Reverses of the Young inequality for matrices and operators
aa r X i v : . [ m a t h . F A ] O c t REVERSES OF THE YOUNG INEQUALITY FOR MATRICES ANDOPERATORS
MOJTABA BAKHERAD , MARIO KRNI ´C AND MOHAMMAD SAL MOSLEHIAN Abstract.
We present some reverse Young-type inequalities for the Hilbert-Schmidt normas well as any unitarily invariant norm. Furthermore, we give some inequalities dealingwith operator means. More precisely, we show that if
A, B ∈ B ( H ) are positive operatorsand r ≥ A ∇ − r B + 2 r ( A ∇ B − A♯B ) ≤ A♯ − r B and prove that equality holds if andonly if A = B . We also establish several reverse Young-type inequalities involving trace,determinant and singular values. In particular, we show that if A, B are positive definitematrices and r ≥
0, then tr((1 + r ) A − rB ) ≤ tr (cid:12)(cid:12) A r B − r (cid:12)(cid:12) − r (cid:16) √ tr A − √ tr B (cid:17) . Introduction and preliminaries
Let H be a Hilbert space and let B ( H ) be the C ∗ -algebra of all bounded linear operatorson H with the operator norm k · k and the identity I H . If dim H = n , then we identify B ( H )with the space M n of all n × n complex matrices and denote the identity matrix by I n . Foran operator A ∈ B ( H ), we write A ≥ A is positive (positive semidefinite for matrices),and A > A is positive invertible (positive definite for matrices). For A, B ∈ B ( H ), wesay A ≥ B if A − B ≥
0. Let B + ( H ) (resp., P n ) denote the set of all positive invertibleoperators (resp., positive definite matrices). A norm ||| . ||| on M n is called unitarily invariantnorm if ||| U AV ||| = ||| A ||| for all A ∈ M n and all unitary matrices U, V ∈ M n . The Hilbert-Schmidt norm is defined by k A k = (cid:16)P nj =1 s j ( A ) (cid:17) / , where s ( A ) = ( s ( A ) , · · · , s n ( A ))denotes the singular values of A , that is, the eigenvalues of the positive semidefinite matrix | A | = ( A ∗ A ) / , arranged in the decreasing order with their multiplicities counted. This normis unitarily invariant. It is known that if A = [ a ij ] ∈ M n , then k A k = (cid:16) P ni,j =1 | a ij | (cid:17) / .The weighted operator arithmetic mean ∇ ν , geometric mean ♯ ν , and harmonic mean ! ν ,for ν ∈ [0 ,
1] and
A, B ∈ B + ( H ), are defined as follows: A ∇ ν B = (1 − ν ) A + νB,A ♯ ν B = A (cid:0) A − BA − (cid:1) ν A ,A ! ν B = (cid:0) (1 − ν ) A − + νB − (cid:1) − . Mathematics Subject Classification.
Key words and phrases.
