Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 5 Issue 2 (2013), Pages 42-60.
ON THE q-BESSEL FOURIER TRANSFORM
(COMMUNICATED BY R.K RAINA)
LAZHAR DHAOUADI
Abstract. In this work, we are interested by the q-Bessel Fourier transform
with a new approach. Many important results of this q-integral transform are
proved with a new constructive demonstrations and we establish in particular
the associated q-Fourier-Neumen expansion which involves the q-little Jacobi
polynomials.
1. Introduction
In the recent mathematical literature one finds many articles which deal with the
theory of q-Fourier analysis associated with the q-Hankel transform. This theory
was elaborated first by Koornwinder and R.F. Swarttouw [12] and then by Fitouhi
and Al [5, 8].
It should be noticed that in [5] we provided the mains results of q-Fourier analysis
in particular that the q-Hankel transform is extended to the Lq,2,ν space like an
isometric operator. Often we use the crucial properties namely the positivity of the
q-Bessel translation operator to prove some results but these last property is not
ensured for any q in the interval ]0, 1[. Thus, we will prove some main results of qFourier analysis without the positivity argument especially the following statments:
- Inversion Formula in the Lq,p,ν spaces with p ≥ 1.
- Plancherel Formula in the Lq,p,ν ∩ Lq,1,ν spaces with p > 2.
- Plancherel Formula in the Lq,2,ν spaces.
Note that in the paper [7] we have proved that the positivity of the q-Bessel translation operator is ensured in all points of the interval ]0, 1[ when ν ≥ 0. In this article
we will try to show in a clear way the part in which the positivity of the q-Bessel
translation operator plays a role in q-Bessel Fourier analysis. In particular, when
we try to prove a q-version of the Young’s inequality for the associated convolution.
Many interesting result about the uncertainty principle for the q-Bessel transform
was proved in the last years. We cite for examples [2, 3, 4, 9]. There are some
differences of the results cited above and our result:
0
2000 Mathematics Subject Classification: Primary 33D15,47A05.
Keywords and phrases. q-Bessel Fourier transform, Heisenberg uncertainty inequality, Hardys
inequality, q-Fourier-Neumen expansion.
Submitted 29 Oct 2012. Published March 29, 2013.
42
ON THE q-BESSEL FOURIER TRANSFORM
43
In this paper the Heisenberg uncertainty inequality is established for functions in
Lq,2,ν space.
The Hardy’s inequality discuss here is a quantitative uncertainty principles which
give an information about how a function and its q-Bessel Fourier transform are
linked.
In the end of this paper we use the remarkable work in [1] to establish a new result
about the q-Fourier-Neumen expansion involving the q-little Jacobi polynomials.
2. The q-Bessel transform
The reader can see the references [10, 11, 16] about q-series theory. The references
[5, 8, 12] are devoted to the q-Bessel Fourier analysis. Throughout this paper, we
consider 0 < q < 1 and ν > −1. We denote by
n
R+
q = {q , n ∈ Z} .
The q-Bessel operator is defined as follows [5]
1 ∆q,ν f (x) = 2 f (q −1 x) − (1 + q 2ν )f (x) + q 2ν f (qx) .
x
The eigenfunction of ∆q,ν associated with the eigenvalue −λ2 is the function x 7→
jν (λx, q 2 ), where jν (., q 2 ) is the normalized q-Bessel function defined by [5, 8, 10,
14, 16]
∞
X
q n(n+1)
2
x2n .
jν (x, q ) =
(−1)n 2ν+2 2
(q
, q )n (q 2 , q 2 )n
n=0
The q-Jackson integral of a function f defined on R+ is
Z ∞
X
f (t)dq t = (1 − q)
q n f (q n ).
0
n∈Z
We denote by Lq,p,ν the space of functions f defined on R+
q such that
Z ∞
1/p
kf kq,p,ν =
|f (x)|p x2ν+1 dq x
exist.
0
We denote by Cq,0 the space of functions defined on R+
q tending to 0 as x → ∞ and
continuous at 0 equipped with the topology of uniform convergence. The space Cq,0
is complete with respect to the norm
kf kq,∞ = sup |f (x)|.
x∈R+
q
The normalized q-Bessel function jν (., q 2 ) satisfies the orthogonality relation
Z ∞
c2q,ν
jν (xt, q 2 )jν (yt, q 2 )t2ν+1 dq t = δq (x, y), ∀x, y ∈ R+
q
0
where
δq (x, y) =
and
cq,ν =
0 if x 6= y
1
(1−q)x2(ν+1)
if x = y
1 (q 2ν+2 , q 2 )∞
.
1 − q (q 2 , q 2 )∞
(1)
44
LAZHAR DHAOUADI
Let f be a function defined on R+
q then
Z ∞
f (y)δq (x, y)y 2ν+1 dq y = f (x).
0
The normalized q-Bessel function jν (., q 2 ) satisfies
1
if n ≥ 0
(−q 2 ; q 2 )∞ (−q 2ν+2 ; q 2 )∞
2
.
|jν (q n , q 2 )| ≤
q n −(2ν+1)n if n < 0
(q 2ν+2 ; q 2 )∞
The q-Bessel Fourier transform Fq,ν is defined by [5, 8, 12]
Z ∞
Fq,ν f (x) = cq,ν
f (t)jν (xt, q 2 )t2ν+1 dq t, ∀x ∈ R+
q .
0
Proposition 1. Let f ∈ Lq,1,ν then Fq,ν f ∈ Cq,0 and we have
kFq,ν (f )kq,∞ ≤ Bq,ν kf kq,1,v
where
Bq,ν =
1 (−q 2 ; q 2 )∞ (−q 2v+2 ; q 2 )∞
.
