This is the last part of a 3 part series on the basics of Littlewood-Paley theory. Today we discuss a couple of applications, that is Marcinkiewicz multiplier theorem and the boundedness of the spherical maximal function (the latter being an application of frequency decompositions in general, and not so much of square functions – though one appears, but only for estimates where one does not need the sophistication of Littlewood-Paley theory).

Part I: frequency projections

Part II: square functions

**7. Applications of Littlewood-Paley theory **

In this section we will present two applications of the Littlewood-Paley theory developed so far. You can find further applications in the exercises (see particularly Exercise 22 and Exercise 23).

** 7.1. Marcinkiewicz multipliers **

Given an function , one can define the operator given by

for all . The operator is called a *multiplier* and the function is called the *symbol* of the multiplier^{1}. Since , Plancherel’s theorem shows that is a linear operator bounded in ; its definition can then be extended to functions (which are dense in ). A natural question to ask is: for which values of in is the operator an bounded operator? When is bounded in a certain space, we say that it is an –*multiplier*.

The operator introduced in Section 1 of the first post in this series is an example of a multiplier, with symbol . It is the linear operator that satisfies the formal identity . We have seen that it cannot be a (euclidean) Calderón-Zygmund operator, and thus in particular it cannot be a Hörmander-Mikhlin multiplier. This can be seen more directly by the fact that any Hörmander-Mikhlin condition of the form is clearly incompatible with the rescaling invariance of the symbol , which satisfies for any . However, the derivatives of actually satisfy some other *superficially similar* conditions that are of interest to us. Indeed, letting for simplicity, we can see for example that . When we can therefore argue that , and similarly when ; this shows that for any with one has

This condition is comparable with the corresponding Hörmander-Mikhlin condition only when , and is vastly different otherwise, being of __product type__ (also notice that the inequality above *is* compatible with the rescaling invariance of , as it should be).

The Littlewood-Paley theory developed in these notes allows us to treat multipliers with symbols that satisfy product-type conditions like the above, and which typically are beyond the reach of Calderón-Zygmund theory. Indeed, we have the following general result, where the conditions assumed of the multiplier are a generalisation of the pointwise ones above.

Theorem 8 (Marcinkiewicz multiplier theorem)Let be a function of that is of class away from the hyperplanes where one of the coordinates is zero. Suppose that satisfies the following conditions:

- ;
- there is a constant such that for every and for every permutation of the set we have
where the are the dyadic Littlewood-Paley rectangles as in § 6 of the second post in this series.

Then the multiplier associated to the symbol satisfies for any the inequality

To be more specific, if and , the rectangle is given by .

A multiplier whose symbol satisfies the conditions of the above theorem is called a *Marcinkiewicz multiplier*. Condition 2., spelled out in words, says that a certain subset of the partial derivatives of (precisely, the derivatives where the components of the multi-index have values in and ) has the property that their integral over any -dimensional dyadic Littlewood-Paley rectangle is uniformly bounded, where the integration is happening in those variables such that .

Remark 1The conditions above are very general but also very cumbersome to spell out, as the above attempt demonstrates. The pointwise conditions, analogous to the one seen before, that imply condition 2. of the theorem are usually easier to check and are as follows:

- exactly as before;
- (pointwise) for any multi-index such that we have

See Exercise 14.

When the dimension is equal to , the statement of the theorem can be superficially generalised and reformulated in the following way, that perhaps clarifies the moral meaning of condition 2.

Theorem 9 (Marcinkiewicz multiplier theorem for )Let be a function of that is ofbounded variationon any interval not containing the origin. Suppose that satisfies the following conditions:

- ;
- there is a constant such that, with the total variation
^{2}of over the interval ,Then the multiplier with symbol satisfies for any the inequality

Condition 2. is now saying that the total variation of on any dyadic Littlewood-Paley interval is bounded. Observe that it is implied by the pointwise condition , but since we are in dimension now this stronger condition would just coincide with the Hörmander-Mikhlin one. There is in general a certain degree of overlap between the Marcinkiewicz multiplier theorem and the Hörmander-Mikhlin theorem in any dimension, but neither implies the other.

We will prove Theorem 9 here, and the proof we will give will already contain all the main ingredients needed for the proof of the full Theorem 8. We leave it to you to extend the proof to higher dimensions in Exercise 15.

