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Let $H$ be a Hilbert space with inner product $\langle\cdot,\cdot\rangle$. We say $A\geq 0$ if $\langle Ax,x\rangle\geq0$ for all $x\in H$ (and similarly $\leq 0$). If $A\geq0$ and $B\leq 0$ do we have $AB\leq 0$ and $BA\leq 0$?

schrodingerscat
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    Up to replacing $B$ by $-B$, this is the same as asking "Does $A\geq 0$ and $B\geq 0$ imply $AB\geq 0$". This is false ($AB$ might not even be symmetric) and I'm sure this has been asked before on this site ... here for example https://math.stackexchange.com/questions/2001516/prove-that-the-product-of-two-positive-linear-operators-is-positive-if-and-only or here https://math.stackexchange.com/questions/3798613/if-a-and-b-are-self-adjoint-and-positive-operators-show-that-also-their-prod?rq=1 – LL 3.14 May 22 '23 at 11:19
  • In fact, $A \geq 0$ does not imply $A^{2} \geq 0$. – geetha290krm May 22 '23 at 11:23
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    @geetha290krm In the complex inner product space $A\ge 0$ does imply $A^2\ge 0.$ – Ryszard Szwarc May 22 '23 at 13:37
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    @RyszardSzwarc True. I made the comment because OP did not specify the scalar field. – geetha290krm May 22 '23 at 15:01

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Much like my other answer I'd like to add to the comments by saying that lack of self-adjointness can be fixed as follows:

Given bounded linear operators $A,B$ on $H$ be given. If $B\leq 0\leq A$, then $$ \sqrt AB\sqrt A\leq 0\qquad\text{and}\qquad \sqrt{-B}(-A)\sqrt{-B}\leq 0 $$ where $\sqrt{(\cdot)}$ is the unique square root.

Using self-adjointness of $\sqrt A,\sqrt{-B}\geq 0$ we compute $$ \langle x,\sqrt AB\sqrt A\rangle=\langle(\sqrt Ax),B(\sqrt Ax)\rangle\leq 0 $$ for all $x\in H$ (and analogously for $\sqrt{-B}(-A)\sqrt{-B}$).

Frederik vom Ende
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