Banach-Mazur distances between parallelograms and other affinely regular even-gons
aa r X i v : . [ m a t h . M G ] A ug Banach-Mazur distance from the parallelogram tothe affine-regular hexagon and other affine-regulareven-gons
Marek Lassak
University of Science and TechnologyKaliskiego 7, 85-789 Bydgoszcz, Polande-mail: [email protected]
Abstract
We show that the Banach-Mazur distance between the parallelogram and the affine-regular hexagon is and we conclude that the diameter of the family of centrally-symmetricplanar convex bodies is just . A proof of this fact does not seem to be published earlier.Asplund announced this without a proof in his paper proving that the Banach-Mazur distanceof any planar centrally-symmetric bodies is at most . Analogously, we deal with the Banach-Mazur distances between the parallelogram and the remaining affine-regular even-gons. Denote by M d the family of all centrally symmetric convex bodies of the Euclidean space E d centered at the origin o of E d . Below always when we say on a homothety, we mean that itscenter is o . For any K ∈ M d and any positive λ , by λK we denote the homothety image of K with ratio λ .The Banach-Mazur distance (or shortly, the BM-distance) of C, D ∈ M d is defined as δ BM ( C, D ) = inf a,λ { λ ; a ( C ) ⊂ D ⊂ λa ( C ) } , where a stands for an affine transformation. This definition is presented by Banach [3] in behalfof him and Mazur. From the role of the affine transformation in this definition we see that adifferent formulation of this definition is formally more precise, namely when we consider theBanach-Mazur distance of the equivalence classes of centrally symmetric bodies with respect toaffine transformations. But in this note it is more convenient to deal with the BM-distance ofconvex bodies from M d . It is well known that the Banach-Mazur distance is a multiplicativemetric, i.e., that log δ BM is a metric. In particular, we have δ BM ( C, D ) = δ BM ( D, C ) and themultiplicative triangle inequality δ BM ( C, D ) · δ BM ( D, E ) ≥ δ BM ( C, E ) for every
C, D, E ∈ M d .1 Marek Lassak
Later the notion of the BM-distance has been generalized to pairs of arbitrary convex bodies.Fora survey of results on the BM-distance we propose the books by Tomczak-Jaegerman [5], Toth[6] (Part 3.2), and Aubrun and Szarek [2] (Chapter 4).By P n we denote the regular n -gon with vertices (cos jn π, sin jn π ) for j = 0 , . . . , n − δ BM ( P , P ) = and to establish all the optimal positionsof the parallelograms a ( P ) with respect to the hexagon P . Recall that the value is stated inTheorem 1 of [1]. It says that δ BM ( P , C ) ≤ for any C ∈ M with equality only for C = P .But the proof of Lemma 3 of [1] by Asplund that δ BM ( P , P ) = is omitted. This statementis quoted on p. 206 by Stromquist [4] and illustrated in Fig. 1 there. Since the hexagon fromthis Fig. 1 is not inscribed in the larger parallelogram, then looking to the second thesis of ourProposition we get further questions. The basic task of [4] is to construct a center R of M (seep. 207 and compare Fig. 2). With the help of Fig. 3 it is shown that both P and P are in thedistance p / R . This implies δ BM ( P , P ) ≤ , but does not imply the equality. Thesedoubts mobilized the writer to present a detailed proof of the equality δ BM ( P , P ) = in thepresent note in order to be sure that the diameter of M is .An additional aim is explained here. The idea of the proof of our Theorem 1 that δ BM ( P , P ) = encouraged the author to consider the analogous task for all the regulareven-gons with more vertices in place of P . Theorem 2 establishes the Banach-Mazur distancesfrom P to all P j and P j +4 . This task for the remaining odd-gons, so to P j +2 and P j +6 appears to be more complicated. We estimate and conjecture the values of these distances. By an inscribed parallelogram in a convex body C we mean a parallelogram with all verticesin the boundary of C and by a circumscribed parallelogram about C we mean a parallelogramcontaining C whose all sides have non-empty intersections with C .By the width width( S ) of a strip S between two parallel hyperplanes of E d we meanthe distance of these hyperplanes. We omit an easy proof of the following lemma, whose two-dimensional version is applied in the proofs of Theorems 1 and 2. Lemma.
