Remarks on planar Blaschke-Santaló inequality
aa r X i v : . [ m a t h . M G ] N ov REMARKS ON PLANAR BLASCHKE-SANTAL ´O INEQUALITY
K. J. B¨or¨oczky, E. Makai, Jr.*
Abstract.
We prove the Blaschke-Santal´o inequality restricted to n -gons: the ex-tremal polygons are the affine regular n -gons. If either the John or the L¨owner ellipseof a planar o -symmetric convex body K is the unit circle about o , then a sharpening ofthe Blaschke-Santal´o inequality holds: even the aritmetic mean ( V ( K ) + V ( K ∗ )) / π . We give stability variants of the Blaschke-Santal´o inequality for theplane. If for some n ≥ K is n -fold rotationally symmetricabout o , then we give the exact maximum of V ( K ∗ ), as a function of V ( K ) and thearea of either the John or the L¨owner ellipse. We introduce some notation that will be used in the paper. For x, y ∈ R , x = y we write aff { x, y } for the line passing through x and y . For compact sets or points X , . . . , X k in R , let [ X , . . . , X k ] denote the convex hull of their union, wherethe X i ’s which are points are replaced by { X i } . For x, y ∈ R we write | xy | forthe length of the segment [ x, y ]. For any point p = o , let p ∗ be the polar line withequation h x, p i = 1. For l = p ∗ , let p = l ∗ . By the intersection point of two parallellines we mean their common point at infinity (in the projective plane). For twonon-negative quantities f, g we write g = Θ( f ) ( g is of exact order f ), if g = O ( f )and f = O ( g ). Optimality of regular n -gons in the o -symmetric case First we treat o -symmetric polygons because the argument is much easier in thiscase. Theorem A.
For even n ≥ , if K is an o -symmetric convex polygon of at most n vertices, then V ( K ) V ( K ∗ ) ≤ n sin πn , with equality if and only if K is a regular n -gon.Proof. Let K maximize the area product among o -symmetric convex polygonsof at most n vertices. Since n sin ( π/n ) is strictly monotonically increasing with n , by induction we may suppose that K has exactly n sides. By compactness of Mathematics Subject Classification . Primary:52A40. Secondary: 52A38, 52A10.
Key words and phrases.
Blaschke-Santal´o inequality, stability, polygons, n -fold rotationalsymmetry. Typeset by AMS -TEX K. J. B ¨OR ¨OCZKY, E. MAKAI, JR. affine equivalence classes of convex bodies such a K exists. In this case, K ∗ is also a maximal body. The vertices of K will be denoted by x , x , . . . , x n , inpositive sense.First we observe that K has exactly n vertices. In fact, if k < n , consider thevertex x of K . Let l be a line such that K ∩ l = { x } . Let ε > l parallelly towards K throgh a distance ε . Then the vertex x iscut off. (Without further mentioning we will make always the same changes at theopposite vertex − x so that K remains o -symmetric.) We write K new for the soobtained new o -symmetric polygon, that still has at most n vertices by hypothesis.Then l cuts off an area Θ( ε ) from K , so V ( K new ) = V ( K ) − Θ( ε ) . Then K ∗ has neighbouring vertices aff { x , x } ∗ and aff { x , x } ∗ but K ∗ new willhave a new vertex l ∗ between these two old vertices. The distance of l ∗ and theside x ∗ of K is Θ( ε ). Hence K ∗ new is obtained from K ∗ by adding to it a smalltriangle of height, and also of area Θ( ε ). Thus V ( K ∗ new ) = V ( K ) + Θ( ε ) . Therefore for the volume products we have V ( K new ) V ( K ∗ new ) = V ( K ) V ( K ∗ ) + Θ( ε ) > V ( K ) V ( K ∗ ) , a contradiction.Next we prove two basic properties of K :(i) If x , x , x are consecutive vertices of K , then o , ( x + x ) / x are collinear.(ii) If y , y , y , y are consecutive vertices of K , and m is the intersection point ofaff { y , y } and aff { y , y } (observe that m is a finite point separated from K byaff { y , y } ), then o , ( y + y ) / m are collinear.Observe that both (i) and (ii) are affine invariant. Then we may supposethat in (i) x , x , x ∈ S , and in (ii) that aff { y , y } , aff { y , y } , aff { y , y } aretangents to S . Then (i) expresses that { x , x , x } is symmetrical to the axis R · ( x + x ) /
2, while (ii) expresses that { aff { y , y } , aff { y , y } , aff { y , y }} issymmetrical to R · m . Otherwise said, in the metric of the o -symmetric ellipse con-taining x , x , x , the rotation carrying x to x carries x to x , and in the metricof the o -symmetric ellipse touched by aff { y , y } , aff { y , y } , aff { y , y } , the rota-tion carrying aff { y , y } to aff { y , y } carries aff { y , y } to aff { y , y } . Then thesetwo symmetry properties, one taken for K , the other taken for K ∗ , clearly implyeach other.Since (i) for K ∗ is equivalent to (ii) for K , therefore it is sufficient to prove (i).We suppose that (i) does not hold, and seek a contradiction with a method goingback to Mahler. (Recall that we assumed x , x , x ∈ S .) We may assume that x and x lie on the same open side of aff { o, ( x + x ) / } . For i = 1 ,
5, 2 .
5, we write y i LANAR BLASCHKE-SANTAL ´O INEQUALITY 3 for the intersection of x ∗ i and x ∗ , and hence y . and y . are consecutive verticesof K ∗ . Let l be the line through x parallel to aff { x , x } . We move x into aposition ˜ x along l towards aff { o, ( x + x ) / } . In other words, ˜ x − x = ε ( x − x )for small ε >
0, where ε is small enough to ensure that x and ˜ x lie on the sameopen side of both aff { o, ( x + x ) / } and the line containing the other side of K with endpoint x (say, aff { x , x } ). (That is, we move toward the symmetricsituation.) Therefore there exists an o -symmetric convex polygon e K obtained from K by removing x and − x from the set of vertices, and adding ˜ x and − ˜ x . Then V ( e K ) = V ( K ).For i = 1 .
5, or i = 2 .
