Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 171, 2006
INEQUALITIES IN q−FOURIER ANALYSIS
LAZHAR DHAOUADI, AHMED FITOUHI, AND J. EL KAMEL
E COLE P RÉPARATOIRE D ’I NGÉNIEUR
B IZERTE , T UNISIA .
lazhardhaouadi@yahoo.fr
FACULTÉ DES S CIENCES DE T UNIS
1060 T UNIS , T UNISIA .
Ahmed.Fitouhi@fst.rnu.tn
Received 17 April, 2006; accepted 28 July, 2006
Communicated by H.M. Srivastava
A BSTRACT. In this paper we introduce the q−Bessel Fourier transform, the q−Bessel translation operator and the q−convolution product. We prove that the q−heat semigroup is contractive
and we establish the q−analogue of Babenko inequalities associated to the q−Bessel Fourier
transform. With applications and finally we enunciate a q−Bessel version of the central limit
theorem.
Key words and phrases: Čebyšev functional, Grüss inequality, Bessel, Beta and Zeta function bounds.
2000 Mathematics Subject Classification. Primary 26D15, 26D20; Secondary 26D10.
1. I NTRODUCTION AND P RELIMINARIES
In introducing q−Bessel Fourier transforms, the q−Bessel translation operator and the q−convolution
product we shall use the standard conventional notation as described in [4]. For further detailed
information on q−derivatives, Jackson q−integrals and basic hypergeometric series we refer
the interested reader to [4], [10], and [8].
The following two propositions will useful for the remainder of the paper.
Proposition 1.1. Consider 0 < q < 1. The series
(w; q)∞1 φ1 (0, w; q; z) =
∞
X
n=0
(−1)n q
n(n−1)
2
(wq n ; q)∞ n
z ,
(q; q)n
defines an entire analytic function in z, w, which is also symmetric in z, w:
(w; q)∞1 φ1 (0, w; q; z) = (z; q)∞1 φ1 (0, z; q; w).
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
115-06
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L AZHAR D HAOUADI , A HMED F ITOUHI , AND J. E L K AMEL
Both sides can be majorized by
|(w; q)∞1 φ1 (0, w; q; z)| ≤ (−|w|; q)∞ (−|z|; q)∞ .
Finally, for all n ∈ N we have
(q 1−n ; q)∞1 φ1 (0, q 1−n ; q; z) = (−z)n q
n(n−1)
2
(q 1+n ; q)∞1 φ1 (0, q 1+n ; q; q n z).
Proof. See [10].
Now we introduce the following functional spaces:
Rq = {∓q n ,
n
R+
q = {q , n ∈ Z}.
n ∈ Z},
Let Dq , Cq,0 and Cq,b denote the spaces of even smooth functions defined on Rq continuous
at 0, which are respectively with compact support, vanishing at infinity and bounded. These
spaces are equipped with the topology of uniform convergence, and by Lq,p,v the space of even
functions f defined on Rq such that
Z ∞
p1
p 2v+1
kf kq,p,v =
|f (x)| x
dq x < ∞.
0
We denote by Sq the q−analogue of the Schwartz space of even function f defined on Rq such
that Dqk f is continuous at 0, and for all n ∈ N there is Cn such that
Cn
, ∀k ∈ N, ∀x ∈ R+
q .
(1 + x2 )n
Ar the end of this section we introduce the q−Bessel operator as follows
1 ∆q,v f (x) = 2 f (q −1 x) − (1 + q 2v )f (x) + q 2v f (qx) .
x
Proposition 1.2. Given two functions f and g in Lq,2,v such that
|Dqk f (x)| ≤
∆q,v f, ∆q,v g ∈ Lq,2,v
then
Z
∞
∆q,v f (x)g(x)x
2v+1
0
Z
dq x =
∞
f (x)∆q,v g(x)x2v+1 dq x.
0
2. T HE N ORMALIZED H AHN -E XTON q−B ESSEL F UNCTION
The normalized Hahn-Exton q−Bessel function of order v is defined as
(q, q)∞ −v (3)
jv (x, q) = v+1
x Jv (x, q) = 1 φ1 (0, q v+1 , q, qx2 ), <(v) > −1,
(q , q)∞
(3)
where Jv (·, q) is the Hahn-Exton q−bessel function, (see [12]).
