18.409 A Algorithist s Toolkit October, 009 Lecture 1 Lecturer: Joatha Keler Scribes: Alex Levi (009) 1 Outlie Today we ll go over soe of the details fro last class ad ake precise ay details that were skipped. We ll the go o to prove Fritz Joh s theore. Fially, we will start discussig the Bru-Mikowski iequality. Separatig Hyperplaes Give a covex body K R ad a poit p, a separatig hyperplae for K ad p is a hyperplae that has K o oe side of it ad p o the other. More forally, for a vector ν, the hyperplae H = {x ν x = 1} is a separatig hyperplae for K ad p if 1. ν x 1 for all x K, ad. ν p 1. Note that if we replace the right had side of both the above coditios by 0 or ay other costat, we get a equivalet forulatio. We call a separatig hyperplae H a strogly separatig hyperplae if the secod iequality is strict. Last tie, we sketched a proof of the followig theore: Theore 1 (Separatig Hyperplae Theore) If K is a covex body ad p is a poit ot cotaied i K, the there exists a hyperplae that strogly separates the. We ll use the above result to show why the polar of the polar of a covex body is the body itself. Recall that for a covex body K, we had defied its polar K to be {p k p 1 k K}. Theore Let K be a covex body. The K = K. Proof We kow that K = {p k p 1 k K}. Siilarly K = {y p y 1 p k }. Let y be ay poit i K. The, by the defiitio of the polar, for all p K we have that p y 1. The defiitio of the polar of K iplies that y K. Sice this is true for every y K, we coclude that K K. The other directio of the proof is the otrivial oe ad we ll have to use the covexity of the body ad the separatig hyperplae theore. Suppose that we ca fid a y K such that y / K. Sice y K, we have that p y 1 for all p K. Sice y K, there exists a strogly separatig hyperplae for y ad K. Let it be H = {x v x = 1}. By the defiitio of separatig hyperplae, we have v k 1 for all k K. Hece, v K. Also, v y > 1 (sice H is a separatig hyperplae), ad we just showed that v K. This cotradicts our assuptio that y K. Hece K K. 3 Baach Mazur Distace Recall fro last tie the defiitio of the Baach Mazur distace betwee two covex bodies: Defiitio 3 Let K ad L be two covex bodies. The Baach Mazur distace d(k, L) is the least positive d R for which there is a liear iage L of L such that L K dl, where dl is the covex body obtaied by ultiplyig every vector i L by the scalar d. 1-1
dl ~ L ~ K Iage by MIT OpeCourseWare. Figure 1: Defiig the distace betwee K ad L. Observe that the above defiitio takes ito cosideratio oly the itrisic shape of the body, ad it is idepedet of ay particular choice of coordiate syste. Also observe that the Baach Mazur distace is syetric i its iput arguets. If L K dl, the by scalig everythig by d, we get that dl dk. Hece K dl dk, which iplies the syetry property. 4 Fritz Joh s Theore Let B deote the -diesioal uit ball. For ay two covex bodies K ad K, let d(k, K ) deote the Baach Mazur distace betwee the. I the rest of this lecture, we will state ad prove Fritz Joh s theore. Theore 4 For ay -diesioal, origi-syetric covex body K, we have d(k, B ). I other words, the theore states that for every origi-syetric covex body K, there exists soe ellipsoid E such that E K E. We will prove that the ellipsoid of axial volue that is cotaied i K will satisfy the above cotaiet. Iforally, the theore says that up to a factor of, every covex body looks like a ball. The above boud of is tight for the cube. If we did t require the coditio that K is origi syetric, the the boud would be, which would be tight for a siplex. The theore ca also be rephrased as the followig: there exists a chage of the coordiate basis for which B K B. 4.1 A slightly stroger versio of the Fritz Joh Theore We will actually state ad prove a ore techical ad slightly stroger versio of the Fritz Joh theore that iplies our previous forulatio. Fro ow o, we assue that all the covex bodies we cosider are origi-syetric. Theore 5 Let K be a origi-syetric covex body. The K cotais a uique ellipsoid of axial volue. Moreover, this largest ellipsoid is B if ad oly if the followig coditios hold: B K There are uit vectors u 1,u,...,u such that o the boudary of K ad positive real ubers c 1,c,...,c 1-
