DOWKER-TYPE THEOREMS FOR DISK-POLYGONS IN NORMED PLANES

. A classical result of Dowker (Bull. Amer. Math. Soc. 50: 120-122, 1944) states that for any plane convex body K in the Euclidean plane, the areas of the maximum (resp. minimum) area convex n -gons inscribed (resp. circumscribed) in K is a concave (resp. convex) sequence. It is known that this theorem remains true if we replace area by perimeter, the Euclidean plane by an arbitrary normed plane, or convex n -gons by disk-n -gons, obtained as the intersection of n closed Euclidean unit disks. The aim of our paper is to investigate these problems for C - n -gons, defined as inter-sections of n translates of the unit disk C of a normed plane. In particular, we show that Dowker’s theorem remains true for the areas and the perimeters of circumscribed C - n -gons, and the perimeters of inscribed C - n -gons. We also show that in the family of origin-symmetric plane convex bodies, for a typical element C with respect to Hausdorff distance, Dowker’s theorem for the areas of inscribed C - n -gons fails.


Introduction
For any integer n ≥ 3 and plane convex body K, let A n (K) (resp.a n (K)) denote the the infimum (resp.supremum) of the areas of the convex n-gons circumscribed about (resp.inscribed in) K. Verifying a conjecture of Kerschner, Dowker [6] proved that for any plane convex body K, the sequences {A n (K)} and {a n (K)} are convex and concave, respectively.It was proved independently by L. Fejes Tóth [12], Molnár [20] and Eggleston [7] that the same statements remain true if we replace area by perimeter, where the last author also showed that these statements are false if we replace area by Hausdorff distance.These results are known to be true also in any normed plane [19].Dowker's theorems have became important in many areas of discrete geometry, in particular in the theory of packing and covering [11,1,13] and are often used even today (see e.g.[21,3]).
Among many variants of Dowker's theorems that have appeared in the literature, we mention only one, which is related to the notion of spindle convexity.This concept goes back to a paper of Mayer [18] who, for any given convex body C in Euclidean space, considered sets X with the property that for any points p, q ∈ X, X contains the intersection of all translates of C containing p, q.He called these sets hyperconvex.His paper led to several papers in this topic in the 1930s and 40s, which, however, seems to have been forgotten by the end of the century.In modern times, a systematic investigation of hyperconvex sets was started in the paper [4] in 2007 for the special case that C is a closed Euclidean ball, and a similar paper [16] appeared in 2013, dealing with any convex body C (see also [15]).Hyperconvex sets have appeared in the literature under several different names: spindle convex, strongly convex or superconvex sets (see e.g.[17,23]), and appear in different areas of mathematics [5,14,17].In this paper, we follow the terminology in [4,16], and call a set satisfying the property in Mayer's paper C-spindle convex, or shortly C-convex, and if C is a closed Euclidean unit ball, we call it spindle convex (see Definition 2).
One of the results related to spindle convex sets is due to G. Fejes Tóth and Fodor [10] who extended Dowker's theorems, together with their variants for perimeter, for spindle convex sets; in these theorems the role of inscribed or circumscribed convex n-gons is played by the so-called disk-n-gons, obtained as the intersections of n closed Euclidean unit disks.They also proved similar theorems in hyperbolic or spherical plane.
Our main goal is to investigate a normed version of the problem in [10].To state our results, recall that the unit ball of any finite dimensional normed space is a convex body symmetric to the origin o, and any such body is the unit ball of a finite dimensional normed space.Thus, in the paper we choose an arbitrary o-symmetric convex disk C in the real normed space R 2 , and work in the normed plane with unit disk C, which we regard as R 2 equipped with the norm ⋅ C of C. In the paper, by a convex disk we mean a compact, convex planar set with nonempty interior.We denote the family of convex disks by K, and the family of o-symmetric convex disks by K o .In the paper we regard K and K o as topological spaces with the topology induced by Hausdorff distance.
