NEW WEIGHTED INTEGRAL INEQUALITIES FOR TWICE DIFFERENTIABLE CONVEX FUNCTIONS

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Volume 40(1) (2016), Pages 15–33.

M. Z. SARIKAYA1 AND S. ERDEN2

Abstract. In this paper, we establish several new weighted inequalities for some twice differentiable mappings that are connected with the celebrated Hermite-Hadamard type and Ostrowski type integral inequalities. Some of the new inequal-ities are Hermite-Hadamard-type inequalinequal-ities involving fractional integrals. The results presented here would provide extensions of those given in earlier works.

1. Introduction

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

f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)

for all x, y ∈ [a, b] and λ ∈ [0, 1]. We say that f is concave if (−f ) is convex.

The following inequality is well known in the literature as the Hermite-Hadamard integral inequality (see, [6]):

f a + b 2 ≤ 1 b − a Z b a f (x)dx ≤ f (a) + f (b) 2 ,

where f : I ⊂ R → R is a convex function on the interval I of real numbers and a, b ∈ I with a < b.

A largely applied inequality for convex functions, due to its geometrical significance, is Hadamard’s inequality, (see [4, 5, 18–23]) which has generated a wide range of directions for extension and a rich mathematical literature.

In 1938, the classical integral inequality established by Ostrowski [8] as follows:

Key words and phrases. Hermite-Hadamard inequality, Ostrowski inequality, Trapezoid inequality, convex function, Hölder inequality, Rieman-Liouville integrals.

2010 Mathematics Subject Classification. Primary: 26D07. Secondary: 26D15. Received: July 15, 2015.

Accepted: November 28, 2015.

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Theorem 1.1. Let f : [a, b]→ R be a differentiable mapping on (a, b) whose derivative f0 _{: (a, b)→ R is bounded on (a, b), i.e., kf}0k_∞ = sup

t∈(a,b) |f0_{(t)| < ∞. Then, the} inequality holds: f (x) − 1 b − a b Z a f (t)dt ≤ " 1 4 + (x − a+b₂ )2 (b − a)2 # (b − a) kf0k_∞

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

Inequality (1.1) has wide applications in numerical analysis and in the theory of some special means; estimating error bounds for some special means, some mid-point, trapezoid and Simpson rules and quadrature rules, etc. Hence inequality (1.1) has attracted considerable attention and interest from mathematicans and researchers. Due to this, over the years, the interested reader is also refered to [1–3, 7, 9–16] for integral inequalities in several independent variables. In addition, the current approach of obtaining the bounds, for a particular quadrature rule, have depended on the use of Peano kernel. The general approach in the past has involved the assumption of bounded derivatives of degree greater than one.

In this study, using functions whose twice derivatives absolute values are convex, we obtain new weighted inequalities that are connected with the celebrated Hermite-Hadamard type and Ostrowski type integral inequalities. In addition, we obtain new inequalities of Hermite-Hadamard type and Ostrowski type involving fractional integrals.The results presented here would provide extensions of those given in earlier works.

2. Main Results

Throughout this section, let us define the S(α; w, f ) operator as follows:

S(α; w, f ) =     x Z a (u − x) w (u) du   α −   x Z b (u − x) w (u) du   α f 0 (x) + α     x Z a (u − x) w (u) du   α−1  x Z a w (u) du   −   x Z b (u − x) w (u) du   α−1  x Z b w (u) du    f (x) + α (α − 1)    x Z a   t Z a (u − t) w (u) du   α−2  t Z a w (u) du   2 f (t)dt

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+ b Z x   t Z b (u − t) w (u) du   α−2  t Z b w (u) du   2 f (t)dt    − α   x Z a   t Z a (u − t) w (u) du   α−1 w (t) f (t) dt + b Z x   t Z b (u − t) w (u) du   α−1 w (t) f (t) dt  .

In order to prove our main results we need the following lemma.

