RECONSTRUCTION OF CONVEX BODIES OF REVOLUTION FROM THE AREAS OF THEIR SHADOWS

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1 RECONSTRUCTION OF CONVEX BODIES OF REVOLUTION FROM THE AREAS OF THEIR SHADOWS D. RYABOGIN AND A. ZVAVITCH Abstract. In this note we reconstruct a convex body of revolution from the areas of its shadows by giving a precise formula for the support function. 1. Introduction The problem of reconstruction of a convex body from the areas of its shadows, goes back to A. D. Aleksandrov Al1], who proved that an origin symmetric convex body K in R n is uniquely defined by the volumes of its projections. Recently R. Gardner and P. Milanfar GM] provided an algorithm for reconstruction of an origin-symmetric convex body K from the volumes of its projections. It is plausible that there exist an explicit formula, connecting the support function of K with volumes of its shadows. The similarities between sections and projections, pointed out in KRZ], suggest that it should exist as a dual version of the formula for sections, proved by A. Koldobsky K] Vol n 1 (K θ ) = 1 ( n+1 K ) (θ), θ S n 1. (1) π(n 1) Note that by inverting the Fourier transform in (1), one can find the direct formula for the norm of K, given the volumes of sections. It was proved in KRZ1] that there is a connection between the volumes of projections of K and the curvature function f K via the Fourier transform: Vol n 1 (K θ ) = 1 π f K (θ), θ S n 1. () Here f K (x) = x n 1 f K (x/ x ), x R n \ {}, is the extension of f K (x), x S n 1, to a homogeneous function of degree n 1. Mathematics Subject Classification. Primary: 5A15, 5A1, 5A38. Key words and phrases. Convex body, geometric tomography, surface area measure, cosine transform, Alexandrov s theorem. 1

2 D. RYABOGIN AND A. ZVAVITCH A. D. Aleksandrov (Al]; Pog], p. 456; Sc], Corollary.5.3) showed that f K is the sum Σ(h K ) of the principal minors of order n 1 of the Hessian matrix of the support function h K. This suggests a dual version of (1) Vol n 1 (K θ ) = 1 π Σ(h K )(θ), θ S n 1, (3) which in the three dimensional case has the form: Vol (K θ ) = 1 ( h xx h xy π h xy h yy + h xx h xz h xz h zz + h yy h zy h zy h zz ) (θ). Unfortunately, to invert the above formulas one needs not only to invert the Fourier transform, but also to solve a nonlinear differential equation. It turns out that in the case of a body of revolution, this differential equation can be considerably simplified. In Section we show how to obtain an expression for h K, given the curvature function f K. In Section 3 we give a simple formula for the curvature function via the projections of K (cf. R], p. 15). All arguments can be generalized to higher dimensions. Let r be a natural number. A real valued function on an open subset U of R 3 is said to be of class C r (cf. Ga3], p. ) if it is r times differentiable, that is all partial derivatives of order r exist and are continuous. We denote this class by C r (U). A function f(σ) on σ S is said to be in C r (S ) if its homogeneous extension (x, y, z) f( x + y + z ) Cr (R 3 \ {}). We say that a convex body K is of class C r (cf. Ga3], p. 3) if K is of class C r as a submanifold of R 3. If k, we say that K is of class C+ k (cf. Ga3], p. 5), if K is of class C k and the Gauss curvature of K at each point is positive. Without loss of generality we may assume that e 3 is the axis of revolution. Our main result is the following Theorem 1. Let (x, y, z) S, and let K be of class C 5 +. Then where h K (x, y, z) = x + y φ(arcsin z ) + z π/ φ(t) = sin(α) f(α)dα, t arcsin z cos tf(t) φ(t) dt,

