International Journal of Analysis and Applications
ISSN 2291-8639
Volume 7, Number 1 (2015), 38-49
http://www.etamaths.com

OPIAL TYPE INTEGRAL INEQUALITIES FOR WIDDER

DERIVATIVES AND LINEAR DIFFERENTIAL OPERATORS

GHULAM FARID1,∗ AND JOSIP PEČARIĆ2

Abstract. In this paper we establish Opial type integral inequalities for Wid-

der derivatives and linear differential operator. Also, for applications we con-

struct some related inequalities as special cases.

1. Introduction

The following inequality established in 1960. by Opial [15] :

Let x(t) ∈ C(1)[0,h] be such that x(0) = x(h) = 0, and x(t) > 0 in
(0,h). Then

(1.1)

∫ h
0

|x(t)x′(t)|dt ≤
h

4

∫ h
0

(x′(t))
2
dt,

where constant h
4

is the best possible.

Over the last 50 years, Opial inequality (1.1) is studied by many mathematicians
and extended, generalized in different ways. It is recognized as fundamental re-
sult in the theory of differential equations (see the monograp[1]). Opial inequality
and its generalizations, extensions and discretizations play a fundamental role in
establishing the existence and uniqueness of initial and boundary value problems
for ordinary and partial differential equations as well as difference equations, (see,
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 18, 19]).
Following theorems by Mitrinović and Pečarić, include such generalizations of Opi-
al’s inequality given in [16, page, 237–238] and for them we need next characteri-
zation:
We say that a function u : [a,b] −→ R belongs to the class U1(v,K) if it admits the
representation

u(x) =

∫ x
a

K(x,t)v(t) dt

where v is a continuous function and K is an arbitrary non-negative kernel such
that v(x) > 0 implies u(x) > 0 for every x ∈ [a,b]. We also assume that all integrals
under consideration exist and are finite.

Theorem 1.1. Let φ : [0,∞) −→ R be a differentiable function such that for
q > 1 the function φ(x

1
q ) is convex and φ(0) = 0. Let u ∈ U1(v,K) where

2010 Mathematics Subject Classification. 26A24, 26D15.
Key words and phrases. Convex functions; Opial–type inequality; Widder derivatives; frac-

tional derivative.

c©2015 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

38



OPIAL TYPE INTEGRAL INEQUALITIES 39

(∫ x
a

(K(x,t))p dt
)1

p ≤ M and 1
p

+ 1
q

= 1. Then

(1.2)∫ b
a

|u(x)|1−q φ′(|u(x)|)|v(x)|q dx ≤
q

Mq (b−a)

∫ b
a

φ
(

(b−a)
1
q M|v(x)|

)
dx.

If the function φ(x
1
q ) is concave, then the reverse inequality holds.

A similar result follows by using another class U2(v,K) of functions
u : [a,b] −→ R which admits representation

u(x) =

∫ b
x

K(x,t)v(t)dt.

Theorem 1.2. Let φ : [0,∞) −→ R be a differentiable function such that for
q > 1 the function φ(x

1
q ) is convex and φ(0) = 0. Let u ∈ U2(v,K) where(∫ b

x
(K(x,t))p dt

)1
p

≤ N and 1
p

+ 1
q

= 1. Then

(1.3)∫ b
a

|u(x)|1−q φ′(|u(x)|)|v(x)|qdx ≤
q

Nq(b−a)

∫ b
a

φ
(

(b−a)
1
q N|v(x)|

)
dx.

If the function φ(x
1
q ) is concave, then reverse inequality holds.

In [5] we gave extensions of above Mitrinović–Pečarić inequalities which are s-
tated in the following theorems.

Theorem 1.3. Let φ : [0,∞) −→ R be a differentiable function such that for
q > 1 the function φ(x

1
q ) is convex and φ(0) = 0. Let u ∈ U1(v,K) where(∫ x

a
(K(x,t))p dt

)1
p ≤ M and 1

p
+ 1

q
= 1. Then∫ b

a

|u(x)|1−q φ′(|u(x)|)|v(x)|q dx ≤
q

Mq
φ
(
M
(∫ b

a

|v(x)|q dx
)1

q
)

≤
q

Mq (b−a)

∫ b
a

φ
(

(b−a)
1
q M|v(x)|

)
dx.(1.4)

If the function φ(x
1
q ) is concave, then reverse inequalities hold.

