Symmetry, Integrability and Geometry: Methods and Applications
SIGMA 5 (2009), 037, 17 pages
Hilbert Transforms Associated
with Dunkl–Hermite Polynomials?
Néjib BEN SALEM and Taha SAMAALI
Department of Mathematics, Faculty of Sciences of Tunis,
Campus Universitaire, 2092 Tunis, Tunisia
E-mail: Nejib.BenSalem@fst.rnu.tn
Received October 14, 2008, in final form March 12, 2009; Published online March 25, 2009
doi:10.3842/SIGMA.2009.037
Abstract. We consider expansions of functions in Lp (R, |x|2k dx), 1 ≤ p < +∞ with respect
to Dunkl–Hermite functions in the rank-one setting. We actually define the heat-diffusion
and Poisson integrals in the one-dimensional Dunkl setting and study their properties. Next,
we define and deal with Hilbert transforms and conjugate Poisson integrals in the same
setting. The formers occur to be Calderón–Zygmund operators and hence their mapping
properties follow from general results.
Key words: Dunkl operator; Dunkl–Hermite functions; Hilbert transforms; conjugate Poisson integrals; Calderón–Zygmund operators
2000 Mathematics Subject Classification: 42A50; 42C10
1
Introduction
The study of Dunkl operators has known a considerable growth during the last two decades due
to their relevance in mathematical physics and since they give the way to built a parallel to
the theory of spherical Harmonics based on finite root systems and depending on a set of real
parameters. In this spirit, the Hilbert transform, a basic tool in signal processing and in Fourier
and harmonic analysis as well, may be defined by means of partial derivatives, so that, since the
commutative algebra of Dunkl operators generalize the one of partial derivatives, it is natural to
extend the study of Hilbert transforms and connected topics as heat diffusion, Poisson integrals
and others to the Dunkl setting. In this work, we start with investigating the rank-one case, that
is why we sketch some facts that are subsequently needed. Let k be a nonnegative parameter
and let Tk be the Dunkl operator acting on smooth functions f as
f (x) − f (−x)
,
f ∈ C 1 (R).
x
To that operator is associated the so-called Dunkl–Hermite operator on R denoted Lk and
defined by
Tk f (x) = f 0 (x) + k
Lk = Tk2 − x2 .
Its spectral decomposition is given by the Dunkl–Hermite functions hkn defined by
hkn (x) = e−
x2
2
Hnk (x),
where Hnk are the generalized Hermite polynomials which we call the Dunkl–Hermite polynomials
as in [3], namely (see [8])
Lk hkn (x) = −(2n + 2k + 1)hkn (x).
?
This paper is a contribution to the Special Issue on Dunkl Operators and Related Topics. The full collection
is available at http://www.emis.de/journals/SIGMA/Dunkl operators.html
2
N. Ben Salem and T. Samaali
Recall also that Hnk were introduced in [7] and studied by Rösler in [8], whence
s
n!
k− 12
k
H2n
(x) = (−1)n
x2 ,
1 Ln
Γ(n + k + 2 )
s
n!
k+ 12
k
x2 ,
H2n+1
(x) = (−1)n
3 xLn
Γ(n + k + 2 )
where Lαn are the Laguerre polynomials of index α ≥ − 21 , given by
1 −α x dn
n+α −x
x e
x
e
.
n!
dxn
Lαn (x) =
2
It is well known that the system {Hnk }n≥0 is complete and orthonormal in L2 (R, e−x |x|2k dx),
therefore the system {hkn }n≥0 is complete and orthonormal in L2 (R, |x|2k dx).
Hereafter, Lp (R, |x|2k dx), 1 ≤ p < +∞ is the space of measurable functions on R satisfying
Z
p
kf kk,p :=
2k
|f (x)| |x| dx
1
p
< +∞,
R
and f belongs to Lp (R, |x|2k dx), 1 ≤ p < +∞, unless mentioned. For a given f , one defines the
heat-diffusion integral Gk (f ) by
Gk (f )(t, x) =
+∞
X
e−t(2n+2k+1) akn (f )hkn (x),
t > 0,
x ∈ R,
n=0
where
akn (f )
Z
=
f (t)hkn (t)|t|2k dt.
R
We shall establish that Gk (f ) possesses the following integral representation
Z
Gk (f )(t, x) =
Pk (t, x, y)f (y)|y|2k dy,
R
where
Pk (t, x, y) =
+∞
X
e−t(2n+2k+1) hkn (x)hkn (y).
n=0
We shall prove that Gk (f )(t, ·), t > 0, satisfies
1
kGk (f )(t, ·)kk,p ≤ (cosh(2t))−(k+ 2 ) kf kk,p .
Next, we define the Poisson integral Fk (f ) by
Fk (f )(t, x) =
+∞
X
√
e−t
2n+2k+1 k
an (f )hkn (x),
t > 0,
x ∈ R.
n=0
We shall establish that Fk (f ) possesses the following integral representation
Z
Fk (f )(t, x) =
f (y)Ak (t, x, y)|y|2k dy,
R
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
3
where
Z
t
Ak (t, x, y) = √
4π
+∞
3
t2
Pk (u, x, y)u− 2 e− 4u du
0
is the Poisson kernel associated with {hkn }n≥0 . We shall show that Fk (f )(t, ·) ∈ Lp (R, |x|2k dx)
and
√
1
kFk (f )(t, ·)kk,p ≤ 2k+ 2 e−t
2k+1
kf kk,p .
