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Schur quadratic concavity of the elliptic Neuman mean and its application

Abstract

For x,y>0 and k[0,1], we prove that the elliptic Neuman mean N k (x,y) is strictly Schur quadratically concave on (0,)×(0,) if and only if k[ 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)= ( x 2 + y 2 ) / 2 are presented.

MSC:26B25, 26E60.

1 Introduction

Let (x,y)(0,)×(0,) and k[0,1]. Then the elliptic Neuman mean N k (x,y), see [1], is defined by

N k (x,y)= { y 2 x 2 c n 1 ( x / y , k ) , x < y , x , x = y , x 2 y 2 n c 1 ( x / y , k ) , y < x ,
(1.1)

where c n 1 (x,k)= x 1 d u ( 1 u 2 ) ( k 2 + k 2 u 2 ) and n c 1 (x,k)= 1 x d u ( u 2 1 ) ( k 2 + k 2 u 2 ) are the inverse functions of Jacobian elliptic functions cn and nc, see [2, 3], respectively, and k = 1 k 2 . In particular, c n 1 (0,k)=K(k)= 0 π / 2 d t 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,)×(0,), and strictly decreasing with respect to k[0,1] for fixed (x,y)(0,)×(0,) with xy. 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) [57] for a real-valued function on (0,)×(0,).

A real-valued function f:(0,)×(0,)R is said to be strictly Schur quadratically convex on (0,)×(0,) 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 )(0,)×(0,) 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,)×(0,). As an application, an inequality between the elliptic Neuman mean N k (x,y) and the quadratic mean Q(x,y)= ( 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,)×(0,)(0,) is a continuous symmetric function. If f is differentiable in (0,)×(0,), then f is strictly Schur quadratically convex on (0,)×(0,) if and only if

(xy) ( y f ( x , y ) x x f ( x , y ) y ) >0
(2.1)

for all x,y(0,) with xy, and f is strictly Schur quadratically concave on (0,)×(0,) if and only if inequality (2.1) is reversed.

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

f k (t)=c n 1 (t,k) ( 1 + t 2 ) 1 t 2 2 t 1 k 2 + k 2 t 2 .
(2.2)

Then f k (t)<0 for all t(0,1) if and only if 2 /2k1, and there exists λ=λ(k)(0,1) such that f k (t)<0 for t(0,λ) and f k (t)>0 for t(λ,1) if k[0, 2 /2).

Proof We distinguish for the proof two cases.

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

f 1 ( t ) = c n 1 ( t , 1 ) ( 1 + t 2 ) 1 t 2 2 t 2 = cosh 1 ( 1 t ) ( 1 + t 2 ) 1 t 2 2 t 2 = log ( 1 + 1 t 2 ) log t ( 1 + t 2 ) 1 t 2 2 t 2 , f 1 ( 1 ) = 0 ,
(2.3)
f (t)= ( 1 t 2 ) ( 2 t 2 ) 2 t 3 1 t 2 >0
(2.4)

for all t(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(0,1).

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

f k ( 0 + ) =,
(2.5)
f k ( 1 ) =0,
(2.6)
f k ( t ) = 1 ( 1 t 2 ) ( 1 k 2 + k 2 t 2 ) k 2 t 6 + ( 3 k 2 2 ) t 4 + ( 1 3 k 2 ) t 2 + k 2 1 2 t 2 1 t 2 ( 1 k 2 + k 2 t 2 ) 3 / 2 = k 2 t 6 + ( 5 k 2 2 ) t 4 + ( 3 5 k 2 ) t 2 + k 2 1 2 t 2 1 t 2 ( 1 k 2 + k 2 t 2 ) 3 / 2 = 1 t 2 [ k 2 t 4 + ( 2 4 k 2 ) t 2 + k 2 1 ] 2 t 2 ( 1 k 2 + k 2 t 2 ) 3 / 2 .
(2.7)

Let

g k (t)= k 2 t 4 + ( 2 4 k 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 ( 2 2 k ) ( 2 2 + k ) ,
(2.10)
g k (t)=4 k 2 t 3 +4 ( 1 2 k 2 ) t,
(2.11)
g k (0)=0,
(2.12)
g k (1)=4(1k)(1+k)>0,
(2.13)
g k (t)=12 k 2 t 2 +4 ( 1 2 k 2 ) ,
(2.14)
g k (0)=8 ( 2 2 k ) ( 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= 2 /2. Then (2.7) leads to the conclusion that

f 2 / 2 (t)= 2 1 t 2 ( 1 t 4 ) 2 t 2 ( 1 + t 2 ) 3 / 2 >0
(2.17)

for all t(0,1).

