How to prove that $-div(\nabla f)$ is positive definite?

Question

How to prove that $-div(\nabla f)$ is positive definite?

I have consulted these two articles:

1. Prove that the Laplacian operator is positive definite

2. The minus Laplacian operator is positive definite

but they don't seem to explain clearly why.

Answer 1

I am asking a general case in $L^2$. Why does the domain matter?

I'll tell you why. Take the function $f(x) = e^x$, considering it as an element of $L^2((-1,1))$. Then $\Delta f(x) = -f''(x) = -e^x$, hence
$$\langle \Delta f, f\rangle = \int_{-1}^1 (-e^x)e^x\,dx < 0$$ So much for being positive definite.

The Laplacian may become positive definite when viewed on a suitable function space, but since you didn't feel like going into such details, I won't go into them either.

Answer 2

Since you tagged this [differential-geometry] but haven't been more specific about your setting, I'll discuss the case of the Laplace-Beltrami operator $$\Delta = \mathrm{div}\circ \nabla : D(\Delta) \subset L^2(M) \to L^2(M)$$ on a compact Riemannian manifold $M$, possibly with boundary. (Here $D(\Delta)$ is the maximal domain of $\Delta$, i.e. the set of all $L^2$ functions $f$ for which $\Delta f$ is an $L^2$ function.) This includes as special cases the classic example where $M$ is a bounded open subset of Euclidean space and $\Delta = \sum_i \partial_i \partial_i$ is the standard Laplacian. (You should be aware that the Laplacian operator has many generalizations - anywhere local averages make sense, there's probably a $\Delta$. This includes on graphs, nice enough metric measure spaces, and a lot more. Thus why it would be nice to know exactly what you're asking about!)

Since $L^2(M)$ is a Hilbert space, the most common definition of positivity of $-\Delta$ is simply that the $L^2$ inner product $(f, -\Delta f)$ is positive for all non-zero $f$, which is the most obvious analog of the finite-dimensional version $x^T A x > 0.$ The basic reason this is something we might expect $-\Delta$ to obey comes from integration by parts: we have $$(f, - \Delta f )= \int_M \langle f, -\Delta f \rangle = -\int_{\partial M} f\nabla_\nu f + \int_M |df|^2, \tag{1}$$ which has a strong tedency to be positive thanks to the last term. In order to make this positivity a universal fact rather than a tendency, however, we need to restrict our domain - the example of the normal human below shows that it's easy to get $(-\Delta f, f)<0$ when you have a boundary. Moreover, even on closed manifolds where the bad term vanishes, the existence of non-zero constant functions means we only have $-\Delta \ge 0,$ not $-\Delta > 0.$ Thus the correct choice of operator domain $H \subset D(\Delta)$ depends on the function domain $M$ you are dealing with:

• When $M$ has boundary, we can impose a zero Dirichlet condition $$H = \{ f \in D(\Delta) : f|_{\partial M} = 0 \}$$ so that the boundary integral in $(1)$ vanishes, and no non-zero constants are admissible.

• Alternatively, in this same situation we could impose a Neumann condition along with an average constraint $$H = \{ f \in D(\Delta) : \nabla_\nu f|_{\partial M} = 0, \int_M f = 0 \},$$ where the first condition makes the boundary integral vanish and the second condition rules out non-zero constants.

• When $M$ is boundaryless, the boundary integral always vanishes, so we can simply choose $$H = \{ f \in D(\Delta) : \int_M f = 0 \}.$$ This can be viewed as the special case of the Neumann condition where $\partial M = \emptyset.$

In each of these cases, it is the restriction $\Delta|_H : H \subset L^2(M) \to L^2(M)$ that is positive-definite.

From a more abstract point of view, the reason $-\Delta$ is positive is because $-\mathrm{div}$ is (once you choose the right $H$!) the $L^2$-adjoint of the gradient operator; so $-\Delta = \nabla^* \circ \nabla$ is positive: we have $$(\nabla^* \nabla f,f) = (\nabla f, \nabla f) > 0$$ so long as $\nabla$ is injective. This should be very familiar from the finite-dimensional version: if $X$ is an invertible matrix then $X^T X$ is always positive definite.

I think all of the above is also true (minus the need to rule out constants) for a noncompact complete manifold(-with-boundary), though I'm not 100% - I know there are a lot of frequently used properties of Sobolev spaces that require some bounded geometry assumption. Even if it's not quite literally the same, the idea will be the same: the Laplacian will be positive-definite once you sensibly restrict the operator domain.