关于满足|f|²≤|g|³的非零整函数f、g的零点性质的推理验证问询

Hi folks, I've been working through a problem involving non-zero entire functions $f$ and $g$ that satisfy $|f|^2 \leq |g|^3$, and here's my line of reasoning so far. I'd really appreciate your feedback on whether it holds water!

• Case 1: Both $f$ and $g$ have zeros
  Suppose $g$ has a zero at $z=a$. Then $f$ must also have a zero at $a$, right? But if we look at a small neighborhood around $a$, $g$ takes values extremely close to 0—and raising that to the third power would make it even smaller. That seems like it would lead to $|g|^3 < |f|^2$, which contradicts our original inequality.

• Case 2: $f$ has zeros but $g$ doesn't
  If $g$ has no zeros, it's either a constant polynomial or another entire function with no roots.

  • If $g$ is constant: If $f$ is a polynomial, $|f|^2$ tends to infinity as $z$ goes to infinity, which breaks the inequality. If $f$ isn't a polynomial, it has an essential singularity at infinity, which also can't satisfy the boundedness condition from the inequality.
  • If $g$ is non-constant: It has an essential singularity at infinity, so we should be able to find some neighborhood where the inequality fails.

• Case 3: $g$ has zeros but $f$ doesn't
  This is an obvious contradiction—if $g$ has a zero at some point, $|g|^3$ would be 0 there, but $f$ doesn't have a zero so $|f|^2 > 0$, which violates $|f|^2 \leq |g|^3$.

• Case 4: Neither $f$ nor $g$ has zeros
  This feels like the only valid scenario to me. But even here, both functions have to be constants, right? If either was non-constant, they'd have an essential singularity at infinity, which would lead to points where the inequality doesn't hold.

Does this reasoning check out? I know I skipped over some details when I said we could find a neighborhood where the inequality fails, but it makes intuitive sense to me. Thanks in advance for your thoughts!

备注：内容来源于stack exchange，提问作者Mateo