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Q-norm inequalities for sequences of Hilbert space operators

This is the Published version of the following publication

Dragomir, Sever S, Moslehian, M and Sandor, J (2009) Q-norm inequalities for sequences of Hilbert space operators. Journal of Mathematical Inequalities, 3 (1). pp. 1-14. ISSN 1846-579X

The publisher’s official version can be found at

http://jmi.ele-math.com/03-01/Q-norm-inequalities-for-sequences-of-Hilbert-space- operators

Note that access to this version may require subscription.

Downloaded from VU Research Repository https://vuir.vu.edu.au/4276/

(2)

Inequalities Volume 3, Number 1 (2009), 1–14

Q–NORM INEQUALITIES FOR SEQUENCES OF HILBERT SPACE OPERATORS

S. S. DRAGOMIR, M. S. MOSLEHIAN ANDJ. S ´ANDOR

(communicated by R. Oinarov)

Abstract. We give some inequalities related to a large class of operator norms, the so-calledQ- norms, for a (not necessary commutative) family of bounded linear operators acting on a Hilbert space that are related to the classical Schwarz inequality. Applications for vector inequalities are also provided.

1. Introduction and preliminaries

Let (H;·,·) be a real or complex Hilbert space,B(H) the Banach algebra of all bounded linear operators acting onH andI denote the identity operator onH .

In many estimates one needs to use upper bounds for the norm of a sum of products A1,...,An,B1,...,BnB(H), where separate information for norms of operators are provided. The special case in whichBiiIiC,1in)and the norms in the question are the operator norm has already investigated in [4]. The celebrated H¨older’s discrete inequality stating

n i=1αiβi

n

i=1

|αi|p

1/p n

i=1

|βi|q 1/q

(1)

1<p, 1<q, 1 p+1

q=1, αi,βiC (1in)

and its variance in the special case p=q=2 (the so-called Cauchy–Bunyakovsky–

Schwarz (CBS) discrete inequality) are of special interest and of larger utility.

To each symmetric gauge functions defined on the sequences of real numbers there corresponds a unitarily invariant norm ||| · ||| defined on a two-sided ideal C|||·||| of B(H) enjoying the invariant property |||UAV|||=|||A||| for all A∈C|||·||| and all unitary operatorsU,V∈B(H). It is known that C|||·||| is a Banach space under the norm||| · |||. If B∈C|||·|||, then |||B|||=|||B||| and |||AB|||A|||B||| for all A∈ B(H), from which we easily conclude that

|||ABC|||A|||B|||C (A,C∈B(H)).

Mathematics subject classification(2000): 47A05, 47A12.

Keywords and phrases: Bounded linear operators, Hilbert spaces, Schwarz inequality, Commuting operators.

c , Zagreb

Paper JMI-03-01 1

(3)

Some well-known examples of unitarily invariant norms are the Schatten p-norms Bp:=tr(|B|p)1/p for 1p<∞, operator norm · , and the Ky-Fan norms; cf.

[1, 6]. A unitarily invariant norm · Q is called aQ-norm if there is a unitarily invari- ant norm · Q such thatBBQ=B2Q.

The aim of the present paper is to establish some upper bounds of interest for the quantityni=1AiBiQ under certain assumptions onAi’s andBi’s. Our results are significant since we do not use any commutative assumption on the families of operators and as well we provide inequalities for a class of norms which are larger than operator norms (compare the results with those of [4]). Applications for vector inequalities are also given.

We need the following lemmas concerning real numbers (the second lemma is obvious):

LEMMA1.1. (Young’s inequality; see[5, p. 30 or p. 49]) If p>1and 1 p+1

q= 1,then

abap p +bq

q for all a,b>0.

LEMMA1.2. If a,b>0, then ab1

4(a+b)21

2(a2+b2).

LEMMA1.3. (Daykin–Eliezer; see[2]) If ak>1,bk>1 and 1 p+1

q <1,then n

k=1

akbk 1p+1q

n

k=1

akp 1p n

k=1

bqk 1q

.