Lemma 2.1. Let f : I ⊂ R → R be twice differentiable function on I◦, a, b ∈ I◦ with a < b, f00 _{is absolutely continuous on [a, b] and let w : [a, b] → R be nonnegative and} continuous on [a, b]. Then the following identity holds:

(2.1) S(α; w, f ) = b Z a Pw(x, t) f00(t) dt, where Pw(x, t) :=            _t R a (u − t) w (u) du α , a ≤ t < x, t R b (u − t) w (u) du α , x ≤ t ≤ b, for α ≥ 1.

Proof. By integration by parts, we have the following identity: b Z a Pw(x, t) f00(t) dt = x Z a   t Z a (u − t) w (u) du   α f00(t) dt + b Z x   t Z b (u − t) w (u) du   α f00(t) dt =   t Z a (u − t) w (u) du   α f0(t) x a + α x Z a   t Z a (u − t) w (u) du   α−1  t Z a w (u) du  f 0 (t) dt

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+   t Z b (u − t) w (u) du   α f0(t) b x + α b Z x   t Z b (u − t) w (u) du   α−1  t Z b w (u) du  f 0 (t) dt =     x Z a (u − t) w (u) du   α −   x Z b (u − t) w (u) du   α f 0 (x) + α        t Z a (u − t) w (u) du   α−1  t Z a w (u) du  f (t) x a + x Z a  (α − 1)   t Z a (u − t) w (u) du   α−2  t Z a w (u) du   2 −   t Z a (u − t) w (u) du   α−1 w (t)  f (t) dt +   t Z b (u − t) w (u) du   α−1  t Z b w (u) du  f (t) b x + b Z x  (α − 1)   t Z b (u − t) w (u) du   α−2  t Z b w (u) du   2 −   t Z b (u − t) w (u) du   α−1 w (t)  f (t) dt    =     x Z a (u − x) w (u) du   α −   x Z b (u − x) w (u) du   α f 0 (x) + α     x Z a (u − x) w (u) du   α−1  x Z a w (u) du   −   x Z b (u − x) w (u) du   α−1  x Z b w (u) du    f (x)

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+ α (α − 1)    x Z a   t Z a (u − t) w (u) du   α−2 ×   t Z a w (u) du   2 f (t)dt + b Z x   t Z b (u − t) w (u) du   α−2  t Z b w (u) du   2 f (t)dt    − α   x Z a   t Z a (u − t) w (u) du   α−1 w (t) f (t) dt + b Z x   t Z b (u − t) w (u) du   α−1 w (t) f (t) dt  ,

which is the required identity in (2.1) and the proof is completed. Remark 2.1. Under the same assumptions of Lemma 2.1 with α = 1, then the following identity holds: S(1; w, f ) =   b Z a (u − x) w (u) du  f0(x) +   b Z a w (u) du  f (x) − b Z a w (t) f (t) dt, which was proved by Sarikaya and Yaldiz in [16].

Definition 2.1. Let f ∈ L1[a, b]. The Riemann-Liouville integrals Ja+α f and Jb−α f of order α > 0 with a ≥ 0 are defined by

J_a+α f (x) = 1 Γ(α) Z x a (x − t)α−1f (t)dt, x > a, and J_b−α f (x) = 1 Γ(α) Z b x (t − x)α−1f (t)dt, x < b,

respectively. Here, Γ(α) is the Gamma function and J_a+0 f (x) = J_b−0 f (x) = f (x). Corollary 2.1. Under the same assumptions of Lemma 2.1 with w(s) = 1, then the following identity holds:

S(α; 1, f ) = −1 2 α h (a − x)2α− (b − x)2αif0(x) + α −1 2 α−1 h (a − x)2α−1− (b − x)2α−1if (x) (2.2)

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+αΓ(2α) 2 −1 2 α−2 h J_x−2α−1f (a) + J_x+2α−1f (b)i. Remark 2.2. If we take x = a+b₂ in (2.2), we get

a+b 2 Z a   t Z a (u − t) du   α f00(t) dt + b Z a+b 2   t Z b (u − t) du   α f00(t) dt = α −1 2 α−1 (b − a)2α−1 22α−2 f ( a + b 2 ) +αΓ(2α) 2 −1 2 α−2 J2α−1 (a+b 2 )− f (a) + J2α−1 (a+b 2 )+ f (b) .

