aa r X i v : . [ m a t h . M G ] S e p A reverse Minkowski-type inequality
Daniel Hug ∗ and K´aroly B¨or¨oczky † Abstract
The famous Minkowski inequality provides a sharp lower bound for the mixedvolume V ( K, M [ n − K, M ⊂ R n in terms of powers of thevolumes of the individual bodies K and M . The special case where K is the unitball yields the isoperimetric inequality. In the plane, Betke and Weil (1991) founda sharp upper bound for the mixed area of K and M in terms of the perimeters of K and M . We extend this result to general dimensions by proving a sharp upperbound for the mixed volume V ( K, M [ n − K andthe surface area of M . The equality case is completely characterized. In addition,we establish a stability improvement of this and related geometric inequalities ofisoperimetric type. Keywords . Geometric inequality, Brunn-Minkowski theory, Minkowski inequality,mean width, surface area, mixed volume, stability result
MSC . Primary: 52A20, 52A38, 52A39, 52A40; secondary: 60D05, 52A22.
Mixed volumes of convex bodies in Euclidean space R n are fundamental functionals whichencode geometric information about the involved convex bodies in a non-trivial way. Let K n denote the space of compact convex subsets of R n . For K, M ∈ K n and α, β ≥
0, thevolume V ( αK + βM ) of the Minkowski sum αK + βM has the polynomial expansion V ( αK + βM ) = n X i =0 (cid:18) ni (cid:19) V ( K [ i ] , M [ n − i ]) α i β n − i , (1.1)by which the coefficients V ( K [ i ] , M [ n − i ]) are uniquely determined. These are specialmixed volumes involving i copies of K and n − i copies of M , for i ∈ { , . . . , n } . We referto [8] for an introduction of more general mixed volumes and a thorough study of theirbasic properties. In the following, we simply write V ( K, M [ n − V ( K, M, . . . , M )if K appears with multiplicity one. In particular, the polynomial expansion (1.1) impliesthat nV ( K, M, . . . , M ) = lim ε → ε ( V ( K + εM ) − V ( K )) . ∗ Karlsruhe Institute of Technology (KIT), D-76128 Karlsruhe, Germany. E-mail: [email protected] † Alfr´ed R´enyi Institute of Mathematics, Hungarian Academy of Sciences, Reltanoda u. 13-15, H-1053Budapest, Hungary, and Department of Mathematics, Central European University, Nador u 9, H-1051,Budapest, Hungary. E-mail: [email protected], Supported in part by NKFIH grants 116451,121649 and 129630. nV ( K, B n , . . . , B n ) is the surface are F ( K ) of K if M = B n is theEuclidean unit ball. The special choice M = B n leads to the intrinsic volumes V i ( K ) = 1 κ n − i (cid:18) ni (cid:19) V ( K [ i ] , B n [ n − i ]) , i ∈ { , . . . , n } , where κ m is the volume of B m in R m . We note that V n = V is the volume functional, V ( K )is proportional to the mean width of K , and equal to the length of K if K is a segment.Furthermore, the intrinsic volume V n − ( K ) = n V ( K, B n [ n − K if int K = ∅ , and V n − ( K ) = H n − ( K ) if dim K = n −
1. Here, we write H i for the i -dimensional Hausdorff-measure, which is normalized in such a way that it coincides withthe Lebesgue measure on R i . In particular, in the Euclidean plane, F ( K ) = 2 V ( K ) is theperimeter of K ∈ K .One of the fundamental results for mixed volumes is Minkowski’s inequality V ( K, M, . . . , M ) n ≥ V ( K ) V ( M ) n − for K, M ∈ K n . (1.2)If int K, int M = ∅ , then equality holds if and only if K and M are homothetic, that is, M = x + λK for some x ∈ R n and λ > K, M ∈ K in terms of the perimeters of K and M . Theorem 1.1 (Betke, Weil (1992)) . If K, M ∈ K , then V ( K, M ) ≤ F ( K ) F ( M ) with equality if and only if K and M are orthogonal (possibly degenerate) segments. We extend this result to general dimensions and thus obtain the following reverseMinkowski-type inequality.
Theorem 1.2. If K, M ∈ K n , then V ( K, M [ n − ≤ n V ( K ) V n − ( M ); if dim ( K ) ≥ and dim ( M ) ≥ n − , then equality holds if and only if K is a segment and M is contained in a hyperplane orthogonal to K . For Minkowski’s inequality various stability versions have been found, the first is dueto Minkowski himself. Here we cite only two such results. Groemer [6] proved that if
K, M ∈ K n with int K, int M = ∅ and ε > V ( K, M, . . . , M ) n ≤ (1 + ε ) V ( K ) V ( M ) n − (1.3)implies that there exist y, z ∈ R n and λ > λ ( K − z ) ⊂ M − y ⊂ (cid:16) γε n +1 (cid:17) λ ( K − z )where γ > n . 2n addition, Figalli, Maggi, Pratelli [3] showed that (1.3) implies that there is some x ∈ R n such that H n ( M ∆( x + λK )) ≤ γ √ ε V ( M )where λ = ( V ( M ) /V ( K )) /n , ∆ stands for the symmetric difference and γ > n .These stability results improve Minkowski’s first inequality provided some informationabout the deviation of the shapes of K and M (up to homothety) is available and atthe same time they provide additional information on how close K and M are if almostequality holds in Minkowski’s inequality.We obtain the following stability version of the reverse Minkowski inequality given inTheorem 1.2. Here and in the following, we write R ( K ) to denote the circumradius of K ∈ K n . Theorem 1.3.
