\(D=\{z\in \mathbb{C}: |z|<1\}\) is the open unit disc centered at the origin.
Let \(\Omega \neq \mathbb{C}\) be a simply connected region. Then \(\Omega\) is conformally equivalent to \(D\). Moreover, the assumption \(\Omega \neq \mathbb{C}\) is necessary.
Remark.
In view of the Liouville theorem, the assumption \(\Omega \neq \mathbb{C}\) is necessary. Indeed, if \(f:\mathbb C\to D\) is a conformal map, then \(f\) is bounded, since \(|f(z)|<1\) for all \(z\in \mathbb C\). Hence, by the Liouville theorem \(f\) must be constant, but then it cannot be injective.
However, \(\mathbb C\) and \(D\) are homeomorphic. The mapping \[\mathbb C\ni z\mapsto \frac{z}{1+|z|}\in D\] is the desired homeomorphism.
Let \(\Omega\) be an open subset of \(\mathbb C\). Then there exists a sequence \((K_{n})_{n \in\mathbb N}\subseteq \Omega\) of compact sets such that \[\Omega=\bigcup_{n=1}^{\infty} K_{n}\quad \text{ and } \quad K_{n} \subseteq \operatorname{int} K_{n+1} \quad \text{ for } \quad n\in\mathbb N,\] where \(\operatorname{int}K_{n+1}\) denotes the interior of \(K_{n+1}\). Further for a compact set \(K\subseteq \Omega\), we have \(K \subseteq K_{n}\) for some \(n \in\mathbb N\).
Proof: For \(n \in\mathbb N\), let \[K_n=\overline{D}(0, n)\cap\{z \in\Omega: d(z, \mathbb{C}\setminus\Omega) \geq 1/n\}.\]
It is clear that \(K_{n} \subseteq \Omega\) and it is bounded.
Each \(K_n\) is closed and consequently compact.
Moreover, \(K_{n} \subseteq K_{n+1}\) for \(n\in\mathbb N\).
We prove that \(K_{n} \subseteq \operatorname{int} K_{n+1}\). Taking \(r=\frac{1}{n}-\frac{1}{n+1}\), it suffices to show \[D(z, r) \subseteq K_{n+1} \quad \text { for } \quad z \in K_{n}.\]
Let \(z \in K_{n}\) and \(\zeta \in D(z, r)\). Then \[|\zeta-z|<\frac{1}{n}-\frac{1}{n+1} \quad \text { and } \quad |z-a| \geq \frac{1}{n} \quad \text { for }\quad a \notin \Omega.\]
Therefore for \(a \notin \Omega\), we have \[|\zeta-a| \geq|z-a|-|\zeta-z| \geq \frac{1}{n}-\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{n+1}.\]
This implies \(\zeta \in K_{n+1}\), since \(|\zeta|<|z|+\frac{1}{n} \leq n+\frac{1}{n} \leq n+1\).
Hence \(K_{n} \subseteq \operatorname{int}{K}_{n+1}\) for \(n \in\mathbb N\) and \(K_{1} \subseteq K_{2} \subseteq K_{3} \subseteq \ldots\).
We observe that \(\bigcup_{n=1}^{\infty} K_{n} \subseteq \Omega\), since each \(K_{n} \subseteq \Omega\).
For the reverse inclusion, let \(z \in \Omega\) and \(M\in\mathbb N\) be so that \(|z| \leq M\). Since \(\Omega\) is open, there exists \(\rho>0\) such that \(D(z, \rho) \subseteq \Omega\). Let \(N\in\mathbb N\) be the least integer greater than or equal to \(\max (M, \rho^{-1})\).
Then \(|z| \leq M \leq N\) and for \(a \notin \Omega\) we have \(|z-a| \geq \rho \geq \frac{1}{N}\). Therefore \(z \in K_{N}\) and thus \(\Omega \subseteq \bigcup_{n=1}^{\infty} K_{n}\), hence \(\Omega=\bigcup_{n=1}^{\infty} K_{n}\).
