IJMMS 2004:8, 407–419
PII. S0161171204212388
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
R. S. PATHAK and S. PATHAK
Received 16 December 2002
The pseudodifferential operator (p.d.o.) A(x, D), associated with the Bessel operator d2 /dx 2
m
+ (1 − 4µ 2 )/4x 2 , is defined. Symbol class Hρ,δ
is introduced. It is shown that the p.d.o.
associated with a symbol belonging to this class is a continuous linear mapping of the
Zemanian space Hµ into itself. An integral representation of p.d.o. is obtained. Using Hankel
p
convolution Lσ ,α -norm continuity of the p.d.o. is proved.
2000 Mathematics Subject Classification: 46F12, 47A05, 47A30, 47G30.
1. Introduction. The theory of the Hankel transformation
hµ f (y) =
∞
0
(xy)1/2 Jµ (xy)f (x)dx,
1
µ≥− ,
2
(1.1)
where 0 < y < ∞ and Jµ is the Bessel function of the first kind and of order µ, has
been extended by Zemanian [14] to distributions belonging to Hµ , the dual of the test
function space Hµ , consisting of all complex-valued smooth functions φ defined on
I = (0, ∞) and satisfying
d k −µ−1/2
µ
φ(x) < ∞
x
γm,k (φ) = sup x m x −1
dx
0<x<∞
(1.2)
for each pair of nonnegative integers m and k. Pseudodifferential operators associated
with a numerical valued symbol a(x, y) were investigated by R. S. Pathak and Pandey
[8, 9] and by R. S. Pathak and S. Pathak [10].
We will use the notation and terminology of [4, 7, 11, 12, 14]. The differential operators Nµ , Mµ , and Sµ are defined by
d
x −µ−1/2 ,
Nµ = Nµ,x = x µ+3/2 x −1
dx
d
x µ+1/2 ,
Mµ = Mµ,x = x −µ+1/2 x −1
dx
d
x −µ−1/2
Sµ = Sµ,x = Mµ Nµ = 2µ + 2 x µ+1/2 x −1
dx
d 2 −µ−1/2
+ x µ+5/2 x −1
x
.
dx
(1.3)
(1.4)
(1.5)
408
R. S. PATHAK AND S. PATHAK
From [14, page 139] and [5, page 948] we know the following relations for any φ ∈ Hµ :
hµ+1 Nµ φ = −yhµ φ,
hµ Sµ φ = −y 2 hµ φ,
k
k
d k −µ−1/2
d v
d k−v −µ−1/2 θφ =
θ x −1
φ ,
x
x
x −1
x −1
v
dx
dx
dx
v=0
r
d r +j −µ−1/2
r
φ(x) =
bj x 2j+µ+1/2 x −1
φ(x) ,
x
Sµ,x
dx
j=0
(1.6)
(1.7)
(1.8)
(1.9)
where the bj are constants depending only on µ.
p
Next, we define the space Lσ (I), 1 ≤ p < ∞, as the space of those real-valued measurable functions on I for which
f Lpσ =
1/p
f (x)p dσ (x)
< ∞,
∞
0
(1.10)
where
−1 2µ+1
dσ (x) = 2µ Γ (µ + 1)
x
dx.
(1.11)
Let (x, y, z) denote the area of a triangle with sides x, y, and z, if such a triangle
exists. For fixed µ ≥ −1/2, set
2
2µ−1
1 −1
D(x, y, z) = 23µ−1/2 (π )−1/2 Γ (µ + 1)
(xyz)−2µ (x, y, z)
,
Γ µ+
2
(1.12)
if exists, and zero otherwise. Then D(x, y, z) ≥ 0 and that D(x, y, z) is symmetric
in x, y, z. From [13, page 411], we know that
j(xt)j(yt) =
∞
j(zt)D(x, y, z)dσ (z),
(1.13)
0
where
j(x) = 2µ Γ (µ + 1)x −µ Jµ (x).
(1.14)
From (1.13) it follows that
−1
Jµ (xt)Jµ (yt) = 2µ Γ (µ + 1)
∞
0
(xyt)µ z−µ Jµ (zt)D(x, y, z)dσ (z).
