An extension of a result by Lonke to intersection bodies
aa r X i v : . [ m a t h . M G ] M a r AN EXTENSION OF A RESULT BY LONKE TO INTERSECTIONBODIES
M. ANGELES ALFONSECA
Abstract.
In this paper we prove that intersection bodies cannot be direct sumsusing Fourier analytic techniques. This extends a result by Lonke. We also prove anecessary regularity condition and a convexity condition for a body of revolution tobe an intersection body of a star body. Introduction
Let
K, L be origin-symmetric star bodies in R n . The body K is called the intersectionbody of L , and denoted by K = IL , if the radius of K in each direction is equal to the( n − L that is perpendicular to thatdirection. In other words, if ρ K ( ξ ) = max { a : aξ ∈ K } is the radial function of K ,then for every ξ ∈ S n − , we have ρ K ( ξ ) = Vo l n − ( L ∩ ξ ⊥ ).The volume of the section of L can be written using the spherical Radon transform R (see [2, 5]): ρ K ( ξ ) = 1 n − Z S n − ρ n − L ( θ ) dθ = 1 n − R ( ρ n − L )( ξ ) , ∀ ξ ∈ S n − . The more general class of intersection bodies is defined as follows. A star body K isan intersection body if its radial function ρ K ( ξ ) = max { a : aξ ∈ K } is the sphericalRadon transform of an even non-negative measure µ , i.e. , for every continuous function g on S n − , Z S n − ρ K ( ξ ) g ( ξ ) dξ = Z S n − Rg ( ξ ) µ ( dξ ) , ∀ ξ ∈ S n − . Work partially supported by the NDSU Advance FORWARD program sponsored by the NationalScience Foundation, HRD-0811239AMS 2010 Subject Classification: 52A30. Keywords: Intersection bodies, bodies of revolution.
Intersection bodies were originally introduced by Lutwak [9] in connection to theBusemann-Petty problem (see, for example, [5]), and were instrumental in findingits complete solution [3].A Fourier-analytic characterization of intersection bodies is due to Koldobsky [5]. If k ξ k K = ρ K ( ξ ) − denotes the norm associated to the body K , then K is an intersectionbody if and only if the Fourier transform of k · k − K is a positive distribution, i.e. itsaction on any non-negative test function gives a non-negative result.The structure and the geometric properties of intersection bodies are hard to under-stand for at least two reasons. First, if n = 2 , ,
4, all origin-symmetric convex bodiesin R n are intersection bodies, and thus their study is only meaningful in dimension n ≥
5. Secondly, Zhang proved that no polytope is an intersection body of a star bodyif n ≥ K is the projection body of L if the width of K in every direction ξ equals twice the ( n − L orthogonal to ξ ).Notice the parallel with the definition of an intersection body of a star body. In acertain sense, it may be expected that the geometric properties of intersection bodiesare dual to the geometric properties of zonoids, since there is a certain duality betweensections and projections (see Section 7.4, Note 4 in [11]). However, this duality doesnot hold in all instances. For example, if Z is a zonoid, then Z ∗ is an intersection body,but the class of intersection bodies is wider than the class of duals of zonoids. A betterunderstanding of intersection bodies could be useful in the solution of the followingopen problem. Problem 1.
