riesz transform characterization of hardy spaces ... - Semantic Scholar
RIESZ TRANSFORM CHARACTERIZATION OF HARDY SPACES ASSOCIATED WITH CERTAIN LAGUERRE EXPANSIONS ´ ski and Gustavo Garrigo ´s Jorge Betancor, Jacek Dziuban
Abstract In this paper we prove Riesz transform characterizations for Hardy spaces associated with certain systems of Laguerre functions.
1
Introduction and statement of the results
Denote the Laguerre polynomials of order α > −1 by d n (e−x xn+α ), n = 0, 1, 2, ... Lαn (x) = (n!)−1 ex x−α dx In this paper we consider the following two systems of Laguerre functions on (0, ∞) √ 2 (1) ϕαn (x) = 2 cn,α e−x /2 xα+1/2 Lαn (x2 ), n = 0, 1, 2 . . . , (2)
Lαn (x) = cn,α e−x/2 xα/2 Lαn (x),
n = 0, 1, 2, . . . ,
where cn,α = (Γ(n + 1)/Γ(n + 1 + α))1/2 . It is well known that, for every α > −1, each α ∞ 2 of the systems {ϕαn }∞ n=0 and {Ln }n=0 is complete and orthonormal on L ((0, ∞), dx). Moreover, these functions are eigenvectors, respectively, of the differential operators d2 x α2 1 d2 1 1 d − 2 + y 2 + 2 α2 − , Lα = − x 2 + − + , Lα = 2 dy y 4 dx dx 4 4x 2000 Mathematics Subject Classification. Primary 42B30; Secondary 42C10. Key words and phrases. Hardy space, Laguerre system, Riesz transform. First author partially supported by MTM2004-05878 and MTM2007-65609. Second author partially 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 Government funds for science - grant N N201 397137, MNiSW. Third author partially supported by MTM2007-60952.
1
satisfying Lα ϕαn = (2n + α + 1)ϕαn
and Lα (Lαn ) = (n + (α + 1)/2) Lαn .
As in [6, 7], the operators Lα and Lα can be factored as Lα = where
1 ∗ α+1 Dα Dα + α + 1 and Lα = δα∗ δα + , 2 2
α + 1/2 d +x− dx x
Dα =
and δα =
√ d 1 √ α x + x− √ , dx 2 x
and where Dα∗ and δα∗ denote, respectively, the formal adjoint operators to Dα and δα in L2 ((0, ∞), dx). Corresponding Riesz transforms are defined in L2 ((0, ∞), dx) by Rα = Dα Lα−1/2
and Rα = δα Lα−1/2 ,
that is, they act on the basis elements by √ √ 2 n n α α+1 α (3) Rα ϕn = − √ ϕn−1 , Rα Ln = − p Lα+1 n−1 . 2n + α + 1 n + (α + 1)/2 There exist kernels Rα (x, y) and Rα (x, y) such that Z ∞ Z Rα f (x) = lim Rα (x, y)f (y) dy, Rα f (x) = lim ε→0
ε→0
0, |x−y|>ε
∞
Rα (x, y)f (y) dy.
0, |x−y|>ε
One can easily deduce from (1), (2) and (3) that these kernels are related by √ √ Rα (x, y) = 2−3/2 (xy)−1/4 Rα ( x, y),
(4)
x, y ∈ (0, ∞).
Riesz tranforms for Laguerre systems were defined and studied by Nowak and Stempak [7], and by Harboure, Torrea and Viviani [6], who proved that Rα for α ≥ −1/2 and Rα for α ≥ 0 extend as bounded linear operators on Lp (0, ∞) when 1 < p < ∞ and are of weak type (1,1). Our goal in the present paper is to characterize the spaces 1 (Lα ) = f ∈ L1 (0, ∞) ; kRα f kL1 < ∞ HRiesz
for α > −1/2,
and 1 HRiesz (Lα ) =
f ∈ L1 (0, ∞) ; kRα f kL1 < ∞
for α > 0.
1 1 In [3], the second-named author considered Hardy spaces Hmax (Lα ) and Hmax (Lα ) defined by means of the maximal functions associated with the semigroups generated by −Lα and −Lα , respectively. To be more precise, if
Wtα (x, y) =
∞ X
e−(2n+α+1)t ϕαn (x)ϕαn (y), Wαt (x, y) =
n=0
∞ X n=0
2
e−t(n+(α+1)/2) Lαn (x)Lαn (y)
denote the integral kernels of the semigroups {e−tLα }t>0 and {e−tLα }t>0 , we say that a 1 function f in (0, ∞) belongs to Hmax (Lα ) when the maximal function Z ∞ α W∗ f (x) = sup Wtα (x, y)f (y)dy t>0
0
α 1 belongs to L1 (0, ∞). Then we set kf kHmax (Lα ) = kW∗ f kL1 . Analogously, we define the 1 1 (Lα ) and the norm k · kHmax maximal function Wα∗ , the space Hmax (Lα ) . It was proved in [3] 1 1 that the spaces Hmax (Lα ), α > −1/2, and Hmax (Lα ), α > 0, admit atomic decompositions. The notion of atom for these spaces depends on the following auxiliary functions
ρLα (x) =
1 1 min(x, 1/x) and ρLα (x) = min(x, 1). 8 8
A measurable function b : (0, ∞) → C is said to be an H 1 (Lα )-atom if there exists a ball B = B(y0 , R) = {|y0 − y| < R} with R ≤ ρLα (y0 ) such that kbk∞ ≤ |B|−1 and Z b(y)dy = 0. if R ≤ ρLα (y0 )/2 then supp b ⊂ B,
1 The space Hat (Lα ) consists of all measurable functions f on (0, ∞) of the form
f=
∞ X
λj bj ,
j=1
where bj are H 1 (Lα )-atoms, λj ∈ C and by
P∞
j=1
1 |λj | < ∞. The norm in Hat (Lα ) is defined
kf kHat1 (Lα ) = inf
∞ X
|λj |,
j=1
P 1 where the infimum is taken over all decompositions f = ∞ j=1 λj bj , where bj are H (Lα )1 atoms and λj ∈ C. Similarly we define the space Hat (Lα ) and the norm k kHat1 (Lα ) , the only difference being that the function ρLα replaces the function ρLα in the definition of H 1 (Lα )-atoms. The main result in [3] was to show that 1 1 1 1 Hmax (Lα ) = Hat (Lα ) for α > −1/2 and Hmax (Lα ) = Hat (Lα ) for α > 0,
with equivalence of the corresponding norms. Our goal in this paper is to characterize these spaces by means of the Riesz transforms Rα and Rα . More precisely, we shall prove the following theorems. 1 1 Theorem 1.1. If α > −1/2, then HRiesz (Lα ) = Hat (Lα ). Moreover, there exists C > 0 such that
(5)
C −1 kf kHat1 (Lα ) ≤ kRα f kL1 + kf kL1 ≤ Ckf kHat1 (Lα ) . 3
1 1 Theorem 1.2. If α > 0, then HRiesz (Lα ) = Hat (Lα ). Moreover, there exists C > 0 such that
C −1 kf kHat1 (Lα ) ≤ kRα f kL1 + kf kL1 ≤ Ckf kHat1 (Lα ) .
(6)
2
Hardy spaces H 1(Lα ) associated with Laguerre operators Lα
In the present section, we shall prove Theorem 1.1. To do this, we recall the equivalence between Riesz and atomic definitions for the Hardy space associated with the Hermite operator, d2 1 H = (− 2 + x2 ), 2 dx which were established in [4]. First we let ρH (y) = (1 + |y|)−1 .
(7)
It is easily seen that there exist constants C, c > 0 such that −1 1/2 (8) cρH (x) 1 + |x − y|/ρH (x) ≤ ρH (y) ≤ CρH (x) 1 + |x − y|/ρH (x) . A function a : R → C is an H 1 (H)-atom if there exists a ball B = B(y0 , R) = {y ∈ R; |y − y0 | < R} with R ≤ ρH (y0 ) such that kakL∞ ≤ |B|−1 and Z if R ≤ ρH (y0 )/2 then a(y) dy = 0.
supp a ⊂ B,
1 The atomic Hardy space Hat (H) and the norm k kHat1 (H) are defined in the standard way. On the other hand, a Riesz transform RH can be defined in L2 (R) by d + x H −1/2 , RH = dx
motivated by the factorization of the Hermite operator d d d i 1 h d H=− +x −x + −x +x . 4 dx dx dx dx To obtain a kernel expression for RH , recall first the Mehler formula for Hermite functions (cf. [10, Lemma 1.1.1]), which asserts that the integral kernel WtH (x, y) of the Hermite semigroup {e−tH }t>0 is given by (9)
WtH (x, y) =
h
i1/2 1 1 + e−2t e−t e−t 2 2 exp − (x + y ) + 2xy π(1 − e−2t ) 2 1 − e−2t 1 − e−2t 4
R∞ when t > 0 and x, y ∈ R. Using the formula H −1/2 = π −1/2 0 e−tH t−1/2 dt, we can express the Riesz transform RH as a principal value singular integral operator of the form Z H R (f )(x) = lim RH (x, y)f (y)dy , ε→0
y∈R : |x−y|>ε
with the kernel given by Z ∞ 1 dt d R (x, y) = √ + x WtH (x, y) √ dx π 0 t Z ∞ Z ∞ 1 1 d H dt dt =√ Wt (x, y) √ + √ xWtH (x, y) √ π 0 dx π 0 t t H
(10)
= R1H (x, y) + R2H (x, y). It is not difficult to prove using (9) and (10) that Z ∞ Z H (11) sup |R2 (x, y)| dx < ∞, sup y∈R
x∈R
−∞
∞
|R2H (x, y)| dy < ∞
−∞
(see Section 4). Therefore, denoting R2H = xH −1/2 , we have kR2H f kL1 (R) ≤ Ckf kL1 (R)
(12)
(see also [2, Theorem 4.5]). It was proved by Thangavelu [9] that the operator RH is bounded on Lp (R) for 1 < p < ∞. Moreover, Theorem 1.2 of Zhong [11] asserts that the operator R1H = (d/dx)H −1/2 is a Calder´on-Zygmund operator, hence it is of weak type (1,1) (see also [8] for a proof based on analysis of the Melher kernel). The above facts could also be deduced from the following lemma. Lemma 2.1. Let ψ ∈ Cc∞ (−2−4 , 2−4 ) be such that ψ(x) = 1 for |x| < 2−5 . Then there exists a constant c0 6= 0 and a kernel h(x, y) such that x−y c0 (13) RH (x, y) = ψ + h(x, y), x−y ρH (x) Z (14)
∞
∞
|h(x, y)|dx + sup
sup y∈R
Z
−∞
x∈R
|h(x, y)|dy < ∞. −∞
This lemma is known, but a self-contained proof based on analysis of the Mehler kernel will be presented in Section 4. We set 1 HRiesz (H) = f ∈ L1 (R) ; kRH f kL1 (R) < ∞ . 1 In view of (12), an L1 -function f belongs to HRiesz (H) if and only if (d/dx)H −1/2 f belongs to L1 (R). From this remark and the results in [4], it follows that 1 1 HRiesz (H) = Hat (H)
5
and there exists a constant C > 0 such that (15)
C −1 kf kHat1 (H) ≤ kRH f kL1 + kf kL1 ≤ Ckf kHat1 (H) .
