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 byand 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 4Let be a bump function supported on that equals one on . For each integer , let be the Fourier multiplier defined for byand 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 5The 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 7By 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 thedecoupling constantto 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 11If is a collection of non-empty open subsets of , show thatfor 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 12If , 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). Thenfor 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 estimatewhenever 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:

*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

*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 :

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 21An alternate arrangement of the above argument is as follows. For any exponent , let denote the claim thatwhenever , 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 22By carefully unpacking the above iterative arguments, establish a bound of the formfor 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 operatorFor 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 diskThen one has

for any .

Clearly Theorem 13 is the case of Theorem 24.

Exercise 25Show 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 diskAssuming 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 constantswhere 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 regionwhere is the map

Then

for any .

Exercise 29Show 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 31Estimates of the form (24) are known asmean 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 estimateExplain 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 33Show that the inequality in (8) is actually an equality, if is the maximal overlap of the .

Exercise 34Show that whenever . (Hint: despite superficial similarity, this is not related to Lemma 14. Instead, adapt the parabolic rescaling argument used to establish Proposition 15.)

## 90 comments

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14 April, 2020 at 3:06 am

mohsenHello professor.

A PURPOSE!

Please put your pdf version of your notes on this website too!

thanks alot!

14 April, 2020 at 4:07 am

AnonymousThis has been asked many times. See for instance this thread: https://terrytao.wordpress.com/2020/03/29/247b-notes-1-restriction-theory/#comment-549677

If you print a blog post to PDF it should be fairly readable.

14 April, 2020 at 4:31 am

IrkutskThere should be an absolute value on the LHS of the first displayed formula.

14 April, 2020 at 5:29 am

shapiroIn Exercise 4 (iv), what kind of counterexamples should I be looking at? I tried a family of dilated bump functions but it is really tough to compute the projections on frequency. I do not even know if this is the right family of function to test. Do you have any hint?

14 April, 2020 at 6:08 am

shapiroWell, it seems that after writing this comment I finally found an answer: condering a function f such that P_[k,k+1] (f)= the same bump function works just fine!

14 April, 2020 at 5:48 am

Trevor WooleyJust a note on nomenclature: in Exercise 30 you refer to the “Vinogradov main conjecture”, which seems to have entered usage in the literature in the past few years in the harmonic analysis community. In the number theory community the name has been different, partly owing to the fact that there are many conjectures associated with the name I. M. Vinogradov, including for example his famous conjecture concerning non-trivial cancellation in short character sums. The phrase “Vinogradov’s mean value theorem” was used to describe any non-trivial estimate (optimal or not) for the mean value of exponential sums (or moment estimate) in question, and the estimate in Exercise 30 was often referred to as the “optimal estimate in Vinogradov’s mean value theorem”, or by me and others over the past decade as the “main conjecture in Vinogradov’s mean value theorem”.

For what it is worth, the “nested efficient congruencing method” can be seen as a p-adic decoupling method, which leads to some unification of all the ideas currently in play.

14 April, 2020 at 7:57 am

Terence TaoThanks for this! I realise that I have indeed mentally contracted “main conjecture for the Vinogradov mean value theorem” as “Vinogradov main conjecture”, and have now corrected this in the text. Also I realise that in my rush to push out these notes I had omitted the historical remarks on this conjecture, which I have now also added.

27 April, 2020 at 9:14 pm

AnonCould you describe what a p-adic decoupling method is (even if the term isn’t strictly defined)?

3 May, 2020 at 5:50 pm

AnonDear Professor Tao, you are also welcome to answer this if you wish.

14 April, 2020 at 6:28 am

Lior SilbermanIn the line after equation (1), it is the that are supported on the annuli, right? (I guess you follow the physics convention where the variable name specifies the coordinate system).

[Corrected, thanks – T.]14 April, 2020 at 6:36 am

Lior SilbermanIn the equation after you mention Hormander–Miklin there is a missing on the LHS and the subscript didn’t come out.

[Corrected, thanks – T.]14 April, 2020 at 5:33 pm

AnonNote:the same small issue also appears at end of proof of Thm1

[Corrected, thanks – T.]14 April, 2020 at 6:38 am

Lior SilbermanIn exercise 2 you need to swap the two balls (so the bump function is supported on the ball of radius 2, equal to 1 on the ball of radius 1).

