Taking the square root is an operator monotone operation. That is, if $A$ and $B$ are positive bounded operators on a Hilbert space $H$, then $A \leq B$ implies $A^{1/2} \leq B^{1/2}$. This is a bit of a pain to prove. One argument proceeds by assuming (without loss of generality) that the operators are invertible, and then making some very clever manipulations. Other strategies rely on the functional calculus and some approximation arguments.
Given a bounded operator $A$ on a Hilbert space $H$, the operator modulus of $A$ is defined to be $|A| = (A^*A)^{1/2}$.
Consider the following proposition and proof:
Proposition: If $A$ and $B$ are bounded operators on a Hilbert space $H$, then $|AB| \leq \|A\| |B|$.
In other words, $B \mapsto AB$ is a "bounded operator" with respect to the operator modulus.
Proof: Since $A^*A \leq \|A\|^2 \cdot \mathrm{id}_H$, we have $(AB)^*(AB) = B^* A^* A B \leq \|A\|^2 B^*B$. Since the square root is operator monotone, the desired inequality is immediate.
I am curious about the following.
Question: Is there a snappy proof of the above proposition which avoids operator monotonicity of the square root? Or are these two results of roughly equal difficulty?
Just a small edit to point out some easy reductions one can make.
Claim: Without loss of generality, $A$ and $B$ are positive invertible operators and $\|A\|=\|B\| = 1$.
With $A$ and $B$ as in the claim, the inequality to be proved is $|AB| \leq B$.
Proof of claim:
- $B$ can be positive. Indeed, let $B = W|B|$ be the polar decomposition of $B$. If the inequality holds for positive $B$ then $|AB| = |(AW)|B|| \leq \|AW\| |B| \leq \|A\| |B|$, and the inequality holds in general.
- $A$ can be positive. Note that $|AB| = \sqrt{BA^*AB}$ and $\|A\| = \sqrt{\|A^*A\|}$ only depend on $A^*A$. Thus, replacing $A$ with $|A|$, we change neither the LHS nor the RHS of $|AB| \leq \|A\| |B|$.
- $A$ and $B$ can be invertible positive operators. The function $(A,B)\mapsto \|A\||B| - |AB|$ is (norm) continuous, and the positive operators are closed, so the set of $A,B$ for which the inequality holds is closed. Since the invertible positive operators are dense in the positive operators, in is enough to check the inequality on those.
- Can assume $\|A\|=\|B\|=1$. Just rescale them.
One consequence of this reduction is that the partial isometry $W$ in the polar decomposition $AB=W|AB|$ becomes a unitary.