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A Hypersurface Containing the Support of a Radon Transform Must Be an Ellipsoid

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A Hypersurface Containing the Support of a Radon Transform Must Be an Ellipsoid The Journal of Geometric Analysis (2021) 31:2726–2741 https://doi.org/10.1007/s12220-020-00372-8 A Hypersurface Containing the Support of a Radon Transform must be an Ellipsoid. I: The Symmetric Case Jan Boman1 Received: 28 October 2019 / Published online: 2 March 2020 © The Author(s) 2020 Abstract If the Radon transform of a compactly supported distribution f = 0inRn is supported on the set of tangent planes to the boundary ∂ D of a bounded convex domain D, then ∂ D must be an ellipsoid. As a corollary of this result we get a new proof of a recent theorem of Koldobsky, Merkurjev, and Yaskin, which settled a special case of a conjecture of Arnold that was motivated by a famous lemma of Newton. Keywords Radon transform supported in hypersurface · Polynomially integrable domain · Arnold’s conjecture Mathematics Subject Classification 44A12 1 Introduction Define a function f0 in the plane by 1 1 2 f0(x) = for x = (x1, x2) ∈ R , |x| < 1, π 1 −|x|2 and f0(x) = 0for|x| > 1. An easy calculation shows that the Radon transform of f0 satisfies 1 Rf0(ω, p) = f0 ds = 1for|p| < 1,ω∈ S , x·ω=p and obviously Rf0(ω, p) = 0for|p|≥1. Define the distribution f by f = f = ∂2 f + ∂2 f . 0 x1 0 x2 0 B Jan Boman [email protected] 1 Stockholm University, Stockholm, Sweden 123 A Hypersurface Containing the Support of a Radon Transform… 2727 ( )(ω, ) = ∂2 (ω, ) = The well-known formula R h p p Rh p with h f0 now shows that Rf(ω, p) = 0for|p| < 1 and all ω. In other words, Rf is a distribution on the manifold of lines in the plane that vanishes in the open set of lines that intersect the open unit disk. Since Rf obviously vanishes on the open set of lines that are disjoint from the closed disk, it follows that the distribution Rf is supported on the set of lines that are tangent to the circle. By means of an affine transformation it is easy to construct a similar example where the circle is replaced by an arbitrary ellipse. For an arbitrary ellipsoidal domain D ⊂ Rn, n > 2, it is also easy to construct examples of distributions f supported in D such that the Radon transform Rf is supported on the set of tangent planes to the boundary of D. However, surprisingly, for other convex domains than ellipsoids such distributions do not exist. Since we will consider arbitrary convex—not necessarily smooth—domains, we have to replace the notion of tangent plane by supporting plane. A supporting plane for D is a hyperplane L such that L ∩ D is non-empty and one of the components of Rn \ L is disjoint from D. Theorem 1 Let D be an open, convex, bounded, and symmetric (that is D =−D) subset of Rn,n ≥ 2, with boundary ∂ D. If there exists a distribution f = 0 with support in D such that the Radon transform of f is supported on the set of supporting planes for D, then ∂ D must be an ellipsoid. The set of tangent planes to an ellipsoid is a quadratic hypersurface in the space of hyperplanes, hence is itself an ellipsoid, as asserted in the title. The more general case when D is not assumed to be symmetric turned out to require different arguments from those given here. This case will therefore be treated in a forthcoming article. Remark Our arguments prove in fact a stronger statement, Theorem 3, which is local in ω but global in p; see Sect. 5. Denote by V (ω, p) the volume of one of the connected components of D\ L(ω, p), where D ⊂ Rn is a convex, bounded domain and L(ω, p) is the hyperplane x · ω = p that we assume intersects D. A famous conjecture of Arnold (Problem 1987-14 in Arnold’s Problems,[3]) asserts that if V (ω, p) is an algebraic function, then n must be odd and D must be an ellipsoid. The background of Arnold’s conjecture is a famous lemma in Newton’s Principia and is described in [4]. The case n even has been settled long ago by Vassiliev [8],seealso[7]. For odd n the question is still open. The special case when n is odd and p → V (ω, p) is assumed to be a polynomial function of degree ≤ N for all ω and some N has also been studied and was settled recently by Koldobsky, Merkurjev, and Yaskin for domains D with smooth boundary [6]; see also [1]. If the domain D is symmetric, then this question is answered by Theorem 1.In[2] the case when p → V (ω, p) is algebraic and satisfies a certain additional condition is reduced to the case p → V (ω, p) is polynomial. 