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volume 6, issue 5, article 129, 2005.

Received 18 April, 2005;

accepted 05 July, 2005.

Communicated by:B. Mond

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Journal of Inequalities in Pure and Applied Mathematics

REVERSES OF THE TRIANGLE INEQUALITY IN BANACH SPACES

S.S. DRAGOMIR

School of Computer Science and Mathematics Victoria University

PO Box 14428, MCMC 8001 VIC, Australia.

EMail:sever@csm.vu.edu.au URL:http://rgmia.vu.edu.au/dragomir/

c

2000Victoria University ISSN (electronic): 1443-5756 123-05

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Reverses of the Triangle Inequality in Banach Spaces

S.S. Dragomir

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J. Ineq. Pure and Appl. Math. 6(5) Art. 129, 2005

http://jipam.vu.edu.au

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.

2000 Mathematics Subject Classification: Primary 46B05, 46C05; Secondary 26D15, 26D10

Key words: Reverse triangle inequality, Hilbert spaces, Banach spaces, Bochner in- tegral.

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]

1. Introduction

The generalised triangle inequality, namely

n

X

i=1

x_i

≤

n

X

i=1

kx_ik,

provided(X,k.k)is a normed linear space over the real or complex fieldK=R, Candx_i, i∈ {1, ..., n}are vectors inXplays 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 a

f(t)dt

≤ Z b

a

kf(t)kdt,

where f : [a, b] ⊂ R → X is a strongly measurable function on the com- pact interval[a, b]with values in the Banach spaceX andkf(·)kis Lebesgue integrable on [a, b], is crucial in the Analysis of vector-valued functions with countless applications in Functional Analysis, Operator Theory, Differential Equations, Semigroups Theory and related fields.

Surprisingly enough, the reverses of these, i.e., inequalities of the following type

n

X

i=1

kx_ik ≤C

n

X

i=1

x_i ,

Z b a

kf(t)kdt≤C

Z b a

f(t)dt ,

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withC ≥1,which we call multiplicative reverses, or

n

X

i=1

kx_ik ≤

n

X

i=1

x_i

+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

|z_k| ≤

n

X

k=1

z_k ,

provided the complex numberszk, k ∈ {1, . . . , n}satisfy the assumption a−θ ≤arg (z_k)≤a+θ, for any k ∈ {1, . . . , n},

wherea∈Randθ ∈ 0,^π₂

was first discovered by M. Petrovich in 1917, [22]

(see [20, p. 492]) and subsequently was rediscovered by other authors, includ- ing J. Karamata [14, p. 300 – 301], H.S. Wilf [23], and in an equivalent form by M. Marden [18]. Marden and Wilf have outlined in their work the important fact that reverses of the generalised triangle inequality may be successfully applied to the location problem for the roots of complex polynomials.

In 1966, J.B. Diaz and F.T. Metcalf [2] proved the following reverse of the triangle inequality in the more general case of inner product spaces:

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Theorem 1.1 (Diaz-Metcalf, 1966). Letabe a unit vector in the inner product space (H;h·,·i) over the real or complex number field K. Suppose that the vectorsx_i ∈H\ {0}, i∈ {1, . . . , n}satisfy

0≤r≤ Rehx_i, ai

kx_ik , 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

x_i =r

n

X

i=1

kx_ik

! a.

A generalisation of this result for orthonormal families is incorporated in the following result [2].

Theorem 1.2 (Diaz-Metcalf, 1966). Let a₁, . . . , a_n be orthonormal vectors in H.Suppose the vectorsx₁, . . . , x_n∈H\ {0}satisfy

0≤r_k ≤ Rehx_i, a_ki

kx_ik , i∈ {1, . . . , n}, k ∈ {1, . . . , m}. Then

m

X

k=1

r_k²

!¹₂ _n X

i=1

kx_ik ≤

n

X

i=1

x_i ,

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where equality holds if and only if

n

X

i=1

x_i =

n

X

i=1

kx_ik

! _m X

k=1

r_ka_k.

