Open Access Published by De Gruyter Open Access April 26, 2018

Generalizations of Steffensen’s inequality via the extension of Montgomery identity

Andrea Aglić Aljinović, Josip Pečarić and Anamarija Perušić Pribanić
From the journal Open Mathematics

Abstract

In this paper, we obtained new generalizations of Steffensen’s inequality for n-convex functions by using extension of Montgomery identity via Taylor’s formula. Since 1-convex functions are nondecreasing functions, new inequalities generalize Stefensen’s inequality. Related Ostrowski type inequalities are also provided. Bounds for the reminders in new identities are given by using the Chebyshev and Grüss type inequalities.

MSC 2010: 26D15; 26A51

1 Introduction

In [1], the authors obtain the following extension of Montgomery identity using Taylor’s formula:

Theorem 1.1

Let f : I → ℝ be suct that f(n − 1)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < b. Then the following identity holds

f ( x ) = 1 b a a b f t d t i = 0 n 2 f i + 1 x b x i + 2 a x i + 2 i + 2 ! b a + 1 ( n 1 ) ! a b T n x , s f n s d s (1)

where

T n ( x , s ) = 1 n b a a s n , a s x ; 1 n b a b s n , x < s b .

Remark 1.2

The last identity holds also for n = 1. In this special case, we assume that i = 0 n 2 is an empty sum. Thus (1) reduces to well-knownMontgomery identity (e.g. [2])

f ( x ) = 1 b a a b f ( t ) d t + a b T 1 ( x , s ) f ( s ) d s

where the Peano kernel is

T 1 ( x , s ) = s a b a , a s x ; s b b a , x < s b .

The aim of this paper is to obtain some new generalizations of Steffensen’s inequality using above extension of Montgomery identity. The Steffensen’s inequality was first given and proved by Steffensen in 1918 ([3]):

Theorem 1.3

Suppose that f is nonincreasing and g is integrable on [a, b] with 0 ≤ g ≤ 1 and λ = a b g ( t ) d t . Then we have

b λ b f ( t ) d t a b f ( t ) g ( t ) d t a a + λ f ( t ) d t . (2)

The inequalities are reversed for f nondecreasing.

In [4] Jakšetić and Pečarić generalized Steffensen’s inequality for positive measures.

Mitrinović stated in [5] that the inequalities in (2) follow from the identities which will be the starting point for our generalizations of Steffensen’s inequality.

First, let us recall the definition of n-convex functions.

Let f be a real-valued function defined on the segment [a, b]. The divided difference of order n of the function f at distinct points x0, …, xn ∈ [a, b], is defined recursively (see [6, 7]) by

f [ x i ] = f ( x i ) , ( i = 0 , , n )

and

f [ x 0 , , x n ] = f [ x 1 , , x n ] f [ x 0 , , x n 1 ] x n x 0 .

The value f[x0, …, xn] is independent of the order of the points x0, …, xn.

The definition may be extended to include the case when some (or all) of the points coincide. Assuming that f(j − 1)(x) exists, we define

f [ x , , x j t i m e s ] = f ( j 1 ) ( x ) ( j 1 ) ! . (3)

The notion of n-convexity goes back to Popoviciu ([8]). We follow the definition given by Karlin ([9]):

Definition 1.4

A function f: [a, b] → ℝ is said to be n-convex on [a, b], n ≥ 0, if for all choices of (n + 1) distinct points in [a, b], nth order divided difference of f satisfies

f [ x 0 , . . . , x n ] 0.

Note that, 1–convex functions are nondecreasing functions. If f(n) exists, then f is n–convex iff f(n) ≥ 0.

The paper is organized as follows. After this Introduction, in Section 2 we obtain new identities related to Steffensen’s inequality. Using these new identities we generalize Steffensen’s inequality for n–convex functions. Further, in Section 3 we give Ostrowski-type inequalities related to our new generalizations. We conclude this paper with some new bounds for our identities, using the Chebyshev and Grüss type inequalities.

Throughout the paper, it is assumed that all integrals under consideration exist and that they are finite.

2 Generalizations of Steffensen’s inequality via the extension of Montgomery identity

In this section we obtain generalizations of Steffensen’s inequality for n-convex functons using identity (1).

