GENERALIZATION OF HADAMARD-TYPE TRAPEZOID INEQUALITIES FOR FRACTIONAL INTEGRAL OPERATORS Текст научной статьи по специальности «Математика»

ISSN 2074-1871 Уфимский математический журнал. Том 13. № 1 (2021). С. 119-130.

GENERALIZATION OF HADAMARD-TYPE TRAPEZOID INEQUALITIES FOR FRACTIONAL INTEGRAL OPERATORS

B. BAYRAKTAR, M. EMIN OZDEMIR

Abstract. The role of convexity theory in applied problems, especially in optimization problems, is well known. The integral Hermite-Hadamard inequality has a special place in this theory since it provides an upper bound for the mean value of a function. In solving applied problems from different fields of science and technology, along with the classical integro-differential calculus, fractional calculus plays an important role. A lot of research is devoted to obtaining an upper bound in the Hermite-Hadamard inequality using operators of fractional calculus.

The article formulates and proves the identity with the participation of the fractional integration operator. Based on this identity, new generalized Hadamard-tvpe integral inequalities are obtained for functions for which the second derivatives are convex and take values at intermediate points of the integration interval. These results are obtained using the convexity property of a function and two classical integral inequalities, the Hermite-Hadamard integral inequality and its other form, the power mean inequality. It is shown that the upper limit of the absolute error of inequality decreases in approximately n2 times, where n is the number of intermediate points. In a particular case, the obtained estimates are consistent with known estimates in the literature. The results obtained in the article can be used in further researches in the integro-differential fractional calculus.

Keywords: convexity, Hermite-Hadamard inequality, Holder inequality, power-mean inequality, Riemann-Liouville fractional Integrals.

Mathematics Subject Classification:26A51, 26D15

1. Introduction

The theory of integral inequalities is being actively developed since it is attracts a lot in interest. In recent decades, by using various methods, inequalities for different classes of convex functions were obtained. These classes of convex functions are based on the definition of the classical convexity of a function. This definition reads as follows (see, for instance, [1] and its references):

Definition 1. The function f : [a,b] ^ R is said to be convex if we have

f (^ + (1 - t) C) ^ tf (£) + (1 - t) f (C) (1.1)

for all E [a, b] and t E [0,1].

One of the most important inequalities in convex analysis is the Hermite-Hadamard integral inequality. This inequality is as follows [2]:

B. Bayraktar, m. Emin Özdemir, Generalization of Hadamard-type trapezoid inequalities for fractional integral operators.

© B. Bayraktar, M. Emin Özdemir. 2021. Submitted April 1, 2020.

Theorem 1.1. Let f : I C R ^ R be a convex function a nd let a,b El with a < b. Then the following double inequality holds:

b

a

A natural extension of classical analysis is the integro-differential calculus. This theory plays a very important role in classical and applied mathematies( [3,4]), In mathematical modeling of complex systems and processes, the application of fractional calculus theory allows one to reflect adequately the properties of real systems and processes in models [4].

Butkovsky et al, [5], along with problems in the theory of fractional calculus, gave examples of real systems in which it is necessary to use that theory.

In the literature, there are various definitions of the fractional integral (see, for example, [6,7]), but the Riemann-Liouville definition is widely used in most applications of fractional calculus.

Definition 2. ( [3]) Let f E L1 [a, b]. The Riemann-Liouville integrals J^+f and JgLf in order a > 0 are defined respectively by

Jaa+f = - t)a-1 f (t) dt, £ > a,

and

b

Jb-f (0 = f^J(t - 0a-1 f (t) dt, £ < b,

oo

where r (a) = f e-Uua-1du is the well known Gamma function. Here for 0

Ja+f (O = J0~f (0 = f(0

a = 0

as a = 1.

Jl+f (0 = Jl-f (0= f (0%

Two classical inequalities are used in the vast majority of studies on the theory of integral inequality. These are Holder inequality and, its other form, the power mean inequality.

Theorem 1.2. (Holder inequality [8]) Let p > 1 and p + ^ = 1. If f(£) and g(£) are real functions defined on [a, b] and if \f\p, \g\q E L[a, b], then

i

b f b \J / b \ i

IP JC I I / C\ \1

\f (0g(0\d£ < I U (0\Pdn I \g (Ofdn (1.3)

with equality holds if and only if A\f(C)\'P = B\g(^)\q almost everywhere, where A and B are constants.

