arXiv:1701.00092v1 [math.FA] 31 Dec 2016

HERMITE-HADAMARD, HERMITE-HADAMARD-FEJE´R, DRAGOMIR-AGARWAL AND PACHPATTE TYPE
INEQUALITIES FOR CONVEX FUNCTIONS VIA FRACTIONAL INTEGRALS
MOKHTAR KIRANE BERIKBOL T. TOREBEK
Abstract. The aim of this paper is to establish Hermite-Hadamard, HermiteHadamard-Fej´er, Dragomir-Agarwal and Pachpatte type inequalities for new fractional integral operators with exponential kernel.

1. Introduction

The inequalities discovered by Hermite and Hadamard for convex functions are very important in the literature (see, e.g.,[PPT92, DP00]). These inequalities state that if [H1883, H1893] u : I  R is a convex function on the interval I  R and a, b  I with b > a, then

b

u

a+b 2



b

1 -

a

u(x)dx



u(a) + 2

u(b) .

a

(1.1)

Both inequalities hold in the reversed direction if u is concave. We note that Hadamard's inequality may be regarded as a refinement of the concept of convexity and it follows easily from Jensen's inequality.
The classical Hermite-Hadamard inequality provides estimates of the mean value of a continuous convex function u : [a, b]  R.
The most well-known inequalities related to the integral mean of a convex function u are the Hermite-Hadamard inequalities or its weighted versions, the so-called Hermite-Hadamard-Fej´er inequalities.
In [F06], Fej´er established the following inequality which is the weighted generalization of Hermite-Hadamard inequality (1.1):
Let u : [a, b]  R be convex function. Then the inequality

b

b

b

u

a+b 2

v(x)dx 

u(x)v(x)dx



u(a)

+ 2

u(b)

v(x)dx

a

a

a

(1.2)

holds;

here

v

:

[a, b]  R

is

nonnegative,

integrable

and

symmetric

to

a+b 2

.

In [DA98], Dragomir and Agarwal proved the following results connected with

the right part of (1.1):

2000 Mathematics Subject Classification. 35A09; 34K06. Key words and phrases. HermiteHadamard inequality, Hermite-Hadamard-Fej´er inequality, Dragomir-Agarwal inequality, Pachpatte inequalities, new fractional integral operator, integral inequalities.
1

2

M. KIRANE AND B. T. TOREBEK

Let u : I  R  R be a differentiable mapping on I, a, b  I. If |u| is convex on [a, b], then the following inequality holds:

b

u(a)

+ 2

u(b)

-

b

1 -

a

u(x)dx



b- 8

a

(|u(a)| +

|u(b)|) .

a

(1.3)

In [P03], Pachpatte established two new Hermite-Hadamard type inequalities for products of convex functions as follows:
Let u and v be nonnegative and convex functions on [a, b]  R, then

b

1 b-a

u(x)v(x)dx

a



u(a)v(a) + u(b)v(b) 3

+

u(a)v(b)

+ 6

u(b)v(a)

and

2u

a+b 2

v

a+b 2

(1.4)

b



b

1 -

a

u(x)v(x)dx

a

(1.5)

+

u(a)v(a)

+ 6

u(b)v(b)

+

u(a)v(b)

+ 3

u(b)v(a) .

Many generalizations and extensions of the Hermite-Hadamard, Hermite-Hadamard-

Fej´er, Dragomir-Agarwal and Pachpatte type inequalities were obtained for vari-

ous classes of functions using fractional integrals; see [SSYB13, WLFZ12, ZW13,

ITM16, JS16, BPP16, C16, HYT14, I16, CK17] and references therein.

Definition 1.1. The function u : [a, b]  R  R, is said to be convex if the following inequality holds

u(µx + (1 - µ)y)  µu(x) + (1 - µ)u(y)

for all x, y  [a, b] and µ  [0, 1]. We say that u is concave if (-u) is convex.

In the following, we will give some necessary definitions and mathematical preliminaries of new fractional integral which are used further in this paper.