We also have a stronger result \(\Omega=\bigcup_{n=1}^{\infty} \operatorname{int}{K}_{n}\).
Indeed, since \(K_{n} \subseteq \operatorname{int}{K}_{n+1}\), we have \[\Omega \subseteq \bigcup_{n=1}^{\infty} K_{n} \subseteq \bigcup_{n=2}^{\infty} \operatorname{int}{K}_{n} \subseteq \bigcup_{n=1}^{\infty} \operatorname{int}{K}_{n}\subseteq \Omega.\]
Let \(K\) be a compact subset of \(\Omega\). Then \[K \subseteq \bigcup_{n=1}^{\infty} \operatorname{int}{K}_{n}\] is an open cover of \(K\). Therefore, \(K \subseteq \bigcup_{n=1}^{P} \operatorname{int}{K}_{n}\) for some \(P\in \mathbb N\).
Thus \[K \subseteq K_{1} \cup K_{2}\cup \ldots \cup K_{P}=K_{P}. \qquad \tag*{$\blacksquare$}\]
Let \(\Omega\subseteq \mathbb C\) be a region and \((f_{n})_{n \in\mathbb N}\) be a sequence such that \(f_{n} \in H(\Omega)\) converges uniformly on compact subsets of \(\Omega\). Assume that \(f_{n}\) is one-to-one for any \(n \in\mathbb N\). Then the limit function \(f=\lim_{n\to\infty}f_n\) is either constant or one-to-one on \(\Omega\).
Proof: Note that \(f \in H(\Omega)\). ( Why?)
Assume that \(f\) is not one-to-one, then there exist \(z_{1}, z_{2} \in \Omega\) such that \(z_{1} \neq z_{2}\) with \(f\left(z_{1}\right)=f\left(z_{2}\right)\).
For \(n \in\mathbb N\), we define \(g_{n} \in H(\Omega)\), and \(g \in H(\Omega)\) by setting \[g_{n}(z)=f_{n}(z)-f_{n}\left(z_{1}\right), \quad \text { and }\quad g(z)=f(z)-f\left(z_{1}\right).\] Then \(g\left(z_{2}\right)=0\) and also \(g_{n}\left(z_{2}\right) \neq 0\), since \(f_{n}\) is one-to-one.
Further, we may suppose that \(g\) is not constant in \(\Omega\) otherwise \(f\) is constant in \(\Omega\) and the assertion follows.
Since the zeros of \(g\) are isolated, there is \(r>0\) so that \(\left|z_{1}-z_{2}\right|>r\) and \(g(z) \neq 0\) whenever \(z\in\Omega\) satisfies \(0<\left|z-z_{2}\right| \leq r\).
Let \(\gamma\) be a circle centered at \(z_{2}\) with radius \(r\). Then there exists \(\delta>0\) such that \(|g(z)|>\delta\) for \(z \in \gamma^{*}\).
By the uniform convergence there exists \(n_0\in\mathbb N\) such that for every \(n \geq n_{0}\), we have \[\left|g_{n}(z)-g(z)\right|<\frac{\delta}{2} \quad \text { uniformly for any }\quad z \in \gamma^{*}.\]
Therefore for \(z \in \gamma^{*}\), we have \[\left|g_{n}(z)\right| \geq|g(z)|-\left|g_{n}(z)-g(z)\right| \geq \delta-\frac{\delta}{2}=\frac{\delta}{2}.\]
Now we see that \(\frac{1}{g_{n}(z)}\) converges uniformly to \(\frac{1}{g(z)}\) on \(\gamma^*\). Also \(g_{n}^{\prime}(z)\) converges uniformly to \(g^{\prime}(z)\) on \(\gamma^*\).
Hence \[\lim _{n \rightarrow \infty} \frac{g_{n}^{\prime}(z)}{g_{n}(z)}=\frac{g^{\prime}(z)}{g(z)}\] uniformly on \(\gamma^*\).
