Part 1:

Write $\psi_a(z) \coloneqq{a-z\over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z}$ for the Blaschke factor of $a$, and define $$\begin{align*} g(z) \coloneqq{f(z) \over \psi_a(z) \psi_{-a}(z)} .\end{align*}$$

In particular, ${\left\lvert {g(0)} \right\rvert} \leq 1$, so $$\begin{align*} 1\geq {\left\lvert {g(0)} \right\rvert} = {{\left\lvert {f(0)} \right\rvert} \over {\left\lvert {B_a(0)} \right\rvert} \cdot {\left\lvert {B_{-a}(0)} \right\rvert}} = {{\left\lvert {f(0)} \right\rvert} \over {\left\lvert {a} \right\rvert}^2} \implies {\left\lvert {a} \right\rvert}^2 \geq {\left\lvert {f(0)} \right\rvert} .\end{align*}$$

Part 2: Applying Schwarz-Pick: $$\begin{align*} {\left\lvert {f'(0)} \right\rvert} \leq {1 - {\left\lvert {f(0)} \right\rvert}^2 \over 1 - {\left\lvert {0} \right\rvert}^2 } = 1-{\left\lvert {a} \right\rvert}^2 < 1 ,\end{align*}$$ using that $a\neq 0$, so $f$ is a contraction.

Can write $f_e(z) \coloneqq{f(a) + f(-a) \over 2}$ to write $f_e(z) = g(z^2)$. Compose with some $\psi_a$ to get $0\to 0$ and apply Schwarz – unclear how to unwind what happens in the case of equality though.

First show the hint: assume $f$ has nonzero zeros. Write $Z(f) \coloneqq f^{-1}(0)$ for the set of zeros in $\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$.

So write $Z(f) = \left\{{z_1,\cdots, z_m}\right\}$ and define $$\begin{align*} g(z) \coloneqq\prod_{1\leq k \leq m} {z-z_k \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5mu z}, \quad h(z) \coloneqq{f(z) \over g(z)} .\end{align*}$$

So we now have $$\begin{align*} f(z) = e^{i\theta} \prod_{1\leq k\leq m} {z-z_k \over 1 -\mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5mu z} ,\end{align*}$$ which has poles at points $z$ for which $\mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5muz=1$ for some $z_k\in Z(f)$. However, since we assumed $f$ was entire, it can have no such poles, which forces $z_k = 0$ for all $k$. But then $$\begin{align*} f(z) = e^{i\theta}\prod_{1\leq k \leq m}{z- 0 \over 1 - 0\cdot z} = e^{i\theta}z^m .\end{align*}$$

Proof due to Swaroop Hegde!

Fix $R, M$ and make a clever choice: define $$\begin{align*} F: {\mathbb{D}}&\to {\mathbb{C}}\\ z &\mapsto {f(Rz) \over M} .\end{align*}$$ Write $a\coloneqq F(0)$ and consider the Blaschke factor $$\begin{align*} \psi_a(z) \coloneqq{a-z \over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z} \in \mathop{\mathrm{Aut}}({\mathbb{D}}) ,\end{align*}$$ and define $$\begin{align*} g: {\mathbb{D}}&\to {\mathbb{D}}\\ z &\mapsto (\psi_a \circ F)(z) .\end{align*}$$ Then $g(0) = 0$ and ${\left\lvert {g(z)} \right\rvert} \leq 1$ for all $z\in {\mathbb{D}}$, so by Schwarz we have ${\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$ for all $z\in {\mathbb{D}}$. Thus for all $z\in {\mathbb{D}}$, $$\begin{align*} &{\left\lvert {g(z)} \right\rvert} \leq z \\ \\ \iff & {\left\lvert {\psi_a(F(z)) } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert { {f(Rz) \over M} - a \over 1 - {\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu f(Rz) \over M} } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert {f(Rz) - f(0) \over 1 - {\mkern 1.5mu\overline{\mkern-1.5muf(0)\mkern-1.5mu}\mkern 1.5mu f(Rz) \over M^2 } } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert {f(Rz) - f(0) \over M^2 - \mkern 1.5mu\overline{\mkern-1.5muf(0)\mkern-1.5mu}\mkern 1.5mu f(Rz) } \right\rvert} \leq {{\left\lvert {z} \right\rvert} \over M} \\ \\ \iff & {\left\lvert {f(w) - f(0) \over M^2 - \mkern 1.5mu\overline{\mkern-1.5muf(0)\mkern-1.5mu}\mkern 1.5mu f(w) } \right\rvert} \leq {{\left\lvert {w} \right\rvert} \over MR} ,\end{align*}$$ which holds for all $w\in {\mathbb{D}}$ by replacing $Rz$ with $w$ (i.e. to show this equality for arbitrary $w\in {\mathbb{D}}$, write $w = Rz$ for some $z\in {\mathbb{D}}$ and run this chain of inequalities backward).

