Partitioning Functional of a Class of Convex Bodies

Abstract

For each n-dimensional real Banach space X, each positive integer m, and each bounded set $A\subseteq X$ with diameter greater than 0, let ${\beta }_X(A,m)$ be the infimum of $\delta \in (0,1]$ such that $A\subseteq X$ can be represented as the union of m subsets of A, whose diameters are not greater than $\delta$ times the diameter of A. Estimating ${\beta }_X(A,m)$ is an important part of Chuanming Zong’s quantitative program for attacking Borsuk’s problem. However, estimating the partitioning functionals of general convex bodies in finite dimensional Banach spaces is challenging, so we will begin with the estimation of partitioning functionals for special convex bodies. In this paper, we prove a series of inequalities about partitioning functionals of convex cones. Several estimations of partitioning functionals of the convex hull of $(A+u)\cup (A-u)$ and $(A+u)\cup (-A-u)$ are also presented, where $A\subseteq {\mathbb{R}}^{n-1}\times \left\{0\right\}$ is a convex body with the origin o in its interior, and $u\in {\mathbb{R}}^n\setminus ({\mathbb{R}}^{n-1}\times \left\{0\right\})$. These results contribute to the study of Borsuk’s problem through Zong’s program.

1. Introduction

Let $X=({\mathbb{R}}^n,\parallel ·\parallel )$ be an n-dimensional real Banach space with origino and unit ball $B_X$. For two distinct points $x,y\in X$, denote by $[x,y]:=\{\alpha x+(1-\alpha )y\mid \alpha \in [0,1\left]\right\}$ the closed segment connecting x and y. For each $A\subseteq X$, denote by $convA$ and $intA$ the convex hull and the interior of A, respectively. Let A be nonempty and bounded; the diameter $\delta \left(A\right)$ of A is defined by

$$sup\{\parallel x-y\parallel \mid x,y\in A\}.$$

A compact convex subset of X with interior points is called a convex body. Let ${\mathcal{K}}^n$ be the set of convex bodies in X. For each $m\in {\mathbb{Z}}^{+}$, set $\left[m\right]:=\{i\in {\mathbb{Z}}^{+}\mid 1\le i\le m\}$.

In 1933, Borsuk [1] posed the following problem:

Problem 1.

(Borsuk’s Problem). Is it possible to partition every bounded set in the n-dimensional Euclidean space into $n+1$ subsets of smaller diameters?

The answer is positive when $n\le 3$ (cf. [2,3,4,5]). In 1993, J. Kahn and G. Kalai gave counterexamples to Borsuk’s problem in high dimensions (cf. [6]). In 2003, dimensions with counterexamples were reduced by A. Hinrichs and C. Richter to $n\ge 298$ (cf. [7]). In 2014, A. Bondarenko presented a 65-dimensional counterexample (cf. [8]). In the same year, T. Jenrich and A. E. Brouwer [9] gave a 64-dimensional one. Up to now, this problem has been open for $4\le n\le 63$. In 2021, C. Zong [10] proposed a quantitative program to attack it.

B. Grünbaum [11] extended Borsuk’s problem to Banach spaces. For a bounded set $A\subseteq X$, let $b_X\left(A\right)$ be the smallest positive integer m such that A can be represented as the union of m sets with smaller diameters. Recently, J. Wang, F. Xue, and C. Zong [12] proved that

$$b_X\left(A\right)\le 2^n(n+1)(log(n+1)+loglog(n+1)+5)$$

holds for every bounded set $A\subseteq X$. For more information about Borsuk’s problem, we refer to [13,14].

In 2021, Y. Lian and S. Wu [15] studied Borsuk’s partition problem in finite dimensional Banach spaces by estimating

$${\beta }_X(A,m)=inf\left\{{\displaystyle \frac{1}{\delta \left(A\right)}}\underset{k\in \left[m\right]}{max}\delta (A_k)|A=\bigcup _{k\in \left[m\right]}{A}_k\right\}$$

for $A\in {\mathcal{B}}^n$ and $m\in {\mathbb{Z}}^{+}$, and

$$\beta (X,m)=sup\{{\beta }_X(A,m)\mid A\in {\mathcal{B}}^n\},$$

where ${\mathcal{B}}^n=\{A\subseteq {\mathbb{R}}^n\mid A\mathrm{is}\mathrm{bounded}\mathrm{and}\delta \left(A\right)>0\}$. They obtained that $\beta ({\mathsf{\ell }}_p^3,8)\le 0.925$, $\forall p\in [1,\infty ]$. Later, L. Zhang, L. Meng, and S. Wu improved this result by showing that $\beta ({\mathsf{\ell }}_p^3,8)\le 0.9$ (cf. [16]). Recently, this result was reduced to $0.88185$ in [17].

The map

$$\begin{array}{cc}\hfill {\beta }_X:{\mathcal{B}}^n\times {\mathbb{Z}}^{+}& \to [0,1]\hfill \\ \hfill (A,m)& \mapsto {\beta }_X(A,m)\hfill \end{array}$$

is called the partitioning functional in X.

Compared to Borsuk’s problem in Euclidean space, Borsuk’s problem in finite dimensional Banach spaces is more difficult due to the influence of the norm of the space on the partitioning functional of bounded sets. In addition, Chuanming Zong proposed a reformulation for Borsuk’s problem, converting the estimation of Borsuk partition numbers into the estimation of partitioning functionals for bounded sets. Therefore, we study the estimation of partitioning functionals for special convex bodies in finite dimensional Banach spaces.

In the sequel, let A be a convex body in ${\mathbb{R}}^{n-1}\times \left\{0\right\}$ with the origin o in its interior, and u be a point in ${\mathbb{R}}^n\setminus ({\mathbb{R}}^{n-1}\times \left\{0\right\})$. Without loss of generality, we assume that $\delta \left(A\right)=1$.

