Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 5, Article 129, 2005
REVERSES OF THE TRIANGLE INEQUALITY IN BANACH SPACES
S.S. DRAGOMIR
S CHOOL OF C OMPUTER S CIENCE AND M ATHEMATICS
V ICTORIA U NIVERSITY
PO B OX 14428, MCMC 8001
VIC, AUSTRALIA .
sever@csm.vu.edu.au
URL: http://rgmia.vu.edu.au/dragomir
Received 18 April, 2005; accepted 05 July, 2005
Communicated by B. Mond
A BSTRACT. 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.
Key words and phrases: Reverse triangle inequality, Hilbert spaces, Banach spaces, Bochner integral.
2000 Mathematics Subject Classification. Primary 46B05, 46C05; Secondary 26D15, 26D10.
1. I NTRODUCTION
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.,
(1.1)
Z b
Z b
f (t) dt
kf (t)k dt,
≤
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
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
This paper is based on the talk given by the author within the “International Conference of Mathematical Inequalities and their Applications,
I”, December 06-08, 2004, Victoria University, Melbourne, Australia [http://rgmia.vu.edu.au/conference].
123-05
2
S.S. D RAGOMIR
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
n
Z b
Z b
n
X X
kf (t)k dt ≤ C f (t) dt
kxi k ≤ C xi ,
,
a
a
i=1
i=1
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,
a
a
i=1
i=1
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
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:
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
, i ∈ {1, . . . , n} .
0≤r≤
kxi k
Then
n
n
X X
r
kxi k ≤ xi ,
i=1
i=1
where equality holds if and only if
n
X
xi = r
i=1
n
X
!
kxi k a.
i=1
A generalisation of this result for orthonormal families is incorporated in the following result
[2].
Theorem 1.2 (Diaz-Metcalf, 1966). Let a1 , . . . , an be orthonormal vectors in H. Suppose the
vectors x1 , . . . , xn ∈ H\ {0} satisfy
0 ≤ rk ≤
Re hxi , ak i
,
kxi k
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
i ∈ {1, . . . , n} , k ∈ {1, . . . , m} .
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R EVERSES OF THE T RIANGLE I NEQUALITY
3
Then
m
X
! 12
rk2
n
X
i=1
k=1
n
X kxi k ≤ xi ,
i=1
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
a
provided
−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. 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
Rb
only if f : [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
(1.2)
kf (t)k dt ≤ K f (t) dt
,
a
a
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
f (t) dt =
kf (t)k dt e.
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.
2. D IAZ -M ETCALF T YPE I NEQUALITIES
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 ) ,
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
i ∈ {1, . . . , n} ;
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4
S.S. D RAGOMIR
then
r
(2.2)
n
X
i=1
n
X
kxi k ≤ xi ,
i=1
where equality holds if and only if both
F
(2.3)
n
X
!
xi
=r
i=1
n
X
kxi k
i=1
and
F
(2.4)
n
X
!
xi
i=1
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
(2.5)
0 ≤ r kxi k ≤ Re hxi , ei ,
i = 1, . . . , n,
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
#
2
m
|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
(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]
Pm 2 12 X
n
n
X
r
k=1 k
(2.8)
kxi k ≤ xi ,
c
i=1
i=1
where equality holds if and only if both
!
n
n
X
X
(2.9)
Fk
x i = rk
kxi k ,
i=1
k = 1, . . . , m
i=1
and
(2.10)
m
X
"
Fk
k=1
n
X
!#2
xi
i=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 ,
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
i = 1, . . . , n, k = 1, . . . , m;
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R EVERSES OF THE T RIANGLE I NEQUALITY
5
implying the following reverse of the generalised triangle inequality
! 12 n
m
n
X
X
X
(2.12)
rk2
kxi k ≤ xi ,
i=1
k=1
i=1
where the equality holds if and only if
n
X
(2.13)
xi =
i=1
n
X
!
kxi k
i=1
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.
3. I NEQUALITIES OF D IAZ -M ETCALF T YPE
FOR
m F UNCTIONALS
3.1. 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 ∈ P
X\ {0} , i ∈ {1, . . . , n} are such that there exists the constants rk ≥ 0, k ∈ {1, . . . , m}
with m
k=1 rk > 0 and
Re Fk (xi ) ≥ rk kxi k
(3.1)
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
P
n
n
X
X
F
k
k m
k
k=1
xi .
(3.2)
kxi k ≤ Pm
k=1 rk i=1
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
kxi k
i=1
and
m
X
(3.4)
!
n
X
Fk
i=1
k=1
!
xi
m
n
X
X
Fk =
xi .
k=1
i=1
Proof. Utilising the hypothesis (3.1) and the properties of the modulus, we have
! n
! " m
! n
!#
m
X
X
X
X
(3.5)
I := Fk
xi ≥ Re
Fk
xi 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
k=1
!
rk
i=1
n
X
k=1 i=1
kxi k .
i=1
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6
S.S. D RAGOMIR
On the other hand, by the continuity property of Fk , k ∈ {1, . . . , m} we obviously have
! n
! m
m
n
X
X X
X
(3.6)
I=
Fk
xi ≤ Fk xi .
i=1
k=1
i=1
k=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
Re Fk (xi ) = rk kxi k
(3.7)
for each i ∈ {1, . . . , n}, k ∈ {1, . . . , m} ;
m
X
(3.8)
Im Fk
n
X
!
xi
=0
i=1
k=1
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
" m
! n
!#
! n
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 3.2. 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:
Pm
n
n
X
X
kF
k
k k=1
x
(3.11)
kxi k ≤ P
i .
m
r
k=1 k
i=1
i=1
3.2. 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 3.3 (Dragomir, 2004). Let (H; h·, ·i) be an inner product space over the real or
complex number fieldPK, ek , xi ∈ H\ {0}, k ∈ {1, . . . , m} , i ∈ {1, . . . , n} . If rk ≥ 0,
k ∈ {1, . . . , m} with m
k=1 rk > 0 satisfy
(3.12)
Re hxi , ek i ≥ rk kxi k
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
P
n
n
X
X
k m
e
k
k (3.13)
kxi k ≤ Pk=1
x
i .
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
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R EVERSES OF THE T RIANGLE I NEQUALITY
7
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
xi , ek
≥
Re
=
k=1
m X
n
X
i=1
m
X
Re hxi , ek i ≥
k=1 i=1
!
rk
n
X
kxi k > 0.
i=1
k=1
P
Observe also that, by (3.15), m
k=1 ek 6= 0.
P
P
On utilising Schwarz’s inequality in the inner product space (H; h·, ·i) for ni=1 xi , m
k=1 ek ,
we have
*
+
n
m
n
m
X
X
X
X
(3.16)
xi ek ≥ xi ,
ek .
i=1
i=1
k=1
k=1
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
m
n
X
(Pm r ) Pn kx k X
i i=1
k=1 k
ek xi =
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)
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
i=1
k=1
k=1
and
Im
(3.19)
* n
X
xi ,
i=1
m
X
k=1
By (3.19) and (3.20), we have
* n
+
m
X X
(3.21)
xi ,
ek =
i=1
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
k=1
ek
= 0.
k=1
From (3.17), on summing over i and k, we get
* n
+
m
X X
(3.20)
Re
xi ,
ek =
i=1
+
m
X
!
rk
k=1
!
rk
kxi k .
i=1
k=1
m
X
n
X
n
X
kxi k .
i=1
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8
S.S. D RAGOMIR
On the other hand, by the use of the following identity in inner product spaces
2
2
2
2
hu, vi v = kuk kvk − |hu, vi| , v 6= 0,
u
−
(3.22)
kvk2 kvk2
the relation (3.18) holds if and only if
P
P
n
m
X
X
h ni=1 xi , m
k=1 ek i
(3.23)
xi =
ek .
