# Schwarz

## Hyperbolic Translations / Blaschke Factors

For $$a\in {\mathbb{D}}$$, the maps \begin{align*} \psi_a \coloneqq{a-z\over 1-\overline{a}z} \end{align*} are hyperbolic translations because they preserve the hyperbolic metric on the Poincaré disc. They’re also commonly called Blaschke factors, and also sometimes taken to be \begin{align*} \phi_a \coloneqq{z-a \over 1-\overline{a} z} = - \psi_a .\end{align*} A rational map of the form \begin{align*} \Psi_{\mathbf{a}}(z) = \lambda \prod_{1\leq k\leq n} \psi_a(z) = \lambda \prod_{1\leq k \leq n} {a_i - z\over 1 - \overline{a_i} z},\qquad \mathbf{a}\coloneqq{\left[ {a_1,\cdots, a_n} \right]} \end{align*} with zeros $$a_i \in {\mathbb{D}}$$ is called a Blaschke product and is a map $${\mathbb{D}}\to {\mathbb{D}}$$ that preserves $$S^1$$.

Some useful properties:

• $$\psi_a \in \mathop{\mathrm{Aut}}({\mathbb{D}})$$
• $$\psi_a(S^1) = S^1$$
• $$\psi_a(0) = a$$ and $$\psi(a) = 0$$
• With this choice of sign, $$\psi_{a} ^{-1}= \psi_{a}$$, so $$\psi_a^2 = \operatorname{id}$$.
• $$\psi_a'(z) = {{\left\lvert {a} \right\rvert}^2 - 1 \over \qty{1-\overline{a} z}^2 }$$
• $$\psi_a(\lambda z) = \lambda\psi_{\overline{\lambda }a}(z)$$

Inverting: set $$f(z) = w$$ and solve for $$z$$: \begin{align*} {a-z \over 1 - \overline{a}z} &= w \\ \implies a-z - w(1-\overline{a} z) &= 0 \\ \implies z&= {w-a \over \overline{a} w - 1} = {a-w\over 1-\overline{a} w} .\end{align*}

Differentiating: the quotient rule \begin{align*} \psi'_a(z) = {-(1-\overline{a} z) + \overline{a}(a-z) \over \qty{1-\overline{a} z}^2} = {-1 + {\left\lvert {a} \right\rvert}^2 \over \qty{1-\overline{a} z}^2} .\end{align*}

Scaling: use a fun trick, insert $$1=\overline{\lambda }\lambda$$ like so \begin{align*} \psi_a(\lambda z) &= {a - \lambda z \over 1 - \overline{a} \lambda z}\\ &= {\lambda\overline{\lambda }a - \lambda z \over 1 - \overline{a} \lambda z} \\ &= \lambda {\overline{\lambda }a - z \over 1 - \overline{\overline{\lambda }a} z} \\ &= \lambda \psi_{\overline{\lambda }a}(z) .\end{align*}

Being an involution: check $$\psi_a(\psi_a(z))$$ satisfies the Schwarz lemma and has two fixed points, forcing it to be the identity.

Every map $$g\in \mathop{\mathrm{Aut}}({\mathbb{D}})$$ is of the form

\begin{align*} \mathop{\mathrm{Aut}}({\mathbb{D}}) = \left\{{ \lambda \psi_a(z) {~\mathrel{\Big\vert}~}a\in {\mathbb{D}}, \lambda \in S^1 }\right\} ,\end{align*} i.e. they are all Blaschke factors and rotations.

• That these maps are biholomorphisms: they’re compositions of $$z\mapsto \lambda z$$ and $$z\mapsto {z-a\over 1-\overline{a} z}$$, which are biholomorphisms.

• Let $$f \in \mathop{\mathrm{BiHol}}(\Delta)$$ be arbitrary, fix $$a\in \Delta$$ with $$f(a) = 0$$

• Write $$M(z) = {z-a\over 1-\overline{a} z}$$, then note that $$M(a) = 0$$ and this is a biholomorphism.

• $$g\coloneqq f\circ M^{-1}\in \mathop{\mathrm{BiHol}}(\Delta)$$ sends $$0\to0$$ and is thus a rotation, so $$g(z) = \lambda z$$.

• Write $$g\circ M = f \circ M \circ M^{-1}= f$$, which exhibits $$f$$ in the desired form.

• Claim: this representation is unique. Consider $$f'(z)$$, this determines $$\operatorname{Arg}(\lambda)$$.

## The Schwarz Lemma

If $$f: {\mathbb{D}}\to {\mathbb{D}}$$ is holomorphic with $$f(0) = 0$$, then

• $${\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$$ for all $$z\in {\mathbb{D}}$$
• $${\left\lvert {f'(0)} \right\rvert} \leq 1$$.

Moreover, if

• $${\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}$$ for any $$z_0\in {\mathbb{D}}\setminus\left\{{0}\right\}$$, or
• $${\left\lvert {f'(0)} \right\rvert} = 1$$,

then $$f$$ is a rotation, i.e. $$f(z) = \lambda z$$ for some $${\left\lvert {\lambda} \right\rvert} = 1$$.

