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ULTRASPHERICAL EXPANSIONS

DARIUSZ BURACZEWSKI, TERESA MARTINEZ, AND JOS´E L. TORREA

Abstract. We define the higher order Riesz transforms and the Littlewood- Paley g-function associated to the differential operator Lλf (θ) = −f′′(θ) − 2λ cot θf(θ)+λ2f (θ). We prove that these operators are Calder´on-Zygmund operators in the homogeneous type space¡(0, π), (sin t)dt¢. Consequently, Lp weighted, H1− L1and L− BMO inequalities are obtained.

1. Introduction intro

B. Muckenhoupt and E. Stein, in[5], defined and studied the versions of someMS objects to classical Fourier Analysis (conjugate functions, maximal functions, g-functions and multipliers) for the system of the ultraspherical polynomials.

Their approach follows the lines of the classical Fourier Analysis in the torus.

In particular, the relationship between Fourier series, analytic functions and harmonic functions plays an essential role. For instance, their definition of the conjugate function was via a boundary value limit of certain “conjugate harmonic function” which satisfies the appropriate Cauchy-Riemann equations.

The technique involved the definition of the “harmonic extension” including a careful analysis of its kernel. Then they built a conjugate harmonic function and proved the existence of a boundary value function, the conjugate function.

They got Lp boundedness for p in the range 1 < p < ∞ and some substitutive inequality in the case p = 1. This method was followed later by different authors when defining classical operators for orthogonal expansions. In MS[5] they did not study the kernel of the conjugate function.

Five years later, there appeared Stein’s celebrated monograph st1[7], where

“maximal functions”, “g-functions”, “Riesz transforms”, and “multipliers” were also defined. He used systematically a point of view based on an analysis of a Laplacian on compact Lie groups. He studied the “heat” and “Poisson”

semigroups associated to the Laplacian and, from them, he derived the rest of the operators by using some spectral formulas.

Date: March 13, 2005.

2000 Mathematics Subject Classification. 42C05,42C15.

Key words and phrases. Ultraspherical polynomials, Calder´on-Zygmund operators.

The authors were partially supported by RTN Harmonic Analysis and Related Problems contract HPRN-CT-2001-00273-HARP. First author was supported in part by KBN grant 1P03A01826. Second and third authors were partially supported by BFM grant 2002-04013- C02-02.

1

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A natural question arising from[5] is studying some other classical operatorsMS in the context of the system of the ultraspherical polynomials. In particular, our aim was to study higher order Riesz transforms. In order to define those operators, it seems that the natural procedure to follow is the one suggested in the monograph st1[7], and we do so in this paper. The method used to prove the boundedness of higher order Riesz transforms could be applied to a more general class of operators. For instance, we present in this paper its application to study the boundedness of the Littlewood-Paley g-function, although more operators fit in this technique (e.g. multipliers of Laplace transform type). More concretely, we find that all these operators are naturally Calder´on-Zygmund operators in a space of homogeneous type. Therefore, we get as a byproduct of the general theory Lp, H1, BM O boundedness and weighted inequalities for them (see Theorem 1.4).

By using a different method, the (first order) Riesz transform was studied in [1]. It was defined followingBMTU [7] and, among other results, the Lst1 p-boundedness for p ∈ (1, ∞) and the weak type (1, 1) were obtained.

We consider the ultraspherical polynomials Pnλ(x), λ > 0, defined as the coefficients in the expansion of the generating function (1 − 2xω + ω2)−λ = P

n=0ωnPnλ(x) (see[8] for further details). It is known that the set {PSzego nλ(cos θ) : n ∈ N} is orthogonal and complete in L2[0, π] with respect to the measure dmλ(θ) = (sin θ)dθ. The functions Pnλ(cos θ) are eigenfunctions of the operator Lλ

op (1.1) Lλf (θ) = −f′′(θ) − 2λ cot θf(θ) + λ2f (θ),

with eigenvalues µn = (n + λ)2. Every function f in L2(dmλ) has the ultra- spherical expansion f (θ) = P

n=0anPnλ(cos θ)kPnλk−1L2(dmλ). Following [7], west1 define its Poisson integral as

our poisson (1.2) P f (e−t, θ) = e−tLλf (θ) = X n=0

ane−t(n+λ)Pnλ(cos θ)kPnλk−1L2(dmλ).

The calculus formula s−a= Γ(a)1 R

0 e−tsta−1dt can be used to define definition Llambda-05 (1.3) L−l/2λ f (θ) = 1

Γ(l) Z

0

e−tLλf (θ)tl−1dt, l ≥ 1, f ∈ L2(dmλ).

On the other hand, it is easy to check that Lλ is formally self-adjoint on the space L2(dmλ) and that it factorizes as Lλf (θ) = (−∂θθ + λ2)f (θ), where

θ = ∂θ+2λ cot θ is formally adjoint to ∂θ, i.e., h∂θf, giL2(mλ) = −hf, ∂θgiL2(mλ). Following[7], the Riesz transform (l = 1) and the higher order Riesz transformsst1 (l > 1) are defined as

Rlλf (θ) = ∂θl(Lλ)−l/2f (θ), l ≥ 1.

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Also, the Littlewood-Paley g-function is defined as Gf(θ) =

µ Z 1 0

r log1

r¡|∂rP f (r, θ)|2+ |∂θP f (r, θ)|2¢ dr

1/2

. For these operators, we have the following result.

theorem principal Theorem 1.4. The operators Rlλf , for any natural l ≥ 1, and Gf are bounded in Lp(w dmλ), 1 < p < ∞ for any weight w in the Muckenhoupt class Ap with respect to the measure dmλ and of weak type (1, 1) with respect to the measure w dmλ for w ∈ A1. Also, they map L into BM O(dmλ) and H1(dmλ) into L1(dmλ) boundedly.

