ADDITIVE REVERSES OF THE GENERALISED TRIANGLE INEQUALITY IN NORMED SPACES

S.S. DRAGOMIR

Abstract. Some additive reverses of the generalised triangle inequality in normed linear spaces are given. Applications for complex numbers are provided as well.

1. Introduction

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

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 ∥·∥ and the vectors x_{1}, …, x_{n} satisfy the condition

(1.1)  0 ≤ r ∥x_{i}∥ ≤ Re F (x_{i}) , i ∈ {1, …, n};

then

(1.2)  r ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

where equality holds if and only if both

(1.3)  F (∑_{i=1}^{n} x_{i}) = r ∑_{i=1}^{n} ∥x_{i}∥

and

(1.4)  F (∑_{i=1}^{n} x_{i}) = ∥∑_{i=1}^{n} x_{i}∥ .

If X = H, (H; ⟨·, ·⟩) is an inner product space and F (x) = ⟨x, e⟩ , ∥e∥ = 1, then the condition (1.1) may be replaced with the simpler assumption

(1.5)  0 ≤ r ∥x_{i}∥ ≤ Re ⟨x_{i}, e⟩ ,  i = 1, …, n,

which implies the reverse of the generalised triangle inequality (1.2). In this case the equality holds in (1.2) if and only if [2]

(1.6)  ∑_{i=1}^{n} x_{i} = r (∑_{i=1}^{n} ∥x_{i}∥) e.

Date: June 21, 2004.

2000 Mathematics Subject Classification. 46B05, 46C05, 26D15, 26D10.

Key words and phrases. Triangle inequality, Reverse inequality, Normed linear spaces, Inner product spaces, Complex numbers.

Let F_{1}, …, F_{m} be linear functionals on X, each of unit norm. Let [2]

c = sup_{x≠0} [\begin{matrix}∑ & {m \atop k=1} & |F_{k} (x)| & 2 \\ ∥x∥ & 2 \end{matrix} ];

it then follows that 1 ≤ c ≤ m. Suppose the vectors x_{1}, …, x_{k} whenever x_{i} ≠ 0, satisfy

(1.7)  0 ≤ r_{k} ∥x_{i}∥ ≤ Re F_{k} (x_{i}) ,  i = 1, …, n, k = 1, …, m.

Then [2]

(1.8)  (\frac{∑_{k=1}^{m} r_{k}^{2}}{c} ) ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

where equality holds if and only if both

(1.9)  F_{k} (∑_{i=1}^{n} x_{i}) = r_{k} ∑_{i=1}^{n} ∥x_{i}∥ ,  k = 1, …, m

and

(1.10)  ∑_{k=1}^{m} [F_{k} (∑_{i=1}^{n} x_{i})]^{2} = c ∥∑_{i=1}^{n} x_{i}∥^{2} .

If X = H, an inner product space, then, for F_{k} (x) = ⟨x, e_{k}⟩ , where \{e_{k}\}_{k=\overline{1,n}} is an orthonormal family in H, i.e., ⟨e_{i}, e_{j}⟩ = δ_{ij}, i, j ∈ {1, …, k} , δ_{ij} is Kronecker delta, the condition (1.7) may be replaced by

(1.11)  0 ≤ r_{k} ∥x_{i}∥ ≤ Re ⟨x_{i}, e⟩ ,  i = 1, …, n, k = 1, …, m.

implying the following reverse of the generalised triangle inequality

(1.12)  (∑_{k=1}^{m} r_{k}^{2})^{\frac{1}{2}} ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

where the equality holds if and only if

(1.13)  ∑_{i=1}^{n} x_{i} = (∑_{i=1}^{n} ∥x_{i}∥) ∑_{k=1}^{m} r_{k}e_{k}.

The main aim of this paper is to provide some new reverse results of the generalised triangle inequality in its additive form, namely, upper bounds for the nonnegative quantity

∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥

under various assumptions for the vectors x_{i}, i ∈ {1, …, n} in a real or complex normed space (X, ∥·∥) . Applications for complex numbers are provided as well.

