Journal of Inequalities in Pure and
Applied Mathematics
REVERSES OF THE TRIANGLE INEQUALITY IN BANACH SPACES
S.S. DRAGOMIR
School of Computer Science and Mathematics
Victoria University
PO Box 14428, MCMC 8001
VIC, Australia.
volume 6, issue 5, article 129,
2005.
Received 18 April, 2005;
accepted 05 July, 2005.
Communicated by: B. Mond
EMail: sever@csm.vu.edu.au
URL: http://rgmia.vu.edu.au/dragomir/
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
Recent reverses for the discrete generalised triangle inequality and its continuous version for vector-valued integrals in Banach spaces are surveyed. New
results are also obtained. Particular instances of interest in Hilbert spaces and
for complex numbers and functions are pointed out as well.
Reverses of the Triangle
Inequality in Banach Spaces
2000 Mathematics Subject Classification: Primary 46B05, 46C05; Secondary
26D15, 26D10
Key words: Reverse triangle inequality, Hilbert spaces, Banach spaces, Bochner integral.
S.S. Dragomir
This paper is based on the talk given by the author within the “International
Conference of Mathematical Inequalities and their Applications, I”, December 0608, 2004, Victoria University, Melbourne, Australia [http://rgmia.vu.edu.au/
conference]
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1
2
3
4
5
6
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Diaz-Metcalf Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . .
Inequalities of Diaz-Metcalf Type for m Functionals . . . . . . .
3.1
The Case of Normed Spaces . . . . . . . . . . . . . . . . . . . . .
3.2
The Case of Inner Product Spaces . . . . . . . . . . . . . . . .
Diaz-Metcalf Inequality for Semi-Inner Products . . . . . . . . . .
Other Multiplicative Reverses for m Functionals . . . . . . . . . .
An Additive Reverse for the Triangle Inequality . . . . . . . . . . .
4
9
12
12
14
24
28
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6.1
The Case of One Functional . . . . . . . . . . . . . . . . . . . . .
6.2
The Case of m Functionals . . . . . . . . . . . . . . . . . . . . . .
6.3
The Case of Inner Product Spaces . . . . . . . . . . . . . . . .
7
Other Additive Reverses for m Functionals . . . . . . . . . . . . . . .
8
Applications for Complex Numbers . . . . . . . . . . . . . . . . . . . . .
9
Karamata Type Inequalities in Hilbert Spaces . . . . . . . . . . . . .
10 Multiplicative Reverses of the Continuous Triangle Inequality
10.1 The Case of One Functional . . . . . . . . . . . . . . . . . . . . .
10.2 The Case of m Functionals . . . . . . . . . . . . . . . . . . . . . .
11 Additive Reverses of the Continuous Triangle Inequality . . . .
11.1 The Case of One Functional . . . . . . . . . . . . . . . . . . . . .
11.2 The Case of m Functionals . . . . . . . . . . . . . . . . . . . . . .
12 Applications for Complex-Valued Functions . . . . . . . . . . . . . .
References
34
37
39
46
50
58
61
61
65
77
77
81
90
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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1.
Introduction
The generalised triangle inequality, namely
n
n
X
X
xi ≤
kxi k ,
i=1
i=1
provided (X, k.k) is a normed linear space over the real or complex field K = R,
C and xi , i ∈ {1, ..., n} are vectors in X plays a fundamental role in establishing
various analytic and geometric properties of such spaces.
With no less importance, the continuous version of it, i.e.,
Z b
Z b
f (t) dt
kf (t)k dt,
≤
(1.1)
a
a
where f : [a, b] ⊂ R → X is a strongly measurable function on the compact interval [a, b] with values in the Banach space X and kf (·)k is Lebesgue
integrable on [a, b] , is crucial in the Analysis of vector-valued functions with
countless applications in Functional Analysis, Operator Theory, Differential
Equations, Semigroups Theory and related fields.
Surprisingly enough, the reverses of these, i.e., inequalities of the following
type
Z b
Z b
n
n
X
X
,
kxi k ≤ C kf (t)k dt ≤ C f
(t)
dt
xi ,
a
a
i=1
i=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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with C ≥ 1, which we call multiplicative reverses, or
Z b
Z b
n
n
X
X
kxi k ≤ xi + M,
kf (t)k dt ≤ f (t) dt
+ M,
i=1
a
i=1
a
with M ≥ 0, which we call additive reverses, under suitable assumptions for
the involved vectors or functions, are far less known in the literature.
It is worth mentioning though, the following reverse of the generalised triangle inequality for complex numbers
n
n
X
X
cos θ
|zk | ≤ zk ,
k=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
k=1
provided the complex numbers zk , k ∈ {1, . . . , n} satisfy the assumption
a − θ ≤ arg (zk ) ≤ a + θ, for any k ∈ {1, . . . , n} ,
where a ∈ R and θ ∈ 0, π2 was first discovered by M. Petrovich in 1917, [22]
(see [20, p. 492]) and subsequently was rediscovered by other authors, including J. Karamata [14, p. 300 – 301], H.S. Wilf [23], and in an equivalent form
by M. Marden [18]. Marden and Wilf have outlined in their work the important fact that reverses of the generalised triangle inequality may be successfully
applied to the location problem for the roots of complex polynomials.
In 1966, J.B. Diaz and F.T. Metcalf [2] proved the following reverse of the
triangle inequality in the more general case of inner product spaces:
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Theorem 1.1 (Diaz-Metcalf, 1966). Let a be a unit vector in the inner product
space (H; h·, ·i) over the real or complex number field K. Suppose that the
vectors xi ∈ H\ {0} , i ∈ {1, . . . , n} satisfy
Re hxi , ai
,
kxi k
0≤r≤
Then
r
n
X
i=1
i ∈ {1, . . . , n} .
n
X
kxi k ≤ xi ,
i=1
where equality holds if and only if
n
X
S.S. Dragomir
xi = r
i=1
n
X
!
kxi k a.
i=1
Theorem 1.2 (Diaz-Metcalf, 1966). Let a1 , . . . , an be orthonormal vectors in
H. Suppose the vectors x1 , . . . , xn ∈ H\ {0} satisfy
Re hxi , ak i
,
kxi k
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A generalisation of this result for orthonormal families is incorporated in the
following result [2].
0 ≤ rk ≤
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Inequality in Banach Spaces
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i ∈ {1, . . . , n} , k ∈ {1, . . . , m} .
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Then
m
X
k=1
! 21
rk2
n
n
X
X
kxi k ≤ xi ,
i=1
i=1
J. Ineq. Pure and Appl. Math. 6(5) Art. 129, 2005
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where equality holds if and only if
n
X
xi =
i=1
n
X
!
kxi k
i=1
m
X
r k ak .
k=1
Similar results valid for semi-inner products may be found in [15], [16] and
[19].
Now, for the scalar continuous case.
It appears, see [20, p. 492], that the first reverse inequality for (1.1) in the
case of complex valued functions was obtained by J. Karamata in his book from
1949, [14]. It can be stated as
Z b
Z b
cos θ
|f (x)| dx ≤ f (x) dx
a
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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a
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provided
−θ ≤ arg f (x) ≤ θ, x ∈ [a, b]
for given θ ∈ 0, π2 .
This result has recently been extended by the author for the case of Bochner
integrable functions with values in a Hilbert space H. If by L ([a, b] ; H) , we
denote the space of Bochner integrable functions with values in a Hilbert space
H, i.e., we recall that f ∈ L ([a, b] ; H) if and only if f : [a, b] → H is strongly
Rb
measurable on [a, b] and the Lebesgue integral a kf (t)k dt is finite, then
Z
(1.2)
a
b
Z b
,
kf (t)k dt ≤ K f
(t)
dt
a
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provided that f satisfies the condition
kf (t)k ≤ K Re hf (t) , ei for a.e. t ∈ [a, b] ,
where e ∈ H, kek = 1 and K ≥ 1 are given. The case of equality holds in (1.2)
if and only if
Z b
Z b
1
kf (t)k dt e.
f (t) dt =
K
a
a
The aim of the present paper is to survey some of the recent results concerning
multiplicative and additive reverses for both the discrete and continuous version
of the triangle inequalities in Banach spaces. New results and applications for
the important case of Hilbert spaces and for complex numbers and complex
functions have been provided as well.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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2.
Diaz-Metcalf Type Inequalities
In [2], Diaz and Metcalf established the following reverse of the generalised
triangle inequality in real or complex normed linear spaces.
Theorem 2.1 (Diaz-Metcalf, 1966). If F : X → K, K = R, C is a linear
functional of a unit norm defined on the normed linear space X endowed with
the norm k·k and the vectors x1 , . . . , xn satisfy the condition
(2.1)
0 ≤ r ≤ Re F (xi ) ,
i ∈ {1, . . . , n} ;
then
(2.2)
r
n
X
i=1
n
X
kxi k ≤ xi ,
i=1
where equality holds if and only if both
!
n
n
X
X
(2.3)
F
xi = r
kxi k
i=1
i=1
and
(2.4)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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F
n
X
i=1
!
xi
n
X
=
xi .
i=1
If X = H, (H; h·, ·i) is an inner product space and F (x) = hx, ei , kek = 1,
then the condition (2.1) may be replaced with the simpler assumption
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(2.5)
0 ≤ r kxi k ≤ Re hxi , ei ,
i = 1, . . . , n,
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which implies the reverse of the generalised triangle inequality (2.2). In this
case the equality holds in (2.2) if and only if [2]
!
n
n
X
X
(2.6)
xi = r
kxi k e.
i=1
i=1
Theorem 2.2 (Diaz-Metcalf, 1966). Let F1 , . . . , Fm be linear functionals on
X, each of unit norm. As in [2], let consider the real number c defined by
"P
#
m
2
|F
(x)|
k
k=1
c = sup
;
kxk2
x6=0
it then follows that 1 ≤ c ≤ m. Suppose the vectors x1 , . . . , xn whenever xi 6=
0, satisfy
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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(2.7)
0 ≤ rk kxi k ≤ Re Fk (xi ) ,
i = 1, . . . , n, k = 1, . . . , m.
Then one has the following reverse of the generalised triangle inequality [2]
n
Pm 2 12 X
n
X
k=1 rk
(2.8)
kxi k ≤ xi ,
c
i=1
where equality holds if and only if both
!
n
n
X
X
(2.9)
Fk
x i = rk
kxi k ,
i=1
i=1
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and
(2.10)
m
X
"
Fk
n
X
!#2
xi
i=1
k=1
2
n
X
= c
xi .
i=1
If X = H, an inner product space, then, for Fk (x) = hx, ek i , where
{ek }k=1,n is an orthonormal family in H, i.e., hei , ej i = δij , i, j ∈ {1, . . . , k} ,
δij is Kronecker delta, the condition (2.7) may be replaced by
(2.11)
0 ≤ rk kxi k ≤ Re hxi , ek i ,
i = 1, . . . , n, k = 1, . . . , m;
implying the following reverse of the generalised triangle inequality
! 21 n
m
n
X
X
X
2
(2.12)
rk
kxi k ≤ xi ,
k=1
i=1
i=1
where the equality holds if and only if
(2.13)
n
X
i=1
xi =
n
X
i=1
!
kxi k
m
X
rk ek .
k=1
The aim of the following sections is to present recent reverses of the triangle
inequality obtained by the author in [5] and [6]. New results are established for
the general case of normed spaces. Their versions in inner product spaces are
analyzed and applications for complex numbers are given as well.
For various classical inequalities related to the triangle inequality, see Chapter XVII of the book [20] and the references therein.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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3.
3.1.
Inequalities of Diaz-Metcalf Type for m
Functionals
The Case of Normed Spaces
The following result may be stated [5].
Theorem 3.1 (Dragomir, 2004). Let (X, k·k) be a normed linear space over the
real or complex number field K and Fk : X → K, k ∈ {1, . . . , m} continuous
linear functionals on X. If xi ∈ X\ {0} , i ∈ {1,
. . . , n} are such that there
Pm
exists the constants rk ≥ 0, k ∈ {1, . . . , m} with k=1 rk > 0 and
S.S. Dragomir
Re Fk (xi ) ≥ rk kxi k
(3.1)
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
P
n
X X
k m
F
k
k xi .
(3.2)
kxi k ≤ Pk=1
m
r
k=1 k
i=1
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i=1
The case of equality holds in (3.2) if both
! n
!
m
X
X
(3.3)
Fk
xi =
i=1
k=1
m
X
k=1
!
rk
n
X
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kxi k
i=1
and
(3.4)
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Inequality in Banach Spaces
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m
X
k=1
!
Fk
n
X
i=1
!
xi
m
n
X X
=
Fk xi .
k=1
i=1
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Proof. Utilising the hypothesis (3.1) and the properties of the modulus, we have
! n
! " m
! n
!#
m
X
X
X
X
I := Fk
xi ≥ Re
Fk
xi (3.5)
i=1
i=1
k=1
k=1
!
m
n
m X
n
X
X
X
≥
Re Fk
xi =
Re Fk (xi )
≥
k=1
m
X
!
rk
i=1
n
X
k=1 i=1
kxi k .
Reverses of the Triangle
Inequality in Banach Spaces
i=1
k=1
On the other hand, by the continuity property of Fk , k ∈ {1, . . . , m} we obviously have
m
n
! n
! m
X
X X
X
(3.6)
I=
Fk
xi ≤ Fk xi .
i=1
k=1
k=1
i=1
Making use of (3.5) and (3.6), we deduce the desired inequality (3.2).
Now, if (3.3) and (3.4) are valid, then, obviously, the case of equality holds
true in the inequality (3.2).
