ON SOME INEQUALITIES IN NORMED ALGEBRAS

S.S. DRAGOMIR

Abstract. Some inequalities in normed algebras that provides lower and upper bounds for the norm of ∑_{j=1}^{n} a_{j}x_{j} are obtained. Applications for estimating the quantities ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ and ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥ for invertible elements x, y in unital normed algebras are also given.

1. Introduction

In [1], in order to provide a generalisation of a norm inequality for n vectors in a normed linear space obtained by Pečarić and Rajić in [2], the author obtained the following result:

(1.1) max_{k∈{1,...,n}} {|α_{k}| ∥∑_{j=1}^{n} x_{j}∥ − ∑_{j=1}^{n} |α_{j} − α_{k}| ∥x_{j}∥}

≤ ∥∑_{j=1}^{n} α_{j}x_{j}∥ ≤ min_{k∈{1,...,n}} {|α_{k}| ∥∑_{j=1}^{n} x_{j}∥ + ∑_{j=1}^{n} |α_{j} − α_{k}| ∥x_{j}∥} ,

where x_{j}, j ∈ {1, …, n} are vectors in the normed linear space (X, ∥·∥) over 𝕂 while α_{j}, j ∈ {1, …, n} are scalars in 𝕂 (𝕂 = ℂ, ℝ) .

For α_{k} = \frac{1}{∥x_{k}∥}, with x_{k} ≠ 0, k ∈ \{1, …, n\} the above inequality produces the following result established by Pečarić and Rajić in [2]:

(1.2) max_{k∈\{1,...,n\}} \{\frac{1}{∥x_{k}∥} [∥∑_{j=1}^{n} x_{j}∥ − ∑_{j=1}^{n} |∥x_{j}∥ − ∥x_{k}∥|]\}

≤ ∥∑_{j=1}^{n} \frac{x_{j}}{∥x_{j}∥}∥ ≤ min_{k∈\{1,...,n\}} \{\frac{1}{∥x_{k}∥} [∥∑_{j=1}^{n} x_{j}∥ + ∑_{j=1}^{n} |∥x_{j}∥ − ∥x_{k}∥|]\} ,

which implies the following refinement and reverse of the generalised triangle inequality due to M. Kato et al. [3]:

(1.3) min_{k∈\{1,...,n\}} \{∥x_{k}∥\} [n − ∥∑_{j=1}^{n} \frac{x_{j}}{∥x_{j}∥}∥]

Date: 31 July, 2007.

2000 Mathematics Subject Classification. Primary 46H05; Secondary 47A10.

Key words and phrases. Normed algebras, Unital algebras, Gneralised triangle inequality.

≤ ∑_{j=1}^{n} ∥x_{j}∥ − ∥∑_{j=1}^{n} x_{j}∥ ≤ max_{k∈\{1,...,n\}} \{∥x_{k}∥\} [n − ∥∑_{j=1}^{n} \frac{x_{j}}{∥x_{j}∥}∥] .

The other natural choice, α_{k} = ∥x_{k}∥ , k ∈ {1, …, n} in (1.1) produces the result

(1.4) max_{k∈{1,...,n}} {∥x_{k}∥ ∥∑_{j=1}^{n} x_{j}∥ − ∑_{j=1}^{n} |∥x_{j}∥ − ∥x_{k}∥| ∥x_{j}∥}

≤ ∥∑_{j=1}^{n} ∥x_{j}∥ x_{j}∥ ≤ min_{k∈{1,...,n}} {∥x_{k}∥ ∥∑_{j=1}^{n} x_{j}∥ + ∑_{j=1}^{n} |∥x_{j}∥ − ∥x_{k}∥| ∥x_{j}∥} ,

which in its turn implies another refinement and reverse of the generalised triangle inequality:

(1.5)  (0 ≤)\frac{∑_{j=1}^{n} ∥x_{j}∥^{2} − ∥∑_{j=1}^{n} ∥x_{j}∥ x_{j}∥}{max_{k∈\{1,...,n\}} \{∥x_{k}∥\}}

≤ ∑_{j=1}^{n} ∥x_{j}∥ − ∥∑_{j=1}^{n} x_{j}∥ ≤ \frac{∑_{j=1}^{n} ∥x_{j}∥^{2} − ∥∑_{j=1}^{n} ∥x_{j}∥ x_{j}∥}{min_{k∈\{1,...,n\}} \{∥x_{k}∥\}},

provided x_{k} ≠ 0, k ∈ {1, …, n} .

