Question about uniform convergence in a proof

[MathLearner123]

TL;DR Summary

Question about uniform convergence in a proof

The below proposition is from David C. Ullrich's "Complex Made Simple" (pages 264-265)

Proposition 14.5. Suppose $D$ is a bounded simply connected open set in the plane, and let $\phi: D \rightarrow \mathbb{D}$ be a conformal equivalence.

(i) If $\zeta$ is a simple boundary point of $D$ then there exists $L \in \partial \mathbb{D}$ such that $\phi\left(z_n\right) \rightarrow L$ whenever $\left(z_n\right)$ is a sequence in $D$ tending to $\zeta$. (In other words, $\phi$ can be extended continuously to the set $D \cup\{\zeta\}$, and the extended function satisfies $\sigma(\zeta) \in \partial \mathbb{D}$.)

(ii) Suppose that $\zeta_1$ and $\zeta_2$ are two simple boundary points; choose corresponding complex numbers $L_1$ and $L_2$ as in (i). If $\zeta_1 \neq \zeta_2$ then $L_1 \neq L_2$.

Proof. First we note that if $\left(z_n\right)$ is any sequence in $D$ tending to a point of the boundary then $\left|\phi\left(z_n\right)\right| \rightarrow 1$ : If $r<1$ then $K=\phi^{-1}(\bar{D}(0, r))$ is a compact subset of $D$ and hence there exists $N$ such that $z_n \notin K$ for all $n>N$; thus $\left|\phi\left(z_n\right)\right|>r$ for every $n>N$.

Now we prove (i). Suppose that $\zeta$ is a simple boundary point of $D$ and that $\left(z_n\right)$ is a sequence of points of $D$ such that $z_n \rightarrow \zeta$. Now the compactness of the closed disk shows that there exists $L \in \overline{\mathbb{D}}$ and a subsequence $\left(z_{n_j}\right)$ such that $\phi\left(z_{n_j}\right) \rightarrow L$ (hence $|L|=1$ by the comments in the previous paragraph). To show that $\phi\left(z_n\right) \rightarrow L$ we need only show that every other convergent subsequence of $\left(\phi\left(z_n\right)\right)$ also tends to $L$.

Suppose not: $\left(\phi\left(z_n\right)\right)$ has two subsequences tending to different limits. Adjusting the notation a bit we may assume that $\phi\left(z_{2 n}\right) \rightarrow L_1$ and $\phi\left(z_{2 n+1}\right) \rightarrow L_2$, where $L_j \in \partial \mathbb{D}$ and $L_1 \neq L_2$. Now, since $\zeta$ is a simple boundary point, there exists a continuous curve $\gamma:[0,1) \rightarrow D$ and a sequence $\left(t_n\right)$ in $[0,1)$ such that $t_n \leq t_{n+1}, t_n \rightarrow 1, \gamma\left(t_n\right)=z_n$, and $\lim _{t \rightarrow 1} \gamma(t)=\zeta$.

Now let $\Gamma=\phi \circ \gamma$. Then $\Gamma:[0,1) \rightarrow \mathbb{D}$, and an argument as above shows that $|\Gamma(t)| \rightarrow 1$ as $t \rightarrow 1$. But the curve $\Gamma$ does not actually tend to a. single point of $\partial \mathbb{D}$, since $\Gamma\left(t_{2 n}\right) \rightarrow L_1$ while $\Gamma\left(t_{2 n+1}\right) \rightarrow L_2$. Thus $\Gamma$ nust actually "cross" some sector $S$ infinitely many times as it tends to $\partial \mathbb{D}$. So, at least for large $n$, there exist nunbers $a_n$ and $b_n$ such that $t_{2 n}<a_n<b_n<t_{2 n+1}, \gamma\left(a_n\right)$ lies on one edge of $S$, $\gamma\left(b_n\right)$ lies on the other edge of $S$, and $\gamma\left(\left(a_n, b_n\right)\right)$ lies in the interior of $S$. Let $\gamma_n=\left.\gamma\right|_{\left[a_n, b_n\right]}$.

Let $f=\phi^{-1}$. Then $f \in H(\mathbb{D})$ and $f$ is bounded. We have $f(\Gamma(t)) \rightarrow \zeta$ as $t \rightarrow 1$, since $f(\Gamma(t))=\gamma(t)$; thus $f \circ \gamma_n \rightarrow \zeta$ uniformly.

I don't understand the last part. How I can prove that uniform convergence? (I think that the author is wrong and it is $\Gamma_n = \Gamma_{|[a_n, b_n]}$ instead of $\gamma_n$ everywhere).

 

[Office_Shredder]

Since the $\gamma_n$ all lie on the curve in $\mathbb{D}$, the $f(\gamma_n)$ all lie on the curve in $D$.

I think a slightly missed step here is that you can pick the $\gamma_n$ such that $|\gamma_n|\to 1$, since the magnitude of the curve in $\mathbb{D}$ approaches 1.

This means that $f(\gamma_n)$ have to lie entirely on the section of the curve in $D$ for which the original $t$ value is very close to 1.

So you have a sequence of functions $g_n=f(\gamma_n)$ for which for every $\epsilon>0$, there exists $M$ such that $\forall t, |g_n(t)-\zeta| \leq \sup_{t>1-\epsilon} |f(\gamma(t))-\zeta|$ as long as $n>M$. And the thing on the right goes to zero as epsilon goes to zero