Cover Page Title: The Problem of Two Sticks 1 Authors: 0 1 Luis A. Caffarelli 0 2 Department of Mathematics, University of Texas at Austin, n Austin, TX, 78712 a J caff[email protected] 8 512-471-3160 2 ] Michael G. Crandall G Department of Mathematics, University of California, D Santa Barbara, Santa Barbara, CA 93106 . h t [email protected] a m 805-964-3256 [ 1 v 6 8 1 5 . 1 0 0 1 : v i X r a 1 MSC 2000 Subject Classification, Primary 46B20,52A21. THE PROBLEM OF TWO STICKS LUIS A. CAFFARELLI MICHAEL G. CRANDALL Abstract. Let l =[l0,l1] be the directed line segment from l0 ∈IRn to l1 ∈IRn. Suppose ¯l=[¯l0,¯l1]isasecondsegmentofequallengthsuchthatl,¯lsatisfythe“twostickscondition”: l1−¯l0 ≥ kl1−l0k, ¯l1−l0 ≥ ¯l1−¯l0 . Here k·k is a norm on IRn. We explore the mannerinwhichl1−¯l1 isthenconstrainedwhenassumptionsaremadeabout“intermediate (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) p(cid:13)oints”(cid:13)l∗ ∈ l,¯l∗ ∈ ¯l. R(cid:13)oughly(cid:13)spea(cid:13)king, o(cid:13)ur most subtle result constructs parallel planes separated by a distance comparable to l∗−¯l∗ such that l1 −¯l1 must lie between these planes, provided that k·k is “geometrically convex” and “balanced”, as defined herein. The (cid:13) (cid:13) standardp-normsareshowntobegeomet(cid:13)ricallyc(cid:13)onvexandbalanced. Otherresultsestimate l1−¯l1 in a Lipschitz or Ho¨lder manner by l∗−¯l∗ . All these results have implications in the theory of eikonal equations, from which this “problem of two sticks” arose. (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Introduction The origin of the “problem of two sticks,” which we are about to describe, lies in the theory of eikonal equations. Roughly speaking, the results of Caffarelli and Crandall [3] rely on knowledge of how the endpoints of “rays” of the distance function to some set, as measuredinanormk·k,thatemanatefrompointsinthesetandpassthroughacommontiny ball in the interior of the region of differentiability of the distance function are constrained. We provide a variety of results that speak to this issue. In particular, the crown jewel of our results, Corollary 5.3 below, implies that the endpoints must lie between parallel planes which are separated by a distance comparable to the radius of the ball. The ingredients of the problem of two sticks are a norm k·k on IRn and two “sticks” ¯ ¯ ¯ l = [l ,l ], l = [l ,l ], 0 1 0 1 where [l ,l ] denotes the directed line segment from l to l ∈ IRn. Sometimes we regard l 0 1 0 1 as a set, as when we write x ∈ l, or x ∈ [l ,l ], but [x,y] has an “initial” point x and a 0 1 “terminal” point y. We assume throughout this paper that the sticks satisfy the “two sticks condition” ¯ ¯ ¯ ¯ (1.1) l −l ≥ kl −l k and l −l ≥ l −l . 1 0 1 0 1 0 1 0 To emphasize our (cid:13)remarks(cid:13)about the ordering o(cid:13)f the en(cid:13)dpoi(cid:13)nts of t(cid:13)he sticks, observe that ¯ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ¯ (cid:13) ¯ if l = l , then (1.1) is satisfied for any choice whatsoever of l ,l . However, if l = l = 0, 0 0 1 1 1 1 ¯ then (1.1) is satisfied iff kl k = l . In particular, in general, interchanging the initial and 0 0 ¯ terminal points of sticks l,l which satisfy (1.1) can lead to sticks which do not satisfy (1.1). (cid:13) (cid:13) (cid:13) (cid:13) Key words and phrases. Minkowski geometry, finite dimensional Banach spaces. 