Let $$f:{\mathbb{D}}\to {\mathbb{D}}$$ be holomorphic with $$f(0) = 0$$. Then either

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

• Idea: apply the maximum modulus principle to $$g(z) \coloneqq f(z)/z$$.
• $${\left\lvert {g(z)} \right\rvert} \leq 1$$:
• Expand $$f$$ at $$z=0$$ as $$\sum_{k\geq 0} c_k z^k$$. Since $$f(0) = c_0$$, we have $$c_0 = 0$$.
• So $$g(z) \coloneqq f(z)/z$$ is holomorphic on $${\mathbb{D}}$$, since the singularity at $$z=0$$ is removable.
• Set $${\left\lvert {z} \right\rvert} = r < 1$$, then $${\left\lvert {g(z)} \right\rvert} = {\left\lvert {f(z)} \right\rvert}/r \leq 1/r$$ since $${\left\lvert {f(z)} \right\rvert} \leq 1$$.
• By MMP, $${\left\lvert {g(z)} \right\rvert} \leq 1/r$$ holds in the entire disc $${\left\lvert {z} \right\rvert} \leq r$$, so take $$r\to 1$$ to get $${\left\lvert {g(z)} \right\rvert} \leq 1$$
• $${\left\lvert {f'(0)} \right\rvert} \leq 1$$ with equality iff $$f$$ is a rotation:
• Note that $$f(0) = 0$$, so we can write $$g(0) = \lim_{z\to 0} {f(z) - f(0) \over z-0} \coloneqq f'(0)$$.
• So $$1 = {\left\lvert {f'(0)} \right\rvert} = {\left\lvert {g(0)} \right\rvert}$$.
• But $${\left\lvert {g(z)} \right\rvert} \leq 1$$ on $${\mathbb{D}}$$ and $$g(z) = 1$$ in the interior, so by MMP this makes $$g$$ constant.
• So again $$f(z) = cz$$ with $${\left\lvert {c} \right\rvert} = 1$$.
• $${\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert}\implies f$$ is a rotation:
• Again $${\left\lvert {g(z)} \right\rvert} \leq 1$$, but $${\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {z_0} \right\rvert} \implies {\left\lvert {g(z_0)} \right\rvert} = 1$$, so $$g$$ attains a maximum on $${\left\lvert {z} \right\rvert}\leq 1$$, making it constant, so $$f(z) = cz$$.
• Then $${\left\lvert {z_0} \right\rvert} = {\left\lvert {f(z_0)} \right\rvert} = {\left\lvert {cz_0} \right\rvert}$$ since $$f(z_0) = z_0$$, so $${\left\lvert {c} \right\rvert} = 1$$ and $$c = e^{i\theta}$$ for some $$\theta$$.

# Exercises

## Blaschke Factors

Let $$f: {\mathbb{D}}\to {\mathbb{D}}$$ with $$\left\{{a_k}\right\}_{k\leq n}$$ the zeros of $$f$$ in $${\mathbb{D}}$$. Show that \begin{align*} {\left\lvert {f(z)} \right\rvert}\leq \prod_{k\leq n} {\left\lvert { \psi_{a_k}(z) } \right\rvert} .\end{align*}

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*}

## Schwarz-Fu

If $$f: {\mathbb{D}}_R(a)\to{\mathbb{D}}_M(0)$$ with $$f(a) = 0$$, then \begin{align*} {\left\lvert {f(z)} \right\rvert}\leq {M\over R}{\left\lvert {z-a} \right\rvert} .\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*}

Let $$\psi_a$$ be a Blaschke factor and use the Schwarz lemma to prove that $$\psi_a \circ \psi_a = \operatorname{id}_{\mathbb{D}}$$.

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.

Prove the following: suppose $${\left\lvert {f(z)} \right\rvert}\leq 1$$, then for all $$z, w\in {\mathbb{D}}$$, \begin{align*} \left|\frac{f(z)-f(w)}{1-\overline{f(w)} f(z)}\right| \leq\left|\frac{z-w}{1-\overline{w} z}\right| \quad\text{ and } \left|f^{\prime}(z)\right| \leq \frac{1-|f(z)|^{2}}{1-|z|^{2}} .\end{align*} If equality holds for some $$z\neq w$$ in either expression, then $$f= \lambda F$$ where $$F$$ is a linear fractional transformation and $${\left\lvert {\lambda} \right\rvert} = 1$$, so $$f\in \mathop{\mathrm{Aut}}({\mathbb{D}})$$.

Note that this does not require $$f(0) = 0$$.

Show that if $$f: \Delta\to\Delta$$ is a biholomorphism with $$f(0) = 0$$ then $$f$$ is a rotation.

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.

Show that if $$f:{\mathbb{H}}\to {\mathbb{D}}$$ is holomorphic and $$f(i) = 0$$ then $${\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z-i\over z+i} \right\rvert}$$.

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 $${\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}$$.

Suppose $$f:{\mathbb{D}}\to {\mathbb{D}}$$ with $$f(0) = 0$$ and that there exists an $$r\in (0, 1)$$ with $$f(r) = f(-r) = 0$$. Show that \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z^2-r^2 \over 1-r^2 z^2} \right\rvert} .\end{align*}

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.