(1.15)
Let f ∈ L1σ (I), then its associated function f (x, y) is defined by
f (x, y) =
∞
f (z)D(x, y, z)dσ (z),
0
0 < x, y < ∞,
(1.16)
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
409
and the Hankel convolution of f and g is defined by
f #g(x) =
∞
f (x, y)g(y)dσ (y),
0
0 < x < ∞.
q
(1.17)
p
From [9, page 102] we know that if f (x) ∈ Lσ (I), g(x) ∈ Lσ (I) with p > 1, q > 1, and
1/p + 1/q > 1, then
f #gr ≤ f q gp ,
(1.18)
1
1 1
= + − 1 > 0.
r
p q
(1.19)
where
Many algebraic and topological properties of the generalized Hankel convolution
have been given by Betancor and his associates [1, 2, 3, 6].
We also note the differentiation formula
x −1
d
dx
m
x −µ Jµ (xt) = (−t)m x −µ−m Jµ+m (xt).
(1.20)
m
In this paper, a general class Hρ,δ
of symbols associated with the differentiation formula
m
is introduced. For ρ = 1/2 and δ = 0, the
(1.20) and similar to the Hörmander class Sρ,δ
m
m
class Hρ,δ reduces to the symbol class H0 studied by R. S. Pathak and Pandey [8]. An
m
.
integral representation for A(x, D) is given when the symbol belongs to the class Hρ,δ
It is shown that A(x, D) is a continuous linear mapping of Zemanian’s space Hµ into
p
p
itself. The space Lσ ,α is defined and an Lσ ,α -boundedness result is also obtained.
2. The pseudodifferential operator A(x, D)
Definition 2.1. Let a(x, ξ) be a complex-valued smooth function belonging to the
space C ∞ (I × I), where I = (0, ∞), and let its derivatives satisfy certain growth conditions such as (2.3). The pseudodifferential operator (p.d.o.) A(x, D), associated with the
symbol a(x, ξ), is defined by
A(x, D)f (x) =
∞
0
(xξ)1/2 Jµ (xξ)a(x, ξ) hµ f (ξ)dξ,
(2.1)
where
hµ f (ξ) =
∞
0
(xξ)1/2 Jµ (xξ)f (x)dx,
1
µ≥− .
2
(2.2)
Definition 2.2. The function a(x, ξ) : C ∞ (I × I) → C belongs to the symbol class
m ∈ R, 0 ≤ δ ≤ ρ ≤ 1, if and only if for all α, β ∈ N0 , there exists Cα,β,m > 0 such
that
α β
m/2−ρα−δβ
−1 d
−1 d
ξ
x
a(x, ξ) ≤ Cα,β,m 1 + ξ 2
.
(2.3)
dξ
dx
m
,
Hρ,δ
410
R. S. PATHAK AND S. PATHAK
m
A few examples of elements of Hρ,δ
are given as
2
2 m/2
, m < 0,
(i) (1 + x + ξ )
(ii) sin x 2 (1 + x 2 + ξ 2 )m/2 , m < 0,
2
(iii) e−x (1 + x 2 + ξ 2 )m/2 , m < 0.
We give a proof for example (i). We have
m/2
a(x, ξ) = 1 + x 2 + ξ 2
,
x −1
d
dx
β
m < 0,
(2.4)
m/2
m/2−β
= Dm,β 1 + x 2 + ξ 2
,
1 + x2 + ξ2
so that
α m/2 d β
−1 d
ξ
x −1
1 + x2 + ξ2
dξ
dx
m/2−β−α m/2−ρα−δβ
≤ Dm,α,β 1 + ξ 2
= Dm,α,β 1 + x 2 + ξ 2
(m < 0)
(2.5)
for 0 ≤ δ ≤ 1 and 0 ≤ ρ ≤ 1.
m
. Then for µ ≥ −1/2, the p.d.o. A(x, D)
Theorem 2.3. Let the symbol a(x, ξ) ∈ Hρ,δ
is a continuous linear mapping of Hµ into Hµ .
