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

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:

©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

, (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

<

< 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

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

(

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

/ ! 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)

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

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)

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

( 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)

<

<

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) ,

< -- / 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 ( -- ) *

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

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) _

_ (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

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

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

— 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

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)

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

< -< 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

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^

<

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)

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)

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

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

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

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

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

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

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)

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) =

= [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

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.

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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,

School of Computer Science and Applied Mathematics

University of the Witwatersrand,

Private Bag 3, Johannesburg 2050, South Africa

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