http://jipam.vu.edu.au/
Volume 6, Issue 5, Article 129, 2005
REVERSES OF THE TRIANGLE INEQUALITY IN BANACH SPACES
S.S. DRAGOMIR
SCHOOL OFCOMPUTERSCIENCE ANDMATHEMATICS
VICTORIAUNIVERSITY
PO BOX14428, 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
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.
Key words and phrases: Reverse triangle inequality, Hilbert spaces, Banach spaces, Bochner integral.
2000 Mathematics Subject Classification. Primary 46B05, 46C05; Secondary 26D15, 26D10.
1. INTRODUCTION
The generalised triangle inequality, namely
n
X
i=1
xi
≤
n
X
i=1
kxik,
provided(X,k.k)is a normed linear space over the real or complex fieldK=R,Candxi, i∈ {1, ..., n}are vectors in X plays a fundamental role in establishing various analytic and geo- metric properties of such spaces.
With no less importance, the continuous version of it, i.e., (1.1)
Z b a
f(t)dt
≤ Z b
a
kf(t)kdt,
where f : [a, b] ⊂ R → X is a strongly measurable function on the compact interval [a, b]
with values in the Banach spaceX andkf(·)kis 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
Analysis of vector-valued functions with countless applications in Functional Analysis, Opera- tor Theory, Differential Equations, Semigroups Theory and related fields.
Surprisingly enough, the reverses of these, i.e., inequalities of the following type
n
X
i=1
kxik ≤C
n
X
i=1
xi ,
Z b a
kf(t)kdt≤C
Z b a
f(t)dt , withC ≥1,which we call multiplicative reverses, or
n
X
i=1
kxik ≤
n
X
i=1
xi
+M, Z b
a
kf(t)kdt≤
Z b a
f(t)dt
+M,
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
cosθ
n
X
k=1
|zk| ≤
n
X
k=1
zk ,
provided the complex numberszk, k ∈ {1, . . . , n}satisfy the assumption a−θ≤arg (zk)≤a+θ, for any k∈ {1, . . . , n}, where a ∈ Rand 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). Letabe a unit vector in the inner product space(H;h·,·i) over the real or complex number fieldK. Suppose that the vectorsxi ∈H\ {0}, i∈ {1, . . . , n}
satisfy
0≤r ≤ Rehxi, ai
kxik , i∈ {1, . . . , n}. Then
r
n
X
i=1
kxik ≤
n
X
i=1
xi ,
where equality holds if and only if
n
X
i=1
xi =r
n
X
i=1
kxik
! a.
A generalisation of this result for orthonormal families is incorporated in the following result [2].
Theorem 1.2 (Diaz-Metcalf, 1966). Leta1, . . . , anbe orthonormal vectors inH.Suppose the vectorsx1, . . . , xn ∈H\ {0}satisfy
0≤rk≤ Rehxi, aki
kxik , i∈ {1, . . . , n}, k ∈ {1, . . . , m}.
Then
m
X
k=1
r2k
!12 n X
i=1
kxik ≤
n
X
i=1
xi ,
where equality holds if and only if
n
X
i=1
xi =
n
X
i=1
kxik
! m X
k=1
rkak.
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
cosθ Z b
a
|f(x)|dx≤
Z b a
f(x)dx provided
−θ≤argf(x)≤θ, x∈[a, b]
for givenθ ∈ 0,π2 .
This result has recently been extended by the author for the case of Bochner integrable func- tions 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 spaceH,i.e., we recall thatf ∈L([a, b] ;H)if and only iff : [a, b]→His strongly measurable on[a, b]and the Lebesgue integralRb
akf(t)kdtis finite, then
(1.2)
Z b a
kf(t)kdt ≤K
Z b a
f(t)dt , provided thatf satisfies the condition
kf(t)k ≤KRehf(t), ei for a.e.t ∈[a, b],
wheree∈H,kek= 1andK ≥1are given. The case of equality holds in (1.2) if and only if Z b
a
f(t)dt = 1 K
Z b a
kf(t)kdt
e.
