Inequalities for the Radon transform on convex sets
aa r X i v : . [ m a t h . M G ] J a n INEQUALITIES FOR THE RADON TRANSFORM ON CONVEXSETS
APOSTOLOS GIANNOPOULOS, ALEXANDER KOLDOBSKY, AND ARTEM ZVAVITCH
Abstract.
Several years ago the authors started looking at some problems of con-vex geometry from a more general point of view, replacing volume by an arbitrarymeasure. This approach led to new general properties of the Radon transform onconvex bodies including an extension of the Busemann-Petty problem and a slicinginequality for arbitrary functions. The latter means that the sup-norm of the Radontransform of any probability density on a convex body of volume one is boundedfrom below by a positive constant depending only on the dimension. In this note,we prove an inequality that serves as an umbrella for these results. Let K and L bestar bodies in R n , let 0 < k < n, and let f, g be non-negative continuous functionson K and L , respectively, so that k g k ∞ = g (0) = 1 . Then R K f (cid:0)R L g (cid:1) n − kn | K | kn ≤ nn − k ( d ovr ( K, BP nk )) k max H R K ∩ H f R L ∩ H g , where | K | stands for volume of proper dimension, C is an absolute constant, maxi-mum is taken over all ( n − k )-dimensional subspaces of R n , and d ovr ( K, BP nk ) is theouter volume ratio distance from K to the class of generalized k -intersection bodiesin R n . This result also implies a mean value inequality for the Radon transform,as follows. If K is an origin-symmetric convex body in R n , < k < n, and f is anintegrable non-negative function on K, then R K f | K | ≤ C k (cid:18)r nk log (cid:16) enk (cid:17)(cid:19) k max H R K ∩ H f | K ∩ H | . Introduction
The Busemann-Petty problem [8] was posed in 1956 and asks the following question.Let K and L be origin-symmetric convex bodies in R n , and suppose that the ( n − K is smaller than thecorresponding one for L, i.e. | K ∩ ξ ⊥ | ≤ | L ∩ ξ ⊥ | for every ξ ∈ S n − . Does itnecessarily follow that the n -dimensional volume of K is smaller than the volumeof L, i.e. | K | ≤ | L | ? Here ξ ⊥ = { x ∈ R n : h x, ξ i = 0 } is the central hyperplaneperpendicular to ξ ∈ S n − , and | K | stands for volume of proper dimension. The Mathematics Subject Classification.
Key words and phrases.
Convex bodies; Sections; Radon transform; Projections, Cosine Trans-form, Intersection body.Part of the work was done when the first named author visited the University of Missouri as aMiller Scholar. The second named author is supported by the Hellenic Foundation for Research andInnovation (Project Number: 1849). The second named author was supported in part by the U.S.National Science Foundation Grant DMS-1700036. The third named author was supported in partby the U.S. National Science Foundation Grant DMS-2000304 and United States - Israel BinationalScience Foundation (BSF). answer is affirmative if the dimension n ≤ , and it is negative when n ≥
5; see[13, 28] for the solution of the problem and its history.Since the answer to the Busemann-Petty problem is negative in most dimensions, itis natural to ask whether the inequality for volumes holds up to an absolute constant.This is known as the isomorphic Busemann-Petty problem introduced in [44]. Doesthere exist an absolute constant C so that for any dimension n and any pair of origin-symmetric convex bodies K and L in R n satisfying | K ∩ ξ ⊥ | ≤ | L ∩ ξ ⊥ | for all ξ ∈ S n − , we have | K | ≤ C | L | ?As pointed out in [44], the isomorphic Busemann-Petty problem is equivalent tothe slicing problem of Bourgain [4, 5]. Does there exist an absolute constant C sothat for any n ∈ N and any origin-symmetric convex body K in R n | K | n − n ≤ C max ξ ∈ S n − | K ∩ ξ ⊥ | ? (1)In other words, is it true that every origin-symmetric convex body K of volume onein R n has a hyperplane section with area greater than an absolute constant, i.e. thereexists ξ ∈ S n − so that | K ∩ ξ ⊥ | > c, where c does not depend on K and n ? Theisomorphic Busemann-Petty problem and the slicing problem are still open. Klartag[25] proved that C ≤ O ( n / ), removing a logarithmic term from an earlier estimateof Bourgain [6].Several years ago the authors decided to look at these problems from a more generalpoint of view, replacing volume by an arbitrary measure. An extension of the slicingproblem was considered in [29, 30, 31, 10, 26, 27]. It was proved in [31] that for any n ∈ N , any star body K in R n and any non-negative continuous function f on K, Z K f ≤ d ovr ( K, I n ) | K | /n max ξ ∈ S n − Z K ∩ ξ ⊥ f. (2)Here d ovr ( K, Ω) = inf ((cid:18) | D || K | (cid:19) /n : K ⊂ D, D ∈ Ω ) (3)is the outer volume ratio distance from K to a fixed class Ω of bodies in R n , in theparticular case of equation (2) that we consider, Ω = I n is the class of intersectionbodies (see definition below).It is interesting to note that we do not need to require the function f to be even,as well as the body K to be symmetric, and additional assumptions are only neededto estimate d ovr ( K, I n ). Since the class of intersection bodies includes ellipsoids, byJohn’s theorem [24], if K is origin-symmetric and convex, then d ovr ( K, I n ) ≤ √ n. Thus, there exists a constant s n ≥ √ n so that for any origin-symmetric convex body K of volume 1 in R n and any integrable non-negative function f on K with R K f = 1 , there exists a direction ξ ∈ S n − for which R K ∩ ξ ⊥ f ≥ s n . In other words, the sup-norm of the Radon transform of any probability density on a convex body of volumeone is bounded from below by a positive constant depending only on the dimension.Note that this estimate was extended later to the case of non-symmetric bodies in[10]. An extension to the derivatives of the Radon transform was obtained in [20].
