International Journal of Analysis and Applications

Volume 17, Number 2 (2019), 244-259

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-17-2019-244

HARDY-COPSON TYPE INEQUALITIES ON TIME SCALES FOR THE FUNCTIONS

OF “n” INDEPENDENT VARIABLES

M. SHAHZAD ASHRAF1,∗, KHURAM ALI KHAN2 AND AMMARA NOSHEEN1

1Department of Mathematics, The University of Lahore (Sargodha Campus), Pakistan

2Department of Mathematics, University of Sargodha, Sargodha, Pakistan

∗Corresponding author: shahzadashraf30@icloud.com

Abstract. The paper consists of some extensions in Hardy and Copson type inequalities on time scales.

The main results are proved by using induction principle, Rules to find derivatives for composition of two

functions, Hölder’s inequality and Fubini’s theorem in time scales settings. The related results and examples

are also investigated in seek of applications.

1. Introduction

Advancements in inequalities had started since the end of 19th century, which laid the theocratical foun-

dations of approximation methods. Generally it is accepted that the classic work “Inequalities [7]”, which

appeared in 1934, gave a systematic discipline to a collection of isolated formulas. This work had played a

role of a strong stem in continuous growth to modern field of Inequalities.

Many researchers had paid attention to these inequalities since invention of these inequalities. A large num-

ber of papers related with extensions, new results, and generalizations can be seen on the topic [5, 8, 10].

Copson in [6] extend weighted Hardy inequality in the following result:

Received 2018-08-04; accepted 2018-10-24; published 2019-03-01.

2010 Mathematics Subject Classification. 26D15, 39A13, 34N05.

Key words and phrases. time Scales; Hardy-Copson inequality; Keller’s chain rule.

c©2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

244

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-244


Int. J. Anal. Appl. 17 (2) (2019) 245

If % > 1, A(l) =

l∑
i=1

a(i)ω(i)

Ω(i)
, then

∞∑
l=1

ω(l)(A(l))% 6 p%
∞∑
l=1

ω(l)a(l)(A(l))%−1 6 p%
∞∑
l=1

ω(l)a%(l). (1.1)

In [12] R. P. Agarwal et al., proved some dynamic Hardy inequalities via time scales calculus. Those

inequalities as special cases contain some integral and discrete inequalities due to Hardy and Copson. One

of their results is stated as:

“Consider T is a time scale with a ∈ [0,∞)T, % >1, ω and g be non-negative functions.

Further assume that

Ω(ς) =

∫ ς
a

ω(r)∆r, Ω(∞) = ∞, ψ(ς) :=
∫ ς
a

ω(r)g(r)∆r, ∀ ς ∈ [a,∞)T

and

∫ ∞
a

(
Ωσ(ς)

Ω(ς)

)%(%−1)
ω(ς)g%(ς)∆ς exists. Then

∫ ∞
a

ω(ς)

(Ωσ(ς))%
(ψσ(ς))%∆ς 6

%

%− 1

∫ ∞
a

ω(ς)g(ς)

Ω%−1(ς)
ψ%−1(ς)∆ς

and ∫ ∞
a

ω(ς)

(Ωσ(ς))%
(ψσ(ς))%∆ς 6

(
%

%− 1

)%∫ ∞
a

(
Ωσ(ς)

Ω(ς)

)p(p−1)
ω(ς)g%(ς)∆ς.”

Taking into account the worth of inequalities, we are motivated to extend the results of [12] for functions of

several variables.

2. Preliminaries

A closed non-empty subset of real numbers is called a time scale. Notation to be used for a time scale

is T. R, Z, N, N0, [a,b] are examples of time scales whereas rationals, irrationals and (a,b) are not closed,

therefore not time scales. Few basic concepts related to time scales theory, are as under (see [2–4]):

Definition 2.1. Let T denotes a time scale. For ς ∈ T, the forward jump operator σ : T → T is defined as

σ(ς) := inf {z ∈ T; z > ς} ,

and the backward jump operator ρ : T → T is defined as

ρ(ς) := sup{z ∈ T; z < ς} .

The point ς ∈ T satisfying σ(ς) > ς is right-scattered whereas the point ς satisfying ρ(ς) < ς is left-

scattered. Points which are simultaneously left and right scattered are isolated. Also, the point ς is called

right-dense if ς < sup T and σ(ς) = ς and is called left-dense if ς > inf T and ρ(ς) = ς. The points that are

left and right dense simultaneously are dense points.



Int. J. Anal. Appl. 17 (2) (2019) 246

Definition 2.2. The Graininess function µ : T → [0,∞) is defined by,

µ(ς) := σ(ς) − ς.

Definition 2.3. Suppose a function g : T → R is satisfying:

(a) g is continuous at all right dense ς ∈ T,

(b) the left hand limits exist and finite at all left dense ς ∈ T

then, g is called right-dense continuous (rd-continuous) in T. The set denoted by Crd(T) contains all rd-

continuous functions.

Remark 2.1. If we exchange the role of left dense points and right dense points in definition of rd-continuous

function, we get left dense continuous functions. If function is continuous with respect to both sided dense

points, it is continuous function ∀t ∈ T.

Definition 2.4. Let g : T → R and ς ∈ T, if g∆(ς) exists with the condition that, for � > 0, there exists a

neighborhood O of ς such that

|[g(σ(ς)) −g(s)] −g∆(ς)(σ(ς) −s)| ≤ �|σ(ς) −s|,

for all s ∈ O, then g is differentiable at ς and is denoted by g∆(ς).