Let
K, M ∈ K n with dim( K ) ≥ and dim( M ) ≥ n − . Suppose that V ( K, M [ n − ≥ (1 − ε ) 1 n V ( K ) V n − ( M ) for some ε ∈ (0 , ε ) . Then there exist e, f ∈ S n − and a segment s of length (2 − γ ε ) R ( K ) parallel to e such that h M ( f ) + h M ( − f ) ≤ γ r ε , h e, f i ≥ − γ √ ε and s ⊂ K ⊂ s + γ R ( K ) √ εB n , where r is the maximal radius of an ( n − -ball in M | e ⊥ , and γ , γ , ε > are constantsdepending on n . Note that the third condition ensures that M is contained in a slab of width at most γ rε and the second condition implies that this slab is almost orthogonal (in a quantitativesense) to the segment s .A key point in proving Theorem 1.2 and Theorem 1.3 is the following result, which isinteresting in its own right. Theorem 1.4.
Let K ∈ K n with diam( K ) ≥ . (i) Then V ( K ) ≥ R ( K ) , with equality if and only if K is a segment. (ii) If V ( K ) ≤ (2 + ε ) R ( K ) for some small ε > , then there exists a segment s of length (2 − γ ε ) R ( K ) such that s ⊂ K ⊂ s + γ R ( K ) √ εB n , where γ , γ > are constantsdepending on n . The inequality between the circumradius and the first intrinsic volume (or the meanwidth) of a convex body, which is stated in Theorem 1.4 (i), is due to J. Linhart [5]. Ourproof for part (i) follows Linhart’s idea, but we introduce several modifications so as tosimplify the discussion of the equality case and prepare for the proof of part (ii). Theproof of the assertion in part (ii) provides a substantial strengthening and refinement ofLinhart’s argument.The order of the error bound in Theorem 1.4 (ii) is √ ε . This is the optimal order, ascan be seen by considering isosceles triangles.Geometric stability results have recently found applications in stochastic geometry, inparticular in the study of shapes of large cells in certain random tessellations. The stabilityresult stated in Theorem 1.4 now leads to the following probabilistic deviation result forstationary and isotropic Poisson hyperplane tessellations in R n . We refer to Section 5 fora brief introduction of the concepts used in the statement of Theorem 1.5. In particular,a suitable choice of a deviation functional ϑ is provided in (5.1).3 heorem 1.5. Let Z denote the zero cell of a stationary and isotropic Poisson hyperplanetessellation in R n with intensity λ > . Then there is a constant c (depending on n ) suchthat the following holds. If ε > and < a < b ≤ ∞ , then P ( ϑ ( Z ) ≥ ε | R ( Z ) ∈ [ a, b )) ≤ c exp { c ε a λ } , where c is a constant which depends on n, ε . We note that Betke and Weil [2] also proved that if K ∈ K , then V ( K, − K ) ≤ √ F ( K ) , (1.4)and under the additional assumption that K is a two-dimensional polygon they showedthat equality holds in (1.4) if and only if K is an equilateral triangle.Betke and Weil [2] suggested as a problem to characterize the equality cases of (1.4)among all planar compact convex sets K ∈ K . This goal is achieved in the forthcomingmanuscript [1].The paper is structured as follows. Some basic notions which are used the followingare introduced in Section 2. Then Theorem 1.4 is proved in Section 3. Our main results,Theorem 1.2 and its stability version Theorem 1.3, are established in Section 4. Finally,the application of Theorem 1.4 to stationary and isotropic Poisson hyperplane tessellationin R n is discussed in Section 5. For the basic notions and results from the Brunn-Minkowski theory which are used in thispaper, we refer to the monograph [8]. We work in Euclidean space R n with scalar product h· , ·i and induced Euclidean norm k · k in R n . The unit ball is denoted by B n , its boundaryis the unit sphere S n − = ∂B n . For a set A in a topological space we denote its closure bycl( A ). If u ∈ S n − , then u ⊥ denotes the linear ( n − u , and we write X | u ⊥ for the orthogonal projection of X ⊂ R n into u ⊥ . The support function of a convexbody K ∈ K n is h K ( x ) = max y ∈ K h x, y i for x ∈ R n . On K n we use the Hausdorff metric d H ( K, M ) = min { r ≥ K ⊂ M + rB n and M ⊂ K + rB n } for K, M ∈ K n .The surface area measure S n − ( K, · ) of K ∈ K n is the (unique) finite Borel measureon S n − such that if M ∈ K n , then V ( M, K [ n − n Z S n − h M ( u ) S n − ( K, du ) . The surface area measure is weakly continuous on K n ; namely, if K m , K ∈ K n and K m → K for m → ∞ (with respect to the Hausdorff metric) and if g : S n − → R is continuous,then lim m →∞ Z S n − g ( u ) S n − ( K m , du ) = Z S n − g ( u ) S n − ( K, du ) . We note that if K ∈ K n and e ∈ S n − , then2 H n − ( K | e ⊥ ) = Z S n − |h e, u i| S n − ( K, du ) . (2.1)4n fact, this holds even if K does not have interior points. We provide some additionalinformation about the surface area measure for a convex body K ∈ K n . If dim K ≤ n − S n − ( K, · ) is the constant zero measure. If dim K = n − K is parallel to u ⊥ for u ∈ S n − , then S n − ( K, · ) is the even measure concentrated on {± u } with S n − ( K, { u } ) = H n − ( { u } ). Now suppose that int K = ∅ . For each x ∈ ∂K , thereexists an exterior unit normal u ∈ S n − such that h K ( u ) = h x, u i . Moreover, for H n − almost all x ∈ ∂K the exterior unit normal of K at x is uniquely determined. In this case, x is called a regular boundary point and the exterior unit normal of K at x is denoted by ν K ( x ). We write ∂ ′ K to denote the set of regular boundary points of K . In particular, if g : S n − → R is a bounded Borel function, then Z S n − g ( u ) S n − ( K, du ) = Z ∂ ′ K g ( ν K ( x )) H n − ( dx ) . If K ∈ K n with int K = ∅ and f ∈ S n − , then S n − ( K, { u ∈ S n − : h u, f i > } ) > . (2.2)Since 2 V n − ( K ) is the surface area F ( K ) of K , we deduce from (2.1) and (2.2) that if e ∈ S n − and K ∈ K n satisfies dim K ≥ n −
1, then H n − ( K | e ⊥ ) ≤ V n − ( K ) , (2.3)with equality if and only if dim M = n − e is normal to M . In addition, whenprojecting a convex body K ∈ K n to e ⊥ for some e ∈ S n − , we have H n − ( K | e ⊥ ) = Z K |h e, u i| H n − ( dx )if dim K = n − u ∈ S n − is normal to K , and H n − ( K | e ⊥ ) = 12 Z ∂ ′ K |h e, ν K ( x ) i| H n − ( dx )if dim K = n . For z ∈ S n − and α ∈ (0 , π ), let B ( z, α ) = { x ∈ S n − : h x, z i ≥ cos α } be the sphericalcap (geodesic ball) centered at z and of radius α . For a spherical set X ⊂ S n − , we writeint s X to denote the interior of X on S n − and ∂ s X for the boundary of X with respectto S n − (and its topology induced by the geodesic metric, which is equal to the subspacetopology of the ambient space). For a point x ∈ S n − , the point − x is the point of S n − which is antipodal to x . We call X ⊂ S n − starshaped with respect to a point x ∈ S n − if x ∈ X , − x X , and for any x ∈ X \ { x } , the spherical geodesic arc connecting x and x is contained in X .The following observation is a key step in proving Theorem 3.3. Lemma 3.1. If α ∈ (0 , π ] , n ≥ , z ∈ S n − and Π ⊂ B ( z, α ) is compact and starshapedwith respect to z , then Z Π h z, u i H n − ( du ) ≥ Z B ( z,α ) h z, u i H n − ( du ) H n − ( B ( z, α )) · H n − (Π) . roof. For the proof, we can assume that H n − (Π) >
0. For u ∈ z ⊥ ∩ S n − , let ϕ ( u ) ∈ [0 , α ]be the “spherical radial function” of Π which is given by ϕ ( u ) = max { t ∈ [0 , α ] : z · cos t + u · sin t ∈ Π } . In addition, let Ξ = { u ∈ z ⊥ ∩ S n − : ϕ ( u ) > } . An application of the transformation formula shows that Z Π h z, u i H n − ( du ) = Z Ξ Z ϕ ( u )0 (cos s )(sin s ) n − ds H n − ( du ) , H n − (Π) = Z Ξ Z ϕ ( u )0 (sin s ) n − ds H n − ( du ) . To shorten the formulas, we set ̺ ( s ) = (sin s ) n − for s ∈ (0 , π ). Since cos s is decreasingin s , for any u ∈ Ξ with ϕ ( u ) < α we have R αϕ ( u ) (cos s ) ̺ ( s ) ds R αϕ ( u ) ̺ ( s ) ds < cos ϕ ( u ) < R ϕ ( u )0 (cos s ) ̺ ( s ) ds R ϕ ( u )0 ̺ ( s ) ds , which in turn yields that R α (cos s ) ̺ ( s ) ds R α ̺ ( s ) ds = R ϕ ( u )0 ̺ ( s ) ds R α ̺ ( s ) ds · R ϕ ( u )0 (cos s ) ̺ ( s ) ds R ϕ ( u )0 ̺ ( s ) ds + R αϕ ( u ) ̺ ( s ) ds R α ̺ ( s ) ds · R αϕ ( u ) (cos s ) ̺ ( s ) ds R αϕ ( u ) ̺ ( s ) ds ≤ R ϕ ( u )0 (cos s ) ̺ ( s ) ds R ϕ ( u )0 ̺ ( s ) ds . This holds in fact for any u ∈ Ξ. Therefore Z Π h z, u i H n − ( du ) ≥ R α (cos s ) ̺ ( s ) ds R α ̺ ( s ) ds Z Ξ Z ϕ ( u )0 ̺ ( s ) ds H n − ( du )= R B ( z,α ) h z, u i H n − ( du ) H n − ( B ( z, α )) · H n − (Π) , which proves the assertion.We note that for any z ∈ S n − , we have R B ( z, π ) h z, u i H n − ( du ) H n − ( B ( z, π )) = 2 κ n − H n − ( S n − ) . (3.1)The following lemma shows how the left side increases when B ( z, π ) is replaced by B ( z, α )and 0 < α ≤ π − ε . Lemma 3.2. If < α ≤ π − ε , ε ∈ [0 , π ] , n ≥ and z ∈ S n − , then Z B ( z,α ) h z, u i H n − ( du ) H n − ( B ( z, α )) ≥ (1 + c ε ) · κ n − H n − ( S n − ) , where c > depends on n . roof. For α ∈ (0 , π ], let f ( α ) = R B ( z,α ) h z, u i H n − ( du ) H n − ( B ( z, α )) = R α (cos s ) ̺ ( s ) ds R α ̺ ( s ) ds , and hence f ′ ( α ) = ̺ ( α ) (cid:0)R α ̺ ( s ) ds (cid:1) (cid:18) cos α · Z α ̺ ( s ) ds − Z α (cos s ) ̺ ( s ) ds (cid:19) < s > cos α for 0 < s < α .Since f is monotone decreasing on (0 , π ], it is sufficient to prove that f ′ ( α ) ≤ − c for α ∈ [ π , π ], where c > n . We observe that Z α (cos s ) ̺ ( s ) ds ≥ cos α · Z α π ̺ ( s ) ds + cos π · Z π ̺ ( s ) ds, and thereforecos α · Z α ̺ ( s ) ds − Z α (cos s ) ̺ ( s ) ds ≤ (cid:16) cos α − cos π (cid:17) Z π ̺ ( s ) ds < . Since α ≥ π > π , we conclude that f ′ ( α ) ≤ (sin π ) n − (cid:16)R π ̺ ( s ) ds (cid:17) · (cid:16) cos π − cos π (cid:17) · Z π ̺ ( s ) ds, which proves Lemma 3.2.Recall that R ( K ) denotes the circumradius of a convex body K , which is the radius ofthe (uniquely determined) smallest ball containing K . We slightly rephrase Theorem 1.4from the introduction as follows. Theorem 3.3.
Let K ∈ K n . (i) Then V ( K ) ≥ R ( K ) , with equality if and only if K is a segment (ii) If V ( K ) ≤ (2 + ε ) R ( K ) for some small ε > , then there exists a vector c ∈ R n anda segment s of length − c ε such that R ( K ) s ⊂ K − c ⊂ R ( K )( s + c √ εB n ) , where c , c > are constants depending on n . Note also that Theorem 1.4 remains true if R ( K ) = 0 (as stated above). In this case K is a point and all assertions hold trivially. If R ( K ) >
0, then the explicit use of thevector c can be avoided by considering a translation of the segment s . Proof of Theorem 3.3.
For the proof, we can assume that R ( K ) >
0. By homogeneityand translation invariance, we can then assume that B n is the circumball of K , and hence R ( K ) = 1. It follows that the origin o is contained in the convex hull of S n − ∩ K . Let k be the minimal number of points of S n − ∩ K whose convex hull contains o , and hence2 ≤ k ≤ n + 1 by Carath´eodory’s theorem. Let x , . . . , x k ∈ S n − ∩ K and λ , . . . , λ k > λ + · · · + λ k = 1 be such that λ x + · · · + λ k x k = o . For i = 1 , . . . , k , we define theDirichlet-Voronoi cell D i = { x ∈ S n − : h x, x i i ≥ h x, x j i for j = 1 , . . . , k } , and hence D i is starshaped with respect to x i and P ki =1 H n − ( D i ) = H n − ( S n − ). In fact,since λ x + · · · + λ k x k = o implies that for any x ∈ S n − , there exists i ∈ { , . . . , k } suchthat h x, x i i ≥
0, it follows that D i ⊂ B (cid:16) x i , π (cid:17) , i = 1 , . . . , k. For x ∈ D i , we have h ([ x , . . . , x k ] , x ) = h x, x i i . Thence, we deduce from Lemma 3.1 and(3.1) that V ( K ) ≥ V ([ x , . . . , x k ]) = nκ n − V ([ x , . . . , x k ] , B n [ n − κ n − k X i =1 Z D i h x, x i i H n − ( dx ) (3.2) ≥ κ n − k X i =1 κ n − H n − ( S n − ) · H n − ( D i ) = 2 . (3.3)If V ( K ) = 2, then equality in (3.2) yields that K = [ x , . . . , x k ]. Moreover, by themonotonicity shown in the first part of the proof of Lemma 3.2, it follows that strictinequality holds in (3.3) if D i = B ( x i , π/
2) for some i ∈ { , . . . , k } . Hence, if equalityholds, we must have k = 2 and K = [ x , x ].To prove the stability statement (ii), we again assume that R ( K ) = 1. First, we showthat there exists a constant η >
