In this final set of notes we will study the Littlewood-Paley decomposition and the Littlewood-Paley inequalities. These consist of very basic tools in analysis which allow us to decompose a function, on the frequency side, to pieces that have almost disjoint frequency supports. These pieces, the Littlewood-Paley pieces of the function, are almost orthogonal to each other, each piece oscillating at a different frequency.
1. The Littlewood-Paley decomposition
We start our analysis with forming a smooth Littlewood-Paley decomposition as follows. Let be a smooth real radial function supported on the closed ball of the frequency plane, which is identically equal to on . We then form the function as
Observing that if and also that if we see that is supported on the annulus .
Now the sequence of functions forms a partition of unity:
To see this first observe that each function is supported on the annulus . Thus for each given there are only finite terms in the previous sum. In particular if , then
Note that we miss the origin in our decomposition of the frequency space as each piece is supported away from . Some attention is needed concerning this point but usually it creates no real difficulty.
Thus we partition the unity in the form and each is smooth and has frequency support on an annulus of the form . Now for let us define the multiplier operators
initially defined for or . The operator frequency cut-off operator is almost a projection to the corresponding frequency annulus . It is not exactly a projection since the function is a smooth approximation of the indicator function , introducing a small tail in the region which is mostly harmless. Similarly, the operator is almost a projection on the ball .
We have the following simple properties of the Littlewood-Paley decomposition:
Proposition 1 (i) For every we have that is in . (ii) For every we have and where the limits are taken in the -sense.
(iii) For every we have that
Remark 1 Property (iii) above holds in a more general sense and for a wider class of functions, for example functions and more generally locally integrable functions that have some decay at infinity. The decomposition fails however if has no decay. Indeed, the function satisfies for all . Observe here that the function has frequency support on which is the point missed in our partition of unity.
Thus, with a Littlewood-Paley decomposition we managed to write any function (and thus any Schwartz function) as a sum of pieces , each piece being well localized in frequency on the annulus .
It is pretty obvious how the operators act on the frequency variable so let us take a look on what the pieces look in the physical space. From the general facts about the Fourier transform (see for example Exercise 2 of Notes 3) we know already that cannot have compact spatial support. Since
and , we have
Here note that . From the discussion that followed the definition of convolutions in Notes 2 we thus see that is an average of around the point at scale . Remembering that is supported on the ball this is also consistent with the uncertainty principle which also implies that the function is essentially constant at scales . Now since a piece has frequency support contained in we get that
Thus is almost constant on scales . On the other hand, since has frequency support on the annulus we have that
As before we can rewrite this as
The previous identity roughly says that the function has zero mean on every ball around of radius .
Similarly, can be viewed as a smooth approximation of the frequency projection
There are however important differences between the rough and smooth versions of these projections. For example, since is a Schwartz function the function is also Schwartz and Young’s inequality shows that
thus is bounded on . Now, consider the rough version given as
Of course is still bounded on because of Plancherel’s theorem. However, the function is no longer in and Young’s inequality cannot be used. In fact, is not bounded on whenever and . This is a deep result of C. Fefferman.
2. Littlewood-Paley Projections and derivatives
Recall the basic relation describing the interaction of derivatives with the Fourier transform:
If has support on some annulus we immediately get
and thus for any function that
In fact the same approximate identity extends to all spaces for .
Proposition 2 For all we have that
We won’t prove this proposition here since it will be covered by a lecture in the student’s seminar.
3. The Littlewood-Paley inequalities
The Littlewood-Paley inequalities quantify the heuristic principle that the pieces , having well separated frequency supports, behave independently of each other, meaning that
in some appropriate sense (for example in ). In this is already an easy consequence of the Plancherel identities. Indeed, note that
Like before observe that for every there are only two terms which don’t vanish, and these add up to . Thus
We can equivalently write this identity in the form
The following theorem extends this approximate identity to all spaces for .
Then for all we have
Proof: Consider the vector valued singular integral operator
and observe that
Observe that is a bounded linear operator from to . Indeed the strong type of follows from the remarks before the theorem. Furthermore, defining
we can verify that is a singular kernel:
Postponing the proof of this lemma for now, we use the vector valued version of the Calderón-Zygmund theorem to show that is bounded from to :
which is one of the estimates in (1). To prove the lower estimate we argue as follows. Let . Then
By vector valued duality and the estimate we conclude that the adjoint operator satisfies
Now we repeat the Littlewood-Paley decomposition but starting with the function
Using exactly the same arguments as before we can show that we also have that
Observe that for we have that and thus for any function with we have that .
and observe that since we already have that . We get
However on the left hand side we have the pointwise identity which shows that
as we wanted to show.
We now go back to the proof of Lemma 4.
