A NOTE ON CHARACTERIZATION OF h-CONVEX FUNCTIONS VIA HERMITE-HADAMARD TYPE INEQUALITY Текст научной статьи по специальности «Математика»

28

Probl. Anal. Issues Anal. Vol. 8(26), No 2, 2019, pp. 28-36

DOI: 10.15393/j3.art.2019.5790

UDC 517.51

M. Rostamian Delavar, S. S. Dragomir, M. De La Sen

A NOTE ON CHARACTERIZATION OF h-CONVEX FUNCTIONS VIA HERMITE-HADAMARD TYPE INEQUALITY

Abstract. A characterization of h-convex function via Hermite-Hadamard inequality related to the h-convex functions is investigated. In fact it is determined that under what conditions a function is h-convex, if it satisfies the h-convex version of Hermite-Hadamard inequality.

Key words: h-convex function, Hermite-Hadamard inequality 2010 Mathematical Subject Classification: 26A51, 26D15, 52A01

1. Introduction. The following result is well-known in the literature:

Theorem 1. [6] A function f : (a,b) C R ^ R is convex if and only if

holds for all x,y E (a, b) with x = y.

Inequality (1) is known as the Hermite-Hadamard integral inequality for convex functions. Note that the left-hand part and the right-hand part of (1) separately are equivalent to the convexity of f (see [5,6]).

In 2006, the concept of h-convex functions related to the nonnegative real functions has been introduced in [9] by S. Varosanec. This class includes a large class of nonnegative functions, such as nonnegative convex functions, Godunova-Levin functions [3], s-convex functions in the second sense [1], and P-functions [2]. In [4], A. Hazy used the following definition of h-convex functions, which is a generalization of convexexity:

© Petrozavodsk State University, 2019

b

(1)

Definition 1. Let h : [0,1] ^ R be a function, such that h ^ 0. We say that f : (a,b) ^ R is an h-convex function, if for all x,y E (a,b), A E [0,1], we have

f (Ax + (1 - A)y) ^ h(A)f (x) + h(1 - A)f (y). (2)

We use this definition for the real functions defined on open intervals (a, b) C R in this paper. The h-convex version of the Hermite-Hadamard inequality was introduced in [8] by Sarikaya et al. as the following:

Theorem 2. Let f : I ^ [0, ro] be an integrable h-convex function. If a, b E I, with a < b, then

f ) ^ l^J f (x)dx ^ [f (a) + f (b)] ( / h(t)dt) .

2h( 1 y \ 2 J b - a

a 0

Motivated by the abovementioned works and results, we, in this paper, reply to the problem of conditions h-convexity of a function that satisfies (3). Since inequality (3) is double, we separate the problem to the right-hand and the left-hand versions, for the sake of convenience.

2. Main results. To achieve our main results about the characterization of an h-convex function via (3), we introduce a primary definition along with an example and then establish a basic lemma related to h-convex functions.

Definition 2. A function h : [0,1] ^ R is said to be self-concave if h(zx + (1 - z)y) ^ h(z)h(x) + h(1 - z)h(y),

for all z E (0,1) and x, y E [0,1].

We can find some simple functions that are self concave. Example. Consider the function h(x) = xn for n E N and x E [0,1]. It is not hard to see that this function is self-concave. In fact, since the function h is nonnegative,

h(Ax + (1 - A)y) = (Ax + (1 - A)y)n = En=o C)(Ax)n-i((1 - A)y) ^ ^ (o)(Ax)n + C)((1 - A)y)n = h(A)h(x) + h(1 - A)h(y).

b

i

1

Now consider the function h(x) = tan(x), for x E (0,1) and z E (0,1). Expanding this function and using the self-concavity of xn for n E N and

x E [0,1], we get

1 3 tan (Ax + (1 - A)y) = (Ax + (1 - A)y) + - (Ax + (1 - A)y) +

3

+15(Ax +(1 -A)y)5 + 37(Ax + (1 -A)y)7 + (Ax + (1 -A)y)9 +... ^

1 r\ i >y r?r\

> Ax + ^(Ax)3 + -(Ax)5 + —(Ax)7 + ^(Ax)9 + ■ ■ ■ + ((1 - A)y)

+1 № - A>y)3№ - A)y)5 + 315 № - A)y)7K1 - A)y)9+- =

= tan(Ax) + tan((1 - A)y) > tan(A) tan(x) + tan(1 - A) tan(y),

which implies the self-concavity of h(x) = tan(x) on (0,1). Note that we have used the fact that tan(xy) > tan(x) tan(y) for all x,y E (0,1).

