This set of notes focuses on the *restriction problem* in Fourier analysis. Introduced by Elias Stein in the 1970s, the restriction problem is a key model problem for understanding more general oscillatory integral operators, and which has turned out to be connected to many questions in geometric measure theory, harmonic analysis, combinatorics, number theory, and PDE. Only partial results on the problem are known, but these partial results have already proven to be very useful or influential in many applications.

We work in a Euclidean space . Recall that is the space of -power integrable functions , quotiented out by almost everywhere equivalence, with the usual modifications when . If then the Fourier transform will be defined in this course by the formula

From the dominated convergence theorem we see that is a continuous function; from the Riemann-Lebesgue lemma we see that it goes to zero at infinity. Thus lies in the space of continuous functions that go to zero at infinity, which is a subspace of . Indeed, from the triangle inequality it is obvious that

If , then Plancherel’s theorem tells us that we have the identity

Because of this, there is a unique way to extend the Fourier transform from to , in such a way that it becomes a unitary map from to itself. By abuse of notation we continue to denote this extension of the Fourier transform by . Strictly speaking, this extension is no longer defined in a pointwise sense by the formula (1) (indeed, the integral on the RHS ceases to be absolutely integrable once leaves ; we will return to the (surprisingly difficult) question of whether pointwise convergence continues to hold (at least in an almost everywhere sense) later in this course, when we discuss Carleson’s theorem. On the other hand, the formula (1) remains valid in the sense of distributions, and in practice most of the identities and inequalities one can show about the Fourier transform of “nice” functions (e.g., functions in , or in the Schwartz class , or test function class ) can be extended to functions in “rough” function spaces such as by standard limiting arguments.

By (2), (3), and the Riesz-Thorin interpolation theorem, we also obtain the Hausdorff-Young inequality

for all and , where is the dual exponent to , defined by the usual formula . (One can improve this inequality by a constant factor, with the optimal constant worked out by Beckner, but the focus in these notes will not be on optimal constants.) As a consequence, the Fourier transform can also be uniquely extended as a continuous linear map from . (The situation with is much worse; see below the fold.)

The *restriction problem* asks, for a given exponent and a subset of , whether it is possible to meaningfully restrict the Fourier transform of a function to the set . If the set has positive Lebesgue measure, then the answer is yes, since lies in and therefore has a meaningful restriction to even though functions in are only defined up to sets of measure zero. But what if has measure zero? If , then is continuous and therefore can be meaningfully restricted to any set . At the other extreme, if and is an arbitrary function in , then by Plancherel’s theorem, is also an arbitrary function in , and thus has no well-defined restriction to any set of measure zero.

It was observed by Stein (as reported in the Ph.D. thesis of Charlie Fefferman) that for certain measure zero subsets of , such as the sphere , one can obtain meaningful restrictions of the Fourier transforms of functions for certain between and , thus demonstrating that the Fourier transform of such functions retains more structure than a typical element of :

Theorem 1 (Preliminary restriction theorem)If and , then one has the estimatefor all Schwartz functions , where denotes surface measure on the sphere . In particular, the restriction can be meaningfully defined by continuous linear extension to an element of .

*Proof:* Fix . We expand out

From (1) and Fubini’s theorem, the right-hand side may be expanded as

where the inverse Fourier transform of the measure is defined by the formula

In other words, we have the identity

using the Hermitian inner product . Since the sphere have bounded measure, we have from the triangle inequality that

Also, from the method of stationary phase (as covered in the previous class 247A), or Bessel function asymptotics, we have the decay

for any (note that the bound already follows from (6) unless ). We remark that the exponent here can be seen geometrically from the following considerations. For , the phase on the sphere is stationary at the two antipodal points of the sphere, and constant on the tangent hyperplanes to the sphere at these points. The wavelength of this phase is proportional to , so the phase would be approximately stationary on a cap formed by intersecting the sphere with a neighbourhood of the tangent hyperplane to one of the stationary points. As the sphere is tangent to second order at these points, this cap will have diameter in the directions of the -dimensional tangent space, so the cap will have surface measure , which leads to the prediction (7). We combine (6), (7) into the unified estimate

where the “Japanese bracket” is defined as . Since lies in precisely when , we conclude that

Applying Young’s convolution inequality, we conclude (after some arithmetic) that

whenever , and the claim now follows from (5) and Hölder’s inequality.

Remark 2By using the Hardy-Littlewood-Sobolev inequality in place of Young’s convolution inequality, one can also establish this result for .

Motivated by this result, given any Radon measure on and any exponents , we use to denote the claim that the *restriction estimate*

for all Schwartz functions ; if is a -dimensional submanifold of (possibly with boundary), we write for where is the -dimensional surface measure on . Thus, for instance, we trivially always have , while Theorem 1 asserts that holds whenever . We will not give a comprehensive survey of restriction theory in these notes, but instead focus on some model results that showcase some of the basic techniques in the field. (I have a more detailed survey on this topic from 2003, but it is somewhat out of date.)

** — 1. Necessary conditions — **

It is relatively easy to find necessary conditions for a restriction estimate to hold, as one simply needs to test the estimate (9) against a suitable family of examples. We begin with the simplest case . The Hausdorff-Young inequality (4) tells us that we have the restriction estimate whenever . These are the only restriction estimates available:

Proposition 3 (Restriction to )Suppose that are such that holds. Then and .

We first establish the necessity of the duality condition . This is easily shown, but we will demonstrate it in three slightly different ways in order to illustrate different perspectives. The first perspective is from scale invariance. Suppose that the estimate holds, thus one has

for all Schwartz functions . For any scaling factor , we define the scaled version of by the formula

Applying (10) with replaced by , we then have

From change of variables, we have

and from the definition of Fourier transform and further change of variables we have

so that

combining all these estimates and rearranging, we conclude that

If is non-zero, then by sending either to zero or infinity we conclude that for all , which is absurd. Thus we must have the necessary condition , or equivalently that .

We now establish the same necessary condition from the perspective of dimensional analysis, which one can view as an abstraction of scale invariance arguments. We give the spatial variable a unit of length. It is not so important what units we assign to the range of the function (it will cancel out of both sides), but let us make it dimensionless for sake of discussion. Then the norm

will have the units of , because integration against -dimensional Lebesgue measure will have the units of (note this conclusion can also be justified in the limiting case ). For similar reasons, the Fourier transform

will have the units of ; also, the frequency variable must have the units of in order to make the exponent appearing in the exponential dimensionless. As such, the norm

has units . In order for the estimate (10) to be dimensionally consistent, we must therefore have , or equivalently that .

Finally, we establish the necessary condition once again using the example of a rescaled bump function, which is basically the same as the first approach but with replaced by a bump function. We will argue at a slightly heuristic level, but it is not difficult to make the arguments below rigorous and we leave this as an exercise to the reader. Given a length scale , let be a bump function adapted to the ball of radius around the origin, thus where is some fixed test function supported on . We refer to this as a bump function *adapted* to ; more generally, given an ellipsoid (or other convex region, such as a cube, tube, or disk) , we define a bump function adapted to to be a function of the form , where is an affine map from (or other fixed convex region) to and is a bump function with all derivatives uniformly bounded. As long as is non-zero, the norm is comparable to (up to constant factors that can depend on but are independent of ). The uncertainty principle then predicts that the Fourier transform will be concentrated in the dual ball , and within this ball (or perhaps a slightly smaller version of this ball) would be expected to be of size comparable to (the phase does not vary enough to cause significant cancellation). From this we expect to be comparable in size to . If (10) held, we would then have

for all , which is only possible if , or equivalently .

