Given bounds on analytic function, finding bounds on its derivative

Question

Let $f$ be analytic on $\Delta$ (unit disc) and satisfy $|f| < 1$. Prove that if $f( \frac{1}{2} ) = f(-\frac{1}{2} ) = 0$, then $|f^{\prime}(0)| \leq \frac{1}{4}$.

Answer 1

So, I think I have a solution: We can factor $f(z)=\frac{z-1/2}{1-1/2 z} \cdot \frac{z+1/2}{1+1/2 z} \cdot f_1(z)$ where $f_1(z)$ is holomorphic; (like Blaschke factorization except that $f_1(z)$ may have other zeroes). Using a limiting argument, we can conclude that $|f_1(z)|\leq 1$ on $|z|=1$. Using Cauchy integral estimates we have $|f_1^{\prime}(0)|\leq 1$. Differentiating $f(z)=\frac{z-1/2}{1-1/2 z} \cdot \frac{z+1/2}{1+1/2 z} \cdot f_1(z)$ and plugging our estimates for $z=0$, gives us the desired result.