Upper Bounds on the Cross-Sectional Volumes of Cubes and Other Problems

Transcription

1 Upper Bounds on the Cross-Sectonal Volumes of Cubes and Other Problems Ben Pooley March 01 1

2 Contents 1 Prelmnares 1 11 Introducton 1 1 Basc Concepts and Notaton Cross-Sectonal Volumes of Cubes (Hyperplane Case) 1 An Exposton of Ball s 1986 Paper Proof of the Brünn-Mnkowsk Inequalty 9 3 An Applcaton to the Busemann-Petty Problem 11 3 The Problem for d-dmensonal Subspaces The Brascamp-Leb Inequalty 15 3 Upper Bounds n the d-dmensonal Case 1 33 A Conjecture on the Best Upper Bound n all Cases 3 4 Some Other Interestng Problems Related to Cubes 5 41 Unversalty of Cross-Sectons 5 4 The Blaschke-Santaló Inequalty and the Mahler Conjecture 7 1 Prelmnares 11 Introducton In ths essay we shall dscuss some problems n convex geometry related to (hyper)cubes n R n Throughout, we shall denote the centred unt cube n R n by Q n := [ 1, 1 ]n Of partcular nterest s the problem of fndng bounds on the cross-sectonal volumes of cubes We say crosssecton to mean the ntersecton between a body (eg Q n ) and an affne subspace of R n The smplcty wth whch ths problem can be descrbed makes t qute attractve It s also puzzlng why there are not easer solutons The key results dscussed wll be those proved by Keth Ball n 1986 and 1989, whch gve upper bounds for the volumes of ntersecton by hyperplanes (affne subspaces of codmenson 1) and for general subspaces respectvely In dong ths we also ntroduce some powerful nequaltes whch are mportant n many areas of mathematcs, namely the Brünn-Mnkowsk nequalty and the Brascamp-Leb nequalty We shall conclude ths part wth a smple conjecture on the maxmum cross-sectonal volumes In the fnal secton, we prove an nterestng result about the ubquty of cross-sectons of cubes Fnally we ntroduce a famous open problem known as the Mahler conjecture, whch clams that cubes have mnmal Mahler volume We am to make ths account readable for Warwck undergraduates n ther fnal year of the MMath course, assumng only a reasonable background n analyss and some geometrc ntuton Before begnnng the essay proper we brefly remnd the reader of some basc notons and set out some notaton for use n the remander of ths account 1

3 1 Basc Concepts and Notaton We treat R n as an nner product space wth the usual product, denoted by, ( wll denote the usual l norm) Unless otherwse stated we let e 1, e,, e n denote the standard orthonormal bass of R n We denote by S n 1 the unt sphere n R n For the orthogonal complement of a set or vector we use the notaton, for example A or v The word subspace should be understood to mean a lnear subspace (contanng the orgn) unless otherwse specfed (eg by descrbng t as affne) For A R n, A means the (dm H (A)-dmensonal) Hausdorff (Lebesgue) measure of A If a dfferent dmenson s ntended we shall wrte A k for the k-dmensonal Hausdorff measure We wll frequently use the notaton of Mnkowsk sums, whch are defned as follows Defnton 11 Let A, B R n, the Mnkowsk sum A + B s defned to be A + B = {a + b : a A, b B} When addng a set to a sngle vector, say A and v respectvely, we may wrte A + v to mean the Mnkowsk sum A + {v} Scalar multples of sets wll denote dlatons and A B := A + ( 1 B) Moreover, products of ntervals and vectors (eg [a, b]v or Rv) should be understood as lne segments, lnes or rays, as approprate Topologcal nteror, closure and boundary of a set A R n wll be denoted by nt(a), A and A respectvely Open Eucldean balls of radus r > 0, centred at v R n wll be denoted by B r (v) and for a set A R n, B r (A) := A + B r (0) means the open r-neghbourhood of A If a set s called a ball then t should be assumed to be a Eucldean ball unless another norm s ndcated Gven a hyperplane H = v + tv R n, where v S n 1 and t R, the (closed) halfspaces nduced by H are H + = {x : x, v t} and H = {x : x, v t} Now we recall some of the key defntons from convex geometry Defnton 1 A set A R n s convex f for all λ [0, 1] and x, y A we have λx + (1 λ)y A A convex body s a compact convex set wth non-empty nteror A set A s (centrally) symmetrc f x A x A The convex hull of a set A (denoted conv(a)) s the ntersecton of all convex sets contanng A If A s convex and x A, a hyperplane H R n s a supportng hyperplane at x f x H and A H + or A H A body s called star-shaped f t s a unon of lne segments that all contan the orgn A convex set s a cone f t contans 0 and s closed under addton Cross-Sectonal Volumes of Cubes (Hyperplane Case) 1 An Exposton of Ball s 1986 Paper We wll now dscuss the proof due to Ball that the volume of any (n 1)-dmensonal crosssecton of Q n s at most regardless of n Moreover f we only consder cross-sectons generated

4 by subspaces, the volume of ntersecton s always at least 1 We wll essentally be followng a paper by Ball from 1986, namely [9] To prove the upper bound ( ) we frst show that t suffces to consder only subspaces Ths requres the Brünn-Mnkowsk nequalty for convex sets whch s the followng (see [18]) Theorem 1 Let K, L R n be convex bodes and λ (0, 1), then λk + (1 λ)l 1/n λ K 1/n + (1 λ) L 1/n The Brünn-Mnkowsk nequalty wll be dscussed n more detal later Applyng Theorem 1 to the problem at hand we see that we can ndeed restrct our attenton to cross-sectons arsng from subspaces In fact the followng more general result holds (we generalse a result from [9] so that t s also relevant later) Lemma Let d Z >0 such that d n and let S be an d dmensonal subspace of R n Suppose H = S + u s a translaton of S by u S, then S Q n H Q n Proof Let H = S u, then by symmetry of Q n and S, we have H Q n = (S u) Q n = ( S u) ( Q n ) = [(S + u) Q n ] = (H Q n ) Hence H Q n = H Q n Let P be the orthogonal projecton map onto S, then as H and H are parallel to S, P preserves the volume of subsets of H and H Now applyng Theorem 1 wth λ = 1 we see that S Q n 1/d 1 P (H Q n) + 1 P ( H Q n ) 1/d 1 P (H Q n) 1/d + 1 P ( H Q n ) 1/d = H Q n 1/d, as requred The frst nequalty holds because, by convexty of Q n, 1 (H Q n+ H Q n ) S Q n Therefore, by lnearty of P and the fact that P S = d S we see that 1 (H Q n + H Q n ) = 1 P (H Q n) + 1 P ( H Q n ) The next step n provng the bounds on cross-sectonal volume wll be to ntroduce a probablty densty functon (pdf) for a specfc random varable so that we can apply some results from probablty theory Ths approach s not partcularly ntutve but lookng at the problem n ths way allows us to use some powerful tools lke the Fourer nverson formula and a handy nequalty relatng: the supremum of a pdf; the p-norm of the assocated random varable and p tself The followng lemma ntroduces the (canddate) pdf, whch we shall call f, and establshes that t s ndeed the requred densty functon Lemma 3 Let X 1, X,, X n be ndependent random varables, each unformly dstrbuted on [ 1, 1 ], furthermore let S be a subspace of Rn and u = (u 1,, u n ) S n 1 a unt normal to S Then the functon f : R R 0 gven by f(r) = (S + ru) Q n s a pdf for the random varable X = n u X (where the event space s taken to be the σ-algebra of Borel sets n R) 3

5 Remarks: Of course, the dea here s that the random vector (X 1,, X n ) nduces the probablty measure on Q n whch agrees wth the Lebesgue measure Therefore, for ε > 0, the probablty P (r ε X r+ε) s the volume of Q n ntersected wth the ε-neghbourhood of S + ru Hence, takng ε 0, t seems natural that f s the pdf of X Ths lemma summarses part of the ntroducton of [9] Note that t s n fact a lttle stronger, n partcular, the proof does not use contnuty of f so the result also holds when u s orthogonal to a face of Q n Proof (of Lemma 3) Fx r R and ε 0 then { } ( ) r+ε n n f(t)dt = x Q n : u x r ε = P u X r ε r ε All we have done here s to rewrte the volume of a certan set usng the probablty measure nduced by the X 1,, X n (see remark above) Of course, we can wrte any bounded open nterval (a, b) R as (r ε, r + ε) for some r R and ε [0, ) Hence for < a < b <, b a f(t)dt = P (X (a, b)) Moreover, snce Q n s bounded, ths s suffcent to show that the equalty stll holds f a or b s Hence by the determnaton of Borel measures 1 from ther values on ntervals we have shown that f(t)dt = P (X E) for any Borel set E R, as requred E We wll later need f to be contnuous at 0 Ths could be checked drectly but we have found the followng more general lemma whch proves an ntutve property of convex bodes We also fnd a further generalsaton whch wll be used n a later secton Lemma 4 Let C R n be a convex body and H R n an (n 1) dmensonal subspace wth unt normal v then the functon g : R R gven by g(x) = (H + xv) C s contnuous at each x such that H + xv contans an nteror pont of C Proof Suppose z (H + xv) s an nteror pont of C Indeed, suppose the open δ-ball B δ (z) s contaned n C Now defne a concal cap K (see Fgure 1) by K := {λ(z + δv) + (1 λ)h : h (H + xv) C, λ [0, 1]} Consder the famly of hyperplanes H ε = H + (x + ε)v parametrsed by ε > 0 Defne an assocated famly of concal caps K ε = {λ(z δv) + (1 λ)h : h H ε C, λ [0, 1]} Now by convexty of C, we have K C and K ε C for all ε > 0 Therefore g(x) = (H + xv) C (H + xv) K ε = ( ) δ n 1 H ε C = δ + ε Smlarly, f ε < δ, we get the followng lower bound on g(x + ε) ( ) δ ε n 1 g(x + ε) = H ε C H ε K = g(x) δ ( ) δ n 1 g(x + ε) δ + ε 1 Some readers mght fnd t useful to thnk of ths argument n terms of the Radon-Nkodym dervatves of measures We use the term concal cap to mean the convex hull of a convex set wth a pont added (the vertex of the concal cap) 4

