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Cent. Eur. J. Math. • 12(5) • 2014 • 778-786 DOI: 10.2478/s11533-013-0369-7

Central European Journal of Mathematics

Characterization of intermediate values of the triangle inequality II Research Article Hiroki Sano1∗ , Tamotsu Izumida2† , Ken-Ichi Mitani3‡ , Tomoyoshi Ohwada4§ , Kichi-Suke Saito5¶

1 Graduate School, Faculty of Science, Shizuoka University, Shizuoka 422-8529, Japan 2 Department of Mathematical Sciences, Graduate School of Science and Technology, Niigata University, Niigata 950-2181, Japan 3 Department of Information and Communication Engineering, Faculty of Computer Science and System Engineering, Okayama Prefectural University, Okayama 719-1197, Japan 4 Faculty of Education, Shizuoka University, Shizuoka 422-8529, Japan 5 Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan

Received 16 May 2013; accepted 24 July 2013 Abstract: In [Mineno K., Nakamura Y., Ohwada T., Characterization of the intermediate values of the triangle inequality, Math. Inequal. Appl., 2012, 15(4), 1019–1035] there was established a norm

Pwhich characterizes all P inequality intermediate values of the triangle inequality, i.e. Cn that satisfy 0 ≤ Cn ≤ nj=1 kxj k − nj=1 xj , x1 , . . . , xn ∈ X . Here we study when this norm inequality attains equality in strictly convex Banach spaces. MSC:

46B20, 46B99, 26D20

Keywords: Triangle inequalities • Strictly convex Banach spaces • Norm inequality

© Versita Sp. z o.o.

∗ † ‡ § ¶

778

E-mail: E-mail: E-mail: E-mail: E-mail:

[email protected] [email protected] [email protected] [email protected] [email protected]

H. Sano et al.

1.

Introduction

The (generalized) triangle inequality, namely,

n n

X

X

xj ≤ kxj k,

j=1

j=1

where (X , k · k) is a normed linear space over the real or complex field K and xj , j ∈ {1, 2, . . . , n}, are vectors in X , plays a fundamental role in establishing various analytic and geometric properties of such spaces. For results related to this inequality and the relevant topics (for example, the Dunkl–Williams inequality) see [2–7, 9, 10, 14–19]. In [8] and [13] two results on intermediate values of the triangle inequality were presented. They were generalized in [11] with a result which characterizes all intermediate values of the triangle inequality. In this paper we slightly modify the statement and present a simpler proof for it. Moreover we study when this new norm inequality attains equality in a strictly convex Banach space. For positive integer n ≥ 2, let Mn ([0, 1]) be the set of all n × n matrices whose all elements belong to the interval [0, 1], and Ln denote the set of all lower triangular matrices of Mn ([0, 1]), i.e.,  Ln = a = (aij ) ∈ Mn ([0, 1]) : aij = 0, i < j . a Let 1 ≤ m ≤ n. For each a = (aij ) in Ln , we set `mj (m) = amj , 1 ≤ j ≤ m, and if 2 ≤ n, then, for each m with 2 ≤ m ≤ n, we put m Y `ija (m) = aij (1 − akj ), 1 ≤ i ≤ m − 1, 1 ≤ j ≤ m. k=i+1

Theorem 1.1. Let n ≥ 2. Take any a = (aij ) in Ln . For any elements x1 , x2 , . . . , xn in a normed linear space X , the following inequality holds: 



n n i i n

X

X X X X



a a  k`ij (n)xj k − `ij (n)xj ≤ kxj k − xj . (1)



i=1

j=1

j=1

j=1

j=1

In [11], the above theorem was established for `1ja (m) = a1j , 1 ≤ j ≤ m, which values do not influence (1). We made a modification in order to consider the equality attainability in (1) in a particular space.

2.

Simple proof of Theorem 1.1

In this section, we present a simpler proof of Theorem 1.1. We would need the following lemma.

Lemma 2.1. Let n ≥ 2 and a = (aij ) ∈ Ln . Take m with 1 ≤ m ≤ n. For each j with m ≤ j ≤ n, the following identity holds: n X i=j

Proof.

a `im (n) +

n Y

(1 − aim ) = 1. i=j

We shall prove this lemma by induction on j. If j = n, then a (1 − anm ) + `nm (n) = (1 − anm ) + anm = 1.

