This is the second part of the series on basic Littlewood-Paley theory, which has been extracted from some lecture notes I wrote for a masterclass. In this part we will prove the Littlewood-Paley inequalities, namely that for any it holds that

This time there are also plenty more exercises, some of which I think are fairly interesting (one of them is a theorem of Rudin in disguise).

Part I: frequency projections.

** 4. Smooth square function **

In this subsection we will consider a variant of the square function appearing at the right-hand side of () where we replace the frequency projections by better behaved ones.

Let denote a smooth function with the properties that is compactly supported in the intervals and is identically equal to on the intervals . We define the *smooth frequency projections* by stipulating

notice that the function is supported in and identically in . The reason why such projections are better behaved resides in the fact that the functions are now smooth, unlike the characteristic functions . Indeed, they are actually Schwartz functions and you can see by Fourier inversion formula that ; the convolution kernel is uniformly in and therefore the operator is trivially bounded for any by Young’s inequality, without having to resort to the boundedness of the Hilbert transform.

We will show that the following smooth analogue of (one half of) () is true (you can study the other half in Exercise 6).

Proposition 3Let denote the square functionThen for any we have that the inequality

We will give two proofs of this fact, to illustrate different techniques. We remark that the boundedness will depend on the smoothness and the support properties of only, and as such extends to a larger class of square functions.

*Proof:* In this proof, we will prove the inequality by seeing it as a -valued to -valued inequality and then applying vector-valued Calderón-Zygmund theory to it.

Consider the vector-valued operator given by

then, since , the inequality we want to prove can be rephrased as

The case is easy to prove: indeed, in this case the left-hand side of (2) is simply , and by Plancherel’s theorem this is equal to

the sum is clearly for any and thus conclude by using Plancherel again.

Next it remains to show is given by convolution with a vector-valued kernel that satisfies a condition analogous to ii) as in the proof of Proposition 1 from the previous notes (we do not need i) and iii) because we have already concluded the boundedness by other means); a final appeal to vector-valued Calderón-Zygmund theory will conclude the proof. It is easy to see that the convolution kernel for is

where denotes . We have to verify the vector-valued Hörmander condition

Recall that this is a consequence of the bound , where the prime denotes the componentwise derivative. Since (prove this), we have

thanks to the smoothness of , we have (or any other large positive exponent) and the estimate follows easily. We let you fill in the details in Exercise 4 below.

For the second proof of the proposition, we will introduce a basic but very important tool – Khintchine’s inequality – that allows one to linearise objects of square function type without resorting to duality.

Lemma 4 (Khintchine’s inequality)Let be a sequence of independent identically distributed random variables over a probability space taking values in the set and such that each value occurs with probability . Then for any we have that for any sequence of complex numbers

where denotes the expectation with respect to .

In other words, by randomising the signs the expression behaves on average simply like the norm of the sequence . See the exercises for some interesting uses of the lemma.

*Proof:* The proof relies on a clever use of the independence assumption.

Let for convenience, and observe that (because ; cf. Exercise 1). First we have two trivial facts: when we have by Hölder’s inequality that , and when we have conversely that by Jensen's inequality. Next, observe that to prove the part of the inequality it suffices to assume that the 's are real-valued. We thus make this assumption and proceed to estimate by expressing the norm as an integral over the superlevel sets,

We split the level set in two and estimate ; observe that for any this is equal to , which we estimate by Markov’s inequality by

Since , it follows by independence^{1} of the ‘s that the integral equals , and each factor is easily evaluated to be . Since , we have shown that the above is controlled by

if we choose , the above is controlled by . Inserting these bounds in the layer-cake representation of we have shown that

which shows for all (and therefore for all when combined with Jensen's inequality, as explained at the beginning of the proof).

Finally, we have to prove as well, and by the initial remarks we only need to do so in the regime . For such a , we take some larger than and find a such that ; by the logarithmic convexity of the norms we have then , and since we have proven above that we can conclude.

Now we are ready to provide a second proof of the proposition.

*Proof:* We claim that it will suffice to prove for every that

with constant independent of . Indeed, if this is the case, then we can raise the above inequality to the exponent and then apply the expectation to both sides. The right-hand side remains because it does not depend on , and the left-hand side becomes by linearity of expectation

which by Khintchine’s inequality is comparable to , thus proving the claim.