1−q
(q 2 ; q 2 )∞
Theorem 1. Let f be a function in the Lq,p,ν space where p ≥ 1 then
2
Fq,ν
f = f.
(2)
Proof. If f ∈ Lq,p,ν then Fq,ν f exist, and we have
Z ∞
2
Fq,ν
f (x) = cq,ν
Fq,ν f (t)jν (xt, q 2 )t2ν+1 dq t
0
Z ∞
Z ∞
=
f (y) c2q,ν
jν (xt, q 2 )jν (yt, q 2 )t2ν+1 dq t y 2ν+1 dq y
0
Z0 ∞
f (y)δq (x, y)y 2ν+1 dq y
=
0
= f (x).
The computations are justified by the Fubuni’s theorem: If p > 1 then we use the
Hölder’s inequality
Z ∞
Z ∞
|f (y)|
|jν (xt, q 2 )jν (yt, q 2 )|t2ν+1 dq t y 2ν+1 dq y
0
≤
0
Z
0
∞
p 2ν+1
|f (y)| y
dq y
1/p
×
Z
∞
p 2ν+1
σ(y) y
dq y
0
The numbers p and p above are conjugates and
Z ∞
σ(y) =
|jν (xt, q 2 )jν (yt, q 2 )|t2ν+1 dq t,
0
then
Z
∞
σ(y)p y 2ν+1 dq y
0
=
Z
0
1
p 2ν+1
σ(y) y
dq y +
Z
1
∞
σy p y 2ν+1 dq y.
1/p
.
ON THE q-BESSEL FOURIER TRANSFORM
45
Note that
1
Z
σ(y)p y 2ν+1 dq y
0
≤ kjν (., q
2
)kpq,∞
1
Z
Z
0
∞
0
2
|jν (xt, q )|t
2ν+1
≤ kjν (., q 2 )kpq,∞ kjν (., q 2 )kpq,1,ν x−2(ν+1)p
and
Z
∞
dq t
Z
1
0
p
y 2ν+1 dq y
y 2ν+1 dq y < ∞,
σ(y)p y 2ν+1 dq y
1
≤ kjν (., q 2 )kpq,∞ kjν (., q 2 )kpq,1,ν
≤ kjν (., q 2 )kpq,∞ kjν (., q 2 )kpq,1,ν
If p = 1 then
Z
∞
kf (y)k
0
Z
∞
∞
Z
Z1 ∞
1
y 2ν+1
dq y
y 2(ν+1)p
1
dq y < ∞.
2(ν+1)(p−1)+1
y
|jν (xt, q 2 )jν (yt, q 2 )|t2ν+1 dq t y 2ν+1 dq y
0
≤ kf kq,1,ν kjν (., q 2 )kq,∞ kjν (., q 2 )kq,1,ν ×
1
x2(ν+1)
.
Theorem 2. Let f be a function in the Lq,1,ν ∩ Lq,p,ν space, where p > 2 then
kFq,ν f kq,2,ν = kf kq,2,ν .
Proof. Let f ∈ Lq,1,ν ∩ Lq,p,ν then by Theorem 1 we see that
2
Fq,ν
f = f.
This implies
Z ∞
Z ∞
Z ∞
Fq,ν f (x)2 x2ν+1 dq x =
Fq,ν f (x) cq,ν
f (t)jν (xt, q 2 )t2ν+1 dq t x2ν+1 dq x
0
0
0
Z ∞
Z ∞
=
f (t) cq,ν
Fq,ν f (x)jν (xt, q 2 )x2ν+1 dq x t2ν+1 dq t
0
Z0 ∞
=
f (t)2 t2ν+1 dq t.
0
The computations are justified by the Fubuni’s theorem
Z ∞
Z ∞
|f (t)| cq,ν
|Fq,ν f (x)||jν (xt, q 2 )|x2ν+1 dq x t2ν+1 dq t
0
≤
where
0
Z
0
∞
p 2ν+1
|f (t)| t
φ(t) = cq,ν
Z
0
dq t
∞
1/p
×
Z
0
∞
p 2ν+1
|φ(t)| t
dq t
1/p
|Fq,ν f (x)||jν (xt, q 2 )|x2ν+1 dq x,
,
46
LAZHAR DHAOUADI
then
kFq,ν f (x)k ≤ cq,ν
This gives
Z
∞
0
≤ cq,ν
Z
≤ cq,ν
Z
|f (y)||jν (xy, q 2 )|y 2ν+1 dq y
∞
p 2ν+1
|f (y)| y
0
∞
0
dq y
|f (y)|p y 2ν+1 dq y
1/p
1/p
×
Z
×
Z
Z
2
p 2ν+1
|jν (xy, q )| y
0
∞
0
−2(ν+1)/p
≤ cq,ν kf kq,p,ν kjν (., q 2 )kq,p,ν x
φ(t) ≤ c2q,ν kf kq,p,ν kjν (., q 2 )kq,p,ν
∞
dq y
|jν (y, q 2 )|p y 2ν+1 dq y
.
1/p
1/p
x−2(ν+1)/p
∞
|jν (xt, q 2 )|x(2ν+1)−2(ν+1)/p dq x
0
Z
∞
2
2
|jν (x, q 2 )|x2(ν+1)/p−1 dq x t−2(ν+1)/p
≤ cq,ν kf kq,p,ν kjν (., q )kq,p,ν
0
and
≤ C1 t−2(ν+1)/p ,
Z ∞
φ(t) = cq,ν
|Fq,ν f (x)||jν (xt, q 2 )|x2ν+1 dq x
0
Z ∞
2
2ν+1
= cq,ν
|Fq,ν f (x/t)||jν (x, q )|x
dq x t−2(ν+1)
0
≤ cq,ν kFq,ν f kq,∞ × kjν (., q 2 )kq,1,ν × t−2(ν+1)
≤ C2 t−2(ν+1) .