ADON TRANSFORM ON CONVEX SETS 3
On the other hand, it was proved in [26] that there exists an origin-symmetricconvex body M in R n and a probability density f on M so that Z M ∩ H f ≤ C √ log log n √ n | M | − /n , for every affine hyperplane H in R n , where C is an absolute constant. The logarithmicterm was later removed in [27], so s n ≤ C/ √ n. Finally, c √ n ≤ s n ≤ c √ n , where c , c > K is a starbody in R n , f is a continuous non-negative function on K , and 1 ≤ k < n, then Z K f ≤ C k ( d ovr ( K, BP nk )) k | K | k/n max H ∈ Gr n − k Z K ∩ H f, (4)where C is an absolute constant, and BP nk is the class of generalized k -intersectionbodies (see definition below). This implies that there exists a constant c n,k > K in R n and any probability density f on K, there existsan ( n − k )-dimensional subspace H in R n so that R K ∩ H f ≥ c n,k . Moreover, applyinginequality (8) we get ( c n,k ) /k ≥ c √ k q n log ( enk ) , where c > c n,k . An extension of the Busemann-Petty problem to arbitrary functions was found in[48, 49]. Suppose that f is an even continuous strictly positive function on R n , and K and L are origin-symmetric convex bodies in R n so that Z K ∩ ξ ⊥ f ≤ Z L ∩ ξ ⊥ f, ∀ ξ ∈ S n − . (5)Does it necessarily follow that R K f ≤ R L f ? The answer is the same as for volume:affirmative if n ≤ n ≥ . An isomorphic version was proved in [37]. For every dimension n, inequalities (5)imply R K f ≤ √ n R L f. In fact, it was proved in [37] that inequalities (5) imply R K f ≤ d BM ( K, I n ) R L f, where d BM ( K, I n ) = inf { a > ∃ D ∈ I n : D ⊂ K ⊂ aD } is the Banach-Mazur distance from K to the class of intersection bodies. Now if K is origin-symmetric and convex, by John’s theorem, d BM ( K, I n ) ≤ √ n, so the √ n estimate follows. It is not known whether the √ n estimate is optimal in this problem.Another version of the isomorphic Busemann-Petty problem was proved in [35]; seeSection 5.In this article we prove a theorem that generalizes these results. Theorem 1.
Let K and L be star bodies in R n , let < k < n, and let f, g be non-negative continuous functions on K and L , respectively, so that k g k ∞ = g (0) = 1 . APOSTOLOS GIANNOPOULOS, ALEXANDER KOLDOBSKY, AND ARTEM ZVAVITCH
Then R K f (cid:0)R L g (cid:1) n − kn | K | kn ≤ nn − k ( d ovr ( K, BP nk )) k max H ∈ Gr n − k R K ∩ H f R L ∩ H g . (6)In Section 5, we deduce various properties of the Radon transform from the latterinequality. For example, if we put K = L and g ≡ R K f | K | ≤ nn − k ( d ovr ( K, BP nk )) k max H ∈ Gr n − k R K ∩ H f | K ∩ H | . (7)The distance d ovr ( K, BP nk ) does not depend on n for some special classes of bodies K. This distance is equal to 1 for intersection bodies K, because I n ⊂ BP nk for all k [22, 43]. It was proved in [31] that for unconditional convex bodies K, d ovr ( K, I n ) ≤ e. Also, this distance is bounded by an absolute constant for the polar bodies of convexbodies with bounded volume ratio [31]. For the unit balls of n -dimensional subspacesof L p , p > , the distance is less than c √ p, where c > L p with 0 < p ≤ K d ovr ( K, BP nk ) ≤ C r nk log (cid:16) enk (cid:17) , (8)where C is an absolute constant. Thus, if K is origin symmetric convex, the meanvalue inequality holds with a constant depending only on n, k : R K f | K | ≤ C k (cid:18)r nk log (cid:16) enk (cid:17)(cid:19) k max H R K ∩ H f | K ∩ H | . In Section 3 we describe an alternative approach to Theorem 1 which is based onBlaschke-Petkantchin formulas and affine isoperimetric inequalities. This approachwas initiated in [10] and leads to a version of (6) which is valid for more general pairsof sets. Below, for any bounded Borel set K in R n we denote by ovr( K ) the outervolume ratio ovr( K ) = d ovr ( K, L ) = inf E (cid:0) |E| / | K | (cid:1) /n , where the infimum is over allorigin symmetric ellipsoids E in R n with K ⊆ E . Theorem 2.
