Convolution estimates and the number of disjoint partitions

TL;DR

This paper provides an upper bound estimate for the number of disjoint partitions within a finite set collection, extending previous results to arbitrary numbers of partitions using convolution estimates.

Contribution

It generalizes the Kane-Tao result for three partitions to any finite number of disjoint partitions with a new bounding function.

Findings

Derived an explicit upper bound for the count of disjoint partitions.

Extended previous specific case to general finite cases.

Provided a mathematical estimate involving the function c(n).

Abstract

Let $X$ be a finite collection of sets. We count the number of ways a disjoint union of $n-1$ subsets in $X$ is a set in $X$, and estimate this number from above by $|X|^{c(n)}$ where $$ c(n)=\left(1-\frac{(n-1)\ln (n-1)}{n\ln n} \right)^{-1}. $$ This extends the recent result of Kane-Tao, corresponding to the case $n=3$ where $c(3)\approx 1.725$, to an arbitrary finite number of disjoint $n-1$ partitions.

Equations

$$c(n)={\left(1-\frac{(n-1)\ln (n-1)}{n\ln n}\right)}^{-1}.$$

$$f_1\ast f_2\ast \dots \ast f_n(1^m):=\sum _{x_j\in \{0,1{\}}^m:x_1+\cdots +x_n=1^m}{f}_1(x_1)\cdots f_n(x_n).$$

$$\Vert f{\Vert }_p:={\left(\sum _{x\in \{0,1{\}}^m}|f(x){|}^p\right)}^{1/p}.$$

$$p_n:=\frac{\ln \frac{n^n}{(n-1{)}^{n-1}}}{\ln n}.$$

$$f_1\ast f_2\ast \dots \ast f_n(1^m)⩽\prod _{j=1}^n\Vert f_j{\Vert }_{p_n}.$$

$$\left|\left\{(A_1,\dots ,A_{n-1},A)\in \underset{n}{\underbrace{X\times \cdots \times X}}:A=\underset{j=1}{\overset{n-1}{⨆}}{A}_j\right\}\right|⩽|X{|}^{\frac{n}{p_n}},$$

$$\sum _{j=1}^n{u}_j\prod _{\begin{array}{c}i=1\\ i\ne j\end{array}}^n{v}_i⩽\prod _{j=1}^n{\left(|u_j{|}^{p_n}+|v_j{|}^{p_n}\right)}^{1/p_n}.$$

$${\left(\sum _{i=1}^n{x}_i^{1/p_n}\right)}^{p_n}⩽\prod _{i=1}^n(1+x_i),$$

$$p\to {\left(\sum _{i=1}^n{x}_i^{1/p}\right)}^p,$$

$$p\ln \left(\sum x_i\right)⩽\sum \ln (1+x_i^p).$$

$$\ln (x)=\underset{b\in \mathbb{R}}{\min }(b+e^{-b}x-1).$$

$$B(x,b):=\sum (b_i+(1+x_i^p)e^{-b_i}-1)-p\ln \left(\sum x_i\right)⩾0,$$

$$x_k^{\ast }=\frac{e^{\frac{b_k}{p-1}}}{{\left({\sum }_i{e}^{\frac{b_i}{p-1}}\right)}^{\frac{1}{p}}}\text{for}k=1,\dots ,n.$$

$$B(x^{\ast },b)=\sum _k(b_k+e^{-b_k})+1-n-(p-1)\ln \left(\sum _k{e}^{\frac{b_k}{p-1}}\right).$$

$$f(y):=1-n+\sum \ln y_i^r+\sum \frac{1}{y_i^r}-r\ln \left(\sum y_i\right)⩾0$$

$$\frac{1}{y_i}-\frac{1}{y_i^{r+1}}=\frac{1}{y_j}-\frac{1}{y_j^{r+1}}=\frac{1}{\sum y_k}.$$

$$\sum \ln y_i-\ln \left(\sum y_i\right)⩾0.$$

$$s\to \frac{1}{s}-\frac{1}{s^{r+1}},s>0,$$

$$\frac{1}{u}-\frac{1}{u^{r+1}}=\frac{1}{v}-\frac{1}{v^{r+1}}=\frac{1}{ku+(n-k)v}.$$

$$v=\frac{u^{r+1}(1-k)+ku}{(u^r-1)(n-k)}.$$

$$\frac{(z-1{)}^r(n-k{)}^{r+1}}{(z(1-k)+k{)}^r}=(n-1)z-k.$$

$$v^r=\frac{z(n-k)}{(n-1)z-k}.$$

$$(n-1)\ln z+(n-k)\ln \frac{n-k}{(n-1)z-k}-r\ln \frac{z}{z-1}⩾0.$$

$$k\to (n-k)\ln \frac{n-k}{(n-1)z-k}$$

$$z\to \frac{\ln \left(1+\frac{1}{z(n-1)-1}\right)}{\ln \left(1+\frac{1}{z-1}\right)}$$