5, let ˜ y i be the intersection of x ∗ and ˜ x ∗ , or x ∗ and ˜ x ∗ ,respectively, and hence ˜ y . and ˜ y . are the two new vertices of e K ∗ replacing thevertices y . and y . of K ∗ . In addition ˜ y . K ∗ and ˜ y . ∈ K ∗ , and l ∗ =[ y . , y . ] ∩ [˜ y . , ˜ y . ], moreover k l ∗ − y . k > k l ∗ − y . k and hence k l ∗ − ˜ y . k > k l ∗ − ˜ y . k , for ε > l ∗ , y . , ˜ y . ] and [ l ∗ , y . , ˜ y . ] havethe same angle at l ∗ , we have V ( ˜ K ∗ ) − V ( K ∗ ) = 2 ( V ([ l ∗ , y . , ˜ y . ]) − V ([ l ∗ , y . , ˜ y . ])) > o -symmetry, an analogous change has to be made at theside opposite to y . y . , that gives the factor 2). This contradiction proves (i). Now we prove Theorem A based on (i) and (ii). Applying a linear transfor-mation, we may assume that x , x , x ∈ S , and hence x lies on the perpendicularbisector of [ x , x ] by (i).Now (ii) yields that aff { x , x } and aff { x , x } are symmetric with respect tothe perpendicular bisector of [ x , x ] (that contains o ). It follows that aff { x , x } and aff { x , x } are symmetric with respect to aff { o, x } . Together with (i), appliedto x , x , x , this yields that x is uniquely determined, and we have x ∈ S , andthe rotation about o taking x to x takes x to x and x to x . Continuing likethis, we conclude that K is a regular n -gon inscribed into B . (cid:4) Area sum for normalized o -symmetric planar convex bodies Recall that in [BMMR] the main tool was the lower estimate for polar sectorsof convex bodies in R . There we had a sector xoy , with vertex at o , and ofangle in (0 , π ), with x, y ∈ ∂K , and we had two supporting lines of K at x and y , respectively, intersecting in the half-plane bounded by aff { x, y } not containing K . Thus the sector contained [ x, o, y ] and was contained in the convex quadranglewith two sides [ o, x ] and [ o, y ], and other two sides lying on the given supportinglines. This quadrangle could be an arbitrary convex quadrangle, up to affinities.(Observe that the only affine invariants of a convex quadrangle are the two ratiosin which the two diagonals bisect each other.) Below we will use a convex deltoid,that is up to affinities characterized by the fact that one diagonal bisects the otherone in its midpoint. However, this will be sufficient for our theorems. In fact wecannot solve the general case about the maximum polar area, this remains an open K. J. B ¨OR ¨OCZKY, E. MAKAI, JR. question.Here we proceed analogously. First we maximize the area product for an angulardomain of angle 2 α ∈ (0 , π ), say ∠ (cos α, − sin α ) o (cos α, sin α ), where we supposethat the convex sector contains the triangle T := [(cos α, − sin α ) , o, (cos α, sin α )],and is contained in the deltoid Q := [(cos α, − sin α ) , o, (cos α, sin α ) , (1 / cos α, S at (cos α, ± sin α ) contain the sides [(cos α, ± sin α ) , (1 / cos α, T K ) ∗ = ( T ∗ ) − K ∗ for a non-singular linear transformation T .For this we need Steiner symmetrization. For an o -symmetric planar convex body K , and a line l through o , translate each chord of K orthogonal to l (including theones that degenerate to points) orthogonally to l in such a way that the midpointof the translated chord should lie on l . The union of these translates is the Steinersymmetral K ′ of K with respect to l . Clearly K ′ is an o -symmetric planar convexbody with V ( K ′ ) = V ( K ). According to K.M. Ball’s PhD thesis [...] (see also M.Meyer, A. Pajor [...]), we have(P1) V (( K ′ ) ∗ ) ≥ V ( K ∗ ) , with equality if and only if K ′ is a linear image of K , by a linear map preservingall straight lines orthogonal to l . (A similar statement holds, without o -symmetry,for V (cid:0) ( C − s ( C )) ∗ (cid:1) , with “linear” replaced by “affine”, cf. ???).For the following proposition we need some notations.Let α ∈ (0 , π/ a := (cos α, − sin α ) , b := (1 / cos α, , c := (cos α, sin α ).Let T := [ o, a, c ] and Q := [ o, a, b, c ]. Then a, c ∈ S and aff { a, b } and aff { b, c } aretangents to S . Let K be an o -symmetric convex body with a, c ∈ ∂K , and let thecounterclockwise arcs I := c ( − a ) and − I = ( − c ) a of S be contained in ∂K . Then[ I, − I ] ⊂ K ⊂ B ∪ [ a, b, c ] ∪ [ − a, − b, − c ] . Then we have also [ I, − I ] ⊂ K ∗ ⊂ B ∪ [ a, b, c ] ∪ [ − a, − b, − c ] . Let C := K ∩ Q and C ∗ := K ∗ ∩ Q . Observe that ∂C ∗ also contains I ∪ ( − I ),and that T ⊂ C ⊂ Q . Further, we have that K or K ∗ is the union of C ∪ ( − C )or C ∗ ∪ ( − C ∗ ), and the two sectors [ ± I, o ] of B . Observe that we have V ( C ) ∈ [cos α sin α, tan α ]. Both the minimal and maximal values of V ( C ) are attained fora unique K : namely for K = [ I, − I ] and for K = B ∪ [ a, b, c ] ∪ [ − a, − b, − c ]. Stillobserve(P2) V ( K ) = 2 V ([ o, I ]) + 2 V ( C ) and V ( K ∗ ) = 2 V ([ o, I ]) + 2 V ( C ∗ ) . Hence maximization of V ( C ∗ ) is equivalent to maximization of V ( K ∗ ). Proposition.
With the above notations, let V ( C ) ∈ (cos α sin α, tan α ) be fixed. LANAR BLASCHKE-SANTAL ´O INEQUALITY 5
Then the maximum of V ( C ∗ ) occurs, e.g., for the following cases.(i) for V ( C ) < α e.g., for ∂C being the union of [ o, a ] ∪ [ o, c ] and an ellipsoidal arc J in B joining a and c (in the positive sense), the ellipse having as centre o ;(ii) for V ( C ) = α e.g., for C = B ∩ Q ;(iii) for V ( C ) > α e.g., for ∂C being the union of [ o, a ] ∪ [ o, c ] and two segmentsof equal length [ a, a ′ ] ⊂ aff { a, b } and [ c, c ′ ] ⊂ aff { c, b } , and an ellipsoidal arc J joining a ′ and c ′ (in the positive sense), the ellipse having as centre o , and havingaff { a, b } and aff { b, c } as supporting lines. Observe that for C of the form (i) or (iii) we have that C ∗ is of the form in (iii)or (i), respectively (for suitable areas). Proof. We begin with Steiner’s symmetrization of K with respect to the x axis,obtaining K ′ . Observe that then K ∩ [ I, − I ] remains invariant, and K ∩ [ a, b, c ] willbe replaced by K ′ ∩ [ a, b, c ], which is just the Steiner symmetral of K ∩ [ a, b, c ] withrespect to the x axis. We introduce the notation C ′ := [ o, ( K ′ ∩ [ a, b, c ])] . Then(P3) (cid:26) V ( K ) = 2 V ([ o, I ]) + 2 V ( C ) = V ( K ′ ) = 2 V ([ o, I ]) + 2 V ( C ′ ) and V ( K ∗ ) = 2 V ([ o, I ]) + 2 V ( C ∗ ) = V (( K ′ ) ∗ ) = 2 V ([ o, I ]) + 2 V (( C ′ ) ∗ ) . Hence maximization of V ( C ∗ ) is equivalent to maximization of V ( K ∗ ). Still observethat by the equality case of (P1) V ( K ∗ ) can attain its maximum only in the casewhen K ′ = K , since the only linear map preserving vertical lines and taking [ I, − I ]to itself is the identity.Therefore we may suppose that K and thus also C is symmetric with respect tothe x axis. In particular,(P4) (cid:26) [ a, b ] ∩ ∂K and [ b, c ] ∩ ∂K will be symmetric images of eachother with respect to the x axis, thus they have the same length . This length can be 0 or positive, but in the second case by the hypothesis V ( C ) ∈ (cos α sin α, tan α ) of the proposition its length is smaller than the length of [ a, b ]. Now let us consider a convex body K satisfying the hypotheses of the Propo-sition for which V ( K ∗ ) is maximal, and the body C corresponding to this K . Thenby both K and C are necessarily symmetric with respect to the x -axis. We havetwo cases.