2. Semi-Inner Products and Diaz-Metcalf Inequality

In 1961, G. Lumer [7] introduced the following concept.

Definition 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.

It is well known that the mapping X ∋ x ⟼ [x, x]^{\frac{1}{2}} ∈ ℝ is a norm on X and for any y ∈ X, the functional X ∋ x ⟼^{φ_{y}} [x, x]^{\frac{1}{2}} ∈ 𝕂 is a continuous linear functional on X endowed with the norm ∥·∥ generated by [·, ·] . Moreover, one has ∥φ_{y}∥ = ∥y∥ (see for instance [3, p. 17]).

Let (X, ∥·∥) 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) = ∥φ∥ ∥x∥ , ∥φ∥ = ∥x∥} , 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 ∥·∥ of the normed linear space (X, ∥·∥) over the real or complex number field 𝕂, is of the form

[x, y] = ⟨J̃(y) , x⟩ for any x, y ∈ X,

where J̃ is a selection of the normalised duality mapping and ⟨φ, x⟩ := 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 1. Let (X, ∥·∥) be a normed linear space, [·, ·] : X ×X → 𝕂 a semi-inner product generating the norm ∥·∥ and e ∈ X, ∥e∥ = 1. If x_{i} ∈ X, i ∈ {1, …, n} and r ≥ 0 such that

(2.1)  r ∥x_{i}∥ ≤ Re [x_{i}, e] for each i ∈ {1, …, n} ,

then we have the inequality

(2.2)  r ∑_{i=1}^{n} ∥x_{i}∥ = ∥∑_{i=1}^{n} x_{i}∥ .

The case of equality holds in (2.2) if and only if both

(2.3)  [∑_{i=1}^{n} x_{i}, e] = r ∑_{i=1}^{n} ∥x_{i}∥

and

(2.4)  [∑_{i=1}^{n} x_{i}, e] = ∥∑_{i=1}^{n} x_{i}∥ .

The proof is obvious from the Diaz-Metcalf theorem [2, Theorem 3] applied for the continuous linear functional F_{e} (x) = [x, e] , x ∈ X.

Before we provide a simpler necessary and sufficient condition of equality in (2.2), we need to recall the concept of strictly convex normed spaces and a classical characterisation of these spaces.

Definition 2. A normed linear space (X, ∥·∥) is said to be strictly convex if for every x, y from X with x ≠ y and ∥x∥ = ∥y∥ = 1, we have ∥λx + (1 − λ) y∥ < 1 for all λ ∈ (0, 1) .

The following characterisation of strictly convex spaces is useful in what follows (see [1], [6], or [3, p. 21]).

Theorem 1. Let (X, ∥·∥) be a normed linear space over 𝕂 and [·, ·] a semi-inner product generating its norm. The following statements are equivalent:

(i) (X, ∥·∥) is strictly convex; (ii) For every x, y ∈ X, x, y ≠ 0 with [x, y] = ∥x∥ ∥y∥ , there exists a λ > 0 such

that x = λy.

The following result may be stated.

Corollary 2. Let (X, ∥·∥) be a strictly convex normed linear space, [·, ·] a semiinner product generating the norm and e, x_{i} (i ∈ {1, …, n}) as in Corollary 1. Then the case of equality holds in (2.2) if and only if

(2.5)  ∑_{i=1}^{n} x_{i} = r (∑_{i=1}^{n} ∥x_{i}∥) e.

Proof. If (2.5) holds true, then, obviously

∥∑_{i=1}^{n} x_{i}∥ = r (∑_{i=1}^{n} ∥x_{i}∥) ∥e∥ = r ∑_{i=1}^{n} ∥x_{i}∥ ,

which is the equality case in (2.2).

Conversely, if the equality holds in (2.2), then by Corollary 1, we have that (2.3) and (2.4) hold true. Utilising Theorem 1, we conclude that there exists a μ > 0 such that

(2.6)  ∑_{i=1}^{n} x_{i} = μe.

