# Schur-harmonic convexity related to co-ordinated harmonically convex functions in plane

## Abstract

In this paper, we investigate Schur-harmonic convexity of some functions which are obtained from the co-ordinated harmonically convex functions on a square in a plane.

## 1 Introduction

Schur-convexity was introduced by Schur in 1923. Since then many researchers have devoted their efforts to it; see for example [6, 8, 12, 17, 19]. Schur-convexity has many important applications in analytic and geometric inequality, combinatorial analysis, numerical analysis, matrix theory, and so on. We recall some definitions.

### Definition 1.1

()

Suppose that $$x=(x_{1}, x_{2}, \ldots, x_{n})$$, $$y=(y_{1}, y_{2}, \ldots, y_{n})\in{\mathbb{R}} ^{n}$$. x is said to be majorized by y (with symbol $$x\prec y$$) if

$$\sum_{i=1}^{k} x_{[i]} \leq \sum _{i=1}^{k} y_{[i]},\quad k=1, 2, \ldots, n-1,$$

and

$$\sum_{i=1}^{n} x_{[i]}= \sum _{i=1}^{n} y_{[i]},$$

where $$x_{[i]}$$, denotes the ith largest component in x.

### Definition 1.2

()

Let $$E\subset{\mathbb{R}}^{n}$$, $$f: E\rightarrow\mathbb{R}$$ is said to be Schur-convex function on E if $$x\prec y$$ on E implies $$f(x) \leq f(y)$$. f is said to be Schur-concave if and only if −f is Schur-convex.

Chu in [4, 5, 7, 18] defined the concept of Schur-harmonically convex function.

### Definition 1.3

()

A set $$E \subset{\mathbb{R}} ^{n}_{+}$$ is said to be harmonically convex if $$( \frac{2x_{1}y_{1}}{x_{1}+y_{1}} , \frac {2x_{2}y_{2}}{x_{2}+y_{2} }, \ldots, \frac{2x_{n}y_{n}}{x_{n}+y_{n} } ) \in E$$, for every $$x=(x_{1}, x_{2}, \ldots, x_{n})$$, $$y=(y_{1}, y_{2}, \ldots, y_{n}) \in E$$.

### Definition 1.4

()

A function $$f: E \rightarrow\mathbb{R_{+}}$$ is said to be Schur-harmonically convex on E, for every $$x, y \in E$$, if $$(\frac{1}{x_{1}},\frac{1}{x_{2}},\ldots, \frac{1}{x_{n}}) \prec (\frac{1}{y_{1}},\frac{1}{y_{2}},\ldots, \frac{1}{y_{n}})$$ implies $$f(x)\leq f(y)$$.

### Definition 1.5

([2, 14])

1. (i)

A set $$E \subset{\mathbb{R}}^{n}$$ is called symmetric, if $$x\in E$$ implies $$Px \in E$$ for every $${n\times n}$$ permutation matrix P.

2. (ii)

A function $$f : E \rightarrow\mathbb{R}$$ is said to be a symmetric function if $$f(Px)=f(x)$$ for every permutation matrix P, and for every $$x\in E$$.

Recall that a $${n\times n}$$ square matrix P is said to be a permutation matrix if each row and column has a single unit entry, and all other entries are zero. The following theorem, called the Schur condition, is very useful for specifying Schur-convexity or Schur-concavity of functions.

### Theorem 1.1

()

Let $$E \subset{\mathbb{R}}^{n}$$be a symmetric convex set with nonempty interior ($$E ^{\circ}$$is the interior ofE), and $$f : E \rightarrow\mathbb{R}$$be a symmetric continuous function onE. Iffis differentiable on $$E ^{\circ}$$, thenfis Schur-convex (Schur-concave) on $$E ^{\circ}$$if and only if

$$( x_{1}-x_{2}) \biggl( \frac{\partial f}{\partial x _{1}} - \frac {\partial f}{\partial x_{2}} \biggr) \geq0 \quad(\leq0),$$

for every $$x=(x_{1}, x_{2}, \ldots, x_{n})\in E ^{\circ}$$.

In  Chu proved the following result, which is useful for determining Schur-harmonic convexity or Schur-harmonic concavity of functions.

### Theorem 1.2

()

Let $$E \subset{\mathbb{R}}_{+}^{n}$$be a symmetric and harmonically convex set with nonempty interior ($$E ^{\circ}$$is the interior ofE), and $$f : E \rightarrow\mathbb{R}_{+}$$is a symmetric continuous function onE. Iffis differentiable on $$E ^{\circ}$$, thenfis Schur-harmonically convex (Schur-harmonically concave) on $$E ^{\circ}$$if and only if

$$( x_{1}-x_{2}) \biggl(x^{2}_{1} \frac{\partial f}{\partial x _{1}} - x^{2}_{2}\frac{\partial f}{\partial x_{2}} \biggr) \geq0 \quad (\leq0),$$

for every $$x=(x_{1}, x_{2}, \ldots, x_{n})\in E ^{\circ}$$.

By Definition 1.4 the following simple fact is obvious.

### Lemma 1.1

()

The function $$f : E \rightarrow\mathbb{R}^{+}$$is Schur-harmonically convex (Schur-harmonically concave ) if and only if $$f( \frac {1}{x_{1}},\frac{1}{x_{2}},\ldots, \frac{1}{x_{n}})$$is Schur-convex (Schur-concave ) on $$\frac{1}{E}= \{\frac{1}{x}: x \in E \}$$.