Theorem 2.1. Assume delta derivatives of g1,g2 : T → R exist at ς ∈ T. Then derivative of the product

g1g2 : T → R exists at ς ∈ T and satisfies

(g1g2)
∆(ς) = g∆1 (ς)g2(ς) + g

σ
1 (ς)g

∆
2 (ς) = g1(ς)g

∆
2 (ς) + g

∆
1 (ς)g

σ
2 (ς). (2.1)

Definition 2.5. Consider g is rd-continuous function. Then for ς0 ∈ T, the function G defined by

G :=

∫ ς
ς0

g(ς)∆ς, for ς ∈ T

is called the antiderivative of g.

Definition 2.6. If α ∈ T, sup T = ∞, and g1 is rd-continuous on the interval [α,∞), then∫ ∞
α

g1(ς)∆ς = lim
β→∞

∫ β
α

g1(ς)∆ς.

Existence of limit gives convergence of the integral elsewise it diverges.

Theorem 2.2. If α,β,γ ∈ T, c ∈ R and u1,u2 ∈ Crd, then

(i)
∫β
α
uσ1 (ς)u

∆
2 (ς)∆ς = |u1u2(ς)|βα −

∫β
α
u∆1 (ς)u2(ς)∆ς,

(ii)
∫β
α
u1(ς)u

∆
2 (ς)∆ς = |u1u2(ς)|βα −

∫β
α
u∆1 (ς)u

σ
2 (ς)∆ς.

These are known as integration by parts formulae.



Int. J. Anal. Appl. 17 (2) (2019) 247

Theorem 2.3 (Chain rule 1). Let u2 : R → R be continuous, u2 : T → R is delta differentiable on Tκ and

suppose that u1 : R → R is continuously differentiable. Then there exists c ∈ [ς,σ(ς)] with

(u1 ◦u2)∆(ς) = u′1(u2(c))u
∆
2 (ς). (2.2)

Theorem 2.4 (Chain rule 2). Assume u1 : R → R is continuously differentiable and suppose u2 : T → R is

delta differentiable. Then u1 ◦u2 : T → R is delta differentiable and

(u1 ◦u2)∆(ς) =
{∫ 1

0

u′1(u2(ς) + hµ(ς)g
∆
2 (ς))dh

}
u∆2 (ς) (2.3)

holds.

Chain rule 2 is given by C. Pötzsche [11], likewise it also appeared in S. Keller’s paper [9].

A simple consequence of Theorem 2.4 is given below:

(uω1 (ς))
∆(ς) = ω

∫ 1
0

{huσ1 + (1 −h)u1(ς)}
ω−1

dhu∆1 (ς). (2.4)

Theorem 2.5. (Hölder’s inequality) Let α,β ∈ T, for rd-continuous functions g1,g2 : [α,β] → R, we have

∫ β
α

|g1(ς)g2(ς)|∆ς ≤

[∫ β
α

|g1(ς)|q∆ς

]1
q
[∫ β

α

|g2(ς)|p∆ς

]1
p

, (2.5)

where p > 1 and q = p/(p− 1).

Next, we present Fubini’s Theorem on time scales [1].

Theorem 2.6. Let T1,T2 be two time scales. Suppose S : T1 × T2 → R is integrable with respect to both

time scales. Define Φ(y2) =
∫
T1
S(y1,y2)∆y1 for a.e y2 ∈ Λ and Ψ(y1) =

∫
T2
S(y1,y2)∆y2 for a.e y1 ∈ T1.

Then Φ and Ψ are both differentiable in time scales settings and∫
T1

∆y1

∫
T2
S(y1,y2)∆y2 =

∫
T2

∆y2

∫
T1
S(y1,y2)∆y1. (2.6)

In the sequel, we denote [a,b)T = [a,b) ∩ T, where T be any time scale. Also partial derivatives for

i ∈ {1, . . . ,n} are denoted by ∂
∆ςi

g(ς1, . . . , ςn) = g
∆i (ς1, . . . , ςn). We also assume that the functions are

nonnegative, rd-continuous and the integrals considered are assumed to exist.

3. Dynamic Hardy & Copson-Type Inequalities via Time Scales for Functions of n

Independent Variables

Theorem 3.1. Let T1, . . . ,Tn denote time scales. For p > 1 and i ∈ {1, . . . ,n}, consider ai ∈ [0,∞)Ti

and g : T1 × . . . × Tn → R+. Let λi : Ti → R+ be such that Λi(ςi) :=
∫ ςi
ai
λi(si)∆si and ψ(ς1, . . . , ςn) =



Int. J. Anal. Appl. 17 (2) (2019) 248

∫ ς1
a1
. . .
∫ ςn
an

∏n
i=1 λi(si)g(s1, . . . ,sn)∆s1 . . . ∆sn.

Assume Λi(∞) = 0 and the delta integrals∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

Λσii (ςi)

[Λi(ςi)]p
λi(ςi)[g(ς1, . . . , ςn)]

p∆ς1 . . . ∆ςn exist. (3.1)

Then

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)

[Λσii (ςi)]
p

[ψ(σ1(ς1), . . . ,σn(ςn))]
p∆ς1 . . . ∆ςn

≤
(

p

p− 1

)np ∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

(
Λσii (ςi)

Λi(ςi)

)p(p−1)
λi(ςi)g

p(ς1, . . . , ςn)∆ς1 . . . ∆ςn, (3.2)

where n is any positive integer.

