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New progress on the operator inequalities involving improved Young’s and its reverse inequalities relating to the Kantorovich constant
Journal of Inequalities and Applications volume 2017, Article number: 69 (2017)
Abstract
The purpose of this paper is to give a survey of the progress, advantages and limitations of various operator inequalities involving improved Young’s and its reverse inequalities related to the Kittaneh-Manasrah inequality. We also present our new progress to the related research topics. New scalar versions of Young’s inequalities are promoted, the operator version and the Hilbert-Schmidt form also get a promotion.
1 Introduction
As is well known, the famous Young’s inequality for real numbers is that
which is called the u-weighted arithmetic-geometric mean inequality.
Recently, research in this area has received continuous attention. Many researches have spent lots of energy on it due to the applications in various fields. In [1, 2], Kittaneh and Manasrah researched Young’s inequality and obtained the following results:
where \(a,b > 0\), \(u \in [ 0,1 ]\) and \(r = \min \{ 1 - u,u \}\), \(s = \max \{ 1 - u,u \}\).
The left inequality of (1.2) can be regarded as a refinement of Young’s inequality and the right one can be regarded as a reverse of Young’s inequality.
In [3], Zhao and Wu deepened inequality (1.2) and showed another refinement of Young’s inequality as follows:
If \(0 \le u \le\frac{1}{2}\), then
If \(\frac{1}{2} < u \le1\), then
where \(a,b \ge0\), \(u \in [ 0,1 ]\) and \(r = \min \{ u,1 - u \}\), \(r_{0} = \min \{ 2r,1 - 2r \}\).
Zhao and Wu also obtained a more precise form of the reverse Young’s inequality as follows:
If \(0 \le u \le\frac{1}{2} \), then
If \(\frac{1}{2} < u \le1 \), then
where \(r = \min \{ u,1 - u \}\), \(r_{0} = \min \{ 2r,1 - 2r \}\) and \(a,b \ge0\), \(u \in [ 0,1 ]\).
In [4], Furuichi obtained Young’s inequality about Specht’s ratio as follows:
where \(r = \min \{ 1 - u,u \}\), \(h = \frac{b}{a}\), and \(a,b > 0\), \(u \in [ 0,1 ]\).
In [5], Tominaga got the reverse Young inequality with the help of Specht’s ratio. He gave the following inequality:
Article [6] pointed out that Specht’s ratio and the Kantorovich constant have the relationship as follows:
where \(t > 0\) and \(0 \le r \le\frac{1}{2}\).
Based on this idea, in the article [6], the authors got the refinement of Young’s inequality:
where \(r = \min \{ 1 - u,u \}\), \(h = \frac{b}{a}\), and \(a,b \ge0\), \(u \in [ 0,1 ]\).
Generally, the Kantorovich constant is written as \(K ( {t},2 ) = \frac{(t + 1)^{2}}{4t}\) for \({t} > 0\), which has properties \(K ( t,2 ) = K ( \frac{1}{t},2 ) \ge 1\) (\(t > 0 \)), and \(K ( t,2 )\) is increasing on \([1,\infty)\) and is decreasing on \(( 0,1 )\).
In [7], Zhao and Wu made a further study about inequality (1.2) with the Kantorovich constant and gave the following results:
where \(r = \min \{ u, 1 - u \}\), \(r' = \min \{ 2r,1 - 2r \}\), \(a,b > 0\), \(u \in [ 0,1 ] - \{ \frac{1}{2} \}\) and \(h = \frac{b}{a}\).
And then Zhao and Wu made a reverse refinement of the second inequality in (1.2)
where \(a,b > 0\), \(u \in [ 0,1 ] - \{ \frac{1}{2} \} \), \(h =\frac{b}{a}\) and \(r = \min \{ u, 1 - u \}\), \(r' = \min \{ 2r,1 - 2r \}\), \(s = \max \{ u, 1 - u \}\).
Let us take a closer look at [8] where Liao et al. made a reverse refinement for Young’s inequality as follows:
where \(r = \min \{ 1 - u,u \}\), \(R' = \max \{ 2r,1 - 2r \}\), \(R = \max \{1 - u,u \}\) and \(a,b > 0\), \(u \in [ 0,1 ]\), \(h = \frac{b}{a}\).
