New inequalities of hermite-hadamard type for convex functions with applications

Abstract

In this paper, some new inequalities of the Hermite-Hadamard type for functions whose modulus of the derivatives are convex and applications for special means are given. Finally, some error estimates for the trapezoidal formula are obtained.

2000 Mathematics Subject Classiffication. 26A51, 26D10, 26D15.[PUBLICATION ABSTRACT]

Kavurmaci et al. Journal of Inequalities and Applications 2011, 2011:86 http://www.journalofinequalitiesandapplications.com/content/2011/1/86

RESEARCH Open Access

New inequalities of hermite-hadamard type for convex functions with applications

Havva Kavurmaci*, Merve Avci and M Emin zdemir

* Correspondence: mailto:[email protected]

Web End [email protected] Aatatrk University, K.K. Education Faculty, Department of Mathematics, 25240, Campus, Erzurum, Turkey

Abstract

In this paper, some new inequalities of the Hermite-Hadamard type for functions whose modulus of the derivatives are convex and applications for special means are given. Finally, some error estimates for the trapezoidal formula are obtained.2000 Mathematics Subject Classiffication. 26A51, 26D10, 26D15.

Keywords: Convex function, Hermite-Hadamard inequality, Hlder inequality, Power-mean inequality, Special means, Trapezoidal formula

1. Introduction

A function f : I is said to be convex function on I if the inequality

f (x + (1 )y) f (x) + (1 )f (y),

holds for all x, y I and a [0,1].

One of the most famous inequality for convex functions is so called Hermite-Hadamards inequality as follows: Let f : I be a convex function defined on the interval I of real numbers and a, b I, with a <b. Then:

f

a + b2 1b a

ba f (x)dx

f (a) + f (b)

2 . (1:1)

In [1], the following theorem which was obtained by Dragomir and Agarwal contains the Hermite-Hadamard type integral inequality.

Theorem 1. Let f : I be a differentiable mapping on I, a, b I with a <b. If |f| is convex on [a, b], then the following inequality holds:

f (a) + f (b)

2

1b a

(b a)(|f (a)| + |f (b)|)

8 . (1:2)

In [2] Kirmaci, Bakula, zdemir and Peari proved the following theorem. Theorem 2. Let f : I , I be a differentiable function on I such that f L [a, b], where a, b I, a <b. If |f|q is concave on [a, b] for some q > 1, then:

ba f (u)du

2011 Kavurmaci et al; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons

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Page 2 of 11

f (a) + f (b)

2

1b a

ba f (u)du

(1:3)

In [3], Kirmaci obtained the following theorem and corollary related to this theorem. Theorem 3. Let f : I be a differentiable mapping on I, a, b I with a <b and let p > 1. If the mapping |f|p is concave on [a, b], then we have

f (ca + (1 c)b)(B A) + f (a)(1 B) + f (b)A

q1 q

f

b a4 q 1 2q 1

a + 3b 4

+

f

3a + b 4

.

1b a

ba f (x)dx

(b a)

K

f

aT + b(K T) K

+ M

f

aN + b(M N) M

where

K = A2 + (c A)2

2 , T =

A3 + c3

3

Ac2

2 , M =

(B c)2 + (1 B)2

2 ,

N = B3 + c3 + 1

3 (1 + c2)

B

2 .

Corollary 1. Under the assumptions of Theorem 3 with A = B = c = 12, we have

f (a) + f (b)

2

1b a

ba f (x)dx

(1:4)

For recent results and generalizations concerning Hermite-Hadamards inequality see[1]-[5] and the references given therein.

2. The New Hermite-Hadamard Type Inequalities

In order to prove our main theorems, we first prove the following lemma:

Lemma 1. Let f : I be a differentiable mapping on I, where a, b I with a <b. If f L [a, b], then the following equality holds:

(b x)f (b) + (x a)f (a)

b a

1b a

(b a)

8

f

5a + b 6

+

f

a + 5b 6

.

ba f (u)du

= (x a)2

b a

10 (t 1)f (tx + (1 t)a)dt +

(b x)2

b a

10 (1 t)f (tx + (1 t)b)dt.

