Научная статья на тему 'Inequalities of Hermite-Hadamard type for HG-convex functions'

Inequalities of Hermite-Hadamard type for HG-convex functions Текст научной статьи по специальности «Математика»

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CONVEX FUNCTIONS / INTEGRAL INEQUALITIES / HG-CONVEX FUNCTIONS

Аннотация научной статьи по математике, автор научной работы — Dragomir S.S.

Some inequalities of Hermite-Hadamard type for HG-convex functions defined on positive intervals are given. Applications for special means are also provided.

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Текст научной работы на тему «Inequalities of Hermite-Hadamard type for HG-convex functions»

Probl. Anal. Issues Anal. Vol. 6(24), No. 2, 2017, pp. 25-41

25

DOI: 10.15393/j3.art.2017.3790

UDC 517.51

S. S. Dragqmir

INEQUALITIES OF HERMITE-HADAMARD TYPE FOR HG-CONVEX FUNCTIONS

Abstract. Some inequalities of Hermite-Hadamard type for HG -convex functions defined on positive intervals are given. Applications for special means are also provided.

Key words: convex functions, integral inequalities, HG-convex functions

2010 Mathematical Subject Classification: 26D15, 25D10

1. Introduction. Following [4] (see also [26]) we say that the function f : I C R \ {0} ^ R is HA-convex if

for all x,y E I and t E [0,1]. If the inequality in (1) is reversed, then f is said to be HA-concave.

If I C (0, to) and f is convex and nondecreasing function then f is HA-convex and if f is HA-convex and nonincreasing function then f is convex.

If [a, 6] C I C (0, oo) and if we consider the function g : [1, 1] ^ R, defined by g (t) = f (1), then we can state the following fact [4]:

Lemma 1. The function f is HA-convex on [a, 6] if and only if g is convex in the usual sense on [1,1].

Therefore, as examples of HA-convex functions we can take f (t) = = g (1), where g is any convex function on [1,1].

In the recent paper [16] we obtained the following characterization result as well:

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©Petrozavodsk State University, 2017

[MglHl

(1)

Lemma 2. Let f, h : [a, b] C (0, to) ^ R be so that h (t) = tf (t) for t E [a, b]. Then f is HA-convex on the interval [a, b] if and only if h is convex on [a, b].

Following [4] (see also [26]) we say that the function f : I C R \ {0} ^ ^ (0, to) is HG-convex if

f (ix+fr^) *f (x)]1-t f (y)]t (2)

for all x,y E I and t E [0,1]. If the inequality in (2) is reversed, then f is said to be HG-concave.

By the geometric-mean - arithmetic mean inequality we have that any HG-convex function is HA-convex. The converse is obviously not true.

We observe that f : I C R \ {0} ^ (0, to) is HG-convex if and only if the function ln f : I C R \ {0} ^ R is HA-convex on I. Using Lemmas 1 and 2 we have:

Theorem 1. Let f : [a, b] C (0, to) ^ (0, to) and define the associated functions Gf : [ 1, a ] ^ R defined by Gf (t) = ln f (1) and Hf : [a, b] C C (0, to) ^ R defined by Hf (t) = t ln f (t). Then the following statements are equivalent:

(i) The function f is HG-convex on [a, 6] ;

(ii) The function Gf is convex on [1, a];

(iii) The function Hf is convex on [a, b].

For a convex function h : [c, d] ^ R, the following inequality is well known in the literature as the Hermite-Hadamard inequality

d

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, (c + d\ 1 ,, , h (c) + h (d)

h(—) * ir-/h (t) dt * ^-V0. (3)

c

For related results, see [1] - [10], [12]-[28].

Motivated by the above results, we establish in this paper some inequalities of Hermite-Hadamard type for HG-convex functions defined on positive intervals. Applications for special means are also provided. 2. Main Results. The following result holds.

