Convex Floating Bodies of Equilibrium
CConvex Floating Bodies of Equilibrium ∗ D.I. Florentin, C. Sch¨utt, E.M. Werner † , N. Zhang Abstract
We study a long standing open problem by Ulam, which is whether theEuclidean ball is the unique body of uniform density which will float in equilib-rium in any direction. We answer this problem in the class of origin symmetric n -dimensional convex bodies whose relative density to water is . For n = 3,this result is due to Falconer. A long standing open problem asked by Ulam in [16] (see also [9], Problem 19), iswhether the Euclidean ball is the unique body of uniform density which floats in aliquid in equilibrium in any direction. We call such a body
Ulam floating body .A two-dimensional counterexample was found for relative density ρ = by Auerbach[2] and for densities ρ (cid:54) = by Wegner [17]. These counterexamples are not originsymmetric. For higher dimension, Wegner obtained results for non-convex bodies(holes in the body are allowed) in [18]. The problem remains largely open in the classof convex bodies in higher dimension. In order to study Ulam floating bodies, weuse the notion of the convex floating body , which was introduced independently byB´ar´any and Larman [3] and by Sch¨utt and Werner [13]. Let K be a convex body in R n and let δ ∈ R , 0 ≤ δ ≤ . Then the convex floating body K δ is defined as K δ = (cid:92) u ∈ S n − H − δ,u . Here H + δ,u is the halfspace with outer unit normal vector u , which “cuts off” a δ proportion of K , i.e. vol n (cid:0) K ∩ H + δ,u (cid:1) = δ vol n ( K ). The convex floating body is a ∗ Keywords: Ulam Problem, floating bodies, 2020 Mathematics Subject Classification: 52A20 † Partially supported by NSF grant DMS-1811146 and by a Simons Fellowship a r X i v : . [ m a t h . M G ] D ec atural variation of Dupin’s floating body K [ δ ] (see [4]). K [ δ ] is this convex bodycontained in K such that every support hyperplane cuts off a set of volume δ vol n ( K )exactly. In general K [ δ ] need not exist. An example is e.g., the simplex in R n . Dupinshowed that if it exists, each supporting hyperplane H touches K [ δ ] at the centroidof K ∩ H . If the floating body K [ δ ] exists, it is equal to the convex floating body K δ .It was shown in [10] that for a symmetric convex body K , one has K [ δ ] = K δ .We recall the relation between the density ρ and the volume δ vol n ( K ) that is cutoff. If the liquid has density 1 and the body K has unit volume and density ρ , thenby Archimedes’ law the submerged volume equals the total mass of the body, i.e. ρ ,and consequently the floating part has volume δ = 1 − ρ .In [7], the authors defined the metronoid M δ ( K ) of a convex body K to be thebody whose boundary consists of centroids of the floating parts of K , i.e. K ∩ H + δ,u .More precisely, denoting X K,δ ( u ) = δ vol n ( K ) (cid:82) K ∩ H + δ,u x dx , they defined M δ ( K ) by ∂M δ ( K ) = (cid:8) X K,δ ( u ) : u ∈ S n − (cid:9) , and showed that X K,δ : S n − → ∂M δ ( K ) is the Gauss map of M δ ( K ), i.e. the normalto M δ ( K ) at X K,δ ( u ) is u . Huang, Slomka and Werner showed that K is an Ulamfloating body if and only if M δ ( K ) is a ball (see [8, Section 2.2] for details). We utilizethis characterization in our proof of Theorem 1.1. We present two results concerning Ulam’s problem. We first establish a relationbetween Ulam floating bodies and a uniform isotropicity property of sections. Oursecond result provides a short proof of a known answer to Ulam’s problem in the classof symmetric convex bodies with relative density ρ = 1 / g ( B ) for the centroid of aset B . Theorem 1.1.