Inserting this in (2.3) we get

μ ∥e∥^{2} = r ∑_{i=1}^{n} ∥x_{i}∥

giving

(2.7)  μ = r ∑_{i=1}^{n} ∥x_{i}∥ .

Finally, by (2.6) and (2.7) we deduce (2.5) and the corollary is proved.

3. An Additive Reverse for the Triangle Inequality

In the following we provide an alternative of the Diaz-Metcalf reverse of the generalised triangle inequality.

Theorem 2. Let (X, ∥·∥) 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)| ≤ ∥x∥ for any x ∈ X (i.e., ∥F ∥ = 1, we say that F is of unit norm). If x_{i} ∈ X, k_{i} ≥ 0, i ∈ {1, …, n} are such that

(3.1)  (0 ≤) ∥x_{i}∥ − Re F (x_{i}) ≤ k_{i} for each i ∈ {1, …, n} ,

then we have the inequality

(3.2)  (0 ≤) ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥ ≤ ∑_{i=1}^{n} k_{i}.

The equality holds in (3.2) if and only if both

(3.3)  F (∑_{i=1}^{n} x_{i}) = ∥∑_{i=1}^{n} x_{i}∥ and F (∑_{i=1}^{n} x_{i}) = ∑_{i=1}^{n} ∥x_{i}∥ − ∑_{i=1}^{n} k_{i}.

Proof. If we sum in (3.1) over i from 1 to n, then we get

(3.4)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ Re [F (∑_{i=1}^{n} x_{i})] + ∑_{i=1}^{n} k_{i}.

Taking into account that |F (x)| ≤ ∥x∥ for each x ∈ X, then we may state that

(3.5)  Re [F (∑_{i=1}^{n} x_{i})] ≤ |Re F (∑_{i=1}^{n} x_{i})| ≤ |F (∑_{i=1}^{n} x_{i})| ≤ ∥∑_{i=1}^{n} x_{i}∥ .

Now, making use of (3.4) and (3.5), we deduce (3.2).

Obviously, if (3.3) is valid, then the case of equality in (3.2) holds true.

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

∑_{i=1}^{n} ∥x_{i}∥ = Re [F (∑_{i=1}^{n} x_{i})] + ∑_{i=1}^{n} k_{i}

and

Re [F (∑_{i=1}^{n} x_{i})] = |F (∑_{i=1}^{n} x_{i})| = ∥∑_{i=1}^{n} x_{i}∥ ,

which imply (3.3).

The following corollary may be stated.

Corollary 3. Let (X, ∥·∥) be a normed linear space, [·, ·] : X × X → 𝕂 a semiinner product generating the norm ∥·∥ and e ∈ X, ∥e∥ = 1. If x_{i} ∈ X, k_{i} ≥ 0, i ∈ {1, …, n} are such that

(3.6)  (0 ≤) ∥x_{i}∥ − Re [x_{i}, e] ≤ k_{i} for each i ∈ {1, …, n} ,

then we have the inequality

(3.7)  (0 ≤) ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥ ≤ ∑_{i=1}^{n} k_{i}.

The equality holds in (3.7) if and only if both

(3.8)  [∑_{i=1}^{n} x_{i}, e] = ∥∑_{i=1}^{n} x_{i}∥ and [∑_{i=1}^{n} x_{i}, e] = ∑_{i=1}^{n} ∥x_{i}∥ − ∑_{i=1}^{n} k_{i}.

Moreover, if (X, ∥·∥) is strictly convex, then the case of equality holds in (3.7) if and only if

(3.9)  ∑_{i=1}^{n} ∥x_{i}∥ ≥ ∑_{i=1}^{n} k_{i}

and

(3.10)  ∑_{i=1}^{n} x_{i} = (∑_{i=1}^{n} ∥x_{i}∥ − ∑_{i=1}^{n} k_{i}) · e.

Proof. The first part of the corollary is obvious by Theorem 2 applied for the continuous linear functional of unit norm F_{e}, F_{e} (x) = [x, e] , x ∈ X. The second part may be shown on utilising a similar argument to the one from the proof of Corollary 2. We omit the details.