Theorem 2.1. Let f : I ⊂ R → R be twice differentiable function on I◦, a, b ∈ I◦ with a < b, f00 _{is absolutely continuous on [a, b] and let w : [a, b] → R be nonnegative} and continuous on [a, b]. If |f00| is convex on [a, b] then for all x ∈ [a, b], the following inequalities hold: |S(α; w, f )| ≤ kwk α [a,x],∞ 2α_{(b − a)} × (b − a)(x − a) 2α+1 2α + 1 − (x − a)2α+2 2α + 2 |f00(a)| + (x − a) 2α+2 2α + 2 |f 00 (b)| + kwk α [x,b],∞ 2α_{(b − a)} (2.3) × (b − x) 2α+2 2α + 2 |f 00 (a)| + (b − a)(b − x) 2α+1 2α + 1 − (b − x)2α+2 2α + 2 |f00(b)| ≤ kwk α [a,b],∞ 2α_{(b − a)} (b − a)(x − a)2α+1 2α + 1 + (b − x)2α+2− (x − a)2α+2 2α + 2 |f00(a)| + (b − a)(b − x) 2α+1 2α + 1 + (x − a)2α+2− (b − x)2α+2 2α + 2 |f00(b)| , where α ≥ 1 and kwk_∞= sup

t∈[a,b]

|w(t)|.

Proof. We take the absolute value of (2.1). Using bounded of the mapping w, we find that |S(α; w, f )| ≤ b Z a |P (x, t)| |f00(t)| dt = x Z a   t Z a (t − u) w (u) du   α |f00(t)| dt

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+ b Z x   b Z t (u − t) w (u) du   α |f00(t)| dt ≤ kwk α [a,x],∞ 2α x Z a (t − a)2α|f00(t)| dt + kwk α [x,b],∞ 2α b Z x (b − t)2α|f00(t)| dt.

Since f00(t) is convex on [a, b] = [a, x] ∪ [x, b], we have (2.4) f00 b − t b − aa + t − a b − ab ≤ b − t b − a f 00 (a) + t − a b − a f 00 (b) . From (2.4), it follows that

|S(α; w, f )| ≤ kwk α [a,x],∞ 2α_{(b − a)} x Z a (t − a)2αh(b − t) f 00 (a) + (t − a) f 00 (b) i dt +kwk α [x,b],∞ 2α_{(b − a)} b Z x (b − t)2αh(b − t) f 00 (a) + (t − a) f 00 (b) i dt = kwk α [a,x],∞ 2α_{(b − a)} (b − a)(x − a)2α+1 2α + 1 − (x − a)2α+2 2α + 2 |f00(a)| +(x − a) 2α+2 2α + 2 |f 00 (b)| +kwk α [x,b],∞ 2α_{(b − a)} (b − x)2α+2 2α + 2 |f 00 (a)| + (b − a)(b − x) 2α+1 2α + 1 − (b − x)2α+2 2α + 2 |f00_(b)|

and because of kwk_[a,x],∞ ≤ kwk_[a,b],∞ and kwk_[x,b],∞ ≤ kwk_[a,b],∞, we obtain |S(α; w, f )| ≤ kwk α [a,b],∞ 2α_{(b − a)} (b − a)(x − a)2α+1 2α + 1 + (b − x)2α+2_{− (x − a)}2α+2 2α + 2 f 00 (a) + (b − a)(b − x) 2α+1 2α + 1 + (x − a)2α+2_{− (b − x)}2α+2 2α + 2 f 00 (b) ,

which completes the proof.