Theorem 2.1

Let f : I → ℝ be suct that f(n − 1)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < b. Let g, p : [a, b] → ℝ be integrable functions such that p is positive and 0 ≤ g ≤ 1. Let a a + λ p ( t ) d t = a b g ( t ) p ( t ) d t and let the function G1be defined by

G 1 ( x ) = a x ( 1 g ( t ) ) p ( t ) d t , x [ a , a + λ ] , x b g ( t ) p ( t ) d t , x [ a + λ , b ] . (4)

Then

a a + λ f ( t ) p ( t ) d t a b f ( t ) g ( t ) p ( t ) d t + a b G 1 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x = 1 ( n 2 ) ! a b a b G 1 ( x ) T n 1 ( x , s ) d x f ( n ) ( s ) d s . (5)

Proof

Using identity

[ a , a + λ ] f ( t ) p ( t ) d t [ a , b ] f ( t ) g ( t ) p ( t ) d t = [ a , a + λ ] f ( t ) ( 1 g ( t ) ) p ( t ) d t ( a + λ , b ] f ( t ) g ( t ) p ( t ) d t

and integration by parts we have

a a + λ f ( t ) p ( t ) d t a b f ( t ) g ( t ) p ( t ) d t = a a + λ [ f ( t ) f ( a + λ ) ] [ 1 g ( t ) ] p ( t ) d t + a + λ b [ f ( a + λ ) f ( t ) ] g ( t ) p ( t ) d t = a a + λ a x ( 1 g ( t ) ) p ( t ) d t d f ( x ) a + λ b x b g ( t ) p ( t ) d t d f ( x ) = a b G 1 ( x ) d f ( x ) = a b G 1 ( x ) f ( x ) d x .

Applying identity (1) to f and replacing n with n − 1 we have

f ( x ) = f ( b ) f ( a ) b a i = 0 n 3 f i + 1 x b x i + 2 a x i + 2 i + 2 ! b a + 1 ( n 2 ) ! a b T n 1 ( x , s ) f ( n ) ( s ) d s .

Now we obtain

a b G 1 ( x ) f ( x ) d x = a b G 1 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x + 1 ( n 2 ) ! a b G 1 ( x ) a b T n 1 ( x , s ) f ( n ) ( s ) d s d x . (6)

After applying Fubini’s theorem on the last term in (6) we obtain (5). □

Theorem 2.2

Let f : I → ℝ be suct that f(n − 1)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < b. Let g, p : [a, b] → ℝ be integrable functions such that p is positive and 0 ≤ g ≤1. Let b λ b p ( t ) d t = a b g ( t ) p ( t ) d t and let the function G2be defined by

G 2 ( x ) = a x g ( t ) p ( t ) d t , x [ a , b λ ] , x b ( 1 g ( t ) ) p ( t ) d t , x [ b λ , b ] . (7)

Then

a b f ( t ) g ( t ) p ( t ) d t b λ b f ( t ) p ( t ) d t + a b G 2 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x = 1 ( n 2 ) ! a b a b G 2 ( x ) T n 1 ( x , s ) d x f ( n ) ( s ) d s . (8)

Proof

Similarly as in the proof of Theorem 2.1, we use the identity

[ a , b ] f ( t ) g ( t ) p ( t ) d t ( b λ , b ] f ( t ) p ( t ) d t = [ a , b λ ] f ( t ) g ( t ) p ( t ) d t ( b λ , b ] f ( t ) ( 1 g ( t ) ) p ( t ) d t .

Now, using the above obtained identites we give generalization of Steffensen’s inequality for n-convex functions.

