The modified complex Busemann-Petty problem on sections of convex bodies

a r X i v :0807.0776v 1 [m a t h .F A ] 4 J u l 2008THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES.MARISA ZYMONOPOULOUAbstract.The complex Busemann-Petty problem asks whether origin symmetric convex bodies in C n with smaller central hyperplane sections necessarily have smaller volume.The answer is affirmative if n ≤3and negative if n ≥4.Since the answer is negative in most dimensions,it is natural to ask what conditions on the (n −1)-dimensional volumes of the central sections of complex convex bodies with complex hyper-planes allow to compare the n -dimensional volumes.In this article we give necessary conditions on the section function in order to obtain an affirmative answer in all dimensions.The result is the complex analogue of [KYY].1.IntroductionThe Busemann-Petty problem was completely solved in the late 90’s as a result of a series of papers of many mathematicians ([LR],[Ba],[Gi],[Bo],[Lu],[Pa],[Ga],[Zh1],[K1],[K2],[Zh2],[GKS];see [K5,p.3]for the history of the solution).The problem asks the following:Suppose K and L are two origin symmetric convex bodies in R n such that for every ξ∈S n −1,Vol n −1 K ∩ξ⊥≤Vol n −1 L ∩ξ⊥ .Does it follow thatVol n K ≤Vol n L ?The problem has an affirmative answer only if n ≤4.Since the answer is negative in most dimensions,it is natural to ask what conditions on the (n −1)-dimensional volumes of central sections do allow to compare the n -dimensional volumes.Such conditions were found in [KYY].The result is as follows.For an origin symmetric convex body K in R n define the section functionS K (ξ)=Vol n −1(K ∩ξ⊥),ξ∈S n −1.Suppose K and L are origin symmetric convex smooth bodies in R n and α∈R with α≥n −4.Then,the inequality−∆α/2S K (ξ)≤ −∆ α/2S L (ξ),ξ∈S n −1implies that Vol n (K )≤Vol n (L ).If α<n −4this is not necessarily true.Here,∆is the Laplace operator on R n .12MARISA ZYMONOPOULOUIn this article we study the complex version of this problem.Forξ∈C n,|ξ|=1we denote byHξ={z∈C n:(z,ξ)=n k=1z kTHE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES3 Main Result.Suppose K and L are two origin symmetric invariant with respect to all Rθconvex bodies in R2n.Suppose thatα∈[2n−6,2n−2),n≥3.If−∆ α/2S CK(ξ)≤ −∆ α/2S CL(ξ),(3)for everyξ∈S2n−1.ThenVol2n(K)≤Vol2n(L).Ifα∈(2n−7,2n−6)then one can construct two convex bodies K and Lthat satisfy(3),but Vol2n(K)>Vol2n(L).This means that one needs to differentiate the section functions at least2n−6times and compare the derivatives in order to obtain the same inequal-ity for the volume of the original bodies.Note that ifα=0the problem coincides with the original complex Busemann-Petty problem.2.The Fourier analytic approachThroughout this paper we use the Fourier transform of distributions.The Schwartz class of the rapidly decreasing infinitely differentiable functions(test functions)in R n is denoted by S(R n),and the space of distributionsover S(R n)by S′(R n).The Fourier transformˆf of a distribution f∈S′(R n)is defined by ˆf,φ = f,ˆφ for every test functionφ.A distribution is called even homogeneous of degree p∈R if f(x),φ(x/α) =|α|n+p f,φ for everytest functionφand everyα∈R,α=0.The Fourier transform of an even homogeneous distribution of degree p is an even homogeneous distributionof degree−n−p.A distribution f is called positive definite if,for every test functionφ, f,φ∗(2π)n|x|α2ˆf(x),(4)where the Fourier transform is considered in the sense of distributions.A compact set K⊂R n is called a star body,if every straight line that passes through