2. Some General Results

The following result containing 9 different inequalities may be stated:

THEOREM2.1. Let A1,...,AnB(H)and B1,...,Bn∈CQ.Then

n i=1

AiBi

2

Q

⎧⎨

A B C

(2)

(4)

where

A:=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1knmaxAk2n

i,j=1

BiBj

Q,

1knmaxAk n

i=1Air 1r

n i=1

n j=1

BiBj

Q

s1s , wherer>1, 1r+1s =1,

1knmaxAkn

i=1Aimax

1in

n j=1

BiBjQ

,

B:=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1inmaxAi n

k=1Akp 1p n

i=1n

j=1

BiBjq

Q

1q ,

n

i=1Ait 1t n

k=1Akp 1p

⎣∑n

i=1

n j=1

BiBjq

Q

uq

1u

, wheret>1, 1t +1u=1,

n i=1Ai

n

k=1Akp 1p

1inmax

⎧⎨

n

j=1

BiBjqQ 1q

,

(3)

for p>1, 1 p+1

q =1; and

C:=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1inmaxAin

k=1Akn

i=1max

1jn

BiBj

Q

, n

i=1Aim m1 n

k=1Ak n

i=1

1maxjn

BiBjQ l1l

, wherem,l>1, m1+1l =1, n

k=1Ak 2

1i,maxjn

BiBj

Q

.

Proof. In the operator partial order ofB(H), we have 0

n i=1

BiAi

n i=1

BiAi

(4)

=

n

i=1

AiBi

n

j=1

BjAj=

n

i,j=1

AiBiBjAj.

(5)

Taking the norm in (4) and noticing thatBBQ=B2Q for anyB∈CQ, we have

n i=1

AiBi

2

Q

=

n i=1

n j=1

AiBiBjAj

Q

n

i=1

n

j=1AiAjBiBj

Q

=

n

i=1Ai n

j=1

AjBiBj

Q

=:M.

Utilizing H¨older’s discrete inequality we have that

n j=1

AjBiBj

Q

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

1knmaxAkn

j=1

BiBj

Q, n

k=1Akp 1p

n j=1

BiBjq

Q

1q , wherep>1, 1p+1q=1,

n

k=1Ak max

1jn

BiBj

Q,

for anyi∈ {1,...,n}.

This provides the following inequalities:

M

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

1knmaxAkn

i=1Ai n

j=1

BiBj

Q

=:M1,

n

k=1Akp 1p n

i=1Ai n

j=1

BiBjq

Q

1q :=Mp, wherep>1, 1p+1q=1,

n

k=1Akn

i=1Ai

1maxjn

BiBj

Q

:=M.

(6)

Utilizing H¨older’s inequality forr,s>1, 1 r+1

s =1,we have:

n i=1

Ai n

j=1

BiBj

Q

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1inmaxAin

i,j=1

BiBj

Q, n

i=1Air 1r

n i=1

n j=1

BiBjQ s1s

, wherer>1, 1r+1s =1,

n

i=1Aimax

1in

n j=1

BiBj

Q

, and thus we can state that

M1

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1knmaxAk2n

i,j=1

BiBjQ,

1knmaxAk n

i=1Air 1r

n i=1

n j=1

BiBjQ s1s

, wherer>1, 1r+1s =1,

1knmaxAkn

i=1Aimax

1in

n j=1

BiBj

Q

, and thefirst part of the theorem is proved.

By H¨older’s inequality we can also have that (for p>1, 1 p+1

q =1)

Mp n

k=1

Akp 1p

×

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1inmaxAin

i=1

n j=1

BiBjqQ 1q

,

n

i=1Ait 1t

⎣∑n

i=1

n j=1

BiBjq

Q

uq

1u

, wheret>1, 1t +1u=1,

n

i=1Aimax

1in

⎧⎨

n

j=1

BiBjq

Q

1q

, and the second part of (2) is proved.

(7)

Finally, we may state that

M

n

k=1Ak ×

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

1inmaxAin

i=1max

1jn

BiBjQ

, n

i=1Aim m1

n i=1

1jnmax

BiBj

Q

l1l , wherem,l>1, m1+1l =1,

n

i=1Ai max

1i,jn

BiBj

Q

, giving the last part of (2).

REMARK2.2. One can obtain various particular inequalities as well. For in- stance, by using the usual Cauchy–Bunyakovsky–Schwarz inequality and norm prop- erties of the spaceCQ we get

n i=1

AiBi

2

Q

n

i=1

AiBiQ 2

n

i=1

AiBiQ 12

n

i=1

Ai2

n

i=1

Bi2Q. (5)

If we consider again the Cauchy–Bunyakovsky–Schwarz inequality

n i=1

AiBi

2

Q

n

i=1Ai2Q

n

i=1Bi2Q (6)

then we can conclude that (5) is a refinement of the (CBS) inequality (6) whenever the operator norm · is majorized by · Q.