Remark 1. If X = H, (H; ⟨·, ·⟩) is an inner product space, then from Corollary 3 we deduce the additive reverse inequality obtained in Theorem 7 of [4]. For further similar results in inner product spaces, see [4] and [5].

4. Reverse Inequalities for m Functionals

The following result generalising Theorem 2 may be stated.

Theorem 3. Let (X, ∥·∥) be a normed linear space over the real or complex number field 𝕂. If F_{k}, k ∈ {1, …, m} are bounded linear functionals defined on X and x_{i} ∈ X, M_{ik} ≥ 0 for i ∈ {1, …, n}, k ∈ {1, …, m} such that

(4.1)  ∥x_{i}∥ − Re F_{k} (x_{i}) ≤ M_{ik} for each i ∈ {1, …, n} , k ∈ {1, …, m} ,

then we have the inequality

(4.2)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥\frac{1}{m} ∑_{k=1}^{m} F_{k}∥ ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}.

The case of equality holds in (4.2) if both

(4.3)  \frac{1}{m} ∑_{k=1}^{m} F_{k} (∑_{i=1}^{n} x_{i}) = ∥\frac{1}{m} ∑_{k=1}^{m} F_{k}∥ ∥∑_{i=1}^{n} x_{i}∥

and

(4.4)  \frac{1}{m} ∑_{k=1}^{m} F_{k} (∑_{i=1}^{n} x_{i}) = ∑_{i=1}^{n} ∥x_{i}∥ − \frac{1}{m} ∑_{k=1}^{m} ∑_{j=1}^{n} M_{ik}.

Proof. If we sum (4.1) over i from 1 to n, then we deduce

∑_{i=1}^{n} ∥x_{i}∥ − Re F_{k} (∑_{i=1}^{n} x_{i}) ≤ ∑_{i=1}^{n} M_{ik}

for each k ∈ {1, …, m} .

Summing these inequalities over k from 1 to m, we deduce

(4.5)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \frac{1}{m} ∑_{k=1}^{m} Re F_{k} (∑_{i=1}^{n} x_{i}) + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}.

Utilising the continuity property of the functionals F_{k} and the properties of the modulus, we have

(4.6)  ∑_{k=1}^{m} Re F_{k} (∑_{i=1}^{n} x_{i}) ≤ |∑_{k=1}^{m} Re F_{k} (∑_{i=1}^{n} x_{i})|

≤ |∑_{k=1}^{m} F_{k} (∑_{i=1}^{n} x_{i})| ≤ ∥∑_{k=1}^{m} F_{k}∥ ∥∑_{i=1}^{n} x_{i}∥ .

Now, by (4.5) and (4.6), we deduce (4.2).

Obviously, if (4.3) and (4.4) hold true, then the case of equality is valid in (4.2). Conversely, if the case of equality holds in (4.2), then it must hold in all the

inequalities used to prove (4.2). Therefore we have

∑_{i=1}^{n} ∥x_{i}∥ = \frac{1}{m} ∑_{k=1}^{m} Re F_{k} (∑_{i=1}^{n} x_{i}) + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik},

∑_{k=1}^{m} Re F_{k} (∑_{i=1}^{n} x_{i}) = ∥∑_{k=1}^{m} F_{k}∥ ∥∑_{i=1}^{n} x_{i}∥

and

∑_{k=1}^{m} Im F_{k} (∑_{i=1}^{n} x_{i}) = 0.

These imply that (4.3) and (4.4) hold true, and the theorem is completely proved.

Remark 2. If F_{k}, k ∈ {1, …, m} are of unit norm, then, from (4.2), we deduce the inequality

(4.7)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik},

which is obviously coarser than (4.2), but perhaps more useful for applications.

The case of inner product spaces, in which we may provide a simpler condition of equality, is of interest in applications.

