weak-type inequalities for higher order riesz–laguerre transforms

Journal of Functional Analysis 256 (2009) 258–274

www.elsevier.com/locate/jfa

Weak-type inequalities forhigher order Riesz–Laguerre transforms

Liliana Forzani a, Emanuela Sasso b,∗, Roberto Scotto a

a Departamento de Matemática and IMAL, Universidad Nacional del Litoral and CONICET, Santa Fe 3000, Argentinab Dipartimento di Matematica, Universitá di Genova. Genova, Via Dodecaneso 35, Genova 16146, Italy

Received 2 April 2008; accepted 5 September 2008

Available online 25 September 2008

Communicated by C. Kenig

Abstract

In this paper we study the weak-type (1,1) boundedness of the higher order Riesz–Laguerre transformsassociated with the Laguerre polynomials. In particular, we obtain the boundedness for the Riesz–Laguerretransforms of order 2 and we find also the sharp polynomial weight ω that makes the Riesz–Laguerretransforms of order greater than two continuous from L1(ω dμα) into L1,∞(dμα), being μα the Laguerremeasure.© 2008 Elsevier Inc. All rights reserved.

Keywords: Riesz transforms; Laguerre; Polynomial expansion; Weak-type

1. Introduction: Riesz–Laguerre transforms of order |m|

The aim of this paper is to study the weak type (1,1) boundedness of Rmα , the mth Riesz–

Laguerre transform, with m ∈ Zd�0, associated with the multidimensional Laguerre operator Lα ,

where α = (α1, . . . , αd) is a multi-index with αi � 0, i = 1, . . . , d .The Laguerre operator Lα is a self-adjoint “Laplacian” on L2(dμα), where μα is the Laguerre

measure of type α = (α1, . . . , αd) with αi > −1, i = 1, . . . , d ; defined on Rd+ = {x ∈ Rd : xi > 0,

* Corresponding author.E-mail addresses: [email protected] (L. Forzani), [email protected] (E. Sasso),

[email protected] (R. Scotto).

0022-1236/$ – see front matter © 2008 Elsevier Inc. All rights reserved.doi:10.1016/j.jfa.2008.09.004

L. Forzani et al. / Journal of Functional Analysis 256 (2009) 258–274 259

for each i = 1, . . . , d}, by

dμα(x) =d∏

i=1

xαi

i e−xi

�(αi + 1)dx.

It is well known that the spectral resolution of Lα is

Lα =∞∑

n=0

nP αn ,

where P αn is the orthogonal projection on the space spanned by Laguerre polynomials of total

degree n and type α in d variables, cf. [4,10]. The operator Lα is the infinitesimal generator ofa “heat” semigroup, called the Laguerre semigroup, {e−t Lα : t � 0}, defined in the spectral senseas

e−t Lα =∞∑

n=0

e−nt P αn .

For any multi-index m = (a1, . . . , ad) ∈ Zd�0, the Riesz–Laguerre transforms Rm

α of order|m| = a1 + · · · + ad are defined by

Rmα = ∇m

α (Lα)−|m|/2 P α0

⊥,

where ∇α is the gradient associated to Lα defined as ∇α = (√

x1 ∂x1 , . . . ,√

xd ∂xd), and P α

0⊥ de-

notes the orthogonal projection onto the orthogonal complement of the eigenspace correspondingto the eigenvalue 0 of Lα .

For every multi-index α such that αi � 0, for all i = 1, . . . , d , we have the following mainresult.

Theorem 1. The second order Riesz–Laguerre transforms map L1(dμα) continuously intoL1,∞(dμα).

This result was proved recently by Graczyk et al. [4] only for half-integer α that includesαi = − 1

2 . The corresponding proof is based on the technique of transference to the Hermitesetting. This technique firstly appears for the Riesz–Laguerre transform of order one in Gutiérrezet al. [5]. The method seems to be inapplicable for any other value of α.

It is known that the first order Riesz–Laguerre transforms are weak-type (1,1), cf. [9]. Fur-thermore, we also know from that same paper that the Riesz–Laguerre transforms of order higherthan 2 need not be weak-type (1,1) with respect to μα . However, we can prove the followingresult that has to do with certain kind of weights we can add on the domain of these transforms tomake them satisfy a weak-type inequality. Let us mention that in the Gaussian context somethingquite similar occur with the higher order Riesz–Gauss transforms. Pérez proved that for |m| > 2,the Riesz–Gauss transforms of order |m|, associated to the Ornstein–Uhlenbeck semigroup, mapL1((1 + |y||m|−2)dγ ) continuously into L1,∞(dγ ), with dγ (x) = e−|x|2 dx (see [6]). RecentlyForzani, Harboure and Scotto in [1] extend this result for a family of singular integral operators

260 L. Forzani et al. / Journal of Functional Analysis 256 (2009) 258–274

related with the Gaussian measure γ . Regarding the weights for the Riesz–Laguerre transformsof order higher than 2 we have

Theorem 2. The Riesz–Laguerre transforms of order |m|, with |m| > 2, map L1(w dμα) contin-uously into L1,∞(dμα), where w(y) = (1 + √|y|)|m|−2.

