# Conditions on an entire function that would make it a polynomial

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 $$M=\max_{|z-a|. Since $$|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!

• "each derivative is bounded, and thus may be estimated by a polynomial, thus š¯‘“ can be too?" - think about k > N. It would mean some derivative is indeed bounded, and entire, which means it is...? Commented Jul 9, 2020 at 2:12
• constant by Louiville's theorem, right? I suppose I was stuck on showing that $\frac{k!(A|z|^N+B)}{r^k}<\infty$. For instance, if $r$ is really small. Commented Jul 9, 2020 at 4:24
• That's right, but since f is entire, it has no poles, so you know it is bounded near r = 0. Commented Jul 10, 2020 at 14:32

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.