Pm
2
k
e
k
k
k=1
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 3.4 (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)
Then we have the inequality
kxk kak2 − r2
(3.25)
21
≤ Re hx, ai
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 .
1
By the second part of (3.24) we have kak2 − r2 2 > 0, therefore, by (3.28), we may state that
(3.28)
(3.29)
0<
kxk2
kak2 − r2
2
2
12 + kak − r
12
≤
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
with equality if and only if α = pq , we may state (for α = kak2 − r2 2 , p = 1, q = kxk2 )
that
1
kxk2
2
2 2
.
(3.30)
2 kxk ≤
1 + kak − r
2
2
2
kak − r
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 only if kx − ak = r and in
1
(3.30) if and only if kak2 − r2 2 = kxk .
The proof is thus completed.
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R EVERSES OF THE T RIANGLE I NEQUALITY
9
We may now state the following corollary [5].
Corollary 3.5. 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
kxi − ek k ≤ ρk < kek k
(3.31)
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
P
n
X
k m
k=1 ek k
(3.32)
kxi k ≤ P
1
m
2
2 2
ke
k
−
ρ
i=1
k
k
k=1
The case of equality holds in (3.32) if and only if
1
Pm
n
2
2 2
X
ke
k
−
ρ
k
k
xi = k=1Pm
2
k k=1 ek k
i=1
n
X
n
X
xi .
i=1
!
kxi k
i=1
m
X
ek .
k=1
Proof. Utilising Lemma 3.4, we have from (3.31) that
1
kxi k kek k2 − ρ2k 2 ≤ Re hxi , ek i
for each k ∈ {1, . . . , m} and i ∈ {1, . . . , n} .
Applying Theorem 3.3 for
1
rk := kek k2 − ρ2k 2 ,
k ∈ {1, . . . , m} ,
we deduce the desired result.
Remark 3.6. If {ek }k∈{1,...,m} are orthogonal, then (3.32) becomes
1
Pm
n
n
2 2
X
X
ke
k
k
k=1
(3.33)
kxi k ≤ P
x
i
1 m
2
ke k − ρ2 2
i=1
k
k=1
i=1
k
with equality if and only if
n
X
Pm
k=1
xi =
kek k2 − ρ2k
Pm
k=1
i=1
12
kek k2
n
X
!
kxi k
i=1
m
X
ek .
k=1
Moreover, if {ek }k∈{1,...,m} is assumed to be orthonormal and
kxi − ek k ≤ ρk for k ∈ {1, . . . , m} , i ∈ {1, . . . , n}
where ρk ∈ [0, 1) for k ∈ {1, . . . , m} , then
n
X
(3.34)
i=1
√
n
X xi kxi k ≤ P
1 m
2 2 i=1
k=1 (1 − ρk )
m
with equality if and only if
n
X
Pm
xi =
i=1
1
2 2
k=1 (1 − ρk )
m
n
X
i=1
!
kxi k
m
X
ek .
k=1
The following lemma may be stated as well [3].
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10
S.S. D RAGOMIR
Lemma 3.7 (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
Re hM y − x, x − myi ≥ 0
(3.35)
or, equivalently,
1
m
+
M
≤ (M − m) kyk ,
x −
y
2
2
(3.36)
then
kxk kyk ≤
(3.37)
1 M +m
· √
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 .
Proof. Obviously,
Re hM y − x, x − myi = (M + m) Re hx, yi − kxk2 − mM kyk2 .
Then (3.35) is clearly equivalent to
(3.39)
√
kxk2
M +m
√
+ mM kyk2 ≤ √
Re hx, yi .
mM
mM
Since, obviously,
(3.40)
√
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.
Finally, we may state the following corollary of Theorem 3.3, see [5].
Corollary 3.8. 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,
1
M
+
µ
k
k
xi −
≤ (Mk − µk ) kek k
e
k
2
2
for each k ∈ {1, . . . , m} and i ∈ {1, . . . , n} , then
n
P
n
X X
k m
e
k
k=1 k
(3.42)
kxi k ≤ Pm 2·√
xi .
µk M k
i=1
k=1 µk +Mk kek k i=1
The case of equality holds in (3.42) if and only if
Pm 2·√µk Mk
n
n
m
X
X
X
k=1 µk +Mk kek k
xi =
kxi k
ek .
P
2
k m
k=1 ek k
i=1
i=1
k=1
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R EVERSES OF THE T RIANGLE I NEQUALITY
11
Proof. Utilising Lemma 3.7, 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 3.3 for
√
2 · µk M k
rk :=
kek k ,
µk + M k
k ∈ {1, . . . , m} ,
we deduce the desired result.
4. D IAZ -M ETCALF I NEQUALITY FOR S EMI -I NNER P RODUCTS
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;
(ii) [λx, y] = λ [x, y] for all x, y ∈ X and λ ∈ K;
(iii) [x, x] ≥ 0 for all x ∈ X and [x, x] = 0 implies x = 0;
(iv) |[x, y]|2 ≤ [x, x] [y, y] for all x, y ∈ X;
(v) [x, λy] = λ̄ [x, y] for all x, y ∈ X and λ ∈ K.
1
It is well known that the mapping X 3 x 7−→ [x, x] 2 ∈ R is a norm on X and for any y ∈ X,
ϕy
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,
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.
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
(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
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
n
X
kxi k ≤ xi .
i=1
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12
S.S. D RAGOMIR
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
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.
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) .
The following characterisation of strictly convex spaces is useful in what follows (see [1],
[13], or [3, p. 21]).
Theorem 4.2. Let (X, k·k) be a normed linear space over K and [·, ·] a semi-inner 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.
The following result may be stated.
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
(4.6)
n
X
xi = µe.
i=1
Inserting this in (4.3) we get
2
µ kek = r
n
X
kxi k
i=1
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R EVERSES OF THE T RIANGLE I NEQUALITY
13
giving
µ=r
(4.7)
n
X
kxi k .
i=1
Finally, by (4.6) and (4.7) we deduce (4.5) and the corollary is proved.
5. OTHER M ULTIPLICATIVE R EVERSES FOR m F UNCTIONALS
Assume that Fk , k ∈ {1, . . . , m} are bounded linear functionals defined on the normed linear
space X.
For p ∈ [1, ∞), define
Pm
p p1
|F
(x)|
k
k=1
(cp )
cp := sup
kxkp
x6=0
and for p = ∞,
c∞ := sup
(c∞ )
max
1≤k≤m
x6=0
|Fk (x)|
kxk
.
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
! p1
m
X
cp ≤
kFk kp
, p≥1
k=1
and
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.
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
i=1
for each k ∈ {1, . . . , m}
i=1
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
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
for any x ∈ X,
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14
S.S. D RAGOMIR
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
1≤k≤m
i=1
i=1
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}
and, from (5.4),
"
n
X
c∞ xi = max Re Fk
1≤k≤m
i=1
n
X
!#
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.