The following lemma plays an important role in obtaining our expected results.

Lemma 1. Let f : (a, b) ^ R be a continuous function and h: [0,1] ^ R be a continuous self-concave function. Suppose that for any x, y E (a, b) with x = y there is a A E (0,1) such that f (Ax + (1 - A)y) ^ h(A)f (x) + + h(1 - A)f (y). Then f is h-convex on (a, b).

Proof. Without loss of generality, consider x,y E (a, b) with x < y. Define Mx,y = {A E [0,1]; f (Ax + (1 - A)y) ^ h(A)f (x) + h(1 -

It is obvious that Mx,y is nonempty. Since f and h are continuous on

their domains, Mx,y is closed in [0,1]. We prove that Mx,y = [0,1]. On

the contrary, suppose that Mx,y is a proper subset of [0,1]; then we can find a, [ E Mx,y such that (a,[) C [0,1] \ Mx,y. Set

w = ax + (1 - a)y , z = [x + (1 - [)y. (4)

From the assumption, there is a A E (0,1) such that

f (Aw + (1 - A)z) ^ h(A)f (w) + h(1 - A)f (z). (5)

Also

f f(w) = /(ax + (1 - a)y) ^ h(a)f (x) + h(1 - a)f(У), (6)

\ f (z) = f ([x + (1 - [)y) ^ h([)f (x) + h(1 - [)f (y). (6)

Set t = Aa + (1 — A)P. It is clear that t E (a, P) and t E Mx,y. Therefore, from the self-concavity of h and relations (4)-(6), we have

f (tx + (1 — t)y) > h(t)f (x) + h(1 — t)f (y) =

= h(Aa + (1 — A)P)f (x) + h(l — (Aa + (1 — A)P ))f (y) = = h(Aa + (1 — A)P)f(x) + h(A(1 — a) + (1 — A)(1 — P))f(y) ^

^ h(A)h(a) + h(1 — A)h(P)] f (x)+ [h(A)h(1 — a) + h(1 — A)h(1 — P)] f (y) =

= h(A) [h(a)f (x) + h(1 — a)f (y)] + h(1 — A) [h(P)f (x) + h(1 — P)f (y)'

^ h(A)f (w) + h(1 — A)f (z) ^ f (Aw + (1 — A)z). On the other hand,

Aw + (1 — A)z = A (ax + (1 — a)y) + (1 — A) (Px + (1 — P)y) =

Aa + (1 - A)P x + A(1 - a) + (1 - A)(1 - P)

y

So,

Aa + (1 - A)P x + 1 - (Aa + (1 - A)P) y = tx + (1 - t)y. f (tx + (1 - t)y) = f (Aw + (1 - A)z) <f (tx + (1 - t)y),

which is a contradiction. It follows that Mx y is not a proper subset of [0,1] and hence Mx,y = [0,1]. Since this happens for any x,y E (a, b) with x < y, we conclude that f is h-convex on (a,b). □

Theorem 3. Let f : (a, b) ^ R be a continuous function. Also suppose that h : [0,1] ^ R is a continuous self-concave function, such that

1

y-x

f (t)dt ^ [f (x) + f (y)] / h(t)dt

for all x, y G (a, b) with x = y. Then f is h-convex on (a, b).