[Corrected, thanks – T.]14 April, 2020 at 11:24 am

extremal010101proof of theorem 1, part (i): why ? I assume . The bump function is supported on , and the frequencies of are not “2-separated”, then may have frequencies not in , or am I missing something?

14 April, 2020 at 11:47 am

Terence TaoIn this part of the argument we are proving (ii), not (i).

17 April, 2020 at 5:16 am

dn1214In Exercise 4 (i), how do you use Parseval for series to get the case? If I introduce the function then I can not compute easily the norm of this function and I’m stuck.

The proof I know goes roughly as follows: for any we write where and then we use the decay of to conclude.

17 April, 2020 at 10:20 am

Terence TaoThe function can be expressed in terms of in a tractable form using either the Poisson summation formula or the Fourier inversion formula for periodic functions, after a certain amount of standard manipulation using the usual Fourier identities. The duality method you indicate would also work.

27 April, 2020 at 12:01 pm

hhy177To apply the Poisson summation formula here, don’t we need to be Schwartz in ? Writing , it seems like is smooth with not enough decay in . After I proceed formally with this calculation (such as taking inverse Fourier transform of 1 in the process), I seem to get something looking more tractable, from which I could derive the boundedness of the square estimate in 4.(i) when .

27 April, 2020 at 12:44 pm

Terence TaoTo convert your formal computations to a rigorous argument, one can either (a) try to work in the language of (tempered) distributions, (b) using some sort of limiting argument, or (c) take the final expression for that you obtained and try to prove it by a different (and fully rigorous) method (e.g., by first establishing a “Fourier transform” of your identity and then applying the Fourier inversion formula).

19 April, 2020 at 5:42 am

dn1214I have two questions about exercise 3:

– In (ii) I am not able to prove the result; since the result is false if and are two fully independent collections of rectangles (pick one collection of disjoint rectangles, and the other to be all the same rectangle), I am wondering how to use that the are translations of ? Does it require to use the boundedness of the disc multiplier?

– I was surprised that without using (ii), one can solve (iv) only using (i) (taking ), is it normal?

There is also a minor typo: at the begining of the exercise, I guess.

Thank you a lot.

19 April, 2020 at 11:56 am

Terence TaoYes, one should use part (i) to establish part (ii).

The definition of encompasses both and , which is why I used a different symbol from either here.

21 April, 2020 at 10:22 am

dn1214Thank you for the clarification!

21 April, 2020 at 12:02 am

AnonymousI’m curious if one should expect to be able to replace the ball of integration on the left in the local decoupling estimates with an integral over a smaller more tubular region?

It seems that when you ultimately rescale and plug in lattice points to prove Vinogrodov-type estimates, there ends up being redundancy in the integral, and that an estimate over a more tubular region would more naturally lead to a the the exact Vinogrodov integral (and be a stronger estimate).

21 April, 2020 at 7:30 am

Terence TaoGood question. There are counterexamples that show that the balls on which decoupling occurs cannot be replaced by smaller balls without significant increase in the decoupling constants, but there may be some possibility of additional useful decoupling estimates involving more eccentric domains than balls. For instance if one is trying to decouple arcs in Fourier space contained in a narrow arc of the paraboloid, then by applying a suitable Galilean transformation and parabolic rescaling, followed by Exercise 23, one obtains a local decoupling estimate involving a certain rectangular region that is a little bit smaller in one dimension than the ball one would get from direct application of Exercise 23. So perhaps there is some “time-frequency” version of decoupling that is a little bit smarter about the spatial localisation, and as you say the redundancy in the Vinogradov integral also points to some possibility of improvement in this direction also.

21 April, 2020 at 10:47 pm

dn1214Aren’t there typos in the proof of Lemma 14? e.g. “assuming that “, since we have , I would expect a sign. And in the proof you write “since contains “: one of the should be a .

22 April, 2020 at 7:57 am

Terence TaoThe additional restriction is intentional; if one can prove the claim in the special case then by transitivity one then obtains the general case as a corollary, because in the latter case one can find a third scale for which .

The statement “ contains ” is stated correctly, but there is an unfortunate placement of a comma immediately afterwards.

27 April, 2020 at 1:02 am

RexIt might be helpful to add a white space just before the comma, so that it doesn’t get confused with a prime. I was also disoriented for a moment by this.

[Reworded -T.]24 April, 2020 at 1:48 pm

AnonymousIn the proof of Theorem 1, how does one get that ?

