Multidimensional lower density versions of Pl\"unnecke's inequality

We investigate the lower asymptotic density of sumsets in $\mathbb{N}^2$ by proving certain Pl\"unnecke type inequalities for various notions of lower density in $\mathbb{N}^2$. More specifically, we introduce a notion of lower tableaux density in $\mathbb{N}^2$ which involves averaging over convex tableaux-shaped regions in $\mathbb{N}^2$ which contain the origin. This generalizes the well known Pl\"unnecke type inequality for the lower asymptotic density of sumsets in $\mathbb{N}$. We also provide a conjectural Pl\"unnecke inequality for the more basic notion of lower rectangular asymtpotic density in $\mathbb{N}^2$ and prove certain partial results.

1. Introduction 1.1. Background. Plünnecke's classical work [6] provided influential techniques for studying the cardinality of sumsets and iterated sumsets. We recall that for subsets A, B of an abelian group (G, +) we can define the sumset and, for positive integers k, the k-fold iterated sumset kA = {a 1 + · · · + a k | a 1 , . . . , a k ∈ A}. Theorem 1.1 (Plünnecke [6]). Suppose that A, B are finite subsets of some abelian group (G, +) and k ≥ 1. Note that the usual definition is σ(A) = inf n≥1 |A∩ [1,n]| n , which is necessary for some applications such as Mann's Theorem, but the results of interest to us remain valid when using our definition. Erdös proved that for A, B ⊂ N such that kB = N for some positive integer k [3]. Plünnecke greatly improved this lower bound by proving the following extension of his cardinality estimate (Theorem 1.1) to Schnirelmann density. A good account of this as well as a proof of Theorem 1.1 and related results can be found in Ruzsa's book [7].
Although Schnirelmann density has played an important role in additive number theory (see, for instance, Schnirelmann's proof that the primes are an asymptotic basis [8]), it lacks many asymptotic features such as translation invariance. From a combinatorial perspective, the lower asymptotic density, given by is a more natural notion of the asymptotic size of a set A ⊂ N. It turns out that Theorem 1.2 is also true with d in place of σ. Theorem 1.3 (See [4], [5] and [7]). Suppose that A, B ⊂ N and k ≥ 1. Then This was first obtained by Ruzsa [7] and an alternative proof, which the author has found insightful, was given by Jin in [4] and [5].

