Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

M is a Riemannian manifold.An end E is said to be a non-parabolic end if it admits a positive Green's function with Neumann boundary conditon on E.Otherwise,it is a parabolic end.If M has more than two nonpara ends,then there is a nonconstant harmonic function on M. What if M has one nonpara end,one para end,is there a nonconstant harmonic function on M?

share|improve this question
    
To clarify the question, what about $R^n$? –  timur Dec 22 '12 at 14:26
    
$\mathbb R^1$ has two parabolic ends; $\mathbb R^n$ for $n>1$ has one end. –  R W Dec 23 '12 at 15:49
    
Do you want your harmonic functions to be bounded? –  John Pardon Dec 24 '12 at 17:24
add comment

2 Answers

The easiest way to answer your question is by using the Brownian motion on the manifold. In these terms parabolicity of an end means its recurrence with respect to the Brownian motion. The reason why presence of at least two non-parabolic ends (not "more than two" as you write) implies existence of non-constant bounded harmonic function (boundedness condition which you don't mention is important - otherwise all these claims fail) is then very simple. The Brownian sample paths escape to infinity along one of non-parabolic ends, so that if there are at least two of them there will be a non-trivial behaviour at infinity, i.e., a non-constant bounded harmonic function.

On the other hand, if there is just one non-parabolic end, then the space of bounded harmonic functions on the whole manifold is the same as just for this end, so that in this case both situations are possible.

share|improve this answer
    
Very interesting. If you don't mind would it be possible to expand a little bit the phrase " there will be a non-trivial behaviour at infinity, i.e., a non-constant bounded harmonic function"? How can one give to this a precise mathematical sense? Is there a reference for this kind of reasoning? –  Dmitri Dec 23 '12 at 12:53
2  
Actually it makes sense for an arbitrary Markov process (chain). The space of bounded harmonic functions (i.e., those which satisfy the mean value property with respect to the transition probabilities) is isometric to the space of bounded functions on the path space measurable with respect to the sigma-algebra of time shift invariant sets. This sigma-algebra describes the "stochastically significant" behaviour of the Markov process at infinity. The key words are "Poisson formula" (this is the above isometry; the name comes from the usual Poisson formula for the disk) and "Poisson boundary". –  R W Dec 23 '12 at 15:46
    
What is an example of a nonconstant bounded harmonic function on $R^1$? –  timur Dec 24 '12 at 16:56
    
I am just trying to understand the meaning of the statement. –  timur Dec 24 '12 at 16:56
    
One should probably add that if we choose a nonparabolic end of $M$, then the function which assigns to $p\in M$ the probability of brownian motion started at $p$ escaping out the chosen end is a harmonic function. If there are at least two nonparabolic ends, then this function will be nonconstant (it approaches 1 deep in the chosen end, and approaches 0 deep in every other nonparabolic end). –  John Pardon Dec 24 '12 at 17:22
show 2 more comments

Any open Riemannian manifold has enough harmonic function to separate points and give local coordinates .This a consequence of a Runge type theorem of Lax and Malgrange independently.In fact Greene and Wu showed that one can harmonically embed the open manifold in Euclidean space. The paper of Greene and Wu is in Annales Institut Fourier vol 25 (1975) 215-235

share|improve this answer
    
Just to clarify, this is a result about possibly unbounded harmonic functions. –  John Pardon Dec 24 '12 at 17:25
    
Yes there is no control over growth of these functions.OP only asked about nonconstant harmonic functions .In the case of parabolic ends one can find a proper harmonic function along that end.This is a theorem of Mitsuru Nakai. –  Mohan Ramachandran Dec 24 '12 at 17:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.