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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.

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  • $\begingroup$ 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
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    $\begingroup$ @Killercamin appologies for my (stupid) comment below. I misread the question totally. $\endgroup$ – Matematleta Mar 22 at 20:50
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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 $|z| \le r$.

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