$$B_n(y_1,\dots ,y_n):=y_1^{1/p_n}\cdots y_n^{1/p_n}.$$

$${\tilde{g}}_j(x_j^{\prime }):=\sum _{{\bar{x}}_j\in \{0,1{\}}^m}{g}_j({\bar{x}}_j,x_j^{\prime })j=1,\dots ,n.$$

$$\sum _{x_j\in \{0,1{\}}^{m+1}:x_1+\cdots +x_n=1^{m+1}.}B(g_1(x_1),\dots ,g_n(x_n))=$$

$$\sum _{x_j^{\prime }\in \{0,1\}:x_1^{\prime }+\cdots +x_n^{\prime }=1.}\sum _{{\bar{x}}_j\in \{0,1{\}}^m:{\bar{x}}_1+\cdots +{\bar{x}}_n=1^m.}B(g_1(x_1),\dots ,g_n(x_n))\stackrel{\mathrm{induction}}{⩽}$$

$$\sum _{x_j^{\prime }\in \{0,1\}:x_1^{\prime }+\cdots +x_n^{\prime }=1.}B({\tilde{g}}_1(x_1^{\prime }),\dots ,{\tilde{g}}_n(x_n^{\prime }))\stackrel{\mathrm{basis}}{⩽}$$

$$B\left(\sum _{x_1\in \{0,1{\}}^{m+1}}{g}_1(x_1),\dots ,\sum _{x_n\in \{0,1{\}}^{m+1}}{g}_n(x_n)\right).$$

$$f_j:\{0,1{\}}^m\to \{0,1\}$$

Full text

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This extends the recent result of Kane–Tao, corresponding to the case $n=3$ where $c(3)\approx 1.725$, to an arbitrary finite number of disjoint $n-1$ partitions.

1 Introduction

Let $\{0,1\}^{m}$ be the Hamming cube of dimension $m\geqslant 1$. Set $1^{m}:=(1,1,\ldots,1)$ to be the corner of $\{0,1\}^{m}$. Take a finite number of functions $f_{1},\ldots,f_{n}:\{0,1\}^{m}\to\mathbb{R}$, and define the convolution at the corner $1^{m}$ as

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Given $f:\{0,1\}^{m}\to\mathbb{R}$ define its $L^{p}$ norm ($p\geqslant 1$) in a standard way

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For $n\in\mathbb{N}$ we set

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Our main result is the following theorem

Theorem 1**.**

For any $n,m\geqslant 1$, and any $f_{1},\ldots,f_{n}:\{0,1\}^{m}\to\mathbb{R}$ we have

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Moreover, for each fixed $n$ exponent $p_{n}$ is the best possible in the sense that it cannot be replaced by any larger number.

As an immediate application we obtain the following corollary (see Section 2.3 below).

Corollary 2**.**

Let $X$ be a finite collection of sets. Then

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where $\bigsqcup$ denotes the disjoint union, and $|X|$ denotes cardinality of the set.

The corollary extends a recent result of Kane–Tao [1], corresponding to the case $n=3$ where $\frac{3}{p_{3}}\approx 1.725$, to an arbitrary finite number $n\geqslant 3$ disjoint partitions.

2 The proof of the theorem

Following [1] the proof goes by induction on the dimension of the cube $\{0,1\}^{m}$. The case $m=1$, which is the most difficult, is the main contribution of the current paper.

2.1 Basis: $m=1$

In this case, set $f_{j}(0)=u_{j}$ and $f_{j}(1)=v_{j}$ for $j=1,\ldots,n$. Then the inequality (1) takes the form

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We do encourage the reader first to try to prove (3) in the case $n=3$, or visit [1], to see what is the obstacle. For example, when $n=3$ equality in (3) is attained at several points. Besides, direct differentiation of (3) reveals many “bad” critical points at which finding the values of (3) would require numerical computations [1]. The number of critical points together with equality cases increases as $n$ becomes larger, therefore one is forced to come up with a different idea. We will overcome this obstacle by looking at (3) in dual coordinates.

Without loss of generality we can assume that $u_{j}$ and $v_{j}$ are nonnegative for $j=1,\ldots,n$. Moreover, we can assume that $v_{j}\neq 0$ for all $j$ otherwise the inequality (3) is trivial.

Let us divide (3) by $\prod_{j=1}^{n}v_{j}$. Denoting $x_{j}:=(u_{j}/v_{j})^{p_{n}}$ we see that it is enough to prove the following lemma.

Lemma 3**.**

For any $n\geqslant 2$ and all $x_{1},\ldots,x_{n}\geqslant 0$ we have

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where $p_{n}=\frac{\ln\frac{n^{n}}{(n-1)^{n-1}}}{\ln n}$

Proof.

