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Few distance sets in p\ell_{p} spaces and p\ell_{p} product spaces

Richard Chen [email protected] Feng Gui [email protected] Jason Tang [email protected] Nathan Xiong [email protected] Lexington High School, Lexington, MA 02421, USA Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA Belmont High School, Belmont, MA 02478, USA Phillips Academy Andover, Andover, MA 01810, USA
Abstract

Kusner asked if n+1n+1 points is the maximum number of points in n\mathbb{R}^{n} such that the p\ell_{p} distance (1<p<)(1<p<\infty) between any two points is 11. We present an improvement to the best known upper bound when pp is large in terms of nn, as well as a generalization of the bound to ss-distance sets. We also study equilateral sets in the p\ell_{p} sums of Euclidean spaces, deriving upper bounds on the size of an equilateral set for when p=p=\infty, pp is even, and for any 1p<1\leq p<\infty.

1 Introduction

1.1 Background

A classic exercise in linear algebra asks for the maximum number of points in n\mathbb{R}^{n} such that the pairwise distances take only two values. One can associate a polynomial to each point in the set such that the polynomials are linearly independent. Then, one can show that the polynomials all lie in a subspace of dimension (n+1)(n+4)/2(n+1)(n+4)/2. Since the number of linearly independent vectors cannot exceed the dimension of the subspace, the cardinality of the set is at most (n+1)(n+4)/2(n+1)(n+4)/2. This beautiful argument was found by Larman, Rogers, and Seidel [10] in 1977, illustrating the power of algebraic techniques in combinatorics.

We can ask the much more general question: “In a metric space XX, what is the maximum number of points such that the pairwise distances take only ss values?” We use es(X)e_{s}(X), or just e(X)e(X) if s=1s=1, to denote the answer to this question (by convention, we do not count 0 as a distance). A set of points SXS\subseteq X satisfying this question’s conditions, i.e., the cardinality of the set {d(x,y):x,yS,xy}\{d(x,y):x,y\in S,x\neq y\} is ss, is called an ss-distance set. A 11-distance set is also known as an equilateral set. Also, we typically restrict the metric space to a normed space, so that the specific distances used do not matter. Thus, we will always assume that the largest of the ss distances is 11.

The posed question has been studied on many different spaces. The most famous result would be the upper bound (n+ss)\binom{n+s}{s} on an ss-distance set in nn-dimensional Euclidean space, found by Bannai, Bannai, and Stanton [3]. This result was also discovered independently by Blokhuis [4]. Another important case is when all points lie on the nn-dimensional sphere. There is strong motivation for this problem as it has many applications in coding theory and design theory [6]. In particular, having tight upper bounds can help us find extremal configurations which satisfy unique properties.

Similar results have been obtained in the hyperbolic space [4], the Hamming space [12], and the Johnson space [12]. Not as much is known in an arbitrary finite-dimensional normed space, known as a Minkowski space, other than Petty’s [13] general bound of e(X)2ne(X)\leq 2^{n}, where n=dimXn=\dim X. Swanepoel [17] then conjectured that es(X)(s+1)ne_{s}(X)\leq(s+1)^{n} for a Minkowski space XX with dimension nn and proved it for the n=2n=2 case. We should mention that equilateral sets in Minkowski spaces have been applied in differential geometry, where they are used to find minimal surfaces [11].

1.2 Definitions

In our paper, we investigate this problem on n\mathbb{R}^{n} with the p\ell_{p} norm, as well as on the p\ell_{p} sum of Euclidean spaces. For a point x=(x1,,xn)nx=(x_{1},\ldots,x_{n})\in\mathbb{R}^{n} and a p1p\geq 1, the p\ell_{p} norm is defined to be

xp=(i=1n|xi|p)1p,\left\lVert x\right\rVert_{p}=\left(\sum_{i=1}^{n}|x_{i}|^{p}\right)^{\frac{1}{p}},

and the \ell_{\infty} norm is defined to be

x=max1in|xi|.\left\lVert x\right\rVert_{\infty}=\max_{1\leq i\leq n}|x_{i}|.

The 1\ell_{1} norm is the well-known “taxicab” norm, and the 2\ell_{2} norm is the standard Euclidean norm. Throughout the paper, we write \left\lVert\cdot\right\rVert instead of 2\left\lVert\cdot\right\rVert_{2} to emphasize that we are using the Euclidean norm. We also use 𝔼n\mathbb{E}^{n} to emphasize that we are in nn-dimensional Euclidean space.

For Euclidean spaces 𝔼a1,,𝔼an\mathbb{E}^{a_{1}},\ldots,\mathbb{E}^{a_{n}}, we define their p\ell_{p} sum as the product space 𝔼a1××𝔼an\mathbb{E}^{a_{1}}\times\cdots\times\mathbb{E}^{a_{n}} equipped with the norm

(x1,,xn)p=(i=1nxip)1p,\left\lVert(x_{1},\ldots,x_{n})\right\rVert_{p}=\left(\sum_{i=1}^{n}\left\lVert x_{i}\right\rVert^{p}\right)^{\frac{1}{p}},

where xi𝔼aix_{i}\in\mathbb{E}^{a_{i}} for i=1,,ni=1,\ldots,n. When p=p=\infty, the norm is

(x1,,xn)=max1inxi.\left\lVert(x_{1},\ldots,x_{n})\right\rVert_{\infty}=\max_{1\leq i\leq n}\left\lVert x_{i}\right\rVert.

Then, for any p[1,]p\in[1,\infty], we let the distance between two points xx and yy be xyp\left\lVert x-y\right\rVert_{p} and use 𝔼a1pp𝔼an\mathbb{E}^{a_{1}}\oplus_{p}\cdots\oplus_{p}\mathbb{E}^{a_{n}} to describe this space.

Although our notation for the norm in p\ell_{p} spaces and in p\ell_{p} sums is the same, the norm being used should be clear from context.

1.3 Previous Work and Our Results

We first study ss-distance sets in n\mathbb{R}^{n} equipped with the p\ell_{p} norm. This space is denoted by pn=(n,p)\ell_{p}^{n}=(\mathbb{R}^{n},\left\lVert\cdot\right\rVert_{p}). The two most famous questions pertaining to this problem are Kusner’s [7] conjectures on equilateral sets.

Conjecture 1 (Kusner).

e(1n)=2ne(\ell_{1}^{n})=2n.

Conjecture 2 (Kusner).

e(pn)=n+1e(\ell_{p}^{n})=n+1 for 1<p<1<p<\infty.

For Conjecture 1, note that the set {±ei:i=1,,n}\{\pm e_{i}:i=1,\ldots,n\}, where eie_{i} is the ii-th standard basis vector, is equilateral in 1n\ell_{1}^{n}, so e(1n)2ne(\ell_{1}^{n})\geq 2n. Currently, the best known upper bound is e(1n)cnlogne(\ell_{1}^{n})\leq cn\log n due to Alon and Pudlák [1]. It is also known that Conjecture 1 holds for n=3n=3 (Bandelt, Chepoi, and Laurent [2]) and n=4n=4 (Koolen, Laurent, and Schrijver [9]).

As for Conjecture 2, note that the set {e1,,en,λi=1nei}\{e_{1},\ldots,e_{n},\lambda\sum_{i=1}^{n}e_{i}\} is equilateral for a suitable choice of λ\lambda, so e(pn)n+1e(\ell_{p}^{n})\geq n+1. For 1<p<21<p<2 and nn large enough, Swanepoel [19] actually showed that e(pn)>n+1e(\ell_{p}^{n})>n+1, disproving Conjecture 2 in this case. The first nontrivial upper bound of e(pn)cpn(p+1)/(p1)e(\ell_{p}^{n})\leq c_{p}n^{(p+1)/(p-1)} was found by Smyth [15] using an approximation argument. This result was later improved by Alon and Pudlák [1] and is currently the best known upper bound on e(pn)e(\ell_{p}^{n}) for arbitrary nn and pp.

