V5B7 — ADVANCED TOPICS IN ANALYSIS: GEOMETRIC FOURIER ANALYSIS

OLLI SAARI

Contents

1. Revision of basic notions 21.1. Lp spaces 21.2. Weak Lp spaces 31.3. Interpolation 31.4. Convolution 42. Maximal functions and differentiation 52.1. General bases 52.2. The Euclidean balls 62.3. Axis parallel rectangles 73. Calderon–Zygmund theory 73.1. Calderon–Zygmund theorem 73.2. Cotlar’s inequality 93.3. Sparse bound 104. Fourier transform 114.1. Basics 114.2. Schwartz class 134.3. Uncertainty 134.4. Classical Fourier multipliers 145. Littlewood–Paley theory 165.1. Khintchine 165.2. Ap weights 175.3. Weighted Littlewood–Paley inequality 185.4. Subdyadic scales 195.5. Extrapolating unweighted bounds 206. Oscillatory integrals 226.1. Principle of stationary phase 226.2. Surface carried measures 236.3. Fourier restriction 247. Waves and spherical means 267.1. Schrodinger equation with Schwartz data 267.2. L2 data 277.3. Wave equation 277.4. Spherical means 297.5. Sparse bound for spherical maximal function 318. Fourier transform and dimension 368.1. Hausdorff dimension 368.2. Riesz energies and capacitary dimension 388.3. Energy integral and Fourier dimension 398.4. Dimension of projections 418.5. Exceptional projections and Sobolev dimension 429. Kakeya problems 449.1. Besicovitch sets 449.2. Nikodym sets 459.3. Furstenberg sets 469.4. Kakeya maximal function 46

Date: January 31, 2020.

1

9.5. The dual Kakeya inequality 479.6. Kakeya maximal bound 499.7. Nikodym maximal function 5110. Conjectures in the Fourier restriction theory 5210.1. Restriction implies Kakeya 5310.2. Square function implies Kakeya 5310.3. Square function implies restriction 5510.4. Remark on other conjectures 59References 59

1. Revision of basic notions

This is a hasted revision of what you should roughly consider as known from the previous courses onreal and harmonic analysis. The most central things are proved again, and for the rest and for moredetails in the proofs, please take a look at Chapter 1 in Grafakos’ Classical Fourier Analysis [10] whereyou can find in detail everything that is sketched below.

1.1. Lp spaces. Several standing assumptions hold throughout these notes. Our setting is harmonicanalysis in the n-dimensional Euclidean space Rn. A function f is always Lebesgue measurable withoutspecific mention, and it is defined on Rn if not otherwise stated. Sets E are measurable sets in Rn. Forp ∈ (0,∞], a function f is in Lp(E) if

∥f∥Lp(E) ∶= (∫E∣f(y)∣p dy)

1/p<∞.

This is called the Lp norm and the class of functions is called Lp space. In several notations, such asfunction spaces and integrals, we often omit E if E = Rn or it does not matter. It is always clear fromthe context which one is the case.

Remember the following

Lp is a Banach space for p ≥ 1. When p ∈ (1,∞), it is reflexive and the dual space is identified

with Lp′

, where p′ = p/(p − 1) and the duality is realized through the pairing

⟨f, g⟩Lp,Lp′ = ∫ fg ≤ ∥f∥p∥g∥p′ .

The last inequality is called Holder’s inequality. Let (X,µ) be a measure space and f ∶ Rn ×X → C measurable in both variables. Then

∥∫Xf(x, ⋅)dµ(x)∥

Lp(Rn)≤ ∫

X∥f(x, ⋅)∥Lp(Rn) dµ(x).

This is Minkowski’s inequality. If ϕ ∶ C→ [0,∞) is convex and E of finite measure, then

ϕ(⨏E∣f ∣) ≤ ⨏

Eϕ(∣f ∣).

This is Jensen’s inequality. If the Lebesgue measure ∣E∣ is positive and finite and f ∈ L1

loc, then

p↦ (⨏E∣f ∣p)

1/p

is increasing. In particular, Lp(E) ⊃ Lq(E) with continuous inclusion when q ≥ p and ∣E∣ <∞. If X = Z and µ = #, we denote Lp(X,µ) = `p. Here `p ⊂ `q when p ≤ q.

The Lp norm as representation through the layer cake formula (Cavalieri’s principle)

∫ ∣f ∣p = p∫∞

0tp−1∣∣f ∣ > t∣dt

Several function spaces of very regular functions are dense in Lp: C∞c , simple function and the Schwartz

class.2

1.2. Weak Lp spaces. We define the quasi-norm

∥f∥Lp,∞(E) ∶= supt>0

t∣x ∈ E ∶ ∣f(x)∣ > t∣1/p.

Endowed with this quasi-norm Lp,∞(E) (functions with finite quasi-norm) becomes a quasi Barnachspaces, that is, triangle inequality does not hold but ∥f +g∥ ≤K(∥f∥+∥f∥) holds instead for some K > 1.When p > 1, the weak Lp can be normed through

(1.1) f ↦ sup0<∣E∣<∞

1

∣E∣1/p′ ∫E∣f(y)∣dy.

By Chebyshev, Lp ⊂ Lp,∞. The inclusion is strict, as power function ∣x∣p never belongs to Lp but itdoes belong to Lp,∞. We will not use the more general Lorentz spaces Lp,q, but the following simpleproposition can also be understood against that background.

Proposition 1.1. Let p > 1 and let T be a linear operator defined on Lp and let T ∗ be its adjoint in thesense that

∫ Tf ⋅ g = ∫ f ⋅ T ∗g.Then T ∶ Lp → Lp,∞ if and only if for all measurable sets of finite measure E

∥T ∗1E∥Lp′ ≲ ∣E∣1/p′

.

Proof. Assume first that the weak type bound holds. Then

∥T ∗1E∥Lp′ = sup∥f∥p≤1

∫ fT ∗1E = sup∥f∥p≤1

( ∣E∣∣E∣

)1/p′

∫ETf

(1.1)

≲ ∣E∣1/p′

sup∥f∥p≤1

∥Tf∥Lp,∞ ≲ ∣E∣1/p′

.

Conversely,

∥Tf∥Lp,∞ = sup0<∣E∣<∞

1

∣E∣1/p′ ∫Tf ⋅ 1E = sup

0<∣E∣<∞

1

∣E∣1/p′ ∫f ⋅ T ∗1E ≤ sup

0<∣E∣<∞∥f∥p.

Note that the constants in the both boundedness properties can obviously be seen to be comparable.

1.3. Interpolation. We recall the real and complex interpolation theorems. The Marcinkiewicz theoremonly needs weak type bounds for the endpoints and works for sublinear operators. It is the prototype ofreal interpolation method.

Theorem 1.2 (Marcinkiewicz interpolation theorem). Let 1 ≤ p1 ≤ p2 ≤∞ and T a subliner operator onLp1,∞ +Lp2,∞. Let

1

p= θ

p1+ (1 − θ)

p2, θ ∈ [0,1].

If T ∶ Lpi → Lpi,∞ is bounded for i ∈ 1,2, then it is bounded Lp → Lp.

Proof. We only do the case p <∞. The other case is slightly simpler, but does not fit the same notation.Fix f ∈ Lp. Take δ, λ > 0. Define

fλ2 = f1f≤δλ ∈ Lp2 , fλ1 = 1 − fλ2 ∈ Lp1 .Then

∫ ∣Tf ∣p ≤ p∫∞

0λp−1∣∣Tfλ2 ∣ + ∣Tfλ2 ∣ > λ∣dλ

≤ p∫∞

0λp−1∣∣Tfλ1 ∣ > λ/2∣dλ + p∫

∞

0λp−1∣∣Tfλ2 ∣ > λ/2∣dλ = I+ II .

By the assumption and Fubini

I ≲ ∫∞

0λp−p1−1 ∫f>δλ

∣f ∣p1 dλ = ∫ ∣f ∣p1 ∫∣f ∣/δ

0λp−p1−1 dλ ≲ (1

δ)p−p1

∫ ∣f ∣p.

The term II has similar bound and the claim follows by optimizing the δ.

The Riesz–Thoring theorem requires strong type endpoints and only works for linear operators, but onthe other hand, it gives usually very good constants in the conclusion and easily applies to off-diagonalcases Lp → Lq. This is the prototype for the complex interpolation method. The idea of complexinterpolation is to find an analytic function with values in Lp0 + Lp1 so that when the real part is zero,its modulus is bounded by Lp0 norm and when the real part is one, the modulus is bounded by Lp1

norm. Finally, for some intermediate value it is supposed to give the Lp norm, and this information can3

be used using complex analysis. See below for the details. An analytic function with values in a Banachspace X means a function so that z ↦ ⟨f(z), x∗⟩ is analytic for all x∗ ∈X∗.

Theorem 1.3 (Riesz–Thorin interpolation theorem). Let 1 ≤ p0, p1, q0, q1 ≤∞, θ ∈ (0,1) and

(1

p,1

q) = (1 − θ) ( 1

p0,

1

q0) + θ ( 1

p1,

1

q1) .

If ∥Tf∥qi ≤ Ai∥f∥pi for i ∈ 0,1, then ∥Tf∥q ≤ A1−θ0 Aθ1∥f∥p.

Proof. By density, it suffices to prove the claim for simple f with compacts support. Further, by dualformulation of the Lp norms we have to show

∣∫ Tf ⋅ g dx∣ ≤ ∥f∥p∥g∥p′

for all simple functions g with compact support. By f being simple, we find pairwise disjoint sets Akand Bk numbers ak, bk, αk, βk ≥ 0 and so that

f =∑k

akeiαk1Ak , g =∑

k

bkeiβk1Bk

where both sums are finite by the compact support property.Let

P (z) = p(1 − zp0

+ z

p1) , Q(z) = q′ (1 − z

q′0+ z

q′1)

andfz =∑

k

aP (z)k eiαk1Ak , gz =∑

k

bQ(z)k eiβk1Bk .

The idea is that ∣fz ∣ transforms ∣f ∣p0 to ∣f ∣p1 when z varies from zero to one, and this transition is analyticwhen z takes the values in the unit strip on the complex plane. In particular, note that ∣fθ ∣ = ∣f ∣p and

∣gθ ∣ = ∣g∣q′

. Then by linearity

F (z) = ∫ Tfz ⋅ g =∑k,j

aP (z)k eiαkb

Q(z)k eiβk ∫ T1Ak1Bk .

This function is analytic in the complex plane and bounded in the unit strip S = z ∈ C ∶ Rez ∈ [0,1].By Hadamard’s three lines lemma from complex analysis

∣F (θ)∣ ≤ ∣F (0)∣1−θ ∣F (1)∣θ

and simplifying this expression completes the proof.

There is a generalization of the Riesz–Thorin theorem to families of operators varying in an analyticfashion along a complex parameter z. The proof of this generalization is similar to the one above butmore technical. We give the statement without proof.

Theorem 1.4 (Stein’s interpolation theorem). Let (pk, qk) for i ∈ 0,1, (p, q) and θ are as in theprevious theorem. Let Tz be a family of operator such that for all pairs of simple functions with compactsupport f and g

z ↦ ∫ Tzf ⋅ g

is analytic and grows at most as eeImz

. If ∥Tk+iy∥pk→qk ≤Mk(y) for k ∈ 0,1 where Mk(y) ≲ eey

, thenTθ ∶ Lp → Lq is bounded.

1.4. Convolution. The following basic properties of the convolution will be useful later. For functionsf and g, we denote

f ∗ g(x) = ∫ f(y)g(x − y)dy

whenever this makes sense (that is, the integral is well defined). Recall the basic algebraic properties

f ∗ g = g ∗ f, f ∗ (g ∗ h) = (f ∗ g) ∗ h, f ∗ (g + h) = f ∗ g + f ∗ h.

If f ∈ L1loc and ϕ ∈ Ckc , then ∂β(f ∗ ϕ) = f ∗ ∂βϕ and in particular, the convolution is Ck smooth. If f

and g are compactly supported, then it is easy to check supp(f ∗ g) ⊂ supp f + supp g. Finally, we have

Proposition 1.5 (Young’s convolution inequality). For numbers r, p, q ∈ [1,∞] with 1+ 1r= 1p+ 1q

it holds

∥f ∗ g∥r ≤ ∥f∥p∥g∥q.4

One important notion best presented via convolutions is that of approximate identity. Let ϕ ∈ L1

have ∫ ϕ = 1 and let ϕε(x) = ε−nϕ(xε−1). Then ϕε form an approximation of identity. In particular,ϕε ∗ f → f in all Lp with p ∈ [1,∞) and ϕε → δ0 (Dirac mass at origin) weakly* as measures.

2. Maximal functions and differentiation

If f is a locally integrable function, then almost every x is its Lebesgue point meaning that the limit

limr→o⨏B(x,r)

f(x)dx

exists. When the Euclidean balls are changed to some sequence of very wild sets, the limit need not existin general. On the other hand, if more regularity is imposed on f the choice of possible differentiationbases becomes easier. In what follows, we only consider regularity in terms of local Lp integrability, andstudy which bases of sets are good. We start with an abstract principle, which will turn out to be usefulalso in other contexts.

Proposition 2.1. Let p, q ≥ 1 and let Ttt>0 be a family of operators acting on Lp. Suppose that themaximal operator T ∗f = supt ∣Ttf ∣ defines a bounded operator Lp → Lq,∞. Then

f ∶ limt→0

Ttf exists almost everywhere.

is a closed subset of Lp.

Proof. Take fj → f so that limt→0 Ttfj exists almost everywhere for every f . By the boundedness of themaximal operator,

∣∣ limt→0

Ttfj − limt→0

Ttfk ∣ > λ∣ ≤ ∣∣ limt→0

(Ttfj − Ttfk)∣ > λ∣ ≤ ∣∣ limt→0

Tt(fj − fk)∣ > λ∣

≤ ∣∣T∗(fj − fk) > λ∣ ≲1

λq∥fj − fk∥qp

so limt→0 Ttfj is Cauchy in Lq,∞. We define T0f as the Lq,∞ limit of this sequence. Then for all k > 0

Sk ∶= ∣lim supt→0

∣Ttf − T0f ∣ >1

k∣ ≤ ∣T∗(f − fj) >

1

2k∣ + ∣∣ lim

t→0Ttfj − T0f ∣ >

1

2k∣

≤ (2k)q∥f − fj∥qp + ∣∣ limt→0

Ttfj − T0f ∣ >1

2k∣Ð→ 0

as j →∞. Then ∣lim supt→0 ∣Ttf − T0f ∣ >∣ ≤ ∑k Sk = 0, and the claim follows.

2.1. General bases. Let B be a collection of open and bounded subsets of Rn. For f ∈ L1loc, we define

MBf(x) = supU∈B

1U ⨏U∣f ∣dy.

If there is a sequence of arbitrarily small sets in terms of the diameter for each point, then the Lebesguedifferentiation theorem holds for continuous functions. As continuous functions are dense in Lp themaximal function above can be studied to find out whether the differentiation theorem holds at almostevery point or not.

The following theorem is the main result to be kept in mind from this discussion. It goes back toCordoba’s and Fefferman’s [6] work on a geometric proof for the strong maximal theorem.

Theorem 2.2. Let B be a collection of open and bounded subsets of Rn. Then MB ∶ Lp → Lp,∞ isbounded if and only if there are constants c,C ∈ (0,∞) such that for any collection R ⊂ B that is finite

and satisfies supdiamR <∞ there exists a subfamily R such that

∣⋃ R∣ ≥ c ∣⋃R∣(2.1)

∫ (∑1R)p′ ≤ C ∣⋃R∣ .(2.2)

Proof. Assume first that the covering properties hold. Take f ≥ 0 and fix λ > 0. Let K ⊂ MBf > λ becompact. For any x ∈ K, there is U ∈ B so that x ∈ U and fU > λ. As K is compact, we can cover it byfinitely many such U . Call the resulting collection R. Let R be the subfamily as in (2.1) and (2.2).Then

∣⋃R∣ ≤ 1

c∣⋃ R∣ ≤ 1

c∑ ∣R∣ ≤ 1

λc∑∫

Rf ≤

∥f∥pcλ

⋅ ∥∑1R∥p′ ≤C1/p′∥f∥p

cλ⋅ ∣⋃R∣1/p

′

.

5

and so

∣K ∣ ≤ ∣⋃R∣1/p = ∣⋃R∣1−1/p′ ≲ 1

λ∥f∥p

taking supremum over compact K as above concludes the proof.To prove the converse, we start with the sequence Rj∞j=1 of open sets whose diameters are uniformly

bounded. We can order the along decreasing diameters. We first define R1 = R1. If R1, . . . , Ri havebeen chosen, we let Ri+1 be the Rj with smallest index j so that its overlap with the previous choices ismoderate in terms of the density:

∣Rj ∩⋃ik=1 Rk ∣∣Rj ∣

≤ 1

2.

If there is no such index, we stop the algorithm. The resulting collection has the following two properties.

Sparsity. Let Ek = Rk ∖⋃k−1j=1 Rj . The sets Ek are disjoint and it holds ∣Ek ∩ Rk ∣ ≥ 1

2∣Rk ∣. Indeed,

if not, then Rk would not have been chosen as the previous rectangles have too high a densityinside it.

Containment. Given any Rj , we know that either Rj has been chosen or the family Rkoverlaps with at least half of it. Hence

⋃j

Rj ⊂ MB1⋃j Rj ≥1

2 .

We define the linear operators adjoint one adjoint to another

Tf =∑k

1EkfRk , T ∗f =∑k

1Rk∣Rk ∣

∫Ekf.

Note in particular that Tf ≤ MBf so that it maps Lp → Lp,∞ by the assumption. By Proposition 1.1,

T ∗ is satisfies the Lp′

→ Lp′

norm inequality when restricted to characteristic functions of sets of finitemeasure. Finally, as Ek has density at least half in Rk, we see that ∑k 1Rk ≤ 2T ∗1⋃k Rk . Altogether

∫ (∑1R)p′ ≲ ∫ (T ∗1⋃k Rk)

p′ ≲ ∣⋃R∣ .

This is (2.2). The remaining (2.1) follows from the containment property and the weak type estimate.

2.2. The Euclidean balls. Verifying the covering properties implying the weak type bounds for themaximal function with respect to the Euclidean balls amounts to proving the following version of theVitali covering lemma.

Theorem 2.3 (Vitali covering theorem). Consider a family of balls B(xi, ri)∞i=1 so that supi ri < ∞.Then there exists a subfamily ik such that B(xik , rik) are disjoint and

∞⋃i=1

B(xi, ri) ⊂∞⋃k=1

B(xik ,5rik).

Proof. Let K = supi ri. Let i1 be the smallest k so that rk ≥ 12K. Let B1 be the collection B(xj , rjj≠k

and K1 = supj≠k rj . Assume that ij has been chosen and Bj and Kj have been defined. Let ij+1 be the

smallest k so that rk ≥ 12Ki and B(xk, rk) ∩B ≠ ∅ for all B ∈ Bi and define Bi+1 and Ki+1 accordingly.

If there is no k to satisfy the choice criterion, the algorithm terminates.The created sequence B(xik , rik) is clearly disjoint. Take a ball B(x, r) that was not chosen. Let

B(y,R) be the largest ball from the chosen sequence that meets B(x, r). Now 2R ≥ r, because otherwiseall the balls meeting B(y, r) would have less than half of its radius and hence B(x, r) would have beenchosen. Now it follows from the triangle inequality that B(x, r) ⊂ B(y,5R).

We can verify that (2.1) and (2.2) are valid for all p > 1 for the Vitali cover, and this shows thatthe Hardy–Littlewood maximal function (MB when B are the Euclidean balls, we denote this by M forbrevity in what follows) is bounded on all weak (and by interpolation also strong) Lp spaces. However,the covering property we get from Vitali is even better. It continues to holds for p = 1. This implies theweak type (1,1) bound. In the next example of rectangles of arbitrary exentricity, the covering lemmawill be strong enough to imply all Lp bounds, but there one will not be able to establish a weak type(1,1) bound.

Corollary 2.4. The Hardy–Littlewood maximal opearator M is bounded Lp → Lp for all p > 1.6

Theorem 2.5. Let µ be a locally finite positive Borel measure. Let f ∈ L1loc(Mµ). Then for all λ > 0

µ(x ∈ Rn ∶Mf(x) > λ ≲ 1

λ∫Rn

∣f(y)∣Mµ(y)dy.

Proof. Take f ≥ 0 and fix λ > 0. Take a compact K ⊂ Mf > λ. Take a Vitali cover of balls where fhas large mean value and note

µ(K) ≤ 5n∑i

µ(5Bi)∣Bi∣∣5Bi∣

≤ 5n

λ∑i∫Bif ⋅Mµ ≤ 5n

λ∫ f ⋅Mµ.

The claim follows by inner regularity of µ.

2.3. Axis parallel rectangles. Let B = ∏i(ai, ai +hi) ∶ ai ∈ R, hi > 0. The related maximal functionis called the strong maximal function. For continuous f , one clearly sees

MBf(x) =M1 ⋯ Mnf(x), Mig(x) ∶= suph>0

1

2h∫

xi+h

xi−h∣g(x1, . . . , xi−1, t, xi+1, . . . , xn)∣dt,

and so the strong maximal function is bounded on every Lp with p > 1 as a consequence of the onedimensional Hardy–Littlewood maximal theorem. This implies that the covering property in (2.1) and(2.2) applies to rectangles. Also in this case, more can be said.

Theorem 2.6. Let R be a finite family of axis parallel rectangles in Rn with n > 1. Then there is asubfamily R′ so that (2.1) holds and

∫⋃R′

exp⎛⎝[ ∑R∈R′

1R]1n−1 ⎞

⎠dx ≲ ∣⋃R′∣ .

We do not prove this theorem here, but the idea is the following: The case n = 1 can be taken as theVitali covering lemma. We proceed by induction on dimension. Suppose the claim holds for dimensionn − 1. Number the rectangles Ri and choose R′

1 = R1. Suppose that R′1, . . . ,R

′i have been chosen. Let

R′i+1 = Rj for the smallest j so that

∣Rj ∩⋃ik=1R′k ∣

∣Rj ∣≤ 1

2.

Write each rectangle as R = T × I where I is an interval. Then

∣Tj ∩⋃ik=1 T′k ∣

∣Tj ∣≥

∣Rj ∩⋃ik=1R′k ∣

∣Rj ∣≥ 1

2

follows from the choice criterion not being valid. This allows one to estimate

⋃T×I∈R

T ⊂ Mn−1(1⋃k T ′k) ≥1

2

slice-wise and reduce the question to the lower dimension. From this point on, the proof is very similarto the proof of Theorem 2.2 except for the duality property from Proposition 1.1 must be replaced by a

more complicated duality of Orlicz spaces L(logL)n−1 and expL1n−1 .

3. Calderon–Zygmund theory

3.1. Calderon–Zygmund theorem. Next we recall the basic class of convolution type Calderon–Zygmund operators, which will serve as a work horse for some later arguments with Fourier multipliers.A good reference for most of the material here is [7]. A function k ∈ L1

loc(Rn ∖ 0) is said to be astandard (convolution) kernel if

∣k(x)∣ ≤K1∣x∣−n, for some constant K1 and all x ∈ Rn and

∣k(x + h) − k(x)∣ ≤ K2∣h∣δ

∣x∣n+δ, for some K2 > 1, δ ∈ (0,1] and all x ∈ Rn ∖ 0 and h ∉ B(0, 1

2∣x∣).

Definition 3.1. A linear operator T acting on L2 is a Calderon–Zygmund operator (CZO) associatedwith the the kernel k if

T ∶ L2 → L2 is bounded.

∫ Tf(x) ⋅ g(x)dx =∬Rn×Rn

f(y)k(x − y)g(x)dxdy for all f, g ∈ L2 with compact disjoint supports.

7

For ε > 0, the operator

T εf(x) = ∫∣x−y∣>εf(y)k(x − y)dy

defined a priory on compactly supported f is called the truncated singular integral. The sublinearoperator

T ∗f(x) = supε>0

∣Tεf(x)∣

is called the maximal truncated singular integral.

At this point it is good to remember several dangerous properties of singular integrals. The kerneldoes not determine the operator uniquely and the truncated singular integrals need not converge to theCZO. Indeed, zero is a standard kernel associated with the identity operator and there the truncatedsingular integrals have nothing to do with the actual value of the operator at any point. A CZO that canbe seen as a limit of its truncated singular integrals is sometimes called a Calderon–Zygmund singularintegral. The main theorem of the Calderon–Zygmund theory is the following.

Theorem 3.2. Let T be a CZO. Then it extends to a bounded operator L1 → L1,∞ with norm onlydepending on dimension, ∥T ∥2→2 and the constant K2.

Proof. Let f ≥ 0 be bounded and compactly supported so that it is in all Lp spaces. Fix λ > 0. Find Q0

containing the support of f so that fQ0 < λ (this is possible as f is in L1). Consider the dyadic subcubesof Q0 (ones obtained through repeated bisection of its sides). Take the maximal ones Qi with respectto the choice criterion fQi > λ. Then Qi are disjoint and their union coincides with

E = Mdf > λ ⊃ f > λ

where Md is the maximal function maximizing over averages of dyadic subcubes of Q0 and the lastinclusion is up to a set of measure zero. Define

g = 1Ef +∑i

1QifQi , b = f − g =∑i

1Qi(f − fQi) =∑i

bi.

Then

∣Tf > λ∣ ≤ ∣⋃i

3Qi∣ + ∣(⋃i

3Qi)c

∩ Tb > λ/2∣ + ∣Tg > λ/2∣ = I+ II+ III .

Here

I ≤∑i

∣3Qi∣ = 3n∑i

∣Qi∣ = 3n∣E∣ ≲ 1

λ∫ f

and

III ≤ 1

λ2 ∫ ∣Tg∣2 ≲ 1

λ2 ∫ ∣g∣2 ≤ 1

λ∫ ∣g∣ ≲ 1

λ∫ f.

Finally

II ≤ 1

λ∑i∫(3Qi)c

∣Tbi∣dx =1

λ∑i

∞∑k=1∫

3k+1Qi∖3kQi∣Tbi∣dx.

Taking a single summand from the last expression and applying the regularity condition of standardkernel as well as the property ∫ bi = 0 we get

∫3k+1Qi∖3kQi

∣Tbi∣dx ≤ ∫3k+1Qi∖3kQi

∣∫Qibi(y)k(x − y)dy∣dx

≤ ∫3k+1Qi∖3kQi

∫Qi

∣bi(y)∣∣k(x − y) − k(x − c(Qi))∣dydx

≲ ∫3k+1Qi∖3kQi

∫Qi

∣bi(y)∣∣y − cQi ∣δ

∣x − y∣n+δdydx ≲ ∣3k+1Qi∣ ⋅

`(Qi)δ

(3k`(Qi))n+δ ∫Qi≲ 3−kδ ∫

Qi∣bi.∣

Summing over i and k, we obtain

II ≲ 1

λ∫ ∣b∣ ≲ 1

λ∫ f

which concludes the proof. 8

3.2. Cotlar’s inequality. Next we prove a bound for maximal truncations of CZO’s. It follows fromCotlar’s inequality. Knowing a weak type bound for the maximal truncations implies that the truncatedsingular integrals do have principal values almsot everywhere. However, it does not imply that theyshould have anything to do this the actual CZO Tf .

Proposition 3.3 (Kolmogorov). Let T be of weak type (1,1). Then for any α ∈ (0,1) and a set E ofpositive and finite measure, the following holds

(⨏E∣Tf ∣α dx)

1/α≲T,α

1

∣E∣ ∫∣f ∣.

The proof is a simple application of the layer cake formula (omitted).

Lemma 3.4 (Cotlar’s inequality). Let p > 1 and f ∈ Lp and let α ∈ (0,1). Then

T ∗f(x) ≲M(T ∗fα)(x)1/α +Mf(x).