In fact, this theorem is a consequence of the general theory for Calder´on- Zygmund operators in spaces of homogeneous type and the following theorems.

principal Riesz Theorem 1.5. For any natural l ≥ 1, the operators Rλl are Calder´on-Zygmund operators in the homogeneous type space ([0, π), dmλ).

For the g-function, we study separately the operators G1f (θ) =

µ Z 1 0

r log1 r

¯¯∂rPrf (θ)¯

¯

2dr

1/2

eq g1 (1.6)

G2f (θ) = µ Z 1

0

r log1 r

¯¯∂θPrf (θ)¯

¯

2dr

1/2

. eq g2 (1.7)

theorem principal g Theorem 1.8. There exist vector valued Calder´on-Zygmund operators in the homogeneous type space ([0, π), dmλ) mapping scalar-valued functions into func- tions with values in L2((0, 1), dr), TGi, i = 1, 2, such that

Gif (θ) = kTGif (θ)kL2((0,1),dr).

The structure of the paper is as follows. In section 2, we state the technical Theorem 2.4, as an intermediate step in the proofs of Theorems 1.5 and 1.8.

In section 3 we verify the hypothesis of Theorem 2.4 for the higher order Riesz transforms, and in section 4 we do the corresponding checking for the g-function.

Let us mention just a word about notation. Throughout the paper, the letter C will denote a constant whose value may vary from line to line, and let us call polynomial function to any finite linear combination of ultraspherical polynomials, that is, any f of the form

eq polynomial (1.9) f =

N

X

n=0

anPnλ

kPnλkL2(dmλ), an= Z π

0

f (θ) Pnλ(cos θ)

kPnλkL2(dmλ)dmλ(θ) 2. Preliminaries and technical tools technical lemmas

Following[3] andC [4], a space of homogeneus type (X, ρ, µ) is a set X togetherCW with a quasimetric ρ and a positive measure µ on X such that for every θ ∈ X and r > 0, µ(B(θ, r)) < ∞, and such that there exists 0 < C < ∞ such that

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for every θ ∈ X and r > 0 µ(B(θ, 2r)) ≤ Cµ(B(θ, r)). In our case, X = [0, π]

with the metric given by the absolute value, and the measure is dmλ, which is doubling, what can be easily checked.

We say that a kernel K : X × X\{x = y} → C is a standard kernel if there exist ε > 0 and C < ∞ such that for all x 6= y ∈ X and z with ρ(x, z) ≤ ερ(x, y),

|K(x, y)| ≤ C

µ(B(x, r)), where r = ρ(x, y), and equation CZ kernel 1 (2.1)

|K(x, y) − K(z, y)| + |K(y, x) − K(y, z)| ≤ ρ(x, z) ρ(x, y)

C µ(B(x, r)). equation CZ kernel 2 (2.2)

Thus, a Calder´on-Zygmund operator (with associated kernel K) is a linear op- erator T bounded in L2 such that, for every f ∈ L2 and x outside the support of f ,

T f (x) = Z

X

K(x, y)f (y) dµ(y).

It is known that any Calder´on-Zygmund operator as above is bounded in Lp(w dµ), for 1 < p < ∞ and any weight w in the Muckenhoupt class Ap with respect to the measure dµ. They also map L1(w dµ) into L1,∞(w dµ) for any weight w in the Muckenhoupt class A1 with respect to the measure dµ, L(µ) boundedly into BM O(dµ) and H1(dµ) into L1(dµ) (see[2]).Cal

For B a Banach space, vector-valued Calder´on-Zygmund operators, T from L2(dµ) into L2B(dµ), are defined in the same way as scalar valued ones, but considering K : X × X\{x = y} → B instead of a scalar valued kernel, and taking B-norms in (2.1) and (2.2) instead of absolute values. The boundedness results mentioned above also hold in the vector valued case (see RT[6]).

drawing1

θ φ

π 2

π 2 π 3

φ = 2θ/3 φ = 3θ/2

The symmetry with respect to π/2 in the kernels of our opera- tors will play an important role in the proofs. Also, it will be use- ful to have in mind the picture of the area where we are placing the variables θ and φ. Our first step is studying in detail the behavior of the measure of B(θ, |θ − φ|).

lema medida bola Lemma 2.3. There exists a constant C > 0 such that for any θ ∈ [0, π/2],

mλ(B(θ, |θ − φ|)) ≤ C

|θ − φ|¡ sin(θ ∨ φ)¢ if φ ∈ [0, π/2]

φ if φ ∈ [π/2, π], φ > 32θ

|θ − φ|¡ sin(θ ∨ φ)¢ if φ ∈ [π/2, π], φ < 32θ.