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
Pn
Pm
1
kxi k
kFk kp p
cp
i=1
k=1
(5.6)
(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
i=1
for each k ∈ {1, . . . , m}
i=1
and
(5.8)
m
X
"
Re Fk
n
X
!#p
xi
i=1
k=1
n
X p
= cpp xi .
i=1
Proof. By the definition of cp , p ≥ 1, we have
cpp
p
kxk ≥
m
X
|Fk (x)|p
for any x ∈ X,
k=1
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R EVERSES OF THE T RIANGLE I NEQUALITY
15
implying that
(5.9)
p
!p
!p
n
m n
m n
X
X
X
X
X
p
cp xi ≥
xi ≥
xi Fk
Re Fk
i=1
i=1
i=1
k=1
k=1
"
!#p
" n
#p
m
n
m
X
X
X
X
≥
Re Fk
xi
=
Re Fk (xi ) .
i=1
k=1
i=1
k=1
Utilising the hypothesis (3.1), we obviously have that
" n
#p
" n
#p
m
m
m
X
X
X
X
X
(5.10)
Re Fk (xi ) ≥
rk kxi k =
rkp
k=1
i=1
k=1
i=1
n
X
!p
kxi k
.
i=1
k=1
Making use of (5.9) and (5.10), we deduce
p
! n
!p
n
m
X
X
X
cpp xi ≥
rkp
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
n
m
n
m
X
X
X
X
p
rk
Re Fk
xi
=
kxi k ,
i=1
k=1
i=1
k=1
which, together with (5.8) shows that the equality case holds true in (5.6).
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
and, from (5.9),
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m}
p
"
n
m
X
X
cpp xi =
Re Fk
i=1
n
X
!#p
xi
,
i=1
k=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.
6. A N A DDITIVE R EVERSE FOR
THE
T RIANGLE I NEQUALITY
6.1. The Case of One Functional. In the following we provide an alternative of the DiazMetcalf 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)
(0 ≤) kxi k − Re F (xi ) ≤ ki for each i ∈ {1, . . . , n} ,
then we have the inequality
(0 ≤)
(6.2)
n
X
i=1
n
n
X
X
kxi k − xi ≤
ki .
The equality holds in (6.2) if and only if both
! n
n
X X
(6.3)
F
xi = xi and F
i=1
i=1
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
i=1
n
X
i=1
i=1
!
xi
=
n
X
i=1
kxi k −
n
X
ki .
i=1
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16
S.S. D RAGOMIR
Proof. If we sum in (6.1) over i from 1 to n, then we get
"
!#
n
n
n
X
X
X
kxi k ≤ Re F
xi
+
ki .
(6.4)
i=1
i=1
i=1
Taking into account that |F (x)| ≤ kxk for each x ∈ X, then we may state that
"
!# !
n
n
X
X
Re F
(6.5)
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
= F
!#
xi
i=1
i=1
! n
X xi = xi ,
n
X
i=1
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
(0 ≤) kxi k − Re [xi , e] ≤ ki for each i ∈ {1, . . . , n} ,
(6.6)
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
xi and
xi , e =
kxi k −
ki .
(6.8)
xi , e = 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
n
X
X
(6.9)
kxi k ≥
ki
i=1
i=1
and
(6.10)
n
X
xi =
i=1
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.
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
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R EVERSES OF THE T RIANGLE I NEQUALITY
17
Remark 6.3. 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].
6.2. The Case of m Functionals. The following result generalising Theorem 6.1 may be stated
[6].
Theorem 6.4 (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
m
n
m X
n
n
X
1 X
1 X
X
Fk xi +
Mik .
(6.12)
kxi k ≤ m
m
i=1
i=1
k=1 i=1
k=1
The case of equality holds in (6.12) if both
m
(6.13)
n
X
1 X
Fk
m k=1
!
xi
i=1
n
m
1 X
X =
Fk xi m
i=1
k=1
and
m
(6.14)
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
for each k ∈ {1, . . . , m} .
Summing these inequalities over k from 1 to m, we deduce
!
n
m
n
m
n
X
X
1 X
1 XX
(6.15)
kxi k ≤
Re Fk
xi +
Mik .
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
(6.16)
Re Fk
xi ≤ Re Fk
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).
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18
S.S. D RAGOMIR
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
!
m
n
m
n
n
X
X
1 X
1 XX
Re Fk
xi +
Mik ,
kxi k =
m k=1
m k=1 i=1
i=1
i=1
m
X
n
X
Re Fk
!
xi
i=1
k=1
m
n
X X
=
Fk xi i=1
k=1
and
m
X
Im Fk
n
X
!
xi
= 0.
i=1
k=1
These imply that (6.13) and (6.14) hold true, and the theorem is completely proved.
Remark 6.5. If Fk , k ∈ {1, . . . , m} are of unit norm, then, from (6.12), we deduce the
inequality
n
n
m X
n
X
1 X
X
(6.17)
kxi k ≤ xi +
Mik ,
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 6.6 (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
kxi k − Re hxi , ek i ≤ Mik
(6.18)
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
(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)
n
X
xi =
m
m
Pm Pn
1
X
M
kx
k
−
i
ik
i=1
k=1
i=1
ek .
Pmm
2
k k=1 ek k
k=1
Pn
i=1
Proof. As in the proof of Theorem 6.4, we have
*
+
n
m
n
m
n
X
1 X X
1 XX
(6.22)
kxi k ≤ Re
ek ,
xi +
Mik ,
m
m
i=1
i=1
k=1
k=1 i=1
Pm
and k=1 ek 6= 0.
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R EVERSES OF THE T RIANGLE I NEQUALITY
19
On utilising the Schwarz inequality in the inner product space (H; h·, ·i) for
we have
*
+
n
m
n
m
X
X
X
X
(6.23)
xi ek ≥ xi ,
ek i=1
k=1
k=1
i=1
* n
+
m
X X
≥ Re
xi ,
ek i=1
k=1
* n
+
m
X X
≥ Re
xi ,
ek .
i=1
Pn
i=1
xi ,
Pm
k=1 ek ,
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 ik
i
i=1
Pmm k=1 i=1
,
xi =
k
e
k
k
k=1
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)
for each i ∈ {1, . . . , n} , k ∈ {1, . . . , m} ,
*
+
n
m
n
m
X
X
X
X
(6.25)
xi ek = xi ,
ek i=1
i=1
k=1
k=1
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
+
m
n
m X
n
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
m
n
X
X
h ni=1 xi , m
k=1 ek i
ek ,
xi =
Pm
2
k
e
k
k
k=1
i=1
k=1
giving, from (6.26) and (6.27), that
P
P Pn
m
n
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
If the inequality holds in (6.19), then obviously (6.20) is valid, and the theorem is proved.
Remark 6.7. 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
X 1 X
X
1 X
2
(6.28)
kxi k ≤
kek k
xi +
Mik .
m
m k=1
i=1
i=1
k=1 i=1
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S.S. D RAGOMIR
Moreover, if {ek }k=1,m is an orthonormal family, then (6.28) becomes
√ n
m
n
n
X
m
X 1 X X
kxi k ≤
xi +
Mik ,
(6.29)
m i=1 m k=1 i=1
i=1
which has been obtained in [12].
Before we provide some natural consequences of Theorem 6.6, we need some preliminary
results concerning another reverse of Schwarz’s inequality in inner product spaces (see for instance [4, p. 27]).
Lemma 6.8 (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
kxk kak − Re hx, ai ≤ r2 .
