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On strengthened form of Copson’s inequality

Abstract

In this paper, the famous Copson inequality has been improved. We obtain some new results by a different method.

MSC:26D15, 25D05.

1 Introduction

Suppose that a k >0, p>1, then we obtain the following Hardy inequality:

( p p 1 ) p n = 1 a n p > n = 1 ( 1 n k = 1 n a k ) p .
(1.1)

Hardy’s inequality plays an important role in the field of analysis; see [14]. In recent decades, some generalizations and strengthening of Hardy’s inequality have been obtained in [15]. We list some previous results as follows.

Copson’s inequality [4, 5]

Suppose that a n >0 (n=1,2,3). If p>1, then

n = 1 ( k = n a k k ) p < p p n = 1 a n p .
(1.2)

If 0<p<1, then

n = 1 ( k = n a k k ) p > p p n = 1 a n p .
(1.3)

And in [4] and [6], the authors pay much attention to the generalization of Copson’s inequality.

In this paper, inequalities (1.2) and (1.3) were strengthened by using a new method.

2 Relational lemmas and definitions

In this section, some relational lemmas and definitions will be introduced.

Theorem A [[7], Th. 1.1]

Suppose that a,bR, a<b, c[a,b], and f: [ a , b ] n R has continuous partial derivatives, and

D i = { ( x 1 , x 2 , , x n 1 , c ) | min 1 k n 1 { x k } c , x i = min 1 k n 1 { x k } c } ,i=1,2,,n1.

If f ( x ) x i >0 holds for all x D i (i=1,2,,n1), then

f( y 1 , y 2 ,, y n 1 ,c)f(c,c,,c,c),

where y i [c,b] (i=1,2,,n1).

Theorem B [[7], Cor. 1.3]

Suppose that a,bR, a<b, and f: [ a , b ] n R has continuous partial derivatives and

D i = { ( x 1 , x 2 , , x n ) | a min 1 k n { x k } < x i = max 1 k n { x k } b } ,i=1,2,,n.

If f ( x ) x i >0 holds for all x D i , where i=1,2,,n, then

f( x 1 , x 2 ,, x n )f( x min , x min ,, x min ),

where x i [a,b] (i=1,2,,n), x min = min 1 k n { x k }.

Definition 1 [1]

Let G R n be a convex set, ϕ:HR be a continuous function. If

ϕ ( α x + ( 1 α ) y ) ()αϕ(x)+(1α)ϕ(y)

holds for all x,yG, α[0,1], then the function φ is convex (concave).

Lemma 1 (Hermite-Hadamard’s inequality)

Let ϕ:[a,b]R be a convex (concave) function. Then

ϕ ( a + b 2 ) () 1 b a a b ϕ(x)dx() ϕ ( a ) + ϕ ( b ) 2 ,
(2.1)

and the equality holds if and only if ϕ is linear.

Lemma 2 Suppose that p>0.

  1. (1)

    If p1, or 0<p 1 2 , then

    2 p p 2 p ;
    (2.2)
  2. (2)

    If 1 2 <p<1, then

    2 p >2 p p ;
    (2.3)
  3. (3)

    If 1 2 <p<1, then

    p p +(p1) 2 p >0.
    (2.4)

Proof Set f 1 :p(0,+)plnp+ln2pln2. Then we have

f 1 (p)=lnp+1ln2=ln ( e p 2 ) .

Obviously, f 1 is monotone decreasing for p(0, 2 e ), f 1 is monotone increasing for p( 2 e ,+), and f 1 (1)=0, f 1 ( 1 2 )= 1 2 ln2+ln2 1 2 ln2=0, then (2.2) and (2.3) hold. Let

f 2 :p(0,1)plnppln2ln(1p).

We get

and

h (p)= 1 p 1 p 2 <0.

Then h is concave for p(0,1). Because h( 1 2 )>0 and lim p 1 h(p)>0, then h(p)>0 and f 2 (p)>0 hold for p[ 1 2 ,1). From f 2 ( 1 2 )=0, we have f 2 (p)>0 for p( 1 2 ,1). Inequality (2.4) is proved. □

Lemma 3

  1. (1)

    If p>1, then the equation

    p p (1x) ( 1 2 x ) p 1 =1
    (2.5)

has only a positive root for x(0, 1 2 ).

  1. (2)

    If 1 2 <p<1, then the equation

    p p (1+x)= ( 1 2 + x ) 1 p
    (2.6)

has only a positive root for x(0, 1 2 ).

