A REVERSE OF THE GENERALISED TRIANGLE INEQUALITY IN NORMED SPACES AND APPLICATIONS

S.S. DRAGOMIR

Abstract. A new reverse of the generalised triangle inequality that complements the classical results of Diaz and Metcalf is obtained. Applications for inner product spaces and for complex numbers are provided.

1. Introduction

In [1], 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 ≤ 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 [1]

(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 [1]

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 [1]

(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 point out a different reverse of the triangle inequality than the one obtained by Diaz-Metcalf in (1.8). Its version in inner product spaces is analysed and applications for complex numbers are given as well.

For various inequalities related to the triangle inequality, see Chapter XVII of the book [3] and the references therein.

2. A New Reverse for m Functionals

The following result may be stated.

Theorem 1. Let (X, ∥·∥) be a normed linear space over the real or complex number field 𝕂 and F_{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 constant r_{k} ≥ 0, k ∈ {1, …, m} with ∑_{k=1}^{m} r_{k} > 0 and

(2.1)  Re F_{k} (x_{i}) ≥ r_{k} ∥x_{i}∥ for each i ∈ {1, …, n} and k ∈ {1, …, m} ,

then

(2.2)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \frac{∥∑_{k=1}^{m} F_{k}∥}{∑_{k=1}^{m} r_{k}} ∥∑_{i=1}^{n} x_{i}∥ .

The case of equality holds in (2.2) if both

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

and

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

Proof. Utilising the hypothesis (2.1) and the properties of the modulus, we have

(2.5)  I := |(∑_{k=1}^{m} F_{k}) (∑_{i=1}^{n} x_{i})| ≥ |Re [(∑_{k=1}^{m} F_{k}) (∑_{i=1}^{n} x_{i})]|

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

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

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

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

Making use of (2.5) and (2.6), we deduce the desired inequality (2.2).

Now, if (2.3) and (2.4) are valid, then, obviously, the case of equality holds true in the inequality (2.2).

Conversely, if the case of equality holds in (2.2), then it must hold in all the inequalities used to prove (2.2). Therefore we have

(2.7)  Re F_{k} (x_{i}) = r_{k} ∥x_{i}∥ for each i ∈ {1, …, n}, k ∈ {1, …, m};

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

and

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

Note that, from (2.7), by summation over i and k, we get

(2.10)  Re [(∑_{k=1}^{m} F_{k}) (∑_{i=1}^{n} x_{i})] = (∑_{k=1}^{m} r_{k}) ∑_{i=1}^{n} ∥x_{i}∥ .

Since (2.8) and (2.10) imply (2.3), while (2.9) and (2.10) imply (2.4) hence the theorem is proved.

Remark 1. If the norms ∥F_{k}∥ , k ∈ {1, …, m} are easier to find, then, from (2.2), one may get the (coarser) inequality that might be more useful in practice:

(2.11)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \frac{∑_{k=1}^{m} ∥F_{k}∥}{∑_{k=1}^{m} r_{k}} ∥∑_{i=1}^{n} x_{i}∥ .

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

Theorem 2. Let (H; ⟨·, ·⟩) 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 r_{k} ≥ 0, k ∈ {1, …, m} with ∑_{k=1}^{m} r_{k} > 0 satisfy

(2.12)  Re ⟨x_{i}, e_{k}⟩ ≥ r_{k} ∥x_{i}∥ for each i ∈ {1, …, n} and k ∈ {1, …, m} ,

then

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

The case of equality holds in (2.13) if and only if

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

Proof. By the properties of inner product and by (2.12), we have

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

= ∑_{k=1}^{m} ∑_{i=1}^{n} Re ⟨x_{i}, e_{k}⟩ ≥ (∑_{k=1}^{m} r_{k}) ∑_{i=1}^{n} ∥x_{i}∥ > 0.

Observe also that, by (2.15), ∑_{k=1}^{m} e_{k} ≠ 0.

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

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

Making use of (2.15) and (2.16), we can conclude that (2.13) holds.

Now, if (2.14) holds true, then, by taking the norm, we have

∥∑_{i=1}^{n} x_{i}∥ = \begin{matrix}(∑_{k=1}^{m} r_{k}) & ∑ & {n \atop i=1} & ∥x_{i}∥ \\ ∥∑_{k=1}^{m} e_{k}∥ & 2 \end{matrix} ∥∑_{k=1}^{m} e_{k}∥

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

i.e., the case of equality holds in (2.13).