123 2728 J. Boman Corollary 1 Let D ⊂ Rn,n≥ 2, be as in Theorem 1, and assume that there exists a number N such that p → V (ω, p) is a polynomial of degree ≤ N for all ω ∈ Sn−1. Then ∂ D must be an ellipsoid. Proof Let χD be the characteristic function of D and choose an integer k such that > ∂2k( χ )(ω, ) = 2k N. The assumption implies that p R D p 0 for all p in the interval inf{x · ω; x ∈ D} < p < sup{x · ω; x ∈ D}, and obviously (RχD)(ω, p) = 0 for all other p. This shows that the distribution ∂2k χ ∂ p R D must be supported on the set of supporting planes to D, a hypersurface in n n k the manifold of hyperplanes in R . Define the distribution f in R by f = χD (k )(ω, ) = ∂2k (ω, ) where denotes the Laplace operator. The formula R h p p Rh p with h = χD now shows that the distribution Rf must be supported on the set of supporting hyperplanes. By Theorem 1 this implies that ∂ D is an ellipsoid. A somewhat related problem is treated in a recent article by Ilmavirta and Paternain [5]. It is proved that the existence of a function in L1(D), D ⊂ Rn, whose X-ray transform (integral over lines) is constant, implies that the boundary of D is a ball. In Sect. 2 we will write down an expression for an arbitrary distribution g on the manifold of hyperplanes that is equal to the Radon transform of some compactly supported distribution and is supported on the submanifold of supporting planes to ∂ D. In Sect. 3 we will use the description of the range of the Radon transform to write down the conditions for g to be the Radon transform of a compactly supported distribution f . Those conditions will be an infinite number of polynomial identities in the supporting function ρ(ω)for D and the densities q j (ω) that define the distribution g. Thereby the problem is transformed to a purely algebraic question. In Sect. 4 we analyze the polynomial identities and prove (Theorem 2) that they imply that the supporting function ρ(ω) must be a quadratic polynomial, which together with the fact that ρ(ω) > 0 implies that ∂ D is an ellipsoid. In Sect. 5 we finish the proof of Theorem 1 and prove the semi-local version Theorem 3. An outline of the proof of Theorem 2 is given in Sect. 4. 2 Distributions on the Manifold of Hyperplanes As is well known the manifold Pn of hyperplanes in Rn can be identified with the manifold (Sn−1 × R)/(±1), the set of pairs (ω, p) ∈ Sn−1 × R, where (ω, p) is identified with (−ω,−p). Thus a function on Pn can be represented as an even function g(ω, p) = g(−ω,−p) on Sn−1 ×R. In this article a distribution on Pn will be a linear ∞( n−1 × R) n−1 × R form on Ce S , the set of smooth even functions on S , and a locally integrable even function h(ω, p) on Sn−1 × R will be identified with the distribution ∞( n−1 × R) ϕ → (ω, )ϕ(ω, ) ω , Ce S h p p d d p R Sn−1 123 A Hypersurface Containing the Support of a Radon Transform… 2729 where dω is area measure on Sn−1. Using the standard definition of R∗, ∗ R ϕ(x)(x) = ϕ(ω, x · ω)dω, Sn−1 we can then define the Radon transform of the compactly supported distribution f on Rn as the linear form ∞( n−1 × R) ϕ → , ∗ϕ . Ce S f R Let D be a bounded, convex subset of Rn with boundary ∂ D. Here we will also assume that D is symmetric with respect to some point, which we may assume to be the origin, so D =−D. We shall denote the supporting function of D by ρ(ω), that is ρ(ω) = sup{x · ω; x ∈ D}. Since D is symmetric, ρ will be an even function, because ρ(−ω) = sup{x · (−ω); x ∈ D}=sup{x · (−ω);−x ∈ D} = sup{(−x) · (−ω); x ∈ D}=ρ(ω). Clearly a hyperplane x · ω = p intersects D if and only if |p| <ρ(ω), and it is a supporting plane to ∂ D if and only if p =±ρ(ω). Weshall consider the hypersurface in Pn that consists of all the supporting planes to ∂ D. Since the origin in Rn is contained in (the interior of) D, none of the supporting planes can contain the origin, hence ρ(ω) > 0 for all ω. A distribution of order 0 that is supported on the set of supporting planes to D can therefore be represented as g(ω, p) = q+(ω)δ(p − ρ(ω)) + q−(ω)δ(p + ρ(ω)) − for some measures q+(ω) and q−(ω) on Sn 1; here δ(·) is the Dirac measure at the origin in R.
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