On Hardy spaces associated with Bessel operators

ON HARDY SPACESASSOCIATED WITH BESSEL OPERATORS

By

JORGE J. BETANCOR∗, JACEK DZIUBANSKI† AND JOSE LUIS TORREA

Abstract. In this paper, we study Hardy spaces associated with two Besseloperators. Two different kind of Hardy spaces appear. These differences aretransparent in the corresponding atomic decompositions.

1 Introduction

In 1965, Muckenhoupt and Stein [17] introduced a notion of conjugacy associatedwith the Bessel operators ∆λ, λ > 0, defined by

∆λf(x) = − d2

dx2f(x) − 2λ

x

d

dxf(x), x > 0.

They developed in this setting a theory parallel to the classical case associated tothe Euclidean Laplacian. In their paper, definitions of Poisson kernels, harmonicfunctions, conjugate functions and fractional integrals associated with ∆λ aregiven. Results parallel to the classical case about Lp((0,∞), x2λdx)-boundedness,1 ≤ p < ∞, for these operators were obtained.

Recently, following a different procedure from that used in [17], Betancor,Buraczewski, Farina, Martınez, and Torrea [2] investigated Lp((0,∞), dx) boundsof the Riesz transforms associated with the Bessel operators Sλ, defined by

Sλf(x) = − d2

dx2f(x) +

λ2 − λ

x2f(x), x > 0.

In the present note, we analyze “the” real H1 Hardy space in the settings of∆λ and Sλ. We recall that in the classical (Euclidean Laplacian) case, it is well-known that the Hardy space (originally defined through the boundedness in L1

∗The first author was partially supported by MTM2004/05878.†The second author was supported by the European Commission Marie Curie Host Fellowship

for the Transfer of Knowledge “Harmonic Analysis, Nonlinear Analysis and Probability” MTKD-CT-2004-013389 and by Polish funds for science in years 2005–2008 (research project 1P03A03029).

JOURNAL D’ANALYSE MATHEMATIQUE, Vol. 107 (2009)

DOI 10.1007/s11854-009-0008-1

195

196 JORGE J. BETANCOR, JACEK DZIUBANSKI AND JOSE LUIS TORREA

of the conjugate function) can also be defined either through the boundedness ofthe maximal operator of the Poisson semigroup or via an atomic decomposition.Our aim is to study these three different characterizations of Hardy spaces in bothsettings ∆λ and Sλ.

Definition 1.1. Let us denote by L either the differential operator ∆λ or thedifferential operator Sλ. We consider the Riesz transform, heat semigroup andPoisson semigroup associated to L. These operators are defined in (1.5) and (1.8)below and are denoted respectively by RL, e−tL and e−t

√L.

We define the following spaces:(a) H1

Riesz(L) = {f : f ∈ L1 and RLf ∈ L1} with norm

‖f‖H1Riesz(L) = ‖f‖L1 + ‖RLf‖L1.

(b) H1max(L) = {f : f ∈ L1 and sup

t>0|e−t

√Lf | ∈ L1} with norm

‖f‖H1max(L) = ‖f‖L1 + ‖ sup

t>0|e−t

√Lf |‖L1.

In contrast with the situation of Lp-bounds, 1 < p < ∞, for the Riesz transformsand maximal functions associated with L, it turns out that although a characteri-zation of the type H1

Riesz(L) ∼ H1max(L) can be proved for both ∆λ and Sλ, the

atomic characterizations are of a different nature. This difference consists in thefact that every atom for the space H1

max(∆λ) satisfies the cancellation condition,while there are some H1

max(Sλ)-atoms which need not have this property.

Operators associated to the operator ∆λ, λ > 0. We review some def-initions and properties useful in the sequel; cf. [2] and [17]. Let Jν denote theBessel function of the first kind and order ν. Then

∆λ,x((xy)−λ+1/2Jλ−1/2(xy)) = y2(xy)−λ+1/2Jλ−1/2(xy), x, y ∈ (0,∞).

The Poisson semigroup {P [λ]t }t>0 generated by the operator −

√∆λ is defined by

P[λ]t f(x) = e−t

√∆λf(x) =

∫ ∞

0

P[λ]t (x, y)f(y)y2λdy,

where

P[λ]t (x, y) =

∫ ∞

0

e−tz(xz)−λ+1/2Jλ−1/2(xz)(yz)−λ+1/2Jλ−1/2(yz)z2λdz(1.1)

=2λt

π

∫ π

0

(sin θ)2λ−1

(x2 + y2 + t2 − 2xy cos θ)λ+1dθ, t, x, y ∈ (0,∞).

HARDY SPACES AND BESSEL OPERATORS 197

The last equality was established by Weinstein [24]. Thus {P [λ]t }t>0 is a

contraction semigroup on Lp((0,∞), x2λdx), for every 1 ≤ p ≤ ∞. Moreover, forevery f ∈ Lp((0,∞), x2λdx), 1 ≤ p < ∞, the Poisson integral u(t, x) = P

[λ]t (f)(x),

satisfies the differential equation

(1.2) ∂2t u(t, x) + ∂2

xu(t, x) +2λ

x∂xu(t, x) = 0, t, x ∈ (0,∞).

If f ∈ Lp((0,∞), x2λdx), 1 ≤ p < ∞, the ∆λ-conjugate of f (or of the Poissonintegral of f ) is defined by

Q[λ]t (f)(x) =

∫ ∞

0

Q[λ]t (x, y)f(y)y2λdy,

where

Q[λ]t (x, y) = −(xy)−λ+1/2

∫ ∞

0

e−tξξJλ+1/2(xξ)Jλ−1/2(ξy) dξ

=−2λ

π

∫ π

0

(x − y cos θ)(sin θ)2λ−1

(x2 + y2 + t2 − 2xy cos θ)λ+1dθ, t, x, y ∈ (0,∞);

(1.3)

see [17, page 84]. For every f ∈ Lp((0,∞), x2λdx), 1 ≤ p < ∞, the pair offunctions

u(t, x) = P[λ]t (f)(x) and v(t, x) = Q

[λ]t (f)(x), t, x ∈ (0,∞),

satisfy the Cauchy-Riemann type equations

(1.4) ∂tv + ∂xu = 0, ∂tv − ∂xu − 2λ

xv = 0.

Moreover, there exists the boundary value function

limt→0

Q[λ]t (f)(x), a.e. x ∈ (0,∞),

which defines the ∆λ-Riesz transform ∂x(∆λ)−1/2f = R∆λf of f . It is known

that R∆λis a bounded operator on Lp((0,∞), x2λdx) for every 1 < p < ∞ and

bounded from L1((0,∞), x2λdx) into L1,∞((0,∞), x2λdx). In [2], it was shownthat if 1 ≤ p < ∞ and f ∈ Lp((0,∞), x2λdx), then

R∆λ(f)(x) = lim

ε→0R∆λ,εf(x)(1.5)

= limε→0

∫ ∞

0,|x−y|>ε

Q[λ]0 (x, y)f(y)y2λdy, a.e. x ∈ (0,∞).


