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Jan 3, 2014 at 3:41 vote accept Jiang
Jun 29, 2011 at 20:06 comment added Matthew Daws @Yemon: Nice Question! In the general case, I don't know. But if $G$ is amenable, then as the trivial rep is contained in the left regular rep, if $f\in C^*(G)$ (with abuse of notation, $f$ is the "Fourier coefficient") is pointwise positive, then $\|f\|_{C^*} = \|f\|_1$. As $\ell^1(G) \not= C^*(G)$ unless $G$ is finite, it cannot be that $f\in C^*(G)$ implies that the absolute value of $f$ is also in $C^*(G)$.
Jun 29, 2011 at 18:59 comment added Yemon Choi Matt, do you know of any example of something in the reduced $C^*$ algebra where replacing each "Fourier coefficient" with its absolute value results in an unbounded operator? (Classical Fourier analysis gives examples of this for the group ${\mathbb Z}$, i.e. you can't always tell if a function on the circle is continuous just by knowing the modulus of each Fourier coefficient.)
Jun 29, 2011 at 17:11 history edited Matthew Daws CC BY-SA 3.0
Better reference.
Jun 29, 2011 at 16:34 history answered Matthew Daws CC BY-SA 3.0