# Prove $|f(z)|\leq M|z|$ inside unit disk if $f(0)=0$ and $|f(z)|\leq M$

The problem goes as follows:

Assume $f(z)$ is analytic inside the unit disk $D=\{z:|z|<1\}$. Also, $f(0)=0$ and $|f(z)|\leq M$ in $D$.

Prove that $|f(z)|\leq M|z|$ in $D$. When does equality occur?

This seems like an easy problem, but I can't seem to come up with a solution.

I have tried using Cauchy's integral formula and the ML-inequality, but I am stuck here:

$|f(z)|=|\frac{1}{2\pi i}\oint_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta|\leq\frac{1}{2\pi}\frac{M}{r-|z|}2\pi r=\frac{Mr}{r-|z|}$

where $C_r$ is a circle of radius $r$ inside $D$.

Any help would be appreciated.

• Try to factor $f$ as $f(z)=zg(z)$. – Yai0Phah Oct 22 '15 at 18:55
• I've considered that, but don't know how to continue from there.. – Anton Lundberg Oct 22 '15 at 19:16
• You can determine that the maximum of the function g is M. Then you are done. You could use maximum modulus to get the bound. – random123 Oct 22 '15 at 20:39

As suggested, write $f(z) = zg(z)$ (where $g$ is also holomorphic on $|z|<1$). Then for $|z|\le r$ and $r < 1$, $$|g(z)| = \frac{|f(z)|}{|z]} \le \frac{M}{r}.$$ Letting $r \to 1^-$, it follows that $$|g(z)| \le M$$ for all $z \in D$. Hence $|f(z)| \le M |z|$ for $z \in D$. (This is a variant of Schwarz' lemma.)
Equality can only happen (at $z=0$ or) if $|g(z)| = M$ for some $z \in D$. But then, the maxmimum modulus principle forces $g(z)$ to be constant on $D$, $g(z) = Me^{i\theta}$ for some real $\theta$, and consequently $f(z) = Mze^{i\theta}$.
HINT: define $$g(z):=\frac{f(z)}M$$ and apply Schwarz Lemma.