Having established the Riesz and atomic characterizations of the Hardy space associated with the Hermite operator, we continue our preparation for the proof of Theorem 1.1. H H H f , where f + R2,loc f = R1,loc For a function f defined on (0, ∞), we denote Rloc Z 2x H Rj,loc f (x) = lim RjH (x, y)f (y) dy, x > 0, j = 1, 2. ε→0
x/2, |x−y|>ε
Proposition 2.2. For f ∈ L1 (0, ∞), let fo denote its odd extension. Then R1H fo ∈ H L1 (R) if and only if R1,loc f is in L1 (0, ∞). Moreover, there exists C > 0 such that H kR1H fo − R1,loc f kL1 (0,∞) ≤ Ckf kL1 (0,∞) .
Proof. Set r = r(t) = e−t ∈ (0, 1). According to (9) and (10), we have (16) Z ∞√ dt r 2ry − (1 + r2 )x 1 1 + r2 2r H 2 2 √ . R1 (x, y) = √ exp − (x + y ) + xy (1 − r2 )3/2 2(1 − r2 ) 1 − r2 π 0 t Note that kR1H fo kL1 (R) = 2kR1H fo kL1 (0,∞) , because R1H fo is an even function. Moreover, Z ∞ H R1 fo (x) = lim R1H (x, y) − R1H (x, −y) f (y)dy, a.e. x ∈ (0, ∞). ε→0
0, |x−y|>ε
Further, R1H fo (x)
−
H R1,loc f (x)
x/2
Z
R1H (x, y) − R1H (x, −y) f (y)dy
= 0
∞
Z
R1H (x, y) − R1H (x, −y) f (y)dy
+ Z2x2x
(17) −
R1H (x, −y)f (y)dy
x/2
=
3 X
Tj (f )(x), a.e. x ∈ (0, ∞).
j=1
It suffices to show that the operators Tj , j = 1, 2, 3, are bounded on L1 ((0, ∞), dx). To deal with T1 and T2 , we estimate the difference DH (x, y) = |R1H (x, y) − R1H (x, −y)| for x, y > 0. By (16) √ Z ∞ rx 2r 2r H exp xy − exp − xy D (x, y) ≤ C (1 − r2 )3/2 1 − r2 1 − r2 0 dt 1 + r2 2 2 √ (18) × exp − (x + y ) 2(1 − r2 ) t √ Z ∞ 2r dt ry 1 + r2 2 2 √ . exp − exp xy +C (x + y ) (1 − r2 )3/2 2(1 − r2 ) 1 − r2 t 0 6
Applying the mean value theorem in the first integral, we can assert that (19) DH (x, y) √ Z ∞ rx2 y r ≤C + y exp − (1 − r2 )3/2 1 − r2 0 √ Z ∞ rx2 y r =C + y exp − (1 − r2 )3/2 1 − r2 0
2r dt 1 + r2 2 2 √ (x + y ) exp xy 2(1 − r2 ) 1 − r2 t 2 1−r dt 1+r (x − y)2 exp − xy √ . 2 2(1 − r ) 1+r t
It is now not difficult to verify using (19) that Cyx−2 (20) DH (x, y) ≤ Cy −1
for x > 2y, for 2x < y.
The estimate (20) easily implies kT1 f kL1 (0,∞) + kT2 f kL1 (0,∞) ≤ Ckf kL1 (0,∞) . Moreover, from (16), we conclude Z ∞ Z 1 1 −cx2 /t C H −cx2 −t e dt ≤ for x/2 < y < 2x. |R1 (x, −y)| ≤ C xe e dt + x 2 y 1 0 t Hence T3 is a bounded operator from L1 (0, ∞) into itself. Proposition 2.3. Let α > −1/2, f ∈ L1 (0, ∞) and fo be the odd extension of f to R. Then Rα f is in L1 (0, ∞) if and only if RH fo is in L1 (R). Moreover, there exists C > 0 such that C −1 (kfo kL1 (R) + kRH fo kL1 (R) ) ≤ kf kL1 (0,∞) + kRα f kL1 (0,∞)
and
H
kf kL1 (0,∞) + kRα f kL1 (0,∞) ≤ C(kfo kL1 (R) + kR fo kL1 (R) ). Proof. According to [1, Lemma 2.13], we have |Rα (x, y)| (21)
≤ Cxα+3/2 y −(α+5/2)
≤ Cy α+1/2 x−(α+3/2) C (xy)1/4 H |Rα (x, y) − R (x, y)| ≤ 1+ y |x − y|1/2 |Rα (x, y)|
for 0 < 2x < y < ∞, for 0 < y < x/2, and for 0 < x/2 < y < 2x.
Each of the Hardy operators Hα (g)(x) = x
x
Z
−α−3/2
y α+1/2 g(y)dy,
x>0
y −α−3/2 g(y)dy,
x>0
0
and α
H (g)(x) = x
α+1/2
Z
∞
x
7
are bounded on L1 (0, ∞) when α > −1/2. Moreover, the operator N defined by Z 2x 1 (xy)1/4 1+ f (y)dy N f (x) = |x − y|1/2 x/2 y is also bounded in L1 (0, ∞). Hence, by (21), (11) and Proposition 2.2, we obtain H H kRα f − RH fo kL1 (0,∞) ≤ kRα f − Rloc f kL1 (0,∞) + kRloc f − RH fo kL1 (0,∞)
≤ C kN |f | kL1 (0,∞) + kH α+1 |f | kL1 (0,∞) + kHα |f | kL1 (0,∞)
H H + kR1,loc f − R1H fo kL1 (0,∞) + kR2,loc f kL1 (0,∞) + kR2H fo kL1 (0,∞)
≤ Ckf kL1 (0,∞) .
The next elementary lemma will be used below. Lemma 2.4. Let b : (0, ∞) → C be an H 1 (Lα )-atom. Then, its odd extension bo satisfies kbo kHat1 (H) ≤ 36. Proof. Let B = B(y, R) ⊂ (0, ∞) be a ball associated with b, that is, R ≤ ρLα (y), R supp b ⊂ B and kbk∞ ≤ |B|−1 . Moreover, b(y)dy = 0 if R ≤ ρLα (y)/2. In this last case, since ρLα (y) ≤ ρH (y)/2, the function b(x) (extended as 0 when x ≤ 0) is an H 1 (H)-atom, and hence so is −b(−x). Thus kbo kHat1 (H) ≤ 2. Suppose now that ρLα (y)/2 < R ≤ ρLα (y). We distinguish two cases. If y ∈ (0, 8/9) then supp bo ⊂ B(0, y + R) ⊂ B(0, 9y/8) ≡ Bo . R Since R bo = 0 and kbo k∞ ≤ ρLα (y)−1 = 18/|Bo |, it follows that bo /18 is an H 1 (H)-atom associated with the ball Bo , and hence kbo kHat1 (H) ≤ 18. In the second case, i.e. y > 8/9, we may regard b/18 as an H 1 (H)-atom associated with the ball B(y, ρH (y)), since supp b ⊂ B(y, ρH (y)) and kbk∞ ≤ (2R)−1 ≤ 18|B(y, ρH (y))|−1 . Similarly, b(−x)/18 is an H 1 (H)-atom associated with the ball B(−y, ρH (−y)). We conclude that kbo kHat1 (H) ≤ 36, establishing the lemma. P 1 Proof of Theorem 1.1. Assume that f is in Hat (Lα ). Then f can be written as j cj bj , P where bj are H 1 (Lα )-atoms and j |cj | ∼ kf kHat1 (Lα ) . By the previous lemma, the odd 1 extension fo of f belongs to Hat (H) and kfo kHat1 (H) ≤ 36kf kHat1 (Lα ) . Applying Proposition 2.3 and using (15), we obtain kRα f kL1 (0,∞) ≤ C(kfo kL1 (R) + kRH fo kL1 (R) ) ≤ C 0 kfo kHat1 (H) ≤ C 00 kf kHat1 (Lα ) . 8
1 To prove the converse, assume that f is in HRiesz (Lα ). Again, using Proposition 2.3 1 1 combined with (15), we obtain fo ∈ HRiesz (H) = Hat (H) and
kfo kHat1 (H) ≤ C(kfo kL1 (R) + kRH fo kL1 (R) ) ≤ C(kf kL1 (0,∞) + kRα f kL1 (0,∞) ). P P Hence fo (x) = j cj aj (x), where aj are H 1 (H)-atoms and j |cj | ∼ kfo kHat1 (H) . Letting 1 (Lα ) and bj = aj (0,∞) , one easily verifies the inequality kbj kHat1 (Lα ) ≤ C. Thus f is in Hat kf kHat1 (Lα ) ≤ C 0 (kf kL1 (0,∞) + kRα f kL1 (0,∞) ). Remark 2.5. Using a similar analysis based on a comparison of the kernels Wtα (x, y) and WtH (x, y) (see [1, Lemma 2.11]), one can prove that W∗H fo belongs to L1 (R) if and only if W∗α f belongs to L1 (0, ∞) and kfo kL1 (R) + kW∗H fo kL1 (R) ∼ kf kL1 (0,∞) + kW∗α f kL1 (0,∞) .