Proposition 2.1. The function
x 7→ jv (λx, q 2 ),
is a solution of the following q−difference equation
∆q,v f (x) = −λ2 f (x)
Proof. See [9].
In the following we put
cq,v =
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
1
(q 2v+2 , q 2 )∞
·
.
1−q
(q 2 , q 2 )∞
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I NEQUALITIES IN q−F OURIER A NALYSIS
3
Proposition 2.2. Let n, m ∈ Z and n 6= m, then we have
Z ∞
q −2n(v+1)
2
cq,v
jv (q n x, q 2 )jv (q m x, q 2 )x2v+1 dq x =
δnm .
1−q
0
Proof. See [10].
Proposition 2.3.
(−q 2 ; q 2 )∞ (−q 2v+2 ; q 2 )∞
|jv (q n , q 2 )| ≤
(q 2v+2 ; q 2 )∞
(
1
2 +(2v+1)n
qn
if
n ≥ 0,
if
n < 0.
Proof. Use Proposition 1.1.
3. q−B ESSEL F OURIER T RANSFORM
The q−Bessel Fourier transform Fq,v is defined as follows
Z ∞
Fq,v (f )(x) = cq,v
f (t)jv (xt, q 2 )t2v+1 dq t.
0
Proposition 3.1. The q−Bessel Fourier transform
Fq,v : Lq,1,v → Cq,0 ,
satisfying
kFq,v (f )kCq,0 ≤ Bq,v kf kq,1,v ,
where
Bq,v =
1 (−q 2 ; q 2 )∞ (−q 2v+2 ; q 2 )∞
.
1−q
(q 2 ; q 2 )∞
Proof. Use Proposition 2.3.
Theorem 3.2. Given f ∈ Lq,1,v then we have
2
Fq,v
(f )(x) = f (x),
∀x ∈ R+
q .
If f ∈ Lq,1,v and Fq,v (f ) ∈ Lq,1,v then
kFq,v (f )kq,2,v = kf kq,2,v .
Proof. Let t, y ∈ R+
q , we put
(
δq,v (t, y) =
It is not hard to see that
Z
1
(1−q)t2v+2
if t = y,
0
if t 6= y.
∞
f (t)δq,v (t, y)t2v+1 dq t = f (y).
0
By Proposition 2.2, we can write
Z ∞
2
cq,v
jv (yx, q 2 )jv (tx, q 2 )x2v+1 dq x = δq,v (t, y),
∀t, y ∈ R+
q ,
0
which leads to the result.
Corollary 3.3. The transformation
Fq,v : Sq → Sq ,
is an isomorphism, and
−1
Fq,v
= Fq,v .
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
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4
L AZHAR D HAOUADI , A HMED F ITOUHI , AND J. E L K AMEL
Proof. The result is deduced from properties of the space Sq .
4. q−B ESSEL T RANSLATION O PERATOR
We introduce the q−Bessel translation operator as follows:
Z ∞
v
Tq,x f (y) = cq,v
Fq,v (f )(t)jv (xt, q 2 )jv (yt, q 2 )t2v+1 dq t,
∀x, y ∈ R+
q , ∀f ∈ Lq,1,v .
0
Proposition 4.1. For any function f ∈ Lq,1,v we have
v
v
f (x),
f (y) = Tq,y
Tq,x
and
v
Tq,x
f (0) = f (x).
Proposition 4.2. For all x, y ∈ R+
q , we have
v
Tq,x
jv (λy, q 2 ) = jv (λx, q 2 )jv (λy, q 2 ).
Proof. Use Proposition 2.2.
Proposition 4.3. Let f ∈ Lq,1,v then
∞
Z
v
Tq,x
f (y)
f (z)Dv (x, y, z)z 2v+1 dq z,
=
0
where
Dv (x, y, z) =
c2q,v
∞
Z
jv (xt, q 2 )jv (yt, q 2 )jv (zt, q 2 )t2v+1 dq t.
0
Proof. Indeed,
v
Tq,x
f (y)
Z
∞
Fq,v (f )(t)jv (xt, q 2 )jv (yt, q 2 )t2v+1 dq t
Z0 ∞ Z ∞
2 2v+1
= cq,v
cq,v
f (z)jv (zt, q )z
dq t jv (xt, q 2 )jv (yt, q 2 )t2v+1 dq t
0
0
Z ∞
Z ∞
2
2
2
2 2v+1
=
f (z) cq,v
jv (xt, q )jv (yt, q )jv (zt, q )t
dq t z 2v+1 dq z,
= cq,v
0
0
which leads to the result.
