Suppose that an entire function $f(z)$ satisfies $\left|f(z)\right|\leq k\left|z\right|^n$ for sufficiently large $\left|z\right|$, where $n\in\mathbb{Z^+}$ and $k>0$ is constant. Show that $f$ is a polynomial of degree at most $n$.

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    $\begingroup$ Do you know Liouville's theorem? en.wikipedia.org/wiki/… $\endgroup$ – Paul May 10 '12 at 13:04
  • $\begingroup$ If $f$ is entire, then so are all of its derivatives. If an entire function is bounded, then it's constant. $\endgroup$ – Gerry Myerson Nov 17 '15 at 23:01
  • $\begingroup$ No I meant the nth derivative is bounded in the plane. And I need to show that f is a polynomial of degree n. @GerryMyerson Thank you for the help. $\endgroup$ – gmath Nov 17 '15 at 23:17
  • $\begingroup$ @user, please take half a minute to work through the logical implications of my earlier comment. Everything you want is there. $\endgroup$ – Gerry Myerson Nov 17 '15 at 23:20
  • $\begingroup$ I am sorry but I don't really see how the function being constant implies that f is a polynomial of degree n. @GerryMyerson $\endgroup$ – gmath Nov 17 '15 at 23:23

Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $\infty$, which must match its Taylor series there.


Since $|f(z)|\leq k|z|^m$, Cauchy's estimate gives

$$|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}$$ for all $|z|=R$. For $n>m$, letting $R\rightarrow\infty$, we see that $|f^{(n)}|=0$. It follows that $f$ is a polynomial of degree $\leq m$.



  • We have by Cauchy's integral formula that $$|f^{(d)}(0)|=\frac{d!}{2\pi R}\left|\int_{C(0,R)}\frac{f(z)}{z^{d+1}}dz\right|.$$
  • What about $f^{(d)}(0)$ if $d\geq n+1$?
  • Use the fact that $f$ is analytic at $0$ to get that $f(z)=\sum_{j=0}^n\frac{f^{(j)}(0)}{j!}z^j$ in a neighborhood of $0$.
  • Show that the last formula is in fact true for all $z\in\Bbb C$.
  • $\begingroup$ shouldn't you have $\frac{d!}{2 \pi i}$? $\endgroup$ – Andres Mejia May 19 '18 at 21:46

A really dirty way to do this:

  • Theorem 1: Jensen's Formula Corollary

    Suppose $f$ has order of growth $\rho$. Then there is a $C$ that for large enough $R$, $n(R) \le C R^{\rho} $ where $n(R)$ is the number of zeros whose magnitude is less than $R$.

  • Theorem 2: Hadamard Factorization Theorem

    Suppose $f$ has order of growth $k \le \rho \lt k+1$ where $k$ is an integer. Then $f(z)$ can be written $z^m e^{g(z)} \prod_n E_k(z/a_n)$ where $E_k$ is the kth canonical Weirerstrass factor and $a_n$ is the nth zero of $f$ and $g(z)$ is a polynomial of degree $k$.

Then note that by assumption $f$ has zero order of growth. Theorem 1 it follows that $f$ has finitely many zeros. From Theorem 2 it follows that $f$ is a polynomial. Then we need to put in a tiny bit of work to show that the degree of this polynomial is the one we need. (Just argue about $|f(z)/z^n|$ as $z$ grows large)


Look at closed discs centered at the origin, use maximum modulus principle to show that the function obtains its maximum value on the boundary, show that if you take a larger disc, you obtain a higher value, and thus use Liouville's Theorem to get that $\lim_{|z| \to \infty} |f(z)| = \infty$. Then show that such a function is a polynomial.


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