Volume 8 (2007), Issue 4, Article 101, 5 pp.
ON SOME WEIGHTED MIXED NORM HARDY-TYPE INTEGRAL INEQUALITY
C.O. IMORU AND A.G. ADEAGBO-SHEIKH
D EPARTMENT OF M ATHEMATICS
O BAFEMI AWOLOWO U NIVERSITY,
I LE -I FE , N IGERIA
cimoru@oauife.edu.ng
adesheikh2000@yahoo.co.uk
Received 07 May, 2007; accepted 24 August, 2007
Communicated by B. Opić
A BSTRACT. In this paper, we establish a weighted mixed norm integral inequality of Hardy’s
type. This inequality features a free constant term and extends earlier results on weighted norm
Hardy-type inequalities. It contains, as special cases, some earlier inequalities established by the
authors and also provides an improvement over them.
Key words and phrases: Hardy-type inequality, Weighted norm.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In a recent paper [2], the authors proved the following result.
Theorem 1.1. Let g be continuous and non-decreasing on [a, b], 0 ≤ a ≤ b ≤ ∞ with g(x) >
0, x > 0, r 6= 1 and let f (x) be non-negative and Lebesgue-Stieltjes integrable with respect to
Rx
Rb
, r 6= 1. Then
g(x) on [a, b]. Suppose Fa (x) = a f (t)dg(t), Fb (x) = x f (t)dg(t) and δ = 1−r
p
Z
b
(1.1)
a
Z
(1.2)
a
149-07
b
1−p
g(x)δ−1 g(x)−δ − g(a)−δ
Fa (x)p dg(x) + K1 (p, δ, a, b)
p Z b
p
≤
g(x)δp−1 [g(x)f (x)]p dg(x),
r−1
a
1−p
g(x)δ−1 g(x)−δ − g(b)−δ
Fb (x)p dg(x) + K2 (p, δ, a, b)
p Z b
p
≤
g(x)δp−1 [g(x)f (x)]p dg(x),
1−r
a
r > 1,
r < 1,
2
C.O. I MORU AND A.G. A DEAGBO -S HEIKH
where
K1 (p, δ, a, b) =
1−p
p
Fa (b)p ,
g(b)δ g(b)−δ − g(a)−δ
r−1
δ < 0, i.e. r > 1
K2 (p, δ, a, b) =
1−p
p
g(a)δ g(a)−δ − g(b)−δ
Fb (a)p ,
1−r
δ > 0, i.e. r < 1.
and
The above result generalizes Imoru [1] and therefore Shum [3]. The purpose of the present
work is to obtain a weighted norm Hardy-type inequality involving mixed norms which contains
the above result as a special case and also provides an improvement over it.
2. M AIN R ESULT
The main result of this paper is the following theorem:
Theorem 2.1. Let g be a continuous function which is non-decreasing on [a, b], 0 ≤ a ≤ b <
∞, with g(x) > 0 for x > 0. Suppose that q ≥ p ≥ 1 and f (x) is non-negative and LebesgueStieltjes integrable with respect to g(x) on [a, b]. Let
Z x
Z x
(2.1)
Fa (x) =
f (t)dg(t), θa (x) =
g(t)(p−1)(1+δ) f (t)p dg(t),
a
Z
Fb (x) =
(2.2)
a
b
Z
f (t)dg(t), θb (x) =
x
and δ =
1−r
,r
p
Z
g(t)(p−1)(1+δ) f (t)p dg(t)
x
6= 1. Then if r > 1, i.e. δ < 0,
b
g(x)
(2.3)
b
δq
−1
p
1q
q
(p−1)
Faq (x)dg(x) + A1 (p, q, a, b, δ)
g(x)−δ − g(a)−δ p
a
Z
b
δp−1
≤ C1 (p, q, δ)
g(x)
p1
[g(x)f (x)] dg(x) ,
p
a
and for r < 1, i.e. δ > 0,
Z
b
g(x)
(2.4)
δq
−1
p
1q
q
−δ
−δ p (p−1) q
g(x) − g(b)
Fb (x)dg(x) + A2 (p, q, a, b, δ)
a
Z
≤ C2 (p, q, δ)
b
p1
g(x)δp−1 [g(x)f (x)]p dg(x) ,
a
where
q
δq
q
p
(−δ) p (1−p)−1 g(b) p θa (b) p ,
q
1q
q
p
(1−p)−1
C1 (p, q, δ) = (−δ) p
,
q
δq
q
p q
A2 (p, q, a, b, δ) = (δ) p (1−p)−1 g(a) p θb (a) p ,
q
1
p pq (1−p)−1 q
C2 (p, q, δ) = δ
.
q
A1 (p, q, a, b, δ) =
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 101, 5 pp.