Now we turn to the other necessary condition . Here one does not use scaling considerations; instead, it is more convenient to work with randomised examples. A useful tool in this regard is Khintchine’s inequality, which encodes the *square root cancellation* heuristic that a sum of numbers or functions with randomised signs (or phases) should have magnitude roughly comparable to the *square function* .

Lemma 4 (Khintchine’s inequality)Let , and let be independent random variables that each take the values with an equal probability of .

- (i) For any complex numbers , one has
- (ii) For any functions on a measure space , one has

*Proof:* We begin with (i). By taking real and imaginary parts we may assume without loss of generality that the are all real, then by normalisation it suffices to show the upper bound

for all , whenever are real numbers with .

When the upper and lower bounds follow by direct calculation (in fact we have equality in this case). By Hölder’s inequality, this yields the upper bound for and the lower bound for . To handle the remaining cases of (11) it is convenient to use the exponential moment method. Let be an arbitrary threshold, and consider the upper tail probability

For any , we see from Markov’s inequality that this quantity is less than or equal to

The expectation here can be computed to equal

By comparing power series we see that for any real , hence by the normalisation we see that

If we set we conclude that

since the random variable is symmetric around the origin, we conclude that

From the Fubini-Tonelli theorem we have

and this then gives the upper bound (11) for any . The claim (12) for then follows from this, Hölder’s inequality (applied in reverse), and the fact that (12) was already established for .

To prove (ii), observe from (i) that for every one has

integrating in and applying the Fubini-Tonelli theorem, we obtain the claim.

Exercise 5For any , let denote the root of the implied constant in (11), that is to saywhere the supremum is over all and all reals with .

- (i) If one analyzes the argument used above to prove (11) carefully, one obtains an upper bound for . How does this upper bound depend on asymptotically in the limit ?
- (ii) Establish (11) for the case of even integers by direct expansion of the left-hand side and some combinatorial calculation. This gives another upper bound on . How does this upper bound compare with that in (i) in the limit .
- (iii) Establish a matching lower bound (up to absolute constants) for the quantity in the limit .

Now we show that the estimate (10) fails in the large regime , even when . Here, the idea is to have “spread out” in physical space (in order to keep the norm low), and also having somewhat spread out in frequency space (in order to prevent the norm from dropping too much). We use the probabilistic method (constructing a random counterexample rather than a deterministic one) in order to exploit Khintchine’s inequality. Let be a non-zero bump function supported on (say) the unit ball , and consider a (random) function of the form

where are the random signs from Lemma 4, and are sufficiently separated points in (all we need for this construction is that for all ); thus is the random sum of bump functions adapted to disjoint balls . In particular, the summands here have disjoint supports and

(note that the signs have no effect on the magnitude of ). If (10) were true, this would give the (deterministic) bound

On the other hand, the Fourier transform of is

so by Khintchine’s inequality

The phases can be deleted, and is not identically zero, so one arrives at

Comparing this with (13) and sending , we obtain a contradiction if . This completes the proof of Proposition 3.

Exercise 6Find a deterministic construction that explains why the estimate (10) fails when and .

Exercise 7 (Marcinkiewicz-Zygmund theorem)Let be measure spaces, let , and suppose is a bounded linear operator with operator norm . Show thatfor any at most countable index set and any functions . Informally, this result asserts that if a linear operator is bounded from scalar-valued functions to scalar-valued functions, then it is automatically bounded from

vector-valuedfunctions to vector-valued functions. (By using gaussians instead of random sums, one can even obtain this bound with the implied constant equal to .)

Exercise 8Let be a bounded open subset of , and let . Show that holds if and only if and . (Note: in order to use either the scale invariance argument or the dimensional analysis argument to get the condition , one should replace with something like a ball of some radius , and allow the estimates to depend on .)

Now we study the restriction problem for two model hypersurfaces:

- (i) The
*paraboloid*equipped with the measure induced from Lebesgue measure in the horizontal variables , thus

(note this is

*not*the same as surface measure on , although it is mutually absolutely continuous with this measure). - (ii) The
*sphere*.

These two hypersurfaces differ from each other in one important respect: the paraboloid is non-compact, while the sphere is compact. Aside from that, though, they behave very similarly; they are both quadric hypersurfaces with everywhere positive curvature. Furthermore, they are also very highly symmetric surfaces. The sphere of course enjoys the rotation symmetry under the orthogonal group . At first glance the paraboloid only enjoys symmetry under the smaller orthogonal group that rotates the variable (leaving the final coordinate unchanged), but it also has a family of Galilean symmetries

for any , which preserves (and also can be seen to preserve the measure , since the horizontal variable is simply translated by ). Furthermore, the paraboloid also enjoys a *parabolic scaling symmetry*

for any , for which the sphere does not have an exact analogue (though morally Taylor expansion suggests that the sphere “behaves like” the paraboloid at small scales, or equivalently that certain parabolically rescaled copies of the sphere behave like the paraboloid in the limit). The following exercise exploits these symmetries:

- (i) Let be a non-empty open subset of , and let . Show that holds if and only if holds.
- (ii) Let be bounded non-empty open subsets of (endowed with the restriction of to ), and let . Show that holds if and only if holds.
- (iii) Suppose that are such that holds. Show that . (
Hint:Any of the three methods of scale invariance, dimensional analysis, or rescaled bump functions will work here.)- (iv) Suppose that are such that holds. Show that . (
Hint:The same three methods still work, but some will be easier to pull off than others.)- (v) Suppose that are such that holds for some bounded non-empty open subset of , and that . Conclude that holds.
- (vi) Suppose that are such that holds, and that . Conclude that holds.

Exercise 10 (No non-trivial restriction estimates for flat hypersurfaces)Let be an open non-empty subset of a hyperplane in , and let . Show that can only hold when .

To obtain a further necessary condition on the restriction estimates or holding, it is convenient to dualise the restriction estimate to an *extension estimate*.

Exercise 11 (Duality)Let be a Radon measure on , let , and let . Show that the following claims are equivalent:

This gives a further necessary condition as follows. Suppose for instance that holds; then by the above exercise, one has

for all . In particular, . However, we have the following stationary phase computation:

for all and some non-zero constants depending only on . Conclude that the estimate can only hold if .

Exercise 13Show that the estimate can only hold if . (Hint:one can explicitly test (15) when is a gaussian; the fact that gaussians are not, strictly speaking, compactly supported can be dealt with by a limiting argument.)

It is conjectured that the necessary conditions claimed above are sufficient. Namely, we have

Conjecture 14 (Restriction conjecture for the sphere)Let . Then we have whenever and .

Conjecture 15 (Restriction conjecture for the paraboloid)Let . Then we have whenever and .

It is also conjectured that Conjecture 14 holds if one replaces the sphere by any bounded open non-empty subset of the paraboloid .

The current status of these conjectures is that they are fully solved in the two-dimensional case (as we will see later in these notes) and partially resolved in higher dimensions. For instance, in one of the strongest results currently is due to Hong Wang, who established for a bounded open non-empty subset of when (conjecturally this should hold for all ); for higher dimensions see this paper of Hickman and Rogers for the most recent results.

We close this section with an important connection between the restriction conjecture and another conjecture known as the *Kakeya maximal function conjecture*. To describe this connection, we first give an alternate derivation of the necessary condition in Conjecture 14, using a basic example known as the *Knapp example* (as described for instance in this article of Strichartz).

Let be a spherical cap in of some small radius , thus for some . Let be a bump function adapted to this cap, say where is a fixed non-zero bump function supported on . We refer to as a *Knapp example* at frequency (and spatially centred at the origin). The cap (or any slightly smaller version of ) has surface measure , thus

for any . We then apply the extension operator to :

The integrand is only non-vanishing if ; since also from the cosine rule we have

we also have . Thus, if lies in the tube

for a sufficiently small absolute constant , then the phase has real part . If we set to be non-negative and not identically zero, and factor out the constant phase from the integral in (16), we conclude that

for . Since the tube has dimensions , its volume is

and thus

for any . By Exercise 11, we thus see that if the estimate holds, then

for all small ; sending to zero, we conclude that

or equivalently that , recovering the second necessary condition in Conjecture 14.