Moreover, in [3] it was proved that R∆λis a Calderon-Zygmund operator

on the space of homogeneous type ((0,∞), ρ, dmλ), where ρ(x, y) = |x − y| anddmλ(x) = x2λdx. For the definition and properties of spaces of homogeneous type,see [4], [5], and [22]. We denote by H1

CW ((0,∞), dmλ) the Hardy space associatedwith ((0,∞), ρ, dmλ) defined in [5]. To be precise; we say that a measurablefunction a is an H1

CW ((0,∞), dmλ)-atom if there exists a bounded interval I ⊂[0,∞) such that suppa ⊂ I, ‖a‖L∞((0,∞), dmλ) ≤ 1/mλ(I), and

∫∞0

a(x)dmλ(x) = 0.A function f ∈ L1((0,∞), dmλ) is in H1

CW ((0,∞), dmλ) if and only if f(x) =∑∞j=1 αjaj(x), in L1((0,∞), dmλ), where for every j, aj is an H1

CW ((0,∞), dmλ)-atom and αj ∈ C, with

∑∞j=1 |αj | < ∞. The norm ‖f‖H1

CW ((0,∞),dmλ) is definedby

(1.6) ‖f‖H1CW ((0,∞),dmλ) = inf

∞∑

j=1

|αj |,

where the infimum is taken over all absolutely summable sequences {αj}j∈N,αj ∈ C, for which f =

∑∞j=1 αjaj , with aj being H1

CW ((0,∞), dmλ)-atoms.Now we can state the theorem about the Hardy space H1 related to the operator

∆λ.

Theorem 1.7. Let λ > 0 and f ∈ L1((0,∞), dmλ). The following assertions

are equivalent.(i) f ∈ H1

CW ((0,∞), dmλ);

(ii) f ∈ H1max(∆λ);

(iii) f ∈ H1Riesz(∆λ).

Moreover, the corresponding norms are equivalent.

The Poisson semigroup {P [λ]t }t>0 can be seen as special case of a Hankel

convolution semigroup (cf. [21]). In Section 2 (see Theorem 2.7), we obtaincharacterizations of H1

max(∆λ) in terms of maximal operators associated with theHankel convolution.

Operators associated to the operator Sλ, λ > 0. In [2], the Riesz trans-form RSλ

associated with the Bessel operator Sλ was investigated. If f is a smoothfunction having compact support on (0,∞), then RSλ

f is defined by

(1.8) RSλ(f)(x) = Mxλ

d

dxMx−λS

−1/2λ f(x),

where Mϕ(x)f(x) = ϕ(x)f(x) and the negative power S−1/2λ is given by the formula

S−1/2λ (f)(x) =

1√π

∫ ∞

0

P[λ]t (f)(x)dt.


Here

P[λ]t (f)(x) =

∫ ∞

0

P[λ]t (x, y)f(y)dy,

is the Poisson semigroup and

P[λ]t (x, y) = (xy)λP

[λ]t (x, y), t, x, y ∈ (0,∞),

with P[λ]t being the kernel defined in (1.1). Thus RSλ

can be extended to abounded operator on Lp((0,∞), dx) when 1 < p < ∞, and from L1((0,∞), dx)

into L1,∞((0,∞), dx). Also, RSλis a principal value Calderon-Zygmund operator.

In [7], Fridli introduced an atomic Hardy type space H1F (0,∞) as follows. A

measurable function a defined on (0,∞) is said to be an F -atom if

(a) a = 1δ χ(0,δ), for some δ > 0, where χ(0,δ) denotes the characteristic function

on the interval (0, δ),or

(b) there exists a bounded interval I⊂(0,∞) such that suppa⊂I,∫

I a(x)dx = 0,and ‖a‖L∞((0,∞), dx) ≤ |I|−1, where |I| denotes the length of I.

Following Fridli, a function f ∈ L1((0,∞), dx) it is said to be in H1F ((0,∞), dx)

if and only if f(x) =∑∞

j=1 αjaj(x), where, for every j ∈ N, aj is an F -atom andαj ∈ C,

∑∞j=1 |αj | < ∞. The norm ‖f‖H1

F (0,∞) of f ∈ H1F (0,∞) is defined by

(1.9) ‖f‖H1F (0,∞) = inf

∞∑

j=1

|αj |,

where the infimum is taken over all absolutely summable sequences {αj}j∈N,αj ∈ C, for which f =

∑∞j=1 αjaj , aj being an F -atom, for every j ∈ N.

Now we can state the result parallel to Theorem 1.7 for the operator Sλ.

Theorem 1.10. Let λ > 0. The following assertions are equivalent:

(i) f ∈ H1F ((0,∞), dx);

(ii) f ∈ H1max(Sλ);

(iii) f ∈ H1Riesz(Sλ).

Moreover, the corresponding norms are equivalent.

In the last section of the paper, we study Hardy type inequalities and trans-plantation and multiplier theorems associated with Hankel transforms on the spaceH1

F (0,∞).


Throughout this paper C denotes a positive constant that can change from oneline to the next.

2 The Hardy space H1

max(∆λ).

In this section, we prove Theorem 1.7. Recall that dmλ(x) = x2λdx. If f and g arein L1((0,∞), dmλ), then their Hankel convolution is defined by

(f#λg)(x) =

∫ ∞

0

f(y)τ [λ]x (g)(y)dmλ(y),

where τ[λ]x g, x ∈ [0,∞), denotes the Hankel translation of g given by

τ [λ]x (g)(y) =

Γ(λ + 1/2)

Γ(λ)√

π

∫ π

0

g(√

(x − y)2 + 2xy(1 − cos θ))(sin θ)2λ−1dθ.(2.1)

It is well-known that for f, g ∈ L1((0,∞), dmλ) ∩ L2((0,∞), dmλ) one has

f#λg = F [λ]((F [λ]f)(F [λ]g)

),

where

(2.2) F [λ]f(ξ) =

∫ ∞

0

(ξy)−λ+1/2Jλ−1/2(ξy)f(y)dmλ(y)

is the Hankel (Fourier-Bessel) transform.The convolution operation associated with the Hankel transform was investi-

gated in [9] and [10]. For every x ∈ [0,∞), the operator τ[λ]x is a contraction on

Lp((0,∞), dmλ), 1 ≤ p ≤ ∞. Moreover, a Young type inequality holds for #λ.Using the Hankel translations and convolution, we have the following repre-

sentations for the Poisson kernels and the Poisson integrals defined in (1.1):

(2.3) P[λ]t (x, y) = τ [λ]

x gt(y), P[λ]t (f)(x) = f#λgt(x), t, x, y ∈ (0,∞),

where

(2.4) g(z) =2λΓ(λ)√

πΓ(λ + 1/2)

1

(1 + z2)λ+1, z ∈ (0,∞),

and gt(z) = t−2λ−1g(z/t), z, t ∈ (0,∞).Let Z [λ] denote the set consisting of all C1-functions on [0,∞) such that φ(0) > 0

and

0 ≤ φ(x) ≤ C(1 + x2)−λ−1, x ≥ 0,(2.5)

|φ′(x)| ≤ Cx(1 + x2)−λ−2, x ≥ 0,(2.6)


with a constant C > 0 which depends on φ. Observe that the function g defined in(2.4) belongs to Z [λ]. Moreover, if h(z) = 2(1−2λ)/2 exp(−z2/2)/Γ(λ + 1/2), thenexp(−t∆λ)f(x) = f#λh√

t(x); cf. [21] . Clearly h ∈ Z [λ].

For φ ∈ Z [λ], we define the maximal function

Mφf(x) = supt>0

|f#λφt(x)|.

The following theorem includes (i)⇐⇒(ii) of Theorem 1.7.

