On some Hadamard type inequalities for MT-convex functions 1 ...

Int. J. Open Problems Compt. Math., Vol. 6, No. 2, June 2013 c ISSN 1998-6262; Copyright °ICSRS Publication, 2013 www.i-csrs.org

On some Hadamard type inequalities for MT-convex functions Mevl¨ ut TUNC ¸ , Yusuf S ¸ UBAS ¸ , Ibrahim KARABAYIR Department of Mathematics, Faculty of Science and Arts, Kilis 7 Aralık University P.O.Box 79000 Kilis, Turkey e-mail:[email protected] e-mail:[email protected] e-mail:[email protected] Abstract In this paper, we establish some new Hadamard type inequalities for MTconvex functions and give few applications for some special means. Keywords: Hadamard’s inequality, M T -convexity, means.

1

Introduction

The following inequality is known in the literature as Hermite-Hadamard inequality. This inequality has the update for a lot of different type convex functions in the mathematics literature. Theorem 1.1 Let f : I ⊆ R → R be a convex function and a < b with a, b ∈ I. Then µ ¶ Z b a+b f (a) + f (b) 1 f f (x) dx ≤ (1) ≤ 2 b−a a 2 Definition 1.2 [3] We say that f : I → R is Godunova-Levin function or that f belongs to the class Q (I) if f is nonnegative and for all x, y ∈ I and t ∈ (0, 1) we have f (tx + (1 − t) y) ≤

f (x) f (y) + . t 1−t

(2)

On some Hadamard type inequalities for ...

103

Definition 1.3 [2] We say that f : I → R is a P -function or that f belongs to the class P (I) if f is nonnegative and for all x, y ∈ I and t ∈ [0, 1] , we have f (tx + (1 − t) y) ≤ f (x) + f (y) . (3)

Obviously, Q(I) ⊃ P (I) and for applications it is important to note that P (I) also consists only of nonnegative monotonic, convex and quasi-convex functions, i.e., nonnegative functions satisfying f (tx + (1 − t) y) ≤ max {f (x) , f (y)} . In [9], Tun¸c and Yıldırım defined the following M T -convex functions class. Definition 1.4 f : I ⊆ R → R nonnegative function provides, with ∀x, y ∈ I and t ∈ (0, 1), √ √ t 1−t f (tx + (1 − t) y) ≤ √ f (x) + √ f (y) 2 1−t 2 t

(4)

f is M T -convex function. This class is shown as M T (I). If (4) change direction, f is M T -concave function. Again in [10], the following inequalities are given by Tun¸c.

Theorem 1.5 Let f, g : [a, b] ⊆ R → R two convex functions and f, g ∈ L1 [a, b]. Then, Z b 1 (b − x) (f (a) g (x) + g (a) f (x)) dx (b − a)2 a Z b 1 (x − a) (f (b) g (x) + g (b) f (x)) dx + (b − a)2 a Z b 1 M (a, b) N (a, b) + + f (x) g (x) dx ≤ 3 6 b−a a where M (a, b) = f (a) g (a) + f (b) g (b), N (a, b) = f (a) g (b) + f (b) g (a) .

104

TUNC ¸ , S¸UBAS¸, KARABAYIR

Theorem 1.6 Let f, g : [a, b] ⊆ R → R two convex functions and f, g ∈ L1 [a, b]. Then, ¶ µ ¶ ¶ Z bµ µ 1 a+b a+b f g (x) + g f (x) dx 2 2 (b − a)2 a Z b 1 M (a, b) N (a, b) ≤ f (x) g (x) + + 2 (b − a) a 12 6 ¶ µ ¶ µ a+b a+b g +f 2 2 where M (a, b) = f (a) g (a) + f (b) g (b), N (a, b) = f (a) g (b) + f (b) g (a) . Definition 1.7 [13]If f, g : X → R functions, for x, y ∈ X provides the following inequality (f (x) + f (y)) (g (x) + g (y)) ≥ 0 it is said that similar sequent (brief ly s.o.) functions for them. In accordance with the above studies, works performed in literature can be looked at no [1]-[13] for references given below. The purpose of this work is to establish some new results on Hermite Hadamard inequality given (1) for M T -convex functions and to apply some basic inequalities for real numbers in numeric integration.