- (i) By considering a random superposition of Knapp examples located at different frequencies , and using Khintchine’s inequality, recover (a weak form of) the first necessary condition of Conjecture 14. (It is possible to recover the strong form of this condition by using the observation of Besicovitch that there exist Kakeya sets of arbitrarily small measure; we leave this as a challenge for the interested reader.)
- (ii) Suppose that holds for some . Establish the estimate
whenever is a collection of tubes – that is to say, sets of the form

whose directions are -separated (thus for any two distinct ), and the are arbitrary real numbers.

- (iii) Establish claims (i) and (ii) with the sphere replaced by a bounded non-empty open subset of the paraboloid .

Using this exercise, we can show that restriction estimates imply assertions about the dimension of Kakeya sets (also known as *Besicovitch sets*.

Exercise 17 (Restriction implies Kakeya)Assume that either Conjecture 14 or Conjecture 15 holds. Define aKakeya setto be a compact subset of that contains a unit line segment in every direction (thus for every , there exists a line segment for some that is contained in ). Show that for any , the -neighbourhood of has Lebesgue measure for any . (This is equivalent to the assertion that has Minkowski dimension .) It is also possible to show that the restriction conjecture implies that all Kakeya sets have Hausdorff dimension , but this is trickier; see this paper of Bourgain. (This can be viewed as a challenge problem for those students who are familiar with the concept of Hausdorff dimension.)

The *Kakeya conjecture* asserts that all Kakeya sets in have Minkowski and Hausdorff dimension equal to . As with the restriction conjecture, this is known to be true in two dimensions (as was first proven by Davies), but only partial results are known in higher dimensions. For instance, in three dimensions, Kakeya sets are known to have (upper) Minkowski dimension at least for some absolute constant (a result of Katz, Laba, and myself), and also more recently for Hausdorff dimension (a result of Katz and Zahl). For the latest results in higher dimensions, see these papers of Hickman-Rogers-Zhang and Zahl.

Much of the modern progress on the restriction conjecture has come from trying to reverse the implication in Exercise 17, and use known partial results towards the Kakeya conjecture (or its relatives) to obtain restriction estimates. We will not give the latest arguments in this direction here, but give an illustrative example (in the multilinear setting) at the end of this set of notes.

** — 2. theory — **

One of the best understood cases of the restriction conjecture is the case. Note that Conjecture 14 asserts that holds whenever , and Conjecture 15 asserts that holds when . Theorem 1 already gave a partial result in this direction. Now we establish the full range of the restriction conjecture, due to Tomas and Stein:

Theorem 18 (Tomas-Stein restriction theorem)Let . Then holds for all , and holds for .

The exponent is sometimes referred to in the literature as the *Tomas-Stein exponent*; though the dual exponent is also referred to by this name.

We first establish the restriction estimate in the non-endpoint case by an interpolation method. Fix . By the identity (5) and Hölder’s inequality, it suffices to establish the inequality

We use the standard technique of *dyadic decomposition*. Let be a bump function supported on that equals on . Then one has the telescoping series

where is a bump function supported on the annulus . We can then decompose the convolution kernel as

so by the triangle inequality it will suffice to establish the bounds

for all and some constant depending only on .

The function is smooth and compactly supported, so (18) is immediate from Young’s inequality (note that when ). So it remains to prove (19). Firstly, we recall from (7) (or (8)) that the kernel is of magnitude . Thus by Young’s inequality we have

We now complement this with an estimate. The Fourier transform of can be computed as

for any , and hence by the triangle inequality and the rapid decay of the Schwartz function we have

By dyadic decomposition we then have

From elementary geometry we have

(basically because the sphere is -dimensional), and then on summing the geometric series we conclude that

From Plancherel’s theorem we conclude that

Applying either the Marcinkiewicz interpolation theorem (or the Riesz-Thorin interpolation theorem) to (20) and (21), we conclude (after some arithmetic) the required estimate (19) with

which is indeed positive when .

At the endpoint the above argument does not quite work; we obtain a decent bound for each dyadic component of , but then we have trouble getting a good bound for the sum. The original argument of Stein got around this problem by using complex interpolation instead of dyadic decomposition, embedding in an analytic family of functions. We present here another approach, which is now popular in PDE applications; the basic inputs (namely, an to estimate similar to (20), an estimate similar to (21), and an interpolation) are the same, but we employ the additional tool of Hardy-Littlewood-Sobolev fractional integration to recover the endpoint.

We turn to the details. Set . We write as and parameterise the frequency variable by with , thus for instance . We split the spatial variable similarly. (One can think of as a “time” variable that we will give a privileged role in the physical domain , with being the dual time-frequency variable.) Let be a non-negative bump function localised to a small neighbourhood of the north pole of . By Exercise 9 it will suffice to show that

for . Squaring as before, it suffices to show that

For each , let denote the function , and let denote the function

Then we have

where on the right-hand side the convolution is now over rather than . By the Fubini-Tonelli theorem and Minkowski’s inequality, we thus have

From Exercise 12 (and the trivial bounds to treat the case when ) we have the bounds

leading to the *dispersive estimate*

for any (the claim is vacuous when vanishes). On the other hand, the -dimensional Fourier transform of can be computed as

which is bounded by , hence by Plancherel we have the *energy estimate*

Interpolating, we conclude after some arithmetic that

Applying the one-dimensional Hardy-Littlewood-Sobolev inequality we conclude (after some more arithmetic) that

and the claim follows.

This latter argument can be adapted for the paraboloid, which in turn leads to some very useful estimates for the Schrödinger equation:

Exercise 19 (Strichartz estimates for the Schrödinger equation)Let .

- (i) By modifying the above arguments, establish the restriction estimate .
- (ii) Let , and let denote the function
(This is the solution to the Schrödinger equation with initial data .) Establish the

Strichartz estimate- (iii) More generally, with the hypotheses as in (ii), establish the bound
whenever are exponents obeying the scaling condition . (The endpoint case of this estimate is also available when , using a more sophisticated interpolation argument; see this paper of Keel and myself.)

The Strichartz estimates in the above exercise were for the linear Schrödinger equation, but Strichartz estimates can also be established by the same method (namely, interpolating between energy and dispersive estimates) for other linear dispersive equations, such as the linear wave equation . Such Strichartz estimates are a fundamental tool in the modern analysis of *nonlinear* dispersive equations, as they often allow one to view such nonlinear equations as perturbations of linear ones. The topic is too vast to survey in these notes, but see for instance my monograph on this topic.

** — 3. Bilinear estimates — **

A restriction estimate such as

are *linear* estimates, asserting the boundedness of either a restriction operator (where denotes the support of ) or an extension operator . In the last ten or twenty years, it has been realised that one should also consider *bilinear* or *multilinear* versions of the extension estimate, both as stepping stones towards making progress on the linear estimate, and also as being of independent interest and application.

In this section we will show how the consideration of bilinear extension estimates can be used to resolve the restriction conjecture for the circle (i.e., the case of Conjecture 14):

Theorem 20 (Restriction conjecture for )One has whenever and .

Note from Exercise 9(vi) that this theorem also implies the case of Conjecture 15. This case of the restriction conjecture was first established by Zygmund; Zygmund’s proof is shorter than the one given here (relying on the Hausdorff-Young inequality (4)), but the arguments here have broader applicability, in particular they are also useful in higher-dimensional settings.

