Suppose $f$ is analytic and $f(a) = f(b) = 0$. Show that $|f(z)| ≤ |{z − a \over 1 − z\bar{a}}| · |{z − b \over 1 − z\bar{b}}|$.

Question

Suppose $f$ is analytic from $D(0, 1)$ to $D(0, 1)$ and $f(a) = f(b) = 0$ for two different numbers $a, b$ in $D(0, 1)$. Show that $\left\vert f(z) \right\vert ≤ \left\vert{z − a \over 1 − z\bar{a} }\right\vert \cdot \left\vert {z − b \over1 − z\bar{b}}\right\vert$.

I'm trying to understand how to show this $\left\vert f(z)\right\vert$ inequality. I'm not sure how to do it. Can I assume that $\left\vert f(z)\right\vert > \left\vert{z − a \over 1 − z\bar{a}}\right\vert \cdot \left\vert{z − b \over 1 − z\bar{b}}\right\vert$ and search for a contradiction? Or is there a better way to show this? Any solutions or help is greatly appreciated.

Answer 1

Hint: Let $g_w(z) = (w-z)/(1-\overline w z).$ Let $m,n$ be the orders of the zeros of $f$ at $a,b.$ Consider $$\frac{f(z)}{g_a(z)^m g_b(z)^n}.$$