Conversely, if the case of equality holds in (3.2), then it must hold in all the
inequalities used to prove (3.2). Therefore we have
(3.7)
Re Fk (xi ) = rk kxi k
for each i ∈ {1, . . . , n}, k ∈ {1, . . . , m} ;
(3.8)
m
X
k=1
Im Fk
n
X
i=1
S.S. Dragomir
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!
xi
=0
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and
(3.9)
m
X
Re Fk
n
X
!
xi
i=1
k=1
m
n
X
X
Fk =
xi .
i=1
k=1
Note that, from (3.7), by summation over i and k, we get
! n
!#
! n
" m
m
X
X
X
X
(3.10)
Re
Fk
xi
=
rk
kxi k .
k=1
i=1
k=1
i=1
Since (3.8) and (3.10) imply (3.3), while (3.9) and (3.10) imply (3.4) hence the
theorem is proved.
Remark 1. If the norms kFk k , k ∈ {1, . . . , m} are easier to find, then, from
(3.2), one may get the (coarser) inequality that might be more useful in practice:
n
Pm
n
X X
kF
k
k k=1
(3.11)
kxi k ≤ P
xi .
m
r
k=1 k
i=1
3.2.
i=1
The Case of Inner Product Spaces
The case of inner product spaces, in which we may provide a simpler condition
for equality, is of interest in applications [5].
Theorem 3.2 (Dragomir, 2004). Let (H; h·, ·i) be an inner product space over
the real or complex number field K, ek , xi P
∈ H\ {0}, k ∈ {1, . . . , m} , i ∈
{1, . . . , n} . If rk ≥ 0, k ∈ {1, . . . , m} with m
k=1 rk > 0 satisfy
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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(3.12)
Re hxi , ek i ≥ rk kxi k
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for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
P
n
X X
k m
e
k
k
(3.13)
kxi k ≤ Pk=1
xi .
m
k=1 rk i=1
i=1
The case of equality holds in (3.13) if and only if
! m
Pm
n
n
X
X
X
r
k
(3.14)
xi = Pmk=1 2
kxi k
ek .
k k=1 ek k
i=1
i=1
k=1
Reverses of the Triangle
Inequality in Banach Spaces
Proof. By the properties of inner product and by (3.12), we have
* n
+ m
* n
+
m
X
X X
X
(3.15) xi ,
ek ≥ Re
xi , ek i=1
i=1
k=1
k=1
*
+
m
n
X
X
≥
Re
xi , ek
=
k=1
m X
n
X
S.S. Dragomir
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i=1
Re hxi , ek i ≥
k=1 i=1
m
X
k=1
!
rk
n
X
kxi k > 0.
i=1
P
Observe also that, by (3.15), m
k=1 ek 6= 0.
PnOn utilising
Pm Schwarz’s inequality in the inner product space (H; h·, ·i) for
i=1 xi ,
k=1 ek , we have
*
+
n
m
n
m
X
X
X
X
(3.16)
ek ≥ ek .
xi xi ,
i=1
k=1
i=1
k=1
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Making use of (3.15) and (3.16), we can conclude that (3.13) holds.
Now, if (3.14) holds true, then, by taking the norm, we have
n
m
(Pm r ) Pn kx k X
X
k
i
k=1
i=1
xi =
ek Pm
2
k k=1 ek k
i=1
k=1
Pm
n
( k=1 rk ) X
P
=
kxi k ,
k m
k=1 ek k i=1
i.e., the case of equality holds in (3.13).
Conversely, if the case of equality holds in (3.13), then it must hold in all the
inequalities used to prove (3.13). Therefore, we have
Re hxi , ek i = rk kxi k
(3.17)
i=1
k=1
S.S. Dragomir
Title Page
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} ,
*
+
n
m
n
m
X
X
X
X
(3.18)
xi ek = xi ,
ek i=1
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Inequality in Banach Spaces
k=1
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Im
(3.19)
* n
X
i=1
xi ,
m
X
+
ek
= 0.
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k=1
From (3.17), on summing over i and k, we get
* n
+
! n
m
m
X
X X
X
(3.20)
Re
xi ,
ek =
rk
kxi k .
i=1
k=1
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i=1
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By (3.19) and (3.20), we have
+
* n
m
X X
ek =
(3.21)
xi ,
i=1
k=1
m
X
!
rk
k=1
n
X
kxi k .
i=1
On the other hand, by the use of the following identity in inner product spaces
2
2
2
2
hu,
vi
v
u −
= kuk kvk − |hu, vi| , v 6= 0,
(3.22)
kvk2 kvk2
the relation (3.18) holds if and only if
P
P
m
n
X
X
h ni=1 xi , m
k=1 ek i
(3.23)
ek .
xi =
P
2
k m
k=1 ek k
i=1
k=1
Finally, on utilising (3.21) and (3.23), we deduce that the condition (3.14) is
necessary for the equality case in (3.13).
Before we give a corollary of the above theorem, we need to state the following lemma that has been basically obtained in [4]. For the sake of completeness,
we provide a short proof here as well.
Lemma 3.3 (Dragomir, 2004). Let (H; h·, ·i) be an inner product space over
the real or complex number field K and x, a ∈ H, r > 0 such that:
kx − ak ≤ r < kak .
(3.24)
Reverses of the Triangle
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Then we have the inequality
(3.25)
kxk kak2 − r2
12
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≤ Re hx, ai
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or, equivalently
kxk2 kak2 − [Re hx, ai]2 ≤ r2 kxk2 .
(3.26)
The case of equality holds in (3.25) (or in (3.26)) if and only if
kx − ak = r and kxk2 + r2 = kak2 .
(3.27)
Proof. From the first part of (3.24), we have
kxk2 + kak2 − r2 ≤ 2 Re hx, ai .
(3.28)
By the second part of (3.24) we have kak2 − r2
we may state that
(3.29)
0<
kxk2
kak2 − r2
2
12 + kak − r
2
12
12
Reverses of the Triangle
Inequality in Banach Spaces
> 0, therefore, by (3.28),
2 Re hx, ai
≤
1 .
kak2 − r2 2
Utilising the elementary inequality
1
√
q + αp ≥ 2 pq, α > 0, p > 0, q ≥ 0;
α
q
1
2
q
2 2
,
with equality if and only if α =
,
we
may
state
(for
α
=
kak
−
r
p
S.S. Dragomir
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2
p = 1, q = kxk ) that
(3.30)
2 kxk ≤
Page 18 of 99
kxk
2
kak2 − r2
2
2
12 + kak − r
12
.
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The inequality (3.25) follows now by (3.29) and (3.30).
From the above argument, it is clear that the equality holds in (3.25) if and
only if it holds in (3.29) and (3.30). However, the equality holds in (3.29) if and
1
only if kx − ak = r and in (3.30) if and only if kak2 − r2 2 = kxk .
The proof is thus completed.
We may now state the following corollary [5].
Corollary 3.4. Let (H; h·, ·i) be an inner product space over the real or complex
number field K, ek , xi ∈ H\ {0}, k ∈ {1, . . . , m} , i ∈ {1, . . . , n} . If ρk ≥ 0,
k ∈ {1, . . . , m} with
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
kxi − ek k ≤ ρk < kek k
(3.31)
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for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
X
(3.32)
i=1
Contents
n
X
k k=1 ek k
kxi k ≤ P
x
.
i
1
m
2
2 2 ke
k
−
ρ
i=1
k
k
k=1
Pm
JJ
J
The case of equality holds in (3.32) if and only if
n
X
i=1
Pm
xi =
2
2
k=1 kek k − ρk
Pm
2
k k=1 ek k
12
n
X
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kxi k
i=1
m
X
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ek .
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k=1
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Proof. Utilising Lemma 3.3, we have from (3.31) that
kxi k kek k2 − ρ2k
12
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for each k ∈ {1, . . . , m} and i ∈ {1, . . . , n} .
Applying Theorem 3.2 for
rk := kek k2 − ρ2k
12
k ∈ {1, . . . , m} ,
,
we deduce the desired result.
Remark 2. If {ek }k∈{1,...,m} are orthogonal, then (3.32) becomes
n
X
(3.33)
i=1
kxi k ≤ P
m
k=1
Reverses of the Triangle
Inequality in Banach Spaces
12
n
X xi 1 2
kek k − ρ2k 2 i=1 Pm
2
k=1 kek k
S.S. Dragomir
with equality if and only if
n
X
Pm
xi =
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k=1
kek k2 − ρ2k
Pm
k=1
i=1
21
kek k2
n
X
i=1
!
kxi k
m
X
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ek .
k=1
Moreover, if {ek }k∈{1,...,m} is assumed to be orthonormal and
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kxi − ek k ≤ ρk for k ∈ {1, . . . , m} , i ∈ {1, . . . , n}
where ρk ∈ [0, 1) for k ∈ {1, . . . , m} , then
(3.34)
n
X
i=1
√
n
X
kxi k ≤ P
x
i
1 m
2 2 i=1
k=1 (1 − ρk )
m
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with equality if and only if
n
X
i=1
Pm
xi =
1
2 2
k=1 (1 − ρk )
m
n
X
!
kxi k
i=1
m
X
ek .
k=1
The following lemma may be stated as well [3].
Lemma 3.5 (Dragomir, 2004). Let (H; h·, ·i) be an inner product space over
the real or complex number field K, x, y ∈ H and M ≥ m > 0. If
(3.35)
Re hM y − x, x − myi ≥ 0
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
or, equivalently,
(3.36)
1
m
+
M
≤ (M − m) kyk ,
x −
y
2
2
then
(3.37)
1 M +m
kxk kyk ≤ · √
Re hx, yi .
2
mM
The equality holds in (3.37) if and only if the case of equality holds in (3.35)
and
√
(3.38)
kxk = mM kyk .
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Proof. Obviously,
Re hM y − x, x − myi = (M + m) Re hx, yi − kxk2 − mM kyk2 .
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Then (3.35) is clearly equivalent to
(3.39)
√
kxk2
M +m
√
+ mM kyk2 ≤ √
Re hx, yi .
mM
mM
Since, obviously,
√
kxk2
2 kxk kyk ≤ √
+ mM kyk2 ,
mM
√
with equality iff kxk = mM kyk , hence (3.39) and (3.40) imply (3.37).
The case of equality is obvious and we omit the details.
(3.40)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
Finally, we may state the following corollary of Theorem 3.2, see [5].
Corollary 3.6. Let (H; h·, ·i) be an inner product space over the real or complex
number field K, ek , xi ∈ H\ {0}, k ∈ {1, . . . , m} , i ∈ {1, . . . , n} . If Mk >
µk > 0, k ∈ {1, . . . , m} are such that either
(3.41)
Re hMk ek − xi , xi − µk ek i ≥ 0
or, equivalently,
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xi − Mk + µk ek ≤ 1 (Mk − µk ) kek k
2
2
for each k ∈ {1, . . . , m} and i ∈ {1, . . . , n} , then
P
n
n
X
X
k m
e
k
k=1 k
(3.42)
kxi k ≤ Pm 2·√
x
i .
µk M k
i=1
k=1 µk +Mk kek k i=1
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The case of equality holds in (3.42) if and only if
n
X
i=1
√
2· µk Mk
n
m
X
X
k=1 µk +Mk kek k
kx
k
ek .
P
i
2
k m
e
k
k
k=1
i=1
k=1
Pm
xi =
Proof. Utilising Lemma 3.5, by (3.41) we deduce
√
2 · µk M k
kxi k kek k ≤ Re hxi , ek i
µk + M k
for each k ∈ {1, . . . , m} and i ∈ {1, . . . , n} .
Applying Theorem 3.2 for
√
2 · µk M k
rk :=
kek k , k ∈ {1, . . . , m} ,
µk + M k
we deduce the desired result.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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4.
Diaz-Metcalf Inequality for Semi-Inner
Products
In 1961, G. Lumer [17] introduced the following concept.
Definition 4.1. Let X be a linear space over the real or complex number field
K. The mapping [·, ·] : X × X → K is called a semi-inner product on X, if the
following properties are satisfied (see also [3, p. 17]):
(i) [x + y, z] = [x, z] + [y, z] for all x, y, z ∈ X;
Reverses of the Triangle
Inequality in Banach Spaces
(ii) [λx, y] = λ [x, y] for all x, y ∈ X and λ ∈ K;
S.S. Dragomir
(iii) [x, x] ≥ 0 for all x ∈ X and [x, x] = 0 implies x = 0;
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(iv) |[x, y]|2 ≤ [x, x] [y, y] for all x, y ∈ X;
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(v) [x, λy] = λ̄ [x, y] for all x, y ∈ X and λ ∈ K.
1
2
It is well known that the mapping X 3 x 7−→ [x, x] ∈ R is a norm on X
ϕy
and for any y ∈ X, the functional X 3 x 7−→ [x, y] ∈ K is a continuous linear
functional on X endowed with the norm k·k generated by [·, ·] . Moreover, one
has kϕy k = kyk (see for instance [3, p. 17]).
Let (X, k·k) be a real or complex normed space. If J : X → 2 X ∗ is the
normalised duality mapping defined on X, i.e., we recall that (see for instance
[3, p. 1])
J (x) = {ϕ ∈ X ∗ |ϕ (x) = kϕk kxk , kϕk = kxk} , x ∈ X,
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then we may state the following representation result (see for instance [3, p.
18]):
Each semi-inner product [·, ·] : X × X → K that generates the norm k·k of
the normed linear space (X, k·k) over the real or complex number field K, is of
the form
D
E
[x, y] = J˜ (y) , x for any x, y ∈ X,
where J˜ is a selection of the normalised duality mapping and hϕ, xi := ϕ (x)
for ϕ ∈ X ∗ and x ∈ X.
Utilising the concept of semi-inner products, we can state the following particular case of the Diaz-Metcalf inequality.