Set $$\begin{align*} g(z) \coloneqq{f(Rz+a) - f(a) \over 2M} ,\end{align*}$$ so that $g(0) = 0$ and $g:{\mathbb{D}}\to {\mathbb{D}}$ so Schwarz applies, $$\begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert { f(Rz+a) - f(a) \over 2M } \right\rvert} &\leq {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert { f(Rz+a) - f(a) } \right\rvert} &\leq 2M {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert { f(w) - f(a) } \right\rvert} &\leq 2M{\left\lvert { w-a\over R} \right\rvert} \\ \implies {\left\lvert {f(w) - f(a) \over w-a} \right\rvert} &\leq {2M \over R} .\end{align*}$$

Define $g:{\mathbb{H}}\to {\mathbb{H}}$ by $g(z) = {1\over 2}iz$, so $g(2) = i$. Then set $F \coloneqq C\circ g \circ f: {\mathbb{D}}\to {\mathbb{D}}$ where $C(z) \coloneqq{z-i\over z+i}$ is the Cayley map.Since $F(0) = C(g(f(0))) = C(g(2)) = C(i) = 0$, Schwarz applies to $F$ and ${\left\lvert {F'(z)} \right\rvert}\leq 1$ for $z\in {\mathbb{D}}$. By the chain rule, $$\begin{align*} F'(z) = f'( (g\circ C) (z))\cdot g'(C(z)) \cdot C'(z) .\end{align*}$$ Setting $g(C(z)) = 0$ yields $z=C^{-1}(g^{-1}(0)) = C^{-1}(0) = i$. $$\begin{align*} F'(i) &= f'(0) \cdot g'(0) \cdot C'(i) \\ \implies {\left\lvert {f'(0)} \right\rvert} &\leq {\left\lvert {F'(i) \over g'(0) C'(i)} \right\rvert} \\ &\leq {1\over {\left\lvert {g'(0)} \right\rvert} \cdot {\left\lvert {C'(i)} \right\rvert} } \\ &= {1\over {\left\lvert {i\over 2} \right\rvert} \cdot {\left\lvert {-{i\over 2} } \right\rvert} } \\ &= 4 .\end{align*}$$

By Schwarz, if ${\left\lvert {F'(z)} \right\rvert} = 1$ for any $z\in {\mathbb{D}}$, we’ll have $F(z) = \lambda z$ for some ${\left\lvert { \lambda} \right\rvert} = 1$. Unwinding this: $$\begin{align*} F(z) &= \lambda z \implies (C\circ g\circ f)(z) = \lambda z \\ \implies f(z) &= g^{-1}(C^{-1}(\lambda z)) = g^{-1}\qty{-i {\lambda z + 1 \over \lambda z - 1}} \\ \implies f(z) &= -2 {\lambda z + 1\over \lambda z - 1} .\end{align*}$$ Moreover $f'(z) = -2\qty{-2\lambda \over (\lambda z - 1)^2}$, so $$\begin{align*} {\left\lvert {f'(0)} \right\rvert} = 4{\left\lvert {\lambda} \right\rvert} = 4 .\end{align*}$$

