Certain geometric structure of $\Lambda $-sequence spaces

Certain geometric structure of Λ-sequence spaces Atanu Manna ∗ Indian Institute of Carpet Technology

arXiv:1612.01519v1 [math.FA] 3 Dec 2016

Chauri Road, Bhadohi- 221401, Uttar Pradesh, India

Abstract The Λ-sequence spaces Λ p for 1 < p ≤ ∞ and its generalization Λ pˆ for 1 < pˆ < ∞,

pˆ = (pn ) is introduced. The James constants and strong n-th James constants of Λ p for 1 < p ≤ ∞ is determined. It is proved that generalized Λ-sequence space Λ pˆ is embedded isometrically in the Nakano sequence space l pˆ (Rn+1 ) of finite dimensional Euclidean

space Rn+1 . Hence it follows that sequence spaces Λ p and Λ pˆ possesses the uniform Opial property, property (β ) of Rolewicz and weak uniform normal structure. Moreover, it is established that Λ pˆ possesses the coordinate wise uniform Kadec-Klee property. Further necessary and sufficient conditions for element x ∈ S(Λ pˆ ) to be an extreme point of B(Λ pˆ )

are derived. Finally, estimation of von Neumann-Jordan and James constants of two di(2)

mensional Λ-sequence space Λ2 is being carried out.

Keywords: Ces`aro sequence space; Nakano sequence space; James constant; von NeumannJordan constant; Extreme point; Kadec-Klee property. 2010 Mathematics Subject Classification: Primary 46B20; Secondary 46A45, 46B45.

1 Introduction There are several important geometric constants of Banach spaces such as von Neumann-Jordan constant, James constant, Dunkl-Williams constant, Khintchine constant, Zb˘aganu, Ptolemy constants and so on. The space is how much close (or far) to a Hilbert space measured by von Neumann-Jordan constant and Dunkl-Williams constant. The uniform non-squareness of a unit ball in a real Banach space measured by James constant (or sometimes called James non-square constant). These constants occupied a prominent place in the study of geometrical properties of Banach spaces, and have investigated recently by many researchers (see e.g. [18], [19], [20]). Maligranda et al. [19] obtained the classical James constant and n-th James constant for the Ces`aro sequence spaces ces p , 1 < p ≤ ∞, which was first introduced and studied by Shiue [31] ∗ Author’s

e-mail: [email protected]/[email protected]

1

and Leibowitz [16]. Let l 0 be the space of all real sequences and N0 be the set of all natural numbers N including 0, i.e., N0 = N ∪ {0} and x = (xn )n∈N0 ∈ l 0 . Throughout the text we shall write x = (xn ) instead of x = (xn )n∈N0 . Then the Ces`aro sequence spaces ces p , 1 < p ≤ ∞ are defined as follows:

 p 1 o n  ∞  1 n p |x | < ∞ for 1 < p < ∞ ces p = x ∈ l 0 : ∑ ∑ k n=0 n + 1 k=0 n o 1 n ces∞ = x ∈ l 0 : sup |x | < ∞ . ∑ k n∈N0 n + 1 k=0 The Ces`aro sequence space ces p is generalized to ces pˆ for pˆ = (pn ), pn > 1, n ∈ N0 ([14]) and defined as n  ∞  1 n  pn  1 o pn ces pˆ = x ∈ l 0 : ∑ |x | < ∞ . ∑ k n=0 n + 1 k=0 Several authors ([5], [6], [8], [11], [21], [23], [24], [27], [29], [30], [32], [33]) studied geometric properties such as Opial property, Kadec-Klee property, property (β ) of Rolewicz, weak uniform normal structure etc. for the spaces ces p and ces pˆ . These constants play very important role in the study of fixed point theory. For example, Opial property has several applications in the Banach fixed point theory, differential equations, integral equations etc. On the other hand, Kadec-Klee property applied to established certain results in the ergodic theory (see [25]). In this present paper, we are willing to study certain geometric structure, as the title said, of Λ-sequence spaces Λ p , 1 < p ≤ ∞ and its generalization Λ pˆ , for 1 < pˆ < ∞, pˆ = (pn ). These spaces Λ p and Λ pˆ are not only gives the sequence spaces ces p and ces pˆ but also include sequence spaces generated by Riesz weighted means, N¨orlund means etc. [2] in special cases. For example, our generalization include the following important results: (i) ces p and ces pˆ has the uniform Opial property (see [6] and [23] respectively), (ii) ces p and ces pˆ has the property (β ) (see [5] and [27] respectively), (iii) ces p has weak uniform normal structure [6] and ces pˆ has the same [27], (iv) ces pˆ possesses the uniform Kadec Klee property [24] and Kadec-Klee property q ([32], [33]), (2)

(v) The James constants J(ces p ) = 2, Jns (ces p ) = n for 1 < p ≤ ∞ and J(ces2 ) = (vi) ces p and ces pˆ has extreme point.

2 + √25 [19],

Therefore studying geometric structure of the spaces Λ p and Λ pˆ is a unified study of geometric structure for the known sequence spaces.