Density Plünnecke inequalites in (semi)groups.
It is natural to ask whether density versions of Plünnecke's inequality hold in other countable abelian (semi)groups with certain notions of asymptotic density. Let us briefly mention some recently established results in this direction. We first recall a standard way of extending the notion of density to other groups that involves replacing the sequence of intervals with a sequence of asymptotically invariant finite sets.
Definition 1.4 (Densities along Følner sequences). Let (G, +) be a countable abelian semigroup. A Følner sequence is a sequence F 1 , F 2 , . . . of finite subsets of G that is asymptotically invariant, i.e., for each g ∈ G we have that lim n→∞ |F n ∩ (g + F n )| |F n | = 1.
Moreover, if A ⊂ G then we define the lower asymptotic density along (F n ) as Similairly, we may define the upper asymptotic density along (F n ) as If F is a collection of Følner sequences in G then we can define the lower and upper densities with respect to this collection as Finally, the lower and upper Banach densities in G may be defined, respectively, as d * = d Følner(G) and d * = d Følner (G) where Følner(G) denotes the collection of all Følner sequences in G.
Theorem 1.5 ( [2]; k = 1 case obtained in [1]). Suppose that (G, +) is a countable abelian group and A, B ⊂ G. Then for integers 0 < k < k we have For the semigroup G = N, the k = 1 cases of these inequalities were obtained by Jin in [5]. We remark that the proofs of Theorem 1.5 (for arbitrary countable abelian groups G) use ergodic theory and it is unclear whether such techniques can be applied to densities associated to smaller classes of Følner sequences, such as lower asymptotic density. The purpose of this paper is to extend the Plünnecke type inequality for lower asymptotic density to the semigroup N 2 .
1.3. Følner sequences and lower asymptotic density in N 2 . A natural candidate for lower asymptotic density in N 2 arises from considering the family as the corresponding lower density is a product density in the sense that of intervals which gives rise to the lower density d in N satisfies the desirable property of being closed under pointwise unions 1 . Unfortunately, Rect does not satisfy this property, so it seems natural to consider which is the smallest subset of Følner(G) that contains Rect and is closed under pointwise unions. The corresponding lower density d Tab has the curious property that it is also a product density as above and hence agrees with d Rect on cartesian products, i.e., . In fact, this property is also satisfied by the collection for all L ∈ N, which is itself a family of Følner sequences that we will be interested in.
1.4. Lower density versions of Plünnecke's inequality in N 2 . The main goal of this paper is to address the following question.
Question 1.6. For A, B ⊂ N 2 , with (0, 0) ∈ B, and positive integers k < k, is it true that Our first partial result is an affirmative answer to this question when A ⊂ N 2 is such that the density d Rect (A) exists, by which we mean that In this case, we denote this common value by d Rect (A).
Then for all B ⊂ N 2 with (0, 0) ∈ B and integers 0 < k < k we have that Remark 1.8. Note that it is very easy to affirmatively answer Question 1.6 up to a constant by using the . For instance, one may use this fact to immediately deduce that for A, B ⊂ N 2 and positive integers k, we have that 1 The pointwise union of two Følner sequences (Fn)n and (Gn)n is the Følner sequence (Fn ∪ Gn)n.
The constant becomes much worse than 1 4 if one applies this method to estimate the lower densities of A + k B for k < k large; more precisely, one gets 1.5. Statements of main results and applications. We are able to affirmatively answer Question 1.6 if we replace Rect with the collection Tab introduced above in (1.2). An element of Tab will be refered to as a tableau Følner sequence and we will refer to the corresponding notion of lower density d Tab (as per Definition 1.4) as the lower tableau density.
In fact, our techniques also give the following partial answer to Question 1.6.
Theorem 1.10. Suppose that 0 < k < k are integers and A, B ⊂ N 2 such that (0, 0) ∈ B, then This enables us to affirmatively answer Question 1.
In particular, since d Tab  in the case where A possesses a certain fractal structure and B is a rectangular asymptotic basis of order k > k (that is, d Rect (kB) = 1). We give an example of such a fractal set before giving a broad definition. {x ∈ N|a ≤ x < b} for a < b ∈ N. Choose a sequence 0 < u 1 < u 2 < . . . in N 2 such that u k > 2u k−1 for k > 1 and lim k→∞ u k u k−1 = ∞. Let A ⊂ N be the set given in Figure 1, more precisely One can generalise the example above to construct more general sets that are asymptotically finite unions of translates of a large square.
Definition 1.14 (Fractal set generated by a pattern). As before, we will use the convention be a set such that (0, 0) ∈ P . We call such a P a pattern. Choose a sequence u 1 , u 2 , . . . of positive integers such that u k > (N + 1)u k−1 and lim k→∞ u k and let where u 0 = 0 (in other words A 1 = P 1 ). Finally, we define A(P, N ) = ∞ k=1 A k to be the fractal set generated by the pattern P of degree N . Figure 2. Definition 1.14 for the case N = 2, P = {(0, 0), (0, 2), (2, 2)}. Note that A k is contained in [0, 3u k ) 2 \ [0, 3u k−1 ) 2 ; while P j and A j are contained in the small square [0, 3u k−1 ) 2 , for all j < k.
Note that the set A given in Example 1.13 is A(P, 1) for P = {(0, 0), (1, 1)}. We now give a combinatorial formula for d Tab and d Rect of fractal sets generated by a pattern. where we define a tableau to be a set of the form Thus if (0, 0) ∈ B ⊂ N 2 and k > 0 is an integer then by Theorem 1.10 we have for integers 0 < k < k. Many such special cases of Question 1.6 may be constructed, which are not covered by Proposition 1.7.

