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LITTLEWOOD-PALEY THEORY AND APPLICATIONSAtelier d’Analyse Harmonique 2019

Harmonic analysis makes a fundamental use of divide-et-impera approaches. Aparticularly fruitful one is the decomposition of a function in terms of the frequen-cies that compose it, which is prominently incarnated in the theory of the Fouriertransform and Fourier series. In many applications however it is not necessary oreven useful to resolve the function f at the level of single frequencies and it sufficesinstead to consider how wildly different frequency components behave instead. Oneexample of this is the (formal) decomposition of functions of R given by

f =∑j∈Z

∆jf,

where ∆jf denotes the operator

∆jf(x) :=

∫ξ∈R:2j≤|ξ|<2j+1

f(ξ)e2πiξ·xdξ,

commonly referred to as a (dyadic) frequency projection (you have encountered asimilar one in the study of Hörmander-Mikhlin multipliers). Thus ∆jf representsthe portion of f with frequencies of magnitude ∼ 2j . The Fourier inversion for-mula can be used to justify the above decomposition if, for example, f ∈ L2(R).Heuristically, since any two ∆jf,∆kf oscillate at significantly different frequencieswhen |j − k| is large, we would expect that for most x’s the different contributionsto the sum cancel out more or less randomly; a probabilistic argument typical ofrandom walks (see Exercise 1) leads to the conjecture that |f | should behave “most

of the time” like(∑

j∈Z |∆jf |2)1/2

(the last expression is an example of a squarefunction). While this is not true in a pointwise sense, we will see in these notesthat the two are indeed interchangeable from the point of view of Lp-norms: moreprecisely, we will show that for any 1 < p <∞ it holds that

‖f‖Lp(R) ∼p∥∥∥(∑

j∈Z|∆jf |2

)1/2∥∥∥Lp(R)

. (†)

This is a result historically due to Littlewood and Paley, which explains the namegiven to the related theory. It is easy to see that the p = 2 case is obvious thanksto Plancherel’s theorem, to which the statement is essentially equivalent. Thereforeone could interpret the above as a substitute for Plancherel’s theorem in generic Lpspaces when p 6= 2.In developing a framework that allows to prove (†) we will encounter some variantsof the square function above, including ones with smoother frequency projectionsthat are useful in a variety of contexts. We will moreover show some applicationsof the above fact and its variants. One of these applications will be a proof ofthe boundedness of the spherical maximal function MSd−1 introduced in a previouslecture.

Notation: We will use A . B to denote the estimate A ≤ CB where C > 0 is some absoluteconstant, and A ∼ B to denote the fact that A . B . A. If the constant C depends on a list ofparameters L we will write A .L B.

1

2 LITTLEWOOD-PALEY THEORY

1. Motivation: estimates for the heat equation with sources

In this section we present certain natural objects that cannot be dealt with by(euclidean) Calderón-Zygmund theory alone, as a way to motivate the study ofLittlewood-Paley theory.Consider the heat equation on Rd

∂tu−∆u = f,

u(x, 0) = f0(x),(1.1)

with f = f(x, t) representing the instantaneous heat generated by the source atpoint x at instant t. If the total heat introduced by the sources is small, we expectto see that the solution u evolves slowly (that is, ∂tu is also small). One sense inwhich this conjecture can be made rigorous is to control the Lp(Rd × R)-norms of∂tu by the corresponding norms of the data f . If we take a Fourier transform inboth x and t we have

∂tu(ξ, τ) = 2πiτ u(ξ, τ),

∆u(ξ, τ) = −4π2|ξ|2u(ξ, τ).

Now, we are trying to relate ∂tu and f on the Fourier side (that is, we are tryingto use spectral calculus). The simplest way to do so is to observe that, by takingthe Fourier transform of both sides of (1.1), we have

∂tu− ∆u = f , that is (2πiτ + 4π2|ξ|2)u(ξ, τ) = f(ξ, τ);

thus we can write with a little algebra

∂tu(ξ, τ) = 2πiτ u = 2πiτ2πiτ + 4π2|ξ|2

2πiτ + 4π2|ξ|2u =

τ

τ − 2πi|ξ|2f .

Let m(ξ, τ) := ττ−2πi|ξ|2 , so that the result can be restated as ∂tu = mf . Observe

that both the real and imaginary parts of m are discontinuous at (0, 0) and nowhereelse. By Fourier inversion we can interpret this equality as defining a linear operatorTm, given by

Tmg(x, t) =

∫Rd×R

m(ξ, τ)g(ξ, τ)ei(ξ·x+tτ)dξdτ ;

thus we have come to realise that the partial derivative ∂tu can be obtained (atleast formally) from the function f by ∂tu = Tmf . To bound ‖∂tu‖Lp(Rd×R) istherefore equivalent to bounding ‖Tmf‖Lp(Rd×R): if Tm is bounded from Lp intoLp then we will have

‖∂tu‖Lp(Rd×R) = ‖Tmf‖Lp(Rd×R) . ‖f‖Lp(Rn×R),

which is the type of control we are looking for.Observe that by the properties of the Fourier transform our operator is actually aconvolution operator: precisely,1

Tmf(x, t) = f ∗ m(x, t) =

∫Rd

∫Rf(x− y, t− s)m(y, s)dyds.

If we knew that the convolution kernel m satisfied ‖m‖L1 <∞ we would be done,since we could appeal to Young’s inequality. Unfortunately, this is not the case. Wehave already encountered this issue in dealing with singular integrals, so the naturalthing to do is to check whether m is a Calderón-Zygmund kernel. However, this

1With a slight abuse of notation we have used m to denote the inverse Fourier transform aswell, in place of the more usual m, motivated by the fact that on Rd the two differ only by areflection. We will continue to do so in the rest of these notes as it will always be clear fromcontext which one is meant.

LITTLEWOOD-PALEY THEORY 3

is also not the case! Indeed, if K is a (euclidean) Calderón-Zygmund convolutionkernel, you have seen that it must satisfy properties such as |∇K(x)| . |x|−d−1.However, m enjoys a certain (parabolic) scaling invariance that makes it incompat-ible with the last inequality: indeed, one can easily see that for any λ > 0

m(y, s) = λd+2m(λy, λ2s)

and deduce from this a similar anisotropic rescaling invariance for ∇m. Since|(y, s)|−d−1 is not invariant with respect to this rescaling, the inequality |∇m(y, s)| .|(y, s)|−d−1 cannot possibly hold for all y, s.One solution to this issue is to develop a theory of parabolic singular integralsthat extends the results of classical Calderón-Zygmund theory to those kernels thatsatisfy anisotropic rescaling invariance identities of the kind above. While this isa viable approach, we will take a different one in these notes. The approach wewill take will not make use of said identities and will thus be more general in na-ture. With the methods of next section we will be able to show that Tm is indeedLp → Lp bounded for all 1 < p <∞.

2. Littlewood-Paley theory

We begin by considering the frequency projections introduced in the beginning.Let I be an interval and define the frequency projection ∆I to be

∆If(x) :=

∫1I(ξ)f(ξ)e2πiξxdξ.

An equivalent way of defining this is to work directly on the frequency side of thingsand stipulate that ∆I is the operator that satisfies

∆If(ξ) = 1I(ξ)f(ξ);

this is well-defined for f ∈ L2 and can be therefore extended to functions f ∈L2∩Lp. We claim that ∆I defines an Lp → Lp bounded operator for any 1 < p <∞and any (possibly semi-infinite) interval I and, importantly, that its Lp → Lp normis bounded independently of I. The point is that the operator ∆I is essentiallya linear combination of two (modulated) Hilbert transforms, that are bounded inthe same stated range. Indeed, observe that the Fourier transform of the Hilberttransform kernel p.v.1/x is −iπsign(ξ); it is easy to show that, if I = (a, b), then

1(a,b)(ξ) =sign(ξ − a)− sign(ξ − b)

2.

Now, the Fourier transform exchanges translations with modulations, and specifi-cally f(ξ + θ) = (f(·)e−2πiθ·)∧(ξ); we see therefore that we can write, with H theHilbert transform,

−iπ∫f(ξ)sign(ξ − a)e2πiξxdξ

= − iπe2πiax

∫f(ξ + a)sign(ξ)e2πiξxdξ

= e2πiaxH(e−2πia·f(·))(x).

If we let Modθ denote the modulation operator Modθf(x) := e−2πiθxf(x), we havetherefore shown that

−iπ∆(a,b) =1

2

(Mod−a H Moda −Mod−b H Modb

),

and from the Lp → Lp boundedness of H it follows that

‖∆If‖Lp .p ‖f‖Lp (2.1)

(since Modθ does not change the Lp norms at all).


At this point, another important thing to notice is that when I, J are disjointintervals, then the operators ∆I ,∆J are orthogonal to each other. Indeed, one hasby Parseval’s identity

〈∆If,∆Jg〉 =⟨

∆If, ∆Jg⟩

=

∫1I(ξ)f(ξ)1J(ξ)g(ξ)dξ

=

∫1I∩J(ξ)f(ξ)g(ξ)dξ = 0.

