Suppose $f$ is an entire function such that $|f(z)|\,|\Im(z)|^2\leq 1$. Then $f$ is identically zero.
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Some things I have thought that are mostly unconclusive: Plugging $w=iz$ one gets $|f(iz)|,|\Re(z)|^2\leq 1$. I tried to combine these two bounds to get one for $f$, unsuccesfully. In hindsight, this is a stupid thing to do since there are non-entire, non-bounded functions satisfying the constraint. – Fernando Martin Oct 16 '16 at 18:03
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Trying to classify the singularity at infinity: no clue how to do this, since I only control how the function behaves on horizontal lines. – Fernando Martin Oct 16 '16 at 18:04
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Trying to use the averaging property of harmonic functions: this may work but it would have to involve a certain amount of clever geometric trickery. – Fernando Martin Oct 16 '16 at 18:06
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One thing that particularly bugs me is the fact that $\Im(z)$ appears squared. I don't immediately see how that helps one to have more control over $f$, and I don't know if the result holds if we drop that exponent or not. – Fernando Martin Oct 16 '16 at 18:10
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Are you sure about this theorem? Does the constant function $f(z) = c$ with $0<|c|<1$ not fulfil the requirements? – Ranc Oct 16 '16 at 18:58
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1@Ranc: It does not. The bound tells you that $f$ goes to zero on every vertical line, so if we knew $f$ was constant then it would have to be zero. – Fernando Martin Oct 16 '16 at 19:06
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@FM, I now notice this is $\Im (z)$ and not $\Im (f(z))$. – Ranc Oct 16 '16 at 20:11
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Have a look at http://math.stackexchange.com/questions/377782 (this is for $|\operatorname{Im}(z)|$, not $|\operatorname{Im}(z)|^2$, but the answers still apply. – mrf Oct 16 '16 at 23:14
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@FM It seems mrf's comment resolves your question. Do you agree? – mathworker21 Jan 17 '20 at 19:53