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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 $a=b$.


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.

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Thanks, I never realized I had to click the check, I only click the up arrow for answers that I liked, I went back, and clicked the check for the good answers just now. – Peter Aug 20 '12 at 1:49
Informally (and incorrectly) speaking, spectral theorem says that $a^2-b^2$ is the multiplication operator $M_x$ acting on some $L^2$ space supported on $\sigma$. Then $c$ is multiplication by $f(x)$. Consequently, $c(a^2-b^2)c$ is multiplication by $xf(x)^2$ (note that these operators commute). The norm of multiplication operator is the supremum of its symbol $xf(x)^2$ on $\sigma$. The supremum is attained at $2\delta$ and is equal to $(2\delta)f(2\delta)^2=2\delta$. – user31373 Aug 20 '12 at 2:37
up vote 2 down vote accepted

$c (a^2 - b^2) c = f(a^2 - b^2)(a^2 - b^2) f(a^2 - b^2) = g(a^2 - b^2)$ where $g(x) = f(x)^2 x$. Now $g(2\delta) = 2 \delta f(2 \delta)^2 = 2 \delta$ and $2 \delta \in \sigma(a^2 - b^2)$ so $\|g(2 \delta)\| = \sup_{z \in \sigma(a^2 - b^2)} |g(z)| = 2 \delta$.

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