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

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

I have consulted these two articles:

but they don't seem to explain clearly why.

• What is your definition of positive-definite? Nov 9, 2017 at 4:57
• @AnthonyCarapetis What are the possible definition of positive-definite? Isn't there only one? Nov 9, 2017 at 6:22
• @Aha I'm fairly sure that there are different definitions depending on which book you're following. I've seen positive semi-definite and positive definite, but I've also seen positive semi-definite being replaced by positive definite and positive definite being replaced by strictly positive definite. Nov 9, 2017 at 6:29
• $x^TAx > 0$ or eigenvalues > 0 Nov 9, 2017 at 6:31
• Those are the usual definitions for a matrix, but the Laplacian is an unbounded operator on an infinite-dimensional space. (You really should specify which space in your question - I'm assuming $L^2$.) The obvious generalizations ($\langle u, A u \rangle > 0$ and $\sigma(A) > 0$ respectively) are often used, but so are stronger variants like $\langle u, Au \rangle \ge \epsilon |u|^2$ and $\sigma(A) \ge \epsilon$ with strictly positive $\epsilon$. Nov 9, 2017 at 6:59

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.

• Sorry for asking in your old answer, but could you explain"When $M$ is boundaryless, the boundary integral always vanishes"? Apr 26, 2018 at 9:57
• @Barabara: the integral of anything over the empty set is zero. Apr 26, 2018 at 9:58
• Indeed, but what if my domain is unbounded, like $\mathbb{R}^n$? I guess it doesn't work that way and I need more assumptions about domain? Apr 26, 2018 at 10:03
• My answer doesn't address the noncompact case, where you need to get a little more functional or harmonic analysis involved. For $\mathbb R^n$, I believe $-\Delta$ is positive-definite when restricted to its maximal domain, which is the set of $L^2$ functions whose distributional Laplacians are in $L^2$; though I can't think of a reference off the top of my head. Apr 26, 2018 at 10:10
• If you're happy to restrict to a more restricted class of functions, like those satisfying $|f(x)| \le C (1+|x|)^{-p}$ for any $C$ and a fixed large $p$, then you should be able to do the integration by parts over a ball of radius $r$ and check that the boundary term vanishes as $r \to \infty.$ Apr 26, 2018 at 10:17

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.

• I use the same sign convention as the OP.
– user357151
Nov 12, 2017 at 4:43