Definition 1.2. Let f  L1(a, b). The fractional integrals Ia and Ib of order   (0, 1) are defined by

x

Iau(x)

=

1 

exp

-

1

- 



(x

-

s)

u(s)ds, x > a

a

and

b

Ibu(x)

=

1 

exp - 1 -  (s - x) u(s)ds, x < b 

x

HERMITE-HADAMARD, HERMITE-HADAMARD-FEJE´ R, DRAGOMIR-AGARWAL AND ... 3

respectively.

If  = 1, then

x

b

lim
1

Iau(x)

=

u(s)ds,

lim
1

Ibu(x)

=

u(s)ds.

a

x

Therefore the operators Ia and Ib are called a fractional integrals of order . Moreover, because

lim 1 exp - 1 -  (x - s) = (x - s),

0 



then

lim
0

Iau(x)

=

u(x),

lim
0

Ibu(x)

=

u(x).

The aim of this paper is to establish some functional inequalities for the above

new fractional integral operators with exponential kernel.

We

henceforth

denote

A

=

1- 

(b

-

a)

2. Hermite-Hadamard type inequality

Theorem 2.1. Let u : [a, b]  R be a positive function with 0  a < b and u  L1(a, b). If u is a convex function on [a, b], then the following inequalities for fractional integrals hold:

u

a+b 2



2 (1

1- - exp (-A))

[Iau(b) +

Ibu(a)]



u(a)

+ 2

u(b)

.

(2.1)

Proof. Since u is a convex function on [a, b], we get for x and y from [a, b] with

µ

=

1 2

u

x+y 2



u(x)

+ 2

u(y) ,

(2.2)

i.e., with x = ta + (1 - t)b, y = (1 - t)a + tb,

2u

a+b 2

 u(ta + (1 - t)b) + u((1 - t)a + tb).

(2.3)

Multiplying both sides of (2.3) by exp (-At) , then integrating the resulting inequality with respect to t over [0, 1], we obtain

2

(1

-

exp A

(-A))

u

a+b 2

1
 exp (-At) [u(ta + (1 - t)b) + u((1 - t)a + tb)] dt

0 1
= exp (-At) u(ta + (1 - t)b)dt

0 1
+ exp (-At) u((1 - t)a + tb)dt

0

4

M. KIRANE AND B. T. TOREBEK

As a results, we obtain

b

=

b

1 -

a

exp - 1 -  (b - s) 

a

b

+

b

1 -

a

exp - 1 -  (s - a) 

a

=

b

 -a

[Iau(b) +

Ibu(a)] .

u(s)ds u(s)ds

2

(1

-

exp A

(-A))

u

a+b 2



b

 -

a

[Iau(b)

+

Ibu(a)] .

The first inequality of (2.1) is proved. For the proof of the second inequality in (2.1) we first note that if u is a convex
function, then, for µ  [0, 1], it yields

u(ta + (1 - t)b)  tu(a) + (1 - t)u(b)

and u((1 - t)a + tb)  (1 - t)u(a) + tu(b).
By adding these inequalities we get

u(ta + (1 - t)b) + u((1 - t)a + tb)  u(a) + u(b).

(2.4)

Then multiplying both sides of (2.4) by exp (-At) and integrating the resulting inequality with respect to t over [0, 1], we obtain

2

(1

-

exp A

(-A))

[u(a)

+

u(b)]

1

 exp (-At) u(ta + (1 - t)b)dt

0 1

+ exp (-At) u((1 - t)a + tb)dt,

0

i.e.

b

 -

a

[Iau(b)

+

Ibu(a)]



2 (1

-

exp (-A)) A

[u(a)

+

u(b)] ,

and the second inequality in (2.1) is proved. The proof of the Theorem 2.1 is

completed.

Corollary 2.2. Let u : [a, b]  R be a positive function with 0  a < b and u  L1(a, b). If u is a concave function on [a, b], then the following inequalities for fractional integrals hold:

u

a+b 2



2 (1

1- - exp (-A))

[Iau(b)

+

Ibu(a)]



u(a)

+ 2

u(b) .

Remark 2.3. For   1, we get

lim
1

2 (1

1- - exp (-A))

=

1 2(b -

a) .