1. c iu i =0, ad. For all vectors x, we have c i x, u i = x. It is ot hard to show that this coditio is T equivalet to the requireet that c iu i u i =Id, where Id is the idetity atrix. Sice the u i are uit vectors, they are poits o the covex body K that also belog to the sphere B Also, the first idetity, i.e. c iu i =0, is actually redudat, sice for origi-syetric bodies it ca be derived fro the secod idetity. This is because for every u i, its reflectio i the origi (aely u i )is also cotaied i K B ; further we ca take the costats i the secod idetity correspodig to u i ad u i to be the sae, ad this establishes the first equatio. The secod idetity says that the cotact poits of the sphere with K act soewhat like a orthooral basis. They ca be weighted so that they are copletely isotropic. I other words, the poits are ot cocetrated ear soe proper subspace, but are pretty evely spread out i all directios. Together they ea that the u i ca be weighted so that their ceter of ass is the origi ad their iertia tesor is the idetity. Also, a siple rak arguet shows that there eed to be at least such cotact poits, sice the secod idetity ca oly hold for x i the spa of the u i. Note that Theore 4 easily follows fro Theore 5. Ideed, assue without loss of geerality that B is the ellipsoid of axial volue cotaied i K. We ca ake this assuptio sice the particular choice of basis is ot iportat for the proof. We eed to show that B K B. Now, for all x K, we have x u i 1 for all i. Hece, x = c i (x u i ) c i. I the course of the proof below, we will see that c i =. This shows that x, ad hece K B. Thus, oce we prove Theore 5, we will have show the existece of a ellipsoid E such that E K E. 4. Proof of Joh s Theore As part of the proof Theore 5, we will prove the followig thigs: 1. If there exist cotact poits {u i } as required i the stateet of Theore 5, the B is the uique ellipsoid of axial volue that is cotaied i K.. If B is the uique ellipsoid of axial volue that is cotaied i K, the there exist poits {u i } such that they satisfy the two idetities i Theore 5. To prove the first stateet, suppose that we are give uit vectors u 1,u,...,u o the boudary of K ad positive real ubers c 1,c,...,c such that c iu i =0, ad for all vectors x, it is the case that c i x, u i = x. We wish to show that B is the uique ellipsoid of axial volue that is cotaied i K. Observe that it suffices to show that aog all axis-aliged ellipsoids cotaied i K, B is the uique ellipsoid of axial volue. This is because what we are tryig to prove does t etio ay basis ad is oly i ters of dot products. Hece, sice the stateet will reai true uder rotatios, provig it for axis-aliged ellipsoids is eough. For each u i it is the case that u i k 1 for all k K, Hece u i K. Let E be ay axis-aliged ellipsoid { such that E K. } The K E. Hece {u 1,u,...,u } E. Sice E is axis-aliged, it is of the for i x x 1. α i We also have that Vol(E)/ Vol(B )= α i. Therefore, to show that Vol(E) < Vol(B ), we ust show that α i < 1 for ay such E that is ot B. ( ) Observe that E = {y α T i y i 1}. Now, sice u i u i =1, we have Tr c iu i u i = c i. Sice Tr(Id )=, this iplies that c i =. Let e j deote the vector which has a1ithe j th coordiate ad 0 i the other coordiates. Clearly u i,e j is the j th coordiate of u i. For 1 i, sice u i E, we get that j=1 α j u i,e j 1. Suig it over all i, weget u i,e j c i =. (1) c i α j j=1. 1-3
Now, switchig the order of suatio o the left-had side of (1) gives us αj c i u i,e j, j=1 ad by the above we kow that this is at ost. Further, by coditio of Theore 5, we kow that c i u i,e j = e j = 1. Therefore, we get j=1 j α. By the AM-GM iequality, we get that ( P ) 1/ i =1 α i α i 1, which iplies that α i 1. Equality oly holds if all the α i are equal. This shows that α i < 1 for ay such E that is ot B, copletig the first part of the proof. For the secod part, assue that we are give that B is the uique ellipsoid of axial volue that is cotaied i K. We wat to show that for soe, there exist c i ad u i for 1 i (as i the stateet of Theore 5), such that for all vectors x, c i x, u i = x. Agai, this is equivalet to showig that T c i u i u i =Id. We already observed that for origi-syetric bodies, the coditio that c iu i = 0 is iplied by the previous requireet. Let U i = u i u T i. Also, observe that we ca view the space of atrices as a vector space of diesio. Hece we ca paraetrize the space of atrices by R. Thus, c iu i u T i =Id for c i > 0 eas that Id / is i the covex hull of the U i (if the idetity holds, we kow that the c i