2
The case of equality holds in (6.30) if and only if
(6.30)
kx − ak = r and kxk = kak .
(6.31)
Proof. The condition kx − ak ≤ r is clearly equivalent to
kxk2 + kak2 ≤ 2 Re hx, ai + r2 .
(6.32)
Since
2 kxk kak ≤ kxk2 + kak2 ,
(6.33)
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.
Utilising the above lemma we may state the following corollary of Theorem 6.6 [6].
Corollary 6.9. Let (H; h·, ·i) , ek , xi be as in Theorem 6.6. 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
(6.35)
n
X
i=1
n
m
m
n
1 X
X 1 XX 2
ek xi +
rik .
kxi k ≤ 2m
m
i=1
k=1
k=1 i=1
The equality holds in (6.35) if and only if
n
X
i=1
m
kxi k ≥
n
1 XX 2
r
2m k=1 i=1 ik
and
n
X
xi =
m
Pm Pn 2 m
1
X
kx
k
−
i
i=1
k=1
i=1 rik
ek .
Pm2m 2
k k=1 ek k
k=1
Pn
i=1
The following lemma may provide another sufficient condition for (6.18) to hold (see also [4,
p. 28]).
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R EVERSES OF THE T RIANGLE I NEQUALITY
21
Lemma 6.10 (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
Re hM y − x, x − myi ≥ 0
(6.36)
or, equivalently,
(6.37)
x − m + M y ≤ 1 (M − m) kyk ,
2
2
holds, then
1 (M − m)2
·
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
(6.38)
kxk kyk − Re hx, yi ≤
kxk =
M +m
kyk .
2
The proof is obvious by Lemma 6.8 for a = M +m
y and r = 12 (M − m) kyk .
2
Finally, the following corollary of Theorem 6.6 may be stated [6].
Corollary 6.11. Assume that (H, h·, ·i) , ek , xi are as in Theorem 6.6. If Mik ≥ mik > 0 satisfy
the condition
Re hMk ek − xi , xi − µk ek i ≥ 0
for each i ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
m
n
m
n
1 X
X
X
1 X X (Mik − mik )2
kxi k ≤ ek xi +
kek k2 .
m
4m
M
+
m
ik
ik
i=1
i=1
k=1
k=1 i=1
7. OTHER A DDITIVE R EVERSES FOR m F UNCTIONALS
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)
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∞ Mik .
xi +
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
X
1 X
X
1
xi +
Mik .
(7.3)
kxi k ≤ 1 cp m p i=1 m k=1 i=1
i=1
Proof. (i) Since
max kFk (x)k ≤ c∞ kxk
1≤k≤m
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
for any x ∈ X,
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22
S.S. D RAGOMIR
then we have
m X
Fk
n
X
i=1
k=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
m
1 X
Re Fk
m k=1
n
X
!
xi
i=1
n
X
xi ,
≤ c∞ i=1
which, together with (6.15) imply the desired inequality (7.2).
(ii) Using the fact that, obviously
! p1
m
X
|Fk (x)|p
≤ cp kxk for any x ∈ X,
k=1
then, by Hölder’s inequality for p > 1, p1 +
m X
Fk
n
X
k=1
i=1
1
q
= 1, we have
!p ! p1
m n
X
X
xi Fk
k=1
i=1
n
1 X
≤ cp m q 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.
Remark 7.2. Since, obviously c∞ ≤ max kFk k , then from (7.2) we have
1≤k≤m
(7.4)
n
X
i=1
n
m X
n
X
1 X
kxi k ≤ max {kFk k} · xi +
Mik .
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
n
n
m X
n
p p1 X
1 X
X
kF
k
k
k=1
(7.5)
kxi k ≤
xi +
Mik .
m
m
i=1
i=1
k=1 i=1
The following corollary for semi-inner products may be stated as well.
Corollary 7.3. 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} .
(i) If
d∞ := sup
x6=0
max1≤k≤n |[x, ek ]|
kxk
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
≤ max kek k ,
1≤k≤n
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R EVERSES OF THE T RIANGLE I NEQUALITY
23
then
n
X
(7.7)
i=1
n
n
m X
X
1 X
kxi k ≤ d∞ xi +
Mik
m
i=1
i=1
n k=1
!
m X
n
X 1 X
≤ max kek k · xi +
Mik ;
1≤k≤n
m
i=1
k=1 i=1
(ii) If
Pm
dp := sup
p p1
|[x, ek ]|
kxkp
k=1
x6=0

≤
m
X
! p1 
kek kp
,
k=1
where p ≥ 1, then
(7.8)
n
X
i=1
n
m X
n
X 1 X
kxi k ≤ 1 dp xi +
Mik
m p i=1 m k=1 i=1
n
!
Pm
m X
n
p p1 X
1 X
ke
k
k
k=1
≤
Mik .
xi +
m
m
1
i=1
k=1 i=1
8. A PPLICATIONS FOR C OMPLEX N UMBERS
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
kF k1 := sup
z6=0
|F (z)|
= |a| ,
|z|1
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 ∈
Pm{1, . . . , m} and j ∈ {1, . . . , n} . If there exist the
constants rk ≥ 0, k ∈ {1, . . . , m} with k=1 rk > 0 and
(8.1)
rk [|Re xj | + |Im xj |] ≤ Re ak · Re xj − Im ak · Im xj
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24
S.S. D RAGOMIR
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
n
#
" n
P
X
X
| m
a
|
k
Re
x
+
Im
x
[|Re xj | + |Im xj |] ≤ Pk=1
.
j
j
m
k=1 rk
j=1
j=1
j=1
n
X
(8.2)
The case of equality holds in (8.2) if both
Re
m
X
!
ak
Re
n
X
!
− Im
xj
j=1
k=1
=
m
X
m
X
!
ak
Im
rk
n
X
!
xj
j=1
k=1
!
n
X
[|Re xj | + |Im xj |]
j=1
k=1
"
#
m
n
n
X
X
X
=
ak Re xj + Im xj .
j=1
k=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
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 √
|d| |z|
|F (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 ∈
Pm{1, . . . , m} and j ∈ {1, . . . , n} . If there exist the
constants rk ≥ 0, k ∈ {1, . . . , m} with 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
max {|Re xj | , |Im xj |} ≤
j=1
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
√
)
( n
Pm
n
X
X
| k=1 ak |
P
2·
max Re xj , Im xj .
m
k=1 rk
j=1
j=1
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R EVERSES OF THE T RIANGLE I NEQUALITY
25
The case of equality holds in (8.3) if both
!
!
!
!
m
n
m
n
X
X
X
X
Re
ak Re
xj − Im
ak Im
xj
j=1
k=1
m
X
=
j=1
k=1
!
rk
n
X
max {|Re xj | , |Im xj |}
j=1
k=1
m
n
)
( n
X
X
√ X
= 2
ak max Re xj , Im xj .
j=1
j=1
k=1
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
=
≤
2
|c| .
1
|z|2p
|Re z|2p + |Im z|2p 2p
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
1
1
showing that F is a bounded linear functional on C, |·|2p , p ≥ 1 and kF k2p = 2 2 − 2p |c| .
If we apply Theorem 3.1, then we can state the following proposition [5].
kF k2p := sup
Proposition 8.3. Let ak , xj ∈ C, k ∈
P {1, . . . , m} and j ∈ {1, . . . , n} . If there exist the
constants rk ≥ 0, k ∈ {1, . . . , m} with m
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
j=1

2p 2p  2p1
Pm
n
n
1 |
1
ak | X
X

Re
x
+
Im
x
.
