A negative answer to Ulam's Problem 19 from the Scottish Book
AA NEGATIVE ANSWER TO ULAM’S PROBLEM 19 FROMTHE SCOTTISH BOOK
DMITRY RYABOGIN
Abstract.
We give a negative answer to Ulam’s Problem 19 from theScottish Book asking, is a solid of uniform density which will float inwater in every position a sphere?
Assuming that the density of wateris 1, we show that there exists a strictly convex body of revolution K ⊂ R of uniform density , which is not a Euclidean ball, yet floatsin equilibrium in every direction. We prove an analogous result in alldimensions d ≥ Introduction
The following intriguing problem was proposed by Ulam [U, Problem 19]:
If a convex body K ⊂ R made of material of uniform density D ∈ (0 , floats in equilibrium in any orientation ( in water, of density , must K bespherical? Schneider [Sch1] and Falconer [Fa] showed that this is true, provided K is centrally symmetric and D = . No results are known for other densities D ∈ (0 ,
1) and no counterexamples have been found so far.The “two-dimensional version” of the problem is also very interesting. Inthis case, we consider floating logs of uniform cross-section, and seek for theones that will float in every orientation with the axis horizontal. If D = ,Auerbach [A] has exhibited logs with non-circular cross-section, both convexand non-convex, whose boundaries are so-called Zindler curves [Zi]. Morerecently, Bracho, Montejano and Oliveros [BMO] showed that for densities D = , , and the answer is affirmative, while Wegner proved that forsome other values of D (cid:54) = the answer is negative, [Weg1], [Weg2]; see alsorelated results of V´arkonyi [V1], [V2]. Overall, the case of general D ∈ (0 , Theorem 1.
Let d ≥ . There exists a strictly convex non-symmetric bodyof revolution K ⊂ R d which floats in equilibrium in every direction at thelevel vol d ( K )2 . This gives
Key words and phrases.
Floating bodies.The author is supported in part by Simons Collaboration Grant for Mathematiciansprogram 638576 and by U.S. National Science Foundation Grant DMS-1600753. a r X i v : . [ m a t h . C A ] F e b D. RYABOGIN
Theorem 2.
The answer to Ulam’s Problem 19 is negative, i.e., there existsa convex body K ⊂ R of density D = , which is not a Euclidean ball, yetfloats in equilibrium in every direction. Our bodies will be small perturbations of the Euclidean ball . We combineour recent results from [R] together with Olovjanischnikoff’s known The-orems [O], and then use the machinery developed together with Nazarovand Zvavitch in [NRZ]. The proofs of Theorem 1 for even and odd d aredifferent. For even d we solve a finite moment problem to obtain our bodyas a local perturbation of the Euclidean ball. The case d ≥ d is more involved. To control the perturbation, we use the properties of the Spherical Radon transform , [He].We refer the reader to [M, pgs. 90-93], [CFG, pgs. 19-20], [Ga, pgs. 376-377], [Sch2, pgs. 560-563] and [G] for an exposition of known results relatedto the problem.This paper is structured as follows. In Section 2, we recall all the nec-essary notions and statements needed to prove the main result. In Section3, we reduce the problem to finding a non-trivial solution to a system oftwo integral equations. In Section 4, we prove Theorem 1 for even d . InSection 5, we give the proof of Theorem 1 for odd d and prove Theorem2. The Appendices contain technical parts of the proofs and some auxiliarystatements. In Appendix A, we present the proof of Theorem 4 given in [O].In Appendix B, we recall some well-known facts about envelopes of curvesand prove Lemma 1.2. Notation and auxiliary results
Let N = { , , . . . , } be the set of natural numbers. A convex body K ⊂ R d , d ≥
2, is a convex compact set with non-empty interior int K . Theboundary of K is denoted by bd K . Let S d − = { ξ ∈ R d : d (cid:80) j =1 ξ j = 1 } be the unit sphere in R d centered at the origin, let B d = { p ∈ R d : d (cid:80) j =1 p j ≤ } be the unit Euclidean ball centered at the origin, and let e j = (0 , . . . , (cid:124)(cid:123)(cid:122)(cid:125) j , . . . ,
0) be the standard basis in R d , j = 1 , . . . , d . Given ξ ∈ S d − , we denote by ξ ⊥ = { p ∈ R d : p · ξ = 0 } the subspace orthogonalto ξ , where p · ξ = p ξ + · · · + p d ξ d is the usual inner product in R d . We saythat a hyperplane H is the supporting hyperplane of a convex body K if K ∩ H (cid:54) = ∅ , but int K ∩ H = ∅ . The center of mass of a non-empty compactconvex set L ⊂ R d will be denoted by C ( L ), C ( L ) = 1vol( L ) (cid:90) L xdx, N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 3 where vol( L ) is the volume of L of the corresponding dimension. For d ≥ m ∈ N we denote by C m ( R d ) the class of functions having continuouspartial derivatives up to order m .Let d ≥
3, let K ⊂ R d be a convex body and let δ ∈ (0 , vol d ( K )) be fixed.Given a direction ξ ∈ S d − and t = t ( ξ ) ∈ R , we call a hyperplane H ( ξ ) = { p ∈ R d : p · ξ = t } the cutting hyperplane of K in the direction ξ , if it cuts out of K the givenvolume δ , i.e., if(1) vol d ( K ∩ H − ( ξ )) = δ, where H − ( ξ ) = { p ∈ R d : p · ξ ≤ t ( ξ ) } , (see Figure 1). l ( ξ ) ξ C ( K ) C δ ( ξ ) Κ Η ( ξ ) Κ Η - ( ξ ) A = L . Figure 1.
A body K , its submerged part K ∩ H − ( ξ ) andthe line (cid:96) ( ξ ) passing through C ( K ) and C δ ( ξ )We recall several well-known facts and definitions (see [DVP, ChapterXXIV], [Zh, Hydrostatics, Part I]). Definition 1.
Let ξ ∈ S d − and let C δ ( ξ ) be the center of mass of thesubmerged part K ∩ H − ( ξ ) satisfying (1). We say that K floats in equilibriumin the direction ξ at the level δ if the line (cid:96) ( ξ ) passing through C ( K ) and C δ ( ξ ) is orthogonal to the “free water surface” H ( ξ ) , i.e., the line (cid:96) ( ξ ) is “vertical” ( parallel to ξ , see Figure 1 ) . Definition 2.
A geometric locus of {C δ ( ξ ) : ξ ∈ S d − } is called the surfaceof centers S = S δ , or the surface of buoyancy. To define the moments of inertia (see [Zh, page 553]), consider a convexbody K and an affine ( d − H ( ξ ) for which (1)holds. Choose any ( d − l ⊂ H ( ξ ) passing D. RYABOGIN through the center of mass C ( K ∩ H ( ξ )) and let η , . . . , η d − , η d − be anorthonormal basis of ξ ⊥ = { p ∈ R d : p · ξ = 0 } such that(2) l = C ( K ∩ H ( ξ )) + span( η , . . . , η d − ) , H ( ξ ) = C ( K ∩ H ( ξ )) + ξ ⊥ . Definition 3.
The moment of inertia I K ∩ H ( ξ ) ( l ) of K ∩ H ( ξ ) with respectto l is calculated by summing r for every “particle” in the set K ∩ H ( ξ ) ,where r is the distance to l , ( see Figure 2 ) , i.e., (3) I K ∩ H ( ξ ) ( l ) = (cid:90) K ∩ H ( ξ ) | r ( v ) | dv = (cid:90) K ∩ H ( ξ ) −C ( K ∩ H ( ξ )) ( u · η d − ) du. K H η η v r ( v ) l O ( ξ ) Figure 2.
Two-dimensional body K ∩ H ( ξ ) with centerof mass at the origin, and a line l parallel to η ; we have | r ( v ) | = | v | − ( v · η ) = ( v · η ) .We will use the following result (see [R, Theorem 1] or [FSWZ, Theorem1.1]). Theorem 3.
Let d ≥ , let K ⊂ R d be a convex body and let δ ∈ (0 , vol d ( K )) .If K floats in equilibrium at the level δ in all directions, then for all ξ ∈ S d − and for all ( d − -dimensional affine subspaces l ⊂ H ( ξ ) passingthrough the center of mass C ( K ∩ H ( ξ )) , the cutting sections K ∩ H ( ξ ) haveequal moments of inertia independent of ξ and l .Conversely, let K be C and let C ( S ) = C ( K ) . If for all cutting hyper-planes H ( ξ ) , ξ ∈ S d − , and for all ( d − -dimensional affine subspaces l ⊂ H ( ξ ) passing through the center of mass C ( K ∩ H ( ξ )) , the cutting sec-tions K ∩ H ( ξ ) have equal moments of inertia independent of ξ and l , then K floats in equilibrium in all directions at the level δ . N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 5
Remark 1.
Let δ = vol d ( K )2 . Since for any ξ ∈ S d − , C ( K ) is the arithmeticaverage of C ( K ∩ H + ( ξ )) and C ( K ∩ H − ( ξ )) , the condition C ( S ) = C ( K ) issatisfied. Now we recall the notion of characteristic points of a family of hyperplanes(cf. [Fikh, page 549]).
Definition 4.