*Proof:*

Let be the multiplier with symbol . Observe that by Theorem 5 of the second part of these notes we have

and thus it will suffice to prove that

by a second application of the boundedness of the Littlewood-Paley square function. Now, is a multiplier with symbol (we discard the negative frequencies for ease of notation), and by condition 2) we see that there exists a complex Borel measure^{3} such that for

By Fubini we see therefore that

where we recall that denotes frequency projection onto the interval . By condition 1), the first term in the right-hand side contributes at most to the left-hand side of (1) overall, and therefore we can safely discard it; let be the second term. We have by Cauchy-Schwarz and condition 2) that

and therefore we have reduced to control the square function

Now comes the tricky part: we want to realise the above expression as the norm in an abstract Hilbert space of a vector-valued frequency projection, as in Proposition 1 of the first part of this series. There are many ways of doing so, and the following is the one we choose. Let , that is the set of elements such that ; observe that a set has necessarily the form for a certain and sets . We define the measure on to be given by

provided the ‘s are -measurable. Thus the Hilbert space consists of those functions such that

is finite. Finally, to any we assign the interval . With these definitions, we see that (2) corresponds to

where in our particular case we can take because . By Proposition 1 of the first part of this series we have for generic vector-valued functions that ; applying this to we see that we have bounded by the norm of

but the integrand does not depend on and thus condition 2) shows that the above is bounded pointwise by . This concludes the proof of (1) and thus the proof of the theorem.

** 7.2. Boundedness of the spherical maximal function **

Recall that the boundedness of the spherical maximal function implies, through the method of rotations, that the Hardy-Littlewood maximal function is bounded for any with constant *independent* of the dimension. In the final part of this lecture we will finally prove the boundedness of (when ).

Theorem 10(Stein, ’76)Let . Then for any the inequality

holds for any .

The range of ‘s stated in the above theorem is sharp (prove it in Exercise 17).

A small caveat: the proof we will give below will yield a constant for the above inequality that depends on the dimension . It is only later – through the method of rotations – that one can remove this dependence on the dimension, showing that if one has a bound for dimension then one also has the bound for dimension .

Let me stress once again that the following is merely a slight re-elaboration of Tao’s excellent post on the subject.

*Proof:* It will suffice to prove the theorem for , since the rest of the exponents can be obtained by Marcinkiewicz interpolation with the trivial estimate. It will also suffice to assume that is a Schwartz function for convenience, as they are dense in .

We will prove the theorem first for a local version of , namely the operator

where denotes the average over the sphere of radius , that is

where is the normalised surface measure on the sphere. Then in Exercise 21 you will conclude the proof for the full case of with little extra effort.

We will use an annular frequency decomposition as the one in Section 6 of the second post in this series, but slightly modified to suit our purposes. We let be a smooth radial function compactly supported in the annulus and such that^{4} for any we have . We let for convenience. Thus, with being given by we have that for all functions we can decompose

Since the radius will be , we will not need to consider the projections to extremely low frequencies separately (see below for an explanation); therefore we define , so that . In view of the above decomposition, to conclude the boundedness of it will suffice by triangle inequality to prove

for some .

Now, the heuristic motivation behind such a decomposition is that is now roughly constant at scale . Indeed, one way to appreciate this informal principle (which is a manifestation of the Uncertainty Principle) is to show for example that is pointwise dominated by its average at scale : since is supported in the ball , we can take a smooth bump function such that is identically on this ball and vanishes outside and see that therefore . The latter means that and the smoothness of means that we have (say); therefore we have by expanding the convolution that

This observation suggests that should just be an average of at unit scale, since and is roughly constant at scales . Indeed, one can realise that is given by convolution with a Schwartz function and thus by Fubini ; it is not hard to show that being Schwartz and being implies bounds such as (or any other exponent really), and therefore we have as seen before

uniformly in , so that the pointwise domination continues to hold after we take the supremum. This immediately implies (3) for any (so even outside our desired range).

Repeating the argument for does not allow us to conclude. Indeed, you can see that in this case is essentially a function of magnitude concentrated in a shell of width and radius (see Exercise 20). A greedy argument, disregarding the width information, shows the rough bound , which in turn implies uniformly in . Therefore this argument gives at best the bound

which blows up as . However, one advantage of this bound is that it holds even for , which would allow for some interpolation if we had a decaying bound for some other exponent to compensate with. A particularly nice exponent is of course, because Plancherel's theorem is available, so we should explore what happens in this case. Consider fixed and observe that

where denotes the Fourier transform of , that is (verify the formula). It turns out that, due to the curvature of the sphere, has a certain amount of decay as gets large – in particular, one has the pointwise bound

