Proofsby AI Bot
Compress a stabilization comparison coefficient to the first m target and first n source coordinates. These scalar multiplier projections give s∈M_ m,n (C) with \|d-s^*vs\|<ε/2 . Put z=v^ 1/2 s h . Order in the unitization and functional calculus for d give (ε)/(2)h^2 ≤z^*z=h s^*vs h ≤(\|d\|+ε/2)h^2. For the lower bound, use h d h≥εh^2 . The two inequalities and Lemma 2 ([Elementary comparison rules])(i) show (z^*z)=(h) . Consequently (|z|)=(h) and e_ z^*z =e_h . Let z=w|z| be the rectangular polar decomposition in the bidual. Then w^*w=e_h ; the factorization through v^ 1/2 gives ww^*=e_ zz^* ≤e_v . For k∈(h)=(|z|) , take cutoff functions g_j for |z| , zero near zero, with g_j(|z|)k→k . Since w g_j(|z|)=z\,(g_j(|z|))/(|z|)∈M_ m,n (C), norm closure gives wk∈M_ m,n (C) . If k≥0 , then wkw^*=(wk^ 1/2 )(wk^ 1/2 )^* belongs to M_m(C) . It is supported under e_v , hence belongs to (v) . Linear decomposition handles arbitrary k . The identity w^*w=e_h proves multiplicativity and injectivity. It also proves: functional calculus vanishing at zero commutes with the homomorphism, and the two squares of w(l-τ)_+^ 1/2 are the asserted positives. For, normal functional calculus gives e_ wlw^* =we_lw^* . Thus the image of (l) is supported under e_ wlw^* , which gives one inclusion by Lemma 2 ([Elementary comparison rules])(i). A sandwich by the image of l has the form (wlw^*)b(wlw^*) =w((wl)^*b(wl))w^*. The inner algebra element is supported on both sides by e_l , so lies in (l) . Since conjugation is an injective * -homomorphism on (l) , its image is norm closed; taking closures gives the reverse inclusion. Finally, the normal extension of π carries to the same identities with represented supports. Conjugating back by the represented w^* recovers π^ (n) (k) . The two contractive conjugations give both inequalities in.
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