6 Fgure 1: Constructng concal caps Hence for ε (0, δ) we get ( ) δ n 1 ( ) δ n 1 g(x + ε) g(x) g(x + ε) (5) δ ε δ + ε The argument s easly modfed for ε ( δ, 0) so g s ndeed contnuous at x Observe that the coeffcents n (5) are ndependent of g (only dependng on C, x and z) therefore we get the followng generalsaton Corollary 6 If C R n s a convex body and S R n s a d-dmensonal subspace then g : S R gven by g(x) = (S +x) C s contnuous at every x such that (S +x) nt(c) Proof (Sketch) Fx δ > 0 and z (S + x) such that B δ (z) C For y S wth y < δ, apply the prevous proof on the (d+1)-dmensonal affne subspace contanng S+x and S+x+y to obtan the followng bounds ( ) δ n 1 ( ) δ n 1 g(x + y) g(x) g(x + y) δ y δ + y The coeffcents depend only on y so contnuty follows Pror to provng the lower bound on the volume of ntersecton of Q n and an (n 1) dmensonal subspace we need the aforementoned lemma concernng the p-norm of a random varable, whch wll enable us to apply Lemma 3 Ths s Lemma 1 from [9] Note that for a ( ) 1/p real random varable Y wth pdf g, Y p denotes g(x) x p (not to be confused wth the L p norm) Lemma 7 Let Y be a real random varable wth pdf g, then for p > 0, Y p g 1 (p + 1) 1/p, where g s of course the essental supremum of g Proof We may assume that Y s symmetrc, for f not we can consder the random varable Ȳ wth pdf ḡ(x) = 1 (g( x) + g(x)) Then Ȳ p = Y p and ḡ g, so showng Ȳ p ḡ 1 (p + 1) 1/p would suffce 5

7 Set G(x) = x 0 g(t)dt for x 0 so G(0) = 0 and G( ) = 1/ by symmetry By constructon G s absolutely contnuous and G (x) = g(x) for almost all x We therefore obtan the followng p = G( ) p+1 = 0 (G(x) p+1 ) dx = (p + 1) 0 (p + 1) g p g(x)g(x) p dx 0 g(x)x p dx Ths last nequalty holds because G(x) x g for all x 0 The proof s completed by takng the p-th root and replacng ( 0 g(x)x p dx ) 1/p by Y p The lower bound now follows easly, ths s Theorem n [9] Proposton 8 Let S R n be a subspace wth dmenson (n 1) then S Q n 1 Proof Fx a unt normal u = (u 1,, u n ) to S and let X, X 1,, X n and f be as above Snce S Q n = f(0) and f s contnuous at 0 (where the maxmum s attaned), t suffces to show that f 1 To apply Lemma 7, consder X = E(X ) Snce the X 1,, X n are ndependent and dentcally dstrbuted, each wth expected value 0, we have E(X X j ) = 0 for all j Moreover, ( n ) 1/ E(X ) = E(X1) = E(X1) = x dx = 1/1 u Hence X = 1 Therefore, by Lemma 7 (wth p = ), f 1 3 = 1 as 3 X requred Next we set about provng the upper bound on f(0) To begn wth we need a lemma to estmate the L p norm of the snc functon, defned by { 1 snc(x) := x sn(x) x 0 1 x = 0 The full proof of ths lemma can be found n [9] but here we only explan an adaptaton of the frst part of the proof whch leads to a weaker upper bound The orgnal verson (lemma 3 n [9]) s as follows 1/ Lemma 9 If p then wth equalty f and only f p = 1 snc(t) p dt π p, However here we only prove the followng Lemma 10 If p then where α = 1 snc(t) p dt α π p, { e log() ( 1030) f p < 4 1 f p 4 6

8 Remark: The essental problem wth boundng ths ntegral s the central peak of the snc functon Snce snc(0) = 1, ts dffcult to fnd a nce bound whch decays suffcently fast as p ncreases for ntegrals of the form ε ε snc(x) p dx In the proof of Lemma 9, Ball fnds a smple bound when p 4 but has to do a lot of work for p < 4 We exhbt the frst part here but apply an nterpolaton 3 to get the weaker bound for smaller p Proof (of Lemma 10) To begn wth suppose p 4 We frst show that for t 36/5, we have 0 snc(t) e t /6 For ths consder the Taylor seres expansons (about 0) Smlarly, snc(t) = 1 t 6 + t4 5! t6 7! + t8 9! < 1 t 6 + t4 5! f t < 9! 7! = 7 e t /6 = 1 t 6 + t4 6! t ! + t ! t ! > 1 t 6 + t4 6! t ! Therefore, f t 36/5 = 6 3 3!/180, we have e t /6 snc(t) f t < 65 5! 6 4 4! = 30 t4 6! t ! t4 5! = t4 180 t ! 0 The fact that snc(t) 0 for t 36/5 follows from the fact that 36/5 < π For the sake of brevty, let c = 36/5 Now we can use e t /6 to estmate the central part of the ntegral n Lemma 10 (e the part on [ c, c]) We fnd a cruder estmate on the rest c snc(t) p dt e pt /6 dt + t p dt < e pt /6 dt + t p dt Snce p 4 we see that = p c R\[ c,c] π 6 p + (p 1) c p 1 = 1 p ( 6π + p 1 3 and cp 1 c 3 = 16/ 15 Therefore p 6π + (p 1) c p 1 6π < π p (p 1) c p 1 Ths gves the requred bound For p < 4, frst notce that we have equalty for p = (t s a well known fact that snc = π) Moreover, by an applcaton of Hölder s nequalty, we deduce that snc p p snc 4 p snc p 4 4 To prove ths, one could consder R snc(x) α snc(x) 4(1 α) dx for a sutable choce of α [0, 1] Now we have already found upper bounds on the terms of the rght hand sde so we obtan the followng snc p L π 4 p p ( π ) p 4 4 ( ) p = π π e log() p The last nequalty s the desred result and follows from elementary calculus (fndng the maxmum of p 4 p/ on the nterval [, 4]) 3 Thanks to Professor James Robnson for suggestng ths trck c ) 7

9 Fnally, we wll use a generalsaton of Hölder s nequalty whch we state here (see [13], page 67) Ths can be obtaned by applyng nducton to the standard -exponent verson Later on we wll dscuss a much more general result, namely the Brascamp-Leb nequalty Lemma 11 Let p > 1 for = 1,, n be constants such that n functons g L p (R) ( = 1,,, n) we have g 1 g g n L 1 (R) and g 1 g n 1 n g p p 1 = 1 Then for any We are now n a poston to prove the upper bound on the volumes of cross-sectons of Q n, ths s Theorem 4 n [9] and s the man result of ths secton Theorem 1 If S s an (n 1) dmensonal subspace of R n then S Q n Proof Let u = (u 1, u,, u n ) be a unt normal to S, then by symmetry of the cube, we may assume that u 0 for = 1,,, n Moreover, f u = 0 for some then S Q n = (u 1,, û,, u n ) Q n 1 and the problem reduces by one dmenson Thus, by nducton, we may assume that u > 0 for all There are two cases n ths proof, the frst s a farly straghtforward geometrcal argument whle the second requres some of the materal prevously dscussed Case 1: Suppose that u 1/ for some, (by symmetry agan, we may assume = 1) Consder the cylnder C wth long axs e 1 = (1, 0,, 0) defned by C = R [ 1, 1 ]n 1 Now clearly Q n C, so S Q n S C Moreover S C s the projecton n the drecton e 1 of {0} [ 1, 1 ]n 1 onto S Fgure : Reducng to the cylnder and projectons of e 1 Ths projecton has the effect of multplyng volumes n e 1 by P (e 1) / R P (e 1 ) were P and R are orthogonal projectons onto S and e 1 respectvely (see Fgure ) To check ths, one mght lke to consder the -plane spanned by e 1 and v It follows that S C = Q n 1 P (e 1 ) / R P (e 1 ) = 1 e 1 e 1, u u / e 1 e 1, u u e 1 (1 e 1, u ) 1 = (1 u u 1 1 u 1 ) + u 1 (u + u u n) u = (1 u u 1 1 u 1 ) + u 1 (1 u 1 ) = 1 = u 1 1 u 1 1 u 1 1 Ths completes the proof of the frst case snce we assumed u 1 1/ 8

