Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f$ 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

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

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 $m$.

share|improve this answer
add comment

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 $|z|=1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.