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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
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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.

$$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$$

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 $\leq m$.

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Hints:

  • 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 $z\in\Bbb C$.
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  • $\begingroup$ shouldn't you have $\frac{d!}{2 \pi i}$? $\endgroup$ – Andres Mejia May 19 '18 at 21:46
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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)

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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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