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Given only that $f(z)$ is analytic and maps the unit disk $|z| < 1$ surjectively to the upper half plane $\Im(z) > 0$, how much can we deduce about $f(z)$? In particular, can we find the radius of convergence of the Taylor series $\sum_{n=0}^\infty a_n z^n$ of $f(z)$?

Of course, the Taylor series will have radius of at least $1$ since $f(z)$ is given as analytic on the unit disk, but can we deduce more?

If we let $Sz = \frac{z-i}{z+i}$, $S$ is a Mobius transformation that maps the upper half plane to the unit disk, so $Sf(z)$ maps the unit disk to itself. If we choose an automorphism $T$ of the disk that maps $Sf(0)$ to $0$, we could apply Schwarz's lemma to find that $|TSf(z)| \leq |z| \Rightarrow |f(z)| \leq |S^{-1}T^{-1}z|$, but does this necessarily tell us anything about $f(z)$? What additional information can we deduce about $f(z)$ under the given conditions?


Here is what I have tried so far.

Let $Sz = \frac{z-i}{z+i}$; $S$ maps the upper half plane to the unit disk, so $S \circ f$ maps the unit disk to itself.

Now, an automorphism $T$ of the unit disk is of the form $Tz = e^{i\theta} \frac{z-a}{1-\overline{a}z}$ where $|a| < 1$. If we choose $a = Sf(0)$, then $TSf(0) = 0$, and we can apply Schwarz's lemma to $TSf(z)$, giving that $|TSf(z)| \leq |z|$ on the unit disk.

If we can find some $z_0$ with $0 < |z_0| < 1$ such that $|TSf(z_0)| = |z_0|$ - or, equivalently, a $z_0$ with $0 < |z_0| < 1$ such that $|f(z_0)| = |S^{-1}T^{-1}(z_0)|$ - then, $TSf(z) = cz$ for some $c \in \mathbb{C}$ such that $|c| = 1$.

$$\begin{align} S^{-1}T^{-1}(z) &= \frac{(i + i\overline{a})e^{-i\theta}z + (i + ia)}{(\overline{a} - 1)e^{-i\theta}z + (1-a)} \\ &= -i\frac{(1 + \overline{a})e^{-i\theta}z + (1 + a)}{(1 - \overline{a})e^{-i\theta}z - (1-a)} \\ \end{align}$$

However, I do not see where to go from here.

My feeling is that an argument along this line of thought will have us end up with $TSf(z) = cz$ for a $c \in \mathbb{C}$ with $|c| = 1$ by Schwartz's lemma, at which point we can say that the radius of convergence of the Taylor series of $f(z)$ will be $\infty$, but I do not currently see a path of logic from the given premises to that potential conclusion.

I've tried a couple calculations but the algebra isn't working out for me. Am I barking up the wrong tree, or am I missing something obvious? (Or, of course, is this entirely the wrong approach?)

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    $\begingroup$ To answer this, it helps to know what the automorphisms of the unit disk are. $\endgroup$ – Lubin Aug 19 '14 at 3:32
  • $\begingroup$ They are of the form $e^{i\theta}\frac{z-a}{1-\overline{a}z}$, where $|a| < 1$. The obvious avenue to pursue is letting $a = Sf(0)$ so as to apply Schwarz's lemma, but I'm not getting anywhere. $\endgroup$ – user169845 Aug 19 '14 at 5:33
  • $\begingroup$ @Lubin Thank you for your comment; I feel as though it pushed me in the right direction, and I have updated my question with what I have tried so far along the suggested lines. However, I have not discovered "the solution"; I would appreciate a further hint, if you wouldn't mind. $\endgroup$ – user169845 Aug 19 '14 at 6:34
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Suppose the radius of convergence were greater than $1$. Then $f$ would map the unit circle to a compact set, in particular, $\lvert f(z)\rvert \leqslant M$ for some $M < \infty$ and all $z$ with $\lvert z\rvert = 1$. Then the maximum principle says that $\lvert f(z)\rvert \leqslant M$ for all $z$ in the unit disk, and $f$ cannot map the unit disk onto the upper half plane (since that is unbounded).

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