Convexity, superquadratic growth, and dot products

Abstract Let P⊂R2 be a point set with cardinality N. We give an improved bound for the number of dot products determined by P, proving that |{p·q:p,q∈P}|≫N2/3+c. A crucial ingredient in the proof of this bound is a new superquadratic expander involving products and shifts. We prove that, for any finite set X⊂R, there exist z,z′∈X such that (zX+1)(2)(z′X+1)(2)(zX+1)(2)(z′X+1)≳|X|5/2. This is derived from a more general result concerning growth of sets defined via convexity and sum sets, and which can be used to prove several other expanders with better than quadratic growth. The proof develops arguments from Hanson, Roche‐Newton, and Rudnev [Combinatorica, to appear], and uses predominantly elementary methods.

1. Introduction 1.1.Dot products.For a finite set P of points in R 2 , let Λ(P ) denote the set of all dot products determined by P , that is Λ(P ) := {p • q : p, q ∈ P }, where the dot product of p = (p 1 , p 2 ) and q = (q 1 , q 2 ) is p • q = p 1 q 1 + p 2 q 2 .In the spirit of the Erdős distinct distance problem, one expects that the set Λ(P ) should be large for any P .The bound (1) |Λ(P follows from an application of the Szemerédi-Trotter Theorem.* The Erdős distance problem was resolved up to logarithmic factors in a remarkable paper of Guth and Katz [4], which proved that the bound ( 2) holds for any finite point set P ⊂ R 2 , where D(P ) is the set of Euclidean distances determined by pairs of points in P .It is widely believed that a bound similar to (2) also holds for Λ(P ).
However, in stark contrast to the distance problem, this question remains wide open, and the apparent connection between these two problems may be somewhat misleading.
For instance, the closely related unit distance conjecture claims that any fixed distance can occur at most N 1+ǫ times among a point set of size N.The bound (3) |{(p, q) ∈ P × P : p − q = 1}| ≪ N 4/3   was established by Spencer, Szemerédi and Trotter [15].This bound is widely believed to be sub-optimal, but there has been no significant improvement to (3) in 37 years.On the other hand, a construction in [6] shows that there exists a point set P ⊂ R 2 such that |{(p, q) ∈ P × P : p • q = 1}| ≫ N 4/3 .
In this paper, we give the following improvement to (1).
Theorem 1.1.For any finite set P ⊂ R 2 We remark that a better bound, with exponent 2 3 + 1 39 , for the case of skew symmetric bilinear forms was obtained in [6].This covers, for instance, the set of wedge/cross products determined by a point set, i.e. p ∧ q = p 1 q 2 − p 2 q 1 .1.2.Superquadratic growth.An essential ingredient in the proof of Theorem 1.1 is a new sum-product type result which shows that a certain expander involving products and shifts gives superquadratic growth.We will prove the following theorem.
Here, A (k) denotes the k-fold product set {a 1 • • • a k : a 1 , . . ., a k ∈ A}.A key quantitative feature of Theorem 1.2 is that the exponent in the right hand side of ( 5) is strictly larger than 2. We refer to the set in Theorem 1.2 as a superquadratic expander.There are many sets defined by a combination of arithmetic operations that, one may suspect, grow in such a way.However, given such a set, it is often difficult to prove that it really is a superquadratic expander, particularly as methods from incidence geometry typically lose their potency in this range.The existence of other superquadratic expanders has been established in [1], [12], [14], [11].
Theorem 1.2 is derived from a more general result, the forthcoming Theorem 2.6.The statement of Theorem 2.6 is a little technical, depending on some further definitions, and we do not state it in full yet.However, Theorem 2.6 can also be used to derive several other new sum-product results.For instance, we will prove the following result about superquadratic growth under addition for cubes of shifts.
Corollary 1.3.Let A ⊂ R be a finite set.Then there exists a, a ′ ∈ A such that Further applications of Theorem 2.6 will be given in the forthcoming Section 4.1.The proof of Theorem 2.6 uses similar elementary methods to an earlier paper of Rudnev and the first two authors [5], which was itself built up from a simple and beautiful observation concerning sumsets of convex sets in a paper of Ruzsa, Shakan, Solymosi and Szemerédi [13].Our first step towards proving Theorem 2.6 and its consequences is to give a simple proof of the following quadratic expander.