Theorem 2. Let f : [a, b] C (0, to) ^ (0, to) be an HG-convex function on the interval [a, b]. Then for any A E [0,1] we have the inequalities

A —\*

Ma + b ' *

j. , 2ab

(1 - A) a + (A + 1) b

1-A

j. , 2ab

(2 - A) a + Ab,

A

<

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< exp

i b

' ab /ln f (t)

b - a J t2

a

dt I <

<1 f

ab

(1 - A) a + Ab

[f (a)]1 [f (b)]A < Vf (a) f (b).

If we take A = 1 in (4), then we get

(4)

f( —1 <a fi —W — I <

a + b < \ Ma + 3b / f V 3a + b

< exp

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ab ba

ln f (t) t2

dt I <

<1 f

2ab a + b

Vf (a) f (b) < Vf (a) f (b).

(5)

The identric mean I (a, 6) for two distinct positive numbers a, 6 is defined by

i

1 / b-a I (a,b);=i(ai

while the logarithmic mean is defined by

T ( n 6 - a

L (a, 6) :=

ln 6 — ln a

Theorem 3. Let f : [a, 6] C (0, to) ^ (0, to) be an HG-convex function on the interval [a, 6]. Then

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(

f (L (a,b)) < exP

1

ba

(L(g,b)-g)b (b-L(g,b))g

ln f (t) dt | < [f (b)] (b a) L(a,b) [f (a)] (b a)L (a,b)

(6)

If we write the classical Hermite-Hadamard inequality for the function Hf that is convex on [a, 6] when f : [a, 6] C (0, to) ^ (0, to) is an HG-convex function on [a, 6] and perform the required calculations, we get:

b

b

Theorem 4. Let f : [a, b] C (0, to) ^ (0, to) be an HG-convex function on the interval [a, b]. Then we have

f

a + b

a + b

— exp

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/ ! Ï

ba

t ln f (t) dtl < V [f (b)]b [f (a)]a. (7)

We have the reverse inequalities as well:

Theorem 5. Let f : [a, b] C (0, to) ^ (0, to) be an HG-convex function on the interval [a, b]. Then we have

exp I ^ I f dt

1 <

f I 2ab f \ a+b

<

(8)

and

< exH 8

f- (b) b2_ f+ (a) a2

Lf (b)

f (a)

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1

Vf (a) f (b)

ex^ baba/

f dt

b — a ab

<

(9)

< exp -

f- (b),2 f+ (a) a2

f (b)

b2

f (a)

b — a ab

The following related result also holds:

Theorem 6. Let f : [a, b] С (0, то) ^ (0, то) be an HG-convex function on the interval [a, b]. Then we have

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1

[f (a)]a [f (b)]b

exp bi/1 ln f (t) dt

<

- Vf (a) У

8 (b-a)

I 1 (b ) f bf- (b) af+ (a)

e !(b—a)lnr— fâT

2

2

and

exp( ft ln f (t) dt

1 < V 1 a+b - <

[f ()]

/ f (6) \ 8 (b"a) /1(6 ) / f (6) af+ (a)\\ (11)

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H/Mi ex\8 (6 — a)l TW — "T(arJJ • (11)

From a different perspective we have:

Theorem 7. Let f : [a, 6] C (0, to) ^ (0, to) be an HG-convex function on the interval [a, 6]. Then

exp

( b _

rtb /"ln f (t) 1 / a(b-x) b(x-a)

1 f (t) dt I < Vf (x) [f (b)] Xb-) [f (a)] (12)

b - a t2

\ a

for any x G [a, b].

If we take in (12), x = a2b, then we get from (12) that

exp

'jk ./ f dt ) < // (^) ff (b)]a+b ff („)] A. (13)

3. Proofs. In [15], in order to improve Iscan's inequality [26] for HA-convex functions g : [a, 6] C (0, to) ^ R,

b

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( 2a6 \ a6 f g (t) , g (a) + g (6) ,„

g brW < ^ rwdt < ^^^ir^, (14)

we obtained the following result:

/ 2a6 \ / 2a6 \

»U + jJ < (1 — A) H (1 — A) a r(A + 1) b) +

b

A / 2a6 \ a6 fa (t) ,

+Ag —-n-— < ,- ^dt <

\ (2 — A) a + A6 y < 6 — a 7 t2 <

1

< -*2

g

ab

(1 - A) a + Ab

+ (1 - A) g (a) + Ag (b)

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<

<

g(a) + g(b) 2

(15)

where A E [0, 1] .

