On Co-ordinated s-Convex Functions 1 Introduction - CiteSeerX

International Mathematical Forum, 3, 2008, no. 40, 1977 - 1989

On Co-ordinated s-Convex Functions Mohammad Alomari

1

and Maslina Darus

2

School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 43600 Selangor, Malaysia

Abstract A new class of s–convex functions in both sense denoted by M W Osi 1 ,s2 (i = 1, 2) is established. The definition of s–convex functions in both sense on the co–ordinates and some of it’s properties is established.

Keywords: s–Hadamard’s inequality, s–convex function, Jensen’s inequality, open problem

1

Introduction

Let f : I ⊆ R → R be a convex mapping defined on the interval I of real numbers and a, b ∈ I, with a < b. The following double inequality: 

f

a+b 2



1 ≤ b−a

b

f (x) dx ≤

a

f (a) + f (b) 2

(1)

is known in the literature as Hadamard’s inequality for convex mappings. For refinements, counterparts and new results on s–convex functions, see the papers [1–8]. In [7], Orlicz introduced two definitions of s–convexity of real valued functions. A function f : R+ → R, where R+ = [0, ∞), is said to be s–convex in the first sense if f (αx + βy) ≤ αs f (x) + β s f (y) 1 2

First author: [email protected] Corresponding author: [email protected]

(2)

1978

M. Alomari and M. Darus

for all x, y ∈ [0, ∞), α, β ≥ 0 with αs + β s = 1 and for some fixed s ∈ (0, 1]. We denote this class of functions by Ks1 . Also, a function f : R+ → R, where R+ = [0, ∞), is said to be s–convex in the second sense if f (αx + βy) ≤ αs f (x) + β s f (y)

(3)

for all x, y ∈ [0, ∞), α, β ≥ 0 with α + β = 1 and for some fixed s ∈ (0, 1]. We denote this class of functions by Ks2 . These definitions of s–convexity, for so called ϕ–functions, was introduced by Orlicz in [7] and was used in the theory of Orlicz spaces (see [5], [6], [8]). A function f : R+ → R+ is said to be ϕ–function if f (0) = 0 and f is non– decreasing and continuous. Its easily to check that the both s–convexity mean just the convexity when s = 1. The concept of s–convex functions and s–convex functions on the co– ordinates in both sense was introduced by Alomari and Darus in [1] and [3]. Definition 1.1 Consider the bidimensional interval Δ := [a, b] × [c, d] in [0, ∞)2 with a < b and c < d. The mapping f : Δ → R is s–convex in the first sense (in the second sense) on Δ if f (αx + βz, αy + βw) ≤ αs f (x, y) + β s f (z, w) , holds for all (x, y), (z, w) ∈ Δ with α, β ≥ 0 with αs + β s = 1 (α + β = 1) and for some fixed s ∈ (0, 1]. We write f ∈ Ξis (i = 1, 2) which means that f is s–convex in the first sense when i = 1, (in the second sense when i = 2). A function f : Δ → R is s–convex in the first sense on Δ is called co–ordinated s–convex in first sense (in the second sense) on Δ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v), are s–convex in the first sense (in the second sense) for all y ∈ [c, d] and x ∈ [a, b] such that s ∈ (0, 1], i.e, the partial mappings fy and fx s–convex with same fixed s ∈ (0, 1].

2

s–Convex Function On The Co–ordinates

In this section we introduce a new class of s–convex functions and s–convex functions on the co–ordinates. Let us start with the following definition.

On co-ordinated s-convex functions

1979

Definition 2.1 Consider the bidimensional interval Δ := [a, b] × [c, d] in [0, ∞)2 with a < b and c < d. A mapping f : Δ → R is called s–convex of 2 first sense on Δ if there exists s1 , s2 ∈ (0, 1] with s = s1 +s , such that 2 f (αx + βz, αy + βw) ≤ αs1 f (x, y) + β s2 f (z, w)