Proof. Let Φ(x) = A(x, D)f (x), f ∈ Hµ . Then using definitions (1.3) and (2.1), we
have
d
x −µ−1/2 Φ(x).
Nµ Φ(x) = x µ+3/2 x −1
dx
(2.6)
By induction, we get
Nµ+k−1 · · · Nµ Φ(x)
d k −µ−1/2
x
Φ(x)
= x µ+k+1/2 x −1
dx
∞
k
d
x −µ−1/2
(xξ)1/2 Jµ (xξ)a(x, ξ) hµ f (ξ)dξ.
= x µ+k+1/2 x −1
dx
0
(2.7)
Using formula (1.8), we get
Nµ+k−1 · · · Nµ Φ(x)
=
∞
k
d k−r −µ
x −1
x Jµ (xξ)
x µ+k+1/2
r
dx
0
r =0
k
d r
a(x, ξ)ξ 1/2 hµ f (ξ)dξ.
× x −1
dx
(2.8)
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
411
Next, applying (1.20), we have
Nµ+k−1 · · · Nµ Φ(x)
∞
k
k
x µ+k+1/2
=
(−ξ)k−r x −µ−k+r Jµ+k−r (xξ)
r
0
r =0
d r
a(x, ξ)ξ 1/2 hµ f (ξ)dξ
× x −1
dx
k
∞
k
x r (xξ)1/2 Jµ+k−r (xξ)
=
r
0
r =0
d r
× x −1
a(x, ξ)(−ξ)k−r hµ f (ξ)dξ.
dx
(2.9)
Using formula (1.6), we get
(−x) Nµ+k−1 · · · Nµ Φ(x)
∞
(xξ)1/2 Jµ+k+1 (xξ)Nµ+k a(x, ξ)(−ξ)k hµ f (ξ) dξ
= A0
0
+ A1
∞
0
x 3/2 ξ 1/2 Jµ+k (xξ)Nµ+k−1
×
+ A2
+ Ak
∞
0
d
a(x, ξ)(−ξ)k−1 hµ f (ξ) dξ
x −1
dx
x 5/2 ξ 1/2 Jµ+k−1 (xξ)Nµ+k−2
d
dx
×
x −1
x
k+1/2 1/2
∞
ξ
0
2
a(x, ξ)(−ξ)k−2 hµ f (ξ) dξ + · · ·
Jµ+1 (xξ)Nµ
x
−1
d
dx
k
a(x, ξ) hµ f (ξ) dξ,
Nµ+k−1 · · · Nµ Φ(x)
∞
(xξ)1/2 Jµ+k+2 (xξ)Nµ+k+1 Nµ+k a(x, ξ)(−ξ)k hµ f (ξ) dξ
= A0
− x2
0
+ A1
∞
0
x 3/2 ξ 1/2 Jµ+k+1 (xξ)Nµ+k Nµ+k−1
×
+ A2
∞
0
x 5/2 ξ 1/2 Jµ+k (xξ)Nµ+k−1 Nµ+k−2
×
+ Ak
∞
x
0
d
a(x, ξ)(−ξ)k−1 hµ f (ξ) dξ
x −1
dx
x
−1
d
dx
k+1/2 1/2
ξ
2
k−2
a(x, ξ)(−ξ)
Jµ+2 (xξ)Nµ+1 Nµ
hµ f (ξ) dξ + · · ·
x
−1
d
dx
k
a(x, ξ) hµ f (ξ) dξ.
(2.10)
412
R. S. PATHAK AND S. PATHAK
By induction, we have
(−x)n Nµ+k−1 · · · Nµ Φ(x)
= A0
∞
0
(xξ)1/2 Jµ+k+n (xξ)
× Nµ+k+n−1 · · · Nµ+k a(x, ξ)(−ξ)k hµ f (ξ) dξ
+ A1
∞
0
x 3/2 ξ 1/2 Jµ+k+n−1 (xξ)Nµ+k+n−2 · · · Nµ+k−1
×
+ Ak
∞
0
x −1
d
dx
a(x, ξ)(−ξ)k−1 hµ f (ξ) dξ + · · ·
(2.11)
x k+1/2 ξ 1/2 Jµ+n (xξ)Nµ+n−1 · · · Nµ
×
x −1
d
dx
k
a(x, ξ) hµ f (ξ) dξ.
