Some Trace Inequalities for Operators in Hilbert Spaces

arXiv:1409.6295v1 [math.FA] 20 Sep 2014

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES S. S. DRAGOMIR1,2 Abstract. Some new trace inequalities for operators in Hilbert spaces are provided. The superadditivity and monotonicity of some associated functionals are investigated and applications for power series of such operators are given. Some trace inequalities for matrices are also derived. Examples for the operator exponential and other similar functions are presented as well.

1. Introduction Let (H, h·, ·i) be a complex Hilbert space and {ei }i∈I an orthonormal basis of H. We say that A ∈ B (H) is a Hilbert-Schmidt operator if X (1.1) kAei k2 < ∞. i∈I

It is well know that, if {ei }i∈I and {fj }j∈J are orthonormal bases for H and A ∈ B (H) then X X X 2 2 2 (1.2) kAei k = kAfj k = kA∗ fj k i∈I

j∈I

j∈I

showing that the definition (1.1) is independent of the orthonormal basis and A is a Hilbert-Schmidt operator iff A∗ is a Hilbert-Schmidt operator. Let B2 (H) the set of Hilbert-Schmidt operators in B (H) . For A ∈ B2 (H) we define !1/2 X 2 (1.3) kAk2 := kAei k i∈I

for {ei }i∈I an orthonormal basis of H. This definition does not depend on the choice of the orthonormal basis. Using the triangle inequality in l2 (I) , one checks that B2 (H) is a vector space and that k·k2 is a norm on B2 (H) , which is usually called in the literature as the Hilbert-Schmidt norm. 1/2 Denote the modulus of an operator A ∈ B (H) by |A| := (A∗ A) . Because k|A| xk = kAxk for all x ∈ H, A is Hilbert-Schmidt iff |A| is HilbertSchmidt and kAk2 = k|A|k2 . From (1.2) we have that if A ∈ B2 (H) , then A∗ ∈ B2 (H) and kAk2 = kA∗ k2 . The following theorem collects some of the most important properties of HilbertSchmidt operators: 1991 Mathematics Subject Classification. 47A63; 47A99. Key words and phrases. Trace class operators, Hilbert-Schmidt operators, Trace, Schwarz inequality, Trace inequalities for matrices, Power series of operators. 1

S. S. DRAGOMIR1,2

2

Theorem 1. We have (i) (B2 (H) , k·k2 ) is a Hilbert space with inner product X X (1.4) hA, Bi2 := hAei , Bei i = hB ∗ Aei , ei i i∈I

i∈I

and the definition does not depend on the choice of the orthonormal basis {ei }i∈I ; (ii) We have the inequalities

(1.5) for any A ∈ B2 (H) and (1.6)

kAk ≤ kAk2 kAT k2 , kT Ak2 ≤ kT k kAk2

for any A ∈ B2 (H) and T ∈ B (H) ; (iii) B2 (H) is an operator ideal in B (H) , i.e.

B (H) B2 (H) B (H) ⊆ B2 (H) ;

H.

(iv) Bf in (H) , the space of operators of finite rank, is a dense subspace of B2 (H) ; (v) B2 (H) ⊆ K (H) , where K (H) denotes the algebra of compact operators on

If {ei }i∈I an orthonormal basis of H, we say that A ∈ B (H) is trace class if X (1.7) kAk1 := h|A| ei , ei i < ∞. i∈I

The definition of kAk1 does not depend on the choice of the orthonormal basis {ei }i∈I . We denote by B1 (H) the set of trace class operators in B (H) . The following proposition holds: Proposition 1. If A ∈ B (H) , then the following are equivalent: (i) A ∈ B1 (H) ; (ii) |A|1/2 ∈ B2 (H) ; (ii) A (or |A|) is the product of two elements of B2 (H) . The following properties are also well known: Theorem 2. With the above notations: (i) We have kAk1 = kA∗ k1 and kAk2 ≤ kAk1

(1.8)

for any A ∈ B1 (H) ; (ii) B1 (H) is an operator ideal in B (H) , i.e. (iii) We have (iv) We have

B (H) B1 (H) B (H) ⊆ B1 (H) ; B2 (H) B2 (H) = B1 (H) ;

kAk1 = sup {hA, Bi2 | B ∈ B2 (H) , kBk ≤ 1} ;

(v) (B1 (H) , k·k1 ) is a Banach space. (iv) We have the following isometric isomorphisms ∗ ∗ B1 (H) ∼ = K (H) and B1 (H) ∼ = B (H) , ∗

∗

where K (H) is the dual space of K (H) and B1 (H) is the dual space of B1 (H) .

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

3

We define the trace of a trace class operator A ∈ B1 (H) to be X (1.9) tr (A) := hAei , ei i i∈I

where {ei }i∈I an orthonormal basis of H. Note that this coincides with the usual definition of the trace if H is finite-dimensional. We observe that the series (1.9) converges absolutely and it is independent from the choice of basis. The following result collects some properties of the trace: Theorem 3. We have (i) If A ∈ B1 (H) then A∗ ∈ B1 (H) and

tr (A∗ ) = tr (A);

(1.10)

(ii) If A ∈ B1 (H) and T ∈ B (H) , then AT, T A ∈ B1 (H) and

(1.11)

tr (AT ) = tr (T A) and |tr (AT )| ≤ kAk1 kT k ;

(iii) tr (·) is a bounded linear functional on B1 (H) with ktrk = 1; (iv) If A, B ∈ B2 (H) then AB, BA ∈ B1 (H) and tr (AB) = tr (BA) ; (v) Bf in (H) is a dense subspace of B1 (H) . Utilising the trace notation we obviously have that   hA, Bi2 = tr (B ∗ A) = tr (AB ∗ ) and kAk22 = tr (A∗ A) = tr |A|2

for any A, B ∈ B2 (H) . Now, for the finite dimensional case, it is well known that the trace functional is submultiplicative, that is, for positive semidefinite matrices A and B in Mn (C), 0 ≤ tr(AB) ≤ tr (A) tr (B) . Therefore k

0 ≤ tr(Ak ) ≤ [tr (A)] , where k is any positive integer. In 2000, Yang [17] proved a matrix trace inequality   (1.12) tr (AB)k ≤ (tr A)k (tr B)k ,

where A and B are positive semidefinite matrices over C of the same order n and k is any positive integer. For related works the reader can refer to [5], [7], [12] and [19], which are continuations of the work of Bellman [2]. If (H, h·, ·i) is a separable infinite-dimensional Hilbert space then the inequality (1.12) is also valid for any positive operators A, B ∈ B1 (H) . This result was obtained by L. Liu in 2007, see [10]. In 2001, Yang et al. [18] improved (1.12) as follows:
1/2
 , (1.13) tr [(AB)m ] ≤ tr A2m tr B 2m

where A and B are positive semidefinite matrices over C of the same order and m is any positive integer. In [14] the authors have proved many trace inequalities for sums and products of matrices. For instance, if A and B are positive semidefinite matrices in Mn (C) then n
o
  k k (1.14) tr (AB)k ≤ min kAk tr B k , kBk tr Ak

