ON SOME INEQUALITIES IN NORMED ALGEBRAS
SEVER S. DRAGOMIR
SCHOOL OFCOMPUTERSCIENCE ANDMATHEMATICS
VICTORIAUNIVERSITY
PO BOX14428, MELBOURNECITYMAILCENTRE
VIC, 8001, AUSTRALIA. sever.dragomir@vu.edu.au URL:http://rgmia.vu.edu.au/dragomir
Received 24 October, 2007; accepted 29 October, 2007 Communicated by S. Saitoh
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, yin unital normed algebras are also given.
Key words and phrases: Normed algebras, Unital algebras, Generalised triangle inequality.
2000 Mathematics Subject Classification. Primary 46H05; Secondary 47A10.
1. INTRODUCTION
In [1], in order to provide a generalisation of a norm inequality for n 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}
(
|αk|
n
X
j=1
xj
−
n
X
j=1
|αj−αk| kxjk )
≤
n
X
j=1
αjxj
≤ min
k∈{1,...,n}
(
|αk|
n
X
j=1
xj
+
n
X
j=1
|αj −αk| kxjk )
,
where xj, j ∈ {1, . . . , n} are vectors in the normed linear space (X,k·k) over K while αj, j ∈ {1, . . . , n}are scalars inK(K=C,R).
324-07
Forαk = kx1
kk, with xk 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 kxkk
"
n
X
j=1
xj
−
n
X
j=1
|kxjk − kxkk|
#)
≤
n
X
j=1
xj kxjk
≤ min
k∈{1,...,n}
( 1 kxkk
"
n
X
j=1
xj
+
n
X
j=1
|kxjk − kxkk|
#) , which implies the following refinement and reverse of the generalised triangle inequality due to M. Kato et al. [3]:
(1.3) min
k∈{1,...,n}{kxkk}
"
n−
n
X
j=1
xj kxjk
#
≤
n
X
j=1
kxjk −
n
X
j=1
xj
≤ max
k∈{1,...,n}{kxkk}
"
n−
n
X
j=1
xj kxjk
# .
The other natural choice,αk =kxkk, k ∈ {1, . . . , n}in (1.1) produces the result (1.4) max
k∈{1,...,n}
( kxkk
n
X
j=1
xj
−
n
X
j=1
|kxjk − kxkk| kxjk )
≤
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=1kxjk2−
Pn
j=1kxjkxj
k∈{1,...,n}max {kxkk}
(1.5)
≤
n
X
j=1
kxjk −
n
X
j=1
xj
≤ Pn
j=1kxjk2−
Pn
j=1kxjkxj
k∈{1,...,n}min {kxkk} , providedxk 6= 0, k ∈ {1, . . . , n}.
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 of Pn
j=1ajxj, where aj, xj, j ∈ {1, . . . , n} are elements in a normed algebra(A,k·k)over the real or complex number fieldK. In the case where(A,k·k)is a unital algebra andx, yare 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.
2. INEQUALITIES FORnPAIRS OFELEMENTS
Let(A,k·k)be a normed algebra over the real or complex number fieldK. Theorem 2.1. If(aj, xj)∈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}
(
ak n
X
j=1
xj
!
−
n
X
j=1
k(aj −ak)xjk )
≤
n
X
j=1
ajxj
≤ min
k∈{1,...,n}
(
ak n
X
j=1
xj
!
+
n
X
j=1
k(aj−ak)xjk )
≤ 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 properties, we have
n
X
j=1
ajxj
≤
ak
n
X
j=1
xj
!
+
n
X
j=1
(aj−ak)xj
≤
ak
n
X
j=1
xj
!
+
n
X
j=1
k(aj−ak)xjk
≤
ak
n
X
j=1
xj
!
+
n
X
j=1
kaj −akk kxjk,
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
ajxj
≥
ak
n
X
j=1
xj
!
−
n
X
j=1
(ak−aj)xj
≥
ak
n
X
j=1
xj
!
−
n
X
j=1
(ak−aj)xj
≥
ak
n
X
j=1
xj
!
−
n
X
j=1
k(ak−aj)xjk
≥
ak
n
X
j=1
xj
!
−
n
X
j=1
kak−ajk kxjk
for anyk∈ {1, . . . , n},which implies the first part in (2.1).
Remark 2.2. If there existsr >0so thatkaj−akk ≤ rkakkfor anyj, k ∈ {1, . . . , n},then, by the second part of (2.1), we have
(2.2)
n
X
j=1
ajxj
≤ min
k∈{1,...,n}{kakk}
"
n
X
j=1
xj
+r
n
X
j=1
kxjk
# .
Corollary 2.3. Ifxj ∈A, j ∈ {1, . . . , n},then
k∈{1,...,n}max (
xk
n
X
j=1
xj
!
