The square root cancellation heuristic, briefly mentioned in the preceding set of notes, predicts that if a collection of complex numbers have phases that are sufficiently “independent” of each other, then
similarly, if are a collection of functions in a Lebesgue space
that oscillate “independently” of each other, then we expect
We have already seen one instance in which this heuristic can be made precise, namely when the phases of are randomised by a random sign, so that Khintchine’s inequality (Lemma 4 from Notes 1) can be applied. There are other contexts in which a square function estimate
or a reverse square function estimate
(or both) are known or conjectured to hold. For instance, the useful Littlewood-Paley inequality implies (among other things) that for any , we have the reverse square function estimate
whenever the Fourier transforms of the
are supported on disjoint annuli
, and we also have the matching square function estimate
if there is some separation between the annuli (for instance if the are
-separated). We recall the proofs of these facts below the fold. In the
case, we of course have Pythagoras’ theorem, which tells us that if the
are all orthogonal elements of
, then
In particular, this identity holds if the have disjoint Fourier supports in the sense that their Fourier transforms
are supported on disjoint sets. For
, the technique of bi-orthogonality can also give square function and reverse square function estimates in some cases, as we shall also see below the fold.
In recent years, it has begun to be realised that in the regime , a variant of reverse square function estimates such as (1) is also useful, namely decoupling estimates such as
(actually in practice we often permit small losses such as on the right-hand side). An estimate such as (2) is weaker than (1) when
(or equal when
), as can be seen by starting with the triangle inequality
and taking the square root of both side to conclude that
However, the flip side of this weakness is that (2) can be easier to prove. One key reason for this is the ability to iterate decoupling estimates such as (2), in a way that does not seem to be possible with reverse square function estimates such as (1). For instance, suppose that one has a decoupling inequality such as (2), and furthermore each can be split further into components
for which one has the decoupling inequalities
Then by inserting these bounds back into (2) we see that we have the combined decoupling inequality
This iterative feature of decoupling inequalities means that such inequalities work well with the method of induction on scales, that we introduced in the previous set of notes.
In fact, decoupling estimates share many features in common with restriction theorems; in addition to induction on scales, there are several other techniques that first emerged in the restriction theory literature, such as wave packet decompositions, rescaling, and bilinear or multilinear reductions, that turned out to also be well suited to proving decoupling estimates. As with restriction, the curvature or transversality of the different Fourier supports of the will be crucial in obtaining non-trivial estimates.
Strikingly, in many important model cases, the optimal decoupling inequalities (except possibly for epsilon losses in the exponents) are now known. These estimates have in turn had a number of important applications, such as establishing certain discrete analogues of the restriction conjecture, or the first proof of the main conjecture for Vinogradov mean value theorems in analytic number theory.
These notes only serve as a brief introduction to decoupling. A systematic exploration of this topic can be found in this recent text of Demeter.
— 1. Square function and reverse square function estimates —
We begin with a form of the Littlewood-Paley inequalities. Given a region , we say that a tempered distribution
on
has Fourier support in
if its distributional Fourier transform
is supported in (the closure of)
.
Theorem 1 (Littlewood-Paley inequalities) Let
, let
be distinct integers, let
, and for each
let
be a function with Fourier support in the annulus
.
- (i) (Reverse square function inequality) One has
- (ii) (Square function inequality) If the
are
-separated (thus
for any
) then
Proof: We begin with (ii). We use a randomisation argument. Let be a bump function supported on the annulus
that equals one on
, and for each
let
be the Fourier multiplier defined by
at least for functions in the Schwartz class. Clearly the operator
is given by convolution with a (
-dependent) Schwartz function, so this multiplier is bounded on every
space. Writing
, we see from the separation property of the
that we have the reproducing formula
Now let be random signs drawn uniformly and independently at random, thus
The operator is a Fourier multiplier with symbol
. This symbol obeys the hypotheses of the Hörmander-Miklin multiplier theorem, uniformly in the choice of signs; since we are in the non-endpoint case
, we thus have
uniformly in the . Taking
power means of this estimate using Khintchine’s inequality (Lemma 4 from Notes 1), we obtain (ii) as desired.
Now we turn to (i). By treating the even and
odd cases separately and using the triangle inequality, we may assume without loss of generality that the
all have the same parity, so in particular are
-separated. (Why are we permitted to use this reduction for part (i) but not for part (ii)?) Now we use the projections
from before in a slightly different way, noting that
for any , and hence
Applying the Hörmander-Mikhlin theorem as before, we conclude that
and on taking power means and using Khintchine’s inequality as before we conclude (i).
