This blog entry is the third in a series which seeks to maximize such that there exist explicit matrices of arbitrarily large aspect ratio which are RIP for . Continuing our study of the breakthrough paper by Bourgain et al., this entry will identify what the set needs to satisfy, how to construct it, and we will also use this set to complete the proof of the main result. In particular, we will prove a general version of the result which allows for optimizing constants, and then we will perform an optimization routine that improves by a few orders of magnitude.

**— Everything you need to know about —**

The main result will use an even number as a free parameter, which in turn determines another parameter . For a given choice of , the proof of the main result requires a subset such that (i)

(i.e., is sufficiently small), and (ii) for any , the only way for to satisfy

is by taking and to be permutations of each other. Here, division (and addition) is taken in the field .

Unfortunately, these requirements on lead to very little intuition compared to our current understanding of . From my perspective, this choice for is merely sufficient to produce the remaining flat-RIP-type cancellations we seek, so there may be room for improvement here. Regardless, we will continue by considering how Bourgain et al. constructs . The following lemma describes their construction and makes a slight improvement to :

**Lemma.** Pick and take and . Then

satisfies (i) and (ii) above if we take

and is a sufficiently large prime.

* Proof:* For (i), we have . For (ii), we claim it suffices to show that for any [note: the paper contains a typo here, namely ], any distinct , and any nonzero integers such that , then

is nonzero (in ). To see this, define and . Then

.

As such, if (ii) fails to hold, then subtracting the right-hand side of (2.4) from the left-hand side produces an example of which is zero, violating our statement. Thus, our statement implies (ii) by the contrapositive.

We now seek to prove our statement. To this end, define and . Note that (1) being nonzero in is equivalent to having not divide as an integer. To prove this, we will show that

(a) does not divide ,

(b) is nonzero, and

(c) .

Indeed, (b) and (c) together imply that does not divide , which combined with (a) implies that does not divide .

For (a), we have since and the ‘s are distinct by assumption, and since and each has size at most , we also have . To complete the proof of (a), it then suffices to have

(a’) ,

which we will verify later.

For (b), we will prove that is nonzero. We first write

.

For each term in the above sum, note that both fractions are integers, and for every , is a factor of . As such, the entire sum is congruent to the first term modulo :

.

Each factor of the form can be further simplified:

,

and so

,

where

.

To prove that is nonzero, it suffices to show that does not divide . To this end, we first note that is nonzero since and the ‘s are distinct by assumption. Next, since and and each is at most , we have , and so it suffices to have

(b’)

since . We will verify this later.

For (c), we have

.

Considering our assumption on the ‘s we then have , and so it suffices to have

(c’) .

Overall, we have shown it suffices to have

(a’) ,

(b’) , and

(c’) .

To satisfy (b’) for sufficiently large , we take . Then , and so for sufficiently large . As such, for (c’), it suffices to take , which also satisfies (a’).

**— The main result —**

Before things get messy, let’s discuss the structure of the proof of the main result (i.e., Lemma 2 in the paper). The goal is to prove flat-RIP-type cancellations, namely that

for subsets . (Actually, we get to assume that these subsets and the ‘s satisfy certain size constraints since we have an extra in the power of ; having this will imply the general case without the .) Thanks to the construction of , Lemma 9 gives sufficient cancellation when each is sufficiently small. In the remaining case, Bourgain et al. prove sufficient cancellation by invoking Lemma 10, which concerns the following quantity:

.

Specifically, Lemma 10 gives that is small provided has sufficient additive structure. In the proof of Lemma 2, they take a maximal subset such that is small, and then they use Lemma 10 show that necessarily has little additive structure. By Lemma 9, this in turn forces to be small, and so (and furthermore ) are also small due to a triangle inequality.

