arxiv:2109.04555v1 [math.ca] 9 sep 2021

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ANALYSIS OF THE AHLFORS-BEURLING TRANSFORM

LECTURE NOTES FOR THE SUMMER SCHOOL AT THE UNIVERSITY OF

SEVILLE, SEPTEMBER 9-13, 2013

OLIVER DRAGICEVIC

Contents

Prologue 2Acknowledgements 21. Notation and main protagonists 31.1. Weak derivatives and Sobolev spaces 41.2. Riesz transforms 61.3. Ahlfors-Beurling transform 71.4. Muckenhoupt weights 82. Motivation 92.1. Some estimates concerning conformal mappings 92.2. Beyond conformality: quasiconformal maps and the Beltrami equation 142.3. Cauchy transform 152.4. Solving the Beltrami equation 192.5. The Iwaniec conjecture 212.6. Weak quasiregularity vs. quasiregularity: weighted estimates of T 222.7. Estimates of T n on Lp 262.8. Weighted estimates of T n 303. Weighted estimate for T : proof of Theorem 2.29. 313.1. Haar functions and martingale transforms 313.2. Main idea 323.3. The averaging 333.4. Estimates for S on L2(w) 423.5. Sharpness 473.6. Bellman function and Wittwer’s theorem 493.7. Unweighted estimates for T n by means of averaging martingale transforms 554. Unweighted estimates for T n 564.1. Lower estimates 564.2. Upper estimates 604.3. Candidates for ‖T n‖p and some necessary conditions 615. Spectral theory for T 626. Open questions 666.1. Hypergeometric functions and de Branges’ theorem 66References 74

1

http://arxiv.org/abs/2109.04555v1

AHLFORS-BEURLING TRANSFORM 2

Prologue

These notes are based on the ten lectures their author held in September of 2013 atthe University of Seville. The purpose of the course was to introduce basic conceptsabout Ahlfors-Beurling operator T and explain a few of the recent (already publishedor known) results and techniques focused around it. The intended level of the coursewas primarily that for beginner graduate students in analysis. In accordance with that,in this text we tried to give proofs of most of the statements characteristic of T , payingattention to details and leaving out as exercises a few of the less demanding proofs.Facts that hold for much larger classes of operators have for the most part only beencited or again left as exercises. Whereever possible, we gave references for statementsthat were not proven.

This text was by no means designed as a comprehensive survey of all the importantresults regarding the subject. In particular, the Bellman-function-heat-flow methodthat was the basis of the works [70, 67, 34] is not presented here. Furthermore, thestochastic approach that has been utilized by Banuelos et al. is not discussed hereeither. It can be found, for example, in [9] where many other interesting issues related tothe Ahlfors-Beurling operator are treated. A brief recount of the probabilistic approachto the Ahlfors-Beurling operator can also be found in [81, Section 4]. For a veryinteresting paper which combines both techniques see [14]. Finally, there has recentlybeen a revival of interest in finding sharp weighted estimates for general Calderon-Zygmund operators. It culminated in the work by Hytonen [51] where the completesolution to the problem was described for the first time. While many results preceedingand following the resolution of the conjecture are beautiful and interesting for their ownsake, they would exceed the scope of this note and are not included either.

Organization. The bulk of this text is devoted to estimates of T and its powers T n

on the usual Lp as well as Lp(w) with w from the so-called Muckenhoupt class Ap. Wewill also briefly discuss spectral theory for T . In Section 1 we introduce the principalobjects we work with and state some of the most basic facts about them. Section 2provides background for the results on T we consider in this text. In particular, wegive a very brief introduction to the theory of quasiconformal maps and explain theconnection with weighted and “unweighted” estimates for T and its powers. For themost recent and very thorough treatise of quasiconformal theory on C the reader isadvised to consider the monograph by Astala, Iwaniec and Martin [5]. Sections 3 and4 are mainly devoted to proofs of results announced in Section 2. Several differenttechniques will be treated in the process: most prominently (i) the Bellman functiontechnique (though without direct association with heat flows) and (ii) the averagingmethod. Finally, in Section 5 we address some spectral properties of T .

Acknowledgements. I am most grateful to Carlos Perez for the generous opportunityto present one of my areas of research in the form of a 10-hour course. Further gratitudegoes to the students and researchers who attended the course – for their interest – andto the University of Seville for its kind hospitality.

Novo mesto, March 2014


1. Notation and main protagonists

Given two quantities A and B, we adopt the convention whereby A.B means thatthere exists an absolute constant C > 0 such that A 6 CB. If both A.B and B .A,then we write A ∼ B. If λ1, . . . , λn is a set of parameters, C(λ1, . . . , λn) denotes aconstant depending only on λ1, . . . , λn. When A 6 C(λ1, . . . , λn)B, we will often writeA. λ1,...,λnB.

For R > 0 and z ∈ C we will write B(z,R) = w ∈ C ; |z − w| < R. The open unitdisc B(0, 1) will also be denoted by ∆.

For any x = (x1, x2) ∈ R2 and p > 0 denote

|x|p = (|x1|p + |x2|p)1/p and |x|∞ = max|x1|, |x2|.Given ψ ∈ R, introduce the rotation matrix on R2,

Oψ =

[cosψ − sinψ

sinψ cosψ

].

For a function f on R2 we will introduce three basic families of transformations (x ∈ R2):

translations : (τtf)(x) = f(x− t) ; t ∈ R2;

dilations : (δaf)(x) = f(x/a) ; a > 0 ;

rotations : (Uψf)(x) = f(O−ψx) ; ψ ∈ R.

(1.1)

Function Ω : C\0 → C which for any r > 0 and ζ ∈ ∂∆ satisfies

• Ω(rζ) = Ω(ζ) is called homogeneous of degree zero;• Ω(rζ) = Ω(r) is called radial;• Ω(rζ) = Ω(r)ζ is called a (radial) stretch function if, in addition, Ω|(0,∞) issctrictly increasing, continuous and extends continuously to zero [5, p. 28].

By m(E) or |E| we shall denote the Lebesgue measure of a Borel set E ⊂ C, whiledx in the integrals means just dm(x). Sometimes even dx will be omitted. We will usethe standard pairing

〈ϕ,ψ〉 =∫

Rnϕψ, (1.2)

where ϕ,ψ are complex functions on Rn such that the above integral makes sense.For complex C1 functions of two real variables x, y denote

∂ =1

2(∂x − i∂y) and ∂ =

1

2(∂x + i∂y) .

We may instead of ∂f, ∂f write ∂f/∂z, ∂f/∂z or ∂zf, ∂zf or simply fz, fz.Let S = S(Rd) be the Schwartz class on Rd. Recall that its topology is given by a

family of seminorms

ρα,β(f) = supx∈Rd

|xα∂βf(x)| ,

with α, β ranging over all Nd0; here N0 := N ∪ 0. Here(x1, . . . , xd)

(α1,...,αd) := xα11 , . . . , xαdd and ∂(β1,...,βd) := ∂β1x1 . . . ∂

βdxd.

For ϕ ∈ S(Rd) and ξ ∈ Rd define the Fourier transform ϕ of ϕ by

ϕ(ξ) =

∫

Rdϕ(x)e−2πi〈x,ξ〉dx.


We will throughout this note work with the obvious identification of C with R2 viax+ iy ≡ (x, y). Accordingly, a function on C will also be thought of as a function onR2 and vice versa. An important notation to bear in mind will be

p∗ = maxp,

p

p− 1

.

Among the standard tools we shall frequently need is the Minkowski’s integral in-equality, e.g. Stein [76, §A.1], Grafakos [44, Exercise 1.1.6], Duoandikoetxea [35, p.xviii]: if 1 6 p <∞ and F is measurable on X×Y , where (X,µ) and (Y, ν) are σ-finitemeasure spaces, then

(∫

Y

∣∣∣∣∫

XF (x, y) dµ(x)

∣∣∣∣p

dν(y)

)1/p

6

∫

X

(∫

Y|F (x, y)|p dν(y)

)1/p

dµ(x) .

In the context of operators, the asterisk ∗ will typically denote their adjoints. Thusif A ∈ B(Lp) then A∗ ∈ B(Lq), where 1/p + 1/q = 1, and 〈Af, g〉 = 〈f,A∗g〉 for allf ∈ Lp, g ∈ Lq.

Finally, for a linear operator A on a vector space X we will define its null space (orkernel) N(A) and range (or image) R(A) by

N(A) := x ∈ X ; Ax = 0;R(A) := Ax ; x ∈ X .

Recall the following basic facts.

Exercise 1.1. Let X be a Banach space and A ∈ B(X). Suppose that both ‖Af‖ & ‖f‖for all f ∈ X and R(A) is dense in X. Then A is surjective, i.e. R(A) = X.

Exercise 1.2. Let X be a Banach space and A ∈ B(X). Then R(A) is dense if andonly if N(A∗) = 0.Exercise 1.3. Suppose X is a σ-finite measure space and

A ∈⋂

p>1

Lp(X) .

Then for every p > 1 we have log ‖A‖p = ϕ(1/p), where ϕ : (0, 1) → R+ is a convexfunction. Consequently, function p 7→ ‖A‖p is continuous on (1,∞).

Exercise 1.4. The set of all C∞c functions on Rn whose average (i.e. integral over

Rn) is zero is dense in Lp(Rn) for any 1 < p <∞. What about p = 1?

Exercise 1.5. Suppose X is a locally compact Hausdorff space and µ a regular measureon X. Take 1 < p < ∞ and a set Ψ := ψα ; 0 < α < 1 ⊂ L2(µ) ∩ Lp(µ). If Ψ isbounded in Lp and ψα → 0 in L2 as α→ 0, then also ψα → 0 weakly in Lp as α→ 0.

Exercise 1.6. Suppose K is a closed convex subset of a Hilbert space H. Prove thatfor any h ∈ H there exists a unique k ∈ K such that d(h,K) = ‖h − k‖. Is the mapH → K, defined by h 7→ k, continuous, uniformly continuous, a contraction?

1.1. Weak derivatives and Sobolev spaces. There are plenty of sources on thismost fundamental class of function spaces. Let us mention only Gilbarg–Trudinger [42,Chapter 7], Evans [37, Chapter 5], Stein [76, V.§2], Hormander [48, Section 7.9] andGrafakos [45, Section 6.2].


Suppose U ⊂ Rn is an open set. Let u ∈ L1loc(U). Function v ∈ L1

loc(U) is called the

αth weak or distributional derivative of u provided that∫

UuDαϕ = (−1)|α|

∫

Uv ϕ ∀ϕ ∈ C |α|

c (U) .

This notion is well-defined, i.e. if the αth weak derivative of a function exists, it isuniquely determined up to a set of measure zero. Its order is by definition |α|. Wewrite v = Dαu.

We say that u is

• weakly differentiable, if all of its first-order weak derivatives exist;• k-times weakly differentiable, if all of its derivatives of orders 1, . . . , k exist.

The vector space of all k-times weakly differentiable functions on U is denoted byW k(U). Clearly, Ck(U) ⊂W k(U).

Exercise 1.7. Suppose D is a homogeneous first-order partial differential operator onRn, i.e.

D =n∑

j=1

aj∂

∂xj

for some a1, . . . , an ∈ C. If h ∈W 1(U) and g ∈ C∞(U) then h · g ∈W 1(U) and

D(h · g) = Dh · g + h ·Dg .Next result is known as (the holomorphic version of)Weyl’s lemma. See, for example,

[5, Lemma A.6.10], [49, Theorem 4.1.6] or [3, p. 45].

Theorem 1.8. Suppose U ⊂ C is open and g ∈ L1loc(U) ∩W 1(U). If gz = 0 weakly,

then also gz = 0 strongly, in the sense that g coincides p.p. U with a holomorphicfunction.

Exercise 1.9. Suppose p > 1 and h ∈ Lp(C) is holomorphic (entire). Then h ≡ 0.

Remark 1.10. Since Lp ⊂ L1loc, Weyl’s lemma implies that Exercise 1.9 holds even if

we only assume that h ∈ Lp(C) is weakly holomorphic, the latter meaning that hz = 0weakly.

Definition 1.11. For k ∈ N0 and p > 1 the corresponding Sobolev space is defined as

W k,p(U) := u ∈W k(U) ; Dαu ∈ Lp(U) for all |α| 6 k .These are clearly vector spaces. On W k,p(U) we define the norm

‖u‖W k,p(U) :=

∫

U

∑

|α|6k

|Dαu|p

1/p

,

which makes it a Banach space. Because of the equivalence of the euclidean norms wesee that

‖u‖W k,p(U) ∼∑

|α|6k

‖Dαu‖p ,

where the implied constants depend on k, p.The special case p = 2 merits its own notation, namely Hk(U) := W k,2(U). Thus in

particular H1(U) =W 1,2(U) and

‖u‖2H1(U) = ‖u‖22 + ‖∇u‖22 . (1.3)


By H10 (U) we denote the closure of C∞

c (U) in H1(U).

The following is a special case of an essential result known as the Sobolev embed-ding theorem, see Stein [76, V.2.2]. It immediately follows from another fundamentaltheorem, the so-called Gagliardo-Nirenberg-Sobolev inequality, see Evans [37, 5.6.1], forexample.

Theorem 1.12. If 1 < p < n then W 1,p(Rn) ⊂ Lp′(Rn), where

1

p′=

1

p− 1

n.

Let us also introduce local Sobolev spaces, by setting [5, p. 646]

W k,ploc (U) :=

⋂

U ′

W k,p(U ′) ,

where U ⊂ C and U ′ runs over all relatively compact subsets of U .

Definition 1.13. Suppose f ∈ W 1(U) for some open set U ∈ C. If f = u + iv, u, vreal, then the Jacobian (determinant) of f is defined as

Jf :=

∣∣∣∣ux uy

vx vy

∣∣∣∣ .

It is immediate thatJf = |fz|2 − |fz|2 . (1.4)

Exercise 1.14. If ϕ ∈ C∞(C) with ℜϕ or ℑϕ compactly supported, and ε > 0, then∫

ϕ 6=0|ϕ|εJϕ = 0 .

Does the statement hold even for ε > −1?

We shall be only interested in functions f with Jf > 0, i.e. those that preserve theorientation. For a given f of this kind define its complex dilatation or the Beltramicoefficient µ = µf by

µf =fzfz.

Clearly |µf | < 1 on U .

1.2. Riesz transforms. [44, Section 4.2] Let Ω be an integrable function on the unitsphere Sd−1 in Rd with zero average. The functional WΩ, defined by

WΩ(ϕ) = p.v.

∫

Rd

Ω(y/|y|)|y|d ϕ(y) dy for ϕ ∈ S(Rd) ,

is a tempered distribution. We denote by TΩ the singular integral that acts on S(Rd)as a convolution with the distribution WΩ:

TΩϕ = ϕ ∗WΩ for ϕ ∈ S(Rd). (1.5)

References for general singular integrals are, for example, [76, 44, 45, 35, 62]. Wewill be almost exclusively concerned with the case d = 2 only. The (first-order) Riesztransforms R1, R2 on R2 are the operators of the form (1.5) in the case of projections

Ω(ζ1, ζ2) =1

2πζj


for j = 1, 2, respectively. Explicitly,

(Rjϕ)(x) =1

2πp.v.

∫

R2

yj|y|3 ϕ(x− y) dy .

It is well known, see [76, III. §1, eq. (8)], [44, Proposition 4.1.14] or [35, eq. (4.8)], thatRj is a Fourier multiplier with the symbol −iξj/|ξ|: for j = 1, 2 and any ϕ ∈ S(R2) wehave

Rjϕ(ξ) = −i ξj|ξ| ϕ(ξ). (1.6)

As a special case of the central result on the Lp boundedness of the Calderon-Zygmund singular integral operators, all operators Rj are bounded on Lp(R2) for1 < p < ∞. More precisely, for p > 1 every Rj admits an extension from S to abounded operator on (entire) Lp. This extension is denoted by the same symbol - Rj.

We will much need the complex Riesz transform R [5, Section 4.2], sometimes alsocalled the “complex Hilbert transform” and denoted HC [55], which is defined by

R = R2 + iR1,

and its integer powersRk = (R2 + iR1)

k. (1.7)

By [5, Section 4.2], the convolution kernel of Rk is given by

p.v.Ωk(z/|z|)

|z|2 , (1.8)

where for ζ ∈ ∂∆,

Ωk(ζ) =i|k||k|2π

ζ−k. (1.9)

On the Fourier side, by (1.6) we have

Rϕ(ξ) =ξ

|ξ| ϕ(ξ). (1.10)

Exercise 1.15. For ψ ∈ R consider the rotation operator Uψ as in (1.1). Prove that

R = eiψU−ψRUψ.

1.3. Ahlfors-Beurling transform. The central object of this note, the Ahlfors-Beurlingtransform T , is obtained by taking k = 2 in (1.7), i.e.

T = H2C = R2 = (R2 + iR1)

2.

Explicitly, for test functions f we have

Tf(z) = − 1

πp.v.

∫

C

f(ζ)

(ζ − z)2dm(ζ) .

Alternatively, it can be introduced in terms of the Fourier transforms:

T f(ξ) =ξ

ξf(ξ) . (1.11)

From here and the Plancherel identity it immediately follows that ‖Tf‖2 = ‖f‖2 forall f ∈ L2.

For any p > 1 and f ∈ Lp define

Sf := Tf .

Since R1, R2 ∈ B(Lp), the same holds for T and thus also for S.


Exercise 1.16. Take p ∈ (1,∞) and let q be its conjugate exponent. The operator T isinvertible on Lp. By a small abuse of notation (since S, T−1 act on Lp, while T ∗ actson Lq) we have S = T−1 = T ∗, in the sense that

i) TSf = STf = f for all f ∈ Lp(R2);ii) 〈Tf, g〉 = 〈f, Sg〉 for all f ∈ Lp(R2) and g ∈ Lq(R2).

As an immediate consequence it follows that for all p > 1,

‖Tf‖p ∼p ‖f‖p and ‖Sf‖p ∼p ‖f‖p . (1.12)

1.4. Muckenhoupt weights. For any locally integrable function f on C and anybounded set Q ⊂ C with positive Lebesgue measure |Q|, denote by 〈f〉Q the average off on Q,

〈f〉Q =1

|Q|

∫

Qf(x) dx .

If w is a positive locally integrable function on C, introduce

[w]p := supQ⊂R2

〈w〉Q⟨w− 1

p−1

⟩p−1

Q,

where the supremum is taken over all squares in C regardless of their orientation.Let Lp(w) be the space of all functions p-integrable with respect to the weight w,

i.e.,∫|f |pw <∞. When w ≡ 1 we will simply write Lp.

From now on we assume that w belongs to the Muckenhoupt class Ap, defined as

w ∈ Apdef⇐⇒ [w]p <∞ .

This class is systematically discussed in the monographs by Garcıa-Cuerva and Rubiode Francia [38], Stein [75, Chapter V], Duoandikoetxea [35, Chapter 7], Grafakos [45,Chapter 9] and Torchinsky [77, Chapter IX].

Both S and C∞c are dense in Lp(w), for any w ∈ Ap(R

d), see [61, Lemma 2.1].

Exercise 1.17. Is the set f ∈ S(Rd) ; f(0) = 0 dense

• in Lp(w)?• in S?

The following sharp version of the Rubio-de-Francia extrapolation theorem wasproven in [30]:

Theorem 1.18. If an operator T satisfies the estimates

‖T‖B(L2(w)) 6 C[w]τA2

for some C, τ > 0 and all w ∈ A2, then it also satisfies the estimates

‖T‖B(Lp(w)) 6 cpC[w](p∗/p)τAp

for some cp > 0 and all p ∈ (1,∞) and w ∈ Ap.

We conclude this introductory section with a terminological remark. Most of this textis dedicated to “weighted estimates” (of the operator A) or the “unweighted estimatesof A”. Let us explain what we mean by this. In the former case we study A as anoperator on Lp(w) and express its norm estimates in terms of [w]Ap , while in the latterwe study A on Lp (the case w ≡ 1), but typically ask for more precise information onthe dependence of the norms on p.


2. Motivation

An important question in mathematical analysis is how mappings transform shapeand size of objects. To make this question precise let us introduce two notions ofdistortion. Take an open Ω ⊂ C and suppose f : Ω → C is an injective continuousfunction. For any z ∈ Ω define the infinitesimal distortion of f at z as [5, Section 2.4]

Hf (z) := lim supr→0

max|h|=r |f(z + h)− f(z)|min|h|=r |f(z + h)− f(z)| .

We are interested in functions of “bounded distortion”, i.e. such f for which Hf < ∞uniformly on Ω. If f is differentiable at z with Jf (z) > 0 then Hf (z) can be simplyexpressed in terms of the Beltrami coefficient [5, (2.27)]. It turns out that findingfunctions of “bounded distortion” corresponds to finding functions with prescribedBeltrami coefficients. We will show how this problem naturally leads to the appearanceof the Ahlfors-Beurling operator T .

Another concept that underlines much of what is discussed here is area distortion [5,Section 13.1], by which we somewhat loosely mean how to control |f(E)| in terms of|E| for all Borel measurable E with finite Lebesgue measure.

As a historical prelude let us consider a special case in which Hf ≡ 1.

2.1. Some estimates concerning conformal mappings. The main source for thissubsection is [23]. For the sake of the reader’s convenience, we nevertheless chose topresent complete proofs of most of the results in this subsection.

Definition 2.1. Suppose Ω ⊂ C is an open set. We say that a function f : Ω → C

is conformal on Ω if it is holomorphic and injective. Such a function is also calledunivalent. Univalent functions f on ∆ which are normalized by f(0) = 0 and f ′(0) = 1are called schlicht.

Remark 2.2. Often, e.g. Rudin [74, Definition 14.1], conformality of f is definedby requiring that f preserve angles (both in terms of size and orientation). This isequivalent to f being holomorphic and that f ′ 6= 0 everywhere on Ω [74, Theorem14.2], which is in turn equivalent to f being holomorphic and locally injective [74,Theorems 10.30 and 10.33]. This however still does not suffice for global injectivitythat is required in our definition, the example being the exponential function.

Our first result on univalent functions is due to Gronwall. The proofs can be foundin [23, Theorem I.1.1] or [74, Theorem 14.13].

Theorem 2.3 (area theorem). Suppose g is univalent on Ω := ∆\0 and has therethe Laurent series expansion

g(z) =1

z+

∞∑

n=0

bnzn .

Then∞∑

n=1

n|bn|2 6 1 .

Proof. Fix 0 < r < 1 and define ∆r = z ∈ ∆ ; 0 < |z| < r and Dr := C\g(∆r). Bythe open mapping theorem [74, Chapter 10], the set Dr is closed. Take w ∈ Ω\∆r and


write α = g(w). Since g was assumed to be injective, the function z 7→ g(z)−α has nozeros on ∆r, therefore we may define G : ∆r → C by

G(z) :=1

g(z) − α=

z

1 + zh(z),

where

h(z) = b0 − α+

∞∑

n=1

bnzn .