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stronger conditions Theorem 2.4. Let B be a Banach space, and T : L2(dmλ) → L2B(dmλ) be an operator given by integration against a kernel K in the Calder´on-Zygmund sense, such that

i) If |θ − φ| ≥ π/6, then

kK(θ, φ)kB ≤ C and

k∂θK(θ, φ)kB+ k∂φK(θ, φ)kB ≤ C.

ii) For every (θ, φ) either belonging to [0, π/2] × [0, π/2] or to [0, π/2] × [π/2, π] ∩ {φ ≤ 32θ}, we have

kK(θ, φ)kB≤ C

|θ − φ|

1

(sin(θ ∨ φ)), and strong kernel 1 (2.5)

k∂θK(θ, φ)kB+ k∂φK(θ, φ)kB≤ C

|θ − φ|2

1

(sin(θ ∨ φ)). strong kernel 2 (2.6)

iii) The kernel K is symmetric (K(π−θ, π−φ) = K(θ, φ)) or antisymmetric (K(π − θ, π − φ) = −K(θ, φ)).

Then, T is a Calder´on-Zygmund operator in (0, π) with ρ(x, y) = |x − y|.

Proof. Lemma 2.3 and (2.5) clearly imply (2.1) in the region stated in ii).

On the other hand, for (θ, φ) ∈ [0, π/2] × [π/2, π] ∩ {φ ≥ 32θ}, we have π/6 ≤

|θ − φ| ≤ π, and therefore, by i), kK(θ, φ)kB ≤ C ≤ φ . Using again Lemma 2.3, (2.5) and the symmetry condition iii), we obtain (2.1) easily.

By using Lemma 2.3 and (2.6), we get equation final kernel 2 (2.7) k∂θK(θ, φ)kB+ k∂φK(θ, φ)kB≤ C

|θ − φ|

1

mλ(B(θ, |θ − φ|))

for θ and φ in the region stated in ii). For (θ, φ) ∈ [0, π/2] × [π/2, π] ∩ {φ ≥ 32θ}, we have π/6 ≤ |θ − φ| ≤ π, and therefore, by i), k∂θK(θ, φ)kB+ k∂φK(θ, φ)kB≤ C ≤ |θ−φ|φC . By Lemma 2.3, we have (2.7) in (θ, φ) ∈ [0, π/2] × [π/2, π] ∩ {φ ≥

3

2θ}. The symmetry condition iii) implies that (2.7) holds for any (θ, φ) ∈ [0, π] × [0, π]. By standard calculations, this inequality implies condition (2.2)

for ε = 1/2 ¤

3. Proof of Theorem 1.5 section Riesz

Given a polynomial function f , we have Rλlf (θ) = ∂θl(Lλ)−l/2f (θ) = ∂θl

à N X

n=0

an

(n + λ)k

Pnλ(cos θ) kPnλkL2(dmλ)

! rt (3.1)

=

N

X

n=0

an

(n + λ)l

θlPnλ(cos θ) kPnλkL2(dmλ).

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The case l = 1 was extensively studied in [1]. There it was proved its bound-BMTU edness in Lp for p ∈ (1, ∞) and its weak type (1, 1), as well as the fact that it is a principal value, the boundedness of the maximal operator, etc.. In order to prove Theorem 1.5, we first prove the boundedness in L2 of the higher order Riesz transforms.

subsection l2 bdd

3.1. L2 boundedness of higher order Riesz transforms. We shall see that for any polynomial function (1.9), we have kRlλf kL2(dmλ) ≤ C kfkL2(dmλ) with a constant independent of N . First of all, let us observe that without loss of generality, we can consider only polynomial functions f such that there only appear Pnλ(cos θ) for n ≥ l + 1 in their expansion. Since any linear operator is bounded on finite dimensional spaces, we have that the restriction of Rλl to the span of {P0λ, . . . , Plλ} is bounded, and it only remains to check the case when n ≥ l + 1.

The case l = 1 was treated in BMTU[1], and it was seen to be bounded in L2(dmλ).

Lets us explore the case l = 2. In this case, for any n ≥ 3 R2λPnλ(cos θ) = 1

(n + λ)2θ2Pnλ(cos θ)

= −2λ

(n + λ)2 cos θPn−1λ+1(cos θ) + 4λ(λ + 1)

(n + λ)2 (sin θ)2Pn−2λ+2(cos θ).

riesz differential (3.2)

On the other hand, from (1.1) we obtain ∂θ2 = −Lλ − 2λ cot θ ∂θ + λ2 and therefore

riesz2 operator (3.3) R2λ = ∂θ2L−1λ = −Id − 2λ cot θ ∂θL−1λ + λ2L−1λ . With this expression,

action riesz operator (3.4) Rλ2Pnλ = −Pnλ+ 4λ2

(n + λ)2 cos θPn−1λ+1+ λ2

(n + λ)2Pnλ. Mixing (3.2) and (3.4) one has

2λ(2λ + 1)

(n + λ)2 cos θPn−1λ+1(cos θ) eq fundamental (3.5)

= Pnλ(cos θ) − λ2

(n + λ)2Pnλ(cos θ) + 4λ(λ + 1)

(n + λ)2 (sin θ)2Pn−2λ+2(cos θ).

In particular, with this calculation we get from (3.4) Rλ2Pnλ(cos θ) = −1

2λ + 1Pnλ(cos θ) + 1 2λ + 1

1

(n + λ)2Pnλ(cos θ) +8λ2(λ + 1)

2λ + 1

1

(n + λ)2(sin θ)2Pn−2λ+2(cos θ),

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and therefore

R2λf (θ) = −1 2λ + 1

N

X

n=3

an

kPnλkL2(dmλ)Pnλ(cos θ)

+ 1

2λ + 1

N

X

n=3

an

kPnλkL2(dmλ)

1

(n + λ)2Pnλ(cos θ) +8λ2(λ + 1)

2λ + 1

N

X

n=3

an

kPnλkL2(dmλ) 1

(n + λ)2(sin θ)2Pn−2λ+2(cos θ), Clearly, the first two sums are operators bounded in L2(dmλ). For the third one, we will use the following lemma.

lemma boundedness Lemma 3.6. For every k ≥ 0, the operator T acting on polynomial functions T f (θ) =

N

X

n=k+1

an kPnλkL2(dmλ)

1

(n + λ)k(sin θ)kPn−kλ+k(cos θ) is bounded in L2(dmλ).

