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Let $M$ be a complete Riemannian manifold, does there exists a positive non-constant harmonic function $f \in L^1(M)$? Who can answer me or give me a counter example? Thank you very much!

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    $L^1$ is an interesting borderline case. For q > 1, the nonexistence result is due to Yau.2012-05-23

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Consider the surface of revolution (so topologically we are dealing with $\mathbb{R}\times\mathbb{S}$) with standard coordinates $(z,\theta)$. Let the metric be $ \mathrm{d}s^2 = \mathrm{d}z^2 + h^2(z) \mathrm{d}\theta^2$ This manifold is clearly complete (it is a warped product of two geodesically complete manifolds).

The Laplace-Beltrami operator associated to it is $ \triangle = \frac{1}{h} \partial_z h \partial_z + \frac{1}{h^2} \partial^2_\theta $ and the volume/area form is $h \mathrm{d}z \mathrm{d}\theta$.

Let $f = f(z)$ be a function. It being $L^1(M)$ is equivalent to $ \int_{-\infty}^\infty |f(z)| h(z) \mathrm{d}z < \infty $ It being harmonic is the same as $ h \partial_z f \equiv c $ for some constant $c$ (which we can assume, WLOG, to be 1). So this implies that we need to find a monotonic function $f$ such that $f / f'$ is absolutely integrable. This requires that $\frac{d}{dz} \log f$ to grow superlinearly in $z$.

So we can consider the following: let $f(z) = \exp (z + z^3)$. Define $h(z) = \frac{1}{(1 + 3z^2) \exp (z + z^3)}$. Then $h f' = 1$ so $f$ is harmonic. On the other hand, $hf = \frac{1}{1+3z^2}$ is integrable in $z$, and hence $f\in L^1(M)$.

Note that the scalar curvature can be computed to be $ - \frac{2}{h} h''$ which has fourth order growth in $z$ and so violates the hypotheses of Li's theorem. (In fact, the order of growth of the scalar curvature will be roughly twice that of the growth of $\frac{d}{dz} \log f$. So in this sense the quadratic growth assumption in Li's theorem is sharp.)

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No.

The second result on Google gives http://intlpress.com/JDG/archive/1984/20-2-447.pdf, "Uniqueness of $L^1$ solutions for the Laplace equation and heat equation on Riemannian manifolds" by Peter Li, J Diff Geo 20 (1984) 447-457. It has the following result:

Theorem 1: If $M$ is a complete noncompact Riemannian manifold without boundary, and if the Ricci curvature of $M$ has a negative quadratic lower bound, then any $L^1$ nonnegative subharmonic function on $M$ is identically constant. In particular, any $L^1$ nonnegative harmonic function on $M$ is identically constant.

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    Oh! just like sometimes a "ring" means a "commutative ring with identity"? (Ok~ just kinding...). Thanks for all your answers, let me think... The last question is... if we don't require any condition of Ricci curvature? (is that similar with the trivial case as Willie said?) Sorry, I am not good at geometry...2012-05-23