Let δ ∈ (cid:0) , (cid:3) and let K ⊂ R n be a convex body such that K δ is C or K δ = K [ δ ] reduces to a point. Then K is an Ulam floating body if and only if thereexists R > such that for all u ∈ S n − and v ∈ S n − ∩ u ⊥ , (cid:90) K ∩ H δ,u (cid:104) x, v (cid:105) − (cid:104) g ( K ∩ H δ,u ) , v (cid:105) dx = δ vol n ( K ) R. (1) In that case, M δ ( K ) is a ball of radius R . Remark.
Note that if K δ reduces to a point, which without loss of generality we canassume to be 0, then the condition (1) reduces to (cid:90) K ∩ H δ,u (cid:104) x, v (cid:105) dx = δ vol n ( K ) R. (2)2e use the characterization given in Theorem 1.1 to give a short proof of the followingresult which was proved in dimension 3 by Falconer [5]. It also follows from a resultin [11]. Theorem 1.2.
Let K ⊂ R n be a symmetric convex body of volume and density .If K is an Ulam floating body, then K is a ball. We collect some definitions and basic results that we use throughout the paper. Forfurther facts in convex geometry we refer the reader to the books by Gardner [6] andSchneider [12].The radial function r K,p : S n − → R + of a convex body K about a point p ∈ R n isdefined by r K,p ( u ) = max { λ ≥ λu ∈ K − p } . If 0 ∈ int( K ), the interior of K , we simply write r K instead of r K, .Let K ⊂ R n be a convex body containing a strictly convex body D in its interior,and let H be a hyperplane supporting D at a point p . If u is the outer unit normalvector at p , we denote the restriction of the radial function r K ∩ H,p to S n − ∩ u ⊥ by r K,D ( u, · ).We denote by B n the Euclidean unit ball centered at 0 and by S n − = ∂B n itsboundary. The spherical Radon transform R : C ( S n − ) → C ( S n − ) is defined by Rf ( u ) = (cid:90) u ⊥ ∩ S n − f ( x ) dx (3)for every f ∈ C ( S n − ). Since δ > implies K δ = ∅ , we restrict our attention to the range δ ∈ (cid:2) , (cid:3) . It wasshown in [14] that there is δ c , 0 < δ c ≤ such that K δ c reduces to a point. It canhappen that δ c < . An example is the simplex.In fact, Helly’s Theorem (and a simple union bound) implies that δ c > n +1 , so wehave δ c ∈ (cid:0) n +1 , (cid:3) .As mentioned above, when Dupin’s floating body K [ δ ] exists, it coincides with theconvex floating body K δ . The following lemma states that existence of K [ δ ] is alsoguaranteed by smoothness of K δ . We use this for Theorem 1.1.3 emma 2.1. If K δ is C , then K [ δ ] exists and K δ = K [ δ ] .Proof. Let x ∈ ∂K δ . By [14], there is at least one hyperplane H through x that cutsoff exactly δ vol n ( K ) from K and this hyperplane touches ∂K δ in the barycenter of H ∩ K . As K δ is C , the hyperplane at every boundary point x ∈ K δ is unique. Thus K [ δ ] exists and K δ = K [ δ ] .We know that K = { } for every centrally symmetric convex body K . The Homo-thety Conjecture [14] (see also [15, 19]), states that the homothety K δ = t ( δ ) K onlyoccurs for ellipsoids. We treat the following conjecture which has a similar flavor. Conjecture 2.2.
Let K ⊂ R n be a convex body and let δ ∈ (cid:0) , (cid:1) . If K δ is aEuclidean ball, then K is a Euclidean ball. We now prove the two dimensional case of the conjecture.
Theorem 2.3.