HERMITE-HADAMARD, HERMITE-HADAMARD-FEJE´ R, DRAGOMIR-AGARWAL AND ... 5
Then the under assumptations of Theorem 2.1 with  = 1, we have HermiteHadamard inequality of (1.1).

3. Hermite-Hadamard-Fej´er type inequality

Theorem 3.1. Let u : [a, b]  R be convex and integrable function with a < b.

If

v

:

[a, b]



R

is

nonnegative,

integrable

and

symmetric

with

respect

to

a+b 2

,

i.e.

v(a + b - x) = v(x), then the following inequalities hold

u

a+b 2

[Iav(b) + Ibv(a)]



[Ia (uv) (b) +

Ib

(uv) (a)]



u(a) + u(b) 2

[Iav(b)

+ Ibv(a)] .

(3.1)

Proof. Since u is a convex function on [a, b], we have for all t  [0; 1] the inequality (2.3). Multiplying both sides of (2.3) by

exp (-At) v ((1 - t)a + tb) ,

(3.2)

then integrating the resulting inequality with respect to t over [0, 1], we obtain

2u

a+b 2

1
exp (-At) v ((1 - t)a + tb) dt

0

1

 exp (-At) u (ta + (1 - t)b) v ((1 - t)a + tb) dt

0 1
+ exp (-At) u ((1 - t)a + tb) v ((1 - t)a + tb) dt

0

b

=

b

1 -

a

exp

a

- 1 -  (s - a) 

u (a + b - s) v(s)ds

b

+

b

1 -

a

exp - 1 -  (s - a) u(s)v(s)ds 

a

b

=

b

1 -

a

exp - 1 -  (b - s) u(s)v (a + b - s) ds 

a

+

b

 -

a Ib

[u(a)v(a)]

=

b

 -

a

[Ia

[u(a)v(a)]

+

Ib

[u(a)v(a)]] ,

i.e.

2u

a+b 2

1
exp (-At) v ((1 - t)a + tb) dt

0



b

 -

a

[Ia

[u(a)v(a)]

+

Ib

[u(a)v(a)]] .

6

M. KIRANE AND B. T. TOREBEK

Since

v

is

symmetric

with

respect

to

a+b 2

,

then

the

following

equalities

hold

Iav(b)

=

Ibv(a)

=

1 2

[Iav(b) +

Ibv(a)] .

Therefore, we have

u

a+b 2

[Iav(b) + Ibv(a)]  Ia [v (b) u(b)] + Ib [v (a) u(a)]

and the first inequality of Theorem 3.1 is proved. For the proof of the second inequality in (3.1) we first note that if u is a convex
function, then, for all t  [0; 1], it yields the inequality (2.4). Then multiplying both sides of (2.3) by (3.2) and integrating the resulting inequality with respect to t over [0; 1], we get

1
exp (-At) u (ta + (1 - t)b) v ((1 - t)a + tb) dt

0 1
+ exp (-At) u ((1 - t)a + tb) v ((1 - t)a + tb) dt

0 1
 [u(a) + u(b)] exp (-At) v ((1 - t)a + tb) dt.

0

As a result, we obtain

Ia

[v

(b) u(b)]

+

Ib

[v

(a) u(a)]



u(a)

+ 2

u(b)

[Iav(b)

+

Ibv(a)] .

Theorem 3.1 is proved

Corollary 3.2. Let u : [a, b]  R be concave and integrable function with a < b. If

v

:

[a, b]  R

is

nonnegative,

integrable

and

symmetric

to

a+b 2

,

i.e.

v(a + b - x) =

v(x), then the following inequalities hold

u

a+b 2

[Iav(b) + Ibv(a)]

 [Ia (uv) (b) + Ib (uv) (a)]



u(a)

+ 2

u(b)

[Iav(b)

+

Ibv(a)] .

Remark 3.3. Under assumptations of Theorem 3.1 with  = 1, we have HermiteHadamard-Fej´er inequality of (1.2).

4. Dragomir-Agarwal type inequality
Theorem 4.1. Let u : I  R  R be a differentiable mapping on I, a, b  I. If |u| is convex on [a, b], then the following inequality holds:

u(a)

+ 2

u(b)

-

2

(1

1- - exp (-A))

[Iau(b)

+

Ibu(a)]



b-a 2A

tanh

A 4

(|u(a)| + |u(b)|) .

