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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.

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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 $c$.)

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  • $\begingroup$ You say "not in this pointwise sense". Do you have any alternative in mind? $\endgroup$
    – Alex M.
    Sep 5 '15 at 19:47
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    $\begingroup$ Well, you already observed that $\int_M \bar{f} \Delta f \,dx \le 0$, so it's true in an "integrated" sense. That integral is usually what people talk about as the "quadratic form" generated by the Laplacian. This is just another way of saying that the Laplacian is negative definite. $\endgroup$ Sep 5 '15 at 19:51

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