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Denote $\mathbb{D}=\left\{ z\,|\,\left|z\right|<1\right\} $ and $\mathbb{H}^{+}=\left\{ z\,|\,\mathfrak{R}\left(z\right)>0\right\}$.

Claim: Suppose $f_{n}:\mathbb{D}\to\mathbb{H}^{+}$ is a sequence of holomorphic functions, then there is a subsequence of $f_{n}$ that converges uniformly on compact subsets of $\mathbb{D}$ either to a holomorphic function or to $\infty$.

I know that if $f_{n}$ are uniformly bounded then there's a subseuqence converging to a holomorphic function uniformly on compact subsets. I'm lost as to how to deal with the case where $f_{n}$ are not uniformly bounded in order to show there's a subsequence converging uniformly to $\infty$. In particular I don't understand the significance of choice of domain and range of the functions to the question.

Note: uniform convergence to $\infty$ means that for all $M>0$ there is an $N$ such that for all $n>N$ it holds that $\left|f_{n}\left(z\right)\right|>M$ for all $z\in\mathbb{D}$ .

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    $\begingroup$ Essentially, a half-plane is conformally equivalent to a disk, so your sequence is equivalent to a uniformly bounded sequence. $\endgroup$ – Daniel Fischer Jan 16 '14 at 10:19
  • $\begingroup$ I read the solution and to be honest I didn't fully understand it. I would appreciate it if I could get an answer specifically on how to show that under these conditions if there's no uniform boundedness there's either a contradiction or a subsequence converging uniformly to $\infty$. $\endgroup$ – Serpahimz Jan 16 '14 at 10:28
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For every $n$, let

$$g_n(z) = \frac{f_n(z)-1}{f_n(z)+1}.$$

Then $(g_n)$ is a sequence of holomorphic functions with values in the unit disk, hence a uniformly bounded sequence, and we know that there is a subsequence $(g_{n_k})$ that is locally uniformly convergent to a holomorphic function $h \colon \mathbb{D}\to \overline{\mathbb{D}}$.

The Möbius transformation

$$T\colon w \mapsto \frac{w-1}{w+1}$$

used to obtain the $g_n$ from the $f_n$ is a homeomorphism of the Riemann sphere, hence so is its inverse

$$T^{-1} \colon w \mapsto \frac{1+w}{1-w},$$

and the locally uniform convergence of $g_{n_k}$ to $h$ is equivalent to the locally uniform convergence of $f_{n_k}$ to $T^{-1}\circ h$.

If $h(z) \equiv 1$, then the $f_{n_k}$ converge locally uniformly to $\infty$. If $h(z) \equiv c \neq 1$, the $f_{n_k}$ converge locally uniformly to the finite constant $T^{-1}(c) = \frac{1+c}{1-c}$, and if $h$ is not constant, by the open mapping theorem its image is actually contained in $\mathbb{D}$ and the $f_{n_k}$ converge locally uniformly to the non-constant function

$$z \mapsto \frac{1+h(z)}{1-h(z)}$$

with values in the right half-plane.

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  • $\begingroup$ How would the approach change, if the original domain were an arbitrary open subset of the complex plane, rather than the unit disk? $\endgroup$ – Ellya Apr 21 '14 at 22:25
  • $\begingroup$ Not at all. We apply the transformation to the range. The family of holomorphic functions $\Omega\to H$, where $\Omega \subset \mathbb{C}$ is an open set and $H$ is any half-plane is normal. By applying a Möbius transformation, we transform it to a bounded family, which we know is normal, and then applying the inverse Möbius transformation keeps locally uniform convergence, since Möbius transformations are homeomorphisms of the Riemann sphere. $\endgroup$ – Daniel Fischer Apr 21 '14 at 22:35
  • $\begingroup$ You have just answered my previous question (math.stackexchange.com/questions/763692/…), and there is a third part, which I am putting in an edit, I have put my approach, and it will use some of the ideas from what is here. the only thing is that mobius transforms etc, are not part of the syllabus I have covered, so I wanted to find a different approach, see my edit please, and Thanks! $\endgroup$ – Ellya Apr 21 '14 at 22:38

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