On polynomially integrable domains in Euclidean spaces 7 1 0 2 Mark L. Agranovsky n a January 24, 2017 J 9 1 ] Abstract A F LetD beaboundeddomaininRn,withsmoothboundary. Denote . V (ω,t), ω ∈ Sn−1,t ∈ R, the Radon transform of the characteristic h D t function χ of the domain D, i.e., the (n−1)− dimensional volume a D m of the intersection D with the hyperplane {x ∈ Rn :< ω,x >= t}. If the domain D is an ellipsoid, then the function V is algebraic and [ D if, in addition, the dimension n is odd, then V(ω,t) is a polynomial 1 v with respect to t. Whether odd-dimensional ellipsoids are the only 1 boundedsmoothdomainswithsuchaproperty? Thearticleisdevoted 5 to partial verification and discussion of that question. 5 5 0 1 . 1 0 7 1 Introduction 1 : v V.A.Vassilievproved[1,9]thatifDisaboundeddomainineven-dimensional i X space Rn, with C∞ boundary, then the two-valued function, evaluated the r a n− dimensional volumes V(cid:98)±(ω,t) of the two complementary portions of D which are cut-off by the section of D by the hyperplane {< ω,x >= t}, is not an algebraic function of the parameters of the hyperplane. That means that no functional equation Q(ω ,··· ,t,ω ,V(cid:98)±(ω,t)) = 0, where Q is a polyno- 1 n mial, is fulfilled. The result of [9] is a multidimensional generalization of a 1MSC 2010: 44A12, 51M99; keywords: Radon transform, Fourier transform, cross- section, polynomial, ellipsoid. 1 celebrated Newton’s Lemma XXVIII in Principia [8] for convex ovals in R2. The smoothness condition is essential. The main object of interest in this article is the section-volume function (cid:90) V (ω,t) = f(x)dV (x), D n−1 <x,ω>=t which is the t− derivative of V(cid:98)±. Contrary to the multi-valued cut-off-volume function V(cid:98) , its derivative D V = dV(cid:98) is single-valued and can be algebraic in any dimension of the D dt D ambientspace(thedifferentiationcan”improve”thealgebraicity, justkeepin mind the algebraic function 1 which is the derivative of the non-algebraic 1+x2 function tanx.) For example, if D is the unit ball in Rn then V (ω,t) = c(1−t2)n−1 D 2 isalgebraicinanydimension. Ifnisoddthen,evenbetter,V isapolynomial D in t. Here and further, c will denote a nonzero constant, which exact value is irrelevant. Ellipsoids in odd-dimensional spaces also have the same property. For instance, if (cid:88)n x2 j E = { = 1} b2 j=1 j then b ···b V(ω,t) = c 1 n(h2(ω)−t2)n−1, 2 hn(ω) where (cid:118) (cid:117) n (cid:117)(cid:88) h(ω) = (cid:116) b2ω2. j j j=1 Definition 1 We will call a domain D polynomially integrable if its section- volume function coincides with a polynomial in t : N (cid:88) V (ω,t) = a (ω)tj D j j−1 in the domain V (ω,t) > 0 of all (ω,t) such that the hyperplane < ω,x >= t D hits the domain D. 2 We assume that the leading coefficient a (ω) is not identical zero and in this N case we call N the degree of the polynomially integrable domain D. Note that in [9] the term ”algebraically integrable” refers to the cut-off volume function V(cid:98) , rather than to V but polynomial integrability with D D respect to either function is the same. It was conjectured in [9] that ellipsoids are the only smoothly bounded algebraically integrable domains in odd-dimensional Euclidean spaces. We are concerned with the similar question about the section-volume function V (ω,t) and the condition is that this function is a polynomial in t. Although D polynomials are definitely algebraic functions, we do not impose conditions on the dependence