Reverses of the Triangle
Inequality in Banach Spaces
Corollary 4.1. Let (X, k·k) be a normed linear space, [·, ·] : X × X → K a
semi-inner product generating the norm k·k and e ∈ X, kek = 1. If xi ∈ X,
i ∈ {1, . . . , n} and r ≥ 0 such that
Title Page
(4.1)
r kxi k ≤ Re [xi , e] for each i ∈ {1, . . . , n} ,
then we have the inequality
(4.2)
r
n
X
i=1
n
X kxi k ≤ xi .
i=1
The case of equality holds in (4.2) if and only if both
" n
#
n
X
X
(4.3)
xi , e = r
kxi k
i=1
i=1
S.S. Dragomir
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and
"
(4.4)
n
X
i=1
n
X xi , e = xi .
#
i=1
The proof is obvious from the Diaz-Metcalf theorem [2, Theorem 3] applied
for the continuous linear functional Fe (x) = [x, e] , x ∈ X.
Before we provide a simpler necessary and sufficient condition of equality
in (4.2), we need to recall the concept of strictly convex normed spaces and a
classical characterisation of these spaces.
Reverses of the Triangle
Inequality in Banach Spaces
Definition 4.2. A normed linear space (X, k·k) is said to be strictly convex if for
every x, y from X with x 6= y and kxk = kyk = 1, we have kλx + (1 − λ) yk <
1 for all λ ∈ (0, 1) .
S.S. Dragomir
The following characterisation of strictly convex spaces is useful in what
follows (see [1], [13], or [3, p. 21]).
Contents
Theorem 4.2. Let (X, k·k) be a normed linear space over K and [·, ·] a semiinner product generating its norm. The following statements are equivalent:
(i) (X, k·k) is strictly convex;
(ii) For every x, y ∈ X, x, y 6= 0 with [x, y] = kxk kyk , there exists a λ > 0
such that x = λy.
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The following result may be stated.
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Corollary 4.3. Let (X, k·k) be a strictly convex normed linear space, [·, ·] a
semi-inner product generating the norm and e, xi (i ∈ {1, . . . , n}) as in Corollary 4.1. Then the case of equality holds in (4.2) if and only if
!
n
n
X
X
(4.5)
xi = r
kxi k e.
i=1
i=1
Proof. If (4.5) holds true, then, obviously
n
!
n
n
X X
X
xi = r
kxi k kek = r
kxi k ,
i=1
i=1
i=1
which is the equality case in (4.2).
Conversely, if the equality holds in (4.2), then by Corollary 4.1, we have that
(4.3) and (4.4) hold true. Utilising Theorem 4.2, we conclude that there exists a
µ > 0 such that
n
X
(4.6)
xi = µe.
i=1
Inserting this in (4.3) we get
2
µ kek = r
n
X
kxi k
i=1
giving
(4.7)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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µ=r
n
X
kxi k .
Page 27 of 99
i=1
Finally, by (4.6) and (4.7) we deduce (4.5) and the corollary is proved.
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5.
Other Multiplicative Reverses for m
Functionals
Assume that Fk , k ∈ {1, . . . , m} are bounded linear functionals defined on the
normed linear space X.
For p ∈ [1, ∞), define
Pm
(cp )
|Fk (x)|p
kxkp
k=1
cp := sup
x6=0
p1
Reverses of the Triangle
Inequality in Banach Spaces
and for p = ∞,
S.S. Dragomir
(c∞ )
c∞ := sup
x6=0
max
1≤k≤m
|Fk (x)|
kxk
.
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Then, by the fact that |Fk (x)| ≤ kFk k kxk for any x ∈ X, where kFk k is the
norm of the functional Fk , we have that
cp ≤
m
X
! p1
kFk kp
,
p≥1
k=1
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c∞ ≤ max kFk k .
1≤k≤m
We may now state and prove a new reverse inequality for the generalised
triangle inequality in normed linear spaces.
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Theorem 5.1. Let xi , rk , Fk , k ∈ {1, . . . , m}, i ∈ {1, . . . , n} be as in the
hypothesis of Theorem 3.1. Then we have the inequalities


Pn
max kFk k
c∞
i=1 kxi k
≤ 1≤k≤m
.
(5.1)
(1 ≤) P
≤
n
k i=1 xi k
max {rk }
max {rk }
1≤k≤m
1≤k≤m
The case of equality holds in (5.1) if and only if
"
!#
n
n
X
X
(5.2)
Re Fk
xi
= rk
kxi k for each k ∈ {1, . . . , m}
i=1
Reverses of the Triangle
Inequality in Banach Spaces
i=1
S.S. Dragomir
and
"
(5.3)
max Re Fk
1≤k≤m
n
X
!#
xi
i=1
n
X
= c∞ xi .
i=1
Proof. Since, by the definition of c∞ , we have
c∞ kxk ≥ max |Fk (x)| ,
1≤k≤m
for any x ∈ X,
P
then we can state, for x = ni=1 xi , that
!
"
!#
n
n
n
X
X
X
(5.4) c∞ xi ≥ max Fk
xi ≥ max Re Fk
xi 1≤k≤m 1≤k≤m i=1
i=1
i=1
#
"
#
"
n
n
X
X
≥ max Re
Fk (xi ) = max
Re Fk (xi ) .
1≤k≤m
i=1
1≤k≤m
i=1
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Utilising the hypothesis (3.1) we obviously have
" n
#
n
X
X
max
Re Fk (xi ) ≥ max {rk } ·
kxi k .
1≤k≤m
1≤k≤m
i=1
i=1
Pn
Also, i=1 xi 6= 0, because, by the initial assumptions, not all rk and xi with
k ∈ {1, . . . , m} and i ∈ {1, . . . , n} are allowed to be zero. Hence the desired
inequality (5.1) is obtained.
Now, if (5.2) is valid, then, taking the maximum over k ∈ {1, . . . , m} in this
equality we get
"
!#
n
n
X
X
max Re Fk
xi
= max {rk } xi ,
1≤k≤m
1≤k≤m
i=1
i=1
which, together with (5.3) provides the equality case in (5.1).
Now, if the equality holds in (5.1), it must hold in all the inequalities used to
prove (5.1), therefore, we have
(5.5) Re Fk (xi ) = rk kxi k
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m}
Reverses of the Triangle
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S.S. Dragomir
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and, from (5.4),
n
"
X xi = max Re Fk
c∞ 1≤k≤m
i=1
n
X
Close
!#
xi
,
i=1
which is (5.3).
From (5.5), on summing over i ∈ {1, . . . , n} , we get (5.2), and the theorem
is proved.
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The following result in normed spaces also holds.
Theorem 5.2. Let xi , rk , Fk , k ∈ {1, . . . , m} , i ∈ {1, . . . , n} be as in the
hypothesis of Theorem 3.1. Then we have the inequality
(5.6)
Pn
Pm
1
kxi k
kFk kp p
cp
i=1
k=1
(1 ≤) Pn
≤ P
≤ Pm p
,
p p1
k i=1 xi k
rk
k=1
( m
r
)
k=1 k
where p ≥ 1.
The case of equality holds in (5.6) if and only if
"
!#
n
n
X
X
(5.7)
Re Fk
xi
= rk
kxi k for each k ∈ {1, . . . , m}
i=1
i=1
S.S. Dragomir
Title Page
and
(5.8)
Reverses of the Triangle
Inequality in Banach Spaces
Contents
m
X
"
Re Fk
n
X
!#p
xi
i=1
k=1
p
n
X
= cpp xi .
i=1
Proof. By the definition of cp , p ≥ 1, we have
cpp kxkp ≥
m
X
k=1
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for any x ∈ X,
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implying that
p
!p
!p
n
m n
m n
X
X
X
X
X
(5.9) cpp xi xi ≥
xi ≥
Fk
Re Fk
i=1
i=1
i=1
k=1
k=1
"
!#
"
#p
p
n
m
n
m
X
X
X
X
Re Fk
xi
=
Re Fk (xi ) .
≥
i=1
k=1
k=1
i=1
Utilising the hypothesis (3.1), we obviously have that
" n
#p
" n
#p
m
m
m
X
X
X
X
X
(5.10)
rkp
Re Fk (xi ) ≥
rk kxi k =
k=1
i=1
k=1
i=1
k=1
n
X
Reverses of the Triangle
Inequality in Banach Spaces
!p
kxi k
.
i=1
Making use of (5.9) and (5.10), we deduce
p
! n
!p
n
m
X
X
X
p
cpp xi ≥
rk
kxi k ,
i=1
i=1
k=1
which implies the desired inequality (5.6).
If (5.7) holds true, then, taking the power p and summing over k ∈ {1, . . . , m} ,
we deduce
" "
!##p
!p
m
n
m
n
X
X
X
X
Re Fk
xi
=
rkp
kxi k ,
k=1
i=1
k=1
S.S. Dragomir
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i=1
which, together with (5.8) shows that the equality case holds true in (5.6).
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Conversely, if the case of equality holds in (5.6), then it must hold in all
inequalities needed to prove (5.6), therefore, we must have:
(5.11) Re Fk (xi ) = rk kxi k
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m}
and, from (5.9),
n
"
m
X p X
cpp xi =
Re Fk
i=1
k=1
n
X
!#p
xi
,
i=1
which is exactly (5.8).
From (5.11), on summing over i from 1 to n, we deduce (5.7), and the theorem is proved.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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6.
6.1.
An Additive Reverse for the Triangle Inequality
The Case of One Functional
In the following we provide an alternative of the Diaz-Metcalf reverse of the
generalised triangle inequality [6].
Theorem 6.1 (Dragomir, 2004). Let (X, k·k) be a normed linear space over
the real or complex number field K and F : X → K a linear functional with the
property that |F (x)| ≤ kxk for any x ∈ X. If xi ∈ X, ki ≥ 0, i ∈ {1, . . . , n}
are such that
(6.1)
S.S. Dragomir
(0 ≤) kxi k − Re F (xi ) ≤ ki for each i ∈ {1, . . . , n} ,
then we have the inequality
Title Page
n
n
n
X
X
X
(0 ≤)
kxi k − xi ≤
ki .
(6.2)
i=1
i=1
Contents
i=1
The equality holds in (6.2) if and only if both
! n
!
n
n
n
n
X X
X
X
X
(6.3)
F
xi = xi and F
xi =
kxi k −
ki .
i=1
i=1
i=1
i=1
Proof. If we sum in (6.1) over i from 1 to n, then we get
"
!#
n
n
n
X
X
X
(6.4)
kxi k ≤ Re F
xi
+
ki .
i=1
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i=1
i=1
i=1
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Taking into account that |F (x)| ≤ kxk for each x ∈ X, then we may state that
"
!# !
n
n
X
X
(6.5)
Re F
xi
≤ Re F
xi i=1
i=1
!
n
n
X
X
≤ F
xi ≤ xi .
i=1
i=1
Now, making use of (6.4) and (6.5), we deduce (6.2).
Obviously, if (6.3) is valid, then the case of equality in (6.2) holds true.
Conversely, if the equality holds in (6.2), then it must hold in all the inequalities used to prove (6.2), therefore we have
"
!#
n
n
n
X
X
X
kxi k = Re F
xi
+
ki
i=1
and
"
Re F
i=1
n
X
i=1
!#
xi
= F
n
X
i=1
i=1
! n
X xi = xi ,
Reverses of the Triangle
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S.S. Dragomir
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i=1
which imply (6.3).
The following corollary may be stated [6].
Corollary 6.2. Let (X, k·k) be a normed linear space, [·, ·] : X × X → K a
semi-inner product generating the norm k·k and e ∈ X, kek = 1. If xi ∈ X,
ki ≥ 0, i ∈ {1, . . . , n} are such that
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(6.6)
(0 ≤) kxi k − Re [xi , e] ≤ ki for each i ∈ {1, . . . , n} ,
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then we have the inequality
(0 ≤)
(6.7)
n
X
i=1
n
n
X
X
kxi k − xi ≤
ki .
i=1
i=1
The equality holds in (6.7) if and only if both
" n
# n
" n
#
n
n
X X
X
X
X
(6.8)
xi , e = xi and
xi , e =
kxi k −
ki .
i=1
i=1
i=1
i=1
i=1
Moreover, if (X, k·k) is strictly convex, then the case of equality holds in (6.7)
if and only if
n
X
(6.9)
kxi k ≥
i=1
n
X
ki
S.S. Dragomir
Title Page
i=1
Contents
and
(6.10)
Reverses of the Triangle
Inequality in Banach Spaces
n
X
i=1
xi =
n
X
i=1
kxi k −
n
X
!
ki
· e.
i=1
Proof. The first part of the corollary is obvious by Theorem 6.1 applied for the
continuous linear functional of unit norm Fe , Fe (x) = [x, e] , x ∈ X. The
second part may be shown on utilising a similar argument to the one from the
proof of Corollary 4.3. We omit the details.
Remark 3. If X = H, (H; h·, ·i) is an inner product space, then from Corollary
6.2 we deduce the additive reverse inequality obtained in Theorem 7 of [12]. For
further similar results in inner product spaces, see [4] and [12].
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6.2.
The Case of m Functionals
The following result generalising Theorem 6.1 may be stated [6].
Theorem 6.3 (Dragomir, 2004). Let (X, k·k) be a normed linear space over
the real or complex number field K. If Fk , k ∈ {1, . . . , m} are bounded linear
functionals defined on X and xi ∈ X, Mik ≥ 0 for i ∈ {1, . . . , n}, k ∈
{1, . . . , m} are such that
kxi k − Re Fk (xi ) ≤ Mik
(6.11)
for each i ∈ {1, . . . , n} , k ∈ {1, . . . , m} , then we have the inequality
n
m
n
m X
n
1 X
X
1 X
X
Fk xi +
Mik .
