Prove
Proposition 24 ([Global profile cost]). For every prescribed profile, its…Lemma 23 ([At most two active incidences]). For two active incidences whose…Proposition
Put m=_ℓs_ℓ and choose ℓ_0 with s_ ℓ_0 =m . Using h=c+m and ∑s_ℓ≤D-c , expand 3C-∑_ℓb(c+s_ℓ)-2b(c+m) , =3(D+1)c-5b(c) -c(∑_ℓs_ℓ+2m) +2∑_ ℓ≠ℓ_0 b(s_ℓ); , ≥b(c)+2∑_ ℓ≠ℓ_0 b(s_ℓ). The last step uses ∑s_ℓ+2m≤3(D-c) ; substituting this upper bound leaves exactly b(c) in the terms involving c . Consequently β_j(h)3+b(c)6 +13∑_ ℓ≠ℓ_0 b(s_ℓ). The gain is zero with zero or one active incidence. With two, it is b( h'/2) for their pair dimension h'≤h . For an integer h≥0 , b( h/2)(h)3, b( h/2)(h+2)8. The first follows separately for h=2k and h=2k+1 ; its differences are k(k-1)/6 and (k+1)(k+2)/6 , respectively, so equality holds exactly for h=0,2 . Combining these inequalities with Equation proves nonnegativity and S_j≥b(h)/3-h(h+2)/8=h^2/24-h/12 . Equality forces c=0 , s_ℓ=0 for ℓ≠ℓ_0 , and h=0 or 2 . When h=2 , the gain must equal 1 , which requires the stated two active endpoints. With at most one active incidence the positive baseline for h>0 excludes equality.