Theorem 4. Let (X, ∥·∥) 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} . If M_{ik} ≥ 0 for i ∈ {1, …, n} , {1, …, n} such that

(4.8)  ∥x_{i}∥ − Re ⟨x_{i}, e_{k}⟩ ≤ M_{ik} for each i ∈ {1, …, n} , k ∈ {1, …, m} ,

then we have the inequality

(4.9)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥\frac{1}{m} ∑_{k=1}^{m} e_{k}∥ ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}.

The case of equality holds in (4.9) if and only if

(4.10)  ∑_{i=1}^{n} ∥x_{i}∥ ≥ \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}

and

(4.11)  ∑_{i=1}^{n} x_{i} = \frac{m (∑_{i=1}^{n} ∥x_{i}∥ − \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik})}{∥∑_{k=1}^{m} e_{k}∥^{2}} ∑_{k=1}^{m} e_{k}.

Proof. As in the proof of Theorem 3, we have

(4.12)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ Re ⟨\frac{1}{m} ∑_{k=1}^{m} e_{k}, ∑_{i=1}^{n} x_{i}⟩ + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik},

and ∑_{k=1}^{m} e_{k} ≠ 0.

On utilising the Schwarz inequality in the inner product space (H; ⟨·, ·⟩) for ∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}, we have

(4.13)  ∥∑_{i=1}^{n} x_{i}∥ ∥∑_{k=1}^{m} e_{k}∥ ≥ |⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩| ≥ |Re ⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩|

≥ Re ⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩ .

By (4.12) and (4.13) we deduce (4.9).

Taking the norm in (4.11) and using (4.10), we have

∥∑_{i=1}^{n} x_{i}∥ = \frac{m (∑_{i=1}^{n} ∥x_{i}∥ − \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik})}{∥∑_{k=1}^{m} e_{k}∥},

showing that the equality holds in (4.9).

Conversely, if the case of equality holds in (4.9), then it must hold in all the inequalities used to prove (4.9). Therefore we have

(4.14)  ∥x_{i}∥ = Re ⟨x_{i}, e_{k}⟩ + M_{ik} for each i ∈ {1, …, n} , k ∈ {1, …, m} ,

(4.15)  ∥∑_{i=1}^{n} x_{i}∥ ∥∑_{k=1}^{m} e_{k}∥ = |⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩|

and

(4.16)  Im ⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩ = 0.

From (4.14), on summing over i and k, we get

(4.17)  Re ⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩ = m ∑_{i=1}^{n} ∥x_{i}∥ − ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}.

On the other hand, by the use of the following identity in inner product spaces,

∥u − \frac{⟨u, v⟩ v}{∥v∥^{2}} ∥^{2} = \frac{∥u∥^{2} ∥v∥^{2} − |⟨u, v⟩|^{2}}{∥v∥^{2}} , v ≠ 0;

the relation (4.15) holds if and only if

∑_{i=1}^{n} x_{i} = \frac{⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩}{∥∑_{k=1}^{m} e_{k}∥^{2}} ∑_{k=1}^{m} e_{k},

giving, from (4.16) and (4.17), that

∑_{i=1}^{n} x_{i} = \frac{m ∑_{i=1}^{n} ∥x_{i}∥ − ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}}{∥∑_{k=1}^{m} e_{k}∥^{2}} ∑_{k=1}^{m} e_{k}.

If the inequality holds in (4.9), then obviously (4.10) is valid, and the theorem is proved.

Remark 3. If in the above theorem the vectors \{e_{k}\}_{k=\overline{1,m}} are assumed to be orthogonal, then (4.9) becomes:

(4.18)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \frac{1}{m} (∑_{k=1}^{m} ∥e_{k}∥^{2})^{\frac{1}{2}} ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik}.

Moreover, if \{e_{k}\}_{k=\overline{1,m}} is an orthonormal family, then (4.18) becomes

(4.19)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \frac{\sqrt{m}}{m} ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{m} ∑_{k=1}^{m} ∑_{i=1}^{n} M_{ik},

which has been obtained in [5].

Before we provide some natural consequences of Theorem 4, we need some preliminary results concerning reverses of Schwarz’s inequality in inner product spaces (see for instance [4, p. 27]).

