Gaussian Measure of Sections of convex bodies

Transcription

1 Gaussian Measure of Sections of convex bodies A. Zvavitch Department of Mathematics, University of Missouri, Columbia, MO 652, USA Abstract In this paper we study properties of sections of convex bodies with respect to the Gaussian measure. We develop a formula connecting the Minkowski functional of a convex symmetric body with the Gaussian measure of its sections. Using this formula we solve an analog of the Busemann-Petty problem for Gaussian measures. ey words: Convex body, Gaussian Measure, Busemann-Petty problem Introduction The standard Gaussian measure on R n is given by its density: γ n () = ( 2π) n e x 2 2 dx, where x denotes the l 2 norm on R n. Let C n denote the collection of convex closed subsets of R n with nonempty interior, which are symmetric about the origin. In this paper we give an answer to the following question: Gaussian Busemann-Petty problem (GBP): Assume,L C n and γ n ( ξ ) γ n (L ξ ), ξ, where ξ denotes the section of by the central hyperplane orthogonal to ξ. Does it follow that γ n () γ n (L)? In Section 3 we show that the answer is affirmative if n 4 and it is negative if n 5. address: zvavitch@math.missouri.edu (A. Zvavitch). Preprint to appear in Advances in Mathematics December 23

2 This problem is a Gaussian analog of the Busemann-Petty problem, posed in 956 (see [3]) and asking the same question with Lebesgue measure in place of the Gaussian measure. The answer is affirmative if n 4 and negative if n 5, The solution appeared as the result of a sequence of papers: [6] n 2, [] n, [9] and [2] n 7, [8] and [4] n 5, [5] n = 3, [22] and [7] n = 4 (we refer to [22], [7] and [4] for more historical details). Let us outline the analytic solution to the Busemann-Petty problem from [7]. The first ingredient is a connection with intersection bodies found by Lutwak [7]. Let and M be symmetric star bodies in R n. We say that is the intersection body of M if the radius of in every direction is equal to the (n )-dimensional volume of the central hyperplane section of L perpendicular to this direction, i.e. for every ξ, ξ = Vol(M ξ ). A more general class of intersection bodies can be defined as the closure of the class of intersection bodies of star bodies in the radial metric. Lutwak [7] proved that if is an intersection body then the answer to the Busemann-Petty problem is affirmative for every L, and, on the other hand, if L not an intersection body, then one can perturb it to construct a body giving together with L a counterexample. Therefore, the answer to the Busemann-Petty problem in R n is affirmative if and only if every symmetric convex body in R n is an intersection body. The second ingredient is the following Fourier analytic characterization of intersection bodies found by oldobsky [2]: an origin symmetric star body in R n is an intersection body if and only if the function represents a positive definite distribution on R n. The characterization is based on the following formula ([], Theorem ) π(n ) ( x n+ ) (ξ) = A,ξ (), () where A,ξ (t) = Vol( {ξ + tξ}). The formula () was generalized by Gardner, oldobsky and Schlumprecht [7] to (i) If k is even ( n+k+ ) (ξ) = ( ) k/2 π(n k )A (k),ξ (); (2) (ii) If k is odd ( n+k+ ) (ξ) = c n,k A,ξ (z) A,ξ ()... A (k ),ξ () zk (k )! dz, (3) z k+ 2

3 where c n,k = ( ) (k+)/2 2(n k )k! and A (k),ξ the order k of the function A,ξ. stands for the derivative of Finally, if the body is convex then, by Brunn s theorem, the central section is maximal among all sections orthogonal to a given direction, so A,ξ() for every ξ, and, therefore, putting n = 4 in (2) and applying Fourier analytic characterization of intersection bodies, we get that every symmetric convex body in R 4 is an intersection body. However, if n = 5 we have to deal with the third derivative of A,ξ which is not controlled by convexity, and one construct symmetric convex bodies in R 5 that are not intersection bodies (see [7]). In the first section of this paper we introduce a Fourier analytic formula for γ n ( ξ ) (see Theorem below). Using this formula and the version of Parseval s identity on the sphere ([3], Lemma 3) we prove that intersection bodies play the same role in GBP as in the original Busemann-Petty problem. In Theorems 3, 4 below we show that if represents a positive definite distribution on R n, then the answer to GBP is affirmative. Finally Theorem 5 below shows that if L is not an intersection body then one can construct a counterexample to GBP. This, together with results about intersection bodies [7], gives affirmative answer to GBP in dimensions 3 and 4 and negative answer in dimension n 5. Some additional remarks are provided in the last section. We show that a counterexample to GBP problem in R n, n 5 can be also constructed directly from the examples to the original Busemann-Petty problem. We also show that this method can be used to give an answer to the projection counterpart of GBP. 2 The main formula In this paper we operate with the Fourier transform of distributions (see [8] for exact definitions and properties). We denote by S the space of rapidly decreasing infinitely differentiable functions (test functions) on R n with values in C. By S we denote the space of distributions over S. The Fourier transform of a distribution f is defined by ˆf, ˆφ = (2π) n f, φ, for every test function φ. The spherical Radon transform is the bounded linear operator on C( ) defined by Rf(ξ) = ξ f(x)dx, f C( ), ξ. 3

