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Show that an entire function bounded by |z|^{10/3} is cubic

Let $f$ be an entire function such that $|f(z)| \le 1 + 2|z|^{10/3}$ for all $z$. Prove that $f$ is a cubic polynomial.

I thought about using the Extended Liouville Theorem:

If f is entire and for some integer $k \ge 0$ there exists positive constants $A$ and $B$ such that $|f(z)| \le A + B|z|^k$ for all $z$, then $f$ is a polynomial of degree $\le k$.

I found a similar problem here, which helped me a lot: if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant?

I replicated the approach to obtain an upper bound. Here's what I got:

Since $f$ is entire, it can be expressed as a power series:

$ f(z)=\sum\limits_{n=0}^\infty a_n z^n $ for all $z\in \mathbb{C}$.

From Cauchy's Integral Formula we have:

$ a_n= \frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz $

So,

$ |a_n| \le |\frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{f(z)}{z^{n+1}}dz| \le \frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{|f(z)|}{|z|^{n+1}}|dz| \le \frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{1 + 2|z|^{10/3}}{|z|^{n+1}}|dz| $

Therefore,

$ |a_n| \le \frac{1}{2\pi i}\int\limits_{\partial B(0,R)}\frac{1 + 2R^{10/3}}{R^n}dz $

And for $n\in\mathbb{N}$ we have $ |a_n|\leq \frac{1}{2\pi i} \lim\limits_{R\to+\infty}\frac{1+2R^{10/3}}{R^n} $ which implies $a_n=0$ for all $n \ge 4$. It follows that $f(z)$ is a polynomial of at most degree 3.

I'm having trouble coming up with a lower bound to show that $f(z)$ is a polynomial of at least degree 3.

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    @PantelisDamianou Thanks for that hint! You're right, it makes the problem much easier.2012-10-15

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