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Sep 20, 2016 at 14:31 comment added Yemon Choi Not that I know of...
Sep 20, 2016 at 14:26 comment added Mikael de la Salle I have often wondered: is there any connection between you and Choi (M-D)?
Sep 20, 2016 at 14:24 comment added Yemon Choi I often feel that Choi (Y) needs to pay closer attention to ideas introduced by Choi (M-D)
Sep 20, 2016 at 11:09 comment added Mikael de la Salle @YemonChoi: sure. $f \colon G \to \mathbf{C}$ being PSD means that the Fourier multiplier $M_f$ with symbol $f$ is completely positive. So if we have such PSD $f$ and $t \in G$ such that $f(t) = f(0)=1$, we have (Choi) that $t$ belongs to the multiplicative domain of $M_f$, and so the whole group generated by $f$ belongs to the multiplicative domain, ie that $f$ is constant equal to $1$ on the group generated by $t$.
Sep 19, 2016 at 13:53 vote accept Yemon Choi
Sep 19, 2016 at 13:53 comment added Yemon Choi Nice! I wonder if this example can be seen as some kind of argument with multiplicative domains or the Schwarz inequality for cp maps... but this is purely idle speculation
Sep 19, 2016 at 8:27 history answered Mikael de la Salle CC BY-SA 3.0