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That is, if a function $f$ is analytic and bounded in all $K$, a $p$-adic field (or more generally a complete non-archimedean field), has to be constant? And does the theorem work for functions on $K^n$, or in $\mathbb{C}^n$?

Daniel Fischer
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iago
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  • "Analytic" means what? Since the space is zero-dimensional, there are certainly sets $U$ that are open and closed, not empty and not the whole space. The characteristic function of such a set is identically constant on a neighborhood of every point. Note that the answer says "entire" and not "analytic"... – GEdgar Jan 14 '12 at 14:46
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    Please do not deface your questions. Others may have the same question too. – Daniel Fischer Aug 19 '18 at 08:55

1 Answers1

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The answer to the first question is yes, Liouville's theorem still holds for valued fields that are algebraically closed (this last part added after Pete Clark's comment below). See, for example, these lecture notes by William Cherry (in particular, see page 16).

Álvaro Lozano-Robledo
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    Note that Cherry's result is for a valued field which is algebraically closed, e.g. $\mathbb{C}_p$. If I remember correctly, there are nonconstant bounded entire functions on $\mathbb{Q}_p$ (or any finite extension thereof), just as there are on $\mathbb{R}$. I believe Alain Robert's GTM has a discussion of this. – Pete L. Clark Jan 14 '12 at 05:01
  • In fact, in any normed field, the nonexistence of nonconstant bounded entire functions implies that every nonconstant polynomial has a root, so algebraic closure is necessary as well as sufficient. – Pete L. Clark Jan 14 '12 at 08:34
  • Thanks, I just need the case of an algebraically closed field. – iago Jan 14 '12 at 12:26
  • Theorem 42.6 in Schikof's Ultrametric calculus seems to be relevant. – Watson Jun 09 '18 at 17:06