Proof. We prove the result by using Principle of mathematical induction.

For n = 1, statement is true by [12, Theorem 2.1]. Assume for 1 ≤ n ≤ k (3.2) holds. To prove the result

for n = k + 1, Take left hand side of (3.2) in the following form

∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

[Λσii (ςi)]
p

{∫ ∞
ak+1

λk+1(ςk+1)

[Λ
σk+1
k+1 (ςk+1)]

p
[ψ(σ1(ς1), . . . ,σk+1(ςk+1))]

p∆ςk+1

}
∆ς1 . . . ∆ςk. (3.3)

Consider

Ik+1 =

∫ ∞
ak+1

λk+1(ςk+1)

[Λ
σk+1
k+1 (ςk+1)]

p
[ψ(σ1(ς1), . . . ,σk+1(ςk+1))]

p∆ςk+1. (3.4)

Apply integration by parts (Theorem 2.2 (i)) on Ik+1 w.r.t ςk+1 ∈ [ak+1,∞) to get

Ik+1 = |uk+1(ςk+1)ψp(σ1(ς1), . . . ,σk(ςk), ςk+1)|∞ak+1

+

∫ ∞
ak+1

−uk+1(ςk+1)[ψp(σ1(ς1), . . . ,σk(ςk), ςk+1)]∆k+1 ∆ςk+1, (3.5)

where

−uk+1(ςk+1) =
∫ ∞
ςk+1

λk+1(sk+1)

[Λ
σk+1
k+1 (sk+1)]

p
∆sk+1 ≤

(
−1
p− 1

)∫ ∞
ςk+1

[Λ
1−p
k+1(sk+1)]

∆k+1 ∆sk+1.

∴−uk+1(ςk+1) ≤
(

1

1 −p

)(
1

Λ
p−1
k+1(ςk+1)

)
. (3.6)

By chain rule (2.4) and for dk+1 ∈ [ςk+1,σk+1(ςk+1)],

[ψp(σ1(ς1), . . . ,σk(ςk), ςk+1)]
∆k+1

= pψp−1(σ1(ς1), . . . ,σk(ςk),dk+1)ψ
∆k+1 (σ1(ς1), . . . ,σk(ςk), ςk+1). (3.7)



Int. J. Anal. Appl. 17 (2) (2019) 249

Since

ψ∆k+1 (σ1(ς1), . . . ,σk(ςk), ςk+1) =
∂

∆ςk+1
ψ(σ1(ς1), . . . ,σk(ςk), ςk+1)

=

∫ σ1(ς1)
a1

. . .

∫ σk(ςk)
ak

k∏
i=1

λi(si)×

{
∂

∆ςk+1

∫ ςk+1
ak+1

λk+1(sk+1)g(s1, . . . ,sk,sk+1)∆sk+1

}
∆s1 . . . ∆sk, (3.8)

also

∂

∆ςk+1

∫ ςk+1
ak+1

λk+1(sk+1)g(σ1(s1), . . . ,σk(sk),sk+1)∆sk+1 ≥ 0

and σk+1(ςk+1) ≥ dk+1. Therefore (3.7) implies

ψ∆k+1 (σ1(ς1), . . . ,σk(ςk), ςk+1)

= λk+1(ςk+1)

∫ σ1(ς1)
a1

. . .

∫ σk(ςk)
ak

k∏
i=1

λi(si)g(s1, . . . ,sk, ςk+1)∆s1 . . . ∆sk.

Thus

ψ∆k+1 (σ1(ς1), . . . ,σk(ςk), ςk+1) = λk+1(ςk+1)ψk(σ1(ς1), . . . ,σk(ςk), ςk+1), (3.9)

where

ψk(σ1(ς1), . . . ,σk(ςk), ςk+1) =

∫ σ1(ς1)
a1

. . .

∫ σk(ςk)
ak

k∏
i=1

λi(si)g(s1, . . . ,sk, ςk+1)∆s1 . . . ∆sk.

Use (3.9) in (3.7) to get,

[ψp(σ1(ς1), . . . ,σk(ςk), ςk+1)]
∆k+1

= pψp−1(σ1(ς1), . . . ,σk(ςk),dk+1)λk+1(ςk+1)ψk(σ1(ς1), . . . ,σk(ςk), ςk+1)

≤ pψp−1(σ1(ς1), . . . ,σk(ςk), ςk+1)λk+1(ςk+1)ψk(σ1(ς1), . . . ,σk(ςk), ςk+1). (3.10)

∵ ψ(ς1, ς2, . . . ,ak+1) = 0 and uk+1(∞) = 0, (3.5) becomes

Ik+1 =

∫ ∞
ak+1

−uk+1(ςk+1)[ψp(σ1(ς1), . . . ,σk(ςk), ςk+1)]∆k+1 ∆ςk+1. (3.11)

Use (3.6) and (3.10) in (3.11) to get

Ik+1 ≤
(

p

p− 1

)∫ ∞
ak+1

λk+1(ςk+1)ψk(σ1(ς1), . . . ,σk(ςk), ςk+1)

Λ
p−1
k+1(ςk+1)

[ψ(σ1(ς1), . . . ,σk+1(ςk+1))]
p−1∆ςk+1. (3.12)