Young inequality; positive operator, operator mean; unitarily invariant norm;determinant; trace. If ν = 1 /
2, we denote arithmetic, geometric and harmonic mean, respectively, by ∇ , ♯ and !,for brevity.The classical Young inequality states that a ν b − ν ≤ νa + (1 − ν ) b, when a, b ≥ ν ∈ [0 , ν = , we obtain the arithmetic-geometric mean inequality √ ab ≤ a + b . An operator Young inequality reads as follows: A ! ν B ≤ A ♯ ν B ≤ A ∇ ν B, ν ∈ [0 , , (1.1)where A, B ∈ B + ( H ) and ν ∈ [0 , s j ( A ν B − ν ) ≤ s j ( νA + (1 − ν ) B ) , in which A, B ∈ M n are positive semidefinite, j = 1 , , . . . , n , and ν ∈ [0 , ||| A ν B − ν ||| ≤ ||| νA + (1 − ν ) B ||| , where A, B ∈ M n are positive semidefinite and 0 ≤ ν ≤
1. Kosaki [13] proved that theinequality k A ν XB − ν k ≤ k νAX + (1 − ν ) XB k (1.2)holds for matrices A, B, X ∈ M n such that A, B are positive semidefinite, and for 0 ≤ ν ≤ ν = inequality (1.2) may not hold for other unitarilyinvariant norms. Hirzallah and Kittaneh [7], gave a refinement of (1.2) by showing that k A ν XB − ν k + r k AX − XB k ≤ k νAX + (1 − ν ) XB k , (1.3)in which A, B, X ∈ M n are such that A, B are positive semidefinite, 0 ≤ ν ≤ r =min { ν, − ν } . A determinant version of the Young inequality is also known (see [9, p. 467]):det( A ν B − ν ) ≤ det( νA + (1 − ν ) B ) , where A, B, X ∈ M n are such that A, B are positive semidefinite and 0 ≤ ν ≤
1. Thisdeterminant inequality was recently improved in [10]. Further, Kittaneh [11], proved that ||| A − ν XB ν ||| ≤ ||| AX ||| − ν ||| XB ||| ν , (1.4)in which ||| . ||| is any unitarily invariant norm, A, B, X ∈ M n are such that A, B are positivesemidefinite and 0 ≤ ν ≤
1. Conde [2], showed that2 ||| A − ν XB ν ||| + (cid:0) ||| AX ||| − ν − ||| XB ||| ν (cid:1) ≤ ||| AX ||| − ν ) + ||| XB ||| ν , EVERSES OF THE YOUNG INEQUALITY 3 where ||| . ||| is unitarily invariant norm, A, B, X ∈ M n are such that A, B are positivesemidefinite and 0 ≤ ν ≤
1. Tominaga [20, 21] employed Specht’s ratio to Young inequality.In addition, some reverses of Young inequality are established in [4].For a, b ∈ R , the number x = νa + (1 − ν ) b belongs to the interval [ a, b ] for all ν ∈ [0 , ν > ν <
0. Exploiting this obvious fact, Fujii [3],showed that if f is an operator concave function on an interval J , then the inequality f ( C ∗ XC − D ∗ Y D ) ≤ | C | f ( V ∗ XV ) | C | − D ∗ f ( Y ) D holds for all self-adjoint operators X, Y and operators
C, D in B ( H ) with spectra in J , suchthat C ∗ C − D ∗ D = I H , σ ( C ∗ XC − D ∗ Y D ) ⊆ J and C = V | C | is the polar decomposition of C .In this direction, by using some numerical inequalities, we obtain reverses of (1.1), (1.2),(1.3) and (1.4) under some mild conditions. We also aim to give some reverses of the Younginequality dealing with operator means of positive operators. Finally, we present some sin-gular value inequalities of Young-type involving trace and determinant.2. Reverses of the Young inequality for the Hilbert-Schmidt norm
In this section we deal with reverses of the Young inequality for the Hilbert-Schmidt norm.To this end, we need some lemmas.
Lemma 2.1.
Let a, b > . If r ≥ or r ≤ − , then (1 + r ) a − rb ≤ a r b − r . (2.1) Proof.
Let f ( t ) = t − r − (1 + r ) + rt , t ∈ (0 , ∞ ). It is easy to see that f ( t ) attains its minimumat t = 1, on the interval (0 , ∞ ). Hence, f ( t ) ≥ f (1) = 0 for all t >
0. Letting t = ba , we getthe desired inequality. (cid:3) Remark . By virtue of Lemma 2.1, it follows that the inequality((1 + r ) a − rb ) ≤ (cid:0) a r b − r (cid:1) (2.2)holds if a ≥ b > r ≥
0, or b ≥ a > r ≤ − Lemma 2.3. [22, Theorem 3.4] (Spectral Decomposition) Let A ∈ M n with eigenvalues λ , λ , . . . , λ n . Then A is normal if and only if there exists a unitary matrix U such that U ∗ AU = diag( λ , λ , · · · , λ n ) . In particular, A is positive definite if and only if λ j > for j = 1 , , . . . , n . Now, our first result reads as follows.
M. BAKHERAD, M. KRNI´C, M.S. MOSLEHIAN
Theorem 2.4.
