A solution to the Fifth and the Eighth Busemann-Petty problems in a small neighborhood of the Euclidean ball
M. Angeles Alfonseca, Fedor Nazarov, Dmitry Ryabogin, Vladyslav Yaskin
AA SOLUTION TO THE FIFTH AND THE EIGHTHBUSEMANN-PETTY PROBLEMS IN A SMALLNEIGHBORHOOD OF THE EUCLIDEAN BALL
M. ANGELES ALFONSECA, FEDOR NAZAROV, DMITRY RYABOGIN,AND VLADYSLAV YASKIN
Abstract.
We show that the fifth and the eighth Busemann-Petty problems have positive solutions for bodies that are suffi-ciently close to the Euclidean ball in the Banach-Mazur distance. Introduction
In 1956, Busemann and Petty [BP] posed ten problems about sym-metric convex bodies, of which only the first one has been solved (see[K]). Their fifth and eighth problems are as follows.
Problem 5.
If for an origin-symmetric convex body K ⊂ R n , n ≥ ,we have (1) h K ( θ )vol n − ( K ∩ θ ⊥ ) = C ∀ θ ∈ S n − , where the constant C is independent of θ , must K be an ellipsoid? Here S n − is the unit sphere in R n , θ ⊥ = { x ∈ R n : (cid:104) x, θ (cid:105) = 0 } is thehyperplane orthogonal to the direction θ ∈ S n − and passing throughthe origin, and h K ( θ ) = max x ∈ K (cid:104) x, θ (cid:105) is the support function of the convexbody K ⊂ R n . Problem 8.
If for an origin-symmetric convex body K ⊂ R n , n ≥ ,we have (2) f K ( θ ) = C (vol n − ( K ∩ θ ⊥ )) n +1 ∀ θ ∈ S n − , where the constant C is independent of θ , must K be an ellipsoid? Key words and phrases.
Projections and sections of convex bodies.The first author is supported in part by the Simons Foundation Grant 711907.The second and the third authors are supported in part by U.S. National ScienceFoundation Grants DMS-1900008 and DMS-1600753. The fourth author is sup-ported by NSERC. a r X i v : . [ m a t h . M G ] J a n M.A. ALFONSECA, F. NAZAROV, D. RYABOGIN, AND V. YASKIN
Here f K is the curvature function of K , which is the reciprocal ofthe Gaussian curvature viewed as a function of the unit normal vector(see [Sch, pg. 419]).The Euclidean ball clearly satisfies (1) and (2). If a body K satisfies(1), then so does T K for any linear transformation T ∈ GL ( n ) (withconstant C · | det T | ). Similarly, if a body K satisfies (2), then sodoes T K for any linear transformation T ∈ GL ( n ) (with constant C · | det T | − n ). Hence, (1) and (2) are satisfied by ellipsoids.In this paper we prove the following result. Theorem.
Let n ≥ . If an origin-symmetric convex body K ⊂ R n satisfies (1) or (2) and is sufficiently close to the Euclidean ball in theBanach-Mazur metric, then K must be an ellipsoid. In dimension 2, there are convex bodies satisfying (1) that are notellipses but, nevertheless, can be arbitrarily close to the unit disc. Thecurve bounding such a body is a so-called
Radon curve , see [D]. Onthe other hand, the only convex bodies satisfying (2) in dimension 2are the ellipses [P, Theorem 5.6].2.
Invariance of Busemann-Petty problems under lineartransformations
Both Busemann-Petty problems are invariant under linear transfor-mations in the sense that if a symmetric convex body K satisfies (1)or (2), then so does T K where T is an invertible linear map from R n to itself.This statement is almost obvious for Problem 5. Indeed, let H beany hyperplane in R n passing through the origin and let H s be a sup-port hyperplane of K parallel to H . Consider any point p ∈ K ∩ H s and the cone C K,H with the base K ∩ H and the vertex p . Notethat due to the symmetry of K and the fact that H s is parallel to H , the volume vol n ( C K,H ) of this cone is independent of the particularchoice of H s and p . Moreover, we clearly have C T K,T H = T ( C K,H ), sovol n ( C T K,T H ) = | det T | vol n ( C K,H ). Since for H = θ ⊥ this volume canbe expressed as vol n ( C K,H ) = n vol n − ( K ∩ θ ⊥ ) h K ( θ ), we see that prop-erty (1) is merely the statement that vol n ( C K,H ) is independent of thechoice of the hyperplane H (this was exactly how the fifth Busemann-Petty problem was originally formulated in [BP]).The invariance of (2) under linear transformations is somewhat lesstransparent. When K has smooth non-degenerate C -boundary withstrictly positive Gaussian curvature at each point, we can restate it asfollows. IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 3
Let, as before, H be an arbitrary hyperplane passing through theorigin, let H s be one of the two supporting hyperplanes of K parallelto H , and let p ∈ K ∩ H s . For t ∈ (0 , H t be the hyperplanebetween H and H s parallel to H such that the distance between H s and H t is t times the distance d between H s and H . Then, for small t , the( n − n − ( K ∩ H t ) is approximately proportialto t n − d n − √ G ( p ) where G ( p ) is the Gaussian curvature of ∂K at p .Note now that vol n − ( K ∩ H t )vol n − ( K ∩ H ) is invariant under linear transformationsand d vol n − ( K ∩ H ) = n vol n ( C K,H ) is multiplied by | det T | when wereplace K by T K and H by T H . Thus, G ( p )vol n − ( K ∩ H ) n +1 equals(up to a constant factor depending on the dimension n only)lim t → t n − vol n ( C K,H ) n − (cid:104) vol n − ( K ∩ H )vol n − ( K ∩ H t ) (cid:105) , and thus is multiplied by | det T | n − when we replace K by T K and H by T H .In general, it is unclear to us what degree of smoothness Buse-mann and Petty assumed when posing Problem 8. We will handlethe most general case, when (2) is understood in the sense that thesurface area measure of K is absolutely continuous with respect tothe ( n − ∂K is smooth and non-degenerate, but K is close to the unit ball onlyin the Banach-Mazur distance and not in C .3. From the Banach-Mazur distance to the Hausdorffone