4 oldobsky ([], Lemma 4) proved that if f(x) is an even homogeneous function of degree n + on R n \ {}, n > so that f L ( ), then Rf(ξ) = π ˆf(ξ), ξ. (4) Let be an origin-symmetric star-shaped body. We denote by x = min{α > : x α} the Minkowski functional on R n generated by. Theorem Let be a symmetric star-shaped body in R n, then γ n ( ξ ) = ( 2π) n π x / x 2 dt (ξ). Proof : If χ is the indicator function of the interval [, ] then, passing to polar coordinates in the hyperplane ξ we get ( 2π) n γ n ( ξ ) = (x,ξ)= χ( x )e x 2 2 dx = ξ θ t n 2 e t2 2 dtdθ. We extend the function under the spherical integral to a homogeneous of degree n + function on R n and apply (4) to get x x x n+ ξ = = π x / x t n 2 e t2 2 dtdx = R x n+ x x t n 2 e t2 2 dt (ξ) 2 dt (ξ). (5) Remark : Note that the function in (5) is homogeneous of degree (with respect to ξ R n ). This gives a natural extension of G (ξ) = γ n ( ξ ) to a homogeneous function of degree, i.e. for ν R n. G (ν) = ν G (ν/ ν ) = ν γ n ( (ν/ ν ) ), 4

5 Remark 2: If n = 3 then γ 2 ( ξ ) = 2π 2 x 2 e 2 x 2 Indeed, from Theorem and the well-known formula for the Fourier transform of powers of the Euclidean norm (see [8], p. 94) we get γ 2 ( ξ ) = x 2 (ξ) x 2 e 2π 2 2π 2 = x 2 e 2π 2 2 x 2 2 x 2 Theorem also implies that a symmetric body is uniquely determined by the Gaussian measure of its sections: Theorem 2 Let and L be star-shaped origin symmetric bodies in R n. If then = L. (ξ). (ξ). γ n ( ξ ) = γ n (L ξ ), ξ, Proof : Remark after Theorem gives that G (ν) = G L (ν), ν R n. Applying the inverse Fourier transform to both sides of the latter equation and using Theorem we get: (ξ) x n+ which gives x / x t n 2 e t2 2 dt = x n+ x / x L t n 2 e t2 2 dt, x R n, so x = x L. L t n 2 e t2 2 dt = x t n 2 e t2 2 dt, x, 3 The Busemann-Petty Problem for Gaussian Measure We start with positive answer to GBP in dimension 3. 5

6 Theorem 3 Let, L C 3. Assume that γ 2 ( ξ ) γ 2 (L ξ ), ξ, (6) then γ 3 () γ 3 (L). Proof : We say that a body R n is infinitely smooth if the restriction of the Minkowski functional to the sphere belongs to the space C ( ). By elementary approximation, it is enough to prove the theorem in the case of infinitely smooth, symmetric, convex bodies and L. Note that then ([7], Theorem or [3], Remark ) and L are continuous functions on, and the same argument can be applied to show that x 2 e 2 x 2 and x 2 e 2 x 2 L are also continuous functions on. From (6) and Remark 2 after Theorem, x 2 e 2 x 2 (ξ) x 2 e 2 x 2 L (ξ), ξ. Every convex symmetric body in R 3 ([2], Theorem ) and x 2 e 2 x 2 is an intersection body ([5], [7]), so (ξ) x 2 e 2 x 2 L (ξ), ξ. In particular, x 2 e 2 x 2 (ξ)dξ x 2 e 2 x 2 L (ξ)dξ. Note that the sum of degrees of homogeneity of functions under both integrals is 3, so we may apply a version of the Parseval identity on ([3], Lemma 4) to get : e 2 x 2 dx e 2 x 2 L dx. (7) 6