1 2 LUIS A. CAFFARELLI MICHAEL G. CRANDALL For further remarks about the nature of the two sticks condition, see Section 2, where we explain its relationship to nearest point mappings and distance functions. Usually we will assume the sticks are of equal length L : ¯ ¯ (1.2) kl −l k = l −l = L. 1 0 1 0 ¯ Assume that a point of l is “close” (to be(cid:13)quant(cid:13)ified) to a point of l, each point being (cid:13) (cid:13) somewhere away from the endpoints of the stick in which it lies. The two sticks problem is ¯ then to obtain information about l −l . In what manner is it constrained? 1 1 For example, suppose that (1.1) holds, (1.2) holds with L = 1 (a normalization), the sticks ¯ intersect at a point z ∈ l ∩l, and z is not an endpoint of either stick. Then one has ¯ ¯ ¯ ¯ (1.3) 1 ≤ l −l ≤ kl −zk+ z −l , 1 ≤ l −l ≤ l −z +kz −l k, 1 0 1 0 1 0 1 0 which, when add(cid:13)ed, giv(cid:13)e (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ¯ ¯ ¯ ¯ 2 ≤ l −l + l −l ≤ kl −zk+kz −l k+ l −z + z −l 1 0 1 0 1 0 1 0 ¯ ¯ (cid:13) (cid:13) (cid:13) (cid:13) = kl1 −l0k+(cid:13) l1 −l(cid:13)0 =(cid:13)2. (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) It follows that each inequality in (1.3) must be an equality. (cid:13)If the n(cid:13)orm is strictly convex (cid:13) (cid:13) (see Section 4), this entails the existence of positive constants α,β such that ¯ ¯ l −z = α(z −l ), l −z = β(z −l ). 1 0 1 0 To continue, since z is an intermediate point of both sticks, each of which has length 1, ¯ ¯ the above implies l −l = l −l . However, a moment’s thought reveals that if the directions 1 0 1 0 of unit length sticks are the same and they have a common intermediate point, they cannot ¯ ¯ satisfy the two sticks condition without being identical, that is l = l , l = l . We go a bit 1 1 0 0 further with this, now allowing, for example, z = l . A picture quickly reveals that the two 1 ¯ ¯ sticks condition then fails unless l = l . We can no longer assert that l = l , but surely 1 1 0 0 ¯ ¯ ¯ l = l still holds. Similarly, if z = l , then l = l , but we can no longer assert that l = l . 1 1 0 0 0 1 1 ¯ In all, l = l holds if the sticks have a common point, so long as that common point is not 1 1 l = ¯l . It follows that given a collection {li,i ∈ I} of sticks of unit length, indexed here by 0 0 some index set I, which pairwise satisfy the two sticks condition, then the mapping from the set of all intermediate points from the family to terminal points of sticks in which they lie is well defined. It is properties of this mapping which are called on in [3]. Using the simple result already noted, straightforward compactness arguments show that if the norm is strictly convex, 0 < ε, and 0 < t ≤ 1, then there is a δ = δ (ε,t) > 0 such 0 0 that ¯ ¯ ¯ ¯ (1.4) l ∈ l,l ∈ l, t ≤ kl −l k, l −l , ∗ ∗ ∗ 0 ∗ 0 and (cid:13) (cid:13) (cid:13) (cid:13) ¯ (1.5) l −l ≤ δ ∗ ∗ 0 ¯ imply l −l ≤ ε. That is, the mapp(cid:13)ing refe(cid:13)rred to in the preceding paragraph is