Proposition 4.4.
lim Dv (x, y, z) = 0
z→∞
and
(1 − q)
X
q (2v+2)s Dv (x, y, q s ) = 1
s∈Z
Proof. To prove the first relation use Proposition 3.1. The second identity is deduced from
v
f = 1.
Proposition 4.2: if f = 1 then Tq,x
Proposition 4.5. Given f ∈ Sq then
v
Tq,x
f (y)
=
∞
X
n=0
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
q n(n+1)
y 2n ∆nq,v f (x).
(q 2 , q 2 )n (q 2v+2 , q 2 )n
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I NEQUALITIES IN q−F OURIER A NALYSIS
5
Proof. By the use of Proposition 2.1 and the fact that
Z ∞
n
n
∆q,v f (x) = (−1) cq,v
Fq,v (f )(t)t2n jv (xt, q 2 )t2v+1 dq t.
0
Proposition 4.6. If v = − 12 then
Dv (q m , q r , q k ) =
q 2(r−m)(k−m)−m 2(r−m)+1
(q
; q)∞1 φ1 (0, q 2(r−m)+1 , q; q 2(k−m)+1 ).
(1 − q)(q; q)∞
Proof. Indeed
∆nq,v
n (k+n)(k+n+1)
q −n(n+1) X
2n
−2kn k
2
=
(−1)k+n q
Λq ,
2n
x
k
+
n
q
k=−n
and use Proposition 4.5.
5. q−C ONVOLUTION P RODUCT
In harmonic analysis the positivity of the translation operator is crucial. It plays a central role
in establishing some useful results, such as the property of the convolution product. Thus it is
v
natural to investigate when this property holds for Tq,x
. In the following we put
v
Tq,x
Qv = {q ∈ [0, 1],
is positive for all
x ∈ R+
q }.
v
v
Recall that Tq,x
is said to be positive if Tq,x
f ≥ 0 for f ≥ 0.
Proposition 5.1. If v = − 12 then
Qv = [0, q0 ],
where q0 is the first zero of the following function:
q 7→ 1 φ1 (0, q, q, q).
v
Proof. The operator Tq,x
is positive if and only if
Dv (x, y, q s ) ≥ 0,
We replace
x
y
∀x, y, q s ∈ R+
q .
by q r , and we can choose r ∈ N, because
v
v
Tq,x
f (y) = Tq,y
f (x),
thus we get
(q
1+2s
, q)∞1 φ1 (0, q
1+2s
, q, q
1+2r
)=
∞
X
Bn (s, r),
∀r, s ∈ N,
n=0
where
2n
∞
Y
q 2r+i Y
q 2r+2n+1
2s+i
2s+2n+1
Bn (s, r) =
(1 − q
) (1 − q
)−
,
1 − q i i=2n+2
1 − q 2n+1
i=1
and
∀n ∈ N∗ ,
∞
Y
q 2r+1
2s+i
2s+1
B0 (s, r) =
(1 − q
) (1 − q
)−
,
1−q
i=2
which leads to the result.
In the rest of this work we choose q ∈ Qv .
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
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6
L AZHAR D HAOUADI , A HMED F ITOUHI , AND J. E L K AMEL
Proposition 5.2. Given f ∈ Lq,1,v then
Z
Z ∞
2v+1
v
dq y =
Tq,x f (y)y
0
∞
f (y)y 2v+1 dq y.
0
The q−convolution product of both functions f, g ∈ Lq,1,v is defined by
Z ∞
v
f ∗q g(x) = cq
Tq,x
f (y)g(y)y 2v+1 dq y.
0
Proposition 5.3. Given two functions f, g ∈ Lq,1,v then
f ∗q g ∈ Lq,1,v ,
and
Fq,v (f ∗q g) = Fq,v (f )Fq,v (g).
Proof. We have
kf ∗q gkq,1,v ≤ kf kq,1,v kgkq,1,v .
On the other hand
∞
Z
Z
∞
v
f (x)Tq,y
jv (λx, q 2 )x2v+1 dq x
Fq,v (f ∗q g)(λ) =
0
g(y)y 2v+1 dq y
0
= Fq,v (f )(λ) Fq,v (g)(λ).