Int. J. Anal. Appl. 17 (2) (2019) 250

Multiply and divide by

(
λk+1(ςk+1)

[Λ
σk+1
k+1 (ςk+1)]

p

)p−1
p

on right hand side of (3.12) and then apply Hölder’s inequality

to get

Ik+1 ≤
(

p

p− 1

)∫ ∞
ak+1


λk+1(ςk+1)Λ1−pk+1(ςk+1)ψk(σ1(ς1), . . . ,σk(ςk), ςk+1)[

[Λ
σk+1
k+1 (ςk+1)]

−pλk+1ςk+1
]p−1

p


p ∆ςk+1




1
p

×

(∫ ∞
ak+1

λk+1(ςk+1)

[Λ
σk+1
k+1 (ςk+1)]

p
[ψ(σ1(ς1), . . . ,σk+1(ςk+1))]

p∆ςk+1

)p−1
p

. (3.13)

Divide both sides by right most term and take power p on both sides. After simplification, we get

Ik+1

≤
(

p

p− 1

)p ∫ ∞
ak+1

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p(p−1)
λk+1(ςk+1) ψ

p
k(σ1(ς1), . . . ,σk(ςk), ςk+1)∆ςk+1. (3.14)

Substitute (3.14) in (3.3) to get

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)

[Λσii (ςi)]
p

[ψ(σ1(ς1), . . . ,σk(ςk),σk+1(ςk+1))]
p∆ς1 . . . ∆ςk∆ςk+1

≤
(

p

p− 1

)p ∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

[Λσii (ςi)]
p

×



∫ ∞
ak+1

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p(p−1)
λk+1(ςk+1)ψ

p
k(σ1(ς1), . . . ,σk(ςk), ςk+1)∆ςk+1


 ∆ς1 . . . ∆ςk. (3.15)

Use Theorem 2.6 “k times” on right hand side of (3.15) to get

∫ ∞
ak+1

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p(p−1)
λk+1(ςk+1)

×

{∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

[Λσii (ςi)]
p
ψ
p
k(σ1(ς1), . . . ,σk(ςk), ςk+1)∆ς1 . . . ∆ςk

}
∆ςk+1. (3.16)

By using induction hypothesis for ψkσ1(ς1), . . . ,σk(ςk), ςk+1) (instead for ψk(σ1(ς1), . . . ,σk(ςk)), ) with fix

tk+1 ∈ Tk+1, in (3.16) and use Fubini’s Theorem (Theorem 2.6) “k times” to get

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)

[Λσii (ςi)]
p

[ψ(σ1(ς1), . . . ,σk+1(ςk+1))]
p∆ς1 . . . ∆ςk+1

≤
(

p

p− 1

)(k+1)p ∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

(
Λσii (ςi)

Λi(ςi)

)p(p−1)
λi(ςi)g

p(ς1, . . . , ςk+1)∆ς1 . . . ∆ςk+1.

Thus by principle of mathematical induction, inequality (3.2) holds for all n ∈ Z+, which completes the

proof. �



Int. J. Anal. Appl. 17 (2) (2019) 251

If we assume

inf
ςi∈[ai,∞)Ti

Λi(ςi)

Λσii (ςi)
= Li > 0 ∀ i ∈{1, . . . ,n}, (3.17)

in Theorem 3.1 (in particular in (3.2)), we obtain the following result.

Corollary 3.1. For p > 1 and i ∈ {1, . . . ,n}, consider Ti be time scales, ai ∈ [0,∞)Ti , λi : Ti → R+ and

g : T1 × . . .×Tn → R+. Let Λi(ςi) and ψ(ς1, . . . , ςn) be defined as in Theorem 3.1 such that (3.17) holds and

the delta integrals ∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

Λσii (ςi)

[Λi(ςi)]p
λi(ςi)[g(ς1, ς2, . . . , ςn)]

p∆ς1∆ς2 . . . ∆ςn exist.

Then (3.2) takes the form

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)

[Λσii (ςi)]
p

[ψ(σ1(ς1), . . . ,σn(ςn))]
p∆ς1 . . . ∆ςn

≤
(

p

p− 1

)np n∏
i=1

(Li)
p(1−p)

∫ ∞
a1

. . .

∫ ∞
an

λi(ςi)g
p(ς1, . . . , ςn)∆ς1 . . . ∆ςn, (3.18)

where “n” is any positive integer.

Remark 3.1. As a special case of Corollary 3.1 when Ti = R, ∀ i ∈ {1, . . . ,n}, generalization of integral

Hardy inequality for the functions of n independent variables (note that when Ti = R, we have σi(ςi) = ςi

and Li = 1) takes the form:

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)

[Λi(ςi)]p

(∫ ς1
a1

. . .

∫ ςn
an

[

n∏
i=1

λi(si)g(s1, . . . ,sn)dsn . . .ds1]
p

)
dςn . . .dς1

≤
(

p

p− 1

)np ∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)g
p(ς1, . . . , ςn)dtn . . .dς1. (3.19)

As a special case of this inequality when λi(ςi) = 1, ∀ i = {1, . . . ,n}, we have

∫ ∞
a1

. . .

∫ ∞
an

1

[
∏n
i=1(ςi −ai)]p

(∫ ς1
a1

. . .