Let K ⊂ R and suppose there is δ ∈ (cid:0) , (cid:1) such that K δ = r B .Then K = R B , for some R > .Proof. We shall prove that the radial function r K : S → R is constant. If thecontinuous function r K is not constant, it must attain some value R > r such thatthe angle θ = arccos (cid:0) rR (cid:1) ∈ (cid:0) , π (cid:1) is not a rational multiple of π . Let u ∈ S bethe point with r K ( u ) = R , and let { u i } ∞ i =1 be the arithmetic progression on S withdifference 2 θ . We claim that r K is constant on { u i } . Indeed, assuming Ru i ∈ ∂K ,we consider the triangle with vertices O, Ru i , Ru i +1 (see the figure below). The edge[ Ru i , Ru i +1 ] is tangent to K δ = rB at its midpoint m i , and since K δ is smooth, thechord on ∂K containing Ru i and m i is bisected by m i , which implies Ru i +1 ∈ ∂K ,i.e., r K ( u i +1 ) = R . Since θ is not a rational multiple of π , the sequence { u i } is densein S . Since r K is constant on a dense set and continuous, it is constant on S , asrequired. 4 Proof of the main theorems
Proof.
We first treat the case n = 2. Also, we first consider when K δ reduces to apoint. Without loss of generality we can assume that this point is 0. Then we havefor all u ∈ S that g ( K ∩ H δ,u ) = 0 by Dupin and thus (cid:104) g ( K ∩ H δ,u ) , v (cid:105) = 0, for all v ∈ u ⊥ ∩ S and the condition reduces to (cid:82) K ∩ H δ,u (cid:104) x, v (cid:105) dx = C . This observation istrue in all dimensions.Let u ∈ S . Let v ∈ u ⊥ ∩ S = H δ,u ∩ S . Then, as H δ,u = span { v } , we get for all v ∈ S , (cid:90) K ∩ H δ,u (cid:104) x, v (cid:105) dx = (cid:90) r K ( v ) − r K ( v ) x dx = 23 r K ( v ) , (4)as r K ( v ) = r K ( − v ). Hence if (cid:82) K ∩ H δ,u (cid:104) x, v (cid:105) dx = C , then we get that for all v ∈ S that r K ( v ) = C , and hence K is a Euclidean ball and therefore also M δ ( K ) is aEuclidean ball.For the other direction, we fix u ∈ S . We may assume that u = (1 , u corresponds to θ = 0 and r ( θ ) = 1. For φ small, let w = (cos φ, sin φ )and define the sets E = H + δ,u ∩ H + δ,w ∩ K, E = H + δ,u ∩ H − δ,w ∩ K, E = H − δ,u ∩ H + δ,w ∩ K.
5n order to compute the derivative of the boundary curve of M δ ( K ) we write δ vol ( K ) · (cid:2) X K,δ ( w ) − X K,δ ( u ) (cid:3) = (cid:90) E ∪ E x dx − (cid:90) E ∪ E x dx = (cid:90) E x dx − (cid:90) E x dx = (cid:90) π + φ π (cid:90) r K ( θ )0 (cos θ, sin θ ) r dr dθ − (cid:90) − π + φ − π (cid:90) r K ( θ )0 (cos θ, sin θ ) r dr dθ = 2 (cid:90) π + φ π (cid:90) r K ( θ )0 (cos θ, sin θ ) r dr dθ = 23 (cid:90) π + φ π r K ( θ ) (cos θ, sin θ ) dθ. Thus ddφ [ X K,δ ( w ) − X K,δ ( u )] = 23 δ vol ( K ) r K (cid:16) φ + π (cid:17) ( − sin φ, cos φ )and hence (cid:12)(cid:12)(cid:12)(cid:12) ddφ [ X K,δ ( w ) − X K,δ ( u )] (cid:12)(cid:12)(cid:12)(cid:12) = 23 δ vol ( K ) r K (cid:16) φ + π (cid:17) , (5)where | · | denotes the Euclidean norm. With z = w ⊥ , we