§8.5 The Great Picard, Bloch, Landau, and Schottky Theorems

Without loss of generality, assume ${𝑧}_0=0$ and that $𝑓$ omits the values $0$ and $1$ (otherwise, consider $𝑧\mapsto \frac{1}{𝛽-𝛼}\left(𝑓\left(𝑧+{𝑧}_0\right)-𝛼\right)$, where $𝛼$ and $𝛽$ are the omitted values). Define the family

of holomorphic functions on ${𝐷}^{\ast }\left(0,𝛿\right)$. Since $𝑓$ omits $0$ and $1$, each element of $\mathrm{ℱ︀}$ does as well. By the Fundamental Normality Test (Theorem 8.4.7$\text{<a href="../../pdf/notes.pdf\#loc-525">Theorem 8.4.7</a>}\text{<a href="../spherical-generalization-of-normal-families/index.html\#thm-theorem-8-4-7">Theorem 8.4.7</a>}$), $\mathrm{ℱ︀}$ is spherically normal. Thus, there exists a subsequence ${\left\{{𝑓}_{{𝑛}_{𝑘}}\right\}}_{𝑘\in \mathbb{N}}\subseteq \mathrm{ℱ︀}$ that converges locally uniformly on ${𝐷}^{\ast }\left(0,𝛿\right)$ in the spherical metric. By Proposition 8.4.5$\text{<a href="../../pdf/notes.pdf\#loc-523">Proposition 8.4.5</a>}\text{<a href="../spherical-generalization-of-normal-families/index.html\#thm-proposition-8-4-5">Proposition 8.4.5</a>}$, this subsequence converges locally uniformly either to a holomorphic function on ${𝐷}^{\ast }\left(0,𝛿\right)$ or to $\infty$ thereon.

1. Suppose ${\left\{{𝑓}_{{𝑛}_{𝑘}}\right\}}_{𝑘\in \mathbb{N}}$ converges locally uniformly to a holomorphic function on ${𝐷}^{\ast }\left(0,𝛿\right)$. Then ${\left\{{𝑓}_{{𝑛}_{𝑘}}\right\}}_{𝑘\in \mathbb{N}}$ is uniformly bounded on $𝜕𝐷\left(0,\frac{𝛿}{2}\right)$. Hence, there exists $𝑀>0$ such that

   $$\left|𝑓\left(\frac{𝑧}{{𝑛}_{𝑘}}\right)\right|=\left|{𝑓}_{{𝑛}_{𝑘}}\left(𝑧\right)\right|<𝑀\forall 𝑧\in 𝜕𝐷\left(0,\frac{𝛿}{2}\right),𝑘\in \mathbb{N}.$$

   In other words, $𝑓$ is bounded by $𝑀$ on every circle $𝜕𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘}}\right)$ for $𝑘\in \mathbb{N}$. By the Maximum Modulus Principle (Theorem 3.4.2$\text{<a href="../../pdf/notes.pdf\#loc-187">Theorem 3.4.2</a>}\text{<a href="../../complex-integration/further-properties-of-holomorphic-functions/index.html\#thm-theorem-3-4-2">Theorem 3.4.2</a>}$), $𝑓$ is then bounded by $𝑀$ on each annulus $\overline{𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘}}\right)}\setminus 𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘+1}}\right)$ for $𝑘\in \mathbb{N}$. As

   $$\bigcup _{𝑘\in \mathbb{N}}\overline{𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘}}\right)}\setminus 𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘+1}}\right)=\overline{𝐷\left(0,\frac{𝛿}{2{𝑛}_1}\right)}\setminus \left\{0\right\},$$
   it follows that $𝑓$ is bounded on ${𝐷}^{\ast }\left(0,\frac{𝛿}{2}\right)$. By Riemann’s Removable Singularity Theorem (Theorem 3.2.9$\text{<a href="../../pdf/notes.pdf\#loc-135">Theorem 3.2.9</a>}\text{<a href="../../complex-integration/analyticity-and-holomorphy/index.html\#thm-theorem-3-2-9">Theorem 3.2.9</a>}$), $𝑓$ therefore extends holomorphically to $0$.