Theorem 2.7. For every φ ∈ Z [λ], there exists a constant C > 0 such that

(2.8) C−1‖Mφf‖L1((0,∞), dmλ) ≤ ‖f‖H1CW (∆λ) ≤ C‖Mφf‖L1((0,∞), dmλ).

Proof. Define (cf. [6])

(2.9) t(x, r) =

rx−2λ for r < x2λ+1,

r1/(2λ+1) for r ≥ x2λ+1.

For φ ∈ Z [λ], we set Φt(x, y) = τ[λ]x φt(y) and introduce the new kernel

(2.10) Φr(x, y) = rΦt(x,r)(x, y), r, x, y ∈ (0,∞).

Clearly,

(2.11) supr>0

∣∣∣∣∫ ∞

0

Φr(x, y)f(y)dmλ(y)

r

∣∣∣∣ = Mφf(x).

Let

dλ(x, y) =

∣∣∣∣∫ y

x

dmλ(t)

∣∣∣∣.

Then ((0,∞), dλ(x, y), dmλ) is a space of homogeneous type (see [5], [22]), and thespaces H1

CW ((0,∞), dλ(x, y), dmλ) and H1CW ((0,∞), |x − y|, dmλ) are exactly the

same. The proof of Theorem 2.7 is completed by applying (2.11) and Corollary 1of [22], once we have established the following proposition. �

Proposition 2.12. There exist constants A > 0 and γ > 0 such that

(i) Φr(x, x) > 1/A, r, x ∈ (0,∞);

(ii) 0 ≤ Φr(x, y) ≤ A(1 + dλ(x,y)

r

)−1−γ, r, x, y ∈ (0,∞);

(iii) for every r, x, y ∈ (0,∞) such that dλ(y, z) ≤ 14A (r + dλ(x, y)),

|Φr(x, y) − Φr(x, z)| ≤ A(dλ(y, z)

r

)γ(1 +

dλ(x, y)

r

)−1−2γ

.


Proof. Note that

t(sx, s2λ+1r) = st(x, r),(2.13)

dλ(sx, sy) = s2λ+1dλ(x, y),(2.14)

Φr(sx, sy) = Φs−2λ−1r(x, y).(2.15)

Proof of (i). According to (2.15), it suffices to prove (i) for x = 1. Sinceφ(0) > 0, there exist constants δ, c > 0 such that φ(x) > c for 0 < x < δ.

If r < 1, then t(1, r) = r. There exists δ′ > 0 such that φ(√

2(1 − cos θ)/r)≥ c

for 0 ≤ θ ≤ δ′r. Hence

(2.16) Φr(1, 1) ≥ r−2λ

∫ δ′r

0

c(sin θ)2λ−1 dθ ≥ A−1.

If r ≥ 1, then t(1, r) = r1/(2λ+1); and for some δ′ > 0, we have φ(√2(1−cos θ)

r1/(2λ+1)

)≥ c

for 0 ≤ θ ≤ δ′. Therefore,

(2.17) Φr(1, 1) ≥ c

∫ δ′

0

(sin θ)2λ−1 dθ ≥ A−1.

Thus we have shown that (i) holds. �

Proof of (ii). We show that (ii) is satisfied with γ = 1/(2λ + 1). Again, by(2.14) and (2.15), it is enough to prove (ii) for x = 1.

Case 1: r < 1. Using (2.1), (2.5), and (2.9), we have

(2.18) Φr(1, y) ≤ Cr2

∫ π

0

(sin θ)2λ−1

(r2 + (1 − y)2 + 2y(1 − cos θ))λ+1dθ.

If r/2 ≤ |1 − y| and 1/2 ≤ y ≤ 2, then dλ(1, y) ∼ |1 − y| and, consequently,

Φr(1, y) ≤ Cr2

(∫ |1−y|

0

(sin θ)2λ−1

|1 − y|2λ+2dθ +

∫ π

|1−y|

(sin θ)2λ−1

θ2λ+2dθ

)

≤ C(1 +

|1 − y|r

)−2

∼ C(1 +

dλ(1, y)

r

)−2

.

(2.19)

If r/2 ≤ |1 − y| and |1 − y| > 1/2, then |1 − y| ∼ dλ(1, y)1/(2λ+1). Hence

Φr(1, y) ≤ Cr2

∫ π

0

(sin θ)2λ−1

|1 − y|2λ+2dθ ≤ Cr2dλ(1, y)−1−1/(2λ+1)

≤ C(1 +

dλ(1, y)

r

)−1−1/(2λ+1)

.

(2.20)


If r/2 > |1 − y|, then y ∼ 1 and |1 − y| ∼ dλ(1, y). Thus

Φr(1, y) ≤ Cr2

∫ r

0

(sin θ)2λ−1

r2λ+2dθ + Cr2

(∫ π/2

r

θ−3 dθ +

∫ π

π/2

(sin θ)2λ−1

θ2λ+2dθ

)

≤ C ≤ C(1 +

dλ(1, y)

r

)−2

.

(2.21)

Case 2: r ≥ 1. Then t(r, 1) = r1/(2λ+1). Applying (2.1) and (2.5), we get(2.22)

Φr(1, y) ≤ Cr(2λ+2)/(2λ+1)

∫ π

0

(sin θ)2λ−1

((1 − y)2 + r2/(2λ+1) + 2y(1 − cos θ))λ+1dθ.

If |1 − y| ≥ r1/(2λ+1), then |1 − y| ∼ dλ(1, y)1/(2λ+1). Hence(2.23)

Φr(1, y) ≤ Cr(2λ+2)/(2λ+1)

∫ π

0

(sin θ)2λ−1

|1 − y|2λ+2dθ ≤ C

(1 +

dλ(1, y)

r

)−1−1/(2λ+1)

.

If |1 − y| < r1/(2λ+1), then dλ(1, y) ≤ Cr. Thus(2.24)

Φr(1, y) ≤ C

∫ π

0

r(2λ+2)/(2λ+1)(sin θ)2λ−1

(r2/(2λ+1) + 2y(1 − cos θ))λ+1dθ ≤ C ≤ C

(1 +

dλ(1, y)

r

)−2

.�

Proof of (iii). First we prove

Lemma 2.25. There exist constants C, γ > 0 such that for every x, y, z ∈(0,∞) and all r > 0, one has

(2.26) |Φr(x, y) − Φr(x, z)| ≤ C(dλ(y, z)

r

)γ

.

Proof. It suffices to prove (2.26) for x = 1 and y < z and dλ(y, z) ≤ r/Cλ,where Cλ is a fixed (large) constant. Set L = |Φr(1, y)− Φr(1, z)|. By (2.1) and themean value theorem we have

L =Γ(λ + 1/2)r√

πΓ(λ)t(r, 1)2λ+1

∣∣∣∣∫ π

0

(y − z)(sin θ)2λ−1((1 − cos θ) − 1 + u)

t(r, 1)2

× φ′(√

(1 − u)2 + 2u(1 − cos θ)/t(r, 1))√(1 − u)2 + 2u(1 − cos θ)/t(r, 1)

dθ

∣∣∣∣,

for some u ∈ (y, z). Applying (2.6), we obtain

(2.27) L ≤ C|y − z|t(r, 1)r

∫ π

0

(sin θ)2λ−1((1 − cos θ) + |u − 1|)((1 − u)2 + t(r, 1)2 + 2u(1 − cos θ))λ+2

dθ.


Set

Ξ(r, u, θ) =((1 − cos θ) + |u − 1|)

((1 − u)2 + t(r, 1)2 + 2u(1 − cos θ))λ+2.

