Let $A,B$ be self-adjoint, bounded, linear operators on a Hilbert space $H$ over $\mathbb{R}$ such that $\langle Ax,x\rangle\geq 0$ and $\langle Bx,x\rangle\geq 0$ for all $x\in H$. Does it also hold that $\langle ABx,x\rangle\geq 0$ for all $x\in H$?
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1$\begingroup$ The answer is no, but the details depend on whether you're talking about Hilbert spaces over $\Bbb R$ or over $\Bbb C$. Over $\Bbb R$, some care is required in constructing $A,B$ such that $\langle ABx ,x \rangle < 0$. $\endgroup$– Ben GrossmannCommented Jan 23 at 15:21
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2$\begingroup$ In the simplest case, just take $H$ to be a two-dimensional space (either over $\mathbb{C}$ or over $\mathbb{R}$), then take $A = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$. $\endgroup$– David GaoCommented Jan 23 at 19:35
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$\begingroup$ You're right, thank you. I mean Hilbert spaces over $\mathbb{R}$ and updated the question. I think for $𝐴𝐵$ to be positive more assumptions are needed, for instance, that $𝐴$ and $𝐵$ commute. Does it maybe also work, if $A$ is strictly positive? $\endgroup$– SuimCommented Jan 24 at 9:54
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$\begingroup$ @Suim Assuming $A$ is strictly positive does not work, just change $A$ to be $\begin{pmatrix} 2 & 0\\0 & 1\end{pmatrix}$ in the example. Similarly, assuming both $A$ and $B$ are strictly positive does not work either. As Disintegrating By Parts pointed out in their answer, this happens if and only if $A$ and $B$ commute. $\endgroup$– David GaoCommented Jan 25 at 18:59
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Assume that $A,B$ are positive operators. A positive operator is self-adjoint. So, if $AB \ge 0$, then it is necessary that $(AB)^*=AB$, or $BA=AB$. In that case $AB \ge 0$ because, you may write $A=\sqrt{A}^2$ where $\sqrt{A}$ is the unique positive square root of $A$ that commutes with everything that commutes with $A$. Then you end up with $\sqrt{A}B=B\sqrt{A}$ and $$ AB=\sqrt{A}\sqrt{A}B=\sqrt{A}B\sqrt{A} \ge 0. $$
answered Jan 25 at 17:47
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$\begingroup$ Double post! SE must have had a server error as you were submitting your answer. $\endgroup$ Commented Jan 25 at 17:50