To prove this conjecture, it suffices to verify it at the endpoint , since from Hölder’s inequality the norm is essentially non-decreasing in , where is arclength measure on . By Exercise 9, we may replace here by (say) the first quadrant of the circle, where is the map ; we let be the arc length measure on that quadrant. (This reduction is technically convenient to avoid having to deal with antipodal points with parallel tangents a little later in the argument.)

By (22) and relabeling, it suffices to show that

whenever , , and (we drop the requirement that is smooth, in order to apply rough cutoffs shortly), and arcs such as are always understood to be endowed with arclength measure.

We now bilinearise this estimate. It is clear that the estimate (23) is equivalent to

for any , since (24) follows from (23) and Hölder’s inequality, and (23) follows from (24) by setting .

Right now, the two functions and are both allowed to occupy the entirety of the arc . However, one can get better estimates if one separates the functions to lie in *transverse* sub-arcs of (where by “transverse” we mean that there is some non-zero separation between the normal vectors of and the normal vectors of . The key estimate is

Proposition 21 (Bilinear estimate)Let be subintervals of such that . Then we havefor an , where denotes the arclength measure on .

*Proof:* To avoid some very minor technicalities involving convolutions of measures, let us approximate the arclength measures . Observe that we have

in the sense of distributions, where is the annular region

Thus we have the pointwise bound

and

and similarly for . Hence by monotone convergence it suffices to show that

for sufficiently small . By Plancherel’s theorem, it thus suffices to show that

for , if is sufficiently small. From Young’s inequality one has

so by interpolation it suffices to show that

But this follows from the pointwise bound

for sufficiently small , whose proof we leave as an exercise.

Exercise 22Establish (25).

Remark 23Higher-dimensional bilinear estimates, involving more complicated manifolds than arcs, play an important role in the modern theory of nonlinear dispersive equations, especially when combined with the formalism of dispersive variants of Sobolev spaces known asspaces, introduced by Bourgain (and independently by Klainerman-Machedon). See for instance this book of mine for further discussion.

From the triangle inequality we have

so by complex interpolation (which works perfectly well for bilinear operators; see for instance the appendix to Muscalu-Schlag) we have

for any . The estimate (26) begins to look rather similar to (24), and we can deduce (24) from (26) as follows. Firstly, it is convenient to use Marcinkiewicz interpolation (using the fact that we have an open range of , and using the Lorentz space form of this theorem due to Hunt) to reduce (23) to proving a restricted strong-type estimate

for any measurable subset of the circle, so to prove (24) it suffices to show that

We can view the expression as a two-dimensional integral

We now perform a Whitney decomposition away from the diagonal of the square to decompose it as rectangles of the form for which Proposition 21 applies. There are many ways to perform this decomposition; here is one such. Define a *dyadic interval* in to be a subinterval of of the form ; then each dyadic interval other than the full interval is contained in a unique parent interval of twice the length. Let us say that two dyadic intervals are *close*, and write , if they are the same length and are disjoint, but whose parents are not disjoint. Observe that if then and almost every pair lies in exactly one pair of the form with . Therefore we can decompose (42) as the sum of

where range over pairs of close dyadic intervals, leading to the decomposition

We perform a triangle inequality based on the length of the dyadic intervals, thus it will suffice to show that

One could apply the triangle inequality a second time to pull out the sum on the left-hand side, but this turns out to be too inefficient for certain ranges of . Instead we need to exploit some further orthogonality of the factors :

Exercise 24Let .

- (i) Show that for each pair with and , the function has Fourier transform supported in a rectangle in with the property that the dilates of these rectangles (the rectangle of twice the sidelengths of but the same center) have bounded overlap (any point in is contained in of these dilates ).
- (ii) Establish the bound
for all and all functions whose Fourier transform is supported in . (

Hint:one needs to apply an interpolation theorem, but one has to take care to make sure the interpolation is rigorous.)

Using the above exercise, we can bound the left-hand side of (28) by

Applying (26), this is bounded by

since , this simplifies to

Bounding and noting that each is close to only intervals and vice versa, we can bound this by

Since and , we can bound this by

and a routine calculation (dividing into the regimes and and using the hypothesis ) shows that the right-hand side is as required. This completes the proof of Theorem 20.

Exercise 25 (Bilinear restriction for paraboloid implies linear restriction)It is a known theorem (first conjectured by Klainerman and Machedon) that one has the bilinear restriction theoremwhenever , , disjoint compact subsets of , and functions , where denotes the measure given by the integral

(The range is known to be sharp for (29) except possibly for the endpoint , which remains open currently.) Assuming this result, show that Conjecture 15 holds for all . (

Hint:one repeats the above arguments, but at one point one will be faced with estimating a bilinear expression involving two “close” regions , which could be very large or very small. The hypothesis (29) does not specify how the implied constants depend on the size or location of , but one can obtain such a dependence by exploiting the parabolic rescaling and Galilean symmetries of the paraboloid.)

** — 4. Multilinear estimates — **

We now turn to multilinear (or more precisely, -linear) Kakeya and restriction estimates, where we happen to have nearly optimal estimates. For instance, we have the following estimate (cf. (17)), first established by Bennett, Carbery, and myself:

Theorem 26 (Multilinear Kakeya estimate)Let , let be sufficiently small, and let . Suppose that are collections of tubes such that each tube in is oriented within of the basis vector . Then we have

Exercise 27Assuming Theorem 26, obtain an estimate for for any in terms of and , and use examples to show that this estimate is optimal in the sense that the exponents for and can only be improved by epsilon factors at best, and that no such estimate is available for .

In the two-dimensional case the estimate is easily established with no epsilon loss. Indeed, in this case we can expand the left-hand side of (30) as

But if is a rectangle oriented near , and is a -rectangle oriented near , then is comparable with , and the claim follows.

The epsilon loss was removed in general dimension by Guth, using the polynomial method. We will not give that argument here, but instead give a simpler proof of Theorem 26, also due to Guth, and based primarily on the method of *induction on scales*. We first treat the case when , that is when all the tubes in each family are completely parallel:

Exercise 28

- (i) (Loomis-Whitney inequality) Let , and for each , let be the linear projection . Establish the inequality
for all . (

Hint:induct on and use Hölder’s inequality and the Fubini-Tonelli theorem.)- (ii) Establish Theorem 26 in the case .

Now we turn to the case of positive . Fix . For any , let denote the best constant in the inequality

whenever are collections of tubes such that each tube in is oriented within of the basis vector . Note that each tube can be covered by tubes whose direction is *exactly* equal to . From this we obtain the crude bound

In particular, is finite, and we have

whenever . Our objective is to show that whenever is sufficiently small, , and . The bound (32) establishes the claim when is large; the strategy is now to use the *induction on scales* method to push this “base case” bound down to smaller values of . The key estimate is

*Proof:* For each , let be a collection of tubes oriented within of . Our objective is to show that

Let be a small constant depending only on . We partition into cubes of sidelength , then the left-hand side of (33) can be decomposed as

Clearly we can restrict the inner sum to those tubes that actually intersect . For small enough, the intersection of with is contained in a tube oriented within of ; such a tube can be viewed as a rescaling by of a tube, also oriented within of . From (31) and rescaling we conclude that

Now let be the tube with the same central axis and center of mass as . For small enough, if then equals on all of , and hence

Combining all these estimates, we can bound the left-hand side of (33) by

But by (31) and rescaling we have

and the claim follows.

We remark that the same argument also gives the more general estimate whenever .