(4)

holds for all (x, y), (z, w) ∈ Δ, α, β ≥ 0 with αs1 + β s2 = 1 and for all fixed s1 , s2 ∈ (0, 1]. We denote this class of functions by MW Os11 ,s2 . Let f : Δ → R be s–convex on Δ, then f is called co–ordinated s–convex of first sense on Δ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v), are s1 –, s2 –convex functions in the first sense for all s1 , s2 ∈ (0, 1], y ∈ [c, d] and x ∈ [a, b]; respectively, with 2 s = s1 +s ∈ (0, 1]. 2 Also, a mapping f : Δ → R is called s–convex of second sense on Δ if there 2 exists s1 , s2 ∈ (0, 1] with s = s1 +s , such that (2.8) holds for all (x, y), (z, w) 2 ∈ Δ, α, β ≥ 0 with α + β = 1 and for all fixed s1 , s2 ∈ (0, 1]. We denote this class of functions by MW Os21 ,s2 . Similarly, we define the s–convex function of second sense on the co– ordinates, i.e., a function f is called co–ordinated s–convex of second sense on Δ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v), are s1 –, s2 –convex functions in the second sense for all 2 s1 , s2 ∈ (0, 1], y ∈ [c, d] and x ∈ [a, b]; respectively, with s = s1 +s ∈ (0, 1]. 2 Remark 2.2 Definition 2.1. implies Definition 1.1. iff s1 = s2 . Theorem 2.3 Let 0 < s1 , s2 ≤ 1. If f ∈ MW Os11 ,s2 , then f is non– decreasing on R2+ and lim + f (x) ≤ f (0), for all x = (x1 , x2 ) ∈ R2 . x→0 Proof. Let 0 < s1 , s2 ≤ 1 and f : R2+ → R, such that f ∈ MW Os11 ,s2 . If for all x, y ∈ R2+ and α, β ≥ 0 with αs1 + β s2 = 1. Then, for x > 0 and λ ∈ [0, 1], we have 

f

1

1

 

λ /s1 + (1 − λ) /s2 x

≤ λf (x) + (1 − λ) f (x) = f (x) .

Notice that, the function 1

1

h (λ) = λ /s1 + (1 − λ) /s2

1980

M. Alomari and M. Darus

is continuous on [0, 1] and differentiable on (0, 1). Since h (0) = h (1) = 1, then by Rolle’s theorem there exists c ∈ (0, 1) such that h (c) = 0. Therefore, h (c) =

1 1 1 1 −1 −1 (c) s1 − (1 − c) s2 = 0. s1 s2

If s1 = s2 , then it’s easy to see that c = if s1 = s2 , then we have two cases:

1 2

(5)

and therefore h (c) = 0. However,

1. If c < 1 − c, then h (c) < 0. 2. If c > 1 − c, then h (c) > 0.





Logically, let  > 0 and c, 1 − c ∈ 12 − , 12 +  , if |c − (1 − c)| < , then h (c) ≈ 0. Thus, h is decreasing on [0, c] and increasing on [c, 1] and h ([0, 1]) = [h (c) , h (1)] = [κ, 1] , 1

1

where, κ = (c) s1 + (1 − c) s2 . This yields that, f (θx) ≤ f (x) , ∀ x > 0, θ ∈ [κ, 1] .

(6)

1

If θ ∈ [κ2 , 1] then θ 2 ∈ [κ, 1]. Therefore, by (6) holds for all x > 0, we get  1  1 

f (θx) = f θ 2 θ 2 x

 1 

≤ f θ 2 x ≤ f (x)

for all x > 0. By induction we obtain that f (θx) ≤ f (x) ∀x, θ ∈ (0, 1] . Therefore, taking 0 < x ≤ y and applying (7) we get   

x y ≤ f (y) y

f (x) = f

which means that f is non–decreasing on R2+ . The second part can be proved as follows, For x > 0 we have f (αx) = f (αx + β0) ≤ αs1 f (x) + β s2 f (0) by taking x → 0+ we get lim f (x) ≤

x→0+

=

lim f (αx)

x→0+

lim f (αx + β0)

x→0+ s1

≤ α

lim f (x) + β s2 f (0)

x→0+

(7)

On co-ordinated s-convex functions

1981

and so, β s2 lim + f (x) ≤ f (0) = f (0) , x→0 1 − α s1

∀s1 , s2 ∈ (0, 1] .

