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On Sharp Triangle Inequalities in Banach Spaces II
Journal of Inequalities and Applications volume 2010, Article number: 323609 (2010)
Abstract
Sharp triangle inequality and its reverse in Banach spaces were recently showed by Mitani et al. (2007). In this paper, we present equality attainedness for these inequalities in strictly convex Banach spaces.
1. Introduction
In recent years, the triangle inequality and its reverse inequality have been treated in [1–5] (see also [6, 7]).Kato et al.[8] presented the following sharp triangle inequality and its reverse inequality with elements in a Banach space .
Theorem 1.1 (see [8]).
For all nonzero elements in a Banach space ,
These inequalities are useful to treat geometrical structure of Banach spaces, such as uniform nonness (see [8]). Moreover,Hsu et al.[9] presented these inequalities for strongly integrable functions with values in a Banach space.
Mitani et al.[10] showed the following inequalities which are sharper than Inequality (1.1) in Theorem 1.1.
Theorem 1.2 (see [10]).
For all nonzero elements in a Banach space with , ,
where .
In this paper we first present a simpler proof of Theorem 1.2. To do this we consider the case as follows.
Theorem 1.3.
For all nonzero elements in a Banach space with
where .
From this result we can easily obtain Theorem 1.2.
Moreover we consider equality attainedness for sharp triangle inequality and its reverse inequality in strictly convex Banach spaces. Namely, we characterize equality attainedness of Inequalities (1.4) and (1.5) in Theorem 1.3.
2. Simple Proofs of Theorems 1.2 and 1.3
Proof of Theorem 1.3.
According to Theorem 1.1 Inequalities (1.4) and (1.5) hold for the case (cf. [3]). Therefore let . We first prove (1.4) by the induction. Assume that (1.4) holds true for all elements in . Let be any elements in with . Let
for all positive numbers with . Then
and . By assumption,
holds, where . Since , from (2.2) and (2.3),
and hence (1.4). Thus (1.4) holds true for all finite elements in .
Next we show Inequality (1.5). Let
Then
and Applying Inequality (1.4) to ,
where . Thus we obtain (1.5). This completes the proof.
Proof of Theorem 1.2.
Let be any nonzero elements in with . For all positive numbers with let
Then . Applying Theorem 1.3 to ,
where for all positive numbers with . As , we have Inequalities (1.4) and (1.5).
3. Equality Attainedness in a Strictly Convex Banach Space
In this section we consider equality attainedness for sharp triangle inequality and its reverse inequality in a strictly convex Banach space. Kato et al. in [8] showed the following.
Theorem 3.1 (see [8]).
Let be a strictly convex Banach space and nonzero elements in . Let and . Let . Then
if and only if either
(a)
or
(b)
Theorem 3.2 (see [8]).
Let be a strictly convex Banach space and nonzero elements in . Let and . Let . Then
if and only if either
(a)
or
(b)
We present equality attainedness for (1.4) and (1.5) in Theorem 1.2. The following lemma given in [8] is quite powerful.
Lemma 3.3 (see [8]).
Let be a strictly convex Banach space. Let be nonzero elements in . Then the following are equivalent.
(i) holds for any positive numbers .
(ii) holds for some positive numbers .
(iii).
Theorem 3.4.
Let be a strictly convex Banach space and nonzero elements in with . Then
if and only if there exists a real number with satisfying .
Proof.
Assume that (3.3) is true. By Theorem 3.1, the equality (3.3) is equivalent to Equality
Put
Then . Since , we obtain . Conversely, if where , then
By , we have (3.4). Thus we get (3.3).
Next we consider the case .
Theorem 3.5.
Let be a strictly convex Banach space and nonzero elements in with . Then
if and only if there exist with satisfying one of the following conditions:
(a)
(b)
Proof.
Assume that (3.7) is true. Put
Then and
Note that . As in the proof of Theorem 1.2 given in [10], we have (3.7) if and only if we have the equalities
By Theorem 3.4, Equality (3.11) implies that
for some with . By (3.8) we have
for some . On the other hand, by Lemma 3.3, Equality (3.10) implies
Hence, by using (3.8), (3.12), and (3.13) we have for some real number . Since , we have . We consider the following two cases
Case 1.
.
Equality (3.14) implies
Hence we have
By and , Equality (3.16) is valid for all real numbers with .
Case 2.
.
Equality (3.14) implies
So we have
Hence . Thus () holds.
Conversely, assume that there exist with satisfying one of the conditions and . Then it is clear that (3.7) holds. Thus we obtain ().
Moreover we consider general cases. For each with , we put For and , we define