In [2], the authors have shown that the case n = 2 in (1.2) produces the Maligranda-Mercer inequality:

(1.6)  \frac{∥x − y∥ − |∥x∥ − ∥y∥|}{min \{∥x∥ , ∥y∥\}} ≤ ∥\frac{x}{∥x∥} − \frac{y}{∥y∥}∥ ≤ \frac{∥x − y∥ + |∥x∥ − ∥y∥|}{max \{∥x∥ , ∥y∥\}},

for any x, y ∈ X\ {0} .

We notice that Maligranda proved the right inequality in [5] while Mercer proved the left inequality in [4].

We have shown in [1] that the following dual result for two vectors is also valid:

(1.7)  (0 ≤)\frac{∥x − y∥}{min \{∥x∥ , ∥y∥\}} − \frac{|∥x∥ − ∥y∥|}{max \{∥x∥ , ∥y∥\}}

≤ ∥\frac{x}{∥y∥} − \frac{y}{∥x∥}∥ ≤ \frac{∥x − y∥}{max \{∥x∥ , ∥y∥\}} + \frac{|∥x∥ − ∥y∥|}{min \{∥x∥ , ∥y∥\}},

for any x, y ∈ X\ {0} .

Motivated by the above results, the aim of the present paper is to establish lower and upper bounds for the norm of ∑_{j=1}^{n} a_{j}x_{j}, where a_{j}, x_{j}, j ∈ {1, …, n} are elements in a normed algebra (A, ∥·∥) over the real or complex number field 𝕂. In the case where (A, ∥·∥) is a unital algebra and x, y are invertible, lower and upper bounds for the quantities

∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ and ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥

are provided as well.

2. Inequalities for n Pairs of Elements

Let (A, ∥·∥) be a normed algebra over the real or complex number field 𝕂.

Theorem 1. If (a_{j}, x_{j}) ∈ A^{2}, j ∈ {1, …, n} , then

max_{k∈{1,...,n}} {∥a_{k} (∑_{j=1}^{n} x_{j})∥ − ∑_{j=1}^{n} ∥a_{j} − a_{k}∥ ∥x_{j}∥} (2.1)

≤ max_{k∈{1,...,n}} {∥a_{k} (∑_{j=1}^{n} x_{j})∥ − ∑_{j=1}^{n} ∥(a_{j} − a_{k}) x_{j}∥}

≤ ∥∑_{j=1}^{n} a_{j}x_{j}∥

≤ min_{k∈{1,...,n}} {∥a_{k} (∑_{j=1}^{n} x_{j})∥ + ∑_{j=1}^{n} ∥(a_{j} − a_{k}) x_{j}∥}

≤ min_{k∈{1,...,n}} {∥a_{k} (∑_{j=1}^{n} x_{j})∥ + ∑_{j=1}^{n} ∥a_{j} − a_{k}∥ ∥x_{j}∥} .

Proof. Observe that for any k ∈ {1, …, n} we have

∑_{j=1}^{n} a_{j}x_{j} = a_{k} (∑_{j=1}^{n} x_{j}) + ∑_{j=1}^{n} (a_{j} − a_{k}) x_{j}.

Taking the norm and utilising the triangle inequality and the normed algebra properties, we have

∥∑_{j=1}^{n} a_{j}x_{j}∥ ≤ ∥a_{k} (∑_{j=1}^{n} x_{j})∥ + ∥∑_{j=1}^{n} (a_{j} − a_{k}) x_{j}∥

≤ ∥a_{k} (∑_{j=1}^{n} x_{j})∥ + ∑_{j=1}^{n} ∥(a_{j} − a_{k}) x_{j}∥

≤ ∥a_{k} (∑_{j=1}^{n} x_{j})∥ + ∑_{j=1}^{n} ∥a_{j} − a_{k}∥ ∥x_{j}∥ ,

for any k ∈ {1, …, n} , which implies the second part in (2.1).

