Let $H$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$. We say $A\geq 0$ if $\langle Ax,x\rangle\geq0$ for all $x\in H$. If $0\leq A\leq B$ and $C\geq0$ then does $0\leq AC\leq BC$?
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schrodingerscat
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Just to double check, $B \geq A$ means $B - A \geq 0$, right? – roundsquare May 22 '23 at 15:41
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4As pointed out by a comment to your other question, it doesn't make sense to ask whether $AC\ge0$, as $AC$ is in general not self-adjoint. – user1551 May 22 '23 at 15:42
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Adding to the comment of user1551, lack of self-adjointness can be fixed as follows:
Let bounded linear operators $A,B,C$ on $H$ be given. If $0\leq A\leq B$ and $C\geq 0$, then $$ 0\leq \sqrt CA\sqrt C\leq\sqrt CB\sqrt C $$ where $\sqrt C$ is the unique square root of $C$.
Because $\sqrt C\geq 0$ it in particular is self-adjoint meaning $$ \langle x,\sqrt CA\sqrt Cx\rangle=\langle(\sqrt Cx),A(\sqrt Cx)\rangle\geq 0 $$ for all $x\in H$ because $A\geq 0$. Analogously one sees $\sqrt C(B-A)\sqrt C\geq 0$.
Frederik vom Ende
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