Is it true that all zonoids whose polars are zonoids tend to the Euclideanball when n → ∞ ? In other words, if d n is the Banach-Mazur distance, is it true that c = lim n →∞ d n ( Z, B n ) = 1? NTERSECTION BODIES 3
This problem is related to an isometric analogue of a well-known theorem of Grothendieck[7], asserting that among infinite dimensional Banach spaces, the ones which are iso-morphic to both a subspace of L and a quotient-space of L ∞ are isomorphic to aHilbert space. R. Schneider [10] used high-order derivatives of the support function toconstruct examples in all dimensions of zonoids whose polars are also zonoids, that arenot ellipsoids. It is known that all his zonoids tend to the Euclidean ball as n goesto infinity. It is not known if there are zonoids whose polars are zonoids that do notconverge to the Euclidean ball.Y. Lonke [8] showed that if K = A + B is a convex body in R n , n ≥
3, such that dim (span A) = n , 1 ≤ dim (span B) ≤ n −
2, then the polar of K is not a zonoid. Thisshows, in particular, that a zonoid whose polar is also zonoid cannot have an ( n − ≤ dim (span B) ≤ n −
1. Onthe other hand, Lonke proved that the barrel Z n = B n + B n − is a zonoid whose polaris also a zonoid in dimensions n = 3 ,
4. Thus, it is possible for a zonoid whose polar isa zonoid to have an ( n − K to be an intersection body.It is an extension of Lonke’s results, proved using the Fourier-analytic characterizationand Koldobsky’s Second Derivative Test [6]. We show that if K ⊂ R n , n ≥ is aconvex body that can be written as a direct sum K = A + B , where dim ( A ) ≥ anddim ( B ) ≥ , then K is not an intersection body . This was previously proved by Zhang,using the Radon transform (see Note 8.1 in [2]). Note that such a result is not truefor non-direct sums, as there exist intersection bodies with low dimensional faces. Forexample, if L is a body of revolution L with a cylindrical part near the equator, then IL has an ( n − R n , if M. ANGELES ALFONSECA the body in R n +2 that has the same 2-dimensional radial function is still an intersectionbody in dimension n + 2. The original motivation for the work in this section was toprove that Lonke’s zonoid Z n is not an intersection body in dimension 5 and higher,thus explaining why its polar is not a zonoid in those dimensions. As it turned out, Z n is an intersection body in dimensions 5 and 6, but not in dimensions 8 and higher.Acknowledgements: The author would like to thank Dmitry Ryabogin and ArtemZvavitch for many conversations and ideas about this paper. The author also thanksthe reviewer for useful suggestions to make the paper accurate and more readable.2. Direct sums are not intersection bodies for n ≥ K be an origin-symmetric convex body in R n , n ≥
5, that can be written as adirect sum K = A ⊕ B , where j = dim ( A ) ≥ n − j = dim ( B ) ≥
4. Note that,because of the direct sum, A and B inherit central symmetry from K . We shall showthat K is not an intersection body. Since a central section of an intersection body isalso an intersection body [4], it is enough to consider the case in which j = 1, i.e. A is a segment. We will write the points z ∈ R n in the form z = ( x, y ), where x ∈ R and y ∈ R n − . Assuming that the segment A has length 2, the norm associated to K canbe written as k ( x, y ) k K = max {| x | , k y k B } . We shall use the second derivative test introduced in [6] to prove that K is notan intersection body. Theorem 1 in [6] requires that the function x → k ( x, y ) k hascontinuous second derivative everywhere on R . However, in our calculations all thederivatives of the norm will be taken in the sense of distributions and we will not needthis regularity.The following proof follows that of Lemma 1 in [6]. For every m ∈ N , we considerthe functions h m ( x ) = m √ π e − x m / and u ( y ) = π ) ( n − / e −k y k / . Lemma 1.
Let k ( x, y ) k be the norm defined by K . For every ǫ > there exists M ∈ N so that, for every m > M , hk ( x, y ) k − , u ( y ) h ′′ m ( x ) i ≥ − ǫ . NTERSECTION BODIES 5
Proof.
Let us define the sets U = { ( x, y ) : | x | > k y k B } ,W = { ( x, y ) : | x | < k y k B } . Then hk ( x, y ) k − , u ( y ) h ′′ m ( x ) i = Z U u ( y ) h ′′ m ( x ) k ( x, y ) k − dx dy + Z W u ( y ) h ′′ m ( x ) k ( x, y ) k − dx dy If ( x, y ) ∈ U , then Z U u ( y ) h ′′ m ( x ) k ( x, y ) k − dx dy =(1) Z R n − \{ } u ( y ) Z | x | > k y k B | x | − h ′′ m ( x ) dx dy =2 Z R n − \{ } u ( y ) Z ∞k y k B h ′′ m ( x ) x dx dy Integrating by parts twice, we obtain(2) 2 Z R n − \{ } u ( y ) (cid:20) − h ′ m ( k y k B ) k y k B − h m ( k y k B ) k y k B + 2 Z ∞k y k B h m ( x ) x dx (cid:21) dy. Since u ( y ) and h m ( x ) are positive functions, and h ′ m ( x ) is negative, the first and thirdintegrals in (2) are positive. Thus, we only need to study the second integral, − Z R n − \{ } u ( y ) h m ( k y k B ) k y k B dy Using the change of variables z = my , we can rewrite this integral as − m − n Z R n − \{ } u (cid:16) zm (cid:17) h ( k z k B ) k z k B dz. Since the integral is well defined and n ≥
5, this term converges to 0 as m goes toinfinity, as we want.Now we turn to the second case. If ( x, y ) ∈ W , then(3) Z W u ( y ) h ′′ m ( x ) k ( x, y ) k − dx dy = Z R n − \{ } u ( y ) Z | x | < k y k B h ′′ m ( x ) k y k B dx dy = M. ANGELES ALFONSECA Z R n − \{ } u ( y ) k y k B Z k y k B h ′′ m ( x ) dx dy =2 Z R n − \{ } u ( y ) k y k B [ h ′ m ( k y k B ) − h ′ m (0)] dy =2 Z R n − \{ } u ( y ) k y k B h ′ m ( k y k B ) dy = − m Z R n − \{ } u ( y ) h m ( k y k B ) dy The change of variables z = my gives − Z R n − \{ } m − n u (cid:16) zm (cid:17) h ( k z k B ) dz which, as before, converges to 0 when m goes to infinity. (cid:3) It follows (cf. [6]) that K is not an intersection body. We write the proof forcompleteness. Theorem 2.
