A GENERALIZATION OF THE BLASCHKE-LEBESGUE PROBLEM TO A KIND OF CONVEX DOMAINS

. In this paper we will introduce for a convex domain K in the Euclidean plane a function Ω n ( K,θ ) which is called by us the biwidth of K , and then try to ﬁnd out the least area convex domain with constant biwidth Λ among all convex domains with the same constant biwidth. When n is an odd integer, it is proved that our problem is just that of Blaschke-Lebesgue, and when n is an even number, we give a lower bound of the area of such constant biwidth domains.


1.
Introduction. If the distance between parallel supporting planes of a convex body in R d is always the same, then the convex body is said to be of constant width. For d = 2 such convex bodies are often called orbiforms, and for d = 3 they are called spheroforms. A well-known example of these bodies is the Reuleaux triangle (see Figure 2(A) below). It has long been known that among all two-demensional convex bodies of constant width w, the Reuleaux triangle with width equal to w has the least area. This fact, to the best of our knowledge, was first proved by W. Blaschke ([5]) and H. Lebesgue ([18,19]), and since then there have been several proofs of this result (see, for example, [7,9,11,13,14,16,20]). However, the Blaschke-Lebesgue problem in dimension d ≥ 3 appears to be very difficult to solve and remains open. For partial results, see Anciaux and Georgiou [1], Anciaux and Guilfoyle [2] and Bayen, Lachand-Robert and Oudet [4] and the literature therein. There are many researches on various variations of the Blaschke-Lesbesgue problem (see [3,8,11,12,23]). In [21], in order to get a generalization of the Chernoff inequality ( [10]), K. Ou and the first author of the present paper introduced for convex domains in the plane a function w k (θ) = p(θ) + p θ + 2π k + · · · + p θ + 2(k − 1)π k to generalize the usual width function, where p(θ) is the Minkowski support function. They also posed a Blaschke-Lebesgue-style problem which motivates us to perform the research of the present paper.

SHENGLIANG PAN, DEYAN ZHANG AND ZHONGJUN CHAO
In this paper we will first introduce a function (which is called by us the biwidth function of K) in association with a convex domain K in the Euclidean plane where w(K, θ) is the usual width function of K, and then consider the similar Blaschke-Lebesgue problem for convex domains with Ω n (K, θ) being constant. In section 2, we will give some basic preliminaries for convex domains in the Euclidean plane. In section 3, we will introduce the function Ω n (K, θ), give the Fourier series expansions of the support functions of K and DK, and construct the regular Reuleaux type polygon with even sides, denote it by R 2m (m ∈ N). In section 4, by the properties of the mixed area, we will prove that when n is an odd integer, the Reuleaux triangle has the least area among K, when n is an even number, the R 2m has the least area among DK and give a new lower bound of the area of K.

Preliminaries.
2.1. Support function and width function of convex domains. A convex body in the Euclidean plane is usual called a convex domain. Let K be a bounded convex domain, its support function is a certain real continuous function and defined by and the width function of K in direction u is defined by For u = e iθ , we can write h(K, θ) instead of h(K, u) and w(K, θ) instead of w(K, u). The supporting line of K in direction u, H(K, u), is defined by The support function From these definitions one can see that h(K, u) is the (signed) distance from the origin O to the supporting line of K in direction u. The support function is nonnegative if and only if the origin is inside K (see Figure 1(A)). The width function w(K, u) is simply the distance between the parallel supporting lines, orthogonal to direction u, of K (see Figure 1(B)). h(K, θ) and w(K, θ) are both periodic functions with periods 2π and π, respectively. It is also easy to see that Denote by L(·) the perimeter of a bounded convex domain, then which is known as Cauchy's formula.
In order to establish most of the results contained in this paper, we need some definitions and known facts about mixed area and some other related notions associated to convex domains. Denote A(K) the area of K, then one can define the mixed area of two convex domains.
Given two bounded convex domains K 1 , K 2 , their Minkowski sum is More generally, for any t 1 , t 2 ∈ R, the Minkowski combination is defined as Definition 2.1. If K 1 , K 2 are bounded convex domains, then their mixed area is defined by Since the mixed area of convex domains has been well-studied (see, for instance, [6,15,22,24]), we state some propositions without proofs for later use.
Proposition 2.2. If K 1 , K 2 , K 3 and K are convex domains, g = e iϕ is a rotation, t 1 , t 2 and t are real numbers, then one gets and w(gK, u) = w(K, g −1 u). (2) Proof. From Proposition 2.2 and Proposition 2.3, one can easily obtain 3. Convex domains with constant biwidth.