Proof. Because of translation invariance, we simplify the notation by assuming x = 0. Let ε > 0 anddecompose f1 = f1B(0,ε), f2 = f − f1. Because 0 ∉ supp f2, we note Tf2(0) = T εT2(0) = T εf(0).

Take z ∈ B(0, ε/2). Then

∣T εf(0)∣ = ∣Tf2(0)∣ ≤ ∣Tf2(0) − Tf2(z)∣ + ∣Tf2(z)∣≤ ∣Tf2(0)∣ ≤ ∣Tf2(0) − Tf2(z)∣ + ∣Tf(z)∣ + ∣Tf1(z)∣ = I+ II+ III .

The first term is dealt with the kernel bound:

I ≤ ∫∣y∣>ε∣k(−y) − k(z − y)∣∣f(y)∣dy ≤

∞∑k=1∫

2k−1ε<∣y∣≤2kε

∣z∣δ

∣z − y∣n+δdy ≲

∞∑k=1

2−kδ ⨏B(0,2kε)

≲Mf(0).

The third term can be bounded using Kolmogorov’s inequality

⨏B(0,ε/2)

IIIα dz ≲ (⨏B(0,ε)

∣f ∣dz)α

Finally,

∣T εf(0)∣ = (⨏B(0,ε/2)

∣T εf(0)∣α dz)1/α

≲Mf(0) + (⨏B(0,ε/2)

∣Tf(z)∣α dz)1/α

+ ⨏B(0,ε/2)

∣T εf(0)∣α dz

whence the claim follows by taking supremum over ε > 0.

Corollary 3.5. Let T be a CZO associated with a kernel k. Then T ∗ is of weak type (1,1).

There is a non-centred variant of the maximal truncation, which first appeared in Lerner’s simplifi-cation of Lacey’s elementary approach to A2 bounds. For Q0 a cube and x ∈ Q0, define the non-centredmaximal truncation of a CZO T as

(3.1) MT,Q0f(x) = supQ∋x

ess supξ∈Q ∣T (f13Q0∖3Q)(ξ)∣.

If Q0 = Rn, we omit if from the notation. This operator is designed to control the non-symmetrictruncations appering in applications. Even this maximal truncation is of weak type (1,1) as follows fromthe next lemma.

Lemma 3.6. Let T be a CZO and f a bounded compactly supported function.

For almost every x ∈ Q0

∣T (f13Q0)(x)∣ ≲ ∣f(x)∣ +MT,Q0f(x).

For all x ∈ Rn

MT f(x) ≲Mf(x) + T ∗f(x).

Proof. Recall first that a measurable function is approximately continuous at almost every point. Thismeans that for almost every x and any ε > 0 the point x is a density point for y ∶ ∣f(y)− f(x)∣ < ε. See1.7.2 in [9]. Hence we can assume that x is a Lebesgue point for f , a point of approximate continuity forT (13Q0f) and an interior point of Q0 without losing more than a null set of points. Let ε > 0 and define

E = y ∈ Rn ∶ ∣T (13Q0f)(x) − T (13Q0f)(y)∣ < ε.9

Let s > 0 be very small and take y ∈ Q(x, s/3) ∩ E (the cube centred at x and diameter 2s). By thedefinition of E

∣T (13Q0f)(x)∣ ≤ ∣T (13Q0f)(y)∣ + ε ≤ ∣T (13Q0∖Q(x,s)f)(y)∣ + ∣T (1Q(x,s)f)(y)∣ + ε≤MT,Q0f(x) + ∣T (1Q(x,s)f)(y)∣ + ε.

Taking infimum over all y as above, we end up estimating

infy∈E∩Q(x,s)

∣T (f1Q(x,s))(y)∣ ≤ (⨏E∩Q(x,s)

∣T (f1Q(x,s))(y)∣1/2 dy)2

≲ 1

∣E ∩Q(x, s)∣ ∫Q(x,s)∣f(y)∣dy → ∣f(x)∣.

The limit as s→ 0 follows because x is a point of density for E and a Lebesgue point for f . Taking ε→ 0,the first claim follows.

To prove the second item, take x, fix Q ∋ x and choose any ξ ∈ Q. Let Bx = B(x,2 diamQ) so that3Q ⊂ Bx. Then

∣T (1(3Q)cf)(ξ)∣ ≤ ∣T (1Bcxf)(ξ) − T (1Bcxf)(x)∣ + ∣T (1Bx∖3Qf)(ξ)∣ + ∣T (1Bcxf)(x)∣ = I+ II+ III .

The first term is controlled by the usual argument using regularity of kernel and decomposition of spaceinto annuli

I ≲ ∫∣y−x∣>2 diamQ∣k(ξ − y) − k(x − y)∣∣f(y)∣dy ≲Mf(x).

The second term is handled using the size estimate of the kernel

II ≲ ∫Bx∖3Q

1

diam(Q)n∣f(y)∣dy ≲Mf(x).

The last term is just controlled by the definition of the maximally truncated singular integral and theproof of the lemma is complete.

Corollary 3.7. Let T be a CZO. Then MT,Q0 is of weak type (1,1) independently of Q0.

3.3. Sparse bound. Weak type (1,1) bound together with the L2 bound and a duality argument implythat Calderon–Zygmund operators extend to bounded operators on all Lp spaces with any p ∈ (1,∞).Next we show that even more is true: there is a sparse bound. Sparse bounds are a modern measure ofsize. They can be seen as an alternative to Lp norms and weighted Lp norms when bounding operators.

Given a family of cubes Q, we say that it is γ ∈ (0,1) sparse if for each Q ∈ Q there is a subset EQ ⊂ Qso that ∣EQ∣ > γ∣Q∣ and for all P,Q ∈ Q either P = Q or EP ∩EQ = ∅. This property implies the Carlesonpacking condition

∑Q⊂Q0

Q∈Q

∣Q∣ ≤ 1

γ∑Q⊂Q0

Q∈Q

∣EQ∣ ≤ ∣Q0∣γ.

In fact, one can show that the packing condition is equivalent to sparseness of a family (we do not dothat here, see Section 6 in [14]).

The sparse bound for Calderon–Zygmund operators is formulated as follows. We already know thatthe norm inequality ∥Tf∥p ≲ ∥f∥p holds. The norm on the left hand side can be replaced using theduality of Lp spaces by the bilinear form

(f, g)↦ ∫ Tf(x) ⋅ g(x)dx.

The idea of sparse domination is to bound this bilinear form by

∑Q

∣Q∣fQgQ

where the sum is over a sparse family. If the sparse family were well structured, say a family of dyadiccubes, then the sum could be understood as an integral over the upper half space of the product ofextensions of f and g (originally functions on Rn, now functions on dyadic cubes) and a Carleson weight(restriction to the sparse family of cubes). This implies all Lp bounds through

≲∑Q

EQfQgQ ≤∑Q∫EQ

Mf ⋅Mg = ∫ Mf ⋅Mg

but the sparse form can in principle be much smaller than the Lp norms.10

The bound by the sparse form follows from the following pointwise result. The proof is the one from[13].

Theorem 3.8. Let f ∈ L1 be compactly supported and T a CZO. Then there exists a sparse family Q(depending on f) so that for almost every x ∈ Rn

∣Tf(x)∣ ≲∑Q

1Q ⨏Q∣f(x)∣dx.

Proof. For notational convenience, assume f ≥ 0. Take Q0 with 110Q0 ⊃ supp f and let

E = x ∈ Q0 ∶ f(x) > cf3Q ∪ x ∈ Q0 ∶MT,Q0f(x) > cf3Q.

We start with the interesting case x ∈ Q0. By the weak type bound for the non-centred maximaltruncation

∣E∣ ≤ C

cf3Q∫ f dx ≤ 1

2n+2∣Q0∣

provided c is chosen large enough (only depending on the dimension and the operator norm for the weaktype bound). Consider then the function 1E . By the bound above, its density in Q0 is less than 2−n−1.Let Pj ⊂ Q0 be the maximal dyadic (relative to Q0) cubes with

1

2n+1<

∣E ∩ Pj ∣∣Pj ∣

.

By maximality, there is an upper bound 12

for the density in the display above. Now ∣E ∖⋃j Pj ∣ = 0. Inparticular, almost every point not meeting any Pj is outside of E. Moreover,

∑j

∣Pj ∣ ≤ 2n+1∑j

∣E ∩ Pj ∣ ≤ 2n+1∣E∣ ≤ 1

2∣Q0∣.

Complement of E is large.Define L(Q0) = 1Q0 ∣T (f13Q0)∣. Then

L(Q0) = 1Q0∖⋃j Pj ∣T (f13Q0)∣ +∑j

1Pj ∣T (f13Q0)∣

≤ 1Q0∖⋃j Pj ∣T (f13Q0)∣ +∑j

1Pj ∣T (f13Q0∖3Pj)∣ +∑j

L(Pj).

Clearly the first term is ≲ 1Q0f3Q0 as it is non-zero only in the complement of E. The same bound holdsfor the second term: For any x ∈ Pj , there is a set of positive measure of ξ ∈ Pj ∖E. For any such ξ,

1Pj(x)∣T (f13Q0∖3Pj)(x)∣ ≤ 1Pj(x)MT,Q0f(ξ) ≤ 1Pjcf3Q0 .

Because Pj are disjoint, we can sum up this estimate in j. Hence

L(Q0) ≲ 1Q0f3Q0 +∑j

1PjL(Pj)

where Pj are disjoint and occupy less than half of measure of Q0. We can set EQ0 = Q0∖⋃j Pj . Iteratingthe argument on Pj gives the construction of the sparse family. To deal with the points x ∉ Q0, it is aneasy exercise to construct an increasing sparse sequence of cubes exhausting Rn and bounding Tf (thesize estimate for the kernel). Finally, note that if Q is a sparse family, so is 3Q.

4. Fourier transform

4.1. Basics. Let µ be a finite Borel measure (e.g. µ ∈ L1(Rn)). We define the Fourier transform throughthe formula

µ(ξ) = Fµ(ξ) = ∫ e−2πix⋅ξ dµ(x).

Because the measure has fintie total variation, the integral above converges absolutely and the definitionmakes sense. The physical motivation is as follows: µ(ξ) is the superposition at ξ of the plane wavesξ ↦ e−2πξ⋅x whose amplitudes are distributed according to µ. Usually we call ξ the frequency and xthe physical variable (wave number and location would be more accurate). This is just the wrong wayaround, but the Fourier transform will turn out to be its own inverse (up to reflection).

11

There are several basic properties that we list without proving. The proofs are not more complicatedthan changes of variables or integration by parts. Before doing that, we set up the following notation.Let f be a measurable function. Define

τhf(x) = f(x − h), Dilpr f(x) =1

rn/pf (x

r) , Moda f(x) = e2πia⋅xf(x).

Proposition 4.1. Let f and g be smooth functions in the intersection of Lp spaces as p ranges over[1,∞).

F(f ∗ g) = f g and fg = f ∗ g and

∫ fg = ∫ fg.

Let α ∈ N (including zero) and denote xα = xα1

1 ⋯xαnn , ∂α = ∂α1

1 ⋯∂αnn and ∣α∣ = α1+⋯+αn. Then

F(∂αf)(ξ) = (2πiξ)f(ξ), ∂αFf(ξ) = Fx[(−2πix)f(x)](ξ).

τω(Ff) = F(Modω f) and Mod−ω Ff = F(τωf).

F Dilpε f = Dilp′

ε−1Ff and for all matrices A that are orthonormal F(f A) = (Ff) A.

The second item of the proposition should be understood as saying that decay of the Fourier transformimplies smoothness of the function and smoothness of the Fourier transfrom implies decay of the Fouriertransform. The third item connects location in the Fourier space to oscillatory phase of in the physicalspace. The last item is an important duality between scalings of the function and its Fourier transform.

Proposition 4.2. Let g(x) = e−π∣x∣2

. Then g is its own Fourier transform.

The proof is left as an exercise. You can consider the first order differential equation solved by theone-dimensional g, take a Fourier transform, and apply the formula for the derivative from the previousproposition. The fact that the normalized Gaussian is its own Fourier transform can be used to provethe Fourier inversion formula.

Proposition 4.3. If f, f ∈ L1, then F2f(x) = f(−x).

Proof. The family gε = Dil1ε gε>0 form an approximation of identity, and clearly F2(gε) = gε andF2τxh = FMod−xFh = τ−xF2h for any h. Let x be such that both x and −x are Lebesgue points of f .Then by even parity of g and the previous propositions

f(−x) = limε∫ f(y)gε(x + y)dy = lim

ε∫ f(y)(F2gε)(x + y)dy

= limε∫ f(y)F2τxgε(y)dy = lim

ε∫ F2f(y)gε(y − x)dy = F2f(x)

as was claimed.

Using the inversion formula, we can prove several norm inequalities that allow us to extend the Fouriertransform to functions for which the defining formula need not make sense as an absolutely convergentintegral.

Proposition 4.4. F extends to a contraction on L2 → L2, L1 → L∞ and Lp → Lp′

whenever p ∈ [1,2].Moreover, it is an isometry on L2.

Proof. Let f ∈ L1 ∩L2. Then

∫ ∣f ∣2 = ∫ ff = ∫ f(y)F2f(−y)dy = ∫ f(y)FFf(y)dy = ∫ f f = ∫ ∣f ∣2.

By density and linearity F has an extension to the whole L2 so that the equality of the norms above isvalid. It is elementary to see that for f ∈ L1

∣f(ξ)∣ = ∣∫ f(y)e−2πix⋅ξ∣ ≤ ∫ ∣f ∣.

Interpolating these two bounds with the Riesz–Thoring interpolation theorem (Theorem 1.3) we get thelast norm inequality claimed.

The fact that the Fourier transform is an isometry on L2 is usually referred as Plancherel’s identity.12

4.2. Schwartz class. Sometimes it is helpful to restrict the considerations to a dense subclass evensmaller than Lp ∩ L1. The correct choice in Fourier analysis is the Schwartz class S. For α,β ∈ Nn, wedefine the Schwartz seminorms

∥f∥Sα,β = supx∈Rn

∣xα∂βf(x)∣.

We say f ∈ S if the seminorms above are finite for all α,β ∈ Nn. This space is endowed with thetopology induced by the countable family of seminorms above (in practise it is enough to remember thatconvergence in Schwartz space topology means convergence in all the Schwartz seminorms separately).Given this structure as a topological vector space, we can define the space of tempered distributions S ′as the topological dual of S (continuous linear functionals).

Any function in f ∈ Lp with p > 1 gives rise to a tempered distribution through ⟨ϕ, f⟩S,S′ = ∫ fϕ. Afunction failing to be locally integrable need not do that, neither does one growing to fast at infinity. Notethat this is different from the usual distributions defined as the dual of smooth and compactly supportedfunctions. The reason for using tempered distributions is exactly here: We cannot hope the Fouriertransform of a compactly supported function to stay compactly supported, but the Fourier transform ofa Schwartz function is a Schwartz function (this is easy to verify using Proposition 4.1). Hence we can

define the Fourier transform of a tempered distribution Λ as Λ(ϕ) = Λ(ϕ). Recall that a sequence ofdistributions is said to converge if it converge weakly*, that is Λj(ϕ) → Λ(ϕ) for all Schwartz functionsϕ.

4.3. Uncertainty. A plane wave e2πixξ with a definite frequency ξ given by a single point in the Fourierspace has no localization in the physical space. The uncertainty principle refers to all manifestations ofthis phenomenon. The first property to note is that not only point support but also compact support ofthe Fourier transform forces the function to be highly delocalized.

Proposition 4.5. Let f be bounded and compactly supported function of one real variable. If f iscompactly supported, then f = 0 identically.

Proof. Let

g(z) = ∫ e−2πizxf(x)dx

for z ∈ C. Because f is compactly supported, this integral is well defined (note that the imaginary part

of z can cause real trouble otherwise). It coincides with f when z is real. It is an analytic function in z.

If f is zero in an interval, so is g. But as g is analytic, this implies it must be identically zero.

The second small proposition in the same spirit is the Heisenberg uncertainty principle.

Proposition 4.6. Let f ∈ S(Rn) and ∥f∥2 = 0. Then

(∫ ∣x∣2∣f(x)∣2 dx)1/2

(∫ ∣ξ∣2∣f(ξ)∣2 dξ)1/2

≥ 1

4π.

The equality is achieved only in case of real Gaussian functions.

Proof. The inequality follows from integration by parts and Plancherel

1 = ∫ ∣f ∣2 ≤ 2∫ x∣f ∣∣f ′∣ ≤ 2(∫ ∣x∣2∣f(x)∣2 dx)1/2

(∫ ∣f ′(x)∣2 dx)1/2

= 4π (∫ ∣x∣2∣f(x)∣2 dx)1/2

(∫ ∣ξ∣2∣f(ξ)∣2 dξ)1/2

.

The case of inequality follows by checking when the inequalities in the proof are saturated (no moredetails here).

Finally, after discussing the phenomenon both f and f cannot have small support, we turn to a partial

converse. If f has support in a bounded set at scale R, then f is blurred out at all scales smaller thanR−1.

Proposition 4.7. Let f be such that supp f ⊂ B(0,R). Then f is locally constant at scale R−1 meaningthat for all M > 0 there is a constant CM and ϕR such that

f(x) = f ∗ ϕR(x), ∣ϕR(x)∣ ≤CMR

n

(1 +R2∣x∣2)M/2

13

Proof. Just take ϕR to be a smooth function so that 1B(0,R) ≤ ϕR ≤ 1B(0,2R) and ∣∇ϕR∣ ≤ 2. Then

f(x) = F−1f = F−1(ϕRf) = ϕR ∗ f

by the elementary properties in Proposition 4.1.

Similar property holds for sets other than balls. For instance, if the Fourier transform of f is supportedin a rectangle R with dimensions (h1, . . . , hn), then the almost constant property holds with the convolv-ing factor decaying away from the dual rectangle with same directions but dimensions (h−1

1 , . . . , h−1n ).

The first application of this property is the following.

Proposition 4.8 (Bernstein’s inequality). Let 1 ≥ p ≥ q < ∞ and f ∈ Lp(Rn). Suppose that supp f ⊂B(ξ0,R). Then

∥f∥Lq ≲ Rn(1p−

1q )∥f∥Lp .

Similarly, for any α ∈ Nn

∥∂αf∥Lp ≲ R∣α∣∥f∥Lp .

Proof. Take M > n and let ϕR be the function given by the Proposition 4.7. By Young’s convolutioninequality

∥f∥Lq = ∥f ∗ ϕR∥Lq ≤ ∥f∥Lp∥ϕR∥L(1+ 1

q−

1p)

−1 .

Here

∫ ∣ϕR∣(1+ 1q −

1p )

−1

≲ ∫Rn(1+ 1

q −1p )

−1

(1 +R2∣x∣2)M/2 dx = Rn(1+ 1

q −1p )

−1−n∫

1

(1 + ∣x∣2)M/2 dx

whence the first claim follows by raising to the power 1 + 1q− 1p. The second claim is proved in the same

way by applying the derivative on the function ϕR.

4.4. Classical Fourier multipliers. Next we study operators that perturb the frequencies of a function.More precicely, we study operators of the form

f ↦ F−1(mFf)

where m is a locally integrable function that is also a tempered distribution. Then its inverse Fouriertransform is a tempered distribution as well, and we denote

k = m, Tmf = k ∗ f = F−1(mFf).

These definitions make sense a priori for Schwartz functions. Indeed, one can define the convolution ofa tempered distribution and a Schwartz function (duality pairing and translation), and the convolutionformula can be verified for this object (not done here, if you are curious about the details, check whathas been done in Chapter 2 of [10] or Chapter 7 in [18]. The second alternative might require a shortmoment of getting used to the style).

We want to know when the Fourier multipliers are bounded operators on Lp spaces. For the casesp ∈ 1,2 there are simple answers.

Proposition 4.9. Let m be as above. Then Tm ∶ L2 → L2 is bounded if and only if m ∈ L∞ and∥m∥∞ = ∥Tm∥L2→L2 .

Proof. Assume first that m is bounded. Then by Plancherel

∫ ∣Tmf ∣2 = ∫ ∣m∣2∣f ∣2 ≤ ∥m∥∞ ∫ ∣f ∣2

so that ∥Tm∥L2→L2 ≤ ∥m∥∞. To prove the converse direction, fix R > 0 and let ϕ = 1B(0,R). Then

∫B(0,R)

∣m∣2 = ∫ ∣Tmϕ∣2 ≤ ∥Tm∥L2→L2 ∫ ∣ϕ∣2 = ∥Tm∥L2→L2 ∣B(0,R)∣.

Dividing the measure on the left and sending R → 0, we conclude the claim by the Lebesgue differentiationtheorem.

Proposition 4.10. The multiplier operator Tm is bounded L1 → L1 if and only if m = µ for a finitecomplex Borel measure.

14

Proof. If m = µ, we take Schwartz f and it follows from Fubini

∫ ∣Tmf ∣ = ∫ ∣µ ∗ f ∣ ≤ ∥f∥L1∥µ∥.

Conversely, if Tm is bounded, we take f(x) = e−π∣x∣2

and let fk = Dil12−k f so that fk ∗ µ→ µ in the senseof (tempered) distributions as k →∞ and each Tmfk is a smooth function. On the other hand,

∥Tmfk∥ ≤ ∥Tm∥L1→L1

so the sequence is uniformly bounded in L1 and hence in total variation. As finite Borel measures arethe dual of continuous functions, it follows from the Banach–Alaoglu theorem that there is a weakly*convergent subsequence of Tmfk. Hence the distributional limit must be a finite Borel measure.

The question for other Lp spaces is much more difficult, and are no characterizations in general. Thefollowing theorem gives a sufficient condition, which is not sharp but it is very useful in applicationslater. The proof is based on a decompostion of the multiplier, which we will also need later. Let φ bea smooth radially decreasing function such that 1B(0,1) ≤ φ ≤ 1B(0,5/4). Denote φ = ϕ(x) − ϕ(2x) and

define ϕ(x) = φ(x)/∑j∈Z φ(2−jx). Note that for any x, there only two non-zero terms in the sum. Inaddition, ϕ(x) = 1 for 5/8 ≤ ∣x∣ ≤ 1 and the dilates ϕj = Dil∞2j form a partition of unity of the puncturedspace Rn ∖ 0.

Theorem 4.11 (Mikhlin). Let m ∈ C∞(Rn ∖ 0) satisfy

∣∂αm(ξ)∣ ≤ Aα∣ξ∣−∣α∣, for all α ∈ Nn.

Then Tm is a CZO and consequently bounded Lp → Lp for p > 1 and L1 → L1,∞.

Proof. First, the condition with α = (0, . . . ,0) means that m is bounded. Hence Tm is a bounded operatoron L2. Next, we study the kernel. Take the partition of the unity constructed before the statement ofthe theorem and denote mj =mϕj . We define the kernels (Schwartz functions) kj = F−1mj and considerthe sum ∑j kj . We prove that this sum converges to a function smooth outside of origin, satisfies thekernel estimates from Definition 3.1 and is associated with the operator Tm in the sense of the samedefinition.

Let M ≥ 0 be a natural number and take a multi-index α ∈ Nn. Then by the L1 → L∞ bound of theFourier (and inverse Fourier) transform as well as Proposition 4.1

∣∣x∣2M∂αkj(x)∣ ≲ ∑β∈Nn

∣β∣=2M

∫ ∣∂βξ [ξαϕj(ξ)m(ξ)]∣dξ ≲ ∫∣ξ∣∼2j

∣ξ∣∣α∣−2M ∼ 2j(∣α∣−2M+n).

Now

∑j≤log2

1∣x∣

∣∂αkj(x)∣ ≲ ∑j≤log2

1∣x∣

2j(∣α∣+n) ≲ ∣x∣−n−∣α∣

∑j>log2

1∣x∣

∣∂αkj(x)∣ ≲ ∑j>log2

1∣x∣

∣x∣2M2j(∣α∣−2M+n) ≲ ∣x∣−n−∣α∣

where 2M > ∣α∣ + n. This implies locally uniform convergence of the sum of all derivatives of the kernelskj , and hence the kernel k is smooth away from the origin and satisfies the size estimate

∣∂αk(x)∣ ≲ ∣x∣−n

by choosing α = (0, . . . ,0). Further for any x ≠ 0 and h ∈ B(0, ∣x∣/2)

∣k(x + h) − k(x)∣ = ∣∫∣h∣

0

h

∣h∣⋅ ∇k(x + th

∣h∣)dt∣ ≲ ∣h∣

∣x∣n+1

so the regularity estimate for Definition 3.1 is valid.To prove that this kernel is associated with the multiplier, take f and g from L2 with compact and

disjoint supports. By dominated convergence (supports and kernel bounds) and Plancherel

∬ f(x)∑j

kj(x − y)g(y)dxdy =∑j∬ f(x)kj(x − y)g(y)dxdy

=∑j∫ Tmjf ⋅ g =∑

j∫ mj f ⋅ g = ∫ mf ⋅ g = ∫ Tmf ⋅ g.

This completes the proof that Tm is CZO associated with ∑j kj . Now the rest of the claim follows fromthe Calderon–Zygmund theorem.

15

5. Littlewood–Paley theory

5.1. Khintchine. Recall the functions ϕjj defined before the statement of Theorem 4.11. Let Pjf =ϕj ∗ f be the Littlewood–Paley operator projecting onto frequencies near 2j . By elementary L2 orthog-onality one sees

∥Sf∥L2 ∶=XXXXXXXXXXXXX

⎛⎝∑j

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXL2

∼ ∥∑j

Pjf∥L2

where the last sum of operators just gives the identity. This inequality, which is geometrically describedas orthogonality in the Hilbert space L2, also remains true in more general Lp spaces. The probabilisticinterpretation of orthogonality is independence, and the square sum in Sf quantifies what would bethe effect of independent random signs. This interpretation carries over to Lp spaces where the L2

orthogonality driven by disjoint Fourier supports no longer has meaning. Also, recall the embeddings ofthe sequence spaces: if p > q, then `q ⊂ `p. Against this background, `1 and random signs give roughlythe effect of `2 norm.

Proposition 5.1 (Khintchine). Let (ai)Ni=1 be complex numbers and ωi independent random variablestaking values −1,1 with equal probability. Let p > 0. Then

(E∣∑i

ωiai∣p)1/p ∼p (∑i

∣ai∣2)1/2

where the hidden constant is in particular independent of N .

Proof. For simplicity, assume ai are real. We start by estimating the expectation. Let f = ∑i ωiai. Forany t > 0, by independence and zero expectation of the random variables

Eetf = E∏i

etωiai =∏i

Eetωiai =∏i

1

2(etai + e−tai) ≤∏

i

e12 t

2a2i = e12 t

2∑i a2i .

Also, P(f > λ) = P(−f > λ) so by Tschebyschev and the preceding remark

P(∣f ∣ > λ) = 2P(f > λ) = P(tf − tλ > 0) = P(etf−tλ > 1) ≤ 2e−tλEetf ≤ 2e−tλe12 t

2∑i a2i = e12λ

2∑i a2i

when we choose t = λ(∑i a2i )−1. Then by Cavalieri principle

E∣f ∣p = p∫∞

0λp−1P(∣f ∣ > λ)dλ ≤ 2p∫

∞

0λp−1e−λ

2/(∑i a2i ) dλ = Cp (∑i

a2i)p/2

which concludes the estimate.For the other direction, we note that for p = 2 the comparability holds as an equality. Also, by Holder’s

inequality for probablitity spaces the case p > 2 is clear. For p ∈ (0,2) take s > 2 and θ ∈ (0,1) such that1/2 = θ/s + (1 − θ)/p. Then (the following are with respect to the probablity measure

∥f∥L2 ≤ ∥f∥1−θLp ∥f∥θLs ≤ Cθs ∥f∥1−θ

Lp ∥f∥θL2

where the last step used the first part of the theorem. Now hiding the L2 norm on the left concludes theproof.

