Lemma 4.4 Sample Clauses

Lemma 4.4. Suppose that less than a 1 -fraction of the parties is dishonest. If Pi and Pj broadcast correctly formed messages (i, v, Li) and (j, vr, Lj), respectively, in ΠHBA at time tr < tout, then v = vr.

Lemma 4.4. Suppose that less than a 1 -fraction of the parties is dishonest. If Pi and Pj broadcast correctly formed messages (i, v, Li) and (j, vj, Lj), respectively, in ΠHBA at time tj < tout, then v = vj. Proof. Let s > 0 and suppose that n( 1 − s) parties are dishonest. By assumption, Li and Lj each contain at least 3n valid signatures on v and vj, respectively. This means, that Li and Lj each contain 3n − n( 1 − s) = n + s signatures from honest parties on v and vj, respectively (since signatures are unforgeable). Since no honest party signs distinct messages v and vj, there must be at least 2( n + s) = n( 1 + 2s) many honest parties. This is a contradiction, since by

Lemma 4.4. 11. Fix p ∈ P0 and t ∈ [0, 1−s ]. There exist uniform constants 0 < C < ∞ and 0 < ρ < 1 such that p,t ǁMln f ǁ0,1 ™ Cρnǁf ǁ0,1 ∫ for all f ∈ C0,1([0, 1]) such that ∫ fdµp,t = 0. Proof. Let f ∈ C0,1([0, 1]) for which fdµp,t = 0. If f is constant, f = 0 since its integral is 0 and thus the result follows trivially. If f is not constant, ǁf ǁ0,1 > 0. Let f1 and f2 be the positive and negative parts of f respectively, so that f = f1 − f2 with f1, f2 “ 0. We can guarantee that they belong to a cone by adding a constant.

Lemma 4.4. 16. For all p ∈ P0 and t ∈ [ δ , δ] and 1 ™ k ™ l, [M

Lemma 4.4. 2. If Gm ≥ 0, then κˆ is non-increasing in r . Proof. Let u1, u2, v be points on µv with 0 < ru1 < ru2 < rv . By Lemma 4.3.14 the ray ξui is the limit of geodesics segments that join ui with points τv(s) as s → ∞. The segments [u1, τv(s)], [u2, τv(s)] only intersect at the endpoint τv(s) for if they intersect at a point z , then z is a cut point for τv(s), so [τv(s), ui] cannot be minimizing. Hence the geodesic triangle with vertices u1 , v, τv(s) contains the geodesic triangle with vertices u2 , v, τv(s). Since Gm ≥ 0, the former triangle has larger total curvature, which is finite as Mm has finite total curvature. As m only vanishes at 0, concavity of m implies that m is non-decreasing. If m is unbounded, Xxxxxxxx’s relation implies that the angles at τv(s) tend to zero as s → ∞. By the Gauss-Bonnet theorem κξ1(0)—κξ2(0) equals the total curvature of the “ideal” triangle with sides ξ1 , ξ2 , [u1, u2]. Thus κˆ(ru1 ) ≥ κˆ(ru2 ) with equality if and only if Gm vanishes on [ru1 , ∞). If m is bounded, then ∫ ∞ m—2 = ∞, so by [Tan92a, Proposition 1.7] the only ray emanating from q is µq so that κˆ = 0 on Mm \ {o}. For future use note that in this case the angle formed by µq = ξq and [q, τq(s)] tends to zero as s → ∞, so Xxxxxxxx’s relation together with boundedness of m imply that the angle at τq(s) in the bigon with sides [q, τq(s)] and τq also tends to zero as s → ∞.

Lemma 4.4. If arg τ > π and |x| ≤ 1 , then ঩(g(τ )) −g(y) ≤ −c2y—1, where c2 is an explicitly given constant depending only on d.