Negativity of a quadratic form on $L^2(M)$

Question

Let $M$ be a compact Riemannian manifold and $V=L^2(M)$. Let $\Delta$ be the negative-definite Laplacian. Let $f \in V$ and $x \in M$ be arbitrary, but fixed.

Is it true that ${\rm Re} \ (\Delta f) (x) \overline {f (x)} \le 0$? (This is the result of a chain of calculations that is not relevant to this question. $\rm{Re}$ is the real part.)

In its support, I have two arguments that lend it some plausibility:

1) if $f \in V$ is an eigenfunction of the Laplacian, the conjecture is true. In fact, it is true for arbitrary complex multiples of such eigenfunctions.

2) if $F(x) = {\rm Re} \ (\Delta f) (x) \overline {f (x)}$, then $\int \limits _M F(x) \ \Bbb d x = {\rm Re} \int \limits _M (\Delta f) (x) \overline {f (x)} \ \Bbb d x = - {\rm Re} \int \limits _M \| \nabla f \| ^2 \ \Bbb d x \le 0$, which suggests that there exist points $x$ where the expression is indeed negative - but are they "many" or "few"?

As a side-note, if the conjecture can be proven for some space other than $L^2(M)$, I shall still be happy.

Answer 1

It's not true in this pointwise sense. Consider $M = S^1$ seen as $[0,2\pi]$ with endpoints identified, so that $\Delta f = f''$. Let $f(x) = \sin(x) + 2$. Then $\bar{f} \Delta f > 0$ everywhere on $(\pi, 2\pi)$. (Notice that $f$ is a sum of two eigenfunctions of $\Delta$, with different eigenvalues.)

(In general, think about what happens to your inequality if you replace $f$ by $f+c$ for a complex constant $c$.)