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Provide a function $g(z)$ that is analytic in a region, bounded throughout the complex plane that is not a constant function (I was thinking $\sin(z)$) The second part of the question asks, does this contradict Liouville's theorem?

I am thinking it does not contradict the theorem as the function need not be entire? Can anyone suggest a function or confirm the one I gave.

Thanks

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    $\begingroup$ The question is unclear to me: "analytic in a region" suggests that the domain is allowed to be chosen, but "bounded throughout the complex plane" suggests that the domain must be all of $\Bbb C$. What is the intended meaning? $\endgroup$ Commented May 31, 2021 at 0:58
  • $\begingroup$ Ya I struggling with that as well. Can you please provide your best interpretation/assumption as that is how the question is written. $\endgroup$
    – Elli
    Commented May 31, 2021 at 1:00
  • $\begingroup$ Perhaps some isolated singularities are allowed. Then the better terminology would be to use the word meromorphic. $\endgroup$ Commented May 31, 2021 at 1:07
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    $\begingroup$ $sin z$ is unbounded, and Liouville's name wasn't Louisville... $\endgroup$ Commented May 31, 2021 at 4:03

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There exists a bounded, non-entire function on the complex plane that is analytic/holomorphic in some region.

One possible example is $g(z) = \max\bigl(0, \frac{|z|-1}{|z|+1}\bigr)$. The verification of properties is left as an exercise.

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    $\begingroup$ Thanks for your suggestion, as a follow-up I don't understand where this would be analytic. $\endgroup$
    – Elli
    Commented May 31, 2021 at 1:28
  • $\begingroup$ It's $0$ on the unit disk. :) $\endgroup$ Commented May 31, 2021 at 2:01
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As meromorphic functions diverge to infinity at the pole, the answer to your question is negative.

Next best I can think of is examples of the following kind:

Define $g(z) = z^n$ inside the unit disc and outside the unit disc as $z^n/|z|^n$. On the boundary of the unit disc both definitions agree and so we get a continuous function on the whole complex plane. This is analytic in the unit disc.

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You can even go further and prescribe an arbitrary region and an arbitrary analytic function thereon (as long as the latter is bounded on the former). If you do not require continuity you can just continue your function by 0 (or any other value you wish for). But you can also work with a smooth cutoff function in the style of Urysohn (or its smooth version).

Suppose you have (what you call) a "region" $A$, then for any closed set $K$ such that $K\subset A^°$ (=interior) you find a smooth function which is constantly 1 on $K$ and constantly 0 on $\mathbb{C}\backslash \bar{A}$. Then you can multiply your "prescribed analytical values" by this function and the result is analytic on $K^°$.

I'm not sure though, whether you can extend this approach to replace $A$ by an open neighborhood of $A$ and $K$ by $\bar{A}$ to receive such a function on all of $A$. Certainly, you have to claim that your prescribed values are analytically continuable to such an open neighborhood, which maybe imposes some niceness properties of the boundary of $A$.

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