problem I already solved to solve the problem at hand.However, I don't see how to use this to solve the above problem. Note, by continuity, $$g(0) = \lim_{z \to 0} f(z) = \lim_{z \to 0} \frac{f(z)-f(0)}{z-0} = f'(0),$$ so if we could show $|g(0)| \le 1/2$, we would be done. But I don't see how to argue this.
I also tried using Mobius transforms of $\Bbb{D}$ of the form $z \mapsto e^{i \theta} \frac{z-z_0}{1-\overline{z_0}z}$, where $\theta \in \Bbb{R}$ and $|z_0| < 1$, and then try applying the Schwarz lemma somehow, but I didn't have any luck.
I could use some help.
So $g : \mathbb{D} \to \mathbb{D}$ satisfies $g(1/2)=0$. Then using your Möbius transformation you can define $h : \mathbb{D} \to \mathbb{D}$ by $h(z) = g((2z+1)/(2+z))$ ($\theta=0$ and $z_0=-1/2$) so that $h(0)=g(1/2)=0$. Note that there is some $w_0$ such that $h(w_0)=g(0)=f'(0)$. So you can get what you want from the Schwarz lemma by solving $w_0$.