δ < 0,
δ>0
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O N S OME W EIGHTED M IXED N ORM H ARDY-T YPE I NTEGRAL I NEQUALITY
3
Proof. For the proof of Theorem 2.1 we will use the following adaptations of Jensen’s inequality
for convex functions,
Z x
1− p1 Z x
p1
Z x
1
1
(2.5)
h(x, t) pq dλ(t) ≤
dλ(t)
h(x, t) q dλ(t)
a
a
a
and
Z
b
Z
1
pq
1− p1 Z b
p1
1
dλ(t)
h(x, t) q dλ(t) ,
b
h(x, t) dλ(t) ≤
(2.6)
x
x
x
where h(x, t) ≥ 0 for x ≥ 0, t ≥ 0, λ is non-decreasing and q ≥ p ≥ 1.
Let
dλ(t) = g(t)−(1+δ) dg(t),
h(x, t) = g(x)δq g(t)pq(1+δ) f (t)pq ,
(2.7)
q
q
∆q1 = (−δ) p (1−p) , if δ < 0 and ∆q2 = (δ) p (1−p) , if δ > 0.
Using (2.7) in (2.5), we get
Z x
δ
p
g(x)
f (t)dg(t)
a
≤ (−δ)
1
(1−p)
p
1 (p−1)
δ
g(x)−δ − g(a)−δ p
g(x) p
x
Z
(p−1)(1+δ)
g(t)
p1
f (t) dg(t) .
p
a
Raising both sides of the above inequalities to power q and using (2.1), we obtain
δq
q
δq
q
g(x) p Fa (x)q ≤ ∆q1 ga (x) p (p−1) g(x) p θa (x) p ,
where ga (x) = g(x)−δ − g(a)−δ .
Integrating over (a, b) with respect to g(x)−1 dg(x) gives
Z b
Z b
δq
q
δq
q
q
−1
(1−p)
q
p
p
(2.8)
g(x)
ga (x)
Fa (x) dg(x) ≤ ∆1
g(x) p −1 θa (x) p dg(x) = J.
a
a
Now integrate the right side of (2.8) by parts to obtain
Z b
q
δq
q
J = ∆1
g(x) p −1 θa (x) p dg(x)
=
a
q
∆1
(δq/p)
q
δq
g(x) p θa (x) p |ba + (−δ −1 )∆q1
Z
×
b
q
δq
g(x) p g(x)(p−1)(1+δ) f (x)p θa (x) p −1 dg(x).
a
However,
Z
b
q
δq
g(x) p g(x)(p−1)(1+δ) f (x)p θap
I=
−1
(x)dg(x)
a
Z
b
δq
p
(p−1)(1+δ)
g(x) g(x)
=
p
Z
f (x)
=
δp+p−1−δ
g(t)
pq −1
f (t) dg(t)
dg(x)
p
a
a
Z
x
b
δp+p−1
g(x)
Z
δ
f (x) g(x)
p
a
x
δp+p−1−δ
g(t)
pq −1
dg(x).
f (t) dg(t)
p
a
Since δ < 0, we have g(x)−δ ≥ g(t)−δ ∀t ∈ [a, x].
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 101, 5 pp.
http://jipam.vu.edu.au/
4
C.O. I MORU AND A.G. A DEAGBO -S HEIKH
Consequently
b
Z
δp+p−1
I≤
g(x)
Z
p
f (t)
δp+p−1
g(t)
a
pq −1
f (t) dg(t)
dg(x)
p
a
Z b Z
x
=
a
p
=
q
x
pq −1
g(t)δp+p−1 f (t)p dg(t)
g(x)δp+p−1 f (t)p dg(x)
a
Z
b
pq
g(x)δp−1 [f (x)g(x)]p dg(x) .
a
Thus (2.8) becomes
Z b
q (1−p)
δq
δq
q
p
(2.9)
g(x) p −1 g(x)−δ − g(a)−δ p
Fa (x)q dg(x) + ∆q1 (−δ −1 )g(b) p θa (b) p
q
a
Z b
pq
p
≤ (−δ −1 )∆q1
g(x)δp−1 [f (x)g(x)]p dg(x) .
q
a
Taking the q th root of both sides yields assertion (2.3) of the theorem.