Let
A, B, X ∈ M n and let m, m ′ be positive scalars. If A ≥ mI n ≥ B > and r ≥ , or B ≥ m ′ I n ≥ A > and r ≤ − , then the following inequality holds: k (1 + r ) AX − rXB k ≤ k A r XB − r k . Proof.
It follows from Lemma 2.3 that there are unitary matrices
U, V ∈ M n such that A = U Λ U ∗ and B = V Γ V ∗ , where Λ = diag( λ , λ , · · · , λ n ), Γ = diag( γ , γ , · · · , γ n ), and λ j , γ j , j = 1 , . . . , n , are positive. If Z = U ∗ XV = (cid:2) z ij (cid:3) , then(1 + r ) AX − rXB = U (cid:16) (1 + r )Λ Z − rZ Γ (cid:17) V ∗ = U h(cid:16) (1 + r ) λ i − rγ j (cid:17) z ij i V ∗ (2.3)and A r XB − r = U Λ r U ∗ XV Γ − r V ∗ = U Λ r Z Γ − r V ∗ = U h(cid:16) λ ri γ − rj (cid:17) z ij i V ∗ . (2.4)Suppose first that A ≥ mI n ≥ B > r ≥
0. Then, it follows that λ i ≥ γ j , ≤ i, j ≤ n, (2.5)so, utilizing (2.3) and (2.4), we have k (1 + r ) AX − rXB k = n X i,j =1 (cid:16) (1 + r ) λ i − rγ j (cid:17) | z ij | ≤ n X i,j =1 (cid:16) λ ri γ − rj (cid:17) | z ij | (by inequality (2.2) and (2.5))= k A r XB − r k . The same conclusion can be drawn for the case of B ≥ m ′ I n ≥ A > r ≤ − (cid:3) Recall that a continuous real valued function f , defined on an interval J , is called operatormonotone if A ≤ B implies f ( A ) ≤ g ( B ), for all A, B ∈ M n with spectra in J . Now, thefollowing result can be accomplished as an immediate consequence of Theorem 2.4. Corollary 2.5.
Suppose that A j , B j , X ∈ M n , ≤ j ≤ n , with spectra in an interval J , andlet m j , m ′ j , ≤ j ≤ n , be positive scalars. If A j ≥ m j I n ≥ B j > , ≤ j ≤ n , and r ≥ , or B j ≥ m ′ j I n ≥ A j > , ≤ j ≤ n , and r ≤ − , then the inequality (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X j =1 (cid:16) (1 + r ) f ( A j ) X − rXf ( B j ) (cid:17)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X j =1 f ( A j ) r X n X j =1 f ( B j ) − r (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) holds for any operator monotone function f defined on interval J .Proof. It suffices to set A = P nj =1 f ( A j ) and B = P nj =1 f ( B j ) in Theorem 2.4 to get thedesired inequality. (cid:3) EVERSES OF THE YOUNG INEQUALITY 5
Generally speaking, Theorem 2.4 does not hold for arbitrary positive definite matrices A and B . The reason for this lies in the fact that the inequality (2.2) is not true for arbitrarypositive numbers a, b . To see this, let a = 1 , b = 4 , r = 2.Our next intention is to derive a result related to Theorem 2.4 which holds for all positivedefinite matrices. Observe that the inequality((1 + r ) a − rb ) − r ( a − b ) = (1 + 2 r ) a − rab ≤ ( a ) r ( ab ) − r = (cid:0) a r b − r (cid:1) yields an appropriate relation instead of (2.2), for arbitrary positive numbers a, b and r ≥ r ≤ − , as follows:((1 + r ) a − rb ) ≤ (cid:0) a r b − r (cid:1) + r ( a − b ) a, b > , r ≥ r ≤ − . Note also that if a = b , then the equality holds.Now, utilizing this inequality and the same argument as in the proof of Theorem 2.4,i.e. the spectral theorem for positive definite matrices, we can accomplish the correspondingresult. Theorem 2.6.