Now let , and let be sufficiently large depending on . If is sufficiently small depending on , then from (32) we have the claim

whenever . On the other hand, from Proposition 29 we see (for large enough) that if (34) holds in some range with then it also holds in the larger range . By induction we then have (34) for all . Combining this with (32), we have shown that

for all , whenever is sufficiently small depending on . This is *almost* what we need to prove Theorem 26, except that we are requiring to be small depending on as well as , whereas Theorem 26 only requires to be sufficiently small depending on and not . We can overcome this (at the cost of worsening the implied constants by an -dependent factor) by the triangle inequality and exploiting affine invariance (somewhat in the spirit of Exercise 9). Namely, suppose that and is only assumed to be small depending on but not on . By what we have previously established, we have

whenever the tubes lie within of , where is a quantity that is sufficiently small depending on . Now we apply a linear transformation to both sides, and also modify slightly, and conclude that for any within of , we still have the bound (35) if the are assumed to lie within (say) of instead of . On the other hand, by compactness (or more precisely, total boundedness), we can find directions that lie within of , such that any other direction that lies within of lies within of one of the . Applying the (quasi-)triangle inequality for , we conclude that

whenever the direction of are merely assumed to lie within of . This concludes the proof of Theorem 26.

Exercise 30By optimising the parameters in the above argument, refine the estimate in Theorem 26 slightly tofor any .

We can use the multilinear Kakeya estimate to prove a multilinear restriction (or more precisely, multilinear extension) estimate:

Theorem 31 (Multilinear restriction estimate)Let , let be sufficiently small, and let . Suppose that are open subsets of that lie within of the basis vector . Then we have

Exercise 32By modifying the arguments used to prove Exercise 16(ii), show that Theorem 31 implies Theorem 26.

Exercise 33Assuming Theorem 31, obtain for each and an estimate of the formwhenever , , and and some exponent , and use examples to show that the exponent you obtain is best possible.

Remark 34In the case, this result with no epsilon loss follows from Proposition 21. It is an open question whether the epsilon can be removed in higher dimensions; see this recent paper of mine for some progress in this direction.

To prove Theorem 31, we again turn to induction on scales; the argument here is a corrected version of one from this paper of Bennett, Carbery and myself, which first appeared in this paper of Bennett. Fix , and let be sufficiently small. For technical reasons it is convenient to replace the subsets of the sphere by annuli. More precisely, for each , let denote the best constant in the inequality

whenever , where is the annular cap

The extra factor of in the upper bound for is technically convenient, as it creates a certain “margin” of room in the nesting that will be useful later on when we perform spatial cutoffs that blur the Fourier support slightly.

Because we have restricted both the Fourier and spatial domains to be compactly supported it is clear that is finite for each , thus

Exercise 35Show that (40) implies Theorem 31. (Hint:starting with , multiply by a suitable weight function that is large on and has Fourier transform supported on , and write this as for a suitable . Then obtain estimates on .)

To establish (40), the key estimate is

Proposition 36 (Induction on scales)For any , one haswhere the multilinear Kakeya constant was defined in (31).

Suppose we can establish this claim. Applying Theorem 26, we conclude that if for a sufficiently large depending on , one has

From (39) one has

for all and some , and then an easy induction then shows that (41) holds for all , giving the claim.

It remains to prove Proposition 36. We will rely on the *wave packet decomposition*. Informally, this decomposes into a sum of “wave packets” that is approximately of the form

where ranges over -tubes in oriented in various directions oriented near , and the coefficients obey an type estimate

(This decomposition is inaccurate in a number of technical ways, for instance the sharp cutoff should be replaced by something smoother, but we ignore these issues for the sake of this informal discussion.) Heuristically speaking, (42) is asserting that behaves like a superposition of various (translated) Knapp examples (16) with .

Let us informally indicate why we would expect the wave packet decomposition to hold, and then why it should imply something like Proposition 36. Geometrically, the annular cap behaves like the union of essentially disjoint -disks , each centred at some point on the unit sphere that is close to , and oriented normal to the direction . Thus should behave like the sum of the components . By the uncertainty principle, each such component should behave like a constant multiple of the plane wave on each translate of the dual region to , which is a tube oriented in the direction . By Plancherel’s theorem, the total norm of should equal . Thus we expect to have a decomposition roughly of the form

where is a collection of parallel and boundedly overlapping tubes oriented in the direction , and the are coefficients with

Summing over and collecting powers of , we (heuristically) obtain the wave packet decomposition (42) with bound (43).

Now we informally explain why the decomposition (42) (and attendant bound (43)) should yield Proposition 36. Our task is to show that

for . We may as well normalise . Applying the wave packet decomposition, one expects to have an approximation of the form

and the are essentially distinct tubes oriented within of . We cover by balls of radius . On each such ball, the cutoffs are morally constant, and so

From the uncertainty principle, the trigonometric polynomial behaves on like the inverse Fourier transform of a function supported on with

and hence by (38) we expect the expression (46) to be bounded by

which is also morally

Averaging in , we thus expect the left-hand side of (44) to be

Applying a rescaled (and weighted) version of Theorem 26, this is bounded by

and the claim now follows from (45).

Now we begin the rigorous argument. We need to prove (44), and we normalise . By Fubini’s theorem we have

Let be a fixed Schwartz function that is bounded away from zero on and has Fourier transform supported on , thus the function is bounded away from zero on and has Fourier transform supported on . In particular we have

We can write

and observe that has Fourier transform supported in . Thus

Thus it remains to establish the bound

We cover by a collection of disks , each one centered at an element that lies within of , and is oriented with normal , with the separated from each other by . A partition of unity then lets us write where each with

The functions then have bounded overlapping supports in the sense that every is contained in at most of these supports. Hence

By Plancherel’s theorem the right-hand side is at most

This is morally bounded by

so one has morally bounded the left-hand side of (47) by

In practice, due to the rapid decay of , one has to add some additional terms involving some translates of the balls , but these can be handled by the same method as the one given below and we omit this technicality for brevity. We can write , where is a Schwartz function adapted to a slight dilate of whose inverse Fourier transform is a bump function adapted to a tube oriented along through the origin, and with

This gives a reproducing-type formula

which by Cauchy-Schwarz (or Jensen’s inequality) gives the pointwise bound

By enlarging slightly, we then have

for all , hence

We have thus bounded the left-hand side of (47) by

which we can rearrange as

Using a rescaled version of (31) (and viewing the convolution here as a limit of Riemann sums) we can bound this by

which by (49), (48) is bounded by

giving (47) as desired.

Exercise 37Show that Theorem 31 continues to hold (but now with implied constants depending on ) if the hypersurfaces are no longer assumed to lie on the sphere , but are compact smooth surfaces with boundary with the property that whenever is a unit normal to a point in for , then the wedge product has magnitude comparable to (that is to say, the determinant of the matrix with rows has magnitude comparable to ). In particular, the hypersurfaces are permitted to be flat; it is not the curvature of these surfaces that induces the multilinear restriction estimate, but rather the transversality between the surfaces.

Exercise 38Let the notation and hypotheses be as in Theorem 31, except that we now assume (in order to keep bounded away from zero).

For further discussion of the multilinear Kakeya and restriction theorems, see this survey of Bennett.

Remark 39In recent years, multilinear restriction estimates have been useful to establish linear ones (although the implication is still not perfectly efficient), starting with this paper of Bourgain and Guth.

## 189 comments

Comments feed for this article

14 April, 2020 at 5:48 am

AnonymousIn the 8th line below (16) there is a missing formula which “does not parse”.

[Corrected, thanks – T.]14 April, 2020 at 6:00 am

AnonymousInstead of the Schwartz class approach, can one define the Fourier transform on via the Riesz representation theorem?

