# an anlaytic function defined in a neighborhood of the unit disk

I have the following question: Let $f$ be analytic in a neighborhood of the unit disk such that $|f(z)|<1$ for $|z|=1$. Show that $f$ has exactly one fixed point $w$ in the unit disk and also show that $|f'(w)|<1$.

I tried to define a new function to apply Schwarz lemma and maximum modulus principle but couldn't get the statement. Any help would be great.

## 2 Answers

Let $D$ be the open unit ball in $\mathbb C$.

First of all, the maximum modulus principle tells you that: $$\forall z\in \overline D, \quad |f(z)|<1 \quad (\star).$$ Thus, $f$ induices a continuous map from the closed unit ball $\overline D$ into it self and due to the Brouwer fixed point theorem, it must have at least one fixed point in $\overline D$.

However, considering $(\star)$, any fixed point has to belong in $D$.

Let $\omega\in D$ be a fixed point of $f$ and $\varphi:D\rightarrow D$ the holomorphic map given by $$\varphi(z)=\dfrac{z-\omega}{1-\overline\omega z}.$$ Actually, $\varphi$ in an automorphism of the unit ball : $\varphi$ is holomorphic, bijective and so is its inverse. Furthermore, one has :$$\varphi(\omega)=0.$$ So the holomorphic map $g=\varphi\circ f\circ \varphi^{-1}$ is well defined on $D$ and satisfies :

1. $g(0)=0$,
2. $|g(z)|<1$,
3. $g'(0)=f'(\omega)$,
4. $g$ and $f$ have the same fixed point on $D$.

Now, applying the Schwarz lemma to $g$ we get that $$\forall z\in D, \quad |g(z)|\leq |z| \quad \text{ and } \quad |g'(0)|\leq 1$$ and moreover if $|g'(0)|=1$ then $g$ is of the form $z\mapsto az$ with $|a|=1$.

But we can't have both the assumption $(\star)$ and $g(z)=az$ on $D$ with $|a|=1$. Indeed, if $g(z)=az$ then $z\mapsto \varphi^{-1}\circ g\circ \varphi(z)$ extends to a holomorphic function on a neighborhood of $\overline D$ that coincides with $f$ on $\overline D$. But you can easily check that $\varphi^{-1}$, $g$ and $\varphi$ send the unit circle to the unit circle so $f$ has to do the same: this is in contradiction to $(\star)$.

So $$\forall z\in D, \quad |g(z)|\leq |z| \quad \text{ and }\quad |f'(\omega)|=|g'(0)|<1.$$

Finally, since $$|[g(z)-z]+z|=|g(z)|\leq |-z|$$ the Rouché theorem applied to $z\mapsto -z$ and $z\mapsto g(z)-z$ on $D$ says that $g$ (and thus $f$) has exactly one fixed point on $D$.

Edit: An other proof of the unicity of the fixed point is given in the comment.

• Second part is identical with the one I wrote. Yesterday. – Gary. Aug 8 '15 at 2:32
• Yes, but in order to apply Rouché's theorem, I need the schwarz lemma to get $|f(z)-z+z|\leq |z|$ and it wasn't clear to me how you processed in your solution. Actually, one could also prove that $g$ has only one fixed point with the Schwarz lemma again: if $f$ has another fixed point than $\omega$ then there exists $z\in D$ such that $|g(z)|=|z|$ and so $g$ has to be of the form $z\mapsto az$ with $|a|=1$. This is a contradiction with the hypothesis $(\star)$. – Bebop Aug 8 '15 at 13:15

For the first part, I think this should work: Using Rouche's theorem with $h(z)=f(z)-z$ and $g(z)=-z$ on $|z|<1$ :

$|h(z)-g(z)|=|f(z)-z-(-z) |=|f(z)|<|f(z)-z|+|z|$ Since $|f(z)|<1$ by assumption and $|z|=1$ on $\partial D$, so that $|f(z)-z|$ and $|-z|$ , both have the same number of zeros in $|z|<1$.

Let me try the second part.

EDIT: We can use Schwarz -Pick lemma http://mathworld.wolfram.com/Schwarz-PickLemma.html to show $|f'(z)| \leq 1$: We are given that $|f((z)| \leq 1$ we can just take $f(w)=w$ , then Schwarz-Pick tells us that $$f'(z) \leq \frac {1-w^2}{1-w^2}=1$$.

So we can conclude $|f'(z)| \leq 1$ . Let me see if we can sharpen it to $|f'(z)|<1$.

• Thanks for help. The rest is as follows: Let $F(z)=\frac{f(z)}{z-w}$ if $z \neq w$ and $f'(w)$ if $z=w$. Then we can apply maximum modulus to get $f'(w)<1$. – delueze Aug 6 '15 at 0:58
• @delueze: Thanks for the follow up. Wouldn't that be $\frac {f(z)-w}{z-w}$ ? – Gary. Aug 6 '15 at 1:29
• Yes, thanks for correction. But now I think my solution is not working, I will try again. – delueze Aug 6 '15 at 1:36
• @delueze: I think Schwrz-Pick lemma may answer the second question. – Gary. Aug 6 '15 at 12:37
• @Bebop: Why we couldnt have both the condition 2. and g(z)=az – delueze Aug 8 '15 at 0:28