24 April, 2020 at 2:53 pm

AnonymousMultiplier equals 1 on the fourier support of f_j and 0 for all the others

24 April, 2020 at 3:24 pm

Terence TaoBy computing Fourier transforms one can see that is equal to when and vanishes when .

24 April, 2020 at 1:54 pm

AnonymousCan one say that “decoupling” is sort of “partition of unity” in the Fourier space? (I may have a very wrong impression.)

24 April, 2020 at 3:27 pm

Terence TaoDecoupling is a property of

somepartitions of unity in Fourier space, but not all partitions of Fourier space have good decoupling estimates (and some partitions only have good decoupling for some exponents and not others). The presence of an decoupling estimate can be viewed as saying that the geometry of that partition behaves somewhat like a Hilbert space geometry in the sense that a (weak) version of the Pythagorean theorem remains valid in this setting.24 April, 2020 at 5:24 pm

AnonymousIn high dimension, if we use a dyadic cubic projection rather than the dyadic ball projection in the note, can the condition that the being 2-separated be removed from Theorem 1(ii)? This would at least avoid encountering the disk multiplier.

[Yes – T.]25 April, 2020 at 3:08 am

RexIn exercise 10.ii (triangle inequality), one of the mathcal Us in the statement of the inequality is missing a ‘

[Corrected, thanks – T.]25 April, 2020 at 6:47 pm

RexDoes this blog entry load correctly for the other people? I am getting something very messy with “formula does not parse”, visible “”, “”, etc. everywhere starting about 10 hours ago. Everything was fine before then. No other blog entries seem to have this issue.

[Corrected, thanks – T.]25 April, 2020 at 7:56 pm

RexMinor omission: in the proof of lemma 6, the A in “By hypothesis, for each frequency {\xi} at most {A} of the {\hat f_j(\xi)} are non-zero, thus by Cauchy-Schwarz” is missing a subscript.

26 April, 2020 at 5:05 am

Xiao-Chuan LiuFor Exercise 3, I can compute and obtain that will take uniform positive value restricted to . Is there some easy way to see this should be true (without computing it)?

26 April, 2020 at 8:44 am

Terence TaoNot really: lower bounds generally are more delicate than upper bounds and usually require some calculation to ensure that one does not have some unexpected cancellation. (In some cases, the function will obey some sort of PDE and one can use tools from the theory of unique continuation to obtain some lower bounds. It’s possible that one could pull off this trick here using the connection between the Hilbert transform and harmonic conjugates, but it is simpler to just directly calculate in this case.)

26 April, 2020 at 6:36 am

RexThe formula

“.”

under Theorem 13 has a \Vert on the left and a parenthesis on the right.

[Corrected, thanks – T.]27 April, 2020 at 12:04 pm

hhy177For Ex 4 (ii), how do one see the boundedness of on from the boundedness of the Hilbert transform? Since both operator is equal (up to a factor of ) on Schwartz functions with Fourier support in , I can conclude is bounded on on Schwartz functions with Fourier support in . How do I see it for arbitrary functions?

27 April, 2020 at 12:45 pm

Terence TaoExpress as the difference of and .

27 April, 2020 at 9:46 pm

AnonymousDear Pro Tao,

Whether you want it or not, Twin prime conjecture is certainly solved in 2020. By any rate, you must succeed. The historic seconds is coming, you have no choice. An only way for you, you can not come back, only go aheah! I think Sir will understand my imply. As you win , it will be useful for many people.

Lovely Pro.Tao,

28 April, 2020 at 9:35 am

AnonymousWhat’s this about?

27 April, 2020 at 1:56 pm

AnonymousDear Professor Tao, in the proof of the Key estimate (Proposition 19), you said “, hence “. I think one needs in order to do so. Otherwise, could you please give me some hints on how it works? Thank you very much.

[Oops, there are some losses of here that I forgot to take into account; however, these turn out to be acceptable losses for this argument, which is now reworded accordingly. -T]28 April, 2020 at 12:02 pm

RexIn the proof of lemma 14:

“(we can extend from {C^\infty_c({\bf R}^2)} to all of {L^6({\bf R}^2)} by a limiting argument).”

I assume that this should be changed to Schwartz functions?

[Corrected, thanks – T.]29 April, 2020 at 11:41 am

AnonymousFor the triangle inequality of decoupling constants, I got the inequality the other way around. I basically used the monotonicity for U and U’ (i.e. ), square both sides then add them, and take square root, I got

How to get the inequality the other way around?