Therefore,
(−x)n Nµ+k−1 · · · Nµ Φ(x)
∞
k
=
x r +1/2 ξ 1/2 Jµ+k−r +n (xξ)Nµ+k−r +n−1 · · · Nµ+k−r
r
0
r =0
k
×
x −1
d
dx
r
(2.12)
a(x, ξ)(−ξ)k−r hµ f (ξ) dξ,
which in view of formula (2.7) can be written as
d k −µ−1/2
x
Φ(x)
(−x)n x µ+k+1/2 x −1
dx
∞
k
k
x r +1/2 ξ 1/2 Jµ+k−r +n (xξ)ξ µ+k−r +n+1/2
=
(−1)k−r
r
0
r =0
n r
−1 d
−µ−1/2
−1 d
× ξ
a(x, ξ) hµ f (ξ) dξ
x
ξ
dξ
dx
(2.13)
∞
k
k
k−r
−µ−k+r
µ+k+1/2
=
(xξ)
Jµ+k−r +n (xξ)x
(−1)
r
0
r =0
d n
× ξ 2(µ+k−r )+n+1 ξ −1
dξ
r
−µ−1/2
−1 d
× ξ
a(x, ξ) hµ f (ξ) dξ.
x
dx
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
413
Putting µ + k − r = λ, we get
d k −µ−1/2
x
Φ(x)
(−x)n x −1
dx
∞
k
k
d n
(xξ)−λ Jλ+n (xξ)ξ 2λ+n+1 ξ −1
=
(−1)k−r
r
dξ
0
r =0
r
d
× ξ −µ−1/2 x −1
a(x, ξ) hµ f (ξ) dξ.
dx
(2.14)
Now an application of formula (1.8) yields
k
n −1 d
x x
x −µ−1/2 Φ(x)
dx
≤
∞
k
(xξ)−λ Jλ+n (xξ)ξ 2λ+n+1
r
0
r =0
v n
n d r
−1 d
ξ
×
a(x, ξ)
x −1
v
dξ
dx
v=0
d n−v −µ−1/2 hµ f (ξ)dξ.
ξ
× ξ −1
dξ
k
(2.15)
Then using inequality (2.3) and the boundedness property of the Bessel function, we
find that the right-hand side of (2.15) is bounded by
∞
m/2−ρv−δr
k
n
ξ 2λ+n+1 1 + ξ 2
Aλ,n Cv,r ,m
r
v
0
r =0 v=0
d n−v −µ−1/2 × ξ −1
ξ
hµ f (ξ)dξ
dξ
∞
k n
λ+n/2+1/2+m/2−ρv−δr
k
n
Aλ,n Cv,r ,m
1 + ξ2
≤
r
v
0
r =0 v=0
n−v
−1 d
−µ−1/2
× ξ
ξ
hµ f (ξ)dξ.
dξ
k n
(2.16)
Let p be a positive integer greater than or equal to µ +k+m/2+n/2+1/2, then we can
write
k
n −1 d
−µ−1/2
x x
x
Φ(x)
dx
k
n
Aλ,n Cv,r ,m
≤
r
v
r =0 v=0
k n
×
p+1 ∞ −1 d n−v −µ−1/2 1 + ξ2
dξ
ξ
f
(ξ)
h
ξ
µ
2
dξ
1
+
ξ
0
414
R. S. PATHAK AND S. PATHAK
≤
k n p+1
k
n
p +1
Aλ,n Cv,r ,m
r
v
s
r =0 v=0 s=0
∞
dξ
d n−v −µ−1/2 .