Conversely, if the case of equality holds in (2.13), then it must hold in all the inequalities used to prove (2.13). Therefore, we have

(2.17)  Re ⟨x_{i}, e_{k}⟩ = r_{k} ∥x_{i}∥ for each i ∈ {1, …, n} and k ∈ {1, …, m} ,

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

and

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

From (2.17), on summing over i and k, we get

(2.20)  Re ⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩ = (∑_{k=1}^{m} r_{k}) ∑_{i=1}^{n} ∥x_{i}∥ .

By (2.19) and (2.20), we have

(2.21)  ⟨∑_{i=1}^{n} x_{i}, ∑_{k=1}^{m} e_{k}⟩ = (∑_{k=1}^{m} r_{k}) ∑_{i=1}^{n} ∥x_{i}∥ .

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 (2.18) holds if and only if

(2.22)  ∑_{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}.

Finally, on utilising (2.21) and (2.22), we deduce that the condition (2.14) is necessary for the equality case in (2.13).

Before we give a corollary of the above theorem, we need to state the following lemma that has been basically obtained in [2]. For the sake of completeness, we provide a short proof here as well.

Lemma 1. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 and x, a ∈ H, r > 0 such that:

(2.23)  ∥x − a∥ ≤ r < ∥a∥ .

Then we have the inequality

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

or, equivalently

(2.25)  ∥x∥^{2} ∥a∥^{2} − [Re ⟨x, a⟩]^{2} ≤ r^{2} ∥x∥^{2} .

The case of equality holds in (2.24) (or in (2.25)) if and only if

(2.26)  ∥x − a∥ = r and ∥x∥^{2} + r^{2} = ∥a∥^{2} .

Proof. From the first part of (2.23), we have

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

By the second part of (2.23) we have (∥a∥^{2} − r^{2})^{\frac{1}{2}} > 0, therefore, by (2.27), we may state that

(2.28)  0 < \begin{matrix}∥x∥ & 2 \\ (∥a∥^{2} − r^{2}) & \frac{1}{2} \end{matrix} + (∥a∥^{2} − r^{2})^{\frac{1}{2}} ≤ \frac{2 Re ⟨x, a⟩}{(∥a∥^{2} − r^{2})^{\frac{1}{2}}} .

Utilising the elementary inequality

\frac{1}{α}q + αp ≥ 2\sqrt{pq}, α > 0, p > 0, q ≥ 0;

with equality if and only if α = \sqrt{\frac{q}{p}}, we may state (for α = (∥a∥^{2} − r^{2})^{\frac{1}{2}} , p = 1,

q = ∥x∥^{2}) that

(2.29)  2 ∥x∥ ≤ \begin{matrix}∥x∥ & 2 \\ (∥a∥^{2} − r^{2}) & \frac{1}{2} \end{matrix} + (∥a∥^{2} − r^{2})^{\frac{1}{2}} .

The inequality (2.24) follows now by (2.28) and (2.29).

From the above argument, it is clear that the equality holds in (2.24) if and only if it holds in (2.28) and (2.29). However, the equality holds in (2.28) if and only if

∥x − a∥ = r and in (2.29) if and only if (∥a∥^{2} − r^{2})^{\frac{1}{2}} = ∥x∥ .

The proof is thus completed.

We may now state the following corollary.

Corollary 1. Let (H; ⟨·, ·⟩) 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 ρ_{k} ≥ 0, k ∈ {1, …, m} with

(2.30) ∥x_{i} − e_{k}∥ ≤ ρ_{k} < ∥e_{k}∥ for each i ∈ {1, …, n} and k ∈ {1, …, m} ,

then

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

The case of equality holds in (2.31) if and only if

∑_{i=1}^{n} x_{i} = \begin{matrix}∑ & {m \atop k=1} & (∥e_{k}∥^{2} − ρ_{k}^{2}) & \frac{1}{2} \\ ∥∑_{k=1}^{m} e_{k}∥ & 2 \end{matrix} (∑_{i=1}^{n} ∥x_{i}∥) ∑_{k=1}^{m} e_{k}.