3 RECONSTRUCTION OF CONVEX BODIES 3 f(α) = f K (, cos α, sin α) = sin α 6 d 1 ds s d ds where α, π/]. ( s(1 s ) 3/ d ds sin α ds (sin α s ) 1 ( Vol (K (, s, 1 s ) ) 1 s ))], We would like to make a remark concerning the smoothness hypothesis in our theorem. It is clear that if the function Vol (K ) : u S Vol (K u ) belongs to C 3 (S ), and is rotation-invariant, then the function Vol (K (,, 1 ) ) : s (, 1) Vol (K (, s, 1 s ) ) belongs to C 3 ((, 1)). Thus, it is enough to assume that K is such that Vol (K ) C 3 (S ). This is true, provided f K C 3 (S ) (see Lemma 4). One can weaken this hypothesis, but this is not our purpose here (see M]). We also remark that in many problems of convexity the bodies of revolution serve as a main source for examples and counterexamples, see for example the Shephard problem P], (Ga3], p. 14), or the Busemann-Petty problem Ga1], Ga], Pa]. Therefore, different types of special formulas for bodies of revolution may lead to the general development of the techniques related to the more general classes of convex bodies (cf. Ga] and GKS]). We hope that the formulas obtained in Lemma 1 and Theorem 1 will help to provide new connections between the volumes of sections (1) and projections () of general convex bodies.. From the curvature to the support function We recall that the curvature function f K is the reciprocal of the Gauss curvature viewed as a function of the unit normal vector (Sc], p. 419). The support function of the convex body K is defined as h K (ξ) = sup{η ξ, η K}. It is proved (see Sc], pp ) that K is of class C+ if and only if h K C and the Gauss curvature of K exists and is positive everywhere. It is enough to consider the case x, y, z (all other cases can be reconstructed by symmetry). We will use the following notation f K (x, y, z) = f K (u, v), u = x + y and v = z, and our goal is to find h K (x, y, z) = h(u, v). We will need three elementary lemmas.

4 4 D. RYABOGIN AND A. ZVAVITCH Lemma 1. Let K be a body of revolution, then the equation f K = h xx h xy + h xx h xz + h yy h zy has the form h xy h yy h xz h zz h zy h zz Proof : f K (u, v) = h u u (h uu + h vv ), u + v = 1. A straightforward computation gives x h x = h u x + y, y x h xx = h uu x + y + h u (x + y ), 3/ y h yy = h uu x + y + h u (x + y ), 3/ xy h xy = h uu x + y h xy u (x + y ). 3/ Observe that due to homogeneity, the Hessian of h(u, v) is zero. Hence h xx h xy = 1 u h uuh u, h xy h yy and h xx h xz h xz h zz + h yy h zy h zy h zz = 1 u h vvh u. Using the homogeneity of f K and h K we may extend the result above to the case (x, y, z) R 3 (or, the same, (u, v) R ): (u + v )f K (u, v) = h u u (h uu + h vv ). (4) Our goal is to solve this differential equation for h. Lemma. Define h(θ) = h(cos θ, sin θ) and f(θ) = f K (cos θ, sin θ), θ, π/]. Then equation (4) can be rewritten as follows Proof : and cos θf(θ) = (h + h ) (h cos θ h sin θ). (5) We pass to polar coordinates in R and use the fact that x h u = cos θh r sin θ r h θ, h uu + h vv = 1 r h rr + 1 r h r + 1 r h θθ.

5 RECONSTRUCTION OF CONVEX BODIES 5 Since h is a homogeneous function of degree one, we have h u = cos θ h(θ) sin θ h θ (θ), (6) and h uu + h vv = 1 r h(θ) + 1 r h θθ(θ). Finally we plug these formulas into (4) and get r r 4 f K (θ) = 1 ( 1 r cos θ (cos θh(θ) sin θh θ(θ)) r h(θ) + 1 ) r h θθ(θ). This gives the desired result. Next we denote φ(θ) = h cos θ h sin θ and observe that φ (θ) = h cos θ h sin θ h sin θ h cos θ = sin θ(h + h ). Equation (5) becomes or sin θ cos θf(θ) = φ(θ)φ (θ), c 1 θ sin(α) f(α)dα = φ (θ). Lemma 3. We have φ(θ) = π sin(α) f(α)dα. Proof : Since φ() = h(), we obtain θ h () θ sin(α) f(α)dα = φ (θ). Now the Cauchy projection formula (Sc], Ga3]), and the fact that the projection of K onto the xy-plane is a disk of radius h(), give πh () = Vol (K e 3 ) = 1 S π/ z f K (x, y, z)dσ(x, y, z) = π sin(α) f(α)dα.