(4.1)

HERMITE-HADAMARD, HERMITE-HADAMARD-FEJE´ R, DRAGOMIR-AGARWAL AND ... 7

Proof. For u  L1(a, b) it is easy to prove the validity of the equality

u(a)

+ 2

u(b)

-

2

(1

1- - exp (-A))

[Ibu(a)

+

Iau(b)]

1

=

2 (1

b-a - exp (-A))

 

exp (-At) u (ta + (1 - t)b) dt

0

1



 - exp (-A(1 - t)) u (ta + (1 - t)b) dt .


0

Then using (4.2) and the convexity of |u|, we find

(4.2)

u(a) + u(b)

2

-

2

(1

1- - exp (-A))

[Ibu(a)

+

Iau(b)]

1



b

- 2

a

|exp

(-At) - exp (-A(1 1 - exp (-A)

-

t))|

|u

(ta

+

(1

-

t)b)|

dt

0

1



b

- 2

a

|exp

(-At) - exp (-A(1 1 - exp (-A)

-

t))|

t

|u

(a)|

dt

0

1

+

b

- 2

a

|exp

(-At) - exp (-A(1 1 - exp (-A)

-

t))|

(1

-

t)

|u

(b)|

dt

0

1

2

=

b-a 2

|u (a)|

exp

(-At) - exp (-A(1 1 - exp (-A)

-

t))

tdt

0

1

+

b

- 2

a

|u

(a)|

exp (-A(1 - t)) - exp (-At)

1 - exp (-A)

tdt

1 2

1

2

+

b

- 2

a

|u

(b)|

exp

(-At) - exp (-A(1 1 - exp (-A)

-

t))

(1

-

t)dt

0

1

+

b

- 2

a

|u

(b)|

exp

(-A(1 - t)) - exp 1 - exp (-A)

(-At)

(1

-

t)dt

1 2

=

2 (1

b-a - exp (-A))

[|u (a)| (I1

+

I2)

+ |u (b)| (I3

+

I4)] .

As a result, we get

u(a)

+ 2

u(b)

-

2

(1

1- - exp (-A))

[Ibu(a)

+

Iau(b)]

8

M. KIRANE AND B. T. TOREBEK



2 (1

b-a - exp (-A))

[|u

(a)| (I1

+

I2)

+

|u

(b)| (I3

+

I4)] .

Calculating I1 we obtain

(4.3)

1 2

I1 = (exp (-At) - exp (-A(1 - t))) tdt

0
Similarly, we find

=

-

exp

- A

A 2

+

1 A2

(1

-

exp

(-A))

.

(4.4)

1

I2 = (exp (-A(1 - t)) - exp (-At)) tdt

1 2

=

1 A

1 - exp

-

A 2

+ exp (-A)

-

1 A2

(1

-

exp

(-A))

,

(4.5)

1 2
I3 = (exp (-At) - exp (-A(1 - t))) (1 - t)dt

0
and

=

-

exp

- A

A 2

+

1 A

(1

+

exp

(-A))

-

1 A2

(1

-

exp

(-A))

(4.6)

1

I2 = (exp (-At) - exp (-A(1 - t))) (1 - t)dt

1 2

=

-

exp

- A

A 2

+

1 A2

(1

-

exp

(-A))

.

(4.7)

Thus if we use (4.4)-(4.7) in (4.3), we obtain the inequality of (4.1). This completes

the proof.

Corollary 4.2. Let u : I  R  R be a differentiable mapping on I, a, b  I. If |u| is concave on [a, b], then the following inequality holds:

u(a)

+ 2

u(b)

-

2

(1

1- - exp (-A))

[Iau(b)

+

Ibu(a)]



b-a 2A

tanh

A 4

(|u(a)| + |u(b)|) .

Remark 4.3. For   1, we get

lim
1

2 (1

1- - exp (-A))

=

1 2(b -

a) ,

HERMITE-HADAMARD, HERMITE-HADAMARD-FEJE´ R, DRAGOMIR-AGARWAL AND ... 9

lim
1

b-a 2A

tanh

A 4

=

b

- 8

a

.