Case 1: r ≥ 1. Then t(r, 1) = r1/(2λ+1).If |1 − u| ≤ r1/(2λ+1), then Ξ(r, u, θ) ≤ Cr−(2λ+3)/(2λ+1). Hence, using (2.27) andthe fact that |y − z| ≤ Cdλ(y, z)1/(2λ+1), we have

(2.28) L ≤ C( |y − z|

r1/(2λ+1)

)≤ C

(dλ(y, z)

r

)1/(2λ+1)

.

If |1 − u| ≥ r1/(2λ+1), then Ξ(r, u, θ) ≤ C|1 − u|−2λ−3. Thus, once again, we have

(2.29) L ≤ C|y − z|

r1/(2λ+1)

(r1/(2λ+1)

|1 − u|)2λ+3

≤ C(dλ(y, z)

r

)1/(2λ+1)

.

Case 2: r < 1. In this case, t(r, 1) = r.If |u − 1| ≥ 1/4, then Ξ(r, u, θ) ≤ C. Hence

(2.30) L ≤ C|y − z|r2 ≤ C(dλ(y, z)

r

)1/(2λ+1)

.

If |u − 1| ≤ 1/4 and |u − 1| > r/4, then

L ≤C|y − z|r2

(∫ |u−1|

0

θ2λ−1(θ2 + |u − 1|)|u − 1|2λ+4

dθ +

∫ π/2

|u−1|

θ2λ−1(θ2 + |u − 1|)θ2λ+4

dθ

(2.31)

+

∫ π

π/2

(sin θ)2λ−1(θ2 + |u − 1|)θ2λ+4

dθ

)

≤C|y − z| r2

|u − 1|3 ≤ C|y − z|

r.

Note that the conditions |u − 1| < 1/4 and dλ(y, z) < r/Cλ imply y ∼ z ∼ 1 and,consequently, dλ(y, z) ∼ |y − z|. Thus (2.31) leads to

(2.32) L ≤ Cdλ(y, z)

r.

Finally, if |u − 1| < r/4, then

L ≤C|y − z|r2

(∫ r

0

θ2λ−1(θ2 + |u − 1|)r2λ+4

dθ +

∫ π/2

r

θ2λ−1(θ2 + |u − 1|)θ2λ+4

dθ(2.33)

+

∫ π

π/2

(sin θ)2λ−1(θ2 + |u − 1|)θ2λ+4

dθ

)

≤C(|y − z| + |y − z||u − 1|

r2+ r2|y − z||u − 1|

)≤ C

|y − z|r

.


Using the same argument as above, we get that (2.32) is also satisfied in this case.The proof of Lemma 2.25 is complete. �

Having proved Lemma 2.25, we are now in a position to finish the proof ofProposition 2.12. Assume that dλ(y, z) ≤ (r + dλ(x, y))/(4A). Then for everyγ′ > 0, one has

(2.34)(1 +

dλ(x, z)

r

)−1−γ′

≤ C(1 +

dλ(x, y)

r

)−1−γ′

.

Using (ii) and (2.34), we get

(2.35) |Φr(x, y) − Φr(x, z)| ≤ C(1 +

dλ(x, y)

r

)−1−1/(2λ+1)

.

Finally, (iii) follows from (2.26) and (2.35). �

As was mentioned in the Introduction, the Riesz transform R∆λis a Calderon-

Zygmund operator on the space of homogeneous type ((0,∞), ρ, dmλ), whereρ(x, y) = |x−y| (see [3]). It follows that R∆λ

is a bounded operator from H1CW (∆λ)

into L1((0,∞), dmλ). Hence Theorem 2.7 gives (ii)=⇒(iii) in Theorem 1.7. Letus mention here that in the case λ ≥ 1

2 , the boundedness of R∆λfrom H1

CW (∆λ)

into L1((0,∞), dmλ) can also be deduced from arguments in [8].

Our next objective is to prove (iii)=⇒(ii) in Theorem 1.7. To do this, we needsome preparation.

Lemma 2.36. Assume that u(t, x) is a C2 function on (0,∞)×(0,∞) satisfying(1.2). Then the following conditions are equivalent:

a) u(t, x) = P[λ]t (f)(x), for some f ∈ H1

max(∆λ);b) u(t, x) can be extended to an even function of x on (0,∞) × R, which is C2

and such that u∗(x) = supt>0 |u(t, x)| is in L1((0,∞), dmλ(x)).

Obviously, in these cases, ‖u∗‖L1((0,∞), dmλ) = ‖f‖H1max(∆λ).

Proof. This proof follows a standard method ([20, Proposition 1, p. 119]).

Let u(t, x) = P[λ]t (f)(x), t, x ∈ (0,∞), for some f ∈ H1

max(∆λ). Then u satisfies(1.2); and, by (1.1), the function u(t, x) has a unique C∞ extension to an evenfunction of x on (0,∞) × R. Trivially, u∗ ∈ L1((0,∞), dmλ). It is clear that

(2.37) supt>0

∫ ∞

0

|u(t, x)|dmλ(x) ≤∫ ∞

0

u∗(x)dmλ(x) < ∞.


Assume now that b) holds. Since the even (with respect to x) extension ofthe function u is of class C2, it satisfies (1.2) on (0,∞) × R. For ε > 0, letfε(x) = u(ε, x) and uε(t, x) = u(t + ε, x), x, t ∈ (0,∞). Note that P

[λ]t (fε)(x) has

a unique continuous extension to an even (with respect to x) function of (t, x) on[0,∞) × R. Therefore, by [17, Lemma 11 and its Corollary, p. 85], we get

uε(t, x) = P[λ]t (fε)(x), t, x, ε > 0.

By (2.37), the set {fε}ε>0 is bounded in L1((0,∞), dmλ); and then {fε(x)dmλ(x)}ε>0

is bounded in the space M([0,∞)) of the complex measures on [0,∞). Hence thereexist a sequence εk > 0, εk → 0, and γ ∈ M([0,∞)) such that fεk

dmλ → dγ ask → ∞ in the weak * topology of M([0,∞)). Then, for every t, x ∈ (0,∞), wehave

u(t + εk, x) = P[λ]t (fεk

)(x) →∫ ∞

0

P[λ]t (x, y)dγ(y), as k → ∞.

On the other hand, for every t, x ∈ (0,∞), u(x, t + εk) → u(x, t), as k → ∞. Hence

u(x, t) =

∫ ∞

0

P[λ]t (x, y)dγ(y), t, x ∈ (0,∞).

Since |fε(x)|x2λ ≤ u∗(x)x2λ and∫∞0 u∗(x)x2λdx < ∞, we easily conclude that

dγ(x) = f(x)dmλ(x) with f ∈ H1max(∆λ). �

Lemma 2.38. Assume that u(t, x) and v(t, x) are, respectively, even and odd(with respect to x) real valued C2 functions on (0,∞) × R satisfying (1.4). Let

F = (u2 + v2)1/2 and suppose that

(2.39) supt>0

∫ ∞

0

F (t, x)dmλ(x) < ∞.

Then u∗ ∈ L1((0,∞), dmλ) and

‖u∗‖L1((0,∞), dmλ) ≤ C supt>0

∫ ∞

0

F (t, x)dmλ(x).