10 Case : If u < 1/ for all = 1,,, n, then consder the random varables X 1,, X n from Lemma 3 For each, let φ be the characterstc functon of the random varable u X e φ (t) = E(e tu X ) Each X s symmetrcally dstrbuted so φ (t) = E(e tu X ) whch s just the Fourer transform of u 1 [ 1/,1/] Here 1 A denotes the ndcator functon (often called the characterstc functon) of A R n Therefore φ (t) = snc( 1 u t) Now snce the X 1,, X n are ndependent (as are the random varables e tu X for fxed t), the characterstc functon φ of X := n u X s gven by φ(t) = n ( ) u t snc Furthermore, by Lemma 3, φ s the Fourer Transform of f(x) = (S + xu) Q n Snce f s an even functon, ths s gven by ˆf(t) = cos(xt)f(x)dx 4 Snce f s contnuous at 0 (by Lemma 4), compactly supported and bounded, we may apply the Fourer nverson formula to see that f(0) = 1 π n sn(u t/) e t 0 dt = 1 u t π n snc(u t)dt Now apply Lemma 11 (generalsed Hölder s) wth p = u (so 1 p = u = 1 and p > ) and g (t) = snc(u t) (clearly snc L p (R) for all p > 1) Ths shows that f(0) 1 π n ( 1/p n ( snc(u t) p 1 1/p dt) = snc(t) p dt) u π By Lemma 9, ths gves the bound f(0) Ths s what we set out to prove ( n u p ) 1/p = n ( ) 1/p = Ths completes our exposton of Ball s 1986 paper We now dscuss the Brünn-Mnkowsk nequalty whch we prevoulsy stated Proof of the Brünn-Mnkowsk Inequalty Recall the statement of the Brünn-Mnkowsk nequalty (Theorem 1) A, B R n and λ [0, 1], we have For convex bodes λa + (1 λ)b 1/n λ A 1/n + (1 λ) B 1/n In some sense ths gves a lower bound on the volume of lnear combnatons of pars of convex bodes, n terms of the volumes of the bodes n queston The proof gven here wll be based on the proofs n [3] pages and [18], however we add several lemmas n order to flesh out the measure-theoretc detal We begn wth a defnton (see [3]) 4 Note that here we have chosen the conventon for the Fourer transform gven by ˆf(ξ) = e ξx f(x)dx as t agrees wth the defnton of charaterstc functons n probablty 9

11 Defnton 13 A set e R n s called elementary f t s the unon of fntely many (nondegenerate, closed) axs-parallel cubods that have parwse dsjont nterors Let the collecton of all such sets be denoted by E n We wll use the followng result from Measure Theory: Lemma 14 If A R n s compact then for any postve ε there exsts e E n wth A e and e < A + ε Moreover gven any δ > 0 we may assume that e B δ (A) Proof From the propertes of the product measure λ n we may choose a countable collecton of open axs-parallel cubods whose unon has volume at most A + ε and contans A By compactness we may assume that ths collecton s n fact fnte We can then take the closure of each cubod and subdvde (f necessary) to get an elementary set contanng A and wth small enough volume For the second clam, gven any elementary set e A, subdvde each cubod nto smaller ones wth dameter at most δ/ then we may remove any that do not ntersect A to get e A wth e B δ (A) and e e Next we verfy that elementary sets can be cut up to reduce the number of cubods 5 and sketch an ntutve result about convex sets Lemma 15 If e E n conssts of k cubods then there exsts an axs-parallel hyperplane H such that e H + and e H both consst of fewer than k cubods Proof It obvously suffces to consder the case k = Suppose e = C 1 C where C 1 = [a 1, b 1 ] [a n, b n ] and C = [c 1, d 1 ] [c n, d n ] are cubods wth mutually dsjont nterors Let P = {(x 1,, x n ) R n : x = a }, and Q = {(x 1,, x n ) R n : x = b }, e the hyperplanes contang the faces of C 1 Now suppose for contradcton that for each, nether P nor Q separate C 1 from C e for each of these hyperplanes, one of the correspondng half-spaces ntersects the nterors of both cubods Then for each we see that (a, b ) (c, d ) Hence nt(c 1 ) nt(c ) Ths s a contradcton, thus for some, ether P or Q has the requred property Lemma 16 Let C R n be a convex set wth nteror pont x, ndeed suppose that B ε (x) C Then for any δ > 0, x + (1 + δ) (C x) contans the open neghbourhood B δ ε (C) of C Proof (Sketch) Gven p C and consder the concal cap K = λ [0,1] λb ε(x) + (1 λ)p By convexty K C It s easy to check that the concal cap K = x + (1 + δ) (K x) contans the ball B δ ε (p) and the result follows Fnally we need a lemma to relate the volumes of lnear combnatons of convex bodes and approxmatng elementary sets Lemma 17 Let A, B R n be convex bodes, then for all ε, δ > 0 there exst elementary sets A A, B B wth A < A + δ, B < B + δ and A + B (1 + ε) A + B 5 Ths result s essentally obvous, but fndng ntuton n hgher dmensons s dffcult so t s reasonable to check t rgorously 10

12 Proof Let x be an nteror pont of A + B and let α > 0 such that B α (x) A + B Now by Lemma 14, there exst A, B E n contanng A and B respectvely, such that B < B + δ, A < A + δ Furthermore, we may assume that B B αr/ (B) and A B αr/ (A) where r = (1 + ε) 1/n 1 Therefore A + B A + B αr/ (0) + B + B αr/ (0) = B αr (A + B) snce Mnkowsk sums are commutatve Now A + B s a convex body, so by Lemma 16, we have that B αr (A + B) s a subset of x + (1 + r) (A + B x) whch has volume (1 + r) n A + B = (1 + ε) A + B Hence A + B (1 + ε) A + B as requred Wth these techncaltes n place we can now prove the Brünn-Mnkowsk nequalty Proof (of Theorem 1) We frst consder the case when A and B are axs-parallel cubods wth respectve sde lenths a,b n the drecton of the th standard-bass vector, = 1,,, n Observe that A + B s a cubod wth sde lengths a + b, hence wth volume n (a + b ) Now by the standard nequalty between geometrc and arthmetc means we have ( n a a + b ) 1 ( n n + b a + b ) 1 n 1 n n a a + b + 1 n n b a + b = 1 Thus A + B 1/n A 1/n + B 1/n as requred Now f A and B are elementary sets consstng of a combned total of k cubods we proceed by nducton on k We have just dealt wth the ntal case (when k = ) so suppose k > We may assume that A contans at least two cubods Snce A s elementary, by Lemma 15 there exsts an axs-parallel hyperplane H, such that A + := A H + and A := A H are both elementary sets wth strctly fewer cubods than A Let µ = A + / A then by translatng B f necessary, we may assume that B = B + B where B ± = B H ± and B + / B = µ Now A + + B + and A + B are subsets of A + B wth dsjont nterors and each made up of at most k 1 cubods Together wth the nductve hypothess ths mples that A + B A + + B + + A + B ( A + 1/n + B + 1/n ) n + ( A 1/n + B 1/n ) n = (µ + 1 µ)( A 1/n + B 1/n ) n Ths concludes the case for elementary sets All that remans s to apply some of the measure theory prepared earler Let A, B R n be convex bodes and ε > 0 Then by Lemma 17 there exst elementary sets A, B contanng A and B respectvely such that A + B 1/n [(1 + ε) A + B ] 1/n Thus, by the prevous case A 1/n + B 1/n A 1/n + B 1/n [(1 + ε) A + B ] 1/n for any postve ε Ths completes the proof (the generalsaton to λa + (1 λ)b s obvous) The Brünn-Mnkowsk nequalty has several generalsatons and applcatons n varous areas of mathematcs For a survey of some of these, as well as a dsscusson of how t relates to other mportant nequaltes (ncludng Brascamp-Leb), one mght lke to access [18] by Gardener 3 An Applcaton to the Busemann-Petty Problem The upper bound n Theorem 1 has a farly straghtforward applcaton to a well-known problem In 1956 Herbert Busemann and Clnton Petty proposed ten problems n convex 11

13 geometry (see [6]) The frst of these has become known as the Busemann-Petty problem One statement of the problem s the followng Problem 18 Let A and B be symmetrc convex bodes n R n Suppose that for any (n 1) dmensonal subspace H we have A H B H, does t follow that A B? The answer n general s, surprsngly, negatve In partcular f n 5, t s not true However for n 4 the answer s ndeed affrmatve (see [5]) The bound on the cross-sectonal volumes of cubes provdes counterexamples for n 10 To prove t, we wll need the followng proposton Later on we shall flesh out the proof that s sketched n the appendx of [10] Proposton 19 The volume of the ntersecton of an (n 1) dmensonal subspace of R n and the (centred) ball wth unt volume n R n s strctly ncreasng (wth respect to n 1) The followng theorem s a consequence of ths proposton Theorem 0 The Busemann-Petty problem has negatve soluton for n 10 Proof Consder the central unt cube Q n n R n and the central ball B n of unt volume n R n As these objects have the same volume, t suffces to prove that the hyperplane (subspace) cross-sectonal volumes of B n are strctly greater than (e the maxmum cross-sectonal volume of Q n ) when n 10 Of course, by Proposton 19 we only need an estmate n the case n = 10 Denote by v n the volume of the unt ball n R n Recall that v n s gven by v n = π n Γ( n + 1) Here Γ(x) = 0 e t t x 1 dt s the usual gamma functon Let a n = B n H for any (n 1) dmensonal subspace H R n, then we have a n = v n 1 v n 1 n n = π n 1 (Γ( n+ n 1 )) n π n (n 1) n Γ( n+1 ) = (Γ( n+ n 1 )) n Γ( n+1 ) Recall that for n Z >0, Γ(n) = (n 1)! and Γ( 1 ) = π More generally, Γ(x + 1) = xγ(x) for x R >0 Hence when n = 10 we get a 10 = Γ(6) 9 10 Γ(5 + 1 ) = 5! π > 140 > Ths s what we wanted to show Furthermore, a smlar calculaton shows that a 9 < 141 < so these partcular convex bodes do not provde counterexamples n lower dmensons We now return to prove Proposton 19 regardng the monotoncty of the volumes of cross-secton of balls wth unt volume One s tempted to try a drect approach, for example by dfferentatng 9 10 a x = ( ( Γ x+ )) x 1 x Γ( x+1 ) 1