1.7.
A remark about higher dimensions. When B is an asymptotic basis of order k (i.e., kB = N 2 ), the natural analogue of Theorem 1.12 (and hence also its consequences) holds in higher dimensions, as the reader may observe in the equation (6.13) found in the proof of Theorem 1.12 in Section 6. However, it is yet unclear whether the higher dimensional analogue of Theorem 1.12 holds in full generality, the main obstacle lies in finding the right higher dimensional extension of Lemma 5.7.
1.8. Organization of the paper. In Section 2 we review some classical Plünnecke inequalities for cardinalities of truncated sumsets (i.e., sets of the form (A + B) \ C), as well as the less well known but crucial δ-heavy Plünnecke inequality. In Section 3 we introduce the main definitions, notations and conventions.
In particular, we introduce the notion of a tableau, the main combinatorial object in this paper, and we establish some basic properties of the lower tableau density d Tab . Sections 4 and 5 include some technical combinatorial lemmata involving tableaux that will be put together in Section 6 to conclude the proof of Theorem 1.12. In Section 7 we state some related open problems; more specifically, some conjectural multidimensional Plünnecke inequalities for various densities. Finally, Appendix A is devoted to proving Proposition 1.15.

Acknowledgement:
The author is grateful for many insightful and encouraging conversations with Alexander Fish.

Plünnecke inequalities for truncated sumsets
One of the key tools used in the proof of our results, as well as Jin's proofs in [4] and [5], is the following Plünnecke inequality for truncated sumsets.
Theorem 2.1 (See [7]). Let A, B, C be finite subsets of an abelian group and define Then D 1/n n is decreasing in n.
We will need (unlike Jin in [4] and [5]) the following δ-heavy version of this inequality. We will include a proof for the sake of completeness as this version, to the best of the author's knowledge, rarely appears in the literature (cf. [1]).
Theorem 2.2. Let A, B, C be finite subsets of an abelian group and let 0 < δ < 1. Then for positive integers k < k, there exists A ⊂ A such that |A | > δ|A| and Proof: Using Theorem 2.1 and the fact that (1−δ) −k/k > 1, take non-empty A ⊂ A of maximal cardinality such that We deduce from this inequality, using the fact that Now by adding the estimates (2.1) and (2.2) we get But since A is non-empty, this contradicts the maximality of A .

Tableaux
We stress that throughout this paper we use (and already have used) the convention tableau is a set of the form where F ⊂ N 2 is a finite set. It will also be convenient to define a tableau region to be a set of the form for some finite F ⊂ N 2 . Thus a tableau is precisely the set of lattice points of some tableau region. {x ∈ R | a ≤ x < b} (very rarely it will mean {x ∈ Z | a ≤ x < b}, in fact only in the definition of fractal sets in the Introduction above and Appendix A).

Important note on notation: If
We note the following simple but useful additive characterization of tableaux.
Lemma 3.1. If T ⊂ N 2 is finite and non-empty, then the following are equivalent: This means that if T is a tableau and B ⊂ N 2 contains (0, 0), then (N 2 \ T ) + B = N 2 \ T . As a consequence, we may apply Theorem 2.2 with C = N 2 \ T to obtain the following crucial proposition.
Proposition 3.2. Let T ⊂ N be a tableau and suppose that A, B ⊂ N 2 with (0, 0) ∈ B. Then for positive integers k < k and 0 < δ < 1, there exists A ⊂ A such that |A | > δ|A| and We now turn to explicating some basic properties of the densities d Tab and d Tab(L) that were introduced above. The following simple lemma will be convenient as it shows that we may, without loss of generality, assume that the side lengths of our rectangles are divisible by a chosen integer.
(ii) The W i,n and H i,n tend to ∞, more precisely remains unchanged. Applying this finitely many times, we can adjust a sequence satisfying (iii) (which exists by a simple diagonalization argument) to one which satisfies the desired properties.
Let us spell out a useful characterization of the lower tableaux density.
. Then for each > 0 and positive integer L, there exists an Moreover, d Tab (A) is the largest choice of α which makes this statement true.

