Volume 9 (2008), Issue 1, Article 5, 10 pp.
ON SOME INEQUALITIES IN NORMED ALGEBRAS
SEVER S. DRAGOMIR
S CHOOL OF C OMPUTER S CIENCE AND M ATHEMATICS
V ICTORIA U NIVERSITY
PO B OX 14428, M ELBOURNE C ITY M AIL C ENTRE
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
A BSTRACT
bounds
for the
Pn. Some inequalities in normed algebras that provides lower and upper
norm of j=1 aj xj are obtained. Applications for estimating the quantities x−1 x ± y −1 y −1 −1 and y x ± x y for invertible elements x, y in 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. I NTRODUCTION
In [1], in order to provide a generalisation of a norm inequality for n vectors in a normed
linear space obtained by Pečarić and Rajić in [2], the author obtained the following result:
n
)
n
X X
|αk | xj −
|αj − αk | kxj k
j=1
j=1
n
n
(
)
n
X
X X
≤
αj xj ≤ min
|αk | xj +
|αj − αk | kxj k ,
k∈{1,...,n}
(
(1.1)
max
k∈{1,...,n}
j=1
j=1
j=1
where xj , j ∈ {1, . . . , n} are vectors in the normed linear space (X, k·k) over K while αj ,
j ∈ {1, . . . , n} are scalars in K (K = C, R) .
324-07
2
S EVER S. D RAGOMIR
For αk = kx1k k , with xk 6= 0, k ∈ {1, . . . , n} the above inequality produces the following
result established by Pečarić and Rajić in [2]:
(
" n
#)
n
X X
1
(1.2)
max
xj −
|kxj k − kxk k|
k∈{1,...,n}
kxk k
j=1
j=1
(
" n
#)
n
n
X X
X x 1
j ≤
xj +
|kxj k − kxk k|
,
≤ min
kxj k k∈{1,...,n} kxk k j=1
j=1
j=1
which implies the following refinement and reverse of the generalised triangle inequality due to
M. Kato et al. [3]:
#
"
n
X
xj (1.3)
min {kxk k} n − k∈{1,...,n}
kxj k j=1
#
"
n
n
n
X
X
X
x
j ≤
kxj k − xj ≤ max {kxk k} n − .
k∈{1,...,n}
kxj k j=1
j=1
j=1
The other natural choice, αk = kxk k , k ∈ {1, . . . , n} in (1.1) produces the result
(
)
n
n
X
X
(1.4)
max
kxk k xj −
|kxj k − kxk k| kxj k
k∈{1,...,n}
j=1
j=1
(
)
n
n
n
X
X
X
≤
kxj k xj ≤ min
kxk k xj +
|kxj k − kxk k| kxj k ,
k∈{1,...,n}
j=1
j=1
j=1
which in its turn implies another refinement and reverse of the generalised triangle inequality:
P
Pn
n
2
kx
k
−
kx
k
x
j=1 j j j
j=1
(1.5)
(0 ≤)
max {kxk k}
k∈{1,...,n}
P
Pn
2
n
n
n
X X
j=1 kxj k − j=1 kxj k xj ≤
kxj k − xj ≤
,
min {kxk k}
j=1
j=1
k∈{1,...,n}
provided xk 6= 0, k ∈ {1, . . . , n} .
In [2], the authors have shown that the case n = 2 in (1.2) produces the Maligranda-Mercer
inequality:
kx − yk + |kxk − kyk|
kx − yk − |kxk − kyk| x
y
(1.6)
≤
−
,
kxk kyk ≤
min {kxk , kyk}
max {kxk , kyk}
for any x, 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:
(1.7)
kx − yk
|kxk − kyk|
(0 ≤)
−
min {kxk , kyk} max {kxk , kyk}
x
y kx − yk
|kxk − kyk|
≤
≤
−
+
,
kyk kxk
max {kxk , kyk} min {kxk , kyk}
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
http://jipam.vu.edu.au/
I NEQUALITIES IN N ORMED A LGEBRAS
3
for any x, y ∈ X\ {0} .
Motivated by the abovePresults, the aim of the present paper is to establish lower and upper
bounds for the norm of nj=1 aj xj , 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
−1 x x ± y −1 y and y −1 x ± x−1 y are provided as well.
2. I NEQUALITIES FOR n PAIRS OF E LEMENTS
Let (A, k·k) be a normed algebra over the real or complex number field K.
Theorem 2.1. If (aj , xj ) ∈ A2 , j ∈ {1, . . . , n} , then
(
!