Theorem 1 (Alon and Pudlák).

For every p1p\geq 1, we have e(pn)cpn(2p+2)/(2p1)e(\ell_{p}^{n})\leq c_{p}n^{(2p+2)/(2p-1)}, where one may take cp=cpc_{p}=cp for an absolute c>0c>0.

Our first result is an improvement of Theorem 1 when pp is large in terms of nn. One can check that c>2c>2, so Theorem 2 is indeed an improvement.

Theorem 2.

Let c>0c>0 be the constant from Theorem 1. If n>1n>1 and pc(nlogn)2p\geq c(n\log n)^{2}, then e(pn)2(p+1)ne(\ell_{p}^{n})\leq 2(p+1)n.

We should mention that when pp satisfies other special properties, Theorem 1 can be strengthened. If pp is an even integer, Swanepoel [19] used a linear independence argument to show that

e(pn){(p21)n+1if p0(mod4),p2n+1if p2(mod4).e(\ell_{p}^{n})\leq\begin{cases}(\frac{p}{2}-1)n+1&\text{if $p\equiv 0\pmod{4}$,}\\ \frac{p}{2}n+1&\text{if $p\equiv 2\pmod{4}$.}\end{cases}

If pp is an odd integer, Alon and Pudlák’s argument for e(1n)e(\ell_{1}^{n}) extends to e(pn)e(\ell_{p}^{n}), giving the bound e(pn)<cpnlogne(\ell_{p}^{n})<c_{p}n\log n in this case.

Our next result is a generalization of Theorem 1 to ss-distance sets. As far as we know, the only literature on ss-distance sets in pn\ell_{p}^{n} is by Smyth [16]. Our theorem below is strictly stronger than Conjecture 5 in [16].

Theorem 3.

If ss is a positive integer and pp is a real number satisfying 2p>s2p>s, then es(pn)cp,sn(2ps+2s)/(2ps)e_{s}(\ell_{p}^{n})\leq c_{p,s}n^{(2ps+2s)/(2p-s)} for a constant cp,sc_{p,s} depending on pp and ss.

Our next three results are on equilateral sets in the p\ell_{p} sum of Euclidean spaces. As far as we know, this problem has not been well studied in this space. Swanepoel [20] showed that e(XY)e(X)bf(Y)e(X\oplus_{\infty}Y)\leq e(X)b_{f}(Y) for normed spaces XX and YY, where bfb_{f} is the finite Borsuk number. However, explicit bounds are still unknown when XX and YY are Euclidean spaces and when we take an p\ell_{p} sum instead of an \ell_{\infty} sum. Our first result in this area almost completely resolves the problem for 𝔼a𝔼b\mathbb{E}^{a}\oplus_{\infty}\mathbb{E}^{b}. Note that we have the obvious lower bound e(𝔼a𝔼b)e(𝔼a)e(𝔼b)=(a+1)(b+1)e(\mathbb{E}^{a}\oplus_{\infty}\mathbb{E}^{b})\geq e(\mathbb{E}^{a})e(\mathbb{E}^{b})=(a+1)(b+1) by taking the Cartesian product of the two equilateral sets. This lower bound is known to meet the upper bound when a=2,3a=2,3 [20].

Theorem 4.

Let 𝔼a\mathbb{E}^{a} and 𝔼b\mathbb{E}^{b} be Euclidean spaces. Then, e(𝔼a𝔼b)(a+1)(b+1)+1e(\mathbb{E}^{a}\oplus_{\infty}\mathbb{E}^{b})\leq(a+1)(b+1)+1.

Our second result in this area provides an upper bound when pp is even.

Theorem 5.

Let 𝔼a\mathbb{E}^{a} and 𝔼b\mathbb{E}^{b} be Euclidean spaces, and let pp be an even integer. Then, e(𝔼ap𝔼b)(a+p/2a)+(b+p/2b)e(\mathbb{E}^{a}\oplus_{p}\mathbb{E}^{b})\leq\binom{a+p/2}{a}+\binom{b+p/2}{b}.

Finally, we extend Alon and Pudlák’s [1] result on equilateral sets in pn\ell_{p}^{n} to certain p\ell_{p} product spaces. We present an upper bound on the p\ell_{p} sum of nn Euclidean spaces for any 1p<1\leq p<\infty. Observe that Theorem 1 is a special case of our theorem below when a1==an=1a_{1}=\cdots=a_{n}=1.

Theorem 6.

Let 𝔼a1,,𝔼an\mathbb{E}^{a_{1}},\ldots,\mathbb{E}^{a_{n}} be Euclidean spaces and set a=max1inai\displaystyle a=\max_{1\leq i\leq n}a_{i}. If 2p>a2p>a, then e(𝔼a1pp𝔼an)cp,an2p+2a2pae(\mathbb{E}^{a_{1}}\oplus_{p}\cdots\oplus_{p}\mathbb{E}^{a_{n}})\leq c_{p,a}n^{\frac{2p+2a}{2p-a}} for a constant cp,ac_{p,a} depending on pp and aa.

Our paper is structured as follows. In Section 2, we introduce two important tools used in our proofs. Then, we prove Theorems 2, 3, 4, 5, and 6 in Sections 3, 4, 5, 6, and 7 respectively.

2 Preliminaries

In this section, we present two famous results that we use later. The first is the Rank Lemma, which allows us to estimate the rank of a symmetric matrix. The second is Jackson’s Theorem, a celebrated result from approximation theory. This theorem allows us to approximate |x|p|x|^{p} by a polynomial with sufficiently small error.

2.1 The Rank Lemma

Lemma 1 (Rank Lemma).

For a real symmetric n×nn\times n nonzero matrix AA,

rankA(i=1naii)2i,j=1naij2.\operatorname{rank}A\geq\frac{(\sum_{i=1}^{n}a_{ii})^{2}}{\sum_{i,j=1}^{n}a_{ij}^{2}}.

We will frequently make use of the following corollary.

Corollary.

Let AA be a real symmetric n×nn\times n matrix with aii=1a_{ii}=1 and |aij|ε|a_{ij}|\leq\varepsilon for all iji\neq j. Then

rankAn1+(n1)ε2.\operatorname{rank}A\geq\frac{n}{1+(n-1)\varepsilon^{2}}.

Choosing ε=n1/2\varepsilon=n^{-1/2} gives rankAn/2\operatorname{rank}A\geq n/2.

2.2 Jackson’s Theorem

Theorem 7 (Jackson, [8]).

For any fCk[1,1]f\in C^{k}[-1,1] and positive integer dd, there exists a polynomial PP with degree at most dd such that

|f(x)P(x)|ck(d+1)kω(f(k),1/d)for all x[1,1],|f(x)-P(x)|\leq\frac{c^{k}}{(d+1)_{k}}\omega(f^{(k)},1/d)\qquad\text{for all }x\in[-1,1],

where c>0c>0 is an absolute constant, ω(f,δ):=sup{|f(x)f(y)|:|xy|δ}\omega(f,\delta):=\sup\{|f(x)-f(y)|:|x-y|\leq\delta\} denotes the modulus of continuity of ff, and (d+1)k=(d+1)d(dk+2)(d+1)_{k}=(d+1)d\cdots(d-k+2) uses falling factorial notation.

From Theorem 7, we can recover the following lemma. This lemma was first given and proved in [14] and later also transmitted in [18].

Lemma 2.

For any p1p\geq 1 and dpd\geq\lceil p\rceil, there exists a polynomial PP with degree at most dd such that

|P(x)|x|p|B(p)dpfor all x[1,1]|P(x)-|x|^{p}|\leq\frac{B(p)}{d^{p}}\qquad\text{for all }x\in[-1,1] (1)

where B(p)=(pp(1+π2/2)p(p)p1)/p!B(p)=(\lceil p\rceil^{p}(1+\pi^{2}/2)^{\lceil p\rceil}(p)_{\lceil p\rceil-1})/\lceil p\rceil!.