Trimming lemma
In this section we formulate and prove the Trimming Lemma, one of the main combinatorial tricks of this paper. It will be most convenient to state and prove it in a rather abstract setting. If X is a set equipped with a measure µ, then we will use the averaging notation If the measure µ is clear, we simply use the shorthand A(f, X). Suppose that is a function and α > 0 is such that for all non-empty tableaux S ⊂ I. Then there exists   ρ ((0, 0)) = α. Now suppose the theorem holds for all tableaux with cardinality strictly less than |I|. Suppose that (iii) fails for some tableau S I with ρ = ρ (otherwise, we may take ρ = ρ). Let S max I be a maximal tableau contained in I such that To define ρ on S max , we use the induction hypothesis as follows. Let ρ 0 = ρ| Smax . Since |S max | < |I| we apply the induction hypothesis to ρ 0 to obtain a map ρ 0 : (c) For all tableau S S max we have We define ρ (x) = ρ 0 (x) for x ∈ S max . Let us now check that ρ : I → [0, 1] satisfies the desired conclusions.

By (4.1) we have
A(ρ , I \ S max ) = α which together with (b) implies that (ii) holds. Now suppose that S I is a tableau. Then we may decompose I \ S = (S max \ (S ∩ S max )) (I \ (S max ∪ S)).

It is enough to show that
A(ρ , X) ≤ α when X is one of these parts. If then, by (c), we do indeed have and consider the following two cases.
Then we have This means that S max S max ∪ S and thus, by the maximality of S max , we have that This verifies (iii) and thus completes the proof.

Q 2 -tilings and approximating subtableaux regions
We now turn to studying tableau regions obtained by subdividing a tableau region. We will consider subdivisions that are equally spaced, thus it is convenient to define following the notion.
be the smallest set that contains S and is in the σ-algebra generated by the partition C 0 . Then Proof: There are at most 2Q elements of C that intersect both S and F \ S, since these elements form a path consisting of right and down steps. In fact, there are at most 2Q − 1. See Figure 4.
for all m ∈ {1, . . . , }. Then we define the Q 2 -tiling  Note that the elements of C are rectangles with both side lengths integers divisible by Q, and thus contain at least Q 2 elements of N 2 .
Naturally, we may identify C with a tableau T = T (F, Q) ⊂ N 2 by constructing a bijection ψ : C → T as follows: (i) ψ −1 (0, 0) is the unique element of C which contains (0, 0).
(ii) ψ −1 (i + 1, j) is the element to the right of ψ −1 (i, j) and ψ −1 (i, j + 1) is the element just above ψ −1 (i, j). (Note: We refined C 0 to C precisely so that the notion of right is well defined.) A set of the form ψ −1 (T ) (we use the notation X = x∈X x), for some tableau T ⊂ T , will be called a C-measurable subtableau region. In general, a union of elements of C will be called a C-measurable set.
Remark 5.4. All C-measurable subtableau regions contain the element of C that contains (0, 0). This element is precisely Hence, each C-measurable subtableau region is a union of rectangles of width at least 1 Q W 1 and height 1 Q H . Apply Lemma 5.2 to S ∩ U m ⊂ U m to get where S ⊃ S is the smallest subset of F that contains S and may be written as a union of elements of C 0 .
Lemma 5.6 (Trimming a set of points). Fix a positive integer Q, a Q 2 -tableaux region F and A ⊂ F ∩ N 2 .
Let C = C(F, Q). Define Then there exists A ⊂ A such that (a) For C-measurable subtableau regions F F we have Proof: T = T (F, Q) ⊂ N 2 be the corresponding subtableau and let ψ : C → T be the bijection constructed in Definition 5.3. Apply the Trimming Lemma (Lemma 4.1) to the tableaux T , the measure given by µ({t}) = |ψ −1 (t)| for t ∈ T and the map ρ : T → [0, 1] given by Since each element of C is a rectangle with both sidelengths multiples of Q, we may find A ⊂ A such that, We use the notation y(a, b) = b.
denote the bottom-left corners of C . Note that the element of E may be ordered vertically: we say E is t 2 ). We now construct the a j recursively.
Choose a 1 ∈ E 1 ∩ A, where E 1 is the highest element of E. Now suppose we have chosen a 1 , . . . , a j with each a j ∈ A ∩ E j for some E j ∈ E (by minimality of S, A ∩ E is non-empty for all E ∈ E). We have that y r ≤ a j < y r+1 for some r = r j ∈ {1, . . . , |Y| − 1} where Y = Y(F, Q) = {y 1 < y 2 < . . . < y |Y| } is as constructed in Definition 5.3. In fact y r and y r+1 are the y ordinates of the corners of the element of E that contains a j . Now let E j+1 be the highest element of E below the horizontal line y = y r−1 and choose a j+1 ∈ E j+1 , and if such E j+1 does not exist then j = J and we are done with our construction. Now let for 1 < j ≤ J . Note that G j is a tableau region translated by a j . Figure 6. The tile E j south-east of E j is an element of E, but it is not E j+1 since it is in the row of C just below E j . This ensures the desired condition that y(a j+1 ) ≤ y(a j ) − Q.