2

Results For M T -Convexity

¢ ¡ 1 Remark 2.1 i) f, g : (1, ∞) → R, f (x) = xp , g (x) = (1 + x)p , p ∈ 0, 1000 , ¡ 1 ¢ m ii) h : [1, 3/2] → R, h(x) = (1 + x2 ) , m ∈ 0, 100 , are M T -convex functions, but they are not convex. All of the positive convex functions is also an M T -convex function, but the√ reverse is not always √ t √ , it is written , (1 − t) ≤ 21−t true. Since f is M T -convex and t ≤ 2√1−t t √ √ t 1−t f (tx + (1 − t) y) ≤ tf (x) + (1 − t) f (y) ≤ √ f (x) + √ f (y) , 2 1−t 2 t this indicates that each positive convex function is a M T -convex function. √√ Remark 2.2 Since 2 t 1 − t ≥ t (1 − t) , for t ∈ (0, 1) , it is written f (ta + (1 − t) b) ≤

tf (a) + (1 − t) f (b) tf (a) + (1 − t) f (b) √√ . ≤ t (1 − t) 2 t 1−t

105


As can be seen from this inequality ,each M T -convex function is Q(I)-GodunovaLevin function. However, M T -convex functions class allows us to obtain a better upper bound than Q(I)-Godunova-Levin function. Obviously, Q(I) ⊃ M T (I). Moreover, for t ∈ [0.2, 0.8] , we have tf (a) + (1 − t) f (b) √√ ≤ f (a) + f (b) . 2 t 1−t

f (ta + (1 − t) b) ≤

Then, we can say that each M T -convex function is P (I) function, on [0.2, 0.8]. So, P (I) ⊃ M T (I).

Theorem 2.3 Let f : [a, b] ⊆ R → R a nonnegative M T -convex function and f ∈ L1 [a, b]. Then, 1 b−a

Z

b

f (x) dx ≤ a

π (f (a) + f (b)) 4

(5)

Proof: i) Because f is a M T -convex function, if we get the integral of √ √ t 1−t f (ta + (1 − t) b) ≤ √ f (a) + √ f (b) 2 1−t 2 t inequality (0, 1) for t, the proof is completed. The above inequality can be proven by another way given below. ii) Since f is a M T -convex function, we can write √ √ t 1−t f (ta + (1 − t) b) ≤ √ f (a) + √ f (b) 2 1−t 2 t √ √ t 1−t f (tb + (1 − t) a) ≤ √ f (b) + √ f (a) 2 1−t 2 t

(6)

(7)

By adding (6) and (7) , we get µ f (ta + (1 − t) b) + f (tb + (1 − t) a) ≤

1 √√ 2 t 1−t

¶ (f (a) + f (b))

(8)

By integrating above inequality (8), according to t over [0, 1], we obtain Z

Z

1

1

(f (ta + (1 − t) b) + f (tb + (1 − t) a)) dt ≤ (f (a) + f (b)) 0

0

1 p dt 2 t (1 − t)

106


By taking into account Z 1 Z f (ta + (1 − t) b) dt = 0

1 0

and

Z

1 0

1 f (tb + (1 − t) a) dt = b−a

Z

b

f (x) dx a

1 1 p dt = π 2 2 t (1 − t)

the proof is completed. Theorem 2.4 Let f, g : [a, b] ⊆ R → R two nonnegative M T -convex functions and f, g ∈ L1 [a, b]. Then, the following inequality holds;

g (a)

p Z (b − x) (b − x) (x − a)

b

3

(b − a) a p Z (b − x) (b − x) (x − a)

f (x) dx + g (b) b

(x − a)

p

(b − x) (x − a)

Z

(9) b

(b − a)3 a p Z (x − a) (b − x) (x − a)