∫ ςn
an

[g(s1, . . . ,sn)dsn . . .ds1]
p

)
dςn . . .dς1

≤
(

p

p− 1

)np ∫ ∞
a1

. . .

∫ ∞
an

gp(ς1, . . . , ςn)dςn . . .dς1, (3.20)

where p > 1.

Remark 3.2. For p > 1 and ∀ i ∈ {1, 2, . . . ,n}, assume that Ti = N with ai = 1 in Corollary 3.1.

Furthermore assume that
∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)g
p(l1, . . . , ln)



Int. J. Anal. Appl. 17 (2) (2019) 252

is convergent. In this case (3.18) becomes the following discrete Copson type inequality for the functions of

“n” independent parameters:

∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)(∑l1
s1=1

λ1(s1) . . .
∑ln
sn=1

λn(sn)
)p
(

l1∑
s1=1

. . .

ln∑
sn=1

λ1(s1) . . .λn(sn)g(s1, . . . ,sn)

)p

≤
(

p

p− 1

)kp n∏
i=1

(Li)
p(1−p)

∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)g
p(l1, . . . , ln). (3.21)

Theorem 3.2. Let T1, . . . ,Tn denote time scales. For p > 1 and i ∈{1, . . . ,n}, consider ai ∈ [0,∞)Ti and

g : T1 × . . .×Tn → R+. Let λi : Ti → R+, Λi(ςi) :=
∫ ςi
ai
λi(si)∆si, Λi(∞) = 0 and the delta integrals∫ ∞

a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)[g(ς1, . . . , ςn)]
p∆ς1 . . . ∆ςn exist. (3.22)

Assume for any ςi ∈ [ai,∞)Ti ,

φ(ς1, . . . , ςn) :=

∫ ∞
ς1

. . .

∫ ∞
ςn

n∏
i=1

(
λi(si)

Λσii (si)

)
g(s1, . . . ,sn)∆s1 . . . ∆sn. (3.23)

Then

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)[φ(ς1, . . . , ςn)]
p∆ς1 . . . ∆ςn

≤ (p)np
∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)g
p(ς1, . . . , ςn)∆ς1 . . . ∆ςn, (3.24)

where n is any positive integer.

Proof. We prove the result by using Principle of mathematical induction.

For n = 1, statement is true by [12, Theorem 2.5]. Assume for 1 ≤ n ≤ k (3.24) holds. To prove result for

n = k + 1, Take left hand side of (3.24) int the following form

∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

{∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk, ςk+1)]
p∆ςk+1

}
∆ς1 . . . ∆ςk. (3.25)

Consider

Ik+1 =

∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1. (3.26)

Apply integration by parts (Theorem 2.2 (i)) on Ik+1 w.r.t ςk+1 ∈ [ak+1,∞) to get

Ik+1 = |φp(ς1, . . . , ςk+1)Λk+1(ςk+1)|∞ak+1 +
∫ ∞
ak+1

[−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

pΛ
σk+1
k+1 (ςk+1)]∆ςk+1. (3.27)

Since φ(ς1, . . . , ςk,∞) = 0 and Λk+1(ak+1) = 0, (3.27) becomes

Ik+1 =

∫ ∞
ak+1

[−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

pΛ
σk+1
k+1 (ςk+1)]∆ςk+1. (3.28)



Int. J. Anal. Appl. 17 (2) (2019) 253

Apply chain rule (2.2) to find upper bound of −
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

p, for dk+1 ∈

[ςk+1,σk+1(ςk+1)], we get

−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

p = −pφp−1(ς1, . . . , ςk,dk+1)
∂

∆ςk+1
φ(ς1, . . . , ςk, ςk+1). (3.29)

Also

∂

∆ςk+1
φ(ς1, . . . , ςk+1) =

∂

∆ςk+1

∫ ∞
ς1

. . .

∫ ∞
ςk+1

k+1∏
i=1

(
λi(si)

Λσii (si)

)
g(s1, . . . ,sk+1)∆s1 . . . ∆sk+1

=

∫ ς1
a1

. . .

∫ ςk
ak

k∏
i=1

(
λi(si)

Λσii (si)

){
∂

∆ςk+1

∫ ∞
ςk+1

λk+1(sk+1)

Λ
σk+1
k+1 (sk+1)

g(s1, . . . ,sk+1)∆sk+1

}
∆s1 . . . ∆sk

=

∫ ς1
a1

. . .

∫ ςk
ak

k∏
i=1

(
λi(si)

Λσii (si)

){
−λk+1(ςk+1)
Λ
σk+1
k+1 (ςk+1)

g(s1, . . . ,sk, ςk+1)

}
∆s1 . . . ∆sk

=

(
−λk+1(ςk+1)
Λ
σk+1
k+1 (ςk+1)

)∫ ς1
a1

. . .

∫ ςk
ak

k∏
i=1

(
λi(si)

Λσii (si)

)
g(s1, . . . ,sk, ςk+1)∆s1 . . . ∆sk. (3.30)

Since dk+1 ≥ ςk+1, use (3.30) in (3.29) to get,

−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

pΛ
σk+1
k+1 (ςk+1) ≤ pλk+1(ςk+1)φ

p−1(ς1, . . . , ςk+1)φk(ς1, . . . , ςk), (3.31)

where

φk(ς1, . . . , ςk) =

∫ ς1
a1

. . .

∫ ςk
ak

k∏
i=1

(
λi(si)

Λσii (si)

)
g(s1, . . . ,sk, ςk+1)∆s1 . . . ∆sk.

