JJ II

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volume 6, issue 2, article 37, 2005.

Received 16 September, 03;

accepted 06 February, 2005.

Communicated by:F. Qi

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Journal of Inequalities in Pure and Applied Mathematics

NEW INEQUALITIES OF HADAMARD’S TYPE FOR LIPSCHITZIAN MAPPINGS

LIANG-CHENG WANG

School of Mathematics Scientia Chongqing Institute of Technology Xingsheng Lu 4, Yangjia Ping 400050 Chongqing City, China.

EMail:wlc@cqit.edu.cn

c

2000Victoria University ISSN (electronic): 1443-5756 123-03

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New Inequalities of Hadamard’s Type for Lipschitzian Mappings

Liang-Cheng Wang

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J. Ineq. Pure and Appl. Math. 6(2) Art. 37, 2005

http://jipam.vu.edu.au

Abstract

In this paper, we study several new inequalities of Hadamard’s type for Lips- chitzian mappings.

2000 Mathematics Subject Classification: Primary 26D07; Secondary 26B25, 26D15.

Key words: Lipschitzian mappings, Hadamard inequality, Convex function.

This author is partially supported by the Key Research Foundation of the Chongqing Institute of Technology under Grant 2004ZD94.

1. Introduction

Let us begin by defining some mappings that we shall need. Iff is a continuous function on a closed interval [a, b] (a < b), and for any t ∈ (0,1), let u = ta+ (1−t)b, then we can define

A(t)=⁴ 1

t(1−t)(b−a)² Z u

a

Z b

u

f(tx+ (1−t)y)dy

dx,

B(t)=⁴ 1 (1−t)(b−a)²

Z u

a

Z b

u

f

(b−y)x+ (y−u)u t(b−a)

dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

f

(u−x)u+ (x−a)y (1−t)(b−a)

dy

dx,

C(t)=⁴ t (1−t)(b−a)

Z u

a

f(x)dx+ 1−t t(b−a)

Z b

u

f(y)dy.

Letf be a continuous convex function on[a, b], then

(1.1) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx≤ f(a) +f(b)

2 .

The inequalities in (1.1) are called Hadamard inequalities [1] – [7]. In [2], the author of this paper gave extensions and refinements of the inequalities in (1.1).

He obtained the following:

(1.2) f(ta+ (1−t)b)≤A(t)≤B(t)≤C(t)≤tf(a) + (1−t)f(b).

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Recently, Dragomir et al. [3], Yang and Tseng [4] and Matic and Peˇcari´c [5]

proved some results for Lipschitzian mappings related to (1.1). In this paper, we will prove some new inequalities for Lipschitzian mappings related to the mappingsA, BandC(or to (1.2)).

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2. Main Results

For the mappingA(t), we have the following theorem:

Theorem 2.1. Let f : [a, b] → R be a M-Lipschitzian mapping. For any t ∈(0,1), then

(2.1) |A(t)−f(ta+ (1−t)b)| ≤ M

3 t(1−t)(b−a).

Proof. Obviously, we have

(2.2) b−u=t(b−a), u−a= (1−t)(b−a).

From integral properties and (2.2), we have (2.3) f(ta+ (1−t)b) = 1

t(1−t)(b−a)² Z u

a

Z b

u

f(ta+ (1−t)b)dy

dx.

Forx∈[a, u], wheny∈

u,−_1−t^t x+ _1−t^u

, then we have (2.4) ta+ (1−t)b=u≥tx+ (1−t)y, ify∈

−_1−t^t x+_1−t^u , b

, then the inequality (2.4) reverses.

Using (2.2) – (2.4) and integral properties, we obtain

|A(t)−f(ta+ (1−t)b)|

≤ 1

t(1−t)(b−a)² Z u

a

Z b

u

|f(tx+ (1−t)y)−f(u)|dy

dx

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≤ M

t(1−t)(b−a)² Z u

a

Z b

u

|tx+ (1−t)y−u|dy

dx

= M

t(1−t)(b−a)² Z u

a

"

Z −_1−t^t x+_1−t^u

u

(u−(tx+ (1−t)y))dy

# dx

+ M

t(1−t)(b−a)² Z u

a

"

Z b

−_1−t^t x+_1−t^u

(tx+ (1−t)y−u)dy

# dx

= M

t(1−t)(b−a)² Z u

a

t²

1−tx²+t

b− 1 +t 1−tu

x + 1 +t²

2(1−t)u²+ 1−t

2 b²−bu

dx

= M

t(b−a)

t²

3(1−t)(u²+ua+a²) + t 2

b−1 +t 1−tu

(u+a) + 1 +t²

2(1−t)u²+ 1−t

2 b²−bu

= M

t(b−a)

2t²−3t+ 3

6(1−t) (ta+ (1−t)b)² +−t(t+ 3)a−3(t²−3t+ 2)b

6(1−t) (ta+ (1−t)b) + t²

3(1−t)a² + t

2ab+ 1−t 2 b²

= M

3 t(1−t)(b−a).

