1
$\begingroup$

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 $C_0(\Omega)$ ?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Maybe there is some misunderstanding, but $\gamma([\mu])$ is linear, while $|\mu|$ is not. $\endgroup$
    – Fan Zheng
    Dec 30, 2015 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, 2015 at 20:46
  • $\begingroup$ @YemonChoi How is the absolute value of a functional on a *-algebra defined? $\endgroup$
    – Fan Zheng
    Dec 30, 2015 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, 2015 at 22:18
  • 2
    $\begingroup$ 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, 2015 at 7:46

1 Answer 1

Reset to default
1
$\begingroup$

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)

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ If you are going to answer your own question, please provide details of how this theorem answers your question $\endgroup$
    – Yemon Choi
    Jan 5, 2016 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.