Exercise 2 (Smooth Littlewood-Paley estimate) Let
and
, and let
be a bump function supported on
that equals
on
. For any integer
, let
denote the Fourier multiplier, defined on Schwartz functions
by
and extended to
functions by continuity. Show that for any
, one has
(in particular, the left-hand side is finite).
We remark that when , the condition that the
be
-separated can be removed from Theorem 1(ii), by using the Marcinkiewicz multiplier theorem in place of the Hörmander-Mikhlin multiplier theorem. But, perhaps surprisingly, the condition cannot be removed in higher dimensions, as a consequence of Fefferman’s surprising result on the unboundedness of the disk multiplier.
Exercise 3 (Unboundedness of the disc multiplier) Let
denote either the disk
or the annulus
. Let
denote the Fourier multiplier defined on Schwartz functions
by
- (i) Show that for any collection
of half-planes in
, and any functions
, that
(Hint: first rescale the set
by a large scaling factor
, apply the Marcinkiewicz-Zygmund theorem (Exercise 7 from Notes 1), exploit the symmetries of the Fourier transform, then take a limit as
.)
- (ii) Let
be a collection of
rectangles for some
, and for each
, let
be a rectangle formed from
by translating by a distance
in the direction of the long axis of
. Use (i) to show that
(Hint: a direct application of (i) will give just one side of this estimate, but then one can use symmetry to obtain the other side.)
- (iii) In Fefferman’s paper, modifying a classic construction of a Besicovitch set, it was shown that for any
, there exists a collection of
rectangles
for some
with
such that all the rectangles
are disjoint, but such that
has measure
. Assuming this fact, conclude that the multiplier estimate (4) fails unless
.
- (iv) Show that Theorem 1(ii) fails when
and the requirement that the
be
-separated is removed.
Exercise 4 Let
be a bump function supported on
that equals one on
. For each integer
, let
be the Fourier multiplier defined for
by
and also define
- (i) For any
, establish the square function estimate
for
. (Hint: interpolate between the
cases, and for the latter use Plancherel’s theorem for Fourier series.)
- (ii) For any
, establish the square function estimate
for
. (Hint: from the boundedness of the Hilbert transform,
is bounded in
. Combine this with the Marcinkiewicz-Zygmund theorem (Exercise 7 from Notes 1), then use the symmetries of the Fourier transform, part (i), and the identity
.)
- (iii) For any
, establish the reverse square function estimate
for
. (Hint: use duality as in the solution to Exercise 2 in this set of notes, or Exercise 11 in Notes 1, and part (ii).)
- (iv) Show that the estimate (ii) fails for
, and similarly the estimate (iii) fails for
.
Remark 5 The inequalities in Exercise 4 have been generalised by replacing the partition
with an arbitrary partition of the real line into intervals; see this paper of Rubio de Francia.
If are functions with disjoint Fourier supports, then as mentioned in the introduction, we have from Pythagoras’ theorem that
We have the following variants of this claim:
Lemma 6 (
and
reverse square function estimates) Let
have Fourier transforms supported on the sets
respectively.
- (i) (Almost orthogonality) If the sets
have overlap at most
(i.e., every
lies in at most
of the
) for some
, then
- (ii) (Almost bi-orthogonality) If the sets
with
have overlap at most
for some
, then
Proof: For (i), observe from Plancherel’s theorem that
By hypothesis, for each frequency at most
of the
are non-zero, thus by Cauchy-Schwarz we have the pointwise estimate
and hence by Fubini’s theorem
The claim then follows by a further application of Plancherel’s theorem and Fubini’s theorem. For (ii), we observe that
and
Since has Fourier support in
, the claim (ii) now follows from (i).
Remark 7 By using
in place of
, one can also establish a variant of Lemma 6(ii) in which the sum set
is replaced by the difference set
. It is also clear how to extend the lemma to other even exponent Lebesgue spaces such as
; see for instance this recent paper of Gressman, Guo, Pierce, Roos, and Yung. However, we will not use these variants here.
