Am I allowed to move $\partial_j$ (partial derivative with regards to $j$th variable) around?
Particularly, because I think that a proof related to Calderon's problem here p. 22 seems to end up with a factorization that has only the term $-\partial_j^2$.
So I think what is done there is that the derivative is treated as an operator?
$$\left(-\partial^2_j + \frac {\partial^2_j\gamma^{1/2}}{\gamma^{1/2}}\right)u$$ means $$-\partial^2_j u + \frac {\partial^2_j\gamma^{1/2}}{\gamma^{1/2}}u$$ You can think of it as just a kind of abbreviation, or you can view $\partial^2_j$ as a map from $C^2(\overline \Omega) \to C^0(\overline \Omega)$, and thus as an object in its own right. If so, then you should also recognize that any $f \in C^0(\overline \Omega)$ can be thought of as another such operator which acts on $C^2(\overline \Omega)$ by multiplication. As the space of operators on $C^2(\overline \Omega)$ is an algebra, any linear combination of them, such as $$\left(-\partial^2_j + \frac {\partial^2_j\gamma^{1/2}}{\gamma^{1/2}}\right)$$ is also an operator.