Function G thus has a removable singularity at zero. We still denote its holomorphicextension to ∆r ∪ 0 by G. By the open mapping theorem and since G(0) = 0, thereexists δ > 0 such that B(0, δ) ⊂ G(∆r ∪ 0). Suppose u ∈ C satisfies |u| > 1/δ. Then0 < |1/u| < δ, therefore 1/u ∈ G(∆r), which means that u ∈ g(∆r)−α. Consequently,Dr ⊂ B(a, 1/δ), so Dr is also bounded.

It can now be shown (e.g. by using the Green’s formula) that

area(Dr) =1

2i

∫

∂Dr

z dz . (2.1)

Let us prove that, as sets, ∂Dr and g(rS1) coincide. We have ∂Dr = ∂(C\g(∆r)

)=

∂g(∆r). We know that g(∆r) is an open set, therefore ∂g(∆r) = g(∆r)\g(∆r). Since

g is continuous, g(∆r) ⊂ g(∆r). Function g is injective, therefore it has an inverse,g−1. By a basic theorem [74, Theorem 10.30], g−1 is again holomorphic, thus contin-

uous. Consequently, g−1(g(∆r)

)⊂ g−1(g(∆r)) = ∆r, which means g(∆r) ⊂ g

(∆r

).

Therefore, in a set-theoretical sense, ∂g(∆r) = g(∆r

)\g(∆r) = g(rS1).

Since g is conformal, it preserves the orientation of angles. As for curves, it re-verts the orientation of rS1, in the following sense: if ωr is the curve rS1 orientedcounterclockwise and a ∈ Dr, then Indgωr(a) = −1 (exercise).

To summarize, one admissible parametrization for the integral in (2.1) is z = g(re−it),t ∈ [0, 2π). We emphasize that the negative value of the winding number above amountsto taking a minus in the exponent. With this choice (and with replacing t by 2π − tafter the first parametrization) one eventually gets

area(Dr) = −r2

∫ 2π

0g(reit) g′(reit) eit dt .

Clearly

g′(z) = − 1

z2+

∞∑

n=1

bnzn−1 .

Expand the integrand above into a double infinite sum of trigonometric monomials,integrate term-by-term (we may do that since the double sum is absolutely integrable)and use that

∫ 2π

0eint dt =

2π ; n = 0

0 ; n 6= 0 .

The outcome is

area(Dr) = π

(1

r2−

∞∑

k=1

k|bk|2r2k).


Since the area is nonnegative, we proved

∞∑

k=1

k|bk|2r2k 61

r2

for all 0 < r < 1. Finally send r ր 1.

Theorem 2.4 (Bieberbach). Suppose

f(z) = z +∞∑

n=2

anzn

is a Taylor series representation of a schlicht function. Then |a2| 6 2.

Proof. Define

h(z) =f(z)

z= 1 +

∞∑

n=1

an+1zn .

Since f is by assumption injective, 0 is its only zero, therefore h has no zeros on ∆.Consequently, there exists ϕ ∈ H(∆) such that h = ϕ2 and ϕ(0) = 1, e.g. [74, Theorem13.11.(j)]. Define also

ψ(z) := zϕ(z2) .

Then

ψ2(z) = z2ϕ2(z2)= z2h

(z2)= f

(z2), (2.2)

i.e. ψ : ∆ → C is a holomorphic square root of the function z 7→ f(z2).We have, for w ∈ ∆,

1

ϕ(w)= 1 +

∞∑

k=1

bkwk ,

where

b1 =

(1

ϕ

)′

(0) = −ϕ′(0)

ϕ(0)2= −ϕ′(0) .

From h = ϕ2 it follows that h′ = 2ϕϕ′, so ϕ′(0) = h′(0)/2 = a2/2, therefore b1 = −a2/2.This means that we can define and expand function g as

g(z) :=1

ψ(z)=

1

zϕ(z2)=

1

z− a2

2z +

∞∑

k=2

bkz2k−1 . (2.3)

Let us show that g is injective. If g(z1) = g(z2) for some z1, z2 ∈ ∆, then ψ(z1) =ψ(z2), thus ψ

2(z1) = ψ2(z2), i.e. f(z21) = f(z22), by (2.2). Since f was injective, z21 = z22 ,

which gives options z2 = ±z1. But g is an odd function, hence g(−z1) = −g(z1), whichequals g(z1) only if g(z1) = 0. However g has no zeros on ∆, which forces z2 = z1. Sog is indeeed injective and thus univalent.

Now we may apply Gronwall’s area theorem and conclude from (2.3) that

∣∣∣a22

∣∣∣2+

∞∑

k=2

(2k − 1)|bk|2k 6 1 .

In particular, |a2| 6 2.


Bieberbach conjectured in 1916 that, in fact, under the assumptions of Theorem 2.4,|an| 6 n for all n ∈ N. This was confirmed many years later in a renowned work by deBranges [20]. It is easy to see that all of these estimates are optimal, the extremals beingthe so-called Koebe functions, cf. Exercise 2.6 below, and its rotations. Proposition6.5 below shows that the converse of de Branges’ theorem is false, in the sense thatif ϕ(z) = z +

∑∞2 bnz

n is holomorphic on ∆ and |bn| 6 n, then ϕ is not necessarilyinjective, even if all bn are strictly positive.

For much more information about this problem see Gong [43].

Theorem 2.5 (Koebe 1/4 theorem). Suppose f is schlicht. Then B(0, 1/4) ⊂ f(∆).

Proof. Take c ∈ C such that f(z) 6= c for any z ∈ ∆. Define

g(z) :=cf(z)

c− f(z)=

c2

c− f(z)− c .

We calculate

g′(z) =c2f ′(z)

(c− f(z)

)2

g′′(z) = c2f ′′(z)

(c− f(z)

)+ 2f ′(z)2

(c− f(z)

)3 .

Therefore g(0) = 0, g′(0) = 1 and g′′(0) = a2 + 1/c, where

f(z) = z +∞∑

k=2

akzk .

Function g is clearly injective. From the Bieberbach’s estimate (Theorem 2.4) appliedto both f and g we get |a2| 6 2 and |a2+1/c| 6 2, respectively. This implies |1/c| 6 4,i.e. |c| > 1/4.

We proved that c 6∈ f(∆) implies |c| > 1/4, which is of course equivalent to |c| < 1/4implying c ∈ f(∆).

The estimate from the Koebe theorem is sharp, meaning that 1/4 appearing in theformulation cannot be replaced by a smaller number:

Exercise 2.6. On ∆ define the Koebe function

K(z) :=z

(1− z)2=

∞∑

n=1

nzn .

Then K(∆) = C\(−∞,−1/4].

Corollary 2.7. For every schlicht function f the following estimate is valid:

1

46 d(0, ∂f(∆)

)6 1 .

Proof. The lower estimate follows immediately from the Koebe theorem.The upper estimate will follow once we prove that ∆ 6⊂ f(∆). Suppose the contrary,

i.e. that ∆ ⊂ f(∆). Then

f−1(∆) ⊂ f−1(∆) ⊂ f−1(f(∆)) = ∆ . (2.4)

The Schwartz lemma implies f(z) = λz for some |λ| = 1. Since f is univalent, we havef ′(0) = 1, therefore λ = 1, i.e. f(z) = z. But this contradicts ∆ ⊂ f(∆).


Observe that the assumption ∆ ⊂ f(∆) was only used at the very end, i.e. (2.4)clearly holds already if ∆ ⊂ f(∆), which can actually happen (e.g. with f = id).

Corollary 2.8. If f is univalent on B(z, δ), then

d(f(z), ∂f(B(z, δ))

)>δ

4|f ′(z)| .

Proof. Define g : ∆ → C by

g(w) =f(δw + z)− f(z)

δf ′(z).

Then g is schlicht, therefore d(0, ∂f(∆)

)> 1/4, i.e.

d(f(z), ∂f(δ∆+ z)

)

δ|f ′(z)| >1

4.

Theorem 2.9. Suppose f is univalent on the region D and z ∈ D. Then

1

46

1

|f ′(z)| ·d(f(z), ∂f(D)

)

d(z, ∂D)6 4 .

Proof. Take δ := d(z, ∂D). Then B(z, δ) ⊂ D, therefore f(B(z, δ)

)⊂ f(D) and

consequently

d(f(z), ∂f(D)

)= d(f(z), f(D)c

)> d(f(z), f(B(z, δ))c

)= d(f(z), ∂

(f(B(z, δ))

))

>|f ′(z)|

4d(z, ∂D) .

This confirms the lower estimate. In order to get the upper one, apply the lowerestimate for

z ! f(z)

f ! f−1

D ! f(D) .

Since

1 = (f−1 f)′(z) =(f−1

)′(f(z)) · f ′(z) ,

it follows that

d(z, ∂D) = d(f−1(f(z)), ∂D

)>

1

4· 1

|f ′(z)| · d(f(z), ∂f(D)

).

This finishes the proof.

Theorem 2.10. If f is schlicht then

d(f(z), ∂f(∆)

)>

1

16

(1− |z|2

)

for all z ∈ ∆.

Proof. See Carleson, Gamelin [23, Theorem I.1.7].


2.2. Beyond conformality: quasiconformal maps and the Beltrami equation.

Let us now start discussing the issues of area distortion for nonholomorphic functions.The simplest nonholomorphic function is probably the linear one, i.e. of the followingform. Take a, b ∈ C\0 and define A : C → C by A(u) = au+ bu.

Exercise 2.11. For A as above, the following properties hold:

• If A is understood in a canonical way as a real 2× 2 matrix, then

detA = |a|2 − |b|2 .• When |a| 6= |b|, we have

A−1u =au− bu

|a|2 − |b|2 .

• A(∂∆) is an ellipse whose semi-axes have lengths |a| + |b| and∣∣|a| − |b|

∣∣. In-clination of the longer axis with respect to the positive half of the real axis is(arg a+ arg b)/2.

• If we define ‖A‖ := max|Au| ; |u| = 1 then ‖A‖ = |a|+ |b|.We proceed with the general case.

Definition 2.12. Let U, V ⊂ C be open sets. Take K > 1 and denote

k :=K − 1

K + 1∈ [0, 1) . (2.5)

We say that a map f : U → V is K-quasiconformal, if

• it is a homeomorphism;• belongs to W 1,2

loc (U);• its distributional derivatives satisfy |fz| 6 k|fz| almost everywhere on U .

Remark 2.13. Let us list a few quick observations:

* The last condition is equivalent to f solving the Beltrami equation

∂f = µ · ∂f , (2.6)

for some µ ∈ L∞(C) such that ‖µ‖∞ 6 k.* When µ = 0 (i.e. K = 1) the Beltrami equation becomes the well-known Cauchy-Riemann system. Weyl’s lemma (Theorem 1.8) implies that any 1-quasiconformalmapping is in fact conformal [49, Corollary 4.1.7], [5, p. 27].

Definition 2.14. A map f is quasiconformal if it is K-quasiconformal for some K > 1.The smallest K with this property is called the quasiconformal constant for f and isdenoted by K(f).

It was shown by Gehring and Lehto [39] that quasiconformality of a homeomorphismis equivalent to its being of bounded distortion, in the sense of Hf < ∞ holdinguniformly.

Suppose now f is K-quasiconformal and also f ∈ C1(U). Then Df(z0) exists in theordinary sense for all z0 ∈ U and we have

[Df(z0)

]u =

∂f

∂z(z0)u+

∂f

∂z(z0) u .

Write a = fz(z0) and b = fz(z0). Then the K-quasiconformality gives |b| 6 k|a| < |a|.Thus, by (1.4) and Exercise 2.11, Df(z0) is nondegenerate and Jf = |a|2 − |b|2 > 0,


meaning that f preserves the orientation. Moreover, the eccentricity of Df(z0) is equalto

|a|+ |b||a| − |b| =

1 + |b|/|a|1− |b|/|a| 6

1 + k

1− k= K .

Again by Exercise 2.11, ‖Df‖ = |fz|+ |fz|, therefore the above inequality could alter-natively be expressed as

‖Df‖2Jf

6 K .

2.2.1. The first objective. We want to solve the Beltrami equation. A very prominentrole in this expedition will be played by the Ahlfors-Beurling operator T .

2.3. Cauchy transform. We will often use the simple fact that

1

ζ∈ Lrloc(C) for all r < 2 . (2.7)

For h ∈ Cc(C) define its (planar) Cauchy transform by

(Ch)(z) := − 1

π

∫

C

h(ζ)

ζ − zdA(ζ) . (2.8)

By (2.7) the above integral converges absolutely for h ∈ Cc(C). Indeed,∫

C

∣∣∣∣h(ζ)

ζ − z

∣∣∣∣ dA(ζ) 6 ‖h‖∞∫

K

dA(η)

|η| <∞ .

Here K := supph − z is a compact set. Observe that the integral also convergesabsolutely under a weaker assumption of h ∈ Lp having compact support.

Exercise 2.15. If h ∈ C1c (C) then Ch ∈ C1(C) and

∂(Ch) = C(∂h)

∂(Ch) = C(∂h) .

In brief, on C1c (C) the operator C commutes with both ∂ and ∂.

This statement has an immediate corollary:

Corollary 2.16. We have C : C∞c (C) → C∞(C) and CD = DC on C∞

c (C) for anylinear partial differential operator D on C∞(C) with constant coefficients.

Recall that T was the Ahlfors-Beurling operator, defined in Section 1.3.

Lemma 2.17. Suppose h ∈ C1c (C). Then Ch ∈ C1 and

(Ch)z = h

(Ch)z = Th .

Proof. Write ζ = α+ iβ and z = x+ iy. Take h ∈ C1c (C). In view of Exercise 2.15 it

suffices to evaluate integrals∫∫

C

∂h

∂ζ(ζ)

1

ζ − zdα dβ and

∫∫

C

∂h

∂ζ(ζ)

1

ζ − zdα dβ .


z

Aε(z)

1/ε

εζ

ν

Figure 1.

We will first do it with ∂α and ∂β in place of ∂ζ = (1/2)(∂α + i∂β) and ∂ζ = (1/2)(∂α−i∂β). We start with

∫∫

C

∂h

∂α(ζ)

1

ζ − zdα dβ = lim

ε→0

∫∫

Aε(z)

∂h

∂α(ζ)

1

ζ − zdα dβ

︸︷︷︸=:Iε(z)

,

where Aε(z) := z + A(ε, 1/ε) = ζ ∈ C ; ε < |ζ − z| < 1/ε. We integrate by parts[37, p.712, Theorem 2], using that for small ε > 0 we have h

∣∣S(z,1/ε)

≡ 0, since h is of

compact support. For ζ ∈ ∂Aε(z) = S(z, ε) ∪ S(z, 1/ε) let ν = ν(ζ) = να + iνβ denotethe outer unit normal vector. We get

Iε(z) = −∫∫

Aε(z)h(ζ)

∂

∂α

( 1

ζ − z

)dαdβ +

∫

S(z,ε)h(ζ) · 1

ζ − z· να dS .

An analogous formula holds for ∂β. Thus for ∂ζ = (1/2)(∂α + i∂β) we get, by taking

into account that ∂ζ[1/(ζ − z)

]= 0,

∫∫

Aε(z)

∂h

∂ζ(ζ)

1

ζ − zdA(ζ) =

∫

S(z,ε)h(ζ) · 1

ζ − z· ν

α + iνβ

2dS .

We see (e.g. by Figure 1) that on S(z, ε) the outer unit normal ν = να + iνβ can beevaluated as

ν = −ζ − z

ε,

therefore

− 1

π

∫∫

Aε(z)

∂h

∂ζ(ζ)

1

ζ − zdA(ζ) =

1

2πε

∫

S(z,ε)hdS = 〈h〉S(z,ε) .


Thus the continuity of h gives

limεց0

− 1

π

∫∫

Aε(z)

∂h

∂ζ(ζ)

1

ζ − zdA(ζ) = h(z) ,

i.e., for h ∈ C1c (C) we indeed have

C(∂zh) = h .

Now let us address C(∂zh). Similarly as before we get∫∫

Aε(z)

∂h

∂ζ(ζ)

1

ζ − zdA(ζ)

= −∫∫

Aε(z)h(ζ)

∂

∂ζ

( 1

ζ − z

)dA(ζ) +

∫

S(z,ε)h(ζ) · 1

ζ − z· ν

α − iνβ

2dS .

Obviously

∂

∂ζ

( 1

ζ − z

)= − 1

(ζ − z)2and να − iνβ = −ζ − z

ε,

therefore

− 1

π

∫∫

Aε(z)

∂h

∂ζ(ζ)

1

ζ − zdA(ζ)

= − 1

π

∫∫

Aε(z)

h(ζ)

(ζ − z)2dA(ζ) +

1

2πε

∫

S(z,ε)h(ζ)

ζ − z

ζ − zdS(ζ)

︸︷︷︸=:Hε(z)

.

We have

Hε(z) =1

2πε

∫

S(0,ε)h(η + z)

η

ηdS(η) .

Since ∫

S(0,ε)

η

ηdS(η) =

∫ 2π

0e−2it dt = 0 ,

by introducing g(η) = h(η + z) we obtain

Hε(z) =1

2πε

∫

S(0,ε)g(η)

η

ηdS(η) =

1

2πε

∫

S(0,ε)

g(η) − g(0)

η· η dS(η) .

The fraction in the integrand above is uniformly bounded in a neighbourhood of 0,since h ∈ C1. Therefore

|Hε(z)| . 〈|η|〉S(0,ε) = ε .

It follows that

− 1

π

∫∫

C

∂h

∂ζ(ζ)

1

ζ − zdA(ζ) = − 1

πlimε→0

∫∫

Aε(z)

h(ζ)

(ζ − z)2dA(ζ) = Th(z) .

Corollary 2.16 now implies its own analogue for T :

Corollary 2.18. We have T : C∞c (C) → C∞(C) and TD = DT on C∞

c (C) for anylinear partial differential operator D on C∞(C) with constant coefficients.

What follows is arguably the most important property of T from the point of viewof complex analysis.


Exercise 2.19. Take p > 1, For h ∈W 1,p(C) we have

T :∂h

∂z7−→ ∂h

∂z.

Hint: first consider smooth h and then approximate.

Exercise 2.20. The set ϕz ;ϕ ∈ C∞c (C) is dense in L2(C).

In view of these two exercises the isometry of T on L2 can be reformulated as Exercise1.14, or vice versa.

2.3.1. Cauchy transform on Lp. We want to have the Cauchy operator available on Lp,p > 2, without the extra assumption of compact support. But we may have integrabilityproblems at infinity if in (2.8) we only assume that h ∈ Lp. Therefore, following Ahlfors[2, Chapter V.A], as a replacement for C we introduce, for p > 2 and h ∈ Lp(C),

(Ph)(z) := − 1

π

∫

C

h(ζ)

(1

ζ − z− 1

ζ

)dA(ζ) . (2.9)

Exercise 2.21. The integral above converges absolutely when h ∈ Lp, i.e. P is welldefined.

Observe that for compactly supported h ∈ Lp we have

Ph(z) = Ch(z)− Ch(0) . (2.10)

Lemma 2.22. If h ∈ Lp then Ph is a continuous function. In addition, it satisfies theuniform Holder condition with exponent 1− 2/p, i.e.

|Ph(z)− Ph(w)| . p ‖h‖p|z − w|1−2/p ∀ z, w ∈ C .

Proof. By the Holder’s inequality we have, for z 6= 0,

|Ph(z)| 6 |z|π

‖h‖p∣∣∣∣∣∣∣∣

1

ζ(ζ − z)

∣∣∣∣∣∣∣∣q

.

By the change of variable ζ = zη we get, for z 6= 0,∫

C

dA(ζ)

|ζ(ζ − z)|q =

∫

C

Jζ(η)

|zη(zη − z)|q dA(η) =∫

C

|z|2|zη(zη − z)|q dA(η)

= |z|2−2q

∫

C

dA(η)

|η(η − 1)|q︸︷︷︸cp

,

therefore

|Ph(z)| 6 |z|π

‖h‖p(cp|z|2−2q

)1/q=

1

π

∣∣∣∣∣∣∣∣

1

η(η − 1)

∣∣∣∣∣∣∣∣Lq(dη)︸︷︷︸

Cp

‖h‖p|z|1−2/p. (2.11)

Fix arbitrary w ∈ C. The function h(z) := h(z + w) of course again belongs toLp(C). We have

Ph(z − w) = − 1

π

∫

C

h(ζ + w)

(1

ζ +w − z− 1

ζ

)dA(ζ)

= − 1

π

∫

C

h(ζ)

(1

ζ − z− 1

ζ − w

)dA(ζ)

= Ph(z) − Ph(w) .


Together with (2.11) this gives

|Ph(z)− Ph(w)| =∣∣∣Ph(z − w)

∣∣∣ 6 Cp‖h‖p|z −w|1−2/p ,

as desired.

This lemma has an important corollary, which can be perceived as a generalizationof Lemma 2.17:

Exercise 2.23. For every h ∈ Lp(C), p > 2, we have

(Ph)z = h

(Ph)z = Th

in the distributional sense.(Recall that this means that

∫

C

Ph · φz = −∫

C

φ · h∫

C

Ph · φz = −∫

C

φ · Th

for all φ ∈ C∞c (C).)

2.4. Solving the Beltrami equation. Exercise 2.19 suggests that T might have animportant role in solving the Beltrami equation (2.6). Indeed this is the case, as weexplain in this section, following Ahlfors [2, Chapter V.B]. We will treat in detail thecase of compactly supported µ.

Theorem 2.24. Suppose µ ∈ L∞(C) is of compact support and k := ‖µ‖∞ < 1. Letp > 2 be such that k‖T‖p < 1. Then there exists a unique f ∈ C(C) ∩W 1(C) thatsatisfies the conditions

∂f = µ · ∂f ;f(0) = 0 ;

fz − 1 ∈ Lp(C) .

(2.12)

Remark 2.25. Since T is an isometry on L2, its norm equals 1, of course. Therefore,by the continuity of Lp norms (Exercise 1.3), the condition k‖T‖p < 1 is indeed fulfilledfor p sufficiently close to 2.

Since µ is compactly supported, the Beltrami equation and the Weyl lemma implythat f is holomorphic outside a large disc.

Proof. First let us prove uniquenness.If f satisfies (2.12), then fz = µ · fz = µ(fz − 1) + µ ∈ Lp(C), so that P (fz) is

well defined and hence so is F := f − P (fz). From Exercise 2.23 we conclude thatFz = fz −

(P (fz)

)z= 0 distributionally. By Lemma 2.22, F is continuous, therefore

also F ∈ L1loc(C). Owing to the Weyl’s lemma (Theorem 1.8), under such conditions F

must be holomorphic. We have, in a weak sense,

Fz − 1 = (fz − 1)−(P (fz)

)z= (fz − 1)− T (fz) ∈ Lp(C) . (2.13)

Certainly, for smooth functions their weak derivative coincides with the strong one.Now Exercise 1.9 implies that Fz − 1 ≡ 0. Consequently (2.13) yields

fz = T (fz) + 1 = T (µfz) + 1 .


Suppose now g is another solution of (2.12). Then gz = T (µgz) + 1 as before andthus

fz − gz = T(µ(fz − gz)

),

therefore ‖fz − gz‖p 6 k‖T‖p‖fz − gz‖p. But k‖T‖p < 1, hence fz − gz = 0 p.p. C.Beltrami’s equation implies fz − gz = 0 p.p. C. By using the Weyl’s lemma again,together with the fact that ϕz = ϕz, we conclude that f − g and f − g are bothholomorphic functions. This is only possible if f − g ≡ c for some c ∈ C, whereuponthe normalization gives f ≡ g.

