Recall that the absolute value of a bounded operator $T$ on a Hilbert space $H$ is the unique positive operator $|T|$ such that $$\||T|x\|=\|Tx\|$$ for all $x\in H$. It can be defined using the continuous functional calculus, or if you have square-roots of positive operators in hand, by $|T|=(T^*T)^{1/2}$. Likewise, this can be defined in any C*-algebra.

With respect to the usual ordering on self-adjoint elements, i.e., $S\leq T$ if $T-S$ is positive, does this absolute value behave like an absolute value? For instance, does it satisfy a triangle inequality like the one below? $$||T|-|S||\leq|T-S|\leq|T|+|S|$$

If $S$ and $T$ are normal and commute, then perhaps one could use the functional calculus on the commutative C*-algebra they generate to show this, but not for general operators.

How about other inequalities with respect to this ordering?

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    $\begingroup$ The identities you mentioned are not necessarily true. See discussion in Reed and Simon Vol-1, chapter 6, exercises 16 and 17. $\endgroup$ – Piyush Grover Jul 8 '14 at 18:29

The answer is no. Loewner partial order is subtle. There are some simple examples given in R. Bhatia, F. Kittaneh, The matrix arithmetic–geometric mean inequality revisited, Linear Algebra and its Applications 428 (2008) 2177–2191


In general, if $p_1$, $p_2$ are non-commuting rank $1$ projectors, then $$p_1 + p_2 \not \succeq |p_1 - p_2|$$ Indeed, may assume $p_1 = \left(\matrix{ 1 &0\\0&0}\right)$, $p_2 = \left(\matrix{ \cos^2 \theta &\cos \theta \sin \theta\\\cos \theta \sin \theta&\sin^2 \theta}\right)$, then one checks $(p_1-p_2)^2 = \sin^2\theta \cdot I_2$, and so $$p_1 + p_2 - |p_1 - p_2| = \left(\matrix{ 1+ \cos^2 \theta - |\sin \theta| &\cos \theta \sin \theta\\\sin \theta \cos \theta&\sin^2 \theta - | \sin \theta|}\right)$$


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