In  harmonical convexity was introduced by Anderson et al. and in  İşcan gave the following definition.

### Definition 1.6

Let $$I\subset{\mathbb{R}}- \lbrace{0}\rbrace$$ be an interval. A function $$f: I\rightarrow\mathbb{R}$$ is said to be HA-convex or harmonically convex, if

$$f \biggl(\frac{xy}{tx+(1-t)y} \biggr) \leq tf(y)+(1-t)f(x),$$
(1)

for every $$x, y \in I$$ and $$t\in[0, 1]$$. If the inequality in (1) is reversed, then f is said to be harmonically concave.

If $$I\subset(0, \infty)$$ and f is a convex and nondecreasing function then f is harmonically convex. If f is an harmonically convex and nonincreasing function then f is convex. If $$[a, b] \subset I\subset(0, \infty)$$ then the function $$g:[\frac {1}{b},\frac{1}{a} ]\rightarrow\mathbb{R}$$, defined by $$g(t)=f(\frac{1}{t})$$, is convex if and only if f is harmonically convex on $$[a, b]$$ (see ).

The following Hermite–Hadamard type inequality for harmonically convex functions was obtained by İşcan  and Dragomir  in different ways.

### Theorem 1.3

Let $$f:I\subset{\mathbb{R}}- \lbrace{0}\rbrace\rightarrow\mathbb {R}$$ba a harmonically convex function and $$a,b \in I$$with $$a< b$$. If $$f\in L[a, b]$$then the following inequalities hold:

$$f \biggl( \frac{2ab}{a+b} \biggr) \leq\frac{ab}{b-a} \int_{a}^{b} f(x)\,dx \leq\frac{f(a)+f(b)}{2}, \quad a,b \in{I}, a< b.$$

The above inequalities are sharp.

In , Dragomir defined convex function on the co-ordinates (or co-ordinated convex functions ) on the set $$[a, b] \times[c, d]$$ in $$\Bbb{R}^{2}$$ with $$a< b$$ and $$c< d$$ as follows.

### Definition 1.7

A function $$f:[a ,b] \times[c, d] \rightarrow\mathbb{R}$$ is said to be convex on the co-ordinates on $$[a ,b] \times[c, d]$$ if for every $$y \in[c , d]$$ and $$x\in[a , b]$$, the partial mappings

$$f_{y} {:[a, b] \rightarrow\mathbb{R}} ,\quad f_{y}(u)=f(u,y),$$

and

$$f_{x}{:[c, d] \rightarrow\mathbb{R}} ,\quad f_{x}(v)=f(x,v),$$

are convex. This means that, for every $$(x, y) , (z , w ) \in[a ,b] \times[c, d]$$ and $$t ,s \in[0, 1]$$,

\begin{aligned} f \bigl(tx+(1-t)z , sy+(1-s)w \bigr) \leq{}& t s f(x, y) +s (1-t) f(z, y) \\ &+ t(1-s)f(x, w)+ (1-t) (1-s)f(z, w). \end{aligned}

Clearly, every convex function is co-ordinated convex. Furthermore, there exist co-ordinated convex functions which are not convex. The following Hermite–Hadamard type inequality for co-ordinated convex functions was also proved in .

### Theorem 1.4

Suppose that $$f:[a ,b] \times[c, d] \rightarrow\mathbb{R}$$is convex on the co-ordinates on $$[a ,b] \times[c, d]$$. Then

\begin{aligned} f \biggl( \frac{a+b}{2}, \frac{c+d}{2} \biggr) \leq{}& \frac{1}{2} \biggl[ \frac{1}{b-a} \int_{a}^{b} f \biggl(x,\frac {c+d}{2} \biggr) \,dx+\frac{1}{d-c} \int_{c}^{d} f \biggl(\frac {a+b}{2},y \biggr) \,dy \biggr] \\ \leq{}& \frac{1}{( b-a) (d-c)} \int_{a} ^{b} \int_{c} ^{d} f(x , y) \,dy \,dx \\ \leq{}& \frac{1}{4} \biggl[\frac{1}{b-a} \int_{a}^{b} f(x, c) \,dx+\frac {1}{b-a} \int_{a}^{b} f(x, d) \,dx \\ &+\frac{1}{d-c} \int_{c}^{d} f(a, y)\,dy+\frac{1}{d-c} \int_{c}^{d} f(b, y)\, dy \biggr] \\ \leq{}& \frac{f(a, c)+f(a, d)+f(b, c)+f(b, d)}{4}. \end{aligned}

The above inequalities are sharp.

In  Set and İşcan defined an harmonically convex and an harmonically convex function on the co-ordinates on the set $$[a, b] \times[c, d]$$ in $${\Bbb{R}}^{2}$$ with $$a< b$$ and $$c< d$$ as follows.