In [9], Liao and Wu further deepened the results of inequality (1.10) and obtained the following results:
If \(0 < u \le\frac{1}{2}\), then
If \(\frac{1}{2} < u < 1\), then
where \(h = \frac{b}{a}\), \(r = \min \{ u,1 - u \}\), \(r_{0} = \min \{ 2r,1 - 2r \}\), \(r_{1} = \min \{ 2r_{0},1 - 2r_{0} \}\).
Similarly, Liao and Wu deepened the reverse results of inequality (1.10) as follows:
If \(0 < u \le\frac{1}{2}\), then
If \(\frac{1}{2} < u < 1\), then
In [10], Hizrallah and Kittaneh showed a square form of the refinement of the first inequality of (1.2) as follows:
where \(a,b > 0\), \(u \in [ 0,1 ]\) and \(r = \min \{ u,1 - u \}\).
In [11] He and Zou presented a reverse of the second inequality of (1.2) as follows:
where \(a,b > 0\), \(u \in [ 0,1 ]\) and \(s = \max \{ u, 1 - u \}\).
For the sake of convenience, we have to introduce a set of symbols. Let \(A,B \in B ( H )\) be two positive and invertible operators, where \(B ( H )\) is regarded as the set of all bounded linear operators on a complex separable Hilbert space H. \(A^{*}\) denotes the conjugate operator of A. Let \(M_{n}\) be the set of \(n \times n\) complex matrices. For \(A = ( a_{ij} ) \in M_{n}\), the Hilbert-Schmidt norm of A is defined by \(\Vert A \Vert _{2} = \sqrt{\sum_{i,j = 1}^{n} a_{ij}^{2}}\), \(\Vert \ \Vert _{2}\) has the unitarily invariant property, that is, \(\Vert UAV \Vert _{2} = \Vert A \Vert _{2}\) for all \(A \in M_{n}\) and all unitary matrices \(U,V \in M_{n}\). We use the following notations:
The operator versions \(A\nabla_{{u}}B\), \(A\#_{{u}}B\), \(A!_{{u}}B\) are called the arithmetic mean, geometric mean and harmonic mean, respectively. When \(u = \frac{1}{2}\), these operators can be written by simplification as \(A\nabla B\), \(A\# B\), \(A!B\).
The article [12] presented the matrix version of (1.1). For any \(B,C \in M_{n}\), if B is a positive definite matrix and C is an invertible matrix, then
where \(A = CC^{*}\) and \(u \in [ 0,1 ]\).
Specially, if \(C = A^{\frac{1}{2}}\), then \(C^{*} ( ( C^{*} )^{ - 1}BC^{ - 1} )^{ - u}C = A\#_{u}B\) and inequality (1.20) can be written as \(A\nabla_{u}B \ge A\#_{u}B\).
Next, we have to introduce the development of operator inequalities. In [2], Kittaneh and Manasrah took a step further and got the matrix version of inequality (1.2)
where \(u \in [ 0,1 ]\), \(r = \min \{ u,1 - u \}\) and \(s = \max \{ u,1 - u \}\).
Based on the idea, articles [7, 8] and [9] also got the corresponding operator forms of inequality (1.10)-(1.19).
In [10], inequality (1.18) for the Hilbert-Schmidt norm form was obtained by Hirzallah and Kittaneh, that is, if \(A,B,X \in M_{n}\), where A, B are positive semi-definite matrices, then
where \(u \in [ 0,1 ]\) and \(r = \min \{ u,1 - u \}\).
The reverse inequality (1.19) for the Hilbert-Schmidt norm form was obtained by Kittaneh and Manasrah in [2] as the following result:
where \(u \in [ 0,1 ]\) and \(s = \max \{ u,1 - u \}\).
We find that there are still limitations in the previous inequalities. This paper continues to research the refinement of Young’s inequality. The difference is that we present Young’s inequalities in four more precise intervals of \([0,1]\) other than two sections of \([0,1]\), and also the operator form and the matrix version get a promotion.
In Section 2, a new scalar version of Young’s inequality and its reverse with the Kantorovich constant are first given. In Section 3, we obtain the corresponding new operator inequalities on the Hilbert space. Finally, in Section 4, we mainly show the matrix version of inequalities for the Hilbert-Schmidt norm.