Proof. Suppose that f is not h-convex on (a, b). Then, by Lemma 1, there are x, y G (a, b) with x < y such that

f (tx + (1 - t)y) > h(t)f (x) + h(1 - t)f (y) Vt G (0,1).

i

y

For such x and y,

y i i

y—xj f (t)dt = / f (tx + (1 - t)y)dt > J[h(t)f (x) + h(1 - t)f (y)]dt

i i h(t)dt)f (x) + h(1 - t)dt)f (y) = [f (x) + f (y)]^J h(t)dt). 0 0 0 This is a contradiction. Hence, f is h-convex on (a, 6). □

The following lemma, along with Lemma 1, are the base for characterization of a h-convex function via the left-hand side of (3).

Lemma 2. (Also see Theorem 1.1.4 in [5].) Suppose that ^ : [a, 6] ^ R is a continuous function such that <^(a) = <^(6) = 0 and <^(t) > 0 for some t E (a, 6). Then there exists an x E (a, 6) such that

<^(x) = max <^(y) and <^(x) > <^(y) for all a ^ y < x.

Proof. From Theorem 4.16 in [7], ^ attains its maximum a in [a, 6]. From the assumption, we have a ^ <^(t) > 0. Set M = {y E [a, 6]; <^(y) = a}. Since is continuous, M is a nonempty compact subset of [a, 6], such that a, 6 E M. If we put x = inf{y; y E M}, then

<^(x) = a = max <^(y),

and f (y) < f (x) for all a ^ y < x. □

In what follows, we assume that the function h : [0,1] ^ R satisfies the conditions

h(A) + h(1 - A) = 1 for all A E (0,1), (7)

h(0) = 0. (7)

Lemma 3. Let h : [0,1] ^ R be a continuous self-concave function. Suppose that f : (a, 6) ^ R is a continuous function and for any x E (a, 6), e > 0, there exist y, z E (a, 6) if (x — e, x + e) with y < x < z such that

f (x) = f (Ay + (1 - A)z) ^ h(A)f (y) + h(1 - A)f (z) for some A E (0,1).

Then f is h-convex on (a, 6).

Proof. If f is not h-convex, then by Lemma 1, there are x^x2 G (a, b) with xi = x2 (assume that xi < x2) such that

f (Axi + (1 - A)x2) > h(A)f (xi) + h(1 - A)f (X2) for all A G (0,1). (8)

Consider the function g : [x1,x2] ^ R defined by

g(y) = g(Axi + (1 - A)x2) =

= f (Axi + (1 - A)x2) - f (xi) - f (x2) - f (xi Vh(A)xi + h(1 - A)x2 - xO.

x2 - xi

It is clear that g is continuous on [xi,x2] and g(xi) = g(x2) = 0. Also, from (7) and (8), we get

g(Axi + (1 - A)x2) = f (Axi + (1 - A)x2) - f (xi)- (9)

- f (x2) - f (xi) /(1 - h(A))x2 - (1 - h(A))xi) =

x2 - xi

= f (Axi + (1 - A)x2) - h(A)f (xi) - h(1 - A)f (x2) > 0. Lemma 2 and (9) imply that there is an x G (xi,x2) such that

g(x) = max g(y) and g(x) > g(y) for xi ^ y < x. (10)

Hence, x = txi + (1 - t)x2 for some 0 < t < 1. Now choose x0, yo G [xi, x2] such that xi ^ x0 < x < y0 ^ x2. Therefore, from (10) for any A G (0,1),

g(x) = [h(A) + h(1 - A)]g(x) > h(A)g(xo) + h(1 - A)g(yo). (11)

f(x) - f(xi) - f (x2) f(xi) (h(A)xo + h(1 - A)yo - xi) > (12) x2 - xi

> h(A) [f(xo) - f(xi) - f(x2) f(xi) (xo - xi)l +

x2 - xi

+h(1 - A) [f (yo) - f (xi) - f (x2) - f (xi) (yo - xi)

x2 - xi

From (7) we deduce, simplifying (12):

f (x) > h(A)f (xo) + h(1 - A)f (yo) for all A G (0,1). (13)

Since x0, y0, and A are arbitrary, (13) contradicts the assumption. Hence, f is an h-convex function on (a, b). □

Using Lemma 3, as an immediate consequence we have two following lemmas. For more details about this kind of results related to the convex functions, see [6].