Let H , H − and H , H − be two pairs of hyperplanes of E d symmetric with respectto o such that all of them are parallel and let L be a straight line through o intersecting them.Assume that H , H , H − , H − intersect L in this order. Denote by ( a i , . . . , a in ) the points ofintersection of L with H i + , where i ∈ { , } . Consider the strips S i = conv( H i + ∪ H i − ) , where i ∈ { , } . We claim that width( S ) / width( S ) = a j /a j for any j ∈ { , . . . , n } . The following proposition is applied later for evaluating particular BM-distances. anach-Mazur distance Proposition.
Let C ∈ M . Assume that P ⊂ C ⊂ µP for a parallelogram P ∈ M and apositive µ . Then there exists a parallelogram P ′ inscribed in C such that µ ′ P ′ is circumscribedabout C for a µ ′ ≤ µ .Proof. Since the thesis is obvious for µ = 1, below we assume that µ > P in the part of P ′ fulfills the thesis, there is nothing to prove. In the opposite case weintend to construct the required P ′ .Of course, there is a homothetic image P α ⊂ P of P such that for least one pair ofopposite sides Z + , Z − of P α the sides µZ + , µZ − of µP α touch C from outside. Denote the otherpair of sides of P α by W + , W − and assume that their order is W + , Z + , W − , Z − when we movecounterclockwise.If µW − , µW + do not touch C , then we lessen P α up to P β by moving W + , W − such thattheir vertices remain in Z + , Z − symmetrically closer to o up to the position F + , F − when both µF + and µF − touch C . The other two sides of the parallelogram P β are denoted by G + , G − .Clearly, G + ⊂ Z + and G − ⊂ Z − . The parallelogram µP β is circumscribed about C . Of course,when we go counterclockwise, then the order of the sides of P β is F + , G + , F − , G − .Clearly P β is a subset of C , but it may be not inscribed in C .Take the homothetic copy P γ ⊂ C of P β such that at least one pair of opposite vertices t, v of P γ is in bd( C ). Denote the other two vertices of P γ by u, w such that t, u, v, w be in thisorder on bd( C ).Let a, b ∈ bd( C ) be the points such that w ∈ ta and w ∈ vb If u, w are not in bd( C ), let us cleverly enlarge P γ up to a parallelogram P δ (not oblig-atory homothetic) inscribed in C such that a homothetic copy of P δ with a ratio at most µ iscircumscribed about C .Here we explain how to provide this task. For every boundary point c of bd( C ) on the arc ⌢ ab provide the straight lines K +1 ( c ) , K +2 ( c ) containing cv and ct , respectively. Let K − ( c ) , K − ( c )be the line symmetric to K +1 ( c ) , K +2 ( c ), respectively. Moreover, for i = 1 , L + i ( c ) and L − i ( c ) supporting C and parallel to the pairs K + i ( c ) and K − i ( c ) such that the order of them is L + i ( c ), K + i ( c ), K − i ( c ), L − i ( c ).Let L i ( c ) be the strip between L + i ( c ) and L − i ( c ) for i = 1 , K i ( c ) be the strip between K + i ( c ) and K − i ( c ) for i = 1 , o and points d + , d − of support of C by L +1 ( c ) and L − ( c ), respectively. Denote by f + , f − its intersections with F + , F − , respectively. Denote by k + , k − its intersections with K + i ( c ) , K − i ( c ), respectively. Denote by h + , h − its intersections with µF + , µF − . Since h + , d + , k + , f +, o , f − , k − , d − , h − are in this order on this straight line, weobtain | h + h − | / | f + f − | ≤ | d + d − | / | k + k − | . Hence by Lemma we obtain that Marek Lassak width( L i ( c )) / width( K i ( c )) ≤ µ for i = 1. Analogously, we conclude this for i = 2.If c is sufficiently close to a , thenwidth( L ( c )) / width( K ( c )) > width( L ( c )) / width( K ( c )) . If c is sufficiently close to b , then we have the opposite inequality. Moreover, observethat when c moves on ⌢ ab from a to b , then the strips L i ( c ) and K i ( c ), where i = 1 ,
2, changecontinuously. Consequently, there is at least one position c of c for whichwidth( L ( c )) / width( K ( c )) = width( L ( c )) / width( K ( c )) . Therefore the thesis of our proposition is true for the parallelograms P ′ = K ( c ) ∩ K ( c ).Still the homothetic one µ ′ P = L ( c ) ∩ L ( c ), where µ ′ ≤ µ , is circumscribed about C . Corollary.