Theorem 2.4 Let 0 < s1 , s2 ≤ 1. If f ∈ MW Os21 ,s2 , then f is non–negative on R2+ . Proof. We have for x ∈ R2+ , 





x x 1 1 + f (x) = f ≤ s1 f (x) + s2 f (x) = 2−s1 + 2−s2 f (x) . 2 2 2 2

Therefore, (2−s1 + 2−s2 ) f (x) ≥ 0 and so f (x) ≥ 0. Remark 2.5 The above two results do not hold in general, in the case of convex functions (s1 = s2 = 1), since a convex function f : R2+ → R, need not be either non–decreasing or non–negative, (see [4]). Remark 2.6 The above two results hold, in the case of s–convex functions such that s = s1 = s2 , (see Definition 1.1). Theorem 2.7 For 0 < s1 , s2 ≤ 1, we have Ξis ≡ MW Osi i ,si , for i = 1, 2 i 2 with s = s1 +s . For instance if s1 = s2 then Ξis ≡ MW Os,s 2 Proof. It’s an immediate consequence of Definition 1.1. and Definition 2.1. Let us take some examples about s–convex function and s–convex function on the co–ordinates in both sense. Example 2.8 Let f : R+ × R+ → R+ be co–ordinated s–convex of second sense and a, b, c ∈ R. For all s1 , s2 ∈ (0, 1] and x, y ∈ R+ , define ⎧ ⎪ ⎨

a , x=y=0 s1 f (x, y) = b (x + y) + c , 0 < x < y ⎪ ⎩ b (x + y)s2 + c , 0 < y < x If x = y = 0, then f ∈ MW Os21 ,s2 , and fy , fx ∈ K2s . Now, if b ≥ 0, 0 ≤ c ≤ a and 0 < x < y then fy , fx ∈ K2s1 for all 0 < x < y, which shows that f is (s = s1 )–convex on the co–ordinates. Moreover, since s2 = s1 for all 0 < x < y, then according to Definition 1.1. we have f ∈ Ξs2 . Similarly, if b ≥ 0 and 0 ≤ c ≤ a and 0 < y < x then fy , fx ∈ K2s for all 0 < y < x, which shows that f is (s = s2 )–convex on the co–ordinates. Moreover, since s1 = s2 for all 0 < y < x, then according to Definition 1.1. we

1982

M. Alomari and M. Darus

have f ∈ Ξs2 . In general, if b ≥ 0 and 0 ≤ c ≤ a and according to Definition 2.1. then, 2 f ∈ / MW Os21 ,s2 , for s = s1 +s ∈ (0, 1]. But, if (x, y) , (z, w) ∈ R+ × R+ , such 2 that 0 < x < y and 0 < w < z, then it’s easy to see that f ∈ MW Os21 ,s2 . Also, for all 0 < x < y, fy ∈ K2s1 and for all 0 < y < x, fx ∈ K2s2 . Example 2.9 In Example 2.8. if b ≥ 0 and c ≤ a, then f ∈ MW Os11 ,s2 . If x = y = 0, then f ∈ MW Os11 ,s2 , and fy , fx ∈ K1s . Now, if b ≥ 0, c ≤ a and 0 < x < y then fy , fx ∈ K1s1 for all 0 < x < y, which shows that f is (s = s1 )–convex on the co–ordinates. Therefore, there are two non–trivial cases: 1. Let x = (x1 , y1 ) and y = (x2 , y2 ) with 0 < x1 < y1 and 0 < x2 < y2 . Then, 0 < αx1 + βy1 < αx2 + βy2 , such that b (αx1 + βx2 + αy1 + βy2 )s1 + c b (α (x1 + y1 ) + β (x2 + y2 ))s1 + c b (αs1 (x1 + y1 )s1 + β s1 (x2 + y2 )s1 ) + c b (αs1 (x1 + y1 )s1 + β s1 (x2 + y2 )s1 ) +c (αs1 + β s1 ) = αs1 (b (x1 + y1 )s1 + c) + β s1 (b (x2 + y2 )s1 + c) = αs1 f (x1 , y1 ) + β s1 f (x2 , y2 ) .

f (αx1 + βx2 , αy1 + βy2 ) = = ≤ =

2. If x2 > x1 = 0 and y2 > y1 = 0 with β > 0. Then, f (α0 + β (x2 , y2 )) = = = = = ≤ =

f (β (x2 , y2)) bβ s1 (x2 + y2 )s1 + c bβ s1 (x2 + y2 )s1 + c (αs1 + β s1 ) αs1 c + β s1 (b (x2 + y2 )s1 + c) αs1 c + β s1 f (x2 , y2) αs1 a + β s1 f (x2 , y2 ) αs1 f (0, 0) + β s1 f (x2 , y2 ) .