6 6 D. RYABOGIN AND A. ZVAVITCH Thus φ (θ) = π/ sin(α) f(α)dα. θ It remains to show that φ is nonnegative. This is a consequence of the fact that φ(θ) = h u (θ) (see (6)), cos θ and the following proposition. Proposition. Let L R n be of class C 1 +. Assume also that if (x 1,, x n ) L, then (ε 1 x 1, ε n x n ) L for any choice of signs ε 1,, ε n. We have u i h L x i (u 1,, u i,, u n ). Proof: It is well known (Sc], p. 4) that max y L u y = u gradh L(u). h Assume that u L i x i (u) < for some 1 i n. To get a contradiction we consider a point y L such that all coordinates of y, with the exception of the ith, are equal to coordinates of gradh L (u), and y i = h L x i (u). But then u y > u gradh L (u). To obtain a formula for h it remains to solve or after differentiation: φ(θ) = h cos θ h sin θ, (7) φ (θ) sin θ = h + h. (8) Note that φ (θ)/ sin θ is a continuous function on, π/]. Using standard method we solve (8) with the initial values π/ h() = sin(α) f(α)dα and h () =, (the last one comes from the fact that K has a smooth boundary, and h(θ) is an even function): π/ θ cos h(θ) = cos θ tf(t)dt sin(α) f(α)dα + sin θ. π/ θ sin(α) f(α)dα t

7 RECONSTRUCTION OF CONVEX BODIES 7 3. From projections to the curvature function The volume of the projection of the convex body K is connected with the curvature function f K via the formula of Cauchy: ) Vol (Ku = 1 u (x, y, z) f K (x, y, z)dσ(x, y, z). (9) S Thus, in view of Section, it is enough to invert (9). Observe that in the case of a body of revolution the functions f K and u Vol (K u ) are invariant under rotations around the axis, so it is enough to invert (9) at the point u = (, s, 1 s ), s, 1]. Denote Then (9) has the form ϕ(s) = 1 1 ϕ(s) = Vol (K (, s, 1 s ) ). 1 1 (1 t ) 1 t 1 z s + z 1 s dt. We substitute t 1 z s = η, and use the evenness of f(z) to get: 1 ϕ(s) = s + 1 s s 1 z s 1 z s 1 z s 1 z (1 z )s η ] 1 (1 z )s η ] 1 = I 1 + I. Observe that z < s implies z 1 s < s 1 z, so s I 1 = s 1 z (1 z )s η ] 1 η + z 1 s dη = η + z 1 s dη = (η + z 1 s ) dη+ s z 1 s z 1 s (1 z )s η ] 1 ( η z ) ] 1 s dη. This gives s I 1 = s 1 z z 1 s s 1 z (1 z )s η ] 1 dη s 1 z

8 8 D. RYABOGIN AND A. ZVAVITCH s z 1 s (1 z )s η ] 1 ηdη s s 1 z z 1 s z 1 s s 1 z (1 z )s η ] 1 dη]. Similarly, 1 z s implies s 1 z z 1 s, so 1 I = Now we have 1 ϕ(s) = I 1 + I = s f(z)z 1 s dz s 1 z s 1 z z 1 s (1 z )s η ] 1 dη. s 1 z (1 z )s η ] 1 dη s z 1 s s 1 z (1 z )s η ] 1 ηdη s In other words, s 1 z z 1 s z 1 s s 1 z π 1 ϕ(s) = 4 1 s z + 4 s (1 z )s η ] 1 dη]. s s z z 1 s f(z) arccos z 1 s s 1 z dz ]. We can simplify this formula, dividing both sides by 1 s and taking the derivative with respect to s (, 1). We get ( ) d ϕ(s) 4 s = s z ds 1 s s(1 s ). 3/ We define g(s) = s(1 s ) 3/ 4 ( ) d ϕ(s), ds 1 s

9 to get the integral equation RECONSTRUCTION OF CONVEX BODIES 9 g(s) = s s z, which can be inverted by standard methods, see for example (H], p. 11, n = 4): f(z) = z ( ) z(z s ) 1 g (s) ds, z 1. 3 s Finally f(z) = 1 6 z z(z s ) 1 ( d 1 ds s d ds ( s(1 s ) 3/ d ( ))) ϕ(s) ds. ds 1 s The following result is well-known (see Sc], p. 431, relation (A.16) and See]). We include it here for the convenience of the reader. Lemma 4. The function Vol (K ) : u S Vol (K u ) belongs to C 3 (S ), provided f K C 3 (S ). Proof : Let n 3, d n (m) = (n+m )(n+m 3)!, and let m!(n )! d n(m) f m,µ Y m,µ (θ), f m,µ = f(σ)y m,µ(σ)dσ, m=1 µ=1 S n 1 be the Fourier-Laplace series of f L (S n 1 ). It is well-known (see See]), that f C r (S n 1 ) implies f m,µ c m r, D j n j + Y m (θ) c m, for all multi-indices j, j = j j n =, 1,,..., D j = j ( x 1 ) j 1...( xn ) jn. Let r > 3(n )/ + j + 1. It follows that the derivatives D j of the Fourier-Laplace series of f converge uniformly and absolutely to D j f(θ). Indeed, d n (m) c m n as m, and we have m=1 d n(m) µ=1 f m,µ D j Y m,µ (θ) C m=1 1 m r 3/(n ) j <.