Proof. We note that

J = (x a)2

b a

10 (t 1)f (tx + (1 t)a)dt

+ (b x)2

b a

10 (1 t)f (tx + (1 t)b)dt.

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Integrating by parts, we get

J = (x a)2

b a

(t 1)f (tx + (1 t)a) x a

1 0

1 0

f (tx + (1 t)a)

x a

dt

+ (b x)2

b a

(1 t)f (tx + (1 t)b) x b

1 0

+

1 0

f (tx + (1 t)b)

x b

dt

= (x a)2

b a

f (a)x a

1

(x a)2

xa f (u)du

f (b)x b+ 1(x b)2 xb f (u)du

= (b x)f (b) + (x a)f (a)

b a

1b a

+ (b x)2

b a

ba f (u)du.

Using the Lemma 1 the following results can be obtained.

Theorem 4. Let f : I be a differentiable mapping on I such that f L [a, b], where a, b I with a <b. If |f| is convex on [a, b], then the following inequality holds:

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

|f (x)| + 2|f (a)|6 + (b x)2 b a

|f (x)| + 2|f (b)| 6

for each x [a, b].

Proof. Using Lemma 1 and taking the modulus, we have

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

10 (1 t)|f (tx + (1 t)a)|dt

+(b x)2

b a

10 (1 t)|f (tx + (1 t)b)|dt.

Since |f| is convex, then we get

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

10 (1 t)[t|f (x)| + (1 t)|f (a)|]dt

+(b x)2

b a

10 (1 t)[t|f (x)| + (1 t)|f (b)|]dt

= (x a)2

b a

|f (x)| + 2|f (a)|6 + (b x)2 b a

|f (x)| + 2|f (b)| 6

which completes the proof.

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Corollary 2. In Theorem 4, if we choose x = a+b2we obtain

f (a) + f (b)

2

1b a

ba f (u)du

b a

12

|f (a)| + f

a + b 2

+ |f (b)|

.

1b a

8 (|f (a)| + |f (b)|)

which is the inequality in (1.2).

Theorem 5. Let f : I be a differentiable mapping on I such that f L [a,

b], where a, b I with a <b. If|f |

p

Remark 1. In Corollary 2, using the convexity of |f| we have

f (a) + f (b)

2

ba f (u)du

b a

p1is convex on [a, b] and for some fixed q > 1, then the following inequality holds:

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

1p + 1

1 p

1 2

1 q

(x a)2

|f (a)|q + |f (x)|q 1q + (b x)2

|f (x)|q + |f (b)|q 1 q

b a

for each x [a, b] and q = p

p1.

Proof. From Lemma 1 and using the well-known Hlder integral inequality, we have

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

10 (1 t)|f (tx + (1 t)a)|dt

+(b x)2

b a

10 (1 t)|f (tx + (1 t)b)|dt

(x a)2

b a

10 (1 t)pdt 1p

1 0

|f (tx + (1 t)a)|qdt 1q

+(b x)2

b a

10 (1 t)pdt 1p

1 0

|f (tx + (1 t)b)|qdt 1q.

|f |

p

Since

p1 is convex, by the Hermite-Hadamards inequality, we have

1 0

|f (tx + (1 t)a)|qdt |

f (a)|q + |f (x)|q

2

and

1 0

|f (tx + (1 t)b)|qdt |

f (b)|q + |f (x)|q

2 ,

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so

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

1p + 1

1 p

1 2

1 q

(x a)2

|f (a)|q + |f (x)|q 1q + (b x)2

|f (x)|q + |f (b)|q 1 q

b a

which completes the proof.