Proof. By using the orthogonality of (sin θ)kPn−kλ+k in L2(dmλ), and kPnλk2L2(dmλ) = 21−2λπΓ(λ)−2Γ(n + 2λ)

(n + λ)n!, (see Szego[8]), we obtain

°

°

°

°

N

X

n=k+1

an

kPnλkL2(dmλ)

1

(n + λ)k(sin θ)kPn−kλ+k

°

°

°

°

2 L2(dmλ)

=

N

X

n=k+1

a2n kPnλk2L2(dmλ)

kPn−kλ+kk2L2(dmλ+k)

(n + λ)2k ≤ C

N

X

n=k+1

a2n= C kfk2L2(dmλ).

¤ We use induction in l to prove the boundedness of Rlλ for any l ≥ 1. Let us assume that Rkλ is bounded in L2(dmλ) for k ≤ l − 1. ¿From (3.3), for l ≥ 3, we have

Rlλ = ∂θlL−l/2λ = ∂θl−2Rλ2L−(l−2)/2λ lgeq2 riesz operator (3.7)

= −∂θl−2L−(l−2)/2λ − 2λ∂θl−2cot θ ∂θL−l/2λ + λ2θl−2L−l/2λ

= −Rl−2λ − 2λ∂θl−2cot θ ∂θLλ−l/2+ λ2Rl−2λ L−1λ .

By the induction hypothesis, first and last operators are bounded in L2(dmλ).

It remains to show that the second term is also bounded.

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lemma derivatives Lemma 3.8. For any k ≥ 0,

θk¡ cos θPn−1λ+1(cos θ)¢ =

k+1

X

a=1

X

A, B:

A+B=a

CAB(cos θ)A(sin θ)BPn−aλ+a(cos θ)

where CAB ∈ R may be zero.

Proof. We will prove the result by induction in k. For k = 0 it is clearly true, with C10 = 1 and C01 = 0. Let us suppose that the formula holds until k − 1.

For k, we have

θk¡ cos θPn−1λ+1(cos θ)¢ = ∂θ k

X

a=1

X

A, B:

A+B=a

CAB(cos θ)A(sin θ)BPn−aλ+a(cos θ)

and it is enough to see that any term ∂θ¡(cos θ)A(sin θ)BPn−aλ+a(cos θ)¢ is a sum of terms of the form (cos θ)A˜(sin θ)B˜Pn−˜aλ+˜a(cos θ) con 1 ≤ ˜a ≤ k + 1, ˜A + ˜B = ˜a.

And this holds, since

θ¡(cos θ)A(sin θ)BPn−aλ+a(cos θ)¢ = A(cos θ)A−1(sin θ)B+1Pn−aλ+a(cos θ) + B(cos θ)A+1(sin θ)B−1Pn−aλ+a(cos θ)

− 2(λ + a)(cos θ)A(sin θ)B+1Pn−(a+1)λ+(a+1)(cos θ).

¤ Thus, for a polynomial function f =PN

n=l+1ankPnλkL−12(dmλ)Pnλ, we can write

θl−2cotθ ∂θL−l/2λ f (θ) =

N

X

n=l+1

an

kPnλkL2(dmλ)

1

(n + λ)lθl−2¡ − 2λ cos θPn−1λ+1(cos θ)¢

= Xl−1

a=1

X

A, B:

A+B=a

CAB N

X

n=l+1

an kPnλkL2(dmλ)

1

(n + λ)l(cos θ)A(sin θ)BPn−aλ+a(cos θ) middle operator 2 (3.9)

=

l−1

X

a=1

X

A, B:

A+B=a

CABTA, B, al−1 f (θ).

Thus, the boundedness of this operator is done with the following lemma.

boundedness TABa Lemma 3.10. For any l ≥ 2, the operators TA, B, al−1 appearing in (3.9) with non zero coefficients are bounded in L2(dmλ).

Proof. We will proceed by induction in A: in the case A = 0, we have to prove that for every a ≤ l − 1,

T0, a al−1 f (θ) =

N

X

n=l+1

an

kPnλkL2(dmλ) 1

(n + λ)l cos θ(sin θ)aPn−aλ+a(cos θ)

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is bounded in L2(dmλ). This holds by Lemma 3.6. Now assume that the induction hypothesis is true until A − 1, that is, for any ˜A ≤ A − 1, we have that for every a ≤ l − 1, the operator TA, a− ˜l−1˜ A, af (θ) is bounded in L2(dmλ). For A ≥ 1, by (3.5) with n − (a − 1) and λ + (a − 1) instead of n and λ, one can write

1

(n + λ)l(cos θ)A(sin θ)BPn−aλ+a = C

(n + λ)l−2(cos θ)A−1(sin θ)B×

× µ

Pn−(a−1)λ+(a−1)+ C

(n + λ)2Pn−(a−1)λ+(a−1)+ C

(n + λ)2(sin θ)2Pn−(a+1)λ+(a+1)

= C

(n + λ)l−2(cos θ)A−1(sin θ)BPn−(a−1)λ+(a−1)+ C

(n + λ)l(cos θ)A−1(sin θ)BPn−(a−1)λ+(a−1)

