THE COMPLEX BUSEMANN-PETTY PROBLEM ON SECTIONS OF CONVEX BODIES

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1 THE COMPEX BUSEMANN-PETTY PROBEM ON SECTIONS OF CONVEX BOIES A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU Abstract. The complex Busemann-Petty problem asks whether origin symmetric convex bodies in C n with smaller central hyperplane sections necessarily have smaller volume. We prove that the answer is affirmative if n 3 and negative if n Introduction The Busemann-Petty problem, posed in 1956 (see [BP]), asks the following question. Suppose that and are origin symmetric convex bodies in R n such that Vol n 1 ( H) Vol n 1 ( H) for every central hyperplane H in R n. oes it follow that Vol n () Vol n ()? The answer is affirmative if n 4 and negative if n 5. The solution was completed in the end of the 90 s as the result of a sequence of papers [R], [Ba], [Gi], [Bo], [u], [Pa], [Ga1], [Ga], [Zh1], [1], [], [Zh], [GS] ; see [10, p. 3] for the history of the solution. In this article we consider the complex version of the problem. For ξ C n, ξ = 1, denote by H ξ = {z C n : (z, ξ) = n z k ξ k = 0} the complex hyperplane perpendicular to ξ. Origin symmetric convex bodies in C n are the unit balls of norms on C n. We denote by the norm corresponding to the body : k=1 = {z C n : z 1}. In order to define volume, we identify C n with R n using the mapping ξ = (ξ 1,..., ξ n ) = (ξ 11 + iξ 1,..., ξ n1 + iξ n ) (ξ 11, ξ 1,..., ξ n1, ξ n ). 1

2 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU Under this mapping the hyperplane H ξ turns into a two-codimensional subspace of R n orthogonal to the vectors ξ = (ξ 11, ξ 1,..., ξ n1, ξ n ) and ξ = ( ξ 1, ξ 11,..., ξ n, ξ n1 ). Since norms on C n satisfy the equality λz = λ z, z C n, λ C, origin symmetric complex convex bodies correspond to those origin symmetric convex bodies in R n that are invariant with respect to any coordinate-wise two-dimensional rotation, namely for each θ [0, π] and each ξ = (ξ 11, ξ 1,..., ξ n1, ξ n ) R n ξ = R θ (ξ 11, ξ 1 ),..., R θ (ξ n1, ξ n ), (1) where R θ stands for the counterclockwise rotation of R by the angle θ with respect to the origin. We shall simply say that is invariant with respect to all R θ if it satisfies the equations (1). Now the complex Busemann-Petty problem can be formulated as follows: suppose and are origin symmetric invariant with respect to all R θ convex bodies in R n such that Vol n ( H ξ ) Vol n ( H ξ ) for each ξ from the unit sphere S n 1 of R n. oes it follow that Vol n () Vol n ()? This formulation reminds of the lower-dimensional Busemann-Petty problem, where one tries to deduce the inequality for n-dimensional volumes of arbitrary origin-symmetric convex bodies from the inequalities for volumes of all (n )-dimensional sections. In the case where n = this amounts to considering two-dimensional sections of fourdimensional bodies, where the answer to the lower dimensional problem is affirmative by the solution to the original Busemann-Petty problem - we first get inequalities for the volumes of all three-dimensional sections and then the inequality for the four-dimensional volumes. However, if n = 3 we get four-dimensional sections of six-dimensional bodies, where the answer to the lower-dimensional problem is negative by a result of Bourgain and Zhang [BZ]. Our problem is different from the lowerdimensional Busemann-Petty problem in two aspects. First, we do not have all (n )-dimensional sections, we only have sections by subspaces coming from complex hyperplanes, which makes the situation worse than for the lower-dimensional problem. Secondly, we consider only those convex bodies in R n that are invariant with respect to all R θ, and we may be able to convert this invariance into affirmative answer in some higher dimensions.

