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Refinements of the Heinz inequalities for matrices

Abstract

This article aims to discuss Heinz inequalities involving unitarily invariant norms. We use a similar method to (Feng in J. Inequal. Appl. 2012:18, 2012; Wang in J. Inequal. Appl. 2013:424, 2013) and we get different refinements of the Heinz inequalities for matrices. Our results are better than some given in (Kittaneh in Integral Equ. Oper. Theory 68:519-527, 2010) and they are different from (Feng in J. Inequal. Appl. 2012:18, 2012; Wang in J. Inequal. Appl. 2013:424, 2013).

1 Introduction

If A, B, X are operators on a complex separable Hilbert space such that A and B are positive, then for every unitarily invariant norm , the function f(v)= A v X B 1 v + A 1 v X B v is convex on the interval [0,1], attains its minimum at v= 1 2 , and attains its maximum at v=0 and v=1. Moreover, f(v)=f(1v) for 0v1. From [1] we know that for every unitarily invariant norm, we have the Heinz inequalities

2 A 1 2 X B 1 2 A v X B 1 v + A 1 v X B v AX+XB.
(1)

In [2], Feng used the following inequalities to get refinements of (1):

f ( a + b 2 ) 1 b a a b f(t)dt 1 4 ( f ( a ) + 2 f ( a + b 2 ) + f ( b ) ) f ( a ) + f ( b ) 2 ,

where f is a real-valued function which is convex on the interval [a,b]. With a similar method, Wang [3] got some new refinements of (1).

In this paper, we use a similar method to [2, 3] and we get different refinements of (1).

When we consider T, we are implicitly assuming that the operator T belongs to the norm ideal associated with . Our results are better than those in [4] and different from [2, 3].

2 Main results

From page 122 of [5], we know the following Hermite-Hadamard integral inequality for convex functions.

Lemma 1 (Hermite-Hadamard integral inequality)

Let f be a real-valued function which is convex on the interval [a,b]. Then

f ( a + b 2 ) 1 b a a b f(t)dt f ( a ) + f ( b ) 2 .

We will use the following lemma.

Lemma 2 Let f be a real-valued function which is convex on the interval [a,b]. Then

f ( a + b 2 ) 1 b a a b f(t)dt 1 8 ( 3 f ( a ) + 2 f ( a + b 2 ) + 3 f ( b ) ) f ( a ) + f ( b ) 2 .

Proof Using the previous lemma, we can easily verify the inequality

1 8 ( 3 f ( a ) + 2 f ( a + b 2 ) + 3 f ( b ) ) f ( a ) + f ( b ) 2 .

Next, we will prove the following inequality:

1 b a a b f(t)dt 1 8 ( 3 f ( a ) + 2 f ( a + b 2 ) + 3 f ( b ) ) .

From the previous lemma, we have

1 b a a b f ( t ) d t = 1 b a ( a a + b 2 f ( t ) d t + a + b 2 b f ( t ) d t ) 1 b a ( f ( a ) + f ( a + b 2 ) 2 b a 2 + f ( a + b 2 ) + f ( b ) 2 b a 2 ) = 1 4 ( f ( a ) + 2 f ( a + b 2 ) + f ( b ) ) = 1 8 ( 2 f ( a ) + 4 f ( a + b 2 ) + 2 f ( b ) ) 1 8 ( 2 f ( a ) + 2 f ( a + b 2 ) + ( f ( a ) + f ( b ) ) + 2 f ( b ) ) = 1 8 ( 3 f ( a ) + 2 f ( a + b 2 ) + 3 f ( b ) ) .

 □

Applying the previous lemma to the function f(v)= A v X B 1 v + A 1 v X B v on the interval [μ,1μ] when 0μ 1 2 , and on the interval [1μ,μ] when 1 2 μ1, we obtain a refinement of the first inequality in (1).