Let d ≥ , let Q be a family of hyperplanes in R d , and let H ∈ Q . We call a point e ∈ H a characteristic point of Q with respect to H if for any sequence of hyperplanes { H ( k ) } ∞ n =1 , H ( k ) ∈ Q , converging to H as k → ∞ , we have e ∈ lim k →∞ H ∩ H ( k ) . We will need the following result from [O] (see Lemma on pages 114-117and Remark 1 on page 117).
Theorem 4.
Let d ≥ , let K ⊂ R d be a convex body and let δ ∈ (0 , vol d ( K )) .The characteristic points of the family of cutting hyperplanes { H ( ξ ) : ξ ∈ S d − } are the centers of mass of the sections { K ∩ H ( ξ ) : ξ ∈ S d − } .Conversely, if the characteristic points of the family of hyperplanes { H ( ξ ) : ξ ∈ S d − } corresponding to the sections { K ∩ H ( ξ ) : ξ ∈ S d − } coincide withthe centers of mass of these sections, then the function ξ → vol d ( K ∩ H − ( ξ )) is constant on S n − . Since the reference [O] is not readily available, for the convenience of thereader we present the proof of Theorem 4 in Appendix A.3.
Reduction to a system of integral equations
Let d ≥
3. We follow the notation from [NRZ]. We will be dealing withbodies of revolution K f = { x ∈ R d : x + x + · · · + x d ≤ f ( x ) } obtained by the rotation of a smooth concave function supported on [ − R , R ]about the x -axis. Let L ( t ) = L s ( t ) = st + h ( s ) be a linear function withslope s ∈ R , and let H ( L ) = { x ∈ R d : x d = L ( x ) } be the corresponding hyperplane. The function h will be chosen later. Nowit is enough to assume that it is infinitely smooth, not identically zero,supported on [1 − τ, − τ ] for some small τ >
0, and it is small togetherwith sufficiently many derivatives. Let − x = − x ( s ) and y = y ( s ) be thefirst coordinates of the points of intersection of ± f and L (see Figure 3).To construct a system of two integral equations we will prove four lemmas.Consider the family of hyperplanes(4) F = { H ( L s ) : s ∈ [0 , ∞ ) } . Lemma 1.
Let E be the set of characteristic points of F . Then, (5) E = { ( − h (cid:48) ( s ) , , . . . , , L ( − h (cid:48) ( s ))) ∈ R d : s ∈ [0 , ∞ ) } . D. RYABOGIN f-f -x y (y, f (y)) L ( -x, -f (-x )) x x d h(s) i . . . / . . I . i Figure 3.
Sections of K f and H ( L ) by the ( x , x d )-plane.For the proof we refer the reader to Appendix B, where the perturbationfunction h is given explicitly. Lemma 2.
Let s > . The condition (6) C ( K f ∩ H ( L s )) = ( − h (cid:48) ( s ) , , . . . , , L ( − h (cid:48) ( s ))) reads as (7) y ( s ) (cid:90) − x ( s ) ( t + h (cid:48) ( s ))( f ( t ) − L ( t )) d − dt = 0 . Let l = { x ∈ H ( L ) : x = − h (cid:48) ( s ) } , l j = { x ∈ H ( L ) : x j = 0 } . be the ( d − -dimensional affine subspaces, j = 2 , . . . , d − . The momentsof inertia conditions I j = I K f ∩ H ( L ) ( l j ) = const, j = 1 , . . . , d − , read as (8) I = v d − (1 + s ) y ( s ) (cid:90) − x ( s ) ( t + h (cid:48) ( s )) ( f ( t ) − L ( t )) d − dt = const, N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 7 (9) I j = (cid:112) s γ d − y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) d dt = const, where v d − = vol d − ( B d − ) , γ d − = (cid:90) B d − p j dp, j = 2 , . . . , d − . Proof.
Fix s >
0. Observe that the slice K f ∩ H ( L ) ∩ H t of the cutting section K f ∩ H ( L ) by the hyperplane H t = { x ∈ R d : x = t } , − x ( s ) < t < y ( s ), isthe ( d − B d − ( r, ( t, , . . . , , L ( t ))) = { ( t, x , . . . , x d − , L ( t )) : x + · · · + x d − ≤ r } of radius r = (cid:112) f ( t ) − L ( t ) centered at ( t, , . . . , , L ( t )). Hence, for thefirst coordinate of the center of mass in (6) we have(10) y ( s ) (cid:90) − x ( s ) ( t + h (cid:48) ( s )) dt (cid:90) B d − ( r, ( t, ,..., ,L ( t ))) dp = v d − y ( s ) (cid:90) − x ( s ) ( t + h (cid:48) ( s ))( f ( t ) − L ( t )) d − dt = 0 . Using the symmetry of K f ∩ H ( L ) with respect to l j , j = 2 , . . . , d −
1, wesee that the x , . . . , x d − -coordinates of C ( K f ∩ H ( L )) are zero. Since L ( t ) − L ( − h (cid:48) ( s )) = s ( t + h (cid:48) ( s )), the condition for the x d -coordinate of C ( K f ∩ H ( L ))is equation (10) multiplied by s . This gives (7).Similarly, since the distance in K f ∩ H ( L ) between ( t, x , . . . , x d ) ∈ K f ∩ H ( L ) ∩ H t and ( − h (cid:48) , x , . . . , x d ) ∈ K f ∩ H ( L ) ∩ H − h (cid:48) is √ s | t + h (cid:48) | , wehave I = (cid:112) s y ( s ) (cid:90) − x ( s ) ( (cid:112) s ( t + h (cid:48) ( s )) dt (cid:90) B d − ( r, ( t, ,..., ,L ( t ))) dp = v d − (1 + s ) y ( s ) (cid:90) − x ( s ) ( t + h (cid:48) ( s )) ( f ( t ) − L ( t )) d − dt. Finally, the expression in the left-hand side of (9) for the other momentscan be obtained as I j = (cid:112) s y ( s ) (cid:90) − x ( s ) dt (cid:90) B d − ( r, ( t, ,..., ,L ( t ))) p j dp = D. RYABOGIN (cid:112) s γ d − y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) d dt. (cid:3) Lemma 3.
Let s o ≥ , let K f be as above and let F be the family ofhyperplanes defined as in (4) for s ≥ s , so that (6) holds for s ≥ s o .Then for all s > s o and for all ( d − -dimensional affine subspaces l ⊂ H ( L s ) passing through the center of mass C ( K f ∩ H ( L s )) , the cutting sections K f ∩ H ( L s ) have equal moments of inertia I K f ∩ H ( L s ) ( l ) independent of s and l , provided (8) and (9) hold with the same constant on the right-hand side,which is independent of s and j = 1 , . . . , d − .Proof. Let s o ≥ s > s o be fixed. If l ⊂ H ( L s ) is any ( d − C s = C ( K f ∩ H ( L s )), then by (3) we have I K f ∩ H ( L s ) ( l ) = (cid:90) K f ∩ H ( L s ) (( u − C s ) · η ) du, where η = η d − is a unit vector in the hyperplane H ( L s ) − C s which isorthogonal to l .Let ι , . . . ι d − be the orthonormal basis in H ( L s ) − C s such that ι ∈ span { e , e d } and ι j = e j for j = 2 , . . . , d −
1. Decomposing η in this basis as d − (cid:80) j =1 η ( j ) ι j , we have I K f ∩ H ( L s ) ( l ) = d − (cid:88) j =1 η j ) (cid:90) K f ∩ H ( L s ) (( u − C s ) · ι j ) du + d − (cid:88) j,l =1 j (cid:54) = l η ( j ) η ( l ) (cid:90) K f ∩ H ( L s ) (( u − C s ) · ι j )(( u − C s ) · ι l ) du = J + J . Using the fact that η is a unit vector, together with (8) and (9), we havethat J is constant.We claim that J = 0. Indeed, if j is equal to 1, then arguing as in theprevious lemma, and using the fact that (cid:82) B d − p l dp = 0 for l = 2 , . . . , d − (cid:90) K f ∩ H ( L s ) (( u − C s ) · ι )(( u − C s ) · ι l ) du = (cid:112) s y ( s ) (cid:90) − x ( s ) ( t + h (cid:48) ( s )) dt (cid:90) B d − ( r, ( t, ,..., ,L ( t ))) p l dp = 0 . N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 9
The case when l = 1 is similar.If j (cid:54) = 1, l (cid:54) = 1, then we use the fact that (cid:82) B d − p j p l dp = 0 for l, j =2 , . . . , d − j (cid:54) = l , to obtain (cid:90) K f ∩ H ( L s ) (( u − C s ) · ι j )(( u − C s ) · ι l ) du = y ( s ) (cid:90) − x ( s ) dt (cid:90) B d − ( r, ( t, ,..., ,L ( t ))) p j p l dp = 0 . Thus, I K f ∩ H ( L s ) ( l ) is a constant independent of s and of the arbitrarilychosen l . The lemma is proved. (cid:3) Lemma 4.