We will always assume that the polynomial PP in Lemma 2 is even and that P(0)=0P(0)=0. If PP is not even, we can take the even part of PP. If P(0)0P(0)\neq 0, we can take the polynomial Q(x)=P(x)P(0)Q(x)=P(x)-P(0) as this only increases the error term by a factor of 22.

3 Bound on equilateral sets in pn\ell_{p}^{n} for large pp

We start with an important lemma about the p\ell_{p} norm.

Lemma 3.

Suppose 1pq1\leq p\leq q. Then for any xnx\in\mathbb{R}^{n},

xqxpn1p1qxq.\left\lVert x\right\rVert_{q}\leq\left\lVert x\right\rVert_{p}\leq n^{\frac{1}{p}-\frac{1}{q}}\left\lVert x\right\rVert_{q}.
Proof.

The left hand inequality is a well-known property of the p\ell_{p} norm. The right hand inequality is the Power Mean Inequality, which can be proven using Jensen’s Inequality. ∎

Suppose our equilateral set is {p1,,pm}\{p_{1},\ldots,p_{m}\}. Let kk be the closest even integer to pp, rounding up if pp is odd. The main idea is that by Lemma 3, we can approximate p\left\lVert\cdot\right\rVert_{p} by k\left\lVert\cdot\right\rVert_{k}. If pp is large enough in terms of nn, the error term is sufficiently small. From there, it suffices to bound the size of an approximately-equilateral set in kn\ell_{k}^{n}, which can be done with a linear algebra argument.

Proof of Theorem 2.

We use the notation explained above. There are two cases to consider depending on whether kk is greater than or less than pp.

Case 1. We have p\lfloor p\rfloor is odd, so p<kp<k.

From our bound on pp and the fact that p2p\geq 2,

m\displaystyle m cpn(2p+2)/(2p1)\displaystyle\leq cpn^{(2p+2)/(2p-1)}
cpn2\displaystyle\leq cpn^{2}
p2(logn)2\displaystyle\leq\frac{p^{2}}{(\log n)^{2}}
(1n1/p)2,\displaystyle\leq(1-n^{-1/p})^{-2},

where the first inequality holds by Theorem 1 and the last inequality by the inequality 11/xlog(x)1-1/x\leq\log(x) valid for x>0x>0. Assuming without loss of generality that m>1m>1, we may rearrange the result above into

plogn(11m)1.-p\log_{n}\left(1-\frac{1}{\sqrt{m}}\right)\geq 1.

Because kk is the smallest even integer greater than pp, we have

p<kp+1pplogn(11m).p<k\leq p+1\leq p-p\log_{n}\left(1-\frac{1}{\sqrt{m}}\right). (2)

Now, embed the mm points into kn\ell_{k}^{n}. Since p<kp<k, Lemma 3 implies

n1k1ppipjk1n^{\frac{1}{k}-\frac{1}{p}}\leq\left\lVert p_{i}-p_{j}\right\rVert_{k}\leq 1 (3)

for all iji\neq j. Consider the mm functions fi:nf_{i}:\mathbb{R}^{n}\to\mathbb{R} given by fi(x)=1pixkkf_{i}(x)=1-\left\lVert p_{i}-x\right\rVert_{k}^{k}, and let AA be the matrix with aij=fi(pj)a_{ij}=f_{i}(p_{j}). Clearly, aii=1a_{ii}=1, and from (2) and (3),

|aij|\displaystyle|a_{ij}| 1n1kp\displaystyle\leq 1-n^{1-\frac{k}{p}}
1n1(1logn(11/m))\displaystyle\leq 1-n^{1-(1-\log_{n}(1-1/{\sqrt{m}}))}
=1m,\displaystyle=\frac{1}{\sqrt{m}},

for all iji\neq j. Since AA is symmetric, Lemma 1 tells us that rankAm/2\operatorname{rank}A\geq m/2. For the upper bound on the rank, note that every fif_{i} lies in the span of the set of polynomials

𝒮={1,x1,,xn,x12,,xn2,,x1k1,,xnk1,t=1nxtk}.\mathcal{S}=\{1,x_{1},\ldots,x_{n},x_{1}^{2},\ldots,x_{n}^{2},\ldots,x_{1}^{k-1},\ldots,x_{n}^{k-1},\sum_{t=1}^{n}x_{t}^{k}\}.

The ii-th row vector of AA is (fi(p1),,fi(pm))(f_{i}(p_{1}),\ldots,f_{i}(p_{m})). Hence, every row vector belongs to the subspace spanned by {(f(p1),,f(pm)):f𝒮}\{(f(p_{1}),\ldots,f(p_{m})):f\in\mathcal{S}\}, which has dimension at most |𝒮|=(k1)n+2kn|\mathcal{S}|=(k-1)n+2\leq kn. It follows that rankAkn(p+1)n\operatorname{rank}A\leq kn\leq(p+1)n. Combining the upper and lower bound, we have

m2(p+1)n,m\leq 2(p+1)n,

as desired.

Case 2. We have p\lfloor p\rfloor is even, so pkp\geq k.

Since c,n2c,n\geq 2, p>4p>4. So, similar to Case 3, we have

m\displaystyle m cpn(2p+2)/(2p1)\displaystyle\leq cpn^{(2p+2)/(2p-1)}
cpn22/p\displaystyle\leq cpn^{2-2/p}
n2/pp2(logn)2\displaystyle\leq n^{-2/p}\cdot\frac{p^{2}}{(\log n)^{2}}
n2/p(1n1/p)2\displaystyle\leq n^{-2/p}(1-n^{-1/p})^{-2}
=(n1/p1)2.\displaystyle=(n^{1/p}-1)^{-2}.

Rearranging the result gives us

plogn(1+1m)1.p\log_{n}\left(1+\frac{1}{\sqrt{m}}\right)\geq 1.

Because kk is the largest even integer less than or equal to pp, we have

pkp1pplogn(1+1m).p\geq k\geq p-1\geq p-p\log_{n}\left(1+\frac{1}{\sqrt{m}}\right). (4)

Embed the mm points into kn\ell_{k}^{n}. This time, Lemma 3 implies

1pipjkn1k1p.1\leq\left\lVert p_{i}-p_{j}\right\rVert_{k}\leq n^{\frac{1}{k}-\frac{1}{p}}. (5)

We define fif_{i} and AA like in Case 3. Again, aii=1a_{ii}=1, and from (4) and (5),

|aij|\displaystyle|a_{ij}| n1kp1\displaystyle\leq n^{1-\frac{k}{p}}-1
n1(1logn(1+1/m))1\displaystyle\leq n^{1-(1-\log_{n}(1+1/{\sqrt{m}}))}-1
=1m,\displaystyle=\frac{1}{\sqrt{m}},

for all iji\neq j. Applying Lemma 1, rankAm/2\operatorname{rank}A\geq m/2. So, similar to Case 3, we have the bound rankAknpn\operatorname{rank}A\leq kn\leq pn. The final result is

m2pn<2(p+1)n.m\leq 2pn<2(p+1)n.

Having considered all cases, the proof is complete. ∎

4 Bound on ss-distance sets in pn\ell_{p}^{n}

Smyth [14], see also [16], first combined Jackson’s theorem with the invertibility of diagonally dominated matrices to prove e(pn)cn(p+1)/(p1)e(\ell_{p}^{n})\leq cn^{(p+1)/(p-1)} for p>1p>1. Alon and Pudlák [1] then improved this bound by substituting the rank lemma for the invertibility lemma to achieve Theorem 1. We would like to extend this method of proof to the ss-distance case. However, as pointed out by Smyth [16], if one of the distances is arbitrarily small, we need arbitrarily high degrees of approximation. So, we want to impose a lower bound on our distances.