Proof. Utilising Lemma 1, we have from (2.30) that

∥x_{i}∥ (∥e_{k}∥^{2} − ρ_{k}^{2})^{\frac{1}{2}} ≤ Re ⟨x_{i}, e_{k}⟩

for each k ∈ {1, …, m} and i ∈ {1, …, n} .

REVERSE OF THE GENERALISED TRIANGLE INEQUALITY 7

Applying Theorem 2 for

r_{k} := (∥e_{k}∥^{2} − ρ_{k}^{2})^{\frac{1}{2}} , k ∈ \{1, …, m\} ,

we deduce the desired result. Remark 2. If {e_{k}}_{k∈{1,...,m}} are orthogonal, then (2.31) becomes

(2.32)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \begin{matrix}(∑_{k=1}^{m} ∥e_{k}∥^{2}) & \frac{1}{2} \\ ∑ & {m \atop k=1} & (∥e_{k}∥^{2} − ρ_{k}^{2}) & \frac{1}{2} \end{matrix} ∥∑_{i=1}^{n} x_{i}∥

with equality if and only if

∑_{i=1}^{n} x_{i} = \begin{matrix}∑ & {m \atop k=1} & (∥e_{k}∥^{2} − ρ_{k}^{2}) & \frac{1}{2} \\ ∑ & {m \atop k=1} & ∥e_{k}∥ & 2 \end{matrix} (∑_{i=1}^{n} ∥x_{i}∥) ∑_{k=1}^{m} e_{k}.

Moreover, if {e_{k}}_{k∈{1,...,m}} is assumed to be orthonormal and

∥x_{i} − e_{k}∥ ≤ ρ_{k} for k ∈ {1, …, m} , i ∈ {1, …, n}

where ρ_{k} ∈ [0, 1) for k ∈ {1, …, m} , then

(2.33) ∑_{i=1}^{n} ∥x_{i}∥ ≤ \begin{matrix}√ & \overline{m} \\ ∑ & {m \atop k=1} & (1 − ρ_{k}^{2}) & \frac{1}{2} \end{matrix} ∥∑_{i=1}^{n} x_{i}∥

with equality if and only if

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

The following lemma may be stated as well [2]. Lemma 2. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x, y ∈ H and M ≥ m > 0. If

(2.34)  Re ⟨My − x, x − my⟩ ≥ 0

or, equivalently,

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

then

(2.36)  ∥x∥ ∥y∥ ≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}} Re ⟨x, y⟩ .

The equality holds in (2.36) if and only if the case of equality holds in (2.34) and

(2.37)  ∥x∥ = \sqrt{mM} ∥y∥ .

Proof. Obviously,

Re ⟨My − x, x − my⟩ = (M + m) Re ⟨x, y⟩ − ∥x∥^{2} − mM ∥y∥^{2} .

Then (2.34) is clearly equivalent to

(2.38)  \begin{matrix}∥x∥ & 2 \\ √ & \overline{mM} \end{matrix} + \sqrt{mM} ∥y∥^{2} ≤ \frac{M + m}{\sqrt{mM}} Re ⟨x, y⟩ .

Since, obviously,

(2.39)  2 ∥x∥ ∥y∥ ≤ \begin{matrix}∥x∥ & 2 \\ √ & \overline{mM} \end{matrix} + \sqrt{mM} ∥y∥^{2} ,

with equality iff ∥x∥ = \sqrt{mM} ∥y∥ , hence (2.38) and (2.39) imply (2.36). The case of equality is obvious and we omit the details.

Finally, we may state the following corollary of Theorem 2.

Corollary 2. Let (H; ⟨·, ·⟩) 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_{k} > μ_{k} > 0, k ∈ {1, …, m} are such that either

(2.40)  Re ⟨M_{k}e_{k} − x_{i}, x_{i} − μ_{k}e_{k}⟩ ≥ 0

or, equivalently,

∥x_{i} − \frac{M_{k} + μ_{k}}{2} e_{k}∥ ≤ \frac{1}{2} (M_{k} − μ_{k}) ∥e_{k}∥

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

(2.41)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \frac{∥∑_{k=1}^{m} e_{k}∥}{∑_{k=1}^{m} ^{\frac{2·\sqrt{μ_{k}M_{k}}}{μ_{k}+M_{k}}} ∥e_{k}∥} ∥∑_{i=1}^{n} x_{i}∥ .