+f (a) g (x) dx + f (b) (b − a)3 a ½ 1 1 1 1 ≤ f (a) g (a) + f (b) g (b) + [f (a) g (b) + f (b) g (a)] 2 3 3 6 ¾ Z b + (b − x) (x − a) f (x) g (x) dx

(b − a)3

a

where M (a, b), N (a, b) are like above. Proof: Since f and g are M T -convex, it is written √ √ t 1−t f (ta + (1 − t) b) ≤ √ f (a) + √ f (b) 2 1−t 2 t √ √ t 1−t g (ta + (1 − t) b) ≤ √ g (a) + √ g (b) 2 1−t 2 t By using er + f p ≤ ep + f r, e, f, r, p ∈ R+ basic inequality, it is written √ µ √ ¶ t 1−t f (ta + (1 − t) b) √ g (a) + √ g (b) 2 1−t 2 t √ ¶ µ √ t 1−t +g (ta + (1 − t) b) √ f (a) + √ f (b) 2 1−t 2 t √ √ · √ ¸· √ ¸ t 1−t t 1−t √ √ ≤ f (a) + √ f (b) g (a) + √ g (b) 2 1−t 2 1−t 2 t 2 t +f (ta + (1 − t) b) g (ta + (1 − t) b)

f (x) dx b

g (x) dx a


107

If this statement is edited, we get √ √ t 1−t g (a) √ f (ta + (1 − t) b) + g (b) √ f (ta + (1 − t) b) 2 1−t 2 t √ √ t 1−t +f (a) √ g (ta + (1 − t) b) + f (b) √ g (ta + (1 − t) b) 2 1−t 2 t t 1−t 1 1 ≤ f (a) g (a) + f (b) g (b) + f (a) g (b) + f (b) g (a) 4 (1 − t) 4t 4 4 +f (ta + (1 − t) b) g (ta + (1 − t) b)

(10)

If both sides of (10) inequality are multiplied by t (1 − t), we get √√ √√ g (a) t t 1 − tf (ta + (1 − t) b) + g (b) (1 − t) t 1 − tf (ta + (1 − t) (11) b) √√ √√ +f (a) t t 1 − tg (ta + (1 − t) b) + f (b) (1 − t) t 1 − tg (ta + (1 − t) b) 1© 2 ≤ t f (a) g (a) + (1 − t)2 f (b) g (b) 2 +t (1 − t) [f (a) g (b) + f (b) g (a)] +t (1 − t) [f (ta + (1 − t) b) + g (ta + (1 − t) b)]} By integrating inequality (11) according to t over [0, 1], Z 1 √ √ g (a) t t 1 − tf (ta + (1 − t) b) dt 0 Z 1 √√ +g (b) (1 − t) t 1 − tf (ta + (1 − t) b) dt Z0 1 √√ +f (a) t t 1 − tg (ta + (1 − t) b) dt Z 01 √√ +f (b) (1 − t) t 1 − tg (ta + (1 − t) b) dt 0 ½ Z 1 Z 1 1 2 f (a) g (a) t dt + f (b) g (b) (1 − t)2 dt ≤ 2 0 0 Z 1 + [f (a) g (b) + f (b) g (a)] t (1 − t) dt 0 ¾ Z 1 + t (1 − t) [f (ta + (1 − t) b) + g (ta + (1 − t) b)] dt

(12)

0

By substituting ta + (1 − t) b = x, (a − b) dt = dx, we get p Z b Z 1 √√ b − x (b − x) (x − a) 1 f (x) dx t t 1 − tf (ta + (1 − t) b) dt = b−a (b − a) b−a a 0 (13)

108 Z

1 0


p

√√ x−a (1 − t) t 1 − tf (ta + (1 − t) b) dt = b−a

(b − x) (x − a) 1 (b − a) b−a

Z

b

f (x) dx a

(14)

similarly Z

1

0

Z

1 0

√√ b−x t t 1 − tg (ta + (1 − t) b) dt = b−a

p

(b − x) (x − a) 1 (b − a) b−a

√√ x−a (1 − t) t 1 − tg (ta + (1 − t) b) dt = b−a

and

Z

1

Z

1

2

t dt = 0

0

1 (1 − t) dt = , 3

Z

p

b

g (x) dx a

(b − x) (x − a) 1 (b − a) b−a

1

2

Z

t (1 − t) dt = 0

(15)