Consider a two-distance set with distances 11 and aa, where aa is very small. Intuitively, this set should look like several “clusters” of points, such that the distance between each cluster is 11. So, if we look globally, this set looks like an equilateral set with distance 11, but if we look locally at each cluster, we have an equilateral set with distance aa. This means that we have essentially reduced the problem from one about two-distance sets to one about equilateral sets. We carry this intuition to ss-distance sets and formalize this argument with an induction.

Proof of Theorem 3.

We use strong induction on ss. The base case is s=1s=1, which was proven by Alon and Pudlák [1]. For the inductive step, assume that the statement holds true for all 1s<k1\leq s<k. We prove that it is true for s=ks=k.

Let SS be a kk-distance set in pn\ell_{p}^{n}. Let our kk distances be 1=a1>a2>>ak1=a_{1}>a_{2}>\cdots>a_{k}. There are two cases to consider depending on whether the kk distances are lower bounded or not.

Case 1. The smallest distance aka_{k} is less than 21k2^{1-k}.

There exists an index 1i<k1\leq i<k such that ai>2ai+1a_{i}>2a_{i+1}. Let SSS^{\prime}\subseteq S be a maximal ii-distance set using the distances a1,,aia_{1},\ldots,a_{i}. Every point pSSp\in S\setminus S^{\prime} is a distance at most ai+1a_{i+1} from some point in SS^{\prime}, otherwise pp can be added to SS^{\prime}, contrary to the maximality of SS^{\prime}. So, if we draw a closed ball of radius ai+1a_{i+1} around every point in SS^{\prime}, every point in SS lies within some ball. These balls are disjoint from the condition ai>2ai+1a_{i}>2a_{i+1}. Furthermore, by the triangle inequality, within each ball, the distance between any two points is at most 2ai+1<ai2a_{i+1}<a_{i}, and thus is at most ai+1a_{i+1}. It follows that within every ball, we have a kk^{\prime}-distance set, where kkik^{\prime}\leq k-i, using kk^{\prime} of the distances from ai+1,,aka_{i+1},\ldots,a_{k}. This implies the bound

ek(pn)ei(pn)max0jkiej(pn).e_{k}(\ell_{p}^{n})\leq e_{i}(\ell_{p}^{n})\cdot\max_{0\leq j\leq k-i}e_{j}(\ell_{p}^{n}).

Applying the inductive hypothesis gives us

ek(pn)\displaystyle e_{k}(\ell_{p}^{n}) cp,in2pi+2i2pimax0jkicp,jn2pj+2j2pj\displaystyle\leq c_{p,i}n^{\frac{2pi+2i}{2p-i}}\cdot\max_{0\leq j\leq k-i}c_{p,j}n^{\frac{2pj+2j}{2p-j}}
cp,in2pi+2i2picp,kin2p(ki)+2(ki)2p(ki)\displaystyle\leq c_{p,i}n^{\frac{2pi+2i}{2p-i}}\cdot c_{p,k-i}n^{\frac{2p(k-i)+2(k-i)}{2p-(k-i)}}
cp,kn2pk+2k2pk,\displaystyle\leq c_{p,k}n^{\frac{2pk+2k}{2p-k}},

as desired. The first inequality follows by analyzing the function

f1(x)=2px+2x2pxf_{1}(x)=\frac{2px+2x}{2p-x}

on the interval [0,2p)[0,2p). Its first derivative is

f1(x)=4p(p+1)(2px)2,f_{1}^{\prime}(x)=\frac{4p(p+1)}{(2p-x)^{2}},

so f1f_{1} is increasing on [0,2p)[0,2p) and hence on [0,ki][0,k-i]. The second inequality follows by analyzing the related function

f2(x)=2px+2x2px+2p(kx)+2(kx)2p(kx)f_{2}(x)=\frac{2px+2x}{2p-x}+\frac{2p(k-x)+2(k-x)}{2p-(k-x)}

on the interval [0,k][0,k]. It is symmetric about x=k2x=\tfrac{k}{2}, and its first derivative is

f2(x)=4p(p+1)(2px)24p(p+1)(2p(kx))2.f_{2}^{\prime}(x)=\frac{4p(p+1)}{(2p-x)^{2}}-\frac{4p(p+1)}{(2p-(k-x))^{2}}.

Since 2p>k2p>k, f2f_{2} is decreasing on [0,k2][0,\tfrac{k}{2}] and increasing on [k2,k][\tfrac{k}{2},k]. Thus, it is maximised at x=0x=0 and x=kx=k.

Case 2. The smallest distance aka_{k} is at least 21k2^{1-k}.

Suppose SS consists of the points {p1,,pm}\{p_{1},\ldots,p_{m}\}. For convenience, let π=a1pa2pakp\pi=a_{1}^{p}a_{2}^{p}\cdots a_{k}^{p}. Fix B(p)B(p) as the constant from Lemma 2. Define cc as

c=max(B(p)k2pk2pk+2k,(21/p1)p).c=\max(B(p)k\cdot 2^{pk^{2}-pk+2k},(2^{1/p}-1)^{-p}).

Then, let dd be a positive integer satisfying

cnm<dp<2cnm,cn\sqrt{m}<d^{p}<2cn\sqrt{m},

which is possible since c(21/p1)pc\geq(2^{1/p}-1)^{-p}.

Lemma 2 allows us to pick an even polynomial PP with P(0)=0P(0)=0 and degree at most dd such that

|P(x)|x|p|B(p)dpfor all x[1,1].|P(x)-|x|^{p}|\leq\frac{B(p)}{d^{p}}\qquad\text{for all }x\in[-1,1].

Consider the mm functions fi:nf_{i}:\mathbb{R}^{n}\to\mathbb{R} given by

fi(x)=1πu=1k(aupt=1n|xtpit|p),f_{i}(x)=\frac{1}{\pi}\prod_{u=1}^{k}\left(a_{u}^{p}-\sum_{t=1}^{n}|x_{t}-p_{it}|^{p}\right),

and their polynomial approximations

gi(x)=1πu=1k(aupt=1nP(xtpit)).g_{i}(x)=\frac{1}{\pi}\prod_{u=1}^{k}\left(a_{u}^{p}-\sum_{t=1}^{n}P(x_{t}-p_{it})\right).

Let AA be the m×mm\times m matrix given by aij=gi(pj)a_{ij}=g_{i}(p_{j}). First, since P(0)=0P(0)=0, aii=1a_{ii}=1 for all ii. We now estimate aija_{ij} for iji\neq j. Expand

fi(x)=1π=0k(1)k+σ()(t=1n|xtpit|p)k,f_{i}(x)=\frac{1}{\pi}\sum_{\ell=0}^{k}(-1)^{k+\ell}\sigma(\ell)\left(\sum_{t=1}^{n}|x_{t}-p_{it}|^{p}\right)^{k-\ell},

and

gi(x)=1π=0k(1)k+σ()(t=1nP(xtpit))k,g_{i}(x)=\frac{1}{\pi}\sum_{\ell=0}^{k}(-1)^{k+\ell}\sigma(\ell)\left(\sum_{t=1}^{n}P(x_{t}-p_{it})\right)^{k-\ell},

where σ()\sigma(\ell) denotes the \ell-th elementary symmetric polynomial in a1p,,akpa_{1}^{p},\ldots,a_{k}^{p}. For convenience, we define Xij=t=1n|pjtpit|pX_{ij}=\sum_{t=1}^{n}|p_{jt}-p_{it}|^{p} and Yij=t=1nP(pjtpit)Y_{ij}=\sum_{t=1}^{n}P(p_{jt}-p_{it}). Applying the triangle inequality with (1), we have

|XijYij|nB(p)dp.|X_{ij}-Y_{ij}|\leq\frac{nB(p)}{d^{p}}.

