Schur-Convexity of the Mean of Convex Functions for Two Variables

Abstract

The results of Schur convexity established by Elezovic and Pecaric for the average of convex functions are generalized relative to the case of the means for two-variable convex functions. As an application, some binary mean inequalities are given.

1. Introduction

Let $\mathbb{R}$ be a set of real numbers, g be a convex function defined on the interval $I\subseteq \mathbb{R}\to \mathbb{R}$ and $c,d\in I$, $c<d$. Then

$$g\left(\frac{d+c}{2}\right)\le \frac{1}{d-c}{\int }_c^{d}g\left(t\right)dt\le \frac{g\left(d\right)+g\left(c\right)}{2}.$$

(1)

This is the famous Hadamard’s inequality for convex functions.

In 2000, utilizing Hadamard’s inequality, Elezovic and Pecaric [1] researched Schur-convexity on the lower and upper limit of the integral for the mean of the convex functions and obtained the following important and profound theorem.

Theorem 1

([1]). Let I be an interval with nonempty interior on $\mathbb{R}$ and g be a continuous function on I. Then,

$$\mathrm{\Phi }(c,d)=\left\{\begin{array}{c}\frac{1}{d-c}{\int }_c^{d}g\left(s\right)ds,c,d\in I,d\ne c\hfill \\ g\left(c\right),d=c\hfill \end{array}\right.$$

is $Schurconvex(Schurconcave,resp.)$ on $I\times I$ iff g is convex (concave, resp.) on I.

In recent years, this result attracted the attention of many scholars (see references [2,3,4,5,6,7,8,9,10,11,12] and Chapter II of the monograph [13] and its references).

In this paper, the result of theorem 1 is generalized to the case of bivariate convex functions, and some bivariate mean inequalities are established.

Theorem 2.

Let I be an interval with non-empty interior on $\mathbb{R}$ and $g(s,t)$ be a continuous function on $I\times I$. If g is convex (or concave resp.) on $I\times I$, then

$$G(u,v)=\left\{\begin{array}{cc}\frac{1}{{(v-u)}^2}{\int }_u^v{\int }_u^{v}g(s,t)dsdt,\hfill & (u,v)\in I\times I,u\ne v\hfill \\ g(u,u),\hfill & (u,v)\in I\times I,u=v\hfill \end{array}\right.$$

(2)

is Schur convex (or Schur concave, resp.) on $I\times I$.

2. Definitions and Lemmas

To prove Theorem 2, we provide the following lemmas and definitions.

Definition 1.

Let $(x_1,x_2)$ and $(y_1,y_2)\in \mathbb{R}\times \mathbb{R}$.

(1)

A set $\mathrm{\Omega }\subset \mathbb{R}\times \mathbb{R}$ is said to be convex if $(x_1,x_2),(y_1,y_2)\in \mathrm{\Omega }$ and $0\le \beta \le 1$ implies

$$(\beta x_1+(1-\beta )y_1,\beta x_2+(1-\beta )y_2)\in \mathrm{\Omega }.$$

(2)

Let $\mathrm{\Omega }\subset \mathbb{R}\times \mathbb{R}$ be convex set. A function ψ: $\mathrm{\Omega }\to \mathbb{R}$ is said to be a convex function on Ω if, for all $\beta \in [0,1]$ and all $(x_1,x_2),(y_1,y_2)\in \mathrm{\Omega }$, inequality

$$\psi (\beta x_1+(1-\beta )y_1,\beta x_2+(1-\beta )y_2)\le \beta \psi (x_1,x_2)+(1-\beta )\psi (y_1,y_2)$$

(3)

holds. If, for all $\beta \in [0,1]$ and all $(x_1,x_2),(y_1,y_2)\in \mathrm{\Omega }$, the strict inequality in (3) holds, then ψ is said to be strictly convex. ψ is called concave (or strictly concave, resp.) iff $-\psi$ is convex (or strictly convex, resp.)

Definition 2.