0, depending only on n , such that if diam K ≤ − η forsome η ∈ [0 , η ], then V ( K ) ≥ c √ η for a constant c > n . (3.4)For η = 0, the assertion holds by (i). Since P ki =1 H n − ( D i ) = nκ n , we may assume that H n − ( D ) ≥ nκ n /k . For i = 2 , . . . , k , we consider the set F i = { x ∈ D : h x, x i = h x, x i i} ,which is contained in the hyper-sphere { x ∈ S n − : h x, x − x i i = 0 } and compact. Inaddition, let C i be the union of all spherical geodesic arcs connecting x to the points of F i , and hence each C i is compact and starshaped with respect x and P ki =2 H n − ( C i ) = H n − ( D ). In particular, since 2 ≤ k ≤ n + 1 we may assume that H n − ( C ) ≥ nκ n k ( k − ≥ κ n n + 1 . Let u ∈ x ⊥ ∩ S n − ∩ lin { x , x } be the vector such that h u, x i >
0, and hence writing β for the angle enclosed by x and x , we have x = x cos β + u sin β . We deduce from F ⊂ B ( x , π ) ∩ ( x − x ) ⊥ that F ⊂ B ( x , π ) ∩ B ( u, π ), and thus C ⊂ B ( x , π ) ∩ B ( u, π ) ∩ B ( x − x k x − x k , π ) . Let Ξ ′ := { x ∈ x ⊥ ∩ S n − : 0 ≤ h x, u i ≤ τ } , τ = τ ( n ), depending only on n , is chosen such that H n − (Ξ ′ ) = H n − ( S n − ) n ( n + 1)Then we connect each point of Ξ ′ to x by a geodesic arc and take the union of all sucharcs to get a compact subset Ξ ⊂ B ( x , π ) which is starshaped with respect to x andsatisfies H n − (Ξ) = H n − ( S n − )2 n ( n + 1) = κ n n + 1) . Now we show that if 0 ≤ η ≤ τ − =: η and γ := τ , then h x, x i ≥ γ √ η for x ∈ C \ Ξ . (3.5)For the proof, we write x in the form x = x cos s + x sin s , where x ∈ x ⊥ and s ∈ [0 , π ].Since x ∈ C \ Ξ, we conclude further that s ∈ (0 , π ] and h x , u i > τ > x − x k x − x k = x sin β − u cos β β k x − x k ≤ − η , since x , x ∈ K and diam K ≤ − η . In addition, we have (cid:18) tan β (cid:19) ≤ (1 − η ) − (1 − η ) ≤ η . Since x ∈ C , we have x ∈ B ( x − x k x − x k , π ), and hence by (3.6) it follows that h x, u i ≤ h x, x i tan β ≤ h x, x i √ √ η . If s ∈ [ π , π ], then h x, u i = h x , u i sin s ≥ τ sin s ≥ √ τ, and hence h x, x i ≥ τ √ η ≥ τ √ η. If s ∈ (0 , π ], then again h x, x i = cos s ≥ ≥ τ √ η, since 0 < η ≤ τ − . This proves the claim.It follows from the construction of C and Ξ that H n − (cl( C \ Ξ)) = H n − ( C \ Ξ) ≥ κ n n + 1) = 12 n ( n + 1) · H n − ( S n − ) . (3.7)We define α ∈ (0 , π ) by cos α = γ √ η ∈ (0 , ], and hence α ≥ π . Then (3.5) implies thatfor x ∈ cl( C \ Ξ) we have x ∈ B ( x , α ) and γ √ η = cos α = sin (cid:16) π − α (cid:17) ≤ π − α,
9o that 0 < α ≤ π − γ √ η and γ √ η ≤ π . Therefore, we can apply Lemma 3.1 tothe topological closure of C \ Ξ, which is starshaped with respect to x , and also useLemma 3.2 and (3.7) to get Z C \ Ξ h x , x i H n − ( dx ) ≥ R B ( x ,α ) h x , x i H n − ( dx ) H n − ( B ( x , α )) · H n − ( C \ Ξ) ≥ (1 + c γ √ η ) · κ n − H n − ( S n − ) · H n − ( C \ Ξ) ≥ κ n − H n − ( S n − ) · H n − ( C \ Ξ) + κ n − c γ n ( n + 1) · √ η. In addition, using again Lemma 3.1 we also have Z C ∩ Ξ h x , x i H n − ( dx ) ≥ κ n − H n − ( S n − ) · H n − ( C ∩ Ξ) , Z C j h x , x i H n − ( dx ) ≥ κ n − H n − ( S n − ) · H n − ( C j ) for j ≥ , Z D i h x i , x i H n − ( dx ) ≥ κ n − H n − ( S n − ) · H n − ( D i ) for i ≥ . Summing up the individual contributions from the subsets, we get V ( K ) ≥ V ([ x , . . . , x k ]) = 1 κ n − k X i =1 Z D i h x, x i i H n − ( dx ) ≥ κ n − c γ n ( n + 1) · √ η + 1 κ n − k X i =1 κ n − H n − ( S n − ) · H n − ( D i )= 2 + κ n − c γ n ( n + 1) · √ η, which completes the proof of (3.4).Let us assume that V ( K ) ≤ ε . If [ y , y ] ⊂ K is a longest segment in K , then k y − y k ≥ − c ε for c = c − by (3.4). For any y ∈ K , writing t for the distance of y from [ y , y ], we have2 + ε ≥ V ( K ) ≥ V ([ y , y , y ]) ≥ k y − y k r k y − y k t ≥ − c ε s(cid:18) − c ε (cid:19) + t . Assuming that ( c ε + 1) ε ≤ s ) ≤ s for s ∈ [0 , t ≤ (cid:18) c (cid:19) ε, which in turn yields that K ⊂ [ y , y ] + p (3 + 3 c ) ε B n . Finally we note that (ii) again implies the equality condition in (i).10
Proofs of Theorem 1.2 and Theorem 1.3
Part (i) of Theorem 3.3 is the main ingredient for the proof of Theorem 1.2.
