Using complex analysis , how to prove that any holomorphic function $f:\overline{D(0;1)} \to \overline{D(0;1)}$ has a fixed point? Using complex analysis , how to prove that any holomorphic function $f:\overline{D(0;1)} \to \overline{D(0;1)}$ has a fixed point ? From this answer Suppose $f(z)$ is analytic in the closed unit disc... I can see that if $|f(z)|<1$ on $|z|=1$ then I can prove it easily ; but what if that condition doesn't hold ? I am stuck . Please help . Thanks in advance 

For every $n \in \mathbb N$ , define $g_n : \overline{D(0;1)} \to \mathbb C $ as $g_n(x)=(1-\dfrac 1n)f(x)$ , then $|g_n(x)|<|f(x)|\le1$ ,  so $g_n: \overline{D(0;1)} \to D(0;1)$ ; then by the question you link , which you say you can easily show , we have that each $g_n$ has a fixed point , hence for every $n$ , $\exists z_n \in \overline {D(0;1)} $ such that $g_n(z_n)=z_n$ i.e. $(1-\dfrac 1n)f(z_n)=z_n , \forall n \in \mathbb N$ ; now as $\overline{D(0;1)}$ is compact so $\{z_n\}$ has a convergent subsequence say $\{z_{r_n}\}$ converging to some $z \in \overline{D(0;1)}$ , then since $f$ is continuous and $\dfrac 1{r_n} \to 0$ , as $n \to \infty$ , so we get $f(z)=z$