Lemma 1. Let (X, ∥·∥) be an inner product space over the real or complex number field 𝕂 and x, a ∈ H, r > 0. If ∥x − a∥ ≤ r, then we have the inequality

(4.20)  ∥x∥ ∥a∥ − Re ⟨x, a⟩ ≤ \frac{1}{2}r^{2}.

The case of equality holds in (4.20) if and only if

(4.21)  ∥x − a∥ = r and ∥x∥ = ∥a∥ .

Proof. The condition ∥x − a∥ ≤ r is clearly equivalent to

(4.22)  ∥x∥^{2} + ∥a∥^{2} ≤ 2 Re ⟨x, a⟩ + r^{2}.

Since

(4.23)  2 ∥x∥ ∥a∥ ≤ ∥x∥^{2} + ∥a∥^{2} ,

with equality if and only if ∥x∥ = ∥a∥ , hence by (4.22) and (4.23) we deduce (4.20). The case of equality is obvious.

Utilising the above lemma we may state the following corollary of Theorem 4.

Corollary 4. Let (H; ⟨·, ·⟩) , e_{k}, x_{i} be as in Theorem 4. If r_{ik} > 0, i ∈ {1, …, n} , k ∈ {1, …, m} such that

(4.24)  ∥x_{i} − e_{k}∥ ≤ r_{ik} for each i ∈ {1, …, n} and k ∈ {1, …, m} ,

then we have the inequality

(4.25)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥\frac{1}{m} ∑_{k=1}^{m} e_{k}∥ ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{2m} ∑_{k=1}^{m} ∑_{i=1}^{n} r_{ik}^{2}.

The equality holds in (4.25) if and only if

∑_{i=1}^{n} ∥x_{i}∥ ≥ \frac{1}{2m} ∑_{k=1}^{m} ∑_{i=1}^{n} r_{ik}^{2}

and

∑_{i=1}^{n} x_{i} = \frac{m (∑_{i=1}^{n} ∥x_{i}∥ − \frac{1}{2m} ∑_{k=1}^{m} ∑_{i=1}^{n} r_{ik}^{2})}{∥∑_{k=1}^{m} e_{k}∥^{2}} ∑_{k=1}^{m} e_{k}.

The following lemma may provide another sufficient condition for (4.8) to hold (see also [4, p. 28]).

Lemma 2. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 and x, y ∈ H, M ≥ m > 0. If either

(4.26)  Re ⟨My − x, x − my⟩ ≥ 0

or, equivalently,

(4.27)  ∥x − \frac{m + M}{2} y∥ ≤ \frac{1}{2} (M − m) ∥y∥ ,

holds, then

(4.28)  ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{4} · \frac{(M − m)^{2}}{m + M} ∥y∥^{2} .

The case of equality holds in (4.28) if and only if the equality case is realised in (4.26) and

∥x∥ = \frac{M + m}{2} ∥y∥ .

The proof is obvious by Lemma 1 for a = \frac{M+m}{2}y and r = \frac{1}{2} (M − m) ∥y∥ .

Finally, the following corollary of Theorem 4 may be stated.

Corollary 5. Assume that (H, ⟨·, ·⟩) , e_{k}, x_{i} are as in Theorem 4. If M_{ik} ≥ m_{ik} > 0 satisfy the condition

Re ⟨M_{k}e_{k} − x_{i}, x_{i} − μ_{k}e_{k}⟩ ≥ 0 for each i ∈ {1, …, n} and k ∈ {1, …, m} ,

then

∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥\frac{1}{m} ∑_{k=1}^{m} e_{k}∥ ∥∑_{i=1}^{n} x_{i}∥ + \frac{1}{4m} ∑_{k=1}^{m} ∑_{i=1}^{n} \frac{(M_{ik} − m_{ik})^{2}}{M_{ik} + m_{ik}} ∥e_{k}∥^{2} .