Let s o ≥ . Assume that (7) is valid for all s > s o . Then (9)holds for all s > s o with the constant independent of s if and only if (8)holds for all s > s o with the constant independent of s .Proof. Let s o ≥ s > s o . We rewrite (9) as y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) d dt = constγ d − √ s and differentiate both sides with respect to s . We have y ( s ) (cid:90) − x ( s ) ( f ( ξ ) − L ( t )) d − ( st + h ( s ))( t + h (cid:48) ( s )) dt = const sdγ d − (1 + s ) . By (7) this is equivalent to s y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) d − ( t + h (cid:48) ( s )) dt = const sdγ d − (1 + s ) . Canceling s and using dγ d − = dd − (cid:90) B d − | p | dp = dd − (cid:90) S d − dσ (cid:90) r d − dr = σ ( S d − ) d − v d − , we have (8).Now we prove the converse statement. We rewrite the first equality in (9)as I j ( s ) γ d − √ s = y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) d dt and differentiate both parts with respect to s . Using (7) and (8), we seethat (cid:16) I j ( s ) √ s (cid:17) (cid:48) = I (cid:48) j ( s )(1 + s ) − sI j ( s )(1 + s ) = − const s (1 + s ) , or I (cid:48) j ( s ) − s s I j ( s ) + const s s = 0 . Solving this linear ODE with an integrating factor √ s , we have I j ( s ) = (cid:112) s (cid:16) const √ s + c (cid:17) = const + c (cid:112) s with some constant c . Since I j is bounded on [ s o , ∞ ), c = 0, and we obtainthe converse part of the lemma. (cid:3) Let f o ( t ) = √ − t , L o ( s, t ) = st , x o ( s ) = y o ( s ) = √ s be the functionscorresponding to the unit Euclidean ball. Our goal is to prove the followingproposition. Proposition 1.
Let n = d . A body K f floats in equilibrium in every direc-tion at the level vol d ( K )2 , provided f is C -smooth and for all s > , (11) y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) n dt = y o ( s ) (cid:90) − x o ( s ) ( f o ( t ) − L o ( t )) n dt = const √ s , (12) y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( t )) n − ∂L s ( t ) ∂s dt = 0 . We remark that (11) and (12) are similar to equations (4) and (5) from[NRZ].
Proof.
Observe that H ( L ) divides K f into two parts of equal volume. Also,(12) is the same as (7) of Lemma 2. Thus, by Lemma 1 and Lemma 2 thecharacteristic points of the family of hyperplanes { H ( L s ), s ∈ [0 , ∞ ) } , areexactly the centers of mass of the sections K ∩ H ( L s ). Hence, we canapply the converse part of Theorem 4 to conclude that they are the cuttinghyperplanes at the level vol d ( K )2 .On the other hand, observing that conditions (11), (12) are the same as(9) and (7), by Lemma 4 condition (8) also holds. Therefore, by Lemma 3,the cutting sections have equal moments of inertia for all ( d − (cid:3) In the sequel, it will be more convenient to write L ( s, t ) = st + h ( s ).In order to construct a counterexample, we will choose the perturbationfunction h with the properties described at the beginning of this section.The convex body corresponding to any such function will be automaticallyasymmetric since not all its sections dividing the volume in half will passthrough a single point. N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 11 The case of even d ≥ n = d ∈ N . Our argument is very similar to theone in Section 3 of [NRZ]. Our body K f will be a local perturbation of theEuclidean ball (see Figure 4). x Figure 4.
The body of revolution K f in the case of even dimensionsThe equations(13) f ( y ( σ )) = L ( σ, y ( σ )) , f ( − x ( σ )) = L ( σ, − x ( σ ))show that to define f , it is enough to define two decreasing functions x ( σ ), y ( σ ) on [0 , + ∞ ). Our functions x ( σ ) and y ( σ ) will coincide with x o and y o for all σ / ∈ [1 − τ, − τ ]. Since the curvature of the semicircle is strictlypositive, the resulting function f will be strictly concave if x and y are closeto x o and y o in C .We shall make our construction in several steps. First, we define x = x o , y = y o on [1 , ∞ ). Second, we will express equations (11), (12) purely interms of x and y (see (16) and (17) below). Then we will use these newequations to extend the functions x and y to [1 − τ, τ and h are sufficiently small. Moreover, the extensions willcoincide with x o and y o on [1 − τ,
1] and will be close to x o and y o up totwo derivatives on [1 − τ, − τ ]. Then, we will show that our extensionsautomatically coincide with x o and y o on [1 − τ, − τ ] as well. This willallow us to put x = x o , y = y o on the remaining interval [0 , − τ ] and geta nice smooth function. At last, we will show that equations (11), (12) willbe satisfied up to s = 0, thus finishing the proof. Step 1 . We put x = x o , y = y o on [1 , ∞ ). Step 2 . To construct x , y on [1 − τ, to obtain a system of four integral equations with four unknown functions x , y , x (cid:48) , y (cid:48) . Next, we will apply Lemma 8 and Remark 2 from [NRZ, pgs.63-66] to show that there exists a solution x , y , x (cid:48) , y (cid:48) of the constructedsystem of integral equations on [1 − τ, x o , y o , dx o ds , dy o ds on [1 − τ, x and y components of thatsolution give a solution of (11), (12) with f defined by (13).Differentiating equation (11) n + 1 times and equation (12) n times, weobtain( − n n ! (cid:104)(cid:16)(cid:16) L ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ( s )) (cid:17) n dxds ( s ) + (cid:16)(cid:16) L ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ( s )) (cid:17) n dyds ( s ) (cid:105) +(14) y ( s ) (cid:90) − x ( s ) (cid:16) ∂∂s (cid:17) n +1 (cid:16) ( f ( t ) − L ( s, t )) n (cid:17) dt = (cid:16) dds (cid:17) n +1 (cid:16) const √ s (cid:17) , and ( − n − ( n − (cid:104)(cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ( s )) dxds ( s ) + (cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ( s )) dyds ( s ) (cid:105) +(15) y ( s ) (cid:90) − x ( s ) (cid:16) ∂∂s (cid:17) n (cid:16) ( f ( t ) − L ( s, t )) n − ∂L∂s ( s, t ) (cid:17) dt = 0 . When s ≤
1, the integral term I in (14) can be split as I = y ( s ) (cid:90) − x ( s ) (cid:16) ∂∂s (cid:17) n +1 (cid:16) ( f ( t ) − L ( s, t )) n (cid:17) dt = (cid:16) − x o (1) (cid:90) − x ( s ) + y ( s ) (cid:90) y o (1) (cid:17) (cid:16) ∂∂s (cid:17) n +1 (cid:16) ( f ( t ) − L ( s, t )) n (cid:17) dt + Ξ ( s ) , where Ξ ( s ) = y o (1) (cid:90) − x o (1) (cid:16) ∂∂s (cid:17) n +1 (cid:16) ( f o ( t ) − L ( s, t )) n (cid:17) dt. Making the change of variables t = − x ( σ ) in the integral (cid:82) − x o (1) − x ( s ) , and t = y ( σ ) in the integral (cid:82) y ( s ) y o (1) , we obtain I = − (cid:90) s (cid:16) ∂∂s (cid:17) n +1 (cid:16) L ( σ, − x ( σ )) − L ( s, − x ( σ )) (cid:17) n dxds ( σ ) dσ − N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 131 (cid:90) s (cid:16) ∂∂s (cid:17) n +1 (cid:16) L ( σ, y ( σ )) − L ( s, y ( σ )) (cid:17) n dyds ( σ ) dσ + Ξ ( s ) . Similarly, we have y ( s ) (cid:90) − x ( s ) (cid:16) ∂∂s (cid:17) n (cid:16) ( f ( t ) − L ( s, t )) n − ∂L∂s ( s, t ) (cid:17) dt = − (cid:90) s (cid:16) ∂∂s (cid:17) n (cid:16)(cid:16) L ( σ, − x ( σ )) − L ( s, − x ( σ )) (cid:17) n − ∂L∂s ( s, − x ( σ )) (cid:17) dxds ( σ ) dσ − (cid:90) s (cid:16) ∂∂s (cid:17) n (cid:16)(cid:16) L ( σ, y ( σ )) − L ( s, y ( σ )) (cid:17) n − ∂L∂s ( s, y ( σ )) (cid:17) dyds ( σ ) dσ + Ξ ( s ) , where Ξ ( s ) = y o (1) (cid:90) − x o (1) (cid:16) ∂∂s (cid:17) n (cid:16) ( f o ( t ) − L ( s, t )) n − ∂L∂s ( s, t ) (cid:17) dt. To reduce the resulting system of integro-differential equations to a puresystem of integral equations we add two independent unknown functions x (cid:48) , y (cid:48) and two new relations x ( s ) = − (cid:90) s x (cid:48) ( σ ) dσ + x o (1) , y ( s ) = − (cid:90) s y (cid:48) ( σ ) dσ + y o (1) . We rewrite our equations (14), (15) as follows:(16) ( − n n ! (cid:104)(cid:16)(cid:16) L ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ( s )) (cid:17) n x (cid:48) ( s ) + (cid:16)(cid:16) L ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ( s )) (cid:17) n y (cid:48) ( s ) (cid:105) − (cid:90) s (cid:16) ∂∂s (cid:17) n +1 (cid:16) L ( σ, − x ( σ )) − L ( s, − x ( σ )) (cid:17) n x (cid:48) ( σ ) dσ − (cid:90) s (cid:16) ∂∂s (cid:17) n +1 (cid:16) L ( σ, y ( σ )) − L ( s, y ( σ )) (cid:17) n y (cid:48) ( σ ) dσ + Ξ ( s ) = (cid:16) dds (cid:17) n +1 (cid:16) const √ s (cid:17) , and(17) ( − n − ( n − (cid:104)(cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ( s )) x (cid:48) ( s ) + (cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ( s )) y (cid:48) ( s ) (cid:105) − (cid:90) s (cid:16) ∂∂s (cid:17) n (cid:16)(cid:16) L ( σ, − x ( σ )) − L ( s, − x ( σ )) (cid:17) n − ∂L∂s ( s, − x ( σ )) (cid:17) x (cid:48) ( σ ) dσ − (cid:90) s (cid:16) ∂∂s (cid:17) n (cid:16)(cid:16) L ( σ, y ( σ )) − L ( s, y ( σ )) (cid:17) n − ∂L∂s ( s, y ( σ )) (cid:17) y (cid:48) ( σ ) dσ +Ξ ( s ) = 0 . Now we rewrite our system in the form(18) G ( s, Z ( s )) = (cid:90) s Θ ( s, σ, Z ( σ )) dσ + Ξ ( s ) . Here Z = xyx (cid:48) y (cid:48) , G ( s, Z ) = xy ( − n n ! (cid:104)(cid:16) L ∂L∂s (cid:12)(cid:12)(cid:12) ( s, − x ) (cid:17) n x (cid:48) + (cid:16) L ∂L∂s (cid:12)(cid:12)(cid:12) ( s,y ) (cid:17) n y (cid:48) (cid:105) ( − n − ( n − (cid:104)(cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ) x (cid:48) + (cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ) y (cid:48) (cid:105) , Θ ( s, σ, Z ) = − x (cid:48) y (cid:48) Θ Θ , whereΘ = − (cid:16) ∂∂s (cid:17) n +1 (cid:16) L ( σ, − x ) − L ( s, − x ) (cid:17) n x (cid:48) − (cid:16) ∂∂s (cid:17) n +1 (cid:16) L ( σ, y ) − L ( s, y ) (cid:17) n y (cid:48) , Θ = − (cid:16) ∂∂s (cid:17) n (cid:16)(cid:16) L ( σ, − x ) − L ( s, − x ) (cid:17) n − ∂L∂s ( s, − x ) (cid:17) x (cid:48) − (cid:16) ∂∂s (cid:17) n (cid:16)(cid:16) L ( σ, y ) − L ( s, y ) (cid:17) n − ∂L∂s ( s, y ) (cid:17) y (cid:48) , and Ξ ( s ) = x o (1) y o (1) − Ξ ( s ) + (cid:16) dds (cid:17) n +1 (cid:16) const √ s (cid:17) − Ξ ( s ) . N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 15
Note that G , Θ , Ξ are well-defined and infinitely smooth for all s, σ ∈ (0 , Z ∈ R . Observe also that D Z G (cid:12)(cid:12)(cid:12) ( s,Z ) = (cid:18) I ∗ A (cid:19) , where I = (cid:18) (cid:19) , A = A ( s, x, y ) = ( − n n ! (cid:16)(cid:16) L ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ) (cid:17) n ( − n n ! (cid:16)(cid:16) L ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ) (cid:17) n ( − n − ( n − (cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s, − x ) ( − n − ( n − (cid:16)(cid:16) L ∂L∂s (cid:17) n − ∂L∂s (cid:17)(cid:12)(cid:12)(cid:12) ( s,y ) . The function Z o ( s ) = x o ( s ) y o ( s ) dx o ds ( s ) dy o ds ( s ) solves the system (18) with G , Θ , Ξ corresponding to h ≡ G o , Θ o , Ξ o ) on [ , (cid:16) D Z G o (cid:12)(cid:12)(cid:12) ( s,Z o ( s )) (cid:17) = det( A o ( s, x o ( s ) , y o ( s ))) (cid:54) = 0 ∀ s ∈ (0 , . Indeed, since the matrix A o ( s, x o ( s ) , y o ( s )) is of the form (cid:18) ( − n n !( sx o ( s )) n ( − n n !( sy o ( s )) n ( − n − ( n − sx o ( s )) n − ( − x o ( s )) ( − n − ( n − sy o ( s )) n − y o ( s ) (cid:19) , its sign pattern is (cid:18) + ++ − (cid:19) , when n is even , and (cid:18) − −− + (cid:19) , when n is odd . Thus, (19) follows. In particular,det (cid:16) D Z G o (cid:12)(cid:12)(cid:12) (1 ,Z o (1)) (cid:17) (cid:54) = 0 . Lemma 8 from [NRZ, page 63] implies then that we can choose some small τ > k ∈ N , construct a solution Z ( s ) of (18) which is C k -close to Z o ( s ) on [1 − τ, G , Θ , Ξ are sufficiently closeto G o , Θ o , Ξ o in C k on certain compact sets. Since G , Θ , Ξ and theirderivatives are some explicit (integrals of) polynomials in Z , s , σ , h ( s ), andthe derivatives of h ( s ), this closeness assumption will hold if h is sufficientlyclose to zero with sufficiently many derivatives. Moreover, since h vanisheson [1 − τ, Z ( s ) = Z o ( s ) on [1 − τ, To prove that the x and y components of the solution we found give asolution of (11), (12) with f defined by (13), we consider the functions F ( s ) := y ( s ) (cid:90) − x ( s ) (cid:16) f ( s, t ) − L ( s, t ) (cid:17) n dt − const √ s ,H ( s ) := y ( s ) (cid:90) − x ( s ) (cid:16) f ( s, t ) − L ( s, t ) (cid:17) n − ∂L∂s ( s, t ) dt. Since equations (16) and (17) of our system (18) were obtained by the dif-ferentiation of equations (11), (12), we have (cid:16) dds (cid:17) n +1 F ( s ) = 0 , (cid:16) dds (cid:17) n H ( s ) = 0on [1 − τ, F and H are polynomials on [1 − τ, h ( s ) = 0, x ( s ) = x o ( s ), y ( s ) = y o ( s ) on [1 − τ, F and H vanish on [1 − τ,
1] and,therefore, identically. Thus, we conclude that the x and y components ofthe solutions of (16), (17) solve (11), (12) on (1 − τ, Step 3 . We claim that x = x o , y = y o on [1 − τ, − τ ], i.e., the perturbedsolution returns to the semicircle. Since h is supported on [1 − τ, − τ ], wehave L = L o = st and ∂∂s L ( s, t ) = t for s ∈ [1 − τ, − τ ]. It follows thatevery time we differentiate equation (11) (with respect to s ) we can dividethe result by s to obtain(20) y ( s ) (cid:90) − x ( s ) ( f ( t ) − L o ( s, t )) n − k t k dt = y o ( s ) (cid:90) − x o ( s ) ( f o ( t ) − L o ( s, t )) n − k t k dt,k ≤ n . If we take k = n in (20), we get(21) y ( s ) (cid:90) − x ( s ) t n dt = y o ( s ) (cid:90) − x o ( s ) t n dt. Similarly, for k ≤ n −
1, equation (12) implies that(22) y ( s ) (cid:90) − x ( s ) ( f ( t ) − L o ( s, t )) n − − k t k +1 dt = y ( s ) (cid:90) − x ( s ) ( f ( t ) − L o ( s, t )) n − − k t k +1 dt = 0 . N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 17
Putting k = n − y ( s ) (cid:90) − x ( s ) t n − dt = 0 = y o ( s ) (cid:90) − x o ( s ) t n − dt. Equation (23) yields x ( s ) = y ( s ), and the symmetry (with respect to 0) of theintervals ( − x o ( s ) , y o ( s )), ( − x ( s ) , y ( s )), together with (21), yield ( − x o ( s ) , y o ( s ))= ( − x ( s ) , y ( s )) for all s ∈ [1 − τ, − τ ]. Step 3 is completed. Step 4 . We put x = x o , y = y o on [0 , − τ ], which will result in afunction f defined on [ − ,
1] and coinciding with f o ( t ) = √ − t outsidesmall intervals around ± √ . It remains to check that (11), (12) are validfor s ∈ [0 , − τ ]. We will prove the validity of (12). The proof for equation(11) is similar and can be found in [NRZ, page 53].Since h ≡ − τ, − τ ), we have L ( s, t ) = st for s ∈ [0 , − τ ],so we need to check that y ( s ) (cid:90) − x ( s ) ( f ( t ) − ( st ) ) n − tdt = y ( s ) (cid:90) − x ( s ) ( f o ( t ) − ( st ) ) n − tdt, ∀ s ∈ [0 , − τ ] . Recall that x = x o and y = y o everywhere on this interval, so we can write x and y instead of x o and y o on the right hand side.Using the binomial formula, we see that it suffices to check that(24) y ( s ) (cid:90) − x ( s ) f j ( t ) t n − − j )+1 dt = y ( s ) (cid:90) − x ( s ) f jo ( t ) t n − − j )+1 dt, ∀ j = 1 , . . . , n − s ∈ [0 , − τ ]. Since f ≡ f o outside [ − x (1 − τ ) , y (1 − τ )], splitting the integrals in (24) into three parts with ranges[ − x ( s ) , − x (1 − τ )], [ − x (1 − τ ) , y (1 − τ )], [ y (1 − τ ) , y ( s )], it is enough tocheck (24) on the middle interval [ − x (1 − τ ) , y (1 − τ )].To this end, we first take s = 1 − τ , k = n − y (1 − τ ) (cid:90) − x (1 − τ ) f ( t ) t n − dt = y (1 − τ ) (cid:90) − x (1 − τ ) f o ( t ) t n − dt, which is (24) for j = 1. Now we go “one step up”, by taking s = 1 − τ , k = n − y (1 − τ ) (cid:90) − x (1 − τ ) ( f ( t ) − ( st ) ) t n − dt = y (1 − τ ) (cid:90) − x (1 − τ ) ( f o ( t ) − ( st ) ) t n − dt. The last equality together with (25) yield y (1 − τ ) (cid:90) − x (1 − τ ) f ( t ) t n − dt = y (1 − τ ) (cid:90) − x (1 − τ ) f o ( t ) t n − dt, which is (24) for j = 2. Proceeding in a similar way we get (24) for j =1 , . . . , n −