We observe that \(\frac{g_{n}^{\prime}(z)}{g_{n}(z)} \in H\left(\overline{D}\left(z_{2}, r\right)\right)\). Now by the Cauchy theorem we obtain that \[0=\frac{1}{2 \pi i} \int_{\gamma} \frac{g_{n}^{\prime}(z)}{g_{n}(z)} d z \quad \text { for }\quad n \in\mathbb N.\]
Consequently, by the argument principle \[0=\lim _{n \rightarrow \infty}\frac{1}{2 \pi i} \int_{\gamma} \frac{g_{n}^{\prime}(z)}{g_{n}(z)} d z=\frac{1}{2 \pi i} \int_{\gamma} \frac{g^{\prime}(z)}{g(z)} d z=N_g \geq 1.\]
This is a contradiction completing the proof of Hurwitz lemma.$$\tag*{$\blacksquare$}$$
Let \(\Omega\subseteq \mathbb C\) be a region and \(\mathcal{F}\) be a family of analytic functions in \(\Omega\). Thus \(\mathcal{F}\) is a sub-family of \(H(\Omega)\). Further \(\mathcal{F}\) is called a normal family if every sequence of elements of \(\mathcal{F}\) contains a subsequence which converges uniformly on compact subsets of \(\Omega\).
\(\mathcal{F}\) is uniformly bounded on compact subsets of \(\Omega\) if for every compact subset \(K\) of \(\Omega\) there exists \(M=M(K)\) such that \[|f(z)| \leq M \quad \text { for } \quad f \in \mathcal{F},\ z \in K.\]
The family \(\mathcal{F}\) is called equicontinuous on compact subsets of \(\Omega\) if for \(\varepsilon>0\) and compact subset \(K \subseteq \Omega\) there exists \(\delta>0\) depending only on \(\varepsilon\) and \(K\) such that \(\left|f\left(z_{1}\right)-f\left(z_{2}\right)\right|<\varepsilon\) for \(f \in \mathcal{F}\) and \(z_{1}, z_{2} \in K\) satisfying \(\left|z_{1}-z_{2}\right|<\delta\).
Remark.
The limit of the subsequence in the above definition (i) belongs to \(H(\Omega)\) but it need not belong to \(\mathcal{F}\).
Let \(\mathcal{F}\) be a family of \(H(\Omega)\) with \(\Omega\subseteq \mathbb C\) region. Assume that \(\mathcal{F}\) is uniformly bounded on compact subsets of \(\Omega\). Then \(\mathcal{F}\) is a normal family.
Proof: We show that \(\mathcal{F}\) is equicontinuous on compact subsets of \(\Omega\).
By the topological lemma, there exists a sequence of compact sets \((K_{n})_{n\in\mathbb N}\subseteq \Omega\) such that \(K_{n} \subseteq \operatorname{int}K_{n+1}\) for \(n\in\mathbb N\) that satisfies \[\Omega=\bigcup_{n=1}^{\infty} K_{n}.\] Also, every compact subset of \(\Omega\) is contained in \(K_{n}\) for some \(n\in\mathbb N\).
Let \(n \in\mathbb N\). Since \(K_{n}\) is compact, \(\left(\operatorname{int}K_{n+1}\right)^{c}\) is closed and \[K_{n} \cap\left(\operatorname{int}K_{n+1}\right)^{c}=\varnothing,\]
We can also find \(\delta_{n}>0\) such that \[\left|z_{1}-z_{2}\right|>2 \delta_{n}\quad \text { for } \quad z_{1} \in K_{n}, \quad z_{2} \notin \operatorname{int}K_{n+1}.\]
Thus \[\overline{D}\left(z, 2 \delta_{n}\right) \subseteq \operatorname{int}K_{n+1} \subseteq K_{n+1} \quad \text { for } \quad z \in K_{n}.\]
Let \(z^{\prime} \in K_{n}\), and \(z^{\prime \prime} \in K_{n}\) with \(\left|z^{\prime}-z^{\prime \prime}\right|<\delta_{n}\).
Let \(\gamma\) be a circle centered at \(z^{\prime}\) and with radius \(2 \delta_{n}\).