6. q−H EAT S EMIGROUP
The q−heat semigroup is defined by:
v
Pq,t
f (x) = Gv (·, t, q 2 ) ∗q f (x)
Z ∞
v
= cq,v
Tq,x
Gv (y, t, q 2 )f (y)y 2v+1 dq y,
∀f ∈ Lq,1,v .
0
v
Gv (·, t, q 2 ) is the q−Gauss kernel of Pq,t
−2v
(−q 2v+2 t, −q −2v /t; q 2 )∞
q
2 2
G (x, t, q ) =
e −
x ,q .
(−t, −q 2 /t; q 2 )∞
t
v
2
and e(·, q) the q-exponential function defined by
e(z, q) =
∞
X
n=0
zn
1
=
,
(q, q)n
(z; q)∞
|z| < 1.
Proposition 6.1. The q−Gauss kernel Gv (·, t, q 2 ) satisfying
Fq,v Gv (·, t, q 2 ) (x) = e(−tx2 , q 2 ),
and
Fq,v e(−ty 2 , q 2 ) (x) = Gv (x, t, q 2 ).
Proof. In [5], the Ramanujan identity was proved
q2
2 2
,
q
,
q
bz,
s
X
bz
z
∞,
=
2s
2
2
q
(bq , q )∞
b, z, b , q 2
s∈Z
∞
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
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I NEQUALITIES IN q−F OURIER A NALYSIS
which implies
Z ∞
7
X (q 2n+2v+2 )s
(−tq 2s , q 2 )∞
s
−2n−2v
−tq 2n+2v+2 , − q t , q 2 , q 2
∞.
= (1 − q)
q2
2n+2v+2
2
−t, q
,− t ,q
e(−ty 2 , q 2 )y 2n y 2v+1 dq y = (1 − q)
0
∞
The following identity leads to the result
(a, q 2 )∞ = (a, q 2 )n (q 2n a, q 2 )∞ ,
and
(aq
−2n
n −n2 +n
2
, q )∞ = (−1) q
a
q2
n q2 2
,q
a
(a, q 2 )∞ .
n
Proposition 6.2. For any functions f ∈ Sq , we have
v
Pq,t
f (x) = e(t∆q,v , q 2 )f (x).
Proof. Indeed, if
Z
cq,v
∞
Gv (y, t, q 2 )y 2n y 2v+1 dq y = (q 2v+2 , q 2 )n q −n(n+n) tn ,
0
then
v
Pq,t
f (x)
=
∞
X
n=0
Z ∞
q n(n+1)
v
2 2n 2v+1
cq,v
G (y, t, q )y y
dq y ∆nq,v f (x).
2
2
2v+2
2
(q , q )n (q
, q )n
0
Theorem 6.3. For f ∈ Lq,p,v and 1 ≤ p < ∞, we have
v
kPq,t
f kq,p,v ≤ kf kq,p,v .
Proof. If p = 1 then
v
kPq,t
f kq,1,v ≤ kGv (·, t, q 2 )kq,1,v kf kq,1,v = kf kq,1,v .
Now let p > 1 and we consider the following function
v
g : y 7→ Tq,x
Gv (y, t; q 2 ).
In addition
v p
Pq,t f ≤ cpq,v
q,p
Z
∞
Z
|f (y)g(y)| y
0
p
∞
2v+1
dq y
x2v+1 dq x.
0
By the use of the Hölder inequality and the fact that kGv (·, t, q 2 )kq,1,v =
immediately.
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
1
,
cq,v
the result follows
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L AZHAR D HAOUADI , A HMED F ITOUHI , AND J. E L K AMEL
7. q−W IENER A LGEBRA
For u ∈ Lq,1,v and λ ∈ R+
q , we introduce the following function
x
1
uλ : x 7→ 2v+2 u
.
λ
λ
Proposition 7.1. Given u ∈ Lq,1,v such that
Z ∞
u(x)x2v+1 dq x = 1,
0
then we have
Z
∞
f (x)uλ (x)x2v+1 dq x = f (0),
lim
λ→0
∀f ∈ Cq,b .
0
Corollary 7.2. The following function
Gvλ : x 7→ cq,v Gv (x, λ2 , q 2 ),
checks the conditions of the preceding proposition.
Proof. Use Proposition 6.1.
Theorem 7.3. Given f ∈ Lq,1,v ∩ Lq,p,v , 1 ≤ p < ∞ and fλ defined by
Z ∞
fλ (x) = cq
Fq,v (f )(y)e(−λ2 y 2 , q 2 )jv (xy, q 2 )y 2v+1 dq y.