get from (4) and (5) that (cid:12)(cid:12)(cid:12)(cid:12) ddφ [ X K,δ ( w ) − X K,δ ( u )] (cid:12)(cid:12)(cid:12)(cid:12) = 1 δ vol ( K ) (cid:90) K ∩ H δ,w (cid:104) x, z (cid:105) dx. (6)If M δ ( K ) is a Euclidean ball with radius R , we write X K,δ (cos φ, sin φ ) = R (cos φ, sin φ )in polar coordinates and get from (6)1 δ vol ( K ) (cid:90) K ∩ H δ,w (cid:104) x, z (cid:105) dx = (cid:12)(cid:12)(cid:12)(cid:12) ddφ [ X K,δ ( w ) − X K,δ ( u )] (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ddφ [ X K,δ ( w )] (cid:12)(cid:12)(cid:12)(cid:12) = R. To treat the case when K δ does not consist of just one point, we introduce the followingcoordinate system for the complement of an open, strictly convex body D ⊂ R withsmooth boundary (see also e.g., [20]). Let γ : [0 , π ] → ∂D be the inverse Gaussmap, and T : [0 , π ] → S be the unit tangent vector to the curve at γ ( θ ), orientedcounterclockwise, i.e., n ( θ ) := n D ( γ ( θ )) = (cid:18) cos θ sin θ (cid:19) , T ( θ ) = γ (cid:48) ( θ ) (cid:12)(cid:12) γ (cid:48) ( θ ) (cid:12)(cid:12) = (cid:18) − sin θ cos θ (cid:19) . The coordinate system F : R × [0 , π ] → R \ D is defined by F ( r, θ ) = γ ( θ ) + rT ( θ ) . (7)6ince ∂F∂r = T and ∂F∂θ = γ (cid:48) − rn , the Jacobian of F is given by | r | .Now we fix 0 < δ < / K δ does not just consist of one point. Wethen set D = int( K δ ), which has smooth boundary by assumption. Without loss ofgenerality, we can assume that 0 ∈ int( K δ ). It was shown in [14] that K δ is strictlyconvex. Let u ∈ S , and assume without loss of generality that u = n (0) = (1 , w = n ( φ ) for an angle φ > H δ,u and H δ,w intersect in the interior of K . Define the sets E = H + δ,u ∩ H + δ,w ∩ K, E = H + δ,u ∩ H − δ,w ∩ K, E = H − δ,u ∩ H + δ,w ∩ K, and let E be the bounded connected component of (cid:0) H − δ,u ∩ H − δ,w (cid:1) \ K δ , see Figure.FigureAgain, in order to compute the derivative of the boundary curve of M δ ( K ) we write δ vol ( K ) · [ X K,δ ( w ) − X K,δ ( u )] = (cid:90) E ∪ E x dx − (cid:90) E ∪ E x dx = (cid:90) E ∪ E x dx − (cid:90) E ∪ E x dx. Now we use the above introduced coordinate system. For n = n ( θ ) and T = T ( θ ), let r K,K δ ( θ ) be such that γ ( θ ) + r K,K δ ( θ ) T ∈ ∂K . As K δ = K [ δ ] , γ ( θ ) is the midpoint of n ( θ ) ⊥ ∩ K by Dupin’s characterization of K [ δ ] . Therefore δ vol ( K ) · [ X K,δ ( w ) − X K,δ ( u )]= (cid:90) φ (cid:90) r K,Kδ ( θ )0 F ( r, θ ) | r | dr dθ − (cid:90) φ (cid:90) − r K,Kδ ( θ ) F ( r, θ ) | r | dr dθ = (cid:90) φ (cid:90) r K,Kδ ( θ )0 F ( r, θ ) r dr dθ + (cid:90) φ (cid:90) − r K,Kδ ( θ ) F ( r, θ ) r dr dθ = (cid:90) φ (cid:90) r K,Kδ ( θ ) − r K,Kδ ( θ ) F ( r, θ ) r dr dθ. F . Thus X K,δ ( w ) − X K,δ ( u ) = 1 δ vol ( K ) (cid:90) φ (cid:90) r K,Kδ ( θ ) − r K,Kδ ( θ ) rγ ( θ ) + r T ( θ ) dr dθ = 1 δ vol ( K ) (cid:90) φ r K,K δ ( θ ) T ( θ ) dθ. As M δ ( K ) is strictly