3 COMPEX BUSEMANN-PETTY PROBEM 3 The latter appears to be the case, as we prove below that the answer to the complex Busemann-Petty problem is affirmative if n 3 and negative if n 4. In 1988 utwak [u] introduced the class of intersection bodies and found a connection between this class and the real Busemann-Petty problem, which played an important role in the solution of the problem. It appears that the complex Busemann-Petty problem is closely related to the class of -intersection bodies introduced in [5, 8], namely the answer to the problem is affirmative if and only if every origin symmetric invariant with respect to all R θ convex body in R n is a - intersection body. We shall prove this connection in Theorem. After that we prove that every origin symmetric invariant with respect to all R θ convex body in R n is a (n 4)-intersection body, but not every such body is a (n 6)-intersection body. Putting n = 3 and then n = 4, one can see how these results imply the solution of the complex Busemann-Petty problem. Our proofs use several results from the recently developed Fourier analytic approach to sections of convex bodies; see [10]. In Section, we collect necessary definitions and results related to this approach. For other results related to the Busemann-Petty problem see [BZ], [BFM], [5], [9], [YY], [Mi], [Ru], [RZ], [Y1], [Y], [Zv1], [Zv].. Elements of the Fourier approach to sections Our main tool is the Fourier transform of distributions. As usual, we denote by S(R n ) the Schwartz space of rapidly decreasing infinitely differentiable functions (test functions) in R n, and S (R n ) is the space of distributions over S(R n ). The Fourier transform ˆf of a distribution f S (R n ) is defined by ˆf, φ = f, ˆφ for every test function φ. A distribution is called even homogeneous of degree p R if f(x), φ(x/α) = α n+p f, φ for every test function φ and every α R, α 0. The Fourier transform of an even homogeneous distribution of degree p is an even homogeneous distribution of degree n p. A distribution f is called positive definite if, for every test function φ, f, φ φ( x) 0. This is equivalent to ˆf being a positive distribution in the sense that ˆf, φ 0 for every non-negative test function φ. A compact set in R n is called a star body if every straight line through the origin crosses the boundary at exactly two points different from the origin, and the boundary of is continuous in the sense that the Minkowski functional of defined by = min{a 0 : x a}

4 4 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU is a continuous function on R n. If in addition is origin symmetric and convex, then the Minkowski functional is a norm on R n. If ξ S n 1, then ρ (ξ) = ξ 1 is the radius of in the direction ξ. A simple calculation in polar coordinates gives the following polar formula for the volume: n Vol n () = n χ( ) dx = R n ξ n S n 1 dξ, where χ is the indicator function of the interval [0, 1]. We say that a star body in R n is k-smooth (infinitely smooth) if the restriction of to the sphere S n 1 belongs to the class C k (S n 1 ) (C (S n 1 )) of k times continuously differentiable (infinitely differentiable) functions on the sphere. It is well-known that one can approximate any convex body in R n in the radial metric d(, ) = sup ρ (ξ) ρ (ξ) ξ S n 1 by a sequence of infinitely smooth convex bodies. This can be proved by a simple convolution argument (see for example [Sch, Th ]). It is also easy to see that any convex body in R n invariant with respect to all R θ can be approximated in the radial metric by a sequence of infinitely smooth convex bodies invariant with respect to all R θ. This follows from the same convolution argument, because invariance with respect to R θ is preserved under convolutions. As proved in [10, emma 3.16], if is an infinitely smooth origin symmetric star body in R n and 0 < p < n then the Fourier transform of the distribution p is a homogeneous function of degree n + p on R n, whose restriction to the sphere is infinitely smooth. We use a version of Parseval s formula on the sphere established in [5] (see also [10, emma 3.]): Proposition 1. et and be infinitely smooth origin symmetric star bodies in R n and 0 < p < n. Then p (ξ) n+p (ξ) dξ = (π) n p n+p dx. S n 1 S n 1 The classes of k-intersection bodies were introduced in [5], [8] as follows. et 1 k < n, and let and be origin symmetric star bodies in R n. We say that is a k-intersection body of if for every (n k)-dimensional subspace H of R n Vol k ( H ) = Vol n k ( H).

