Arrangements of homothets of a convex body II

TL;DR

This paper establishes upper bounds on the size of arrangements of homothets of convex bodies in Euclidean space, improving previous results and providing new bounds for specific configurations.

Contribution

It improves bounds on the maximum size of Minkowski arrangements and related configurations of homothets of convex bodies, using novel geometric arguments.

Findings

Any pairwise intersecting Minkowski arrangement has at most 2·3^d members.

Sequences of homothets with centers on boundaries have at most O(3^d d) members.

The results generalize and strengthen previous bounds by Polyanskii.

Abstract

A family of homothets of an o-symmetric convex body K in d-dimensional Euclidean space is called a Minkowski arrangement if no homothet contains the center of any other homothet in its interior. We show that any pairwise intersecting Minkowski arrangement of a d-dimensional convex body has at most $2\cdot 3^d$ members. This improves a result of Polyanskii (arXiv:1610.04400). Using similar ideas, we also give a proof the following result of Polyanskii: Let $K_1,\dots,K_n$ be a sequence of homothets of the o-symmetric convex body $K$, such that for any $i<j$, the center of $K_j$ lies on the boundary of $K_i$. Then $n\leq O(3^d d)$.

Equations

$$\left|f_{ij}(x_k)\right|\le \left|f_{ij}(x_i)-f_{ij}(x_j)\right|\text{for all k∈[n].}$$

$$\left|f_{ij}(x_k)\right|\le \frac{\lambda }{2}\left|f_{ij}(x_i)-f_{ij}(x_j)\right|\text{for all k∈[n].}$$

$$n\le \frac{(1+2\alpha {)}^{D-1}(1+3\alpha )}{2{\alpha }^D}.$$

$$\phi (v_j)-{\lambda }_j\le \alpha \le \phi (v_i)+{\lambda }_i.$$

$$\phi (v_j-v_i)={\Vert v_j-v_i\Vert }_K\ge \max \{{\lambda }_i,{\lambda }_j\}.$$

$$\phi (v_i)\le \alpha \le \phi (v_j).$$

$$f(x_j-x_i)$$

$$=\frac{\phi (v_j)-\alpha }{{\lambda }_j}+\frac{\alpha -\phi (v_i)}{{\lambda }_i}$$

$$\stackrel{\text{eqref}eq2}{\ge }\frac{\phi (v_j)-\alpha +\alpha -\phi (v_i)}{\max \{{\lambda }_i,{\lambda }_j\}}$$

$$=\frac{{\Vert v_j-v_i\Vert }_K}{\max \{{\lambda }_i,{\lambda }_j\}}\ge 1.$$

$$\left|f(x_k)\right|=\left|\phi ({\lambda }_k^{-1}{v}_k)-\alpha {\lambda }_k^{-1}\right|=\left|\phi (x)\right|\le {\Vert x\Vert }_K\le 1.$$

$$X_m=\{x_i:\lfloor N{\log }_2{\lambda }_i\rfloor \equiv m(\mathrm{mod}N+1)\}.$$

$$\phi (x_j-x_i)=\Vert v_j-v_i\Vert ={\lambda }_i.$$

$$\phi (v_j)-{\lambda }_j<\alpha <\phi (v_i)+{\lambda }_i.$$

$$-{\lambda }_i<\phi (v_i)-\alpha <{\lambda }_j-{\lambda }_i.$$

$$n\le (N+1)(1+\lambda /2)(1+\lambda {)}^d.$$

$$\lfloor N{\log }_2{\lambda }_j\rfloor -\lfloor N{\log }_2{\lambda }_i\rfloor =kN\text{for some k∈Z.}$$

$${\lambda }_i=\Vert v_i-v_j\Vert \ge \Vert v_j-v_n\Vert -\Vert v_n-v_i\Vert ={\lambda }_j-{\lambda }_i,$$

$$\frac{{\lambda }_j}{{\lambda }_i}<2^{1/N}.$$

$$f(x_j-x_i)$$

$$=\frac{\phi (v_j)-\alpha }{{\lambda }_j}+\frac{\alpha -\phi (v_i)}{{\lambda }_i}$$

$$=\frac{\phi (v_i)+{\lambda }_i-\alpha }{{\lambda }_j}+\frac{\alpha -\phi (v_i)}{{\lambda }_i}$$