Denote the Banach-Mazur distance between C ∈ M and P by µ . Assume that P ⊂ C ⊂ µP for a particular affine image P ∈ M of P . Then the parallelogram P is inscribedin C and µP is circumscribed about C . The author does not know if Proposition holds true in higher dimensions for the paral-lelotope or the cross-polytope in place of the parallelogram.
Theorem 1.
We have δ BM ( P , P ) = .Proof. The parallelogram with the four vertices at ( ± ,
0) and (0 , ± √
3) is contained in P andits homothetic image with ratio contains P . Consequently, δ BM ( P , P ) ≤ .Having in mind Proposition, in order to show that δ BM ( P , P ) ≥ , it is sufficient toconsider only any parallelogram P = a ( P ) inscribed in P such that a positive homothetic copy λP of P is circumscribed about P , and to show that this homothety ratio λ is at least .Denote by P the class of all such parallelograms P .Consider a parallelogram P = pqrs from P . Some two consecutive vertices of P must bein two consecutive sides of P . The reason is that in the opposite case no positive homotheticimage of our P is circumscribed about P in contradiction to P ∈ P . In order to fix attention,thanks to the symmetries of P , we do not make our considerations narrower assuming that p ∈ v v and q ∈ v v . For the same reason, we may additionally assume that p ∈ v m , where m denotes the middle of the side v v . anach-Mazur distance P ∈ P and h ( a ) P with respect to the hexagonTake the line y = bx passing through p . It is easy to show that p has the form ( √ b + √ , √ bb + √ ),where b belongs to the interval [0 , √ ]. Here b = 0 generates v , while b = √ generates m .Our q is the intersection of the segment v v with a straight line x = cy , where c ∈ [ − √ , √ ]. So q has the form ( √ c, √ ). An easy calculation shows that the directional coefficientof the straight line containing pq is σ = b −√ − bc −√ c and that the directional coefficient of thestraight line containing sp is ς = b + √ bc + √ c .Denote by S the strip between the straight lines containing pq and rs , and by S + thenarrowest strip parallel to S which contains P . Denote by T the strip between the straightlines containing qr and sp , and by T + the narrowest strip parallel to T which contains P .We see that P = S ∩ T . Of course, S + ∩ T + is the parallelogram with sides parallel tothe sides of P which is circumscribed about P . Since we are looking only for parallelograms P ∈ P , the parallelogram S + ∩ T + should be a positive homothetic copy of P .Now, our task is to describe such parallelograms P .We find the first coordinate of the intersection of the straight line through pq with y = √ x . It is x = √ · − cσ √ − σ . Next we evaluate the ratio of the first coordinate of v to x which is √ · √ − σ − cσ . By Lemma this ratio equals to width( S + ) / width( S ). By the substitu-tion of σ we get width( S + ) / width( S ) = √ · √ −√ bc − c − b − bc . In a similar way we obtain thatwidth( T + ) / width( T ) = √ · √ ς − cς = √ · √ b +3 c + √ bc − bc . Marek Lassak
