Schur quadratic concavity of the elliptic Neuman mean and its application

Abstract

For $x,y>0$ and $k\in [0,1]$, we prove that the elliptic Neuman mean $N_k(x,y)$ is strictly Schur quadratically concave on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if and only if $k\in [\sqrt{2}/2,1]$. As an application, the bounds for elliptic Neuman mean $N_k(x,y)$ in terms of the quadratic mean $Q(x,y)=\sqrt{(x^2+y^2)/2}$ are presented.

MSC:26B25, 26E60.

1 Introduction

Let $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$ and $k\in [0,1]$. Then the elliptic Neuman mean $N_k(x,y)$, see [1], is defined by

$$N_k(x,y)=\{\begin{array}{ll}\frac{\sqrt{y^2-x^2}}{c{n}^{-1}(x/y,k)},& x<y,\\ x,& x=y,\\ \frac{\sqrt{x^2-y^2}}{n{c}^{-1}(x/y,k)},& y<x,\end{array}$$

(1.1)

where $c{n}^{-1}(x,k)={\int }_x^1\frac{du}{\sqrt{(1-u^2)({k^{\prime }}^2+k^2{u}^2)}}$ and $n{c}^{-1}(x,k)={\int }_1^x\frac{du}{\sqrt{(u^2-1)(k^2+{k^{\prime }}^2{u}^2)}}$ are the inverse functions of Jacobian elliptic functions cn and nc, see [2, 3], respectively, and $k^{\prime }=\sqrt{1-k^2}$. In particular, $c{n}^{-1}(0,k)=\mathcal{K}(k)={\int }_0^{\pi /2}\frac{dt}{\sqrt{1-k^2{sin}^{2}t}}$ is the well-known complete elliptic integral of the first kind.

In [1] Neuman proved that $N_k(x,y)$ is symmetric and homogeneous on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$, and strictly decreasing with respect to $k\in [0,1]$ for fixed $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$ with $x\ne y$. In this context let us note that if a mean is homogeneous, then the order of its homogeneity must be 1; see [4].

Let us recall the notion of Schur quadratic convexity (concavity) [5–7] for a real-valued function on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$.

A real-valued function $f:(0,\mathrm{\infty })\times (0,\mathrm{\infty })\to \mathbb{R}$ is said to be strictly Schur quadratically convex on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if $f(x_1,x_2)<f(y_1,y_2)$ for each pair of 2-tuples $(x_1,x_2),(y_1,y_2)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$ with $max\{x_1,x_2\}<max\{y_1,y_2\}$ and $x_1^2+x_2^2=y_1^2+y_2^2$. f is said to be strictly Schur quadratically concave if −f is strictly Schur quadratically convex.

The main purpose of this paper is to present the range of k such that the elliptic Neuman mean $N_k(x,y)$ is strictly Schur quadratically concave on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$. As an application, an inequality between the elliptic Neuman mean $N_k(x,y)$ and the quadratic mean $Q(x,y)=\sqrt{(x^2+y^2)/2}$ is also given.

2 Two lemmas

In order to prove our main results we need two lemmas, which we present in this section.

Lemma 2.1 (See [[5], Corollary 2.1], [[6], Corollary 1], [[7], Corollary 1])

Suppose that $f:(0,\mathrm{\infty })\times (0,\mathrm{\infty })\to (0,\mathrm{\infty })$ is a continuous symmetric function. If f is differentiable in $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$, then f is strictly Schur quadratically convex on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if and only if

$$(x-y)(y\frac{\partial f(x,y)}{\partial x}-x\frac{\partial f(x,y)}{\partial y})>0$$

(2.1)

for all $x,y\in (0,\mathrm{\infty })$ with $x\ne y$, and f is strictly Schur quadratically concave on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if and only if inequality (2.1) is reversed.

Lemma 2.2 Let $t\in (0,1)$, $k\in [0,1]$, and

$$f_k(t)=c{n}^{-1}(t,k)-\frac{(1+t^2)\sqrt{1-t^2}}{2t\sqrt{1-k^2+k^2{t}^2}}.$$

(2.2)

Then $f_k(t)<0$ for all $t\in (0,1)$ if and only if $\sqrt{2}/2\le k\le 1$, and there exists $\lambda =\lambda (k)\in (0,1)$ such that $f_k(t)<0$ for $t\in (0,\lambda )$ and $f_k(t)>0$ for $t\in (\lambda ,1)$ if $k\in [0,\sqrt{2}/2)$.