The remainder of this section mimics the proofs of Lemmas 2 and 10, but instead of arbitrarily picking certain constants, we leave them as variables which must satisfy certain constraints. We then solve the corresponding constrained optimization in the next section. What follows is a generalized version of the statement of Lemma 10, which we then use to prove a generalized version of Lemma 2. Later, we will prove a generalized version of Lemma 10 (which says that certain constraints imply the following statement, and thus Lemma 2):

**Definition.** Let denote the following statement about the construction of and : For every , there exists such that for every prime the following holds:

Take such that

and for which there exist powers of two such that

and

for and for every . Then for every such that

and

we have

with for every .

**Lemma 2′.** Pick and an even integer . According to the construction of , define . According to the construction of , take , and let denote the solution to

.

Pick such that L10 holds with

Then there exists such that the following holds for every prime :

Take for which there exist powers of two such that

and

for and for every . Then .

* Proof: *First note that implies that

by (2.9a), and by (2.9b), we also have

.

As such, the triangle inequality gives that

,

where the last step uses the fact that

Thus, we can assume , and so (2.3) gives

Applying (2.9a) and (6.1) then gives

,

where the last step uses the fact that

Note that we can redo all of the preceding analysis by interchanging indices 1 and 2. As such, we also have . (This will enable us to use Corollary 4.) At this point, we bound

using Lemma 9:

.

Next, since holds and , Corollary 4 with gives

.

At this point, the triangle inequality gives

which can be further bounded using (2.9a) and (2.9b):

Thus, if , then

,

where the last step uses the fact that

As such, we may assume that either or is . Without loss of generality, we assume

(This will enable us to use L10.) At this point, take to be a maximal subset satisfying (6.5) for , and denote . Then the triangle inequality gives

,

and then Lemma 9 gives

.

This can be bounded further by applying , (2.9a), (2.9b) and (2.3):

At this point, we claim that . To see this, suppose otherwise. Then , implying

and by (2.9a), we also have

.

Thus, Corollary 1 with produces a subset such that

where the second and third inequalities follow from () and (6.1), respectively, and

,

where the last step follows from the fact that

As such, satisfies (6.3) and (6.4), implying that satisfies (6.5) by L10. By the triangle inequality, also satisfies (6.5), contradicting ‘s maximality. We conclude that , and continuing () gives

.

Now we apply (6.5) to and combine with this to get

.

Applying by (2.9a) then gives

.

Now we apply the triangle inequality to get

and applying (2.9b) and (2.3) then gives

,

where the last steps use the facts that

and

This completes the proof.

Now we wish to determine when L10 is true:

**Lemma 10′.** L10 is true with , and provided

and

* Proof:* [The proof is identical to the original until immediately after equation (6.14).] At this point, we have

.

Thus,

,

where the last step uses (6.2). Next, (6.3) and (6.4) together give

and

.

Thus,

,

where the last step follows from the fact that

So, by (6.12) and (6.14) [following the original proof], we have

,

and subsequent application of (6.9), (6.10), and (6.11) gives

By Lemma 4 (which states that ), condition (6.4) implies that

.

We now use this with (2.9b), (6.2) and (6.4) to bound ():

Next, the left-hand inequality of (6.3) gives that , leading to the following bound:

.

Overall, we have

,

since and

Thus, (6.8) gives

.

Finally, taking square roots produces the result.

**— Maximizing —**

In this last section, we optimize the constants from the previous section to make as large as possible. If we ignore the ‘s, then the inequalities — can be simplified quite a bit. In fact, the existence of and satisfying these is equivalent to having

Note that determines and , and so for each , the above system determines a polytope over which we seek to maximize the (linear) objective

.

I performed this optimization for various values of , and for , I got . [I wanted to upload the Mathematica notebook file I used to simplify the system of inequalities and optimize, but it doesn’t seem to be an allowed file type for WordPress…] Dividing by 2 then gives a new world record:

.

While this optimization did make a substantial improvement (this is over 300 times larger than the previous record, and over 8,000 times larger than the original record of Bourgain et al.), the constant is still tiny! For this particular construction of and , I think the remaining bottlenecks lie at the very foundations of additive combinatorics. For example, if , then taking leads to being on the order of .