S. S. DRAGOMIR1,2

4

for any positive integer k. Also, if A, B ∈ Mn (C) then for r ≥ 1 and p, q > 1 with 1 1 p + q = 1 we have the following Young type inequality  p q r  |A| |B| r (1.15) tr (|AB ∗ | ) ≤ tr . + p q

Ando [1] proved a very strong form of Young’s inequality - it was shown that if A and B are in Mn (C), then there is a unitary matrix U such that   1 1 p q ∗ |A| + |B| U ∗ , |AB | ≤ U p q where p, q > 1 with

1 p

+

1 q

= 1, which immediately gives the trace inequality

1 1 p q tr (|A| ) + tr (|B| ) . p q This inequality can also be obtained from (1.15) by taking r = 1. Another H¨ older type inequality has been proved by Manjegani in [11] and can be stated as follows: tr (|AB ∗ |) ≤

(1.16)

(1.17) 1 p

tr(AB) ≤ [tr(Ap )]

1/p

[tr(B q )]

1/q

,

1 q

where p, q > 1 with + = 1 and A and B are positive semidefinite matrices. For the theory of trace functionals and their applications the reader is referred to [15]. For other trace inequalities see [3], [5], [8], [9], [13] and [16]. In this paper, motivated by the above results, some new inequalities for HilbertSchmidt operators in B (H) are provided. The superadditivity and monotonicity of some associated functionals are investigated and applications for power series of such operators are given. Some trace inequalities for matrices are also derived. Examples for the operator exponential and other similar functions are presented as well. 2. Some Trace Inequalities Via Hermitian Forms Let P a selfadjoint operator with P ≥ 0. For A ∈ B2 (H) and {ei }i∈I an orthonormal basis of H we have X X 2 2 2 kAk2,P := tr (A∗ P A) = hP Aei , Aei i ≤ kP k kAei k = kP k kAk2 , i∈I

i∈I

which shows that h·, ·i2,P defined by

hA, Bi2,P := tr (B ∗ P A) =

X i∈I

hP Aei , Bei i =

X i∈I

hB ∗ P Aei , ei i

is a nonnegative Hermitian form on B2 (H) , i.e. h·, ·i2,P satisfies the properties: (h) hA, Ai2,P ≥ 0 for any A ∈ B2 (H) ; (hh) h·, ·i2,P is linear in the first variable; (hhh) hB, Ai2,P = hA, Bi2,P for any A, B ∈ B2 (H) . Using the properties of the trace we also have the following representations     2 2 2 kAk2,P := tr P |A∗ | = tr |A∗ | P

and

hA, Bi2,P := tr (P AB ∗ ) = tr (AB ∗ P ) = tr (B ∗ P A)

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

5

for any A, B ∈ B2 (H) . We start with the following result: Theorem 4. Let P a selfadjoint operator with P ≥ 0, i.e. hP x, xi ≥ 0 for any x ∈ H. (i) For any A, B ∈ B2 (H) we have     (2.1) |tr (P AB ∗ )|2 ≤ tr P |A∗ |2 tr P |B ∗ |2

and

(2.2)

i1/2   h  2 2 tr P |A∗ | + 2 Re tr (P AB ∗ ) + tr P |B ∗ | i1/2 h  i1/2 h  2 2 ≤ tr P |A∗ | + tr P |B ∗ | ;

(ii) For any A, B, C ∈ B2 (H) we have 2   2 (2.3) tr (P AB ∗ ) tr P |C ∗ | − tr (P AC ∗ ) tr (P CB ∗ ) i    h  ≤ tr P |A∗ |2 tr P |C ∗ |2 − |tr (P AC ∗ )|2 i    h  2 2 2 × tr P |B ∗ | tr P |C ∗ | − |tr (P BC ∗ )| , (2.4)

and (2.5)

  2 |tr (P AB ∗ )| tr P |C ∗ |   ≤ tr (P AB ∗ ) tr P |C ∗ |2 − tr (P AC ∗ ) tr (P CB ∗ ) + |tr (P AC ∗ ) tr (P CB ∗ )| h  i1/2 h  i1/2   2 2 2 ≤ tr P |A∗ | tr P |B ∗ | tr P |C ∗ |

|tr (P AC ∗ ) tr (P CB ∗ )|    i1/2  i1/2 h  1 h  2 ∗ 2 ∗ ∗ 2 ≤ tr P |B | + |tr (P AB )| tr P |C ∗ | . tr P |A | 2

Proof. (i) Making use of the Schwarz inequality for the nonnegative hermitian form h·, ·i2,P we have 2 hA, Bi2,P ≤ hA, Ai2,P hB, Bi2,P for any A, B ∈ B2 (H) and the inequality (2.1) is proved. We observe that k·k2,P is a seminorm on B2 (H) and by the triangle inequality we have kA + Bk2,P ≤ kAk2,P + kBk2,P for any A, B ∈ B2 (H) and the inequality (2.2) is proved. (ii) Let C ∈ B2 (H) , C 6= 0. Define the mapping [·, ·]2,P,C : B2 (H) × B2 (H) → C by 2

[A, B]2,P,C := hA, Bi2,P kCk2,P − hA, Ci2,P hC, Bi2,P .

6

S. S. DRAGOMIR1,2

Observe that [·, ·]2,P,C is a nonnegative Hermitian form on B2 (H) and by Schwarz inequality we have 2 2 (2.6) hA, Bi2,P kCk2,P − hA, Ci2,P hC, Bi2,P  2   2  2 2 2 2 ≤ kAk2,P kCk2,P − hA, Ci2,P kBk2,P kCk2,P − hB, Ci2,P for any A, B ∈ B2 (H) , which proves (2.3). The case C = 0 is obvious. Utilising the elementary inequality for real numbers m, n, p, q