Then the under assumptations of Theorem 4.1 with  = 1, we have DragomirAgarwal inequality of (1.3).

5. Pachpatte type inequalities

Theorem 5.1. Let u and v be real-valued, nonnegative and convex functions on [a, b]. Then the following inequalities hold

 2(b -

a)

[Ia

(u(b)v(b))

+

Ib

(u(a)v(a))]

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

 [u(a)v(a) + u(b)v(b)]

2A3

+

[u(a)v(b)

+

u(b)v(a)]

A

-

2

+

exp (-A) A3

(A

+

2)

,

(5.1)

2u

a+b 2

v

a+b 2



2 (1

1- - exp (-A))

[Iau(b)v(b)

+

Ibu(a)v(a)]

+

[u(a)v(a)

+

u(b)v(b)]

A

- 2 + exp (-A) (A + A2 (1 - exp (-A))

2)

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

+ [u(a)v(b) + u(b)v(a)]

2A2 (1 - exp (-A))

. (5.2)

Proof. Since u and v are convex on [a, b], then for t  [0, 1] from definition 1.1, we get

u (ta + (1 - t)b) v (ta + (1 - t)b)  t2u(a)v(a) + (1 - t)2u(b)v(b) + t(1 - t) [u(a)v(b) + u(b)v(a)] .

Similarly, we have

u ((1 - t)a + tb) v ((1 - t)a + tb)  (1 - t)2u(a)v(a) + t2u(b)v(b) + t(1 - t) [u(a)v(b) + u(b)v(a)] .

Consequently u (ta + (1 - t)b) v (ta + (1 - t)b) + u ((1 - t)a + tb) v ((1 - t)a + tb)  (2t2 - 2t + 1) [u(a)v(a) + u(b)v(b)] + 2t(1 - t) [u(a)v(b) + u(b)v(a)] .

(5.3)

Multiplying both sides of inequality (5.3) by exp (-At) , then integrating the resulting inequality with respect to t  [0, 1], we obtain

10

M. KIRANE AND B. T. TOREBEK

1
exp (-At) u (ta + (1 - t)b) v (ta + (1 - t)b) dt

0 1

+ exp (-At) u ((1 - t)a + tb) v ((1 - t)a + tb) dt

0

=

b

 -

a

[Ia

(u(b)v(b))

+

Ib

(u(a)v(a))]

1

 [u(a)v(a) + u(b)v(b)] exp (-At) (2t2 - 2t + 1)dt

0 1

+ [u(a)v(b) + u(b)v(a)] exp (-At) 2t(1 - t)dt

0

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

= [u(a)v(a) + u(b)v(b)]

A3

+

2

[u(a)v(b)

+

u(b)v(a)]

A

-

2

+

exp (-A) A3

(A

+

2)

.

So

 2(b -

a)

[Ia

(u(b)v(b))

+

Ib

(u(a)v(a))]

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

 [u(a)v(a) + u(b)v(b)]

2A3

+

[u(a)v(b)

+

u(b)v(a)]

A

-

2

+

exp (-A) A3

(A

+

2)

,

which completes the proof of (5.1). Now let us prove the inequality (5.2). The functions u and v are convex on [a, b],
then we obtain

u

a+b 2

v

a+b 2

=u

ta

+

(1 2

-

t)b

+

(1

-

t)a 2

+

tb

×

×v

ta

+

(1 2

-

t)b

+

(1

-

t)a 2

+

tb



u (ta

+

(1

-

t)b)

+ 2

u ((1

-

t)a

+

tb) ×

×

v

(ta

+

(1

-

t)b)

+ 2

v

((1

-

t)a

+

tb)



u (ta

+

(1

-

t)b) v 4

(ta

+

(1

-

t)b)

+

u ((1

-

t)a

+

tb) v 4

((1

-

t)a

+

tb)

+

t(1 - 2

t)

[u(a)v(a)

+

u(b)v(b)]

HERMITE-HADAMARD, HERMITE-HADAMARD-FEJE´ R, DRAGOMIR-AGARWAL AND ... 11

That is

u

a+b 2

v

a+b 2

+

(2t2

- 2t 4

+

1)

[u(a)v(b)

+

u(b)v(a)]

.



u (ta + (1

-

t)b) v (ta + (1 4

-

t)b)

+

u

((1

-

t)a

+

tb) v 4

((1

-

t)a

+

tb)

(5.4)

+

t(1

- 2

t)

[u(a)v(a)

+

u(b)v(b)]

+

(2t2

- 2t 4

+

1)

[u(a)v(b)

+

u(b)v(a)]

.