Theorem 1.4.Let A ⊂ R be a finite set and let f be a strictly convex or concave function.Then there exists a, a ′ ∈ A such that Theorem 1.4 gives the optimal order of growth.This can be seen by considering the case A further example illustrating the optimality of Theorem 1.4 is given in Section 2.
1.3.Sketch of the proof of Theorem 1.1.The proof of Theorem 1.1 is rather long, and so we sketch it here in an attempt to convey some of the main ideas behind the proof to the reader.In this sketch we are using Theorem 1.2 as a black box.
That is, we make the simplifying assumption that z = z ′ in Theorem 1.2.After several applications of Plünnecke-Ruzsa type inequalities, it would then follow that, for any Y ⊂ R, Quantitatively, the additional 1/128 in the exponent is important here.Later in the sketch, we will apply this bound in a case whereby |Y | = N 2/3 and |X| = N 1/3 , in which case (6) yields the non-trivial bound ( 7) To prove Theorem 1.1, we begin with an application of the Szemerédi-Trotter Theorem which tells us that we obtain an exponent better than 2/3 unless we are in a very specific situation.Roughly speaking, this allows us to assume that P , our given set of N points, is supported on exactly N 1/3 lines through the origin, each containing N 2/3 points from P .Furthermore, since the problem is rotation invariant, we may assume that one of these lines is the x-axis.This in fact implies that the set of x-coordinates involved in P also has size around N 2/3 , otherwise we have too many dot products between a fixed point from P on the x-axis and the remaining elements of P .Pigeonholing, we deduce that there is some vertical line ℓ 0 , with equation x = x 0 , which supports around N 1/3 points of P .Simplifying slightly, let's assume that all of the N 1/3 lines through the origin contain a point on P ∩ ℓ 0 .
Let S denote the set of slopes of these N 1/3 lines, each containing a point from P ∩ ℓ 0 and N 2/3 points from P in total.Now apply (6) for the set S. It follows that there is some slope s ∈ S such that the bound holds for any Y ⊂ R.However, if we consider the set of dot products between points of P ∩ {y = sx} and P ∩ ℓ 0 , this is equal to the set x 0 Y (sS + 1), where Y is the set of the N 2/3 x-coordinates of P ∩ {y = sx}.Recalling the calculation in (7), we obtain an exponent better than 2/3, as required.
If we remove the simplifying assumption that z = z ′ , things become a bit trickier when it comes to constructing the set that plays the role of Y .We need to choose a set of xcoordinates that works with respect to two lines y = sx and y = s ′ x simultaneously, and it is not immediately obvious that these lines share many x-coordinates.
However, an extra piece of information comes to our aid here.We may consider the dot products between P ∩ {y = sx} and P ∩ {y = s ′ x}.This turns out to be essentially a product set of the two sets of x-coordinates.Each set has size N 2/3 , and their product set has approximately the same size, otherwise we are done.We can use this information, and some fairly basic additive combinatorial trickery, to deduce the existence of a good set Y which works simultaneously for both of the fixed lines.
1.4.Structure of this paper.We begin Section 2 by proving Theorem 1.4.The main goal of the section is to prove our general result on superquadratic growth, which is Theorem 2.6.We then apply this bound to prove our main result on growth of products with shifts, Theorem 1.2.In Section 3 we use Theorem 1.2, or rather a slightly refined statement containing some extra information, to prove our main result, Theorem 1.1.In Section 4.1, we give several other applications of Theorem 2.6.This includes the proof of Corollary 1.3.

