Define \(\Psi(z) \coloneqq\prod_{k\leq n} \psi_{a_k}(z)\) and \(g(z) \coloneqq f(z)/\Psi(z)\). The claim is that \({\left\lvert {g(z)} \right\rvert} \leq 1\), which implies the result directly. Note that \({\left\lvert {\Psi(z)} \right\rvert} = 1\) for \({\left\lvert {z} \right\rvert} = 1\), so \(\lim_{r\to 1^-} {\left\lvert {\Psi(re^{it})} \right\rvert} = 1\) along any ray. Now for \(z_r \coloneqq re^{it}\) for \(r< 1\), \begin{align*} {\left\lvert {g(z_r)} \right\rvert} = { {\left\lvert {f(z_r)} \right\rvert} \over {\left\lvert {\Psi(z_r)} \right\rvert} } \leq {1\over {\left\lvert {\Psi(z_r)} \right\rvert} } \leq {1\over \displaystyle\sup_{t\in [0, 2\pi]} {\left\lvert {\Psi(z_r) } \right\rvert} } \overset{r\to 1^-}\longrightarrow 1 .\end{align*}

Set \(g(z) \coloneqq{f(Rz + a) \over M}\), then \(g: {\mathbb{D}}\to {\mathbb{D}}\) with \(g(0) = f(a)/M = 0\), so unwinding Schwarz yields \begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {f(Rz+a)\over M} \right\rvert} &\leq {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert {f(Rz+a)} \right\rvert} &\leq M{\left\lvert {z} \right\rvert} \\ \implies {\left\lvert {f(w)} \right\rvert} &\leq M{\left\lvert {w-a\over R} \right\rvert} \qquad w = Rz+a\implies z={w-a\over R} .\end{align*}

First, \(\psi_a\) maps \({\mathbb{D}}\to {\mathbb{D}}\). Fix \(z\in S^1\), then \begin{align*} {\left\lvert {\psi_a(z)} \right\rvert} \coloneqq {\left\lvert {a-z\over 1-\overline{a} z} \right\rvert} &= {\left\lvert {a-z\over 1-\overline{a} z} \right\rvert} \cdot {\left\lvert {\overline{z}} \right\rvert}^{-1}\\ &= {\left\lvert {a-z\over \overline{a} - \overline{z}} \right\rvert} \\ &= {\left\lvert {a-z \over \overline{a-z} } \right\rvert} \\ &\coloneqq{\left\lvert {w \over \overline{w} } \right\rvert} \\ &= {\left\lvert {w \over {\left\lvert {w} \right\rvert}^2/w } \right\rvert}\\ &= {\left\lvert {w^2 \over {\left\lvert {w} \right\rvert}^2} \right\rvert} \\ &= 1 .\end{align*} \(\psi_a\) is holomorphic on \({\mathbb{D}}\) since it has a simple pole at \(z=1/\overline{a}\), but \({\left\lvert {a} \right\rvert}<1\) implies \({\left\lvert {1/\overline{a}} \right\rvert} > 1\). Let \(g\coloneqq\psi_a\circ \psi_a\), then \(g(0) = 0\) so Schwarz applies. Since \(g(a) = a\) with \(a\neq 0\), the 2nd part of Schwarz also applies since \({\left\lvert {g(a)} \right\rvert} = {\left\lvert {a} \right\rvert}\) and \(g(z) = \lambda z\) is a rotation. Since \(a = g(a) = \lambda a\), this forces \(\lambda = 1\), so \(g\) is the identity.

By Schwarz, \({\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\), and if we write \(h \coloneqq f^{-1}\) then \(h(0) = 0\) and \({\left\lvert {h(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\) as well. Schwarz says \(f\) will be a rotation if there is any \(z_0\) such that \({\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\). Write \(f(z) = w\), we’ll then show that in fact \({\left\lvert {f(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) for all \(z\in {\mathbb{D}}\). \begin{align*} {\left\lvert {z} \right\rvert} = {\left\lvert {(h\circ f)(z)} \right\rvert} = {\left\lvert {h(w)} \right\rvert} \leq {\left\lvert {w} \right\rvert} = {\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert} .\end{align*}

By Schwarz, \({\left\lvert {f'(0)} \right\rvert} \leq 1\) – the claim is that we have equality, so that by Schwarz \(f(z) = \lambda z\) for some \({\left\lvert {\lambda} \right\rvert} = 1\). Use that \(f'(0) \neq 0\) since \(f\) is a bijection near zero and if \(g\coloneqq f^{-1}\) then \(g'(0) = 1/f'(0)\). Moreover Schwarz applies to \(g\), so \(1 \geq {\left\lvert {g'(0)} \right\rvert} \geq {1\over {\left\lvert {f'(0)} \right\rvert} }\), forcing \({\left\lvert { f'(0) } \right\rvert} = 1\). By the equality clause in the Schwarz lemma for \(f\), \(f\) is rotation.