(i) [ a, b ] ∩ ∂K (and then also [ b, c ] ∩ ∂K ) has a positive length.(ii) [ a, b ] ∩ ∂K (and then also [ b, c ] ∩ ∂K ) is a point.For this K and C let us inscribe to [( ∂C ) \ T ] ∪ { a, c } a convex polygonal arc x . . . x n , in positive orientation, in the following way. We have x = a and x n = c .In case (i) the segment [ a, b ] ∩ ∂K (and [ b, c ] ∩ ∂K ) should be [ x , x ] (or [ x n − x n ],respectively). The further points x , . . . , x n − in case (i) or x , . . . , x n − in case (ii)are placed so that they divide the counterclockwise arc \ x . . . x n − or \ x . . . x n K. J. B ¨OR ¨OCZKY, E. MAKAI, JR. of ∂C to subarcs of equal length. This guarantees symmetry of our polygonalarc with respect to the x axis. Since the length of the arc [ ∂C \ T ] ∪ { a, c } is atmost the length of the arc c abc , therefore the length of the subarcs \ x i x i +1 , excepting [ x x and \ x n − x n in case (i), is O (1 /n ).We write (cid:26) K n := [ I, − I ] ∪ conv { x , . . . , x n } ∪ conv {− x , . . . , − x n } and C n := [ o, K n ∩ [ a, b, c ]] . Then K n → K and C n → C in the Hausdorff metric in both cases (i) and (ii). Infact, we have on one hand K n ⊂ K and C n ⊂ C . On the other hand, if some p ∈ C is separated from C n by a side x i x i +1 (in case (i) this cannot be x x or x n − x n ),then min {| x i p | , | x i +1 p |} ≤ ( | x i p | + | x i +1 p | ) / ≤ | \ x i x i +1 | / O (1 /n ) . Hence the Hausdorff distance of K and K n , as well as that of C and C n is O (1 /n ).Now we define ˜ K n := [ I, − I, J, − J ], where J is a strictly convex polygonal arc J := ˜ x , . . . , ˜ x k which has the following properties. We have k ≤ n . Further, J liesin [ a, b, c ] and satisfies x = a and x k = c , and in case (i) still [ x , x ] ⊃ [ a, b ] ∩ ∂K and [ x k − , x k ] ⊃ [ b, c ] ∩ ∂K , and V ([˜ x , . . . , ˜ x k ]) = V ([ x , . . . , x n ]). Lastly, J issuch a one among all polygonal arcs satisfying all above listed properties, for which V (cid:16) ( ˜ K n ) ∗ (cid:17) attains its maximal value. (By a (strictly) convex polygonal arc wemean one for which the sides following each other turn (strictly) counterclockwise.)Observe that till now we could not exclude [ ˜ x , ˜ x ] = [ a, b ] ∩ ∂K and [˜ x k − , ˜ x k ] =[ b, c ] ∩ ∂K in either case (i) or case (ii), and also we do not yet know k = n .Then V (( K n ) ∗ ) ≤ V (cid:16) ( ˜ K n ) ∗ (cid:17) . Then for some subsequence ˜ K n ( i ) of the ˜ K n ’sthere exists a limit convex body, and also the equal lengths of the intersections˜ K n ( i ) ∩ [ a, b ] and ˜ K n ( i ) ∩ [ b, c ] are convergent. Then also V (cid:16) (lim ˜ K n ( i ) ) ∗ (cid:17) will havea maximal value among the considered convex bodies K . Now we want to apply the method of proof of Theorem A. Its part consistedof three main steps. First we showed k = n . Then we showed (i) and (ii) from theproof of Theorem A.Now, rather than (i) and (ii) from we have to distinguish between the followingcases.(i’) ˜ x ∈ ( a, b ) and ˜ x k − ∈ ( b, c ), and(ii’) ˜ x ( a, b ) and ˜ x k − ( b, c ). Observe that by strict convexity, in case (i’) wehave ˜ x , . . . , ˜ x n − ∈ int [ a, b, c ] , while in case (ii’) we have ˜ x , . . . , ˜ x n − ∈ int [ a, b, c ] . As soon as some vertex ˜ x i lies in int [ a, b, c ], it is freely movable till some smalldistance. Therefore the statements in the proof of Theorem A, , which used smallmovements of such x i ’s, remain valid also here. LANAR BLASCHKE-SANTAL ´O INEQUALITY 7
In particular, (i) of Theorem A, remains true for any three consecutive vertices˜ x i , ˜ x i +1 , ˜ x i +2 , for 2 ≤ i ≤ k − ≤ i ≤ k − k = n and (ii) in the proof of Theorem A, the areaof K was not preserved, so we have to take more care in the proof here. We begin with the proof of k = n . In the proof of Theorem A, we cut offfrom K a part of area Θ( ε ), so for keeping the area constant, we have to give thisarea back at some other place.Namely, supposing k < n consider a straight line l such that ˜ K n ∩ l = { ˜ x i +1 } .Then push parallelly l towards o through a small distance ε = ε ( ˜ K n ) >
0, obtain-ing l ′ . Then consider the o -symmetric convex body ( ˜ K n ) ′ ( ⊂ ˜ K n ) obtained from˜ K n by cutting ˜ x i +1 and − ˜ x i +1 from ˜ K n by l ′ and by − l ′ . Then ( ˜ K n ) ′ has in theclosed angular domain ∠ aoc k + 1 ≤ n vertices, and V ( ˜ K n ) − V (cid:16) ( ˜ K n ) ′ (cid:17) = Θ( ε ).Moreover, ( ˜ K n ) ′ has vertices ˜ x i , ˜ x ′ i +1 , ˜ x ′′ i +1 , ˜ x i +2 in positive sense. Now fix ˜ x i andaff { ˜ x ′ i +1 , ˜ x ′′ i +2 } , and rotate the side line aff { ˜ x i , ˜ x ′ i +1 } about ˜ x i outwards throughan angle Θ( ε ) (and making the analogous change also at − ˜ x i ) so that for the new o -symmetric convex body ( ˜ K n ) ′′ , that has also k + 1 vertices in the closed angulardomain ∠ aoc we should have ( ˜ K n ) ′′ ⊃ ( ˜ K n ) ′ and V (cid:16) ( ˜ K n ) ′′ (cid:17) = V ( ˜ K n ). Now weturn to the polar bodies. By ˜ K n ⊃ ( ˜ K n ) ′ ⊂ ( ˜ K n ) ′′ we have ( ˜ K n ) ∗ ⊂ (cid:16) ( ˜ K n ) ′ (cid:17) ∗ ⊃ (cid:16) ( ˜ K n ) ′′ (cid:17) ∗ . Then ( ˜ K n ) ∗ has neighbouring vertices (˜ x i ˜ x i +1 ) ∗ and (˜ x i +1 ˜ x i +2 ) ∗ , con-nected by the side on the line ˜ x ∗ i +1 . Then (cid:16) ( ˜ K n ) ′ (cid:17) ∗ will have a new vertex( l ′ ) ∗ , at a distance Θ( ε ) from ˜ x ∗ i +1 , hence V (cid:16)(cid:16) ( ˜ K n ) ′ (cid:17) ∗ (cid:17) − V (cid:16) ( ˜ K n ) ∗ (cid:17) = Θ( ε ).The rotation of the side line aff { ˜ x i , ˜ x ′ i +1 } about ˜ x i through an angle O ( ε ) im-plies motion of (aff { ˜ x i , ˜ x i +1 } ) ∗ on the line (˜ x i ) ∗ through a distance O ( ε ), hence V (cid:16)(cid:16) ( ˜ K n ) ′ (cid:17) ∗ (cid:17) − V (cid:16)(cid:16) ( ˜ K n ) ′′ (cid:17) ∗ (cid:17) = O ( ε ). Therefore, V (cid:16)(cid:16) ( ˜ K n ) ′′ (cid:17) ∗ (cid:17) = V (cid:16) ( ˜ K n ) ∗ (cid:17) + Θ( ε ) − O ( ε ) > V (cid:16) ( ˜ K n ) ∗ (cid:17) , for ε > k = n . For the proof of (ii) in the proof of Theorem A, (by dualizing (i) in the proofof Theorem A), the area of K was not preserved, even just conversely, the area ofthe polar was preserved. Therefore we have to give a new proof, which preservesthe area of K .So we consider four consecutive vertices ˜ x i , ˜ x i +1 , ˜ x i +2 , ˜ x i +3 of ˜ K n , and m is theintersection point of aff { ˜ x i , ˜ x i +1 } and aff { ˜ x i +2 , ˜ x i +3 } . Then m is a finite point,separated from ˜ K n by aff { ˜ x i +1 , ˜ x i +2 } .We may suppose that the lines aff { ˜ x i , ˜ x i +1 } , aff { ˜ x i +1 , ˜ x i +2 } and aff { ˜ x i +2 , ˜ x i +3 } are tangent to S , and also that m lies on the positive x axis. So aff { o, m } isthe x axis. Then aff { ˜ x i , ˜ x i +1 } and aff { ˜ x i +2 , ˜ x i +3 } are symmetric with respect to K. J. B ¨OR ¨OCZKY, E. MAKAI, JR. the x axis. Then of course, also their polars (aff { ˜ x i , ˜ x i +1 } ) ∗ and (aff { ˜ x i +2 , ˜ x i +3 } )are symmetric with respect to the x axis, thus(P5) (cid:26) the line connecting (aff { ˜ x i , ˜ x i +1 } ) ∗ and (aff { ˜ x i +2 , ˜ x i +3 } ) ∗ is vertical . We may assume, for contradiction, that (˜ x i +1 + ˜ x i +2 ) / x axis.We begin with rotating the side line aff { ˜ x i +1 , ˜ x i +2 } about the midpoint (˜ x i +1 +˜ x i +2 ) / x i +1 , ˜ x i +2 ], through some small angle ε > of theproof of Theorem A.) The new positions of ˜ x i +1 and ˜ x i +2 are denoted by (˜ x i +1 ) ′ and (˜ x i +2 ) ′ . The body obtained by this change is denoted by ( ˜ K n ) ′ . Then we have V (cid:16) ( ˜ K n ) ′ (cid:17) − V ( ˜ K n ) = O ( ε ).Of course we have to still to restore the original area. This is done by a parallelreplacement of the already rotated side line through a distance O ( ε ). The newpositions of (˜ x i +1 ) ′ and (˜ x i +2 ) ′ will be denoted by (˜ x i +1 ) ′′ and (˜ x i +2 ) ′′ , and thethus obtained body will be denoted by ( ˜ K n ) ′′ .Observe that the centre of rotation (˜ x i +1 + ˜ x i +2 ) / m, s, t ] \ B ⊂ [ m, s, t ], where s and t are the points of tangency of S with the sidelines aff { ˜ x i , ˜ x i +1 } and aff { ˜ x i +2 , ˜ x i +3 } . Therefore (˜ x i +1 + ˜ x i +2 ) / x >
0. By hypothesis, it also lies in the open lower half-plane given by y <
0. So it lies in the open fourth coordinate quadrant. Therefore(P6) the slope of the polar line [(˜ x i +1 + ˜ x i +2 ) / ∗ lies in (0 , ∞ ) . Now we consider the polars. We begin with the first part of the motion, i.e.,with the rotation of the side line about the side midpoint. Rotation of the side lineaff { ˜ x i +1 , ˜ x i +2 } about the midpoint (˜ x i +1 + ˜ x i +2 ) / x i +1 , ˜ x i +2 ] impliesmoving the polar point (aff { ˜ x i +1 , ˜ x i +2 } ) ∗ on the polar line [(˜ x i +1 + ˜ x i +2 ) / ∗ , sothat this point moves counterclockwise, when looked upon from o . Therefore by(P5) its x coordinate increases, by a quantity Θ( ε ).The polar body ( ˜ K n ) ′ has consecutive vertices (aff { ˜ x i , ˜ x i +1 } ) ∗ , (aff { (˜ x i +1 ) ′ , (˜ x i +2 ) ′ } ) ∗ and (aff { ˜ x i +2 , ˜ x i +3 } ). The first and third of these vertices are fixed,and their connecting line is vertical, by (P5). The second vertex lies on the righthand side of this vertical line, and in its new position ((aff { ˜ x i +1 , ˜ x i +2 } ) ′ ) ∗ its x coordinate is greater than in its original position (aff { ˜ x i +1 , ˜ x i +2 } ) ∗ , namely thedifference is Θ( ε ). Therefore(P7) V ( K ′ ) ∗ ) − V ( K ∗ ) = Θ( ε ) . Now we consider the second part of this motion, i.e., the parallel displace-ment of the already rotated side, through a distance O ( ε ). Then the vertices(aff { ˜ x i , ˜ x i +1 } ) ∗ , and (aff { ˜ x i +2 , ˜ x i +3 } ) ∗ remain fixed, but the vertex (aff { (˜ x i +1 ) ′ , LANAR BLASCHKE-SANTAL ´O INEQUALITY 9 (˜ x i +2 ) ′ } ) ∗ moves to its new position (aff { (˜ x i +1 ) ′′ , (˜ x i +2 ) ′′ } ) ∗ , and the distanceof these last two points is O ( ε ). Hence(P8) V (( K ′′ ) ∗ ) − V (( K ′ ) ∗ ) = O ( ε )Then (P7) and (P8) imply(P9) V (( K ′′ ) ∗ ) = V ( K ∗ ) + Θ( ε ) + O ( ε ) = V ( K ∗ ) + Θ( ε ) > V ( K ∗ ) , that is a contradiction. This proves (ii) in the proof of Theorem A, . By , and we have that the extremal polygonal line x . . . x n has exactly n vertices, and both (i) and (ii) from the proof of Theorem A, hold.Recall the cases (i’) and (ii’) from . From , in case (i’) we have˜ x , . . . , ˜ x n − ∈ int [ a, b, c ] , while in case (ii’) we have ˜ x , . . . , ˜ x n − ∈ int [ a, b, c ] . Then, like in the proof of Theorem A, we obtain that in case (i’) ˜ x . . . ˜ x n − ,while in case (ii) ˜ x . . . ˜ x n is inscribed to an elliptical arc, where the respectiveellipse E n has centre o , and, in the metric of E n , the rotation carrying x to x in case (i), or carrying x to x in case (ii) also carries all x i to x i +1 , for3 ≤ i ≤ n −
2, or 2 ≤ i ≤ n −
2, respectively. Since any two metrics on R areequivalent, therefore changing the original metric of R to the metric determined by E n preserves lengths and areas, up to an at most constant positive factor. Thereforewe may use henceforward for fixed n the metric of E n .Now let us replace the polygonal arc ˜ x . . . ˜ x n − in case (i’), and ˜ x . . . ˜ x n − incase (ii’) by the respective arc of E n (and similarly for their mirror images withrespect to o ). Then in both cases we obtain a convex body ˜ L n , since the chords[˜ x , ˜ x ] and [˜ x n − , ˜ x n − ] in case (i’), and [˜ x , ˜ x ] and [˜ x n − , ˜ x n ] in case (ii’) spanlines intersecting in [ a, b, c ].Then we have ˜ K n ⊂ L n , and the Hausdorff distance of ˜ K n and L n is O (1 /n ) = o (1). For a suitable subsequence n ( i ) (cf. ) we have that(a) ˜ K n is convergent, and then necessarily ˜ L n ( i ) has the same limit, andwe have case (i’) for all n ( i ) or case (ii’) for all n ( i ), and(b) in case (i’) the points ˜ x ∈ ( a, b ) and ˜ x n − ∈ ( b, c ) are convergent, and also(c) E n ( i ) is convergent.Therefore we claim that we may use the metric of any E n , or the metric of E :=lim i →∞ E n ( i ) , the changes in lengths and areas are still bounded by an at mostconstant factor.For this we have to show that all E n ’s lie in a compact family. All of them passthrough all four vertices ( ± cos α, ± sin α of an axis-parallel rectangle, hence haveaxisparallel axes themselves. The positive horizontal semiaxes have endpoints inint [ a, b, c ], hence are bounded from below and from above. The positive vertical semiaxes are bounded from below by the y -coordinate of the vertex ˜ x in case(i’) and of the vertex ˜ x in case (ii’). In case (i’) this is fixed. Also in case (ii’)these cannot be arbitrarily small, since then the areas of [˜ x , . . . , ˜ x n ] would tendto V ([ a, b, c ]), which was excluded. Also they cannot be arbitrarily large. Namelythen the chords of E n from the endpoints of their vertical semiaxes with positive(negative) y -coordinates to the endpoints with positive x -coordinates and positive(negative) y -coordinates of the above considered elliptical arcs span lines on whoseleft hand side lies [˜ x , . . . , ˜ x n ]. Then the areas of [˜ x , . . . , ˜ x n ] would tend to 