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This question already has an answer here:

I need to prove the following question

Question

Let $g(z)$ be an entire function such that there exists an $\alpha > 0$ such that $\left|\operatorname{Im}(g(z))\right| \le \alpha$. Prove that $g(z)$ is a constant function.


My first thought was to use Liouville's theorem. As that states, for an entire function $g$, if $g$ is bounded, then $g$ is constant.

So if we could prove that $g$ is bounded, then by Liouville's theorem, $g$ is constant.

I have attempted to prove that $g$ is bounded, but I am struggling. Some help would be much appreciated.

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marked as duplicate by Martin R, Did, carmichael561, Rolf Hoyer, user223391 Nov 19 '17 at 20:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This question is about an entire function with bounded imaginary part, but the same arguments apply. Or apply the above to $ig(z)$. $\endgroup$ – Martin R Nov 19 '17 at 16:14
  • $\begingroup$ Please avoid titles both misleading and incomplete. $\endgroup$ – Did Nov 19 '17 at 16:15
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Hint: compose it with the function $z\mapsto e^{iz}$

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  • $\begingroup$ $$g(z) = exp(-y) [cos(x) + i sin(x)] $$ Then as $$|im(g(z))| \le \alpha$$ $$\implies exp(-y)sin(x)\le \alpha$$ I know $-1 \le sin(x) \le 1$. How does this prove that $g$ is bounded? Apologies if I'm completely off track $\endgroup$ – Ben Nov 19 '17 at 15:09
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    $\begingroup$ @Ben : Standard usage is $\exp(-y),$ not $exp(-y)$ and $\sin(x),$ not $sin(x).$ They're coded as \exp and \sin. That not only prevents italicization but also results in proper spacing in things like $a\sin b$ and $a\sin(b). \qquad$ $\endgroup$ – Michael Hardy Nov 19 '17 at 15:11
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    $\begingroup$ @Ben - Let $f(z) = e^{iz}$. What boon heT is suggesting is looking at the function $f(g(z))$. Since both $f$ and $g$ are entire, so must their composition be. Now, what can you figure out about the boundedness of that composition? $\endgroup$ – Paul Sinclair Nov 19 '17 at 18:55
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This is little picard theorem: https://en.wikipedia.org/wiki/Picard_theorem

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  • $\begingroup$ Isn't Little Picard theorem for a non-constant entire function? $\endgroup$ – Ben Nov 19 '17 at 14:28
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    $\begingroup$ Admittedly, bazookas will kill flies, but I still recommend a fly-swatter. It is easy enough, and more informative IMO, to prove this from simple principles instead of pulling in really high-powered tools like Picard's Theorem. $\endgroup$ – Paul Sinclair Nov 19 '17 at 19:04
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The Little Picard Theorem states that if $f$ is an entire, non-constant function from $\mathbb C$ to $\mathbb C$, then $f$ takes on either every value in $\mathbb C$ or every value but one.

There is more than one value that your function doesn’t take on, so cannot be both entire and non-constant.

The “or every value but one” is necessary here because $e^z$ is an entire, non-constant function that is never $0$.

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  • $\begingroup$ May I suggest that there is an important difference between "Your function doesn't take on more than one value" to "There is more than one value that your function doesn't take on", which you probably meant to say? $\endgroup$ – Marie. P. Nov 19 '17 at 20:53
  • $\begingroup$ @Marie.P. thanks, I fixed that. $\endgroup$ – Stella Biderman Nov 19 '17 at 22:38

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