Now, if f : [a, b] C (0, to) ^ (0, to) is an HG-convex function on the interval [a, b], then g := ln f is HA-convex on [a, b], and by (15) we get

2ab 2ab

flaTb) * (1 - A)lnf((1 - A)a + (A +1)b

+

1

< -*2

+A ln f ln f

2ab

(2 - A) a + Ab J ~ b - a ab

ab f ln f (t) ,

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< -- / dt <

t2

(1 - A) a + Ab

ln f (a) + ln f (b)

+ (1 - A) ln f (a) + A ln f (b)

<

<

2

that is equivalent to

ln

f

2ab

(1 - A) a + (A + 1) b

2ab

ln f ( -- ) *

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f

a+b

1-A

2ab

(2 - A) a + Ab

(16)

<

* f dt * W /

a

ab

(1 - A) a + Ab

[f (a)]1-A [f (b)]A *

* ln Vf (a) f (b),

and by taking the exponential we get the desired result (4). We have the following result for HA-convex functions [15]:

Lemma 3. Let g : [a, b] C (0, to) ^ R be an HA-convex function on the interval [a, b]. Then

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g (L (a, b)) *

1

ba

g (x) dx *

(L (a, b) - a) bg (b) + (b - L (a, b)) ag (a)

(b - a) L (a, b)

b

A

b

b

If f : [a, b] С (0, то) ^ (0, to) is an HG-convex function on the interval [a, b], then g := ln f is HA-convex on [a, b], and by (17) we have

b

ln f (L (a, b)) t ln f (x) dx <

ba

(L (a, b) - a) b ln f (b) + (b - L (g,b)) a ln f (a) _

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_ (b — a) L (a, b)

((L(a,b)-a)b (b-L(a,b))a \

[f (b)] (b a)L(a,b) [f (a)] (b a) L(a,b) \ . (1g)

By taking the exponential in (18) we get the desired result (6).

We use the following results obtained by the author in [10] and [11]:

Lemma 4. Let h : [a, ß] — R be a convex function on [a, ß]. Then we have the inequalities

ß

0 < h (a) + h (ß) — ß—^/h (t) dt < 1 [hi. (ß) — h+ (a)] (ß — a) (19)

a

and

ß

0 < / h (t) dt — h^^) < g [hi (ß) — h+ (a)] (ß — a). (20)

a

The constant 8 is best possible in (19) and (20).

If i : [a, b] C (0, to) —> R is an HA-convex function on the interval [a, b], then the function g : [1, 1] — R, g (s) = i (1) , is convex on [1,1] . Now, by (19) and (20) we have

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0 - g Ш + g (bi — — Ig (t) dt -2 a — ь J

1

< -< 8

11

a — gn b

a b> (21)

a

b

and

0 <

— 1

a

g (t) dt - g( i+i I —

We also have

and then

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1

< -— 8

g- i - g+n

11

g± « =^ (1) (-h)

g'_ Q) = -t+ (a) a2 and g^ (i) = -t- (b) b2.