We can use this lemma to establish the following reverse square function estimate for the circle:
Exercise 8 (Square function estimate for circle and parabola) Let
, let
be a
-separated subset of the unit circle
, and for each
, let
have Fourier support in the rectangle
- (i) Use Lemma 6(ii) to establish the reverse square function estimate
- (ii) If the elements of
are
-separated for a sufficiently large absolute constant
, establish the matching square function estimate
- (iii) Obtain analogous claims to (i), (ii) in which
for some
-separated subset
of
, where
is the graphing function
, and to each
one uses the parallelogram
in place of
.
- (iv) (Optional, as it was added after this exercise was first assigned as homework) Show that in (i) one cannot replace the
norms on both sides by
for any given
. (Hint: use a Knapp type example for each
and ensure that there is enough constructive interference in
near the origin.) On the other hand, using Exercise 4 show that the
norm in (ii) can be replaced by an
norm for any
.
For a more sophisticated estimate along these lines, using sectors of the plane rather than rectangles near the unit circle, see this paper of Cordóba. An analogous reverse square function estimate is also conjectured in higher dimensions (with replaced by the endpoint restriction exponent
), but this remains open, and in fact is at least as hard as the restriction and Kakeya conjectures; see this paper of Carbery.
— 2. Decoupling estimates —
We now turn to decoupling estimates. We begin with a general definition.
Definition 9 (Decoupling constant) Let
be a finite collection of non-empty open subsets of
for some
(we permit repetitions, so
may be a multi-set rather than a set), and let
. We define the decoupling constant
to be the smallest constant for which one has the inequality
We have the trivial upper and lower bounds
with the lower bound arising from restricting to the case when all but one of the vanish, and the upper bound following from the triangle inequality and Cauchy-Schwarz. In the literature, decoupling inequalities are also considered with the
summation of the
norms replaced by other summations (for instance, the original decoupling inequality of Wolff used
norms) but we will focus only on
decoupling estimates in this post. In the literature it is common to restrict attention to the case when the sets
are disjoint, but for minor technical reasons we will not impose this extra condition in our definition.
Exercise 10 (Elementary properties of decoupling constants) Let
and
.
- (i) (Monotonicity) Show that
whenever
are non-empty open subsets of
with
for
.
- (ii) (Triangle inequality) Show that
for any finite non-empty collections
of open non-empty subsets of
.
- (iii) (Affine invariance) Show that
whenever
are open non-empty and
is an invertible affine transformation.
- (iv) (Interpolation) Suppose that
for some
and
, and suppose also that
is a non-empty collection of open non-empty subsets of
for which one has the projection bounds
for all
,
, and
, where the Fourier multiplier
is defined by
Show that
- (v) (Multiplicativity) Suppose that
is a family of open non-empty subsets of
, with each
containing further open non-empty subsets
for
. Show that
- (vi) (Adding dimensions) Suppose that
is a family of disjoint open non-empty subsets of
and that
. Show that for any
, one has
where the right-hand side is a decoupling constant in
.
The most useful decoupling inequalites in practice turn out to be those where the decoupling constant is close to the lower bound of
, for instance if one has the sub-polynomial bounds
for every . We informally say that the collection
of sets exhibits decoupling in
when this is the case.
For , Lemma 6 (and (3)) already gives some decoupling estimates: one has
if the sets have an overlap of at most
, and similarly
when the sets ,
have an overlap of at most
.
For , it is not possible to exhibit decoupling in the limit
:
Exercise 11 If
is a collection of non-empty open subsets of
, show that
for any
. (Hint: select the
to be concentrated in widely separated large balls.)
Henceforth we now focus on the regime . By (8),
decoupling is easily obtained if the regions
are of bounded overlap. For
larger than
, bounded overlap is insufficient by itself; the arrangement of the regions
must also exhibit some “curvature”, as the following example shows.
Exercise 12 If
, and
, show that
(Hint: for the upper bound, use a variant of Exercise 24(ii) from Notes 1, or adapt the interpolation arguemnt used to establish that exercise.)
Now we establish a significantly more non-trivial decoupling theorem:
Theorem 13 (Decoupling for the parabola) Let
, let
for some
-separated subset
of
, where
, and to each
let
be the parallelogram (5). Then
for any
.
This result was first established by Bourgain and Demeter; our arguments here will loosely follow an argument of Li, that is based in turn on the efficient congruencing methods of Wooley, as recounted for instance in this exposition of Pierce.
