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.