A strengthening of Freiman's 3k−4$3k-4$ theorem

In its usual form, Freiman's 3k−4$3k-4$ theorem states that if A$A$ and B$B$ are subsets of Z${\mathbb {Z}}$ of size k$k$ with small sumset (of size close to 2k$2k$ ), then they are very close to arithmetic progressions. Our aim in this paper is to strengthen this by allowing only a bounded number of possible summands from one of the sets. We show that if A$A$ and B$B$ are subsets of Z${\mathbb {Z}}$ of size k$k$ such that for any four‐element subset X$X$ of B$B$ the sumset A+X$A+X$ has size not much more than 2k$2k$ , then already this implies that A$A$ and B$B$ are very close to arithmetic progressions.


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BOLLOBÁS, LEADER and TIBA Let  and  be finite subsets of ℤ with || = || =  such that | + | = 2 − 1 + , where  ⩽  − 3. Then there are arithmetic progressions  and  of the same common difference, containing  and , respectively, such that each has size  + .
A result of this kind is often called an 'inverse' theorem, as it describes what happens when an inequality is close to being tight.(These are also often known as 'stability' results.) What about a 'bounded sumset' version of Freiman's 3 − 4 theorem?Our aim in this paper is to show that indeed we can weaken the condition that  +  is small to just a condition that all sumsets of  with a bounded number of terms of  are small.This is really rather surprising.One could view this as a kind of inverse result to Theorem 1.
The key point in this result is that the sizes of the progressions  and  are 'about' (1 + ): the precise form of the error term in this, as quadratic in , is not so important.Note that the dependence of the error terms (the sizes of |Δ| and |Δ|), as linear functions of , is best possible.This may be seen by taking  to be an interval of length  and  to consist of an interval of length (1 − 2) together with, on either side of it, random half-sized subsets of the adjacent intervals of length 2.
We remark that the condition that  has small symmetric difference with  cannot be strengthened to insist that  lies inside .This is because one may always have a few 'rogue' points in , far from the rest of .The same would not apply to , because adding a few faraway points to  is a great help in selecting the four points.
It would be extremely interesting to decide whether this result remains valid if we consider three elements instead of four.To elaborate on this, the way that Theorem 1 is proved is by taking the three points of  as follows: we first take the smallest and largest points of  (the reader will see that we have to take those points, as may be seen from the case when the two sets are intervals), and then choosing the third point at random from the remaining points of .It is very surprising that these two different ingredients mesh together so well, to give exactly the lower bound required.However, and this is the key point, this approach cannot be used to prove a three-element version of Theorem 2. Indeed, this may be seen by taking  = [0, ] and  =  ∪ ( + ) where  is a random set of density 1∕2 in an interval of length .Perhaps an easier way to see this is by taking  =  ⋅ [0, ] and  = [0, ].Then for any  ∈ [1,  − 1], we have | + {0, , }| = 2|| + 1.Therefore, on the one hand, all these sumsets are quite small.On the other hand, it is clear that we cannot find arithmetic progressions  and  with the same difference such that |Δ| and |Δ| are small.
We mention that the  0 produced in our proof is fairly small.It would be interesting to know how large  0 may be taken; in particular, can we take  0 close to 1?
The plan of the paper is as follows.In Section 2, we mention some background results from [1] that we will need.We do give a brief summary of the proof of Theorem 1, to help make the paper self-contained and also because the methods there form a good 'toy case' of the arguments in our main result.Indeed, those background results from [1] that we do not prove here may be proved by methods very similar to those in the proof of Theorem 1. Then in Section 3, we present our main result.
Our notation is standard.To make our paper more readable, we often omit integer-part signs when these do not affect the argument.
Sometimes we write ' mod ' as shorthand for the infinite arithmetic progression { ∈ ℤ ∶  ≡  mod }, and refer to it as a fibre mod .When  is a subset of ℤ we often write   for the intersection of this fibre with  -when the value of  is clear.(We sometimes write   as    when we want to stress the value of .)Thus,   =  ∩  −1 (), where  =   denotes the natural projection from ℤ to ℤ  .We also write S for   ().
When we write a probability or an expectation over a finite set, we always assume that the elements of the set are being sampled uniformly.Thus, for example, for a finite set  ⊂ ℤ we denote the expectation and probability when we sample uniformly over all  ∈  by, respectively,  ∈ and ℙ ∈ .
For more general background on sumsets, see the survey of Breuillard, Green and Tao [2] or the books of Nathanson [11] or Tao and Vu [13].For extensions of Freiman's 3 − 4 theorem to other settings, such as ℤ  in particular, see Grynkiewicz [8].
To end the introduction, we give a brief overview of the proof of Theorem 2. With  having first element 0 and last element , we start by showing that (unless we are done)  must have a large projection onto ℤ  .This is accomplished by a careful analysis of what happens when we take our four points of  to be 0,  and two random points: it is the interaction of the two random points that is critical here.From this we find that  is contained in a quite short arithmetic progression, and in turn this will imply that  is also relatively close to an arithmetic progression.However, these progressions are not small enough to give Theorem 2, and so we need an argument that 'boosts' this.This is a very delicate analysis of, again, how the two random translates interact with each other and with the two fixed translates.In fact, it is rather surprising that this boosting argument actually works: in length it is most of the proof.(It is Theorem 9.)

