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!
positive non-constant harmonic function $f $ in $L^1(M)$ on a complete Riemannian manifold
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0$L^1$ is an interesting borderline case. For $q > 1$, the nonexistence result is due to Yau. – 2012-05-23
2 Answers
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.)
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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1I was just about to post... see also Yau's paper _Harmonic Functions on Complete Riemannian Manifolds_. – 2012-05-23
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0I 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! – 2012-05-23
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2The 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. – 2012-05-23
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0@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. – 2012-05-23
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1More 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 http://math.uci.edu/~pli/ ) from 2008, which I think captures more or less the state of the art about harmonic functions on complete Riemannian manifolds. – 2012-05-23
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0Oh! 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