In Section 2, we obtain a series of results concerning partitioning functionals of convex cones (the convex hull of the union of A and $\left\{u\right\}$). In Section 3, we show that

$${\beta }_X(C,2m)\le {\displaystyle \frac{{\beta }_X(A,m)}{2}}+{\displaystyle \frac{1}{2}},$$

where C is the sum of A and $[-u,u]$. In Section 4, we show that ${\beta }_X(T,2m)\le {\displaystyle \frac{{\beta }_X(A,m)}{2}}+{\displaystyle \frac{1}{2}}$, where T is the convex hull of $A+u$ and $-A-u$.

2. Partitioning Functionals of Convex Cones

For $0\le \alpha \le \beta \le 1$, and each subset B of A, we put

$$D_{\alpha ,\beta }\left(B\right)=\{\mu x+(1-\mu )u\mid x\in B,\mu \in [\alpha ,\beta ]\}.$$

Lemma 1.

Let $0\le \alpha \le \beta \le 1$. Then,

$$\delta \left(D_{\alpha ,\beta }\left(B\right)\right)\le max\left\{\beta \delta \left(B\right),\alpha \delta \left(B\right)+(\beta -\alpha )\underset{y\in B}{sup}\parallel u-y\parallel \right\}.$$

Proof. 

For every pair of points $x,y\in D_{\alpha ,\beta }\left(B\right)$, there exist two numbers ${\mu }_1,{\mu }_2\in [\alpha ,\beta ]$ and two points $x_1,y_1\in B$ such that

$$x={\mu }_1{x}_1+(1-{\mu }_1)u\mathrm{and}y={\mu }_2{y}_1+(1-{\mu }_2)u.$$

Assume, without loss of generality, that ${\mu }_2\ge {\mu }_1$. Then,

$$\begin{array}{cc}\hfill \parallel x-y\parallel & = \parallel {\mu }_1{x}_1-{\mu }_2{y}_1+({\mu }_2-{\mu }_1)u\parallel \hfill \\ \hfill & = \parallel {\mu }_1(x_1-y_1)+({\mu }_2-{\mu }_1)(u-y_1)\parallel \hfill \\ \hfill & \le {\mu }_1\parallel x_1-y_1\parallel +({\mu }_2-{\mu }_1)\parallel u-y_1\parallel \hfill \\ \hfill & \le {\mu }_1\delta \left(B\right)+({\mu }_2-{\mu }_1)\underset{y\in B}{sup}\parallel u-y\parallel \hfill \\ \hfill & \le {\mu }_1\delta \left(B\right)+(\beta -{\mu }_1)\underset{y\in B}{sup}\parallel u-y\parallel \hfill \\ \hfill & =\beta \underset{y\in B}{sup}\parallel u-y\parallel +{\mu }_1\left(\delta \left(B\right)-\underset{y\in B}{sup}\parallel u-y\parallel \right).\hfill \end{array}$$

If $\delta \left(B\right)\ge {sup}_{y\in B}\parallel u-y\parallel$, then $\delta \left(D_{\alpha ,\beta }\left(B\right)\right)\le \beta \delta \left(B\right)$.

If $\delta \left(B\right)\le {sup}_{y\in B}\parallel u-y\parallel$, then

$$\delta \left(D_{\alpha ,\beta }\left(B\right)\right)\le \alpha \delta \left(B\right)+(\beta -\alpha )\underset{y\in B}{sup}\parallel u-y\parallel .$$

Therefore,

$$\delta \left(D_{\alpha ,\beta }\left(B\right)\right)\le max\left\{\beta \delta \left(B\right),\alpha \delta \left(B\right)+(\beta -\alpha )\underset{y\in B}{sup}\parallel u-y\parallel \right\}.$$

□

Lemma 2.

Let $\alpha \in (0,1]$. Then,

$$\delta \left(D_{0,\alpha }\left(B\right)\right)=max\left\{\alpha \delta \left(B\right),\alpha \underset{y\in B}{sup}\parallel u-y\parallel \right\}.$$

Proof. 

By Lemma 1,

$$\delta \left(D_{0,\alpha }\left(B\right)\right)\le max\left\{\alpha \delta \left(B\right),\alpha \underset{y\in B}{sup}\parallel u-y\parallel \right\}.$$

Since $\alpha B+(1-\alpha )u\subseteq D_{0,\alpha }\left(B\right)$ and

$$\delta \left(D_{0,\alpha }\left(B\right)\right)\ge \parallel \alpha y+(1-\alpha )u-u\parallel =\alpha \parallel y-u\parallel ,\forall y\in B,$$

we have $\delta \left(D_{0,\alpha }\left(B\right)\right)\ge \alpha \delta \left(B\right)$ and $\delta \left(D_{0,\alpha }\left(B\right)\right)\ge \alpha {sup}_{y\in B}\parallel y-u\parallel$. It follows that

$$\delta \left(D_{0,\alpha }\left(B\right)\right)\ge max\left\{\alpha \delta \left(B\right),\alpha \underset{y\in B}{sup}\parallel y-u\parallel \right\}.$$

□

Remark 1.

Lemma 2 provides the diameter of convex cones, which is an indispensable step for estimating partitioning functionals of convex cones.

Put $D=conv(A\cup \{u\left\}\right)$; then, D is a cone with apex u and base A.

Theorem 1.

Let m be a positive integer satisfying ${\beta }_X(A,m)<1$, and $\lambda ={sup}_{y\in A}\parallel y-u\parallel$. Then,

$${\beta }_X(D,m+1)\le \left\{\begin{array}{cc}{\displaystyle \frac{\lambda }{2\lambda -{\beta }_X(A,m)}},\hfill & \lambda \ge 1,\hfill \\ {\displaystyle \frac{\lambda }{1+\lambda -{\beta }_X(A,m)}},\hfill & {\beta }_X(A,m)<\lambda <1,\hfill \\ {\beta }_X(A,m),\hfill & \lambda \le {\beta }_X(A,m).\hfill \end{array}\right.$$

Proof. 