Theorem 1.3. (Power mean inequality [8]) Let q^ 1 and p + ^ = 1. If f(£) and g(£) are real functions defined on [a, b] and if \ f\p, \g\q E L[a, b], then

1 1 1

b / b \1- W b \q

\f (0g(0\dC ^ II \f (OKI If\f (OMy (£)rd£| . (1.4)

b

In recent years, many authors (see [7-18] and references therein) studied Hermite-Hadamard-tvpe inequalities for improvements and generalizations. In these papers, new inequalities for functions from various convexity classes were obtained.

In the studies [14-16], generalizations of an Hadamard-type integral inequality are given for functions with convex first derivatives taking values at intermediate points. In this study, by using the formulated identity, we obtain some new generalizations of Hadamard-type inequalities for functions such that the absolute values of the second derivatives are convex and take values at intermediate points of the interval. In addition, some applications to special means are given.

2. Main results

Let n G N and a,b G R and a < b. The interval [a, b] with a uniform step h = ^ is divided

n

into n subintervals: [a, b] = (J [£k-i, ], where ^ = a + ih, i = 0,1, 2,... ,n.

k=1

To avoid some repetitions, we use the following notation:

A(.,.) is the arithmetic means of real numbers,

/(&-1) + f(60 Vk = —

2

a

~T l J I " t~ J \>>K—1J I \ J \>>KJ l

h V ?fc-l J \ Sfc-1 s

The following theorem can be easily proved

u = ate^) + JVf(ik—i)) - (JZln&) + JTj(ik—i)), k = 1,2,...,rn-

Theorem 2.1. Let f : [a, b] ^ R be a differentiate function on [a, b]. If f G L[a, b], then for all a > 0 the following inequality holds:

A(xi,x2,... ,xn) ^ ^^ 1) A( s i, S2,..., sn) ^ A( yi, y2,..., yn), (2.1)

where

x> = i (^, .. = (jS-,f(&) + ^ns-i)).

n = 1

by (2.1), we get:

1 ) « + «

Lemma 2.1. Let f : I C R ^ R be a twice differentiate mapping on 1°, where 1° is the interior of I. If f" g L[a, b], where a,b el, then for all a > 2, the following identity holds

k=i k=i k=i

where

hk = J ita—i - e) f"(tCk—i + (i -1)&)dt, 0

i

hk = J (ta—i - ta) f((1 - t)Ck—i +t^k)dt. 0

i

Proof. By taking into account that h = — £k-1 and integrating by parts twice integral I1k, we obtain

'-1k

J (ta-1 — ta) f"(l&-1 + (1 — t)^)dt

j- a—1 _ 4- a 1

-h— f'(t^k-1 + (1 — t)^)|0 + ((a — 1) ta-2 — ata-1) f(tik-i + (1 — t)^)dt

h

(

1( (a — 1) ta-2 — ata-1

h

- f(t Ck-1 + (1 — t) Zk )\0

+h (a — 1) I ((a — 2) ta-3 — ata-2) f(t&-1 + (1 — t)&)dt

f(ik-1) + ^^ I (a — 2) I ta-3f(t^-1 + (1 — t)£k)dt

h2 h2 1

,a-2

—a ta-2f (t^-1 + (1 — t)a)dt I .

In the obtained integrals, we change the variables t£k-1 + (1 — t)= x and by taking into account the properties of the gamma function (sT(s) = T(s + 1)), we obtain

f(tk-1) (a — 1)

11k =-VT,--+

h2

h°

dk

dk

a

(a — 2) l( & — x)a-if (x)dx — hi (Zk — x)a-2f (x)dx

dk~

f(£k-1) , r(a) I ja—2r/£\ - Ta-

h2

+

J

a

ha \ «+-

h «+-

1 f(ik )) .

dk-

Similarlv, for the other integral

f(Ck) , r(a)

hk = J-^ + f(^-1)— ^ f( ^

By summing these identities, we obtain

11k + I 2 k

f(Ck-1) + m) V(aUT ( JX1 f(Ck) + J?-1/(Ck-1)

h2

ha \h\

)

or

— (v2 mk) + Jaf2f dk-1)

r + r =№-1)±fM r(a) TT

11k + 12k = -To---" Uk.

h2

h°

(2.3)

By multiplying both sides of the equality (2.3) by the expression t2 and taking the sum over k, we obtain (2.2). This completes the proof. □

Theorem 2.2. Let f : I C R ^ R be twice differentiable function on 1°. If f E L[a, b], where a,b El and \ f"\ is a convex function, then for all a > 2 the following inequality holds:

£ f(&-1) + f(&) — r(a) y Uk

k=1

2ha-2 ^

k=1

<

h2

¡a^ZM"^* \f" & )\) (M k=1

1

1

1

1

Proof. Bv Lemma 2,1 and the triangle inequality, we obtain

^ f(&-i) + f(&) r(g)

Z^ 2 Oha-2 2^tUk

k=i

k=1

u2 n

^ Y e (i ^ i + 112k i) k=1

(2.5)

Since |/''| is convex with the account of inequality (1.1), we can write

i hk i

(t*-1 - r) f^k-! + (1 - t)Ck)dt

< If''(&-i)i / t (ta-1 - ta) dt + If'' (a)I (ta-1 - ta) (1 - t)dt

(a + 1) (a + 2)

If''(Ck-i)I +

a ( a + 1) ( a + 2)

If''(Ck )I

Hence, for the integral Iik we obtain

i hk i ^

( +1)1( +2) (mCk-i)I + - i f(tk)i

( a + 1) ( a + 2) a

)

Similarly, for the second integral 112k | we can write

i hk I <

( a + 1) ( a + 2) a

(

2

- If''(Ck-i)I + If"(£k)I

)

By adding the last two inequalities, we get

i hk I + I hk I < -fO-rrs (If'(Zk-i)I + If (&)i) a ( a + 1)

(2.6)

By multiplying both sides of inequality (2.6) by the expression ^ and taking into account inequality (2.5), we obtain (2.4). The proof is complete. □

a = 2

A( yi, V2,..., Vn) T- f(x)dx

<

(b -a) 24n2

A(Zi, Z2, . . ., Zn),

(2.7)

where

Zk = If(Ck-i)I + If(Ck )i , fc = 1, 2,... ,n.

i

i

i

1

2

1

Proof. Since h = — £k-1 = ^ and a = 2, for the second expression under the absolute value in the right hand side of inequality (2,4) we have:

r(a) 2ha-2

EU =2$ £ (h & > + JVfM— (C &) +

k=1

2h

1 2

1 2

1 2

k=1

n _

£ Ti{Jk-1 ^k) + ^ m-1)) — m) +J- m-1)

k=1

2 n b — a

i dfc

E

k=1

dk

\

4n

J f(x)dx + J f(x)dx \^k-1 dk-1 J

( d, \

E ( f( & ) + f( Ck-1))

k=1

b — a 1 / 4n

E

k=1

( x) d x

e( № k ) + f(t k-1))

/

k=1

2 1 b — a b

\dfc-1

f(x)dx — £ ( /(£k) + /(Zk-1))

k=1

2n ff(x)dx — 2 £ ( ) + f(Zk-1)).

b — a

k=1

Thus, we can rewrite inequality (2,4) as

^ m-)±m) _ rial ^TT

Z^ 2 2h»-2 2^Uk

k=1

k=1

E

k=1

f( Ck-1) + f( Ck ) 2

2 n 1 n n 1 f(x)dx — K ik ) + f( tk-1))

£ ( f( Ck-1) + /( Ck))

k=1

b — a

2 n b — a

a

k=1

f(x)dx

£ (\f''(Ck-1)\ + \f''(Ck )\)

12 n2

k=1

or

E

k=1

f( Ck-1) + f( Ck) n

f(x)dx

(h — )2 n

£ (\f "(C k-1)\ + \f ''(£ k )\)

<

24n2

k=1

Bv dividing both sides of the last inequality by n, we obtain (2,7), The proof is completed, □

n = 1

f(a) + f(b) 1

b — a

f(x)dx

^ (1—T (\f"(a)\ + \f"m.

(2.8)

This inequality for convex functions was obtained, by M. Sankaya and N. Aktan, see [18, Prop. 2], and in [11, Cor. 3.1].

2

2

Proposition 1. Let £ G (0,1) and n G N. In order to satisfy the inequality,

A(y u У2,..., yn) -

1

a

( x) d x

< e,

it is sufficient to

where ||/''|| = sup |/''(x)|

x€(a,b)

n >

(2.9)

Proof. From the right hand side of the inequality (2.7), we have

it-fA( ^ ^..., *») = ^J2 £ (if &—i)i + if & )i)

24n3

(2.10)

k=i

c (2n uni)-(b -a

24n3

12 n2

|

and since

n

The proof is complete.

(b - a)2

12 n2

b - a

n >

'|| C e,

□

Remark 2. Let R( f, h) be the estimation error of the Trapezoidal rule of numerical integration, then by (2.7) we obtain the estimate for the error known in the literature:

R( f,h)=b--1 h2 I If I.