### Definition 1.8

Let $$\Delta=[a, b]\times[c, d]\subset(0, \infty)\times(0, \infty )$$ with $$a< b$$ and $$c< d$$. A function $$f: \Delta\rightarrow \mathbb{R}$$ is said to be harmonically convex on Δ if the following inequality holds:

\begin{aligned}[b] f \biggl(\frac{xz}{tz+(1-t)x}, \frac{yw}{tw+(1-t)y} \biggr) = {}& f \biggl(\frac{1}{\frac{t}{x}+\frac{1-t}{z}}, \frac{1}{\frac{t}{y}+\frac {1-t}{w}} \biggr) \\ \leq{}& t f(x,y) +(1-t) f(z, w), \end{aligned}
(2)

for every $$(x, y), (z, w) \in\Delta$$ and $$t\in[0, 1]$$. If the inequality in (2) is reversed, then f is said to be harmonically concave on Δ.

### Definition 1.9

Let $$\Delta=[a, b]\times[c, d]\subset(0, \infty)\times(0, \infty )$$ with $$a< b$$ and $$c< d$$. A function $$f: \Delta\rightarrow \mathbb{R}$$ is said to be harmonically convex on the co-ordinates on Δ if for every $$y \in[c , d]$$ and $$x\in[a , b]$$, the partial mappings,

$$f_{y} {:[a, b] \rightarrow\mathbb{R}} ,\quad f_{y}(u)=f(u,y),$$

and

$$f_{x}{:[c, d] \rightarrow\mathbb{R}} ,\quad f_{x}(v)=f(x,v),$$

are harmonically convex.

Clearly, every harmonically convex function is harmonically convex on the co-ordinates. Furthermore, there exist co-ordinated harmonically convex functions which are not harmonically convex. Note that if $$f_{x}$$ and $$f_{y}$$ are convex and nondecreasing functions then $$f_{x}$$ and $$f_{y}$$ are harmonically convex. The following Hermite–Hadamard type inequality for harmonically co-ordinated convex functions was also proved in .

### Theorem 1.5

Let $$f:\Delta=[a, b]\times[c, d]\subset(0, \infty)\times(0, \infty)\rightarrow\mathbb{R}$$is harmonically convex on the co-ordinates on Δ. Then

\begin{aligned} f \biggl( \frac{2ab}{a+b}, \frac{2cd}{c+d} \biggr) \leq{}& \frac{1}{2} \biggl[ \frac{ab}{b-a} \int_{a}^{b}\frac{f ( x,\frac{ 2cd}{c+d} ) }{x^{2}}\,dx + \frac{cd}{d-c} \int_{c}^{d} \frac{f ( \frac{2ab}{a+b},y ) }{y^{2}}\,dy \biggr] \\ \leq{}& \frac{abcd}{( b-a) (d-c)} \int_{a} ^{b} \int_{c} ^{d} \frac{f(x , y)}{(xy)^{2}} \,dy \,dx \\ \leq{}& \frac{1}{4} \biggl[\frac{ab}{b-a} \int_{a}^{b}\frac{ f(x, c)}{x^{2}} \,dx+ \frac{ab}{b-a} \int_{a}^{b}\frac{ f(x, d)}{x^{2}} \,dx \\ &+\frac{cd}{d-c} \int_{c}^{d}\frac{ f(a, y)}{y^{2}}\,dy+ \frac {cd}{d-c} \int_{c}^{d} \frac{f(b, y)}{y^{2}}\,dy \biggr] \\ \leq{}& \frac{f(a, c)+f(a, d)+f(b, c)+f(b, d)}{4}. \end{aligned}

The above inequalities are sharp.

In  Elezović and Pečarić investigated the Schur-convexity on the upper and the lower limit of the integral for the mean of convex function and proved the following important result by using the Hermite–Hadamard inequality.

### Theorem 1.6

Letfbe a continuous function on an intervalI, and

\begin{aligned} F(x, y)= \textstyle\begin{cases} \frac{1}{y-x}\int_{x}^{y}f(t)\,dt, & x, y\in I, x \neq y,\\ f(x),&x=y\in I. \end{cases}\displaystyle \end{aligned}

Then $$F(x, y)$$is Schur-convex (Schur-concave) on $$I^{2}$$if and only iffis convex (concave) onI.

Let $$I\subset\mathbb{R}$$ be an open interval and $$f\in{C^{2}(I)}$$. In  Chu et al. proved the following theorems.

### Theorem 1.7

Let $$f:I\rightarrow\mathbb{R}$$be a continuous function. The function

\begin{aligned} F(x, y)= \textstyle\begin{cases} \frac{1}{y-x}\int_{x}^{y}f(t)\,dt-{f(\frac{x+y}{2})},&x, y\in I, x \neq y,\\ 0, &x=y\in I, \end{cases}\displaystyle \end{aligned}

is Schur-convex (Schur-concave) on $$I^{2}$$if and only iffis convex (concave) onI.

### Theorem 1.8

Let $$f:I\rightarrow\mathbb{R}$$be a continuous function. The function

\begin{aligned} F(x, y)= \textstyle\begin{cases} \frac{f(x)+f(y)}{2} -\frac{1}{y-x}\int_{x}^{y}f(t)\,dt,&x, y\in I, x \neq y,\\ 0, &x=y\in I, \end{cases}\displaystyle \end{aligned}

is Schur-convex (Schur-concave) on $$I^{2}$$if and only iffis convex (concave) onI.

We recall the following lemma from , which is known as Leibniz’s formula.