the origin crosses the boundary of the set at exactly two points and the boundary of K is continuous in the sense that the Minkowski functional of K,defined byx K=min{α≥0:x∈αK}is a continuous function on R ing polar coordinates it is possible to obtain the following polar formula of the volume of the body:4MARISA ZYMONOPOULOUVol n(K)= R nχ( x K)dx=1THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES5In addition,if we consider the action of the distribution|u|−q−22/Γ(−q/2)on A D,H,p we may apply a standard regularization argument(see[GS,p.71-74])and define the functionq−→ |u|−q−222),A D,H,p(u) .(6)For q∈C with Re q≤2n−p−3,the function is an analytic function of q.If q<0|u|−q−222),A D,H,p(u) =1Γ(−q22d d!∆d A f,D,H(0),(8) where∆= 2i=1∂2/∂u2i is the2-dimensional Laplace operator(see[GS, p.71-74]).Note that the function(6)is equal,up to a constant,to the fractional power of∆q/2A D,H,p.(see[KKZ,p.6-7]or[K4,p.6-7]for complete definition).If the body D is origin symmetric the function A D,H,p is even and for0< q<2we have(see also[K5,p.39])|u|−q−222),A D,H,p(u)=12) 2π0∞0A D,H,p(tθ)−A D,H,p(0)2=Γ(−q/2)Γ(−q πΓ q+26MARISA ZYMONOPOULOUAlso,for every d ∈N ∪0,d <n −1∆dA D,H,p (0)=(−1)dΓ(−qΓ(−q/2)R 2|u |−q −22A D,H,p (u )duUsing the expression (5)for the function A D,H,p ,writing the integral in polarcoordinates and then using (10),we see that the right-hand side of the latter equation is equal to12)Rn(x,e 1)2+(x,e 2)2)−q −2Γ(−q2θ −n +q +2Ddθ=12)π12Γ(−q −12(n −q −p −2)×2π0S n −1θ −n +q +p +2D(u 1e 1+u 2e 2,θ)−q −2dθdu.(13)Let us show that the function under the integral over [0,2π]is the Fouriertransform of x −n +q +p +2D |x |−p2at the point u 1e 1+u 2e 2.For any even test function φ∈S (R n ),using the well-known connection between the Fourier and Radon transforms (see [K5,p.27])and the expression for the Fouriertransform of the distribution |z |q −12(see [K5,p.38]),we get( x −n +q +p +2D |x |−p 2)∧,φ =R nx −n +q +p +2D |x |−p 2ˆφ(x )dx =S n −1θ −n +q +p +2D ∞r q +1ˆφ(rθ)dr dθ=1πΓ((q +2)/2)πΓ((q +2)/2)THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES 7Since φis an arbitrary test function,this proves that,for every y ∈R n \{0},x −n +q +p +2D |x |−p 2∧(y )=2q +2√2Γ((−q −1)/2)S n −1|(θ,y )|−q −2 θ −n +q +p +2D dθ.Together with (13),the latter equality shows that|u |−q −22Γ((q +2)/2)(n −q −p −2)S n −1∩H ⊥x −n +q +p +2D |x |−p 2 ∧(θ)dθ,(14)because in our notation S n −1∩H ⊥=[0,2π].We have proved (14)under the assumption that q ∈(−2,−1).However,both sides of (14)are analytic functions of q ∈C in the domain where −2<Re q <2n −2.This implies that the equality (14)holds for every q from this domain (see [K5,p.61]for the details of a similar argument).Putting q =2m,m ∈N ∪{0},m <n −1in (14)and applying (8)and the fact that Γ(x +1)=x Γ(x ),we get the second formula. Brunn’s theorem (see for example [K5,Theorem 2.3])states that for an origin symmetric convex body and a fixed direction,the central hyperplane section has the maximal volume among all the hyperplane sections per-pendicular to the given direction.As a consequence we have the following generalization proved in [KKZ,Lemma 1]for p =0.Proposition 3.Suppose D is a 2-smooth origin symmetric convex body in R 2n ,then the function A D,H,p is twice differentiable at the origin and∆A D,H,p (0)≤0.Moreover,for any q ∈(0,2),|u |−q −222),A D,H,p (u ) ≥0.Proof.Differentiability follows from the same argument as in [K5,Lemma2.4].The body D is origin symmetric and convex,so,to prove the first inequal-ity we need to observe that the function u −→A D,H,p (u ),u ∈R 2,attains its