Let K be a direct sum of a segment and an ( n − -dimensional body, n ≥ . Then K is not an intersection body. Proof:
Suppose that K is an intersection body. Then k ( x, y ) k − K is a positive definitedistribution (by Theorem 4.1 of [5]). This implies, by Lemma 2.24 and Corollary 2.26in [5], that there exists a finite Borel measure µ on S n − such that, for every even testfunction φ ,(4) Z R n k ( x, y ) k − K φ ( x, y ) dx dy = Z S n − (cid:18)Z ∞ b φ ( tξ ) dt (cid:19) dµ ( ξ ) . In our case, (4) becomes(5) Z R n k ( x, y ) k − K ∂ φ∂x = − Z S n − ξ dµ ( ξ ) Z ∞ t b φ ( tξ ) dt. Let φ ( x, y ) = h m ( x ) u ( y ), where h m ( x ) and u ( y ) are the functions defined before Lemma1. Then, d φ ( ξ ) = e − ξ / m e − ( ξ + ··· + ξ n ) / . NTERSECTION BODIES 7
With this choice of φ , (5) becomes − ǫ ≤ h ( k ( x, y ) k − K ) , h m ( x ) ′′ · u ( y ) i = − p π/ Z S n − ξ (cid:18) ξ m + ξ + . . . + ξ n (cid:19) − / dµ ( ξ ) ≤ − p π/ Z S n − ξ dµ ( ξ ) ≤ . In the first inequality we have used Lemma 1. Thus, the measure µ is supported on S n − ∩ { ξ = 0 } , which is a contradiction with the fact that K is an n -dimensionalbody. (cid:3) Remark 3.
Theorem 2 proves that any direct sum in n ≥ n ≥
5, there is only one case left to consider: a direct sum of a 2-dimensionaland a 3-dimensional body in R . Example 4.
The cylinder B n − (0 , × [0 , e n is not an intersection body for any n ≥ Example 5.
A slight variation of the proof of Lemma 1 allows us to prove a version ofthe Second Derivative Test for bodies of revolution in R n with an ( n − n − K ∈ R n be a body of revolution with an ( n − R n as ( x, y ), with x ∈ R and y ∈ R n − . Assume that the axis of M. ANGELES ALFONSECA revolution of K is the x -axis, and that the face perpendicular to it has radius 1 and isplaced at height 1. Then the norm k · k K can be written as k ( x, y ) k K = | x | | x | > k y k g ( x, k y k ) | x | < k y k where g ( x, r ) is positive, convex, homogeneous of degree 1 and even with respect toeach variable. In particular, for every r = 0 fixed, g ( x, r ) has a positive minimum at x = 0. Proposition 6.
Let K be a body of revolution with a face and assume that:(1) For every fixed r = 0 , the function x → g ( x, r ) has continuous second derivativeon | x | < | r | and ∂ g∂x (0 , r ) = 0 .(2) lim x → ∂ g∂x ( x, r ) = 0 uniformly on r .Then K is not an intersection body. Proof:
With u ( y ) and h m ( x ) defined as in Lemma 1, we consider the integral hk ( x, y ) k − K , u ( y ) h ′′ m ( x ) i = Z R n − \{ } u ( y ) Z | x | > k y k h ′′ m ( x ) | x | − dx dy + Z R n − \{ } u ( y ) Z | x | < k y k h ′′ m ( x ) ( g ( x, k y k )) − dx dy. The first of these two integrals is exactly the same as the integral in (1). As we showedin (2), it is well defined and converges to 0 as m goes to infinity if n ≥ g is homogeneous of degree1 and that g (1 ,
1) = 1 (from the definition of the norm of K ), obtaining2 Z R n − \{ } u ( y ) " h ′ m ( k y k ) k y k + Z k y k h ′ m ( x ) ∂g∂x ( x, k y k ) 1 g ( x, k y k ) dx dy = I + II.