Thus \(z^{\prime \prime}\) lies inside the circle \(\gamma\) and \(\gamma^{*} \subseteq K_{n+1}\).
Let \(f \in \mathcal{F}\), then by the Cauchy integral formula \[\begin{aligned} f\left(z^{\prime}\right)-f\left(z^{\prime \prime}\right) & =\frac{1}{2 \pi i} \int_{\gamma} f(\zeta)\left(\frac{1}{\zeta-z^{\prime}}-\frac{1}{\zeta-z^{\prime \prime}}\right) d \zeta \\ & =\frac{z^{\prime}-z^{\prime \prime}}{2 \pi i} \int_{\gamma} \frac{f(\zeta)}{\left(\zeta-z^{\prime}\right)\left(\zeta-z^{\prime \prime}\right)} d \zeta. \end{aligned}\]
If \(\left|z^{\prime}-z^{\prime \prime}\right|<\delta_{n}\), we observe that \[\left|\zeta-z^{\prime}\right|=2 \delta_{n} \quad \text { and } \quad \left|\zeta-z^{\prime \prime}\right|=\left|\zeta-z^{\prime}+z^{\prime}-z^{\prime \prime}\right|>2 \delta_{n}-\delta_{n}=\delta_{n}.\]
Since \(\mathcal{F}\) is uniformly bounded on compact subsets of \(\Omega\), there exists a constant \(M\left(K_{n+1}\right)\) depending only on \(K_{n+1}\) such that \[\left|f\left(z^{\prime}\right)-f\left(z^{\prime \prime}\right)\right|<\frac{\left|z^{\prime}-z^{\prime \prime}\right|}{2 \pi} \frac{4\pi \delta_n}{2 \delta_{n}^2 } M\left(K_{n+1}\right)=\frac{M\left(K_{n+1}\right)}{\delta_{n}}\left|z^{\prime}-z^{\prime \prime}\right|.\]
The above inequality holds for all \(f \in \mathcal{F}\) and \(z^{\prime}, z^{\prime \prime} \in K_{n}\) whenever \(\left|z^{\prime}-z^{\prime \prime}\right|<\delta_{n}\).
Let \(\varepsilon>0\) and set \[\delta=\delta(n)=\frac{\varepsilon \delta_{n}}{\varepsilon+M\left(K_{n+1}\right)}<\delta_{n}.\]
Then for \(\left|z^{\prime}-z^{\prime \prime}\right|<\delta\), we have \[\frac{M\left(K_{n+1}\right)}{\delta_{n}}\left|z^{\prime}-z^{\prime \prime}\right|<\frac{M\left(K_{n+1}\right)}{\delta_{n}} \delta=\frac{\varepsilon M\left(K_{n+1}\right)}{\varepsilon+M\left(K_{n+1}\right)}<\varepsilon.\]
Therefore we have \[\left|f\left(z^{\prime}\right)-f\left(z^{\prime \prime}\right)\right|<\varepsilon \tag{*}\] for \(f \in \mathcal{F}\) and \(z^{\prime}, z^{\prime \prime} \in K_{n}\) and \(\left|z^{\prime}-z^{\prime \prime}\right|<\delta\).
Thus the family \(\mathcal{F}\) is equicontinuous on compact subsets of \(\Omega\), since every compact subset \(K\) of \(\Omega\) is contained in \(K_{n}\) for some \(n\in\mathbb N\), and therefore (*) holds for all \(f \in \mathcal{F}\) and \(z^{\prime}, z^{\prime \prime} \in K\) so that \(\left|z^{\prime}-z^{\prime \prime}\right|<\delta\).
Let \((f_{m})_{m\in\mathbb N}\subseteq\mathcal{F}\). We show that it has a subsequence which converges uniformly on compact subsets of \(\Omega\).
Let \(E\) be a countable dense set of \(\Omega\).
For example, we take \(E\) to be the set of all points of \(\Omega\) with rational coordinates. We arrange the elements of \(E\) as \(w_{1}, w_{2}, w_3, w_4, \ldots\).