2. Suppose ${\left\{{𝑓}_{{𝑛}_{𝑘}}\right\}}_{𝑘\in \mathbb{N}}$ converges locally uniformly to $\infty$ on ${𝐷}^{\ast }\left(0,𝛿\right)$. Then, for every $𝜀>0$, there exists $𝑁\in \mathbb{N}$ such that, for all $𝑘>𝑁$,

   $$\left|\frac{1}{𝑓\left(\frac{𝑧}{{𝑛}_{𝑘}}\right)}\right|=\left|\frac{1}{{𝑓}_{{𝑛}_{𝑘}}\left(𝑧\right)}\right|<𝜀\forall 𝑧\in 𝜕𝐷\left(0,\frac{𝛿}{2}\right).$$

   By the same reasoning as in the previous case, $\left|\frac{1}{𝑓}\right|<𝜀$ on

   $$\bigcup _{𝑘>𝑁}\overline{𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘}}\right)}\setminus 𝐷\left(0,\frac{𝛿}{2{𝑛}_{𝑘+1}}\right)\supset {𝐷}^{\ast }\left(0,\frac{𝛿}{2{𝑛}_{𝑁+1}}\right).$$
   Thus, by the definition of the limit, ${lim}_{𝑧\to 0}\frac{1}{𝑓\left(𝑧\right)}=0$, so $𝑓$ has a pole at $0$.

In either case, we have derived a meromorphic continuation of $𝑓$ to $0$, contradicting the assumption that $0$ is an essential singularity of $𝑓$. ⁠$\square$

Let $𝑔\left(𝑧\right)=𝑓\left(\frac{1}{𝑧}\right)$ with an isolated singularity at $0$ and a removable singularity at $\infty$. By Picard’s Great Theorem (Theorem 8.5.1$\text{<a href="../../pdf/notes.pdf\#loc-529">Theorem 8.5.1</a>}\text{<a href="\#thm-theorem-8-5-1">Theorem 8.5.1</a>}$), $𝑔\left(𝑧\right)$ has a meromorphic extension to $𝑧=0$. If $𝑧=0$ is removable, by virtue of Proposition 4.3.1$\text{<a href="../../pdf/notes.pdf\#loc-241">Proposition 4.3.1</a>}\text{<a href="../../weierstrass-theory/entireness-and-meromorphy/index.html\#thm-proposition-4-3-1">Proposition 4.3.1</a>}$ and Theorem 3.2.5$\text{<a href="../../pdf/notes.pdf\#loc-129">Theorem 3.2.5</a>}\text{<a href="../../complex-integration/analyticity-and-holomorphy/index.html\#thm-theorem-3-2-5">Theorem 3.2.5</a>}$, the constancy of $𝑔$ and $𝑓$ follows.

If instead $𝑧=0$ is a pole of $𝑔$, then $𝑧=\infty$ is a pole of $𝑓$, and hence $𝑓$ is a polynomial. Assume, for the sake of contradiction that $𝑓$ is non-constant. Then $\forall 𝑤\in ℂ$, the Fundamental Theorem of Algebra (Theorem 3.3.1$\text{<a href="../../pdf/notes.pdf\#loc-173">Theorem 3.3.1</a>}\text{<a href="../../complex-integration/zeros-of-a-holomorphic-function/index.html\#thm-theorem-3-3-1">Theorem 3.3.1</a>}$) gives the existence of some $𝑧\in ℂ$ such that $𝑓\left(𝑧\right)=𝑤$. Hence, $𝑓$ attains every value $𝑤\in ℂ$. This contradicts the statement and hence $𝑓$ is constant. ⁠$\square$

For $𝑤\in 𝑓\left(𝔻\right)$, let $𝜙\left(𝑤\right)$ denote the radius of the largest schlicht disk in $𝑓\left(𝔻\right)$ centered at $𝑤$ (it is mapped to univalently by $𝑓$ on some subdomain). Trivially, $𝜙$ is ${𝐶}^0$ and vanishes toward the boundary of $𝑓\left(𝔻\right)$.

Define the metric

for $𝑤\in 𝑓\left(𝔻\right)$ and $𝜌\left(𝑤\right)\ne 0$, where ${𝐴}^2>{sup}_{𝑤}𝜙\left(𝑤\right)$ is a constant. We may assume that ${sup}_{𝑤}𝜙\left(𝑤\right)$ is finite, since otherwise the theorem is already proved for $𝑓$.

