# Poincaré inequality on compact manifolds without boundary

The question arises from a paper The heat flow for the full bosonic string, Ann. Global Anal. Geom. 50 (2016).

In line 4, Page 362, the author claimed the following inequality which looks similar to Poincaré inequality: Let $$(M,g)$$ be a closed Riemannian manifold (i.e. compact and without boundary). Then for a nonconstant smooth function $$f:M\longrightarrow \mathbb{R}$$, there holds $$\int_{M}|\nabla f|^{2}\geq \lambda_{1}\int_{M}f^{2}.\label{1}\tag{\ast}$$ As far as I know, if we have zero mean value, i.e., $$\int_M f=0$$, then the above inequality follows from Poincaré inequality immediately. However, in that paper the author applied the above inequality \eqref{1} to a function which cannot have zero mean value. So I have some confusions now :

1. Is the inequality \eqref{1} true? In my view, I suspect this is wrong, but I cannot give countexamples except for constant functions.

2. If the inequality \eqref{1} is true, how can we prove it ?

I have emailed the author. But the reasons he gave still cannot convince me. Here are the reasons the author gave :

You are of course right, one needs to have vanishing mean value in order to apply the Poincare inequality, I did not state this correctly in my paper. What I did here instead is the following: I can expand $$f$$ in terms of eigenfunctions of the Laplace operator and as $$f$$ cannot be constant the eigenvalue zero does not appear. Then, I get the inequality that looks similar to the Poincare inequality.

On a closed Riemannian manifold all harmonic functions (which correspond to the eigenvalue zero) are constant as a consequence of the maximum principle. Hence, if $$f$$ is assumed to be non-constant, then there cannot be an eigenvalue zero in the expansion in terms of eigenfunctions.

1. Why the non-constant condition implies no term related to the eigenfunction corresponding to eigenvalue zero in the expansion of $$f$$ ?
• Is the $f$ on p.362 a generic $f$ (say, smooth or Sobolev space) or is some specific $f$ that was introduced earlier ? Commented Jul 25 at 10:35
• A non-constant function can have a non-zero coefficient for the eigenvalue zero. Just take any non-constant function and shift it by an eigenvalue. Commented Jul 25 at 10:36
• The $f$ in the original paper is the kinetic energy, which is non-negative and smooth (since we are doing some a priori estimates). There are no other special features. @shuhalo Commented Jul 26 at 6:07

Without going into the cited paper, the reasoning as such is not correct.

Let $$f : M \rightarrow \mathbb R$$ be any function for which the Poincare inequality is satisfied. Such a function is not constant.

Take any $$c \in \mathbb R$$ and let $$g := f + c$$ be a shift of $$f$$ by some constant. Still, $$g$$ itself is not constant. By the argument of the author, we still have

$$\int_M |\nabla g|^2 = \int_M |\nabla f|^2 \geq \lambda_1 \int_M |g|^2 = \lambda_1 \int_M | f + c |^2$$

When picking $$c$$ large enough, that inequality is simply no longer true.

Possibly remedies: you essentially need some feature of $$f$$ that distinguishes it from the constants. Alternatively, the left-hand side needs some additional continuous functional that is non-zero for constants. Your best hope is that the specific $$f$$ in the paper has got some additional feature that can be exploited.

On the positive side, there is an interest in the literature for Poincare-type inequalities. If the author has got such a new variant, that should be publicized more explicitly. I'd be curious to see what Kato's inequality can accomplish.

• I think you are right. There must be something wrong in the original paper. The $f$ in the original paper is the kinetic energy, which is non-negative and smooth (since we are doing some a priori estimates). There are no other special features. Commented Jul 26 at 6:08
• These inequalities are usually associated with names such as Poincare, Friedrichs, or Wirtinger, among others, and come in different variations. As a rule of thumb, one can find such inequalities if (a) you restrict the functions so that no non-zero constant is present (b) you add a continuous functional on the left-hand side that is non-zero for the constants. For example, depending on the specific situation, you can require the boundary values to be zero or you can add the boundary trace norm on the left. Similar with point evaluations, etc.... Commented Jul 26 at 21:10