on ω. Still, our conjecture is that the odd-dimensional ellipsoids are the only bounded polynomially integrable domains in odd- dimensional Euclidean spaces. The present article contains some partial results and observations on the conjecture. Namely, in Sections 1-4 we prove that there are no bounded polynomi- ally integrable domains with smooth boundaries in Euclidean spaces of even dimensions (Theorem 2), while in odd dimensions the following is true: 2) polynomially integrable bounded domains with smooth boundaries are con- vex (Theorem 5), 3) there are no polynomially integrable bounded domains inRn,withsmoothboundaries, ofdegreestrictlylessthann−1(Theorem7), 4) polynomially integrable bounded domains in Rn, with smooth boundaries, of degree ≤ n−1 are only ellipsoids (Theorem 8, 4) polynomially integrable bounded domains in R3, with smooth boundaries, having axial and central symmetry are only ellipsoids (Theorem 11). In Section 7 a relation of polyno- mial integrability and stationary phase expansions is discussed. Some open questions are formulated in Section 8. 2 Preliminaries 2.1 Support functions For any bounded domain D define the support functions h+(ω) = sup < ω,x >, D x∈D (1) h−(ω) = inf < ω,x >, D x∈D 3 where ω is a vector in the unit sphere Sn−1 and (cid:104),(cid:105) is the standard inner product in Rn The two support functions are related by h+(−ω) = −h−(ω). D D The relation with translations is given by h± (ω) = h±(ω)+ < a,ω > . (2) D+a D If t ∈/ [h−(ω),h+(ω)] then the < ω,x >= t is disjoint from the closed D D domain D (and therefore V (ω,t) = 0 ). D 2.2 Radon transform We will be using several standard facts about the Radon transform. They can be found in many books or articles on Radon transform, see e.g., [6],[7]. 4 The Radon transform of a continuous compactly supported function f in Rn is defined as the integral (cid:90) Rf(ω,t) = f(x)dV (x) n−1 <ω,x>=t of f over the (n−1)− dimensional plane < ω,x >= t, with the unit normal vector ω, on the distance t from the origin, against the (n−1)− dimensional volume element dV . n−1 There is a nice relation, called Projection-Slice Formula, between Fourier and Radon transforms: f(cid:98)(rω) = F Rf(ω,t), (3) t→r where f(cid:98) is the n− dimensional Fourier transform and F is the one- t→r dimensional Fourier transform with respect to the variable t. The formula followsimmediatelybywritingtheintegralasadoubleintegralagainstdV n−1 and dt. The Radon transform can be inverted in different ways. The following formula is called back-projection inversion formula ([7], Thm 2.13) : (cid:90) n−1 f(x) = c∆ Rf(ω,< ω,x >)dω, 2 ω∈Sn−1 where f is sufficiently smooth compactly supported function. There is a crucial difference between the inversion formulas in even and odd dimensions. If n is even then the exponent n−1 is fractional and ∆n−1 2 2 becomes an integral operator, so that the inversion formula is not local. However, when n is odd then n−1 is integer and the operator in front of the 2 integral is differential so that the formula is local. Due to the relation (cid:100)∂f ∂f(cid:98) (ω,t) = ω (ω,t), (4) i ∂x dt i the inversion formula can be rewritten in odd dimensions as: (cid:90) (cid:18) dn−1 (cid:19) f(x) = c Rf (ω,< ω,x >)dω, (5) dtn−1 ω∈Sn−1 5 where dω is the Lebesgue measure on the unit sphere Sn−1. The Plancherel formula for the Radon transform in odd dimensions is: (cid:90) (cid:90) (cid:18) dn−1 (cid:19) c f(x)g(x)dV(x) = Rf (ω,t)Rg(ω,t)dωdt. (6) dtn−1 Rn