Suppose $f$ is a function which is holomorphic on $\mathbb{C}\setminus A$ where $A$ is the set of points where $f$ has a singularity. Suppose that all of the points in $A$ are removable singularities of $f$. Here is my question: does this imply that $f$ is itself entire? I understand that by Riemann Extension Theorem, $f$ can be extended to an entire $F$, but my question has to do with whether we can say that $f$ itself is entire. I have seen a few other questions on this site which make such a statement, e.g. Removable singularities and an entire function and I am not sure if they are just slurring notation or if I am missing something.
The context in which this arose: I am trying to show that if two entire functions $f,g$ are such that $|f|\leq |g|$, then one is a multiple of the other. Obviously the strategy is to take quotient, and show that each singularity is removable. I have been able to do this, but then after that I am lost. I know I am supposed to use liouville to show that bounded and entire implies constant, but I am not sure if $|f|/|g|$ is itself entire. Is it not supposed to be some extended function which is supposed to be entire? With such an extended function, indeed we would have bounded and entire, but then I am not sure how to show that $f$ and $g$ are multiples of one another on all of $\mathbb{C}$, since things get strange around the singularities.
I would appreciate anything which clarifies my understanding.
Best Answer
Regarding your context:
Let $S$ be the set of zeroes of $g$.
By the inequality, the set of zeroes of $f$ is also $S$.
The domain of an entire function is necessarily $\mathbb C$ by definition. Therefore, barring any sort of ‘extensions’, the largest possible domain of $\frac fg$ is $\mathbb C\setminus S$, due to the fact that $\frac fg=\frac 00$ on $S$ and $\frac 00$ is not well-defined. Thus, $\frac fg$ cannot be entire.
The statement you want to prove is ‘$f$ and $g$ are multiple of each other.’ Mathematically, this can be restated as $f=cg$ for some universal, non-zero constant $c$.
This statement is trivially true on $S$, what remains is proving it on $\mathbb C\setminus S$.
You may proceed like this:
Let $S$ be the set of zeroes of $g$.
By the inequality, $$\left\vert\frac fg\right\vert \le 1\text{ for }\mathbb C\setminus S$$
Let $h=\frac fg$. Since the zeroes of $g$ is isolated, there exist a neighbourhood $N$ of every element of $S$, such that $N\in\mathbb C\setminus S$ and thus $|h|\le 1$ holds on $N$.
By Riemann’s removable singularity theorem, $h$ can be extended to an entire $H$.
Then, by Liouville theorem $H=c$ on $\mathbb C$.
Recall that $H=h$ on $\mathbb C\setminus S$. Hence $h=c$ on $\mathbb C\setminus S$.
Therefore, you can conclude $f=cg$ on $\mathbb C\setminus S$.
A few final words: Your first question regarding $f$ cannot be answered because you did not specify how $f$ is defined on $A$.
Whenever you ask whether a function $f$ is entire, always think about where did you define it. A function is always defined along with a domain, and $f$ can be entire only if its domain is $\mathbb C$.
If you define $\sin z :[0,1]$, it can never be entire. If you have $f$ holomorphic on $\mathbb C\setminus A$, before you ask whether it is entire, ask yourself how $f$ is defined on $A$. If for $a\in A$, $f(a)$ does not return a complex number but a set, or a function, or $\text{Donald Trump}$, then there is no point to discuss about being entire or not.
It turns out that it is the same case in your context: without any extensions, $\frac fg$ cannot be defined on $S$ because we don’t know how to define $\frac 00$. Discussion on entireness immediately ends. Of course, if you define $\frac fg$ on $S$ by its continuous extension, then by Riemann’s removable singularity theorem continuous extension is same as holomorphic extension, hence $\frac fg$ is holomorphic on $S$ too.