Theorem 1.4. Let φ : [0,∞) −→ R be a differentiable function such that for
q > 1 the function φ(x

1
q ) is convex and φ(0) = 0. Let u ∈ U2(v,K) where(∫ b

x
(K(x,t))p dt

)1
p

≤ M and 1
p

+ 1
q

= 1. Then∫ b
a

|u(x)|1−qφ′(|u(x)|)|v(x)|qdx ≤
q

Mq
φ
(
M
(∫ b

a

|v(x)|q dx
)1

q
)

≤
q

Mq(b−a)

∫ b
a

φ
(

(b−a)
1
q M|v(x)|

)
dx.(1.5)

If the function φ(x
1
q ) is concave, then reverse inequalities hold.

In view of the great importance of Taylor’s formula in analysis, it may be regard-
ed as extremely surprising that so few attempts at generalization have been made.



40 FARID AND PEČARIĆ

The problem of the representation of an arbitrary function by means of linear com-
binations of prescribed functions has received no small amount of attention (see
the introduction of [17]). Therefore, in 1927 Widder gave generalization of Taylor’s
formula [17].
In this paper we are interested to give Opial type integral inequalities by Mitrinović
and Pečarić for Widder derivatives and linear differential operators. Also, we give
their extensions and as applications discuss their special cases and examples.

2. Mitrinović-Pečarić Inequalities for Widder Derivatives

In this section we give Opial type integral inequalities for Widder derivatives. We
extend these inequalities, also give their special cases and provide some examples.
The following are taken from [17].
Let f,u0,u1, ...,un ∈ Cn+1([a,b]),n ≥ 0, and the Wronskians

(2.1) Wi(x) := W [u0(x),u1(x), ...,un(x)] :=

∣∣∣∣∣∣∣∣∣∣∣∣

u0(x) u1(x) . . . ui(x)

u
′

0(x) u
′

1(x) . . . u
′

i(x)
.
.
.

ui0(x) u
i
1(x) . . . u

i
i(x)

∣∣∣∣∣∣∣∣∣∣∣∣
,

i = 1, ...,n. Here W0(x) = u0(x). Assume W0(x) > 0 over [a,b], i = 1, ...,n.
For i ≥ 0, the differential operator of order i (Widder derivative):

(2.2) Lif(x) :=
W [u0(x),u1(x), ...,ui−1(x),f(x)]

Wi−1(x)

i = 1, ...,n + 1; L0f(x) := f(x),∀x ∈ [a,b]. Consider also

(2.3) gi(x,t) :=
1

Wi(t)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

u0(t) u1(t) . . . ui(t)

u
′

0(t) u
′

1(t) . . . u
′

i(t)
.
.
.

ui−10 (t) u
i−1
1 (t) . . . u

i
i−1(x)

u0(x) u1(x) . . . ui(x)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣
,

i = 1, ...,n; g0(x,t) :=
u0(x)
u0(t)

,∀x,t ∈ [a,b].

Example 2.1. Sets of the form {u0,u1, ...,un} are {u0,u1, ...,un},
{1,sinx,−cosx,−sin2x,cos2x,..., (−1)nsinnx, (−1)ncosnx}, etc.

We also mention the generalized Widder-Talylor’s formula, see, [17] also [3].

Theorem 2.2. Let the functions f,u0,u1, ...,un ∈ Cn+1([a,b]), and the Wron-
skians W0(x),W1(x), ...,Wn(x) > 0 on [a,b], x ∈ [a,b]. Then for t ∈ [a,b] we have

(2.4) f(x) = y(t)
u0(x)

u0(t)
+ L1f(t)g1(x,t) + ... + Lnf(t)gn(x,t) + Rn(x),

where

Rn(x) :=

∫ x
t

gn(x,s)Ln+1f(s)ds



OPIAL TYPE INTEGRAL INEQUALITIES 41

For example [17] one could take u0(x) = c > 0. If ui(x) = x
i, i = 0, 1, ...,n,

defined on [a,b], then

Liy(t) = f
i(t) and gi(x,t) =

(x−t)i
i!

, t ∈ [a,b].

We need the following result.

Corollary 2.3. By additionally assuming for fixed x0 ∈ [a,b] that Lif(x0) = 0, i =
0, 1, ...,n, we get that

(2.5) f(x) =

∫ x
x0

gn(x,s)Ln+1f(s)ds.

Note that all results of this section are under the assumptions of Theorem 2.2
and Corollary 2.3.