Applying an appropriate linear transformation, we can assume thatthe constants in (1) and (2) are equal to those for the unit ball B n andthat (1 − ε ) rB n ⊂ K ⊂ (1 + ε ) rB n for some r > ε > r must be close to 1, i.e., K mustbe close to the unit Euclidean ball B n in the Hausdorff metric. Wehave(3) (1 − ε ) rh B n ≤ h K ≤ (1 + ε ) rh B n M.A. ALFONSECA, F. NAZAROV, D. RYABOGIN, AND V. YASKIN and(4) (1 − ε ) n − r n − vol n − ( B n ∩ θ ⊥ ) ≤ vol n − ( K ∩ θ ⊥ ) ≤ (1 + ε ) n − r n − vol n − ( B n ∩ θ ⊥ ) . In the case of (1), combining (3) and (4) with the equation h K ( θ )vol n − ( K ∩ θ ⊥ ) = h B n ( θ )vol n − ( B n ∩ θ ⊥ ) , we obtain (1 − ε ) n r n ≤ ≤ (1 + ε ) n r n , i.e., ε ≤ r ≤ − ε .In the case of (2), we can integrate both sides with respect to the( n − S n − to conclude (see [Sch],Section 5.3.1) that(5) Σ( K ) = (cid:90) S n − f K ( θ ) dm n − ( θ ) = c n (cid:90) S n − (cid:0) vol n − ( K ∩ θ ⊥ ) (cid:1) n +1 dm n − ( θ ) , where Σ( K ) is the surface area of ∂K and c n is defined byΣ( B n ) = c n (cid:90) S n − (cid:0) vol n − ( B n ∩ θ ⊥ ) (cid:1) n +1 dm n − ( θ ) . From our assumption (1 − ε ) rB n ⊂ K ⊂ (1 + ε ) rB n , it follows that(1 − ε ) n − r n − Σ( B n ) ≤ Σ( K ) ≤ (1 + ε ) n − r n − Σ( B n ) , which, together with (4) and (5), gives(1 − ε ) n − r n − ≤ (1 + ε ) ( n − n +1) r ( n − n +1) and (1 + ε ) n − r n − ≥ (1 − ε ) ( n − n +1) r ( n − n +1) , i.e., 1 − ε (1 + ε ) n +1 ≤ r n ≤ ε (1 − ε ) n +1 . The isotropic position
We have seen in the previous section that, without loss of generality,we may assume that (1 − ε ) B n ⊂ K ⊂ (1 + ε ) B n . However, thisrequirement still leaves some freedom as to what affine image of K tochoose. In this section we will reduce this freedom even further byputting K into the so-called isotropic position, i.e., the position where (cid:90) K (cid:104) x, y (cid:105) dy = c | x | ∀ x ∈ R n . IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 5
The existence of such a position is well known and easy to derive (see[BGVV], Section 2.3.2). Indeed, for an arbitrary symmetric convexbody K , the mapping R n (cid:51) x (cid:55)→ (cid:90) K (cid:104) x, y (cid:105) dy = (cid:88) i,j (cid:16) (cid:90) K y i y j dy (cid:17) x i x j is a positive-definite quadratic form. Thus, it can be written as (cid:104) Sx, x (cid:105) ,where S is a self-adjoint positive definite operator on R n .Moreover, if K = B n , then S = c n I for some c n >
0. If (1 − ε ) B n ⊂ K , then (cid:104) Sx, x (cid:105) = (cid:90) K (cid:104) x, y (cid:105) dy ≥ (cid:90) (1 − ε ) B n (cid:104) x, y (cid:105) dy = (1 − ε ) n +2 c n | x | and, similarly, if K ⊂ (1 + ε ) B n , then (cid:104) Sx, x (cid:105) ≤ (1 + ε ) n +2 c n | x | . Thus, setting (cid:101) S = c − n S , we have(1 − ε ) n +2 | x | ≤ (cid:104) (cid:101) Sx, x (cid:105) ≤ (1 + ε ) n +2 | x | . It follows that (1 − ε ) n ( n +2) ≤ det( (cid:101) S ) ≤ (1 + ε ) n ( n +2) and (cid:107) (cid:101) S (cid:107) ≤ (1 + ε ) n +2 , (cid:107) (cid:101) S − (cid:107) ≤ (1 − ε ) − ( n +2) , whence the operator T = (cid:113) det( (cid:101) S ) − n (cid:101) S satisfiesdet T = 1 , (cid:107) T (cid:107) , (cid:107) T − (cid:107) ≤ (cid:16) ε − ε (cid:17) n +22 , and T − ST − is a multiple of the identity.The body (cid:101) K = T − K satisfies (cid:90) (cid:101) K (cid:104) x, y (cid:105) dy = (cid:90) K (cid:104) x, T − y (cid:105) dy = (cid:90) K (cid:104) T − x, y (cid:105) dy = (cid:104) ST − x, T − x (cid:105) = (cid:104) T − ST − x, x (cid:105) = c | x | for some c >
0, while we also have(1 − ε ) (cid:16) − ε ε (cid:17) n +22 B n ⊂ T − (1 − ε ) B n ⊂ T − K ⊂ T − (1 + ε ) B n ⊂ (1 + ε ) (cid:16) ε − ε (cid:17) n +22 B n . M.A. ALFONSECA, F. NAZAROV, D. RYABOGIN, AND V. YASKIN The analytic reformulation
Let ρ K , h K : S n − → R be the radial and the support functions ofthe convex body K respectively, i.e., ρ K ( θ ) = max { t > tθ ∈ K } and h K ( θ ) = max {(cid:104) x, θ (cid:105) : x ∈ K } . The ( n − K ∩ θ ⊥ is given byvol n − ( K ∩ θ ⊥ ) = c n R (cid:104) ρ n − K (cid:105) , where c n is a positive constant depending on the dimension n only and R is the Radon transform on S n − , i.e., R f ( θ ) = (cid:90) S n − ∩ θ ⊥ f ( ξ ) dσ ( ξ )with σ being the ( n − S n − ∩ θ ⊥ normalized by the condition σ ( S n − ∩ θ ⊥ ) = 1, i.e., R h K R (cid:104) ρ n − K (cid:105) = C , where, due tothe normalization made at the beginning of Section 3, the constant C should be the same as for the unit ball B n , i.e., C = 1. So, we arriveat the equation(6) h K = (cid:16) R (cid:104) ρ n − K (cid:105)(cid:17) − . Rewriting (2) in terms of h K and ρ K is trickier. The right-hand sidepresents no problem: it is just proportional to (cid:16) R (cid:104) ρ n − K (cid:105)(cid:17) n +1 . So, theequation becomes f K = C (cid:16) R (cid:104) ρ n − K (cid:105)(cid:17) n +1 . Due to the normalizationmade at the beginning of Section 3, the constant C should be thesame as for the unit ball B n , i.e., C = 1. However, f K can be readilyexpressed in terms of h K only if h K is C and we have made no suchassumption.The expression for f K in the C -case can be written as f K = Ah K where the operator A is defined as follows. For a function h ∈ C ( S n − )denote by H ( x ) its degree 1 homogeneous extension to the entire space(i.e., H ( x ) = | x | h ( x | x | ) for x (cid:54) = 0). Let (cid:98) H = ( H x i x j ) ni,j =1 be the Hessianof H and let (cid:98) H j be the matrix obtained from H by removing the j -throw and the j -th column. Let Ah be the restriction of n (cid:80) j =1 det (cid:98) H j to theunit sphere S n − (see [Sch], Corollary 2.5.3). IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 7
We shall show that when ρ K is close to 1, we can solve the equation Ah = (cid:16) R (cid:104) ρ n − K (cid:105)(cid:17) n +1 with h close to 1 in C . This h will determine a convex body L thatsatisfies f L = f K . By the uniqueness theorem (see [Sch], Theorem8.1.1) we will then conclude that K = L , so h K = h and the smoothnessof h K will be justified a posteriori. Thus, it will be possible to rewrite(2) as(7) Ah K = (cid:16) R (cid:104) ρ n − K (cid:105)(cid:17) n +1 . Maximal function
For e ∈ S n − , ϑ ∈ (0 , π ], let S ϑ ( e ) = { e (cid:48) ∈ S n − , (cid:104) e, e (cid:48) (cid:105) ≥ cos ϑ } denote the spherical cap centered at e with spherical radius ϑ . Thespherical Hardy-Littlewood maximal function is defined by M f ( e ) = max ϑ ∈ (0 ,π ] σ ( S ϑ ( e )) (cid:90) S ϑ ( e ) | f ( x ) | dσ ( x ) , f ∈ L ( S n − ) , where σ is the surface measure on S n − normalized by the condition σ ( S n − ) = 1. It is well known that M is bounded as an operator from L ( S n − ) to itself (see [Kn]). Lemma 1.
Let K be a -dimensional origin-symmetric convex bodyand let R be a positive real number. Let h K = R + ω be the supportfunction of K and let e ∈ S be a unit vector. Assume that h K ( e ) ≤ R cos ϑ for some ϑ ∈ (0 , π ) . Denote by e (cid:48) ( t ) the unit vector situatedclockwise from e and making an angle t with e . Then | ω ( e ) | ≤ ϑ (cid:90) ϑ ϑ | ω ( e (cid:48) ( t )) | dt. Proof.