Remark 2.3. Under the same assumptions of Theorem 2.1 with α = 1, then the following inequality holds:

|S(1; w, f )| ≤ kwk[a,b],∞ 2 (b − a) (b − a)(x − a)3 3 + (b − x)4− (x − a)4 4 f 00 (a) + (b − a)(b − x) 3 3 + (x − a)4 − (b − x)4 4 f 00 (b) , which was proved by Sarikaya and Yaldiz in [16].

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Corollary 2.2. Under the same assumptions of Theorem 2.1 with w(s) = 1, then the following inequality holds:

|S(α; 1, f )| ≤ 1 2α_{(b − a)} (b − a)(x − a)2α+1 2α + 1 + (b − x)2α+2_{− (x − a)}2α+2 2α + 2 f 00 (a) + (b − a)(b − x) 2α+1 2α + 1 + (x − a)2α+2− (b − x)2α+2 2α + 2 f 00 (b) . (2.5)

Remark 2.4. If we take α = 1 in (2.5), we get a + b 2 − x f0(x) + f (x) − 1 b − a b Z a f (t) dt ≤ 1 2 (b − a)2 " (b − a)(x − a) 3 3 + (b − x)4− (x − a)4 4 ! f 00 (a) + (x − a) 4_{− (b − x)}4 4 + (b − a) (b − x)3 3 ! f 00 (b) # , which is proved by Sarikaya and Yaldiz in [16].

Corollary 2.3. Under the same assumptions of Theorem 2.1 with w(s) = 1 and x = a+b₂ , then we have

α −1 2 α−1(b − a) 2α−1 22α−2 f ( a + b 2 ) + αΓ(2α) 2 −1 2 α−2 h J2α−1a+b 2 − f (a) + J2α−12+b 2 + f (b)i ≤ (b − a) 2α+1 (2α + 1)23α+1 h f 00 (a) + f 00 (b) i . (2.6)

Remark 2.5. If we take α = 1 in (2.6), we obtain f a + b 2 − 1 b − a b Z a f (t) dt ≤ (b − a) 2 24 f00(a)+ f00(b) 2 ! , which is proved by Sarikaya and Yildirim in [17].

Corollary 2.4. Under the same assumptions of Theorem 2.1 with f00(a)= f00(b) in (2.3), then the following inequalities hold:

|S(α; w, f )| ≤ 1 (2α + 1)2α h kwkα_[a,x],∞(x − a)2α+1+ kwkα_[x,b],∞(b − x)2α+1i f 00 (a) ≤ kwk α [a,b],∞ (2α + 1)2α (x − a) 2α+1 + (b − x)2α+1 f 00 (a) .

Theorem 2.2. Let f : I ⊂ R → R be twice differentiable function on I◦, a, b ∈ I◦ with a < b, f00 _{is absolutely continuous on [a, b] and let w : [a, b] → R be nonnegative}

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and continuous on [a, b]. If |f00|q is convex on [a, b], q > 1, then for all x ∈ [a, b], the following inequalities hold:

|S(α; w, f )| ≤ kwk α [a,x],∞(x − a) 2α+1_p 2α_{(b − a)}1q _{(2αp + 1)} 1 p × " (b − a)2− (b − x)2 2 f 00 (a) q + (x − a) 2 2 f 00 (b) q# 1 q + kwk α [x,b],∞(b − x) 2α+1_p 2α_{(b − a)}1q _{(2αp + 1)} 1 p × " (b − x)2 2 f 00 (a) q +(b − a) 2_{− (x − a)}2 2 f 00 (b) q# 1 q ≤ kwk α [a,b],∞ 2α_{(b − a)}1q _{(2αp + 1)} 1 p ×    (x − a)2α+p1 " (b − a)2− (b − x)2 2 f 00 (a) q + (x − a) 2 2 f 00 (b) q# 1 q + (b − x)2α+1p " (b − x)2 2 f 00 (a) q +(b − a) 2_{− (x − a)}2 2 f 00 (b) q# 1 q    ,

where α ≥ 1, 1_p + 1_q = 1, and kwk_∞= sup t∈[a,b]

|w(t)|.

