$z_{0}$ is a zero of order $m$. Prove that $|z_{0}|^m\geq|f(0)|$ where $f$ is analytic in the unit disc

Question

$f$ is an analytic function in the unit disc, so that $|f(z)|\leq1$.

Let $z_{0}$ be a zero of order $m$. Prove that $|z_{0}|^m\geq|f(0)|$

My approach:

We can write: $$(1) \ \ \ f(z)=(z-z_0)^mg(z)$$

where $g(z_0)\neq0$

Then we define the automorphism on the unit disc: $$\varphi(z)=\frac{z-z_0}{1-\bar{z_0}z}$$

Then we have,

$$f\circ\varphi^{-1}(0)=0$$

And we can apply Schwarz Lemma on $f\circ\varphi^{-1}(z)$:

$$|f\circ\varphi^{-1}(z)|=|(\varphi^{-1}(z)-z_0)^mg(\varphi^{-1}(z))|\leq|z|$$

Then we choose $z=\varphi(0)=-z_0$:

$$|z_0|^m|g(0)|\leq|z_0|$$

and by (1) we only get:

$$|f(0)|=|z_0|^m|g(0)|\leq|z_0|$$

I've noticed that I don't "really" use the fact that $z_0$ is of order $m$.

Any ideas?

Thanks

Answer 1

Your idea is good, but instead of applying directly Schwarz lemma, you would better use (and prove) the following generalization:

If $f\colon D \to D$ is an holomorphic function with zero of order $m\geq 1$ at $z_0=0$, then $|f(z)|\leq |z|^m$ for every $z \in D$.

Hint:

Induction on $m$.

Answer 2

A quicker way to the goal is to write $$f(z) = (\varphi(z))^m g(z) = \left( \frac{z-z_0}{1-\bar z_0 z} \right)^{\!m} g(z)$$ and exploit the maximum modulus principle and the fact that $|\varphi(z)| = 1$ on $|z|=1$.