how many nonparabolic ends guarantee a nonconstant harmonic function on Riemannian manifold? 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?
 A: 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. 
A: 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