Note ${\left\lvert {f(z)} \right\rvert}\leq 1$, and $g\coloneqq f(z)/z^k$ has a removable singularity at zero since $g$ is bounded on ${\mathbb{D}}$: fixing ${\left\lvert {z} \right\rvert} = r < 1$, $$\begin{align*} {\left\lvert {g(z)} \right\rvert} = {\left\lvert {f(z)\over z^k} \right\rvert} = {\left\lvert {f(z)} \right\rvert}r^{-k}\leq r^{-k}\overset{r\to 1}\longrightarrow 1 .\end{align*}$$ So $g:{\mathbb{D}}\to {\mathbb{D}}$ since ${\left\lvert {g(z)} \right\rvert}\leq 1$ on ${\mathbb{D}}$ by the MMP. Since $g$ has no zeros on ${\mathbb{D}}$, by the MMP ${\left\lvert {g} \right\rvert} \geq 1$ on ${\mathbb{D}}$, so ${\left\lvert {g} \right\rvert} = 1$ is constant, making $g(z) = \lambda z$ a rotation. Then $f(z) = \lambda z^n$.

Alternative to MMP: if $g$ has no zeros in ${\mathbb{D}}$, $g$ admits a conjugate reflection through ${\mathbb{D}}$ by $z\mapsto 1/\mkern 1.5mu\overline{\mkern-1.5muf(1/\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu$. This is bounded and entire, thus constant, making $g$ constant.

Since $f$ is injective, it has a left-inverse $f^{-1}$, and $F\coloneqq f^{-1}g$ is well-defined. Since $F:{\mathbb{D}}\to {\mathbb{D}}$ and $F(0) = 0$, Schwarz applies and ${\left\lvert {F(z)} \right\rvert} \leq z$ on ${\mathbb{D}}$. Unwinding: $$\begin{align*} {\left\lvert {(f^{-1}\circ g)(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {f(z)} \right\rvert} \qquad \forall {\mathbb{D}}\in {\mathbb{Z}} .\end{align*}$$ This says that $g({\mathbb{D}}) \subseteq f({\mathbb{D}})$, and in particular this holds on all ${\mathbb{D}}_r(0)$, so $g({\mathbb{D}}_r(0)) \subseteq f({\mathbb{D}}_r(0))$.

Define a function $$\begin{align*} F(z) \coloneqq \begin{cases} f(z) & \Im(z)\geq 0 \\ \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu & \Im(z) < 0 \end{cases} .\end{align*}$$ Then $F$ is holomorphic on $\tilde S\coloneqq\left\{{z\in {\mathbb{D}}{~\mathrel{\Big\vert}~}\Im(z) < 0}\right\}$ – write $w_0\in \tilde S$ as $w_0 = \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu$ for some $z_0\in S$, then $$\begin{align*} f(z) &= \sum_{k\geq 0}c_k (z-z_0)^k \\ \implies \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu &= \sum_{k\geq 0}\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5mu\qty{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - z_0}^k\mkern-1.5mu}\mkern 1.5mu \\ &= \sum_{k\geq 0}\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \qty{z - \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu}^k \\ &= \sum_{k\geq 0}\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \qty{z - w_0}^k ,\end{align*}$$ which yields a power series expansion of $F$ about $w_0$. So $f$ is analytic at every point in $\tilde S$ and thus holomorphic. Since $\mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu = f(z)$ for $z, f(z)\in {\mathbb{R}}$, $F$ is a continuous extension of $f$ to ${\mathbb{D}}$. By the symmetry principle, $F$ is holomorphic, and ${ \left.{{F}} \right|_{{S}} } = f$.

This is the Schwarz–Pick lemma.

• Fix $z_1$ and let $w_1 = f(z_1)$. Define $$\begin{align*} \psi_{a}(z) \coloneqq{a-z \over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5muz} \in \mathop{\mathrm{Aut}}({\mathbb{D}}) .\end{align*}$$

  • Note that inequality now reads $$\begin{align*} {\left\lvert {\psi_{f(w)}(f(z)) } \right\rvert} \leq {\left\lvert {\psi_w(z)} \right\rvert} .\end{align*}$$ Moreover $\psi_a$ is an involution that swaps $a$ and $0$.