To prove the boundedness of the operators we appeal to a well-known result, namely the Hörmander-Mikhlin multiplier theorem. Indeed, is given by convolution with the kernel , whose Fourier transform is simply

For any , there are *at most* 3 terms in the sum above that are non-zero; this and the other assumptions on readily imply that, uniformly in ,

therefore is indeed a Hörmander-Mikhlin multiplier, and by the Hörmander-Mikhlin multiplier theorem we have that is bounded for with constant uniform in , and we are done.

** 5. Littlewood-Paley square function **

With the material developed in the previous subsections we are now ready to prove (). We restate it properly:

Theorem 5Let and let denote the Littlewood-Paley square functionThen for all functions it holds that

*Proof:* A first observation is that it will suffice to prove the part of the statement, thanks to duality. Indeed, since

we can write

where we have used the orthogonality of the projections in the third equality and then Cauchy-Schwarz inequality. Now Hölder’s inequality shows that

and by the assumed boundedness of we can bound the right hand side by , thus concluding that as well. We could also have gone the opposite direction, namely if one can prove then one can also deduce , but duality alone does not suffice – see Exercise 11 for details.

Now observe the following fundamental fact: if denotes the smooth frequency projection as defined in Section 4, then we have for any

indeed, recall that and that is identically equal to on the intervals .

Using this fact, we can argue by boundedness of vector-valued frequency projections (that is, by Corollary 2 in the previous part of this series) that

but the right-hand side is now the smooth Littlewood-Paley square function , and by Proposition 3 it is bounded by , thus concluding the proof.

** 6. Higher dimensional variants **

So far we have essentially worked only in dimension (except for Proposition 1 from the previous part). There are however some generalisations to higher dimensions of the theorems above, whose proofs follow from similar arguments.

Let . First of all, with as in Section 4, define the (smooth) *annular frequency projections*^{2} by

(there is no need for these functions to be exactly radial however, as long as they are uniformly smooth and compactly supported in the annula ). Then we have that the corresponding square function satisfies the analogue of Proposition 3:

Proposition 6Let denote the square functionThen for any we have for any the inequality

Either of the proofs given for Proposition 3 generalises effortlessly to this case.

Interestingly, the non-smooth analogue of does not satisfy the analogue of Theorem 5! Indeed, the reason behind this fact is that the operator taking on the rôle of the Hilbert transform would be the so called *ball multiplier*, given by

however, it is a celebrated result of Charles Fefferman that the operator is only bounded when , and unbounded otherwise. This result is very deep and interesting but we do not discuss it in these notes. One of the ingredients needed in the proof is nevertheless presented in Exercise 7, if you are interested.

Another way in which one can generalise Littlewood-Paley theory to higher dimensions is to take products of the dyadic intervals . That is, let for convenience denote the *dyadic Littlewood-Paley interval*

then for one defines the “rectangle” to be

and defines the square function to be

It is easy to see that the above rectangles are all disjoint and they tile . Then we have the rectangular analogue of Theorem 5

Theorem 7For any and all functions it holds that

*Proof:* We will deduce the theorem from the one-dimensional case, that is from Theorem 5. With the same argument given there, one can see that it will suffice to establish the part of the inequalities.

First of all, with the same notation as in Lemma 4, when we define the operators

These operators are bounded for any with constant independent of , as a consequence of Theorem 5 (prove this in Exercise 12).

Next, when , define . By Khintchine’s inequality, it will suffice to prove that the operator

is bounded independently of (you are invited to check that this is indeed enough). However, it is easy to see that

where denotes the operator above applied to the variable. The boundedness of thus follows from that of by integrating in one variable at a time.

In the next part of this series we will use the theory developed so far to prove the Marcinkiewicz multiplier theorem and Stein’s theorem on the spherical maximal function.

**Exercises **

Exercise 4Fill in the missing details in the first proof of Proposition 3 (that is, the pointwise bound for ).

Exercise 5Look at the proof of Khintchine’s inequality (Lemma 4). What is the asymptotic behaviour as of the constant in the inequality

obtained in that proof?

Exercise 6Let be a linear operator that is bounded for some . Show, using Khintchine's inequality and the linearity of expectation, that for any vector-valued function in we have

In other words, Khintchine’s inequality provides us with an upgrade to vector-valued estimates, free of charge.