This completes the proof of Theorem2.1.

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For the mappingB(t), we have the following theorem:

Theorem 2.2. Letf be defined as in Theorem2.1. For anyt∈(0,1), then

(2.5) |B(t)−A(t)| ≤ M

2 t(1−t)(b−a), and

(2.6) |B(t)−f(ta+ (1−t)b)| ≤ M

2 t(1−t)(b−a).

Proof. From _t(1−t)¹ = ¹_t +_1−t¹ , we have (2.7) A(t) =

1 t + 1

1−t

1 (b−a)²

Z u

a

Z b

u

f(tx+ (1−t)y)dy

dx.

LetF be defined by

F(y) = tx+ (1−t)y− (b−y)x+ (y−u)u t(b−a) .

Obviously, ifF(y)is the function of first degree fory, thenF(y)is monotone with y on [u, b]. And for any x ∈ [a, u], F(u) = tx + (1− t)u− x ≥ 0, F(b) = tx+ (1−t)b−u ≥ 0. Hence for x ∈ [a, u] andy ∈ [u, b], we get F(y)≥0, i.e.,

(2.8) tx+ (1−t)y≥ (b−y)x+ (y−u)u t(b−a) .

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LetGbe defined by

G(x) = (u−x)u+ (x−a)y

(1−t)(b−a) −(tx+ (1−t)y). UsingG(x)and the same method as in the proof of (2.8), we obtain

(2.9) (u−x)u+ (x−a)y

(1−t)(b−a) ≥tx+ (1−t)y, where,x∈[a, u]andy∈[u, b].

Using (2.7) – (2.9) and (2.2), we obtain

|B(t)−A(t)|

≤ 1

(1−t)(b−a)² Z u

a

Z b

u

f

−f(tx+ (1−t)y)

dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

f

(u−x)u+ (x−a)y (1−t)(b−a)

−f(tx+ (1−t)y)

dy

dx

≤ M

(1−t)(b−a)² Z u

a

Z b

u

(b−y)x+ (y−u)u

t(b−a) −(tx+ (1−t)y)

dy

dx

+ M

t(b−a)² Z u

a

Z b

u

(u−x)u+ (x−a)y

(1−t)(b−a) −(tx+ (1−t)y)

dy

dx

= M

(1−t)(b−a)² Z u

a

Z b

u

tx+ (1−t)y− (b−y)x+ (y−u)u t(b−a)

dy

dx

+ M

t(b−a)² Z u

a

Z b

u

(u−x)u+ (x−a)y

(1−t)(b−a) −(tx+ (1−t)y)

dy

dx

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= M

t(1−t)(b−a)³ Z u

a

Z b

u

(b−a)(2t−1) (tx+ (1−t)y) + 2xy−(b+u)x−(a+u)y+ 2u²

dy

dx

= M

(1−t)(b−a)² Z u

a

1

2((b−a)(2t−1)(1−t)−(a+u)) (b+u) + (b−a)(2t−1)tx+ 2u²

dx

= M

2(b−a)

((b−a)(2t−1)(1−t)−(a+u)) (b+u) + (b−a)(2t−1)t(u+a) + 4u²

= M

2 t(1−t)(b−a).

This completes the proof of inequality (2.5).

From (2.3) and _t(1−t)¹ = ¹_t +_1−t¹ , we have (2.10) f(ta+ (1−t)b)

= 1

t + 1 1−t

1 (b−a)²

Z u

a

Z b

u

f(ta+ (1−t)b)dy

dx.

Ifx∈[a, u]andy ∈[u, b], then we have

ta+ (1−t)b=u≥ (b−y)x+ (y−u)u t(b−a) , (2.11)

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(u−x)u+ (x−a)y

(1−t)(b−a) ≥ta+ (1−t)b =u.