Theorem 5.2 (Littlewood–Paley theorem). Let Pj be the Littlewood–Paley pieces. If f ∈ Lp with p > 1,then

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp∼ ∥f∥Lp .

Proof. By Fatou, Khintchine, and Theorem 4.11,

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp≤ lim inf

N→∞

XXXXXXXXXXXXX

⎛⎝ ∑∣j∣≤N

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp≲ lim inf

N→∞(E

XXXXXXXXXXXX∑

∣j∣≤NωiPjf

XXXXXXXXXXXX

p

Lp

)1/p ≲ lim infN→∞

(E ∥f∥pLp)1/p

= ∥f∥Lp .

Conversely, we know that Pj = ∑j+100k=j−100 PkPj so set Fj = ∑j+100

k=j−100 Pk. Let ⟨⋅⟩ be the L2 inner product

and let g be a Schwartz function. The Littlewood–Paley decomposition converges as, say, tempered16

distribution so the following holds

⟨f, g⟩ = ⟨∑j

FjPjf, g⟩ = limN

⟨ ∑∣j∣≤N

Pjf,Fjg⟩ ≤ ∫⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2⎛⎝∑j∈Z

∣Fjg∣2⎞⎠

1/2

≤XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp

XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Fjg∣2⎞⎠

1/2XXXXXXXXXXXXXLp′≤XXXXXXXXXXXXX

⎛⎝∑j∈Z

∣Pjf ∣2⎞⎠

1/2XXXXXXXXXXXXXLp∥g∥Lp′ .

5.2. Ap weights. We only use Ap weights as a tool even if one could study them as a subject of itsown interest. Hence many proofs and results here are highly unoptimal when it comes to finer aspectsof the theory (see [7] for more). We call a locally integrable function a weight if it is non-negative. Sucha weight w is understood as a measure wdx, and for E a measurable set, we define w(E) = ∫E wdx.

Definition 5.3. Let p ∈ (1,∞). Then w ∈ Ap if

[w]Ap ∶= supQ⨏Qw ⋅ (⨏

Qw1−p′)

p−1

<∞.

We say w ∈ A1 ifMw ≤ Cw

almost everywhere. The smallest C possible is denoted by [w]A1 .

Proposition 5.4. The following hold:

If p ≤ q, then Ap ⊂ Aq. In particular, A1 ⊂ Ap for all p ≥ 1.

Let σ = w1−p′ . Then w ∈ Ap if and only if σ ∈ Ap′ If w ∈ Ap, then M ∶ Lq(w)→ Lq(w) for all q > p.

Proof. The first item follows from Holder’s inequality. The second one is clear by writing down thedefinitions. The third one follows by interpolation: we only prove Lp(w) → Lp,∞ and then use the firstitem together with Marcinkiewicz interpolation theorem.

Let λ > 0 and K ⊂ Mf > λ be compact. Form a Vitali cover 5Qi so that Qi are disjoint andfQi > λ. Then

w(K) ≤∑i

w(5Qi) ≤∑i

w(5Qi) (1

λ⨏Qi

∣f ∣dx)p

= λ−p∑i

w(5Qi)∣Qi∣

(∫Qi

∣f ∣w1/pw−1/p dx)p

≤ λ−p∑i

w(5Qi)σ(Qi)p−1

∣Qi∣p ∫Qi

∣f ∣pwdx ≲ ∫ ∣f ∣pw

and the claim follows by exhausting the level set by compact sets and using inner regularity of theweighted measure.

Remark 5.5. The last item extends to w ∈ Ap implying M ∶ Lp(w)→ Lp(w), but we will not use that.

Theorem 5.6. Let T be a CZO and w ∈ A1. Then T is bounded on Lp(w) for any p ∈ (1,∞).

Proof. By assumption and the previous proposition w,σ ∈ Aq for all q and consequently M is bounded

both Lp(w) and Lp′

(σ). By density, assume that f is compactly supported and bounded. Let g ∈ Lp′

(w).By sparse domination 3.8

∣∫ Tf ⋅ g wdx∣ ≤∑Q

∣Q∣⟨f⟩Q⟨gw⟩Q ≤∑Q

∣E∣ infQMf ⋅ inf

QM(gw) ≤ ∫ w1/pMf ⋅w−1/pM(gw)⋅

≤ ∥Mf∥Lp(w)∥M(gw)∥Lp′(σ) ≲ ∥f∥Lp(w)∥g∥Lp′(w)

whence the claim follows by taking supremum over all g with norm at most one.

The next question is whether the theorem above is of any use. The following proposition shows thatall reasonable functions are controlled by A1 weights of essentially equal size in terms of all Lp norms.Indeed, denoting Mpf = (M(fp))1/p we see that for f ∈ Lp and p > 1 this functions is of the form claimedin the following proposition.

Proposition 5.7. Let w be a weight such that Mw < ∞ almost everywhere. Let δ ∈ [0,1). Then[(Mw)δ]A1 ∼n,δ 1.

17

Proof. We use the uncentred maximal function M in the definition of A1 (it can always be chosen). FixQ. Let w1 = w12Q and w2 = w −w1. By Kolmogorov

⨏QMwδ1 ≲ (⨏

2Qw)

δ

≤ infx∈Q

Mw(x)δ.

Suppose a cube R meets x ∈ Q and ∫Rw2 ≠ 0. Then R ∩ (2Q)c ≠ ∅ and consequently ∣R∣ ≥ 2n∣Q∣. Itfollows that Q ⊂ 100R and hence

⨏QMwδ2 ≤ sup

x∈Q

⎛⎜⎜⎝

supR∋x

R∩(2Q)c≠∅⨏Rw

⎞⎟⎟⎠

δ

≲ infx∈Q

Mw(x)δ.

This concludes the proof.

5.3. Weighted Littlewood–Paley inequality. Next we see that the orthogonality of the Littlewood–Paley projections almost carries over to weighted L2 spaces (exactly as it carried over to the unweightedLp spaces).

Theorem 5.8. Let w be a weight and α > 1. Let f ∈ L2(Mαw) and Sf be the Littlewood–Paley squarefunction. Then

∥Sf∥L2(w) ≲ ∥f∥L2(Mαw), ∥f∥L2(w) ≲ ∥Sf∥L2(Mαw).

Proof. Let (εj) be a sequence of random signs as in Khintchine’s inequality. By independence

E∣∑j

εjPjf ∣2 = E∑i,j

εiεjPifPjf =∑i

∣Pif ∣2 = ∣Sf ∣2.

On the other hand, Mαw ≥ w and Mαw ∈ A1 provided it is almost everywhere finite (and otherwise thetheorem is trivially true). Now ∑i εiPi are Mikhlin multipliers uniformly in the choice of the randomsigns and hence by the weighted CZ theorem 5.6

∥Sf∥2L2(w) ≤ E∥∑

i

εiPif∥2L2(Mαw) ≲ E∥f∥2

L2(Mαw) = ∥f∥2L2(Mαw).

It remains to prove the converse direction. We factor the identity multiplier as follows. Let ϕj be the

symbol of Pj , that is, Pjf = ϕj f . Denote ψj = ∑j+100i=j−100 ϕi. Then

1 =∑j

ϕj =1000

∑k=1

∑j∈1000Z

ε2jϕk+jψk+j

for any sequence of signs (εj)j . Fix k ∈ 1, . . . ,1000. By disjoint supports

∑j∈1000Z

ε2jϕk+jψk+j =⎛⎝ ∑j∈1000Z

εjϕj+k⎞⎠⎛⎝ ∑j∈1000Z

εjψj+k⎞⎠.

Let Fjf = ψj ∗ f . Then by the weighted CZ Theorem and Mikhlin multiplier theorem

∥f∥L2(w) ≤1000

∑k=1

XXXXXXXXXXX

⎛⎝ ∑j∈1000Z

εjPj+k⎞⎠⎛⎝ ∑j∈1000Z

εjFj+k⎞⎠fXXXXXXXXXXXL2(w)

≲1000

∑k=1

XXXXXXXXXXX∑

j∈1000ZεjPj+kf

XXXXXXXXXXXL2(Mαw).

Taking the expectation over the random sign sequences εj and applying Khintchine, we get

∥f∥2L2(w) ≲

XXXXXXXXXXXERRRRRRRRRRR∑

j∈1000ZεjPj+kf

RRRRRRRRRRR

XXXXXXXXXXX

2

L2(Mαw)≲XXXXXXXXXXXXX

⎛⎝ ∑j∈1000Z

∣Pj+kf ∣2⎞⎠

1/2XXXXXXXXXXXXX

2

L2(Mαw)

≤ ∥Sf∥2L2(Mαw)

18

5.4. Subdyadic scales. Next we use the weighted inequalities to study multipliers beyond CZ theory.A typical Mikhlin multiplier is essentially constant at dyadic scales. Indeed, if ε∣ξ∣ ∼ ε∣ξ′∣ ∼ ∣ξ − ξ′∣ ∼ R,then the condition assumed in Theorem 4.11 implies

∣m(ξ) −m(ξ′)∣ ≤ ∫∣ξ−ξ′∣

0∣∇m(ξ + ξ′ − ξ

∣ξ′ − ξ∣t)∣dt ≲ ∫

∣ξ−ξ′∣

0

1

R≲ 1.

Next we consider multipliers, which have more complicated structure. Let ρ ∈ (0,1]. We let the multiplier

have non-trivial variation down to ρ-subdyadic scales identified as the regions ∣ξ0∣ ∼ ∣ξ∣ ∼ ∣ξ0 − ξ∣1/ρ withξ0 ∈ Rn by asking a derivative in the frequency variable to cause only decay according to the power ρ.The largest sets where the multiplier can then be regarded as almost constant are the ρ-subdyadic ballsthat are given by

diam(B) ∼ dist(B,0)ρ.These balls are smaller than the dyadic balls with diam(B) ∼ dist(0,B) only if diam(B) is large.

Definition 5.9. Let s ∈ R, ρ ∈ (0,1] and m ∈ C∞. We say m ∈ Ssρ,0 if for all α ∈ Nn it holds

∣∂αm(ξ)∣ ≲α (1 + ∣ξ∣)s−ρ∣α∣

This is a special case of more general pseudodifferential operators where we would allow the symbolm to depend on the physical variable x and impose limited growth with respect to spaces derivatives.The following weighted inequalities for the multipliers can be skimmed from the papers [2, 3], whereasthe case of more general pseudodifferential operators is treated in [1]. We prove a weighted L2 boundand deduce Lp bounds from that. The interplay with the weigths and the structural conditions on theFourier multiplier is the key aspect of to keep eyes on in what follows.

Definition 5.10 (Approach region). Let ρ ∈ (0,1]. Define Λ(x) = (y, t) ∈ Rn × (0,1] ∶ ∣x − y∣ ≤ tρ. Letalso s ∈ R. We define

Mρ,sf(x) = sup(y,r)∈Λ(x)

r−2s ⨏B(y,r)

∣f(z)∣dz.

Theorem 5.11. Let ρ ∈ (0,1], s ∈ R and m ∈ Ssρ,0. For any weight w and index q > 1, we have that

∫ ∣Tmf ∣2w ≲ ∫ ∣f ∣2MqMMρ,sMqw.

Proof. We can assume for the time being that w is bounded and compactly supported. The general casethen follows by monotone convergence. Let ψ0 be a smooth bump supported in the unit ball and ψjpartition of unity adapted to the annuli 2j ∼ ∣ξ∣. Denote fj = ψj ∗ f . By weighted Littlewood–Paleytheory

∫ ∣Tmf ∣2w ≲∑j≥1∫ ∣Tmfj ∣2Mqw + ∫ ∣Tmf0∣2w.

We fix a frequency annulus with j ≥ 0. As it were, the multiplier operator is given through a convolution,which can be controlled by the Hardy–Littlewood maximal function provided we manage to decouplethe frequency space annulus to subdyadic balls.

First note that because supp Tmfj ⊂ B(0,2j+1), it follows that

suppF(∣Tmfj ∣2) = supp(FTmfj ∗FTmfj) ⊂ suppFTmfj +FTmfj ⊂ B(0,2j+1) +B(0,2j+1) ⊂ B(0,2j+2).

Let Ψ0 be a smooth function such that 1B(0,1) ≤ Ψ0 ≲ 1B(0,4) such that Ψ0 is positive. Indeed, we can

choose Ψ0 to be radial and vanishing outside B(0,2). Its inverse Fourier transform is real so ( ˇΨ0)2 is

positive and Ψ0 = Ψ0 ∗ Ψ0 ≲ 1B(0,4). Then let Ψj = Dil∞2j+2 Ψ0. Now

∫ ∣Tmfj ∣2Mqw = ∫ F(∣Tmfj ∣2) ⋅FMqw = ∫ F(∣Tmfj ∣2) ⋅ΨjFMqw = ∫ ∣Tmfj ∣2Ψj ∗Mqw.

Next we want to shrink the Fourier support of the weight further to a ball B(0,2jρ). This is doneby an additional Fourier cut-off (convolution), but this time we have to pay a price. Discarding effect offrequencies with ∣ξ∣ ≳ 2ρj means passing to averages at scale 2−ρj . Maximizing over translations at thatscale then pays back the loss of the high frequency part: Define

Wj(x) = ϕj ∗ ( sup∣z−⋅∣≤2−ρj

Ψj ∗Mqw(z))(x)

where ϕj = Dil2ρj Ψ0. Then

suppFWj ⊂ B(0,2ρj+2) by the choice of ϕj .19

Wj ≳ Ψj ∗Mqw because

Wj(x) ≳ ⨏B(x,2−ρj−10)

sup∣z−y∣≤2−ρj

Ψj ∗Mqw(z) ≥ Ψj ∗Mqw(x).

Next let B = B be a boundedly overlapping cover of the frequencies around 2j by subdyadic ballsand let χB form the partition of unity adapted to it. Note that one can choose it so that

∣∂αχB(ξ)∣ ≲ 2−ρj∣α∣

for all α ∈ Nn and all B in the covering. Further, for each B define the set

S(B) = B′ ∈ B ∶ dist(B′,B) ≤ 10 diam(B)

(satellites) and we know that #S(B) ≤ Cn. Now

∫ ∣Tmfj ∣2Ψj ∗Mqw ≲ ∫ ∣Tmfj ∣2Wj = ∫ ∣∑B∈BF−1(χBmfj)∣

2

Wj

= ∫ ∑B,B′∈B

F−1(χBmfj)[F−1(χB′mfj)Wj]

= ∑B∈B

∑B′∈S(B′)

∫ [F−1(χBmfj)][F−1(χB′mfj)Wj].(5.1)

The last equality used that if B′ ∉ S(B), then 10B′ ∩B = ∅ and the two square bracketed expressionshave disjoint Fourier support and hence their product must have integral zero.

Next we use the symbol estimates for the multiplier. Set kB = F−1(χBm) and let fB be a smoothrestriction of f to a dilate of B. Then

∣F−1(χBmf)(x)∣ = ∣kB ∗ fj(x)∣ ≤ ∫ ∣kB(x − y)∣∣fB(y)∣dy ≤ ∥kB∥2L1 (∫ ∣fB(y)∣2∣kB(x − y)∣dy)

1/2.

(5.2)

By the symbol estimates and scaling of B

∣xβkB ∣ ≤ ∫ ∣∂β(χBm)∣ ≲ ∫4B

2−j(∣β∣ρ−s) ∼ 2−j(∣β∣ρ−s−ρn)

so

∣kB(x)∣ ≲ 2js ⋅ 2jρn(1 + (2jρ∣x∣)2)−N ∶=KN(x).This estimate is uniform for all B, and ∥KN∥ ≲ 2sj so (5.1) can be estimated using (5.2) and the fact2∣ab∣ ≤ ∣a∣2 + ∣b∣2 through

∥KN∥2L1 ∑

B∈B∑

B′∈S(B′)∫ (∫ ∣fB(y)∣2∣KN(x − y)∣dy)

1/2(∫ ∣fB(y)∣2KN(x − y)dy)

1/2Wj(x)dx

≲ 22sj ∑B∈B

∑B′∈S(B′)

∬ ∣fB(y)∣2KN(x − y)Wj(x)dxdy

= 22sj ∑B∈B

∑B′∈S(B′)

∫ ∣fB(y)∣2KN ∗Wj(x)dy ≲ 22sj ∫ ∣fj(y)∣2KN ∗Wj(x)dy

≲ ∫ ∣fj(y)∣2MMρ,sMqw(y)dy.

Finally, summing over all possible j, we conclude the proof by the weighted Littlewood–Paley theory.

5.5. Extrapolating unweighted bounds. The L2 bounds with very general weights imply unweightedLp bounds by a simple duality argument, which we study next. This can be seen as a simple instanceof more general extrapolation of weighted inequalities. Typically weighted Lp bounds imply bounds forseveral different values of p even if the initial claim were for a single value of p.

Proposition 5.12. Let 2 ≤ p ≤ q and assume that S ∶ L(q/2)′ → L(p/2)′ is bounded. If T is a sublinearoperator such that there exists a dense X ⊂ Lp so that Tf ∈ Lq for all f ∈X and the weighted inequality

∫ ∣Tf ∣2w ≤ C ∫ ∣f ∣2Sw

holds for all weights w, then T ∶ Lp → Lq is bounded.20

Proof. It suffices to prove the norm inequality for f ∈X. Denote g = ∣Tf ∣q−2. Then

∫ ∣Tf ∣q = ∫ ∣Tf ∣2g ≲ ∫ ∣f ∣2Sg ≤ ∥f∥2Lp∥Sg∥L(p/2)′ ≲ ∥f∥2

Lp∥g∥L(q/2)′ = ∥f∥Lp∥Tf∥q−2Lq .

As the second factor on the right hand side is finite by the fact f ∈ X, we can divide by ∥Tf∥q−2Lq and

take the square root to conclude the proof.

Once this proposition is knwon, the role of the maximal functions appearing in the right hand side ofthe conclusion of Theorem 5.11 is clear. The Hardy–Littlewood maximal function together with its Lq

variants with small q are bounded on all Lp spaces p > q so the nature of the multiplier is captured bythe maximal function Mρ,s.

Proposition 5.13. Let 1 < p ≤ q <∞, s ∈ R and ρ ∈ [0,1]. If

−2s

n≥ (1 − ρ)

qn+ (1

p− 1

q) ,

then Mρ,s is bounded Lp → Lq.

Proof. This is only sketch of the proof, which is based on interpolation. There are three Lp → Lq estimateswhere one can hope to easily prove a sharp estimate. The case of averaging operator L∞ → L∞, thecase of convolution with a bounded function L1 → L∞, and the case of non-tangential maximal functionH1 → L1. The last one requires the use of Hardy spaces, which suggests we have to control the maximalfunction by another one with smooth kernel. Moreover, as the proof proceeds via analytic interpolation,we have to linearize the whole setup in order to be able to apply that.

First, we can easily find non-negative ϕ ≳ 1B(0,1) compactly supported in B(0,2) so that we can replacethe rough average by a smooth one in the definition of Mρ,s. Second, it suffices to take supremum overrational points in Γ(x) = (x,0) + Γ(0). Finally, for any simple function f ≥ 0, we can find a measurableγ ∶ Rn → Γ(0) so that

Mρ,sf(x) ≲ t−2sϕt ∗ f(y) =∶ Pγs f(x)

where (y, t) = γ(x). This operator is linear, and it can be seen to be analytic in s. Once this reductionhas been done, one establishes for all simple functions g that the operator Pγs defined using f satisfiesthe bounds

∣Pγ0 g(x)∣ ≲ ∥g∥L∞∥ϕ∥L1(5.3)

∣Pγ−n/2g(x)∣ ≲ ∥g∥L1∥ϕ∥L∞(5.4)

∣Pγ−n(1−ρ)/2g(x)∣ ≲ ∥g∥L1∥ϕ∥L∞ .(5.5)

The first two are clear by inspection. First we do not take fractional part so we end up having anaveraging operator. Then we put so much weight on the fractional part that we do not average anymorebut we just integrate. The third alternative corresponds to increasing the radius in the integral from r torρ and balancing the ratio of measures ∣B(0, rρ)∣/∣B(0, r)∣ = rn(ρ−1) by the fractional part so that we endup having a non-tangential maximal function (one whose approach region is a cone). It then remains tobound this operator from the Hardy space to L1.

The atomic Hardy space is defined as follows. We call a measurable a with

suppa ⊂ B, ∫ a = 0, ∥a∥Lp ≤ ∣B∣−1/p′

an p-atom adapted to B. If h is a finite linear combination of 2-atoms, we norm it by

∥h∥H1 = inf∑j

∣λj ∣ ∶ h =∑j

λjaj

where aj are atoms. We could close this space with respect to the norm above, but even that isnot necessary. From this definition one already sees that the space is very well suited for complexinterpolation. From this point on, it remains to estimate the action of Pγ−n(1−ρ)/2 on a single atom and

use linearity. This is a computation that we omit. All the bounds are uniform in f determining thelinear operator so the desired Lp → Lq bounds follow for all linearizations, and hence for any given simplefunction f , we can find a linearization that beats the original maximal function and apply the boundsfor that linearization.

21

6. Oscillatory integrals

All the content of this section can be found in Chapter 8 of [17]. If you can, read that and skip thissection of the lecture notes. An integral over a product of a smooth function and a highly oscillatingfactor is typically much smaller than what can be said without taking into account the effect of thecancellation. A typical example of this fact is the sequence of functions sin(2πnx) on the unit intervalconverging to zero in the sense of distributions while being bounded away from zero in all Lp norms withp ∈ [1,∞].

This phenomenon is very similar to the definition of Fourier transform and in particular its values forhigh values of the frequency parameter. To verify certain (important) features of Fourier transforms ofmeasures supported on smooth and curved hypersurfaces, we study in a modular way the following typeof oscillatory integrals. Let ψ ∈ C∞

c (Rn) be a complex valued amplitude function and ϕ ∈ C∞(Rn) a realvalued phase function. An oscillatory integral characterized by the two is an expression

I(λ) = ∫ eiλϕ(x)ψ(x)dx, λ > 0.

As ϕ is real, the exponent is purely imaginary and the exponential factor is bounded and oscillating(Euler’s identity). Because ψ is assumed to be compactly supported, the oscillatory integral convergesabsolutely. We aim at studying its decay along λ→∞.

6.1. Principle of stationary phase. Principle of the stationary phase is the following. Consider thephase ϕ as a graph of a function on Rn. The oscillatory integral is an integral over the surface presentedas the graph. The parameter λ gives a slicing of the upper half space to slabs of height π/λ. Thefunction t ↦ eiλt has its real part single signed in each of the slabs and the signs of the neighboringslabs alternate. If ϕ does not have a stationary point (e.g. local maximum), the contributions of the slabscancel each other and the oscillatory integral decays fast. If the phase function has a stationary point,the slab containing that one contains a large area of the graph of ϕ, and the (single signed) contributionof that area weights more than all the rest as the slicing gets finer and finer.

Theorem 6.1 (Unstationary phase). Assume ∇ϕ ≠ 0 in suppψ. Then I(λ) ≲N,ϕ,ψ λ−N for any N > 0.

Proof. For smooth f , define

Lf = 1

iλ

∇ϕ∣∇ϕ∣

⋅ ∇f, Ltf = − 1

iλ∇ ⋅ ( ∇ϕ

∣∇ϕ∣f) .

Integration by parts then shows

∣I(λ)∣ = ∣∫ (LNeiλϕ(x))ψ(x)dx∣ = ∣∫ eiλϕ(x)(Lt)Nψ(x)dx∣ ≲ λ−N

where the last step used the fact that all derivatives of the phase function are bounded and that thegradient is uniformly bounded away from zero in the support of ψ.

Theorem 6.2 (Stationary phase). Assume det∇2ϕ ≠ 0 in suppψ. Then I(λ) ≲ϕ,ψ λ−n/2.

Proof. Let ε > 0 be a small number to be fixed later depending on the phase. By a partition of unity, wecan assume diam(suppψ) < ε. Then

I(λ)I(λ) =∬ eiλ(ϕ(x)−ϕ(y))ψ(x)ψ(y)dxdy =∬ eiλ(ϕ(y+u)−ϕ(y))ψ(y + u)ψ(y)dxdy = ∫ J(y)dy

where

J(u) = ∫ eiλ(ϕ(y+u)−ϕ(y))β(y, u)dy, β(y, u) = ψ(y + u)ψ(y).

We prove a bound for J(u) first. Define the operator

Lf = 1

iλa ⋅ ∇f, a = ∇(ϕ(y + u) − ϕ(y))

∣∇(ϕ(y + u) − ϕ(y))∣2=∶ b

∣b∣2.

Because all derivatives of ϕ are bounded, we see that for all α ∈ Nn

∣∂αb∣ = ∣∇∂αϕ(y + u) −∇∂αϕ(y)∣ = ∣∫∣u∣

0(∇2∂αϕ)(y + t u

∣u∣) u∣u∣dt∣ ≲ ∣u∣.

On the other hand, as det∇2ϕ ≠ 0, it follows from compactness and continuity that all the eigenvaluesof ∇2ϕ are uniformly bounded away from zero. Hence ∇2ϕ(x) is invertible and ∣∇2ϕ(x)u∣ ≳ ∣u∣ for allx ∈ suppψ and u ∈ Rn. Finally, as

(∇2ϕ)(y + t u∣u∣

) u∣u∣

22

is a smooth function in the variable t, we conclude that it cannot change sign provided the support of ψis small enough. Hence ∣b∣ ∼ ∣u∣. These estimates together imply ∣∂αa∣ ≲ ∣u∣−1 and further

∣(Lt)Nβ(y, u)∣ ≲ (λ∣u∣)−1

and

∣J(y)∣ = ∣∫ (LNeiλ(ϕ(y+u)−ϕ(y)))β(y, u)dy∣ = ∣∫ eiλ(ϕ(y+u)−ϕ(y))(Lt)Nβ(y, u)dy∣ ≲ (λ∣u∣)−N .

Hence

∣I(λ)∣2 ≤ ∫ ∣J(u)∣du ≲ ∫du

(1 + λ∣u∣)n+1≲ λ−n.

Taking the square root then concludes the proof.

Finally, in dimension one it is possible to prove decay estimates without the assumption on compactsupport of the amplitude function.

Theorem 6.3 (Van der Corput lemma). Let ϕ ∈ Ck(R) and assume that ∣ϕk(x)∣ ≥ 1 for all x ∈ [a, b].Then for all λ > 0

∣∫b

aeiλϕ(x) dx∣ ≤ Ckλ−1/k

where the constant only depends on k (and not on a and b or anything else) provided that k ≥ 2 or thatϕ′ is monotone.

The case when k = 1 follows by a simple integration by parts and the higher values of k can be dealtwith using induction on k. The proof is omitted here.

6.2. Surface carried measures. The main application of decay estimates for oscillatory integrals isin the study Fourier transforms of measures supported on hypersurfaces. We will assume throughoutthis subsection that Ω ⊂ Rn is a bounded open set so that there is a finite collection of balls Bi so that∂Ω ⊂ ⋃Ni=1Bi and for each i there is a smooth function ϕi such that

∂Ω ∩Bi = (x′, xn) ∈ Bi ∶ xn = ϕi(x′)for a suitably rotated coordinate system. We will usually restrict the attention to one such ball Bi andomit the index i from the notation. We call Ω a smooth domain and M = ∂Ω a smooth hypersurface.

The most central geometric quantities related to the hypersurface are the following. The definingfunction ρ(x) = ϕ(xn)−xn gives the piece of the hypersurface as its zero set in a ball. Hence the normalof the surface is given by N = ∇ρ/∣∇ρ∣. For u and v tangent to the surface, we can note that the shapeoperator of the surface is Lv = ∇Nv and the second fundamental form II(u, v) = Lu ⋅ v = ∣∇ρ∣−1∇2u ⋅ v.The Gaussian curvature is the determinant of ∇2ρ (one can show it does not depend on the choice ofthe defining function ρ) and the eigenvalues of ∇2ρ are called the principal curvatures.