Suppose that
A, B ∈ P n and X ∈ M n . Then the inequality k (1 + r ) AX − rXB k ≤ (cid:13)(cid:13) A r XB − r (cid:13)(cid:13) + r k AX − XB k (2.6) holds for r ≥ or r ≤ − . Reverse Young-type inequalities involving unitarily invariant norms
It has been shown in [8] that the inequality k A r XB r k ≥ k X k − r k AXB k r (3.1)holds for A, B ∈ P n , 0 = X ∈ M n and r ≥
0. Applying inequality (3.1) yields the relation k A r XB − r k ≥ k AX k r k XB k − r , (3.2)where r ≥ A, B ∈ P n and X ∈ M n with X = 0.Our next intention is to show that inequality (3.2) holds for every unitarily invariant norm.This can be done by virtue of inequality (1.4). In fact, the following result is, in some way,complementary to inequality (1.4). Lemma 3.1.
Suppose that
A, B ∈ P n , X ∈ M n are such that X = 0 . If r ≥ or r ≤ − ,then the inequality ||| AX ||| r ||| XB ||| − r ≤ ||| A r XB − r ||| holds for any unitarily invariant norm ||| . ||| . M. BAKHERAD, M. KRNI´C, M.S. MOSLEHIAN
Proof.
First, let r ≥
0. Set α = r + 1. Utilizing inequality (1.4), it follows that ||| AX ||| = ||| ( A α ) α ( XB − α )( B α ) α − α ||| ≤ ||| A α XB − α ||| α ||| XB − α B α ||| α − α = ||| A α XB − α ||| α ||| XB ||| α − α , that is, ||| AX ||| ||| XB ||| − αα ≤ ||| A α XB − α ||| α . Hence, ||| AX ||| α ||| XB ||| − α ≤ ||| A α XB − α ||| , whence ||| AX ||| r ||| XB ||| − r ≤ ||| A r XB − r ||| . On the other hand, if r ≤ −
1, set α = − r . By a similar argument, we get the desiredresult. (cid:3) Applying Lemmas 2.1 and 3.1 yields the Young-type inequality(1 + r ) ||| AX ||| − r ||| XB ||| ≤ ||| A r XB − r ||| , (3.3)which holds for matrices A, B ∈ P n , X ∈ M n such that X = 0 and r ≥ r ≤ −
1. It isinteresting that the inequality (3.3) can be improved. But first we have to improve the scalarinequality (2.1).
Lemma 3.2.
Let a, b > and r ≥ or r ≤ − . Then, (1 + r ) a − rb + r ( √ a − √ b ) ≤ a r b − r . (3.4) Proof.
Due to Lemma 2.1, it follows that(1 + r ) a − rb + r ( √ a − √ b ) = − r √ ab + (1 + 2 r ) a ≤ ( √ ab ) − r a r = a r b − r . (cid:3) Obviously, if r ≥
0, inequality (3.4) represents an improvement of inequality (2.1). Finally,we give now an improvement of matrix inequality (3.3).
Theorem 3.3.
Let
A, B ∈ P n , X ∈ M n be such that X = 0 and let r ≥ . Then theinequality (1 + r ) ||| AX ||| − r ||| XB ||| + r ( p ||| AX ||| − p ||| XB ||| ) ≤ ||| A r XB − r ||| holds for any unitarily invariant norm ||| . ||| . EVERSES OF THE YOUNG INEQUALITY 7
Proof. (1 + r ) ||| AX ||| − r ||| XB ||| + r ( p ||| AX ||| − p ||| XB ||| ) ≤ ||| AX ||| r ||| XB ||| − r (by Lemma 3 . ≤ ||| A r XB − r ||| (by Lemma 3 . . (cid:3) Remark . It should be noticed here that the Theorem 3.3 is also true in the case of r ≤ − .However, in this case, the corresponding inequality is less precise than the relation (3.3) anddoes not represent its refinement.4. Reverse Young-type inequalities related to operator means
The matrix Young inequality can be considered in a more general setting. Namely, thisinequality holds also for self-adjoint operators on a Hilbert space. The main objective ofthis section is to derive inequalities which are complementary to mean inequalities in (1.1),presented in the Introduction.The main tool in obtaining inequalities for self-adjoint operators on Hilbert spaces, is thefollowing monotonicity property for operator functions: If X is a self-adjoint operator withthe spectrum sp( X ), then f ( t ) ≥ g ( t ) , t ∈ sp( X ) = ⇒ f ( X ) ≥ g ( X ) . (4.1)For more details about this property the reader is referred to [18].Since A, B ∈ B + ( H ), the expressions A ∇ ν B and A♯ ν B are also well-defined when ν ∈ R \ [0 , Theorem 4.1. If A, B ∈ B + ( H ) and r ≥ or r ≤ − , then A ∇ − r B ≤ A♯ − r B. (4.2) Proof.