5 COMPEX BUSEMANN-PETTY PROBEM 5 More generally, we say that an origin symmetric star body in R n is a k-intersection body if there exists a finite Borel measure µ on S n 1 so that for every even test function φ S(R n ), ( ) k φ(x) dx = t k 1 ˆφ(tξ) dt dµ(ξ). R n S n 1 0 Note that k-intersection bodies of star bodies are those k-intersection bodies for which the measure µ has a continuous strictly positive density; see [8] or [10, p. 77]. When k = 1 we get the class of intersection bodies introduced by utwak in [u]. A more general concept of embedding in p was introduced in [7]. et be an origin symmetric star body in R n, and X = (R n, ). For 0 < p < n, we say that X embeds in p if there exists a finite Borel measure µ on S n 1 so that, for every even test function φ R n p φ(x) dx = S n 1 ( ) z p 1 ˆφ(zθ) dz dµ(θ). R Obviously, an origin symmetric star body in R n is a k-intersection body if and only if the space (R n, ) embeds in k. In this article we use embeddings in p only to state some results in continuous form; for more applications of this concept, see [10, Ch. 6]. Embeddings in p and k-intersection bodies admit a Fourier analytic characterization that we are going to use throughout this article: Proposition. ([8], [10, Th. 6.16]) et be an origin symmetric star body in R n, 0 < p < n. The space (R n, ) embeds in p if and only if the function p represents a positive definite distribution on R n. In particular, is a k-intersection body if and only if k is a positive definite distribution on R n. It was proved in [6] (see also [10, Corollary 4.9]) that every n- dimensional normed space embeds in p for each p [n 3, n). In particular, every origin symmetric convex body in R n is a k-intersection body for k = n 3, n, n 1. On the other hand, the spaces l n q, q > do not embed in p if 0 < p < n 3, hence, the unit balls of these spaces are not k-intersection bodies if k < n 3; see [3], [10, Theorem 4.13]. We are going to use a generalization of the latter result, the so-called second derivative test for k-intersection bodies and embeddings in p, which was first proved for intersection bodies in [4] and then generalized in [10, Theorems 4.19, 4.1]. Recall that for normed spaces X and Y and q R, q 1, the q-sum (X Y ) q of X and Y is defined as the space of pairs {(x, y) : x X, y Y } with

6 6 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU the norm (x, y) = ( q X + y q Y )1/q. Proposition 3. et n 3, k N {0}, q > and let Y be a finite dimensional normed space of dimension greater or equal to n. Then the q-sum of R and Y does not embed in p with 0 < p < n. In particular, this direct sum is not a k-intersection body for any 1 k < n. et 1 k < n and let H be an (n k)-dimensional subspace of R n. Fix any orthonormal basis e 1,..., e k in the orthogonal subspace H. For a convex body in R n, define the (n k)-dimensional parallel section function A,H as a function on R k such that A,H (u) = Vol n k ( {H + u 1 e u k e k }) = χ( ) dx, u R k. () {x R n :(x,e 1 )=u 1,...,(x,e k )=u k } et be the Euclidean norm on R k. For every q C, the value of the distribution u q k /Γ( q/) on a test function φ S(R k ) can be defined in the usual way (see [GS, p.71]) and represents an entire function of q C. If is infinitely smooth, the function A,H is infinitely differentiable at the origin (see [10, emma.4]), and the same regularization procedure can be applied to define the action of these distributions on the function A,H. The function u q k q Γ( q/), A,H(u) (3) is an entire function of q C. In particular, if q < 0 u q k Γ( q/), A 1,H(u) = u q k A,H (u) du. Γ( q/) R k If q = m, m N {0}, then u q k, A,H (u) Γ( q/) q=m = ( 1) m S k 1 m+1 k(k + )...(k + m ) m A,H (0), (4) where S k 1 = π k/ /Γ(k/) is the surface area of the unit sphere S k 1 in R k, and = k i=1 / u i is the k-dimensional aplace operator (for details, see [GS, p.71-74]). Since the body is origin-symmetric, the

7 COMPEX BUSEMANN-PETTY PROBEM 7 function A,H is even, and for 0 < q < we have (see also [10, p. 49]) u q k Γ( q/), A,H(u) ( 1 ) A,H (tθ) A,H (0) = dt dθ. (5) Γ( q/) S n 1 0 t 1+q Note that the function (3) is equal (up to a constant) to the fractional power of the aplacian q/ A,H. The following proposition was proved in [8, Th. ]. We reproduce the proof here for the sake of completeness. We use a well-known formula (see for example [GS, p. 76]): for any v R k and q < k + 1, (v vk) ( q k)/ Γ( q/) = (v, u) q k du. (6) Γ(( q k + 1)/)π (k 1)/ S k 1 Proposition 4. et be an infinitely smooth origin symmetric convex body in R n and 1 k < n. Then for every (n k)-dimensional subspace H of R n and any q R, k < q < n k, u q k Γ( q/), A,H(u) q k π k/ = n+q+k (θ) dθ. (7) Γ((q + k)/)(n q k) S n 1 H Also for every m N {0}, m < (n k)/, m A,H (0) = ( 1) m k π k (n m k) where, as before, is the aplacian on R k. ( n+m+k S n 1 H ) (η) dη, (8) Proof : et first q ( k, k + 1). Then, u q k Γ( q/), A 1,H(u) = u q k A,H (u) du. Γ( q/) R k Using the expression () for the function A,H, writing the integral in polar coordinates and then using (6), we see that the right-hand side of the latter equation is equal to 1 ( (x, Γ( q ) e1 ) (x, e k ) ) ( q k)/ χ( ) dx = R n