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 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 thatf ∈L([a, b] ;H)if and only iff : [a, b]→H is strongly measurable on[a, b]and the Lebesgue integralRb

a kf(t)kdtis finite, then (1.2)

Z b a

kf(t)kdt≤K

Z b a

f(t)dt ,

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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.

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2. Diaz-Metcalf Type Inequalities

In [2], Diaz and Metcalf established the following reverse of the generalised triangle inequality in real or complex normed linear spaces.

Theorem 2.1 (Diaz-Metcalf, 1966). If F : X → K, K = R,C is a linear functional of a unit norm defined on the normed linear space X endowed with the normk·kand the vectorsx₁, . . . , x_nsatisfy the condition

(2.1) 0≤r≤ReF (x_i), i∈ {1, . . . , n}; then

(2.2) r

n

X

i=1

kx_ik ≤

n

X

i=1

x_i ,

where equality holds if and only if both

(2.3) F

n

X

i=1

x_i

!

=r

n

X

i=1

kx_ik

and

(2.4) F

n

X

i=1

x_i

!

=

n

X

i=1

x_i .

IfX =H,(H;h·,·i)is an inner product space andF (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,

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which implies the reverse of the generalised triangle inequality (2.2). In this case the equality holds in (2.2) if and only if [2]

(2.6)

n

X

i=1

x_i =r

n

X

i=1

kx_ik

! 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|F_k(x)|² kxk²

#

;

it then follows that1 ≤c ≤m.Suppose the vectorsx₁, . . . , x_n wheneverx_i 6=

0,satisfy

(2.7) 0≤r_kkx_ik ≤ReF_k(x_i), i= 1, . . . , n, k = 1, . . . , m.

Then one has the following reverse of the generalised triangle inequality [2]

(2.8)

Pm k=1r²_k

c

¹₂ ⁿ X

i=1

kx_ik ≤

n

X

i=1

x_i ,

where equality holds if and only if both

(2.9) F_k

n

X

i=1

x_i

!

=r_k

n

X

i=1

kx_ik, k = 1, . . . , m

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and (2.10)

m

X

k=1

"

F_k

n

X

i=1

x_i

!#2

=c

n

X

i=1

x_i

2

.

If X = H, an inner product space, then, for Fk(x) = hx, eki, where {e_k}_k=1,n is an orthonormal family in H,i.e., he_i, e_ji =δ_ij, i, j ∈ {1, . . . , k}, δ_ij 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

r_k²

!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

x_i =

n

X

i=1

kx_ik

! _m X

k=1

r_ke_k.

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 Chap- ter XVII of the book [20] and the references therein.

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3. Inequalities of Diaz-Metcalf Type for m Functionals

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 fieldKandF_k :X → K,k ∈ {1, . . . , m}continuous linear functionals on X. If x_i ∈ X\ {0}, i ∈ {1, . . . , n} are such that there exists the constantsr_k ≥0,k ∈ {1, . . . , m}withPm

k=1r_k >0and (3.1) ReF_k(x_i)≥r_kkx_ik

for each i∈ {1, . . . , n}andk ∈ {1, . . . , m},then

(3.2)

n

X

i=1

kx_ik ≤ kPm k=1F_kk Pm

k=1r_k

n

X

i=1

x_i .

The case of equality holds in (3.2) if both (3.3)

m

X

k=1

F_k

! _n X

i=1

x_i

!

=

m

X

k=1

r_k

! _n X

i=1

kx_ik

and (3.4)

m

X

k=1

Fk

! _n X

i=1

xi

!

=

m

X

k=1

Fk

n

X

i=1

xi

.

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Proof. Utilising the hypothesis (3.1) and the properties of the modulus, we have I :=

m

X

k=1

F_k

! _n X

i=1

x_i

!

≥

Re

" _m X

k=1

F_k

! _n X

i=1

x_i

!#

(3.5)

≥

m

X

k=1

ReF_k

n

X

i=1

x_i

!

=

m

X

k=1 n

X

i=1

ReF_k(x_i)

≥

m

X

k=1

rk

! _n X

i=1

kxik.