Proof. By [17, Lemma 5], for every p > 2λ/(2λ + 1), the function F p issatisfies

∂2t (F p)(t, x) + ∂2

x(F p)(t, x) +2λ

x∂x(F p)(t, x) ≥ 0

on the region where F > 0. Let 2λ/(2λ + 1) < p < 1, and set r = 1/p. Forevery ε > 0, we write Fε(x) = F (ε, x), x ∈ (0,∞). According to (2.39), F p

ε ∈Lr((0,∞), dmλ). Then P

[λ]0 (F p

ε )(x) = F pε (x); and, applying [17, Lemma 11], we

get

(2.40) F p(t + ε, x) ≤ P[λ]t (F p

ε )(x).


By (2.39), there exists h ∈ Lr((0,∞), dmλ) and a sequence εk → 0 suchthat F p

εkconverges weakly to h in Lr((0,∞), dmλ), and ‖h‖r

Lr((0,∞), dmλ) ≤supt>0

∫∞0

F (t, x)dmλ(x). Hence, by (2.40), we obtain

F (t, x)p ≤ P[λ]t (h)(x), x, t ∈ (0,∞).

Then, u∗(x)p ≤ (F p)∗(x) ≤ supt>0 P[λ]t (h)(x), x ∈ (0,∞); and from [17, (c), p. 86],

we conclude

‖u∗‖L1((0,∞), dmλ) ≤ C‖h‖rLr((0,∞), dmλ) ≤ C sup

t>0

∫ ∞

0

F (x, t)dmλ(x).�

Before we turn to the proof of (iii)=⇒(ii) in Theorem 1.7, we need to make pre-cise in which sense the action of R∆λ

on L1((0,∞), dmλ)-functions is understood.We say that ϕ belongs to the set Dλ of test functions if ϕ ∈ C(0,∞),

∫ 1

0

|ϕ(x)|dx

x< ∞,

and there exist C, δ, ε > 0 such that

|ϕ(x)| ≤ C(1 + x)−2λ−1−ε,

|ϕ(x) − ϕ(y)| ≤ C|x − y|δ(1 + x)−2λ−1−ε for |x − y| < x/2.(2.41)

Lemma 2.42. Assume that ϕ ∈ Dλ. Then the limit

limε→0

R∆λ,εϕ(y) = limε→0

∫ ∞

0, |x−y|>ε

Q[λ]0 (x, y)ϕ(x) dmλ(x)

is uniform on (0,∞) and defines a bounded continuous function R∆λϕ(y). More-

over, there exists a constant Cϕ such that

supε>0

|R∆λ,εϕ(y)| ≤ Cϕ < ∞.

Proof. The kernel Q[λ]0 (x, y) satisfies the estimates (cf. [2], [17])

(2.43)

Q[λ]0 (x, y) =

O(y−2λ−1) for 0 < x < y/2,

1π

(xy)−λ

x−y + O(x−2λ−1(1 + log+

xy(x−y)2 )

)for y/2 ≤ x ≤ 3y/2,

O(x−2λ−1) for 3y/2 < x.

The lemma now follows from (2.43) by standard arguments. �


Lemma 2.44. For every t, x > 0 the function (0,∞) 3 y 7→ xyP[λ+1]t (x, y)

belongs to Dλ.

Proof. This follows from (1.1). �

Let f ∈ L1((0,∞), dmλ). We say that R∆λf belongs to L1((0,∞), dmλ) in

the sense of distributions if there exists F ∈ L1((0,∞), dmλ) such that for everyϕ ∈ Dλ,

〈R∆λf, ϕ〉λ :=

∫ ∞

0

f(x)R ∆λϕ(x) dmλ(x) =

∫ ∞

0

F (x)ϕ(x)dmλ(x);

cf. Lemma 2.42. Then we set R∆λf = F . Clearly, limε→0 R∆λ,εf = F in the sense

of distributions.

We are now in a position to complete the proof of Theorem 1.7.

Proof of (iii)=⇒(ii) in Theorem 1.7. Suppose that f ∈ L1((0,∞), dmλ)

and that R∆λ(f) ∈ L1((0,∞), dmλ) in the sense described above. We define the

functions

u(t, x) = P[λ]t (f)(x),

v(t, x) = xP[λ+1]t (( · )−1R∆λ

(f)( · ))(x) =

∫ ∞

0

xyP[λ+1]t (x, y)R∆λ

(f)(y)dmλ(y).

One can check, using Hankel transforms (see (2.2)), (1.3), and Lemma 2.44, that

(2.45) xP[λ+1]t (( · )−1R∆λ

(f)( · ))(x) = Q[λ]t (f)(x), t, x ∈ (0,∞).

Then u and v can be extended to even and odd functions with respect to x on(0,∞) × R, respectively. The extensions of u and v are C2-functions, and theysatisfy (1.4). Moreover,

(2.46) supt>0

∫ ∞

0

|u(t, x)|dmλ(x) ≤ ‖f‖L1((0,∞), dmλ) < ∞,

and

(2.47) supt>0

∫ ∞

0

|v(t, x)|dmλ(x) ≤ C‖R∆λf‖L1((0,∞), dmλ) < ∞.

Indeed, to see (2.46), it is sufficient to take into account that∫∞0

P[λ]t (x, y)dmλ(x)=1


for every t > 0. To prove (2.47), we use [17, (b), p. 86] and get∫ ∞

0

xyP[λ+1]t (x, y)dmλ(x)

=

(∫ y/2

0

+

∫ ∞

y/2

)yP

[λ+1]t (x, y)x2λ+1dx

≤ C

(y−λ

∫ y/2

0

xλt

t2 + (x − y)2dx +

∫ ∞

y/2

y

xP

[λ+1]t (x, y)x2λ+2dx

)

≤ C

(ty

t2 + y2+

∫ ∞

0

P[λ+1]t (x, y)x2λ+2dx

)

≤ C,

independently of y, t > 0. Now (2.47) follows because R∆λf ∈ L1((0,∞), dmλ).

Finally, from Lemmas 2.38 and 2.36, we conclude that f ∈ H1max(∆λ) and

‖f‖H1max(∆λ) ≤ C(‖f‖L1((0,∞), dmλ) + ‖R∆λ

(f)‖L1((0,∞), dmλ)). �

It is worth noting that in the particular cases λ = n−12 , n = 2, 3, . . . , which corre-

spond to the Hardy spaces of radial functions on Rn, the inequality ‖f‖H1CW (∆λ) ≤

C(‖f‖L1((0,∞), dmλ) + ‖R∆λ(f)‖L1((0,∞), dmλ)) can be deduced by using arguments

of [8].

3 The Hardy space H1

max(Sλ).

In this section we study the Hardy space H1max(Sλ). This space, initially defined

by means of the maximal function associated with the Poisson semigroup {Pt}t>0,is characterized here in terms of special atomic decompositions, the maximalfunction associated with the heat semigroup {Wt}t>0, and the Riesz transform RSλ

(see Theorem 1.10).The Telyakovskii transform, also called local Hilbert transform (cf. [1]), is

defined by

T(f)(x) = PV

∫ 3x/2

x/2

f(y)

x − ydy.

Fridli [7] studied Hardy spaces generated by the Telyakovskii transform and intro-duced a class of atoms and the corresponding Hardy space H1

F (0,∞); cf. 1.9.The following useful result was established in [7, Theorem 2.1].

Theorem 3.1. Let f ∈ L1((0,∞), dx). The following assertions are equiva-lent:

(i) T(f) ∈ L1((0,∞), dx);(ii) f ∈ H1

F (0,∞);


(iii) the odd extension fo of f to R is in the classical Hardy space H1(R).

