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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 $a=b$.

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.

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Have you thought of enhancing your accept rate? As it is it might disuade people to write down proposal of solution as it seems like you don't like the answers you get here. –  DonAntonio Aug 20 '12 at 1:41
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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
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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
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1 Answer 1

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$c (a^2 - b^2) c = f(a^2 - b^2)(a^2 - b^2) f(a^2 - b^2) = g(a^2 - b^2)$ 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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