Also, we define the Hilbert transforms associated with the Dunkl–Hermite operators formally by
1
Hk± = (Tk ± x)(−Lk )− 2 .
We write f ∼
+∞
P
n=0
akn (f )hkn , to say that the last series represents the expansions of f with respect
to {hkn }n≥0 . Note, that if f ∼
Hk+ f ∼
+∞
X
akn (f ) √
n=0
where
+∞
P
n=0
akn (f )hkn , then again formally,
θ(n, k)
hkn−1 ,
2n + 2k + 1
Hk− f ∼ −
+∞
X
θ(n + 1, k) k
hn+1 ,
akn (f ) √
2n + 2k + 1
n=0
√
θ(n, k) =
2n
if n is even,
√
θ(n, k) = 2n + 4k if n is odd;
here and also later on, we use the convention that hkn−1 = 0 if n = 0.
We shall see that
Z
±
f (y)Rk± (x, y)|y|2k dy,
Hk f (x) = lim
−→0 |x−y|>
exist for almost every x, where Rk± (x, y) are appropriate kernels. Next, we shall prove that the
operators Hk± are bounded on Lp (R, |x|2k dx).
Finally, we use the Dunkl–Hermite functions to define the conjugate Poisson integrals fk± (t, x)
by
fk+ (t, x)
=
fk− (t, x) =
+∞
X
n=0
+∞
X
n=0
√
e−t
√
e−t
2n+2k+1 k
an (f ) √
θ(n, k)
hkn−1 (x),
2n + 2k + 1
θ(n
2n+2k+1 k
an (f ) √
+ 1, k) k
hn+1 (x).
2n + 2k + 1
We shall establish that fk± (t, x) possesses the integral representations
Z
Z
fk+ (t, x) =
Qk (t, x, y)f (y)|y|2k dy,
fk− (t, x) =
Mk (t, x, y)f (y)|y|2k dy,
R
R
where Qk (t, x, ·) and Mk (t, x, ·) are kernels expressed in terms of the Dunkl kernel Ek (x, y) which
is the eigenfunction of the Dunkl operator Tk .
We point out that recently (see [6]) A. Nowak and K. Stempak have studied Riesz transforms
for the Dunkl harmonic oscillator in the rank-one case.
We conclude this introduction by giving the organization of this paper. In the second section,
we define the heat-diffusion integral Gk (f ) and the Poisson integral Fk (f ) and give the integral
representations of Gk (f ) and Fk (f ). In the third section, we deal with the Hilbert transforms Hk±
and prove that these operators are of the strong type (p, p), 1 < p < +∞. The remaining part
is concerned with the study of the conjugate Poisson.
4
2
2.1
N. Ben Salem and T. Samaali
Heat-dif fusion and Poisson integrals
Heat-dif fusion
With the help of the Dunkl–Hermite functions introduced in the previous section, we define and
study the heat-diffusion in the Dunkl setting. As the Dunkl–Hermite polynomials are expressed
in terms of Laguerre polynomials, using Lemma 1.5.4 in [12], we have the following limiting
behavior of khkn kk,p , with respect to n.
Proposition 1. For 1 ≤ p ≤ 4, we have
(
− 1 + 1 +k( p1 − 21 )
, if k(p − 2) < 1,
n 4 2p
k
kh2n kk,p ∼
1
+k( 12 − p1 )
− 41 − 2p
, if k(p − 2) > 1,
n
(
1
− 41 + 2p
+k( p1 − 21 )
n
, if k(p − 2) < 1,
khk2n+1 kk,p ∼
1
+k( 12 − p1 )
− 41 − 2p
, if k(p − 2) > 1.
n
For p > 4, we have

1
1
1
1

 n− 12 − 6p +k( p − 2 ) , if
khk2n kk,p ∼
1
1
1
1

 n− 4 − 2p +k( 2 − p ) , if

1
1
1
1

 n− 12 − 6p +k( p − 2 ) ,
khk2n+1 kk,p ∼
1
1
1
1

 n− 4 − 2p +k( 2 − p ) ,
1 p
+ ,
3 6
1 p
k(p − 2) > + ,
3 6
1 p
if k(p − 2) ≤ + ,
3 6
1 p
if k(p − 2) > + .
3 6
Proposition 2. There exists a positive constant C such that
k
k(p − 2) ≤
1
khkn k∞ ≤ Cn 2 − 12 .
Proof . Using the fact that
q
n!Γ(n + k + 21 )
k
H2n =
Vk (H2n )
(2n)!Γ(k + 12 )
(see [7]), where {H2n }n≥0 is the set of classical Hermite polynomials, and Vk is the intertwining
d
operator between Tk and the usual derivative dx
given by
Z
k−1
2−2k Γ(2k + 1) 1
Vk (f )(x) =
f (xt) 1 − t2
(1 + t)dt,
Γ(k)Γ(k + 1) −1
we deduce that
khk2n k∞
q
2n n!Γ(n + k + 12 )
kh2n k∞ ,
≤C p
(2n)!Γ(k + 12 )
where {h2n }n≥0 is the classical Hermite-functions. In view of the following estimate
1
khn k∞ ≤ Cn− 12
given in [4, Lemma 2.1], using Stirling’s formula, we can deduce easily the result.
