Entire function Problem

Question

I am new to complex analysis and I am facing difficulty in the following problem:

Let $f(z)$ be an entire function satisfying $|f(z)| \leq k|z|^2$ for some +ve constant k and all z. Show that $ f(z) = az^2$ for some constant a.

Which concepts will be used here?

Does this answer your question? Show that if $|f(z)| \leq M |z|^n$ then $f$ is a polynomial max degree n — The Amplitwist, Dec 10 '21 at 12:32
Does this answer your question? Entire function bounded by a polynomial is a polynomial — Arctic Char, Feb 04 '22 at 21:22

Spock · Accepted Answer · 2014-04-12T10:32:16.087

4

This is an application of Liouville's theorem.

$|f(z)|\leq|(\sqrt{k}z)^2|\implies|\frac{f(z)}{(\sqrt{k}z)^2}|\leq1$ Notice that $f(z)$ and $(\sqrt{k}z)^2$ are both entire functions thus $\frac{f(z)}{(\sqrt{k}z)^2}$ is entire and it's also bounded therefore it's constant and the result follows.

edited Apr 12 '14 at 10:32

answered Apr 12 '14 at 08:17

Spock

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To the answer above, observe that the obtained function is not necessarily entire. However, it can be extended to an entire function, for it is bounded in some neighbourhood of 0, which is its only singularity point. Liouville's theorem provides a constant that works for the new function, and so for the old one, as well. – Apr 14 '14 at 15:42

Entire function Problem

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