+ C

(n + λ)l(cos θ)A−1(sin θ)B+2Pn−(a+1)λ+(a+1). The first two terms give rise to operators bounded in L2(dmλ), by the induction hypothesis. For the third factor, if a+1 ≤ l−1, then by the same reason it is also a bounded operator. But we want to prove the boundedness for any a ≤ l − 1, and thus it remains to prove the case a = l − 1. Since ∂θl−2cot θ ∂θL−l/2λ Pnλ =

−2λ∂θl−2¡ cos θPn−1λ+1¢, an operator TA,B,l−1l−1 in which there appear terms with Pn−(l−1)λ+(l−1) in the left-hand of expression (3.9), necessarily comes from the factors in which the derivatives ∂θl−2 act on the polynomial Pn−1λ+1 and not in any other term (otherwise we would obtain Pn−aλ+a with a < l − 1). Therefore, the only terms with non-zero coefficients coming by applying (3.5) in the right-hand of (3.9) must sum up a constant times the operator

T1, l−2, l−1l−1 f (θ) =

N

X

n=l+1

an kPnλkL2(dmλ)

1

(n + λ)l cos θ(sin θ)l−2Pn−(l−1)λ+(l+1)(cos θ).

By using (3.5) with n − (l − 2) and λ + (l − 2) instead of n and λ, we can write 1

(n + λ)lcos θ(sin θ)l−2Pn−(l−1)λ+(l+1)

= C(sin θ)l−2 (n + λ)l−2

µ

Pn−(l−2)λ+(l−2)+ C

(n + λ)2Pn−(l−2)λ+(l−2)+ C

(n + λ)2(sin θ)2Pn−lλ+l

= C(sin θ)l−2

(n + λ)l−2Pn−(l−2)λ+(l−2)+ C(sin θ)l−2

(n + λ)l Pn−(l−2)λ+(l−2)+ C(sin θ)l (n + λ)lPn−lλ+l. By Lemma 3.6, the operators are bounded in L2(dmλ).

kernel of the riesz

3.2. Kernel of the Riesz transfom. Observe that by (1.2), P f (e−t, θ) = e−tλf (e−t, θ), where f (r, θ) = P

n=0anrnPnλ(cos θ)kPnλk−1L2(dmλ) is defined by Muckenhoupt and Stein in MS[5]. They compute explicitly the kernel P (r, θ, φ) of

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f (r, θ):

p1 (3.11) P f (r, θ) = rλ Z π

0

P (r, θ, φ)f (φ)dmλ(φ), θ ∈ [0, π]

where r = e−t and prtf (3.12) P (r, θ, φ) = λ

π Z π

0

(1 − r2) sin2λ−1t

¡1 − 2r(cos θ cos φ + sin θ sin φ cos t) + r2¢λ+1dt.

Before continuing further, let us state some useful notation.

notation (3.13)

σ = sin θ sin φ, a = cos θ cos φ + σ cos t = cos(θ − φ) − σ(1 − cos t),

r = 1 − 2r cos(θ − φ) + r2 = (1 − r)2+ 2r(1 − cos(θ − φ)),

∆ = ∆1, Dr = 1 − 2ra + r2 = ∆r+ 2rσ(1 − cos t).

lemma kernels Lemma 3.14 (Lemma 2 in[1]). Given f ∈ LBMTU 1(dmλ) and l ≥ 1, for almost every θ ∈ [0, π], we have that

equation Wlambda (3.15)

(Lλ)2lf (θ) = Z π

0

Wλl(θ, φ)f (φ) dmλ(φ), where Wλl(θ, φ) = 1

Γ(l) Z 1

0

rλ−1¡ log1 r

¢l−1

P (r, θ, φ) dr.

Given f ∈ L1(dmλ), l ≥ 1 and θ outside the support of f, we have that Rlλf (θ) =

Z π 0

Rlλ(θ, φ)f (φ) dmλ(φ), where Rlambda (3.16)

Rλl(θ, φ) = 1 Γ(l)

Z 1 0

rλ−1¡ log1 r

¢l−1λ

π(1 − r2) Z π

0

(sin t)2λ−1θl¡ 1 Drλ+1¢ dt.

Proof. L−1/2λ is defined in (1.3). Then, to get (3.15) it is enough to apply Fubini’s theorem. To prove (3.16), it is enough to justify the differentiation inside the integral sign. See BMTU[1] for the details. ¤ Now we shall see that the kernel Rlλ(θ, φ) satisfies the hypothesis of The- orem 2.4. Since P (r, π − θ, π − φ) = P (r, θ, φ), then Rlλ(π − θ, π − φ) = (−1)lRlλ(θ, φ). Therefore, Rlλ(θ, φ) satisfies condition iii) in Theorem 2.4. Ob- serve that ∂θl+1(D−λ−1r ) and ∂φθl(D−λ−1r ) are quotients with a bounded func- tion in the numerator and a certain power of Dr in the denominator. Since for r ∈ (0, 1/2), Dr ≥ C and for r ∈ (1/2, 1) and |θ − φ| > π/6, Dr ≥ C, we get

|Rlλ(θ, φ)| + |∂θRlλ(θ, φ)| + |∂φRlλ(θ, φ)|

≤ C Z 1/2

0

rλ−1¡ log1 r

¢l−1

dr + C Z 1

1/2

rλ−1¡ log1 r

¢l−1

dr ≤ C.

