Using complex analysis , how to prove that any holomorphic function $f:\overline{D(0;1)} \to \overline{D(0;1)}$ has a fixed point?

Question

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

Answer 1

Only "using comple analysis".

Take a sequence $(p_n)$ where $0<p_n<p_{n+1}<1$ and $p_n\to 1$ as $n\to \infty$. Let $f_n(z)=p_nf(z)$. Then $f_n(z)$ satisfies $|f_n(z)|<1$, so $f_n(z)$ has a fixed point $z_n$ $(|z_n|<1)$. Since $\overline{D(0;1)}$ is compact, there is a subsequence $(z_{n_k})$ which converges to some point $ z_0\in \overline{D(0;1)}$. Since $(f_n(z))$ converges uniformly to $f(z)$ on $\overline{D(0;1)}$, $$f(z_0)-z_0=\lim_{k\to \infty} \left(f_{n_k}(z_{n_k})-z_{n_k}\right)=0.$$ Note: $z_0$ may lie on $|z|=1.$
Example: $f(z)=\frac{2z-1}{2-z}.$ This function satisfies $|f(z)|\le 1$ on $|z|\le 1$ and has two fixed points $z=1, -1$, but $f$ has no fixed point in $|z|<1$.

Answer 2

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 $f(z)=z$

Answer 3

This is loosely modeled on a portion of chp 11, section 3 of Beardon's Complex Analysis and only requires standard complex analysis knowledge of winding numbers, e.g. written as an integral. With a slightly better knowledge of winding numbers we can see this proof works identically for merely continuous $f$.

Let analytic $f:\overline D(0,1)\longrightarrow \mathbb C$ with $f\big(\overline D(0,1)\big)\subseteq \overline D(0,1)$ and define
$g(z):= f(z)-z$
$\gamma(t):=\exp\big(2\pi i \cdot t\big)$ for $t\in [0,1]$

Suppose $f$ does not have a fixed point. Then we may calculate $n\big(g\circ \gamma,0\big)$ two different ways:
(i.) $n\big(g\circ \gamma,0\big)=0$
since $0\not \in g\big(\overline D(0,1)\big)\implies g\Big(p\cdot 0 + (1-p)\cdot z\Big)$ for $p\in [0,1]$ gives homotopy to a point

$h(z):=\frac{ f(z)}{z}-1$ so $g(z)=z\cdot h(z)$
(ii.) $n\big(g\circ \gamma,0\big)=n\big((\gamma)\cdot (h\circ \gamma),0\big)=n\big(\gamma,0\big) + n\big(h\circ \gamma,0\big)=1+n\big(h\circ \gamma,0\big)=1$
$\implies 0=1$.

justification of $n\big(h\circ \gamma,0\big)=0$
$h$ maps the image of $\gamma$ into $\overline D(-1,1)$ i.e. $h(z)=\frac{ f(z)}{z}-1$ and for $\vert z \vert =1$ we know $\big \vert \frac{ f(z)}{z} \big\vert \leq 1$. By assumption of no fixed point we know for $\vert z\vert =1$ that $h(z)\neq 0$ thus $n\big(h\circ \gamma,0\big)$ is well defined and the image of $h\circ\gamma$ does not meet the ray on the real half line, i.e. $[0,+\infty)$ so the winding number about $0$ is $0$.