Suppose that $f(z)$ is analytic for $|z<1|$ and satisfies $|f(z)|<1, f(0)=0, $and $|f'(0)<1|.$ Let $r<1$. Show that there is a constant $c<1$ sucht that $|f(z)| \leq c|z|$ for $|z|\leq r$.

I'm struggling with this, but this is what I have right now:

Using the Schwarz Lemma, then we can assume $z$ is analytic thus we can factor $f(z)=z \cdot h(z)$, where c is analytic. Given $r<1$. If $|z|=r$, then $|h(z)|=\frac{|f(z)|}{r} \leq \frac{1}{r}$. Using the maximum principle then $|h(z)| \leq\frac{1}{r} \forall z$ satisfying $|z| \leq r$.

By letting $r \rightarrow 1$, then $|h(z)|<1 \forall |z|<1$. This implies that $|f(z)|=|z||h(z)| \leq |z|$. Since $|f(z_o)|=|z_o|$ for some $z_o \neq 0$ then $|h(z_o)|=1$ for some $z_o \neq 0$ and using the strict maximum principle h(z) is constant. We can say that $h(z)= \lambda$. Then $f(z)= \lambda z$.

What am I missing or did I miss the picture completely? Any hints would be appreciated. I tried using the Schwarz lemma since I feel it hits most of the points.

  • $\begingroup$ Do you mean $|f(z)|\le c|z|$? $\endgroup$ – Henning Makholm Mar 22 at 14:42
  • $\begingroup$ Yes my bad, already corrected it. $\endgroup$ – Killercamin Mar 22 at 14:42
  • $\begingroup$ It also seems like you're using $c$ in your attempt to mean a function whereas in the problem it is a constant. That's rather confusing at best. $\endgroup$ – Henning Makholm Mar 22 at 14:43
  • $\begingroup$ Then I would need to change c into a function and then show that the function itself is constant? $\endgroup$ – Killercamin Mar 22 at 14:46
  • 1
    $\begingroup$ @Killercamin appologies for my (stupid) comment below. I misread the question totally. $\endgroup$ – Matematleta Mar 22 at 20:50

The function $h: \Bbb D \to \Bbb C$ defined by $$ h(z) = \begin{cases} \dfrac{f(z)}{z} & \text{ for } z \ne 0 \\ f'(0) & \text{ for } z = 0 \end{cases} $$ is holomorphic in the unit disk $\Bbb D$ with $|h(z)| \le 1$ according to the Schwarz Lemma.

If $h$ is constant then we are done: $$ |f(z)| = c |z| $$ for all $z\in \Bbb D$, with $c = |f'(0)| < 1$.

Now assume that $h$ is not constant, and fix $0 < r < 1$. Then $$ c = \max \{ |h(z)| : |z| \le r \} $$ must satisfy $c < 1$, because $h$ can not have a maximum in the interior of the unit disk (the maximum modulus principle). If follows that $$ |f(z)| \le c |z| $$ for $|z| \le r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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