Since \((f_{m}\left(w_{1}\right))_{m\in\mathbb N}\) is a bounded sequence by assumption on \(\mathcal F\), the sequence \((f_{m})_{m\in\mathbb N}\) has a subsequence \((f_{m1})_{m\in\mathbb N}\) so that \((f_{m 1}\left(w_{1}\right))_{m\in\mathbb N}\) converges. Here, we have used the Bolzano–Weierstrass theorem.
Similarly, the sequence \((f_{m1})_{m\in\mathbb N}\) has a subsequence \((f_{m2})_{m\in\mathbb N}\) such that \((f_{m 2}\left(w_{2}\right))_{m\in\mathbb N}\) converges.
Proceeding recursively, we see that for \(i \in\mathbb N\) there is a subsequence \((f_{m i})_{i\in\mathbb N}\) of \((f_{m, i-1})_{i\in\mathbb N}\) such that \((f_{m i}\left(w_{i}\right))_{i\in\mathbb N}\) converges.
Here we write \(f_{m 0}\) for \(f_{m}\).
Now the diagonal sequence \((f_{mm})_{m\in\mathbb N}\) converges at all \(w\in E\).
We show that \((f_{mm})_{m\in\mathbb N}\) converges uniformly on \(K_{n}\) for any \(n \in\mathbb N\).
Then the diagonal sequence \((f_{mm})_{m\in\mathbb N}\) converges uniformly on all compact subsets \(K\) of \(\Omega\) since \(K \subseteq K_{n}\) for some \(n\in\mathbb N\).
For \(n \in \mathbb N\) and \(\delta=\delta(n)\) as above, we have \[K_{n} \subseteq \bigcup_{z \in K_{n}} D(z, \delta).\]
Then \[K_{n} \subseteq \bigcup_{z \in E \cap K_{n}} D(z, \delta),\] since \(E \cap K_{n}\) is dense in \(K_{n}\). Since \(K_{n}\) is compact, we observe that the above open cover admits a finite subcover.
Thus there exist \(z_{1}, z_{2}, \ldots, z_{p} \in E \cap K_{n}\) such that \[K_{n} \subseteq D\left(z_{1}, \delta\right) \cup D\left(z_{2}, \delta\right)\cup \cdots \cup D\left(z_{p}, \delta\right).\]
For \(\varepsilon>0\), there exists \(M\) depending only on \(\varepsilon\) and \(K_{n}\) such that \[\left|f_{r r}\left(z_{i}\right)-f_{s s}\left(z_{i}\right)\right| \leq \varepsilon\] for \(r \geq M, s \geq M\) and \(1 \leq i \leq p\).
Let \(z \in K_{n}\). Then \(z \in D\left(z_{i}, \delta\right)\) for some \(i\) with \(1 \leq i \leq p\).
Further, by the equicontinuity, we have \[\begin{aligned} |f_{r r}(z)&-f_{s s}(z)| \\ & \leq\left|f_{r r}(z)-f_{r r}\left(z_{i}\right)\right|+\left|f_{r r}\left(z_{i}\right)-f_{s s}\left(z_{i}\right)\right|+\left|f_{s s}(z)-f_{s s}\left(z_{i}\right)\right| \\ & <\varepsilon+\varepsilon+\varepsilon=3 \varepsilon \end{aligned}\] wherever \(r \geq M, s \geq M\). Thus \((f_{mm})_{m\in\mathbb N}\) converges uniformly on \(K_{n}\).
Since every compact subset of \(\Omega\) is contained in \(K_{n}\) for some \(n\in\mathbb N\), we conclude that \((f_{mm})_{m\in\mathbb N}\) converges uniformly on compact subsets of \(\Omega\). This completes the proof of Montels’ lemma. $$\tag*{$\blacksquare$}$$
Proof: Assume that \(\Omega\neq \mathbb C\) is a simply connected region. Let \(z_{0} \in \Omega\) and define \[\Sigma=\{\psi \in H(\Omega): \psi \text{ is one-to-one on } \Omega \text{ and } \psi(\Omega) \subseteq D\}.\] Our aim is to prove that \(\Sigma\) contains an element which is onto.