Moreover,

Theorem 3. The weight w is the optimal polynomial weight needed to get the weak type (1,1)

inequality for the Riesz–Laguerre transforms of order |m|.

It should be noted that there is another proof of Theorem 2 for multi-indices of half-integertype (even for the negative one). Indeed, take f w as the function f in Lemma 2.1 of the paperby Gutiérrez et al. [5] after using Corollary 4.8 of the reference [4].

The paper is organized as follows. Section 2 contains basic facts and notation needed in the se-quel and the proof of Theorem 1. In particular, in order to exploiting the well-known relationshipwith the Ornstein–Uhlenbeck context, we introduce the “modified” Riesz–Laguerre transformsrelated to the “modified” Laguerre measure. In Section 3, we prove auxiliary propositions andTheorems 2 and 3.

2. Some preliminaries and proof of Theorem 1

In order to use the well-known relationship with the Ornstein-Uhlenbeck context, but not toomuch exploited in the weak-type inequalities, we are going to perform a change of coordinatesin Rd+.

If x = (x1, . . . , xd) is a vector in Rd+, then x2 will denote the vector (x21 , . . . , x2

d). LetΨ : Rd+ → Rd+ be defined as Ψ (x) = x2 and let dμ̃α = dμα ◦ Ψ −1 be the pull-back measurefrom dμα . Then the modified Laguerre measure dμ̃α is the probability measure

dμ̃α(x) = 2d

d∏i=1

x2αi+1i e−x2

i

�(αi + 1)dx = 2d

d∏i=1

x2αi+1i

�(αi + 1)e−|x|2 dx, (2.1)

on Rd+.The map f → UΨ f = f ◦ Ψ is an isometry from Lq(dμα) onto Lq(dμ̃α) and from

Lq,∞(dμα) onto Lq,∞(dμ̃α), for every q in [1,∞]. So we may reduce the problem of study-ing the weak-type boundedness of Rm

α to the study of the same boundedness for the modifiedRiesz–Laguerre transforms R̃m

α = UΨ Rmα U −1

Ψ with respect to the measure dμ̃α .Observe that R̃m

α coincides, up to a multiplicative constant, with ∇m(L̃α)−|m|P̃ α ⊥0 , being

L̃α = UΨ Lα U −1Ψ , P̃ α ⊥

0 = UΨ P α ⊥0 U −1

Ψ and ∇ the gradient on Rd associated to the Laplacianoperator (for more details, see [8, Section 2]).

For the sequel it is convenient to express the kernel of R̃mα with respect to the polynomial

measure mα defined on Rd+ as

dmα(x) = e|x|2 dμ̃α(x). (2.2)


According to [6,9], for αi > −1/2, i = 1, . . . , d , the kernel of the modified Riesz–Laguerretransforms of order |m| with respect to the polynomial measure mα is defined, off the diagonal,as

Km(x, y) =∫

[−1,1]dKm(x, y, s)Πα(s)ds

with

Km(x, y, s) =1∫

0

(√

r)|m|−2(

− log r

1 − r

) |m|−22

d∏i=1

Hai

(√rxi − yisi√

1 − r

)e− q−(rx2,y2,s)

1−r

(1 − r)|α|+d+1dr (2.3)

where Haiis the Hermite polynomial of degree ai and

q±(x, y, s) =d∑

i=1

(xi + yi ± 2(xiyi)

1/2si),

Πα(s) =d∏

i=1

�(αi + 1)

�(αi + 1/2)√

π

(1 − s2

i

)αi−1/2,

cos θ = cos θ(x, y, s) =∑d

i=1 xiyisi

|x||y| ,

sin θ = sin θ(x, y, s) = (1 − cos2 θ

)1/2 =(

1 −(∑d

i=1 xiyisi

|x||y|)2)1/2

.

Observe that cos θ (and also sin θ ) does not depend on |x| nor on |y|.Throughout this paper the symbol a � b means a � C b where C is a constant that may be

different on each occurrence. And we write a ∼ b whenever a � b and b � a.The proof of Theorem 1 follows from the three propositions below whose proofs can be found

in Section 3.