([14,15]). Let $\mathrm{\Omega }\subseteq \mathbb{R}\times \mathbb{R},(x_1,x_2)$ and $(y_1,y_2)\in \mathrm{\Omega }$, and let $\phi :\mathrm{\Omega }\to \mathbb{R}$:

(1)

$(x_1,x_2)$ is said to be majorized by $(y_1,y_2)$ (in symbols $(x_1,x_2)\prec (y_1,y_2)$) if $max\{x_1,x_2\}$$\le max\{y_1,y_2\}$ and $x_1+x_2=y_1+y_2$.

(2)

ψ is said to be a Schur-convex function on Ω if $(x_1,x_2)\prec (y_1,y_2)$ on $\mathrm{\Omega }$ implies $\psi (x_1,x_2)\prec \psi (y_1,y_2)$, and ψ is said to be a Schur-concave function on Ω iff $-\psi$ is a Schur-convex function.

Lemma 1

([14] (p. 5)). Let $(x_1,x_2)\in \mathbb{R}\times \mathbb{R}$. Then

$$\left(\frac{x_1+x_2}{2},\frac{x_1+x_2}{2}\right)\prec (x_1,x_2).$$

Lemma 2

([14] (p. 5)). Let $\mathrm{\Omega }\subseteq \mathbb{R}\times \mathbb{R}$ be symmetric set with a nonempty interior ${\mathrm{\Omega }}^{\circ }$. $\psi :\mathrm{\Omega }\to \mathbb{R}$ is continuous on Ω and differentiable in ${\mathrm{\Omega }}^{\circ }$. Then, function ψ is Schur convex (or Schur concave, resp.) iff ψ is symmetric on Ω and

$$\left(x_1-x_2\right)\left(\frac{\partial \psi }{\partial x_1}-\frac{\partial \psi }{\partial x_2}\right)\ge 0(or\le 0,resp.)$$

holds for any $\left(x_1,x_2\right)\in {\mathrm{\Omega }}^{\circ }$.

Lemma 3

([16]). Let $\phi \left(x,w\right)$ and $\frac{\partial \phi \left(x,w\right)}{\partial w}$ be continuous on

$$D=\left\{(x,w):a\le x\le b,c\le w\le d\right\};let$$

$a\left(w\right),b\left(w\right)$ and their derivatives be continuous on $[c,d]$; $v\in [c,d]$ implies $a\left(w\right),b\left(w\right)\in [a,b]$. Then,

$$\frac{d}{dw}{\int }_{a\left(w\right)}^{b\left(w\right)}\phi (x,w)dx={\int }_{a\left(w\right)}^{b\left(w\right)}\frac{\partial \phi (x,w)}{\partial w}dx+\phi (b\left(w\right),u)b^{\prime }\left(w\right)-\phi (a\left(w\right),w)a^{\prime }\left(w\right).$$

(4)

Lemma 4.

Let $g(s,t)$ be continuous on rectangle $[a,p;a,q]$, $G\left(c,d\right)={\int }_c^d{\int }_c^{d}g(s,t)dsdt$. If $c=c\left(b\right)$ and $d=d\left(b\right)$ are differentiable with b, $a\le c\left(b\right)\le p$ and $a\le d\left(b\right)\le q$, then

$$\begin{array}{cc}\hfill \frac{\partial G}{\partial b}=& {\int }_c^{d}g(s,d)d^{\prime }\left(b\right)ds-{\int }_c^{d}g(s,c)c^{\prime }\left(b\right)ds\hfill \\ \hfill +& d^{\prime }\left(b\right){\int }_c^{d}g(d,t)dt-c^{\prime }\left(b\right){\int }_c^{d}g(c,t)dt.\hfill \end{array}$$

(5)

Proof. 

Let $\phi (s,b)={\int }_c^{d}g(s,t)dt$. Then,

$$\frac{\partial \phi (s,b)}{\partial b}=g(s,d)d^{\prime }\left(b\right)-g(s,c)c^{\prime }\left(b\right).$$

By Lemma 3, we have

$$\begin{array}{cc}\hfill \frac{\partial G}{\partial b}& =\frac{d}{db}{\int }_c^d\phi (s,b)ds\hfill \\ \hfill & ={\int }_c^d\frac{\partial \phi (s,b)}{\partial b}ds+\phi (d,b)d^{\prime }\left(b\right)-\phi (c,b)c^{\prime }\left(b\right)\hfill \\ \hfill & ={\int }_c^{d}g(s,d)d^{\prime }\left(b\right)ds-{\int }_c^{d}g(s,c)c^{\prime }\left(b\right)ds\hfill \\ \hfill & +d^{\prime }\left(b\right){\int }_c^{d}g(d,s)ds-c^{\prime }\left(b\right){\int }_c^{d}g(c,s)ds.\hfill \end{array}$$

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Remark 1.