0
then we have
lim kf − fλ kq,p,v = 0.
λ→0
Proof. We have
f
∗q Gvλ (x)
∞
Z
Fq,v (f )(t)e(−λ2 t2 , q 2 )jv (tx, q 2 )t2v+1 dq t.
= cq,v
0
In addition, for all ε > 0, there exists a function h ∈ Lq,p,v with compact support in [q k , q −k ]
such that
kf − hkq,p,v < ε,
however
kGvλ ∗q f − f kq,p,v ≤ kGvλ ∗q (f − h)kq,p,v + kGvλ ∗q h − hkq,p,v + kf − hkq,p,v .
By Theorem 6.3 we get
kGvλ ∗q (f − h)kq,p,v ≤ kf − hkq,p,v .
Now, we will prove that
lim kGvλ ∗q h − hkq,p,v = 0.
λ→0
Indeed, by the use of Corollary 7.2 we get
Z 1
lim
|Gvλ ∗q h(x) − h(x)|p x2v+1 dq x = 0.
λ→0
0
On the other hand the following function is decreasing on the interval [1, ∞[:
u 7→ u2v+2 Gv (u).
If λ < 1, then we deduce that
v
v
v
Tq,q
i Gλ (x) ≤ Tq,q i G(x).
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I NEQUALITIES IN q−F OURIER A NALYSIS
9
We can use the dominated convergence theorem to prove that
Z ∞
lim
|Gvλ ∗q h(x) − h(x)|p x2v+1 dq x = 0.
λ→0
1
Corollary 7.4. Given f ∈ Lq,1,v then
Z ∞
f (x) = cq,v
Fq,v (f )(y)jv (xy, q 2 )y 2v+1 dq y,
∀x ∈ R+
q .
0
Proof. The result is deduced by Theorem 7.3 and the following relation
(1 − q)x2v+2 |f (x) − fλ (x)| ≤ kf − fλ kq,1,v
∀x ∈ R+
q .
Now we attempt to study the q−Wiener algebra denoted by
Aq,v = {f ∈ Lq,1,v ,
Fq,v (f ) ∈ Lq,1,v } .
Proposition 7.5. For 1 ≤ p ≤ ∞, we have
(1) Aq,v ⊂ Lq,p,v and Aq,v = Lq,p,v .
(2) Aq,v ⊂ Cq,0 and Aq,v = Cq,0 .
Proof. 1. Given h ∈ Lq,p,v with compact support, and we put hn = h ∗q Gvqn . The function
hn ∈ Aq,v and by Theorem 7.3 we get
lim kh − hn kq,p,v = 0.
n→∞
2. If f ∈ Cq,0 , then there exist h ∈ Cq,0 with compact support on [q k , q −k ], such that
kf − hkCq,0 < ε,
and by Corollary 7.4 we prove that
"
#
sup |h(x) − hn (x)| = 0.
lim
n→∞
x∈R+
q
Theorem 7.6. For f ∈ Lq,2,v ∩ Lq,1,v , we have
kFq,v (f )kq,2,v = kf kq,2,v .
Proof. We put
fn = f ∗q Gvqn ,
which implies
Fq,v (fn )(t) = e(−q 2n t2 , q 2 )Fq,v (f )(t),
by Corollary 7.4 we get
Z
fn (x) = cq
∞
Fq,v (fn )(t)jv (tx, q 2 )t2v+1 dq t.
0
On the other hand
Z ∞
f (x)fn (x)x
2v+1
Z
dq x =
0
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
∞
Fq,v (f )(x)Fq,v (fn )(x)x2v+1 dq x.
0
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10
L AZHAR D HAOUADI , A HMED F ITOUHI ,
AND
J. E L K AMEL
Theorem 7.3 implies
Z
lim
n→∞
∞
Fq,v (f )(x)2 e(−q 2n x2 , q 2 )x2v+1 dq x = kf k2q,2,v .
0
2n 2
The sequence e(−q x , q 2 ) is increasing. By the use of the Fatou-Beppo-Levi theorem we
deduce the result.
Theorem 7.7.
(1) The q−cosine Fourier transform Fq,v possesses an extension
U : Lq,2,v → Lq,2,v .
(2) For f ∈ Lq,2,v , we have
kU (f )kq,2,v = kf kq,2,v .