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. DIAZ-METCALFTYPEINEQUALITIES
In [2], Diaz and Metcalf established the following reverse of the generalised triangle inequal- ity 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 spaceX endowed with the norm k·k and the vectors x1, . . . , xnsatisfy the condition
(2.1) 0≤r≤ReF (xi), i∈ {1, . . . , n};
then
(2.2) r
n
X
i=1
kxik ≤
n
X
i=1
xi , where equality holds if and only if both
(2.3) F
n
X
i=1
xi
!
=r
n
X
i=1
kxik
and
(2.4) F
n
X
i=1
xi
!
=
n
X
i=1
xi .
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≤rkxik ≤Rehxi, 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]
(2.6)
n
X
i=1
xi =r
n
X
i=1
kxik
! e.
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 numbercdefined by
c= sup
x6=0
"
Pm
k=1|Fk(x)|2 kxk2
#
;
it then follows that1≤c≤m.Suppose the vectorsx1, . . . , xnwheneverxi 6= 0,satisfy (2.7) 0≤rkkxik ≤ReFk(xi), i= 1, . . . , n, k = 1, . . . , m.
Then one has the following reverse of the generalised triangle inequality [2]
(2.8)
Pm k=1rk2
c
12 n X
i=1
kxik ≤
n
X
i=1
xi
, where equality holds if and only if both
(2.9) Fk
n
X
i=1
xi
!
=rk
n
X
i=1
kxik, k = 1, . . . , m
and (2.10)
m
X
k=1
"
Fk n
X
i=1
xi
!#2
=c
n
X
i=1
xi
2
.
IfX =H,an inner product space, then, forFk(x) =hx, eki,where{ek}k=1,nis an orthonor- mal family inH,i.e.,hei, eji=δij, i, j ∈ {1, . . . , k}, δij is Kronecker delta, the condition (2.7) may be replaced by
(2.11) 0≤rkkxik ≤Rehxi, eki, i= 1, . . . , n, k = 1, . . . , m;
implying the following reverse of the generalised triangle inequality (2.12)
m
X
k=1
r2k
!12 n X
i=1
kxik ≤
n
X
i=1
xi ,
where the equality holds if and only if (2.13)
n
X
i=1
xi =
n
X
i=1
kxik
! m X
k=1
rkek.
The aim of the following sections is to present recent reverses of the triangle inequality ob- tained 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 num- bers 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. INEQUALITIES OFDIAZ-METCALFTYPE FORmFUNCTIONALS
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 ∈ X\ {0}, i ∈ {1, . . . , n}are such that there exists the constantsrk ≥ 0, k ∈ {1, . . . , m}
withPm
k=1rk >0and
(3.1) ReFk(xi)≥rkkxik
for each i∈ {1, . . . , n}andk∈ {1, . . . , m},then
(3.2)
n
X
i=1
kxik ≤ kPm k=1Fkk Pm
k=1rk
n
X
i=1
xi . The case of equality holds in (3.2) if both
(3.3)
m
X
k=1
Fk
! n X
i=1
xi
!
=
m
X
k=1
rk
! n X
i=1
kxik
and (3.4)
m
X
k=1
Fk
! n X
i=1
xi
!
=
m
X
k=1
Fk
n
X
i=1
xi .
Proof. Utilising the hypothesis (3.1) and the properties of the modulus, we have I :=
m
X
k=1
Fk
! n X
i=1
xi
!
≥
Re
" m X
k=1
Fk
! n X
i=1
xi
!#
(3.5)
≥
m
X
k=1
ReFk
n
X
i=1
xi
!
=
m
X
k=1 n
X
i=1
ReFk(xi)
≥
m
X
k=1
rk
! n X
i=1
kxik.
On the other hand, by the continuity property ofFk, k ∈ {1, . . . , m}we obviously have
(3.6) I =
m
X
k=1
Fk
! n X
i=1
xi
!
≤
m
X
k=1
Fk
n
X
i=1
xi
. 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 in- equality (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) ReFk(xi) = rkkxik
for each i∈ {1, . . . , n},k ∈ {1, . . . , m}; (3.8)
m
X
k=1
ImFk n
X
i=1
xi
!
= 0
and (3.9)
m
X
k=1
ReFk
n
X
i=1
xi
!
=
m
X
k=1
Fk
n
X
i=1
xi .
Note that, from (3.7), by summation overiandk,we get
(3.10) Re
" m X
k=1
Fk
! n X
i=1
xi
!#
=
m
X
k=1
rk
! n X
i=1
kxik.