Note that the hypothesis h K ( e ) ≤ R cos ϑ implies that ω ( e ) < ω ( e (cid:48) ( t )) ≥ | ω ( e ) | for all t ∈ [ ϑ , ϑ ], the inequality clearly holds.Assume now that ω ( e (cid:48) ( t )) < | ω ( e ) | for some t ∈ [ ϑ , ϑ ]. Let p bethe intersection point of the lines (cid:104) x, e (cid:105) = R + ω ( e ) and (cid:104) x, e (cid:48) ( t ) (cid:105) = R + | ω ( e ) | . Then p lies clockwise from e and, since | p | > R and (cid:104) p, e (cid:105) = h K ( e ) < R cos ϑ , the angle α between p and e is at least ϑ (seeFigure 1). Also, since (cid:104) p, e (cid:105) = h K ( e ) >
0, we have α < π .Since K is contained entirely in the angle (cid:104) x, e (cid:105) ≤ R + ω ( e ), (cid:104) x, e (cid:48) ( t ) (cid:105) ≤ R + | ω ( e ) | with vertex p , we have h K ( e (cid:48) ( t )) ≤ (cid:104) p, e (cid:48) ( t ) (cid:105) = | p | cos( α − t )for all t ∈ [0 , t ].We shall now use the following elementary property of the cosinefunction: if γ , δ> γ − δ, γ + δ ] ⊂ [0 , π ], then cos β ≤ γ − δ )+cos( γ + δ )4 M.A. ALFONSECA, F. NAZAROV, D. RYABOGIN, AND V. YASKIN pR (cid:16) R + | ω ( e ) | (cid:17) e ( t ) h ( e ) e = ( R + ω ( e )) et ϑ K Figure 1.
The body K , the lines (cid:104) x, e (cid:105) = R + ω ( e ), (cid:104) x, e (cid:48) ( t ) (cid:105) = R + | ω ( e ) | , and the point p for all β ∈ [ γ, γ + δ ]. Indeed, since cos β ≤ cos γ , it suffices to showthat cos γ ≤ γ − δ ) + cos( γ + δ )4 = cos γ cos δ + 12 sin γ sin δ. Rewriting this as cos γ (1 − cos δ ) ≤ sin γ sin δ and using the iden-tity 1 − cos δ = − cos δ δ = sin δ δ , we see that we need to prove that cos γ δ sin δ ≤ sin γ sin δ . However, since 0 ≤ δ ≤ γ ≤ π , we havecos γ ≤ cos δ ≤ γ ≥ sin δ , so the left hand side is at most sin δ and the right hand side is at least that.Applying this property to the interval [ α − t , α ], i.e., with γ = α − t , δ = t , we conclude that h K ( e (cid:48) ( t )) ≤ (cid:16) R + 17 | ω ( e ) | (cid:17) + 14 ( R + ω ( e )) = R − | ω ( e ) | for every t ∈ [0 , t ] ⊃ [ ϑ , ϑ ] and the conclusion of the lemma followsagain. (cid:3) IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 9
Corollary 1.
Let K be a convex body in R n and let R > . Let h K = R + ω be the support function of K and let e ∈ S n − be a unitvector. Assume that h K ( e ) ≤ R cos ϑ for some ϑ ∈ (0 , π ) . Then | ω ( e ) | ≤ C σ ( S ϑ ( e )) (cid:90) S ϑ ( e ) | ω ( e (cid:48) ) | dσ ( e (cid:48) ) . Proof.
We will use the parametrization e (cid:48) = e (cid:48) ( t, v ) ∈ S n − where t is the angle between e and e (cid:48) , and v ∈ S n − ∩ e ⊥ is such that e (cid:48) = e cos t + v sin t .Note that dσ n − ( e (cid:48) ) = c n (sin t ) n − dtdσ n − ( v ). It follows from thelemma applied to the projection of K to the plane spanned by e and v that | ω ( e ) | ≤ ϑ (cid:90) ϑ ϑ | ω ( e (cid:48) ( t, v )) | dt ≤ ϑ (cid:0) sin( ϑ ) (cid:1) n − (cid:90) ϑ | ω ( e (cid:48) ( t, v )) | (sin t ) n − dt. Integrating this inequality with respect to v and observing that σ ( S ϑ ( e )) (cid:16) ϑ n − (cid:16) ϑ (sin( ϑ )) n − , we get the statement of the corollary. (cid:3) Lemma 2.
Assume that a symmetric convex body K is very close tothe unit ball and l ∈ N . Let h = h K and ρ = ρ K be the support and theradial functions of K respectively. Trivially, ρ ≤ h . Let h = ∞ (cid:80) m =0 h m bethe decomposition of h into spherical harmonics (since h is even, only h m with even m are not identically ). Put η = l (cid:80) m =1 h m , ν = ∞ (cid:80) m = l +1 h m .We claim that for every ε, l > there exists δ = δ ( ε, l ) such thatwhenever (cid:107) h − (cid:107) ∞ ≤ δ , the inequality h − ρ ≤ ε (cid:107) η (cid:107) L + CM ν holds, where C is an absolute constant and M is the spherical Hardy-Littlewood maximal function.Proof. We have ρ ( e ) = inf { e (cid:48) ∈ S n − , (cid:104) e,e (cid:48) (cid:105) > } h ( e (cid:48) ) (cid:104) e, e (cid:48) (cid:105) . Note that the admissible range of e (cid:48) can be further restricted to | e − e (cid:48) | < δ with arbitrarily small δ >
0, provided that δ is chosen smallenough. Indeed, since h ( e (cid:48) ) ≥ − δ δ h ( e ), e (cid:48) can compete with e only if (cid:104) e, e (cid:48) (cid:105) ≥ − δ δ , so | e − e (cid:48) | = 2(1 − (cid:104) e, e (cid:48) (cid:105) ) ≤ δ δ < δ if δ > l onthe unit sphere are equivalent, and that any semi-norm is dominatedby any norm, whence (cid:107) η (cid:107) C ( S n − ) ≤ C ( l ) (cid:107) η (cid:107) L ( S n − ) and (cid:107)∇ η (cid:107) C ( S n − ) ≤ C ( l ) (cid:107) η (cid:107) L ( S n − ) . In particular, if | e − e (cid:48)(cid:48) | < δ , we get | η ( e ) − η ( e (cid:48)(cid:48) ) | ≤ (cid:107)∇ η (cid:107) C ( S n − ) δ ≤ C ( l ) δ (cid:107) η (cid:107) L ( S n − ) . Let us now assume that e (cid:48) ∈ S n − , with | e − e (cid:48) | < δ , is a competitor,so h ( e (cid:48) ) (cid:104) e,e (cid:48) (cid:105) ≤ h ( e ). Then, if ϑ is the angle between e and e (cid:48) , we have h ( e (cid:48) ) ≤ h ( e ) cos ϑ , so we can apply Corollary 1 to the vector e (cid:48) with R = h ( e ) and conclude that h ( e ) − h ( e (cid:48) ) (cid:104) e, e (cid:48) (cid:105) ≤ h ( e ) − h ( e (cid:48) ) ≤ Cσ ( S ϑ ( e (cid:48) )) (cid:90) S ϑ ( e (cid:48) ) | h ( e ) − h ( e (cid:48)(cid:48) ) | dσ ( e (cid:48)(cid:48) ) ≤ C (cid:48) σ ( S ϑ ( e )) (cid:90) S ϑ ( e ) | h ( e ) − h ( e (cid:48)(cid:48) ) | dσ ( e (cid:48)(cid:48) ) . However, | h ( e ) − h ( e (cid:48)(cid:48) ) | ≤ | η ( e ) − η ( e (cid:48)(cid:48) ) | + | ν ( e ) | + | ν ( e (cid:48)(cid:48) ) | , and | η ( e ) − η ( e (cid:48)(cid:48) ) | ≤ C ( l ) δ (cid:107) η (cid:107) L ( S n − ) , while | ν ( e ) | ≤ M ν ( e ) and 1 σ ( S ϑ ( e )) (cid:90) S θ ( e ) | ν ( e (cid:48)(cid:48) ) | dσ ( e (cid:48)(cid:48) ) ≤ M ν ( e ) , so the desired statement follows if we choose δ > C (cid:48) C ( l ) δ <ε . (cid:3) Contraction
Let M be a bounded linear operator on L = L ( S n − ) such that M is proportional to the identity on every space H m of spherical harmonicsof degree m , i.e., for some µ m ∈ R , M f = (cid:88) m ≥ µ m f m where f = (cid:88) m ≥ f m and f m ∈ H m is the spherical harmonic decomposition of f ∈ L ( S n − ). We say that M is a strong contraction ifmax m ≥ | µ m | < , and lim m →∞ µ m = 0 . IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 11
Lemma 3.