2 New progress of Young’s and its reverse inequalities
In this section, we mainly present the improved scalar Young and its reverse inequalities relating to the Kantorovich constant.
Theorem 2.1
Let \(a,b > 0\), \(u \in ( 0, 1 )\), we have
-
(1)
If \(0 < {u} < \frac{1}{4}\), then
$$\begin{aligned} ( 1 - u )a + ub \ge& u ( \sqrt{a} - \sqrt{b} )^{2} + 2u \bigl( \sqrt[4]{ab} - \sqrt{a} \bigr)^{2} + r \bigl( \sqrt{a} - \sqrt[8]{a^{3}b} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.1)where \(h = \frac{b}{a}\), \(r = \min \{ 1 - 4u,4u \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(2)
If \(\frac{1}{4} \le u < \frac{1}{2}\), then
$$\begin{aligned} ( 1 - u )a + ub \ge& u ( \sqrt{a} - \sqrt{b} )^{2} + ( 1 - 2u ) \bigl( \sqrt[4]{ab} - \sqrt{a} \bigr)^{2} + r \bigl( \sqrt[4]{ab} - \sqrt[8]{a^{3}b} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.2)where \(h = \frac{b}{a}\), \(r = \min \{ 2 - 4u,4u - 1 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(3)
If \(\frac{1}{2} \le u < \frac{3}{4}\), then
$$\begin{aligned} ( 1 - u )a + ub \ge& ( 1 - u ) ( \sqrt{a} - \sqrt{b} )^{2} + ( 2u - 1 ) \bigl( \sqrt[4]{ab} - \sqrt{b} \bigr)^{2} + r \bigl( \sqrt[4]{ab} - \sqrt[8]{ab^{3}} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.3)where \(h = \frac{b}{a}\), \(r = \min \{ 3 - 4u,4u - 2 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(4)
If \(\frac{3}{4} \le u < 1\), then
$$\begin{aligned} ( 1 - u )a + ub \ge& ( 1 - u ) ( \sqrt{a} - \sqrt{b} )^{2} + ( 2 - 2u ) \bigl( \sqrt[4]{ab} - \sqrt{b} \bigr)^{2} + r \bigl( \sqrt{b} - \sqrt[8]{ab^{3}} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.4)where \(h = \frac{b}{a}\), \(r = \min \{ 4 - 4u,4u - 3 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
Proof
The process of the proof of inequalities (2.3) and (2.4) is similar to that of inequalities (2.1) and (2.2), so we only need to prove inequalities (2.1) and (2.2).
We first consider the case \(0 < {u} \le\frac{1}{4}\). By inequality (1.10) and inequality (1.14), we have
Then we can conclude the inequality
Then, considering the case \(\frac{1}{4} \le u < \frac{1}{2}\), we have
We conclude the inequality
So the proof is completed. □
It is not hard to see that the results above are superior to inequalities (1.14) and (1.15) through more precise subsections.
And the following result is superior to inequalities (1.16) and (1.17).
Theorem 2.2
Let \(a,b > 0\), \(u \in ( 0,1 )\), we have the following.