Moreover, the norms (‖f‖L1((0,∞), dx) + ‖T(f)‖L1((0,∞), dx)), ‖f‖H1F (0,∞), and

‖fo‖H1(R) are equivalent.

Thus, the space H1F (0,∞) is the subspace of L1((0,∞), dx) that consists of all

those functions f satisfying the conditions in Theorem 3.1.

The proof of Theorem 1.10 consists of Propositions 3.7, 3.8 and 3.9.

First we characterize the space H1F (0,∞) by means of the maximal operators

P[λ]∗ f(x) = supt>0 |P

[λ]t f(x)| and W

[λ]∗ f(x) = supt>0 |W

[λ]t f(x)| associated with the

Poisson and heat semigroups, respectively. To do this, we use the estimates for thekernels which are stated in Lemmas 3.3 and 3.6. The heat semigroup {W

[λ]t }t>0

generated by −Sλ is defined by

W[λ]t (f)(x) =

∫ ∞

0

W[λ]t (x, y)f(y)dy,

where

(3.2) W[λ]t (x, y) =

(xy)1/2

2tIλ−1/2

(xy

2t

)exp

(− x2 + y2

4t

), t, x, y ∈ (0,∞).

Here Iν represents the modified Bessel function of the first kind and order ν.

Lemma 3.3. Let λ > 0. There exists C > 0 such that for t > 0 and x, y ∈(0,∞),

(i) 0 < W[λ]t (x, y) ≤ Cyλ/xλ+1, 0 < y < x/2;

(ii) 0 < W[λ]t (x, y) ≤ Cxλ/yλ+1, 2x < y;

(iii) |W[λ]t (x, y) − Wt(x, y)| ≤ C/x, x/2 < y < 2x, where Wt denotes the classical

heat kernel, i.e.,

Wt(x, y) =1√4πt

e−|x−y|2/4t.

Proof. It follows from (3.2) that

s−1/2W

[λ]t (x/

√s, y/

√s) = W

[λ]ts (x, y).

Thus it suffices to prove (i)-(iii) for x = 1. Using (3.2), we rewrite

W[λ]t (1, y) =

( y

2t

)1/2

Iλ−1/2

( y

2t

)exp

(− y

2t

) 1√2t

exp(− |1 − y|2

4t

).

It is well-known, (cf. [23]), that

(3.4)1

Cxν ≤ Iν(x) ≤ Cxν for 0 < x ≤ 1,


(3.5) Iν(x) =ex

√2πx

+ Ψν(x), where |Ψν(x)| ≤ Cexx−3/2 for x > 1/4

for some C > 0. If y ≤ 1/2, then (3.4) and (3.5) imply

Wt(1, y) ≤ C(y

t

)λ

t−1/2 exp(− 1

16t

)≤ Cyλ.

If y ≥ 2, then |y − 1| ≥ y/2; and, by (3.4) and (3.5), we obtain

Wt(1, y) ≤ C(y

t

)λ

t−1/2 exp(− y2

16t

)≤ Cy−1−λ.

Clearly,

|W[λ]t (1, y) − Wt(1, y)|

=1√2t

exp(− |1 − y|2

4t

)∣∣∣( y

2t

)1/2

Iλ−1/2

( y

2t

)exp

(− y

2t

)− 1√

2π

∣∣∣.

Therefore, if 1/2 < y < 2 and t ≥ 1, we get, using (3.4), |W[λ]t (1, y)−Wt(1, y)| ≤ C.

Finally, if 1/2 < y < 2 and t < 1, then applying (3.5), we get∣∣W[λ]

t (1, y) − Wt(1, y)∣∣ ≤ Ct−1/2(t/y) ≤ C.

The proof of the lemma is complete. �

Lemma 3.6. Let λ > 0. Then there exist constants C such that(i) 0 < P

[λ]t (x, y) ≤ C yλ

xλ+1 , for t > 0 and 0 < y < x/2;

(ii) 0 < P[λ]t (x, y) ≤ C xλ

yλ+1 , for t > 0 and 0 < 2x < y;

(iii) |P[λ]t (x, y) − 1

πt

(x−y)2+t2 | ≤ C/x, for t > 0 and 0 < x/2 < y ≤ 2x.

Proof. By the principle of subordination; see, e.g., [19, p. 60],

P[λ]t (x, y) =

1

2√

π

∫ ∞

0

e−1/(4s)s−3/2W

[λ]t2s(x, y) ds,

1

π

t

(x − y)2 + t2=

1

2√

π

∫ ∞

0

e−1/(4s)s−3/2Wt2s(x, y) ds.

Therefore Lemma 3.6 is a direct consequence of Lemma 3.3. �

The following proposition asserts that H1F (0,∞) = H1

max(Sλ).

Proposition 3.7. Let λ > 0. A function f ∈ L1((0,∞), dx) is in H1F (0,∞) if

and only if P[λ]∗ (f) ∈ L1((0,∞), dx). Moreover, there exists C > 0 such that for

every f ∈ H1F (0,∞),

1

C‖P

[λ]∗ (f)‖L1((0,∞), dx) ≤ ‖f‖H1

F (0,∞) ≤ C‖P[λ]∗ (f)‖L1((0,∞), dx).


Proof. Let f ∈ L1((0,∞), dx). Then

P[λ]t (f)(x) =

(∫ x/2

0

+

∫ 2x

x/2

+

∫ ∞

2x

)P

[λ]t (x, y)f(y)dy

= A1(f)(t, x) + A2(f)(t, x) + A3(f)(t, x).

Setting R(t, x, y) = P[λ]t (x, y) − π−1t

((x − y)2 + t2)−1, we write

A2(f)(t, x) =1

π

∫ 2x

x/2

t

(x − y)2 + t2f(y)dy +

∫ 2x

x/2

R(t, x, y)f(y)dy

= A2,1(f)(t, x) + A2,2(f)(t, x).

It follows from Lemma 3.6 that the maximal operators

A∗1(f) = sup

t>0|A1(f)(t, x)|, A

∗3(f) = sup

t>0|A1(f)(t, x)|,

and A∗2,2(f) = sup

t>0|A2,2(f)(t, x)|

are bounded on L1((0,∞), dx). Therefore, supt>0 |P[λ]t (f)(x)| ∈ L1((0,∞), dx) if

and only if

A∗2,1f(x) = sup

t>0

∣∣∣∣1

π

∫ 2x

x/2

t

(x − y)2 + t2f(y)dy

∣∣∣∣ ∈ L1((0,∞), dx).

For f ∈ L1((0,∞), dx) let fo be the odd extension of f to R. For t > 0 and x ∈ R,we write

∫ ∞

−∞

t

(x − y)2 + t2fo(y)dy

=

∫ ∞

0

t

(x − y)2 + t2f(y)dy −

∫ 0

−∞

t

(x − y)2 + t2f(−y)dy

=

∫ ∞

0

f(y)( t

(x − y)2 + t2− t

(x + y)2 + t2

)dy.

Note that this function is odd in x ∈ R. Then, for t, x > 0,∫ ∞

−∞

t

(x − y)2 + t2fo(y)dy −

∫ 2x

x/2

t

(x − y)2 + t2f(y) dy

=

∫ x/2

0

f(y)4xyt

((x − y)2 + t2)((x + y)2 + t2)dy

+

∫ ∞

2x

f(y)4xyt

((x − y)2 + t2)((x + y)2 + t2)dy −

∫ 2x

x/2

t

(x + y)2 + t2f(y) dy

=H1(f)(t, x) + H2(f)(t, x) − H3(f)(t, x), t, x ∈ (0,∞).


