1. Fundamental Principles
Let indicate the set of all bounded linear operators on a complex separable Hilbert space H, and let indicate the two-sided ideal of compact operators in. If, the singular values of T, denoted by are the eigenvalues of the positive operator ordered as and repeated according to multiplicity. It is well known that for. It follows by Weyl’s monotonicity principle (see, e.g., [1] , p. 63 or [2] , p. 26) that if
are positive and, then for. Moreover, for if and only if
for. Here, we use the direct sum notation
for the block-diagonal operator defined on. The sin- gular values of and are the same, and they consist of those of
S together with those of T.
Bhatia and Kittaneh have proved in [3] that if such that is self-adjoint, , and, then
(1.1)
for.
Audeh and Kittaneh in [4] prove inequality which is equivalent to inequality (1.1):
If such that, then
(1.2)
for.
The well-known arithmetic-geometric mean inequality for singular values, due to Bhatia and Kittaneh [5] , says that if, then
(1.3)
for. On the other hand, Zhan has proved in [6] that if are positive, then
(1.4)
for. Moreover, Tao has proved in [7] that if such
that, then
(1.5)
for.
Audeh and Kittaneh have proved in [4] that:
If such that is self-adjoint, , and, then
(1.6)
for.
It has been pointed out in [4] that the four inequalities (1.3)-(1.6) are equi- valent.
Moreover, Tao in [7] uses inequality (1.3) to prove that if and are positive operators in,. Then
(1.7)
for.
2. Introduction
In this study, we will present several new inequalities, and prove that they are equivalent to arithmetic-geometric mean inequality.
The following are the proved inequalities in this study:
Let and be operators in where, and arbitrary operators. Then
(2.1)
for.
Let and be arbitrary operators in. Then we have
(2.2)
for.
Let be operators in. Then
(2.3)
for.
If and are operators in. Then
(2.4)
for.
Let be positive operators in Then
(2.5)
for.
3. Main Results
Our first singular value inequality needs the following lemma.
Lemma 1: Let be a positive operator in, be an arbitrary operator in. Then we have
(3.1)
Now we will prove the first Theorem which is equivalent to arithmetic- geometric mean inequality.
Theorem 3.1 Let and be operators in where, and arbitrary operators. Then
for.
Proof. Let (because by assumption), and let
. Then we have
From (1.5) we have
for.
Now we will prove that Theorem (3.1) is equivalent to arithmetic-geometric mean inequality.
Theorem 3.2 The following statements are equivalent:
1) Let, then
for.
2) Let and be operators in where, and arbitrary operators. Then
for.
Proof. 1) ® 2) Let
Now apply arithmetic-geometric mean inequality to get
for. But
The above steps implies that
for.
2) ® 1) The matrix can be factorized as
, but it is well known that
for. So
for, from (2) we have
(3.2)
for. Now let in Inequality (3.2) we get
(3.3)
for, which is the arithmetic-geometric mean inequality.
The following lemma which was proved by Bhatia [1] is essential to prove the next theorem.
Lemma 2 Let be arbitrary operator in. Then
(3.4)
Now we will prove the following theorem which is more general than Theo- rem (3.1) and equivalent to arithmetic-geometric mean inequality.
Theorem 3.3 Let and be arbitrary operators in. Then we have
for.
Proof. Applying Lemma (2) gives for an arbitrary ope- rator. Let by using Inequality (3.1) we have
Hence using Inequality (1.5) gives
.
Remark 1 Theorem (3.3) is generalization of Theorem (3.1) because here X is arbitrary operator but there A should be positive operator.
Remark 2 Inequality (2.2) is equivalent to arithmetic-geometric mean inequality. We can prove this equivalent by similar steps used to prove Theorem (3.2).
The following theorem is a generalization of Theorem (3.1) and Theorem (3.3).
Theorem 3.4 Let and be arbitrary operators in. Then we have
for.
Proof. Let Then
Hence
use Inequality (1.5) to get the required result.
Remark 3 Replace B, D by 0 in Inequality (2.4) will gives Inequality (2.1).
Remark 4 Replace A, C by 0 in Inequality (2.4) will also gives Inequality (2.1).
Now we will use Inequality (1.3) to prove the following theorem, then we will show that they are equivalent.
Theorem 3.5 Let be operators in. Then
for.
Proof. Let Then and
Now use Inequality (1.3) we get
for.
Now we will prove that Inequality (2.3) is equivalent to Inequality (1.3).
Theorem 3.6 The following statements are equivalent:
1) Let. Then
for.
2) Let be operators in. Then
for.
Proof. 1) ® 2) It is the proof of Theorem (3.5).
2) ® 1) By replacing and in Inequality (2.3), we
get From this we reach to
which implies that for.
In the rest of this paper, we will prove new inequality which is equivalent to Inequality (1.7).
Theorem 3.7 Let be positive operators in, n is an even integer,. Then
(3.5)
for.
Proof. Let Then we have
and Now apply
Inequality (1.7) we get the result.
We will prove that Inequality (1.7) is equivalent to Inequality (3.5).
Theorem 3.8 The following statements are equivalent:
1) Let and be positive operators in,. Then
for.
2) Let be positive operators in, n is even integer,. Then
for.
Proof. 1) ® 2) This implication follows from the proof of Theorem 3.7.
2) ® 1) Let in Inequality (3.5) to get
for. But and for
.
If and only if, this gives
for, replace by, by in this inequality we will get
for.
4. Conclusion
Since this study has been completed, we can conclude that several singular value inequalities for compact operators are equivalent to arithmetic-geometric mean inequality, which in turns have many crucial applications in operator theory, and from this point we advise interested authors to join these results with results in other studies to make connection between several branches in operator theory.
Acknowledgements
The author is grateful to the University of Petra for its Support. The Author is grateful to the referee for his comments and suggestions.