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Characterize the set of holomorphic functions $g:\mathbb{C}\setminus\{0\}\to\mathbb{C}$ that are bounded away from zero with $|g(z)|>|z|^{-7/3}$ for all $z\in\mathbb{C}\setminus\{0\}$

I understand that from the inequality, it has a pole at $0$.

But other than that I really don't see a method to obtain the mentioned characterization.

Appreciate your help

Charith
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1 Answers1

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Let $f(z)=\frac{1}{z}$. Then consider functions $h=f \circ g$. Then $|h(z)|<|z|^{\frac{7}{3}}$ Then, $h(z)$ is similar to a quadratic polynomial. This in turn characterises $g(z)$.

Ciarán Ó Raghaillaigh
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  • Thank you but why did you say that $h$ is similar to a quadratic polynomial? – Charith Dec 31 '19 at 21:38
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    @gune : Because it has a removable singularity at $z=0$ and must thus be an entire function. Now entire functions with such bounds are actually polynomials of the same or less degree,... – Lutz Lehmann Dec 31 '19 at 21:40
  • Oh I see. Removable singularity is there because $h$ is bounded. Right? And the polynomial property follows from https://math.stackexchange.com/questions/86772/show-that-if-fz-leq-m-zn-then-f-is-a-polynomial-max-degree-n

    Am I correct

     – Charith Dec 31 '19 at 21:45
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    As I mentioned, $\frac{1}{z}$ does not work as working for all $z$ without a constant to balance it out like in that link is very strict. – Ciarán Ó Raghaillaigh Dec 31 '19 at 21:59
  • Thank you I edited the question – Charith Dec 31 '19 at 22:00
  • And I need a small clarification.
    So according to the obtained condition, $g(z)=\frac{1}{z^2}$ should work. Isn't it? (Reciprocal of a quadratic function)
    But wouldn't that also have the same problem of satisfying the given inequality for $z$ close to $0$     – Charith Dec 31 '19 at 22:06
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    Indeed. I believe the set of such functions is empty. I should have clarified that my answer was only a hint. You just need show no quadratic will work. – Ciarán Ó Raghaillaigh Dec 31 '19 at 22:12
  • Than you very much! – Charith Dec 31 '19 at 22:33