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Prove $f(z)$ is a polynomial if $f(z)$ is entire and $|f(z)| \leq (1 + |z|)^n$ $\forall z \in C$.

Here is what I wrote for my proof: $f(z)$ can be represented as a power series $\sum\limits_{n=0}^\infty a_n z^n$ where $a_n = \frac{f^n(0)}{n!}$ if we choose $|z| = r$. Then by Cauchy's Estimates, we have that $|a_m| = \frac{|f^{(m)} (0)|}{m!} \leq \frac{(1+r)^n}{r^m}$ where $m > n$, so $a_m \to 0$ and $f(z)$ is a polynomial of degree $\leq n$.

My solution was marked incorrect, but I didn't have any other ideas on how to approach this. What is the proper solution?

mr eyeglasses
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    Well, I would conclude that $a_m = 0$ (as opposed to $a_m \to 0$) and you have $a_n = {1 \over n!} f^{(n)}(0)$ regardless of $z$, but, other than that, it is a reasonable approach. – copper.hat May 19 '15 at 17:47
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    Perhaps you forgot to include the $20 bill in your answer? – copper.hat May 19 '15 at 17:48

1 Answers1

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Seems fine to me... perhaps apart from saying "$|a_m|\to 0$, since those coefficients are constants, so can't "go" anywhere. But this is minor...

paul garrett
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