Application of Schwarz's Lemma to $f(a)=f(-a)=0$

Question

Suppose $f : D\rightarrow D$ is analytic function where $D$ is open unit disk. Suppose there exists $a\in D\backslash\{0\}$ such that $f(a)=f(-a)=0$. I want to show that $|f(0)|\leq |a|^2$. So I consider $g(z)=\begin{cases} f(z)/\left(\dfrac{z-a}{1-\overline{a}z}\right), \text{if }z\neq a\\ (1-|a|^2)f'(a),\text{if }z=a \end{cases}$. Similarly for $g$ but I change $a$ to $-a$. Then $g$ and $h$ are analytic on unit disk and bounded by $1$, so we can use Schwarz's lemma. By multiplying the functions, I got $|f(0)|\leq |a|$. How can I improve this so $|f(0)|\leq |a|^2$?

Answer 1

Let $B(z)=\frac{z-a}{1-\bar a z}\frac{z+a}{1+\bar a z}$ the Blaschke product that satisfies $B(\pm a)=0$

Let $h(z)=\frac{f(z)}{B(z)}$ which is analytic on the unit disc by hypothesis (extended by continuity at $\pm a$) and since $|B(z)|=1, |z|=1$ it follows that $|h(z)| \le 1$ so $|h(0)|=|\frac{f(0)}{a^2}| \le 1$ and we are done!