14 However, dong so does not seem to gve the requred result, even usng some recently proved bounds on Γ and Ψ = Γ /Γ (the dgamma functon) such as those from [16] Instead we elaborate on the proof sketched by Ball It s qute elementary but the choce of steps s not obvous so we spend some tme on the detals The frst step uses the followng smple result about ntegratng convex functons Lemma 1 If f : [a, b] R s a C convex functon on a bounded nterval (e f (x) 0 n (a, b)) then ( ) a + b f 1 b f(a) + f(b) f(x)dx b a a Proof As f s convex, the regon above the graph of f, e U := {(x, y) [a, b] R : y f(x)} s a convex set Hence U s contaned n the upper half-plane generated by any tangent In partcular U {(x, y) : y f ((a+b)/)x+c}, where C = f((a+b)/) f ((a+b)/)(a+b)/ Thus f(x) f ((a + b)/)x + C on (a, b), so b a f(x)dx b a f ( a + b ) + (b a)f ( ) a + b b a ( ) a + b f as requred Ths proves the left-hand nequalty For the rght hand sde, observe that by convexty of U, f(x) 1 b a [f(a)(b x)+f(b)(x a)] Integratng gves (b a) b Ths proves the rght-hand nequalty a f(x)dx b a (f(b) f(a)) + (b a)(f(a)b f(b)a) = (b a) (f(b) + f(a)) Lemma The functon f : (0, ) (0, ) gven by f(x) := ( x+1 x the functon g : (1, ) (0, ) gven by g(x) := ( x x 1) x 1 1/x s ncreasng Proof The dervatve of f s gven by [ ( ) x + 1 f (x) = log x + 1 ] f(x), x x(x + 1) so t suffces to show that ( log ) x + 1 x x(x + 1) ) x+1 s decreasng and Now the functon x x 1 s convex Therefore by Lemma 1 on the nterval (x, x + 1), log(x + 1) log(x) = x+1 as requred Smlarly, the dervatve of g s g = [( + 1x ) ( x log x 1 [ x ( + 1 x = x log x 1 x 1 x dx 1 x + 1 (x + 1) = x + 1 x(x + 1) ) ( x 1 1 ) x 1 x x(x 1) ) x + 1 x ] g(x) ] g(x) 13

15 So we only need to show that ( ) x log x + 1 x 1 x + 1 In fact, more s true If we apply the frst nequalty of Lemma 1 as before on the nterval (x 1, x), we get as requred log(x) log(x 1) 1 x 1 = x 1 x + 1 x + 1, These gve rse to the followng result (whch s what we actually need) Corollary 3 If n 3 then ( ) n 3 n 1 n n 5 ( n + 1 n ) n n 1 < n n 1 Proof For n 3, by Lemma we have (by the frst part) Moreover, by the second part ( n ) n n 1 n+1 ( n 1 n ( n ) (n+1)(n 1) (n 3) n 1 ) n n 1 n(n 1)(n )(n 5) ( n + 1 n So the result follows from the easly checked fact that, for n 3, ) n 3 n 5 (n + 1)(n 1) (n 3) n(n 1)(n )(n 5) + n n + 1 < 1 Now we can prove the proposton Proof (of Proposton 19) For n 0 let I n = π/ π/ cosn xdx and note that I n s strctly decreasng and moreover I n+ = n+1 n+ I n Usng the same notaton as above, a n = v n 1 vn (n 1)/n s the sequence we clam s strctly ncreasng It s known that v n = I n v n 1, so a n = vn 1/n /I n and we only need to show that for n 1 1 < a n a n 1 = ( In 1 I n ) n 1 v n 1 n n v n 1 1 n 1 = [ (In 1 I n ) n 1 1 a n ] 1 n 1 (4) ( ) n 1 Let u n = In 1 I n then un > 1 for n 1 thus, n partcular u 1 > a 1 and u > a 1 Moreover, from the above observatons, one can easly obtan that a n = u 1 n a n 1 n n 1 It follows that f (u n) s a strctly ncreasng sequence and u n > a n 1 then u n+1 > u n > a n In ths case u n > a n for all n 1 by nducton, whch would complete the proof as ths s exactly (4) 14

16 ( ) 1 Now we show that (u n ) s ndeed strctly ncreasng Let x n = un+1 n 1 u n then x n = I n ( I n n+1 I n 1 n 1 ) 1 n 1 = ( In I n 1 ) ( n + 1 n ) n n 1 Snce 1 > In I n 1 > In I we must have x n n 1 as n Furthermore we shall show that x n < x n for all n 3 Hence x n > 1 for n 1 so (u n ) s strctly ncreasng as clamed The fact that x n < x n follows drectly from Corollary 3 wth the followng observaton = n 1 n Ths completes the proof x n x n = ( ) ( In I n 1 n 1 I n 3 I n n ( ) ( n n 1 = n 1 n ( ) [ ( n n 1 = n 1 n ) n n 5 ( ) n (n 5) n 5 ) n 3 n 5 ( n + 1 n n n + 1 ( n n + 1 ) n n 1 ) n n 1 ) n n 1 ] 1 3 The Problem for d-dmensonal Subspaces So far we have only dealt wth cross-sectons correspondng to hyperplanes, however we may equally thnk about cross-sectons of lower dmenson In ths secton we exhbt two upper bounds (due to Ball) on the volume of d-dmensonal cross-sectons of Q n (for d n) whch depend only on n and d In many cases one or other of these bounds s optmal and we present a conjecture on the best upper bound n the remanng cases The proof of each bound requres the Brascamp-Leb nequalty, whch we frst prove 31 The Brascamp-Leb Inequalty Ths result s a powerful generalsaton of the nequaltes of both Hölder and Young It was orgnally due to Herm Brascamp and Ellott Leb (see [7]) We only requre a specal case (the geometrc verson), but because of the mportance of ths result we prove t n the orgnal generalty (t can also be generalsed beyond the orgnal statement) In 1997 an elegant proof was presented by Barthe n [1] We shall now commence an exposton and translaton of ths statement and ts proof (wth a lttle smplfcaton) One of nterestng thngs about t s that t reles on some convex geometry, though the result tself s essentally analytc Theorem 31 Fx n Z >0 also fx an nteger m n Let c 1, c,, c m R >0 such that m c = n Suppose v 1, v,, v m are vectors n R n and defne D = { det( m c γ v v ) m γc } : γ > 0 for all We have used the followng notaton, for a, b R n, a b s a lnear map on R n defned by 15

17 a b(x) = a, x b Now defne the followng sets: { m R F = n (f ( v, x )) c } dx m f c : each f s non-negatve and ntegrable, 1 { m R F g = n exp( c γ v, x } )dx m ( R exp( γ x )dx) c : γ > 0, { R sup ({ m Ẽ g = n exp( c γ θ ) : x = m c } ) } θ v {0} dx m ( R exp( γ x )dx) c : γ > 0 Furthermore let F = sup( F ), F g = sup( F g ), E g = nf(ẽg) and D = nf( D) Then the concluson of the theorem s that F = F g = 1 D Note that F g s the subset of F correspondng to the restrcton that each f s a (centered) Gaussan The {0} terms appearng n the defntons of Ẽg s smply to prevent undesrable behavour when the {v } m does not span Rn, arsng from the fact that sup = The proof has three parts We frst show that F g = 1 D, then that E g D F (t s clear by the remark after the theorem that F F g ) Fnally we shall show that E g = D F g to complete the proof Lemma 3 F g = 1 D If D = 0 then F g = Proof We show that for every x F 1 g, D The same argument also shows the converse x (e x D 1 x F g ) Fx γ 1, γ,, γ m > 0 then the correspondng element of F g s m R y = n exp( c γ v, x )dx m ( R exp( γ x )dx) c The denomnator s the product of Gaussan ntegrals and so s easly calculated ( c ( ) π exp( γ x )dx) c π = n = R γ m γc Furthermore the numerator can be calculated by a change of varables (usng the fact that Q(x) := x, ( m c γ v v ) (x) s a postve-semdefnte quadratc form and thus has a real matrx-square-root) exp( c γ v, x )dx = exp( Q(x))dx R n R n π n = exp( πx π )dx = n det Q R n det m c γ v v Hence 1/y s the element n D correspondng to {γ } m The result follows easly Lemma 33 If D 0, then E g F D F g D 16