From (21) and (22) we have

(22)

and

£(a)±M_ft(i dS<

b — a

— 1 [/_ (b)b2 - (a) a2](

ab 1 2ab

0 — t- t - ds - N -r I —

b — a / \s J \a ± b

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— 1 (b) b2 - 4 (a) a2](

(23)

(24)

If we change the variable 1 = u, then ds = — tr and (23) and (24) can be written as

b

0< I (a) + I (6) a6 j £Jt) dt <

b - a ./ t2

ba

— 1 ['- (b) b2 - (a) a2](

(25)

1

b

b

a

and

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0 <

ab [£ (t) / 2ab

b - a J t2

a

a + b, ba

<

< 1 [C- (b)b2 - 4 (a) a2](

(26)

If f : [a, b] C (0, to) — (0, to) is an HG-convex function on the interval [a, b], then i := ln f is HA-convex on [a, b], and by (25) and (26) we have

o < in vrwrn - f dt <

1

< -< 8

f- (b) b2_ f+ (a) a2 f (b) b f (a)

b — a ab

(27)

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and

0 /fdt - inff Jit I <

b - a J t2

a

1

< -< 8

f- (b) b2 _ f+ (a) a2

Lf (b)

f (a)

a + b

b - a ab

(28)

and the Theorem 5 is proved.

If f : [a, b] C (0, to) ^ (0, to) is an HG-convex function on the interval [a, b], then Hf is convex on [a, b] and by (19) and (20) we have after appropriate calculations

1

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< -< 8

0 < in Vf (a)]a [f (b)]b - b-^/t in f (t) dt <

a

bf 1 (b) , , , , af+ (a)

in f (b) +

f (b)

(b - a) =

= m( fi)

8 (b-a)

1

f (a)

bf- (b) af; (a)

+ 8 (b - aH f (b) f (a)

b

and

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0

6a

t ln f (t) dt — ln

a + 6

a + b -2

<

( f (6) x 1 (b-g) +1 / 6f- (6) af; (a)

< htmj r1(6—a)

f (6) f (a)

These inequalities are equivalent to

(

0 ln

V[f (a)]1 [f (6)]b

expi b—ftlnf (t) dtj^

<

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ln

and

ln

(¿(6)x 8 (b"a) exp (1 (6 - a) ( f-^ - af±ja) Vf (a J exH8(6 f(6) f (a)

0 ln

exp 1 ln f (t) dt

)] a+b

g+bM

[f №)]

V

<

/

/f (6)A Vf (a)7

8 (b-g)

I 1(h ^ f (6) af; (a)

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e 8(6—a)bw~— fa)"

and by taking the exponential we get the desired results (10) and (11). The following lemma is of interest in itself:

Lemma 5. Let g : [a, 6] C (0, to) ^ R be a HA-convex function on the interval [a, 6]. Then

1 fg (6) a (6 — x)+ g (a) 6 (x — a)+ xg ^ ^^/g^dy (29)

2x

6a

6 - a y2

for any x E [a, 6] •

b

1

2

Proof. Since h (t) = tg (t) for t E [a, 6] is convex, then by the gradient inequality for convex functions we have

xg (x) — yg (y) > (g (y) + yg- (y^ (x — y)

for any x, y E (a, 6) • This is equivalent to

xg (x) — xg (y) > yg- (y) (x — y) (30)

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for any x, y E (a, 6) •

From (30) we have, by division with xy2 > 0, that

~2 o (x) — ^ o (y) > ^ (1 — ^)

y2 y2 y x

for any x, y E (a, 6) •

Taking the integral mean over y we have

b b

g (x) —^ / 4 — / ^ dy >

b — a J y2 b — a J y2

a

b / ь

>7-^ / g-(y)dy - / g- (y) dy

b — a J У xb — a J

that is equivalent to

b

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g(x) 1 f g (y) , . ~ar — b—~aj dy>

>

ba

g(b) g(a) b

+

g (y)

dy

1 g(b) — g(a)

x

ba

ba

g(b) g(a) b

a

+

ba

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g (yK 1 g(b) — g(a)

-dy---Г-

y2 x b — a

b

1

2

У

b

1

1

for any x E (a, b). This can be written as

b

1 g(b) -g(a) 1 f g(b) g(aA > 2 j" ^Mdy - g(x)

— - ■ - —

x b - a b - a b a b - a y2 ab

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a

or as

b

2ba

b x x a

g(b) —r- + g(a)-

xb ax

+> > ¡m. dy.

ab b - a y2

This is equivalent to the desired result (29). □

If f : [a, b] C (0, to) ^ (0, to) is an HG-convex function on the interval [a, b], then g := ln f is HA-convex on [a, b], and by (29) we have

1 fa(b-x)lnf (b) + b(x-a)lnf (a) + xlnf (xA >

2x b a

b

> a^ ¡fri dy

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b - a J y2

for any x E [a, b].