Assume that M as above is a strong contraction. Then,there exists δ ∈ (0 , such that for any symmetric convex body K andany c ∈ (1 − δ, δ ) , the conditions − δ ≤ ρ K ≤ δ, (cid:107) ( h K − h ) − c M ( ρ K − r ) (cid:107) L ≤ δ (cid:107) ρ K − r (cid:107) L , imply h K = ρ K = const . Here, h and r are the constant terms of thespherical harmonic decomposition of h K and ρ K , respectively.Proof. Fix a large l and consider the decompositions h K = h + η + ν and ρ K = r + ϕ + ψ, where h , r are the constant terms, η and ϕ are the parts correspond-ing to the harmonics of degrees 1 to l and ν and ψ are the partscorresponding to the harmonics of degrees greater than l .Fix ε >
0. Since the projection to any sum of spaces of sphericalharmonics in L has norm 1, we have(8) (cid:107) η − c M ϕ (cid:107) L ≤ (cid:107) h K − h − c M ( ρ K − r ) (cid:107) L ≤ δ (cid:107) ρ K − r (cid:107) L ≤ δ ( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L ) . Similarly,(9) (cid:107) ν − c M ψ (cid:107) L ≤ δ ( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L ) . From (9) we obtain that (cid:107) ν (cid:107) L ≤ c (cid:107) M ψ (cid:107) L + δ ( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L ) ≤ (10) (1 + δ )(max m>l | µ m | ) (cid:107) ψ (cid:107) L + δ ( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L ) ≤ ε ( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L )if l is large enough (recall that lim m →∞ µ m = 0) and δ is small enough.The same computation for η , using in this case that max m ≥ | µ m | < (cid:107) η (cid:107) L ≤ (1 + 2 δ )( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L ) . On the other hand, by Lemma 2 and the boundedness of the maximalfunction in L , we have (cid:107) h K − ρ K (cid:107) L ≤ ε (cid:107) η (cid:107) L + C (cid:107) ν (cid:107) L , which implies(12) (cid:107) η − ϕ (cid:107) L ≤ ε (cid:107) η (cid:107) L + C (cid:107) ν (cid:107) L and (cid:107) ν − ψ (cid:107) L ≤ ε (cid:107) η (cid:107) L + C (cid:107) ν (cid:107) L . Combining (8), (9), (10), (11), and (12), we obtain (cid:107) ϕ − c M ϕ (cid:107) L + (cid:107) ψ − c M ψ (cid:107) L ≤ (cid:107) ϕ − η (cid:107) L + (cid:107) η − c M ϕ (cid:107) L + (cid:107) ψ − ν (cid:107) L + (cid:107) ν − c M ψ (cid:107) L ≤ C ( δ + ε )( (cid:107) ϕ (cid:107) L + (cid:107) ψ (cid:107) L ) . On the other hand, for any function χ ∈ L ( S n − ), we have (cid:107) χ − c M χ (cid:107) L ≥ (1 − (1 + δ ) max m ≥ | µ m | ) (cid:107) χ (cid:107) L , so we can conclude that ϕ = 0, ψ = 0 if C ( δ + ε ) < − (1+ δ ) max m ≥ | µ m | . (cid:3) Remark 1.
Note that ( h K − h ) − c M ( ρ K − r ) is orthogonal to con-stants and, therefore, its L -norm does not exceed (cid:107) ( h K − λ ) − c M ( ρ K − r ) (cid:107) L for any λ ∈ R . Thus, to verify the conditions of the lemma itsuffices to check that (cid:107) ( h K − λ ) − c M ( ρ K − r ) (cid:107) L ≤ δ (cid:107) ρ K − r (cid:107) L with any λ ∈ R of our choice. Properties of the function ( R [ ρ αK ]) β when ρ K is close to K be a symmetric convex body in the isotropic position suchthat 1 − δ ≤ ρ K ≤ δ for some small δ >
0. Let α, β ∈ R . We wantto derive several useful properties of the function ( R [ ρ αK ]) β .The first observation is that ρ K is Lipschitz with Lipschitz constant5 √ δ . Indeed, let x, y ∈ S n − . If | x − y | ≥ √ δ , then we have | ρ K ( x ) − ρ K ( y ) | ≤ δ ≤ √ δ | x − y | , so we may assume that 0 < | x − y | < √ δ . Without loss of generality, ρ K ( x ) ≥ ρ K ( y ). Let us denote X = ρ K ( x ) x , Y = ρ K ( y ) y , where X, Y ∈ ∂K . By the convexity of K , every point on the line Y − t ( X − Y )with t ≥ K and, therefore, outside (1 − δ ) B n as well.Hence,(1 − δ ) ≤ | Y − t ( X − Y ) | = | Y | − t (cid:104) X − Y, Y (cid:105) + t | X − Y | . Since | Y | ≤ (1 + δ ) , we conclude that, for all t ≥ t (cid:104) X − Y, Y (cid:105) − t | X − Y | ≤ δ. From (13) it follows that(14) (cid:104) X − Y, Y (cid:105) ≤ √ δ | X − Y | . Indeed, if (cid:104) X − Y, Y (cid:105) ≤
0, the inequality is obvious. Otherwise, we canplug t = (cid:104) X − Y,Y (cid:105)| X − Y | into (13), obtaining (cid:104) X − Y,Y (cid:105) | X − Y | ≤ δ , which is equivalentto (14). Now, equation (14) can be rewritten as | X || Y |(cid:104) x, y (cid:105) − | Y | ≤ √ δ | X − Y | , IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 13 or, equivalently, | Y | ( | X | − | Y | ) ≤ √ δ | X − Y | + | X || Y | (1 − (cid:104) x, y (cid:105) ) . Observe that 1 − (cid:104) x, y (cid:105) = | x − y | ≤ √ δ | x − y | , while | X − Y | ≤| X | − | Y | + | Y || x − y | . Hence, | X | − | Y | ≤ √ δ | Y | ( | X | − | Y | ) + (cid:18) | X | (cid:19) √ δ | x − y | . Now, if δ ∈ (0 , / √ δ | Y | ≤ √ δ − δ ≤ / − / < , and we conclude that | X |−| Y | ≤ (cid:18) | X | (cid:19) √ δ | x − y | ≤ (cid:18) δ (cid:19) √ δ | x − y | ≤ √ δ | x − y | , as required.Since the mapping t (cid:55)→ t p is Lipschitz on any compact subset of(0 , + ∞ ) and the Radon transform does not increase the Lipschitz con-stant of the function, we immediately conclude that ( R [ ρ αK ]) β has Lip-schitz constant at most C α,β √ δ .Let now r = (cid:82) S n − ρ K dσ be the mean value of ρ K on the unit sphere.Clearly, | r − | ≤ δ , so | ρ K − r | ≤ δ and, thereby, R| ρ K − r | ≤ δ as well. Now, using the fact that t (cid:55)→ t p is C on any compact subsetof (0 , + ∞ ) and linearizing, we successively derive that | ρ αK − ( r α + αr α − ( ρ K − r )) | ≤ C α δ | ρ K − r | , |R [ ρ αK ] − ( r α + αr α − R ( ρ K − r )) | ≤ C α δ R| ρ K − r | , | ( R [ ρ αK ]) β − ( r α + αr α − R ( ρ K − r )) β | ≤ C α,β δ R| ρ K − r | , | ( R [ ρ αK ]) β − ( r αβ + αβr αβ − R ( ρ K − r )) | ≤ C α,β δ R| ρ K − r | . Thus, ( R [ ρ αK ]) β = r αβ + γ , where | γ − αβr αβ − R ( ρ K − r ) | ≤ C α,β