18 Proof As mentoned above, the rght-hand nequalty s obvous from the defnton We prove the left-hand nequalty usng a change of varables Let f 1,, f m, g 1,, g m be strctly postve, ntegrable and contnuous real valued functons on R We may make such assumptons about {g } m as these wll later correspond to Gaussans, for {f } m (whch later correspond to any non-negatve ntegrable functons) we nclude an approxmaton argument at the end of the proof For each defne T by the equaton T (t) g (x)dx = R g t f (x)dx R f Observe that f and g have ant-dervatves whch are ncreasng and dfferentable Hence each T : R R s well defned and bjectve Moreover, the rght-hand sde of the above s dfferentable and the ant-dervatve of g has a dfferentable nverse on (0, ) (by the nverse functon theorem) Precomposng by ths nverse we see that T s dfferentable Indeed we have T (x)g (T (x)) = R g f (x) Moreover each T s strctly postve snce T s strctly ncreasng Now the change of varables wll be gven by Θ(x) = m c T ( x, v )v Notce that the drectonal dervatves are gven by x j Θ(x) = R f m c T ( x, v )v j v Where v = (v 1,, vn ) It follows that the dfferental s dθ(x) = m c T ( x, v )v v Notce that (as n the proof of Lemma 3) dθ(x) corresponds to a postve-defnte quadratc form on R n for all x (because D 0 so {v } m spans Rn ) Hence Θ s njectve Usng the drect substtuton x = Θ(y), then the change of varables, we get the followng nequaltes sup x= (g (θ )) c dx m c θ v ( m [g (T ( y, v ))] c det R n D = D R n R n [g (T ( y, v )) T ( y, v )] c dy ( R g ) c R f R n [f ( y, v )] c dy sup Θ(R n ) Θ(y)= m c θ v c T ( y, v )v v ) dy (g (θ )) c dθ(y) To complete the proof, dvde by m g c 1 then take the nfmum of the left-hand sde wth respect to {g } m (over all centered Gaussans) and the supremum of the rght-hand sde wth respect to {f } m (over postve contnuous, ntegrable functons) All that remans s to justfy the assumpton that, for each, f s strctly postve and contnuous We adapt an approach whch s used n another paper by Barthe (see []) 17

19 We show that n calculatng F we only need to consder postve and contnuous functons Let {f } m,{v } m and {c } m be as they are n the defnton of F It suffces to show that we may assume f 1 to be postve and contnuous By monotone convergence theorem, we only need to consder functons whch are bounded above by centred Gaussans More precsely f G(x) = e πx and f 1 s non-negatve and ntegrable then ζ k (x) := mn(f 1 (x), kg(x)) s ncreasng (wth respect to k Z >0 ) and converges pontwse to f 1 as k Hence ζ k 1 f 1 1 and moreover R n ζ k ( x, v 1 ) c 1 f ( x, v ) c dx = R n f ( x, v ) c dx We can now assume that for some Gaussan G, f 1 (x) G(x) for all x R For postvty and contnuty, let G k (x) = kg(kx) and defne η k (x) := mn(f 1 G k (x), G(x)) Then η k (x) > 0 for all x (we may assume that f 1 > 0) and s contnuous (by the propertes of convolutons) Furthermore t can be shown that lm k f 1 G k f 1 1 = 0 (usng Mnkowsk s ntegral nequalty) and so lm k η k f 1 1 = 0 Hence, passng to a subsequence f necessary, we may assume that η k converges pontwse to f 1 almost everywhere By domnated convergence theorem (we use G to fnd the domnatng functon), we now have η k ( x, v 1 ) c 1 f ( x, v ) c dx f ( x, v ) c dx R n R n = Ths s unless G( x, v 1 ) c 1 m = f ( x, v ) c s not ntegrable, n whch case F = anyway Hence wthout loss of generalty, we may ndeed assume that f 1 s postve and contnuous The fnal part of Barthe s proof of the Brascamp Leb nequalty uses the followng facts and defntons from convex geometry The frst two defntons can be found n [19], (pages 33 and 37 respectvely) the thrd s a quantty dscussed n [8] Defnton 34 Let C R n be a symmetrc convex body then: 1 the polar body of C s C = {x R n : x, y 1 for all y C} the support functon of C s the map h(c) : R n R gven by h(c)(x) = sup y C x, y 3 the Mahler volume (or volume product) of C s defned to be M(C) = C C Lemma 35 If C R n s a symmetrc convex body and T : R n R n s a non-sngular lnear map, then: 1 C s unquely determned by h(c) (T (C)) = (T 1 ) (C ) 3 M(T (C)) = M(C) 4 If C = B r (0) for r > 0 then C = B 1/r (0) 5 C s a symmetrc convex body 6 C = C (Dualty) 18

20 Proof 1) Suppose that C and D are convex bodes and h(c) = h(d) Now for any y S n 1, h(c)(y) s the dstance from the orgn to a supportng hyperplane parallel to y Hence x C f and only f x, y h(c)(y) for all y S n 1 (snce a convex body s the ntersecton of all half-spaces whch contan t) Smlarly, x D f and only f x, y h(d)(y) = h(c)(y) for all y S n 1 So C = D as requred ) x (T (C)) f and only f x, T (y) 1 for all y C An equvalent condton s of course T (x) C e x (T 1 ) (C ) 3) By defnton and part we have M(T C) = det(t ) C det(t 1 ) C = M(C) as requred x 4) By part t suffces to consder C = B 1 (0) If x / C then x, x = x > 1 so x / C The converse s by Cauchy-Schwarz So C = C as requred 5) Clearly C s bounded as C has non-empty nteror and hence contans a closed ball, B ε (0) say, so x / B 1/ε (0) would mply that x, ε x x > 1 e x / C C = (h(c)) 1 ([0, 1]) and so s closed by contnuty of h(c) Symmetry follows drectly from the defnton C has non-empty nteror snce C s bounded, say C B r (0) for some r > 0 so f x < 1 r then h(c)(x) 1 e x C C s convex, for f λ [0, 1] and x, y C then for all z C, (1 λ)x + λy, z 1 by lnearty Hence (1 λ)x + λy C So ndeed C s compact, symmetrc and convex wth non-empty nteror e a symmetrc convex body as requred 6) It s clear from the defnton that C C For the converse suppose that C C (e strctly contaned), then by takng the polar bodes 6 we would get C C However, just as C C, we have C C whch s a contradcton By Lemmas 3 and 33, to complete the proof of the Brascamp-Leb nequalty we only need to show the followng Lemma 36 If E g and F g are fnte then E g F g = 1 If F g = then the E g = 0 and n any case E g must be fnte Proof Begnnng wth the last pont, E g s the nfmum of a non-empty set of non-negatve reals and so cannot be nfnte We now fnd alternatve expressons for the elements of F g and Ẽg Let γ 1,, γ m > 0 and for x R n defne N(x) = m c γ x, v If {v } spans R n then N s clearly a norm (the trangle nequlty follows from the trangle nequalty for the Eucldean norm) In ths case the closed unt ball wth respect to N s a symmetrc convex body (n fact an ellpsod) In any case (even f {v } m does not span Rn ), the trangle nequalty holds for N so the set F = {x R n : N(x) 1} s convex It s also clearly symmetrc, therefore F s star-shaped Hence, just as wth spheres and balls, we have the relaton F = F n (or both sdes are nfnte) Therefore, consderng the element of F g correspondng to the choce of γ 1,, γ m, we have m R n exp( c γ x, v )dx m ( R exp( γ x )dx) c = R e n N(x) dx = π c γ c m 0 F r n 1 e r dr m π n = F n 0 r n e r d r m = F Γ( n + 1) π n π n c γ c γ c γ = F B 1 (0) 6 For symmetrc convex bodes A and B, the fact that A B A B follows from the defnton and convexty c γ 19

21 Note that the ntroducton of r was just a change of varable If {v } does not span R n then F = so, by the above calculaton, we have F g = In ths case we also have (from the defnton) that E g = 0, as requred Hence we may now assume that {v } = Consder the element of Ẽg correspondng to the constants 1 1 γ 1,, γ m Agan we defne a norm on R n, ths tme gven by M(x) = nf m θ c : x = γ m c θ v wth closed unt ball E (a symmetrc convex body) Then, as above, we have the followng { m R sup n exp( c θ γ ) : x = } m c θ v dx ( m ) R e c = n M(x) dx x m R exp( γ )dx (πγ ) c = E c B 1 (0) γ Thus, as we have a 1-to-1 correspondance between Ẽg and F g arsng from the choce of γ 1,, γ m t suffces to show that F E = B 1 (0) for any such choce Here we can apply the prevously dscussed geometry In fact we shall see that E s the polar body of F Ths mmedately gves the requred result by Lemma 35 and the fact that F s an ellpsod (e the mage of B 1 (0) under a non-sngular lnear operator) Consderng the support functon of F, we see that h(f )(x) = N(x) Ths follows from dualty More precsely, f x R n x \{0} then N(x) F = F so h(f )( x N(x) ) = 1 whch s enough by homogenety of h(f ) Now by Cauchy-Schwarz appled to c γ x, v and c γ θ, we have that N(x) = m c γ x, v sup c γ θ 1 m c θ x, v In partcular, settng θ = γ x,v N(x) gves equalty Now applyng the defnton of E, we see that ths s exactly N(x) = sup y E x, y = h(e)(x) Therefore, snce support functons unquelly determne symmetrc convex bodes, we have F = E as requred We have proved Brascamp-Leb nequalty n ts orgnal generalty, however we shall only use the followng specal case (often called the geometrc Brascamp-Leb nequalty) Corollary 37 Fx m n and suppose {c } m R >0 and {v } m Rn wth and m c = n (38) m c v v = I n (39) 0