Now we let
As desired, we have that (cf. Figure 6) It now remains to estimate |S \ G|. To this end, let G denote the smallest set that contains G and may be written as a union of elements of C. We have by Lemma 5.5 that One may argue (see Figure 7 and its caption below) that Combining (5.1) with (5.2) we get that We finish this section with a simple lemma which will allow us to remove a negligible set of integral points which lie too closely to the boundary of a tableaux region.
We now apply the techniques developed in Section 5 to the tableau regions F n . Note that the C(F n , Q)measurable subtableau regions of F n are a union of at most Q 2 L 2 (and thus a bounded function of n) rectangles with sidelengths tending to ∞ as n → ∞ (see Remark 5.4), we thus have that Thus there existsÑ (which depends on Q and the sequence F n , which we have fixed) such that α n > 3 4 α for all n >Ñ .
From here on, we fix n >Ñ and let = (n), W m = W n,m , H m = H n,m for m ∈ {1, . . . , }. Applying Lemma 5.6 to F n we obtain a subset A ⊂ A ∩ F n such that and |A | |F n | ≥ α n .

Now let
be the smallest set that contains A 0 and is the complement (in F n ) of a tableau region contained in F n . We have by Lemma 5.5 that where S n is the smallest C-measurable set that contains S n . Now note that by (6.5). Combining these two estimates gives Also notice that S n is the complement of a C(F n , Q)-measurable tableau, and thus we may apply (6.3) to Now applying the inequalities (6.6), followed by (6.4), followed by (6.10) and then finally (6.9) we obtain which, by letting λ(n, Q) denote the factor before |(A 0 +kB)∩Fn| |Sn| , we rewrite as Using lim inf n→∞ α n ≥ α (justified in (6.2)) it is an easy calculation to show that 2 We now wish to show that Note that this holds trivially in the case that kB = N 2 , as in this case we have that |(A 0 +kB)∩F n | = |S n ∩Z 2 | and thus |(A 0 + kB) ∩ F n | |S n | = 1. (6.13) We now return to the general case. Apply Lemma 5.7 to A 0 and S n to obtain a 1 , a 2 , . . . , a J ∈ A with 3 y(a j ) ≤ y(a j−1 ) − Q such that the set for 1 < j ≤ J. Note that G j is a tableau region translated by a j (cf. Figure 6 in the proof of Lemma 5.7).
Claim: Each G j is a union of at most L rectangles with bottom corner a j , each with sidelengths at least Q.
Proof of Claim: Writing a j = (x, y), we can write (see Figure 6) where the union is over m ∈ {1, . . . , } such that x < W m and y < H m (for j = 1, we omit the y(a j−1 )). For such m, we have that W m − x ≥ Q and H m − y ≥ Q since a j avoids BadRow m and BadCol m , respectively.
We also have y(a j−1 ) − y ≥ Q by construction of the a j . This completes the proof of the Claim.
This claim implies that is a map (it depends on kB, but not n) such that 4 lim Q→∞ δ(Q) = 0. We thus have that Now combining (6.14) with (6.8) gives and so we have that So in summary, we have (see (6.11)) that Finally, this completes the proof as from which we deduce Theorem 1.12 by letting Q → ∞ and using (6.12).