### Lemma 1.2

Suppose that $${ f:\Delta=[a, b]\times[c, d]\rightarrow\mathbb{R}}$$and $${ \frac{\partial f}{\partial t} :[a, b]\times[c, d]\rightarrow \mathbb{R}}$$are continuous and $$\alpha_{1} , \alpha_{2}:[c,d]\rightarrow[a, b]$$are differentiable functions. Then the function $$\varphi :[c,d]\rightarrow\mathbb{R}$$defined by

$$\varphi(t)= \int_{{\alpha}_{1} (t)}^{{\alpha}_{2} (t)} f(x,t)\,dx,$$

has a derivative for each $$t\in[c, d]$$, which is given by

$$\varphi^{\prime}(t)=f\bigl(\alpha_{2} (t), t\bigr) \alpha_{2}^{\prime }(t)-f\bigl(\alpha_{1} (t), t\bigr) \alpha_{1}^{\prime}(t)+ \int_{{\alpha}_{1} (t)}^{{\alpha}_{2} (t)}\frac{\partial f}{\partial t}(x, t)\,dx.$$

Moreover, in  we proved the following lemma which will be useful in the sequel. A version of the following lemma was proved in .

### Lemma 1.3

Let $$F (u , v ) =\int_{u} ^{v} \int_{u} ^{v} f(x, y) \,dx \,dy$$, where $$f(x, y)$$is continuous on the rectangle $$[a,p]\times[ a, q]$$, $$u =u(b)$$and $$v= v (b)$$are differentiable with $$a\leq u(b)\leq p$$and $$a\leq v (b)\leq q$$. Then

\begin{aligned} \frac{\partial F}{\partial b} ={} & \biggl( \int_{u} ^{v} f(x, v) \,dx + \int_{u} ^{v} f(v, y) \,dy \biggr)v^{\prime} (b) - \biggl( \int_{u} ^{v} f(x,u)\,dx+ \int_{u} ^{v} f(u,y) \,dy \biggr)u^{\prime} (b). \end{aligned}

## 2 Main result

In this section we prove new theorems like Theorem 1.6 and Theorems 1.7 and 1.8 for harmonically convex functions and co-ordinated harmonically convex functions.

### Theorem 2.1

Let $$I \subset(0, \infty)$$be an open interval, and the function $$f:I\rightarrow\mathbb{R_{+}}$$be continuously differentiable onI. Suppose that the function $$F:I^{2}\rightarrow\mathbb{R_{+}}$$is defined by

$$F(x, y) := \textstyle\begin{cases} \frac{xy}{y-x}\int_{x}^{y}\frac{ f(t)}{t^{2}} \,dt , & x, y\in I, x \neq y,\\ f(x),&x=y\in I. \end{cases}$$
(3)

ThenFis Schur-harmonically convex on $$I^{2}$$if and only iffis harmonically convex onI.

### Proof

According to Lemma 1.1 it is sufficient to show that the function $$F(\frac{1}{x} , \frac{1}{y})$$ is Schur-convex on $$\frac {1}{I}\times\frac{1}{I}$$. From (3) we have

$$F\biggl(\frac{1}{x},\frac{1}{y}\biggr)= \frac{\frac{1}{xy}}{\frac{1}{y} - \frac{1}{x}} \int_{\frac{1}{x}}^{\frac{1}{y}} \frac{f(t)}{t^{2}}\, dt= \frac{1}{x-y} \int_{\frac{1}{x}}^{\frac{1}{y}}\frac {f(t)}{t^{2}}\,dt,$$

for every $$x, y\in I$$, with $$x \neq y$$. Using the change of variable $$s=\frac{1}{t}$$, then $$F(\frac{1}{x} , \frac{1}{y})=\frac{1}{y-x}\int_{x}^{y}f(\frac {1}{s})\,ds$$. Thus by Theorem 1.6 the function

\begin{aligned} F\biggl(\frac{1}{x}, \frac{1}{y} \biggr)= \textstyle\begin{cases} \frac{1}{y-x}\int_{x}^{y}f(\frac{1}{t})\,dt , & x, y\in I, x \neq y,\\ f(\frac{1}{x}),&x=y\in I, \end{cases}\displaystyle \end{aligned}

is Schur-convex if and only if the function $$f(\frac{1}{t})$$ is convex on $$\frac{1}{I}$$. This implies that the function $$f(t)$$ is harmonically convex on I. Therefore by Theorem 1.6 the result follows. □

The proofs of the following two theorems are similar to the one for Theorem 2.1, hence we omit them.

### Theorem 2.2

Let $$I \subset(0, \infty)$$be an open interval, and the function $$f:I\rightarrow\mathbb{R}$$has continuous second order derivatives onI. Suppose that the function $$G:I^{2}\rightarrow\mathbb{R_{+}}$$is defined by

\begin{aligned} G(x, y) := \textstyle\begin{cases} \frac{xy}{y-x}\int_{x}^{y}\frac{ f(t)}{t^{2}} \,dt -f(\frac {2xy}{x+y}), & x, y\in I, x \neq y,\\ 0,&x=y\in I. \end{cases}\displaystyle \end{aligned}

ThenGis Schur-harmonically convex on $$I^{2}$$if and only iffis harmonically convex onI.