Making the change of variables w = my , I is equal to I = − m n − Z R n − \{ } u (cid:16) wm (cid:17) h ( k w k ) dw which tends to 0 as m goes to infinity. NTERSECTION BODIES 9
As for II , another integration by parts gives II = 2 ∂g∂x (1 , Z R n − \{ } u ( y ) h m ( k y k ) 1 k y k dy − Z R n − \{ } u ( y ) Z k y k h m ( x ) ∂ g∂x ( x, k y k ) 1 g ( x, k y k ) + (cid:18) ∂g∂x ( x, k y k ) (cid:19) − g ( x, k y k ) ! dx dy Here, ∂g∂x (1 ,
1) is defined as lim x → − ∂g∂x ( x, − Z R n − \{ } u ( y ) Z k y k h m ( x ) ∂ g∂x ( x, k y k ) 1 g ( x, k y k ) dx dy. Since h m ( x ) is an approximate identity and ∂ g∂x ( x, k y k ) converges to zero uniformly in y as x goes to zero, this term tends to zero as m goes to infinity and Lemma 1 holds. (cid:3) Remark 7.
A similar proof shows that any centrally symmetric body (not necessarilyof revolution) that has a cylindrical part is not an intersection body.3.
Regularity and convexity conditions for an intersection body ofrevolution to be the intersection body of a star body
Let L be a centered star body of revolution about the x n -axis in R n , and let ρ L beits radial function, which we assume to be continuous. The radial function ρ L may beconsidered as a function of the angle ϕ from the x n -axis.Let K be the intersection body of L , defined as the body whose radial function ρ K is the spherical Radon transform of ρ n − L / ( n − ρ K ( ϕ ) = 2 ω n − ( n −
1) sin ϕ Z π/ π/ − ϕ ρ L ( ψ ) n − (cid:18) − cos ψ sin ϕ (cid:19) ( n − / sin ψ dψ, if 0 < ϕ ≤ π/
2, and ρ K (0) = κ n − ρ L ( π/ ( n − . Here, ω n denotes the surface area ofthe unit ball in R n . A derivation of this formula can be found in [2], Theorem C.2.9.If we substitute x = sin ϕ , t = cos ψ in (6), we obtain(7) ρ K (arcsin x ) = 2 ω n − ( n − x n − Z x ρ L (arccos t ) n − ( x − t ) ( n − / dt, for 0 < x ≤
1. When ρ L is continuous, this formula can be inverted:(8) ρ L (arccos t ) n − = 1( n − ω n − t (cid:18) t ddt (cid:19) n − Z t ρ K (arcsin x ) x n − ( t − x ) ( n − / dx, for 0 < t ≤ ρ K and ρ L ,thus providing a necessary regularity condition for a body of revolution to be an inter-section body of a star body. Proposition 8.
Let L be a star body of revolution in R n , where n ≥ is an even num-ber. Assume that its radial function ρ L is of class C m ( S n − ) . Let K be the intersectionbody of L , with radial function ρ K ( ϕ ) given by (6). Then ρ K ( ϕ ) is of class C m + n − for < ϕ < π/ , of class C m at ϕ = 0 , and of class C m + n − at ϕ = π/ . Corollary 9.