Using (2.10)-(2.11) and (2.2), we obtain

|B(t)−f(ta+ (1−t)b)|

≤ 1

(1−t)(b−a)² Z u

a

Z b

u

f

−f(u)

dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

f

(u−x)u+ (x−a)y (1−t)(b−a)

−f(u)

dy

dx

≤ M

(1−t)(b−a)² Z u

a

Z b

u

(b−y)x+ (y−u)u t(b−a) −u

dy

dx

+ M

t(b−a)² Z u

a

Z b

u

(u−x)u+ (x−a)y (1−t)(b−a) −u

dy

dx

= M

(1−t)(b−a)² Z u

a

Z b

u

u−(b−y)x+ (y−u)u t(b−a)

dy

dx

+ M

t(b−a)² Z u

a

Z b

u

(u−x)u+ (x−a)y (1−t)(b−a) −u

dy

dx

= M

t(1−t)(b−a)³ Z u

a

Z b

u

(b−a)(2t−1)u

+ (2x−u−a)y−(b+u)x+ 2u² dy

dx

= M

(1−t)(b−a)² Z u

a

−1

2(u+a)(b+u) + (b−a)(2t−1)u+ 2u²

dx

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= M

b−a

−1

2(u+a)(b+u) + (b−a)(2t−1)u+ 2u²

= M

2 t(1−t)(b−a).

This completes the proof of Theorem (2.2).

For the mappingC(t), we have the following theorem:

Theorem 2.3. Letf be defined as in Theorem2.1. For anyt∈(0,1), then

(2.12) |C(t)−B(t)| ≤ M

2 t(1−t)(b−a), and

(2.13) |C(t)−(tf(a) + (1−t)f(b))| ≤M t(1−t)(b−a).

Proof. From the integral property and (2.2), we have

(2.14) C(t) = 1

(1−t)(b−a)² Z u

a

Z b

u

f(x)dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

f(y)dy

dx.

Ifx∈[a, u]andy ∈[u, b], then we have (2.15) x≤ (b−y)x+ (y−u)u

t(b−a) , (u−x)u+ (x−a)y (1−t)(b−a) ≤y.

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Using (2.14) – (2.15) and (2.2), we obtain

|C(t)−B(t)|

≤ 1

(1−t)(b−a)² Z u

a

Z b

u

f

−f(x)

dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

f

(u−x)u+ (x−a)y (1−t)(b−a)

−f(y)

dy

dx

≤ M

(1−t)(b−a)² Z u

a

Z b

u

(b−y)x+ (y−u)u t(b−a) −x

dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

(u−x)u+ (x−a)y (1−t)(b−a) −y

dy

dx

= M

(1−t)(b−a)² Z u

a

Z b

u

(b−y)x+ (y−u)u t(b−a) −x

dy

dx

+ 1

t(b−a)² Z u

a

Z b

u

y−(u−x)u+ (x−a)y (1−t)(b−a)

dy

dx

= M

t(1−t)(b−a)³ Z u

a

Z b

u

(b−a) ((1−t)y−tx)

−2xy+ (b+u)x+ (a+u)y−2u² dy

dx

= M

(1−t)(b−a)² Z u

a

1

2((b−a)(1−t) + (a+u)) (b+u)

−(b−a)tx−2u²

dx

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= M

b−a 1

2(((b−a)(1−t) + (a+u)) (b+u)−(b−a)t(u+a))−2u²

= M

2 t(1−t)(b−a).

From integral property and (2.2), we have (2.16) tf(a)+(1−t)f(b) = t

(1−t)(b−a) Z u

a

f(a)dx+ 1−t t(b−a)

Z b

u

f(b)dy.

Using (2.16) and (2.2), we obtain

|C(t)−(tf(a) + (1−t)f(b))|

≤ 1 b−a

t 1−t

Z u

a

|f(x)−f(a)|dx+ 1−t t

Z b

u

|f(y)−f(b)|dy

≤ M b−a

t 1−t

Z u

a

|x−a|dx+1−t t

Z b

u

|y−b|dy

= M

b−a t

1−t Z u

a

(x−a)dx+ 1−t t

Z b

u

(b−y)dy

= M

b−a t

1−t 1

2(u²−a²)−a(u−a)

+1−t t

b(b−u)− 1

2(b² −u²)

=M t(1−t)(b−a).

This completes the proof of Theorem2.3.

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Corollary 2.4. Letf be defined as in Theorem2.1. For anyt∈(0,1), then (2.17) |C(t)−f(ta+ (1−t)b)| ≤M t(1−t)(b−a),

(2.18) |A(t)−C(t)| ≤M t(1−t)(b−a),

(2.19) |A(t)−(tf(a) + (1−t)f(b))| ≤2M t(1−t)(b−a),

(2.20) |B(t)−(tf(a) + (1−t)f(b))| ≤ 3M

2 t(1−t)(b−a), and

(2.21) |f(ta+ (1−t)b)−(tf(a) + (1−t)f(b))| ≤2M t(1−t)(b−a).

Proof. Using (2.5) and (2.12), we get inequality (2.18):

Using the same method as the proof of (2.18), from (2.6) and (2.12), (2.13) and (2.18), (2.12) and (2.13), and (2.13) and (2.17), we get (2.17), (2.19), (2.20) and (2.21), respectively.