0, whichwas also excluded.Therefore the limit of the ˜ L n ( i ) ’s exists. It is o -symmetric. In the closed angulardomains [ α, π − α ] and its mirror image with respect to o it is bounded by therespective arcs of S . In the closed angular domains [ − α, α ] it is bounded in case (i’)by the segment [lim ˜ x , lim ˜ x ], its mirror image w.r.t. the x -axis, and an ellipticalarc symmetric w.r.t. the x -axis, while in case (ii’) only with the hyperbola arcsymmetric w.r.t. the x -axis. In case (i’) possibly lim ˜ x = lim ˜ x — this case wecount to case (ii’).In case (ii’) the area of the convex hull of the arc of ellipse (that is equal to V ([˜ x , . . . , ˜ x n ]) for each n ) uniquely determines the arc of ellipse. Namely theboundary of the ellipse passes through the four points ( ± cos α, ± sin α ). Thereforeany two of them does not have any further intersection points, so their parts inthe open angular domain ( − α, α ) are disjoint, hence they are linearly ordered byinclusion. The arc giving the minimal area is [ a, c ], which however cannot beattained by the restriction on the area. We assert that the maximal of them is thearc of S in this angular domain. Namely any two such ellipses have a transversalintersection point at a (and c ). Else they would have eight common points withmultiplicities, hence would be equal. In particular, the arc of S in this angulardomain and any elliptical arc passing through the four points ( ± cos α, ± sin α )have different tangents at a . In case of strict inclusion of the sector of B and therespective sector of the elliptical arc the tangent of the elliptical arc at a wouldpoint outside from the quadrangle Q , that is impossible, since it lies in Q .In case (i’) the situation is more complicated. Then we have on the boundaryof ˜ L segments of equal positive lengths [˜ x , ˜ x ] and [˜ x n − , ˜ x n ]. On ∂L , at ˜ x andits mirror image w.r.t. the x -axis there joins to these segments an arc of an ellipse,also symmetric w.r.t. the x -axis. Now we have two parameters: the positive lengthof the segment [˜ x , ˜ x ], and one more parameter that distinguishes the ellipticalarc among all those elliptical arcs that pass through the four points ± ˜ x and theirsymmetric images w.r.t. the x -axis.However, here is still one restriction. The body L cannot have a non-smoothpoint at ˜ x . This can be proved similarly as formerly already several times. Namely,we put a line l that is a supporting line of L at ˜ x , but is not equal to any of thehalf-tangents at ˜ x . Now let ε > l inward to L (and similarly at all three symmetric images of ˜ x ). The loss of area is Θ( ε ). Sincethe area of K must remain constant, we have to give this loss of area somewhereback. For this consider the elliptical arc in question, and close to the midpoint ofthis elliptical arc we choose a point p on the positive x -axis outside of L , having adistance δ from L . We consider the convex hull [ L, p − p ]. This has area LANAR BLASCHKE-SANTAL ´O INEQUALITY 11 V ( L )+Θ( δ / ). Similarly, [ L, p − p ] ∗ has an area V ( L ∗ )+Θ( δ / ). Then choosing δ suitably, the added area will be exactly equal to the lost area, so the area of K willnot change by performing these two operations. At the same time, δ / = Θ( ε ).By cutting off K with l (and its three symmetric images) K ∗ will change to theconvex hull of the original K ∗ and l ∗ and its three symetric images. However, thismeans an increase in the area by Θ( ε ). Then cutting the new K ∗ by the line p ∗ (andits three symmetric images) causes the area to decrease with Θ( δ / ) = Θ( ε ). Thatis, the original V ( K ∗ ) changed to V ( K ∗ ) + Θ( ε ) + Θ( ε ) = V ( K ∗ ) + Θ( ε ) > V ( K ∗ )a contradiction.This contradiction implies that the extremal body L is smooth at ˜ x (and itsthree symmetric images). This means that it is already uiniquely determined (bypassing through the point ˜ x and having given tangents there — namely two suchellipses would have eight common points with multiplicities, which is a contradic-tion). Observe that the polars of such arcs, consisting of two symmetrically placedsegments and an arc of a hyperbola, are just elliptical arcs of the form in case (ii’),hence the corresponding (polar) sectors are totally ordered by inclusion. Therefore,for a given area of the sector of K in the closed angular domain aob the part of theboundary in this angular domain is uniquely determined. By the upper inequalityon the area of the sector of K in this angular domain the degenerate position, whenthe sector of K would coincide with the deltoid [ o, a, b, c ], cannot be attained.We obtain case (ii) of the Proposition when both (i’) and (ii’) hold. (cid:4) The
John ellipse E i ( K ), or the L¨owner ellipse E o ( K ) of an o -symmetric convexbody K ⊂ R d is a the o -symmetric ellipsoid of maximal volume contained in K ,or of minimal volume containing K , respectively. Both of them are unique, andthe polar of the John ellipsoid of K is the L¨owner ellipsoid of K ∗ (using duality).By [Be], B is the John ellipse or the L¨owner ellipse of K ⊂ R if and only if itis contained in K , or contains K , and ( ∂K ) ∩ S contains the vertices of a squarewith vertices p , ..., p in positive cyclic order, inscribed to S , or the vertices of an o -symmetric convex hexagon with vertices p , ..., p , in positive cyclic order, where ∠ p i op i +1 < π/ Theorem B. If B is the John ellipse or the L¨owner ellipse of a planar o-sym-metric convex body K , then V ( K ) + V ( K ∗ ) ≤ π. Remark.
This inequality is specific to o -symmetric planar convex bodies. If ei-ther K is a regular triangle inscribed into S , or or a regular cross-polytope cir-cumscribed about S d − (for the John ellipsoid) or a cube inscribed to S d − (forthe L¨owner ellipsoid) of dimension d ≥
3, the analogous statement (i.e., when π isreplaced by the volume of the unit ball B d in R d ) does not hold.In fact, for K a regular triangle inscribed to S we have V ( K ) + V ( K ∗ ) =15 √ / . ... > π . Further, observe that the John ellipsoid of a regularcross-polytope D d , circumscribed about B d , is B d (by the criterion of John [John]).Then, by polarity, the L¨owner ellipsoid of a cube C d , inscribed to B d , is B d . For d = 3,we have V ( C ) + V ( D ) = 44 √ / . ... > V ( B ) = 8 . ... . For d = 4 , V ( D ) = 10 . ... > . ... = 2 V ( B ) and V ( D ) = 14 . ... > . ... = 2 V ( B ), so V ( D d ) > V ( B d ). Further, using the recursion formulas,one easily sees that V ( D d ) / (cid:0) V ( B d ) (cid:1) is increasing, separately for even, and for odd d ≥
4, that proves our claim. (For comparing the cases of dimensions d and d + 2,the increasing property is equivalent to (1 + 2 /d ) d/ · [( d + 2) / ( d + 1)] · (2 /π ) > Proof of Theorem B.