COROLLARY2.3. Let A1,...,AnB(H)and B1,...,Bn∈CQ so that BiBj=0 with i=j.Then

n i=1

AiBi

2

Q

⎧⎨

A˜ B˜ C˜

(7)

(8)

where

A˜:=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1knmax Ak2n

i=1Bi2Q,

1knmax Ak n

i=1Air 1r n

i=1Bi2sQ 1s

, wherer>1, 1r+1s =1,

1knmax Akn

i=1Aimax

1in

Bi2Q ,

B˜:=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1inmaxAi n

k=1Akp 1p n

i=1Bi2Q, n

i=1Ait 1t n

k=1Akp 1p n

i=1Bi2uQ 1u

, wheret>1, 1t +1u=1,

n i=1Ai

n

k=1Akp 1p

1inmax

Bi2Q , where p>1and

C˜:=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

1inmaxAin

k=1Akn

i=1Bi2Q, n

i=1Aim m1 n

k=1Ak n

i=1Bi2lQ 1l

, wherem,l>1, m1+1l =1, n

k=1Ak 2

1i,maxjn

Bi2Q . As in the proof of the Theorem 2.1 one has

n i=1

AiBi

2

Q

n

i=1Ai n

j=1AjBiBjQ

=:M

Using H¨older’s generalized inequality from Lemma 1.3, and assuming thatAj>

1, BiBjQ>1,we get n

j=1AjBiBjQ

n

k=1

Akp

pq+q n

j=1BiBjqQ pP+q

. Thus the following result is true:

(9)

THEOREM2.4. Let A1,...,AnB(H)and B1,...,Bn∈CQ. Assume thatAj

>1andBiBjQ>1 for all i,j∈ {1,2,...,n}. Then

n i=1

AiBi

2

Q

n

k=1

Akp pq

+q n i=1

Ai

n

j=1BiBjqQ pp

+q

,

where 1 p+1

q<1.

3. Other Results

A different approach is embodied in the following theorem:

THEOREM3.1. If A1,...,AnB(H)and B1,...,Bn∈CQ,then

n i=1

AiBi

2

Q

n

i=1Ai2

n

j=1

BiBj

Q (8)

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

n

i=1Ai2max

1in

n j=1

BiBjQ

,

n

i=1Ai2p 1p

n i=1

n j=1

BiBj

Q

q1q , wherep>1, 1p+1q=1,

1inmaxAi2n

i,j=1

BiBj

Q.

Proof. From the proof of Theorem 2.1 we have that

n i=1

AiBi

2

Q

n

i=1

n j=1

AiAjBiBj

Q.

Using the simple observation that AiAj1

2

Ai2+Aj2

, i,j∈ {1,...,n},

(10)

we get

n i=1

n

j=1AiAjBiBj

Q1

2

n i=1

n j=1

Ai2+Aj2BiBj

Q

=1 2

n i=1

n j=1

Ai2BiBj

Q+Aj2BjBi

Q

=

n

i=1Ai2

n

j=1

BiBj

Q, which proves thefirst inequality in (8).

The second part follows by H¨older’s inequality and the details are omitted.

REMARK3.2. If in (8) we chooseA1=...=An=I,then we get

n i=1

Bi

Q

n

i,

j=1

BiBj

Q

1/2

n

i,

j=1BiQBj

Q

1/2

=

n i=1

Bi

Q

,

which is a refinement for the generalized triangle inequality for · Q, whenever · Q

is majorized by · Q.

The following corollary may be stated:

COROLLARY3.3. If A1,...,AnB(H)and B1,...,Bn∈CQ are such that BiBj

=0for i=j, i,j∈ {1,...,n},then

n i=1

AiBi

2

Q

n

i=1Ai2Bi2Q (9)

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

n

i=1Ai2max

1inBi2Q, n

i=1Ai2p 1p

n j=1Bi2qQ

1q , wherep>1, 1p+1q=1,

1inmaxAi2n

i=1Bi2Q.

(11)

THEOREM3.4. If A1,...,AnB(H)and B1,...,Bn∈CQ,then

n i=1

AiBi

2

Q

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

1inmaxAi2n

i,j=1

BiBj

Q, n

i=1Aip 2p

n i,j=1

BiBjqQ 1q

, wherep>1, 1p+1q=1, n

i=1Ai 2

1i,maxjn

BiBjQ

.