1. This finishes the proof of Theorem 1 in even dimensions. (cid:3) The case of odd d ≥ x Figure 5.
The body of revolution K f in the case of odd dimensionsNote that n = q + , q ∈ N . Then (11) and (12) take the form(26) y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( s, t )) q + dt = y o ( s ) (cid:90) − x o ( s ) ( f o ( t ) − L o ( s, t )) q + dt = const √ s , (27) y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( s, t )) q − ∂L∂s ( s, t ) dt = 0 , where f o ( t ) = √ − t , L o ( s, t ) = tξ , and y o ( s ) = x o ( s ) = √ s .Our argument is similar to the one in [NRZ, Section 4]. Our body ofrevolution K f will be constructed as a perturbation of the Euclidean ball.We remark that in the case of odd dimensions, the perturbation will not belocal (see Figure 5).We shall make our construction in several steps corresponding to the sloperanges s ∈ [1 , ∞ ), s ∈ [1 − τ, s ∈ (0 , − τ ]. We will use differentways to describe the boundary of K f within those ranges. We will define N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 19 f ( t ) = f o ( t ) for t ∈ (cid:104) − √ , √ (cid:105) . We will differentiate (26), (27) and rewritethe resulting equations in terms of x and y , to extend x and y to [1 − τ, f is related to x and y by (13).Finally, we will change the point of view and define the remaining part of f in terms of the radial functions R ( α ) and r ( α ), related to f by(28) f ( R ( α ) cos α ) = R ( α ) sin α, f ( − r ( α ) cos α ) = r ( α ) sin α, α ∈ [0 , π ] . Note that the radial function ρ K ( w ) = sup { t > tw ∈ K } of the resultingbody K satisfies(29) ρ K ( w ) = (cid:40) R ( α ) if w > ,r ( α ) if w < , where w = ( w , . . . , w d ) ∈ S d − and α ∈ [0 , π ], cos α = | w | . Step 1 . We put x = x o , y = y o on [1 , ∞ ), which is equivalent to putting f ( t ) = √ − t for t ∈ [ − √ , √ ]. Step 2 . Differentiating equation (26) q + 1 times, we obtain(30) (cid:16) ∂∂s (cid:17) q +1 y ( s ) (cid:90) − x ( s ) ( f ( t ) − L ( s, t )) q + dt = (cid:16) − x o (1) (cid:90) − x ( s ) + y ( s ) (cid:90) y o (1) (cid:17)(cid:16) ∂∂s (cid:17) q +1 (cid:16) ( f ( t ) − L ( s, t )) q + (cid:17) dt + E ( s ) = (cid:16) dds (cid:17) q +1 const √ s , where E ( s ) = y o (1) (cid:90) − x o (1) (cid:16) ∂∂s (cid:17) q +1 (cid:16) ( f o ( t ) − L ( s, t )) q + (cid:17) dt. Note that, unlike it was for the function Ξ in the even-dimensional case,the function E is well-defined only for s ≤ (cid:107) h (cid:107) C is muchsmaller than 1. Also, even with these assumptions, E ( s ) is C ∞ on [0 , (cid:16) ∂∂s (cid:17) q +1 (cid:16) ( f ( t ) − L ( s, t )) q + (cid:17) = J ( s, t, f ( t )) (cid:112) f ( t ) − L ( t ) , where J ( s, t, f ) is some polynomial expression in s , t , f , h ( s ), and thederivatives of h at s . Making the change of variables t = − x ( σ ) in the integral (cid:82) − x o (1) − x ( s ) , and t = y ( σ ) in the integral (cid:82) y ( s ) y o (1) , we can rewrite the sum of integrals on theleft hand side of (30) as − (cid:90) s (cid:104) J ( s, − x ( σ ) , L ( σ, − x ( σ ))) (cid:112) L ( σ, − x ( σ )) − L ( s, − x ( σ )) dxds ( σ ) ++ J ( s, y ( σ ) , L ( σ, y ( σ ))) (cid:112) L ( σ, y ( σ )) − L ( s, y ( σ )) dyds ( σ ) (cid:105) dσ. Now write L ( σ, t ) − L ( s, t ) = ( L ( σ, t ) − L ( s, t ))( L ( σ, t ) + L ( s, t )) , and L ( σ, t ) − L ( s, t ) = σt + h ( σ ) − st − h ( s ) = ( σ − s )( t + H ( s, σ )) , where H ( s, σ ) = h ( σ ) − h ( s ) σ − s = (cid:90) h (cid:48) ( s + ( σ − s ) τ ) dτ is an infinitely smooth function of s and σ . Denote K ( s, σ, t ) = J ( s, t, L ( σ, t )) (cid:112) ( t + H ( s, σ ))( L ( σ, t ) + L ( s, t )) . The function K is well-defined and infinitely smooth for all s , σ , t satisfying( t + H ( s, σ ))( L ( σ, t ) + L ( s, t )) >
0. If (cid:107) h (cid:107) C is small enough, this conditionis fulfilled whenever s , σ ∈ [ ,
1] and | t | > .Now we can rewrite equation (30) in the form(31) − (cid:90) s (cid:16) K ( s, σ, − x ( σ )) dxds ( σ ) + K ( s, σ, y ( σ )) dyds ( σ ) (cid:17) dσ √ σ − s = − E ( s ) + (cid:16) dds (cid:17) q +1 const √ s . Similarly, we can differentiate (27) q times and transform the resultingequation into(32) − (cid:90) s (cid:16) K ( s, σ, − x ( σ )) dxds ( σ ) + K ( s, σ, y ( σ )) dyds ( σ ) (cid:17) dσ √ σ − s == − E ( s ) , where K is well-defined and infinitely smooth in the same range as K . N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 21
The function E on the right hand side of (32) is given by E ( s ) = y o (1) (cid:90) − x o (1) (cid:16) ∂∂s (cid:17) q (cid:16) ( f o ( t ) − L ( s, t )) q − ∂L∂s ( s, t ) (cid:17) dt, and everything that we said about E applies to E as well.Equations (31) and (32) together can be written in the form(33) (cid:90) s K ( s, σ, z ( σ ) , dzds ( σ )) √ σ − s dσ = Q ( s ) , where, for z = (cid:18) xy (cid:19) , z (cid:48) = (cid:18) x (cid:48) y (cid:48) (cid:19) ∈ R , K ( s, σ, z, z (cid:48) ) = − (cid:18) K ( s, σ, − x ) x (cid:48) + K ( s, σ, y ) y (cid:48) K ( s, σ, − x ) x (cid:48) + K ( s, σ, y ) y (cid:48) (cid:19) ,Q ( s ) = − E ( s ) + (cid:16) dds (cid:17) q +1 const √ s − E ( s ) . By Lemma 8 in [NRZ, page 63] with b = 1, equation (33) is equivalent to(34) − G ( s, s, z, z (cid:48) ) + (cid:90) s ∂∂s G ( s, σ, z ( σ ) , dzds ( σ )) dσ = (cid:101) Q ( s ) , where G ( s, σ, z, z (cid:48) ) = (cid:90) K ( s + τ ( σ − s ) , σ, z, z (cid:48) ) (cid:112) τ (1 − τ ) dτ, (cid:101) Q ( s ) = dds (cid:90) s Q ( s (cid:48) ) √ s (cid:48) − s ds (cid:48) . Note that G ( s, s, z, z (cid:48) ) = C · K ( s, s, z, z (cid:48) ) , C = (cid:90) dτ (cid:112) τ (1 − τ ) . To reduce the resulting system of integro-differential equations to a puresystem of integral equations we add two independent unknown functions x (cid:48) , y (cid:48) , denote z (cid:48) = (cid:18) x (cid:48) y (cid:48) (cid:19) , z o ( s ) = (cid:18) x o ( s ) y o ( s ) (cid:19) , and add two new relations z ( s ) = − (cid:90) s z (cid:48) ( σ ) dσ + z o (1) . Together with (34), they lead to the system(35) G ( s, Z ( s )) = (cid:90) s Θ ( s, σ, Z ( σ )) dσ + Ξ ( s ) , Z = (cid:18) zz (cid:48) (cid:19) = xyx (cid:48) y (cid:48) . Here G ( s, Z ) = (cid:18) z − G ( s, s, z, z (cid:48) ) (cid:19) , Θ ( s, σ, Z ) = − (cid:18) z (cid:48) ∂∂s G ( s, σ, z, z (cid:48) ) (cid:19) , and Ξ ( s ) = (cid:18) z o (1) (cid:101) Q ( s ) (cid:19) . In what follows, we will choose h so that (cid:107) h (cid:107) C is much smaller than 1. Inthis case, G , Θ are well-defined and infinitely smooth whenever s , σ ∈ [ , | x | , | y | > , z (cid:48) ∈ R , and Ξ is well-defined and infinitely smooth on [ , D Z G (cid:12)(cid:12)(cid:12) ( s,Z ( s )) = (cid:18) I ∗ A (cid:19) , where I = (cid:18) (cid:19) , A ( s, z ) = C · E ( s, z ) , and E ( s, z ) = (cid:18) K ( s, s, − x ) K ( s, s, y ) K ( s, s, − x ) K ( s, s, y ) (cid:19) . The function Z o ( s ) = (cid:32) z o ( s ) dz o ds ( s ) (cid:33) = x o ( s ) y o ( s ) dx o ds ( s ) dy o ds ( s ) solves the system (35) with G , Θ , Ξ corresponding to h ≡ G o , Θ o , Ξ o ) on [ , (cid:16) D Z G o (cid:12)(cid:12)(cid:12) ( s,Z o ( s )) (cid:17) = det( A o ( s, z o ( s ))) (cid:54) = 0 for all s ∈ [ , . Indeed, since K , ( s, s, t ) have the same signs as J , ( s, ξ, L ( s, t )) and since J ( s, t, L ( s, t )) = (2 q + 1)!! (cid:16) − L ( s, t ) ∂L∂s ( s, t ) (cid:17) q +1 ,J ( s, t, L ( s, t )) = (2 q − (cid:16) − L ( s, t ) ∂L∂s ( s, t ) (cid:17) q ∂L∂s ( s, t ) , we conclude that the matrix A o ( s, z o ( s )) has the same sign pattern as thematrix (cid:18) ( − q +1 ( − q +1 ( − q ( − x o ( s )) ( − q y o ( s ) (cid:19) , N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 23 i.e., the signs in the first row are the same, and the signs in the second oneare opposite.Thus, (36) follows. In particular,det (cid:16) D Z G o (cid:12)(cid:12)(cid:12) (1 ,Z o (1)) (cid:17) (cid:54) = 0 . Lemma 8 from [NRZ, page 63] implies then that we can choose some small τ > C k -close to Z o ( s ) solution Z ( s ) of (35) on [1 − τ, G , Θ , Ξ are sufficiently close to G o , Θ o , Ξ o in C k on certaincompact sets. Since G , Θ , Ξ and their derivatives are (integrals of) someexplicit elementary expressions in Z , s , σ , h ( s ), and the derivatives of h ( s ),this closeness assumption will hold if h is sufficiently close to zero along withsufficiently many derivatives. Moreover, since h vanishes on [1 − τ, Z ( s ) = Z o ( s ) on [1 − τ, x and y components of Z solve the equations obtained by differenti-ating (26) and (27). The passage to (26), (27) is now exactly the same as inthe even case. Step 3 . From now on, we change the point of view and switch to thefunctions R ( α ) and r ( α ), α ∈ (0 , π ), related to f by (28). The functions x and y , which we have already constructed, implicitly define C ∞ -functions R h ( α ) and r h ( α ) for all α with tan α > − τ .Instead of parameterizing hyperplanes by the slopes s of the correspondinglinear functions, we will parameterize them by the angles β they make withthe x -axis, where β is related to s by tan β = s .Our next task will be to derive the equations that will ensure that all central sections corresponding to angles β with tan β < − τ are the cut-ting sections with equal moments with respect to any ( d − β ∈ (1 − τ, − τ ).It will be convenient to rewrite conditions (7), (8) and (9) in terms of the Spherical Radon Transform (see [He]), defined as R f ( ξ ) = (cid:90) S d − ∩ ξ ⊥ f ( w ) dw, f ∈ C ( S d − ) , ξ ∈ S d − . We will use the following proposition.
Proposition 2.
Let K be a convex body of revolution about the x -axiscontaining the origin in its interior and let ξ = ( ± sin α, , . . . , , ∓ cos α ) ∈ S d − be the unit vector corresponding to the angle α ∈ [0 , π ) . Then thecenter of mass of the central section K ∩ ξ ⊥ is at the origin if and only if (37) ( R ( w j ρ dK ( w ))( ξ ) = 0 , j = 1 , . . . , d − . Also, the moments of inertia of the central section K ∩ ξ ⊥ with respect toany ( d − -dimensional subspace l are constant independent of l if and onlyif (38) ( R ( w ρ d +1 K ( w ))( ξ ) = const ( d + 1)(1 − ξ ) , (39) ( R ( w j ρ d +1 K ( w ))( ξ ) = const ( d + 1) f or all j = 2 , . . . , d − , and (40) ( R ( w j w l ρ d +1 K ( w ))( ξ ) = 0 , j, l = 1 , . . . , d − , j (cid:54) = l. Proof.
If the center of mass of K ∩ ξ ⊥ is at the origin, we have1vol d − ( K ∩ ξ ⊥ ) (cid:90) K ∩ ξ ⊥ xdx = 0 . Passing to the polar coordinates in ξ ⊥ and taking into account the fact thatfor w ∈ ξ ⊥ we have w d = w tan α , we obtain the first statement of thelemma.Let l be any ( d − ξ ⊥ and let u = u d − be a unitvector in ξ ⊥ orthogonal to l . By (3) the condition on the moments reads as(41) I K ∩ ξ ⊥ ( l ) = (cid:90) K ∩ ξ ⊥ ( x · u ) dx = const ∀ u ∈ S d − ∩ ξ ⊥ . Denote by ι , . . . ι d − the orthonormal basis in ξ ⊥ such that ι = cos αe +sin αe d and ι j = e j for j = 2 , . . . , d −
1. Passing to polar coordinates anddecomposing u in the basis { τ j } d − j =1 , we see that the moments of inertia ofthe central section K ∩ ξ ⊥ with respect to any ( d − R (( w · τ ) ρ d +1 K ( w ))( ξ ) = const ( d + 1) , (39) holds, and(43) ( R (( w · τ j )( w · τ l ) ρ d +1 K ( w ))( ξ ) = 0 , j, l = 1 , . . . , d − , j (cid:54) = l, (see the proof of Theorem 1 in [R]). Since w · τ = w cos α + w d sin α and w d = w tan α , we see that (42) and (43) are equivalent to (38) and (40).This gives the second statement and the lemma is proved. (cid:3) We remark that for any body of revolution around the x -axis, (37) holdsfor j = 2 , . . . , d −
1. Taking u = τ j in the integral in (41), by rotationinvariance we obtain that the moments in (39) are equal for j = 2 , . . . , d − s o = 1 − τ and Proposition2 with K = K f , when K f is the body of revolution we are constructing,equations (37), (38), (39) and (40) hold if tan α ∈ (1 − τ, − τ ) with the N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 25 constant in (38), (39) independent of ξ . Also, the left hand sides of (37),(38) and (39) are already defined on the cap U τ = { ξ ∈ S d − : ξ = ± sin α, α ∈ [0 , π , tan α ≥ − τ } and are smooth even rotation invariant functions there.Assume for a moment that we have constructed a smooth body K f sothat conditions(44) ( R ( w ρ d +1 K f ( w ))( ξ ) = const ( d + 1)(1 − ξ ) , ( R ( w ρ dK f ( w ))( ξ ) = 0 , hold for all unit vectors ξ ∈ S d − with ξ = ± sin α , corresponding to theangles α ∈ [0 , π ] such that tan α < − τ . Then by the above remarks,Proposition 2 and the converse part of Lemma 4 with s o = 0, conditions (11),(12) of Proposition 1 are satisfied for all s > K f floats in equilibriumin every direction at the level vol d ( K )2 .Thus, it remains to construct the part of K f so that (44) holds for all unitvectors ξ corresponding to the angles α ∈ [0 , − τ ]. To this end, denoteby ϕ h and ψ h the left-hand sides of (44) defined on U τ . We put ϕ h ( ξ ) = const ( d + 1)(1 − ξ ) and ψ h ( ξ ) = 0 for ξ ∈ S d − such that ξ = ± sin α andtan α ∈ [0 , − τ ]. This definition agrees with the one we already have whentan α ∈ [1 − τ, − τ ], so ϕ h and ψ h are even rotation invariant infinitelysmooth functions on the entire sphere.Recall that the values of R g ( ξ ) for all ξ ∈ S d − such that ξ = ± sin α and tan α > − τ are completely determined by the values of the evenfunction g ( w ) for all w ∈ S d − satisfying w = ± cos α and tan α > − τ .Moreover, for bodies of revolution (but not in general) the converse is alsotrue (see the explicit inversion formula in [Ga, page 433, formula (C.17)]).Since the equation R g = (cid:101) g with even C ∞ right hand side (cid:101) g is equivalentto g ( ξ ) + g ( − ξ )2 = R − (cid:101) g ( ξ ) , we can rewrite the equations in (44) as(45) w ( ρ d +1 K ( w ) + ρ d +1 K ( − w )) = 2( R − ϕ h )( w )and(46) w ( ρ dK ( w ) − ρ dK ( − w )) = 2( R − ψ h )( w ) . The already constructed part of ρ K satisfies these equations for the vectors w ∈ S d − such that w = ± cos α and tan α > − τ .Since the Spherical Radon Transform commutes with rotations and ourinitial ρ K was rotation invariant, the even functions 2 R − ϕ h ( w ), 2 R − ψ h ( w )are rotation invariant as well and can be written as Φ h ( α ) and Ψ h ( α ), where w ∈ S d − is such that w = ± cos α and α ∈ [0 , π ]. Note that the mappings h (cid:55)→ Φ h , h (cid:55)→ Ψ h are continuous from C k + d to C k , say. Thus, for all h sufficiently close to zero in C k + d , Φ h and Ψ h will be close to Φ ≡ w andΨ ≡ C k .We will be looking for a rotation invariant solution ρ K of (45) and (46),which will be described in terms of the two functions R ( α ) and r ( α ) relatedto it by (29). Equations (45) and (46) translate into(47) R d +1 ( α ) + r d +1 ( α ) = Φ h ( α )cos α , R d ( α ) − r d ( α ) = Ψ h ( α )cos α . Equations (47), together with the conditions R ( α ) > r ( α ) >
0, de-termine R ( α ) and r ( α ) uniquely, and they coincide with the functions R h and r h obtained in Step 2 for all α ∈ [0 , π ] with tan α ≥ − τ . Thus, anysolution R , r of this system will satisfy R ( α ) = R h ( α ), r ( α ) = r h ( α ) in thisrange.If h along with several derivatives are small enough, the functions Φ h − w and Ψ h are close to zero uniformly with several derivatives. Since the map D : ( R, r ) (cid:55)→ ( R d +1 + r d +1 , R d − r d )is smoothly invertible near the point (1 ,
1) by the inverse function theo-rem, the functions R , r exist in this case on the entire interval [0 , π ], andare close to 1 in C . Moreover, R (cid:48) (0) = r (cid:48) (0) = 0, because Φ (cid:48) h (0) = 0,Ψ (cid:48) h (0) = 0, (otherwise the functions R − ϕ h , R − ψ h would not be smoothat (1 , , . . . , R and r is convex and corresponds to some strictly concave function f defined on[ − r (0) , R (0)].This completes the proof of Theorem 1 in the case of odd dimensions. (cid:3) It remains to prove Theorem 2. Assume that a body K ⊂ R have density D and volume V . If K is submerged in liquid of density D (cid:48) and V (cid:48) is thevolume of a submerged part, then, by Archimedes law, D V = D (cid:48) V (cid:48) , [Zh,page 657]. Taking D (cid:48) = 1 and V (cid:48) = V , we obtain the result. (cid:3) Appendix A: proof of Theorem 4 from [O]6.1.
The “if ” part.
We begin with several auxiliary Lemmas. Let P H bethe orthogonal projection onto a hyperplane H . For a small ε > ε = P H ( { x ∈ bd K : dist( x, H ) < ε } ) . Let D be the length of a diameter of K and let µ = D d vol d ( K ∩ H − ( ξ )) . We put(48) Σ µε = { x ∈ H ( ξ ) : dist( x, bd K ∩ H ( ξ )) < µε } , where H ( ξ ) is a hyperplane for which (1) holds. Lemma 5.
We have Ξ ε ⊂ Σ µε , and vol d − (Σ µε ) < c d µD d − ε → as ε → .Proof. Consider a hyperplane G ( ξ ) ∈ H − ( ξ ) which is parallel to H ( ξ ) andsuch that dist( H ( ξ ) , G ( ξ )) = ε for ε > T containing any two corresponding parallel ( d − N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 27 planes which are the supporting planes to K ∩ H ( ξ ) and K ∩ G ( ξ ). In thehalf-space H − ( ξ ) containing these sections choose an angle γ between T and H ( ξ ) which is not obtuse (see Figure 6, cf. Figure 1 in [O]). T ε Ψ H( ξ )G( ξ ) Κ γ V X ( Figure 6.
The hyperplanes H ( ξ ), G ( ξ ), and T .Denote by Ψ the maximal distance between H ( ξ ) and any point in K ∩ H − ( ξ ). ThenΨ ≤ D sin γ, vol d ( K ∩ H − ( ξ )) < D d − Ψ ≤ D d sin γ. On the other hand, if λ = vol d ( K ∩ H − ( ξ ))vol d ( K ) , thenvol d ( K ∩ H − ( ξ )) ≥ λ λ vol d ( K ) ≥ λ vol d ( K ) , which yields sin γ > λ vol d ( K )2 D d , | cot γ | < D d λ vol d ( K ) = µ. Since the distance between the corresponding ( d − K ∩ H ( ξ ) and P H ( ξ ) ( K ∩ G ( ξ )) is ε | cotγ | < µε , we see that Ξ ε isa subset of Σ µε .Let P be the ( d − K ∩ H ( ξ ). Thenvol d − (Σ µε ) ≤ µεP < µεc d D d − → , as ε → . These estimates hold if bd K ∩ H ( ξ ) is a polytope, and the general case canbe obtained by passing to the limit; in the second inequality we used the factthat the surface area measure of bd K ∩ H ( ξ ) does not exceed c d D d − , where c d is some constant depending on the dimension, (it follows, for example,from inequality (7) in [CSG, Theorem 1]). (cid:3) Consider now a family W of hyperplanes H satisfying (1) which are par-allel to some ( d − l . Each such hyperplane is de-termined by the angle θ it makes with some fixed H ∈ W (we take theorientation into account). We will denote by H ( θ ) and H ( θ + ∆ θ ) the hy-perplanes in W making angles θ and θ + ∆ θ with H . Lemma 6.
For sufficiently small ∆ θ the ( d − -dimensional affine subspace H ( θ ) ∩ H ( θ + ∆ θ ) passes through K .Proof. Observe at first that for ∆ θ small enough, the convex bodies K ∩ H − ( θ ) and K ∩ H − ( θ + ∆ θ ) have a common point in the interior of H − ( θ ).Indeed, let β be the smallest angle between H ( θ ) and the supporting hyper-planes to K at points in bd K ∩ H ( θ ). As in the proof of Lemma 5, one canshow that β > sin β > λ vol d ( K )2 D d = 1 µ . Therefore, any supporting hyperplane to K making a positive angle with H ( θ ) which is less than µ , must also be supporting to K ∩ H − ( θ ). Let H (cid:48) ( θ + ∆ θ ) be the supporting hyperplane to K ∩ H − ( θ + ∆ θ ) parallel to H ( θ +∆ θ ). Then H (cid:48) ( θ +∆ θ ) is also the supporting hyperplane to K ∩ H − ( θ ),provided ∆ θ < µ . This proves the observation.Using the observation, we see that if H ( θ ) ∩ H ( θ + ∆ θ ) does not passthrough K , then K ∩ H − ( θ ) and K ∩ H − ( θ + ∆ θ ) are contained in oneanother. This contradicts the fact that they have the same volume and theresult follows. (cid:3) Now choose a “moving” system of coordinates in which the ( d − H ( θ ) ∩ H ( θ +∆ θ ) is the x x . . . x d − -coordinate plane,the axis x d − is in H ( θ ) and the axis x d is orthogonal to H ( θ ). We can as-sume that ∆ θ is acute and is less than µ .The next lemma is a direct consequence of the fact that all hyperplanesin W satisfy (1). Denote by A & B the symmetric difference of two sets A and B , i.e., A & B = ( A \ B ) ∪ ( B \ A ). Lemma 7.
Let
Λ = ( K ∩ H ( θ )) & P H ( θ ) ( K ∩ H ( θ + ∆ θ )) . Then (49) ∆ V = vol d ( K ∩ H − ( θ )) − vol d ( K ∩ H − ( θ + ∆ θ )) = (cid:90) K ∩ H ( θ ) x d − tan ∆ θ d vol d − ( x ) − (cid:90) Λ ζ d d vol d − ( x ) = 0 , where x d − = x d − ( θ, ∆ θ ) and ζ d = ζ d ( θ, ∆ θ ) is an error of x d = x d − tan θ in Λ which is obtained during the computation of ∆ V using the first integralabove ( see Figure 7 ) . We are ready to finish the proof of the “ if ” part of Theorem 4. Let x d − ( C ( K ∩ H ( θ ))) be the ( d − C ( K ∩ H ( θ )) with respect N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 29 H ( θ + Δθ ) Η ( θ ) Δθ x d-1 ζ d ζ d K : i Figure 7.