Fig 2. Two best positions of P ∈ P and h ( a ) P with respect to the hexagonSolving the equation width( S + ) / width( S ) = width( T + ) / width( T ), we conclude that onlyfor c = − b √ b +3 its both sides are equal. We see that q ∈ nv , where n is the midpoint of v v . Sinceour P is a function of b , we denote it by P ( b ). Substituting c = − b √ b +3 into width( S + ) / width( S )we obtain that the common value of w ( S + ) /w ( S ) and width( T + ) / width( T ) is h ( b ) = b +4 √ b +94 b +2 √ b +6 ,where h ( b ) stands in place of h ( P ( b )). This ends our task.Every P ( b ) is inscribed in P and every h ( b ) P ( b ) is circumscribed about P (see Fig. 1).The vertices of h ( b ) P ( b ) being the images of vertices p, q, r, s of P ( b ) are denoted by p ′ , q ′ , t ′ , u ′ ,respectively. Of course, the straight line containing the side p ′ q ′ of h ( b ) P ( b ) supports P at v .We are considering here only the situation when the straight line containing the side s ′ p ′ of h ( b ) P ( b ) supports P at v . This holds true if and only if the directional coefficient ς of thestraight line containing s ′ p ′ is at most the directional coefficient of the straight line containing v v , so when it is at most √
3. Recall that ς = b + √ bc + √ c , which after substituting c = − b √ b +3 gives ς = √ b +32( √ − b ) . Solving √ b +32( √ − b ) = √ b = √ .So the straight line containing the side h ( b ) ur of h ( b ) P ( b ) supports P at v if and onlyif b is in the interval [0 , √ ]. In other words, if and only if r ∈ v k , where k is the point ofintersection of y = √ x with v v .The derivative of the function h ( b ) is − √ b − b +3 √ b + √ b +6) . An evaluation shows that thisderivative is 0 if and only if b = ( − √ ± √ anach-Mazur distance b = ( − √ √ ≈ . , √ ]. The value of h ( b ) for this b is approximately 1 . . Since h (0) = and h ( √ ) = , we conclude that the globalminimum of h ( b ) in the interval [0 , √ ] is and that it is attained only for b = 0 and b = √ .In Fig. 2 we see the two pairs P ( b ) , P ( b ) for these two values of b .It remains to explain what happens for b ∈ [ √ , √ ], so when p ∈ km . Observe that thenthe boundary of the smallest positive homothetic copy of P containing P touches it at points v , v , v , v . Hence this situation is symmetric to the preceding one with respect to the straightline through o perpendicular to v v . Considering b only in the interval [0 , √ ] is sufficient sincewe have in mind rotations of P ( b ) by 60 ◦ and 120 ◦ and the axial symmetries with respect to thelines containing v v , v v , v v .Let us add that when b changes from 0 to √ in the paragraph before the last of the proof,the point p changes from v = (0 ,
1) to k = ( , √ ) beating of the unit. Simultaneously q changes from (0 , √ ) to ( − , √ ) beating of the unit. Remark.
The only two positions of P ⊂ P such that P ⊂ P (besides their rotations by60 ◦ and 120 ◦ and axial symmetries with respect to the lines containing v v , v v , v v ) are theparallelograms with vertices (1 , , √ ), ( − , , − √ ) and with vertices ( , √ ), ( − √ , √ ), ( − , − √ ), ( √ , − √ ). Again see Fig. 2. In this section we consider the BM-distances from P to the regular even-gons with more thansix vertices. In Theorem 2 we find these distances to P j and P j +4 , and we present the estimatesfrom above from P to P j +2 and P j +6 . Next we conjecture that the values of these two upperestimates are just the BM-distances from P to P j +2 and P j +6 . Theorem 2.
