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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

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    $\begingroup$ The identity doesn't work? $\endgroup$ – Neal Apr 5 '13 at 2:32
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    $\begingroup$ @Neal I assume "proper" doesn't imply bounded (an open strip seems to satisfy the hypothesis of $\Omega$), so the identity may not be bounded. $\endgroup$ – Matt Apr 5 '13 at 2:36
  • $\begingroup$ How are you going to do this without using Riemann? That's like the point of the theorem. $\endgroup$ – Euler....IS_ALIVE Apr 5 '13 at 3:05
  • $\begingroup$ @Euler....IS_ALIVE This is weaker. The poster doesn't need a biholomorphism to the unit disc. $\endgroup$ – Potato Apr 5 '13 at 3:06
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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.

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  • $\begingroup$ @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. $\endgroup$ – Neal Apr 5 '13 at 3:31
  • $\begingroup$ Hmm, there is a problem here. A branch of the square root doesn't not always map to a half plane. $\endgroup$ – Potato Apr 5 '13 at 21:48
  • $\begingroup$ You need to justify why taking the square root of the region is not dense, I think. It's not immediately obvious to me. $\endgroup$ – Potato Apr 5 '13 at 21:49
  • $\begingroup$ See: math.stackexchange.com/questions/352553/… $\endgroup$ – Potato Apr 5 '13 at 21:53
  • $\begingroup$ @Potato The intuition seemed clear to me, but your solution below seems to do a nice job formalizing it. $\endgroup$ – Neal Apr 8 '13 at 0:06
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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.

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