Let us now address the existence. First we will find a natural candidate for thesolution and then prove that this candidate is indeed the right one.

Suppose f is a solution to (2.12). Note that f 6∈W 1,p(C), because fz ∈ 1+Lp. Hencefor g := f − z we have gz = fz−1 ∈ Lp and gz = fz ∈ Lp, by the Beltrami equation. Inother words, g ∈W 1,p(C). It follows from Exercise 2.19 that gz = µ(gz+1) = µ(Tgz)+µand so µ = (I − µT )gz. The assumed relation between µ and p means that µT isinvertible on Lp, by the Neumann series, therefore gz = (I − µT )−1µ. Exercise 2.23offered a solution to the (distributional) ∂-equation with data from Lp, namely bymeans of the operator P . That is, g = P

((I − µT )−1µ

). Therefore we have our

candidate for a solution of (2.12):

f = z + P((I − µT )−1µ

). (2.14)

It remains to verify that this function indeed has all the required properties.

• We saw before that (I −µT )−1 is invertible on Lp, therefore (I −µT )−1µ ∈ Lp,for µ ∈ L∞

c ⊂ Lp. By Lemma 2.22, P((I−µT )−1µ

)is a well-defined continuous

function on C, therefore so is our f .• Since Ph(0) = 0 for any h ∈ Lp, we get f(0) = 0.• Exercise 2.23 gives, in a distributional sense,

fz = (I − µT )−1µ

fz = 1 + T((I − µT )−1µ

).

(2.15)

Since (I − µT )−1µ ∈ Lp and T ∈ B(Lp), we conclude that fz, fz − 1 ∈ Lp.• Finally,

µfz = µ+ µT((I − µT )−1µ

)= (I − µT )(I − µT )−1µ+ µT

((I − µT )−1µ

)

=[(I − µT ) + µT

]((I − µT )−1µ

)= (I − µT )−1µ

= fz .

Thus all the properties of (2.12) hold, as claimed.

Function f defined in (2.14) is called the normal solution to the Beltrami equation(2.6). It is known [2, Theorem V.B.2] that normal solutions are not only continuous,but in fact homeomorphisms.

Remark 2.26. The formula (2.14) is essentially due to Bojarski [13].From (2.14) and (2.15) we see that f = z + P (fz).As our final – yet very imporant – remark in this section let us mention that in

Theorem 2.17 one can remove that assumption about µ having compact support [2,Theorem V.B.3].


2.5. The Iwaniec conjecture. We also see from (2.15) that f − z ∈ W 1,p, providedthat ‖µT‖p < 1. Hence the Sobolev integrability of quasiconformal maps self-improves,

meaning that the a priori assumption f ∈ W 1,2loc (C) when coupled with the Beltrami

equation improves to f ∈ W 1,ploc (C) for some p > 2. This fact is known as Bojarski’s

theorem, see [12] or [5, Theorem 5.4.2]. For a long time a major question in the areawas to determine the best (i.e. largest) p that can be attained if one knows ‖µ‖∞ =: k.The question was solved by K. Astala [4] who showed that the answer is 1+1/k−. Oneimmediately sees that the condition ‖µT‖p < 1 would be satisfied for all p ∈ [2, 1+1/k)if we knew that ‖T‖p = p−1 for p > 2. This is however a notoriously difficult problem,open at least since 1982 when it was formulated by T. Iwaniec [54]:

Conjecture 2.27 (Iwaniec). For every p ∈ (1,∞),

‖T‖p = p∗ − 1 . (2.16)

It is not very difficult to prove that ‖T‖p > p∗ − 1, see [58] or [8]. As to the upperestimates, following a series of results by different authors, the best to-date result,due to Banuelos and Janakiraman, is

√2p(p − 1) for p > 2, see [10] and the references

there. In the asymptotical sense, the strongest estimate known today is due to Borichev,Janakiraman, Volberg [14] who obtained ‖T‖p 6 1·3922(p∗ − 1) for p∗ → ∞.

Apart from immediately implying the optimal Sobolev integrability of quasiconfor-mal maps, the confirmation of the Iwaniec conjecture would have many other deepconsequences. See e.g. [8] for a related discussion. In particular, the validity of (2.16)is closely related to the calculus of variations and the Morrey’s problem [8, 7].

Another implication that (2.16) would bring regards the issue we are well familiarwith by now – distortion of area under the action of quasiconformal mappings.

2.5.1. Area distortion by quasiconformal mappings. The proof of Theorem 2.24 givesthe following important consequence [23, Theorem I.7.2].

Theorem 2.28. Take R > 0 and µ ∈ L∞ so that suppµ ⊂ R∆. Write k := ‖µ‖∞ < 1.Suppose f is the normal solution to the Beltrami equation (2.12). If p > 2 is such thatk‖T‖p < 1, then for every |E| ⊂ R∆ we have

|f(E)| .R,k,p |E|1−2/p .

Proof. We start by recalling (1.4) and calculating

|f(E)| =∫

f(E)1 =

∫

EJf 6

∫

E|fz|2 . (2.17)

By using (2.15) and writing ϕ := T((I − µT )−1µ

), so that fz = 1 + ϕ, we obtain

∫

E|fz|2 = ‖fz‖2L2(E)

. ‖1‖2L2(E) + ‖ϕ‖2L2(E)

= |E|+∣∣∣∣ϕ2

∣∣∣∣L1(E)

.

(2.18)

The last term can be estimated by applying the Holder’s inequality to the pair p/2,(p/2)′ = p/(p− 2) to get

∣∣∣∣ϕ2∣∣∣∣L1(E)

6∣∣∣∣ϕ2

∣∣∣∣Lp/2(E)

‖1‖L(p/2)′ (E) = ||ϕ||2Lp(E) |E|1−2/p .


But

||ϕ||Lp(E) 6 ||ϕ||p 6 ‖T‖p ·1

1− k‖T‖p· ‖µ‖p 6

k‖T‖p1− k‖T‖p

· (πR2)1/p <∞ ,

therefore ∣∣∣∣ϕ2∣∣∣∣L1(E)

.R,k,p |E|1−2/p . (2.19)

On the other hand, |E|/R 6 1 implies |E|/R 6 (|E|/R)1−2/p, i.e.

|E| 6 R2/p|E|1−2/p . (2.20)

Finally merging (2.17), (2.18), (2.19) and (2.20) finishes the proof.

It was conjectured by Gehring and Reich [40], see also [8], that the best integrabilityexponent in the theorem above should be 1/K, where K is inferred from k through

(2.5). This exponent is attained in the case of the radial stretch function z 7→ z|z|1/K−1.The Gehring-Reich conjecture has attracted a lot of attention as the central problemin the area of planar quasiconformal mappings. It was eventually solved by Astala [4].Eremenko and Hamilton [36] gave a simplified proof of Astala’s theorem.

One sees that the extremal case 1− 2/p = 1/K holds precisely when

1

k=K + 1

K − 1=

1 + 1/K

1− 1/K=

2− 2/p

2/p= p− 1 .

In view of Theorem 2.28 this small observation shows that one would immediately re-cover Astala’s theorem if – again – one knew (2.16). At this point one may guess thatthe Gehring-Reich conjecture is essentially equivalent to the optimal Sobolev integra-bility of solutions to the Beltrami equation, see [5, Chapter 13].

2.6. Weak quasiregularity vs. quasiregularity: weighted estimates of T . InSection 2.5 we mentioned the question of the optimal Sobolev integrability that an apriori W 1,2

loc function is pushed to by satisfying the Beltrami equation. One can pose a“dual” version of this question:

What is the smallest p− ∈ (1, 2) such that f ∈W 1,p−loc which satisfies the Beltrami equa-

tion with ‖µ‖ = k, automatically belongs to W 1,2loc (and therefore, by Astala’s theorem,

to W1,1+1/k−loc )?

The aim of this section is to answer this question. We will show that it reduces toproving sharp estimates of the Ahlfors-Beurling operator on weighted Lp spaces withweights from the Muckenhoupt class, as specified in the following fundamental theorem.It was first proven by Petermichl and Volberg [70]. In this note we will give a proof asin [33]. We will devote Section 3 to this purpose.

Theorem 2.29 ([70, 33]). For any p > 1 and any w ∈ Ap we have

‖T‖B(Lp(w)) . p [w]p∗/pAp

.

Recently there have been remarkable efforts by many mathematicians to extend thisresult to all Calderon-Zygmund operators. The final step towards the proof of thatextension was done by T. Hytonen [51].

Assuming Theorem 2.29 until the end of this section, let us proceed to its conse-quences. It will be directly used in the proof of Theorem 2.35 which will in turn implythe answer to the question posed above.


Denote, for k ∈ (0, 1),

Ik := (1 + k, 1 + 1/k) .

Lemma 2.30. Let µ ∈ L∞(C) be such that k := ‖µ‖∞ < 1. Suppose that for all p ∈ Ikand g ∈ Lp we have ‖(I − µT )g‖p & p,k ‖g‖p. Then I − µT is invertible on Lp for allp ∈ Ik.

Proof. By Exercises 1.1, 1.2 it suffices to show that (I − µT )∗ is injective. Recall alsoExercise 1.16. Then we have on Lq, compare with [5, p. 371],

(I − µT )∗ = I − Sµ = S(I − µS)T .

Clearly (I − µS)h = (I − µT )h. Consequently, we get from (1.12) and the assumptionon I − µT

‖(I − µT )∗g‖q & ‖(I − µS)Tg‖q = ‖(I − µT )Tg‖q & ‖Tg‖q & ‖g‖q .

The next result is stated in [81, p. 4].

Exercise 2.31. Let µ ∈ L∞(C) be such that k := ‖µ‖∞ ∈ (0, 1). Take p ∈ Ik. If theIwaniec conjecture (2.16) holds, then I − µT is invertible on Lp and

∣∣∣∣(I − µT )−1∣∣∣∣p6

1/k

d(p, Ick).

Actually, the invertibility of I−µT is in fact true, regardless of the validity of (2.16).More precisely, we have this result [6, Theorem 3], [81, Theorem 1.4]:

Theorem 2.32. Suppose ‖µ‖∞ =: k < 1. Take p ∈ Ik. Then I − Tµ and I − µT areinvertible in Lp.

Before proceeding to the proof of this theorem let us quote another beautiful resultof Astala, Iwaniec and Saksman [6, Theorem 12] which is crucial for establishing con-nection between quasiconformal maps and the theory of Ap weights. The formulationpresented here is not the most general one, but instead one which just suffices for ourpurposes. For a more general statement see [5, Theorem 13.4.2 and estimate (13.58)].

Theorem 2.33. Choose K > 1 and let k be as in (2.5). Let p ∈ [2, 1 + 1/k). Suppose

f : C → C is a K-quasiconformal map. Define w := J1−p/2f . Then w ∈ A2 and

[w]A2 .K1

1 + 1/k − p.

Proof of Theorem 2.32. We have I − Tµ = T (I − µT )S and since T, S are bothinvertible in any Lp, cf. Exercise 1.16, it is enough to verify the statement for I − µT .Lemma 2.30 reduces the theorem to proving

‖(I − µT )g‖p & p,k ‖g‖p for all p ∈ Ik and g ∈ Lp . (2.21)

We follow [6, proof of Lemma 14]. By Exercise 1.4 we may take g ∈ C∞c (C) with zero

average, i.e.∫Cg = 0.

Define ϕ := Cg, with C being the planar Cauchy transform. Lemma 2.17 givesϕz = g. Let h = (I − µT )g. Therefore our goal is to prove

‖ϕz‖p . k,p ‖h‖p . (2.22)


We know that there is a K−qasiconformal homeomorphism f such that fz − µfz = 0,cf. Remark 2.26. Set u = ϕ f−1, that is, ϕ = u f . The chain rule reads

ϕz = (uz f) · ∂zf + (uz f) · ∂z fϕz = (uz f) · ∂zf + (uz f) · ∂z f .

(2.23)

We will start with

|ϕz|p . p |uz f |p|∂zf |p + |uz f |p|∂z f |p . (2.24)

From (2.23) it follows, through Exercise 2.19, that

h = ϕz − µϕz = 0 + (uz f) ·(∂z f − µ · ∂z f

).

Since ∂z f = ∂zf and ∂z f = ∂zf = µ · ∂zf = µ · ∂zf , we obtain

h =(1− |µ|2

)· (uz f) · ∂zf ,

which together with Jf = |fz|2 − |fz|2 6 |fz|2 implies

|uz f |p · Jp/2−1f · Jf 6 |uz f |p · |fz|p =

|h|p(1− |µ|2

)p . k |h|p .

Integrate, change the variable into z′ = f(z) and use Jf f−1 = (Jf−1)−1 to get∫

|uz|p · J1−p/2f−1 . k

∫|h|p . (2.25)

On the other hand, the definition of f gives |fz| 6 k|fz|, therefore Jf > (1/k2−1)|fz|2and so

|uz f |p|∂zf |p 6k2

1− k2|uz f |p · Jp/2−1

f · Jf ,therefore ∫

|uz f |p|∂zf |p . k

∫|uz|p · J1−p/2

f−1 =

∫|Tuz|p · J1−p/2

f−1 .

Write

w = J1−p/2f−1 .

Let us summarize our findings so far:

•∫

|uz f |p|∂zf |p 6k2

1− k2

∫|Tuz|pw

•∫

|uz|pw 6

∫|uz f |p · |fz|p 6

1

(1− k2)p

∫|h|p .

In fact, the top chain of inequalities can be (up to a multiplicative constant) continuedby the bottom one:

It is a well-known fact that every Calderon-Zygmund operator (such as T ) is boundedon Lr(w) for any Ar weight w, e.g. [35, Theorem 7.11]. Since f is K-quasiconformal,so is f−1 [5, Theorem 3.7.7]. Theorem 2.33 implies that w ∈ A2.

If p > 2 then also w ∈ Ap [35, Proposition 7.2]. Now suppose 1 < p < 2. Thenw ∈ Ap if and only if w1−q ∈ Aq, where q = p/(p− 1) ∈ [2, 1 + 1/k). Now

w1−q = J(1−p/2)(1−q)f−1 = J

1−q/2f−1 ,

which by Theorem 2.33 belongs to A2 and thus also Aq.This shows that, after integration, both of the summands in (2.24) are controlled by∫|h|p, which finishes the proof.


Remark 2.34. The proof above shows that if∫|Tuz|pw . F

([w]A2

) ∫|uz |pw

for some increasing function F , then∣∣∣∣(I − µT )−1∣∣∣∣p→p

.K 1 + F([w]A2

),

where, as before, w = J1−p/2f−1 and f is K-quasiconformal.

We are nearing the main objective of this section. The next result can be found in[5, Theorem 14.4.1].

Theorem 2.35. Suppose that µ ∈ L∞(C) with ‖µ‖∞ =: k < 1. Let q := 1 + k. ThenI − µT and I − Tµ are injective on L1+k(C).

Proof. By Exercise 1.2 it suffices to prove that I − µT and I − Tµ have dense rangeson L1+1/k(C). Now Exercise 1.16 and the identity I − Tµ = T (I − µT )S imply that itis enough to prove this for I − µT . Since R(I − µT ) is a convex set, it suffices to provethat it is weakly dense in L1+1/k(C), see Rudin [73, Theorem 3.12].

Choose h ∈ C∞c (C) and 0 < ε < 1. Denote by p the conjugate exponent of q, i.e.

p = 1 + 1/k. Since p ∈ I(1−ε)k, Theorem 2.32 says that

φε :=[I − (1− ε)µT

]−1h ∈ Lp(C) .

Let us write this formula as

(I − µT )φε = h− µT (εφε) .

Now invoking again Exercise 1.16 ii) reduces our problem to proving that εφε → 0weakly in Lq as ε→ 0.

At this point we use our weighted estimate for T – Theorem 2.42. For together withRemark 2.34 and Theorem 2.33 it gives the estimate

∣∣∣∣∣∣[I − (1− ε)µT

]−1∣∣∣∣∣∣B(Lp(C))

.K [w]A2 61

ε.

This immediately implies ‖εφε‖p .K‖h‖p. On the other hand, by applying in L2 theNeumann series to the definition of φε we see that

‖φε‖2 61

1− (1− ε)k‖h‖2 .

Hence εφε → 0 in L2 as ε→ 0. Now Exercise 1.5 implies that εφε → 0 weakly in Lp asε→ 0, which had to be proven.

The next result summarizes our efforts in this section. It appeared in [6, Lemma 19and Proposition 19], [5, Corollary 14.4.3], [81, Theorem 1.5].

Theorem 2.36. Suppose µ ∈ L∞(C) and ‖µ‖∞ =: k < 1. If a solution of Fz−µFz = 0

lies in W 1,1+kloc (C), then it automatically also lies in W 1,2

loc (C). Therefore it even lies in

W1,1+1/k−loc (C).

Proof. The case k = 0 follows from Weyl’s lemma. Now assume k ∈ (0, 1). Write

q = 1 + k. Suppose F ∈ W 1,qloc (C) satisfies Fz − µFz = 0. Take φ ∈ C∞

c (C) and define

G = φF . Then G ∈W 1,q(C). Let ω ∈ C∞c (C) be such that ω

∣∣supp φ

≡ 1. Then clearly

Gz − µGz = (φz − µφz)F = (φz − µφz)(ωF ) =: F .


Since F ∈W 1,qloc (C) it follows that ωF ∈W 1,q(C), therefore, by the Sobolev embedding

(Theorem 1.12), ωF ∈ L2K , where K > 1 is as in (2.5). Since F contains φz − µφz as

a factor, it is compactly supported, therefore F ∈ L2 ∩ L2K .Let us now try to solve the equation

Hz − µHz = F (2.26)

but for H whose derivatives belong to L2(C) instead of Lq(C). By Exercise 2.19 this isthe same as

(I − µT )Hz = F .

Since F ∈ Lr with r 6 2K, 2K > 2 and I − µT is invertible on L2+ε for small ε > 0(Neumann series), this is the same as

Hz = (I − µT )−1F ∈ L2+ε .

By Exercise 2.23 the solution of this ∂-problem is given by the Cauchy transform of

(I−µT )−1F . But obviously also (I−µT )−1F ∈ L2, therefore Hz,Hz = THz ∈ L2. Wemay assume that µ has a compact support, otherwise we may consider λ := µχsuppφ

or λ = ωµ instead. Therefore Hz = F + µHz has compact support and hence belongs

to Lq, since F does. Then also Hz = THz ∈ Lq.

So we got two solutions to (2.26) on Lq, namelyG andH. ThusGz−µGz = Hz−µHz.Since Gz = TGz and Hz = THz we obtained (I − µT )(Gz − Hz) = 0. We sawthat Gz ,Hz ∈ Lq. But Theorem 2.35 says that I − µT is injective on Lq, thereforeGz = Hz ∈ L2 and consequently also Gz = Hz ∈ L2. Thus G−H is a constant. SinceH is of the form Ph for some h ∈ L2+ε, it is continuous, by Lemma 2.22. ThereforeH ∈ L2

loc and hence also G ∈ L2loc. Since G has compact support, it follows that G ∈ L2.

Thus for G = φF we got G,Gz , Gz ∈ L2. Since φ ∈ C∞c (C) was arbitrary this means

that F ∈W 1,2loc .

2.7. Estimates of T n on Lp. In the article [55] Iwaniec and Martin study singularintegrals that appear in regularity theory of nonlinear PDE in arbitrary dimensions.For example, they compute Lp norms of scalar Riesz transforms on Rn, thus extendinga well-known result of Pichorides [71]. One of the key features of their work is that theyreduce the estimates of vector-valued operators on Rn (such as combinations of Riesztransforms, complex Riesz transforms, certain differential operators, etc.) to those ofscalar-valued operators on C. They succeed in doing that by developing a so-calledcomplex method of rotations [55, Section 6]. We summarize this technique in Section2.7.1 below. As it emerges from their approach, the crucial role is played by the Ahlfors-Beurling operator T , its “square root” HC (recall it was defined in Section 1.2) andtheir powers Hk

C. Throughout their article they extensively work with them; most of

their estimates are expressed in terms of the norm of HkC on Lp(C), which they denote

by Hp(k). However, no estimate on Hp(k) itself is given. In a subsequent paper byIwaniec and Sbordone [56] it was noticed that for odd k one can resort to the methodof rotations, developed in the 1956 paper [22] of Calderon and Zygmund, see also [35,Section 4.3] or [76, Chapter II], which yields

Hp(2n − 1) 6π

2(2n− 1) cot

π

2p∗. np∗, ∀n ∈ N . (2.27)


On the other hand, the same approach applied in the case of even k does not give sucha linear estimate. Recall that Hp(2n) = ‖T n‖p. Obviously

Hp(2n) 6 Hp(2n − 1)Hp(1). np∗2 . (2.28)

The slight difference lies in the fact that the kernel of H2nC = T n is even. Calderon and

Zygmund [22] derive a method for operators with even kernels as well, but that methodyields the same quadratic estimate in p as (2.28), namely

Hp(2n). np∗2 .

In [31] yet another method of rotation was presented, which works very well exactlyfor even kernels. Applying it to T n = H2n

C returned an estimate Hp(2n). np∗, whichis the best possible estimate of the type φ(n)ψ(p), but still not the overall optimalestimate of Hp(2n) simultaneously in n and p. The optimal result is stated in Theorem2.39.

Finally, the estimates of Hp(k) for any k ∈ Z are also considered in [5, Section 4.5].The estimate in Corollary 4.5.1 there is

Hp(k) 6 C(p)(1 + k2) ∀ k ∈ Z . (2.29)

This is improved in our Theorem 2.38 below.

Section 4 is devoted to the estimates ofHp(k). In Theorems 2.38 and 2.39 we improve(2.27), (2.28) and (2.29), and in the case of Hp(2n) our estimate is asymptotically sharpin terms of both n and p. We also conjecture the exact value of Hp(2n). See Section2.7.2 for precise formulations of our results.

2.7.1. Iwaniec and Martin’s complex method of rotations. Here we summarize [55, Sec-tion 6].

Let E and F be finite-dimensional complex vector spaces, each of them endowedwith its own Hermitian inner product. Denote by L(E,F ) the vector space of all linearmappings from E to F . The norm on L(E,F ) is introduced by

‖Λ‖ := suph∈E\0

‖Λh‖F‖h‖E

.

We additionally assume that m ∈ Z\0 and p ∈ (1,∞) are given.

Theorem 2.37. Suppose Ω : Cn → L(E,F ) is a measurable function satisfying

Ω(λζ) =

(λ

|λ|

)−m Ω(ζ)

|λ|2n (2.30)

for all (λ, ζ) ∈ C × S2n−1 and suppose that the restriction Ω∣∣S2n−1 is integrable. Then

the singular integral operator T : Lp(Cn, E) → Lp(Cn, F ), defined by

(Tf)(z) =

∫

CnΩ(z − w)f(w) dw ,

is bounded and admits the following norm estimate:

‖T‖p 6Hp(m)

|m|

∫

S2n−1

‖Ω(ζ)‖ dζ .


