On some Hadamard-type inequalities for product of two s-convex functions on the co-ordinates

Abstract

In this article, Hadamard-type inequalities for product of s-convex in the second sense on the co-ordinates in a rectangle from the plane are established.

1. Introduction

A function f : I → ℝ, I ⊆ ℝ is an interval, is said to be a convex function on I if

$$f\left(tx+\left(1-t\right)y\right)\le tf\left(x\right)+\left(1-t\right)f\left(y\right)$$

(1.1)

holds for all x, y ∈ I and t ∈ [0, 1]. If the reversed inequality in (1.1) holds, then f is concave. Let f : I ⊆ ℝ → ℝ be a convex function and a, b ∈ I with a < b. Then the following double inequality:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{2}$$

(1.2)

is known as Hermite-Hadamard inequality for convex mappings. For particular choice of the function f in (1.2) yields some classical inequalities of means. Both inequalities in (1.2) hold in reversed direction if f is concave.

Some basic definitions can be given as followings:

Definition 1. (See [1], [2, p. 410]) We say that f : I → ℝ is a Godunova-Levin function or that f belongs to the class Q(I) if f is non-negative and for all x, y ∈ I and t ∈ (0, 1) we have

$$f\left(tx+\left(1-t\right)y\right)\le \frac{f\left(x\right)}{t}+\frac{f\left(y\right)}{1-t}.$$

Definition 2. (See [3]) We say that f : I ⊆ ℝ → ℝ is a P-function or that f belongs to the class P (I) if f is non-negative and for all x, y ∈ I and t ∈ [0, 1], we have

$$f\left(tx+\left(1-t\right)y\right)\le f\left(x\right)+f\left(y\right).$$

Definition 3. (See [4]) Let s ∈ (0, 1]. A function f : [0, ∞) → [0, ∞) is said to be s-convex in the second sense if

$$f\left(tx+\left(1-t\right)y\right)\le t^{s}f\left(x\right)+{\left(1-t\right)}^{s}f\left(y\right)$$

for all x, y ∈ (0, b) and t ∈ [0, 1].

In [5], Hudzik and Maligranda considered among others the class of functions which are s-convex in the first sense. This class is defined in the following way:

Definition 4. A function f : ℝ⁺ → ℝ, where ℝ⁺ = [0, ∞), is said to be s-convex in the first sense if

$$f\left(\alpha x+\beta y\right)\le {\alpha }^{s}f\left(x\right)+{\beta }^{s}f\left(y\right)$$

for all x, y ∈ [0, ∞), α, β ≥ 0 with α^s + β^s = 1 and for some fixed s ∈ (0, 1]. We denote by $K_s^1$ the class of all s-convex functions.

In 1978, Breckner introduced s-convex functions as a generalization of convex functions in [4]. Also, he proved the important fact that the set valued map is s-convex only if the associated support function is s-convex function [6]. Of course, s-convexity means just convexity when s = 1. The definition of s-convexity of real valued functions are very important for Orlicz spaces and Banach normed spaces (see [7–9]). A number of properties of s-convex functions are discussed in articles [5, 10–13].

In article [14] the following generalization of the previously described functions was given.

Definition 5. (See [14, 15]) Let I, J be intervals ℝ, (0, 1) ⊆ J and let h : J → ℝ. A function f : I → ℝ is called an h-convex function, or that f belongs to the class SX(h, I), if for all x, y ∈ I and t ∈ (0, 1) we have

$$f\left(tx+\left(1-t\right)y\right)\le h\left(t\right)f\left(x\right)+h\left(1-t\right)f\left(y\right).$$

(1.3)

If inequality in (1.3) is reversed, then f is said to be h-concave.

Obviously, if h(t) = t, for all t ∈ [0, 1] ⊆ J, then all convex functions belong to SX (h, I) and all concave functions are h-concave; if $h\left(t\right)=\frac{1}{t}$, for all t ∈ (0, 1), then SX(h, I) = Q(I); if h(t) = 1, SX(h, I) ⊇ P(I); and if h(t) = t^s, where s ∈ (0, 1), then $SX\left(h,I\right)\supseteq K_s^2$. For some recent results about h-convex functions we refer the reader to articles [15–18].

In [10], Dragomir and Fitzpatrick proved the following variant of Hermite-Hadamard inequality which hold for s-convex functions in the second sense:

Theorem 1. Suppose that f : [0, ∞) → [0, ∞) is an s-convex function in the second sense, where s ∈ (0, 1) and let a, b ∈ [0, ∞), a < b. If f ∈ L₁([a, b]), then the following inequalities hold:

$$2^{s-1}f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{s+1}$$

(1.4)

The constant $k=\frac{1}{s+1}$ is the best possible in the second inequality in (1.4).