Theorem 1 Let A, B, X be operators such that A, B are positive. Then for 0μ1 and for every unitarily invariant norm, we have

2 A 1 2 X B 1 2 1 | 1 2 μ | | μ 1 μ A v X B 1 v + A 1 v X B v d v | 1 4 ( 3 A μ X B 1 μ + A 1 μ X B μ + 2 A 1 2 X B 1 2 ) A μ X B 1 μ + A 1 μ X B μ .
(2)

Proof First assume that 0μ 1 2 . Then it follows by the previous lemma that

f ( 1 μ + μ 2 ) 1 1 2 μ μ 1 μ f ( t ) d t 1 8 ( 3 f ( μ ) + 2 f ( 1 μ + μ 2 ) + 3 f ( 1 μ ) ) f ( μ ) + f ( 1 μ ) 2 ,

and so

f ( 1 2 ) 1 1 2 μ μ 1 μ f ( t ) d t 1 4 ( 3 f ( μ ) + f ( 1 2 ) ) f ( μ ) .

Thus,

2 A 1 2 X B 1 2 1 1 2 μ μ 1 μ A v X B 1 v + A 1 v X B v d v 1 4 ( 3 A μ X B 1 μ + A 1 μ X B μ + 2 A 1 2 X B 1 2 ) A μ X B 1 μ + A 1 μ X B μ .
(3)

Now, assume that 1 2 μ1. Then by applying (3) to 1μ, it follows that

2 A 1 2 X B 1 2 1 2 μ 1 1 μ μ A v X B 1 v + A 1 v X B v d v 1 4 ( 3 A μ X B 1 μ + A 1 μ X B μ + 2 A 1 2 X B 1 2 ) A μ X B 1 μ + A 1 μ X B μ .
(4)

Since

lim μ 1 2 1 | 1 2 μ | | μ 1 μ A v X B 1 v + A 1 v X B v d v | = lim μ 1 2 1 4 ( 3 A μ X B 1 μ + A 1 μ X B μ + 2 A 1 2 X B 1 2 ) = 2 A 1 2 X B 1 2 ,

the inequalities in (2) follow by combining (3) and (4). □

Applying the previous lemma to the function f(v)= A v X B 1 v + A 1 v X B v on the interval [μ, 1 2 ] when 0μ 1 2 , and on the interval [ 1 2 ,μ] when 1 2 μ1, we obtain the following.

Theorem 2 Let A, B, X be operators such that A, B are positive. Then for 0μ1 and for every unitarily invariant norm, we have

A 2 μ + 1 4 X B 3 2 μ 4 + A 3 2 μ 4 X B 2 μ + 1 4 2 | 1 2 μ | | μ 1 2 A v X B 1 v + A 1 v X B v d v | 1 8 ( 3 A μ X B 1 μ + A 1 μ X B μ + 2 A 2 μ + 1 4 X B 3 2 μ 4 + A 3 2 μ 4 X B 2 μ + 1 4 + 6 A 1 2 X B 1 2 ) 1 2 ( A μ X B 1 μ + A 1 μ X B μ + 2 A 1 2 X B 1 2 ) .
(5)

The inequality (5) and the first inequality in (1) yield the following refinement of the first inequality in (1).

Corollary 1 Let A, B, X be operators such that A, B are positive. Then for 0μ1 and for every unitarily invariant norm, we have

2 A 1 2 X B 1 2 A 2 μ + 1 4 X B 3 2 μ 4 + A 3 2 μ 4 X B 2 μ + 1 4 2 | 1 2 μ | | μ 1 2 A v X B 1 v + A 1 v X B v d v | 1 8 ( 3 A μ X B 1 μ + A 1 μ X B μ + 2 A 2 μ + 1 4 X B 3 2 μ 4 + A 3 2 μ 4 X B 2 μ + 1 4 + 6 A 1 2 X B 1 2 ) 1 2 ( A μ X B 1 μ + A 1 μ X B μ + 2 A 1 2 X B 1 2 ) A μ X B 1 μ + A 1 μ X B μ .
(6)

Applying the previous lemma to the function f(v)= A v X B 1 v + A 1 v X B v on the interval [0,μ] when 0μ 1 2 , and on the interval [μ,1] when 1 2 μ1, we obtain the following theorem.