On the other hand, by the continuity property of F_k, k ∈ {1, . . . , m}we obviously have

(3.6) I =

m

X

k=1

F_k

! _n X

i=1

x_i

!

≤

m

X

k=1

F_k

n

X

i=1

x_i .

Making use of (3.5) and (3.6), we deduce the desired inequality (3.2).

Now, if (3.3) and (3.4) are valid, then, obviously, the case of equality holds true in the inequality (3.2).

Conversely, if the case of equality holds in (3.2), then it must hold in all the inequalities used to prove (3.2). Therefore we have

(3.7) ReF_k(x_i) =r_kkx_ik for each i∈ {1, . . . , n},k ∈ {1, . . . , m}; (3.8)

m

X

k=1

ImF_k

n

X

i=1

x_i

!

= 0

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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 1. If the norms kF_kk, 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

kx_ik ≤ Pm

k=1kF_kk Pm

k=1r_k

n

X

i=1

x_i .

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 3.2 (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}.Ifr_k≥0, k ∈ {1, . . . , m}withPm

k=1r_k >0satisfy (3.12) Rehxi, eki ≥rkkxik

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for each i∈ {1, . . . , n}andk∈ {1, . . . , m},then

(3.13)

n

X

i=1

kx_ik ≤ kPm k=1e_kk Pm

k=1r_k

n

X

i=1

x_i .

The case of equality holds in (3.13) if and only if (3.14)

n

X

i=1

xi = Pm

k=1r_k kPm

k=1e_kk²

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

x_i,

m

X

k=1

e_k +

≥

m

X

k=1

Re

* _n X

i=1

x_i, e_k +

(3.15)

≥

m

X

k=1

Re

* _n X

i=1

xi, ek

+

=

m

X

k=1 n

X

i=1

Rehx_i, e_ki ≥

m

X

k=1

r_k

! _n X

i=1

kx_ik>0.

Observe also that, by (3.15),Pm

k=1e_k 6= 0.

On utilising Schwarz’s inequality in the inner product space (H;h·,·i)for Pn

i=1x_i,Pm

k=1e_k,we have (3.16)

n

X

i=1

x_i

m

X

k=1

e_k

≥

* _n X

i=1

x_i,

m

X

k=1

e_k +

.

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Making use of (3.15) and (3.16), we can conclude that (3.13) holds.

Now, if (3.14) holds true, then, by taking the norm, we have

n

X

i=1

x_i

= (Pm

k=1r_k)Pn i=1kx_ik kPm

k=1e_kk²

m

X

k=1

e_k

= (Pm k=1r_k) kPm

k=1e_kk

n

X

i=1

kx_ik,

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) Rehx_i, e_ki=r_kkx_ik for each i∈ {1, . . . , n}andk ∈ {1, . . . , m}, (3.18)

n

X

i=1

x_i

m

X

k=1

e_k

=

* _n X

i=1

x_i,

m

X

k=1

e_k +

and

(3.19) Im

* _n X

i=1

x_i,

m

X

k=1

e_k +

= 0.

From (3.17), on summing overiandk,we get

(3.20) Re

* _n X

i=1

x_i,

m

X

k=1

e_k +

=

m

X

k=1

r_k

! _n X

i=1

kx_ik.

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By (3.19) and (3.20), we have (3.21)

* _n X

i=1

x_i,

m

X

k=1

e_k +

=

m

X

k=1

r_k

! _n X

i=1

kx_ik.

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

x_i = hPn

i=1xi,Pm k=1eki kPm

k=1e_kk²

m

X

k=1

e_k.

Finally, on utilising (3.21) and (3.23), we deduce that the condition (3.14) is necessary for the equality case in (3.13).

Before we give a corollary of the above theorem, we need to state the following lemma that has been basically obtained in [4]. For the sake of completeness, we provide a short proof here as well.

Lemma 3.3 (Dragomir, 2004). Let (H;h·,·i)be an inner product space over the real or complex number fieldKandx, a∈H, r > 0such that:

(3.24) kx−ak ≤r <kak.