Sketch of the proof. For each complex “direction” ζ ∈ S2n−1 we introduce the op-erator Aζ : L

p(Cn, E) → Lp(Cn, E) by

(Aζf)(z) =|m|2π

∫

C

(λ

|λ|

)−m f(z − λζ)

|λ|2 dλ .

Iwaniec and Martin call it “directional complex Hilbert transform”; strictly speakingthough, it looks more like “directional m-th power of the complex Hilbert transform”.What matters more is that they prove the following:

‖Aζ‖p = Hp(m), (2.31)

which is not so surprising, in view of (1.8) and (1.9).The second key calculation in their proof, and the one we present here, shows how

to pass from Aζ to T. Compare with the classical method of rotation by Calderon andZygmund.

Fix z ∈ Cn and consider a test function Φ = Φz : S2n−1 → F , defined by

Φ(ζ) = Ω(ζ)Aζf(z) .

By the definition of Aζ we get

Φ(ζ) =|m|2π

Ω(ζ)

∫

C

(λ

|λ|

)−m f(z − λζ)

|λ|2 dλ .

It is easy to verify that for Λ ∈ L(E,F ) and g : C → E we have Λ∫Cg =

∫CΛ g.

Apply this simple fact to Λ = Ω(ζ) and then use property (2.30). The outcome is:

Φ(ζ) =|m|2π

∫

C

|λ|2n−2Ω(λζ)f(z − λζ) dλ . (2.32)

At this point we need the following identity: for even m ∈ N we have

2π

∫

RmF (x) dx =

∫

Sm−1

∫

C

F (λσ)|λ|m−2 dλ dσ. (2.33)

To prove this start with the well-known formula [37, Appendix C.3, Theorem 4]∫

RmF (x) dx =

∫ ∞

0

∫

rSm−1

F dσ dr.

By using spherical coordinates we get∫

RmF (x) dx =

∫

Sm−1

∫ ∞

0F (rσ) rm−1 dr dσ.

We integrate as∫ 2π0 dϕ and obtain

2π

∫

RmF (x) dx =

∫ ∞

0

∫ 2π

0

∫

Sm−1

F (rσ) rm−1 dσ dϕdr.

Since m is even, we may think of σ ∈ Rm as σ ∈ Cn, where of course m = 2n. Thenit makes sense, for any fixed ϕ, to introduce the new variable σ′ = e−iϕσ, of courseunderstood as a tensor product of rotations applied to a vector in Rm. We get

2π

∫

RmF (x) dx =

∫ ∞

0

∫ 2π

0

∫

Sm−1

F (reiϕσ′) dσ′ dϕ rm−1 dr

=

∫

Sm−1

∫ 2π

0

∫ ∞

0F (reiϕσ)rm−2r dr dϕdσ ,

which we recognize as (2.33).


By recalling that |ζ| = 1, we may apply (2.33) to (2.32) and arrive at∫

S2n−1

Φz(ζ) dζ = |m|∫

CnΩ(ζ)f(z − ζ) dζ = |m|(Tf)(z) .

Finally we apply Minkowski’s integral inequality:

‖Tf‖p 61

|m|

∫

S2n−1

‖Φ·(ζ)‖p dζ

61

|m|

∫

S2n−1

‖Ω(ζ)‖‖Aζf‖p dζ

6Hp(m)

|m|

(∫

S2n−1

‖Ω(ζ)‖ dζ)‖f‖p .

For the last inequality we applied (2.31).

2.7.2. Our results. The following estimates give partial answers to the questions dis-cussed above. They will be proven in Section 4.

Theorem 2.38 ([31]). We have Hp(k) . |k|1−2/p∗(p∗−1), where k ∈ Z\0 and p > 1.

When k is even the one-sided inequality from the above theorem is mirrored by thesame estimate from below, thus establishing the correct asymptotic behaviour of ‖T n‖psimultaneously in n and p:

Theorem 2.39 ([31, 28]). We have ‖T n‖p ∼ n1−2/p∗(p∗ − 1), for n ∈ N and p > 1.

Our proving Theorem 2.39 in Section 4 will eventually lead us to conjecture that

‖T n‖p =(1/q)n(1/p)n

for all p > 2, q = p/(p − 1) and n ∈ N. See Conjecture 4.10 below. Here

(a)n =Γ(a+ n)

Γ(a)= a (a+ 1) · . . . · (a+ n− 1) (2.34)

is the Pochhammer symbol.

Let us explain why the case of odd k is more difficult. Recall that, by definition,Hp(k) = ‖Rk‖B(Lp(C)).

If k is even we have R2l = T l. The operator T is characterized by ∂zf 7→ ∂zf(Exercise 2.19). This property is used in proving lower estimate of Hp(k) for even k,see Proposition 4.2.

If k is odd we have R2l+1 = RT l. Now, R has a characterization in terms of(−∆)1/2f 7→ 2∂zf , however (−∆)1/2 is not a differential operator, which makes thecalculations less explicit. See [27, Example 3.7.5] for details regarding (−∆)1/2.

Still, we believe that Hp(k) ∼ |k|1−2/p∗(p∗ − 1) might hold for all k ∈ Z\0,not only the even ones which Theorem 2.39 addresses. At the moment however thisquestion is still open. We have trivial inequalities Hp(2l + 2) 6 Hp(2l + 1)Hp(1) andHp(4l+2) 6 Hp(2l+1)2. It follows from [55, Theorem 1.1] that Hp(1). p

∗. Thereforewe obtain, as a simple consequence of Theorems 2.38 and 2.39, the following estimateof Hp(k) from below in the case when k is odd:

Corollary 2.40. Hp(2l + 1)& maxl1−2/p∗ , l1/2−1/p∗

√p∗ − 1

.


We conclude by a weak-type (1,1) result analogous to Theorem 2.39.

Theorem 2.41. We have ‖T n‖1→1,∞ ∼ n, for n ∈ N.

2.8. Weighted estimates of T n. Based on the results announced in this section(estimates of T n on Lp and of T on Lp(w)) we find it natural to consider estimatingpowers of T on weighted spaces. It turns out an adequate modification of the techniquesfrom [33] and [31] yields the following generalization of Theorem 2.29:

Theorem 2.42 ([29]). For every p > 1 there is C(p) > 0 such that for every n ∈ Z\0and w ∈ Ap we have

‖T n‖B(Lp(w)) 6 C(p) |n|3 [w]p∗/pp . (2.35)

The proof will not be presented in this note. The theorem above has an improvementdue to Hytonen [50, Corollary 7.5] who proved, for every n ∈ N, p > 1 and ε > 0, theestimate

‖T n‖B(Lp(w)) 6 C(p, ε) |n|1+ε [w]p∗/pp . (2.36)

More recently [52] he observed that one may improve this to

‖T n‖B(Lp(w)) 6 C(p) |n|(1 + log |n|)2 [w]p∗/pp .

Even more recently, Hytonen, Roncal and Tapiola [53, Corollary 4.2] improved thisestimate to

‖T n‖B(Lp(w)) 6 C(p) |n|(1 + log |n|) [w]p∗/pp . (2.37)

On the other hand, Hytonen also noticed (personal communication, 2011) that, forany ε > 0, the estimate

‖T n‖B(L2(w)) 6 C(ε)n1−ε [w]2

is not valid for all n ∈ N and w ∈ A2. Indeed, suppose the contrary. Then, by Theorem1.18, we would get

‖T n‖B(Lp(w)) 6 C(p, ε)n1−ε

for any n ∈ N, p > 2 and all w ∈ Ap. In particular, in the unweighted case (w ≡ 1) wewould have

‖T n‖B(Lp(C)) 6 C(p, ε)n1−ε,

for all n ∈ N and p > 2. However, Theorem 2.39 gives

‖T n‖B(Lp(C)) > C(p)n1−2/p,

therefore the combination of the last two inequalities would lead to

n1−2/p 6 C(p, ε)n1−ε

for all n ∈ N and p > 2, which is clearly impossible.

It remains an open question whether one can get rid of ε in (2.36) or, equivalently,of the logarithmic term in (2.37). See [52, Section 6.2] and [53, Conjecture 4.6].

Hytonen’s observation in [50, Corollary 7.5] arrives from a much stronger result –the A2 theorem for general Calderon-Zygmund operators – whose original proof, alsodue to Hytonen [51], is way more involved than our proof of Theorem 2.42, while onthe other hand it also uses a (far subtler) version of the averaging technique.


3. Weighted estimate for T : proof of Theorem 2.29.

In view of the above-cited extrapolation result (Theorem 1.18), in order to proveTheorem 2.29 it suffices to consider the case p = 2 (recall that

[w]2 = supQ⊂R2

〈w〉Q⟨w−1

⟩Q,

where the supremum is taken over all squares in C regardless of their orientation):

Theorem 3.1 ([70, 33]). For any w ∈ A2 we have

‖T‖B(L2(w)) . [w]A2 .

This estimate is sharp.

This theorem was first proven by Petermichl and Volberg [70] by means of the Bell-man function technique. Here we present another approach, essentially from [33]. Thekey result, and one of the inspirations for the proof in [33], is an inequality by Wittwer[82] (see Theorem 3.28 below), saying, roughly, that martingale transforms admit onL2(w) linear estimates in terms of [w]A2 , i.e. just the kind of estimates we want forT . Its proof is an example of the Bellman function technique and is modelled on thepaper by Nazarov, Treil and Volberg [65]. We present it in Section 3.6.

Another source of inspiration was a paper by Petermichl, Treil and Volberg [69],where (first-order) Riesz transforms were represented as averages of so-called dyadicshifts. That result, in its turn, was an upgrade of the theorem by Petermichl [71], wherethe Hilbert transform was represented as an average of dyadic shifts. While first-orderRiesz transforms are singular operators with odd kernels, the Ahlfors-Beurling operatorT has an even kernel and thus dyadic shifts cannot be turned into T by the averagingmethod. It turns out that an appropriate replacement are precisely the martingaletransforms.

3.1. Haar functions and martingale transforms. For each interval I ⊂ R let I−, I+be its left and its right half, respectively. Denote by χI its characteristic function andby hI its Haar function, which is to say hI = |I|−1/2(χI+ − χI−). An interval in R iscalled dyadic if it is of the form [k2m, (k+1)2m) for some integers k,m. The collectionhI ; I ⊂ [0, 1] dyadic is called the (standard) Haar system on [0, 1]. Together with theconstant function 1 it forms a complete orthogonal system in L2([0, 1]); see Muscalu-Schlag [62, Section 8.4] or Grafakos [44, Section 5.4].

Typically Haar functions are defined on intervals in R. By working in R2 one is ledto define them on squares [33]. Let L be the standard dyadic lattice in R2, i.e., thecollection of squares Q = I × J , where both I and J are dyadic intervals of the samelength. The corresponding Haar functions are then

h0Q =1√|I|

χI ⊗ hJ , h+Q =

√2

|J | hI ⊗ χJ+ , h−Q =

√2

|J | hI ⊗ χJ− .

The constants in the front are chosen so that the functions are normalized in (the usual)L2.

We define martingale transforms Mσ to be the operators

Mσf(ζ) =∑

Q∈L

∗∈0,+,−

σ∗Q〈f, h∗Q〉h∗Q(ζ) , (3.1)

where σ∗Q ∈ ∆ and h∗Q are Haar functions.


Wittwer [82] originally addressed weighted estimates of martingale transforms on theline and associated with the standard dyadic lattice in R. Her result (here Theorem3.28) is formulated and proven in Section 3.6. A careful reading of the proof showsthat the same holds for the “planar” operators Mσ defined above:

Theorem 3.2 (Wittwer [82]). There exists C > 0 such that for any family of coeffi-cients σ and any w ∈ A2,

‖Mσ‖B(L2(w)) 6 C[w]A2 .

3.2. Main idea. Briefly, we represent T as an “average” of operators that on B(L2(w))admit linear estimates in terms of [w]A2 .

Instead of a dyadic lattice let us for a moment consider a unit grid G of squares. Thisis a family of squares I × J , where I and J are dyadic intervals of unit length.

Our plan is to consecutively average 0−type projections (i.e. operators of the form

f 7→∑

Q∈F

〈f, h0Q〉h0Q,

where F is some dyadic collection of squares) over (ever wider) families of F’s comprising

• translated unit grids• translated grids of arbitrary (fixed) size• translated and rotated grids of arbitrary (fixed) size• all lattices of a fixed calibre• all lattices.

More precisely, introduce, for f ∈ S,

P0f :=∑

Q∈G

〈f, h0Q〉h0Q .

Note that, at least in the unweighted case, P0 is simply the orthogonal projection ontothe subspaces generated by all Haar functions of type 0 having as supports dyadicsquares of unit size. Note also that we have one sole set of coefficients, namely σ0Q = 1

and σ+Q = σ−Q = 0 for all such Q.This P0 is our starting operator, the one we will conjugate by translations, dilations,

rotations, and then average. In the process we will take care of:

• approaching T , meaning that after the final average our resulting operator willequal the operator T , up to a nonzero multiplicative constant;

• justifying the name “averaging”, i.e. preserving at any phase the same estimatesthat the original operator, P0, enjoys on L

2(w), w ∈ A2, by Theorem 3.2.

The first part (the averaging) will be carried out in Section 3.3, while the second one(estimates) in Section 3.4.

The idea presented here comes from the simple observation that T is a (p.v.) in-tegral operator whose kernel k(z, w) = (z − w)−2 is characterized, up to a nonzeromultiplicative constant, by the following properties:

• translation invariance (convolution kernel): k(z, w) = k(z+u,w+u), ∀z, w, u ∈C;

• “weighted dilation invariance” (homogeneity of order -2): k(λz, λw) = λ−2k(z, w)for any λ > 0;

• “weighted rotation invariance”: k(eiψz, eiψw) = e−2iψk(z, w) for any ψ ∈ R.


3.3. The averaging. We start by carrying out the plan announced above.

Averaging over translated unit grids. For t ∈ R2 define Pt as the conjugation of P0 by

the translation τt, explicitly,

Pt = τt P0 τ−1t . (3.2)

It is straightforward that, for f ∈ S,

Ptf =∑

Q∈G

〈f, τth0Q〉τth0Q .

Observe that τth0Q = h0Q+t, therefore we may write

Ptf =∑

Q∈Gt

〈f, h0Q〉h0Q , (3.3)

where Gt := G+ t is the grid of unit squares such that one of them contains the pointt as one of its vertices.

The family Ω := Gt ; t ∈ R2 of all unit grids naturally corresponds to the torusR2/Z2, which is of course in one-to-one correspondence with the square [0, 1)2. Thuswe are able to regard Ω as a probability space [0, 1)2 where the probability measureequals the Lebesgue measure.

Consider the “mathematical expectation” of the “random variable” P : for f ∈ S

and x ∈ R2 define

(EPf)(x) =

∫

Ω(Ptf)(x) dt .

It makes good sense to call the process of passing from all Pt to EP the averaging. Thestructure of the operator EP is revealed in the following proposition.

Proposition 3.3. The operator EP is a convolution operator with the kernel −β ⊗ α,where α = h0 ∗ h0 and β = χ0 ∗ χ0. Here χ0 and h0 stand (respectively) for thecharacteristic and Haar function of the interval (−1/2, 1/2]. That is, we have

EPf = f ∗ F,where f ∈ S(R2) and F (x, y) = −β(x)α(y).

Proof. Choose t = (t1, t2) ∈ R2 and Q = I × J ∈ Gt. Then

〈f, h0Q〉 =∫

R

∫

R

f(s1, s2)h0Q(s1, s2) ds1 ds2 =

∫

J

∫

If(s1, s2)χI(s1)hJ (s2) ds1 ds2, .

Thus for (fixed) f ∈ S(R2) and x = (x1, x2) ∈ R2 we have, by (3.3),

(Ptf)(x) =∑

Q∈Gt

∫

R

∫

R

f(s1, s2)χI(s1)hJ (s2) ds1 ds2 · h0Q(x).

The expression under the summation sign in the last row is nonzero for exactly oneQ ∈ Gt; namely, one such that h0Q(x) 6= 0. This means that x = (x1, x2) ∈ Q and hence

x1 ∈ I and x2 ∈ J . Thus h0Q(x) = hJ(x2). We thus obtain

(Ptf)(x) =

∫

R

∫

R

f(s1, s2)χI(s1)hJ (s2)hJ(x2) ds1 ds2 . (3.4)


Because Gt does not change if we increase or decrease any component of t by 1, wemay assume that I = [t1 − 1, t1) and J = [t2 − 1, t2). Denoting I0 = [−1/2, 1/2), thisassumption implies

I = t1 −1

2+ I0 and J = t2 −

1

2+ I0 .

Now let χ0 and h0 be as in the formulation of the proposition and let k0 := −h0. ThenχI(s) = χ0(t1 − 1/2− s)

hI(s) = k0(t1 − 1/2− s)

for all s ∈ R2. The analogue pair of identities is valid also for J , of course.The point here is that we modified the expressions on the left to look more like a

part of a convolution integral with t1 and t2 as integration variables and s as a centerof convolving.

Together with (3.4), the last two equalities imply

(Ptf)(x) =

∫

R

∫

R

f(s1, s2)χ0(t1−1/2−s1)k0(t2−1/2−s2)k0(t2−1/2−x2) ds1 ds2. (3.5)

Recall that x1 ∈ I = [t1 − 1, t1) and x2 ∈ J = [t2 − 1, t2). Hence xj < tj 6 xj + 1 forj = 1, 2. Averaging in our case means integrating over all admissible tj. Therefore,

(EPf)(x) =

∫ x2+1

x2

∫ x1+1

x1

(P(t1,t2)f)(x) dt1 dt2 .

By using the most recent expression for (Ptf)(x), changing variables (from tj to tj−1/2)and applying Fubini’s theorem, we obtain

(EPf)(x1, x2) =

∫

R

∫

R

f(s1, s2)

∫ x2+1/2

x2−1/2

∫ x1+1/2

x1−1/2[A] dt1 dt2 ds1 ds2 , (3.6)

where

A = k0(t2 − x2)χ0(t1 − s1) k0(t2 − s2).

Let us take a closer look at the inner integral. We have∫ x2+1/2

x2−1/2

∫ x1+1/2

x1−1/2[A] dt1 dt2

=

∫ x1+1/2

x1−1/2χ0(t1 − s1) dt1 ·

∫ x2+1/2

x2−1/2k0(t2 − x2) k0(t2 − s2) dt2

=

∫ 1/2

−1/2χ0(y1 − u1) du1 ·

∫ 1/2

−1/2k0(−u2) k0(y2 − u2) du2,

where in the last integral we introduced a new pair of variables, namely

uj = xj − tj

yj = xj − sj(3.7)

for j = 1, 2. Clearly, this is the same as −(χ0 ∗ χ0)(y1)(k0 ∗ k0)(y2). Observe thatk0 ∗ k0 = h0 ∗ h0 and recall that yj ’s were given in (3.7). Hence we proved

∫ x2+1/2

x2−1/2

∫ x1+1/2

x1−1/2[A] dt1 dt2 = F (x1 − s1, x2 − s2),

where F is as in the formulation of the proposition. By (3.6), this finishes the proof.


12

12

−12

x

1

−1

Figure 2. Graph of α

x

11

1

Figure 3. Graph of β

Graphs of functions α and β are shown as Figures 2 and 3, respectively.

Averaging over translated grids of arbitrary (fixed) size. Instead of the unit grid we

may consider a grid of squares with sides of an arbitrary length ρ > 0. Denote such agrid by G

ρt if t ∈ R2 is a vertex of one of its members. Henceforth we will call ρ the size

of the grid and t its reference point. We introduce another family of operators, definedby

Pρt f :=

∑

Q∈Gρt

〈f, h0Q〉h0Q .

Since Gρt = ρG+ t, it follows that, for any Q ∈ G

ρt ,

h0Q =1

ρ(τt δρ)h0Q,


where Q is some (unique) square belonging to G. From here we get a generalization of(3.2), namely

Pρt = Λρ,t P0 Λ−1

ρ,t (3.8)

where Λρ,t := τt δρ. So Pρt is obtained by conjugating our initial operator P0 by a

composition of a translation and dilation.In a similar way, by writing G

ρt = ρGt/ρ, we may derive the representation

Pρt = δρ Pt/ρ δ−1

ρ . (3.9)

Let us fix ρ > 0 and average operators Pρt over all admissible t. This means studyingthe operator EPρ, given on S by

EPρf :=1

ρ2

∫

ΩρPρt f dt,

where Ωρ := [0, ρ) × [0, ρ).We have the following generalization of Proposition 3.3.

Exercise 3.4. Fix ρ > 0. Then the operator EPρ can be on S(R2) expressed as

EPρf = f ∗ F ρ,where

F ρ(x, y) :=1

ρ2F

(x

ρ,y

ρ

)

and F is as in Proposition 3.3.

Averaging over translated and rotated grids of arbitrary (fixed) size. At this step we

bring rotations into play. Recall the notation (1.1). For ρ > 0 define

Gρ(ξ) =1

2π

∫ 2π

0(UψF

ρ)(ξ)e−2iψdψ (3.10)

and G = G1. Observe thatsupp Gρ ⊂ K2(0, ρ

√2). (3.11)

Indeed, if |ξ|2 > ρ√2, then |O−ψξ|2 > ρ

√2 for every ψ ∈ R, hence |O−ψξ|∞ > ρ,

therefore F ρ(O−ψξ) = 0 for every ψ ∈ R.Exercise 3.4 helps to establish the following formula.

Exercise 3.5. For every ρ > 0, f ∈ S(R2) and x ∈ R2 we have

(f ∗Gρ)(x) = 1

2π

∫ 2π

0[(Uψ EPρ U−1

ψ )f ](x) e−2iψ dψ.

Remark 3.6. From the definition of EPρ it follows that

[(Uψ EPρ U−1ψ )f ](x) =

1

ρ2

∫

Ωρ[(Uψ Pρt U−1

ψ )f ](x) dt.

Now Exercise 3.5 and (3.8) imply

f ∗Gρ = 1

2π

∫ 2π

0

1

ρ2

∫

Ωρ

(Λρ,t,ψ P0 Λ−1

ρ,t,ψ

)f dt e−2iψdψ.

where Λρ,t,ψ := Uψ τt δρ. Therefore we could restate Exercise 3.5 by saying thataveraging the operators

Λρ,t,ψ P0 Λ−1ρ,t,ψ : f 7−→

∑

Q∈Gρt

〈f, Uψh0Q〉Uψh0Q


where ρ > 0 is fixed and t, ψ run over all admissible values, returns a convolutionoperator whose (convolution) kernel is Gρ. Note that the operators Λρ,t,ψ P0 Λ−1

ρ,t,ψ

are all martingale transforms, and that the supports of the “rotated” Haar functionsare squares from the rotated grid OψG

ρt .

The following property of Gρ is simple yet important:

Lemma 3.7. Gρ has zero average on every circle centered at the origin.

Proof. Here it is convenient to resort to the “complex” notation. For every R > 0,∫ 2π

0Gρ(Reiϕ) dϕ =

1

2π

∫ 2π

0e−2iψ

∫ 2π

0F ρ(Rei(ϕ−ψ)) dϕdψ

=1

2π

∫ 2π

0e−2iψ dψ

∫ 2π

0F ρ(Reiλ) dλ

= 0.