Sn−1×R The following conditions characterize the range of Radon transform in class of Schwartz functions: 1. g(−ω,−t) = g(ω,t), (cid:82) 2. The k− moment g(ω,t)tkdt,k = 0,1,··· extends from the unit Rn sphere Sn−1 to Rn as a polynomial of degree at most k. N The immediate corollary of property 1 is that if V (ω,t) = (cid:80) a (ω)tk then D k k=0 a (−ω) = (−1)ka (ω), k k i.e. a is an even function on Sn−1 when k is even and is an odd function k when k is odd. In the sequel, the above facts will be used for the section-volume function V which is just the Radon transform of the characteristic function of the D domain D : V (ω,t) = (Rχ )(ω,t), t ∈ [h−(ω),h+(ω)]. D D D D All domains under consideration will be assumed bounded, with C∞ boundary, although some statements are true even under weaker smooth assumptions. As it is shown in [9],[1], smoothness plays an important role in that circle of questions. 3 There are no polynomially integrable do- mains in even dimensions The condition of algebraic integrability in [9] involves both variables ω and t, while the polynomial integrability imposes a condition with respect to t only, so that the statement formulated in the title on this section is not a straightforward corollary of the result in [9]. However, this statement easily follows from the asymptotic behavior of the section-volume function near the boundary points of D : 6 Theorem 2 There are no polynomially integrable domain with C2− smooth boundary in Rn with even n. Proof Let D be such a domain. We will show that V (ω,t) can not behave D polynomially when the section of D by the hyperplanes < x,ω >= t shrink to an elliptical point on ∂D. The elliptical point of the surface ∂D is a point at which the principal curvatures are all nonzero and of the same sign. To find such a point, consider the maximally distant, from the origin, boundary point a ∈ ∂D : |a| = max |x|. x∈∂D Using rotation and translation, we can move a to 0 and make the tangent plane at 0 the coordinate plane T (0) = {x = 0} ∂D n so that the domain D is in the upper half-plane x ≥ 0. Let b the image of n the point a under those transformations Then b is located on the x − axis n and D ⊂ B(b,|b|), where B(b,r) is the ball of radius r with center b. The boundary surface ∂D can be represented, in a neighborhood of 0 ∈ ∂D, as the graph ∂D∩U = {x = f(x(cid:48)),x(cid:48) = (x ,...,x )}. n 1 n−1 Since the tangent plane at 0 is x = 0 the first differential vanishes at the n origin, df = 0. Using rotations around the x axis we can diagonalize the 0 n second differential and then the equation of ∂D near 0 becomes n−1 1 (cid:88) ∂D∩U = {x = λ x2 +S(x(cid:48))}, n 2 j j j=1 where S(x(cid:48)) = o(|x(cid:48)|2),x(cid:48) → 0. The coefficients λ are the principal curvatures of the surface ∂D and j since D ⊂ B(b,|b|) all of them are not less than the curvature 1 of the |b| sphere S(b,|b|) : 1 λ ≥ > 0,j = 1,...,n−1. j |b| 7 Indeed, the equation of the sphere S(b,|b|) is (cid:112) x = |b|− |b|2 −|x(cid:48)|2 n and since D ⊂ B(b,|b|) we have n−1 1 (cid:88) (cid:112) λ x2 +o(|x(cid:48)|2) ≥ |b|− |b|2 −|x(cid:48)|2 = 2 j j j=1 |x(cid:48)|2 = . (cid:112) |b|+ |b|2 −|x(cid:48)|2 and then the inequalities follow by the passage to the limit. LetusturnfurthertothefunctionV .ThevalueV (e ,t),e = (0,...,0,1),t > D D n n 0,isthe(n−1)−dimensionalvolumeofthesectionD∩{x = t}andthelead- n ing term of the asymptotic when t → 0 is defined by the leading, quadratic, term in the decomposition of f. This