I think so - but I'd rather ask the MSE community too.
Say I am given the bound |f(z)| < $|z|^3$, and that f is entire. Show f must be a polynomial.
I used Cauchy's Integral Formula for derivatives and showed that for n>3, all of the derivatives are zero, when we let the closed contour grow to infinity.
But, I have only used CIF at the point z = 0 - I didn't keep z arbitrary.
Then the CIF shows that, at z = 0, the Taylor series has only finitely many terms and so f(z) must be a polynomial.
So I feel there's actually no need to apply the CIF to an arbitrary z.
We have enough information on f(z) just from looking at z=0.
Thanks,
The condition $|f(z)| < |z|^3$ should be $|f(z)| \le |z|^3$, otherwise we get $0\le |f(0)| < |0^3| = 0$.
The condition implies that $0$ is a zero of $f$ of order at least $3$ (see below) and so $f(z)=z^3 g(z)$ with $g$ entire and bounded. Hence $g$ is constant by Liouville's theorem.
Indeed, $|f(z)| \le |z|^3$ implies $|f(0)|\le0$ and so $f(0)=0$. Write $f(z)=z f_1(z)$ with $f_1$ entire. Then $|f_1(z)| \le |z|^2$ and as before this implies $f_1(0)=0$. Repeat the argument twice and conclude that $f(z)=z^3 g(z)$ with $g$ entire and $|g(z)|\le1$.