contin- 1 1 (cid:13) (cid:13) uous. (cid:13) (cid:13) (cid:13) (cid:13) THE PROBLEM OF TWO STICKS 3 These remarks are not strong enough for our intended applications to eikonal equations, owing to the general behavior of δ as a function of ε. Thus we prove a hierarchy of variants 0 under additional conditions. Indeed, in the case of the Euclidean norm on IRn, when the two sticks and equal length conditions are satisfied as well as (1.4), the mapping associated ¯ ¯ with (1.4) here, that is l 7→ l ,l 7→ l , is Lipschitz continuous; in fact, Corollary 3.3 below ∗ 1 ∗ 1 implies that then 2 ¯ ¯ l −l ≤ l −l . 1 1 ∗ ∗ t (cid:13) (cid:13) (cid:13) (cid:13) ThisLipschitzcontinuityalsoholdsfornormswhichare“2uniformlysmoothand2uniformly (cid:13) (cid:13) (cid:13) (cid:13) convex” (seeSection 3.2forthedefinition). This isaspecial caseofthemainresult ofSection 3.2, which states that if the norm is p-uniformly convex and q-uniformly smooth, then the mapping is H¨older continuous with exponent q/p. The results of Section 3.2 apply to the p-norms on IRn, that is k·k = k·k , where p n 1/p (1.6) kxk := |x |p , p i ! i=1 X in the range 1 < p < ∞. Indeed, k · k is 2-uniformly smooth and p-uniformly convex for p 2 ≤ p < ∞, and it is 2-uniformly convex and p-uniformly smooth for 1 < p ≤ 2. In Section 7 we provide examples to show that the H¨older continuity established for the k·k cases is p asymptotically an optimal modulus of continuity, up to constants. However, the H¨older continuity obtained in the k·k case is not always sufficient for the p purposes of [3], even if the modulus is optimal. This deficiency led us to the notion of norms which are “geometrically convex,” as introduced in Section 4.1. For geometrically convex norms, which are also “balanced”, it is shown in Section 5 that, roughly speaking, if δ is 0 ¯ sufficiently small, then (1.4), (1.5) imply that l −l is confined between two parallel planes 1 1 which are separated by a distance which is an estimable multiple of δ . This is, of course, 0 not a “modulus of continuity” result; it is more subtle. It is another task to verify that the p-norms are geometrically convex and balanced, and this we do in Section 6. We begin with the Euclidean case, after some remarks about the two sticks problem and distance functions. In this regard, it is clear that the problem of two sticks is related to properties of nearest point mappings onto convex sets, and we recognized that the results of Section 3.2 were likely to hold via papers concerning this issue. These include, for example, B. Bj¨ornestal [2], Y. Alber [1] and C. Li, X. Wang and W. Yang [4]. However, our Section 3.2 is short and self-contained; correspondingly, our constants are not sharp. In contrast, the results and notions of Sections 4.1, 4.2, 5 and 6 are not suggested by other literature of which we are aware. As this entire paper could be made essentially self-contained, we have done so. Thus in the first part of Section 4 we have presented some well-known elementary material with perhaps a different spirit than is usual; in particular, we do not use dual spaces or dual norms explicitly anywhere in this work. 4 LUIS A. CAFFARELLI