With this we obtain condition i) in Theorem 2.4. In order to prove condition ii), we need a careful analysis of ∂θl³

1 Drλ+1

´. Recall our notation (3.13) and let

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us call

b = ∂θa = − sin(θ − φ) − cos θ sin φ(1 − cos t), and observe ∂θb = a. We have the following lemma to describe ∂θl³

1 Dλr+1

´ sum more precisely:

Lemma 3.17.

lθ³ 1 Dλ+1r

´=X cl,k,i,j

ri+jaibj Drλ+1+k and cl,k,i,j 6= 0 only if:

conditions derivative (3.18) k = 1, ..., l, i + j = k, j ≥ 2k − l.

Proof. For l = 1, ∂θl³

1 Dλr+1

´ = Crb

Drλ+2 and therefore only c1,1,0,1 is nonzero. As- sume that the lemma is true for l. Since

derivative Mlijk (3.19) ∂θ

³ri+jaibj Dλ+1+kr

´= c1

ri+jai−1bj+1 Dλ+1+kr + c2

ri+jai+1bj−1 Dλ+1+kr + c3

ri+j+1aibj+1 Dλ+1+(k+1)r

(we assume c1 = 0 if i = 0 and c2 = 0 if j = 0), we easily see that all these

expressions satisfy (3.18) for l + 1. ¤

As a consequence we may write: Rlλ(θ, φ) = P cl,k,i,jMn,k,i,j(θ, φ), where cl,k,i,j are as in the last lemma and

Ml,k,i,j(θ, φ) = Z 1

0

Z π 0

rλ+i+j−1¡ log1 r

¢l−1

(1 − r2)aibjsin2λ−1t Drλ+1+k dtdr.

The next step is checking that each of these terms verifies condition ii) of The- orem 2.4. This will be done in two lemmas.

Lemma 3.20. There exists a constant C > 0 such that, for every θ ∈ [0, π/2]

and 2θ/3 ≤ φ ≤ 3θ/2, we have

|Ml,k,i,j(θ, φ)| ≤ C

|θ − φ|

1 (sin φ), equation r5 (3.21)

|∂θMl,k,i,j(θ, φ)| + |∂φMl,k,i,j(θ, φ)| ≤ C

|θ − φ|2 1 (sin φ). equation r6 (3.22)

Proof. Let us start with the first inequality. Since for r ∈ (0, 1/2), Dr ≥ C, and for r ∈ (1/2, 1), log1r ≤ C(1 − r), and also by using that |a| ≤ C, |b| ≤

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sin |θ − φ| + sin φ(1 − cos t), we have

|Ml,k,i,j(θ, φ)| ≤ C Z 1/2

0

rλ+k−1¡ log1 r

¢l−1

dr + C

Z 1 1/2

Z π

0 (1 − r)l(sin |θ − φ| + sin φ(1 − cos t))j

Drλ+k+1 (sin t)2λ−1dt dr eq m sin derivar (3.23)

≤ C + C Z 1

1/2

Z π

0 (1 − r)l(sin |θ − φ|)j(sin t)2λ−1 Dλ+k+1r dt dr + C

Z 1 1/2

Z π/2

0 (1 − r)l(sin φ(1 − cos t))j(sin t)2λ−1

Drλ+k+1 dt dr

+ C Z 1

1/2

Z π

π/2(1 − r)l(sin φ)j(sin t)2λ−1

Dλ+k+1r dt dr = C + I + II + III.

For the first term, we use that j ≥ 2k − l, and thus l ≥ 2k − j, and then we perform successively the changes of variables x = p σ

rt and u = 1−r.

I ≤ C Z 1

1/2(1 − r)2k−j(sin |θ − φ|)j Z π/2

0

t2λ−1

(∆r+ σt2)λ+k+1 dt dr

= C Z 1

1/2(1 − r)2k−j(sin |θ − φ|)j

Z π2σ

∆r

0

¡q

r

σ2λ−1q

r σ dx

λ+k+1r (1 + x2)λ+k+1 dr

≤ (sin |θ − φ|)jC Z 1

1/2

(1 − r)2k−jdr σλk+1r

≤ C(sin |θ − φ|)j σλ

Z 2√∆1

0

(√

∆u)2k−j

∆ du

k+1(1 + u2)k+1 dr

≤ C(sin |θ − φ|)j

σλj/2+1/2 ≤ C

σλ1/2 ≤ C

|θ − φ|

1

(sin(θ ∨ φ)),

where in the last two inequalities we have used that for 2θ/3 ≤ φ ≤ 3θ/2,

equivalences local (3.24) 1 − cos(θ − φ) ∼ (sin |θ − φ|)2, sin θ ∼ sin φ, | sin(θ − φ)| ≤ C sin φ.

The proof of these inequalities is trivial, although the argument differs when φ ∈ [0, π/2] than when φ ∈ [π/2, π]. Analogously we may estimate II and III.

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By (3.19) and analogous arguments as in (3.23), we have

|∂θMl,k,i,j(θ, φ)| ≤ C + C Z 1

1/2

Z π

0 (1 − r)l |b|j+1

Drλ+k+1(sin t)2λ−1dt dr +C

Z 1 1/2

Z π

0 (1 − r)l |b|j−1

Drλ+k+1(sin t)2λ−1dt dr +C

Z 1 1/2

Z π

0 (1 − r)l |b|j+1

Drλ+k+2(sin t)2λ−1dt dr

= C + A + B + D.