29 April, 2020 at 2:24 pm

Lior SilbermanApply the 2d C–S inequality to the dot product where (resp. ) have Fourier support in (resp. ).

29 April, 2020 at 2:56 pm

Terence TaoOne trick that can help conceptually with trying to prove estimates of the form or is to deconstruct the degenerate version of these inequalities, namely that “If vanishes (or is very small), then must also vanish (or is very small)”. Suppose you know that vanishes (or is very small). What does this tell you about ? If one approaches this question literally it is vacuous due to the fact that decoupling constants are necessarily bounded from below by 1, but try to ignore this fact for the purposes of performing a mental exercise. Your answer to this exercise will help suggest how to actually approach the inequality at hand.

29 April, 2020 at 2:26 pm

Lior SilbermanQuestion about Ex. 10(v): where do we use the disjointness? It seems to me that the argument works whenever are any sets such that (up to sets of measure zero) .

[Good point; I’ve reworded the exercise accordingly. -T]29 April, 2020 at 3:19 pm

AnonymousIn the explanation of significance of , how do you get the factor of for the norm of , ie.

(the 2nd equation after Theorem 13)? Also, I am not sure how do you get in the 3rd equation after we restrict to a small ball of radius .

29 April, 2020 at 6:26 pm

Terence TaoBoth of these follow from the previously mentioned fact that has cardinality .

1 May, 2020 at 6:49 am

RexIn proof of Prop 18:

“(ii) (Broad case) There exist distinct intervals {I,J \in {\mathcal I}} with {\mathrm{dist}(I,J) \geq \nu} such that

\displaystyle |\sum_{\omega \in \phi(\Sigma)} f_\omega(x)| \lesssim \nu^{-O(1)} |\sum_{\omega \in \phi(\Sigma_I)} f_\omega(x)|, \nu^{-O(1)} |\sum_{\omega \in \phi(\Sigma_I)} f_\omega(x)|.”

One of the subscripts should be a _J instead of a _I, I assume.

Could you say more about how this pigeonhole argument works? I’m not seeing why the broad case should follow if the narrow case doesn’t hold.

[More details added – T.]1 May, 2020 at 8:48 am

RexMore in the same proof:

“From (13) we have

\displaystyle \int_{{\bf R}^2} |\sum_{\omega \in \phi(\Sigma_I)} f_\omega|^2 |\sum_{\omega \in \phi(\Sigma_I)} f_\omega|^4 \lesssim M_{2,4}(\delta,\nu,\nu,\nu)

while”

Another subscript here should be _J, and the constant M should be M^6

[Corrected, thanks – T.]1 May, 2020 at 11:27 am

RexIn the same proof:

“If we immediately apply insert (14) into the above lemma, we obtain a useless inequality due to the loss of {\nu^{-O(1)})} in the main term on the…”

has an extra parenthesis.

[Corrected, thanks – T.]1 May, 2020 at 7:51 am

AnonymousIs the subset really supposed to be a subset of and of cardinality O(1)? Shouldn’t it be and the cardinality isn’t used in the rest of the proof.

[Corrected, thanks. The cardinality bound is needed for a minor reason, namely to ensure that the polygons formed by taking various boolean combinations of the parallelograms involved only have sides. -T.]1 May, 2020 at 11:04 am

AnonymousShould the RHS of eq.(13) in the definition of bilinear decoupling constant be instead of ?

1 May, 2020 at 11:11 am

AnonymousSorry, it is right.

1 May, 2020 at 12:24 pm

Alan ChangIn the proof of Lemma 14, why do you assume instead of just ?

2 May, 2020 at 8:32 am

Terence TaoHuh, I think that was a relic from an earlier version of the argument. You are right, the extra safety margin of is not needed here.

1 May, 2020 at 2:41 pm

Lior SilbermanIn the second equation after (21), I would replace with .

[Corrected, thanks – T.]3 May, 2020 at 1:37 am

RexIn the proof of Prop 19:

“we have the normalisation

\displaystyle \sum_{\omega_1 \in \phi(\Sigma_1)} \|f_{\omega_1}\|_{L^6({\bf R}^2)} = \sum_{\omega_2 \in \phi(\Sigma_2)} \|g_{\omega_2}\|_{L^6({\bf R}^2)}^2 = 1.