1.1 Sequence spaces Λ p , 1 < p ≤ ∞ The notions of Λ-strong convergence was first introduced by M´oricz [22] and is behind the genesis of Λ-sequence spaces. Let Λ = {λk : k = 0, 1, 2, . . .} be a non-decreasing sequence of λ positive numbers tending to ∞, i.e., 0 < λ0 < λ1 < λ2 < . . ., λk → ∞ with λk+1 → 1 as k → ∞ k and x = (xk ) ∈ l 0 . Define sequence spaces Λ p , for 1 < p ≤ ∞ as

2

 Λ p = x = (xk ) ∈ l 0 : kxk p < ∞ for 1 < p < ∞  Λ∞ = x = (xk ) ∈ l 0 : kxk∞ < ∞ for p = ∞,

where k.k p , 1 < p < ∞ and kxk∞ are defined by kxk p =



∞

∑

n=0


p  1p 1 n − )|x | ( λ λ ∑ k k−1 k λn k=0

1 n ∑ (λk − λk−1 )|xk |. n∈N0 λn k=0

and kxk∞ = sup

It is a routine work to established that these spaces (Λ p , k.k p) for 1 < p < ∞ and (Λ∞ , k.k∞ ) are Banach spaces.

1.2 Sequence spaces Λ pˆ , pˆ = (pn ), pn > 1 Let pˆ = (pn ) be a bounded sequence of positive real numbers such that pn > 1 for each n ∈  pn ∞  1 n N0 . Define a convex modular σ (x) on l 0 as σ (x) = ∑ − )|x | ( λ λ ∑ k k−1 k and denote n=0 λn k=0

Λx(n) =

1 λn

n

∑ (λk − λk−1 )|xk |.

k=0

Then we define the following set  Λ pˆ = x = (xk ) ∈ l 0 : σ (rx) < ∞ for some r > 0 ,

which is a normed linear space equipped with the Luxemberg norm o x n ≤1 . kxk pˆ = inf r > 0 : σ r Indeed (Λ pˆ , kxk pˆ ) becomes a Banach space.

Remark 1. In particular, (i) if we put λn = n + 1, then sequence spaces Λ p and Λ pˆ reduces to ces p and ces pˆ respectively ([14], [16] and [31]). (ii) choose q = (qk ) = (λk − λk−1 ) and Qn =

n

∑ qk = λn , then sequence spaces Λ p and Λ pˆ

k=0

reduces to ces[p, q] and ces[ p, ˆ q] ˆ respectively [14]. (iii) when pn−k = (λk − λk−1 ) for k = 0, 1, . . ., n and N =

n

∑ qk = λn, then we get absolute

k=0

type N¨orlund sequence spaces. The N¨orlund sequence spaces of non-absolute type is studied by Wang [34].

2 James constants of Λ p for 1 < p ≤ ∞ The unit ball of a normed linear space is uniformly non-square if and only if there is a positive number δ such that there do not exist members x and y of the unit ball for which k 21 (x + y)k >

1 − δ and k 21 (x − y)k > 1 − δ (see [13]).

3

The James constant (or measure of uniform non-squareness) of a real Banach space (X , k · k) with dim(X ) ≥ 2 is denoted by J(X ) and is defined as ([9], [12]) J(X ) = sup{min(kx + yk, kx − yk) : x, y ∈ X , kxk = 1, kyk = 1}. We begin with finding the James constants of the sequence spaces Λ p for 1 < p ≤ ∞. Theorem 1. The James constants of Λ-sequence spaces Λ p for 1 < p ≤ ∞ is 2. In notation

J(Λ p ) = 2, 1 < p ≤ ∞.

Proof. First we discuss the case 1 < p < ∞. For each m = 0, 1, 2, . . ., we denote em = (emk ) = (0, 0, . . ., 0, 1, 0, . . .), where 1 is at the m-th position. Note that 1 n ∑ (λk − λk−1 )emk = λn k=0

(

0 λm −λm−1 λn

Let xm = (xmk ) and ym = (ymk ), where xm = Then kxm k p = 1 and kym k p = 1.

if n < m, if n ≥ m. em kem k p

and ym =

em+1 kem+1 k p

for each m = 0, 1, 2, . . ..

Now we find the value of kxm ± ym k p for m = 0, 1, 2, . . .. We have kxm ± ym k pp p ∞  1 n λ λ − )|x ± y | ( =∑ ∑ k k−1 mk mk n=0 λn k=0 e ∞  e(m+1)k  p 1 n mk =∑ ∑ (λk − λk−1 ) kem k p ± kem+1k p n=0 λn k=0 λ − λ  ∞  1  λm − λm−1 λm+1 − λm  p m m−1 p = + ∑ + λm kem k p kem k p kem+1 k p n=m+1 λn    ∞ 1 λm − λm−1 λm+1 − λm p + ≥ ∑ kem k p kem+1 k p n=m+1 λn   λm − λm−1 kem+1 k p p = 1+ . . λm+1 − λm kem k p

(1)

Therefore by removing the power p, we get kxm ± ym k p ≥ 1 + λλm −λ−m−1 λm . m+1

kem+1 k p kem k p .

We claim that

λm − λm−1 kem+1 k p . = 1. m→∞ λm+1 − λm kem k p lim

Since ∞

∞

p  λ − λ m m−1 p kem+1 k p . λm+1 − λm kem k pp

=

1 ∑ λnp n=m+1 ∞

1 ∑ λnp n=m

1 p n=m+1 λn

∑

= 1 p λm

4

∞

1 + ∑ p n=m+1 λn

=

1+

∞

1 λmp

1 ∑ λnp n=m+1

!−1

,

by the discrete version of Bernoulli-de l’ Hospital rule, we have

lim

m→∞

∞

1 p λm

1 ∑ λnp n=m+1

= lim

1 λmp

− λ p1

m+1

1

m→∞

p λm+1

1 λm p = lim 1 m→∞ p λm+1

− 1 = lim

m→∞

λ

m+1

λm

p

−1 = 0

(2)

Eqn (2) proves our claim. Therefore from Eqn (1), we have kxm ± ym k p ≥ 2 as m → ∞. Since we always know that kxm ± ym k p ≤ 2, so J(Λ p ) = 2 for 1 < p < ∞.

