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Banach lattices which are N -order uniformly noncreasy
J. Math. Anal. Appl. 399 (2013) 459–471
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Banach lattices which are N-order uniformly noncreasy Anna Betiuk-Pilarska, Stanisław Prus ∗ Institute of Mathematics, M. Curie-Skłodowska University, 20-031 Lublin, Poland
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info
Article history: Received 30 June 2012 Available online 22 October 2012 Submitted by T.D. Benavides
abstract An N-dimensional Riesz angle for a Banach lattice and the class of N-order uniformly noncreasy Banach lattices are introduced and studied. A fixed point theorem for these lattices is proved. © 2012 Elsevier Inc. All rights reserved.
Keywords: Uniform monotonicity Order uniform smoothness N-dimensional Riesz angle N-order uniform noncreasiness Mapping with condition (C) Weak fixed point property
1. Introduction One of the main problems of metric fixed point theory concerns existence of fixed points of nonexpansive mappings. In [1], Maurey applied an ultrapower technique to this problem (see also [2]). His approach turned out to be very fruitful. In particular it enabled many authors to prove fixed point theorems for nonexpansive mappings in Banach spaces satisfying various geometric conditions (see [3,4]). One of them was introduced in [5] as a combination of two classical geometric properties: uniform convexity and uniform smoothness. Banach spaces with this property are called uniformly noncreasy. In [6] a new property which is related to order was introduced. Banach lattices with this property are said to be order uniformly noncreasy. This definition was inspired by Kurc [7] who considered uniform monotonicity and order uniform smoothness of Banach lattices and showed a duality theorem for these properties. Order uniform noncreasiness can be seen as a combination of uniform monotonicity and order uniform smoothness and it is essentially weaker than each of them. Modulus of order uniform smoothness is strictly related to the Riesz angle which was introduced by Borwein and Sims [8] to prove a fixed point theorem for nonexpansive mappings. In [6] their result was extended to the class of order uniformly noncreasy lattices. Later on a further extension was given by S. Dhompongsa and A. Kaewcharoen in [9]. Their result covers the class of continuous mappings satisfying condition (C), which is larger than the class of nonexpansive mappings. In this paper we introduce a new coefficient αN (X ) for a Banach lattice X which we call the N-dimensional Riesz angle. For N = 2 it coincides with the Riesz angle considered by Borwein and Sims. We give estimates for αN (X ) and find the exact value of this coefficient for some lattices, including Lp spaces. We show relations between αN (X ) and some geometric lattice properties. Following an idea from [6], we introduce a new class of lattices: N-order uniformly noncreasy Banach lattices. Then we extend the fixed point result from [9] to this class. 2. Geometric properties of Banach lattices Let X be a Banach space. By BX we denote the unit ball of X . For notation and terminology concerning Banach lattices we refer the reader to [10]. Let us recall that if an inequality involves lattice and algebraic operations, then it is enough to
∗
Corresponding author. E-mail addresses: [email protected] (A. Betiuk-Pilarska), [email protected] (S. Prus).
0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.10.025
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check its validity for real numbers to be sure that it holds for vectors in an arbitrary Banach lattice (see [10, p. 1]). In the next lemma we collect some inequalities which will be used in the sequel. Lemma 2.1. Let X be a Banach lattice. Then (i) 0 ≤ |x| − (|x| − |y|)+ ≤ |y|, (ii) |z | ≤ |x − z | ∨ |y − z | + |x| ∧ |y|, (iii) |x| + |y| − 2(|x| ∧ |y|) ≤ |x − y| ≤ |x| + |y| for all x, y, z ∈ X . Given a Banach lattice X and ε ∈ [0, 1], we put
δm,X (ε) = inf{1 − ∥x − y∥ : 0 ≤ y ≤ x, ∥x∥ ≤ 1, ∥y∥ ≥ ε}. We say that X is uniformly monotone if δm,X (ε) > 0 for all ε ∈ (0, 1]. The coefficient
ε0,m (X ) = sup{ε ∈ [0, 1) : δm,X (ε) = 0} is called the characteristic of monotonicity of the lattice X . In [7], the modulus of order smoothness of a Banach lattice X was introduced as follows
ρm,X (t ) = sup{∥x ∨ ty∥ − 1 : x, y ∈ BX , x, y ≥ 0} where t ∈ [0, 1]. A Banach lattice X is called order uniformly smooth if lim
ρm,X (t )
t →0
t
= 0.