Proof. We take absolute value of (2.1). Using Holder’s inequality, we find that

|S(α; w, f )| ≤ x Z a   t Z a (t − u) w (u) du   α |f00(t)| dt + b Z x   b Z t (u − t) w (u) du   α |f00(t)| dt ≤ kwk α [a,x],∞ 2α x Z a (t − a)2α|f00_{(t)| dt +} kwk α [x,b],∞ 2α b Z x (b − t)2α|f00_{(t)| dt} ≤ kwk α [a,x],∞ 2α   x Z a (t − a)2αpdt   1 p  x Z a |f00(t)|qdt   1 q

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+kwk α [x,b],∞ 2α   b Z x (b − t)2αpdt   1 p  b Z x |f00(t)|qdt   1 q . Since f00(t) q

is convex on [a, b] = [a, x] ∪ [x, b], we have (2.7) f00 b − t b − aa + t − a b − ab q ≤ b − t b − a f 00 (a) q + t − a b − a f 00 (b) q . From (2.7), it follows that

|S(α; w, f )| ≤ kwk α [a,x],∞ 2α (x − a)2α+1p (2αp + 1)1p   x Z a b − t b − a f 00 (a) q + t − a b − a f 00 (b) q dt   1 q +kwk α [x,b],∞ 2α (b − x)2α+1p (2αp + 1)1p   b Z x b − t b − a f 00 (a) q + t − a b − a f 00 (b) q dt   1 q = kwk α [a,x],∞ 2α_{(b − a)}1q (x − a)2α+1p (2αp + 1)1p × " (b − a)2− (b − x)2 2 f 00 (a) q +(x − a) 2 2 f 00 (b) q# 1 q + kwk α [x,b],∞ 2α_{(b − a)}1q (b − x)2α+1p (2αp + 1)1p × " (b − x)2 2 f 00 (a) q +(b − a) 2 _{− (x − a)}2 2 f 00 (b) q# 1 q

and because of kwk_[a,x],∞≤ kwk_[a,b],∞ and kwk_[x,b],∞ ≤ kwk_[a,b],∞. Hence, the proof is

complete.

|S(1; w, f )| ≤ kwk[a,b],∞ 2 (b − a)1q _{(2p + 1)} 1 p ×    (x − a)2+1p " (b − a)2− (b − x)2 2 f 00 (a) q + (x − a) 2 2 f 00 (b) q# 1 q + (b − x)2+1p " (b − x)2 2 f 00 (a) q + (b − a) 2 − (x − a)2 2 f 00 (b) q# 1 q    ,

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which is given by Sarikaya and Yaldiz in [16].

|S(α; 1, f )| ≤ 1 2α_{(b − a)}1q _{(2αp + 1)} 1 p ×    (x − a)2α+p1 " (b − a)2− (b − x)2 2 f 00 (a) q + (x − a) 2 2 f 00 (b) q# 1 q (2.8) + (b − x)2α+1p " (b − x)2 2 f 00 (a) q +(b − a) 2 − (x − a)2 2 f 00 (b) q# 1 q    .

Remark 2.7. If we take α = 1 in (2.8), we get a + b 2 − x f0(x) + f (x) − 1 b − a b Z a f (t) dt ≤ 1 2 (b − a)1+1q _{(2p + 1)} 1 p ×    (x − a)2+p1 " (b − a)2− (b − x)2 2 f 00 (a) q + (x − a) 2 2 f 00 (b) q# 1 q + (b − x)2+1p " (b − x)2 2 f 00 (a) q +(b − a) 2 − (x − a)2 2 f 00 (b) q# 1 q    , which is given by Sarikaya and Yaldiz in [16].

