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.

  • $\begingroup$ possible duplicate of if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant? In particular, see the comment to the accepted answer by Robert Israel. $\endgroup$ – EuYu Oct 14 '12 at 7:10
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    $\begingroup$ 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. $\endgroup$ – uncookedfalcon Oct 14 '12 at 7:12
  • $\begingroup$ 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. $\endgroup$ – ABC Bach Oct 14 '12 at 7:21
  • $\begingroup$ 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 $|f(z)| \ge |-1-2|z|^{10/3}|$. $\endgroup$ – Christopher A. Wong Oct 14 '12 at 7:57
  • $\begingroup$ @Wong Thank you for the feedback. I had not realized that this was an error. $\endgroup$ – ABC Bach Oct 14 '12 at 8:05

The general form of this classic problem is the following proposition:

If $f$ is an entire function satisfying $|f(z)| \le A + B|z|^k$ for some positive constants $A,B$ and some nonnegative integer $k$, then $f$ is a polynomial of degree at most $k$.

I'll prove this proposition in a moment, by induction on $k$. Note that if the proposition is true, then it implies the following theorem:

If $f$ is an entire function satisfying $|f(z)| \le A + B|z|^\gamma$ for some positive constants $A,B,\gamma$, then $f$ is a polynomial of degree at most $\lfloor\gamma\rfloor$.

(Justification: the inequality with $\gamma$ implies the inequality for $\lfloor\gamma\rfloor+1$, possibly with different constants $A,B$. So $f$ is a polynomial of degree at most $\lfloor\gamma\rfloor+1$. But if it truly had degree $\lfloor\gamma\rfloor+1$, then it would grow faster than $|z|^\gamma$, violating the given inequality; so it actually has degree at most $\lfloor\gamma\rfloor$.)

The base case $k=0$ of the Proposition is simply Liouville's Theorem: a bounded entire function is constant. Suppose the proposition is true for $k-1$, and let $f$ satisfy $|f(z)| \le A+B|z|^k$. Let $g(z) = (f(z)-f(0))/z$, which is also an entire function (its simgularity at $z=0$ is removable). It's easy to check that $|g(z)| \le C + D|z|^{k-1}$ for some positive constants $C,D$ (check separately on $|z|\le1$ and $|z|\ge1$). By the induction hypothesis, $g$ is a polynomial of degree at most $k-1$; since $f(z) = zg(z)+f(0)$, we see that $f$ is a polynomial of degree at most $k$.


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