)
n
n
X
X
(2.1)
max
xj −
kaj − ak k kxj k
ak
k∈{1,...,n} j=1
j=1
(
!
)
n
n
X
X
≤ max
xj −
k(aj − ak ) xj k
ak
k∈{1,...,n} j=1
j=1
n
X
≤
aj x j j=1
(
!
)
n
n
X
X
xj +
k(aj − ak ) xj k
≤ min
ak
k∈{1,...,n} j=1
j=1
(
!
)
n
n
X
X
≤ min
xj +
kaj − ak k kxj k .
ak
k∈{1,...,n} j=1
j=1
Proof. Observe that for any k ∈ {1, . . . , n} we have
!
n
n
n
X
X
X
aj x j = ak
xj +
(aj − ak ) xj .
j=1
j=1
j=1
Taking the norm and utilising the triangle inequality and the normed algebra properties, we have
n
! n
n
X
X
X
aj xj ≤ ak
xj + (aj − ak ) xj j=1
j=1
j=1
!
n
n
X
X
≤ ak
xj +
k(aj − ak ) xj k
j=1
j=1
!
n
n
X
X
≤ ak
xj +
kaj − ak k kxj k ,
j=1
j=1
for any k ∈ {1, . . . , n} , which implies the second part in (2.1).
Observing that
!
n
n
n
X
X
X
aj x j = ak
xj −
(ak − aj ) xj
j=1
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
j=1
j=1
http://jipam.vu.edu.au/
4
S EVER S. D RAGOMIR
and utilising the continuity of the norm, we have
!
n
n
n
X
X
X
a
x
≥
x
−
(a
−
a
)
x
a
k
j j
j
k
j
j j=1
j=1
j=1
! n
n
X
X
≥ ak
xj − (ak − aj ) xj j=1
j=1
!
n
n
X
X
≥ ak
xj −
k(ak − aj ) xj k
j=1
j=1
!
n
n
X
X
≥ ak
xj −
kak − aj k kxj k
j=1
j=1
for any k ∈ {1, . . . , n} , which implies the first part in (2.1).
Remark 2.2. If there exists r > 0 so that kaj − ak k ≤ r kak k for any j, k ∈ {1, . . . , n} , then,
by the second part of (2.1), we have
n
" n
#
n
X
X X
(2.2)
aj xj ≤ min {kak k} xj + r
kxj k .
k∈{1,...,n}
j=1
j=1
j=1
Corollary 2.3. If xj ∈ A, j ∈ {1, . . . , n} , then
(
!
)
n
n
X
X
(2.3)
max
xj −
kxj − xk k kxj k
xk
k∈{1,...,n} j=1
j=1
(
!
) n
n
n
X
X X
≤ max
xj −
k(xj − xk ) xj k ≤ x2j xk
k∈{1,...,n} j=1
j=1
j=1
(
!
)
n
n
X
X
≤ min
xj +
k(xj − xk ) xj k
xk
k∈{1,...,n} j=1
j=1
(
!
)
n
n
X
X
≤ min
xj +
kxj − xk k kxj k .
xk
k∈{1,...,n} j=1
j=1
Corollary 2.4. Assume that A is a unital normed algebra. If xj ∈ A are invertible for any
j ∈ {1, . . . , n} , then
#
" n
n
X
−1 X
(2.4)
min
xk
kxj k − xj k∈{1,...,n}
j=1
j=1
n
n
X
X
−1 −1 x kxj k − x xj ≤
j
j
j=1
j=1
#
" n
n
X
−1 X
≤ max xk
kxj k − xj .
k∈{1,...,n}
j=1
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
j=1
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I NEQUALITIES IN N ORMED A LGEBRAS
5
· 1 in (2.1) we get
Proof. If 1 ∈ A is the unity, then on choosing ak = x−1
k
(
)
n
n
X
X
−1 x x−1 − x−1 kxj k
(2.5)
max
xj −
j
k
k
k∈{1,...,n}
j=1
j=1
n
X
−1 ≤
xj xj j=1
(
)
n
n
X
X
−1 x x−1 − x−1 kxj k .
xj +
≤ min
j
k
k
k∈{1,...,n}
j=1
Now, assume that
min
k∈{1,...,n}
j=1
−1 −1 x = x . Then
k
k0
n
n
X
X
−1 −1 −1 x x − x kxj k
x
+
j
j
k0
k0
j=1
j=1
!
n
n
n
X
X
−1 X
−1 x kxj k .