maximum at the origin:If p =0then it follows immediately from Brunn’s theorem (see [K5,Theorem 2.3]and [KKZ,Lemma 1].)8MARISA ZYMONOPOULOULet p>0.Since|x|−p2=p ∞0χ(z|x|2)z q−1dz,we have that for any u∈R2 A D,H,p(u)= D∩H u|x|−p2dx=p D∩H u ∞0χ(z|x|2)z q−1dzdx=p ∞0z q−1 D∩H uχ(z|x|2)dxdz=p ∞0z q−1 B(1/z)∩H uχ( x D)dxdz,where B(1/z)is the unit ball of radius1/z,.Applying Brunn’s theorem to the body B(1/z)∩D,we have that the latter integral is≤p ∞0z q−1 Hχ( x B(1/z)∩D)dxdz=A D,H,p(0).If q∈(0,2)thenΓ(−q/2)<0.Hence,for the second inequality we use(9)to get that |u|−q−222),A D,H,p(u)=12) 2π0∞0A D,H,p(tθ)−A D,H,p(0)THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES 9Lemma 2.Let D be an origin symmetric invariant with respect to all R θconvex body in R 2n ,n ≥3.If q ∈(−2,2]and 0≤p <2n −q −3then|x |−p 2 x −2n +p +q +2D is a positive definite distribution.Proof.If p =0then by [KKZ,Theorem 3], x −2n +q +2D∧≥0,since 2n −q −2∈[2n −4,2n ).Let p >0.If q ∈(−2,0)then by equation (7)and Proposition 2(formula (11))we have that2−q −22(2n −q −p −2) S 2n −1∩H ⊥x −2n +q +p +2D |x |−p2 ∧(θ)dθ=12)>0,Γ(−q4π(n −1)|x |−p 2 x −2n +p +q +2D∧(ξ)≥0.For the case where q ∈(0,2)we use Proposition 2and the Remark to getthat |u |−q −222),A D,H,p (u )=2−q −12(2n −q −p −2) x −2n +q +p +2D |x |−p2 ∧(ξ).Then,by the generalization of Brunn’s theorem,Proposition 3,the desired follows.Lastly,if q =2,(12)and (15)imply that∆A D,H,p (0)=−110MARISA ZYMONOPOULOUBefore we prove the main result of this article we need the following:Lemma3.Let D be an infinitely smooth origin symmetric invariant with respect to all Rθconvex body in R2n andα∈R.Then−∆ α/2S CD(ξ)=14π(n−1) x −2n+2D ∧(ξ).(17) By the definition of the section function of D,and equation(17)we obtain the following formula:1S CD(ξ)=THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES 11in (20)by |x |−α2 x −2K∧and integrate over the unit sphere S 2n −1.Then we can apply Parseval’s spherical version,Proposition 1,to get thatS 2n −1x −2nK dx ≤S 2n −1x −2K x −2n +2L .(21)Then,by a simple application of H¨o lder’s inequality on formula (21)and the polar formula of the bodies (see Section 2.)we obtain the affirmative answer to the problem,since2n Vol 2n (K )≤ 2n Vol 2n (K ) 1/n 2n Vol 2n (L ) (n −1)/n.To prove the negative part we need the following lemma.Lemma 4.Let α∈(2n −7,2n −6).There exists an infinitely smooth originsymmetric convex body L with positive curvature,so that |x |−α2 x −2Lis not a positive definite distribution.We postpone the proof of Lemma 4until the end of this section to showthat the existence of such a body gives a negative answer to the problem.Theorem 2.(NEGATIVE PART)Suppose there exists an infinitely smooth,origin symmetric convex body L for which |x |−α2 x −2L is not a positive definite distribution.Then one can construct an origin symmetric convex body K in R 2n ,n ≥3,so that together with L they satisfy (19),for every ξ∈S 2n −1butVol 2n (K )>Vol 2n (L ).Proof.The body L is infinitely smooth,so by [K5,Lemma 3.16]the Fouriertransform of the distribution |x |−α2 x −2L is a continuous function on the unit sphere S 2n −1.Moreover there exists an open subset Ωof S 2n −1in which|x |−α2 x −2L∧<0.Since L is invariant with respect to all R θwe may assume that Ωis also invariant we respect to rotations R θ.We use a standard perturbation procedure for convex bodies,see for ex-ample [K5,p.96](similar argument was used in [KKZ,Lemma 5]).Consider a non-negative infinitely differentiable even function g supported on Ωthat is also invariant with respect to rotations R θ.We extend it to a