If $A,B$ are self-adjoint and $A+B\ge 0$, then $A_+-B_-,\ B_+-A_-\ge 0$?

Question

Let $0\le X\in B(H)$ ($X$ is a positive bounded linear operator on a Hilbert $H$). Assume that there are self-adjoint $A,B\in B(H)$ such that $X=A+B$. I am thinking how to decompose $X$ into some positive operators depending on $A,B$. My idea is $$X=A+B=A_+-A_- +B_+-B_- =(A_+-B_-)+ (B_+-A_-).$$ However, I don't know how to show that $A_+-B_-$, $B_+-A_-\ge 0$.

Answer 1

This is not true. Define $$ B = \frac12\pmatrix{1 & 1 \\ -1 & 1} \pmatrix{1&0\\0&-1}\pmatrix{1 & -1 \\ 1 & 1} = \pmatrix{0 & -1\\ -1 & 0} $$ with $$ B^- = - \frac12\pmatrix{1 & 1 \\ -1 & 1} \pmatrix{0&0\\0&-1}\pmatrix{1 & -1 \\ 1 & 1} = -\pmatrix{ -1 & -1\\ -1 & -1}. $$ Set $$ A=\pmatrix{3&0\\0&1} = A^+. $$ Then $$ A+B =\pmatrix{3&-1\\-1&1} $$ is positive definite, $$ A^+-B^-=\pmatrix{2&-1\\-1&0} $$ is indefinite.