A consequence of this fact, together with Plancherel’s theorem, is the p = 2 caseof the result that we mentioned in the introduction, that is inequality (†); andactually, one has more, namely that for any collection I of disjoint intervals thatpartition R one has ∥∥∥(∑

I∈I

|∆If |2)1/2∥∥∥

L2(R)= ‖f‖L2 . (2.2)

Prove this in Exercise 2 to familiarise yourself with frequency projections.

2.1. Vector-valued estimates. As a step towards proving (†) (and also importantfor the applications), it will be useful to study vector-valued analogues of theseinequalities, which are easier to prove in general. We change the setting to Rd towork in more generality: so, if R is an axis-parallel rectangle (that is, of the formI1× . . .× Id with Ij intervals, possibly semi-infinite) in Rd, we define the frequencyprojection ∆R to be the operator that satisfies for any f ∈ L2(Rd)

∆Rf(ξ) = 1R(ξ)f(ξ).

The following is an abstract vector-valued analogue of (2.1).

Proposition 2.1 (Vector-valued square functions). Let (Γ, dµ) be a measure spacewith dµ a σ-finite measure, and let H denote the Hilbert space L2(Γ, dµ), that isthe space of functions G : Γ → C such that ‖G‖H := (

∫|G(γ)|2dµ(γ))1/2 < ∞.

We let (Rγ)γ∈Γ be a measurable collection of arbitrary rectangles of Rd (that is, themapping R : γ 7→ Rγ is a measurable function2).Given any vector-valued function f(x) = (fγ(x))γ∈Γ ∈ L2(Rd; H ) (that is, a func-tion of Rd that takes values in H ), we can define its vector-valued frequency pro-jection ∆R by

∆Rf := (∆Rγfγ)γ∈Γ.

Then we have that for any 1 < p < ∞ it holds for all functions f = (fγ)γ∈Γ ∈Lp(Rd; H ) that

‖∆Rf‖Lp(Rd;H ) .p ‖f‖Lp(Rd;H ).

In particular, the constant is independent of the collection of rectangles and evenof the measure space (Γ, dµ).

The level of abstraction in the previous statement is quite high, and it mighttake the reader a while to fully unpack its meaning. The following special case ofProposition 2.1 should help:

Corollary 2.2. Let (Rj)j∈N be a collection of arbitrary axis-parallel rectangles inRd. If f = (fj)j∈N is a sequence of functions such that (

∑j |fj |2)1/2 = ‖f‖`2(N) is

in Lp(Rd), we have∥∥∥(∑j

|∆Rjfj |2)1/2∥∥∥

Lp(Rd).p∥∥∥(∑

j

|fj |2)1/2∥∥∥

Lp(Rd),

2Notice the space of rectangles of Rd can be identified with Rd × Rd.


with constant independent of the collection of rectangles.

To prove (†) we will only need Corollary 2.2, but for the application of Littlewood-Paley theory we will give in Section 3.1 we will need the full power of Proposition2.1. Prove in Exercise 3 that the corollary is indeed just a special case of theproposition, before looking at the proof below.

Remark 1. It is important to notice that there is no assumption on the family ofrectangles other than the fact that they are all axis-parallel - in particular, they arenot necessarily pairwise disjoint or even distinct.

Proof of Proposition 2.1. The proof is a vector-valued generalisation of what wesaid for frequency projections over intervals at the beginning of this section.First consider d = 1 and introduce the vector-valued Hilbert transform H given by

Hf := (Hfγ)γ∈Γ;

we claim that this operator is Lp(R; H )→ Lp(R; H ) bounded. Indeed, H is givenby convolution with the vector-valued3 kernelK(x) := (p.v.1/x)γ∈Γ, and this kernelsatisfies (as you can easily verify)i) ‖K(x)‖B(H ,H ) . |x|−1,ii)∫|x|>2|y| ‖K(x− y)−K(x)‖B(H ,H )dx . 1,

iii)∫r<|x|<RK(x)dx = 0 for any 0 < r < R.

Then i)-ii)-iii) imply Lp(R; H ) boundedness simply by vector-valued Calderón-Zygmund theory.Next, observe that if d = 1 and the intervals I = (Iγ)γ∈Γ are arbitrary, we canuse the same trick used to prove (2.1) (rewriting ∆Iγ as sum of Hilbert transformsconjugated with a modulation) to deduce boundedness of the corresponding vector-valued operator ∆I from the boundedness of H above.Finally, in the case of general d > 1 one should observe that if R = I1 × . . . × Idthen ∆R factorises as4

∆R = ∆(1)I1 . . . ∆

(d)Id,

where ∆(k)I denotes frequency projection in the xk variable. Therefore, writing

Rγ = I1γ ×R′γ and x = (x1, x

′) ∈ R× Rd−1, we have by the one-dimensional resultapplied to x1 that

‖∆Rf‖pLp(Rd;H )

=

∫∫ ∥∥∥(∆Rγfγ(x1, x′))γ∈Γ

∥∥∥pHdx1dx

′

=

∫∫ ∥∥∥(∆(1)I1γ

∆R′γfγ(x1, x

′))γ∈Γ

∥∥∥pHdx1dx

′

.p

∫∫ ∥∥∥(∆R′γfγ(x1, x

′))γ∈Γ

∥∥∥pHdx1dx

′;

iterating the argument for each variable, we obtain the claimed boundedness forgeneric d.

2.2. Smooth square function. In this subsection we will consider a variant ofthe square function appearing at the right-hand side of (†) where we replace thefrequency projections ∆j by better behaved ones.

Let ψ denote a smooth function with the properties that ψ is compactly sup-ported in the intervals [−4,−1/2] ∪ [1/2, 4] and is identically equal to 1 on the

3Taking values in B(H ,H ), the space of bounded linear operators in H . Specifically, K(x)

is the operator that multiplies f(x) componentwise by 1/x.4The identity can be proved by appealing to the fact that it is trivial for tensor products and

tensor products of single variable functions are dense in Lp(Rd; H ).


intervals [−2,−1] ∪ [1, 2]. We define the smooth frequency projections ∆j by stipu-lating

∆jf(ξ) := ψ(2−jξ)f(ξ);

notice that the function ψ(2−jξ) is supported in [−2j+2,−2j−1] ∪ [2j−1, 2j+2] andidentically 1 in [−2j+1,−2j ]∪[2j , 2j+1]. The reason why such projections are betterbehaved resides in the fact that the functions ψ(2−jξ) are now smooth, unlike thecharacteristic functions 1[2j ,2j+1]. Indeed, they are actually Schwartz functions andyou can see by Fourier inversion formula that ∆jf = f ∗ (2jψ(2j ·)); the convolutionkernel 2jψ(2j ·) is uniformly in L1 and therefore the operator is trivially Lp → Lp

bounded for any 1 ≤ p ≤ ∞ by Young’s inequality, without having to resort to theboundedness of the Hilbert transform.We will show that the following smooth analogue of (one half of) (†) is true (youcan study the other half in Exercise 9).

Proposition 2.3. Let S denote the square function

Sf :=(∑j∈Z

∣∣∆jf∣∣2)1/2

.

Then for any 1 < p <∞ we have that the inequality∥∥Sf∥∥Lp(R)

.p ‖f‖Lp(R) (2.3)

holds for any f ∈ Lp(R).

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

First proof of Proposition 2.3. In this proof, we will prove the inequality by see-ing it as a C-valued to `2(Z)-valued inequality and then applying vector-valuedCalderón-Zygmund theory to it.Consider the vector-valued operator S given by

Sf(x) :=(∆jf(x)

)j∈Z;

then, since ‖Sf‖`2(Z) = Sf , the inequality we want to prove can be rephrased as∥∥Sf∥∥Lp(R;`2(Z))

.p ‖f‖Lp(R). (2.4)

The p = 2 case is easy to prove: indeed, in this case the left-hand side of (2.4) issimply

∑j ‖∆jf‖2L2(R), and by Plancherel’s theorem this is equal to∫

R|f(ξ)|2

∑j∈Z|ψ(2−jξ)|2dξ;

the sum is clearly . 1 for any ξ and thus conclude by using Plancherel again.Next it remains to show S is given by convolution with a vector-valued kernel thatsatisfies a condition analogous to ii) as in the proof of Proposition 2.1 (we do notneed i) and iii) because we have already concluded the L2 boundedness by othermeans); a final appeal to vector-valued Calderón-Zygmund theory will conclude theproof. It is easy to see that the convolution kernel for S is

K(x) := (ψj(x))j∈Z,

where ψj(x) denotes 2jψ(2jx). We have to verify the vector-valued Hörmandercondition ∫

|x|>2|y|‖K(x− y)− K(x)‖B(C,`2(Z))dx . 1.


Recall that this is a consequence of the bound ‖K′(x)‖B(C,`2(Z)) . |x|−2, where theprime ′ denotes the componentwise derivative. Since ‖K(x)‖B(C,`2(Z)) = (

∑j |ψj(x)|2)1/2

(prove this), we have

‖K′(x)‖2B(C,`2(Z)) =∑j∈Z

24j |ψ ′(2jx)|2;

thanks to the smoothness of ψ, we have |ψ ′(x)| . (1 + |x|)−100 (or any other largepositive exponent) and the estimate follows easily. We let you fill in the details inExercise 4.

