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Let $\Omega$ be a locally compact and Hausdorff topological space. The Riesz representation theorem says that $C_0(\Omega)^*$ , dual of the commutative C*-algebra $C_0(\Omega)$, is just the space of complex Radon measures $M(\Omega)$.

$$\gamma: M(\Omega)\simeq C_0(\Omega)^* : \gamma(\mu)(f)=\int f d\mu$$

Let $\mu$ be a complex Radon measure and denote $[\mu]$, by the total variation of $\mu$.

Question: It seems that $\gamma([\mu])=|\gamma(\mu)|$, where $|\gamma(\mu|)$ is the absolute value of the bounded linear functional $\gamma(\mu)$ on $C_0(\Omega)$ ?

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    $\begingroup$ Maybe there is some misunderstanding, but $\gamma([\mu])$ is linear, while $|\mu|$ is not. $\endgroup$ – Fan Zheng Dec 30 '15 at 17:09
  • $\begingroup$ @FanZheng From the surrounding context I think $|\mu|$ should be $|\gamma(\mu)|$ where the latter is understood as the absolute value of a functional on a $*$-algebra, hence is itself a linear functional $\endgroup$ – Yemon Choi Dec 30 '15 at 20:46
  • $\begingroup$ @YemonChoi How is the absolute value of a functional on a *-algebra defined? $\endgroup$ – Fan Zheng Dec 30 '15 at 21:49
  • $\begingroup$ @FanZheng Well, perhaps this only works for normal functionals on a von Neumann algebra, I admit I'm a bit sketchy on the details. But for such a functional $f$ it can be factorized as $f=u\cdot g$ where $h$ is a positive normal functional and $u$ is a partial isometry in the (von Neumann) algebra $\endgroup$ – Yemon Choi Dec 30 '15 at 22:18
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    $\begingroup$ It would be helpful if you could define more explicitly the objects you are talking about, so that people don't have to guess about the notation. $\endgroup$ – Nate Eldredge Dec 31 '15 at 7:46
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The answer is directly obtained by the following interesting fact:

Theorem. Let $M$ be a von neumann algebra and $f$ be a normal functional on $M$. Assume $a$ is in the unit ball of $M$ with $||af||=|f||$ and $af$ is positive. Then $af$ is just the absolute value of $|f|$. (see theorem 3.2 in Order ideals in a C*-algebra and its dual by E. Effros 1962)

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    $\begingroup$ If you are going to answer your own question, please provide details of how this theorem answers your question $\endgroup$ – Yemon Choi Jan 5 '16 at 14:13

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