(3.12)

Averaging over all lattices of a fixed calibre. Only now, owing to Lemma 3.7, will it

make real sense to introduce the “kernel for averages over whole lattices”. Define

kr(x) :=

∞∑

n=−∞

G2nr(x), (3.13)

in the sense that

kr(x) = limM→∞

krM (x), (3.14)

where

krM (x) :=

M∑

n=−∞

G2nr(x). (3.15)

The fact that kr∗ is a sum of operators, obtained by averaging over grids of size r ·2n,hints at kr∗ itself being an average, this time over unions of these grids, i.e. latticesof calibre r. While it is not clear what could be a probability space corresponding toall lattices of a fixed calibre, we define the above-said average as a limit of averages oftruncated lattices. Then the statement makes sense and holds, as will be shown.

For every x 6= 0 and M ∈ Z, the sum krM (x) is finite (i.e. has only finitely manynonzero terms). Indeed, from (3.11) it emerges that G2nr(x) 6= 0 implies

n > log2|x|2r

− 1

2=: C(x, r) > −∞.

Therefore, with N(x, r) := [C(x, r)], where [y] is the integer part of y ∈ R,

krM (x) =

M∑

n=N(x,r)

G2nr(x).

Consequently,

kr(x) = limM→∞

M∑

n=N(x,r)

G2nr(x). (3.16)


Remark 3.8. Since N(x, r) rises as |x| rises, this shows that we also have

krM (y) =

M∑

n=N(x,r)

G2nr(y)

for any y ∈ R2 such that |y| > |x|. Consequently, krM is continuous on R2\0 for anyM ∈ Z and r > 0, since on any set of the form |y| > ε it is a finite sum of continuousfunctions G2nr.

Lemma 3.9. The sum in (3.14) converges absolutely and uniformly on every comple-ment of a neighbourhood of 0 and satisfies the estimate |kr(x)|. |x|−2, uniformly inr > 0. By Remark 3.8, this implies that kr is continuous on R2\0.Proof. Take x = (x1, x2) ∈ R2\0. Therefore, by (3.16),

kr(x) =1

r2

∞∑

n=N(x,r)

1

4nG( x

2nr

).

Since |F | 6 1, and thus |G| 6 1, this implies

|kr(x)| 6 1

r2(1/4)N(x,r) 1

1− 1/46

4

3r2(1/4)C(x,r) =

8

3· 1

|x|22.

Remark 3.10. Actually, the above proof shows that, for any nonzero x ∈ R2 and anysubset Z of Z, ∑

n∈Z

|G2nr(x)|. |x|−2,

with the implied constant independent of x, r and Z.

Lemma 3.11. The average of kr over any circle centered at the origin is zero. Thatis, for every R, r > 0 we have

∫ 2π

0kr(Reiψ) dψ = 0.

Proof. We have, by (3.16),

∫ 2π

0kr(Reiψ) dψ =

∫ 2π

0limM→∞

M∑

n=N(Reiψ ,r)

G2nr(Reiψ) dψ .

Because N(Reiψ, r) = N(R, r) for every ψ ∈ R, and by Remark 3.10, we may applythe dominated convergence theorem and conclude

∫ 2π

0kr(Reiψ) dψ = lim

M→∞

M∑

n=N(R,r)

∫ 2π

0G2nr(Reiψ) dψ.

By Lemma 3.7, this is equal to zero.

Proposition 3.12. For every r > 0, the function kr defines a tempered distribution,understood in the sense

Wkr(φ) = limε→0

∫

|x|>εkr(x)φ(x) dx.


Proof. We start with the splitting∫

|x|>εkr(x)φ(x) dx =

∫

ε<|x|<1+

∫

|x|>1.

The second integral obviously converges, because φ ∈ S and kr decays (quadratically),by Lemma 3.9. As for the first integral, by Lemma 3.11 we may write

∫

ε<|x|<1kr(x)φ(x) dx =

∫

ε<|x|<1kr(x)|x| φ(x)− φ(0)

|x| dx. (3.17)

By Lemma 3.9, the function kr(x)|x| is integrable on 0 < |x| < 1, while φ(x)−φ(0)|x| is

bounded. Indeed, since φ(y)− φ(0) = 〈∇φ(ty), y〉 for some t ∈ (0, 1), it follows that

|φ(y) − φ(0)||y| 6 ‖∂y1φ‖∞ + ‖∂y2φ‖∞. (3.18)

Thus the limit ε → 0 of (3.17) exists. We proved, for any r > 0 and φ ∈ S, thatWkr(φ) is well defined and that

Wkr(φ) =

∫

|x|<1φ(x)kr(x)|x| dx +

∫

|x|>1φ(x)kr(x) dx, (3.19)

where

φ(x) :=

φ(x)− φ(0)

|x| ; x 6= 0;

0 ; x = 0.

Now it is easy to show that Wkr defines a tempered distribution: by combining(3.19), Lemma 3.9 and (3.18) we get

|Wkr(φ)|. ‖φ‖∞∫

|x|<1

dx

|x| + supy∈R2

|φ(y)||y|∫

|x|>1

dx

|x|3. ‖∂y1φ‖∞ + ‖∂y2φ‖∞ + sup

y∈R2

|φ(y)||y|.

This means, e.g. [44, Proposition 2.3.4.(b)], that indeed Wkr ∈ S′(R2).

Remark 3.13. As before, we see that

WkrM(φ) = lim

ε→0

∫

|x|>εkrM (x)φ(x) dx

also defines a tempered distribution and that

WkrM(φ) =

∫

|x|<1φ(x)krM (x)|x| dx+

∫

|x|>1φ(x)krM (x) dx. (3.20)

Moreover, WkrM→ Wkr in the sense of tempered distributions as M → ∞, i.e.

WkrM(φ) →Wkr(φ) for every φ ∈ S, e.g. [44, p. 110]. Indeed,

Wkr(φ)−WkrM(φ) =

∫

|x|<1φ(x)[kr(x)− krM (x)]|x| dx +

∫

|x|>1φ(x)[kr(x)− krM (x)] dx

=

∫

|x|<1φ(x)

∞∑

n=M+1

G2nr(x)|x| dx+

∫

|x|>1φ(x)

∞∑

n=M+1

G2nr(x) dx.

Because of Lemma 3.9 and the fact that φ ∈ S, we may use the dominated convergencetheorem and show that the above expression tends to zero as M → ∞, for any x ∈ R2.


The operators we are interested in are the convolutions with the distributions WκrMand Wκr :

KrMf := f ∗WkrM

.

Krf := f ∗Wkr .

It is well known [44, Theorem 2.3.20] that a convolution of a Schwartz function ϕ anda tempred distribution u is actually a (C∞) function and [44, eq. (2.3.21)] that it isgiven by

x 7→ u(ϕ(x− ·)).Therefore for f ∈ S we have, by Proposition 3.12 and Remark 3.13,

(Krf)(x) = limε→0

∫

|y|>εkr(y)f(x− y) dy

and

(KrMf)(x) = lim

ε→0

∫

|y|>εkrM (y)f(x− y) dy. (3.21)

Averaging over all lattices. The final step is to average over dilations, in other words,

over all calibres r. The family of all calibres (of dyadic lattices) is represented by anyinterval of the form [a, 2a). For our purpose, the most appropriate measure turns outto be dr/r. This makes all such intervals have the same measure (log 2). Thereforewhen averaging over all (translated, rotated, dilated) lattices we may take, for instance,[1, 2) and define, for f ∈ S and x ∈ R2,

(Sf)(x) :=

∫ 2

1(Krf)(x)

dr

r.

This integral is well defined.

The operator S is the final outcome of our averaging process. We needed to sumover n and integrate over dt, dψ and dr. The problem was to do this rigorously and inthe correct order.

As explained above, we need to show that with S we “hit the target” - T .

Proposition 3.14. We have S = cT , where c ∈ R is nonzero.

Proof. We have

(Krf)(x) =

∫

|y|<1

f(x− y)− f(x)

|y| kr(y)|y| dy +∫

|y|>1f(x− y)kr(y) dy.

Now from Lemma 3.9 it emerges that

Sf = p.v. f ∗ k,

where

k(x) =

∫ 2

1kr(x)

dr

r.


We need to prove that k is a (nonzero) constant multiple of the (p.v.) convolutionkernel of T . We have, by (3.13), Lemma 3.9 and (3.10) that

k(x) =

∫ 2

1

∞∑

n=−∞

G2nr(x)dr

r

=∞∑

n=−∞

∫ 2

1G2nr(x)

dr

r

=

∞∑

n=−∞

∫ 2n+1

2nGs(x)

ds

s

=

∫ ∞

0Gs(x)

ds

s

=1

2π

∫ ∞

0

∫ 2π

0F s(O−ψx)e

−2iψdψds

s.

Therefore, in polar coordinates,

k(reiϕ) =1

2π

∫ ∞

0

∫ 2π

0s−2F (e−iψreiϕ/s)e−2iψdψ

ds

s

=e−2iϕ

2π

∫ ∞

0

∫ 2π

0F (reit/s)e2itdt

ds

s3

=e−2iϕ

πr2· 12

∫ ∞

0

∫ 2π

0F (ueit)e2itdt u du .

Since the (p.v.) convolution kernel of T is equal to

reiϕ 7→ −e−2iϕ

πr2,

we indeed obtain S = cT , where, by the evenness of F ,

c = −1

2

∫ ∞

0

∫ 2π

0F (ueit) cos 2t dt u du.

It remains to be verified that c 6= 0. In cartesian coordinates, we can express c as

c = −1

2

∫

R

∫

R

F (x, y)x2 − y2

x2 + y2dx dy .

In Proposition 3.3 we saw that F (x, y) = −β(x)α(y), which leads to

c = −1

2

∫

R

∫

R

α(x)β(y)x2 − y2

x2 + y2dx dy .

Since α and β are even functions, supported on the interval [−1, 1], we obtain

c = −2

∫ 1

0

∫ 1

0α(x)β(y)

x2 − y2

x2 + y2dx dy .

Figure 3 shows that β(y) = 1− y on [0, 1]. Combine this with the identity

x2 − y2

x2 + y2= 1− 2y2

x2 + y2


and the observation (following from Figure 2) that∫ 10 α(x) dx = 0 to get

c = 4

∫ 1

0(1− y)y2

∫ 1

0

α(x)

x2 + y2dx dy .

For y > 0, computation returns

C(y) : =

∫ 1

0

α(x)

x2 + y2dx

=1

y

(arctan

1

y− 2 arctan

1

2y

)+ 2 log(4y2 + 1)− 1

2log(y2 + 1)− 3 log y − 4 log 2 .

We now need only evaluate the integral∫ 1

0(1− y)y2C(y) dy .

We can directly calculate this integral to find that it equals

1

12

(arctan 2− 4 arctan

1

2+

15

8log 5− 4 log 2

),

which is approximately −0·042 .

3.4. Estimates for S on L2(w).

Remark 3.15. Recall that, by (3.21) and (3.15),

KrMf = p.v. f ∗

(M∑

n=−∞

G2nr

).

Let us show that we also have

KrMf =

M∑

n=−∞

(f ∗G2nr

). (3.22)

Indeed, this follows from

(KrMf)(x) =

∫

|y|<1

f(x− y)− f(x)

|y| krM (y)|y| dy +∫

|y|>1f(x− y)krM (y) dy,

Lemma 3.9 and the dominated convergence theorem.

By (3.22) and Exercise 3.5,

(K1Mf)(x) =

M∑

n=−∞

1

2π

∫ 2π

0[(Uψ EP2n U−1

ψ )f ](x) e−2iψdψ

=

M∑

n=−∞

1

2π

∫ 2π

0

1

4n

∫

[0,2n)2P2nt (U−1

ψ f)(O−ψx) dt e−2iψdψ .

As implied by (3.5) and (3.9), for every f, x, n, the function

(t, ψ) 7→ P2nt (U−1

ψ f)(O−ψx) e−2iψ

is bounded on [0, 2n)2 × [0, 2π), therefore it belongs to L1 of that space and hence wecan apply Fubini’s theorem. With

(Rρt f)(x) =1

2π

∫ 2π

0[(Uψ Pρt U−1

ψ )f ](x) e−2iψdψ, (3.23)


we get

(K1Mf)(x) =

M∑

n=−∞

1

4n

∫

[0,2n)2(R2n

t f)(x) dt.

By using the fact that Pρ· is periodic and that n 6M , we can continue as

(K1Mf)(x) =

M∑

n=−∞

1

4M

∫

[0,2M )2(R2n

t f)(x) dt. (3.24)

Let us define, for N ∈ ZM := k ∈ Z ; k 6M,

(K1M,Nf)(x) =

M∑

n=N

1

4M

∫

[0,2M )2(R2n

t f)(x) dt.

The following result is straightforward yet useful.

Lemma 3.16. Let A be an invertible affine transformation of R2, i.e. Ax = Λx + cfor x ∈ R2, where Λ is linear and invertible on R2 and c ∈ R2. Define the associatedoperator C = CA on S(R2) by Cg = g A. Then

‖Cg‖L2(Sw) =1√detΛ

‖g‖L2(w)

for any w ∈ A2.

Proposition 3.17. For every f ∈ S, M ∈ Z and w ∈ A2, the sequence (K1M,Nf)N∈ZM

is a Cauchy sequence in L2(w).

Proof. We have, for N1 < N2 ∈ ZM ,

‖K1M,N1

f −K1M,N2

f‖L2(w) =

∫

R2

∣∣∣ 1

4M

∫

[0,2M )2

N2−1∑

n=N1

(R2n

t f)(x) dt∣∣∣2w(x) dx

1/2

.

By the Minkowski’s integral inequality,

‖K1M,N1

f −K1M,N2

f‖L2(w) 61

4M

∫

[0,2M )2

∫

R2

∣∣∣N2−1∑

n=N1

(R2nt f)(x)

∣∣∣2w(x) dx

1/2

dt.

From (3.23) and by using the Minkowski’s integral inequality again we get

(∫

R2

∣∣∣N2−1∑

n=N1

(R2nt f)(x)

∣∣∣2w(x) dx

1/2

61

2π

∫ 2π

0

∫

R2

∣∣∣∣[(Uψ

N2−1∑

n=N1

P2nt U−1

ψ )f](x)

∣∣∣∣2

w(x) dx

1/2

dψ,


where(. . .

)1/2

=

∫

R2

∣∣∣∣[ N2−1∑

n=N1

P2n

t (U−1ψ f)

](y)

∣∣∣∣2

w(Oψy) dy

1/2

=

∣∣∣∣∣∣

∣∣∣∣∣∣

N2−1∑

n=N1

P2nt (U−ψf)

∣∣∣∣∣∣

∣∣∣∣∣∣L2(U−ψw)

=

∣∣∣∣∣∣

∣∣∣∣∣∣

(τt

N2−1∑

n=N1

P2n

0 τ−t)(U−ψf)

∣∣∣∣∣∣

∣∣∣∣∣∣L2(U−ψw)

=

∣∣∣∣∣∣

∣∣∣∣∣∣

( N2−1∑

n=N1

P2n

0

)(τ−tUψf)

∣∣∣∣∣∣

∣∣∣∣∣∣L2(τ−tUψw)

.

The last inequality is due to Lemma 3.16. To summarize, we so far proved that

‖K1M,N1

f −K1M,N2

f‖L2(w)

61

4M

∫

[0,2M )2

1

2π

∫ 2π

0

∣∣∣∣∣∣

∣∣∣∣∣∣

( N2−1∑

n=N1

P2n

0

)(τ−tUψf)

∣∣∣∣∣∣

∣∣∣∣∣∣L2(τ−tUψw)

dψ dt.(3.25)

Verifying the Cauchy condition means proving that

limN2→−∞

supN1∈ZN2

‖K1M,N1

f −K1M,N2

f‖L2(w) = 0.

Since for any integrable sequence (fk)k on any space we clearly have sup∫|fk| 6∫

sup |fk|, we get from (3.25) that

supN1∈ZN2

‖K1M,N1

f −K1M,N2

f‖L2(w)

61

4M

∫

[0,2M )2

1

2π

∫ 2π

0sup

N1∈ZN2

∣∣∣∣∣∣

∣∣∣∣∣∣

( N2−1∑

n=N1

P2n0

)(τ−tUψf)

∣∣∣∣∣∣

∣∣∣∣∣∣L2(τ−tUψw)

dψ dt.(3.26)

Certainly, we would now like to use the dominated convergence theorem and putlimN2→−∞ inside the integral. For that purpose we need to know that functions

γN2(ψ, t) := supN1∈ZN2

∣∣∣∣∣∣

∣∣∣∣∣∣

( N2−1∑

n=N1

P2n0

)(τ−tUψf)

∣∣∣∣∣∣

∣∣∣∣∣∣L2(τ−tUψw)

have a L1(dψ dt) majorant which is independent on N2. In order to show that weinvoke a result by Treil and Volberg [78, Theorem 5.2] stating, roughly, that a (one-dimensional) Haar system is a Riesz basis in L2(w) when w ∈ A2. Recall that acomplete system of vectors en in a Hilbert space is called a Riesz basis if

∣∣∣∣∣

∣∣∣∣∣∑

n

λnen

∣∣∣∣∣

∣∣∣∣∣ ∼(∑

n

|λn|2‖en‖2)1/2

for all finitely supported sequences λn of coefficients.


Lemma 3.18. Let w ∈ A2, g ∈ S and

gN1,N2 =( N2−1∑

n=N1

P2n0

)g.

Then

limN2→−∞

supN1∈ZN2

‖gN1,N2‖L2(w) = 0.

Proof. Since∑N2−1

n=N1P2n0 is a classical martingale transform (in R2), we can apply (a

two-dimensional analogue of) the Wittwer’s result [82], see Theorems 3.28 and 3.2, andconclude that

‖gN1,N2‖L2(w). [w]A2‖g‖L2(w), (3.27)

where the implied constants are independent of N1, N2.On the other hand, we can write

gN1,N2 =

N2−1∑

n=N1

∑

Q∈G2n0

〈g, h0Q〉h0Q.

Assume for a moment that g is of compact support. Then the sum above is finite and

we may use the above-noted fact that Haar system hjQQ,j is a Riesz basis. We get

‖gN1,N2‖L2(w) &

N2−1∑

n=N1

∑

Q∈G2n0

|〈g, h0Q〉|2‖h0Q‖2L2(w)

1/2

. (3.28)

Together (3.27) and (3.28) yield convergence of the series

∞∑

n=−∞

∑

Q∈G2n0

|〈g, h0Q〉|2‖h0Q‖2L2(w),

which thus satisfies the Cauchy condition, in particular

limN2→−∞

supN1∈ZN2

N2−1∑

n=N1

∑

Q∈G2n0

|〈g, h0Q〉|2‖h0Q‖2L2(w) = 0. (3.29)

Finally, Haar system’s being a Riesz basis also means that

‖gN1,N2‖L2(w) .

N2−1∑

n=N1

∑

Q∈G2n0

|〈g, h0Q〉|2‖h0Q‖2L2(w)

1/2

,

which in view of (3.29) finishes the proof of the lemma in the case of compactly sup-ported g.

When g ∈ S, use that C∞c is a dense subspace of L2(w), cf. [61, p. 93 bottom], and

combine with (3.27): start with gN1,N2 = (g − f)N1,N2 + fN1,N2 . Then

limN2→−∞

supN1∈ZN2

‖gN1,N2‖L2(w) 6 ε+ 0

for every ε > 0.


We continue with the proof of Proposition 3.17. From (3.27) we get

γN2(ψ, t). [τ−tUψw]A2‖τ−tUψf‖L2(τ−tUψw)

. [w]A2‖f‖L2(w).

Therefore we may use the dominated convergence theorem and conclude from (3.26)that

limN2→−∞

supN1∈ZN2

‖K1M,N1

f −K1M,N2

f‖L2(w)

61

4M

∫

[0,2M )2

1

2π

∫ 2π

0lim

N2→−∞γN2(ψ, t)dψ dt.

By the previous lemma, this expression is equal to zero.

It is straightforward to show that if (ϕN )N is Cauchy sequence in L2 and convergesto ϕ pointwise, then it also converges to ϕ in L2. [True: ϕN Cauchy in L2, thereforeϕN → ψ in L2, therefore (ϕNk)k → ψ almost everywhere. At the same time we know(ϕNk)k → ϕ. Hence ψ = ϕ almost everywhere.] This gives the following.

Corollary 3.19. For every M ∈ Z and f ∈ S we have limN→−∞K1M,Nf = K1

Mf in

L2(w).

The preceeding proof also shows that for any N1, N2 ∈ Z, N1 > N2,

‖K1N1,N2

f‖L2(w) . [w]A2‖f‖L2(w) ,

with the implied constant independent of M,N, f,w. Indeed, with any M > N1 wemay write

K1N1,N2

= K1M,N2

−K1M,N1+1.

Now (3.25) and Lemma 3.16 imply

‖K1N1,N2

f‖L2(w) 6

∣∣∣∣∣∣

∣∣∣∣∣∣

( N2∑

n=N1

P2n0

)f

∣∣∣∣∣∣

∣∣∣∣∣∣L2(w)

,

which is dominated by C[w]A2‖f‖L2(w), according to Theorem 3.2.Hence the previous corollary immediately yields

Corollary 3.20. For every M ∈ Z and f ∈ S we have

‖K1Mf‖L2(w) . [w]A2‖f‖L2(w) .

Because the “Cauchy differences” of the sequence (K1M,Nf)N∈ZM look exactly as the

“Cauchy differences” of the sequence (K1Mf)M∈ZM , effectively the same proof works for

K1:

Corollary 3.21. For every f ∈ S we have limM→−∞K1Mf = K1f in L2(w) and

therefore‖K1f‖L2(w) . [w]A2‖f‖L2(w) .

And the same can be done for any r > 0:

Corollary 3.22. For every r > 0, f ∈ S and w ∈ A2 we have

‖Krf‖L2(w). [w]A2‖f‖L2(w) .

where the implied constants do not depend on r, f, w.

Finally, this yields the weighted estimate of S:


Corollary 3.23. For any w ∈ A2 we have

‖S‖B(L2(w)) . [w]A2 .

Proof. For every f ∈ S, another application of Minkowski’s integral inequality gives

‖Sf‖L2(w) 6

∫ 2

1‖Krf‖L2(w)

dr

r.

The previous corollary implies ‖Sf‖L2(w) . [w]A2‖f‖L2(w). Since S is dense in L2(w)

[61], this allows an extension of S onto the whole of L2(w) with the same bound.

Now Proposition 3.14 implies Theorem 3.1 and thus Theorem 2.29, as explained atthe beginning of this section.

3.5. Sharpness. We still need to show that this estimate is sharp, in the same sensethe Wittwer’s estimate (Theorem 3.2) was. For that purpose we shall need the followingauxiliary result, which can be proved by direct simple calculations.

Exercise 3.24. Let α ∈ (−2, 2) and define wα : R2 → [0,∞) by wα(x) = |x|α . Then

[wα]A2 ∼ 1

2− |α| .

Now we are ready to prove the sharpness. Calculations of this kind have alreadybeen made for other singular integral operators (e.g. [21]) but not for T , so we includethis calculation for the sake of completeness.