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 normskFkk, k ∈ {1, . . . , m}are easier to find, then, from (3.2), one may get the (coarser) inequality that might be more useful in practice:
(3.11)
n
X
i=1
kxik ≤ Pm
k=1kFkk Pm
k=1rk
n
X
i=1
xi .
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 field K, ek, xi ∈ H\ {0}, k ∈ {1, . . . , m}, i ∈ {1, . . . , n}. If rk ≥ 0, k ∈ {1, . . . , m}withPm
k=1rk>0satisfy
(3.12) Rehxi, eki ≥rkkxik
for each i∈ {1, . . . , n}andk ∈ {1, . . . , m},then
(3.13)
n
X
i=1
kxik ≤ kPm k=1ekk Pm
k=1rk
n
X
i=1
xi . The case of equality holds in (3.13) if and only if
(3.14)
n
X
i=1
xi = Pm
k=1rk
kPm k=1ekk2
n
X
i=1
kxik
! m X
k=1
ek.
Proof. By the properties of inner product and by (3.12), we have
* n X
i=1
xi,
m
X
k=1
ek
+
≥
m
X
k=1
Re
* n X
i=1
xi, ek
+ (3.15)
≥
m
X
k=1
Re
* n X
i=1
xi, ek +
=
m
X
k=1 n
X
i=1
Rehxi, eki ≥
m
X
k=1
rk
! n X
i=1
kxik>0.
Observe also that, by (3.15),Pm
k=1ek 6= 0.
On utilising Schwarz’s inequality in the inner product space(H;h·,·i)forPn
i=1xi,Pm k=1ek, we have
(3.16)
n
X
i=1
xi
m
X
k=1
ek
≥
* n X
i=1
xi,
m
X
k=1
ek +
. 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
X
i=1
xi
= (Pm
k=1rk)Pn i=1kxik kPm
k=1ekk2
m
X
k=1
ek
= (Pm k=1rk) kPm
k=1ekk
n
X
i=1
kxik,
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
(3.17) Rehxi, eki=rkkxik
for each i∈ {1, . . . , n}andk∈ {1, . . . , m}, (3.18)
n
X
i=1
xi
m
X
k=1
ek
=
* n X
i=1
xi,
m
X
k=1
ek +
and
(3.19) Im
* n X
i=1
xi,
m
X
k=1
ek +
= 0.
From (3.17), on summing overiandk,we get
(3.20) Re
* n X
i=1
xi,
m
X
k=1
ek +
=
m
X
k=1
rk
! n X
i=1
kxik.
By (3.19) and (3.20), we have (3.21)
* n X
i=1
xi,
m
X
k=1
ek +
=
m
X
k=1
rk
! n X
i=1
kxik.
On the other hand, by the use of the following identity in inner product spaces (3.22)
u−hu, viv kvk2
2
= kuk2kvk2− |hu, vi|2
kvk2 , v 6= 0, the relation (3.18) holds if and only if
(3.23)
n
X
i=1
xi = hPn
i=1xi,Pm k=1eki kPm
k=1ekk2
m
X
k=1
ek.
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 fieldKandx, a∈H, r > 0such that:
(3.24) kx−ak ≤r <kak.
Then we have the inequality
(3.25) kxk kak2−r212
≤Rehx, ai or, equivalently
(3.26) kxk2kak2−[Rehx, ai]2 ≤r2kxk2. The case of equality holds in (3.25) (or in (3.26)) if and only if (3.27) kx−ak=r and kxk2+r2 =kak2. Proof. From the first part of (3.24), we have
(3.28) kxk2+kak2−r2 ≤2 Rehx, ai. By the second part of (3.24) we have kak2−r212
>0,therefore, by (3.28), we may state that
(3.29) 0< kxk2
kak2−r212
+ kak2−r212
≤ 2 Rehx, ai kak2−r212
. Utilising the elementary inequality
1
αq+αp ≥2√
pq, α >0, p >0, q ≥0;
with equality if and only ifα = qq
p,we may state (forα = kak2−r212
, p= 1, q =kxk2) that
(3.30) 2kxk ≤ kxk2
kak2−r212
+ kak2−r212 . 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 (3.30) if and only if kak2−r212
=kxk.