Finally, the surface measure (which coincides with the n − 1 dimensional Hausdorff measure) can berecovered as a weak limit of normalized Lebesgue measures around the surface: For each f continuouson the surface there is an extension f continuous in a neughborhood of the surface. We set

∫Mf dσ ∶= lim

ε→0

1

2ε∫

dist(x,M)<εf(x)dx = ∫ f(x′, ϕ(x′))(1 + ∣∇ϕ(x′)∣2)1/2 dx′

for all f = f ∣M with f continuous. The last equality is just the familiar computation of surface integralfrom calculus. Using this formula, we can do precise computations about the Fourier transforms ofmeasures of the form f dσ.

Theorem 6.4. Let M be as above and k ∈ 1, . . . , n−1. If M has k non-vanishing principal curavtures,

then for all ψ ∈ C∞c (Rn) we have ψσ(ξ) ≲ψ ∣ξ∣−k/2.

Proof. We only prove the claim for the case k = n. The remaining cases can be deduced from thatby applying the theorem in lower dimensional subspaces of Rn. As discussed above, we can cover thehypersurface by balls inside of which it can be represented as a graph of a smooth function. We canrestrict the attention to a single piece. Further, as the constant in the theorem is allowed to depend onthe surface, we can use a partition of unity inside of the chosen piece to assume that diam(suppψ) < εfor some small ε > 0. By rotation and translation, we assume the ball we study is centred at origin,∇ϕ(0) = 0 and N(0) = (0, . . . ,1). After these reductions, the Fourier transform to be estimated becomes

ψσ(ξ) = ∫Rn−1

e2πi(x′⋅ξ′+ξnϕ(x′))ψ(x′)dx′.23

Without loss of generality, we assume that ξn ≥ 0.We restrict the study to two cases. In case 1, we assume that ∣ξ′∣δ < ξn and in case 2, we assume that

∣ξ′∣δ ≥ ξn. These correspond to frequencies almost normal to the surface and the complementary case.This is where the principle of the stationary phase can be applied. Consider first the case 1. Here we set

λ = 2πξn and Φ(x′) = −ϕ(x′) − x′⋅ξ′ξn

. By assumption

det∇2Φ = det∇2ϕ ≠ 0

so that Theorem 6.2 implies

∣ψσ(ξ)∣ ≲ ξ−n−12

n ≲ ∣ξ∣−n−12 .

Consider then the case 2. Here we set λ = 2π∣ξ′∣ and Φ(x′) = −ϕ(x′) ξn∣ξ′∣ −ξ′⋅x′∣ξ′∣ . Because ∇ϕ(0) = 0 and

∣x′∣ ≤ ε by the support assumption on ψ, we see that

∣∇Φ(x′) = ∣ξn∇ϕ∣ξ′∣

− ξ′

∣ξ′∣∣ ≥ 1 − ∣ξn∣∣∇ϕ∣

∣ξ′∣≥ 1

2.

Hence it follows from Theorem 6.1∣ψσ(ξ)∣ ≲ ∣ξ′∣−N ≲ ∣ξ∣−N .

This concludes the proof.

With a little additional work, one can prove the following corollary stating that a characteristicfunction of the solid set has one order faster Fourier decay than its surface measure. Intuitively this isclear. Dirac mass is the derivative of the characteristic function of an interval on real line and similarlysurface measure is normal derivative of the characteristic function of the set.

Corollary 6.5. Let Ω be a smooth bounded domain so that ∂Ω has nonvanishing Gaussian curvature.Then

1Ω(ξ)∣ ≲ ∣ξ∣−n+12

The proof is omitted here, but it is easy to reconstruct from the previous one. It can also be found in[17].

6.3. Fourier restriction. As we saw in Section 4 of these notes, the Fourier transform is an isometryof L2 into itself. Hence the Fourier transform of a generic L2 function is not better than a generic L2

function. On the other hand, Fourier transform of an L1 function is always continuous. Fourier restrictiontheorems deal with the slightly mysterious phenomenon that for some values p ∈ (1,2), one can definerestriction of the Fourier transform onto a lower dimensional surface, something that is obviously trivialfor p = 1 and obviously impossible for p = 2. However, this does not work for any surface, but the surfacehas to be curved. From point of view of restricting the Fourier transform, there is not too much intuitionto be used, but dualizing the problem in a suitable way show an intimate connection to decay estimatesof the Fourier transforms just discussed.

For this section, fix a compact hypersurface M as in the previous section. Let σ be its surface measure.We define the extension operator and its adjoint

Ef(x) = ∫ e2πix⋅ξf dσ, E∗f(ξ) = f ∣M(ξ).

The extension operator maps functions defined on the surface to functions defined in the whole ambientspace and the restriction operator maps functions defined in the whole ambient spaces to ones definedon the surface. The following proposition shows the relation between these two operators.

Proposition 6.6. Let 1 < p, q <∞. The following are equivalent

∥Ef∥Lq(Rn) ≤ C1∥f∥Lp(σ)(6.1)

∥E∗f∥Lp′(σ) ≤ C1∥f∥Lq′(Rn)(6.2)

∥EE∗f∥Lq(Rn ≤ C1∥f∥Lq′(Rn) and p = 2.(6.3)

Proof. This follows immediately by writing down the definitions and using Fubini and Holder.

The main result of this section is the following.

Theorem 6.7 (Tomas–Stein theorem). Let M be a compact hypersurface with non-vanishing gaussiancurvature, E the related extension operator and σ the surface measure. Then

∥Ef∥L

2(n+1)n−1 (Rn)

≲ ∥f∥L2(σ).

24

Proof. By taking a partition of unity, it suffices to study the case when f is compactly supported in acube Q ⊂ Rn such that there is smooth ϕ and suitably chosen coordinate system with

M ∩Q = (x′, xn) ∈ Q ∶ xn = ϕ(x′).

We also define the set Ω = (x,x′) ∈ Q ∶ xn < ϕ(x′). We study the kernel of EE∗ denoted by k0 = ψσwhere ψ is a smooth function supported in Q. To prove an Lq

′

→ Lq bound, it is possible to interpolateL1 → L∞ bound with an L2 → L2 bound. The decay of the Fourier transform of a surface carried measure

given by Theorem 6.4 shows that (1 + ∣ξ∣)n−12 k0(ξ) is bounded. Note that multiplying by this weightcorresponds to differentiating the Fourier transform of the kernel up to order (n − 1)/2. On the otherhand, as we know that the Fourier transform of k0 is ψσ, we are lead to guess that “integrating” theFourier transform once to the normal direction of the surface will yield something reminiscent of thecharacteristic function of Ω, which is a bounded function. This suggests an analytic family of kernelsalong the number of derivatives applied to the Fourier transform.

Let

Is(ρ) = (−1)N+1 ∫∞

0us+N∂N+1

u (F (u)e−2πiuρ)du

where N is a large natural number to be chose and F is a smooth function compactly supported in theinterval (−2,2). The following hold.

The map s↦ Is(ρ) can be continued as an analytic function to s ∈ C ∶ −(N + 1) < Re s. For a constant C bounded by the CN+1 norm of F , we have

∣Is(ρ)∣ ≤ C(1 + ∣s∣)N+1(1 + ∣ρ∣)−Re s

whenever −(N + 1) < Re s ≤ 1. I0(ρ) = N !F (0).

The first two items in the list are clear. For the third one, we note that for ∣ρ∣ ≤ 1 the claim is clear soit suffices to deal with the case ρ > 1. Let η be a smooth function such that 1[− 1

2 ,12 ] ≤ η ≤ 1[−1,1]. Then

∣∫∞

0us+N∂N+1

u (F (u)η(uρ)e−2πiuρ)du∣ ≲ ρN+1 ∫1/ρ

0uRe s+N du ≲ ρ−Re s.

On the other hand, integrating by parts, we see

∣∫∞

0us+N∂N+1

u (F (u)(1 − η(uρ))e−2πiuρ)du∣ = ∣∫∞

0us−1F (u)(1 − η(uρ)) ∂u

(−2πiρ)e−2πiuρ du∣

≲ 1

ρ∫

2

1/(2ρ)∣∂uus−1F (u)(1 − η(uρ))∣du ≲ ρ−Re s.

Note that the factor 1 − η is non-zero only on and interval of length ∼ 1/ρ. This concludes the proof forthe third item on the list. Finally, integrating by parts N times in the definition of I0 shows the lastitem claimed to be true.

Using the properties of Is(ρ), it is easy to conlcude the proof. Indeed, define

ks(ξ) = es2

∫Rn−1

∫∞

ϕ(x′)(xn − ϕ(x′))s+N∂N+1

xn (ψ(x′)e−2πi(x′⋅ξ′+xnξn)) dxndx′ = e2Is(ξn)

where the function F in the definition of Is is given by

F (u) = ∫Rn−1

e−2πi(x′⋅ξ′+ϕ(x′)u)ψ(x′, u + ϕ(x′))dx′.

Then it follows from the stationary phase computations (Theorem 6.2) that F together with all its

derivatives is bounded by (1 + ∣ξ∣)−(n−1)/2, and by the properties of ψ, it has support in the interval[−2,2]. Define the operator Tsf = ks ∗ f , and now we can conclude the following.

It holds∣ks(ξ)∣ ≲ (1 + ∣s∣)n(1 + ∣ξ∣)−Re s(1 + ∣ξ∣)−

n−12 .

In particular, ∣ks(ξ)∣ is uniformly bounded when Re s ∈ [−n−12,1]. In particular, Ts maps L1 →

L∞ boundedly when s = −(n − 1)/2.

k1 = 1Ωcψ ∈ L∞. In particular, T1 maps L2 → L2 boundedly. The family Ts is analytic.

The boundedness of T0 from Lq′

to Lq with q = 2(n+1)n−1

follows from Theorem 1.4.

An important corollary of the above proof is the following generalization to surface carried measuressatisfying an decay estimate of the Fourier transform.

25

Corollary 6.8. Let M be a local smooth hypersurface such that there is δ > 0 so that for all µ = ψσ withψ ∈ C∞

c it holds

∣µ(ξ)∣ ≲ (1 + ∣ξ∣)−δ.Then ∥Eµf∥Lq(Rn) ≲ ∥f∥L2(µ) for q = 2(δ+1)

δalso holds for all such measures.

Next we briefly discuss the possible improvements of the restriction theorem. For simplicity, we letM to be the unit sphere. We derive necessary conditions for an restriction (or extension) theorem tohold. Take δ > 0 and consider the cap Cδ = x ∈ Sn−1 ∶ 1 − x ⋅ en ≤ δ2. Here en is the unit vector to thedirection of the nth coordinate and the cap consists of the points whose position vectors make angle atmost δ to it. Let f = 1Cδ . Clearly

∣1Cδσ = ∣∫Cδe−2πiξ⋅(x−en) dσ(x)∣ ≥ ∫

Cδcos(2πiξ ⋅ (x − en)) ≥

1

2σ(Cδ)

whenever ∣2πiξ ⋅ (x − en)∣ ≤ π/3. As x − en is almost orthogonal to nth coordinate, one can move δ−2 tothat direction while keeping the quantity inside the cosine bounded by a numerical constant. Similarly,to all other directions, the vectors x− en can be as parallel to coordinate axes as possible, and hence onecan only move ∼ δ−1 without increasing the argument of the cosine. As a result, the lower bound in thedisplay above holds in a box Rδ of dimensions δ−1 ×⋯−1 × δ−2.

Suppose the extension operator E ∶ Lp → Lq boundedly. Then

∣Rδ ∣1/qσ(Cδ) ≲ ∥Ef∥Lq ≲ ∥f∥Lp(σ) = σ(Cδ)1/p

and

1 ≲ δ(n−1)( 1p−1)+n+1q .

As this must hold for all δ > 0, we see that

n − 1

n + 1(1 − 1

p) ≥ 1

q.

This condition is called the scaling line for the restriction estimate. Using a more careful stationary

phase analysis of the Fourier transform of the surface measure, one can show that σ ≳ (1 + ∣ξ∣)−n−12 andconsequently q > 2n

n−1is a necessary condition for the extension operator to map bounded functions to

Lq.

Conjecture 6.9 (Restriction). Let E be the extension operator of the unit sphere in Rn. Then E is abounded operator Lp(σ)→ Lq(Rn) for all

n − 1

n + 1(1 − 1

p) = 1

q< n − 1

2n.

This conjecture has only been verified in dimension 2.

7. Waves and spherical means

7.1. Schrodinger equation with Schwartz data. Next we discuss the linear Schrodinger equation. Itcan be understood as a description of the time evolution of the state vector of a certain simple quantumsystem. From this point of view one expects the L2 norm (total probability density) to be preserved astime runs. Given a function f , we call u a solution to the Cauchy problem with initial data f if

ut − i∆u = 0, in Rn+1

u(x,0) = f(x), in Rn.

We will first consider classical solutions with u ∈ C2 and the initial data being attained as pointwiselimit. Later we study the L2 version where the equation is understood in the sense of distributions andthe initial data as an L2 limit.

Suppose that the initial data f is Schwartz. Then we can write down an explicit formula for thesolution, and there is no difficulty in seeing that this is indeed the only solution with the given initialdata. Let

u(x, t) = eit∆f(x) = F−1(e−it(2π∣ξ∣)2

f(ξ)) = Eµfwhere µ = (1 + 16π2∣ξ∣2)−1/2dσ is the measure on the paraboloid (ξ,−4π2∣ξ∣2) that makes the equalityabove true. Eµ is the Fourier extension operator relative to the measure µ. The following are easy toverify

eit∆ maps Schwartz class into itself.26

u(x, t) = eit∆f is smooth in (x, t) whenever f is Schwartz.

eit∆ has kernel Kt(x) = (4πit)−n/2e−∣x∣2/(4it).

It holds ∥eit∆f∥L∞(Rn) ≤ (4π∣t∣)−n/2∥f∥L1(Rn).

Two first items are clear, the third one follows from the formula of the Fourier transform of a realGaussian via a detour where one continues the parameter t in the exponent to the right half of thecomplex plane and takes limits. The last item is and immediate consequence of the third one.

The last item shows that the solution spreads out in space and loses its L∞ norm. This can be seenas a consequence of the dispersive nature of the equation. Given a plane wave with complex exponentialprofile ei(x⋅k+ωt), one sees that it is a solution if ∣k∣2 = ω, that is, the wave number squared equalsfrequency (in the correct physical terminology, usually we do not follows the physical definitions). Aparticle located at a point stays at the plane of constant phase if

d

dt(x ⋅ k + ωt) = 0,

dx

dt⋅ k = ω.

Hence one can say that the phase travels at velocity ω/∣k∣ to the direction of the wave number vector.In particular, a plane wave obeying Schrodinger’s equation has phase velocity ∣k∣. Plane waves withdifferent magnitude of wave numbers (that is, different Fourier modes) travel at different velocities, andthe evolution separates them. The wave disperses.

7.2. L2 data. Next we study a solution to the Schrodinger equation with L2 data. We ask the initialdata to be attained as an L2 limit when t→ 0, and the equation is satisfied in the sense of distributions.We try to find solutions u,ut ∈ L2(Rn) for all t. The following proposition is easy to prove and the proofis not recorded here.

Proposition 7.1. The following hold.

The Schrodinger propagator eit∆ is a unitary operator on L2(Rn). The mapping t↦ eit∆ is continuous R→ L2(Rn). For f ∈ L2(Rn), the function u(x, t) = eit∆f(x) is a solution in the sense above.

The first item in the list shows that the time decay driven by the dispersion of the solutions does notextend to L2 norms on constant times. On the other hand, this was to be expected from the quantummechanical interpretation. However, there is a weaker form of a decay estimate the holds for L2 initialdata. An estimate where the space–time Lp norm is controlled by a norm of the initial data is called aStrichartz estimate.

Theorem 7.2 (Strichartz estimate). Let u be a solution with L2 initial data f . Then

∥u∥L

2(n+2)n (Rn+1)

≲n ∥f∥L2(Rn).

Proof. That any solution is given by the Schrodinger propagator defined as the extension operator canbe proved but is left as an exercise. Assume that f is supported in the unit ball. Then the claimedinequality follows from the Tomas–Stein Theorem 6.7. Then let R > 0 and consider f with Fouriertransform supported in the ball B(0,R). Define F (x) = f(R−1x) so that F (ξ) = Rnf(Rξ) is supportedin the unit ball. A change of variables shows that

U(x, t) = eit∆F (x) = (eiR−2t∆f)(R−1x) = u(R−1x,R−2t).

Now F has Fourier transform supported in the unit ball so two more changes of variables show

∥u∥Lq(Rn+1) = R− 1q (n+2)∥U∥Lq(Rn+1) ≲ R− 1

q (n+2)∥F ∥L2(Rn) = R− 1q (n+2)+n2 ∥f∥L2(Rn) = ∥f∥L2(Rn).

The final claim follows by approximating an arbitrary L2 function by one that has compact Fouriersupport.

7.3. Wave equation. Another equation related to restriction theory is the wave equation describing forinstance propagation of light or an acoustic wave. Many of its characteristic features differ remarkablyfrom those of the Schrodinger equation, and we will see what is the interpretation of this in terms of theFourier restriction. For f, g ∈ S, we attempt to find u ∈ C2 such that

utt −∆u = 0, on Rn+1,

u(x,0) = f(x)ut(x,0) = g(x).

27

Relative to this problem, we define the half wave propagator

eit√−∆f(x) = F−1

ξ (eit⋅2π∣ξ∣f(ξ)).

This can be also seen as a Fourier extension from the surface M = (ξ, ∣ξ∣) ∶ ξ ∈ Rn with a suitablemeasure. Note that there are two obvious differences to the paraboloid. There is a singular point atorigin and there is one vanishing principal curvature. In general, the initial value problem for the waveequation with Schwartz data can always be solved via the representation formula

u = u+ + u−, u± =1

2e±it

√−∆(f ∓ i(−∆)−1/2g).

It is easy to see that there cannot be other (at least smooth) solutions.

Proposition 7.3. Given f, (−∆)−1/2g ∈ L2(Rn)∩C2(Rn). Then the solution with these initial values isunique.

Proof. If u, v are solutions with the same initial data, then w = u − v solves the equation with initialvalue 0. It suffices to consider real valued solutions. Define E(t) = 1

2 ∫ (∣∇w∣2 −w2t )dx. Differentiating in

t shows that E is constant and hence w must be zero as was claimed.

Proposition 7.4. Let C(x0, t0) = (x, t) ∶ 0 ≤ t ≤ t0, ∣x − x0∣ < t − t0. If u is C2 and u = ut = 0 inB(x0, t0), then u = 0 in C(x0, t0).

Proof. As before, define e(t) = 12 ∫B(x0,t)(∣∇u∣

2 − u2t )dx. Now

d

dte(t) = ∫

B(x0,t)(∇u ⋅ ∇ut − uttut)dx −

1

2∫∂B(x0,t)

(∣∇u∣2 − u2t )dσ(x)

= ∫B(x0,t)

ut(∆u − utt)dx + ∫∂B(x0,t)

∂u

∂νut dσ(x) −

1

2∫∂B(x0,t)

(∣∇u∣2 − u2t )dσ(x).

The first integral is zero and the second one bounded by the negative of the third one so the claim followsas in the previous proposition.

Similarly to the case of teh Scrodinger equation, it is easy to see that the solutions preserve the L2

norm of the initial the. On the other hand, it is not so easy to write down a formula for the convolutionkernel of the half wave propagator. Indeed, the previous proposition suggests the kernel to be verylocal (check 2.4 of [8] for exact formulas for the solutions; they can be computed, but not with Fourier

techniques). Moreover, a simple study of the plane wave ei(x⋅k+ωt) shows that the phase velocity of theplane wave satisfying the wave equation is ±1. There is no dispersion in the same sense as with theSchrodinger case. However, we can still prove a Strichartz estimate, as there is some cancellation drivenby different directions of the phases.

Theorem 7.5 (Strichartz estimate for the wave). Let q = 2(n+1)n−1

and f, g ∈ S. Let u be the solution ofthe wave equation with these initial data. Then

∥eit√−∆f∥Lq(Rn+1) ≲n ∥(−∆)1/4f∥L2(Rn)

∥u∥Lq(Rn+1) ≲n ∥(−∆)1/4f∥L2(Rn) + ∥(−∆)−1/4g∥L2(Rn)

where the fractional powers of the Laplacian are defined as Fourier multipliers.

Proof. It clearly suffices to prove the claim for the half wave propagator and apply it to the representationformula of the solution to conclude the second claim. The proof proceeds in three steps. Assume first fis supported in a unit annulus B(0,10) ∖B(0,1). Choose µ = ϕ0dσ so that

eit√−∆f = Eµf.

This is possible as we removed the singularity in the origin by restricting the attention to the functionsf with Fourier support in an annulus. Then the claim follows immediately from the generalized versionof the Tomas–Stein theorem given in Corollary 6.8.

Assume then f is supported in B(0,10R) ∖ B(0,R). Define F (x) = f(x/R) so that F has Fouriersupport in B(0,10) ∖B(0,1). Now

eit√−∆F (x) = ∫ e2πi(x⋅ξ+t∣ξ∣)f(Rξ)Rn dξ = Eµf ( x

R,t

R) =∶ U(x, t).

28

By change of variables (indeed after all)

∥eit√−∆f∥Lq(Rn+1) = R− 1

q (n+1)∥U∥Lq(Rn+1) ≲ R− 1q (n+1)∥F ∥L2(Rn)

= R− 1q (n+1)+n2 ∥f∥L2(Rn) ≲ R− 1

q (n+1)+n2 −12 ∥∣ ⋅ ∣

12 f∥L2(Rn) ∼ ∥(−∆)1/4f∥L2(Rn).

This concludes the proof for f with Fourier support in B(0,10R) ∖B(0,R).Finally, form the Littlewood–Paley decomposition of f into pieces fj so that the Fourier transforms

are supported around 2j . Denote uj = eit√−∆fj . By the Littlewood–Paley theorem and the previous

case,

∥eit√−∆f∥Lq(Rn+1) ∼

⎛⎜⎝∫

XXXXXXXXXXXXX

⎛⎝∑j

∣uj ∣2⎞⎠

1/2XXXXXXXXXXXXX

q

Lq(Rn+1)

dt⎞⎟⎠

1/q

≤⎛⎝∑j

∥uj∥2Lq(Rn+1)

⎞⎠

1/2

≲⎛⎝∑j

∥(−∆)1/4fj∥2L2(Rn)

⎞⎠

1/2

∼ ∥(−∆)1/4f∥L2(Rn).

7.4. Spherical means. Let σ be the normalized measure on the unit sphere. Define the sphericalaverages at scale t > 0 as

Atf(x) = (σt ∗ f)(x) = ∫∂B(0,1)

f(x + ty)σ(y).

The spherical means are well defined for apriori for continuous functions, but the following propositionshows that the definition extends to Lp. What is more, the spherical convolution is a smoothing operator.

Proposition 7.6. A1 extends to a bounded operator Lp → Lq for all p and q so that (1/p,1/q) are inthe closed convex hull (without (0,0)) of (0,0), (1,1) and (n/(n + 1),1/(n + 1)). Moreover, for k beingthe smallest integer at least (n − 1)/2, ∂αA1 maps L2 → L2.

The proof is an interpolation of analytic families similar to the proof of the Tomas–Stein theorem.Next we try are intersted in recovering a function from its spherical averages Atf by means of a limitt → 0. The almost everywhere convergence of these means is equivalent to Lp boundedness of thespherical maximal function. We will see that such an estimate holds, but not in all Lp spaces. As anintermediate result, we show that the single scale maximal function

Sf(x) = supt∈[1,2)

∣Atf(x)∣.

The following theorem goes back to Schlag and Sogge, but we mostly follow [12]

Theorem 7.7 (Schlag and Sogge). Let n ≥ 3, R > 1 and let f be a Schwartz function such that supp f ⊂B(0,10R) ∖B(0,R). Let

P = (0,0), Q = (n − 1

n,n − 1

n) , R = (n − 1

n,

1

n) , S = (n

2 − nn2 + 1

,n − 1

n2 + 1)

and suppose that (1/p,1/q) is in the interior of the convex hull of P,Q,R,S or p = q > n/(n−1). Thenthere exists ε > 0 so that

∥Sf∥Lq ≲ R−ε∥f∥Lp .

Proof. We first prove bounds when (1/p,1/q) is A = (1,1), B = (1,0), C = (1/2,1/2) and D = (1/2, (n −1)/(2(n + 1)). We start with A because, f has compact Fourier support, everything is smooth and forany s ∈ (1,2)

supt

∣Atf(x)∣ = supt

∣Asf(x) + ∫t

s∂τAτf(x)dτ ∣ ≤ ∣Asf(x)∣ + ∫

2

1∣∂τAτf(x)∣dτ.

Integrating over s ∈ (1,2) and x ∈ Rn, we get

∥Sf∥L1 ≤∬Rn×(1,2)

Asf(x)dxds +∬Rn×(1,2)

∣∂sAsf(x)∣ds ≤ ∥f∥L1 +∬Rn×(1,2)

∣∂sAsf(x)∣ds.29

For the second term, note that

∂sAsf(x) = ∂sF−1(σ(sξ)f(ξ))(x)

= ∂s ∫Rne2πiξ⋅x ∫

∂B(0,1)e−2πsiξ⋅y dσ(y)f(ξ)dξ

= ∫Rne2πiξ⋅x ∫

∂B(0,1)ye−2πsiξ⋅y dσ(y) ⋅ (−2πiξ)f(ξ)dξ

= ∫∂B(0,1)

y ⋅ ∇f(x + sy)dσ(y).

By Young’s convolution the L1 norm of thi expression is bounded by R∥f∥L1 and so

(7.1) ∥Sf∥L1 ≲ (1 +R)∥f∥L1 .

This is the bound at A.Next we treat the bound at B. As f has compact Fourier support, we can find a Schwartz function ψ

whose Fourier transform is identically one in B(0,10R) and vanishes outside B(0,20R). Then f = f ∗ψand

supt

∣Atf(x)∣ = supt∈(1,2)

∣(ψ ∗ σt) ∗ f ∣ ≤ ∥f∥L1∥ψ ∗ σt∥L∞ .

Because ψ is Schwartz, for any N > 0

∣ψ ∗ σt∣ ≲ ∫∂B(0,1)

Rn

(1 + ∣x + ty∣R)Ndσ(y).

Now for any given x, decompose the sphere regions Θk where ∣ty + x∣ ∼ 2kR−1 as k ≥ 1 and the finalregion Θ0 where ∣ty + x∣ ≤ R−1. Then

∫∂B(0,1)

Rn

(1 + ∣x + ty∣R)Ndσ(y) = ∑

k≥0∫

Θk

Rn

(1 + ∣x + ty∣R)Ndσ(y) ≤ R∑

k≥0

2k(n−1) 1

(1 + 2k)N≲ R

provided N ≥ n. Hence

(7.2) ∥Sf∥L∞ ≲ R∥f∥L1 .

This is the bound at B.We turn to the bound at C. Here the usual trick with the fundamental theorem of calculus gives

supt∈(1,2)

∣Atf ∣2 ≤ ∣Asf ∣2 + (∫2

1∣Aτf ∣2 dτ)

1/2(∫

2

1∣∂τAτf ∣2 dτ)

1/2.

Integrating in x and s, we can use the fact that the Fourier transform of f is supported around ∼ R andbounds for the multiplier corresponding to the Fourier transform of the spherical measure to see

(7.3) ∥Sf∥L2 ≲ (R−n−12 + (R−n−12 R−n−32 )1/2)∥f∥L2 ≲ R−n−22 ∥f∥L2 .