Construction of convex domain with constant biwidth.
In this section, we will construct a class of convex domains with constant biwidth.
Definition 3.1. Let K be a convex domain in the plane. We call the following Ω n (K, θ) the biwidth function of K: It is easy to see that Ω n (K, θ) is a periodic function in θ with period π. We are more interested in the case when Ω n (K, θ) is a constant. We call a convex domain K satisfying Ω n (K, θ) = Λ (a constant) a convex domain with constant biwidth, and denote Proof. By Definition 3.1, we obtain which, together with the fact that K ∈ C n (Λ), gives us Ω n (DK, θ) = 2Λ, i.e., DK ∈ C n (2Λ). (w(K, θ) + w(K, θ + π/n)) dθ = πΛ, and then L(K) = πΛ 2 .
Theorem 3.3 is a Barbier-style result which tells us that all convex domains with constant biwidth have the same length. Among all domains with the same length L, the classical isoperimetric inequality implies that the circular disc with circumference L has the greatest area. For convex domains of constant width w 0 , the elegant Blaschke-Lebesgue theorem claims that the Reuleaux triangle with diameter w 0 has the least area. Now, we consider the Blaschke-Lebesgue-Style problem for convex domains with constant biwidth, that is, which member of the set C n (Λ) = {K|Ω n (K, θ) = Λ} has the least area?
3.2. Fourier series expansion. Since the support function of a given convex domain is always continuous, bounded and 2π-periodic, it can be expressed in the Fourier series form where The width function of K can be written as and thus a 2k cos 2kθ + cos 2k θ + π n + b 2k sin 2kθ + sin 2k θ + π n .
Now, using (4)- (7), we can obtain and that of K can be given by