×
ξ 2s ξ −1
ξ
hµ f (ξ) 1 + ξ2
dξ
0
(2.17)
Thus
µ
γn,k (Φ) ≤
n p+1
k µ
Dλ,n,k,m γ2s,n−v hµ f ,
(2.18)
r =0 v=0 s=0
where Dλ,n,k,m is a positive constant. From (2.18), the continuity of A(x, D) follows.
3. An integral representation. The function ax (ξ), associated with the symbol a(x, ξ)
and defined by
ax (ξ) =
∞
0
(ηξ)1/2 Jµ (ηξ) (xη)1/2 Jµ (xη)a(x, η) dη,
(3.1)
will play a fundamental role in our investigation. An estimate for ax (ξ) is given by the
following lemma.
m
. Then the function ax (ξ) defined by (3.1)
Lemma 3.1. Let the symbol a(x, η) ∈ Hρ,δ
satisfies the inequality
ax (ξ) ≤ Eµ,m,t 1 + ξ 2 −t (1 + x)µ+4t+1/2 ,
(3.2)
where Eµ,m,t is a positive constant, µ ≥ −1/2, and m < −µ − 3/2 − t.
Proof. Using property (1.7), we can write
t
1 − Sµ,η 1/2
(ηξ)1/2 Jµ (ηξ) t (xη) Jµ (xη)a(x, η) dη
2
0
1+ξ
∞
t
r
Sµ,η
t
1/2
r
1/2
=
(ηξ) Jµ (ηξ)
(−1)
t (xη) Jµ (xη)a(x, η) dη.
2
r
0
1+ξ
r =0
ax (ξ) =
∞
In view of (1.9), we have
ax (ξ) =
t
(−1)r
r =0
×
∞
0
t
1
t
r
1 + ξ2
(ηξ)1/2 Jµ (ηξ)
d r +j
bj η2j+µ+1/2 η−1
dη
j=0
r
× η−µ−1/2 a(x, η)(xη)1/2 Jµ (xη) dη
(3.3)
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
=
r
t (−1)r
r =0 j=0
×
∞
0
415
t
1
bj t
r
1 + ξ2
(ηξ)1/2 Jµ (ηξ)η2j+µ+1/2
r +j r +j d i
d r +j−i
×
a(x, η) η−1
η−1
i
dη
dη
i=0
× η−µ Jµ (xη)x 1/2 dη.
(3.4)
Again applying formula (1.20), we have
t
r +j
1
(−1)
ax (ξ) =
bj t
r
i
1 + ξ2
r =0 j=0 i=0
+j
t r r
×
r
∞
0
d i
(ηξ)1/2 Jµ (ηξ)η2j+µ+1/2 η−1
dη
(3.5)
× a(x, η)(−1)r +j−i x r +j−i η−µ−r −j+i
× Jµ+r +j−i (xη)x 1/2 dη.
Therefore,
+j t r r
µ+1/2+2(r +j−i)
t
r +j ax (ξ) ≤
bj x t
r
i
1 + ξ2
r =0 j=0 i=0
∞
d i
(ηξ)1/2 Jµ (ηξ)η2j+µ+1/2 η−1
×
a(x, η)
dη
0
× (xη)−µ−r −j+i Jµ+r +j−i (xη)dη.
(3.6)
Then, using inequality (2.3) and the boundedness property of the Bessel function, we
have
+j t r r
µ+1/2+2(r +j−i)
r +j t
bj x ax (ξ) ≤
t
i
r
1 + ξ2
r =0 j=0 i=0
× Aµ Bµ,i,j,r Ci,m
∞
0
η2j+µ+1/2
−m/2+ρi dη
1 + η2
+j t r r
t
r +j bj Aµ Bµ,i,j,r Ci,m x µ+2r +2j−2i+1/2
≤
r
i
r =0 j=0 i=0
−t
µ 3 m µ 3
× 1 + ξ2
B j + + , − − − + ρi − j
2 4
2
2 4
(3.7)
416
R. S. PATHAK AND S. PATHAK
for m < −µ − 3/2 − t. Therefore there exists a constant Eµ,m,t such that
ax (ξ) ≤ Eµ,m,t 1 + ξ 2 −t (1 + x)µ+4t+1/2 .