Note that
−1 < −2(ν + 1) pp + 2v + 1
0 < −2(ν + 1)(p − 2)
⇔
⇔ 1 < p < 2 ⇔ p > 2.
−2(ν + 1)(p − 1) < 0
−2(ν + 1)p + 2v + 1 < −1
Hence
Z
Z ∞
|φ(t)|p t2ν+1 dq t =
1
0
0
≤ C1
which prove the result.
|φ(t)|p t2ν+1 dq t +
Z
1
t
Z
∞
1
−2(v+1)p/p 2ν+1
t
|φ(t)|p t2ν+1 dq t
dq t + C2
Z
∞
1
0
t−2(v+1)p t2ν+1 dq t < ∞,
Theorem 3. Let f be a function in the Lq,2,ν space then
kFq,ν f kq,2,ν = kf kq,2,ν .
Proof. We introduce the function ψx as follows
ψx (t) = cq,ν jν (tx, q 2 ).
The inner product h, i in the Hilbert space Lq,2,ν is defined by
Z ∞
f, g ∈ Lq,2,ν ⇒ hf, gi =
f (t)g(t)t2ν+1 dq t.
0
Using (1) we write
x 6= y ⇒ hψx , ψy i = 0
(3)
ON THE q-BESSEL FOURIER TRANSFORM
kψx k2q,2,ν =
We have
47
1
x−2(ν+1) .
1−q
Fq,ν f (x) = hf, ψx i,
and by Theorem 1
2
f ∈ Lq,2,ν ⇒ Fq,ν
f = f,
then
+
hf, ψx i = 0, ∀x ∈ R+
q ⇒Fq,ν f (x) = 0, ∀x ∈ Rq ⇒f = 0.
Hence, {ψx , x ∈ R+
q } form an orthogonal basis of the Hilbert space Lq,2,ν and we
have
{ψx , ∀x ∈ R+
q } = Lq,2,ν .
Now
X
1
hf, ψx iψx ,
f ∈ Lq,2,ν ⇒ f =
2
kψ
k
x q,2,ν
+
x∈Rq
and then
kf k2q,2,ν =
X
x∈R+
q
X
1
x2(ν+1) Fq,ν f (x)2 = kFq,ν f k2q,2,ν ,
hf, ψx i2 = (1 − q)
2
kψx kq,2,ν
+
x∈Rq
which achieve the proof.
Proposition 2. Let f ∈ Lq,p,ν where p ≥ 1 then Fq,ν f ∈ Lq,p,ν . If 1 ≤ p ≤ 2 then
2
−1
p
kf kq,p,ν .
kFq,ν f kq,p,ν ≤ Bq,ν
(4)
Proof. This is an immediate consequence of Proposition 1, Theorem 3, the RieszThorin theorem and the inversion formula (2).
The q-translation operator is given as follow
Z ∞
ν
Tq,x
f (y) = cq,ν
Fq,ν f (t)jν (yt, q 2 )jν (xt, q 2 )t2ν+1 dq t.
0
Let us now introduce
Qν = q ∈]0, 1[,
ν
Tq,x
is positive for all x ∈ R+
q
ν
ν
v
the set of the positivity of Tq,x
. We recall that Tq,x
is called positive if Tq,x
f ≥0
′
for f ≥ 0. In a recent paper [6] it was proved that if −1 < ν < ν then Qν ⊂ Qν ′ .
As a consequence :
-: If 0 ≤ ν then Qν =]0, 1[.
-: If − 21 ≤ ν < 0 then ]0, q0 ] ⊂ Q− 21 ⊂ Qν (]0, 1[, q0 ≃ 0.43.
-: If −1 < ν ≤ − 12 then Qν ⊂ Q− 12 .
ν
Theorem 4. Let f ∈ Lq,p,ν then Tq,x
f exists and we have
Z ∞
Z ∞
ν
Tq,x
f (y)y 2ν+1 dq y =
f (y)y 2ν+1 dq y.
0
0
and
ν
Tq,x
f (y)
=
Z
0
∞
f (z)Dν (x, y, z)z 2ν+1 dq z,
48
LAZHAR DHAOUADI
where
Dν (x, y, z) =
∞
Z
c2q,ν
jν (xs, q 2 jν (ys, q 2 jν (zs, q 2 )s2ν+1 dq s.
0
ν
If we suppose that Tq,x
is a positive operator then for all p ≥ 1 we have
ν
kTq,x
f kq,p,ν ≤ kf kq,p,ν .
(5)
ν
Proof. We write the operator Tq,x
in the following form
ν
Tq,x
f (y) =
0
=
So we have
Z ∞
ν
Tq,x
f (y)y 2ν+1 dq y
∞
Z
Fq,v f (z)jv (xz, q 2 )jv (yz, q 2 )z 2v+1 dq z
Fq,ν Fq,ν f (z)jν (xz, q 2 ) (y).
cq,ν
=
0
=
=
=
=
Z
∞
Fq,ν Fq,ν f (z)jν (xz, q 2 ) (y)y 2ν+1 dq y
0
Z ∞
1
cq,ν
Fq,ν Fq,ν f (z)jν (xz, q 2 ) (y)jν (0, q 2 )y 2ν+1 dq y
cq,ν
0
1 2 Fq,ν Fq,ν f (z)jν (xz, q 2 ) (0)
cq,ν
1
Fq,ν f (0)
cq,ν
Z ∞
f (y)y 2v+1 dq y.