Corollary 2.6. Under the same assumptions of Theorem 2.2 with w(s) = 1 and x = a+b₂ , then, we have

α −1 2 α−1 (b − a)2α−1 22α−2 f a + b 2 + αΓ(2α) 2 −1 2 α−2 J2α−1 (a+b 2 )− f (a) + J2α−1 (a+b 2 )+ f (b) ≤ (b − a) 2α+1 (2αp + 1)1p₂3α+1 (2.9) ×    " 3f00(a) q +f00(b) q 4 #1_q + " f00(a) q + 3f00(b) q 4 #1_q   .

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Remark 2.8. If we take α = 1 in (2.9), we obtain f a + b 2 − 1 b − a b Z a f (t) dt ≤ (b − a) 2 16 (2p + 1)p1    " 3f00(a) q +f00(b) q 4 #1_q + " f00(a) q + 3f00(b) q 4 #1_q   ,

which is proved by Sarikaya and Yildirim in [17].

Theorem 2.3. Let f : I ⊂ R → R be twice differentiable function on I◦, a, b ∈ I◦ with a < b, f00 _{is absolutely continuous on [a, b] and let w : [a, b] → R be nonnegative} and continuous on [a, b]. If |f00|q is convex on [a, b], q > 1, then for all x ∈ [a, b], the following inequalities hold:

|S(α; w, f )| ≤ (b − a) 1 q 2α_{(2αp + 1)}1p h kwkαp_[a,x],∞(x − a)2αp+1+ kwkαp_[x,b],∞(b − x)2αp+1i 1 p × " f00(a) q +f00(b) q 2 #1_q ≤ kwk α [a,b],∞(b − a) 1 q 2α_{(2αp + 1)}1p (x − a)2αp+1 + (b − x)2αp+11p (2.10) × " f00(a) q +f00(b) q 2 #1_q ,

|w(t)|.

Proof. We take absolute value of (2.1). Using Holder’s inequality, we find that

|S(α; w, f )| ≤ b Z a |P (x, t)| |f00(t)| dt ≤   x Z a   t Z a (t − u) w (u) du   αp dt + b Z x   b Z t (u − t) w (u) du   αp dt   1 p ×   b Z a |f00(t)|qdt   1 q

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≤" kwk αp [a,x],∞ 2αp (x − a)2αp+1 2αp + 1 + kwkαp_[x,b],∞ 2αp (b − x)2αp+1 2αp + 1 #1_p ×   b Z a |f00(t)|qdt   1 q .

From (2.7), it follows that

|S(α; w, f )| ≤" kwk αp [a,x],∞ 2αp (x − a)2αp+1 2αp + 1 + kwkαp_[x,b],∞ 2αp (b − x)2αp+1 2αp + 1 #1_p ×   b Z a b − t b − a f 00 (a) q + t − a b − a f 00 (b) q dt   1 q = (b − a) 1 q 2α_{(2αp + 1)}p1 h kwkαp_[a,x],∞(x − a)2αp+1+ kwkαp_[x,b],∞(b − x)2αp+1i 1 p × " f00(a) q +f00(b) q 2 #1_q

and because of kwk_[a,x],∞≤ kwk_[a,b],∞ and kwk_[x,b],∞ ≤ kwk_[a,b],∞. Hence, the proof is

complete.

|S(1; w, f )| ≤ kwk[a,b],∞(b − a) 1 q 2 (2p + 1)1p (x − a)2p+1 + (b − x)2p+1 1 p × " f00(a) q +f00(b) q 2 #1_q

which is proved by Sarikaya and Yaldiz in [16].

|S(α; 1, f )| ≤ (b − a) 1 q 2α_{(2αp + 1)}1p (x − a)2αp+1 + (b − x)2αp+1 1 p × " f00(a) q +f00(b) q 2 #1_q . (2.11)

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Remark 2.10. If we take α = 1 in (2.11), we get a + b 2 − x f0(x) + f (x) − 1 b − a b Z a f (t) dt ≤ (x − a) 2p+1 + (b − x)2p+1 1 p 2 (b − a)1p_{(2p + 1)} 1 p " f00(a) q +f00(b) q 2 #1_q .