Let K be a body of revolution in R n , with n ≥ even. A necessarycondition for K to be an intersection body of a star body is that its radial function ρ K ( ϕ ) is of class C n − for < ϕ < π/ , and of class C n − at ϕ = π/ . Some immediate applications of Corollary 9 are the following: • Any body of revolution that is not C , such as the cylinder, or a cylinder withtwo conical caps, is not an intersection body of a star body in dimensions 4 andhigher. • A double cone is not an intersection body of a star body in dimensions 4 andhigher, because its radial function is not C at ϕ = π/ Proof of Proposition 8:Part 1:
We consider first the case 0 < ϕ < π/
2. It will be more convenient to useequation (7), with 0 < x <
1. Let us denote r ( t ) = ρ L (arccos t ) n − , and(9) F ( x ) = x − n +3 Z x r ( t )( x − t ) ( n − / dt. Thus, ρ K (arcsin x ) = ω n − ( n − F ( x ), and we have to prove that if r ( t ) ∈ C m , then F ( x ) ∈ C m + n − . NTERSECTION BODIES 11
The first derivative of F is( − n + 3) x − n +2 Z x r ( t )( x − t ) ( n − / dt + x − n +4 ( n − Z x r ( t )( x − t ) ( n − / dt. Looking at the second term, we see that the exponent of ( x − t ) has decreased byone. Continuing this process, the k -th derivative of F will thus contain a term in whichthe exponent of ( x − t ) is ( n − − k ) /
2. Hence, the ( n − / F will be the first to contain a term without integral, which is the term with the lowestregularity. It is equal to ( n − x − n/ r ( x ) . Thus, if r ∈ C m , then F ∈ C m +( n − / . Part 2:
To study the regularity at the point x = 0, we extend F evenly (since it isthe radial function of a body of revolution). At x = 0, the value of F must be ( n − / X j =0 (cid:18) n − j (cid:19) ( − j r (0)2 j + 1 , so that F is continuous. This is easy to see by expanding the term ( x − t ) ( n − / inEquation (9) and applying L’Hˆopital’s rule.We will show that, for every natural number k ,(10) ˜ F ( k ) (0+) = ( n − / X j =0 (cid:18) n − j (cid:19) ( − j lim x → r ( k ) ( x )2 j + k + 1 . Hence the regularity of F ( x ) at x = 0 is the same as the regularity of r ( x ) at x = 0.We proceed by induction. Assume that (10) holds for every k ≤ k . We will showthat the formula is true for k + 1. Expanding the binomial inside equation (9), we canwrite F as F ( x ) = ( n − / X j =0 (cid:18) n − j (cid:19) ( − j g j ( x ) I j ( x ) , where g j ( x ) = 1 x j +1 and I j ( x ) = Z x r ( t ) t j dt ≡ Z x r j ( t ) dt Then,(11) F ( k ) ( x ) = ( n − / X j =0 (cid:18) n − j (cid:19) ( − j " k X i =0 (cid:18) ki (cid:19) g ( i ) j ( x ) I ( k − i ) j ( x ) , and F ( k +1) (0+) = lim x → F ( k ) ( x ) − F ( k ) (0) x == lim x →
0+ ( n − / X j =0 (cid:18) n − j (cid:19) ( − j x " k X i =0 (cid:18) k i (cid:19) g ( i ) j ( x ) I ( k − i ) j ( x ) − r ( k ) (0+)2 j + k + 1 . Let us fix j and consider the term T j ( x ) = 1 x " k X i =0 (cid:18) k i (cid:19) g ( i ) j ( x ) I ( k − i ) j ( x ) − r ( k ) (0+)2 j + k + 1 . We have to prove that(12) lim x → T j ( x ) = r ( k +1) (0+)(2 j + k + 2) . Observing that g ( i ) j ( x ) = (2 j + i )!(2 j )! ( − i x − (2 j + i +1) , and, if k − i ≥ I ( k − i ) j ( x ) = r ( k − i − j ( x ) , and multiplying both the numerator and denominator of T j ( x ) by x j + k +1 , T j ( x ) canbe rewritten as T j ( x ) = 1 x j + k +2 " k − X i =0 (cid:18) k i (cid:19) (2 j + i )!(2 j )! ( − i x k − i r ( k − i − j ( x )+(2 j + k )!(2 j )! ( − k Z x r j ( t ) dt − r ( k ) (0+)2 j + k + 1 x j + k +1 (cid:21) . By L’Hˆopital’s rule,lim x → T j ( x ) = lim x → j + k + 2) x j + k +1 " k − X i =0 (cid:18) k i (cid:19) (2 j + i )!(2 j )! ( − i (cid:16) ( k − i ) x k − i − r ( k − i − j ( x )+(13) x k − i r ( k − i ) j ( x ) (cid:17) + (2 j + k )!(2 j )! ( − k r j ( x ) − r ( k ) (0+) x j + k (cid:21) NTERSECTION BODIES 13
