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

([2])

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

([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

([5])

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

([5])

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])

• (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.

• (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

([2])

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

Theorem 1.2

([5])

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

([5])

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 [1] harmonical convexity was introduced by Anderson et al. and in [13] 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 [10]).

The following Hermite–Hadamard type inequality for harmonically convex functions was obtained by İşcan [13] and Dragomir [10] 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 [9], 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 [9].

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 [16] 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 [16].

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 [11] 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 [6] 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 [3], 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 [15] we proved the following lemma which will be useful in the sequel. A version of the following lemma was proved in [17].

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.