This question is about the proof of lemma 14 in 1, which deals with the decomposition of an unbounded self-adjoint operator $A$ in positive and negative parts: $A = A^+ – A^-$. I have trouble understanding the proof given at the bottom of page 27 that $A^- \geq 0$.
Indeed, the author writes that $(A^+ + A^-)Ex = |A|Ex$, where $x \in \mathcal{D}(A)$ and $E$ is the orthogonal projector onto $\mathcal{N}(A^+)$. This seems to assume that $Ex \in \mathcal{D}(A)$ but I don't see where this would come from. Any idea?
EDIT: ~~In fact, I wonder whether one could not start to prove that $|A| – A \geq 0$. Since $A^-$ is the closure of this operator, the result would follow, right?~~
Best Answer
So stupid of me! I had simply forgotten that $A = A^+ - A^-$ by point i), hence in particular $\mathcal{D}(A^+) \cap \mathcal{D}(A^-) = \mathcal{D}(A)$.