Proof of Theorem 1.2.
We can assume that the circumball of K has its centre at the origin.Then V ( K, M [ n − n Z S n − h K ( u ) S n − ( M, du ) ≤ n R ( K ) F ( M ) (4.1) ≤ n V ( K )2 V n − ( M ) = 1 n V ( K ) V n − ( M ) , where we used Theorem 3.3 for the second inequality. If equality holds, then equalityholds in Theorem 3.3, since V n − ( M ) >
0, and therefore K = [ − Re, Re ] with R = R ( K )and for some e ∈ S n − . Moreover, we then also have equality in the first inequality, whichyields Z S n − |h u, e i| S n − ( M, du ) = S n − ( M, S n − ) . This implies that the area measure of M is concentrated in {− e, e } , hence M is containedin a hyperplane orthogonal to e .We now start to build the argument leading to the stability version Theorem 1.3 ofTheorem 1.2. Recall that if M is an at least ( n − R n and e ∈ S n − , then 2 H n − ( M | e ⊥ ) = Z S n − |h e, u i| S n − ( M, du )= Z ∂ ′ M |h e, ν M ( x ) i| H n − ( dx ) . Moreover, Z S n − |h e, u i| S n − ( M, du ) ≤ V n − ( M ) , (4.2)with equality if and only if M ⊂ e ⊥ .In the following proposition, we explore what can be said about M if the integral onthe left side of (4.2) is ε -close to the upper bound. Proposition 4.1.
Let ε ∈ (0 , (cid:0) n (cid:1) n ) and e ∈ S n − . Suppose that M is an at least ( n − -dimensional compact convex set in R n such that Z S n − |h e, u i| S n − ( M, du ) ≥ (1 − ε )2 V n − ( M ) . (4.3) Then there is some f ∈ S n − such that h M ( f ) + h M ( − f ) ≤ c r √ ε and h e, f i ≥ − c ε ,where c ≤ n √ n , c ≤ (10 n ) (2 n ) n and r is the maximal radius of an ( n − -ball in M | e ⊥ . Remark
The lemma is essentially optimal for n ≥
3, in the sense that one cannotconclude in general that h M ( e ) + h M ( − e ) ≤ c V n − ( M ) n − √ ε . To show this, let f ∈ S n − with h e, f i = 1 − ε , and let f , . . . , f n be an orthonormal basis such that f = f and e ∈ lin { f , f } . For large λ , we define M = (cid:2) ±√ εf , ± λf , ± nf , . . . , ± nf n (cid:3) , which satisfies H n − ( M | e ⊥ ) ≥ (1 − ε ) V n − ( M ) if λ > ε > roof of Proposition 4.1. For the proof, we can assume that M is n -dimensional. Thisfollows from an approximation argument (which will require adjustments of M , ε and r ).The main idea is as follows. Let us consider some n -dimensional convex body C with V n − ( C ) ≤ V n − ( M ) and H n − ( C | e ⊥ ) = H n − ( M | e ⊥ ), and let ∂ + C = { y ∈ ∂ ′ C : h e, ν C ( y ) i > } ,∂ − C = { y ∈ ∂ ′ C : h e, ν C ( y ) i < } . Claim
Suppose there exist η, γ > X ⊂ C | e ⊥ with H n − ( X ) = γ H n − ( C | e ⊥ ) such that any y ∈ ∂ + C with y | e ⊥ ∈ X satisfies tan ∠ ( e, ν C ( y )) ≥ η . Then η ≤ √ ε √ γ provided ε < γ . (4.4)To prove the Claim, let Y denote the set of all y ∈ ∂ + C with y | e ⊥ ∈ X . For any y ∈ Y ,we have 0 < h e, ν C ( y ) i = cos ∠ ( e, ν C ( y )) = s
11 + (tan ∠ ( e, ν C ( y ))) ≤ r
11 + η . It follows that γ H n − ( C | e ⊥ ) = H n − ( X ) = Z Y |h e, ν C ( y ) i| H n − ( dy ) ≤ r