-
(1)
If \(0 < u < \frac{1}{4}\), then
$$\begin{aligned} ( 1 - u )a + ub \le&(1 - u) ( \sqrt{a} - \sqrt{b} )^{2} - 2u \bigl( \sqrt[4]{ab} - \sqrt{b} \bigr)^{2} - r \bigl( \sqrt[8]{ab^{3}} - \sqrt{b} \bigr) \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.5)where \(h = \frac{b}{a}\), \(r = \min \{ 4u,1 - 4u \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(2)
If \(\frac{1}{4} \le u < \frac{1}{2}\), then
$$\begin{aligned} ( 1 - u )a + ub \le&(1 - u) ( \sqrt{a} - \sqrt{b} )^{2} - ( 1 - 2u ) \bigl( \sqrt[4]{ab} - \sqrt{b} \bigr)^{2} - r \bigl( \sqrt[8]{ab^{3}} - \sqrt[4]{ab} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.6)where \(h = \frac{b}{a}\), \(r = \min \{ 2 - 4u,4u - 1 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(3)
If \(\frac{1}{2} \le u < \frac{3}{4}\), then
$$\begin{aligned} ( 1 - u )a + ub \le& u ( \sqrt{a} - \sqrt{b} )^{2} - ( 2u - 1 ) \bigl( \sqrt[4]{ab} - \sqrt{a} \bigr)^{2} - r \bigl( \sqrt[8]{a^{3}b} - \sqrt[4]{ab} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.7)where \(h = \frac{b}{a}\), \(r = \min \{ 3 - 4u,4u - 2 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(4)
If \(\frac{3}{4} \le u < 1\), then
$$\begin{aligned} ( 1 - u )a + ub \le& u ( \sqrt{a} - \sqrt{b} )^{2} - ( 2 - 2u ) \bigl( \sqrt[4]{ab} - \sqrt{a} \bigr)^{2} - r \bigl( \sqrt[8]{a^{3}b} - \sqrt{a} \bigr)^{2} \\ &{}+ K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}a^{1 - u}b^{u}, \end{aligned}$$(2.8)where \(h = \frac{b}{a}\), \(r = \min \{ 4 - 4u,4u - 3 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
Proof
The process of the proof of inequalities (2.7) and (2.8) is similar to that of inequalities (2.5) and (2.6), so we only need to prove inequalities (2.5) and (2.6).
We consider the first case \(0 < u < \frac{1}{4}\). By inequality (1.14) obtained by Liao and Wu, we have
That is to say,
So the proof is completed. □
Remark 2.1
By the property of the Kantorovich constant, the inequalities in Theorems 2.1 and 2.2 are the improved results of Liao and Wu in [9].
Next we are going to deduce another form of reverse ratio Young’s inequality by virtue of inequality (1.13).
Theorem 2.3
Let \(a,b > 0\), \(u \in ( 0,1 )\).
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
where \(h = \frac{b}{a}\), \(r = \min \{ 2u,1 - 2u \}\), \(R = \max \{2r,1 - 2r \}\).
Proof
The process of the proof of the second inequality is similar to that of the first one, we only need to investigate the first one. By inequality (1.13), we have
Then we conclude the inequality
So the proof is completed. □
Remark 2.2
It is worth to mention that we can obtain a new reverse Young’s inequality about Specht’s ratio on the basis of inequalities (2.9) and (2.10). That is:
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
When \(0 < u < \frac{1}{2}\), it is easy to see that the right-hand side of the inequality above and the corresponding side of the inequality in Theorem 2.3 cannot be compared because the value of \(K(\sqrt[4]{h},2)^{R}\) will change with R; neither of them is uniformly better than the other. The other case when \(\frac{1}{2} \le u < 1\) has the same conclusion analysis, so we omit it here.
Remark 2.3
If we use \(a^{2}\), \(b^{2}\) to take the place of a, b in inequalities (2.9) and (2.10), we can easily obtain the following results:
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
Note that we can easily obtain the square form of reverse Young’s inequality on the basis of the above inequalities (2.11) and (2.12) which are more excellent than the inequality in [8].
Theorem 2.4
Let \(a,b > 0\), \(u \in ( 0,1 )\). We have the following.
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
where \(h = \frac{b}{a}\), \(r = \min \{ 2u,1 - 2u \}\), \(R = \min \{2r,1 - 2r \}\).
Proof
The process of proving the second inequality is similar to that of proving the first one, so we omit it and only investigate the first one. By inequalities (2.11) and (2.12), we can deduce that
And then we can conclude the following inequality:
□
At the end of the section, we are going to consider a new square form based on Theorem 2.3 which is more creative than the inequality in [12].
Theorem 2.5
Let \(a,b > 0\), \(u \in ( 0,1 )\). We have the following.
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
where \(h = \frac{b}{a}\), \(r = \min \{ 2u,1 - 2u \}\) and \(R = \max \{ 1 - 2r,2r \}\).
Proof
It is to mention that the process always has the similarity due to the properties of the problem. So we only need to investigate the first inequality. By inequality (2.9), we have
Then we can conclude the inequality
So the proof is completed. □
3 New operator versions of Young-type inequalities
In the section, we will give some more excellent versions of Young-type operator inequalities and their reverse by the monotonic property of operator functions.
First, we present the monotonic property of operator function, which is the basis of the following discussion.