2 m2 − n2 p2 − q 2 ≤ (mp − nq) ,

we can easily see that  2  2   2 2 2 2 (2.7a) kBk2,P kCk2,P − hB, Ci2,P kAk2,P kCk2,P − hA, Ci2,P 2  ≤ kAk2,P kBk2,P kCk22,P − hA, Ci2,P hB, Ci2,P

for any A, B, C ∈ B2 (H) . Since, by Schwarz’s inequality we have

and

kAk2,P kCk2,P ≥ hA, Ci2,P

kBk2,P kCk2,P ≥ hB, Ci2,P ,

then by multiplying these inequalities we have 2 kAk2,P kBk2,P kCk2,P ≥ hA, Ci2,P hB, Ci2,P

for any A, B, C ∈ B2 (H) . Utilizing the inequalities (2.6) and (2.7a) and taking the square root we get 2 (2.8) hA, Bi2,P kCk2,P − hA, Ci2,P hC, Bi2,P 2 ≤ kAk2,P kBk2,P kCk2,P − hA, Ci2,P hB, Ci2,P for any A, B, C ∈ B2 (H) , which proves the second inequality in (2.4). The first inequality is obvious by the modulus properties. By the triangle inequality for modulus we also have 2 (2.9) hA, Ci2,P hC, Bi2,P − hA, Bi2,P kCk2,P ≤ hA, Bi2,P kCk22,P − hA, Ci2,P hC, Bi2,P for any A, B, C ∈ B2 (H) . On making use of (2.8) and (2.9) we have 2 hA, Ci2,P hC, Bi2,P − hA, Bi2,P kCk2,P ≤ kAk2,P kBk2,P kCk22,P − hA, Ci2,P hB, Ci2,P , which is equivalent to the desired inequality (2.5).



SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

7

Remark 1. By the triangle inequality for the hermitian form [·, ·]2,P,C : B2 (H) × B2 (H) → C, 2

[A, B]2,P,C := hA, Bi2,P kCk2,P − hA, Ci2,P hC, Bi2,P we have  2 1/2 2 2 kA + Bk2,P kCk2,P − hA + B, Ci2,P ≤

 2 1/2  2 1/2 2 2 2 2 kAk2,P kCk2,P − hA, Ci2,P + kBk2,P kCk2,P − hB, Ci2,P ,

which can be written as 1/2  i   h 2 2 ∗ 2 (2.10) tr P (A + B) tr P |C ∗ | − |tr [P (A + B) C ∗ ]| 1/2      ≤ tr P |A∗ |2 tr P |C ∗ |2 − |tr (P AC ∗ )|2 1/2      2 2 2 + tr P |B ∗ | tr P |C ∗ | − |tr (P BC ∗ )| for any A, B, C ∈ B2 (H) .

Remark 2. If we take B = λC in (2.10), then we get  i  h 2 2 ∗ 2 (2.11) 0 ≤ tr P (A + λC) tr P |C ∗ | − |tr [P (A + λC) C ∗ ]|     2 2 2 ≤ tr P |A∗ | tr P |C ∗ | − |tr (C ∗ P A)|

for any λ ∈ C and A, C ∈ B2 (H) . Therefore, we have the bound o  i  n h 2 2 ∗ 2 (2.12) sup tr P (A + λC) tr P |C ∗ | − |tr [P (A + λC) C ∗ ]| λ∈C     = tr P |A∗ |2 tr P |C ∗ |2 − |tr (P AC ∗ )|2 . We also have the inequalities  h 2 i  (2.13) 0 ≤ tr P (A ± C)∗ tr P |C ∗ |2 − |tr [P (A ± C) C ∗ ]|2     2 2 2 ≤ tr P |A∗ | tr P |C ∗ | − |tr (P AC ∗ )|

for any A, C ∈ B2 (H) .

Remark 3. We observe that, by replacing A∗ with A, B ∗ with B etc...above, we can get the dual inequalities, like, for instance (2.14)

|tr (P A∗ C) tr (P C ∗ B)|    i1/2  i1/2 h  1 h  2 2 2 ∗ ≤ tr P |B| + |tr (P A B)| tr P |C| , tr P |A| 2

that holds for any A, B, C ∈ B2 (H) . Since  ∗  |tr (P A∗ C)| = tr (P A∗ C) = tr (P A∗ C) = |tr (C ∗ AP )| = |tr (P C ∗ A)| , |tr (P C ∗ B)| = |tr (P B ∗ C)|

S. S. DRAGOMIR1,2

8

and |tr (P A∗ B)| = |tr (P B ∗ A)| then the inequality (2.14) can be also written as (2.15)

|tr (P C ∗ A) tr (P B ∗ C)|    i1/2  i1/2 h  1 h  tr P |B|2 + |tr (P B ∗ A)| tr P |C|2 , tr P |A|2 ≤ 2

that holds for any A, B, C ∈ B2 (H) . If we take in (2.15) B = A∗ then we get the following inequality (2.16)

|tr (P C ∗ A) tr (P AC)|     i1/2 i1/2 h 
1 h  2 2 ∗ 2 2 tr P |C| , tr P |A | + tr P A tr P |A| ≤ 2

for any A, B, C ∈ B2 (H) . If A is a normal operator, i.e. |A|2 = |A∗ |2 then we have from (2.16) that  
i  1h  2 2 (2.17) |tr (P C ∗ A) tr (P AC)| ≤ tr P |A| + tr P A2 tr P |C| , 2 In particular, if C is selfadjoint and C ∈ B2 (H) , then 
i
1h  2 2 tr P |A| + tr P A2 tr P C 2 , (2.18) |tr (P AC)| ≤ 2 for any A ∈ B2 (H) a normal operator. We notice that (2.18) is a trace operator version of de Bruijn inequality obtained in 1960 in [4], which gives the following refinement of the Cauchy-BunyakovskySchwarz inequality: # 2 " n n n n X X 1X 2 X 2 2 zi , |zi | + ai ai z i ≤ (2.19) 2 i=1

i=1

i=1

i=1

provided that ai are real numbers while zi are complex for each i ∈ {1, ..., n} . We notice that, if P ∈ B1 (H) , P ≥ 0 and A, B ∈ B (H) , then

hA, Bi2,P := tr (P AB ∗ ) = tr (AB ∗ P ) = tr (B ∗ P A)

is a nonnegative Hermitian form on B (H) and all the inequalities above will hold for A, B, C ∈ B (H) . The details are left to the reader. 3. Some Functional Properties We consider now the convex cone B+ (H) of nonnegative operators on the complex Hilbert space H and, for A, B ∈ B2 (H) define the functional σ A,B : B+ (H) → [0, ∞) by h  i1/2 h  i1/2 2 2 (3.1) σ A,B (P ) := tr P |A| tr P |B| − |tr (P A∗ B)| (≥ 0) .

The following theorem collects some fundamental properties of this functional. Theorem 5. Let A, B ∈ B2 (H) . (i) For any P, Q ∈ B+ (H) we have (3.2)

σ A,B (P + Q) ≥ σ A,B (P ) + σ A,B (Q) (≥ 0) ,

namely, σ A,B is a superadditive functional on B+ (H) ;

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

9

(ii) For any P, Q ∈ B+ (H) with P ≥ Q we have (3.3)

σ A,B (P ) ≥ σ A,B (Q) (≥ 0) ,

namely, σ A,B is a monotonic nondecreasing functional on B+ (H) ; (iii) If P, Q ∈ B+ (H) and there exist the constants M > m > 0 such that M Q ≥ P ≥ mQ then (3.4)

M σ A,B (Q) ≥ σ A,B (P ) ≥ mσ A,B (Q) (≥ 0) .

