Let the $f(z)$ be analytic and $\vert f(z) \vert \leq \vert \sin({1\over z})\vert$ on $\mathbb{C}^\# (= \mathbb{C} \setminus \{0\})$[There are no more information aobut the $f$]
Put $f(\frac{1}{z}) =g(z)$
In my lecture's note, he said $h(z)$ is a entire function by defining new function $h$(Here the $n\in \mathbb{Z}$)
$$h(z) = \begin{cases}
g(z) \over \sin z & \text{$z \neq n\pi$} \\
\lim\limits_{z \to n\pi} \frac{g(z)}{\sin z} & \text{$z = n\pi$}
\end{cases}$$
And he drew the conclusion that $h$ is a constant on $\mathbb{C}$(By Liouville thm)
But I have a doubt about his solution. Let me suggest Why I do think like that. At least I've known, To claim the $h$ is a entire(or applying the Riemann thm), $\frac{g}{\sin z}$ should have a removable singularity at $0$. Then, owing to the $z=0$ is a zero for $\sin z$, $g$ does be analytic or removable at $0$. But the since $g$ is a analytic on $\mathbb{C}^\#$, this function is having a removable at $z = 0$. But we can't sure what the type of the singularity is at $0$ for $g(z)$. For example, if the $g(z)$ has a Essential singularity at $0$, then $0$ is not removable for $g \over \sin z$.(I guess the $\frac{g(z)}{\sin z}$ has the Essential singularity at $0$, we can't define the limits at $0$) Therfore, we can't apply the Riemann's thm for $0$ in that case.(I.e. $h$ can have a singularity at $0$[Not entire] )
Is my thought right? Or His solution is right?
Thanks.
Best Answer
I have no idea how to type # in mathematical mode, so I will write $\mathbb{C}^*=\{z\in\mathbb{C};z\ne 0\}$.
Your hyphotesis $|f(z)| \le |\sin(\frac{1}{z})|, \forall z \in \mathbb{C}^*$ means $|g(z)|=|f(\frac{1}{z})| \le |\sin(z)|, \forall z \in \mathbb{C}^*$. As $|g(z)| \ge 0$ you can easily use sandwich rule to conclude that $\lim_{z \to 0}g(z) = 0$, so $g(z)$ has a removable singularity at $z=0$.
Well, I hope it helps (I wrote this in a hurry and my English sucks).