# Must every holomorphic function $f: \mathbb{D} \to \mathbb{D}$ have a fixed point?

(a) Prove that if $$f: \mathbb{D} \to \mathbb{D}$$ is analytic and has two distinct fixed points, then $$f$$ is identity.

(b) Must every holomorphic function $$f: \mathbb{D} \to \mathbb{D}$$ have a fixed point?

Notation. $$\mathbb{D}$$ is the open unit disk.

My attempt.

(a) Let $$\displaystyle \psi_{z_{1}} = \frac{z_{1} - z}{1 - \bar{z_{1}}z}$$ and $$z_{1},z_{2}$$ the fixed points of $$f$$, and define $$g: \mathbb{D} \to \mathbb{D}$$ by $$g(z) = (\psi_{z_{1}} \circ f \circ \psi_{z_{1}}^{-1})(z)$$. Since $$\psi_{z_{1}}$$ is holomorphic and an automorphism of $$\mathbb{D}$$ (such that $$\psi_{z_{1}}^{2} = id$$), $$g$$ maps $$\mathbb{D}$$ into itself. Note that $$g(0) = (\psi_{z_{1}} \circ f \circ \psi_{z_{1}}^{-1})(0) = (\psi_{z_{1}} \circ f)(\psi_{z_{1}}(0)) = \psi_{z_{1}}(f(z_{1})) = 0.$$ Since $$\psi_{z_{1}}$$ is bijective, there is $$\alpha$$ such that $$\psi_{z_{1}}(\alpha) = z_{2}$$, moreover, $$\alpha = \psi_{z_{1}}^{2}(\alpha) = \psi_{z_{1}}(z_{2})$$. Then, $$g(\alpha) = (\psi_{z_{1}} \circ f \circ \psi_{z_{1}}^{-1})(\alpha) = (\psi_{z_{1}} \circ f)(\psi_{z_{1}}(\alpha)) = \psi_{z_{1}}(f(z_{2})) = \alpha.$$ Also, if $$\alpha = 0$$, $$z_{1} = z_{2}$$, a contradiction. By Schwarz lemma, $$g(z) = cz$$ where $$c = e^{i\theta}$$. But, since $$g(\alpha) = \alpha$$, $$c\alpha = \alpha$$ and so, $$c=1$$.

(b) I know that, with tha same ideia that I use in (a), if the answer is yes, I can always apply the Schwarz lemma in every holomorphic function from $$\mathbb{D}$$ to $$\mathbb{D}$$. So, I think the answer is no, but I could not find a counterexample. I'm trying to get an example where Schwarz lemma fails.

$$f(z)=\dfrac{z+1}{2}$$
The unit disc $$\Bbb D$$ is conformally equivalent to the upper half-plane $$\Bbb H$$. That has a holomorphic map $$\Bbb H\to \Bbb H$$ without fixed points, for instance $$z\mapsto z+1$$. So if $$h:\Bbb D\to\Bbb H$$ is a holomorphic map, with a holomorphic inverse, then you can take $$f(z)=h^{-1}(h(z)+1)$$. I'll leave it to you to find such an $$h$$.
• $h(z) = i\frac{1-z}{1+z}$ is a holomorphic map from $\mathbb{D}$ to $\mathbb{H}$ (since $\mathrm{Im}(h(z)) > 0$) with inverse $g(z) = \frac{i-z}{i+z}$, right? Nov 22 '18 at 21:34