PREREQUISITES
To give a feel of how the arguments proceed, we start by including the short proof of Theorem 1 presented in [1].
As explained above, the main idea is to fix two elements as the first and last elements of  and then to select the remaining element at random.
If we choose an element  ∈  ⧵ {}, uniformly at random, then To see this, note that   ∶  ⧵ {} →  is a bijection.Also, for  ∈  we have Therefore, ) .
It is easy to check that if || ⩾ ||, then Combining the last inequality with (1) and ( 2), gives the following stronger version of Theorem 1.Let  and  be finite non-empty subsets of integers with || ⩾ || and  having smallest element 0 and greatest element .Then we have We now pass to some related results from [1].These are proved along broadly similar lines to the above results.The first one is about the case of equality.Theorem 3. Let  and  be finite non-empty subsets of ℤ with || = ||, min() = 0 and max() = .Suppose that when we choose an element  of  ⧵ {0, } uniformly at random we have Then  and  are arithmetic progressions with the same common difference.
In particular, suppose that when we choose any three elements  1 ,  2 ,  3 of  we have Then  and  are arithmetic progressions with the same common difference.
Then we need another version of Theorem 1.
We note also a simple variant of the ideas above.
The next result follows easily from Lemma 5.
Kneser's inequality [9] states the following.A simple corollary of Kneser's inequality is the following.

Corollary 9.
If  is a subset of an abelian group , such that | + | < 1.5||, then  +  is a coset.Moreover, if 0 ∈ , then  +  is a subgroup containing .

THE MAIN RESULT
In this section, we prove Theorem 2, our stability version of Theorem 1.As we mentioned earlier, the dependence of the error terms on , being linear, is of best possible order.It will turn out that a key ingredient is actually the following weaker version of Theorem 2.
We will also need the following technical result.
Assume that min() = 0 and max() = .In particular, we have Take any subset  1 ⊂  with  1 = | 1 | such that min( 1 ) = 0 and max( 1 ) = .(We used here that  ⩾  1 ⩾ 2.) On the one hand, by construction we have On the other hand, by (7) we have The last two relations imply Lemma 6 and ( 9) tell us that As  ⩽ 0.5 by hypothesis, the last two inequalities imply By (3), we have The last two inequalities imply As  ⩽ 3∕16 by hypothesis, by ( 7) and ( 12) we have For the rest of the proof, we focus on the sets  1 and  1 and thus we may assume without loss of generality that  = 1.
The construction of  ′ .As we supposed without loss of generality  = 1, by translating  1 , we may assume that the partition  1 =   ⊔   ⊔   given by for some  ⩾ 0 and  ∈ {0, 1}.
Combining the last two inequalities, we get We now break the argument into two cases.Case 1: In this case, by (20) we get Let  =  ∩ [−, 0).We may assume without loss of generality that Indeed, otherwise replace the sets  and  with the sets  ′ = − − 1 −  and  ′ =  − .As showing the result for  and  is equivalent to showing the result for  ′ and  ′ , this can be done without loss of generality.Note that sets  ′ and  ′ also satisfy {0, } ⊂ , ) .
Here the last inequality comes from (18).This concludes the proof of Claim A. □ We now turn to the second case.Case 2: We may assume without loss of generality Indeed, otherwise replace the sets  and  with the sets  ′ = − − 1 −  and  ′ =  − .As showing the result for  and  is equivalent to showing the result for  ′ and  ′ , this can be done without loss of generality.Note that sets  ′ and  ′ also satisfy {0, } ⊂  and . By ( 17) and ( 27), we get Claim B.
This finishes the proof of Theorem 11. □ We now prove Theorem 10.
Proof.The proof of this lemma is based on two claims.Suppose for a contradiction that Claim A. There exists  1 ∈  such that Proof.By Theorem 7, we have By hypothesis, we have Define the random variable  ∶  ⧵ {} → ℝ by By the two lines above, we get that () ⩾ 0 ∀ ∈  ⧵ {} and  ∈⧵{} () ⩽ .
This establishes Claim A. □ Claim B.
Proof.From the first part of Claim A, we get From the second part of Claim A and Lemma 5, we get where the last inequality uses  1∕2  ⩾ 4. Combining the inequalities above, we obtain □ Our final lemma is as follows.