For the first integral, we have that A ≤ C

Z 1 1/2

Z π

0 (1 − r)l |b|j+1

Drλ+k+1(sin t)2λ−1dt dr ≤ C

|θ − φ|2

1 (sin(θ ∨ φ)) as it was proven by the estimates of I, II and III above. Observe that if j = 0, the second term would not appear. So, we may proceed to estimate B assuming j ≥ 1, and obtaining

B ≤ C Z 1

1/2(1 − r)l Z π

0

(sin |θ − φ| + sin φ(1 − cos t))j−1

Dλ+k+1r (sin t)2λ−1dt dr

≤ C Z 1

1/2

Z π/2

0 (1 − r)l(sin |θ − φ|)j−1(sin t)2λ−1 Drλ+k+1 dt dr +C

Z 1 1/2

Z π/2

0 (1 − r)l(sin φ(1 − cos t))j−1(sin t)2λ−1

Drλ+k+1 dt dr

+C Z 1

1/2

Z π

π/2(1 − r)l(sin φ)j−1(sin t)2λ−1 Dλ+k+1r dt dr

≤ C

σλ∆ ≤ C

|θ − φ|2

1

(sin(θ ∨ φ)),

where the last two inequalities follows as in I, II and III above. Observe that D ≤ C

Z 1

1/2(1 − r)l Z π

0

(sin |θ − φ| + sin φ(1 − cos t))j+1

Drλ+k+2 (sin t)2λ−1dt dr The same arguments drive to the bound

D ≤ C

|θ − φ|2

1

(sin(θ ∨ φ)).

The proof of (3.22) for ∂φ follows the same lines. ¤ parte global riesz Lemma 3.25. For every θ, φ ∈ [0, π/2] and φ outside the region 2θ/3 ≤ φ ≤

3θ/2, we also have (3.21) and (3.22).

Proof. In the following calculations we will use that for θ, φ ∈ [0, π/2], then 1 − cos(θ − φ) ∼ |θ − φ|2 ∼ (sin |θ − φ|)2. Also, for φ outside the region

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2θ/3 ≤ φ ≤ 3θ/2, sin |θ − φ| ∼ sin(θ ∨ φ) and |b| ≤ C sin |θ − φ|. Therefore, by the same techniques applied in (3.23), for Ml,k,i,j(θ, φ) we have

|Ml,k,i,j(θ, φ)| ≤ C + C Z 1

1/2(1 − r)2k−j |b|j

λ+k+1r dr

≤ C + C(sin |θ − φ|)j Z 1

1/2

(1 − r)2k−j

λ+k+1r dr

≤ C + C(sin |θ − φ|)j Z 2√∆1

0

(√

∆u)2k−j

∆ du

λ+k+1(1 + u2)λ+k+1 dr

≤ C + C(sin |θ − φ|)j

λ+j/2+1/2 ≤ C

|θ − φ|

1

(sin(θ ∨ φ)). We proceed analogously for the derivative

|∂θMl,k,i,j(θ, φ)| ≤ C + C Z 1

1/2

Z π

0 (1 − r)l |b|j+1

Dλ+k+1r (sin t)2λ−1dt dr + C

Z 1 1/2

Z π

0 (1 − r)l |b|j−1

Drλ+k+1(sin t)2λ−1dt dr + C

Z 1 1/2

Z π

0 (1 − r)l |b|j+1

Drλ+k+2(sin t)2λ−1dt dr ≤ C

|θ − φ|2

1

(sin(θ ∨ φ)).

¤ 4. Proof of Theorem 1.8

section g function r

Let us write Gif (θ) = kTGif (θ)kL2((0,1),dr), i = 1, 2 where TGi is the operator mapping a scalar valued functions into L2(dr)-valued functions given by equation Tg 1 (4.1) TG1f (θ) =

r r log1

r ∂rP f (r, θ), TG2f (θ) = r

r log1

r ∂θP f (r, θ).

subsection bddness of g

4.1. Boundedness in L2(dmλ) of the g-functions. For any polynomial func- tion f ∈ L2(dmλ), the operator ∂rP f =PN

n=0(n+λ)rn+λ−1anPnλkPnλk−1 gives a well defined function in L2(dmλ), since for each fixed r ∈ (0, 1), |(n + λ)rn+λ| ≤ C for any n ≥ 0. Thus, we can write

TG1f =

X

n=0

r r log1

r (n + λ)rn+λ−1anPnλkPnλk−1 =

X

n=0

g1(n)anPnλkPnλk−1,

where g1(n) = q

r log1r (n + λ)rn+λ−1 belongs to L2(dr) uniformly in n. In particular, this implies that G1 is bounded in L2. To get the boundedness of G2 in L2 we proceed similarly. Observe that for any polynomial f ∈ L2(dmλ),

θP f =

N

X

n=1

rn+λan(−2λ) sin θPn−1λ+1kPnλk−1

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We can write TG2f =

N

X

n=1

r r log1

r rn+λ an

n + λ(−2λ) sin θPn−1λ+1kPnλk−1

= X n=0

g2(n) an

n + λ(−2λ) sin θPn−1λ+1kPnλk−1 = Tg2(Rλf ),

where Rλ is the Riesz transform operator (see [1]) and TBMTU g2 is the multiplier associated to the orthogonal system in L2(dmλ) given by the functions hn(θ) = sin θPn−1λ+1(cos θ) with coefficients g2(n) = q

r log1r (n + λ)rn+λ−1. The coeffi- cients are uniformly bounded in L2(dr). This, together with the L2-boundedness of the Riesz transform, see BMTU[1] for the details, give that G2 is bounded in L2(dmλ).