5. Applications for Complex Numbers

Let ℂ be the field of complex numbers. If z = Re z + i Im z, then by |·|_{p} : ℂ → [0, ∞), p ∈ [1, ∞] we define the p−modulus of z as

|z|_{p} := \{ \begin{matrix}max \{|Re z| , |Im z|\} & if p = ∞, \\ (|Re z|^{p} + |Im z|^{p})^{\frac{1}{p}} & if p ∈ [1, ∞), \end{matrix}

where |a| , a ∈ ℝ is the usual modulus of the real number a. Obviously, for p = 2, we recapture the usual modulus of a complex number.

It is well known that (ℂ, |·|_{p}) , p ∈ [1, ∞] is a Banach space over the complex number field ℂ.

Consider the Banach space (ℂ, |·|_{1}) and F : ℂ → ℂ, F (z) = az with a ∈ ℂ, a ≠ 0. Obviously, F is linear on ℂ. For z ≠ 0, we have

\frac{|F (z)|}{|z|_{1}} = \begin{matrix}|a| & |z| \\ |z| & 1 \end{matrix} = \frac{|a| \sqrt{|Re z|^{2} + |Im z|^{2}}}{|Re z| + |Im z|} ≤ |a| .

Since, for z_{0} = 1, we have |F (z_{0})| = |a| and |z_{0}|_{1} = 1, hence

∥F ∥_{1} := sup_{z≠0} \frac{|F (z)|}{|z|_{1}} = |a| ,

showing that F is a bounded linear functional on (ℂ, |·|_{1}) and ∥F ∥_{1} = |a| .

We can apply Theorem 3 to state the following reverse of the generalised triangle inequality for complex numbers.

Proposition 1. Let a_{k}, x_{j} ∈ ℂ, k ∈ {1, …, m} and j ∈ {1, …, n} . If there exist the constants M_{jk} ≥ 0, k ∈ {1, …, m} , j ∈ {1, …, n} such that

(5.1)  |Re x_{j}| + |Im x_{j}| ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j} + M_{jk}

for each j ∈ {1, …, n} and k ∈ {1, …, m} , then

(5.2)  ∑_{j=1}^{n} [|Re x_{j}| + |Im x_{j}|]

≤ \frac{1}{m} |∑_{k=1}^{m} a_{k}| [|∑_{j=1}^{n} Re x_{j}| + |∑_{j=1}^{n} Im x_{j}|] + \frac{1}{m} ∑_{k=1}^{m} ∑_{j=1}^{n} M_{jk}.

The proof follows by Theorem 3 applied for the Banach space (ℂ, |·|_{1}) and F_{k} (z) = a_{k}z, k ∈ {1, …, m} on taking into account that:

∥∑_{k=1}^{m} F_{k}∥_{1} = |∑_{k=1}^{m} a_{k}| .

Now, consider the Banach space (ℂ, |·|_{∞}) . If F (z) = dz, then for z ≠ 0 we have

\frac{|F (z)|}{|z|_{∞}} = \begin{matrix}|d| & |z| \\ |z| & ∞ \end{matrix} = |d| \sqrt{|Re z|^{2} + |Im z|^{2}} ; max \{|Re z| , |Im z|\} ≤ \sqrt{2} |d| .

Since, for z_{0} = 1 + i, we have |F (z_{0})| = \sqrt{2} |d| , |z_{0}|_{∞} = 1, hence

∥F ∥_{∞} := sup_{z≠0} \frac{|F (z)|}{|z|_{∞}} = \sqrt{2} |d| ,

showing that F is a bounded linear functional on (ℂ, |·|_{∞}) and ∥F ∥_{∞} = \sqrt{2} |d| .

If we apply Theorem 3, then we can state the following reverse of the generalised triangle inequality for complex numbers.

