Let M be a complete Riemannian manifold.Suppose there are two non-parabolic ends on M with respect to $M\backslash {B_p}\left( {{R_0}} \right)$Then there is a harmonic function f on M.Is it right that $\int_{{B_p}\left( R \right)} {f \le CR} $ for $R \ge {R_0}$ and a constant C independent of R?
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What made you expect this? Since the ends are non-parabolic, their volume growth is at least quadratic. On the other hand, the harmonic functions arising from two non-parabolic ends are linear combinations of the constant and the function whose value at a point $x$ is the probability that the Brownian motion started at $x$ will end up at infinity in a fixed end. Therefore these functions converge to constants along ends, so that the growth of their integrals along balls will be at least quadratic (unless the function is zero). The simplest example of this kind is provided by two copies of $\mathbb R^3$ joined with a "bridge".
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$\begingroup$ If one parabolic end,one non-parabolic end,the inequality holds? $\endgroup$ Commented Mar 14, 2013 at 12:01
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$\begingroup$ If only one end is non-parabolic, then you may have no harmonic functions to talk about (I presume you are interested in bounded or at least positive ones). $\endgroup$– R WCommented Mar 14, 2013 at 13:21