Then we have the inequality

(3.25) kxk kak² −r²¹₂

≤Rehx, ai

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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 if α = q_q

p, we may state (for α = kak²−r²¹₂ , p= 1, q=kxk²) that

(3.30) 2kxk ≤ kxk²

kak²−r²¹₂

+ kak²−r²¹₂ .

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The inequality (3.25) follows now by (3.29) and (3.30).

From the above argument, it is clear that the equality holds in (3.25) if and only if it holds in (3.29) and (3.30). However, the equality holds in (3.29) if and only ifkx−ak=rand in (3.30) if and only if kak²−r²¹₂

=kxk. The proof is thus completed.

We may now state the following corollary [5].

Corollary 3.4. Let(H;h·,·i)be an inner product space over the real or complex number fieldK, e_k, x_i ∈ H\ {0},k ∈ {1, . . . , m}, i ∈ {1, . . . , n}.Ifρ_k ≥ 0, k ∈ {1, . . . , m}with

(3.31) kx_i−e_kk ≤ρ_k <ke_kk for each i∈ {1, . . . , n}andk∈ {1, . . . , m},then

(3.32)

n

X

i=1

k=1 ke_kk² −ρ²_k¹₂

n

X

i=1

x_i .

The case of equality holds in (3.32) if and only if

n

X

i=1

x_i = Pm

k=1 ke_kk²−ρ²_k¹₂ kPm

k=1e_kk²

n

X

i=1

kx_ik

! _m X

k=1

e_k.

Proof. Utilising Lemma3.3, we have from (3.31) that kxik kekk²−ρ²_k¹₂

≤Rehxi, eki

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for eachk ∈ {1, . . . , m}andi∈ {1, . . . , n}. Applying Theorem3.2for

r_k := ke_kk²−ρ²_k¹₂

, k∈ {1, . . . , m},

we deduce the desired result.

Remark 2. If{e_k}k∈{1,...,m}are orthogonal, then (3.32) becomes

(3.33)

n

X

i=1

kx_ik ≤

Pm

k=1ke_kk²¹₂ Pm

k=1 kekk²−ρ²_k¹₂

n

X

i=1

x_i with equality if and only if

n

X

i=1

x_i = Pm

k=1 ke_kk²−ρ²_k¹₂ Pm

k=1kekk²

n

X

i=1

kx_ik

! _m X

k=1

e_k.

Moreover, if{e_k}k∈{1,...,m} is assumed to be orthonormal and kxi−ekk ≤ρk fork ∈ {1, . . . , m}, i ∈ {1, . . . , n}

whereρ_k ∈[0,1)fork∈ {1, . . . , m},then

(3.34)

n

X

i=1

kx_ik ≤

√m

Pm

k=1(1−ρ²_k)¹²

n

X

i=1

xi

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with equality if and only if

n

X

i=1

x_i = Pm

k=1(1−ρ²_k)

1 2

m

n

X

i=1

kx_ik

! _m X

k=1

e_k.

The following lemma may be stated as well [3].

Lemma 3.5 (Dragomir, 2004). Let (H;h·,·i)be an inner product space over the real or complex number fieldK,x, y ∈H andM ≥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−mMkyk².

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Then (3.35) is clearly equivalent to

(3.39) kxk²

√mM +√

mM kyk² ≤ M+m

√mM Rehx, yi.

Since, obviously,

(3.40) 2kxk kyk ≤ kxk²

√mM +√

mMkyk²,

with equality iffkxk=√

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 Theorem3.2, see [5].

Corollary 3.6. Let(H;h·,·i)be an inner product space over the real or complex number field K, e_k, x_i ∈ H\ {0}, k ∈ {1, . . . , m}, i ∈ {1, . . . , n}. IfM_k >

µ_k >0, k ∈ {1, . . . , m}are such that either

(3.41) RehM_ke_k−x_i, x_i−µ_ke_ki ≥0 or, equivalently,

x_i− M_k+µ_k 2 e_k

≤ 1

2(M_k−µ_k)ke_kk for eachk ∈ {1, . . . , m}andi∈ {1, . . . , n},then

(3.42)

n

X

i=1

k=1 2·√

µkMk

µk+Mk kekk

n

X

i=1

x_i .