We may assume that B is the L¨owner ellipse of K . It issufficient to prove that if p := p i , q := p i +1 ∈ S ∩ ∂K , and the angle of the vectors p and q is 2 α ∈ (0 , π/ V ( K ∩ S ) + V ( K ∗ ∩ S ) ≤ α for the convex cone S := { tp + sq : t, s ≥ } . Namely, summing all four, or all sixsuch inequalities, we obtain the statement of the theorem.We may suppose p = (1 , r := (0 , ∈ S , and let e S = { tr + sq : t, s ≥ } .For e K pq = [ ± ( S ∩ K ) , ± ( e S ∩ B )] , we have e K ∗ pq = [ ± ( S ∩ K ∗ ) , ± ( e S ∩ B ) , ±√ r − p )] . Therefore (*) is equivalent to V ( e K pq ) + V ( e K ∗ pq ) ≤ π . Let E be an o -symmetric ellipse such that p, r ∈ ∂E , and for the part M betweenthe chords [ − r, p ] and [ − p, r ], we have V ( M ) = V ( e K pq ) ≤ π/
2. By ± p, ± r ∈ ∂E we have V ( E ) ≥ π . It follows from the Proposition that V ( e K ∗ pq ) ≤ V ( M ∗ ), andhence (*) follows from(**) V ( M ∩ S ) + V ( M ∗ ∩ S ) ≤ π/ , where S := S ∪ e S = { tp + sr : t, s ≥ } and V ( A ∩ S ) ≤ π/
4. Thus it suffices toshow ( ∗∗ ), which we are going to do.Let a ≤ b ≥ E . For E = B ( ∗∗ ) is fulfilled, thereforewe may assume a < < b . We choose a new orthonormal system of coordinateswith basic vectors (1 / √ , / √ , ( − / √ , / √ E = Φ B ,for a diagonal matrix Φ = (cid:18) a b (cid:19) . In particular, E ∩ S = Φ σ , where σ ⊂ S is a sector of B , having an acute angle2 α LANAR BLASCHKE-SANTAL ´O INEQUALITY 13 at 0, and being symmetric with respect to the perpendicular bisector of [ p, r ].(For 2 α < π/ V ( E ) > π , and hence V ( E ∩ S ) ≤ π/ < V ( E ) / α, sin α ) = ( a cos α, b sin α ) = (1 / √ , / √ a = 1 / ( √ α ) and b = 1 / ( √ α ), hence tan( α ) = a/b and M ∗ ∩ S = [ p, r, τ ] , where τ is the sector of the polar ellipse E ∗ corresponding to the sector σ of E bypolarity. Hence V ( M ∩ S ) + V ( M ∗ ∩ S ) = abα + αab + 1 − tan α α . In particular, V ( M ∩ S )+ V ( M ∗ ∩ S ) = f ( α ) for α ∈ (0 , π/ f ( π/
4) = π/ f ( α ) = α (sin(2 α ) + 1 / sin(2 α )) + cos(2 α ) . We write β := 2 α and g ( β ) := f ( β/ β ∈ (0 , π/ g ′ ( β ) > β < tan β , for β ∈ (0 , π/ V ( M ∩ S ) + V ( M ∗ ∩ S ) = g ( β ) ≤ g ( π/
2) = π/ , completing the proof of (**), and in turn the proof of Theorem B. (cid:4) Stability of the Blaschke-Santal´o inequality
Theorem D.
Let K be a planar convex body, and s ( K ) the Santal´o point of K .If the Banach-Mazur distance of K from the ellipses is at least ε , where ε > ,then V ( K ) V ( K − s ( K )) ∗ ) ≤ π (1 − cε ) , where c is a positive absolute constant. If additionally K is -symmetric, we have V ( K ) V ( K − s ( K )) ∗ ) = V ( K ) V ( K ∗ ) ≤ π (cid:0) − (8 + o (1)) ε (cid:1) , for ε → .Proof. It follows by [B¨o], Theorem 1.4, together with [MR06], Theorem 1 (and The-orem 13), that there exists an o -symmetric planar convex body e K with V ( e K ) V ( e K ∗ ) ≥ V ( K ) V (cid:0) ( K − s ( K )) ∗ (cid:1) , whose Banach-Mazur distance from the ellipses is at least1 + c ε , where c is a positive absolute constant ???. We may assume for the Johnellipse of ˜ K that E i ( e K ) = B . Then there is a point p of e K of distance at least1 + c ε from o . We deduce that V ( e K ) ≥ π + c / (cid:0) √ /
3) + o (1) (cid:1) ε , for ε → K ) ∗ has a supporting line p ∗ at distance at most 1 − ( c + o (1)) ε from o , hence V (cid:16) ˜ K ∗ (cid:17) ≤ π − c / (cid:0) √ /
3) + o (1) (cid:1) ε . Therefore Theorem B yields V ( ˜ K ) V (cid:16) ( ˜ K ) ∗ (cid:17) = (cid:16) V ( ˜ K ) + V (cid:16) ( ˜ K ) ∗ (cid:17)(cid:17) − (cid:16) V ( ˜ K ) − V (cid:16) ( ˜ K ) ∗ (cid:17)(cid:17) ≤ π − (cid:16)(cid:16) √ c / + o (1) (cid:17) ε (cid:17) . The o -symmetric case follows similarly. (cid:4) Remark.
A somewhat weaker result is given by K. M. Ball, K. J. B¨or¨oczky [BB],Theorem 2.2, case of R (observe that there n ≥ n ≥ | log ε | can in fact bedoubled at the application of (3) and (4) there, for the o -symmetric case.) Underthe conditions of Theorem D, except that K ⊂ R n , we have V ( K ) V ( K ∗ ) ≤ κ n (1 − const n ε n +3 | log ε | − ) , and for the o -symmetric case V ( K ) V ( K ∗ ) ≤ κ n (1 − const n ε (3 n +3) / | log ε | − ) . For n = 2 these give V ( K ) V ( K ∗ ) ≤ π (1 − const ε | log ε | − ) , and V ( K ) V ( K ∗ ) ≤ π (1 − const ε / | log ε | − ) . As stated in [BB], we cannot have a better upper estimate, even in the o -symmetriccase, than κ n (1 − const n ε ( n +1) / ).For the planar o -symmetric case we can let K := { ( x, y ) ∈ B | | x | ≤ − ε .Then the John ellipse E i ( K ) of K has, by Behrends’ characterization, the equa-tion ( x/ (1 − ε )) + y = 1. Possibly the Banach-Mazur distance is attained for E i ( K ) and that inflated copy λE i ( K ) of it that passes through the four non-smooth points ( ± (1 − ε, ±√ ε − ε of K . If this were true, we would have λ =((1 − ε ) / (1 − ε )) +( √ ε − ε ) , hence λ − ∼ ε . Also we have V ( K ) V ( K ∗ ) = π − (cid:0) π · √ o (1) (cid:1) ε / /
3. Observe that the same Banach-Mazur distance is attainedif we consider the intersection K ′ of B with an o -symmetric square Q of sidelengths1 − ε , that follows from the fact that the John and L¨owner ellipses of K ′ are the incir-cle of Q and B , respectively. But ( V ( K ′ ) V ( K ′ ) ∗ ) = π − (cid:0) π · √ o (1) (cid:1) ε / / K ′ is worse than K . Similarly, for the general case the intersection of B witha regular triangle of sides at a distance 1 − ε from o is probably worse than thebody K := { ( x, y ) ∈ B | x ≤ − ε . Whether K and K could be conjectured asoptimal, is not clear: possibly some truncation of B with some other curve(s) canbe better. About the stability of the Santal´o point
LANAR BLASCHKE-SANTAL ´O INEQUALITY 15
Let us review some basic formulas about polar bodies, due to Santal´o and Meyer-Pajor. Let M be a convex body in R d . The support function of M is h M ( u ) = max x ∈ M h u, x i . In particular, if z ∈ int M , then the support function of M − z is h M ( u ) − h u, z i ,and the radial function of ( M − z ) ∗ at u ∈ S d − is ( h M ( u ) − h u, z i ) − . It followsthat (with V ( · ) denoting volume) V (( M − z ) ∗ ) = d − Z S d − ( h M ( u ) − h z, u i ) − d du. Here V (( M − z ) ∗ ) is a strictly convex analytic function of z ∈ int M , that tendsto infinity as dist ( z, ∂M ) →
0. In fact, its second differential is a positive definitequadratic form. E.g., (cid:18) ∂∂x (cid:19) V (( M − z ) ∗ ) = ( d + 1) Z S d − u ( h M ( u ) − h z, u i ) − d du > , and the respective foirmula holds for thecsecind derivetive along any direction.Hence it has a unique minimum at some s ( M ) ∈ int M , which is called the Santal´opoint s ( M ) of M . Differentiation yields that Z S d − u ( h M ( u ) − h s ( M ) , u i ) d +1 du = o . Thus R ( M − s ( M )) ∗ y dy = o , and hence o is the center of mass of ( M − s ( M )) ∗ . Thisimplies that − ( M − s ( M )) ∗ ⊂ d ( M − s ( M )) ∗ , hence(0) − ( M − s ( M )) ⊂ d ( M − s ( M )) . There are known several statements about stability of the Santal´o point, orbehaviour of V ( K − z ) ∗ for z close to the Santal´o point s ( K ) of K . See e.g.,Santal´o [San], , pp. 156-157, Kim-Reisner [KR], Propositions 1 and 2, and thefirst part of this paper [BMMR], Lemma 11, first statement (about c ( K ) — thesecond statement, about c ( K ), will not be needed here).Now we cite [BMMR], Lemma 11, first part. [BMMR], Lemma 11 becomes cor-rect.) This is a more explicit version of Kim-Reisner [KR], Proposition 1, inasmuchthe constants are explicitly given. This will render it possible to give estimates withabsolute constants (for fixed dimension). Lemma E’. ( [BMMR] , Lemma 11, first part) Let d ≥ be an integer, K ⊂ R d be a convex body, and let < ε ≤ ε ( K ) := min { / , − d − (cid:0) κ d − / ( dκ d ) (cid:1) ·| K | / (diam K ) d } . Let K ⊂ R d be a convex body, and let (1 − ε ) K + a ⊂ K ⊂ (1 + ε ) K + b, where a, b ∈ R d . Additionally, let (E2) from above hold. Then (E2.5) k s ( K ) − s ( K ) k ≤ c ( K ) · ε , where (E3) c ( K ) := (diam K ) ( d +1) | K | − d − · d ( dκ d /κ d − ) d +2 . Theorem E.