The coefficient
ρm,X (1) + 1 = sup{∥x ∨ y∥ : x, y ∈ BX , x, y ≥ 0} is called the Riesz angle of X and denoted by α(X ) (see [8]). Recall that 1 ≤ α(X ) ≤ 2 and α(X ) = 1 if and only if X is an abstract M space. Moreover, if X is an abstract L1 space, then α(X ) = 2. In particular, α(c0 ) = α(c ) = 1 and α(l1 ) = 2. Kurc [7] proved the duality formulas
ρm,X ∗ (t ) = sup (ε t − δm,X (ε)) 0≤ε≤1
δm,X (ε) = sup (εt − ρm,X ∗ (t )) 0≤t ≤1
for all ε, t ∈ [0, 1]. They show that a Banach lattice X is uniformly monotone (resp. order uniformly smooth) if and only if the dual lattice X ∗ is order uniformly smooth (resp. uniformly monotone). Using the above formulas it is also easy to see that ρm,X ∗ (1) < 1 if and only if ε0,m (X ) < 1. In [6] it was proved that if ρm,X (1) < 1 and ε0,m (X ) < 1, then X is superreflexive. In particular, if a Banach lattice X is order uniformly smooth and uniformly monotone, then X is superreflexive. Now we introduce a generalized Riesz angle. Definition. Let X be a Banach lattice and N ∈ N, N ≥ 2. The N-dimensional Riesz angle of X is defined as
αN (X ) = sup (xi ∨ xj ) : x1 , . . . , xN ∈ BX , x1 , . . . , xN ≥ 0 . N i,j=i1̸=,..., j Of course αN (X ) ≥ 1. Moreover, α2 (X ) = α(X ) and αN (X ) ≤ αN −1 (X ) for every N ≥ 3. Hence αN (X ) ≤ α(X ). This estimate is not sharp. To get a better upper bound for αN (X ) we need the following lemma. Lemma 2.2. For N ∈ N, N ≥ 2 and for all nonnegative scalars x1 , . . . , xN ∈ R the following inequality holds
(xi ∨ xj ) ≤
i,j=1,...,N i̸=j
1
N − 1 i=1
Proof. We have
i,j=1,...,N i̸=j
N
(xi ∨ xj ) = xk
xi .
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461
for some k. Let xm = min{xi : i = 1, . . . , N }. Then xk ≤ xi ∨ xm = xi for every i ̸= m. Hence xk ≤
N
1 N −1
xi ≤
i=1 i̸=m
N
1
N − 1 i=1
xi .
Proposition 2.3. Let X be a Banach lattice and N ∈ N, N ≥ 2. Then
αN (X ) ≤
N N −1
.
Proof. Let x1 , . . . , xN ∈ BX , x1 , . . . , xN ≥ 0. Lemma 2.2 leads to the inequality
z :=
(xi ∨ xj ) ≤
i,j=1,...,N i̸=j
Hence ∥z ∥ ≤
1 N −1
N
i =1
N
1
N − 1 i=1
∥x i ∥ ≤
N , N −1
xi .
which proves that αN (X ) ≤
N . N −1
We will show relations between αN (X ) and some geometric lattice properties. Let us recall that a Banach lattice X is said to satisfy an upper p-estimate, where 1 ≤ p < +∞, if there exists a constant M > 0 such that
1p n n p ∥x i ∥ x≤M i=1 i i =1 for all pairwise disjoint elements x1 , . . . , xn in X . Our next result generalizes Theorem 2.3(i) in [6]. Theorem 2.4. Let X be a Banach lattice and N ∈ N, N ≥ 2. If αN (X ) < p ∈ (1, +∞).
