# Schwarz Lemma

## Fall 2020.4 (Schwarz double root) #stuck#complex/qual/completed

Let $$\mathbb{D}:=\{z:|z|<1\}$$ denote the open unit disk. Suppose that $$f(z): \mathbb{D} \rightarrow \mathbb{D}$$ is holomorphic, and that there exists $$a \in \mathbb{D} \backslash\{0\}$$ such that $$f(a)=f(-a)=0$$.

• Prove that $$|f(0)| \leq|a|^{2}$$.

• What can you conclude when $$|f(0)|=|a|^{2} ?$$

Part 1:

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

$${\left\lvert {g(z)} \right\rvert}\leq 1$$ on $${\mathbb{D}}$$.

$${\left\lvert {\psi_a(z)} \right\rvert} = 1$$ on $${{\partial}}{\mathbb{D}}$$, so $$\lim_{r\to 1}\psi_a(re^{it}) = 1$$ for any fixed $$t$$. Then for any $$f$$ with $${\left\lvert {f} \right\rvert} \leq 1$$ in $${\mathbb{D}}$$, \begin{align*} {\left\lvert {f(re^{it} ) \over \psi_a(re^{it} ) } \right\rvert} \leq {1\over \psi_a(re^{it})} \leq {1\over \sup_{t} \psi_a(re^{it}) } \overset{r\to 1}\longrightarrow 1 .\end{align*} So apply this to $$f=g$$ and $$f={g\over \psi_a}$$ to get it for $${f\over \psi_a \psi_{-a}}.$$

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.

## Fall 2021.5 #complex/qual/completed

Assume $$f$$ is an entire function such that $$|f(z)|=1$$ on $$|z|=1$$. Prove that $$f(z)=e^{i \theta} z^{n}$$, where $$\theta$$ is a real number and $$n$$ a non-negative integer.

Suggestion: First use the maximum and minimum modulus theorem to show \begin{align*} f(z)=e^{i \theta} \prod_{k=1}^{n} \frac{z-z_{k}}{1-\overline{z_{k}} z} \end{align*} if $$f$$ has zeros.

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

If we assume $$f$$ is continuous on $${\mathbb{D}}$$, then $${\sharp}Z(f) < \infty$$

Suppose $${\sharp}Z(f) = \infty$$, then by compactness of $$\overline{{\mathbb{D}}}$$ there is a limit point $$z_0$$. If $$z_0 \in {\mathbb{D}}$$, then there is a sequence $$\left\{{z_k}\right\}\to z_0$$ with $$f(z_k) = 0$$ for every $$k$$, so $$f$$ is zero on a set $$S\coloneqq\left\{{z_k}\right\}_{k\geq 1} \cup\left\{{z_0}\right\}$$ with an accumulation point and this forces $$f\equiv 0$$ on $$\overline{{\mathbb{D}}}$$ by the identity principle, contradicting $${\left\lvert {f} \right\rvert} = 1$$ on $${{\partial}}{\mathbb{D}}$$>

Otherwise, if $$z_0\in {{\partial}}{\mathbb{D}}$$, using continuity of $$f$$ we have $$f(z_k) = 0$$ for all $$k$$ and $$z_k\to z_0$$ so $$f(z_0) = 0$$, again contradiicting $${\left\lvert {f} \right\rvert} = 1$$ on $${{\partial}}{\mathbb{D}}$$.

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 - \overline{z_k} z}, \quad h(z) \coloneqq{f(z) \over g(z)} .\end{align*}

$$h(z) \equiv 1$$ is constant on $$\overline{{\mathbb{D}}}$$, so that $$f = \lambda g$$ for some $$\lambda \in S^1$$, i.e. $$\lambda = e^{i\theta}$$ for some $$\theta$$.

Note that $$h$$ cancels all zeros of $$f$$, so $$h$$ is nonzero and holomorphic on $$\overline{{\mathbb{D}}}$$. Moreover $${\left\lvert {g(z)} \right\rvert} \leq 1$$ on $$\overline{{\mathbb{D}}}$$ since these are well-known to be in $$\mathop{\mathrm{Aut}}({\mathbb{D}})$$. It’s also well-known that $${\left\lvert {g(z)} \right\rvert} = 1$$ on $${{\partial}}{\mathbb{D}}$$. Thus $${\left\lvert {h(z)} \right\rvert} = 1$$ and $${\left\lvert {1\over h(z)} \right\rvert} =1$$ on $${{\partial}}{\mathbb{D}}$$, and by the maximum modulus principle, \begin{align*} {\left\lvert {h(z)} \right\rvert} \leq 1 \quad\text{ and }\quad {\left\lvert {1\over h(z)} \right\rvert}\leq 1 \quad \text{ on } {\mathbb{D}} ,\end{align*} forcing $${\left\lvert {h(z)} \right\rvert}\equiv 1$$ and thus $$h(z) = e^{i\theta}$$ for some $$\theta$$.