14 April, 2020 at 7:56 am

Terence TaoIn principle yes, through the formal identity , but when and it is not immediately obvious that the inner product is absolutely convergent, unless one already has proven enough of Plancherel’s theorem that one can establish that is square-integrable. I would imagine that a purely Riesz representation theorem approach is going to be nontrivial, as it would also likely extend to other locally compact abelian groups, and the construction of the Fourier transform at this level of generality seems to require some machinery (e.g., Bochner’s theorem).

14 April, 2020 at 1:15 pm

AnonymousHow do you think about why the crude covering of each by tubes automatically gives the bound ?

14 April, 2020 at 3:38 pm

Terence TaoThis covering lets one dominate pointwise by where is a collection of tubes oriented parallel to with . If one then applies Exercise 28(ii) to these families of tubes, we obtain the claim.

15 April, 2020 at 2:01 pm

Lior SilbermanTypos in the proof of multilinear Kakeya (after equation (33)): you switch between tubes to tubes and -tubes (it should be the reverse in the latter cases). Also, when you count tubes meeting a cube using norms of characteristic functions, you get several times which I think should be

[Corrected, thanks – T.]16 April, 2020 at 1:04 am

dn1214I’m trying to prove inequality (25) (which you left as an exercise), I thought it would be easy but that it seems more involved. I guess that one has to use the separation hypothesis efficiently. Usually this is done by making a change of variable directly in the integral, just like in the proof of Bourgain’s bilinear estimates in the dispersive pde’s context. But here, you ask for a pointwise bound. However is not easy to compute explicitely. If I start by writing

then how can I bound this by the required ? I can see that there will be an coming from the integral in with brutal estimates. My guess is that the inner integral should be bounded by but I do not see how.

Can you give me a hint or some help?

16 April, 2020 at 7:42 am

Terence TaoThis problem is much easier to understand if you approach it geometrically rather than purely analytically. A convolution of two indicator functions evaluated at a point is nothing more than the area of the intersection of with a reflected translate of , so the main issue is to understand the shape and size of the intersection . Drawing pictures should convince you that this intersection (when it is non-empty and does not occur on the angular boundary of the annular sectors involved) resembles a parallelogram whose dimensions one can work out, and this should lead you to a heuristic bound of . One can then make this rigorous by proving that the intersection lies in an actual parallelogram (or other object whose area is easy to compute, such as a rectangle).

It may also be worth revisiting the proof of the change of variables formula in several dimensions to see exactly where the Jacobian term in that formula comes from. In particular the geometric fact that a determinant can be viewed as the area of a parallelogram or parallelopiped, when applied to certain infinitesimal approximate parallelopipeds, is the geometric essence of the change of variables formula, and that fact is also the essence of the computation here. (It is also possible to establish the desired claim from the change of variables formula and a duality argument, working primarily with the integration over the angular variables; this is also a good exercise for you to work out.)

More generally, this exercise is an illustration of the substantial gain in effectiveness one gains when one transitions from the “rigorous” approach to mathematics to the “post-rigorous” one, as I discuss in https://terrytao.wordpress.com/career-advice/theres-more-to-mathematics-than-rigour-and-proofs/

16 April, 2020 at 8:36 am

dn1214Thank you for this geometric insight, it is really helpful! I think that your geometrical proof can be turned into a rigorous proof (without change of variables), I’ll try to also write a proof with a change of variable as you suggested. It seems that I am still at the “rigorous” stage of my mathematical development.

16 April, 2020 at 11:38 am

AnonymousFor the interpretation of multilinear kakeya estimate, equation (30), you wrote , what does stands for?

16 April, 2020 at 12:59 pm

Terence Taowas an arbitrary natural number parameter (well, it shouldn’t exceed , but otherwise it is arbitrary) used to construct a certain testing example for (30) (in which each family of tube occupied a tube).

16 April, 2020 at 5:49 pm

hhy177On the interpolation in Ex 24(ii), if A denote the index set one can show boundedness from , , and . One concern is that the space in which to evaluate functions in , i.e. the in , is changing as well. How do one justify interpolation in this case? Thanks!

16 April, 2020 at 7:13 pm

Terence TaoThe proof of the Riesz-Thorin interpolation theorem can be adapted to this context.

16 April, 2020 at 10:41 pm

dn1214In Exercise 30 I’m not able to recover the but I have instead a , what am I missing?

Here is how I get it: let , then Proposition 29 tells us that . Now we already have the "good estimate" in the range so for one seeks for such that which is done by taking and then the estimate follows. However, proving the bound with the square root is more challenging, and I feel like the induction Proposition can not do better than that.

17 April, 2020 at 8:52 am

Terence TaoA bound of is in fact worse than the qualitative version of Theorem 26. One should stop using Proposition 29 at a smaller value of , e.g., , and use the arguments near (35) and (36) to conclude, rather than (32).

18 April, 2020 at 1:05 am

dn1214You are right my estimates was worse than what was obtained before. Thank you for the hint, it is greatly effective.

17 April, 2020 at 2:33 pm

AnonymousYou mentioned in class that rotation symmetry is not strong enough because it preserves angles, we should use symmetry. Doesn’t rotation symmetry also have the properties of symmetry?, i.e. determinant =1, … How does symmetry remove the angle preservation problem?

17 April, 2020 at 3:12 pm

Terence TaoMatrices in are not required to have determinant 1, and furthermore determinant 1 is not enough to preserve angles (it preserves volume instead). For instance the shear transformation in the plane is an element of of determinant 1, but does not preserve angles.

19 April, 2020 at 2:24 pm

AnonymousDear Prof. Tao:

Wonder if you can target the audience towards graduate students more. It seems that the only people ask questions during lectures are professional mathematicians who are familiar with the subject already. It is too fast for graduate students to grasp the material the first time. Or do you intend the course to be like an overview of the subject?

20 April, 2020 at 10:02 am

AnonymousProf. Tao:

Could you post some solutions to the HWs?

20 April, 2020 at 1:51 pm

Lior SilbermanThe fourth displayed equation after (46) should be because we have the squared -norm.

20 April, 2020 at 2:00 pm

Lior SilbermanSorry — this is wrong (I missed the point where you took the square root). However, three equations down (right after the mention of Theorem 26) you are missing the left parenthesis that matches the right parenthesis at the end of the equation (presumably it goes to the left of the sum over the ).

[Corrected, thanks – T.]20 April, 2020 at 3:12 pm

AnonymousFor the proof of multilinear restriction estimate, you mentioned again that the important thing is transversality, not curvature. The first time you introduced the concept of transversality is in Proposition 21. I am not sure I understand (or fully appreciate) the importance of transversality as you indicated. In equation (37), for flat surface, which makes (37) trivial. Back to Proposition 21, transversality comes in equation (25). if no transversality, which makes (25) obviously true. It seems that transversality becomes important because you set up the problem (Prop. 25) to make it important, not because it is intrinsically important. I feel that curvature and transversality are two sides of the same coin. ???

20 April, 2020 at 4:13 pm

Terence TaoThe parameter in (37) is not directly related to the curvature of the surfaces; (37) is indeed simpler to prove in the case of completely flat surfaces (it follows from the Loomis-Whitney inequality and Plancherel’s theorem, with no loss) but it is not entirely trivial. But the point is that the inequality (37) continues to hold for flat transverse hypersurfaces (Exercise 37), in contrast to linear restriction estimates that fail in the flat case (Exercise 10). In the linear theory curvature and transversality are essentially equivalent hypotheses (a surface is curved if different portions of it tend to be transversely arranged), but in the multilinear theory there is a distinction between the hypotheses (both flat and curved surfaces can be transverse to other flat or curved surfaces).

20 April, 2020 at 11:20 pm

AnonymousI have two questions. First, in Exercise 27, we are asked to obtain an estimate for all p>0, but in your paper given before Theorem 26, it seems to me that the estimates were proved for p>=1/(d-1) only. I was wondering how we can prove the estimate for p<1/(d-1), and whether there is a typo here.