In the term inside square brackets, we group together the terms with derivatives of r j ( x ) of the same order, obtaining k X i =0 (2 j + i − j − − i r ( k − i ) j ( x ) x k − i ! − r ( k ) (0+) x j + k . Recalling now that r j ( x ) = r ( x ) x j , this sum equals(14) 2 j k − X l =0 (cid:18) k l (cid:19) r ( l ) ( x ) x j + l k − l X i =0 ( − i (cid:18) k − li (cid:19) (2 j + i − j + i − k + l )! ! +( r ( k ) ( x ) − r ( k ) (0+)) x j + k . The inner sum appearing in (14) is zero for every value of k and l , because it is equalto ∆ m P ( x ) evaluated at x = 2 j , where P ( x ) = ( x + 1)( x + 2) · · · ( x + m −
1) and ∆ isthe difference operator ∆ P ( x ) = P ( x − − P ( x ). But if we apply ∆ m to a polynomialof degree m −
1, we obtain zero. Thus, (14) equals( r ( k ) ( x ) − r ( k ) (0+)) x j + k . Writing this instead of the square brackets in (13), we now havelim x → T j ( x ) = 1(2 j + k + 2) lim x → ( r ( k ) ( x ) − r ( k ) (0+)) x = r ( k +1) (0+)(2 j + k + 2) . This proves (12) and hence the regularity of F at 0 is the same as the regularity of r at 0. Part 3:
Finally, we will show that ρ K ( ϕ ) ∈ C m + n − at ϕ = π/
2. If we set u = π/ − ψ and s ( u ) = ρ L ( π/ − u ) n − (cid:16) − sin u sin ϕ (cid:17) ( n − / , equation (6) becomes, disregarding theconstants, ρ K ( ϕ ) = csc ϕ Z ϕ s ( u ) cos u du, . for 0 ≤ ϕ ≤ π/
2. Since K is centrally symmetric, for π/ < ϕ ≤ π its radial function is ρ K ( π − ϕ ). In particular, all even derivatives of ρ K are continuous at π/
2, and the oddderivatives are continuous if and only if their value at π/ the body L and thus for s ( u ). Our hypothesis says that s ∈ C m , and we will assumethat s ( m +1) is not continuous at π/ I ( ϕ ) = R ϕ s ( u ) cos u du . Observe that I ′ ( π/
2) = 0, and that all the odd deriva-tives of g ( ϕ ) = csc ϕ are 0 at ϕ = π/
2. Hence, ρ ′ K is continuous at π/
2, and for k ≥ ρ ( k ) K (cid:16) π (cid:17) = ( k − / X i =0 (cid:18) k i (cid:19) g (2 i ) (cid:16) π (cid:17) I ( k − i ) (cid:16) π (cid:17) . The highest order derivative of s ( u ) will appear in the term I ( k ) . Since s ∈ C m \ C m +1 at ϕ = π/
2, we have to find the first value of k for which ρ ( k ) K contains a term with s ( m ) multiplied by a function which is non-zero at π/
2. That value of k will give us theregularity of ρ K .Since s ( u ) = ρ L ( π/ − u ) n − (cid:16) − sin u sin ϕ (cid:17) ( n − / , we need to differentiate the integral I ( ϕ ) ( n − / s to appear outside of an integral. Indeed, I (( n − / contains a term of the form (cos ϕ ) ( n − / (sin ϕ ) ( n − / s ( ϕ ) . Let f ( ϕ ) = (cos ϕ ) ( n − / (sin ϕ ) ( n − / . Then, f ( l ) ( π/
2) = 0 for 0 ≤ l ≤ ( n − /
2, and f (( n − / ( π/ =0. Thus, I ( n − m ) is the first of the derivatives of I containing the term s ( m ) ( ϕ ) f (( n − / ( ϕ ).This shows that ρ K has regularity C m + n − \ C m + n − at π/ (cid:3) Theorem 8.1.13 in [2] proves that if a body K ⊂ R is axis-convex and its radialfunction ρ K ( ϕ ) is C for 0 < ϕ < π/ C at ϕ = π/
2, then K is an intersectionbody of a star body. (The statement of theorem actually asks for ρ K to be in C , butin its proof the continuity of the second derivative is only used at the point π/ ρ K , as inProposition 8. The axis-convexity guarantees that the inverse Radon transform of ρ K is non-negative. It is well known that, in dimensions 5 and higher, there exist infinitelysmooth convex bodies that are not intersection bodies of star bodies [2]. Hence, wecannot expect a result similar to Theorem 8.1.13 in dimension n ≥
5. However, the
NTERSECTION BODIES 15 following theorem shows that if a body of revolution K ⊂ R n is an intersection bodyof a star body that verifies an additional axis-convexity-type property, then K is alsoan intersection body of a star body in dimension 2 n + 2.Before stating the theorem, we will introduce some notation. We say that a bodyof revolution L ⊂ R n is equator-convex if its intersection with every plane parallel tothe equator of L is convex. If we consider L to be 2-dimensional, L is equator-convexif every line parallel to the x-axis intersects the body in a line segment.Given a radial function ρ K ( ϕ ), 0 ≤ ϕ ≤ π/
2, we will denote by K n the body ofrevolution in R n whose radial function is ρ K . Theorem 10.