For p = ∞, we choose xm =

λm λm −λm−1 em

and ym =

λm λm+1 −λm em+1 .

Then it is easy to verify that

kxm k∞ = 1, kym k∞ = 1 and kxm ± ym k∞ = 1 + λλm → 2 as m → ∞. Hence J(Λ∞ ) = 2. m+1

Corollary 1. (i) Choose λn = n + 1, then J(ces p ) = 2 for 1 < p ≤ ∞ ([19]). n

(ii) If Qn =

∑ qk = λn, where q = (qk ) = (λk − λk−1 ) then J(ces[p, q]) = 2 for 1 < p ≤ ∞.

k=0

(1)

if there is δ ∈ (0, 1) such that for any

n−1 x0 , x1 , x2 , . . ., xn−1 from the unit ball of X , we have min ∑ εk xk ≤ n(1 − δ ). A Banach space is said to be uniformly non-ln

εk =±1

k=0

(1)

This definition leads to the notion of n-th James constant (or the measure of uniformly non-ln ) Jn (X ), n ∈ N of a Banach space X is defined as Jn (X ) = sup

n

o

n−1

min ∑ εk xk : xk ∈ X , kxk k ≤ 1, k = 0, 1, 2, . . ., n − 1 [9].

εk =±1

k=0

When we restrict to unit sphere of a real Banach space X then the James constants (or n-th strong James constants) are denoted by Jns (X ), n ∈ N and defined by Jns (X ) =

sup

n

n−1

o

min ∑ ε j x j : kx j k = 1, j = 0, 1, 2, . . ., n − 1 .

ε j =±1

j=0

It is to be noted that Jns (X ) ≤ Jn (X ) ≤ n and J2s (X ) = J2 (X ) = J(X ) [19].

Theorem 2. The strong n-th James constant Jns (Λ p ) = n for 1 < p < ∞, and Jns (Λ∞ ) = n. Proof. In previous theorem, we have established the result for the case when n = 2. Therefore we deduce this result for n ≥ 3. Let m ∈ N0 , n ∈ N such that n ≥ 3 and put, x j,m =

em+ j , j = 0, 1, 2, . . ., n − 1. kem+ j k p

With this setting it is clear that kx j,m k p = 1 for each j = 0, 1, 2, . . ., n − 1. Since x j1 ,m and x j2 ,m have disjoint supports for j1 6= j2 , so it implies that

n−1

n−1



min ∑ ε j x j,m = ∑ x j,m .

ε j =±1

p

j=0

5

j=0

p

λi − λi−1 , l = 0, 1, 2, . . ., n − 1, then we have i=m kei k p

m+l

Choose bm,l =

∑

n−1

p m+n−1 e p

j

∑ x j,m = ∑

p j=m ke j k p p j=0

  p 1 1 1

, ,..., , 0, 0, . . . = 0, 0, . . ., 0, kem k p kem+1 k p kem+n−1 k p p  b p  b p b p  b p m,0 m,1 m,n−1 m,n−1 = + +...+ + +... λm λm+1 λm+n−1 λm+n  b ∞ m,n−1 p ≥ ∑ λk k=m+n−1  1 p ∞ = (bm,n−1 ) p ∑ k=m+n−1 λk p  m+n−1 λ − λ  p  ke m+n−1 k p i i−1 = ∑ keik p . λm+n−1 − λm+n−2 i=m p λ − λ   ke m+n−1 k p m m−1 p ≥ np . kem k p λm+n−1 − λm+n−2 Consequently, we have

n−1

n ≥ ∑ x j,m ≥ n p

j=0

kem+n−1 k p λm − λm−1 . λm+n−1 − λm+n−2 kem k p

(3)

kem+n−1 k p λm − λm−1 . = 1. Denote am+n−1 := m→∞ λm+n−1 − λm+n−2 kem k p

We claim that lim Then we have

∞

1 p. j=m+n−1 λ j

∑

∞



p  p ke λm − λm−1 m+n−1 k p = . λm+n−1 − λm+n−2 kem k pp

1 p j=m+n−1 λ j

∑ ∞

1

∑ λp

j=m

 = 1+

=

1 λmp

1

+ λp

m+1

am+n−1 +...+ λp 1

m+n−2

+ am+n−1

j

−1 1 1 + p +...+ p . λmp am+n−1 λm+1 λm+n−2 am+n−1 am+n−1 1

1

Now we find out the limits lim

p m→∞ λ m+i am+n−1

for i = 0, 1, . . ., n − 1. When i = 0, applying

Bernoulli-de l’ Hospital rule, we get

lim

m→∞

∞

1 p λm

1 ∑ λp j=m+n−1 j

= lim

m→∞

1 λmp

− λ p1

m+1

1

p λm+n−1

= lim

m→∞

6

λ

m+n−1

λm

p

− lim

m→∞

λ

m+n−1

λm+1

p

= 0.