This completes the proof of Corollary2.4.

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Remark 1. If we lett= ¹₂, then (2.13), (2.17) and (2.21) reduce to (2.22)

1 b−a

Z b

a

f(x)dx− f(a) +f(b) 2

≤ M

4 (b−a),

(2.23)

1 b−a

Z b

a

f(x)dx−f

a+b 2

≤ M

4 (b−a),

(2.24)

f

a+b 2

− f(a) +f(b) 2

≤ M

2 (b−a).

(2.22) is better than (2.2) in [3]. (2.23) and (2.24) are (2.1) and (2.4) in [3], respectively.

Corollary 2.5. Let f be a convex mapping on[a, b], withf₊⁰ (a)andf₋⁰(b)ex- isting. For anyt∈(0,1), then we have

0≤A(t)−f(ta+ (1−t)b) (2.25)

≤ max{|f₊⁰ (a)|,|f₋⁰ (b)|}

3 t(1−t)(b−a), 0≤B(t)−A(t)

(2.26)

≤ max{|f₊⁰ (a)|,|f₋⁰ (b)|}

2 t(1−t)(b−a),

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0≤B(t)−f(ta+ (1−t)b) (2.27)

≤ max{|f₊⁰ (a)|,|f₋⁰ (b)|}

2 t(1−t)(b−a), 0≤C(t)−B(t)

(2.28)

≤ max{|f₊⁰ (a)|,|f₋⁰ (b)|}

2 t(1−t)(b−a), 0≤tf(a) + (1−t)f(b)−C(t)

(2.29)

≤max{|f₊⁰ (a)|,|f₋⁰ (b)|}t(1−t)(b−a),

0≤C(t)−f(ta+ (1−t)b) (2.30)

≤max{|f₊⁰ (a)|,|f₋⁰ (b)|}t(1−t)(b−a),

0≤C(t)−A(t) (2.31)

≤max{|f₊⁰ (a)|,|f₋⁰ (b)|}t(1−t)(b−a),

0≤tf(a) + (1−t)f(b)−A(t) (2.32)

≤2 max{|f₊⁰(a)|,|f₋⁰(b)|}t(1−t)(b−a),

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0≤tf(a) + (1−t)f(b)−B(t) (2.33)

≤ 3

2max{|f₊⁰(a)|,|f₋⁰(b)|}t(1−t)(b−a), and

0≤tf(a) + (1−t)f(b)−f(ta+ (1−t)b) (2.34)

≤2 max{|f₊⁰ (a)|,|f₋⁰ (b)|}t(1−t)(b−a).

Proof. For any x, y ∈ [a, b], from the properties of convex functions, we have the followingmax{|f₊⁰(a)|,|f₋⁰(b)|}-Lipschitzian inequality (see [7]):

(2.35) |f(x)−f(y)| ≤max{|f₊⁰ (a)|,|f₋⁰ (b)|}|x−y|.

Using (1.2), (2.35), Theorem2.1 – 2.2 and Corollary 2.4, we obtain (2.25) – (2.34).

This completes the proof of Corollary (2.5).

Remark 2. The condition in Corollary2.5is better than that in Corollary 2.2, Corollary 4.2 and Theorem 3.3 of [3]. Actually, it is that f is a differentiable convex mapping on[a, b]withM = sup

t∈[a,b]

|f⁰(t)|<∞.

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References

[1] J. HADAMARD, Etude sur les propriétés des fonctions entières et en par- ticulier d’une fonction considérée par Riemann, J. Math. Pures Appl., 58 (1893), 171–215.

[2] L.-C. WANG, On extensions and refinements of Hermite-Hadamard in- equalities for convex functions, Math. Ineq. Appl., 6(4) (2003), 659–666.

[3] S.S. DRAGOMIR, Y.J.CHO AND S.S. KIM, Inequalities of Hadamard’s type for Lipschitzian mappings and their applications, J. Math. Anal. Appl., 245 (2000), 489–501.

[4] G.-S. YANGAND K.-L. TSENG, Inequalities of Hadamard’s type for Lip- schitzian mappings, J. Math. Anal. Appl., 260 (2001), 230–238.

[5] M. MATICANDJ. PE ˇCARI ´C, Note on inequalities of Hadamard’s type for Lipschitzian mappings, Tamkang J. Math., 32(2) (2001), 127–130.

[6] L.-C. WANG, On some extentions of Hadamard inequalities of convex function, Math. in Practice and Theory, 32 (2002), 1027–1030. (In Chi- nese).

[7] L.-C. WANG, Convex Functions and Their Inequalities, Sichuan University Press, Chengdu, China, 2001, pp.55. (In Chinese).