• Now set up a situation where Schwarz’s lemma will apply: $$\begin{align*} 0 \xrightarrow{\psi_{z_1}} z_1 \xrightarrow{f} f(z) \xrightarrow{\psi_{f(z_1)}} 0 ,\end{align*}$$ so $F\coloneqq\psi_{f(z_1)} \circ f \circ \psi_{z_1} \in \mathop{\mathrm{Aut}}({\mathbb{D}})$ and $F(0) = 0$.

• Apply Schwarz we get ${\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$ for all $z$, so $$\begin{align*} {\left\lvert {F(z)} \right\rvert} &\leq {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert { f(z_1) - (f\circ \psi_{z_1})(z) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(z_1)\mkern-1.5mu}\mkern 1.5mu \cdot (f\circ \psi_{z_1}) (z) } \right\rvert} &\leq {\left\lvert { z} \right\rvert} \\ \implies {\left\lvert {f(z_1) - f(w) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(z_1)\mkern-1.5mu}\mkern 1.5mu\cdot f(w) } \right\rvert} &\leq {\left\lvert {\psi_{z_1}(z)} \right\rvert} && w\coloneqq\psi_{z_1}(z) \\ \implies {\left\lvert {f(z_1) - f(w) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(z_1)\mkern-1.5mu}\mkern 1.5mu\cdot f(w) } \right\rvert} &\leq {\left\lvert {z_1 - z \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz_1\mkern-1.5mu}\mkern 1.5mu z } \right\rvert} .\end{align*}$$

• Since $z_1$ was arbitrary and fixed and $w$ was a free variable, this holds for all $z,w\in {\mathbb{D}}$.

• Strictness: suppose equality holds, we’ll show that $f(z) = {az+b\over cz+d}$

• By Schwarz, $F(z) = \lambda z$ for $\lambda \in S^1$. Thus $$\begin{align*} (\psi_{f(z_1)} \circ f \circ \psi_{z_1}) (z) &= \lambda z \\ \implies (f \circ \psi_{z_1}) (z) &= \psi_{f(z_1)}^{-1}(\lambda z ) \\ \implies f(w) &= \psi_{f(z_1)}^{-1}(\lambda \psi_{z_1}^{-1}(w) ) && w\coloneqq\psi_{z_1}(z) \\ &= \psi_{f(z_1)} \qty{\lambda \psi_{z_1}(w)} \\ &= \lambda \psi_{\mkern 1.5mu\overline{\mkern-1.5mu\lambda \mkern-1.5mu}\mkern 1.5muf(z_1)} \qty{\psi_{z_1}(w)} \\ &\coloneqq\lambda \psi_a(\psi_b(w)) \\ &=\lambda\qty{ a- \psi_b(w) \over 1 - \mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu \psi_b(w) } \\ &= \quad \vdots \\ &= -\lambda \qty{ \frac{{\left(a \overline{b} - 1\right)} z - a + b}{{\left(\overline{a} - \overline{b}\right)}z - b \overline{a} + 1} } \\ &= \qty{ \frac{-\lambda {\left(a \overline{b} - 1\right)} z + \lambda( a - b)}{{\left(\overline{a} - \overline{b}\right)}z + (- b \overline{a} + 1)} } ,\end{align*}$$ which is evidently a linear fractional transformation.