(4)

Therefore Eqn (4) proved our claim. Similarly, we get lim

m→∞ λ

1 p m+i am+n−1

= 0 for i = 1, . . ., n−1.

n−1

Hence by Eqn (3), we get ∑ x j,m ≥ n as m → ∞. p

j=0

n−1

Since ∑ x j,m ≤ n, so by definition we obtain Jns (Λ p ) = n for 1 < p < ∞. p

j=0

For the case p = ∞, we choose x j,m =

λm+ j em+ j , for j = 0, 1, . . ., n − 1. λm+ j − λm+ j−1

It is easy to find that kx j,m k∞ = 1 and for any fixed n ∈ N

n−1

n−1



∑ ε j x j,m = ∑ x j,m = 1 + ∞

j=0

j=0

∞

λm λm+n−1

+

λm+1 λm+n−2 +...+ λm+n−1 λm+n−1

→ n as m → ∞

λm+i

(Since lim

m→∞ λm+n−1

= 1 for each i = 0, 1, 2, . . ., n − 2).

Again by definition, we have Jns (Λ∞ ) = n and thus the theorem is proved. Corollary 2. (i) Choose λn = n + 1, then Jns (ces p ) = n for 1 < p ≤ ∞ ([19]). n

(ii) If Qn =

∑ qk = λn, where q = (qk ) = (λk − λk−1 ) then Jns (ces[p, q]) = n for 1 < p ≤ ∞.

k=0

3 Geometric properties of Λ pˆ Let (X , k.k) be a Banach space and being a subspace of l 0 . As usual, we denote S(X ) and B(X ) for the unit sphere and closed unit ball respectively. A point x ∈ S(X ) is said to be an extreme point of B(X ) if there does not exist two distinct points y, z ∈ B(X ) such that 2x = y+z.

The concept of extreme point plays an important role in the study of Krein-Milman theorem, Choquet integral representation theorem etc. Foralewski [10] introduced the notion of coordinatewise Kadec-Klee property of a Banach

space and is denoted by (Hc ). X is said to possess the property (Hc ), if x ∈ X and every sequence (xl ) ⊂ X such that kxl k → kxk and xli → xi as l → ∞ for each i, then kxl − xk → 0. If for every ε > 0 there exists a δ > 0 such that (xl ) ⊂ B(X ), sep(xl ) ≥ ε , kxl k → kxk and xli → xi for each i implies kxk ≤ 1 − δ , where sep(xl ) = inf{kxl −xm k : l 6= m}, then we say X has the coordinatewise uniformly KadecKlee property and is denoted by X ∈ (U KKc) [35]. For any Banach space X , (U KKc) ⇒ (Hc). 7

X is said to have the uniform Opial property ((U OP), in short) if for each ε > 0 there exists µ > 0 such that 1 + µ ≤ lim inf kxl + xk l→∞

for any weakly null sequence (xl ) in S(X ) and x ∈ X with kxk ≥ ε (see [7], [25]).

A Banach space X has the property (β ) if and only if, for every ε > 0, there exists δ > 0 such that, for each element x ∈ B(X ) and each sequence (xl ) ∈ B(X ) with sep(xl ) ≥ ε , there is an index k such that



x+xk

2 ≤ 1 − δ [15].

The weakly uniform normal structure of a Banach space X (WU NS(X ), in short) is determined by the weakly convergent sequence coefficient of X (WCS(X ), in short) ([1], [3]) is defined as WCS(X ) = inf

n

lim sup kxn − xm k k n,m≥k

inf{lim sup kxn − yk n

: y ∈ Conv( xn )}

o ,

where infimum is taken over all weakly convergent sequence (xn ) which is not norm convergent. If WCS(X ) > 1 then Banach space X has WU NS. A Banach space X has WU NS if it possesses (U OP) [17]. Let (Xn , k.kn) be Banach spaces for each n ∈ N0 . Then the Nakano sequence spaces l pˆ (Xn) is defined as o n ∞ l pˆ (Xn ) = x = (xn )n=0 : xn ∈ Xn for each n ∈ N0 and ρ (rx) < ∞ for some r > 0 , where convex modular ρ is defined as ρ (x) =

∞

∑ kxn knpn . It is easy to show that the sequence

n=0

space l pˆ (Xn ) is a Banach space equipped with the Luxemberg norm o n x kxk = inf r > 0 : ρ ( r ) ≤ 1 .

Saejung [27] proved that Ces`aro sequence spaces ces p for 1 < p < ∞ are isometrically embedded in the infinite l p -sum l p(Rn ) of finite dimensional spaces Rn . Here we present similar result for the sequence space Λ pˆ . Lemma 3.1. The sequence space Λ pˆ is isometrically embedded in the Nakano sequence spaces l pˆ (Rn+1 ), where Rn+1 is the (n + 1)-dimensional Euclidean space equipped with the following norm: n

k(α0 , α1 , . . . , αn )k = ∑ |αi | for (α0 , α1 , . . . , αn ) ∈ Rn+1 . i=0

Proof. For all x = (xi ) ∈ Λ pˆ , we define the following linear isometry T : Λ pˆ → l pˆ (Rn+1 ) by       T ((xi )) = x0 , λλ0 x0 , (λ1λ−λ0 ) x1 , . . ., λλ0n x0 , (λ1λ−nλ0 ) x1 , . . . , (λn −λλnn−1 ) xn , . . . . 1

1

8

Indeed kT ((xi ))kl pˆ (Rn+1 ) = kT (x0 , x1 , . . . , xi , . . .)kl pˆ (Rn+1 )