are positive ad su to ). If we caot fid c i,u i such that c iu i u T i =Id, it eas that Id / is ot i the covex hull of the U i. Hece, there ust be a hyperplae i the space of atrices that separates Id / fro the covex hull of the U i. For two atrices A ad B, let A B deote their dot product i R, i.e. A B = i,j A ij B ij. Thus, the separatig hyperplae gives a atrix H such that A H 1 for all A cov(u i ) ad (Id /) H < 1. Let t = Tr(H) = H Id. Let H = H (t/)(id ). The (Id /) H = (Id /) (H (t/)id )= t/ ((Id /) (t/)id ) = 0. Siilarly, sice Tr(A) = 1 for all A i cov(u i ), we get that A H > 0. Hece, H is such that: 1. Tr(H )=0, ad. H T (u i u i ) > 0 for all i. { } Now, let E δ = x R x T (Id +δh )x 1. For all i, we have u T (Id +δh T i )u i =1+ δu i H u i, which is greater tha 1 sice u T i H u i > 0 = H (u i u T i ) > 0. Hece u i E δ. Also, sice H (u i u T i ) > 0 for all i, by cotiuity, there exists ɛ > 0 such that for all vectors w i the ɛ-eighborhood of the set of all u i satisfy H (ww T ) > 0. Hece, by the previous arguet, ay such w is ot cotaied i E δ. Note that whe δ = 0, we get the uit ball B. For every δ > 0 we have that all w i the ɛ-eighborhood of the cotact poits of B are ot cotaied i E δ. Hece, as we icrease δ cotiuously startig fro 0, the cotiuity of the trasforatio of E δ iplies that for sufficietly sall δ, boudary(k) E δ =. Hece ɛ > 0suchthat(1 + ɛ )E δ K. Therefore, to coclude the proof, it suffices to show that Vol(E δ ) Vol(B ), which gives give us a cotradictio (as (1 + ɛ )E δ is a ellipse of volue larger tha B cotaied i K). Let λ 1,λ,...,λ be the eigevalues of Id +δh. Sice Vol(E δ )=( λ i) 1, to show that Vol(E δ ) Vol(B ), we eed to show that λ i 1. However we kow that λ i = Tr(Id +δh ) = Tr(Id )=. By the AM-GM iequality, ( λ i) 1/ ( λ i)/ = 1. Hece λ i 1. This cocludes the proof of part. 1-4
5 Sketch of a Sipler Proof If we just wish to prove the existece of a ellipse E that satisfies the coditios of Fritz Joh s Theore without actually characterizig it, the the picture below suggests a alterative ad possibly sipler proof of the result. If ay poit of K is ore tha distace away fro the origi, the we ca fid a ellipse of larger volue tha B that is cotaied i K. 1 S L M C E P Iage by MIT OpeCourseWare. Figure : A sipler proof of the Roudig result. 6 The Bru-Mikowski Iequality Defiitio 6 For A, B R, the Mikowski su A B is give by A B = {a + b a A, b B}. The Mikowski su ca be defied for ay subsets of R, but it is icely behaved if A ad B are covex. Ituitively, the Mikowski su is obtaied by ovig oe of the sets aroud the boudary of the other oe. The Bru-Mikowski iequality, which relates the volue of A B to the volues of A ad B, iplies ay iportat theores i covex geoetry. The goal is to boud Vol(A B) i ters of Vol(A) ad Vol(B). The followig are soe loose bouds that ca be siply verified. Fact 7 Vol(A B) ax{vol(a), Vol(B)} Proof Let a A. We have {a} B A B, by defiitio. Hece, Vol(A B) Vol({a} B) =Vol(B). Siilarly, Vol(A B) Vol(B). Fact 8 Vol(A B) Vol(A)+Vol(B) Proof By ovig oe of the sets aroud the other oe (suig the extree poits), we ca get disjoit copies of A ad B i A B. The boud give by Fact 8 is loose. To see that, cosider the case that A = B. I this case, A A = A ad hece, Vol(A A) = Vol(A). So the volue of A A grows expoetially with, while the lower boud give i the above fact do ot. This suggests takig the -th roots ad still get a valid boud. Let us first prove it for boxes. 1-5
Lea 9 Let A ad B be boxes i R. The Vol(A B) 1/ Vol(A) 1/ + Vol(B) 1/. Proof Let A have sides of legth a 1,...,a ad B have sides of legth b 1,...,b. It directly follows fro the defiitio of Mikowski sus that A B has sides of legth a 1 + b 1,...,a + b. We just eed to show the followig: We ca rewrite the left-had side of () as Vol(A) 1/ + Vol(B) 1/ 1. () Vol(A B) 1/ b i ) 1/ ( a i) 1/ +( = ( a i) 1/ + ( b i ) 1/ ( (a i + b i )) 1/ ( (a i + b i )) 1/ ( (a i + b i )) 1/ ( ) 1/ ( ) 1/ a i b i = + a i + b i a i + b i where the iequality is just a applicatio of AM-GM. 1 a i 1 b i + =1 a i + b i a i + b i Next tie, we will prove the Bru-Mikowski iequality for ore geeral bodies, ad study soe of its applicatios. 1-6
MIT OpeCourseWare http://ocw.it.edu 18.409 Topics i Theoretical Coputer Sciece: A Algorithist's Toolkit Fall 009 For iforatio about citig these aterials or our Ters of Use, visit: http://ocw.it.edu/ters.