We first explain the significance of the exponent in Theorem 13. Let
be a maximal
-separated subset
for some small
, so that
has cardinality
. For each
, choose
so that
is a non-negative bump function (not identically zero) adapted to the parallelogram
, which is comparable to a
rectangle. From the Fourier inversion formula,
will then have magnitude
on a dual rectangle of dimensions comparable to
, and is rapidly decreasing away from that rectangle, so we have
for all and
. In particular
On the other hand, we have for
if
is a sufficiently small absolute constant, hence
Comparing this with (6), we conclude that
so Theorem 13 cannot hold if the exponent is replaced by any larger exponent. On the other direction, by using Exercise 10(iv) and the trivial
decoupling from (8), we see that we also have decoupling in
for any
. (Note from the boundedness of the Hilbert transform that a Fourier projection to any polygon of boundedly many sides will be bounded in
for any
with norm
.) Note that reverse square function estimates in Exercise 8 only give decoupling in the smaller range
; the
version of Lemma 6 is not strong enough to extend the decoupling estimates to larger ranges because the triple sums
have too much overlap.
For any , let
denote the supremum of the decoupling constants
over all
-separated subsets
of
. From (7) we have the trivial bound
for any .
We first make a minor observation on the stability of that is not absolutely essential for the arguments, but is convenient for cleaning up the notation slightly (otherwise we would have to replace various scales
that appear in later arguments by comparable scales
).
Proof: Without loss of generality we may assume that . We first show that
. We need to show that
whenever is a
-separated subset of the
. By partitioning
into
pieces and using Exercise 10(ii) we may assume without loss of generality that
is in fact
-separated. In particular
The claim now follows from Exercise 10(i) and the inclusion
Conversely, we need to show that
whenever is
-separated, or equivalently that
when (we can extend from
to all of
by a limiting argument). From elementary geometry we see that for each
we can find a subset
of
of cardinality
, such that the parallelograms
with
an integer,
, and
a sufficiently small absolute constant, cover
. In particular, using Fourier projections to polygons with
sides, one can split
where each has Fourier support in
and
Now the collection can be partitioned into
subcollections, each of which is
-separated. From this and Exercise 10(ii), (iii) we see that
and thus
Applying (10), (11) we obtain the claim.
More importantly, we can use the symmetries of the parabola to control decoupling constants for parallelograms in a set of diameter
in terms of a coarser scale decoupling constant
:
Proposition 15 (Parabolic rescaling) Let
, and let
be a
-separated subset of an interval
of length
. Then
Proof: We can assume that for a small absolute constant
, since when
the claim follows from Lemma 14. Write
. Applying the Galilean transform
(which preserves the parabola, and maps parallelograms to
) and Exercise 10(iii), we may normalise
, so
.
Now let be the parabolic rescaling map
Observe that maps
to
for any
. From Exercise 10(iii) again, we can write the left-hand side of (12) as
since is
-separated, the claim then follows.
The multiplicativity property in Exercise 16 suggests that an induction on scales approach could be fruitful to establish (9). Interestingly, it does not seem possible to induct directly on ; all the known proofs of this decoupling estimate proceed by introducing some auxiliary variant of
that looks more complicated (in particular, involving additional scale parameters than just the base scale
), but which obey some inequalities somewhat reminiscent of the one in Exercise 16 for which an induction on scale argument can be profitably executed. It is yet not well understood exactly what choices of auxiliary quantity work best, but we will use the following choice of Li of a certain “asymmetric bilinear” variant of the decoupling constant:
Definition 17 (Bilinear decoupling constant) Let
. Define
to be the best constant for which one has the estimate
whenever
are
-separated subsets of intervals
of length
respectively with
, and for each
,
has Fourier support in
, and similarly for each
,
has Fourier support in
.
The scale is present for technical reasons and the reader may wish to think of it as essentially being comparable to
. Rather than inducting in
, we shall mostly keep
fixed and primarily induct instead on
. As we shall see later, the asymmetric splitting of the sixth power exponent as
is in order to exploit
orthogonality in the first factor.