$$\stackrel{\text{eqref}2.1,\text{eqref}2.2}{>}\frac{2^{-1/N}(\phi (v_i)+{\lambda }_i-\alpha )+\alpha -\phi (v_i)}{{\lambda }_i}$$

$$=2^{-1/N}+\frac{(1-2^{-1/N})(\alpha -\phi (v_i))}{{\lambda }_i}$$

$$\stackrel{\text{eqref}2.1}{>}{2}^{-1/N}+\frac{(1-2^{-1/N})({\lambda }_i-{\lambda }_j)}{{\lambda }_i}$$

$$=1-(1-2^{-1/N})\frac{{\lambda }_j}{{\lambda }_i}$$

$$\stackrel{\text{eqref}2.1}{>}1-(1-2^{-1/N})2^{1/N}$$

$$=2-2^{1/N}.$$

$$\left|f(x_k)\right|=\left|\phi ({\lambda }_k^{-1}{v}_k)-\alpha {\lambda }_k^{-1}\right|=\left|\phi (x)\right|\le {\Vert x\Vert }_K\le 1.$$

$$w\le \left|f(t_i)-f(t_j)\right|.$$

$$F(y):=\frac{1}{f(y)}\underset{0}{\overset{y}{\int }}f(x)dx$$

$$\underset{-\alpha }{\overset{1+2\alpha }{\int }}{V}_{D-1}\left(\left(\bigcup _{i=1}^n\alpha K+t_i\right)\cap h(x)\right)dx=n{\alpha }^D{V}_D(K).$$

$$\underset{0}{\overset{1+2\alpha }{\int }}{V}_{D-1}\left(\left(\bigcup _{i=1}^n\alpha K+t_i\right)\cap h(x)\right)dx$$

$$\le \underset{0}{\overset{1+2\alpha }{\int }}{V}_{D-1}\left((1+2\alpha )K\cap h(x)\right)dx=\frac{(1+2\alpha {)}^d}{2}{V}_D(K).$$

$$\underset{-\alpha }{\overset{0}{\int }}{V}_{D-1}\left(\left(\bigcup _{i=1}^n(\alpha K+t_i)\right)\cap h(x)\right)dx=\sum _{i=1}^m\underset{0}{\overset{\alpha -a_i}{\int }}f(x)dx$$

Full text

1. Introduction

We use the notation $[n]=\{1,2,\dots,n\}$. A convex body $K$ in the $d$-dimensional Euclidean space $\mathbb{R}^{d}$ is a compact convex set with non-empty interior, and is $o$-symmetric if $K=-K$. A (positive) homothet of $K$ is a set of the form $\lambda K+v:=\{\lambda k+v:k\in K\}$, where $\lambda>0$ is the homothety ratio, and $v\in\mathbb{R}^{d}$ is a translation vector. If $K$ is $o$-symmetric, we also call $v$ the center of the homothet $\lambda K+v$. An arrangement of homothets of $K$ is a collection $\{\lambda_{i}K+v_{i}:i\in[n]\}$. A Minkowski arrangement of an $o$-symmetric convex body $K$ is a family $\{v_{i}+\lambda_{i}K\}$ of homothets of $K$ such that none of the homothets contains the center of any other homothet in its interior. This notion was introduced by L. Fejes Tóth [3] in the context of Minkowski’s fundamental theorem on the minimal determinant of a packing lattice for a symmetric convex body, and was further studied by him in [4, 5], by Böröczky and Szabó in [2], and in connection with the Besicovitch covering theorem by Füredi and Loeb [6]. Recently, Minkowski arrangements have been used to study a problem arising in the design of wireless networks [10]. In [9] it was shown that the largest cardinality of a pairwise intersecting Minkowski arrangement of homothets of an $o$-symmetric convex body in $\mathbb{R}^{d}$ is $O(3^{d}d\log d)$. This was improved to $3^{d+1}$ by Polyanskii [11]. We make the following slight improvement.

Theorem 1**.**

For any $o$-symmetric convex body $K$ in $\mathbb{R}^{d}$, a pairwise intersecting Minkowski arrangement has at most $2\cdot 3^{d}$ members.