Proposition 3.25. Let φ : R → (0,∞) grow at infinity more slowly than a linearfunction, i.e.

limx→∞

φ(x)

x= 0 .

Then there is a weight w ∈ A2 such that

‖T‖B(L2(w)) > φ([w]A2) . (3.30)

Proof. For α ∈ (−2, 2) define w(z) = wα(z) = |z|α as in Exercise 3.24. Denote

E = reiϕ ; 0 < r < 1, 0 < ϕ < π/2,X = −E, and f(z) = |z|−αχE(z).

We begin by

‖f‖2L2(w) =

∫∫

R2

|f(x, y)|2 w(x, y) dx dy

=

∫∫

E(x2 + y2)−α(x2 + y2)α/2 dx dy

=

∫ π/2

0

∫ 1

0r−2α+α r dr dϕ

=π

2· 1

2− α.

(3.31)

In particular, f ∈ L2(w) and therefore also Tf ∈ L2(w). Let us estimate its norm frombelow:

‖Tf‖2L2(w) =

∫

R2

|(Tf)(ξ)|2 w(ξ) dξ >∫

X|(Tf)(ξ)|2 w(ξ) dξ . (3.32)


When z ∈ X we have

(Tf)(z) = − 1

π

∫

E

f(u)

(z − u)2du ,

i.e. the integral converges in the ordinary sense (not only as a principal value integral),because z ∈ X is on a positive distance from supp f ⊂ E. Therefore (3.32) implies

‖Tf‖2L2(w) >

∫∫

X

∣∣∣∣∣1

π

∫∫

E

(s2 + t2)−α/2

[(x− s) + i(y − t)]2ds dt

∣∣∣∣∣

2

(x2 + y2)α/2 dx dy .

We use the identity1

[a+ ib]2=

a2 − b2

[a2 + b2]2− i

2ab

[a2 + b2]2

to estimate the square of modulus of the inner integral (i.e. the one over E) from belowby the square of its imaginary part, that is, by

(∫∫

E

2(x− s)(y − t)

[(x− s)2 + (y − t)2]2(s2 + t2)−α/2 ds dt

)2

.

Since (x, y) ∈ X and (s, t) ∈ E, we have

(x− s)(y − t) > xy .

The bound for the denominator comes from the triangle inequality:

[(x− s)2 + (y − t)2]2 = |(x, y)− (s, t)|4 6 (|(x, y)|+ |(s, t)|)4 .Therefore,

∣∣∣∣∣

∫∫

E

(s2 + t2)−α/2

[(x− s) + i(y − t)]2ds dt

∣∣∣∣∣

2

> 4x2y2

(∫∫

E

(s2 + t2)−α/2

(|(x, y)| + |(s, t)|)4 ds dt)2

= π2x2y2(∫ 1

0

r−α

(|(x, y)| + r)4r dr

)2

.

By taking u = r/|(x, y)| we can continue with

π2x2y2(|(x, y)|−α−2

∫ 1/|(x,y)|

0

u1−α

(1 + u)4du

)2

>

>π2x2y2

(x2 + y2)α+2

(∫ 1

0

u1−α

(1 + 1)4du

)2

=π2

256(2 − α)2· x2y2

(x2 + y2)α+2.

We proved that

‖Tf‖2L2(w) >1

256 (2 − α)2

∫∫

Xx2y2 (x2 + y2)α/2−α−2 dx dy =

Cl(2− α)3

, (3.33)

where

Cl =1

256

∫ π/2

0cos2 ϕ sin2 ϕdϕ =

1

256· 12B

(3

2,3

2

)=

π

4096.

Thus we summarize (3.33) and (3.31) as

‖T‖B(L2(w)) >‖Tf‖L2(w)

‖f‖L2(w)&

1

2− α.

In order to finish the proof we need to find α that satisfies the inequality

1

2− α> C φ([wα]A2)


for a given constant C. It suffices to show that

limα→2−

(2− α)φ([wα]A2) = 0 ,

which however follows from Exercise 3.24 and the assumption on φ.

3.6. Bellman function and Wittwer’s theorem. Let D = D[0, 1] be the collectionof all dyadic intervals in [0, 1). As remarked in Section 3.1, the set 1 ∪ hI ; I ∈ Dis a complete orthonormal system in L2([0, 1]) [44, Theorem 5.4.6]. Let σ = (σI)I∈D bea sequence in ∆. Define on S(R) the operator

Tσf =∑

I∈D

σI〈f, hI〉hI .

Define also a dyadic A2 weight: this is a positive function w ∈ L1([0, 1]), such that

[w]dyadicA2:= sup

I∈D〈w〉I〈w−1〉I

is finite.The following theorem was proven by Wittwer [82] who based her approach on the

paper by Nazarov, Treil and Volberg [65]. We will give another proof, following Volberg[81, Theorem 2.10]. A key element will be a concrete Bellman function.

First we need the following simple lemma introducing “Haar functions” which aremore suitable to the weighted space L2(w) (see [64, Section 2.1]). Denote

dλ = w dm .

Exercise 3.26. For any interval I and any positive w ∈ L1loc(R) there exists a unique

function (called hwI ) satisfying the properties:

(i) hwI is a linear combination of functions χI− and χI+;(ii) ∫

IhwI dλ = 0 ;

(iii) ∫

I|hwI |2 dλ = 1 ;

(iv)

hwI

∣∣∣I+> 0 .

Functions hwI ; I ∈ D form an orthonormal system in L2(λ).

Since hwI and χI are (linearly independent) linear combinations of functions χI− ,χI+, and so is hI , there exist unique real constants αI , βI such that

hI = αIhwI + βI

χI√|I|

. (3.34)

Observe that αI cannot be zero, since otherwise hI would be constant on I. Certainly,we can extract αI , βI by using Exercise 3.26:

• compute the L2(w) norm of (3.34):

〈w〉I = ‖hI‖2L2(w) = α2I + β2I 〈w〉I ;


• divide (3.34) by√|I| and integrate with respect to w dm to get

1

2∆Iw = βI〈w〉I .

Here ∆Iw := 〈w〉I+ − 〈w〉I− .Let us summarize the above findings.

Lemma 3.27. The constants αI , βI from (3.34) obey estimates

0 < |αI | 6√

〈w〉I and |βI | 6|∆Iw|〈w〉I

.

Theorem 3.28.

supσ

‖Tσ‖B(L2(w)) . [w]dyadicA2.

Proof. We want to prove ‖Tσf‖L2(w) . [w]dyadicA2‖f‖L2(w) for f ∈ L2(w) and w ∈ A2.

Replacing f by fw−1 enables us to rephrase this inequality as

‖Tσ(fw−1)‖L2(w). [w]dyadicA2‖f‖L2(w−1)

for f ∈ L2(w−1) and w ∈ A2, which is (take w−1 in place of w) the same as

‖Tσ(fw)‖L2(w−1) . [w]dyadicA2‖f‖L2(w) for f ∈ L2(w), w ∈ A2 .

Since the duality in L2(w−1) means that

‖ϕ‖L2(w−1) = sup‖g‖L2(w−1)

|〈ϕ, g〉L2(w−1)| ,

we see that the theorem is equivalent to proving

|〈Tσ(fw), gw−1〉|. [w]dyadicA2‖f‖L2(w)‖g‖L2(w−1) .

We emphasize that in the last inequality we take the usual (unweighted) scalar product,i.e. with respect to the Lebesgue measure.

Since functions hI ; I ∈ D are orthonormal in (the unweighted) L2, we get

〈Tσ(fw), gw−1〉 =∑

I

σI〈fw, hI〉〈gw−1, hI〉 .

By using the assumption on σI , we see that our task will be completed once we prove

∑

I∈D

|〈fw, hI〉| |〈gw−1, hI〉| . [w]dyadicA2‖f‖L2(w)‖g‖L2(w−1) .

We use (3.34) and Lemma 3.27 to estimate

∑

I∈D

|〈fw, hI 〉| |〈gw−1, hI〉| 6 I + II + III + IV,

where


I =∑

I∈D

|〈fw, hwI 〉|√

〈w〉I |〈gw−1, hw−1

I 〉|√

〈w−1〉I

II =∑

I∈D

|〈fw〉I ||∆Iw|〈w〉I

√|I| · |〈gw−1, hw

−1

I 〉|√

〈w−1〉I

III =∑

I∈D

|〈fw, hwI 〉|√

〈w〉I · |〈gw−1〉I ||∆Iw

−1|〈w−1〉I

√|I|

IV =∑

I∈D


√|I| · |〈gw−1〉I |

|∆Iw−1|

〈w−1〉I√

|I| .

Estimate of I. We obviously have

I 6

√[w]dyadicA2

√∑

I∈D

|〈fw, hwI 〉|2√∑

I∈D

|〈gw−1, hw−1

I 〉|2

=√

[w]dyadicA2

√∑

I∈D

|〈f, hwI 〉L2(w)|2√∑

I∈D

|〈g, hw−1

I 〉L2(w−1)|2

6

√[w]dyadicA2

‖f‖L2(w)‖g‖L2(w−1).

For the second inequality we applied Exercise 3.26.

Estimate of II, III, IV . Since III is symmetric to II, we only consider II and IV .

Fix α ∈ (0, 1/2) and introduce

µI := 〈w〉αI 〈w−1〉αI( |∆Iw|2

〈w〉2I+

|∆Iw−1|2

〈w−1〉2I

)|I| .

Observe that

max

|∆Iw|〈w〉I

√|I| , |∆Iw

−1|〈w−1〉I

√|I|

6 〈w〉−α/2I 〈w−1〉−α/2I

√µI . (3.35)

Now suppose that p ∈ (1, 2). Then Holder’s inequality gives

|〈fw〉I | 6 〈|f |pw〉1/pI 〈w〉1−1/pI . (3.36)

Introduce measures dλ = w dm and dλ∗ = w−1dm. Then (3.36) can be rewritten as

|〈fw〉I | 6 〈w〉I(

1

λ(I)

∫

I|f |p dλ

)1/p

6 〈w〉I infx∈I

(Mdλ |f |p(x)

)1/p, (3.37)

where for any positive measure µ on [0, 1] we define its dyadic (weighted) maximalfunction by

Mdµϕ(x) := sup

I∈D

I∋x

1

µ(I)

∫

I|ϕ| dµ .

We summarize (3.35) and (3.37) into


√|I| 6 〈w〉1−α/2I 〈w−1〉−α/2I inf

I

(Mdλ |f |p

)1/p√µI .


Consequently

II 6∑

I∈D

〈w〉1−α/2I 〈w−1〉1−α/2I

infI(Mdλ |f |p

)1/p

〈w−1〉1/2I

√µI · |〈gw−1, hw

−1

I 〉|

6([w]dyadicA2

)1−α/2∑

I∈D

infI(Mdλ |f |p

)1/p

〈w−1〉1/2I

√µI · |〈gw−1, hw

−1

I 〉|

6([w]dyadicA2

)1−α/2√√√√∑

I∈D

infI(Mdλ |f |p

)2/p

〈w−1〉IµI

√∑

I∈D

|〈gw−1, hw−1

I 〉|2

6([w]dyadicA2

)1−α/2‖g‖L2(w−1)

√√√√∑

I∈D

infI(Mdλ |f |p

)2/p

〈w−1〉IµI

and similarly

IV 6([w]dyadicA2

)1−α∑

I∈D

infI

(Mdλ |f |p

)1/pinfI

(Mdλ∗ |g|p

)1/pµI

6([w]dyadicA2

)1−α∑

I∈D

infI

(Mdλ |f |p

)1/p (Mdλ∗ |g|p

)1/pµI .

Lemma 3.29. The sequence µII∈D is a Carleson sequence with the Carleson constant

bounded by .

([w]dyadicA2

)α=: Cα(w). This means that, for every I ∈ D,

∑

J∈D

J⊂I

µJ . αCα(w)|I|.

This lemma will be proven at the end of the section. We couple it with another result,which follows directly from the dyadic version of the Carleson embedding theorem, see[63], [65] (for the weighted version and a nicely written proof) and [66] for a proofestablishing the optimal constant.

Lemma 3.30. Suppose (αJ)J∈D defines a Carleson measure with Carleson constantbounded by B > 0. If F,G are positive measurable functions on (0, 1), then

∑

J∈D

(infJF )αJ . B

∫

(0,1)F

∑

J∈D

infJ G

〈w〉JαJ . B

∫

(0,1)

G

w.

We apply the above lemmas with F =(Mdλ |f |p

)1/p (Mdλ∗ |g|p

)1/pandG =

(Mdλ |f |p

)2/p.

We get

II 6([w]dyadicA2

)1−α/2 ([w]dyadicA2

)α/2‖g‖L2(w−1)

√∫

(0,1)

(Mdλ |f |p

)2/pdλ

= [w]dyadicA2‖g‖L2(w−1)‖Md

λ |f |p‖1/p

L2/p(w)

. [w]dyadicA2‖g‖L2(w−1)‖|f |p‖1/pL2/p(w)

= [w]dyadicA2‖g‖L2(w−1)‖f‖L2(w).


Similarly,

IV 6([w]dyadicA2

)1−α ([w]dyadicA2

)α ∫

(0,1)

(Mdλ |f |p

)1/p (Mdλ∗ |g|p

)1/pdm

6 [w]dyadicA2

√∫

(0,1)

(Mdλ |f |p

)2/pdλ

√∫

(0,1)

(Mdλ∗ |g|p

)2/pdλ∗

. [w]dyadicA2‖f‖L2(w).‖g‖L2(w−1)

At the end we used that p < 2, i.e. 2/p > 1 and that the weighted maximal functionMdλ is bounded on Lr(λ) for any r > 1 and with a constant that does not depend on w.

See [44, Exercises 2.1.2, 2.1.12, 1.3.3]. (As acknowledged in [72], this fact was learnedfrom [26].) Note that dλ = w dm is a doubling measure when w belongs to the (usual)A2: by using the ad hoc notation 2I = (x − 2ε, x + 2ε) for I = (x − ε, x + ε), we seethat

λ(2I) = 2|I|〈w〉2I 6 2|I| [w]A2

〈w−1〉2I= 4|I|2 [w]A2∫

2I w−1

6 4|I|2 [w]A2∫I w

−1= 4|I| [w]A2

〈w−1〉I6 4|I|[w]A2〈w〉I = 4[w]A2λ(I) .

We still need to prove Lemma 3.29. We do this by means of a Bellman function.

Proof of Lemma 3.29. Let Ω = (0,∞)× (0,∞) . Choose α ∈ (0, 1/2) and define thefunction b = bα : Ω → (0,∞) by b(x, y) = (xy)α. We have, at (x, y) ∈ Ω, its Hessianmatrix d2b(x, y):

−d2b(x, y) = αxαyα

[(1− α)x−2 −αx−1y−1

−αx−1y−1 (1− α)y−2

].

Obviously

−d2b(x, y) > α(1− 2α)(xy)α((dx)2

x2+

(dy)2

y2

),

meaning that for any u, v ∈ R we have⟨−d2b(x, y)

[u

v

],

[u

v

]⟩> α(1− 2α)(xy)α

(u2

x2+v2

y2

). (3.38)

Owing to the choice of α, the last expression is positive for any (x, y) ∈ Ω and anyu, v ∈ R. Moreover, if Q > 1 and xy 6 Q, then obviously b(x, y) 6 Qα.

Now fix an interval I. Denote σ = w−1. Let

a = (〈w〉I , 〈σ〉I)aj = (〈w〉Ij , 〈σ〉Ij ) ,

where j ∈ −,+ and I−, I+ are the left and the right half of I, respectively.Consider

cj : [0, 1] −→ Ω

t 7−→ (1− t)a+ taj .

and

qj = b cj : [0, 1] −→ (0,∞)

t 7−→ b((1 − t)a+ taj) .


Since qj is of order C2, we have

qj(0) − qj(1) = −q′j(0)−∫ 1

0(1− t)q′′j (t) dt . (3.39)

The chain rule gives

q′j(t) = 〈∇b(cj(t)), aj − a〉q′′j (t) =

⟨d2b(cj(t))(aj − a), aj − a

⟩.

(3.40)

Hence from (3.40) and (3.38) it follows that −q′′j (t) > 0, therefore (3.39) gives

qj(0) − qj(1) > −q′j(0)−1

2

∫ 1/2

0q′′j (t) dt . (3.41)

Now estimate −q′′j (t) again, but in a more delicate way. Namely, we use again (3.40)

and (3.38), but now observe that for t away from 1, say, if 0 6 t 6 1/2 as in the integralabove, the first component of cj(t) cannot be smaller than its own half, i.e. (1/2)〈w〉I :indeed, either draw a picture or estimate

(1− t)〈w〉I + t〈w〉Ij > (1− t)〈w〉I >1

2〈w〉I .

On the other hand, since

〈w〉I =〈w〉I− + 〈w〉I+

2, (3.42)

it is always true that the first component of cj(t) is not larger than 2〈w〉I . To summa-rize, for 0 6 t 6 1/2, the first component of cj(t) is comparable with 〈w〉I , while thesecond one is comparable with 〈σ〉I .

Therefore, when 0 6 t 6 1/2 we get from (3.40) and (3.38)

−q′′j (t) & α(1− 2α) (〈w〉I〈σ〉I)α((〈w〉I − 〈w〉Ij )2

〈w〉2I+

(〈σ〉I − 〈σ〉Ij )2〈σ〉2I

)

& α(1− 2α)b(a)

( |∆Iw|2〈w〉2I

+|∆Iσ|2〈σ〉2I

).

For the last inequality we used again (3.42). Now we get from here and (3.41) that

qj(0)− qj(1) > −q′j(0)− Cα(1− 2α)b(a)

( |∆Iw|2〈w〉2I

+|∆Iσ|2〈σ〉2I

). (3.43)

We want to add these inequalities for j ∈ −,+. Since (a+ − a) + (a− − a) = 0, wesee from (3.40) that what remains is

q+(0)− q+(1) + q−(0) − q−(1) & α(1− 2α)b(a)

( |∆Iw|2〈w〉2I

+|∆Iσ|2〈σ〉2I

),

that is,

b(aI)−b(aI+) + b(aI−)

2& α(1− 2α) b(aI )

( |∆Iw|2〈w〉2I

+|∆Iσ|2〈σ〉2I

).

Multiply this by |I| and write temporarily γJ = |J |b(aJ ) to get

γI > γI+ + γI− + Cα(1− 2α)µI .


Finally iterate this inequality. After the first iteration we get

γI > γI++ + γI+− + γI−+ + γI−︸︷︷︸>0

+Cα(1− 2α)[µI + µI+ + µI− ] .

Similarly, after n iterations we obtain

γI & α(1− 2α)∑

J∈D

J⊆I

|J|>2−n|I|

µJ .

This is valid for all n ∈ N, therefore we conclude that

γI & α(1 − 2α)∑

J∈D

J⊆I

µJ .

Since γI = |I|b(aI) 6 |I|Cα(w), where Cα(w) is as in the formulation of the lemma,this finishes the proof.

3.7. Unweighted estimates for T n by means of averaging martingale trans-

forms. Let us briefly discuss an approach to unweighted estimates of T or T n bymodifying the averaging technique from Section 3.3.

When using Proposition 3.14 to estimate T on a nonweighted space (cf. the problemof Iwaniec), one needs to consider a type of estimates of martingale transforms that isdifferent from Wittwer’s. For on the unweighted space the problem is very specific - oneneeds as small numerical constant as possible in front of the norms of the martingaletransforms; the Iwaniec conjecture comprises an equality, not an inequality, namely‖T‖B(Lp) = p∗ − 1. Fortunately, there is a sharp estimate of martingale transforms.This is a famous result of Burkholder [16, 17, 19], see also [62, Theorem 8.18]:

supσ

‖Tσ‖B(Lp) = p∗ − 1 . (3.44)

Actually, for martingale transforms on the plane we need a more general version, re-garding Hilbert-space-valued differentially subordinate pairs of martingales. Again itexists and was proven by Burkholder in [18]. It is very useful due to its generality andsharpness. We present it here for the convenience of the reader.

Lemma 3.31. Let (W,F, P ) be a probability space, Fn ; n ∈ N a filtration in F andH a separable Hilbert space. Furthermore, let (Xn,Fn, P ) and (Yn,Fn, P ) be H-valuedmartingales satisfying

‖Y0(ω)‖H 6 ‖X0(ω)‖H and ‖Yn(ω)− Yn−1(ω)‖H 6 ‖Xn(ω)−Xn−1(ω)‖H (3.45)

for all n ∈ N and almost every ω ∈ W. Then for any p ∈ (1,∞)

‖Yn‖p 6 (p∗ − 1)‖Xn‖p .The constant p∗ − 1 is sharp.

The property (3.45) is called differential subordination.

Theorem 3.32. For any Q ∈ L and ∗ ∈ 0,+,− let σ∗Q be an arbitrary unimodularcomplex number. Define the operator

Tσf :=∑

Q∈L

[σ0Q〈f, h0Q〉h0Q + σ+Q〈f, h+Q〉h+Q + σ−Q〈f, h−Q〉h−Q

].

Then ‖Tσ‖p 6 p∗ − 1. This estimate is sharp.


As we saw in the weighted case, when dealing with averaging processes it is desirableto make the objects of averaging have the same norm estimate. In the unweightedcase [31] this was achieved by a construction which permitted a uniform applicationof the sharp Burkholder’s result on the norms of general differentially subordinatedmartingales. Doing so all the martingale transforms we considered were, regardless oftheir symbols, on Lp bounded by p∗ − 1.

It turns out, however, that the averaging method adjusted for powers T n also callsfor Haar systems defined on rectangles. When using it to obtain as precise estimatesas possible of T on Lp, we can further extend and sharpen our method by consideringdifferent sets of coefficients, and also different Haar systems - ones built on rectangles,parallelograms, triangles and yet more exotic shapes (such as “L-shaped” tiles). See[31, 32]. The problem we encounter is that a certain amount of precision is lost duringthe averaging process. The best constant obtained so far in this way is approximately2·007 [32, p. 4]. That was at the time (2004) when the best known constant was 2.

Another problem might be the fact that the Burkholder’s estimate is sharp whenconsidering all martingale transforms. Yet we typically average just one operator, andits translations, dilations and rotations should all obey pretty much the same estimates(as outlined in Section 3.2).

However, by merging the averaging method and the Burkholder estimate we managedto prove the following result [31]:

Theorem 3.33. There is N0 ∈ N such that for all n > N0 and 1 < p <∞,

‖T n‖p 6 2.716n(p∗ − 1) . (3.46)

One may compare this estimate to the estimate of κn(p) in Proposition 4.3 and withConjecture 4.10.

It seems to be quite difficult to derive the numerical value of ‖T n‖L1→L1,∞ in [24],[47]. Estimate (3.46) hints that these norms may be bounded by en.

4. Unweighted estimates for T n

The main objective of this section is to prove the results announced in Section 2.7.2.We will of course separately treat estimates from below and above.

4.1. Lower estimates. Take n ∈ N, p > 2, α < 1/p and define f = fn,α by

fn,α(z) = zn|z|−2αgn(|z|) .Here gn : [0,∞) → [0, 1] is a sufficiently smooth, rapidly decaying function satisfyinggn(x) = 1 for x ∈ [0, 1]. Its primary role is to extend zn|z|−2α from the unit disc to thewhole plane in a way such that the outcome lies in W n,p. For example, one can take

gn(x) =

1 ; 0 6 x 6 1

exp(−(x− 1)n) ; x > 1.