Note that \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {g(z)} \right\rvert} \impliedby {\left\lvert {(f\circ g^{-1})(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} ,\end{align*} so one can use the Schwarz lemma on \(F \coloneqq f\circ g^{-1}\). Noting that \(g(z) \coloneqq{z-i\over z+i}: {\mathbb{H}}\to {\mathbb{D}}\) is the Cayley map, the inverse is \(g^{-1}(z) = i{1-z\over 1+z}: {\mathbb{D}}\to {\mathbb{H}}\). Then \(F(0) = f(g^{-1}(0)) = f(i) = 0\) by assumption, so Schwarz yield \({\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\).

The key observation is that this factors: \begin{align*} {z^2 - r^2 \over 1-r^2 z^2} = {r-z\over 1+rz}{-r-z\over 1-rz} = \psi_{r}(z) \psi_{-r}(z) ,\end{align*} so this inequality will follow from Schwarz on \(g(z) \coloneqq f(z)/\psi_a(z)\psi_{-a}(z)\). Schwarz does apply since \({\left\lvert {f} \right\rvert}\leq 1\) in \({\mathbb{D}}\) and \({\left\lvert {\psi_a(z)} \right\rvert} = 1\) on \(S^1\), so \({\left\lvert {g(z)} \right\rvert} \leq 1\) on \(S^1\) and by the MMP this inequality holds in all of \({\mathbb{D}}\). So \({\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\) and unwinding gives the desired inequality.

Suppose \(f\) is entire and bounded, we’ll show \(f\) is constant. If \(f\) is bounded by \(M\), then \(f({\mathbf{C}}) \subseteq {\mathbb{D}}_M(0)\). Without loss of generality, replace \(f\) with \(g(z) \coloneqq f(z) - f(0)\), so \(g(0) = 0\) and is still bounded by \(M' \coloneqq M + {\left\lvert {f(0)} \right\rvert}\) by the triangle inequality. This is still finite since \(0\) is not a singularity since \(f\) is entire.

By the radius \(R\) variant of the Schwarz lemma, for every \({\mathbb{D}}_R(0)\), \begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {M\over R}{\left\lvert {z} \right\rvert} \qquad \text{for } z\in {\mathbb{D}}_R(0) .\end{align*} Using that \(g(0) = 0\), \begin{align*} {\left\lvert {g(z) - g(0) \over z} \right\rvert} \leq {M\over R}\overset{R\to\infty}\longrightarrow 0 ,\end{align*} where dividing by \(z\) is not an issue since \(z=0\) is a zero of \(g\) of at least order one. This forces \(g(z) = g(0) = 0\) for all \(z\), so \(f(z) = f(z_0)\) is constant.

By Schwarz, \({\left\lvert {f(z)} \right\rvert}\leq z\). Write \(g(z) \coloneqq f(z)/z\), then \({\left\lvert {g(z)} \right\rvert}\leq 1\) and \(g:{\mathbb{D}}\to {\mathbb{D}}\) is holomorphic since \(f\) has a zero of order at least one at \(0\). Note that \(f(z) = c_2z^2 + { \mathsf{O}}(z^3)\), where \(c_0 = 0\) since \(f(0) = 0\) and \(c_1 = 0\) since \(f'(0) = 0\). Thus \(g(z) = c_2z + { \mathsf{O}}(z^2)\), and \(g(0) = 0\), so Schwarz applies and

• \({\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\)
• \({\left\lvert {g'(0)} \right\rvert} \leq 1\)

We have \(g'(z) = c_2 + { \mathsf{O}}(z)\) so \(g'(0) = c_2 = {f^{(2)}(0) \over 2!}\). \begin{align*} 1 \geq {\left\lvert {g'(0)} \right\rvert} = {\left\lvert {c_2} \right\rvert} = {\left\lvert {f^{(2)}(0) \over 2} \right\rvert} \implies {\left\lvert { f^{(2)}(0) } \right\rvert} \leq 2! .\end{align*}

Suppose this is an equality – then Schwarz on \(g\) shows \(g\) is a rotation, so \begin{align*} {f(z)\over z} = g(z) = \lambda z \implies f(z) = \lambda z^2 \qquad \lambda \in S^1 .\end{align*}

Note: if \({\left\lvert {z} \right\rvert}\leq 1\), we have \(g({\mathbb{D}}) \subseteq \overline{{\mathbb{D}}}\), but the open mapping theorem always guarantees \(g({\mathbb{D}}) = {\mathbb{D}}\).