Proof. (i) Let P, Q ∈ B+ (H). On utilizing the elementary inequality
1/2
1/2 2 ≥ ac + bd, a, b, c, d ≥ 0 c + d2 a2 + b 2 and the triangle inequality for the modulus, we have

σ A,B (P + Q) i1/2 i1/2 h  h  tr (P + Q) |B|2 − |tr ((P + Q) A∗ B)| = tr (P + Q) |A|2 i1/2 i1/2 h  h  2 2 2 2 tr P |B| + Q |B| = tr P |A| + Q |A| − |tr (P A∗ B + QA∗ B)| i1/2   i1/2 h    h  tr P |B|2 + tr Q |B|2 = tr P |A|2 + tr Q |A|2

− |tr (P A∗ B) + tr (QA∗ B)| h  i1/2 h  i1/2 h  i1/2 h  i1/2 2 2 2 2 ≥ tr P |A| tr P |B| + tr Q |A| tr Q |B| − |tr (P A∗ B)| − |tr (QA∗ B)| = σ A,B (P ) + σ A,B (Q)

and the inequality (3.2) is proved. (ii) Let P, Q ∈ B+ (H) with P ≥ Q. Utilising the superadditivity property we have σ A,B (P ) = ≥

σ A,B ((P − Q) + Q) ≥ σ A,B (P − Q) + σ A,B (Q) σ A,B (Q)

and the inequality (3.3) is obtained. (iii) From the monotonicity property we have σ A,B (P ) ≥ σ A,B (mQ) = mσ A,B (Q) and a similar inequality for M, which prove the desired result (3.4).



Corollary 1. Let A, B ∈ B2 (H) and P ∈ B (H) such that there exist the constants M > m > 0 with M 1H ≥ P ≥ m1H . Then we have h   i1/2 h  i1/2 2 2 M tr |A| tr |B| (3.5) − |tr (A∗ B)| i1/2 i1/2 h  h  2 2 tr P |B| − |tr (P A∗ B)| ≥ tr P |A|  h  i1/2 i1/2 h  2 2 ∗ tr |B| − |tr (A B)| . ≥ m tr |A|

S. S. DRAGOMIR1,2

10 2

Let P = |V | with V ∈ B (H) . If A, B ∈ B2 (H) then i1/2  i1/2 h    h   tr |V |2 |B|2 − tr |V |2 A∗ B σ A,B |V |2 = tr |V |2 |A|2 = [tr (V ∗ V A∗ A)]

1/2

[tr (V ∗ V B ∗ B)]

1/2

− |tr (V ∗ V A∗ B)|

= [tr (V A∗ AV ∗ )]1/2 [tr (V B ∗ BV ∗ )]1/2 − |tr (V A∗ BV ∗ )|

1/2
1/2   − tr (AV ∗ )∗ BV ∗ tr (BV ∗ )∗ BV ∗ = tr (AV ∗ )∗ AV ∗ i1/2 i1/2 h  h 
2 ∗ 2 tr |BV ∗ | − tr (AV ∗ ) BV ∗ . = tr |AV ∗ |

On utilizing the property (3.2) for P = |V |2 , Q = |U |2 with V, U ∈ B (H) , then we have for any A, B ∈ B2 (H) the following trace inequality (3.6)

i1/2 i1/2 h  h  2 2 2 2 tr |BV ∗ | + |BU ∗ | tr |AV ∗ | + |AU ∗ |
∗ ∗ − tr (AV ∗ ) BV ∗ + (AU ∗ ) BU ∗ i1/2 i1/2 h  h 
2 ∗ 2 tr |BV ∗ | − tr (AV ∗ ) BV ∗ ≥ tr |AV ∗ | i1/2 i1/2 h  h 
tr |BU ∗ |2 − tr (AU ∗ )∗ BU ∗ (≥ 0) . + tr |AU ∗ |2 2

2

Also, if |V | ≥ |U | with V, U ∈ B (H) , then we have for any A, B ∈ B2 (H) that i1/2 i1/2 h  h 
2 ∗ 2 (3.7) tr |BV ∗ | − tr (AV ∗ ) BV ∗ tr |AV ∗ | i1/2 i1/2 h  h 
tr |BU ∗ |2 − tr (AU ∗ )∗ BU ∗ (≥ 0) . ≥ tr |AU ∗ |2 If U ∈ B (H) is invertible, then

1 kxk ≤ kU xk ≤ kU k kxk for any x ∈ H, kU −1 k which implies that 1

1H kU −1 k2

≤ |U |2 ≤ kU k2 1H .

By making use of (3.5) we have the following trace inequality  h  i1/2 i1/2 h  2 2 2 ∗ (3.8) tr |B| − |tr (A B)| tr |A| kU k

i1/2 i1/2 h  h 
2 ∗ 2 tr |BU ∗ | − tr (AU ∗ ) BU ∗ ≥ tr |AU ∗ |  h  i1/2 i1/2 h  1 2 2 ∗ ≥ tr |B| − |tr (A B)| tr |A| kU −1 k2

for any A, B ∈ B2 (H) . Similar results may be stated for P ∈ B1 (H) , P ≥ 0 and A, B ∈ B (H) . The details are omitted.

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

11

4. Inequalities for Sequences of Operators For n ≥ 2, define the Cartesian products B (n) (H) := B (H) × ... × B (H) , (n) (H) := B2 (H) × ... × B2 (H) and B+ (H) := B+ (H) × ... × B+ (H) where B+ (H) denotes the convex cone of nonnegative selfadjoint operators on H, i.e. P ∈ B+ (H) if hP x, xi ≥ 0 for any x ∈ H. (n) B2

(n)

Proposition 2. Let P = (P1 , ..., Pn ) ∈ B+ (H) and A = (A1 , ..., An ) , B = (n) (B1 , ..., Bn ) ∈ B2 (H) and z = (z1 , ..., zn ) ∈ Cn with n ≥ 2. Then ! 2 ! ! n n n X X X 2 2 ∗ (4.1) zk Pk Ak Bk ≤ tr |zk | Pk |Ak | tr |zk | Pk |Bk | . tr k=1

k=1

k=1

Proof. Using the properties of modulus and the inequality (2.1) we have ! n n X X zk tr (Pk A∗k Bk ) zk Pk A∗k Bk = tr k=1

k=1

≤

≤

n X

k=1 n X

k=1

|zk | |tr (Pk A∗k Bk )|

i1/2 i1/2 h  h  2 2 tr Pk |Bk | . |zk | tr Pk |Ak |

Utilizing the weighted discrete Cauchy-Bunyakovsky-Schwarz inequality we also have n i1/2 i1/2 h  h  X tr Pk |Bk |2 |zk | tr Pk |Ak |2 k=1

≤ =

n X

k=1

n X

k=1

=

tr

! h  i1/2 2 1/2 2 |zk | tr Pk |Ak | 

|zk | tr Pk |Ak | n X

k=1

2

|zk | Pk |Ak |



2

!1/2

!!1/2

n X

k=1

tr

n X

k=1



! h  i1/2 2 1/2 2 |zk | tr Pk |Bk |

|zk | tr Pk |Bk | n X

k=1

2

!  1/2

|zk | Pk |Bk |

2

!!1/2

,

which is equivalent to the desired result (4.1).



