# How much does the absolute value of an operator behave like an absolute value?

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?

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

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 $$p_1 + p_2 - |p_1 - p_2| = \left(\matrix{ 1+ \cos^2 \theta - |\sin \theta| &\cos \theta \sin \theta\\\sin \theta \cos \theta&\sin^2 \theta - | \sin \theta|}\right)$$