Moreover, since s2 = s1 for all 0 < x < y, then according to Definition 1.1. we have f ∈ Ξs1 . Similarly, if b ≥ 0, c ≤ a and 0 < y < x then fy , fx ∈ K1s2 for all 0 < y < x, which shows that f is (s = s2 )–convex on the co–ordinates. Moreover, since

On co-ordinated s-convex functions

1983

s1 = s2 for all 0 < y < x, then according to Definition 1.1. we have f ∈ Ξs1 . In general, if b ≥ 0 and c ≤ a and according to Definition 2.1. then, 2 f ∈ / MW Os11 ,s2 , for s = s1 +s ∈ (0, 1]. But, if (x, y) , (z, w) ∈ R+ × R+ , such 2 that 0 < x < y and 0 < w < z, then it’s easy to see that f ∈ MW Os11 ,s2 . Therefore, there are two non–trivial cases: 1. Let x = (x, y) and y = (z, w) with 0 < x < y and 0 < w < z. Then, either 0 < αx + βy < αz + βw , or αx + βy > αz + βw > 0, without loss of generality, if αs1 + β s2 = 1, then we have f (αx + βz, αy + βw) = = ≤ = = =

b (αx + βz + αy + βw)s1 + c b (α (x + y) + β (z + w))s1 + c b (αs1 (x + y)s1 + β s1 (z + w)s1 ) + c b (αs1 (x + y)s1 + β s1 (z + w)s1 ) + c (αs1 + β s1 ) αs1 (b (x + y)s1 + c) + β s1 (b (z + w)s1 + c) αs1 f (x, y) + β s1 f (z, w) .

Also, if αs2 + β s1 = 1, then we have f (αx + βz, αy + βw) ≤ αs2 f (x, y) + β s2 f (z, w). Therefore, f (αx + βz, αy + βw) ≤ 2f (αx + βz, αy + βw) ≤ αs1 f (x, y) + β s1 f (z, w) + αs2 f (x, y) + β s2 f (z, w) = [αs1 f (x, y) + β s2 f (z, w)] + [β s1 f (z, w) + αs2 f (x, y)] , which shows that f ∈ MW Os11 ,s2 . 2. If z > x = 0 and w > y = 0 with β > 0. Then, f (α0 + β (z, w)) = = = = = ≤ =

f (β (z, w)) bβ s1 (z, w)s1 + c bβ s1 (z, w)s1 + c (αs1 + β s1 ) αs1 c + β s1 (b (z, w)s1 + c) αs1 c + β s1 f (z, w) αs1 a + β s1 f (z, w) αs1 f (0, 0) + β s1 f (z, w) .

Also, f (α0 + β (z, w)) ≤ αs2 f (0, 0) + β s2 f (z, w), therefore f (α0 + β (z, w)) ≤ 2f (α0 + β (z, w)) ≤ αs1 f (0, 0) + β s1 f (z, w) + αs2 f (0, 0) + β s2 f (z, w) = [αs1 f (0, 0) + β s2 f (z, w)] + [αs2 f (0, 0) + β s1 f (z, w)] ,