Since iji\neq j, recall that Xij2(1k)pX_{ij}\geq 2^{(1-k)p}. Combining this with the fact that c>B(p)2(k1)pc>B(p)\cdot 2^{(k-1)p} implies that YijY_{ij} is positive. Now, we are ready to upper bound |aij||a_{ij}|. Since fi(pj)=0f_{i}(p_{j})=0 for iji\neq j, we have aij=gi(pj)=gi(pj)fi(pj)a_{ij}=g_{i}(p_{j})=g_{i}(p_{j})-f_{i}(p_{j}). Thus

|aij|\displaystyle|a_{ij}| =1π|=0k(1)k+σ()(YijkXijk)|\displaystyle=\frac{1}{\pi}\left|\sum_{\ell=0}^{k}(-1)^{k+\ell}\sigma(\ell)(Y_{ij}^{k-\ell}-X_{ij}^{k-\ell})\right|
1πnB(p)dp=0k1σ()r=1k|XijkrYijr1|\displaystyle\leq\frac{1}{\pi}\cdot\frac{nB(p)}{d^{p}}\sum_{\ell=0}^{k-1}\sigma(\ell)\sum_{r=1}^{k-\ell}|X_{ij}^{k-\ell-r}Y_{ij}^{r-1}|
1πnB(p)dp=0k1σ()r=1k|(kr)Xij+(r1)Yijk1|k1\displaystyle\leq\frac{1}{\pi}\cdot\frac{nB(p)}{d^{p}}\sum_{\ell=0}^{k-1}\sigma(\ell)\sum_{r=1}^{k-\ell}\left|\frac{(k-\ell-r)X_{ij}+(r-1)Y_{ij}}{k-\ell-1}\right|^{k-\ell-1}
=1πnB(p)dp=0k1σ()r=1k|(r1)(YijXij)k1+Xij|k1\displaystyle=\frac{1}{\pi}\cdot\frac{nB(p)}{d^{p}}\sum_{\ell=0}^{k-1}\sigma(\ell)\sum_{r=1}^{k-\ell}\left|\frac{(r-1)(Y_{ij}-X_{ij})}{k-\ell-1}+X_{ij}\right|^{k-\ell-1}
1πnB(p)dp=0k1σ()(k)(nB(p)dp+1)k1\displaystyle\leq\frac{1}{\pi}\cdot\frac{nB(p)}{d^{p}}\sum_{\ell=0}^{k-1}\sigma(\ell)(k-\ell)\left(\frac{nB(p)}{d^{p}}+1\right)^{k-\ell-1}
1πnB(p)kdp2k=0k1σ().\displaystyle\leq\frac{1}{\pi}\cdot\frac{nB(p)k}{d^{p}}\cdot 2^{k}\sum_{\ell=0}^{k-1}\sigma(\ell).

The first inequality is achieved by the triangle inequality, the factorization xryr=(xy)i=0r1xriyix^{r}-y^{r}=(x-y)\sum_{i=0}^{r-1}x^{r-i}y^{i}, and the upper bound on |XijYij||X_{ij}-Y_{ij}|. The second inequality follows by the AM-GM inequality. The last inequality holds because nB(p)/dp<B(p)/c<1nB(p)/d^{p}<B(p)/c<1 by our choice of cc and dd, whence k2k(kl)(1+nB(p)/dp)k1k2^{k}\geq(k-l)(1+nB(p)/d^{p})^{k-\ell-1}.

Now, since au21ka_{u}\geq 2^{1-k}, we can lower bound π\pi with

π=a1pa2pakp(2(1k)p)k=2pkpk2.\pi=a_{1}^{p}a_{2}^{p}\cdots a_{k}^{p}\geq(2^{(1-k)p})^{k}=2^{pk-pk^{2}}.

On the other hand, since at1a_{t}\leq 1, we can upper bound σ()\sigma(\ell) with

σ()=1j1<j2<<jkaj1paj2pajp(k).\sigma(\ell)=\sum_{1\leq j_{1}<j_{2}<\cdots<j_{\ell}\leq k}a_{j_{1}}^{p}a_{j_{2}}^{p}\cdots a_{j_{\ell}}^{p}\leq\binom{k}{\ell}.

This gives us =0k1σ()<2k\sum_{\ell=0}^{k-1}\sigma(\ell)<2^{k}. Putting everything together, we have

|aij|\displaystyle|a_{ij}| <2pk2pk+2knB(p)kdp\displaystyle<2^{pk^{2}-pk+2k}\cdot\frac{nB(p)k}{d^{p}}
<1m\displaystyle<\frac{1}{\sqrt{m}}

from our choice of cc and dd.

Recall that PP is even, so the matrix AA is symmetric. Lemma 1 then tells us that rankAm/2\operatorname{rank}A\geq m/2. We now find an upper bound for the rank.

Note that the polynomials aupt=1nP(xtpit)a_{u}^{p}-\sum_{t=1}^{n}P(x_{t}-p_{it}) belong to the span of the set

{1,x1,,xn,x12,,xn2,,x1d1,,xnd1,t=1nxtd}.\{1,x_{1},\ldots,x_{n},x_{1}^{2},\ldots,x_{n}^{2},\ldots,x_{1}^{d-1},\ldots,x_{n}^{d-1},\sum_{t=1}^{n}x_{t}^{d}\}.

This set consists of (d1)n+2dn(d-1)n+2\leq dn polynomials. Thus, all the gig_{i} belong to the span of a set 𝒮\mathcal{S} of at most (dn)k(dn)^{k} polynomials. The ii-th row vector of AA is (gi(p1),,gi(pm))(g_{i}(p_{1}),\ldots,g_{i}(p_{m})). Hence, every row vector belongs to the subspace spanned by

{(f(p1),,f(pm)):f𝒮},\{(f(p_{1}),\ldots,f(p_{m})):f\in\mathcal{S}\},

which has dimension at most |𝒮|(dn)k|\mathcal{S}|\leq(dn)^{k}. This implies rankA(dn)k\operatorname{rank}A\leq(dn)^{k}.

The upper and lower bounds combine to give (dn)km/2(dn)^{k}\geq m/2. Using the upper bound on dpd^{p} and the condition 2p>k2p>k, we can rearrange the inequality to obtain

mcp,kn(2pk+2k)/(2pk).m\leq c_{p,k}n^{(2pk+2k)/(2p-k)}.

This completes the induction. ∎

5 Bound on equilateral sets in \ell_{\infty} product spaces

Proof of Theorem 4.

Write x=(x~1,x~2)x=(\widetilde{x}_{1},\widetilde{x}_{2}) for each point x𝔼a𝔼bx\in\mathbb{E}^{a}\oplus_{\infty}\mathbb{E}^{b}, where x~1𝔼a\widetilde{x}_{1}\in\mathbb{E}^{a} and x~2𝔼b\widetilde{x}_{2}\in\mathbb{E}^{b}. Let SS be our equilateral set with cardinality mm. Consider the mm functions fu:a+bf_{u}:\mathbb{R}^{a+b}\to\mathbb{R} defined by

fu(x)=(1x~1u~12)(1x~2u~22),f_{u}(x)=\left(1-\left\lVert\widetilde{x}_{1}-\widetilde{u}_{1}\right\rVert^{2}\right)\left(1-\left\lVert\widetilde{x}_{2}-\widetilde{u}_{2}\right\rVert^{2}\right),

for all uSu\in S. Note that fu(v)=δuvf_{u}(v)=\delta_{uv} for all u,vSu,v\in S, so the fuf_{u} are linearly independent.