Superquadratic expansion
2.1.An elementary proof of a quadratic expander.The basic argument underlying this section is contained in the forthcoming proof of Theorem 1.4.In fact we record the following more general result, which will prove useful.
be a finite set of real numbers, and let f be a strictly convex or strictly concave function.Suppose that h < h ′ are parameters and In particular, there exists a, a ′ ∈ A such that Proof.Observe that the intervals all have length h ′ − h and do not overlap because of the spacing condition.Therefore, by convexity of f , the intervals are increasing (or decreasing, if f is concave) in length as i increases, and they also do not overlap.This enables us to squeeze the smaller intervals into the bigger ones.In particular, for all 1 ≤ i < j ≤ N we have This shows the existence of at least j − 1 elements of f . Summing over all intervals, it therefore follows that For the final statement of the theorem, let a and a ′ be two elements of A such that a ′ − a is the minimum positive element of A − A.
As we mentioned in the introduction, Theorem 1.4 is optimal in the case when This bound is also optimal, again in the case B = [N].In fact, a closer look at the proof shows that we obtain the bound (9) |bB A very similar bound to (9) was obtained by Garaev using a different elementary argument.He made the observation that, if B is a set of positive integers with maximal element b max and with d min = b ′ − b being the smallest positive element in B − B, then as there are no repetitions in this sum set.
There is some flexibility in Theorem 1.4, hinted at in Theorem 2.1, as to the choice of the elements a and a ′ .Indeed, we chose them so that a ′ − a is a minimal positive element of A−A, but we could also have taken another close pair of elements, with minimal modification of the proof.In particular, the same proof works if we replace a, a ′ with a second nearest consecutive pair b, b ′ ∈ A. That is, we choose b, b ′ to be consecutive elements of A such that We then have the following refined version of the statement of Theorem 2.1, which will be useful later.
Theorem 2.2.Let A ⊂ R be a finite set and let f be a strictly convex or concave function.Let a < a ′ be a nearest consecutive pair of elements in A and let b < b ′ be a second nearest consecutive pair.Then (10) |f and Proof.The proof of Theorem 2.1 immediately implies (10).To prove (11), define A ′ ⊂ A to be the set consisting of every second element of and so the spacing condition ( 8) is satisfied.Therefore as required.
In this paper, we are particularly interested in the expander consisting of products of shifts of products, and so we record the following corollary.

and with the additional properties
• log z 1 < log z ′ 1 is a nearest consecutive pair of elements in log X, and Proof.Apply Theorem 2.2 with A = log X and f (t) = log(e t + 1).
2.2.More variables, more growth.We would like to do better than Theorem 2.1 using differences of differences, in the spirit of [5].We begin to work towards this goal by proving the following lemma.
Lemma 2.4.Let X = {x 1 < . . .< x N } be a finite set of real numbers and suppose that F is a function which is strictly increasing and strictly convex (or concave) on the interval Proof.This argument is taken from [5].We have We will estimate the number of elements in the appropriate interval coming from of each factor on the right separately. Let there is some integer L and set D ′ ⊆ D, say of size m, such that (12) mL ≫ N log N , There are at least L/2 such intervals, they are visibly disjoint, and each has a distinct length in view of the convexity of F .Thus if and by considering such contributions across all d ∈ D ′ , we deduce Finally, combining the estimates ( 12) and ( 13) concludes the proof.
The following is merely a two-dimensional extension of Theorem 2.1, obtained by applying the one-dimensional result in each coordinate independently, and stated in a way that is convenient to our remaining arguments.The lemma extends with no difficulty to arbitrary cartesian products in any dimension.

Lemma 2.5. Suppose that we have a real numbers h
is a finite set of reals which satisfies the spacing condition Suppose f 1 and f 2 are strictly increasing and strictly convex functions.Then, for any contains all points of the form Proof.From the spacing condition, we deduce that the boxes are disjoint.That the boxes B k 1 ,k 2 are disjoint follows from this and the fact that each of f 1 and f 2 is strictly increasing.To conclude the proof, note that for each l ∈ {1, 2} we have j l ≤ k l .The fact that f l is strictly increasing and strictly convex yields Lemma 2.5 will produce for us a basic amount of expansion, and can be applied rather generally.From there, we will work locally, but in order to find further expansion we will require additional convexity.Such convexity will be found in a pair of discrete derivatives of the function f , provided f is itself sufficiently convex.To that end, let f 1 , f 2 : [u, ∞) → R be continuous functions defined on some interval, which are each strictly increasing and strictly convex.For a non-negative integer W , we say the quadruple (h is the graph of a strictly convex or strictly concave function.This definition just allows us to pass to a large subset of a curve which is convex, should the full curve fail to be so.With it in hand, we are ready to state and prove our main result about superquadratic expanders in its most general form.Again, we remark that the spacing condition serves only to provide additional freedom in our future choice of parameters h 1 , h 2 , h ′ 1 , and h ′ 2 .