Corollary 3. In Theorem 5, if we choose x = a+b2we obtain

f (a) + f (b)

2

1b a

ba f (u)du

b a

4

1p + 1

1 p

1 2

1 q

|f (a)|q + f

a + b 2

q

1q

+

|f (b)|q + f

a + b 2

q

1q

1q (|f (a)| + |f (b)|).

The second inequality is obtained using the following fact:

nk=1 (ak + bk)s

nk=1 (ak)s +

b a

2

1p + 1

1 p

1 2

nk=1 (bk)sfor (0 s < 1), a1, a2, a3,..., an 0; b1, b2, b3,..., bn 0 with 0 p1p < 1, for p > 1.

Theorem 6. Let f : I be a differentiable mapping on I such that f L [a, b], where a, b I with a <b. If |f|q is concave on [a, b], for some fixed q > 1, then the following inequality holds:

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

f

q 1

2q 1

q1 q

a+x

2 | + (b x)2|f

b+x

2

b a

for each x [a, b].

Proof. As in Theorem 5, using Lemma 1 and the well-known Hlder integral inequal

ity for q > 1 and p = q

q1, we have

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(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

10 (1 t)|f (tx + (1 t)a)|dt

+(b x)2

b a

10 (1 t)|f (tx + (1 t)b)|dt

(x a)2

b a

10 (1 t)

q

q1 dt

q1q

1 0

|f (tx + (1 t)a)|qdt 1q

+(b x)2

b a

10 (1 t)

q

q1 dt

q1q

1 0

|f (tx + (1 t)b)|qdt

1 q .

Since |f|q is concave on [a, b], we can use the Jensens integral inequality to obtain:

1 0

|f (tx + (1 t)a)|qdt =

10 t0|f (tx + (1 t)a)|qdt

10 t0dt

f

1

0 t0dt

10 (tx+(1 t)a)dt

q

1

=

f

a + x 2

q

Analogously,

1 0

|f (tx + (1 t)b)|qdt f

b + x 2

q

.

Combining all the obtained inequalities, we get

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

q 1

2q 1

(x a)2|f

q1 q

a+x

2 | + (b x)2|f

b+x

2

|

b a

which completes the proof.

Remark 2. In Theorem 6, if we choose x = a+b2we have

f (a) + f (b)

2

1b a

ba f (u)du

q1 q

b a4 f

q 1

2q 1

3a + b 4

+

f

a + 3b 4

which is the inequality in (1.3).

Theorem 7. Let f : I be a differentiable mapping on I such that f L [a, b], where a, b I with a <b. If |f|q is convex on [a, b] and for some fixed q 1, then the following inequality holds:

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(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

1

2

1 3

1 q

(x a)2

|f (x)|q + 2|f (a)|q 1q + (b x)2

|f (x)|q + 2|f (b)|q 1 q

b a

for each x [a, b].

Proof. Suppose that q 1. From Lemma 1 and using the well-known power-mean inequality, we have

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

10 (1 t)|f (tx + (1 t)a)|dt

+(b x)2

b a

10 (1 t)|f (tx + (1 t)b)|dt

(x a)2

b a

10 (1 t)dt 11q

10 (1 t)|f (tx + (1 t)a)|qdt 1q

+(b x)2

b a

10 (1 t)dt 11q

10 (1 t)|f (tx + (1 t)b)|qdt 1q

.

Since |f|q is convex, therefore we have

10 (1 t)|f (tx + (1 t)a)|qdt

10 (1 t)

t|f (x)|q + (1 t)|f (a)|q

dt

= |f (x)|q + 2|f (a)|q

6

Analogously,

f (x)|q + 2|f (b)|q

6 .

Combining all the above inequalities gives the desired result.

Corollary 4. In Theorem 7, choosing x = a+b2and then using the convexity of |f|q we have

f (a) + f (b)

2

10 (1 t)|f (tx + (1 t)b)|qdt |

1b a

ba f (u)du

1 q

2|f (a)|q + f

b a 8

1 3

a + b 2

q

1q+

2|f (b)|q + f

1 q

a + b 2

q

31

1

q

(b a)(|f (a)| + |f (b)|).