Theorem 2.3

Let f : I → ℝ be suct that f(n − 1)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < b. Let g, p : [a, b] → ℝ be integrable functions such that p is positive and 0 ≤ g ≤1. Let a a + λ p ( t ) d t = a b g ( t ) p ( t ) d t and let the function G1be defined by (4). If f is nconvex and

a b G 1 ( x ) T n 1 ( x , s ) d x 0 , s [ a , b ] , (9)

then

a b f ( t ) g ( t ) p ( t ) d t a a + λ f ( t ) p ( t ) d t + a b G 1 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x . (10)

Proof

If the function f is n-convex, without loss of generality we can assume that f is n–times differentiable and f(n) ≥ 0 see [7, p. 16 and p. 293]. Now we can apply Theorem 2.1 to obtain (10). □

Theorem 2.4

Let f : I → ℝ be suct that f(n − 1)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < b. Let g, p : [a, b] → ℝ be integrable functions such that p is positive and 0 ≤ g ≤1. Let b λ b p ( t ) d t = a b g ( t ) p ( t ) d t and let the function G2be defined by (7). If f is nconvex and

a b G 2 ( x ) T n 1 ( x , s ) d x 0 , s [ a , b ] , (11)

then

a b f ( t ) g ( t ) p ( t ) d t b λ b f ( t ) p ( t ) d t a b G 2 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x . (12)

Proof

Similar to the proof of Theorem 2.3. □

Remark 2.5

If the integrals in (9) and (11) are nonpositive, then the reverse inequalities in (10) and (12) hold. Note that in this case for some odd n ≥ 3, functions Gi, i = 1, 2 are nonnegative so integrals in (9) and (11) are nonpositive. Hence, inequalities (10) and (12) are reversed.

3 Ostrowski-type inequalities

In this section we give the Ostrowski-type inequalities related to generalizations obtained in the previous section.

Here, the symbol Lp [a, b] (1 ≤ p < ∞) denotes the space of p-power integrable functions on the interval [a, b] equipped with the norm

f p = a b f t p d t 1 p

and L[a, b] denotes the space of essentially bounded functions on [a, b] with the norm

f = ess sup t a , b f t .

Theorem 3.1

Suppose that all assumptions of Theorem 2.1 hold. Assume also that (p, q) is a pair of conjugate exponents, that is 1 ≤ p, q ≤ ∞, 1/p+1/q = 1 andf(n)Lp[a, b] for some n ≥ 2. Then we have

a a + λ f ( t ) p ( t ) d t a b f ( t ) g ( t ) p ( t ) d t + a b G 1 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x 1 ( n 2 ) ! f n p a b G 1 ( x ) T n 1 ( x , ) d x q . (13)

The constant on the right-hand side of (13) is sharp for 1 < p ≤ ∞ and the best possible for p = 1.

Proof

Let’s denote

C ( s ) = 1 ( n 2 ) ! a b G 1 ( x ) T n 1 ( x , s ) d x .

By taking the modulus of (5) and applying Hölder’s inequality we obtain

a a + λ f ( t ) p ( t ) d t a b f ( t ) g ( t ) p ( t ) d t + a b G 1 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x = a b C ( s ) f ( n ) ( s ) d s f n p C q .

For the proof of the sharpness of the constant ∥Cq let us find a function f for which the equality in (13) is obtained.

For 1 < p < ∞ take f to be such that

f ( n ) ( s ) = sgn  C ( s ) C ( s ) 1 p 1 .

For p = ∞ take f(n)(s) = sgn C(s).

For p = 1 we prove that

a b C ( s ) f ( n ) ( s ) d t max t [ a , b ] C ( t ) a b f ( n ) ( s ) d s (14)

is the best possible inequality. C(⋅) is a continuous function on [a, b] and so is |C(⋅)| . Suppose that |C(⋅)| attains its maximum at s0 ∈ [a, b]. First we assume that C(s0) > 0. For ε > 0 small enough we define fε(s) by

f ε ( s ) = 0 , a s s 0 , 1 ε n ! ( s s 0 ) n , s 0 s s 0 + ε , 1 n ! ( s s 0 ) n 1 , s 0 + ε s b .

Then

a b C ( s ) f ε ( n ) ( s ) d s = s 0 s 0 + ε C ( s ) 1 ε d s = 1 ε s 0 s 0 + ε C ( s ) d s .

Now from the inequality (14) we have

1 ε s 0 s 0 + ε C ( s ) d s 1 ε C ( s 0 ) s 0 s 0 + ε d s = C ( s 0 ) .