Again in [10], Dragomir and Fitzpatrick also proved the following Hadamard-type inequality for s-convex functions in the first sense:

Theorem 2. Suppose that f : [0, ∞) → [0, ∞) is an s-convex function in the first sense, where s ∈ (0, 1) and let a, b ∈ [0, ∞). If f ∈ L₁([a, b]) then the following inequalities hold:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le \frac{f\left(a\right)+sf\left(b\right)}{s+1}$$

(1.5)

The above inequalities are sharp.

A modification for convex functions which is also known as co-ordinated convex functions was introduced as following by Dragomir [19]:

Let us consider a bidimensional interval Δ =: [a, b] × [c, d] in ℝ² with a < b and c < d. A mapping f : Δ → ℝ is said to be convex on Δ if the following inequality:

$$f\left(\alpha x+\left(1-\alpha \right)z,\alpha y+\left(1-\alpha \right)w\right)\le \alpha f\left(x,y\right)+\left(1-\alpha \right)f\left(z,w\right)$$

holds, for all (x, y), (z, w) ∈ Δ and α ∈ [0, 1].

A function f : Δ → ℝ is said to be convex on the co-ordinates on Δ if the partial mappings f _y : [a, b] → ℝ, f _y (u) = f(u, y) and f _x : [c, d] → ℝ, f _x (v) = f(x, v) are convex where defined for all x ∈ [a, b], y ∈ [c, d].

In the same article, Dragomir established the following Hadamard-type inequalities for convex functions on the co-ordinates in a rectangle from the plane ℝ²:

Theorem 3. Suppose f : Δ = [a, b] × [c, d] ⊆ [0, ∞) → ℝ is convex function on the co-ordinates on Δ. Then one has the inequalities:

$$\begin{array}{ll}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dydx\\ \le \frac{f\left(a,c\right)+f\left(b,c\right)+f\left(a,d\right)+f\left(b,d\right)}{4}\end{array}$$

(1.6)

The concept of s-convex functions on the co-ordinates in both sense was introduced by Alomari and Darus [20, 21]:

Definition 6. 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

$$f\left(\alpha x+\beta z,\alpha y+\beta w\right)\le {\alpha }^{s}f\left(x,y\right)+{\beta }^{s}f\left(z,w\right)$$

holds for all (x, y), (z, w) ∈ Δ, α, β ≥ 0 with α^s + β^s = 1 (α + β = 1) and for some fixed s ∈ (0, 1]. We write $f\in K_s^i\left(i=1,2\right)$ 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 =: [a, b] × [c, d] ⊆ [0, ∞)² → ℝ is called s-convex in first sense (in the second sense) on the co-ordinates on Δ if the partial mappings f _y : [a, b] → ℝ, f _y (u) = f(u, y) and f _x : [c, d] → ℝ, f _x (v) = f (x, v), are s-convex in the first sense (in the second sense) for all y ∈ [c, d], x ∈ [a, b], and s ∈ (0, 1], i.e, the partial mappings f _y and f _x are s-convex in the first sense (second sense) with some fixed s ∈ (0, 1].

In [20], Alomari and Darus proved the following inequalities for s-convex functions (in the second sense) on the co-ordinates in a rectangle from the plane ℝ²:

Theorem 4. Suppose f : Δ = [a, b] × [c, d] ⊆ [0, ∞) → ℝ is s-convex function (in the second sense) on the co-ordinates on Δ. Then one has the inequalities:

$$\begin{array}{ll}\hfill 4^{s-1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dydx\\ \le \frac{f\left(a,c\right)+f\left(b,c\right)+f\left(a,d\right)+f\left(b,d\right)}{{\left(s+1\right)}^2}\end{array}$$

(1.7)

Also in [21] (see also [22]), Alomari and Darus established the following inequalities for s-convex functions (in the first sense) on the co-ordinates in a rectangle from the plane ℝ²:

Theorem 5. Suppose f : Δ = [a, b] × [c, d] ⊆ [0, ∞) → ℝ is s-convex function (in the first sense) on the co-ordinates on Δ. Then one has the inequalities:

$$\begin{array}{ll}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dydx\\ \le \frac{f\left(a,c\right)+sf\left(b,c\right)+sf\left(a,d\right)+s^{2}f\left(b,d\right)}{{\left(s+1\right)}^2}\end{array}$$

(1.8)

The above inequalities are sharp.