### Theorem 2.3

Let $$I \subset(0, \infty)$$be an open interval, and the function $$f:I\rightarrow\mathbb{R}$$has continuous second order derivatives onI. Suppose that the function $$H:I^{2}\rightarrow\mathbb{R_{+}}$$is defined by

\begin{aligned} H(x, y) := \textstyle\begin{cases} \frac{f(x)+f(y)}{2}-\frac{xy}{y-x}\int_{x}^{y}\frac{ f(t)}{t^{2}} \, dt , & x, y\in I, x \neq y,\\ 0,&x=y\in I. \end{cases}\displaystyle \end{aligned}

ThenHis Schur-harmonically convex on $$I^{2}$$if and only iffis harmonically convex onI.

To reach our main results, we need the following two lemmas.

### Lemma 2.1

Let $$D=[a_{1}, b_{1}]\times[a_{1}, b_{1}]$$be a square in $${\Bbb {R}}^{2}- \lbrace( 0, 0) \rbrace$$with $$a_{1}< b_{1}$$, and the function $$f :D\rightarrow\Bbb{R}$$be continuous, and have continuous second order partial derivatives on $$D^{\circ}$$. Choose $$a, b \in(a_{1}, b_{1})$$, with $$a< b$$, and let $$\Delta=[a, b]\times [a, b]$$. Suppose that the function $$F:\Delta\rightarrow\Bbb{R}$$is defined by

$$F(x,y) := \textstyle\begin{cases} \frac{x^{2}y^{2}}{(y-x)^{2}} {\int_{x}^{y} \int_{x} ^{y} \frac{ f(t, s)}{t^{2}s^{2}} \,dt \,ds },& x\neq y, x, y\in[a, b], \\ f(x , x) , & x=y, x, y\in[a, b]. \end{cases}$$

Then

\begin{aligned}[b] \frac{\partial F}{\partial x}\bigg|_{(t_{0 } , t _{0} )} &= \frac {\partial F}{\partial y}\bigg|_{(t_{0 } , t _{0} )} \\ &=\frac{1}{6} \biggl[ 2t_{0} \bigl( g(t_{0}, t_{0}) +h(t_{0}, t_{0}) \bigr)\\&\quad+ t^{2}_{0} \biggl(g_{1}(t_{0},t_{0})+h_{1}(t_{0},t_{0}) +\frac{\partial g}{\partial t}(t,t)\bigg|_{t_{0}}+\frac{\partial h}{\partial t}(t,t)\bigg|_{t_{0}} \biggr) \biggr] , \end{aligned}
(4)

for all $$t_{0}\in[a, b]$$, where

$$g(u,t_{0}+t )=\frac{f(u,t_{0}+t) }{u^{2}} , \quad h(t_{0}+t,v)= \frac {f(t_{0}+t,v) }{v^{2}} ,$$

and

$$g_{1}(u , t_{0}+t)=\frac{\partial g}{\partial t}( u, t_{0}+ t ),\qquad h_{1}(t_{0}+t ,v)= \frac{\partial h}{\partial t}(t_{0}+ t,v ).$$

### Proof

Fix $$t_{0} \in[a, b]$$. By using L’Hopital’s rule and Lemmas 1.2, and 1.3 we see that

\begin{aligned}[b] \frac{\partial F}{\partial x}\bigg| _{(t_{0 } , t _{0})} &= \lim_{t\rightarrow0}\frac{F(t_{0}+t, t_{0}) -F(t_{0}, t_{0})}{t} \\ &=\lim_{t\rightarrow0}\frac{1}{t^{3}} \biggl[ t_{0}^{2}(t_{0}+t)^{2}{ \int_{t_{0}}^{ t_{0}+t} \int_{t_{0} } ^{ t_{0}+t}\frac{f(u, v)}{u^{2}v^{2}}\,du \,dv -t^{2} f(t_{0}, t_{0})} \biggr] \\ &=\lim_{t\rightarrow0}\frac{1}{3t^{2}} \biggl[2t_{0}^{2}(t_{0}+t) \int_{t_{0}}^{ t_{0}+t} \int_{t_{0} } ^{ t_{0}+t}\frac {f(u,v)}{{u}^{2}v^{2}}\,du \,dv \\ &\quad+t_{0}^{2} \biggl( \int_{t_{0}}^{ t_{0}+t} \frac{f(u, t_{0}+t)}{u^{2}} \,du + \int_{t_{0}}^{ t_{0}+t} \frac{f(t_{0}+t,v)}{v^{2}} \,dv \biggr) -2t f(t_{0}, t_{0}) \biggr]. \end{aligned}
(5)

Again, using L’Hopital’s rule we obtain

\begin{aligned} \frac{\partial F}{\partial x}={}&\lim_{t\rightarrow0}\frac {1}{6t} \biggl[ 2t_{0}^{2} \int_{t_{0}}^{ t_{0}+t} \int_{t_{0}}^{ t_{0}+t}\frac{f(u,v)}{{u}^{2}v^{2}}\,du \,dv+ \frac {2t_{0}^{2}}{t_{0}+t} \biggl( \int_{t_{0}}^{ t_{0}+t} \frac{f(u, t_{0}+t)}{u^{2}}\,du \\ &+ \int_{t_{0}}^{ t_{0}+t}\frac{f(t_{0}+t,v)}{v^{2}} \,dv \biggr)+t_{0}^{2} \biggl(g(t_{0}+t , t_{0}+t)+ \int_{t_{0}}^{ t_{0}+t}\frac {\partial g}{\partial t}(u, t_{0}+t ) \,du \\ &+h(t_{0}+t , t_{0}+t)+ \int_{t_{0}}^{ t_{0}+t}\frac{\partial h}{\partial t} (t_{0}+t, v ) \,dv \biggr) -2f(t_{0}, t_{0}) \biggr] . \end{aligned}