22 Then for any non-negatve ntegrable functons f 1, f,, f m on R, we have [ [f ( x, v )] c dx R n c f (x)dx] R Condtons (38) and (39) may be refered to as the Frtz John condtons Ths corollary follows from the followng lemma whch s proved n [1], but we shall not prove t here Lemma 310 Wth {v } m and {c } m as n Corollary 37 and D from the statement of the Brascamp-Leb nequalty, we have D 1 We now swtch our attenton back to the problem of boundng the cross-sectonal volumes of cubes 3 Upper Bounds n the d-dmensonal Case In ths secton we dscuss two upper bounds on the d-dmensonal cross-sectonal volumes of Q n R n where d may be less than n 1 The followng theorem s the frst of these bounds Wth the use of the Brascamp-Leb nequalty the proof s qute straghtforward For ths result we follow the proof n [11] Theorem 311 If S R n s a subspace wth dmenson d > 0 then moreover ths bound s attaned f d n S Q n ( n d ) d, Proof Fx a d-dmensonal subspace S, then f P denotes orthogonal projecton onto S, we see that P (e 1 ),, P (e n ) forms a bass of S Moreover, for x S, x, e = x, P (e ) for each Therefore S Q n s the set {x R n : x, P (e ) 1 for = 1,, n} Clearly the restrcton of P to S s the dentty and s also gven by P = n e P (e ) = n P (e ) P (e ) We may assume that for some m wth d m n we have P (e ) 0 m Then gven an sometry ψ : S R d, the vectors v = ψ(p (e )) P (e ) and constants c = P (e ) > 0 ( = 1,, m) satsfy (39) Furthermore, { ψ(s Q n ) = x R d 1 : x, v P (e ) = 1 } for = 1,, m c Notce also that {c } m satsfy (38), e m c = n P (e ) = d It s easy to see ths by wrtng each P (e ) n terms of an orthonormal bass of S and usng the fact that x, e = x, P (e ) for all x S 1

23 Therefore, lettng f = 1 [ 1/( c ),1/( c )] for each we conclude, usng the Brascamp-Leb nequalty that S Q n = [ [f ( x, v )] c dx R d c f (x)dx] = R c c We shall shortly sketch a proof that a product of the form n xx s mnmsed (subject to each x beng postve and n x = r beng constant) when x 1 = x = = x n Hence c m c ( ) d m d ( ) d n d as requred We also clamed that f n s an nteger multple of d then ths bound s the best possble Indeed n ths case, let k = n d and consder the subspace H Rn spanned by ẽ := e k+1 + e k+ + + e (+1)k, for = 0, 1,, d 1 Thnkng of {ẽ } =0 d 1 as an orthogonal bass for Rd wth ẽ = k t s not dffcult to see that H Q n s sometrc to the set [ k/, k/] d n R d Hence H Q n = ( k) d = ( n d ) d As mentoned n the prevous proof, we sketch a proof of the followng lemma Lemma 31 If x 1,, x n R >0 and n x = r then n xx ( r r n) Proof (Sketch) Let R n be the (open) smplex defned by x = r (and each x postve) Then let f : R >0 be defned by f(x) = n xx The tangent plane of s spanned by vectors of the form e e j Gven x we sketch an algorthm to construct a path n, from x to ( r n,, r n ) on whch f s always decreasng We argue by nducton Fx k Z >0, k < n and suppose that x 1 = x = = x k then we shall construct a path from x to x where x 1 = = x k = x k+1 and f( x) f(x) Let α = 1 k and assume that x 1 x k+1 (f not then replace t wth t where approprate n the followng argument) Dfferentatng f wth respect to the drecton v = ( α,, α, 1, 0, 0) (k terms are α) we get ( ) d xk+1 + s dt f(x + tv) = log f(x + sv) t=s x 1 αs Ths s non-postve for s s := x 1 x k+1 1+α In partcular, f(x) f(x+ sv) and lettng x = x+ sv, we clearly have x 1 = = x k+1 as requred We shall now dscuss the second upper bound of ths secton Ths s a theorem from the same paper (e [11]) whch gves the best upper bound n the case where d n The proof of optmalty s agan constructve and wll be dscussed later as part of a more general set of examples The proof s essentally a generalsaton of the method used n the hyperplane case (Theorem 1), so we may only sketch some parts Theorem 313 As before, let S R n be a d-dmensonal subspace (d > 0) then S Q n ( ) n d

24 Proof (Sketch) Assume for nducton that the result holds for any cross-secton of Q n 1 As before there are two parts to the proof Case 1: We frst consder the case where S contans a drecton that s suffcently close to a standard bass drecton n R n The case d = n s obvous so we also assume that d n 1 Suppose that there exsts a unt vector v S wth some component larger than 1 Wthout loss of generalty we assume that ths s the frst component, e v 1 1 Denote by C n the cylnder R [ 1, 1 ]n 1 and let P denote orthogonal projecton onto e 1 Then (as P S s non-sngular) we have, by nducton that P (S C n ) ( ) n 1 d In other words, P (S C n ) s a subspace of e 1 wth dmenson d Let R be the orthogonal projecton onto v, then P R(e 1 ) / R(e 1 ) = v 1 (see the calculatons n the proof of Theorem 1) Moreover, snce we can fnd an orthogonal bass of v contaned n {R(e 1 )} e 1, we have that for any x v, P (x) / x v 1 In partcular, by consderng an orthogonal bass of S v wth at least d 1 vectors n e 1 we see that P (S C n ) S C n v 1 Hence S Q n ( ) n d as requred Case : Suppose that for any unt vector n S 1, each component s at most (n absolute value) Let T be the orthogonal projecton onto S and defne c = T (e ) and w = T (e ) c for = 1,, n We may assume (by nducton) that for each, c > 0, moreover, by the hypothesese of the case, c < 1 Notce that as n the proof of Theorem 311 we have constructed vectors and constants satsfyng the Frtz John condtons (on S ) Once agan, consder ndependent random varables X 1,, X n, each unformly dstrbuted on [ 1, 1 ] and let X = (X 1,, X n ) Observe that the probablty densty functon of T (X) s gven by f : S R >0 where f(x) = (S + x) Q n By consderng the characterstc functon of T (X) (e the Fourer transform of f); usng the propertes of expectaons and applyng the Fourer nverson formula, we obtan the followng (cf the hyperplane case) S Q n = f(0) = 1 π n d S n snc( c w, ξ )dξ Thus, takng absolute values and applyng the geometrc Brascamp-Leb nequalty, we obtan S Q n 1 n ( π n d snc(x ) c c ) 1 c dx R = 1 π n d 1 π n d n n ( 1 c R ) c snc(x) 1 c dx ( π c c ) c = ( ) n d The last nequalty s just Lemma 9, appled n tmes, wth p = 1 c to show Ths s what we wanted 33 A Conjecture on the Best Upper Bound n all Cases We have seen some good upper bounds on the d-dmensonal cross-sectonal volumes of n- dmensonal cubes whch are ndeed attaned n many cases However, by consderng the form of the worst examples when the optmal bounds are known, we can fnd examples of large crosssectons n every case and conjecture that these are ndeed the worst We frst dscuss such examples 3

25 Fx d n and let r be the remander from the dvson of n by d Also let k = n r d, then defne v 1,, v d R n as follows v +1 = e k+1 + e k+ + + e k(+1) for 0 < d r v d r+ = e (d r)k+(k+1)+1 + e (d r)k+(k+1)+ + + e (d r)k+(k+1)(+1) for 0 < r It s easy to check that the v 1,, v d are orthogonal and hence span a d-dmensonal subspace, call t S, of R n Moreover, the ntersecton S Q n s the set { d } S Q n = α v : 1 α 1 for each Hence S Q n = d ( n r v = d ) d r ( n r + d One can check that ths agrees wth the result of Theorem 311 when d n (r = 0) and wth bound n Theorem 313 when d n (n r = d) In partcular these examples verfy our clams about optmalty n Theorem 313 After consderng ths, the followng conjecture seems natural d ) r Conjecture 314 Let S be a d-dmensonal subspace of R n and r as above, then ( n r S Q n d ) d r ( n r + d We shall now dscuss one of the apparent dffcultes n fndng a proof for ths result As a frst attack on the problem, one mght look at the applcaton of the Brascamp-Leb nequalty n Theorem 311 as ths result looks smlar Furthermore, we may ask whether choosng c 1,, c n to satsfy the Frtz John condtons n that proof s too restrctve After all, the full statement of Brascamp-Leb gves the followng (usng the notaton of Theorem 311) R d n [f ( x, v )] c dx n ( ) c f (sup R d ) r { n }) 1 γc det( n c γ v v ) : γ > 0 Here we have only assumed that n c = d ( and c > 0 for each ) Now the left-hand sde does not depend on the {c } n as each f s an ndcator functons Hence we would lke to take the nfmum of the rght-hand sde wth respect to {c } n Unfortunately ths approach wll probably not work Indeed Stefán Valdmarsson has recently proved (see [1]) that 1 s the best Brascamp-Leb constant e the optmal constant n the Brascamp-Leb 7 nequalty More precsely, for {c } n and {v } n Rd as above, we have ( { n }) 1 sup γc det( n c γ v v ) : γ > 0 1, wth equalty only n the geometrc case Therefore the only hope left for ths strategy s f, for a sutable choce of {c } n, n ( R f c ) can be reduced wthout substantally ncreasng the Brascamp-Leb constant 7 Valdmarsson actually proves ths result for a more general verson of the Brascamp-Leb nequalty 4