In the same way, we have
q
n!Γ(n + k + 32 )
k
H2n+1 =
Vk (H2n+1 ),
(2n + 1)!Γ(k + 12 )
where {H2n+1 }n≥0 is the classical Hermite polynomials, we obtain the result as above.
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
5
Let
akn (f )
Z
=
f (t)hkn (t)|t|2k dt.
R
0
Using the fact that hkn ∈ Lp (R, |x|2k dx), where p0 is the conjugate exponent of p, and Hölder’s
inequality, we deduce that
|akn (f )| ≤ kf kk,p khkn kk,p0 .
In view of Propositions 1 and 2, we have the following.
Proposition 3. The series
+∞
X
e−t(2n+2k+1) akn (f )hkn (x),
t > 0,
n=0
converges uniformly in x ∈ R.
Definition 1. We define the heat-diffusion integral of f by
Gk (f )(t, x) =
+∞
X
e−t(2n+2k+1) akn (f )hkn (x),
t > 0,
x ∈ R.
n=0
Proposition 4. The heat-diffusion integral Gk (f ) possesses the following integral representation
Z
Pk (t, x, y)f (y)|y|2k dy,
Gk (f )(t, x) =
R
where
Pk (t, x, y) =
+∞
X
e−t(2n+2k+1) hkn (x)hkn (y).
n=0
Proof . We obtain an integral form of Gk (f ) by writing
Gk (f )(t, x) =
=
+∞
X
e−t(2n+2k+1) hkn (x)
n=0
Z X
+∞
Z
f (y)hkn (y)|y|2k dy
R
e−t(2n+2k+1) hkn (x)hkn (y)f (y)|y|2k dy
R n=0
Z
=
Pk (t, x, y)f (y)|y|2k dy.
R
Interchanging the order of summation and integration is justified by Lebesgue’s dominated
convergence theorem since
+∞
X
−t(2n+2k+1)
Z
e
|hkn (x)hkn (y)f (y)||y|2k dy
R
n=0
≤
+∞
X
n=0
e−t(2n+2k+1) khkn k∞ khkn kk,p0 kf kk,p < +∞.
6
N. Ben Salem and T. Samaali
Mehler’s formula established by Margit Rösler for Dunkl–Hermite polynomials (see [8]),
adapted to Dunkl–Hermite functions {hkn }n≥0 reads
+∞
X
rn hkn (y)hkn (z) =
n=0
2
ck
− 12 ( 1+r 2 )(y 2 +z 2 )
1 e
1−r
(1 − r2 )k+ 2
Ek
2ry
,
z
,
1 − r2
0 < r < 1,
(1)
where ck is the constant defined by
Z
−1
−x2
2k
ck =
e |x| dx
R
and Ek is the Dunkl kernel expressed in terms of the normalized Bessel function
Ek (x, y) = jk− 1 (ixy) +
2
xy
j 1 (ixy),
2k + 1 k+ 2
where
jα (z) = Γ(α + 1)
+∞
X
n=0
(−1)n ( z2 )2n
,
n!Γ(n + α + 1)
1
α≥− .
2
Set
+∞
X
Uk (r, y, z) :=
rn hkn (y)hkn (z),
0 < r < 1.
n=0
Proposition 5. The kernel Uk satisfies the following properties
Uk (r, y, z) ≥ 0,
(i)
(ii)
Uk (r, y, z) = Uk (r, z, y),
k+ 1
2
2
− 21
e
(iii) kUk (r, y, ·)kk,1 =
1 + r2
(2)
1−r 2
1+r 2
y2
.
(3)
Proof . (i) and (ii) are obvious, let us therefore prove (iii).
Z
2
ck
2ry
− 12 1+r 2 (y 2 +z 2 )
1−r
kUk (r, y, ·)kk,1 =
e
Ek
, z |z|2k dz
2
k+ 12 R
2
1
−
r
(1 − r )
Z
2
2
ck
2ry
− 21 1+r 2 z 2
− 12 1+r 2 y 2
1−r
1−r
=
e
e
E
,
z
|z|2k dz.
k
2
k+ 12
2
1
−
r
R
(1 − r )
q
2
Performing the change of variables u = 1+r
z, and using the following identity (see [2])
1−r 2
Z
Ek (x, y)e−
x2
2
1
|x|2k dx = 2k+ 2 c−1
k e
y2
2
R
we are done.
Proposition 6. The heat-diffusion integral Gk (f ) is a C ∞ function on R+ × R satisfying the
differential-difference equation
∂
Lk,x −
Gk (f )(t, x) = 0,
(4)
∂t
(here Lk,x means that the operator Lk acts on the variable x).
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
7
Proof . On the one hand, one has for all m ∈ N
+∞
X
∂m
(−1)m (2n + 2k + 1)m e−t(2n+2k+1) akn (f )hkn (x).
m Gk (f )(t, x) =
∂t
(5)
n=0
On the other hand, it is easy to see that
(hkn )0 (x) = e−
x2
2
∂ k
H (x) − xhkn (x),
∂x n
thus for fixed t, the series (5) can be differentiated termwisely with respect to x. A similar
argument holds for higher derivatives and then Gk (f ) is C ∞ on R+ × R. Differentiating term
by term shows that Gk (f )(t, x) satisfies (4).