  λ  λ (λ1 − λ0 )  (λ1 − λ0 ) (λn − λn−1 ) 

0 0 = x0 , x0 , x1 , . . ., x0 , x1 , . . ., xn , . . . n+1 l pˆ (R ) λ1 λ1 λn λn λn  o n ∞  n pn 1 ≤1 = inf r > 0 : ∑ (λk − λk−1 )|xk | ∑ r λ n n=0 o  x  k=0 n ≤1 = inf r > 0 : σ r = k(xi )k pˆ Hence the lemma. Instead of studying geometric properties of Λ pˆ it is enough to study geometric properties of l pˆ (Rn+1 ) and if such geometric properties are inherited by subspaces then Λ pˆ will have the same property. Certain geometric structure of finite dimensional Banach spaces l p (Xn ), p > 1 is also investigated by Rolewicz [26]. Saejung in his recent work ([27], Theorem 11 (2), p.535) established the following important result. Proposition 1. Suppose each Xn is finite dimensional. Then the space l pˆ (Xn) has property (β ) and uniform Opial property if and only if lim sup pn < ∞. n→∞

Now we have the following new result. Theorem 3. The sequence space Λ pˆ has property (β ) and uniform Opial property if and only if lim sup pn < ∞. n→∞

Proof. Since Rn+1 is finite dimensional, Proposition 1 implies that the space l pˆ (Rn+1 ) possesses the property (β ) and uniform Opial property if and only if lim sup pn < ∞. Since propn→∞

erty (β ) and uniform Opial property are inherited by subspaces, so by Lemma 3.1 we obtain the result. Corollary 3. (i) The space Λ pˆ has WU NS if lim sup pn < ∞. n→∞

(ii) If lim sup pn < ∞, and λn = n + 1 then ces pˆ has property (β ) and uniform Opial property n→∞

([27], Theorem 11 (2)). Theorem 4. The sequence space Λ pˆ possesses coordinate-wise uniform Kadec-Klee property. Proof. Let ε ∈ (0, 1) and take η = ( ε4 ) p , where p∗ = sup pn . We choose δ ∈ (0, 1) such that ∗

∗ (1 − δ ) p

n

> 1 − η . Suppose (xl ) ⊂ B(Λ pˆ ), sep(xl ) ≥ ε , kxl k pˆ → kxk pˆ , xil → xi as l → ∞ and 9

for all i ∈ N0 . We show that there exists a δ > 0 such that kxk pˆ ≤ 1 − δ . In contradict, we suppose that kxk pˆ > 1 − δ . Then we can select a finite set I = {1, 2, . . ., N − 1} on which kx|I k pˆ > 1 − δ . Since xil → xi for each i ∈ N0 , therefore xl → x uniformly on I. Consequently, since kxl k pˆ → kxk pˆ , there exists lN ∈ N such that

kxl |I k pˆ > 1 − δ and k(xl − xm )|I k pˆ ≤

ε 2

∗

for all l, m ≥ lN . ∗

From the first inequality, we simply get σ (xl |I ) ≥ kxl |I k ppˆ > (1 − δ ) p > 1 − η for l ≥ lN . Since

sep(xl ) ≥ ε , i.e., kxl − xm k pˆ ≥ ε , so the second inequality implies that k(xl − xm )|N−I k pˆ ≥ ε2 for l, m ≥ lN , l 6= m. Hence for N ∈ N there exists a lN such that kxlN |N−I k pˆ ≥ ε4 . Without loss of generality, we may assume that kxl |N−I k pˆ ≥ ε4 for all l, N ∈ N. Therefore from the relation p∗

between norm and modular, we have σ (xl |N−I ) ≥ kxl |N−I k pˆ ≥ ( ε4 ) p = η . The convexity of the function f (t) = |t| pn for each n ∈ N0 , we have for any γ ∈ [0, 1] and u ∈ R ∗

that f (γ u) = f (γ u+(1− γ )0) ≤ γ f (u). Therefore, if 0 ≤ u < v < ∞, then f (u) = f ( uv v) ≤ uv f (v), f (v) f (u) which means that u ≤ v for each n ∈ N0 . Assuming now that 0 ≤ u, v < ∞, u + v > 0, we get f (u + v) = u

f (u+v) u+v

+v

f (u+v) u+v

≥u

f (u) u

+v

f (v) v

= f (u) + f (v)

for each n ∈ N0 . Since xl = xl |I + xl |N−I , applying the above fact we get σ (xl |I ) + σ (xl |N−I ) ≤ σ (xl ) ≤ 1. It implies that σ (xl |N−I ) ≤ 1 − σ (xl |I ) < 1 −(1 − η ) = η , i.e., σ (xl |N−I ) < η , which contradicts the inequality σΦ (xl |N−I ) ≥ η and this contradiction completes the proof.