We will take this for granted here (it is a well known fact) but if you want to see a proof, prove the decay yourself in Exercise 19. Now, combining the decay information above with the frequency localisation of , we see by Plancherel that, since ,

a bound nicely decaying in . However, this bound is not directly useful to us as we still have to take the supremum in ! To deal with this issue we will do something a bit unconventional: we will replace the supremum by the integral of the derivative in the parameter! Indeed, observe that the fact that is approximately constant at scale suggests that cannot change by much if we change by ; that is, we expect to be somewhat “smooth” in the parameter . We could take advantage of this heuristic as follows: since for a generic function we have by the Fundamental Theorem of Calculus that

we could replace the supremum by the right-hand side. However, if we proceeded with this approach we would get bad bounds in the end – the second term at the right-hand side would contribute much more than the other one. We need therefore some more sophisticated inequality which allows us to optimally balance the two contributions, and the following will do. Observe that, by the Fundamental Theorem of Calculus and Cauchy-Schwarz, if we have

and by the trivial inequality we have therefore

for any , which *crucially* we are free to choose, and which will therefore allow us to balance two distinct contributions equally. We take and estimate the terms on the right-hand side separately. The bound (6) shows that, since the expressions commute,

(and similarly for ). For the other term, we need to evaluate (the term disappears), and this is easy on the Fourier side, giving

The gradient has the same decay as , namely (and this can be proven in the same way). Combining this decay information with the frequency localisation of we obtain by Plancherel’s theorem that

Putting all this information together, we see that inequality (7) gives us a bound of for , which is optimised if we choose , resulting in

(observe that when this expression gives no decay at all, thus explaining why the proof does not work in that case). For a given we can then interpolate between estimates (5) and (8) to obtain a constant for the norm of that is at most for any small (do the calculation; you will need to interpolate between the exponent and an exponent extremely close to but not – hence the ). The expression is only positive when , and therefore in this regime we have that (4) holds for some , allowing us to sum in and thus to conclude that is bounded.

** Exercises: **

Exercise 13Assume that function satisfies and that it has bounded variation over all of (that is ). Show,without appealing to the Marcinkiewicz multiplier theorem, that defines a multiplier which is bounded for all .

[hint: simplify the proof of Theorem 9 as much as you can.]

Exercise 14Show that the pointwise condition 2 in Remark 1 implies condition 2 in Theorem 8.

Exercise 15In this exercise you will prove Theorem 8 in dimensions . Actually, the notation required becomes nightmarish pretty quickly, and thus we will content ourselves with proving the case , since it already contains the full generality of the argument. Let and satisfy the conditions stated in the theorem. The proof is essentially a repetition of the argument given for Theorem 9.

- Show that, by Theorem 7 of the second part of these notes, it will suffice to show
under the assumption that is supported in the first quadrant .

- Show that in each rectangle we can write

- Show that by ii) and Fubini we have

- Use condition 1. of the theorem to dispense with the first term at the right-hand side of the last expression.
- Use condition 2. and Cauchy-Schwarz to show that the (square of) the remaining terms is bounded by

- Find measure spaces and collections of rectangles or intervals for such that each of the terms above can be treated by Proposition 1 of the first part of this series, as in the proof of the case.
- Conclude using condition 2. one last time.

Exercise 16Show that the multiplier given by symbol

where and , is a Marcinkiewicz multiplier.

Exercise 17Show that the range for the boundedness of the spherical maximal function is sharp, in the sense that is not bounded on for any (find a counterexample).

Exercise 18Let the dimension be and let denote the spherical average , where is the normalised surface measure on . Show that solves the wave equationwith , ; here we assume . This is an instance of

Huygens’ principle, and more in general if the initial data becomes with , the complete solution is . Similar formulas hold in higher dimensions, but only when is odd they involve only the spherical averages . When is even, one needs to average over balls instead (that is, the classical Huygens’ principle fails in even dimensions).

- Using Stokes theorem, show that
- Using polar coordinates, show that
- Use i)-ii) to show that
- Show that for any twice differentiable it holds that , and combine this with iii) to show that the wave equation is satisfied.
- It remains to check the initial conditions. Using Theorem 10 argue that for every , and then use this fact together with i) to argue that the initial conditions are indeed satisfied in the limit (and are continuous in ).

Exercise 19Let be a smooth function compactly supported in the ball of . In this exercise you will show thatfor any . It is a matter of doing a smooth partition of unity on and a change of variables with a suitable diffeomorphism to show that the above implies ; however, we will content ourselves with the object above (you can see it as the Fourier transform of a measure supported on the elliptic paraboloid parametrised by , instead of the sphere; but the two surfaces are locally the same).

