0
$\begingroup$

Let $f$ be a conformal map from unit disk $|z| <1$ to square $D=\{x+iy \in \mathbb{C}:|x|<1,|y|<1\}.$ Could anyone advise me how to prove $f$ cannot be extended to holomorphic function defined on disk $|z| <R, \forall R>1 \ ?$

Suppose $f$ can be extended to a holomorphic function on $|z|<R,$ where $R>1.$

Consider $z$ where $|z|=1.$ Then $f(z) \not\in D. $

But how do I continue from here? Hints will suffice, thank you very much.

$\endgroup$
1
$\begingroup$

Do you know that $f$ can be continuously extended to a homeomorphism from the closed disk to the closed square? What would happen at the points being mapped to the vertices of the square if an analytic extension existed?

Let $z_0 \in \partial\mathbb{D}$ a point that is mapped to the vertex $1+i$ of the square $\overline{D}$ by the continuous extension of $f$ to $\overline{\mathbb{D}}$. Since that continuous extension is a homeomorphism, the two boundary arcs of $\mathbb{D}$ meeting in $z_0$ - $A_- = \{z_0 e^{i\varphi} : -\delta < \varphi < 0\}$ and $A_+ = \{z_0 e^{i\varphi} : 0 < \varphi < \delta\}$ - are mapped to the two boundary segments $B_- = \{1 + ti : 0 < t < 1\}$ and $B_+ = \{(1-t) + i : 0 < t < 1\}$ of $D$ [for small enough $\delta > 0$], $A_-$ being mapped into $B_-$ and $A_+$ into $B_+$, since $f$ is orientation-preserving.

The two arcs $A_-$ and $A_+$ meet at an angle of $\pi$ (the unit circle is smooth), and the two segments $B_-$ and $B_+$ meet at an angle of $\frac{\pi}{2}$ (or $\frac{3\pi}{2}$ if you look at the exterior angle).

Now, if $f$ had a holomorphic continuation $\tilde{f}$ to a neighbourhood of $z_0$, then $\tilde{f}$ would attain the value $1+i$ with multiplicity $m$ in $z_0$.

Hence $\tilde{f}$ would map the two arcs $A_-$ and $A_+$ to two curves meeting at an angle of - what?

$\endgroup$
  • $\begingroup$ Thanks for the advice. I'm still unsure of how to proceed. Here is my attempt: Consider $B(z,r) \subset D(0,R),$ where $|z|=1$ and $R>1. $By Opening Mapping theorem, $f[(B(z,r)]$ is open in $\mathbb{C}. \ $ Also, $f[B(z,r)]$ contains points that do not lie within and on the square. $\endgroup$ – Alexy Vincenzo Sep 28 '14 at 13:16
  • $\begingroup$ Look at the behaviour of $f$ at $z_0$, where $z_0$ is a (the) point of $\partial \mathbb{D}$ that is mapped to the vertex $1+i$. What does $f$ do to the boundary curve (the unit circle)? Is that compatible with $f$ being holomorphic in a neighbourhood of $z_0$? $\endgroup$ – Daniel Fischer Sep 28 '14 at 13:20
  • $\begingroup$ $f$ maps the boundary curve of the unit circle to the boundary of the square. $\endgroup$ – Alexy Vincenzo Sep 28 '14 at 13:40
  • $\begingroup$ Yes, and at $z_0$ that means that what happens? $\endgroup$ – Daniel Fischer Sep 28 '14 at 13:42
  • $\begingroup$ I apologize for being unable to make full use of your hints yet. I am thinking of using Open Mapping theorem but the set of boundary points of the circle is obviously not open. $\endgroup$ – Alexy Vincenzo Sep 28 '14 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.