We have (I) δ BM ( P , P j ) = √ , (II) δ BM ( P , P j +2 ) ≤ sec j j +2 π + cos j j +2 π , (III) δ BM ( P , P j +4 ) = √ j +4 π , (IV) δ BM ( P , P j +6 ) ≤ sin j +28 j +6 π · csc j +28 j +6 π + cos j +28 j +6 π .Proof. The inequalities showing that the left sides are at most the right sides result from positionsof the inscribed parallelograms whose vertices are the intersections of the coordinate axes withthe boundaries of P j , P j +2 , P j +4 , P j +6 , respectively. Evaluating the ratio of the smallesthomothetic copy of such an inscribed parallelogram which contains our polygon from amongst P j , P j +2 , P j +4 , P j +6 , we find each value at the right sides of (I)–(IV). Marek Lassak
Let us show the opposite inequalities for (I) and (III).Ad (I). To prove that δ BM ( P , P j ) ≥ √
2, we have to show that for any parallelogram P ∈ M contained in P j such that P j ⊂ λP , where λ >
0, the inequality λ ≥ √ δ BM ( P , P j ) ≥ √
2, it is sufficient to consider only any P = a ( P ) inscribed in P j such that λP is circumscribed about P j , and to show that λ ≥ √ P ⊂ P j . This is realized in the following Parts ( α ) and ( β ).( α ) If a parallelogram P ∈ M is inscribed in P j and its homothetic image is circum-scribed, then P is a square. Take into account a parallelogram P ∈ M inscribed in P j , which is not a square. Denoteits successive vertices by a, b, c, d , when we go counterclockwise.First let us explain that they must be in some 2 j -th sides of P j . In the opposite case,some of them, say a and b does not fulfill this. We may assume that a ∈ v v and b v j v j +1 (besides when a = v or b = v j +1 ). Provide the straight line parallel to ab which supports P j at a vertex v k with 1 < k < j . Provide the straight line parallel to bc which supports P j ata vertex v l with 2 j < l < j . Then the distances from v k to ab and from v l to bc are different.This means that no homothetic image of abcd has a chance to be circumscribed about P j .Hence consider the situation when succesive vertices of P are at every 2 j -th side of P j .Clearly, a, c are symmetric with respect to o and c is in different distances from a and b . Say,let a ∈ v v and b ∈ v j v j +1 . From the symmetry we see that | v a | = | v j c | . Let b ′ ∈ v v fulfill | v o a | = | v j b ′ | and let d ′ be opposite to b ′ . Then ab ′ cd ′ is a square inscribed in P j Clearly an enlarged homothetic image of it is circumscribed about P j . Thus the half-lineswith origin at o through v and v j +1 intersect the sides ab ′ and b ′ c , respectively, at points w +1 , w − j +1 , respectively, in equal distances from o . These half-lines intersect the sides ab and cb at points z +1 , z − j +1 , respectively. Since b = b ′ , then P is not a square [[?rhombus]] and thus | oz | 6 = | oz j +1 | . Since | ov | = | ov j +1 | , we get | z +1 z − | / | w +1 w − | 6 = | z +2 j +1 z − j +1 | / | w +2 j +1 w − j +1 | where w − i , z − i denote the points symmetric to w + i , z + i . From Lemma we conclude that the stripbetween the straight lines through v and v j +1 , parallel to ab has a different width than thestrip between the straight lines through v j +1 and v j +1 , parallel to bc . Hence no homotheticimage of abcd is circumscribed about P j .Consequently, P must be a square.( β ) If a square P is inscribed in P j and λP is circumscribed about P j , then λ ≥ √ , and λ = √ if and only if the vertices of P are at every j -th vertex of P j or at the middle of every j -th side of P j . Having in mind the axial symmetries of P j , we may limit our considerations to the squares P inscribed in P j whose one vertex denoted by p is in v m , where m is the midpoint of v v .Denote by k the directional factor sin( π/ j )cos( π/ j ) − of the straight line containing the side v v . anach-Mazur distance y = k ( x −
