New Properties of m-Convex Functions - Hikari

International Journal of Mathematical Analysis Vol. 9, 2015, no. 15, 735 - 742 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ijma.2015.412389

New Properties of m-Convex Functions 1 Teodoro Lara, Edgar Rosales Departamento de F´ısica y Matem´aticas Universidad de los Andes, N. U. Rafael Rangel, Trujillo, Venezuela Jos´ e L. S´ anchez Universidad Central de Venezuela Escuela de Matem´aticas. Caracas. Venezuela c 2014 Teodoro Lara, Edgar Rosales and Jos´e L. S´anchez. This is an open Copyright access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this research we present some properties for m- convex functions, algebraic, inequalities of Fej´er type, a sandwich like theorem and a result on Hyers-Ulam stability.

Mathematics Subject Classification: 26A51, 39B62 Keywords: m-convex function, strongly convex function, Fej´er type inequalities

1. Introduction The concept of m-convex function was introduced in ([4, 7]) and since then many properties, especially inequalities, have been obtained for them. In this paper we establish and prove some additional properties of different type, namely algebraic properties, inequalities and a sandwich type theorem. 1This research has been partially supported by Central Bank of Venezuela.

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Definition 1. A function f : [0, b] → R is called m-convex, 0 ≤ m ≤ 1, if for any x, y ∈ [0, b] and t ∈ [0, 1] we have f (tx + m(1 − t)y) ≤ tf (x) + m(1 − t)f (y).

(1.1)

Remark 2. It is important to point out that the above definition is equivalent to f (mtx + (1 − t)y) ≤ mtf (x) + (1 − t)f (y), with x, y and t as before. Remark 3. If f is a m-convex function, with m ∈ [0, 1) and x = y = 0 in (1.1) then f (0) ≤ 0. Geometrically a function f : [0, b] → R is m- convex if for any x, y ∈ [0, b], say x ≤ y, the segment between the points (x, f (x)) and (my, mf (y)) is above the graph of f in [x, my]. 2. Main Results We begin by setting some algebraic and topological properties for m-convex functions. The incoming propositions and corresponding proofs follow ideas from [6]. Proposition 4. Let f, g : [a, b] → R, a ≥ 0. If f is m1 -convex and g is m2 convex, with m1 ≤ m2 , then f + g and αf , α ≥ 0 a constant, are m1 -convex. Proof. Since g is m2 -convex and m1 ≤ m2 , then g is m1 -convex ([3], Lemma 2). Thus, for x, y ∈ [a, b] and t ∈ [0, 1], (f + g)(tx + m1 (1 − t)y) = f (tx + m1 (1 − t)y) + g(tx + m1 (1 − t)y) ≤ t(f + g)(x) + m1 (1 − t)(f + g)(y). Also, (αf )(tx+m1 (1−t)y) ≤ α[tf (x)+m1 (1−t)f (y)] = t(αf )(x)+m1 (1−t)(αf )(y).  Proposition 5. Let f : [0, b] → R, g : [0, b1 ] → R, with range(f ) ⊆ [0, b1 ]. If f and g are m-convex and g is increasing, then g ◦ f is m-convex on [0, b]. Proof. For x, y ∈ [0, b] and t ∈ [0, 1], (g ◦ f )(tx + m(1 − t)y) = g(f (tx + m(1 − t)y)) ≤ tg(f (x)) + m(1 − t)g(f (y)) = t(g ◦ f )(x) + m(1 − t)(g ◦ f )(y).  Proposition 6. If f, g : [0, b] → R are both nonnegative, increasing and mconvex, then f g is m-convex.

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Proof. If x ≤ y (the case y ≤ x runs in the same fashion) then f (x) − f (y) ≤ 0 and g(y) − g(x) ≥ 0 which implies (2.1)

f (x)g(y) + f (y)g(x) ≤ f (x)g(x) + f (y)g(y).