(10)

Proof. We know that

n i=1

AiBi

2

Q

n

i=1

n

j=1AiAjBiBj

Q=:P. Firstly, we obviously have that

P max

1i,jn AiAj!

n

i,j=1

BiBj

Q= max

1inAi2

n

i,j=1

BiBj

Q. Secondly, by the H¨older inequality for double sums, we obtain

P n

i,j=1

"

AiAj#p1p

n i,j=1

BiBjq

Q

1q

= n

i=1

Aip

n

j=1

Ajp1p

n i,j=1

BiBjq

Q

1q

= n

i=1

Aip 2p

n i,j=1

BiBjq

Q

1q , wherep>1, 1

p+1 q =1. Finally, we have

P max

1i,jn

BiBj

Q

n

i,j=1

AiAj

= n

i=1

Ai 2

1i,maxjn

BiBj

Q

and the theorem is proved.

(12)

COROLLARY3.5. If A1,...,AnB(H)and B1,...,Bn∈CQ are such that BiBj

=0for i,j∈ {1,...,n}with i=j, then

n i=1

AiBi

2

Q

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

1inmaxAi2n

i=1Bi2Q, n

i=1Aip 2p n

i=1Bi2qQ 1q

, wherep>1, 1p+1q=1, n

i=1Ai 2

1inmax

Bi2Q .

(11)

THEOREM3.6. If 1 p+1

q <1 andAi>1 andBiBj>1 for all 1i,jn

then

n i=1

AiBi

2

Q

n

i=1

Aip

p2q+q n

i,

j=1BiBjqQ p+pq

.

Proof. Using Lemma 1.3 for double sums, we have, as in the proof of Theorem 3.4, that

p=

n

i=1

n

j=1AiAjBiBjQ n

i,

j=1(AiAj)p p+qq

n

i,j=1BiBjqQ p+qp

, if we assume thatAi>1 and BiBj>1 for all 1i,jn.

Thus, we get

p n

i=1

Aip p+q2q

n

i,j=1BiBjqQ p+qp

, if 1

p+1 q<1.

Finally, the following result may be stated as well:

THEOREM3.7. If 1 p+1

q =1, A1,...,AnB(H )and B1,...,Bn∈CQ, then

n i=1

AiBi

2

Q

1

p

n

i=1Aip+1 q

n i=1Aiq

1inmax n

j=1BiBjQ

.

(13)

Proof. As in the proof of Theorem 3.1 we have

n i=

AiBi

2

Q

n

i=1

n j=1

AiAjBiBjQ.

Now, using Lemma 1.1, we can write AiAj 1

pAip+1 qAjq. Thus, we have

n i=1

n

j=1AiAjBiBjQ

n

i=1

n j=1

1

pAip+1 qAjq

BiBjQ

= 1 p

n i=1

n

j=1AipBiBjQ+1 q

n i=1

n

j=1AjqBiBjQ

= 1 p

n i=1

Aip

n

j=1

BiBjQ+1 q

n j=1

Ajq

n

i=1

BiBjQ

1 p

n

i=1Aip+1 q

n j=1Ajq

1inmax n

j=1BiBjQ

.

REMARK3.8. Another variant may be obtained via Lemma 1.2:

AiAj1

4(Ai+Aj)2,so

n i=1

n

j=1AiAjBiBjQ1 4

n i=1

n

j=1(Ai+Aj)2BiBjQ

sup

1in

n j=1

BiBjQ n

i,

j=1

(Ai+Aj)2

.

4. Applications

Throughout this section we assume · Q (and hence · Q) to be the operator norm. If byM(B,A)we denote any of the bounds provided by (2), (5), (8) or (10) for the quantity

n

i=1AiBi

2, then we may state the following general fact:

Under the assumptions of Theorem 2.1, we have:

n i=1

AiBix

2

x2M(B,A). (12)

(14)

for any x∈H and

n

i=1AiBix,y

2

x2y2M(B,A). (13)

for any x,y∈H, respectively.

The proof follows by the Schwarz inequality in the Hilbert space (H,.,.). Now, we consider the non zero vectorsy1,...,yn∈H . Define Bi:H →H by Bi=yyi⊗yii (1in). Then

BiBj=yj,yiyi⊗yj

yiyj =yi,yj (1i,jn).

If (yi)1in is an orthogonal family on H, then Bi=1 and BiBj=0 for i,j∈ {1,...,n}, i=j.

Now, utilizing, for instance, the inequalities in Theorem 3.1 we may state that:

n i=1

Aix,yi yi yi

2

x2

n

i=1Ai2

n

j=1

$yi,yj% (14)

x2×

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩

n

i=1Ai2max

1in

n j=1

$yi,yj% ,

n

i=1Ai2p 1p

n i=1

n j=1

$yi,yj%q1q , wherep>1, 1p+1q=1,

1inmaxAi2n

i,j=1

$yi,yj%. for anyx,y1,...,yn

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