MICHAEL G. CRANDALL Contents 2. Two Sticks and the Distance Function 4 3. Cases with Lipschitz or H¨older Continuity 5 3.1. Two sticks in the Euclidean case 5 3.2. Two Sticks in the p-Uniformly Convex, q-Uniformly Smooth Case 7 4. More General Norms: Preliminaries 10 4.1. Geometric Convexity 12 4.2. Some Consequences of Geometric Convexity 13 5. The Problem of Two Sticks and Geometric Convexity 16 6. Verifying Geometric Convexity, etc. 22 6.1. Notions “in the Tangent Plane” 22 6.2. From Tangent Plane Estimates to Full Estimates 24 6.3. The Norm k·k in the Tangent Plane 28 p 7. Examples in the Case of the p-norm 30 References 32 2. Two Sticks and the Distance Function Suppose that C ⊂ IRn and l ,¯l ∈ IRn. Suppose that l ,¯l ∈ C and 1 1 0 0 ¯ ¯ ¯ (2.1) kl −l k ≤ kl −xk, l −l ≤ l −x for x ∈ C. 1 0 1 1 0 1 ¯ Then l is a point of C which is as close(cid:13)to l as(cid:13)any(cid:13)other p(cid:13)oint of C, etc. Choosing x = l in 0 1 0 (cid:13) (cid:13) (cid:13) (cid:13) ¯ ¯ ¯ the first inequality of (2.1) and x = l in the second, we see that l = [l ,l ], l = [l ,l ] satisfy 0 0 1 0 1 ¯ ¯ the two sticks condition. Conversely, if l,l satisfy the two sticks condition and C = {l ,l }, 0 0 we have (2.1). Moreover, we have, in both cases, ¯ ¯ ¯ (2.2) kl −l k = dist(l ,C), l −l = dist(l ,C), 1 0 1 1 0 1 where dist(x,C) is the distance, as measured(cid:13) by k·(cid:13)k, from x to C. If we add the equal (cid:13) (cid:13) length condition, we are assuming these distances are equal. Thus the study of the two sticks problem is a kind of atomization of the study of “rays” of distance functions, wherein lies its connection to Hamilon-Jacobi equations. Continuing in this line, the notation ¯ ¯ ¯ (2.3) l := (1−t)l +tl , l := (1−t)l +tl , t 0 1 t 0 1 is used in the next remarks. Note that we use A := B to indicate that A is defined to be B. ¯ ¯ ¯ First, if l,l satisfy the two sticks condition, then so do [l ,l ],[l ,l ] for 0 ≤ t ≤ 1. To see 0 t 0 1 this, merely note that ¯ kl −l k = kl −l k+kl −l k ≤ l −l 1 0 1 t t 0 1 0 implies (cid:13) (cid:13) (cid:13) (cid:13) ¯ ¯ (2.4) kl −l k ≤ l −l −kl −l k ≤ l −l . t 0 1 0 1 t t 0 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) THE PROBLEM OF TWO STICKS 5 ¯ ¯ Iterating this remark, if 0 ≤ s,t ≤ 1, then [l ,l ],[l ,l ] also satisfy the two sticks condition. 0 t 0 s When we add an equal length condition, say ¯ ¯ kl −l k = l −l = L, 1 0 1 0 there is an additional symmetry. Observe the(cid:13)n that(cid:13) (cid:13) (cid:13) ¯ ¯ ¯ ¯ kl −l k ≤ l −l =⇒ l −l ≤ l −l , 1 0 1 0 1 0 1 0 ¯ ¯ ¯ ¯ l1 −l0 ≤(cid:13)l1 −l0(cid:13) =⇒ (cid:13)kl1 −l0(cid:13)k ≤ (cid:13)l1 −l0(cid:13); (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ¯ ¯ that is the sticks [l ,l(cid:13)],[l ,l ](cid:13)obt(cid:13)ained by(cid:13) switching initial a(cid:13)nd term(cid:13)inal points also satisfy 1 0(cid:13) 1 0(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) thetwo sticks condition. It follows fromthisthat the equal lengthcondition andtheprevious discussion guarantee that each line below implies the next when 0 ≤ t,s ≤ 1 : ¯ ¯ (i) [l ,l ],[l ,l ] satisfy the two sticks and equal length conditions. 