3.3.
Construction of the regular Reuleaux type polygon with even sides. A polygon built by circular arcs is called a curvilinear polygon. A Reuleaux polygon is a curvilinear polygon where the center of each circular arc is its opposite vertex. That is why a Reuleaux polygon necessarily has odd number of sides. If all sides of a Reuleaux polygon are of equal length, then it is called regular. In this paper, we use the symbol R 2m+1 , m ∈ N, to denote a regular Reuleaux polygon of 2m + 1 sides. Figure 2 below shows some examples of regular Reuleaux polygons. In the following, we will describe how to construct a regular Reuleaux type polygon with even number of sides. We denote it by R 2m . For the sake of simplicity, we assume that the radius of curvature of R 2m is 4. Set α = π 2m , r 2m = 2 cos(α/2) .
Draw a circle with radius r 2m and centered at the origin O (see Figure 3). We will build a 2m-gon inscribed in the circle. Let AB be one side of the 2m-gon, draw a circular arc connecting A and B with radius equal to 4 and centered at O which is the midpoint of the circular arc CD, where C and D are chosen such that AB CD. Similarly, other sides of the 2m-gon can be built and we can obtain the regular Reuleaux type polygon R 2m . Obviously, R 2m is centrosymmetric. Some examples are given in Figure 5. Lemma 3.5. R 2m + gR 2m = 4B, where g = e i π 2m . Furthermore, R 2m ∈ C 2m (8).
Proof. Let ∂K denote the boundary curve of the convex domain K. Recall that the form of the sum of two domains(or curves) does not depend upon the choice of the origin and is not changed by parallel displacement of the summands, and under these circumstances the sum undergoes only a parallel displacement. So we can choose the center of R 2m as the origin in Figure 4. We rotate R 2m with a rotation of angle π 2m and translate it to the regular Reuleaux 2m−gon on the right side in Figure 4, for the sake of simplicity, it is still denoted by gR 2m .
It is well known that the sum R 2m + gR 2m is equivalent to a convex domain which is enclosed by the sum curve ∂R 2m + ∂(gR 2m ). However, the sum curve ∂R 2m +∂(gR 2m ) is the locus of all points which are the sum of corresponding points of ∂R 2m and ∂(gR 2m )(Let l 1 and l 2 be parallel and similarly oriented supporting lines of the curves ∂R 2m and ∂(gR 2m ) respectively. Let P and Q be the points of l 1 and l 2 with ∂R 2m and ∂(gR 2m ) respectively. We say that the points P and Q are corresponding points of the curves ∂R 2m and ∂(gR 2m )).
In Figure 4, it is easy to see that B on ∂R 2m and arbitrary point on CD in ∂(gR 2m ) are corresponding points, and the sum B + CD is F G according to the parallelogram law. Arbitrary point on AB in ∂R 2m and C on ∂(gR 2m ) are corresponding points , and the sum AB + C is EF . Continuing this process, the closed convex curve ∂R 2m + ∂(gR 2m ) is consisted of 4m circular arcs.
On the other hand, CD and I (the center of CD) can be translated to F G and O along − − → CF respectively, so O is the center of F G. From Figure 4, it is easy to Regular Reuleaux type polygons with even number of sides have been constructed, they are strictly convex domains, and their support functions are C 1,1 , (see [17]), so the radius of curvature of the boundary ∂R 2m exists a.e., and Since R 2m in Figure 3 is centrosymmetric, we get immediately Lemma 3.6. For any regular Reuleaux type polygon R 2m in Figure 3, there is a unique regular Reuleaux-type 2m-polygonR 2m such that DR 2m = R 2m .
From Lemmata 3.5 and 3.6 it follows obviously that Lemma 3.7. If K is a convex domain and DK = R 2m , then K ∈ C 2m (4).
Next we will calculate the area of R 2m .

4.
The main results. In this section, without loss of generality, let Λ = 4, and denote P 4m the regular 4m−gon circumscribing a circle of radius 2. If K ∈ C n (4), B is a unit disc, set q n (K) = min From the Blaschke selection theorem and the classical isoperimetric inequality it is easy to see that 0 < q n (K) ≤ 1, 0 < q n (DK) ≤ 1. They are hypothetically invariant, like the isoperimetric ratio, they can measure the area and the shape of convex domains in C n (4). Proof. Observe that w(K, θ) is continuous. Suppose w(K, θ) ≡ 2, that is to say that there exists a θ 0 such that w(K, θ 0 ) = 2. Without loss of generality, we may assume that w(K, θ 0 ) > 2. By Definition 3.1, we get Hence w(K, θ 0 ) = w(K, θ 0 + π) = w K, θ 0 This is a contradiction. Therefore w(K, θ) ≡ 2.
Theorem 4.1 tells us that C 2m−1 (4) are consists of convex domains with constant width 2, and thus one can obtain the area of the difference domains.
In this case, our problem is just that of Blaschke-Lebesgue. According to the Blaschke-Lebesgue theorem, the Reuleaux triangle has the least area, that is where R 3 is a Reuleaux triangle with width 2, and by Corollary 4.2, we can obtain We have considered the case when n is an odd integer. Next, we will deal with the case when n is an even integer.
where the rotation g = e i π 2m .

Remark 1.
Here, it is worth mentioning that L 4m in the proof above is not certainly regular, see for instance the octagon in Figure 8, formed by two squares with the same edges.