Proposition 2. Let a_{k}, x_{j} ∈ ℂ, k ∈ {1, …, m} and j ∈ {1, …, n} . If there exist the constants M_{jk} ≥ 0, k ∈ {1, …, m} , j ∈ {1, …, n} such that

max {|Re x_{j}| , |Im x_{j}|} ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j} + M_{jk}

for each j ∈ {1, …, n} and k ∈ {1, …, m} , then

(5.3) ∑_{j=1}^{n} max {|Re x_{j}| , |Im x_{j}|}

≤ \frac{\sqrt{2}}{m} |∑_{k=1}^{m} a_{k}| max \{|∑_{j=1}^{n} Re x_{j}| , |∑_{j=1}^{n} Im x_{j}|\} + \frac{1}{m} ∑_{k=1}^{m} ∑_{j=1}^{n} M_{jk}.

Finally, consider the Banach space (ℂ, |·|_{2p}) with p ≥ 1.

Let F : ℂ → ℂ, F (z) = cz. By Hölder’s inequality, we have

\frac{|F (z)|}{|z|_{2p}} = |c| \sqrt{|Re z|^{2} + |Im z|^{2}} ; (|Re z|^{2p} + |Im z|^{2p})^{\frac{1}{2p}} ≤ 2^{\frac{1}{2} − \frac{1}{2p}} |c| .

Since, for z_{0} = 1 + i we have |F (z_{0})| = 2^{\frac{1}{2}} |c| , |z_{0}| = 2^{\frac{1}{2p}} (p ≥ 1) , hence

∥F ∥_{2p} := sup_{z≠0} \frac{|F (z)|}{|z|_{2p}} = 2^{\frac{1}{2} − \frac{1}{2p}} |c| ,

showing that F is a bounded linear functional on (ℂ, |·|_{2p}) , p ≥ 1 and ∥F ∥_{2p} =

2^{\frac{1}{2} − \frac{1}{2p}} |c| .

If we apply Theorem 3, then we can state the following proposition.

Proposition 3. Let a_{k}, x_{j}, M_{jk} be as in Proposition 2. If

[|Re x_{j}|^{2p} + |Im x_{j}|^{2p}]^{\frac{1}{2p}} ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j} + M_{jk}

for each j ∈ {1, …, n} and k ∈ {1, …, m} , then

(5.4) ∑_{j=1}^{n} [|Re x_{j}|^{2p} + |Im x_{j}|^{2p}]^{\frac{1}{2p}}

≤ \frac{2^{\frac{1}{2} − \frac{1}{2p}}}{m} |∑_{k=1}^{m} a_{k}| [|∑_{j=1}^{n} Re x_{j}|^{2p} + |∑_{j=1}^{n} Im x_{j}|^{2p}]^{\frac{1}{2p}} + \frac{1}{m} ∑_{k=1}^{m} ∑_{j=1}^{n} M_{jk}.

where p ≥ 1.

Remark 4. If in the above proposition we choose p = 1, then we have the following reverse of the generalised triangle inequality for complex numbers

∑_{j=1}^{n} |x_{j}| ≤ |\frac{1}{m} ∑_{k=1}^{m} a_{k}| |∑_{j=1}^{n} x_{j}| + \frac{1}{m} ∑_{k=1}^{m} ∑_{j=1}^{n} M_{jk}

provided x_{j}, a_{k}, j ∈ {1, …, n}, k ∈ {1, …, m} satisfy the assumption

|x_{j}| ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j} + M_{jk}

for each j ∈ {1, …, n}, k ∈ {1, …, m} . Here |·| is the usual modulus of a complex number and M_{jk} > 0, j ∈ {1, …, n}, k ∈ {1, …, m} are given.

References

[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, Preprint [Online http://front.math.ucdavis.edu/math.FA/0309354].

[5] 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].

[6] S. GUDDER and D. STRAWTHER, Strictly convex normed linear spaces, Proc. Amer. Math. Soc., 59(2) (1976), 263-267.

[7] G. LUMER, Semi-inner product spaces, Trans. Amer. Math. Soc., 100 (1961), 29-43.

[8] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

School of Computer Science and Mathematics, Victoria University of Technology, PO Box 14428, MCMC 8001, Victoria, Australia.

   E-mail address: sever.dragomir@vu.edu.au
   URL: http://rgmia.vu.edu.au/SSDragomirWeb.html