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Denote the class of Schlicht functions (injective, holomorphic on the unit disk, with $f(0)=0$ and $f'(0)=1$) by $\mathcal{S}$. And let $\gamma :[0,\infty )\rightarrow \mathbb{C}$ be a simple curve (continuous and injective) and such that $\gamma (0)=0$ and $\lim _{t\to \infty}\gamma (t)=\infty$.

I have to show that, for each such $\gamma$, there is a unique $t>0$ so that $\mathbb{C} \backslash \gamma \left( [t,\infty )\right)$ is the image of some $f\in \mathcal{S}$.

Here's my idea so far:

For each $t$, via the Riemann Maping Theorem, I can construct a bijective, holomorphic function $f_t:\mathbb{C} \backslash \gamma \left( [t,\infty )\right) \rightarrow \mathbb{D}$ ($\mathbb{D}$ is the unit disk) such that $f_t(0)=0$. Furthermore, such an $f_t$ is unique up to a choice of the argument of $f_t'(0)$. If I could somehow show that there must be some $t>0$ such that $\left| f_t'(0)\right| =1$ I would be done, but I am stuck as how to do this.

Anyone have any ideas? In the same direction or not, all suggestions are welcome.

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1 Answer 1

up vote 2 down vote accepted

Here is a simple idea, unfortunately not finished (unlike I thought a minute ago; I'm posting it anyway:) It shows that there is at most one such $t$, but not existence (for existence I give only an unfinished idea).

Let $F_t:\mathbb{C}-\gamma([t,\infty))\to\mathbb{D}$ be the holomorphic bijection such that $F_t(0)=0$ and $F_t'(0)>0$. If $t_1<t_2$ then $F_{12}=F_{t_2}\circ F_{t_1}^{-1}:\mathbb{D}\to\mathbb{D}$ is injective but not bijective and $F_{12}(0)=0$, so by Schwarz lemma $F_{12}'(0)<1$, i.e. $F_{t_1}'(0)>F_{t_2}'(0)$. So there is at most one such $t$ that $F_t'(0)=1$.

For existence: If $t$ and $a>1$ are such that $\gamma(t')>a$ for all $t'>t$ (i.e. if $t$ is big enough) then $z\mapsto F_t(z/a)$ maps $\mathbb{D}$ injectively to $\mathbb{D}$, so $F_t'(0)\leq1/a<1$ (again by Schwarz lemma). If $F_t'(0)>1$ for some (small) $t$ and if $F_t'(0)$ is a continuous function of $t$ then we're done.

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