On the other hand for x, y ∈ [0, b] and t ∈ [0, 1], (f g)(tx + m(1 − t)y) = f (tx + m(1 − t)y) g(tx + (1 − t)y) ≤ [tf (x) + m(1 − t)f (y)] [tg(x) + m(1 − t)g(y)] = t2 f (x)g(x) + m2 (1 − t)2 f (y)g(y) + mt(1 − t)[f (x)g(y) + f (y)g(x)]. Using now (2.1), we obtain (f g)(tx + m(1 − t)y) ≤ t2 f (x)g(x) + m2 (1 − t)2 f (y)g(y) + mt(1 − t)[f (x)g(x) + f (y)g(y)] = t[t + m(1 − t)]f (x)g(x) + m(1 − t)[t + m(1 − t)]f (y)g(y). But t + m(1 − t) ≤ 1, therefore (f g)(tx+m(1−t)y) ≤ tf (x)g(x)+m(1−t)f (y)g(y) = t(f g)(x)+m(1−t)(f g)(y).   b Proposition 7. Let f : [0, +∞) → R a finite function on a, m ⊂ [0, +∞), m-convex with m ∈ (0, 1]. Then f is bounded on any closed interval [a, b]. Proof. Let M = max{f (a), mf ( mb )}, so for any z = ta + (1 − t)b in [a, b],   b f (z) = f (ta + (1 − t)b) =f ta + m(1 − t) m   b ≤ tf (a) + m(1 − t)f m ≤ M. Thus, f is upper bounded in [a, b]. Now we notice that any z ∈ [a, b] can be written as a+b + t for |t| ≤ 2 hence        a+b 1 a+b 1 a+b f =f +t + −t 2 2 2 2 2 !   a+b − t 1 a+b m 2 ≤ f +t + f . 2 2 2 m In other words,     a+b a+b f + t ≥ 2f − mf 2 2

a+b 2

−t m

!

 ≥ 2f

a+b 2

 − M,

b−a , 2

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and since

a+b 2

+ t is arbitrary in [a, b], f is also bounded below in [a, b].



The following result gives necessary and sufficient conditions under which a pair of functions may be separated by an m- convex one, we call it a sandwich like theorem and is inspired in [1]. Theorem 8. Let b > 0, m ∈ [0, 1] and f, g : [0, b] → R functions with g nonnegative. Then there is an m-convex function h : [0, b] → R such that f ≤ h on [0, mb] and h ≤ g on [0, b] if and only if (2.2)

f (tx + m(1 − t)y) ≤ tg(x) + m(1 − t)g(y); x, y ∈ [0, b], t ∈ [0, 1].

Proof. Let us assume that functions f and g satisfy (2.2) and call A the convex hull of the epigraph of g, that is, A = hull{(x, y) ∈ [0, b] × R : g(x) ≤ y}. Let (x, y) ∈ A, then by using Caratheodory’s theorem ([1, 6]), it belongs to the interior of a 2- simplex S in A which in turn is a triangle whose vertices are in epigraph of g. Let y0 be y0 = inf{z ∈ R : (x, z) ∈ S}, notice that y0 exists because S is a triangle therefore bounded. So, y ≥ y0 and (x, y0 ) is in the boundary of S; therefore (x, y0 ) = t(x1 , y1 ) + (1 − t)(x2 , y2 ) for some t ∈ [0, 1] and (x1 , y1 ), (x2 , y2 ) vertices of S, that is in the epigraph of g. From here, and using g ≥ 0, we get y ≥ y0 ≥ tg(x1 )+m(1−t)g(x2 ) ≥ f (tx1 +m(1−t)x2 ) = f (x), with x ∈ [x1 , mx2 ]. It is worth to point out that expression tx1 + m(1 − t)x2 , t ∈ [0, 1] is as parametrization of [x1 , mx2 ] (or [x2 , mx1 ]). We now define h : [0, b] → R as h(x) = inf{y ∈ R : (x, y) ∈ A}, this infimun exists because g ≥ 0. Actually f ≤ h on [0, mb], moreover, because (x, g(x)) ∈ A for any x ∈ [0, b] we conclude that h ≤ g on [0, b]. Now we show h is m-convex, let us choose x1 , x2 ∈ [0, b] arbitrary and t ∈ [0, 1]; so for any y1 , y2 such that (x1 , y1 ), (mx2 , my2 ) ∈ A which is convex, then (tx1 + (1 − t)mx2 , ty1 + (1 − t)my2 ) ∈ A and consequently h(tx1 + (1 − t)mx2 ) ≤ ty1 + (1 − t)my2 , now by passing to the infimun the desired result shows up. For the converse we take x, y ∈ [0, b] and t ∈ [0, 1], then f (tx + m(1 − t)y) ≤ th(x) + m(1 − t)h(y) ≤ tg(x) + m(1 − t)g(y). 

