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 $z\in\mathbb{D}$
.
Best Answer
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.