First suppose \(a=0\) – then Schwarz applies, and since \({\left\lvert {f'(a)} \right\rvert} < 1\) is strict, \(f\) is not a rotation.

With such an \(r\) and \(C<1\) in hand,

\begin{align*} {\left\lvert {z_k} \right\rvert} = {\left\lvert {f(z_{k-1})} \right\rvert} \leq C{\left\lvert {z_{k-1}} \right\rvert} = C{\left\lvert {f(z_{k-2})} \right\rvert} \leq C^2 {\left\lvert {z_{k-2}} \right\rvert} \cdots \implies {\left\lvert {z_k} \right\rvert} \leq C^k{\left\lvert {z_0} \right\rvert} \overset{k\to\infty}\longrightarrow 0 ,\end{align*} which proves the \(a=0\) case.

For \(a\neq 0\), take a Blaschke factor \(\psi_a(z)\) and consider \(F \coloneqq\psi_a^{-1}\circ f\circ \psi_a\). The claim is that this reduces to the case \(a=0\).

Note \(F(0) = 0\), so \(0\) is a fixed point of \(F\). Moreover, a clever calculation shows \begin{align*} F'(0) &= (\psi_a^{-1})'(f(\psi_a(0))) \cdot f'(\psi_a(0)) \cdot \psi_a'(0) \\ \\ &= (\psi_a^{-1})'(f( a )) \cdot f'(a ) \cdot \psi_a'(0) \\ \\ &= (\psi_a^{-1})'(a) \cdot f'(a ) \cdot \psi_a'(0) \qquad \text{since } f(a) = a \\ \\ &= (\psi_a)'(a) \cdot \psi_a'(0) \cdot f'(a) \qquad \text{since } \psi_a^{-1}= \psi_a \\ \\ &= (\psi_a)'( \psi_a(0) ) \cdot \psi_a'(0) \cdot f'(a) \\ \\ &= (\psi_a \circ \psi_a)'(0) \cdot f'(a) \\ &= 1 \cdot f'(a) ,\end{align*} so \({\left\lvert {F'(0)} \right\rvert} = {\left\lvert {f'(a)} \right\rvert} < 1\). Now setting \(w_k \coloneqq\psi_a(z_n)\) and writing \(f = \psi_a \circ F \circ \psi_a^{-1}\), by continuity we have \begin{align*} f(z_k) = \psi_a(F(w_k)) \overset{k\to\infty}\longrightarrow \psi_a(0) = a .\end{align*}

Note that without loss of generality we can assume \(a=0\) so \(f(0) = 0\) and \(b\neq 0\). If not, if \(a,b\neq 0\) then let \(F\coloneqq\psi_a \circ f \circ \psi_a\), then \(F(0) = 0\) and \(F(b') = b'\) for \(b\coloneqq\psi_a(b)\).

Since \(f(0) = 0\), Schwarz applies, so \({\left\lvert {f(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) with equality attained because \({\left\lvert {f(b)} \right\rvert} = {\left\lvert {b} \right\rvert}\), and \(f(z) = \lambda z\) must be a rotation. Since \(b = f(b) = \lambda b\), we have \(\lambda = 1\).

Set \(f(a) = a\) in Schwarz-Pick: \begin{align*} \left|f^{\prime}(a)\right| \leq \frac{1-|f(a)|^{2}}{1-|a|^{2}} \implies {\left\lvert {f'(a)} \right\rvert} \leq {1 - {\left\lvert {a} \right\rvert}^2 \over 1 - {\left\lvert {a} \right\rvert}^2} \leq 1 .\end{align*} Set \(a=0\): \begin{align*} \left|f^{\prime}(0)\right| \leq \frac{1-|f(0)|^{2}}{1-|0|^{2}} \implies {\left\lvert {f'(0)} \right\rvert}^2 \leq 1 - {\left\lvert {f(0)} \right\rvert}^2 .\end{align*}

Apply Schwarz-Pick: \begin{align*} {\left\lvert {f'\qty{1\over 2} } \right\rvert} \leq {1 - {\left\lvert {f\qty{1\over 2}} \right\rvert}^2 \over 1 - {\left\lvert {1\over 2} \right\rvert}^2 } = {7\over 2}< {2\over 3} ,\end{align*} so this is not possible.