8

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Page 8 of 11

Theorem 8. Let f : I be a differentiable mapping on I such that f L [a, b], where a, b I with a <b. If |f|q is concave on [a, b], for some fixed q 1, then the following inequality holds:

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

1

2

(x a)2|f

x+2a

3 | + (b x)2|f

x+2b

3

|

.

b a

Proof. First, we note that by the concavity of |f|q and the power-mean inequality, we have

|f (tx + (1 t)a)|q t|f (x)|q + (1 t)|f (a)|q.

Hence,

|f (tx + (1 t)a)| t|f (x)| + (1 t)|f (a)|,so |f| is also concave.

Accordingly, using Lemma 1 and the Jensen integral inequality, we have

(b x)f (b) + (x a)f (a)

b a

1b a

ba f (u)du

(x a)2

b a

10 (1 t)|f (tx + (1 t)a)|dt

+(b x)2

b a

10 (1 t)|f (tx + (1 t)b)|dt

(x a)2

b a

10 (1 t)dt

f

1

0 (1 t)(tx + (1 t)a)dt

1

0 (1 t)dt

+(b x)2

b a

10 (1 t)dt

f

1

0 (1 t)(tx + (1 t)b)dt

1

0 (1 t)dt

1

2

(x a)2

f

x+2a

3

+ (b x)2

f

x+2b 3

.

b a

Remark 3. In Theorem 8, if we choose x = a+b2we have

f (a) + f (b)

2

1b a

ba f (u)du

b a

8

f

5a + b 6

+

f

a + 5b 6

which is the inequality in (1.4).

3. Applications to Special Means

Recall the following means which could be considered extensions of arithmetic, logarithmic and generalized logarithmic from positive to real numbers.

Kavurmaci et al. Journal of Inequalities and Applications 2011, 2011:86 http://www.journalofinequalitiesandapplications.com/content/2011/1/86

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(1) The arithmetic mean:

A = A(a, b) = a + b

2 ; a, b R

(2) The logarithmic mean:

L(a, b) = b a

ln |b| ln |a|

; |a| = |b|, ab = 0, a, b R

(3) The generalized logarithmic mean:

Ln(a, b) =

bn+1 an+1 (b a)(n + 1)

1n; n Z\{1, 0}, a, b R, a = b

Now using the results of Section 2, we give some applications to special means of real numbers.

Proposition 1. Let a, b , a <b, 0 [a, b] and n , |n| 2. Then, for all p > 1(a)

|A(an, bn) Lnn(a, b)| |n|(b a)

1p + 1

1 p

1 2

1q A

|a|n1, |b|n1

(3:1)

and(b)

|A(an, bn) Lnn(a, b)| |n|(b a)

31

1

q

4 A

|a|n1, |b|n1

. (3:2)

Proof. The assertion follows from Corollary 3 and 4 for f (x) = xn, x , n , |n| 2.

Proposition 2. Let a, b , a <b, 0 [a, b]. Then, for all q 1,(a)

|A(a1, b1) L1(a, b)| (b a)

1p + 1

1 p

1 2

1q A

|a|2, |b|2

(3:3)

and(b)

|A(a1, b1) L1(a, b)| (b a)

31

1

q

A

|a|2, |b|2

. (3:4)

Proof. The assertion follows from Corollary 3 and 4 for f (x) = 1x.