0 1 0 1 ¯ ¯ (ii) [l ,l ],[l ,l ] satisfy the two sticks and equal length conditions. 1 0 1 0 ¯ ¯ (2.5) (iii) [l ,l ],[l ,l ] satisfy the two sticks and equal length conditions. 1 t 1 t ¯ ¯ (iv) [l ,l ],[l ,l ] satisfy the two sticks and equal length conditions. t 1 t 1 ¯ ¯ (v) [l ,l ],[l ,l ] satisfy the two sticks and equal length conditions. t s t s ˜ ˆ ˜ Indeed, note that if 0 ≤ t ≤ s ≤ 1, and l = [l ,l ],l = [l ,l ],l = [l ,l ], then l = 0 1 1 0 t 1 1−t ˆ l , l = l . t (s−t)/(1−t) s In particular, we note for later use that, via (2.5) (v), ¯ ¯ ¯ ¯ (2.6) l −l ≥ l −l , l −l ≥ kl −l k for 0 ≤ t,s ≤ 1. s t s t s t s t (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 3. C(cid:13)ase(cid:13)s with(cid:13)Li(cid:13)pschit(cid:13)z or Ho¨lder Continuity In this section we first treat the Euclidean case. Then we turn to the “p-uniformly con- vex, q-uniformly smooth” case. The Euclidean (or, more generally, Hilbert) case is also an example in which p = q = 2. However, as is usual, it is clean and elegant in comparison to its generalization, and deserves to be singled out. 3.1. Two sticks in the Euclidean case. We will denote the Euclidean norm by |·|; (3.1) |x| := hx,xi, where p n (3.2) hx,yi := x y j j j=1 X is the Euclidean inner-product. ¯ We begin with estimates valid for sticks l,l which satisfy the two sticks condition (1.1), but which do not necessarily have the same length. The notation (2.3) is employed. The next result is well known. 6 LUIS A. CAFFARELLI MICHAEL G. CRANDALL Proposition 3.1. Let l,¯l satisfy the two sticks condition (1.1). Then ¯ ¯ (3.3) l −l ,l −l ≥ 0. 1 1 0 0 In consequence, for 0, (cid:10) (cid:11) (3.4) (1−t)2|l −¯l |2 +t2|l −¯l |2 ≤ |l −¯l |2. 0 0 1 1 t t Proof. The relation (3.3) follows from adding the extremes in the relations |l −l |2 +|l −¯l |2 +2 l −l ,l −¯l =|(l −l )+(l −¯l )|2 = |l −¯l |2 ≥ |l −l |2, 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 |¯l −¯l |2 +|l −¯l |2 +2 ¯l −¯l ,¯l −l =|(¯l −¯l )+(¯l −l )|2 = |¯l −l |2 ≥ |¯l −¯l |2, 1 0 0 0 (cid:10) 1 0 0 0(cid:11) 1 0 0 0 1 0 1 0 and simplifying the result. (cid:10) (cid:11) To verify (3.4), we use (3.3) to deduce |l −¯l |2 = |(1−t)(l −¯l )+t(l −¯l )|2 t t 0 0 1 1 = (1−t)2|l −¯l |2 +2t(1−t) l −¯l ,l −¯l +t2|l −¯l |2 0 0 0 0 1 1 1 1 ≥ (1−t)2|l −¯l |2 +t2|l −¯l |2, 0 0 1 1(cid:10) (cid:11) (cid:3) which is (3.4). Remark 3.2. The relation (3.4) shows that the terminal point l is a Lipschitz continuous 1 function of the intermediate point l , 0 < t ≤ 1, with Lipschitz constant 1/t, in any family t ¯ of sticks which pairwise satisfy the two sticks condition. Note again that if l = l then the 0 0 two sticks condition is always satisfied, so 0 < t is necessary to have Lipschitz continuity. If we add the equal length assumption (1.2), the Lipschitz continuity may be extended to ¯ intermediate points l ,l , where s 6= t. s t Corollary 3.3. Let l,¯l satisfy (1.1) and (1.2). Then 2 ¯ ¯ (3.5) |l −l | ≤ |l −l | for 0 < t ≤ s ≤ 1. 1 1 s t t Proof. Using (3.4) 1 ¯ ¯ (3.6) |l −l | ≤ |l −l |. 