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Let $f$ be an entire function such that $|f(z)|\leq A+B|z|^k$ for all $z$, where $A$, $B$, $k$ are positive numbers.

Prove that $f$ is polynomial.

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By "full" I'm guessing you mean "entire". Please consider adding some remarks on what you've tried for this problem. –  Dylan Moreland Jul 2 '12 at 19:35
    
Hint: show that the n'th derivatives of $f$ begin to vanish at a certain point, using Cauchy's theorem –  Alex R. Jul 2 '12 at 19:39
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3 Answers

Since $f(z)$ is entire, from Cauchy integral formula, we have $$f^{(n)}(0) = \dfrac{n!}{2 \pi i}\oint_{C_r} \dfrac{f(z)}{z^{n+1}} dz$$ On $C_r$, the integrand is bounded by $$\left \vert \dfrac{f(z)}{z^{n+1}} \right \vert \leq \dfrac{A}{\left \vert z^{n+1} \right \vert} + \dfrac{B}{\left \vert z^{n+1-k} \right \vert} = \dfrac{A}{r^{n+1}} + \dfrac{B}{r^{n+1-k}}$$ Now argue out why $f^{(n)}(0) = 0$ for $n > k$ by letting $r \to \infty$ and looking at what happens to the integral.

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Note that this is nothing but a minor refinement of the usual proof of Liouville's theorem –  t.b. Jul 2 '12 at 20:11
    
@DylanMoreland Ah Yes. I have changed it now. –  user17762 Jul 2 '12 at 20:26
    
+1 Very nice and focused answer. –  DonAntonio Jul 2 '12 at 21:57
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Hint: start with $f(z)/z^k$ (assuming $k$ is an integer) and subtract the principal part of the pole at $0$. What could be left?

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Show that $f$ has (at worst) a pole at infinity.

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