Let K and L be two bounded Borel sets in R n . Let f and g be twobounded non-negative measurable functions on K and L , respectively, and assumethat k g k > and k g k ∞ = 1 . For every ≤ k ≤ n − we have that R K f (cid:0)R L g (cid:1) n − kn | K | kn ≤ ( C · ovr( K )) k max H ∈ Gr n − k R K ∩ H f R L ∩ H g , (9) where C > is an absolute constant. It should be noted that even for origin-symmetric convex bodies K in R n the outervolume ratio ovr( K ) can be as large as √ n . This is a disadvantage of Theorem 2which does not provide estimates depending on k in contrast to Theorem 1. Regardingthis comparison, we mention that besides Blaschke-Petkantchin formulas and affineisoperimetric inequalities, the proof of Theorem 2 exploits a well-known result of ADON TRANSFORM ON CONVEX SETS 5
Bar´any and F¨uredi [3] which may be stated as follows: if E is an ellipsoid in R m , s ≥ m + 1 and w , . . . , w m ∈ E , then (cid:18) | conv( w , . . . , w m ) ||E| (cid:19) /m ≤ C p log(1 + s/m ) /m where C > w , . . . , w m are chosen from a body L ∈BP nk .In Section 4 we obtain a generalization of the isomorphic version of the Shephardproblem due to Ball [1, 2]. Ball has proved that if K and L are origin-symmetricconvex bodies in R n such that | K | ξ ⊥ | ≤ | L | ξ ⊥ | , ∀ ξ ∈ S n − , then | K | ≤ d vr ( L, Π n ) | L | . In this statement, K | ξ ⊥ denotes the orthogonal projection of K in the direction of ξ and d vr ( L, Π n ), defined in (3), is the volume ratio distance from L to the classΠ n of projection bodies. Replacing Π n by the class Π p,n of p -projection bodies (seeSection 4 for background information) we obtain the following: Theorem 3.
Fix p ≥ and let K and L be convex bodies in R n , then (cid:18) | K || L | (cid:19) n − ppn ≤ d vr ( L, Π p,n ) max ξ ∈ S n − h Π p K ( ξ ) h Π p L ( ξ ) . Throughout the paper, we write a ≃ b if c b ≤ a ≤ c b, where c , c > Proof of Theorem 1
We need several definitions and facts. A closed bounded set K in R n is called a star body if every straight line passing through the origin crosses the boundary of K at exactly two points different from the origin, the origin is an interior point of K, and the Minkowski functional of K defined by k x k K = min { a ≥ x ∈ aK } is a continuous function on R n . We use the polar formula for the volume | K | of a starbody K : | K | = 1 n Z S n − k θ k − nK dθ. (10)If f is an integrable function on K , then Z K f = Z S n − Z k θ k − K r n − f ( rθ ) dr ! dθ. (11)Let Gr n − k be the Grassmanian of ( n − k )-dimensional subspaces of R n . For 1 ≤ k ≤ n − , the ( n − k ) -dimensional spherical Radon transform R n − k : C ( S n − ) → C ( Gr n − k )is a linear operator defined by R n − k g ( H ) = Z S n − ∩ H g ( x ) dx, ∀ H ∈ Gr n − k APOSTOLOS GIANNOPOULOS, ALEXANDER KOLDOBSKY, AND ARTEM ZVAVITCH for every function g ∈ C ( S n − ) . For every H ∈ Gr n − k , the ( n − k )-dimensional volume of the section of a star body K by H can be written as | K ∩ H | = 1 n − k R n − k ( k · k − n + kK )( H ) . (12)More generally, for an integrable function f and any H ∈ Gr n − k , Z K ∩ H f = R n − k Z k·k − K r n − k − f ( r · ) dr ! ( H ) . (13)The class of intersection bodies I n was introduced by Lutwak [38]. We consider ageneralization of this concept due to Zhang [47]. We say that an origin symmetricstar body D in R n is a generalized k -intersection body , and write D ∈ BP nk , if thereexists a finite Borel non-negative measure ν D on Gr n − k so that for every g ∈ C ( S n − ) Z S n − k x k − kD g ( x ) dx = Z Gr n − k R n − k g ( H ) dν D ( H ) . (14)When k = 1 we get the original Lutwak’s class of intersection bodies BP n = I n . Proof of Theorem 1.