From Hölder’s inequality, the left-hand side of (13) is bounded by
from which we conclude the bound
When are at their maximal size
we can use these bilinear decoupling constants
to recover control on the decoupling constants
, thanks to parabolic rescaling:
Proposition 18 (Bilinear reduction) If
, then
Proof: Let be a
-separated subset of
, and for each
let
be Fourier supported in
. We may normalise
. It will then suffice to show that
We partition into disjoint components
, each of which is supported in a subinterval
of
of length
, with the family
of intervals
having bounded overlap, so in particular
has cardinality
. Then for any
, we of course have
From the pigeonhole principle, this implies at least one of the following statements needs to hold for each given :
- (i) (Narrow case) There exists
such that
- (ii) (Broad case) There exist distinct intervals
with
such that
(The reason for this is as follows. Write and
, then
. Let
be the number of intervals
, then
, hence
. If there are only
intervals
for which
, then by the pigeonhole principle we have
for one of these
and we are in the narrow case (i); otherwise, if there are sufficiently many
for which
, one can find two such
with
, and we are in the broad case (ii).) This implies the pointwise bound
(We remark that more advanced versions of this “narrow-broad decomposition” in higher dimensions, taking into account more of the geometry of the various frequencies that arise in such sums, are useful in both restriction and decoupling theory; see this paper of Guth for more discussion.) From (13) we have
while from Proposition 15 we have
and hence
Combining all these estimates, we obtain the claim.
In practice the term here will be negligible as long as
is just slightly smaller than
(e.g.
for some small
). Thus, the above bilinear reduction is asserting that up to powers of
(which will be an acceptable loss in practice), the quantity
is basically comparable to
.
If we immediately apply insert (14) into the above lemma, we obtain a useless inequality due to the loss of in the main term on the right-hand side. To get an improved estimate, we will need a recursive inequality that allows one to slowly gain additional powers of
at the cost of decreasing the size of
factors (but as long as
is much larger than
, we will have enough “room” to iterate this inequality repeatedly). The key tool for doing this (and the main reason why we make the rather odd choice of splitting the exponents as
) is
Proposition 19 (Key estimate) If
with
then
Proof: It will suffice to show that
whenever are
-separated subsets of intervals
of length
respectively with
, and
have Fourier support on
respectvely for
, and we have the normalisation
We can partition as
, where
is a collection of intervals
of length
that have bounded overlap, and
is a
-separated subset of
. We can then rewrite the left-hand side of (16) as
where
and
From (13) we have
so it will suffice to prove the almost orthogonality estimate
By Lemma 6(i), it suffices to show that the Fourier supports of have overlap
.
Applying a Galilean transformation, we may normalise the interval to be centered at the origin, thus
, and
is now at a distance
from the origin. (Strictly speaking this may push
out to now lie in
rather than
, but this will not make a significant impact to the arguments.) In particular, all the rectangles
,
, now lie in a rectangle of the form
, and hence
and
have Fourier support in such a rectangle also (after enlarging the implied constants in the
notation appropriately). Meanwhile, if
is centered at
, then (since the map
has Lipschitz constant
when
and
) the parallelogram
is supported in the strip
for any
, hence
will also be supported in such a strip. Since
,
is supported in a similar strip (with slightly different implied constants in the
notation). Thus, if
and
have overlapping Fourier supports for
, then
, hence (since
)
. Since the intervals
have length
and bounded overlap, we thus see that each
has at most
intervals
for which
and
have overlapping Fourier supports, and the claim follows.
The final ingredient needed is a simple application of Hölder’s inequality to allow one to (partially) swap and
:
Exercise 20 For any
, establish the inequality
Now we have enough inequalities to establish the claim (9). Let be the least exponent for which we have the bound
for all and
; equivalently, we have
Another equivalent formulation is that is the least exponent for which we have the bound
as where
denotes a quantity that goes to zero as
. Clearly
; our task is to show that
.
Suppose for contradiction that . We will establish the bound
as for some
, which will give the desired contradiction.
Let for some small exponent
(independent of
, but depending on
) to be chosen later. From Proposition 18 and (17) we have
Since , the second term on the right-hand side is already of the desired form; it remains to get a sufficiently good bound on the first term. Note that a direct application of (14), (17) bounds this term by
; we need to improve this bound by a large multiple of
to conclude. To obtain this improvement we will repeatedly use Proposition 19 and Exercise 20. Firstly, from Proposition 19 we have
if is small enough. To control the right-hand side, we more generally consider expressions of the form
for various ; this quantity is well-defined if
is small enough depending on
. From Exercise 20 and (17) we have
and then by Proposition 19
if is small enough depending on
. We rearrange this as
The crucial fact here is that we gain a small power of on the right-hand side when
is large. Iterating this inequality
times, we see that
for any given , if
is small enough depending on
, and
denotes a quantity that goes to zero in the limit
holding
fixed. Now we can afford to apply (14), (17) and conclude that
which when inserted back into (20), (19) gives
If we then choose large enough depending on
, and
small enough depending on
, we obtain the desired improved bound (18).