The case of equality holds in (2.41) if and only if

∑_{i=1}^{n} x_{i} = \begin{matrix}∑ & {m \atop k=1} & \frac{2·\sqrt{μ_{k}M_{k}}}{μ_{k}+M_{k}} & ∥e_{k}∥ \\ ∥∑_{k=1}^{m} e_{k}∥ & 2 \end{matrix} ∑_{i=1}^{n} ∥x_{i}∥ ∑_{k=1}^{m} e_{k}.

Proof. Utilising Lemma 2, by (2.40) we deduce

∥x_{i}∥ \frac{2 · \sqrt{μ_{k}M_{k}}}{μ_{k} + M_{k}} ∥e_{k}∥ ≤ Re ⟨x_{i}, e_{k}⟩

for each k ∈ {1, …, m} and i ∈ {1, …, n} .

Applying Theorem 2 for

r_{k} := \frac{2 · \sqrt{μ_{k}M_{k}}}{μ_{k} + M_{k}} ∥e_{k}∥ , k ∈ \{1, …, m\} ,

we deduce the desired result.

3. 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.

For p = 2, we recapture the usual modulus of a complex number, i.e.,

|z|_{2} = \sqrt{|Re z|^{2} + |Im z|^{2}} = |z| , z ∈ ℂ.

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 1 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 r_{k} ≥ 0, k ∈ {1, …, m} with ∑_{k=1}^{m} r_{k} > 0 and

(3.1)  r_{k} [|Re x_{j}| + |Im x_{j}|] ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j}

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

(3.2)  ∑_{j=1}^{n} [|Re x_{j}| + |Im x_{j}|] ≤ \frac{|∑_{k=1}^{m} a_{k}|}{∑_{k=1}^{m} r_{k}} [|∑_{j=1}^{n} Re x_{j}| + |∑_{j=1}^{n} Im x_{j}|] .

The case of equality holds in (3.2) if both

Re (∑_{k=1}^{m} a_{k}) Re (∑_{j=1}^{n} x_{j}) − Im (∑_{k=1}^{m} a_{k}) Im (∑_{j=1}^{n} x_{j})

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

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

The proof follows by Theorem 1 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 1, 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 r_{k} ≥ 0, k ∈ {1, …, m} with ∑_{k=1}^{m} r_{k} > 0 and

r_{k} max {|Re x_{j}| , |Im x_{j}|} ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j}

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

(3.3) ∑_{j=1}^{n} max \{|Re x_{j}| , |Im x_{j}|\} ≤ \sqrt{2}· \frac{|∑_{k=1}^{m} a_{k}|}{∑_{k=1}^{m} r_{k}} max \{|∑_{j=1}^{n} Re x_{j}| , |∑_{j=1}^{n} Im x_{j}|\} .

The case of equality holds in (3.3) if both

Re (∑_{k=1}^{m} a_{k}) Re (∑_{j=1}^{n} x_{j}) − Im (∑_{k=1}^{m} a_{k}) Im (∑_{j=1}^{n} x_{j})

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

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

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 1, then we can state the following proposition.

Proposition 3. Let a_{k}, x_{j} ∈ ℂ, k ∈ {1, …, m} and j ∈ {1, …, n} . If there exist the constants r_{k} ≥ 0, k ∈ {1, …, m} with ∑_{k=1}^{m} r_{k} > 0 and

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

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

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

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

The case of equality holds in (3.4) if both:

Re (∑_{k=1}^{m} a_{k}) Re (∑_{j=1}^{n} x_{j}) − Im (∑_{k=1}^{m} a_{k}) Im (∑_{j=1}^{n} x_{j})

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

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

Remark 3. 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{|∑_{k=1}^{m} a_{k}|}{∑_{k=1}^{m} r_{k}} |∑_{j=1}^{n} x_{j}|

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

r_{k} |x_{j}| ≤ Re a_{k} · Re x_{j} − Im a_{k} · Im x_{j}

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

References

[1] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces, Proceedings Amer. Math. Soc., 17(1) (1966), 88-97.

[2] 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].

[3] 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