So we now have \begin{align*} f(z) = e^{i\theta} \prod_{1\leq k\leq m} {z-z_k \over 1 -\overline{z_k} z} ,\end{align*} which has poles at points $$z$$ for which $$\overline{z_k}z=1$$ 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*}

## Fall 2021.6 (Schwarz manipulation) #complex/qual/completed

Show that if $$f: D(0, R) \rightarrow \mathbb{C}$$ is holomorphic, with $$|f(z)| \leq M$$ for some $$M>0$$, then \begin{align*} \left|\frac{f(z)-f(0)}{M^{2}-\overline{f(0)} f(z)}\right| \leq \frac{|z|}{M R} . \end{align*}

The strategy:

• Write the RHS as $$a$$. Note that we need to get rid of the $$M^2$$ on the LHS, so keep the $$M$$ around and write $$a \coloneqq z/R$$ so $$z = aR$$.
• Make the substitution to get \begin{align*} {\left\lvert {f(aR) - f(0) \over M^2 - \overline{f(0)} f(aR) } \right\rvert} \leq M^{-1}{\left\lvert {a} \right\rvert} \\ \implies {\left\lvert {M\qty{ f(aR) - f(0)} \over M^2 - \overline{f(0)} f(aR) } \right\rvert} \leq {\left\lvert {a} \right\rvert} \\ {\left\lvert { f(aR)/M - f(0)/M \over 1 - \overline{f(0)} f(aR)/M^2 } \right\rvert} \leq {\left\lvert {a} \right\rvert} .\end{align*}
• Recognize the LHS as $$\psi_w(g(a))$$ for $$w\coloneqq f(0)/M$$ and $$g(a) \coloneqq f(aR)/M$$.

Proof due to Swaroop Hegde!

Fix $$R, M$$ and make a clever choice: define \begin{align*} F: {\mathbb{D}}&\to {\mathbf{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-\overline{a} 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 - {\overline{a} f(Rz) \over M} } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert {f(Rz) - f(0) \over 1 - {\overline{f(0)} f(Rz) \over M^2 } } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert {f(Rz) - f(0) \over M^2 - \overline{f(0)} f(Rz) } \right\rvert} \leq {{\left\lvert {z} \right\rvert} \over M} \\ \\ \iff & {\left\lvert {f(w) - f(0) \over M^2 - \overline{f(0)} 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).

## Scaling Schwarz #complex/exercise/completed

Let $$\overline{B}(a, r)$$ denote the closed disc of radius $$r$$ about $$a\in {\mathbf{C}}$$. Let $$f$$ be holomorphic on an open set containing $$\overline{B}(a, r)$$ and let \begin{align*} M \coloneqq\sup_{z\in \overline{B}(a, r)} {\left\lvert {f(z)} \right\rvert} .\end{align*}

Prove that \begin{align*} z\in \overline{B}\qty{a, {r\over 2}},\,z\neq a, \qquad {{\left\lvert { f(z) - f(a)} \right\rvert} \over {\left\lvert {z-a} \right\rvert}} \leq {2M \over r} .\end{align*}

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*}

## Bounding derivatives #complex/exercise/completed

Suppose $$f: {\mathbb{D}}\to {\mathbb{H}}$$ is analytic and satisfies $$f(0) = 2$$. Find a sharp upper bound for $${\left\lvert {f'(0)} \right\rvert}$$, and prove it is sharp by example.

Some useful facts about the Cayley map:

• $$C(z) \coloneqq{z-i\over z+i}$$ maps $${\mathbb{H}}\to {\mathbb{D}}$$ sending $$i\to 0$$.
• $$C^{-1}(z) \coloneqq-i {z+1\over z-1}$$ maps $${\mathbb{D}}\to{\mathbb{H}}$$ sending $$0\to i$$.
• $$C'(z) = {2i\over (z+i)^2}$$ and $$C'(i) = -{1\over 2}i$$.
• $$(C^{-1})'(z) = {2i\over (z-1)^2}$$ and $$C'(0) = 2i$$.
• A mistake that’s useful to know: $$\psi_w'(z) = {1-{\left\lvert {w} \right\rvert}^2 \over (1-\overline{w}z )^2}$$ and $$\psi_w'(w) \to \infty$$.