By virtue of Lemma 2.1, it follows that f ( x ) = x − r + rx − (1+ r ) ≥ x >
0. Moreover,since B ∈ B + ( H ), it follows that A − BA − ∈ B + ( H ), that is, sp( A − BA − ) ∈ (0 , ∞ ).Thus, applying the monotonicity property (4.1) to the above function f , we have that (cid:16) A − BA − (cid:17) − r + rA − BA − − (1 + r ) I H ≥ . Finally, multiplying both sides of this relation by A , we have A (cid:16) A − BA − (cid:17) − r A + rB − (1 + r ) A ≥ , and the proof is completed. (cid:3) M. BAKHERAD, M. KRNI´C, M.S. MOSLEHIAN If A, B ∈ B + ( H ) are such that A ≤ B , the expression A ! − r B is well defined for r ≥ h ( x ) = − x on (0 , ∞ ) (for more details,see [18]), A ≤ B implies that B − ≤ A − , so that ( r + 1) A − − rB − ∈ B + ( H ). Therefore,the operator A ! − r B = (cid:0) ( r + 1) A − − rB − (cid:1) − is well-defined for r ≥ Corollary 4.2.
Let
A, B ∈ B + ( H ) be such that A ≤ B . If r ≥ , then A♯ − r B ≤ A ! − r B .Proof. Theorem 4.1 with operators A and B replaced by A − and B − , respectively, followsthat A − ∇ − r B − ≤ A − ♯ − r B − . (4.3)Now, applying operator monotonicity of the function h ( x ) = − x , x ∈ (0 , ∞ ), to relation(4.3), we have that (cid:0) A − ♯ − r B − (cid:1) − ≤ (cid:0) A − ∇ − r B − (cid:1) − . Finally, the result follows since (cid:0) A − ♯ − r B − (cid:1) − = A♯ − r B . (cid:3) Kittaneh et . al . obtained in [12] the following relation (see also [14]):2 max { ν, − ν } ( A ∇ B − A♯B ) ≥ A ∇ ν B − A♯ ν B ≥ { ν, − ν } ( A ∇ B − A♯B ) . (4.4)Clearly, the left inequality in (4.4) represents the converse, while the right inequality repre-sents the refinement of arithmetic-geometric mean operator inequality in (1.1).Our next goal is to derive refinement of inequality (4.2) which is, in some way, comple-mentary to above relations in (4.4). Clearly, this will be carried out by virtue of Lemma3.2. Theorem 4.3. If A, B ∈ B + ( H ) and r ≥ , then the following inequality holds A ∇ − r B + 2 r ( A ∇ B − A♯B ) ≤ A♯ − r B. (4.5) Proof.
By virtue of Lemma 3.2, it follows that(1 + r ) − rx + r ( x − √ x + 1) ≤ x − r (4.6)holds for all x >
0. Now, applying the functional calculus, i.e. the property (4.1) to thisscalar inequality, we have(1 + r ) I H − rA − BA − + r ( A − BA − − (cid:16) A − BA − (cid:17) + I H ) ≤ (cid:16) A − BA − (cid:17) − r . Finally, multiplying both sides of this operator inequality by A , we obtain (4.5). (cid:3) Corollary 4.4.