Consider \(g(z) \coloneqq f(z)/e^z\) – since \(g(0) = 0\) and \(g: {\mathbb{D}}\to {\mathbb{D}}\), Schwarz applies and \begin{align*} {\left\lvert {g(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {ze^z} \right\rvert} \leq e{\left\lvert {z} \right\rvert} .\end{align*} So \begin{align*} {\left\lvert {f\qty{1+i\over 2}} \right\rvert} \leq e{\left\lvert {1+i\over 2} \right\rvert}= {e\sqrt{2} \over 2} .\end{align*}

Compose with the inverse Cayley map \(g(z) \coloneqq i{1+z\over 1-z}\) so \(g: {\mathbb{D}}\to {\mathbb{H}}\) to get \(F\coloneqq f\circ g:{\mathbb{D}}\to {\mathbb{D}}\), where \(F(0) =f(g(0))=f(i) = 0\). So Schwarz applies and \({\left\lvert {F(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\). Now a small trick: \begin{align*} {\left\lvert {f(2i)} \right\rvert} = {\left\lvert {(f\circ g\circ g^{-1})(2i)} \right\rvert} = {\left\lvert { F \qty{z+i\over z-i}\Big|_{z=2i} } \right\rvert} = {\left\lvert {F \qty{1\over 3} } \right\rvert} \leq {1\over 3} .\end{align*}

First note that the image of \(f\) is in fact \({\mathbb{D}}\) rather than \(\overline{{\mathbb{D}}}\), using the open mapping theorem and that the domain \({\mathbb{D}}\) is open. So Schwarz applies to \(f\). Write \(g(z) \coloneqq f(z) - (2z)^m\), the claim is that \(g\) has \(m\) zeros. Toward applying Rouché, identify

• The big part of \(g\): \(M(z) \coloneqq-(2z)^m\)
• The small part of \(g\): \(m(z) \coloneqq g(z) - M(z) = f(z)\).

Now \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} = {1\over 2} < {\left\lvert {m(z)} \right\rvert} = {\left\lvert {2z} \right\rvert}^m = 1 \qquad \text{on } {\left\lvert {z} \right\rvert} = {1\over 2} ,\end{align*} so Rouché applies: \({\sharp}Z_m = {\sharp}Z_M\) on \({\left\lvert {z} \right\rvert} < {1\over 2}\), where \(m(z) = f(z)\) and \(M(z) = -(2z)^m\) which has exactly \(m\) zeros.

Note that this works with \(g(z) \coloneqq(cz)^m\) and \(R = {1\over c}\).

Use the conformal map \(g: z\mapsto -1{z+1\over z-1}\) to map \(\Re(z)>0\) to \({\mathbb{D}}\). Composing \(F: {\mathbb{D}}\xrightarrow{g} -i{\mathbb{H}}\xrightarrow{f} {\mathbb{D}}\) yields a map \(F = f\circ g:{\mathbb{D}}\to {\mathbb{D}}\). Since \(F(0) = f(g(0)) = f(1) = 0\), Schwarz applies and \({\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\). Using the standard trick, \begin{align*} {\left\lvert {f(2)} \right\rvert} = {\left\lvert {(f\circ \circ g^{-1})(2)} \right\rvert} = {\left\lvert {F(g^{-1}(2))} \right\rvert} = {\left\lvert {F\qty{z-1\over z+1}\Big|_{z=2}} \right\rvert} = {\left\lvert {F\qty{1\over 3}} \right\rvert} \leq {\left\lvert {1\over 3} \right\rvert} .\end{align*}

Define \(F(z) \coloneqq{ f(z) + f(-z) \over 2 z^2}\), then \(F(0) = 0\) and thus Schwarz applies and yields the desired inequality. If \({\left\lvert {F(z)} \right\rvert} = {\left\lvert {z} \right\rvert}\) for any \(z\), then \(F(z) = \lambda z^2\) is a rotation and \(f(z) + f(-z) = \lambda z\). Noting that \(F\) is an even function, \(f(z) = \lambda z^2 + f_o(z)\) for \(f_o\) some odd function.

Given this, the result follows immediately since \(f(z) = \lambda z^2\).