Theorem 3 Let A, B, X be operators such that A, B are positive. Then

  1. (1)

    for 0μ 1 2 and for every unitarily norm,

    A μ 2 X B 1 μ 2 + A 1 μ 2 X B μ 2 1 μ 0 μ A v X B 1 v + A 1 v X B v d v 1 8 ( 3 A X + X B + 2 A μ 2 X B 1 μ 2 + A 1 μ 2 X B μ 2 + 3 A μ X B 1 μ + A 1 μ X B μ ) 1 2 ( A X + X B + A μ X B 1 μ + A 1 μ X B μ ) ;
    (7)
  2. (2)

    for 1 2 μ1 and for every unitarily norm,

    A 1 + μ 2 X B 1 μ 2 + A 1 μ 2 X B 1 + μ 2 1 1 μ μ 1 A v X B 1 v + A 1 v X B v d v 1 8 ( 3 A X + X B + 2 A 1 + μ 2 X B 1 μ 2 + A 1 μ 2 X B 1 + μ 2 + 3 A μ X B 1 μ + A 1 μ X B μ ) 1 2 ( A X + X B + A μ X B 1 μ + A 1 μ X B μ ) .
    (8)

Since the function f(v)= A v X B 1 v + A 1 v X B v is decreasing on the interval [0, 1 2 ] and increasing on the interval [ 1 2 ,1], and using the inequalities (7) and (8), we obtain the refinement of the second inequality in (1).

Corollary 2 Let A, B, X be operators such that A, B are positive. Then for 0μ1 and for every unitarily invariant norm, we have the following.

  1. (1)

    For 0μ 1 2 and for every unitarily norm,

    A μ X B 1 μ + A 1 μ X B μ A μ 2 X B 1 μ 2 + A 1 μ 2 X B μ 2 1 μ 0 μ A v X B 1 v + A 1 v X B v d v 1 8 ( 3 A X + X B + 2 A μ 2 X B 1 μ 2 + A 1 μ 2 X B μ 2 + 3 A μ X B 1 μ + A 1 μ X B μ ) 1 2 ( A X + X B + A μ X B 1 μ + A 1 μ X B μ ) A X + X B .
    (9)
  2. (2)

    For 1 2 μ1 and for every unitarily norm,

    A μ X B 1 μ + A 1 μ X B μ A 1 + μ 2 X B 1 μ 2 + A 1 μ 2 X B 1 + μ 2 1 1 μ μ 1 A v X B 1 v + A 1 v X B v d v 1 8 ( 3 A X + X B + 2 A 1 + μ 2 X B 1 μ 2 + A 1 μ 2 X B 1 + μ 2 + 3 A μ X B 1 μ + A 1 μ X B μ ) 1 2 ( A X + X B + A μ X B 1 μ + A 1 μ X B μ ) A X + X B .
    (10)

It should be noticed that in the inequalities (7) to (10), we have

lim μ 0 1 μ 0 μ A v X B 1 v + A 1 v X B v d v lim μ 1 1 1 μ μ 1 A v X B 1 v + A 1 v X B v d v = A X + X B .

References

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Acknowledgements

This work is supported by NSF of China (Grant Nos. 11171364 and 11271301).

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Correspondence to Guiyun Chen.

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The authors declare that they have no competing interests.

Authors’ contributions

YY carried out convex function. YF carried out unitarily invariant norm. GC carried out the calculation. All authors read and approved the final manuscript.

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Yan, Y., Feng, Y. & Chen, G. Refinements of the Heinz inequalities for matrices. J Inequal Appl 2014, 50 (2014). https://doi.org/10.1186/1029-242X-2014-50

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Keywords

  • convex function
  • Heinz inequality
  • Hermite-Hadamard inequality
  • unitarily invariant norm