Hardy spaces H 1(Lα ) associated with Laguerre op-
3
erators Lα . In this section we prove Theorem 1.2. The proof is based on the following estimates for the kernel Rα (x, y). Proposition 3.1. Let ψ be as in Lemma 2.1. Then, for every α > 0, there exists a kernel K(x, y) such that x−y c0 (22) Rα (x, y) = √ + K(x, y), x, y ∈ (0, ∞), ψ 2(x − y) ρLα (x) Z (23)
|K(x, y)|dx < ∞,
sup y>0
∞
0
where c0 is the constant from (13). Proof. Set (24)
K(x, y) = Rα (x, y) − √
x−y c0 ψ . 2(x − y) ρLα (x)
If x < y/4 or y < x/4, then K(x, y) = Rα (x, y). From (4) and (21), we conclude ( Cx(α+1)/2 y −(α+3)/2 if 4x < y < ∞ (25) |K(x, y)| ≤ Cy α/2 x−(α+2)/2 if 0 < y < x/4. Hence (26)
sup y>0
Z
y/4
Z
∞
|K(x, y)| dx +
0
4y
9
|K(x, y)| dx < ∞.
In order to deal with the kernel K(x, y) in the local part y/4 ≤ x ≤ 4y, we set √ √ E(x, y) = Rα (x, y) − 2−3/2 (xy)−1/4 RH ( x, y), √x − √y c0 2c0 x − y −3/2 −1/4 √ √ G(x, y) = 2 (xy) − ψ . √ ψ x−y ρLα (x) x− y ρH ( x) Then, by (4) and Lemma 2.1, we have √ √ K(x, y) = E(x, y) + 2−3/2 (xy)−1/4 h( x, y) + G(x, y).
(27)
According to (21), we get (28)
√ (xy)−1/4 1 (xy)1/8 x |E(x, y)| ≤ C √ 1+ √ √ 1/2 ≤ C 1 + y y |x − y|1/2 | x − y|
for y/4 ≤ x ≤ 4y. Trivially, using (28) and (14), we obtain Z 4y √ √ (29) |E(x, y)| + (xy)−1/4 |h( x, y)| dx ≤ C. y/4
The proof will be complete if we show the inequality Z 4y |G(x, y)| dx ≤ C. (30) y/4
Let us note that (31)
√ √ x − y i 2−3/2 c0 h x + y x−y √ √ G(x, y) = ψ . − 2ψ √ x−y (xy)1/4 ρLa (x) ( x + y)ρH ( x)
If y > 10, y/4 ≤ x ≤ 4y and |x − y| > 1, then G(x, y) = 0. If y > 10, y/4 < x < 4y and |x − y| ≤ 1, then, by the mean value theorem, |G(x, y)| ≤ C. Thus (30) is satisfied for y > 10. If 0 < y ≤ 10 and y/4 ≤ x ≤ 4y, then applying the mean value theorem we deduce |G(x, y)| ≤ Cy −1 and, consequently, (30) holds. Before we turn to the proof of Theorem 1.2, we state some results from the theory of local Hardy spaces [5]. Fix l > 0 . We say that a function b is an atom for the local Hardy space h1l (R) if there exists a ball B(y0 , R) with R < l such that supp b ⊂ R B(y0 , R), kbk∞ ≤ (2R)−1 , and if R ≤ l /2 , then b(y) dy = 0. A function f belongs to P the space h1l if there exist a sequence bj of h1l -atoms and λj ∈ C with j |λj | < ∞ such that (32)
f=
X j
10
λj bj .
P The atomic norm in h1l is defined in a standard way, that is, kf kh1l = inf j |λj |, where the infimum is taken over all decompositions (32). Moreover, if f ∈ h1l and supp f ⊂ B(y0 , l ), then there exists decomposition (32) of f such that supp bj ⊂ B(y0 , 10l /9 ) and P j |λj | ≤ Ckf kh1l . We define a local Hilbert transform Z x − y c √ 0 Hl f (x) = lim ψ f (y) dy, ε→0 |x−y|>ε l 2(x − y) where c0 and ψ are as in Lemma 2.1. The following result was actually proved in [5]. There exists a constant C > 0 independent of l such that (33)
C −1 kf kh1l ≤ kHl f kL1 + kf kL1 ≤ Ckf kh1l .
Proof of Theorem 1.2. Since Rα maps continuously L1 (0, ∞) into the space of distributions, to prove the second inequality in (6), it suffices to verify that there exists a constant C > 0 such that, for every H 1 (Lα )-atom b, one has kRα bkL1 ≤ C.
(34)
Let b be an H 1 (Lα )-atom with associated ball B(y0 , R). Clearly, letting l = ρLα (y0 ), we see that b is also an h1l -atom. By Proposition 3.1, Z Rα b(x) = K(x, y)b(y) dy + Hl b(x) Z (35) x − y x−y c0 √ −ψ + lim ψ χB(y0 ,l) (y)b(y) dy. ε→0 |x−y|>ε ρLα (x) l 2(x − y) The kernel U (x, y) = √
x−y x − y c0 ψ −ψ χB(y0 ,l) (y), ρLα (x) l 2(x − y)
as a function of (x, y), is supported by B(y0 , 3l ) × B (y0 , l ). Moreover, |U (x, y)| ≤ Cl −1 , R which implies supy>0 |U (x, y)|dx < ∞. Therefore, (34) holds by applying (23) and (33). We now turn to prove the first inequality in (6). We define the intervals {Ij }j∈Z , Ij = (βj , βj+1 ), βj = (9/8)j for j ≤ 1, and βj = 1 + j/8 for j ≥ 1. Set lj = ρLα (βj ). Let ηj be a family of smooth functions such that d X ∗ ηj (x ) = 1 for x > 0 , (36) 0 ≤ ηj ≤ 1, supp ηj ⊂ Ij , ηj (x) ≤ Clj−1 , dx j P where Ij∗ = [βj−1 , βj+2 ]. Set Ij∗∗ = [βj−2 , βj+3 ]. Then j χIj∗∗ ≤ 5. Fix f ∈ L1 (0, ∞) such that kRα f kL1 < ∞. We shall verify that X (37) kHlj (ηj f )kL1 ≤ C(kRα f kL1 + kf kL1 ) j
11
with a constant C > 0 independent of f . To this end, note that (38) x − y c0 ηj (y) − ηj (x) √ ψ f (y) dy + ηj (x)Hlj f (x) lj 2(x − y) |x−y|>ε
Z Hlj (ηj f )(x) = lim
ε→0
= Ξj f (x) + ηj (x)Hlj f (x). Observe that the kernel c0 x − y ψ ηj (y) − ηj (x) √ , lj 2(x − y) as a function of (x, y), is supported by Ij∗∗ × Ij∗∗ and bounded by Clj−1 . Since each y > 0 belongs to at most 5 intervals Ij∗∗ , and |Ij∗∗ | ∼ lj , we can easily obtain XZ (39) |Ξj f (x)| dx ≤ Ckf kL1 . j
Now we shall deal with ηj (x)Hlj f (x), defined by Z h x − y x − y i c √ 0 ηj (x)Hlj f (x) = ηj (x) ψ f (y) dy −ψ lj ρLα (x) 2(x − y) (40) Z + ηj (x)Rα f (x) − ηj (x) K(x, y)f (y) dy. The integral kernel h x − y x − y i c0 √ −ψ ηj (x) ψ , lj ρLα (x) 2(x − y) as a function of (x, y), is supported by Ij∗ × Ij∗∗ and bounded by Clj−1 . Hence Z ∞X x − y x − y c √ 0 (41) sup −ψ ηj (x) ψ dx < ∞. lj ρLα (x) y>0 0 2(x − y) j Using (40), (41), we obtain (42)
X
kηj Hlj f kL1 ≤ C(kf kL1 + kRα f kL1 ),
j
which combined with (38), (39) and (36) gives (37). Having (37) already proved, we are in a position to complete the proof of the first inequality in (6). Applying (37) together with the results from the theory of local Hardy spaces stated in this section, we have X XX (43) f= (ηj f ) = λij aij , j
j
12
i
P where aij are h1lj -atoms supported by Ij∗∗ , and ij |λij | ≤ C(kRα f kL1 +kf kL1 ). The proof will be complete once we observe that each of these atoms is either an H 1 (Lα )-atom, or can be written as a sum of at most 20 such atoms. Indeed, fix an h1lj -atom a supported in Ij∗∗ . Then, for some 0 < R0 < lj and y0 ∈ Ij∗∗ we have supp a ⊂ B(y0 , R0 ) ⊂ Ij∗∗ , R kak∞ ≤ (2R0 )−1 , and if R0 ≤ lj /2 then also a(x)dx = 0. Notice that, by construction, ρLα (y) ≤ 2ρLα (y 0 ), for all y, y 0 ∈ Ij∗∗ = [βj−2 , βj+3 ]. R If R0 ≤ lj /2 = ρLα (βj )/2 then a = 0 and R0 ≤ ρLα (y0 ), and therefore a is also an H 1 (Lα )-atom. If R0 > lj /2, then Ij∗∗
=
4 [
Ij−2+k
with |Ij−2+k | = ρLα (βj−2+k ) ,
k=0
and using again ρLα (βj+2 ) ≤ 2ρLα (βj ) we see that kaχIj−2+k k∞ ≤ (2R0 )−1 ≤ ρLα (βj )−1 ≤ 2|Ij−2+k |−1 . Hence, each piece aχIj−2+k /4 is an H 1 (Lα )-atom for the ball B(βj−2+k , ρLα (βj−2+k )) and, consequently, kakHat1 (Lα ) ≤ 20.