Let
A, B ∈ B + ( H ) and r > . Then, A ∇ − r B = A♯ − r B if and only if A = B .Proof. It follows from Theorem 4.3 and the fact that A ∇ B = A♯B if and only if A = B . (cid:3) EVERSES OF THE YOUNG INEQUALITY 9
Remark . Having in mind that scalar inequality (4.6) holds also for r ≤ − (see Lemma3.2), it follows that inequality (4.5) holds also for r ≤ − . However, if r < −
1, relation (4.5)is less precise than the original inequality (4.2) and does not represent its refinement. Onthe other hand, it is interesting to consider the case when − ≤ r ≤ − . Namely, denoting ν = − r , where ≤ ν ≤
1, (4.5) reduces to A ∇ ν B − ν ( A ∇ B − A♯B ) ≤ A♯ ν B, and this relation coincides with the converse of the arithmetic-geometric mean inequality,that is, with the left inequality in (4.4). Remark . In [12], the authors considered operator version of the classical Heinz mean, i.e.,the operator H ν ( A, B ) =
A ♯ ν B + A ♯ − ν B , (4.7)where A, B ∈ B + ( H ), and ν ∈ [0 , A ♯ B ≤ H ν ( A, B ) ≤ A ∇ B. (4.8)On the other hand, since A, B ∈ B + ( H ), the expression (4.7) is also well-defined for ν ∈ R \ [0 , H − r ( A, B ) = A♯ − r B + A♯ r B ≥ A ∇ − r B + A ∇ r B A ∇ B, r ≥ r ≤ − , complementary to (4.8).In order to conclude this section, we mention yet another inequality closely connected tothe Young inequality. Namely, in [6], it has been shown the equivalence between the Younginequality and the H¨older-McCarthy inequality which asserts that h Ax, x i − r ≤ h A − r x, x i , x ∈ H , k x k = 1 , (4.9)holds for all A ∈ B + ( H ) and r > r < −
1. If − < r <
0, then the sign of inequality in(4.9) is reversed.Now, we give a refinement of the H¨older-McCarthy, once again by exploiting Lemma 3.2.
Theorem 4.7.
Let A ∈ B + ( H ) and r > . Then the inequality ≤ r (cid:16) − h A x, x ih Ax, x i − (cid:17) ≤ h A − r x, x ih Ax, x i r − holds for any unit vector x ∈ H .Proof. By virtue of (4.6), it follows that the inequality 2 r (1 − √ x ) ≤ x − r − x >
0. Now, applying the functional calculus to this inequality and the positive operator λ r A , λ >
0, we have 2 r (cid:16) I H − λ r A (cid:17) ≤ λ − A − r − I H . Further, fix a unit vector x ∈ H . Then we have2 r (cid:16) − λ r h A x, x i (cid:17) ≤ λ − h A − r x, x i − . Finally, putting λ = h Ax, x i − r in the last inequality, we obtain second inequality in (4.10).Clearly, the first inequality sign in (4.10) holds due to (4.9) since h A x, x i ≤ h Ax, x i . (cid:3) Remark . Since relation (4.6) holds for r ≤ − , it follows that the second inequality in(4.10) holds also for r ≤ − . Clearly, the case of r < − − < r < − yields a converse of (4.9).5. Reverse Young-type inequalities for the trace and the determinant
In this section we derive some Young-type inequalities for the trace and the determinantof a matrix. The starting point for this direction is already used Lemma 3.2.In [10], Kittaneh and Manasrah obtained the inequalitytr (cid:12)(cid:12) A ν B − ν (cid:12)(cid:12) + r (cid:16) √ tr A − √ tr B (cid:17) ≤ tr ( νA + (1 − ν ) B ) , (5.1)which holds for positive semidefinite matrices A, B ∈ M n , 0 ≤ ν ≤
1, and r = min { ν, − ν } .By virtue of Lemma 3.2, we can accomplish the inequality complementary to (5.1). To dothis, we also need the following inequality regarding singular values of complex matrices: n X j =1 s j ( A ) s n − j +1 ( B ) ≤ n X j =1 s j ( AB ) ≤ n X j =1 s j ( A ) s j ( B ) . (5.2)Now, we have the following result: Theorem 5.1. If A, B ∈ P n and r ≥ , then the following inequality holds: tr((1 + r ) A − rB ) ≤ tr (cid:12)(cid:12) A r B − r (cid:12)(cid:12) − r (cid:16) √ tr A − √ tr B (cid:17) . (5.3) Proof.