11 + η · H n − ( Y ) . (4.5)Furthermore, we have(2 − γ ) H n − ( C | e ⊥ ) = H n − ( C | e ⊥ ) + H n − (cid:0) ( C | e ⊥ ) \ X (cid:1) = Z ( ∂ ′ C ) \ Y |h e, ν C ( y ) i| H n − ( dy ) ≤ H n − (( ∂C ) \ Y ) . (4.6)From (4.3), (4.5) and (4.6), we deduce that(1 + 2 ε ) · H n − ( C | e ⊥ ) = (1 + 2 ε ) · H n − ( M | e ⊥ ) ≥ (1 + 2 ε )(1 − ε ) · H n − ( ∂M ) ≥ H n − ( ∂M ) ≥ H n − ( ∂C )= H n − (( ∂C ) \ Y ) + H n − ( Y ) ≥ (cid:16) − γ + γ p η (cid:17) H n − ( C | e ⊥ ) , and hence 4 ε ≥ γ ( p η − s ) ≤ s for s ∈ (0 , t = max (cid:8) H (( x + R e ) ∩ M ) : x ∈ M | e ⊥ (cid:9) r to denote the maximal radius of ( n − M | e ⊥ . Possibly after atranslation of M , there exists an origin symmetric ellipsoid E such that E ⊂ M ⊂ nE (4.7)according to John’s theorem. By changing the orientation of e ∈ S n − , if necessary, thereis a positive τ such that τ e ∈ ∂E and τ ≤ t ≤ nτ , and hence τ ∈ [ t n , t ]. Let f ∈ S n − bethe exterior unit normal at τ e to E . It follows that h M ( f ) + h M ( − f ) ≤ h nE ( f ) + h nE ( − f ) = 2 nh E ( f )= 2 n h τ e, f i = 2 nτ h e, f i ≤ nτ, and thus h M ( f ) + h M ( − f ) ≤ nt. (4.8)We prove Lemma 4.1 in two steps. First, we bound t from above, then we establish alower bound for |h e, f i| . Step 1
We show that t ≤ c r √ ε with a constant c ≤ n √ n − .Let w ∈ S n − ∩ e ⊥ be such that h E ( w ) = min { h E ( u ) : u ∈ S n − ∩ e ⊥ } , which equalsthe inradius of E | e ⊥ , and hence h E ( w ) ≤ r . In turn, for y ∈ M we deduce from (4.7) that |h y, w i| ≤ max { h M ( w ) , h M ( − w ) } ≤ h nE ( w ) = nh E ( w ) ≤ nr, that is, |h y, w i| ≤ nr for y ∈ M. (4.9)We may choose an orthonormal basis e , . . . , e n of R n such that e = e and e = w ,and let M ′ be the convex body resulting from M via successive Steiner symmetrizationswith respect to e ⊥ , . . . , e ⊥ n . Then M ′ satisfies |h y, w i| ≤ nr for y ∈ M ′ , ± te ∈ ∂M ′ , ( rB n ∩ e ⊥ ) ⊂ M ′ ,M ′ is symmetric with respect to the coordinate subspaces e ⊥ , . . . , e ⊥ n , and hence in par-ticular M ′ is centrally symmetric, H n − ( M ′ | e ⊥ ) = H n − ( M | e ⊥ ), M ′ | e ⊥ = M ′ ∩ e ⊥ and V n − ( M ′ ) ≤ V n − ( M ). Finally, we consider the double cone f M = conv (cid:26) te, − te, M ′ ∩ e ⊥ (cid:27) , which satisfies |h y, w i| ≤ nr for y ∈ f M , ± te ∈ ∂ f M , ( rB n ∩ e ⊥ ) ⊂ f M , f M is symmetric with respect to the coordinate subspaces e ⊥ , . . . , e ⊥ n (and hence centrallysymmetric), H n − ( f M | e ⊥ ) = H n − ( M | e ⊥ ) and V n − ( f M ) ≤ V n − ( M ).For ̺ = h f M ( w ), we have ̺w ∈ ∂ f M , and hence ̺ ≤ nr . To prepare an application of(4.4), we consider X = 56 ̺w + 16 (cid:16) f M | e ⊥ (cid:17) ⊂ f M | e ⊥ . L = e ⊥ ∩ w ⊥ . Let Y denote the set of all y ∈ ∂ + f M such that y | e ⊥ ∈ X . For y ∈ Y , wewrite α for the angle of ν f M ( y ) and e , thus y = se + pw + v and ν f M ( y ) = e cos α + qw + v ,where v ∈ (cid:16) f M ∩ L (cid:17) , v ∈ L, s ≥ , ̺ ≤ p ≤ ̺, q ≤ sin α. Since f M is a double cone, there exists z ∈ e ⊥ ∩ ∂ f M such that y ∈ [ z, te ] and y | e ⊥ ∈ [ z, o ].We deduce that s t = k z − ( y | e ⊥ ) kk z k = h z − ( y | e ⊥ ) , w ih z, w i ≤ , thus s ≤ t . As te ∈ f M , we have0 ≤ (cid:10) y − te, ν f M ( y ) (cid:11) = ( s − t ) cos α + pq + h v , v i , which yields −h v , v i ≤ ( s − t ) cos α + pq. Therefore, since 2 v ∈ f M we obtain0 ≤ h y − v , ν f M ( y ) i = s · cos α + pq − h v , v i ≤ (2 s − t ) cos α + 2 pq ≤ − t cos α + 2 ̺ sin α ≤ − t cos α + 2 nr sin α, which implies that tan α ≥ t nr . Now an application of (4.4) proves the estimate of Step 1. Step 2