Proposition 4. For |m| = 2, ∣∣Km(x, y, s)∣∣ � e|x|2 K∗(x, y, s)e−|y|2

on the global region

G = Rd+ × [−1,1]d \ R0 = R1 ∪ R2 ∪ R3 ∪ R4,

with

R0 ={(y, s) ∈ Rd+ × [−1,1]d : q−

(x2, y2, s

)1/2 � C

1 + |x|},

R1 = {(y, s) /∈ R0: cos θ < 0

},


R2 = {(y, s) /∈ R0: cos θ � 0, |y| � |x|},

R3 = {(y, s) /∈ R0: cos θ � 0, |x| � |y| � 2|x|},

R4 = {(y, s) /∈ R0: cos θ � 0, |y| � 2|x|},

and

K∗(x, y, s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

e−|x|2 , (y, s) ∈ R1, (1)

|x|2|α|+2de−c|x|q1/2− (x2,y2,s), (y, s) ∈ R2, (2)

|x|2|α|+2d

(1 ∧ e

− c|x|4 sin2 θ

|y|2−|x|2+sin θ |x|2

(|y|2 − |x|2 + sin θ |x|2) 2|α|+2d−12

), (y, s) ∈ R3, (3)

(1 + |x|)e− sin2 θ |x|2 , (y, s) ∈ R4. (4)

Proposition 5. The operator K∗ defined as

K∗f (x) = e|x|2∫

Rd+

∫[−1,1]d

χG(y, s)K∗(x, y, s)Πα(s)ds∣∣f (y)

∣∣dμ̃α(y),

is of weak type (1,1) with respect to the measure μ̃α .

Proposition 6. For all m, the operator

T m0 f (x) = p.v.

∫Rd+

∫[−1,1]d

χR0(y, s)Km(x, y, s)Πα(s)ds f (y)dmα(y),

which is the modified Riesz–Laguerre transform restricted to the local region R0, is of weak type(1,1) with respect to the measure μ̃α .

Proof of Theorem 1. The result follows by splitting the modified Riesz–Laguerre transforms ofsecond order into a local operator and a global one. Let us observe that for a simple coveringlemma, we may pass from estimates with respect to the measure mα on the local part R0 toestimates with respect to the modified Laguerre measure μ̃α . Therefore the local operator isequivalent to T m

0 defined in Proposition 6 for |m| = 2. On the other hand, the global operator,taking into account Proposition 4, is bounded by the operator defined in Proposition 5. Both areweak type (1,1) and therefore so are the second order modified Riesz–Laguerre transforms. �3. Proofs of propositions and Theorems 2 and 3

From [6] and [9], it is known that an upper bound for |Km(x, y, s)| on G is

K̃m(x, y, s) =⎧⎨⎩ (|x|2 + |y|2) |m|−2

2 e−|y|2 if cos θ < 0,

(q+q−)|m|−2

4 (q+q

)|α|+d

2 (1 + (q+q−)1/4 |x||y|2 2 )e−u0 if cos θ � 0,

(3.1)
− |x| +|y|


with

u0 = |y|2 − |x|22

+ (q+(x2, y2, s)q−(x2, y2, s))1/2

2.

Proof of Proposition 4. In this proposition, |m| = 2.If cos θ < 0, it is immediate that∣∣Km(x, y, s)

∣∣ � e|x|2 K∗(x, y, s)e−|y|2 .

Let us then assume that cos θ � 0.

(1) First let us consider |x| > |y|.Since cos θ � 0, q

1/2+ � |x| and since |x| > |y|, then q

1/2+ � 2|x|. Therefore q

1/2+ ∼ |x|. On

the other hand, since q1/2− � C

1+|x| then |x| � c. Thus

∣∣Km(x, y, s)∣∣ �

[(q+q−

) |α|+d2 +

(q+q−

) |α|+d2

(q+q−)1/4]e−u0

�[|x||α|+d

(1 + |x|)|α|+d + (q

1/2+ )|α|+d+1/2

(q1/2− )|α|+d−1/2

]e−u0

� |x|2|α|+2de−u0 = e|x|2 |x|2|α|+2de− (q+q−)1/2

2 e|y|2−|x|2

2 e−|y|2

� e|x|2 |x|2|α|+2de− |x|q1/2− (x2,y2,s)

2 e−|y|2 = e|x|2 K∗(x, y, s)e−|y|2 .