For a small δ > , let D ∈ BP nk be a body such that K ⊂ D and | D | n ≤ (1 + δ ) d ovr ( K, BP nk ) | K | n , (15)and let ν D be the measure on Gr n − k corresponding to D by the definition (14).Let ε be such that Z K ∩ H f ≤ ε Z L ∩ H g, ∀ H ∈ Gr n − k . By (13), we have R n − k Z k·k − K r n − k − f ( r · ) dr ! ( H ) ≤ ε R n − k Z k·k − L r n − k − g ( r · ) dr ! ( H )for every H ∈ Gr n − k . Integrating both sides of the latter inequality with respect to ν D and using the definition (14), we get Z S n − k x k − kD Z k x k − K r n − k − f ( rx ) dr ! dx (16) ≤ ε Z S n − k x k − kD Z k x k − L r n − k − g ( rx ) dr ! dx, which is equivalent to Z K k x k − kD f ( x ) dx ≤ ε Z L k x k − kD g ( x ) dx. (17)Since K ⊂ D, we have 1 ≥ k x k K ≥ k x k D for every x ∈ K. Therefore, Z K k x k − kD f ( x ) dx ≥ Z K k x k − kK f ( x ) dx ≥ Z K f. ADON TRANSFORM ON CONVEX SETS 7
On the other hand, by [44, Lemma 2.1], R L k x k − kD g ( x ) dx R D k x k − kD dx ! / ( n − k ) ≤ (cid:18) R L g ( x ) dx R D dx (cid:19) /n . Since R D k x k − kD dx = nn − k | D | , we can estimate the right-hand side of (17) by Z L k x k − kD g ( x ) dx ≤ ε nn − k (cid:18)Z L g (cid:19) n − kn | D | kn . Applying (15) and sending δ to zero, we see that the latter inequality in conjunctionwith (17) implies Z K f ≤ ε nn − k ( d ovr ( K, BP nk )) k | K | kn . Now put ε = max H ∈ Gr n − k R K ∩ H f R L ∩ H g . ✷ If f ≡ , g ≡ , we can get a slightly sharper inequality than what Theorem 1gives in this case. Theorem 4.
Let
K, L be star bodies in R n and < k < n, then (cid:18) | K || L | (cid:19) n − kn ≤ ( d ovr ( K, BP nk )) k max H ∈ Gr n − k | K ∩ H || L ∩ H | . Proof:
Let ε be such that | K ∩ H | ≤ ε | L ∩ H | for all H ∈ Gr n − k , and let D be asin the proof of Theorem 6. By (12), for all H R n − k ( k · k − n + kK )( H ) ≤ ε R n − k ( k · k − n + kL )( H ) . Integrating both sides with respect to ν D and using the definition (14) we get Z S n − k x k − kD k x k − n + kK dx ≤ ε Z S n − k x k − kD k x k − n + kL dx. Since K ⊂ D, we have 1 ≥ k x k K ≥ k x k D , and by (10) the left-hand side is greaterthan n | K | . Using this and H¨older’s inequality, n | K | ≤ ε Z S n − k x k − kD k x k − n + kL dx ≤ ε (cid:18)Z S n − k x k − nD dx (cid:19) kn (cid:18)Z S n − k x k − nL dx (cid:19) n − kn = ε n | D | kn | L | n − kn ≤ ε n (1 + δ ) k ( d ovr ( K, BP nk )) k | K | kn | L | n − kn . Sending δ to zero and setting ε = max H ∈ Gr n − k | K ∩ H || L ∩ H | , we get the result. ✷ APOSTOLOS GIANNOPOULOS, ALEXANDER KOLDOBSKY, AND ARTEM ZVAVITCH Proof of Theorem 2
Our first tool will be a Blaschke-Petkantchin formula (see [46, Chapter 7.2] and[14, Lemma 5.1]).
