If $0 \leq A \leq B$ and $A$ and $B$ commute, then $A^n \leq B^n$

Question

I am dealing with a problem from an old exam paper which might be simple but it turns out to be difficult for me to show.

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Here is the problem: Suppose that $0\leq A\leq B$ for self-adjoint elements $A,B$ in a $C^\ast$-algebra. Then show that if $0\leq A\leq B$ and both $A$ and $B$ commute, then $A^n\leq B^n$ for every positive integer $n$. More generally, show that if there are positive elements $C_j$, $1\leq j\leq k$ with $0\leq A\leq C_1\leq C_2\leq \cdots \leq C_k\leq B$ so that any two neighbors in this list commute, then $A^n\leq B^n$ for every positive integer $n$.

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I don't want the solution for this but maybe a sketch of strategy/sketch of proof so I can work the details out by my self. If you feel you would like to share a example of a solution then feel free. I just want to prepare for my exam in January. Thanks in advance.

Answer 1

Hint: If $A$ and $B$ commute, $$B^n - A^n = (B-A) (B^{n-1} + B^{n-2} A + \ldots + B A^{n-2} + A^{n-1})$$

Answer 2

Your claim immediately follows from functional calculus.

Consider the unital $C^*$-algebra $\mathscr{C}$ generated (inside a unitisation of the $C^*$-algebra if the $C^*$-algebra is not-unital) by $A,B$. This $C^*$-algebra is commutative, since $A$ and $B$ commute, so by Gelfand theory, $\mathscr{C}\cong C(X)$ for a compact Hausdorff space $X$. Hence, you may asume that $A,B$ are continuous functions on $X$. But then the claim is trivially true.

If you know basic $C^*$-algebra theory, you can prove this without working. After a while, you can do these kinds of arguments in your head. In this case, this approach is overkill, but in more complicated situations this is the way to go.