(2) Now let us assume |y| � |x| and rewrite u0 in the following way:

u0 = |y|2 − |x|22

+ (q+(x2, y2, s)q−(x2, y2, s))1/2

2

= |y|2 − |x|2 + (q+(x2, y2, s)q−(x2, y2, s))1/2 − (|y|2 − |x|2)2

= |y|2 − |x|2 + q+q− − (|y|2 − |x|2)2

2(|y|2 − |x|2 + (q+q−)1/2)

= |y|2 − |x|2 + 2 sin2 θ |x|2|y|2|y|2 − |x|2 + (q+q−)1/2

. (3.2)

Since

q+q− = (|x|2 + |y|2)2 − 4|x|2|y|2 cos2 θ = (|y|2 − |x|2)2 + 4|x|2|y|2 sin2 θ,

and taking into account that sin θ is non-negative, we obtain that

(q+q−)1/2 ∼ |y|2 − |x|2 + |x||y| sin θ � |y|2 − |x|2 + |x|2 sin θ. (3.3)


Thus, from (3.2) together with (3.3) we get

u0 � |y|2 − |x|2 + c|x|4 sin2 θ

|y|2 − |x|2 + |x||y| sin θ. (3.4)

Claim. max(|y|2 − |x|2, |x|2 sin θ) � 1.

Proof. If sin θ � 1|x|2 , the inequality is immediate.

If sin θ � 1|x|2 , then |y|2 � |x|2 + 1. This inequality is immediate when |x| � 1 by adjusting

conveniently the constant C in the definition of the global zone and it is also immediate for d = 1and |x| > 1. Now let us assume that d � 2 and |x| > 1.

(C/2)2

|x|2 � C2

(1 + |x|)2� q−

(x2, y2, s

) = |x|2 + |y|2 − 2|x||y|√

1 − sin2 θ

� |x|2 + |y|2 − 2|x||y|√

1 − 1

|x|4 .

Hence

|y|2 − 2|x|√

1 − 1

|x|4 |y| + |x|2 − (C/2)2

|x|2 � 0

for all |y| � |x|, then

|y| � |x|√

1 − 1

|x|4 +√

(C/2)2 − 1

|x|

which implies that

|y|2 � |x|2 + 2√

(C/2)2 − 1

√1 − 1

|x|4 + (C/2)2 − 2

|x|2 � |x|2 + 1. �Therefore by applying this claim to inequality (3.3) we obtain that q+q− � c in this context.

If |x| � |y| � 2|x|, by taking into account (3.4), we get

u0 � |y|2 − |x|2 + c|x|4 sin2 θ

|y|2 − |x|2 + |x|2 sin θ, (3.5)

then

∣∣Km(x, y, s)∣∣ �

(q+q−

) |α|+d2

(q+q−)1/4e−u0

� q|α|+d+

2|α|+2d(q+q−)1/4e−u0

(q+q−) 4


� |x|2|α|+2d

[(q+q−)1/2] 2|α|+2d−12

e−u0

� e|x|2 |x|2|α|+2d

(|y|2 − |x|2 + sin θ |x|2) 2|α|+2d−12

e− c|x|4 sin2 θ

|y|2−|x|2+|x|2 sin θ e−|y|2 .

To get the last inequality we have used (3.3) and (3.5).On the other hand, since q+q− � c, it is immediate the following inequality∣∣Km(x, y, s)

∣∣ � |x|2|α|+2de|x|2−|y|2 .

Thus ∣∣Km(x, y, s)∣∣ � e|x|2 K∗(x, y, s)e−|y|2 .

Now if |y| � 2|x|, then (q+q−)1/4 � (|x|2 + |y|2)1/2 � |y|, and thus(1 + (q+q−)1/4 |x||y|

|x|2 + |y|2)

�(1 + |x|).

Besides q− � (|y| − |x|)2 � c|y|2 and q+ � C|y|2 therefore q+q− � C. On the other hand, from

(3.2) together with (q+q−)1/2 � |y|2 − |x|2 + 2|x||y| sin θ and |y|2 − |x|2 + |x||y| sin θ � 2|y|2we get

u0 � |y|2 − |x|2 + sin2 θ |x|2|y|2|y|2 − |x|2 + |x||y| sin θ

� |y|2 − |x|2 + sin2 θ

2|x|2.

Therefore ∣∣Km(x, y, s)∣∣ � e|x|2 K∗(x, y, s)e−|y|2 . �

Proof of Proposition 5. The method of proof we use here is an adaptation to our context of thetechniques developed in [1–3], which allows us to get rid of the classical one called “forbiddenregions technique.”

The kernels (1) and (2) define strong type (1,1) operators. Indeed,

e|x|2∫

Rd+

∫[−1,1]d

χR1(y, s)e−|x|2Πα(s)ds∣∣f (y)

∣∣dμ̃α(y) � C‖f ‖1.

Moreover, for semi-integer values of the parameter α, by [6],

|x|2|α|+2de−c|x|q1/2− (x2,y2,s) (3.6)

is in L1(dmα) uniformly in y and s and so the operator is of strong type with respect to μ̃α on R2.Finally the result for the other values of α is obtained via the multidimensional Stein’s complexinterpolation theorem (the same argument is used at the end of this proof).