4
Proof of (11) and Lemma 2.1
During the proof we set r = e−t ∈ (0, 1). We can rewrite (9) as √ 1 1 + r2 1−r r H 2 exp − |x − y| exp − xy , (44) Wt (x, y) = p 2 1 − r2 1+r π(1 − r2 ) for all x, y ∈ R. A simple computation using (44) or (9) gives √ 1 1 + r2 r H 2 (45) Wt (x, y) ≤ p exp − |x − y| . 4 1 − r2 π(1 − r2 ) Let us note that, for every N > 0, there exists a constant CN such that e−t/3 t −N (46) WtH (x, y) ≤ CN p 1+ . ρH (x)2 (1 − r2 ) Indeed, if |x − y| > |x|/2, then 1 1 + r2 e−t/2 e−t/3 t −N 2 p (47) WtH (x, y) ≤ p exp − x ≤ C 1 + . N 8 1 − r2 ρH (x)2 π(1 − r2 ) (1 − r2 ) If |x − y| ≤ |x|/2, then xy ∼ x2 and, using (44), we get (48)
e−t/2 e−t/3 t −N WtH (x, y) ≤ C √ 1+ . exp − c(1 − r)x2 ≤ CN √ ρH (x)2 1 − r2 1 − r2 13
Applying (45) and (46) combined with the fact that WtH (x, y) = WtH (y, x), we obtain e−t/3 t −N t −N |x − y|2 (49) WtH (x, y) ≤ CN √ 1 + 1 + exp − . 12(1 − e−2t ) ρ(x)2 ρ(y)2 1 − e−2t We are now in a position to prove (11). If |x − y| ≤ CρH (y), then by (10) and (49) we have (50) |R2H (x, y)|
Z
|x−y|2
N dt t + ≤ CN |x − y|2 t 0 Cρ (y) H ≤ CN |x| + |x| ln . |x − y| |x|
Z
C 2 ρH (y)2
|x−y|2
dt |x| + t
∞
ρ (y)2 N dt H |x| t t C 2 ρH (y)2
Z
If |x − y| ≥ CρH (y), then we use again (49) and get (51) |R2H (x, y)|
N dt Z |x−y|2 N t −2N dt t t ≤ CN + |x| |x − y|2 t |x − y|2 ρH (y)2 t 0 C 2 ρH (y)2 Z ∞ ρ (y)2 N dt H + |x| t t |x−y|2 Z
C 2 ρH (y)2
|x|
|x|ρH (y)2N |x − y|2N |x − y|ρ (y)2N |y|ρH (y)2N H + . ≤ CN |x − y|2N |x − y|2N
≤ CN
Now the first inequality in (11) is a consequence of (50) and (51). Similarly to (50) and (51), we also conclude that C |x| + |x| ln CρH (x)/|x − y| for |x − y| ≤ CρH (x) (52) |R2H (x, y)| ≤ C |x|ρ (x)N /|x − y|N for |x − y| > Cρ (x), N
H
H
from which we easily obtain the second inequality in (11). Having (11) already established, we now turn to prove Lemma 2.1. By (44), (53) √ 1 1 + r2 1−r ∂ H r 1 + r2 2 W (x, y) = − p (x − y) exp − |x − y| exp − xy ∂x t 2 1 − r2 1+r π(1 − r2 ) 1 − r2 √ 1 1 + r2 1−r 1−r r 2 p −y exp − |x − y| exp − xy . 1 + r π(1 − r2 ) 2 1 − r2 1+r From (53) we deduce that, for |x − y| > CρH (y), we have (54) ∂ 1 e−t/3 |x − y|2 t −N H √ (x, y) − 1 + . W ≤ C + |y|(1 − r) exp N ∂x t |x − y| 12(1 − r2 ) ρH (y)2 1 − r2 14
Proceeding as in (51), we obtain Z ∞ ∂ 1 ρ (y)2N dt H H (55) Wt (x, y) √ ≤ CN + |y| for |x − y| > CρH (y), ∂x |x − y| |x − y|2N t 0 which leads to Z (56)
|R1H (x, y)| dx ≤ C.
sup y∈R
|x−y|>CρH (y)
Our next step is to estimate R1H (x, y) for |x − y| ≤ CρH (y). Note that (53) implies (57) ∂ e−t/3 1 + r2 |x − y|2 t −N −1 |x − y| exp − 1+ Wt (x, y) ≤ CN √ ∂x 12(1 − r2 ) ρH (y)2 1 − r2 1 − r2 t −N −1 |x − y|2 e−t/3 1 + + CN √ |y|(1 − r) exp − 12(1 − r2 ) ρH (y)2 1 − r2 t −N e−t/4 1 + . ≤ CN 1 − r2 ρH (y)2 Consequently, using (57) we get Z ∞ dt ∂ (58) Wt (x, y) √ ≤ CρH (y)−1 . t ρH (y)2 ∂x In order to investigate the integral Z ρH (y)2 0
∂ dt Wt (x, y) √ , ∂x t
we study first the difference ρH (y)2
Z Q(x, y) = 0
dt ∂ H Wt (x, y) − Pt (x − y) √ , ∂x t
where Pt (x) = (2πt)−1/2 exp(−x2 /2t) is the classical Gauss-Weierstrass kernel. The perturbation formula asserts that Z 2 Z t Z ∞ 1 ρH (y) ∂ dt Q(x, y) = − Pt−s (x − z)z 2 WsH (z, y)dz ds √ . 2 0 t 0 −∞ ∂x Therefore, (59) Z |Q(x, y)|dx
J= |x−y| CρH (x), |x − y|N
and consequently, Z (61)
|R1H (x, y)| dy < ∞.
sup x∈R
|x−y|>CρH (x)
A similar procedure to that employed to estimate J gives Z (62) sup |Q(x, y)|dy ≤ C. x∈R
|x−y|≤CρH (x)
Finally, our analysis of the kernel R1H (x, y) is reduced to the integral Z (63)
0
ρH (y)2
ρH (y)2
x−y 1 dt √ exp(−|x − y|2 /2t) √ t t 2πt 0 2 |x − y| 2 exp − . = −√ 2ρH (y)2 2π(x − y)
∂ dt Pt (x − y) √ = − ∂x t
Z
Taking into account (10), (55), (58), (60), (61), (62) and (63), we get √ |x − y|2 2 H (64) R1 (x, y) = − exp − + h1 (x, y) π(x − y) 2ρH (y)2 with Z (65)
∞
∞
|h1 (x, y)|dx + sup
sup y∈R
Z
−∞
x∈R
|h1 (x, y)|dy < ∞. −∞
To complete the proof, take any ψ ∈ Cc∞ (R) as in the statement of Lemma 2.1. Define a function h2 (x, y) by √ √ x−y |x − y|2 2 2 ψ − exp − , x, y ∈ R. h2 (x, y) = π(x − y) ρH (x) π(x − y) 2ρH (y)2 By (10), (64), (65) and (11), the lemma will be established once we show that, for some C > 0 we have Z Z (66) sup |h2 (x, y)| dy ≤ C and sup |h2 (x, y)| dx ≤ C. x∈R
y∈R
16
Set A = {(x, y) ∈ R2 ; |x − y| > ρH (x)}, B = {(x, y) ∈ R2 ; |x − y| ≤ ρH (x)}. Then (67)
|x − y|2 1 C |x − y| |h2 (x, y)| ≤ exp − χA (x, y) + C + χB (x, y), |x − y| 2ρH (y)2 ρH (x) ρH (y)2
where the last summand is obtained by applying the mean value theorem. Using (8), we see that ρ(y)2 ≤ cρ(x)|x − y| when (x, y) ∈ A, and therefore Z Z |x − y|2 |x − y| C C exp − χA (x, y) dy ≤ exp − c χA (x, y) dy |x − y| 2ρH (y)2 |x − y| ρH (x) Z (68) du ≤ exp(−c|u|) ≤ C. |u| |u|>1 On the other hand, ρH (x) ∼ ρH (y) when (x, y) ∈ B (again by (8)), so we have Z C C|x − y| + χB (x, y) dy ≤ C, ρH (x) ρH (y)2 which together with (68) implies the first inequality in (66). From (8) we also see that e = {(x, y) ∈ R2 ; |x − y| > ερH (y)} and B ⊂ B e = {(x, y) ∈ R2 ; |x − y| ≤ ρH (y)/ε} A⊂A for some ε > 0. Using this fact, the second inequality in (66) follows by similar arguments. This completes the proof of Lemma 2.1.
References [1] J. J. Betancor, J. C. Fari˜ na, L. Rodr´ıguez-Mesa, A. Sanabria and J. L. Torrea, Transference between Laguerre and Hermite settings, J. Funct. Anal. 254 (2008), 826–850. [2] J. Dziuba´ nski, A note on Schr¨ odinger operators with polynomial potentials, Colloq. Math. 78 (1998), 149–161. [3] J. Dziuba´ nski, Hardy spaces for Laguerre expansions, Constr. Approx. 27 (2008), 269–287. [4] J. Dziuba´ nski and J. Zienkiewicz, Hardy spaces associated with some Schr¨ odinger operators, Studia Math. 126 (2) (1997), 149–160. [5] D. Goldberg, A local version of real Hardy spaces, Duke Math. J. 46 (1979), 27–42. [6] E. Harboure, J. L. Torrea and B. Viviani, Riesz transforms for Laguerre expansions, Indiana Univ. Math. J. 55 (2006), 999–1014. [7] A. Nowak and K. Stempak, Riesz transforms and conjugacy for Laguerre function expansions of Hermite type, J. Funct. Anal. 244 (2007), 399–443. [8] K. Stempak and J. L. Torrea, Poisson integrals and Riesz transforms for Hermite function expansions with weights, J. Funct. Anal. 202 (2003), 443–472.