This is the bound at C.To prove the remaining case at point D, we need (unfortunately) some additional estimates on os-

cillatory integrals whose significance was overlooked back in the previous section. Namely, one canwrite

σ(ξ) = a+(ξ)e2πi∣ξ∣ + a−(ξ)e−2πi∣ξ∣

where∣∂αξ a±(ξ)∣ ≲ (1 + ∣ξ∣)−

n−12 −∣α∣

for α ∈ Nn. Compare to e.g. Section 6 in [20]. This implies in particular that the multipliers (1 +∣ξ∣)n−12 a±(ξ) are Mikhlin multipliers (Theorem 4.11) and hence induce Lp bounded operators. Using thefundamental theorem of calculus as before, we see that

∥Sf∥Lq ≲ ∥Atf∥Lq(Rn×(1,2)) + ∥Atf∥q−1q

Lq(Rn×(1,2))∥∂tAtf∥1q

Lq(Rn×(1,2))

≲ R−n−12 (∥e−it√−∆f∥Lq(Rn×(1,2)) +R

1q ∥e−it

√−∆f∥Lq(Rn×(1,2)))

≲ R−n−12 + 1q +

12 ∥f∥L2(Rn)

where the last inequality used the Strichartz estimate for the wave propagator and the fact that theFourier support of f is in annulus around R. Hence

(7.4) ∥Sf∥L

2(n+1)n−1

≲ R−n2−2n−2

2(n+1) ∥f∥L2 .

30

Interpolating the bounds at A and B, we see that the exponent of R is negative provided p is larger thann/(n − 1). This gives the point Q in the quadrilateral. Interpolating both A and D with C gives thepoints R and S. Finally, a bound without Fourier decay is immediate at (0,0) so the proof is complete.Note that there is slight technical complication in interpolation as the operator is not linear. However,this can be circumvented by, for instance, a linearization procedure like the one described in the proofof Proposition 5.13.

Next we see how the single scale maximal operator bounds can be upgraded to bounds for the maximalfunction defined through the full supremum

f ↦ supt>0

∣Atf ∣.

This means gluing together all the scales instead of restricting the attention to a single scale (singlefrequency band). We will do this by two alternative methods, first one of which is the use of Littlewood–Paley theory. This is a very short proof, but on the other hand it is only available when p ≥ 2. Part ofthe restriction comes from the unoptimal formulation of the previous theorem. Some of the bounds therehold for the full supremum too (the proof for (1,1) and (2,2) for instance). However, the second proofwhich is based on sparse bound elegantly avoids these problems. We start with the Littlewood–Paleytheory.

Theorem 7.8. Let α ∈ (0, n − 1) and (1/p,1/q) in the interior of the convex hull of P,Q,R,S suchthat q ≥ 2. Let α = n(1/p − 1/q). Define Sαf(x) = supt>0 t

α∣Atf(x)∣. Then

∥Sαf∥Lq ≲ ∥f∥Lp .

Proof. Decompose f into frequency localized pieces so that f0 has Fourier support in the unit ball andeach fj in an annulus around 2j when j ≥ 1. If ψ0 is the frequency cut-off associated with f0, then we seethat the bound ∥Sαf0∥Lq ≲ ∥f∥Lp is an easy consequence of the Hardy–Littlewood–Sobolev inequality.This will be left as an exercise.

Consider then the other pieces. Note that for any R > 0 and t ∈ (1,2)

ARtf(x) = σRt ∗ f(x) = ∫∂B(0,1)

f (R( xR− ty)) = 1

RnAtDil1R−1 f ( x

R) .

Hence by a change of variables and Theorem 7.7

∥ supt∈(2k,2k+1)

∣Atfj ∣∥Lq = ∥ supt∈(1,2)

∣A2ktfj ∣∥Lq = 2kn( 1

q −1)∥ supt∈(1,2)

∣AtDil1R−1 fj ∣∥Lq

≲ 2kn( 1

q −1)2−ε(j+k)∥Dil1R−1 fj∥Lp = 2

kn( 1q −

1p )−ε(j+k)∥fj∥Lp .

Recalling the definition of α and applying the inequality with j − k in place of j, we have

∥ supt∈(2k,2k+1)

tα∣Atfj−k ∣∥Lq ≲ 2−εj∥fj−k∥Lp .

Then

∥ supt∈(1,∞)

tα∣Atf≥1∣∥Lq = ∥ supk∈Z+

supt∈(2k,2k+1)

tα∣Atf≥1∣∥Lq ≤⎛⎝ ∑k∈Z+

∥ supt∈(2k,2k+1)

tα∣Atf≥1∣∥qLq⎞⎠

1/q

≤∞∑j=1

⎛⎝ ∑k∈Z+

∥ supt∈(2k,2k+1)

tα∣Atfj−k ∣∥qLq⎞⎠

1/q

≲∞∑j=1

2−εj (∑k∈Z

∥fj−k ∣∥qLp)1/q

≲∞∑j=1

2−εjXXXXXXXXXXXX(∑k∈Z

∣fj−k ∣q)1/qXXXXXXXXXXXXLp

≲∞∑j=1

2−εj∥f∥Lp ≲ ∥f∥Lp

where the last line required q ≥ 2 in order to controll the `q sum by an `2 sum in order to apply theLittlewood–Paley theorem. Finally, the theorem follows by scaling from the above inequality.

7.5. Sparse bound for spherical maximal function. Next we show how the sparse bound givesan alternative to the Littlewood–Paley theory. In a certain sense, the sparse bound encodes moreinformation. It implies various weighted bounds that would be more difficult to derive from the otherapproach. The whole subsection follows quite closely [11]. We start by reformulating the Theorem 7.7in a Fourier–free way.

31

Proposition 7.9. Let (1/p,1/q) be as in Theorem 7.7. Let τhf(x) = f(x − h). Then there exists ε > 0so that for R > ∣h∣

∥ supt∈(R,2R)

∣Atf − τhAtf ∣∥Lq ≲ ( ∣h∣R

)ε

Rn( 1

q −1p )∥f∥Lp .

Proof. Fix k0 so that 1/R ∈ [2k0 ,2k0+1). Let f0 be the piece of f with Fourier support in the ballB(0,2k0), f1 = f − f0, and let fj be the pieces with Fourier support in annuli around each 2j with j ≥ 1.

Now f0 = ψ0 ∗ f where ψ0 is an L1 normalized bump function at scale 2−k0 ∼ R. Then

∥ supt∈(R,2R)

∣Atf0 − τhAtf0∣∥Lq ≲ Rn(1q −

1p )∥f0 − τhf0∥Lp ≲ Rn(

1q −

1p )∥f∥Lp∥ψ0 − τhψ0∥L1 ≲ ∣h∣

RRn( 1

q −1p )∥f∥Lp .

For the remaining pieces we use the epsilon gain in the Fourier variable in Theorem 7.7. Indeed,

∥ supt∈(R,2R)

∣Atfj − τhAtfj ∣∥Lq ≲∞∑j=k0

(2−j

R)ε′

Rn( 1

q −1p )∥fj − τhfj∥Lp .

Consider first the parts ∣h∣ ≤ 2−j . Let ψj be the multiplier of the Littlewood–Paley projection around 2j .Then by elementary geomtery

∫ ∣1 − e2πiξ⋅h∣ψj ≲ 2jε∣h∣ε2nj

and further

∣∫ ∂α (1 − e2πiξ⋅h)ψj∣ ≲ 2jε∣h∣ε2nj max(∣h∣∣α∣,2−j∣α∣) ≲ 2jn−∣α∣j .

Note that there is a potential problem with all derivatives hitting the exponential factor and killing theconstant one. However, because ψj decays fast, we can integrate by parts once to make sure that at leastone derivative sits on ψj and make the 2εj ∣h∣ε appear as

∫ ψj∂α(1 − e2πiξ⋅h) = −∫ ∂α−α

′

ψj∂α′(1 − e2πiξ⋅h) = (2πih)α

′

∫ (∂α−α′

ψj)e2πiξ⋅h

= (2πih)α′

∫ (∂α−α′

ψj)(e2πiξ⋅h − 1)

where the last equality used the mean zero property of the derivative.By the L1 → L∞ bound for the Fourier transform

∣ψj − τhψj ∣ ≲ (2j ∣h∣)ε 2jn

(1 + 22j ∣x∣2)n+1

and hence the bound

∥fj − τhfj∥Lp ≲ (2j ∣h∣)ε∥fj∥Lp

holds for ∣h∣ ≤ 2−j . Let k1 be the index so that 2k1 ≤ ∣h∣−1 < 2k1+1. Then

k1

∑j=k0

(2−j

R)ε′

Rn( 1

q −1p )∥fj − τhfj∥Lp ≲ ( ∣h∣

R)ε k1

∑j=k0

(2−j

R)ε′−ε

Rn( 1

q −1p )∥f∥Lp

and trivially also

∞∑j=k1

(2−j

R)ε′

Rn( 1

q −1p )∥fj − τhfj∥Lp ≲ ( ∣h∣

R)ε ∞∑j=k1

(2−j

R)ε′−ε

Rn( 1

q −1p )∥f∥Lp

which completes the proof.

Next we control the fractional spherical maximal function Sαf = supt>0 tα∣Atf ∣ by a dyadic operator.

Here we need a useful fact we did not prove at any point earlier.

Theorem 7.10 (Theorem 3.1 in [14]). For every dyadic lattice D = 2k([a, b)n+(b−a)j) ∶ j ∈ Zn, k ∈ Zlattice there exists 3n dyadic lattices Dj such that

3D =3n

⋃j=1

Dj .

32

This is called the three lattice theorem, and we use it as follows. Fix a scale R and consider a cubeQ with sidelength R. Find k so that R ∈ [2k,2k+1). We can cover Q with 2n dyadic cubes from thestandard dyadic grid. Call any one of them Q0 so that Q ⊂ 3Q0. By the three lattice theorem, there arethree dyadic lattices so that 3Q0 is in one of them. Iterating, we find 9n dyadic grids so that 3 ⋅ 3Q0 isin one of them. In particular, we have found out that the original cube Q is in 1

3P where ∣P ∣ ∼ ∣Q∣ and

the latter cube is in a dyadic grid. Enumerate the dyadic grids as Dα with α = 1, . . . ,9n. Then

supt>0

tα∣Atf ∣ ≲ supk∈Z

2kα supt∈(2k,2k+1)

∣Atf ∣ ≤ supk∈Z

2kα9n

∑α=1

supQ∈Dα

`(Q)≥2k+5

supt∈(2k,2k+1)

1Q∣At(1 13Qf)∣

≲9n

∑α=1

supQ∈Dα

1Q`(Q)αAQf

where

AQf(x) = supt∈(2−10`(Q),2−2`(Q))

∣At(1 13Qf)(x)∣.

Hence it suffices to study the dyadic Sαd f = supQ∈Dα 1Q`(Q)αAQf for a single dyadic grid and sum theestimates.

Lemma 7.11. Let p and q be as in Theorem 7.7. Let f be a constant multiple of a characteristic

function, let g be an arbitrary Lq′

function, and let Q0 be a dyadic cube and E a collection of its dyadicsubcubes so that

supQ′∈E

supQ′⊂Q⊂Q0

(⟨f⟩Q,p⟨f⟩Q0,p

+⟨g⟩Q,q′⟨g⟩Q0,q′

) ≤ C0

for some constant C0 (only depending on p, q and the dimension). Then

∫ supQ∈E

AQf ⋅ g dx ≲ ∣Q0∣⟨f⟩Q0,p⟨g⟩Q0,q′ .

Proof. By setting f/⟨f⟩Q0,p, we can normalize ⟨f⟩Q0,p = ⟨g⟩Q0,q′ = 1. We can linearize the supremumover dyadic cubes: for Q0, let E0 be the points x where AQ0f(x) is at least half of the supremum.Recursively, define the sets EQ from the points x ∉ E0 where Q is at least half of the supremum. Thenwe get a collection EQ of pairwise disjoint sets indexed along the cubes in E so that

supQ∈E

AQf ≤ 2 ∑Q∈E

1EQAQf.

We are led to estimate

∑Q∫ AQf ⋅GQ

where GQ = 1EQg.Run the Calderon–Zygmund construction with the stopping condition

⟨f⟩P,p + ⟨g⟩P,q′ > 2Cn.

Collect the maximal cubes satisfying this to B and denote Bk = P ∈ B ∶ `(P ) = 2−k`(Q0). Further,write the decomposition

f = F +B, F =∑P

1P ⟨f⟩P,1 + 1Q0∖⋃P , B =∞∑k=1

Bk =∞∑k=1

∑P ∈Bk

bP =∞∑k=1

∑P ∈Bk

1P (f − ⟨f⟩P,1).

Note that Q ∩ P ≠ ∅ only if P ⊊ Q.We first notice the claim is clear for F . Indeed, by stopping time construction of F and disjoint

supports

∑Q∫ AQF ⋅GQ ≲∑

Q∫ GQ ≤ ∣Q0∣.

It remains to study the piece B. Take a cube Q with side length 2i`(Q0) and consider all P ∈ Bi+kcontained in it. Let P0 be the cube with side length 2i+k`(Q0) that is centred at origin, and let cP be

33

the center of P . Let τhf(x) = f(x − h). Then

AQBi+k = AQ⎛⎝ ∑P ∈Bi+k

1P∩Q ⨏P0

(f − τhf(cP ))dh⎞⎠

≤ ⨏P0

AQ⎛⎝ ∑P ∈Bi+k

1P∩Q(f − τhf(cP ))⎞⎠dh

≲ ⨏2P0

AQ(Bi+k − τhBi+k)dh.

Consequently

∫ AQBi+k ⋅GQ ≤ ∥AQBi+k∥q∥GQ∥q′ ≲ ⨏2P0

∥AQ(Bi+k − τhBi+k)∥q∥GQ∥q′ dh

≲ 2−kε∣Q∣⟨Bi+k⟩Q,p⟨GQ⟩Q,q′(7.5)

where we applied Proposition 7.9 with ∣h∣ ∼ 2−i−k`(Q0) in the cube Q with side length 2i`(Q0).Recall that we aim at estimating

(7.6) ∑Q∈E

∑k≥0

∣∫ AQBi+k ⋅GQ∣ .

The trick is to sum first in Q. Now

∣Bi∣ ≤ ∑P ∈Bi

(1P ∣f ∣ + ⟨∣f ∣⟩P,1) ≲ ∑P ∈Bi

1P ∣f ∣ +∑P

1P ⟨∣f ∣⟩Q0,p = Ci +Di.

In particular, the functions on the right hand side are supported in the union of the maximal cubes fromthe Calderon–Zygmund decomposition having side length 2−i`(Q0). The Ci with distinct i have disjointsupport, they are constant multiples of a characteristic functions and so is their sum over i. The samegoes for Di. If k is kept fixed, then 1QCi+k are pairwise disjoint when Q varies. This shows that thebounds

∑Q

∣Q∣⟨Ci+k⟩Q,1⟨GQ⟩Q,q′ ≲ ∣Q0∣⟨C⟩Q0,1

∑Q

∣Q∣⟨Ci+k⟩Q,r⟨GQ⟩Q,r′ ≲ ∣Q0∣⟨C⟩Q0,r⟨g⟩Q0,r′

hold whenever r ∈ [1,∞). The same holds with D in place of C. Here C is the sum over i of all Ci.It remains to interpolate. Because q ≥ p, then there exists θ ∈ [0,1] such that

1

p= θ + 1 − θ

q.

Because Ci+k is a constant multiple of a characteristic function (as is also Di+k), we do the arithmeticswith the exponents and apply Holder to see

∑Q

∣Q∣⟨Ci+k⟩Q,p⟨GQ⟩Q,q′ =∑Q

∣Q∣⟨Ci+kf⟩θQ,1⟨Ci+k⟩1−θQ,q⟨GQ⟩θQ,q′⟨GQ⟩1−θQ,q′

≤⎛⎝∑Q

∣Q∣⟨Ci+k⟩Q,1⟨GQ⟩Q,q′⎞⎠

θ⎛⎝∑Q

∣Q∣⟨Ci+k⟩Q,q⟨GQ⟩Q,q′⎞⎠

1−θ

≤ ∣Q0∣⟨C⟩θQ0,1⟨C⟩1−θQ0,q = ∣Q0∣⟨C⟩Q0,p ≤ ∣Q0∣⟨f⟩Q0,p ≲ ∣Q0∣

Here we also used the fact ⟨GQ⟩Q,q′ ≤ 1. The same bound applies for Di+k, so using the fact ∣Bi+k ∣ ≤Ci+k+Di+k, we can use (7.5) on (7.6), estimate the Q sum by the estimate above and sum in k to conludethe proof.

Theorem 7.12 (Lacey [11]). Let f and g be bounded compactly supported functions and α ∈ [0, n − 1).Then there exists a sparse family S such that

∫ Sαf ⋅ g dx ≲ ∑Q∈S

∣Q∣1+α/n⟨f⟩Q,p⟨g1FQ⟩Q,q′

where FQ are pairwise disjoint and (1/p,1/q) as in Theorem 7.7.34

Proof. It suffices to deal with the dyadic operator Sαd . Moreover, we can assume f and g are supportedin a single cube Q0. We let Q0 to be the first cube in the sparse family (eventually we will include allits parents too, but this preserves sparsity). Let M be the maximal dyadic subcubes with respect to

either ⟨f⟩P,p > Cn⟨f⟩Q0,p or ⟨g⟩P,q′ > Cn⟨g⟩Q0,q′ or both.

Let E = ∪M. For Cn large enough, ∣E∣ ≤ ∣Q0∣/2 (weak type bounds). Let N be the cubes not containedin E and hence not contained in any cube in M. Assume first that f is a characteristic function abounded set. These cubes satisfy the assumption of lemma 7.11, and hence

(7.7) ∫Q0∖E

Sαd f ⋅ g dx ≤ `(Q0)α ∫Q0∖E

supQ∈N

AQf ⋅ g dx ≤ ∣Q0∣1+α/n⟨f⟩Q0,p⟨g1FQ0⟩Q0,q′

where FQ0 is the set of points in Q0 where Sd coincides with the supremum over N . It remains toestimate the integral inside E. However, we can recurse the argument on each of the (disjoint) maximalcubes, and the claim follows for all characteristic functions f .

Suppose then that f is a general bounded and non-negative function. We choose the stopping cubesM and outsider cubes N as before, and it remains to verify (7.7). We decompose f by setting Fk = x ∈Q0 ∶ 2k < f(x) < 2k+1 for all k ∈ Z. Then

f ≤∑k

2k+11Fk .

Next we use sublinearity of Sd and apply the result for characteristic functions with the pair of exponentsr < p and q′ and α = 0 (this is possible as the range of exponents we are proving the claim for is open)to get

∫Q0∖E

supQ∈N

AQf ⋅ g dx ≤∑k

2k ∫ supQ∈N

AQ1Fk ⋅ g1Q0∖E dx

≲∑k

2k ∑Q∈Sk

∣Q∣⟨1Fk⟩Q,p⟨g1FQ⟩Q,q′

≲ ⟨g1Q0∖E⟩Q0,q′∑k

2k ∑Q∈Sk

∣Q∣⟨1Fk⟩Q,p.

Here we applied that the (not modified) function g1Q0∖E is still subject to the stopping condition in allcubes in the sparse families (note that this would not hold without the restriction to Q0 ∖E.

Letting ˜p ∈ (p, p), we can use sparsity and a maximal function bound to estimate

∑Q∈Sk

∣Q∣⟨1Fk⟩Q,p ≲ ∫Q0

Mp(1Fk∩Q0) ≤ ∣Q0∣ (⨏Q0

Mp(1Fk∩Q0)˜p)

1/ ˜p

≲ ∣Q0∣⟨1Fk⟩Q0, ˜p.

Finally, noting (by cutting the integral in two pieces)

∑k

2k⟨1Fk⟩Q0, ˜p≲ ∫

∞

0∣Q0 ∩ f > t∣1/ ˜p dt ≲ ⟨1Fk⟩Q0,p,

we conclude

∫Q0∖E

supQ∈N

AQf ⋅ g dx ≲ ∣Q0∣⟨f⟩Q0,p⟨g1FQ0⟩Q0,q′ .

This is (7.7).

Remark 7.13. Lacey [11] did not have the case α > 0 in his paper, but as this extension was actuallytrivial and required no additional work, I included it in his theorem.

Remark 7.14. The Lp bounds for the spherical maximal function follow in the usual way from thesparse bound. Fix p > r > n/(n − 1) and q > p > r so that (1/r,1/q) are as in the previous theorem. Notethat as q > p, then q′ < p′. Now

∑Q∈S

∣Q∣⟨f⟩Q,r⟨g1FQ⟩Q,q′ ≲ ∫ Mrf ⋅Mq′g ≤ ∥Mrf∥p∥Mq′∥p′ ≲ ∥f∥p∥g∥p′

where the first inequality used sparsity and the last one used p > r and p′ > q′. This was in the case α = 0.The other cases follow by similar argument but using the fractional version of the Hardy–Littlewoodmaximal theorem instead of the usual one.

35

Remark 7.15. One could also derive weighted bounds with Ap type weights as in the Calderon–Zygmundcase in Theorem 3.8, but now the conditions on weights would be more complicated, and we do not enterthat topic. In general, it is not believed that the Muckenhoupt weights are the correct tool to understandfiner properties of the spherical maximal function. Finally, note that unlike in Theorem 7.8, we werenot compelled to work under the (artificial) restriction q ≥ 2 when using the sparse method instead ofLittlewood–Paley theory.

8. Fourier transform and dimension

Next we study Fourier transforms of measures that are more irregular than the ones seen before. Theymight be, for instance, supported in sets wilder than the surfaces considered so far. As was the case withoscillatory integrals, the absolute size of the set plays an important role when it comes to the decay ofthe Fourier transform of measures supported in that set. Indeed, 1Ω had one order faster Fourier decaythan the surface measure on the boundary. On the other hand, we will see that the finer structure ofthe sets in question also has an effect in the rough setting.

We will use the notation

M(A) = µ Borel ∶ suppµ ⊂ A, 0 < µ(A) <∞

where A is a set in Rn for measures supported in A. Most of this section is from [15, 16].

8.1. Hausdorff dimension. Let E ⊂ Rn be any set. We define the s-Hausdorff δ > 0 content by

Hsδ(E) = inf∑i

diam(Ai)s ∶ diam(Ai) ≤ δ, E ⊂⋃i

Ai

where the collections Aii are otherwise arbitrary. The Hausdorff content does not take into accountthe local structure of the set. For instance, n − 1-Hausdorff δ content of a δ/2 sphere is δ/2 which doesnot coincide with the surface measure as we would like it to do. We define the Hausdorff measure (orHausdorff outer measure, but following the traditional language in geometric measure theory, we callouter measures just measures) as

Hs(E) = supδ>0Hsδ(E) = lim

δ→0Hsδ(E).

The basics of Hausdorff measure can be found for instance in Section 2 of [9]. One can show thatHs is Borel regular (all Borel sets are Hs-measurable and every set E has a Borel set B ⊃ E such thatHs(B) = Hs(E)). It also coincides with the surface measure when it comes to smooth surfaces andLebegue measure when it comes to Euclidean subspaces. We will mostly be interested when the measureis zero. For this question it would be enough to consider a measure defined through a covering only usingballs instead of general sets. The resulting measures give a comparable size (up to a constant 2) to everyset.

To define the Hausdorff dimension, we need the following simple fact.

Proposition 8.1. Let t < s. If Ht(E) <∞, then Hs(E) = 0.

Proof. Let δ > 0. Take a cover of E with sets Ai of diameter less than δ. Then

Hsδ(E) ≤∑i

diam(Ai)s ≤ δs−t∑i

diam(Ai)t.

Taking infimum of the right hand side over all such collections, we reach

Hsδ(E) ≤ δs−tHtδ(E)

and the claim follows from sending δ → 0.

This is refinement of the fact that a line has zero area. We define the dimension of the set to be theborderline case of s where the s-Hausdorff measure starts seeing the set.

Definition 8.2. Let E be a set. We define

dim(E) = inft > 0 ∶Ht(E) = 0 = supt > 0 ∶Ht(E) =∞.

That the infimum and supremum coincide requires a proof, of course, but we omit it. Even if theHausdorff content is not a good measure, it is very useful for studying the dimension.

Proposition 8.3. Suppose δ ∈ (0,∞]. Then Hs(E) = 0 if and only if Hsδ(E) = 0.36

Proof. If Hs(E) = 0, then clearly Hsδ(E) ≤ Hs(E) ≤ 0 so it remains to establish the other implication.Assume Hsδ(E) = 0. Take ε > 0. Take a cover of E with sets having diameters ri < δ and ∑i rsi < εs. Thenri < ε and

Hsε(E) ≤∑i

rsi < εs.

Taking the limit ε→ 0 finishes the proof.

Theorem 8.4 (Frostman’s lemma). Let B be closed. Then Hs(B) > 0 if and only if there existsµ ∈M(B) such that µ(B(x, r)) ≤ rs for all x ∈ Rn and r > 0.

Proof. Suppose µ exists. Let Bi be a covering of B. Then

0 < µ(B) ≤∑i

µ(Bi) ≲∑i

rsi

so that Hs∞(B) ≥ µ(B) > 0. Hence the same holds for the Hausdorff measure.Conversely, assume B has positive measure. Assume first B is compact. Let Q0 be a cube that

contains B. Let Di be the collection of cubes obtained by partitioning Q0 into 2ni cubes with side length2−il(Q0). We construct a sequence of measures indexed along the smallest non-trivial scale. Let m > 0.For Q ∈ Dm set

µmm∣Q =⎧⎪⎪⎨⎪⎪⎩

2−ms dx∣Q∣ , if B ∩Q ≠ ∅0, otherwise.

Next we force the correct behaviour on the parent cubes through a recursion. For k ∈ [1,m], if µmk hasbeen defined, we let for Q ∈ Dk−1

µmk−1∣Q =⎧⎪⎪⎨⎪⎪⎩

µmk ∣Q, if µmk (Q) ≤ 2−(k−1)s

2−(k−1)s µmk ∣Qµmk

(Q) .

Denote µm = µm0 . For any k ≤m and Q ∈ Dm−k it clearly holds µm(Q) ≤ 2−(m−k)s.Next we use a compactness argument to prove that there is a weak* limit of the sequence approximating

measures. For each x ∈ B, we can find k and x ∈ Q ∈ Dm−k so that µm(Q) = 2−(m−k)s. Of all such cubescollect a subfamily maximal with respect to inclusion and call it Qi. By the lower bound on theHausdorff content of B

µm(Rn) =∑i

µm(Qi) ∼∑i

l(Qi)s ≳ b.

As the lower bound is non-zero and independent of m, we can normalize νm = µm(Rn)−1µm so that ν isa probability measure.

The sequence νm is bounded in total mass so there is a subsequence converging weakly* to a Radonmeasure ν (to see it is non-zero apply on a continuous function equal to one in Q0). Similarly, we cancheck it is supported in B. Moreover, for each ball B(x, r), there are 2n cubes with side length ∼ r fromthe dyadic system generated from Q0 so that we get

ν(B(x, r)) ≤ ν(U) ≤ lim infi

νi(U) ≲ rs

for their union U . This concludes the proof.

A measure as the one whose existence is guaranteed by Frostman’s lemma is usually called a Frostmanmeasure. Because the theorem characterizes the existence of the Frostman measure as equivalent toHausdorff measure being non-zero, it is typically only information one needs to take into account whenderiving further properties of sets with large Hausdorff measure.

Example 8.5 (Cantor sets). Let d ∈ (0,1/2). Take the interval I0 = [0,1] to form I0. Let Ik+1 be theintervals obtained by cutting away the middles dI from all I ∈ Ik. Define the Cantor set

Cd =∞⋂k=0

⋃I∈Ik

I.