Z

b

g (x) dx a

1 6

(16) (17)

If these statements (13)-(17)can be used in (12), the proof is completed. Theorem 2.5 Let f, g ∈ [a, b] → R two nonnegative M T -convex functions and f, g ∈ L1 [a, b]. Then, 8 f 3

µ

a+b 2

¶ ≤ M (a, b) + N (a, b)

(18)

where M and N are like above. Proof: Since f and g are M T -convex functions, we can write µ f

a+b 2

¶

µ = f

ta + (1 − t) b (1 − t) a + tb + 2 2

¶ (19)

1 [f (ta + (1 − t) b) + f ((1 − t) a + tb)] 2µ √ √ ¶ 1 t 1−t √ ≤ (f (a) + f (b)) + √ 2 2 1−t 2 t

≤

and µ g

a+b 2

¶

µ

¶ ta + (1 − t) b (1 − t) a + tb + = g 2 2 √ ¶ µ √ t 1−t 1 √ + √ (g (a) + g (b)) ≤ 2 2 1−t 2 t

(20)


By multiplying (19) and (20) we get µ ¶ a+b fg 2 √ · µ √ ¶¸2 1 t 1−t √ ≤ + √ (f (a) + f (b)) (g (a) + g (b)) 4 1−t t µ ¶ 1 t 1−t = + + 2 (f (a) + f (b)) (g (a) + g (b)) 16 1 − t t µ ¶ 1 1 (f (a) + f (b)) (g (a) + g (b)) = 16 t (1 − t) If both sides of (21) inequality are multiplied by t (1 − t) , we get µ ¶ a+b 1 t (1 − t) f g ≤ (f (a) + f (b)) (g (a) + g (b)) 2 16

109

(21)

(22)

By integrating inequality (22) according to t over [0, 1], the proof is completed. Corollary 2.6 In Theorem 2.5, if f and g are s.o., we obtain µ ¶ 4 a+b fg ≤ M (a, b) 3 2 Theorem 2.7 Let f, g : [a, b] ⊆ R → R two nonnegative M T -convex functions and f, g ∈ L1 [a, b]. Then, we have µ ¶ µ ¶ a+b a+b f (g (a) + g (b)) + g (f (a) + f (b)) (23) 2 2 µ ¶ a+b 16 ≤ fg + 2 (f (a) + f (b)) (g (a) + g (b)) 3π 2 Proof: Since f and g are M T -convex functions, we can write √ µ ¶ µ √ ¶ a+b t 1−t 1 √ f ≤ + √ (f (a) + f (b)) 2 2 2 1−t 2 t √ ¶ µ ¶ µ √ a+b 1 t 1−t √ (g (a) + g (b)) g ≤ + √ 2 2 2 1−t 2 t By using er + f p ≤ ep + f r, e, f, r, p ∈ R+ basic inequality, we obtain, √ µ ¶µ √ ¶ a+b 1 t 1−t √ f + √ (g (a) + g (b)) 2 2 2 1−t 2 t √ µ ¶ ¶µ √ a+b t 1−t 1 √ (f (a) + f (b)) + g + √ 2 2 2 1−t 2 t √ µ ¶¶2 ¶ µ µ √ a+b t 1−t 1 √ ≤ fg + √ (f (a) + f (b)) (g (a) + g (b)) + 2 4 1−t t

110


If this statement is edited, we get √ µ ¶ ¶ µ √ 1 a+b t 1−t f + √ (g (a) + g (b)) √ 4 2 1−t t √ µ ¶ µ √ ¶ t 1−t 1 a+b + √ + g (f (a) + f (b)) √ 4 2 1−t t ¶ µ 1 a+b 1 + ≤ fg (f (a) + f (b)) (g (a) + g (b)) 2 16 t (1 − t)

(24)