So to get the weak-type (1,1) inequality for the operator K∗ it suffices to prove that theoperators

Sif (x) = e|x|2∫

Rd+

∫[−1,1]d

χRi+3(y, s)K∗(x, y, s)Πα(s)ds∣∣f (y)

∣∣dμ̃α(y), i = 0,1,

map L1(dμ̃α) continuously into L1,∞(dμ̃α).Without loss of generality, we may assume that f � 0. Fix λ > 0 and let Ei = {x ∈ Rd+:

Sif (x) > λ}, for i = 0,1. We must prove that μ̃α(Ei) � C‖f ‖1

λ. Let r0 and r1 be the positive

roots of the equations

r2|α|+2d0 er2

0 ‖f ‖1 = λ and r1er21 ‖f ‖1 = λ.

We may observe that Ei ∩ {x ∈ Rd+: |x| < ri} = ∅: indeed, if |x| < ri , we have

S0f (x) � |x|2|α|+2de|x|2‖f ‖1 < λ,

S1f (x) � |x|e|x|2‖f ‖1 < λ.

On the other hand, we may take λ > K‖f ‖1 and by choosing K large enough we may assumethat both r0 and r1 are larger that one. Hence

μ̃α

{x ∈ Rd+: |x| > 2ri

}�

∫|x|>2ri

d∏j=1

x2αj +1j e−|x|2 dx

� r2|α|i e−4r2

i � C‖f ‖1

λ.

Thus we only need to estimate μ̃α{x ∈ Rd+: ri � |x| � 2ri}.We let E′

i denote the set of x′ ∈ Sd−1 for which there exists a ρ ∈ [ri ,2ri] with ρx′ ∈ E. Foreach x′ ∈ E′

i we let ρi(x′) be the smallest such ρ. Observe that sin θ(x, y, s) = sin θ(x′, y, s) =

sin θ . Then Sif (ρi(x′)x′) = λ, by continuity. This implies for i = 0 and x′ ∈ E′

0,

λ = S0f(ρ0(x

′)x′)=

∫Rd+

∫[−1,1]d

χR3e|x|2 |x|2|α|+2d

(1 ∧ e

− c|x|4 sin2 θ

|y|2−|x|2+sin θ |x|2

(|y|2 − |x|2 + sin θ |x|2) 2|α|+2d−12

)

× Πα(s)ds f (y)dμ̃α(y)

� eρ20 (x′)2

r2|α|+2d0

∫Rd+

∫[−1,1]d

χ{|y|�r0}(y)

(1 ∧ e

− cr40 sin2 θ

|y|2−r20 +sin θr2

0

(|y|2 − r20 + sin θr2

0 )2|α|+2d−1

2

)

× Πα(s)ds f (y)dμ̃α(y), (3.7)


and for i = 1 and x′ ∈ E′1,

λ = S1f(ρi(x

′)x′)=

∫Rd+

∫[−1,1]d

χR5(y, s)e|x|2(1 + |x|)e− sin2 θ |x|2Πα(s)ds f (y)dμ̃α(y)

� eρ21 (x′)2

r1

∫Rd+

∫[−1,1]d

χ{|y|>r1}(y)e−c sin2 θr21 Πα(s)ds f (y)dμ̃α(y). (3.8)

Clearly, since r0 and r1 are greater than one, we have

μ̃α

{x ∈ Ei : ri � |x| � 2ri

}�

∫E′

i

dσ(x′)2ri∫

ρi (x′)

e−ρ2ρ2|α|+2d−1 dρ

�∫E′

i

e−ρi(x′)2

r2|α|+2d−2i dσ(x′).

Combining this estimate for i = 0 with (3.7), we get

μ̃α

{x ∈ Ei : ri � |x| � 2ri

}� C

λ

∫E′

0

r4|α|+4d−20 dσ(x′)(I0 + II0), (3.9)

with

I0 =∫

[−1,1]d

∫{|y|�r0,sin θr2

0 �c}

e− cr4

0 sin2 θ


0

(|y|2 − r20 + sin θr2

0 )2|α|+2d−1

2

f (y)dμ̃α(y)Πα(s)ds,

and

II0 =∫

[−1,1]d

∫{sin θr2

0 �c}f (y)dμ̃α(y)Πα(s)ds.

Similarly for i = 1 with (3.8), we obtain

μ̃α

{x ∈ E1: r1 � |x| � 2r1

}� C

λ

∫E′

1

r2|α|+2d−11 dσ(x′)(I1 + II1), (3.10)

with

I1 =∫

[−1,1]d

∫{|y|�r ,sin θr2�c}

e−c sin2 θr21 f (y)dμ̃α(y)Πα(s)ds,

1 1


and

II1 =∫

[−1,1]d

∫{sin θr2

1 �c}f (y)dμ̃α(y)Πα(s)ds.