779

Characterization of intermediate values of the triangle inequality II

Next let j = n − 1. Then n Y

n X

(1 − aim ) + i=n−1

a a a `im (n) = (1 − an−1,m )(1 − anm ) + `n−1,m (n) + `nm (n)

i=n−1

= (1 − an−1,m )(1 − anm ) + an−1,m (1 − anm ) + anm = (1 − an−1,m )(1 − anm ) + {1 − (1 − an−1,m )}(1 − anm ) + anm = (1 − anm ) + anm = 1.

Now if we assume that, for each j with m + 1 ≤ j ≤ n, n n Y X a (1 − aim ) + `im (n) = 1, i=j

i=j

then we have n Y

n X

(1 − aim ) + i=j−1

a `im (n) =

i=j−1

n Y

a (1 − aim ) + `j−1,m (n) +

a `jm (n)

i=j

i=j−1 n Y

=

n X

n Y

(1 − aim ) + aj−1,m

n X

(1 − aim ) + i=j

i=j−1

i=j

n Y

=

n Y

(1 − aim ) + {1 − (1 − aj−1,m )}

n X

(1 − aim ) + i=j

i=j−1 n Y

=

a `jm (n)

n X

(1 − aim ) + i=j

a `jm (n)

i=j

a `jm (n) = 1.

i=j

Thus, by induction on j, we obtain the lemma.

Proof of Theorem 1.1. n X

n X

xj = j=1

For each x1 , . . . , xn ∈ X and a = (aij ) ∈ Ln , the following equations hold: n X

1− j=1

! `ija (n)

n X n X

xj +

i=j

j=1 i=j

`ija (n)xj

n X

=

n X

1− j=1

! `ija (n)

i=j

n X i X

xj +

`ija (n)xj .

i=1 j=1

By Lemma 2.1, we see that n Y

0≤

n X

(1 − aij ) = 1 −

`ija (n).

i=j

i=j

Moreover, 0 ≤ `ija (n) ≤ 1 for all i, j ∈ {1, . . . , n}. Thus, applying the (generalized) triangle inequality, we have



! n n n n X i

X

X

X X



a a xj = 1− `ij (n) xj + `ij (n)xj



i=j j=1 j=1 i=1 j=1



i

! ! n n n X i n n n X

X

X

X X X X



a a a a ≤ `ij (n) xj + `ij (n)xj = 1− `ij (n) kxj k + `ij (n)xj

1−





i=j i=j j=1 i=1 j=1 j=1 i=1 j=1



! n n n X i n n X n n X i

X

X X X X

a

a

X



a a

`ij (n)xj +

`ij (n)xj + = kxj k − `ij (n)xj = kxj k − `ij (n)xj



i=j j=1 i=1 j=1 j=1 j=1 i=j i=1 j=1



! n n X i n X i n n i i

X

X X X X

a

X

a



X a

a



= kxj k − `ij (n)xj + `ij (n)xj = kxj k − `ij (n)xj − `ij (n)xj .



j=1

780

i=1 j=1

i=1

j=1

j=1

i=1

j=1

j=1

(2)



H. Sano et al.

3.

Equality attainability in a strictly convex Banach space

In this section, we consider inequality (1) in a strictly convex Banach space.

Lemma 3.1 (cf. [1, Problem 11.1]). Let (X , k · k) be a strictly convex Banach space. For x1 , . . . , xn ∈ X , the following assertions are equivalent:

n n

X X

(i) xj = kxj k;

j=1

j=1

(ii) kxj kxi = kxj kxi for all i, j ∈ {1, . . . , n}.

For a = (aij ) ∈ Ln and x1 , . . . , xn ∈ X , put ( J=

n X

j ∈ {1, . . . , n} :

) `ija (n)

(

6= 1 ,

I=

i X

i ∈ {1, . . . , n} :

i=j

) `ija (n)xj

6= 0 .

j=1

Lemma 3.2. Take a = (aij ) ∈ Ln . Then (i) J = {1, . . . , n} if and only if aij ∈ [0, 1) for all i, j ∈ {1, . . . , n}; and (ii) J = ∅ if and only if, for each j ∈ {1, . . . , n}, there exists i with j ≤ i ≤ n such that aij = 1.