Given this, there’s just a clever rearrangement to obtain the stated result: $$\begin{align*} {\left\lvert {f(w) - f(z) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(w)\mkern-1.5mu}\mkern 1.5muf(z)} \right\rvert} &\leq {\left\lvert {w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} \\ \implies {\left\lvert { 1\over 1-\mkern 1.5mu\overline{\mkern-1.5muf(w)\mkern-1.5mu}\mkern 1.5muf(z) } \right\rvert} \cdot {\left\lvert {f(z) - f(w) \over z-w} \right\rvert} &\leq {\left\lvert {1\over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5muz} \right\rvert} \\ ,\end{align*}$$ and taking $z\to w$ on both sides yields $$\begin{align*} {\left\lvert {1\over 1 - {\left\lvert {f(w)} \right\rvert}^2 } \right\rvert} {\left\lvert {f'(w)} \right\rvert} \leq {1\over {\left\lvert {w} \right\rvert}^2} \implies {\left\lvert {f'(w)} \right\rvert} \leq {1-{\left\lvert {f(w)} \right\rvert}^2\over 1-{\left\lvert {w} \right\rvert}^2 } .\end{align*}$$

Part 1: use the reflection principle to define $$\begin{align*} F(z) \coloneqq \begin{cases} f(z) & {\left\lvert {z} \right\rvert} \leq 1 \\ {1\over \mkern 1.5mu\overline{\mkern-1.5muf\qty{1/\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\mkern-1.5mu}\mkern 1.5mu } & {\left\lvert {z} \right\rvert} \geq 1 \end{cases} .\end{align*}$$

Now $F:{\mathbb{CP}}^1\to {\mathbb{CP}}^1$ is holomorphic and all such functions are rational. As a consequence, $f$ is rational.

Part 2: As in the proof of Schwarz, define $g(z) \coloneqq{f(z)\over z^n}$ where $n = {\operatorname{Ord}}_{f}(0)$. Then $g$ is holomorphic on ${\mathbb{D}}$ since the singularity at $z=0$ is removable. On ${\left\lvert {z} \right\rvert} = r<1$, $$\begin{align*} {\left\lvert {g(z)} \right\rvert} = { {\left\lvert {f(z)} \right\rvert} \over {\left\lvert {z} \right\rvert} } = {{\left\lvert {f(z)} \right\rvert} \over r} \leq {1\over r} \overset{r\to 1^-}\longrightarrow 1 ,\end{align*}$$ using that ${\left\lvert {f} \right\rvert} \leq 1$ on ${\mathbb{D}}$. By the MMP, ${\left\lvert {g} \right\rvert} \leq 1$ on all of ${\mathbb{D}}$. Note that ${\left\lvert {g} \right\rvert} = 1$ when ${\left\lvert {z} \right\rvert}=1$, so ${\left\lvert {1/g} \right\rvert}\leq 1$ in ${\mathbb{D}}$ by the MMP, forcing ${\left\lvert {g} \right\rvert} = 1$. Unwinding this, ${\left\lvert {f} \right\rvert} = {\left\lvert {z} \right\rvert}^n$, go $f(z) = \lambda z^n$ for some ${\left\lvert {\lambda} \right\rvert} = 1$.

Part 3: Define $\Psi(z) \coloneqq\prod_{k\leq n} \psi_{a_k}(z)$ where $\psi_a(z) \coloneqq{a-z\over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z}$. Set $g(z) \coloneqq{f(z) \over \Psi(z)}$, then by the same argument as above, ${\left\lvert {g} \right\rvert} \leq 1$ and ${\left\lvert {g} \right\rvert} = 1$ on ${\left\lvert {z} \right\rvert} = 1$. Then $g$ has no zeros, since they’ve all been divided out, and no poles since $f$ is holomorphic on ${\mathbb{D}}$, so $1/g$ is holomorphic on ${\mathbb{D}}$. Since ${\left\lvert {1/g} \right\rvert} = 1$ on $S^1$, this forces $g$ to be constant. Equality in the Schwarz lemma implies $g(z) = \lambda z$ is a rotation, and unwinding this yields $f(z) = \lambda \Psi(z)$.

Use Rouché: if ${\left\lvert {f(z)} \right\rvert} < 1$ is strict when ${\left\lvert {z} \right\rvert} = 1$, then consider $F(z) \coloneqq f(z) - z$. Write the big part as $M(z) = z$ and the small as $m(z) = f(z)$, then on ${\left\lvert {z} \right\rvert} = 1$ $$\begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {f(z)} \right\rvert} < 1 = {\left\lvert {z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*}$$ so $M(z)$ and $m(z) + M(z) = f(z) - z$ have the same number of zeros in ${\mathbb{D}}$ – precisely one.

