Is there a $p$-adic version of Liouville theorem?

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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 $mathbbC^n$?




analytic-number-theory p-adic-number-theory analyticity




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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






  • 1




    Please do not deface your questions. Others may have the same question too.
    – Daniel Fischer♦
    Aug 19 at 8:55














up vote
5
down vote

favorite
2












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 $mathbbC^n$?




analytic-number-theory p-adic-number-theory analyticity




share|cite|improve this question






















  • "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






  • 1




    Please do not deface your questions. Others may have the same question too.
    – Daniel Fischer♦
    Aug 19 at 8:55












up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2





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 $mathbbC^n$?




analytic-number-theory p-adic-number-theory analyticity




share|cite|improve this question














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 $mathbbC^n$?





analytic-number-theory p-adic-number-theory analyticity





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 19 at 8:54









Daniel Fischer♦

172k16155276




172k16155276










asked Jan 14 '12 at 0:38









iago

416213




416213











  • "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






  • 1




    Please do not deface your questions. Others may have the same question too.
    – Daniel Fischer♦
    Aug 19 at 8:55
















  • "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






  • 1




    Please do not deface your questions. Others may have the same question too.
    – Daniel Fischer♦
    Aug 19 at 8:55















"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




"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




1




1




Please do not deface your questions. Others may have the same question too.
– Daniel Fischer♦
Aug 19 at 8:55




Please do not deface your questions. Others may have the same question too.
– Daniel Fischer♦
Aug 19 at 8:55










1 Answer
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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).






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  • 2




    Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
    – Pete L. Clark
    Jan 14 '12 at 5:01











  • 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 8:34










  • 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 9 at 17:06










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










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).






share|cite|improve this answer


















  • 2




    Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
    – Pete L. Clark
    Jan 14 '12 at 5:01











  • 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 8:34










  • 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 9 at 17:06














up vote
3
down vote



accepted










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).






share|cite|improve this answer


















  • 2




    Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
    – Pete L. Clark
    Jan 14 '12 at 5:01











  • 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 8:34










  • 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 9 at 17:06












up vote
3
down vote



accepted







up vote
3
down vote



accepted






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).






share|cite|improve this answer














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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 '12 at 14:18

























answered Jan 14 '12 at 2:14









Álvaro Lozano-Robledo

12.2k21837




12.2k21837







  • 2




    Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
    – Pete L. Clark
    Jan 14 '12 at 5:01











  • 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 8:34










  • 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 9 at 17:06












  • 2




    Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
    – Pete L. Clark
    Jan 14 '12 at 5:01











  • 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 8:34










  • 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 9 at 17:06







2




2




Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
– Pete L. Clark
Jan 14 '12 at 5:01





Note that Cherry's result is for a valued field which is algebraically closed, e.g. $mathbbC_p$. If I remember correctly, there are nonconstant bounded entire functions on $mathbbQ_p$ (or any finite extension thereof), just as there are on $mathbbR$. I believe Alain Robert's GTM has a discussion of this.
– Pete L. Clark
Jan 14 '12 at 5:01













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 8:34




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 8:34












Thanks, I just need the case of an algebraically closed field.
– iago
Jan 14 '12 at 12:26




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 9 at 17:06




Theorem 42.6 in Schikof's Ultrametric calculus seems to be relevant.
– Watson
Jun 9 at 17:06












 

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