4. The Trapezoidal Formula

Let d be a division a = x0 <x1 < ... <xn - 1 <xn = b of the interval [a, b] and consider the quadrature formula

ba f (x)dx = T(f , d) + E(f , d)

4

(4:1)

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Page 10 of 11

where

i=0f (xi) + f (xi+1)2 (xi+1 xi)

for the trapezoidal version and E (f, d) denotes the associated approximation error. Proposition 3. Let f : I be a differentiable mapping on I such that f L

[a, b], where a, b I with a <b and|f |

p

T(f , d) =

n1

p1is convex on [a, b], where p > 1. Then in(4.1), for every division d of [a, b], the trapezoidal error estimate satisfies

|E(f , d)|

1p + 1

1 p

1 2

1q n1

i=0(xi+1 xi)22 (|f (xi)| + |f (xi+1)|).

Proof. On applying Corollary 3 on the subinterval [xi, xi+1] (i = 0, 1, 2,..., n - 1) of the division, we have

f (xi) + f (xi+1)2

1 xi+1 xi

xi+1xi f (x)dx

(xi+1 xi) 2

1p + 1

1 p

1 2

1q (|f (xi)| + |f (xi+1)|).

Hence in (4.1) we have

ba f (x)dx T(f , d)

=

n1

i=0

xi+1xi f (x)dx

f (xi) + f (xi+1)2 (xi+1 xi)

n1

i=0

xi+1xi f (x)dx

f (xi) + f (xi+1)

2 (xi+1 xi)

i=0(xi+1 xi)22 (|f (xi)| + |f (xi+1)|)

which completes the proof.

Proposition 4. Let f : I be a differentiable mapping on I such that f L [a, b], where a, b I with a <b. If |f|q is concave on [a, b], for some fixed q > 1, Then in (4.1), for every division d of [a, b], the trapezoidal error estimate satisfies

|E(f , d)|

q 1

2q 1

1p + 1

1 p

1 2

1q n1

f

q1q n1

i=0(xi+1 xi)2 4

3xi + xi+1 4

+

f

xi + 3xi+1 4

.

Proof. The proof is similar to that of Proposition 3 and using Remark 2. Proposition 5. Let f : I be a differentiable mapping on I such that f L [a, b], where a, b I with a <b. If |f|q is concave on [a, b], for some fixed q 1, Then in (4.1), for every division d of [a, b], the trapezoidal error estimate satisfies

|E(f , d)|

1 8

i=0(xi+1 xi)2 f

5xi + xi+1 6

+

f

xi + 5xi+1 6

.

n1

Proof. The proof is similar to that of Proposition 3 and using Remark 3.

Kavurmaci et al. Journal of Inequalities and Applications 2011, 2011:86 http://www.journalofinequalitiesandapplications.com/content/2011/1/86

Authors contributions

HK and MA carried out the design of the study and performed the analysis.

MEO (adviser) participated in its design and coordination. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Received: 11 December 2010 Accepted: 13 October 2011 Published: 13 October 2011

References1. Dragomir, SS, Agarwal, RP: Two inequalities for differentiable mappings and applications to special means of real numbers and to trapezoidal formula. Appl Math Lett. 11(5), 9195 (1998). doi:10.1016/S0893-9659(98)00086-X

2. Kirmaci, US, Klarii Bakula, M, zdemir, ME, Peari, J: Hadamard-type inequalities for s-convex functions. Appl Math Comput. 193(1), 2635 (2007). doi:10.1016/j.amc.2007.03.030

3. Kirmaci, US: Improvement and further generalization of inequalities for differentiable mappings and applications. Computers and Mathematics with Applications. 55, 485493 (2008). doi:10.1016/j.camwa.2007.05.004

4. Pearce, CEM, Peari, J: Inequalities for differentiable mappings with application to special means and quadrature formula. Appl Math Lett. 13(2), 5155 (2000). doi:10.1016/S0893-9659(99)00164-0

5. Peari, JE, Proschan, F, Tong, YL: Convex Functions, Partial Ordering and Statistical Applications. Academic Press, New York (1991)

doi:10.1186/1029-242X-2011-86Cite this article as: Kavurmaci et al.: New inequalities of hermite-hadamard type for convex functions with applications. Journal of Inequalities and Applications 2011 2011:86.

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