This is clearly equivalent to

b

/ / a(b-x) b(x-a) .-\ ab i ln f (y)

ln^V[f (b)] ^ [f (a)]« f)J > — j dy (31)

a

for any x E [a, b] .

If we take the exponential in (31), then we get the desired result (12). 4. Applications. Consider the function f : [a, b] C (0, to) ^ (0, to), f (t) = t. Using the geometric mean - harmonic mean inequality, we have

f (-1 =-x^^ * x1-ty* = [f (x)]1-t [f (y)]* ,

J Vtx + (1 - t) y) tx + (1 - t) y * y L,/ WJ u vyyj '

which shows that f is HG-convex on [a, b]. We need the following integrals

b b

1 ln f (t) dt = —^ I ln tdt = ln I (a, b),

b a b a

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1

1

b b 11

t ln f (t) dt = -- t ln tdt =

b — a b — a

aa

1

= - A (a, b) ln I (a2, b2) = ln I/ (a2, b2 )1 A(a,b)

2 L

and

b b

ln f (t) , /" ln t ! b — a , ^ ( 1 , 1 )т -1

"F"1dt = ~Fdt = "¡S" Г'(a 1-1)1

J _ d = I -2dt =— 'n I1 (a-1,6

g

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giving that

b

a /fdt = ln [i (a-1,6-1)]-1 •

6 — a 7 t2 L v '

g

Now, if we write the inequality (4) for the function f : [a, 6] C (0, to) ^ ^ (0, to), f (t) = t, we get

( 2a6 \1-A ( 2a6 \A

H (a,6) <V(1 — A) a + (A + 1) 67 V(2 — A) a r A6 J <

< [I (a-1,6-1)]-1 < ^((1 — A^a + A*) ^* < G (a, 6), (32)

where H (a, 6) := g+b is the harmonic mean.

If we use the inequality (6) for f (t) = t, then we have

(L(a,b)-a)b (b-L(a,b))a

(L (a, 6) <) I (a, 6) < 6 (b a) L(a,b) a (b-a)L(a,b) . (33)

If we use the inequality (7) for f (t) = t, t E [a, 6], then we also get

[A (a, 6)] A(g'b) < I (a2, 62) 1 A(g,b) < G (6b, ag0 . (34)

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From (8) and (9) for f (t) = t we have

1 < tlw1 <exp (^ l (35)

and

1 < w^ < • (36)

From (10) and (11) we also have

G aa,bb) (b) i(b-a) , x

1 *-V \[, ,, * - (37)

* [I(a2,62)]2A(ab) *Va) ( )

and [ ( )]

[I (a2,b2)] 2Aab) (b) 8(b-a) (38)

1 * [A M)]A<-b) * UJ ■ (38)

Finally, from (13) we obtain

[I (a-1, b-1)]-1 * \JA (a, b) ba^. (39)

Now consider the function f : [a, b] C (0, to) ^ (0, to), f (t) = exp (t). Using the harmonic mean-arithmetic mean inequality we have

f(tx + (7- t) J = exp (tx + (1y- t) y) * exp ((1 - t) x + ty) =

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= [exp (x)]1-t [exp (y)]* = [f (x)]1-t [f (y)]*

for any x, y E [a, b] and t E [0,1].