4.2. Kernel of the g-function. In the next lemma, we find the vector-valued kernel in the Calder´on-Zygmund sense of TGi, i = 1, 2.

kernel in CZ sense Lemma 4.2. For every f ∈ L1(dmλ) and θ outside the support of f , we have equation Tg 2 (4.3) TGif (θ) =

Z π 0

τri(θ, φ)f (φ) dmλ(φ), i = 1, 2, where

τr1(θ, φ) = r

r log1 r

λ π

·

λrλ−1P (r, θ, φ) − 2rλ+1 Z π

0

(sin t)2λ−1 Drλ+1 dt

− (λ + 1)rλ(1 − r2) Z π

0

(sin t)2λ−1rDr

Dλ+2r dt

¸ equation Tg1 kernel (4.4)

τr2(θ, φ) = r

r log1 r

λ(λ + 1)

π rλ(1 − r2) Z π

0

(sin t)2λ−1θDr

Drλ+2 dt equation Tg2 kernel (4.5)

are kernels taking values in L2(dr).

Proof. By the definition of TGi, in order to get (4.3) we only need to put the derivative inside the integrals in the expressions (4.1). With our usual notation (3.13), we write

partial P (4.6) ∂rP (r, θ, φ) = −λ π2r

Z π 0

(sin t)2λ−1

Dλ+1r dt + λ

π(1 − r2)∂r

Z π 0

(sin t)2λ−1 Drλ+1 dt For negative a, Dr ≥ 1, and for positive a, a ≤ cos(θ − φ). Therefore, if θ does not belong to the support of f , we have that Dr ≥ 1 − cos(θ − φ)2 ≥ C.

This implies that for each r the remaining integrands in the right hand of the equations in (4.1) belong to L1(dmλ × dt) for f ∈ L1(dmλ). This shows (4.4). Also, if θ does not belong to the support of f , we have that |τr1(θ, φ)| ≤ Cq

r log1r rλ−1 and |τr2(θ, φ)| ≤ C, which are functions in L2(dr). ¤

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Our aim is to prove that the kernels of TG1 verify the hypothesis of Theorem 2.4. We have already seen that they are bounded in L2(dmλ). Observe that since P (r, π − θ, π − φ) = P (r, θ, φ), then the symmetry condition iii) holds.

On the other hand, for |θ − φ| > π/6 and r ∈ (1/2, 1), then Dr ≥ 1 − (cos(θ − φ))2 ≥ C. For r ∈ (0, 1/2), clearly Dr ≥ C. Then kτr1(θ, φ)kL2(dr) ≤ Ckq

r log1r (rλ−1+ 1)kL2(dr) = C and kτr2(θ, φ)kL2(dr) ≤ C, as can be easily seen from (4.4) and (4.5). Thus condition i) in Theorem 2.4 also holds.

Checking condition ii) requires a bit more of work.

local estimates Lemma 4.7. There exists a constant C > 0 such that, for every θ ∈ [0, π/2]

and 2θ/3 ≤ φ ≤ 3θ/2, we have (for i = 1, 2)

ri(θ, φ)kL2(dr) ≤ C

|θ − φ|

1

(sin(θ ∨ φ)), local estimate g 1 (4.8)

k∂θτri(θ, φ)kL2(dr)+ k∂φτri(θ, φ)]kL2(dr) ≤ C

|θ − φ|2

1

(sin(θ ∨ φ)). local estimate g 2 (4.9)

Proof. When r ∈ (0, 1/2), Dr ≥ (1 − r)2 ≥ C and therefore |τr1(θ, φ)| ≤ Cq

r log1r (rλ−1+C) and |τr2(θ, φ)| ≤ C. For r ∈ (1/2, 1), let us observe that, by (4.4), we can split |τr1(θ, φ)| according to the following sum |∂r(rλP (r, θ, φ))| ≤ P3

i=1Ni(r, θ, φ), where N1(r, θ, φ) = C

Z π/2 0

(sin t)2λ−1

Drλ+1 dt, N2(r, θ, φ) = C Z π

π/2

(sin t)2λ−1 Drλ+1 dt, N3(r, θ, φ) = C(1 − r)

Z π 0

(sin t)2λ−1|∂rDr| Drλ+2 dt.

We will use the following estimate ialphabeta (4.10) Iλ+12λ−1 =

Z π2

0

t2λ−1

(∆r+ rσt2)λ+1dt ≤ C

rrλσλ, which can be easily obtained by the change of variables t = ∆

1

r2(rσ)12u. By using that sin x ∼ x, 1 − cos x ∼ x2 and 1 + cos x ≥ Cx2 for x ∈ [0, π/2], it is not difficult to obtain the following estimate,

N1(r, θ, φ) + N2(r, θ, φ) ≤ CIλ+12λ−1,

where in the case of N2 we have firstly made the change of variables π − x = t.

For the term N3, observe that for r ∈ [1/2, 1]

partial dr (4.11) (1−r)|∂rDr| ≤ C(1−r)[|1−r|+|1−cos(θ −φ)|+|σ(1−cos t)|] ≤ C Dr. Thus, after applying the same change of variables π − x = t used above for N2, we get that N3(r, θ, φ) ≤ C Iλ+12λ−1. This gives (and observe that with the only restriction r ∈ [1/2, 1]) that

bound for partial p (4.12) |∂r(rλP (r, θ, φ))| ≤ C Iλ+12λ−1.

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