Proof of Lemma 4: Remember that the kernel is given as
Let so that
For (2) we write
On the one hand we have that
On the other hand for any positive integer we have
3.1. A rough version for -dimensional dyadic intervals
So far we carried out the Littlewood-Paley decomposition based on a smooth partition of unity. The use of smooth functions to form the Littlewood-Paley decomposition has many advantages since then the projections are bounded multiplier operators. On the other hand, Remark 2 shows that in dimensions , the multiplier associated with a Euclidean ball is not bounded on . This means that the Littlewood-Paley inequalities based on the projections
will fail in any dimension .
The previous discussion leaves the one-dimensional case open. In fact we will see now that one can form the Littlewood-Paley decomposition in one dimension based on the rough partition of unity
and still have the Littlewood-Paley inequalities. So let us define to be the exact frequency projection defined by (4). We have the following.
Proof: Writing in the form
For let us define the vector valued analogue
Using the fact that is a CZO and the representation (5) of in terms of we can see that is a vector valued Calderón-Zygmund operator, thus is bounded from to . Applying this property to the function
Now observe that since we have the identity . Thus the previous estimate implies that
By Theorem 3 we get one of the inequalities in the statement of the theorem:
To prove the opposite inequality, we write the dual estimate that was obtained in proof of Theorem 3:
Now take and observe that
so the previous estimate implies
which gives the other inequality in the theorem.
Hint: Consider the vector valued operator
The problem reduces to showing that is bounded from to . Observe that is associated with the kernel
where is the identity from to and is the (scalar) kernel associated with . You can assume a Banach space version of the vector valued Calderón-Zygmund theorem.
Exercise 2 Let be a sequence of bounded or unbounded intervals on the real line, where is a finite or countably infinite index set. Define the frequency projections
We have already remarked (see remark 2) that Theorem 5 does not generalize to annuli in the -dimensional Euclidean space if we insist on using the rough projections . However, there is a generalization of the `rough’ Littlewood-Paley theorem to dimensions . This is based on decomposing the frequency space to a union of disjoint dyadic `intervals’, that is, -dimensional rectangles with axes parallel to the coordinate axes, where every side of the rectangle is an interval of the form or . This allows for `tensoring’ Theorem 5 to several dimensions without great difficulty. This is done as follows. For we set
where each is the one-dimensional projection previously defined acting only on the -th variable. For we have
The corresponding square function is defined as
This leads to
Theorem 6 For we have
We omit the proof of this theorem as it is mostly technical, based on induction and starting from the one dimensional version of the theorem already proved. You can find the proof for example in [D] or [S].
4. Two theorems on multipliers
We now go back to multiplier operators and reconsider them from the point of view of Calderón-Zygmund theory. We have already seen that a multiplier operator is the linear operator with for some . This definition automatically implies that is bounded on with norm . Alternatively, the discussion from Paragraph 8.1 of Notes 4 reveals that these are all the bounded linear operators on that commute with translations and can be realized in the form
where is the unique tempered distribution such that .
If the operator extends to a bounded linear operator on we say that is an -multiplier and write . We set
The previous remarks then show that . It turns out that the space is a Banach space but we will not dwell on this issue here. We also have the following easy proposition:
and in this case we have that
(ii) For all we have
Proof: This is a consequence of the following obvious identity; for we have
That is, is the adjoint of . Thus
since and have the same norm. To prove the second assertion assume that otherwise there is nothing to prove. By (i), the linear operator is of strong type and with the same operator norm. By the Riesz-Thorin interpolation theorem we get that
which proves .
Remark 3 Observation (ii) above shows that multipliers are necessarily bounded functions. The opposite however is not true. Another easy consequence of the discussion above is the following. We always have
where and as observed above. The problem with this representation is that we don’t know whether is actually a function that can give meaning to the formula
If however it happens that then Young’s inequality readily applies to yield that
The main problem in the theory of multipliers is to get away from the case and place suitable conditions on so that we can conclude that . The previous generalities easily imply that if then since in this case. A similar result with weaker hypothesis is the following.
Suppose that for some . Then and .
Remark 4 Observe that for any tempered distribution we have that
If is an even integer we can write
Thus, at least when is an even integer, the Sobolev space is the space of tempered distributions such that
where makes sense as a partial differentiable operator since is an integer. Similarly one can define the Sobolev spaces to be the space of tempered distributions such that
In fact one can take one step further and define the space for any real number . In the case this presents no difficulty since one has a direct interpretation of as a Fourier integral operator. In particular, is a pseudo-differential operator. Although this sounds a bit cryptic at the moment, we want to make the point here that for example is a condition that imposes decay on derivatives of .
Exercise 3 Prove Proposition 8 above.
The general flavor of the previous results is that if a function has no local singularities and, together with its derivatives, decays fast enough at infinity, then is an multiplier for all . Besides a (controllable) singularity at infinity, one can also allow for a singularity at the origin.