0
On the other hand
Z ∞
ν
Tq,x f (y) = cq,ν
Fq,ν f (z)jν (xz, q 2 )jν (yz, q 2 )z 2ν+1 dq z
0
Z ∞
Z ∞
cq,ν
f (t)jν (tz, q 2 )t2ν+1 dq t jν (xz, q 2 )jν (yz, q 2 )z 2v+1 dq z
= cq,ν
0
0
Z ∞
Z ∞
2
=
cq,ν
jν (xz, q 2 )jν (yz, q 2 )jν (tz, q 2 )z 2ν+1 dq z f (t)t2ν+1 dq t
0
Z0 ∞
=
Dq,ν (x, y, t)f (t)t2ν+1 dq t.
0
The computations are justified by the Fubuni’s theorem
Z
0
∞
Z
∞
0
Z
2 2ν+1
|f (t)| jν (tz, q ) t
dq t jv (xz, q 2 ) jν (yz, q 2 ) z 2v+1 dq z
∞
Z
∞
jν (tz, q 2 )p t2ν+1 dq t
≤
kf kq,p,ν
≤
kf kq,p,ν jν (., q 2 )q,p,ν
0
0
Z
0
∞
p1
jν (xz, q 2 ) jv (yz, q 2 ) z 2ν+1 dq z
jν (xz, q 2 ) jν (yz, q 2 ) z 2(ν+1)(1− p1 )−1 dq z.
ON THE q-BESSEL FOURIER TRANSFORM
49
ν
Now suppose that Tq,x
is positive. Given a function f ∈ Cq,0 we obtains
Z ∞
ν
2ν+1
T f (y) = Dq,ν (x, y, t)f (t)t
dq t
q,x
Z 0∞
≤
|Dq,ν (x, y, t)| |f (t)| t2v+1 dq t
0
Z ∞
2v+1
≤
Dq,ν (x, y, t)t
dq t kf kq,∞ = kf kq,∞
0
which implies
ν Tq,x f ≤ kf kq,∞ .
q,∞
If the function f ∈ Lq,1,ν then we obtains
Z ∞
ν ν
T f T f (y) y 2ν+1 dq y
=
q,x
q,x
q,1,ν
Z0 ∞ Z ∞
2ν+1
|Dq,ν (x, y, t)| |f (t)| t
dq t y 2ν+1 dq y
≤
0
0
Z ∞ Z ∞
2ν+1
Dq,ν (x, y, t)y
dq y |f (t)| t2ν+1 dq t
≤
0
0
Z ∞
≤
|f (t)| t2ν+1 dq t = kf kq,1,ν .
0
The result is a consequence of the Riesz-Thorin theorem.
Notice that the kernel Dq,ν (x, y, t) can be written as follows
Z ∞
2
Dq,ν (x, y, t) = cq,ν
jν (xz, q 2 )jν (yz, q 2 )jν (tz, q 2 )z 2ν+1 dq z
0
= cq,ν Fq,ν jν (xz, q 2 )jν (yz, q 2 ) (t),
which implies
Z ∞
Dq,ν (x, y, t)t2ν+1 dq t
0
Z
∞
Fq,ν jν (xz, q 2 )jν (yz, q 2 ) (t)t2ν+1 dq t
0
2
jν (xz, q 2 )jν (yz, q 2 ) (0) = 1.
= Fq,ν
= cq,ν
The q-convolution product is defined by
f ∗q g = Fq,ν [Fq,ν f × Fq,ν g] .
Theorem 5. Let 1 ≤ p, r, s such that
1
1 1
+ −1=
p r
s
Given two functions f ∈ Lq,p,ν and g ∈ Lq,r,ν then f ∗q g exists and we have
Z ∞
ν
f ∗q g(x) = cq,ν
Tq,x
f (y)g(y)y 2ν+1 dq y.
0
and
f ∗q g ∈ Lq,s,ν .
Fq,ν (f ∗q g) = Fq,ν (f ) × Fq,ν (g).
50
LAZHAR DHAOUADI
If s ≥ 2 then
If we suppose that
ν
Tq,x
kf ∗q gkq,s,ν ≤ Bq,ν kf kq,p,ν kgkq,r,ν .
(6)
is a positive operator then
kf ∗q gkq,s,ν ≤ cq,ν kf kq,p,ν kgkq,r,ν .
(7)
Proof. We have
f ∗q g(x)
= Fq,ν [Fq,ν f × Fq,ν g] (x)
Z ∞
= cq,ν
Fq,ν f (y) × Fq,ν g(y)jν (xy, q 2 )y 2ν+1 dq y
0
Z ∞
Z ∞
= cq,ν
Fq,ν f (y) × cq,ν
g(z)jν (zy, q 2 )z 2ν+1 dq z jv (xy, q 2 )y 2ν+1 dq y
0
0
Z ∞
Z ∞
= cq,ν
cq,ν
Fq,ν f (y)jv (zy, q 2 )jν (xy, q 2 )y 2ν+1 dq y g(z)z 2ν+1 dq z
0
Z0 ∞
ν
= cq,ν
Tq,x f (z)g(z)z 2ν+1dq z.
0
The computations are justified by the Fubuni’s theorem
Z ∞
Z ∞
2 2ν+1
|Fq,ν f (y)| ×
|g(z)| × jν (zy, q ) z
dq z jν (xy, q 2 ) y 2ν+1 dq y
0
≤
≤
≤
≤
0
Z
∞
Z
∞
jν (zy, q 2 )r z 2ν+1 dq z
r1
jν (xy, q 2 ) y 2ν+1 dq y
|Fq,ν f (y)| ×
0
Z ∞0
h
2ν+2 i
2
|Fq,ν f (y)| × jv (xy, q 2 ) y − r y 2ν+1 dq y
kgkq,r,ν jν (., q )q,r,ν
kgkq,r,ν
0
kgkq,r,ν jν (., q 2 )q,r,ν kFq,ν f kq,p,v
kgkq,r,ν jν (., q 2 )q,r,ν kFq,ν f kq,p,ν
∞
Z
0
Z
∞
0
From Proposition 2 we deduce that
p1
ip
h
jν (xy, q 2 ) y − 2ν+2
r
y 2ν+1 dq y
p
jν (xy, q 2 )p y 2(ν+1)(1− r )−1 dq y
Fq,ν f ∈ Lq,p,ν and Fq,ν g ∈ Lq,r,ν .