Before presenting our results, recall the well-known fact that any finite dimensional real normed space can be equipped with a Haar measure, and that this measure is unique up to multiplication of the standard Lebesgue measure by a scalar (cf.e.g.[22]).This scalar does not play a role in our investigation and in the paper area(⋅) denotes 2-dimensional Lebesgue measure.Definition 1.For any C ∈ K o and convex polygon Q, we define the C-perimeter of Q as the sum of the lengths of the sides of Q, measured in the norm generated by C. The C-perimeter of a convex disk K ⊂ R 2 , denoted by perim C (K), is the supremum of the C-perimeters of all convex polygons inscribed in K.
We note that, moving its vertices one by one to the boundary of K in a suitable direction, for any convex polygon Q contained in K one can find a convex polygon Q ′ inscribed in K with perim C (Q) ≤ perim C (Q ′ ).This shows, in particular, that for any two plane convex bodies K ⊆ L ⊂ R 2 , we have perim C (K) ≤ perim C (L), with equality if and only if K = L (see also [19]).Furthermore, it is worth observing that a straightforward modification of Definition 1 can be used to define the C-length of a rectifiable curve Γ ⊂ R 2 , denoted by arclength C (Γ).
Our next definition can be found in [16] and its origin goes back to [18].
Definition 2. Let C ∈ K o and consider two (not necessarily distinct) points p, q ∈ R 2 such that a translate of C contains both p and q.Then the C-spindle (denoted as [p, q] C ) of p and q is the intersection of all translates of C that contain p and q.If no translate of C contains p and q, we set [p, q] C = R 2 .We call a set K ⊂ R 2 C-spindle convex (or shortly C-convex), if for any p, q ∈ K, we have [p, q] C ⊆ K.
We recall from [16,Corollary 3.13] that a closed set in R 2 different from R 2 is C-convex if and only if it is the intersection of some translates of C. Definition 3. The intersection of n translates of C is called a C-n-gon for n ≥ 3.
In our next definition and throughout the paper, area(⋅) denotes standard Lebesgue measure.
Theorem 1.For any C ∈ K o and C-convex disk K, the sequences { ÂC n (K)}, { P C n (K)} are convex, and the sequence {p C n (K)} is concave.That is, for any n ≥ 4, we have As a consequence of Theorem 1, we prove Theorem 2, and recall that similar statements have been derived in [9] for the Euclidean areas of inscribed and circumscribed polygons from the classical results of Dowker in [6] (for their spindle convex variants, see [10]).
Theorem 2. Let n ≥ 3 and k ≥ 2. Assume that k is a divisor of n and both K and C have k-fold rotational symmetry.Then there are C-n-gons Q A , Q P circumscribed about K which have k-fold rotational symmetry, and area(Q A ) = ÂC n (K) and perim C (Q P ) = P C n (K).Similarly, there is a C-n-gon Q p inscribed in K which has k-fold rotational symmetry, and perim C (Q p ) = pC n (K).
Before our next theorem, we remark that in a topological space F, a subset is called residual if it is a countable intersection of sets each of which has dense interior in F. The elements of a residual subset of F are called typical.Our next result shows that Dowker's theorem for the sequence {A C n (K)} fails in a strong sense.Theorem 3. A typical element C of K o satisfies the property that for every n ≥ 4, there is a C-convex disk K with The structure of the paper is as follows.In Section 2, we present the necessary notation and prove some lemmas.Then in Sections 3 and 4 we prove Theorems 1 and 2, and Theorem 3, respectively.Finally, in Section 5, we collect our additional remarks and propose some open problems.

Preliminaries
In the paper, for simplicity, for any x, y ∈ R 2 , we denote by [x, y] the closed segment with endpoints x, y.We equip R 2 also with a Euclidean norm, which we denote by ⋅ , and use the notation B 2 for the Euclidean closed unit disk centered at o. Recall that the Euclidean diameter of a compact set X ⊂ R 2 is the Euclidean distance of a farthest pair of points in X.If we replace Euclidean distance by distance measured in the norm of C, we obtain the C-diameter of X.