Since,

lim ε 0 1 ε s 0 s 0 + ε C ( s ) d s = C ( s 0 )

the statement follows. In the case C(s0) < 0, we define fε(s) by

f ε ( s ) = 1 n ! ( s s 0 ε ) n 1 , a s s 0 , 1 ε n ! ( s s 0 ε ) n , s 0 s s 0 + ε , 0 , s 0 + ε s b ,

and the rest of the proof is the same as above. □

Theorem 3.2

Suppose that all assumptions of Theorem 2.2 hold. Assume also (p, q) is a pair of conjugate exponents, that is 1 ≤ p, q ≤ ∞, 1/p+1/q = 1. Letf(n)Lp[a, b] for some n ≥ 2. Then we have

a b f ( t ) g ( t ) p ( t ) d t b λ b f ( t ) p ( t ) d t + a b G 2 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x 1 ( n 2 ) ! f n p a b G 2 ( x ) T n 1 ( x , ) d x q . (15)

The constant on the right-hand side of (15) is sharp for 1 < p ≤ ∞ and the best possible for p = 1.

Proof

Similar to the proof of Theorem 3.1. □

4 Generalizations related to the bounds for the Chebyshev functional

Let f, h : [a, b] → ℝ be Lebesgue integrable functions. We define the Chebyshev functional T(f, h) by

T ( f , h ) := 1 b a a b f ( t ) h ( t ) d t 1 b a a b f ( t ) d t 1 b a a b h ( t ) d t .

In 1882 Chebyshev proved that

T f , h 1 12 f h b a 2 ,

provided that f, h exist and are continuous on [a, b] and ∥f = supt ∈ [a, b] | f(t) |. It also holds if f, h : [a, b] → are absolutely continuous and f, gL[a, b] while ∥f = ess supt ∈ [a, b]|f(t) |.

In 1934, Grüss in his paper [10] proved that

T f , h 1 4 M m N n ,

provided that there exist the real numbers m, M, n, N such that

m f t M , n h t N

for a.e. t ∈ [a, b]. The constant 1/4 is the best possible.

In [11] Cerone and Dragomir proved the following theorems:

Theorem 4.1

Let f : [a, b] → ℝ be a Lebesgue integrable function and h : [a, b] → ℝ be an absolutely continuous function with (⋅ −a)(b – ⋅)[h]2L1[a, b]. Then we have the inequality

| T ( f , h ) | 1 2 [ T ( f , f ) ] 1 2 1 b a a b ( x a ) ( b x ) [ h ( x ) ] 2 d x 1 2 . (16)

The constant 1 2 in (16) is the best possible.

Theorem 4.2

Assume that h : [a, b] → ℝ is monotonic nondecreasing on [a, b] and f : [a, b] → ℝ is absolutely continuous with fL[a, b]. Then we have the inequality

| T ( f , h ) | 1 2 ( b a ) f a b ( x a ) ( b x ) d h ( x ) .1 (17)

The constant 1 2 in (17) is the best possible.

In the sequel we use the above theorems to obtain some new bounds for integrals on the left hand side in the perturbed version of identities (5) and (8).

Firstly, let us denote

Ω i ( s ) = a b G i ( x ) T n 1 ( x , s ) d x , i = 1 , 2. (18)

Theorem 4.3

Let f : I → ℝ be suct that f(n)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < b and (⋅ –a)(b – ⋅)[f(n+1)]2L1[a, b]. Let g, p : [a, b] → ℝ be integrable functions such that p is positive and 0 ≤ g ≤ 1. Let a a + λ p ( t ) d t = a b g ( t ) p ( t ) d t and let the functions G1and Ω1be defined by (4) and (18). Then

a a + λ f ( t ) p ( t ) d t a b f ( t ) g ( t ) p ( t ) d t + a b G 1 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x + f ( n 1 ) ( b ) f ( n 1 ) ( a ) ( b a ) ( n 2 ) ! a b Ω 1 ( s ) d s = S n 1 ( f ; a , b ) , (19)

where the remainder S n 1 ( f ; a , b ) satisfies the estimation

S n 1 ( f ; a , b ) b a 2 ( n 2 ) ! T ( Ω 1 , Ω 1 ) 1 2 a b ( s a ) ( b s ) [ f ( n + 1 ) ( s ) ] 2 d s 1 2 . (20)