We can partition {\Sigma_1} as {\sum_{I’ \in {\mathcal I}’} \Sigma_{1,I’}}, where {{\mathcal I}’} is a collection of intervals”

The left side of the displayed formula is missing an exponent of 2 (sum of squares of norms rather than norms).

[Corrected, thanks – T.]3 May, 2020 at 12:01 pm

RexIn the same proof:

“We can partition {\Sigma_1} as {\sum_{I’ \in {\mathcal I}’} \Sigma_{1,I’}}, where {{\mathcal I}’} is a collection of intervals {I’ \subset [-1,1]} of length {\rho’_1} that have bounded overlap,”

I think here you mean to write union rather than sum over the I’.

[Corrected, thanks – T.]3 May, 2020 at 12:04 pm

Rex(or better yet, \coprod since this is a disjoint union)

3 May, 2020 at 12:26 pm

RexAlso

“here

\displaystyle F_{I’} := \sum_{\omega_1 \in \phi(\Sigma_{1,I})} f_{\omega_1}

and”

has a missing ‘ in the subscript

[Corrected, thanks – T.]3 May, 2020 at 7:59 am

Alan ChangIn the Knapp-type example to show that decoupling cannot hold for , is there a good way to understand why Khintchine cannot be used to obtain a good lower bound for ?

In the notes, we used the following pointwise bound: on a ball of radius

However, by randomization, we could instead try to use the following pointwise bound: on a ball of radius .

To me, it is not clear at first which of these two estimates will do better. After some computations, I see that the random estimate doesn’t give anything useful about (since all the s cancel out), but is there a way to see this in advance?

3 May, 2020 at 8:34 am

Terence TaoKhintchine tells us that random sums always decouple on the average (for ):

So if one wants to create a good counterexample to decoupling one has to produce constructive interference (in which is significantly larger in magnitude than the square function on some non-empty region of space) rather than random-type interference.

3 May, 2020 at 2:18 pm

John MangualWhat is an example of “independent” phases ? is a complex number. The relation could possibly be replaced with or …

Also is there any way to visualize Fourier series with disjoint annuli? In position space or momentum space.

3 May, 2020 at 5:45 pm

Alan ChangIn the centered equation preceding (21), should be ?

[Corrected, thanks – T.]4 May, 2020 at 8:05 am

Alan ChangActually, now I think that in the displayed equation preceding (21), the should be and that this change should propogate down through the displayed equations that follow.

Also, two displayed equations after (21), should be .

[Corrected, thanks – T.]4 May, 2020 at 10:40 am

AnonymousTo prove Prop.18, why is it sufficient to prove equation (14)?

5 May, 2020 at 9:48 am

Terence TaoI assume you mean (15) instead of (14). From (15) and the definition of we conclude that

(noting that in the definition of decoupling constants we are free to normalise without loss of generality), which then gives Proposition 18.

4 May, 2020 at 1:19 pm

Lior SilbermanIn the counter-example after the statement of Thm 13 (significance of the exponent 6) shouldn’t be ?

[Corrected, thanks – T.]5 May, 2020 at 11:29 am

AnonymousWhat is Exercise 27(ii) supposed to say? I think it’s incomplete as stated. There’s also a typo in the third line after Exercise 20 “for all 00”.

[Corrected, thanks – T.]5 May, 2020 at 5:50 pm

Alan ChangIn the centered equation above “so it will suffice to prove the almost orthogonality estimate,” the M_{2,4} constant should be raised to the 6th power.

Also, in the last paragraph of that proof, “(with slightly different implied constants in the O() notation” is missing a closing parenthesis.

[Corrected, thanks – T.]6 May, 2020 at 11:13 am

GaroWhat are some of the heuristic explanations for the gap between the best possible decoupling results for curves with torsion and the conjectured bounds on exponential sums? For example, we get decoupling on for but expect the estimate to hold for exponential sums up to .

7 May, 2020 at 10:08 am

AnonOne direct explanation is the lack of self-similarity in the phase, or the lack of a ‘Galilean’ invariance symmetry. For the parabola notice that an affine change of variables of the form preserves the parabola extension operator after a simple change of variables; this is no longer true if the phase is changed to . This self-similarity plays an important role in the proof of decoupling inequalities via parabolic rescaling.

8 May, 2020 at 7:37 am

AnonymousI think an easier way to do the upper bound in Exercise 12 is to interpolate the p = 2 bound from Plancherel with the trivial bound.