Theorem 5. Let lim sup pn < ∞. Then a point x ∈ S(Λ pˆ ) is an extreme point of B(Λ pˆ ) if and n→∞

only if (i) σ (x) = 1 and (ii) Card(Ax) ≤ 1,

where Ax = {n ∈ Supp x : Λx(n), Λx(n+1), . . . , Λx(m−1) are different from zero and belongs to the interior of affine interval of f (t) = |t| pn for each n ∈ N0 , where m is the smallest number such that m > n and m ∈ Supp x.} Proof. We first prove the necessary part of the theorem. Let x ∈ S(Λ pˆ ) be an extreme point and condition (i) is not true. Put ε = 1 − σ (x) > 0 and consider the following two sequences: y = (yk ) = (x0 , x1 , . . . , xn0 , 0, 0, . . .) z = (zk ) = (x0 , x1 , . . . , xn0 , 2xn0 +1 , 2xn0 +2 , . . .). Then it is clear that 2x = y + z and y 6= z. But n0   pn 1 n − )|x | ( σ (y) = ∑ λ λ ∑ k k−1 k ≤ σ (x) = 1 − ε < 1, and λ n n=0 k=0

10

σ (z) ≤

1 n   pn  pn ∞ n pn 1 λ λ λ λ 2 − )|x | − )|x | + ( ( ∑ ∑ ∑ k k−1 k ∑ k k−1 k λn k=0 n=0 λn k=0 n=n0 +1 n0

(5)

Since lim sup pn < ∞ and x ∈ Λ pˆ , so ∃ a natural number n0 and a constant M > 0 such that n→∞  pn 1 n ∞ ε − )|x | < M. ( for all n > n0 , we have 2 pn ≤ 2M and for every ε > 0, ∑ λ λ k k−1 k ∑ 2 n=n0 +1 λn k=0 Hence from Eqn. (5), we obtain σ (z) < σ (x) + ε = 1. The relation between norm and modular implies that kyk pˆ ≤ 1 and kzk pˆ ≤ 1, which contradicts to the assumption that x is an extreme point. Therefore σ (x) = 1, i.e., condition (i) is proved. To prove condition (ii), if possible, we assume that Ax has at least two elements and show that x is not an extreme point. Let n1 , n2 ∈ Ax and n1 6= n2 . We denote by m1 (respectively m2 ) the smallest number in Supp x such that m1 > n1 ( respectively m2 > n2 ) and assume that m1 ≤ n2 . Therefore by assumption, we have Λx(n1 ), Λx(n1 + 1), . . . , Λx(m1 − 1) and Λx(n2 ), Λx(n2 + 1), . . ., Λx(m2 − 1) are different from zero and belongs to interior of the affine intervals of f (t) =

|t| pn ,

these intervals by the formula f (t) = Denote

ε1 =

n

∑

(λk − λk−1 )|xk |. The affine function defined on k=0 a j t + b j , where j = n1 , n1 + 1, . . . , m1 − 1, n2 , . . . , m2 − 1.

n ∈ N0 . Recall that Λx(n) =

am −1 an1 an1 +1 + +...+ 1 λn1 λn1 +1 λm1 −1

1 λn

and ε2 =

am −1 an2 an2 +1 + +...+ 2 . λn2 λn2 +1 λm2 −1

Choose δ1 6= 0 and δ2 6= 0 such that ε1 δ1 = ε2 δ2 and

Λx(i) ± δi1 , i = n1 , n1 + 1, . . ., m1 − 1 and Λx( j) ± δj2 , j = n2 , . . ., m2 − 1

are all belongs to affine intervals of f . Consider two sequences y = (yn ) and z = (zn ) defined as follows when m1 < n2 : sgn xn1 sgn xm1 δ1 , xn1 +1 , . . ., xm1 −1 , xm1 − δ1 , xm1 +1 , λn1 − λn1 −1 λm1 − λm1 −1 sgn xm2 sgn xn2 . . . , xn2 −1 , xn2 − δ2 , xn2 +1 , . . . , xm2 −1 , xm2 + δ2 , xm2 +1 , . . .) λn2 − λn2 −1 λm2 − λm2 −1 sgn xn1 sgn xm1 z = (x0 , . . ., xn1 −1 , xn1 − δ1 , xn1 +1 , . . . , xm1 −1 , xm1 + δ1 , xm1 +1 , λn1 − λn1 −1 λm1 − λm1 −1 sgn xn2 sgn xm2 . . . , xn2 −1 , xn2 + δ2 , xn2 +1 , . . . , xm2 −1 , xm2 − δ2 , xm2 +1 , . . .), λn2 − λn2 −1 λm2 − λm2 −1 y = (x0 , . . . , xn1 −1 , xn1 +

where sgn denotes the signum function. The case when m1 = n2 , we define y and z similarly, where ym1 = xm1 −

sgn xm1 (δ1 + δ2 ) λm1 − λm1 −1

and zm1 = xm1 +

For the above two sequences y and z, we have y + z = 2x and

11

sgn xm1 (δ1 + δ2 ). λm1 − λm1 −1

m1 −1

∑ (Λy(n))

pn

=

m1 −1

∑ (Λx(n))

m2 −1

∑ (Λy(n))

+ ε1 δ1 ,

m1 −1

m1 −1

n=n1

n=n1

(Λz(n)) pn =

∑

(Λx(n)) pn − ε1 δ1 ,

pn

=

n=n2

n=n1

n=n1

∑

pn

m2 −1

∑ (Λx(n)) pn − ε2δ2

n=n2

m2 −1

m2 −1

n=n2

n=n2

∑

(Λz(n)) pn =

∑ (Λx(n)) pn + ε2δ2 .

Hence σ (y) = σ (z) = σ (x) = 1 and relation between norm and modular implies that kxk = 1,

kyk = 1 and kzk = 1. Therefore x is not an extreme point and our assumption that Ax has atleast two elements is wrong. So Card(Ax) ≤ 1, i.e, condition (ii) is proved.

