Polynomial Growth on a Region The current question is related to this past question.
Let $f(z)$ be an entire function. If $|f(z)|<|z|^n$ for $|\text{Im}(z)|>1$, does it follow that $f(z)$ is a polynomial?
I know how to prove the statement is true if we haven't restricted the domain where the inequality holds; in fact, we could then say it's a polynomial of degree at most $n$. This makes me lean toward assuming the statement here is true. When I asked my professor, he wasn't as convinced, so I thought I'd come here.
Does anyone have any proofs or counterexample to the above?
 A: A counterexample is provided on p. 165 of the book Complex Analysis by Joseph Bak and Donald Newman; their function is bounded by a constant on $|\text{Im}(z)|\geq \pi$, but obviously you can adjust accordingly.
I'll paraphrase the idea, though you can view it yourself on the Google preview of the book. Put
$$
f(z) = \int_0^\infty \frac{e^{zt}}{t^t}\,dt;
$$
the integral converges absolutely for every $z$, and by applying Fubini's and Morera's theorems one sees that $f$ is entire. Likewise, it is clear from the formula that $\overline{f(z)} = f(\bar{z})$, so $f$ is symmetric about the real axis and we need only show that $|f(z)| \leq C$ for some constant $C$ on $\text{Im}(z)>\pi$. To prove this bound, they use the analyticity of the integrand (in the integral defining $f$) in the variable $t$ to replace the integral over the real line with the limit of two contour integrals:
$$
\int_0^\infty \frac{e^{zt}}{t^t}\,dt = \int_{I}-\lim_{r\to \infty} \int_{c_r} \frac{e^{zt}}{t^t}\,dt;
$$
here $I$ is the imaginary axis and $c_r$ is the quarter circle of radius $r$ in the upper right quadrant. They then parametrize both curves and bound the integrands straightforwardly to show that $f$ is bounded when $\text{Im}(z)>\pi$.
