[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.
1.6. Truncation
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