The proof is thus completed.
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}.Ifρk ≥0, k ∈ {1, . . . , m}with
(3.31) kxi−ekk ≤ρk<kekk
for each i∈ {1, . . . , n}andk ∈ {1, . . . , m},then
(3.32)
n
X
i=1
kxik ≤ kPm k=1ekk Pm
k=1 kekk2−ρ2k12
n
X
i=1
xi .
The case of equality holds in (3.32) if and only if
n
X
i=1
xi = Pm
k=1 kekk2−ρ2k12 kPm
k=1ekk2
n
X
i=1
kxik
! m X
k=1
ek.
Proof. Utilising Lemma 3.4, we have from (3.31) that kxik kekk2−ρ2k12
≤Rehxi, eki for eachk∈ {1, . . . , m}andi∈ {1, . . . , n}.
Applying Theorem 3.3 for
rk := kekk2 −ρ2k12
, k ∈ {1, . . . , m},
we deduce the desired result.
Remark 3.6. If{ek}k∈{1,...,m}are orthogonal, then (3.32) becomes (3.33)
n
X
i=1
kxik ≤
Pm
k=1kekk212 Pm
k=1 kekk2−ρ2k12
n
X
i=1
xi with equality if and only if
n
X
i=1
xi = Pm
k=1 kekk2−ρ2k12 Pm
k=1kekk2
n
X
i=1
kxik
! m X
k=1
ek.
Moreover, if{ek}k∈{1,...,m}is assumed to be orthonormal and
kxi−ekk ≤ρk fork∈ {1, . . . , m}, i∈ {1, . . . , n}
whereρk ∈[0,1)fork ∈ {1, . . . , m},then (3.34)
n
X
i=1
kxik ≤
√m
Pm
k=1(1−ρ2k)12
n
X
i=1
xi with equality if and only if
n
X
i=1
xi = Pm
k=1(1−ρ2k)12 m
n
X
i=1
kxik
! m X
k=1
ek. The following lemma may be stated as well [3].
Lemma 3.7 (Dragomir, 2004). Let(H;h·,·i)be an inner product space over the real or complex number fieldK,x, y ∈HandM ≥m >0.If
(3.35) RehM y−x, x−myi ≥0
or, equivalently, (3.36)
x− m+M 2 y
≤ 1
2(M −m)kyk, then
(3.37) kxk kyk ≤ 1
2· M +m
√mM Rehx, yi.
The equality holds in (3.37) if and only if the case of equality holds in (3.35) and
(3.38) kxk=√
mMkyk. Proof. Obviously,
RehM y−x, x−myi= (M +m) Rehx, yi − kxk2−mM kyk2. Then (3.35) is clearly equivalent to
(3.39) kxk2
√mM +√
mMkyk2 ≤ M +m
√mM Rehx, yi.
Since, obviously,
(3.40) 2kxk kyk ≤ kxk2
√mM +√
mMkyk2,
with equality iffkxk=√
mMkyk,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) RehMkek−xi, xi−µkeki ≥0 or, equivalently,
xi− Mk+µk 2 ek
≤ 1
2(Mk−µk)kekk for eachk ∈ {1, . . . , m}andi∈ {1, . . . , n},then
(3.42)
n
X
i=1
kxik ≤ kPm k=1ekk Pm
k=1 2·√
µkMk
µk+Mk kekk
n
X
i=1
xi .
The case of equality holds in (3.42) if and only if
n
X
i=1
xi = Pm
k=1 2·√
µkMk µk+Mk kekk kPm
k=1ekk2
n
X
i=1
kxik
m
X
k=1
ek.
Proof. Utilising Lemma 3.7, by (3.41) we deduce 2·√
µkMk
µk+Mk kxik kekk ≤Rehxi, eki for eachk∈ {1, . . . , m}andi∈ {1, . . . , n}.
Applying Theorem 3.3 for
rk := 2·√ µkMk
µk+Mk kekk, k ∈ {1, . . . , m},
we deduce the desired result.
4. DIAZ-METCALFINEQUALITY FORSEMI-INNER PRODUCTS
In 1961, G. Lumer [17] introduced the following concept.