Then, using the Hölder inequality and the fact that
1 1
1
+ =
p r
s
to conclude that
Fq,ν f × Fq,ν g ∈ Lq,s,ν .
Which implies that
f ∗q g = Fq,ν [Fq,ν f × Fq,ν g] ∈ Lq,s,ν
and by the inversion formula (2) we obtain
Fq,ν (f ∗q g) = Fq,ν f × Fq,ν g.
p1
.
ON THE q-BESSEL FOURIER TRANSFORM
51
Suppose that s ≥ 2, so 1 ≤ s ≤ 2 and we can write
kf ∗q gkq,s,ν
ν
Tq,x
=
kFq,ν [Fq,ν f × Fq,ν g]kq,s,ν
≤
s
Bq,ν
kFq,ν f kq,p,ν kFq,ν gkq,r,ν
≤
p
s
r
Bq,ν
Bq,ν
kf kq,p,ν kgkq,r,ν
Bq,ν
≤
2
−1
2
−1
2
−1
2
−1
Bq,ν kf kq,p,ν kgkq,r,ν .
Now suppose that
is a positive operator.
We introduce the operator Kf as follows
Z ∞
ν
Kf g(x) = cq,ν
Tq,x
f (z)g(z)z 2ν+1 dq z.
0
By the Hölder inequality and (5) we get
kKf gkq,∞ ≤ cq,ν kf kq,p,ν kgkq,p,ν .
The Minkowski inequality leads to
kKf gkq,p,ν ≤ cq,ν kf kq,p,ν kgkq,1,ν .
Hence we have
Kf : Lq,p,ν → Cq,0 ,
Then the operator Kf satisfies
Kf : Lq,1,ν → Lq,p,ν .
Kf : Lq,r,ν → Lq,s,ν
and
kf ∗q gkq,s,ν = kKf gkq,s,ν ≤ cq,ν kf kq,p,ν kgkq,r,ν .
Remark 1. We discuss here the sharp results for the Hausdorf-Young inequality
provided above. An inequality already sharper than (6) is given in formula (7). In
fact we have cq,ν < Bq,ν .
To obtained (7) without the positivity argument, we can do by using which is a
q-Riemann-Liouville fractional integral generalizing the q-Mehler integral representation for the q-Bessel function jν (., q 2 ) which can be proved in a straightforward
way [8]
Z 1 2 2 2
(q t , q )∞
j0 (λt, q 2 )tdq t
jν (λ, q 2 ) = [2ν]q
2ν t2 , q 2 )
(q
∞
0
together with the inequalities for the q-Bessel function which is given as formula
(24) in the paper [4]
|j0 (x; q 2 )| ≤ 1, ∀x ∈ R+
q .
Combine this formulas we arrive at
|jν (x; q 2 )| ≤ 1,
∀x ∈ R+
q ,
ν ≥ 0.
Then the inequalities (4) can be written as follows
2
−1
p
kFq,ν f kq,p,ν ≤ cq,ν
kf kq,p,ν .
This should give the sharpest version of (6) in the cases ν ≥ 0. Unfortunately the
ν
positivity of the operator Tq,x
is satisfied in this case.
52
LAZHAR DHAOUADI
In fact we can prove that if we are in the positivity cases then
jν (., q 2 )
≤ 1.
q,∞
To prove this recalling that
ν
Tq,x
jν (y, q 2 ) = jν (x, q 2 )jν (y, q 2 ).
So we have
Z
∞
Dq,ν (x, y, t)jv (t, q 2 )t2v+1 dq t = jν (x, q 2 )jν (y, q 2 ).
0
We obtains for all x, y ∈ R+
q
jν (x, q 2 ) × jν (y, q 2 )
Z
∞
Dq,ν (x, y, t) jν (t, q 2 ) t2ν+1 dq t
0
Z
∞
2ν+1
≤
Dq,ν (x, y, t)t
dq t jν (., q 2 )q,∞ .
≤
0
The fact that
Z
∞
Dq,ν (x, y, t)t2ν+1 dq t = 1
0
implies
which gives the result.
jν (., q 2 )2 ≤ jν (., q 2 )
q,∞
q,∞
3. Uncertainty principle
We introduce two q-difference operators
∂q f (x) =
f (q −1 x) − f (x)
x
and
∂q∗ f (x) =
f (x) − q 2ν+1 f (qx)
.
x
Then we have
∂q ∂q∗ f (x) = ∂q∗ ∂q f (x) = ∆q,ν f (x).
Proposition 3. If h∂q f, gi exist and lim a2ν+1 f (q −1 a)g(a) = 0 then
a→∞
h∂q f, gi = − f, ∂q∗ g .
ON THE q-BESSEL FOURIER TRANSFORM
53
Proof. The following computation
Z a
∂q f (x)g(x)x2ν+1 dq x
0
Z a
f (q −1 x) − f (x)
g(x)x2ν+1 dq x
=
x
0
Z a
Z a
f (q −1 x)
f (x)
2ν+1
=
g(x)x
dq x −
g(x)x2ν+1 dq x
x
x
0
0
Z a
Z q−1 a
f (x)
f (x)
g(qx)x2ν+1 dq x −
∂q g(x)x2ν+1 dq x
= q 2ν+1
x
x
0
0
Z a
Z a
f (x)
f (x)
= q 2ν+1
∂q g(qx)x2ν+1 dq x −
g(x)x2v+1 dq x + a2ν+1 f (q −1 a)g(a)
x
x
0
0
Z a
g(x) − q 2ν+1 g(qx) 2ν+1
x
dq x + a2ν+1 f (q −1 a)g(a)
= −
f (x)
x
0
Z a
= =−
f (x)∂q∗ g(x)x2ν+1 dq x + a2ν+1 f (q −1 a)g(a)
0
leads to the result.
