ON SOME INEQUALITIES IN NORMED ALGEBRAS

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Inequalities in Normed Algebras Sever S. Dragomir vol. 9, iss. 1, art. 5, 2008

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ON SOME INEQUALITIES IN NORMED ALGEBRAS

SEVER S. DRAGOMIR

School of Computer Science and Mathematics Victoria University

PO Box 14428, Melbourne City 8001, VIC, Australia

EMail:sever.dragomir@vu.edu.au URL:http://rgmia.vu.edu.au/dragomir

Received: 24 October, 2007 Accepted: 29 October, 2007 Communicated by: S. Saitoh

2000 AMS Sub. Class.: Primary 46H05; Secondary 47A10.

Key words: Normed algebras, Unital algebras, Generalised triangle inequality.

Abstract: Some inequalities in normed algebras that provides lower and upper bounds for the norm ofPn

j=1ajxjare obtained. Applications for estimating the quantities

x⁻¹

x± y⁻¹

y and

y⁻¹

x± x⁻¹

y

for invertible elementsx, y in unital normed algebras are also given.

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1. Introduction

In [1], in order to provide a generalisation of a norm inequality forn vectors in a normed linear space obtained by Peˇcari´c and Raji´c in [2], the author obtained the following result:

(1.1) max

k∈{1,...,n}

(

|α_k|

n

X

j=1

x_j

−

n

X

j=1

|α_j −α_k| kx_jk )

≤

n

X

j=1

α_jx_j

≤ min

k∈{1,...,n}

(

|α_k|

n

X

j=1

x_j

+

n

X

j=1

|α_j−α_k| kx_jk )

,

where x_j, j ∈ {1, . . . , n} are vectors in the normed linear space (X,k·k) over K whileα_j, j∈ {1, . . . , n}are scalars inK(K=C,R).

Forα_k = _kx¹

kk, withx_k 6= 0, k ∈ {1, . . . , n} the above inequality produces the following result established by Peˇcari´c and Raji´c in [2]:

(1.2) max

k∈{1,...,n}

( 1 kx_kk

"

n

X

j=1

x_j

−

n

X

j=1

|kx_jk − kx_kk|

#)

≤

n

X

j=1

x_j kx_jk

≤ min

k∈{1,...,n}

( 1 kx_kk

"

n

X

j=1

x_j

+

n

X

j=1

|kx_jk − kx_kk|

#) ,

which implies the following refinement and reverse of the generalised triangle in-

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equality due to M. Kato et al. [3]:

(1.3) min

k∈{1,...,n}{kxkk}

"

n−

n

X

j=1

x_j kx_jk

#

≤

n

X

j=1

kx_jk −

n

X

j=1

x_j

≤ max

k∈{1,...,n}{kx_kk}

"

n−

n

X

j=1

x_j kx_jk

# .

The other natural choice,αk=kxkk, k ∈ {1, . . . , n}in (1.1) produces the result

(1.4) max

k∈{1,...,n}

( kx_kk

n

X

j=1

x_j

−

n

X

j=1

|kx_jk − kx_kk| kx_jk )

≤

n

X

j=1

kxjkxj

≤ min

k∈{1,...,n}

( kxkk

n

X

j=1

xj

+

n

X

j=1

|kxjk − kxkk| kxjk )

,

which in its turn implies another refinement and reverse of the generalised triangle inequality:

(0≤) Pn

j=1kx_jk²−

Pn

j=1kx_jkx_j

k∈{1,...,n}max {kx_kk}

(1.5)

≤

n

X

j=1

kx_jk −

n

X

j=1

x_j

≤ Pn

j=1kx_jk²−

Pn

j=1kx_jkx_j min

k∈{1,...,n}{kx_kk} , providedx_k6= 0, k ∈ {1, . . . , n}.

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In [2], the authors have shown that the casen = 2in (1.2) produces the Maligranda- Mercer inequality:

(1.6) kx−yk − |kxk − kyk|

min{kxk,kyk} ≤

x

kxk − y kyk

≤ kx−yk+|kxk − kyk|

max{kxk,kyk} , for anyx, y ∈X\ {0}.

We notice that Maligranda proved the right inequality in [5] while Mercer proved the left inequality in [4].