The function ζ d to the moving coordinate system. By (49), we have x d − ( C ( K ∩ H ( θ ))) = (cid:82) K ∩ H ( θ ) x d − d vol d − ( x )vol d − ( K ∩ H ( θ )) = (cid:82) Λ ζ d d vol d − ( x )vol d − ( K ∩ H ( θ )) tan ∆ θ . Since for every x ∈ Λ there exists y ∈ Ξ D sin ∆ θ such that P H ( θ ) y = x ,applying Lemma 5 we see thatvol d − (Λ) ≤ vol d − (Ξ D sin ∆ θ ) ≤ vol d − (Σ µD sin ∆ θ ) ≤ c d µD d − ∆ θ → θ →
0. Using the estimate | ζ d | ≤ D tan ∆ θ , the previous inequalitiesand the fact that Λ ⊂ Σ µD sin ∆ θ , we obtain | x d − ( C ( K ∩ H ( θ ))) | ≤ D tan ∆ θ vol d − (Λ)vol d − ( K ∩ H ( θ )) tan ∆ θ → θ → θ →
0, the ( d − H ( θ ) ∩ H ( θ + ∆ θ ) tends to a limiting position that passes through the center ofmass of K ∩ H ( θ ). Since the affine subspace l and the angle θ were chosenarbitrarily, we obtain the proof of the “ if ” part of the theorem.6.2. Proof of the converse part of Theorem 4.
Let l be an arbitrary( d − W be defined as above. Also, as above,choose an arbitrary angle θ , the hyperplanes H ( θ ) and H ( θ + ∆ θ ) in W and a “moving” coordinate system. We can assume that x d − ( C ( K ∩ H ( θ ))) → θ → V ∆ θ = tan ∆ θ ∆ θ (cid:90) K ∩ H ( θ ) x d − d vol d − ( x ) − (cid:90) Λ ζ d ∆ θ d vol d − ( x ) . Since C ( K ∩ H ( θ +∆ θ )) → C ( K ∩ H ( θ )) and bd K ∩ H ( θ +∆ θ ) → bd K ∩ H ( θ )as ∆ θ →
0, the set Λ defined in Lemma 7 satisfies vol d − (Λ) → θ → | ζ d | ≤ D tan ∆ θ we see that both summandsin the right-hand side of the above identity tend to 0 as ∆ θ →
0. This giveslim ∆ θ → V ∆ θ = 0.Now consider the function ξ (cid:55)→ q ( ξ ) := vol d ( K ∩ H − ( ξ )) on S n − , where H ( ξ ) is the hyperplane from our family. Since l and θ we chosen arbitrarily,writing q in terms of the spherical angles ϕ , . . . , ϕ d − , ϕ j ∈ [0 , π ), j = 1, . . . , d − ϕ d − ∈ [0 , π ), we obtain that ∂∂ϕ j q ( ϕ , . . . , ϕ d − ) = 0 for all ϕ , . . . , ϕ d − . Hence, q is constant on S d − and the proof of the conversepart is complete.This finishes the proof of Theorem 4. (cid:3) Appendix B: proof of Lemma 1
We recall the definition of an envelope of a family of curves and necessaryand sufficient conditions for its existence (see [Za, pgs. 9-17]).
Definition 5.
Let a, b, κ, θ ∈ R . A family Γ of parametrized locally simplecurves in R is a map r ( t, s ) = ( ζ ( t, s ) , ϑ ( t, s )) , ζ, ϑ ∈ C , t ∈ ( a, b ) , s ∈ ( κ, θ ) , such that for every fixed parameter s ∈ ( κ, θ ) the function r ( t, s ) defines alocally simple curve on R . Definition 6.
A regular curve (cid:37) ( λ ) = ( ζ ( λ ) , ϑ ( λ )) , ζ, ϑ ∈ C , (cid:37) λ (cid:54) = 0 , λ ∈ ( α, β ) , is a piece of an envelope of Γ if for every value of the parameter λ , (cid:37) istangent to at least one of the curves in the family Γ , and the correspondencebetween λ and the parameters t, s of the corresponding curve at the point oftangency can be expressed as continuous functions t ( λ ) , s ( λ ) , where s ( λ ) isnot constant on any interval.In other words,(1) r ( t ( λ ) , s ( λ )) = (cid:37) ( λ ) for every λ ∈ ( α, β ) ,(2) for every λ , the curve r ( t, s ( λ )) is tangent to (cid:37) ( λ ) at (cid:37) ( λ ) ,(3) s ( λ ) is not constant on ( α, β ) .The envelope is the union of all the pieces of an envelope. N BODIES FLOATING IN EQUILIBRIUM IN EVERY DIRECTION 31
Theorem 5.
Let Γ have an envelope, and let t = t ( λ ) , s = s ( λ ) . If apoint on the envelope corresponds to ( t , s ) , then ( t , s ) belongs to one ofthe two following sets:1) the collection of points ( t, s ) , such that r ( t, s ) ∈ C in some neighbor-hood of ( t, s ) ;2) the collection of points ( t, s ) for which (50) g := ζ t ϑ s − ϑ t ζ s = 0 . Theorem 6.
Let Γ be a family of C -curves such that (51) g = 0 , r t (cid:54) = 0 , g t (cid:54) = 0 , r s g t − r t g s (cid:54) = 0 . Then there exists a neighborhood of the point ( t , s ) with t ∈ ( a, b ) , s ∈ ( κ, θ ) , such that the envelope exists and is given by r ( t, s ) , where g ( t, s ) = 0 in the neighborhood. On this envelope, one can take s as the parameter λ and we have that t ( s ) ∈ C , t ( s ) = t . Moreover, the envelope is uniqueand every curve in Γ is tangent to the envelope at a unique point, called thecharacteristic point. Finally we are ready to prove Lemma 1. We remark that the choice of theperturbation function is rather wide. It is enough, for example, to requirethat h is infinitely smooth, supported by [1 − τ, − τ ] for sufficiently small τ >
0, small along with sufficiently many derivatives, and such that h (cid:48)(cid:48) hasfinitely many zeros on [1 − τ, − τ ].We make the simplest choice. For sufficiently small τ >
0, we denote c = 1 − τ and r = τ . Consider an infinitely smooth Mexican hat function h ( s ) = (cid:40) b exp (cid:16) s − c ) − r (cid:17) , if | s − c | < r , , if | s − c | ≥ r , where we choose b small enough so that h is small together with sufficientlymany derivatives.Since h (cid:48)(cid:48) ( s ) = (cid:40) b P ( s )(( s − c ) − r ) exp (cid:16) s − c ) − r (cid:17) , if | s − c | < r , , if | s − c | ≥ r , where P ( s ) = 3( s − c ) + 2(1 − r )( s − c ) − r , we have h (cid:48)(cid:48) ( s ) = 0 for s = s ± := c ± (cid:16) ((1 − r ) + 3 r ) − (1 − r )3 (cid:17) . Instead of the family F of hyperplanes given in (4) it is enough to considerthe family of lines H = { (cid:96) s : s ∈ [0 , ∞ ) } , (cid:96) s = H ( L s ) ∩ Π d = { ( t, ϑ ) : ϑ = st + h ( s ) } , where Π d is the x x d -plane. This family is given parametrically as r ( t, s ) = ( t, st + h ( s )) , t ∈ R , s ∈ [0 , ∞ ) . Hence, the sufficient conditions in (51) read as g = t + h (cid:48) ( s ) , r t ( t, s ) = (1 , s ) , g t = 1 , r s g t − r t g s = ( − h (cid:48)(cid:48) , t + h (cid:48) − sh (cid:48)(cid:48) ) , and are satisfied provided t = − h (cid:48) ( s ) and h (cid:48)(cid:48) ( s ) (cid:54) = 0.Using (50), we see that (cid:37) ( s ) = ( − h (cid:48) ( s ) , − sh (cid:48) ( s )+ h ( s )) describes the piecesof the envelope of H when s belongs to the intervals (1 − τ, s − ), ( s − , s + ),( s + , − τ ). Note that this parametrization is not regular at s ± , 1 − τ ,1 − τ , for, t (cid:48) ( s ) = − h (cid:48)(cid:48) ( s ), ϑ (cid:48) ( s ) = − sh (cid:48)(cid:48) ( s ), and h (cid:48)(cid:48) (1 − τ ) = h (cid:48)(cid:48) (1 − τ ) = h (cid:48)(cid:48) ( s ± ) = 0. By Theorem 6, the lines in H are tangent to the pieces of theenvelope for s ∈ (1 − τ, s − ) ∪ ( s − , s + ) ∪ ( s + , − τ ). Hence, the points oftangency are the characteristic ones for the corresponding tangent lines in H when s ∈ (1 − τ, s − ) ∪ ( s − , s + ) ∪ ( s + , − τ ).Let { (cid:96) s ( k ) } ∞ k =1 be a converging sequence of lines in H corresponding tothe points (cid:37) ( s ( k ) ), s ( k ) ∈ (0 , − τ ) ∪ (1 − τ, − τ ) ∪ (1 − τ, ∞ ), where s ( k ) isconverging to one of the points 1 − τ, − τ or s ± as k → ∞ . Passing to thelimit, we see that the points (cid:37) (1 − τ ) = (cid:37) (1 − τ ) = (0 ,
0) and (cid:37) ( s ± ) are alsothe characteristic points for the lines (cid:96) − τ , (cid:96) − τ , (cid:96) s ± . This gives (5) and thelemma is proved. (cid:3) Acknowledment . The author is very thankful to Mariangel Alfonseca,Alexander Fish, Fedor Nazarov, Alina Stancu, Peter V´arkonyi and VladYaskin for their invaluable help and very useful discussions.
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