(3) The application U is bijective and
U −1 = U.
Proof. Let the maps
u : Aq,v → Aq,v ,
f 7→ Fq,v (f ).
Theorem 3.2 implies
ku(f )kq,2,v = kf kq,2,v .
The map u is uniformly continuous, with values in complete space Lq,2,v . It has a prolongation
U on Aq,v = Lq,2,v .
Proposition 7.8. Given 1 < p ≤ 2 and
1
p
+
1
p0
= 1, if f ∈ Lq,p,v , then Fq,v (f ) ∈ Lq,p0 ,v ,
kFq,v (f )kq,p0 ,v ≤ Bp,q,v kf kq,p,v ,
where
( 2 −1)
Bp,q,v = Bq,vp
.
Proof. The result is a consequence of Proposition 3.1, Theorem 7.7 and the Riesz-Thorin theorem, see [13].
As an immediate consequence of Proposition 7.8, we have the following theorem:
Theorem 7.9. Given 1 < p, p0 , r ≤ 2 and
1
1
1
+ 0 −1= ,
p p
r
if f ∈ Lq,p,v and g ∈ Lq,p0 ,v , then
f ∗q g ∈ Lq,r,v ,
and
kf ∗q gkq,r,v ≤ Bq,p,v Bq,p0 ,v Bq,r0 ,v kf kq,p,v kgkq,p0 ,v ,
where
1
1
+ 0 = 1.
r r
Proof. We can write
f ∗q g = Fq,v {Fq,v (f )Fq,v (g)} ,
the use of Proposition 7.8 and the Hölder inequality leads to the result.
v
Now we are in a position to establish the hypercontractivity of the q−heat semigroup Pq,t
.
For more information about this notion, the reader can consult ([1, 2, 3]).
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I NEQUALITIES IN q−F OURIER A NALYSIS
11
Proposition 7.10. For f ∈ Lq,p0 ,v and t ∈ R+
q , we have
v
f kq,p,v ≤ Bq,p0 ,v Bq,p1 ,v c(r, q, v)t−
kPq,t
v+1
r
kf kq,p0 ,v ,
where
1
1
+
= 1,
p p1
1 < p0 < p ≤ 2,
1
1
1
= 0− ,
r
p
p
and
c(r, q, v) = ke(−x2 , q 2 )kq,r,v .
Proof. The result is deduced by the following relations
Fq,v Gv (·, t, q 2 ) (x) = e(−tx2 , q 2 ),
and
v+1
kFq,v Gv (·, t, q 2 ) kq,r,v = c(r, q, v)t− r .
8. q−C ENTRAL L IMIT T HEOREM
In this section we study the analogoue of the well known central limit theorem with the aid
of the q−Bessel Fourier transform.
+
For this, we consider the set M+
q of positive and bounded measures on Rq . The q-cosine
Fourier transform of ξ ∈ M+
q is defined by
Z ∞
Fq,v (ξ)(x) =
jv (tx, q 2 )t2v+1 dq ξ(t).
0
The q−convolution product of two measures ξ, ρ ∈ M+
q is given by
Z ∞
v
ξ ∗q ρ(f ) =
Tq,x
f (t)t2v+1 dq ξ(x)dq ρ(t),
0
and we have
Fq,v (ξ ∗q ρ) = Fq,v (ξ)Fq,v (ρ).
We begin by showing the following result
Proposition 8.1. For f ∈ Aq,v and ξ ∈ M+
q , we have
Z ∞
Z ∞
2v+1
f (x)x
dq ξ(x) = cq,v
Fq,v (f )(x)Fq,v (ξ)(x)x2v+1 dq x.
0
0
As a direct consequence we may state
Corollary 8.2. Given ξ, ξ 0 ∈ M+
q such that
Fq,v (ξ) = Fq,v (ξ 0 ),
then ξ = ξ 0 .
Proof. By Proposition 8.1, we have
Z
Z ∞
2v+1
f (x)x
dq ξ(x) =
0
∞
f (x)x2v+1 dq ξ 0 (x),
from the assertion (2) of Proposition 7.5, we conclude that ξ = ξ 0 .
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
∀f ∈ Aq,v .
