I am looking for all analytic functions $f : \Bbb{C}\longrightarrow\Bbb{C}$ such that for all $z$ satisfy $|f(z)|^5 \leq |z|^6$. Any suggestion would be helpful.
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From the condition you get $f(0)=0$, hence either $f=0$ or $f(z)=z^k g(z)$ for some $k>0$ and $g$ entire with $g(0)\ne 0$. Now $|g(z)|^5\le |z|^{6-5k}$ for all $z\ne 0$. If $k>1$, this implies that $g$ is bounded, hence constant, but $f(z)=cz^k$ is no solution if $k>1$, $c\ne 0$. Therefore $k=1$ and $|g(z)|^5\le|z|$ for $z\ne0$ and by continuity also for $z=0$, contradicting $g(0)\ne 0$.
Therefore $f(z)=0$ is the only solution.
Hagen von Eitzen
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How do you get the implication that "either $f=0$ or $f(z) = z^kg(z)$ for some $k >0$ ?? – Drop Jul 27 '13 at 11:11
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Analytic implies $f$ is the sum of a Taylor series at zero. – GEdgar Jul 27 '13 at 11:47
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Your problem hinges on the following result
If $f(z)$ is an entire function such that $|f(z)|\leq B|z|^n$, $B>0,\, n\in\mathbb{Z^{+}}$, then $f(z)$ is a polynomial of degree at most $n$.
Mhenni Benghorbal
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