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.

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 $n$.

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.

if two symmetric matrices commute they are simultaneously diagonalizable through an orthogonal base chang — Exodd, Dec 08 '22 at 23:09
I would try induction and $$ A^{n+1} = \sqrt{A} A^n \sqrt{A}. $$ Then I would try to show that if $A,B$ commute, that also $\sqrt{A}, B$ commute. If that does not work, I'd still try to do induction, but $A^{n+2} = A A^n A$ (one would then need to consider the case $n=2$ separately). — Severin Schraven, Dec 08 '22 at 23:10

score 5 · Answer 1 · answered Dec 08 '22 at 23:47

5

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

answered Dec 08 '22 at 23:47

Robert Israel

470,583

So the idea is to make a induction proof? – NabbKitha Dec 09 '22 at 00:13
I guess this https://math.stackexchange.com/questions/2001516/prove-that-the-product-of-two-positive-linear-operators-is-positive-if-and-only would help too? – Severin Schraven Dec 09 '22 at 00:37

score 1 · Answer 2 · answered Dec 15 '22 at 12:22

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.

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

2 Answers2