We consider the functional for n-tuples of nonnegative operators as follows: " !#1/2 " !#1/2 n n X X 2 2 σ A,B (P) := tr tr Pk |Ak | (4.2) Pk |Bk | k=1

k=1

! n X ∗ − tr Pk Ak Bk . k=1

Utilising a similar argument to the one in Theorem 5 we can state: (n)

Proposition 3. Let A = (A1 , ..., An ) , B = (B1 , ..., Bn ) ∈ B2 (H) .

S. S. DRAGOMIR1,2

12 (n)

(i) For any P, Q ∈ B+ (H) we have

(4.3)

σ A,B (P + Q) ≥ σ A,B (P) + σ A,B (Q) (≥ 0) , (n)

namely, σ A,B is a superadditive functional on B+ (H) ; (n) (ii) For any P, Q ∈ B+ (H) with P ≥ Q, namely Pk ≥ Qk for all k ∈ {1, ..., n} we have (4.4)

σ A,B (P) ≥ σ A,B (Q) (≥ 0) ,

(n)

namely, σ A,B is a monotonic nondecreasing functional on B+ (H) ; (n) (iii) If P, Q ∈ B+ (H) and there exist the constants M > m > 0 such that M Q ≥ P ≥ mQ then (4.5)

M σ A,B (Q) ≥ σ A,B (P) ≥ mσ A,B (Q) (≥ 0) .

If P = (p1 1H , ..., pn 1H ) with pk ≥ 0, k ∈ {1, ..., n} then the functional of nonnegative weights p = (p1 , ..., pn ) defined by " !#1/2 " !#1/2 n n X X 2 2 σ A,B (p) := tr tr pk |Ak | (4.6) pk |Bk | k=1 ! n X − tr pk A∗k Bk .

k=1

k=1

has the same properties as in (4.3)-(4.5). Moreover, we have the simple bounds: " !#1/2 " !#1/2 n n  X X 2 2 max {pk } tr tr |Ak | (4.7) |Bk |  k∈{1,...,n} k=1 k=1 ! ) n X ∗ − tr Ak Bk k=1 " !#1/2 " !#1/2 ! n n n X X X 2 2 ≥ tr tr pk |Ak | pk |Bk | − tr pk A∗k Bk k=1 k=1 k=1 " !#1/2 " !#1/2 n n  X X 2 2 ≥ min {pk } tr tr |Ak | |Bk |  k∈{1,...,n} k=1 k=1 ! ) n X ∗ − tr Ak Bk . k=1

5. Inequalities for Power Series of Operators

Denote by: D(0, R) = and consider the functions:



{z ∈ C : |z| < R}, C,

λ 7→ f (λ) : D(0, R) → C, f (λ) :=

if R < ∞ if R = ∞, ∞ X

n=0

αn λn

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

and λ 7→ fa (λ) : D(0, R) → C, fa (λ) :=

∞ X

n=0

13

|αn | λn .

As some natural examples that are useful for applications, we can point out that, if (5.1)

f (λ) =

∞ n X 1 (−1) n λ = ln , λ ∈ D (0, 1) ; n 1 + λ n=1

∞ n X (−1) 2n g (λ) = λ = cos λ, λ ∈ C; (2n)! n=0

h (λ) = l (λ) =

∞ n X (−1) λ2n+1 = sin λ, λ ∈ C; (2n + 1)! n=0 ∞ X

n

(−1) λn =

n=0

1 , λ ∈ D (0, 1) ; 1+λ

then the corresponding functions constructed by the use of the absolute values of the coefficients are ∞ X 1 1 n (5.2) λ = ln , λ ∈ D (0, 1) ; fa (λ) = n 1−λ n=1 ∞ X

ga (λ) =

1 λ2n = cosh λ, λ ∈ C; (2n)! n=0

ha (λ) =

∞ X

1 λ2n+1 = sinh λ, λ ∈ C; (2n + 1)! n=0

la (λ) =

∞ X

λn =

n=0

1 , λ ∈ D (0, 1) . 1−λ

Other important examples of functions as power series representations with nonnegative coefficients are: (5.3)

∞ X 1 n exp (λ) = λ λ ∈ C, n! n=0  X  ∞ 1 1 1+λ = ln λ2n−1 , λ ∈ D (0, 1) ; 2 1−λ 2n − 1 n=1
∞ X Γ n + 12 −1 √ sin (λ) = λ2n+1 , λ ∈ D (0, 1) ; π (2n + 1) n! n=0

tanh−1 (λ) = 2 F1

(α, β, γ, λ) =

∞ X

1 λ2n−1 , 2n − 1 n=1

λ ∈ D (0, 1)

∞ X Γ (n + α) Γ (n + β) Γ (γ) n λ , α, β, γ > 0, n!Γ (α) Γ (β) Γ (n + γ) n=0

λ ∈ D (0, 1) ; where Γ is Gamma function.

S. S. DRAGOMIR1,2

14

P∞ Proposition 4. Let f (λ) := n=0 αn λn be a power series with complex coefficients and convergent on the open disk D (0, R) , R > 0. If (H, h·, ·i) is a separable infinitedimensional Hilbert space and A, B ∈ B1 (H) are positive operators with tr (A) , tr (B) < R1/2 , then

(5.4) |tr (f (AB))|2 ≤ fa2 (tr A tr B) ≤ fa (tr A)2 fa (tr B)2 .

Proof. By the inequality (1.12) for the positive operators A, B ∈ B1 (H) we have " n # n X X   (5.5) αk tr (AB)k αk (AB)k = tr k=0

k=0

≤

≤

n X

k=0 n X

k=0

n     X |αk | tr (AB)k |αk | tr (AB)k = k=0 n X

|αk | (tr A)k (tr B)k =

k=0

|αk | (tr A tr B)k .