17
[9] S. Thangavelu, Riesz transforms and wave equation for the Hermite operator, Comm. Partial Differential Equations 15(8) (1990), 1199–1215. [10] S. Thangavelu, Lectures on Hermite and Laguerre expansions, Mathematical Notes, 42, Princeton University Press, Princeton, NJ, 1993. [11] J. Zhong, Harmonic analysis for some Schr¨ odinger operators, Ph. D. Thesis, Princeton Univ., 1993. J. Betancor ´ lisis Matema ´ tico Departamento de Ana Universidad de la Laguna Campus de Anchieta, Avda. Astrof´ısico ´ nchez, s/n Francisco Sa 38271 La Laguna (Sta. Cruz de Tenerife) Spain E-mail address: [email protected] ´ ski J. Dziuban Instytut Matematyczny Uniwersytet Wroclawski 50-384 Wroclaw, PL. Grunwaldzki 2/4 Poland E-mail address: [email protected] ´s G. Garrigo ´ ticas Departamento de Matema Facultad de Ciencias ´ noma de Madrid Universidad Auto 28049 Madrid Spain E-mail address: [email protected]
18
Abstract In this paper we prove Riesz transform characterizations for Hardy spaces associated with certain systems of Laguerre functions.
1
Introduction and statement of the results
Denote the Laguerre polynomials of order α > −1 by d n (e−x xn+α ), n = 0, 1, 2, ... Lαn (x) = (n!)−1 ex x−α dx In this paper we consider the following two systems of Laguerre functions on (0, ∞) √ 2 (1) ϕαn (x) = 2 cn,α e−x /2 xα+1/2 Lαn (x2 ), n = 0, 1, 2 . . . , (2)
Lαn (x) = cn,α e−x/2 xα/2 Lαn (x),
n = 0, 1, 2, . . . ,
where cn,α = (Γ(n + 1)/Γ(n + 1 + α))1/2 . It is well known that, for every α > −1, each α ∞ 2 of the systems {ϕαn }∞ n=0 and {Ln }n=0 is complete and orthonormal on L ((0, ∞), dx). Moreover, these functions are eigenvectors, respectively, of the differential operators d2 x α2 1 d2 1 1 d − 2 + y 2 + 2 α2 − , Lα = − x 2 + − + , Lα = 2 dy y 4 dx dx 4 4x 2000 Mathematics Subject Classification. Primary 42B30; Secondary 42C10. Key words and phrases. Hardy space, Laguerre system, Riesz transform. First author partially supported by MTM2004-05878 and MTM2007-65609. Second author partially 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 Government funds for science - grant N N201 397137, MNiSW. Third author partially supported by MTM2007-60952.
1
satisfying Lα ϕαn = (2n + α + 1)ϕαn
and Lα (Lαn ) = (n + (α + 1)/2) Lαn .
As in [6, 7], the operators Lα and Lα can be factored as Lα = where
1 ∗ α+1 Dα Dα + α + 1 and Lα = δα∗ δα + , 2 2
α + 1/2 d +x− dx x
Dα =
and δα =
√ d 1 √ α x + x− √ , dx 2 x
and where Dα∗ and δα∗ denote, respectively, the formal adjoint operators to Dα and δα in L2 ((0, ∞), dx). Corresponding Riesz transforms are defined in L2 ((0, ∞), dx) by Rα = Dα Lα−1/2
and Rα = δα Lα−1/2 ,
that is, they act on the basis elements by √ √ 2 n n α α+1 α (3) Rα ϕn = − √ ϕn−1 , Rα Ln = − p Lα+1 n−1 . 2n + α + 1 n + (α + 1)/2 There exist kernels Rα (x, y) and Rα (x, y) such that Z ∞ Z Rα f (x) = lim Rα (x, y)f (y) dy, Rα f (x) = lim ε→0
ε→0
0, |x−y|>ε
∞
Rα (x, y)f (y) dy.
0, |x−y|>ε
One can easily deduce from (1), (2) and (3) that these kernels are related by √ √ Rα (x, y) = 2−3/2 (xy)−1/4 Rα ( x, y),
(4)
x, y ∈ (0, ∞).
Riesz tranforms for Laguerre systems were defined and studied by Nowak and Stempak [7], and by Harboure, Torrea and Viviani [6], who proved that Rα for α ≥ −1/2 and Rα for α ≥ 0 extend as bounded linear operators on Lp (0, ∞) when 1 < p < ∞ and are of weak type (1,1). Our goal in the present paper is to characterize the spaces 1 (Lα ) = f ∈ L1 (0, ∞) ; kRα f kL1 < ∞ HRiesz
for α > −1/2,
and 1 HRiesz (Lα ) =
f ∈ L1 (0, ∞) ; kRα f kL1 < ∞
for α > 0.
1 1 In [3], the second-named author considered Hardy spaces Hmax (Lα ) and Hmax (Lα ) defined by means of the maximal functions associated with the semigroups generated by −Lα and −Lα , respectively. To be more precise, if
Wtα (x, y) =
∞ X
e−(2n+α+1)t ϕαn (x)ϕαn (y), Wαt (x, y) =
n=0
∞ X n=0
2
e−t(n+(α+1)/2) Lαn (x)Lαn (y)
denote the integral kernels of the semigroups {e−tLα }t>0 and {e−tLα }t>0 , we say that a 1 function f in (0, ∞) belongs to Hmax (Lα ) when the maximal function Z ∞ α W∗ f (x) = sup Wtα (x, y)f (y)dy t>0
0
α 1 belongs to L1 (0, ∞). Then we set kf kHmax (Lα ) = kW∗ f kL1 . Analogously, we define the 1 1 (Lα ) and the norm k · kHmax maximal function Wα∗ , the space Hmax (Lα ) . It was proved in [3] 1 1 that the spaces Hmax (Lα ), α > −1/2, and Hmax (Lα ), α > 0, admit atomic decompositions. The notion of atom for these spaces depends on the following auxiliary functions
ρLα (x) =
1 1 min(x, 1/x) and ρLα (x) = min(x, 1). 8 8
A measurable function b : (0, ∞) → C is said to be an H 1 (Lα )-atom if there exists a ball B = B(y0 , R) = {|y0 − y| < R} with R ≤ ρLα (y0 ) such that kbk∞ ≤ |B|−1 and Z b(y)dy = 0. if R ≤ ρLα (y0 )/2 then supp b ⊂ B,
1 The space Hat (Lα ) consists of all measurable functions f on (0, ∞) of the form
f=
∞ X
λj bj ,
j=1
where bj are H 1 (Lα )-atoms, λj ∈ C and by
P∞
j=1
1 |λj | < ∞. The norm in Hat (Lα ) is defined
kf kHat1 (Lα ) = inf
∞ X
|λj |,
j=1
P 1 where the infimum is taken over all decompositions f = ∞ j=1 λj bj , where bj are H (Lα )1 atoms and λj ∈ C. Similarly we define the space Hat (Lα ) and the norm k kHat1 (Lα ) , the only difference being that the function ρLα replaces the function ρLα in the definition of H 1 (Lα )-atoms. The main result in [3] was to show that 1 1 1 1 Hmax (Lα ) = Hat (Lα ) for α > −1/2 and Hmax (Lα ) = Hat (Lα ) for α > 0,
with equivalence of the corresponding norms. Our goal in this paper is to characterize these spaces by means of the Riesz transforms Rα and Rα . More precisely, we shall prove the following theorems. 1 1 Theorem 1.1. If α > −1/2, then HRiesz (Lα ) = Hat (Lα ). Moreover, there exists C > 0 such that
(5)
C −1 kf kHat1 (Lα ) ≤ kRα f kL1 + kf kL1 ≤ Ckf kHat1 (Lα ) . 3
1 1 Theorem 1.2. If α > 0, then HRiesz (Lα ) = Hat (Lα ). Moreover, there exists C > 0 such that
C −1 kf kHat1 (Lα ) ≤ kRα f kL1 + kf kL1 ≤ Ckf kHat1 (Lα ) .
(6)
2
Hardy spaces H 1(Lα ) associated with Laguerre operators Lα
In the present section, we shall prove Theorem 1.1. To do this, we recall the equivalence between Riesz and atomic definitions for the Hardy space associated with the Hermite operator, d2 1 H = (− 2 + x2 ), 2 dx which were established in [4]. First we let ρH (y) = (1 + |y|)−1 .
(7)
It is easily seen that there exist constants C, c > 0 such that −1 1/2 (8) cρH (x) 1 + |x − y|/ρH (x) ≤ ρH (y) ≤ CρH (x) 1 + |x − y|/ρH (x) . A function a : R → C is an H 1 (H)-atom if there exists a ball B = B(y0 , R) = {y ∈ R; |y − y0 | < R} with R ≤ ρH (y0 ) such that kakL∞ ≤ |B|−1 and Z if R ≤ ρH (y0 )/2 then a(y) dy = 0.
