I'm stuck with the following question:

Let $\Delta$ be the unit disk and let $H$ be the upper half-plane. Show that any sequence of holomorphic functions $f_n:\Delta \rightarrow H$ either has a subsequence converging normally or a subsequence converging to $\infty$ at every point in $\Delta$.

I found a theorem in Bruce Palka's textbook (Theorem 4.6 in chapter 7) which states that this is indeed the case for a family of analytic functions on a domain $D$ provided that all the functions in the family miss two complex values. It is not proven there because further tools are needed.

Is there an easy proof in my simpler context?

  • 1
    $\begingroup$ Sure. Let $T\colon z \mapsto \frac{z-i}{z+i}$ and consider the sequence $g_n = T\circ f_n$. $\endgroup$ – Daniel Fischer Aug 17 '14 at 16:11
  • $\begingroup$ Composing with $T$ yields a sequence $g_n:\Delta \rightarrow \Delta$, so in particular $g_n$ is locally bounded and hence by Monetl's theorem it has a subsequence which converges uniformly on compact subets. When is there a subsequence converging to $\infty$ at every point in $\Delta$ then? $\endgroup$ – user54631 Aug 17 '14 at 16:31
  • $\begingroup$ When you have the compactly convergent subsequence of the $g_n$ (without loss of generality, the entire sequence), then you look at $f_n = T^{-1}\circ g_n$. Then $f_n \to \infty$ if and only if $g_n$ converges to which function? $\endgroup$ – Daniel Fischer Aug 17 '14 at 16:35
  • $\begingroup$ I see. $T^{-1}(w)=i(w+1)/(w-1)$ so if the sequence $g_n$ converges to $g\equiv 1$, then $f_n=T^{-1}\circ g_n$ diverges at every point. Thank you very much for your hints. $\endgroup$ – user54631 Aug 17 '14 at 16:53

If $G\subset \mathbb{C}$ is an open subset that is not dense, then the normality of the family $\mathscr{O}(U,G)$ of holomorphic functions $f\colon U\to G$, where $U\subset \mathbb{C}$ is a connected open set, in the sense that every sequence $(f_n)$ in $\mathscr{O}(U,G)$ has a subsequence converging compactly to a holomorphic function on $U$ or to the constant $\infty$ is an easy consequence of the classical Montel theorem.

Suppose we have $D_r(z_0) = \{ z\in\mathbb{C} : \lvert z-z_0\rvert < r\}$ in the complement of $G$. The Möbius transformation $\mu \colon z \mapsto \frac{1}{z-z_0}$ maps $\hat{\mathbb{C}}\setminus D_r(z_0)$ biholomorphically to $\overline{D_{1/r}(0)}$, in particular $\mu(G)\subset D_{1/r}(0)$, so the family $\{\mu\circ f : f \in \mathscr{O}(U,G)\}$ is uniformly bounded, and by the classical Montel theorem the sequence $(\mu\circ f_n)$ has a subsequence that converges compactly to a holomorphic function $h\colon U \to \overline{D_{1/r}(0)}$. By Hurwitz's theorem, either $h$ has no zeros, or $h\equiv 0$, since $h_n = \mu\circ f_n$ is zero-free for all $n$.

Since Möbius transformations are automorphisms of the sphere $\hat{\mathbb{C}}$, it follows that

$$f_n = \mu^{-1}\circ h_n \to \mu^{-1}\circ h = \frac{1}{h} + z_0$$

uniformly on compact subsets of $U$, and that is either a holomorphic function (if $h$ has no zeros) or the constant $\infty$ (if $h\equiv 0$).

The mentioned theorem that the family $\mathscr{O}(U,G)$ is normal (in the above sense) if $G = \mathbb{C}\setminus \{a,b\}$ is the so-called big theorem of Montel, which is indeed much harder to prove.


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