Epistemic status: some proofs here may well contain a small bug.
The Laplacian of a function from n-dimensional space to the real line is a generalization of the second derivative of a normal function: specifically, it’s the sum of second derivatives of the function along each dimension. One day, I recall coming across a theorem that said that the Laplacian of a function was equal to something like the limit of the average function value on a shell around a point, minus the function value at that point, divided by the squared radius of the shell. This is a nice theorem, because it connects the Laplacian to one of my favourite mathematical objects, the graph Laplacian. However, I later couldn’t find this theorem anywhere. Yesterday, I decided to re-derive the theorem, and today I decided to write it up. You can see the theorem and its proof here. Do let me know if there are any errors I should fix.
Nice theorem and write up. Already the one dimensional case is something interesting called “The second symmetric derivative”. And I think it might be used to prove the general case directly: If you add up that result for n orthogonal directions then the left hand side is the Laplacian. The right hand side is a sum of a n limits that at first glance seems to depend heavily on the picked direction, but they can’t as the left hand side is independent of the picked directions. We are free to pick arbitrary many different directions and take the average. In the limit it becomes the average over a sphere but the order of limits is wrong, some standard theorem might apply to resolve that?
Ooh, good idea - you're usually allowed to exchange integrals/sums and limits when you end up wanting to do that, so something like this should probably work.
Nice. My immediate instinct is to do this with the divergence theorem. The Laplacian is the divergence of the gradient, so as soon as you say "average value on a shell" I'm immediately getting flashbacks to Physics 260 and thinking "well the integral of the Laplacian inside the shell is equal to the integral of the gradient through the shell - is there some way to get the normal component of the gradient out of the difference between the value at the center minus at the surface?"
You made a mistake in the third from bottom line of page four: there is one Γ(n/2) too many in a denominator. One should go into a numerator so it can cancel on the next line.
Thanks for this by the way, it is quite cool. Sort of reminds me of the interpretation of the Ricci Tensor interpretation, so I'm guessing this generalises to other geometeries without too much trouble.
Before the second Γ(n/2) there's a slash that's meant to indicate that you're dividing by it, not multiplying.
By the way: I don't actually know what the interpretation of the Ricci tensor is, so if you were to write a blog post about it (or link it) I'd probably appreciate that.