Recall that for any set X ⊆ R 2 , the C-convex hull, or shortly C-hull is the intersection of all C-convex sets that contain C. We denote it by conv C (X), and note that it is Cconvex, and if X is closed, then it coincides with the intersection of all translates of C containing X [16].
In the following list we collect some elementary properties of C-spindles and C-n-gons that we are going to use frequently in the paper.
Remark 1.We have the following.
(a) For any x, y ∈ R We observe that the statement in Remark 2 does not necessarily hold if x−y C = 2.As an example, we can choose C as a parallelogram, x m = x and y m = y as the midpoints of two opposite sides S 1 , S 2 of C, and {C m } as a sequence of o-symmetric hexagons inscribed in C whose elements intersect S 1 and S 2 only in x and y, respectively.
For any n ≥ 4, let K n a denote the subfamily of the elements C of K 0 satisfying the Dowker-type inequality âC A , K n p and K n P similarly.Our first lemma describes the topological properties of these families.
Lemma 1.For any n ≥ 4, K n a , K n A , K n p and K n P are closed.
Proof.We prove the assertion only for K n a , as for the other quantities the proof is analogous.Let C ∉ K n a , and suppose for contradiction that there is a sequence , which yields that âj (K) = area(K) for any j ≥ 3. Thus, according to our assumptions, K does not contain points at C-distance equal to 2, i.e its C-diameter is strictly less than 2. On the other hand, since K is C-convex, K is the intersection of the translates of C that contain it.Thus, there is a set By the properties of Hausdorff distance, the C m -diameter of K m is strictly less than 2 if m is sufficiently large.Then we can apply Remark 2, and obtain that area(Q . By compactness, we may assume that Lemma 1 readily yields Corollary 1, since the intersection of arbitrarily many closed sets is closed. and let x, y be points with x − y C ≤ 2. Then the arc-distance ρ C (x, y) of x, y with respect to C (or shortly, C-arc-distance of x and y) is the minimum of the C-length of the arcs, with endpoints x, y, that are contained in z + bd(C) for some y ∈ R 2 .
We recall the following version of the triangle inequality from [16,Theorem 6].
Lemma 2 (Lángi, Naszódi, Talata).Let C ∈ K 0 , and let x, y, z be points such that each pair has a C-arc-distance.
We start with a consequence of this inequality.
Lemma 3. Let p, q, r, s ∈ R 2 be distinct points contained in a translate of the smooth o-symmetric convex disk C, and assume that bd conv C {p, q, r, s} contains all of them and in this counterclockwise order.Then Proof.Note that according to our conditions, the two C-arcs in the boundary of [p, r] C intersect both C-arcs consisting of the boundary of [q, s] C .Let s ′ denote the intersection point of one of the C-arcs in bd[p, r] C and one of the C-arcs in bd[q, s], where the arcs are chosen to satisfy s ′ ∈ bd conv C {p, q, r} and s ′ ∈ conv C {p, r, s}.Then s ′ ∉ [p, q] C and , it is easy to see that p, q, s ′ , and also r, s, s ′ are in C-convex position.Thus, by Lemma 2, we have ρ C (p, q) ≤ ρ C (p, s ′ ) + ρ C (q, s ′ ) and ρ C (r, s) ≤ ρ C (r, s ′ ) + ρ C (s, s ′ ), implying the assertion.
In the following lemma, let S 1 denote the Euclidean unit circle centered at the origin.For simplicity, if x, y ∈ S 1 , we denote by xy the Euclidean closed circle arc obtained as the orbit of x when it is rotated around o in counterclockwise direction until it reaches y.Let S denote the family of closed circle arcs xy of S. Furthermore, we say that a function f ∶ S → R has a k-fold rotational symmetry for some positive integer k, if for any S, S ′ ∈ S, where S ′ is a rotated copy of S in counterclockwise direction with angle 2π k , we have f (S) = f (S ′ ).Lemma 4 can be regarded as a functional form of Dowker's theorems.
then the sequence {M n } is concave.Furthermore, if in addition, there is some positive integer k such that k n and f has k-fold rotational symmetry, and there is an n-element tiling X of S 1 such that M n = ∑ S∈X f (S) then there is an n-element tiling X ′ of S 1 with k-fold rotational symmetry such that Before the proof, we remark that X ⊂ S is called an m-tiling of S 1 for some positive integer m if every point of S 1 belongs to at least m members of X, and to the interiors of at most m members of X.