Corollary 1. If f ∈ Lq,2,ν such that xFq,ν f ∈ Lq,2,ν then
k∂q f k2 = kxFq,ν f k2 .
Proof. In fact we have
k∂q f k22
=
=
=
=
=
h∂q f, ∂q f i = − f, ∂q∗ ∂q f
− hf, ∆q,ν f i
− hFq,ν f, Fq,ν ∆q,ν f i
Fq,ν f, x2 Fq,ν f
kxFq,ν f k22 ,
which prove the result.
Theorem 6. Assume that f belongs to the space Lq,2,ν . Then the q-Bessel transform satisfies the following uncertainty principal
2
kf k2 ≤ kq,v kxf k2 kxFq,ν f k2
where
kq,ν
Proof. In fact
√
1 + q × q ν+1
.
=
1 − q 2(ν+1)
∂q∗ xf = f (x) − q 2ν+2 f (qx)
x∂q f = f (q −1 x) − f (x).
We introduce the following operator
Λq f (x) = f (qx),
then
hΛq f, gi = q −2(ν+1) f, Λ−1
q g .
54
LAZHAR DHAOUADI
So
∗
1
∂q xf (x) − q 2ν+2 Λq x∂q f (x) = f (x)
2(ν+1)
1−q
Assume that xf and xFq,ν f belongs to the space Lq,2,ν . Then we have
hf, f i = −
1
1
hxf, ∂q f i −
∂q f, xΛ−1
q f .
2(ν+1)
2(ν+1)
1−q
1−q
By Cauchy-Schwartz inequality we get
hf, f i ≤
1
1
kxf k2 k∂q f k2 +
k∂q f k2 xΛ−1
q f 2.
2(ν+1)
2(ν+1)
1−q
1−q
On the other hand
−1 xΛq f = √q × q ν+1 kxf k ,
2
2
Corollary 1 leads to the result.
4. Hardy’s theorem
The following Lemma from complex analysis is crucial for the proof of our main
theorem.
Lemma 1. For every p ∈ N, there exist σp > 0 for which
|z|2p |jν (z, q 2 )| < σp e|z| ,
∀z ∈ C.
Proof. In fact
|z|2p |jν (z, q 2 )| ≤
≤
∞
X
1
q n(n−1) |z|2n+2p
(q 2 , q 2 )∞ (q 2ν+2 , q 2 )∞ n=0
∞
X
q p(p+1)
q n(n−2p−1) |z|2n .
(q 2 , q 2 )∞ (q 2ν+2 , q 2 )∞ n=p
Now using the Stirling’s formula
n! ∼
√
2πn
nn
,
en
we see that there exist an entire n0 ≥ p such that
q n(n−2p−1) <
1
,
(2n)!
∀n ≥ n0 ,
which implies
∞
X
n=n0
q n(n−2p−1) |z|2n <
Finally there exist σp > 0 such that
∞
X
n=n0
|z|2p |jν (z, q 2 )|
< σp ,
e|z|
This complete the proof.
1
|z|2n < e|z| .
(2n)!
∀z ∈ C
ON THE q-BESSEL FOURIER TRANSFORM
55
Lemma 2. Let h be an entire function on C such that
2
|h(z)| ≤ Cea|z| ,
z ∈ C,
2
|h(x)| ≤ Ce−ax ,
x ∈ R,
for some positive constants a and C. Then there exist C ∗ ∈ R such
2
h(x) = C ∗ e−ax .
The reader can see the reference [17] for the proof.
Now we are in a position to state and prove the q-analogue of the Hardy’s theorem
Theorem 7. Suppose f ∈ Lq,1,ν satisfying the following estimates
1
2
|f (x)| ≤ Ce− 2 x ,
1
∀x ∈ R+
q ,
2
|Fq,ν f (x)| ≤ Ce− 2 x ,
(8)
∀x ∈ R,
where C is a positive constant. Then there exist A ∈ R such that
1 2
f (z) = Acq,ν Fq,ν e− 2 x (z), ∀z ∈ C.
Proof. We claim that Fq,ν f is an analytic function and there exist C ′ > 0 such that
1
2
|Fq,ν f (z)| ≤ C ′ e 2 |z| ,
We have
|Fq,ν f (z)| ≤ cq,ν
Z
∞
0
∀z ∈ C.
|f (x)||jν (zx, q 2 )|x2ν+1 dq x.
From the Lemma 1, if |z| > 1 then there exist σ1 > 0 such that
x2ν+1 |jν (zx, q 2 )| =
Then we obtain
|Fq,ν f (z)| ≤ Cσ1 cq,ν
1
2ν+1
|jν (zx, q 2 )|
2ν+1 (|z| x)
|z|
"Z
1
∞
e− 2 (x−|z|)
2
1 + |z|2 x2
0
#
<
σ1
2
1 + |z| x2
2
1
dq x e 2 |z| < Cσ1 cq,ν
Now, if |z| ≤ 1 then there exist σ2 > 0 such that
x2ν+1 |jν (zx, q 2 )| ≤ σ2 ex ,
Z
ex|z| ,
∞
0
∀x ∈ R+
q .
2
1
1
d
x
e 2 |z| .
q
2
1+x
∀x ∈ R+
q .