= − x k0
kxj k − xj +
j
j=1
j=1
j=1
Utilising the second inequality in (2.5), we deduce
!
n
n
n
n
X
X
X
−1 X
−1 −1 x kx
k
−
x
x
kx
k
−
x
≤
x
j
j
j
j
j
j
k0
j=1
j=1
j=1
j=1
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:
P Pn −1 n
−1 x kxj k − j=1 xj xj j
j=1
(2.6)
max x−1 k∈{1,...,n}
≤
k
n
X
j=1
P Pn n
−1 x−1 kxj k − n
X
j=1 xj xj j
j=1
kxj k − xj ≤
,
min x−1
k
j=1
k∈{1,...,n}
which provides both a refinement and a reverse inequality for the generalised triangle inequality.
3. I NEQUALITIES FOR T WO PAIRS OF E LEMENTS
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}
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
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6
S EVER S. D RAGOMIR
or, equivalently,
1
(3.2)
{ka (x ± y)k + kb (x ± y)k − [k(b − a) yk + k(b − a) xk]}
2
1
+ |ka (x ± y)k − kb (x ± y)k + k(b − a) yk − k(b − a) xk|
2
≤ kax ± byk
1
≤ {ka (x ± y)k + kb (x ± y)k + [k(b − a) yk + k(b − a) xk]}
2
1
− |ka (x ± y)k + kb (x ± y)k − k(b − a) yk − k(b − a) xk| .
2
Proof. The inequality (3.1) follows from Theorem 2.1 for n = 2, a1 = a, a2 = b, x1 = x and
x2 = ±y.
Utilising the properties of real numbers,
1
1
min {α, β} = [α + β − |α − β|] , max {α, β} = [α + β + |α − β|] ; α, β ∈ R;
2
2
the inequality (3.1) is clearly equivalent with (3.2).
The following result contains some upper bounds for kax ± byk that are perhaps more useful
for applications.
Theorem 3.2. If (a, b) , (x, y) ∈ A2 , then
(3.3)
kax ± byk ≤ min {ka (x ± y)k , kb (x ± y)k} + kb − ak max {kxk , kyk}
≤ kx ± yk min {kak , kbk} + kb − ak max {kxk , kyk}
and
(3.4)
kax ± byk ≤ kx ± yk max {kak , kbk} + min {k(b − a) xk , k(b − a) yk}
≤ kx ± yk max {kak , kbk} + kb − ak min {kxk , kyk} .
Proof. Observe that k(b − a) xk ≤ kb − ak kxk and k(b − a) yk ≤ kb − ak kyk, and then
k(b − a) xk , k(b − a) yk ≤ kb − ak max {kxk , kyk} ,
(3.5)
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 − ak max {kxk , kyk}
≤ kx ± yk min {kak , kbk} + kb − ak max {kxk , kyk} .
Utilising the second inequality in (3.1), we deduce (3.3).
Also, since ka (x ± y)k ≤ kak kx ± yk and kb (x ± y)k ≤ kbk kx ± yk , hence
ka (x ± y)k , kb (x ± y)k ≤ kx ± yk max {kak , kbk} ,
which implies that
min {ka (x ± y)k + k(b − a) yk , kb (x ± y)k + k(b − a) xk}
≤ kx ± yk max {kak , kbk} + min {k(b − a) xk , k(b − a) yk}
≤ kx ± yk max {kak , kbk} + kb − ak min {kxk , kyk} ,
and the inequality (3.4) is also proved.
The following corollary may be more useful for applications.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
http://jipam.vu.edu.au/
I NEQUALITIES IN N ORMED A LGEBRAS
7
Corollary 3.3. If (a, b) , (x, y) ∈ A2 , then
kak + kbk
kxk + kyk
+ kb − ak ·
.
2
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 that min {α, β} + max {α, β} = α + β, α, β ∈ R.
(3.6)
kax ± byk ≤ kx ± yk ·
The following lower bounds for kax ± byk can be stated as well:
Theorem 3.4. For any (a, b) and (x, y) ∈ A2 , we have:
(3.7)
max {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak max {kxk , kyk}
≤ max {ka (x ± y)k , kb (x ± y)k} − kb − ak max {kxk , kyk}
≤ kax ± byk
and
(3.8)
min {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak min {kxk , kyk}
≤ 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 − ak max {kxk , kyk}
≥ max {|kaxk − kayk| , |kbxk − kbyk|} − kb − ak max {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 − ak min {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
1
kxk + kyk
· [|kaxk − kayk| + |kbxk − kbyk|] − kb − ak ·
≤ kax ± byk .