We now study Hj for j = 1, 2, 3. For x ∈ (0,∞) and t > 0, we have

|H1(f)(t, x)| ≤ Ct

∫ x/2

0

|f(y)| xy

(x2 + t2)((x + y)2 + t2)dy ≤ C

∫ x/2

0

|f(y)| y

x2dy.

Hence

‖ supt>0

|H1(f)(t, x)|‖L1((0,∞), dx) ≤ C

∫ ∞

0

1

x2

∫ x/2

0

|f(y)|ydydx ≤ C‖f‖L1((0,∞), dx).

Similarly,

|H2(f)(t, x)| ≤ C

∫ ∞

2x

|f(y)| x

y2dy for t, x > 0;

and, consequently,

‖ supt>0

|H2(f)(t, x)|‖L1((0,∞), dx) ≤ C‖f‖L1((0,∞), dx).

Since supt>0 t((x + y)2 + t2)−1 ≤ (x + y)−1 for x, y > 0, we deduce

∥∥∥ supt>0

|H3(f)(t, x)∥∥∥

L1((0,∞), dx)≤ C

∥∥∥∥∫ 2x

x/2

|f(y)|x + y

dy

∥∥∥∥L1((0,∞), dx)

≤ C‖f‖L1((0,∞), dx).

Thus, we have proved that for f ∈ L1((0,∞), dx),

supt>0

∣∣∣∣∫ 2x

x/2

t

(x − y)2 + t2f(y)dy

∣∣∣∣ ∈ L1((0,∞), dx)

if and only if

supt>0

∣∣∣∣∫ ∞

−∞

t

(x − y)2 + t2fo(y)dy

∣∣∣∣ ∈ L1((0,∞), dx).

Finally, we conclude by invoking [7, Theorem 2.1] (see also Theorem 3.1) thatP

[λ]∗ f ∈ L1((0,∞), dx) if and only if f ∈ H1

F (0,∞), and the norm ‖P[λ]∗ f‖L1((0,∞), dx)

is comparable to ‖f‖H1F (0,∞). �


and only if W[λ]∗ (f) ∈ L1((0,∞), dx). Moreover, there exists C > 0 such that for


C−1‖W[λ]∗ (f)‖L1((0,∞), dx) ≤ ‖f‖H1

F (0,∞) ≤ C‖W[λ]∗ (f)‖L1((0,∞), dx).

Proof. The proof of the proposition is similar to that of Proposition 3.7 anduses Lemma 3.3. We omit the details. �


Finally, we characterize the Hardy space H1F (0,∞) = H1

max(Sλ) by using theRiesz transform RSλ

.


and only if RSλ(f) ∈ L1((0,∞), dx). Moreover, there exists C > 0 such that for


C−1‖f‖H1F (0,∞) ≤ ‖f‖L1((0,∞), dx) + ‖RSλ

(f)‖L1((0,∞), dx) ≤ C‖f‖H1F (0,∞).

Proof. We write

RSλ(f)(x) =

∫ x/2

0

RSλ(x, y)f(y)dy + PV

∫ 3x/2

x/2

RSλ(x, y)f(y)dy

+

∫ ∞

3x/2

RSλ(x, y)f(y)dy

=A1(f)(x) + A2(f)(x) + A3(f)(x).

Using estimates for the kernel RSλ(x, y) (cf. [2, (3.15)]), we have

‖A1(f)‖L1((0,∞), dx) ≤ C

∫ ∞

0

1

xλ+1

∫ x/2

0

yλ|f(y)|dydx ≤ C‖f‖L1((0,∞), dx).

Similarly, by (3.16) of [2], we get

‖A3(f)‖L1((0,∞), dx) ≤ C

∫ ∞

0

xλ+1

∫ ∞

2x

1

yλ+2|f(y)|dydx ≤ C‖f‖L1((0,∞), dx).

Moreover, estimates for |RSλ(x, y) − (π(x − y))−1| (cf. [2, (3.17)]) imply

|A2(f)(x) − 1

πT(f)(x)| ≤ CA4(|f |)(x),

where

A4(g)(x) =

∫ 3x/2

x/2

1

y

(1 + log

(1 +

√xy

|x − y|))

g(y)dy.

Since A4 is a bounded operator on L1((0,∞), dx), we conclude that RSλ(f) ∈

L1((0,∞), dx) if and only if T(f) ∈ L1((0,∞), dx). Thus the proof is finished byapplying Theorem 3.1. �

4 Hardy type inequalities, transplantation and multi-pliers for H1

F.

Let Hλ denote the Hankel transform defined by

Hλ(f)(x) =

∫ ∞

0

√xyJλ−1/2(xy)f(y)dy, x > 0.


It is known that HλHλ = I for λ ≥ 0 and that the following Parseval formula holds

(4.1)∫ ∞

0

f(x)g(x) dx =

∫ ∞

0

Hλf(x)Hλg(x) dx.

Kanjin [12] established a Hardy type inequality for Hλ. He proved that thereexists a constant Cλ such that if f is in the classical Hardy space H1(R) andsuppf ⊂ [0,∞), then

(4.2)∫ ∞

0

|Hλ(f)(y)|y

dy ≤ Cλ‖f‖H1(R).

Set H1(0,∞) = {h|(0,∞) : h ∈ H1(R), supp h ⊂ [0,∞)}. The space H1(0,∞)

admits atomic decomposition. A function a is an H1(0,∞)–atom if there exists aninterval I ⊂ [0,∞) such that suppa ⊂ I, ‖a‖L∞ ≤ |I|−1,

∫a(x) dx = 0. The space

H1(0,∞) is contained in H1F (0,∞). We now extend (4.2) to the space H1

F (0,∞).

Proposition 4.3. Let λ > 0. There exists C > 0 such that, for every f ∈H1

F (0,∞),

(4.4)∫ ∞

0

|Hλ(f)(y)|y

dy ≤ C‖f‖H1F (0,∞).

Proof. Assume that f ∈ H1F (0,∞). Then f =

∑∞j=1 λjaj , where for each

j ∈ N, aj is an F -atom and λj ∈ C,∑∞

j=1 |λj | < ∞. The series defining f

converges in L1((0,∞), dx). Since√

xJλ−1/2(x) is a bounded continuous functionon (0,∞) provided that λ > 0, Hλf(x) is a continuous function on (0,∞) and

Hλ(f)(x) =

∞∑

j=1

λjHλ(aj)(x), x ∈ (0,∞).

Hence, to prove (4.4), it suffices to show that there exists C > 0 such that

(4.5)∫ ∞

0

|Hλ(a)(y)|y

dy ≤ C,

for every F -atom a.

Let a be an F -atom. If there exists an interval I ⊂ (0,∞) for which supp a ⊂ I,∫I a(x)dx = 0 and ‖a‖L∞((0,a), dx) ≤ |I|−1, then [12, Lemma 2] implies that (4.5)

holds with C > 0 independent of a.