10 1 D. RYABOGIN AND A. ZVAVITCH Let f K C r (S n 1 ), r > j + n 7/. Then the Cauchy formula and the Fourier-Laplace decomposition of the Cosine transform imply ( 1 ) D j Vol n 1 (K θ ) = D j θ x f K (x)dx = = m=1 d n(m) µ=1 S n 1 γ m (f K ) m,µ D j Y m,µ (θ), where γ m c m (3+n)/ as m. Moreover, m=1 d n(m) µ=1 γ m (f K ) m,µ D j Y m,µ (θ) C 1 <. mr 3/(n ) j +(3+n)/ m=1 Thus, the function Vol (K ) : u S Vol (K u ) is three times continuously differentiable, provided f K C r (S ), r 3. Acknowledgments: We wish to thank Alexander Koldobsky for fruitful discussions and for a series of suggestions that improved the paper. We thank Alex Iosevich for pointing to the book of A. Pogorelov Pog]. References Al1] A. D. Aleksandrov, On the theory of mixed volumes of convex bodies. I. Extensions of certain concepts in the theory of convex bodies (in Russian). Mat. Sbornik N. S. (1937), Al] A. D. Aleksandrov, Almost everywhere existence of the second differential of a convex function and some properties of convex surfaces connected with it. Uchen. Zap. Leningrad. Gos. Univ. 37 Mat. 3 (1939), Ga1] R. J. Gardner, Intersection bodies and the Busemann-Petty problem. Trans. Amer. Math. Soc. 34 (1994), no. 1, pp Ga] R. J. Gardner, A positive answer to the Busemann-Petty problem in three dimensions. Ann. of Math. () 14 (1994), pp Ga3] R. J. Gardner, Geometric tomography. Encyclopedia of Mathematics and its Applications, 58. Cambridge University Press, Cambridge, (1995). GKS] R. J. Gardner, A. Koldobsky, Th. Schlumprecht, An analytic solution to the Busemann-Petty problem on sections of convex bodies, Ann. of Math. () 149 (1999), GM] R. J. Gardner, P. Milanfar, Reconstruction of convex bodies from brightness functions. Discrete Comput. Geom. 9 (3), Vol., H] S. Helgason, The Radon Transform. Progress in Mathematics, 5. Birkhäuser Boston, Inc., Boston, MA, K] A. Koldobsky, An application of the Fourier transform to sections of star bodies, Israel J. Math. 16 (1998), KRZ1] A. Koldobsky, D. Ryabogin and A. Zvavitch, Projections of convex bodies and the Fourier transform, to appear in Israel J. Math..

11 RECONSTRUCTION OF CONVEX BODIES 11 KRZ] A. Koldobsky, D. Ryabogin and A. Zvavitch, Unified Fourier analytic approach to the volume of projections and sections of convex bodies, to appear in Proceedings of Workshop on Convexity and Fourier analysis. M] Y. Martinez-Maure, Hedgehogs and Zonoids, Adv. Math. Vol. 158 (1), no. 1, Pa] M. Papadimitrakis, On the Busemann-Petty problem about convex, centrally symmetric bodies in R n, Mathematika 39 (199), P] C. M. Petty, Projection bodies, Proc. Coll. Convexity (Copenhagen 1965), Kobenhavns Univ. Mat. Inst., Pog] A. V. Pogorelov, Extrinsic Geometry of Convex Surfaces, Volume Thirtyfive of Translations of Mathematical Monographs, AMS, Providence, R] B. Rubin, Inversion and characterization of the hemispherical transform, J. Anal. Math. 77 (1999), Sc] R. Schneider, Convex bodies: the Brunn-Minkowski theory, Cambridge University Press, Cambridge, (1993). See] R. T. Seeley, Spherical harmonics. Amer. Math. Monthly 73 (1966) no. 4, part II, Dmitry Ryabogin, Department of Mathematics, Kansas State University, Manhattan, KS , USA address: ryabs@math.ksu.edu Artem Zvavitch, Department of Mathematics, University of Missouri, Columbia, MO 6511, USA address: zvavitch@math.missouri.edu