Theorem 1. The heat-diffusion integral Gk (f )(t, ·), t > 0, satisfies
1
kGk (f )(t, ·)kk,p ≤ (cosh(2t))−(k+ 2 ) kf kk,p .
Proof . Using
Pk (t, x, y) = e−t(2k+1) Uk e−2t , x, y
and (3), we obtain
Z
1
1
2
Pk (t, x, y)|y|2k dy = (cosh(2t))−(k+ 2 ) e− 2 tanh(2t)x .
R
By Hölder’s inequality, it follows that
−
|Gk (f )(t, x)|p ≤ (cosh(2t))
p(k+ 1
2)
p0
Z
|f (y)|p |Pk (t, x, y)||y|2k dy,
R
where
yield
p0
is the conjugate exponent of p. Integration with respect to x and using Fubini’s Theorem
kGk (f )(t, ·)kk,p ≤ (cosh(2t))−
2.2
k+ 12
kf kk,p .
Poisson integral
In this subsection, we introduce the Poisson integral and we give its Lp boundedness.
Definition 2. The Poisson integral Fk (f ) of f is defined by
Fk (f )(t, x) =
+∞
X
√
e−t
2n+2k+1 k
an (f )hkn (x),
t > 0,
x ∈ R.
n=0
Again the defining series is convergent by Propositions 1 and 2.
Proposition 7. Fk (f ) possesses the following integral representation
Z
Fk (f )(t, x) =
f (y)Ak (t, x, y)|y|2k dy,
R
where
t
Ak (t, x, y) = √
4π
Z
+∞
3
t2
Pk (u, x, y)u− 2 e− 4u du.
0
Ak is called the Poisson kernel associated with {hkn }n≥0 .
(6)
8
N. Ben Salem and T. Samaali
Proof . By using the subordination formula
Z +∞
β2
3
β
−β
e =√
e−s s− 2 e− 4s ds,
β > 0,
4π 0
(7)
we obtain an integral form of Fk (f )(t, x) by writing
Z +∞
Z
+∞ √
X
t2 (2n+2k+1)
3
t 2n + 2k + 1 k
4s
√
hn (x)
e−s s− 2 e−
ds f (y)hkn (y)|y|2k dy
Fk (f )(t, x) =
4π
0
R
n=0
Z
Z
+∞
+∞
3
t2
t X k
=√
hn (x)
u− 2 e−u(2n+2k+1) e− 4u du f (y)hkn (y)|y|2k dy
4π n=0
0
R
Z +∞ X
Z
+∞
3
t2
t
f (y)
e−(2n+2k+1)u hkn (x)hkn (y)u− 2 e− 4u du|y|2k dy
=√
4π R
0
n=0
Z
Z +∞
Z
t2
t
− 23 − 4u
2k
=√
f (y)
Pk (u, x, y)u e
du|y| dy =
f (y)Ak (t, x, y)|y|2k dy.
4π R
0
R
The same argument used for the heat-diffusion integral implies that Fk (f ) is C ∞ on R+ × R
and satisfies
∂2
Lk,x + 2 Fk (f )(t, x) = 0.
∂t
Theorem 2. Fk (f )(t, ·) ∈ Lp (R, |x|2k dx) and
√
1
kFk (f )(t, ·)kk,p ≤ 2k+ 2 e−t
2k+1
kf kk,p .
Proof . One has
Z Z +∞
3
t2
t
Ak (t, x, y)f (y)|y|2k dy = √
Pk (u, x, y)u− 2 e− 4u duf (y)|y|2k dy
4π R 0
R
Z +∞ Z
3
t2
t
2k
=√
Pk (u, x, y)f (y)|y| dy u− 2 e− 4u du
4π 0
R
Z +∞
3
t2
t
=√
Gk (f )(u, x)u− 2 e− 4u du.
4π 0
Z
Fk (f )(t, x) =
It follows that
t
kFk (f )(t, ·)kk,p ≤ √
4π
3
+∞
Z
3
t2
1
√
kGk (f )(u, ·)kk,p u− 2 e− 4u du ≤ 2k+ 2 e−t
2k+1
kf kk,p .
0
Hilbert transforms
The operator (−Lk ) is positive and symmetric in L2 (R, |x|2k dx) on the domain Cc∞ (R). It may
be easily checked that the operator (−Lk ) given by
! +∞
+∞
X
X
k
k
(−Lk )
an (f )hn =
(2n + 2k + 1)akn (f )hkn
n=0
n=0
on the domain
(
Dom(−Lk ) =
f ∈ L2 (R, |x|2k dx) :
+∞
X
n=0
)
|(2n + 2k + 1)akn (f )|2 < +∞
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
9
is a self-adjoint extension of (−Lk ), has the spectrum {2n + 2k + 1} and admits the spectral
decomposition
(−Lk )f =
+∞
X
(2n + 2k + 1)akn (f )hkn ,
f ∈ Dom(−Lk ).
n=0
Following [10, p. 57] the Hilbert transforms associated with the Dunkl–Hermite operator are
formally given by
1
Hk ± = (Tk ± x)(−Lk )− 2 .