This is a somewhat natural extension of the example considered by Lehto [58] whengiving the lower Lp estimates on the Ahlfors-Beurling operator T ; see also [8]. Notehowever that f is not a radial stretch function, since it involves a power of z. We useit to prove lower estimates for the Lp norms of powers of T .

Exercise 4.1. Function f belongs to the Sobolev space W n,p(C).


This ensures that T n(∂nf) = ∂nf . Denote by q the conjugate exponent of p, i.e.q = p/(p − 1). Recall that the Pochhammer symbol (a)n = a (a + 1) · . . . · (a + n − 1)was introduced in (2.34). Define

κn(p) =(1/q)n(1/p)n

. (4.1)

Proposition 4.2 ([28]). We have ‖T n‖p > κn(p), more precisely,

limα→1/p

‖∂nfn,α‖Lp(C)‖∂nfn,α‖Lp(C)

= κn(p) .

Proof. Write

‖∂nfn,α‖pLp(C)‖∂nfn,α‖pLp(C)

=

∫

∆|∂nf |p +

∫

∆c|∂nf |p

∫

∆|∂nf |p +

∫

∆c|∂nf |p

.

The integrals over ∆ can be explicitly calculated. Recall that gn(|z|) = 1 for z ∈ ∆.Consequently, for z ∈ ∆\0,

∂nfn,α(z) = (n− α)(n − 1− α) . . . (1− α)|z|−2α

and

∂nfn,α(z) = (−α)(−α − 1) . . . (−α− n+ 1)(zz

)n|z|−2α .

Hence ∫

∆|∂nfn,α(z)|pdA(z) =

n∏

k=1

(k − α)p ·∫

∆|z|−2αp dA(z)

=

n∏

k=1

(k − α)p · 2π∫ 1

0r1−2αp dr

=π

1− αp

n∏

k=1

(k − α)p

and ∫

∆|∂nfn,α(z)|pdA(z) =

π

1− αp

n−1∏

k=0

(k + α)p .

Therefore

‖∂nfn,α‖pLp(C)‖∂nfn,α‖pLp(C)

=

πn∏

k=1

(k − α)p + (1− αp)

∫

∆c|∂nfn,α|p

π

n−1∏

k=0

(k + α)p + (1− αp)

∫

∆c|∂nfn,α|p

. (4.2)

As for the integrals over ∆c, it suffices to notice that they stay bounded as α→ 1/p.Indeed, for D ∈ ∂, ∂ and z ∈ ∆c we have

|Dnfn,α(z)| 6 P (n, α)|z|M e−(|z|−1)n .

Here P is a polynomial in two variables, whereasM > 0 is independent of α. Thus whenwe raise this to the p−th power we can estimate the integrands from above uniformlyfor α close to 1/p. The majorant is integrable. As a result we can apply the dominatedconvergence theorem and send α → 1/p. The integrals over ∆c will stay away from


infinity. Hence, owing to the factor 1 − αp in front of them, they will disappear from(4.2). So the result will be

n−1∏

k=0

(k + 1− 1/p

k + 1/p

)p.

By taking the p−th root we get κn(p).

The asymptotic behaviour of κn(p) is described by the following proposition.

Proposition 4.3 ([31, 28]). We have κn(p) ∼ n1−2/p(p− 1). More precisely, for everyn ∈ N and p > 2,

0·964 6κn(p)

n1−2/p(p− 1)6 1 , (4.3)

the upper estimate being sharp.

Proof. Denote δ = 1− 2/p. Then

κn(p) =

n−1∏

k=0

2k + 1 + δ

2k + 1− δ,

therefore (4.3) is equivalent to

0·964 6 n−δn−1∏

k=1

2k + 1 + δ

2k + 1− δ6 1 , ∀δ ∈ [0, 1),∀n ∈ N\1 . (4.4)

First let us prove the lower estimate. So our aim is to find the best (i.e. largest)constant C in the inequality

logC 6

n−1∑

k=1

log2k + 1 + δ

2k + 1− δ− δ log n .

Fix k ∈ N. The function

g(δ) = log2k + 1 + δ

2k + 1− δ

is convex on (0, 2k + 1): indeed,

g′(δ) =2(2k + 1)

(2k + 1)2 − δ2

obviously increases with δ. Therefore g(δ) > g(0) + δg′(0), that is,

g(δ) >2δ

2k + 1.

Hence C > C1, where C1 is the best (i.e. largest) constant in the inequality

logC1 6 δ

( n−1∑

k=1

2

2k + 1− log n

), 0 6 δ < 1, n ∈ N\1 . (4.5)

Denote the expression in the parentheses above by ψ(n).

Exercise 4.4. Function ψ is decreasing and negative on N\1.


Therefore, since ψ is negative on N\1, we can in our search for the best C1 in (4.5)get rid of δ by replacing it with the supremum of all of its admissible values (i.e. 1).And since ψ is also decreasing, we conclude

logC1 = infn∈N\1

ψ(n) = limn→∞

ψ(n) . (4.6)

Now introduce

an =

n∑

k=1

1

k

and bn = an − log n . Then

ψ(n) = 2

(1

3+

1

5+

1

7+ . . .+

1

2n− 1

)− log n

= 2

(a2n−1 − 1− an−1

2

)− log n

= 2b2n−1 − bn−1 + log

(4 +

1

n(n− 1)

)− 2 .

It is a classical fact that limn→∞ bn = γ, where γ ≈ 0·5772 . . . is the Euler–Mascheroniconstant, therefore limn→∞ ψ(n) = γ + 2 log 2− 2. By (4.6), C1 = 4eγ−2 ≈ 0·964.

Now let us turn to the upper estimate in (4.4). Because

nδ =

n−1∏

k=1

(k + 1

k

)δ,

it is enough to show that

2k + 1 + δ

2k + 1− δ6(k + 1

k

)δ, ∀ k ∈ N , δ ∈ [0, 1) .

When δ = 0 the inequality clearly holds, so we may assume that δ ∈ (0, 1). Write2k + 1 = δ/w for some w ∈ (0, δ). Then the equality to prove becomes

1 + w

1− w6(δ + w

δ − w

)δ, ∀ 0 < w < δ < 1 .

Verifying this is left as an exercise.

Exercise 4.5. If 0 < w < δ < 1 then

1− w

1 + w

(δ + w

δ − w

)δ> 1 .

Remark 4.6. Recently D. Kalaj (personal communication, November 2016) calculatedthe optimal lower constant in (4.3). It turns out to be

minx∈[0,1/2]

Γ(3/2− x)

Γ(3/2 + x)

which is approximately 0·985796.

Remark 4.7. Proposition 4.3 can be formulated in a slightly more symmetric way,namely

(a)n(b)n

∼ ana

bnb,

where a = 1/q and b = 1/p. Note that “without fractions” this is far from true, i.e.clearly (a)n 6∼ ana.


As an immediate consequence of Propositions 4.2 and 4.3, we have the followingestimate:

Corollary 4.8. We have ‖T n‖p & n1−2/p∗(p∗ − 1).

4.2. Upper estimates. Here we will show (Theorem 2.38) that the lower asymptoticestimate from Corollary 4.8 has its upper counterpart of the same order, which is validnot only for T n but for all Hk

C, i.e. even when k is odd. Our plan is to single out

constants in the weak (1,1) and strong (2,2) inequalities for HkC and then interpolate.

Weak-type (1,1) estimate for singular integrals is one of the cornerstones of theCalderon-Zygmund theory. Many papers were devoted to establishing such resultsunder various hypotheses. See e.g. [35, 44, 62] for the corresponding references. Weare interested in a) precise and “simultaneous” Lp estimates of a concrete sequence ofoperators and b) the conditions that allow such estimates in our case. For example,given the oscillation factor ζ−k in our kernels (see (1.9) below), usual gradient (orLipschitz) conditions would add another factor of k and thus prevent us from gettingsharp Lp results in terms of k; see e.g. [62, Proposition 7.4] or [44, Theorem 4.3.3].A theorem adequate for our purpose, Theorem 4.9 below, was proven independentlyby Christ, Rubio de Francia [24] and Hofmann [47]. (It should be acknowledged thatboth of these references were brought to the attention of the authors of [31] by MichaelChrist.) The formulation in [47] is explicit about the behaviour of the estimates. Wepresent it here for the reader’s convenience. Note that it is valid for kernels far moregeneral (“rough”) than ours, since no “smoothness” condition is assumed.

Theorem 4.9 ([47]). Suppose Ω ∈ Lq(S1) for some q > 1 and∫S1 Ω = 0. For any

ε > 0 define the operator Tε associated with Ω and ε by

Tεf(z) =

∫

|ζ|>εf(z − ζ)

Ω(ζ/|ζ|)|ζ|2 dm(ζ). (4.7)

Then, for any α, ε > 0 and f ∈ S,

m|Tεf | > α .‖Ω‖qα

‖f‖1.

Proof of Theorem 2.38. Fix k ∈ Z\0. First consider the case when 1 < p 6 2.By (1.10), each Hk

C is an isometry on L2.

Now let us address the weak (1,1) inequality. Fix also α, ε > 0 and f ∈ L1. Let HkC,ε

be the “ε-truncated” version of HkC, in the sense of (4.7). Denote Mε = |Hk

C,εf | > αand M = |Hk

Cf | > α.A well-known result on the almost everywhere convergence of homogeneous singular

integrals (see, for example, [62, Corollary 7.11]) implies that HkCf = limε→0H

kC,εf

almost everywhere. This leads to χM 6 lim infε→0 χMε . Consequently, by Fatou’slemma,

m(M) =

∫

C

χM dm 6

∫

C

lim infε→0

χMε dm 6 lim infε→0

∫

C

χMε dm.‖Ωk‖2α

‖f‖1.

In the last inequality we applied Proposition 4.9, in particular the fact that the estimatesthere are independent of ε. So we proved

m|Hk

Cf | > α

.|k|α

‖f‖1. (4.8)


Now everything is set for interpolation. We actually do it as in [44, Exercise 1.3.2].Choose p ∈ (1, 2) and r ∈ (1, p). Let N1,k be the weak (1,1) constant for Hk

C. By theMarcinkiewicz interpolation theorem, applied to 1 < r < 2,

‖HkC‖r 6 2r1/r

(1

r − 1+

1

2− r

)1/r

N2−rr

1,k . (4.9)

The Riesz-Thorin interpolation theorem we apply to r < p < 2 and obtain

‖HkC‖p 6 ‖Hk

C‖r

2−r· 2−pp

r (4.10)

Together (4.9) and (4.10) give

‖HkC‖p .

[1

(r − 1)(2 − r)

] 1p

N2p−1

1,k .

By choosing r = (p + 1)/2 and using that (p − 1)1/p ∼ p − 1 for p ∈ (1, 2), we finallyarrive at

‖HkC‖p .

N2p−1

1,k

p− 1. (4.11)

Recall that, according to (4.8), we have N1,k . |k|. This proves the theorem for p ∈(1, 2).

When p > 2 use the behaviour of HkC in the Fourier domain [5, Section 4.2] and

duality: with mk(ζ) = (|ζ|/ζ)k we have

〈HkCf, g〉 = 〈Hk

Cf, g〉 = 〈mkf , g〉 = 〈f ,mkg〉 = 〈f ,m−kg〉 = 〈f,H−k

Cg〉 .

This completes the proof of Theorem 2.38.

Proof of Theorem 2.39. In the special case when k is even we get the upper estimatefor (integer) powers of the Ahlfors-Beurling operator: ‖T n‖p . n1−2/p∗(p∗−1), for anyn ∈ Z\0 and p > 1. Together with Corollary 4.8 this constitutes Theorem 2.39.

Proof of Theorem 2.41. When k = 2n, by combining (4.8) with Corollary 4.8 and(4.11) we also get Theorem 2.41.

4.3. Candidates for ‖T n‖p and some necessary conditions. Based on the abovefindings one may cautiously make the following supposition:

Conjecture 4.10 ([31, 28]). For p > 2 and all n ∈ N,

‖T n‖p = κn(p) =(1/q)n(1/p)n

. (4.12)

As mentioned earlier (Conjecture 2.27), this question is open even for n = 1, in whichcase it is well-known since 1982 as the Iwaniec conjecture.

Since ‖Tm+n‖p 6 ‖Tm‖p‖T n‖p the same must be true for κn(p), if (4.12) is to hold.Indeed, such is the case, which slightly reinforces our belief in (4.12):

Exercise 4.11. For m,n ∈ N and p > 2 we have

κm+n(p) 6 κm(p)κn(p) .

with the equality only when p = 2.

We have seen in Proposition 4.3 that the function (n, p) 7→ κn(p) tightly fits to the

function (n, p) 7→ n1−2/p(p− 1), which makes the latter worth investigating further.


Exercise 4.12. The analogue of Exercise 4.11 also holds for the function

γp(s) = s1−2/p(p − 1).

Actually, we have γp(s+ t) 6 γp(s)γp(t) for p > 2 and any s, t > 1.

Let us consider further necessary conditions, arising from varying not n as above,but p.

Exercise 4.13. Suppose (Ω, µ) is a measurable space and Λ a bounded operator onLr(Ω, µ) for any r > 2. Assume also that ‖Λ‖2→2 6 1. Then the function

p 7→ ‖Λ‖p/(p−2)p→p

is increasing on (2,∞).

Exercise 4.14. For u > 0 define

ϕ(u) :=

u+ 1

u− 1log u ; u 6= 1

2 ; u = 1.

Verify that:

• ϕ(1/u) = ϕ(u);• ϕ ∈ C1(0,∞);• ϕ

∣∣[1,∞)

is strictly increasing.

Compare with Exercise 5.4.It follows immediately that the function

p 7→ (p− 1)p/(p−2)


In view of Exercise 4.13, if (4.12) is to hold, then the following must also hold:

Exercise 4.15. Suppose n ∈ N, p > 2, 1/p + 1/q = 1. Then the function

p 7→[(1/q)n(1/p)n

]p/(p−2)


Finally observe that by Exercise 4.12 and the last statement of Exercise 4.14, onemay argue that γp(n) could also be a candidate for ‖T n‖p when p > 2. Note, however,that it is not known whether ‖T n‖p > γp(n), in contrast with Proposition 4.2.

5. Spectral theory for T

The estimates of T on Lp and Lp(w), and of T n on Lp, were all motivated byvarious aspects of the quasiconformal theory, see [54, 6, 55] or Sections 2.5, 2.6 and 2.7,respectively. As indicated in Theorem 2.42, the averaging method developed in [33]and [31] can be equally successfully applied for estimating T n on Lp(w) for arbitraryn ∈ Z, p > 1, w ∈ Ap. Thus to complete the picture one might want to pose a questionin the “reverse” direction, i.e., to ask about possible complex-analytic implications ofan estimate such as (2.35).

Since powers of an operator feature prominently in the spectral radius formula, weare led to inquire whether Theorem 2.42 offers any meaningful information about thespectrum of T on Lp(w). Let us first recall a few basic notions.


Let X be a Banach space, Λ ∈ B(X), and I the identity on X. We denote thespectrum, approximate point spectrum and continuous spectrum, respectively, by

σ(Λ) = λ ∈ C ; Λ− λI not invertible in Xσap(Λ) = λ ∈ C ; ∃ (fn)n∈N ⊂ X : ‖fn‖X = 1 and (Λ− λI)fn → 0 in Xσc(Λ) = λ ∈ σ(Λ) ; Λ− λI has trivial kernel and dense range in X.

It is well known that σc(Λ) ⊂ σap(Λ) ⊂ σ(Λ) [1, Theorem 6.17]. Furthermore, letρ(T ) = C\σ(T ) be the resolvent set and r(T ) = max|λ| ; λ ∈ σ(T ) the spectral

radius of T . Recall that r(T ) = limn→∞ ‖T n‖1/n.Fix p > 1 and w ∈ Ap. We have the following result.

Corollary 5.1. For any p > 1 and w ∈ Ap, we have σ(T ) = ∂∆ on Lp(w).

In the unweighted case this result already appeared in [6, Proposition 2] and [5,Theorem 14.1.1]. In order to prove it the authors first showed (by a method differentfrom the one in this paper) that (a) σ(T ) ⊂ ∂∆, and then argued that (b) any λ ∈ ∂∆is an eigenvalue for T . Yet the claim (b) cannot hold, which can for p = 2 be seen eitherby taking the Fourier transform or by recalling that any normal bounded operator on aseparable Hilbert space has an at most countable point spectrum. A closer inspectionof the proofs in [6] and [5] reveals that their candidate for an eigenvector of T − ζI isactually a zero function if |ζ| = 1. It is true, however, that σ(T ) consists of approximate

eigenvalues, not only in Lp but in all Lp(w). This is because of a simple fact that forgeneral Λ as above, every boundary point of σ(Λ) is an approximate eigenvalue of Λ[25, Chapter VII, Proposition 6.7].

It should be emphasized that the proofs from [6] and [5] have the advantage ofbringing up an explicit formula for the inverse of T − ζI; see [5, identity (14.9)]. Ourproof does not entail that.

These remarks motivate us to classify σ(T ) in the unweighted case, following theusual way of decomposing spectra of bounded operators on Banach spaces.

Theorem 5.2. On any Lp(R2), p > 1, we have σ(T ) = σc(T ).

Recall that S denotes the Schwartz class on C. We are still working in Lp(w) withp > 1 and w ∈ Ap fixed.

Proof of Corollary 5.1. We get straight from Theorem 2.42 that maxr(T ), r(T−1) 6

1. Hence σ(T ) ∪ σ(T−1) ⊂ ∆. By the spectral mapping theorem, σ(T−1) = [σ(T )]−1.Thus we conclude that σ(T ) ⊂ ∂∆.

Now let us verify the inclusion in the opposite direction. The author is indebted toMichael Cowling for showing him how to do that.

Take λ ∈ ∂∆ and write λ = e−2iϑ. Our intention is to show that λ is an approximateeigenvalue for T on Lp(w). For n ∈ Z let Sn be the operator of multiplication by the

function z 7→ e2nπiℜ(e−iϑz), i.e.

(Sng)(z) = e2nπiℜ(e−iϑz)g(z)

for any function g : C → C and any z ∈ C. Take f ∈ S such that ‖f‖Lp(w) = 1 and

supp f ⊂ ∆. We claim that limn→∞(T − λI)(Snf) = 0 in Lp(w) . This is equivalent to

limn→∞

(Tn − λI)f = 0 in Lp(w) , (5.1)

where Tn = S−1n TSn.


It is known that [61, p. 94] that w ∈ Ap implies∫R2 w(x)/(1+ |x|)k dx <∞ for some

sufficiently large k > 0. Consequently, (5.1) would follow if we knew that

|(Tn − λI)f |p(x)(1 + |x|)k → 0 in L∞(C) as n→ ∞,

or|(Tn − λI)f |(x)(1 + |x|)k/p → 0 in L∞(C) as n→ ∞. (5.2)

By using (1.11) and a well-known property of the Fourier transform, e.g. [44, Propo-sition 2.2.11.(7)], one calculates

Tnf(ξ) = Φ(ξ/n+ eiϑ)f(ξ) , (5.3)

where

Φ(ζ) =ζ

ζfor ζ 6= 0.

The assumption on f implies that ξ/n + eiϑ stays away from zero for any n ∈ Z\0and ξ ∈ supp f . Therefore (5.3) shows that Tnf ∈ S. Now (5.2) can be rephrased as a

special case of Tnf → λf in S. This is in turn equivalent to Tnf → λf in S. Note thatλ = Φ(eiϑ). So we are proving, for g = f ∈ C∞

c (∆) and ζ = eiϑ ∈ ∂∆,

limn→∞

Φ(·/n + ζ)g = Φ(ζ)g in S.

Owing to the assumption on supp f it is now enough to verify that Φ(·/n + ζ) tendsto Φ(ζ) in C∞(∆), which emerges after a computation (for “everything happens” in asmall neighbourhood of ζ, i.e. in a ball centered at ζ and of radius 1/n).

Remark 5.3. With Uζϕ(z) = ϕ(ζz) for ζ ∈ ∂∆ and z ∈ C we have

T − ζ2I = ζ2Uζ(T − I)U−1ζ . (5.4)

When w is rotation invariant this identity quickly implies that σ(T ) is also rotation-invariant, hence σ(T ) ⊂ ∂∆ immediately gives σ(T ) = ∂∆. Kari Astala pointed out(personal communication, 2009) that a similar argument is valid in many other functionspaces, e.g., Besov, Holder, Triebel-Lizorkin.

Now we turn to proving Theorem 5.2. First we need a few simple auxiliary lemmas.

Exercise 5.4. Suppose ϕ ∈ C∞(R) is such that ϕ(0) = 0. Define

ψ(x) =

ϕ(x)

x; x 6= 0

ϕ′(0) ; x = 0 .

Then ψ ∈ C∞(R) and for all n ∈ N,

ψ(n)(0) =ϕ(n+1)(0)

n+ 1. (5.5)

Before stating the next result we need to invoke a few standard notions.If G ∈ S′(R2) then by ξ2G we denote the product of the function (projection)

(x1, x2) 7→ x2 and the tempered distribution G. For necessary definitions see, forexample, [44, Definition 2.3.15] or [48, Chapter III]. Basic facts about convolution of afunction and a tempered distribution can be found, for example, in [44, Section 2.3] and[48, Chapter IV]. Furthermore, δ is the Dirac delta distribution (evaluation at zero),

while by 1 we mean the distribution φ 7→∫Rφ. It is straightforward that 1 = δ and

δ = 1.


Exercise 5.5. Suppose u ∈ S′(R2) and that u(φ) = 0 for any φ ∈ S(R) ⊗ S(R). Thenu = 0.

Exercise 5.6. Suppose that G ∈ S′(R2) and that ξ2G = 0. Then there exists a tempered

distribution w ∈ S′(R) such that G = w ⊗ δ. Consequently, G = w ⊗ 1.

Exercise 5.7. Suppose g is a Holder continuous function on R2 ≡ C such that, in thesense of tempered distributions, g = W ⊗ 1 for some W ∈ S′(R). Then g(z) = g(ℜz)for all z ∈ C.

Proof of Theorem 5.2. It suffices to consider the case p > 2. Indeed, since the sameproofs as for T also work for T ∗ (cf. Exercise 1.16) and since the spaces Lp and Lq aremutually dual, the case 1 < p < 2 follows from Exercise 1.2. Furthermore, by (5.4) itis enough to see that 1 ∈ σc(T ).

Thus assume p > 2 and write T1 = T − I. First we will show that KerT1 = 0 onLp. If p = 2 this follows directly from (1.11). Now suppose p > 2. For every h ∈ Lp wedefine Ph as in (2.9). By Exercises 2.21 and 2.23, if p > 2, then g = Ph is well definedon Lp and satisfies

h = ∂zg and Th = ∂zg (5.6)

in the distributional sense, with test functions from C1c . Here ∂z = (∂x + i∂y)/2 and

∂z = (∂x − i∂y)/2, as usual. If we additionally assume that h ∈ KerT1, then ∂yg = 0in the same distributional sense. We want to show that ∂yg exists and is equal to zeroin the usual sense, as well. Note the similarity with the Weyl lemma.