Definition 4.1. LetXbe a linear space over the real or complex number fieldK. The mapping [·,·] :X×X →Kis called a semi-inner product onX,if the following properties are satisfied (see also [3, p. 17]):
(i) [x+y, z] = [x, z] + [y, z]for allx, y, z∈X;
(ii) [λx, y] =λ[x, y]for allx, y ∈X andλ∈K;
(iii) [x, x]≥0for allx∈Xand[x, x] = 0impliesx= 0;
(iv) |[x, y]|2 ≤[x, x] [y, y]for allx, y ∈X;
(v) [x, λy] = ¯λ[x, y]for allx, y ∈X andλ∈K.
It is well known that the mappingX 3x7−→[x, x]12 ∈Ris a norm onXand for anyy∈X, the functionalX 3 x7−→ϕy [x, y]∈ Kis a continuous linear functional onX endowed with the normk·kgenerated by[·,·].Moreover, one haskϕyk=kyk(see for instance [3, p. 17]).
Let(X,k·k)be a real or complex normed space. IfJ :X → 2X∗ is the normalised duality mapping defined onX,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 fieldK, is of the form
[x, y] =D
J˜(y), xE
for any x, y ∈X,
whereJ˜is a selection of the normalised duality mapping and hϕ, xi :=ϕ(x)forϕ ∈ 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 →Ka semi-inner product generating the normk·kande∈X,kek= 1.Ifxi ∈X, i∈ {1, . . . , n}andr≥0such that (4.1) rkxik ≤Re [xi, e] for each i∈ {1, . . . , n},
then we have the inequality
(4.2) r
n
X
i=1
kxik ≤
n
X
i=1
xi .
The case of equality holds in (4.2) if and only if both (4.3)
" n X
i=1
xi, e
#
=r
n
X
i=1
kxik
and (4.4)
" n X
i=1
xi, e
#
=
n
X
i=1
xi .
The proof is obvious from the Diaz-Metcalf theorem [2, Theorem 3] applied for the continu- ous linear functionalFe(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 fromX withx6=yandkxk=kyk= 1,we havekλx+ (1−λ)yk<1for allλ∈(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 ande, xi (i∈ {1, . . . , n})as in Corollary 4.1. Then the case of equality holds in (4.2) if and only if
(4.5)
n
X
i=1
xi =r
n
X
i=1
kxik
! e.
Proof. If (4.5) holds true, then, obviously
n
X
i=1
xi
=r
n
X
i=1
kxik
!
kek=r
n
X
i=1
kxik,
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 aµ >0such that
(4.6)
n
X
i=1
xi =µe.
Inserting this in (4.3) we get
µkek2 =r
n
X
i=1
kxik
giving
(4.7) µ=r
n
X
i=1
kxik.
Finally, by (4.6) and (4.7) we deduce (4.5) and the corollary is proved.
5. OTHER MULTIPLICATIVE REVERSES FORm FUNCTIONALS
Assume thatFk, k∈ {1, . . . , m}are bounded linear functionals defined on the normed linear spaceX.
Forp∈[1,∞),define
(cp) cp := sup
x6=0
Pm
k=1|Fk(x)|p kxkp
1p
and forp=∞,
(c∞) c∞ := sup
x6=0
1≤k≤mmax
|Fk(x)|
kxk
.
Then, by the fact that |Fk(x)| ≤ kFkk kxk for any x ∈ X, where kFkk is the norm of the functionalFk,we have that
cp ≤
m
X
k=1
kFkkp
!1p
, p≥1
and
c∞ ≤ max
1≤k≤mkFkk.
We may now state and prove a new reverse inequality for the generalised triangle inequality in normed linear spaces.
Theorem 5.1. Letxi, rk, Fk, k ∈ {1, . . . , m},i∈ {1, . . . , n}be as in the hypothesis of Theorem 3.1. Then we have the inequalities
(5.1) (1≤)
Pn i=1kxik kPn
i=1xik ≤ c∞
1≤k≤mmax {rk}
≤
1≤k≤mmax kFkk
1≤k≤mmax {rk}
.
The case of equality holds in (5.1) if and only if
(5.2) Re
"
Fk
n
X
i=1
xi
!#
=rk
n
X
i=1
kxik for each k∈ {1, . . . , m}
and
(5.3) max
1≤k≤mRe
"
Fk
n
X
i=1
xi
!#
=c∞
n
X
i=1
xi .
