$f,g$ entire, both $\neq 0$ such that $|f|^2\leq |g|^3$
My reasoning is the following:
- 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.
- $f$ 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.
- $g$ can have zeroes but $f$ can't
Obvious contradiction
- 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.
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.