The following problem is an assignment of my complex analysis course, which seems to be a converse of the Carathéodory's theorem that a biholomorphism between two Jordan regions can be extended to a homeomorphism between their closures.

Suppose that $G$ and $\Omega$ are Jordan regions and $f$ is a continuous function on the closure $\overline G$ of $G$ such that $f$ is analytic on $G$ and $f(G)\subset\Omega$. If $f$ maps $\partial G$ onto $\partial\Omega$ homeomorphically, then $f(G)=\Omega$ and $f$ is univalent.

To be honest it is a bit difficult to prove and I have worked on it for days. By Riemann mapping theorem and its extension to the boundary, we can take both $G$ and $\Omega$ in this problem as the unit disk $\Delta$ and $f(0)=0$. The image of $f$ is a region since $f$ here cannot be constant in $G$, and then we can consider a formally simpler but in fact equivalent problem as follows.

Suppose that $f\colon\overline\Delta\to\overline\Delta$ is continuous. If $f\in\mathcal O(\Delta)$ and $f(0)=0$ and $f\colon\partial\Delta\to\partial\Delta$ is a homeomorphism, then $f\in\mathrm{Aut}\,(\Delta)$, namely $f(z)=e^{i\theta}z$ where $\theta\in\mathbb R$.

For what happens next I completely have no idea. So may I ask how to prove such an $f$ is a bijection? Thanks in advance..

ps. I found it an exercise in J. B. Conway's Functions of One Complex Variable II, GTM159. It is the Excercise 10 of Section 14.5, while a preceding one (Excercise 8, for which a proof is here) says that if $f$ is continuous over $\overline\Delta$ and $f\in\mathcal O(\Delta)$, then there is a sequence of polynomial $P_n(z)$ such that $P_n$ converges uniformly to $f$ on $\overline\Delta$. I think this may help since if we managed to prove the moduli of coefficients $a_n$ of $z$ in $P_n$ tend to $1$ then by Schwarz's lemma we can conclude that $f$ is biholomorphic. However, I don't know what the use of $f$ being homeomorphic on $S^1=\partial\Delta$ is and maybe it is the bottleneck of this proof. Now I would like to ask if I am in a right way, and what to do next...


Starting with your formulation of $f:\bar\Delta\to\bar\Delta$, for each $w\in\Delta$ and each $1\geq t>\sup\{|z|:f(z)=w\}$ consider the curve $\gamma_{t,w}:\partial \Delta\to\mathbb C\setminus\{0\}$ defined by $$\gamma_{t,w}(z) = f(tz)-w.$$

By the argument principle the number of zeroes $N(w)$ is the winding number $$\pi_1(\gamma_{t,w})=\frac{1}{2\pi i}\oint_{\gamma_{t,w}}\frac{dz}{z}$$ at least for $t<1$. For $t=1$, using the homeomorphism assumption, the curve $\gamma_{1,w}$ can be shown to be rectifiable, so the integral above is Riemann-Stieltjes integrable. (This isn't really needed. I don't know if this is discussed in Conway's book, but Stewart and Tall's Complex Analysis book has a chapter "Homotopy versions of Cauchy's Theorem" which explains that analytic functions can be "integrated" along an arbitrary closed path.) More importantly, it is homotopy invariant so takes the same value wherever it is defined. Homotopy invariance for rectifiable curves is in Conway's book.

Given that the winding number is a homotopy invariant, the rest is easy. A homeomorphism $\partial\Delta\to\partial\Delta$ must have winding number $\leq 1$. One way to see this is that winding number is multiplicative - $\pi_1(\phi\circ \psi)=\pi_1(\phi)\pi_1(\psi)$ for continuous functions $\phi,\psi:\partial\Delta\to\partial\Delta$ - and a homeomorphism has an inverse. A more complex-analytic argument would be to consider the primitive of $1/2\pi i z$ defined on $\mathbb C\setminus\mathbb R_{\geq 0}$ which takes values in $(0,1)$ on the unit circle, and argue that the winding number is a limit of differences between two values in $(0,1)$, so must be at most $1$.

So $N(0)$ must be $\leq 1$, and all the values $N(w)$ are the same by homotopy invariance, and $N(f(0))>0$, which gives $N(w)=1$ for all $w$.

  • $\begingroup$ Thanks but I have a little question: now that $f$ on the boundary curve is merely homeomorphic, how can we use the argument principle if we don't know whether $f$ is analytic on $\partial\Delta$? $\endgroup$ – josephz Oct 3 '17 at 2:44
  • $\begingroup$ @josephz: I've added details for the argument I had in mind. I realize now that I was assumping homotopy invariance of winding numbers, which might not be in Conway's book. $\endgroup$ – Dap Oct 3 '17 at 6:10
  • $\begingroup$ So I guess in this way the proof is made somewhat like below $N(w)=\oint_{\{z:|z|<t\}}\frac{f(z)-w}{f(z)-w}dz=\oint_{\gamma_{t, w}}\frac{dz}{z}=\mathrm{deg}(f)=\pm 1$ where $\mathrm{deg}$ is the degree of the continous map $f$. $N(0)=1$ implies $N(w)=1$. Thus $f$ in $\Delta$ is bijective. $\endgroup$ – josephz Oct 3 '17 at 10:26
  • $\begingroup$ @josephz: yeah, though I wasn't sure how much to assume about the degree of continuous maps being well-defined and homotopy invariant, i.e. if you are happy to use the computation of the fundamental group of the circle. $\endgroup$ – Dap Oct 3 '17 at 10:35
  • $\begingroup$ Wow, thank you for answering so meticulously! So afterall I guess the easiest way of proving it is to use homotopy and homology?(I just searched and found that winding number is actually the degree of a continuous map. For continuous $f\colon S^1\to S^1$, it induces a homomorphism $f_*\colon H_1(S^1)\to H_1(S^1)$ while $H_1(S^1)\cong \mathbb Z$ and thus $f\colon x\mapsto ax$. This $a$ is defined as the degree of $f$ or the winding number. In this way it seems easier to explain why $N(w)=\pm 1$.) $\endgroup$ – josephz Oct 3 '17 at 10:47

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