Exercise 7Recall that the Hausdorff-Young inequality says that for any it holds thatCrucially, the inequality cannot hold for (notice that when one has ) and this can be seen in many ways. In this exercise you will show this fact using Khintchine’s inequality – this is an example of the technique known as

randomisation, which is useful to construct counterexamples.

- Let be a smooth function with compact support in the unit ball and let be an integer. Choose vectors for such that the translated functions have pairwise disjoint supports and define the function
Show that

for any .

- Show, using Khintchine’s inequality, that
- Deduce that there is an (equivalently, a choice of signs in ) such that , and conclude that the Hausdorff-Young inequality cannot hold when by taking sufficiently large.

Exercise 8Consider the circle and notice that Hölder’s inequality shows that when we have for all functions . Here we describe a class of functions for which the reverse inequality holds (which in particular implies that all the norms are comparable).

Let be a function on such that is supported in the set ; in particular, the Fourier series of is given by . You will show that for such functions one hasfor any . The proof again rests on a

randomisationtrick enabled by Khintchine’s inequality.

- Let be as in the statement of Lemma 4. Assume that there esist Borel measures such that for any and such that uniformly in . Show that for functions as above we can write
where has Fourier series .

- Show that, always under the assumption that the measures exist, the above identity together with Young’s and Khintchine’s inequalities implies (5). The constant comes from Exercise 5.
- It remains to show that the measures really exist, and you will do so by constructing them explicitely. Let be the collection of trigonometric polynomials given by
and consider the limit given by

Show that this limit exists in the weak sense, that is for any trigonometric polynomial the limit exists and is unique.

- Show that if an integer admits an expression of the form
for some integers and some choice of signs, then such an expression is necessarily unique.

- Show that, thanks to 4) above, , as desired.
- Show that to conclude the proof.
- Let be a lacunary sequence, that is ( is called the
lacunarity constantof the sequence). Show that if the proof above still works (step 4) in particular).- If , show that one can decompose into sequences of lacunarity constant at least 3. Conclude that (5) holds for any function supported on a lacunary sequence, although the constant depends on the lacunarity constant of the sequence.
- Show that there exists a constant such that the functions of the above kind satisfy the following interesting exponential integrability property:
[hint: Taylor-expand the exponential.]

Exercise 9Show that if is chosen in such a way thatfor all , then the converse of (1) is also true; that is, show that for any

Finally, construct a function such that the above condition holds.

[hint: see the proof of Theorem 5.]

Exercise 10Let denote the ball multiplier as in (4), that is the operator defined by . To keep things simple, we let the dimension be . Assume that is bounded for a certain (in reality it is not, unless , but Fefferman's proof of this fact proceeds by contradiction, so assume away). You will show that the following inequality would then also be true. Let be a collection of unit vectors in and denote by the half-plane ; finally, let be the operator given by (you can see is essentially a Hilbert transform in direction ). If is bounded, then for any vector-valued function we haveThe connection between and the ‘s is that, morally speaking, a very large ball looks like a half-plane near its boundary.

- Argue that the functions such that is smooth and compactly supported are dense in .
- Let denote the ball of radius and center (thus is tangent to the boundary of at the origin for any ) and let be the operator given by . Show, using the Fourier inversion formula, that for functions as in i) it holds that
- Argue by Fatou’s lemma that
and conclude that, by i), it will suffice to bound the right-hand side uniformly in .

- Let be the operator given by with the ball of radius centred at the origin (a rescaled ball multiplier); show that has the same -norm of .
- Show that , where .
- Combine 5) with Exercise 6 to conclude.

Exercise 11Let be the Littlewood-Paley square function. Show that the inequality that would imply by a duality argument is

and not as one could naively expect. Next, assume that holds and use it to prove the inequality above.

[hint: use Proposition 1 of part I.]

Exercise 12Show, using both and inequalities of (), that the operators

are bounded for , with constant independent of . Notice that, unlike the operators , the operators are not Calderón-Zygmund operators in general (prove this for some special choice of signs), and therefore we really need to use Littlewood-Paley theory to prove they are bounded.

**Footnotes:**

^{1}: Technically, we can only argue so for finite products; but we can assume that at most coefficients are non-zero and then take the limit at the end of the argument.

^{2}: We call them “annular” because is smoothly supported in the annulus .