Let n ≥ even, and let K n be a body of revolution with radial function ρ K . Assume that K n is the intersection body of a star body L ⊂ R n . If L is equator-convex and ρ L ∈ C , then K n +2 is also an intersection body of a star body. If L is notequator-convex, then K n +2 is not an intersection body. Proof:
Since K n is the intersection body of L , and denoting f ( t ) = ρ L (arccos t ), wehave by (7):(15) ρ K ( x ) = 2 ω n − ( n − x n − Z x f ( t ) n − ( x − t ) ( n − / dt. Let e ρ ( t ) be the ( n + 2)-inverse Radon transform of ρ K . To show that K n +2 is anintersection body of a star body, we need to check that e ρ ( t ) is a non-negative continuousfunction. By (8) and (15), e ρ ( t ) = 1( n − ω n +1 t (cid:18) t ddt (cid:19) n Z t ρ K (arcsin x ) x n ( t − x ) ( n − / dx (16) = c n t (cid:18) t ddt (cid:19) n Z t x ( t − x ) ( n − / (cid:18)Z x f ( u ) n − ( x − u ) ( n − / du (cid:19) dx, where c n = ω n − ( n − ω n +1 ( n − . After ( n ) applications of the operator (cid:0) t ddt (cid:1) , equation (16)becomes(17) ( n − c n t (cid:18) t ddt (cid:19) n t Z t f ( u ) n − ( t − u ) ( n − / du. The next application of (cid:0) t ddt (cid:1) results in two terms,( n − c n t (cid:18) t ddt (cid:19) n − (cid:20) Z t f ( u ) n − ( t − u ) ( n − / du +( n − t Z t f ( u ) n − ( t − u ) ( n − / du (cid:21) . Let us call A = ( n − c n t (cid:18) t ddt (cid:19) n − Z t f ( u ) n − ( t − u ) ( n − / du and B = ( n − c n t (cid:18) t ddt (cid:19) n − ( n − t Z t f ( u ) n − ( t − u ) ( n − / du It is not hard to see that A = 2 c n ( n − n − f ( t ) n − . As for B , it is the same as equation (17), with n replaced by n −
2. Hence, after ( n − A + B = c n ( n − n − " ( n − f ( t ) n − + t (cid:18) ddt (cid:19) t Z t f ( u ) n − du = c n ( n − n − (cid:20) ( n − f ( t ) n − + ddt (cid:18) Z t f ( u ) n − du + tf ( t ) n − (cid:19)(cid:21) = c n ( n − n − (cid:2) ( n − f ( t ) n − + ( n − tf ( t ) n − f ′ ( t ) (cid:3) = 12 π f ( t ) n − (cid:20) tf ′ ( t ) f ( t ) (cid:21) . Thus, e ρ ( t ) = π f ( t ) n − [ f ( t ) + tf ′ ( t )]. Since f ∈ C , e ρ is continuous. Since L isequator-convex, ( tf ( t )) is increasing, which means that f ( t ) + tf ′ ( t ) and e ρ are non-negative. We conclude that K n +2 is an intersection body of the body whose radialfunction is (( n + 1) e ρ ( t )) / ( n +1) .On the other hand, if L is not equator-convex, then e ρ takes negative values and K n +2 is not an intersection body (with the general definition). In this case, the assumption f ∈ C may be relaxed. (cid:3) NTERSECTION BODIES 17
Example 11.
Our first application of Theorem 10 will be the construction of a body that is anintersection body of a star body up to dimension 2 n , and is not an intersection bodystarting from dimension 2 n + 2. Consider the star body of revolution L whose radialfunction is ρ L ( ψ ) = ψ + sin ψ ψ − , ≤ ψ < π/ ψ, π/ < ψ ≤ π/ . The function ρ L is just continuous, and L is not equator-convex (see Figure 1). Let K n be the intersection body of L in dimension n = 2 n , and ρ K its radial function.Notice that, although ρ L is not C , its derivative is piecewise continuous. Hence, indimension 2 n + 2, ρ K is the Radon transform of a piecewise continuous, sign-changingfunction., and this means that K n +2 is not an intersection body. Figure 2 shows thecross-section of the intersection body of L in dimension 4. - - - - Figure 1.
Cross-section of the body L in Example 11. - - - - Figure 2.