By Lemma 3.2, we have(1 + r ) s j ( A ) − rs n − j +1 ( B ) ≤ s rj ( A ) s − rn − j +1 ( B ) − r (cid:18)q s j ( A ) − q s n − j +1 ( B ) (cid:19) , for j = 1 , , . . . , n . EVERSES OF THE YOUNG INEQUALITY 11
Now, utilizing the above inequality and (5.2), as well as the properties of the trace func-tional, it follows thattr((1 + r ) A − rB ) = (1 + r )tr A − r tr B = n X j =1 ((1 + r ) s j ( A ) − rs n − j +1 ( B )) ≤ n X j =1 s rj ( A ) s − rn − j +1 ( B ) − r n X j =1 (cid:18) s j ( A ) + s n − j +1 ( B ) − q s j ( A ) s n − j +1 ( B ) (cid:19) = n X j =1 s j ( A r ) s n − j +1 ( B − r ) − r tr A + tr B − n X j =1 q s j ( A ) s n − j +1 ( B ) ≤ n X j =1 s j ( A r B − r ) − r tr A + tr B − n X j =1 q s j ( A ) s n − j +1 ( B ) . Moreover, by virtue of the well-known Cauchy-Schwarz inequality, we have n X j =1 q s j ( A ) s n − j +1 ( B ) ≤ n X j =1 s j ( A ) n X j =1 s n − j +1 ( B ) = √ tr A tr B, so that tr((1 + r ) A − rB ) ≤ tr (cid:12)(cid:12) A r B − r (cid:12)(cid:12) − r (cid:16) tr A + tr B − √ tr A tr B (cid:17) . This completes the proof. (cid:3)
Remark . Although the proof of Theorem 5.1 seems to be very interesting, it can be ac-complished in a much simpler way, if we take into account Theorem 3.3. Namely, consideringTheorem 3.3 with X = I n and with the trace norm k · k , that is, k A k = P ni =1 s j ( A ) = tr | A | ,it follows that (1 + r ) k A k − r k B k + r ( p k A k − p k B k ) ≤ k A r B − r k . Now, since
A, B ∈ P n , it follows that k A k = tr A and k B k = tr B , that is, (1 + r ) k A k − r k B k = tr((1 + r ) A − rB ), so we retain the inequality (5.3).Our next intention is to obtain an analogous reverse relation for the determinant of amatrix. In [10], the authors obtained inequalitydet( A ν B − ν ) + r n det(2 A ∇ B − A♯B ) ≤ det( νA + (1 − ν ) B ) , where 0 ≤ ν ≤ r = min { ν, − ν } , and A, B are positive definite matrices. The corre-sponding complementary result can also be established by virtue of Lemma 3.2.
Theorem 5.3.
Let r ≥ and let A, B ∈ P n be such that A ≥ rr +1 B . Then the followinginequality holds: det ((1 + r ) A − rB ) ≤ det (cid:0) A r +1 B − r (cid:1) − r n det (2 A ∇ B − A♯B ) . (5.4) Proof.