Let d ≥ be an integer. Then there exists an ε ( d ) > such that forall ε ∈ [0 , ε ( d )) the following holds. Let K , K ⊂ R d be convex bodies, and let (1 − ε ) K + a ⊂ K ⊂ (1 + ε ) K + b, where a, b ∈ R d and [((1 − ε ) K + a ) + ((1 + ε ) K + b )] / K . Then we have s ( K ) ∈ int K , and ( V (( K − s ( K )) ∗ ) − V (( K − s ( K )) ∗ ) ≤ d +4 d +1 d d +6 d +9 κ − d − d − κ d ( d + 1) d +2 · ε . Proof.
We proceed on the lines of Santal´o [San] and Kim-Reisner [KR], Proposition2 and [BMMR], Lemma 11, second part.Observe that the statement of Theorem E is affine invariant. Therefore we maysuppose, by John’s theorem, that B d ⊂ K ⊂ dB d . Then (E3) remains valid if inthe expression of c ( K ) in Lemma E’ we replace | K | with its lower bound κ d anddiam K by its upper bound 2 d , obtaining(E4) k s ( K ) − s ( K ) k ≤ c · ε , where(E5) c = 2 ( d +1) d d +3 d +4 κ − d − d − . The minimum of f ( z ) := V (( K − x ) ∗ ) for z ∈ int K is attained for z = s ( K ),the Santal´o point of K . The function f ( z ) is analytic, hence has a power series LANAR BLASCHKE-SANTAL ´O INEQUALITY 17 expansion at s ( K ), convergent in any open ball of centre s ( K ) and containedin int K . Let us suppose that s ( K ) − s ( K ) is non-zero (else we have nothingto prove), and it points, say, to the direction of the positive x d -axis. The firstderivatives of f ( z ) vanish at z = s ( K ), hence(E6) V (( K − s ( K )) ∗ ) − V (( K − s ( K )) ∗ ) = f ( s ( K )) − f ( s ( K )) = (1 / k s ( K ) − s ( K ) k · ( ∂/∂x d ) f ( x ) =(1 / k s ( K ) − s ( K ) k · ( d + 1) R S d − u d ( h K − h u, x i ) − d − du, where x lies in the relative interior of the segment [ s ( K ) , s ( K )], and hence k x − s ( K ) k < k s ( K ) − s ( K ) k . We estimate ( ∂/∂x d ) f ( x ) in (E6). We estimate u d from above by 1. Then still we have to estimate h K − h u, x i from below by somepositive number. We have(E7) h K ( u ) − h u, x i = ( h K ( u ) − h u, s ( K ) i ) + ( h u, s ( K ) i − h u, x i ) . So we have to estimate from below ( h K ( u ) − h u, s ( K ) i ) and to estimate fromabove |h u, s ( K ) i − h u, x i| .For the first estimate we recall (0): K − s ( K ) ⊂ − d ( K − s ( K )) , or, equivalently,(E8) h K − s ( K ) ≤ dh s ( K ) − K . This last inequality can be interpreted geometrically as follows. Take any support-ing parallel strip of K . Then for the (positive) distances of s ( K ) from the twoboundary hyperplanes of the parallel strip we have that their quotient is at most d , or, otherwise said, the distance of s ( K ) to any of these boundary hyperplanescannot be less than 1 / ( d + 1) times the width of the parallel strip. We apply thisfor K . Thus the distance of s ( K ) to any supporting hyperplane of K cannot beless than 1 / ( d + 1) times the minimal width of K , which width is by K ⊃ B d atleast 2. Returning to the original formulation, we have that(E9) h K − h s ( K ) , u i ≥ / ( d + 1) . For the second estimate we have, by k u k = 1,(E10) (cid:26) |h u, x i − h u, s ( K ) i| = |h u, x − s ( K ) i| ≤ k x − s ( K ) k < k s ( K ) − s ( K ) k ≤ c ε < c ε ( d ) . Suppose that(E11) c ε ( d ) ≤ / ( d + 1) . Then, using (E7), (E9) and (E10), we get(E12) (cid:26) h K ( u ) − h u, x i = ( h K ( u ) − h u, s ( K ) i ) +( h u, s ( K ) i − h u, x i ) ≥ / ( d + 1) − / ( d + 1) = 1 / ( d + 1) . This implies, by (E6), (E12) and (E4), (E5), that(E13) V [( K − s ( K )) ∗ ] − V [( K − s ( K )) ∗ = (1 / k s ( K ) − s ( K ) k · R S d − u d ( h K ( u ) − h u, x i ) − d − du ≤ (1 / c ε · dκ d [1 / ( d + 1)] − d − , and this is equivalent to the inequality of the Theorem. Optimality of regular n -gons We introduce some notation that will be used in the next section, as well. For x = y ∈ R , we write aff { x, y } for the line passing through x and y . For compactsets X , . . . , X k in R , let [ X , . . . , X k ] denote the convex hull of their union. Forany point p = o , let p ∗ be the polar line with equation h x, p i = 1. For 0 l = p ∗ ,let p = l ∗ . Theorem F.
