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Find all entire functions such that $|f(z)|\leq |z^2-1|$ for all $z\in\mathbb C$.

For large $z$ we have $$|f(z)|\leq 2|z|^2$$ so $f$ is a polynomial of degree $\leq 2$. But how to continue? Could someone give me a hint?

Jolien
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    What can you say about $f(1)$ and $f(-1)$? – Hagen von Eitzen May 27 '14 at 14:53
  • We have $|f(1)|=|f(-1)|=0$, so $f(z) = C(z+1)(z-1)$. Wow, that's beautiful. – Jolien May 27 '14 at 14:54
  • @Jolien How do you stat that $f$ is a polynomial of degree$\leq$ 2? Are you using some theorem? – ajotatxe May 27 '14 at 15:01
  • @ajotatxe Yes, I use a theorem from my book: "Extended version of Liouvilles Theorem": Let $f$ be entire with $|f(z)|\leq M|z|^m$ for large $z$, for some $M$ and some $m>0$. Then $f$ is a polynomial of degree $\leq m$. – Jolien May 27 '14 at 15:05
  • @Jolien : Did you finally find an answer for this ? can you please post it as an answer below if you did. – Aman Mittal Jun 14 '14 at 15:49

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Since $|f(z)|\leq M |z|^2$ we know that $f$ has to be a polynomial of degree $\leq 2$. (Extended version of Liouvilles Theorem). Furthermore, we have $$ f(1) = 0 $$ and $$ f(-1)=0 $$ so therefore $f(z) = C(z-1)(z+1)=C(z^2-1)$ for some constant $-1\leq C\leq 1$.

Jolien
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