Corollary 1. Let h : [0,1] ^ R be a continuous self-concave function. Suppose that f : (a, b) ^ R is a continuous function and for any x E (a, b), £ > 0, there exists a 5 E (0,£) such that

f (x) ^ h(1/2)[f (x - 5) + f (x + 5)].

Then f is h-convex on (a, b).

Proof. In Lemma 3, take y = x — 5, z = x + 5 and A = 1/2. □

Lemma 4. Let h : [0,1] ^ R be a continuous self-concave function. Suppose that f : (a, b) ^ R is a continuous function and for any x E (a, b), £ > 0, there exists 5 E (0,£) such that

x+S

f (x) ^ J f (u)du.

x—S

Then f is h-convex on (a, b).

Proof. Suppose that f is not h-convex on (a,b). From Corollary 1, there are x E (a, b) and £ > 0 such that a<x — £<x + £<b and

f (x) > h(1/2) [f (x — 5) + f (x + 5)] for any 0 < 5 < £.

Integrating with respect to 5 in the above inequality, we get

s S S

/ f (x)dt > J f (x — t)dt + J f (x + t)dt =

0 0

x—S x+S x+S

= — J f (w)dw + J f (w)dw = J f (w)dw.

x x x—S

So,

x+S

f (x) ■ 5 ^ h(1/2^ y f (u)du.

x—S

This contradicts the assumption and, hence, f is h-convex on (a, b). □

Now, using Lemma 4, we can obtain a characterization-type theorem for h-convex functions via the left-hand side of (3).

Theorem 4. Let h : [0,1] ^ R be a continuous self-concave function. Suppose that f : (a, b) ^ R is a continuous function and for all y, z G (a, b) with y = z we have

1 fi^l i' f (u)du; (14)

2h(1/2)'V 2 J z - y„ y

then f is h-convex on (a, b).

Proof. Suppose that f is not h-convex on (a,b). From Lemma 4, there exist x G (a, b) and e > 0 such that for all 6 G (0, e)

f (x) > / f (u)du.

x—&

Now, if we choose 6 < e and y, z G (a, b) with y < z such that

x 2 y ^^ 2

x — y = z — x = 6,

then we have

z

f (^) > I f <u)du.

y

This contradicts (14). Thus, f is h-convex on (a, b). □

References

[1] Breckner W. W. Stetigkeitsaussagen für eine Klasse verallgemeinerter konvexer funktionen in topologischen linearen Rüumen. Publ. Inst. Math., 1978, vol. 23, pp. 13-20.

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[4] Hazy A. Bernstein-Doetsch type results for h-convex functions. Math. Ineq. Appl., 2011, vol. 14, no. 3, pp. 499-508. DOI: https://doi .org/10.7153/ mia-14-42.

[5] Niculescu C. P., Persson L. E. Convex functions and their applications: a contemporary approach. Springer, Berlin, CMS Books in Mathematics, 2006.

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[7] Rudin W., Principles of mathematical analysis. McGraw-Hill, 1976.

[8] Sarikaya M. Z., Saglam A., Yildirim H. On Some Hadamard-Type Inequalities for h-Convex Functions. J. Math. Ineq., 2008, vol. 2, no. 3, pp. 335341. DOI: https://doi.org/10.7153/jmi-02-30.

[9] Varosanec S. On h-convexity. J. Math. Anal. Appl., 2007, vol. 326, pp. 303311. DOI: https://doi.org/10.1016/j.jmaa.2006.02.086.

Received December 18, 2018.

In revised form, March 11, 2019.

Accepted March 12, 2019.

Published online April 1, 2019.

M. Rostamian Delavar

Department of Mathematics, Faculty of Basic Sciences, University of Bojnord,

Bojnord, Iran

E-mail: [email protected]

S. S. Dragomir

Mathematics, College of Engineering & Science, Victoria University,

PO Box 14428, Melbourne City, MC 8001, Australia

E-mail: [email protected]

M. De La Sen

Institute of Research and Development of Processes, University of Basque

Country, Campus of Leioa (Bizkaia) - Aptdo. 644 - Bilbao, Bilbao, 48080,

Spain

E-mail: [email protected]