1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

216 6. Amenable Traces

for all finite-rank h, k. Letting hk = ulhkl be the polar decomposition, one
checks that

which completes the proof.

We will need the Powers-Stf}rmer inequality:

Proposition 6.2.4. For any 0 :S: h, k E S2(1i), one has

llh - kJI~ :s: llh^2 - k^2 ll1 :s: llh + kl/2llh - kll2·

In particular, if u E IBl(7-i) is a unitary and h 2:: 0 has finite rank, then

lluh^1 /^2 - h^1 /^2 uJl2 = lluh^112 u - h^1 /^2 1/2 :S: lluhu - hJli^12.

0

Proof. We may assume that h and k are finite-rank. (Why?) The right
inequality follows from the previous lemma and the identity

h^2 - k^2 = ((h + k)(h - k) + (h-k)(h + k))/2.

If h and k commute, then the operator inequality lh - kl^2 :S: lh^2 - k^2 j
implies the left inequality. Unfortunately, this operator inequality doesn't
hold in the noncommutative case. To circumvent this problem, we exploit
the fact that 'I'r(xy) 2:: 0 for any x, y 2:: 0, so long as either x or y is finite-
rank. Now, let e = X[o,oo)(h-k) be the spectral projection and let e..l = 1-e.
Since (h-k)e 2:: 0 and (k-h)e..l 2:: 0, one has

Tr((h - k)^2 ) = Tr((h - k)(h - k)e + (k - h)(k - h)e..l)

:::; Tr((h + k)(h - k)e + (k + h)(k - h)e..l)

= Tr((h^2 - k^2 )e + (k^2 - h^2 )e..l)

:S: Tr(lh^2 - k^2 le + lk^2 - h^2 1e..l)

= Tr(jh^2 - k^2 j),

where the third line equality follows from

Tr((h + k)(h - k)e) = Tr((h + k)e(h - k)) = Tr((h - k)(h + k)e)

and a similar identity for e..l.^6 0

Our next lemma is the main technical result of this section - you'll need
lots of scratch paper. Seriously, lots. ·

(^6) we thank Sorin Popa for sharing this elegant proof with us.