Lecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)


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1 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the geeral versio of the BruMikowski iequality ad the move o to applicatios, icludig the Isoperimetric iequality ad Grubaum s theorem. 2 The BruMikowski iequality Theorem 1 Let A, B R be compact measurable sets. The (Vol(A B)) 1/ (Vol(A)) 1/ + (Vol(B)) 1/. (1) The equality holds whe A is a traslatio of a dilatio of B (up to zeromeasure sets). Proof A equivalet versio of BruMikowski iquality is give by 1/ Vol(λA (1 λ)b) λ(vol(a)) 1/ +(1 λ)(vol(b)) 1/, λ [0, 1]. (2) The equivalece of (1) ad (2) follows from the fact that Vol(λA) = λ Vol(A): 1/ Vol(λA (1 λ)b) Vol(λA) 1/ + Vol((1 λ)b) 1/ = λ Vol(A) 1/ + (1 λ) Vol(B) 1/ = λ Vol(A) 1/ +(1 λ) Vol(B) 1/. (3) The iequality (2) implies that the th root of the volume fuctio is cocave with respect to the Mikowski sum. Here, we sketch the proof for Theorem 1 by provig (1) for ay set costructed from a fiite collectio of boxes. The proof ca be geeralized to ay measurable set by approximatig the set with a sequece of fiite collectios of boxes ad takig the limit. We omit the aalysis details here. Let A ad B be fiite collectios of boxes i R. We prove (1) by iductio o the umber of boxes i A B. Defie the followig subsets of R : A + = A {x R x 0}, A = A {x R x 0}, B + = B {x R x 0}, B = B {x R x 0}. (4) Traslate A ad B such that the followig coditios hold: 1. A has some pair of boxes separated by the hyperplae {x R x 1 =0}. i.e. there exists a box that lies completely i the halfspace {x R x 1 0} ad there is some other box that lies i its complemet halfspace (see figure 1). (If there s o such box i that directio we ca chage coordiates.) 2. It holds that Vol(A + ) Vol(B + ) =. (5) Vol(A) Vol(B) 131
2 Note that traslatio of A or B just traslates A B, so ay statemet about the traslated sets holds for the origial oes. Sice A + ad A are strict subsets of A, we kow that A + B + ad A B have fewer boxes tha A B. Therefore, (1) is true for them by the iductio hypothesis. Moreover, A + B + ad A B are disjoit because they differ i sig of the x 1 coordiate. Hece, we have Vol(A B) Vol(A + B + )+Vol(A B ) Vol(A + ) 1/ + Vol(B + ) 1/ + Vol(A ) 1/ + Vol(B ) 1/ 1/ 1/ = Vol(A + Vol(B + ) ) 1+ + Vol(A Vol(B ) ) 1+ Vol(A + ) Vol(A ) 1/ Vol(B) = Vol(A + )+Vol(A ) 1+ Vol(A) = Vol(A) 1/ + Vol(B) 1/, (6) where the secod iequality follows from the iductio hypothesis, ad the secod equality is implied by (5). Figure 1: A + ad B + as defied i the proof of Theorem 1. 3 Applicatios of BruMikowski Iequality I this sectio, we demostrate the power of BruMikowski iequality by usig it to prove some importat theorems i covex geometry. 3.1 Volumes of Parallel Slices Let K R be a covex body. A parallel slice, deoted by K t, is defied as a itersectio of the body with a hyperplae, i.e. K t = K {x R x 1 = t}. (7) 132
3 Defie the volume of the parallel slice K t, deoted by v K (t), to be its ( 1)dimesioal volume. v K (t) =Vol (K t ). (8) We are iterested i the behavior of the fuctio v K (t), ad i particular, i whether it is cocave. Cosider the Euclidea ball i R. The followig plots of v K (t) for differet suggest that except for = 2, the fuctio v K (t) is ot cocave i t. As aother example, cosider a circular coe i R 3. The volume of a parallel slice is proportioal to t 2,so v K (t) is ot cocave. More geerally, v K (t) is proportioal to t for a circular coe i R. This suggests that the ( 1) th root of v K is a cocave fuctio. This guess is verified by Bru s theorem. Theorem 2 (Bru s Theorem) Let K be a covex body, ad let v K (t) be defied as i (8). The the 1 fuctio v K (t) is cocave. Proof Let s, r, t R with s =(1 λ)r + λt for some λ [0, 1]. Defie the ( 1)dimesioal slices K r,k s,k t as i (7). First, we claim that (1 λ)a r λa t A s. (9) We show this by provig that for ay x A r,y A t, we have z = (1 λ)x λy A s, as follows. Coect the poits (r, x) ad (t, y) with a straight lie (see figure 2). By covexity of K, the lie lies completely i the body. I particular, the poit (s, z), which is a covex combiatio of (r, x) ad (t, y), lies i A s. Therefore, z A s ad the claim i (9) is true. Now, by applyig the versio of BruMikowski iequality i (2), we have Vol(A s ) 1 (1 λ)vol(a r ) 1 + λvol(a t ) 1 v K (s) 1 (1 λ)v K (r) 1 + λv K (t) 1 (10) 3.2 Isoperimetric Iequality Figure by MIT OpeCourseWare. Figure 2: dimesioal covex body K i Theorem 2. A few lectures ago, we asked the questio of fidig the body of a give volume with the smallest surface area. The aswer, amely the Euclidea ball, is a direct cosequece of the Isoperimetric iequality. Before statig the theorem, let us defie the surface area of a body usig the Mikowski sum. Defiitio 3 Let K be a body. The surface area of K is defied as the differetial rate of volume icrease as we add a small Euclidea ball to the body, i.e., Vol(K ɛb 2 ) Vol(K) S(K) = Vol( K) = lim. (11) e 0 ɛ 133
4 Now we state the theorem: Theorem 4 (Isoperimetric iequality) For ay covex body K, with dimesioal volume ad surface area S(K), 1 1/ S(K) (12) V (B 2 S(B 2 ) Proof By applyig the BruMikowski iequality, we have the followig: V (K ɛb 2 ) 1/ + ɛv (B 2 ) 1/ 1/ V (B 2 ) = 1+ ɛ V (B 2 ) 1+ ɛ (13) where the secod iequality is obtaied by keepig the first two terms of the Taylor expasio of (1 + x). Now, the defiitio of surface area i (11) implies: S(K) = V ( K) V ( + ɛ V ( ɛ 1/ V (B = 2 ) B K 2 ) For a dimesioal uit ball, we have S(B 2 )= V (B 2 ). Therefore, ) 1/ = V (B 2 ) 1/. (14) ) 1/ S(K) V (B 2 S(B 2 ) 1 1 S(K) V (B 2 ) 1/ S(B 2 ) V (B 2 ) 1/ = (15) V (B 2 ) 3.3 Grubaum s Theorem Give a highdimesioal covex body, we would like to pick a poit x such that for ay cut of the body by a hyperplae, the piece cotaiig x is big. A reasoable choice for x is the cetroid, i.e. 1 x = ydy. Vol(K) y K This choice guaratees to get at least half of the volume for ay origi symmetric body, such as a cube or a ball. The questio is how much we are guarateed to get for a geeral covex body, ad i particular, what body gives the worst case. Do we get a costat fractio of the body, or does the guaratee deped o dimesio? 134
5 Let us first cosider the simple example of a circular dimesioal coe (figure 3). Suppose we cut the coe C by the hyperplae {x 1 = x 1 } at its cetroid, where 1 h tr x 1 = t Vol dt = h. (16) Vol(C) t=0 h +1 Grubaum s theorem states that the circular coe is ideed the worst case if we choose the cetroid. First we ll eed the followig lemma: Figure 3: dimesioal circular coe. Lemma 5 Let L = C {x 1 x 1} by the left side of the coe (which is x 1 aliged with vertex at the origi). The 1 2 V (L) V (C) 1 e. Proof V (L) V ( = +1 C) = V (C) V (C) e Theorem 6 (Grubaum s Theorem) Let K be a covex body, ad divide it ito K 1 ad K 2 usig a hyperplae. If K 1 cotais the cetroid of K, the Vol(K 1 ) Vol(K) 1 e. (17) I particular, the hyperplae through the cetroid divides the volume ito almost equal pieces, ad the worst case ratio is approximately 0.37:0.63. Proof WLOG, chage coordiates with a affie trasformatio so that the cetroid is the origi ad the hyperplae H used to cut is x 1 = 0. The perform the followig operatios: 1. Replace every ( 1)dimesioal slice K t with a ( 1)dimesioal ball with the same volume to get K, which is covex per Lemma 7 below. 2. Tur K ito a coe, such that the ratio gets smaller per Lemma 8 below. Lemma 7 K is covex. Proof Let K t = K {x 1 = t} be a parallel slice i the modified body. The radius of K t is proportioal 1 1 to V (K t ). By applyig BruMikowski iequality, we get that V (K t ) is a cocave fuctio i t. Thus K is covex. 135
6 Lemma 8 We ca tur K ito a coe while decreasig the ratio. Proof Let K + = K {x 1 0},K = K {x 1 0}. Make a coe yq 0 by pickig y havig x 1 coordiate positive o the x 1 axis, ad V (yq 0)= V (K +. Exted the code i the {x 1 0} regio, so that the volume of the exteded part equals V (K ); ame this code C. Now by Lemma 5, the cetroid of C must lie i yq 0.Let H be the traslatio of H alog the x 1 axis so that it cotais the cetroid of C. The r(k, H) = r(c,h) r(c,h ) 1/e. This completes the proof of Grubaum s theorem. 4 Next Time Next time, we will discuss approximatig the volume of a covex body. 136
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