7 Passing to polar coordinates and integrating by parts we get the following expression for γ 3 (): γ 3 () = ( 2π) 3 e 2 x 2 dx + ( 2π) 3 e t2 2 dtdx. Then, to show that γ 3 () γ 3 (L) it is enough to prove e 2 x 2 dx L e 2 x 2 L dx + L e t2 2 dtdx. (8) We may apply the inequality (7) to claim that (8) follows from e 2 x 2 L dx L e 2 x 2 L dx + L e t2 2 dtdx. If we denote = a, L (b a)e b2 /2 = b the latter inequality follows from b a e t2 2 dt, a, b. Theorem 4 (General Case) Consider origin symmetric convex bodies and L in R n, n 4, and suppose that be an intersection body. Then from γ n ( ξ ) γ n (L ξ ), ξ, (9) it follows that γ n () γ n (L). Proof : As in the proof of Theorem 6, it is enough to prove Theorem 4 in the case of infinitely smooth, bodies and L. Then, again,, L, and x / x 2 dt, are continuous functions on. x / x L 2 dt 7

8 We apply Theorem to rewrite (9) as x / x 2 dt (ξ) x / x L 2 dt Multiplying both sides of the above equation by (ξ), integrating over and applying Parseval s identity in the form of [3], Lemma 4 (note the sum of degrees of homogeneity is n) we get: t n 2 e t2 2 dt dx Next, integration by parts gives: L (ξ). t n 2 e t2 2 dt dx. x n+2 e 2 x 2 dx x n+3 L e 2 x 2 L +(n 3) L t n 4 e t2 2 dt dx. () Passing to polar coordinates in R n γ n () γ n (L) is equivalent to and integrating by parts we get that x n+2 e 2 x 2 dx x n+2 L e 2 x 2 L dx + (n 2) L Note that by (), the latter inequality will follow from x n+3 L e 2 x 2 L + (n 3) L t n 3 e t2 2 dtdx. t n 4 e t2 2 dt dx x n+2 L e 2 x 2 L dx + (n 2) L t n 3 e t2 2 dtdx. 8

9 Denote = a, L = b, then the latter inequality follows from an elementary inequality for a, b : b n 3 (b a)e b2 2 b + (n 3)a a b t n 4 e t2 2 dt (n 2) a t n 3 e t2 2 dt. () To verify () we fix a > and define a function h(b): b h(b) = (n 2) a b t n 3 e t2 2 dt (b n 2 b n 3 a)e b2 /2 (n 3)a Then h (b) = e b2 /2 b n 2 (b a) and h attains its minimum at b = a. a t n 4 e t2 2 dt. The affirmative solution to GBP in dimension 4 follows directly from Theorem 4 and the fact that every symmetric convex body in R 4 is an intersection body (see [7], [22]). Corollary Let,L C 4 and assume that γ 3 ( ξ ) γ 3 (L ξ ), then γ 4 () γ 4 (L). Next we give a negative answer to GBP in dimensions n 5. Theorem 5 If L is an infinitely smooth, origin symmetric, convex body in R n with positive curvature, and L is not a positive definite distribution, then there exists a convex body D in R n such that γ n (D ξ ) γ n (L ξ ), ξ, but γ n (D) > γ n (L). Proof : Note that L is a continuous function on (see [7], Theorem or [3], Remark ). By our assumption, this function is negative on some open symmetric subset Ω of. Let f C ( ) be any non-negative even function supported in Ω. Extend f to a homogeneous function f(θ)r of degree on R n. Then the Fourier transform of f is a homogeneous function of degree n+: f(θ)r = g(θ)r n+, where g is an infinitely smooth function 9