We have shown in [1] that the following dual result for two vectors is also valid:

(0≤) kx−yk

min{kxk,kyk} − |kxk − kyk|

max{kxk,kyk}

(1.7)

≤

x

kyk − y kxk

≤ kx−yk

max{kxk,kyk}+ |kxk − kyk|

min{kxk,kyk}, for anyx, y ∈X\ {0}.

Motivated by the above results, the aim of the present paper is to establish lower and upper bounds for the norm ofPn

j=1ajxj, where aj, xj, j ∈ {1, . . . , n}are elements in a normed algebra (A,k·k) over the real or complex number field K. In the case where(A,k·k)is a unital algebra and x, y are invertible, lower and upper bounds for the quantities

x⁻¹ x±

y⁻¹ y

and

y⁻¹ x±

x⁻¹ y

are provided as well.

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2. Inequalities for n Pairs of Elements

Let(A,k·k)be a normed algebra over the real or complex number fieldK. Theorem 2.1. If(a_j, x_j)∈A²,j ∈ {1, . . . , n},then

k∈{1,...,n}max (

ak n

X

j=1

xj

!

−

n

X

j=1

kaj −akk kxjk ) (2.1)

≤ max

k∈{1,...,n}

(

a_k

n

X

j=1

x_j

!

−

n

X

j=1

k(a_j −a_k)x_jk )

≤

n

X

j=1

a_jx_j

≤ min

k∈{1,...,n}

(

a_k

n

X

j=1

x_j

!

+

n

X

j=1

k(a_j−a_k)x_jk )

≤ min

k∈{1,...,n}

(

ak n

X

j=1

xj

!

+

n

X

j=1

kaj−akk kxjk )

.

Proof. Observe that for anyk ∈ {1, . . . , n}we have

n

X

j=1

ajxj =ak n

X

j=1

xj

! +

n

X

j=1

(aj −ak)xj.

Taking the norm and utilising the triangle inequality and the normed algebra proper-

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ties, we have

n

X

j=1

a_jx_j

≤

a_k

n

X

j=1

x_j

!

+

n

X

j=1

(a_j−a_k)x_j

≤

a_k

n

X

j=1

x_j

!

+

n

X

j=1

k(a_j−a_k)x_jk

≤

a_k

n

X

j=1

x_j

!

+

n

X

j=1

ka_j −a_kk kx_jk,

for anyk ∈ {1, . . . , n},which implies the second part in (2.1).

Observing that

n

X

j=1

ajxj =ak n

X

j=1

xj

!

−

n

X

j=1

(ak−aj)xj

and utilising the continuity of the norm, we have

n

X

j=1

a_jx_j

≥

a_k

n

X

j=1

x_j

!

−

n

X

j=1

(a_k−a_j)x_j

≥

a_k

n

X

j=1

x_j

!

−

n

X

j=1

(a_k−a_j)x_j

≥

a_k

n

X

j=1

x_j

!

−

n

X

j=1

k(a_k−a_j)x_jk

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≥

a_k

n

X

j=1

x_j

!

−

n

X

j=1

ka_k−a_jk kx_jk

for anyk ∈ {1, . . . , n},which implies the first part in (2.1).

Remark 1. If there existsr >0so thatka_j −a_kk ≤rka_kkfor anyj, k∈ {1, . . . , n}, then, by the second part of (2.1), we have

(2.2)

n

X

j=1

a_jx_j

≤ min

k∈{1,...,n}{ka_kk}

"

n

X

j=1

x_j

+r

n

X

j=1

kx_jk

# .

Corollary 2.2. Ifx_j ∈A, j ∈ {1, . . . , n},then

k∈{1,...,n}max (

x_k

n

X

j=1

x_j

!

−

n

X

j=1

kx_j−x_kk kx_jk ) (2.3)

≤ max

k∈{1,...,n}

(

x_k

n

X

j=1

x_j

!

−

n

X

j=1

k(x_j−x_k)x_jk )

≤

n

X

j=1

x²_j

≤ min

k∈{1,...,n}

(

x_k

n

X

j=1

x_j

!

+

n

X

j=1

k(x_j −x_k)x_jk )

≤ min

k∈{1,...,n}

(

x_k

n

X

j=1

x_j

!

+

n

X

j=1

kx_j−x_kk kx_jk )

.

