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If $Ω$ is a simply connected proper sub-domain of $\mathbb{C}$, then show that there is a one-one bounded analytic function on $Ω$ without and using riemann mapping theorem

can anyone help me please to solve the problem

damuru
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2 Answers2

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Potato points out that since $\Omega$ a proper subset, you can translate so that $0$ is in the complement of $\Omega$. Since $\Omega$ is 1-connected, you can choose a path from $0$ to $\infty$. Take a branch of the square root on the complement of the path. This maps to a half-plane. Now conformally map the half-plane to the unit disk, scale if necessary, and translate if necessary to a subset of $\Omega$. All functions involved are injective holomorphic, hence their composition is injective holomorphic.

My original answer is below.

If $\Omega$ is bounded, restrict the identity map to $\Omega$.

If $\Omega$ is not dense in $\mathbb{C}$, translate so that the complement of $\Omega$ has $\mathbb{C}$ as an interior point. Now invert. The result is bounded. Scale if necessary. Now translate back to a subset of $\Omega$.

If $\Omega$ is dense in $\mathbb{C}$, translate so that $0$ is in the complement of $\Omega$. By simple-connectedness, we can connect $0$ to $\infty$ by a single line in the complement. Choose this line as a branch cut for the square root. The image of this is not dense in $\mathbb{C}$, so apply the operations of the previous case.

In each of these cases, we have used analytic injective functions (translation, inversion, dilation, branch of square root), so their composition is analytic and injective.

Neal
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  • @Potato Because I'm out of practice. Your solution is much more elegant; I hope you don't mind that I have edited it into my answer. – Neal Apr 05 '13 at 03:31
  • Hmm, there is a problem here. A branch of the square root doesn't not always map to a half plane. – Potato Apr 05 '13 at 21:48
  • You need to justify why taking the square root of the region is not dense, I think. It's not immediately obvious to me. – Potato Apr 05 '13 at 21:49
  • See: http://math.stackexchange.com/questions/352553/the-range-of-an-arbitrary-branch-of-logz/352560?noredirect=1#352560 – Potato Apr 05 '13 at 21:53
  • @Potato The intuition seemed clear to me, but your solution below seems to do a nice job formalizing it. – Neal Apr 08 '13 at 00:06
  • Yes, in retrospect it actually is obvious. I just got flustered because my original proposal was wrong. – Potato Apr 08 '13 at 00:44
  • @Potato I guess the logarithm doesn't work because exp is $\infty$-to-one, while $z^2$ is just a double-cover. – Neal Apr 08 '13 at 03:42
  • Yeah, but since $\sqrt z$ is defined via $\log z$, knowing that $\sqrt z$ maps to a "nice" region requires knowing the same of $\log z$. – Potato Apr 08 '13 at 04:11
  • @Potato Maybe one could argue that $z\mapsto z^2$ is a double cover of $C^*$ and local sections of covers cannot have dense images. – Neal Apr 10 '13 at 18:34
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I believe there is an issue with the other answer, so here's the solution given in Ahlfors' text. There is some point $a\notin \Omega$. Since the region is simply connected, we can define a branch of $\sqrt{z-a}$. Call this $h(z)$. This in injective and does not take "opposite values": $w$ and $-w$ cannot both be in the range. Fix a point $z_0\in \Omega$. Then the image $h(\Omega)$ covers a disc of radius $r$ at $h(z_0)$ for some $r>0$, so by the opposite value remark, it does not meet the disc of radius $r$ centered at $-h(z_0)$.

You can now compose the above with a translation of $-h(z_0)$ to $0$ and then take $1/z$. The resulting region is bounded because there is an open neighborhood around $0$ before you take the reciprocal, so there will be an open neighborhood around infinity after.

Potato
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