It is immediate to verify that

r4|α|+4d−20

∫[−1,1]d

∫{x′: sin θr2

0 �c}dσ(x′)Πα(s)ds � C

and

r2|α|+2d−11

∫[−1,1]d

∫{x′: sin θr2

0 �c}dσ(x′)Πα(s)ds � C

(for more details, see [8, p. 254]), which give, after changing the order of integration in (3.9) and(3.10), the desired estimate for the terms involving II0 and II1, respectively.

Now let us prove that for |y| � r0

r4|α|+4d−20

∫[−1,1]d

∫{x′: sin θr2

0 �c}

e− cr4

0 sin2 θ


0

(|y|2 − r20 + sin θr2

0 )2|α|+2d−1

2

dσ(x′)Πα(s)ds � C

and for |y| � r1

r2|α|+2d−11

∫[−1,1]d

∫{x′: sin θr2

1 �c}e−c sin2 θr2

1 dσ(x′)Πα(s)ds � C.

Firstly, one considers the case where α = ( n12 − 1, . . . ,

nd

2 − 1), with ni ∈ N and ni > 1 foreach i = 1, . . . , d . In this case the inner integrals can be interpreted as integrals over S|n|−1 withrespect to the Lebesgue measure, expressed in polyradial coordinates. For these cases, the desired

estimates can be found in [2, p. 13]. The same estimates are obtained also for α ∈ Nd

2 − 1 + iRd .Finally the result for the other values of α are obtained via the multidimensional Stein’s complexinterpolation theorem. Indeed, let F : Cd → C the function defined by

F(ξ) = r4ξ+4d−20

∫{sin θr2

0 �c}

e− cr4

0 sin2 θ


0

(|y|2 − r20 + sin θr2

0 )2|ξ |+2d−1

2

Πξ(s)ds.

We have seen that |F(n2 − 1)| � C and it is easy to prove that |F(n

2 − 1 + iζ )| � |F(n2 − 1)|,

whenever n is a integer vector and ζ ∈ Rd . �


Proof of Proposition 6. The proof of this result follows the same steps like the proof of theweak-type boundedness on the local zone of the first order Riesz–Laguerre transforms donein [9]. Indeed the only two things left to prove are Lemma 3.3 of [9] for the kernel of a mod-ified Riesz–Laguerre transform of any order and the L2(dμ̃α)-boundedness of this transformover the global region G. For the former we have the Calderón–Zygmund-type estimates for thekernel Km. �Lemma 7. There exists a constant C such that∣∣Km(x, y, s)

∣∣ϕ(x, y, s) � Cq−(x2, y2, s

)−|α|−d,∣∣∇(x,y)

(Km(x, y, s)ϕ(x, y, s)

)∣∣ � Cq−(x2, y2, s

)−|α|−d−1/2

being ϕ(x, y, s) a cut-off function defined in [9] and (y, s) ∈ R0.

For the latter we have the following theorem regarding the Lp(dμ̃α)-boundedness for 1 <

p < ∞ of the modified Riesz–Laguerre transform of any order on G.

Theorem 8. The operator

Rmg f (x) =

∫Rd+

∫[−1,1]d

χG(y, s)Km(x, y, s) Πα(s)ds f (y)dmα(y)

is strong-type (p,p) for 1 < p < ∞ with respect to the measure μ̃α .

Proof of Lemma 7. Since |√rxi − yisi | � q1/2− (rx2, y2, s), then∣∣∣∣∣

d∏i=1

Hai

(√rxi − yisi√

1 − r

)∣∣∣∣∣e− q−(rx2,y2,s)

1−r �|m|∑k=0

(q−(rx2, y2, s)

1 − r

)k/2

e− q−(rx2,y2,s)

1−r

� e− q−(rx2,y2,s)

2(1−r) � e−cq−(x2,y2,s)

1−r ,

where last inequality follows from this one:

q−(rx2, y2, s

)� q−

(x2, y2, s

) − 2C(1 − r1/2)

when (y, s) ∈ R0 (see [7]). Thus on R0,

∣∣Km(x, y, s)∣∣ϕ(x, y, s) �

1∫0

(√

r)|m|−2(

− log r

1 − r

) |m|−22 e−c

q−(x2,y2,s)

1−r

(1 − r)|α|+d+1dr

�1/2∫0

(√

r)|m|−2(− log r)|m|−2

2 dr +1∫

1/2

e−cq−(x2,y2,s)

1−r

(1 − r)|α|+d+1dr

� 1 + q−(x2, y2, s

)−|α|−d.


