I am working through a problem where I am given some $ h \in \mathcal{H}(\mathbb{C})$ to be an entire function, with a pole of order $m$ at the point $\infty$.
I know this is equivalent to $\hat{h}(z)=h(\frac{1}{z})$ having a pole of order $m$ at $z=0$.
I have also shown previously that if there exists constants $\alpha, \beta$ and $m \in \mathbb{Z}$ such that $|h(z)| \leq \alpha |z|^m + \beta$ then $h$ is at most a polynomial of degree $m$.
The question is asking me to prove that $h$ is at most a polynomial of degree $m$, but I am struggling to see how the existence of the pole would imply that there exists a bound of the form $\alpha|z|^m + \beta$.
I tried using the Laurent series for $\hat{h}(z)$, and using the fact that all coefficients with indices $<-m$ will be $0$, but I don't see how to apply this further.
Any advice?
EDIT - Current Attempt
If $\hat{h}$ has a pole of order $m$ at $0$, then there is an entire function $g(z)$ with $g(0) \neq 0$ such that $\hat{h}(z) = z^{-m}g(z)$.
Since $g$ is entire, it has power series representation $g(z) = \sum_{n\geq 0} c_nz^n$. So on some punctured disk $D'_r(0)$ we have $$\hat{h}(z) = z^{-m}\sum_{n \geq 0} c_nz^n = \sum_{n \geq 0}c_nz^{n - m}$$
Inverting this gives $$h(z) = \sum_{n \geq 0} c_n z^{m-n} = \sum_{n=0}^{m}c_nz^{m-n} + \sum_{n \geq 0}c_{m+n}z^{-n}$$ on the set $\{z : |z| > \frac{1}{r} \}$
The second sum of course goes to $0$ for large $|z|$, but how do I obtain the bound of $|z|^m + 1$ from the first sum?
Let $$ \hat{h}(z)=\sum_{k=-m}^\infty c_kz^k $$ be the Laurent-series of $\hat{h}$ in $0$. Then $c:=c_{-m}=\lim_{z \to 0} z^m \hat{h}(z)$. Thus, there is some $\delta > 0$ such that $$ |z|^m |\hat{h}(z)| <|c|+1 \quad (0<|z| < \delta), $$ so $$ \frac{1}{|z|^m} |h(z)| < |c|+1=:\alpha \quad (|z| > \frac{1}{\delta}). $$ Since $\{z:|z| \le 1/\delta\}$ is compact, the maximum $\beta:=\max_{|z| \le 1/\delta}|h(z)|$ exists. Now, $$ |h(z)| \le \alpha |z|^m + \beta \quad (z \in \mathbb{C}). $$
Edit: We have (by continuity of power series) $$ \lim_{z \to 0} z^m \hat{h}(z)= \lim_{z \to 0}\sum_{k=-m}^\infty c_kz^{k+m}=c_{-m}=:c, $$ that is (by definition of the limit) to each $\varepsilon > 0$ there is some $\delta > 0$ such that $$ 0<|z| < \delta ~ \Rightarrow ~ |z^m \hat{h}(z) - c| < \varepsilon. $$ We apply this to $\varepsilon=1$. Then, for $0<|z| < \delta$ we get (using the reverse triangle inequality) $$ |z^m \hat{h}(z)| - |c| \le |z^m \hat{h}(z) - c| < 1, $$ so $$ |z|^m |\hat{h}(z)| <|c|+1. $$ Replacing $z$ by $1/z$ leads to $$ \frac{1}{|z|^m}|\hat{h}(1/z)| = \frac{1}{|z|^m} |h(z)| < |c|+1 $$ for $|z|> 1/\delta$. Note that the $c_k$ above differ from the $c_k$ in your representation of $\hat{h}$ just by an index shift. Both is correct but I wanted to avoid a reformulation of the original answer. In your notation $c=c_0$.