Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
$L^1$ is an interesting borderline case. For $q > 1$, the nonexistence result is due to Yau. – Willie Wong May 23 '12 at 16:23
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.)

share|cite|improve this answer


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.

share|cite|improve this answer
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

Your Answer


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.