In [23], Sarikaya et al. proved some Hadamard-type inequalities for co-ordinated convex functions as followings:

Theorem 6. Let f : Δ ⊂ ℝ² → ℝ be a partial differentiable mapping on Δ := [a, b] × [c, d] in ℝ² with a < b and c < d. If $\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$ is a convex function on the co-ordinates on Δ, then one has the inequalities:

$$\begin{array}{c}\left|\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}\right.\\ \left.\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{16}\left(\frac{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(a,c\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(a,d\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(b,c\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(b,d\right)}{4}\right)\end{array}$$

where

$$A=\frac{1}{2}\left[\frac{1}{\left(b-a\right)}{\int }_a^b\left[f\left(x,c\right)+f\left(x,d\right)\right]dx+\frac{1}{\left(d-c\right)}{\int }_c^d\left[f\left(a,y\right)dy+f\left(b,y\right)\right]dy\right].$$

Theorem 7. Let f : Δ ⊂ ℝ² → ℝ be a partial differentiable mapping on Δ := [a, b] × [c, d] in ℝ² with a < b and c < d. If ${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q$, q > 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:

$$\begin{array}{c}\left|\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}\right.\\ \left.\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^d\left(x,y\right)dxdy-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}{\left(\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(b,d\right)}{4}\right)}^{\frac{1}{q}}\end{array}$$

where

$$A=\frac{1}{2}\left[\frac{1}{\left(b-a\right)}{\int }_a^b\left[f\left(x,c\right)+f\left(x,d\right)\right]dx+\frac{1}{\left(d-c\right)}{\int }_c^d\left[f\left(a,y\right)dy+f\left(b,y\right)\right]dy\right]$$

and $\frac{1}{p}+\frac{1}{q}=1.$

Theorem 8. Let f : Δ ⊂ ℝ² → ℝ be a partial differentiable mapping on Δ := [a, b] × [c, d] in ℝ² with a < b and c < d. If ${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q$, q ≥ 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:

$$\begin{array}{c}\left|\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}\right.\\ \left.\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{16}{\left(\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^q\left(b,d\right)}{4}\right)}^{\frac{1}{q}}\end{array}$$

where

$$A=\frac{1}{2}\left[\frac{1}{\left(b-a\right)}{\int }_a^b\left[f\left(x,c\right)+f\left(x,d\right)\right]dx+\frac{1}{\left(d-c\right)}{\int }_c^d\left[f\left(a,y\right)dy+f\left(b,y\right)\right]dy\right].$$

For refinements, counterparts, generalizations and new Hadamard-type inequalities see the articles [3, 5, 10–13, 19–28]. In [25] (see also [22]), Alomari and Darus introduced new classes of s-convex functions on the co-ordinates as following:

Definition 7. 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 on Δ if there exist s₁, s₂ ∈ (0, 1] such that $s=\frac{s_1+s_2}{2}$,

$$f\left(\alpha x+\beta z,\alpha y+\beta w\right)\le {\alpha }^{s_1}f\left(x,y\right)+{\beta }^{s_2}f\left(z,w\right)$$

holds for all (x, y), (z, w) ∈ Δ with α, β ≥ 0 with ${\alpha }^{s_1}+{\beta }^{s_2}=1$ and for some fixed s₁, s₂ ∈ (0, 1]. We denote this class of functions by $MW{O}_{s_1,s_2}^1$.

Definition 8. 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 second sense on Δ if there exist s₁, s₂ ∈ (0, 1] such that $s=\frac{s_1+s_2}{2}$,

$$f\left(\alpha x+\beta z,\alpha y+\beta w\right)\le {\alpha }^{s_1}f\left(x,y\right)+{\beta }^{s_2}f\left(z,w\right)$$

holds for all (x, y), (z, w) ∈ Δ with α, β ≥ 0 with α + β = 1 and for all fixed s₁, s₂ ∈ (0, 1]. We denote this class of functions by $MW{O}_{s_1,s_2}^2$.

In [26], Pachpatte established some inequalities for product of convex functions as follows:

Theorem 9. Let f, g : [a, b] ⊆ ℝ → [0, ∞) be convex functions on [a, b], a < b.

Then

$$\frac{1}{b-a}{\int }_a^{b}f\left(x\right)g\left(x\right)dx\le \frac{1}{3}M\left(a,b\right)+\frac{1}{6}N\left(a,b\right)$$

(1.9)

and

$$2f\left(\frac{a+b}{2}\right)g\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)g\left(x\right)dx+\frac{1}{6}M\left(a,b\right)+\frac{1}{3}N\left(a,b\right)$$

(1.10)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

Similar results for s-convex functions is due to Kirmaci et al. [13] as follows:

Theorem 10. Let f, g : [a, b] ⊆ ℝ → ℝ a, b ∈ [0, ∞), a < b, be functions such that g and fg are in L₁([a, b]). If f is convex and non-negative on [a, b] and if g is s-convex on [a, b], for some s ∈ (0, 1), then

$$\begin{array}{c}{2}^{s}f\left(\frac{a+b}{2}\right)g\left(\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f\left(x\right)g\left(x\right)dx\\ \le \frac{1}{\left(s+1\right)\left(s+2\right)}M\left(a,b\right)+\frac{1}{s+2}N\left(a,b\right)\end{array}$$

(1.11)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

Theorem 11. Let f, g : [a, b] ⊆ ℝ → ℝ a, b ∈ [0, ∞), a < b, be functions such that g and fg are in L₁([a, b]). If f is convex and non-negative on [a, b] and if g is s-convex on [a, b] for some s ∈ (0, 1), then

$$\frac{1}{b-a}{\int }_a^{b}f\left(x\right)g\left(x\right)dx\le \frac{1}{s+2}M\left(a,b\right)+\frac{1}{\left(s+1\right)\left(s+2\right)}N\left(a,b\right)$$

(1.12)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

Theorem 12. Let f, g : [a, b] ⊆ ℝ → ℝ a, b ∈ [0, ∞), a < b, be functions such that f, g and fg are in L₁([a, b]). If f is s₁-convex and g is s₂-convex on [a, b] for some fixed s₁, s₂ ∈ (0, 1), then

$$\begin{array}{ll}\hfill \frac{1}{b-a}{\int }_a^{b}f\left(x\right)g\left(x\right)dx& \le \frac{1}{s_1+s_2+1}M\left(a,b\right)+B\left(s_1+1,s_2+1\right)N\left(a,b\right)\\ =\frac{1}{s_1+s_2+1}\left[M\left(a,b\right)+s_1{s}_2\frac{\Gamma \left(s_1\right)\Gamma \left(s_2\right)}{\Gamma \left(s_1+s_2+1\right)}N\left(a,b\right)\right]\end{array}$$

(1.13)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

In the last theorem Beta function of Euler type, defined by

$$B\left(x,y\right)={\int }_0^t{t}^{x-1}{\left(1-t\right)}^{y-1}dt=\frac{\mathrm{\Gamma }\left(x\right)\mathrm{\Gamma }\left(y\right)}{\mathrm{\Gamma }\left(x+y\right)}$$

has been used.

The main purpose of the present article is to establish new Hadamard-type inequalities similar to the above inequalities, but now for product of s-convex functions (in the second sense) on the co-ordinates in a rectangle from the plane ℝ².

2. Main results

We will start with the following theorem.

Theorem 13. Let f, g : Δ = [a, b] × [c, d] ⊆ [0, ∞)² → ℝ, a < b, c < d, be functions such that f, g and fg are in L²(Δ). If f is non-negative and convex on the coordinates on Δ and if g is s-convex in the second sense on the co-ordinates on Δ, for all s₁, s₂ ∈ (0, 1), such that $s=\frac{s_1+s_2}{2}$, then one has the inequality;

$$\begin{array}{c}\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ \le \frac{1}{2}\left(p^2+r^2\right)L\left(a,b,c,d\right)+\frac{1}{2}\left(pq+rt\right)M\left(a,b,c,d\right)+\frac{1}{2}\left(q^2+t^2\right)N\left(a,b,c,d\right)\end{array}$$

(2.1)

where

$$\begin{array}{ll}\hfill L\left(a,b,c,d\right)& =f\left(a,c\right)g\left(a,c\right)+f\left(b,c\right)g\left(b,c\right)+f\left(a,d\right)g\left(a,d\right)+f\left(b,d\right)g\left(b,d\right),\\ \hfill M\left(a,b,c,d\right)& =f\left(a,c\right)g\left(a,d\right)+f\left(a,d\right)g\left(a,c\right)+f\left(b,c\right)g\left(b,d\right)+f\left(b,d\right)g\left(b,c\right)\\ +f\left(b,c\right)g\left(a,c\right)+f\left(b,d\right)g\left(a,d\right)+f\left(a,c\right)g\left(b,c\right)+f\left(a,d\right)g\left(b,d\right),\\ \hfill N\left(a,b,c,d\right)& =f\left(b,c\right)g\left(a,d\right)+f\left(b,d\right)g\left(a,c\right)+f\left(a,c\right)g\left(b,d\right)+f\left(a,d\right)g\left(b,c\right)\end{array}$$

and

$$p=\frac{1}{s_2+2},q=\frac{1}{\left(s_2+1\right)\left(s_2+2\right)},r=\frac{1}{s_1+2},t=\frac{1}{\left(s_1+1\right)\left(s_1+2\right)}.$$