Among convex polygons K of at most n vertices whose Santal´o pointis o , the ones maximizing the area product V ( K ) V ( K ∗ ) are the non-singular linearimages of regular n -gons with centre at o .Proof. As all triangles are affine images of each other, we may assume n ≥
4. ByTheorem ? we may assume even n ≥
5. Observe that the area product of a regular n -gon, i.e., n sin ( π/n ) is strictly increasing with n . Therefore, using inductionw.r.t. n , with base of induction n = 4, we may restrict our attention to convexpolygons with exactly n sides. By the intersection point of two distinct parallellines, we mean their common point of infinity (in the projective plane).Let K maximize the area product among convex polygons of at most n vertices,whose Santal´o point is o . We prove two basic properties of K :(i) If x , x , x are consecutive vertices of K , then o , ( x + x ) and x are collinear.(ii) If x , x , x , x are consecutive vertices of K , and m is the intersection pointof aff { x , x } and aff { x , x } , then o , ( x + x ) / m are collinear. Here m ismeant as a point of the projective plane.To prove these claims, we show with a method going back to Mahler that ifeither (i) or (ii) does not hold, then K can be deformed into a convex polygon withthe same number of vertices but with a larger area product. We write γ , γ , . . . LANAR BLASCHKE-SANTAL ´O INEQUALITY 19 for positive constants that depend on K , but not on its deformations .First we suppose that (i) does not hold, and seek a contradiction. For i = 1 , y i/ for the intersection of x ∗ i and x ∗ , and hence y . and y . areconsecutive vertices of K ∗ . Let l be the line through x parallel to aff { x , x } , andhence l ∗ ∈ [ y . , y . ]. We consider two cases depending on the position of o withrespect to the triangle [ x , x , x ]. Case 1. o int [ x , x , x ]We may assume that x and x lie on the same side of aff { o, ( x + x ) / } if o [ x , x , x ] (and hence o [ x x x ]), or k x k < k x k if o ∈ [ x , x ]. Inparticular, k l ∗ − y . k > k l ∗ − y . k . We move x into a position ˜ x where ˜ x − x = ε ( x − x ) for small ε > ε is small enough to ensure that x and ˜ x lie on the same side of the linecontaining the other side of K with endpoint x , and if o [ x , x , x ], then still x and ˜ x lie on the same side of aff { o, ( x + x ) / } . Therefore there exists a convexpolygon e K obtained from K by removing x from the set of vertices, and adding˜ x . Clearly V ( e K ) = V ( K ).For i = 1 ,
3, let ˜ y ( i +2) / be the intersection of x ∗ i and ˜ x ∗ , and hence ˜ y . and ˜ y . are the two new vertices of e K ∗ replacing the vertices y . and y . of K ∗ . In addition,˜ y . K ∗ and ˜ y . ∈ K ∗ , because o int [ x , x , x ]. Writing α := ∠ x o ˜ x for theangle of x and ˜ x ???, we have γ ε < α < γ ε . Since V ( ˜ K ∗ ) − V ( K ∗ ) = V ([ l ∗ , y . , ˜ y . ]) − V ([ l ∗ , y . , ˜ y . ]) , and α is the angle of both triangles [ l ∗ , y . , ˜ y . ] and [ l ∗ , y , ˜ y ] at l ∗ ???, we have V ( ˜ K ∗ ) − V ( K ∗ ) > ( k l ∗ − y . k − k l ∗ − y . k ) sin α − γ α > γ ε . It follows by Theorem E that V (( ˜ K − s ( ˜ K )) ∗ ) > V ( ˜ K ∗ ) − γ ε . Therefore if ε > V (cid:16)(cid:16) ˜ K − s ( ˜ K ) (cid:17) ∗ (cid:17) > V ( K ∗ ) + γ ε > V ( K ∗ ), which is acontradiction, and thus we have proved (i) in Case 1. Case 2. o ∈ int [ x , x , x ]We may assume that x and x lie on the same side of aff { o, ( x + x ) / } , andhence k l ∗ − y k < k l ∗ − y k . We move x into a position ˜ x where ˜ x − x = ε ( x − x ) for small ε >
0, and ε is small enough to ensure that there exists a convex polygon e K obtained from K by removing x from the set of vertices, and adding ˜ x . Clearly V ( e K ) = V ( K ).For i = 1 ,
3, let again ˜ y ( i +2) / be the intersection of x ∗ i and ˜ x ∗ , and hence ˜ y . and ˜ y . are the two new vertices of e K ∗ replacing the vertices y . and y . of K ∗ .In this case ˜ y . ∈ K ∗ and ˜ y . K ∗ , thus V ( ˜ K ∗ ) − V ( K ∗ ) = V ([ l ∗ , y . , ˜ y . ]) − V ([ l ∗ , y . , ˜ y . ]) . Now the argument can be finished as in Case 1, completing the proof of (i).Next we suppose that (ii) does not hold. We may assume that(*) (cid:26) [ x , x ] and p = ( x + x ) / o and m . There are three cases. The point m can be a finite point, lying on the other sideof aff { x , x } as K , or can be a point at infinity, or can be a finite point lying onthe same side of aff { x , x } as K . Let l . = aff { x , x } , l . = aff { x , x } and l = aff { x , x } . We observe that if m ∈ R , then h m, l ∗ . i = h m, l ∗ . i = 1, and also h , l ∗ . i = h , l ∗ . i = 0. Therefore the assumption on p yields that(**) h p, l ∗ . i > h p, l ∗ . i holds independently whether m ∈ R , or m is a point at infinity.We rotate l about p into a new position ˜ l through a small angle ε ∈ (0 , π/ l intersects [ x , x ] \ { x } in a point ˜ x , but intersects l . in a point˜ x lying outside of K . In particular, k ˜ l ∗ − l ∗ k > γ ε . Let e K be the convex polygon obtained from K by removing x and x from the setof vertices, and adding ˜ x and ˜ x . Then V ( ˜ K ) = V ( K ) + V ([ p, x , ˜ x ]) − V ([ p, x , ˜ x ]) > V ( K ) − γ ε . Now e K ∗ is obtained from K ∗ by removing l ∗ from the set of vertices, and adding ˜ l ∗ .Here l ∗ , ˜ l ∗ ∈ p ∗ , in all the three cases for m , i.e., if it is a point at infinity, or it is afinite point, lying on either side of aff { x , x } = l . We have three different figures,but the following calculations are valid for each of these three figures. Supposingthat the vertices of K follow each other in the positive sense, then assumption (*)on p yields h ( l ∗ . − l ∗ . ) × (˜ l ∗ − l ∗ ) , (0 , , i >
0, hence by (**) actually h ( l ∗ . − l ∗ . ) × (˜ l ∗ − l ∗ ) , (0 , , i > γ ε (observe that the angle of m ∗ and p ∗ only depends on K , but not on ε ). Therefore LANAR BLASCHKE-SANTAL ´O INEQUALITY 21 V ( ˜ K ∗ ) − V ( K ∗ ) = h ( l ∗ . − l ∗ . ) × (˜ l ∗ − l ∗ ) , (0 , , i / > γ ε . Furthermore, V (( ˜ K − s ( ˜ K )) ∗ ) > V ( ˜ K ∗ ) − γ ε by Theorem E. We conclude that V ( ˜ K ) V (( ˜ K − s ( ˜ K )) ∗ ) > (cid:0) V ( K ) − γ ε (cid:1) (cid:0) V ( K ∗ ) − γ ε (cid:1) > (cid:0) V ( K ) − γ ε (cid:1) (cid:0) V ( K ∗ ) + γ ε − γ ε (cid:1) > V ( K ) V ( K ∗ ) + γ ε > V ( K ) V ( K ∗ ) , provided ε ∈ (0 , π/
2) is small enough. This contradiction proves (ii).Now we prove Theorem F based on (i) and (ii). Let x , . . . , x k , with 3 ≤ k ≤ n be the vertices of K in this order. By the beginning of the proof of this theorem wehave k = n ≥
5. By n ≥ ∠ x i − ox i + ∠ x i ox i +1 is4 π/n < π (indices meant cyclically). Let, e.g., ∠ x ox + ∠ x ox < π . (Observe thatthis property is invariant under linear maps.) Applying a linear transformation, wemay assume that x , x , x ∈ S and hence x lies on the perpendicular bisector of[ x , x ] by (i). Applying another linear transformation, we may also assume that x , x , x ∈ S — actually, they lie in an open half-circle.Now (ii) yields that aff { x , x } and aff { x , x } are symmetric with respect tothe perpendicular bisector of [ x , x ], and m is a finite point separated from K by l .It follows that aff { x , x } and aff { x , x } are symmetric with respect to aff { o, x } .Therefore (i) yields that x ∈ S , with k x − x k = k x − x k = k x − x k .Continuing like this, we conclude that K is a regular k -gon inscribed into B .From the first paragraph of the proof we have k = n . (cid:4) References [BB] Ball, K. M., B¨or¨oczky, K. J.,
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Alfr´ed R´enyi Mathematical Institute,Hungarian Academy of Sciences,H-1364 Budapest, Pf. 127,HUNGARY
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