Then the Hsd measure of Cd is positive and finite for sd = − log 2/ log d.37

8.2. Riesz energies and capacitary dimension. We define the Riesz kernel of order s ∈ (0, n) asks(x) = ∣x∣s−n and the corresponding convoluton operator (the Riesz potential) as Isf = ks ∗ f . Here fcan be a locally integrable bounded function or any locally integrable function that decays sufficientlyrapidly at infinity. The importance of the Riesz kernel comes from its relation to Frostman measuresbut also from the fact it inverts certain fractional differentiation operator.

We start by computing the Fourier transform of ks.

Proposition 8.6. Let s ∈ (0, n). Then ks = C(s, n)kn−s.

Proof. Recall that a tempered distribution Λ is radial if Λ(ϕ A) = Λ(ϕ) for all orthogonal matrices A.

It is homogeneous of degree α if Λ(Dil1s ϕ) = sαΛ(ϕ) for all positive s. Here Dil1s ϕ(x) = s−1ϕ(xs−1). Nowfor any distribution Λ homogeneous of degree α

Λ(Dil1s ϕ) = Λ(Dil∞s−1 ϕ) = s−nΛ(Dil1s−1 ϕ) = s−n−αΛ(ϕ) = s−n−αΛ(ϕ).

We see that the Fourier transform of a distribution homogeneous of degree α is homogeneous of degree−n + α.

Clearly ks is radial and homogeneous of degree s − n. Then its Fourier transform is radial andhomogeneous of degree −n − (s − n) = −s. Suppose s ∈ (0, n/2) so that ks ∈ L1 + L2. Then the Fourier

transform must be in L∞ + L2. In particular, ∣x∣sks(x) must be a constant function. Consider then

s > n/2. Then ks = c(s, n)kn−s. Taking one more Fourier transform (as distributions) then proves theclaim. Finally, to deal with the remaining case one has to take limit s → n/2. Here one needs precise

information about how the constant c(n, s) behaves, but this can be computed by testing ks against aGaussian.

The proposition shows that ks inverts (−∆)s/2 up to an additive constant. This motivates the followingterminology. Given a measure µ, we call

Es(µ) = ∫ In−sµdµ

the s-energy of µ. If s = 2, the potential operator inside the integral is just the solution of

−∆u = µ,

an electrostatic potential in physical terms, and the integral represents the energy of the electric fieldgiven by the charge distribution µ.

For A a closed set, we define

Cs(A) = supEsµ−1 ∶ µ ∈M(A), µ(Rn) = 1.

Using the capacity, we define the capacitary dimension as

dimcA ∶= sups ∶ Cs(A) > 0 = infs ∶ Cs(A) = 0= sups ∶ ∃µ ∈M(A), µ(Br) ≤ rs ∀r > 0 = inft ∶ ∃µ ∈M(A),Etµ <∞.

The first equality is the definition and the remaining ones are simple exercises to verify.

Theorem 8.7. Let A be closed. If Λs(A) < ∞, then Cs(A) = 0. If Cs(A) = 0, then Ht(A) = 0 for allt > s. In particular, dimc(A) = dim(A).

Proof. For the first claim, suppose that Cs(A) > 0. Then there is µ ∈ M(A) such that Esµ < ∞.In particular, In−sµ(x) < ∞ for µ-almost every x. Consequently, µ(x) = 0 for any such x and bydominated convergence theorem

limr→0

∫B(x,r)

∣x − y∣−s dµ(y) = 0.

Choose compact B ⊂ A so that µ(B) > 1/2. Let ε > 0. In a compact set, pointwise convergence isuniform, and hence there is δ so that for every x ∈ B and r ∈ (0, δ] it holds

µ(B(x, r)) ≤ ∫ ( r

∣x − y∣)s

dµ(y) ≤ εrs.

Finally, fix a covering Bi of B by sets with diameter ri ≤ δ/2 so that Hs(A)+ 1 ≥ ∑i rsi . Then for eachi there is xi ∈ B ∩Bi so that B(xi,2ri) ⊃ B ∩Bi and

1

2< µ(B) ≤∑

i

µ(B(xi,2ri)) ≲ ε∑i

rsi ≲ ε(Hs(A) + 1).

38

As ε can be taken arbitrarily small, this can be true only if Hs(A) = ∞. This concludes the proof forthis case.

For the second one, suppose Ht(A) > 0. By Frostman, there is a measure µ ∈M(A) with µ(Br) ≤ rtfor all balls with radius r. Now

Esµ = ∫ (∫∞

0µ(y ∶ ∣x − y∣−s > λ)dλ) dµ(x)

= ∫ (∫∞

0µ(B(x,λ−1/s))dλ) dµ(x)

≤ ∫ (µ(Rn) + ∫∞

1λ−t/s dλ) dµ(x) <∞

so that Cs(A) > 0.

8.3. Energy integral and Fourier dimension. The energy of a general Borel measure can be ef-fectively computed using the Fourier transform. The advantage of this approach is that the Fouriertransform of a typical measure is a function and hence easier to handle.

Proposition 8.8. Let µ be a non-negative finite Borel measure and s ∈ (0, n). Then

Esµ = C(s, n)∫ ∣µ(ξ)∣2∣ξ∣s−n dξ.

Proof. We first assume that µ is absolutely continuous and has a density that is a Schwartz function.Later we approximate general measures by more regular ones. Let f ∈ S be the density of µ. Then byProposition

Esµ = ∫ In−sf ⋅ f dx = ∫ In−sf ⋅ f dξ = C ∫ ∣f(ξ)∣2∣ξ∣s−n dξ.

This proves the claim for measures with Schwartz density.Let then µ be a general Borel measure. Let ϕ be a smooth bump function which has integral one and

support contained in the unit ball. We let ϕε be the L1 dilate with support in the ε ball and assume inaddition ϕ to be positive. Then

µε(x) = ϕε ∗ µ(x) = ∫1

εnϕ(x − y

ε) dµ(y)

are non-negative Borel measures with Schwartz density, and they converge to µ weakly* as ε → 0. ByTonelli’s theorem

Lε = Esµε =∬ (∬ ϕε(x − z)∣x − y∣−sϕε(y −w)dxdy) dµ(z)dµ(w).

We also denote

Rε = ∫ ∣µ(ξ)∣2∣ϕ(εξ)∣2∣ξ∣s−n dξ.

We prove the proposition in two complementary cases. Assume first Esµ(x) <∞. Because

∬ ϕε(x − z)∣x − y∣−sϕε(y −w)dxdy ≲ ∣z −w∣−s

uniformly in ε (this requires a small computation which we omit), we can apply dominated convergenceto see

Esµ = limε→0

Lε.

We know from the case already handled that Lε = Rε. Moreover, again by dominated convergence

limε→0

Rε = limε→0

∫ ∣µ(ξ)∣2∣ϕ(εξ)∣2∣ξ∣s−n dξ = ∫ ∣µ(ξ)∣2∣ξ∣s−n dξ

where we used ∣ϕ(εξ)∣2 → 1 as ε→ 0 as well as ∣ϕ(εξ)∣2 ≤ 10.It remains to study the case Eµ =∞. Now by Fatou

∞ = Esµ ≤ lim infε

Lε = lim infε

Rε.

Suppose

∫ lim infε

∣µ(ξ)∣2∣ϕ(εξ)∣2∣ξ∣s−n dξ <∞.

As ∣ϕ(εξ)∣2 → 1, there would be an integrable majorant and we could apply dominated convergencegiving a contradiction. This finishes the proof.

39

The expression of the energy integral in terms of the Fourier transform shows that existence of ameasure decaying as

∣µ(ξ)∣ ≲ (1 + ∣ξ∣)−s/2

implies the Hausdorff dimension to be at least s. The reverse implication is of course not true. It sufficesfor the function to decay in average to make an integral converge. However, this motivates the notion ofFourier dimension:

dimF A = sups ∈ [0, n] ∶ ∃µ ∈M(A), ∣µ(ξ)∣ ≲ (1 + ∣ξ∣)−s/2.

It is immediate from the coincidence of Hausdorff and capacitary dimensions that the Fourier dimensionis always at most the Hausdorff dimension. Conversely, it can be much smaller and it depends on theambient space with respect to which the Fourier transform is understood. Indeed, a subset of a k-planehas always Fourier dimension zero, because the measurse supported there do not decay to any normaldirection. A set whose Hausdorff and Fourier dimensions coincide is called a Salem set. The unit sphereis an example of n − 1 dimensional Salem set. More generally, Salem sets are tricky to construct. Theprototype examples of Salem sets with fractional dimension come from measures supported on trajectoriesof Brownian paths and from number theoretic constructions. We just state the following two exmplesand do not discuss the (lenghty) details of the constructions

Example 8.9. Let α > 0 and define Eα to be the set of real numbers in the unit interval so that thereare infinitely many rationals p/q so that ∣x − p/q∣ ≤ Cq−2−α. Then Eα is Salem and the dimension (bothFourier and Hausdorff) is 2/(2 + α).

Recall the Dirichlet theorem on rational approximation. It states that any irrational number hasinfinitely many rationals p/q so that the distance is less than q−2. Hence the set with α = 0 is just the setof irrationals, but the set Eα consist of number that have better approximability. The tedious part of theproof is the lower bound for the Fourier (and hence Hausdorff) dimension. It boils down to constructinga measure with suitable decay using the prime number theorem. This is discussed in [20].

Example 8.10. Even more briefly, a piece of a trajectory of a Brownian motion is almost surely Salem.There is a chapter devoted to this fact in [16].

Now that we know that there are sets with intersting fractional Fourier dimensions, the followingproposition has some content. For a set A ⊂ Rn, define the sum set

Ak = A +⋯ +A = k

∑i=1

xi ∶ ∀i ∶ xi ∈ A.

Similarly, we define the k fold convolution. If µ is a finite Borel measure, then

µk = µ ∗⋯ ∗ µ

acts on continuous functions through

⟨µk, ϕ⟩ = ∫ ⋯∫ ϕ(x1 +⋯ + xk)dµ(x1)⋯dµ(xk).

The product formula for the Fourier transform extends to convolution of measures with no additionalpain.

Proposition 8.11. Let A ⊂ R be a bounded Borel set and let k > 0 be an integer.

If dimF A > 1/k, then ∣Ak ∣ > 0 (Lebesgues measure). If dimF A > 2/k, then intAk ≠ ∅.

Proof. Let s < dimF A. By definition, there exists a measure µ ∈M(A) such that ∣µ(ξ)∣ ≲ (1 + ∣ξ∣)−s/2.

Clearly µk is supported in Ak and it has positive and finite mass. On the other hand, µk = (µ)k. Because

∣µk(ξ)∣ ≲ (1+∣ξ∣)ks/2, we conclude µk ∈ L2. By Fourier inversion, µk ∈ L2 and as it is non-zero, we conclude0 < ∣ suppµk ∣ ≤ ∣Ak ∣.

To prove the second assertion, just note that now we can find a measure whose decay is of order

ks/2 > 1 and ∣µk ∣ ∈ L1. By Fourier inversion, µk is then a continuous function and support, if non-empty,contains an interior point. However, this is the case by assumptions.

40

8.4. Dimension of projections. Next we discuss how the dimension behaves under projections. Weonly deal with projections onto lines, but most of what follows has its counterpart in the setting of pro-jections on k-spaces, see [16] for higher dimensional projections. That is not too much more complicated,but notationally considerably more annoying.

We consider the unit sphere Sn−1 as a set of directions. The letter e is reserved for this notation, andalmost every e means with respect to the surface measure unless something else has been said explicitly.For a unit vector e ∈ Sn−1, we define the projection Pex = e ⋅ x. This is Lipschitz continuous, and hencedimPeA ≤ dimA for any Borel set A. The questions is, how small the projections can be (very small)and how small they are typically (very large).

Throughout the section, we denote the image measure by

Peµ(A) = µ(P −1e A).

This is a measure on subsets of the real line, and its energy integrals are hence understood to be onedimensional whenever we encounter them. The formula

Peµ(r) = ∫Re−2πirx dPeµ(x) = ∫

Re−2πir(y⋅e) dµ(y) = µ(re)

is also very useful.

Theorem 8.12. Let A ⊂ Rn be a Borel set and denote s = dimA ≤ n.

If s ≤ 1, then dim(PeA) = s for almost every e ∈ Sn−1. If s > 1, then H1(PeA) > 0 for almost every e ∈ Sn−1. If dimA > 2, then PeA has non-empty interior for almsot every e ∈ Sn−1.

Proof. Assume first s ≤ 1. Because dimA = s, for every t < s the set has positive t-Riesz capacity andthere exists µ ∈M(A) such that Etµ <∞. Then using Proposition 8.8 in one dimension, we compute

∫Sn−1

EtPeµdσ(e) = C ∫Sn−1

∫R∣Peµ(r)∣2∣r∣t−1 drdσ(e) = 2C ∫

Sn−1∫

∞

0∣µ(er)∣2rt−1 drdσ(e)

= C ∫Rn

∣µ(x)∣2∣x∣t−n dx = CEtµ <∞.

Hence for almost every e we have found a measure in M(PeA) with finite t energy. ConsequentlydimPeA ≥ t. Taking now a sequence ti → s, we have dimPeA ≥ ti for all e ∉ Ni where Ni is of measurezero. Outside ⋃iNi, the dimension of projections is at least s and the first claim has been proved.

Assume then s > 1. Again, as dimA > 1, there exists a measure µ ∈ M(A) so that E1µ < ∞.Consequently

∫Sn−1

∫R∣Peµ(r)∣2 drdσ(e) = CE1µ <∞.

Hence Peµ ∈ L2(R) for almost every e. By inverting the Fourier transform, Peµ is also in L2 and hencethe measure is absolutely continuous with respect to the Lebesgue measure. As 0 < µ(A) = (Peµ)(PeA)the set PeA must have positive Lebesgue measure too. This concludes the proof for the second item.

Finally, to prove the last item take t such that 2 < t < dimA. Then there exists µ such that Etµ <∞.Now

∫Sn−1

∫R∣Peµ(r)∣drdσ(e) ≲ 1 + ∫

Sn−1∫

∞

1∣µ(er)∣drdσ(e)

≤ 1 + ∫Sn−1

∫∞

1∣µ(er)∣2rt−nrn−1 drdσ(e) + ∫

Sn−1∫

∞

1r1−t drdσ(e)

≲ 1 +Etµ <∞.

Consequently, for almost every e the Fourier transform of the image measure is in L1. By Fourierinversion, the image measure itself is then a continuous function. As it is non-zero, its positivity set hasnon-empty interior.

The first item in the theorem is not so much about Fourier transform, although the proof used theFourier transform. It is just an application of polar coordinates to pass information about finite energyin Rn to finite energy at almost every ray. The second and third items are more interesting relative toour theme. To know that the set has positive projection, we show that it supports a measure that is inL2 and we de so by finding a measure whose Fourier transform is in L2. Similary, the non-empty interiorfollows by continuous density supported in the projection, and continuity is connected with the Fouriertransform being in L1. The proof is of course very inprecise. The properties Fourier transform in L2 andL1 are much more than what is eventually claimed.

41

8.5. Exceptional projections and Sobolev dimension. Next we refine the estimates in the previoussubsection by quantifying what is the dimension of the set of projections where PeA is having dimensionsmaller than t. In order to do so, we introduce the notion of Sobolev dimension of measure. We definethe modified energy quantity

Esµ = ∫ ∣µ(ξ)∣2(1 + ∣ξ∣)s−n dx

for all s ∈ R. When s ∈ (0, n), this is essentially Esµ, but for technical reasons we cut away the singularityat origin. Moreover, by setting up the definition on Fourier side, we see that the expression makes sensefor s ∉ (0, n). In that range, Esµ does not coincide with the corresponding energy integral formallywritten using the Riesz kernels.

For µ a non-negative Borel measure, we define the Sobolev dimension

dimS µ = sups ∈ R ∶ Esµ <∞.Of course, if µ = f dx and f is a test function, then Esf = ∥f∥

Hs−n2,2 and the Sobolev dimension tells

the borderline smoothness so that the function still manages to belong to the relevant Sobolev (Besselpotential) space. By virtue of being a finite Borel measure, any µ has Sobolev dimension at least 0.On the other hand, having larger Sobolev dimension means being smoother. Absolute continuity withrespect to the Lebesgue measure comes at dimension n, continuity comes at 2n, and smooth densityimplies infinite Sobolev dimension.

The Sobolev dimension of the measure is one more intermediate step when estimating Hausdorffmeasures of sets. If the set has Hausdorff dimension greater than s, then there exists a measure withfinite s-energy and hence the set supports measures with Sobolev dimension larger than s. Small setsonly support rough measures. We summarize the following properties.

Proposition 8.13. Let µ ∈M(Rn). Then

If dimS µ > n, then µ has L2 density with respect to the Lebesgue measure. If dimS µ > 2n, then µ has continuous density with respect to the Lebesgue measure.

Proof. The proof is clear from computations similar to the ones in the proof of Theorem 8.12. Thefirst one follows from the Fourier transform of the measure being in L2 and the second one is shown byproving µ ∈ L1. The dimension tells what kind of weight one can divide and multiply into the definingintegral before applying Cauchy–Schwarz.

Now we are in position to improve the bounds in Theorem 8.12. All three cases (dimension less thanone, less than two and the rest) have slightly different arguments. The first proposition is useful whenwe study objects with dimension greater than one.

Proposition 8.14. Let µ ∈M(Rn) be compactly supported and let t > 0. Then

dim(e ∈ Sn−1 ∶ dimS Peµ < t) ≤ max(0, n − 1 + t − dimS µ)

Proof. Denote e ∈ Sn−1 ∶ dimS Peµ < t = St. This is a Borel set (exercise). Suppose for contradictionthe dimension is more than the claimed. Then there exists τ ∈ (n − 1 + t − s,dimSt) so that Hτ(St) > 0.By Frostman’s lemma, there exists ν ∈M(St) so that ν(Br) ≲ rτ for all balls with radius r. We showthat

∫Sn−1

EtPeµdν(e) <∞.

That will imply ν = 0 or dimS Peµ ≥ t in St, which are both contradictions.Let ϕ be a Schwartz function identically one in the support of µ. Then

∣µ∣2 = ∣ϕ ∗ µ∣2 ≲ ∣ϕ∣ ∗ ∣µ∣2.Now by Fubini

∫Sn−1

EtPeµdν(e) = ∫Sn−1

∫R∣Peµ(u)∣2(1 + ∣u∣)t−1 dudν(e)

≲ ∫Sn−1

∫R∣ϕ∣ ∗ ∣µ∣2(ue)(1 + ∣u∣)t−1 dudν(e)

= ∫Rn

∣µ(x)∣2 (∫Sn−1

∫R∣ϕ(ue − x)∣(1 + ∣u∣)t−1 dudν(e)) dx.

Because ϕ, and hence also its Fourier transform, is Schwartz, we can bound the quantity in the parenthesisby

∫Sn−1

∫R(1 + ∣ue − x∣)−N(1 + ∣u∣)t−1 dudν(e).

42

This is an explicit function, and so we are in position to use the knowledge of the growth of ν.Denote Le = ue ∶ u ∈ R. The set e ∈ Sn−1 ∶ d(x,Le) ≤ r is contained in a cap of radius ∼ r/∣x∣.

Hence its ν measure is ≲ (r/∣x∣)τ . Using this, we show that the display above is bounded by (1+ ∣x∣)t−1−τ .Indeed,

∫Sn−1

∫R(1 + ∣ue − x∣)−N(1 + ∣u∣)t−1 dudν(e) ≲ (1 + ∣x∣)t−1∬∣ue−x∣≤2−1

(1 + ∣ue − x∣)−N dudν(e)

+⎛⎝ ∑

2j+1≤∣x∣+ ∑

∣x∣≤2j

⎞⎠∬2j−1<∣ue−x∣≤2j

(1 + ∣ue − x∣)−N(1 + ∣u∣)t−1 dudν(e) = I+ II+ III .

In the first term we used ∣x − ue∣ ∈ (0,2−1).Further, as

ν(e ∈ Sn−1 ∶ ∣ue − x∣ ≤ 2−1) ≤ ν(e ∈ Sn−1 ∶ d(x,Le) ≤ 2−1) ≲ (1 + ∣x∣)−τ ,

the term I satisfies the desired bound. Similarly, in the second term ∣ue − x∣ ∼ 2j ≤ 2−1∣x∣ so

II ≤ (1 + ∣x∣)t−1 ∑2j+1≤∣x∣

2j2−Njν(e ∈ Sn−1 ∶ d(x,Le) ≤ 2j) ≲ ∣x∣t−1−τ .

Finally,

III ≲ ∑∣x∣≤2j

2−(N−1)j ≲ ∣x∣−N+1.

Using these three estimates and the trivial case when ∣x∣ ≤ 1, we obtain

∫Sn−1

EtPeµdν(e) ≲ ∫Rn

∣µ(x)∣2(1 + ∣x∣)t−1−τ ≤ Esµ <∞

because t − 1 − τ ≤ s − n. This is the desired contradition as discussed in the first paragraph of theproof.

Proposition 8.15. Let µ ∈M(Rn) be compactly supported, 0 < s ≤ 1 and Esµ <∞. Then for t ∈ (0, s]we have

dime ∈ Sn−1 ∶ dimS Peµ < t ≤ n − 1 + (t − 1).

Proof. The Hausdorff dimension is countably stable in the sense that

limk→∞

dim(k

⋃i=1

Ai) = dim(k

⋃i=1

Ai).

Hence it suffices to prove the upper bound on dimension for all

e ∈ Sn−1 ∶ dimS Peµ < σ ⊂ e ∈ Sn−1 ∶ EσPeµ =∞

with σ < t. Let the set on the right hand side be S∞. This is again Borel, but we omit the proof. Assume,for contradiction, than dim(S∞) > n−2+ t. Then Hn−2+t(S∞) > 0 and by Frostman’s lemma there existsν ∈M(S∞) such that ν(Br) ≤ rn−2+t for all balls Br. We will show that either ν is zero or EσPe <∞ inS∞ which will be a contradiction.

The set e ∈ Sn−1 ∶ ∣Pe(x)∣ ≤ δ is covered by ∼ (δ/∣x∣)2−n balls of radius δ/∣x∣ when δ/∣x∣ is very small.Indeed, because Pe(x) = ∣x∣ cos∡(e, x), we see that ∣Pe(x)∣ ≲ δ when π/2 − ∡(e, x) ≲ δ/∣x∣. Hence thereis n− 2 dimensional great circle on n− 1 dimensional sphere whose ∼ δ/∣x∣ neighborhood contains all therelevant points. The claimed covering property follows. Consequently

ν(e ∈ Sn−1 ∶ ∣Pe(x)∣ ≤ δ) ≤∑i

ν(Bi) ≤ #Bi ⋅ (δ∣x∣−1)n−2+t ≲ (δ∣x∣−1)t.

Now we can estimate the ν average of energies of projections to different directions. By Fubini

∫Sn−1

EσPeµdν(e) = ∫Sn−1

(∬ ∣Pe(x − y)∣−σ dµ(x)dµ(y)) dν(e)

=∬ (∫Sn−1

∣Pe(x − y)∣−σ dν(e)) dµ(x)dµ(y).

43

Applying the Cavalieri principle to the inner integral, we estimate

∫∞

0ν(e ∈ Sn−1 ∶ ∣Pe(x − y)∣−σ > λ)dλ = ∫

∞

0ν(e ∈ Sn−1 ∶ ∣Pe(x − y)∣ < λ−1/σ)dλ

≲ T + ∫∞

T( λ

−1/σ

∣x − y∣)t

dt

≲ T + σ

t − σ⋅ T

−t/σ+1

∣x − y∣t

whenever T −1/σ > 10∣x − y∣. Setting T ∼ ∣x − y∣−σ, we reach the upper bound

∫Sn−1

EσPeµdν(e) ≲ Eσµ <∞

as σ < s. This is a contradiction to either ν ∈M(S∞) or EσPeµ =∞ in S∞.

Putting together Propositions 8.14, 8.15 and 8.13, we now have the following refinement of Theorem8.12.

Theorem 8.16. Let A ⊂ Rn be a bounded Borel set.

If dimA ≤ 1 and t ∈ (0,dimA], then

dime ∈ Sn−1 ∶ dimPeA < t ≤ n − 2 + t. If dimA > 1, then

dime ∈ Sn−1 ∶H1(PeA) = 0 ≤ n − dimA.

If s > 2, thendime ∈ Sn−1 ∶ intPeA = ∅ ≤ n + 1 − dimA.

9. Kakeya problems

The original Kakeya problem asks the following: Suppose K ⊂ R2 is such that a line segment of unitlength can be continuously rotated by the angle π. What is the smallest possible area of such a K? Theanswer is: arbitrarily small. Another variant of the question is how large K must be if it contains a unitline segment in all directions. Such a set can even have area zero. The first problem was posed by Kakeyaand both answers were provided by Besicovitch a hundred years ago. The next natural question is, whatis the Hausdorff dimension of a set containing line segments to all directions. The Kakeya conjecturesays it should be n in Rn. This has only been verified when n = 2. Most of the following is from [16], therest comes from [4].

9.1. Besicovitch sets. A Borel set K ⊂ Rn such that for all e ∈ Sn−1 there is a ∈ Rn with

a + te ∶ t ∈ [0,1] ⊂Kand ∣K ∣ = 0 is called a Besicovitch set. We denote K(δ) = x ∈ Rn ∶ d(x,K) < δ.

Theorem 9.1. There exists a compact Besicovitch set K ⊂ Rn. For 0 < δ < 1 and a numerical constantc ≥ 1, it holds

∣K(δ)∣ ≤ − c

log δ.

Proof. We first prove the claim for n = 2. Consider the dyadic rationals dn = j2−n and enumerate themas

(ak)∞k=0 = (0,0,1, 1

2,0,

1

4,2

4,3

4,1,

7

8, . . .)

that is, a0 = 0 and for k ∈ [2n,2n+1)

ak =⎧⎪⎪⎨⎪⎪⎩

2−n(k − 2n) − 1, n even

1 − 2−n(k − 2n), n odd.

We also set ε(k) = ∣ak+1 − ak ∣ so that ε(k) = 2−n for k ∈ [2n,2n+1). Finally, we define a function through

g(t) = t − ⌊t⌋

fk(t) =k

∑m=1

am−1 − am2m

g(2mt)

f(t) = limk→∞

fk(t)

44

and set K = (a, f(t)+at) ∶ a, t ∈ [0,1]. We will show that four rotates of K form a Besicovitch set withthe desired properties. Indeed, it is clear that K contains line segments with all slopes in [0,1].

To estimate the size, we cover the set with suitable squares. Consider first a fixed a and defineF (t) = f(t) + at. For fixed k, it holds f ′k(t) = −ak for all t ∈ [0,1] ∖ 2−kN. Denote the connected

components of this set by U ik where i = 1, . . . ,2k. Choose k such that ∣a−ak ∣ < ε(k). Fix one U ik and takeu, v ∈ U ik. Then

∣F (u) − F (v)∣ ≤ ∣f(u) − f(v)∣ + ∣fk(u) − fk(v) + a(u − v)∣ + ∣fk(v) − f(v)∣ = I+ II+ III .

Here

I+ III ≤ 2 supt∈[0,1]

∑m=k+1

∣am−1 − am∣2m

g(2mt) ≤∞∑

m=k+1

ε(k)2−m ≲ ε(k)2−k

and

II = ∣(u − v)(a − ak)∣ ≤ 2−kε(k).Consequently diam(U ik) ≲ 2−kε(k).

If we now let a vary inside an interval J with length 2−kε(k), we see that

(a, f(t) + at) ∶ a ∈ J, t ∈ Uki ⊂ Q(J, i).

The set on the right hand side is a square with side-length δ = 2−kε(k). As there are 2k sets Uki to coverall t ∈ [0,1], and for each of them 2k/ε(k) intervals J to cover all a ∈ [0,1], we see that K is covered by22k/ε(k) squares of area ε(k)22−2k. Dilating these squares by three, we see that

∣K(2−k−n)∣ = ∣K(2−kε(k))∣ ≲ 2−n.