Therefore, f 2 / 2 (t)<0 for all t(0,1) follows from (2.6) and (2.17).

Subcase 2.2. 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 λ 0 (0,1) such that g k is strictly decreasing on (0, λ 0 ] and strictly increasing on [ λ 0 ,1).

From (2.12) and (2.13) together with the piecewise monotonicity of g k we clearly see that there exists λ 1 ( λ 0 ,1) such that g k is strictly decreasing on (0, λ 1 ] and strictly increasing on [ λ 1 ,1). Then (2.9) and (2.18) lead to the conclusion that

g k (t)<0
(2.20)

for all t(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(0,1) follows easily from (2.6) and the monotonicity of f k .

Subcase 2.3. 0k< 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 μ 0 (0,1) such that g k (t)>0 for t(0, μ 0 ) and g k (t)<0 for t( μ 0 ,1). Then (2.7) and (2.8) lead to the conclusion that f k is strictly increasing on (0, μ 0 ] and strictly decreasing on [ μ 0 ,1). Therefore, there exists λ=λ(k)(0, μ 0 )(0,1) such that f k (t)<0 for t(0,λ) and f k (t)>0 for t(λ,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,)×(0,) if and only if k[ 2 /2,1], and N k (x,y) is not Schur quadratically convex on (0,)×(0,) if k[0, 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(0,1), then

N k (x,y)=y N k (t,1), t y = x y 2 , t x = 1 y ,
(3.1)
N k ( x , y ) y = N k (t,1)t d N k ( t , 1 ) d t , N k ( x , y ) x = d N k ( t , 1 ) d t .
(3.2)

Note that

d N k ( t , 1 ) d t = t 1 t 2 c n 1 ( t , k ) + 1 ( c n 1 ( t , k ) ) 2 1 k 2 + k 2 t 2 .
(3.3)

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

( y x ) ( x N k ( x , y ) y y N k ( x , y ) x ) = x ( y x ) [ N k ( t , 1 ) ( t + 1 t ) d N k ( t , 1 ) d t ] = 2 x ( y x ) 1 t 2 ( c n 1 ( t , k ) ) 2 ( c n 1 ( t , k ) ( 1 + t 2 ) 1 t 2 2 t 1 k 2 + k 2 t 2 ) = 2 x ( y x ) 1 t 2 ( c n 1 ( t , k ) ) 2 f k ( t ) ,
(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,)×(0,) 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)= ( 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(0,1), then from (3.1) and (3.2) together with (3.4) we get

d ( N k ( t , 1 ) / Q ( t , 1 ) ) d t = 2 t ( t 2 + 1 ) 3 / 2 ( N k ( t , 1 ) ( t + 1 t ) d N k ( t , 1 ) d t ) = 2 y ( y x ) ( t 2 + 1 ) 3 / 2 ( y x ) ( x N k ( x , y ) y y N k ( x , y ) x ) .
(3.5)

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

Theorem 3.3 The inequalities

Q(x,y)> N 2 / 2 (x,y)
(3.6)

and

Q(x,y)< 2 K ( k ) 2 N k (x,y)
(3.7)

hold for all x,y>0 with xy, and k[0,1], and N 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(0,1) and L k (t)= N k (t,1)/Q(t,1). Then

L k (t)= N k ( x , y ) Q ( x , y ) ,
(3.8)
L k (0)= 2 K ( k ) ,
(3.9)
L k (1)=1.
(3.10)

We distinguish for the proof two cases.

Case 1. k[ 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

2 K ( k ) < N k ( x , y ) Q ( x , y ) <1.
(3.11)

In particular, for k= 2 /2 we have

N 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[0, 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 λ=λ(k)(0,1) such that L k (t)>0 for t(0,λ) and L k (t)<0 for t(λ,1), hence L k is strictly increasing on (0,λ] and strictly decreasing on [λ,1). Therefore, N k (x,y)>Q(x,y) for all x,y>0 with x/y(λ,1) follows from (3.8) and (3.10) together with the monotonicity of L k on [λ,1), and the optimality of inequality (3.6) follows.

Note that

L k (0)= 2 K ( k ) < 2 K ( 0 ) = 2 2 π =0.90031<1.
(3.13)

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

N k ( x , y ) Q ( x , y ) = L k (t)> L k (0)= 2 K ( k ) .
(3.14)

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

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Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11171307 and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

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Correspondence to Yu-Ming Chu.

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Chu, Y., Zhang, Y. & Qiu, S. Schur quadratic concavity of the elliptic Neuman mean and its application. J Inequal Appl 2014, 397 (2014). https://doi.org/10.1186/1029-242X-2014-397

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Keywords

  • elliptic Neuman mean
  • Schur quadratically concave
  • quadratic mean