Note, that if f ∼
+∞
P
n=0
Hk+ f ∼
+∞
X
akn (f )hkn , then again formally,
akn (f ) √
n=0
θ(n, k)
hkn−1 ,
2n + 2k + 1
Hk− f ∼ −
+∞
X
θ(n + 1, k) k
hn+1 ,
akn (f ) √
2n
+
2k
+
1
n=0
where
θ(n, k) =
√
if n is even,
√2n
2n + 4k if n is odd.
We use the convention that hkn−1 = 0 if n = 0. It is clear that Hk± are defined on L2 (R, |x|2k dx) by
Hk+ f =
+∞
X
akn (f ) √
n=0
θ(n, k)
hkn−1 ,
2n + 2k + 1
Hk− f = −
+∞
X
θ(n + 1, k) k
akn (f ) √
hn+1 .
2n + 2k + 1
n=0
To proceed to a deeper analysis of these definitions, in particular to consider Hk± on a wider
class of functions, we define the kernels Rk± (x, y) by
Rk± (x, y)
1
= √ (Tk,x ± x)
π
+∞
Z
Pk (t, x, y)t
− 21
0
1
dt = √
π
Z
+∞
1
(Tk,x ± x)Pk (t, x, y)t− 2 dt. (8)
0
It will be shown in Proposition 8 that the second integral in (8) converges.
We have
Pk (t, x, y) =
+∞
X
e−t(2n+2k+1) hkn (x)hkn (y)
n=0
=
ck
1
1
2k+ 2 (sinh(2t))k+ 2
− 12 coth(2t)(x2 +y 2 )
e
Ek
x
,y .
sinh(2t)
1+s
The change of variables 2t = log( 1−s
) furnishes a useful variant of (8):
Rk± (x, y)
√ Z 1
− 1
2
2
1+s
ds
=√
(Tk,x ± x)Ks (x, y) log
,
1−s
1 − s2
π 0
where
Ks (x, y) = ck
1 − s2
4s
k+ 21
− 14 s+ 1s (x2 +y 2 )
e
Ek
1 − s2
2s
x, y .
10
N. Ben Salem and T. Samaali
We then write
Rk± (x, y)
√
2
= √ [Rk,1 (x, y) ± Rk,2 (x, y)],
π
where
1
Z
Tk,x Ks (x, y) log
Rk,1 (x, y) =
0
1+s
1−s
− 1
2
ds
,
1 − s2
and
1
Z
xKs (x, y) log
Rk,2 (x, y) =
0
1+s
1−s
− 1
2
ds
.
1 − s2
Proposition 8. There exists a positive constant C such that for (x, y) ∈ ∆c = {(x, y) ∈ R2 :
x 6= y}, the kernels Rk,1 (x, y) and Rk,2 (x, y) satisfy
C
,
|x − y|
C
.
|Rk,2 (x, y)| ≤
|x − y|
|Rk,1 (x, y)| ≤
(9)
(10)
Proof . We start with proving (10). We have
Z 1
1 − s2
− 41 s+ 1s (x2 +y 2 )
|Rk,2 (x, y)| ≤ C
β(s)|x|e
Ek
x, y ds,
2s
0
where we let
k− 12 − k+ 21
β(s) = (1 − s)
s
log
1+s
1−s
− 1
2
.
Using the following estimate (see [9])
1−s2
1 − s2
|xy|
2s
x, y ≤ e
Ek
,
2s
the same reasoning as in [11, pp. 460–461] in the classical case gives the result. In order to
estimate Rk,1 , write
1
1
Tk,x Ks (x, y) = − s(x + y) + (x − y) Ks (x, y)
2
s
to see that
Z
|Rk,1 (x, y)| ≤ C
0
1
1
1
β(s) s|x + y| + |x − y| e− 4
s
s+ 1s (x2 +y 2 )
and use the same above arguments used to get the bound for Rk,1 .
Ek
1 − s2
2s
x, y ds,
Proposition 9. There exists a positive constant C such that for (x, y) ∈ ∆c = {(x, y) ∈ R2 :
x 6= y}, if |x − y| ≥ 2|x − x0 |, then
C|x − x0 |
,
|x − y|2
C|x − x0 |
|Rk,2 (x, y) − Rk,2 (x0 , y)| ≤
.
|x − y|2
|Rk,1 (x, y) − Rk,1 (x0 , y)| ≤
(11)
(12)
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
11
Proof . Start with
− 1
2
1+s
ds
0
0
|Rk,2 (x, y) − Rk,2 (x , y)| ≤
xKs (x, y) − x Ks (x , y) log
1
−
s
1
−
s2
0
Z 1
1
1
2
2
1 − s2
β(s)xe− 4 s+ s (x +y ) Ek
≤C
x, y
2s
0
2
1
1
02
2
1
−
s
x0 , y ds.
− x0 e− 4 s+ s (x +y ) Ek
2s
0
Z
1
Using the following estimates (see [9])
∂
1 − s2
1 − s2
1 − s2
x, y
|y|Ek
x, y ,
∂x Ek
≤
2s
2s
2s
1−s2
1 − s2
|xy|
2s
Ek
x, y ≤ e
,
2s
then the same reasoning as in [11, pp. 461–463] in the classical case gives the result.