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n

X

i=1

xi = Pm

k=1 2·√

µkMk

µ_k+M_k ke_kk kPm

k=1e_kk²

n

X

i=1

kxik

m

X

k=1

ek.

Proof. Utilising Lemma3.5, by (3.41) we deduce 2·√

µ_kM_k

µ_k+M_k kx_ik ke_kk ≤Rehx_i, e_ki for eachk ∈ {1, . . . , m}andi∈ {1, . . . , n}.

Applying Theorem3.2for rk := 2·√

µ_kM_k

µ_k+M_k kekk, k ∈ {1, . . . , m}, we deduce the desired result.

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4. Diaz-Metcalf Inequality for Semi-Inner Products

In 1961, G. Lumer [17] introduced the following concept.

Definition 4.1. Let X be a linear space over the real or complex number field K. The mapping[·,·] :X×X →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 ∈Xandλ∈K;

(iii) [x, x]≥0for allx∈X and[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 ∈Xandλ∈K.

It is well known that the mappingX 3 x 7−→ [x, x]¹² ∈ Ris a norm onX and for anyy∈ X,the functionalX 3x7−→^ϕ^y [x, y]∈Kis a continuous linear functional onX endowed with the normk·kgenerated by[·,·].Moreover, one haskϕ_yk=kyk(see for instance [3, p. 17]).

Let (X,k·k) be a real or complex normed space. If J : X → ₂X^∗ is the normalised duality mapping defined on X,i.e., we recall that (see for instance [3, p. 1])

J(x) = {ϕ∈X^∗|ϕ(x) =kϕk kxk, kϕk=kxk}, x∈X,

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then we may state the following representation result (see for instance [3, p.

18]):

Each semi-inner product[·,·] :X×X →K that generates the normk·kof the normed linear space(X,k·k)over the real or complex number field K, is of the form

[x, y] =

DJ˜(y), x E

for any x, y ∈X,

where J˜is a selection of the normalised duality mapping andhϕ, xi := ϕ(x) forϕ ∈X^∗andx∈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 norm k·kand e ∈ X,kek = 1.If x_i ∈ 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

kx_ik ≤

n

X

i=1

x_i .

The case of equality holds in (4.2) if and only if both

(4.3)

" _n X

i=1

x_i, e

#

=r

n

X

i=1

kx_ik

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and (4.4)

" _n X

i=1

x_i, e

#

=

n

X

i=1

x_i .

The proof is obvious from the Diaz-Metcalf theorem [2, Theorem 3] applied for the continuous linear functionalF_e(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 everyx, yfromXwithx6=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 overKand [·,·]a semi- inner product generating its norm. The following statements are equivalent:

(i) (X,k·k)is strictly convex;

(ii) For everyx, y ∈ X, x, y 6= 0 with[x, y] =kxk kyk,there exists aλ > 0 such thatx=λy.

The following result may be stated.

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Corollary 4.3. Let (X,k·k) be a strictly convex normed linear space, [·,·] a semi-inner product generating the norm ande, xi(i∈ {1, . . . , n})as in Corol- lary4.1. Then the case of equality holds in (4.2) if and only if

(4.5)

n

X

i=1

x_i =r

n

X

i=1

kx_ik

! e.

Proof. If (4.5) holds true, then, obviously

n

X

i=1

x_i

=r

n

X

i=1

kx_ik

!

kek=r

n

X

i=1

kx_ik,

which is the equality case in (4.2).

Conversely, if the equality holds in (4.2), then by Corollary4.1, we have that (4.3) and (4.4) hold true. Utilising Theorem4.2, we conclude that there exists a µ > 0such that

(4.6)

n

X

i=1

x_i =µe.