(6.12)
kxi k ≤ m
m
i=1
i=1
k=1
k=1 i=1
The case of equality holds in (6.12) if both
n
! m
n
m
X
X
X
X
1
1
Fk
xi = Fk (6.13)
xi m
m k=1
i=1
i=1
k=1
and
(6.14)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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m
1 X
Fk
m k=1
n
X
i=1
!
xi
=
n
X
i=1
m
n
1 XX
kxi k −
Mik .
m k=1 j=1
Proof. If we sum (6.11) over i from 1 to n, then we deduce
!
n
n
n
X
X
X
kxi k − Re Fk
xi ≤
Mik
i=1
i=1
i=1
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for each k ∈ {1, . . . , m} .
Summing these inequalities over k from 1 to m, we deduce
!
n
n
m
m
n
X
X
1 XX
1 X
Re Fk
(6.15)
xi +
Mik .
kxi k ≤
m k=1
m k=1 i=1
i=1
i=1
Utilising the continuity property of the functionals Fk and the properties of the
modulus, we have
! m
!
m
n
n
X
X
X
X
Re Fk
Re Fk
(6.16)
xi ≤ xi i=1
i=1
k=1
k=1
!
m
n
X
X
≤
Fk
xi k=1
i=1
m
n
X
X
≤
Fk xi .
k=1
i=1
Now, by (6.15) and (6.16), we deduce (6.12).
Obviously, if (6.13) and (6.14) hold true, then the case of equality is valid in
(6.12).
Conversely, if the case of equality holds in (6.12), then it must hold in all the
inequalities used to prove (6.12). Therefore we have
!
n
m
n
m
n
X
X
1 X
1 XX
kxi k =
Re Fk
xi +
Mik ,
m k=1
m k=1 i=1
i=1
i=1
Reverses of the Triangle
Inequality in Banach Spaces
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m
X
Re Fk
!
xi
i=1
k=1
and
n
X
m
X
k=1
Im Fk
m
n
X
X
=
Fk xi i=1
k=1
n
X
!
xi
= 0.
i=1
These imply that (6.13) and (6.14) hold true, and the theorem is completely
proved.
Remark 4. If Fk , k ∈ {1, . . . , m} are of unit norm, then, from (6.12), we deduce
the inequality
n
m X
n
n
X 1 X
X
Mik ,
(6.17)
kxi k ≤ xi +
m
i=1
i=1
k=1 i=1
which is obviously coarser than (6.12), but perhaps more useful for applications.
6.3.
The Case of Inner Product Spaces
The case of inner product spaces, in which we may provide a simpler condition
of equality, is of interest in applications [6].
Theorem 6.4 (Dragomir, 2004). Let (X, k·k) be an inner product space over
the real or complex number field K, ek , xi ∈ H\ {0} , k ∈ {1, . . . , m} , i ∈
{1, . . . , n} . If Mik ≥ 0 for i ∈ {1, . . . , n} , {1, . . . , n} such that
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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(6.18)
kxi k − Re hxi , ek i ≤ Mik
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for each i ∈ {1, . . . , n} , k ∈ {1, . . . , m} , then we have the inequality
m
n
m X
n
n
X
1 X
1 X
X
(6.19)
kxi k ≤ ek xi +
Mik .
m
m
i=1
i=1
k=1
k=1 i=1
The case of equality holds in (6.19) if and only if
n
X
(6.20)
i=1
m
n
1 XX
kxi k ≥
Mik
m k=1 i=1
and
(6.21)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
n
X
i=1
xi =
m
Pn
i=1
m
P Pn
X
kxi k − m1 m
k=1
i=1 Mik
ek .
Pm
2
k k=1 ek k
k=1
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Proof. As in the proof of Theorem 6.3, we have
*
+
n
m
n
m
n
X
X
X
1
1 XX
(6.22)
kxi k ≤ Re
ek ,
xi +
Mik ,
m k=1
m k=1 i=1
i=1
i=1
P
and m
k=1 ek 6= 0.
On utilising the Schwarz inequality in the inner product space (H; h·, ·i) for
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Pn
i=1
xi ,
(6.23)
Pm
k=1 ek ,
we have
*
+
n
m
n
m
X
X
X
X
xi ek ≥ xi ,
ek i=1
k=1
i=1
* n k=1 m +
X X
≥ Re
xi ,
ek k=1
+
* i=1
m
n
X
X
≥ Re
xi ,
ek .
i=1
k=1
By (6.22) and (6.23) we deduce (6.19).
Taking the norm in (6.21) and using (6.20), we have
n
X
m Pn kx k − 1 Pm Pn M i
ik
i=1
Pmm k=1 i=1
xi =
,
k k=1 ek k
i=1
showing that the equality holds in (6.19).
Conversely, if the case of equality holds in (6.19), then it must hold in all the
inequalities used to prove (6.19). Therefore we have
kxi k = Re hxi , ek i + Mik
(6.24)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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for each i ∈ {1, . . . , n} , k ∈ {1, . . . , m} ,
n
m * n
+
m
X X
X X
(6.25)
xi ek = xi ,
ek i=1
k=1
i=1
k=1
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and
Im
(6.26)
* n
X
i=1
xi ,
m
X
+
ek
= 0.
k=1
From (6.24), on summing over i and k, we get
+
* n
n
m X
n
m
X
X
X X
(6.27)
Re
xi ,
ek = m
kxi k −
Mik .
i=1
k=1
i=1
k=1 i=1
On the other hand, by the use of the identity (3.22), the relation (6.25) holds if
and only if
P
P
n
m
X
X
h ni=1 xi , m
k=1 ek i
xi =
ek ,
P
2
k m
k=1 ek k
i=1
k=1
giving, from (6.26) and (6.27), that
P
P Pn
n
m
X
X
m ni=1 kxi k − m
k=1
i=1 Mik
xi =
ek .
Pm
2
k
e
k
k
k=1
i=1
k=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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If the inequality holds in (6.19), then obviously (6.20) is valid, and the theorem
is proved.
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Remark 5. If in the above theorem the vectors {ek }k=1,m are assumed to be
orthogonal, then (6.19) becomes:
! 12 n
n
m
m X
n
1 X
X
X
X
1
2
kek k
Mik .
(6.28)
kxi k ≤
xi +
m
m k=1
i=1
i=1
k=1 i=1
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Moreover, if {ek }k=1,m is an orthonormal family, then (6.28) becomes
√ n
n
n
m X
1 X
X
X
m
(6.29)
kxi k ≤
x
+
Mik ,
i
m
m
i=1
i=1
k=1 i=1
which has been obtained in [12].
Before we provide some natural consequences of Theorem 6.4, we need
some preliminary results concerning another reverse of Schwarz’s inequality
in inner product spaces (see for instance [4, p. 27]).
Lemma 6.5 (Dragomir, 2004). Let (X, k·k) be an inner product space over the
real or complex number field K and x, a ∈ H, r > 0. If kx − ak ≤ r, then we
have the inequality
1
(6.30)
kxk kak − Re hx, ai ≤ r2 .
2
The case of equality holds in (6.30) if and only if
(6.31)
kx − ak = r and kxk = kak .
Proof. The condition kx − ak ≤ r is clearly equivalent to
(6.32)
kxk2 + kak2 ≤ 2 Re hx, ai + r2 .
S.S. Dragomir
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Since
(6.33)
Reverses of the Triangle
Inequality in Banach Spaces
2 kxk kak ≤ kxk2 + kak2 ,
with equality if and only if kxk = kak , hence by (6.32) and (6.33) we deduce
(6.30).
The case of equality is obvious.
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Utilising the above lemma we may state the following corollary of Theorem
6.4 [6].
Corollary 6.6. Let (H; h·, ·i) , ek , xi be as in Theorem 6.4. If rik > 0, i ∈
{1, . . . , n} , k ∈ {1, . . . , m} such that
(6.34)
kxi − ek k ≤ rik for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} ,
then we have the inequality
m
n
m
n
n
X
1 X
X
1 XX 2
(6.35)
kxi k ≤ ek xi +
rik .
2m
m
i=1
i=1
k=1
k=1 i=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
The equality holds in (6.35) if and only if
n
X
i=1
and
n
X
i=1
xi =
m
Pn
m
n
1 XX 2
kxi k ≥
r
2m k=1 i=1 ik
i=1
Pm Pn 2 m
1
X
kxi k − 2m
k=1
i=1 rik
ek .
Pm
2
k k=1 ek k
k=1
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The following lemma may provide another sufficient condition for (6.18) to
hold (see also [4, p. 28]).
Close
Lemma 6.7 (Dragomir, 2004). Let (H; h·, ·i) be an inner product space over
the real or complex number field K and x, y ∈ H, M ≥ m > 0. If either
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(6.36)
Re hM y − x, x − myi ≥ 0
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or, equivalently,
(6.37)
x − m + M y ≤ 1 (M − m) kyk ,
2
2
holds, then
(6.38)
1 (M − m)2
kxk kyk − Re hx, yi ≤ ·
kyk2 .
4
m+M
The case of equality holds in (6.38) if and only if the equality case is realised in
(6.36) and
M +m
kxk =
kyk .
2
The proof is obvious by Lemma 6.5 for a = M +m
y and r = 12 (M − m) kyk .
2
Finally, the following corollary of Theorem 6.4 may be stated [6].
Corollary 6.8. Assume that (H, h·, ·i) , ek , xi are as in Theorem 6.4. If Mik ≥
mik > 0 satisfy the condition
Re hMk ek − xi , xi − µk ek i ≥ 0
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
n
m
m
n
1 X
X X
1 X X (Mik − mik )2
kxi k ≤ ek kek k2 .
xi +
m
4m
M
+
m
ik
ik
i=1
i=1
k=1
k=1 i=1
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7.
Other Additive Reverses for m Functionals
A different approach in obtaining other additive reverses for the generalised
triangle inequality is incorporated in the following new result:
Theorem 7.1. Let (X, k·k) be a normed linear space over the real or complex
number field K. Assume Fk , k ∈ {1, . . . , m} , are bounded linear functionals
on the normed linear space X and xi ∈ X, i ∈ {1, . . . , n} , Mik ≥ 0, i ∈
{1, . . . , n} , k ∈ {1, . . . , m} are such that
kxi k − Re Fk (xi ) ≤ Mik
(7.1)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} .
(i) If c∞ is defined by (c∞ ), then we have the inequality
n
n
m X
n
X
1 X
X
(7.2)
kxi k ≤ c∞ xi +
Mik .
m
i=1
i=1
k=1 i=1
(ii) If cp is defined by (cp ) for p ≥ 1, then we have the inequality:
n
n
m X
n
1 X
X
X
1
(7.3)
kxi k ≤ 1 cp xi +
Mik .
m p i=1 m k=1 i=1
i=1
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Proof. (i) Since
max kFk (x)k ≤ c∞ kxk
1≤k≤m
for any x ∈ X,
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then we have
m X
Fk
n
X
k=1
i=1
!
xi ≤ m max Fk
1≤k≤m n
X
i=1
n
!
X xi ≤ mc∞ xi .
i=1
Using (6.16), we may state that
n
X
m
1 X
Re Fk
m k=1
!
xi
i=1
n
X ≤ c∞ xi ,
i=1
which, together with (6.15) imply the desired inequality (7.2).
(ii) Using the fact that, obviously
m
X
! p1
p
|Fk (x)|
≤ cp kxk
then, by Hölder’s inequality for p > 1, p1 +
k=1
n
X
i=1
S.S. Dragomir
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for any x ∈ X,
k=1
m X
Fk
Reverses of the Triangle
Inequality in Banach Spaces
1
q
= 1, we have
!p ! p1
m n
X
X
xi Fk
k=1
i=1
n
1 X
q
≤ cp m xi ,
!
1
xi ≤ m q
i=1
which, combined with (6.15) and (6.16) will give the desired inequality (7.3).
The case p = 1 goes likewise and we omit the details.
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Remark 6. Since, obviously c∞ ≤ max kFk k , then from (7.2) we have
1≤k≤m
(7.4)
n
X
i=1
m X
n
n
X
1 X
kxi k ≤ max {kFk k} · Mik .
xi +
1≤k≤m
m
i=1
k=1 i=1
P
p p1
Finally, since cp ≤ ( m
k=1 kFk k ) , p ≥ 1, hence by (7.3) we have
Pm
m X
n
n
n
p p1 X
1 X
X
kF
k
k
k=1
xi +
Mik .
(7.5)
kxi k ≤
m
m
i=1
i=1
k=1 i=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
The following corollary for semi-inner products may be stated as well.
Corollary 7.2. Let (X, k·k) be a real or complex normed space and [·, ·] : X ×
X → K a semi-inner product generating the norm k·k . Assume ek , xi ∈ H and
Mik ≥ 0, i ∈ {1, . . . , n} , k ∈ {1, . . . , m} are such that
kxi k − Re [xi , ek ] ≤ Mik ,
(7.6)
for any i ∈ {1, . . . , n} , k ∈ {1, . . . , m} .
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(i) If
d∞ := sup
x6=0
max1≤k≤n |[x, ek ]|
≤ max kek k ,
1≤k≤n
kxk
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then
(7.7)
n
X
i=1
n
n
m X
X 1 X
Mik
kxi k ≤ d∞ xi +
m
i=1
k=1 i=1
n
!
n
m X
X 1 X
≤ max kek k · xi +
Mik ;
1≤k≤n
m
i=1
k=1 i=1
(ii) If
Pm
dp := sup
x6=0
p p1
|[x, ek ]|
kxkp
k=1

≤
m
X
! p1 
kek kp
,
S.S. Dragomir
k=1
where p ≥ 1, then
(7.8)
Reverses of the Triangle
Inequality in Banach Spaces
n
n
m X
n
X
1 X
X
1
kxi k ≤ 1 dp xi +
Mik
m
p
m
i=1
i=1
k=1 i=1
!
Pm
n
m X
n
p p1 X
1 X
ke
k
k
k=1
≤
xi +
Mik .
m
m
i=1
k=1 i=1
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8.