δ R| ρ K − r | . In particular, we conclude that | γ | ≤ C α,β δ , which, together with theabove observation about the Lipschitz constant, implies that (cid:107) γ (cid:107) C = max S n − | γ | + sup x,y ∈ S n − ,x (cid:54) = y | γ ( x ) − γ ( y ) || x − y | ≤ C α,β √ δ. Now let ρ K − r = Y + Y + . . . be the spherical harmonic decompo-sition of ρ K − r . It follows from the definition of the isotropic positionthat 0 = (cid:90) K p ( x ) dx = c n (cid:90) S n − ρ n +2 K ( x ) p ( x ) dσ ( x )for all quadratic polynomials p ( x ) = (cid:80) i,j a ij x i x j with n (cid:80) i =1 a ii = 0. Inother words, ρ n +2 K has no second order term in its spherical harmonicdecomposition.On the other hand, | ρ n +2 K − ( r n +2 + ( n + 2) r n +1 ( ρ K − r )) | ≤ Cδ | ρ K − r | . Taking the second order component in the spherical harmonic decom-position of the expression under the absolute value sign on the left handside, we get( n + 2) r n +1 (cid:107) Y (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) , so (cid:107) Y (cid:107) L ( S n − ) ≤ C (cid:48) δ (cid:107) ρ K − r (cid:107) L ( S n − ) . A solution to the fifth Busemann-Petty problem in asmall neighborhood of the Euclidean ball
Recall that for the fifth Busemann-Petty problem we have the equa-tion h K = ( R [ ρ n − K ]) − . By the results of the previous section, the righthand side can be written as r − n +1 − ( n − r − n R ( ρ K − r ) + γ (cid:48) , where | γ (cid:48) | ≤ Cδ R| ρ K − r | , so (cid:107) γ (cid:48) (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) .Let M be the linear operator that maps every m -th order sphericalharmonic Z m to − ( n − R Z m = − ( n − − m · · . . . · ( m − n − n + 1) · . . . · ( n + m − Z m for even m ≥ m . Then M is a strong contractionand (cid:107) ( h K − r − n +1 ) − r − n M ( ρ K − r ) (cid:107) L ( S n − ) ≤ r − n (cid:107) Y (cid:107) L ( S n − ) + (cid:107) γ (cid:48) (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) , so Lemma 3 and Remark 1 yield h K = ρ K = const , i.e., K is a ball,provided that δ is small enough. IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 15
A solution to the eighth Busemann-Petty problem ina small neighborhood of the Euclidean ball
We now turn to the equation Ah K = ( R [ ρ n − K ]) n +1 (see Section 5).Below we will use several standard results about A and the Laplaceoperator which, for completeness, are proven in the Appendices.By the results of Section 8, ( R [ ρ n − K ]) n +1 can be rewritten as r ( n − n +1) + γ , where (cid:107) γ (cid:107) C ≤ C √ δ and γ = ( n − n + 1) r ( n − n +1) − R ( ρ K − r ) + γ (cid:48) , (cid:107) γ (cid:48) (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) . Then
A h K r n +1 = 1 + r − ( n − n +1) γ and, provided that δ > h K r n +1 = 1 + ϕ (cid:48) + ϕ (cid:48)(cid:48) , where(15) (cid:101) ∆ ϕ (cid:48) = r − ( n − n +1) γ (see Appendix II for the definition of (cid:101) ∆) and (cid:107) ϕ (cid:48)(cid:48) (cid:107) L ( S n − ) ≤ εr − ( n − n +1) (cid:107) γ (cid:107) L ( S n − ) ≤ Cε (cid:107) ρ K − r (cid:107) L ( S n − ) with as small ε > ϕ (cid:48) + ϕ (cid:48) where ϕ (cid:48) solves (cid:101) ∆ ϕ (cid:48) = ( n − n + 1) r − R ( ρ K − r ) , and ϕ (cid:48) solves (cid:101) ∆ ϕ (cid:48) = r − ( n − n +1) γ (cid:48) .The norm of ϕ (cid:48) can be estimated immediately: (cid:107) ϕ (cid:48) (cid:107) L ( S n − ) ≤ C (cid:107) γ (cid:48) (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) . As to ϕ (cid:48) , it is equal to (see the end of Appendix I) r − (cid:88) m ≥ m even µ m Y m , where ρ K = r + (cid:80) m ≥ m even Y m is the spherical harmonic decomposition of ρ K and µ m = ( n − n + 1)(1 − m )( m + n −
1) ( − m · · . . . · ( m − n − n + 1) · . . . · ( n + m − , so µ = 1 and | µ m | < m ≥ µ m → m → ∞ . Since (cid:107) Y (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) , we conclude that (cid:107) ϕ (cid:48) − r − M ( ρ K − r ) (cid:107) L ( S n − ) ≤ Cδ (cid:107) ρ K − r (cid:107) L ( S n − ) , with the strong contraction M given by Z m (cid:55)→ µ m Z m , m even, m ≥ Z m (cid:55)→ m .Putting all these estimates together, we conclude that (cid:107) ( h K − r n +1 ) − r n M ( ρ K − r ) (cid:107) L ( S n − ) ≤ ε (cid:107) ρ K − r (cid:107) L ( S n − ) , with as small ε > δ > h K = ρ K = const , so K is a ball.11. Appendix I. Solving the Laplace equation
Below we shall use the following notation. For a function f : S n − → R and α ∈ (0 , (cid:107) f (cid:107) C α = (cid:107) f (cid:107) C α ( S n − ) = max S n − | f | + sup x,y ∈ S n − , x (cid:54) = y | f ( x ) − f ( y ) || x − y | α , (cid:107) f (cid:107) C α = (cid:107) f (cid:107) C α ( S n − ) =max S n − | f | + max x ∈ S n − , i =1 ,...,n | F x i ( x ) | + max i,j =1 ,...,n (cid:107) F x i x j (cid:107) C α ( S n − ) , where F ( x ) = | x | f ( x | x | ) is the 1-homogeneous extension of f to R n \ { } (we assume that it is at least C in R n \ { } ).Let g : S n − → R be an even C α function on the unit sphere S n − with some α ∈ (0 , G be the ( − g to R n \ { } , i.e., G ( x ) = | x | − g ( x | x | ) for x (cid:54) = 0. We will show that thereexists a unique 1-homogeneous even function F : R n → R of class L loc such that ∆ F = G in R n in the sense of generalized functions.Moreover, F ∈ C α ( S n − ) and for all i, j = 1 , . . . , n , we have(16) (cid:107) F x i x j (cid:107) L ( S n − ) ≤ C (cid:107) g (cid:107) L ( S n − ) , (cid:107) F (cid:107) C α ( S n − ) ≤ C (cid:107) g (cid:107) C α ( S n − ) , with some C = C ( n, α ) > Uniqueness.
If we have two even 1-homogeneous functions F , F such that ∆ F = ∆ F = G in R n , then F − F is an even 1-homogeneous harmonic function, but the only such function is 0. IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 17
Existence.