1984

M. Alomari and M. Darus

which shows that f ∈ MW Os11 ,s2 . Also, for all 0 < x < y we have, fy ∈ K1s1 and for all 0 < y < x we have, fx ∈ K1s2 . Example 2.10 In Example 2.8. if b > 0 and c < 0, then f ∈ / MW Os21 ,s2 , which it follows from Theorem 2.2. since for sufficiently small x the function f is being negative. Remark 2.11 If 0 < s1 , s2 < 1, then f ∈ MW Os11 ,s2 is non–decreasing on (0, ∞)2 but not necessarily on [0, ∞)2 . Example 2.12 In Example 2.8. if b ≥ 0 and c < a, then f non–decreasing on (0, ∞)2 but not on [0, ∞)2 . Theorem 2.13 Let 0 < s1 , s2 ≤ 1. If f, g ∈ MW Os11 ,s2 , then f + g ∈ MW Os11 ,s2 , |f | ∈ MW Os11 ,s2 and max (f, g) ∈ MW Os11 ,s2 . Proof. Let 0 < s1 , s2 ≤ 1 and f, g : R2+ → R, such that f, g ∈ MW Os11 ,s2 . If for all x, y ∈ R2+ and α, β ≥ 0 with αs1 + β s2 = 1. Then f (αx + βy) ≤ αs1 f (x) + β s2 f (y) and

g (αx + βy) ≤ αs1 g (x) + β s2 g (y) .

Therefore, (f + g) (αx + βy) = ≤ = =

f (αx + βy) + g (αx + βy) αs1 f (x) + β s2 f (y) + αs1 g (x) + β s2 g (y) αs1 {f (x) + g (x)} + β s2 {f (y) + g (y)} αs1 (f + g) (x) + β s2 (f + g) (y)

which shows that f + g ∈ MW Os11 ,s2 . Also, |f | (αx + βy) ≤ |αs1 f (x) + β s2 f (y)| ≤ αs1 |f (x)| + β s2 |f (y)| = αs1 |f | (x) + β s2 |f | (y) which shows that |f | ∈ MW Os11 ,s2 . Finally, by previous two statements one can show that max (f, g) ∈ MW Os11 ,s2 by writing max (f, g) =

(f + g) + |f − g| ∈ MW Os11 ,s2 . 2

It’s important to know when the condition αs1 + β s2 = 1 (α + β = 1) in the definition of MW Os11 ,s2 (MW Os21 ,s2 ) can be equivalently replaced by the condition αs1 + β s2 ≤ 1 (α + β ≤ 1, respectively).

On co-ordinated s-convex functions

1985

Theorem 2.14 Let f ∈ MW Os11 ,s2 , then inequality (4) holds for all x, y ∈ R2+ and all α, β with αs1 + β s2 ≤ 1 iff f (0) ≤ 0. Proof. Let f ∈ MW Os11 ,s2 . Firstly, take x = y = 0, and α = β = 0, we obtain f (0) ≤ 0. Now, let x, y ∈ R2+ and α, β ≥ 0 with 0 < τ = αs1 + β s2 < 1. Put a = ατ

− s1

1

and b = βτ 

− s1

2

. Then, as1 + bs2 =

1

1



f (αx + βy) = f aτ s1 x + bτ s2 y 

1





1

1

+

β s2 τ

= 1, and so



1

≤ as1 f τ s1 x + bs2 f τ s2 y 

αs1 τ





1

1



= as1 f τ s1 x + (1 − τ ) s1 0 + bs2 f τ s2 y + (1 − τ ) s2 0

≤ as1 [τ f (x) + (1 − τ ) f (0)] + bs2 [τ f (y) + (1 − τ ) f (0)] = as1 τ f (x) + bs2 τ f (y) + (1 − ζ) f (0) ≤ αs1 f (x) + β s2 f (y) . Therefore, inequality (4) holds. Theorem 2.15 Let f ∈ MW Os21 ,s2 , then inequality (4) holds for all x, y ∈ R2+ and all α, β with α + β ≤ 1 iff f (0) = 0. Proof. Let f ∈ MW Os21 ,s2 . Firstly, take x = y = 0, and α = β = 0, we obtain f (0) ≤ 0 and by using Theorem 2.3. we get f (0) ≥ 0, therefore, f (0) = 0. Now, let x = y ∈ R2+ and α, β ≥ 0 with 0 < ζ = α + β < 1. Put α = aζ and β = bζ. Then, αζ + βζ = a + b = 1, and so f (αx + βy) = ≤ = ≤

f (aζx + bζy) as1 f (ζx) + bs2 f (ζy) as1 f (ζx + (1 − ζ) 0) + bs2 f (ζy + (1 − ζ) 0) as1 [ζ s1 f (x) + (1 − ζ)s1 f (0)] +bs2 [ζ s2 f (y) + (1 − ζ)s2 f (0)] = as1 ζ s1 f (x) + bs2 ζ s2 f (y) + [(1 − ζ)s1 + (1 − ζ)s2 ] f (0) = αs1 f (x) + β s2 f (y) .

