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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.

mika

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    See also: http://math.stackexchange.com/questions/183486/inversion-in-a-unital-c-algebra2012-12-15

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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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    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.2012-01-02