≤ 2 2 − 2p Pk=1
j
j
m
r
k=1 k
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
1
1
= 2 2 − 2p
j=1

2p 2p  2p1
m
n
n
X
X
X

ak Re xj + Im xj  .
k=1
j=1
j=1
Remark 8.4. If in the above proposition we choose p = 1, then we have the following reverse
of the generalised triangle inequality for complex numbers
Pm
n
n
X
X
| k=1 ak | P
|xj | ≤
xj m
k=1 rk
j=1
j=1
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26
S.S. D RAGOMIR
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 6.4 to state the following reverse of the generalised triangle inequality
for complex numbers [6].
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
|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
[|Re xj | + |Im xj |]
j=1
"
#
m
n
n
m
n
X
X
X
1 XX
1 ak Re xj + Im xj +
Mjk .
≤
m k=1 j=1
m k=1 j=1
j=1
The proof follows by Theorem 6.4 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 6.4 for the Banach space (C, |·|∞ ), then we can state the following
reverse of the generalised triangle inequality for complex numbers [6].
Proposition 8.6. 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
max {|Re xj | , |Im xj |} ≤ Re ak · Re xj − Im ak · Im xj + Mjk
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
(8.7)
n
X
j=1
max {|Re xj | , |Im xj |}
)
( n
√ m
n
m
n
X
X
2 X 1 XX
≤
ak max Re xj , Im xj +
Mjk .
m k=1 m
j=1
j=1
k=1 j=1
Finally, if we apply Theorem 6.4, for the Banach space C, |·|2p with p ≥ 1, then we can
state the following proposition [6].
Proposition 8.7. Let ak , xj , Mjk be as in Proposition 8.6. If
1
|Re xj |2p + |Im xj |2p 2p ≤ Re ak · Re xj − Im ak · Im xj + Mjk
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R EVERSES OF THE T RIANGLE I NEQUALITY
27
for each j ∈ {1, . . . , n} and k ∈ {1, . . . , m} , then
(8.8)
n
X
1
|Re xj |2p + |Im xj |2p 2p
j=1
≤
2
1
1
− 2p
2
m
m
 n
2p n
2p  2p1
m
n
X X
X
1 XX


ak Re xj + Im xj +
Mjk .
m k=1 j=1
j=1
j=1
k=1
where p ≥ 1.
Remark 8.8. If in the above proposition we choose p = 1, then we have the following reverse
of the generalised triangle inequality for complex numbers
n
m
n
m X
n
1 X
X
1 X
X
|xj | ≤ ak xj +
Mjk
m
m
j=1
j=1
k=1
k=1 j=1
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.
9. K ARAMATA T YPE I NEQUALITIES IN H ILBERT S PACES
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
−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 : [a, b] → H is strongly
Rb
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
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.
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28
S.S. D RAGOMIR
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)
then
p
(9.7)
Z
ρ2
1−
a
with equality if and only if
Z
b
Z b
kf (t)k dt ≤ f (t) dt
a
b
f (t) dt =
a
p
b
Z
1 − ρ2
kf (t)k dt · e.
a
Also, for e as above and if M ≥ m > 0, f ∈ L ([a, b] ; H) such that either
Re hM e − f (t) , f (t) − mei ≥ 0
(9.8)
or, equivalently,
1
M
+
m
f (t) −
e
≤ 2 (M − m)
2
(9.9)
for a.e. t ∈ [a, b] , then
Z
(9.10)
a
with equality if and only if
Z
b
Z b
M +m ,
kf (t)k dt ≤ √
f
(t)
dt
2 mM a
√
Z b
2 mM
f (t) dt =
kf (t)k dt · e.
M +m
a
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.
b
10. M ULTIPLICATIVE R EVERSES OF THE C ONTINUOUS T RIANGLE I NEQUALITY
10.1. 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)
r kf (t)k ≤ Re F [f (t)] for a.e. t ∈ [a, b] .
Then
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
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R EVERSES OF THE T RIANGLE I NEQUALITY
29
and
b
Z
(10.4)
F
a
Z b
.
f (t) dt = f
(t)
dt
a
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
(10.5)
f (t) dt
f (t) dt ≥ F
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] ,
Z
Im F
(10.8)
b
f (t) dt = 0
a
and
(10.9)
Z b
Z b
f (t) dt .
f (t) dt
= Re F
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 semi-inner 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
Z
(10.12)
b
r
a
Z b
f (t) dt
kf (t)k dt ≤ a
where equality holds in (10.12) if and only if both
Z b
Z b
f (t) dt, e = r
kf (t)k dt
(10.13)
a
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a
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30
S.S. D RAGOMIR
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
kf (t)k dt,
kf (t)k dt kek = r
f (t) dt = r
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
Replacing this in (10.13), we get
Z
2
b
µ kek = r
kf (t)k dt,
a
giving
Z
kf (t)k dt.
µ=r
(10.17)
b
a
Utilising (10.16) and (10.17) we deduce (10.15) and the proof is completed.
10.2. The Case of m Functionals. The following result may be stated [8]:
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,
Pb] → X is a Bochner integrable function on [a, b] and there exists rk ≥ 0, k ∈ {1, . . . , m}
with m
k=1 rk > 0 and
rk kf (t)k ≤ Re Fk [f (t)]
(10.18)
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
k
a
a
k=1
The case of equality holds in (10.19) if both
! Z
m
b
X
(10.20)
Fk
f (t) dt =
k=1
a
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
m
X
k=1
!Z
b
kf (t)k dt
rk
a
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R EVERSES OF THE T RIANGLE I NEQUALITY
31
and
m
X
(10.21)
! Z
b
Fk
a
k=1
Z
m
X
b
f (t) dt = f (t) dt
Fk .
a
k=1
Proof. Utilising the hypothesis (10.18), we have
"X
Z b
#
Z b
m
m
X
(10.22)
f (t) dt ≥ Re
Fk
f (t) dt I := Fk
a
a
k=1
k=1
#
" m
m Z b
X Z b
X
Fk
f (t) dt
=
Re Fk f (t) dt
≥ Re
a
k=1
≥
k=1
a
! Z
m
b
X
rk ·
kf (t)k dt.
a
k=1
On the other hand, by the continuity property of Fk , k ∈ {1, . . . , m} , we obviously have
Z
m
! Z
m
X
b
b
X f (t) dt ≤ Fk f (t) dt
(10.23)
I=
Fk
.
a
a
k=1
k=1
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
f (t) dt = 0,
(10.25)
Im
Fk
k=1
(10.26)
Re
m
X
! Z
Fk
k=1
a
b
a
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
f (t) dt =
Fk
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.
The following new results may be stated as well:
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] P
→ X is a Bochner integrable function on [a, b] and there exists rk ≥ 0, k ∈ {1, . . . , m}
with m
k=1 rk > 0 and
rk kf (t)k ≤ Re Fk [f (t)]
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
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32
S.S. D RAGOMIR
(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
b
Z
Re (Fk )
Z b
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
a
k=1
≤ P
(1 ≤) ≤ Pm p
R b
p p1
k=1 rk
( m
r
)
a f (t) dt
k=1 k
with equality if and only if
b
Z
Re (Fk )
a
Z b
f (t) dt = rk
kf (t)k dt
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
a
k=1
a
where p ≥ 1.