convex and C by [8], we can differentiate with respect to φ andget, dX K,δ ( n ( φ )) dφ = 2 r K,K δ ( φ )3 δ vol ( K ) T ( φ ) . (8)On the other hand, for any θ ∈ [0 , π ], (cid:90) K ∩ H δ,n ( θ ) (cid:104) x, T ( θ ) (cid:105) dx = (cid:90) r K,Kδ ( θ ) − r K,Kδ ( θ ) (cid:104) γ ( θ ) + rT ( θ ) , T ( θ ) (cid:105) dr = 2 r K,K δ ( θ ) (cid:104) γ ( θ ) , T ( θ ) (cid:105) + 2 (cid:104) γ ( θ ) , T ( θ ) (cid:105) (cid:90) r K,Kδ ( θ ) − r K,Kδ ( θ ) rdr + (cid:90) r K,Kδ ( θ ) − r K,Kδ ( θ ) r dr = (cid:104) g ( K ∩ H δ,n ( θ ) ) , T ( θ ) (cid:105) vol ( K ∩ H δ,n ( θ ) ) + 2 r K,K δ ( θ )3 , (9)since γ ( θ ) is the centroid g ( K ∩ H δ,n ( θ ) ) of K ∩ H δ,n ( θ ) . Combining (8) and (9), we getthat for θ = φ , (cid:12)(cid:12)(cid:12)(cid:12) dX K,δ ( n ( φ )) dφ (cid:12)(cid:12)(cid:12)(cid:12) = 1 δ vol ( K ) (cid:90) K ∩ H δ,n ( φ ) (cid:104) x, T ( φ ) (cid:105) − (cid:104) g ( K ∩ H δ,n ( φ ) ) , T ( φ ) (cid:105) dx. By [7, 8], the normal to ∂M δ ( K ) at X K,δ ( n ( φ )) is n ( φ ) = (cos φ, sin φ ) and as M δ ( K )is strictly convex and C by [8], ξ ( φ ) = X K,δ ( n ( φ )) is a parametrization of ∂M δ ( K )with respect to the angle of the normal. The curvature is given by dφds where s isthe arc length along the curve. Since dξds is a unit vector, we get by the chain rule dξds = dξdφ dφds that the radius of curvature is given by R ( φ ) = (cid:12)(cid:12)(cid:12)(cid:12) dX K,δ ( n ( φ )) dφ (cid:12)(cid:12)(cid:12)(cid:12) = 1 δ vol ( K ) (cid:90) K ∩ H δ,n ( φ ) (cid:104) x, T ( φ ) (cid:105) − (cid:104) g ( K ∩ H δ,n ( φ ) ) , T ( φ ) (cid:105) dx. Since M δ ( K ) is a disk if and only if its radius of curvature is constant, the theoremfollows.Let now n ≥ u ∈ S n − be arbitrary, but fixed, and let v ∈ S n − ∩ u ⊥ . We denote by W =span { u, v } the span of u and v and by W ⊥ the ( n − W . ¯ K = K | W is the orthogonal projection of the convex8ody K onto the 2-dimensional subspace W . For a small φ , let w = cos φ u + sin φ v. We define ¯ E , ¯ E and ¯ E as follows,¯ E = ( H + δ,u ∩ H + δ,w ) (cid:12)(cid:12) W, ¯ E = ( H + δ,u ∩ H − δ,η ) (cid:12)(cid:12) W, ¯ E = ( H − δ,u ∩ H + δ,w ) (cid:12)(cid:12) W and ¯ E is the curvilinear triangle enclosed by H δ,u | W , H δ,w (cid:12)(cid:12) W , and the boundary of¯ K δ = K δ | W . Then the picture is identical to the previous Figure. We let E i = ¯ E i × W ⊥ , for i = 1 , , , . When K δ reduces to a point, we can assume without loss of generality that K δ = { } .As noted before, the condition then reduces to (cid:82) K ∩ H δ,u (cid:104) x, v (cid:105) dx = C . In this case¯ E = ∅ and the proof continues along the same lines as below. Alternatively, one canreplace the coordinate system (7) by the usual polar coordinates in W as it was donein the case n = 2.In the general case we thus have that δ vol n ( K ) [ X K,δ ( w ) − X K,δ ( u )] = (cid:90) K ∩ H + δ,w x dx − (cid:90) K ∩ H + δ,u