Applications for Complex Numbers
Let C be the field of complex numbers. If z = Re z + i Im z, then by |·|p : C →
[0, ∞), p ∈ [1, ∞] we define the p−modulus of z as

 max {|Re z| , |Im z|} if p = ∞,
|z|p :=
1

(|Re z|p + |Im z|p ) p if p ∈ [1, ∞),
where |a| , a ∈ R is the usual modulus of the real number a.
For p = 2, we recapture the usual modulus of a complex number, i.e.,
q
|z|2 = |Re z|2 + |Im z|2 = |z| , z ∈ C.
It is well known that C, |·|p , p ∈ [1, ∞] is a Banach space over the real
number field R.
Consider the Banach space (C, |·|1 ) and F : C → C, F (z) = az with a ∈ C,
a 6= 0. Obviously, F is linear on C. For z 6= 0, we have
q
|a| |Re z|2 + |Im z|2
|F (z)|
|a| |z|
=
=
≤ |a| .
|z|1
|z|1
|Re z| + |Im z|
Since, for z0 = 1, we have |F (z0 )| = |a| and |z0 |1 = 1, hence
|F (z)|
kF k1 := sup
= |a| ,
|z|1
z6=0
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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showing that F is a bounded linear functional on (C, |·|1 ) and kF k1 = |a| .
We can apply Theorem 3.1 to state the following reverse of the generalised
triangle inequality for complex numbers [5].
Proposition 8.1. Let ak , xj ∈ C, k ∈ {1, . . . , m}
j ∈ {1, . . . , n} . If there
Pand
m
exist the constants rk ≥ 0, k ∈ {1, . . . , m} with k=1 rk > 0 and
rk [|Re xj | + |Im xj |] ≤ Re ak · Re xj − Im ak · Im xj
(8.1)
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
#
" n
P
n
X
X
X
| m
a
|
k
Re
x
+
Im
x
(8.2)
[|Re xj | + |Im xj |] ≤ Pk=1
j
j .
m
r
k=1 k
j=1
j=1
j=1
The case of equality holds in (8.2) if both
!
!
!
!
m
n
m
n
X
X
X
X
Re
ak Re
xj − Im
ak Im
xj
j=1
k=1
=
m
X
k=1
j=1
k=1
!
rk
n
X
[|Re xj | + |Im xj |]
j=1
m
" n
n
#
X X
X
=
ak Re xj + Im xj .
k=1
j=1
j=1
The proof follows by Theorem 3.1 applied for the Banach space (C, |·|1 ) and
Fk (z) = ak z, k ∈ {1, . . . , m} on taking into account that:
m
m
X X Fk = ak .
k=1
1
k=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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Now, consider the Banach space (C, |·|∞ ) . If F (z) = dz, then for z 6= 0 we
have
q
|d| |Re z|2 + |Im z|2 √
|F (z)|
|d| |z|
=
=
≤ 2 |d| .
|z|∞
|z|∞
max {|Re z| , |Im z|}
√
Since, for z0 = 1 + i, we have |F (z0 )| = 2 |d| , |z0 |∞ = 1, hence
kF k∞ := sup
z6=0
|F (z)| √
= 2 |d| ,
|z|∞
√
showing that F is a bounded linear functional on (C, |·|∞ ) and kF k∞ = 2 |d| .
If we apply Theorem 3.1, then we can state the following reverse of the
generalised triangle inequality for complex numbers [5].
Proposition 8.2. Let ak , xj ∈ C, k ∈ {1, . . . , m}
Pand j ∈ {1, . . . , n} . If there
exist the constants rk ≥ 0, k ∈ {1, . . . , m} with m
k=1 rk > 0 and
rk max {|Re xj | , |Im xj |} ≤ Re ak · Re xj − Im ak · Im xj
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
(8.3)
n
X
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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max {|Re xj | , |Im xj |}
Quit
j=1
≤
√
)
( n
P
n
X
X
| m
a
|
k
2 · Pk=1
max
Re
x
,
Im
x
j j .
m
k=1 rk
j=1
j=1
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The case of equality holds in (8.3) if both
!
!
!
!
n
m
n
m
X
X
X
X
Re
ak Re
xj − Im
ak Im
xj
j=1
k=1
=
m
X
k=1
!
rk
k=1
n
X
j=1
max {|Re xj | , |Im xj |}
j=1
m
n
)
( n
X
X
√ X
ak max Re xj , Im xj .
= 2
j=1
k=1
j=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
Finally, consider the Banach space C, |·|2p with p ≥ 1.
Let F : C → C, F (z) = cz. By Hölder’s inequality, we have
q
|c| |Re z|2 + |Im z|2
1
1
|F (z)|
− 2p
2
=
|c| .
1 ≤ 2
2p
2p 2p
|z|2p
|Re z| + |Im z|
1
1
Since, for z0 = 1 + i we have |F (z0 )| = 2 2 |c| , |z0 |2p = 2 2p (p ≥ 1) , hence
1
1
|F (z)|
= 2 2 − 2p |c| ,
z6=0 |z|2p
showing that F is a bounded linear functional on C, |·|2p , p ≥ 1 and kF k2p =
kF k2p := sup
1
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1
2 2 − 2p |c| .
If we apply Theorem 3.1, then we can state the following proposition [5].
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Proposition 8.3. Let ak , xj ∈ C, k ∈ {1, . . . , m}
j ∈ {1, . . . , n} . If there
Pand
m
exist the constants rk ≥ 0, k ∈ {1, . . . , m} with k=1 rk > 0 and
1
rk |Re xj |2p + |Im xj |2p 2p ≤ Re ak · Re xj − Im ak · Im xj
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
(8.4)
n
X
1
|Re xj |2p + |Im xj |2p 2p
Reverses of the Triangle
Inequality in Banach Spaces
j=1

2p 2p  2p1
Pm
n
n
X
X
1 |
1
ak | 

Re
x
+
.
≤ 2 2 − 2p Pk=1
Im
x
j
j
m
k=1 rk
j=1
j=1
The case of equality holds in (8.4) if both:
!
!
!
!
m
n
m
n
X
X
X
X
Re
ak Re
xj − Im
ak Im
xj
j=1
k=1
=
m
X
k=1
!
rk
n
X
1
|Re xj |2p + |Im xj |2p 2p
j=1
k=1
=2
1
1
− 2p
2
j=1

2p 2p  2p1
m
n
n
X
X
X

ak Re xj + Im xj  .
k=1
j=1
S.S. Dragomir
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Remark 7. If in the above proposition we choose p = 1, then we have the
following reverse of the generalised triangle inequality for complex numbers
n
P
n
X X
| m
a
|
k |xj | ≤ Pk=1
xj m
r
k=1 k
j=1
j=1
provided xj , ak , j ∈ {1, . . . , n}, k ∈ {1, . . . , m} satisfy the assumption
rk |xj | ≤ Re ak · Re xj − Im ak · Im xj
for each j ∈ {1, . . . , n}, k ∈ {1, . . . , m} . Here |·| is the usual modulus of a
complex number and rk > 0, k ∈ {1, . . . , m} are given.
We can apply Theorem 6.3 to state the following reverse of the generalised
triangle inequality for complex numbers [6].
Proposition 8.4. Let ak , xj ∈ C, k ∈ {1, . . . , m} and j ∈ {1, . . . , n} . If there
exist the constants Mjk ≥ 0, k ∈ {1, . . . , m} , j ∈ {1, . . . , n} such that
|Re xj | + |Im xj | ≤ Re ak · Re xj − Im ak · Im xj + Mjk
(8.5)
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
(8.6)
n
X
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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[|Re xj | + |Im xj |]
j=1
m
" n
n
#
m
n
X
1 XX
1 X X
ak Mjk .
≤
Re xj + Im xj +
m k=1 j=1
m k=1 j=1
j=1
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The proof follows by Theorem 6.3 applied for the Banach space (C, |·|1 ) and
Fk (z) = ak z, k ∈ {1, . . . , m} on taking into account that:
m
m
X
X
Fk = ak .
k=1
1
k=1
If we apply Theorem 6.3 for the Banach space (C, |·|∞ ), then we can state
the following reverse of the generalised triangle inequality for complex numbers
[6].
Reverses of the Triangle
Inequality in Banach Spaces
Proposition 8.5. Let ak , xj ∈ C, k ∈ {1, . . . , m} and j ∈ {1, . . . , n} . If there
exist the constants Mjk ≥ 0, k ∈ {1, . . . , m} , j ∈ {1, . . . , n} such that
S.S. Dragomir
max {|Re xj | , |Im xj |} ≤ Re ak · Re xj − Im ak · Im xj + Mjk
Title Page
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
(8.7)
n
X
max {|Re xj | , |Im xj |}
j=1
n
)
( n
√ m
m
n
X
X
2 X 1 XX
ak max Re xj , Mjk .
Im xj +
≤
m k=1 m
j=1
j=1
k=1 j=1
Finally, if we apply Theorem 6.3, for the Banach space C, |·|2p with p ≥ 1,
then we can state the following proposition [6].
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Proposition 8.6. Let ak , xj , Mjk be as in Proposition 8.5. If
1
|Re xj |2p + |Im xj |2p 2p ≤ Re ak · Re xj − Im ak · Im xj + Mjk
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
X
1
|Re xj |2p + |Im xj |2p 2p
(8.8)
j=1
≤
2
1
1
− 2p
2
m
2p 2p  2p1

m
n
n
m
n
X
X
X
1 XX


Re
x
+
Im
x
a
+
Mjk .
k j
j
m
j=1
j=1
k=1
k=1 j=1
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
where p ≥ 1.
Title Page
Remark 8. If in the above proposition we choose p = 1, then we have the
following reverse of the generalised triangle inequality for complex numbers
n
n
m
m X
n
1 X
X 1 X
X
ak Mjk
|xj | ≤ xj +
m
m
j=1
j=1
k=1
k=1 j=1
Contents
provided xj , ak , j ∈ {1, . . . , n}, k ∈ {1, . . . , m} satisfy the assumption
|xj | ≤ Re ak · Re xj − Im ak · Im xj + Mjk
for each j ∈ {1, . . . , n}, k ∈ {1, . . . , m} . Here |·| is the usual modulus of a
complex number and Mjk > 0, j ∈ {1, . . . , n}, k ∈ {1, . . . , m} are given.
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9.
Karamata Type Inequalities in Hilbert Spaces
Let f : [a, b] → K, K = C or R be a Lebesgue integrable function. The
following inequality, which is the continuous version of the triangle inequality
Z b
Z b
(9.1)
f (x) dx ≤
|f (x)| dx,
a
a
plays a fundamental role in Mathematical Analysis and its applications.
It appears, see [20, p. 492], that the first reverse inequality for (9.1) was
obtained by J. Karamata in his book from 1949, [14]. It can be stated as
Z b
Z b
(9.2)
cos θ
|f (x)| dx ≤ f (x) dx
a
a
provided
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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−θ ≤ arg f (x) ≤ θ, x ∈ [a, b]
π
for given θ ∈ 0, 2 .
This result has recently been extended by the author for the case of Bochner
integrable functions with values in a Hilbert space H (see also [10]):
Theorem 9.1 (Dragomir, 2004). If f ∈ L ([a, b] ; H) (this means that f :
Rb
[a, b] → H is strongly measurable on [a, b] and the Lebesgue integral a kf (t)k dt
is finite), then
Z b
Z b
,
(9.3)
kf (t)k dt ≤ K f
(t)
dt
a
a
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provided that f satisfies the condition
(9.4)
kf (t)k ≤ K Re hf (t) , ei for a.e. t ∈ [a, b] ,
where e ∈ H, kek = 1 and K ≥ 1 are given.
The case of equality holds in (9.4) if and only if
Z b
Z b
1
(9.5)
f (t) dt =
kf (t)k dt e.
K
a
a
As some natural consequences of the above results, we have noticed in [10]
that, if ρ ∈ [0, 1) and f ∈ L ([a, b] ; H) are such that
kf (t) − ek ≤ ρ for a.e. t ∈ [a, b] ,
(9.6)
p
Z
1 − ρ2
a
b
Z b
kf (t)k dt ≤ f
(t)
dt
a
with equality if and only if
Z b
Z b
p
f (t) dt = 1 − ρ2
kf (t)k dt · e.
a
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a
Also, for e as above and if M ≥ m > 0, f ∈ L ([a, b] ; H) such that either
(9.8)
S.S. Dragomir
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then
(9.7)
Reverses of the Triangle
Inequality in Banach Spaces
Re hM e − f (t) , f (t) − mei ≥ 0
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or, equivalently,
1
M
+
m
f (t) −
e
≤ 2 (M − m)
2
(9.9)
for a.e. t ∈ [a, b] , then
b
Z
(9.10)
a
Z b
M +m f (t) dt
kf (t)k dt ≤ √
,
2 mM a
with equality if and only if
Z
a
b
√
Z b
2 mM
f (t) dt =
kf (t)k dt · e.
M +m
a
The main aim of the following sections is to extend the integral inequalities
mentioned above for the case of Banach spaces. Applications for Hilbert spaces
and for complex-valued functions are given as well.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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10.
10.1.
Multiplicative Reverses of the Continuous
Triangle Inequality
The Case of One Functional
Let (X, k·k) be a Banach space over the real or complex number field. Then
one has the following reverse of the continuous triangle inequality [11].
Theorem 10.1 (Dragomir, 2004). Let F be a continuous linear functional of
unit norm on X. Suppose that the function f : [a, b] → X is Bochner integrable
on [a, b] and there exists a r ≥ 0 such that
(10.1)
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
r kf (t)k ≤ Re F [f (t)] for a.e. t ∈ [a, b] .