8 8 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU 1 Γ( q )(n q k) θ n+q+k S n 1 S k 1 ( S n 1 ( (θ, e1 ) (θ, e k ) ) ( q k)/ θ n+q+k dθ = Γ( q k+1 1 )π k 1 (n q k) ( k ( u i e i, θ) ) q k du dθ = S k 1 i=1 1 Γ( q k+1 )π k 1 (n q k) θ n+q+k S n 1 k ( u i e i, θ) ) q k dθ du. (9) et us show that the function under the integral over S k 1 is the Fourier transform of n+q+k at the point u i e i. For any even test function φ S(R n ), using the well-known connection between the Fourier and Radon transforms (see [10, p. 7]) and the expression for the Fourier transform of the distribution z q+k 1 (see [10, p. 38]), we get i=1 ( n+q+k ), φ = n+q+k, ˆφ = ( θ n+q+k S n 1 1 q+k π Γ((q + k)/) Γ(( q k + 1)/) 0 θ n+q+k S n 1 θ n+q+k S n 1 ( n+q+k R n z q+k 1 ˆφ(zθ) dz ) dθ = z q+k 1, ˆφ(zθ) dθ = ˆφ(x) dx = t q k, φ(y) dy dθ = (y,θ)=t q+k πγ((q + k)/) ) (θ, y) q k θ n+q+k dθ φ(y) dy. Γ(( q k + 1)/) R n S n 1 Since φ is an arbitrary test function, this proves that, for every y R n \ {0}, n+q+k (y) q+k πγ((q + k)/) = dθ. Γ(( q k + 1)/) Together with (9), the latter equality shows that u q k Γ( q/), A,H(u) (θ, y) q k θ n+q+k S n 1 (10)

9 COMPEX BUSEMANN-PETTY PROBEM 9 q k π k/ = n+q+k (θ) dθ, Γ((q + k)/)(n q k) S n 1 H because in our notation S k 1 = S n 1 H. We have proved (10) under the assumption that q ( k, k + 1). However, both sides of (10) are analytic functions of q C in the domain where k < Re(q) < n k. This implies that the equality (10) holds for every q from this domain (see [10, p. 61] for the details of a similar argument). Putting q = m, m N {0}, m < (n k)/ in (10) and applying (4) and the fact that Γ(x + 1) = xγ(x), we get the second formula. Brunn s theorem (see for example [10, Th..3]) states that the central hyperplane section of an origin symmetric convex body has maximal (n 1)-dimensional volume among all hyperplane sections perpendicular to a given direction. This implies the following emma 1. If is a -smooth origin symmetric convex body in R n, then the function A,H is twice differentiable at the origin and A,H (0) 0. Besides that for any q (0, ), u q k Γ( q/), A,H(u) 0. Proof : ifferentiability follows from [10, emma.4]. Applying Brunn s theorem to the bodies span(h, θ), θ S n 1 H, we see that the function t A,H (tθ) has maximum at zero. Therefore, the interior integral in (5) is negative, but Γ( q/) < 0 for q (0, ), which implies the second statement. The first inequality also follows from the fact that each of the functions t A,H (te j ), j = 1,..., k has maximum at the origin. We often use emma 4.10 from [10] for the purpose of approximation by infinitely smooth bodies. For convenience, let us formulate this lemma: emma. ([10, emma 4.10]) et 1 k < n. Suppose that is an origin-symmetric convex body in R n that is not a k-intersection body. Then there exists a sequence m of origin-symmetric convex bodies so that m converges to in the radial metric, each m is infinitely smooth, has strictly positive curvature and each m is not a k-intersection body.