Lemma 3.1
[13]
Let \(T \in B ( H )\) be self-adjoint. If f and g are both continuous functions with \(f ( t ) \ge g ( t )\) for \(t \in \operatorname{Sp} ( T )\) (where the sign \(\operatorname{Sp} ( T )\) denotes the spectrum of operator T), then \(f ( T ) \ge g ( T )\).
Next we present our main results on the basis of inequalities (2.1)-(2.10). By Lemma 3.1, we have the following.
Theorem 3.1
Let \(A,B \in B ( H )\) be positive invertible operators, I is the identity operator and \(u \in ( 0,1 )\). If all positive numbers m, \(m'\) and M, \(M'\) satisfy either of the conditions \(0 < mI \le A \le m'I < M'I \le B \le MI\) or \(0 < mI \le B \le m'I < M'I \le A \le MI\), then:
-
(1)
If \(0 < {u} < \frac{1}{4}\), then
$$\begin{aligned} A\nabla_{u}B \ge&2u ( A\nabla B - A\# B ) + 2u ( A\# B + A - 2A \#_{\frac{1}{4}}B ) \\ &{}+ r ( A + A\#_{\frac{1}{4}}B - 2A\#_{\frac{1}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}A\#_{u}B, \end{aligned}$$(3.1)where \(h = \frac{M}{m}\), \(r = \min \{ 1 - 4u,4u \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(2)
If \(\frac{1}{4} \le u < \frac{1}{2}\), then
$$\begin{aligned} A\nabla_{u}B \ge&2u ( A\nabla B - A\# B ) + ( 1 - 2u ) ( A\# B + A - 2A\#_{\frac{1}{4}}B ) \\ &{}+ r ( A\# B + A\#_{\frac{1}{4}}B - 2A\#_{\frac{3}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}A\#_{u}B, \end{aligned}$$(3.2)where \(h = \frac{M}{m}\), \(r = \min \{ 2 - 4u,4u - 1 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(3)
If \(\frac{1}{2} \le u < \frac{3}{4}\), then
$$\begin{aligned} A\nabla_{u}B \ge&2 ( 1 - u ) ( A\nabla B - A\# B ) + ( 2u - 1 ) ( A\# B + B - 2A\#_{\frac{1}{4}}B ) \\ &{}+ r ( A\# B + A\#_{\frac{3}{4}}B - 2A\#_{\frac{5}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}A\#_{u}B, \end{aligned}$$(3.3)where \(h = \frac{M}{m}\), \(r = \min \{ 3 - 4u,4u - 2 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(4)
If \(\frac{3}{4} \le u < 1\), then
$$\begin{aligned} A\nabla_{u}B \ge&2 ( 1 - u ) ( A\nabla B - A\# B ) + ( 2 - 2u ) ( A\# B + B - 2A\#_{\frac{1}{4}}B ) \\ &{}+ r ( B + A\#_{\frac{3}{4}}B - 2A\#_{\frac{7}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{r_{1}}A\#_{u}B, \end{aligned}$$(3.4)where \(h = \frac{M}{m}\), \(r = \min \{ 4 - 4u,4u - 3 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
And these equalities hold if and only if \(A = B\) and \(m = M\).
Proof
We only need to investigate inequality (3.1) due to the similarity of the process of proof.
If \(0 < {u} < \frac{1}{4}\), by inequality (2.1), for any \(x > 0\), we have
For \(X = A^{ - \frac{1}{2}}BA^{ - \frac{1}{2}}\), under the first condition, we get \(I \le hI = \frac{M}{m}I \le X \le h'I = \frac{M'}{m'}\), and then \(\operatorname{Sp} ( X ) \subseteq [ h,h' ] \subseteq ( 1, +\infty )\).