Theorem 2.4. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≥ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x

x0
|gn(x,t)|pdt

)1
p

≤ M, 1
p

+ 1
q

= 1. Then

(2.6)∫ b
x0

|f(x)|1−qφ′(|f(x)||Ln+1f(x)|q dx ≤
q

Mq
φ


M

(∫ b
x0

|Ln+1f(t)|qdt

)1
q


 .

If the function φ(x
1
q ) is concave, then the reverse inequality holds.

Proof. As f has representation

f(x) =

∫ x
x0

gn(x,t)Ln+1f(t)dt.

By applying Holder’s inequality we have

|f(x)| =
∣∣∣∣
∫ x
x0

gn(x,t)Ln+1f(t)dt

∣∣∣∣
≤

(∫ x
x0

|gn(x,t)|pdt
)1

p
(∫ x

x0

|Ln+1f(t)|qdt
)1

q

≤ M
(∫ x

x0

|Ln+1f(t)|qdt
)1

q

(2.7)

Let

p(x) =

∫ x
x0

|Ln+1f(t)|qdt.

Then p′(x) = |Ln+1f(x)|q. Now from (2.7) we can have |f(x)| ≤ M (p(x))
1
q . From

the convexity of φ(x
1
q ) it follows that the function x1−qφ′(x) is increasing. Thus



42 FARID AND PEČARIĆ

we have∫ b
x0

|f(x)|1−qφ′(|f(x)|)|Ln+1f(x)|qdx ≤
∫ b
x0

M1−q (p(x))
1
q
−1
φ′
(
M (p(x))

1
q

)
p′(x)dx

=
q

Mq

∫ b
x0

φ′
(
M(p(x))

1
q

)
d(M(p(x)

1
q )

=
q

Mq
φ
(
M (p(b))

1
q

)
=

q

Mq
φ


M

(∫ b
x0

|Ln+1f(t)|qdt

)1
q


 .

�

Next we give extension of above theorem.

Theorem 2.5. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≥ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x

x0
|gn(x,t)|pdt

)1
p

≤ M, 1
p

+ 1
q

= 1. Then

∫ b
x0

|f(x)|1−qφ′(|f(x)||Ln+1f(x)|q dx ≤
q

Mq
φ


M

(∫ b
x0

|Ln+1f(t)|qdt

)1
q




≤
q

Mq (b−x0)

∫ b
x0

φ
(

(b−x0)
1
q M|Ln+1f(t)|

)
dt.(2.8)

If the function φ(x
1
q ) is concave, then reverse inequalities hold.

Proof. Inequality (2.6) holds by Theorem 2.4. Since φ(x
1
q ) is convex, the following

Jensen’s inequality holds

(2.9) φ
(( 1

b−a

∫ b
x0

g(t) dt
)1

q
)
≤

1

b−a

∫ b
x0

φ
(
g

1
q (t)

)
dt.

Applying (2.9) on (2.6) we get (2.8). �

The counter part of above theorem is given in the following.

Theorem 2.6. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≤ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x0

x
|gn(x,t)|pdt

)1
p ≤ M, 1

p
+ 1

q
= 1. Then

(2.10)∫ x0
a

|f(x)|1−qφ′(|f(x)||Ln+1f(x)|q dx ≤
q

Mq
φ

(
M

(∫ x0
a

|Ln+1f(t)|qdt
)1

q

)
.

If the function φ(x
1
q ) is concave, then the reverse inequality holds.

Proof. As f has representation

f(x) =

∫ x
x0

gn(x,t)Ln+1f(t)dt.



OPIAL TYPE INTEGRAL INEQUALITIES 43

By applying Holder’s inequality we have

|f(x)| =
∣∣∣∣
∫ x0
x

gn(x,t)Ln+1f(t)dt

∣∣∣∣
≤
(∫ x0

x

|gn(x,t)|pdt
)1

p
(∫ x0

x

|Ln+1f(t)|qdt
)1

q

≤ M
(∫ x0

x

|Ln+1f(t)|qdt
)1

q

(2.11)

Let

p(x) =

∫ x0
x

|Ln+1f(t)|qdt.

Then −p′(x) = |Ln+1f(x)|q ≥ 0. Now from (2.11) we can have |f(x)| ≤ M (p(x))
1
q .

From the convexity of φ(x
1
q ) it follows that the function x1−qφ′(x) is increasing.

