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.

  • 1
    $\begingroup$ Try to factor $f$ as $f(z)=zg(z)$. $\endgroup$ – Yai0Phah Oct 22 '15 at 18:55
  • $\begingroup$ I've considered that, but don't know how to continue from there.. $\endgroup$ – Anton Lundberg Oct 22 '15 at 19:16
  • $\begingroup$ 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. $\endgroup$ – 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 $f(z) = Mze^{i\theta}$.

  • $\begingroup$ Really, the question to have asked the OP is: Do you know the Schwarz Lemma? If not, are they asking the OP to prove the SL? That would be a little weird. But if so, why is eveyone reproving the SL? $\endgroup$ – zhw. Oct 22 '15 at 23:48

HINT: define $$ g(z):=\frac{f(z)}M $$ and apply Schwarz Lemma.


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