$\textbf{Problem.}$ Suppose $f$ is holomorphic on the unit disk $\mathbb{D}$. Show there are points $a_n\in \mathbb{D}$, $a\in \partial \mathbb{D}$, and $b\in \mathbb{C}$ such that $a_n\to a$ and $f(a_n)\to b$, as $n\to \infty.$
$\textbf{Attempt at a Solution:}$
First, if $f$ has an essential singularity at some point $a\in \partial\mathbb{D}$, I think the statement follows easily.
To contradiction suppose that the statement is false. That is, suppose that $f(a_n)\to \infty$ for all sequences $\{a_n\}$ in $\mathbb{D}$ where $a_n\to a$, $a\in \partial\mathbb{D}.$
Then $$\frac{1}{f(z)}\to 0, \hspace{1cm} \mbox{as} \hspace{1cm}|z|\to 1.$$
$\textbf{ Case 1}$: If $f$ is nowhere vanishing on $\mathbb{D}$, then $1/f(z)$ would be holomorphic in $\mathbb{D}$ and can be extended continuously on $\bar{\mathbb{D}}$. Then, by the maximum principle this would imply that $1/f(z)\equiv 0$ on all of $\mathbb{D}$. But this contradicts the fact that $f$ is holomorphic (non-constant) in $\mathbb{D}.$
$\textbf{Case 2}$: Suppose $f$ has a finite number of zeros in $\mathbb{D}$. That is, let $\{b_1,…,b_n\}$ be points of $\mathbb{D}$ such that $f(b_i)=0.$
Let $r=\min \{|z-b_i| \forall z\in \partial\mathbb{D}\}$, and consider the annulus $$A=\{z\in\mathbb{C}: r<r_1<|z|<1\}$$
Then, $1/f(z)$ is holomorphic on $A$, and can be continuously extended to $\bar A$. (Note that, $1/f(z)$ is already holomorphic for $|z|=r_1$.
Let $$g(z)=\frac{1}{f(z)}\frac{1}{f(r_1/z)}.$$ Then $g(z)=0$ for all $z\in \partial \bar{A}$. By the maximum principle it follows that $g(z)\equiv 0$ on $A$, and by the identity theorem it follows that $1/f(z)\equiv 0$ on $A$. But this, again, contradicts the fact that $f$ is holomorphic on $\mathbb{D}$ and thus on $A$ as well.
$\textbf{Case 3:}$ This is the case where i'm not so sure what i'm doing is correct.
Suppose, now, that $f$ has an infinite number of zeros in $\mathbb{D}$. That is, let $E=\{b_i\in \mathbb{D}: g(b_i)=0\}$. Then, $E$ cannot have an accumulation point in $\mathbb{D}.$
My idea was to take neighborhoods, $\mathbb{B}_{r_i}(b_i)$, and consider $$g(z)=\frac{1}{f(z)}\prod_{w\in A_w}\frac{1}{f(z/w)},$$ where $A_w=\{w\in \mathbb{D}: |w-b_i|=r_i\},$ for all $z\in \overline{\mathbb{D}\setminus\bigcup \mathbb{B}_{r_i}(b_i)}.$
Then, I would argue similarly as in Case 2.
There is probably a much shorter method.
I'd appreciate any thoughts or suggestions!!
Best Answer
That is inaccurate, a pole is by definition an isolated singularity.
That doesn't affect the argument, though.
Note that case 3. cannot occur. If $f$ has infinitely many zeros in $\mathbb{D}$, since the zeros cannot have an accumulation point in $\mathbb{D}$ without $f$ being identically $0$, you have a seuqence $(z_k)$ of points in $\mathbb{D}$ converging to a boundary point $z_\ast$ with $f(z_k) = 0$ for all $k$, hence $f(z_k) \to 0$.
To shorten the argument further, treat cases 1. and 2. together, consider
$$g(z) = \frac{\prod(z-z_k)^{n_k}}{f(z)},$$
where the $z_k$ are the finitely many zeros of $f$ in $\mathbb{D}$ (possibly none) with multiplicities $n_k$. Then you still have $g(z) \to 0$ for $z\to \partial \mathbb{D}$, hence $g \equiv 0$, contradicting the assumption.