218 6. Amenable Traces
and, by the Powers-St0rmer inequality,
1 1 1 1/2 1 2 1
l -Tr(h2uh2u*) = 2JJh -uh2u*Jb :S 2JJh-uhu*J[1.
We use here and there the fact that Tr(ST) = Tr(S^112 TS^112 ) 2: 0 for any
positive operators S, T with S finite-rank. Now, let's compute tr(cp( u)cp(u*)):
tr ( cp ( u ) cp ( u )) = Tr(P(u@ Tr(P) l)P(u@ l)P)
l:~j=l Tr(QiuQju*) Tr(PiPj)
q
= t min{pi,Pj} Tr(QiuQju*)
i,J= ..^1 q
k
2: Tr(h~uh~u*) - L [Pi; Pj[ Tr(QiuQju*)
i,J=.. 1 q
since min{pi,Pj} = ~(Pi+ Pj - IPi - Pjl) 2: y'PiPj-~[Pi - Pjf· We have
further that
k k k k
L Pi Tr(QiuQju*) = L Pi Tr(Qiu(L Qj)u*)::::; L Pi Tr(Qi) = 1
i,j=l q i=l q j=l i=l q
and hence
k
L [Pi; Pjl Tr(QiuQju*)
i,j=l q
~ ,tl ( 1Pf 2:;J I Tr( Q,uQ;u') l) (IP! ;JI Tr( Q,uQ ;u') l)
1 1
::::; (t IPf ;P}I' Tr(Q,uQ;u')) 2 (t Jpf -pJJ2 Tr(QiuQju*)) 2
i,J= ..^1 q i,J= ..^1 q
s: (2+2Tr(Zluhlu•J )' ( 2 -2Tr(hluhlu•))l
:S [[h-uhu*JJi^12 ,
by the previous computations. Finally, combining everything, we get
k
tr(cp(uu) - cp(u)cp(u)) ::::; 1 -Tr(h^112 uh^112 u) + L IPi; Pj[ Tr(QiuQju)
i,j=l q