Proof.To prove the assertion for {M n }, we need to show that M n−1 + M n+1 ≤ 2M n is satisfied for any n ≥ 4. In other words, we need to show that for any tilings Note that the union A 0 of the two tilings is a 2-tiling of S 1 .Assume that x 1 , x 2 , . . ., x n−1 , and y 1 , y 2 , . . ., y n+1 are in this counterclockwise order in S 1 , and that y 1 ∈ x 1 x 2 .Due to the possible existence of coinciding points in the above two sequences, we unite these sequences as a single sequence v 1 , v 2 , . . ., v 2n in such a way that the points are in this counterclockwise order in S 1 , v 1 = x 1 , and removing the x i (resp.y j ) from this sequence we obtain the sequence y 1 , . . ., y n+1 (resp.x 1 , . . ., x n−1 ).In the proof we regard this sequence as a cyclic sequence, where the indices are determined mod 2n, and, with a little abuse of notation, we say that Our main goal will be to modify the 2-tiling A 0 in such a way that the value of f does not decrease but the number of covering pairs strictly decreases.
Note that since A 0 is the union of two tilings consisting of (n − 1) and (n − 1) arcs, respectively, A 0 contains covering pairs.Assume that v i v j covers v k v l .Then let A 1 denote the 2-tiling of S 1 in which v i v j and v k v l are replaced by v i v l and v k v j .According to our conditions, ∑ S∈A 0 f (S) ≤ ∑ S∈A 1 f (S), and the number of covering pairs in A 1 is strictly less than in A 0 .Repeating this procedure we obtain a 2-tiling A t of S 1 for which ∑ S∈A 0 f (S) ≤ ∑ S∈At f (S) and which does not contain covering pairs.Then, A t decomposes into the two tilings { v 1 , v 3 , v 3 v 5 , . . ., v 2n−1 v 1 } and { v 2 , v 4 , v 4 v 6 , . . ., v 2n v 2 }, each of which contains exactly n arcs.This proves the assertion for {M n }.
Now we prove the second part.Let X be an n-element tiling of S 1 such that M n = ∑ S∈X f (S).Assume that X does not have k-fold rotational symmetries.For i = 1, 2, . . ., k, let X i denote the rotated copy of X by 2iπ k in counterclockwise direction.Then Y = ⋃ k i=1 X i is a k-fold tiling of S 1 with k-fold rotational symmetry, and Since X has no k-fold rotational symmetry, Y contains covering pairs, and we may apply the argument in the previous paragraph.
We remark that an analogous proof yields Lemma 5, the proof of which we leave to the reader.Lemma 5. Let f ∶ S → R be a bounded function with f ( pp) = 0 for all p ∈ S 1 .For any integer n ≥ 3, let If for any x 2 x 3 ⊂ x 1 x 4 , we have then the sequence {m n } is convex.Furthermore, if in addition, there is some positive integer k such that k n, and f has k-fold rotational symmetry, and there is an n-element tiling X of S 1 such that m n = ∑ S∈X f (S) then there is a tiling X ′ of S 1 with k-fold rotational symmetry such that m n = ∑ S∈X ′ f (S).