Observing that

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

and utilising the continuity of the norm, we have

∥∑_{j=1}^{n} a_{j}x_{j}∥ ≥ |∥a_{k} (∑_{j=1}^{n} x_{j}) − ∑_{j=1}^{n} (a_{k} − a_{j}) x_{j}∥|

≥ ∥a_{k} (∑_{j=1}^{n} x_{j})∥ − ∥∑_{j=1}^{n} (a_{k} − a_{j}) x_{j}∥

≥ ∥a_{k} (∑_{j=1}^{n} x_{j})∥ − ∑_{j=1}^{n} ∥(a_{k} − a_{j}) x_{j}∥

≥ ∥a_{k} (∑_{j=1}^{n} x_{j})∥ − ∑_{j=1}^{n} ∥a_{k} − a_{j}∥ ∥x_{j}∥

for any k ∈ {1, …, n} , which implies the first part in (2.1).

Remark 1. If there exists r > 0 so that ∥a_{j} − a_{k}∥ ≤ r ∥a_{k}∥ for any j, k ∈ {1, …, n} , then, by the second part of (2.1), we have

(2.2)  ∥∑_{j=1}^{n} a_{j}x_{j}∥ ≤ min_{k∈{1,...,n}} {∥a_{k}∥} [∥∑_{j=1}^{n} x_{j}∥ + r ∑_{j=1}^{n} ∥x_{j}∥] .

Corollary 1. If x_{j} ∈ A, j ∈ {1, …, n} , then

max_{k∈{1,...,n}} {∥x_{k} (∑_{j=1}^{n} x_{j})∥ − ∑_{j=1}^{n} ∥x_{j} − x_{k}∥ ∥x_{j}∥} (2.3)

≤ max_{k∈{1,...,n}} {∥x_{k} (∑_{j=1}^{n} x_{j})∥ − ∑_{j=1}^{n} ∥(x_{j} − x_{k}) x_{j}∥} ≤ ∥∑_{j=1}^{n} x_{j}^{2}∥

≤ min_{k∈{1,...,n}} {∥x_{k} (∑_{j=1}^{n} x_{j})∥ + ∑_{j=1}^{n} ∥(x_{j} − x_{k}) x_{j}∥}

≤ min_{k∈{1,...,n}} {∥x_{k} (∑_{j=1}^{n} x_{j})∥ + ∑_{j=1}^{n} ∥x_{j} − x_{k}∥ ∥x_{j}∥} .

Corollary 2. Assume that A is a unital normed algebra. If x_{j} ∈ A are invertible for any j ∈ {1, …, n} , then

(2.4)  min_{k∈{1,...,n}} ∥x_{k}^{−1}∥ [∑_{j=1}^{n} ∥x_{j}∥ − ∥∑_{j=1}^{n} x_{j}∥]

≤ ∑_{j=1}^{n} ∥x_{j}^{−1}∥ ∥x_{j}∥ − ∥∑_{j=1}^{n} ∥x_{j}^{−1}∥ x_{j}∥

≤ max_{k∈{1,...,n}} ∥x_{k}^{−1}∥ [∑_{j=1}^{n} ∥x_{j}∥ − ∥∑_{j=1}^{n} x_{j}∥] .

Proof. If 1 ∈ A is the unity, then on choosing a_{k} = ∥x_{k}^{−1}∥ · 1 in (2.1) we get

max_{k∈{1,...,n}} {∥x_{k}^{−1}∥ ∥∑_{j=1}^{n} x_{j}∥ − ∑_{j=1}^{n} |∥x_{j}^{−1}∥ − ∥x_{k}^{−1}∥| ∥x_{j}∥} (2.5)

≤ ∥∑_{j=1}^{n} ∥x_{j}^{−1}∥ x_{j}∥

≤ min_{k∈{1,...,n}} {∥x_{k}^{−1}∥ ∥∑_{j=1}^{n} x_{j}∥ + ∑_{j=1}^{n} |∥x_{j}^{−1}∥ − ∥x_{k}^{−1}∥| ∥x_{j}∥} .