Utilising the weighted Cauchy-Bunyakovsky-Schwarz inequality for sums we have !1/2 !1/2 n n n X X X k k 2k 2k (5.6) |αk | (tr A) (tr B) ≤ . |αk | (tr B) |αk | (tr A) k=0

k=0

k=0

Then by (5.5) and (5.6) we have " n #2 # 2 " n X X (5.7) |αk | (tr A tr B)k αk (AB)k ≤ tr k=0

k=0

≤

n X

k=0

n k X  k  |αk | (tr A)2 |αk | (tr B)2 k=0

for n ≥ 1. Since 0 ≤ tr (A) , tr (B) < R1/2 , the numerical series ∞ X

k=0

k

|αk | (tr A tr B) ,

∞ X

k=0



 2 k

|αk | (tr A)

and

∞ X

k=0

k  |αk | (tr B)2

are convergent. P k Also, since 0 ≤ tr(AB) ≤ tr (A) tr (B) < R, the operator series ∞ k=0 αk (AB) is convergent in B1 (H) . Letting n → ∞ in (5.7) and utilizing the continuity property of tr (·) on B1 (H) we get the desired result (5.4).  Example 1. a) If we take in (5.4) f (λ) = (1 ± λ)−1 , |λ| < 1 then we get the inequality  −1  2  −1  2 −1 2 1 − (tr B) (5.8) tr (1H ± AB) ≤ 1 − (tr A)

for any A, B ∈ B1 (H) positive operators with tr (A) , tr (B) < 1. −1 b) If we take in (5.4) f (λ) = ln (1 ± λ) , |λ| < 1, then we get the inequality   2 −1  −1  −1 2 2 (5.9) ln 1 − (tr B) tr ln (1H ± AB) ≤ ln 1 − (tr A) for any A, B ∈ B1 (H) positive operators with tr (A) , tr (B) < 1.

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

15

We have the following result as well: P∞ Theorem 6. Let f (λ) := n=0 αn λn be a power series with complex coefficients and convergent on the open disk D (0,R) , R  > 0.If A, B ∈ B2 (H) are normal op2 2 ∗ ∗ erators with A B = BA and tr |A| , tr |B| < R then we have the inequality       2 2 2 . tr fa |B| (5.10) |tr (f (A∗ B))| ≤ tr fa |A|

Proof. From the inequality (4.1) we have ! 2 ! ! n n n X X X k 2 k 2 ∗ k k (5.11) αk (A ) B ≤ tr . tr |αk | B |αk | A tr k=0 k=0 k=0 2 2 2k 2k Since A, B are normal operators, then we have Ak = |A| and B k = |B| k k for any k ≥ 0. Also, since A∗ B = BA∗ then we also have (A∗ ) B k = (A∗ B) for any k ≥ 0.     2 2 Due to the fact that A, B ∈ B2 (H) and tr |A| , tr |B| < R, it follows that tr (A∗ B) ≤ R and the operator series ∞ ∞ ∞ X X X k 2k 2k αk (A∗ B) , |αk | |A| and |αk | |B| k=0

k=0

k=0

are convergent in the Banach space B1 (H) . Taking the limit over n → ∞ in (5.11) and using the continuity of the tr (·) on B1 (H) we deduce the desired result (5.10).  −1

Example 2. a) If we take in (5.10) f (λ) = (1 ± λ) , |λ| < 1 then we get the inequality   −1   2 −1   2 −1 2 ∗ tr 1 − |B| (5.12) (1 ± A B) ≤ tr 1 − |A| tr H     2 2 for any A, B ∈ B2 (H) normal operators with A∗ B = BA∗ and tr |A| , tr |B| < 1. b) If we take in (5.10) f (λ) = exp (λ), λ ∈ C then we get the inequality       2 2 2 (5.13) |tr (exp (A∗ B))| ≤ tr exp |A| tr exp |B| for any A, B ∈ B2 (H) normal operators with A∗ B = BA∗ . P∞ P∞ Theorem 7. Let f (z) := j=0 pj z j and g (z) := j=0 qj z j be two power series with nonnegative coefficients and convergent on the open disk D (0, R) , R> 0. If T and V are two normal and commuting operators from B2 (H) with tr |T |2 ,   2 tr |V | < R, then i1/2   i1/2 h     h   2 2 2 2 (5.14) tr f |V | + g |V | tr f |T | + g |T | − |tr (f (T ∗ V ) + g (T ∗ V ))| i1/2 i1/2 h   h   2 2 tr f |V | − |tr (f (T ∗ V ))| ≥ tr f |T | h   i1/2 h   i1/2 2 2 + tr g |T | tr g |V | − |tr (g (T ∗ V ))| (≥ 0) .

S. S. DRAGOMIR1,2

16

Moreover, if pj ≥ qj for any j ∈ N, then, with the above assumptions on T and V, we have i1/2 i1/2 h   h   2 2 (5.15) tr f |V | − |tr (f (T ∗ V ))| tr f |T | i1/2 i1/2 h   h   tr g |V |2 − |tr (g (T ∗ V ))| (≥ 0) . ≥ tr g |T |2

Proof. Utilising the superadditivity property of the functional σ A,B (·) above as a function of weights p and the fact that T and V are two normal and commuting operators we can state that " !#1/2 " !#1/2 n n X X 2k 2k tr tr (pk + qk ) |T | (pk + qk ) |V | (5.16) k=0

k=0

! n X k ∗ − tr (pk + qk ) (T V ) k=0 " !# " !#1/2 ! 1/2 n n n X X X 2k 2k k ∗ ≥ tr tr pk |T | pk |V | − tr pk (T V ) k=0 k=0 k=0 " !# " !# ! 1/2 1/2 n n n X X X 2k 2k k + tr tr qk |T | qk |V | − tr qk (T ∗ V ) k=0

k=0

k=0

for any n ≥ 1. Since all the series whose partial sums are involved in (5.16) are convergent in B1 (H) , by letting n → ∞ in (5.16) we get (5.14). The inequality (5.15) follows by the monotonicity property of σ A,B (·) and the details are omitted.  Example 3. Now, observe that if we take f (λ) = sinh λ = and

∞ X

1 λ2n+1 (2n + 1)! n=0

g (λ) = cosh λ = then

∞ X

1 λ2n (2n)! n=0

f (λ) + g (λ) = exp λ =

∞ X 1 n λ n! n=0

for any λ ∈ C. If T and V are two normal and commuting operators from B2 (H), then by (2.11) we have i1/2 i1/2 h   h   2 2 (5.17) tr exp |V | − |tr (exp (T ∗ V ))| tr exp |T | i1/2 i1/2 h   h   2 2 tr sinh |V | − |tr (sinh (T ∗ V ))| ≥ tr sinh |T | h   i1/2 h   i1/2 2 2 + tr cosh |T | tr cosh |V | − |tr (cosh (T ∗ V ))| (≥ 0) .