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*}

## Schwarz for higher order zeros #complex/exercise/completed

Suppose $$f:{\mathbb{D}}\to{\mathbb{D}}$$ is analytic, has a single zero of order $$k$$ at $$z=0$$, and satisfies $$\lim_{{\left\lvert {z} \right\rvert} \to 1} {\left\lvert {f(z)} \right\rvert} = 1$$. Give with proof a formula for $$f(z)$$.

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 $$z\mapsto 1/\overline{f(1/\overline{z})}$$. This is bounded and entire, thus constant, making $$g$$ constant.

## Schwarz with an injective function #complex/exercise/completed

Suppose $$f, g: {\mathbb{D}}\to \Omega$$ are holomorphic with $$f$$ injective and $$f(0) = g(0)$$.

Show that \begin{align*} \mathop{\mathrm{\forall}}0 < r < 1,\qquad g\qty{\left\{{{\left\lvert {z} \right\rvert} < r}\right\}} \subseteq f\qty{\left\{{{\left\lvert {z} \right\rvert} < r}\right\}} .\end{align*}

The first part of this problem asks for a statement of the Schwarz lemma.

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 {\mathbf{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 $$g({\mathbb{D}}_r(0)) \subseteq f({\mathbb{D}}_r(0))$$.

## Reflection principle #complex/exercise/completed

Let $$S\coloneqq\left\{{z\in {\mathbb{D}}{~\mathrel{\Big\vert}~}\Im(z) \geq 0}\right\}$$. Suppose $$f:S\to {\mathbf{C}}$$ is continuous on $$S$$, real on $$S\cap{\mathbf{R}}$$, and holomorphic on $$S^\circ$$.

Prove that $$f$$ is the restriction of a holomorphic function on $${\mathbb{D}}$$.

Define a function \begin{align*} F(z) \coloneqq \begin{cases} f(z) & \Im(z)\geq 0 \\ \overline{f(\overline{z})} & \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 $$w_0 = \overline{z_0}$$ for some $$z_0\in S$$, then \begin{align*} f(z) &= \sum_{k\geq 0}c_k (z-z_0)^k \\ \implies \overline{f(\overline{z})} &= \sum_{k\geq 0}\overline{c_k} \overline{\qty{\overline{z} - z_0}^k} \\ &= \sum_{k\geq 0}\overline{c_k} \qty{z - \overline{z_0}}^k \\ &= \sum_{k\geq 0}\overline{c_k} \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 $$\overline{f(\overline{z})} = f(z)$$ for $$z, f(z)\in {\mathbf{R}}$$, $$F$$ is a continuous extension of $$f$$ to $${\mathbb{D}}$$. By the symmetry principle, $$F$$ is holomorphic, and $${ \left.{{F}} \right|_{{S}} } = f$$.

# Blaschke Factors

## Spring 2019.5, Spring 2021.5 (Blaschke contraction) #complex/qual/completed

Let $$f$$ be a holomorphic map of the open unit disc $${\mathbb{D}}$$ to itself. Show that for any $$z, w\in {\mathbb{D}}$$, \begin{align*} \left|\frac{f(w)-f(z)}{1-\overline{f(w)} f(z)}\right| \leq\left|\frac{w-z}{1-\overline{w} z}\right| .\end{align*} Show that this inequality is strict for $$z\neq w$$ except when $$f$$ is a linear fractional transformation from $${\mathbb{D}}$$ to itself.

The Schwarz conjugation trick:

Write the RHS as $$a$$, we then want something in the form $${\left\lvert {F(a)} \right\rvert}\leq {\left\lvert {a} \right\rvert}$$. The choice $$a=\psi_w(z)$$ is forced, so $$z= \psi_w^{-1}(a)$$. This forces the choice for the LHS \begin{align*} { f(w) - (f\circ \psi_w^{-1})(a) \over 1 - \overline{f(w)} (f\circ \psi_w^{-1})(a) } = (\psi_{f(w)} \circ f \circ \psi_w^{-1})(a) \coloneqq F(a) .\end{align*}

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-\overline{a}z} \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 - \overline{f(z_1)} \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 - \overline{f(z_1)}\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 - \overline{f(z_1)}\cdot f(w) } \right\rvert} &\leq {\left\lvert {z_1 - z \over 1 - \overline{z_1} 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_{\overline{\lambda }f(z_1)} \qty{\psi_{z_1}(w)} \\ &\coloneqq\lambda \psi_a(\psi_b(w)) \\ &=\lambda\qty{ a- \psi_b(w) \over 1 - \overline{a} \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.

