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Related papers: A Sum-Product Estimate for Well Spaced Sets

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Let $A\subset [1, 2]$ be a $(\delta, \sigma)$-set with measure $|A|=\delta^{1-\sigma}$ in the sense of Katz and Tao. For $\sigma\in (1/2, 1)$ we show that $$ |A+A|+|AA|\gtrapprox \delta^{-c}|A|, $$ for…

Combinatorics · Mathematics 2020-02-26 Changhao Chen

Given $d \in \mathbb{N}$, we establish sum-product estimates for finite, non-empty subsets of $\mathbb{R}^d$. This is equivalent to a sum-product result for sets of diagonal matrices. In particular, let $A$ be a finite, non-empty set of $d…

Combinatorics · Mathematics 2021-01-27 Akshat Mudgal

We give a new proof of the discretized ring theorem for sets of real numbers. As a special case, we show that if $A\subset\mathbb{R}$ is a $(\delta,1/2)_1$-set in the sense of Katz and Tao, then either $A+A$ or $A.A$ must have measure at…

Classical Analysis and ODEs · Mathematics 2023-08-24 Larry Guth , Nets Hawk Katz , Joshua Zahl

This paper considers various formulations of the sum-product problem. It is shown that, for a finite set $A\subset{\mathbb{R}}$, $$|A(A+A)|\gg{|A|^{\frac{3}{2}+\frac{1}{178}}},$$ giving a partial answer to a conjecture of Balog. In a…

Combinatorics · Mathematics 2014-01-09 Brendan Murphy , Oliver Roche-Newton , Ilya D. Shkredov

By making use of arithmetic information inequalities, we give a strong quantitative bound for the discretised ring theorem. In particular, we show that if $A \subset [1,2]$ is a $(\delta,\sigma)$-set, with $|A| = \delta^{-\sigma},$ then…

Classical Analysis and ODEs · Mathematics 2025-11-19 András Máthé , William O'Regan

Let $\mathbb{F}_p$ be a finite field of prime order $p$ and let $A \subset \mathbb{F}_p$ be a subset. In the dense regime when $|A| \geq \alpha p$ for some $\alpha \in (0,1)$, we determine the optimal constant $f(\alpha)$ in the inequality…

Number Theory · Mathematics 2026-04-21 Xuancheng Shao

In this paper we prove some results on sum-product estimates over arbitrary finite fields. More precisely, we show that for sufficiently small sets $A\subset \mathbb{F}_q$ we have \[|(A-A)^2+(A-A)^2|\gg |A|^{1+\frac{1}{21}}.\] This can be…

Number Theory · Mathematics 2018-07-17 Doowon Koh , Sujin Lee , Thang Pham , Chun-Yen Shen

A variation on the sum-product problem seeks to show that a set which is defined by additive and multiplicative operations will always be large. In this paper, we prove new results of this type. In particular, we show that for any finite…

Combinatorics · Mathematics 2014-02-25 Antal Balog , Oliver Roche-Newton

I prove two variants of the $ABC$ sum-product theorem for $\delta$-separated sets $A,B,C \subset [0,1]$ satisfying Katz-Tao spacing conditions. The main novelty is that the cardinality of the sets $B,C$ need not match their…

Combinatorics · Mathematics 2025-11-10 Tuomas Orponen

Let $0 < \beta \leq \alpha < 1$ and $\kappa > 0$. I prove that there exists $\eta > 0$ such that the following holds for every pair of Borel sets $A,B \subset \mathbb{R}$ with $\dim_{\mathrm{H}} A = \alpha$ and $\dim_{\mathrm{H}} B =…

Combinatorics · Mathematics 2023-11-13 Tuomas Orponen

Besides various asymptotic results on the concept of sum-product bases in $\mathbb{N}_0$, we consider by probabilistic arguments the existence of thin sets $A,A'$ of integers such that $AA+A=\mathbb{N}_0$ and $A'A'+A'A'=\mathbb{N}_0$.

Number Theory · Mathematics 2019-04-15 Francois Hennecart , Gyan Prakash , E. Pramod

Let $\F_q$ be a finite field of order $q$ and $P$ be a polynomial in $\F_q[x_1, x_2]$. For a set $A \subset \F_q$, define $P(A):=\{P(x_1, x_2) | x_i \in A \}$. Using certain constructions of expanders, we characterize all polynomials $P$…

Combinatorics · Mathematics 2007-05-23 Van Vu

We establish several sum-product estimates over finite fields that involve polynomials and rational functions. First, |f(A)+f(A)|+|AA| is substantially larger than |A| for an arbitrary polynomial f over F_p. Second, a characterization is…

Combinatorics · Mathematics 2014-02-26 Boris Bukh , Jacob Tsimerman

The lambda-dilate of a set A is lambda*A={lambda a : a \in A}. We give an asymptotically sharp lower bound on the size of sumsets of the form lambda_1*A+...+lambda_k*A for arbitrary integers lambda_1,...,lambda_k and integer sets A. We also…

Number Theory · Mathematics 2008-04-03 Boris Bukh

Let $B$ be a Blaschke product with zeros $\{a_n\}$. If $B' \in A^p_{\alpha}$ for certain $p$ and $\alpha$, it is shown that $\sum_n (1 - |a_n|)^{\beta} < \infty$ for appropriate values of $\beta$. Also, if $\{a_n\}$ is uniformly discrete…

Complex Variables · Mathematics 2010-09-29 David Protas

Let $A$ be a subset of a finite field $F := \Z/q\Z$ for some prime $q$. If $|F|^\delta < |A| < |F|^{1-\delta}$ for some $\delta > 0$, then we prove the estimate $|A+A| + |A.A| \geq c(\delta) |A|^{1+\eps}$ for some $\eps = \eps(\delta) > 0$.…

Combinatorics · Mathematics 2007-05-23 Jean Bourgain , Nets Katz , Terence Tao

An achievement set of a series is a set of all its subsums. We study the properties of achievement sets of conditionally convergent series in finite dimensional spaces. The purpose of the paper is to answer some of the open problems…

Functional Analysis · Mathematics 2018-03-01 Jacek Marchwicki , Vaclav Vlasak

We improve a previous sum--products estimates in R, namely, we obtain that max{|A+A|,|AA|} \gg |A|^{4/3+c}, where c any number less than 5/9813. New lower bounds for sums of sets with small the product set are found. Also we prove some pure…

Combinatorics · Mathematics 2016-02-11 Sergei Konyagin , Ilya D. Shkredov

Let $R$ be a finite valuation ring of order $q^r.$ Using a point-plane incidence estimate in $R^3$, we obtain sum-product type estimates for subsets of $R$. In particular, we prove that for $A\subset R$, $$|AA+A|\gg \min\left\{q^{r},…

Combinatorics · Mathematics 2017-01-30 Esen Aksoy Yazici

We estimate the sizes of the sumset A + A and the productset A $\cdot$ A in the special case that A = S (x, y), the set of positive integers n less than or equal to x, free of prime factors exceeding y.

Number Theory · Mathematics 2010-10-19 William D. Banks , David Covert
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