10 on ([3], Lemma 5). Choosing a small ε >, we define a body D by x n+ x / x D t n 2 e t2 2 dt = x n+ x / x L t n 2 e t2 2 dt εg(x/ x ) x n+. Then, one can choose a small enough ε so that the body D is convex. Indeed, for small enough ε we may define a function α ε (x) such that then or x / x L x / x L αε(x/ x ) t n 2 e t2 2 dt εg(x/ x ) = t n 2 e t2 2 dt, x x D = D x x L α ε (x/ x ), = L α ε (x/ x ) x. Moreover α ε (θ) and its derivatives converge uniformly to (for θ ). Using that L is convex with positive curvature, one can choose a small enough ε so that the body D is convex. Now we are ready to finish the proof: γ n (D ξ ) = = ( 2π) n π ( 2π) n π x / x L x / x D 2 dt γ n (L ξ ). 2 dt (ξ) (ξ) (2π)n εf(ξ) ( 2π) n π On the other hand, the function f is positive only where x L = x L x L x / x L x L x / x D 2 dt x / x L 2 dt (ξ) is negative so (ξ) (2π) n x L (ξ)εf(ξ) 2 dt (ξ). Now the same computations as in the previous theorem give γ n (D) > γ n (L).

11 It was proved in [7] (see also [4]) that, for each fixed n 5, there exists an infinitely smooth, symmetric, convex body L R n, so that L is not positive definite. Therefore, Corollary 2 For any n 5 there exist convex symmetric bodies D and L in R n such that γ n (D ξ ) γ n (L ξ ), ξ, but γ n (D) > γ n (L). 4 Remarks Remark : One can present a different proof of Corollary 2, based on the original counterexample for the Busemann-Petty problem and the flatness of the Gaussian density near the origin. Note that e x 2 /2 for any x R n and e x 2 /2 e δ2 /2 for any x δb 2. Then, for any δb 2 : so Vol() = dx e x 2 2 dx = ( 2π) n γ n () e δ2 2 dx = e δ2 2 Vol(), ( 2π) n γ n () Vol() e δ2 2 (2) and ( 2π) n γ n ( ξ ) Vol( ξ ) e δ2 2, ξ. (3) Now, for n 5, we may consider convex, origin symmetric sets and L in R n, which give a counterexample to the Busemann-Petty problem. Clearly, any dilations α and αl will also provide a counterexample. Consider a function α(δ) = max{α : α, αl δb 2 }. Note that if Vol(L) < Vol() then there is a δ such that for all δ < δ : Then for δ < δ : γ n (α(δ)l) Vol(L) < e δ2 2 Vol(). ( Vol(α(δ)L) < e 2π) n δ 2 2 ( 2π) n Vol(α(δ)) γ n(α(δ)).

12 Using that and L are smooth convex bodies and Vol( ξ ) < Vol(L ξ ), ξ, we may choose a δ such that for any δ < δ : Vol(α(δ) ξ )) < e δ2 2 Vol(α(δ)L ξ )), ξ, and so γ n (α(δ) ξ ) < γ n (α(δ)l ξ ), ξ. Finally, choosing δ < min{δ, δ } we get a counterexample to GBP in R n, n 5. Remark 2: The method described in Remark, can be used to solve the following dual version of GBP, which is a Gaussian analog of the Shephard problem (see [2] or [6], p. 4): Assume,L C n and γ n ( ξ ) γ n (L ξ ), ξ, where ξ denotes the projection of to the hyperplane orthogonal to ξ. Does it follow that γ n () γ n (L)? The answer to this problem is affirmative if n = 2. This can easily be shown using monotonic properties of the Gaussian measure. On the other hand, the answer is negative if n 3. To show this one needs to apply the argument of Remark, using a counterexample to the original Shephard problem (see [2], [9] or [5]). Remark 3: One can use a similar method to provide counterexamples to the Busemann-Petty problem in R n, n 5 for a more general class of measures: γ n,p () = C n,p e x p/p dx, p, where x p denotes the standard l p norm on R n. We also note that γ n,p (p 2) is not rotation invariant, so the proofs of Theorems, 3, 4 can not be immediately generalized, and the Busemann-Petty problem for γ n,p, p 2 is open in R 3 and R 4. References []. Ball, Cube slicing in R n, Proc. Amer. Math. Soc. 97 (986), [2] J. Bourgain, On the Busemann-Petty problem for perturbations of the ball, Geom. Funct. Anal. (99), -3. 2

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