## Schwarz-Pick derivative #complex/exercise/completed

Suppose $$f:{\mathbb{D}}\to {\mathbb{D}}$$ is analytic. Prove that \begin{align*} \forall a\in {\mathbb{D}}, \qquad {{\left\lvert {f'(a)} \right\rvert} \over 1 - {\left\lvert {f(a)} \right\rvert}^2 } \leq {1 \over 1 - {\left\lvert {a} \right\rvert}^2} .\end{align*}

Holomorphic maps on $${\mathbb{D}}$$ contract Blaschke factors: \begin{align*} {\left\lvert { \psi_w(z) } \right\rvert} \geq {\left\lvert {\psi_{f(w)}(f(z)) } \right\rvert} ,\end{align*} i.e.  \begin{align*} {\left\lvert {f(w) - f(z) \over 1 - \overline{f(w)}f(z)} \right\rvert} \leq {\left\lvert {w-z \over 1-\overline{w} z} \right\rvert} .\end{align*}

Make a change of variables $$a\coloneqq\psi_w(z)$$ so $$z=\psi_w^{-1}(a) = \psi_w(a)$$, then the desired inequality follows if we can show \begin{align*} {\left\lvert { \psi_{f(w)}(f(\psi_w(a))) } \right\rvert} \leq {\left\lvert {a} \right\rvert} .\end{align*}

So define $$F \coloneqq\psi_{f(w)} \circ f \circ \psi_w$$, then since $$\psi_w(0) = w$$, \begin{align*} F(0) = \psi_{f(w)}(f(w)) = 0 .\end{align*} Moreover $${\left\lvert {F(z)} \right\rvert}\leq 1$$ since each constituent is a map $${\mathbb{D}}\to {\mathbb{D}}$$. So $$F$$ satisfies Schwarz and the claim follows.

Given this, there’s just a clever rearrangement to obtain the stated result: \begin{align*} {\left\lvert {f(w) - f(z) \over 1 - \overline{f(w)}f(z)} \right\rvert} &\leq {\left\lvert {w-z \over 1-\overline{w} z} \right\rvert} \\ \implies {\left\lvert { 1\over 1-\overline{f(w)}f(z) } \right\rvert} \cdot {\left\lvert {f(z) - f(w) \over z-w} \right\rvert} &\leq {\left\lvert {1\over 1-\overline{w}z} \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*}

## Schwarz and Blaschke products #complex/exercise/completed

Suppose $$f:{\mathbb{D}}\to{\mathbb{D}}$$ is analytic and admits a continuous extension $$\tilde f: \overline{{\mathbb{D}}}\to \overline{{\mathbb{D}}}$$ such that $${\left\lvert {z} \right\rvert} = 1 \implies {\left\lvert {f(z)} \right\rvert} = 1$$.

• Prove that $$f$$ is a rational function.

• Suppose that $$z=0$$ is the unique zero of $$f$$. Show that \begin{align*} \exists n\in {\mathbb{N}}, \lambda \in S^1 {\quad \operatorname{ such that } \quad}f(z) = \lambda z^n .\end{align*}

• Suppose that $$a_1, \cdots, a_n \in {\mathbb{D}}$$ are the zeros of $$f$$ and prove that \begin{align*} \exists \lambda \in S^1 {\quad \operatorname{such that} \quad} f(z) = \lambda \prod_{j=1}^n {z - a_j \over 1 - \overline{a_j} z} .\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 \overline{f\qty{1/\overline{z}}} } & {\left\lvert {z} \right\rvert} \geq 1 \end{cases} .\end{align*}

Now $$F:{\mathbf{CP}}^1\to {\mathbf{CP}}^1$$ 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 $$\psi_a(z) \coloneqq{a-z\over 1-\overline{a} z}$$. 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 $$f(z) = \lambda \Psi(z)$$.

### Tie’s Extra Questions: Fall 2009 #complex/exercise/completed

Let $$g$$ be analytic for $$|z|\leq 1$$ and $$|g(z)| < 1$$ for $$|z| = 1$$.

• Show that $$g$$ has a unique fixed point in $$|z| < 1$$.

• What happens if we replace $$|g(z)| < 1$$ with $$|g(z)|\leq 1$$ for $$|z|=1$$? Give an example if (a) is not true or give an proof if (a) is still true.

• What happens if we simply assume that $$f$$ is analytic for $$|z| < 1$$ and $$|f(z)| < 1$$ for $$|z| < 1$$? Suppose that $$f(z) \not\equiv z$$. Can f have more than one fixed point in $$|z| < 1$$?

