Suppose Clause Samples

Suppose. F : X → X {u ∈ X|u = λF [u] for some λ with 0 ≤ λ ≤ 1} The next theorem is the well-posedness of the discrete problem.

Suppose κˆ(rq) = π , so that γ = γq , and suppose that Gm vanishes along γ . We show that q is a soul by showing that every point in Mm is a soul. Our strategy is twofold: First we show that o must be in the horoball of γq . Using this fact, we then show that if we choose basepoint q appropriately, any point in Mm can be rendered a soul. By rotational symmetry Gm = 0 for r ≥ rq , so m(r) = ar + m(0) for r ≥ rq where a > 0, as m only vanishes at 0. The turn angle of γ can be computed explicitly as q 1 ∫ ∞ dr ∫ ∞ dt 1 √ 2 π m(r)2 m(x)2 = a t √t2 — 1 = — a arccot( t — 1) 1 = 2a 2 where x := rq . Since γ is a ray, we deduce that a ≥ 1 , for if a < 1 , then the turn angle of γ would be greater than π, implying that γ intersects τq . Let z ≤ x be the smallest number such that mr|[z,∞) = a; thus there is no neighborhood of z in (0, ∞) on which Gm is identically zero. Note that m(r) = a(r — z) +m(z) for r ≥ z , so the surface Mm – B(o, z) is isometric to C – B(o¯, m(rq ) ) where C is the cone with apex o¯ such that cutting C along the meridian from o¯ gives a sector in R2 of angle 2πa with the portion inside the radius m(rq ) removed. Since γq is a ray, Lemma 6.0.2 implies the existence of a neighborhood Uq of q such that each point in Uq \ [o, q] lies in a horoball for a ray from q . We now check that o lies in the horoball of γq . Concavity of m implies that the graph of m lies below its tangent line at z , so evaluating the tan- gent line at r = 0 and using m(0) = 0 gives m(z) > z . The Pythagorean theorem in the sector in R2 of angle 2πa implies that Mm a a d (γ (s), o) = rs2 + (x — z + m(z))2 + z — m(z) which is < s for large s, implying that o is in the horoball of γq . In the second phase of our proof, we show that every point of Mm is a soul. To realize q as a soul, we need to look at the soul construction with arbitrary basepoint v, which starts by considering the complement in Mm of the union of horoballs for all rays from v, which by the above is either v or a segment [u, v] contained in (o, v], where u is uniquely determined by v. It will be convenient to allow for degenerate segments for which u = v; with this convention, the soul is the midpoint of [u, v]. Since z is the smallest such that Gm|[z,∞) = 0, the focal point argument of Case 2 shows that u = v when 0 < rv < z . Set y := rv , and let e(y) := ru ; note that 0 < e(y) ≤ y, and the midpoint of [u, v] has r-coordinate h(y) := y+e(y) . To realize each point of Mm as a soul, it suffices to show that each ...