Proof We distinguish for the proof two cases.

Case 1. $k=1$. Then from (2.2) one has

$$\begin{array}{c}\begin{array}{rl}{f}_1(t)& =c{n}^{-1}(t,1)-\frac{(1+t^2)\sqrt{1-t^2}}{2t^2}\\ ={cosh}^{-1}\left(\frac{1}{t}\right)-\frac{(1+t^2)\sqrt{1-t^2}}{2t^2}\\ =log(1+\sqrt{1-t^2})-logt-\frac{(1+t^2)\sqrt{1-t^2}}{2t^2},\end{array}\hfill \\ f_1\left(1^{-}\right)=0,\hfill \end{array}$$

(2.3)

$$f^{\prime }(t)=\frac{(1-t^2)(2-t^2)}{2t^3\sqrt{1-t^2}}>0$$

(2.4)

for all $t\in (0,1)$. (Here and in the sequel, $f(t^{-})$ and $f(t^{+})$ denote, respectively, the left and right limit of f at t.)

From (2.3) and (2.4) we clearly see that $f_1(t)<0$ for all $t\in (0,1)$.

Case 2. $0\le k<1$. Then (2.2) leads to

$$f_k\left(0^{+}\right)=-\mathrm{\infty },$$

(2.5)

$$f_k\left(1^{-}\right)=0,$$

(2.6)

$$\begin{array}{rl}{f}_k^{\prime }(t)& =-\frac{1}{\sqrt{(1-t^2)(1-k^2+k^2{t}^2)}}-\frac{-k^2{t}^6+(3k^2-2)t^4+(1-3k^2)t^2+k^2-1}{2t^2\sqrt{1-t^2}{(1-k^2+k^2{t}^2)}^{3/2}}\\ =-\frac{-k^2{t}^6+(5k^2-2)t^4+(3-5k^2)t^2+k^2-1}{2t^2\sqrt{1-t^2}{(1-k^2+k^2{t}^2)}^{3/2}}\\ =-\frac{\sqrt{1-t^2}[k^2{t}^4+(2-4k^2)t^2+k^2-1]}{2t^2{(1-k^2+k^2{t}^2)}^{3/2}}.\end{array}$$

(2.7)

Let

$$g_k(t)=k^2{t}^4+(2-4k^2)t^2+k^2-1.$$

(2.8)

Then simple computations lead to

$$g_k(0)=k^2-1<0,$$

(2.9)

$$g_k(1)=2(\frac{\sqrt{2}}{2}-k)(\frac{\sqrt{2}}{2}+k),$$

(2.10)

$$g_k^{\prime }(t)=4k^2{t}^3+4(1-2k^2)t,$$

(2.11)

$$g_k^{\prime }(0)=0,$$

(2.12)

$$g_k^{\prime }(1)=4(1-k)(1+k)>0,$$

(2.13)

$$g_k^{″}(t)=12k^2{t}^2+4(1-2k^2),$$

(2.14)

$$g_k^{″}(0)=8(\frac{\sqrt{2}}{2}-k)(\frac{\sqrt{2}}{2}+k),$$

(2.15)

$$g_k^{″}(1)=4(1+k^2)>0.$$

(2.16)

We distinguish for the proof three subcases.

Subcase 2.1. $k=\sqrt{2}/2$. Then (2.7) leads to the conclusion that

$$f_{\sqrt{2}/2}^{\prime }(t)=\frac{\sqrt{2}\sqrt{1-t^2}(1-t^4)}{2t^2{(1+t^2)}^{3/2}}>0$$

(2.17)

for all $t\in (0,1)$.

Therefore, $f_{\sqrt{2}/2}(t)<0$ for all $t\in (0,1)$ follows from (2.6) and (2.17).

Subcase 2.2. $\sqrt{2}/2<k<1$. Then (2.10) and (2.15) lead to

$$g_k(1)<0,$$

(2.18)

$$g_k^{″}(0)<0.$$

(2.19)

It follows from (2.14) that $g_k^{″}$ is strictly increasing on $(0,1)$, then (2.16) and (2.19) lead to the conclusion that there exists ${\lambda }_0\in (0,1)$ such that $g_k^{\prime }$ is strictly decreasing on $(0,{\lambda }_0]$ and strictly increasing on $[{\lambda }_0,1)$.