supp a ⊂ B,
1 The atomic Hardy space Hat (H) and the norm k kHat1 (H) are defined in the standard way. On the other hand, a Riesz transform RH can be defined in L2 (R) by d + x H −1/2 , RH = dx
motivated by the factorization of the Hermite operator d d d i 1 h d H=− +x −x + −x +x . 4 dx dx dx dx To obtain a kernel expression for RH , recall first the Mehler formula for Hermite functions (cf. [10, Lemma 1.1.1]), which asserts that the integral kernel WtH (x, y) of the Hermite semigroup {e−tH }t>0 is given by (9)
WtH (x, y) =
h
i1/2 1 1 + e−2t e−t e−t 2 2 exp − (x + y ) + 2xy π(1 − e−2t ) 2 1 − e−2t 1 − e−2t 4
R∞ when t > 0 and x, y ∈ R. Using the formula H −1/2 = π −1/2 0 e−tH t−1/2 dt, we can express the Riesz transform RH as a principal value singular integral operator of the form Z H R (f )(x) = lim RH (x, y)f (y)dy , ε→0
y∈R : |x−y|>ε
with the kernel given by Z ∞ 1 dt d R (x, y) = √ + x WtH (x, y) √ dx π 0 t Z ∞ Z ∞ 1 1 d H dt dt =√ Wt (x, y) √ + √ xWtH (x, y) √ π 0 dx π 0 t t H
(10)
= R1H (x, y) + R2H (x, y). It is not difficult to prove using (9) and (10) that Z ∞ Z H (11) sup |R2 (x, y)| dx < ∞, sup y∈R
x∈R
−∞
∞
|R2H (x, y)| dy < ∞
−∞
(see Section 4). Therefore, denoting R2H = xH −1/2 , we have kR2H f kL1 (R) ≤ Ckf kL1 (R)
(12)
(see also [2, Theorem 4.5]). It was proved by Thangavelu [9] that the operator RH is bounded on Lp (R) for 1 < p < ∞. Moreover, Theorem 1.2 of Zhong [11] asserts that the operator R1H = (d/dx)H −1/2 is a Calder´on-Zygmund operator, hence it is of weak type (1,1) (see also [8] for a proof based on analysis of the Melher kernel). The above facts could also be deduced from the following lemma. Lemma 2.1. Let ψ ∈ Cc∞ (−2−4 , 2−4 ) be such that ψ(x) = 1 for |x| < 2−5 . Then there exists a constant c0 6= 0 and a kernel h(x, y) such that x−y c0 (13) RH (x, y) = ψ + h(x, y), x−y ρH (x) Z (14)
∞
∞
|h(x, y)|dx + sup
sup y∈R
Z
−∞
x∈R
|h(x, y)|dy < ∞. −∞
This lemma is known, but a self-contained proof based on analysis of the Mehler kernel will be presented in Section 4. We set 1 HRiesz (H) = f ∈ L1 (R) ; kRH f kL1 (R) < ∞ . 1 In view of (12), an L1 -function f belongs to HRiesz (H) if and only if (d/dx)H −1/2 f belongs to L1 (R). From this remark and the results in [4], it follows that 1 1 HRiesz (H) = Hat (H)
5
and there exists a constant C > 0 such that (15)
C −1 kf kHat1 (H) ≤ kRH f kL1 + kf kL1 ≤ Ckf kHat1 (H) .
Having established the Riesz and atomic characterizations of the Hardy space associated with the Hermite operator, we continue our preparation for the proof of Theorem 1.1. H H H f , where f + R2,loc f = R1,loc For a function f defined on (0, ∞), we denote Rloc Z 2x H Rj,loc f (x) = lim RjH (x, y)f (y) dy, x > 0, j = 1, 2. ε→0
x/2, |x−y|>ε
Proposition 2.2. For f ∈ L1 (0, ∞), let fo denote its odd extension. Then R1H fo ∈ H L1 (R) if and only if R1,loc f is in L1 (0, ∞). Moreover, there exists C > 0 such that H kR1H fo − R1,loc f kL1 (0,∞) ≤ Ckf kL1 (0,∞) .
Proof. Set r = r(t) = e−t ∈ (0, 1). According to (9) and (10), we have (16) Z ∞√ dt r 2ry − (1 + r2 )x 1 1 + r2 2r H 2 2 √ . R1 (x, y) = √ exp − (x + y ) + xy (1 − r2 )3/2 2(1 − r2 ) 1 − r2 π 0 t Note that kR1H fo kL1 (R) = 2kR1H fo kL1 (0,∞) , because R1H fo is an even function. Moreover, Z ∞ H R1 fo (x) = lim R1H (x, y) − R1H (x, −y) f (y)dy, a.e. x ∈ (0, ∞). ε→0
0, |x−y|>ε
Further, R1H fo (x)
−
H R1,loc f (x)
x/2
Z
R1H (x, y) − R1H (x, −y) f (y)dy
= 0
∞
Z
R1H (x, y) − R1H (x, −y) f (y)dy
+ Z2x2x
(17) −
R1H (x, −y)f (y)dy
x/2
=
3 X
Tj (f )(x), a.e. x ∈ (0, ∞).
j=1
It suffices to show that the operators Tj , j = 1, 2, 3, are bounded on L1 ((0, ∞), dx). To deal with T1 and T2 , we estimate the difference DH (x, y) = |R1H (x, y) − R1H (x, −y)| for x, y > 0. By (16) √ Z ∞ rx 2r 2r H exp xy − exp − xy D (x, y) ≤ C (1 − r2 )3/2 1 − r2 1 − r2 0 dt 1 + r2 2 2 √ (18) × exp − (x + y ) 2(1 − r2 ) t √ Z ∞ 2r dt ry 1 + r2 2 2 √ . exp − exp xy +C (x + y ) (1 − r2 )3/2 2(1 − r2 ) 1 − r2 t 0 6
Applying the mean value theorem in the first integral, we can assert that (19) DH (x, y) √ Z ∞ rx2 y r ≤C + y exp − (1 − r2 )3/2 1 − r2 0 √ Z ∞ rx2 y r =C + y exp − (1 − r2 )3/2 1 − r2 0
2r dt 1 + r2 2 2 √ (x + y ) exp xy 2(1 − r2 ) 1 − r2 t 2 1−r dt 1+r (x − y)2 exp − xy √ . 2 2(1 − r ) 1+r t
It is now not difficult to verify using (19) that Cyx−2 (20) DH (x, y) ≤ Cy −1
for x > 2y, for 2x < y.
The estimate (20) easily implies kT1 f kL1 (0,∞) + kT2 f kL1 (0,∞) ≤ Ckf kL1 (0,∞) . Moreover, from (16), we conclude Z ∞ Z 1 1 −cx2 /t C H −cx2 −t e dt ≤ for x/2 < y < 2x. |R1 (x, −y)| ≤ C xe e dt + x 2 y 1 0 t Hence T3 is a bounded operator from L1 (0, ∞) into itself. Proposition 2.3. Let α > −1/2, f ∈ L1 (0, ∞) and fo be the odd extension of f to R. Then Rα f is in L1 (0, ∞) if and only if RH fo is in L1 (R). Moreover, there exists C > 0 such that C −1 (kfo kL1 (R) + kRH fo kL1 (R) ) ≤ kf kL1 (0,∞) + kRα f kL1 (0,∞)
and
H
kf kL1 (0,∞) + kRα f kL1 (0,∞) ≤ C(kfo kL1 (R) + kR fo kL1 (R) ). Proof. According to [1, Lemma 2.13], we have |Rα (x, y)| (21)
≤ Cxα+3/2 y −(α+5/2)
≤ Cy α+1/2 x−(α+3/2) C (xy)1/4 H |Rα (x, y) − R (x, y)| ≤ 1+ y |x − y|1/2 |Rα (x, y)|
for 0 < 2x < y < ∞, for 0 < y < x/2, and for 0 < x/2 < y < 2x.
Each of the Hardy operators Hα (g)(x) = x
x
Z
−α−3/2
y α+1/2 g(y)dy,
x>0
y −α−3/2 g(y)dy,
x>0
0
and α
H (g)(x) = x
α+1/2
Z
∞
x
7
are bounded on L1 (0, ∞) when α > −1/2. Moreover, the operator N defined by Z 2x 1 (xy)1/4 1+ f (y)dy N f (x) = |x − y|1/2 x/2 y is also bounded in L1 (0, ∞). Hence, by (21), (11) and Proposition 2.2, we obtain H H kRα f − RH fo kL1 (0,∞) ≤ kRα f − Rloc f kL1 (0,∞) + kRloc f − RH fo kL1 (0,∞)
≤ C kN |f | kL1 (0,∞) + kH α+1 |f | kL1 (0,∞) + kHα |f | kL1 (0,∞)
H H + kR1,loc f − R1H fo kL1 (0,∞) + kR2,loc f kL1 (0,∞) + kR2H fo kL1 (0,∞)
≤ Ckf kL1 (0,∞) .
The next elementary lemma will be used below. Lemma 2.4. Let b : (0, ∞) → C be an H 1 (Lα )-atom. Then, its odd extension bo satisfies kbo kHat1 (H) ≤ 36. Proof. Let B = B(y, R) ⊂ (0, ∞) be a ball associated with b, that is, R ≤ ρLα (y), R supp b ⊂ B and kbk∞ ≤ |B|−1 . Moreover, b(y)dy = 0 if R ≤ ρLα (y)/2. In this last case, since ρLα (y) ≤ ρH (y)/2, the function b(x) (extended as 0 when x ≤ 0) is an H 1 (H)-atom, and hence so is −b(−x). Thus kbo kHat1 (H) ≤ 2. Suppose now that ρLα (y)/2 < R ≤ ρLα (y). We distinguish two cases. If y ∈ (0, 8/9) then supp bo ⊂ B(0, y + R) ⊂ B(0, 9y/8) ≡ Bo . R Since R bo = 0 and kbo k∞ ≤ ρLα (y)−1 = 18/|Bo |, it follows that bo /18 is an H 1 (H)-atom associated with the ball Bo , and hence kbo kHat1 (H) ≤ 18. In the second case, i.e. y > 8/9, we may regard b/18 as an H 1 (H)-atom associated with the ball B(y, ρH (y)), since supp b ⊂ B(y, ρH (y)) and kbk∞ ≤ (2R)−1 ≤ 18|B(y, ρH (y))|−1 . Similarly, b(−x)/18 is an H 1 (H)-atom associated with the ball B(−y, ρH (−y)). We conclude that kbo kHat1 (H) ≤ 36, establishing the lemma. P 1 Proof of Theorem 1.1. Assume that f is in Hat (Lα ). Then f can be written as j cj bj , P where bj are H 1 (Lα )-atoms and j |cj | ∼ kf kHat1 (Lα ) . By the previous lemma, the odd 1 extension fo of f belongs to Hat (H) and kfo kHat1 (H) ≤ 36kf kHat1 (Lα ) . Applying Proposition 2.3 and using (15), we obtain kRα f kL1 (0,∞) ≤ C(kfo kL1 (R) + kRH fo kL1 (R) ) ≤ C 0 kfo kHat1 (H) ≤ C 00 kf kHat1 (Lα ) . 8
1 To prove the converse, assume that f is in HRiesz (Lα ). Again, using Proposition 2.3 1 1 combined with (15), we obtain fo ∈ HRiesz (H) = Hat (H) and
kfo kHat1 (H) ≤ C(kfo kL1 (R) + kRH fo kL1 (R) ) ≤ C(kf kL1 (0,∞) + kRα f kL1 (0,∞) ). P P Hence fo (x) = j cj aj (x), where aj are H 1 (H)-atoms and j |cj | ∼ kfo kHat1 (H) . Letting 1 (Lα ) and bj = aj (0,∞) , one easily verifies the inequality kbj kHat1 (Lα ) ≤ C. Thus f is in Hat kf kHat1 (Lα ) ≤ C 0 (kf kL1 (0,∞) + kRα f kL1 (0,∞) ). Remark 2.5. Using a similar analysis based on a comparison of the kernels Wtα (x, y) and WtH (x, y) (see [1, Lemma 2.11]), one can prove that W∗H fo belongs to L1 (R) if and only if W∗α f belongs to L1 (0, ∞) and kfo kL1 (R) + kW∗H fo kL1 (R) ∼ kf kL1 (0,∞) + kW∗α f kL1 (0,∞) .
