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$f,g$ entire, both $\neq 0$ such that $|f|^2\leq |g|^3$

My reasoning is the following:

  1. Both can have zeroes

If $g$ has a zero at $z=a$, then $f$ must also have it, but we could find a neighborhood where the values are very close to $0$, so probably raising to the third power will make it even smaller, so $|g|^3<|f|^2$, thus a contradiction.

  1. $f$ can have zeroes but $g$ can't

Then $g$ is either a constant polynomial or something else, and $f$ is a polynomial of degree>0 or something else. $g$ can't be $cte$ because then, $f$ being a polynomial, going to $\infty$ the inequality doesn't hold, and if $f$ isn't a polynomial, then it has an essential singularity at $\infty$, which doesn't hold the inequality. If $g$ is anything else, at $\infty$ will have an essential singularity, so we could probably find some neighborhood where the inequality doesn't hold.

  1. $g$ can have zeroes but $f$ can't

Obvious contradiction

  1. None can have zeroes

Seems to me the last and only answer, in which both functions have to be $cte$, otherwise we are in the same position as before (both having essential singularities at $\infty$)

Any thoughts? Is it fine what I've done? I know I haven't elaborate the cases where I say that we could find a neighborhood... but intuitively seems to work out fine. Thanks in advance.

Mateo
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  • I do not understand your reasoning (and what does “cte” mean?). But $f$ and $g$ obviously can have zeros, e.g. $f(z) = z^3$ and $g(z) = z^2$, or more generally $f(z) = c_1 h(z)^3$ and $g(z) = c_2 h(z)^2$ where $c_1, c_2$ are constants and $h$ is an arbitrary entire function (with or without zeros). – Martin R Jan 10 '24 at 15:18
  • My fault, I was using the abbreviation of constant in spanish ("cte" for "constante"). And you are right, I didn't think about that example, thanks! – Mateo Jan 10 '24 at 16:28
  • I was thinking that if you had 2 distinct functions, with no relation, that means they have different rates when approaching zero, so I thought it is always the case that because $g$ is raised to the third power, even though $|f|\leq|g|$, there is some point near the zero where the inequality $|f|^2\leq|g|^3$ wouldn't hold. Is that thought correct? – Mateo Jan 10 '24 at 16:54

1 Answers1

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By Entire function dominated by another entire function is a constant multiple if $f$ and $g$ are non-zero the functions have the same zeroes.

Kenta S
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  • Thank you!, So it must be the case that $f^2=\alpha g^3$ for some $\alpha\leq1$. Instead of what @Martin R said, would it be wrong to say that $f=\sqrt{\alpha g^3}$? This should give us an entire function right? (Because normally we should have to select a banch cut...) – Mateo Jan 10 '24 at 16:43