Now, assume that min_{k∈{1,...,n}} {∥x_{k}^{−1}∥} = ∥x_{k_{0}}^{−1}∥ . Then

∥x_{k_{0}}^{−1}∥ ∥∑_{j=1}^{n} x_{j}∥ + ∑_{j=1}^{n} |∥x_{j}^{−1}∥ − ∥x_{k_{0}}^{−1}∥| ∥x_{j}∥

= − ∥x_{k_{0}}^{−1}∥ (∑_{j=1}^{n} ∥x_{j}∥ − ∥∑_{j=1}^{n} x_{j}∥) + ∑_{j=1}^{n} ∥x_{j}^{−1}∥ ∥x_{j}∥ .

Utilising the second inequality in (2.5), we deduce

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

and the first inequality in (2.4) is proved.

The second part of (2.4) can be proved in a similar manner, however, the details are omitted.

Remark 2. An equivalent form of (2.4) is:

(2.6) \frac{∑_{j=1}^{n} ∥x_{j}^{−1}∥ ∥x_{j}∥ − ∥∑_{j=1}^{n} ∥x_{j}^{−1}∥ x_{j}∥}{max_{k∈\{1,...,n\}} ∥x_{k}^{−1}∥}

≤ ∑_{j=1}^{n} ∥x_{j}∥ − ∥∑_{j=1}^{n} x_{j}∥ ≤ \frac{∑_{j=1}^{n} ∥x_{j}^{−1}∥ ∥x_{j}∥ − ∥∑_{j=1}^{n} ∥x_{j}^{−1}∥ x_{j}∥}{min_{k∈\{1,...,n\}} ∥x_{k}^{−1}∥},

which provides both a refinement and a reverse inequality for the generalised triangle inequality.

3. Inequalities for Two Pairs of Elements

The following particular case of Theorem 1 is of interest for applications.

Lemma 1. If (a, b) , (x, y) ∈ A^{2}, then

(3.1) max {∥a (x ± y)∥ − ∥(b − a) y∥ , ∥b (x ± y)∥ − ∥(b − a) x∥}

≤ ∥ax ± by∥ ≤ min {∥a (x ± y)∥ + ∥(b − a) y∥ , ∥b (x ± y)∥ + ∥(b − a) x∥} or, equivalently,

\frac{1}{2} \{∥a (x ± y)∥ + ∥b (x ± y)∥ − [∥(b − a) y∥ + ∥(b − a) x∥]\} (3.2)

+ \frac{1}{2} |∥a (x ± y)∥ − ∥b (x ± y)∥ + ∥(b − a) y∥ − ∥(b − a) x∥|

≤ ∥ax ± by∥

≤ \frac{1}{2} \{∥a (x ± y)∥ + ∥b (x ± y)∥ + [∥(b − a) y∥ + ∥(b − a) x∥]\}

− \frac{1}{2} |∥a (x ± y)∥ + ∥b (x ± y)∥ − ∥(b − a) y∥ − ∥(b − a) x∥| .

Proof. The inequality (3.1) follows from Theorem 1 for n = 2, a_{1} = a, a_{2} = b, x_{1} = x and x_{2} = ±y.

Utilising the properties of real numbers,

min \{α, β\} = \frac{1}{2} [α + β − |α − β|] , max \{α, β\} = \frac{1}{2} [α + β + |α − β|] ; α, β ∈ ℝ;

the inequality (3.1) is clearly equivalent with (3.2).

The following result contains some upper bounds for ∥ax ± by∥ that are perhaps more useful for applications.