In computing the gradient of the kernel with respect to y we are going to have integrals such as

Km(x, y, s)∂xjϕ(x, y, s),

1∫0

(√

r)|m|−1(

− log r

1 − r

) |m|−22 ∏

i �=j

Hai

(√rxi − yisi√

1 − r

)

× Haj −1

(√rxj − yj sj√

1 − r

)e− q−(rx2,y2,s)

1−r

(1 − r)|α|+d+3/2dr,

with H−1 ≡ 0 and

1∫0

(√

r)|m|−1(

− log r

1 − r

) |m|−22

d∏i=1

Hai

(√rxi − yisi√

1 − r

)

×√

rxj − yj sj√1 − r

e− q−(rx2,y2,s)

1−r

(1 − r)|α|+d+3/2dr.

In order to estimate these three integrals we use the same estimates described at the beginning ofthis proof. For the first one we have to use also that

∣∣∇xϕ(x, y, s)∣∣ + ∣∣∇yϕ(x, y, s)

∣∣ � 1

q1/2− (x2, y2, s)

.

The gradient with respect to x is treated similarly. �Proof of Theorem 8. The proof of this result is an adaptation to our context of the same re-sult for the higher order Riesz–Gauss transforms done in [6]. Taking into account that on G,q+(x2, y2, s)q−(x2, y2, s) � c when cos θ � 0, by (3.1) an upper bound for |Km(x, y, s)| is

˜̃K m(x, y, s) =⎧⎨⎩ (|x|2 + |y|2) |m|−2

2 e−|y|2 if cos θ < 0,

q|α|+d+ (q+q−)

|m|−14 e−(

|y|2−|x|22 + (q+q−)1/2

2 ) if cos θ � 0.

Thus, ∫Rd+

∣∣∣∣ ∫G∩{cos θ<0}

Km(x, y, s)Πα(s)ds f (y)dmα(y)

∣∣∣∣p dμ̃α(x)

�∫

Rd+

( ∫G∩{cos θ<0}

˜̃Km

(x, y, s)Πα(s)ds∣∣f (y)

∣∣dmα(y)

)p

dμ̃α(x)

�∫

Rd

( ∫Rd

(|x|2 + |y|2) p′(|m|−2)2 dμ̃α(y)

)p−1

dμ̃α(x)‖f ‖p

Lp(dμ̃α).

+ +


For the region G ∩ {cos θ � 0} we are going to use the following estimates:

|x|2 + |y|2 � q+ � |x + y|2, q− � |x − y|, (q+q−)1/2 �∣∣|y|2 − |x|2∣∣,(

1

p− 1

2

)(|y|2 − |x|2) �∣∣∣∣ 1

p− 1

2

∣∣∣∣(q+q−)1/2 <(q+q−)1/2

2since p > 1,∫

Rd+

∣∣∣∣ ∫G∩{cos θ�0}

Km(x, y, s)Πα(s)ds f (y)dmα(y)

∣∣∣∣p dμ̃α(x)

�∫

Rd+

( ∫G∩{cos θ�0}

˜̃Km

(x, y, s)Πα(s)ds∣∣f (y)

∣∣dmα(y)

)p

dμ̃α(x)

�∫

Rd+

( ∫G∩{cos θ�0}

|x + y|2|α|+2d(q+q−)m−1

4 e− |y|2−|x|2

2 + |y|2−|x|2p

− (q+q−)1/2

2 Πα(s)ds

× ∣∣f (y)∣∣e− |y|2

p dmα(y)

)p

dmα(x)

�∫

Rd+

( ∫G∩{cos θ�0}

|x + y|2|α|+2d(q+q−)m−1

4 e−( 1

2 −| 1p

− 12 |)(q+q−)1/2

Πα(s)ds

× ∣∣f (y)∣∣e− |y|2

p dmα(y)

)p

dmα(x)

�∫

Rd+

( ∫G∩{cos θ�0}

|x + y|2|α|+2de−cp(q+q−)1/2Πα(s)ds

∣∣f (y)∣∣e− |y|2

p dmα(y)

)p

dmα(x).

To finish the proof we just need to check that the kernel

H(x,y, s) := |x + y|2|α|+2de−cp(q+q−)1/2χG∩{cos θ�0},

for G = {(x, y, s): q1/2− (x2, y2, s) � C

1+|x|+|y| }, is in L1(dmα(y)) and L1(dmα(x)) indepen-dently of the remaining variables. Due to the symmetry of the kernel we are going to check onlythe first claim:∫

Rd+

H(x,y, s)dmα(y) �∫

|x−y|�1

|y|2|α|+2de−cp |y|q1/2− dmα(y)

+∫

|x−y|>1

(|x|2|α|+2d + |y|2|α|+2d)e−c̃p(|x|+|y|) dmα(y).

It is clear that the second integral is bounded independently of x and s, for the first one see (3.6)with x substituted by y. �


Proof of Theorem 2. As we mention in the preliminaries to prove this theorem is equivalentto prove that the modified Riesz–Laguerre transforms of order higher than 2 map L1(w̃ dμ̃α)

continuously into L1,∞(dμ̃α), with w̃(y) = (1 + |y|)|m|−2.For each x ∈ Rd+, let us write

Rd+ × [−1,1]d =4⋃

i=0

Ri.

