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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. – Willie Wong May 23 '12 at 16:23

2 Answers 2

up vote 3 down vote accepted

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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The second result on Google gives, "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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I was just about to post... see also Yau's paper Harmonic Functions on Complete Riemannian Manifolds. – Henry T. Horton May 23 '12 at 16:01
I have a little question, the theorem above, require that $M$ is non-compact without boundary. How about $M$ is just a complete Riemannian manifold? Thank you very much! – Peter Hu May 23 '12 at 16:08
The compact case trivially follows from the maximum principle. If $M$ is complete it cannot have boundary. You should read "complete" as modifying "noncompact Riemannian manifold without boundary", since the latter can be incomplete as a Riemannian manifold and the theorem won't apply. – Willie Wong May 23 '12 at 16:12
@PeterHu, in the OP you asked for all general complete Riemannian manifold. This result provides a class of complete Riemannian manifolds which admit no $L^1$ positive non-constant harmonic functions. The paper also discusses some manifolds which do admit nonnegative nonconstant harmonic functions. – Neal May 23 '12 at 16:18
More interesting is the condition that $M$ requires a curvature bound. As @Henry mentioned, if one just requires non-negative curvature, the result is already contained in Yau's paper. You may also want to take a look at Peter Li's survey article (available on his webpage ) from 2008, which I think captures more or less the state of the art about harmonic functions on complete Riemannian manifolds. – Willie Wong May 23 '12 at 16:21

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