Note that the $d$-cube has $3^{d}$ pairwise intersecting translates that form a Minkowski arrangement. The proof uses ideas from [8] and [7].

In [9], bounds on pairwise intersecting Minkowski arrangements were used to give an upper bound of $O(6^{d}d^{2}\log d)$ on the length of a sequence of homothets $v_{i}+\lambda_{i}K$ of an $o$-symmetric convex body $K$ such that $v_{j}\in\operatorname{bd}(v_{i}+\lambda_{i}K)$ whenever $j>i$. This bound was improved to $O(3^{d}d)$ by Polyanskii [11]. We use some similar ideas to the proof of Theorem 1 to give a short proof of this result of Polyanskii.

Theorem 2** (Polyanskii [11]).**

Let $K$ be an $o$-symmetric convex body, and $v_{1},v_{2},\ldots,v_{n}\in\mathbb{R}^{d}$. Let $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}>0$, and assume that for any $1\leq i<j\leq n$ we have $v_{j}\in\operatorname{bd}(v_{i}+\lambda_{i}K)$. Then $n\leq O(3^{d}d)$.

The interest in this result is that it gives the upper bound $k^{O(3^{d}d)}$ to the cardinality of a set in a $d$-dimensional normed space in which only $k$ non-zero distances occur between pairs of points. This is currently the best known upper bound if $k=\Omega(3^{d}d)$ (see [12] for a survey of this problem).

2. Proof of Theorem 1

Theorem 3**.**

Let $d\geq 1$. Suppose that there exists an $o$-symmetric convex body $K$ in $\mathbb{R}^{d}$ which has a pairwise intersecting Minkowski arrangement of $n$ homothets. Then there exists a set $\{x_{1},\dots,x_{n}\}$ of $n$ points in $\mathbb{R}^{d+1}$ such that $o\notin\operatorname{conv}\{x_{1},\ldots,x_{n}\}$, and for any distinct $i,j\in[n]$, $i<j$, there exists a non-zero linear functional $f_{ij}\colon\mathbb{R}^{d+1}\to\mathbb{R}$ with

[TABLE]

We remark that the converse of the above theorem does not hold. For a simple counterexample, let $\{x_{1},\dots,x_{5}\}$ be the vertex set of a regular pentagon, with $o$ just outside the pentagon, close to the midpoint of an edge. It is easy to see that for any pair $x_{i},x_{j}$ of vertices there is a line through $o$ such that the projections $\pi(x_{k})$ of the vertices onto the line are all within distance $\left\lvert\pi(x_{i})-\pi(x_{j})\right\rvert$ of $o$. On the other hand, it is also easy to see that a pairwise intersecting Minkowski arrangement of intervals in $\mathbb{R}$ can have at most two members.

The above remark is to be contrasted with the equivalence in the following result, which generalizes part of Theorem 1.4 of [7].

Theorem 4**.**

Given $\lambda\geq 1$, and $D\in\mathbb{Z},D\geq 1$. Then the following statements are equivalent.

1. (i)

There exists a set $\{x_{1},\ldots,x_{n}\}$ of $n$ points in $\mathbb{R}^{D}$, such that $o\notin\operatorname{conv}\{x_{1},\ldots,x_{n}\}$, and for any distinct $i,j\in[n],i<j$ there exists a non-zero linear functional $f_{ij}:\mathbb{R}^{D}\to\mathbb{R}$ with

[TABLE] 2. (ii)

There is an $o$-symmetric convex set $L$ in $\mathbb{R}^{D}$ that has $n$ non-overlapping translates $L+t_{1},\ldots,L+t_{n}$, each intersecting $(\lambda-1)L$, with $o\notin\operatorname{conv}\{t_{1},\ldots,t_{n}\}$.

We note that the equivalence between (ii) and (iv) of Theorem 1.4 in [7] is exactly the above theorem in the case $\lambda=1$.