Because k ≥ 2n, the area bound follows. Clearly K is bounded so K is compact.To prove the claim in dimensions n ≥ 3, let Q be the unit cube in Rn−2. Then K ×Q is an example of

the high dimensional Besicovitch set.

Next we consider the dimension bound which matches the conjecture only in dimension n = 2. In factwe prove something more. Even the Fourier dimension of a Besicovitch set is at least two.

Theorem 9.2. Let K ⊂ Rn be a compact Besicovitch set. Then dimF K ≥ 2. In particular, dimK ≥ 2.

Proof. For e ∈ Sn−1, let ae be a point such that ae + te ∶ t ∈ [0,1] ⊂K. Let ϕ be a non-negative smoothfunction with compact support in [0,1] and integral one. Define µ ∈ M(K) as the weak limit of µkwhere

∫ g dµk = ∫Sn−1

∫1

0g(ae + te)ϕ(t)dtdσk(e)

and each σk is a weighted sum of Dirac masses so that σk σ (this is to avoid problems of measurabilitywhen reagarding the integrand as function of e).

Let α ∈ (0,1) and ξ ∈ Rn ∖B(0,1). Then

∣µ(ξ)∣ = ∣∫Sn−1

e−2πiξ⋅(ae+te)ϕ(t)dtdσk(e)∣ ≤ ∫Sn−1

∣ϕ(e ⋅ ξ)∣dσk(e).

Let ε > 0. For k large enough (depending on ε), σk(e ∈ Sn−1 ∶ ∣ξ ⋅ e∣ ≤ ε∣ξ∣) ≲ ε (angular ε deviation fromthe equator of the sphere gives a band with thickness ε and radius 1). As ϕ is Schwartz,

∣µk(ξ)∣ ≲ ε + (ε∣ξ∣)−N .

Choosing ε = ∣ξ∣−α and N with −N(1 − α) = −α proves ∣µk(ξ)∣ ≲ ∣ξ∣−α for k > k∣ξ∣. Taking k → ∞ thenproves that µ has the same decay. Existence of a measure on K with such a decay means dimF K ≥ 2.

9.2. Nikodym sets. Original Nikodym sets were defined as sets N ⊂ [0,1]2 such that for every x ∈ Nthere is a line L with L ∩N = x such that ∣N ∣ = 1. Sets more like complements of the ones above areusually called Nikodym sets: A Borel set N is Nikodym if for every x there is a line L ∋ x such thatL∩N contains a unit line segment and ∣N ∣ = 0. Such sets exists and the Nikodym conjecture hopes thatdimN = n for all Nikodym sets in Rn. The Nikodym conjecture follows from the Kakeya conjecture, butthe converse is not known.

Proposition 9.3. Let 1 ≤ s ≤ n. If there exists a Nikodym set N with dimN = s, then there exists aBesicovitch set B with dimB = s.

45

Proof. Define F (x′, xn) = x−1n (x′,1) for xn ≠ 0. Then

te + (a,0) ∶ t ∈ [1,2]↦ (ten)−1(te′ + a,1) ∶ t ∈ [1,2] = e−1n (e′,0) + (ent)(a,1) ∶ t ∈ [1,2].

A line through (a,0) is mapped to a line with direction (a,1). Applying a few rotated copies of F showsthat a Nikodym set can be used to construct a Besicovitch set.

9.3. Furstenberg sets. One more variant of a Besicovitch set is a Furstenberg set. A Borel set on theplane is β-Furstenberg with β ∈ [0,1] is for every e ∈ Sn−1 there is a line le so that dim le ∩K ≥ β. It isan open problem what is the dimension of Furstenberg sets. Let d(β) be the infimum of dimensions ofβ Furstenberg sets. It is known that

β + 1/2 ≤ d(β) ≤ 3/2β + 1/2, d(1/2) > 1 .

9.4. Kakeya maximal function. Kakeya maximal function conjecture is a stronger variant of theKakeya conjecture. We discuss two equivalent formulations and prove it in the plane. For a ∈ Rn,e ∈ Sn−1 and δ ∈ (0,1), define the δ-tube at a to direction e as

T δe (a) = x ∈ Rn ∶ ∣(x − a) ⋅ e∣ ≤ 1

2, ∣(x − a) − e[(x − a) ⋅ e]∣ ≤ δ.

For a function f ∈ L1loc(Rn), define the Kakeya maximal function

Kδf(e) = supa∈Rn

⨏T δe (a)

∣f(y)∣dy

where e ∈ Sn−1. Note that the Kakeya maximal function is a function on the sphere whereas the inputfunction lives in the full space. This is very different from the maximal functions we have seen so far.The supremum is over translated tubes in the space. A maximal function defined using rectangular tubes(dimensions δ × ⋯ × 1) is comparable to the one above, but for the time being we stick to the circularcylinders. The following proposition is an immediate consequence of the definition:

Proposition 9.4. The following hold

∥Kδf∥L∞(Sn−1) ≤ ∥f∥L∞(Rn)

∥Kδf∥L∞(Sn−1) ≲ δ1−n∥f∥L1(Rn)

No other Lp → Lq bounds hold uniformly in δ! The reason for this is the Besicovitch set. Indeed,let K be the set from Theorem 9.1. Let K(δ) be its δ neighborhood which we proved to have theproperty ∣K(δ)∣ → 0 as δ → 0. Let f = 1K(δ). For every e ∈ Sn−1 there is a tube T δe (a) ⊂ K(δ). So if

Kδ ∶ Lp(Rn)→ Lq(Sn−1), then

1 ∼ infe∈Sn−1

Kδf(e) ≤ ∥Kδf∥Lq(Sn−1) ≤ C∥f∥Lp(Rn) = ∣K(δ)∣1/p

which cannot hold for δ small enough as the right hand side tends to zero. However, if we allow C ≳ δ−ε,the Besicovitch set stops being a problem.

There is also another limiting example, namely the characteristic function of a ball f = 1B(0,δ). Withthis example we see that

Kδf ∼ δ, ∥f∥Lp(Rn) = ∣B(0, δ)∣1/p ∼ δn/p.

This also leads to contradiction with small δ in the Lp → Lp bounds if p < n. The Kakeya maximalconjecture says that these are the only limitations that occur.

Conjecture 9.5 (Kakeya maximal conjecture). Let n ≥ 2. Then for all ε > 0 and δ ∈ (0,1/2) it holds

∥Kδ∥Lp(Rn)→Lp(Sn−1) ≲ δ−ε.

There are also conjectured Lp → Lq bounds that one guesses by interpolating the conjecture with thetrivial L1 → L∞ bound.

46

9.5. The dual Kakeya inequality. Next we derive an equivalent formulation of the Kakeya maximalinequality. This is split in several propositions.

Proposition 9.6. Let 1 < p <∞, 0 < δ < 1 and 0 <M <∞. Suppose that

∫ (∑k

1Tktk)p

≤M

whenever tk ≥ 0 are a sequence with ∥(tk)∥`p′ ≤ δ(1−n)/p′

and Tk = T δek(a) are δ separated in the sensethat ∣ek − ej ∣ > δ whenever k ≠ j (δ separated δ tubes in what follows). Then

∥Kδf∥Lp(Sn−1) ≤ C(n)M∥f∥Lp(Rn)for all f ∈ Lp(Rn).

Proof. Let e1, . . . , em be a maximal δ separated subset of the unit sphere. If e ∈ B(ek, δ), then anytube T δe (a) is covered by a finite number of δ tubes to the direction Tek . This number only depends onthe dimension. Hence Kδf(e) ≲ Kδf(ek) for such a direction e. Now

∫Sn−1Kδf(e)p dσe ≤

m

∑k=1∫B(0,ek)

Kδf(e)p dσ(e) ≲m

∑k=1

Kδf(ek)pδ1−n ≲ δ1−n (m

∑k=1

tkKδf(ek))p

for some sequence tk with `p′

norm at most one. The last step used the dual representation of the `p

norm. We did this in order to linearize in Kδf .For each k there exists a δ tube Tk to direction ek so that

Kδf(ek) ∼ ⨏Tk

∣f ∣.

Hence

m

∑k=1

tkKδf(ek) ∼ δ1−n ∫ (m

∑k=1

1Tk(y)tk) f(y)dy

≤ δ1−n∥f∥Lp(Rn) ∥m

∑k=1

1Tktk∥Lp′(Rn)

≤Mδ(1−n)(1−1/p′)∥f∥Lp(Rn)

so∥Kδf∥Lp(Sn−1) ≤ C(n)M∥f∥Lp(Rn)

as 1/p′ − 1 = −1/p.

Proposition 9.7. Let ε > 0, 0 < δ < 1 and M > 1. Then

∥Kδf∥Lp(Sn−1) ≲Mδ−ε∥f∥Lp(Rn)for all f ∈ Lp(Rn) if and only if

∥m

∑k=1

1Tk∥Lp′(Rn)

≲Mδ−ε(mδn−1)1/p′

for all collections Tkmk=1 of δ separated δ tubes.

Proof. Assume first the maximal function inequality. Let ekmk=1 be δ separated directions and Tk =T δek(ak) for some ak ∈ Rn. Let g have Lp

′

(Rn) norm one. Then

∫m

∑k=1

1Tkg =∑k∫Tkg ≲∑

k

δn−1Kδg(ek) ≲ ∫⋃mk=1B(ek,δ)

Kδf(e)dσ(e)

≤ ∥Kδg∥Lp(Sn−1)σ (m

⋃k=1

B(ek, δ))1/p′

≲ δ−εM(mδn−1)1/p′ .

Taking supremum over g with norm at most one, the inequality in the second display of the theoremfollows.

To prove the other direction of the asserted equivalence, we invoke the previous proposition. Let tkbe non-negative numbers so that ∑k t

p′

k ≤ δ1−n, and let Tk be a collection of δ separated δ tubes. It is

obvious that tk ≤ δ(1−n)/p′

for all k. On the other hand,XXXXXXXXXXX∑

tk≤δn−1tk1Tk

XXXXXXXXXXXLp′(Rn)≤ (mδn−1)1/p′ ,

47

so it suffices to estimate the sum over tk ∈ (δn−1, δ(1−n)/p′

].Denote

Ij = k ∶ 2j−1 ≤ tk < 2j.These collections are non-empty only for j such that

2j−1 ≤ δn−1 and δ(1−n)/p′

≤ 2j .

There are C log(1/δ) ≲ δ−ε/2 such numbers j. Then by the assumption

XXXXXXXXXXXX∑

k∶δn−1≤tk≤δ(1−n)/p′tk1Tk

XXXXXXXXXXXXLp′(Rn)≤∑

j

2jXXXXXXXXXXXX∑k∈Ij

1Tk

XXXXXXXXXXXXLp′(Rn)≲∑

j

2jMδε/2(#Ijδn−1)1/p′ .

But here

2j(#Ijδn−1)1/p′ ≲ (∑k

δn−1tp′

k )1/p′

≤ 1

by definition so the claim follows by the counting estimate

#j ∶ Ij ≠ ∅ ≲ δ−ε/2.

This concludes the proof.

Proposition 9.8. If

∥m

∑k=1

1Tk∥Lp(Rn)

≲p,n,ε Mδ−ε

for all collections Tkmk=1 of δ separated δ tubes, then

∥m

∑k=1

1Tk∥Lp′(Rn)

≲p,n,ε Mδ−ε(mδn−1)1/p′

for all collections Tkmk=1 of δ separated δ tubes.

Proof. We only sketch the proof. For fixed m, let c(m) be the smallest constant controlling

∥m

∑k=1

1Tk∥Lp(Rn)

for all δ separated families of m elements. Suppose Tk is a family so that

∥m

∑k=1

1Tk∥Lp(Rn)

= c(m).

Denote the directions by S. As the left hand side is invariant under rotations, for any rotation g ∈ O(n)

∥m

∑k=1

1Tk + 1g(Tk)∥p

Lp(Rn)≥ ∥

m

∑k=1

1Tk∥p

Lp(Rn)+ ∥

m

∑k=1

1g(Tk)∥p

Lp(Rn)= 2c(m)p.

On the other hand, we can find a rotation g so that S ∪ g(S) is a union of a δ separated family and avery small family.

Let θn be a probability measure on O(n). Denote

a(S, g) = #(e, e′) ∈ S × g(S) ∶ ∣e − e′∣ ≤ δ

∫ a(S, g)dθn(g) = ∑e,e′∈S

f(e, e′)

f(e, e′) = ∫ 1B(0,δ)(e − g(e′))dθn(g).

The last quantity is constant in the variable e′ so by Fubini

f(e, e′) = ⨏ f(e, e′)dσ(e′) ≤ bδn−1.

Now

∫ a(S, g)dθn(g) ≤ bm2δn−1

48

and hence there must exist at least one g such that a(S, g) ≤ bm2δn−1. Define

S2 = e ∈ S ∶ ∃e′ ∈ g(S) ∶ ∣e − e′∣ ≤ δS1 = (S ∖ S2) ∪ g(S).

Now S2 is δ separated as a subset of S, and by the estimates above #S2 ≤ bm2δn−1. By definition, S1 isδ separated. It holds #S1 ≤ 2m. Then by Minkowski and the estimate in the beginning of the proof

21/qc(m) ≤ c(2m) + c(bm2δn−1).The claim can be derived from this inequality, but we omit the details. See Proposition 22.7 in [16].

Corollary 9.9. Let p > 1 and δ ∈ (0,1). Then

∥Kδf∥Lp(Sn−1) ≲n,ε δ−ε∥f∥Lp(Rn)for all ε if and only if

∥m

∑k=1

1Tk∥Lp(Rn)

≲p,n,ε δ−ε

for all families Tk of δ separated δ tubes and all ε > 0.

Before moving on, we consider two important configurations at the critical exponent p = n. Thereare at most δ1−n tubes in the sum by angular separation. If all the tubes on the left are disjoint, wehave δ1−n tubes, each with measure δn−1 and the inequality holds. If we violate the angular separation,things go wrong. If we have δ1−n identical tubes, the left hand side has overlap of δ1−n. This gives

δ((1−n)n′+(n−1))/n′ = δ−n−1n , which is a lot. In some sense, Kakeya is claiming that angularly separated

tubes overlap negligibly little.

9.6. Kakeya maximal bound. Next we prove a bound for the Kakeya maximal function on L2. Thisbound is enough to imply the Kakeya conjecture in the plane. In higher dimensions, the conjectureasks about behaviour around Ln, and hence the L2 bound is not that interesting. However, it can beinterpolated with the trivial L∞ bound to get some bounds around Ln.

Theorem 9.10. Let n ≥ 2. For all δ ∈ (0,1) and all f ∈ L2(Rn), the following is valid:

∥Kδf∥L2(Sn−1) ≤ C(n)⎧⎪⎪⎨⎪⎪⎩

√log(1/δ)∥f∥L2(Rn), n = 2

δ(2−n)/2∥f∥L2(Rn), n ≥ 3

Proof. Let ϕ ∈ S be such that supp ϕ ⊂ [−1,1], ϕ ≥ 0 and ϕ(x) ≥ 1 for ∣x∣ ≤ 1. Let

ψ(x) = 1

δn−1ϕ(x1)ϕ( ∣x′∣

δ) , x = (x1, x

′) ∈ Rn.

Then ψ(ξ1, ξ′) = ϕ(ξ1)ϕ(δ∣ξ′∣) so that

supp ψ ⊂ [−1,1] ×Bn−1(0, δ−1),that is, the Fourier transform is supported in a 1-fat δ−1 disk. Moreover, from the lower bound for ϕ isfollows that δ1−n1T δe1(0)

≤ ψ. This is a function for a canonical tube. Next we rotate it to obtain functions

for other tubes. Let e ∈ Sn−1. Let ge be a rotation of Rn so that ge(e1) = e. Let ψe = ψ g−1e . Now

supp ψe ⊂ Ce ∶= ge([−1,1] ×Bn−1(0, δ−1), δ1−n1T δe ≤ ψe.Now

Kδf(e) = Cn supa∈Rn

∫ δ1−n1T δe (a)(y)f(y)dy = Cn supa∈Rn

∫ δ1−n1T δe (0)(a − y)f(y)dy

≤ Cn supa∈Rn

∫ ψ(a − y)f(y)dy = Cn supa∈Rn

ψe ∗ f(a) = Cn∥ψe ∗ f∥L∞(Rn) ≤ ∥ψef∥L1(Rn).

Then

∫ ∣ψe(ξ)f(ξ)∣dξ ≤ (∫ ∣ψe(ξ)∣∣f(ξ)∣2(1 + ∣ξ∣)dξ)1/2

(∫∣ψe(ξ)∣1 + ∣ξ∣

dξ)1/2

= I ⋅ II .

Clearly

II2 ≤ ∫Ce

1

1 + ∣ξ∣dξ ∼ ∫

B(0,1/δ)

1

1 + ∣ξ′∣dξ′ ∼M ∶=

⎧⎪⎪⎨⎪⎪⎩

log(1/δ), n = 2

δ2−n, n ≥ 2.

49

Hence

∫Sn−1Kδf(e)2 σ(e) ≲M ∫

Sn−1I2 dσ(e) = ∫

Rn(∫

Sn−1∣ψe(ξ)∣dσ(e)) (1 + ∣ξ∣)∣f(ξ)∣2 dξ.

The quantity in the parenthesis is estimated by

σ(e ∶ ψe(ξ) ≠ 0) ≤ σ(e ∶ ξ ∈ Ce) ≲1

1 + ∣ξ∣.

Indeed, if the disk Ce meets ξ, a rotation of 1/∣ξ∣ to the thin direction might still keep ξ inside, whereasn − 1 full rotations that keep the Ce fixed certainly do that. Hence the set on the sphere is a band ofangular width 1/∣ξ∣, and the Theorem follows from the surface area estimate by Plancherel.

Next we see that the Kakeya maximal inequality implies a lower bound for the Hausdorff dimensionof any Besicovitch set, the maximal conjecture implying the Kakeya conjecture.

Theorem 9.11. Let 1 < p <∞ and β > 0 such that n − βp > 0. Assume that ∥Kδ∥Lp(Sn−1)→Lp(Rn) ≲ δ−β.Then Hausdorff dimension of any Besicovitch set is at least n − βp.

Proof. It suffices to deal with compact Besicovitch set so we assume K is a compact Besicovitch. Bydefinition, for each e ∈ Sn−1 there is ae so that Ie = ae + te ∶ t ∈ [0,1] ⊂ K. To compute a lower boundfor the Hausdorff dimension, let α ∈ (0, n− βp) and let Bjj be a cover of K with balls Bj having radiirj ≤ 1. Once we show ∑j rαj ≥ c, for c independent of the collection, we can take infimum over familiesBj to bound the α Hausdorff content, and hence the Hausdorff measure, away from zero. That impliesa lower bound for the dimension.

Suppose all the balls were of the same size rj ∼ k. Then

∑j

rαj ∼ #Bjkα ≳ #Bjkn−pβ ∼ k−pβ ∫K(k)

1(⋃j Bj)(k)(x)dx.

This is bounded below by the Kakeya maximal function at scale k which in turn is uniformly boundedfrom below by virtue of K being Besicovitch: for each e the tube T ke (ae) ⊂ K(k) and the claim wouldfollow. However, the balls Bj need not be of the same size, and this makes it impossible to relate theunion of balls in the cover to the neighborhood of the set to be covered. To overcome this issue, we haveto decompose the cover to single scaled “bubble sets”.

For k ≥ 1, denote Jk = j ∶ 2−k ≤ rj < 21−k. Let

Fk = ⋃j∈Jk

Bj , fk = 1Fk .

Next we try to find cover the whole Sn−1 by the sets where Kδkfk are large. If Ie goes through only very

few balls in Fk, then ∣T 2−k

e (ak)∩Fk ∣ ≲ 2−k ∣T 2−k

e (ak)∣ (intersection is basically a ball with measure ≳ 2−kn

whereas the tube has measure 2−k(n−1)). Hence we set

Sk = e ∈ Sn−1 ∶H1(Ie ∩ Fk) ≥1

2k2.

The exact lower bound is not so important. We want it to be larger than 2−k in order to only considerdirections with non-transverse intersection of Ie and Fk, but we also want it to sum up like ∑k 1/(2k2) < 1so that there cannot be e ∈ Sn−1 ∖⋃∞k=1. Indeed, suppose there is e with e ∉ Sk holding for all k. Thenby K being Besicovitch

1 ≤H1(Ie ∩K) ≤∞∑k=1

H1(Ie ∩ Fk) <∞∑k=1

1

2k2< 1

which is a contradiction. No such e can exist and consequently

Sn−1 =∞⋃k=1

Sk.

Then

σ(Sk) ≤ (2k2)p ∫SkK2−kfk(e)p dσ(e) ≲ k2p2kβp∣Fk ∣ ≲ k2p2kβp2−kn#Jk ≲ 2−kα#Jk.

Summing over k we see

1 ≤ σ(Sn−1) ≤∑k

σ(Sk) ≤∑k

2−kα#Jk ≲∑j

rα.

This completes the proof. 50

9.7. Nikodym maximal function. The Nikodym set conjecture also has its maximal function vari-ant, although we do not discuss the proof of Nikodym maximal conjecture implying the Nikowdym setconjecture. However, we study the proof that Nikodym maximal function conjecture is equivalent withthe Kakeya maximal function conjecture. This is remarkable, since the set conjectures are not known tobe equivalent.

Let δ > 0 and x ∈ Rn. We define the Nikodym maximal function of f ∈ L1loc(Rn) at x as

Nδf(x) = supe∈Sn−1

⨏T δe (x)

∣f(y)∣dy.

There are essentially equivalent variants by only asking the tube to contain x instead of being centredthere.

Theorem 9.12. Let α,β ≥ 0, 1 < p <∞ and n ≥ 2. Consider the inequalities

∥Kδ∥Lp(Rn)→Lp(Rn) ≲ δ−α(9.1)

∥Nδ∥Lp(Rn)→Lp(Rn) ≲ δ−β(9.2)

If (9.1) holds and β + 1 > n/p, then (9.2) also holds with β ≤ α.If (9.2) holds, then (9.1) also holds with α ≤ n/p − 1 − 2β.

Proof. We start with the first mentioned assertion. We use the transformation F (x′, xn) = x−1n (x′,1),

defined for xn ≠ 0. Suppose first supp f ⊂ B(0,1). Assume that f(x′, xn) = 0 unless 12≤ xn ≤ 1. Fix

x′ ∈ Rn−1 with ∣x′∣ ≤ 2. Recalling

F (te + (a,0) ∶ t > 0) = e−1n (e′,0) + (ent)−1(a,1)

and using the fact that F is well-behaved on tubes through B(0,1) ∩ 1/2 ≤ xn ≤ 2 with small angle toxn axis. Take e ∈ spr. Then

F (T δe (x′) ∩ 1/2 ≤ xn ≤ 1) =⊂ T cδ(x′,1)

∣(x′,1)∣

(a).

The set on the left is a union of line segments

e−1n (e′,0) + (ent)−1(a,1) ∶ t ∈ [1/8,1], ∣a − x′∣ ≲ δ

from which the containment can be seen.Now

∫Rn−1Nδf(x′,0)p dx′ ≲ ∫

Sn−1Kδ(f F −1)(e)p dσ(e) ≲ δ−αp ∫ ∣f F −1(x)∣p dx ≲ δ−αp ∫ ∣f(x)∣p dx

where we used the support properties of f to keep all changes of variables reasonable.Next we want to establish the same estimate for f with support in B(0,1) ∩ 2−k−1 ≤ xn ≤ 2−k. To

do this we, first define a restricted Nikodym maximal function

Nδf(x) = sup∣e−en∣≤1

⨏T δe (x)

∣f(y)∣dy

Denote f(x,xn) = f(x,2−kxn) so that f(x,xn) is supported in the strip (x′, xn) ∶ 1/2 ≤ xn ≤ 1. Leta0 = (a,0) ∈ Rn and let T δe (a0) be a tube that meets the support of f . Assume ∣e − en∣ ≤ 1. Then bychange of variable

⨏T δe (a0)

∣f(x)∣dx = ⨏2kT δe (a0)

∣f(2−kx)∣dx ≲ 2−kNδ f(2ka0).

Now

∫Rn−1Nδf(x′,0)p dx′ = ∫

B(0,2)Nδf(x′,0)p dx′ ≲ 2−kp ∫

B(0,2)Nδ f(2kx′,0)p dx′

= 2−kp−(n−1)k ∫B(0,2k+1)

Nδ f(x′,0)p dx′ ≤ 2−kp−(n−1)k ∑y∶B(y,1)⊂B(0,2k)

∫B(y,1)

Nδ f(x′,0)p dx′.

Since f is supported in a unit ball intersected with a unit strip (at least f as much as it affects N f inB(y,1)), we can apply the case already proved to see

∥Nδf(⋅,0)∥pLp(Rn−1) ≲ 2−kp−(n−1)k ∑y∶B(y,1)⊂B(0,2k)

(2kδ)−β∥f∥pLp

= 2−kp−(n−1)k ∑y∶B(y,1)⊂B(0,2k)

(2kδ)−βp2nk∥f∥pLp ≲ 2−βkp−(p−n)kδ−β∥f∥pLp .

51

Finally, taking a generic f supported in the unit ball and denoting fk = 12−k−1≤xn≤2−kf , we see

∥Nδf(⋅,0)∥Lp(Rn−1) ≤∞∑k=0

∥Nδfk(⋅,0)∥Lp(Rn−1) ≲∞∑k=0

2−βk−(1−n/p)kδ−β∥f∥Lp ≲ ∥f∥Lp

provided β + 1 > n/p.As the norm inequality holds for all functions in the unit ball, it also holds for translations of f .

Consequently, we can use Fubini to conclude

∫2

−2∥Nδf(⋅, t)∥pLp(Rn−1) dt = ∫

2

−2∥Nδ(τtf)(⋅,0)∥pLp(Rn−1) dt ≲ δ

−β∥f∥Lp(Rn).

By dilations, the claim holds for compactly supported functions. By density, it holds for all Ln functions.To control the full Nδ, we can represent it as a sum of finitely many Nδ operators corresponding todifferent directions.

Next we study the converse. Assume that Nikodym maximal bound holds, that is, (9.2) holds. Takef supported in the unit ball. Fix δ. Take e. To bound the Kakeya maximal function, take a tube T δe (a)meeting the unit ball. Let fδ = f(⋅/δ). Then supp fδ ⊂ B(0, δ). We have

⨏T δe (a)

f(y)dy = ⨏δT δe (a)

fδ(y)dy = cnδ1−2n ∫δT δe (a)

fδ.

Now δT δe (a) is a tube of length δ and thickness δ2. It meets B(0, δ) Let b ∈ δT δe (a) ∩ B(0, δ) andt ∈ (1/2,1). Let

l1 = b(1 − θ)θte ∶ θ ∈ [0,1], l2 = θte ∶ θ ∈ [0,1].

These lines make an angle ∼ ∣b−0∣/1 ≲ δ. Because δT δe (a) is δ long, meets l1 and makes an angle ≲ δ withit, we conclude

δT δe (a) ⊂ l1(cδ2).

Consequently

⨏T δe (a)

f(y)dy ≲ δ−1Ncδ2f(te)

and

∥Kδf∥pLp(Sn−1) = 2∫1

1/2 ∫Sn−1Kδf(e)p σ(e) ≲ 2δ−p ∫

1

1/2 ∫Sn−1Ncδ2fδ(te)p σ(e)

≲ δ−p ∫B(0,1)

Ncδ2fδ(x)p dx ≲ δ−p+n−2βp ∫ fp

The claim for more general f follows as in the previous assertion.

