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Suppose that $f$ is an entire function and has the property that for all $z ∈ \mathbb{C} \backslash \mathbb{R}$, $|f(z)| \le |1/|Im(z)|$. I want to show that $f ≡ 0$.

I think I probably want to use Liouville's Theorem to show that $f$ is in fact bounded on on $\mathbb{C}$, and then show that the constant $c$ that $f$ is equal to is actually $0$, but I'm getting lost in showing this.

Any hints or suggestions would be appreciated.

tor
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  • I'm not sure that's correct. Take $f(z)=\frac{1}{\text{Im}(z)}$. It's neither constant nor zero. –  Dec 14 '14 at 10:13
  • Perhaps it's a good thing I wasn't able to solve this problem then. Thank you!

    If you are wondering, the problem was copied directly from #8 in this old final complex analysis exam: https://math.berkeley.edu/sites/default/files/pages/F03_Final_Exam-D.Geba__0.pdf

    edit: Actually, isn't your function not entire?

     – tor Dec 14 '14 at 10:17
  • hmm. on $\mathbb{C}$? no. on the other hand $\Im(z)=0\Leftrightarrow z\in\mathbb{R}$ and it's outside of your domain. –  Dec 14 '14 at 10:34
  • @DanisFischer But your choice for $f$ isn't entire is it? – Cameron Williams Dec 14 '14 at 20:53
  • This question has already been posted. It found several answers at http://math.stackexchange.com/questions/377782/application-of-liouvilles-theorem?rq=1 The key to one solution is Jensen's inequality for entire functions. – Jo Wehler Dec 17 '14 at 21:59

2 Answers2

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On the lines $z=i$ and $z=-i$ we have $|f(z)|\leq 1 $ hence from Hadamard Three Lines Theorem $|f(z)|\leq 1 $ on whole strip $|\mbox{Im}(z) |\leq 1 .$ If $|\mbox{Im}(z) |\geq 1 $ then $|f(z)|\leq \frac{1}{|\mbox{Im}(z) |}\leq 1$ hence $f$ is bounded on whole plane.

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    Hadamard three-lines theorem requires boundedness of $f(x)$ on the strip $-1 \leq \operatorname{Im} z \leq 1$ – uranix Dec 14 '14 at 11:36
  • Yes, you,re right. –  Dec 14 '14 at 18:28
  • I've never heard of this theorem -- Perhaps I don't yet have the machinery I need to solve this, then. Thanks for the input, though. – tor Dec 14 '14 at 18:43
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If, $|f(z)|\leq \frac{1}{|\mbox{Im}(z) |}$ then $|\frac{1}{f(z) }| \leq |z|.$ Hence $\frac{1}{f(z)} =az$ but this is impossible. So $f$ must be identicaly zero.