26 In many cases ths s not possble Indeed, suppose we can fnd a d-dmensonal subspace S R n such that P (e ) = P (e 1 ) for n, where P s the orthogonal projecton onto S In ths case, constructng {f } n as n Theorem 311, we have f 1 = f = = f n and therefore n ( c f (x)dx) = R ( d f 1 (x)dx) = R ( n d ) d The fnal equalty holds because as before, the constants ( R f ) satsfy the Frtz John condtons wth correspondng vectors {P (e )/ P (e ) } n It would follow that we could not mprove on Theorem 311 wth ths method It s easy to check that such subspaces exst when d dvdes n (and consequently when n d dvdes n) by explct constructons For example n the case of a hyperplane (n d = 1) we can take the subspace (1, 1,, 1) However these cases happen to be examples where ether Theorem 311 or Theorem 313 already provde optmal bounds It does not seem clear whether such subspaces exst n general, but at least n one case of nterest, they can be constructed 8 Namely for d = Ths follows easly from the fact that the Frtz John condtons 9 are not only necessary but also suffcent for a set of vectors to be the mage under projecton of an orthogonal set of vectors n some superspace Ths follows from a smple result about sets of vectors n Hlbert spaces whch was proved for example, by Stenberg (see [0]) To fnd subspaces wth the requred propertes, t therefore suffces to fnd vectors wth equal lengths that satsfy the Frtz John condtons Ths can obvously be done when d =, for example consder the normal drectons of the edges of a regular n-gon In summary of ths dscusson, t seems that f Conjecture 314 holds then a proof should take a dfferent approach to the proof of Theorem 311 (at least for d = ) However, at present t does not seem clear what ths should be Ths concludes our nvestgaton of the cross-sectonal volumes of cubes We now dscuss two other nterestng problems from convex geometry whch nvolve cubes n some way 4 Some Other Interestng Problems Related to Cubes 41 Unversalty of Cross-Sectons If one were to wrte a computer programme to generate mages of cross-sectons of cubes, one would see that a remarkable range of polytopes can be attaned (for example see Fgures 3 and 4) Ths s no concdence Indeed, we now dscuss a result proved n 1967 by Epfanov (see [4]) that any polytope occurs as a cross-secton of a cube The proof begns wth the followng lemma Lemma 41 Fx n Z >0 and real numbers b,j for all 1 < j n + 1 Then there exst vectors v 1,, v n+1 R n wth v,, v n lnearly ndependant and v, v j = b,j for < j Proof (Sketch) If n = 1 ths s obvous Otherwse we argue by nducton Assume we can fnd such vectors {v } n+1 wth respect to an n-dmensonal subspace, S, of Rn+1 Then, choosng a unt vector u S and replacng v n+1 wth v n+1 + u, we obtan a bass for R n+1 There s then a unque choce of v n+ that has the prescrbed nner product wth v 1,, v n and v n+1 + u 8 Thanks to Professor Keth Ball for pontng ths out 9 When we dscuss the Frtz John condtons and only menton the vectors, {u } n say, t s meant that c = u and v = u / c satsfy the (38) and (39) 5

27 Fgure 3: A few hyperplane cross-sectons of Q 3 (not to scale) correspondng to the planes {x : x, v = a} where v and a are (from left to rght): v = (1, 1, 0), a = 0; v = (1, 1, 1), a = 0; v = (1, 1, 1), a = 05 and v = (1, 1, 05), a = 05 Fgure 4: A few hyperplane cross-sectons of Q 4 (projected nto R 3 and not to scale) correspondng to the hyperplanes {x : x, v = a} where v and a are (from left to rght): v = (1, 1, 1, 1), a = 0; v = (1, 1, 1, 0), a = 0; v = (1, 1, 05, 05), a = 0 and v = (1, 1, 1, 1), a = 1 The followng result and proof are based on Epfanov s deas but we have made the argument more explct Epfanov mentons cubes but essentally proves somethng more general, whch means that the dmenson of the cube constructed s not optmal (see the dscusson after the proof) Though we do not mprove on ths method, we am to make the constructon of the cube clearer Proposton 4 Suppose P R n s a polytope wth non-empty nteror and m (n 1) dmensonal faces Then there exsts an n-dmensonal affne space S R n+m 1 and r > 0 such that P s sometrc to S rq n+m 1 Proof Let T R n+m 1 be an n-dmensonal subspace and let ψ : R n T be an sometry such that 0 ψ(nt(p )) Also let ν 1,, ν m R n be unt vectors n the normal drectons of the (n 1) dmensonal faces of P Defne ν = ψ( ν ) ψ(0) for each 1 m (e the mage of these drectons under ψ) By the lemma, there are vectors k 1,, k m T such that k, k j = ν, ν j for all 1 < j m In ths case one can easly check that the vectors {k + ν } m are mutually orthogonal Moreover T (k + ν ) = T ν s parallel to the face of ψ(p ) correspondng to ν For 1 m let h = ν + k and extend to an orthogonal bass {h } n+m 1 of R n+m 1 Wthout loss of generalty assume that there exsts M wth m M n + m 1 for whch h T f and only f > M Furthermore, for each and x R, defne H (x) R n+m 1 to be the halfspace, contanng 0, nduced by the hyperplane h + xh Then there exst nonzero constants a 1,, a M such that ψ(p ) = T ( ( m H M ) (a )) = T H (a ) We may assume that (h + a h ) ψ(p ) e each a s mnmal (n absolute value) wth respect to 6

28 the prevous equaltes (ths only affects the choce of a m+1,, a M ) Let r = max{max x ψ(p ) x, h mn x ψ(p ) x, h : 1 M}, e the maxmum wdth of ( n+m 1 ) ψ(p ) n any of the drectons h We now also see that ψ(p ) T H (b ) where b = a (1 r a M < n + m 1 ) For those hyperplanes parallel to T, defne a = r/ and b = r/ for Then Q = n+m 1 (H (a ) H (b )) s a cube of sde length r wth T Q = ψ(p ) Thus there s an sometry φ of R n+m 1 so that rq n+m 1 = φ(q) and f S := φ(t ) we get φ ψ(p ) = rq n+m 1 S as requred We now brefly hghlght a shortcomng of ths result and ponder possble mprovements In the proof we were only really consderng cross-sectons of cones generated by ntersectng halfspaces Therefore the cube we found wll n general be n a hgher dmenson than necessary For example, a hexagon can be generated as a -dmensonal cross-secton of a 3-cube However, the above constructon would nstead fnd a hexagonal cross-secton of a 7-dmensonal cube The heurstc reason for ths s that a cross-secton of a cube may ntersect faces that are opposte to one another, however n the cone there are no pars of parallel faces Ths leads us to the queston of fndng the mnmum dmenson for a cube wth a gven polytope as one of ts cross-sectons By adaptng the above method, t seems lkely that we can fnd an (n 1 + N) dmensonal cubod wth the correct cross-secton, where N s the number of normal drectons to faces of the polytope, e the maxmum number non-parallel faces However, ths s stll not optmal (consder the hexagon agan) and moreover, t s not clear whether somethng smlar can be done for cubes Ths concludes our dscusson of Epfanov s result We now arrve at the fnal part of ths essay, n whch we ntroduce a well known open problem 4 The Blaschke-Santaló Inequalty and the Mahler Conjecture In ths secton we contnue the dscusson of the Mahler volume, (see Defnton 34) In partcular we consder the bodes that extremse the Mahler volume We shall prove the Blaschke- Santaló nequalty whch shows that ellpsods have maxmal Mahler volume and dscuss a conjecture on the bodes whch mnmse ths quantty Recall that the Mahler volume s nvarant under lnear transformatons Hence when we say for example, that ellpsods have maxmal Mahler volume, ths really does mean that the Mahler volume of any (centred) ellpsod s maxmal n the set of all symmetrc convex bodes Theorem 43 (Blaschke-Santaló) For any symmetrc convex body C, M(C) M(B 1 (0)) It s also known that equalty s only acheved n the case of an ellpsod Ths s dscussed n [15], where more general statements for non-symmetrc convex bodes are also proved For the proof, we wll follow the approach of Meyer and Pajor (see [14] or [15]) Pror to ths however, we need to dscuss an operaton known as Stener symmetrzaton, the ntuton for whch s to make a gven convex body more sphercal To begn wth recall that the Hausdorff metrc on K n (the set of non-empty compact subsets of R n ) s gven by d(a, B) = nf{ε > 0 : A B + B ε (0) and B A + B ε (0)} When we dscuss convergence of sets, ths s the metrc we wll use unless stated otherwse 7