Proposition 3.3. Let n ≥ 2 and a = (aij ) in Ln . For x1 , . . . , xn in a strictly convex Banach space X , the equality  n X i=1



i X

k`ija (n)xj k

j=1

i



n n

X

X

 X

a − `ij (n)xj = kxj k − xj



j=1

j=1

(3)

j=1

holds if and only if either I = ∅ and J = ∅ or, for each g, h, k, m ∈ {1, . . . , n}, the following assertions hold: (i) if J 6= ∅, then, for each g, h, k ∈ J and m ∈ {1, . . . , n}, kxh kxg = kxg kxh ,

m

m

X

X

a a `mj (n)xj xk = kxk k `mj (n)xj ;

j=1

(4) (5)

j=1

(ii) if J = ∅, then, for each k, m ∈ {1, . . . , n},

m

k

k

m

X

X

X

X



a a a a `mj (n)xj `kj (n)xj = `kj (n)xj `mj (n)xj .



j=1

Proof.

j=1

j=1

(6)

j=1

Let I = ∅ and J = ∅. In this case, for each i, j ∈ {1, . . . , n}, i X j=1

`ija (n)xj = 0

n X

and

`ija (n) = 1.

i=j

781

Characterization of intermediate values of the triangle inequality II

Hence n X i X

0=

`ija (n)xj =

n X n X

n X

`ija (n)xj =

xj .

j=1 i=j

i=1 j=1

j=1

Therefore we have  n X



i=1

i X

k`ija (n)xj k

j=1

i



n n i n X i n n

X

X X X X

 X X a

a a − `ij (n)xj = `ij (n)kxj k = `ij (n)kxj k = kxj k = kxj k − xj .



j=1

j=1 i=j

i=1 j=1

j=1

j=1

j=1

Next let J 6= ∅. If (3) holds, then equality (2) also holds. Thus, by Lemma 3.1, the following equalities hold for any g, h, k, m ∈ {1, . . . , n}:

! ! ! ! n n n n



X X X X



a a a a `ih (n) xh 1 − `ig (n) xg = 1 − `ig (n) xg 1 − `ih (n) xh ,

1−



i=g i=g i=h i=h

m

! ! m n n

X

X X X



a a a a `mj (n)xj 1 − `ik (n) xk = 1 − `ik (n) xk `mj (n)xj .



j=1

i=k

i=k

j=1

P By Lemma 2.1, we see that 0 < 1 − ni=j `ija (n) for all j ∈ J. Thus the above equalities are equivalent to kxh kxg = kxg kxh , g, h ∈ J, and

m m

X

X

a a `mj (n)xj xk = kxk k `mj (n)xj , k ∈ J, m ∈ {1, . . . , n},

j=1

j=1

respectively. The converse is clear. Finally, let J = ∅ and I 6= ∅. In this case if (3) holds, then, for each k, m ∈ I, we have

m

k

k

m

X

X

X

X



a a a a `mj (n)xj `kj (n)xj = `kj (n)xj `mj (n)xj .



j=1

j=1

j=1

j=1

The converse is clear, and the proof is completed.

Theorem 3.4. Let n ≥ 2 and a = (aij ) in Ln . Assume that J 6= ∅ and put j0 = minj∈J j. For nonzero elements x1 , . . . , xn in a strictly convex Banach space X , equality (3) holds if and only if there exist positive real numbers αj , j ∈ J, such that xj = αj xj0 , and, for each i ∈ {1, . . . , n}, there are real numbers βi such that X 0≤

X

αj `ija (n) + βi

and

`ija (n)xj = βi xj0 ,

j∈Jic

j∈Ji

where Ji = {j ∈ J : j ≤ i} and Jic is a complement of Ji in {1, . . . , n}.

Proof.

(⇒) Assume that (3) holds. By Proposition 3.3 (i), if we put αj = kxj k/kxj0 k, j ∈ J, then αj > 0 and xj = αj xj0 , j ∈ J. Moreover, for each i ∈ {1, . . . , n}, i X j=1

`ija (n)xj

i

X

x

j a = `ij (n)xj 0 .

kxj0 k j=1

Since i X j=1

782

`ija (n)xj =

X j∈Ji

`ija (n)xj +

X j∈Jic

`ija (n)xj =

X j∈Ji

αj `ija (n)xj0 +

X j∈Jic

`ija (n)xj ,

i ∈ {1, . . . , n},

H. Sano et al.

we have





i

X

X 1

`ija (n)xj =  ` a (n)xj − αj `ija (n) xj0 .

kxj0 k j=1 ij c j∈J

X j∈Ji

i

Put

i

X 1

X a

βi = `ij (n)xj − αj `ija (n).

kxj0 k j=1 j∈J i

Then we see that 0 ≤

P

a j∈Ji αj `ij (n)

+ βi and X

`ija (n)xj = βi xj0 ,

i ∈ {1, . . . , n}.