Proof. Since, by the definition ofc∞,we have c∞kxk ≥ max
1≤k≤m|Fk(x)|, for any x∈X,
then we can state, forx=Pn
i=1xi,that c∞
n
X
i=1
xi
≥ max
1≤k≤m
Fk
n
X
i=1
xi
!
≥ max
1≤k≤m
"
ReFk
n
X
i=1
xi
!
# (5.4)
≥ max
1≤k≤m
"
Re
n
X
i=1
Fk(xi)
#
= max
1≤k≤m
" n X
i=1
ReFk(xi)
# .
Utilising the hypothesis (3.1) we obviously have
1≤k≤mmax
" n X
i=1
ReFk(xi)
#
≥ max
1≤k≤m{rk} ·
n
X
i=1
kxik.
Also,Pn
i=1xi 6= 0,because, by the initial assumptions, not allrk andxi withk ∈ {1, . . . , m}
andi∈ {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 overk ∈ {1, . . . , m}in this equality we get
1≤k≤mmax Re
"
Fk n
X
i=1
xi
!#
= max
1≤k≤m{rk}
n
X
i=1
xi
, 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) ReFk(xi) =rkkxik for each i∈ {1, . . . , n} and k ∈ {1, . . . , m}
and, from (5.4),
c∞
n
X
i=1
xi
= max
1≤k≤mRe
"
Fk n
X
i=1
xi
!#
, which is (5.3).
From (5.5), on summing overi∈ {1, . . . , n},we get (5.2), and the theorem is proved.
The following result in normed spaces also holds.
Theorem 5.2. Letxi, rk, Fk, k ∈ {1, . . . , m}, i∈ {1, . . . , n}be as in the hypothesis of Theorem 3.1. Then we have the inequality
(5.6) (1≤)
Pn i=1kxik kPn
i=1xik ≤ cp (Pm
k=1rkp)1p
≤ Pm
k=1kFkkp Pm
k=1rpk 1p
,
wherep≥1.
The case of equality holds in (5.6) if and only if
(5.7) Re
"
Fk
n
X
i=1
xi
!#
=rk
n
X
i=1
kxik for each k∈ {1, . . . , m}
and (5.8)
m
X
k=1
"
ReFk
n
X
i=1
xi
!#p
=cpp
n
X
i=1
xi
p
.
Proof. By the definition ofcp, p≥1,we have cppkxkp ≥
m
X
k=1
|Fk(x)|p for any x∈X,
implying that cpp
n
X
i=1
xi
p
≥
m
X
k=1
Fk
n
X
i=1
xi
!
p
≥
m
X
k=1
ReFk
n
X
i=1
xi
!
p
(5.9)
≥
m
X
k=1
"
ReFk
n
X
i=1
xi
!#p
=
m
X
k=1
" n X
i=1
ReFk(xi)
#p
.
Utilising the hypothesis (3.1), we obviously have that (5.10)
m
X
k=1
" n X
i=1
ReFk(xi)
#p
≥
m
X
k=1
" n X
i=1
rkkxik
#p
=
m
X
k=1
rkp
n
X
i=1
kxik
!p
.
Making use of (5.9) and (5.10), we deduce cpp
n
X
i=1
xi
p
≥
m
X
k=1
rpk
! n X
i=1
kxik
!p
, which implies the desired inequality (5.6).
If (5.7) holds true, then, taking the powerpand summing overk∈ {1, . . . , m},we deduce
m
X
k=1
"
Re
"
Fk
n
X
i=1
xi
!##p
=
m
X
k=1
rkp
n
X
i=1
kxik
!p
, 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) ReFk(xi) =rkkxik for each i∈ {1, . . . , n} and k ∈ {1, . . . , m}
and, from (5.9),
cpp
n
X
i=1
xi
p
=
m
X
k=1
"
ReFk n
X
i=1
xi
!#p
, which is exactly (5.8).
From (5.11), on summing overifrom1ton,we deduce (5.7), and the theorem is proved.