Proof

  1. (1)

    Let g 1 :x[0, 1 2 ] p p (1x) ( 1 2 x ) p 1 1. Then g 1 is monotone decreasing. According to inequality (2.2), we have

    g 1 (0)= p p ( 1 2 ) p 1 1= ( 1 2 ) p 1 [ p p 2 p 1 ] >0

and g 1 ( 1 2 )=1. So, equation (2.5) has only a positive root for x(0, 1 2 ).

  1. (2)

    Let g 2 :x[0, 1 2 ] p p (1+x) ( 1 2 + x ) 1 p . Thus,

    g 2 (x)= p p (1p) ( 1 2 + x ) p > ( 1 2 + x ) p [ 1 2 p p p ( 1 p ) ] .

By inequality (2.4), g 2 is monotone increasing. According to inequality (2.3), we get

lim x 0 + g 2 (x)= p p 1 2 1 p <0

and

lim x ( 1 / 2 ) + g 2 (x)= 3 2 p p 1 3 2 ( 1 2 ) 1 2 1>0.

Therefore, equation (2.6) has only a positive root for x(0, 1 2 ).

 □

Lemma 4 If p>1, m>0, mN and c(0, 1 2 ) is the only one positive root of equation (2.5), then

n = 1 m [ k = n 1 ( k c ) 1 + 1 p ] p 1 < p p ( m c ) 1 p
(2.7)

and

n = 1 m [ k = n m 1 ( k c ) 1 + 1 / p ] p < p p n = 1 m n p ( n c ) p + 1 .
(2.8)

Proof (1) If m=1, by Lemma 1, we get

n = 1 m [ k = n 1 ( k c ) 1 + 1 / p ] p 1 = [ k = 1 1 ( k c ) 1 + 1 / p ] p 1 < [ 1 2 1 ( x c ) 1 + 1 / p d x ] p 1 = p p 1 ( 1 2 c ) ( p 1 ) / p .

If m2, by Lemma 1, we get

n = 1 m [ k = n 1 ( k c ) 1 + 1 / p ] p 1 < n = 1 m [ n 1 2 1 ( x c ) 1 + 1 / p ] p 1 = p p 1 n = 1 m ( n 1 2 c ) ( p 1 ) / p = p p 1 [ ( 1 2 c ) ( p 1 ) / p + n = 2 m ( n 1 2 c ) ( p 1 ) / p ] < p p 1 [ ( 1 2 c ) ( p 1 ) / p + 3 2 m + 1 2 ( x 1 2 c ) ( p 1 ) / p d x ] = p p 1 [ ( 1 2 c ) ( p 1 ) / p + p ( m c ) 1 / p p ( 1 c ) 1 / p ] .

So,

n = 1 m [ k = n 1 ( k c ) 1 + 1 / p ] p 1 < p p 1 [ ( 1 2 c ) ( p 1 ) / p + p ( m c ) 1 / p p ( 1 c ) 1 / p ]
(2.9)

holds for every m>0 and mN. Since inequalities (2.9), (2.10) and

( 1 2 c ) ( p 1 ) / p =p ( 1 c ) 1 / p ,
(2.10)

inequality (2.7) holds.

  1. (2)

Let q>1 and 1/p+1/q=1. Using Hölder’s inequality, we have

Since

[ n = 1 m ( k = n m 1 ( k c ) 1 + 1 / p ) p ] 1 / p <p [ n = 1 m n p ( n c ) p + 1 ] 1 / p ,

inequality (2.8) holds. □

Lemma 5 If 1 2 <p<1, m>0, mN and d(0, 1 2 ) is the only one positive root of equation (2.6), then

n = 1 m [ k = n m 1 ( k + d ) 1 + 1 / p ] p > p p n = 1 m n p ( n + d ) p + 1 .
(2.11)

Proof

(2.12)

By Hölder’s inequality, we have

And by using inequality (2.12), we obtain

(2.13)

From inequality (2.12) and inequality (2.13), we get

n = 1 m ( k = n m 1 ( k + d ) 1 + 1 / p ) p >p [ n = 1 m n p ( n + d ) p + 1 ] 1 / p { n = 1 m [ k = n m 1 ( k + d ) 1 + 1 / p ] p } 1 p p

and

{ n = 1 m [ k = n m 1 ( k + d ) 1 + 1 / p ] p } 1 / p >p [ n = 1 m n p ( n + d ) p + 1 ] 1 / p .

