# Show that there is a constant in an analytic function.

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.

• Do you mean $|f(z)|\le c|z|$? – Henning Makholm Mar 22 at 14:42
• Yes my bad, already corrected it. – Killercamin Mar 22 at 14:42
• 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. – Henning Makholm Mar 22 at 14:43
• Then I would need to change c into a function and then show that the function itself is constant? – Killercamin Mar 22 at 14:46
• @Killercamin appologies for my (stupid) comment below. I misread the question totally. – 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 $$|z| \le r$$.