(3.8)
m
Theorem 3.2. For any symbol a(x, ξ) ∈ Hρ,δ
, the associated operator A(x, D) can
be represented by
A(x, D)f (x) =
∞∞
0
0
(ηξ)1/2 Jµ (ηξ)ax (ξ) hµ f (η)dηdξ,
f ∈ Hµ (I),
(3.9)
where all the involved integrals are convergent for µ ≥ −1/2 and m < −µ − 3/2 − t with
t > 1/2.
∞
Proof. Since ax (ξ) = 0 (ηξ)1/2 Jµ (ηξ)[(xη)1/2 Jµ (xη)a(x, η)]dη, by inversion, formally, we have
∞
0
(ηξ)1/2 Jµ (ηξ)ax (ξ)dξ = (xη)1/2 Jµ (xη)a(x, η).
(3.10)
Therefore,
∞
(xη)1/2 Jµ (xη)a(x, η) hµ f (η)dη
0
∞
∞
1/2
=
hµ f (η)
(ηξ) Jµ (ηξ)ax (ξ)dξ dη
A(x, D)f (x) =
0
=
∞∞
0
0
(3.11)
0
(ηξ)1/2 Jµ (ηξ)ax (ξ) hµ f (η)dη dξ,
so that
A(x, D)f (x) ≤ Aµ
∞
0
ax (ξ)
∞
0
hµ f (η)dη dξ.
(3.12)
Since (hµ f )(η) ∈ Hµ (I), we have
hµ f (η) ≤ Cηµ+1/2 (1 + η)−l
∀l > 0.
(3.13)
Now using the above estimate and (3.2), we obtain
A(x, D)f (x)
∞∞
−t
Eµ,m,t 1 + ξ 2
(1 + x)µ+4t+1/2 Cηµ+1/2 (1 + η)−l dη dξ
≤ Aµ
0
0
∞
∞
−t
1
≤ Eµ,m,t
1 + ξ2
(1 + x)µ+4t+1/2
dξ
(1 + η)µ+1/2−l dη.
0
(3.14)
0
The above integrals are convergent since µ ≥ −1/2, and t can be chosen greater than
1/2 and l sufficiently large.
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
417
p
4. Lσ ,α -boundedness of A(x, D). In the proof of the boundedness result, we will
need the following estimate for the Hankel transform of ηµ+1/2 a(x, η). We write
Ax (z) = hµ ηµ+1/2 a(x, η) (z).
(4.1)
Lemma 4.1. For 1/2 < ρ < 1 and µ + m + 3/2 < 0, there exists a constant Aµ,m,ρ > 0
such that
Ax (z) ≤ Aµ,m,ρ 1 + z2 −1 .
(4.2)
Proof. We have
Ax (z) =
∞
0
(zη)1/2 Jµ (zη)ηµ+1/2 a(x, η)dη.
(4.3)
Then using the property (1.7), we can write
Ax (z) =
∞
0
1 − Sµ,η µ+1/2
η
(zη)1/2 Jµ (zη) a(x, η)dη.
1 + z2
(4.4)
Using (1.5), we get
1
Ax (z) = 1 + z2
∞
0
(zη)1/2 Jµ (zη)ηµ+1/2 a(x, η)dη
∞
d
a(x, η)dη
(zη)1/2 Jµ (zη)ηµ+1/2 η−1
dη
0
∞
2
1/2
µ+5/2
−1 d
−
η
(zη) Jµ (zη)η
a(x, η)dη ,
dη
0
− (2µ + 2)
(4.5)
so that
Ax (z) ≤ 1
1 + z2
∞
0
m/2
Aµ ηµ+1/2 1 + η2
dη
∞
m/2−ρ
Aµ ηµ+1/2 1 + η2
dη
0
∞
m/2−2ρ
Aµ ηµ+5/2 1 + η2
dη .
+
+ (2µ + 2)
(4.6)
0
Since all the three integrals are convergent for µ+m+3/2 < 0 and 1/2 < ρ < 1, therefore
there exists a constant Aµ,m,ρ such that
Ax (z) ≤ Aµ,m,ρ 1 + z2 −1 .