Considering Rk,1 , we have
− 1
2
1+s
ds
0
Tk,x Ks (x, y) − Tk,x Ks (x , y) log
|Rk,1 (x, y) − Rk,1 (x , y)| ≤
1
−
s
1
−
s2
0
Z 1
1
1
2
2
1
1 − s2
x, y
≤C
β(s) s(x + y) + (x − y) e− 4 s+ s (x +y ) Ek
s
2s
0
2
1
1
02
2
1
1
−
s
x0 , y ds.
− s(x0 + y) + (x0 − y) e− 4 s+ s (x +y ) Ek
s
2s
0
Z
1
The proof of (11) follows the same steps of the one of (12).
Proposition 10. There exists a positive constant C such that for (x, y) ∈ ∆c = {(x, y) ∈ R2 :
x 6= y}, if |x − y| ≥ 2|y − y 0 | then
C|y − y 0 |
,
|x − y|2
C|y − y 0 |
|Rk,1 (x, y) − Rk,1 (x, y 0 )| ≤
.
|x − y|2
|Rk,2 (x, y) − Rk,2 (x, y 0 )| ≤
Proof . The proofs of (13) and (14) are quite similar to the ones of (11) and (12).
(13)
(14)
Lemma 1. Given m, m = 1, 2, . . . , and f ∈ Cc∞ (R) there exists C = Cm,f > 0 such that
|hf, hkn i| ≤ C(2n + 2k + 1)−m ,
where
Z
hf, gi =
f (x)g(x)|x|2k dx.
R
Proof .
|hf, hkn i|
Z
k
2k
k −m
= f (t)hn (t)|t| dt = (−(2n + 2k + 1))−m hLm
. k f, hn i ≤ C(2n + 2k + 1)
R
12
N. Ben Salem and T. Samaali
Theorem 3. Let f, g ∈ Cc∞ (R) with disjoint supports. Then
Z Z
±
Rk± (x, y)f (y)g(x)|y|2k dy|x|2k dx.
hHk f, gi =
R
(15)
R
Proof . We first consider Hk+ . Let
+∞
X
f=
akn (f )hkn
n=0
and
g=
+∞
X
bkn (g)hkn .
n=0
Then
Hk+ f =
+∞
X
n=0
√
θ(n, k)
akn (f )hkn−1 .
2n + 2k + 1
The convergence of the three series is in L2 (R, |x|2k dx)) and by Parseval’s identity
hHk+ f, gi =
+∞
X
n=0
√
θ(n, k)
akn (f )bkn−1 (g).
2n + 2k + 1
(16)
We will show that the right sides in (15) and (16) coincide. Note that we can see as in Proposition 8 that
Z +∞
1
C
|(Tk,x + x)Pk (t, x, y)|t− 2 dt ≤
.
|x − y|
0
This result and the assumption made on the supports of f and g show that
Z Z Z +∞
1
|(Tk,x + x)Pk (t, x, y)|t− 2 dt|g(x)f (y)||y|2k dy|x|2k dx < +∞.
R
R
(17)
0
Now,
Z Z
R
=
=
=
=
=
Rk+ (x, y)f (y)g(x)|y|2k dy|x|2k dx
Z Z Z +∞
1
1
√
{(Tk,x + x)Pk (t, x, y)}t− 2 dtf (y)g(x)|y|2k dy|x|2k dx
π R R 0
!)
Z +∞Z Z (
+∞
X
1
1
√
(Tk,x + x)
e−t(2n+2k+1) hkn (x)hkn (y) g(x)f (y)|x|2k dx|y|2k dyt− 2 dt
π 0
R R
n=0
)
Z +∞ Z Z (X
+∞
1
1
−t(2n+2k+1)
k
k
√
e
θ(n, k)hn−1 (x)hn (y) g(x)f (y)|x|2k dx|y|2k dyt− 2 dt
π 0
R R n=0
)
Z +∞ Z (X
+∞
1
1
√
e−t(2n+2k+1) θ(n, k)bkn−1 (g)hkn (y) f (y)|y|2k dyt− 2 dt
π 0
R n=0
)
Z +∞ (X
+∞
1
1
√
e−t(2n+2k+1) θ(n, k)akn (f )bkn−1 (g) t− 2 dt = hHk+ f, gi.
π 0
n=0
R
Note that Fubini’s theorem is justified by (17). Recall that
Hk− f
+∞
X
θ(n + 1, k) k
√
=−
an (f )hkn+1 ,
2n
+
2k
+
1
n=0
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
13
then one gets similarly
Z Z
Rk− (x, y)f (y)g(x)|y|2k dy|x|2k dx
R R
Z Z Z +∞
1
1
=√
{(Tk,x − x)Pk (t, x, y)}t− 2 dtf (y)g(x)|y|2k dy|x|2k dx
π R R 0
+∞
X
θ(n + 1, k) k
√
=−
an (f )bkn+1 (g) = hHk− f, gi.
2n
+
2k
+
1
n=0
Theorem 4. For almost every x in R, the Hilbert transforms are given by
Z
±
f (y)Rk± (x, y)|y|2k dy.