For a finite set , the cardinal number of is denoted by .
Lemma 3.6.
Let . If for all with , then
Proof.
Let
where and . By the assumption, we have . We first show for all with . It is clear that . Assume that for all with . We will show . Suppose that . By , we have
Hence we have which is a contradiction. Therefore we have . Namely, for all with . From this result, we obtain . Hence
Theorem 3.7.
Let be a strictly convex Banach space and nonzero elements in with . Then
if and only if there exists with such that
for every with .
Proof.
(): According to Theorems 3.4 and 3.5, Theorem 3.7 is valid for the cases . Therefore let . We will prove Theorem 3.7 by the induction. Assume that this theorem holds true for all nonzero elements in less than . Let and assume that Equality (3.24) holds. Let
for positive integer with . As in the proof of Theorem 1.3, Equality (3.24) holds if and only if
hold, where . Hence, by assumption, there exists with such that
Since by Lemma 3.6, we have
Since
by the definition of , we have
where
Since
we have
for all with . By Lemma 3.3, Equality (3.28) implies
Hence there exists such that . Also,
Since , we have from (3.37) and (3.38),
which implies
If , then it is clear that .
If , then, by (3.41), we have . Hence . Thus we obtain .
(): Let with satisfying (3.25) and (3.26), and let . From (3.25) we have
By Lemma 3.6,
Let , where . From (3.26) and we have
Thus we obtain (3.24). This completes the proof.
In what follows, we characterize the equality condition of Inequality (1.3) in Theorem 1.2. For and positive integer with we define , and
Lemma 3.8.
Let with . If
for all positive integers with , then one has
Proof.
Let and , where and . As in the proof of Lemma 3.6, we have for all . So for all . Hence
This completes the proof.
Theorem 3.9.
Let be a strictly convex Banach space and nonzero elements in with . Then
holds if and only if there exists with , for all positive integers with satisfying
for all positive integers with and .
Proof.
: Let and
For positive integers with we put
Note that and
Then Equality (3.48) holds if and only if
Thus, by the equality condition of sharp triangle inequality with elements, there exists such that
for all positive integers with ,
for all positive integers with . Since
for each with , we have , where
Note that and . By (3.53),
Hence there exists such that . We also have and
Hence we have
From (3.3), we obtain . Thus we have ().
(): Assume that there exists with , for all positive integers with satisfying (3.49) for all positive integers with and . By Theorem 3.7, we have (3.54). From the assumption,
By and Lemma 3.8, we obtain (3.53). Thus we have (). This completes the proof.
If , then we have the following corollary.
Corollary 3.10.
Let be a strictly convex Banach space and nonzero elements in with . Then
if and only if there exists a real number with such that .
If , then we have the following corollary.
Corollary 3.11.
Let be a strictly convex Banach space and nonzero elements in with . Then
if and only if there exist with such that and .
4. Remark
In this section we consider equality attainedness for sharp triangle inequality in a more general case, that is, the case without the assumption that . Let us consider the case .
Proposition 4.1.
Let be a strictly convex Banach space, and nonzero elements in .
(i)If , then the equality
always holds.
(ii)If , then the equality (4.1) holds if and only if there exists a real number satisfying and .
(iii)If , then the equality (4.1) holds if and only if there exists a real number satisfying and .
Proof.

(i)
Is clear.

(ii)
Assume that (4.1) holds. By , (4.1) implies
(4.2)
From Theorem 3.1, we have
Hence for some . The following
implies
Hence and so The converse is clear.

(iii)
Assume that (4.1) holds. Put and . As in the proof of Theorem 3.5, we have
(4.6)
Since , we have for some . The following
implies
Hence . The converse is clear.
Conjecture 1.
What is the necessary and sufficient condition when Equality (3.24) (resp. Equality (3.48)) holds for elements with in Theorem 3.7 (resp. Theorem 3.9)?
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Acknowledgment
The second author was supported in part by GrantsinAid for Scientific Research (no. 20540158), Japan Society for the Promotion of Science.
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Mitani, KI., Saito, KS. On Sharp Triangle Inequalities in Banach Spaces II. J Inequal Appl 2010, 323609 (2010). https://doi.org/10.1155/2010/323609
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DOI: https://doi.org/10.1155/2010/323609
Keywords
 Banach Space
 Positive Integer
 Equality Condition
 Kato
 Integrable Function