1) + 1. Clearly, p is in the intersection of the segment v m with a ray y = bx , where x ≥ ≤ b ≤ tan π j . We omit an easy calculation showing that p = ( kk − b , kbk − b ). Since the diagonals of P are orthogonal, the successive vertex q of P is theintersection of the rotated by 90 ◦ ray y = − b x , where x ≥
0, with the side v j v j +1 . We easilyestablish that q = ( − kbk − b , kk − b ).We provide the straight line containing pq . Its equation is y = b − b +1 x + k ( b +1)( k − b )( b +1) . Considerits intersection point u with the line y = x . An easy calculation shows that both coordinates of u are equal to k · b +1 k − b . Consequently, the ratio of the first coordinate of v j to the first coordinateof u equals to h ( b ) = √ : ( k · b +1 k − b ) = √ · k − bk ( b +1) . By Proposition, the homothetic square h ( b ) P is circumscribed about P j .Clearly, we consider the function h ( b ) in the interval [0 , tan π j ]. Observe that h (0) = √ h (tan π j ). Our aim is to show that h ( b ) ≥ √ b from this interval. We findthe derivative h ′ ( b ) = √ k · b − bk − b +1) . An easy calculation shows that h ′ ( b ) = 0 in the interval[0 , tan π j ] if and only if b = k + √ k + 1. We have h ( k + √ k + 1) = − √ k · √ k +1( k + √ k +1) +1 .Applying the fact that k < √
2. Hence h ( b ) ≥ √ b ∈ [0 , tan π j ] with equality only for b = 0 and b = tan π j .Consequently, the thesis of Part ( β ) holds true.We obtain δ BM ( P , P j ) ≥ √
2, which confirms the required opposite inequality of (I).Ad (III). The vertices v , v h +1 , v h +2 , v h +3 of P j +4 are the vertices of P . Observe thatevery side of P is parallel to every h -th side of P j +4 . So the side v v h +1 of P is parallel tothe side v j v j +1 of P j +4 , and so on. An evaluation shows that the ratio of the first coordinatesof the points of the intersection of the line y = (tan π ) x with these two sides (the points arecenters of these two sides) is cos j +4 π · sec π . Analogous is true for every side of P and thecorresponding parallel side of P j +4 . Consequently, by Lemma the homothetic copy of P withratio cos j +4 π · sec π contains P j +4 . Consequently, δ BM ( P , P j +4 ) ≤ cos j +4 π · sec π .From Part ( β ) we see the only positions of P for which δ BM ( P , P j ) = √ P for which δ BM ( P , P j +4 ) is realized are these with thevertices at every (2 j + 1)-th vertex of P j +4 .Generalizing (III) we observe that δ BM ( P n , P hn ) = cos hn π · sec n π for every even n ≥ h ≥ δ BM ( P , P j +2 ) = sec j j +2 π +cos j j +2 π and δ BM ( P , P j +6 ) = sin j +28 j +6 π · csc j +28 j +6 π + cos j +28 j +6 π .We also do not know the BM-distances between P and the regular odd-gons. Besidessome special cases, the task of finding the distances δ BM ( P m , P n ) seems to be very complicated.0 Marek Lassak
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Comparison between plane symmetric convex bodies and parallelograms , Math.Scand. (1960), 171-180.[2] G. Aubrun and S. J. Szarek, Alice and Bob Meet Banach, The interface of asymptotic geo-metric analysis and quantum information theory. Mathematical Surveys and Monographs,223. American Mathematical Society, Providence, RI, 2017.[3] S. Banach, Th´eorie des op´erations lin´eaires, Monogr. Mat. . Warszawa (1932). [Englishtranslation: Theory of linear operations. Translated from the French by F. Jellett. Withcomments by A. Pe lczy´nski and Cz. Bessaga. North-Holland Mathematical Library, 38.North-Holland Publishing Co., Amsterdam, 1987.][4] W. Stromquist, The maximum distance between two-dimensional Banach spaces , Math.Scand.48