10 10 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU If in addition is invariant with respect to R θ, one can choose m with the same property. 3. Connection with intersection bodies We now return to the complex case. The following simple observation is crucial for applications of the Fourier methods to convex bodies in the complex case: emma 3. Suppose that is an origin-symmetric infinitely smooth invariant with respect to all R θ star body in R n. Then for every 0 < p < n and ξ S n 1 the Fourier transform of the distribution p is a constant function on S n 1 Hξ. Proof : By [10, emma 3.16], the Fourier transform of p is a continuous function outside of the origin in R n. The function is invariant with respect to all R θ, so by the connection between the Fourier transform of distributions and linear transformations, the Fourier transform of p is also invariant with respect to all R θ. Recall that the two-dimensional space Hξ is spanned by vectors ξ and ξ (see the Introduction). Every vector in S n 1 Hξ is the image of ξ under one of the coordinate-wise rotations R θ, so the Fourier transform of p is a constant function on Sn 1 Hξ. Of course, this argument also applies to the Fourier transform of any distribution of the form h( ). Similarly to the real case (see [1], [10, Theorem 3.8]), one can express the volume of hyperplane sections in terms of the Fourier transform. Theorem 1. et be an infinitely smooth origin symmetric invariant with respect to R θ convex body in R n, n. For every ξ S n 1, we have Vol n ( H ξ ) = 1 4π(n 1) n+ (ξ). Proof : et us fix ξ S n 1. We apply formula (8) with n in place of n, H = H ξ, k =, m = 0. We get 1 Vol n ( H ξ ) = A,Hξ (0) = n+ 8π (η) dη. (n 1) S n 1 H ξ By emma 3, the function under the integral in the right hand side is constant on the circle S n 1 Hξ. Since ξ H ξ, the integral is equal to π n+ (ξ).

11 COMPEX BUSEMANN-PETTY PROBEM 11 The connection between the complex Busemann-Petty problem and intersection bodies is as follows: Theorem. The answer to the complex Busemann-Petty problem in C n is affirmative if and only if every origin symmetric invariant with respect to all R θ convex body in R n is a -intersection body. This theorem will follow from the next two lemmas. Note that, since we can approximate the body in the radial metric from inside by infinitely smooth convex bodies invariant with respect to all R θ, and also approximate from outside in the same way, we can argue that if the answer to the complex Busemann-Petty problem is affirmative for infinitely smooth bodies and then it is affirmative in general. emma 4. et and be infinitely smooth invariant with respect to R θ convex bodies in R n so that is a -intersection body and, for every ξ S n 1, Then Vol n ( H ξ ) Vol n ( H ξ ). Vol n () Vol n (). Proof : By [10, emma 3.16], the Fourier transforms of the distributions n+, n+ and are continuous functions outside of the origin in R n. By Theorem 1 and Proposition, the conditions of the lemma imply that for every ξ S n 1, n+ (ξ) n+ (ξ) and Therefore, ( n+ S n 1 ( n+ S n 1 (ξ) 0. ) (ξ) (ξ) dξ ) (ξ) (ξ) dξ. Now we apply Parseval s formula on the sphere, Proposition 1, to remove the Fourier transforms in the latter inequality and then use the polar formula for the volume and Hölder s inequality: n Vol n () = ( n S n 1 n S n 1 ) n 1 n dx dx n+ S n 1 1 n dx n Sn 1 dx

12 1 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU which gives the result. = (n Vol n ()) n 1 n (n Voln ()) 1 n, emma 5. Suppose that there exists an origin symmetric invariant with respect to all R θ convex body in R n which is not a -intersection body. Then one can perturb twice to construct other origin symmetric invariant with respect to R θ convex bodies and in R n such that for every ξ S n 1, but Vol n ( H ξ ) Vol n ( H ξ ), Vol n () > Vol n ( ). Proof : We can assume that the body is infinitely smooth and has strictly positive curvature. In fact, approximating in the radial metric by infinitely smooth invariant with respect to all R θ convex bodies with strictly positive curvature, we get by emma that approximating bodies can not all be -intersection bodies. So there exists an infinitely smooth invariant with respect to all R θ convex body with strictly positive curvature that is not a -intersection body. Now as is infinitely smooth, by [10, emma 3.16], the Fourier transform of is a continuous function outside of the origin in R n. The body is not a -intersection body, so by Proposition, the Fourier transform is negative on some open subset Ω of the sphere S n 1. Since is invariant with respect to rotations R θ, we can assume that the set Ω is also invariant with respect to rotations R θ. This allows us to choose an even non-negative invariant with respect to rotations R θ function f C (S n 1 ) which is supported in Ω. Extend f to an even homogeneous function f(x/ x ) x of degree - on R n. By [10, emma 3.16], the Fourier transform of this extension is an even homogeneous function of degree -n+ on R n, whose restriction to the sphere is infinitely smooth: f(x/ x ) x (y) = g(y/ y ) y n+, where g C (S n 1 ). By the connection between the Fourier transform and linear transformations, the function g is also invariant with respect to rotations R θ. efine a body in R n by n+ = n+ ɛg(x/ x ) x n+. (11)