For each $ϵ\in (0,1-{\beta }_X(A,m))$, there exists a collection $\{A_k\mid k\in \left[m\right]\}$ of subsets of A such that

$$A=\bigcup \{A_k\mid k\in \left[m\right]\}$$

and

$$\underset{k\in \left[m\right]}{max}\delta \left(A_k\right)\le {\beta }_X(A,m)+ϵ.$$

Let $\alpha \in (0,1]$. It is not difficult to verify that $D=D_{0,1}\left(A\right)=D_{0,\alpha }\left(A\right)\cup D_{\alpha ,1}\left(A\right)$. Moreover,

$$\begin{array}{cc}\hfill D_{\alpha ,1}\left(A\right)& =\{\mu x+(1-\mu )u\mid x\in A,\mu \in [\alpha ,1\left]\right\}\hfill \\ \hfill & =\left\{\mu x+(1-\mu )u\mid x\in \bigcup _{k\in \left[m\right]}{A}_k,\mu \in [\alpha ,1]\right\}\hfill \\ \hfill & =\bigcup _{k\in \left[m\right]}\{\mu x+(1-\mu )u\mid x\in A_k,\mu \in [\alpha ,1]\}\hfill \\ \hfill & =\bigcup _{k\in \left[m\right]}{D}_{\alpha ,1}\left(A_k\right).\hfill \end{array}$$

By Lemmas 1 and 2,

$$\delta \left(D\right)=max\left\{1,\underset{y\in A}{sup}\parallel y-u\parallel \right\},\delta \left(D_{0,\alpha }\left(A\right)\right)=\alpha \delta \left(D\right),$$

and

$$\begin{array}{cc}\hfill \delta \left(D_{\alpha ,1}\left(A_k\right)\right)& \le max\left\{\delta \left(A_k\right),\alpha \delta \left(A_k\right)+(1-\alpha )\underset{y\in A_k}{sup}\parallel y-u\parallel \right\}\hfill \\ \hfill & \le max\left\{{\beta }_X(A,m)+ϵ,\alpha ({\beta }_X(A,m)+ϵ)+(1-\alpha )\underset{y\in A}{sup}\parallel y-u\parallel \right\}.\hfill \end{array}$$

We consider the following three cases.

Case 1: $\lambda \ge 1$. Put $\alpha ={\displaystyle \frac{\lambda }{2\lambda -{\beta }_X(A,m)-ϵ}}$. Then, $\delta \left(D\right)={sup}_{y\in A}\parallel y-u\parallel =\lambda$, and

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\delta \left(D\right)}}max\{max\{\delta \left(D_{\alpha ,1}\left(A_k\right)\right)\mid k\in \left[m\right]\},\delta \left(D_{0,\alpha }\left(A\right)\right)\}\\ \le max\left\{max\left\{{\displaystyle \frac{{\beta }_X(A,m)+ϵ}{\lambda }},\alpha {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{\lambda }}+1-\alpha \right\},\alpha \right\}\hfill \\ =max\left\{\alpha {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{\lambda }}+1-\alpha ,\alpha \right\}\hfill \\ ={\displaystyle \frac{\lambda }{2\lambda -{\beta }_X(A,m)-ϵ}}.\hfill \end{array}$$

It follows that

$${\beta }_X(D,m+1)\le {\displaystyle \frac{\lambda }{2\lambda -{\beta }_X(A,m)}}.$$

(1)

Case 2: ${\beta }_X(A,m)<\lambda <1$. In this case, we require further that $ϵ\in (0,\lambda -{\beta }_X(A,m))$. Put $\alpha ={\displaystyle \frac{\lambda }{1+\lambda -{\beta }_X(A,m)-ϵ}}$. Then, $\delta \left(D\right)=\delta \left(A\right)=1$ and

$$\begin{array}{c}\hspace{1em}{\displaystyle \frac{1}{\delta \left(D\right)}}max\{max\{\delta \left(D_{\alpha ,1}\left(A_k\right)\right)\mid k\in \left[m\right]\},\delta \left(D_{0,\alpha }\left(A\right)\right)\}\hfill \\ \le max\left\{max\left\{{\beta }_X(A,m)+ϵ,\alpha ({\beta }_X(A,m)+ϵ)+(1-\alpha )\lambda \right\},\alpha \right\}\hfill \\ =max\left\{\alpha ({\beta }_X(A,m)+ϵ)+(1-\alpha )\lambda ,\alpha \right\}\hfill \\ ={\displaystyle \frac{\lambda }{1+\lambda -{\beta }_X(A,m)-ϵ}}.\hfill \end{array}$$

It follows that

$${\beta }_X(D,m+1)\le {\displaystyle \frac{\lambda }{1+\lambda -{\beta }_X(A,m)}}.$$

Case 3: $\lambda \le {\beta }_X(A,m)<1$. Put $\alpha ={\beta }_X(A,m)+ϵ$. Then, $\delta \left(D\right)=\delta \left(A\right)=1$ and

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\delta \left(D\right)}}max\{max\{\delta \left(D_{\alpha ,1}\left(A_k\right)\right)\mid k\in \left[m\right]\},\delta \left(D_{0,\alpha }\left(A\right)\right)\}\\ \le max\{{\beta }_X(A,m)+ϵ,\alpha \}={\beta }_X(A,m)+ϵ.\hfill \end{array}$$

It follows that

$${\beta }_X(D,m+1)\le {\beta }_X(A,m).$$

This completes the proof. □

Remark 2.