In passing, it is pointed out that (9) in Lemma 5 of reference [2] is incorrect and should be replaced by (4) of this paper.

Lemma 5.

Let I be an interval with nonempty interior on $\mathbb{R}$ and $g(s,t)$ be a continuous function on $I\times I$. For $(u,v)\in I\times I,u\ne v$, let $G\left(u,v\right)={\int }_u^v{\int }_u^{v}g(s,t)dsdt$. Then,

$$\frac{\partial G}{\partial v}={\int }_u^{v}g(s,v)ds+{\int }_u^{v}g(v,t)dt,$$

(6)

$$\frac{\partial G}{\partial u}=-\left({\int }_u^{v}g(s,u)ds+{\int }_u^{v}g(u,t)dt\right).$$

(7)

Proof. 

By taking $c\left(b\right)=a$ and $d\left(b\right)=b$, we have $c^{\prime }\left(b\right)=0$ and $d^{\prime }\left(b\right)=1$. By (5) in Lemma 4, we obtain (6).

Notice that $G\left(u,v\right)={\int }_v^u{\int }_v^{u}g(s,t)dsdt$; from (5), we have

$$\begin{array}{c}\hfill \frac{\partial G}{\partial u}={\int }_v^{u}g(s,u)ds+{\int }_v^{u}g(u,t)dt=-\left({\int }_u^{v}g(s,u)ds+{\int }_u^{v}g(u,t)dt\right).\end{array}$$

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Lemma 6

([14] (p. 38, Proposition 4.3) and [15] (p. 644, B.3.d)). Let $\mathrm{\Omega }\subset \mathbb{R}\times \mathbb{R}$ be an open convex set and let $\psi (x,y):\mathrm{\Omega }\to \mathbb{R}$ be twice differentiable. Then, ψ is convex on Ω iff the Hessian matrix

$$H(x,y)=\left(\begin{array}{cc}\frac{{\partial }^2\psi }{\partial x\partial x}& \frac{{\partial }^2\psi }{\partial x\partial y}\\ \frac{{\partial }^2\psi }{\partial y\partial x}& \frac{{\partial }^2\psi }{\partial y\partial y}\end{array}\right)$$

is non-negative definite on Ω. If $H\left(\mathbf{x}\right)$ is positive definite on Ω, then ψ is strictly convex on Ω.

3. Proofs of Main Results

Proof of Theorem 2.

Let $g(s,t)$ be convex on $I\times I$. $G(u,v)$ is evidently symmetric. By Lemma 5, we have

$$\frac{\partial G(u,v)}{\partial v}=\frac{-2}{{(v-u)}^3}{\int }_u^v{\int }_u^{v}g(s,t)dsdt+\frac{1}{{(v-u)}^2}\left({\int }_u^{v}g(s,v)ds+{\int }_u^{v}g(v,t)dt\right).$$

$$\frac{\partial G(u,v)}{\partial u}=\frac{2}{{(v-u)}^3}{\int }_u^v{\int }_u^{v}g(s,t)dsdt-\frac{1}{{(v-u)}^2}\left({\int }_u^{v}g(s,u)ds+{\int }_u^{v}g(u,t)dt\right).$$

$$\begin{array}{cc}\hfill \mathrm{\Delta }:=& (v-u)\left(\frac{\partial G(u,v)}{\partial v}-\frac{\partial G(u,v)}{\partial u}\right)=-\frac{4}{{(v-u)}^2}{\int }_u^v{\int }_u^{v}g(s,t)dsdt\hfill \\ \hfill & +\frac{1}{v-u}{\int }_u^v(g(s,v)+g(s,u))ds+\frac{1}{v-u}{\int }_u^v(g(u,t)+g(v,t))dt\hfill \end{array}$$

