A Trace Class Operator Inequality ✩

J. Math. Anal. Appl. 328 (2007) 1484–1486 www.elsevier.com/locate/jmaa

Note A trace class operator inequality ✩

Lei Liu

College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, PR China Received 7 March 2006 Available online 22 June 2006 Submitted by J.H. Shapiro

Abstract In this note, we generalize the inequality about the trace of positive semidefinite matrix tr(AB)k  (tr A)k(tr B)k to Hilbert space, and obtain a relevant inequality about positive trace class operator. © 2006 Elsevier Inc. All rights reserved.

Keywords: Hilbert space; Positive trace class operator; Inequality

1. Introduction

Recently, Yang [1] has proved a matrix trace inequality

tr(AB)k  (tr A)k(tr B)k, (1) where A and B are positive semidefinite matrices over C of the same order and k is any positive integer. For related works the reader can refer to [2–7], which are continuations of the work of Bellman [2]. In this paper, we generalize the inequality about the trace of matrix to Hilbert space, and obtain a relevant inequality about positive trace class operator. Let B(H) be the set of all bound linear operators of an infinite separable Hilbert space H .LetA ∈ B(H) be a compact ∗ 1/2 operator. The eigenvalues of |A|=(A A) are called singular values of A and denoted by   ··· ∞ ∞ s1(A), s2(A),...,sn(A), . . . , and s1(A) s2(A) sn(A) . . . .If i=1 si(A) < , then A is called a trace class operator. The trace of A is denoted by tr Aand the set of all trace class ∈   = ∞ operators is denoted by B1(H ).LetA B1(H ), then A 1 i=1 si(A) is called the trace

✩ This work is partially supported by NNSF of China (No. 10571114), NSBRP of Shaanxi Province (No. 2004A17), and Innovation Foundation for Postgraduate of Shaanxi Normal University. E-mail address: [email protected].

0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2006.04.092 L. Liu / J. Math. Anal. Appl. 328 (2007) 1484–1486 1485  = ∞ ⊗ norm of A. From [8] we know that if A is positive in B1(H ), then A i=1 si(A)ei ei , where ei is the relevant eigenvector of si(A).

2. Main result

The following lemma can be found in [9].

Lemma 1. Let S ∈ B(H) and T ∈ B1(H ). Then TS1  T 1S and ST1  ST 1.

k k k Theorem 1. If A,B ∈ B1(H ) are positive, then tr(AB)  (tr A) (tr B) for all natural num- bers k.   = ∞ ⊗ = ∞ ⊗ Proof. Let A i=1 si(A)ei ei and B i=1 si(B)gi gi . We write n n An = si(A)ei ⊗ ei,Bn = si(B)gi ⊗ gi i=1 i=1 ˆ and Mn = span{e1,...,en,g1,...,gn}.LetPn be the projection from H to Mn, and An = ˆ ˆ ˆ PnAPn, Bn = PnBPn. Clearly, An, Bn are positive semidefinite matrices over C and ˆ ˆ An  A, Bn  B. Since ei ⊗ ei1 = 1, we have from Lemma 1 that

A − Aˆ   A − A  +A − Aˆ  n 1  n 1 n  n 1   ∞   ∞        =  ⊗  +  ⊗   si(A)ei ei  si(A)Pn(ei ei)Pn i=n+1 1 i=n+1 1 ∞ ∞  si(A) + si(A) i=n+1 i=n+1 ∞ = 2 si(A) → 0 (n →∞). i=n+1 Similarly, we can show that ˆ B − Bn1 → 0 (n →∞). ˆ ˆ ˆ ˆ Since |tr A − tr An|  A − An1 and |tr B − tr Bn|  B − Bn1,wehave ˆ |tr A − tr An|→0 (n →∞) and ˆ |tr B − tr Bn|→0 (n →∞). Hence    k k ˆ k ˆ k (tr A) (tr B) − (tr An) (tr Bn)          k ˆ k  k  ˆ k  k ˆ k  (tr A) − (tr An) · (tr B) + (tr An) · (tr B) − (tr Bn) 1486 L. Liu / J. Math. Anal. Appl. 328 (2007) 1484–1486    −  k 1    =| − ˆ |· k−i ˆ i−1 ·  k tr A tr An  (tr A) (tr An)  (tr B) = i 1    −    k 1  +  ˆ k ·| − ˆ |· k−i ˆ i−1 → →∞ (tr An) tr B tr Bn  (tr B) (tr Bn)  0 (n ). i=1 k ˆ ˆ k Next, we will prove (AB) − (AnBn) 1 → 0 (n →∞) for any positive integer k. When k = 1, we see that ˆ ˆ ˆ ˆ AB − AnBn1  A·B − Bn1 +A − An1 ·B→0 (n →∞). t ˆ ˆ t Suppose that (AB) − (AnBn) 1 → 0 (n →∞) for k = t. In the case when k = t + 1, we have     + + + (AB)t 1 − (Aˆ Bˆ )t 1  (AB)t (Aˆ Bˆ ) − (Aˆ Bˆ )t 1 n n 1  n n n n 1 + + (AB)t 1 − (AB)t (Aˆ Bˆ )   n n 1  (AB)t − (Aˆ Bˆ )t  ·Aˆ Bˆ    n n 1 n n  t  ˆ ˆ + (AB) ·AB − AnBn1 → 0 (n →∞). k ˆ ˆ k Hence for any positive integer k,wehave(AB) − (AnBn) 1 → 0 (n →∞). That is,      tr(AB)k − tr(Aˆ Bˆ )k = tr (AB)k − (Aˆ Bˆ )k  n n  n n   k − ˆ ˆ k → →∞ (AB) (AnBn) 1 0 (n ). From the above result, we conclude that ˆ k ˆ k k k (tr An) (tr Bn) → (tr A) (tr B) (n →∞) (2) and ˆ ˆ k k tr(AnBn) → tr(AB) (n →∞). (3) ˆ ˆ k ˆ k ˆ k It follows from (1) that tr(AnBn)  (tr An) (tr Bn) for all n ∈ N. Hence by (2) and (3), we have tr(AB)k  (tr A)k(tr B)k. The proof is complete. 2

References

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