Then inequality (2.11) holds. □

3 Strengthened Copson’s inequality (p>1)

Theorem 1 Assume that p>1, m>0, mN, a n >0 (n=1,2,,m), c(0, 1 2 ) is the only one positive root of equation (2.5) and B m = min 1 n m { ( n c ) 1 / p a n }. Then

(3.1)

Proof Set b n = ( n c ) 1 / p a n (n=1,2,,m). Then inequality (3.1) is equivalent to

(3.2)

where B m = min 1 n N { b n }. Let

f:b=( b 1 , b 2 ,, b m ) [ 0 , + ) m p p n = 1 m b n p n c n = 1 m ( k = n m b k ( k c ) 1 + 1 / p ) p

and

D i = { ( b 1 , b 2 , , b n ) | 0 min 1 n m { b n } < b i = max 1 n m { b n } } .

If ( b 1 , b 2 ,, b n ) D i , then

f b i = p p p b i p 1 i c p ( i c ) 1 + 1 / p n = 1 i ( k = n m b k ( k c ) 1 + 1 / p ) p 1 > p b i p 1 ( i c ) 1 + 1 p [ p p ( i c ) 1 / p n = 1 i ( k = n m 1 ( k c ) 1 + 1 / p ) p 1 ] > p b i p 1 ( i c ) 1 + 1 p [ p p ( i c ) 1 / p n = 1 i ( k = n 1 ( k c ) 1 + 1 / p ) p 1 ] .

By inequality (2.7), we know f b i >0. By Theorem B, inequality (3.2) holds, the proof is completed. □

Corollary 1 If p>1, m>0, mN, a n >0 (n=1,2,,m), c(0, 1 2 ) is the only one positive root of equation (2.5), and B m = min 1 n m { ( n c ) 1 / p a n }, then

p p n = 1 m a n p n = 1 m ( k = n m a k k c ) p > p p B m p n = 1 m n p ( n c ) p ( n c ) p + 1 .
(3.3)

Proof By (3.1) and (2.8), we can obtain

p p n = 1 m a n p n = 1 m ( k = n m a k k c ) p > p p B m p [ n = 1 m 1 n c n = 1 m n p ( n c ) p + 1 ] = p p B m p n = 1 m n p ( n c ) p ( n c ) p + 1 .

 □

Corollary 2 If p>1, a n >0 (n=1,2,), n = 1 a n p <+ and c(0, 1 2 ) is the only one positive root of equation (2.5), then

n = 1 ( k = n a k k c ) p p p n = 1 a n p .
(3.4)

Proof Because of n = 1 a n p <+, the infimum of { ( n c ) 1 / p a n } n = 1 is zero. Then there exists a sequence such that { ( m i c ) 1 / p a m i } i = 1 decrease to zero. Since (3.3), we have

p p n = 1 m i a n p n = 1 m i ( k = n m i a k k c ) p > p p [ ( m i c ) 1 / p a m i ] p n = 1 m i n p ( n c ) p + 1 ( n c ) p + 1 .
(3.5)

Let i+ in inequality (3.5), we have m i + and

lim i + [ ( m i c ) 1 / p a m i ] p n = 1 m i n p ( n c ) p + 1 ( n c ) p + 1 =0.

Then by (3.5), we can obtain

p p n = 1 m i a n p n = 1 m i ( k = n m i a k k c ) p 0.

Therefore, inequality (3.4) holds. □

Remark Obviously, inequality (3.4) strengthens inequality (1.2).

4 Strengthened Copson’s inequality (1/2<p<1)

Theorem 2 If 1 2 <p<1, m>0, mN, a n >0 (n=1,2,,m), d(0, 1 2 ) is the only one positive root of equation (2.6) and B m = min 1 n m { ( n + d ) 1 / p a n }. Then

n = 1 m ( k = n m a k k + d ) p p p n = 1 m a n p B m p [ n = 1 m ( k = n m 1 ( k + d ) 1 + 1 / p ) p p p n = 1 m 1 n + d ] .
(4.1)

Proof Let b n = ( n + d ) 1 / p a n (n=1,2,,m). Then inequality (4.1) is equivalent to

n = 1 m ( k = n m b k ( k + d ) 1 + 1 / p ) p p p n = 1 m b n p n + d B m p [ p p n = 1 m 1 n n = 1 m ( k = n m 1 k 1 + 1 / p ) p ] ,
(4.2)

where B m = min 1 n m { b n }. Set

f:b ( 0 , + ) m n = 1 m ( k = n m b k ( k + d ) 1 + 1 / p ) p p p n = 1 m b n p n + d

and D i ={( b 1 , b 2 ,, b n )|0 min 1 n m { b n }< b i = max 1 n m { b n }}. If ( b 1 , b 2 ,, b n ) D i , then