10. Conjectures in the Fourier restriction theory

Next we discuss the restriction conjecture and some of its variants. In particular, we shall see thatall these conjectures imply the Kakeya maximal conjecture. The connection of the Kakey maximalconjecture with the restriction theory is best seen via the dual Kakeya inequality. We first note that theProposition 9.6 can be reformulated as follows:

Proposition 10.1. Let δ ∈ (0,1) and C > 1. If for every δ-separated set ek ⊂ Sn−1, set of pointsak ⊂ Rn and for all sequences of non-negative numbers (tk)k

∥∑k

1Tktk∥Lp′

≤ Cδn−1p′

−β (∑k

tp′

k )1p′

then

∥Kδ∥Lp(Rn)→Lp(Sn−1) ≲n,p Cδ−β .

Using this formulation, we show that the content of the Kakeya conjecture is contained in the restric-tion conjecture.

52

10.1. Restriction implies Kakeya. The characteristic functions of the tubes can be bounded above bysuitable wave packets whose frequencies live on the sphere (direction and scaling). Such wave packets canbe glued together by Khintchine to give rise to a function that is Fourier extension of another functionon the sphere, which is the setup of the restriction conjecture.

Theorem 10.2. Let q ∈ (2n/(n − 1),∞) and let E be the extension operator relative to the surfacemeasure of the unit sphere in Rn. If

∥Ef∥Lq(Rn) ≲ ∥f∥Lq(Sn−1)for all f ∈ C(Sn−1), then

∥Kδf∥Lp(Sn−1) ≲ δ4nq −2(n−1)∥f∥Lp(Rn), p = q

q − 2

for all f ∈ Lp(Rn).

Proof. Let ek ⊂ Sn−1 be a δ-separated set, tk ≥ 0 and Tk = T δek(δ2ak) with ak ∈ Rn. Let τk = δ−2Tk and

Sk = e ∈ Sn−1 ∶ 1 − e ⋅ ek ≤ (δ/C)2. Define fk = e−2πix⋅ak1Sk . By a stationary phase argument (Knappexample, compare to the discussion just before 6.9),

∣Efk(ξ)∣ = ∣E1Sk(ξ − ak)∣ ≳ δn−11τk .

This is almost all we need. Let ωk be random signs and define

fω =∑k

√tkωkfk.

Then

∫ (∑k

tk1τkδ2(n−1))

q/2stat. phase

≲ ∫ (∑k

tk ∣Efk ∣2)q/2

Khintchine∼ E∫ ∣Efω ∣q

hypothesis≲ ∫ ∑

k

∣fk ∣qtq/2k = ∫ ∑k

1Sktq/2k ∼ δn−1∑

k

tq/2k .

Changing variable x δ−2y so that x ∈ τk if and only if y ∈ Tk and rearranging the factors, we get

∥∑k

1Tktk∥Lp′

≤ Cδn−1p′

+ 4nq −2(n−1) (∑

k

tp′

k )1p′

for p′ = q/2, that is, p = q/(q − 2), which implies the claim by Proposition 10.1.

Corollary 10.3. The restriction conjecture implies the Kakeya maximal conjecture.

Proof. This is immediate from applying the theorem above to q = c 2nn−1

for c > 1. On can choose ε > 0in the Kakeya conjecture and then find c small enough so that ∥Kδf∥Lp(Rn)→Lp(Sn−1) ≲ δ−ε. Then p < n,and this bound can be interpolated with the trivial L∞ bound for the Kakeya maximal function.

10.2. Square function implies Kakeya. To discuss the next inequality, we change the point of view abit. Instead of the Fourier extension of a function supported on a surface, we study functions supportedin a thin neighborhood of the surface. We denote A(δ) = B(0,1+ δ)∖B(0,1− δ). Let ek be a maximal

δ1/2-separated set on the unit sphere and let Sk = e ∈ ∂B(0,1) ∶ ∣1 − e ⋅ ek ∣ ≤ δ. Let ϕk be an associated

smooth partition of unity on the unit sphere and let θk(re) = ϕk(e). Suppose f ∈ Cc(A(δ)). Denote

fk = f θk .

Conjecture 10.4. For all δ ∈ (0,1), ε > 0, all balls Bδ−1 and all f with f ∈ Cc(A(δ)) it holds

∥f∥Lp(Bδ−1) ≲ε δ−ε

XXXXXXXXXXXX(∑k

∣fk ∣2)1/2XXXXXXXXXXXXLp(Rn)

.

This is reminiscent to one direction of the Littlewood–Paley inequality, but here the pieces are notgiven by dydic scales but separated directions. The content is similar though. Disjoint Fourier supportsmean orthogonality in L2 but not necessarily in Lp (where a Pythagoraean type identity only followsfrom disjoint physical supports). Square function estimate is the best substitute, and the conjecturehopes it indeed be the case. From this point of view the Kakeya connection is immediate: Kakeya saysany separated tube family behaves as if it were disjoint, the square function conjecture says functionsconsisting of transversal wave packets behave almost as if they had disjoint supports.

The next proof follows the presentation in [5].53

Theorem 10.5. The square function conjecture implies Kakeya maximal and hence Nikodym maximal.

Proof. We start by restating the dual Kakeya inequality in several rescaled forms. Passing from oneformulation to another will play a role in the proof. The conjecture is about p = n, but we state theinequalitites with general p in order to better track the parameters in the scalings.

Let Tk be δ-separated δ ×⋯ × δ × 1 tubes. Then for all δ > 0 and ε > 0

∥∑k

tk1Tk∥p′

≲ δn−1p′

−ε (∑k

tp′

k )1p′

.

(Change of variable). Let Tk be δ-separated λδ×⋯×λδ×λ tubes. Then for all δ > 0 and ε > 0

∥∑k

tk1Tk∥p′

≲ (λδ)np′ δ

− 1p′−ε (∑

k

tp′

k )1p′

.

(Set λ = δ−2). Let Tk be δ-separated δ−1 ×⋯ × δ−1 × δ−2 tubes. Then for all δ > 0 and ε > 0

∥∑k

tk1Tk∥p′

≲ δ−np′ δ

− 1p′−ε (∑

k

tp′

k )1p′

.

(Set δ = δ1/2). Let Tk be δ1/2-separated δ−1/2 × ⋯ × δ−1/2 × δ−1 tubes. Then for all δ > 0 andε > 0

∥∑k

tk1Tk∥p′

≲ δ−n+12p′

−ε (∑k

tp′

k )1p′

.

We start the proof by fixing a δ1/2 separated collection of δ−1/2 ×⋯× δ−1/2 × δ−1 tubes Tk = Tek(ak) anda sequence of non-negative numbers tk as in the last formulation of Kakeya. As we can cover the Rn bya countable union of balls with radius δ−1, it suffices to restrict the attention to one such ball. Indeed,a tube meets at most Cn such balls. We can associate to each ball all the tubes that meet it and allcoefficients tk corresponding to these tubes. If we prove Kakeya for the Ln

′

norm restricted to a singleball and the tubes meeting it, we can eventually sum the estimates up to reach the claimed inequalityin the whole space. Hence, assume all the tubes meet B(0, δ−1).

Next we define geometric creatures living on or inspired by the unit sphere of the frequency space.Let χk be a smooth partition of unity on the unit sphere adapted to the cover by δ1/2 caps centred atek (each function supported in the cap in question). Let χik be a smooth partition of χk adapted to δ

caps centred at eik. Let T ik = Teik(ak) be δ−1−ε ×⋯ × δ−1−ε × δ−2−ε tubes.

The geometric objects are converted to wave functions as follows. Let f ik = e2πiak ⋅ξχiksk. This is theFourier transform of a wave packet. There is a coefficient sk, a frequency location χik to make it live in

δ−1 × ⋯ × δ−1 ×∞ tube and a modulating factor e2πiak ⋅ξ to make this tube centred at ak. Further, wedefine fk = ∑i fk and f = ∑k fk.

Recall we were attempting to prove the dual Kakeya inequality in the δ−1 ball centred at origin. Letϕ ≥ 1B(0,δ−2) be a Schwartz function such that ϕ ⊂ B(0, δ2) and ϕ ≲ 1. In particular, multiplication by

ϕ equals convolving the Fourier transform with ϕ. This causes uncertainty of δ2 in the frequency andmakes Ef ik a function with smooth Fourier transform compactly supported in the δ2 neighbourhood ofthe δ cap centred at eik. Then

by the Knapp type stationary phase argument ∣Efk ∣ ≳ δn−12 sk1Tk ;

by the uncertainty principle heuristics above,

ϕ∣Ef ik ∣ = ∣F−1(ϕ ∗ (f ikdσ))∣ ≲M δn−11T ik+ δMsk

for any M > 1.54

Let ωk be random signs with zero expectation. ThenXXXXXXXXXXXX(∑k

s2k1Tk)

1/2XXXXXXXXXXXXL2p′(Bδ−1)

Knapp≲

XXXXXXXXXXXX(∑k

δ1−n∣Efk ∣2)1/2XXXXXXXXXXXXL2p′(Bδ−2)

Khintchine∼ δ(1−n)/2E2p′ ∥∑k

ωkEfk∥L2p′(Bδ−2)

Sq. f. conj.≲ δ(1−n)/2−ε

XXXXXXXXXXXX(∑k

∑i

∣ϕEf ik ∣2)1/2XXXXXXXXXXXXL2p′(Rn)

uncert. pr.≲ δ(1−n)/2−ε

⎛⎜⎝δn−1

XXXXXXXXXXXX(∑k

∑i

1T iks2k)

1/2XXXXXXXXXXXXL2p′(Rn)

+⎛⎝∑k,i

s2k

⎞⎠

1/2

∥ϕ∥Lp′(Bc2δ−2−ε

⎞⎟⎠

= I+ II .

By the Schwartz decay of ϕ, we have

II ≲M δM (∑k

s2p′

k )1/p′

for any M > 1 so this can be regarded as an error term.To deal with the remaining term I, we take advantage of the main geometric insight of the proof.

The square function conjecture helped us to cut the Fourier supports smaller. The resulting tubes Tare correspondingly fatter. Because every ak is δ−1 close to origin and each T has smaller side δ−1−ε,we realise that all the tubes contain origin. Next, we show that the Kakeya inequality holds for such afamily of tubes (called a bush). By rescaling, we can prove this fact in our preferred formulation, andwe choose that of 1×⋯× 1× δ−1 tubes with δ−1 = N (the choice λ = δ−1 in the list of formulations in thebeginning of the proof).

Suppose τk are N−1 separated tubes meeting at the origin. Let B(0,N) = ⋃Nj=1B(0, j) ∖B(0, j − 1) =⋃Nj=1Bj . We decompose each annulus further as Bj = ⋃αBαj where Bαj = te ∶ t ∈ [j −1, j],1−e ⋅eα < j−2and eα is a suitable 1/j separated set on the unit sphere. It holds ∣Bαj ∣ ∼ 1. Now

∫Bj

(∑k

tk1τk)p′

≲∑α∫Bαj

(∑k

tp′

k )( 1/j1/N

)(n−1)p′

p

≤ Nj−1.

Summing over j we obtain

⎛⎝∫

(∑k

tk1τk)p′⎞⎠

1/p′

≲ (N logN)1/p′ (∑k

tp′

k )1/p′

where we used p = n. This is the desired upper bound for a family of tubes all meeting at one point.Next we apply the bound above to the term I. By the Kakeya estimate for the bush and the fact that

for each k there are (δ−1/2/δ−1)n−1 indices i, we get

I ≲⎛⎝δ−n+1p′

−εδ−n−1

2p′ (∑k

s2p′

k )1/p′⎞

⎠

1/2

.

Squaring the total estiamte so far, we haveXXXXXXXXXXXX(∑k

s2k1Tk)

1/2XXXXXXXXXXXXLp′(Bδ−1)≲ δ(n−1)−(n+1

p′+n−1

2p′)−ε∥s2

k∥`p′ = δ−n+1

2p′−ε∥s2

k∥`p′ .

Along with the discussion in the beginning, how to extend the inequality to the full space, this concludesthe proof.

10.3. Square function implies restriction. We saw that the square function conjeecture 10.4 impliesthe Kakeya maximal conjecture. Next we see that it also implies restriction. Briefly, the square functionestimate captures the cancellation between the the pieces of the function with δ1/2 separated Fouriersupports. These pieces can be viewed as wave packets essentially constant on tubes, and the overlapof the tubes can controlled by a the Kakeya estimate, which is also implied by the square functionconjecture. The following is an averaged version of the restriction estimate in local form.

55

Lemma 10.6. Assume the square function conjecture 10.4 holds. Let f ∈ Cc(A(δ)). Then

∥F−1f∥Lp(Rn) ≲ δ−ε+1− 1p ∥f∥Lp(A(δ))

Proof. Let T = [−δ−1/2, δ−1/2]n−1 × [−δ−1, δ−1], L = δ−1/2Zn−1 × δ−1Z and 1 12T

≤ ϕ ≤ 12T be a smooth

function so that ∑`∈L ϕ(x − `) = 1 for all x ∈ Rn. Let eθ be a maximal δ1/2 separated set of directions

and θ a smooth partition of unity on the sphere apadted to the δ1/2 caps around eθ. We can extend itto a partition of unity of Rn away from origin by ignoring the radial coordinate. Take a rotation gθ suchthat gθ(en) = eθ. Define ϕ`(⋅ − `) and ϕ`θ = ϕ` g−1

θ .Now f = ∑θ θf and by the hypothesis of the square function conjecture holding, we obtain

∥F−1f∥Lp(Rn)SFC≲ δ−ε

XXXXXXXXXXXX(∑θ

∣F−1(fθ)∣2)1/2XXXXXXXXXXXXLp(Rn)

≤ δ−ε∑`∈L

XXXXXXXXXXXX(∑θ

∣ϕ`θF−1(fθ)∣2)1/2XXXXXXXXXXXXLp(Rn)

= δ−ε∑`∈L

XXXXXXXXXXXX(∑θ

∣ϕ`θF `θ ∣2)1/2XXXXXXXXXXXXLp(Rn)

Kakeya≲ δ−εδ−

n+14n′ ∑

`∈L(∑θ

∣F `θ ∣2n′

)1/(2n′)

where

F `θ ∶= supx∈T

θ`

∣F−1(θf)∣.

Letting ψθ be a smooth compactly supported function such that ψ = 1 on the support of θf , we canestimate

F `θ = supx∈T `

θ

∣ψθ ∗ θ ∗ f(x)∣ ≤ supy∈Rn

∣θ ∗ f(y)∣ ⋅ supx∈4T 0

θ

∫ ∣ψθ(x − gθ(`))∣dy.

As ψθ is is an L1 normalized Schwartz function at scale of a δ−1/2 ×⋯× δ−1/2 × δ−1 rectangle, we see thatthe second term obeys a bound by (1 + ∣`∣)−n−1 (or any negative power, but this one at least sums up).On the other hand,

supy∈Rn

∣θ ∗ f(y)∣ ≤ ∫ ∣θ(ξ)f(ξ)∣dξ ≲ (δn+12 )1−n−12n ∥fθ∥

L2nn−1 (Rn)

,

so we conclude the bound

∥F−1f∥Lp(Rn) ≲ δ−εδ−(n+1)(n−1)

4n (δn+12 )1−n−12n ∥f∥Lp(Rn) = δ−εδ1−n−12n ∥f∥Lp(Rn).

This concludes the proof.

To pass from the local averaged form to the actual form of Conjecture 6.9 requires some additionalwork. Morally, we want to send δ → 0, but the ε loss in the exponent shows that such an idea isnot possible to implemented as such, and indeed, the restriction estimate fails at the critical exponent.However, the local localized estimate with an epsilon loss at the endpoint exponent implies the globalestimate for all larger exponents. This is the content of the following argument from [19], which isreferred to as ‘epsilon removal lemma’.

Lemma 10.7. Suppose that ∥f∥Lp(A(R−1)) ≲ RεR−1/p∥f∥Lp(BR) for all ε > 0 and all R balls with R > 1.

Here p ∈ [1,2). Then ∥f∥Lq(σ) ≲ ∥f∥Lq(Rn) whenever supp f ⊂ ⋃Ni=1B(xi,R) where ∣xi − xj ∣ ≥ (RN)C fora dimensional constant C (here j ≠ i).

Proof. Let B(xi,R) be the sparse family, let ϕ be a smooth bump adapted to the unit ball and let

ϕi = ϕ(R−1(x − xi)). Then f = ∑i ϕif and f(x) = ∑i(ϕi ∗ (1A(10/R)ϕif))(x) for all x ∈ Sn−1. We denote

Fi = 1A(10/R)ϕif and attempt to estimate the quantity

∥f∥Lp(σ) = ∥∑i

Fi ∗ ϕi∥Lp(σ)

.

The first part of the proof is a decoupling of the terms. We consider a general sequence of functionsFi and prove by interpolation

∥∑i

Fi ∗ ϕi∥Lp(σ)

≲ R1/p (∑i

∥Fi∥pLp(Rn))1/p

.

56

The bound when p = 1 follows by Fubini:

∥∑i

Fi ∗ ϕi∥L1p(σ)

≤∑i∫Rn

∣Fi(y)∣∫ ∣ϕi(x − y)∣dσ(x)dy ≲ RnR1−n∥f∥L1 = R∥f∥L1 .

Here we used that ϕi is supported in a R−1 ball and it is L1 normalized.Next we prove an L2 bound starting from the estimate

∥∑i

Fi ∗ ϕi∥2

L2(σ)= ∥∑

i

F(Fiϕi)∥2

L2(σ)= ∥∑

i

Fiϕi∥2

L2(σ)

= ∫ ∑i

∣(Fiϕi)EE∗(∑j

Fjϕj)∣ ≤ (∑i

∥Fi∥2L2)

1/2 ⎛⎝∑i

∥ϕiEE∗(∑j

Fjϕj)∥2L2

⎞⎠

1/2

≤ (∑i

∥Fi∥2L2)

1/2 ⎛⎜⎝∑i

⎛⎝∑j

∥ϕiEE∗(Fjϕj)∥L2

⎞⎠

2⎞⎟⎠

1/2

.

Consider first the off-diagonal terms i ≠ j. Then the kernel of EE∗ is σ that satisfies ∣σ(ξ)∣ ≲ (1 +∣ξ∣)−(n−1)/2 by Theorem 6.4. Hence

∫B(xi,NR3/2)

∣ϕiEE∗(Fjϕj)∣2 = ∫B(xi,NR3/2)

ϕi(x)∣∫ σ(x − y)ϕj(y)(1B(xj ,NR3/2) + 1B(xj ,NR3/2)c)Fi(y)∣.

For the first term, we use the bound

∣σ(x − y)∣ ≤ 1

∣x − y∣n−12

≲ 1

(RN)C n−12

.

For the second term, we use the Schwartz decay of ϕj to bound

∣σ(x − y)ϕj(y)1B(xj ,NR3/2)c ∣ ≲ (NR)−M .

There estimates show that all off-diagonal terms are bounded by ∥Fi∥2L2(RN)−M so crude summation

gives the desired right hand side.To dealt with the diagonal terms, we need a local restriction estimate in L2. Recall the duality of

restriction and extension operators:

∥E∗(Fiϕi)∥L2(Sn−1) ∶= ∥F(Fiϕi)∥L2(σ) = supg∫ Fi(x) ⋅ ϕi(x)Eg(x)dx ≤ ∥Fi∥L2(Rn) sup

g∥ϕiEg∥L2(Rn).

Here

∥ϕiEg∥2L2(Rn) = ∥F−1(ϕi ∗ gdσ)∥2

L2(Rn) = ∥ϕi ∗ gdσ∥2L2(Rn)

=∬(x,y)∈Sn−1∶∣x−y∣≤10R−1 ∫Rn∣g(x)g(y)ϕi(z − y)ϕi(z − x)∣dzdσ × σ(x, y)

≲ R2n−n∬(x,y)∈Sn−1∶∣x−y∣≤10R−1∣g(x)g(y)∣dzd(σ × σ)(x, y) ≤ Rn∥g∥2

L2(σ)R−(n−1).

Hence TF = E∗(Fϕi) has L2 operator norm bounded by R1/2. But now

∥ϕjEE∗(Fjϕj)∥L2 = ∥TT ∗ Fj∥L2 ≤ ∥T ∥2∥Fj∥ ≲ R∥Fj∥.

Plugging in both the off-diagonal and the diagonal estimates to (??), we conclude the proof of

∥∑i

Fi ∗ ϕi∥Lp(σ)

≲ R1/p (∑i

∥Fi∥pLp(Rn))1/p

with p = 2, and by interpolation the inequality holds for p ∈ (1,2). In particular, it holds for Fi =1A(10/R)ϕif , and hence

∥f∥Lp(σ) = ∥∑i

Fi ∗ ϕi∥Lp(σ)

≲ R1/p (∑i

∥Fi∥pLp(Rn))1/p

≲ Rε (∑i

∥ϕif∥pLp(Rn))1/p

≲ ∥f∥Lp(Rn)

where the penultimate inequality used the local restriction estimate that was the hypothesis of thetheorem.

57

Lemma 10.8. Let K be a compact set with ∣K ∣ > 1. Fix a large integers N and C. Then there exist

∼ N ∣K ∣1/N collections of balls of radius ∼ ∣K ∣CN

so that these collections are sparse in the sense of Lemma10.7 and they cover K.

Proof. Let R0 = 1 and define recursively Rk+1 = ∣K ∣CRCk . Let K0 = ∅, and define recursively

Kk+1 = x ∈K ∖k

⋃i=0

Ki ∶ ∣E ∩B(x,Rk+1)∣ ≤ ∣K ∣(k+1)/N.

Now K = ⋃Ni=1Ki. Indeed, K∖KN = ⋃N−1i=1 Ki is obvious from the definition. We cover each Ki separately,

and as there are N many of them, it suffices to do it with ∣K ∣1/N sparse collections. We proceed by fixingthe radius Ri and covering Rn by boundedly overlapping collection of Ri balls. This we can furtherdivide into cn subcollections, whose balls are Ri separated. Given any such ball B, we inspect B ∩Ki.Now every x ∈ B ∩Ki is in K ∖Ki−1 and satisfies

∣B(x,Ri−1) ∩K ∣ > ∣K ∣i−1N .

Finding a maximal Ri−1/2 separated set of such points xα, we can divide it into cn subcollections thatare Ri−1 separated. Let xα be any of them. Then

∣Ki ∩B∣ ≥∑α

∣B(xα,Ri−1) ∩K ∣ >Ki−1N #xα.

On the other hand, B has radius Ri so we can find cn points y in Ki ∩B so that B(y,Ri) cover it. Then

∣Ki ∩B∣ ≲ ∣K ∣i/N .

In total, #xα ≲ ∣K ∣1/N . As this holds for all Ri−1/2 separated families, we see that Ki ∩ B can be

covered by ∣K ∣1/N balls of radius Ri.

Putting all these balls to different families, we have covered Ki by cn∣K ∣1/N families balls of radius

Ri−1 so that inside one family, all balls are Ri = ∣K ∣CRCi−1 separated. Noting that Rk ≤ ∣K ∣CN

, the proofis complete.

Lemma 10.9. Assume the local restriction estimate as in the hyposthesis of Lemma 10.7 holds. Let Kbe a compact set of measure at least 1. Then for any N large enough and all ε > 0,

∥1K∥Lp(σ) ≲ε ∣K ∣1/p+C/ log(1/ε)

Proof. Just apply the previous lemma and Lemma 10.7 to obtain

∥1K∥Lp(σ) ≲ε N ∣K ∣1/N ∣K ∣CN ε∣K ∣1/p.

The claim follows from setting N = C−1 log(1/ε).

Theorem 10.10. Let p ∈ (2,∞) and assume that

∥F−1f∥Lp(BR) ≲ Rε−1+1/p∥f∥LpRn

holds for all balls BR, all ε and all f ∈ Cc(A(R−1)). Then for all q > p,

∥E∥Lq(σ)→Lq(Rn) ≤ cn.

Proof. Recall that E ∶ L1 → L∞ is trivially bounded. Hence it suffices to prove E ∶ Lp → Lq,∞ bound forany q > p and invoke the Marcinkiewicz interpolation theorem. The assumption of the theorem can berestated in the dual form as

∥f1A(R−1)∥Lp′(Rn) ≲ Rε−1/p′∥f∥Lp′(BR)

holding for all balls. By Lemma 10.9, this implies

(10.1) ∥1K∥Lp′(σ) ≲ε ∣K ∣1/p′+C/ log(1/ε)

for all compact sets K with measure greater than one.Now

∥Ef∥Lq,∞(Rn) = sup0<∣K∣<∞

1

∣K ∣1/q′ ∫∣Ef ∣1K dx ≤ sup

0<∣K∣<∞

4

∣K ∣1/p′∣∫ Ef1K dx∣ .

The last step used the freedom to choose the set K. Indeed, we can restrict it to a part where Ef doesnot change its sign. This part is smaller than the original set, and there is one part for each sign andboth real and complex part, for instance.

58

Next, because Ef is Fourier supported on the unit sphere, we can assume ∣K ∣ > 1. Indeed, for ∣K ∣ ≤ 1we just estimate

∫K

∣Ef ∣dx ≤ ∣K ∣∫ ∣f(y)∣dσ ≤ ∣K ∣∥f∥Lp(σ).

For ∣K ∣ > 1, we can use (10.1) to obtain

∣∫ Ef1K dx∣ = ∣∫ f 1K dσ∣ ≤ ∥f∥Lp(σ)∥1K∥Lp′ ≲ ∥f∥Lp(σ)∣K ∣1/p′+C/ log(1/ε).

Because ∣K ∣ > 1 and q > p so that 1/q′ > 1/p′, we see

∥Ef∥Lq,∞(Rn) ≲ ∥f∥Lp(σ) sup1<∣K∣<∞

∣K ∣1p′− 1q′+ C

log(1/ε) .

For any q > p, the exponent can be made negative by choosing ε small enough. The proof is complete.

10.4. Remark on other conjectures. The previous implications represent a relatively appealing setof ideas relating Kakeya problems, Fourier analysis and restriction. However, the story neither startshere nor ends here. Historically, one can point out the ball multiplier problem as the origin. The problemif F−1(1B(0,1)f) is bounded in Lp was solved in negative by Fefferman for all p ≠ 2.

The ball multiplier can be approximated by smoother Fourier multipliers

(1 − ∣ξ∣)α+where α ∈ (0,1). These symbols do give bounded operators on Lp with p close enough to 2 and therelation of α and the Lp spaces is the content of the Bochner–Riesz conjecture, which is equivalent torestriction (see [19]). Note that Kakeya and restriction problems involve change from Fourier to physicalspace, change from the unit sphere parametrizing directions to Rn parametrizing translations, which isnot the case in the Bochner–Riesz multiplier or Nikodym maximal function problems.

The next, harder, question is about maximal function defined using the Bochner–Riesz means, muchin the spirit of the spherical maximal function. As in the discussion in 7, the maximal estimate can beseen as a Fourier analytic problem in higher dimension (through Sobolev embedding). This gives rise tothe cone multiplier problem and the local smoothing conjecture, which are the most precise formulationsof what one hopes to be true. The local smoothing conjecture claims that

(∫2

1∥eit

√−∆f∥p

Lp(Rn) dt)1/p

≲ ∥f∥Hp,s(Rn), s = (n − 1) (1

2− 1

p) − 1

p

for all p ≥ 2nn−1

, ε > 0, and it is the most quantitative statement here. If true, it would imply maximalBochner-Riesz and consequently everything else. The numerology matches with an interesting guessregarding the wave propagator: the first term in s is how much the solution of the wave equation is moresingular than the initial data (fixed time estimate). Given a function f on Rn, one can regard it as atrace of a Sobolev function F on Rn+1. Such an extension can be at most 1/p derivatives smoother, andthe local smoothing conjecture alleges that this maximal amount of smoothness gain would indeed bethe case when dealing with solutions to the wave equation.

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