Theorem 5**.**

Let $K$ be an $o$-symmetric convex set in $\mathbb{R}^{D}$ with $D\geq 2$, and let $\alpha K+t_{1},\ldots,\alpha K+t_{n}$ be $n$ non-overlapping translates of $\alpha K$ with $\alpha>0$ such that each translate intersects $K$, and $o\notin\operatorname{int}(\operatorname{conv}\{t_{1},\ldots,t_{n}\})$. Then

[TABLE]

This theorem is a slight modification of Theorem 1.5 of [7]. There the translates of $\alpha K$ touch $K$, whereas here they may overlap with $K$. Theorem 5 is sharp for $\alpha=1$. Indeed, let $K$ be the cube $[-1,1]^{D}$, and consider the $2\cdot 3^{D-1}$ translation vectors $\{t\in\{-2,0,2\}^{D}\;:\;t^{(1)}\geq t^{(2)}\}$.

Combining Theorems 3, 4 and 5 (with $\lambda=2$, $K=(\lambda-1)L=L$, $\alpha=\frac{1}{\lambda-1}=1$), we immediately obtain Theorem 1.

3. Proof of Theorem 3

Let the Minkowski arrangement by $\{v_{i}+\lambda_{i}K:i\in[n]\}$, where $\lambda_{i}>0$ and $v_{i}\in\mathbb{R}^{d}$ for each $i\in[n]$. Let $x_{i}=(\lambda_{i}^{-1}v_{i},\lambda_{i}^{-1})\in\mathbb{R}^{d}\times\mathbb{R}$, $i\in[n]$. Fix distinct $i,j\in\{1,\dots,n\}$. We will find a linear $f\colon\mathbb{R}^{d}\times\mathbb{R}\to\mathbb{R}$ that satisfies (1). Let $\varphi\colon\mathbb{R}^{d}\to\mathbb{R}$ be a linear functional such that $\varphi(x)\leq\left\lVert x\right\rVert_{K}$ for all $x\in\mathbb{R}^{d}$ and $\varphi(v_{j}-v_{i})=\left\lVert v_{j}-v_{i}\right\rVert_{K}$. (Thus, $\varphi^{-1}(1)$ is a hyperplane that supports $K$ at $\left\lVert v_{j}-v_{i}\right\rVert_{K}^{-1}(v_{j}-v_{i})$.)

Since any two homothets $v_{k}+\lambda_{k}K$ and $v_{\ell}+\lambda_{\ell}K$ intersect, any two of the compact intervals $\varphi(v_{k}+\lambda_{k}K)$ and $\varphi(v_{\ell}+\lambda_{\ell}K)$ intersect in $\mathbb{R}$. By Helly’s Theorem in $\mathbb{R}$, there exists $\alpha\in\bigcap_{t=1}^{n}\varphi(v_{t}+\lambda_{t}K)$. Since $\varphi(v_{i}+\lambda_{i}K)=[\varphi(v_{i})-\lambda_{i},\varphi(v_{i})+\lambda_{i}]$ and $\varphi(v_{j}+\lambda_{j}K)=[\varphi(v_{j})-\lambda_{j},\varphi(v_{j})+\lambda_{j}]$, we have

[TABLE]

By the Minkowski property,

[TABLE]

It follows that

[TABLE]

We set $f=(\varphi,-\alpha)\in(\mathbb{R}^{d}\times\mathbb{R})^{*}$, that is, define $f(x)=\varphi(v)-\alpha\mu$, where $x=(v,\mu)\in\mathbb{R}^{d}\times\mathbb{R}$. We show that $f(x_{j}-x_{i})\geq 1$, and $\left\lvert f(x_{k})\right\rvert\leq 1$ for all $k\in\{1,\dots,n\}$. This will show that (1) is satisfied, which will finish the proof.

[TABLE]

Since $\alpha\in\varphi(v_{k}+\lambda_{k}K)$, there exists $x\in K$ such that $\varphi(v_{k}+\lambda_{k}x)=\alpha$. Therefore,

[TABLE]

∎

4. Proof of Theorem 2

The following proof is very similar to the proof of Theorem 3.