Then (3.28) becomes,

Ik+1 ≤ p
∫ ∞
ak+1

λk+1(ςk+1)φ
p−1(ς1, . . . , ςk+1)φk(ς1, . . . , ςk). (3.32)

Multiply and divide by [λk+1(ςk+1)]
p−1
p on R.H.S of (3.32) then apply Hölder’s inequality on R.H.S, we get

Ik+1 ≤ p

(∫ ∞
ak+1

λk+1(ςk+1)[φk(ς1, . . . , ςk, ςk+1)]
p∆ςk+1

)1
p

×

(∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1

)p−1
p

. (3.33)

Divide both sides by right most term then take power p on both sides and after simplification, we get

∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1 ≤ pp

∫ ∞
ak+1

λk+1(ςk+1)[φk(ς1, . . . , ςk, ςk+1)]
p∆ςk+1. (3.34)



Int. J. Anal. Appl. 17 (2) (2019) 254

Substitute (3.34) in (3.25), we get

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)[φ(ς1, . . . , ςk+1)]
p∆ς1 . . . ∆ςk+1

≤ pp
∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

{∫ ∞
ak+1

λk+1(ςk+1)[φk(ς1, . . . , ςk+1)]
p∆ςk+1

}
∆ς1 . . . ∆ςk. (3.35)

Use Fubini’s Theorem (Theorem 2.6) “k times” on right hand side of (3.35), we get

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)[φ(ς1, . . . , ςk+1)]
p∆ς1 . . . ∆ςk+1

≤ pp
∫ ∞
ak+1

λk+1(ςk+1)

{∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)φ
p
k(ς1, . . . , ςk+1)∆ς1 . . . ∆ςk

}
∆ςk+1. (3.36)

By using induction hypothesis for φk(ς1, . . . , ςk+1) (instead φk(ς1, . . . , ςk+1), with fix tk+1 ∈ Tk+1, in (3.16)

and again apply Theorem 2.6 “k times” on right hand side to get

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)[φ(ς1, . . . , ςk+1)]
p∆ς1 . . . ∆ςk+1

≤ p(k+1)p
∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)g
p(ς1, . . . , ςk+1)∆ς1 . . . ∆ςk+1. (3.37)

Thus by principle of mathematical induction (3.24) holds for all positive integers n, which completes the

proof. �

Remark 3.3. As a special case of Theorem 3.2 when Ti = R, ∀ i ∈{1, . . . ,n}, we have following general-

ization of integral inequality of Copson-type for the functions of “n” independent variables (note that when

Ti = R, we have φ(σ(ς1), . . . ,σ(ςn)) = φ(ς1, . . . , ςn), Λσii (ςi) = Λi(ςi) and µ(ςi) = 0) and (3.24) takes the

form

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)

(∫ ∞
ς1

. . .

∫ ∞
ςn

n∏
i=1

λi(si)

Λi(si)
g(s1, . . . ,sn)dsn . . .ds1

)p
dςn . . .dς1

≤ pnp
∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)g
p(ς1, . . . , ςn)dςn . . .dς1. (3.38)

As a special case of (3.38) when λi(ςi) = 1, ∀ i ∈{1, . . . ,n}, we have

∫ ∞
a1

. . .

∫ ∞
an

(∫ ∞
ς1

. . .

∫ ∞
ςn

n∏
i=1

λi(si)

si
g(s1, . . . ,sn)dsn . . .ds1

)p
dςn . . .dς1

≤ pnp
∫ ∞
a1

. . .

∫ ∞
an

gp(ς1, . . . , ςn)dςn . . .dς1, (3.39)

where p > 1.



Int. J. Anal. Appl. 17 (2) (2019) 255

Remark 3.4. For p > 1 and ∀ i ∈ {1, 2, . . . ,n}, assume that Ti = N with ai = 1 in Theorem 3.2.

Furthermore assume that
∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)g
p(l1, . . . ln)

is convergent. In this case (3.24) becomes the following discrete Copson type inequality for the functions of

“n” independent parameters:

∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)

(
∞∑

s1=l1

. . .

∞∑
sn=ln

λ1(s1) . . .λn(sn)∑l1
s1=1

λ1(s1) . . .
∑ln
sn=1

λn(sn)
g(s1, . . . ,sn)

)p

≤ pkp
∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)g
p(l1, . . . , ln). (3.40)

Theorem 3.3. Let T1, . . . ,Tn denote time scales. For p > 1 and i ∈{1, . . . ,n}, consider ai ∈ [0,∞)Ti and

g : T1 × . . .×Tn → R+. Let λi : Ti → R+, Λi(ςi) :=
∫ ςi
ai
λi(si)∆si, Λi(∞) = 0 and the delta integrals∫ ∞

a1

. . .

∫ ∞
an

n∏
i=1

(
λi(ςi)

Λi(ςi)

)
[g(ς1, . . . , ςn)]

p∆ς1 . . . ∆ςn exist. (3.41)

Assume for any ςi ∈ [ai,∞)Ti

φ(ς1, . . . , ςn) :=

∫ ∞
ς1

. . .

∫ ∞
ςn

n∏
i=1

λi(si)

Λi(si)
g(s1, . . . ,sn)∆s1 . . . ∆sn. (3.42)

Then

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)[φ(ς1, . . . , ςn)]
p∆ς1 . . . ∆ςn

≤ (p)np
∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)

(
Λσii (si)

Λi(si)

)p
gp(ς1, . . . , ςn)∆ς1 . . . ∆ςn, (3.43)

where n is any positive integer.