Therefore, inequality (4) holds. Corollary 2.16 Let 0 < s ≤ 1. If f ∈ MW Os21 ,s2 and f (0) = 0 then f ∈ MW Os11 ,s2 . In other words, the s–convexity in the second since implies s–convexity in the first sense if f (0) = 0.

1986

M. Alomari and M. Darus

Proof. Assume that f ∈ MW Os21 ,s2 and f (0) = 0. For x, y ∈ R2+ and α, β ≥ 0 with αs1 + β s2 = 1, we have α + β ≤ αs1 + β s2 = 1, and by Theorem 2.4 we obtain that f (αx + βy) ≤ αs1 f (x) + β s2 f (y) , which means that f ∈ MW Os11 ,s2 . Theorem 2.17 Let 0 < si ≤ 1 , (i = 1, 2, 3, 4). Let g ∈ MW Os11 ,s3 and f ∈ MW Os12 ,s4 . 1. If f is a non–decreasing function and g is non–negative function such that f (0) ≤ 0 = g (0). Then, the composition f ◦ g ∈ MW Ot11 ,t2 , where t1 = s1 s2 and t2 = s3 s4 . 2. Assume that 0 < si ≤ 1 , (i = 1, 2, 3, 4). If f and g are non–negative functions such that either f (0) = 0 and g (0+ ) = g (0) or g (0) = 0 and f (0+ ) = f (0). Then, the products f g ∈ MW Ot11 ,t2 , where t1 = min (s1 , s2 ) and t2 = min (s3 , s4 ). Proof. 1. Let x, y ∈ R2+ and α, β ≥ 0 with αt1 + β t2 = 1, where t1 = s1 s2 and t2 = s3 s4 . Since αt1 + β t2 = αs1 s2 + β s3 s4 = 1. Therefore, by assumptions and according to Theorems 2.3–4, we have f ◦ g (αx + βy) = ≤ ≤ =

f (g (αx + βy)) f (αs1 g (x) + β s3 g (y)) αs1 s2 f (g (x)) + β s3 s4 f (g (y)) αt1 f ◦ g (x) + β t2 f ◦ g (y)

which means that f ◦ g ∈ MW Ot11 ,t2 . 2. According to Theorem 2.3, both functions f and g are non–decreasing on (0, ∞)2 . Therefore (f (x) − f (y)) (g (y) − g (x)) ≤ 0 or equivalent, f (x) g (y) + f (y) g (x) ≤ f (x) g (x) + f (y) g (y)

(8)

for all y ≥ x > 0. If y > x = 0, then inequality (8) is still true because f, g are non–negative and either and either f (0) = 0 and

On co-ordinated s-convex functions

1987

g (0+ ) = g (0) or g (0) = 0 and f (0+ ) = f (0). Now let x, y ∈ R+ and α, β ≥ 0 with αt1 +β t2 = 1, where t1 = min (s1 , s2 ) and t2 = min (s3 , s4 ). Without loss of generality, let 0 < s1 < s2 < 1 and 0 < s3 < s4 < 1, with αs1 + β s3 = 1 and αs2 + β s4 = 1. Then, αs1 + β s2 ≤ αt1 + β t2 = 1 and αs3 + β s4 ≤ αt1 + β t2 = 1. Also, since 0 < s1 < s2 < 1 and 0 < s3 < s4 < 1, then we have, αs1 +s2 ≤ α ≤ αs2 ≤ αs1 and β s3 +s4 ≤ β ≤ β s4 ≤ β s3 and by Theorems 2.3–4 and inequality (8) we have f (αx + βy) g (αx + βy) ≤ (αs2 f (x) + β s4 f (y)) (αs1 g (x) + β s3 g (y)) = αs1 +s2 f (x) g (x) + αs2 β s3 f (x) g (y) + αs1 β s4 f (y) g (x) +β s3+s4 f (y) g (y) ≤ αs1 +s2 f (x) g (x) + β s3 +s4 f (y) g (y) +αs2 β s3 f (y) g (y) + αs1 β s4 f (y) g (y)