f b i = p ( i + d ) 1 + 1 / p n = 1 i ( k = n m b k ( k + d ) 1 + 1 / p ) p 1 p p + 1 b i p 1 i + d = p b i p 1 ( i + d ) 1 + 1 / p [ n = 1 i ( k = n m b k ( k + d ) 1 + 1 / p b i ) ( 1 p ) p p ( i + d ) 1 / p ] > p b i p 1 ( i + d ) 1 + 1 / p [ n = 1 i ( k = n m 1 ( k + d ) 1 + 1 / p ) ( 1 p ) p p ( i + d ) 1 / p ] > p b i p 1 ( i + d ) 1 + 1 / p [ n = 1 i ( k = n 1 ( k + d ) 1 + 1 / p ) ( 1 p ) p p ( i + d ) 1 / p ] .

By Lemma 1, we have

f b i > p b i p 1 ( i + d ) 1 + 1 / p [ n = 1 i ( n 1 2 1 ( x + d ) 1 + 1 / p d x ) ( 1 p ) p p ( i + d ) 1 / p ] = p b i p 1 ( i + d ) 1 + 1 / p [ p ( 1 p ) n = 1 i ( n 1 2 + d ) ( 1 p ) / p p p ( i + d ) 1 / p ] .

As i=1, by the definition of d, we have

f b 1 > p b 1 p 1 ( 1 + d ) 1 + 1 / p [ p ( 1 p ) ( 1 2 + d ) ( 1 p ) / p p p ( 1 + d ) 1 / p ] =0.

As 2im, because 1 2 <p<1, 0< p 1 p 1 and g:x(0,+) x ( 1 p ) / p is concave, we have

f b i > p b i p 1 ( i + d ) 1 + 1 / p [ p ( 1 p ) ( ( 1 2 + d ) ( 1 p ) / p + n = 2 i ( n 1 2 + d ) ( 1 p ) / p ) p p ( i + d ) 1 / p ] > p b i p 1 ( i + d ) 1 + 1 / p [ p ( 1 p ) ( ( 1 2 + d ) ( 1 p ) / p + 3 2 i + 1 2 ( x 1 2 + d ) ( 1 p ) / p d x ) p p ( i + d ) 1 / p ] = p b i p 1 ( i + d ) 1 + 1 / p [ p ( 1 p ) ( ( 1 2 + d ) ( 1 p ) / p + p ( i + d ) 1 / p p ( 1 + d ) 1 / p ) p p ( i + d ) 1 / p ] = p b i p 1 ( i + d ) 1 + 1 / p [ p ( 1 p ) p ( i + d ) 1 / p p p ( i + d ) 1 / p ] = 0 .

Thus, for every D i , f b i >0. By Theorem B, inequality (4.2) holds. □

Corollary 3 If 1 2 <p<1, m>0, mN, a n >0 (n=1,2,,m), d(0, 1 2 ) is the only one positive root of equation (2.6) and B m = min 1 n m { ( n + d ) 1 / p a n }. Then

n = 1 m ( k = n m a k k + d ) p p p n = 1 m a n p p p B m p n = 1 m n p ( n + d ) p ( n + d ) p + 1 .
(4.3)

Proof From Theorem 2 and Lemma 5, we have

n = 1 m ( k = n m a k k + d ) p p p n = 1 m a n p B m p [ p p n = 1 m n p ( n + d ) p + 1 p p n = 1 m 1 n + d ] .

Then inequality (4.3) holds. □

Corollary 4 If 1 2 <p<1, a n >0 (n=1,2,), d(0, 1 2 ) is the only one positive root of equation (2.6) and series n = 1 ( k = n a k k + d ) p <+. Then

n = 1 ( k = n a k k + d ) p p p n = 1 a n p .
(4.4)

Proof According to inequality (4.3), we obtain

n = 1 m ( k = n m a k k + d ) p + p p B m p n = 1 m ( n + d ) p n p ( n + d ) p + 1 p p n = 1 m a n p .

The following proof is the same as the relevant proof for Corollary 2, omitted here. □

Remark For 1 2 <p<1, there is no doubt that inequality (4.4) strengthens inequality (1.3).

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Acknowledgements

The research is supported by the Nature Science Foundation of China (No. 110771069) and the NS Foundation of the Educational Committee of Zhejiang Province under Grant Y201223283.

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Xu, Q. On strengthened form of Copson’s inequality. J Inequal Appl 2012, 305 (2012). https://doi.org/10.1186/1029-242X-2012-305

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Keywords

  • inequality
  • Copson’s inequality
  • analytic inequalities