Further questions
We now list some related open problems. We start by recalling the main motivating question of this paper.
Question 7.1. For A, B ⊂ N 2 and positive integers k < k, is it true that We note that the M = 0 case is precisely Plünnecke's classical inequality for Schnirelmann density in N (Theorem 1.2). Even the following special case is not clear.
What is the answer to Question 7.2 in the case M = 1?
Appendix A. Densities of fractal sets We will now prove our formulae for d Tab (A) and d Rect (A) (Proposition 1.15) for fractal sets A. We will use the notation [a, b) = {x ∈ Z|a ≤ x < b}. So let us fix in this section A = A(P, N ) a fractal set, where N ∈ N and P ⊂ {0, 1, . . . , N } 2 is a pattern. We also fix the data A k ⊂ P k and u k ∈ N given in Definition 1.14.
Define the core of F n to be the set which will imply the formula for rectangular density. The key ingredient is the following Perturbation Lemma. Proof: Since F \ [0, (N + 1)u k ) 2 ⊂ A, it suffices to consider the case where W, H ≤ (N + 1)u k . Write H = qu k + r where r ∈ {0, 1, . . . u k − 1} and q ∈ {1, . . . , (N + 1)}. If q = N + 1 then we must have r = 0, which means that we can setH := H = (N + 1)u k and we are done with the proof, so let us now assume all have the same projection onto the x-axis (see Figure 9). We denote this constant by C. It will be convenient to use the notation where X, Y are finite sets. Hence if we define, for t = 0, 1, . . . u k , the expression then we can rewrite it as a convex combination Now notice that g(t) is a monotonic function of λ, and thus a monotonic function of t. Hence the minimum of g(t) occurs at either t = 0 or t = u k . If it occurs at t = 0, we setH = qu k while if it occurs at t = u k , we setH = (q + 1)u k . We thus get that Remark A.2. In Lemma A.1 one may drop the hypothesis that u k+1 ≥ H if one replaces A with P k . The reason is that this hypothesis was only used to show that the cross section density C was constant with respect to t, which is not an issue if one replaces A with P k . This variation of the lemma will also be useful.
Now if we are in the degenerate case where H n ≤ u kn+1 then in fact F n \ F o n = ∅ and so we are done. If however H n > u kn+1 then |Fn\F o n | |Fn| ≥ 1 − n , for some n → 0, which means that we may replace F n \ F o n with F n in (A.3) to get a logically equivalent statement. So we apply Lemma A.1 (with k being the largest integer such that H n ≥ u k and F = F n ) and argue as before.
Remark A.4. We did not really need to consider the definition of a core, but this notion will be useful in establishing the formula for Tableaux density in the next subsection.
A.2. Tableaux density. We will now prove the formula for d Tab (A) for the fractal set A = A(N, P ). In this subsection we fix an integer L ≥ 1 and a Følner sequence We wish to show that where H n,j is decreasing and W n,j is increasing in j, for each fixed n.
Exactly as in Section A.1, we define the core of F n to be the set where k n is the largest positive integer such that [0, u kn ) 2 ⊂ F n . As before, we start with a perturbation lemma. We note again that (since u k u k−1 → ∞) However ifH 1 < H 2 then we see, by the same convexity argument as in Lemma A.1, that and so we are done (in this case) by applying the induction hypothesis to F . We constructH 2 , . . .H L by continuing in this way (for example, to getH 2 one applies the same technique to [W 1 , W 2 ) × [0, H 2 )).
The formula for the tableaux density of A now follows by applying Lemma A.5 to the core of F n (with k = k n ) and arguing as we did in the case of rectangular density.

Appendix B. Density of cartesian products
We now prove the property and so we focus on proving the reverse inequality. We do this by induction on L. The L = 1 case is clear, so let us suppose that L > 0 and that the property holds for Tab(L − 1). Let A, B ⊂ N and let be a sequence (where W n,1 < · · · < W n,L and H n,1 > · · · > H n,L are positive integers with W n,i , H n,i → ∞ as n → ∞ for each i) such that  which completes the induction step in this case.