In addition, in the estimate of L^2 norm of G_{i,x_0} (the first displayed equation after (46)), I was wondering whether the power of sqrt(R) should be d instead of d+1. In the third displayed formula after (46) (right after "which is also morally"), if we assume 1_{T_i}(x)\approx 1_{T_i}(x_0), then we get a factor sqrt(R)^{d(d-1)/2}, which doesn't seem to match the power of sqrt(R) in the formula right above this one.

Thank you in advance for your time and help!

21 April, 2020 at 7:20 am

Terence TaoI’ve edited the problem slightly: estimates are in fact available for all , with the range being deducible from the case by localising to balls or cubes of scale and using H\"older's inequality. There are counterexamples that show that no estimate is possible below .

The exponent of has now been fixed.

21 April, 2020 at 5:36 am

Alan ChangIn the two centered equations following “From the uncertainty principle, the trigonometric polynomial behaves…,” should the exponent be d instead of d+1?

This is what I get when I apply the uncertainty principle. Also, an exponent of d seems to agree with the centered equation following “which is also morally.”

[Corrected, thanks – T.]21 April, 2020 at 5:37 am

Alan ChangLaTeX typos: Starting with the centered equation after “This gives a reproducing-type formula,” there are three instances of “vee” instead of “\vee.”

[Corrected, thanks – T.]21 April, 2020 at 6:01 am

Alan ChangSome other typos: There are several instances of when I think you mean :

in Proposition 36

in line (44)

in line (47)

in the two centered equations following “Using a rescaled version of (31)”

[Corrected, thanks – T.]22 April, 2020 at 8:11 am

Alan ChangI believe the centered equation following “Applying a rescaled (and weighted) version of Theorem 26, this is bounded by” is missing two square roots. It should be:

[Corrected, thanks – T.]22 April, 2020 at 12:19 pm

Seungly OhIn Exercise 24, I am confused about the stated definition of “bounded overlap.” Is this sort of eluding to the almost disjoint (or finite overlap) in Fourier support for a product of two functions, like paraproducts?

Also, a little further down, you say that is controlled by because I is close to J. Could you explain why this is the case?

Thank you!

22 April, 2020 at 12:51 pm

Terence TaoA collection of sets is said to have overlap at most if any given point lies in at most of the . In particular, bounded overlap means that any given point lies in of the . In this particular case the (which are called in the exercise) have an interpretation as a container for the Fourier support of a certain product (with some room to spare), but this is not essential to the definition of bounded overlap.

The quantities and do not control each other. However, by symmetry and the fact that each is close to and vice versa, the sums and are both controlled by .

22 April, 2020 at 5:07 pm

AnonYour notation is defined in other articles on this blog, but it doesn’t appear to be defined in this one.

23 April, 2020 at 7:32 am

Terence TaoThis symbol (as well as the symbol) is not being assigned a precise interpretation; it is only used in the heuristic portions of the disucssion.

23 April, 2020 at 6:28 am

Xiao-Chuan LiuRegarding the statement of Theorem 26, should it be that the choice of \delta is arbitrarily small after one fixes the set of families, or at least the directions of tubes? Here I am a bit confused with the order of choices.

23 April, 2020 at 7:51 am

Terence TaoThe family of tubes will have to depend on , since each tube in these collections have thickness . The set of directions that these tubes point in is also permitted to depend on . The key point though is that the implied constant in the conclusion (30) is uniform in .

24 April, 2020 at 6:40 am

Jiqiang ZhengDear Prof. Tao:

From Exercise 9(vi), we know that the restriction estimate for the sphere implies the restriction estimate for the paraboloid. My question is “How about the reverse?” Is it possible to use the restriction estimate for the paraboloid to deduce the restriction estimate for the sphere? Thank you for your reply.

24 April, 2020 at 3:17 pm

Terence TaoCurrently there is no known satisfactory implication of this form. (The restriction conjecture for the paraboloid does imply the Kakeya conjecture, which can be used to improve slightly the known restriction estimates for the sphere, but at our current level of understanding we cannot recover the full restriction conjecture for the sphere in this fashion.)

6 May, 2020 at 11:44 am

Fourier multipliers: examples on the torus – Zeros and Ones[…] question what happens in higher dimensions when . Bourgain made some progress. See also these lecture notes on Euclidean space . Such type of estimates are called Fourier Restriction […]

7 May, 2020 at 6:40 pm

rajeshd007an interesting non computability theorem, https://www.sciencedirect.com/science/article/pii/S0021904519301121

14 May, 2020 at 7:26 am

247B, Notes 4: almost everywhere convergence of Fourier series | What's new[…] of two. By performing a suitable Whitney type decomposition (similar to that used in Section 3 of Notes 1), establish the pointwise […]

15 May, 2020 at 11:06 am

AnonymousThank you for this blog post.

I want to remark on the proof of Lemma 4 (Khintchine’s inequality).

I believe the trick with and is employed to obtain a sharper constant in the inequality. Since the constant is hidden, a simpler exposition is possible: Estimate

The p-norm should still be estimable. I am remarking this because that trick always confused me, and only now I realized it is probably not necessary.

15 May, 2020 at 11:48 am

AnonymousTo reply to my own comment, I now realize the constant is looked at in Exercise 5.

I also solved the confusion with and : If you follow the method I posted, you put “all” of in the denominator (of Chebyshev inequality right hand side); you may elect to put it in the numerator instead, or some power in the numerator and its complementary in the denominator. The optimization in the power results in the analysis that is presented in this blog.

18 May, 2020 at 10:49 am

Anonymous“we first give an alternate derivation of the necessary condition in Conjecture 14″

I think this should be , the signs are reversed.

[Corrected, thanks – T.]21 May, 2020 at 3:19 pm

Anonymous“we have the bounds”

should be

[I am using the notational convention that if (thus the absolute values are automatically applied) -T.]and

” Applying the one-dimensional Hardy-Littlewood-Sobolev inequality we conclude (after some more arithmetic) that”

should have a , so

[Corrected, thanks – T.]24 May, 2020 at 12:04 pm

AnonymousA question, I am grateful for help:

I am trying to follow the reduction of

into that of

I took a look at the Marcinkiewicz theorem for Lorentz spaces; I have not worked with Lorentz spaces before. Still, I don’t understand how to do that reduction.

24 May, 2020 at 8:32 pm

AnonymousI think I figured it out, but not sure yet. I will post tomorrow with my findings. One thing I missed is that there was a relabeling of what and were.

25 May, 2020 at 8:05 am

AnonymousHere it is: a relabeling of given in the blog text gives and . As mentioned, may be taken since is a finite measure, thus . It is then immediate that (23) implies the “restricted strong-type estimate”.

Then to establish (23) for a given pair , we establish it for two pairs for characteristic functions (i.e. the restricted strong-type estimate) where and use Marcinkiewicz interpolation. The openness simply allows us to always find such .

This "distributional inequality" is also mentioned in Bourgain's paper ( Geom. Funct. Anal. 1 (1991), see (1.21) there) which is linked to in Exercise 17. I didn't understand it at the time I encountered it.

28 May, 2020 at 6:41 pm

Xiao-Chuan LiuFor Exercise 13, I took a (d-1) gaussian g and try to compute following the hint. However, I only get the estimate of this function of (x’,x_d) with the order , which is not enough for me. I might not really understand how to use gaussian as hinted.

29 May, 2020 at 12:54 pm

Terence TaoThis is the right order of decay in the variable. Of course, to get necessary conditions we need lower bounds on rather than upper bounds, and good lower bounds will be available in the parabolic region for which will give the required claim.