Corollary 2.8. Under the same assumptions of Theorem 2.3 with w(s) = 1 and x = a+b₂ , then we have

α −1 2 α−1 (b − a)2α−1 22α−2 f a + b 2 + αΓ(2α) 2 −1 2 α−2 J2α−1 (a+b 2 )− f (a) + J2α−1 (a+b 2 )+ f (b) (2.12) ≤ (b − a) 2α+1 (2αp + 1)1p₂3α " f00(a) q +f00(b) q 2 # . Remark 2.11. If we take α = 1 in (2.12), we obtain

f a + b 2 − 1 b − a b Z a f (t) dt ≤ (b − a) 2 8 (2p + 1)1p " f00(a) q +f00(b) q 2 #1_q , which is proved by Sarikaya and Yildirim in [17].

Corollary 2.9. Under the same assumptions of Theorem 2.3 with f00(a) q

=f00(b) q in (2.10), then the following inequality holds:

|S(α; w, f )| ≤ (b − a) 1 q _f00_(a) 2α_{(2αp + 1)}1p h kwkαp_[a,x],∞(x − a)2αp+1+ kwkαp_[x,b],∞(b − x)2αp+1i 1 p ≤ kwk α [a,b],∞(b − a) 1 q 2α_{(2αp + 1)}1p f 00 (a) (x − a) 2αp+1 + (b − x)2αp+11p_.

Theorem 2.4. Let f : I ⊂ R → R be twice differentiable function on I◦, a, b ∈ I◦ with a < b, f00 _{is absolutely continuous on [a, b] and let w : [a, b] → R be nonnegative} and continuous on [a, b]. If |f00|q is convex on [a, b], q ≥ 1, then for all x ∈ [a, b], the following inequalities hold:

|S(α; w, f )| ≤ kwk α [a,x],∞(x − a) 2α+1 22α_{(2α + 1)}1p_{(b − a)} 1 q × (b − a) + (2α + 1) (b − x) (2α + 1) (2α + 2) f 00 (a) q +(x − a) 2α + 2 f 00 (b) q1_q

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+ kwk α [x,b],∞(b − x) 2α+1 22α_{(2α + 1)}1p _{(b − a)} 1 q × (b − a) + (2α + 1) (x − a) (2α + 1) (2α + 2) f 00 (b) q +(b − x) 2α + 2 f 00 (a) q1_q ≤ kwk α [a,b],∞ 22α_{(2α + 1)}1p_{(b − a)} 1 q (2.13) × (x − a)2α+1 (b − a) + (2α + 1) (b − x) (2α + 1) (2α + 2) f 00 (a) q +(x − a) 2α + 2 f 00 (b) q1_q + (b − x)2α+1 × (b − a) + (2α + 1) (x − a) (2α + 1) (2α + 2) f 00 (b) q +(b − x) 2α + 2 f 00 (a) q1_q) ,

|w(t)|.

Proof. We take absolute value of (2.1). Because of 1_p + 1

q = 1, α 1 p + 1 q can be written instead of α. Using Holder’s inequality and the convexity of f0