2
2
The proof follows from Theorem 3.4 by adding (3.7) to (3.8). The details are omitted.
(3.9)
4. A PPLICATIONS FOR T WO I NVERTIBLE E LEMENTS
In this section we assume that A is a unital algebra with the unity 1. The following results
provide some upper bounds for the quantity kkx−1 k x ± ky −1 k yk , where x and y are invertible
in A.
Proposition 4.1. If (x, y) ∈ A2 are invertible, then
(4.1) x−1 x ± y −1 y ≤ kx ± yk min x−1 , y −1 + x−1 − y −1 max {kxk , kyk}
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
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8
S EVER S. D RAGOMIR
and
(4.2)
−1 x x ± y −1 y ≤ kx ± yk max x−1 , y −1 + x−1 − y −1 min {kxk , kyk} .
Proof. Follows by Theorem 3.2 on choosing a = kx−1 k · 1 and b = ky −1 k · 1.
Corollary 4.2. With the above assumption for x and y, we have
(4.3)
−1
−1
−1 x x ± y −1 y ≤ kx ± yk · kx k + ky k + x−1 − y −1 · kxk + kyk .
2
2
Lower bounds for kkx−1 k x ± ky −1 k yk are provided below:
Proposition 4.3. If (x, y) ∈ A2 are invertible, then
(4.4) kx ± yk max x−1 , y −1 − x−1 − y −1 max {kxk , kyk}
≤ x−1 x ± y −1 y and
(4.5)
kx ± yk min 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−1 k · 1 and b = ky −1 k · 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 choose a = kx−1 k · 1 and b = ky −1 k · 1, then we get
−1 x x ± y −1 y ≥ max x−1 kx ± yk − x−1 − y −1 kyk , y −1 kx ± yk − x−1 − y −1 kxk
≥ kx ± yk min 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
kxk + kyk −1 kx−1 k + ky −1 k − x−1 − y −1 ·
≤ x x ± y −1 y .
2
2
Remark 4.5. We observe that the inequalities (4.3) and (4.7) are in fact equivalent with:
−1 −1 kxk + kyk
kx−1 k + ky −1 k (4.8) x
x± y
y − kx ± yk ·
≤ x−1 − y −1 ·
.
2
2
(4.7)
kx ± yk ·
Now we consider the dual expansion kky −1 k x ± kx−1 k yk, for which the following upper
bounds can be stated.
Proposition 4.6. If (x, y) are invertible in A, then
(4.9) y −1 x ± x−1 y ≤ kx ± yk min x−1 , y −1 + x−1 − y −1 max {kxk , kyk}
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
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I NEQUALITIES IN N ORMED A LGEBRAS
9
and
(4.10)
−1 y x ± x−1 y ≤ kx ± yk max x−1 , y −1 + x−1 − y −1 min {kxk , kyk} .
In particular,
(4.11) y −1 x ± x−1 y kxk + kyk
kx−1 k + ky −1 k + x−1 − y −1 ·
.
2
2
The proof follows from Theorem 3.2 on choosing a = ky −1 k · 1 and b = kx−1 k · 1.
The lower bounds for the quantity kky −1 k x ± kx−1 k yk are incorporated in:
≤ kx ± yk ·
Proposition 4.7. If (x, y) are invertible in A, then
(4.12) kx ± yk max x−1 , y −1 − x−1 − y −1 max {kxk , kyk}
≤ y −1 x ± x−1 y and
(4.13)
kx ± yk min x−1 , y −1 − x−1 − y −1 min {kxk , kyk}
≤ y −1 x ± x−1 y .
In particular,
(4.14)
kx ± yk ·
kxk + kyk
kx−1 k + ky −1 k − x−1 − y −1 ·
2
2
≤ y −1 x ± x−1 y .
Remark 4.8. We observe that the inequalities (4.11) and (4.14) are equivalent with
−1
−1 −1 −1 kx
k
+
ky
k
(4.15) y x ± x y − kx ± yk ·
2
kxk + kyk
≤ x−1 − y −1 ·
.
2
R EFERENCES
[1] S.S. DRAGOMIR, A generalisation of the Pečarić-Rajić 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. PEČARIĆ AND R. RAJIĆ, 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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J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 5, 10 pp.
http://jipam.vu.edu.au/