Proof. Since f is convex and g is s-convex in the second sense on the co-ordinates on Δ. Therefore the partial mappings

$$f_y:\left[a,b\right]\to \left[0,\mathrm{\infty }\right),f_y\left(x\right)=f\left(x,y\right)$$

and

$$f_x:\left[c,d\right]\to \left[0,\mathrm{\infty }\right),f_x\left(y\right)=f\left(x,y\right)$$

are convex and non-negative on [a, b] and [c, d], respectively. The partial mappings

$$g_y:\left[a,b\right]\to \left[0,\mathrm{\infty }\right),g_y\left(x\right)=g\left(x,y\right)$$

and

$$g_x:\left[c,d\right]\to \left[0,\mathrm{\infty }\right),g_x\left(y\right)=g\left(x,y\right)$$

are s₁-, s₂-convex on [a, b] and [c, d], respectively, for all x ∈ [a, b], y ∈ [c, d], for all s₁, s₂ ∈ (0, 1], such that $s=\frac{s_1+s_2}{2}$. Now by applying f _x (y)g _x (y) to (1.12) on [c, d], we get

$$\begin{array}{ll}\hfill \frac{1}{d-c}{\int }_c^d{f}_x\left(y\right)g_x\left(y\right)dy& \le p\left[f_x\left(c\right)g_x\left(c\right)+f_x\left(d\right)g_x\left(d\right)\right]\\ +q\left[f_x\left(c\right)g_x\left(d\right)+f_x\left(d\right)g_x\left(c\right)\right].\end{array}$$

That is

$$\begin{array}{ll}\hfill \frac{1}{d-c}{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dy& \le p\left[f\left(x,c\right)g\left(x,c\right)+f\left(x,d\right)g\left(x,d\right)\right]\\ +q\left[f\left(x,c\right)g\left(x,d\right)+f\left(x,d\right)g\left(x,c\right)\right].\end{array}$$

Integrating over [a, b] with respect to x and dividing both sides by b - a, we have

$$\begin{array}{c}\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ \le p\left[\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx+\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx\right]\\ +q\left[\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx+\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx\right].\end{array}$$

(2.2)

Now by applying (1.12) to each integral on right-hand side of (2.2) again, we get

$$\begin{array}{ll}\hfill \frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx& \le p\left[f\left(a,c\right)g\left(a,c\right)+f\left(b,c\right)g\left(b,c\right)\right]\\ +q\left[f\left(a,c\right)g\left(b,c\right)+f\left(b,c\right)g\left(a,c\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill \frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx& \le p\left[f\left(a,d\right)g\left(a,d\right)+f\left(b,d\right)g\left(b,d\right)\right]\\ +q\left[f\left(a,d\right)g\left(b,d\right)+f\left(b,d\right)g\left(a,d\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill \frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx& \le p\left[f\left(a,c\right)g\left(a,d\right)+f\left(b,c\right)g\left(b,d\right)\right]\\ +q\left[f\left(a,c\right)g\left(b,d\right)+f\left(b,c\right)g\left(a,d\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill \frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx& \le p\left[f\left(a,d\right)g\left(a,c\right)+f\left(b,d\right)g\left(b,c\right)\right]\\ +q\left[f\left(a,d\right)g\left(b,c\right)+f\left(b,d\right)g\left(a,c\right)\right].\end{array}$$

On substitution of these inequalities in (2.2), we obtain

$$\begin{array}{c}\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ \le p^{2}L\left(a,b,c,d\right)+pqM\left(a,b,c,d\right)+q^{2}N\left(a,b,c,d\right).\end{array}$$

(2.3)

Similarly, if we apply f _y (x)g _y (x) to (1.12) on [a, b], we get the following result:

$$\begin{array}{c}\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ \le r^{2}L\left(a,b,c,d\right)+rtM\left(a,b,c,d\right)+t^{2}N\left(a,b,c,d\right)\end{array}$$

(2.4)

Adding the inequalities (2.3), (2.4) and dividing by 2, we get (2.1).   □

Theorem 14. Let f, g : Δ = [a, b] × [c, d] ⊆ ℝ² → ℝ, a < b, c < d, be functions such that f, g, and fg are in L²(Δ). If f is non-negative and convex on the co-ordinates on Δ and if g is s-convex on the co-ordinates on Δ, for all s₁, s₂ ∈ (0, 1), such that $s=\frac{s_1+s_2}{2}$, then one has the inequality;

$$\begin{array}{ll}\hfill 2^{s_1+s_2}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ +\frac{1}{2}\left(q^2+t^2+2pt+2qr\right)L\left(a,b,c,d\right)\\ +\frac{1}{2}\left(pq+rt+2qt+2rp\right)M\left(a,b,c,d\right)\\ +\frac{1}{2}\left(p^2+r^2+2pt+2rq\right)N\left(a,b,c,d\right)\end{array}$$

(2.5)

where L(a, b, c, d), M(a, b, c, d), N(a, b, c, d), p, q, r and t as in Theorem 13.

