# Inverse operator of Laplacian

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.

• Strictly speaking the Laplacian is not invertable as in there is not injective. But leaving some detail, I think you are forgetting some boundary conditions to state? Because in pde we can use the divergence theorem and usually using some boundary conditions we can kill one term and get rid of the Laplacian and obtain a $-u$.
– Kori
Apr 21, 2017 at 5:00
• @Kori Good point! 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 in my question. Apr 21, 2017 at 5:16
• Can you edit your question to add what you stated in the comment? That'd make it easier to read. The context really helps make more sense of the question. Apr 21, 2017 at 5:32
• @JoonasIlmavirta Sure! Apr 21, 2017 at 5:35
• Thanks! I made a little polishing edit (feel free to re-edit or roll back!) and gave an answer. I hope it clarifies things. Apr 21, 2017 at 5:48

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!