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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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matt
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  • Surely $f(z)$ doesn't need to have degree at least 3. If $f(z) \equiv 0$, then $f$ satisfies the inequality and is entire. – Christopher A. Wong Oct 14 '12 at 07:49
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    @matt here is a duplicate post with a tentative solution http://math.stackexchange.com/q/213491/43176 – ABC Bach Oct 14 '12 at 07:52
  • ah! sorry, I didn't see it because it was a new post. thanks! – matt Oct 14 '12 at 08:02
  • You can save a lot of work by using Cauchy Estimates: $$|a_n| \le \frac{n!M(R)}{R^n} .$$ In this case for $n\ge 4$ the right hand side goes to zero for big $R$. Here $M(R)$ is the maximum value of the modulus of $f$ on a circle with center at 0 and radius $R$. – PAD Oct 14 '12 at 08:31
  • @PantelisDamianou Thanks for that hint! You're right, it makes the problem much easier. – matt Oct 15 '12 at 01:14

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