13 COMPEX BUSEMANN-PETTY PROBEM 13 For small enough ɛ the body is convex. This essentially follows from a simple two-dimensional argument: if h is a strictly concave function on an interval [a, b] and u is a twice differentiable function on [a, b], then for small ɛ the function h + ɛu is also concave. Note that here we use the condition that has strictly positive curvature. Besides that, the body is invariant with respect to rotations R θ because so are the body and the function g. We can now choose ɛ so that is an origin symmetric invariant with respect to all R θ convex body in R n. et us prove that the bodies and provide the necessary counterexample. We apply the Fourier transform to both sides of (11). By definition of the function g and since f is non-negative, we get that for every ξ S n 1 ( n+ ) (ξ) = ( n+ By Theorem 1, this means that for every ξ ) (ξ) (π) n ɛf(ξ) n+ (ξ). Vol n ( H ξ ) Vol n ( H ξ ). On the other hand, the function f is positive only where is negative, so n+ (ξ) (ξ) dξ S n 1 = ( n+ S n 1 (π) n ɛ > ( n+ S n 1 ) (ξ) (ξ) dξ ( S n 1 ) (ξ)f(ξ) dξ ) (ξ) (ξ) dξ. The end of the proof is similar to that of the previous lemma - we apply Parseval s formula to remove Fourier transforms and then use Hölder s inequality and the polar formula for the volume to get Vol n () > Vol n (). 4. The solution of the problem It is known (see [6] or [10, Corollary 4.9] plus Proposition ) that for every origin symmetric convex body in R n, n the space (R n, ) embeds in p for each p [n 3, n), or, in other words, every origin-symmetric convex body in R n is a (n 3)-, (n )- and (n 1)-intersection body. On the other hand, for q > the unit ball of the real space l n q is not a (n 4)-intersection body, and, moreover,

14 14 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU R n provided with the norm of this space does not embed in p with p < n 3 (see [3] or [10, Th. 4.13]). Now we have to find out what happens if we consider convex bodies invariant with respect to all R θ. It immediately follows from the second derivative test ([10, Th and 4.1] ; see Corollary 4 below) that for q > the complex space l n q does not embed in p with p < n 4, which means that the unit ball Bq n of this space (which is invariant with respect to all R θ ) is not a k-intersection body with k < n 4. The only question that remains open is what happens in the interval p [n 4, n 3). The following result answers this question. Theorem 3. et n 3. Every origin symmetric invariant with respect to R θ convex body in R n is a (n 4)-intersection body. Moreover, the space (R n, ) embeds in p for every p [n 4, n). If n = the space (R n, ) embeds in p for every p (0, 4). Proof : By emma, it is enough to prove the result in the case where is infinitely smooth. Fix ξ S n 1. et n 3. Applying formula (8) and then emma 3 with H = H ξ, m = 1, k = and dimension n instead of n, we get 1 A,Hξ (0) = ( n+4 8π ) (η) dη (n ) = π 8π (n ) S n 1 H ξ n+4 (ξ). ) (ξ) 0 for every By Brunn s theorem (see emma 1), ( n+4 ξ S n 1, so n+4 is a positive definite distribution on R n. By Proposition, is a (n 4)-intersection body. Now let n. For 0 < q <, formula (7) and emma 1 imply that n+q+ (ξ) 0. By Proposition, the space (R n, ) embeds in n+q+, and, using the range of q, every such space embeds in p, p (n 4, n ). As mentioned before, these spaces also embed in p, p [n 3, n), because so does any n-dimensional normed space. We now give an example of an origin symmetric invariant with respect to all R θ convex body in R n which is not a k-intersection body for any 1 k < n 4. enote by Bn q the unit ball of the complex space l n q considered as a subset of R n : B n q = {ξ R n : ξ q = ( (ξ 11 + ξ 1) q/ ( ξ n1 + ξ n) q/ ) 1/q 1}.