In Theorem 1, a partition method for convex cones is presented, from which an estimate of the partitioning functional of convex cones is derived. It is evident from Theorem 1 that the value of the partitioning functional of a bounded set in a Banach space is closely related to the properties of the set itself.

Corollary 1.

Let m be a positive integer satisfying ${\beta }_X(A,m)<1$. Then,

$${\beta }_X(D,m+1)\le {\displaystyle \frac{1}{2-{\beta }_X(A,m)}}.$$

Proof. 

Set

$$f\left(\lambda \right)={\displaystyle \frac{\lambda }{2\lambda -{\beta }_X(A,m)}},\lambda \ge 1,$$

and

$$g\left(\lambda \right)={\displaystyle \frac{\lambda }{1+\lambda -{\beta }_X(A,m)}},{\beta }_X(A,m)<\lambda <1.$$

Then, $f\left(\lambda \right)$ is decreasing on $[1,\infty )$ and $g\left(\lambda \right)$ is increasing on $({\beta }_X(A,m),1)$. Hence,

$$f\left(\lambda \right)\le f\left(1\right)={\displaystyle \frac{1}{2-{\beta }_X(A,m)}}\mathrm{and}g\left(\lambda \right)\le \underset{\lambda \to 1^{-}}{lim}g\left(\lambda \right)={\displaystyle \frac{1}{2-{\beta }_X(A,m)}}.$$

By Theorem 1,

$${\beta }_X(D,m+1)\le max\left\{{\displaystyle \frac{1}{2-{\beta }_X(A,m)}},{\beta }_X(A,m)\right\}.$$

Since ${\beta }_X(A,m)<1$, ${\beta }_X(D,m+1)\le {\displaystyle \frac{1}{2-{\beta }_X(A,m)}}$. □

Proposition 1.

Let ${\mathsf{\ell }}_1^3=({\mathbb{R}}^3{,\parallel ·\parallel }_1)$, $A=B_{{\mathsf{\ell }}_1^2}\times \left\{0\right\}$, $u=(0,0,1)$, and $D=conv(A\cup \{u\left\}\right)$. Then,

$${\beta }_{{\mathsf{\ell }}_1^3}(D,5)={\displaystyle \frac{1}{2-{\beta }_{{\mathsf{\ell }}_1^3}(A,4)}}={\displaystyle \frac{2}{3}}.$$

Proof. 

It is clear that $\delta \left(D\right)=\delta \left(A\right)={max}_{y\in A}\parallel y-u\parallel =2$. It follows from Proposition 4 in [15] that ${\beta }_{{\mathsf{\ell }}_1^3}(A,4)={\displaystyle \frac{1}{2}}$. By Theorem 1,

$${\beta }_{{\mathsf{\ell }}_1^3}(D,5)\le {\displaystyle \frac{1}{2-{\beta }_{{\mathsf{\ell }}_1^3}(A,4)}}={\displaystyle \frac{2}{3}}.$$

Suppose the contrary that ${\beta }_{{\mathsf{\ell }}_1^3}(D,5)<{\displaystyle \frac{2}{3}}$. Then, there exist five subsets $D_1$, $D_2$, $D_3$, $D_4$, and $D_5$ of D such that $D={\bigcup }_{i\in \left[5\right]}{D}_i$ and $max\{\delta \left(D_i\right)\mid i\in \left[5\right]\}<{\displaystyle \frac{4}{3}}$. Denote by $\{x_1,x_2,x_3,x_4\}$ the set of extreme points of A. We may assume that $u\in D_5$ and $x_i\in D_i$, $\forall i\in \left[4\right]$. The partition is shown in Figure 1. Since $p=\left({\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{3}}\right)\in D$,

$$\parallel p-x_i\parallel \ge {\displaystyle \frac{4}{3}},\forall i\in \left[4\right],$$

and $\parallel p-u\parallel ={\displaystyle \frac{4}{3}}$, we have $p\notin {\bigcup }_{i\in \left[5\right]}{D}_i$, a contradiction. Therefore, ${\beta }_{{\mathsf{\ell }}_1^3}(D,5)={\displaystyle \frac{2}{3}}$. □

It follows from Theorem 1 and Proposition 1 that estimations of partitioning functionals of convex cones, which are the best possible for certain pairs of m and D, are obtained.

Furthermore, when m and $\lambda$ are sufficiently large, by (1), the above result is not optimal. Proposition 2 will give a better estimation in this situation.

For $\alpha <\beta$, and each subset B of A, we put

$$C_{\alpha ,\beta }\left(B\right)=B+[\alpha u,\beta u].$$

Lemma 3.

Let $\alpha <\beta$. Then,

$$\delta \left(C_{\alpha ,\beta }\left(B\right)\right)=\underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel .$$

Proof. 

For every pair of points $x,y\in C_{\alpha ,\beta }\left(B\right)$, there exist two numbers ${\mu }_1,{\mu }_2\in [0,1]$ and two points $x_1,y_1\in B$ such that

$$x=x_1+{\mu }_1\alpha u+(1-{\mu }_1)\beta u$$

and

$$y=y_1+{\mu }_2\alpha u+(1-{\mu }_2)\beta u.$$

Assume, without loss of generality, that ${\mu }_2\ge {\mu }_1$. Then,

$$\begin{array}{cc}\hfill \parallel x-y\parallel & = \parallel x_1-y_1+({\mu }_2-{\mu }_1)(\beta -\alpha )u\parallel \hfill \\ \hfill & = \parallel (1-{\mu }_2+{\mu }_1)(x_1-y_1)+({\mu }_2-{\mu }_1)(x_1-y_1+(\beta -\alpha )u)\parallel \hfill \\ \hfill & \le (1-{\mu }_2+{\mu }_1)\parallel x_1-y_1\parallel +({\mu }_2-{\mu }_1)\parallel x_1-y_1+(\beta -\alpha )u\parallel \hfill \\ \hfill & \le max\{\parallel x_1-y_1\parallel ,\parallel x_1-y_1+(\beta -\alpha )u\parallel \}\hfill \\ \hfill & \le max\left\{\delta \left(B\right),\underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel \right\}.\hfill \end{array}$$