Proof

Applying Theorem 4.1 for fΩ1 and hf(n) we obtain

1 b a a b Ω 1 ( s ) f ( n ) ( s ) d s 1 b a a b Ω 1 ( s ) d s 1 b a a b f ( n ) ( s ) d s 1 2 T ( Ω 1 , Ω 1 ) 1 2 1 b a a b ( s a ) ( b s ) [ f ( n + 1 ) ( s ) ] 2 d t 1 2 . (21)

Now if we add

1 ( b a ) ( n 2 ) ! a b Ω 1 ( s ) d s a b f ( n ) ( s ) d s = f ( n 1 ) ( b ) f ( n 1 ) ( a ) ( b a ) ( n 2 ) ! a b Ω 1 ( s ) d s

to both sides of identity (5) and use inequality (21), we obtain representation (19) and bound (20). □

Similarly, using identity (8) we obtain the following result:

Theorem 4.4

Let f : I → ℝ be suct that f(n)is absolutely continuous for some n ≥ 2, I ⊂ ℝ an open interval, a, bI, a < band (⋅ –a)(b – ⋅)[f(n+1)]2L1[a, b]. Let g, p : [a, b] → ℝ be integrable functions such that p is positive and 0 ≤ g ≤ 1. Let b λ b p ( t ) d t = a b g ( t ) p ( t ) d t and let the functions G2andΩ2be defined by (7) and (18). Then

a b f ( t ) g ( t ) p ( t ) d t b λ b f ( t ) p ( t ) d t + a b G 2 ( x ) f ( b ) f ( a ) b a i = 0 n 3 f ( i + 2 ) ( x ) b x i + 2 a x i + 2 i + 2 ! b a d x + f ( n 1 ) ( b ) f ( n 1 ) ( a ) ( b a ) ( n 2 ) ! a b Ω 2 ( s ) d s = S n 2 ( f ; a , b ) , (22)

where the remainder S n 2 ( f ; a , b ) satisfies the estimation

S n 2 ( f ; a , b ) b a 2 ( n 2 ) ! T ( Ω 2 , Ω 2 ) 1 2 a b ( s a ) ( b s ) [ f ( n + 1 ) ( s ) ] 2 d s 1 2 .

Proof

Similar to the proof of Theorem 4.3. □

The following Grüss-type inequalities also hold.

Theorem 4.5

Let f : I → ℝ be suct that f(n)is absolutely continuous for some n ≥ 2 and f(n+1) ≥ 0 on [a, b]. Let functions Ωi, i = 1, 2 be defined by (18).

  1. Let a a + λ p ( t ) d t = a b g ( t ) p ( t ) d t . Then we have representation (19) and the remainder S n 1 ( f ; a , b ) satisfies the bound

    S n 1 ( f ; a , b ) b a ( n 2 ) ! Ω 1 f ( n 1 ) ( b ) + f ( n 1 ) ( a ) 2 f ( n 2 ) ( b ) f ( n 2 ) ( a ) b a . (23)

  2. Let b λ b p ( t ) d t = a b g ( t ) p ( t ) d t . . Then we have representation (22) and the remainder S n 2 ( f ; a , b ) satisfies the bound

    S n 2 ( f ; a , b ) b a ( n 2 ) ! Ω 2 f ( n 1 ) ( b ) + f ( n 1 ) ( a ) 2 f ( n 2 ) ( b ) f ( n 2 ) ( a ) b a .

Proof

  1. Applying Theorem 4.2 for fΩ1, hf(n) we obtain

    1 b a a b Ω 1 ( s ) f ( n ) ( s ) d s 1 b a a b Ω 1 ( s ) d s 1 b a a b f ( n ) ( s ) d s 1 2 ( b a ) Ω 1 a b ( s a ) ( b s ) f ( n + 1 ) ( s ) d s . (24)

    Since

    a b ( s a ) ( b s ) f ( n + 1 ) ( s ) d s = a b [ 2 s ( a + b ) ] f ( n ) ( s ) d s = ( b a ) f ( n 1 ) ( b ) + f ( n 1 ) ( a ) 2 f ( n 2 ) ( b ) f ( n 2 ) ( a )

    Using representation (5) and inequality (24) we deduce (23).

  2. Similar to the (a)-part.

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Received: 2016-10-13
Accepted: 2018-02-07
Published Online: 2018-04-26

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