Lemma 1 (Blaschke-Petkantschin) . Let ≤ s ≤ n − . There exists a constant p ( n, s ) > such that, for every non-negative bounded Borel measurable function F :( R n ) s → R , Z R n · · · Z R n F ( x , . . . , x s ) dx s · · · dx (18)= p ( n, s ) Z G n,s Z H · · · Z F f ( x , . . . , x s ) | conv(0 , x , . . . , x s ) | n − s dx s · · · dx dν n,s ( H ) , where ν n,s is the Haar probability measure on Gr s . The exact value of the constant p ( n, s ) is p ( n, s ) = ( s !) n − s ( nω n ) · · · (( n − s + 1) ω n − s +1 )( sω s ) · · · (2 ω ) ω , (19) where ω n is the volume of the unit Euclidean ball B n . We shall also use the next inequality, proved independently by Busemann andStraus [9], and Grinberg [21].
Lemma 2 (Busemann-Straus, Grinberg) . Let K be a bounded Borel set of volume in R n . For any ≤ k ≤ n − and T ∈ SL ( n ) we have Z Gr n − k | K ∩ H | n dν n,n − k ( H ) ≤ Z Gr n − k | B n ∩ H | n dν n,n − k ( H ) , (20) where B n is the Euclidean ball of volume . Our next tool will be a theorem of Dann, Paouris and Pivovarov from [11]; the proofof this fact combines Blaschke-Petkantschin formulas with rearrangement inequalities.
Lemma 3 (Dann-Paouris-Pivovarov) . Let g be a non-negative, bounded integrablefunction on R n with k g k > . For every ≤ k ≤ n − we have Z Gr n − k k g | H k k ∞ (cid:18)Z H g ( x ) dx (cid:19) n dν n,n − k ( H ) ≤ γ − nn,k (cid:18)Z R n g ( x ) dx (cid:19) n − k , (21) where γ n,k = ω n − kn n /ω n − k . It is checked in [10] that for every 1 ≤ k ≤ n − e − k/ < γ n,k < γ − nn,k p ( n, n − k )] k ( n − k ) ≃ √ n − k. (22)Finally, we need a well-known theorem of B´ar´any and F¨uredi [3]: if s ≥ m + 1 and w j ∈ R m satisfy k w j k ≤ j = 1 , . . . , s , then | conv( w , . . . , w s ) | /m ≤ C p log(1 + s/m ) m . Equivalently, this says that if w j ∈ B m , 1 ≤ j ≤ s , then the volume radius of theirconvex hull is bounded by C p log(1 + s/m ) /m . By affine invariance we obtain: ADON TRANSFORM ON CONVEX SETS 9
Lemma 4.
There exists an absolute constant
C > such that if E is an ellipsoid in R m , s ≥ m + 1 and w , . . . , w m ∈ E , then (cid:18) | conv( w , . . . , w m ) ||E| (cid:19) /m ≤ C p log(1 + s/m ) /m. Proof of Theorem 2.
Let E be a centered ellipsoid such that K ⊆ E andovr( K ) = ( |E| / | K | ) /n . We shall use the next consequence of Lemma 4: if F ∈ Gr n − k and x , . . . , x n − k ∈ K ∩ H ⊆ E ∩ H then conv(0 , x , . . . , x n − k ) ⊆ E ∩ H , and since E ∩ H is an ( n − k )-dimensional centered ellipsoid we must have | conv(0 , x , . . . , x n − k ) | ≤ C p log(1 + ( n − k + 1) / ( n − k ) √ n − k ! k |E ∩ H | (23) ≤ (cid:18) C √ n − k (cid:19) k |E ∩ H | . Applying Lemma 1 for the function F ( x , . . . , x n − k ) = Q n − ki =1 f ( x i ) K ( x i ), with s = n − k , we get (cid:18)Z K f ( x ) dx (cid:19) n − k = Z R n · · · Z R n F ( x , . . . , x n − k ) dx n − k · · · dx = p ( n, n − k ) Z Gr n − k Z K ∩ H · · · Z K ∩ H g ( x ) · · · g ( x n − k ) × | conv(0 , x , . . . , x n − k ) | k dx n − k · · · dx dν n,n − k ( H ) . Let M := max H ∈ Gr n − k R K ∩ H f R L ∩ H g . Then, (23) shows that (cid:18)Z K f ( x ) dx (cid:19) n − k ≤ p ( n, n − k ) (cid:18) C √ n − k (cid:19) k ( n − k ) Z Gr n − k Z K ∩ H · · · Z K ∩ H |E ∩ H | k × f ( x ) · · · f ( x n − k ) dx n − k · · · dx dν n,n − k ( H )= p ( n, n − k ) (cid:18) C √ n − k (cid:19) k ( n − k ) Z Gr n − k |E ∩ H | k (cid:16) Z K ∩ H f ( x ) dx (cid:17) n − k dν n,n − k ( H ) ≤ p ( n, n − k ) (cid:18) C √ n − k (cid:19) k ( n − k ) M n − k Z Gr n − k |E ∩ H | k × (cid:16) Z L ∩ H g ( x ) dx (cid:17) n − k dν n,n − k ( H ) . Now, by H¨older’s inequality and Grinberg’s inequality (20) we get Z Gr n − k |E ∩ H | k (cid:18)Z L ∩ H g ( x ) dx (cid:19) n − k dν n,n − k ( H ) ≤ Z Gr n − k |E ∩ H | n dν n,n − k ( H ) ! kn Z Gr n − k (cid:18)Z L ∩ H g ( x ) dx (cid:19) n dν n,n − k ( H ) ! n − kn ≤ γ − nn,k |E| k ( n − k ) n Z Gr n − k (cid:18)Z L ∩ H g ( x ) dx (cid:19) n dν n,n − k ( H ) ! n − kn = γ − nn,k | K | k ( n − k ) n ovr( K ) k ( n − k ) Z Gr n − k (cid:18)Z L ∩ H g ( x ) dx (cid:19) n dν n,n − k ( H ) ! n − kn . Finally, since k g | H k ∞ ≤ k g k ∞ = 1 for all H ∈ Gr n − k , we may apply (21) to get Z Gr n − k (cid:18)Z L ∩ H g ( x ) dx (cid:19) n dν n,n − k ( H ) ≤ γ − nn,k (cid:18)Z L g ( x ) dx (cid:19) n − k . (24)Combining the above we get (cid:18)Z K f ( x ) dx (cid:19) n − k ≤ h γ − nn,k p ( n, n − k )( C / √ n − k ) k ( n − k ) i (25) × M n − k | K | k ( n − k ) n ovr( K ) k ( n − k ) (cid:18)Z L g ( x ) dx (cid:19) ( n − k )2 n . Note that, by (22), h γ − nn,k p ( n, n − k ) i n − k (cid:18) C √ n − k (cid:19) k ≤ C k for some absolute constant C >
0. Then, the result follows from (25). ✷ Proof of Theorem 3
The proof of Theorem 3 requires several additional definitions and facts from convexgeometry. We refer the reader to [45] for details.The support function of a convex body K in R n is defined by h K ( x ) = max ξ ∈ K h x, ξ i , x ∈ R n . If K is origin-symmetric, then h K is a norm on R n . One of the crucial properties ofthe support function is its relation to the Minkowski sum of convex bodies: h K + L ( x ) = h K ( x ) + h L ( x ) . (26)The surface area measure S ( K, · ) of a convex body K in R n is defined as follows: forevery Borel set E ⊂ S n − , S ( K, E ) is equal to Lebesgue measure of the part of the
ADON TRANSFORM ON CONVEX SETS 11 boundary of K where normal vectors belong to E. The volume of a convex body canbe expressed in terms of its support function and surface area measure: | K | = 1 n Z S n − h K ( x ) dS ( K, x ) . (27)If K and L are two convex bodies in R n , the mixed volume V ( K, L ) is equal to V ( K, L ) = 1 n lim ε → +0 | K + ǫL | − | K | ε . (28)We shall use the first Minkowski inequality: for any pair of convex bodies K, L in R n ,V ( K, L ) ≥ | K | n − n | L | /n . (29)The mixed volume V ( K, L ) can also be expressed in terms of the support functionand surface area measure: V ( K, L ) = 1 n Z S n − h L ( x ) dS ( K, x ) . (30)For a convex body K in R n and ξ ∈ S n − , denote by K | ξ ⊥ the orthogonal projectionof K to the central hyperplane ξ ⊥ . The Cauchy formula states that | K | ξ ⊥ | = 12 Z S n − |h x, ξ i| dS ( K, x ) . (31)Let K be a convex body in R n . The projection body Π K of K is defined as an origin-symmetric convex body in R n whose support function in every direction is equal tothe volume of the orthogonal projection of K to this direction: for every θ ∈ S n − ,h Π K ( ξ ) = | K | ξ ⊥ | . (32)We denote by Π n the class of projection bodies of convex bodies and if D ∈ Π n wesimply say that D is a projection body. By Cauchy’s formula (31), for every projectionbody D there exists a finite measure ν D on S n − such that h D ( x ) = Z S n − |h x, ξ i| dν D ( ξ ) , ∀ x ∈ S n − . (33)Let K denote the class of convex bodies containing the origin in their interior. Firey[12] extended the concept of Minkowski sum (26), and introduced for each real p ≥ p -sum: h pαK + p βL ( x ) = αh pK ( x ) + βh pL ( x ) . Here