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

Lemma 2.4 (Khintchine’s inequality). Let (εk(ω))k∈N be a sequence of independentidentically distributed random variables over a probability space (Ω,P) taking valuesin the set +1,−1 and such that each value occurs with probability 1/2. Then forany 0 < p <∞ we have that for any sequence of complex numbers (ak)k∈N(

Eω[∣∣∣∑k∈N

εk(ω)ak

∣∣∣p])1/p

∼p(∑k∈N|ak|2

)1/2

,

where Eω denotes the expectation with respect to P.

In other words, by randomising the signs the expression∑k ±ak behaves on

average simply like the `2 norm of the sequence (ak)k. See the exercises for someinteresting uses of the lemma.

Proof of Lemma 2.4. The proof relies on a clever use of the independence assump-tion.Let Ep :=

(Eω[∣∣∣∑k∈N εk(ω)ak

∣∣∣p])1/p

for convenience, and observe that E2 =(∑k∈N |ak|2

)1/2

(because Eω[εk(ω)εk′(ω)] = δk,k′ ; cf. Exercise 1). First we havetwo trivial facts: when p > 2 we have by Hölder’s inequality that E2 ≤ Ep, andwhen 0 < p < 2 we have conversely that E2 ≥ Ep by Jensen’s inequality. Next,observe that to prove the .p part of the inequality it suffices to assume that theak’s are real-valued. We thus make this assumption and proceed to estimate Ep byexpressing the Lp norm as an integral over the superlevel sets,

Epp =

∫ ∣∣∣∑k∈N

εk(ω)ak

∣∣∣pdP(ω) = p

∫ ∞0

λp−1P(∣∣∣∑

k∈Nεk(ω)ak

∣∣∣ > λ)dλ.

We split the level set in two and estimate P(∑

k∈N εk(ω)ak > λ); observe that for

any t > 0 this is equal to P(et

∑k εk(ω)ak > etλ

), which we estimate by Markov’s

inequality by

e−tλ∫et

∑k εk(ω)akdP(ω).

Since et∑k εk(ω)ak =

∏k e

tεk(ω)ak , it follows by independence5 of the εk’s thatthe integral equals

∏k

∫etεk(ω)akdP(ω), and each factor is easily evaluated to be

cosh(tak). Since cosh(x) ≤ ex2/2, we have shown that the above is controlled by

e−tλ∏k∈N

e12 t

2a2k = e−tλ+ 12 t

2 ∑k a

2k ;

5Technically, we can only argue so for finite products; but we can assume that at most Ncoefficients ak are non-zero and then take the limit N →∞ at the end of the argument.


if we choose t = λ/‖a‖2`2 , the above is controlled by exp(− λ2

2‖a‖2`2

). Inserting these

bounds in the layer-cake representation of Epp we have shown that

Epp ≤ 2p

∫ ∞0

λp−1e− λ2

2‖a‖2`2 dλ = ‖a‖p`22p/2p

∫ ∞0

sp/2−1e−sds,

which shows Ep .p ‖a‖`2(Z) = E2 for all 0 < p <∞.Finally, we have to prove E2 .p Ep as well, and by the initial remarks we only needto do so in the regime 0 < p < 2. For such a p, take an exponent q > 2 and find0 < θ < 1 such that 1

2 = 1−θp + θ

q ; by the logarithmic convexity of the Lp norms wehave then E2 ≤ E1−θ

p Eθq , and since we have proven above that Eq .q E2 we canconclude.

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

Second proof of Proposition 2.3. We claim that it will suffice to prove for every ωthat ∥∥∥∑

j∈Zεj(ω)∆jf

∥∥∥Lp(R)

.p ‖f‖Lp

with constant independent of ω. Indeed, if this is the case, then we can raise theabove inequality to the exponent p and then apply the expectation Eω to bothsides. The right-hand side remains ‖f‖pLp because it does not depend on ω, and theleft-hand side becomes by linearity of expectation∫

REω[∣∣∣∑j∈Z

εj(ω)∆jf(x)∣∣∣p]dx,

which by Khintchine’s inequality is comparable to∫|Sf(x)|pdx, thus proving the

claim.To prove the Lp boundedness of the operators Tωf(x) :=

∑j∈Z εj(ω)∆jf we ap-

peal to a result encountered in a previous lecture, namely the Hörmander-Mikhlinmultiplier theorem. Indeed, Tω is given by convolution with the kernel Kω :=∑j∈Z εj(ω)ψj , whose Fourier transform is simply

Kω(ξ) =∑j∈Z

εj(ω)ψ(2−jξ).

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

|Kω(ξ)| . 1,∣∣∣ ddξKω(ξ)

∣∣∣ . |ξ|−1;

therefore Kω is indeed a Hörmander-Mikhlin multiplier, and by the Hörmander-Mikhlin multiplier theorem we have that Tω is Lp → Lp bounded for 1 < p < ∞with constant uniform in ω, and we are done.

2.3. Littlewood-Paley square function. With the material developed in theprevious subsections we are now ready to prove (†). We restate it properly:

Theorem 2.5. Let 1 < p < ∞ and let Sf denote the Littlewood-Paley squarefunction

Sf :=(∑

j

|∆jf |2)1/2

.

Then for all functions f ∈ Lp(R) it holds that

‖Sf‖Lp(R) ∼p ‖f‖Lp(R). (†)


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

‖f‖Lp = supg∈Lp

′:

‖g‖p′=1

∫fg dx,

we can write∫fg dx =

∫ (∑j

∆jf)(∑

k

∆kg)dx =

∑j,k

∫∆jf∆kg dx =

∫ ∑j

∆jf∆jg dx

≤∫ (∑

j

|∆jf |2)1/2(∑

j

|∆jg|2)1/2

dx =

∫SfSg dx,

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

fg dx ≤ ‖Sf‖Lp‖Sg‖Lp′ ,

and by the assumed Lp′ → Lp

′boundedness of S we can bound the right hand side

by .p ‖Sf‖Lp‖g‖Lp′ = ‖Sf‖Lp , thus concluding that ‖f‖Lp .p ‖Sf‖Lp as well.(See Exercise 11 for why going the opposite direction would not work).Now observe the following fundamental fact: if ∆j denotes the smooth frequencyprojection as defined in Section 2.2, then we have for any j

∆j∆j = ∆j ;

indeed, recall that ∆jf(ξ) = ψ(2−jξ)f(ξ) and that ψ(2−jξ) is identically equal to1 on the intervals [2j , 2j+1] ∪ [−2j+1,−2j ].Using this fact, we can argue by Corollary 2.2 that∥∥∥(∑

j

|∆jf |2)1/2∥∥∥

Lp(R)=∥∥∥(∑

j

|∆j∆jf |2)1/2∥∥∥

Lp(R).p∥∥∥(∑

j

|∆jf |2)1/2∥∥∥

Lp(R);

but the right-hand side is now the smooth Littlewood-Paley square function Sf ,and by Proposition 2.3 it is bounded by .p ‖f‖Lp , thus concluding the proof.

2.4. Higher dimensional variants. So far we have essentially worked only indimension d = 1 (except for Proposition 2.1). There are however some generalisa-tions to higher dimensions of the theorems above, whose proofs follow from similararguments.

Let d > 1. First of all, with ψ as in Section 2.2, define the (smooth) annularfrequency projections6 Pj by

Pjf(ξ) := ψ(2−j |ξ|)f(ξ).

Then we have that the corresponding square function satisfies the analogue ofProposition 2.3:

Proposition 2.6. Let S denote the square function

Sf =(∑j∈Z

∣∣Pjf ∣∣2)1/2

.

Then for any 1 < p <∞ we have for any f ∈ Lp(Rd) the inequality∥∥Sf∥∥Lp(Rd)

.p,d ‖f‖Lp(Rd). (2.5)

6We call them “annular” because ψ(2−j |ξ|) is smoothly supported in the annulus ξ ∈ Rd :2j−1 ≤ |ξ| ≤ 2j+2.


Either of the proofs given for Proposition 2.3 generalises effortlessly to this case.Interestingly, the non-smooth analogue of S does not satisfy the analogue of Theo-rem 2.5! Indeed, the reason behind this fact is that the operator taking on the rôleof the Hilbert transform H would be the so called ball multiplier, given by

TBf(x) :=

∫Rd

1B(0,1)(ξ)f(ξ)e2πiξ·xdξ; (2.6)

however, as seen before, it is a celebrated result of Charles Fefferman that theoperator TB is only bounded when p = 2, and unbounded otherwise. This result isvery deep and we do not have enough room to discuss it in these notes. One of theingredients needed in the proof is nevertheless presented in Exercise 10, if you areinterested.

Another way in which one can generalise Littlewood-Paley theory to higher di-mensions is to take products of the dyadic intervals [2k, 2k+1]. That is, let forconvenience Ik denote the dyadic Littlewood-Paley interval

Ik := [2k, 2k+1] ∪ [−2k+1,−2k];

then for k = (k1, . . . , kd) ∈ Zd one defines the “rectangle” Rk to be

Rk = Ik1 × . . .× Ikdand defines the square function Srect to be

Srectf :=( ∑k∈Zd

|∆Rkf |2)1/2

.