By Hadamards inequality, we have

$$\begin{array}{cc}\hfill & \frac{2}{{(v-u)}^2}{\int }_u^v{\int }_u^{v}g(s,t)dsdt=\frac{2}{v-u}{\int }_u^v\left(\frac{1}{v-u}{\int }_u^{v}g(s,t)ds\right)dt\hfill \\ \hfill \le & \frac{2}{v-ua}{\int }_u^v\frac{g(u,t)+g(v,t)}{2}dt=\frac{1}{v-u}{\int }_u^{v}a(g(u,t)+g(v,t))dt\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \frac{2}{{(v-u)}^2}{\int }_u^v{\int }_u^{v}g(s,t)dsdt=\frac{2}{v-u}{\int }_u^v\left(\frac{1}{v-u}{\int }_u^{v}g(s,t)dt\right)ds\hfill \\ \hfill \le & \frac{2}{v-u}{\int }_u^v\frac{g(s,u)+g(s,v)}{2}ds=\frac{1}{v-u}{\int }_u^v(g(s,u)+g(s,v))ds.\hfill \end{array}$$

Moreover, we have

$$\begin{array}{cc}\hfill & \frac{4}{{(v-u)}^2}{\int }_u^v{\int }_u^{v}g(s,t)dsdt\hfill \\ \hfill \le & \frac{1}{v-u}{\int }_u^v(g(s,v)+g(s,u))ds+\frac{1}{v-u}{\int }_u^v(g(u,t)+g(v,t))dt.\hfill \end{array}$$

Therefore, $\mathrm{\Delta }\ge 0$, so $G(u,v)$ is Schur-convex on $I\times I$.

When $g(s,t)$ is a concave function on $I\times I$, it can be proved with similar methods. □

4. Application on Binary Mean

Theorem 3.

Let $c>0$ and $d>0$. If $c\ne d,0<s<1$, then

$$A(d,c)\ge S_{s+1}^s(d,c)S_s^{s-1}(d,c)\ge \frac{{(c+d)}^{2s-1}}{s(s+1)},$$

(8)

where $A(d,c)=\frac{c+d}{2}$ and $S_s(d,c)={\left(\frac{d^s-c^s}{s(d-c)}\right)}^{\frac{1}{s-1}}$ are the arithmetic mean and the s-order Stolarsky mean of positive numbers c and d, respectively.

Proof. 

Let $x>0,y>0$ and $0<s<1$. From Theorem 4 in the reference [17], we know that $g(x,y)=x^s{y}^{1-s}$ is concave on $(0,+\infty )\times (0,+\infty )$. For $c\ne d$, by Theorem 2, from $(\frac{d+c}{2},\frac{d+c}{2})\prec (c,d)\prec (d+c,0)$, it follows that

$$\begin{array}{cc}\hfill G(d+c,0)=& \frac{1}{{(d+c-0)}^2}{\int }_c^d{\int }_0^{d+c}{x}^s{y}^{1-s}dxdy\hfill \\ \hfill =& \frac{1}{{(d+c)}^2}{\int }_0^{d+c}{x}^{s}dx{\int }_0^{d+c}{y}^{1-s}dy\hfill \\ \hfill =& \frac{1}{{(d+c)}^2}\frac{{(c+d)}^{s+1}}{s+1}\frac{{(c+d)}^s}{s}=\frac{{(c+d)}^{2s-1}}{s(s+1)}\hfill \\ \hfill \le & G(c,d)=\frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^d{x}^s{y}^{1-s}dxdy\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}{\int }_c^d{x}^{s}dx{\int }_c^d{y}^{1-s}dy\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}\frac{d^{s+1}-c^{s+1}}{s+1}\frac{d^s-c^s}{s}\hfill \\ \hfill \le & G\left(\frac{d+c}{2},\frac{d+c}{2}\right)=\frac{d+c}{2},\hfill \end{array}$$

That is, we obtain the following.

$$\frac{{(c+d)}^{2s-1}}{s(s+1)}\le S_{s+1}^s(d,c)S_s^{s-1}(d,c)=\frac{d^{s+1}-c^{s+1}}{(s+1)(d-c)}·\frac{d^s-c^s}{s(d-c)}\le \frac{d+c}{2}=A(d,c).$$

□

Theorem 4.