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

17

P∞ P∞ 1 1 = n=0 λn , λ ∈ D (0, 1) and ln 1−λ = n=1 n1 λn , Now, consider the series 1−λ λ ∈ D (0, 1) and define pn = 1, n ≥ 0, q0 = 0, qn = n1 , n ≥ 1, then we observe that for any n ≥ 0 we have pn ≥ qn .   2 If T and V are two normal and commuting operators from B2 (H) with tr |T | ,   tr |V |2 < 1, then by (2.12) we have (5.18)   −1 1/2   −1 1/2   2 2 −1 tr tr 1H − |T | 1H − |V | − tr (1H − T ∗ V )    −1 1/2 −1 1/2    2 2 tr ln 1H − |V | ≥ tr ln 1H − |T |   −1 − tr ln (1H − T ∗ V ) (≥ 0) . 6. Inequalities for Matrices We have the following result for matrices. P n Proposition 5. Let f (λ) := ∞ n=0 αn λ be a power series with complex coefficients and convergent on the open disk D (0,
R) , R >
0. If A and B are positive semidefinite matrices in Mn (C) with tr A2 , tr B 2 < R, then we have the inequality

   2 |tr [f (AB)]| ≤ tr fa A2 tr fa B 2 .

(6.1)

√ R, then also

If tr (A) , tr (B) < (6.2)

|tr [f (AB)]| ≤ min {tr (fa (kAk B)) , tr (fa (kBk A))} .

Proof. We observe that (1.13) holds for m = 0 with equality. By utilizing the generalized triangle inequality for the modulus and the inequality (1.13) we have

(6.3)

" m # X n αn (AB) tr n=0 m m X X n αn tr [(AB) ] ≤ |αn | |tr [(AB)n ]| = n=0

=

m X

n=0

for any m ≥ 1.

|αn | tr [(AB)n ] ≤

n=0

m X

n=0


1/2
1/2   , tr B 2n |αn | tr A2n

S. S. DRAGOMIR1,2

18

Applying the weighted Cauchy-Bunyakowsky-Schwarz discrete inequality we also have m X
1/2
1/2   (6.4) tr B 2n |αn | tr A2n n=0

m X

≤

n=0

=

m X

n=0

"

= tr


1/2 2 |αn | tr A2n 2n



|αn | tr A m X

n=0

2n

|αn | A

!1/2

!1/2


m X

n=0

!#1/2 "

n=0

|αn |



tr B


  |αn | tr B 2n

m X

tr

m X

n=0

|αn | B

2n

2n

!1/2


1/2 2

!1/2

!#1/2

for any m ≥ 1. Therefore, by (6.3) and (6.4) we get " m # 2 ! ! m m X X X n 2n 2n (6.5) tr αn (AB) ≤ tr |αn | B |αn | A tr n=0

n=0

n=0

for any m ≥ 1.
p
Since tr A2 , tr B 2 < R, then tr (AB) ≤ tr (A2 ) tr (B 2 ) < R and the series ∞ X

n=0

αn (AB)n ,

∞ X

n=0

|αn | A2n and

∞ X

n=0

|αn | B 2n

are convergent in Mn (C) . Taking the limit over m → ∞ in (6.5) and utilizing the continuity property of tr (·) on Mn (C) we get (6.1). The inequality (6.2) follows from (1.14) in a similar way and the details are omitted.  −1

Example 4. a) If we take f (λ) = (1 ± λ) , |λ| < 1 then we get the inequality h  
i
i h tr (In ± AB)−1 2 ≤ tr In − A2 −1 tr In − B 2 −1 (6.6)

for any A and B positive semidefinite matrices in Mn (C) with tr A2 , tr B 2 < 1. Here In is the identity matrix in Mn (C) . We also have the inequality o   n    tr (In ± AB)−1 ≤ min tr (In − kAk B)−1 , tr (In − kBk A)−1 (6.7) for any A and B positive semidefinite matrices in Mn (C) with tr (A) , tr (B) < 1. b) If we take f (λ) = exp λ, then we have the inequalities

   2 (6.8) (tr [exp(AB)]) ≤ tr exp A2 tr exp B 2 and

(6.9)

tr [exp(AB)] ≤ min {tr (exp (kAk B)) , tr (exp (kBk A))}

for any A and B positive semidefinite matrices in Mn (C) .

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

19

P∞ Proposition 6. Let f (λ) := n=0 αn λn be a power series with complex coefficients and convergent on the open disk D (0, R) , R > 0. If A and B are matrices in Mn (C) with tr (|A|p ) , tr (|B|q ) < R, where p, q > 1 with p1 + 1q = 1, then   p  |A| |B|q |tr (f (|AB ∗ |))| ≤ tr fa (6.10) + p q   1 1 p q ≤ tr fa (|A| ) + fa (|B| ) . p q Proof. The inequality (1.15) holds with equality for r = 0. By utilizing the generalized triangle inequality for the modulus and the inequality (1.15) we have ! m m X X ∗ n ∗ n αn |AB | = (6.11) αn tr (|AB | ) tr n=0

n=0

≤

≤

m X

n=0 m X

n

|αn | |tr (|AB ∗ | )| = |αn | tr

n=0

= tr

"

m X

n=0

p

m X

n=0 q n 



|B| |A| + p q



|B| |A| + p q

|αn |

p

n

|αn | tr (|AB ∗ | )

q n

#

for any m ≥ 1 and p, q > 1 with p1 + 1q = 1. It is know that if f : [0, ∞) → R is a convex function, then tr f (·) is convex on the cone Mn+ (C) of positive semidefinite matrices in Mn (C). Therefore, for n ≥ 1 we have  p q n  1 |B| 1 |A| pn qn ≤ tr (|A| ) + tr (|B| ) + (6.12) tr p q p q

where p, q > 1 with p1 + 1q = 1. The inequality reduces to equality if n = 0. Then we have  p   m m q n  X X |A| 1 |B| 1 |αn | tr |αn | ≤ (6.13) + tr (|A|pn ) + tr (|B|qn ) p q p q n=0 n=0 # " m m 1X 1X pn qn |αn | |A| + |αn | |B| = tr p n=0 q n=0 for any m ≥ 1 and p, q > 1 with p1 + 1q = 1. From (6.11) and (6.13) we get ! "m n #  p m X X |B|q |A| ∗ r (6.14) αn |AB | ≤ tr + |αn | tr p q n=0 n=0 # " m m 1X 1X pn qn ≤ tr |αn | |A| + |αn | |B| p n=0 q n=0

for any m ≥ 1 and p, q > 1 with

1 p

+

1 q

= 1.

