- #1
MathLearner123
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- 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).
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).