Thus we have∫ x0
a

|f(x)|1−qφ′(|f(x)|)|Ln+1f(x)|qdx ≤−
∫ x0
a

M1−q (p(x))
1
q
−1
φ′
(
M (p(x))

1
q

)
p′(x)dx

= −
q

Mq

∫ x0
a

φ′
(
M(p(x))

1
q

)
d(M(p(x)

1
q )

=
q

Mq
φ
(
M (p(a))

1
q

)
=

q

Mq
φ

(
M

(∫ x0
a

|Ln+1f(t)|qdt
)1

q

)
.

�

Next we give extension of above theorem.

Theorem 2.7. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≤ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x0

x
|gn(x,t)|pdt

)1
p ≤ M, 1

p
+ 1

q
= 1. Then∫ x0

a

|f(x)|1−qφ′(|f(x)||Ln+1f(x)|q dx ≤
q

Mq
φ

(
M

(∫ x0
a

|Ln+1f(t)|qdt
)1

q

)

≤
q

Mq (b−x0)

∫ x0
a

φ
(

(b−x0)
1
q M|Ln+1f(t)|

)
dt.(2.12)

If the function φ(x
1
q ) is concave, then reverse inequalities hold.

Proof. As in proof of the Theorem 2.7, inequalities follow from Theorem 2.6 and
Jensen’s inequality (2.9). �

Remark 2.8. If directly we replace u by f, v by Ln+1f and general kernal K(x,t)
by gn(x,t) in (1.2), (1.3), (1.4) and (1.5) we can see above results.

Next we discuss extreme cases.

Theorem 2.9. Let p = 1,q = ∞ and x ≥ x0 ∈ [a,b]. Then we have
1

b−x0

∫ b
x0

|f(x)||Ln+1f(x)|dx ≤ M||Ln+1f||2∞.(2.13)



44 FARID AND PEČARIĆ

Proof. As f has representation

f(x) =

∫ x
x0

gn(x,t)Ln+1f(t)dt.

From this we get

|f(x)| ≤
(∫ x

x0

|gn(x,t)|dt
)
||Ln+1f||∞

≤ M||Ln+1f||∞
and using |Ln+1f(t)| ≤ ||Ln+1f||∞ we have

|f(x)||Ln+1f(t)| ≤ M||Ln+1f||2∞.

Now integrating the last inequality on [x0,b], we get (2.13). �

Theorem 2.10. Let p = 1,q = ∞ and x ≤ x0 ∈ [a,b]. Then we have
1

x0 −a

∫ x0
a

|f(x)||Ln+1f(x)|dx ≤ M||Ln+1f||2∞.(2.14)

Proof. As f has representation

f(x) =

∫ x0
x

gn(x,t)Ln+1f(t)dt.

From this we get

|f(x)| ≤
(∫ x

x0

|gn(x,t)|dt
)
||Ln+1f||∞

≤ M||Ln+1f||∞
and using |Ln+1f(t)| ≤ ||Ln+1f||∞ we have

|f(x)||Ln+1f(t)| ≤ M||Ln+1f(t)||2∞.

Now integrating the last inequality on [a,x0], we get (2.14). �

Corollary 2.11. Let p,q ∈ (1,∞) such that 1
p

+ 1
q

= 1. Then we have∫ x
x0

|f(x)|q|Ln+1f(x)|q dx ≤
Mq

2

(∫ x
x0

|Ln+1f(t)|qdt
)2

.(2.15)

Proof. By setting φ(t) = t2q and b = x ≥ x0 in Theorem 2.4 one can get (2.15). �

Corollary 2.12. Let p,q ∈ (1,∞) such that 1
p

+ 1
q

= 1. Then we have∫ x0
x

|f(x)|q|Ln+1f(x)|q dx ≤
Mq

2

(∫ x0
x

|Ln+1f(t)|qdt
)2

.(2.16)

Proof. By setting φ(t) = t2q and a = x ≤ x0 in Theorem 2.5 one can get (2.16). �

Corollary 2.13. Let p,q ∈ (1,∞) such that 1
p

+ 1
q

= 1. Then we have∣∣∣∣
∫ x
x0

|f(x)|q|Ln+1f(x)|q dx
∣∣∣∣ ≤ Mq2

(∫ x
x0

|Ln+1f(t)|qdt
)2

.(2.17)

Proof. From (2.15) and (2.16) one can easily get (2.17). �



OPIAL TYPE INTEGRAL INEQUALITIES 45

Example 2.14. If we take u0(x) = c > 0 and un(x) = x
n,n = 0, 1, 2, ...,n defined

on [a,b], then Lnf(x) = f
n(x) and gn(x,t) =

(x−t)n
n!