From (2.7), it follows that

|S(α; w, f )| ≤ kwk α [a,x],∞ 22α (x − a)2α+1 2α + 1 !1_p

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×   x Z a (t − a)2α b − t b − a f 00 (a) q + t − a b − a f 00 (b) q dt   1 q +kwk α [x,b],∞ 22α (b − x)2α+1 2α + 1 !1_p ×   b Z x (b − t)2α b − t b − a f 00 (a) q + t − a b − a f 00 (b) q dt   1 q = kwk α [a,x],∞ 22α_{(b − a)}1q (x − a)2α+1 2α + 1 !_p1 × " (b − a)(x − a) 2α+1 2α + 1 − (x − a)2α+2 2α + 2 ! f 00 (a) q +(x − a) 2α+2 2α + 2 f 00 (b) q# 1 q + kwk α [x,b],∞ 22α_{(b − a)}1q (b − x)2α+1 2α + 1 !1_p × " (b − a)(b − x) 2α+1 2α + 1 − (b − x)2α+2 2α + 2 ! f 00 (b) q +(b − x) 2α+2 2α + 2 f 00 (a) q# 1 q = kwk α [a,x],∞(x − a) 2α+1 22α_{(2α + 1)}p1 _{(b − a)} 1 q × (b − a) 2α + 1 − (x − a) 2α + 2 f 00 (a) q + (x − a) 2α + 2 f 00 (b) q1_q + kwk α [x,b],∞(b − x) 2α+1 22α_{(2α + 1)}1p_{(b − a)} 1 q × (b − a) 2α + 1 − (b − x) 2α + 2 f 00 (b) q +(b − x) 2α + 2 f 00 (a) q1_q

and because of kwk_[a,x],∞ ≤ kwk_[a,b],∞ and kwk_[x,b],∞ ≤ kwk_[a,b],∞, therefore, the proof

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Corollary 2.10. Under the same assumptions of Theorem 2.4 with α = 1, then the following inequality holds:

|S(1; w, f )| ≤ kwk[a,b],∞ 4 · 31p(b − a) 1 q ( (x − a)3 4b − a − 3x 12 f 00 (a) q +(x − a) 4 f 00 (b) q1_q + (b − x)3 3x − 4a + b 12 f 00 (b) q +(b − x) 4 f 00 (a) q1_q) .

|S(α; 1, f )| ≤ 1 22α_{(2α + 1)}1p_{(b − a)} 1 q × ( (x − a)2α+1 (b − a) + (2α + 1) (b − x) (2α + 1) (2α + 2) f 00 (a) q + (x − a) 2α + 2 f 00 (b) q1_q (2.14) + (b − x)2α+1 (b − a) + (2α + 1) (x − a) (2α + 1) (2α + 2) f 00 (b) q +(b − x) 2α + 2 f 00 (a) q1_q ) . Corollary 2.12. If we take α = 1 in (2.14), we get

a + b 2 − x f0(x) + f (x) − 1 b − a b Z a f (t) dt ≤ 1 4 · 31p_{(b − a)} 1 q+1 ( (x − a)3 4b − a − 3x 12 f 00 (a) q + (x − a) 4 f 00 (b) q1_q + (b − x)3 3x − 4a + b 12 f 00 (b) q +(b − x) 4 f 00 (a) q1_q) .

Corollary 2.13. Under the same assumptions of Theorem 2.4 with w(s) = 1 and x = a+b₂ , we obtain α −1 2 α−1 (b − a)2α−1 22α−2 f a + b 2 + αΓ(2α) 2 −1 2 α−2 J2α−1 (a+b 2 )− f (a) + J2α−1 (a+b 2 )+ f (b)

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≤ (b − a) 2α+1 (2α + 1)1p₂4α+1+ 1 q ( (2α + 3) (2α + 1) (2α + 2) f 00 (a) q + 1 2α + 2 f 00 (b) q1_q + (2α + 3) (2α + 1) (2α + 2) f 00 (b) q + 1 2α + 2 f 00 (a) q1_q) .

Corollary 2.14. Under the same assumptions of Corollary 2.13 with α = 1, then we have f a + b 2 − 1 b − a b Z a f (t) dt ≤ (b − a) 2 21q+5× 3 1 p    " 5f00(a) q + 3f00(b) q 12 #1_q + " 3f00(a) q + 5f00(b) q 12 #1_q   . References

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1

Department of Mathematics, Faculty of Science and Arts, University of Düzce

E-mail address: sarikayamz@gmail.com

2

Department of Mathematics, Faculty of Science,

University of Bartın