Now we will show that the function F defined by F ( x ) = c n (cid:90) R n (cid:104) | x − y | n − − | y | n − (cid:105) G ( y ) dy is a well-defined 1-homogeneous function on R n satisfying ∆ F = G and estimates (16). Here, c n is chosen so that ∆ c n | x | n − = δ (the Diracdelta measure) in the sense of generalized functions, and the integralis understood as lim R →∞ (cid:82) B (0 ,R ) .In order to show the convergence of the integral, we note that1 | x − y | n − − | y | n − = ( n − (cid:104) x, y (cid:105)| y | n + O (cid:16) | y | n (cid:17) as y → ∞ uniformly on compact sets in x .Since G is even, the integral of (cid:104) x,y (cid:105)| y | n G ( y ) over each sphere centered atthe origin vanishes. Since G is ( − | y | n G ( y ) = O ( (cid:107) g (cid:107) C | y | n +1 ) as y → ∞ , which is integrable at ∞ .The singularities at x ∈ S n − and 0 are of degrees − ( n −
2) and − ( n −
1) respectively, so the local integrability there presents no problemeither, and we get the estimate (cid:107) F (cid:107) C ( S n − ) ≤ C (cid:107) g (cid:107) C ( S n − ) . The change of variable y (cid:55)→ − y and the identity G ( y ) = G ( − y ) implythat F is even.To show the 1-homogeneity of F , take t > y (cid:55)→ ty to write F ( tx ) = c n lim R →∞ (cid:90) B (0 ,R ) (cid:16) | tx − y | n − − | y | n − (cid:17) G ( y ) dy = c n lim R →∞ (cid:90) B (0 , Rt ) (cid:16) | tx − ty | n − − | ty | n − (cid:17) G ( ty ) d ( ty ) = c n t lim R →∞ (cid:90) B (0 , Rt ) (cid:16) | x − y | n − − | y | n − (cid:17) G ( y ) dy = tF ( x ) , (we used that | tz | n − = t n − | z | n − , G ( ty ) = t − G ( y ), and d ( ty ) = t n dy ).To estimate (cid:107) F (cid:107) C α ( S n − ) , we split the integral defining F into 3parts. Let ξ , ξ , ξ : [0 , + ∞ ) → [0 ,
1] be as on Figure 2, so ξ i areLipschitz with constant 4, and ξ + ξ + ξ = 1.
14 12 ξ
14 12 ξ
14 12 ξ Figure 2.
The functions ξ i , i = 1 , , G i ( x ) = G ( x ) ξ i ( | x | ) and F i ( x ) = c n (cid:90) R n (cid:104) | x − y | n − − | y | n − (cid:105) G i ( y ) dy, so G = G + G + G and F = F + F + F .Our first observation is that G ( y ) is an α -H¨older, compactly sup-ported function on R n , with C α -norm bounded by C (cid:107) g (cid:107) C α ( S n − ) .Indeed, we clearly have max R n |G | ≤ S n − | g | . On the other hand, |G ( x ) − G ( y ) | = (cid:12)(cid:12)(cid:12) ξ ( | x | ) | x | − g (cid:16) x | x | (cid:17) − ξ ( | y | ) | y | − g (cid:16) y | y | (cid:17)(cid:12)(cid:12)(cid:12) . Since (cid:101) ξ ( t ) = ξ ( t ) t − is a compactly supported Lipschitz function on[0 , + ∞ ), it is also α -H¨older for any α ∈ (0 , | (cid:101) ξ ( t ) − (cid:101) ξ ( s ) | ≤ C | t − s | α for all t, s ≥ . Thus, if x, y ∈ B (0 , \ B (0 , ), then (cid:12)(cid:12)(cid:12)(cid:12)(cid:101) ξ ( | x | ) g (cid:18) x | x | (cid:19) − (cid:101) ξ ( | y | ) g (cid:18) y | y | (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤| (cid:101) ξ ( | x | ) − (cid:101) ξ ( | y | ) | (cid:12)(cid:12)(cid:12)(cid:12) g (cid:18) x | x | (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + | (cid:101) ξ ( | y | ) | (cid:12)(cid:12)(cid:12)(cid:12) g (cid:18) x | x | (cid:19) − g (cid:18) y | y | (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12) | x | − | y | (cid:12)(cid:12)(cid:12) α max S n − | g | + 4 (cid:12)(cid:12)(cid:12) x | x | − y | y | (cid:12)(cid:12)(cid:12) α (cid:107) g (cid:107) C α ( S n − ) ≤ C (cid:107) g (cid:107) C α ( S n − ) (cid:16) | x − y | α + (cid:12)(cid:12)(cid:12) x | x | − y | y | (cid:12)(cid:12)(cid:12) α (cid:17) ≤ C (cid:107) g (cid:107) C α ( S n − ) | x − y | α , because the mapping x (cid:55)→ x | x | is C and, thereby, Lipschitz on B (0 , \ B (0 , ).If x, y / ∈ B (0 , \ B (0 , ), then G ( x ) = G ( y ) = 0, so the inequality |G ( x ) − G ( y ) | ≤ C (cid:107) g (cid:107) C α ( S n − ) | x − y | α IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 19 holds trivially. Finally, if x ∈ B (0 , \ B (0 , ) but y / ∈ B (0 , \ B (0 , ),then the segment [ x, y ] intersects the boundary of B (0 , \ B (0 , ) atsome point y (cid:48) , so G ( y ) = G ( y (cid:48) ) = 0 and |G ( x ) − G ( y ) | = |G ( x ) − G ( y (cid:48) ) | ≤ C (cid:107) g (cid:107) C α ( S n − ) | x − y (cid:48) | α ≤ C (cid:107) g (cid:107) C α ( S n − ) | x − y | α . The functions G and G are supported on B (0 , ) and R n \ B (0 , ),respectively, and satisfy the bound |G ( y ) | , |G ( y ) | ≤ | y | max S n − | g | . Now we are ready to estimate (cid:107) F (cid:107) C α ( S n − ) . Consider x with ≤| x | ≤ . Note that x (cid:55)→ | x − y | n − is a C -function (in x ) in this domainwith uniformly bounded (in y ) C -norm as long as y ∈ B (0 , ). Hence, F ∈ C ( B (0 , ) \ B (0 , )) and (cid:107) F (cid:107) C ( B (0 , ) \ B (0 , )) ≤ C (cid:107) g (cid:107) L ( S n − ) (the constant term (cid:82) R n | y | n − G ( y ) dy is also bounded by C (cid:107) g (cid:107) L ( S n − ) ).To estimate F , note that for | x | ≤ and | y | ≥ , we have (cid:12)(cid:12)(cid:12) | x − y | n − − | y | n − − ( n − (cid:104) x, y (cid:105)| y | n (cid:12)(cid:12)(cid:12) ≤ C | y | n , (cid:12)(cid:12)(cid:12) ∂∂x i | x − y | n − − ( n − y i | y | n (cid:12)(cid:12)(cid:12) ≤ C | y | n , (cid:12)(cid:12)(cid:12) ∂ ∂x i ∂x j | x − y | n − (cid:12)(cid:12)(cid:12) ≤ C | y | n , and (cid:12)(cid:12)(cid:12) ∂ ∂x i ∂x j ∂x k | x − y | n − (cid:12)(cid:12)(cid:12) ≤ C | y | n +1 . Since y (cid:55)→ (cid:104) x,y (cid:105)| y | n and y (cid:55)→ y i | y | n are odd functions, their integrals againstthe even function G ( y ) over any sphere centered at the origin are 0 and,therefore, | F | , |∇ F | , |∇ F | ≤ C (cid:90) R n \ B (0 , ) | y | − n |G ( y ) | dy ≤ C (cid:107) g (cid:107) L ( S n − ) and |∇ F | ≤ C (cid:90) R n \ B (0 , ) | y | − n − |G ( y ) | dy ≤ C (cid:107) g (cid:107) L ( S n − ) , so (cid:107) F (cid:107) C ( B (0 , )) ≤ C (cid:107) g (cid:107) L ( S n − ) . It remains to estimate F . We clearly have | F ( x ) | ≤ C (cid:107) g (cid:107) C ( S n − ) (cid:90) B (0 , \ B (0 , ) (cid:12)(cid:12)(cid:12) | x − y | n − − | y | n − (cid:12)(cid:12)(cid:12) dy ≤ C (cid:107) g (cid:107) C ( S n − ) and |∇ F ( x ) | ≤ C (cid:107) g (cid:107) C ( S n − ) (cid:90) B (0 , \ B (0 , ) | x − y | n − dy ≤ C (cid:107) g (cid:107) C ( S n − ) . As for ( F ) x i x j , these partial derivatives are images of G under cer-tain Calder´on-Zygmund singular integral operators (see [GT], Lemma4.4 and Theorem 9.9), so, since G ∈ C α ( R n ) and has fixed compactsupport, we obtain that (cid:107) ( F ) x i x j (cid:107) C α ( R n ) ≤ C (cid:107)G (cid:107) C α ( R n ) ≤ C (cid:107) g (cid:107) C α ( S n − ) and (cid:107) ( F ) x i x j (cid:107) L ( R n ) ≤ C (cid:107)G (cid:107) L ( R n ) ≤ C (cid:107) g (cid:107) L ( S n − ) . The final conclusion is that (cid:107) F (cid:107) C α ( S n − ) ≤ C (cid:107) F (cid:107) C α ( B (0 , ) \ B (0 , )) ≤ C (cid:107) g (cid:107) C α ( S n − ) and (cid:107) ( F ) x i x j (cid:107) L ( S n − ) ≤ C (cid:107) ( F ) x i x j (cid:107) L ( B (0 , ) \ B (0 , )) ≤ C (cid:107) g (cid:107) L ( S n − ) (we used the ( − F ) x i x j here).The desired equality ∆ F = G follows from the fact that the mapping x (cid:55)→ | x − y | n − − | y | n − is harmonic in x for | x | ≤ | y | ≥ . This impliesthat ∆ F = 0 in B (0 , F + F differs by a constant fromthe classical Newton potential of the compactly supported L function G + G = G in B (0 , F = G in B (0 , R n by homogeneity.We shall also need the relation between the spherical harmonic de-compositions of F | S n − and g . To this end, we will start with the fol-lowing computation. Let P m be a homogeneous harmonic polynomialof degree m , so that Y m = P m | S n − is a spherical harmonic of degree m .The 1-homogeneous extension of Y m is (cid:101) Y m ( x ) = | x | − m P m ( x ). Then∆ (cid:101) Y m ( x ) = ∆( | x | − m ) P m ( x ) + 2 (cid:104)∇ ( | x | − m ) , ∇ P m ( x ) (cid:105) =(1 − m )( − m − n ) | x | − m − P m ( x ) + 2(1 − m ) | x | − m ∂∂r P m ( x ) = IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 21 | x | − − m (1 − m )( − m − n + 2 m ) P m ( x ) =(1 − m )( m + n − | x | − − m P m ( x ) . Thus, if g ∈ L ( S n − ) and g = (cid:80) m ≥ m even Y m on S n − , the series F = (cid:88) m ≥ m even − m )( m + n − (cid:101) Y m converges in L loc (the series is orthogonal on every ball B (0 , R ) and (cid:107) (cid:101) Y m (cid:107) L ( B (0 ,R )) ≤ C R (cid:107) Y m (cid:107) L ( S n − ) ) and formally solves ∆ F = G . Toshow that it is a true solution, it suffices to observe that we have∆ F ( l ) = G ( l ) for the partial sums F ( l ) and G ( l ) of the correspondingseries and F ( l ) → F in L loc , G ( l ) → G in L loc as l → ∞ . Thus, ∆ F = G in the sense of generalized functions. If g ∈ C α ( S n − ), then, by theuniqueness part, this solution has to coincide with the explicit solutionconstructed above, so the spherical harmonic decomposition of F | S n − is (cid:80) m ≥ m even 1(1 − m )( m + n − Y m . In particular, the decomposition implies that (cid:107) F (cid:107) L ( S n − ) ≤ (cid:107) g (cid:107) L ( S n − ) . Appendix II. Solution of Monge-Ampere equation
For a function f : S n − → R , we denote by F its 1-homogeneousextension to R n . By Af we will denote the restriction of n (cid:80) k =1 det (cid:98) F k to the unit sphere where (cid:98) F k is the matrix obtained from the Hessian (cid:98) F = ( F x i x j ) ni,j =1 by deleting the k -th row and the k -th column.We now turn to the solution of the equation Af = g where g is closeto 1. Note that A A commutes with the rotationsof the sphere, we can check this identity at the point (1 , , . . . , | x | , so the Hessian is (cid:16) δ ij | x | − x i x j | x | (cid:17) ni,j =1 ,which at the point (1 , , . . . ,
0) turns into . . . . . . . . . . . . . . . . The rotation invariance also allows us to compute the linear partof Af (meaning the linear terms in Φ x i x j ) for f = 1 + ϕ , where Φ is the 1-homogeneous extension of ϕ . Again, computing the Hessian at(1 , , . . . , (cid:98) F = Φ x x Φ x x . . . Φ x x n Φ x x x x . . . Φ x x n ... ... . . . ...Φ x n x Φ x n x . . . x n x n , so n (cid:88) i =1 det (cid:98) F i = det (cid:98) F + n (cid:88) i =2 det (cid:98) F i =1 + n (cid:88) i =2 Φ x i x i + ( n − x x + P (Φ) , where P (Φ) is some linear combination of products of two or more sec-ond partial derivatives of Φ. Note now that, since Φ is 1-homogeneous,the mapping t (cid:55)→ Φ( t, , . . . ,
0) is linear and, thereby, Φ x x (1 , , . . . ,
0) =0. Thus we can just as well write n (cid:80) i =2 Φ x i x i + ( n − x x at (1 , , . . . , , , . . . , n (cid:88) i =1 det (cid:98) F i = 1 + ∆Φ + P (Φ)in general, though P (Φ) will now be a sum of products of at least twosecond partial derivatives of Φ and some fixed functions of x that aresmooth near the unit sphere.Using identities of the type a a . . . a m − b b . . . b m = ( a − b ) a . . . a m + b ( a − b ) a . . . a m + · · · + b . . . b m − ( a m − − b m − ) a m + b . . . b m − ( a m − b m ) , we see that for any 1-homogeneous C -functions Ψ (cid:48) , Ψ (cid:48)(cid:48) satisfyingmax i,j (cid:107) Ψ (cid:48) x i x j (cid:107) C α ( S n − ) ≤ , max i,j (cid:107) Ψ (cid:48)(cid:48) x i x j (cid:107) C α ( S n − ) ≤ , we have(17) (cid:107) P (Ψ (cid:48) ) − P (Ψ (cid:48)(cid:48) ) (cid:107) L ( S n − ) ≤ C max i,j (cid:107) Ψ (cid:48) x i x j − Ψ (cid:48)(cid:48) x i x j (cid:107) L ( S n − ) max i,j (cid:16) (cid:107) Ψ (cid:48) x i x j (cid:107) C ( S n − ) + (cid:107) Ψ (cid:48)(cid:48) x i x j (cid:107) C ( S n − ) (cid:17) and(18) (cid:107) P (Ψ (cid:48) ) − P (Ψ (cid:48)(cid:48) ) (cid:107) C α ( S n − ) ≤ IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 23 C max i,j (cid:107) Ψ (cid:48) x i x j − Ψ (cid:48)(cid:48) x i x j (cid:107) C α ( S n − ) max i,j (cid:16) (cid:107) Ψ (cid:48) x i x j (cid:107) C α ( S n − ) + (cid:107) Ψ (cid:48)(cid:48) x i x j (cid:107) C α ( S n − ) (cid:17) . This will enable us to solve the equation Af = g with g = 1 + γ byiterations if (cid:107) γ (cid:107) C α ( S n − ) is small enough.By (cid:101) ∆ f we shall denote the restriction of the Laplacian ∆ F of F to theunit sphere. Note that the Laplacian ∆ F itself is a ( − R n \ { } (assuming again that F is twice continuouslydifferentiable away from the origin). Lemma 4.