Use a version of the Schwarz lemma to prove Liouville’s theorem.

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.

Let $$f:{\mathbb{D}}\to {\mathbb{D}}$$ with $$f(0) = f'(0) = 0$$. Show that $${\left\lvert {f''(0)} \right\rvert} \leq 2$$ and describe all $$f$$ for which this is an equality.

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 $$g({\mathbb{D}}) = {\mathbb{D}}$$.

Suppose $$f:{\mathbb{D}}\to {\mathbb{D}}$$ with $$f(a) = a$$ a fixed point where $${\left\lvert {f'(a)} \right\rvert} < 1$$. Show that for any initial point $$z_0$$, the sequence $$z_k \coloneqq f(z_{k-1})$$ converges to $$a$$.

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

For any choice of $$z_0\in {\mathbb{D}}$$, there is an $$r$$ with $$0< {\left\lvert {z_0} \right\rvert} < r < 1$$ and a constant $$C<1$$ such that $${\left\lvert {f(z)} \right\rvert} \leq C{\left\lvert {z} \right\rvert}$$ for $${\left\lvert {z} \right\rvert} < r$$.

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.

The claim is that for any given $$r$$, the constant $$C\coloneqq M/r$$ works, where $$M\coloneqq\max_{{\left\lvert {z} \right\rvert} = r} {\left\lvert {f(z)} \right\rvert}>0$$. The scaled Schwarz lemma gives $${\left\lvert {f(z)} \right\rvert}\leq {M\over r}{\left\lvert {z} \right\rvert} = C{\left\lvert {z} \right\rvert}$$, and $${\left\lvert {C} \right\rvert} \leq 1$$ since $${\left\lvert {M} \right\rvert} \leq r$$, which follows because $${\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}$$ on $${\mathbb{D}}$$ itself.

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*}

Show that the only holomorphic map $$f:{\mathbb{D}}\to {\mathbb{D}}$$ two distinct fixed points $$a\neq b$$ is the identity.

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 $$\lambda = 1$$.

## Estimating

Let $$f\in \mathop{\mathrm{Hol}}({\mathbb{D}})$$. Show that if $$f$$ has a fixed point $$a$$ then $${\left\lvert {f'(a)} \right\rvert} \leq 1$$, and that \begin{align*} {\left\lvert {f(0)} \right\rvert}^2 + {\left\lvert {f'(0)} \right\rvert}^2 \leq 1 .\end{align*}

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*}

Does there exist a map $$f: {\mathbb{D}}\to {\mathbb{D}}$$ with

• $$f\qty{1\over 2} = {3\over 4}$$
• $$f'\qty{1\over 2} = {2\over 3}$$

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.

Suppose $$f: {\mathbb{D}}\to {\mathbb{D}}$$ with $$f(0) = 0$$ and $${\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {e^z} \right\rvert}$$ when $${\left\lvert {z} \right\rvert} = 1$$. Find an upper bound for $$f\qty{1+i\over 2}$$.

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*}

Suppose $$f:{\mathbb{H}}\to {\mathbf{C}}$$ with $${\left\lvert {f(z)} \right\rvert}< 1$$ and $$f(i) = 0$$. Find an upper bound for $$f(2i)$$.

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*}

Use Rouché’s theorem and the Schwarz lemma to prove the following: if $$f:{\mathbb{D}}\to \overline{ {\mathbb{D}}}$$ is holomorphic with $$f(0) = 0$$, then there are exactly $$m$$ solutions (counted with multiplicity) to $$f(z) = (2z)^m$$ in the disc $${\left\lvert {z} \right\rvert} < 1/2$$.

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 $$R = {1\over c}$$.

Suppose $$f$$ is holomorphic and $${\left\lvert {f(z)} \right\rvert}\leq 1$$ for $$\Re(z) > 0$$ with $$f(1) = 0$$. Find an upper bound for $$f(2)$$.

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*}

Suppose $$f:{\mathbb{D}}\to {\mathbb{D}}$$ and $$f(0) = 0$$. Show that \begin{align*} {\left\lvert {f(z) + f(-z)} \right\rvert} \leq 2{\left\lvert {z} \right\rvert}^2 ,\end{align*} with equality only if $$f(z) = \lambda z^2$$ for some $${\left\lvert {\lambda} \right\rvert} = 1$$.

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.

$$f_0$$ is identically zero.

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

Note that on the LHS, $${\left\lvert { f(re^{it})} \right\rvert} \to 1$$ as $$r\to 1$$ for any $$t$$. So this must be true on the RHS, and the first term $$\lambda {\left\lvert { re^{it}} \right\rvert}^2 \to 1$$, forcing $${\left\lvert {h(re^{it})} \right\rvert} \to 0$$. So $$h\equiv 0$$ on $${\left\lvert {z} \right\rvert} = 1$$, and by the MMP, $${\left\lvert {h} \right\rvert} = 0$$ in $${\mathbb{D}}$$, making $$h\equiv 0$$ on $${\mathbb{D}}$$.