Hint: The map $$\displaystyle{\psi_{\alpha}(z)=\frac{\alpha-z}{1-\overline{\alpha}z}}$$ may be useful.

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 $$\overline{{\mathbb{D}}}$$, it is in particular continuous. By the Brouwer fixed point theorem, every continuous map $$\overline{{\mathbb{D}}} \to \overline{{\mathbb{D}}}$$ 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 $$\overline{{\mathbb{D}}}$$. 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.

### Tie’s Extra Questions: Fall 2015 (Blaschke factor properties) #complex/exercises/completed

• Let $$z, w$$ be complex numbers, such that $$\overline{z} w \neq 1$$. Prove that \begin{align*}{\left\lvert {\frac{w - z}{1 - \overline{w} z}} \right\rvert} < 1 \; \; \; \mbox{if} \; |z| < 1 \; \mbox{and}\; |w| < 1,\end{align*} and also that \begin{align*}{\left\lvert {\frac{w - z}{1 - \overline{w} z}} \right\rvert} = 1 \; \; \; \mbox{if} \; |z| = 1 \; \mbox{or}\; |w| = 1.\end{align*}

• Prove that for fixed $$w$$ in the unit disk $$\mathbb D$$, the mapping \begin{align*}F: z \mapsto \frac{w - z}{1 - \overline{w} z}\end{align*} satisfies the following conditions:

• $$F$$ maps $$\mathbb D$$ to itself and is holomorphic.

• $$F$$ interchanges $$0$$ and $$w$$, namely, $$F(0) = w$$ and $$F(w) = 0$$.

• $${\left\lvert {F(z)} \right\rvert} = 1$$ if $$|z| = 1$$.

• $$F: {\mathbb D} \mapsto {\mathbb D}$$ is bijective.

Hint: Calculate $$F \circ F$$.

## Tie’s Extra Questions: Spring 2015 #complex/exercise/completed

Suppose $$f$$ is analytic in an open set containing the unit disc $$\mathbb D$$ and $$|f(z)| =1$$ when $$|z|$$=1. Show that either $$f(z) = e^{i \theta}$$ for some $$\theta \in \mathbb R$$ or there are finite number of $$z_k \in \mathbb D$$, $$k \leq n$$ and $$\theta \in \mathbb R$$ such that \begin{align*} \displaystyle f(z) = e^{i\theta} \prod_{k=1}^n \frac{z-z_k}{1 - \overline{z}_k z } \, . .\end{align*}

Also cf. Stein et al, 1.4.7, 3.8.17

## Tie’s Extra Questions: Spring 2015 (Equality of modulus) #complex/exercise/completed

Let $$f$$ and $$g$$ be non-zero analytic functions on a region $$\Omega$$. Assume $$|f(z)| = |g(z)|$$ for all $$z$$ in $$\Omega$$. Show that $$f(z) = e^{i \theta} g(z)$$ in $$\Omega$$ for some $$0 \leq \theta < 2 \pi$$.

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

$$F$$ is holomorphic on $$\Omega$$.

Note that $$g(a) = 0$$ iff $$f(a) = 0$$, so $$F$$ has no poles. If $$F$$ has a singularity at $$z_0$$, noting that $${\left\lvert {F(z_0)} \right\rvert} = 1$$, $$F$$ is bounded in a neighborhood of $$z_0$$ and thus the singularity must be removable. By Riemann’s removable singularity theorem, $$F$$ extends to a holomorphic function.

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 $${\mathbf{C}}$$ 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 $$f(z) = \lambda g(z)$$.

# Fixed Points

## Fall 2020.7 #complex/qual/completed

Suppose that $$f: \mathbb{D} \rightarrow \mathbb{D}$$ is holomorphic and $$f(0)=0$$. Let $$n \geq 1$$, and define the function $$f_{n}(z)$$ to be the $$n$$-th composition of $$f$$ with itself; more precisely, let

\begin{align*} f_{1}(z):=f(z), f_{2}(z):=f(f(z)), \text { in general } f_{n}(z):=f\left(f_{n-1}(z)\right) . \end{align*}

Suppose that for each $$z \in \mathbb{D}, \lim _{n \rightarrow \infty} f_{n}(z)$$ exists and equals to $$g(z)$$. Prove that either $$g(z) \equiv 0$$ or $$g(z)=z$$ for all $$z \in D$$.

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.

#stuck #complex/qual/completed #complex/exercise/completed #complex/exercises/completed