I am having trouble with understanding part of a proof, I wrote the entire proof out, and I marked the part which I do not understand. I am hoping that someone could explain it to me. I guess I don't understand functional calculus that well :( .
Suppose that $a$ and $b$ are positive elements of a $C^*$-algebra $A$ such that $\| ac \| = \| bc \| $ for all $c\in A$. Then $a=b$.
Proof:
Without loss of generality suppose $a$ and $b$ have norm $1$. It suffices to show that $a^2=b^2$.
Assume $a^2-b^2 \not = 0$. Let $\delta = \frac{1}{2} \sup \sigma (a^2-b^2)$. We can suppose $\delta > 0$ (by excanging $a$ and $b$ if necessary.
Let $f$ be a countinuous real-valued function on $\sigma (a^2-b^2)$ such that $ 0\leq f(\lambda ) \leq 1 \ (\lambda \in \sigma (a^2 - b^2)), $ $ f(\lambda )=0 \ (\lambda \leq \delta ), \text{ and } f(2\delta )=1. $
The following part is what I don't understand
Let $c= f(a^2 - b^2) \in A$ and by applying functional calculus, we can get $ \| c(a^2-b^2)c \| = 2\delta . $
I understand the rest of the proof
Since $\lambda > \delta $ whenever $f(\lambda )\not = 0$, it follows that $c(a^2-b^2)c \geq \delta c^2$.
Let $\rho$ be a state of $A$ such that $\rho (cb^2c) = \| cb^2 c \| $. Since $\| b \| ^2 = 1$, we get that $\rho (cb^2c)\leq \rho (c^2)$, and so $ \rho (ca^2c)\geq \rho (cb^2c + \delta c^2) \geq (1+\delta )\rho (cb^2c). $ Thus $\| ca^2c \| \geq (1+\delta )\| cb^2c \| $. This implies that $\| cb^2 c \| > \| cb^2 c \| $ if $cb^2c \not = 0$. If $cb^2c=0$ then we have $ca^2c \not = 0$ (by the equality $\| c(a^2-b^2)c \| = 2\delta $) and hence $\| ca^2 c \| > \| cb^2c \| $ for all $c$. This implies that $\| ab \| > \| bc \| $, a contradiction.