Inserting this in (4.3) we get

µkek² =r

n

X

i=1

kxik

giving

(4.7) µ=r

n

X

i=1

kx_ik.

Finally, by (4.6) and (4.7) we deduce (4.5) and the corollary is proved.

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5. Other Multiplicative Reverses for m Functionals

Assume that F_k, k∈ {1, . . . , m}are bounded linear functionals defined on the normed linear spaceX.

Forp∈[1,∞),define

(cp) cp := sup

x6=0

Pm

k=1|F_k(x)|^p kxk^p

¹_p

and forp=∞,

(c∞) c∞:= sup

x6=0

1≤k≤mmax

|F_k(x)|

kxk

.

Then, by the fact that|F_k(x)| ≤ kF_kk kxk for anyx ∈ X, wherekF_kkis the norm of the functionalF_k,we have that

c_p ≤

m

X

k=1

kF_kk^p

!¹_p

, p≥1

and

c∞≤ max

1≤k≤mkF_kk.

We may now state and prove a new reverse inequality for the generalised triangle inequality in normed linear spaces.

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Theorem 5.1. Let x_i, r_k, F_k, k ∈ {1, . . . , m}, i ∈ {1, . . . , n} be as in the hypothesis of Theorem3.1. Then we have the inequalities

(5.1) (1≤)

Pn i=1kx_ik kPn

i=1x_ik ≤ c_∞

1≤k≤mmax {r_k}



≤

1≤k≤mmax kF_kk

1≤k≤mmax {r_k}



.

(5.2) Re

"

F_k

n

X

i=1

x_i

!#

=r_k

n

X

i=1

kx_ik for each k∈ {1, . . . , m}

and

(5.3) max

1≤k≤mRe

"

F_k

n

X

i=1

x_i

!#

=c∞

n

X

i=1

x_i .

Proof. Since, by the definition ofc∞,we have c∞kxk ≥ max

1≤k≤m|F_k(x)|, for any x∈X, then we can state, forx=Pn

i=1x_i,that c∞

n

X

i=1

x_i

≥ max

1≤k≤m

F_k

n

X

i=1

x_i

!

≥ max

1≤k≤m

"

ReF_k

n

X

i=1

x_i

!

# (5.4)

≥ max

1≤k≤m

"

Re

n

X

i=1

F_k(x_i)

#

= max

1≤k≤m

" _n X

i=1

ReF_k(x_i)

# .

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Utilising the hypothesis (3.1) we obviously have

1≤k≤mmax

" _n X

i=1

ReF_k(x_i)

#

≥ max

1≤k≤m{r_k} ·

n

X

i=1

kx_ik.

Also, Pn

i=1x_i 6= 0,because, by the initial assumptions, not all r_k andx_i with k ∈ {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

"

F_k

n

X

i=1

x_i

!#

= max

1≤k≤m{r_k}

n

X

i=1

x_i ,

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) ReF_k(x_i) =r_kkx_ik for each i∈ {1, . . . , n} and k ∈ {1, . . . , m}

and, from (5.4), c∞

n

X

i=1

x_i

= max

1≤k≤mRe

"

F_k

n

X

i=1

x_i

!#

,

which is (5.3).

From (5.5), on summing overi∈ {1, . . . , n},we get (5.2), and the theorem is proved.

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The following result in normed spaces also holds.

Theorem 5.2. Let xi, rk, Fk, k ∈ {1, . . . , m}, i ∈ {1, . . . , n} be as in the hypothesis of Theorem3.1. Then we have the inequality

(5.6) (1≤)

Pn i=1kx_ik kPn

i=1xik ≤ c_p (Pm

k=1r^p_k)¹^p

≤ Pm

k=1kF_kk^p Pm

k=1r_k^p ¹_p

,

wherep≥1.

(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

"

ReF_k

n

X

i=1

x_i

!#p

=c^p_p

n

X

i=1

x_i

p

.

Proof. By the definition ofc_p, p≥1,we have c^p_pkxk^p ≥

m

X

k=1

|F_k(x)|^p for any x∈X,