Hardy spaces H 1(Lα ) associated with Laguerre op-
3
erators Lα . In this section we prove Theorem 1.2. The proof is based on the following estimates for the kernel Rα (x, y). Proposition 3.1. Let ψ be as in Lemma 2.1. Then, for every α > 0, there exists a kernel K(x, y) such that x−y c0 (22) Rα (x, y) = √ + K(x, y), x, y ∈ (0, ∞), ψ 2(x − y) ρLα (x) Z (23)
|K(x, y)|dx < ∞,
sup y>0
∞
0
where c0 is the constant from (13). Proof. Set (24)
K(x, y) = Rα (x, y) − √
x−y c0 ψ . 2(x − y) ρLα (x)
If x < y/4 or y < x/4, then K(x, y) = Rα (x, y). From (4) and (21), we conclude ( Cx(α+1)/2 y −(α+3)/2 if 4x < y < ∞ (25) |K(x, y)| ≤ Cy α/2 x−(α+2)/2 if 0 < y < x/4. Hence (26)
sup y>0
Z
y/4
Z
∞
|K(x, y)| dx +
0
4y
9
|K(x, y)| dx < ∞.
In order to deal with the kernel K(x, y) in the local part y/4 ≤ x ≤ 4y, we set √ √ E(x, y) = Rα (x, y) − 2−3/2 (xy)−1/4 RH ( x, y), √x − √y c0 2c0 x − y −3/2 −1/4 √ √ G(x, y) = 2 (xy) − ψ . √ ψ x−y ρLα (x) x− y ρH ( x) Then, by (4) and Lemma 2.1, we have √ √ K(x, y) = E(x, y) + 2−3/2 (xy)−1/4 h( x, y) + G(x, y).
(27)
According to (21), we get (28)
√ (xy)−1/4 1 (xy)1/8 x |E(x, y)| ≤ C √ 1+ √ √ 1/2 ≤ C 1 + y y |x − y|1/2 | x − y|
for y/4 ≤ x ≤ 4y. Trivially, using (28) and (14), we obtain Z 4y √ √ (29) |E(x, y)| + (xy)−1/4 |h( x, y)| dx ≤ C. y/4
The proof will be complete if we show the inequality Z 4y |G(x, y)| dx ≤ C. (30) y/4
Let us note that (31)
√ √ x − y i 2−3/2 c0 h x + y x−y √ √ G(x, y) = ψ . − 2ψ √ x−y (xy)1/4 ρLa (x) ( x + y)ρH ( x)
If y > 10, y/4 ≤ x ≤ 4y and |x − y| > 1, then G(x, y) = 0. If y > 10, y/4 < x < 4y and |x − y| ≤ 1, then, by the mean value theorem, |G(x, y)| ≤ C. Thus (30) is satisfied for y > 10. If 0 < y ≤ 10 and y/4 ≤ x ≤ 4y, then applying the mean value theorem we deduce |G(x, y)| ≤ Cy −1 and, consequently, (30) holds. Before we turn to the proof of Theorem 1.2, we state some results from the theory of local Hardy spaces [5]. Fix l > 0 . We say that a function b is an atom for the local Hardy space h1l (R) if there exists a ball B(y0 , R) with R < l such that supp b ⊂ R B(y0 , R), kbk∞ ≤ (2R)−1 , and if R ≤ l /2 , then b(y) dy = 0. A function f belongs to P the space h1l if there exist a sequence bj of h1l -atoms and λj ∈ C with j |λj | < ∞ such that (32)
f=
X j
10
λj bj .
P The atomic norm in h1l is defined in a standard way, that is, kf kh1l = inf j |λj |, where the infimum is taken over all decompositions (32). Moreover, if f ∈ h1l and supp f ⊂ B(y0 , l ), then there exists decomposition (32) of f such that supp bj ⊂ B(y0 , 10l /9 ) and P j |λj | ≤ Ckf kh1l . We define a local Hilbert transform Z x − y c √ 0 Hl f (x) = lim ψ f (y) dy, ε→0 |x−y|>ε l 2(x − y) where c0 and ψ are as in Lemma 2.1. The following result was actually proved in [5]. There exists a constant C > 0 independent of l such that (33)
C −1 kf kh1l ≤ kHl f kL1 + kf kL1 ≤ Ckf kh1l .
Proof of Theorem 1.2. Since Rα maps continuously L1 (0, ∞) into the space of distributions, to prove the second inequality in (6), it suffices to verify that there exists a constant C > 0 such that, for every H 1 (Lα )-atom b, one has kRα bkL1 ≤ C.
(34)
Let b be an H 1 (Lα )-atom with associated ball B(y0 , R). Clearly, letting l = ρLα (y0 ), we see that b is also an h1l -atom. By Proposition 3.1, Z Rα b(x) = K(x, y)b(y) dy + Hl b(x) Z (35) x − y x−y c0 √ −ψ + lim ψ χB(y0 ,l) (y)b(y) dy. ε→0 |x−y|>ε ρLα (x) l 2(x − y) The kernel U (x, y) = √
x−y x − y c0 ψ −ψ χB(y0 ,l) (y), ρLα (x) l 2(x − y)
as a function of (x, y), is supported by B(y0 , 3l ) × B (y0 , l ). Moreover, |U (x, y)| ≤ Cl −1 , R which implies supy>0 |U (x, y)|dx < ∞. Therefore, (34) holds by applying (23) and (33). We now turn to prove the first inequality in (6). We define the intervals {Ij }j∈Z , Ij = (βj , βj+1 ), βj = (9/8)j for j ≤ 1, and βj = 1 + j/8 for j ≥ 1. Set lj = ρLα (βj ). Let ηj be a family of smooth functions such that d X ∗ ηj (x ) = 1 for x > 0 , (36) 0 ≤ ηj ≤ 1, supp ηj ⊂ Ij , ηj (x) ≤ Clj−1 , dx j P where Ij∗ = [βj−1 , βj+2 ]. Set Ij∗∗ = [βj−2 , βj+3 ]. Then j χIj∗∗ ≤ 5. Fix f ∈ L1 (0, ∞) such that kRα f kL1 < ∞. We shall verify that X (37) kHlj (ηj f )kL1 ≤ C(kRα f kL1 + kf kL1 ) j
11
with a constant C > 0 independent of f . To this end, note that (38) x − y c0 ηj (y) − ηj (x) √ ψ f (y) dy + ηj (x)Hlj f (x) lj 2(x − y) |x−y|>ε
Z Hlj (ηj f )(x) = lim
ε→0
= Ξj f (x) + ηj (x)Hlj f (x). Observe that the kernel c0 x − y ψ ηj (y) − ηj (x) √ , lj 2(x − y) as a function of (x, y), is supported by Ij∗∗ × Ij∗∗ and bounded by Clj−1 . Since each y > 0 belongs to at most 5 intervals Ij∗∗ , and |Ij∗∗ | ∼ lj , we can easily obtain XZ (39) |Ξj f (x)| dx ≤ Ckf kL1 . j
Now we shall deal with ηj (x)Hlj f (x), defined by Z h x − y x − y i c √ 0 ηj (x)Hlj f (x) = ηj (x) ψ f (y) dy −ψ lj ρLα (x) 2(x − y) (40) Z + ηj (x)Rα f (x) − ηj (x) K(x, y)f (y) dy. The integral kernel h x − y x − y i c0 √ −ψ ηj (x) ψ , lj ρLα (x) 2(x − y) as a function of (x, y), is supported by Ij∗ × Ij∗∗ and bounded by Clj−1 . Hence Z ∞X x − y x − y c √ 0 (41) sup −ψ ηj (x) ψ dx < ∞. lj ρLα (x) y>0 0 2(x − y) j Using (40), (41), we obtain (42)
X
kηj Hlj f kL1 ≤ C(kf kL1 + kRα f kL1 ),
j
which combined with (38), (39) and (36) gives (37). Having (37) already proved, we are in a position to complete the proof of the first inequality in (6). Applying (37) together with the results from the theory of local Hardy spaces stated in this section, we have X XX (43) f= (ηj f ) = λij aij , j
j
12
i
P where aij are h1lj -atoms supported by Ij∗∗ , and ij |λij | ≤ C(kRα f kL1 +kf kL1 ). The proof will be complete once we observe that each of these atoms is either an H 1 (Lα )-atom, or can be written as a sum of at most 20 such atoms. Indeed, fix an h1lj -atom a supported in Ij∗∗ . Then, for some 0 < R0 < lj and y0 ∈ Ij∗∗ we have supp a ⊂ B(y0 , R0 ) ⊂ Ij∗∗ , R kak∞ ≤ (2R0 )−1 , and if R0 ≤ lj /2 then also a(x)dx = 0. Notice that, by construction, ρLα (y) ≤ 2ρLα (y 0 ), for all y, y 0 ∈ Ij∗∗ = [βj−2 , βj+3 ]. R If R0 ≤ lj /2 = ρLα (βj )/2 then a = 0 and R0 ≤ ρLα (y0 ), and therefore a is also an H 1 (Lα )-atom. If R0 > lj /2, then Ij∗∗
=
4 [
Ij−2+k
with |Ij−2+k | = ρLα (βj−2+k ) ,
k=0
and using again ρLα (βj+2 ) ≤ 2ρLα (βj ) we see that kaχIj−2+k k∞ ≤ (2R0 )−1 ≤ ρLα (βj )−1 ≤ 2|Ij−2+k |−1 . Hence, each piece aχIj−2+k /4 is an H 1 (Lα )-atom for the ball B(βj−2+k , ρLα (βj−2+k )) and, consequently, kakHat1 (Lα ) ≤ 20.