Remark 3. //||/"|| = sup if(x)i, then from inequality (2.8), it follows that

x£[a, b)

f(a) + f(b)

1

a

f(x)dx

<

(b - a)2 12~

(2.11)

and from (2.7), by taken into account inequality (2.10), we get

A (yu y2,..., yn) -

1

( x) x

<

(b - a)2 12 n2

|

(2.12)

n

n2

Theorem 2.3. Let f : I C R ^ R be a twice differentiate function on 1°. If f G L [a, b] , where a,b G I and | ff is a convex function on [a, b], then for all a > 2, q > 1, and

i + i = 1 the following inequality holds

2 2h»—2/.^Uk

k=i

k=i

<

h2

W

2 a

(2.13)

k=i

|

2

where

= (r(a + 1)T(q + 1) \ 1 — V T(a + q + 2) ) ,

Dk = (a\f''(£k-1)\q + (q + 1)\f''(£k-1)\q)1

+ ((Q + 1) \f''(tk-1)\q + a \f''(ik-1)\q)1. Proof. By Lemma 2,1 and the triangle inequality, we obtain

2^ 2 2h»-2 2sUk

k=1 k=1

By using the Holder integral inequality (1,3) and since \ f'\q is a convex function, we have

(b - a)2

^ (\hk\ + |hk|). (2.14)

i hu \

i

V1 (1 -t) f"(tCu-i + (1 -1)&)dt

0

1

a — 1 a — 1

< J t^t^r(1 — t) \f"(t£k-1 + (1 — t)&)\dt

o

^ IJ ta-1dij IJ ta-1(1 — t)q (t\f"(ik-1)\q + (1 — t) \f(ik )\q)dt

= a-i (\f"( Ck-1)\q J ta(1 — t)qdt + \f"( £k-1)\q j ta-1(1 — t) q+1dt

\ o o = a-? (B (a + 1,q + 1) \f''(^-1)\q + \f''()\9 B (a,q + 2)) ,

B ( ., . )

By the Euler functions properties

T(a + 1)T(q+1) aT(a)T(q+1)

B (a + 1,q + 1) B (a, q + 2)

r(a + q + 2) r(a + q + 2) '

r(a)r(q + 2) _ (q + 1)r(a)r(q + 1)

r(a + q + 2) r(a + q + 2) '

for the first integral, we get

m (ar1 ()1 (pwMr+b+wMf)1

= 1 (T(a + 1)r(^+ 1) ^ 1

a \ r(a

— 1

- ■ (a\f''(£k-1)\q + (q+1)\f''(Ck-1)\q)1 .

a

2 k

— 1

\ I-k\ i —— ■ ((« + 1) + a\f"(Sk-^)1

(r<Z +>r + +)i})1 (ai ^) i 5 + + 1) i i")1 i hull

Bv adding the last inequalities, we get

w

\hk\ + \hk\ ^ - ■Dk. (2.15)

a

Q

By multiplying both sides of inequality (2,15) by the expression and taking into account (2,14), we obtain inequality (2,13), The proof is complete, □

a = 2,

(b - a)2 w(q)

A(yu V2,..., yn) -

1

a

f(x)dx

<

8 n2

A(Di,D2,...,Dn),

(2.16)

where

W( )

2

(q+1)(q + 2)(q + 3)

Dk = (2 | f'(ik—i )r + (q + 1) in&)iq)« + {(q + 1) \f"(Zk—i)\q + 2 \f"(^ )|

a = 2

f(Ck—i) + f(Ck) r(a) =

k=i

E-

2

2ha—2

E ( f(&) + f( ^—i))

k=i k=i For the right hand side of inequality (2,13), we get

' T(a + 1)T(q + 1) \ 1 _ (T(3)T(q+1) \ 1

2n

a

( x) x

W( ) =

r(a + q + 2) 2qT(q)

(q + 3)(q + 2)(q + 1)qT(q) Hence, inequality (2,13) becomes

V r(q + 4) )1=(

2

2 n

^f(&) + f(Ck—i)^ - 1 J

(q + 3)(q + 2)(q + 1)

h2 W( )

)

( x) x

4

ZD,.

k=i

2 n

n = 1

(b - a)2

f(a) + Kb) 1

a

f(x)dx

<

w(q) • Di,

where

W( )

2

(q+1)(q + 2)(q + 3)

)

Di = (2 if "(a)\q + (q + 1) \f(b)\q )i + {(q + 1) \f(a)\q + 2 \f

Since the

and

we hence have

G

2

lim . . . . . . . q^i+\ (q + 1)(q + 2)(q + 3)

2

G

lim , . . . . . . (q+1)(q + 2)(q + 3)

3

) )

« _ 1 = 12

1

— <

(

12^ \(q + 1)(q + 2)(q + 3)

)

<1

(2.17)

□

(2.18)

9

Q

2

9

1

Q

for all q > 1. For q ^ 1+, by (2.18) we get (2.8).