N , N −1
then X satisfies an upper p-estimate for some
Proof. Suppose that X does not satisfy an upper p-estimate for any p ∈ (1, +∞). By [10, Theorem 1.f.12], for every ε > 0 there exist disjoint elements x1 , . . . , xN ∈ X such that
(1 − ε)
N i=1
N N ai x i ≤ |a | |ai | ≤ i=1 i=1 i
(1)
for all scalars a1 , . . . , aN . Put yk =
N
1 N −1
|xi |
i=1 i̸=k
for k = 1, . . . , N. By (1) we have ∥xk ∥ ≤ 1 and consequently, ∥yk ∥ ≤ 1. Moreover, since x1 , . . . , xN are pairwise disjoint, yk ∨ yl =
1
N
N − 1 i =1
| xi |
whenever k ̸= l. Substituting ai =
1 N −1
into (1), we obtain
N N N 1 1 (1 − ε) ≤ xi ≤ |xi | = (yk ∨ yl ) . N − 1 i=1 N − 1 i=1 k,l=1,...,N N −1 k̸=l This and Proposition 2.3 show that αN (X ) =
N . N −1
Let us now recall that a Banach lattice X is said to be p-convex, where 1 ≤ p < +∞, if there exists a constant M > 0 such that
1p 1p n n p p ≤M |xi | ∥x i ∥ i=1 i=1 for all vectors x1 , . . . , xn ∈ X . If a Banach lattice X is p-convex, then X satisfies the upper p-estimate and if X satisfies an upper r-estimate for some 1 < r < +∞, then X is p-convex for every p ∈ [1, r ) (see [10, p. 85]). We give an estimate for αN (X ) for a p-convex Köthe function space X (see [10, p. 28] for the definition).
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Theorem 2.5. Let N ∈ N, N ≥ 2. If a Köthe function space X is p-convex for some p ≥ 1 with the constant M = 1, then
αN (X ) ≤
1p
N N −1
.
Proof. Our space X is a space of locally integrable functions on some set Ω with a σ -finite measure µ. Let f1 , . . . , fN ∈ BX be nonnegative. We put
z ( x) =
(fi (x) ∨ fj (x))
i,j=1,...,N i̸=j
for every x ∈ Ω . Let Ak = {x ∈ Ω : fk (x) < z (x)} , k = 1, . . . , N. The sets A1 , . . . , AN are pairwise disjoint. Indeed, if x ∈ Ai ∩ Aj for some i ̸= j, then fi (x) < z (x) and fj (x) < z (x). But z (x) ≤ fi (x) ∨ fj (x), which is a contradiction. For a nonempty set D ⊂ Ω by 1D we denote the characteristic function of D. In addition, we put 1∅ = 0. We have z p − z p 1Ak = z p 1Ac for every k = 1, . . . , N, where Ack = Ω \ Ak . Adding these equalities, we obtain k
N
Nz p −
z p 1Ak =
k=1
N
z p 1Ac . k
k=1
But N
z p 1Ak = z p 1A ≤ z p ,
k=1
where A =
N
k=1
Ak , so we have
(N − 1)z p ≤
N
z p 1Ac . k
k=1
By our assumption of p-convexity we therefore get
1p 1p 1p N N N 1 p p p ∥z1Ack ∥ (z1Ack ) ∥f k ∥ ≤ ≤ Np. (N − 1) ∥z ∥ ≤ ≤ k=1 k=1 k=1 1 p
This clearly gives us the estimate αN (X ) ≤
N N −1
1p
.
Consider some special cases. Let us start with the trivial one, when Ω = {1, . . . , N − 1} with the counting measure, i.e. X = RN −1 endowed with a lattice norm. Following the first part of the proof of Theorem 2.5, we obtain pairwise disjoint subsets A1 , . . . , AN of the set Ω of N − 1 elements. Consequently, Ak = ∅ for some k, which means that z ≤ fk and consequently, ∥z ∥ ≤ 1. This shows that αN (X ) = 1. Consider now the case when X = Lp (Ω ) and Ω has N pairwise disjoint subsets A1 , . . . , AN with finite positive measure. We put gk = ((N − 1)µ(Ak )) fi =
N
− 1p
− 1p
1Ak for k = 1, . . . , N. Then ∥gk ∥ = (N − 1)
. Let
gk
k=1 k̸=i
for i = 1, . . . , N. We have ∥fi ∥ = 1 and z :=
(fi ∨ fj ) =
i,j=1,...,N i̸=j
N
gk .
k=1
Hence
∥z ∥p =
N k=1
∥gk ∥p =
N N −1
.
In view of Theorem 2.5, this shows that αN (Lp (Ω )) = and lp .
