[Added in post: One might consult this post for a simpler/more general (and slightly more correct) version of the results presented here.]
In our last post regarding Chang’s Lemma, let us visit a version due to Thomas Bloom. We will offer a new proof using entropy maximization. In particular, we will again use only boundedness of the Fourier characters.
There are two new (and elementary) techniques here: (1) using a trade-off between entropy maximization and accuracy and (2) truncating the Taylor expansion of .
We use the notation from our previous post: for some prime and is the uniform measure on . For and , we define . We also use to denote the set of all densities with respect to .
Theorem 1 (Bloom) There is a constant such that for every and every density , there is a subset such that and is contained in some subspace of dimension at most
Note that we only bound the dimension of a subset of the large spectrum, but the bound on the dimension improves by a factor of . Bloom uses this as the key step in his proof of what (at the time of writing) constitutes the best asymptotic bounds in Roth’s theorem on three-term arithmetic progressions:
Theorem 2 If a subset contains no non-trivial three-term arithmetic progressions, then
This represents a modest improvement over the breakthrough of Sanders achieving , but the proof is somewhat different.
1.1. A stronger version
In fact, we will prove a stronger theorem.
Theorem 3 For every and every density , there is a random subset such that almost surely
and for every , it holds that
This clearly yields Theorem 1 by averaging.
1.2. The same polytope
To prove Theorem 3, we use the same polytope we saw before. Recall the class of test functionals
We defined by
Let us consider a slightly different convex optimization:
Here, is a constant that we will set soon. On the other hand, is now intended as an additional variable over which to optimize. We allow the optimization to trade off the entropy term and the accuracy . The constant represents how much we value one vs. the other.
Notice that, since , this convex program satisfies Slater’s condition (there is a feasible point in the relative interior), meaning that strong duality holds (see Section 5.2.3).
1.3. The optimal solution
As in our first post on this topic, we can set the gradient of the Lagrangian equal to zero to obtain the form of the optimal solution: For some dual variables ,
Furthermore, corresponding to our new variable , there is a new constraint on the dual variables:
Observe now that if we put then we can bound (the error in the optimal solution): Since is a feasible solution with , we have
which implies that since .
To summarize: By setting appropriately, we obtain of the form (2) and such that
Note that one can arrive at the same conclusion using the algorithm from our previous post: The version unconcerned with sparsity finds a feasible point after time . Setting yields the same result without using duality.
1.4. A Taylor expansion
Let us slightly rewrite by multiplying the numerator and denominator by . This yields:
The point of this transformation is that now the exponent is a sum of positive terms (using ), and furthermore by (3), the exponent is always bounded by
Let us now Taylor expand . Applying this to the numerator, we arrive at an expression
where , , and each is a density. Here, ranges over all finite sequences of elements from and
where we use to denote the length of the sequence .
1.5. The random subset
We now define a random function by taking with probability .
Consider some . Since , we know that . Thus
But we also have . This implies that .
Equivalently, for any , it holds that . We would be done with the proof of Theorem 3 if we also knew that were supported on functions for which because . This is not necessarily true, but we can simply truncate the Taylor expansion to ensure it.
Let denote the Taylor expansion of to degree . Since the exponent in is always bounded by (recall (4)), we have
By standard estimates, we can choose to make the latter quantity at most .
Since , a union bound combined with our previous argument immediately implies that for , we have
This completes the proof of Theorem 3.
1.7. Prologue: A structure theorem
Generalizing the preceding argument a bit, one can prove the following.
Let be a finite abelian group and use to denote the dual group. Let denote the uniform measure on . For every , let denote the corresponding character. Let us define a degree- Reisz product to be a function of the form
for some and and .
Theorem 4 For every , the following holds. For every with , there exists a with such that and is a convex combination of degree- Reisz products where
1.8. A prologue’s prologue
To indicate the lack of algebraic structure required for the preceding statement, we can set things up in somewhat greater generality.
For simplicity, let be a finite set equipped with a probability measure . Recall that is the Hilbert space of real-valued functions on equipped with the inner product . Let be a set of functionals with the property that for .
Define a degree- -Riesz product as a function of the form
for some functions . Define also the (semi-) norm .
Theorem 5 For every , the following holds. For every with , there exists a with such that and is a convex combination of degree- -Riesz products where