For every ε > , there exists δ > such that for every even g = 1 + γ with (cid:107) γ (cid:107) C α ( S n − ) ≤ δ , there exists f = 1 + ϕ solving Af = g and such that (cid:107) ϕ (cid:107) C α ( S n − ) ≤ ε and, moreover, ϕ = ϕ (cid:48) + ϕ (cid:48)(cid:48) , where (cid:101) ∆ ϕ (cid:48) = γ , while (cid:107) ϕ (cid:48)(cid:48) (cid:107) L ( S n − ) ≤ ε (cid:107) γ (cid:107) L ( S n − ) .Proof. Define the sequence ϕ m as follows: (cid:101) ∆ ϕ = γ , (cid:101) ∆ ϕ = γ − P (Φ ), (cid:101) ∆ ϕ = γ − P (Φ ), etc., where as before, Φ m is the 1-homogeneousextension of ϕ m . Recall that by the results of Appendix I, for everyeven function χ ∈ C α ( S n − ), there exists a unique solution ψ of theequation (cid:101) ∆ ψ = χ and we have the estimates (cid:107) ψ (cid:107) C α ( S n − ) ≤ K (cid:107) χ (cid:107) C α ( S n − ) , max i,j (cid:107) Ψ x i x j (cid:107) L ( S n − ) ≤ K (cid:107) χ (cid:107) L ( S n − ) with some constant K >
0. So, all ϕ m are well-defined.Let δ > (cid:107) γ (cid:107) C α ( S n − ) ≤ δ , we have (cid:107) ϕ (cid:107) C α ( S n − ) ≤ K (cid:107) γ (cid:107) C α ( S n − ) ≤ Kδ.
Fix κ ∈ (0 , K ). It follows from (18) that as long as (cid:107) ψ (cid:48) (cid:107) C α ( S n − ) , (cid:107) ψ (cid:48)(cid:48) (cid:107) C α ( S n − ) ≤ κ C , we have (cid:107) P (Ψ (cid:48) ) − P (Ψ (cid:48)(cid:48) ) (cid:107) C α ( S n − ) ≤ κ (cid:107) ψ (cid:48) − ψ (cid:48)(cid:48) (cid:107) C α ( S n − ) . If δ is small enough, so that Kδ < κ C , we obtain (cid:107) P (Φ ) (cid:107) C α ( S n − ) = (cid:107) P (Φ ) − P (0) (cid:107) C α ( S n − ) ≤ κKδ. Hence, from (cid:101) ∆( ϕ − ϕ ) = − P (Φ ), we conclude that (cid:107) ϕ − ϕ (cid:107) C α ( S n − ) ≤ K ( κK ) δ, so (cid:107) ϕ (cid:107) C α ( S n − ) ≤ K (1 + κK ) δ. If this value is still less that κ C , we can continue and write (cid:107) P (Φ ) − P (Φ ) (cid:107) C α ( S n − ) ≤ κ (cid:107) ϕ − ϕ (cid:107) C α ( S n − ) ≤ ( κK ) δ. Thus, from (cid:101) ∆( ϕ − ϕ ) = − ( P (Φ ) − P (Φ )), we get (cid:107) ϕ − ϕ (cid:107) C α ( S n − ) ≤ K ( κK ) δ, (cid:107) ϕ (cid:107) C α ( S n − ) ≤ K (1 + κK + ( κK ) ) δ, and so on. We can continue this chain of estimates as long as K (1 + κK + ( κK ) + · · · + ( κK ) m ) δ < κ C , which is forever if Kδ − κK < κ C .The outcome is that (cid:107) ϕ m +1 − ϕ m (cid:107) C α ( S n − ) ≤ K ( κK ) m +1 δ, (cid:107) P (Φ m +1 ) − P (Φ m ) (cid:107) C α ( S n − ) ≤ ( κK ) m +2 δ. It follows that the sequence ϕ m converges in C α ( S n − ) to some func-tion ϕ ∈ C α ( S n − ) with (cid:107) ϕ (cid:107) C α ( S n − ) ≤ Kδ − κK < ε if δ > ϕ will solve the equation (cid:101) ∆ ϕ = γ − P (Φ), i.e., the function f = 1 + ϕ will solve Af = g .We put ϕ (cid:48) = ϕ and ϕ (cid:48)(cid:48) = ϕ − ϕ . It remains to estimate (cid:107) ϕ (cid:48)(cid:48) (cid:107) L ( S n − ) = (cid:107) ϕ − ϕ (cid:107) L ( S n − ) . To this end, we shall use (17) instead of (18) to obtain (cid:107) P (Φ ) (cid:107) L ( S n − ) = (cid:107) P (Φ ) − P (0) (cid:107) L ( S n − ) ≤≤ κ max i,j (cid:107) (Φ ) x i x j (cid:107) L ( S n − ) ≤ κK (cid:107) γ (cid:107) L ( S n − ) , so from the equation (cid:101) ∆( ϕ − ϕ ) = − P (Φ ), we obtain (cid:107) ϕ − ϕ (cid:107) L ( S n − ) ≤ (cid:107) P (Φ ) (cid:107) L ( S n − ) ≤ κK (cid:107) γ (cid:107) L ( S n − ) and (cid:107) (Φ ) x i x j − (Φ ) x i x j (cid:107) L ( S n − ) ≤ K (cid:107) P (Φ ) (cid:107) L ( S n − ) ≤ K ( κK ) (cid:107) γ (cid:107) L ( S n − ) . Then (cid:107) P (Φ ) − P (Φ ) (cid:107) L ( S n − ) ≤ ( κK ) (cid:107) γ (cid:107) L ( S n − ) , and we can continue as above to get inductively the inequalities (cid:107) ϕ m +1 − ϕ m (cid:107) L ( S n − ) ≤ ( κK ) m +1 (cid:107) γ (cid:107) L ( S n − ) , (cid:107) (Φ m +1 ) x i x j − (Φ m ) x i x j (cid:107) L ( S n − ) ≤ K ( κK ) m +1 (cid:107) γ (cid:107) L ( S n − ) (that requires the estimate max i,j (cid:107) (Φ m ) x i x j (cid:107) C ( S n − ) ≤ κ C , but we havealready obtained that bound even for the C α -norm of ϕ m ). IFTH AND EIGHTH BUSEMANN-PETTY PROBLEMS NEAR THE BALL 25
Adding these estimates up, we get (cid:107) ϕ − ϕ (cid:107) L ( S n − ) ≤ ∞ (cid:88) m =0 (cid:107) ϕ m +1 − ϕ m (cid:107) L ( S n − ) ≤ ∞ (cid:88) m =0 ( κK ) m +1 (cid:107) γ (cid:107) L ( S n − ) = κK − κK (cid:107) γ (cid:107) L ( S n − ) , and it remains to choose κ > κK − κK < ε . (cid:3) References [BGVV]
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E. Lutwak , Extended affine surface area , Adv. Math., (1991), 39-68.3[P] C. Petty , On the geometrie of the Minkowski plane , Rivista Mat. Univ.Parma, (1955), 269-292. 2[Sch] R. Schneider , Convex Bodies: The Brunn-Minkowski theory , Encyclope-dia of Mathematics and its Applications, , Cambridge University Press,Cambridge, 2014. 2, 4, 6, 7, 15 Department of Mathematics, North Dakota State University, Fargo,ND 58108, USA
Email address : [email protected] Department of Mathematical Sciences, Kent State University, Kent,OH 44242, USA
Email address : [email protected] Department of Mathematical Sciences, Kent State University, Kent,OH 44242, USA
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