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 :

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

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.

- Why the non-constant condition implies no term related to the eigenfunction corresponding to eigenvalue zero in the expansion of $f$ ?