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Then
Contents
Z
(10.2)
b
r
a
Z b
,
kf (t)k dt ≤ f
(t)
dt
a
where equality holds in (10.2) if and only if both
Z b
Z b
(10.3)
F
f (t) dt = r
kf (t)k dt
a
a
and
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Z
(10.4)
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F
a
Z b
b
f (t) dt = f (t) dt
.
a
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Proof. Since the norm of F is one, then
|F (x)| ≤ kxk for any x ∈ X.
Rb
Applying this inequality for the vector a f (t) dt, we get
Z b
Z b
f
(t)
dt
≥
F
f
(t)
dt
(10.5)
a
a
Z b
Z b
≥ Re F
f (t) dt = Re F (f (t)) dt .
a
a
Now, by integration of (10.1), we obtain
Z b
Z b
(10.6)
Re F (f (t)) dt ≥ r
kf (t)k dt,
a
a
and by (10.5) and (10.6) we deduce the desired inequality (10.2).
Obviously, if (10.3) and (10.4) hold true, then the equality case holds in
(10.2).
Conversely, if the case of equality holds in (10.2), then it must hold in all the
inequalities used before in proving this inequality. Therefore, we must have
(10.7)
r kf (t)k = Re F (f (t)) for a.e. t ∈ [a, b] ,
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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Z
(10.8)
Im F
a
b
f (t) dt = 0
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and
(10.9)
Z b
Z b
f (t) dt = Re F
f (t) dt .
a
a
Integrating (10.7) on [a, b] , we get
Z b
Z b
(10.10)
r
kf (t)k dt = Re F
f (t) dt .
a
a
On utilising (10.10) and (10.8), we deduce (10.3) while (10.9) and (10.10)
would imply (10.4), and the theorem is proved.
Corollary 10.2. Let (X, k·k) be a Banach space, [·, ·] : X × X → R a semiinner product generating the norm k·k and e ∈ X, kek = 1. Suppose that the
function f : [a, b] → X is Bochner integrable on [a, b] and there exists a r ≥ 0
such that
(10.11)
r kf (t)k ≤ Re [f (t) , e] for a.e. t ∈ [a, b] .
Then
S.S. Dragomir
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(10.12)
Reverses of the Triangle
Inequality in Banach Spaces
r
a
Z b
b
f (t) dt
kf (t)k dt ≤ a
where equality holds in (10.12) if and only if both
Z b
Z b
(10.13)
f (t) dt, e = r
kf (t)k dt
a
a
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and
Z
(10.14)
a
b
Z b
f (t) dt, e = f (t) dt
.
a
The proof follows from Theorem 10.1 for the continuous linear functional
F (x) = [x, e] , x ∈ X, and we omit the details.
The following corollary of Theorem 10.1 may be stated [8].
Corollary 10.3. Let (X, k·k) be a strictly convex Banach space, [·, ·] : X ×
X → K a semi-inner product generating the norm k·k and e ∈ X, kek = 1. If
f : [a, b] → X is Bochner integrable on [a, b] and there exists a r ≥ 0 such that
(10.11) holds true, then (10.12) is valid. The case of equality holds in (10.12) if
and only if
Z b
Z b
(10.15)
f (t) dt = r
kf (t)k dt e.
a
a
Proof. If (10.15) holds true, then, obviously
Z b
Z b
Z b
f (t) dt = r
kf (t)k dt kek = r
kf (t)k dt,
a
a
a
which is the equality case in (10.12).
Conversely, if the equality holds in (10.12), then, by Corollary 10.2, we must
have (10.13) and (10.14). Utilising Theorem 4.2, by (10.14) we can conclude
that there exists a µ > 0 such that
Z b
(10.16)
f (t) dt = µe.
a
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Replacing this in (10.13), we get
Z
2
µ kek = r
b
kf (t)k dt,
a
giving
Z
(10.17)
b
kf (t)k dt.
µ=r
a
Utilising (10.16) and (10.17) we deduce (10.15) and the proof is completed.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
10.2.
The Case of m Functionals
The following result may be stated [8]:
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Theorem 10.4 (Dragomir, 2004). Let (X, k·k) be a Banach space over the real
or complex number field K and Fk : X → K, k ∈ {1, . . . , m} continuous linear
functionals on X. If f : [a, b] → X is a Bochner
P integrable function on [a, b]
and there exists rk ≥ 0, k ∈ {1, . . . , m} with m
k=1 rk > 0 and
(10.18)
rk kf (t)k ≤ Re Fk [f (t)]
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] , then
Z b
P
Z b
k m
Fk k k=1
.
(10.19)
kf (t)k dt ≤ Pm
f
(t)
dt
r
a
a
k=1 k
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The case of equality holds in (10.19) if both
! Z
m
b
X
(10.20)
Fk
f (t) dt =
a
k=1
m
X
!Z
b
kf (t)k dt
rk
a
k=1
and
m
X
(10.21)
! Z
Fk
a
k=1
b
Z
m
X
b
f (t) dt = Fk f (t) dt
.
a
k=1
Proof. Utilising the hypothesis (10.18), we have
Z b
"X
Z b
#
m
m
X
(10.22)
I := Fk
f (t) dt ≥ Re
Fk
f (t) dt a
a
k=1
k=1
#
" m
m Z b
X Z b
X
≥ Re
Fk
f (t) dt
=
Re Fk f (t) dt
a
k=1
≥
m
X
k=1
a
! Z
b
rk ·
kf (t)k dt.
k=1
Reverses of the Triangle
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On the other hand, by the continuity property of Fk , k ∈ {1, . . . , m} , we obviously have
m
Z
! Z
X
m
b
b
X (10.23)
I=
Fk
f (t) dt ≤ Fk f (t) dt
.
a
a
k=1
k=1
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Making use of (10.22) and (10.23), we deduce (10.19).
Now, obviously, if (10.20) and (10.21) are valid, then the case of equality
holds true in (10.19).
Conversely, if the equality holds in the inequality (10.19), then it must hold
in all the inequalities used to prove (10.19), therefore we have
rk kf (t)k = Re Fk [f (t)]
(10.24)
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] ,
! Z
m
b
X
(10.25)
Im
Fk
f (t) dt = 0,
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
a
k=1
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(10.26)
Re
m
X
k=1
! Z
Fk
a
b
Z
m
X
b
f (t) dt = Fk f (t) dt
.
a
k=1
Note that, by (10.24), on integrating on [a, b] and summing over k ∈ {1, . . . , m} ,
we get
! Z
!Z
m
m
b
b
X
X
(10.27)
Re
Fk
f (t) dt =
rk
kf (t)k dt.
k=1
a
k=1
a
Now, (10.25) and (10.27) imply (10.20) while (10.25) and (10.26) imply (10.21),
therefore the theorem is proved.
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The following new results may be stated as well:
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Theorem 10.5. Let (X, k·k) be a Banach space over the real or complex number field K and Fk : X → K, k ∈ {1, . . . , m} continuous linear functionals on
X. Also, assume that f : [a, b] → X is a Bochner
Pm integrable function on [a, b]
and there exists rk ≥ 0, k ∈ {1, . . . , m} with k=1 rk > 0 and
rk kf (t)k ≤ Re Fk [f (t)]
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
(i) If c∞ is defined by (c∞ ), then we have the inequality
Rb
kf (t)k dt
c∞
max1≤k≤m kFk k
a
≤
(10.28)
(1 ≤) ≤
R b
max1≤k≤m {rk }
max1≤k≤m {rk }
a f (t) dt
with equality if and only if
Z b
Z b
Re (Fk )
f (t) dt = rk
kf (t)k dt
a
a
for each k ∈ {1, . . . , m} and
Z b
Z b
f (t) dt
= c∞
kf (t)k dt.
max Re (Fk )
1≤k≤m
a
a
(ii) If cp , p ≥ 1, is defined by (cp ) , then we have the inequality
Rb
Pm
1
kf (t)k dt
kFk kp p
cp
k=1
a
≤ P
≤ Pm p
(1 ≤) R b
m
p p1
k=1 rk
(
r
)
f
(t)
dt
a
k=1 k
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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with equality if and only if
Z b
Z b
Re (Fk )
f (t) dt = rk
kf (t)k dt
a
a
for each k ∈ {1, . . . , m} and
Z b
p
Z b
p
m X
p
Re Fk
f (t) dt
= cp f (t) dt
k=1
a
a
where p ≥ 1.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
The proof is similar to the ones from Theorems 5.1, 5.2 and 10.4 and we
omit the details.
The case of Hilbert spaces for Theorem 10.4, which provides a simpler condition for equality, is of interest for applications [8].
Theorem 10.6 (Dragomir, 2004). Let (X, k·k) be a Hilbert space over the
real or complex number field K and ek ∈ H\ {0} , k ∈ {1, . . . , m} . If f :
[a,
Pmb] → H is a Bochner integrable function and rk ≥ 0, k ∈ {1, . . . , m} and
k=1 rk > 0 satisfy
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(10.29)
rk kf (t)k ≤ Re hf (t) , ek i
for each k ∈ {1, . . . , m} and for a.e. t ∈ [a, b] , then
Z b
P
Z b
k m
ek k k=1
.
(10.30)
kf (t)k dt ≤ Pm
f
(t)
dt
a
a
k=1 rk
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The case of equality holds in (10.30) for f 6= 0 a.e. on [a, b] if and only if
Z
(10.31)
a
b
Rb
P
m
X
( m
k=1 rk ) a kf (t)k dt
ek .
f (t) dt =
Pm
2
k k=1 ek k
k=1
Proof. Utilising the hypothesis (10.29) and the modulus properties, we have
*Z
+
m
b
X
(10.32)
f
(t)
dt,
e
k a
k=1
Z b
X
Z b
m
m
X
≥
Re
f (t) dt, ek ≥
Re
f (t) dt, ek
a
a
k=1
k=1
!Z
m
m Z b
b
X
X
=
Re hf (t) , ek i dt ≥
rk
kf (t)k dt.
k=1
a
k=1
a
Rb
P
By Schwarz’s inequality in Hilbert spaces applied for a f (t) dt and m
k=1 ek ,
we have
*Z
+
Z b
m
m
b
X
X
(10.33)
ek ≥ ek .
f (t) dt f (t) dt,
a
a
k=1
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S.S. Dragomir
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k=1
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Making use of (10.32) and (10.33), we deduce (10.30).
Rb
P
Now, if f 6= 0 a.e. on [a, b] , then a kf (t)k dt 6= 0 and by (10.32) m
k=1 ek 6=
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0. Obviously, if (10.31) is valid, then taking the norm we have
Rb
P
Z b
m
X
( m
k=1 rk ) a kf (t)k dt f (t) dt =
ek Pm
2
k k=1 ek k
a
k=1
Pm
Z b
k=1 rk
= Pm
kf (t)k dt,
k k=1 ek k a
i.e., the case of equality holds true in (10.30).
Conversely, if the equality case holds true in (10.30), then it must hold in all
the inequalities used to prove (10.30), therefore we have
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
Re hf (t) , ek i = rk kf (t)k
(10.34)
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] ,
*Z
+
Z b
m
m
b
X
X
e
(10.35)
f
(t)
dt
=
f
(t)
dt,
e
k
k ,
a
a
k=1
k=1
and
*Z
Im
(10.36)
b
f (t) dt,
a
m
X
= 0.
k=1
k=1
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From (10.34) on integrating on [a, b] and summing over k from 1 to m, we get
+
!Z
*Z
m
m
b
b
X
X
(10.37)
Re
f (t) dt,
ek =
rk
kf (t)k dt,
a
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ek
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a
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and then, by (10.36) and (10.37), we have
*Z
+
m
b
X
(10.38)
f (t) dt,
ek =
a
k=1
m
X
!Z
b
kf (t)k dt.
rk
a
k=1
On the other hand, by the use of the identity (3.22), the relation (10.35) holds
true if and only if
DR
E
Pm
b
Z b
m
f
(t)
dt,
e
X
k
k=1
a
P
ek .
(10.39)
f (t) dt =
k m
a
k=1 ek k
k=1
Finally, by (10.38) and (10.39) we deduce that (10.31) is also necessary for the
equality case in (10.30) and the theorem is proved.
Remark 9. If {ek }k∈{1,...,m} are orthogonal, then (10.30) can be replaced by
1
Pm
2 2
b
ke
k
k
k=1
P
kf (t)k dt ≤
m
a
k=1 rk
Z
(10.40)
Z b
,
f
(t)
dt
a
with equality if and only if
Rb
P
Z b
m
X
( m
k=1 rk ) a kf (t)k dt
(10.41)
f (t) dt =
ek .
Pm
2
a
k=1 kek k
k=1
Moreover, if {ek }k∈{1,...,m} are orthonormal, then (10.40) becomes
Z b
√
Z b
m ,
(10.42)
kf (t)k dt ≤ Pm
f
(t)
dt
a
a
k=1 rk
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with equality if and only if
Z
(10.43)
a
b
1
f (t) dt =
m
m
X
! Z
b
rk
X
m
ek .
kf (t)k dt
a
k=1
k=1
The following corollary of Theorem 10.6 may be stated as well [8].
Corollary 10.7. Let (H; h·, ·i) be a Hilbert space over the real or complex number field K and ek ∈ H\ {0} , k ∈ {1, . . . , m} . If f : [a, b] → H is a Bochner
integrable function on [a, b] and ρk > 0, k ∈ {1, . . . , m} with
S.S. Dragomir
kf (t) − ek k ≤ ρk < kek k
(10.44)
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] , then
P
Z b
k m
k=1 ek k
(10.45)
kf (t)k dt ≤ P
1
m
2
2 2
a
k=1 kek k − ρk
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Z b
f (t) dt
.
a
The case of equality holds in (10.45) if and only if
Z
Pm
b
f (t) dt =
(10.46)
a
2
2
k=1 kek k − ρk
P
2
k m
k=1 ek k
12 Z
12
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b
X
m
kf (t)k dt
ek .
a
Proof. Utilising Lemma 3.3, we have from (10.44) that
kf (t)k kek k2 − ρ2k
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≤ Re hf (t) , ek i
k=1
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for any k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
Applying Theorem 10.6 for
1
rk := kek k2 − ρ2k 2 , k ∈ {1, . . . , m} ,
we deduce the desired result.
