推演由 AI Bot
If q=0 , take r=0 ; if Y=0 , apply Proposition 35 to (h,z,a)=(q,q,tq) . Scaling Y and the tolerance reduces to 0≤Y≤q . All functions below are evaluated at Y in the unital corner qDq . Choose a large integer m . Let F_0=0 , F_m=1 ; for 1≤j<m let F_j be 1 up to (j-1/2)/m , decrease linearly to 0 at j/m , and vanish thereafter. Let G_j be 1 up to j/m , decrease to 0 by (j+1/3)/m , and vanish thereafter for j<m ; put G_m=1 . These functions satisfy G_jF_j=F_j, F_ j+1 G_j=G_j. Construct orthogonal projections r_j recursively with G_jr_j=r_j, τ(r_j)=tτ(F_j-F_ j-1 ). Indeed, if R=∑_ i<j r_i , then F_jR=G_jR=R and τ(R)=tτ(F_ j-1 ) . Apply Proposition 35 to h=G_j-R, z=F_j-R, a=t(F_j-F_ j-1 ). They are positive contractions, hz=z , and τ(z)-τ(a)=(1-t)τ(F_j)≥0 . The selected projection is orthogonal to R , lies under q , and is fixed by G_j . Now r=∑_jr_j has trace tτ(q) . The spectral support conditions give r_jYr_j≤(j+1/3)/(m)r_j, whereas F_j-F_ j-1 is supported at values at least (j-3/2)/m . The difference of these thresholds is 11/(6m)<2/m . Summing therefore gives τ(rY)≤t∑_j(j+1/3)/(m)τ(F_j-F_ j-1 ) ≤t(τ(Y)+(2/m)τ(q)). Take 2/m<ε .
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