Now we proof sufficiency of the theorem, that is if x ∈ S(Λ pˆ ) then x is an extreme point of B(Λ pˆ ). Let the assumptions (i) − (ii) holds. We assume that there exists y, z ∈ S(Λ pˆ ) such that x = y+z 2 but y 6= z. Let n1 ∈ N be the smallest number such that yn1 6= zn1 . Then we

show that n1 ∈ Suppx. Suppose in contradict, we assume that n1 ∈ / Suppx, that is xn1 = 0. Then we immediately have |yn1 | = |zn1 | > 0. Let l be the smallest number in Suppx such that

n1 < l. With out loss of generality, we assume that |xl | ≤ |yl | and we have 0 = |xn | ≤ |yn | for n = n1 + 1, . . ., l − 1. Now Λy(l) 1 = (λ0 |y0 | + . . . + (λn1 − λn1 −1 )|yn1 | + (λn1 +1 − λn1 )|yn1 +1 | + . . . + (λl − λl−1 )|yl |) λl 1 = (λ0 |x0 | + . . . + (λn1 − λn1 −1 )|yn1 | + (λn1 +1 − λn1 )|yn1 +1 | + . . . + (λl − λl−1 )|yl |) λl 1 ≥ (λ0 |x0 | + . . . + (λn1 − λn1 −1 )|yn1 | + (λn1 +1 − λn1 )|xn1 +1 | + . . . + (λl − λl−1 )|xl |) λl 1 > (λ0 |x0 | + . . . + (λn1 − λn1 −1 )|xn1 | + (λn1 +1 − λn1 )|xn1 +1 | + . . . + (λl − λl−1 )|xl |) λl = Λx(l),

which implies that σ (y) > σ (x) = 1, a contradiction with the assumption that y ∈ S(Λ pˆ ). There-

fore n1 ∈ Suppx. Since

∞

y + z  pn 1 n k k =∑ 1 = σ (x) = σ (λk − λk−1 ) ∑ 2 2 n=0 λn k=0  pn ∞  1 n 1 n (λk − λk−1 )|yk | + (λk − λk−1 )|zk | ≤∑ ∑ ∑ 2λn k=0 n=0 2λn k=0   pn  pn 1 ∞ 1 n 1 ∞ 1 n ≤ ∑ ∑ (λk − λk−1 )|yk | + 2 ∑ λn ∑ (λk − λk−1 )|zk| 2 n=0 λn k=0 n=0 k=0
1 = σ (y) + σ (z) = 1. 2 y + z

12

Therefore for each n ∈ N0 , we have ∞

y + z  pn 1 n k k − ) ( λ λ ∑ λn ∑ k k−1 2 n=0 k=0 =

 pn 1 ∞  1 n  pn 1 ∞ 1 n λ λ λ λ − )|y | − )|z | + ( ( ∑ λn ∑ k k−1 k ∑ λn ∑ k k−1 k 2 n=0 2 n=0 k=0 k=0

(6)

With out loss of generality, we assume that 0 ≤ |yn1 | < |xn1 | < |zn1 |. Using the strict convexity of the function f (t) = |t| pn , n ∈ N0 , we have (Λy(n1 )) pn1 < (Λz(n1)) pn1 .

(7)

Let m1 ∈ N0 be the smallest number such that m1 > n1 and m1 ∈ Supp x. If xn = 0 for every n > n1 then by Eqn (7), we arrived at a contradiction σ (y) < σ (z). For n = n1 , n1 +1, . . . , m1 −1,

we have Λy(n) < Λz(n). If there exists n ∈ {n1 , n1 + 1, . . . , m1 − 1} such that   1 p p p n n n (Λx(n)) < 2 (Λy(n)) + (Λz(n))

then we get a contradiction with Eqn (6). Therefore for every n ∈ {n1 , n1 + 1, . . . , m1 − 1} we

assume that

(Λx(n)) pn =

1 2



 (Λy(n)) pn + (Λz(n)) pn .

Then by assumption, we have n1 ∈ Ax . Since σ (y) = 1, σ (z) = 1, so by Eqn (7), there exists n ≥ m1 such that Λy(n) > Λz(n). Let n2 be the such a smallest number by which Λz(n2 ) < Λy(n2 ). Then |zn2 | < |yn2 | and so n2 ∈ Supp x. If xn = 0 for every n > m2 then we get a contradiction with σ (z) < σ (y). Let m2 be the smallest number in Supp x such that m2 > n2 . By assumption n2 ∈ / Ax , so there

exists n ∈ {n2 , n2 + 1, . . ., m2 − 1} such that   (Λx(n)) pn < 12 (Λy(n)) pn + (Λz(n)) pn ,

which contradicts the Eqn (6). Thus we must have x = y = z, i.e., x is an extreme point. (2)

4 Von Neumann-Jordan constant of Λ2 (2)

For two dimensional sequence space Λ p , norm k(u, v)k p is given by   λ |u| + (λ − λ )|v|  p  1 p 0 1 0 p k(u, v)k p = |u| + . λ1 The von Neumann-Jordan constant CNJ (X ) of a Banach space X was introduced by Clarkson [4] and is defined as follows:

13

CNJ (X ) = sup

n

kx+yk2 +kx−yk2 2(kxk2 +kyk2 )

Now we begin with the following result.

o : x, y ∈ X and kxk + kyk = 6 0 . (2)

Theorem 6. The von Neumann-Jordan constant CNJ (Λ2 ) = 1 + √

λ0 . λ02 +λ12

(2)