In the next lemma, by the partial derivatives (∂ p f )( p 0 q 0 ) (resp.(∂ q f )( p 0 q 0 )) of the function f ( pq) at p 0 q 0 , we mean the derivative of the function f ( p(t)q 0 ) (resp.f ( q(t)p 0 )) at t = 0, where p(t) (resp.q(t)) is the rotated copy of p 0 (resp.q 0 ) around o by angle t in counterclockwise direction.Lemma 6.Let f ∶ S → R be a bounded function with f ( pp) = 0 for all p ∈ S 1 .Assume that for any p 0 q 0 ∈ S 1 , where p 0 ≠ q 0 , (∂ p ∂ q f )( p 0 q 0 ) is a continuous function of p 0 q 0 in both variables.Then, for any x 1 , x 2 , x 3 , x 4 ∈ S 1 in this counterclockwise order, we have if and only if (∂ p ∂ q f )( p 0 q 0 ) ≥ 0 for all p 0 ≠ q 0 .Similarly, for any x 1 , x 2 , x 3 , x 4 ∈ S 1 in this counterclockwise order, we have if and only if (∂ p ∂ q f )( p 0 q 0 ) ≤ 0 for all p 0 ≠ q 0 .Proof.We prove only the first part.Assume that (∂ p ∂ q f )( p 0 q 0 ) ≥ 0 for all p 0 ≠ q 0 .Let x 2 x 3 ⊂ x 1 x 4 .Then, by the Newton-Leibniz Theorem we have Furthermore, if we have (∂ p ∂ q f )( p 0 q 0 ) < 0 for some p 0 ≠ q 0 , then, by continuity and the same argument, there are some points x 1 , x 2 and x 3 , x 4 sufficiently close to p 0 and q 0 , respectively, such that x 2 x 3 ⊂ x 1 x 4 , and

Proof of Theorems 1 and 2
Note that by Lemma 1 and Corollary 1, it is sufficient to prove Theorem 1 for any everywhere dense subset of K o , and applying a similar consideration, we have the same for Theorem 2. Thus, we may assume that C has C ∞ -class boundary and strictly positive curvature.Under this condition, the quantities defined in Definition 4 are continuous functions of K for any fixed value of n, and thus, we may assume that K has C ∞ -class boundary, and the curvature of bd(K) at any point p is strictly greater than the curvature of bd(C) at the point q with the same outer unit normal as p. Remark 4.Under the above conditions, for any points p, q ∈ bd(K), [p, q] C ∖ {p, q} ⊂ int(K).
We define the following regions: (2)

The proof of Theorems 1 and 2 for ÂC
Then it can be seen directly that for any ϕ 1 < ϕ 2 < ϕ 3 < ϕ 4 < ϕ 1 + 2π, the function has nonnegative values at every point.Thus, the conditions of Lemma 5 are satisfied, implying the statement.

The proof of Theorems 1 and 2 for pC
Thus, the conditions of Lemma 4 are satisfied, implying our statement.

3.3.
The proof of Theorems 1 and 2 for P C n (K).By Lemmas 5 and 6, it is sufficient to prove that for any ϕ 1 < ϕ 2 < ϕ 1 + π, the function ∂ ϕ 1 ∂ ϕ 2 P is a continuous nonpositive function.In the remaining part of the subsection we prove this property.
For i = 1, 2, let v i denote the tangent vector of C(ϕ i ) at p pointing 'towards' q i in its boundary, and let w i denote the tangent vector of bd K at Γ(ϕ i ) pointing towards p in bd(C(ϕ i )).
Notation for the proof of Theorems 1 and 2 for P C n (K).
x(ϕ+∆)−x(ϕ) = ±v for any value of ϕ, where v is the unit tangent vector of bd(K) at Γ(ϕ) pointing in the positive direction.
By Remark 1, C(ϕ 1 ) ∩ C(ϕ 2 ) is the C-spindle of p and another point, which we denote by p ′ .By convexity, the tangent vectors of bd(C(ϕ 1 )) pointing in counterclockwise direction, turn in counterclockwise direction from p to p ′ .Thus, the directions of the vectors v 2 , w 1 , v 1 are in this order in counterclockwise orientation, and the same holds for the vectors v 2 , w 2 , v 1 .
Then, by Lemma 7, if ∆ i is sufficiently small, we have that the vectors y 1 , y 2 are between v 1 and v 2 according to counterclockwise orientation.