Theorem 2.6.Let f 1 , f 2 : [a 1 , ∞) → R be continuous, strictly increasing and strictly convex functions, and suppose (h , for some W . Suppose A = {a 1 < . . .< a N } is a finite set of positive real numbers satisfying the spacing condition Then Proof.Apply Lemma 2.5 to deduce that the disjoint boxes B k 1 ,k 2 contain all points of the form Observe that for fixed values of k 1 and k 2 , the set of such points can be re-written as the cartesian product where . By W -goodness, there is an interval I such that is the graph of a strictly convex (or concave) function, say F , and such that A ′′ = I ∩ A ′ has size at least N/3W .Define which is the projection of these points to the x-axis, so that Apply Lemma 2.4 to set X and the function F , in order to deduce that and The boxes B k 1 ,k 2 are disjoint, so we conclude the proof by adding the contributions from each choice of k 1 and k 2 .
2.3.Proof of Theorem 1.2.We will now use Theorem 2.6 to prove Theorem 1.2.We instead prove it in the form of the following statement, which contains some useful extra information.
Corollary 2.7.Let X be a set of positive reals.Suppose that z

and with the additional properties
1 is a nearest consecutive pair of elements in log X, and 2 is a second nearest consecutive pair of elements in log X.
In particular, there exist z, z ′ ∈ X such that Proof.Let A = log(X) and write Let A ′ be the set of all elements of A with even indices, relabelled as A ′ = {b 1 , b 2 , . . .b ⌊N/2⌋ } and observe that A ′ satisfies the spacing condition . Similarly, .
With an appropriate value of W , we have from Theorem 2.6 that We turn to estimating W .The curve in question can be parameterized as a shift of (log(z ′ 1 t + 1) − log(z 1 t + 1), log(z ′ 2 t + 1) − log(z 2 t + 1)) , t > 0 and so , which is not identically zero for positive values of z 1 , z ′ 1 , z 2 and z ′ 2 unless z 1 = z 2 and As long as this is guaranteed, there are at most 4 points at which this curve can change convexity, and so we can take W = 5.This completes the proof.record one more important corollary here, which is that the pair z, z ′ satisfying (15) also satisfies the bound (zX + 1)(z ′ X + 1) (zX + 1) ≫ |X| 2 by Corollary 2.3.We also make a note here of the obvious fact that z = z ′ .
Corollary 2.8.Let X be a set of positive reals.Then there exists z, z ′ ∈ X, with z = z ′ , such that both hold.

2.4.
The Plünnecke-Ruzsa inequality and applications.Before we are ready to start the proof of Theorem 1.1, we need to manipulate Plünnecke's inequality in order to deduce from Corollary 2.8 a sum-product type theorem which is tailored for the dot product problem.We will prove Corollary 2.13, the key feature of which is that we use only product sets (as opposed to including ratio sets) and still obtain better than quadratic growth.We believe it is of independent interest.We will use the standard Plünnecke-Ruzsa inequality.See Petridis [8] for a modern proof of the statement.Theorem 2.9.Let X be a finite set in an additive abelian group.Then for any positive integers k and l The Ruzsa Triangle Inequality will be used later in the paper.This is the following result.
Theorem 2.10.Let X, Y, Z be finite sets in an additive abelian group.Then A quick observation is that Theorem 2.10 gives the bound We will also use the following consequence of the Plünnecke-Ruzsa inequality, which was observed by Garaev [2].
Theorem 2.11.Let X 1 , . .X k and Y be finite subsets in an additive abelian group.Then We also need a non-trivial bound for the size of A(A + 1).The following result of Garaev and Shen [3] is sufficient for our purposes.