Theorem 2. If (a, b) , (x, y) ∈ A^{2}, then

(3.3)  ∥ax ± by∥ ≤ min {∥a (x ± y)∥ , ∥b (x ± y)∥} + ∥b − a∥ max {∥x∥ , ∥y∥}
≤ ∥x ± y∥ min {∥a∥ , ∥b∥} + ∥b − a∥ max {∥x∥ , ∥y∥}

and

(3.4)  ∥ax ± by∥ ≤ ∥x ± y∥ max {∥a∥ , ∥b∥} + min {∥(b − a) x∥ , ∥(b − a) y∥}
≤ ∥x ± y∥ max {∥a∥ , ∥b∥} + ∥b − a∥ min {∥x∥ , ∥y∥} .

Proof. Observe that ∥(b − a) x∥ ≤ ∥b − a∥ ∥x∥ and ∥(b − a) y∥ ≤ ∥b − a∥ ∥y∥, and then

(3.5)  ∥(b − a) x∥ , ∥(b − a) y∥ ≤ ∥b − a∥ max {∥x∥ , ∥y∥} ,

which implies that

min {∥a (x ± y)∥ + ∥(b − a) y∥ , ∥b (x ± y)∥ + ∥(b − a) x∥}
≤ min {∥a (x ± y)∥ , ∥b (x ± y)∥} + ∥b − a∥ max {∥x∥ , ∥y∥}
≤ ∥x ± y∥ min {∥a∥ , ∥b∥} + ∥b − a∥ max {∥x∥ , ∥y∥} .

Utilising the second inequality in (3.1), we deduce (3.3).

Also, since ∥a (x ± y)∥ ≤ ∥a∥ ∥x ± y∥ and ∥b (x ± y)∥ ≤ ∥b∥ ∥x ± y∥ , hence

∥a (x ± y)∥ , ∥b (x ± y)∥ ≤ ∥x ± y∥ max {∥a∥ , ∥b∥} ,

which implies that

min {∥a (x ± y)∥ + ∥(b − a) y∥ , ∥b (x ± y)∥ + ∥(b − a) x∥}
≤ ∥x ± y∥ max {∥a∥ , ∥b∥} + min {∥(b − a) x∥ , ∥(b − a) y∥}
≤ ∥x ± y∥ max {∥a∥ , ∥b∥} + ∥b − a∥ min {∥x∥ , ∥y∥} ,

and the inequality (3.4) is also proved.

The following corollary may be more useful for applications.

Corollary 3. If (a, b) , (x, y) ∈ A^{2}, then

(3.6)  ∥ax ± by∥ ≤ ∥x ± y∥ · \frac{∥a∥ + ∥b∥}{2} + ∥b − a∥ · \frac{∥x∥ + ∥y∥}{2}.

Proof. Follows from Theorem 2 by adding the last inequality in (3.3) to the last inequality (3.4) and utilising the property that min {α, β} + max {α, β} = α + β, α, β ∈ ℝ.

The following lower bounds for ∥ax ± by∥ can be stated as well:

Theorem 3. For any (a, b) and (x, y) ∈ A^{2}, we have:

max {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} − ∥b − a∥ max {∥x∥ , ∥y∥} (3.7)

≤ max {∥a (x ± y)∥ , ∥b (x ± y)∥} − ∥b − a∥ max {∥x∥ , ∥y∥}
≤ ∥ax ± by∥

and

min {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} − ∥b − a∥ min {∥x∥ , ∥y∥} (3.8)

≤ min {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} − min {∥(b − a) x∥ , ∥(b − a) y∥}
≤ ∥ax ± by∥ .