Remark 10. If {ek }k∈{1,...,m} are orthogonal, then (10.45) becomes
1
Pm
Z b
Z b
2 2
ke
k
k
k=1
(10.47)
kf (t)k dt ≤ P
f (t) dt
1 ,
m
2
2 2
a
a
k=1 kek k − ρk
with equality if and only if
1
Pm
X
Z b
m
2
2 2 Z b
k=1 kek k − ρk
(10.48)
f (t) dt =
kf (t)k dt
ek .
Pm
2
a
a
k=1 kek k
k=1
Moreover, if {ek }k∈{1,...,m} is assumed to be orthonormal and
kf (t) − ek k ≤ ρk
for a.e. t ∈ [a, b] ,
where ρk ∈ [0, 1), k ∈ {1, . . . , m}, then
√
Z b
m
(10.49)
kf (t)k dt ≤ P
1
m
2 2
a
k=1 (1 − ρk )
Z b
,
f
(t)
dt
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S.S. Dragomir
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a
with equality iff
1 Z
Pm
X
Z b
m
b
2 2
(1
−
ρ
)
k
k=1
(10.50)
f (t) dt =
kf (t)k dt
ek .
m
a
a
k=1
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Finally, we may state the following corollary of Theorem 10.6 [11].
Corollary 10.8. Let (H; h·, ·i) be a Hilbert space over the real or complex number field K and ek ∈ H\ {0} , k ∈ {1, . . . , m} . If f : [a, b] → H is a Bochner
integrable function on [a, b] and Mk ≥ µk > 0, k ∈ {1, . . . , m} are such that
either
(10.51)
Re hMk ek − f (t) , f (t) − µk ek i ≥ 0
or, equivalently,
(10.52)
Reverses of the Triangle
Inequality in Banach Spaces
1
M
+
µ
k
k
f (t) −
ek ≤ 2 (Mk − µk ) kek k
2
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] , then
Z b
P
Z b
k m
k=1 ek k
(10.53)
kf (t)k dt ≤ Pm 2·√µ M
f (t) dt
.
k k
ke
k
a
a
k
k=1 µk +Mk
The case of equality holds if and only if
Pm 2·√µk Mk
Z b
X
Z b
m
k=1 µk +Mk kek k
f (t) dt =
kf (t)k dt ·
ek .
P
2
k m
a
a
k=1 ek k
k=1
Proof. Utilising Lemma 3.5, by (10.51) we deduce
√
2 · µk M k
kf (t)k
kek k ≤ Re hf (t) , ek i
µk + M k
S.S. Dragomir
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for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
Applying Theorem 10.6 for
√
2 · µk M k
kek k , k ∈ {1, . . . , m}
rk :=
µk + M k
we deduce the desired result.
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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11.
11.1.
Additive Reverses of the Continuous Triangle
Inequality
The Case of One Functional
The aim of this section is to provide a different approach to the problem of reversing the continuous triangle inequality. Namely, we are interested in finding
upper bounds for the positive difference
Z b
Z b
kf (t)k dt − f (t) dt
a
a
under various assumptions for the Bochner integrable function f : [a, b] → X.
In the following we provide an additive reverse for the continuous triangle
inequality that has been established in [8].
Theorem 11.1 (Dragomir, 2004). Let (X, k·k) be a Banach space over the real
or complex number field K and F : X → K be a continuous linear functional of
unit norm on X. Suppose that the function f : [a, b] → X is Bochner integrable
on [a, b] and there exists a Lebesgue integrable function k : [a, b] → [0, ∞) such
that
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Inequality in Banach Spaces
S.S. Dragomir
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kf (t)k − Re F [f (t)] ≤ k (t)
(11.1)
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for a.e. t ∈ [a, b] . Then we have the inequality
Z b
Z b
Z b
≤
(11.2)
(0 ≤)
kf (t)k dt − f
(t)
dt
k (t) dt.
a
a
a
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The equality holds in (11.2) if and only if both
Z b
Z b
(11.3)
F
f (t) dt = f
(t)
dt
a
a
and
Z
(11.4)
F
b
Z b
Z b
f (t) dt =
kf (t)k dt −
k (t) dt.
a
a
a
Reverses of the Triangle
Inequality in Banach Spaces
Proof. Since the norm of F is unity, then
S.S. Dragomir
|F (x)| ≤ kxk for any x ∈ X.
Applying this inequality for the vector
(11.5)
Rb
a
f (t) dt, we get
Z b
Z b
Z b
f (t) dt
f (t) dt ≥ Re F
f (t) dt ≥ F
a
a
Z b a
Z b
= Re F [f (t)] dt ≥
Re F [f (t)] dt.
a
a
Integrating (11.1), we have
Z b
Z b
Z b
(11.6)
kf (t)k dt − Re F
f (t) dt ≤
k (t) dt.
a
a
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a
Now, making use of (11.5) and (11.6), we deduce (11.2).
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Obviously, if the equality hold in (11.3) and (11.4), then it holds in (11.2)
as well. Conversely, if the equality holds in (11.2), then it must hold in all the
inequalities used to prove (11.2). Therefore, we have
Z b
Z b
Z b
kf (t)k dt = Re F
f (t) dt +
k (t) dt.
a
a
a
and
Z b
Z b
Z b
Re F
f (t) dt
= F
f (t) dt = f (t) dt
a
a
a
S.S. Dragomir
which imply (11.3) and (11.4).
Corollary 11.2. Let (X, k·k) be a Banach space, [·, ·] : X × X → K a semiinner product which generates its norm. If e ∈ X is such that kek = 1, f :
[a, b] → X is Bochner integrable on [a, b] and there exists a Lebesgue integrable
function k : [a, b] → [0, ∞) such that
(11.7)
(0 ≤) kf (t)k − Re [f (t) , e] ≤ k (t) ,
for a.e. t ∈ [a, b] , then
Z b
Z b
Z b
(11.8)
(0 ≤)
kf (t)k dt − f (t) dt
k (t) dt,
≤
a
a
a
where equality holds in (11.8) if and only if both
Z b
Z b
(11.9)
f (t) dt, e = f
(t)
dt
a
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and
b
Z
(11.10)
a
Z b
Z b
−
f (t) dt, e = f
(t)
dt
k (t) dt.
a
a
The proof is obvious by Theorem 11.1 applied for the continuous linear functional of unit norm Fe : X → K, Fe (x) = [x, e] .
The following corollary may be stated.
Corollary 11.3. Let (X, k·k) be a strictly convex Banach space, and [·, ·] , e, f,
k as in Corollary 11.2. Then the case of equality holds in (11.8) if and only if
Z b
Z b
(11.11)
kf (t)k dt ≥
k (t) dt
a
S.S. Dragomir
a
and
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Z
b
Z
b
Z
kf (t)k dt −
f (t) dt =
(11.12)
a
a
b
k (t) dt e.
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a
Proof. Suppose that (11.11) and (11.12) are valid. Taking the norm on (11.12)
we have
Z b
Z b
Z b
Z b
Z b
f (t) dt = kf (t)k dt −
k (t) dt kek =
kf (t)k dt−
k (t) dt,
a
Reverses of the Triangle
Inequality in Banach Spaces
a
a
a
a
and the case of equality holds true in (11.8).
Now, if the equality case holds in (11.8), then obviously (11.11) is valid, and
by Corollary 11.2,
Z b
Z b
kek .
f (t) dt, e = f
(t)
dt
a
a
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Utilising Theorem 4.2, we get
Z b
(11.13)
f (t) dt = λe with λ > 0.
a
Replacing
Rb
a
f (t) dt with λe in the second equation of (11.9) we deduce
Z
(11.14)
b
Z
kf (t)k dt −
λ=
a
b
k (t) dt,
a
and by (11.13) and (11.14) we deduce (11.12).
Remark 11. If X = H, (H; h·, ·i) is a Hilbert space, then from Corollary 11.3
we deduce the additive reverse inequality obtained in [7]. For further similar
results in Hilbert spaces, see [7] and [9].
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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11.2.
The Case of m Functionals
The following result may be stated [8]:
Theorem 11.4 (Dragomir, 2004). Let (X, k·k) be a Banach space over the real
or complex number field K and Fk : X → K, k ∈ {1, . . . , m} continuous linear
functionals on X. If f : [a, b] → X is a Bochner integrable function on [a, b]
and Mk : [a, b] → [0, ∞), k ∈ {1, . . . , m} are Lebesgue integrable functions
such that
(11.15)
kf (t)k − Re Fk [f (t)] ≤ Mk (t)
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for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] , then
Z
Z b
m
m Z b
b
1 X
1 X
(11.16)
kf (t)k dt ≤ Fk f (t) dt
Mk (t) dt.
+ m
a
m
a
a
k=1
k=1
The case of equality holds in (11.16) if and only if both
Z
Z b
m
m
1 X
b
1 X
Fk
Fk f (t) dt
(11.17)
f (t) dt = m
a
m
a
k=1
k=1
and
S.S. Dragomir
m
(11.18)
Reverses of the Triangle
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1 X
Fk
m k=1
Z
b
a
Z b
m Z
1 X b
f (t) dt =
kf (t)k dt −
Mk (t) dt.
m k=1 a
a
Proof. If we integrate on [a, b] and sum over k from 1 to m, we deduce
Z
b
kf (t)k dt
(11.19)
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a
Z b
m
m Z
1 X
1 X b
≤
Re Fk
f (t) dt +
Mk (t) dt.
m k=1
m k=1 a
a
Utilising the continuity property of the functionals Fk and the properties of the
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modulus, we have:
(11.20)
m
X
Z
Re Fk
a
k=1
b
X
Z b
m
f (t) dt ≤ f (t) dt Re Fk
a
k=1 Z
m
X
b
≤
Fk
f (t) dt a
k=1
Z
m
X
b
.
≤
Fk f
(t)
dt
a
k=1
Now, by (11.19) and (11.20) we deduce (11.16).
Obviously, if (11.17) and (11.18) hold true, then the case of equality is valid
in (11.16).
Conversely, if the case of equality holds in (11.16), then it must hold in all
the inequalities used to prove (11.16). Therefore, we have
Z b
Z b
m
m Z
1 X b
1 X
Re Fk
f (t) dt +
Mk (t) dt,
kf (t)k dt =
m k=1
m k=1 a
a
a
Z b
Z b
m
m
X
X
Re Fk
f (t) dt
=
f
(t)
dt
F
k
a
a
k=1
and
k=1
m
X
k=1
Z b
Im Fk
f (t) dt
= 0.
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S.S. Dragomir
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a
These imply that (11.17) and (11.18) hold true, and the theorem is completely
proved.
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Remark 12. If Fk , k ∈ {1, . . . , m} are of unit norm, then, from (11.16) we
deduce the inequality
Z b
Z b
m Z b
1 X
(11.21)
kf (t)k dt ≤ f (t) dt
Mk (t) dt,
+ m
a
a
k=1 a
which is obviously coarser than (11.16) but, perhaps more useful for applications.
The following new result may be stated as well:
Theorem 11.5. Let (X, k·k) be a Banach space over the real or complex number field K and Fk : X → K, k ∈ {1, . . . , m} continuous linear functionals on
X. Assume also that f : [a, b] → X is a Bochner integrable function on [a, b]
and Mk : [a, b] → [0, ∞), k ∈ {1, . . . , m} are Lebesgue integrable functions
such that
(11.22)
kf (t)k − Re Fk [f (t)] ≤ Mk (t)
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
(i) If c∞ is defined by (c∞ ), then we have the inequality
Z b
Z b
m Z b
1 X
(11.23)
kf (t)k dt ≤ c∞ Mk (t) dt.
f (t) dt +
m k=1 a
a
a
(ii) If cp , p ≥ 1, is defined by (cp ) , then we have the inequality
Z b
Z b
m Z b
1 X
cp kf (t)k dt ≤ 1/p f (t) dt
Mk (t) dt.
+ m
m a
a
a
k=1
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S.S. Dragomir
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The proof is similar to the ones from Theorem 7.1 and 11.4 and we omit the
details.
The case of Hilbert spaces for Theorem 11.4, in which one may provide a
simpler condition for equality, is of interest in applications [8].
Theorem 11.6 (Dragomir, 2004). Let (H, h·, ·i) be a Hilbert space over the
real or complex number field K and ek ∈ H, k ∈ {1, . . . , m} . If f : [a, b] → H
is a Bochner integrable function on [a, b] , f (t) 6= 0 for a.e. t ∈ [a, b] and
Mk : [a, b] → [0, ∞), k ∈ {1, . . . , m} is a Lebesgue integrable function such
that
kf (t)k − Re hf (t) , ek i ≤ Mk (t)
(11.24)
for each k ∈ {1, . . . , m} and for a.e. t ∈ [a, b] , then
Z
Z b
m
m Z b
1 X
b
1 X
(11.25)
kf (t)k dt ≤ ek f (t) dt +
Mk (t) dt.
a
m
m
a
a
k=1
k=1
The case of equality holds in (11.25) if and only if
Z b
m Z
1 X b
Mk (t) dt
(11.26)
kf (t)k dt ≥
m k=1 a
a
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and
Z
m
b
f (t) dt =
(11.27)
a
R
b
a
P Rb
m
kf (t)k dt − m1 m
M
(t)
dt
X
k
k=1 a
ek .