Proof. This result is proved by using Principle of mathematical induction. For n = 1, statement is true

by [12, Theorem 2.8]. Assume for 1 ≤ n ≤ k (3.43) holds. To prove the result for n = k + 1, Take left hand

side of (3.43) in the following form

∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

{∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1

}
∆ς1 . . . ∆ςk. (3.44)

Consider

Ik+1 =

∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1. (3.45)

Use integration by parts Theorem 2.2 (i) with φ(ς1, . . . , ςk,∞) = 0 and Λk+1(ςk+1) = 0 to get

∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1 =

∫ ∞
ak+1

−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

pΛ
σk+1
k+1 (ςk+1)∆ςk+1. (3.46)



Int. J. Anal. Appl. 17 (2) (2019) 256

By chain rule (2.4) for dk+1 ∈ [ςk+1, ςk+1],

−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

p = −pφp−1(ς1, . . . , ςk,dk+1)
∂

∆ςk+1
φ(ς1, . . . , ςk+1), (3.47)

and

∂

∆ςk+1
φ(ς1, . . . , ςk+1)

=

∫ ∞
ς1

. . .

∫ ∞
ςk

k∏
i=1

λi(si)

Λi(si)

{
∂

∆ςk+1

∫ ∞
ak+1

λk+1(sk+1)

Λk+1(sk+1)
g(s1, . . . ,sk+1)∆sk+1

}
∆s1 . . . ∆sk. (3.48)

Also

∂

∆ςk+1

∫ ∞
ak+1

λk+1(sk+1)

Λk+1(sk+1)
g(s1, . . . ,sk+1)∆sk+1 = −

λk+1(ςk+1)

Λk+1(ςk+1)
g(s1, . . . , ςk+1) ≤ 0

and dk+1 ≥ ςk+1. So after simplifications (3.48) implies

∂

∆ςk+1
φ(ς1, . . . , ςk+1) = −

λk+1(ςk+1)

Λk+1(ςk+1)

∫ ∞
ς1

. . .

∫ ∞
ςk

k∏
i=1

λi(si)

Λi(si)
g(s1, . . . , ςk+1)∆s1 . . . ∆sk. (3.49)

Substitute (3.49) in (3.47),

−
∂

∆ςk+1
[φ(ς1, . . . , ςk+1)]

p

= pφp−1(ς1, . . . , ςk,dk+1)
λk+1(ςk+1)

Λk+1(ςk+1)

∫ ∞
ς1

. . .

∫ ∞
ςk

k∏
i=1

λi(si)

Λi(si)
g(s1, . . . , ςk+1)∆s1 . . . ∆sk

≤ pφp−1(ς1, . . . , ςk+1)
λk+1(ςk+1)

Λk+1(ςk+1)
φk(ς1, . . . , ςk+1) , (3.50)

where

φk(ς1, . . . , ςk+1) =

∫ ∞
ς1

. . .

∫ ∞
ςk

k∏
i=1

λi(si)

Λi(si)
g(s1, . . . , ςk+1)∆s1 . . . ∆sk.

Put (3.49) in (3.46),

∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1

≤ p
∫ ∞
ak+1

λk+1(ςk+1)
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)
φp−1(ς1, . . . , ςk+1)φk(ς1, . . . , ςk+1)∆ςk+1. (3.51)

Multiply and divide by [λk+1(ςk+1)]
p−1
p on right hand side of (3.51) and apply Hölder’s inequality to get

Ik+1 ≤ p

(∫ ∞
ak+1

λk+1(ςk+1)

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p
φ
p
k(ς1, . . . , ςk+1)∆ςk+1

)1
p

×

(∫ ∞
ak+1

λk+1(ςk+1)φ
p(ς1, . . . , ςk+1)∆ςk+1

)p−1
p

. (3.52)



Int. J. Anal. Appl. 17 (2) (2019) 257

Divide both sides by right most term then take power p on both sides and after simplifying, we get

∫ ∞
ak+1

λk+1(ςk+1)[φ(ς1, . . . , ςk+1)]
p∆ςk+1

≤ (p)p
∫ ∞
ak+1

λk+1(ςk+1)

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p
φ
p
k(ς1, . . . , ςk+1)∆ςk+1. (3.53)

Substitute (3.53) in (3.52),

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)[φ(ς1, . . . , ςk+1)]
p∆ς1 . . . ∆ςk+1

≤ (p)p
∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)

{∫ ∞
ak+1

λk+1(ςk+1)

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p
φ
p
k(ς1, . . . , ςk+1)∆ςk+1

}
∆ς1 . . . ∆ςk.

(3.54)

Use Theorem 2.6 “k times” on right hand side of (3.54), to get

= (p)
p
∫ ∞
ak+1

λk+1(ςk+1)

(
Λ
σk+1
k+1 (ςk+1)

Λk+1(ςk+1)

)p
{
∫ ∞
a1

. . .