Consider the translate C ′ i of C(ϕ i ) by q i − p.The boundary of this translate contains q i , and v i is a tangent vector of C ′ i at q i .Thus, if q ′ = q 1 + q 2 − p (i.e.q ′ is the unique point for which p, q 1 , q ′ , q 2 are the vertices of a parallelogram in this counterclockwise order), then q ′ lies in the boundary of both C ′ 1 and C ′ 2 .On the other hand, by our observation about the tangent lines, if ∆ i are sufficiently small, then q ′ is contain in Q.By symmetry, ρ C (p, q 1 ) = ρ C (q ′ , q 1 ) and ρ C (p, q 2 ) = ρ C (q ′ , q 2 ), and thus, the required inequality follows from the remark after Definition 1.
The hexagon H and the octagon K 1 .In the illustration, t = 10.
We define C 1 as an o-symmetric convex 14-gon with vertices x 1 , x 2 , . . ., x 14 in counterclockwise order such that (a) x 1 and x 8 are on the negative and the positive half of the y-axis, respectively; and [p 3 , p 4 ], respectively; (d) we have Note that by our conditions, for any two point u, v ∈ G, each of the two C 1 -arcs in the boundary of [u, v] C 1 consists of translates of subsets of at most two consecutive sides of We estimate area([p 1 , q, p 6 ] C 1 ) for any q ∈ G with nonnegative y-coordinate.In the following p = (0, t + 2) denotes the midpoint of [p 3 , p 4 ].
Observe that if t is sufficiently large, in this case area([p 1 , q, p 6 ] C 1 ) is maximal if the xcoordinate of q is equal to t + 1.A similar consideration shows that if the x-coordinate of q is at most t + 1, then area([p 1 , q, p 6 ] C 1 ) is maximal if q = p 5 .Thus, in Case 2 we have Then bd([p 1 , q, p 6 ] C 1 ) consists of G ′ , a segment parallel to [q 2 , p 6 ] and ending at q, a segment containing [p 1 , q 1 ] as a subset, and a translate of [p 3 , p 4 ].Thus, in this case area([p 1 , q, p 6 ] C 1 ) is maximal if q = p 5 , and we have Combining our results, if t is sufficiently large, for any q, q ′ ∈ G (2) area([p 1 , q, p 6 ] C 1 ) + area([p 1 , q ′ , p 6 ] C 1 ) ≤ area(H) + area(K where we used the observation that [p 1 , p 2 , p 5 , p 6 ] C 1 = K 1 . In the remaining part of the construction, we fix t in such a way that (2) is satisfied.
Step 2. In the next step, based on Step 1, we construct some . Let p 7 = (−s, 0), where s is sufficiently large, and set K 2 = conv(K 1 ∪ {p 7 }) (see Figure 4).Let D 2 denote the Euclidean diameter of K 2 , and let C + 1 (resp.C − 1 ) denotes the set of the points of bd(C 1 ) with nonnegative (resp.nonpositive) x-coordinates.We define C 2 as follows: We remark that if s is sufficiently large, then there is some C 2 ∈ K o satisfying the above conditions, and In the following, let Let us define the points v 1 and v 6 as the images of p 1 and p 6 , respectively, under the homothety with center p 7 and homothety ratio 1 s is sufficiently large, and bd(C 2 ) contains two vertical segments of length 2, we may assume that In other words, we may assume that there is a translate of C that contains ))−2; a contradiction.Consequently, in the following we may assume that z Let v ′ 5 and v ′ 7 be the images of p 5 and p 7 , respectively, under the homothety with center p 6 and ratio , and, as in the previous paragraph, if ]).On the other hand, we have p 6 − p 7 > s and that the length of the corresponding height of [p 5 , p 6 , p 7 ] is greater than 0.1 by the definition of p 5 .Thus, area( > 0.1 3 √ s, implying that since area(Q 4 ) ≥ area(Q), which otherwise by our inequalities does not hold if s is sufficiently large, we may assume that some z i , say z 3 , is an element of [v 1 , p 7 , v 6 ].
