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.