Corollary 2.3. Assume thatAis a unital normed algebra. Ifx_j ∈ Aare invertible

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for anyj ∈ {1, . . . , n},then

k∈{1,...,n}min x⁻¹_k

" _n X

j=1

kx_jk −

n

X

j=1

x_j

# (2.4)

≤

n

X

j=1

x⁻¹_j

kx_jk −

n

X

j=1

x⁻¹_j x_j

≤ max

k∈{1,...,n}

x⁻¹_k

" _n X

j=1

kx_jk −

n

X

j=1

x_j

# .

Proof. If1∈Ais the unity, then on choosinga_k = x⁻¹_k

·1in (2.1) we get

k∈{1,...,n}max (

x⁻¹_k

n

X

j=1

x_j

−

n

X

j=1

x⁻¹_j −

x⁻¹_k kx_jk

) (2.5)

≤

n

X

j=1

x⁻¹_j x_j

≤ min

k∈{1,...,n}

( x⁻¹_k

n

X

j=1

x_j

+

n

X

j=1

x⁻¹_j −

x⁻¹_k kx_jk

) .

Now, assume that min

k∈{1,...,n}

x⁻¹_k =

x⁻¹_k

0

.Then

x⁻¹_k

0

n

X

j=1

x_j

+

n

X

j=1

x⁻¹_j −

x⁻¹_k

0

kx_jk

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=− x⁻¹_k

0

n

X

j=1

kx_jk −

n

X

j=1

x_j

! +

n

X

j=1

x⁻¹_j kx_jk. Utilising the second inequality in (2.5), we deduce

x⁻¹_k

0

n

X

j=1

kx_jk −

n

X

j=1

x_j

!

≤

n

X

j=1

x⁻¹_j

kx_jk −

n

X

j=1

x⁻¹_j x_j

and the first inequality in (2.4) is proved.

The second part of (2.4) can be proved in a similar manner, however, the details are omitted.

Remark 2. An equivalent form of (2.4) is:

(2.6) Pn

j=1

x⁻¹_j

kxjk −

Pn j=1

x⁻¹_j xj

k∈{1,...,n}max x⁻¹_k

≤

n

X

j=1

kxjk −

n

X

j=1

xj

≤ Pn

j=1

x⁻¹_j

kx_jk −

Pn j=1

x⁻¹_j x_j

k∈{1,...,n}min x⁻¹_k

,

which provides both a refinement and a reverse inequality for the generalised triangle inequality.

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3. Inequalities for Two Pairs of Elements

The following particular case of Theorem2.1is of interest for applications.

Lemma 3.1. If(a, b),(x, y)∈A²,then

(3.1) max{ka(x±y)k − k(b−a)yk,kb(x±y)k − k(b−a)xk}

≤ kax±byk ≤min{ka(x±y)k+k(b−a)yk,kb(x±y)k+k(b−a)xk}

or, equivalently, 1

2{ka(x±y)k+kb(x±y)k −[k(b−a)yk+k(b−a)xk]}

(3.2)

+ 1

2|ka(x±y)k − kb(x±y)k+k(b−a)yk − k(b−a)xk|

≤ kax±byk

≤ 1

2{ka(x±y)k+kb(x±y)k+ [k(b−a)yk+k(b−a)xk]}

− 1

2|ka(x±y)k+kb(x±y)k − k(b−a)yk − k(b−a)xk|. Proof. The inequality (3.1) follows from Theorem 2.1 forn = 2, a1 = a, a2 = b, x1 =xandx2 =±y.

Utilising the properties of real numbers, min{α, β}= 1

2[α+β− |α−β|], max{α, β}= 1

2[α+β+|α−β|] ; α, β ∈R; the inequality (3.1) is clearly equivalent with (3.2).

The following result contains some upper bounds forkax±bykthat are perhaps more useful for applications.

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Theorem 3.2. If(a, b),(x, y)∈A²,then

kax±byk ≤min{ka(x±y)k,kb(x±y)k}+kb−akmax{kxk,kyk}

(3.3)

≤ kx±ykmin{kak,kbk}+kb−akmax{kxk,kyk}

and

kax±byk ≤ kx±ykmax{kak,kbk}+ min{k(b−a)xk,k(b−a)yk}

(3.4)

≤ kx±ykmax{kak,kbk}+kb−akmin{kxk,kyk}.