In fact, we will show that for \(\psi \in \Sigma\), which is not onto \(D\), there exists \(\psi_{1} \in \Sigma\) such that \[\left|\psi_{1}^{\prime}\left(z_{0}\right)\right|>\left|\psi^{\prime}\left(z_{0}\right)\right|. \tag{*}\]
Next we consider \[\eta=\sup \left\{\left|\psi^{\prime}\left(z_{0}\right)\right|: \psi \in \Sigma\right\},\tag{**}\] and we will prove that the supremum is assumed for some \(\psi_{0} \in \Sigma\).
Then it will be clear that \(\psi_{0}\) has to be an onto function. Otherwise, if \(\psi_{0}\) is not onto \(D\) then there is \(\psi_1\in\Sigma\) satisfying (*). Hence, by (**), we have \(\eta=\left|\psi_0^{\prime}\left(z_{0}\right)\right|<\left|\psi_{1}^{\prime}\left(z_{0}\right)\right|\le \eta\), which is impossible.
The proof will consist of a few steps.
Step 1. We first prove that \(\Sigma\neq\varnothing\).
From Lecture 8 we know that if \(\Omega\subseteq \mathbb C\) is a simply connected region, then for every closed curve \(\Gamma\) in \(\Omega\) we have \[\operatorname{Ind}_{\Gamma}(\alpha)=0 \quad \text{ whenever }\quad \alpha\not\in \Omega.\]
From Lecture 9, we know that the latter is equivalent to the statement that for every \(f\in H(\Omega)\) satisfying \(1/f\in H(\Omega)\) there exists \(g\in H(\Omega)\) such that \(f=g^2\).
Since \(\Omega \neq \mathbb{C}\), let \(w_{0} \in \mathbb{C}\) such that \(w_{0} \notin \Omega\). Consider \[f(z)=z-w_{0}.\]
We observe that \(f \in H(\Omega)\) and \(f\) has no zero in \(\Omega\) since \(w_{0} \notin \Omega\).
Thus \(1 / f \in H(\Omega)\). Therefore, there exists \(g \in H(\Omega)\) such that \[f(z)=g^{2}(z) \quad \text { for }\quad z \in \Omega.\]
Let \(a \in \Omega\). By open mapping theorem, we observe that \(g(\Omega)\) is open containing \(g(a)\).
Therefore there exists \(r>0\) such that \[D(g(a), r) \subseteq g(\Omega).\]
Now we show that \(D(-g(a), r) \cap g(\Omega)=\varnothing\). Suppose there exists \(z_{1} \in \Omega\) such that \(g\left(z_{1}\right) \in D(-g(a), r)\). Thus \[\left|g\left(z_{1}\right)+g(a)\right|<r \quad \iff \quad \left|-g\left(z_{1}\right)-g(a)\right|<r\]
Then \[-g\left(z_{1}\right) \in D(g(a), r) \subseteq g(\Omega) .\]
Therefore there exists \(z_{2} \in \Omega\) such that \(-g\left(z_{1}\right)=g\left(z_{2}\right)\).
By squaring both sides of this equation, we derive that \[z_{1}-w_{0}=f(z_1)=\left(-g\left(z_{1}\right)\right)^{2}=\left(g\left(z_{2}\right)\right)^{2}=f(z_2)=z_{2}-w_{0}\] implying \(z_{1}=z_{2}\). Thus \(g\left(z_{1}\right)=0\), and consequently \(z_{1}-w_{0}=0\).