Since g is Holder continuous with the exponent 1 − 2/p, it defines a tempered dis-tribution [44, Example 2.3.5.6], denoted by the same letter. The space C∞

c is dense inS, equipped with the usual topology induced by the Schwartz seminorms [48, Lemma7.1.8]; therefore ∂yg = 0 is also valid for test functions from S. By taking the Fouriertransform we get ξ2 g = 0, see [44, Proposition 2.3.22.(8)], where (ξ1, ξ2) are coordi-nates in the Fourier domain and ξ2 is also a projection onto the second coordinate.From Exercise 5.6 we conclude that g = w ⊗ δ, where w ∈ S′(R) and δ is the Diracdistribution, i.e., the evaluation at 0. Repeated application of the Fourier transformyields g = w⊗1 for some w ∈ S′(R). Exercise 5.7 implies that g, now again understoodas a function, only depends on the real part, i.e., g(z) = g(Re z) for all z ∈ C, asdesired. In particular, Qh ≡ 0, where (Qh)(z) = g(z)− g(z + i) . By means of (5.6) weobtain [∂z(Qh)](z) = h(z) − h(z + i) in the distributional sense. Thus we proved thath(z) = h(z + i) p.p. z ∈ C, which for h ∈ Lp is only possible if h ≡ 0 p.p. C. Thisconfirms that KerT1 is trivial.

We are left with proving that ImT1 is dense in Lp. Take f ∈ Lp and ε > 0. Ourgoal is to find g ∈ S such that ‖f − T1g‖p < ε. We can assume that f ∈ S. Takeany ϕ ∈ S such that ‖ϕ‖q = 1. By using (1.11) and the Plancherel identity compute,

for a generic g which is to be determined later, 〈f − T1g, ϕ〉 = 〈f − Hg, ϕ〉 , whereH(ξ) = e−2i arg ξ − 1 for ξ ∈ C\0. Since 1 < q 6 2, the Hausdorff-Young and the

Cauchy-Schwarz inequalities imply that |〈f − T1g, ϕ〉| 6 ‖f −Hg‖q . Therefore, givenF ∈ S, our problem reduces to finding G ∈ S such that ‖F − HG‖q < ε . We mayassume that F ∈ C∞

c (C). Choose R > 0 so that suppF ⊂ z ∈ C ; |z| 6 R =: KR.For small δ > 0 define Dδ := (x, y) ∈ KR ; |y| > δ and Eδ := KR\Dδ . There existsan Urysohn function uδ ∈ C∞

c (C) such that

(i) uδ ∈ [0, 1] everywhere on C,(ii) uδ ≡ 1 on Dδ,


(iii) uδ ≡ 0 on Eδ/2.

By letting Gδ(ξ) := 0 for ξ ∈ R and Gδ := F · uδ/H otherwise, we see that Gδ ≡ 0 onEδ/2 and so Gδ ∈ C∞

c (C). Now clearly ‖F −HGδ‖q → 0 as δ → 0.

6. Open questions

Let us conclude this presentation by summarizing a few problems encountered inour study of the Ahlfors-Beurling operator. Most likely, in terms of difficulty thesequestions vary greatly, the first three being extremely difficult.

Question 1. Is it true that ‖T‖p 6 p − 1 for p > 2? This is the well-known Iwaniecconjecture, see Section 2.5. It is a special case of the following question:

Question 2. Is it true that ‖T n‖p 6 n1−2/p(p − 1) for p > 2 and n ∈ N? In view ofProposition 4.3, an affirmative answer to this question would follow if we could confirmthe following one:

Question 3. Is it true that

‖T n‖p 6(1/q)n(1/p)n

(6.1)

for p > 2, where 1/p + 1/q = 1 and (a)n is defined in (2.34)? See Conjecture 4.10.

Question 4. Is it true that Hp(2k + 1) & |k|1−2/p(p − 1) for all k ∈ Z and p > 2?Recall that Hp(m) was defined, following [55], in Section 2.7, which is also where theproblem’s background was presented.

Question 5. Is it true that the optimal weighted estimates of powers T n on Lp(w) arelinear in n? That is, does for every p > 1 exist C(p) > 0 such that for every n ∈ Z\0and w ∈ Ap we have

‖T n‖B(Lp(w)) 6 C(p) |n| [w]p∗/pp ?

See Section 2.8 for a related discussion.

Question 6. Is it true that the spectrum of T on Lp(w) is continuous for all p > 1 andw ∈ Ap? This is known to be true in the unweighted case (w ≡ 1), see Theorem 5.2.

6.1. Hypergeometric functions and de Branges’ theorem. Here we present apoint of view of the Iwaniec conjecture which was first published in [28].

So far all the attempts to prove (2.16) relied heavily on the Burkholder’s sharp Lp es-timate for martingale transforms (3.44). Although this theorem proved to be extremelyimportant for obtaining very good estimates for ‖T‖p, see [10] and the references listedthere, up to now there has not been a definite proof of the Iwaniec conjecture, as far aswe are aware. One of the obstacles in applying Burkholder’s theorem to that purpose isthat one is indirectly prompted to work with the real and the imaginary part of T (see[34], for example). Nazarov and Volberg [67] proved that ReT and ImT have Lp normsbounded from above by p∗ − 1. Recently Geiss, Montgomery-Smith and Saksman [41]and then Boros, Szekelyhidi Jr. and Volberg [15] gave the same estimate from below,thus proving ‖ReT‖p = ‖ImT‖p = p∗ − 1. This additionally hints that one should trynot to treat the real and the imaginary part of T separately when determining ‖T‖p.

Here we bring up a perspective of the p−1 problem different from the ones pursued sofar in that it is entirely “complex” and does not involve Burkholder’s theorem. Instead,we construct some kind of a generating function whose Taylor coefficients involve T n


and then discuss possibilities of applying other sharp estimates, such as de Branges’theorem about schlicht functions. The latter has, to our knowledge, so far not beenconnected to the p− 1 problem.

Caveat lector: This section should be viewed merely as a short discussion. Theapproach presented here has so far brought no definitive results regarding the operatorT and it may likely be that it actually cannot. Still, we decided to include it, in case itmight raise some of the readers’ attention and motivate thoughts in this direction thatwould reach beyond the level of simple remarks that can be found here.

Remember that κn(p) was defined in (4.1). The fact that it is a quotient of Pochham-mer symbols makes us recall the standard hypergeometric function

F (a, b, c; z) = 1 +

∞∑

n=1

(a)n(b)n(c)n n!

zn .

Thus (4.12) is the same as to say that the Lp norms of T n are the coefficients of thehypergeometric function F (1, q−1, p−1; z). This suggests considering a sort of a powerseries involving quotients of T n and κn(p), and then trying to estimate its coefficients.In such a manner we might think about approaching (6.1), which would in view ofProposition 4.2 immediately establish the Lp norms of T n, i.e. (4.12). To summarize,the idea is to take all powers at the same time and to think of them as coefficientsof some generating function which are then to be estimated by means of some sharptheorem.

All said invokes the following celebrated theorem, which was already mentioned inSection 2.1.

Theorem 6.1 (de Branges [20]). Suppose the function

g(z) =

∞∑

n=1

bnzn

is holomorphic and injective in ∆. Then |bn| 6 n|b1| and this estimate is sharp.

The sharpness part is equally important for us, because we can hardly hope toobtain another sharp result, i.e. (6.1), by applying an unsharp one. The sharpness inthe theorem above is obtained (only) by considering the Koebe function or its rotations,explicitly,

gβ(z) =β2z

(β − z)2,

where β ∈ ∂∆. Functions gβ somewhat resemble the integral kernel of T , which isanother – though very circumstantial – hint that there might be some connection.

Thus we wonder whether Theorem 6.1 (or some other sharp estimate of Taylor coef-ficients) can be applied in this context. Take a real sequence a = (an)n∈N, functions f, gfrom the Schwartz class S such that ‖f‖p = ‖g‖q = 1 and define function Ψ = Ψa,p,f,g

as

Ψ(z) = z +

∞∑

n=2

eian〈T n−1f, g〉κn−1(p)

nzn.


We wonder whether it is injective for some sequence a. Alternatively, by targetingQuestion 2 instead of Question 3 we may define

Ψ(z) = (p − 1)z +

∞∑

n=2

eian〈T n−1f, g〉n2/pzn .

Cauchy-Hadamard formula and Theorem 2.39 imply that every such Ψ (or Ψ) is holo-morphic in ∆.

Theorem 6.1 immediately yields:

Proposition 6.2. Let f, g ∈ S satisfy ‖f‖p = ‖g‖q = 1. If there exists a real sequencea = (an)n∈N such that the function Ψ = Ψa,f,g is injective on ∆, then |〈T nf, g〉| 6 κn(p)for all n ∈ N.

Surely, the problem with verifying (4.12) through this approach is that it is difficultto check whether the function Ψ is injective on ∆.

Question 7. By de Branges’ theorem, we know that if the function

z 7−→ z +

∞∑

n=2

bn zn

is injective on ∆, then |bn| 6 n for all n ∈ N. We also know that the converse is notnecessarily true, even if all bn are strictly positive (Proposition 6.5).

Suppose 0 < |bn| 6 n for all n ∈ N. Does there exist a real sequence a = (an)n∈Nsuch that the function

z 7−→ z +

∞∑

n=2

eianbn zn

is injective on ∆?

If the answer is affirmative, then we have an alternative (equivalent) formulation ofConjecture 4.10, in the sense that for any p > 2 and test functions f, g of unit norm inLp and Lq, respectively, we have a real sequence a such that Ψa,f,g is injective.

Certainly this scheme can be repeated in a much more general setting. Assumewe are given operators Sn on a Banach space X and positive numbers cn such that‖Sn‖ 6Mcn for some absolute M > 0 and all n ∈ N. Suppose we want to improve thisestimate to ‖Sn‖ 6 cn. Surely we may consider the series

z +

∞∑

n=2

〈Snx, ϕ〉cn

nzn ,

where x ∈ X and ϕ ∈ X∗ are such that ‖x‖ = ‖ϕ‖ = 1. Yet it is but impossibleto expect that such an approach would work in such generality. One must take intoaccount the special structure of the operator(s) involved (in our case, T ).

Our situation, however, is somewhat special, for:

• we dealing with powers of a single operator, T , which should somehow naturallycorrespond to power series;

• the conjectured sequence of norms – (1/q)n/(1/p)n – reminds one of power(hypergeometric) functions, as noted earlier;

• we also have the property of T being a Fourier multiplier (1.11).


Then one can apply the Plancherel theorem for an alternative description of Ψ involvinghypergeometric functions, namely

Ψa,p,f,g(z) = z(1 + 〈Gz · f , g〉L2(C)

),

where

Gz(ζ) = Gz,a,p(ζ) =∞∑

n=1

eian(1/p)n(1/q)n

(n + 1)

(ζ

ζz

)n.

Remark 6.3. One may remark that neither the Bellman-function-heat-flow methodnor the stochastic methods, that both yield quite good asymptotic estimates for T ,reproduce the simple fact that ‖T‖2 = 1. So one may test the observation discussed

now first by taking p = 2 in the definition of Ψ (or Ψ) and ask whether for given f, g ∈ S

with L2-norm equal to one there exists a real sequence (an)n such that

Ψ(z) = z +∞∑

n=2

eian〈T n−1f, g〉nzn

is injective. Certainly, in this case the corresponding Gz simplifies to

Gz(ζ) = Gz,a,p(ζ) =∞∑

n=1

eian(n+ 1)

(ζ

ζz

)n.

Hence by defining h := f g we have

Ψa,p,f,g(z) = z

(1 +

∫

C

Gz(ζ)h(ζ) dA(ζ)

)

= z

(1 +

∞∑

n=1

eian(n+ 1)

∫

C

(ζ

ζz

)nh(ζ) dA(ζ)

).

However, injectivity still turns out to be an elusive property in general.

Lack of injectivity. Indeed, assume f, g ∈ S satisfy

‖f‖p = ‖g‖q = 1, 〈Tf, g〉 6= 0, but 〈T nf, g〉 = 0 for all n ∈ N\1.Then Ψ(z) = z + 2az2, where

a =〈Tf, g〉p− 1

.

It is trivial to verify that polynomials of the form z + 2az2 are injective on ∆ (if and)only if |a| 6 1/4, which is to say that if |a| ∈ (1/4, 1] then the conclusion of Theorem6.1 is still true, but the assumption is not.

Example 6.4. Let us construct a concrete Ψ which is not injective.

For z = (x, y) ≡ x + iy define f(z) = g(z) = ∂xe−π|z|2 = −2πxe−π|z|

2. Then

f(z) = 2πixe−π|z|2. Compute

〈T nf, g〉 = 〈T nf, g〉 =∫

C

e−2ni arg z · 2πixe−π|z|2 · (−2πi)xe−π|z|2dA(z)

= 4π2∫ ∞

0r3e−2πr2 dr

∫ 2π

0e−2niϕ cos2 ϕdϕ

=1

2

∫ 2π

0e−2niϕ cos2 ϕdϕ .


By writing cosϕ as the average of eiϕ and e−iϕ we immediately see that the only n ∈ N

for which the last integral is nonzero is n = 1. In that case 〈Tf, g〉 = π4 . We still have

to normalize functions f and g in Lp and Lq, respectively. We calculate

‖f‖pp =2pπ

p−12 Γ(p+1

2

)

p1+p2

.

It suffices to choose p = q = 2. Then ‖f‖22 = ‖g‖22 = π2 . Consequently, for f1 = f/‖f‖2

and g1 = g/‖g‖2 ,

a = 〈Tf1, g1〉 =1

2>

1

4.

Hence the corresponding function Ψ is not injective.

The situation described above, namely when g belongs to the intersection of almostall annihilators (T nf)⊥, seems rather odd. It might still be that for sufficiently manypairs f, g the function Ψf,g is injective.

6.1.1. More on de Branges’ theorem and injectivity. Here we present an example show-ing that the function z+

∑∞2 bnz

n may have coefficients satisfying 0 < bn 6 n, yet notbe injective.

Proposition 6.5. Let K be the Koebe function from Exercise 2.6. If α ∈ [0, 1] thenthe convex combination (1− α)z + αK is injective if and only if α ∈ 0 ∪ [2/3, 1].

Before proving the proposition we have to confirm a few simple auxiliary statements.

Lemma 6.6. If ℜa,ℜb > 1/2, then∣∣∣∣

ab

1− a− b

∣∣∣∣ >1

2.

Proof. Set

η :=1− a− b

ab=

1

a· 1b− a

a· 1b− b

b· 1a.

By writing u = 1/a and v = 1/b, we see that

η = uv − u

u· v − v

v· u .

Multiply this expression by (u/u)(v/v). We obtain

ηu

u

v

v= uv − v − u = uv − u− v = (u− 1)(v − 1)− 1 .

Since ∣∣∣∣1

z− 1

∣∣∣∣2

= 1− 2

|z|2(ℜz − 1

2

)(6.2)

for z ∈ C\0, we have u, v ∈ 1 + ∆ and hence

|η| =∣∣∣∣ηu

u

v

v

∣∣∣∣ = | (u− 1)︸︷︷︸∈∆

(v − 1)︸︷︷︸∈∆

−1| 6 |u− 1||v − 1|+ 1 < 2 .

This finishes the proof of the lemma.


Set P := ζ ∈ C ; ℜζ > 1/2 and, for γ, λ ∈ C,

Mγ = (a, b) ∈ P × P ; a 6= b and ab(1− a− b) = γNλ = (a, b) ∈ P × P ; a 6= b and ab = λ(1− a− b) .

The two sets (or equations) are related via the identity

ab(1− a− b) =ab

1− a− b· |1− a− b|2 . (6.3)

Lemma 6.7. For all λ > 1/2 we have Nλ 6= ∅. If (a, b) ∈ Nλ then |1− a− b| > 1.

Proof. Fix λ > 1/2. We will describe all the elements in Nλ and for each of themverify the last inequality.

We are solving the equation

ab = λ(1− a− b) ; ℜa,ℜb > 1/2 and a 6= b . (6.4)

Writea = x1 + y1i

b = x2 + y2i

for some x1, x2 > 1/2 and y1, y2 ∈ R to be determined, and plug into (6.4). Byconsidering separately the real and imaginary part we get the system

y1y2 = x1x2 + λ(x1 + x2 − 1)

(x2 − λ)y1 + (x1 − λ)y2 = 0 .(6.5)

One obvious family of solutions is obtained by taking x1 = x2 = λ, whereupon the firstequation means that (y1, y2) lie on the curve y1y2 = 3λ2 − λ and off the diagonal, i.e.

y1 = c√

3λ2 − λ

y2 = c−1√

3λ2 − λ

for some c ∈ R\0, 1. Then

|a+ b− 1|2 = (2λ− 1)2 + (c2 + 2 + c−2)(3λ2 − λ)

> (2λ− 1)2 + 4(3λ2 − λ)

= (4λ− 1)2

> 1 .

If, alternatively, x1 − λ and x2 − λ are both nonzero, they must be of different sign,since the first equation in (6.5) implies that y1, y2 must be of the same (nonzero) sign.By symmetry of (6.4) we may thus choose 1/2 < x1 < λ < x2 and write

δ :=

√λ− x1x2 − λ

and η :=√x1x2 + λ(x1 + x2 − 1) .

Now (6.5) reduces to

y1y2 = η2

y1/y2 = δ2 ,


which has obvious solutions, namely (y1, y2) = ±η(δ, 1/δ). In this case

|a+ b− 1|2 = (x1 + x2 − 1)2 + η2(δ2 + 2 + δ−2)

> 0 + (x1x2 + λ(x1 + x2 − 1)) · 4> 4x1x2

> 1 .

This finishes the proof.

Corollary 6.8. If ℜa,ℜb > 1/2 and ab(1− a− b) > 0, then ab(1− a− b) > 1/2.

Proof. From (6.3) and from our assumption we get that

λ :=ab

1− a− b> 0 .

By Lemma 6.6 it follows λ > 1/2. Since (a, b) ∈ Nλ by the definition of λ, Lemma 6.7and (6.3) finish the proof.

Lemma 6.9. For every γ > 1/2 we have Mγ 6= ∅.Proof. We saw that some pairs from Nλ have the same real parts. As we know by(6.3), the sets Nλ and Mγ are related, therefore let us try finding (a, b) ∈ Mγ with

a = x+ y1i

b = x+ y2i

where x > 1/2 and y1, y2 ∈ R, y1 6= y2, are the quantities we are looking for. Byconsidering in ab(1 − a − b) = γ separately the real and imaginary part we get thesystem

(2x− 1)(y1y2 − x2) + xy2 = γ

y[3x2 − x− y1y2

]= 0 ,

(6.6)

where y = y1 + y2. If y = 0 then the first equation gives −(2x− 1)(x2 + y21) = γ, whichis impossible, since 2x− 1 > 0. Therefore y 6= 0, thus 3x2−x− y1y2 = 0. Now the firstequation in (6.6) can be expressed as

x[(2x− 1)2 + y2

]= γ . (6.7)

Since we require that y1 6= y2, the inequality between arithmetic and geometric meangives (y1 + y2)/2 >

√y1y2, i.e. y

2 > 4(3x2 − x), i.e.

y2 = 4(3x2 − x) + ε (6.8)

for some ε > 0. Then (6.7) reads

x[(4x− 1)2 + ε

]= γ . (6.9)

Observe that infx(4x− 1)2 ; x > 1/2 = 1/2 < γ.So we may finally construct a solution to our problem. First choose x > 1/2 such

that x(4x−1)2 < γ and then ε > 0 so that (6.9) is valid. In view of (6.8) this determinesy (up to a sign). Again by (6.8) we see that the intersection of the curves y1 + y2 = yand y1y2 = 3x2 − x is nonvoid and lies off the diagonal in the plane (y1, y2).

Proof of Proposition 6.5. Fix α ∈ (0, 1) and write g(z) = (1−α)z+αK(z). Clearlyg is holomorphic on ∆ and g(0) = 0, g′(0) = 1. Thus if g were injective, it would be


schlicht. We aim to show this is not the case precisely for α ∈ (0, 2/3). We have, forz, w ∈ ∆,

g(z) − g(w) = (1− α)(z − w) + α

(z

(1− z)2− w

(1− w)2

)

= (z − w)

[(1− α) + α

1− zw

(1− z)2(1− w)2

].

We are wondering whether g(z) − g(w) = 0 for some z, w ∈ ∆ such that z 6= w. Thuswe want to find different z, w ∈ ∆ for which

(1− z)2(1− w)2 + β(1 − zw) = 0 , (6.10)

whereβ :=

α

1− α∈ (0,∞) .

Write u = 1− z and v = 1− w. Then u, v ∈ 1 + ∆ and (6.10) becomes

(uv)2 + β(u+ v − uv) = 0 .

Since u, v 6= 0 we may divide by uv and get

uv + β

(1

u+

1

v− 1

)= 0 . (6.11)

Now write a = 1/u, b = 1/v. Then (6.2) gives a, b ∈ 1/(1+∆) = ζ ∈ C ; ℜζ > 1/2 =P and the equation (6.11) becomes 1/(ab) + β(a+ b− 1) = 0, or

ab(1− a− b) =1

β.

In other words, g is not injective if and only if M1/β 6= ∅. However by Corollary 6.8and Lemma 6.9 this happens if and only if β < 2, i.e. α < 2/3.

6.1.2. Few other conjectures on Taylor coefficients of holomorphic functions in ∆. Itmight be interesting to ask about other possible sets of assumptions on Ψ (other thaninjectivity) which would imply sharp upper estimates of its Taylor coefficients. We lista few of them.

Rogosinski conjecture [43, §1.3, p. 29]. If

g(z) =

∞∑

n=1

bnzn

is a holomorphic function in ∆ whose image is contained in the image of some schlichtfunction, then |bn| 6 n for all n ∈ N.

Clearly the Rogosinski Conjecture implies the Bieberbach Conjecture.

Krzyz conjecture [57]. If

f(z) =

∞∑

n=0

anzn

is holomorphic in ∆ and maps ∆ → ∆\0, then for all n ∈ N we have |an| 6 2/e.The extremals are precisely the functions of the form αF (βzn) where α, β ∈ ∂∆, n ∈ N

and

F (z) = exp

(z − 1

z + 1

).


The conjecture was presented in 1967 in Krakow [83]. See also a very recent paperby Martın, Sawyer, Uriarte-Tuero and Vukotic [60].

Zalcman conjecture. Any schlicht function f(z) = z +∑∞

2 anzn on ∆ satisfies

|a2n − a2n−1| 6 (n− 1)2

for all n ∈ N\1. The equality holds if and only if f is the Koebe function.

The Zalcman conjecture implies the Bieberbach conjecture.

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Oliver Dragicevic, Department of Mathematics, Faculty of Mathematics and Physics,University of Ljubljana, Jadranska 21, SI-1000 Ljubljana, Slovenia

Email address: [email protected]

http://lanl.arxiv.org/abs/1203.2291

http://lanl.arxiv.org/abs/1106.3899

arxiv:2109.04555v1 [math.ca] 9 sep 2021

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