15 COMPEX BUSEMANN-PETTY PROBEM 15 If q 1 then Bq n is an origin symmetric invariant with respect to R θ convex body in R n. The next theorem immediately follows from Proposition 3. Theorem 4. If q > then the space (R n, q ) does not embed in p with 0 < p < n 4. In particular, the body Bq n is not a k-intersection body for any 1 k < n 4. Proof : The space (R n, q ) contains as a subspace the q-sum of R and a (n )-dimensional subspace (R n, q ). This q-sum does not embed in p, 0 < p < n 4 by Proposition 3. By a result of E.Milman [Mi], the larger space cannot embed in p, 0 < p < n 4 either (the proof in [Mi] is only for integers p, but it is exactly the same for non-integers; note that for the complex Busemann-Petty problem we need only the second statement of the corollary, where p is an integer). We are now ready to prove the main result of this article: Theorem 5. The solution to the complex Busemann-Petty problem in C n is affirmative if n 3 and it is negative if n 4. Proof : By Theorem 3, every origin symmetric invariant with respect to R θ convex body in R 6 (where n = 3) is a n 4 = -intersection body, and in R 4 (where n = ) it is a n = -intersection body. The affirmative answers for n = 3 and n = follow now from Theorem. If n 4 then n 4 >, so by Theorem 4 the body Bq n is not a -intersection body. The negative answer follows from Theorem. Remark 1. The transition between the dimensions n = 3 and n = 4 is due to the fact that convexity controls only derivatives of the second order. To see this let us look again at formula (8), which we apply with k =. We want to get information about the Fourier transform of, so we need to choose m so that n+m+ =. If n = 3 then m = 1, but when n = 4 we need m =. This means that for n = 3 we consider A,H (0), which is always negative by convexity, but when n = 4 we look at A,H (0), which is not controlled by convexity and can be sign-changing. One can construct a counterexample in dimension n = 4 using this argument, similarly to how it was done for the real Busemann-Petty problem; see [10, Corollary 4.4].

16 16 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU Remark. Applying Theorem 3 to n = we get that every two-dimensional complex normed space (which is a 4-dimensional real normed space) embeds in p for every p [ 1, 0). By [YY, Th. 6.4], this implies that every such space embeds isometrically in 0. The concept of embedding in 0 was introduced in [YY]: a normed space (R n, ) embeds in 0 if there exist a probability measure µ on S n 1 and a constant C so that for every x R n, x 0 log = log (x, ξ) dµ(ξ) + C. S n 1 We have Theorem 6. Every two-dimensional complex normed space embeds in 0. On the other hand, there exist two-dimensional complex normed spaces that do not embed isometrically in any p, p > 0. An example supporting the second claim is the complex space l q with q >. This follows from a version of the second derivative test proved in [] (see also [10, Theorem 6.11]). Recall that every twodimensional real normed space embeds isometrically in 1 (see [Fe], [He], [i] or [10, p. 10]), but the real space l q does not embed isometrically in any p, 1 < p, as proved by or [o]; see also [10, p. 14]. Acknowledgments: The first named author was supported in part by the NSF grants MS and MS The third named author was partially supported by the NSF grant MS and by the European Network PH, FP6 Marie Curie Actions, Contract MCRN References [Ba]. Ball, Some remarks on the geometry of convex sets, Geometric aspects of functional analysis (1986/87), ecture Notes in Math. 1317, Springer-Verlag, Berlin-Heidelberg-New York, 1988, [BFM] F. Barthe, M. Fradelizi and B. Maurey, A short solution to the Busemann- Petty problem, Positivity 3 (1999), [Bo] J. Bourgain, On the Busemann-Petty problem for perturbations of the ball, Geom. Funct. Anal. 1 (1991), [BZ] J. Bourgain and Gaoyong Zhang, On a generalization of the Busemann-Petty problem, Convex geometric analysis (Berkeley, CA, 1996), 65 76, Math. Sci. Res. Inst. Publ., 34, Cambridge Univ. Press, Cambridge, [BP] H. Busemann and C. M. Petty, Problems on convex bodies, Math. Scand. 4 (1956), [o]. or, Potentials and isometric embeddings in 1, Israel J. Math. 4 (1976),