1 1 t t t First we assume that 1 ¯ (3.7) |l −l | ≤ |l −l |. t s t t 2 Then, using (3.6), 1 1 t ¯ ¯ ¯ ¯ ¯ ¯ (3.8) |l −l | ≥ |l −l |−|l −l | ≥ |l −l |− |l −l | = |l −l | ≥ |l −l |, s t t t t s t t t t t t 1 1 2 2 2 so (3.5) holds. If (3.7) does not hold, then we use (2.6), (3.4), to again conclude that 1 t ¯ ¯ ¯ (3.9) |l −l | ≥ |l −l | ≥ |l −l | ≥ |l −l |. s t t s t t 1 1 2 2 (cid:3) THE PROBLEM OF TWO STICKS 7 Remark 3.4. If the equal length assumption is not satisfied, there is no Lipschitz estimate ¯ quite like (3.5). To see this, let n = 1 and take l = [0,1],l = [0,2],s = 1,t = 1/2. 3.2. Two Sticks in the p-Uniformly Convex, q-Uniformly Smooth Case. In this section, k·k is a norm for which there are constants 0 < A,B,q,p, with 1 < q ≤ p, such that (3.10) Ake−e¯kp ≤ 2−ke+e¯k for kek = ke¯k = 1, that is, k·k is “p-uniformly convex,” and B (3.11) kx+yk+kx−yk−2kxk ≤ kykq for x 6= 0, kxkq−1 that is, k·k is “q-uniformly smooth.” Remark 3.5. Note that (3.11) holds in general if it holds for kxk = 1. Moreover, if kxk = 1 and (3.11) holds for small kyk, then it holds (with a different B) for all y, as the left hand side is at most 2kyk and q ≥ 1. Remark 3.6. If 2 ≤ p < ∞,the p-normk·k is 2-uniformly smooth and p-uniformly convex, p while if 1 < p ≤ 2, it is 2-uniformly convex, and p-uniformly smooth. The first assertion was proved by Clarkson [5] and the second by Hanner [6]. Regarding the more general and precise notions of “modulus of convexity” and “modulus of smoothness” and relations between them, see Lindenstrauss [7]. Proposition 3.7. Let (3.10), (3.11) hold and R > 0. Then there is a constant C = ¯ C(A,B,R,p,q) such that if l,l, satisfy the two sticks condition, have unit length, and satisfy ¯ (3.12) l −l ≤ R, 1 1 then (cid:13) (cid:13) (cid:13) (cid:13) 1 ¯ ¯ q/p (3.13) l −l ≤ C l −l for 0 < t ≤ 1. 1 1 t t t In consequence, if 0 < t(cid:13)≤ s ≤ 1(cid:13), (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 1 (3.14) l −¯l ≤ 2q/pC l −¯l q/p. 1 1 t s t Remark 3.8. The unit length(cid:13)condit(cid:13)ion and (3.1(cid:13)2) impl(cid:13)y that (cid:13) (cid:13) (cid:13) (cid:13) ¯ ¯ ¯ ¯ l −l = (l −l )−(l −l )+l −l ≤ 2+R, 0 0 1 0 1 0 1 1 and then (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ¯(cid:13) (cid:13) ¯ ¯ (cid:13) l −l ≤ (1−t) l −l +t l −l ≤ 2+R. t t 0 0 1 1 For this reason we relabel 2+R as simply “R” and simply assume hereafter that (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ¯ (cid:13) (cid:13) (cid:13) (cid:13) (3.15) l −l ≤ R for 0 ≤ t ≤ 1. t t Further note that then (cid:13) (cid:13) (cid:13) (cid:13) (3.16) l −¯l ≤ R1−q/p l −¯l q/p, t t t t as q ≤ p. (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 8 LUIS A. CAFFARELLI MICHAEL G. CRANDALL Proof. First we establish (3.13) for t = 1/2. Put ¯ ¯ (3.17) e := l −l , e¯:= l −l . 1 0 1 0 Noting that 1 1 1 1 ¯ ¯ ¯ ¯ l = l + e, l = l − e, l = l + e¯, l = l − e¯, 1 1/2 0 1/2 1 1/2 0 1/2 2 2 2 2 the two sticks and equal length conditions are 1 ¯ ¯ 1 = kek ≤ l −l = (e+e¯)+l −l , 1 0 1/2 1/2 2 (3.18) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)1 (cid:13) (cid:13)¯ (cid:13) (cid:13) ¯ (cid:13) 1 = ke¯k ≤ l −l = (e+e¯)+l −l . 