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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.
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 :
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.
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 $|f'(z)|<1$.
