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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