Let $c>0,d>0$. Then,

$$log{\left(\frac{A(d,c)}{B(d,c)}\right)}^2\ge {\left(\frac{c-d}{d+c}\right)}^2,$$

(9)

where $B(d,c)=\sqrt{dc}$ is the geometric mean of of positive numbers c and d.

Proof. 

From reference [17], we know that the function $g(x,y)=\frac{1}{{(x+y)}^2}$ is convex on $(0,+\infty )\times (0,+\infty )$. For $c>0,d>0$ and $d\ne c$, by Theorem 2, from $(\frac{d+c}{2},\frac{d+c}{2})\prec (d,c)$, it follows that

$$\begin{array}{cc}\hfill G(c,d)=& \frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^d\frac{1}{{(x+y)}^2}dxdy\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}{\int }_c^d\left(\frac{1}{c+y}-\frac{1}{d+y}\right)dy\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}[(log(d+c)-log\left(2c\right))-(log\left(2d\right)-log(d+c))]\hfill \\ \hfill \ge & G\left(\frac{d+c}{2},\frac{d+c}{2}\right)=\frac{1}{{(d+c)}^2},\hfill \end{array}$$

That is, we obtain the following.

$$log{\left(\frac{A(d,c)}{B(d,c)}\right)}^2=log\frac{{(d+c)}^2}{4dc}\ge {\left(\frac{c-d}{d+c}\right)}^2.$$

□

Theorem 5.

Let $c>0,d>0$. Then,

$$H_e(c^2,d^2)\ge A^2(c,d),$$

(10)

where $H_e(c,d)=\frac{c+\sqrt{cd}+d}{3}$ is the Heronian mean of positive numbers c and d.

Proof. 

From reference [18], we know that the function of two variables

$$\psi (x,y)=\frac{x^2}{2r^2}+\frac{y^2}{2s^2}$$

is a convex function on $(0,+\infty )\times (0,+\infty )$, where $s>0$ and $r>0$. For $d>0,c>0$, and $c\ne d$, by Theorem 2, from $(\frac{d+c}{2},\frac{d+c}{2})\prec (d,c)$, it follows that

$$\begin{array}{cc}\hfill G(c,d)=& \frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^d\left(\frac{x^2}{2r^2}+\frac{y^2}{2s^2}\right)dxdy\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}{\int }_c^d\left(\frac{d^3-c^3}{6r^2}+\frac{y^2(d-c)}{2s^2}\right)dy\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}\left(\frac{(d^3-c^3)(d-c)}{6r^2}+\frac{(d^3-c^3)(d-c)}{6s^2}\right)\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}·\frac{(d^3-c^3)(d-c)}{6}\left(\frac{1}{r^2}+\frac{1}{s^2}\right)\hfill \\ \hfill \ge & G\left(\frac{d+c}{2},\frac{d+c}{2}\right)=\frac{{(c+d)}^2}{8}\left(\frac{1}{r^2}+\frac{1}{s^2}\right),\hfill \end{array}$$

namely

$$H_e(c^2,d^2)=\frac{c^2+cd+d^2}{3}=\frac{(d^3-c^3)}{3(d-c)}\ge \frac{{(d+c)}^2}{4}=A^2(d,c).$$

□

Theorem 6.

Let $c>0,d>0$. We have

$$H_e(c^2,d^2)\ge L(d,c)A(d,c),$$

(11)

where $L(d,c)=\frac{d-c}{logd-logc}$ is the logarithmic mean of positive numbers c and d.

Proof. 