Suppose now that a = δ−1χ(0,δ), with δ > 0. Then, using an asymptotic


behaviour of the Bessel function ([23, p. 199], see also [15, p. 461–463]), we get∫ ∞

0

|Hλ(a)(y)|y

dy

=

∫ ∞

0

1

δx

∣∣∣∣∫ δ

0

(xy)1/2Jλ−1/2(xy)dy

∣∣∣∣dx

≤∫ ∞

0

1

v2

∣∣∣∣∫ v

0

u1/2Jλ−1/2(u)du

∣∣∣∣dv

≤C

(∫ 1

0

1

v2

∣∣∣∣∫ v

0

uλdu

∣∣∣∣dv +

∫ ∞

1

1

v2

∣∣∣∣∫ v

0

u1/2Jλ−1/2(u)du

∣∣∣∣dv

)

≤C

(∫ 1

0

vλ−1dv +

∫ ∞

1

1

v2

∣∣∣( ∫ 1

0

+

∫ v

1

)u1/2Jλ−1/2(u)du

∣∣∣dv

)

≤C

(1 +

∫ ∞

1

1

v2dv +

∫ ∞

1

1

v2

∣∣∣∣∫ v

1

(cos(u − σ) + O(u−1))du

∣∣∣∣dv

)

≤C

with C > 0 independent of a. Here σ = π(λ − 1/4)/2. �

In [13], it was proved that if µ, ν > 0, then

(4.6) ‖HµHνf‖H1(0,∞) ≤ C‖f‖H1(0,∞) for f ∈ H1(0,∞) ∩ L2((0,∞), dx).

We now use the methods of [13] to prove that for every µ, ν > 0 there exists aconstant C > 0 such that

(4.7) ‖HµHνf‖H1F (0,∞) ≤ C‖f‖H1

F (0,∞) for f ∈ H1F (0,∞) ∩ L2((0,∞), dx).

Then, having (4.7), we can easily deduce a multiplier theorem for the Hankeltransforms on H1

F (0,∞) (see Theorem 4.11).Since HνHµ is a continuous operator from D(0,∞) = C∞

c (0,∞) into L∞(0,∞),HµHν maps H1

F (0,∞) continuously into D′(0,∞). Therefore, it suffices to show(4.7) for f an H1

F (0,∞) atom. By (4.6), it is enough to verify (4.7) for f(x) =

δ−1χ(0,δ)(x), δ > 0. This can be done, using the identity HµHν = HµHν+2Hν+2Hν ,once we prove the following two lemmas.

Lemma 4.8. For every ν > 0 there exists a constant C > 0 such that

‖Hν+2Hνf‖H1F (0,∞) ≤ C for f(x) = δ−1χ(0,δ)(x), δ > 0.

Lemma 4.9. If ν ≥ 1 and µ > 0 then there exists a constant C > 0 such that

‖HµHνf‖H1F (0,∞) ≤ C for f(x) = δ−1χ(0,δ)(x), δ > 0.


Proof of Lemma 4.8. Take f = δ−1χ(0,δ) with δ > 0. Clearly,

‖(Hν+2Hνf)χ(0,2δ)‖L2 ≤ C‖f‖L2 = Cδ−1/2.

Thus (Hν+2Hνf)χ(0,2δ) ∈ H1F (0,∞) and

‖(Hν+2Hνf)χ(0,2δ)‖H1F (0,∞) ≤ C.

In order to estimate Hν+2Hνf(x) for x ≥ 2δ, we use the identity

Hν+2Hνf = (2ν + 1)V(ν)f − f, where V(ν)f(x) =1

x

∫ x

0

(t/x)νf(t)dt,

(cf. [13, Section 3, p. 237]). Hence, for some cν ∈ R,

(4.10) Hν+2Hνf(x) = cνδνx−ν−1 = cν

∞∑

j=1

δνx−ν−1χ[2jδ,2j+1δ](x) for x ≥ 2δ.

Observing that ‖δνx−ν−1χ[2jδ,2j+1δ](x)‖H1F (0,∞) ≤ C2−jν completes the proof of

the lemma. �

Proof of Lemma 4.9. Let f = δ−1χ(0,δ), δ > 0. Obviously,

‖(HµHνf)χ(0,2δ)‖H1F (0,∞) ≤ Cδ1/2‖(HµHνf)χ(0,2δ)‖L2 ≤ C.

Using Schindler’s integral representation of HµHν [18] (see also [13, (2)]), we get

HµHνf(x) = δ−1

∫ δ

0

Iµ,ν(x, y) dy for x > 2δ,

where Iν,µ(x, y) = Cµ,ν

(yx

)ν( 1x−y + 1

x+y

)F(

ν−µ2 , ν+µ−1

2 ; ν + 12 ; y2

x2

), F is the hy-

pergeometric function and Cµ,ν ∈ R. The function F(

ν−µ2 , ν+µ−1

2 ; ν + 12 ; y2

x2

)is

bounded for 0 < y < x (see [14, (9.3.4)]). Hence |HµHνf(x)| ≤ Cµ,νδνx−ν−1 forx > 2δ, which is the same estimate we got in (4.10). �

Theorem 4.11. Let m : (0,∞) → C be a bounded continuous function that

satisfiessupt>0

‖η( · )m(t · )‖L2s(R) < ∞

for some s > 1/2, where η ∈ C∞c (0,∞) is a fixed nonzero auxiliary function and

‖ · ‖L2s(R) is the Sobolev norm of order s. Then for every µ, ν > 0, there exists a

constant C > 0 such that

‖HµMmHνf‖H1F (0,∞) ≤ C‖f‖H1

F (0,∞) for f ∈ H1F (0,∞) ∩ L2((0,∞), dx),

where Mmf(ξ) = m(ξ)f(ξ).


Proof. Since HµMmHν = HµH1H1MmH1H1Hν , it suffices by (4.7) to verifythat

‖H1MmH1f‖H1F (0,∞) ≤ C‖f‖H1

F (0,∞) for f ∈ H1F (0,∞) ∩ L2((0,∞), dx).

Recall that for ξ > 0,

H1f(ξ) =

√2

π

∫ ∞

0

f(t) sin(ξt) dt =i√2π

∫ ∞

−∞fo(t)e

−iξt dt =i√2π

(Ffo)(ξ),

where fo is the odd extension of f on R. There is the natural one-to-one corre-spondence between the spaces H1

F (0,∞) and {f ∈ H1(R) : f is an odd function}.Moreover, ‖f‖H1

F (0,∞) ∼ ‖fo‖H1(R); cf. Theorem 3.1. Let m(ξ) be the even exten-sion of m on R. Then

H1MmH1f =1

2πF−1MmFfo on (0,∞).

Using the well-known multiplier theorem for the classical Hardy space H1(R) ([20,Chapter III, Section 5.25] combined with [11]), we get

‖F−1MmFfo‖H1(R) ≤ C‖fo‖H1(R).

This, together with the fact that F−1MmFfo is an odd function, completes theproof of the theorem. �

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Jorge J. BetancorDEPARTAMENTO DE ANALISIS MATEMATICO

UNIVERSIDAD DE LA LAGUNA

CAMPUS DE ANCHIETA, AVDA. ASTROFISICO FRANCISCO SANCHEZ, S/N38271 LA LAGUNA (STA. CRUZ DE TENERIFE), SPAIN

email: [email protected]

Jacek DziubanskiINSTITUTE OF MATHEMATICS

UNIVERSITY OF WROCLAW

PL. GRUNWALDZKI 2/450-384 WROCLAW, POLAND


Jose Luis TorreaDEPARTAMENTO DE MATEMATICAS

FACULTAD DE CIENCIAS

UNIVERSIDAD AUTONOMA DE MADRID

28049 MADRID, SPAIN


(Received October 24, 2007)

On Hardy spaces associated with Bessel operators

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