We can expand fuf_{u} as

fu(x)\displaystyle f_{u}(x) =(1t=1a(x~1tu~1t)2)(1t=1b(x~2tu~2t)2)\displaystyle=\left(1-\sum_{t=1}^{a}(\widetilde{x}_{1t}-\widetilde{u}_{1t})^{2}\right)\left(1-\sum_{t=1}^{b}(\widetilde{x}_{2t}-\widetilde{u}_{2t})^{2}\right)
=(1u~12x~12+2t=1ax~1tu~1t)(1u~22x~22+2t=1bx~2tu~2t)\displaystyle=\left(1-\left\lVert\widetilde{u}_{1}\right\rVert^{2}-\left\lVert\widetilde{x}_{1}\right\rVert^{2}+2\sum_{t=1}^{a}\widetilde{x}_{1t}\widetilde{u}_{1t}\right)\left(1-\left\lVert\widetilde{u}_{2}\right\rVert^{2}-\left\lVert\widetilde{x}_{2}\right\rVert^{2}+2\sum_{t=1}^{b}\widetilde{x}_{2t}\widetilde{u}_{2t}\right)

So, the fuf_{u} are all spanned by the following set of (a+2)(b+2)(a+2)(b+2) polynomials

{1,x~1i,x~2j,x~1ix~2j,x~12,x~22,x~22x~1i,x~12x~2j,x~12x~22:1ia,1jb}.\{1,\widetilde{x}_{1i},\widetilde{x}_{2j},\widetilde{x}_{1i}\widetilde{x}_{2j},\left\lVert\widetilde{x}_{1}\right\rVert^{2},\left\lVert\widetilde{x}_{2}\right\rVert^{2},\left\lVert\widetilde{x}_{2}\right\rVert^{2}\widetilde{x}_{1i},\left\lVert\widetilde{x}_{1}\right\rVert^{2}\widetilde{x}_{2j},\left\lVert\widetilde{x}_{1}\right\rVert^{2}\left\lVert\widetilde{x}_{2}\right\rVert^{2}:1\leq i\leq a,1\leq j\leq b\}.

This implies the bound m(a+2)(b+2)m\leq(a+2)(b+2).

We will now prove that the set of polynomials {fu,1,xk,x~12:uS,1ka+b}\{f_{u},1,x_{k},\left\lVert\widetilde{x}_{1}\right\rVert^{2}:u\in S,1\leq k\leq a+b\} is linearly independent. Assume for the sake of contradiction that we have a dependence

uSαufu+k=1a+bβkxk+γx~12+δ=0.\sum_{u\in S}\alpha_{u}f_{u}+\sum_{k=1}^{a+b}\beta_{k}x_{k}+\gamma\left\lVert\widetilde{x}_{1}\right\rVert^{2}+\delta=0. (6)

The left hand side of (6) is a polynomial in the xkx_{k} that is identically zero. Thus, extracting the coefficient of x~12x~22\left\lVert\widetilde{x}_{1}\right\rVert^{2}\left\lVert\widetilde{x}_{2}\right\rVert^{2}, we have

uSαu=0.\sum_{u\in S}\alpha_{u}=0. (7)

Extracting the coefficient of x~r2xk\left\lVert\widetilde{x}_{r}\right\rVert^{2}x_{k}, for the appropriate r{1,2}r\in\{1,2\}, we have

uSαuuk=0.\sum_{u\in S}\alpha_{u}u_{k}=0. (8)

Extracting the coefficient of x~22\left\lVert\widetilde{x}_{2}\right\rVert^{2} and applying (7), we have

uSαuu~12=0.\sum_{u\in S}\alpha_{u}\left\lVert\widetilde{u}_{1}\right\rVert^{2}=0. (9)

Now, plug uu into (6), multiply both sides by αu\alpha_{u}, and sum over all uSu\in S.

uSαu2+k=1a+bβkuSαuuk+γuSαuu~12+δuSαu=0.\sum_{u\in S}\alpha_{u}^{2}+\sum_{k=1}^{a+b}\beta_{k}\sum_{u\in S}\alpha_{u}u_{k}+\gamma\sum_{u\in S}\alpha_{u}\left\lVert\widetilde{u}_{1}\right\rVert^{2}+\delta\sum_{u\in S}\alpha_{u}=0.

Applying (7), (8), and (9) implies αu=0\alpha_{u}=0 for all uSu\in S. It easily follows that all other coefficients are zero, as desired.

Now, we know that m+a+b+2(a+2)(b+2)m+a+b+2\leq(a+2)(b+2). This rearranges into m(a+1)(b+1)+1m\leq(a+1)(b+1)+1. ∎

6 Bound on equilateral sets in p\ell_{p} product spaces for even pp

Proof of Theorem 5.

Let SS be an equilateral set in 𝔼ap𝔼b\mathbb{E}^{a}\oplus_{p}\mathbb{E}^{b} with mm points. For every uSu\in S, define the function fu:a+bf_{u}:\mathbb{R}^{a+b}\to\mathbb{R} by

fu(x)=1x~1u~1px~2u~2pf_{u}(x)=1-\left\lVert\widetilde{x}_{1}-\widetilde{u}_{1}\right\rVert^{p}-\left\lVert\widetilde{x}_{2}-\widetilde{u}_{2}\right\rVert^{p}

for all x=(x1~,x2~)𝔼ap𝔼bx=(\widetilde{x_{1}},\widetilde{x_{2}})\in\mathbb{E}^{a}\oplus_{p}\mathbb{E}^{b} so that fu(v)=δuvf_{u}(v)=\delta_{uv} for all u,vSu,v\in S. It follows that the mm polynomials are linearly independent. Now, we can expand fuf_{u} as

fu(x)\displaystyle f_{u}(x) =1(t=1a(x~1tu~1t)2)p/2(t=1b(x~2tu~2t)2)p/2\displaystyle=1-\left(\sum_{t=1}^{a}(\widetilde{x}_{1t}-\widetilde{u}_{1t})^{2}\right)^{p/2}-\left(\sum_{t=1}^{b}(\widetilde{x}_{2t}-\widetilde{u}_{2t})^{2}\right)^{p/2}
=1(x~12+u~122t=1ax~1tu~1t)p/2(x~22+u~222t=1bx~2tu~2t)p/2\displaystyle=1-\left(\left\lVert\widetilde{x}_{1}\right\rVert^{2}+\left\lVert\widetilde{u}_{1}\right\rVert^{2}-2\sum_{t=1}^{a}\widetilde{x}_{1t}\widetilde{u}_{1t}\right)^{p/2}-\left(\left\lVert\widetilde{x}_{2}\right\rVert^{2}+\left\lVert\widetilde{u}_{2}\right\rVert^{2}-2\sum_{t=1}^{b}\widetilde{x}_{2t}\widetilde{u}_{2t}\right)^{p/2}
=1fu~1(x)fu~2(x)\displaystyle=1-\widetilde{f_{u}}_{1}(x)-\widetilde{f_{u}}_{2}(x)

Utilizing multi-index notation, we can expand

fu~1(x)=ε,gε+γp/2(2)γ(γg)(p/2ε,γ,p/2εγ)u~1gu~1p2ε2γx~1gx~12ε.\widetilde{f_{u}}_{1}(x)=\sum_{\begin{subarray}{c}\varepsilon,g\\ \varepsilon+\gamma\leq p/2\end{subarray}}(-2)^{\gamma}{\gamma\choose g}{p/2\choose\varepsilon,\gamma,p/2-\varepsilon-\gamma}\widetilde{u}_{1}^{g}\left\lVert\widetilde{u}_{1}\right\rVert^{p-2\varepsilon-2\gamma}\widetilde{x}_{1}^{g}\left\lVert\widetilde{x}_{1}\right\rVert^{2\varepsilon}.

The sum is taken over all non-negative integers ε\varepsilon and non-negative integer vectors gg with aa entries such that ε+γp/2\varepsilon+\gamma\leq p/2, where γ\gamma is the sum of the components of gg. Similarly, we can expand

fu~2(x)=ε,gε+γp/2(2)γ(γg)(p/2ε,γ,p/2εγ)u~2gu~2p2ε2γx~2gx~22ε.\widetilde{f_{u}}_{2}(x)=\sum_{\begin{subarray}{c}\varepsilon,g\\ \varepsilon+\gamma\leq p/2\end{subarray}}(-2)^{\gamma}{\gamma\choose g}{p/2\choose\varepsilon,\gamma,p/2-\varepsilon-\gamma}\widetilde{u}_{2}^{g}\left\lVert\widetilde{u}_{2}\right\rVert^{p-2\varepsilon-2\gamma}\widetilde{x}_{2}^{g}\left\lVert\widetilde{x}_{2}\right\rVert^{2\varepsilon}.

