$A,B$ are bounded linear operators, $AB\geq 0$ implies ($A\geq 0$ and $B\geq 0$) or ($-A\geq 0$ and $-B\geq 0$)?

Question

Let $F$ be a complex Hilbert space $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

Assume that

Let $A,B\in\mathcal{B}(F)$ be such that

• $AB\geq 0$ (i.e. $\langle ABx\;, \;x\rangle\geq 0$ for all $x\in F$).

• $\|AB\|=\|BA\|$.

I claim that ($A\geq 0$ and $B\geq 0$) or ($-A\geq 0$ and $-B\geq 0$). Is it possible to prove the claim?

Answer 1

If you don't assume anything else about the operators then I think it is false. Define $A(x)=ix$ and $B(x)=-ix$. Obviously none of them is positive and none of them is negative. But $\langle AB(x),x\rangle=\langle x,x\rangle\geq 0$ for all $x\in F$. And we also have $||AB||=||BA||$.

Answer 2

Hint: consider $2 \times 2$ diagonal matrices.

Answer 3

If the dimension is at least two, the answer is "no", even in the case of real spaces. Take $$ A=B=\begin{bmatrix}1&0\\0&-1\end{bmatrix}. $$ Then $AB=A^2=BA=I\geq0$, but neither $A\geq0$ nor $-A\geq0$.