Let $g(x,y)=y^2{x}^{-1},x>0,y>0$. Then,

$$g_{xx}=2x^{-3}{y}^2,g_{xy}=-2x^{-2}y=g_{yx},g_{yy}=2x^{-1}.$$

The Hesse matrix of $g(x,y)$ is

$$H=\left(\begin{array}{cc}2x^{-3}{y}^2& -2x^{-2}y\\ -2x^{-2}y& 2x^{-1}\end{array}\right).$$

$$det(H-\lambda I)=det\left(\begin{array}{cc}2x^{-3}{y}^2-\lambda & -2x^{-2}y\\ -2x^{-2}y& 2x^{-1}-\lambda \end{array}\right)=0$$

$$\Rightarrow \lambda (\lambda -2x^{-3}{y}^2-2x^{-1})=0\Rightarrow {\lambda }_1=0,{\lambda }_2=2x^{-3}{y}^2+2x^{-1}>0.$$

Therefore, matrix H is positive semidefinite, so it is known that $g(x,y)$ is a convex function on $(0,+\infty )\times (0,+\infty )$. For $d>0,c>0$ and $d\ne c$, by Theorem 2, from $(\frac{d+c}{2},\frac{d+c}{2})\prec (d,c)$, it follows that

$$\begin{array}{cc}\hfill G(c,d)=& \frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^d{y}^2{x}^{-1}dxdy\hfill \\ \hfill =& \frac{logd-logc}{d-c}·\frac{d^2+cd+c^2}{3}\ge \frac{{\left(\frac{d+c}{2}\right)}^2}{\frac{c+c}{2}}=\frac{d+c}{2},\hfill \end{array}$$

which is

$$H_e(c^2,d^2)\ge L(d,c)A(d,c).$$

□

Theorem 7.

Let $d>0,c>0,d\ne c$. Then

$$\tilde{E}(d,c)\le A(d,c)e^{(d+c)}{\left(\frac{d-c}{e^d-e^c}\right)}^2\le A(d,c),$$

(12)

where

$$\tilde{E}(d,c)=\left\{\begin{array}{c}\frac{c{e}^d-d{e}^c}{e^d-e^c}+1,d,c\in I,d\ne c\hfill \\ c,c=d\hfill \end{array}\right.$$

is exponent type mean of positive numbers c and d (see [13] (p. 134)).

Proof. 

Let $g(x,y)=x{e}^{-(x+y)},y>0,x>0$. From reference [19], we know that function $g(x,y)$ is convex on $\mathbb{R}\times \mathbb{R}$. For $d>0,c>0$, and $d\ne c$ by Theorem 2 from $(\frac{d+c}{2},\frac{d+c}{2})\prec (d,c)$, it follows that

$$\begin{array}{cc}\hfill G(c,d)=& \frac{1}{{(c-d)}^2}{\int }_c^d{\int }_c^{d}x{e}^{-x-y}dxdy\hfill \\ \hfill =& \frac{1}{{(c-d)}^2}{\int }_c^{d}x{e}^{-x}dx{\int }_c^d{e}^{-y}dy\hfill \\ \hfill =& \frac{1}{{(c-d)}^2}\left(\frac{c+1}{e^c}-\frac{d+1}{e^d}\right)·\left(\frac{1}{e^c}-\frac{1}{e^d}\right)\hfill \\ \hfill =& \frac{1}{{(d-c)}^2}\frac{(c{e}^d-d{e}^c)+(e^d-e^c)}{e^{(c+d)}}·\frac{e^d-e^c}{e^{(c+d)}}\hfill \\ \hfill \le & G\left(\frac{d+c}{2},\frac{d+c}{2}\right)=\frac{c+d}{2}\frac{1}{e^{(d+c)}},\hfill \end{array}$$

which is

$$\frac{c{e}^d-d{e}^c}{e^d-e^c}+1\le \frac{d+c}{2}{e}^{(d+c)}{\left(\frac{d-c}{e^d-e^c}\right)}^2.$$

For the rest, we only need to prove that

$$e^{(c+d)}{\left(\frac{d-c}{e^d-e^c}\right)}^2\le 1.$$

(13)

We write $e^d=u$ and $e^c=v$; then, the above inequality is equivalent to the well-known log-geometric mean inequality.

$$L(v,u)=\frac{v-u}{logv-logu}\ge \sqrt{vu}=B(v,u).$$

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