For $n=2$ the lemma is trivial. By induction on $n$, monotonicity of the map

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and the fact that $p_{n}$ is decreasing, we can assume that all $x_{i}$ are strictly positive. For convenience we set $p:=p_{n}$. Introducing new variables we rewrite (4) as follows

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Concavity of the function $\ln(x)$ provides us with a simple representation of the logarithmic function

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Therefore we are left to show that for all $x_{i}>0$ and all $b_{i}\in\mathbb{R}$ we have

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where $x=(x_{1},\ldots,x_{n})$ and $b=(b_{1},\ldots,b_{n})$. Notice that given a vector $b\in\mathbb{R}^{n}$, the infimum of $B(x,b)$ in $x$ cannot be reached at infinity because of the slow growth of the logarithmic function. Therefore, we look at critical points of $B$ in $x$

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Notice that $\sum x^{*}_{i}=\left(\sum_{i}e^{\frac{b_{i}}{p-1}}\right)^{\frac{p-1}{p}}$. Therefore

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Setting $r:=p-1>0$, and introducing new variables again we are left to show that

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for all $y_{i}>0$. It is straightforward to check that $f(y)\geqslant 0$ on the diagonal, i.e., when $y_{1}=y_{2}=\ldots=y_{n}$.

In general, we notice that critical points of $f(y)$ satisfy the equation

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Equation (5) gives the identity $\sum y_{i}^{-r}=n-1$, and so at critical points (5) we are only left to show

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Since the mapping

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is increasing on $(0,(1+r)^{1/r})$ and decreasing on the remaining part of the ray, we can assume without loss of generality that $k$ numbers of $x_{i}$ equal to $u\geqslant(1+r)^{1/r}$, and the remaining $n-k$ numbers of $x_{i}$ equal to $v\leqslant(1+r)^{1/r}$. Moreover, we can assume that $0<k<n$ otherwise the statement is already proved. From (5), we have

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From the equality of the first and the third expressions in (7) it follows that

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In order $v$ to be positive we assume that the numerator of (8) is non negative. If we plug the expression for $v$ from (8) into the first equality of (7) then after some simplifications we obtain the following equation in the variable $z:=u^{r}\geqslant 1+r$

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It follows from (7) that $(ku+(n-k)v)^{r}=\left(\frac{z}{z-1}\right)^{r}z$, and so using (9) we obtain

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Therefore at critical points (6) simplifies as follows

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Now it is pretty straightforward to show that (10) is non negative even under the assumption $z\geqslant 1+r$ for $r=p-1=\frac{(n-1)\ln\frac{n}{n-1}}{\ln n}$. Indeed, notice that $z>\frac{n}{n-1}$, and the map

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is increasing on $[1,n-1]$. Therefore it is enough to check nonnegativity of (10) when $k=1$, in which case the inequality follows again using $z>\frac{n}{n-1}$, and the fact that the map

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is increasing for $z\geqslant\frac{n}{n-1}$. ∎

Remark 4*.*

Choice $x_{1}=\ldots=x_{n}=\frac{1}{n-1}$ gives equality in (4), and this confirms the fact that $p_{n}$ is the best possible in Theorem 1.

2.2 Inductive step

Inductive step is the same as in [1] without any modifications. This is a standard argument for obtaining estimates on the Hamming cube (see for example [2]). In order to make the paper self contained we decided to repeat the argument.

Suppose (1) is true on the Hamming cube of dimension $m$. Without loss of generality assume $f_{j}\geqslant 0$, and set $g_{j}:=f_{j}^{p}$ for all $j$. Define

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For $x_{j}\in\{0,1\}^{m+1}$, let $x_{j}=(\bar{x}_{j},x^{\prime}_{j})$ where $\bar{x}_{j}$ is the vector consisting of the first $m$ coordinates of $x_{j}$, and number $x^{\prime}_{j}$ denotes the last $m+1$ coordinate of $x_{j}$. Set

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We have

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2.3 The proof of Corollary 2

Without loss of generality we may assume that all the sets $A$ in $X$ are subsets of $\{1,\ldots,m\}$ with some natural $m\geqslant 1$ (see [1]). For $j=1,\ldots,n$ define functions

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as follows:

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if the set $\{1\leqslant i\leqslant m\;:\;a_{i}=1\}$ lies in $X$, and $f_{j}=0$ otherwise. Finally we define

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if the set $\{1\leqslant i\leqslant m\;:\;a_{i}=0\}$ lies in $X$, and $f_{n}=0$ otherwise. Notice that in this case inequality (1) becomes (2).

Acknowledgements

I would like to thank an anonymous referee, and Benjamin Jaye for helpful comments.