It is easy to see that the above rectangles are all disjoint and they tile Rd. Thenwe have the rectangular analogue of Theorem 2.5

Theorem 2.7. For any 1 < p <∞ and all functions f ∈ Lp(Rd) it holds that

‖Srectf‖Lp(Rd) ∼p,d ‖f‖Lp(Rd).

Proof. We will deduce the theorem from the one-dimensional case, that is fromTheorem 2.5. With the same argument given there, one can see that it will sufficeto establish the .p,d part of the inequalities.First of all, with the same notation as in Lemma 2.4, when d = 1 we define theoperators

Tωf :=∑j∈Z

εj(ω)∆jf.

These operators are Lp(R) → Lp(R) bounded for any 1 < p < ∞ with constantindependent of ω, as a consequence of Theorem 2.5 (prove this in Exercise 12).Next, when d > 1, define εk(ω) := εk1(ω) · . . . · εkd(ω). By Khintchine’s inequality,it will suffice to prove that the operator

Tωf :=∑k∈Zd

εk(ω)∆Rkf

is Lp → Lp bounded independently of ω (you are invited to check that this is indeedenough). However, it is easy to see that

Tω = T (1)ω . . . T (d)

ω ,

where T (k)ω denotes the operator Tω above applied to the xk variable. The bounded-

ness of Tω thus follows from that of Tω by integrating in one variable at a time.


3. Applications

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

3.1. Marcinkiewicz multipliers. Given an L∞(Rd) function m, one can definethe operator Tm given by

Tmf(ξ) := m(ξ)f(ξ)

for all f ∈ L2(Rd). The operator Tm is called a multiplier and the function m iscalled the symbol of the multiplier7. Since m ∈ L∞, Plancherel’s theorem showsthat Tm is a linear operator bounded in L2; its definition can then be extended toL2 ∩ Lp functions (which are dense in Lp). A natural question to ask is: for whichvalues of p in 1 ≤ p ≤ ∞ is the operator Tm an Lp → Lp bounded operator? WhenTm is bounded in a certain Lp space, we say that it is an Lp-multiplier.

The operator Tm introduced in Section 1 is an example of a multiplier, withsymbol m(ξ, τ) = τ/(τ − 2πi|ξ|2). 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 |∂αm(ξ, τ)| .α |(ξ, τ)|−|α| = (|ξ|2 + τ2)−|α|/2 isclearly incompatible with the rescaling invariance of the symbol m, which satis-fies m(λξ, λ2τ) = m(ξ, τ) for any λ 6= 0. However, the derivatives of m actuallysatisfy some other superficially similar conditions that are of interest to us. In-deed, letting (ξ, τ) ∈ R2 for simplicity, we can see for example that ∂ξ∂τm(ξ, τ) =λ3∂ξ∂τm(λξ, λ2τ). When |τ | . |ξ|2 we can therefore argue that |∂ξ∂τm(ξ, τ)| =|ξ|−3|∂ξ∂τm(1, τ |ξ|−2)| . |ξ|−1|τ |−1 sup|η|.1 |∂ξ∂τm(1, η)|, and similarly when |τ | &|ξ|2; this shows that for any (ξ, τ) with ξ, τ 6= 0 one has

|∂ξ∂τm(ξ, τ)| . |ξ|−1|τ |−1.

This condition is comparable with the corresponding Hörmander-Mikhlin conditiononly when |ξ| ∼ |τ |, and is vastly different otherwise, being of product type (alsonotice that the inequality above is compatible with the rescaling invariance of m,as it should be).

The Littlewood-Paley theory developed in these notes allows us to treat multi-pliers with symbols that satisfy product-type conditions like the above, and whichtypically are beyond the reach of Calderón-Zygmund theory. Indeed, we have thefollowing general result, where the conditions assumed of the multiplier are a gen-eralisation of the pointwise ones above.

Theorem 3.1 (Marcinkiewicz multiplier theorem). Let m be a function of Rd thatis of class Cd away from the hyperplanes where one of the coordinates is zero.Suppose that m satisfies the following conditions:i) m ∈ L∞;ii) there is a constant C > 0 such that for every 0 < ` ≤ d and for every permu-

tation j1, . . . , j`, j`+1, . . . , jd of the set 1, . . . , d we have

supk∈Z`

supξj`+1

,...,ξjd 6=0

∫Rk

∣∣∂ξj1 · · · ∂ξj`m(ξ)∣∣dξj1 . . . dξj` ≤ C,

where the Rk are the dyadic Littlewood-Paley rectangles as in §2.4.Then the multiplier Tm associated to the symbol m satisfies for any 1 < p <∞ theinequality

‖Tmf‖Lp(Rd) .p ‖f‖Lp(Rd).

7The etymology of the term “multiplier” should be clear from the definition.


A multiplier whose symbol satisfies the conditions of the above theorem is calleda Marcinkiewicz multiplier. Condition ii), spelled out in words, says that a certainsubset of the partial derivatives ∂αm of m (precisely, the derivatives ∂α where thecomponents of the multi-index α have values in 0, 1 and 0 < |α| ≤ d) has theproperty that their integral over any |α|-dimensional dyadic Littlewood-Paley rec-tangle is uniformly bounded, where the integration is happening in those variablesξj such that αj 6= 0.

Remark 2. The conditions above are very general but also very cumbersome tospell out, as the above attempt demonstrates. The pointwise conditions, analogousto the one seen before, that imply condition ii) of the theorem are usually easier tocheck and are as follows:ii’) for any multi-index α ∈ 0, 1d such that 0 < |α| ≤ d we have

|∂αm(ξ)| .α |ξ1|−α1 · . . . · |ξd|−αd .See Exercise 14.

When the dimension d is equal to 1, the statement of the theorem can be super-ficially generalised and reformulated in the following way, that perhaps clarifies themoral meaning of condition ii).

Theorem 3.2 (Marcinkiewicz multiplier theorem for d = 1). Let m be a function ofR that is of bounded variation on any interval not containing the origin. Supposethat m satisfies the following conditions:1) m ∈ L∞;2) there is a constant C > 0 such that, with

∫I|dm| the total variation8 of m over

the interval I,

supk∈Z

∫[2k,2k+1]∪[−2k+1,−2k]

|dm|(ξ) ≤ C.

Then the multiplier Tm with symbol m satisfies for any 1 < p <∞ the inequality

‖Tmf‖Lp(R) .p ‖f‖Lp(R).

Condition ii) is now saying that the total variation ofm on any dyadic Littlewood-Paley interval is bounded. Observe that it is implied by the pointwise condition|m′(ξ)| . |ξ|−1, but since we are in dimension d = 1 now this stronger condi-tion would just coincide with the Hörmander-Mikhlin one. There is in general acertain degree of overlap between the Marcinkiewicz multiplier theorem and theHörmander-Mikhlin theorem in any dimension, but neither implies the other.

We will prove Theorem 3.2 here, and the proof we will give will already containall the main ingredients needed for the proof of the full Theorem 3.1. We leave itto you to extend the proof to higher dimensions in Exercise 15.

Proof of Theorem 3.2. Let T = Tm be the multiplier with symbol m. Observe thatby (†) (Theorem 2.5) we have

‖Tf‖Lp .p∥∥∥(∑

j∈Z|∆jTf |2

)1/2∥∥∥Lp,

and thus it will suffice to prove that∥∥∥(∑j∈Z|∆jTf |2

)1/2∥∥∥Lp.p∥∥∥(∑

j∈Z|∆jf |2

)1/2∥∥∥Lp

(3.1)

8Recall that the total variation of a function f on the interval [a, b] is defined assupN supa≤ξ1<...<ξN≤b

∑N−1j=1 |f(ξj+1)− f(ξj)|; if f has finite total variation on [a, b] then there

exists a complex Borel measure µ such that f(x) = f(a) +∫ xa dµ.


(again by (†)). Now, ∆jT is a multiplier with symbol m(ξ)1[2j ,2j+1](ξ) (we discardthe negative frequencies for ease of notation), and by condition 2) we see that thereexists a complex Borel measure9 dm such that for ξ ∈ [2j , 2j+1]

m(ξ) = m(2j) +

∫ ξ

2jdm(η).