Hk f (x) = lim
−→0 |x−y|>
Proof . We have
Z
Z
±
2k
f (y)Rk (x, y)|y| dy =
2k
p
f (y)|y| |y|
|x−y|>
|x−y|>
2k
p0
Rk± (x, y)dy
Z
=
g(y)Wk± (x, y)dy,
|x−y|>
where p0 is the conjugate exponent of p,
2k
g ∈ Lp (R, dx),
g(y) = f (y)|y| p ,
2k
Wk± (x, y) = |y| p0 Rk± (x, y),
Wk± (x, y) are Calderón–Zygmund kernels (see Propositions 8, 9 and 10). It follows that
Z
Z
±
2k
g(y)Wk± (x, y)dy
lim
f (y)Rk (x, y)|y| dy = lim
−→0 |x−y|>
−→0 |x−y|>
exist for almost every x (see [5, p. 55]).
Remark 1. For f ∈ L2 (R, |x|2k dx), we have
Fk (Hk± f )(x) = ±iHk± (Fk f )(x),
where Fk is the Plancherel Dunkl transform, (see [1]).
Theorem 5. The operators Hk± are bounded on Lp (R, |x|2k dx), 1 < p < +∞.
Proof . Consider the truncated operators
Z
±
H,k f (x) =
f (y)Rk± (x, y)|y|2k dy,
|x−y|>
Z
Z
±
kHk± f kpk,p =
|Hk± f (x)|p |x|2k dx =
| lim H,k
f (x)|p |x|2k dx
−→0
R Z
ZR
±
±
p
2k
=
lim |H,k f (x)| |x| dx =
lim inf |H,k
f (x)|p |x|2k dx
−→0
−→0
R
R
Z
Z
2k ±
±
p
2k
|x| p H f (x)p dx.
≤ lim inf
|H,k f (x)| |x| dx = lim inf
,k
−→0
−→0
R
R
Yet,
2k
±
|x| p H,k
f (x) =
Z
|x−y|>
2k
f (y)|x| p Rk± (x, y)|y|2k dy =
Z
|x−y|>
g(y)Zk± (x, y)dy,
14
N. Ben Salem and T. Samaali
where
2k
g(y) = f (y)|y| p , g ∈ Lp (R, dx),
2k
2k
Zk± (x, y) = |x| p |y| p0 Rk± (x, y).
Zk± (x, y) are Calderón–Zygmund kernels (see Propositions 8, 9 and 10). Let
Z
Sg± (x) =
g(y)Zk± (x, y)dy.
R
The operators Sg± are Calderón–Zygmund type associated with the Calderón–Zygmund kernels
Zk± (x, y) (see [5, p. 48]), then
Z
sup g(y)Zk± (x, y)dy >0 |x−y|>
are bounded on Lp (R, dx) for 1 < p < +∞ (see [5, p. 56]). Consequently, there exists a positive
constant C = Cp such that if f ∈ Lp (R, |x|2k dx) then
kHk± f kk,p ≤ Ckf kk,p .
Lemma 2. There exists a positive constant C such that for f ∈ L1 (R, |x|2k dx), λ > 0, we have
Z
C
dx ≤ kf kk,1 .
±
λ
{x∈R:sup>0 |H,k
f (x)|>λ}
Proof . We have
±
f (x)
H,k
Z
=
|x−y|>
f (y)Rk± (x, y)|y|2k dy
Z
=
|x−y|>
g(y)Wk± (x, y)dy,
where
2k
g(y) = f (y)|y| p
2k
Wk± (x, y) = |y| p0 Rk± (x, y).
and
Let
Sg± (x)
Z
=
g(y)Wk± (x, y)dy.
R
The operators Sg± are Calderón–Zygmund operators associated with the Calderón–Zygmund kernels Wk± (x, y) then there exists a positive constant C such that for λ > 0 and f ∈ L1 (R, |x|2k dx),
we have
Z
C
dx ≤ kf kk,1 .
±
λ
{x∈R:sup>0 |H,k
f (x)|>λ}
As a by-product, we have the following.
Theorem 6. There exists a positive constant C such that for f ∈ L1 (R, |x|2k dx), we have
!
Z
sup λ
λ>0
{x∈R:|Hk± f (x)|>λ}
dx
≤ Ckf kk,1 .
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
4
15
Conjugate Poisson integrals
Dunkl–Hermite functions allow to define the conjugate Poisson integrals.
Definition 3. The conjugate Poisson integrals fk± (t, x) of f are defined by
fk+ (t, x)
=
fk− (t, x) =
+∞
X
√
e−t
n=0
+∞
X
√
e−t
2n+2k+1 k
an (f ) √
θ(n, k)
hkn−1 (x),
2n + 2k + 1
θ(n
2n+2k+1 k
an (f ) √
+ 1, k) k
hn+1 (x).
2n + 2k + 1
n=0
Remark 2. The same arguments used for the heat-diffusion integral show that fk± (t, x) ∈
C ∞ (R+ × R) and satisfy the differential-difference equations
∂2
(18)
(i)
Lk,x + 2 fk± (t, x) = ±2fk± (t, x),
∂t
∂
(ii) (Tk,x ± x)Fk (f )(t, x) = ∓ fk± (t, x),
(19)
∂t
where Fk (f )(t, x) is the Poisson integral of f .