From (2.12) and (2.13) together with the piecewise monotonicity of $g_k^{\prime }$ we clearly see that there exists ${\lambda }_1\in ({\lambda }_0,1)$ such that $g_k$ is strictly decreasing on $(0,{\lambda }_1]$ and strictly increasing on $[{\lambda }_1,1)$. Then (2.9) and (2.18) lead to the conclusion that

$$g_k(t)<0$$

(2.20)

for all $t\in (0,1)$.

It follows from (2.7) and (2.8) together with (2.20) that $f_k$ is strictly increasing on $(0,1)$. Therefore, $f_k(t)<0$ for all $t\in (0,1)$ follows easily from (2.6) and the monotonicity of $f_k$.

Subcase 2.3. $0\le k<\sqrt{2}/2$. Then from (2.10) and (2.11) we know that $g_k$ is strictly increasing on $(0,1)$ and

$$g_k(1)>0.$$

(2.21)

It follows from (2.9) and (2.21) together with the monotonicity of $g_k$ that there exists ${\mu }_0\in (0,1)$ such that $g_k(t)>0$ for $t\in (0,{\mu }_0)$ and $g_k(t)<0$ for $t\in ({\mu }_0,1)$. Then (2.7) and (2.8) lead to the conclusion that $f_k$ is strictly increasing on $(0,{\mu }_0]$ and strictly decreasing on $[{\mu }_0,1)$. Therefore, there exists $\lambda =\lambda (k)\in (0,{\mu }_0)\subset (0,1)$ such that $f_k(t)<0$ for $t\in (0,\lambda )$ and $f_k(t)>0$ for $t\in (\lambda ,1)$ follows from (2.5) and (2.6) together with the piecewise monotonicity of $f_k$. □

3 Main results

Theorem 3.1 The elliptic Neuman mean $N_k(x,y)$ is strictly Schur quadratically concave on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if and only if $k\in [\sqrt{2}/2,1]$, and $N_k(x,y)$ is not Schur quadratically convex on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if $k\in [0,\sqrt{2}/2)$.

Proof Since $N_k(x,y)$ is symmetric and homogeneous of degree 1, without loss of generality, we assume that $x<y$. Let $t=x/y\in (0,1)$, then

$$N_k(x,y)=y{N}_k(t,1),\frac{\partial t}{\partial y}=-\frac{x}{y^2},\frac{\partial t}{\partial x}=\frac{1}{y},$$

(3.1)

$$\frac{\partial N_k(x,y)}{\partial y}=N_k(t,1)-t\frac{d{N}_k(t,1)}{dt},\frac{\partial N_k(x,y)}{\partial x}=\frac{d{N}_k(t,1)}{dt}.$$

(3.2)

Note that

$$\frac{d{N}_k(t,1)}{dt}=-\frac{t}{\sqrt{1-t^2}c{n}^{-1}(t,k)}+\frac{1}{{(c{n}^{-1}(t,k))}^2\sqrt{1-k^2+k^2{t}^2}}.$$

(3.3)

It follows from (1.1) and (3.2) together with (3.3) that

$$\begin{array}{r}(y-x)(x\frac{\partial N_k(x,y)}{\partial y}-y\frac{\partial N_k(x,y)}{\partial x})\\ =x(y-x)[N_k(t,1)-(t+\frac{1}{t})\frac{d{N}_k(t,1)}{dt}]\\ =\frac{2x(y-x)}{\sqrt{1-t^2}{(c{n}^{-1}(t,k))}^2}(c{n}^{-1}(t,k)-\frac{(1+t^2)\sqrt{1-t^2}}{2t\sqrt{1-k^2+k^2{t}^2}})\\ =\frac{2x(y-x)}{\sqrt{1-t^2}{(c{n}^{-1}(t,k))}^2}{f}_k(t),\end{array}$$

(3.4)

where $f_k(t)$ is defined as in Lemma 2.2.