By Fubini we see therefore that

∆jTf(x) = m(2j)∆jf(x) +

∫ 2j+1

2j∆[η,2j+1]f(x) dm(η),

where we recall that ∆[η,2j+1] denotes frequency projection onto the interval [η, 2j+1].By condition 1), the first term in the right-hand side contributes at most ‖m‖L∞Sfto the left-hand side of (3.1) overall, and therefore we can safely discard it; letTjf(x) be the second term. We have by Cauchy-Schwarz and condition 2) that

|Tjf(x)|2 ≤ C∫ 2j+1

2j|∆[η,2j+1]f(x)|2 |dm|(η),

and therefore we have reduced to control the square function(∑j∈Z

∫ 2j+1

2j|∆[η,2j+1]f(x)|2 |dm|(η)

)1/2

. (3.2)

Now comes the tricky part: we want to realise the above expression as the norm inan abstract Hilbert space H of a vector-valued frequency projection, as in Propo-sition 2.1. There are many ways of doing so, and the following is the one wechoose. Let Γ :=

⋃j∈Z(2j , 2j+1] × j, that is the set of elements (η, j) such that

2j < η ≤ 2j+1; observe that a set E ⊆ Γ has necessarily the form E =⋃j∈AEj×j

for a certain A ⊆ Z and sets Ej ⊆ (2j , 2j+1]. We define the measure dµ on Γ to begiven by

µ(E) :=∑j∈A

∫Ej

|dm|(η),

provided the Ej ’s are |dm|-measurable. Thus the Hilbert space H = L2(Γ, dµ)consists of those functions G(η, j) such that∑

j∈Z

∫ 2j+1

2j|G(η, j)|2|dm|(η)

is finite. Finally, to any (η, j) ∈ Γ we assign the interval Iη,j := [η, 2j+1]. Withthese definitions, we see that (3.2) corresponds to

‖(∆Iη,jgη,j(x))(η,j)∈Γ‖H ,

where in our particular case we can take gη,j(x) = ∆jf(x) because ∆Iη,j∆j = ∆Iη,j .By Proposition 2.1 we have for generic vector-valued functions (gη,j)(η,j)∈Γ that‖(∆Iη,jgη,j)(η,j)∈Γ‖Lp(R;H ) .p ‖(gη,j)(η,j)∈Γ‖Lp(R;H ); applying this to gη,j = ∆jf

we see that we have bounded ‖(∑j |Tjf |2)1/2‖Lp by the Lp(R) norm of(∑

j∈Z

∫ 2j+1

2j|∆jf(x)|2|dm|(η)

)1/2

,

but the integrand does not depend on η and thus condition 2) shows that the aboveis bounded pointwise by C Sf(x). This concludes the proof of (3.1) and thus theproof of the theorem.

9If m is absolutely continuous we simply have dm(η) = m′(η)dη.


3.2. Boundedness of the spherical maximal function. Recall that the bound-edness of the spherical maximal function MSd−1 implies, through the method ofrotations, that the Hardy-Littlewood maximal function M is Lq(Rd) → Lq(Rd)bounded for any 1 < q < ∞ with constant independent of the dimension. In thefinal part of this lecture we will finally prove the boundedness of MSd−1 (whend ≥ 3).

Theorem 3.3. Let d ≥ 3. Then for any dd−1 < p ≤ ∞ the inequality

‖MSd−1f‖Lp(Rd) .p ‖f‖Lp(Rd)

holds for any f ∈ Lp(Rd).

The range of p’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 Cd,p for the aboveinequality that depends on the dimension d. It is only later - through the methodof rotations - that one can remove this dependence on the dimension, showing thatif one has a bound ‖MSd−1‖Lp(Rd)→Lp(Rd) ≤ A for dimension d then one also hasthe bound ‖MSd‖Lp(Rd+1)→Lp(Rd+1) ≤ A for dimension d+ 1.

Proof. It will suffice to prove the theorem for dd−1 < p < 2, since the rest of

the exponents can be obtained by Marcinkiewicz interpolation with the trivial L∞estimate. It will also suffice to assume that f is a Schwartz function for convenience,as they are dense in Lp.We will prove the theorem first for a local version of MSd−1 , namely the operator

M locSd−1f(x) := sup

1≤r≤2|Arf(x)|,

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

Arf(x) :=

∫Sd−1

f(x− ry)dσ(y),

where dσ is the normalised surface measure on the sphere. Then in Exercise 21you will conclude the proof for the full case of MSd−1 with little extra effort.We will use an annular frequency decomposition as the one in Section 2.4, butslightly modified to suit our purposes. We let ψ be a smooth radial function com-pactly supported in the annulus ξ ∈ Rd : 1/2 < |ξ| < 2 and such that10 for anyξ 6= 0 we have

∑j∈Z ψ(2−jξ) = 1. We let ψj(ξ) := ψ(2−jξ) for convenience. Thus,

with Pj being given by Pjf(ξ) = ψj(ξ)f(ξ) we have that for all functions f ∈ L2

we can decomposef =

∑j∈Z

Pjf.

Since the radius r will be ∼ 1, we will not need to consider the projections toextremely low frequencies separately; therefore we define Plowf :=

∑j<0 Pjf , so

that f = Plowf +∑j∈N Pjf . In view of the above decomposition, to conclude the

Lp → Lp boundedness of M locSd−1 it will suffice by triangle inequality to prove∥∥∥ sup

1≤r≤2|ArPlowf |

∥∥∥Lp.p ‖f‖Lp , (3.3)∥∥∥ sup

1≤r≤2|ArPjf |

∥∥∥Lp.p 2−δj‖f‖Lp , (3.4)

for some δ = δp,d > 0.Now, the heuristic motivation behind such a decomposition is that |Pjf | is now

10Recall that such a function exists, and can be realised for example by taking ψ(ξ) = ϕ(ξ)−ϕ(ξ/2) with ϕ smooth compactly supported in [−2, 2] and identically 1 on [−1, 1].


roughly constant at scale 2−j . Indeed, one way to appreciate this informal principle(which is a manifestation of the Uncertainty Principle) is to show for example that|Pjf(x)| is pointwise dominated by its average at scale 2−j : since Pjf is supportedin the ball B(0, 2j+1), we can take a smooth bump function φ such that φ(2−jξ)is identically 1 on this ball and vanishes outside B(0, 2j+2) and see that thereforePjf(ξ) = φ(2−jξ)Pjf(ξ). The latter means that Pjf = Pjf ∗ 2jdφ(2j ·) and thesmoothness of φ means that we have |φ(x)| . (1 + |x|)−10d (say); therefore we haveby expanding the convolution that

|Pjf(x)| .∫Rd|Pjf(x− 2−jy)| dy

(1 + |y|)10d.

This observation suggests that ArPlowf should just be an average of f at unit scale,since r ∼ 1 and Plowf is roughly constant at scales . 1. Indeed, one can realisethat Plow is given by convolution with a Schwartz function ϕ and thus by FubiniArPlowf = f ∗Arϕ; it is not hard to show that ϕ being Schwartz and r being ∼ 1implies bounds such as |Arϕ(x)| . (1+ |x|)−10d (or any other exponent really), andtherefore we have as seen before

|ArPlowf(x)| .Mf(x)

uniformly in r, so that the pointwise domination continues to hold after we takethe supremum. This immediately implies (3.3) for any 1 < p <∞ (so even outsideour desired range).Repeating the argument for Pjf does not allow us to conclude. Indeed, you can seethat in this case |Arψj | is essentially a function of magnitude ∼ 2j concentratedin a shell of width ∼ 2−j and radius r (see Exercise 20). A greedy argument,disregarding the width information, shows the rough bound |Arψj(x)| . 2j(1 +|x|)−10d, which in turn implies |ArPjf | . 2jMf uniformly in r ∼ 1. Therefore thisargument gives at best the bound∥∥∥ sup

1≤r≤2|ArPjf |

∥∥∥Lp.p 2j‖f‖Lp , (3.5)

which blows up as j → ∞. However, one advantage of this bound is that it holdseven for 1 < p ≤ d/(d − 1), which would allow for some interpolation if we had adecaying bound for some other exponent to compensate with. A particularly niceexponent is p = 2 of course, because Plancherel’s theorem is available, so we shouldexplore what happens in this case. Consider r fixed and observe that

Arf(ξ) = dσ(rξ)f(ξ),

where dσ denotes the Fourier transform of dσ, that is dσ(ξ) =∫Sd−1 e

−2πiy·ξdσ(y)

(verify the formula). It turns out that, due to the curvature of the sphere, dσ hasa certain amount of decay as ξ gets large - in particular, as you will see in a futurelecture, one has the pointwise bound

|dσ(ξ)| . (1 + |ξ|)−(d−1)/2.

We will take this for granted here, but if you cannot wait you can prove the decayin Exercise 19. Now, combining the decay information above with the frequencylocalisation of Pjf , we see by Plancherel that, since r ∼ 1,

‖ArPjf‖L2 . 2−j(d−1)/2‖f‖L2 , (3.6)

a bound nicely decaying in j. However, this bound is not directly useful to us as westill have to take the supremum in r! To deal with this issue we will do something


a bit unconventional: since for a generic function φ(r) we have by the FundamentalTheorem of Calculus that

sup1≤r≤2

|φ(r)| ≤ |φ(1)|+∫ 2

1

|φ′(r)|dr,

we would like to replace the supremum by the right-hand side. This is usually agood idea when the supremum is taken over a controlled interval and the derivativeis a nice object (as will be the case for us); however, if we proceeded with thisapproach we would get bad bounds in the end - the second term at the right-handside would contribute much more than the other one. We need therefore some moresophisticated inequality which allows us to optimally balance the two contributions,and the following will do. Observe that, by the Fundamental Theorem of Calculusand Cauchy-Schwarz, if φ(1) = 0 we have

1

2sup

1≤r≤2|φ(r)|2 ≤

∫ 2

1

|φ(r)||φ′(r)|dr

≤(∫ 2

1

|φ(r)|2dr)1/2(∫ 2

1

|φ′(r)|2dr)1/2

,

and by the trivial inequality xy . Bx2 +B−1y2 we have therefore

sup1≤r≤2

|φ(r)| . B(∫ 2

1

|φ(r)|2dr)1/2

+B−1(∫ 2

1

|φ′(r)|2dr)1/2

(3.7)

for any B > 0. We take φ(r) = ArPjf − A1Pjf and estimate the terms on theright-hand side separately. The L2 bound (3.6) shows that, since the L2 expressionscommute, ∥∥∥(∫ 2

1

|ArPjf(x)|2dr)1/2∥∥∥

L2. 2−j(d−1)/2‖f‖L2

(and similarly for A1Pjf). For the other term, we need to evaluate ddrArPjf (the

A1Pjf term disappears), and this is easy on the Fourier side, giving

ddrArPjf(ξ) = ξ · ∇dσ(rξ)Pjf(ξ).