We want to count the number of monomials in each polynomial. Since fu~1\widetilde{f_{u}}_{1} is a polynomial in x~11,,x~1a\widetilde{x}_{11},\ldots,\widetilde{x}_{1a} and fu~2\widetilde{f_{u}}_{2} is a polynomial in x~21,,x~2b\widetilde{x}_{21},\ldots,\widetilde{x}_{2b}, the two sets of monomials are disjoint, except for the constant monomial. Let us consider fu~1\widetilde{f_{u}}_{1} first. By choosing ε=0\varepsilon=0, we must count all the monomials with degree at most p/2p/2. There are (a+p/2a){a+p/2\choose a} of these. If the degree is greater than p/2p/2, say p/2+cp/2+c, we only need to count the monomials formed when ε=c\varepsilon=c and γ=p/2c\gamma=p/2-c. Hence, the total number of monomials in fu~1\widetilde{f_{u}}_{1} is

(a+p/2a)+(a+p/22a1)+(a+p/23a1)++(a1a1)=(a+p/2a)+(a+p/21a).{a+p/2\choose a}+{a+p/2-2\choose a-1}+{a+p/2-3\choose a-1}+\cdots+{a-1\choose a-1}={a+p/2\choose a}+{a+p/2-1\choose a}.

Similarly, there are (b+p/2b)+(b+p/21b){b+p/2\choose b}+{b+p/2-1\choose b} monomials in fu~2\widetilde{f_{u}}_{2}. In total, fuf_{u} has (a+p/2a)+(a+p/21a)+(b+p/2b)+(b+p/21b)1{a+p/2\choose a}+{a+p/2-1\choose a}+{b+p/2\choose b}+{b+p/2-1\choose b}-1 monomials (we subtract 11 as the constant monomial is counted twice). This gives an upper bound on mm.

By finding a larger linearly independent set of polynomials, a trick first used by Blokhuis [5], we can lower this bound to (a+p/2a)+(b+p/2b)\binom{a+p/2}{a}+\binom{b+p/2}{b}. We prove that the set of polynomials

{fu,x~1m,x~2n,1:uS,0<μ<p/2,0<ν<p/2}\{f_{u},\widetilde{x}_{1}^{m},\widetilde{x}_{2}^{n},1:u\in S,0<\mu<p/2,0<\nu<p/2\}

where m,nm,n are integer vectors with a,ba,b entries and sums of components μ,ν\mu,\nu respectively, is linearly independent. The details are very similar to those in Blokhuis’s [4] bound for the ss-distance set in n\mathbb{R}^{n}.

Suppose we have a dependence

uSaufu(x)+0<μ<p/2amx~1m+0<ν<p/2anx~2n+δ=0.\sum_{u\in S}a_{u}f_{u}(x)+\sum_{0<\mu<p/2}a_{m}\widetilde{x}_{1}^{m}+\sum_{0<\nu<p/2}a_{n}\widetilde{x}_{2}^{n}+\delta=0. (10)

The main claim is the following lemma.

Lemma 4.

For all mm with μ<p/2\mu<p/2,

uSauu~1m=0.\sum_{u\in S}a_{u}\widetilde{u}_{1}^{m}=0.

Similarly, for all nn with ν<p/2\nu<p/2,

uSauu~2n=0.\sum_{u\in S}a_{u}\widetilde{u}_{2}^{n}=0.
Proof.

We only prove the first statement. The argument for the second statement is identical.

Suppose μ=t<p/2\mu=t<p/2. Consider the part of the left hand side of (10) that is homogeneous in x~11,,x~1a\widetilde{x}_{11},\ldots,\widetilde{x}_{1a} with degree ptp-t. Note that the monomials x~1m\widetilde{x}_{1}^{m} do not contribute to this, so we only have to look at the part from the fuf_{u}. Using our expansion of fuf_{u} above, the part of fuf_{u} homogeneous in the aa variables with degree ptp-t is

ε,g2ε+γ=ptε+γp/2(2)γ(γg)(p/2ε,γ,p/2εγ)u~1gu~1p2ε2γx~1gx~12ε.-\sum_{\begin{subarray}{c}\varepsilon,g\\ 2\varepsilon+\gamma=p-t\\ \varepsilon+\gamma\leq p/2\end{subarray}}(-2)^{\gamma}{\gamma\choose g}{p/2\choose\varepsilon,\gamma,p/2-\varepsilon-\gamma}\widetilde{u}_{1}^{g}\left\lVert\widetilde{u}_{1}\right\rVert^{p-2\varepsilon-2\gamma}\widetilde{x}_{1}^{g}\left\lVert\widetilde{x}_{1}\right\rVert^{2\varepsilon}.

The left hand side of (10) is a polynomial in the xix_{i} which is identically zero. Thus, we have

ε,g2ε+γ=ptε+γp/2(2)γ(γg)(p/2ε,γ,p/2εγ)uSauu~1gu~1p2ε2γx~1gx~12ε=0.\sum_{\begin{subarray}{c}\varepsilon,g\\ 2\varepsilon+\gamma=p-t\\ \varepsilon+\gamma\leq p/2\end{subarray}}(-2)^{\gamma}{\gamma\choose g}{p/2\choose\varepsilon,\gamma,p/2-\varepsilon-\gamma}\sum_{u\in S}a_{u}\widetilde{u}_{1}^{g}\left\lVert\widetilde{u}_{1}\right\rVert^{p-2\varepsilon-2\gamma}\widetilde{x}_{1}^{g}\left\lVert\widetilde{x}_{1}\right\rVert^{2\varepsilon}=0.

Now, substitute x=vx=v, multiply by avv~12tpa_{v}\left\lVert\widetilde{v}_{1}\right\rVert^{2t-p}, and sum over all vSv\in S.

ε,g2ε+γ=ptε+γp/2(2)γ(γg)(p/2ε,γ,p/2εγ)(uSauu~1p2ε2γu~1g)2=0.\sum_{\begin{subarray}{c}\varepsilon,g\\ 2\varepsilon+\gamma=p-t\\ \varepsilon+\gamma\leq p/2\end{subarray}}(-2)^{\gamma}{\gamma\choose g}{p/2\choose\varepsilon,\gamma,p/2-\varepsilon-\gamma}\left(\sum_{u\in S}a_{u}\left\lVert\widetilde{u}_{1}\right\rVert^{p-2\varepsilon-2\gamma}\widetilde{u}_{1}^{g}\right)^{2}=0.

This is a sum of squares where all coefficients have the same sign. So,

uSauu~1p2ε2γu~1g=0.\sum_{u\in S}a_{u}\left\lVert\widetilde{u}_{1}\right\rVert^{p-2\varepsilon-2\gamma}\widetilde{u}_{1}^{g}=0.

Plugging in γ=t\gamma=t proves the lemma. ∎

Now, plug in uu into (10), multiply by aua_{u} and sum over all uSu\in S,

uSau2+0<μ<p/2amuSauu~1m+0<ν<p/2anuSauu~2n+δuSau=0.\sum_{u\in S}a_{u}^{2}+\sum_{0<\mu<p/2}a_{m}\sum_{u\in S}a_{u}\widetilde{u}_{1}^{m}+\sum_{0<\nu<p/2}a_{n}\sum_{u\in S}a_{u}\widetilde{u}_{2}^{n}+\delta\sum_{u\in S}a_{u}=0.

Applying Lemma 4, we obtain au=0a_{u}=0 for all uu. We already know that the other polynomials in our set are linearly independent, so all coefficients vanish, as desired. This implies m(a+p/2a)+(b+p/2b)m\leq{a+p/2\choose a}+{b+p/2\choose b}. ∎

Remark.

This proof can be easily extended to the p\ell_{p} sum of nn Euclidean spaces. If we consider the product space 𝔼a1pp𝔼an\mathbb{E}^{a_{1}}\oplus_{p}\cdots\oplus_{p}\mathbb{E}^{a_{n}}, our bound is just (a1+p/2a1)++(an+p/2an){a_{1}+p/2\choose a_{1}}+\cdots+{a_{n}+p/2\choose a_{n}}.

7 Bound on equilateral sets in p\ell_{p} product spaces for 1p<1\leq p<\infty

Proof of Theorem 6.

Let {p1,,pm}\{p_{1},\ldots,p_{m}\} be an equilateral set in 𝔼a1pp𝔼an\mathbb{E}^{a_{1}}\oplus_{p}\cdots\oplus_{p}\mathbb{E}^{a_{n}}, and write pi=(pi~1,,pi~n)p_{i}=(\widetilde{p_{i}}_{1},\ldots,\widetilde{p_{i}}_{n}) where pi~k𝔼ak\widetilde{p_{i}}_{k}\in\mathbb{E}^{a_{k}} for k=1,,nk=1,\ldots,n.

Let B(p)B(p) be the constant from Lemma 2. Take c=max(B(p),(21/p1)p)c=\max(B(p),(2^{1/p}-1)^{-p}) and set dd to be a positive integer satisfying

cnm<dp<2cnm.cn\sqrt{m}<d^{p}<2cn\sqrt{m}.

By Lemma 2, there exists an even polynomial PP with P(0)=0P(0)=0 and degree at most dd that approximates |x|p|x|^{p}. Thus, for iji\neq j and k=1,,nk=1,\ldots,n,

|P(pi~kpj~k)pi~kpj~kp|B(p)dp.\left|P(\left\lVert\widetilde{p_{i}}_{k}-\widetilde{p_{j}}_{k}\right\rVert)-\left\lVert\widetilde{p_{i}}_{k}-\widetilde{p_{j}}_{k}\right\rVert^{p}\right|\leq\frac{B(p)}{d^{p}}.

Next, for i=1,,mi=1,\ldots,m define the functions fi:a1++anf_{i}:\mathbb{R}^{a_{1}+\cdots+a_{n}}\to\mathbb{R} by

fi(x)=1k=1nP(pi~kx~k).f_{i}(x)=1-\sum_{k=1}^{n}P(\left\lVert\widetilde{p_{i}}_{k}-\widetilde{x}_{k}\right\rVert).

Let MM be the m×mm\times m matrix given by mij=fi(pj)m_{ij}=f_{i}(p_{j}). Since P(0)=0P(0)=0, we have mii=1m_{ii}=1. For iji\neq j,

|mij|\displaystyle|m_{ij}| =|k=1npi~kpj~kpk=1nP(pi~kpj~k)|\displaystyle=\left|\sum_{k=1}^{n}\left\lVert\widetilde{p_{i}}_{k}-\widetilde{p_{j}}_{k}\right\rVert^{p}-\sum_{k=1}^{n}P(\left\lVert\widetilde{p_{i}}_{k}-\widetilde{p_{j}}_{k}\right\rVert)\right|
k=1n|pi~kpj~kpP(pi~kpj~k)|\displaystyle\leq\sum_{k=1}^{n}\left|\left\lVert\widetilde{p_{i}}_{k}-\widetilde{p_{j}}_{k}\right\rVert^{p}-P(\left\lVert\widetilde{p_{i}}_{k}-\widetilde{p_{j}}_{k}\right\rVert)\right|
nB(p)dp\displaystyle\leq\frac{nB(p)}{d^{p}}
<1m,\displaystyle<\frac{1}{\sqrt{m}},

with the last inequality following from our conditions on cc and dd. Because MM is symmetric, we can apply Lemma 1 to get rankMm/2\operatorname{rank}M\geq m/2.

Now, we look for an upper bound on the rank. We can write any point x𝔼a1pp𝔼anx\in\mathbb{E}^{a_{1}}\oplus_{p}\cdots\oplus_{p}\mathbb{E}^{a_{n}} as

x=(x~1(1),x~1(2),,x~1(a1),x~2(1),x~2(2),,x~2(a2),,x~n(1),x~n(2),,x~n(an)).x=(\widetilde{x}_{1}^{(1)},\widetilde{x}_{1}^{(2)},\ldots,\widetilde{x}_{1}^{(a_{1})},\widetilde{x}_{2}^{(1)},\widetilde{x}_{2}^{(2)},\ldots,\widetilde{x}_{2}^{(a_{2})},\ldots,\widetilde{x}_{n}^{(1)},\widetilde{x}_{n}^{(2)},\ldots,\widetilde{x}_{n}^{(a_{n})}).

Because PP is even, all terms in the expansion of P(pi~kx~k)P(\left\lVert\widetilde{p_{i}}_{k}-\widetilde{x}_{k}\right\rVert) have integer exponent, i.e., fif_{i} is actually a polynomial. Additionally, since PP has degree at most dd, each fif_{i} is in the span of the set

𝒮={1,k=1nx~kd}𝒮1𝒮2𝒮n,\mathcal{S}=\{1,\sum_{k=1}^{n}\left\lVert\widetilde{x}_{k}\right\rVert^{d}\}\cup\mathcal{S}_{1}\cup\mathcal{S}_{2}\cup\cdots\cup\mathcal{S}_{n},

where each set 𝒮i\mathcal{S}_{i} consists of all the monomials with degree less than dd formed by x~i(1),,x~i(ai)\widetilde{x}_{i}^{(1)},\ldots,\widetilde{x}_{i}^{(a_{i})}. Thus, the fif_{i} are spanned by at most 2+k=1n(ak+d1ak)n2+\sum_{k=1}^{n}\binom{a_{k}+d-1}{a_{k}}-n polynomials. Because every row vector of MM, each of the form (fi(p1),,fi(pm))(f_{i}(p_{1}),\ldots,f_{i}(p_{m})), belongs to the subspace spanned by the set {(f(p1),,f(pm)):f𝒮}\{(f(p_{1}),\ldots,f(p_{m})):f\in\mathcal{S}\}, we have

rankM\displaystyle\operatorname{rank}M 2+k=1n(ak+d1ak)n\displaystyle\leq 2+\sum_{k=1}^{n}\binom{a_{k}+d-1}{a_{k}}-n
k=1n(ak+d1ak)\displaystyle\leq\sum_{k=1}^{n}\binom{a_{k}+d-1}{a_{k}}
n(a+d1a)\displaystyle\leq n\binom{a+d-1}{a}
<nea(1+da)a,\displaystyle<ne^{a}\left(1+\frac{d}{a}\right)^{a},

where we let a=max1inai\displaystyle a=\max_{1\leq i\leq n}a_{i} and use the well-known bound (nk)(en/k)k{n\choose k}\leq(en/k)^{k} for 1kn1\leq k\leq n. Now, recall that dB(p)1/ppd\geq B(p)^{1/p}\geq p and 2p>a2p>a, so d>a/2d>a/2. Thus,

m2rankM<n(3eda)a.\frac{m}{2}\leq\operatorname{rank}M<n\left(\frac{3ed}{a}\right)^{a}.

Combining this inequality with the condition dp<2cnmd^{p}<2cn\sqrt{m} yields

m<cp,an2p+2a2pa,m<c_{p,a}n^{\frac{2p+2a}{2p-a}},

as desired. ∎

8 Acknowledgements

The authors would like to thank the MIT PRIMES program for making this research possible. The authors would also like to thank Larry Guth for introducing them to the problem, as well as Yufei Zhao for some helpful suggestions.

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