homogeneous function of degree −α−2on R 2n .By [K5,Lemma 3.16]its Fourier trans-form is an even homogeneous function of degree −2n +α+2on R 2n ,whose restriction to the sphere is infinitely smooth: g (x/|x |2)|x |−α−22∧(y )=h (y/|y |2)|y |−2n +α+22,where h ∈C ∞(S 2n −1).We define a body K so thatx −2n +2K= x −2n +2L +ε|x |−2n +22h x12MARISA ZYMONOPOULOUfor small enough ε>0so that the body K is strictly convex.Note that K isalso invariant with respect to all R θ.We multiply both sides by14π(n −1)|x |−α−22g x4π(n −1)S 2n −1|x |−α2 x −2L∧(θ)g (θ)dθ>S 2n −1|x |−α2 x −2L ∧(θ) −∆ α/2S CL (θ)dθ,since|x |−α2 x −2L∧<0on the support of ing equation (16)and the spherical version of Parseval’s identity,the latter becomesS 2n −1x −2L x −2n +2K> S 2n −1x −2n L dx.As in Theorem 1we apply H¨o lder’s inequality and the polar representationof the volume to obtain the desired inequality for the volumes of the bodies.Proof of Lemma 4.The construction of the body follows similar steps as in [KYY].We put q =2n −α−4,so q ∈(2,3).From the definition of the fractional derivatives,Proposition 2and the Remark we see that for a ξ∈S 2n −1we need to construct a convex body D so that2π∞0t−q −1A D,H ξ,α(tθ)−A D,H ξ,α(0)−∆A D,H ξ,α(0)t 22)>0for q ∈(2,3).We define the functionf (|u |)=(1−|u |22−N |u |42)1THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES 13n−1 i =1j =1,2x 2ij1/2≤f (|¯x |2),where a N is the first positive root of the equation f (t )=0.From its defini-tion,the body D is strictly convex with an infinitely smooth boundary.Wechoose ξ∈S 2n −1in the direction of ¯x .For u ∈R 2with |u |2∈[0,a N ],we write equation (5)in polar coordinates and get thatA D,H ξ,α(u )= S 2n −3uf (|u |2)(r 2+|u |22)−α2r 2n −3dr,where |S 2n −3u|is the volume of the (2n −3)-dimensional unit sphere.Note that if |u |2>αN then A D,H ξ,α(u )=0.Moreover,if u =tθ,t ∈[0,∞)and θ∈S 1,the parallel section function A D,H ξ,α(tθ)is independent of θsinceA D,H ξ,α(tθ)=|S 2n −3t| f (t )(r 2+t 2)−α2dt <0.(24)Note that the condition |u |2∈[0,αN ]is now equivalent to t ∈[0,αN ].Inorder to prove the above we first computeA D,H ξ,α(0)=|S 2n −3|2n −α−2+α4.Also,for every t ∈[0,αN ],f (t )>0and f (t )≥t if and only if t ∈[0,βN ].For the first part,the interval [0,βN ],since f (t )≥t ,the 2-dimensional parallel section function A D,H ξ,αcan be easily estimated if we split it into two integrals.For the second we use the inequality (1+x )−γ≤1−γx +γ(γ+1)2r2n −3dr ≤tr −α+2n −3dr =t14MARISA ZYMONOPOULOUandf(t)t(r2+t2)−α2t24t42n−α−2−α2n−α−4+α(α+2)2n−α−6f(t)t=f2n−α−2(t)2t2f2n−α−4(t)4t4f2n−α−6(t)2n−α−2−p4(2n−α−6)>0,since n≥3andα∈(2n−7,2n−6).We now use the definition of the function f and the inequality(1−x)γ≥1−γx(1−x)γ−1,for0<γ<1and0<x<1.We then write=1−t2−Nt42t2(1−t2−Nt4)2n−α−42n−α−4+α(α+2)2n−α−22n−α−2−αt22(2n−α−2)(1−t2−Nt4)24(2n−α−6)−α(α+2)(1−t2−Nt4)42 dt= βN0t−q−1 Ct2n−α−2−Dt4+E t2(t2+Nt4)2n−α−2−F t4(t2+Nt4)2n−α−2 dt,(25)where E=α2n−α−2>0and D=N4(2n−α−6)>0,for N large enough.Now,in order to obtain an upper bound for(25)we need to estimate four different integrals.Thefirst one simply gives C2,and theTHE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES15second D4,for large N.For the third one,we make a change of variables,u=N1(1−t2−Nt4)24βN N12+u2)2−u4)2 4,sinceβN N1(1−u4)24t,for the last integral and find that it is comparable to N q(1−t2−Nt4)44−1βN N12+u4)2−u4)44−→1as N→∞.So,we can roughly bound the integral from below by a pos-itive constant and have that equation(26)is greater than F1N q2 dt ≤ αNβN t−q−1 12(2n−α−4) t2dt<A αNβN t−q−1dt. Recall thatαN andβN are the positive