N N −1
1p
. In particular, this formula is true for the spaces Lp ([0, 1])
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Definition. Let r ∈ (0, 1]. A Banach lattice X is said to be r-N-order uniformly noncreasy (r-N-OUNC) if for all u1 , . . . , uN ∈ N −1 BX such that ∥ui − uj ∥ ≤ 1 we have either N
(|ui | ∨ |uj |) ≤ r i,j=1,...,N i̸=j or there exist i ̸= j such that for every y ∈ X the conditions |y| ≤ |ui − uj |, ∥y∥ ≥ r imply ∥ |ui − uj | − |y| ∥ ≤ r. A Banach lattice X is N-order uniformly noncreasy (N-OUNC) if it is r-N-OUNC for some r ∈ (0, 1). For N = 2 this definition coincides with the definition of an order uniformly noncreasy Banach lattice given in [6]. Of course, each Banach lattice X is r-N-OUNC with r = NN−1 αN (X ). Therefore, if αN (X ) < NN−1 , then X is N-OUNC. It is also easy to see that X is r-N-OUNC where r = max ε, 1 − δm,X (ε) for any ε ∈ (0, 1). It follows that if ε0,m (X ) < 1, then
X is N-OUNC. The class of all N-OUNC Banach lattices contains therefore all Banach lattices X with αN (X ) < NN−1 and all uniformly monotone lattices. The following example shows that for different values of N the classes of N-OUNC Banach lattices are essentially different. N −1 Let N ∈ N, N ≥ 3 and X = R ⊕ i=1 c0 . Elements of X are of the form x = (ξ , x1 , . . . , xN −1 ), where ξ ∈ R and x1 , . . . , xN −1 ∈ c0 . We define the norm of such vector x by the formula
∥x∥ = |ξ | +
N −1
∥x i ∥0 ,
i=1
where ∥·∥0 is the standard norm in c0 . In the sequel, given a vector y ∈ c0 we write y = (y(n)). We will show that X is N-OUNC, but it is not M-OUNC for any M < N. We begin by proving that X is r-N-OUNC where r = 1 − N (N1+2) . Let u1 , . . . , uN ∈ X be such that ∥ui ∥ ≤ NN−1 and ∥ui − uj ∥ ≤ 1 for any i, j = 1, . . . , N. We put
z=
(|ui | ∨ |uj |).
i,j=1,...,N i̸=j
Then z = (ζ , z1 , . . . , zN −1 ) for some ζ ∈ R and some z1 , . . . , zN −1 ∈ c0 . For every k = 1, . . . , N − 1 we choose nk ∈ N such that ∥zk ∥0 = zk (nk ). Given x = (ξ , x1 , . . . , xN −1 ), where ξ ∈ R and x1 , . . . , xN −1 ∈ c0 , we put P 1 ( x) = ξ ,
Pk+1 (x) = xk (nk )
for k = 1, . . . , N − 1. Consider the sets Ai = {k : |Pk (ui )| < Pk (z )}, i = 1, . . . , N. It easy to show that they are pairwise disjoint. If Ai = ∅ for some i, then Pk (z ) ≤ |Pk (ui )| for every k = 1, . . . , N and hence
∥z ∥ =
N
Pk (z ) ≤
k=1
N
|Pk (ui )| ≤ ∥ui ∥ ≤
k=1
N −1 N
< r.
Now suppose that sets A1 , . . . , AN are nonempty. Then we can assume that Ai = {i} for every i = 1, . . . , N. Therefore |Pi (ui )| < Pi (z ) and Pk (z ) ≤ |Pk (ui )| for all k ̸= i. Thus
(N − 1)Pk (z ) ≤
N
|Pk (ui )|
i=1 i̸=k
for k = 1, . . . , N, and from this we obtain
∥z ∥ =
N
Pk (z ) ≤
k=1
=
1 N −1
1
N N
N − 1 k=1
N N
|Pk (ui )| −
i=1 k=1
|Pk (ui )| =
i=1 i̸=k
N k=1
1
N N
N − 1 k=1
i=1
|Pk (uk )| ≤
1 N −1
|Pk (ui )| − |Pk (uk )|
N
∥ ui ∥ − s ≤ 1 −
i =1
s N −1
,
|Pk (uk )|. Consequently, if s ≥ N (NN−+12) , then ∥z ∥ ≤ r. Now let us consider the case when s < N (NN−+12) . Of course it suffices to examine the case ∥z ∥ > r. We have
where s =
N
k=1
r < ∥z ∥ =
N k=1
Pk ( z ) ≤ P1 ( z ) +
N k=2
|Pk (u1 )| ≤ P1 (z ) + ∥u1 ∥ ≤ P1 (z ) +
N −1 N
,
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hence N −1
P1 ( z ) ≥ r −
N
.
