Liouville's theorem of complex analysis states that a bounded entire function is constant. This means that for a non-constant entire function $f$ there exists a sequence $z_n \in \mathbb{C}$ such that $|z_n| \to \infty$ and $f(z_n) \to \infty$. I am trying to understand if a sort of converse holds in the following sense: consider a closed set $S \subset \mathbb{C}$ ($S$ may be unbounded) such that $\mathbb{C} \setminus S$ is unbounded. Can one construct a non-constant entire function $f$ such that $f$ is bounded on $S$ ?

Comment: Cross-posted on MO here.

  • $\begingroup$ Try the Maximum modulus principle. But see math.stackexchange.com/questions/894304/…. $\endgroup$ – lhf Oct 28 '16 at 17:14
  • 3
    $\begingroup$ What if $S= \{0\}?$ $\endgroup$ – zhw. Oct 28 '16 at 17:21
  • $\begingroup$ @lhf What if the maximum is attained at some boundary point of $S$? $\endgroup$ – user346998 Oct 28 '16 at 17:53
  • 1
    $\begingroup$ Yes, for every curve $\gamma$ such that $\gamma(t) \to \infty$, there exists a function bounded everywhere except in the neighborhood of the curve, search for the Alice Roth's theorems on entire functions $\endgroup$ – reuns Oct 30 '16 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.