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Corollary 9. If f and g are as before but now m ∈ [0, 1), and they satisfy (2.2), then f (0) ≤ 0. An important consequence of the Theorem 8 is the following Hyers-Ulamtype stability result for m-convex functions. Next definition is similar to the one given for convex functions ([2]). Definition 10. Let  > 0 and m ∈ [0, 1]. A function f : [0, b] → R is called -m-convex if for any x, y ∈ [0, b] and t ∈ [0, 1] we have f (tx + m(1 − t)y) ≤ tf (x) + m(1 − t)f (y) + . Corollary 11. Let  > 0 and m ∈ (0, 1]. If f : [0, b] → [− m , +∞) is -mconvex, then there exists a function h : [0, b] → R m-convex, such that  , x ∈ [0, mb]. |f (x) − h(x)| ≤ 2m Proof. We set g = f + m ; thus g is nonnegative; by hypothesis on f and t ≤ 0, it is clear that f and g satisfy (2.2) and from Theorem because t − m 8 there exists an m-convex function ϕ : [0, b] → R such that f ≤ ϕ on [0, mb] and ϕ ≤ g = f + m on [0, b]. Therefore, f ≤ ϕ ≤ f + m on [0, mb]. Putting  h = ϕ − 2m , h is m-convex and even more  |f (x) − h(x)| ≤ , x ∈ [0, mb]. 2m  The following result follows ideas from [5]. Theorem 12. Let f : [0, b] → R, twice differentiable function such that there are k1 , k2 ∈ R so that k1 ≤ f 00 ≤ k2 . Then for m ∈ [0, 1] fixed, a ∈ [0, b] and t ∈ [0, 1] arbitrary (2.3) k1 (mb − a)2 t(1 − t) k2 (mb − a)2 t(1 − t) ≤ tf (a)+(1−t)f (mb)−f (ta+m(1−t)b) ≤ . 2 2 Proof. We define G(t) = tf (a) + (1 − t)f (mb) − f (ta + m(1 − t)b) −

k1 (mb − a)2 t(1 − t) , t ∈ [0, 1] 2

then G00 (t) = (mb − a)2 [k1 − f 00 (ta + m(1 − t)b)] ≤ 0, that is, G is concave function on [0, 1], moreover G(0) = G(1) = 0. Therefore G(t) ≥ 0, t ∈ [0, 1] so left hand side of inequality holds. For the second half of mentioned inequality we define, as before, H(t) = tf (a)+(1−t)f (mb)−f (ta+m(1−t)b)−