x dx = (cid:90) K ∩ ( E ∪ E ) x dx − (cid:90) K ∩ ( E ∪ E ) x dx = (cid:90) K ∩ E x dx − (cid:90) K ∩ E x dx = (cid:90) K ∩ ( E ∪ E ) x dx − (cid:90) K ∩ ( E ∪ E ) x dx. For x ∈ W , we consider the following parallel section function, A K,W ( x ) = vol n − (cid:0) K ∩ { W ⊥ + x } (cid:1) (10)and observe that by Fubini, δ vol n ( K ) [ X K,δ ( w )] = (cid:90) K ∩ H + δ,w z dz = (cid:90) ¯ K (cid:32)(cid:90) ( x + W ⊥ ) ∩ K ∩ H + δ,w y dy (cid:33) dx. We denote by g ( B ) = n ( B ) (cid:82) B y dy the centroid of the set B . Then we get δ ( X K,δ ( w )) (cid:12)(cid:12) W = (cid:18)(cid:90) K ∩ ( E ∪ E ) x dx (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) W = (cid:18)(cid:90) ¯ K ∩ ( ¯ E ∪ ¯ E ) (cid:18)(cid:90) ( x + W ⊥ ) ∩ K y dy (cid:19) dx (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) W = (cid:18)(cid:90) ¯ K ∩ ( ¯ E ∪ ¯ E ) A K,W ( x ) g (( x + W ⊥ ) ∩ K ) dx (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) W = (cid:90) ¯ K ∩ ( ¯ E ∪ ¯ E ) A K,W ( x ) (cid:0) g (( x + W ⊥ ) ∩ K ) (cid:1) (cid:12)(cid:12) W dx = (cid:90) ¯ K ∩ ( ¯ E ∪ ¯ E ) A K,W ( x ) x dx, δ vol n ( K )( X K,δ ( u )) (cid:12)(cid:12) W . Now we will use the coordinate system F : R × [0 , π ] → R \ int (cid:0) ¯ K δ (cid:1) , introduced earlier in (7), F ( r, θ ) = γ ( θ ) + rT ( θ ) . We can assume that n (0) = u . Then T (0) = v . We recall that the Jacobian of F is given by | r | . We abbreviate n = n ( θ ) and T = T ( θ ). We let r ¯ K, ¯ K δ ( n ( θ )) , T ( θ )) = r ¯ K, ¯ K δ ( n, T ) > γ ( θ ) + r ¯ K, ¯ K δ ( n, T ) T ( θ ) ∈ ∂ ¯ K , and r ¯ K, ¯ K δ ( n, − T ) > γ ( θ ) + r ¯ K, ¯ K δ ( n, T ) ( − T ( θ )) ∈ ∂ ¯ K . We get δ vol n ( K ) ( X K,δ ( w ) − X K,δ ( u )) (cid:12)(cid:12) W = (cid:90) φ (cid:90) r ¯ K, ¯ Kδ ( n,T )0 F ( r, θ ) A K,W ( F ( r, θ )) | r | dr dθ − (cid:90) φ (cid:90) − r ¯ K, ¯ Kδ ( n, − T ) F ( r, θ ) A K,W ( F ( r, θ )) | r | dr dθ = (cid:90) φ (cid:90) r ¯ K, ¯ Kδ ( n,T )0 F ( r, w ) A K,W ( F ( r, θ )) r dr dθ + (cid:90) φ (cid:90) − r ¯ K, ¯ Kδ ( n, − T ) F ( r, θ ) A K,W ( F ( r, θ )) r dr dθ = (cid:90) φ (cid:90) r ¯ K, ¯ Kδ ( n,T ) − r ¯ K, ¯ Kδ ( n, − T ) F ( r, θ ) A K,W ( F ( r, θ )) r dr dθ. We differentiate with respect to φ , δ vol n ( K ) ddφ (cid:0) ( X K,δ ( w ) − X K,δ ( u )) (cid:12)(cid:12) W (cid:1) = δ vol n ( K ) ddφ (cid:0) ( X K,δ ( w )) (cid:12)(cid:12) W (cid:1) = (cid:90) r ¯ K, ¯ Kδ ( n ( φ ) ,T ( φ )) − r ¯ K, ¯ Kδ ( n ( φ ) , − T ( φ )) F ( r, φ ) A K,W ( F ( r, φ )) r dr. Putting φ = 0, results in δ vol n ( K ) ddφ (cid:0) ( X K,δ ( w )) (cid:12)(cid:12) W (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ =0 = (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) F ( r, A K,W ( F ( r, r dr = (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) [ γ (0) + rv ] A K,W ( F ( r, r dr = (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) γ (0) A K,W ( F ( r, r dr + (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) r v A K,W ( F ( r, dr = v (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) r A K,W ( F ( r, dr, (11)where the last equality holds by Dupin since H δ,u ∩ K δ is the centroid of H δ,u ∩ K , i.e. (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) γ (0) A K,W ( F ( r, r dr = 0 . (12)10ndeed, in the coordinate system (7), the centroid of H δ,u ∩ K is γ (0). Thus, withcoordinate system (7) we get as abovevol n − ( K ∩ H δ,u ) (cid:104) v, γ (0) (cid:105) = (cid:90) K ∩ H δ,u (cid:104) v, x (cid:105) dx = (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) (cid:104) γ (0) + rv, v (cid:105) A K,W ( F ( r, dr = (cid:104) γ (0) , v (cid:105) (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) A K,W ( F ( r, dr + (cid:104) v, v (cid:105) (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) r A K,W ( F ( r, dr = (cid:104) γ (0) , v (cid:105) vol n − ( K ∩ H δ,u ) + (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) r A K,W ( F ( r, dr. On the other hand, again in the coordinate system (7), and also using (12), (cid:90) K ∩ H δ,u (cid:104) x, v (cid:105) dx = (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) (cid:104) γ (0) + rv, v (cid:105) A K,W ( F ( r, dr = (cid:104) γ (0) , v (cid:105) (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) A K,W ( F ( r, dr + (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) r A K,W ( F ( r, dr = (cid:104) γ (0) , v (cid:105) vol n − ( K ∩ H δ,u ) + (cid:90) r ¯ K, ¯ Kδ ( u,v ) − r ¯ K, ¯ Kδ ( u, − v ) r A K,W ( F ( r, dr. (13)As w = cos φ u + sin φ v , it follows from (11) and (13) that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ddφ (cid:16) X K,δ (cos φ u + sin φ v ) (cid:12)(cid:12)(cid:12) W (cid:17) (cid:12)(cid:12)(cid:12) φ =0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 δ vol n ( K ) (cid:90) K ∩ H δ,u (cid:0) (cid:104) x, v (cid:105) − (cid:104) γ (0) , v (cid:105) (cid:1) dx = 1 δ vol n ( K ) (cid:90) K ∩ H δ,u (cid:0) (cid:104) x, v (cid:105) − (cid:104) g ( K ∩ H δ,u ) , v (cid:105) (cid:1) dx. Observe that in the case when K δ = { } , (cid:104) g ( K ∩ H δ,u ) , v (cid:105) = 0. We have that w = n ( φ ) = cos φ u + sin φ v ∈ W is the outer unit normal to M δ ( K ) in X δ,K ( w ).Therefore, w is the outer unit normal to M δ ( K ) (cid:12)(cid:12) W in X δ,K ( w ) (cid:12)(cid:12) W . Again, as M δ ( K )and therefore M δ ( K ) (cid:12)(cid:12) W is strictly convex and C by [8], X δ,K ( n ( φ )) is a parametriza-tion of the boundary of M δ ( K ) (cid:12)(cid:12) W with respect to the angle of the normal. Thus thecurvature of M δ ( K ) (cid:12)(cid:12) W is constant, which implies that M δ ( K ) (cid:12)(cid:12) W is a disk. Since W is arbitrary, we get that every two dimensional projection of M δ is a disk, and itfollows that M δ ( K ) is a Euclidean ball ([6], Corollary 3.1.6).11 .2 Proof of Theorem 1.2 Proof.