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,
P. . . , m} . If f : [a, b] → H is a Bochner integrable
function and rk ≥ 0, k ∈ {1, . . . , m} and m
k=1 rk > 0 satisfy
(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
The case of equality holds in (10.30) for f 6= 0 a.e. on [a, b] if and only if
Rb
P
Z b
m
( m
r
)
kf (t)k dt X
k
k=1
a
(10.31)
f (t) dt =
ek .
P
2
k m
a
k=1 ek k
k=1
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R EVERSES OF THE T RIANGLE I NEQUALITY
33
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 Z b
m
b
X
X
=
Re hf (t) , ek i dt ≥
rk
kf (t)k dt.
a
k=1
a
k=1
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)
f (t) dt ek ≥ f (t) dt,
ek .
a
a
k=1
k=1
Making use of (10.32) and (10.33), we
R b deduce (10.30).
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= 0. Obviously,
if (10.31) is valid, then taking the norm we have
Rb
P
Z b
m
X
( m
r
)
kf
(t)k
dt
k
k=1
a
=
e
f
(t)
dt
P
k
2
m
k k=1 ek k
a
k=1
Pm
Z b
k=1 rk
P
kf (t)k dt,
=
k m
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
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
f
(t)
dt
e
=
f
(t)
dt,
e
(10.35)
,
k
k
a
a
k=1
k=1
and
*Z
Im
(10.36)
b
f (t) dt,
a
m
X
+
ek
= 0.
k=1
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
k=1
and then, by (10.36) and (10.37), we have
+
*Z
m
b
X
(10.38)
f (t) dt,
ek =
a
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
k=1
a
k=1
m
X
k=1
!Z
b
kf (t)k dt.
rk
a
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34
S.S. D RAGOMIR
On the other hand, by the use of the identity (3.22), the relation (10.35) holds true if and only
if
DR
b
Z
f (t) dt =
(10.39)
b
a
a
E
Pm
m
f (t) dt, k=1 ek X
P
ek .
k m
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 10.7. If {ek }k∈{1,...,m} are orthogonal, then (10.30) can be replaced by
1
Pm
Z b
2 2 Z b
ke
k
k
k=1
Pm
(10.40)
kf (t)k dt ≤
f (t) dt
,
a
a
k=1 rk
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
ke
k
a
k
k=1
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
r
k
a
a
k=1
with equality if and only if
Z b
1
(10.43)
f (t) dt =
m
a
m
X
! Z
rk
b
X
m
kf (t)k dt
ek .
a
k=1
k=1
The following corollary of Theorem 10.6 may be stated as well [8].
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
ρk > 0, k ∈ {1, . . . , m} with
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
Z b
.
f
(t)
dt
a
The case of equality holds in (10.45) if and only if
1
Pm
X
Z b
m
2
2 2 Z b
ke
k
−
ρ
k
k
k=1
(10.46)
f (t) dt =
kf
(t)k
dt
ek .
P
2
k m
e
k
a
a
k
k=1
k=1
Proof. Utilising Lemma 3.4, we have from (10.44) that
1
kf (t)k kek k2 − ρ2k 2 ≤ Re hf (t) , ek i
for any k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
Applying Theorem 10.6 for
rk := kek k2 − ρ2k
we deduce the desired result.
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
12
, k ∈ {1, . . . , m} ,
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R EVERSES OF THE T RIANGLE I NEQUALITY
35
Remark 10.9. 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
12 m
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
ke
k
−
ρ
k
k
k=1
(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
where ρk ∈ [0, 1), k ∈ {1, . . . , m}, then
Z b
(10.49)
kf (t)k dt ≤ P
m
k=1
a
for a.e. t ∈ [a, b] ,
√
m
1
(1 − ρ2k ) 2
Z b
,
f
(t)
dt
a
with equality iff
Z
Pm
b
f (t) dt =
(10.50)
a
1
2 2
k=1 (1 − ρk )
m
Z
b
X
m
kf (t)k dt
ek .
a
k=1
Finally, we may state the following corollary of Theorem 10.6 [11].
Corollary 10.10. 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)
f (t) − Mk + µk ek ≤ 1 (Mk − µk ) kek k
2
2
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] , then
Z b
P
Z b
k m
e
k
k
k=1
.
(10.53)
kf (t)k dt ≤ Pm 2·√
f
(t)
dt
µk M 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
kf (t)k dt ·
ek .
f (t) dt =
P
2
k m
e
k
a
a
k
k=1
k=1
Proof. Utilising Lemma 3.7, by (10.51) we deduce
√
2 · µk M k
kf (t)k
kek k ≤ Re hf (t) , ek i
µk + M k
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] .
Applying Theorem 10.6 for
√
2 · µk M k
rk :=
kek k , k ∈ {1, . . . , m}
µk + M k
we deduce the desired result.
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S.S. D RAGOMIR
11. A DDITIVE R EVERSES OF THE C ONTINUOUS T RIANGLE I NEQUALITY
11.1. 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
kf (t)k − Re F [f (t)] ≤ k (t)
(11.1)
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
The equality holds in (11.2) if and only if both
Z b
Z b
f (t) dt = f
(t)
dt
(11.3)
F
a
a
and
Z
F
(11.4)
b
Z b
Z b
f (t) dt =
kf (t)k dt −
k (t) dt.
a
a
a
Proof. Since the norm of F is unity, 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
Z b
(11.5)
f (t) dt f (t) dt ≥ Re F
f (t) dt ≥ F
a
a
a
Z b
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
k (t) dt.
kf (t)k dt − Re F
f (t) dt ≤
(11.6)
a
a
a
Now, making use of (11.5) and (11.6), we deduce (11.2).
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
k (t) dt.
kf (t)k dt = Re F
f (t) dt +
a
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R EVERSES OF THE T RIANGLE I NEQUALITY
and
37
Z b
Z b
Z b
Re F
f (t) dt
= F
f (t) dt = f (t) dt
a
a
a
which imply (11.3) and (11.4).
Corollary 11.2. Let (X, k·k) be a Banach space, [·, ·] : X × X → K a semi-inner 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
(0 ≤) kf (t)k − Re [f (t) , e] ≤ k (t) ,
(11.7)
for a.e. t ∈ [a, b] , then
Z
(0 ≤)
(11.8)
a
b
Z b
Z b
kf (t)k dt − f (t) dt
k (t) dt,
≤
a
a
where equality holds in (11.8) if and only if both
Z b
Z b
f
(t)
dt
(11.9)
f (t) dt, e = a
a
and
Z
(11.10)
a
b
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
a
and
b
Z
Z
b
kf (t)k dt −
f (t) dt =
(11.12)
a
Z
a
b
k (t) dt e.
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
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
f (t) dt, e = f (t) dt
kek .
a
a
Utilising Theorem 4.2, we get
Z
b
f (t) dt = λe with λ > 0.
(11.13)
a
Replacing
(11.14)
Rb
a
f (t) dt with λe in the second equation of (11.9) we deduce
Z b
Z b
k (t) dt,
kf (t)k dt −
λ=
a
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38
S.S. D RAGOMIR
and by (11.13) and (11.14) we deduce (11.12).
Remark 11.4. 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].