Proof. Observe thatk(b−a)xk ≤ kb−ak kxkandk(b−a)yk ≤ kb−ak kyk, and then

(3.5) k(b−a)xk,k(b−a)yk ≤ kb−akmax{kxk,kyk}, which implies that

min{ka(x±y)k+k(b−a)yk,kb(x±y)k+k(b−a)xk}

≤min{ka(x±y)k,kb(x±y)k}+kb−akmax{kxk,kyk}

≤ kx±ykmin{kak,kbk}+kb−akmax{kxk,kyk}. Utilising the second inequality in (3.1), we deduce (3.3).

Also, sinceka(x±y)k ≤ kak kx±ykandkb(x±y)k ≤ kbk kx±yk,hence ka(x±y)k,kb(x±y)k ≤ kx±ykmax{kak,kbk},

which implies that

min{ka(x±y)k+k(b−a)yk,kb(x±y)k+k(b−a)xk}

≤ kx±ykmax{kak,kbk}+ min{k(b−a)xk,k(b−a)yk}

≤ kx±ykmax{kak,kbk}+kb−akmin{kxk,kyk}, and the inequality (3.4) is also proved.

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The following corollary may be more useful for applications.

Corollary 3.3. If(a, b),(x, y)∈A²,then (3.6) kax±byk ≤ kx±yk · kak+kbk

2 +kb−ak · kxk+kyk

2 .

Proof. Follows from Theorem 3.2 by adding the last inequality in (3.3) to the last inequality (3.4) and utilising the property thatmin{α, β}+ max{α, β} = α+β, α, β ∈R.

The following lower bounds forkax±bykcan be stated as well:

Theorem 3.4. For any(a, b)and(x, y)∈A²,we have:

max{|kaxk − kayk|,|kbxk − kbyk|} − kb−akmax{kxk,kyk}

(3.7)

≤max{ka(x±y)k,kb(x±y)k} − kb−akmax{kxk,kyk}

≤ kax±byk and

min{|kaxk − kayk|,|kbxk − kbyk|} − kb−akmin{kxk,kyk}

(3.8)

≤min{|kaxk − kayk|,|kbxk − kbyk|} −min{k(b−a)xk,k(b−a)yk}

≤ kax±byk.

Proof. Observe that, by (3.5) we have that

max{ka(x±y)k − k(b−a)yk,kb(x±y)k − k(b−a)xk}

≥max{kax±ayk,kbx±byk} − kb−akmax{kxk,kyk}

≥max{|kaxk − kayk|,|kbxk − kbyk|} − kb−akmax{kxk,kyk}

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and on utilising the first inequality in (3.1), the inequality (3.7) is proved.

Observe also that, since

ka(x±y)k,kb(x±y)k ≥min{|kaxk − kayk|,|kbxk − kbyk|}, then

max{ka(x±y)k − k(b−a)yk,kb(x±y)k − k(b−a)xk}

≥min{|kaxk − kayk|,|kbxk − kbyk|} −min{k(b−a)xk,k(b−a)yk}

≥min{|kaxk − kayk|,|kbxk − kbyk|} − kb−akmin{kxk,kyk}. Then, by the first inequality in (3.1), we deduce (3.8).

Corollary 3.5. For any(a, b),(x, y)∈A², we have (3.9) 1

2·[|kaxk − kayk|+|kbxk − kbyk|]− kb−ak · kxk+kyk

2 ≤ kax±byk. The proof follows from Theorem 3.4 by adding (3.7) to (3.8). The details are omitted.

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4. Applications for Two Invertible Elements

In this section we assume thatAis a unital algebra with the unity 1.The following results provide some upper bounds for the quantitykkx⁻¹kx± ky⁻¹kyk,wherex andyare invertible inA.

Proposition 4.1. If(x, y)∈A² are invertible, then (4.1)

x⁻¹

x± y⁻¹

y

≤ kx±ykmin x⁻¹

, y⁻¹

+

x⁻¹ −

y⁻¹

max{kxk,kyk}

and (4.2)

x⁻¹

x± y⁻¹

y

≤ kx±ykmax x⁻¹

, y⁻¹

+

x⁻¹ −

y⁻¹

min{kxk,kyk}. Proof. Follows by Theorem3.2on choosinga=kx⁻¹k ·1andb=ky⁻¹k ·1.