Without loss of generality, $\min_{i}\lambda_{i}=1$. Denote the unit ball of $\left\lVert\cdot\right\rVert$ by $K$. Let $x_{i}=(\lambda_{i}^{-1}v_{i},\lambda_{i}^{-1})\in\mathbb{R}^{d}\times\mathbb{R}$, $i=0,\dots,n-1$. Let $N\geq 1$, to be fixed later. For each $m=0,\dots,N$, let

[TABLE]

Then $X_{0},\dots,X_{N}$ partition $\{x_{0},\dots,x_{n-1}\}$ into $N+1$ parts. Fix $i,j\in X_{m}$ such that $0\leq i<j<n$. We will find a linear $f\colon\mathbb{R}^{d}\times\mathbb{R}\to\mathbb{R}$ such that (2) is satisfied for all $x_{k}\in X_{m}$ and $\lambda=2-2^{1/N}$. Let $\varphi\colon\mathbb{R}^{d}\to\mathbb{R}$ be a linear functional such that $\varphi(x)\leq\left\lVert x\right\rVert$ for all $x\in\mathbb{R}^{d}$ and

[TABLE]

(Thus, $\varphi^{-1}(1)$ is a hyperplane that supports $K$ at $\left\lVert v_{j}-v_{i}\right\rVert_{K}^{-1}(v_{j}-v_{i})$.)

Since any two homothets $v_{k}+\lambda_{k}K$ and $v_{\ell}+\lambda_{\ell}K$ intersect in their interiors, any two of the open intervals $\varphi(v_{k}+\lambda_{k}\operatorname{int}K)$ and $\varphi(v_{\ell}+\lambda_{\ell}\operatorname{int}K)$ intersect in $\mathbb{R}$. By Helly’s Theorem in $\mathbb{R}$, there exists $\alpha\in\bigcap_{t=1}^{n}\varphi(v_{t}+\lambda_{t}\operatorname{int}K)$. Since $\varphi(v_{i}+\lambda_{i}\operatorname{int}K)=(\varphi(v_{i})-\lambda_{i},\varphi(v_{i})+\lambda_{i})$ and $\varphi(v_{j}+\lambda_{j}\operatorname{int}K)=(\varphi(v_{j})-\lambda_{j},\varphi(v_{j})+\lambda_{j})$, we have

[TABLE]

By (5), we can rewrite this as

[TABLE]

We set $f=(\varphi,-\alpha)\in(\mathbb{R}^{d}\times\mathbb{R})^{*}$, that is, for $x=(v,\mu)\in\mathbb{R}^{d}\times\mathbb{R}$, we let $f(x)=\varphi(v)-\alpha\mu$. It remains to show that $f(x_{j}-x_{i})>2-2^{1/N}$, and $\left\lvert f(x_{k})\right\rvert\leq 1$ for all $k\in\{0,\dots,n\}$, since this will show that (2) is satisfied with $\lambda=2-2^{1/N}$. By applying Theorems 4 and 5 with $\lambda=2/(2-2^{1/N})=2+\frac{\log 4}{N}+O(N^{-2})$, $K=(\lambda-1)L$ and $\alpha=1/(\lambda-1)=2^{1-1/N}-1$, we obtain $\left\lvert X_{m}\right\rvert\leq(1+\lambda/2)(1+\lambda)^{d}$, and it follows that

[TABLE]

If we choose $N=d$, we obtain $\lambda=2+\frac{\log 4}{d}+O(d^{-2})$ and $n=3^{d}O(d)$, which would finish the proof.

By definition of $X_{m}$,

[TABLE]

If $k\geq 1$, then $N\log_{2}\lambda_{j}-N\log_{2}\lambda_{i}>N$, hence $\lambda_{j}/\lambda_{i}>2$. However, we also have

[TABLE]

a contradiction. Therefore, $k\leq 0$, that is, $\lfloor N\log_{2}\lambda_{j}\rfloor-\lfloor N\log_{2}\lambda_{i}\rfloor\leq 0$. This gives $N\log_{2}\lambda_{j}-N\log_{2}\lambda_{i}<1$ and

[TABLE]

It follows that

[TABLE]

Since $\alpha\in\varphi(v_{k}+\lambda_{k}K)$, there exists $x\in K$ such that $\varphi(v_{k}+\lambda_{k}x)=\alpha$. Therefore,

[TABLE]

