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Let $f : \Bbb C \longrightarrow \Bbb C$ be an entire function i.e., analytic everywhere in $\Bbb C.$ Suppose $$\lim\limits_{\left \lvert z \right \rvert \to \infty} \dfrac {f(z)} {z} = 0.$$ Prove that $f$ is a constant function.

My attempt $:$ By the given condition it follows that $$\lim\limits_{z \to 0} z\ f \left (\dfrac {1} {z} \right ) = 0.$$

That means the function $h(z) = f \left (\dfrac {1} {z} \right )$ has a removable singularity at the origin. Hence $h$ is also an entire function. So there exist $M_1,M_2 \gt 0$ such that for all $z \in \Bbb C$ with $\left \lvert z \right \rvert \leq 1$ we have $\left \lvert f(z) \right \rvert \leq M_1$ and $\left \lvert h(z) \right \rvert \leq M_2.$ Which in turn implies that for all $z \in \Bbb C$ we have $\left \lvert f(z) \right \rvert \leq M,$ where $M = \max\ \{M_1,M_2 \}.$ Now Liouville's theorem finishes the proof.

Is my attempt correct? Please verify it. I am also eager to know other alternative methods (if available) by using power series expansion of the entire function $f$ over the entire complex plane (this is the hint given with the problem). Any help in this regard will be highly appreciated.

Thanks in advance.

Anacardium
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  • For alternative methods you might want to look at https://math.stackexchange.com/questions/1745437/if-f-is-entire-and-lim-limitsz-to-infty-fracfzz-0-show-that-f – offline Nov 01 '20 at 19:27
  • @offline I wonder that my process is different from all the processes in the link you provided. – Anacardium Nov 01 '20 at 19:30
  • Your proof seems fine. Another way, which allows you to prove a more general statement, is to use Cauchy's estimates. You can prove, without much additional effort, that an entire function which is asymptotically bounded by a polynomial must be itself a polynomial. Try using Cauchy's estimates to show that an entire function bounded by $z^n$ must have a finite series expansion. – EuYu Nov 01 '20 at 19:32
  • Do you mean extended Liouville's theorem? – Anacardium Nov 01 '20 at 19:45

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