Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Entire function bounded by a polynomial is a polynomial

Let $f \colon \mathbb{C}\to\mathbb{C}$ be an entire function (i.e. $f$ is analytic on $\mathbb{C}$) that satisfies $|f(z)|\leq M|z|^n$ where $n$ is a positive integer and $M \geq 0$ is a constant. By considering the estimates in the proof of Liouville's Theorem, deduce that $f$ has to be a polynomial of degree at most $n$.

share|improve this question
    
Use the hint. Try to estimate derivatives of $f$ of order $> n$. –  mrf Nov 2 '12 at 23:08
add comment

marked as duplicate by Cameron Buie, tomasz, froggie, Martin Sleziak, Noah Snyder Nov 3 '12 at 11:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

up vote 0 down vote accepted

$f$ is entire so it has an everywhere convergent taylor expansion $f(z) = \displaystyle\sum_{n=1}^{\infty }.$ For $k>n,$ $|a_k | = \frac{ f^{(k)} (0) }{k!} | = |\frac{1}{2\pi i} \displaystyle\int_{C_r} \frac{ f(\zeta )}{\zeta^{k+1} } | \le \frac{1}{2\pi } \frac{Mr^n}{2\pi r^{k+1} } 2\pi r = \frac{M}{r^{n-k} } $ - this follows from some basic integral inequalities and the Cauchy Integral Formula. Letting $r$ tend toward $\infty ,$ we have that $a_k=0.$

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.