Theorem 2.4. Let f : I C R ^ R be a twice differentiate fun ction on 1° .If f G L [a, b], where a,b G I and \ f"\q is a convex function on [a, b], then for all a > 2 and q^ 1 the following inequality holds

2 2ha—2/-^Uk

k=i

where

§

k=i

c —,

2

(2.19)

k=i

( — )9

\a + 2J

a(a + 1) \a

vfc={\f'(^—i)\q+a \r (^ )\9)9 +( a \f'(^—i)\q+\f & )\*)9.

Proof. Owing to identity (2.2) and the triangle inequality, we can write:

(b - a)2

2 2ha—2Z^Uk

(\hk \ + \hk\) .

(2.20)

2 2ha—2

k=i k=i

Bv using the well-known power mean integral inequality (1.4) and since \ f\q is a convex function, we have

\ hk \

ta—i (1 - t) f(t^—i + (1 - t)a)dt

i-1

C \ j ta—i (1 - t)dt

ta—i (1 - t) (t\f''(Zk—i)r + (1 - t) \f"(6)\q)dt

or

(a(a + 1))

i-1

(->—v—u

\a(a + 1)J V

i i

\f(Ck—i)\q j (1 - t)tadt + \f''(Ck )r I ta—i(1 - t)2dt\ 00

2

1

a(a + 1) 1

(a + 1)(a + 2)

V—! (

\a(a + 1)J \

\f "(Z k—i)\q +

\f''(a )\9

a(a + 1)(a + 2)

( f1) i 9( ( +u( +2))q (\f"(^—i)\q+a a''(Ck )\q

a(a + 1) J \(a + 1)(a + 2) J \ a

)

\ hk \ C

1

a( a + 1) a + 2 Similarly, for the second integral, we can write

a 2 \ t \f(^—i)\q + 2 \f"(&)r

a

\ hk \ c

1

a

va+2

a 2 \f(Ck—i)\q + \f"(Ck )!*)'

a( a + 1) a + 2 a By adding side by side the last inequalites (2.21) and (2.22), we get

1

\ hk \ + \ hk \ C

a( a + 1) a

(-Y

\a + 2J

Vk.

(2.21)

(2.22)

(2.23)

By multiplying both sides of inequality (2.23) by the expression ^r and taking into account (2.20), we obtain inequality (2.19). The proof is completed. □

1

i

9

9

9

Corollary 4. For a = 2, from Theorem 2-4, we get

A( Vn) -

1

b — a

f(x)dx

<

(b -a)2 12 n2

A(v 1, V2, ... , Vn),

where

vk

(

If ''( îk-1 )r + If ''( îk )r V

2

k =1, 2,... ,n.

a = 2

1

f a \ 1 = 1 (1\ 1

U + 2J =6\2j

a (a + 1) [a + 2J 6 \2

vk=(mck-i)iq+a if (^)r)9 +(a if(^+if" (^)9

= 2 (if k-i)iq + if ''( & )Iq )«

and, in view of (2,17), we can write

£ ( f( Ck ) + f( Ck-i))

k=1

2 n

a

f(x)dx

<

(b -a)2 ( 1

12 n2 2

k 6 n2

®; t

v / k=1

(2.24)

2 n

k.

k=1

Bv dividing both sides of the last inequality by 2n, we get (2,24), Remark 5. If we choose n = 1, from (2.24), we 9et

f(a) + f(b) 1

2

( x) x

(b-a)2 ( if"(a)iq + If»™*

<

12

(

2

)1

□

(2.25)

For q = 1, from (2.25), we get (2.8).

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Bahtiyar Bavraktar,

Bursa Uludag University,

Faculty of Education,

Gorukle Campus,

16059, Bursa, Turkey

E-mail: bbayraktar@uludag.edu.tr

Muhammet Emin Özdemir,

Bursa Uludag University,

Faculty of Education,

Gorukle Campus,

16059, Bursa, Turkey

E-mail: eminozdemirSuludag. edu. tr