By Lemma 3.1, we have
Since the Kantorovich constant \(K ( t,2 ) = \frac{ ( 1 + t )^{2}}{4t}\) is an increasing function on \(( 1, \infty )\), then
In a similar way, under the second condition, we have \(I \le\frac{1}{h}I = \frac{m}{M}I \le X \le\frac{1}{h'}I = \frac{m'}{M'}\), and \(\operatorname{Sp} ( X ) \subseteq [ \frac{1}{h},\frac{1}{h'} ] \subseteq ( 0,1 )\). By Lemma 3.1, we have
Since the Kantorovich constant \(K ( t,2 ) = \frac{ ( 1+ t )^{2}}{4t}\) is a decreasing function on \(( 0,1 )\), then
Then multiplying inequalities (3.5) and (3.6) by \(A^{\frac{1}{2}}\) on the left-hand side and on the right-hand side, we can deduce the required inequality (3.1). □
Theorem 3.2
Let \(A,B \in B ( H )\) be positive invertible operators, I is the identity operator and \(u \in ( 0,1 )\). If all positive numbers m, \(m'\) and M, \(M'\) satisfy either of the conditions \(0 < mI \le A \le m'I < M'I \le B \le MI\) or \(0 < mI \le B \le m'I < M'I \le A \le MI\), then:
-
(1)
If \(0 < {u} < \frac{1}{4}\), then
$$\begin{aligned} A\nabla_{u}B \le&2 ( 1 - u ) ( A\nabla B - A\# B ) - 2u ( A\# B + B - 2A\#_{\frac{1}{4}}B ) \\ &{}- r ( B + A\#_{\frac{3}{4}}B - 2A\#_{\frac{7}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}A\#_{u}B, \end{aligned}$$(3.7)where \(h = \frac{M}{m}\), \(r = \min \{ 1 - 4u,4u \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(2)
If \(\frac{1}{4} \le u < \frac{1}{2}\), then
$$\begin{aligned} A\nabla_{u}B \le&2 ( 1 - u ) ( A\nabla B - A\# B ) - ( 1 - 2u ) ( A\# B + B - 2A\#_{\frac{1}{4}}B ) \\ &{}- r ( A\# B + A\#_{\frac{3}{4}}B - 2A\#_{\frac{5}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}A\#_{u}B, \end{aligned}$$(3.8)where \(h = \frac{M}{m}\), \(r = \min \{ 2 - 4u,4u - 1 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(3)
If \(\frac{1}{2} \le u < \frac{3}{4}\), then
$$\begin{aligned} A\nabla_{u}B \le&2u ( A\nabla B - A\# B ) - ( 2u - 1 ) ( A\# B + A - 2A\#_{\frac{1}{4}}B ) \\ &{}- r ( A\# B + A\#_{\frac{1}{4}}B - 2A\#_{\frac{3}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}A\#_{u}B, \end{aligned}$$(3.9)where \(h = \frac{M}{m}\), \(r = \min \{ 3 - 4u,4u - 2 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
-
(4)
If \(\frac{3}{4} \le u < 1\), then
$$\begin{aligned} A\nabla_{u}B \le&2u ( A\nabla B - A\# B ) - ( 2 - 2u ) ( A\# B + A - 2A\#_{\frac{1}{4}}B ) \\ &{}- r ( A + A\#_{\frac{1}{4}}B - 2A\#_{\frac{1}{8}}B ) + K \bigl( \sqrt[8]{h},2 \bigr)^{ - r_{1}}A\#_{u}B, \end{aligned}$$(3.10)where \(h = \frac{M}{m}\), \(r = \min \{ 4 - 4u,4u - 3 \}\), \(r_{1} = \min \{ 2r,1 - 2r \}\).
And these equalities hold if and only if \(A = B\) and \(m = M\).
Proof
The process of the proof is analogous to that of Theorem 3.1, so we omit it here. □
Theorem 3.3
Let \(A,B \in B ( H )\) be positive invertible operators, I is the identity operator and \(u \in ( 0,1 )\). If all positive numbers m, \(m'\) and M, \(M'\) satisfy either of the conditions \(0 < mI \le A \le m'I < M'I \le B \le MI\) or \(0 < mI \le B \le m'I < M'I \le A \le MI\), then:
-
(1)
If \(0 < u < \frac{1}{2}\), then
$$A\nabla B \le2u ( A\nabla B - A\# B ) + r ( A\# B + A - 2A\#_{\frac{1}{4}}B ) + K ( \sqrt{h},2 )^{R}A\#_{u}B; $$ -
(2)
If \(\frac{1}{2} \le u < 1\), then
$$A\nabla B \le2 ( 1 - u ) ( A\nabla B - A\# B ) + r ( A\# B + B - 2A\#_{\frac{1}{4}}B ) + K ( \sqrt{h},2 )^{R}A\#_{u}B, $$
where \(r = \min \{ 2u,1 - 2u \}\), \(R = \max \{ 2r,1 - 2r \}\) and \(h = \frac{M}{m}\).