Proof. Applying $2^{s_1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)$ to (1.11) and multiplying both sides by $2^{s_2}$, we get

$$\begin{array}{c}{2}^{s_1+s_2}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{2^{s_2}}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)g\left(x,\frac{c+d}{2}\right)dx\\ +t\left[2^{s_2}f\left(a,\frac{c+d}{2}\right)g\left(a,\frac{c+d}{2}\right)+2^{s_2}f\left(b,\frac{c+d}{2}\right)g\left(b,\frac{c+d}{2}\right)\right]\\ +r\left[2^{s_2}f\left(a,\frac{c+d}{2}\right)g\left(b,\frac{c+d}{2}\right)+2^{s_2}f\left(b,\frac{c+d}{2}\right)g\left(a,\frac{c+d}{2}\right)\right].\end{array}$$

(2.6)

Similarly, by applying $2^{s_2}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)$ to (1.11) and multiplying both sides by $2^{s_1}$, we get

$$\begin{array}{c}{2}^{s_1+s_2}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{2^{s_1}}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)g\left(\frac{a+b}{2},y\right)dy\\ +q\left[2^{s_1}f\left(\frac{a+b}{2},c\right)g\left(\frac{a+b}{2},c\right)+2^{s_1}f\left(\frac{a+b}{2},d\right)g\left(\frac{a+b}{2},d\right)\right]\\ +p\left[2^{s_1}f\left(\frac{a+b}{2},c\right)g\left(\frac{a+b}{2},d\right)+2^{s_1}f\left(\frac{a+b}{2},d\right)g\left(\frac{a+b}{2},c\right)\right].\end{array}$$

(2.7)

Adding (2.6) and (2.7), we have

$$\begin{array}{c}{2}^{s_1+s_2+1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{2^{s_2}}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)g\left(x,\frac{c+d}{2}\right)dx+\frac{2^{s_1}}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)g\left(\frac{a+b}{2},y\right)dy\\ +t\left[2^{s_2}f\left(a,\frac{c+d}{2}\right)g\left(a,\frac{c+d}{2}\right)+2^{s_2}f\left(b,\frac{c+d}{2}\right)g\left(b,\frac{c+d}{2}\right)\right]\\ +r\left[2^{s_2}f\left(a,\frac{c+d}{2}\right)g\left(b,\frac{c+d}{2}\right)+2^{s_2}f\left(b,\frac{c+d}{2}\right)g\left(a,\frac{c+d}{2}\right)\right]\\ +q\left[2^{s_1}f\left(\frac{a+b}{2},c\right)g\left(\frac{a+b}{2},c\right)+2^{s_1}f\left(\frac{a+b}{2},d\right)g\left(\frac{a+b}{2},d\right)\right]\\ +p\left[2^{s_1}f\left(\frac{a+b}{2},c\right)g\left(\frac{a+b}{2},d\right)+2^{s_1}f\left(\frac{a+b}{2},d\right)g\left(\frac{a+b}{2},c\right)\right].\end{array}$$

(2.8)