The gradient ∇dσ has the same decay as dσ, namely |∇dσ(ξ)| . (1 + |ξ|)−(d−1)/2

(and this can be proven in the same way). Combining this decay information withthe frequency localisation of Pjf we obtain by Plancherel’s theorem that∥∥∥(∫ 2

1

∣∣∣ ddrArPjf

∣∣∣2dr)1/2∥∥∥L2. 2j2−j(d−1)/2‖f‖L2 = 2−j(d−3)/2‖f‖L2 .

Putting all this information together, we see that inequality (3.7) gives us a bound of(B2−j(d−1)/2+B−12−j(d−3)/2))‖f‖L2 for ‖ sup1≤r≤2 |ArPjf |‖L2 , which is optimisedif we choose B = 2j/2, resulting in

‖ sup1≤r≤2

|ArPjf |‖L2 . 2−j(d−2)/2‖f‖L2 (3.8)

(observe that when d = 2 this expression gives no decay at all, thus explainingwhy the proof does not work in that case). For a given 1 < p < 2 we can theninterpolate between estimates (3.5) and (3.8) to obtain a constant for the Lp → Lp

norm of sup1≤r≤2 |ArPjf | that is at most .ε 2−j(d−1−d/p)+εj for any small ε > 0(do the calculation; you will need to interpolate between the p = 2 exponent and anexponent extremely close to 1 but not 1 - hence the ε). The expression d− 1− d/pis only positive when p > d/(d− 1), and therefore in this regime we have that (3.4)holds for some δ = δp,d > 0, allowing us to sum in j ∈ N and thus to conclude thatM loc

Sd−1 is bounded.


Exercises

Exercises that require a bit more work are labeled by a F.

Exercise 1. Let (εk)k∈N be independent random variables taking values in +1,−1such that εk takes each value with equal probability. Show that

E[∣∣∣ N∑k=1

εk

∣∣∣2] = N.

Generalise this to the case where the sum is∑Nk=1 εkak with ak some complex

coefficients.

Exercise 2. Show that (2.2) holds for any collection I of disjoint intervals.

Exercise 3. Show that Corollary 2.2 is a special case of Proposition 2.1. Whatis the measure space (Γ, dµ) that gives the corollary? Go to the proof of theproposition and spell out all the Hilbert norms in the argument in terms of themeasure space you found.

Exercise 4. Fill in the missing details in the first proof of Proposition 2.3 (thatis, the pointwise bound for ‖K′(x)‖B(C,`2(Z))).

Exercise 5. Look at the proof of Khintchine’s inequality (Lemma 2.4). What isthe asymptotic behaviour as p→∞ of the constant Cp in the inequality(

Eω[∣∣∣∑k∈N

εk(ω)ak

∣∣∣p])1/p

≤ Cp(∑k∈N|ak|2

)1/2

obtained in that proof?

Exercise 6. Let T be a linear operator that is Lp(Rd)→ Lp(Rd) bounded for some1 ≤ p < ∞. Show, using Khintchine’s inequality and the linearity of expectation,that for any vector-valued function (fj)j∈Z in Lp(Rd; `2(Z)) we have∥∥∥(∑

j∈Z|Tfj |2

)1/2∥∥∥Lp(Rd)

.∥∥∥(∑

j∈Z|fj |2

)1/2∥∥∥Lp(Rd)

.

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

Exercise 7. Recall that the Hausdorff-Young inequality says that for any 1 ≤ p ≤ 2it holds that ∥∥f∥∥

Lp′≤ ‖f‖Lp .

Crucially, the inequality cannot hold for p > 2 (notice that when p > 2 one hasp > 2 > p′) and this can be seen in many ways. In this exercise you will show thisfact using Khintchine’s inequality - this is an example of the technique known asrandomisation, which is useful to construct counterexamples.i) Let ϕ be a smooth function with compact support in the unit ball B(0, 1) and

let N > 0 be an integer. Choose vectors xj ∈ Rd for j = 1, . . . , N such thatthe translated functions (ϕ(·−xj))j=1,...,N have pairwise disjoint supports anddefine the function

Φω(x) :=

N∑j=1

εj(ω)ϕ(x− xj).

Show that‖Φω‖Lp ∼ N1/p

for any ω.


ii) Show, using Khintchine’s inequality, that

Eω[∥∥Φω

∥∥p′Lp′

]. Np′/2.

iii) Deduce that there is an ω (equivalently, a choice of signs in∑Nj=1±ϕ(· − xj))

such that∥∥Φω

∥∥Lp′∼ N1/2, and conclude that the Hausdorff-Young inequality

cannot hold when p > 2 by taking N sufficiently large.

Exercise 8. (F) Consider the circle T = R/Z and notice that Hölder’s inequalityshows that when p ≥ 2 we have ‖f‖L2(T) ≤ ‖f‖Lp(T) for all functions f . Herewe describe a class of functions for which the reverse inequality holds (which inparticular implies that all the Lp norms are comparable).Let f be a function on T such that f is supported in the set 3k : k ∈ N; inparticular, the Fourier series of f is given by

∑k∈N f(3k)e2πi3kx. You will show

that for such functions one has

‖f‖Lp(T) . p1/2‖f‖L2(T) (3.9)

for any p ≥ 2. The proof again rests on a randomisation trick enabled by Khint-chine’s inequality.i) Let (εk(ω))k∈N be as in the statement of Lemma 2.4. Assume that there esist

Borel measures (µω)ω∈Ω such that µω(3k) = εk(ω) for any k ∈ N and such that‖µω‖ . 1 uniformly in ω. Show that for functions f as above we can write

f = fω ∗ µω,

where fω has Fourier series∑k f(3k)εk(ω)e2πi3kx.

ii) Show that, always under the assumption that the measures (µω)ω∈Ω exist,the above identity together with Young’s and Khintchine’s inequalities implies(3.9). The p1/2 constant comes from Exercise 5.

iii) It remains to show that the measures (µω)ω∈Ω really exist, and you will do so byconstructing them explicitely. Let (pωk (θ))k∈N be the collection of trigonometricpolynomials given by

pωk (θ) := 1 +εk(ω)

2(e2πi3kθ + e−2πi3kθ)

and consider the limit µω given by

µω(θ) = limK→∞

K∏k=0

pωk (θ).

Show that this limit exists in the weak sense, that is for any trigonometricpolynomial q(θ) the limit limK〈q,

∏Kk=0 p

ωk (θ)〉 exists and is unique.

iv) Show that if an integer m ∈ Z admits an expression of the form

m = 3n1 ± . . .± 3n`

for some integers 0 ≤ n1 < . . . < n` and some choice of signs, then such anexpression is necessarily unique.

v) Show that, thanks to iv), µω(3k) = εk(ω), as desired.vi) Show that ‖µω‖ =

∫|µω| = 1 to conclude the proof.

vii) Let (Nk)k∈N ⊂ N be a lacunary sequence, that is lim infk→∞Nk+1/Nk =: ρ > 1(ρ is called the lacunarity constant of the sequence). Show that if ρ ≥ 3 theproof above still works (step iv) in particular).

viii) If 3 > ρ > 1, show that one can decompose (Nk)k∈N into O((log3 ρ)−1) se-quences of lacunarity constant at least 3. Conclude that (3.9) holds for anyfunction supported on a lacunary sequence, although the constant depends onthe lacunarity constant of the sequence.


ix) Show that there exists a constant C > 0 such that the functions f of the abovekind satisfy the following interesting exponential integrability property:∫

TeC|f(θ)|2/‖f‖2

L2dθ . 1.

[hint: Taylor-expand the exponential.]

Exercise 9. Show that if ψ is chosen in such a way that∑j∈Z|ψ(2−jξ)|2 = c

for all ξ 6= 0, then the converse of (2.3) is also true; that is, show that for any1 < p <∞

‖f‖Lp .p ‖Sf‖Lp .

Finally, construct a function ψ such that the above condition holds.[hint: see the proof of Theorem 2.5.]

Exercise 10. (F) Let TB denote the ball multiplier as in (2.6), that is the operatordefined by TBf = 1B(0,1)f . To keep things simple, we let the dimension be d = 2.Assume that TB is Lp(R2)→ Lp(R2) bounded for a certain 1 < p <∞ (in reality itis not, unless p = 2, but Fefferman’s proof of this fact proceeds by contradiction, soassume away). You will show that the following inequality would then also be true.Let (vj)j∈N be a collection of unit vectors in R and denote by Hj the half-planex ∈ R2 : x · vj ≥ 0; finally, let Hj be the operator given by Hjf(ξ) = 1Hj (ξ)f(ξ)

(you can seeHj is essentially a Hilbert transform in direction vj). If TB is Lp(R2)→Lp(R2) bounded, then for any vector-valued function (fj)j∈N ∈ Lp(R2; `2(N)) wehave ∥∥∥(∑

j∈N|Hjfj |2

)1/2∥∥∥Lp(R2)

.∥∥∥(∑

j∈N|fj |2

)1/2∥∥∥Lp(R2)

.