31 May, 2020 at 11:19 am

Xiao-Chuan LiuFor the challenge problem in exercise 16(i), could you please give a hint? I suppose the endpoint exponent and try to repeat the Khintchine inequality with the superposition of Knapp’s examples with no success..

6 June, 2020 at 11:02 am

James LengI’m also a bit stuck on this. I tried applying a phase shift to rearrange these delta^{-1} \times \delta^{-2} tunes to be an approximate Kakeya set. Then I applied Holder’s inequality with that and the characteristic function of the approximate Kekaya set while varying the L^r on the left hand side (e.g. varying r so that 1 \le r \le 2d/(d + 1) or something similar to that). However, whatever benefit that the Kakeya set gives you gets cancelled out if you let p = r (since you work with the L^infty norm with the characteristic function) leading you to only get the condition p \le 2d/(d + 1). So I don’t think the Holder inequality approach I used is correct…

8 April, 2021 at 10:07 am

Giacomo Del NinSuperimpose Knapp examples, so that . After rescalings, the non-strict inequality boils down to the following: given -separated tubes of size , then , which forces . The rightmost inequality can be proved by Holder for any disposition of the tubes (the worst case is when they are disjoint). To prove that can not happen, it is sufficient to improve the rightmost inequality from to . To do this you have to phase-shift the tubes in an approximate Kakeya set (i.e. with total area ). Then the estimate follows again by Holder’s inequality (and because ).

2 June, 2020 at 12:37 am

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

3 June, 2020 at 2:03 pm

AnonymousSome minor corrections:

In (43), the sum should be over instead of and in the right hand side the space should be over instead of .

“and observe that has Fourier transform”, there it should be , the inverse Fourier transform is missing.

(48) can’t be an equality, only comparable, I think. It doesn’t affect the rest of the proof.

[Corrected, thanks. An equality can be obtained by defining each to be the restriction of to . -T]17 May, 2023 at 9:13 am

AnonymousBut if you define $G_i$ to be the restriction like that, it is no longer smooth.

[This is not a problem in practice, since one can smoothly approximate to arbitrary accuracy in various norms. -T.]29 August, 2020 at 1:32 pm

Strichartz estimates, nonendpoint and homogeneous – Mathssww's Math Blog[…] estimates for the Schordinger equations follow from the restriction estimates of paraboloid. See this note by […]

10 November, 2020 at 8:52 pm

luyufDear Tao: thanks a lot for the note. I have some problems with the proof of proposition 36. Firstly, why we can just control the cutoff function by the character function in

I try to control it by the sum of the character function of the ball with radius , but if so, the convolution of the character function of the tube below that equation may not with the same scale and cannot use the multilinear Kakeya estimate. Secondly, why can we see the convolution of the character function as the limit of Riemann sum so that to use the multilinear Kakeya.

14 November, 2020 at 5:35 pm

Terence TaoThis is only morally true, as the text indicates. In practice one has to instead replace the expression by a more complicated expression such as .

A convolution can be written as the limit of Riemann sums if say are Schwartz functions. (The indicator functions are not quite Schwartz functions, but one can approximate them by such functions without much difficulty.)

21 April, 2021 at 10:30 am

AnonymousKlainerman and Machedon’s conjecture was motivated by a nonlinear PDE kind of problem to which regular Strichartz estimates were not well suited. I wonder if the something similar happens in the $k$-linear world: in other words, if you have $k$ transversal caps on a paraboloid (or cone) and you manage to prove the correct $k$-linear restriction estimate, would that help studying regularity of some nonlinear PDE to which not even bilinear estimates are well suited?

26 April, 2021 at 8:34 am

Terence TaoIn principle one could concoct an artificial nonlinear PDE designed specifically to take advantage of such a multilinear estimate, but there is no a priori reason to expect that there is a “natural” PDE (e.g., one arising from physics or geometry) that would be able to benefit from these estimates. (I once worked with a wave maps equation with a trilinear nonlinearity that required a somewhat delicate trilinear estimate that did not follow immediately from bilinear estimates. One could say that this estimate was a sort of trilinear restriction estimate, but the function spaces involved were customised to the wave maps equation and were not exactly the standard type spaces appearing in the usual restriction theory.)

26 May, 2021 at 11:21 am

boningdiDear Prof. Tao,

In the proof of Prop. 36, more precisely in (48), you say that “cover by disks whose centers are at around , and the centers are separated from each other by ”. I’m a little confused about this. As far as I understand, due to the curvature of the sphere and , the disks should be if the centers are at . Here the disks with sizes are only able to cover rather than .

Besides, I don’t really understand why we need the conditiont “centers are separated from each other by ”. I wonder if the bounded overlapping condition (i.e. Heine-Borel Theorem) is enough here in this proof.

Thanks in advance for your help!

[This was a typo: I have corrected to . Bounded overlap would be a satisfactory substitute for separation if desired. -T]17 May, 2023 at 9:14 am

AnonymousI think the typo has not been corrected?

[Corrected (again), thanks – T.]1 June, 2021 at 6:48 pm

luyufDear Prof.Tao,

I have some questions about Hausdorff-Young’s inequality. First, we know that it is sharp for , my question is that if we know that for all function f which is supported on the unit cube, what is the sufficient and necessary condition for this inequality? I guess the result is ,and . By scaling, we can have , but I do not know how to prove .

Thanks for your time.

[Hint: experiment with random Fourier series using Khintchine’s inequality. -T]15 January, 2022 at 12:27 am

Boning DiDear Prof. Tao:

Thank you very much for this note! It’s very helpful. In Exercise 25, for the implied constant in bilinear restriction for the paraboloid, you mentioned that we could get the dependence of subsets and by parabolic rescaling and Galilean symmetries. And I do get this by following the hint there. But I wonder what about other surfaces such as ? In this case, can we still get a similar dependence of the distance ? I’m a little confused since there will be some extra terms when . Is this possible to obtain this dependence?

Thanks in advance for your reply!

15 January, 2022 at 12:32 am

Boning DiMaybe the example of surface should be which is better…

20 January, 2022 at 5:20 pm

Terence TaoFor surfaces without an exact parabolic rescaling, such as the surface you mention, one can still use rescaling to obtain quantitative dependence on as long as the base bilinear restriction estimate used (when are uniformly separated) holds uniformly for all pairs of surfaces obeying various smoothness and curvature bounds. See for instance Section 2 of my paper https://arxiv.org/abs/math/9807163 with Vargas and Vega for an example of this.

11 June, 2022 at 11:26 am

AnonymousIs there a higher dimensional version of Proposition 21 where the measures of the caps appear on the RHS instead of the measure of I?

11 June, 2022 at 3:52 pm

Terence TaoYes, there are similar estimates in any dimension (though with possibly different numerology regarding the dependence on the size of the cap), with similar proofs; among other things, they form the backbone of the bilinear estimates of Bourgain, Klainerman-Machedon, Kenig-Ponce-Vega, and others, which are of importance in nonlinear dispersive equations.

11 June, 2022 at 5:45 pm

AnonymousThanks! Can you point out some references in the literature?

[See for instance this paper of mine. – T.]20 May, 2023 at 6:42 pm

wbshuDear Prof. TAO:

If and then its fourier transform is defined in your formula (1), I wonder if can infer that . i.e., Do the plancherel inverse fourier transform of have the same L2 norm with ? If , this is trivial. The problem is that how about if we only know .

21 May, 2023 at 8:22 am

Terence TaoYes, this is a consequence of Plancherel’s theorem and the uniqueness of the Fourier transform (the latter can for instance be derived from the theory of the Fourier transform on (tempered) distributions).

21 May, 2023 at 6:31 pm

wbshuThanks very much.