Proof. Choose (a, b), (c, d) ∈ Λ2 . Then k(a, b) ± (c, d)k2 = k(a ± c, b ± d)k2   λ |a ± c| + (λ − λ )|b ± d| 2  0 1 0 = |a ± c|2 + λ1  λ2 2λ 0 ( λ 1 − λ 0 ) (λ1 − λ0 )2 |a ± c||b ± d| + |b ± d|2 . = 1 + 02 |a ± c|2 + 2 2 λ1 λ1 λ1 Now q λ02 + λ12

o (λ1 − λ0 ) 2λ 0 ( λ 1 − λ 0 ) λ0 q |a ± c| |b ± d| 2 |a ± c||b ± d| = λ1 λ1 λ12 λ02 + λ12 nλ2 +λ2 o λ0 (λ1 − λ0 )2 2 2 0 1 ≤q . |a ± c| + |b ± d| λ12 λ12 λ02 + λ12 n

Therefore k(a, b) + (c, d)k2 + k(a, b) − (c, d)k2 n λ 2 + λ 2 o  (λ1 − λ0 )2 λ0 2 2 2 2 0 1 (|a + c| + |a − c| ) + (|b + d| + |b − d| ) ≤ 1+ q λ12 λ12 λ02 + λ12  n λ 2 + λ 2 o λ0 (λ1 − λ0 )2 2 2 2 2 0 1 = 2 1+ q (|a| + |c| ) + (|b| + |d| ) λ12 λ12 λ02 + λ12  n λ 2 + λ 2 λ0 2λ 0 ( λ 1 − λ 0 ) (λ1 − λ0 )2 2 2 0 1 ≤ 2 1+ q |a| + |a||b| + |b| 2 2 2 λ λ λ 2 2 1 1 1 λ0 + λ1 λ2 +λ2 2λ 0 ( λ 1 − λ 0 ) (λ1 − λ0 )2 2 o + 0 2 1 |c|2 + |c||d| + |d| λ1 λ12 λ12   λ0 = 1+ q 2(k(a, b)k2 + k(c, d)k2 ) . λ02 + λ12 (2)

Hence CNJ (Λ2 ) ≤ 1 + √

λ0 . λ02 +λ12

14

Consider (a, b) = (λ1 − λ0 , 0) and (c, d) = (0, (2) CNJ (Λ2 ) ≥

q λ02 + λ12 ). Then from the definition, we have

k(a, b) + (c, d)k2 + k(a, b) − (c, d)k2 λ0 . = 1+ q 2 2 2(k(a, b)k + k(c, d)k ) λ2 +λ2 0

(2)

Combining last two inequalities, we obtain CNJ (Λ2 ) = 1 + √ Corollary 4. The James constant

(2) J(Λ2 ) =

r

2 + √ 22λ0

λ0 +λ12

1

λ0 . λ02 +λ12

.

(2)
2 (2) (2) Proof. It is well-known that 21 J(Λ2 ) ≤ CNJ (Λ2 ) = 1 + √ λ20 2 . Therefore J(Λ2 ) ≤ λ0 +λ1 r     2 + √ 22λ0 2 . Equality occurs when x = √ λ21 2 , 0 and y = 0, λ λ−1λ .

λ0 +λ1

λ0 +λ1

1

0

(2)

(2)

Corollary 5. If λ0 = 1 and λ1 = 2 then CNJ (ces2 ) = 1 + √1 and J(ces2 ) = 5 [19], [27]).

q 2 + √2 (see 5

Corollary 6. If λ0 = q0 and λ1 = q0 + q1 then CNJ (ces[2, q](2) ) = 1 + √ 2 q0 and 2q0 +2q0 q1 +q21 r J(ces[2, q](2) ) = 2 + √ 2 2q0 . 2 2q0 +2q0 q1 +q1

The following theorem is a counter part of a theorem presented in ([27], Theorem 15, p.536) (2) for the sequence space Λ p . We repeat here for the sake of completeness.

ψ (t) 2 , 1 < p ≤ 2 where 0≤t≤1 ψ2 (t) q  p  p 1  p λ1 (1−t) p λ0 (1−t) and ψ2 (t) = ψ (t) = λ p +λ p + (λ p +λ p )1/p + t (1 − t)2 + t 2 . (2)

Theorem 7. The von Neumann-Jordan constant CNJ (Λ p ) =

0

1

0



sup

1

Proof. For (x, y) ∈ R2 , we define a norm on R2 as

 

|(x, y)| = (λ p +λλ1 xp )1/p , (λ λ−1 yλ ) 1 0 0

1

(2)

Λp

.

It is easy to verify that |(x, y)| = |(|x|, |y|)| and |(1, 0)| = 1 = |(0, 1)|, that is |(x, y)| defines an (2) absolute and normalized norm. Additionally, choose a map T : (R2 , |(, )|) → Λ p defined as 
 T (x, y) = (λ p +λλ1 xp )1/p , (λ λ−1 yλ ) . 0

1

1

0

(2)

It is easy to show that T is an isometric isomorphism, i.e, sequence spaces Λ p are isometrically isomorphic to (R2 , |(, )|). Using Saito et al. result ([28], Theorem 1, p. 521) it is enough to prove that ψ ≥ ψ2 and observe that p  λ0p (1−t) p λ0 (1−t) + t ≥ + t p, p p 1/p λ p +λ p (λ +λ ) 0

0

1

1 p

1

  1 which implies ψ (t) ≥ (1 − t) p + t p ≥ (1 − t)2 + t 2 2 = ψ2 (t). Hence the result. 15

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