The connection between TB and the Hj ’s is that, morally speaking, a very largeball looks like a half-plane near its boundary.

i) Argue that the functions f such that f is smooth and compactly supportedare dense in Lp.

ii) Let Brj denote the ball of radius r and center rvj (thus Brj is tangent to theboundary of Hj at the origin for any r > 0) and let T rj be the operator givenby T rj f = 1Brj f . Show, using the Fourier inversion formula, that for functionsf as in i) it holds that

limr→∞

|Hjf(x)− T rj f(x)| = 0 for all x.

iii) Argue by Fatou’s lemma that∥∥∥(∑j∈N|Hjfj |2

)1/2∥∥∥Lp(R2)

≤ lim infr→∞

∥∥∥(∑j∈N|T rj fj |2

)1/2∥∥∥Lp(R2)

and conclude that, by i), it will suffice to bound the right-hand side uniformlyin r.

iv) Let TBr be the operator given by TBrf = 1Br f (a rescaled ball multiplier);show that TBr has the same Lp(R2)→ Lp(R2)-norm of TB = TB1 .

v) Show that T rj = Mod−rvjTBrModrvj , where Modξf(x) := e−2πiξ·xf(x).vi) Combine v) with Exercise 6 to conclude.


Exercise 11. Let S be the Littlewood-Paley square function. Show that the in-equality that would imply ‖Sf‖Lp .p ‖f‖Lp by a duality argument is∥∥∥∑

j∈Z∆jgj

∥∥∥Lp′.p′

∥∥∥(∑j∈Z|gj |2

)1/2∥∥∥Lp′

and not ‖f‖Lp′ .p′ ‖Sf‖Lp′ as one could naively expect.

Exercise 12. Show, using both .p and &p inequalities of (†), that the operators

Tωf :=∑j∈Z

εj(ω)∆jf

are Lp(R) → Lp(R) bounded for 1 < p < ∞, with constant independent of ω.Notice that, unlike the operators Tω, the operators Tω are not Calderón-Zygmundoperators in general (prove this for some special choice of signs), and therefore wereally need to use Littlewood-Paley theory to prove they are bounded.

Exercise 13. Assume that function m(ξ) satisfies ‖m‖L∞ < ∞ and that it hasbounded variation over all of R (that is

∫R |dm|(ξ) <∞). Show, without appealing to

the Marcinkiewicz multiplier theorem, that m defines a multiplier which is Lp(R)→Lp(R) bounded for all 1 < p <∞.[hint: simplify the proof of Theorem 3.2 as much as you can.]

Exercise 14. Show that the pointwise condition ii’) in Remark 2 implies conditionii) in Theorem 3.1.

Exercise 15. (F) In this exercise you will prove Theorem 3.1 in dimensions d > 1.Actually, the notation required becomes nightmarish pretty quickly, and thus wewill content ourselves with proving the case d = 2, since it already contains the fullgenerality of the argument. Let T = Tm and m satisfy the conditions stated in thetheorem. The proof is essentially a repetition of the argument given for Theorem3.2.i) Show that, by Theorem 2.7, it will suffice to show∥∥∥( ∑

k∈Z2

|∆RkTf |2)1/2∥∥∥

Lp.p∥∥∥( ∑

k∈Z2

|∆Rkf |2)1/2∥∥∥

Lp

under the assumption that f is supported in the first quadrant [0,∞)× [0,∞).ii) Show that in each rectangle Rk1,k2 we can write

m(ξ, η) =m(2k1 , 2k2) +

∫ ξ

2k1∂ξm(ζ1, 2

k2)dζ1

+

∫ η

2k2∂ηm(2k1 , ζ2)dζ2 +

∫ η

2k2

∫ ξ

2k1∂ξ∂ηm(ζ1, ζ2)dζ1dζ2.

iii) Show that by ii) and Fubini we have

∆Rk1,k2Tf =m(2k1 , 2k2)∆Rk1,k2

f +

∫ 2k1+1

2k1∂ξm(ζ1, 2

k2)∆(1)

[ζ1,2k1+1]fdζ1

+

∫ 2k2+1

2k2∂ηm(2k1 , ζ2)∆

(2)

[ζ2,2k2+1]fdζ2

+

∫ 2k2+1

2k2

∫ 2k1+1

2k1∂ξ∂ηm(ζ1, ζ2)∆[ζ1,2k1+1]×[ζ2,2k2+1]fdζ1dζ2.

iv) Use condition i) of the theorem to dispense with the first term at the right-handside of the last expression.


v) Use condition ii) and Cauchy-Schwarz to show that the (square of) the remain-ing terms is bounded by

C(∫ 2k1+1

2k1

∣∣∆(1)

[ζ1,2k1+1]f∣∣2 |∂ξm(ζ1, 2

k2)|dζ1

+

∫ 2k2+1

2k2

∣∣∆(2)

[ζ2,2k2+1]f |2 |∂ηm(2k1 , ζ2)

∣∣dζ2+

∫ 2k2+1

2k2

∫ 2k1+1

2k1

∣∣∆[ζ1,2k1+1]×[ζ2,2k2+1]f∣∣2 |∂ξ∂ηm(ζ1, ζ2)|dζ1dζ2

).

vi) Find measure spaces (Γj , dµj) and collections of rectangles or intervals forj = 1, 2, 3 such that each of the 3 terms above can be treated by Proposition2.1, as in the proof of the d = 1 case.

vii) Conclude using condition ii) one last time.

Exercise 16. Show that the multiplier given by

m(ξ1, . . . , ξd) :=|ξ1|α1 · . . . · |ξd|αd

(ξ21 + . . .+ ξ2

d)|α|/2,

where αj > 0 and |α| := α1 + . . .+ αd, is a Marcinkiewicz multiplier.

Exercise 17. Show that the range dd−1 < p ≤ ∞ for the boundedness of the

spherical maximal function MSd−1 is sharp, in the sense that MSd−1 is not boundedon Lp(Rd) for any 1 ≤ p ≤ d

d−1 (find a counterexample).

Exercise 18. Let the dimension be d = 3 and let Ar denote the spherical averageArf(x) :=

∫S2 f(x − ty)dσ(y), where dσ is the normalised surface measure on S2.

Show that u(x, t) := tAtg(x) solves the wave equation∂2t u−∆u = 0,

u(x, 0) = 0,

∂tu(x, 0) = g(x),

with x ∈ R3, t ≥ 0; here we assume g ∈ C3(R3) ∩ L2(R3). This is an instance ofHuygens’ principle, and more in general if the initial data becomes u(x, 0) = f(x)with f ∈ C2, the complete solution is tAtg + d

dt (tAtf). Similar formulas hold inhigher dimensions, but only when d is odd they involve only the spherical averagesAt. When d is even, one needs to average over balls instead (that is, the classicalHuygens’ principle fails in even dimensions).i) Using Stokes theorem, show that

d

dtAtg(x) =

1

t2∆(∫|y|≤t

g(x− y)dy).

ii) Using polar coordinates, show that∫|y|≤t

g(x− y)dy =

∫ t

0

s2Asg(x)ds.

iii) Use i)-ii) to show that

d

dt

(t2d

dtAtg(x)

)= t2∆Atg(x).

iv) Show that for any F twice differentiable it holds that 1tddt

(t2 ddtF (t)

)=(ddt

)2

(tF (t)),and combine this with iii) to show that the wave equation is satisfied.


v) It remains to check the initial conditions. Using Theorem 3.3 argue thatlimt→0Atg(x) = g(x) for every x, and then use this fact together with i)to argue that the initial conditions are indeed satisfied in the limit t→ 0 (andu, ∂tu are continuous in t ≥ 0).

Exercise 19. Let ψ be a smooth function compactly supported in the ball B(0, 1)of Rd−1. In this exercise you will show that∣∣∣ ∫

Rd−1

ei(x′,xd)·(ξ,|ξ|2)ψ(ξ)dξ

∣∣∣ . (1 + |x′|+ |xd|)−(d−1)/2 (3.10)

for any (x′, xd) ∈ Rd−1 × R = Rd. It is a matter of doing a smooth partition ofunity on Sd−1 and a change of variables with a suitable diffeomorphism to showthat the above implies |dσ(ξ)| + |∇dσ(ξ)| . (1 + |ξ|)−(d−1)/2; however, we willcontent ourselves with the object above (you can see it as the Fourier transform ofa measure supported on the elliptic paraboloid parametrised by (ξ, |ξ|2), instead ofthe sphere; but the two surfaces are locally the same).

(1) Square the left-hand side of (3.10) and show by a change of variables thatthe result equals

I =

∫∫Rd−1×Rd−1

ei(x′·η1+xdη1·η2)Ψ(η1, η2)dη1dη2

for some smooth function Ψ compactly supported in B(0, 2)×B(0, 2).(2) You will show (3.10) in two complementary cases. First let |x′| < 200|xd|.

