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(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.

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Another Counterexample :

$f(z)=\dfrac{z+1}{2}$

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    $\begingroup$ Not sure why this wasn't the accepted answer. It is the simplest one. $\endgroup$
    – zhw.
    Jun 17 '20 at 15:11
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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 $h$.

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  • $\begingroup$ $h(z) = i\frac{1-z}{1+z}$ 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? $\endgroup$
    – Lucas
    Nov 22 '18 at 21:34
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    $\begingroup$ @LucasCorrêa That looks very plausible. $\endgroup$ Nov 22 '18 at 21:43

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