Inverse operator of Laplacian

Question

Suppose $u$ is a vector-valued function in $\mathbb R^n$, i.e. $u(x)=(u_1 (x), u_2 (x), \cdots,u_n (x))$ for $x \in \mathbb R^n$. How to derive: $$\partial_j \partial_k u = \partial_j \partial_k (-\Delta)^{-1}\Delta u$$

So the theorem goes like this: "Let $\Omega$ be $\mathbb R^3$ and assume $-\Delta u = f$ in $\Omega$ and that $f$ belongs to $L^p (\Omega)$ for some $p \in (0, \infty)$. Then all second derivatives of $u$ can be bounded in $L^p$ in terms of $f$, namely $$\|\partial_j \partial_k u\|_p \leq C_p \|f\|_p".$$ Then the proof goes using the identity above.

For me, it means the inverse operator of Laplacian is $(-\Delta)^{-1}$ (but why with the negative sign?) or maybe I understand it wrong.

Answer 1

From my point of view the question is not about the inverse of the Laplacian. One can indeed define the inverse of the Laplacian in suitable spaces, but I don't think that's a useful point of view here.

First, the equation $-\Delta u=f$ implies that the norm of $\sum_i\partial_i\partial_i u$ is the same as that of $f$. That is, you can estimate the sum of the diagonals of the Hessian $D^2u$ by $f$ in any space you like because they are the same thing. The big surprise is that in fact you can also estimate any matrix element of the Hessian — any $\partial_i\partial_ju$ — with $f$ as well.

Very roughly, the theorem says: If you know something about the Laplacian of $u$, then you know something about all second order partial derivatives of $u$.

Also, let me explain the negative sign. The Laplace operator $\Delta$ is a "negative operator" in the sense that all eigenvalues are necessarily negative (or zero). This negativity has many incarnations, and I mention two:

1. If $\Delta g=\lambda g$ for $g\in W^{1,2}_0(\Omega)$, then $\lambda\leq0$.
2. As a Fourier multiplies $\Delta=-|\xi|^2$ which is always non-positive.

Ask for more details in separate questions if you are interested. It follows that $-\Delta$ is a positive operator, and that makes taking powers more convenient. This is why things are often phrased in terms of $-\Delta$ rather than $\Delta$. In fact, some define the Laplacian to be $\Delta=-\sum_i\partial_i\partial_i$ to make it positive — beware of varying sign conventions!