Show that|I| ≤

∫|η1|≤2

|F2Ψ(η1, xdη1)|dη1,

where F2 denotes the Fourier transform in the second variable. Argue that|F (2)Ψ(η1, xdη1)| .N (1 + |η1|+ |xd||η1|)−N for any arbitrarily large N > 0and conclude that this shows |I| . (1+ |xd|)−(d−1) ∼ (1+ |x′|+ |xd|)−(d−1),thus proving (3.10) in this regime.

(3) Let now |x′| ≥ 200|xd|. Show that

|I| ≤∫|η2|≤2

|F1Ψ(x′ + xdη2, η2)|dη2;

next argue that |F (1)Ψ(x′ + xdη2, η2)| .N (1 + |x′ + xdη2| + |η2|)−N .N(1 + |x′|)−N for any N > 0 and show that this proves (3.10) in this case aswell (even with arbitrarily large exponents).

Exercise 20. Let ψj be as in the proof of Theorem 3.3. Show that, thanks to thefact that |ψ(x)| .N (1 + |x|)−N for any N > 0, one has the rough bound

|Arψj(x)| . 2j(1 + |x|)−10d

for any r ∼ 1 and any x ∈ Rd.

Exercise 21. (F) In this exercise you will finish the proof of Theorem 3.3. Hereyou will need to use a smooth square function at some point, but other than thatthe proof is just a repetition of the one given for M loc

Sd−1 .i) We see MSd−1 as

supk∈Z

sup2k≤r≤2k+1

|Arf |.

For a fixed k ∈ Z let P≤k =∑j≤k Pj be the operator that will take on the rôle

of Plow, so thatf = P≤kf +

∑j∈N

Pj+kf.


Show that it suffices to prove for any d/(d− 1) < p < 2

‖ supk∈Z

sup2−k≤r≤2−k+1

|ArP≤kf |‖Lp .p ‖f‖Lp , (3.11)

‖ supk∈Z


|ArPj+kf |‖Lp .p 2−δj‖f‖Lp , (3.12)

for some δ = δp,d > 0.ii) Show that for r ∼ 2−k one has

|ArP≤kf | .Mf

uniformly in r, k and conclude (3.11). You can simply rescale the argumentgiven for the analogous part of M loc

Sd−1 , that is for k = 0.iii) Show that for r ∼ 2−k and j > 0 one has

|ArPj+kf | . 2jMf

uniformly in r and k (again, just rescale the argument used when k = 0).Conclude that

‖ supk∈Z


|ArPj+kf |‖Lq .p 2j‖f‖Lq (3.13)

for any q > 1.iv) Show, using the decay of dσ, that when r ∼ 2−k

‖ArPj+kf‖L2 . 2−j(d−1)/2‖f‖L2 .

v) Show, using the decay of ∇dσ, that when r ∼ 2−k∥∥∥ ddrArPj+kf

∥∥∥L2. 2k2−j(d−3)/2‖f‖L2 .

vi) Show, using a suitably rescaled inequality (3.7), that for any k ∈ Z

‖ sup2−k≤r≤2−k+1

|ArPj+kf |‖L2 . 2−j(d/2−1)‖f‖L2 . (3.14)

vii) Let ψ be a smooth function compactly supported in the annulus ξ ∈ Rd :1/4 ≤ |ξ| ≤ 4 and identically equal to 1 in the annulus ξ ∈ Rd : 1/2 ≤ |ξ| ≤2; finally, let Pj be the annular projection given by Pjf(ξ) = ψ(2−jξ)f(ξ)(thus they are just a slight modification of the projections in Section 2.4).Show that in (3.14) we can replace ‖f‖L2 in the right-hand side by ‖Pj+kf‖L2 .

viii) Replace the supremum in k ∈ Z by the `2(Z) sum and show that, by Proposition2.6, we have

‖ supk∈Z


|ArPj+kf |‖L2 . 2−j(d/2−1)‖f‖L2 . (3.15)

ix) Interpolate between (3.13) and (3.15) to show that (3.12) holds. This concludesthe proof.

Exercise 22. (F) The Carleson operator for R (already mentioned in previouslectures for T) is given by

Cf(x) := supN>0

∣∣∣ ∫ 1[−N,N ](ξ)f(ξ)e2πiξxdξ∣∣∣;

recall that its Lp → Lp,∞ boundedness for a certain p implies the a.e. convergenceof the integral above to f(x) as N →∞ when f ∈ Lp. The methods developed inthese notes are not powerful enough to show the boundedness of C, but they are


enough for a simplified version of it. Indeed, in this exercise you will prove the Lpboundedness of the lacunary Carleson operator

Clacf(x) := supk∈Z

∣∣∣ ∫ 1[−2k,2k](ξ)f(ξ)e2πiξxdξ∣∣∣.

Fix p such that 1 < p <∞.i) First, consider the additional simplification given by replacing the character-

istic function 1[−N,N ](ξ) by the smooth function ϕ(ξ/N) where ϕ is a non-negative smooth function compactly supported in [−1, 1] and identically equalto 1 on [−1/2, 1/2]. Show that the associated maximal operator

Cf(x) := supN>0

∣∣∣ ∫ ϕ(ξ/N)f(ξ)e2πiξxdξ∣∣∣

is bounded pointwise by a constant multiple of the Hardy-Littlewood maximalfunction Mf(x) and conclude Lp → Lp boundedness of C.

ii) Show that it suffices to prove the Lp boundedness of the operator

Clacf(x) := supk∈Z

∣∣∣ ∫ (1[−2k,2k](ξ)− ϕ(2−kξ))f(ξ)e2πiξxdξ∣∣∣

to conclude that of Clac.iii) Show that there is a smooth non-negative function θ compactly supported in

[1/2, 2] ∪ [−2,−1/2] and such that

1[−2k,2k](ξ)− ϕ(2−kξ) = 1[2k−1,2k]∪[−2k,−2k−1](ξ) · θ(2−kξ).

iv) Let Θk denote the frequency projections associated to θ, that is Θkf(ξ) =

θ(2−kξ)f(ξ). Show that it suffices to show that the square function

Sf :=(∑k∈Z|∆kΘkf |2

)1/2

is Lp → Lp bounded to conclude the same for Clac.v) Show that S is indeed Lp → Lp bounded.vi) Argue that the argument generalises to treat the operators

C′lacf(x) := supk∈N

∣∣∣ ∫ 1[−Nk,Nk](ξ)f(ξ)e2πiξxdξ∣∣∣

where (Nk)k∈N is any fixed lacunary sequence - that is, a sequence that satisfieslim infk→∞Nk+1/Nk =: ρ > 1. However, the constant produced by the proofwill end up depending on ρ.

Exercise 23. (F) Recall that in the lecture on Calderón-Zygmund theory youhave seen how, thanks to the boundedness of the Riesz transforms (defined byRjf(ξ) = (ξj/|ξ|)f(ξ)), we can control the Lp norms of all the partial derivativesof order 2 by the Laplacian alone: for any i, j ∈ 1, . . . , d

‖∂xi∂xju‖Lp .p ‖∆u‖Lp for all 1 < p <∞.

With a little spherical harmonics theory, this can be generalised to generic homo-geneous elliptic differential operators, always within the framework of Calderón-Zygmund theory. More precisely, let Q(X) ∈ C[X1, . . . , Xd] be a homogeneouselliptic polynomial of degree k, that is1) Q(λX) = λkQ(X) for all λ 6= 0;2) the zero set Z(Q) of Q(X) consists only of the origin.


If Q(X) =∑α∈Nd:|α|=k cαX

α, the homogeneous elliptic differential operator Q(∂)

is defined asQ(∂) :=

∑α∈Nd:|α|=k

cα∂α.

Then one can show that for all multi-indices α ∈ Nd such that |α| = k and (atleast) for all the functions u of class Ck and compact support, one has

‖∂αu‖Lp .p,Q ‖Q(∂)u‖Lp for all 1 < p <∞.The point is that the operator T such that ∂α = T Q(∂) exists and is a boundedCalderón-Zygmund singular integral operator (or a composition of such operators).Now, as a further application of Littlewood-Paley theory, you will show that theresult above generalises even further. Indeed, you will show that if P (X) ∈C[X1, . . . , Xd] is a polynomial of degree k whose top-degree component is ellip-tic, then for all multi-indices α ∈ Nd such that |α| ≤ k and (at least) for all thefunctions u of class Ck and compact support, one has

‖∂αu‖Lp .p,P ‖P (∂)u‖Lp + ‖u‖Lp for all 1 < p <∞.i) Let Z(P ) denote the zero set of P . Show that Z(P ) is contained in a ballB(0, R) for some large R depending on P .

ii) Let ϕ be a smooth function compactly supported in a neighbourhood of Z(P )and that vanishes outside B(0, 2R).

iii) With ξα = ξα11 · . . . · ξ

αdd , show that m1(ξ) := ξαϕ(ξ) is a Schwartz function

and deduce that Tm1 is Lp → Lp bounded for 1 < p <∞.iv) Show that m2(ξ) := (1−ϕ(ξ)) ξα

P (ξ) is a Marcinkiewicz multiplier and thereforeTm2 is Lp → Lp bounded for 1 < p <∞.

v) Decompose ξαf(ξ) = m1(ξ)f(ξ) +m2(ξ)P (ξ)f(ξ) and use this to conclude theinequality.

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