Theorem 2.12.For any finite set A ⊂ R and any non-zero λ ∈ R, We remark that improvements to Theorem 2.12 are known, and a recent result of Stevens and Warren [16] gives the better exponent 49/38.We use Theorem 2.12 in the proof of Theorem 1.1 in order to simplify the exposition slightly.If we instead apply the result of Stevens and Warren, we obtain a further small improvement, (namely c = 1/2739 instead of 1/3057).
We are now ready to derive the following consequence of Corollary 2.8.
Corollary 2.13.Let X be a set of positive reals.Then there exists z, z ′ ∈ X such that Proof.Let z, z ′ ∈ X be elements given by Corollary 2.8 which satisfy the bounds
(18) again and (20), We can lower bound |T | using (21).After doing this and rearranging, it follows that .Moreover, should it be the case that P is incident to at least |P | 1/3+c lines through the origin, the exponent 2/3 can be improved.
Lemma 3.1.Let P ⊂ R 2 be a finite point set.Let L denote the set of all lines through the origin incident to P .Then there exists p ∈ P such that Proof.We may assume without loss of generality that each line in L has the form y = λx, with λ = 0. Otherwise, we can rotate the whole point set so that it avoids the co-ordinate axes, since all of the quantities involved are rotation invariant.This is just done in order to simplify some notation.Let ℓ λ denote the line y = λx.For each ℓ λ ∈ L, we choose a point p λ ∈ P on the line ℓ λ arbitrarily.Consider the set {p λ • q : q ∈ P } of all dot products determined by p λ .Note that p λ • q = p λ • q ′ if and only if q and q ′ lie on a line which is perpendicular to ℓ λ .Let S λ denote the set of all lines with slope −1/λ which are incident to P .It therefore follows that Define S to be the set of lines S = Note that this union is disjoint, since elements of S λ and S λ ′ have distinct slopes for λ = λ ′ .By the Szemerédi-Trotter Theorem as well, and so (25) holds in either case.Because S is a disjoint union, It then follows from (23) that there is some p λ ∈ P such that We will take this lemma as a launch point for our theorem.The lemma cannot itself be improved without additional information.Indeed, if P = [N] × [N] then L has size of order N 2 .Meanwhile, any single point (m, n) makes at most 2N 2 distinct dot products with the rest of P , so this result is optimal.The key will be to use a few different lines from L, whose slopes will introduce different arithmetic constraints and allow us to invoke an expansion result.This lemma puts us in a regime where this approach is viable.

Proof of Theorem 1.1.
Proof.We repeat some notation from the proof of the previous subsection: L denotes the set of lines through the origin covering P , and ℓ λ ∈ L is the line with equation y = λx.Fix c = 1 3057 .The proof is presented in such a way that c can be viewed as a parameter, and we will calculate the optimal choice for c at the end of the proof.
The idea of the proof is to use the assumption of few dot products in order to deduce arithmetic combinatorial constraints associated to the set of points P .We do this by considering dot products arising from various subsets of P , each of which is chosen to impose a different constraint.The resulting combination of constraints will combine in such a way as to violate the expansion results proved in the previous section.
Step 1.There is a large subset, P ′ , of points lying on a small set, L ′ , of rich lines through the origin.The x-axis is in L ′ , and the other lines in L ′ have positive slope.
Lemma 3.1 implies the desired bound (4) if we can henceforth assume that (26) We perform a dyadic decomposition of the line set L according to the number of points from P they contain.Note that For the remaining part of the claim, observe first that at least half of the slopes of the lines in L ′ have the same sign.If the most popular sign is negative, we rotate the whole point set by 90 degrees so that at least half of the lines in L ′ have positive slope.We can then make a further rotation so that the line with smallest positive slope goes to the x-axis.Abusing notation slightly, we redefine the remaining set of lines as L ′ .Let P ′ denote the set of points from P lying on the lines of L ′ .We have Step 2. There is a large subset, L ′′ , of lines which intersect the vertical line x = x 0 in P ′ .Fix a point p ∈ P ′ lying on the x-axis.The set of dot products {p • q : q ∈ P ′ } is in bijection with L 0 , the set of vertical lines incident to P ′ .If we denote by X the set of all x-coordinates appearing in P ′ , then |X| = |L 0 |, and we may therefore assume that |X| ≤ |P | 2/3+c .For each x 0 ∈ X, define and observe that It therefore follows by the pigeonhole principle that there exists x 0 ∈ X such that Let L ′′ := L(x 0 ) be the set satisfying (29).