−
n
X
j=1
kxj−xkk kxjk ) (2.3)
≤ max
k∈{1,...,n}
(
xk
n
X
j=1
xj
!
−
n
X
j=1
k(xj −xk)xjk )
≤
n
X
j=1
x2j
≤ min
k∈{1,...,n}
(
xk
n
X
j=1
xj
!
+
n
X
j=1
k(xj −xk)xjk )
≤ min
k∈{1,...,n}
(
xk
n
X
j=1
xj
!
+
n
X
j=1
kxj −xkk kxjk )
.
Corollary 2.4. Assume that A is a unital normed algebra. If xj ∈ A are invertible for any j ∈ {1, . . . , n},then
k∈{1,...,n}min x−1k
" n X
j=1
kxjk −
n
X
j=1
xj
# (2.4)
≤
n
X
j=1
x−1j
kxjk −
n
X
j=1
x−1j xj
≤ max
k∈{1,...,n}
x−1k
" n X
j=1
kxjk −
n
X
j=1
xj
# .
Proof. If1∈Ais the unity, then on choosingak = x−1k
·1in (2.1) we get
k∈{1,...,n}max (
x−1k
n
X
j=1
xj
−
n
X
j=1
x−1j −
x−1k kxjk
) (2.5)
≤
n
X
j=1
x−1j xj
≤ min
k∈{1,...,n}
( x−1k
n
X
j=1
xj
+
n
X
j=1
x−1j −
x−1k kxjk
) .
Now, assume that min
k∈{1,...,n}
x−1k =
x−1k
0
.Then
x−1k
0
n
X
j=1
xj
+
n
X
j=1
x−1j −
x−1k
0
kxjk
=− x−1k
0
n
X
j=1
kxjk −
n
X
j=1
xj
! +
n
X
j=1
x−1j kxjk. Utilising the second inequality in (2.5), we deduce
x−1k
0
n
X
j=1
kxjk −
n
X
j=1
xj
!
≤
n
X
j=1
x−1j
kxjk −
n
X
j=1
x−1j xj
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 2.5. An equivalent form of (2.4) is:
(2.6) Pn
j=1
x−1j
kxjk −
Pn j=1
x−1j xj
max
k∈{1,...,n}
x−1k
≤
n
X
j=1
kxjk −
n
X
j=1
xj
≤ Pn
j=1
x−1j
kxjk −
Pn j=1
x−1j xj
min
k∈{1,...,n}
x−1k
,
which provides both a refinement and a reverse inequality for the generalised triangle inequality.
3. INEQUALITIES FORTWO PAIRS OFELEMENTS
The following particular case of Theorem 2.1 is 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 =xand x2 =±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.
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.
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}
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.
4. APPLICATIONS FORTWOINVERTIBLE ELEMENTS
In this section we assume that A is a unital algebra with the unity1.The following results provide some upper bounds for the quantitykkx−1kx± ky−1kyk,wherexandyare 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 Theorem 3.2 on 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
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 choosing a=kx−1k ·1andb =ky−1k ·1.
We know from the proof of Theorem 3.4 that
(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
≥max x−1
kx±yk −
x−1 −
y−1
kyk, y−1
kx±yk −
x−1 −
y−1 kxk
≥ 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 . Remark 4.5. 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 expansion kky−1kx± kx−1kyk, for which the following upper bounds can be stated.
Proposition 4.6. 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 Theorem 3.2 on choosinga=ky−1k ·1andb =kx−1k ·1.
The lower bounds for the quantitykky−1kx± kx−1kykare incorporated in:
Proposition 4.7. 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 4.8. 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 .
REFERENCES
[1] S.S. DRAGOMIR, A generalisation of the Peˇcari´c-Raji´c inequality in normed linear spaces, Preprint.
RGMIA Res. Rep. Coll., 10(3) (2007), Art. 3. [ONLINE:http://rgmia.vu.edu.au/v10n3.
html].
[2] J. PE ˇCARI ´C AND R. RAJI ´C, The Dunkl-Williams inequality with n elements in normed linear spaces, Math. Ineq. & Appl., 10(2) (2007), 461–470.
[3] M. KATO, K.-S. SAITO AND T. TAMURA, Sharp triangle inequality and its reverses in Banach spaces, Math. Ineq. & Appl., 10(3) (2007).
[4] P.R. MERCER, The Dunkl-Williams inequality in an inner product space, Math. Ineq. & Appl., 10(2) (2007), 447–450.
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