Therefore,

$$\delta \left(C_{\alpha ,\beta }\left(B\right)\right)\le max\left\{\delta \left(B\right),\underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel \right\}.$$

Clearly, $\delta \left(C_{\alpha ,\beta }\left(B\right)\right)\ge \delta \left(B\right)$. Since $x+\beta u\in C_{\alpha ,\beta }\left(B\right)$ and $y+\alpha u\in C_{\alpha ,\beta }\left(B\right)$ for any two points $x,y\in B$, $\delta \left(C_{\alpha ,\beta }\left(B\right)\right)\ge \parallel x-y+(\beta -\alpha )u\parallel$. Thus,

$$\delta \left(C_{\alpha ,\beta }\left(B\right)\right)\ge max\left\{\delta \left(B\right),\underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel \right\}.$$

Since, for arbitrary two points $x,y\in B$,

$$\begin{array}{cc}\hfill \parallel x-y\parallel & =∥{\displaystyle \frac{x-y}{2}}+{\displaystyle \frac{\beta -\alpha }{2}}u+{\displaystyle \frac{x-y}{2}}-{\displaystyle \frac{\beta -\alpha }{2}}u∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\parallel x-y+(\beta -\alpha )u\parallel +{\displaystyle \frac{1}{2}}\parallel y-x+(\beta -\alpha )u\parallel \hfill \\ \hfill & \le \underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel ,\hfill \end{array}$$

we have

$$\delta \left(B\right)\le \underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel .$$

It follows that

$$\begin{array}{cc}\hfill \delta \left(C_{\alpha ,\beta }\left(B\right)\right)& =max\left\{\delta \left(B\right),\underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel \right\}\hfill \\ \hfill & =\underset{x,y\in B}{sup}\parallel x-y+(\beta -\alpha )u\parallel .\hfill \end{array}$$

□

Proposition 2.

Let m be a positive integer satisfying ${\beta }_X(A,m)<1$, and $\lambda ={sup}_{y\in A}\parallel y-u\parallel >1$. Then,

$${\beta }_X(D,m)\le {\displaystyle \frac{1}{m}}+{\displaystyle \frac{1}{\lambda }}.$$

Proof. 

It is clear that

$$\begin{array}{cc}\hfill D& =conv(A\cup \{u\left\}\right)\subseteq A+[0,u]\hfill \\ \hfill & =\bigcup _{i=1}^m\left(A+\left[{\displaystyle \frac{i-1}{m}}u,{\displaystyle \frac{i}{m}}u\right]\right)\hfill \\ \hfill & =\bigcup _{i=1}^m{C}_{\frac{i-1}{m},\frac{i}{m}}\left(A\right).\hfill \end{array}$$

By Lemma 3, for each $i\in \left[m\right]$,

$$\delta \left(C_{\frac{i-1}{m},\frac{i}{m}}\left(A\right)\right)=\underset{x,y\in A}{sup}∥x-y+{\displaystyle \frac{1}{m}}u∥\le \delta \left(A\right)+{\displaystyle \frac{1}{m}}\parallel u\parallel .$$

Then,

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{\delta \left(D\right)}}\underset{i\in \left[m\right]}{max}\delta \left(C_{{\displaystyle \frac{i-1}{m}},{\displaystyle \frac{i}{m}}}(A)\right)& \le {\displaystyle \frac{1}{\delta \left(D\right)}}\left(\delta \left(A\right)+{\displaystyle \frac{1}{m}}\parallel u\parallel \right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{\lambda }}\left(1+{\displaystyle \frac{1}{m}}\lambda \right)={\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{1}{m}}.\hfill \end{array}$$

Therefore,

$${\beta }_X(D,m)\le {\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{1}{m}}.$$

□

When $\lambda$ is sufficiently small, we have the following result.

Proposition 3.

Let m be a positive integer satisfying ${\beta }_X(A,m)<1$. If $\lambda ={max}_{y\in A}\parallel y-u\parallel \le {\beta }_X(A,m)$, then

$${\beta }_X(D,m)={\beta }_X(A,m).$$

Proof. 

Since ${max}_{y\in A}\parallel y-u\parallel \le {\beta }_X(A,m)$, $\delta \left(D\right)=\delta \left(A\right)=1$. For each $ϵ>0$, there exists a collection $\{D_i\mid i\in \left[m\right]\}$ of subsets of D such that

$$D=\bigcup \{D_i\mid i\in \left[m\right]\}\mathrm{and}\underset{i\in \left[m\right]}{max}\delta \left(D_i\right)\le {\beta }_X(D,m)+ϵ.$$

Then, $A=A\cap D={\bigcup }_{i\in \left[m\right]}(A\cap D_i)$. Thus,

$${\beta }_X(A,m)\le max\{\delta (A\cap D_i)\mid i\in \left[m\right]\}\le {\beta }_X(D,m)+ϵ.$$

It follows that ${\beta }_X(A,m)\le {\beta }_X(D,m)$.

