Let $\varepsilon >0$ and let $f:B(0,1+\varepsilon )\rightarrow B(0,1)$ be a holomorphic function. Show that there exists a unique $z_0\in B(0,1)$ s.t $f(z_0)=z_0$.
For this problem, I don't want to use the Earle-Hamilton fixed-point Theorem because my complex-analysis textbook doesn't have it, which implies the uniqueness of $z_0$ straightforwardly.
However, what I'm considering instead is the maximum modulus principle (or Schwarz Lemma or Rouche's Theorem).
Here's my idea.
Since $B(0,1)$ is bounded, $\exists \alpha \in B(0,1+\varepsilon )$ s.t $|f(z)| \leq |f(\alpha)| <1$ for any $z \in B(0,1+\varepsilon )$.
Then, by the Maximum modulus principle, $f$ is constant.
So, $f(z)=z_0$ for any $z\in B(0,1+\varepsilon )$ and for some $z_0 \in B(0,1)$.
Since, $z_0 \in B(0,1) \subset B(0,1\varepsilon )$, $f(z_0)=z_0$.
Lastly, the uniqueness of $z_0$ is obvious because for any $z \in B(0,1) \setminus \{z_0\}$, $z \neq z_0$. $\blacksquare $
How does this sound like?
I have a little doubt due to the fact that $f$ is constant.
Another question is this. If we want to use the Schwarz Lemma for this problem, then we would need the condition $f(0)=0$. But, since $f$ might not satisfy this, I wonder how we should modify this Lemma? I think we can let $g(z)=f(z)-f(0)$ so that the Lemma can be used. Would this method work?
Update:
I'm also trying to use Rouche's Theorem, comparing $f(z)$ and $g(z)=z$ on $\partial B(0,1)$.
But, I am stuck at showing $|f(z_0)-z_0| \neq |f(z_0)|+1$ for any $z_0 \in \partial B(0,1)$. How this could be shown?
Update:
I finally could get a simple solution to this problem using Rouche's Theorem.
Note that $\overline{B(0,1)}\subseteq B(0,1+\varepsilon )$.
Now, we are going to compare $|f(z)|$ and $|-z|$ on $\partial B(0,1)$.
As $f(z) \in B(0,1)$, $|f(z)|<1$ for any $z \in \partial B(0,1)$. Also, $|-z|=1$ for any $z \in \partial B(0,1)$.
So, $|f(z)|<|-z|$ on $\partial B(0,1)$.
By a Corollary derived from Rouche's Theorem, we can say that $-z$ and $f(z)-z$ have the same number of zeros in $B(0,1)$.
Then, this implies that $f(z)-z = 0$ at a unique $z_0 \in B(0,1)$.
Hence, we conclude that there exists unique $z_0 \in B(0,1)$ such that $f(z_0)=z_0$.
However, I still want to know if it is possible to use Schwarz Lemma for this problem.
Any help would be really appreciated.