By a similar computation it follows that

\begin{aligned} \frac{\partial F}{\partial x}={}&\lim_{t\rightarrow0}\frac{1}{6} \biggl[ \frac{2t_{0}^{2}}{(t_{0}+t)^{2}} \biggl( \int_{t_{0}}^{ t_{0}+t} \frac {f(u, t_{0}+t)}{u^{2}}\,du + \int_{t_{0}}^{ t_{0}+t}\frac {f(t_{0}+t,v)}{v^{2}} \,dv \biggr) \\ &+\frac{2t_{0}^{2}}{t_{0}+t} \biggl( g(t_{0}+t , t_{0}+t)+ \int _{t_{0}}^{ t_{0}+t}\frac{\partial g}{\partial t}(u, t_{0}+t ) \, du+h(t_{0}+t , t_{0}+t) \\ &+ \int_{t_{0}}^{ t_{0}+t}\frac{\partial h}{\partial t} (t_{0}+t, v ) \,dv \biggr)+ t_{0}^{2} \biggl(\frac{\partial g}{\partial t}( t_{0}+ t, t_{0}+ t )+ g_{1}(t_{0}+t , t_{0}+t) \\ &+ \int_{t_{0}}^{ t_{0}+t}\frac{\partial g_{1}}{\partial t}(u, t_{0}+t)\,du+\frac{\partial h}{\partial t}( t_{0}+ t, t_{0}+ t )+h_{1}(t_{0}+t , t_{0}+t) \\ &+ \int_{t_{0}}^{ t_{0}+t}\frac{\partial h_{1}}{\partial t}(t_{0}+t,v )\,dv \biggr) \biggr] \\ ={}&\frac{1}{6} \biggl[{ 2t_{0} \bigl( g(t_{0}, t_{0}) +h(t_{0}, t_{0}) \bigr)+ t^{2}_{0} \biggl(g_{1}(t_{0},t_{0})+h_{1}(t_{0},t_{0}) +\frac{\partial g}{\partial t}(t,t)\bigg|_{t_{0}}+\frac{\partial h}{\partial t}(t,t)\bigg|_{t_{0}}} \biggr) \biggr]. \end{aligned}

By changing the role of x by y in (5), we obtain the required results in (4). □

The proof of the following lemma is similar to the one in Lemma 2.1, hence we omit it.

### Lemma 2.2

Let $$D = [a_{1}, b_{1}]\times[a_{1}, b_{1}]$$be a square in $${\Bbb {R}}^{2}- \lbrace(0, 0) \rbrace$$with $$a_{1}< b_{1}$$, and the function $$f :D\rightarrow\Bbb{R}$$be continuous, and have continuous third order partial derivatives on $$D^{\circ}$$. Choose $$a, b \in(a_{1}, b_{1})$$, with $$a< b$$, and let $$\Delta=[a, b]\times[a, b]$$. Suppose that the function $$G:\Delta\rightarrow\mathbb{R}$$is defined by

$$G(x,y) : = \textstyle\begin{cases} \frac{x^{2}y^{2}}{(y-x)^{2}} {\int_{x}^{y} \int_{x} ^{y} \frac{ f(t, s)}{t^{2}s^{2}} \,dt \,ds }-f(\frac{2xy}{x+y} , \frac{2xy}{x+y}) ,& x\neq y, x, y \in[a, b], \\ 0 , & x=y, x, y \in[a,b]. \end{cases}$$

Then

\begin{aligned} \frac{\partial G}{\partial x}\bigg| _{(t_{0 } , t _{0} )} =& \frac{\partial G}{\partial y}\bigg| _{(t_{0 } , t _{0} )} \\ ={}& \frac{1}{6} \biggl[ \frac{4f(t_{0}, t_{0})}{t_{0}}\\ &+t_{0}^{2} \biggl(g_{1}(t_{0}, t_{0}) + h_{1}(t_{0}, t_{0}) +\frac{\partial g}{\partial t}(t,t)\bigg|_{t_{0}} + \frac{\partial h}{\partial t}(t,t)\bigg|_{t_{0}} \biggr) - 6 \frac{\partial f}{\partial t}(t,t)\bigg|_{t_{0}} \biggr], \end{aligned}

for all $$t_{0} \in[a, b]$$, where

$$g(u,t_{0}+t )=\frac{f(u,t_{0}+t) }{u^{2}} , \qquad h(t_{0}+t,v)= \frac {f(t_{0}+t,v) }{v^{2}} ,$$

and

$$g_{1}(u , t_{0}+t)=\frac{\partial g}{\partial t}( u, t_{0}+ t ),\qquad h_{1}(t_{0}+t ,v)= \frac{\partial h}{\partial t}(t_{0}+ t,v ).$$

We now derive the next results for co-ordinated harmonically convex functions.