For each $ϵ>0$, there exists a collection $\{A_i\mid i\in \left[m\right]\}$ of subsets of A such that

$$A=\bigcup \{A_i\mid i\in \left[m\right]\}\mathrm{and}\underset{i\in \left[m\right]}{max}\delta \left(A_i\right)\le {\beta }_X(A,m)+ϵ.$$

Then,

$$\begin{array}{cc}\hfill D& =D_{0,1}\left(A\right)=\{\mu x+(1-\mu )u\mid x\in A,\mu \in [0,1]\}\hfill \\ \hfill & =\left\{\mu x+(1-\mu )u\mid x\in \bigcup _{i\in \left[m\right]}{A}_i,\mu \in [0,1]\right\}\hfill \\ \hfill & =\bigcup _{i\in \left[m\right]}\{\mu x+(1-\mu )u\mid x\in A_i,\mu \in [0,1]\}\hfill \\ \hfill & =\bigcup _{i\in \left[m\right]}{D}_{0,1}\left(A_i\right).\hfill \end{array}$$

Thus, by Lemma 2,

$$\begin{array}{cc}\hfill {\beta }_X(D,m)& \le \underset{i\in \left[m\right]}{max}\delta \left(D_{0,1}\left(A_i\right)\right)\hfill \\ \hfill & =\underset{i\in \left[m\right]}{max}\left\{max\left\{\delta \left(A_i\right),\underset{y\in A_i}{sup}\parallel y-u\parallel \right\}\right\}\hfill \\ \hfill & \le {\beta }_X(A,m)+ϵ.\hfill \end{array}$$

It follows that ${\beta }_X(D,m)\le {\beta }_X(A,m)$. □

3. Estimations of ${\mathit{\beta }}_{\mathit{X}}(\mathit{A}+[-\mathit{u},\mathit{u}],\mathbf{2}\mathit{m})$

Theorem 2.

Let m be a positive integer satisfying ${\beta }_X(A,m)<1$. Then,

$${\beta }_X(A+[-u,u],2m)\le {\displaystyle \frac{{\beta }_X(A,m)}{2}}+{\displaystyle \frac{1}{2}}.$$

Proof. 

For each $ϵ\in (0,1-{\beta }_X(A,m))$, there exists a collection $\{A_i\mid i\in \left[m\right]\}$ of subsets of A such that

$$A=\bigcup \{A_i\mid i\in \left[m\right]\}$$

and

$$\underset{i\in \left[m\right]}{max}\delta \left(A_i\right)\le {\beta }_X(A,m)+ϵ.$$

Put $C=C_{-1,1}\left(A\right)=A+[-u,u]$. Then,

$$\begin{array}{cc}\hfill C& =A+\left(\right[-u,o]\cup [o,u\left]\right)=(A+[-u,o\left]\right)\cup (A+[o,u\left]\right)\hfill \\ \hfill & =\left(\left(\bigcup _{i\in \left[m\right]}{A}_i\right)+[-u,o]\right)\bigcup \left(\left(\bigcup _{i\in \left[m\right]}{A}_i\right)+[o,u]\right)\hfill \\ \hfill & =\left(\bigcup _{i\in \left[m\right]}(A_i+[-u,o])\right)\bigcup \left(\bigcup _{i\in \left[m\right]}(A_i+[o,u])\right)\hfill \\ \hfill & =\left(\bigcup _{i\in \left[m\right]}{C}_{-1,0}\left(A_i\right)\right)\bigcup \left(\bigcup _{i\in \left[m\right]}{C}_{0,1}\left(A_i\right)\right),\hfill \end{array}$$

By Lemma 3,

$$\delta \left(C\right)=\underset{x,y\in A}{sup}\parallel x-y+2u\parallel ,$$

and

$$\delta \left(C_{-1,0}\left(A_i\right)\right)=\delta \left(C_{0,1}\left(A_i\right)\right)=\underset{x,y\in A_i}{sup}\parallel x-y+u\parallel .$$

Since, for two arbitrary points $x,y\in A_i$,

$$\begin{array}{cc}\hfill \parallel x-y+u\parallel & ={\displaystyle \frac{1}{2}}\parallel x-y+x-y+2u\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\parallel x-y\parallel +{\displaystyle \frac{1}{2}}\parallel x-y+2u\parallel \hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\delta \left(A_i\right)+{\displaystyle \frac{1}{2}}\underset{x,y\in A_i}{sup}\parallel x+2u-y\parallel ,\hfill \end{array}$$

we have

$$\underset{x,y\in A_i}{sup}\parallel x-y+u\parallel \le {\displaystyle \frac{1}{2}}\delta \left(A_i\right)+{\displaystyle \frac{1}{2}}\underset{x,y\in A_i}{sup}\parallel x+2u-y\parallel .$$

Put $\lambda ={sup}_{x,y\in A}\parallel x+2u-y\parallel$. Then, $\lambda \ge 1$ and

$$\begin{array}{cc}\hfill {\displaystyle \frac{\delta \left(C_{0,1}\left(A_i\right)\right)}{\delta \left(C\right)}}& \le {\displaystyle \frac{\delta \left(A_i\right)}{2\lambda }}+{\displaystyle \frac{{sup}_{x,y\in A_i}\parallel x+2u-y\parallel }{2\lambda }}\hfill \\ \hfill & \le {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{2\lambda }}+{\displaystyle \frac{1}{2}},i\in \left[m\right].\hfill \end{array}$$

Since $\delta \left(C_{-1,0}\left(A_i\right)\right)=\delta \left(C_{0,1}\left(A_i\right)\right)$, we have

$${\displaystyle \frac{\delta \left(C_{-1,0}\left(A_i\right)\right)}{\delta \left(C\right)}}\le {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{2\lambda }}+{\displaystyle \frac{1}{2}},i\in \left[m\right].$$

Therefore,

$${\beta }_X(C,2m)\le {\displaystyle \frac{{\beta }_X(A,m)}{2}}+{\displaystyle \frac{1}{2}}.$$

This completes the proof. □

4. Estimations of ${\mathit{\beta }}_{\mathit{X}}(conv((\mathit{A}+\mathit{u})\cup (-\mathit{A}-\mathit{u})),\mathbf{2}\mathit{m})$

Put

$$T=conv\left(\right(A+u)\cup (-A-u\left)\right).$$

It is clear that T is a centrally symmetric convex body in ${\mathbb{R}}^n$.