≤ αs1 f (x) g (x) + β s3 +s4 + αs2 β s3 + αs1 β s4 f (y) g (y) = ≤ ≤ =

αs1 f (x) g (x) + (β s3 (β s4 + αs2 ) + αs1 β s4 ) f (y) g (y) αs1 f (x) g (x) + (β s3 + (β s3 + αs1 )) f (y) g (y) αs1 f (x) g (x) + β s3 f (y) g (y) αt1 f (x) g (x) + β t2 f (y) g (y) .

Similarly, one can show that if 0 < s2 < s1 < 1 and 0 < s4 < s3 < 1, then we have, αs1 +s2 ≤ α ≤ αs1 ≤ αs2 and β s3 +s4 ≤ β ≤ β s3 ≤ β s4 then f (αx + βy) g (αx + βy) ≤ αt1 f (x) g (x) + β t2 f (y) g (y) which means that f g ∈ MW Ot11 ,t2 . Remark 2.18 From the above proof it is also easy to see that, if f is a non–decreasing function in MW Os21 ,s2 and g is a non–negative convex function on R2+ , then the composition f ◦ g of f with g belongs to MW Os21 ,s2 , where 0 < s1 , s2 ≤ 1. Remark 2.19 Convex on R2+ need not be monotonic. However, if f and g are non–negative, convex and either both are non–decreasing or both are non– increasing on R2+ then the product f g is also a convex function. Remark 2.20 In Theorem 2.3, we consider the function h defined on [0, 1] such as 1

1

h (λ) = λ /s1 + (1 − λ) /s2 .

1988

M. Alomari and M. Darus

By Rolle’s theorem we proved that h has a zero in (0, 1), i.e., there exists c ∈ (0, 1) such that h (c) = 0. Therefore, h (c) =

1 1 1 1 −1 −1 (c) s1 − (1 − c) s2 = 0. s1 s2

Indeed, we have unsolved problem and it is being very hard to get such

c ∈ (0, 1). 1 1 As in Theorem 3.3 we proved that such c belongs to 2 + , 2 −  whenever  > 0. By determining the value of c, the intervals of where h is increasing or decreasing will be known and the proof of Theorem 3.3 will completely finish, so we state this problem as follows: Unsolved Problem 2.21 Solve the following equation for x, 1 1 1 1 −1 −1 (x) s1 − (1 − x) s2 = 0 , s1 s2

where, x ∈ (0, 1) and 0 < s1 , s2 ≤ 1 with s1 = s2 .

ACKNOWLEDGEMENTS. The work here is supported by the Grant: UKM–GUP–TMK–07–02–107.

References [1] M. Alomari and M. Darus, The Hadamard’s inequality for s–convex function of 2–variables On The co–ordinates, Int. Journal of Math. Analysis, 2 (13) (2008), 629–638. [2] M. Alomari and M. Darus, The Hadamard’s inequality for s–convex function, Int. Journal of Math. Analysis, 2 (13) (2008), 639–646. [3] M. Alomari and M. Darus, Co–ordinates s–convex function in the first sense with some Hadamard–type inequalities, Int. J. Contemp. Math. Sci., submitted. [4] H. Hudzik, L. Maligranda, Some remarks on s–convex functions, Aequationes Math., 48 (1994) 100–111. [5] W. Matuszewska and W. Orlicz, A note on the theorey of s–normed spaces of ϕ–integrable functions, studia Math., 21 (1981), 107–115. [6] J. Musielak, Orlicz spaces and modular spaces, Lecture Notes in Mathematics, Vol. 1034, Springer–Verlag, New York / Berlin, 1983.

On co-ordinated s-convex functions

1989

[7] W. Orlicz, A note on modular spaces, I, Bull. Acad. Polon. Sci. Math. Astronom. Phys., 9 (1961), 157–162. [8] S. Rolewicz, Metric Linear Spaces, 2nd ed., PW N, Warsaw, 1984. Received: May 22, 2008