Here is the question:
Suppose $f$ is entire.
a) Suppose $|f(z)|\leq A|z|^N+B$ $\forall z\in\mathbb{C}$ where $A,B<\infty$ are constants. Show $f$ is a polynomial of degree $\leq N$.
b) Suppose $f$ satisfies $f(z_n)\rightarrow\infty$ whenever $z_n\rightarrow\infty$. Show $f$ is a polynomial.
Here is my idea:
for (a), since $f$ is entire we can consider $f$ on a circle centered at $a$ of radius $r$. By Cauchy, we can estimate the $k^{th}$ derivative by $|f^{(k)}(z)|\leq\frac{k!M}{r^k}$ where $M=\max_{|z-a|<r}|f(z)|$. Since $|f(z)|\leq A|z|^N+B$, we know that $M\leq A|z|^N+B$. Thus, $|f^{(k)}(z)|\leq\frac{k!(A|z|^N+B)}{r^k}$. So could this lead me to an argument where I could say that each derivative is bounded, and thus may be estimated by a polynomial, thus $f$ can be too? Moreover, since the highest power of any derivative is $N$, then can we say that the highest power of $f$ is $N$ too? The trouble is that what if the $k^{th}$ derivative is of power $N$, then $f$ would have power $N+1$… so something is wrong here…
I'm not too sure where to start with (b). I was going to try and piggy-back off my argument in (a) but I wasn't seeing any way that it could be done.
Any hints, ideas, thoughts, etc. are greatly appreciated! Thank you so much!
Best Answer
For (b), that condition implies that $f$ extends to a meromorphic function on the Riemann sphere, with a pole at $\infty$ and no poles anywhere else. The meromorphic functions on the Riemann sphere are precisely the rational functions, so $f(x) = p(x) / q(x)$ for some coprime polynomials $p$ and $q$. Since $f$ has no poles on $\mathbb{C}$, $q(x)$ cannot have any roots so must be constant.