Lemma 4.

Let $T\in {\mathcal{K}}^{n+1}$, defined as above. Then,

$$\delta \left(T\right)=\underset{x,y\in A}{sup}\parallel x+y+2u\parallel .$$

Proof. 

For every pair of points $x,y\in T$, there exist two numbers $\alpha ,\beta \in [0,1]$ and four points $x_1,x_2,y_1,y_2\in A$ such that

$$x=(1-\alpha )(x_1+u)+\alpha (-x_2-u)$$

and

$$y=(1-\beta )(y_1+u)+\beta (-y_2-u).$$

Without loss of generality, we assume that $\alpha \le \beta$. Then,

$$\begin{array}{cc}\hfill \parallel x-y\parallel & =\parallel x_1-y_1+\beta y_1-\alpha x_1-\alpha x_2+\beta y_2+2(\beta -\alpha )u\parallel \hfill \\ \hfill & =\parallel (1-\beta )(x_1-y_1)+(\beta -\alpha )(x_1+y_2+2u)+\alpha (y_2-x_2)\parallel \hfill \\ \hfill & \le (1-\beta )\parallel x_1-y_1\parallel +(\beta -\alpha )\parallel x_1+y_2+2u\parallel +\alpha \parallel x_2-y_2\parallel \hfill \\ \hfill & \le (1-\beta )\delta \left(A\right)+(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel +\alpha \delta \left(A\right)\hfill \\ \hfill & =(1+\alpha -\beta )\delta \left(A\right)+(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \hfill \\ \hfill & \le max\left\{\delta \left(A\right),\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \right\}.\hfill \end{array}$$

Therefore,

$$\delta \left(T\right)\le max\left\{\delta \left(A\right),\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \right\}.$$

Since $A+u\subseteq T$ and $\parallel x+u-(-y-u)\parallel \le \delta \left(T\right)$ for all $x,y\in A$, $\delta \left(A\right)\le \delta \left(T\right)$ and ${sup}_{x,y\in A}\parallel x+y+2u\parallel \le \delta \left(T\right)$. Thus,

$$\delta \left(T\right)\ge max\left\{\delta \left(A\right),\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \right\}.$$

Suppose that ${sup}_{x,y\in A}\parallel x+y+2u\parallel <\delta \left(A\right)$. Since, for each $x\in A$,

$$\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \ge 2\parallel x+u\parallel .$$

we have $\parallel x+u\parallel <{\displaystyle \frac{\delta \left(A\right)}{2}}$. It follows that $A+u\subseteq int\left({\displaystyle \frac{\delta \left(A\right)}{2}}{B}_X\right)$. Thus,

$$\delta \left(A\right)=\delta (A+u)<\delta \left(A\right),$$

which is impossible. Therefore,

$$\delta \left(T\right)=max\left\{\delta \left(A\right),\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \right\}=\underset{x,y\in A}{sup}\parallel x+y+2u\parallel .$$

□

Theorem 3.

Let $T\in {\mathcal{K}}^n$, defined as above, and m be a positive integer such that ${\beta }_X(A,m)<1$. Then,

$${\beta }_X(T,2m)\le {\displaystyle \frac{{\beta }_X(A,m)}{2}}+{\displaystyle \frac{1}{2}}.$$

Proof. 

For each $ϵ\in (0,1-{\beta }_X(A,m))$, there exists a collection $\{A_i\mid i\in \left[m\right]\}$ of subsets of A such that

$$A=\bigcup \{A_i\mid i\in \left[m\right]\},$$

and

$$\underset{i\in \left[m\right]}{max}\delta \left(A_i\right)\le {\beta }_X(A,m)+ϵ.$$

Then, $A+u=\bigcup \{A_i+u\mid i\in \left[m\right]\}=\bigcup \{S_i\mid i\in \left[m\right]\}$ and $-A-u=\bigcup \{-A_i-u\mid i\in \left[m\right]\}=\bigcup \{-S_i\mid i\in \left[m\right]\}$, where $S_i=A_i+u$. Put

$$H^{+}=\{(x_1,x_2,\cdots ,x_n)\mid x_n\ge 0\},$$

and

$$H^{-}=\{(x_1,x_2,\cdots ,x_n)\mid x_n\le 0\}.$$

Without loss of generality, we assume that $u\in H^{+}$. Then,

$$\begin{array}{cc}\hfill T& =(conv((A+u)\cup (-A-u))\cap H^{+})\cup (conv(A+u\cup (-A-u))\cap H^{-})\hfill \\ \hfill & =\left(\left(conv\left(\left(\bigcup _{i\in \left[m\right]}{S}_i\right)\cup (-A-u)\right)\right)\cap H^{+}\right)\bigcup \left(conv\left((A+u)\cup \bigcup _{i\in \left[m\right]}(-S_i)\right)\cap H^{-}\right)\hfill \\ \hfill & =\left(\bigcup _{i\in \left[m\right]}((conv(S_i\cup (-A-u)))\cap H^{+})\right)\bigcup \left(\bigcup _{i\in \left[m\right]}((conv((-S_i)\cup (A+u))\cap H^{-})\right)\hfill \\ \hfill & =\left(\bigcup _{i\in \left[m\right]}{C}_i\right)\bigcup \left(\bigcup _{i\in \left[m\right]}{D}_i\right),\hfill \end{array}$$

where

$$C_i=(conv((A_i+u)\cup (-A-u)))\cap H^{+},i\in \left[m\right],$$

and

$$D_i=(conv((-A_i-u)\cup (A+u)))\cap H^{-},i\in \left[m\right].$$

For every pair of points $x,y\in C_i$, there exist two numbers $\alpha ,\beta \in [0,{\displaystyle \frac{1}{2}}]$, two points $x_1,y_1\in conv{A}_i$, and two points $x_2,y_2\in A$ such that