Cross-section of the body K = IL in Example 11. In dimension6, this body is not an intersection body. We can also start with a body that has higher regularity. For example, let ρ e L ( ψ ) =(2 − ψ + 5 cos ψ ) / . This function is C ∞ , but e L is not equator-convex (seeFigure 3). As before, let ρ e K be given by (6) with ρ L = ρ e L and n = 2 n . Then,by Theorem 10, e K n is an intersection body of a star body, but e K n +2 is not anintersection body. Example 12.
We will now study the barrel B = B n + B n − ⊂ R n , where B n is the unit ball in R n .See Figure 4. This body was introduced by Lonke in the paper [8], where he provedthat, in dimensions 3 and 4, B is a zonoid whose dual is a zonoid (and, in particular, B is an intersection body). Lonke’s proof that the dual of B is a zonoid does not workin dimensions 5 and higher, and we are interested in studying if the reason can be that B no longer is an intersection body. - - - - Figure 3.
Cross-section of the body e L in Example 11. - - - - Figure 4.
Cross-section of Lonke’s barrel zonoid (see Example 12).
NTERSECTION BODIES 19
Its radial function is ρ B ( ϕ ) = sec ϕ, ≤ ϕ < π/
42 sin ϕ, π/ < ϕ ≤ π/ , and thus ρ B ( ϕ ) is C at the point ϕ = π/
4, and C ∞ everywhere else. By Proposition 8, B is not an intersection body of a star body in dimensions 6 and higher. In dimension4, we use the inversion formula (8) to obtain that B is the intersection body of thebody L whose radial function is ρ L ( ϕ ) = (cid:18) ψπ (cid:19) / , ≤ ψ < π/ (cid:18) π (cid:19) / csc ψ, π/ < ψ ≤ π/ , Note that ρ ′ L is piecewise continuous and that the body L is equator-convex (see Figure5). Although B is not an intersection body of a star body in dimension 6, it is anintersection body in R , since ρ B is the Radon transform of a non-negative piecewisecontinuous function.Proposition 8 shows that, for a given function ρ K , increasing the dimension by twounits decreases the regularity of its inverse Radon transform by 1. Since the inverseRadon transform of ρ B ( ϕ ) is piecewise continuous in dimension 6, we expect thatin dimension 8 it will contain delta functions. Indeed, in the sense of distributions - - - - - Figure 5.
Cross-section of the body whose intersection body in R isLonke’s barrel zonoid (see Example 12). ρ B (arcsin x ) equals, up to a constant, the Radon transform of ρ L (arccos t ) − δ / √ ( t ),where δ / √ is the Dirac measure supported at the point t = 1 / √
2, and ρ L (arccos t ) = − t ) / , < t < / √ t, / √ ≤ t < . Since this measure is negative at the point t = 1 / √ B is not an intersection bodyin dimensions 8 and higher. Thus, the dual of Lonke’s barrel’s is not a zonoid indimensions 8 and higher. Only in dimensions 5, 6 and 7 the question is still unanswered. References [1] E. D. Bolker,
A class of convex bodies,
Trans. Amer. Math. Soc. 145 (1969), 323–345. 2[2] R. J. Gardner,
Geometric Tomography, 2nd edition , Cambridge University Press, 2006. 1, 3, 9,10, 14[3] R. J. Gardner, A. Koldobsky, T. Schlumprecht,
An analytic solution to the Busemann-Pettyproblem on sections of convex bodies,
Ann. Math. 149 (1999), 691–703. 2[4] P. Goodey, W. Weil,
Intersection bodies and ellipsoids , Mathematika 42 (1995), 295–304. 4[5] A. Koldobsky,
Fourier Analysis in Convex Geometry,
Math. Surveys and Monographs, AMS(2005). 1, 2, 6[6] A. Koldobsky,
Second derivative test for intersection bodies,
Adv. Math. 136 (1998), 15–25. 3, 4,6[7] J. Lindenstrauss, A. Pelczy´nski,
Absolutely summing operators in L p spaces and their applications, Studia Math. 29 (1968), 257–326. 3[8] Y. Lonke,
On zonoids whose polars are zonoids,
Israel J. Math. 102 (1997), 1–12. 3, 18[9] E. Lutwak,
Intersection bodies and dual mixed volumes,
Advances in Math. 71 (1988), 232–261.2[10] R. Schneider,
Zonoids whose polars are zonoids,
Proc. AMS 50 (1974), 365–368. 3[11] R. Schneider,
Convex bodies: The Brunn-Minkowski Theory,
Cambridge University Press, 1993.2[12] G. Zhang,
Intersection bodies and polytopes,
Mathematika, 46 (1999), 29–34. 2
Department of Mathematics, North Dakota State University
E-mail address ::