S. S. DRAGOMIR1,2

20 p

q

Since tr (|A| ) , tr (|B| ) < R, then all the series whose partial sums are involved in (6.14) are convergent, then by letting m → ∞ in (6.14) we deduce the desired inequality (6.10). 

Example 5. a) If we take f (λ) = (1 ± λ)−1 , |λ| < 1 then we get the inequalities   (6.15) tr (In ± |AB ∗ |)−1

!   p q −1 |B| |A| + ≤ tr In − p q   1 1 p −1 q −1 , (In − |A| ) + (In − |B| ) ≤ tr p q p

q

where A and B are matrices in Mn (C) with tr (|A| ) , tr (|B| ) < 1 and p, q > 1 with 1p + 1q = 1. b) If we take f (λ) = exp λ, then we have the inequalities

(6.16)

∗

tr (exp (|AB |)) ≤ ≤

  p q  |A| |B| tr exp + p q   1 1 p q tr exp (|A| ) + exp (|B| ) , p q

where A and B are matrices in Mn (C) and p, q > 1 with

1 p

+

1 q

= 1.

Finally, we can state the following result: P∞ Proposition 7. Let f (λ) := n=0 αn λn be a power series with complex coefficients and convergent on the open disk D (0, R) , R > 0. If A and B are commuting positive semidefinite matrices in Mn (C) with tr (Ap ) , tr (B q ) < R, where p, q > 1 with p1 + 1q = 1, then we have the inequality

(6.17)

|tr (f (AB))| ≤ [tr(fa (Ap ))]

1/p

1/q

[tr(fa (B q ))]

.

Proof. Since A and B are commuting positive semidefinite matrices in Mn (C) , then by (1.17) we have for any natural number n including n = 0 that

(6.18)

n

tr((AB) ) = tr(An B n ) ≤ [tr(Anp )]

where p, q > 1 with

1 p

+

1 q

= 1.

1/p

[tr(B nq )]

1/q

,

SOME TRACE INEQUALITIES FOR OPERATORS IN HILBERT SPACES

21

By (6.18) and the weighted H¨ older discrete inequality we have ! m m m X X X n n n αn (AB) = αn tr(A B ) ≤ |αn | |tr(An B n )| tr n=0

n=0

≤

≤ × =

m X

n=0

n=0

|αn | [tr(Anp )]

m X

n=0 m X

n=0 m X

n=0

=

tr(

1/p

[tr(B nq )]



np

1/p



nq

1/q

|αn | [tr(A )] |αn | [tr(B )] np

!1/p

|αn | tr(A )

m X

n=0

!1/p

|αn | Anp )

p

q

1/q

!1/p

!1/q

m X

n=0

tr(

nq

!1/q

|αn | tr(B )

m X

n=0

!1/q

|αn | B nq )

where p, q > 1 with p1 + 1q = 1. The proof follows now in a similar way with the ones from above and the details are omitted.  −1

Example 6. a) If we take f (λ) = (1 ± λ) , |λ| < 1 then we get the inequality   h i1/p h i1/q −1 −1 (6.19) tr((In − B q )−1 ) , tr (In ± AB) ≤ tr((In − Ap ) )

for any A and B commuting positive semidefinite matrices in Mn (C) with tr (Ap ) , tr (B q ) < 1, where p, q > 1 with p1 + 1q = 1. b) If we take f (λ) = exp λ, then we have the inequality

(6.20)

tr (exp (AB)) ≤ [tr(exp (Ap ))]

1/p

[tr(exp (B q ))]

1/q

,

for any A and B commuting positive semidefinite matrices in Mn (C) and p, q > 1 with p1 + 1q = 1. References [1] T. Ando, Matrix Young inequalities, Oper. Theory Adv. Appl. 75 (1995), 33–38. [2] R. Bellman, Some inequalities for positive definite matrices, in: E.F. Beckenbach (Ed.), General Inequalities 2, Proceedings of the 2nd International Conference on General Inequalities, Birkh¨ auser, Basel, 1980, pp. 89–90. [3] E. V. Belmega, M. Jungers and S. Lasaulce, A generalization of a trace inequality for positive definite matrices. Aust. J. Math. Anal. Appl. 7 (2010), no. 2, Art. 26, 5 pp. [4] N. G. de Bruijn, Problem 12, Wisk. Opgaven, 21 (1960), 12-14. [5] D. Chang, A matrix trace inequality for products of Hermitian matrices, J. Math. Anal. Appl. 237 (1999) 721–725. [6] L. Chen and C. Wong, Inequalities for singular values and traces, Linear Algebra Appl. 171 (1992), 109–120. [7] I. D. Coop, On matrix trace inequalities and related topics for products of Hermitian matrix, J. Math. Anal. Appl. 188 (1994) 999–1001. [8] S. Furuichi and M. Lin, Refinements of the trace inequality of Belmega, Lasaulce and Debbah. Aust. J. Math. Anal. Appl. 7 (2010), no. 2, Art. 23, 4 pp.

22

S. S. DRAGOMIR1,2

[9] H. D. Lee, On some matrix inequalities, Korean J. Math. 16 (2008), No. 4, pp. 565-571. [10] L. Liu, A trace class operator inequality, J. Math. Anal. Appl. 328 (2007) 1484–1486. [11] S. Manjegani, H¨ older and Young inequalities for the trace of operators, Positivity 11 (2007), 239–250. [12] H. Neudecker, A matrix trace inequality, J. Math. Anal. Appl. 166 (1992) 302–303. [13] K. Shebrawi and H. Albadawi, Operator norm inequalities of Minkowski type, J. Inequal. Pure Appl. Math. 9(1) (2008), 1–10, article 26. [14] K. Shebrawi and H. Albadawi, Trace inequalities for matrices, Bull. Aust. Math. Soc. 87 (2013), 139–148. [15] B. Simon, Trace Ideals and Their Applications, Cambridge University Press, Cambridge, 1979. [16] Z. Uluk¨ ok and R. T¨ urkmen, On some matrix trace inequalities. J. Inequal. Appl. 2010, Art. ID 201486, 8 pp. [17] X. Yang, A matrix trace inequality, J. Math. Anal. Appl. 250 (2000) 372–374. [18] X. M. Yang, X. Q. Yang and K. L. Teo, A matrix trace inequality, J. Math. Anal. Appl. 263 (2001), 327–331. [19] Y. Yang, A matrix trace inequality, J. Math. Anal. Appl. 133 (1988) 573–574. 1 Mathematics, School of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia. E-mail address: [email protected] URL: http://rgmia.org/dragomir 2 School of Computational & Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa