why is $f(xx^*)$ is defined and $xf(x^*x)=f(xx^*)x$?

Question

Let $A$ be a $c^*$algebra, $x\in A$ and $f:\sigma(x^*x)\to\mathbb{C}$ continuous and $f(0)=0$ ($\sigma(x^*x)$ is the spectrum of $x^*x$). Why is $f(xx^*)$ is defined and $xf(x^*x)=f(xx^*)x$?

It is $x^*x$ and $xx^*$ selfadjoint and therefore you can consider the continuous functional-calculus $\phi:C(\sigma(x^*x))\to C^*(x^*x,1)\subset A, f\mapsto f(x^*x)$ and the same for $xx^*$: $\phi':C(\sigma(xx^*))\to C^*(xx^*,1)\subset A , f\mapsto f(xx^*)$ (here I'm not sure if it is correct because A maybe don't need to be unital!) . I'm not sure why $f(0)=0$ is important (and if it is important that $x^*x$ is not invertible) and what the answer of the question should be. Could you help me to answer the question why $f(xx^*)$ is defined and $xf(x^*x)=f(xx^*)x$?? Is the continuos functional-calculus useful? Regards

Edit: T.A.E gave me the hint to show the identity first for polynomials: Let $p(x^*x)=\sum\limits_{k=0}^na_k(x^*)^kx^k$, $a_k\in\mathbb{C}$. $xp(x^*x)=x\sum\limits_{k=0}^na_k(x^*)^kx^k=\sum\limits_{k=0}^na_kx(x^*)^kx^k$ and $p(xx^*)x=\sum\limits_{k=0}^na_kx^k(x^*)^kx$. The problem reduces that I need to know, why $x(x^*x)^k=(xx^*)^kx$ for every natural number k is true. I'm not sure how to justify this (do we need that $x^*x$ is self-adjoint now)?

Answer 1

There is an error in your edited material. In general, $ (x x^{*})^{k} \neq x^{k} (x^{*})^{k} $ unless we know that $ x $ commutes with $ x^{*} $.

The justification that you seek is simple.

• When $ k = 0 $, you get $ x = x $, which is true.
• When $ k = 1 $, you get $ x (x^{*} x) = (x x^{*}) x $, which is also true as $ A $ is an associative algebra.
• When $ k = 2 $, you get $ x (x^{*} x) (x^{*} x) = (x x^{*}) (x x^{*}) x $, which is true yet again by associativity.
• Perform mathematical induction to prove that $ x (x^{*} x)^{k} = (x x^{*})^{k} x $ for all $ k \in \mathbb{N} $. This step is purely algebraic manipulation in an associative algebra.