I am new to this but I kept on looking answer about this problem
Suppose $\displaystyle{f: \overline{B_1(0)} \rightarrow \overline{B_1(0)}}$ is continuous and $f$ is analytic in $\displaystyle{B_1(0)}$. Prove that $f(z)$ has a fixed point in $\displaystyle{\overline{B_1(0)}}$.
I don't even know how to start with this to be honest. Can you help me guide through it? I would appreciate it. Thanks
On
If we consider the Poincare distance on the unit disk $$d(z_1, z_2) = 2\sinh^{-1} \frac{|z_1-z_2|}{\sqrt{(1-|z_1|^2)(1-|z_2|^2)}}$$ then any holomorphic map from $D(0,1)$ to itself does not increase the distance between points. This is called the Schwarz-Pick Theorem
Now, if $0<\rho < 1$, the one checks right away that $$d(\rho w_1, \rho w_2) < d(w_1, w_2)$$ for every $z_1\ne z_2$ in $D(0,1)$.
We conclude that $g =\rho f\colon D[0,\rho]\to D[0,\rho]$, continuous, and decreasing the distance between points. It follows that $g$ has a fixed point. Indeed, otherwise consider a point $z$ such that $d(z, g(z))$ is smallest and $>0$. But then $d(g(z), g^2(z)) < d(z, g(z))$, contradiction.
Now take a sequence $\rho_n \to 1$. There exists $z_n \in D[0, \rho_n]$ such that $\rho_n f(z_n) = z_n$. The sequence $z_n$ has a subsequence convergent to a point $z\in B[0,1]$. We have $f(z) = z$.
Note that it is possible that $f$ has a fixed point only on $S^1$. For example, consider $$f(z) = \frac{3+4 i}{5}\cdot \frac{z -\frac{1+ 2 i}{5}}{ 1- \frac{1-2i}{5} z}$$
with unique fixed point $z=1$.
Edit: When $f$ is a self-mapping of the boundary circle, then you may also want to know that $f$ is a finite Blaschke product (see https://en.m.wikipedia.org/wiki/Blaschke_product); that is, $f$ can be written as $$f(z)=e^{i\theta}\prod_{k=1}^n\frac{z-a_k}{1-\overline{a_k}z}\,$$ where $\{a_k\}_k$ is a sequence of complex numbers (with possible repetitions) lying entirely in $B_1(0,1)$, and $\theta$ is a real number. Note that $f$ is defined for all $z\in\mathbb{C}\setminus\{1/\overline{a_k}\}_k$, and that we have the identity $$\overline{f(z)}f(\frac{1}{\overline{z}})=1\,,$$ for all $z\in\mathbb{C}\setminus\{1/\overline{a_k}\}_k$ . It follows that if $z\ne 0$ and $f(z)=z$, then we also have that $f(\frac{1}{\overline{z}})= \frac{1}{\overline{z}}$. Thus it suffices to show that $f$ has at least one fixed point $z$ in $\mathbb{C}\setminus\{1/\overline{a_k}\}_k$ then it would follow that either $z$ or $\frac{1}{\overline{z}}$ lies in $\overline{B_1(0,1)}$. But observe that the fixed point equation $f(z)=z$ in $\mathbb{C}\setminus\{1/\overline{a_k}\}_k$ is equivalent to solving for the zeroes of the polynomial $$g(z):=e^{i\theta}\prod_{k=1}^n(z-a_k)-z\prod_{k=1}^n(1-\overline{a_k}z)\,,$$ which exist thanks to the Fundamental Theorem of Algebra.