11.2. The Case of m Functionals. The following result may be stated [8]:
Theorem 11.5 (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
kf (t)k − Re Fk [f (t)] ≤ Mk (t)
(11.15)
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b] , then
Z
Z b
m
m Z b
1 X
b
1 X
(11.16)
kf (t)k dt ≤ Fk f (t) dt +
Mk (t) dt.
a
m
m k=1 a
a
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
(11.17)
Fk
f (t) dt = Fk f (t) dt
a
m
m
a
k=1
k=1
and
m
(11.18)
1 X
Fk
m k=1
Z
a
b
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
Z b
m
m Z
1 X
1 X b
Re Fk
f (t) dt +
Mk (t) dt.
(11.19)
kf (t)k dt ≤
m k=1
m k=1 a
a
a
Utilising the continuity property of the functionals Fk and the properties of the modulus, we
have:
Z b
X
Z b
m
m
X
Re Fk
(11.20)
f (t) dt ≤ Re Fk
f (t) dt a
a
k=1
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
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R EVERSES OF THE T RIANGLE I NEQUALITY
m
X
k=1
and
39
Z b
Z b
m
X
Re Fk
f (t) dt
=
f (t) dt Fk a
a
k=1
m
X
Z b
Im Fk
f (t) dt
= 0.
k=1
a
These imply that (11.17) and (11.18) hold true, and the theorem is completely proved.
Remark 11.6. 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 k=1 a
a
a
which is obviously coarser than (11.16) but, perhaps more useful for applications.
The following new result may be stated as well:
Theorem 11.7. 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∞ f (t) dt +
Mk (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 k=1 a
a
a
The proof is similar to the ones from Theorem 7.1 and 11.5 and we omit the details.
The case of Hilbert spaces for Theorem 11.5, in which one may provide a simpler condition
for equality, is of interest in applications [8].
Theorem 11.8 (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
(11.24)
kf (t)k − Re hf (t) , ek i ≤ Mk (t)
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.
m
a
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
(11.26)
kf (t)k dt ≥
Mk (t) dt
m k=1 a
a
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40
S.S. D RAGOMIR
and
Z
m
b
R
f (t) dt =
(11.27)
b
a
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
Proof. As in the proof of Theorem 11.5, we have
*
+
Z b
Z b
m
m Z
1 X
1 X b
(11.28)
kf (t)k dt ≤ Re
ek ,
f (t) dt +
Mk (t) dt
m k=1
m k=1 a
a
a
P
and m
k=1 ek 6= 0.
Rb
P
On utilising Schwarz’s inequality in Hilbert space (H, h·, ·i) for a f (t) dt and m
k=1 ek , we
have
*Z
+
Z b
m
m
b
X
X
(11.29)
f
(t)
dt
e
≥
f
(t)
dt,
e
k
k a
a
k=1
k=1
*Z
+
m
b
X
≥ Re
f (t) dt,
ek a
k=1
*Z
+
m
b
X
≥ Re
f (t) dt,
ek .
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 k=1 ek k
a
showing that the equality holds in (11.25).
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
e
=
f
(t)
dt,
e
k
k a
a
k=1
k=1
and
*Z
Im
(11.32)
b
f (t) dt,
a
m
X
+
ek
= 0.
k=1
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.
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R EVERSES OF THE T RIANGLE I NEQUALITY
41
If the equality holds in (11.25), then obviously (11.26) is valid and the theorem is proved. Remark 11.9. If in the above theorem, the vectors {ek }k∈{1,...,m} are assumed to be orthogonal,
then (11.25) becomes
! 12 Z
Z b
m Z b
m
b
1 X
1 X
2
f (t) dt +
Mk (t) dt.
(11.34)
kf (t)k dt ≤
kek k
m k=1 a
m k=1
a
a
Moreover, if {ek }k∈{1,...,m} is an orthonormal family, then (11.34) becomes
Z b
Z b
m Z b
1 X
1 Mk (t) dt
(11.35)
kf (t)k dt ≤ √ f (t) dt +
m k=1 a
m a
a
which has been obtained in [4].
The following corollaries are of interest.
Corollary 11.10. Let (H; h·, ·i), ek , k ∈ {1, . . . , m} and f be as in Theorem 11.8. If
rk : [a, b] → [0, ∞), k ∈ {1, . . . , m} are such that rk ∈ L2 [a, b] , k ∈ {1, . . . , m} and
kf (t) − ek k ≤ rk (t) ,
(11.36)
for each k ∈ {1, . . . , m} and a.e. t ∈ [a, b], then
Z
Z b
m
m Z
1 X
b
1 X b 2
(11.37)
kf (t)k dt ≤ ek f (t) dt +
r (t) dt.
m
a
2m k=1 a k
a
k=1
The case of equality holds in (11.37) if and only if
Z b
m Z
1 X b 2
kf (t)k dt ≥
r (t) dt
2m k=1 a k
a
and
Z
m
b
f (t) dt =
R
b
a
a
Pm R b 2
1
m
kf (t)k dt − 2m
r
(t)
dt
X
k=1 a k
ek .
P
2
k m
e
k
k
k=1
k=1
Finally, the following corollary may be stated.
Corollary 11.11. Let (H; h·, ·i), ek , k ∈ {1, . . . , m} and f be as in Theorem 11.8. If Mk , µk :
2
k −µk )
∈ L [a, b] and
[a, b] → R are such that Mk ≥ µk > 0 a.e. on [a, b] , (M
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
Z
Z b
Z b
m
m
1 X
b
X
1
[Mk (t) − µk (t)]2
2
kf (t)k dt ≤ ek f (t) dt
+
ke
k
dt.
k
4m
m
a
M
k (t) + µk (t)
a
a
k=1
k=1
12. A PPLICATIONS FOR C OMPLEX -VALUED F UNCTIONS
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|
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42
S.S. D RAGOMIR
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 complex-valued functions.
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
r [|Re f (t)| + |Im f (t)|] ≤ α Re f (t) − β Im f (t)
(12.1)
for a.e. t ∈ [a, b] . Then
Z b
Z b
Z b
Z b
(12.2)
r
|Re f (t)| dt +
|Im f (t)| dt ≤ Re f (t) dt + Im f (t) dt .
a
a
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
b
Z
Z
Re f (t) dt − β
α
a
a
b
Z b
Z b
Im f (t) dt .
Im f (t) dt = Re f (t) dt + a
a
√
Now, consider the Banach space (C, |·|∞ ) . If F (z) = dz with d = γ + iδ and |d| =
2
γ + δ 2 = 12 , then F is linear on C. For z 6= 0 we have
q
√
|Re z|2 + |Im z|2
|F (z)|
|d| |z|
2
=
=
≤ 1.
·
|z|∞
|z|∞
2 max {|Re z| , |Im z|}
2
,
2
i.e.,
Since, for z0 = 1 + i, we have |F (z0 )| = 1, |z0 |∞ = 1, hence
kF k∞ := sup
z6=0
|F (z)|
= 1,
|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.
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 b
Z b
Z b
(12.3) r
max {|Re f (t)| , |Im f (t)|} dt ≤ max Re f (t) dt , Im f (t) dt .
a
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
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R EVERSES OF THE T RIANGLE I NEQUALITY
43
and
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
a
a
a
Now, consider the Banach space C, |·|2p with p ≥ 1. Let F : C → C, F (z) = cz with
Z
1
Z
b
1
|c| = 2 2p − 2 (p ≥ 1) . Obviously, F is linear and by Hölder’s inequality
q
1
− 12
2p
2
|Re z|2 + |Im z|2
|F (z)|
=
≤ 1.