Therefore
|Fq,ν f (z)| ≤ Cσ2 cq,ν
Z
∞
1
e− 2 x
2
+x
0
Z
dq x ≤ Cσ2 cq,ν
∞
1
e− 2 x
0
2
+x
2
1
dq x e 2 |z| ,
which leads to the estimate (8). Using Lemma 2, we obtain
1
2
Fq,ν f (z) = const.e− 2 z ,
∀z ∈ C,
and by Theorem 1, we conclude that
1 2
f (z) = const.Fq,ν e− 2 t (z),
∀z ∈ C.
56
LAZHAR DHAOUADI
Corollary 2. Suppose f ∈ Lq,1,ν satisfying the following estimates
2
|f (x)| ≤ Ce−px ,
∀x ∈ R+
q ,
2
|Fq,ν f (x)| ≤ Ce−σx ,
∀x ∈ R,
where C, p, σ are a positive constant and pσ = 14 . We suppose that there exist
1
2
a ∈ R+
q such that a p = 2 . Then there exist A ∈ R such that
2
f (z) = Acq,ν Fq,ν e−σt (z), ∀z ∈ C.
Proof. Let a ∈ R+
q , and put
fa (x) = f (ax),
then
1
Fq,ν f (x/a).
a2ν+2
In the end, applying Theorem 7 to the function fa .
Fq,ν fa (x) =
Corollary 3. Suppose f ∈ Lq,1,ν satisfying the following estimates
2
|f (x)| ≤ Ce−px ,
∀x ∈ R+
q ,
2
|Fq,ν f (x)| ≤ Ce−σx ,
where C, p, σ are a positive constant and pσ >
1
2
a ∈ R+
q such that a p = 2 . Then f ≡ 0.
∀x ∈ R,
1
4.
(9)
We suppose that there exist
Proof. In fact there exists σ ′ < σ such that pσ ′ = 14 . Then the function f satisfying
the estimates of Corollary 2, if we replacing σ by σ ′ . Which implies
′
2
Fq,ν f (x) = const.e−σ x ,
∀x ∈ R.
On the other hand, f satisfying the estimates (9), then
′ 2
2
const.e−σ x ≤ Ce−σx , ∀x ∈ R.
This implies Fq,ν f ≡ 0, and by Theorem 1 we conclude that f ≡ 0.
5. The q-Fourier-Neumann Expansions
The little q-Jacobi polynomials are defined for ν, β > −1 by [15]
n+ν+β+1 −n q
, q ν β
q;
qx
.
pn (x; q , q ; q) = 2 φ1
q ν+1
We define the functions
Pν,n (x; q 2 ) = σq,ν (n)q −n(ν+1)
(q 2+2n , q 2ν+2 ; q 2 )∞
pn (x2 ; q 2ν , 1; q 2 )
(q 2+2n+2ν , q 2 ; q 2 )∞
and
Jν,n (x; q 2 ) = σq,ν (n)
where
σq,ν (n) =
s
Jν+2n+1 (q n x; q 2 )
,
xν+1
1 − q 2ν+4n+2
.
1−q
ON THE q-BESSEL FOURIER TRANSFORM
57
Consider Lνq,2 as an Hilbert space with the inner product
Z 1
hf |gi =
f (x)g(x)x2ν+1 dq x.
0
The q-Paley-Wiener space is defined by
Z
P Wqν = f ∈ Lq,2,ν : f (x) = cq,ν
1
u(t)jν (xt, q 2 )t2ν+1 dq t,
0
u ∈ Lνq,2 .
Proposition 4. P Wqν is a closed subspace of Lq,2,ν and with the inner product
given in (3) is an Hilbert space.
Proof. In fact, given f ∈ Lq,2,ν and let {fn }n∈N be a sequence of element of P Wqν
which converge to f in L2 -norm. For n ∈ N, there exist un ∈ Lνq,2 such that
Z 1
fn (x) = cq,ν
un (t)jν (xt, q 2 )t2ν+1 dq t.
0
Moreover
lim kfn − f kq,2,ν = 0.
n→∞
This give
lim kFq,ν fn − Fq,ν f kq,2,ν = 0,
n→∞
and then
Z
lim
n→∞
0
1
2 2ν+1
|Fq,ν fn (x) − Fq,ν f (x)| x
dq x +
Z
∞
1
|Fq,ν f (x)| x
which implies
Z ∞
|Fq,ν f (x)|2 x2ν+1 dq x = 0 ⇒ Fq,ν f (x) = 0,
1
2 2ν+1
dq x = 0,
∀x ∈ R+
q ∩]1, +∞[.
Then f ∈ P Wqν .
Proposition 5. We have
As a consequence
Fq,ν (Jν,n )(x) = Pν,n (x; q 2 )χ[0,1] (x),
Z
1
∀x ∈ R+
q .
Pν,n (x; q 2 )Pν,m (x; q 2 )x2ν+1 dq x = δn,m .
0
Proof. The following proof is identical to the proof of Lemma 1 in [1]. Using an
identity established in [12, 13]
Z ∞
t−λ Jµ (q m t; q 2 )Jθ (q n t; q 2 )dq t
0
(q 1+λ+θ−µ , q 2µ+2 ; q 2 )∞
= (1 − q)q n(λ−1)+(m−n)µ
(q 1−λ+θ+µ , q 2 ; q 2 )∞
1−λ+µ+θ 1−λ+µ−θ 2 2m−2n+1+λ+θ−µ
q
,q
, (10)
q
;
q
× 2 φ1
q 2µ+2
where n, m ∈ Z and θ, µ, λ ∈ C such that Re(1 − λ + θ + µ) > 0, θ, µ are not equal
to a negative integer and
(λ + θ + 1 − µ)/2,
m − n + (λ + θ + 1 − µ)/2
58
LAZHAR DHAOUADI
are not a non-positive integer [13].