Proof. Observe that, by (3.5) we have that

max {∥a (x ± y)∥ − ∥(b − a) y∥ , ∥b (x ± y)∥ − ∥(b − a) x∥}
≥ max {∥ax ± ay∥ , ∥bx ± by∥} − ∥b − a∥ max {∥x∥ , ∥y∥}
≥ max {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} − ∥b − a∥ max {∥x∥ , ∥y∥}

and on utilising the first inequality in (3.1), the inequality (3.7) is proved. Observe also that, since

∥a (x ± y)∥ , ∥b (x ± y)∥ ≥ min {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} ,

then

max {∥a (x ± y)∥ − ∥(b − a) y∥ , ∥b (x ± y)∥ − ∥(b − a) x∥}
≥ min {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} − min {∥(b − a) x∥ , ∥(b − a) y∥}
≥ min {|∥ax∥ − ∥ay∥| , |∥bx∥ − ∥by∥|} − ∥b − a∥ min {∥x∥ , ∥y∥} .

Then, by the first inequality in (3.1), we deduce (3.8).

Corollary 4. For any (a, b) , (x, y) ∈ A^{2}, we have

(3.9) \frac{1}{2} · [|∥ax∥ − ∥ay∥| + |∥bx∥ − ∥by∥|] − ∥b − a∥ · \frac{∥x∥ + ∥y∥}{2} ≤ ∥ax ± by∥ .

The proof follows from Theorem 3 by adding (3.7) to (3.8). The details are omitted.

4. Applications for Two Invertible Elements

In this section we assume that A is a unital algebra with the unity 1. The following results provide some upper bounds for the quantity ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ , where x and y are invertible in A.

8 S.S. DRAGOMIR

Proposition 1. If (x, y) ∈ A^{2} are invertible, then

(4.1) ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥
≤ ∥x ± y∥ min {∥x^{−1}∥ , ∥y^{−1}∥} + |∥x^{−1}∥ − ∥y^{−1}∥| max {∥x∥ , ∥y∥}

and

(4.2) ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥
≤ ∥x ± y∥ max {∥x^{−1}∥ , ∥y^{−1}∥} + |∥x^{−1}∥ − ∥y^{−1}∥| min {∥x∥ , ∥y∥} .

Proof. Follows by Theorem 2 on choosing a = ∥x^{−1}∥ · 1 and b = ∥y^{−1}∥ · 1. Corollary 5. With the above assumption for x and y, we have

(4.3) ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥

≤ ∥x ± y∥ · \frac{∥x^{−1}∥ + ∥y^{−1}∥}{2} + |∥x^{−1}∥ − ∥y^{−1}∥| · \frac{∥x∥ + ∥y∥}{2}.

Lower bounds for ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ are provided below: Proposition 2. If (x, y) ∈ A^{2} are invertible, then (4.4) ∥x ± y∥ max {∥x^{−1}∥ , ∥y^{−1}∥} − |∥x^{−1}∥ − ∥y^{−1}∥| max {∥x∥ , ∥y∥}

≤ ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥

and (4.5) ∥x ± y∥ min {∥x^{−1}∥ , ∥y^{−1}∥} − |∥x^{−1}∥ − ∥y^{−1}∥| min {∥x∥ , ∥y∥}

≤ ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ .

Proof. The first inequality in (4.4) follows from the second inequality in (3.7) on choosing a = ∥x^{−1}∥ · 1 and b = ∥y^{−1}∥ · 1.

We know from the proof of Theorem 3 that

(4.6) max {∥a (x ± y)∥ − ∥(b − a) y∥ , ∥b (x ± y)∥ − ∥(b − a) x∥} ≤ ∥ax ± by∥ .

If in this inequality we choose a = ∥x^{−1}∥ · 1 and b = ∥y^{−1}∥ · 1, then we get

∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ ≥ max {∥x^{−1}∥ ∥x ± y∥ − |∥x^{−1}∥ − ∥y^{−1}∥| ∥y∥ , ∥y^{−1}∥ ∥x ± y∥ − |∥x^{−1}∥ − ∥y^{−1}∥| ∥x∥}

≥ ∥x ± y∥ min {∥x^{−1}∥ , ∥y^{−1}∥} − |∥x^{−1}∥ − ∥y^{−1}∥| min {∥x∥ , ∥y∥}

and the inequality (4.5) is obtained. Corollary 6. If (x, y) ∈ A^{2} are invertible, then

(4.7) ∥x ± y∥ · \frac{∥x^{−1}∥ + ∥y^{−1}∥}{2} − |∥x^{−1}∥ − ∥y^{−1}∥| · \frac{∥x∥ + ∥y∥}{2}

≤ ∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ .

Remark 3. We observe that the inequalities (4.3) and (4.7) are in fact equivalent with:

(4.8) |∥∥x^{−1}∥ x ± ∥y^{−1}∥ y∥ − ∥x ± y∥ · \frac{∥x^{−1}∥ + ∥y^{−1}∥}{2}|

≤ |∥x^{−1}∥ − ∥y^{−1}∥| · \frac{∥x∥ + ∥y∥}{2}.

Now we consider the dual expansion ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥, for which the following upper bounds can be stated.

Proposition 3. If (x, y) are invertible in A, then

(4.9) ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥
≤ ∥x ± y∥ min {∥x^{−1}∥ , ∥y^{−1}∥} + |∥x^{−1}∥ − ∥y^{−1}∥| max {∥x∥ , ∥y∥}

and

(4.10) ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥
≤ ∥x ± y∥ max {∥x^{−1}∥ , ∥y^{−1}∥} + |∥x^{−1}∥ − ∥y^{−1}∥| min {∥x∥ , ∥y∥} .

In particular,

(4.11) ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥

≤ ∥x ± y∥ · \frac{∥x^{−1}∥ + ∥y^{−1}∥}{2} + |∥x^{−1}∥ − ∥y^{−1}∥| · \frac{∥x∥ + ∥y∥}{2}.

The proof follows from Theorem 2 on choosing a = ∥y^{−1}∥ · 1 and b = ∥x^{−1}∥ · 1. The lower bounds for the quantity ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥ are incorporated in:

Proposition 4. If (x, y) are invertible in A, then

(4.12) ∥x ± y∥ max {∥x^{−1}∥ , ∥y^{−1}∥} − |∥x^{−1}∥ − ∥y^{−1}∥| max {∥x∥ , ∥y∥}

≤ ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥

and

(4.13) ∥x ± y∥ min {∥x^{−1}∥ , ∥y^{−1}∥} − |∥x^{−1}∥ − ∥y^{−1}∥| min {∥x∥ , ∥y∥}

≤ ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥ .

In particular,

(4.14) ∥x ± y∥ · \frac{∥x^{−1}∥ + ∥y^{−1}∥}{2} − |∥x^{−1}∥ − ∥y^{−1}∥| · \frac{∥x∥ + ∥y∥}{2}

≤ ∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥ .

Remark 4. We observe that the inequalities (4.11) and (4.14) are equivalent with

(4.15) |∥∥y^{−1}∥ x ± ∥x^{−1}∥ y∥ − ∥x ± y∥ · \frac{∥x^{−1}∥ + ∥y^{−1}∥}{2}|

≤ |∥x^{−1}∥ − ∥y^{−1}∥| · \frac{∥x∥ + ∥y∥}{2}.

References

[1] S.S. DRAGOMIR, A generalisation of the Pečarić-Rajić inequality in normed linear spaces, Preprint. RGMIA Res. Rep. Coll., 10(3) (2007), Art. 3. [ONLINE:

   http://rgmia.vu.edu.au/v10n3.html].

[2] J. PEČARIĆ and R. RAJIĆ, The Dunkl-Williams inequality with n elements in normed linear spaces, Math. Ineq. & Appl., 10(2) (2007), 461-470.

[3] M. KATO, K.-S. SAITO and T. TAMURA, Sharp triangle inequality and its reverses in Banach spaces, Math. Ineq. & Appl., 10(3) (2007).

[4] P.R. MERCER, The Dunkl-Williams inequality in an inner product space, Math. Ineq. & Appl., 10(2) (2007), 447-450.

[5] L. MALIGRANDA, Simple norm inequalities, Amer. Math. Monthly, 113 (2006), 256-260.

School of Computer Science and Mathematics, Victoria University, PO Box 14428, Melbourne City Mail Centre, VIC, 8001, Australia.

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