(7)

j∈Jic

P (⇐) Assume that there exist positive real numbers αj and real numbers βi such that xj = αj xj0 , j ∈ J, 0 ≤ j∈Ji αj `ija (n)+ βi and (7) holds. Since kxj k = αj kxj0 k for all j ∈ J, it is clear that (4) is valid. So, by Proposition 3.3 (i), we need only to prove (5). Since ! i X X a a `ij (n)xj = αj `ij (n) + βi xj0 , i ∈ {1, . . . , n}, j=1

we have

j∈Ji

i

X

a `ij (n)xj =

j=1

Thus i X j=1

`ija (n)xj

X

! αj `ija (n)

+ βi kxj0 k,

i ∈ {1, . . . , n}.

j∈Ji

i

1

X a

`ij (n)xj xj0 , =

kxj0 k j=1

i ∈ {1, . . . , n},

and the proof is completed. By Lemma 3.2, for a = (aij ) ∈ Ln , recall that aij ∈ [0, 1), i, j ∈ {1, . . . , n}, is equivalent to J = {1, . . . , n}. Thus we immediately have the following corollary.

Corollary 3.5. Let n ≥ 2 and a = (aij ) in Ln . For nonzero elements x1 , . . . , xn in a strictly convex Banach space X , if aij ∈ [0, 1) for all i, j ∈ {1, . . . , n}, then equality (3) holds if and only if there exist positive real numbers αj , j ∈ {1, . . . , n}, such that xj = αj x1 , j ∈ {1, . . . , n}.

Next, we consider the case J = ∅ for a = (aij ) ∈ Ln . In this case we note that ann = 1. Take a = (aij ) ∈ Ln with anj = 1 for all j ∈ {1, . . . , n}. Then we see that 

0 .  .. (`ija (n)) =   0 1

... .. . ... ...

0 .. . 0 1

 0 ..   ..  0 1

Thus it is clear that, for each x1 , . . . , xn in a normed linear space X , (3) always holds. Thus we are interested in the case anj 6= 1 for some j ∈ {1, . . . , n − 1}. However, generally it is complicated. We give the following special case.

Theorem 3.6. Let n ≥ 2 and a = (aij ) in Ln . For nonzero elements x1 , . . . , xn in a strictly convex Banach space X , if aij ∈ [0, 1) for all i > j and aii = 1 for all i ∈ {1, . . . , n}, then equality (3) holds if and only if there exist real numbers αj , j ∈ {1, . . . , n}, P such that xj = αj x1 and ij=1 αj `ija (n) ≥ 0, i ∈ {1, . . . , n}.

783

Characterization of intermediate values of the triangle inequality II

P (⇒) Since aii = 1 for all i ∈ {1, . . . , n}, by Lemma 2.1, ni=k `ika (n) = 1, k ∈ {1, . . . , n}, and we note that a `11 (n) > 0. If (3) holds, then, by Proposition 3.3, for all i ∈ {1, . . . , n},

Proof.

i

i

X

a

X

a a

` (n)x `ija (n)xj .

j `11 (n)x1 = `11 (n)x1 ij

j=1

(8)

j=1

P a a a Let i = 2 and assume that 2j=1 `2ja (n)xj = 0. Since `22 (n) > 0, if we put α2 = −`21 (n)/`22 (n), then x2 = α2 x1 and P2 P 2 a a α ` (n) = 0. Next assume that ` (n)x = 6 0. By (8), we have j j=1 j 2j j=1 2j  

2

def 1  1

X a

a x2 = a ` (n)xj − α1 `21 (n) x1 = α2 x1 .

`22 (n) kx1 k j=1 2j We see that

2

2

2 X

X

x X

1 a a αj `2j (n) x1 = `2j (n)xj = `2ja (n)xj =

kx1 k j=1

Thus

P2 j=1

j=1

2 X

j=1

! αj `2ja (n)

x1 .

j=1

αj `2ja (n) > 0, and hence in any case we have 2 X

αj `2ja (n) ≥ 0.