Applying (1.11) to each term within the brackets, we get

$$\begin{array}{ll}\hfill 2^{s_2}f\left(a,\frac{c+d}{2}\right)g\left(a,\frac{c+d}{2}\right)& \le \frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(a,y\right)dy\\ +t\left[f\left(a,c\right)g\left(a,c\right)+f\left(a,d\right)g\left(a,d\right)\right]\\ +r\left[f\left(a,c\right)g\left(a,d\right)+f\left(a,d\right)g\left(a,c\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_2}f\left(b,\frac{c+d}{2}\right)g\left(b,\frac{c+d}{2}\right)& \le \frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(b,y\right)dy\\ +t\left[f\left(b,c\right)g\left(b,c\right)+f\left(b,d\right)g\left(b,d\right)\right]\\ +r\left[f\left(b,c\right)g\left(b,d\right)+f\left(b,d\right)g\left(b,c\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_2}f\left(a,\frac{c+d}{2}\right)g\left(b,\frac{c+d}{2}\right)& \le \frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(b,y\right)dy\\ +t\left[f\left(a,c\right)g\left(b,c\right)+f\left(a,d\right)g\left(b,d\right)\right]\\ +r\left[f\left(a,c\right)g\left(b,d\right)+f\left(a,d\right)g\left(b,c\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_2}f\left(b,\frac{c+d}{2}\right)g\left(a,\frac{c+d}{2}\right)& \le \frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(a,y\right)dy\\ +t\left[f\left(b,c\right)g\left(a,c\right)+f\left(b,d\right)g\left(a,d\right)\right]\\ +r\left[f\left(b,c\right)g\left(a,d\right)+f\left(b,d\right)g\left(a,c\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_1}f\left(\frac{a+b}{2},c\right)g\left(\frac{a+b}{2},c\right)& \le \frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx\\ +q\left[f\left(a,c\right)g\left(a,c\right)+f\left(b,c\right)g\left(b,c\right)\right]\\ +p\left[f\left(a,c\right)g\left(b,c\right)+f\left(b,c\right)g\left(a,c\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_1}f\left(\frac{a+b}{2},d\right)g\left(\frac{a+b}{2},d\right)& \le \frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx\\ +q\left[f\left(a,d\right)g\left(a,d\right)+f\left(b,d\right)g\left(b,d\right)\right]\\ +p\left[f\left(a,d\right)g\left(b,d\right)+f\left(b,d\right)g\left(a,d\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_1}f\left(\frac{a+b}{2},c\right)g\left(\frac{a+b}{2},d\right)& \le \frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx\\ +q\left[f\left(a,c\right)g\left(a,d\right)+f\left(b,c\right)g\left(b,d\right)\right]\\ +p\left[f\left(a,c\right)g\left(b,d\right)+f\left(b,c\right)g\left(a,d\right)\right].\end{array}$$

$$\begin{array}{ll}\hfill 2^{s_1}f\left(\frac{a+b}{2},d\right)g\left(\frac{a+b}{2},c\right)& \le \frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx\\ +q\left[f\left(a,d\right)g\left(a,c\right)+f\left(b,d\right)g\left(b,c\right)\right]\\ +p\left[f\left(a,d\right)g\left(b,c\right)+f\left(b,d\right)g\left(a,c\right)\right].\end{array}$$

Substituting these inequalities in (2.8) and simplifying, we obtain

$$\begin{array}{c}{2}^{s_1+s_2+1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{2^{s_2}}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)g\left(x,\frac{c+d}{2}\right)dx\\ +\frac{2^{s_1}}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)g\left(\frac{a+b}{2},y\right)dy\\ +t\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(a,y\right)dy+t\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(b,y\right)dy\\ +r\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(b,y\right)dy+r\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(a,y\right)dy\\ +q\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx+q\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx\\ +p\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx+p\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx\\ +\left(q^2+t^2\right)L\left(a,b,c,d\right)+\left(pq+rt\right)M\left(a,b,c,d\right)+\left(p^2+r^2\right)N\left(a,b,c,d\right).\end{array}$$

(2.9)

Now by applying $2^{s_1}f\left(\frac{a+b}{2},y\right)g\left(\frac{a+b}{2},y\right)$ to (1.11), integrating over [c, d] and dividing both sides by d - c, we get

$$\begin{array}{c}\frac{2^{s_1}}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)g\left(\frac{a+b}{2},y\right)dy\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dxdy\\ +t\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(a,y\right)dy+t\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(b,y\right)dy\\ +r\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(b,y\right)dy+r\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(a,y\right)dy.\end{array}$$

(2.10)

Again by applying $2^{s_2}f\left(x,\frac{c+d}{2}\right)g\left(x,\frac{c+d}{2}\right)$ to (1.11), integrating over [a, b] and dividing both sides by b - a, we get

$$\begin{array}{c}\frac{2^{s_2}}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)g\left(x,\frac{c+d}{2}\right)dx\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ +q\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx+q\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx\\ +p\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx+p\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx.\end{array}$$

(2.11)

Adding (2.10) and (2.11), we have

$$\begin{array}{c}\frac{2^{s_2}}{b-a}{\int }_c^{d}f\left(x,\frac{c+d}{2}\right)g\left(x,\frac{c+d}{2}\right)dx+\frac{2^{s_1}}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)g\left(\frac{a+b}{2},y\right)dy\\ \le \frac{2}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ +t\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(a,y\right)dy+t\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(b,y\right)dy\\ +r\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(b,y\right)dy+r\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(a,y\right)dy\\ +q\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx+q\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx\\ +p\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx+p\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx.\end{array}$$

(2.12)