K, L ∈ K and α, β are positive real numbers. In a series of papers Lutwak[39, 40] showed that the Firey sums lead to a Brunn-Minkowski theory for each p ≥ p -mixed volume, V p ( K, L ), p ≥ V p ( K, L ) = pn lim ε → V ( K + p ε L ) − V ( K ) ε , for all K, L ∈ K . Lutwak proved that for each K ∈ K , there exists a positive Borelmeasure S p ( K, · ) on S n − so that V p ( K, L ) = 1 n Z S n − h pL ( u ) dS p ( K, u ) for all L ∈ K . It turns out that the measure S p ( K, · ) is absolutely continuous withrespect to S ( K, · ), with Radon-Nikodym derivative dS p ( K, · ) dS ( K, · ) = h ( K, · ) − p . Lutwak [39] generalized the Brunn-Minkowski inequality to the case of p -mixed vol-umes as follows: V p ( K, L ) n ≥ | K | n − p | L | p , p > . (34)We will also use the concept of a p -projection body, introduced by Lutwak [40, 41].Let Π p K , p ≥ h Π p K ( ξ ) p = 12 n Z S n − |h x, ξ i| p d S p ( K, x ) , ξ ∈ S n − . (35)We note that Π K = n Π K . Moreover, for some fixed p ≥
1, we say that D is a p -projection body if D is the p -projection body of some convex body. By (35), forevery p -projection body D there exists a finite measure ν D on S n − such that h pD ( x ) = Z S n − |h x, ξ i| p dν D ( ξ ) , x ∈ R n . (36)Let us denote by Π p,n the class of all p -projection bodies in R n . We may pass now to the proof of Theorem 3.
Proof of Theorem 3.
Let ε > ξ ∈ S n − h p Π p K ( ξ ) ≤ εh p Π p L ( ξ ) . (37)By (35), the condition (37) is equivalent to Z S n − |h x, ξ i| p dS p ( K, x ) ≤ ε Z S n − |h x, ξ i| p dS p ( L, x ) , ∀ ξ ∈ S n − . (38)For small δ > , let D ∈ Π p,n be such that D ⊂ L and | L | n ≤ (1 + δ ) d vr ( L, Π p,n ) | D | n , (39)and let ν D be the measure on S n − corresponding to D by (36). Integrating bothsides of (38) with respect to dν D ( ξ ) , we get Z S n − Z S n − |h x, ξ i| p dS p ( K, x ) dν D ( ξ ) ≤ ε Z S n − Z S n − |h x, ξ i| p dS p ( L, x ) dν D ( ξ ) , for all ξ ∈ S n − . Applying Fubini’s theorem on S n − together with (35) we get Z S n − h pD ( x ) dS p ( K, x ) ≤ ε Z S n − h pD ( x ) dS p ( L, x ) . (40)Since D ⊂ L, we have h D ( x ) ≤ h L ( x ) for every x ∈ S n − , so the right-hand side of(40) can be estimated from above by ε Z S n − h pD ( x ) dS p ( L, x ) ≤ ε Z S n − h pL ( x ) dS p ( L, x ) = εn | L | . ADON TRANSFORM ON CONVEX SETS 13
By (30), (34), (29) and (39), the left-hand side of (40) can be estimated from belowby Z S n − h pD ( x ) dS p ( K, x ) = nV p ( K, L ) ≥ | K | n − pn | D | pn ≥ n (1 + δ ) p d p vr ( L, Π p,n ) | K | n − pn | L | pn . Combining these estimates we see that n (1 + δ ) p d p vr ( L, Π p,n ) | K | n − pn | L | pn ≤ εn | L | . Sending δ → , we get (cid:18) | K || L | (cid:19) n − pn ≤ d p vr ( L, Π p,n ) ε. Now, putting ε = max ξ ∈ S n − h p Π p K ( ξ ) h p Π p L ( ξ ) , we get the result. ✷ A mixed version of the Busemann-Petty and Shephard problems was posed byMilman and solved in [15]. Namely, if K is a convex body in R n , D is a compactsubset of R n and 1 ≤ k ≤ n −
1, then the inequalities | K | H | ≤ | D ∩ H | for all H ∈ Gr n − k imply | K | ≤ | D | . Here K | H is the orthogonal projection of K onto H. One can easily modify the proof from [15] to get a slightly stronger version of thisresult.
Theorem 5.
Let K be a convex body in R n , let D be a compact set in R n , and ≤ k ≤ n − . Then (cid:18) | K || L | (cid:19) n − kn ≤ max H ∈ Gr n − k | K | H || L ∩ H | . Applications
Comparison theorem for the Radon transform.