Since K is symmetric and has volume 1 and density ρ = , we have that δ = ,as noted above. Therefore, K = K [ ] = { } . Since K is an Ulam floating body, theremark after Theorem 1.1 implies that for any u ∈ S n − and v ∈ u ⊥ ∩ S n − (cid:90) u ⊥ ∩ K (cid:104) x, v (cid:105) dx = C, (14)for some constant C . Let u ∈ S n − be arbitrary, but fixed. We pass to polar coordi-nates in u ⊥ and get for all v ∈ u ⊥ ∩ S n − , C = (cid:90) u ⊥ ∩ S n − (cid:90) r K ( ξ ) t =0 t n (cid:104) ξ, v (cid:105) dt dσ ( ξ ) = 1 n + 1 (cid:90) u ⊥ ∩ S n − r K ( ξ ) n +1 (cid:104) ξ, v (cid:105) dσ ( ξ ) . Now we integrate over all v ∈ u ⊥ ∩ S n − = S n − w.r. to the normalized Haar measure µ on S n − . We use that (cid:82) S n − (cid:104) ξ, v (cid:105) dµ ( v ) = c n (cid:107) ξ (cid:107) = c n , where c n = 2 vol n − ( B n − ) vol n − ( S n − ) andget that ( n + 1) Cc n = (cid:90) u ⊥ ∩ S n − r K ( ξ ) n +1 dσ ( ξ ) = R r n +1 K ( u ) , where R is the spherical Radon transform (3). We rewrite this equation as (cid:90) u ⊥ ∩ S n − dσ ( ξ ) = (cid:90) u ⊥ ∩ S n − n − ( B n − )( n + 1) C r K ( ξ ) n +1 dσ ( ξ ) , or 0 = (cid:90) u ⊥ ∩ S n − (cid:18) n − ( B n − )( n + 1) C r K ( ξ ) n +1 − (cid:19) dσ ( ξ ) . As u was arbitrary and as r K is even, it then follows from e.g., Theorem C.2.4 of [6]that r K = const. for σ almost all u and as r K is continuous, r K = const. on S n − .Thus K is a ball. References [1] D. Amir,
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Non-central Sections of Convex Bodies . Israel J. Math. , 763–790 (2017).Dan I. FlorentinDepartment of MathematicsBar-Ilan University, Israel e-mail : danfl[email protected] Sch¨uttMathematisches SeminarChristian-Albrechts University of KielLudewig-Meyn-Strasse 4, 24098 Kiel, Germany e-mail : [email protected] M. WernerDepartment of Mathematics Universit´e de Lille 1Case Western Reserve University UFR de Math´ematiqueCleveland, Ohio 44106, U. S. A. 59655 Villeneuve d’Ascq, France [email protected]
Ning ZhangSchool of Mathematics and StatisticsHuazhong University of Science and Technology1037 Luoyu Road, Wuhan, Hubei 430074, China e-maile-mail