17 COMPEX BUSEMANN-PETTY PROBEM 17 [Fe] T. S. Ferguson, A representation of the symmetric bivariate Cauchy distributions, Ann. Math. Stat. 33 (196), [Ga1] R. J. Gardner, Intersection bodies and the Busemann-Petty problem, Trans. Amer. Math. Soc. 34 (1994), [Ga] R. J. Gardner, A positive answer to the Busemann-Petty problem in three dimensions, Annals of Math. 140 (1994), [GS] R.J. Gardner, A. oldobsky, and Th. Schlumprecht, An analytic solution of the Busemann-Petty problem on sections of convex bodies, Annals of Math. 149(1999), [GS] I.M. Gelfand and G.E. Shilov, Generalized functions, vol. 1. Properties and operations, Academic Press, New York, (1964). [Gi] A. Giannopoulos, A note on a problem of H. Busemann and C. M. Petty concerning sections of symmetric convex bodies, Mathematika 37 (1990), [Ha] H. Hadwiger, Radialpotenzintegrale zentralsymmetrischer rotationkörper und ungleichheitsaussagen Busemannscher art, Math. Scand. 3 (1968), [He] C. Herz, A class of negative definite functions, Proc. Amer. Math. Soc. 14 (1963), [YY] N. J. alton, A. oldobsky, V. Yaskin and M. Yaskina, The geometry of 0, Canad. J. Math., to appear. [1] A. oldobsky, An application of the Fourier transform to sections of star bodies, Israel J. Math 106(1998), [] A. oldobsky, Intersection bodies, positive definite distributions and the Busemann-Petty problem, Amer. J. Math. 10 (1998), [3] A. oldobsky, Intersection bodies in R 4, Adv. Math. 136 (1998), [4] A. oldobsky, Second derivative test for intersection bodies, Adv. Math. 136 (1998), [5] A. oldobsky, A generalization of the Busemann-Petty problem on sections of convex bodies, Israel J. Math. 110 (1999), [6] A. oldobsky, A correlation inequality for stable random vectors, Advances in stochastic inequalities (Atlanta, GA, 1997), 11 14, Contemp. Math. 34, Amer. Math. Soc., Providence, RI, [7] A. oldobsky, Positive definite distributions and subspaces of p with applications to stable processes, Canad. Math. Bull. 4 (1999), [8] A. oldobsky, A functional analytic approach to intersection bodies, Geom. Funct. Anal. 10 (000), [9] A. oldobsky, The Busemann-Petty problem via spherical harmonics, Adv. Math. 177 (003), [10] A. oldobsky, Fourier analysis in convex geometry, Amer. Math. Soc., Providence RI, 005. [] A. oldobsky and Y. onke, A short proof of Schoenberg s conjecture on positive definite functions, Bull. ondon Math. Soc. 31 (1999), [YY] A. oldobsky, V. Yaskin and M. Yaskina, Modified Busemann-Petty problem on sections of convex bodies, Israel J. Math. 154 (006), [R]. G. arman and C. A. Rogers, The existence of a centrally symmetric convex body with central sections that are unexpectedly small, Mathematika (1975),

18 18 A. OOBSY, H. ÖNIG, AN M. ZYMONOPOUOU [i] J. indenstrauss, On the extension of operators with finite dimensional range, Illinois J. Math. 8 (1964), [u] E. utwak, Intersection bodies and dual mixed volumes, Adv. Math. 71 (1988), [Mi] E. Milman, Generalized intersection bodies, J. Funct. Anal. 40 (006), [Pa] M. Papadimitrakis, On the Busemann-Petty problem about convex, centrally symmetric bodies in R n, Mathematika 39 (199), [Ru] B. Rubin, The lower dimensional Busemann-Petty problem for bodies with generalized axial symmetry, preprint. [RZ] B. Rubin and Gaoyong Zhang, Generalizations of the Busemann-Petty problem for sections of convex bodies, J. Funct. Anal. 13 (004), [Sch] R. Schneider, Convex bodies: the Brunn-Minkowski theory, Cambridge University Press, Cambridge, [Y1] V. Yaskin, The Busemann-Petty problem in spherical and hyperbolic spaces, Adv. Math. 03 (006), [Y] V. Yaskin, A solution to the lower dimensional Busemann-Petty problem in the hyperbolic space, J. Geom. Anal. 16 (006), [Zh1] Gaoyong Zhang, Intersection bodies and Busemann-Petty inequalities in R 4, Annals of Math. 140 (1994), [Zh] Gaoyong Zhang, A positive answer to the Busemann-Petty problem in four dimensions, Annals of Math. 149 (1999), [Zv1] A. Zvavitch, Gaussian measure of sections of convex bodies, Adv. Math. 188 (004), [Zv] A. Zvavitch, The Busemann-Petty problem for arbitrary measures, Math. Ann. 331 (005), Alexander oldobsky, epartment of Mathematics, University of Missouri, Columbia, MO 6511, USA address: koldobsk@math.missouri.edu Hermann önig, Mathematisches Seminar, Christian-Albrechts-Universitt iel, udewig-meyn Str. 4, iel address: hkoenig@math.uni-kiel.de Marisa Zymonopoulou, epartment of Mathematics, University of Missouri, Columbia, MO 6511, USA address: marisa@math.missouri.edu