If both sides of (24) are multiplied by t (1 − t) , we get ¶ µ ³√√ ´ √√ 1 a+b f (g (a) + g (b)) t t 1 − t + (1 − t) t 1 − t (25) 4 2 µ ¶ ³√√ ´ √√ 1 a+b + g (f (a) + f (b)) t t 1 − t + (1 − t) t 1 − t 4 2 µ ¶ a+b 1 (f (a) + f (b)) (g (a) + g (b)) ≤ fg t (1 − t) + 2 16 By integrating of inequality (25) , according to t over [0, 1], we get µ ¶ µZ 1 ¶ Z 1 √√ √√ a+b 1 t t 1 − tdt + f (g (a) + g (b)) (1 − t) t 1 − tdt 4 2 0 0 µ ¶ µZ 1 ¶ Z 1 √√ √√ 1 a+b + g t t 1 − tdt + (f (a) + f (b)) (1 − t) t 1 − tdt 4 2 0 0 µ ¶Z 1 a+b 1 t (1 − t) dt + ≤ fg (f (a) + f (b)) (g (a) + g (b)) . 2 16 0 By accounting the following equalities Z 1 √ Z √ t t 1 − tdt = 0

Z

1

1 0

√√ (1 − t) t 1 − tdt =

0

1

3

t 2 (1 − t) 2 dt = Z

1

1

3

π 16

t 2 (1 − t) 2 dt = 0

π 16

the proof is completed.

3

Applications to some special means

We now consider the applications of our Theorems to the following special means

111


, a, b ≥ 0, The arithmetic mean: A = A (a, b) := a+b √2 The geometric mean: G = G (a, b)  := ab, a, b ≥ 0,  a if a = b 1 ³ b ´ b−a The identric mean: I = I (a, b) := , a, b ≥ 0,  1e aba if a 6= b ( h i1/p bp+1 −ap+1 if a 6= b , (p+1)(b−a) The p-logarithmic mean: Lp = Lp (a, b) := a if a = b p ∈ R\ {−1, 0} ; a, b > 0. The following propositions holds: Proposition 3.1 Let 0 < a < b. Then one has the inequality π

I (a, b) ≥ G 2 (a, b) .

(26)

Proof: If we choose in Theorem 2.3, applied to the M T -convex function f : (0, 1] → R, f (x) = − ln x, we obtain Z b 1 π − ln xdx ≤ − (ln a + ln b) b−a a 4 π ≤ − ln (ab) 4 ¢ π ¡ 2 ≤ − ln G (a, b) 4 ¡ ¢ π ≤ − ln G 2 (a, b) which gives us

¡ π ¢ − ln I (a, b) ≤ − ln G 2 (a, b)

and the inequality is proved. ¡ ¢ 1 , then we have Proposition 3.2 Let 1 < a < b, p ∈ 0, 1000 Lpp (a, b) ≤

π A (ap , bp ) . 2

(27)

Proof: The inequality follows ¡ from ¢(5) applied to the M T -convex function 1 f : (1, ∞) → R, f (x) = xp , p ∈ 0, 1000 . The details are omitted. ¡ ¢ 1 Proposition 3.3 Let 1 < a < b, p ∈ 0, 1000 , then, we have p (b − x) (x − a) p Lp (a, b) {ap (b − x) + bp (x − a)} (b − a)2 ½ 1 1 2p 1 2p 1 ≤ a + b + (ab)p (28) 4 3 3 3 ª 2p+2 2p − (b − a) L2p+2 (a, b) − L2p+1 2p+1 (a, b) − ab (b − a) L2p (a, b)

112


Proof: The inequality follows from to the M T -convex functions ¡ (9)1applied ¢ p f, g : R → R, f (x) = g (x) = x , p ∈ 0, 1000 . The details are omitted.

4

Open Problem

It is a well-known fact that if f is a convex function on the interval I ⊂ R, then the Hadamard’s inequality retains for the convex functions. As a matter of fact, it has been demonstrated lots of this type inequalities for several convex functions. Therefore, there is one questions as follows: How can be set up the general versions of the inequalities (5), (9), (18) and (23) including several differentiable M T -convex function on I.

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