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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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Do you know Liouville's theorem? en.wikipedia.org/wiki/… –  Paul May 10 '12 at 13:04
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3 Answers 3

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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.

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

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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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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