(4.7)
We will use the following subspace of Hµ .
p
Definition 4.2. The space Lσ ,α (0, ∞), α ∈ R, 1 ≤ p ≤ ∞, is defined as the set of all
those elements f ∈ Hµ (I) which satisfy
f Lpσ ,α = ξ −α f (ξ)Lpσ .
(4.8)
418
R. S. PATHAK AND S. PATHAK
Theorem 4.3. Pseudodifferential operator A(x, D) is a continuous linear mapping
p
from Lσ , µ+1/2 into Lrσ , µ+1/2 , where 1 ≤ p ≤ ∞, q > (2µ + 2)/(µ + 5/2), and 1/r = 1/p +
1/q − 1.
Proof. From (3.9), we have
A(x, D)f (x) =
=
∞
0
∞
ax (ξ)
(ηξ)1/2 Jµ (ηξ) hµ f (η)dη dξ
0
∞
0
(4.9)
ax (ξ)f (ξ)dξ.
Using (3.1), this can be written as
A(x, D)f (x) =
∞ ∞
0
0
(ηξ)1/2 Jµ (ηξ)(xη)1/2 Jµ (xη)a(x, η)dη f (ξ)dξ.
(4.10)
Next using relation (1.15), we can express it in the form
A(x, D)f (x)
=
∞
∞
0
0
−1
(ηξ)1/2 (xη)1/2 2µ Γ (µ + 1)
×
=
0
µ
(xξη) z
0
∞ ∞ ∞
0
∞
0
−µ
Jµ (zη)D(x, ξ, z)dσ (z)a(x, η)dη f (ξ)dξ
(4.11)
(zη)1/2 Jµ (zη)ηµ+1/2 a(x, η)dη
× x µ+1/2 ξ −µ−1/2 f (ξ)z−µ−1/2 D(x, ξ, z)dσ (ξ)dσ (z)
= x µ+1/2
∞∞
0
0
Ax (z)ξ −µ−1/2 f (ξ)z−µ−1/2 D(x, ξ, z)dσ (ξ)dσ (z).
An application of inequality (4.2) yields
−µ−1/2
x
A(x, D)f (x)
∞∞
(4.12)
−µ−1/2
−1
ξ
f (ξ)z−µ−1/2 1 + z2
D(x, ξ, z)dσ (ξ)dσ (z).
≤ Aµ,m,ρ
0
0
In view of definitions (1.16) and (1.17), the last expression can be expressed as a convolution, and for F (ξ) = |ξ −µ−1/2 f (ξ)| and G(z) = |z−µ−1/2 |(1 + z2 )−1 , we have
−µ−1/2
x
A(x, D)f (x) ≤ Aµ,m,ρ (F #G)(x),
(4.13)
so that
∞
0
1/r
1/r
∞
−µ−1/2
r
x
A(x, D)f (x) dσ (x)
≤ Aµ,m,ρ
(F #G)r (x)dσ (x)
.
0
(4.14)
419
THE PSEUDODIFFERENTIAL OPERATOR A(x, D)
q
p
We note that for q > (2µ +2)/(µ +5/2), G(z) ∈ Lσ (I), and F (z) ∈ Lσ (I), 1 ≤ p ≤ ∞, and
using (1.18) finally we obtain
−µ−1/2
x
A(x, D)f (x)
Lrσ
≤ F #GLrσ ≤ F Lpσ GLqσ ,
(4.15)
where 1/r = 1/p + 1/q − 1 > 0. Therefore, in view of Definition 4.2, we get
A(x, D)f (x)
Lrσ , µ+1/2
−1 ≤ f Lpσ , µ+1/2 1 + z2
q
Lσ , µ+1/2
.
(4.16)
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R. S. Pathak: Department of Mathematics, Banaras Hindu University, Varanasi 221005, India
E-mail address: rspathak@banaras.ernet.in
S. Pathak: Department of Mathematics, Banaras Hindu University, Varanasi 221005, India