, t ∈ [a,b],. Here we can have

M =
(b−x0)

np+1
p

n!(np+1)
1
p

and the inequality (2.6) becomes

∫ b
x0

|f(x)|1−qφ′
(
|f(x)||fn+1(x)|q

)
dx

≤
n!q(np + 1)

q
p

(b−x0)
q(np+1)

p

φ


(b−x0) np+1p
n!(np + 1)

1
p

(∫ b
x0

|fn+1(t)|qdt

)1
q


 .

Also extension of (2.6) becomes∫ b
x0

|f(x)|1−qφ′
(
|f(x)||fn+1(x)|q

)
dx

≤
n!q(np + 1)

q
p

(b−x0)
q(np+1)

p

φ


(b−x0) np+1p
n!(np + 1)

1
p

(∫ b
x0

|fn+1(t)|qdt

)1
q




≤
n!q(np + 1)

q
p

(b−x0)q(n+1)

∫ b
x0

φ

(
(b−x0)n+1

n!(np + 1)
1
p

|Ln+1f(t)|

)
dt.

Remark 2.15. Examples similar to Example 2.14 can be obtained by using other
inequalites given in above results. We omit here such examples.

3. Mitrinović-Pečarić Inequalities for Linear differential operators

In this section we give Opial type integral inequalities for Linear differential
operators. We extend these inequalities, also give some special cases.

Here we follow [12, page, 145–154].
Let I be a closed interval of R. Let ai(x), i = 0, 1, ...,n−1(n ∈ N),h(x) be continuous
functions on I and let L = Dn + an−1(x)D

n−1 + ... + a0(x) be a fixed linear
differential operator on Cn(I). Let y1(x), ...,yn(x) be a set of linear independent
solutions to Ly = 0. Here the associated Green’s function for L is

(3.1) H(x,t) :=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

y1(t) . . . yn(t)

y
′

1(t) . . . y
′

n(t)
.
.
.

yn−21 (t) . . . y
n−2
n (t)

y1(x) . . . yn(x)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

/
∣∣∣∣∣∣∣∣∣∣∣∣∣∣

y1(t) . . . yn(t)

y
′

1(t) . . . y
′

n(t)
.
.
.

yn−21 (t) . . . y
n−2
n (t)

y1(t) . . . yn(t)

∣∣∣∣∣∣∣∣∣∣∣∣∣∣
,

which is a continuous function on I2. Consider fixed x0 ∈ I, then

(3.2) y(x) =

∫ x
x0

H(x,t)h(t)dt,x ∈ I

is the unique solution to the initial value problem

(3.3) Ly = h; y(i)(x0) = 0, i = 0, 1, ...,n− 1.



46 FARID AND PEČARIĆ

Theorem 3.1. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≥ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x

x0
(H(x,t))pdt

)1
p

≤ M, 1
p

+ 1
q

= 1. Then

∫ b
x0

|y(x)|1−qφ′(|y(x)||(Ly)(x)|q dx ≤
q

Mq
φ


M

(∫ b
x0

|(Ly)(t)|qdt

)1
q


 .(3.4)

If the function φ(x
1
q ) is concave, then the reverse inequality holds.

Proof. Here y has representation

y(x) =

∫ x
x0

H(x,t)h(t)dt.

rest of proof follows from the proof of Theorem 2.4. �

Next we give extension of above theorem.

Theorem 3.2. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≥ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x

x0
(H(x,t))pdt

)1
p

≤ M, 1
p

+ 1
q

= 1. Then

∫ b
x0

|y(x)|1−qφ′(|y(x)||(Ly)(x)|q dx ≤
q

Mq
φ


M

(∫ b
x0

|(Ly)(t)|qdt

)1
q




≤
q

Mq (b−x0)

∫ b
x0

φ
(

(b−x0)
1
q M|(Ly)(t)|

)
dt.

If the function φ(x
1
q ) is concave, then reverse inequalities hold.

Proof. Proof is similar to the proof of Theorem 2.5. �

The counter part of above theorem is given in the following.

Theorem 3.3. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≤ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x0

x
(H(x,t))pdt

)1
p ≤ M, 1

p
+ 1

q
= 1. Then

(3.5)∫ x0
a

|y(x)|1−qφ′(|y(x)||(Ly)(x)|q dx ≤
q

Mq
φ

(
M

(∫ x0
a

|(Ly)(t)|qdt
)1

q

)
.