To prove (2.4), we start with inequality (2.6) and use (2.7) with (2.2) to obtain
δ
1
1
δ
1
g(x) p Fb (x) ≤ (−δ) p (1−p) gb (x) p (p−1) g(x) p θb (x) p
1
1
δ
1
= (δ −1 ) p (p−1) (−gb (x)) p (p−1) g(x) p θb (x) p ,
where gb (x) = g(b)−δ − g(x)−δ .
On rearranging and raising to power q and then integrating both sides over [a, b] with respect
to g(x)−1 dg(x), we obtain
Z b
Z b
q
δq
δq
q
−1
−δ
−δ p (1−p)
q
p
(2.10)
g(x)
g(x) − g(b)
Fb (x) dg(x) ≤ ∆2
g(x) p −1 θb (x)dg(x).
a
a
We denote the right side of (2.10) by H, integrate it by parts and use the fact that for δ > 0,
g(x)δ ≤ g(t)δ ∀t ∈ [x, b] to obtain
Z b
δp
q
p q
b
−1 q
q
p
H ≤ ∆2 g(x) θb (x) |a + (δq/p) ∆2
g(x)δp−1 [f (x)g(x)]p dg(x).
δq
a
Using this in (2.10) we obtain
Z b
q (1−p)
δq
q
q
p
(2.11)
g(x) p −1 g(x)−δ − g(b)−δ p
Fb (x)q dg(x) + δ −1 ∆q2 g(a) p θb (a) p
q
a
Z b
pq
p −1 q
≤ δ ∆2
g(x)δp−1 [f (x)g(x)]p dg(x) .
q
a
We take the q th root of both sides to obtain assertion (2.4) of the theorem.
Remark 2.2. Let p = q, and δ =
Z
(2.12)
a
b
1−r
p
< 0, i.e., r > 1, then (2.3) reduces to
p−1
g(x)δ−1 g(x)−δ − g(a)−δ
Fa (x)p dg(x) + A1 (p, p, a, b, δ)
Z b
p
δp−1
≤ C1 (p, p, δ)
g(x)
[f (x)g(x)] dg(x) ,
a
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O N S OME W EIGHTED M IXED N ORM H ARDY-T YPE I NTEGRAL I NEQUALITY
5
where
(2.13)
A1 (p, p, a, b, δ) = (−δ)−p g(b)δ θa (b),
δ<0
and
(2.14)
−p
C1 (p, p, δ) = (−δ)
p
=
r−1
p
.
Now from (2.18) in [2] we have that, for δ < 0
(2.15)
1−p
g(b)δ θa (b) ≥ (−δ −1 )1−p g(b)δ g(b)−δ − g(a)−δ
Fa (b)p .
Thus, from (2.13) and (2.15), using notations in (1.1), we have
(2.16)
A1 (p, p, a, b, δ) = (−δ)−p g(b)δ θa (b)
1−p
≥ (−δ)−p (−δ −1 )1−p g(b)δ g(b)−δ − g(a)−δ
Fa (b)p
1−p
= (−δ)−1 g(b)δ g(b)−δ − g(a)−δ
Fa (b)p
1−p
p
=
g(b)δ g(b)−δ − g(a)−δ
Fa (b)p
r−1
= K1 (p, δ, a, b),
i.e., A1 (p, p, a, b, δ) = K1 (p, δ, a, b) + B1 for some B1 ≥ 0.
Thus we can write (2.12), using (2.14), as
Z b
p−1
(2.17)
g(x)δ−1 g(x)−δ − g(a)−δ
Fa (x)p dg(x) + K1 (p, δ, a, b) + B1
a
p Z b
p
p
δp−1
g(x)
[f (x)g(x)] dg(x) .
≤
r−1
a
So, when B1 = 0, (2.17) reduces to (1.1). When B1 6= 0, i.e., B1 > 0, (2.17) is an improvement
of (1.1). Similarly with notations in (1.2) and (2.4) in this paper we use (2.19) in [2] to prove
that
A2 (p, p, a, b, δ) = K2 (p, δ, a, b) + B2
for some B2 ≥ 0.
Thus, when p = q, (2.4) reduces to (1.2) if B2 = 0 and is an improvement of (1.2) when B2 6= 0,
i. e., when B2 > 0.
R EFERENCES
[1] C.O. IMORU, On some integral inequalities related to Hardy’s, Canadian Math. Bull., 20(3) (1977),
307–312.
[2] C.O. IMORU AND A.G. ADEAGBO-SHEIKH, On an integral inequality of the Hardy-type, (accepted) Austral. J. Math. Anal. and Applics.
[3] D.T. SHUM, On integral inequalities related to Hardy’s, Canadian Math. Bull., 14(2) (1971), 225–
230.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 101, 5 pp.
http://jipam.vu.edu.au/