This is a contradiction since \(z_{1} \in \Omega\) and \(w_{0} \notin \Omega\). Hence, \[D(-g(a), r) \cap g(\Omega)=\varnothing.\]
Now we consider \[\psi(z)=\frac{r}{g(z)+g(a)} \quad \text { for }\quad z \in \Omega.\]
We observe that \(\psi(z) \in H(\Omega)\) and \(|\psi(z)| \leq 1\) for \(z \in \Omega\), since \[|g(z)+g(a)| \geq r\quad \text{ for } \quad z \in \Omega.\]
In fact \(|\psi(z)|<1\) for \(z \in \Omega\) by the maximum modulus principle.
Further \(\psi \in H(\Omega)\) and is one-to-one on \(\Omega\), since \(\psi\left(z^{\prime}\right)=\psi\left(z^{\prime \prime}\right)\) implies \(g^{2}\left(z^{\prime}\right)=g^{2}\left(z^{\prime \prime}\right)\), and therefore \(z^{\prime}=z^{\prime \prime}\).
Hence \(\psi \in \Sigma\) as desired.
Step 2. We will show that if \(\psi \in \Sigma\) and \(\psi(\Omega)\) is a proper subset of \(D\), then there exists \(\psi_{1} \in \Sigma\) satisfying \[\left|\psi_{1}^{\prime}\left(z_{0}\right)\right|>\left|\psi^{\prime}\left(z_{0}\right)\right|. \tag{*}\]
Fix \(\psi \in \Sigma\). Since \(\psi(\Omega)\) is a proper subset of \(D\), there exists \(\alpha \in D\) such that \(\alpha \notin \psi(\Omega)\). We consider \(\phi_{\alpha} \circ \psi\), where \[\phi_{\alpha}(w)=\frac{w-\alpha}{1-\overline{\alpha} w}.\]
We recall that \(\phi_{\alpha}\) is an automorphism of \(D\). For \(z \in \Omega\), observe that \[\phi_{\alpha} \circ \psi(z)=\phi_{\alpha}(\psi(z))=\frac{\psi(z)-\alpha}{1-\overline{\alpha} \psi(z)}=0\] only when \(\psi(z)=\alpha\) which is not the case since \(\alpha \notin \psi(\Omega)\).
Therefore \(\phi_{\alpha} \circ \psi \in H(\Omega)\) has no zero in \(\Omega\). Then, there exists \(g \in H(\Omega)\) such that \[g^{2}(z)=\phi_{\alpha} \circ \psi(z) \quad \text { for }\quad z \in \Omega.\]
By writing \(s(w)=w^{2}\) for \(w \in D\), we rewrite the last equation as \[s \circ g=\phi_{\alpha} \circ \psi\quad \text { in } \quad \Omega.\]
If \(g\left(z_{1}\right)=g\left(z_{2}\right)\) for \(z_{1},z_{2}\in\Omega\), then \(\psi\left(z_{1}\right)=\psi\left(z_{2}\right)\), since \(\phi_{\alpha}\) is one-to-one. Therefore, \(g\) is one-to-one, and \(g \in \Sigma\).
Let \[\psi_{1}=\phi_{\beta} \circ g \quad \text { with }\quad g\left(z_{0}\right)=\beta.\]
Observe that \[\psi_{1}\left(z_{0}\right)=\phi_{\beta}\left(g\left(z_{0}\right)\right)=\phi_{\beta}(\beta)=0.\]
Hence, \[\psi=\phi_{-\alpha} \circ s \circ g=\phi_{-\alpha} \circ s \circ \phi_{-\beta} \circ \psi_{1}=F \circ \psi_{1},\] where \[F=\phi_{-\alpha} \circ s \circ \phi_{-\beta}.\]
By the chain rule and \(\psi_{1}\left(z_{0}\right)=0\), we have \[\psi^{\prime}\left(z_{0}\right)=F^{\prime}\left(\psi_{1}\left(z_{0}\right)\right) \psi_{1}^{\prime}\left(z_{0}\right)=F^{\prime}(0) \psi_{1}^{\prime}\left(z_{0}\right).\]
Inequality (*) will follow if we show that \(\left|F^{\prime}(0)\right|<1\), since \[\left|\psi^{\prime}\left(z_{0}\right)\right|=\left|F^{\prime}(0)\right|\left|\psi_{1}^{\prime}\left(z_{0}\right)\right|.\]
Recall the following lemma from the previous lecture.