Proof
We only need to investigate the first inequality due to the similarity of the process of proof.
By inequality (2.9), for any \(x > 0\), we have
For \(X = A^{ - \frac{1}{2}}BA^{ - \frac{1}{2}}\), under the first condition, we get \(I \le hI = \frac{M}{m}I \le X \le h'I = \frac{M'}{m'}\).
By Lemma 3.1 and the property of the Kantorovich constant, we have
Then multiplying the above inequality by \(A^{\frac{1}{2}}\) on the left-hand side and right-hand side respectively, we can deduce the required inequality. The investigation under the other condition is similar to the above, so we omit it here. □
Remark 3.1
We get the promotion of the inequality by Liao et al. in [8]. Since the Kantorovich constant \(K ( t,2 ) = \frac { ( t + 1 )^{2}}{4t} > 1\) for \(t > 0\) and \({t} \ne1\), combining (3.1) and (3.7) for \(0 < u < \frac{1}{4}\), we have
A similar analysis can be performed in other cases, so we omit them here.
4 New matrix versions of Young’s inequalities for the Hilbert-Schmidt norm
In the last part, we focus on the matrix version of Young’s inequality for the Hilbert-Schmidt norm. As is well known, every positive semi-definite matrix can be unitarily diagonalizable. Then, for any positive semi-definite matrices A and B, there exist two unitary matrices U and V such that \(A = U\operatorname{diag} ( \lambda_{1},\lambda_{2}, \ldots,\lambda_{n} )U^{*}\) and \(B = V\operatorname{diag} ( \mu_{1},\mu_{2}, \ldots,\mu_{n} )V^{*}\) (\(\lambda_{i} \ge0\), \(\mu_{i} \ge0\), \(i = 1,2,\ldots,n\)). The spectrum of A and B is denoted by \(\operatorname{Sp} ( A ) = \{ \lambda_{1},\lambda_{2},\ldots,\lambda_{n} \}\) and \(\operatorname{Sp} ( B ) = \{ \mu_{1},\mu_{2}, \ldots,\mu_{n} \}\), respectively.
Theorem 4.1
Suppose \(A,B,X \in M_{n}\) and A, B are positive semi-definite matrices, I is the identity operator, and \(0 < mI \le A,B \le MI\) and \(u \in ( 0,1 )\). Then we have the following:
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
where \(h = \frac{b}{a}\), \(r = \min \{ 2u,1 - 2u \}\), \(R = \max \{ 2r,1 - 2r \}\).
Proof
The proof of the second inequality is similar to that of the first one. Thus we only need to investigate the first one. Since A and B are positive matrices, by the spectral decomposition theorem, there exist unitary matrices \(U,V \in M_{n}\) satisfying \(A = U\Lambda _{1}U^{*}\), \(B = V\Lambda_{2}V^{*}\), where \(\Lambda_{1} = \operatorname{diag} ( \lambda_{1},\lambda_{2}, \ldots,\lambda_{n} )\), \(\Lambda_{2} = \operatorname{diag} ( \mu_{1},\mu_{2}, \ldots,\mu_{n} )\) (\(\lambda_{i} \ge0\), \(\mu_{i} \ge0\), \(i = 1,2,\ldots,n\)).
Suppose \(Y = U^{*}XV = [y_{ij}]\), we have
By Theorem 2.4 and the unitarily invariant property of the Hilbert-Schmidt norm, we can see that
where \(t_{ij} = \frac{\lambda_{i}}{\mu_{j}}\).
Utilizing the condition \(0 < mI \le A,B \le MI\), \(\frac{m}{M} = \frac{1}{h} \le t_{ij} = \frac{\lambda_{i}}{\mu_{j}} \le h = \frac{M}{m}\) and the property of the Kantorovich constant, we have
So the proof is completed. □
Theorem 4.2
Let \(A,B \in M_{n}\) such that A, B are positive invertible operators, I is the identity operator, \(0 < mI \le A,B \le MI\) and \(u \in ( 0,1 )\). We have the following:
If \(0 < u < \frac{1}{2}\), then
If \(\frac{1}{2} \le u < 1\), then
where \(h = \frac{M}{m}\), \(r = \min \{ 2u,1 - 2u \}\), \(R = \min \{2r,1 - 2r \}\).