j=1

Now we assume that, for each i = m with 2 ≤ m ≤ n − 1, (3) holds. Then there exist real numbers αj , j ∈ {1, . . . , m}, P P a such that xj = αj x1 , j ∈ {1, . . . , m}, and ij=1 αj `ija (n) ≥ 0, i ∈ {1, . . . , m}. If m+1 j=1 `m+1 j (n)xj = 0, then we see that Pm+1 P Pm+1 a a a a xm+1 = αm+1 x1 and j=1 αj `m+1,j (n) = 0 for αm+1 = − m j=1 αj `m+1,j (n)/`m+1,m+1 (n). In the case that j=1 `m+1,j (n)xj 6= 0, if we put  

m+1

X

1 1

a a  αm+1 = a `m+1,j (n)xj − α1 `m+1,1 (n),

`m+1,m+1 (n) kx1 k j=1

then we see that xm+1 = αm+1 x1 and, by (8), m+1

m+1

m+1 X

X

x X

1 a a a αj `m+1,j (n) x1 = `m+1,j (n)xj = `m+1,j (n)xj =

kx1 k j=1

Thus

Pm+1 j=1

j=1

j=1

m+1 X

! a αj `m+1,j (n)

x1 .

j=1

a αj `m+1,j (n) > 0, and hence we have

m+1 X

a αj `m+1,j (n) ≥ 0.

j=1

By using induction on i, we have desired result. (⇐) Suppose that there exist real numbers αj , j ∈ {1, . . . , n}, such that xj = αj x1 , j ∈ {1, . . . , n}, and i ∈ {1, . . . , n}, where α1 = 1. If m 6∈ I, then, for each k ∈ {1, . . . , n},

m

m

X

X

a a `mj (n)xj xk = 0 = kxk k `mj (n)xj .

j=1

784

j=1

Pi j=1

αj `ija (n) ≥ 0,

H. Sano et al.

On the other hand, since

i

i

X

X

a `ij (n)xj = kx1 k αj `ija (n)

j=1

j=1

for each i ∈ I, we have i X

`ija (n)xj

j=1

i

X

x

1 a = `ij (n)xj .

kx1 k j=1

Therefore, by Proposition 3.3 (ii), the proof is completed. As a corollary of Theorem 3.6, we can get [12, Theorem 3.7].

Corollary 3.7 ([12, Theorem 3.7]). For all nonzero elements x1 , . . . , xn in a strictly convex Banach space X with kx1 k > kx2 k > . . . > kxn k, the equality  n X i=1

i



n n

X x

X

X

j

i − kxj k − xj

 (kxi k − kxi+1 k) =

kx jk j=1 j=1 j=1

(9)

holds if and only if there exist real numbers αj , j ∈ {1, . . . , n}, with 1 = α1 > |α2 | > . . . > |αn | such that xj = αj x1 , j ∈ {1, . . . , n}, and |Im+ (α)| ≥ |Im− (α)| for every m with 1 ≤ m ≤ n, where |Im+ (α)| and |Im− (α)| are the cardinal numbers of Im+ (α) = {j ∈ {1, . . . , m} : αj > 0} and Im− (α) = {j ∈ {1, . . . , m} : αj < 0}, respectively.

Proof.

As in the proof of [11, Corollary 3.5], if we put

aij =

kxi k − kxi+1 k , kxj k − kxi+1 k

i, j ∈ {1, . . . , n},

then a = (aij ) ∈ Dn and `ija (n) =

kxi k − kxi+1 k , kxj k

i, j ∈ {1, . . . , n}.

In this case, (8) is equivalent to (9). Thus, (6), if (9) holds, then there exist αj ∈ R such that m X

xj = α j x1

and

a αj `mj (n) ≥ 0,

j, m ∈ {1, . . . , n}.

j=1

Since a `mj (n) =

we see that

m X

0≤

kxm k − kxm+1 k |αm | − |αm+1 | = , kxj k |αj |

a αj `mj (n) = (|αm | − |αm+1 |)

j=1

m X αj . |α j| j=1

Thus we have

0 ≤

m m m m m X X X X X αj αj αj = + = 1+ (−1) = |Im+ (α)| − |Im− (α)|, |α |α |α j| j| j| + + − − j=1 j∈Im (α)

j∈Im (α)

j∈Im (α)

j∈Im (α)

m ∈ {1, . . . , n}. The converse is clear, and the proof is completed.

785

Characterization of intermediate values of the triangle inequality II

Acknowledgements The authors wish to thank the referee for his/her comments on the manuscript. Kichi-Suke Saito is partly supported by the Grant-in-Aid for Scientific Research(C), Japan Society for the Promotion of Science (23540189).

References

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