Step 4. The sets, X(s) and X(s ′ ), of x-coordinates from ℓ s ∩ P ′ and ℓ s ′ ∩ P ′ respectively, have multiplicative structure controlled by Λ(P ).In particular, there is a non-zero a ∈ R, for which the intersection Z = X(s) ∩ aX(s ′ ) is large: Recall that all lines in L ′′ have approximately M points from P ′ on them, and so Now consider the subset of Λ(P ) consisting of all dot products between points of P ′ ∩{y = sx} and P ′ ∩ {y = s ′ x}.This is the set On the other hand Therefore, by (28), there exists a ∈ X(s)/X(s ′ ) such that Step 5.The multiplicative structure of Z is controlled by Λ(P ): the following subset of Λ(P ): This is indeed a subset of Λ(P ), since (x, sx) and (x 0 , s 1 x 0 ) are elements of P .Furthermore, The first claimed estimate in (33) follows immediately.Similarly, the set is a subset of Λ(P ) and has cardinality This shows the second claimed estimate.
Step 6.A second appeal to expansion -the sets (sS + 1)Z and (s ′ S + 1)Z cannot both be too small.
Observe first that the other hand, we can still use Theorem 2.6 to deduce results about growth which concerning sums and differences of (A + A) 2 .For instance, we can deduce the following result.
Corollary 4.2.For any A ⊂ R, there are elements a, a ′ ∈ A such that In particular, Proof.We apply Theorem 2.6 with f 1 (t) = t 2 and f 2 (t) = t 3 .This time we take (h 1 , h ′ 1 ) = (h 2 , h ′ 2 ) equal to any pair of nearest neighbours in A, which will play the role of (a, a ′ ).In this case, the curve in question is parameterized by which is a parabola, and hence W = 1 is permissible.
One can interpret the estimate (36) as saying that, if the set 4(A + A) 2 − 3(A + A) 2 attains its minimal possible value O(N 2 ), then we obtain a cubic order of growth for the set |4(A + A) 3 − 3(A + A) 3 |, which is better than that given by Corollary 4.1.The estimate (36) is seen to be best possible in view of the case X = [N].

4.2.
A superquadratic bound for three products and shifts.We conclude the paper by recording the following consequence of the work in Section 2. In order to prove this, we will need the fact that (37) |(AA + 1)(AA + 1)| |A| 2 holds for any A ⊂ R.This follows from a simple application of the Szemerédi-Trotter Theorem.The proof of (37) is a very small modification of the proof of the main result in [9], and also exists implicitly in earlier work of Jones [7].We give the full statement and proof for completeness., which was proven in [10].We omit the details to avoid repetition.

3 . 1 .
Controlling the number of slopes with the Szemerédi-Trotter Theorem.The next lemma uses the Szemerédi-Trotter Theorem to give a lower bound for the size of Λ(P ) in terms of the number of lines through the origin.The result is well-known, and can be used to quickly deduce the threshold bound |Λ(P )| ≫ |P | 2/3 (24) |L||P | = ℓ λ ∈L I(P, S λ ) = I(P, S) ≪ |P | 2/3 |S| 2/3 + |P | + |S| If |P | dominates the right side of (24), then we must have |L| ≪ 1.In this case, there is some line ℓ ∈ L such that |ℓ ∩ P | ≫ |P |.This set determines Ω(|P |) dot products, since the set of dot products determined by set of points on a line is in one-to-one correspondence with the product set of the set of magnitudes.If the first term on the right hand side of (24) is dominant then a rearrangement gives (25) |S| ≫ |P | 1/2 |L| 3/2 , while if |S| dominates the right hand side of (24) then we notice that by definition |P | ≥ |L|, so we have |S| ≫ |P ||L| ≥ |P | 1/2 |L| 3/2