Remark 21 An alternate arrangement of the above argument is as follows. For any exponent
, let
denote the claim that
whenever
,
with
sufficiently small depending on
, and
is sent to zero holding
fixed. The bounds (14), (17) give the claim
. On the other hand, the bound (21) shows that
implies
for any given
. Thus if
, we can establish
for arbitrarily large
, and for
large enough we can insert the bound (22) (with
sufficiently large depending on
) into (19), (20) to obtain the required claim (18). See this blog post for a further elaboration of this approach, which allows one to systematically determine the optimal exponents one can conclude from a system of inequalities of the type one sees in Proposition 19 or Exercise 20 (it boils down to computing the Perron-Frobenius eigenvalue of certain matrices).
Exercise 22 By carefully unpacking the above iterative arguments, establish a bound of the form
for all sufficiently small
. (This bound was first established by Li.)
Exercise 23 (Localised decoupling) Let
, and let
be a family of boundedly overlapping intervals
in
of length
. For each
, let
be an integrable function supported on
, and let
denote the extension operator
For any ball
of radius
, use Theorem 13 to establish the local decoupling inequality
for any
, where
is the weight function
The decoupling theorem for the parabola has been extended in a number of directions. Bourgain and Demeter obtained the analogous decoupling theorem for the paraboloid:
Theorem 24 (Decoupling for the paraboloid) Let
, let
, let
for some
-separated subset
of
, where
is the map
, and to each
let
be the disk
Then one has
for any
.
Clearly Theorem 13 is the case of Theorem 24.
Exercise 25 Show that the exponent
in Theorem 24 cannot be replaced by any larger exponent.
We will not prove Theorem 24 here; the proof in Bourgain-Demeter shares some features in common with the one given above (for instance, it focuses on a -linear formulation of the decoupling problem, though not one that corresponds precisely to the bilinear formulation given above), but also involves some additional ingredients, such as the wave packet decomposition and the multilinear restriction theorem from Notes 1.
Somewhat analogously to how the multilinear Kakeya conjecture could be used in Notes 1 to establish the multilinear restriction conjecture (up to some epsilon losses) by an induction on scales argument, the decoupling theorem for the paraboloid can be used to establish decoupling theorems for other surfaces, such as the sphere:
Exercise 26 (Decoupling for the sphere) Let
, let
, and let
be a
-separated subset of the sphere
. To each
, let
be the disk
Assuming Theorem 24, establish the bound
for any
. (Hint: if one lets
denote the supremum over all expressions of the form of the left-hand side of (23), use Exercise 10 and Theorem 24 to establish a bound of the form
, taking advantage of the fact that a sphere resembles a paraboloid at small scales. This argument can also be found in the above-mentioned paper of Bourgain and Demeter.)
An induction on scales argument (somewhat similar to the one used to establish the multilinear Kakeya estimate in Notes 1) can similarly be used to establish decoupling theorems for the cone
from the decoupling theorem for the parabola (Theorem 13). It will be convenient to rewrite the equation for the cone as , then perform a linear change of variables to work with the tilted cone
which can be viewed as a projective version of the parabola .
Exercise 27 (Decoupling for the cone) For
, let
denote the supremum of the decoupling constants
where
ranges over
-separated subsets of
, and
denotes the sector
More generally, if
, let
denote the supremum of the decoupling constants
where
ranges over
-separated subsets of
, and
denotes the shortened sector
- (i) For any
, show that
.
- (ii) For any
, show that
for any
. (Hint: use Theorem 13 and various parts of Exercise 10, exploiting the geometric fact that thin slices of the tilted cone resemble the Cartesian product of a parabola and a short interval.)
- (iii) For any
, show that
. (Hint: adapt the argument used to establish Exercise 16, taking advantage of the invariance of the tilted light cone under projective parabolic rescaling
and projective Galilean transformations
; these maps can also be viewed as tilted (conformal) Lorentz transformations ).
- (iv) Show that
for any
and
.
- (v) State and prove a generalisation of (iv) to higher dimensions, using Theorem 24 in place of Theorem 13.
This argument can also be found in the above-mentioned paper of Bourgain and Demeter.