We obtain similarly that if s is sufficiently large, some z i , say , where v ′′ 7 and v ′′ 2 are the images of p 7 and p 2 , respectively, under the homothety with center p 1 and ratio 1 3 √ s .These observations, the consideration in Step 1, and the inequality area(Q 4 ) ≥ area(Q) yield that as s → ∞, we have z 1 → p 1 , z 3 → p 6 and z 4 ∈ [v 1 , p 7 , v 6 ], and min{ z 2 − p 2 , z 2 − p 5 } → 0, implying that in this case area(Q 4 ) → area(Q).This shows that if s is sufficiently large, then area(H ′ ) + area(K 2 ) > 2 area(Q 4 ).
Before proceeding to the final step, we make two important observations that we are going to use.Here, by C + 2 and C − 2 , we denote the parts of bd(C 2 ) contained in the closed half planes {x ≥ 0} and {x ≤ 0}, respectively.
(1) A straightforward modification of the construction in Step 2 yields, for any n ≥ 4, the existence of some C n ∈ K 0 and a C n -convex disk K n such that âCn n−1 (K n ) + âCn n+1 (K n ) > 2â Step 3. Now we prove Theorem 3. Let n ≥ 4. Recall that K n a denotes the elements C of K o such that for any C-convex disk K, we have âC n−1 (K)+â C n+1 (K) ≤ 2â C n (K), and set K n a = K o ∖K n a .Observe that by Lemma 1, K n a is open.We show that it is everywhere dense in K o .Let C be an arbitrary element of K o and let ε > 0. Note that for any nondegenerate linear transformation h ∶ R 2 → R 2 , K is C-convex if and only if h(K) is h(C)-convex, and for any n ≥ 4, if K is C-convex, then âC n (K) = âh(C) n (h(K)).Thus, without loss of generality, we may assume that there are vertical supporting lines of C meeting bd(C) at some points ±p of the x-axis.We choose our notation such that p is on the positive half of the axis.

Lemma 4 .
Let f ∶ S → R be a bounded function with f ( xx) = 0 for all x ∈ S 1 .For any integer n ≥ 3, let

(a) C 2
is symmetric to both coordinate axes.(b) bd(C 2 ) contains some translates u+C + 1 and −u+C − 1 , where u points in the direction of the positive half of the x-axis.We set w 3 = u + x 1 .(c) In addition to the above two translates, bd(C 2 ) consists of segments [w 1 , w 2 ], [w 2 , w 3 ] and their reflections about one or both of the coordinate axes, such that [w 1 , w 2 ], [w 2 , w 3 ] are parallel to [p 6 , p 7 ] and [p 5 , p 7 ], respectively, and w 1 −w 2 , w 2 − w 3 > D 2 .

3 √s
. An elementary computation shows that then

Remark 6 .
For C ∈ K o , K ∈ K and positive integer n ≥ 3, let (4) P C n (K) = inf{perim C (Q) ∶ Q is a convex n − gon circumscribed about K}; 2 with x − y C ≤ 2, [x, y] C is the intersection of at most two translates of C, and if [x, y] C is a translate of C, then x − y C = 2. (b) Conversely, a nonempty intersection of at most two translates of C is the C-spindle of two (not necessarily distinct) points.(c) For any x, y ∈ R 2 , [x, y] C = [x, y] if and only if a translate of C contains [x, y] in Let x, y ∈ C ∈ K o , with x−y C < 2.Then, for any sequences 1 ) − 4 < < area(H) + area(K 1 ) = area([p 1 , p 6 ] C 1 ) + area([p 1 , p 2 , p 5 , p 6 ] C 1 ), Cn n (K n ).(2) To guarantee the required inequalities in Steps 1 and 2, we used the properties of the arcs of C 2 entirely contained in C + 2 or C − 2 .Thus, if C ′ 2 is an o-symmetric plane convex body containing C + 2 and C − 2 in its boundary, then we have âC ′ We combine these two observations in the following remark.Remark 5.For any n ≥ 4, there is some C n ∈ K o and a C n -convex disk K n such that if any C ′ n ∈ K o contains C + n and C − n in its boundary, where by C + n and C − n , we denote the parts of bd(C n ) contained in the closed half planes {x ≥ 0} and {x ≤ 0}, respectively, then K n is C ′ n -convex, and âC