### Theorem 2.4

Let $$D = [a_{1}, b_{1}]\times[a_{1}, b_{1}]$$be a square in $${\Bbb {R}}^{2}_{+}$$with $$a_{1}< b_{1}$$, and the function $$f :D\rightarrow \Bbb{R_{+}}$$be continuous, and have continuous second order partial derivatives on $$D^{\circ}$$. Choose $$a, b \in(a_{1}, b_{1})$$, with $$a< b$$, and let $$\Delta=[a, b]\times[a, b]$$. Suppose thatfis harmonically convex on the co-ordinates on Δ, then the function $$F:\Delta\rightarrow\Bbb{R_{+}}$$defined by

$$F(x,y) := \textstyle\begin{cases} \frac{x^{2}y^{2}}{(y-x)^{2}} {\int_{x}^{y} \int_{x} ^{y} \frac{ f(t, s)}{t^{2}s^{2}} \,dt \,ds },& x\neq y, x, y \in[a, b],\\ f(x , x) , & x=y, x,y\in[a, b], \end{cases}$$
(6)

is Schur-harmonically convex on Δ.

### Proof

Case 1: if $$x,y \in[a, b]$$, with $$x= y$$. Then Lemma 2.1 implies that

$$(y-x) \biggl( y^{2}\frac{\partial F}{\partial y} -x^{2} \frac{\partial F}{\partial x} \biggr)=0.$$

Case 2: if $$x, y\in[a, b]$$, with $$x\neq y$$. Then by Lemma 1.3 we have

\begin{aligned} \frac{\partial F}{\partial y}={}&\frac{-2x^{3}y}{(y-x)^{3}} \int _{x}^{y} \int_{x}^{y} \frac{ f(t,s)}{t^{2}s^{2}}\,dt \,ds \\ &+\frac{x^{2}}{(y-x)^{2}} \biggl( \int_{x} ^{y}\frac{f(t,y)}{t^{2}}\,dt + \int_{x}^{y} \frac{ f(y, s)}{s^{2}}\,ds \biggr) \end{aligned}

and

\begin{aligned} \frac{\partial F}{\partial x}={}&\frac{2xy^{3}}{(y-x)^{3}} \int _{x}^{y} \int_{x}^{y} \frac{ f(t,s)}{t^{2}s^{2}}\,dt \,ds \\ &-\frac{y^{2}}{(y-x)^{2}} \biggl( \int_{x} ^{y} \frac{f(t,x)}{t^{2}}\,dt + \int_{x}^{y} \frac{f(x, s)}{s^{2}}\,ds \biggr). \end{aligned}

Thus,

\begin{aligned} \biggl( y^{2}\frac{\partial F}{\partial y} -x^{2} \frac{\partial F}{\partial x} \biggr) ={}&\frac{-4x^{3}y^{3}}{(y-x)^{3}} \int_{x}^{y} \int_{x}^{y} \frac {f(t,s)}{t^{2}s^{2}}\,dt \,ds \\ &+\frac{x^{2}y^{2}}{(y-x)^{2}} \biggl( \int_{x} ^{y}\frac{f(t,x)+f(t, y)}{t^{2}}\,dt \\ & + \int_{x}^{y}\frac{f(x, s)+ f(y, s)}{s^{2}} \,ds \biggr). \end{aligned}

Then $$(y-x) (y^{2} \frac{\partial F}{\partial y} -x^{2} \frac {\partial F}{\partial x} )$$ is nonnegative if

\begin{aligned}& \frac{xy}{(y-x)^{2}} \int_{x}^{y} \int_{x}^{y} f(t,s)\,dt \,ds \\& \quad\leq \frac{1}{ 4 (y-x) } \biggl( \int_{x}^{y} \frac{ f(t, y)+f(t, x)}{t^{2}}\,dt+ \frac{f(y, s)+f(x, s)}{s^{2}}\,ds \biggr). \end{aligned}

The last inequality follows from Theorem 1.5. Therefore by Theorem 1.2 the function F is Schur-harmonically convex. □

The following theorem also holds.

### Theorem 2.5

Let $$D = [a_{1}, b_{1}]\times[a_{1}, b_{1}]$$be a square in $${\Bbb {R}}^{2}_{+}$$with $$a_{1}< b_{1}$$, and the function $$f :D\rightarrow \Bbb{R}$$be continuous, and have continuous third order partial derivatives on $$D^{\circ}$$. Choose $$a, b \in(a_{1}, b_{1})$$, with $$a< b$$, and let $$\Delta=[a, b]\times[a, b]$$. Suppose thatfis harmonically convex on the co-ordinates on Δ, then the function $$G:\Delta\rightarrow\Bbb{R}_{+}$$defined by

$$G(x,y):= \textstyle\begin{cases} \frac{x^{2}y^{2}}{(y-x)^{2}} {\int_{x}^{y} \int_{x} ^{y} \frac{f(t, s)}{t^{2}s^{2}} \,dt \,ds }-f( \frac{2xy}{x+y}, \frac{2xy}{x+y}) ,& x\neq y, x, y \in[a, b], \\ 0 , & x=y, x, y\in[a, b], \end{cases}$$
(7)

is Schur-harmonically convex for Δ.