Multiplying both sides of (5.4) by exp (-At) , then integrating the resulting in-

equality with respect to t  [0, 1], we have

1

-

exp A

(-A)

u

a+b 2

v

a+b 2

1



exp

(-At)

u

(ta

+

(1

-

t)b) v 4

(ta

+

(1

-

t)b)

dt

0

1

+

exp

(-At)

u

((1

-

t)a

+

tb) v 4

((1

-

t)a

+

tb)

dt

0

1

+

exp

(-At)

t(1

- 2

t)

[u(a)v(a)

+

u(b)v(b)]

dt

0

1

+

exp

(-At)

2t2

- 2t 4

+

1

[u(a)v(b)

+

u(b)v(a)]

dt

0

=

 4(b -

a)

[Iau(b)v(b) + Ibu(a)v(a)]

That is, we have

+

[u(a)v(a)

+

u(b)v(b)]

A

-

2

+

exp (-A) 2A3

(A

+

2)

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

+ [u(a)v(b) + u(b)v(a)]

4A3

.

u

a+b 2

v

a+b 2



4 (1

1- - exp (-A))

[Iau(b)v(b)

+

Ibu(a)v(a)]

+

[u(a)v(a)

+

u(b)v(b)]

A

- 2 + exp 2A2 (1 -

(-A) (A + exp (-A))

2)

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

+ [u(a)v(b) + u(b)v(a)]

4A2 (1 - exp (-A))

.

12

M. KIRANE AND B. T. TOREBEK

This ends the proof.

Corollary 5.2. Let u and v be real-valued, nonnegative and concave functions on [a, b]. Then the following inequalities hold

 2(b -

a)

[Ia

(u(b)v(b))

+

Ib

(u(a)v(a))]

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

 [u(a)v(a) + u(b)v(b)]

2A3

+

[u(a)v(b)

+

u(b)v(a)]

A

-

2

+

exp (-A) A3

(A

+

2)

,

2u

a+b 2

v

a+b 2



2 (1

1- - exp (-A))

[Iau(b)v(b)

+

Ibu(a)v(a)]

+

[u(a)v(a)

+

u(b)v(b)]

A

- 2 + exp (-A) (A + A2 (1 - exp (-A))

2)

A2 - 2A + 4 - A2 + 2A + 4 exp (-A)

+ [u(a)v(b) + u(b)v(a)]

2A2 (1 - exp (-A))

.

Remark 5.3. For   1, we get

lim
1

2 (1

1- - exp (-A))

=

1 2(b -

a) ,

lim
1

A

- 2 + exp (-A) (A + A3

2)

=

1 6

,

lim
1

A2

-

2A

+ 4 - A2 + 2A + 4 2A2 (1 - exp (-A))

exp (-A)

=

1 3

.

The the under assumptations of Theorem 5.1 with  = 1, we have Pachpatte inequalities of (1.4) and (1.5).

Acknowledgements
The second named author is supported by the target program 0085/PTSF-14 from the Ministry of Science and Education of the Republic of Kazakhstan.

[PPT92] [DP00] [H1883] [H1893] [F06] [DA98]

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Mokhtar Kirane LaSIE, Facult´e des Sciences, Pole Sciences et Technologies, Universit´e de La Rochelle, Avenue M. Crepeau, 17042 La Rochelle Cedex, France NAAM Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
E-mail address: mkirane@univ-lr.fr

Berikbol T. Torebek Department of Differential Equations, Institute of Mathematics and Mathematical Modeling. 125 Pushkin str., 050010 Almaty, Kazakhistan
E-mail address: torebek@math.kz