If the function φ(x
1
q ) is concave, then the reverse inequality holds.

Proof. Proof is similar to the proof of Theorem 2.6. �

Next we give extension of above theorem.



OPIAL TYPE INTEGRAL INEQUALITIES 47

Theorem 3.4. Let φ : [0,∞) → R be a differentiable function such that for q > 1
the function φ(x

1
q ) is convex and φ(0) = 0. Let x ≤ x0 ∈ [a,b] and p ∈ (1,∞) such

that
(∫ x0

x
(H(x,t))pdt

)1
p ≤ M, 1

p
+ 1

q
= 1. Then∫ x0

a

|y(x)|1−qφ′(|y(x)||(Ly)(x)|q dx ≤
q

Mq
φ

(
M

(∫ x0
a

|(Ly)(t)|qdt
)1

q

)

≤
q

Mq (b−x0)

∫ x0
a

φ
(

(b−x0)
1
q M|(Ly)(t)|

)
dt.(3.6)

If the function φ(x
1
q ) is concave, then reverse inequalities hold.

Proof. Proof is similar to the proof of Theorem 2.7. �

Remark 3.5. If directly we replace u by y, v by Ly, and general kernal K(x,t) by
H(x,t) in (1.2), (1.3), (1.4) and (1.5) we can see above results.

Next we discuss extreme cases.

Theorem 3.6. Let p = 1,q = ∞ and x ≥ x0 ∈ [a,b]. Then we have
1

b−x0

∫ b
x0

|y(x)||(Ly)(x)|dx ≤ M||Ly||2∞.(3.7)

Proof. As f has representation

y(x) =

∫ x
x0

H(x,t)(Ly)(t)dt.

From this we get

|y(x)| ≤
(∫ x

x0

|H(x,t)|dt
)
||Ly||∞

≤ M||Ly||∞
and using |(Ly)(t)| ≤ ||Ly||∞ we have

|y(x)||(Ly)(t)| ≤ M||Ly||2∞.
Now integrating the last inequality on [x0,b], we get (3.7). �

Theorem 3.7. Let p = 1,q = ∞ and x ≤ x0 ∈ [a,b]. Then we have
1

x0 −a

∫ x0
a

|y(x)||(Ly)(x)|dx ≤ M||Ly||2∞.(3.8)

Proof. As y has representation

y(x) =

∫ x0
x

H(x,t)(Ly)(t)dt.

From this we get

|y(x)| ≤
(∫ x

x0

|H(x,t)|dt
)
||Ly||∞

≤ M||Ly||∞
and using |(Ly)(t)| ≤ ||Ly||∞ we have

|y(x)||(Ly)(t)| ≤ M||Ly||2∞.
Now integrating the last inequality on [a,x0], we get (3.8). �



48 FARID AND PEČARIĆ

Corollary 3.8. Let p,q ∈ (1,∞) such that 1
p

+ 1
q

= 1. Then we have∫ x
x0

|y(x)|q|(Ly)(x)|q dx ≤
Mq

2

(∫ x
x0

|(Ly)(t)|qdt
)2

.(3.9)

Proof. By setting φ(t) = t2q and b = x ≥ x0 in Theorem 3.1 one can get (3.9). �

Corollary 3.9. Let p,q ∈ (1,∞) such that 1
p

+ 1
q

= 1. Then we have∫ x0
x

|y(x)|q|(Ly)(x)|q dx ≤
Mq

2

(∫ x0
x

|(Ly)(t)|qdt
)2

.(3.10)

Proof. By setting φ(t) = t2q and a = x ≤ x0 in Theorem 3.2 one can get (3.10). �

Corollary 3.10. Let p,q ∈ (1,∞) such that 1
p

+ 1
q

= 1. Then we have∣∣∣∣
∫ x
x0

|y(x)|q|(Ly)(x)|q dx
∣∣∣∣ ≤ Mq2

(∫ x
x0

|(Ly)(t)|qdt
)2

.(3.11)

Proof. From (3.9) and (3.10) one can easily get (3.11). �

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OPIAL TYPE INTEGRAL INEQUALITIES 49

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1Department of Mathematics, COMSATS Institute of Information Technology, At-

tock Campus, Pakistan

2Faculty of Textile Technology, University of Zagreb, Prilaz baruna Filipovića

28A, 10000 Zagreb, Croatia

∗Corresponding author