- Square the left-hand side of (9) and show by a change of variables that the result equals
for some smooth function compactly supported in .

- You will show (9) in two complementary cases. First let . Show that
where denotes the Fourier transform in the second variable. Argue that for any arbitrarily large and conclude that this shows , thus proving (9) in this regime.

- Let now . Show that
next argue that for any and show that this proves (9) in this case as well (even with arbitrarily large exponents).

Exercise 20Let be as in the proof of Theorem 10. Show that, thanks to the fact that for any , one has the rough bound

for any and any .

Exercise 21In this exercise you will finish the proof of Theorem 10. Here you will need to use a smooth square function at some point, but other than that the proof is just a repetition of the one given for .

- We see as
For a fixed let be the operator that will take on the rôle of , so that

Show that it suffices to prove for any

- Show that for one has
uniformly in and conclude (10). You can simply rescale the argument given for the analogous part of , that is for .

- Show that for and one has
uniformly in and (again, just rescale the argument used when ). Conclude that

- Show, using the decay of , that when
- Show, using the decay of , that when
- Show, using a suitably rescaled inequality (7), that for any
- Let be a smooth function compactly supported in the annulus and identically equal to in the annulus ; finally, let be the annular projection given by (thus they are just a slight modification of the projections in Section 6 of the second post). Show that in (13) we can replace in the right-hand side by .
- Replace the supremum in by the sum and show that, by Proposition 6 of the second part in this series, we have
- Interpolate between (12) and (14) to show that (11) holds. This concludes the proof.

Exercise 22The Carleson operator for is given byrecall that its boundedness for a certain implies the a.e. convergence of the integral above to as when . The methods developed in these notes are not powerful enough to show the boundedness of , but they are enough for a simplified version of it. Indeed, in this exercise you will prove the boundedness of the

lacunary Carleson operatorFix such that .

- First, consider the additional simplification given by replacing the characteristic function by the smooth function where is a non-negative smooth function compactly supported in and identically equal to on . Show that the associated maximal operator
is bounded pointwise by a constant multiple of the Hardy-Littlewood maximal function and conclude boundedness of .

- Show that it suffices to prove the boundedness of the operator
to conclude that of .

- Show that there is a smooth non-negative function compactly supported in and such that
- Let denote the frequency projections associated to , that is . Show that it suffices to show that the square function
is bounded to conclude the same for .

- Show that is indeed bounded.
- Argue that the argument generalises to treat the operators
where is any fixed lacunary sequence – that is, a sequence that satisfies . However, the constant produced by the proof will end up depending on .

Exercise 23There is a known application of Riesz transforms that provides important motivation for singular integrals. The application is the following: the Riesz transforms are defined by , and working on the Fourier transform side one can verify that

(modulus some constants); thanks to the above identity and the boundedness of the Riesz transforms, one can therefore control the norms of all the partial derivatives of order by the Laplacian alone: for any

With a little spherical harmonics theory, this can be generalised to generic homogeneous elliptic differential operators, always within the framework of Calderón-Zygmund theory. More precisely, let be a homogeneous elliptic polynomial of degree , that is

- for all ;
- the zero set of consists only of the origin.
If , the homogeneous elliptic differential operator is defined as

Then one can show that for all multi-indices such that and (at least) for all the functions of class and compact support, one has

The point is that the operator such that exists and is a bounded Calderón-Zygmund singular integral operator with homogeneous kernel, that is of the form with (or a composition of such operators).

Now, as a further application of Littlewood-Paley theory, you will show that the result above generalises even further. Indeed, you will show that if is a polynomial of degree whose top-degree component is elliptic, then for all multi-indices such that and (at least) for all the functions of class and compact support, one has

- Let denote the zero set of . Show that is contained in a ball for some large depending on .
- Let be a smooth function compactly supported in a neighbourhood of and that vanishes outside .
- With , show that is a Schwartz function and deduce that is bounded for .
- Show that is a Marcinkiewicz multiplier and therefore is bounded for .
- Decompose and use this to conclude the inequality.

** Footnotes: **

^{1}: The etymology of the term “multiplier” should be clear from the definition.

^{2}: Recall that the total variation of a function on the interval is defined as ; if has finite total variation on then there exists a complex Borel measure such that .

^{3}: If is absolutely continuous we simply have .

^{4}: Recall that such a function exists, and can be realised for example by taking with smooth compactly supported in and identically on .