$$x=(1-\alpha )(x_1+u)+\alpha (-x_2-u)$$

and

$$y=(1-\beta )(y_1+u)+\beta (-y_2-u).$$

Without loss of generality, we assume that $\alpha \le \beta$. Then,

$$\begin{array}{cc}\hfill \parallel x-y\parallel & =\parallel x_1-y_1+\beta y_1-\alpha x_1-\alpha x_2+\beta y_2+2(\beta -\alpha )u\parallel \hfill \\ \hfill & =\parallel (1-\beta )(x_1-y_1)+(\beta -\alpha )(x_1+y_2+2u)+\alpha (y_2-x_2)\parallel \hfill \\ \hfill & \le (1-\beta )\parallel x_1-y_1\parallel +(\beta -\alpha )\parallel x_1+y_2+2u\parallel +\alpha \parallel x_2-y_2\parallel \hfill \\ \hfill & \le (1-\beta )\delta (conv{A}_i)+(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel +\alpha \delta \left(A\right)\hfill \\ \hfill & =(1-\beta )\delta \left(A_i\right)+(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel +\alpha \hfill \\ \hfill & \le (1-\beta )({\beta }_X(A,m)+ϵ)+(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel +\alpha .\hfill \end{array}$$

By Lemma 4,

$$\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \ge \delta \left(A\right)=1>{\beta }_X(A,m)+ϵ.$$

Then, for each $i\in \left[m\right]$,

$$\begin{array}{cc}\hfill \delta \left(C_i\right)& \le (1-\beta )({\beta }_X(A,m)+ϵ)+\alpha +(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \hfill \\ \hfill & ={\beta }_X(A,m)+ϵ+\beta \left(\underset{x,y\in A}{sup}\parallel x+y+2u\parallel -{\beta }_X(A,m)-ϵ\right)\hfill \\ \hfill & +\alpha \left(1-\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \right)\hfill \\ \hfill & \le {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{2}}+{\displaystyle \frac{{sup}_{x,y\in A}\parallel x+y+2u\parallel }{2}}.\hfill \end{array}$$

By Lemma 4, $\delta \left(T\right)={sup}_{x,y\in A}\parallel x+y+2u\parallel \ge 1$. Put $\lambda ={sup}_{x,y\in A}\parallel x+y+2u\parallel$. Then, for $i\in \left[m\right]$,

$$\begin{array}{cc}\hfill {\displaystyle \frac{\delta \left(C_i\right)}{\delta \left(T\right)}}& \le {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{2\lambda }}+{\displaystyle \frac{1}{2}}.\hfill \end{array}$$

For every pair of points $x,y\in D_i$, there exist two numbers $\alpha ,\beta \in [{\displaystyle \frac{1}{2}},1]$, two points $x_1,y_1\in conv{A}_i$, and two points $x_2,y_2\in A$ such that

$$x=\alpha (-x_1-u)+(1-\alpha )(x_2+u)$$

and

$$y=\beta (-y_1-u)+(1-\beta )(y_2+u).$$

Without loss of generality, we assume that $\alpha \le \beta$. Then,

$$\begin{array}{cc}\hfill \parallel x-y\parallel & =\parallel x_2-y_2+\beta y_1-\alpha x_1-\alpha x_2+\beta y_2+2(\beta -\alpha )u\parallel \hfill \\ \hfill & =\parallel (1-\beta )(x_2-y_2)+(\beta -\alpha )(x_2+y_1+2u)+\alpha (y_1-x_1)\parallel \hfill \\ \hfill & \le (1-\beta )\parallel x_2-y_2\parallel +(\beta -\alpha )\parallel x_2+y_1+2u\parallel +\alpha \parallel y_1-x_1\parallel \hfill \\ \hfill & \le 1-\beta +(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel +\alpha \delta \left(A_i\right)\hfill \\ \hfill & \le 1-\beta +(\beta -\alpha )\underset{x,y\in A}{sup}\parallel x+y+2u\parallel +\alpha ({\beta }_X(A,m)+ϵ).\hfill \end{array}$$

By Lemma 4,

$$\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \ge \delta \left(A\right)=1>{\beta }_X(A,m)+ϵ.$$

Then, for each $i\in \left[m\right]$,

$$\begin{array}{cc}\hfill \delta \left(D_i\right)& \le 1+\beta \left(\underset{x,y\in A}{sup}\parallel x+y+2u\parallel -1\right)+\alpha \left({\beta }_X(A,m)+ϵ-\underset{x,y\in A}{sup}\parallel x+y+2u\parallel \right)\hfill \\ \hfill & \le {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{2}}+{\displaystyle \frac{{sup}_{x,y\in A}\parallel x+y+2u\parallel }{2}}.\hfill \end{array}$$

Thus,

$${\displaystyle \frac{\delta \left(D_i\right)}{\delta \left(T\right)}}\le {\displaystyle \frac{{\beta }_X(A,m)+ϵ}{2\lambda }}+{\displaystyle \frac{1}{2}},i\in \left[m\right].$$

It follows that ${\beta }_X(T,2m)\le {\displaystyle \frac{{\beta }_X(A,m)}{2\lambda }}+{\displaystyle \frac{1}{2}}\le {\displaystyle \frac{{\beta }_X(A,m)}{2}}+{\displaystyle \frac{1}{2}}$. This completes the proof. □

5. Discussion

Estimating partitioning functionals of special convex bodies plays an important role in Chuanming Zong’s proof program to attack Borsuk’s problem. In this paper, when a convex body D is constructed from a convex body A, for example, $D\subseteq {\mathbb{R}}^n$ is a convex cone with $A\subseteq {\mathbb{R}}^{n-1}\times \left\{0\right\}$ as the base, using a partition of A, we obtain a partition of D; it follows that a relationship between the partitioning functional of D and partitioning functional of A is given.