Now, if we use the inequality (4) for the HG-convex function f : [a, b] C C (0, to) ^ (0, to), f (t) = exp (t), then we get, after suitable calculations, that

2 (1 - A) ab 2Aab

H (a b) * (1 - A) a + (A +1) b + (2 - A) a + Ab *

G2 (a, b) 1 ( ab ,„ ., w \ . , ,,

* r, ' * - ^-7t-ry + (1 - A) a + Ab * A (a, b),

" L (a,b) " 2 V(1 - A) a + Ab v y J ~ v J

for any A E [0, 1] .

If we use the inequalities (8) and (9) for the HG-convex function f : [a, b] C (0, to) ^ (0, to), f (t) = exp (t), then, by performing the required calculations, we get

G2 (a,b) 1 A (a, b)

0 < ~ /T/ — H (a, b) < ^ ^Vr?\ (b — a)2 (40)

L (a, b) 4 G2 (a, b)

and

, / G2 (a, b) 1 A (a, b) ,, x2

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0 < A(a,b) — T^ < 4GWb-a)2- (41)

From the inequality (13) we also have

G2 M) ^ 1 ,

L (a, b) _ 2

< - (A (a, b) + b^a^^ . (42)

Acknowledgement. The author would like to thank the anonymous referees for valuable comments that have been implemented in the final version of the paper.

References

[1] Alomari M., Darus M. The Hadamard's inequality for s-convex function. Int. J. Math. Anal., 2008, vol. 2, no. 13-16, pp. 639-646.

[2] Alomari M., Darus M. Hadamard-type inequalities for s-convex functions. Int. Math. Forum, 2008, vol. 3, no. 37-40, pp. 1965-1975.

[3] Anastassiou G. A. Univariate Ostrowski inequalities, revisited. Monatsh. Math., 2002, vol. 135, no. 3, pp. 175-189.

[4] Anderson G. D., Vamanamurthy M. K., Vuorinen M. Generalized convexity and inequalities. J. Math. Anal. Appl., 2007, vol. 335, pp. 1294-1308.

[5] Barnett N. S., Cerone P., Dragomir S. S., Pinheiro M. R., Sofo A. Ostrowski type inequalities for functions whose modulus of the derivatives are convex and applications. Inequality Theory and Applications, vol. 2 (Chinju/Masan, 2001), 19-32, Nova Sci. Publ., Hauppauge, NY, 2003. Preprint: RGMIA Res. Rep. Coll., 2002, vol. 5, no. 2, art. 1 [Online http://rgmia.org/papers/v5n2/Paperwapp2q.pdf].

[6] Beckenbach E. F. Convex functions. Bull. Amer. Math. Soc., 1948, vol. 54, pp. 439-460.

[7] Bombardelli M., Varosanec S. Properties of h-convex functions related to the Hermite-Hadamard-Fejer inequalities. Comput. Math. Appl., 2009, vol. 58, no. 9, pp. 1869-1877.

[8] Cristescu G. Hadamard type inequalities for convolution of h-convex functions. Ann. Tiberiu Popoviciu Semin. Funct. Equ. Approx. Convexity, 2010, vol. 8, pp. 3-11.

iНе можете найти то, что вам нужно? Попробуйте сервис подбора литературы.

[9] Dragomir S. S. On the Ostrowski's inequality for mappings of bounded variation and applications. Math. Ineq. & Appl., 2001, vol. 4, no. 1, pp. 33-40.

[10] Dragomir S. S. An inequality improving the first Hermite-Had,a,m,ard inequality for convex functions defined on linear spaces and applications for semi-inner products. J. Inequal. Pure Appl. Math., 2002, vol. 3, no. 2, art. 31, 8 p.

[11] Dragomir S. S. An inequality improving the second Hermite-Hadamard inequality for convex functions defined on linear spaces and applications for semi-inner products. J. Inequal. Pure Appl. Math., 2002, vol. 3, no. 3, art. 35.

[12] Dragomir S. S. An Ostrowski like inequality for convex functions and applications. Revista Math. Complutense, 2003, vol. 16, no. 2, pp. 373-382.