A separate generalization of Theorem 13, to the moment curve
was obtained by Bourgain, Demeter, and Guth:
Theorem 28 (Decoupling for the moment curve) Let
, let
, and let
be a
-separated subset of
. For each
, let
denote the region
where
is the map
Then
for any
.
Exercise 29 Show that the exponent
in Theorem 28 cannot be replaced by any higher exponent.
It is not difficult to use Exercise 10 to deduce Theorem 13 from the case of Theorem 28 (the only issue being that the regions
are not quite the same the parallelograms
appearing in Theorem 13).
The original proof of Theorem 28 by Bourgain-Demeter-Guth was rather intricate, using for instance a version of the multilinear Kakeya estimate from Notes 1. A shorter proof, similar to the one used to prove Theorem 13 in these notes, was recently given by Guo, Li, Yung, and Zorin-Kranich, adapting the “nested efficient congruencing” method of Wooley, which we will not discuss here, save to say that this method can be viewed as a -adic counterpart to decoupling techniques. See also this paper of Wooley for an alternate approach to (a slightly specialised version of) Theorem 28.
Perhaps the most striking application of Theorem 28 is the following conjecture of Vinogradov:
Exercise 30 (Main conjecture for the Vinogradov mean value theorem) Let
. For any
and any
, let
denote the quantity
where
.
- (i) If
is a natural number, show that
is equal to the number of tuples
of natural numbers between
and
obeying the system of equations
for
.
- (ii) Using Theorem 28, establish the bound
for all
. (Hint: set
and
, and apply the decoupling inequality to functions
that are adapted to a small ball around
.)
- (iii) More generally, establish the bound
for any
and
. Show that this bound is best possible up to the implied constant and the loss of
factors.
Remark 31 Estimates of the form (24) are known as mean value theorems, and were first established by Vinogradov in 1937 in the case when
was sufficiently large (and by Hua when
was sufficiently small). These estimates in turn had several applications in analytic number theory, most notably the Waring problem and in establishing zero-free regions for the Riemann zeta function; see these previous lecture notes for more discussion. The ranges of
for which (24) was established was improved over the years, with much recent progress by Wooley using his method of efficient congruencing; see this survey of Pierce for a detailed history. In particular, these methods can supply an alternate proof of (24); see this paper of Wooley.
Exercise 32 (Discrete restriction) Let
and
, and let
be a
-separated subset of either the unit sphere
or the paraboloid
. Using Theorem 24 and Exercise 26, show that for any radius
and any complex numbers
, one has the discrete restriction estimate
Explain why the exponent
here cannot be replaced by any larger exponent, and also explain why the exponent
in the condition
cannot be lowered.
For further applications of decoupling estimates, such as restriction and Strichartz estimates on tori, and application to combinatorial incidence geometry, see the text of Demeter.
[These exercises will be moved to a more appropriate location at the end of the course, but are placed here for now so as not to affect numbering of existing exercises.]
Exercise 33 Show that the inequality in (8) is actually an equality, if
is the maximal overlap of the
.
Exercise 34 Show that
whenever
. (Hint: despite superficial similarity, this is not related to Lemma 14. Instead, adapt the parabolic rescaling argument used to establish Proposition 15.)
94 comments
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6 November, 2021 at 10:44 pm
Dmitrii
Do I understand correctly that Proposition 19 is the only part where the proof breaks for a line (which does not exhibit strong decoupling ofc)?
Also I wonder if in general the number of solutions to
where
lie on a parabola should be dominated by the diagonal term even if the points are not well-separated?
7 November, 2021 at 8:08 pm
Terence Tao
Yes, this is the key proposition that crucially exploits the curvature of the parabola.
It is tentatively conjectured that L^6 decoupling for the parabola should hold for arbitrary collections of N disjoint arcs (not necessarily adjacent and of equal size), with losses of
type, which would imply a positive answer to your last question (again with
type losses), but this has not yet been established.
8 January, 2022 at 11:46 am
Dmitrii
Thanks. A related question: can one similarly prove “one-dimensional” L_4 decoupling for the y-axis projections and thus show that the set of squares
has additive energy
without resorting to the divisor bound? Curious if any such a proof is known.
20 January, 2022 at 3:59 pm
Terence Tao
I don’t know of any decoupling-based proof of this fact. But any such argument would probably have to use some number theory in addition to pure decoupling, as decoupling estimates alone wouldn’t be able to detect if one replaced the set of squares with some perturbation of that set, and I would imagine that there are perturbations in which the additive energy becomes significantly larger than
.