Therefore, Theorem 3.1 follows easily from Lemmas 2.1 and 2.2 together with (3.4). □

Theorem 3.2 The elliptic Neuman mean $N_k(x,y)$ is strictly Schur quadratically concave (or convex, respectively) on $(0,\mathrm{\infty })\times (0,\mathrm{\infty })$ if and only if the function $N_k(t,1)/Q(t,1)$ is strictly increasing (or decreasing, respectively) in $(0,1)$, where $Q(x,y)=\sqrt{(x^2+y^2)/2}$ is the quadratic mean of x and y.

Proof Without loss of generality, we assume that $x<y$. Let $t=x/y\in (0,1)$, then from (3.1) and (3.2) together with (3.4) we get

$$\begin{array}{rl}\frac{d(N_k(t,1)/Q(t,1))}{dt}& =-\frac{\sqrt{2}t}{{(t^2+1)}^{3/2}}(N_k(t,1)-(t+\frac{1}{t})\frac{d{N}_k(t,1)}{dt})\\ =-\frac{\sqrt{2}}{y(y-x){(t^2+1)}^{3/2}}(y-x)(x\frac{\partial N_k(x,y)}{\partial y}-y\frac{\partial N_k(x,y)}{\partial x}).\end{array}$$

(3.5)

Therefore, Theorem 3.2 follows easily from Lemma 2.1 and (3.5). □

Theorem 3.3 The inequalities

$$Q(x,y)>N_{\sqrt{2}/2}(x,y)$$

(3.6)

and

$$Q(x,y)<\frac{\sqrt{2}\mathcal{K}(k)}{2}{N}_k(x,y)$$

(3.7)

hold for all $x,y>0$ with $x\ne y$, and $k\in [0,1]$, and $N_{\sqrt{2}/2}(x,y)$ is the best possible lower elliptic Neuman mean bound for the quadratic mean $Q(x,y)$.

Proof Without loss of generality, we assume that $y>x>0$. Let $t=x/y\in (0,1)$ and $L_k(t)=N_k(t,1)/Q(t,1)$. Then

$$L_k(t)=\frac{N_k(x,y)}{Q(x,y)},$$

(3.8)

$$L_k(0)=\frac{\sqrt{2}}{\mathcal{K}(k)},$$

(3.9)

$$L_k(1)=1.$$

(3.10)

We distinguish for the proof two cases.

Case 1. $k\in [\sqrt{2}/2,1]$. Then from Theorems 3.1 and 3.2 we clearly see that $L_k$ is strictly increasing on $(0,1)$. Then (3.8)-(3.10) lead to the conclusion that

$$\frac{\sqrt{2}}{\mathcal{K}(k)}<\frac{N_k(x,y)}{Q(x,y)}<1.$$

(3.11)

In particular, for $k=\sqrt{2}/2$ we have

$$\frac{N_{\sqrt{2}/2}(x,y)}{Q(x,y)}<1.$$

(3.12)

Therefore, inequalities (3.6) and (3.7) follow from (3.11) and (3.12).

Case 2. $k\in [0,\sqrt{2}/2)$. Then (3.4) and (3.5) together with the Subcase 2.3 in Lemma 2.2 lead to the conclusion that there exists $\lambda =\lambda (k)\in (0,1)$ such that $L_k^{\prime }(t)>0$ for $t\in (0,\lambda )$ and $L_k^{\prime }(t)<0$ for $t\in (\lambda ,1)$, hence $L_k$ is strictly increasing on $(0,\lambda ]$ and strictly decreasing on $[\lambda ,1)$. Therefore, $N_k(x,y)>Q(x,y)$ for all $x,y>0$ with $x/y\in (\lambda ,1)$ follows from (3.8) and (3.10) together with the monotonicity of $L_k$ on $[\lambda ,1)$, and the optimality of inequality (3.6) follows.

Note that

$$L_k(0)=\frac{\sqrt{2}}{\mathcal{K}(k)}<\frac{\sqrt{2}}{\mathcal{K}(0)}=\frac{2\sqrt{2}}{\pi }=0.90031\cdots <1.$$

(3.13)

From (3.8), (3.9), (3.13), and the piecewise monotonicity of $L_k$ we clearly see that

$$\frac{N_k(x,y)}{Q(x,y)}=L_k(t)>L_k(0)=\frac{\sqrt{2}}{\mathcal{K}(k)}.$$

(3.14)

Therefore, inequality (3.7) follows from (3.14). □