leading term of V (e ,t) is the volume D n of the ellipsoid n−1 1 (cid:88) λ x2 = t, 2 j j j=1 n−1 n−1 which is c t 2 ,c = (2π) 2 . Therefore, λ1···λn−1 Γ(n+21) tn−1 2 n−1 V (e ,t) = c +o(t ),t → 0+. D n 2 λ ···λ 1 n−1 Thus, the function V (e ,t) has zero at t = 0 of order n−1. If this function is D n 2 a polynomial then the order must be integer, which is not the case when n is even. Thus, V (e ,t) is not a polynomial in t. In fact, all we have used is D n that a is an elliptical point of the convex hypersurface ∂D. Since the points of ∂D, sufficiently close to a, are elliptical as well, we conclude that, moreover, for an open set of directions ω, the function V (ω,t) is not a polynomial in D t. This proves the theorem. 8 4 Polynomially integrable domains in R2m+1 are convex Lemma 3 Suppose that the section-volume function V coincides with a D polynomial in t : V (ω,t) = a (ω)+a (ω)t+...+a (ω)tN, D 0 1 N when t ∈ [h−(ω),h+(ω)]. Then D D (cid:90) a (ω)p(ω)dA(ω) = 0 k |ω|=1 for any polynomial p of deg p ≤ k −n+1 and any k > n−1. Proof Regarding the function V as the Radon transform of the charac- D teristic function χ of the domain D, we can write by the inversion formula D (5): (cid:90) dn−1 χ (x) = V (ω,< x,ω >)dA(ω) D drn−1 D |ω|=1 N (cid:90) (cid:88) k! = a (ω) < x,ω >k−(n−1) dA(ω). k (k −n+1)! k=n−1 |ω|=1 Since χ (x) = 1 for x ∈ D, the power series in x in the left hand side D equals 1 identically and hence each term with k −(n−1) > 0 vanishes: (cid:90) a (ω) < x,ω >k−(n−1) dA(ω) = 0,x ∈ D. k |ω|=1 Since x is taken from an open set, the functions y →< x,y >k−n+1 span the space of all homogeneous polynomials of degree at most k − n + 1. Those polynomials, being restricted on the unit sphere, generate restrictions of all polynomials p, degp ≤ k −n+1. Lemma is proved. Remark 4 The assertion of Lemma remains true also under the assumption that V expands as an infinite power series in t : D ∞ (cid:88) V (ω,t) = a (ω)tk, D j k=0 9 in a neighborhood of [h−(ω),h+(ω)]. Theorem 5 If a smoothly bounded domain D in Rn, with n odd, is polyno- mially integrable then it is convex. Proof Let D be such a domain and denote D(cid:98) the convex hull of D. If D (cid:54)= D(cid:98) then there is an open portion Γ of the boundary ∂D which is disjointfromaneighborhoodof∂D(cid:98).Thenonecanconstructafunctionwhich vanishes near ∂D(cid:98) and behaves in a prescribed way on Γ. In particular, one can construct a compactly supported C∞ function ψ in Rn with the following properties: 1. supp ψ ⊂ D(cid:98), (cid:82) 2. ψ(x)ν (x)dS(x) (cid:54)= 0,whereν(x),x ∈ ∂D,istheexternalunitnormal 1 ∂D field on the boundary of D. Denote Ψ(ω,t) the Radon transform of ψ : (cid:90) Ψ(ω,t) = (Rψ)(ω,t) = ψ(x)dV (x). n−1 <ω,x>=t Formula 4 implies Ψ (ω,t) = ω Ψ(cid:48)(ω,t). (7) 1 1 t Using (7), write the Plancherel formula (6) for the pair of functions: χ and ∂ψ . Although formula (6) works for Schwartz functions, we can D ∂x1 extend the equality to our case approximating χ by smooth functions, if D the differentiation in t in the left hand side applies to the second, smooth, factor Ψ = Rψ. We obtain (cid:90) (cid:90) ∂ψ ∂ψ (x)dV(x) = χ (x) (x)dV(x) = D ∂x ∂x D 1 1 Rn (cid:90) (cid:90) (8) = Rχ (ω,t)(Ψ (ω,t))(n−1)dAω)dt. D 1 t R |ω|=1 By the condition, the Radon transform Rχ (ω,t) of the characteristic D function of D coincides with the polynomial V (ω,t) as long as the hyper- D plane < ω,x >= t hits D. The latter occurs, if and only if it hits D(cid:98). In turn, 10