### Proof

Case 1: If $$x,y \in[a, b]$$, with $$x= y$$. Then Lemma 2.2 implies that

$$(y-x) \biggl( y^{2}\frac{\partial G}{\partial y} - x^{2} \frac{\partial G}{\partial x} \biggr)=0.$$

Case 2: If $$x, y\in[a, b]$$, with $$x\neq y$$. Then by Lemma 1.2 we have

$$(y-x) \biggl(y^{2} \frac{\partial G}{\partial y} - x^{2} \frac{\partial G}{\partial x} \biggr)\geq0,$$

if

\begin{aligned}& \frac{xy}{(y-x)^{2}} \int_{x}^{y} \int_{x}^{y} f(t,s)\,dt \,ds \\& \quad\leq \frac{1}{ 4 (y-x) } \biggl( \int_{x}^{y} \frac{ f(t, y)+f(t, x)}{t^{2}}\,dt+ \frac{f(y, s)+f(x, s)}{s^{2}}\,ds \biggr). \end{aligned}

The result follows from Theorem 1.2 and Theorem 1.5. □

In the following examples we show that the converses of Theorems 2.4 and 2.5 are not true in general.

### Example 2.1

Consider the non-harmonically co-ordinated convex function,

$$f(t, s) : =t^{2}-\frac{1}{3}s^{2}, \quad t, s\in[1, 2].$$

It is easy to see that for the function F as defined in (6) we have $$F(x, x)=\frac{2}{3}x^{2}$$, for every $$x \in[1, 2]$$, and

$$F(x, y)=\frac{x^{2}y^{2}}{(y-x)^{2}} \int_{x}^{y} \int_{x}^{y} \frac {t^{2}-\frac{1}{3} s^{2}}{t^{2}s^{2}}\,dt \,ds= \frac{2}{3}xy,$$

for every $$x, y\in[1, 2]$$, with $$x\neq y$$. Thus,

$$F(x, y)=\frac{2}{3}xy,$$

for every $$x, y\in[1, 2]$$. Clearly F is symmetric, continuous and differentiable on $$[1, 2]\times[1, 2]$$.

If $$x,y \in[1,2]$$, with $$x \neq y$$, we have

$$(y-x) \biggl( y^{2}\frac{\partial F}{\partial y} - x^{2} \frac{\partial F}{\partial x} \biggr) =\frac{2}{3}xy(y-x)^{2}\geq0.$$

Therefore by Theorem 1.2 the function F is Schur-harmonically convex.

### Remark 2.1

It is easy to see that for the function f as defined in Example 2.1 we have

\begin{aligned} f \biggl( \frac{2xy}{x+y}, \frac{2xy}{x+y} \biggr) \leq{}& \frac{1}{2} \biggl[ \frac{xy}{y-x} \int_{x}^{y}\frac{f ( t,\frac{ 2xy}{x+y} ) }{t^{2}}\,dt + \frac{xy}{y-x} \int_{x}^{y} \frac{f ( \frac{2xy}{x+y},s ) }{s^{2}}\,ds \biggr] \\ \leq{}& \frac{x^{2}y^{2}}{( y-x)^{2} } \int_{x} ^{y} \int_{x}^{y} \frac{f(t , s)}{t^{2}s^{2}} \,dt \,ds \\ \leq{}& \frac{1}{4} \biggl[\frac{xy}{y-x} \int_{x}^{y}\frac{ f(t, x)}{t^{2}} \,dt+ \frac{xy}{y-x} \int_{x}^{y}\frac{ f(t, y)}{t^{2}} \,dt \\ &+\frac{xy}{y-x} \int_{x}^{y}\frac{ f(x, s)}{s^{2}}\,ds+ \frac {xy}{y-x} \int_{x}^{y} \frac{f(y, s)}{s^{2}}\,ds \biggr] \\ \leq{}& \frac{f(x, x)+f(x, y)+f(y,x)+f(y, y)}{4}, \end{aligned}

for every $$x, y \in[1, 2]$$, with $$x\neq y$$. This means that each of the inequalities in Theorem 1.5 is valid while f is not harmonically convex on co-ordinates.

### Example 2.2

Consider the non-harmonically co-ordinated convex function:

$$f(t, s) : =2t^{2}-s^{2},\quad t, s \in[1, 2].$$

It is easy to see that for the function G as defined in (7) we have $$G(x, x)=0$$, for every $$x \in[1, 2]$$, and

\begin{aligned} G(x, y)&=\frac{1}{(y-x)^{2}} \int_{x}^{y} \int_{x}^{y} \frac{2 t^{2}- s^{2}}{t^{2}s^{2}}\,dt \,ds- \biggl( \frac{2xy}{x+y} \biggr)^{2} \\ &=xy- \biggl(\frac{2xy}{x+y} \biggr)^{2}, \end{aligned}

for every $$x\neq y$$, with $$x, y\in[1, 2]$$. Thus,

$$G(x, y)= xy- \biggl(\frac{2xy}{x+y} \biggr)^{2},$$

for every $$x, y\in[1, 2]$$. Clearly G is symmetric, continuous and differentiable on $$[1, 2]\times[1, 2]$$.

If $$x, y \in[1, 2]$$, we have

$$(y-x) \biggl(y^{2} \frac{\partial G}{\partial y} - x^{2} \frac{\partial G}{\partial x} \biggr) =xy(y-x)^{2}\geq0.$$

Therefore by Theorem 1.2 the function G is Schur-harmonically convex.

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