1 0 (cid:13) 1/2 1/2(cid:13) 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) The desired estimate (3.13) for t (cid:13)= 1/2 is(cid:13)of t(cid:13)he form (cid:13) (cid:13) (cid:13) 1 ¯ ¯ ¯ q/p (3.19) l −l = l −l + (e¯−e) ≤ C l −l , 1 1 1/2 1/2 1/2 1/2 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) where the meaning(cid:13)of C v(cid:13)aries(cid:13)according to need. (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Thus there are only three vectors to be concerned about, e,e¯, and ¯ (3.20) m := l −l ; 1/2 1/2 the notation is a mnemonic for “middle”. In these terms, we want 1 1 (3.21) 1 = kek ≤ m+ (e+e¯) , 1 = ke¯k ≤ −m+ (e+e¯) , 2 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) to imply (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 1 (3.22) m+ (e−e¯) ≤ Ckmkq/p. 2 (cid:13) (cid:13) (cid:13) (cid:13) If we show, instead, that (3.21)(cid:13)implies (cid:13) (cid:13) (cid:13) (3.23) ke−e¯k ≤ Ckmkq/p, with some other constant C, then 1 (3.24) m+ (e−e¯) ≤ Ckmkq/p +kmk, 2 (cid:13) (cid:13) (cid:13) (cid:13) and Remark 3.8 takes us bac(cid:13)k to the form ((cid:13)3.22). For the moment, we will obtain the bound (cid:13) (cid:13) (3.23) and leave the resulting (3.24) in “raw” form. From the two sticks condition (3.21) and the uniform smoothness assumption (3.11), we have 1 1 2q−1B 2 ≤ m+ (e+e¯) + −m+ (e+e¯) ≤ ke+e¯k+ kmkq, 2 2 ke+e¯kq−1 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) THE PROBLEM OF TWO STICKS 9 or 1 1 2q−1B 2−ke+e¯k ≤ m+ (e+e¯) + −m+ (e+e¯) −ke+e¯k ≤ kmkq. 2 2 ke+e¯kq−1 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Thisestimatedeteri(cid:13)orateswhenke(cid:13)+e¯k(cid:13)issmall. Tohan(cid:13)dlethis, wenoteagain, asinRemark (cid:13) (cid:13) (cid:13) (cid:13) 3.5, that the intermediate term above is never more that 2kmk. Thus we consider cases as follows: 2q−1Bkmkq if ke+e¯k ≥ 1, (3.25) 2−ke+e¯k ≤ 2kmk if ke+e¯k ≤ 1. ( Combining (3.10) and (3.25), we find: 2q−1Bkmkq if ke+e¯k ≥ 1, (3.26) Ake−e¯kp ≤ 2kmk if ke+e¯k ≤ 1. ( Next note that if ke+e¯k ≤ 1, then ke−e¯k = ke+e¯−2e¯k ≥ 2ke¯k−ke+e¯k ≥ 2−1 = 1. Therefore, in this case, we use q ≥ 1 to find 2 2q 1 ≤ ke−e¯kp ≤ kmk =⇒ ke−e¯kp ≤ kmkq. A Aq Therefore, choosing C(A,B,p,q) appropriately, (3.26) implies the estimate (3.27) ke−e¯k ≤ C(A,B,p,q)kmkq/p. Recalling what we were about, we have established (3.24) with C as above, or ¯ ¯ q/p ¯ (3.28) l −l ≤ C(A,B,p,q) l −l + l −l . 1 1 1/2 1/2 1/2 1/2 Next let us obse(cid:13)rve tha(cid:13)t if sticks l∗,˜l hav(cid:13)e equal len(cid:13)gths L(cid:13) ≤ 1, whi(cid:13)ch is not necessarily (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) 1, and satisfy the two sticks condition, we may apply (3.28) to l∗/L,˜l/L (with the obvious meaning) to find q/p l∗ −˜l ≤ C(A,B,p,q)L1−q/p l∗ −˜l + l∗ −˜l 1 1 1/2 1/2 1/2 1/2 (3.29) (cid:13) (cid:13) (cid:13) q/p (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ≤ C(A,B,p,q) l∗ −(cid:13)˜l +(cid:13)l∗ −(cid:13)˜l , (cid:13) (cid:13) (cid:13) 1/2 (cid:13)1/2 (cid:13) 1/2 (cid:13)1/2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) where we used q ≤ p. (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) To proceed, we next treat the case 0 < t ≤ 1/2. With this assumption, we note that 1−t 1−2t 1−t 1−2t ¯ ¯ ¯ l = l − l , l = l − l , 1 2t t 1 2t t t t t t and, from this, 1−t 1−2t ¯ ¯ ¯ l −l = (l −l )− (l −l ), 1 1 2t 2t t t t t