∎

5. Proof of Theorem 4

Assume that (i) holds. Let $C:=\bigcap_{i\neq j}S_{ij}$ be the intersection of the $o$-symmetric slabs $S_{ij}:=\left\{p\in\mathbb{R}^{D}:\left\lvert f_{ij}(p)\right\rvert\leq\frac{\lambda}{2}\left\lvert f_{ij}(x_{i})-f_{ij}(x_{j})\right\rvert\right\}$. By assumption, $C\supseteq\{x_{1},\ldots,x_{n}\}$. For each $i\in[n]$, let $C_{i}:=\frac{\lambda x_{i}+C}{\lambda+1}$ be the homothetic copy of $C$ with center of homothety $x_{i}$, and of ratio $\frac{1}{\lambda+1}$. It is an easy exercise that the $C_{i}$s are non-overlapping. Moreover, by the symmetry of $C$, we have $\frac{\lambda-1}{\lambda+1}x_{i}\in C_{i}\cap\frac{\lambda-1}{\lambda+1}C$. Thus, for $L:=\frac{1}{\lambda+1}C$, and $t_{i}:=\frac{\lambda}{\lambda+1}x_{i}$, (ii) holds as promised.

Next, assume that (ii) holds. Fix $i,j\in[n],i\neq j$. Since $L+t_{i}$ and $L+t_{j}$ are non-overlapping, there is a linear functional $f$ such that the two real intervals $s_{i}:=f(L+t_{i})$ and $s_{j}:=f(L+t_{i})$ do not overlap. These two intervals are of equal length, which we denote by $w$. Thus, we have

[TABLE]

On the other hand, $s_{k}:=f(L+t_{k})$ is also a real interval of length $w$ for any $k\in[n]$; and $s_{0}:=f((\lambda-1)L)$ is a 0-symmetric real interval of length $(\lambda-1)w$, which intersects each $s_{k}$. Thus, for the center $f(t_{k})$ of $s_{k}$, we have $\left\lvert f(t_{k})\right\rvert\leq\frac{(\lambda-1)w}{2}+\frac{w}{2}=\frac{\lambda w}{2}$. Now, (8) yields $\left\lvert f(t_{k})\right\rvert\leq\frac{\lambda}{2}\left\lvert f(t_{i})-f(t_{j})\right\rvert$. Thus, we may set $f_{ij}:=f$. This argument is valid for any $i$ and $j$, thus, with $x_{i}:=t_{i}$, we obtain (i).

6. Proof of Theorem 5

The proof is an almost verbatim copy of the proof of Theorem 1.5 of [7]. There are two points of difference, which we will note.

We recall Lemma 3.1. of [7], which is a slightly more general version of the Lemma of [1].

Lemma 1**.**

Let $f$ be a function on $[0,1]$ with the properties $f(0)\geq 0$, $f$ is positive and monotone increasing on $(0,1]$, and $f(x)=(g(x))^{k}$ for some concave function $g$ and $k>0$. Then

[TABLE]

is strictly increasing on $(0,1]$.

Proof of Theorem 5..

Clearly, we may assume that $K$ is bounded, otherwise, by a projection, we can reduce the dimension. Let $\alpha K+t_{1}$, $\alpha K+t_{2},\ldots,\alpha K+t_{n}$ be pairwise non-overlapping translates of $\alpha K$ that intersect $K$. By the assumptions of the theorem, there is a non-zero vector $v\in\mathbb{R}^{D}$ such that $a_{i}:=\langle t_{i},v\rangle\geq 0$ for $i\in[n]$. Set $h(x):=\{p\in\mathbb{R}^{D}:\langle p,v\rangle=x\}$. Without loss of generality, we may assume that $h(-1)$ and $h(1)$ are supporting hyperplanes of $K$.

Clearly, $\alpha K+t_{i}$ is between $h(-\alpha)$ and $h(1+2\alpha)$, and it is contained in $(1+2\alpha)K$, for $i\in[n]$.

[TABLE]

[TABLE]

[TABLE]

We note that this was the first point of difference from the proof in [7]: here, we do not subtract the contribution of $K$ in the total volume on the right hand side of the inequality.

Set $f(x):=\operatorname{V}_{D-1}\left(\alpha K\cap h(x-\alpha)\right)$, and observe that the conditions of Lemma 1 are satisfied by $f$ (with $k=D-1$, by the Brunn–Minkowski inequality). We may assume that $a_{1},\dots,a_{m}\leq\alpha<a_{m+1},\dots,a_{n}$. By Lemma 1,

[TABLE]

We note that this was the second point of difference from the proof in [7]: again, the contribution of $K$ to the volume is not subtracted.

This inequality, combined with (9) and (10), yields (3).

∎