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Question: Let $f$ be an entire function such that $|f(z)|\leq1+2|z|^{10/3}$ for all z. Prove that $f$ is a cubic polynomial

Thoughts so far: Using a corollary of Liouville's theorem, we know that we want to show that $|f(z)|\leq a+b|z|^3$ and $|f(z)|\geq a+b|z|^3$ for some constants a and b. We know that within the unit circle $|f(z)|\leq 1+2|z|^{10/3} < 1+2|z|^3$ which gives us an upper bound, while outside of the unit circle we know that $-|f(z)|\geq -1-2|z|^{10/3} \implies |f(x)| \geq |-1-2|z|^{10/3}| = |2|z|^{10/3}--1|$ (by triangle inequality) $\geq 2|z|^{10/3}-1 > 2|z|^3-1$, which provides an lower bound of three, which by the corollary of Liouville's theorem implies that f(z) must be cubic. However, this proof makes me quesy because I feel that the upper and lower limits were chosen arbitrarily and could be any such function with a power less than $\frac{10}{3}%$, which makes me feel rather frustrated. Furthermore, this also leads me to believe that this is not a constructive line of thought for this problem.

Thank you in advance for any help that you may provide.

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    possible duplicate of [if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant?](http://math.stackexchange.com/questions/151700/if-f-is-entire-and-fz-leq-1z1-2-why-must-f-be-constant) In particular, see the comment to the accepted answer by Robert Israel.2012-10-14
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    hmm note that say the function $f \equiv \frac{1}{2}$ satisfies our condition, so I think by cubic we mean $f = a_0 + a_1z + a_2z^2 + a_3z^3$, for $a_i \in \mathbb{C}$ (and possibly zero). With that, for $|z|$ large enough, your condition reads $|f| \leqslant C|z|^{10/3}$ for your favorite $C >> 0$. From here, by a corollary of Liouville (Cauchy's inequality I think it's called), we deduce $\frac{d^4}{dz^4}f$ vanishes identically, as desired.2012-10-14
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    Thank you for your responses. I think I understand. We can show that the third derivative is a constant (see linked problem) and thus the fourth derivative vanishes, which means that f is at most a cubic function. So the question is badly worded then, since the function may be at most a cubic function rather than must be a cubic function.2012-10-14
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    As mentioned above, one can solve this problem using the integral formula. However, I would like to point out the error in your argument, which is that if $-|f(z)| \ge -1 - 2|z|^{10/3}$, it is *false* that $|f(z)| \ge |-1-2|z|^{10/3}|$.2012-10-14
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    @Wong Thank you for the feedback. I had not realized that this was an error.2012-10-14

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