P
2
k m
e
k
k
k=1
k=1
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Proof. As in the proof of Theorem 11.4, we have
*
+
Z b
Z b
m
m Z
1 X
1 X b
(11.28)
ek ,
kf (t)k dt ≤ Re
f (t) dt +
Mk (t) dt
m k=1
m k=1 a
a
a
P
and m
k=1 ek 6= 0.
Rb
On
utilising Schwarz’s inequality in Hilbert space (H, h·, ·i) for a f (t) dt
Pm
and k=1 ek , we have
*Z
+
Z b
m
m
b
X
X
f (t) dt (11.29)
ek ≥ f (t) dt,
ek a
a
k=1
k=1
*Z
+
m
b
X
≥ Re
f (t) dt,
ek a
k=1
*Z
+
m
b
X
f (t) dt,
ek .
≥ Re
a
k=1
By (11.28) and (11.29), we deduce (11.25).
Taking the norm on (11.27) and using (11.26), we have
Z b
m R b kf (t)k dt − 1 Pm R b M (t) dt
k
k=1 a
m
a
Pm
f (t) dt
,
=
k
ek k
a
showing that the equality holds in (11.25).
k=1
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Conversely, if the equality case holds in (11.25), then it must hold in all the
inequalities used to prove (11.25). Therefore we have
kf (t)k = Re hf (t) , ek i + Mk (t)
(11.30)
for each k ∈ {1, . . . , m} and for a.e. t ∈ [a, b] ,
*Z
+
Z b
m
m
b
X
X
(11.31)
f (t) dt ek = f (t) dt,
ek a
a
k=1
k=1
Reverses of the Triangle
Inequality in Banach Spaces
and
*Z
Im
(11.32)
b
f (t) dt,
a
m
X
+
ek
S.S. Dragomir
= 0.
k=1
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From (11.30) on integrating on [a, b] and summing over k, we get
*Z
+
Z b
m
m Z b
b
X
X
(11.33) Re
f (t) dt,
ek = m
kf (t)k dt −
Mk (t) dt.
a
k=1
a
k=1
a
On the other hand, by the use of the identity (3.22), the relation (11.31) holds if
and only if
DR
E
Pm
b
Z b
m
f
(t)
dt,
e
k=1 k X
a
f (t) dt =
ek ,
P
2
k m
e
k
a
k
k=1
k=1
giving, from (11.32) and (11.33), that (11.27) holds true.
If the equality holds in (11.25), then obviously (11.26) is valid and the theorem is proved.
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Remark 13. If in the above theorem, the vectors {ek }k∈{1,...,m} are assumed to
be orthogonal, then (11.25) becomes
b
Z
kf (t)k dt
(11.34)
a
1
≤
m
! 21 Z
m Z b
b
1 X
2
kek k
Mk (t) dt.
f (t) dt
+ m
a
a
k=1
k=1
m
X
Moreover, if {ek }k∈{1,...,m} is an orthonormal family, then (11.34) becomes
Z
(11.35)
a
b
Z b
m Z b
1 X
1 f (t) dt
Mk (t) dt
kf (t)k dt ≤ √ + m
m a
a
k=1
which has been obtained in [4].
The following corollaries are of interest.
Corollary 11.7. Let (H; h·, ·i), ek , k ∈ {1, . . . , m} and f be as in Theorem
11.6. If rk : [a, b] → [0, ∞), k ∈ {1, . . . , m} are such that rk ∈ L2 [a, b] ,
k ∈ {1, . . . , m} and
(11.36)
kf (t) − ek k ≤ rk (t) ,
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b], then
Z
Z b
m
m Z
b
1 X
1 X b 2
ek f (t) dt +
r (t) dt.
(11.37)
kf (t)k dt ≤ a
m
2m k=1 a k
a
k=1
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The case of equality holds in (11.37) if and only if
Z
b
a
m
1 X
kf (t)k dt ≥
2m k=1
Z
b
rk2 (t) dt
a
and
Z
m
b
f (t) dt =
a
R
b
a
1
2m
Pm R b
r2
a k
kf (t)k dt −
k=1
Pm
2
k k=1 ek k
m
(t) dt X
ek .
k=1
S.S. Dragomir
Finally, the following corollary may be stated.
Corollary 11.8. Let (H; h·, ·i), ek , k ∈ {1, . . . , m} and f be as in Theorem
11.6. If Mk , µk : [a, b] → R are such that Mk ≥ µk > 0 a.e. on [a, b] ,
(Mk −µk )2
∈ L [a, b] and
Mk +µk
Re hMk (t) ek − f (t) , f (t) − µk (t) ek i ≥ 0
for each k ∈ {1, . . . , m} and for a.e. t ∈ [a, b] , then
m
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Z
Z b
m
1 X
b
kf (t)k dt ≤ ek f (t) dt
m
a
a
k=1
1 X
+
kek k2
4m k=1
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Z
a
b
[Mk (t) − µk (t)]2
dt.
Mk (t) + µk (t)
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12.
Applications for Complex-Valued Functions
We now give some examples of inequalities for complex-valued functions that
are Lebesgue integrable on using the general result obtained in Section 10.
Consider the Banach space (C, |·|1 ) over the real field R and F : C → C,
F (z) = ez with e = α + iβ and |e|2 = α2 + β 2 = 1, then F is linear on C. For
z 6= 0, we have
q
|Re z|2 + |Im z|2
|F (z)|
|e| |z|
=
=
≤ 1.
|z|1
|z|1
|Re z| + |Im z|
Since, for z0 = 1, we have |F (z0 )| = 1 and |z0 |1 = 1, hence
kF k1 := sup
z6=0
|F (z)|
= 1,
|z|1
showing that F is a bounded linear functional on (C, |·|1 ).
Therefore we can apply Theorem 10.1 to state the following result for complexvalued functions.
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Proposition 12.1. Let α, β ∈ R with α2 + β 2 = 1, f : [a, b] → C be a Lebesgue
integrable function on [a, b] and r ≥ 0 such that
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r [|Re f (t)| + |Im f (t)|] ≤ α Re f (t) − β Im f (t)
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(12.1)
for a.e. t ∈ [a, b] . Then
Z b
Z b
(12.2)
r
|Re f (t)| dt +
|Im f (t)| dt
a
a
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Z b
Z b
Im f (t) dt .
≤
Re f (t) dt + a
a
The equality holds in (12.2) if and only if both
Z b
Z b
Z b
Z b
α
Re f (t) dt − β
Im f (t) dt = r
|Re f (t)| dt +
|Im f (t)| dt
a
a
a
a
and
Z
b
Z
Re f (t) dt − β
α
a
a
b
Z b
Z b
Im f (t) dt = Re f (t) dt + Im f (t) dt .
a
a
Now, consider
the Banach space (C, |·|∞ ) . If F (z) = dz with d = γ + iδ
√
2
and |d| = 2 , i.e., γ 2 + δ 2 = 12 , then F is linear on C. For z 6= 0 we have
√
|Re z|2 + |Im z|2
|d| |z|
2
|F (z)|
=
=
·
≤ 1.
|z|∞
|z|∞
2 max {|Re z| , |Im z|}
q
Since, for z0 = 1 + i, we have |F (z0 )| = 1, |z0 |∞ = 1, hence
kF k∞
|F (z)|
= 1,
:= sup
z6=0 |z|∞
showing that F is a bounded linear functional of unit norm on (C, |·|∞ ).
Therefore, we can apply Theorem 10.1, to state the following result for
complex-valued functions.
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Proposition 12.2. Let γ, δ ∈ R with γ 2 + δ 2 = 12 , f : [a, b] → C be a Lebesgue
integrable function on [a, b] and r ≥ 0 such that
r max {|Re f (t)| , |Im f (t)|} ≤ γ Re f (t) − δ Im f (t)
for a.e. t ∈ [a, b] . Then
Z
(12.3)
r
a
b
max {|Re f (t)| , |Im f (t)|} dt
Z b
Z b
≤ max Re f (t) dt , Im f (t) dt .
a
a
The equality holds in (12.3) if and only if both
Z b
Z b
Z b
γ
Re f (t) dt − δ
Im f (t) dt = r
max {|Re f (t)| , |Im f (t)|} dt
a
a
a
a
a
Now, consider the Banach space C, |·|2p with p ≥ 1. Let F : C → C,
1
1
F (z) = cz with |c| = 2 2p − 2 (p ≥ 1) . Obviously, F is linear and by Hölder’s
inequality
q
1
− 12
2p
2
S.S. Dragomir
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a
and
Z b
Z b
Z b
Z b
γ
Re f (t) dt−δ
Im f (t) dt = max Re f (t) dt , Im f (t) dt .
a
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2
2
|Re z| + |Im z|
|F (z)|
=
1 ≤ 1.
|z|2p
|Re z|2p + |Im z|2p 2p
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1
1
Since, for z0 = 1 + i we have |F (z0 )| = 2 p , |z0 |2p = 2 2p (p ≥ 1) , hence
kF k2p := sup
z6=0
|F (z)|
= 1,
|z|2p
showing that F is a bounded linear functional of unit norm on C, |·|2p , (p ≥ 1) .
Therefore on using Theorem 10.1, we may state the following result.
1
1
Proposition 12.3. Let ϕ, φ ∈ R with ϕ2 + φ2 = 2 2p − 2 (p ≥ 1) , f : [a, b] → C
be a Lebesgue integrable function on [a, b] and r ≥ 0 such that
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S.S. Dragomir
1
r |Re f (t)|2p + |Im f (t)|2p 2p ≤ ϕ Re f (t) − φ Im f (t)
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for a.e. t ∈ [a, b] , then
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Z
(12.4) r
b
1
|Re f (t)|2p + |Im f (t)|2p 2p dt
a
"Z
2p Z b
2p # 2p1
b
≤ Re f (t) dt + Im f (t) dt
, (p ≥ 1)
a
a
where equality holds in (12.4) if and only if both
Z b
Z b
Z b
1
ϕ
Re f (t) dt − φ
Im f (t) dt = r
|Re f (t)|2p + |Im f (t)|2p 2p dt
a
a
a
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and
Z
b
Z
Re f (t) dt − φ
ϕ
a
b
Im f (t) dt
a
"Z
2p Z b
2p # 2p1
b
= Re f (t) dt + Im f (t) dt
.
a
a
Remark 14. If p = 1 above, and
r |f (t)| ≤ ϕ Re f (t) − ψ Im f (t) for a.e. t ∈ [a, b] ,
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
2
2
provided ϕ, ψ ∈ R and ϕ + ψ = 1, r ≥ 0, then we have a reverse of the
classical continuous triangle inequality for modulus:
Z b
Z b
r
|f (t)| dt ≤ f (t) dt ,
a
a
with equality iff
Z b
Z b
Z b
ϕ
Re f (t) dt − ψ
Im f (t) dt = r
|f (t)| dt
a
and
Z
a
b
Z
Re f (t) dt − ψ
ϕ
a
a
a
b
Z b
Im f (t) dt = f (t) dt .
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a
If we apply Theorem 11.1, then, in a similar manner we can prove the following result for complex-valued functions.
J. Ineq. Pure and Appl. Math. 6(5) Art. 129, 2005
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Proposition 12.4. Let α, β ∈ R with α2 + β 2 = 1, f, k : [a, b] → C Lebesgue
integrable functions such that
|Re f (t)| + |Im f (t)| ≤ α Re f (t) − β Im f (t) + k (t)
for a.e. t ∈ [a, b] . Then
Z
b
b
Z
|Re f (t)| dt +
|Im f (t)| dt
a
Z b
Z b
Z b
− Re f (t) dt + Im f (t) dt ≤
k (t) dt.
(0 ≤)
a
a
a
a
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
Applying Theorem 11.1, for (C, |·|∞ ) we may state:
Proposition 12.5. Let γ, δ ∈ R with γ 2 + δ 2 = 12 , f, k : [a, b] → C Lebesgue
integrable functions on [a, b] such that
max {|Re f (t)| , |Im f (t)|} ≤ γ Re f (t) − δ Im f (t) + k (t)
for a.e. t ∈ [a, b] . Then
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Z
(0 ≤)
a
b
max {|Re f (t)| , |Im f (t)|} dt
Z b
Z b
Z b
− max Re f (t) dt , Im f (t) dt ≤
k (t) dt.
a
a
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Finally, utilising Theorem 11.1, for C, |·|2p with p ≥ 1, we may state that:
J. Ineq. Pure and Appl. Math. 6(5) Art. 129, 2005
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1
1
Proposition 12.6. Let ϕ, φ ∈ R with ϕ2 + φ2 = 2 2p − 2 (p ≥ 1) , f, k : [a, b] →
C be Lebesgue integrable functions such that
1
|Re f (t)|2p + |Im f (t)|2p 2p ≤ ϕ Re f (t) − φ Im f (t) + k (t)
for a.e. t ∈ [a, b] . Then
Z
(0 ≤)
b
1
|Re f (t)|2p + |Im f (t)|2p 2p dt
a
"Z
2p Z b
2p # 2p1
Z b
b
≤
k (t) dt.
− Re f (t) dt + Im f (t) dt
a
Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
a
a
Remark 15. If p = 1 in the above proposition, then, from
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|f (t)| ≤ ϕ Re f (t) − ψ Im f (t) + k (t) for a.e. t ∈ [a, b] ,
2
2
provided ϕ, ψ ∈ R and ϕ +ψ = 1, we have the additive reverse of the classical
continuous triangle inequality
Z b
Z b
Z b
(0 ≤)
|f (t)| dt − f (t) dt ≤
k (t) dt.
a
a
a
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References
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Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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Reverses of the Triangle
Inequality in Banach Spaces
S.S. Dragomir
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Inequality in Banach Spaces
S.S. Dragomir
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[23] H.S. WILF, Some applications of the inequality of arithmetic and geometric means to polynomial equations, Proceedings Amer. Math. Soc., 14
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