Let \(f\) be non-constant and analytic in \(D\), and satisfy \(|f(z)|<1\) for \(z\in D\). Let \(w \in D\) with \(f(w)=a\). Then \[\left|f^{\prime}(w)\right| \leq \frac{1-|a|^{2}}{1-|w|^{2}} .\] Moreover equality occurs only when \[f=\phi_{-a} \circ\left(c \phi_{w}\right) \text { in } D,\] for some constant \(c\) whose absolute value is \(1\).
We observe that \(F(D) \subseteq D\), and let \(F(0)=a\). Then by the lemma with \(w=0\), we have \[\left|F^{\prime}(0)\right| \leq 1-|a|^{2}.\]
Suppose that \(\left|F^{\prime}(0)\right|=1\). Then \(a=0\) and by the second part of the previous lemma, we conclude with \(w=0\) that \(F(z)=\lambda z\) for \(z \in D\) where \(\lambda\) is a constant of absolute value \(1\).
This is not possible since \(F\) is not one-to-one as \(s(w)=w^{2}\) is not one-to-one. Hence \(\left|F^{\prime}(0)\right|<1\) and the proof of (*) is complete.
Step 3. In this step we finish the proof. In Step 1, we have proved that \(\Sigma\neq\varnothing\), hence we can define \[\eta=\sup \left\{\left|\psi^{\prime}\left(z_{0}\right)\right|: \psi \in \Sigma\right\},\tag{**}\]
By the inverse mapping theorem we have \(\left|\psi^{\prime}\left(z_{0}\right)\right|>0\) for every \(\psi \in \Sigma\). Hence \(\eta>0\).
There exists a sequence \((\psi_{n})_{n\in\mathbb N}\subseteq\Sigma\) such that \[\lim _{n \rightarrow \infty}\left|\psi_{n}^{\prime}\left(z_{0}\right)\right|=\eta>0.\]
We observe that \(|\psi(z)|<1\) for \(\psi \in \Sigma\).
In particular, \(\Sigma\) is uniformly bounded on compact subsets of \(\Omega\).
Therefore \(\Sigma\) is a normal family by Montel’s lemma.
Hence the above sequence \((\psi_{n})_{n\in\mathbb N}\subseteq\Sigma\) has a subsequence which we denote again by \((\psi_{n})_{n\in\mathbb N}\subseteq\Sigma\), and which converges uniformly on compact subsets of \(\Omega\) satisfying \[\lim _{n \rightarrow \infty}\left|\psi_{n}^{\prime}\left(z_{0}\right)\right|=\eta>0.\]
Let \[\lim _{n \rightarrow \infty} \psi_{n}(z)=h(z)\quad \text { for } \quad z \in \Omega\] converge uniformly on compact subsets of \(\Omega\). Then \(h \in H(\Omega)\).
Further \[\lim _{n \rightarrow \infty} \psi_{n}^{\prime}(z)=h^{\prime}(z)\quad \text { for }\quad z \in \Omega\] converges uniformly on compact subsets \(\Omega\).
Therefore \[\lim _{n \rightarrow \infty}\left|\psi_{n}^{\prime}\left(z_{0}\right)\right|=\left|h^{\prime}\left(z_{0}\right)\right|,\] which implies \[\left|h^{\prime}\left(z_{0}\right)\right|=\eta>0.\]
Hence \(h\) cannot be constant, otherwise we would have \(\eta=0\).
Therefore \(|h(z)|<1\) for \(z \in \Omega\) by the maximum modulus principle, in other words \[h(\Omega)\subseteq D.\]
By the Hurwitz lemma \(h\) must be also one-to-one, thus \(h\in\Sigma\) and is the maximizer for (**) as desired.
Hence \(h(\Omega)=D\) as it was explained by combining (*) and (**) at the beginning of the proof.
This completes the proof of the Riemann mapping theorem. $$\tag*{$\blacksquare$}$$