k2 (mb − a)2 t(1 − t) , t ∈ [0, 1]. 2

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So H is convex on [0, 1], H(0) = H(1) = 0 and consequently H(t) ≤ 0, t ∈ [0, 1].  Corollary 13. Under the same assumption as in Theorem 12, with a 6= mb, the following inequalities hold Z mb f (a) + f (mb) 1 k2 (mb − a)2 k1 (mb − a)2 ≤ − f (x)dx ≤ . (2.4) 12 2 mb − a a 12 Moreover, if f is m-convex, then the following inequalities of Hermite-Hadamard type take place (2.5) Z mb f (a) + f (mb) k2 (mb − a)2 1 f (a) + mf (b) k1 (mb − a)2 − ≤ − . f (x)dx ≤ 2 12 mb − a a 2 12 Proof. (2.4) is obtened by integrating each term of (2.3) on [0, 1] with respect to t, and by the change of variable x = at + m(1 − t)b. The proof of (2.5) is similar, but considering now f (mb) ≤ mf (b), since f is m-convex.  Corollary 14. Preserving the same conditions of Theorem 12, with a ≤ mb, we obtain k1 (1 − 2t)2 (mb − a)2 f (ta + m(1 − t)b) + f ((1 − t)a + mtb) ≤ 8 2   a + mb k2 (1 − 2t)2 (mb − a)2 −f ≤ 2 8 for all t ∈ [0, 1]. Theorem 15. Let f : [0, b] → R, twice differentiable function such that there are k1 , k2 ∈ R so that k1 ≤ f 00 ≤ k2 . Let g : [a, b] → R+ , with a ≥ 0, integrable (i.e. g(x) = g(a + mb − x)), where m ∈ [0, 1]. Then, and symmetric about a+mb 2 the following inequalities hold Z Z k1 mb f (a) + f (mb) mb (x − a)(mb − x)g(x)dx ≤ g(x)dx− 2 a 2 a Z mb Z k2 mb f (x)g(x)dx ≤ (x − a)(mb − x)g(x)dx 2 a a Proof. We use mainly inequality (2.3), notice that for a = mb, the result is clear. Let a 6= mb, multiply each member of (2.3) by g(ta + m(1 − t)b) and integrate the result on [0, 1], we shall work each member of this resulting inequality separately. First we notice that (2.6) Z Z mb k1 (mb − a)2 1 k1 t(1−t)g(ta+m(1−t)b)dt = (x−a)(mb−x)g(x)dx, 2 2(mb − a) a 0

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and (2.7) Z Z mb k2 k2 (mb − a)2 1 t(1−t)g(ta+m(1−t)b)dt = (x−a)(mb−x)g(x)dx. 2 2(mb − a) a 0 The middle term of the mention resulting inequality is treated as follows, Z 1 Z mb f (a) (2.8) tf (a)g(ta + m(1 − t)b)dt = (mb − x)g(x)dx, (mb − a)2 a 0 Z 1 Z mb f (mb) (1 − t)f (mb)g(ta + m(1 − t)b)dt = (2.9) (x − a)g(x)dx. (mb − a)2 a 0 and Z 1 Z mb 1 f (ta+m(1−t)b)g(ta+m(1−t)b)dt = − (2.10) − f (x)g(x)dx. mb − a a 0 Hence, summing up (2.8), (2.9) and (2.10), and the fact that Z Z mb a + mb mb g(x)dx, xg(x)dx = 2 a a which is a consequence of of symmetry of g, shows that middle term of the mention inequality coming from (2.3) is Z Z mb f (a) + f (mb) mb 1 (2.11) g(x)dx − f (x)g(x)dx. 2(mb − a) mb − a a a Result follows from (2.6), (2.7) and (2.11).



Corollary 16. If f and g are as in Theorem 15 and f is m-convex, then the following inequalities of Fej´er type hold Z mb Z Z k2 mb f (a) + f (mb) mb f (x)g(x)dx g(x)dx − (x − a)(mb − x)g(x)dx ≤ 2 2 a a a f (a) + mf (b) ≤ 2

Z a

mb

k1 g(x)dx − 2

Z

mb

(x − a)(mb − x)g(x)dx. a

References [1] K. Baron and J. Matkowski and K. Nikodem, Sandwich with convexity, Math. Pannonica vol.5, 1 (1994), 139-144. [2] D. H. Hyers and S. M. Ulam, Approximately convex functions, Proc. Amer. Math. Soc. 3, (1952), 821–828. http://dx.doi.org/10.1090/s0002-9939-1952-0049962-5 [3] S. S. Dragomir, On Some New inequalities of Hermite-Hadamrd Type for m-Convex Functions, Tamkang J. of Math., vol. 33, 1 (2002), 45–55. [4] S. S. Dragomir and G. H. Toader, Some Inequalities for m-Convex Functions, Studia Univ. Babes-Bolyai, Math., vol.38, 1, (1993), 21–28.

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[5] N. Minculete and F. C. Mitroi, Fej´er Type Inequalities, arXiv: 1 105.5778v2, 1 (2011) 1–9. [6] A. W. Roberts and D. E. Varberg, Convex Functions. Academic Pres. NY. 1973. [7] G. H. Toader Some generalizations of the Convexity, Proc. Colloq. Approx. Optim. Cluj-Naploca (Romania) (1984), 329–338.

Received: December 17, 2014; Published: March 15, 2015