6. AN ADDITIVEREVERSE FOR THETRIANGLE INEQUALITY
6.1. 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 fieldK andF : X → K a linear functional with the property that|F (x)| ≤ kxkfor anyx∈X. Ifxi ∈X, ki ≥0, i∈ {1, . . . , n}are such that
(6.1) (0≤)kxik −ReF (xi)≤ki for each i∈ {1, . . . , n}, then we have the inequality
(6.2) (0≤)
n
X
i=1
kxik −
n
X
i=1
xi
≤
n
X
i=1
ki. The equality holds in (6.2) if and only if both
(6.3) F
n
X
i=1
xi
!
=
n
X
i=1
xi
and F
n
X
i=1
xi
!
=
n
X
i=1
kxik −
n
X
i=1
ki.
Proof. If we sum in (6.1) overifrom1ton,then we get (6.4)
n
X
i=1
kxik ≤Re
"
F
n
X
i=1
xi
!#
+
n
X
i=1
ki.
Taking into account that|F (x)| ≤ kxkfor eachx∈X,then we may state that Re
"
F
n
X
i=1
xi
!#
≤
ReF
n
X
i=1
xi
! (6.5)
≤ F
n
X
i=1
xi
!
≤
n
X
i=1
xi . 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
X
i=1
kxik= Re
"
F
n
X
i=1
xi
!#
+
n
X
i=1
ki and
Re
"
F
n
X
i=1
xi
!#
= F
n
X
i=1
xi
!
=
n
X
i=1
xi ,
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 →Ka semi-inner product generating the normk·kande∈X,kek= 1.Ifxi ∈X, ki ≥0, i∈ {1, . . . , n}are such that (6.6) (0≤)kxik −Re [xi, e]≤ki for each i∈ {1, . . . , n},
then we have the inequality
(6.7) (0≤)
n
X
i=1
kxik −
n
X
i=1
xi
≤
n
X
i=1
ki. The equality holds in (6.7) if and only if both
(6.8)
" n X
i=1
xi, e
#
=
n
X
i=1
xi
and
" n X
i=1
xi, e
#
=
n
X
i=1
kxik −
n
X
i=1
ki.
Moreover, if(X,k·k)is strictly convex, then the case of equality holds in (6.7) if and only if
(6.9)
n
X
i=1
kxik ≥
n
X
i=1
ki and
(6.10)
n
X
i=1
xi =
n
X
i=1
kxik −
n
X
i=1
ki
!
·e.
Proof. The first part of the corollary is obvious by Theorem 6.1 applied for the continuous linear functional of unit normFe, 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 6.3. IfX =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 ofmFunctionals. 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 fieldK. IfFk,k ∈ {1, . . . , m}are bounded linear functionals defined onXandxi ∈X, Mik ≥0fori∈ {1, . . . , n},k ∈ {1, . . . , m}are such that
(6.11) kxik −ReFk(xi)≤Mik
for each i∈ {1, . . . , n}, k ∈ {1, . . . , m},then we have the inequality
(6.12)
n
X
i=1
kxik ≤
1 m
m
X
k=1
Fk
n
X
i=1
xi
+ 1 m
m
X
k=1 n
X
i=1
Mik. The case of equality holds in (6.12) if both
(6.13) 1
m
m
X
k=1
Fk
n
X
i=1
xi
!
=
1 m
m
X
k=1
Fk
n
X
i=1
xi and
(6.14) 1
m
m
X
k=1
Fk n
X
i=1
xi
!
=
n
X
i=1
kxik − 1 m
m
X
k=1 n
X
j=1
Mik.
Proof. If we sum (6.11) overifrom1ton,then we deduce
n
X
i=1
kxik −ReFk
n
X
i=1
xi
!
≤
n
X
i=1
Mik for eachk∈ {1, . . . , m}.
Summing these inequalities overk from1tom,we deduce (6.15)
n
X
i=1
kxik ≤ 1 m
m
X
k=1
ReFk
n
X
i=1
xi
! + 1
m
m
X
k=1 n
X
i=1
Mik.
Utilising the continuity property of the functionals Fk and the properties of the modulus, we have
m
X
k=1
ReFk
n
X
i=1
xi
!
≤
m
X
k=1
ReFk
n
X
i=1
xi
! (6.16)
≤
m
X
k=1
Fk
n
X
i=1
xi
!
≤
m
X
k=1
Fk
n
X
i=1
xi . 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).