Therefore, in order to get the result, it will be enough to prove that each of the followingoperators

T mi f (x) =

∫Rd+

∫[−1,1]d

χRi(y, s)Km(x, y, s)Πα(s)ds f (y)d(mα),

for i = 0, . . . ,4, maps L1(w̃ dμ̃α) continuously into L1,∞(dμ̃α).Observe that according to Proposition 6, for all m the operator T m

0 is weak-type (1,1) withrespect to μ̃α .

On the other hand, for the ‘global parts’: R1, R2, R3, and R4, we have the following estimatefor the kernel Km:

∣∣Km(x, y, s)∣∣ �

{(|x|2 + |y|2) |m|−2

2 e|x|2 K∗(x, y, s)e−|y|2 , cos θ < 0,

(q+q−)|m|−2

4 e|x|2 K∗(x, y, s)e−|y|2 , cos θ � 0

(for more details see kernel (3.1) together with proof of Proposition 4 or papers [6,9]).If (y, s) ∈ R1, |Km(x, y, s)| is controlled by C(1 + max{|x|, |y|})|m|−2e−|y|2 and therefore it

is immediate to prove that T m1 maps L1(w̃ dμ̃α) into L1(dμ̃α).

Now if (y, s) ∈ Ri , with i = 2,3,4, we claim that∣∣Km(x, y, s)∣∣ � w̃(y)e|x|2 K∗(x, y, s)e−|y|2 .

Assuming the claim, the conclusion follows from Proposition 5. We must finally prove the claim.If (y, s) ∈ R2, since q+ � (|x| + |y|)2 � |x|2, then

∣∣Km(x, y, s)∣∣ � (q+q−)

|m|−24 e|x|2e−c|x|q1/2

− e−|y|2

�(|x|q1/2

−) |m|−2

2 e|x|2e−c|x|q1/2− e−|y|2

� w̃(y)e|x|2e−c|x|q1/2− e−|y|2 .

If (y, s) ∈ R3 ∪ R4,

q+q− = (|x|2 + |y|2)2 − 4|x|2|y|2 sin2 θ

�(|x|2 + |y|2)2 � |y|4.


Thus ∣∣Km(x, y, s)∣∣ � (q+q−)

|m|−24 e|x|2 K∗(x, y, s)e−|y|2

� w̃(y)e|x|2 K∗(x, y, s)e−|y|2 ,

and this concludes the proof of the theorem. �Proof of Theorem 3. This proof follows essentially from the proof of Theorem 4.1 in [9]. Withthe notation of that theorem, one takes η ∈ R

d+ with |η| sufficiently large, away from the axis andobtains the following lower bound for Km(x,η)

Km(x,η) = C

∫[−1,1]d

Km(x,η, s)Πα(s)ds � C|η||m|−2|α|−d−1eξ2−|η|2 (3.11)

for x ∈ J = {ξ η|η| + v: v ⊥ η, |v| < 1, 1

2 |η| < ξ < 32 |η|}.

Now if we assume that the Riesz–Laguerre transforms of order |m| > 2 map L1(w̃ε dμ̃α)

continuously into L1,∞(dμ̃α) with w̃ε(y) = (1 + |y|)ε and 0 < ε < |m| − 2 then by takingf � 0 in L1(w̃ε dμ̃α) close to an approximation of a point mass at η, with ‖f ‖L1(w̃ε dμ̃α) = 1 we

have that Rmα f (x) is close to e|η|2 Km(x,η)|η|−ε and by applying inequality (3.11) we get that

e|η|2 Km(x,η)|η|−ε � |η||m|−2|α|−d−1−εe(|η|2 )2

. Therefore by setting λ = c|η||m|−2|α|−d−1−εe(|η|2 )2

we obtain

e−(|η|2 )2 |η|2|α|+d−1 � μ̃α(J )

� μ̃α

{x ∈ Rd+: Rm

α f (x) > λ}

� 1

λ= C|η|2|α|+d−|m|+1+εe−(

|η|2 )2

.

Hence |η||m|−2−ε must be bounded which is a contradiction. Therefore the conclusion of Theo-rem 3 holds. �Acknowledgment

We thank the referee on pointing us out Gutiérrez et al. [5] and the alternative proof of Theo-rem 2 for half-integer multi-indices.

References

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[7] E. Sasso, Spectral multipliers of Laplace transform type for the Laguerre operator, Bull. Austral. Math. Soc. 69 (2)(2004) 255–266.

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weak-type inequalities for higher order riesz–laguerre transforms

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