Corollary 4.2. With the above assumption forxandy,we have

(4.3)

x⁻¹

x± y⁻¹

y

≤ kx±yk · kx⁻¹k+ky⁻¹k

2 +

x⁻¹

− y⁻¹

·kxk+kyk

2 .

Lower bounds forkkx⁻¹kx± ky⁻¹kykare provided below:

Proposition 4.3. If(x, y)∈A² are invertible, then (4.4) kx±ykmax

x⁻¹ ,

y⁻¹ −

x⁻¹

− y⁻¹

max{kxk,kyk}

≤

x⁻¹ x±

y⁻¹ y

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and

(4.5) kx±ykmin x⁻¹

, y⁻¹

−

x⁻¹ −

y⁻¹

min{kxk,kyk}

≤

x⁻¹ x±

y⁻¹ y

. Proof. The first inequality in (4.4) follows from the second inequality in (3.7) on choosinga=kx⁻¹k ·1andb =ky⁻¹k ·1.

We know from the proof of Theorem3.4that

(4.6) max{ka(x±y)k − k(b−a)yk,kb(x±y)k − k(b−a)xk}

≤ kax±byk. If in this inequality we choosea=kx⁻¹k ·1andb =ky⁻¹k ·1,then we get

x⁻¹ x±

y⁻¹ y

≥maxn x⁻¹

kx±yk −

x⁻¹ −

y⁻¹ kyk, y⁻¹

kx±yk −

x⁻¹ −

y⁻¹ kxko

≥ kx±ykmin x⁻¹

, y⁻¹

−

x⁻¹ −

y⁻¹

min{kxk,kyk}

and the inequality (4.5) is obtained.

Corollary 4.4. If(x, y)∈A² are invertible, then

(4.7) kx±yk · kx⁻¹k+ky⁻¹k

2 −

x⁻¹

− y⁻¹

· kxk+kyk 2

≤

x⁻¹ x±

y⁻¹ y

.

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Remark 3. We observe that the inequalities (4.3) and (4.7) are in fact equivalent with:

(4.8)

x⁻¹ x±

y⁻¹ y

− kx±yk · kx⁻¹k+ky⁻¹k 2

≤

x⁻¹ −

y⁻¹

·kxk+kyk

2 .

Now we consider the dual expansionkky⁻¹kx± kx⁻¹kyk, for which the following upper bounds can be stated.

Proposition 4.5. If(x, y)are invertible inA,then (4.9)

y⁻¹

x± x⁻¹

y

≤ kx±ykmin x⁻¹

, y⁻¹

+

x⁻¹ −

y⁻¹

max{kxk,kyk}

and (4.10)

y⁻¹

x± x⁻¹

y

≤ kx±ykmax x⁻¹

, y⁻¹

+

x⁻¹ −

y⁻¹

min{kxk,kyk}. In particular,

(4.11)

y⁻¹ x±

x⁻¹ y

≤ kx±yk · kx⁻¹k+ky⁻¹k

2 +

x⁻¹

− y⁻¹

·kxk+kyk

2 .

The proof follows from Theorem3.2on choosinga=ky⁻¹k·1andb=kx⁻¹k·1.

The lower bounds for the quantitykky⁻¹kx± kx⁻¹kykare incorporated in:

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Proposition 4.6. If(x, y)are invertible inA,then (4.12) kx±ykmax

x⁻¹ ,

y⁻¹ −

x⁻¹

− y⁻¹

max{kxk,kyk}

≤

y⁻¹ x±

x⁻¹ y

and

(4.13) kx±ykmin x⁻¹

, y⁻¹

−

x⁻¹ −

y⁻¹

min{kxk,kyk}

≤

y⁻¹ x±

x⁻¹ y

. In particular,

(4.14) kx±yk · kx⁻¹k+ky⁻¹k

2 −

x⁻¹

− y⁻¹

· kxk+kyk 2

≤

y⁻¹ x±

x⁻¹ y

. Remark 4. We observe that the inequalities (4.11) and (4.14) are equivalent with (4.15)

y⁻¹ x±

x⁻¹ y

− kx±yk · kx⁻¹k+ky⁻¹k 2

≤

x⁻¹ −

y⁻¹

·kxk+kyk

2 .

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References

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