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I need the following result. I think it's quite obvious but I don't know how to prove that: Let $C, T : \mathcal{H} \rightarrow \mathcal{H}$ be two positive, bounded, self-adjoint, invertible operators on a Hilbert space $\mathcal{H}$ such that $C \leq T$. Then it follows also that $T^{-1} \leq C^{-1}$. Or maybe this is even not true? If this is not true can somebody give me a counter-example or if it is true some strategy how to solve this? I would be very thankful.


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It is not obvious if you don't know how to show it. Note for example that $0\leq C\leq T$ does not imply $C^2\leq T^2$. What this result shows is that $x\mapsto -1/x$ is operator monotone on $(0,\infty)$, which is a much stronger condition than just being monotone. – Jonas Meyer Jan 2 '12 at 21:41
See also:… – Jonas Meyer Dec 15 '12 at 5:39
up vote 4 down vote accepted

Note that $T \ge C$ iff $C^{-1/2} T C^{-1/2} \ge I$ iff $C^{1/2} T^{-1} C^{1/2} = (C^{-1/2} T C^{-1/2})^{-1} \le I$ iff $T^{-1} \le C^{-1}$

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Nice, and yes, I upvoted this. But (no offense intended) note that in order to be able to note this, you need to note that $C$ has a root which, you have to note in addition, is invertible. This is in fact known given the assumptions the OP stated, but still involves quite some machinery to prove. At least this is way beyond what is needed to formulate the question. – user20266 Jan 2 '12 at 22:07
That a positive operator has a unique positive square root is one of the basic facts of operator theory, and not hard to prove. That any square root of an invertible operator must be invertible is very easy. – Robert Israel Jan 2 '12 at 23:45

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