For every point ${𝑤}_0\in 𝑓\left(𝔻\right)$, the bounding circle corresponding to $𝜙\left({𝑤}_0\right)$ passes through a (at least one) boundary point, denoted by $𝑏={𝑏}_{{𝑤}_0}$. Let ${𝜙}_{{𝑤}_0}\left(𝑤\right)=\left|𝑤-𝑏\right|$ and let

By the definition of $𝜙$, we have ${𝜙}_{{𝑤}_0}\ge 𝜙$ everywhere in this neighborhood. Since ${𝜌}_{{𝑤}_0}={\left({𝜙}_{{𝑤}_0}\mapsto \sqrt{{𝜙}_{{𝑤}_0}}\right)}^{\ast }\left(𝜑\mapsto \frac{2𝐴}{{𝐴}^2-{𝜑}^2}\right)$ is the pullback of the hyperbolic metric in (8.3.4)$\text{<a href="../../pdf/notes.pdf\#loc-511">(8.3.4)</a>}\text{<a href="../schwarz-pick-ahlfors/index.html\#loc-4">(8.3.4)</a>}$, the metric ${𝜌}_{{𝑤}_0}\left(𝑤\right)$ has the constant negative curvature of $-1$.

Our goal is to construct ${𝜌}_{{𝑤}_0}$ so that it is the function of a supporting metric for $𝜌$ (satisfies the criteria for Theorem 8.3.3$\text{<a href="../../pdf/notes.pdf\#loc-510">Theorem 8.3.3</a>}\text{<a href="../schwarz-pick-ahlfors/index.html\#thm-theorem-8-3-3">Theorem 8.3.3</a>}$). For ${𝜌}_{{𝑤}_0}\le 𝜌$ to be satisfied, we consider

In particular, we want

to be increasing on $\left[0,𝜙\left({𝑤}_0\right)+𝛿\right]$ for arbitrary $𝛿>0$. The function itself can be calculated to be increasing for $𝑡\le \frac{{𝐴}^2}{3}$ by elementary methods (using derivative tests). Therefore, the conditions for a supporting metric are satisfied if $\frac{{𝐴}^2}{3}\ge {sup}_{𝑤\in 𝑓\left(𝔻\right)}𝜙\left(𝑤\right)+𝛿\ge 𝜙\left({𝑤}_0\right)+𝛿$. Without loss of generality we let $𝛿\to 0^{+}$ and thus, under the condition that $\frac{{𝐴}^2}{3}>{sup}_{𝑤}𝜙\left(𝑤\right)$, Theorem 8.3.3$\text{<a href="../../pdf/notes.pdf\#loc-510">Theorem 8.3.3</a>}\text{<a href="../schwarz-pick-ahlfors/index.html\#thm-theorem-8-3-3">Theorem 8.3.3</a>}$ gives that

Let $𝑧=0$, $𝑤=𝑓\left(0\right)$, so that by the theorem conditions, $\left|\frac{\mathrm{d}𝑤}{\mathrm{d}𝑧}\right|=1$, and therefore

By the previous assumptions on $𝐴$, the corresponding function on the right-hand side is increasing, and since $𝜙\left(𝑓\left(0\right)\right)\le {sup}_{𝑤}𝜙\left(𝑤\right)$, we have

As $\frac{{𝐴}^2}{3}\to {sup}_{𝑤}{𝜙\left(𝑤\right)}^{+}$ ($𝐴$ was chosen arbitarily, so this is valid), it follows that

Consider conformal map ${𝜁}_1\left(𝑤\right):ℂ\setminus \left[0,1\right]\to ℂ\setminus \overline{𝔻}$, which extends to $0$, $1$, and $\infty$ continuously such that ${𝜁}_1\left(\infty \right)=\infty$, ${𝜁}_1\left(1\right)=1$, ${𝜁}_1\left(0\right)=-1$. Explicitly, we have the relationship

as an affine transformation of the inverse Joukowski transform (inverse of $𝑧\mapsto 𝑧+{𝑧}^{-1}$). The solution is given by