[Hint reworded – T.]9 May, 2020 at 8:40 am

Alan ChangIn the proof of Proposition 19, I see four instances of L^2(R^d) which should be L^2(R^2).

[Corrected, thanks – T.]2 June, 2020 at 12:37 am

MATH 247B: Modern Real-Variable Harmonic Analysis – Countable Infinity[…] Decoupling Estimates […]

4 June, 2020 at 2:34 pm

AnonymousA small comment (please don’t take my words for granted):

For the Littlewood-Paley inequalities, once

is shown, it follows (by choosing optimizing values for ) that the minimum and maximum (over all sign choices) of the expression

are comparable, and therefore we in fact have

After that, Khintchine will yield both directions of the proof.

This may help clarify the second bit of the proof; it is the same idea expressed differently, i.e. one may also prove

and then use Khintchine on either side.

7 June, 2020 at 2:26 pm

Alan ChangI have two questions about Remark 21.

1. “On the other hand, the bound (21) shows that implies for any given .”

In the sentence quoted above, should be ?

2. “Thus if , we can establish for arbitrarily large , and for large enough we can insert the bound (22) (with {a=1}) into (19), (20) to obtain the required claim (18).”

From (22) with a = 1, we gain a factor , but this is not necessarily an improvement, since the constant in the exponent can depend on theta. It seems like we need the to actually be ?

8 June, 2020 at 8:39 am

Terence TaoOops, it isn’t enough to take , one has to take sufficiently large depending on in order to overcome both the losses, and the losses when one converts the estimate back to control on .

Thanks for the corrections!

8 June, 2020 at 9:26 am

Alan ChangThanks!

What do you mean by “convert the estimate back to control on “? If we don’t take in (22), how do we use (19) and (20)?

8 June, 2020 at 12:06 pm

Terence TaoThere is an easy triangle inequality argument that shows that . Granted, this makes this version of the argument slightly different from the one in the notes, which required more precise bookkeeping of what in this formulation is the error, but it hopefully illustrates that that bookkeeping is not an essential part of the argument.

29 September, 2020 at 6:45 pm

zorichDear Tao ,

I have a small question, you said:

“Note from the boundedness of the Hilbert transform that a Fourier projection to any polygon of bounded many sides will be bounded in ."

But I think polygons can be used to construct the same counter-example as balls.

Use the same construction in Fefferman's example, if (U is the polygon) is bounded in ,then for any dilation and rigid motion L is also bounded with the same norm.Then rescale U based on a point on its edges (but not vertices) to exhaust the whole half plane.By Fatou's lemma, it implies the boundness of projection to half plane…

1 October, 2020 at 7:51 am

Terence TaoThe square function over any bounded number of half-planes is still bounded. Only the square function over an unbounded number of half-planes becomes unbounded. Similarly, the multiplier to a polygon with boundedly many sides remains bounded in , but as the number of sides goes to infinity, the operator norm will also go to infinity (assuming the orientations of the sides are suitably well distributed in space).

4 October, 2020 at 5:06 am

zorichThanks a lot! For polygons, to get half plane pointing different directions, we must use rotations, but can’t be controlled by If is different rotations.but for spheres, we only need to use translation(multiplied by a phase) and dilation

4 February, 2021 at 8:05 am

Small typoSmall typo on the index of the random signs:

“…with symbol {\sum_{j=1}^n \epsilon_n \psi(\xi/2^{k_j})}.”, should be “…with symbol {\sum_{j=1}^n \epsilon_j\psi(\xi/2^{k_j})}.”

[Corrected, thanks – T.]26 July, 2021 at 2:24 pm

BI’m a little confused as to how to make use of the boundedness of the Fourier projection operators in Exercise 10(iv); any advice? I’m guessing it must be needed to get some sort of “reverse interpolation inequality”, but I’m not sure how to make this precise.

[In order to apply an interpolation theorem, such as Riesz-Thorin, the estimates one is interpolating must be allowed to range freely the entirety of an space; one cannot automatically impose a restriction on these functions, such as a requirement that the functions have specified Fourier support, and still expect to have an interpolation theorem. However, one can get around this for this particular problem by inserting Fourier projections into the decoupling inequality to be proven, which permit the functions involved to now have arbitrary Fourier support. -T]16 September, 2021 at 8:57 am

AnonymousIn the statememt of Ex 3, $\Omega$ should be $D$ right? Or am I missing smth?

[I am defining the projections here for arbitrary sets , such as the half-planes in part (i). -T.]