∫ ∞
ak

k∏
i=1

λi(ςi)φ
p
k(ς1, . . . , ςk+1)∆ς1 . . . ∆ςk}∆ςk+1. (3.55)

By using induction hypothesis for φk(ς1, . . . , ςk+1) (instead φk(ς1, . . . , ςk+1), with fix tk+1 ∈ Tk+1, in (3.55)

and again by applying Theorem 2.6 on right hand side, we get

∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)[φ(ς1, . . . , ςk+1)]
p∆ς1 . . . ∆ςk+1

≤ (p)(k+1)p
∫ ∞
a1

. . .

∫ ∞
ak+1

k+1∏
i=1

λi(ςi)

(
Λσii (ςi)

Λi(ςi)

)p
gp(ς1, . . . , ςk+1)∆ς1 . . . ∆ςk+1. (3.56)

Thus by principle of mathematical induction (3.43) holds for all n ∈ Z+, which completes the proof. �

From Theorem 3.3 with condition (3.17), we obtain following result.

Corollary 3.2. For p > 1 and i ∈ {1, . . . ,n}, consider Ti be time scales, ai ∈ [0,∞)Ti , λi : Ti → R+ and

g : T1 × . . .×Tn → R+. Let Λi(ςi) and ψ(ς1, . . . , ςn) be defined as in Theorem 3.1 such that (3.17) holds and

the delta integrals ∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

(
λi(ςi)

Λi(ςi)

)
[g(ς1, . . . , ςn)]

p∆ς1 . . . ∆ςn exists.

Then

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)[φ(ς1, . . . , ςn)]
p∆ςn . . . ∆ς1

≤ pnp
n∏
i=1

(
1

Li

)p ∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)g
p(ς1, . . . , ςn)∆ςn . . . ∆ς1. (3.57)



Int. J. Anal. Appl. 17 (2) (2019) 258

Remark 3.5. As a special case of Corollary 3.2 when Ti = R, integral inequality of Copson-type for the

functions of “n” independent variables is

∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)

(∫ ∞
ς1

. . .

∫ ∞
ςn

n∏
i=1

λi(si)

Λi(si)
g(s1, . . . ,sn)dsn . . .ds1

)p
dςn . . .dς1

≤ pnp
∫ ∞
a1

. . .

∫ ∞
an

n∏
i=1

λi(ςi)g
p(ς1, . . . , ςn)dςn . . .dς1, (3.58)

where p > 1.

Remark 3.6. Assume that Ti = N in Theorem 3.3, p > 1, ai = 1,∀ i ∈{1, 2, . . . ,n}. Furthermore assume

that
∑∞
s1=1

. . .
∑∞
sn=1

λ1(s1) . . .λn(sn)
Λ1(s1+1)

Λ1(s1)
. . .

Λn(sn+1)
Λn(sn)

gp(s1, . . . ,sn) is convergent. Then (3.43) becomes

the following discrete inequality of Copson’s type for the functions of “n” independent variables:

∞∑
l1=1

. . .

∞∑
ln=1

λ1(l1) . . .λn(ln)

(
∞∑

s1=l1

. . .

∞∑
s2=ln

λ1(s1) . . .λn(sn)

Λ1(s1) . . . Λn(sn)
g(s1, . . . ,sn)

)p

≤ pkp
∞∑
l1=1

. . .

∞∑
ln=1

Λ1(l1 + 1) . . . Λn(ln + 1)

Λ1(l1) . . . Λn(ln)
λ1(l1) . . .λn(ln)g

p(l1, . . . , ln),

where Λi(si) =

si−1∑
ai=1

λi(si), ∀i = 1, 2, . . . ,n.

References

[1] M. Bohner, A. Nosheen, J. Peĉarić, A. Younus, Some dynamic Hardy-type inequalities with general kernel, J. Math.

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2017.

[3] M. Bohner and A. Peterson, Dynamic Equations on time scales: an introduction with Applications, Birkhauser, Boston,

MA, 2001.

[4] M. Bohner and A. Peterson, Advances in Dynamic Equations on time scales: an introduction with Applications, Birkhauser,

Boston, MA, 2003.

[5] L. Y. Chan, Some extensions of Hardy’s inequality, Canad. Math. Bull. 22 (1979), 165-169.

[6] E. T. Copson, Note on series of positive terms, J. London. Math. Soc. 2 (1927), 9-12.

[7] G. H. Hardy, J. E. Littlewood and G. Pólya, Inequalities, 2nd ed., Cambridge University Press, Cambridge, 1952.

[8] S. Kaijser, L. Nikolova, L. E. Persson and A. Wedestig, Hardy-type inequalities via convexity, Math. Inequal. Appl. 8

(2005), 403-417.

[9] S. Keller, Asymotitishces Verhalten invarianter Faserbündel bei Diskretisierung und Mittelwertbildung im Rahmen der

Analysis auf Zeitskalen, PhD thesis, Universität Augsburg (1999).

[10] J. Pan, Geometric character of domain and the Harnack inequality of solution of a singular parabolic equation, Math.

Slovaca, 62 (2012), 721-734.

[11] C. Pötzsche, Chain rule and invariance principle on measure chains, J. Comput. Appl. Math. 141 (1-2) (2002), 249-254.



Int. J. Anal. Appl. 17 (2) (2019) 259

[12] S. H. Saker, D. O’Regan and R. P. Agarwal, Dynamic inequalities of Hardy and Copson type on time scales, Analysis, 34

(4) (2014), 391-402.


	1. Introduction
	2. Preliminaries
	3. Dynamic Hardy & Copson-Type Inequalities via Time Scales for Functions of n Independent Variables
	References