solutions of the equations f(t)=0 and1−t2−Nt4=t q+1respectively,and that for large N,αN≃N−1(αN+βN)(1+N(α2N+β2N))≃AN−12 dt= ∞αN −t−q−12n−α−2+α2 dt=−A1α−q N+A2α−q+2≃−A1N q4,16MARISA ZYMONOPOULOUwhere A 1,A 2>0.Combining all the above estimations,for N large enough,we obtain the following upper bound for the integral in (24),∞t−q −1A D,H ξ,α(tθ)−A D,H ξ,α(0)−∆A D,H ξ,α(0)t 22+D 1Nq −44−F 1Nq4−A 1Nq4,which clearly shows that it is negative since all the constants are positive and q ∈(2,3).Acknowledgments:The author was partially supported by the NSF grant DMS-0652571.Also,part of this work was carried out when the author was visiting the Department of Mathematics of University of Crete,which the author thanks for its hospitality.References[Ba]K.Ball,Some remarks on the geometry of convex sets ,Geometric aspects of functional analysis (1986/87),Lecture Notes in Math.1317,Springer-Verlag,Berlin-Heindelberg-New York,1988,224–231.[Bo]J.Bourgain,On the Busemann-Petty problem for perturbations of the ball ,Geom.Funct.Anal.1(1991),1–13.[BP]H.Busemann and C.M.Petty,Problems on convex bodies ,Math.Scand.4(1956),88–94.[Ga1]R.J.Gardner,Intersection bodies and the Busemann-Petty problem ,Trans.Amer.Soc.342(1994),435–445.[Ga2]R.J.Gardner,A positive answer to the Busemann-Petty problem in three dimen-sions ,Annals of Math.140(1994),435–447.[Ga3]R.J.Gardner,Geometric tomography ,Cambridge University Press,New York,1995.[GKS]R.J.Gardner,A.Koldobsky and T.Schlumprecht,An analytic solution to the Busemann-Petty problem on sections of convex bodies ,Annals of Math.149(1999),691–703.[GS]I.M.Gelfand and G.E.Shilov,Generalized function,vol.1.Properties and oper-ations ,Academic Press,New York,(1964).[Gi] A.Giannopoulos,A note on a problem of H.Busemann and C.M.Petty concern-ing sections of symmetric convex bodies ,Mathematika 37(1990),239–244.[KKZ]A.Koldobsky,H.K¨o nig and M.Zymonopoulou,The complex Busemann-Petty problem on sections of convex bodies ,Advances in Mathematics,218(2008),352–367.[KYY] A.Koldobsky,V.Yaskin and M.Yaskina,Modified Busemann-Petty problem on sections of convex bodies ,Israel J.Math.154(2006),191–207.[K1] A.Koldobsky,An application of the Fourier transform to sections of star bodies ,Israel J.Math.106(1998),157–164.[K2] A.Koldobsky,Intersection bodies,positive definite distributions and the Busemann-Petty problem ,Amer.J.Math.120(1998),827–840.[K3]A.Koldobsky,A generalization of the Busemann-Petty problem on sections of convex bodies ,Israel J.Math.110(1990),75–91.THE MODIFIED COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES17 [K4] A.Koldobsky,A functional analytic approach to intersection bodies,Geom.Funct.Anal.,10(2000),no.4,1507–1526.[K5] A.Koldobsky,Fourier analysis in convex geometry,Amer.Math.Soc.,Providence RI,2005.[LR] rman and C.A.Rogers,The existence of a centrally symmetric convex body with central sections that are unexpectedly small,Mathematika22(1975),164–175.[Lu] E.Lutwak,Intersection bodies and dual mixed volumes,Adv.Math.71(1988), 232–261.[Pa]M.Papadimitrakis,On the Busemann-Petty problem about convex,centrally symmetric convex bodies in R n,Mathematika39(1992),258–266.[Sch]R.Schneider,Convex:the Brunn-Minkowski theory,Cambridge University Press,Cambridge,1993.[Zh1]G.Zhang,Intersection bodies and Busemann-Petty inequalities in R4,Annals of Math.140(1994),331–346.[Zh2]G.Zhang,A positive answer to the Busemann-Petty problem in four dimensions, Annals of Math.149(1999),535–543.Marisa Zymonopoulou,Department of Mathematics,University of Missouri, Columbia,MO65211,USAE-mail address:marisa@@。