Proof
The proof of the second inequality is similar to that of the first one. Thus we only need to investigate the first one. Since A and B are positive matrices, by the spectral decomposition theorem, there exist unitary matrices \(U,V \in M_{n}\) satisfying \(A = U\Lambda _{1}U^{*}\), \(B = V\Lambda_{2}V^{*}\), where \(\Lambda_{1} = \operatorname{diag} ( \lambda_{1},\lambda_{2}, \ldots,\lambda_{n} )\), \(\Lambda_{2} = \operatorname{diag} ( \mu_{1},\mu_{2}, \ldots,\mu_{n} )\) (\(\lambda_{i} \ge0\), \(\mu_{i} \ge0\), \(i = 1,2,\ldots,n\)).
Suppose \(Y = U^{*}XV = [y_{ij}]\), we have
And so
By Theorem 2.5 and the unitarily invariant property of the Hilbert-Schmidt norm, we can see that
where \(t_{ij} = \frac{\lambda_{i}}{\mu_{j}}\). Under the condition \(0 < mI \le A,B \le MI\), \(\frac{m}{M} = \frac{1}{h} \le t_{ij} = \frac{\lambda_{i}}{\mu_{j}} \le h = \frac{M}{m}\) and by the property of the Kantorovich constant, we have
So the proof is completed. □
References
Kittaneh, F, Manasrah, Y: Improved Young and Heinz inequalities for matrix. J. Math. Anal. Appl. 361(1), 262-269 (2010)
Kittaneh, F, Manasrah, Y: Improved Young and Heinz inequalities for matrices. Linear Multilinear Algebra 59(9), 1031-1037 (2011)
Zhao, J, Wu, J: Operator inequalities involving improved Young and its reverse inequalities. J. Math. Anal. Appl. 421(2), 1779-1789 (2015)
Furuichi, S: Refined Young inequalities with Specht’s ratio. J. Egypt. Math. Soc. 20(1), 46-49 (2011)
Tominaga, M: Specht’s ratio in the Young inequality. Sci. Math. Jpn. 55, 583-588 (2002)
Zuo, H, Shi, G, Fujii, M: Refined Young inequality with Kantorovich constant. J. Math. Inequal. 5(4), 551-556 (2011)
Wu, J, Zhao, J: Operator inequalities and reverse inequalities related to the Kittaneh-Manasrah inequalities. Linear Multilinear Algebra 62(7), 884-894 (2014)
Liao, W, Wu, J, Zhao, J: New versions of reverse Young and Heinz mean inequalities with the Kantorovich constant. Taiwan. J. Math. 19(2), 467-479 (2015)
Liao, W, Wu, J: Improved Young and Heinz inequalities with the Kantorovich constant. J. Math. inequalities 10(2), 559-570 (2015)
Hirzallah, O, Kittaneh, F: Matrix Young inequalities for the Hilbert-Schmidt norm. Linear Algebra Appl. 308(1-3), 77-84 (2000)
He, C, Zou, L: Some inequalities involving unitarily invariant norms. Math. Inequal. Appl. 15(4), 767-776 (2012)
Merris, R, Pierce, S: Monotonicity of positive semidefinite Hermitian matrices. Proc. Am. Math. Soc. 31(2), 437-440 (1972)
Pečarić, J, Furuta, T, Mićić, T, Seo, Y: Mond-Pečarić Method in Operator Inequalities. ELEMENT, Zagreb (2005)
Acknowledgements
The authors would like to thank the editor for their helpful and detailed feedback and for revising the manuscript. The authors are also grateful to Dr. Haisong Cao for fruitful discussion and revising the manuscript. This work is supported by the National Natural Science Foundation of China (Grant No. 70872123).
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Zhang, J., Wu, J. New progress on the operator inequalities involving improved Young’s and its reverse inequalities relating to the Kantorovich constant. J Inequal Appl 2017, 69 (2017). https://doi.org/10.1186/s13660-017-1344-9
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DOI: https://doi.org/10.1186/s13660-017-1344-9