Therefore from (2.9) and (2.12), we obtain

$$\begin{array}{c}{2}^{s_1+s_2+1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{2}{\left(b-a\right)\left(d-c\right)}{\int }_a^{b}f\left(x,y\right)g\left(x,y\right)dxdy\\ +2t\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(a,y\right)dy+2t\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(b,y\right)dy\\ +2r\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)g\left(b,y\right)dy+2r\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)g\left(a,y\right)dy\\ +2q\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,c\right)dx+2q\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,d\right)dx\\ +2p\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)g\left(x,d\right)dx+2p\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)g\left(x,c\right)dx\\ +\left(q^2+t^2\right)L\left(a,b,c,d\right)+\left(pq+rt\right)M\left(a,b,c,d\right)+\left(p^2+r^2\right)N\left(a,b,c,d\right).\end{array}$$

(2.13)

By applying (1.12) to each of the integral in (2.13) and simplifying, we get

$$\begin{array}{c}{2}^{s_1+s_2+1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{2}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ +\left(q^2+t^2+2pt+2qr\right)L\left(a,b,c,d\right)+\left(pq+rt+2qt+2rp\right)M\left(a,b,c,d\right)\\ +\left(p^2+r^2+2pt+2rq\right)N\left(a,b,c,d\right)\end{array}$$

Dividing both sides by 2, we get required result.   □

Theorem 15. Let f, g : Δ = [a, b] × [c, d] ⊆ ℝ² → ℝ, a < b, c < d, be functions such that f, g, and fg are in L₂(Δ). If f is s₁-convex on the co-ordinates on Δ and g is s₂-convex on the co-ordinates on Δ, for some fixed s₁₁, s₁₂, s₂₁, s₂₂ ∈ (0, 1), such that $s_1=\frac{s_{11}+s_{12}}{2},s_2=\frac{s_{21}+s_{22}}{2}$, then one has the following inequality;

$$\begin{array}{c}\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ \le \frac{1}{2}\left(p_1^2+r_1^2\right)L\left(a,b,c,d\right)+\frac{1}{2}\left(p_1{q}_1+r_1{t}_1\right)M\left(a,b,c,d\right)+\frac{1}{2}\left(q_1^2+t_1^2\right)N\left(a,b,c,d\right)\end{array}$$

(2.14)

where L(a, b, c, d), M(a, b, c, d), and N(a, b, c, d) as defined in Theorem 13 and $p_1=\frac{1}{s_{12}+s_{22}+1}$, q₁ = B(s₁₂ + 1, s₂₂ + 1),$r_1=\frac{1}{s_{11}+s_{21}+1}$, t₁ = B(s₁₁ + 1, s₂₁ + 1).

Proof. By a similar way to Theorem 13 with $p_1=\frac{1}{s_{12}+s_{22}+1}$, q₁ = B(s₁₂+1, s₂₂+1), $r_1=\frac{1}{s_{11}+s_{21}+1}$, t₁ = B(s₁₁ + 1, s₂₁ + 1) and thus (2.14) is established.   □

Theorem 16. Let f, g : Δ = [a, b] × [c, d] ⊆ ℝ² → ℝ, a < b, c < d, be functions such that f, g, and fg are in L₂(Δ). If f is s₁-convex on the co-ordinates on Δ and g is s₂-convex on the co-ordinates on Δ, for some fixed s₁₁, s₁₂, s₂₁, s₂₂ ∈ (0, 1), such that $s_1=\frac{s_{11}+s_{12}}{2},s_2=\frac{s_{21}+s_{22}}{2}$, then one has the following inequality;

$$\begin{array}{c}{2}^{s_{11}+s_{21}+s_{12}+s_{22}-2}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)g\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)g\left(x,y\right)dydx\\ +\frac{1}{2}\left(q_1^2+t_1^2+2p_1{t}_1+2q_1{r}_1\right)L\left(a,b,c,d\right)\\ +\frac{1}{2}\left(p_1{q}_1+r_1{t}_1+2q_1{t}_1+2r_1{p}_1\right)M\left(a,b,c,d\right)\\ +\frac{1}{2}\left(p_1^2+r_1^2+2p_1{t}_1+2r_1{q}_1\right)N\left(a,b,c,d\right)\end{array}$$

(2.15)

where L(a, b, c, d), M(a, b, c, d), and N(a, b, c, d) as defined in Theorem 13 and $p_1=\frac{1}{s_{12}+s_{22}+1}$, q₁ = B(s₁₂ + 1, s₂₂ + 1),$r_1=\frac{1}{s_{11}+s_{21}+1}$, t₁ = B(s₁₁ + 1, s₂₁ + 1).

Proof. By a similar way to Theorem 13 with $p_1=\frac{1}{s_{12}+s_{22}+1}$, q₁ = B(s₁₂+1, s₂₂+1), $r_1=\frac{1}{s_{11}+s_{21}+1}$, t₁ = B(s₁₁ + 1, s₂₁ + 1) the proof is completed   □