29 Defnton 44 For a bounded measurable set S R n and a unt vector v S n 1 the Stener symmetrzaton st v (S) s gven by st v (S) = { [ x + v λ, λ ] } : λ = (x + vr) S, x P (S) where P s the orthogonal projecton onto v In partcular suppose S s convex We can thnk of S as the unon of lne-segments n the drecton v (one for each x P (S)) then st v (S) s the unon of translated copes of those segments such that each s bsected by v Fgure 5: Stener symmetrzaton We wll need the followng propertes of the Stener symmetrzaton (based on pages of [17] and [14]) Lemma 45 Let C, D be symmetrc convex bodes n R n such that C D, and let v S n 1 then we have the followng: 1 If C s a ball then st v (C) = C st v (C) st v (D) 3 st v (C) s a symmetrc convex body 4 st v (C) = C 5 st v (C) C Proof Let K be the mage of C under the orthogonal projecton onto v 1 & ) are obvous from the defnton 3) (Sketch) Snce C s bounded, (1) and () together mply that st v (C) s bounded Let f : K R be gven by x (x + vr) C Closedness follows from the contnuty of f (see Corollary 6 and thnk of st v (C) locally as a graph of 1 f) For symmetry, observe that K s symmetrc and f( x) = f(x) by symmetry of C In two dmensons, wth C a trapezum and v parallel to two of the sdes of C, st v (C) s stll a trapezum, hence convex To deduce convexty n general, let x, y st v (C), then the convex hull H = conv{[(y + vr) C] [(x + vr) C]} C s a trapezum wth parallel sdes n the drecton v Thus st v (C) st v (H), the latter beng a convex set contanng x and y as requred (see Fgure 6) 8

30 Fgure 6: Stener symmetrzaton preserves convexty Fnally, st v (C) has non-empty nteror snce 0 nt(c) thus 0 nt(st v (C)) by (1) 4) Preservaton of volume s a consequence of Fubn s Theorem 5) (from [14]) Assume, wthout loss of generalty, that v = e n then we can rewrte st v (C) and C as follows: C = { (Y, y) R n 1 R : Y, X + xy 1 for all X K and x C (X + vr)}, { st v (C) = (Y, y) R n 1 R : Y, X + y x 1 x 1 for all X K, } x 1, x C (X + vr) The second equaton follows from the observaton that 1 {x 1 x : x 1, x C (X + vr)} = st v (C) (X + vr) For t R, let H t denote the hyperplane v + tv = R n 1 {t} Then from the equatons above, we deduce that, for each t R H t st v (C) 1 (H t C + H t C ) = 1 (H t C H t C ) The equalty follows from the symmetry of C Now an applcaton of the Brünn-Mnkowsk nequalty shows that the rght hand sde of the above has volume at least H t C Therefore we have st v (C) = H t st v (C) dt H t C dt = C Ths s the requred result R Combnng (4) and (5) we see that Stener symmetrzaton does not decrease the Mahler volume of a convex body Ths s the frst of two observatons whch together gve Theorem 43 as a corollary The second s the followng (whch can be found n [17] pages ) Lemma 46 Let C be a symmetrc convex body n R n Then there exsts a sequence of vectors v 1, v, R n such that the sequence of convex bodes C k = st vk st vk 1 st v1 (C) converges to a ball B r (0) (where r s determned by C ) Ths wll be proved shortly We shall also use wthout proof the followng consequence of the Arzelà-Ascol theorem For a proof, see [17], pages R 9

31 Lemma 47 Let C 1, C, be a sequence of compact convex sets n R n, each contaned n some ball B r (0) (e a bounded sequence n some sense) Then the sequence has an accumulaton pont (wth respect to the Hausdorff metrc) whch s a compact convex set Corollary 48 Wth C 1, C as above, suppose addtonally that each C s a symmetrc convex body and C = C 1 Then the sequence has an accumulaton pont whch s also a symmetrc convex body Proof Let K be an accumulaton pont from the lemma To show symmetry, suppose for contradcton that there s ε (0, 1) and x K wth x := (ε 1)x K Let v S n 1 and t R be such that v + tv s the supportng hyperplane at x Then there exsts N > 0 such that k > N C k {y R n : y, v t + εt (1 ε) } Now by symmetry of C k we also have C k {y R n : y, v t + εt x (1 ε) }, but x, v = 1 ε, v = t 1 ε = t + εt 1 ε Hence x / C k + B εt (0) whch s a contadcton, as C k was supposed to converge to K (1 ε) For non-emptness of the nteror of K we suppose that K contans no ball around the orgn Passng to a subsequence f necessary, we may assume that for any ε > 0 there exsts N ε Z >0 such that f > N ε then C does not contan B ε (0) By symmetry and convexty, t would follow that, for > N ε, each C s contaned n some (symmetrc) plank of wdth at most ε (e there s a unt vector v such that y, v < ε for all y C ) Moreover, snce {C } s unformly bounded, we have > N ε C Dε for some D ndependent of and ε Now ε was arbtrary so we cannot have C = C 1 > 0 for all, whch s the requred contradcton We can now prove Lemma 46 The proof of Theorem 43 follows drectly Proof (of Lemma 46) For any non-empty compact set D K n, defne ρ(d) = nf{r > 0 : D B r (0)} Then wth C as n the statement, let F be the set of convex bodes whch can be obtaned from C by applyng fntely many Stener symmetrzatons and defne σ = nf{ρ(d) : D F } Snce C s bounded (and by (1) & () of Lemma 45) we have that any sequences n F are bounded (n the sense of Lemma 47) Thus by the corollary, there s a symmetrc convex body K and a sequence C 1, C, F such that C k K and ρ(c k ) σ By the defnton of the Hausdorff metrc t s easy to check that ρ s contnuous on K n and so ρ(k) = σ It now suffces to show that K = B σ (0) By defnton of ρ, we have K B σ (0) Suppose for contradcton that equalty does not hold By convexty of K we must have K B σ (0)\S where S s a sphercal cap (the ntersecton of B σ (0) and a halfspace) By compactness of B σ (0) there are fntely many unt vectors v 1,, v m so that B σ (0) s covered by the mages of S B under reflecton n the hyperplanes v From the defnton of st v, one can check that f T B σ (0)\K and T s the reflecton of T n v then st v (K) (T T ) = (see Fgure 7) It follows that st vm st v1 (K) B σ (0) = Hence ρ(k) = σ s not mnmal n F, ths s the requred contradcton Proof (of Theorem 43) By Lemma 46, gven a convex body C, there s a sequence of convex bodes C = C 1, C, C 3,, convergng to a ball, wth C k+1 = st vk (C k ) for some v k It s farly straghtforward to show that the Mahler volume s contnuous wth respect to the Hausdorff metrc (at least when deallng wth symmetrc convex bodes snce neghbourhoods are gven by dlatons) Therefore M(C k ) M(B σ (0)) = M(B 1 (0)) Furthermore, the sequence M(C k ) s ncreasng as a consequence of Lemma 45 It follows that M(C) M(B 1 (0)), as requred 30

32 Fgure 7: st v (K) contans less of the boundary of B σ (0) than K We conclude ths secton wth a bref dscusson of a problem whch s currently open, known as the Mahler conjecture We state t here n ts orgnal form (due to Mahler) Conjecture 49 (Mahler) For a symmetrc convex body C R n, M(C) M(Q n ) That s to say, the mnmum Mahler volume of a symmetrc convex body s attaned by the cube (and therefore by ts polar body- known as the cross-polytope) The (unt) cross-polytope s the unt ball n the l 1 norm on R n and t s easy to check that ths s the polar body of the cube (wth sde length ) We shall not dscuss attempts to prove ths conjecture n detal, but nstead pont out a few results For example, t s known that the conjecture holds true n the case of zonods (e sets that are the lmts of fnte sums of lne-segments) For a proof of ths see [4] One recent paper about ths problem (see [3]) shows that cubes are ndeed local mnmsers of the Mahler volume Here the word local s n the sense of the Banach-Mazur dstance, whch n the aforementoned paper s defned by { } b d BM (K, L) = nf : ak T (L) bk for some T GL(n), a for symmetrc convex bodes K and L Note that n ths form, the Banach-Mazur dstance s not a metrc but heurstcally behaves lke the mnmum Hausdorff dstance between mages of the gven sets under lnear maps It therefore seems lke a natural noton of dstance when consderng the Mahler volume Another recent developement uses a generalsaton of the Gauss curvature to show that a convex body wth mnmum Mahler volume must have a curvature of zero on almost all of ts boundary (see []) Ths at least ponts to the mnmsers beng polytopes References [1] FBarthe, Inégaltés de Brascamp-Leb et convexté, Comptes Rendus de l Académe des Scences 34 (1997), no 8, [], Optmal Young s Inequalty and ts Converse: a Smple Proof, Geometrc and Functonal Analyss 8 (1998), no, 34 4 [3] FNazarov, FPetrov, DRyabogn, and AZvavtch, A Remark on the Mahler Conjecture: Local Mnmalty of the Unt Cube, Duke Mathematcal Journal 154 (010), no 3, [4] GVEpfanov, Unversalty of the Sectons of Cubes, Mathematcal Notes (1967), no 1,