0
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12
L AZHAR D HAOUADI , A HMED F ITOUHI ,
AND
J. E L K AMEL
Theorem 8.3. Let (ξn )n≥0 be a sequence of probability measures of M+
q such that
lim Fq,v (ξn )(t) = ψ(t),
n→∞
then there exists ξ ∈ M+
q such that the sequence ξn converges strongly toward ξ, and
Fq,v (ξ) = ψ.
Proof. We consider the map In defined by
Z ∞
In (u) =
u(x)x2v+1 dq ξn (x),
∀f ∈ Cq,0 .
0
By the following inequality
|In (u)| ≤ kukCq,0 ,
and by Proposition 8.1, we get
Z ∞
In (f ) = cq,v
Fq,v (f )(x)Fq,v (ξn )(x)x2v+1 dq x,
∀f ∈ Aq,v ,
0
which implies
∞
Z
Fq,v (f )(x)ψ(x)x2v+1 dq x,
lim In (f ) =
n→∞
∀f ∈ Aq,v .
0
On the other hand, by assertion (2) of Proposition 7.5, and by the use of the Ascoli theorem (see
[11]):
Consider a sequence of equicontinuous linear forms on Cq,0 which converge on a dense part
Aq,v then converge on the entire Cq,0 . We get
Z ∞
Fq,v (u)(x)ψ(x)x2v+1 dq x, ∀u ∈ Cq,0 .
lim In (u) =
n→∞
0
M+
q
such that
Finally there exist ξ ∈
Z ∞
Z
2v+1
lim
u(x)x
dq ξn (x) =
n→∞
0
∞
u(x)x2v+1 dq ξ(x),
∀u ∈ Cq,0 .
0
On the other hand
Fq,v (Aq,v ) = Aq,v ,
and
Z
∞
Z
Fq,v (f )(x)Fq,v (ξ)(x)dq x =
∞
Fq,v (f )(x)ψ(x)x2v+1 dq x,
∀f ∈ Aq,v ,
0
0
which implies
Fq,v (ξ) = ψ.
Proposition 8.4. Given ξ ∈ M+
q , and supposing that
Z ∞
t2 t2v+1 dq ξ(t) < ∞,
σ=
0
then
Fq,v (ξ)(x) = 1 −
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
q2σ
x2 + o(x2 ).
(q 2 , q 2 )1 (q 2v+2 , q 2 )1
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I NEQUALITIES IN q−F OURIER A NALYSIS
13
Proof. We write
jv (tx, q 2 ) = 1 −
q 2 t2
x2 + x2 θ(tx)t2 ,
(q 2 , q 2 )1 (q 2v+2 , q 2 )1
where
lim θ(x) = 0,
x→0
then
q2σ
Fq,v (ξ)(x) = 1 − 2 2
x2 +
(q , q )1 (q 2v+2 , q 2 )1
Z
∞
2
2v+1
t θ(tx)t
dq ξ(t) x2 .
0
Now we are in a position to present the q−central limit theorem.
Theorem 8.5. Let (ξn )n≥0 be a sequence of probability measures of M+
q of total mass 1, satisfying
Z ∞
lim nσn = σ, where σn =
t2 t2v+1 dq ξn (t),
n→∞
0
and
Z
lim ne
σn = 0,
σ
en =
where
n→∞
0
∞
t4 2v+1
t
dq ξn (t),
1 + t2
then ξn∗n converge strongly toward a measure ξ defined by
2σ
− 2 2 q 2v+2
t2
2
dq ξ(x) = cq,v Fq,v e (q ,q )1 (q ,q )1
(x)dq x.
Proof. We have
Fq,v (ξn∗n ) = (Fq,v (ξn ))n ,
and
Fq,v (ξn )(x) = 1 −
q 2 σn
x2 + θn (x)x2 ,
(q 2 , q 2 )1 (q 2v+2 , q 2 )1
where
Z
θn (x) =
∞
t2 θ(tx)t2v+1 dq ξn (t).
0
Consequently
(Fq,v (ξn )) (x) = exp n log 1 −
n
q 2 σn
x2 + θn (x)x2
(q 2 , q 2 )1 (q 2v+2 , q 2 )1
.
By the following inequality
|t2 θ(tx)| ≤ Cx
t4
,
1 + t2
∀t ∈ R+
q ,
where Cx is some constant, the result follows immediately.
J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
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14
L AZHAR D HAOUADI , A HMED F ITOUHI ,
AND
J. E L K AMEL
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J. Inequal. Pure and Appl. Math., 7(5) Art. 171, 2006
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