There is still a unique fixed point. Use the Brouwer fixed point theorem: since $g$ is holomorphic on $\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$, it is in particular continuous. By the Brouwer fixed point theorem, every continuous map $\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu \to \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$ has a fixed point. If $g$ is nonconstant, then the fixed point is unique by Schwarz: without loss of generality one can assume $f(0) = 0$ by composing with a Blaschke factor. Apply Schwarz to $f$, then if $f(a) = a$ we have the equality clause and $f(z) = \lambda z$. Since $a = f(a) = \lambda a$, $\lambda = 1$ and $f$ is the identity. If $g$ is constant, then ${\left\lvert {g(z)} \right\rvert} < 1$ on ${\left\lvert {z} \right\rvert} = 1$ forces $g\equiv 0$.

Note that there is a major difference between self maps to ${\mathbb{D}}$ versus $\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu$. By the argument in part 2, if $f(z)$ is not the identity then $f$ can have at most one fixed point. Moreover, not every map $f:{\mathbb{D}}\to{\mathbb{D}}$ need have a fixed point: consider $$\begin{align*} g: {\mathbb{H}}&\to {\mathbb{H}}\\ z &\mapsto z+1 .\end{align*}$$ Now conjugate with the Cayley map $C:{\mathbb{H}}\to {\mathbb{D}}$ to define $f\coloneqq CgC^{-1}:{\mathbb{D}}\to {\mathbb{D}}$ which has no fixed points at all.

Define $F(z) \coloneqq{f(z) \over g(z)}$.

Given this, note that ${\left\lvert {F(z)} \right\rvert} = 1$ for all $z$, so $F(\Omega) \subseteq S^1$, which is codimension 1 in ${\mathbb{C}}$ and not open. By the open mapping theorem, $F$ must be constant, so $F(z) = \lambda$, and in particular since ${\left\lvert {F(z)} \right\rvert} = 1$, $\lambda = e^{it}\in S^1$ for some $t$. Then $f(z) = \lambda g(z)$.

Note that there is a unique fixed point. We have $f(0) = 0$, so there is at least one, so suppose $a$ is another fixed point with $f(a) = a$. By Schwarz, ${\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}$ with equality at any nonzero point implying $f$ is a rotation, and $f(a) = a\implies {\left\lvert {f(a)} \right\rvert} = {\left\lvert {a} \right\rvert}$, so write $f(z) = e^{i\theta}z$. Now $f(a) = a = e^{i\theta }a$ forces $\theta = 0$, so $f(z) = z$ is the identity.

Since $f(0) = 0$, the Schwarz lemma applies and either

• $f(z) = e^{i\theta} z$ is a rotation, or
• ${\left\lvert {f'(0)} \right\rvert} < 1$ and ${\left\lvert {f(z)} \right\rvert} < z$ for all $z\in {\mathbb{D}}$.

Supposing the latter, $f$ is a contraction, and ${\left\lvert {f_{n+1}(z)} \right\rvert} < {\left\lvert {f_{n}(z)} \right\rvert}$ for all $n$ and all $z$, so ${\left\lvert {f_n(z)} \right\rvert} \overset{n\to\infty}\longrightarrow 0$ for all $z$. Since $f_n\to g$ pointwise, this means $g(z) = 0$ for all $z$, making $g\equiv 0$.

Otherwise, suppose $f$ is a rotation. Then if $f(z) = e^{i\theta}z$, $f_n(z) = e^{in\theta}z$. The pointwise limit $\lim_{n\to\infty}e^{in\theta}z$ can only exist if $\theta = 0$, otherwise this is periodic when $\theta$ is rational or the points $e^{i\theta}z, e^{2i\theta }z,\cdots$ form form a countably infinite set of distinct points. So $f(z) = z$, making $\lim_{n\to \infty}f_n(z) = z$ as well.