We can recover the iso-morphic Busemann-Petty theorem for the Radon transform established in [35], asfollows. If, in addition to the conditions of Theorem 1, we assume that Z K ∩ H f ≤ Z L ∩ H g, ∀ H ∈ Gr n − k , then we get Z K f ≤ nn − k ( d ovr ( K, BP nk )) k | K | kn (cid:18)Z L g (cid:19) n − kn . A lower estimate for the sup-norm of the Radon transform.
Theorem 1with L = B n and g ≡ Z K f ≤ nn − k | B n | n − kn | B n − k | ( d ovr ( K, BP nk )) k | K | kn max H Z K ∩ H f. Note that the constant | B n | n − kn | B n − k | is less than 1, and nn − k ≤ e k . Mean value inequality for the Radon transform.
Let K = L, and g ≡ . Then, as we have mentioned in the Introduction, R K f | K | ≤ nn − k ( d ovr ( K, BP nk )) k max H R K ∩ H f | K ∩ H | . The isomorphic Busemann-Petty problem for sections of proportionaldimensions.
Theorem 4 and inequality (8) imply the following result from [33],which solves the isomorphic Busemann-Petty problem in affirmative for sections ofproportional dimensions. If
K, L are origin-symmetric convex bodies in R n and k ≥ λn, where 0 < λ < , so that | K ∩ H | ≤ | L ∩ H | for every H ∈ Gr n − k , then | K | n − kn ≤ ( C ( λ )) | L | n − kn , where the constant C ( λ ) depends only on λ. Inequalities for projections.
Theorem 3 immediately implies an isomorphicversion of the Shephard problem first established by Ball; it immediately follows from[1, 2].
Corollary 1.
Let K and L be origin-symmetric convex bodies in R n such that | K | ξ ⊥ | ≤ | L | ξ ⊥ | , ∀ ξ ∈ S n − , then | K | ≤ d vr ( L, Π n ) | L | . Corollary 2.
Let L be an origin-symmetric convex body in R n . Then min ξ ∈ S n − | L | ξ ⊥ | ≤ √ e d vr ( L, Π n ) | L | n − n . Proof:
Apply Theorem 3 to K = B n and L, and then use the fact that c n, = | B n − | / | B n | n − n ≤ √ e. ✷ By John’s theorem [24] and the fact that ellipsoids are projection bodies (see, forexample [28, 45]), we have d vr ( L, Π n ) ≤ √ n for any origin-symmetric convex body L in R n . On the other hand, Ball [1] proved that there exists an absolute constant c > n there exists an origin-symmetric convex body L n of volume1 in R n satisfying | L | ξ ⊥ | ≥ c √ n for all ξ ∈ R n . Combined with Corollary 2, theseestimates show that c √ n ≤ max L d vr ( L, Π n ) ≤ √ n, (41)where c is an absolute constant, and the maximum is taken over all origin-symmetricconvex bodies in R n . This estimate was first established by Ball; it immediately followsfrom [2, Example 2].
ADON TRANSFORM ON CONVEX SETS 15
Note that the distance d vr ( L, Π n ) has been studied by several authors. It wasintroduced in [2] and was proved to be equivalent to the weak-right-hand-Gordon-Lewis constant of L . Also it was connected to the random unconditional constantof the dual space (see Theorem 5 and Proposition 6 in [2]). In [16] this distancewas called zonoid ratio, and it was proved that it is bounded from above by theprojection constant of the space. In the same paper the zonoid ratio was computedfor several classical spaces. We refer the interested reader to [16], [18], [17], [19] formore information.5.6. Milman’s estimate for the isotropic constant.
We say that a compact set K with volume 1 in R n is in isotropic position if for each ξ ∈ S n − Z K h x, ξ i dx = L K where L K is a constant that is called the isotropic constant of K. In the case where K is origin-symmetric convex, the slicing problem of Bourgain is equivalent to provingthat L K is bounded by an absolute constant.Hensley [23] has proved that there exist absolute constants c , c > K in R n in isotropic position and any ξ ∈ S n − c L K ≤ | K ∩ ξ ⊥ | ≤ c L K . The following inequality was proved by Milman [42]. We present a simpler proof.
Theorem 6.
There exists an absolute constant C so that for any origin-symmetricisotropic convex body K in R n L K ≤ C d ovr ( K, I n ) . Proof:
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Math. Ann. 331, no. 4 (2005),867–887.(Apostolos Giannopoulos)
Department of Mathematics, National and Kapodistrian Uni-versity of Athens, Panepistimiopolis 157-84, Athens, Greece
Email address : [email protected] (Alexander Koldobsky) Department of Mathematics, University of Missouri, Columbia,MO 65211, USA
Email address : [email protected] (Artem Zvavitch) Department of Mathematical Sciences, Kent State University, Kent,OH USA
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