[13] Dragomir S. S. Operator Inequalities of Ostrowski and Trapezoidal Type. Springer Briefs in Mathematics. Springer, New York, 2012. x+112 pp. ISBN: 978-1-4614-1778-1

[14] Dragomir S. S. Some new inequalities of Hermite-Hadamard type for GA-convex functions. Preprint RGMIA Res. Rep. Coll., 2015, vol. 18, art. 30. [http://rgmia.org/papers/v18/v18a30.pdf].

[15] Dragomir S. S. Inequalities of Hermite-Hadamard type for HA-convex functions. Preprint RGMIA Res. Rep. Coll., 2015, vol. 18, art. 38. [http://rgmia.org/papers/v18/v18a38.pdf].

[16] Dragomir S. S. New inequalities of Hermite-Hadamard type for HA-convex functions. Preprint RGMIA Res. Rep. Coll., 2015, vol. 18, art. 41. [http://rgmia.org/papers/v18/v18a41.pdf].

[17] Dragomir S. S., Cerone P., Roumeliotis J., Wang S. A weighted version of Ostrowski inequality for mappings of Holder type and applications in numerical analysis. Bull. Math. Soc. Sci. Math. Romanie, 1999, vol. 42(90), no. 4, pp. 301-314.

[18] Dragomir S. S., Fitzpatrick S. The Hadamard inequalities for s-convex functions in the second sense. Demonstratio Math., 1999, vol. 32, no. 4, pp. 687-696.

[19] Dragomir S. S., Fitzpatrick S. The Jensen inequality for s-Breckner convex functions in linear spaces. Demonstratio Math., 2000, vol. 33, no. 1, pp. 43-49.

[20] Dragomir S. S., Mond B. On Hadamard's inequality for a class of functions of Godunova and Levin. Indian J. Math., 1997, vol. 39, no. 1, pp. 1-9.

[21] Dragomir S. S., Pearce C. E. M. On Jensen's inequality for a class of functions of Godunova and Levin. Period. Math. Hungar., 1996, vol. 33, no. 2, pp. 93-100.

[22] Dragomir S. S., Pearce C. E. M. Quasi-convex functions and Hadamard's inequality. Bull. Austral. Math. Soc., 1998, vol. 57, pp. 377-385.

[23] Dragomir S. S., Pecaric J., Persson L.-E. Some inequalities of Hadamard type. Soochow J. Math., 1995, vol. 21, no. 3, pp. 335-341.

iНе можете найти то, что вам нужно? Попробуйте сервис подбора литературы.

[24] Dragomir S. S., Rassias Th. M. (Eds) Ostrowski Type Inequalities and Applications in Numerical Integration. Kluwer Academic Publisher, 2002.

[25] El Farissi A. Simple proof and refinement of Hermite-Hadamard, inequality. J. Math. Ineq., 2010, vol. 4, no. 3, pp. 365-369.

[26] Iscan I. Hermite-Hadamard type inequalities for harmonically convex functions. Hacettepe Journal of Mathematics and Statistics, 2014, vol. 43, no. 6, pp. 935-942.

[27] Kikianty E., Dragomir S. S. Hermite-Hadamard's inequality and the p-HH-norm on the Cartesian 'product of two copies of a normed space. Math. Inequal. Appl., 2010, vol. 13, no. 1, pp. 1-32.

[28] Pecaric J. E., Dragomir S. S. On an inequality of Godunova-Levin and some refinements of Jensen integral inequality. Itinerant Seminar on Functional Equations, Approximation and Convexity (Cluj-Napoca, 1989), pp. 263268, Preprint, 89-6, Univ. "Babes-Bolyai", Cluj-Napoca, 1989.

Received April 19, 2017. In revised form, July 11, 2017. Accepted July 11, 2017. Published online August 16, 2017.

Mathematics, College of Engineering & Science Victoria University, PO Box 14428 Melbourne City, MC 8001, Australia;

DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences,

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E-mail: sever.dragomir@vu.edu.au