1
2p
2p 2p
|z|2p
|Re z| + |Im z|
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
1
r |Re f (t)|2p + |Im f (t)|2p 2p ≤ ϕ Re f (t) − φ Im f (t)
for a.e. t ∈ [a, b] , then
Z b
1
(12.4) r
|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
and
Z
b
Z
Re f (t) dt − φ
ϕ
a
a
b
"Z
2p Z b
2p # 2p1
b
Im f (t) dt = Re f (t) dt + Im f (t) dt
.
a
a
Remark 12.4. If p = 1 above, and
r |f (t)| ≤ ϕ Re f (t) − ψ Im f (t) for a.e. t ∈ [a, b] ,
provided ϕ, ψ ∈ R and ϕ2 + ψ 2 = 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
Re f (t) dt − ψ
ϕ
a
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
b
Z
a
b
|f (t)| dt
Im f (t) dt = r
a
http://jipam.vu.edu.au/
44
S.S. D RAGOMIR
and
Z
b
Z
Re f (t) dt − ψ
ϕ
a
a
b
Z b
Im f (t) dt = f (t) dt .
a
If we apply Theorem 11.1, then, in a similar manner we can prove the following result for
complex-valued functions.
Proposition 12.5. 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
Z b
Z b
Z b
(0 ≤)
|Re f (t)| dt +
|Im f (t)| dt − Re f (t) dt + Im f (t) dt
a
a
a
a
Z
b
≤
k (t) dt.
a
Applying Theorem 11.1, for (C, |·|∞ ) we may state:
1
,
2
Proposition 12.6. Let γ, δ ∈ R with γ 2 + δ 2 =
functions on [a, b] such that
f, k : [a, b] → C Lebesgue integrable
max {|Re f (t)| , |Im f (t)|} ≤ γ Re f (t) − δ Im f (t) + k (t)
for a.e. t ∈ [a, b] . Then
Z b
Z b
Z b
(0 ≤)
max {|Re f (t)| , |Im f (t)|} dt − max Re f (t) dt , Im f (t) dt
a
a
a
Z
≤
b
k (t) dt.
a
Finally, utilising Theorem 11.1, for C, |·|2p with p ≥ 1, we may state that:
1
1
Proposition 12.7. 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 b
1
(0 ≤)
|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
a
a
Remark 12.8. If p = 1 in the above proposition, then, from
|f (t)| ≤ ϕ Re f (t) − ψ Im f (t) + k (t) for a.e. t ∈ [a, b] ,
provided ϕ, ψ ∈ R and ϕ2 + ψ 2 = 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
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
a
a
http://jipam.vu.edu.au/
R EVERSES OF THE T RIANGLE I NEQUALITY
45
R EFERENCES
[1] E. BERKSON, Some types of Banach spaces, Hermitian systems and Bade functionals, Trans.
Amer. Math. Soc., 106 (1965), 376–385.
[2] J.B. DIAZ AND F.T. METCALF, A complementary triangle inequality in Hilbert and Banach
spaces, Proc. Amer. Math. Soc., 17(1) (1966), 88–97.
[3] S.S. DRAGOMIR, Semi-Inner Products and Applications, Nova Science Publishers Inc., New
York, 2004, pp. 222.
[4] S.S. DRAGOMIR, Advances in Inequalities of the Schwarz, Grüss and Bessel Type in Inner Product
Spaces, Nova Science Publishers Inc., New York, 2005, pp. 249.
[5] S.S. DRAGOMIR, A reverse of the generalised triangle inequality in normed spaces and applications, RGMIA Res. Rep. Coll., 7(2004), Supplement, Article 15. [ONLINE: http://rgmia.
vu.edu.au/v7(E).html].
[6] S.S. DRAGOMIR, Additive reverses of the generalised triangle inequality in normed spaces,
RGMIA Res. Rep. Coll., 7(2004), Supplement, Article 17. [ONLINE: http://rgmia.vu.
edu.au/v7(E).html].
[7] S.S. DRAGOMIR, Additive reverses of the continuous triangle inequality for Bochner integral of
vector-valued functions in Hilbert spaces, RGMIA Res. Rep. Coll., 7(2004), Supplement, Article
12. [ONLINE: http://rgmia.vu.edu.au/v7(E).html].
[8] S.S. DRAGOMIR, Additive reverses of the continuous triangle inequality for Bochner integral of
vector-valued functions in Banach spaces, RGMIA Res. Rep. Coll., 7(2004), Supplement, Article
16[ONLINE: http://rgmia.vu.edu.au/v7(E).html].
[9] S.S. DRAGOMIR, Quadratic reverses of the continuous triangle inequality for Bochner integral of
vector-valued functions in Hilbert spaces, RGMIA Res. Rep. Coll., 7(2004), Supplement, Article
10. [ONLINE: http://rgmia.vu.edu.au/v7(E).html].
[10] S.S. DRAGOMIR, Reverses of the continuous triangle inequality for Bochner integral of vectorvalued functions in Hilbert spaces, RGMIA Res. Rep. Coll., 7(2004), Supplement, Article 11. [ONLINE: http://rgmia.vu.edu.au/v7(E).html].
[11] S.S. DRAGOMIR, Reverses of the continuous triangle inequality for Bochner integral of vectorvalued functions in Banach spaces, RGMIA Res. Rep. Coll., 7(2004), Supplement, Article 14 [ONLINE: http://rgmia.vu.edu.au/v7(E).html].
[12] S.S. DRAGOMIR, Reverses of the triangle inequality in inner product spaces, RGMIA Res.
Rep. Coll., 7(2004), Supplement, Article 7. [ONLINE: http://rgmia.vu.edu.au/v7(E)
.html].
[13] S. GUDDER AND D. STRAWTHER, Strictly convex normed linear spaces, Proc. Amer. Math.
Soc., 59(2) (1976), 263–267.
[14] J. KARAMATA, Stieltjes Integral, Theory and Practice (Serbo-Croatian) SANU, Posebna Izdonija,
154, Beograd, 1949.
[15] S.M. KHALEELULA, On Diaz-Metcalf’s complementary triangle inequality, Kyungpook Math. J.,
15 (1975), 9–11.
[16] S. KUREPA, On an inequality, Glasnik Math., 3(23)(1968), 193–196.
[17] G. LUMER, Semi-inner product spaces, Trans. Amer. Math. Soc., 100 (1961), 29–43.
[18] M. MARDEN, The Geometry of the Zeros of a Polynomial in a Complex Variable, Amer. Math.
Soc. Math. Surveys, 3, New York, 1949.
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
http://jipam.vu.edu.au/
46
S.S. D RAGOMIR
[19] P.M. MILIČIĆ, On a complementary inequality of the triangle inequality (French), Mat. Vesnik
41(2) (1989), 83–88.
[20] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis,
Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.
[21] B. NATH, On a generalisation of semi-inner product spaces, Math. J. Okoyama Univ., 15(10)
(1971), 1–6.
[22] M. PETROVICH, Module d’une somme, L’ Ensignement Mathématique, 19 (1917), 53–56.
[23] H.S. WILF, Some applications of the inequality of arithmetic and geometric means to polynomial
equations, Proceedings Amer. Math. Soc., 14 (1963), 263–265.
J. Inequal. Pure and Appl. Math., 6(5) Art. 129, 2005
http://jipam.vu.edu.au/