4
Proof of (11) and Lemma 2.1
During the proof we set r = e−t ∈ (0, 1). We can rewrite (9) as √ 1 1 + r2 1−r r H 2 exp − |x − y| exp − xy , (44) Wt (x, y) = p 2 1 − r2 1+r π(1 − r2 ) for all x, y ∈ R. A simple computation using (44) or (9) gives √ 1 1 + r2 r H 2 (45) Wt (x, y) ≤ p exp − |x − y| . 4 1 − r2 π(1 − r2 ) Let us note that, for every N > 0, there exists a constant CN such that e−t/3 t −N (46) WtH (x, y) ≤ CN p 1+ . ρH (x)2 (1 − r2 ) Indeed, if |x − y| > |x|/2, then 1 1 + r2 e−t/2 e−t/3 t −N 2 p (47) WtH (x, y) ≤ p exp − x ≤ C 1 + . N 8 1 − r2 ρH (x)2 π(1 − r2 ) (1 − r2 ) If |x − y| ≤ |x|/2, then xy ∼ x2 and, using (44), we get (48)
e−t/2 e−t/3 t −N WtH (x, y) ≤ C √ 1+ . exp − c(1 − r)x2 ≤ CN √ ρH (x)2 1 − r2 1 − r2 13
Applying (45) and (46) combined with the fact that WtH (x, y) = WtH (y, x), we obtain e−t/3 t −N t −N |x − y|2 (49) WtH (x, y) ≤ CN √ 1 + 1 + exp − . 12(1 − e−2t ) ρ(x)2 ρ(y)2 1 − e−2t We are now in a position to prove (11). If |x − y| ≤ CρH (y), then by (10) and (49) we have (50) |R2H (x, y)|
Z
|x−y|2
N dt t + ≤ CN |x − y|2 t 0 Cρ (y) H ≤ CN |x| + |x| ln . |x − y| |x|
Z
C 2 ρH (y)2
|x−y|2
dt |x| + t
∞
ρ (y)2 N dt H |x| t t C 2 ρH (y)2
Z
If |x − y| ≥ CρH (y), then we use again (49) and get (51) |R2H (x, y)|
N dt Z |x−y|2 N t −2N dt t t ≤ CN + |x| |x − y|2 t |x − y|2 ρH (y)2 t 0 C 2 ρH (y)2 Z ∞ ρ (y)2 N dt H + |x| t t |x−y|2 Z
C 2 ρH (y)2
|x|
|x|ρH (y)2N |x − y|2N |x − y|ρ (y)2N |y|ρH (y)2N H + . ≤ CN |x − y|2N |x − y|2N
≤ CN
Now the first inequality in (11) is a consequence of (50) and (51). Similarly to (50) and (51), we also conclude that C |x| + |x| ln CρH (x)/|x − y| for |x − y| ≤ CρH (x) (52) |R2H (x, y)| ≤ C |x|ρ (x)N /|x − y|N for |x − y| > Cρ (x), N
H
H
from which we easily obtain the second inequality in (11). Having (11) already established, we now turn to prove Lemma 2.1. By (44), (53) √ 1 1 + r2 1−r ∂ H r 1 + r2 2 W (x, y) = − p (x − y) exp − |x − y| exp − xy ∂x t 2 1 − r2 1+r π(1 − r2 ) 1 − r2 √ 1 1 + r2 1−r 1−r r 2 p −y exp − |x − y| exp − xy . 1 + r π(1 − r2 ) 2 1 − r2 1+r From (53) we deduce that, for |x − y| > CρH (y), we have (54) ∂ 1 e−t/3 |x − y|2 t −N H √ (x, y) − 1 + . W ≤ C + |y|(1 − r) exp N ∂x t |x − y| 12(1 − r2 ) ρH (y)2 1 − r2 14
Proceeding as in (51), we obtain Z ∞ ∂ 1 ρ (y)2N dt H H (55) Wt (x, y) √ ≤ CN + |y| for |x − y| > CρH (y), ∂x |x − y| |x − y|2N t 0 which leads to Z (56)
|R1H (x, y)| dx ≤ C.
sup y∈R
|x−y|>CρH (y)
Our next step is to estimate R1H (x, y) for |x − y| ≤ CρH (y). Note that (53) implies (57) ∂ e−t/3 1 + r2 |x − y|2 t −N −1 |x − y| exp − 1+ Wt (x, y) ≤ CN √ ∂x 12(1 − r2 ) ρH (y)2 1 − r2 1 − r2 t −N −1 |x − y|2 e−t/3 1 + + CN √ |y|(1 − r) exp − 12(1 − r2 ) ρH (y)2 1 − r2 t −N e−t/4 1 + . ≤ CN 1 − r2 ρH (y)2 Consequently, using (57) we get Z ∞ dt ∂ (58) Wt (x, y) √ ≤ CρH (y)−1 . t ρH (y)2 ∂x In order to investigate the integral Z ρH (y)2 0
∂ dt Wt (x, y) √ , ∂x t
we study first the difference ρH (y)2
Z Q(x, y) = 0
dt ∂ H Wt (x, y) − Pt (x − y) √ , ∂x t
where Pt (x) = (2πt)−1/2 exp(−x2 /2t) is the classical Gauss-Weierstrass kernel. The perturbation formula asserts that Z 2 Z t Z ∞ 1 ρH (y) ∂ dt Q(x, y) = − Pt−s (x − z)z 2 WsH (z, y)dz ds √ . 2 0 t 0 −∞ ∂x Therefore, (59) Z |Q(x, y)|dx
J= |x−y| CρH (x), |x − y|N
and consequently, Z (61)
|R1H (x, y)| dy < ∞.
sup x∈R
|x−y|>CρH (x)
A similar procedure to that employed to estimate J gives Z (62) sup |Q(x, y)|dy ≤ C. x∈R
|x−y|≤CρH (x)
Finally, our analysis of the kernel R1H (x, y) is reduced to the integral Z (63)
0
ρH (y)2
ρH (y)2
x−y 1 dt √ exp(−|x − y|2 /2t) √ t t 2πt 0 2 |x − y| 2 exp − . = −√ 2ρH (y)2 2π(x − y)
∂ dt Pt (x − y) √ = − ∂x t
Z
Taking into account (10), (55), (58), (60), (61), (62) and (63), we get √ |x − y|2 2 H (64) R1 (x, y) = − exp − + h1 (x, y) π(x − y) 2ρH (y)2 with Z (65)
∞
∞
|h1 (x, y)|dx + sup
sup y∈R
Z
−∞
x∈R
|h1 (x, y)|dy < ∞. −∞
To complete the proof, take any ψ ∈ Cc∞ (R) as in the statement of Lemma 2.1. Define a function h2 (x, y) by √ √ x−y |x − y|2 2 2 ψ − exp − , x, y ∈ R. h2 (x, y) = π(x − y) ρH (x) π(x − y) 2ρH (y)2 By (10), (64), (65) and (11), the lemma will be established once we show that, for some C > 0 we have Z Z (66) sup |h2 (x, y)| dy ≤ C and sup |h2 (x, y)| dx ≤ C. x∈R
y∈R
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Set A = {(x, y) ∈ R2 ; |x − y| > ρH (x)}, B = {(x, y) ∈ R2 ; |x − y| ≤ ρH (x)}. Then (67)
|x − y|2 1 C |x − y| |h2 (x, y)| ≤ exp − χA (x, y) + C + χB (x, y), |x − y| 2ρH (y)2 ρH (x) ρH (y)2
where the last summand is obtained by applying the mean value theorem. Using (8), we see that ρ(y)2 ≤ cρ(x)|x − y| when (x, y) ∈ A, and therefore Z Z |x − y|2 |x − y| C C exp − χA (x, y) dy ≤ exp − c χA (x, y) dy |x − y| 2ρH (y)2 |x − y| ρH (x) Z (68) du ≤ exp(−c|u|) ≤ C. |u| |u|>1 On the other hand, ρH (x) ∼ ρH (y) when (x, y) ∈ B (again by (8)), so we have Z C C|x − y| + χB (x, y) dy ≤ C, ρH (x) ρH (y)2 which together with (68) implies the first inequality in (66). From (8) we also see that e = {(x, y) ∈ R2 ; |x − y| > ερH (y)} and B ⊂ B e = {(x, y) ∈ R2 ; |x − y| ≤ ρH (y)/ε} A⊂A for some ε > 0. Using this fact, the second inequality in (66) follows by similar arguments. This completes the proof of Lemma 2.1.
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[9] S. Thangavelu, Riesz transforms and wave equation for the Hermite operator, Comm. Partial Differential Equations 15(8) (1990), 1199–1215. [10] S. Thangavelu, Lectures on Hermite and Laguerre expansions, Mathematical Notes, 42, Princeton University Press, Princeton, NJ, 1993. [11] J. Zhong, Harmonic analysis for some Schr¨ odinger operators, Ph. D. Thesis, Princeton Univ., 1993. J. Betancor ´ lisis Matema ´ tico Departamento de Ana Universidad de la Laguna Campus de Anchieta, Avda. Astrof´ısico ´ nchez, s/n Francisco Sa 38271 La Laguna (Sta. Cruz de Tenerife) Spain E-mail address: [email protected] ´ ski J. Dziuban Instytut Matematyczny Uniwersytet Wroclawski 50-384 Wroclaw, PL. Grunwaldzki 2/4 Poland E-mail address: [email protected] ´s G. Garrigo ´ ticas Departamento de Matema Facultad de Ciencias ´ noma de Madrid Universidad Auto 28049 Madrid Spain E-mail address: [email protected]
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