We show that h e, f i ≥ − c ε .Let β = ∠ ( f, e ) ∈ [0 , π ), and let e w ∈ S n − ∩ e ⊥ be such that f = e cos β + e w sin β .Since the shadow boundary of E in direction e lies in a hyperplane and by [7, Theorem1], it follows from the definition of f that E | e ⊥ = ( E ∩ f ⊥ ) | e ⊥ . Hence (4.7) yields that( E ∩ f ⊥ ) | e ⊥ ⊂ ± M | e ⊥ ⊂ n ( E ∩ f ⊥ ) | e ⊥ . (4.10)Since there is some a ∈ e ⊥ such that ( rB n ∩ e ⊥ ) + a ⊂ M | e ⊥ it follows that rn ( B n ∩ e ⊥ ) ± an ⊂ ± n ( M | e ⊥ ) ⊂ E | e ⊥ , which shows that rn ( B n ∩ e ⊥ ) ⊂ ( E ∩ f ⊥ ) | e ⊥ , and we deduce that ̺ = max (cid:8) ξ > ξ e w ∈ ( E ∩ f ⊥ ) | e ⊥ (cid:9) ≥ rn , (4.11)In order to apply (4.4), we choose C = M and the set X ⊂ M | e ⊥ is chosen as X = 12 ( E | e ⊥ ) = 12 ( E ∩ f ⊥ ) | e ⊥ ⊂ M | e ⊥ , which satifies H n − ( X ) = e γ H n − ( M | e ⊥ ) , e γ ≥ (2 n ) − ( n − , according to (4.10). 14s before, we define Y as the set of all y ∈ ∂ + M with y | e ⊥ ∈ X . Then, for y ∈ Y wehave x = y | e ⊥ ∈
12 ( E ∩ f ⊥ ) | e ⊥ . It follows that x − ̺ e w ∈ ( E ∩ f ⊥ ) | e ⊥ , and hence x − ̺ e w = z | e ⊥ for a z ∈ E ∩ f ⊥ ⊂ M . In addition, the definition of t impliesthe existence of an s ∈ [0 , t ] such that y − se ∈ E ∩ f ⊥ . Since f = e cos β + e w sin β , wededuce that z − ( y − se ) = − ̺ e w + e · ̺ tan β, thus y − z = 12 ̺ e w + e · (cid:18) s − ̺ tan β (cid:19) . We set α = ∠ ( e, ν M ( y )) ∈ [0 , π ), and hence ν M ( y ) = e cos α + e p e w + e v, where e v ∈ (lin { e w, e } ) ⊥ and e p + k e v k = (sin α ) .We deduce from Step 1, 0 ≤ s ≤ t and ̺ ≥ r/n (compare (4.11)) that0 ≤ h y − z, ν M ( y ) i = (cid:28) ̺ e w + e · (cid:18) s − ̺ tan β (cid:19) , e cos α + e p e w + e v (cid:29) = (cid:18) s − ̺ tan β (cid:19) cos α + e p · ̺ ≤ (cid:18) c n̺ √ ε − ̺ tan β (cid:19) cos α + sin α · ̺, thus tan α ≥ tan β − c n √ ε . If tan β − c n √ ε >
0, we conclude from (4.4) thattan β − c n √ ε ≤ n ) n − √ ε, which in turn yields β ≤ tan β ≤ c √ ε with c ≤ n √ n − + 4 √ n n − ≤ (10 n ) √ n n .If tan β ≤ c n √ ε , we directly arrive at the same conclusion. Therefore, in any case h e, f i = cos β ≥ − c ε , completing Step 2.Finally, we combine the stability estimates above to derive (Theorem 1.3 in the follow-ing form. Theorem 4.2.
Let
K, M ∈ K n with dim( K ) ≥ and dim( M ) ≥ n − . Suppose that V ( K, M [ n − ≥ (1 − ε ) 1 n V ( K ) V n − ( M ) (4.12) for some sufficiently small ε ∈ (0 , / . Then there is a segment s of length (2 − c ε ) anddirection e ∈ S n − such that R ( K ) s ⊂ K ⊂ R ( K )( s + c √ εB m ) , and there is a vector f ∈ S n − such that h e, f i ≥ − c √ ε and h M ( f ) + h M ( − f ) ≤ c rε , where r is themaximal radius of ( n − -balls in M | e ⊥ and c , c , c , c are constants depending on n . roof. Suppose that (4.12) is satisfied for some ε ∈ (0 , ε ), where ε ∈ (0 , ) is sufficientlysmall. A combination of (4.1) and (4.12) yields1 n R ( K ) F ( M ) ≥ V ( K, M [ n − ≥ (1 − ε ) 1 n V ( K ) V n − ( M ) , and hence V ( K ) ≤ − ε R ( K ) ≤ (1 + 2 ε )2 R ( K ) = (2 + 4 ε ) R ( K ) . By Theorem 3.3, there is a segment s of length 2 − c ε such that R ( K ) s ⊂ K ⊂ R ( K )( s + c √ εB n ). In particular, V ( K ) ≥ V ( s ) R ( K ) ≥ (2 − c ε ) R ( K ) . If s has direction e ∈ S n − , then again by the assumption we obtain(1 − ε )(2 − c ε ) 1 n R ( K ) V n − ( M ) ≤ (1 − ε ) 1 n V ( K ) V n − ( M ) ≤ V ( K, M [ n − n Z S n − h K ( u ) S n − ( M, du ) ≤ n Z S n − (cid:0) h s ( u ) + c √ ε (cid:1) R ( K ) S n − ( M, du )= R ( K ) n V ( s ) Z S n − |h u, e i| S n − ( M, du ) + c √ ε R ( K ) n V n − ( M ) , and hence Z S n − |h u, e i| S n − ( M, du ) ≥ (1 − ε )2 V n − ( M ) − c c √ ε − c ε V n − ( M ) ≥ V n − ( M )(1 − c √ ε ) . An application of Proposition 4.1 completes the proof.