We now use√(19) to find an integral formula for fk± (t, x). Using the subordination formula (7),
taking β = t 2n + 2k + 1, making the change of variables s −→ (2n + 2k + 1)u, and then
substituting r = e−2u leads to the formula
√
e−t
2n+2k+1
1
Z
1
L(t, r)rn+k+ 2 dr,
=
0
where
t2
te 2 log r
L(t, r) =
1
3
.
(2π) 2 r(− log r) 2
Then if Ak (t, x, y) denotes the Poisson kernel (6), we have
Ak (t, x, y) =
+∞
X
√
e−t 2n+2k+1 hkn (x)hkn (y)
=
+∞
1X
hkn (x)hkn (y)
1
rn hkn (x)hkn (y)L(t, r)rk+ 2 dr =
0 n=0
Z
1
1
L(t, r)rn+k+ 2 dr
0
n=0
n=0
Z
=
+∞
X
Z
1
1
L(t, r)Uk (r, x, y)rk+ 2 dr.
0
Combining this and (1) we obtain
(Tk,x + x)Ak (t, x, y)
2 2
2 2
√
t2
Z 1
− r x +r2 y
2 log r e
1−r
1
1
2
(y
−
rx)te
2rx
2
2
= √ ck e− 2 (x +y )
Ek (
, y)rk+ 2 dr.
3
3
2
k+
1−r
π
0
(− log r) 2 (1 − r2 ) 2
(20)
Now
Z
(Tk,x + x)Fk (f )(t, x) =
R
(Tk,x + x)Ak (t, x, y)f (y)|y|2k dy.
(21)
16
N. Ben Salem and T. Samaali
From Propositions 1 and 2, it is easy to check that fk+ (t, x) −→ 0 as t −→ +∞ and so
Z +∞
∂ +
+
fk (t, x) = −
f (u, x)du.
∂t k
t
Using (21), (20) and (19) we find after integration
Z
+
Qk (t, x, y)f (y)|y|2k dy,
fk (t, x) =
R
where
1
2 +y 2 )
Qk (t, x, y) = e− 2 (x
Q1,k (t, x, y)
and
1
Z
Q1,k (t, x, y) =
0
2 +r 2 y 2
(y − rx) − r2 x1−r
2
e
Ek
(1 − r2 )k+2
2rx
, y W1,k (t, r)dr
1 − r2
with
√
1
t2
1
2
1 − r2 2 2 log
r r k+ 2 .
W1,k (t, r) = √ ck
e
− log r
π
We now use (19) to find an integral formula for fk− (t, x)
(Tk,x − x)Ak (t, x, y)
2 2
2 2
√
t2
Z 1
− r x +r2 y
2 log r e
1−r
1
1
2
(ry
−
x)te
2rx
2
2
− 2 (x +y )
= √ ck e
, y rk− 2 dr.
Ek
3
3
2
1−r
π
0
(− log r) 2 (1 − r2 )k+ 2
(22)
Now
Z
(Tk,x − x)Fk (f )(t, x) =
(Tk,x − x)Ak (t, x, y)f (y)|y|2k dy.
R
The same reasoning as above gives fk− (t, x) −→ 0 as t −→ +∞ and so
Z +∞
∂ −
−
fk (t, x) = −
f (u, x)du.
∂t k
t
Using (23), (22) and (19) we find after integration
Z
−
fk (t, x) =
Mk (t, x, y)f (y)|y|2k dy,
R
where
1
Mk (t, x, y) = e− 2 (x
2 +y 2 )
M1,k (t, x, y)
and
Z
M1,k (t, x, y) =
0
1
−r
2 x2 +r 2 y 2
1−r 2
(x − ry)e
(1 − r2 )k+2
Ek
with
√
1
t2
1
2
1 − r2 2 2 log
r r k− 2 .
Yk (t, r) = √ ck
e
− log r
π
2rx
, y Yk (t, r)dr,
1 − r2
(23)
Hilbert Transforms Associated with Dunkl–Hermite Polynomials
17
Theorem 7. There exists a positive constant C such that for 1 < p < +∞, f ∈ Lp (R, |x|2k dx),
we have
√
kfk± (t, ·)kk,p ≤ Ce−t
2k+1
kf kk,p .
Proof . We have
√
kfk± (t, ·)kk,p = k ± Hk± Fk (f )(t, ·)kk,p ≤ CkFk (f )(t, ·)kk,p ≤ Ce−t
2k+1
kf kk,p .
Theorem 8. There exists a positive constant C such that for f ∈ L1 (R, |x|2k dx), we have
!
Z
√
sup λ
λ>0
{x∈R:|fk± (t,x)|>λ}
Proof . We have
Z
sup λ
λ>0
{x∈R:|fk± (t,x)|>λ}
≤ Ce−t
dx
2k+1
!
dx
kf kk,1 .
Z
= sup λ
!
dx
{x∈R:|±Hk± Fk (f )(t,x)|>λ}
√
CkFk (f )(t, ·)kk,1 ≤ Ce−t 2k+1 kf kk,1 .
λ>0
≤
Acknowledgments
The authors thank the anonymous referees for their careful reading of the manuscript and their
valuable suggestions to improve the style of this paper.
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