To evaluate Fq,ν (Jν,n )(x) when x = q m ≤ 1, we take in (10)
q m = x, µ = ν, θ = ν + 2n + 1, λ = 0
then
Fq,ν (Jν,n )(x)
=
=
=
x−ν
1−q
Z
∞
Jν (xt; q 2 )Jν+2n+1 (q n t; q 2 ) dq t
0
2+2ν+2n −2n 2+2n 2ν+2 2
2 2 2
,q
; q )∞
q
,q
−n(ν+1) (q
q
;
q
x
φ
σq,ν (n)q
2 1
q 2ν+2
(q 2+2n+2ν , q 2 ; q 2 )∞
σq,ν (n)
Pν,n (x; q 2 ).
To evaluate Fq,ν (Jν,m )(x) when x = q n > 1, we consider in (10)
q n = x, µ = ν + 2m + 1, θ = ν, λ = 0
In this way, 1 + λ + θ − µ = −2m. This gives for m ∈ N a factor
(q −2m ; q 2 )∞ = 0
on the numerator and then
Fq,ν (Jν,m )(x) = 0,
x>1
By setting λ = 1, θ = ν + 2n + 1, and µ = ν + 2m + 1 in , it is clear that, for
n, m = 0, 1, 2, . . . ,
Z ∞
dq x
1
Jν+2n+1 (q n x; q 2 )Jν+2m+1 (q m x; q 2 )
δ
=
2 n,m
x
σ
q,ν (n)
0
and then
Z
0
∞
Jν,n (x; q 2 )Jν,m (x; q 2 ) x2ν+1 dq x = δn,m .
Now we use the arguments of q-Bessel Fourier analysis provided in this paper to
show that
Pν,n χ[0,1] , Pν,m χ[0,1] = hFq,ν (Jv,n ), Fq,ν (Jv,m )i = hJν,n , Jν,m i = δn,m . (11)
Another proof of the orthogonality of the little q-Jacobi polynomials can be found
in [15]
Proposition 6. The systems
{Jν,n }∞
n=0 ,
{Pν,n }∞
n=0
forme two orthonormals basis respectively of the Hilbert spaces P Wqv and Lνq,2 .
Proof. From (11) we derive the orthonormality. To prove that the system {Jν,n }∞
n=0
is complet in P Wqν , given a function f ∈ P Wqν such that
Then
hf, Jν,n i = 0,
∀n ∈ N.
hFq,ν (f ), Fq,ν (Jν,n )i = 0,
∀n ∈ N,
which implies
Fq,ν (f ), Pν,n χ[0,1] = Fq,ν (f )χ[0,1] , Pν,n = hFq,ν (f ), Pν,n i = 0,
From the definition of the polynomial Pν,n we conclude that
Fq,ν (f ), t2n = 0, ∀n ∈ N.
∀n ∈ N.
ON THE q-BESSEL FOURIER TRANSFORM
59
Then
cq,ν
∞
X
(−1)n
n=0
q n(n+1)
Fq,ν (f ), t2n x2n = 0,
(q 2 , q 2 )n (q 2ν+2 , q 2 )n
∀x ∈ R+
q ,
which can be written as
2
Fq,ν
(f )(x) = 0,
∀x ∈ R+
q .
By the inversion formula (2) we conclude that f = 0. From (11) we derive the
ν
orthonormality. To prove that the system {Pν,n }∞
n=0 is complet in Lq,2 , given a
function f ∈ Lνq,2 such that
Then
Which leads to the result.
hf |Pν,n i = 0,
∀n ∈ N
hf |t2n i = 0,
∀n ∈ N.
Proposition 7. Let λ ∈ R+
q then
cq,ν jν (λx; q 2 ) =
∞
X
n=0
Jn,ν (λ; q 2 )Pn,ν (x),
As a consequence we have
∞
X
2
x−2(ν+1)
Pn,ν (x; q 2 ) =
,
1−q
n=0
∀x ∈ [0, 1] ∩ R+
q .
∀x ∈ [0, 1] ∩ R+
q
and for all λ ∈ R+
q
∞
X
2
Jn,ν (λ; q 2 ) = −
n=0
×
qν
2(1 − q)λ1+2ν
λ
′
Jν+1 (λ; q 2 )Jν′ (λ/q; q 2 ) − Jν+1 (λ; q 2 )Jν (λ/q; q 2 ) − Jν+1
(λ; q 2 )Jν (λ/q; q 2 ) .
q
Proof. Let λ ∈ R and consider the function
ψλ : [0, 1] ∩ R+
q → R,
Then ψλ ∈ Lνq,2 and we can write
ψλ (x) =
∞
X
n=0
Note that
x 7→ cq,ν jν (λx; q 2 ).
hψλ |Pn,ν i Pn,ν (x),
∀x ∈ [0, 1] ∩ R+
q .
(12)
2
hψλ |Pn,ν i = ψλ , Pn,ν χ[0,1] = hψλ , Fq,ν (Jn,ν )i = Fq,ν
(Jn,ν )(λ) = Jn,ν (λ; q 2 ).
Then we deduce the result. Using the Parseval’s theorem and (12) we obtain
∞
X
2
x−2(ν+1)
.
Pn,ν (x; q 2 ) = kψx k2q,2,ν =
1−q
n=0
The second identity is deduced also from the Parseval’s theorem
∞
X
2
2
(ψλ ),
Jn,ν (λ; q 2 ) = Nq,ν,2
n=0
60
LAZHAR DHAOUADI
and the following relation proved in [14]
Z 1
2
(1 − q)q ν−1
Jν (aqt; q 2 ) tdq t = −
2a
0
h
i
2
′
2
′
× aJν+1 (aq; q )Jν (a; q ) − Jν+1 (aq; q 2 )Jν (a; q 2 ) − Jν+1
(aq; q 2 )Jν (a; q 2 ) .
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