(2)
For x = (ξ , x1 , . . . , xN −1 ), where ξ ∈ R and x1 , . . . , xN −1 ∈ c0 we put Q (x) = (0, x1 , . . . , xN −1 ). Applying inequality (2), we obtain 1 ≥ ∥u2 − u1 ∥ = |P1 (u2 − u1 )| + ∥Q (u2 − u1 )∥
≥ |P1 (u2 )| − |P1 (u1 )| + ∥Q (u2 − u1 )∥ ≥ P1 (z ) − s + ∥Q (u2 − u1 )∥ N −1 ≥r− − s + ∥Q (u2 − u1 )∥. N
Therefore, ∥Q (u2 − u1 )∥ ≤ 2 + s − N1 − r. Let y ∈ X be such that 0 ≤ y ≤ |u2 − u1 | and ∥y∥ ≥ r. Then Q (y) ≤ Q (|u2 − u1 |), so r ≤ ∥y∥ = P1 (y) + ∥Q (y)∥ ≤ P1 (y) + ∥Q (u2 − u1 )∥ ≤ P1 (y) + 2 + s − Hence P1 (y) ≥ 2r − 2 − s +
1 N
1 N
− r.
and we obtain
∥ |u2 − u1 | − y∥ = |P1 (u2 − u1 )| − P1 (y) + ∥Q (|u2 − u1 | − y)∥ ≤ |P1 (u2 − u1 )| − P1 (y) + ∥Q (u2 − u1 )∥ = ∥u2 − u1 ∥ − P1 (y) ≤ 3 − 2r + s − But s <
N −1 , N (N +2)
1 N
.
which gives us the inequality
∥ |u2 − u1 | − y∥ < 3 − 2r +
N −1 N (N + 2)
−
1 N
= r.
This proves that X is r-N-OUNC. Now we show that X is not M-OUNC for any M < N. Let (en ) be the standard basis of c0 and for k = 1, . . . , M let ekn = (0, x1 , . . . , xN −1 ) be the element of X such that xk = en and xi = (0, 0, . . .) for i ̸= k. Put
vk =
M 2k 1
M
eij
j =1
i=1 i̸=k
for k = 1, . . . , M. Of course
M 2k M −1 1 i ej = . ∥vk ∥ = M i=1 j=1 M 0
i̸=k
Assume that 1 ≤ l < k ≤ M. Then
2k M 2k 2l 1 elj + |vk − vl | = eij + ekj , M
j =1
i=1 j=2l+1 i̸=k,l
j =1
so ∥vk − vl ∥ = 1. Moreover, for
M
1 i y= e2k + ek2l M i=1 i̸=k
we have 0 ≤ y ≤ |vk − vl |, ∥y∥ = 1 and
M 2k −1 2l−1 1 |vk − vl | − y = elj + eij + ekj , 2k−1
M
j =1
i=1 j=2l+1 i̸=k,l
which shows that ∥ |vk − vl | − y∥ = 1.
j=1
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465
Finally we see that
M 2k 2l M 2 1 1 i i k vk ∨ vl = ej + ej ≥ ej M
i=1 i̸=k
j=1
M
j =1
i=1 j=1
and hence
M 2 1 i ej = 1. (vk ∨ vl ) ≥ k,l=1,...,M M i=1 j=1 k̸=l Note also that X contains a copy of c0 , so X is not uniformly monotone. Moreover, X contains a copy of the lattice RN with the l1 -norm and consequently, αN (X ) = NN−1 . To prove our fixed point theorem we will use the ultrapower method (see [4]). Let us therefore briefly recall the construction of the ultrapower. Given a Banach space X , by l∞ (X ) we denote the space of all bounded sequences in X with the supremum norm. Let U be a free ultrafilter on N. The ultrapower (X )U is the quotient space l∞ (X )/N where
N = (xn ) ∈ l∞ (X ) : lim ∥xn ∥ = 0 . n→U
Here limn→U ∥xn ∥ stands for the limit over the ultrafilter U (see [11, p. 13]). The equivalence class of the sequence (xn ) ∈ l∞ (X ) will be denoted by (xn )U . It is easy to see that the quotient norm is given by the formula (see [11, p. 23])
∥(xn )U ∥ = lim ∥xn ∥. n→U
If X is a Banach lattice, then (X )U is also a Banach lattice with the nonnegative cone consisting of all equivalent classes of sequences (xn ) such that xn ≥ 0 for every n. Our next result generalizes Theorem 3.3 in [6]. Theorem 2.6. