Large Convex sets in Difference sets

We give a construction of a convex set $A \subset \mathbb R$ with cardinality $n$ such that $A-A$ contains a convex subset with cardinality $\Omega (n^2)$. We also consider the following variant of this problem: given a convex set $A$, what is the size of the largest matching $M \subset A \times A$ such that the set \[ \{ a-b : (a,b) \in M \} \] is convex? We prove that there always exists such an $M$ with $|M| \geq \sqrt n$, and that this lower bound is best possible, up a multiplicative constant.


Introduction
This paper is concerned with the existence or non-existence of convex sets in difference sets.A set A ⊂ R is said to be convex if its consecutive differences are strictly increasing.That is, writing holds for all i = 2, . . ., n − 1.Most research on convex sets comes in the context of sumproduct theory, and one may think of the notion of a convex set as a generalisation of a set with multiplicative structure.For instance, it is known than convex sets determine many distinct sums and differences.In particular, it was proven in [8] that the bound 1|A − A| ≫ |A| 8/5−o (1)   holds for any convex set A. Here, A − A := {a − b : a, b ∈ A} denotes the difference set determined by A. This results captures the vague notion that convex sets cannot be additively structured, and there has been considerable effort expended to quantify and apply this idea in various way, see for instance [1], [3] and [9].
Given a finite set B ⊂ R, define

C(B) := max
That is, C(B) denotes the size of the largest convex subset of B. The first question that we consider is the following: given a convex set A ⊂ R, what can we say about the possible value of C(A − A)?A first observation is that as can be seen by considering the convex set A − a ⊂ A − A, where a is an arbitrary element of A. There are some simple constructions showing that the lower bound (1) is optimal up to a multiplicative constant; for instance, we can take A = {i 2 : 1 ≤ i ≤ n}.
In this paper, we give a construction of a convex set A whose difference set contains a very large convex set.
Theorem 1.1.For all n ∈ N, there exists a convex set A ⊂ R with |A| = n such that A − A contains a convex subset S with cardinality |S| ≫ n 2 .
Using the notation introduced earlier, Theorem 1.1 states that there exists a convex set A such that C(A−A) ≫ |A| 2 .This result shares some similarities with main result of [6], where it was established that there exists a set A ⊂ R such that A + A contains a convex subset with cardinality Ω(|A| 2 ).The main qualitative difference is that we have the additional restriction that the set A is also assumed to be convex.Also, Theorem 1.1 provides a convex subset of the difference set, rather than the sum set.
The simple construction giving rise to the lower bound (1) feels like something of a cheat, and so we consider a variant of this problem where we make a further restriction concerning the origin of the convex subset of a difference set.A set M ⊂ A × A is a matching if the elements of M are pairwise disjoint.Given a matching M ⊂ A × A, define the restricted difference set

CM(A) := max
M ⊂A×A:M is a matching and A− M A is convex.

|M|.
That is, CM(A) denotes the size of the largest matching on A which gives rise to a convex subset of A − A. Now, we ask a similar question for this quantity: given a convex set A ⊂ R, what can we say about the size of CM(A)?In particular, how small can this quantity be?Should we expect an analogue of the bound (1) if we rule out this simple construction?In this paper we answer this question by giving the following two complimentary results, showing that CM(A) ≥ |A| and that this bound is optimal up to a multiplicative constant.Theorem 1.2.Let n ∈ N be sufficiently large and suppose that A ⊂ R is a convex set with cardinality n.Then there exists a matching

Proof of Theorem 1.1
Define where Assume that n is a sufficiently large multiple of 100.This assumption is made only to simplify the notation slightly, and can be easily removed at the price of introducing some floor and ceiling functions to the calculations.Define A = {a 1 < • • • < a n }.Observe that A is convex.Indeed, the sequence a i+1 − a i is increasing, as can be seen by calculating that For each integer k ∈ [0.009n, 0.01n] define D k to be the set of kth differences increases with i.This can be seen by observing that (2) d and hence increases with i.
We will find a large convex subset of A − A by efficiently gluing together consecutive convex sets D k .We will make use of the following observation from [6].
Before we get into the details of the proof of Theorem 1.1, which involves some rather tedious calculations, let us take a moment to try and explain the idea behind it, with the help of some pictures.
Firstly, we note that, although the sets D k are convex, they are only slightly convex, in the sense that, if we zoom out and take a look at D k , it appears to resemble an arithmetic progression with common difference 2c 1 k.Note also that this common difference increases slightly as k increases.Figure 1.This picture shows the first three elements of D 10 after setting n = 10000.The three elements form a convex set, but to the naked eye they appear to be arranged in an arithmetic progression.
The other important feature of this construction is that we have chosen the parameters in such a way that the D k have convenient overlapping properties.In particular, each D k has diameter approximately 3  2 , and starts at k.In particular, this means that neighbouring D k have a significant overlap, but also that each D k takes sole ownership of a section of the real line.We can use this setup to form a convex set by gluing together consecutive D k .In the region where D k and D k+1 overlap, the elements of D k are slightly more dense (because the common difference of the approximate arithmetic progression is smaller).This ensures that there exist two consecutive elements of D k in this region, which allows for an application of Lemma 2.1.Meanwhile, the existence of the non-overlapping region ensures that this glued set contains many elements of D k for each k.
Now we come to the formal details.We use the notation d max ].We will prove the following two facts about the intersection properties of these intervals.
Claim 2.3.For each k ∈ [0.009n, 0.01n], there are at least Cn elements of D k in the interval (d min ), where C > 0 is an absolute constant.
Once we have proved these two claims, the proof will be finished.Indeed, we can use Claim 2.2 together with Lemma 2.1 to glue together consecutive convex sets D k to form a set Claim 2.3 guarantees that, for each k ∈ [0.009n, 0.01n], there at least Cn elements in D k ∩ S that do not appear in D j ∩ S for any j = k.This implies that It remains to prove the two claims.
Proof of Claim 2.2.We will show that the interval max ] contains at least two more elements of D k than it does of D k+1 .It then follows that there must exist two consecutive elements of D k in this interval, which then implies the existence of the claimed configuration We begin by establishing a lower bound for |D k ∩ I|, namely We need an upper bound for |{i ∈ N : We will compare the bound in ( 6) with an upper bound for |D k+1 ∩ I|, which we deduce now.Observe that Note that the term which involves the multiple of c 2 in the previous step is at most 1/n 2 .Therefore, this term is less than c 1 , which allows us to write It remains to show that the lower bound given in ( 6) is at least as big as the upper bound given in (7), plus two.That is, we need to show that , and eventually Since we have 0.009n ≤ k ≤ 0.01n, it would be sufficient to prove that Substituting in the definition of c 1 , the previous inequality becomes  max ]|, we make use of (7) and deduce that Next, we present a lower bound for The last inequality above uses the fact that the term c 2 (3i 2 k + 3ik 2 ) is at most 0.01, provided that n is sufficiently large.Therefore, By combining this inequality with (8), it follows that

Matchings
Proof of Theorem 1.2.Again, write For convenience, we the shorthand use k := ⌈ √ n⌉.The matching M is given by The matching M has cardinality k ≥ √ n, and the choice of parameters ensures that it is indeed a well-defined subset of A × A, provided that n is sufficiently large.To verify this, we just need to check that k + 1 + k(k+1)

It remains to check that
denote the ith element of A − M A. We need to check that e i+1 − e i > e i − e i−1 holds for all 2 ≤ i ≤ k − 1.A telescoping argument gives and therefore It then follows that There are i + 2 terms with a positive sign and i + 1 with a negative sign.We can pair off the i + 1 largest positive terms with smaller (in absolute value) negative terms to conclude the proof, as follows: Proof of Theorem 1.3.For each j = 1, . . ., n, define The set A = {a j : 1 ≤ j ≤ n} is a convex set.Indeed, the consecutive differences of A are given by a sequence which is strictly increasing.
Let M ⊂ A × A be a matching such that A − M A is convex.Our goal is to prove that |M| ≪ √ n.Let k ≤ n − 1 be an integer.Repeating notation used earlier in the paper, set d We calculate that 1) .
An important feature of this construction is that the diameter of the components D k , which is approximately (2n) n−2 , is significantly smaller than the gaps between consecutive components, which is approximately (2n) n .This allows us to conclude that, with at most one exception, a convex set can have at most one representative from each D k .This is formalised in the following claim.
Proof.The first sentence of the claim follows from the second, and so it is sufficient to prove only the second sentence.Suppose for a contradiction that is not the first element of S, and since S also contains a larger element of D k 2 , we also know that d .By the convexity of S, On the other hand The second inequality uses the fact that k 2 ≤ n, while the third inequality is an application of the inequality (12) (2n) j + (2n which is valid for all j, n ∈ N.
Using the same basic fact about the blocks D k again, namely that the gap between consecutive blocks is significantly larger than their individual diameters, we now show that the blocks which contain elements of a convex set must occur in a weakly convex form.A set holds for all i = 2, . . ., n − 1.For a given set S ⊂ A − A, we define Proof.Suppose for a contradiction that K(S) is not weakly convex.Then there exists three consecutive elements The difference between d and d Meanwhile, the next difference can be bounded by However, by the convexity of S, we also have d Combining this with the previous two inequalities and applying (14) yields We then once again use inequality (12) to obtain Finally, note that k 3 + k 1 < 2n.This holds because k 3 , k 1 ≤ n − 1, as the sets D k i are only defined within this range.Plugging this into (15), we obtain the contradiction Another useful feature of this construction is that the consecutive differences within the components D k shrink rapidly, which makes it difficult to find a large convex sets in A − A.   In this picture, we zoom in to take a closer look at the way the elements of D k are distributed (here we consider the set D 1 with n = 5).Crucially, the gaps between consecutive elements of D k shrink rapidly, with the conseutive differences resembling a geometric progression with a small common ratio.This picture can be used for a sketchy justification of Claims 3.3 and 3.4 We use this fact in the following claim to establish that a convex set cannot contain more than two elements from any D k .Proof.Suppose for a contradiction that there exist three consecutive elements of S belonging to the same block D k .In particular, we have d On the other hand, Combining the previous two inequalities with (16), we obtain the intended contradiction By proving Claims 3.1 and 3.3, we have essentially proved that each block of D k in A − A can contain at most one element of a convex set S ⊂ A − A. We can be a little more precise; taking the potential exceptional block into account, we have the bound It remains to upper bound the size of the indexing set K(S).We need one more claim to allow us to achieve this goal.Note that the following claim represents the first time in the proof where we use the fact that the convex set S is derived from a matching.
Claim 3.4.Suppose that M ⊂ A × A is a matching and that S = A − M A is a convex set.Then the indexing set K(S) does not contain four consecutive elements which form an arithmetic progression.
Proof.Suppose for a contradiction that four consecutive elements of K(S) form an arithmetic progression.It then follows from Claim 3.1 that there exist four consecutive elements d and d in S, for some positive integers k, t such that k + 3t ≤ n − 1.Since S is derived from a matching, it must be the case that the j i are pairwise distinct.Write Since S is convex, we have e 1 < e 2 < e 3 .
We will now show that it must be the case that j 2 < j 1 .Suppose for a contradiction that this is not true, and so j 2 > j 1 .Then On the other hand It follows from the previous two bounds that This is a contradiction, and we have thus established that j 2 < j 1 .The exact same argument implies that j 3 < j 2 .Now, since j 2 < j 1 , we have Similarly, since j 3 < j 2 , it follows that Combining the previous two inequalities, and again making use of the fact that j 3 < j 2 , we have This contradicts the fact that e 1 < e 2 and completes the proof of the claim.
When we combine Claim 3.2 and Claim 3.4, we see that the set K is a weakly convex subset of {1, . . ., n} which does not contain four consecutive terms in an arithmetic progression.It follows that Combining this with (17), the proof is complete.

Concluding remarks; sums instead of differences
The problems considered in this paper were partly motivated by a potential application to a problem in discrete geometry concerning the minimum number of angles determined by a set of points in the plane in general position.This problem was considered recently in [2], and similar problems can be traced back to the work of Pach and Sharir [5].We found that progress on this question could be given by a solution to the following problem: given a convex set A ⊂ R estimate the size of the largest matching on A which gives rise to a convex set in the image set f (A, A), where f : R × R → R is a specific bivariate function whose rather complicated formula is omitted here.Theorems 1.2 and 1.3 of this paper solve this problem for this simplified case when f (x, y) = x − y.
With the potential application to the distinct angles problem in mind, an interesting future research direction could be to generalise the problems considered in this paper by considering an arbitrary f : R × R → R in place of the function f (x, y) = x − y.We conclude this paper with some remarks about the most natural case, whereby f (x, y) = x + y.
It is interesting to see that we can quite easily obtain an optimal result, giving a significant quantitative improvement to Theorem 1.2, if we consider sums instead of differences, as follows.
Theorem 4.1.Let n ∈ N and suppose that A ⊂ R is a convex set with cardinality n.Then there exists a matching M ⊂ A × A such that |M| ≥ ⌊ n 2 ⌋ and A + M A is convex.
Then the set A + M A = {a n/2+k + a k : k ∈ {1, . . ., n/2}} is convex.If n is odd then we omit a n and use the same argument as above.
In particular, it follows from Theorem 4.1 that an analogue of the construction in the proof of Theorem 1.3 is not possible if we take sums instead of differences.There are other cases in which problems concerning additive properties of convex sets are sensitive to sums and differences.For instance, a construction in [4] (see also [7]) shows that there exists a convex set A ⊂ R and for any convex A ⊂ R and x ∈ A + A.
We would also be interested to know whether Theorem 1.1 is still valid when A − A is replaced by A + A. We were unable to prove anything non-trivial for this question.

Figure 2 .
Figure 2.This diagram illustrates the intersection pattern of the sets D k .
min for the smallest element of D k and d (k) max for the largest element of D k .Note that

3 Figure 3 .
Figure 3.This diagram illustrates how the gaps between the consecutive D i are significantly larger than the diameters of the individual D i .This is the heuristic reason why Claims 3.1 and 3.2 are valid.

(k 2
) j is not the last element of S. Let x be the element of S preceding d (k 2 ) j and let y be the element of S following d (k 2 ) j d

Figure 4 .
Figure 4.In this picture, we zoom in to take a closer look at the way the elements of D k are distributed (here we consider the set D 1 with n = 5).Crucially, the gaps between consecutive elements of D k shrink rapidly, with the conseutive differences resembling a geometric progression with a small common ratio.This picture can be used for a sketchy justification of Claims 3.3 and 3.4