The starting point is Lemma 3.2 with a = s j (cid:16) B − AB − (cid:17) and b = 1, i.e. theinequality s r +1 j (cid:16) B − AB − (cid:17) ≥ (1 + r ) s j (cid:16) B − AB − (cid:17) − r + r (cid:18) s j (cid:16) B − AB − (cid:17) − (cid:19) . Furthermore, since A ≥ rr +1 B , it follows that B − AB − ≥ rr +1 I n , which means that s j (cid:16) B − AB − (cid:17) ≥ rr +1 . Consequently, we have that(1 + r ) s j (cid:16) B − AB − (cid:17) − r ≥ . Hence, by virtue of the above two relations and the well-known determinant properties, wehave det (cid:16) B − AB − (cid:17) r +1 = n Y j =1 s r +1 j (cid:16) B − AB − (cid:17) ≥ n Y j =1 " (1 + r ) s j (cid:16) B − AB − (cid:17) − r + r (cid:18) s j (cid:16) B − AB − (cid:17) − (cid:19) ≥ n Y j =1 h (1 + r ) s j (cid:16) B − AB − (cid:17) − r i + r n n Y j =1 "(cid:18) s j (cid:16) B − AB − (cid:17) − (cid:19) = det (cid:16) (1 + r ) B − AB − − rI n (cid:17) + r n det (cid:18)(cid:16) B − AB − (cid:17) − I n (cid:19) . Finally, multiplying both sides of the obtained inequality by det( B ) and utilizing the well-known Binet-Cauchy theorem, we obtain (5.4), as claimed. (cid:3) EVERSES OF THE YOUNG INEQUALITY 13 Reverses of the Young inequality dealing with singular values
Let x = ( x , . . . , x n ) , y = ( y , . . . , y n ) ∈ R n be such that 0 ≤ x ≤ · · · ≤ x n and 0 ≤ y ≤· · · ≤ y n . Then x is said to be log majorized by y , and denoted by x ≺ log y , if k Y j =1 x j ≤ k Y j =1 y j (1 ≤ k < n ) and n Y j =1 x j = n Y j =1 y j . For X ∈ M n and k = 1 , . . . , n , the k -th compound of X is defined as the (cid:0) nk (cid:1) × (cid:0) nk (cid:1) complexmatrix C k ( X ), whose entries are defined by C k ( X ) r,s = det X [( r , r , · · · , r k ) | ( s , s , · · · , s k )],where ( r , r , · · · , r k ) , ( s , s , · · · , s k ) ∈ P k,n = { ( x , · · · , x k ) | ≤ x < · · · < x k ≤ n } arearranged in a lexicographical order and ( r , r , · · · , r k ) and ( s , s , · · · , s k ) are the r -th and s -th element in P k,n , respectively. X [ r, s ] is the k × k matrix that contains the elements inthe intersection of rows ( r , r , · · · , r k ) ∈ P k,n and columns ( s , s , · · · , s k ) ∈ P k,n (for moredetails, see [17]). For example, if n = 3 and k = 2, then (1 , , (1 ,
3) and (2 ,
3) are the first,the second and the third element of P k,n , respectively. So, C ( X ) = det X [1 , | ,
2] det X [1 , | ,
3] det X [1 , | , X [1 , | ,
2] det X [1 , | ,
3] det X [1 , | , X [2 , | ,
2] det X [2 , | ,
3] det X [2 , | , . In a general case, for
A, B ∈ M n , we have C k ( AB ) = C k ( A ) C k ( B ) and s ( C k ( A )) = k Y j =1 s j ( A ) (1 ≤ k ≤ n ) . (6.1)Finally we use the corresponding ideas from [19] to present our last result. Theorem 6.1.
Suppose that
A, B ∈ P n and X ∈ M n . If r ≥ , then (i) s ( A r XB r ) ≻ log s r ( AXB ) s − r ( X ) , (ii) s ( A r XB − r ) ≻ log s r ( AX ) s − r ( XB ) . Proof. (i) Let C k ( X ) ∈ C ( nk ) × ( nk ) denote the k -th component of X , 1 ≤ k ≤ n . Then, we have k Y i =1 s i ( A r XB r ) = s ( C k ( A r XB r )) by (6.1)= s ( C k ( A ) r C k ( X ) C k ( B ) r ) by (6.1) ≥ s − r ( C k ( X )) s r ( C k ( AXB )) (by inequality (3.1))= k Y i =1 s − ri ( X ) k Y i =1 s ri ( AXB ) . Moreover, if k = n , we have n Y i =1 s i ( A r XB r ) = | det( A r XB r ) | = (det A ) r | det X | (det B ) r and n Y i =1 s i ( X ) − r n Y i =1 s i ( AXB ) r = | det X − r | | det( AXB ) r | = (det A ) r | det X | (det B ) r . (ii) The second conclusion can be accomplished by a similar argument as in (i) and by utilizingthe inequality (3.2). (cid:3) References
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EVERSES OF THE YOUNG INEQUALITY 15 Department of Pure Mathematics, Ferdowsi University of Mashhad, P. O. Box 1159, Mash-had 91775, Iran.
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