We present two instances of this principle, usually referred to as the Hörmander multiplier theorem. We start with an `easy’ version where the function is bounded, to assure the hypothesis is satisfied, away from the origin and its derivatives decay at least as fast as their order.
for all multi-indices . Then agrees with a function away from the origin and satisfies
for all multi-indices . In particular, is an multiplier for all with .
Proof: Using the Littlewood-Paley decomposition we can write
whenever . Each piece is supported on the annulus and as a product of smooth functions so it makes sense to define
Furthermore, from our hypotheses on we can get some good estimates on each together with its derivatives. Indeed since by our hypothesis (with the zero multi-index ) we have
On the other hand for every multi-index we have
for every non-negative integer . Integrating by parts times to pass the derivatives to the term , using Leibniz’s rule and the hypothesis on the derivatives we get the estimate
for all multi-indices and non-negative integers . Using (6) for we have
On the other hand, using (6) for we get
Now since the series converges absolutely and uniformly in (when ) for every multi-index we conclude that the series converges in to some function which also satisfies the estimate
for all multi-indices . On the other hand converges to in we conclude that when . In particular,
whenever has compact support and since then . However, satisfies
by taking the zero multi-index and furthermore
by considering multi-indices with . These estimates are enough to assure that and thus is a singular kernel so is a CZO associated with . However this means that and we are done.
Observe that what we really used in order to show that is the estimates with of the derivatives of which in turn required a control of the derivatives of up to order . Thus we have the following corollary.
Corollary 10 Let be a function such that
for all multi-indices with . Then for all .
Remark 5 The hypothesis of the previous theorem is not optimal as one can get away with less derivatives of . However it already applies to many practical case. For example for any multi-index of order , consider the operator with symbol
Observe that falls into the scope of Theorem 9 since
for all multi-indices . So for all . Now observe that for (say) we have
which shows in particular that
for all multi-indices of order , whenever . Thus all partial derivatives of order are control by the Laplacian in .
Now consider the space to be the space of functions such that all the partial derivatives of order up to are in and equip this space with the norm
By the remarks above this norm is equivalent to
Similar conclusions hold for any even integer and the space . Thus the two definitions of the Sobolev space , the one given here and then one given in Remark 3 coincide whenever is an even integer:
We now give a sharper form of the multiplier theorem which requires control only on derivatives of .
Theorem 11 (Hörmander-Mikhlin multiplier theorem version II) (i) Let be the smallest integer and suppose that the multiplier is of class with
for all multi-indices with . Then agrees with a function away from the origin which is locally integrable away from the origin and satisfies
for all .
(ii) Under the assumptions of (i) we have that .
Proof: As in the proof of Theorem 9 it will be enough to control the pieces . For this, let be a multi-index. We have
For this implies that
Now for any we have
where . Choosing and these estimates imply that
We will now prove a similar estimate for the derivatives of using a very similar approach. Indeed, we start from the identity
Now for and using the Leibniz rule we get
Thus we have
Choosing and combining the last two estimates we conclude
This estimate for together with the mean value theorem implies that
We now have for all
On the other hand
by (7). Using now that converges in to some locally integrable function for every compact set that doesn’t contain we conclude that coincides with a locally integrable function away from and satisfies
Now since away from the origin we have that
whenever in and has compact support and . Furthermore, by the assumption we automatically get that is bounded on . Here condition (8) is enough to substitute the conditions given in the definition of a singular kernel and show that is a CZO with playing the role of the kernel. Indeed, the type of can be used to treat the bad part in the Calderón-Zygmund decomposition of a function . On the other hand, if is a bad piece supported on a dyadic cube with center and is the cube with the same center and twice the side-length, we have
Now if and we have that . Thus for we have from (8) that
This treats the bad part of the Calderón-Zygmund decomposition of so we conclude the proof that is of weak type as in the general case of a CZO. Interpolating between this bound and the strong bound we get that for . By Proposition 7 or using the symmetry of in and , we also get the range with .
Exercise 4 The purpose of this exercise is to clear out some of the calculation in the proofs of the two versions of Hörmander’s theorem. (i) Prove the identity
for any positive integer . Here the meaning of the symbol is
Let and be two multi-indices in . We write of for all . With this notation the Leibniz rule says that for any multi-index and functions which are say smooth, we have
Here the generalized binomial coefficients are defined as
Alternatively we use the notation
(ii) For any two multi-indices show that
(iii) Let satisfy the estimate
for some and , be as in the Littlewood-Paley decomposition. Show that satisfies the same estimates, that is,
with different implied constants of course. Remember that and thus is supported on .
(iv) Let satisfy the estimate
for some and , be as in the Littlewood-Paley decomposition. Set . Show that for any multi-index of order we have
(v) Let be a smooth function which is supported on . Show that
and by iterating that
for all positive integers .
Exercise 5 Let be such that . Furthermore suppose that satisfies the mean regularity condition
Show that .
Hint: Briefly describe the key elements of the proof showing that is of weak type . Argue why this implies that for . You get the complementary interval for free (why?).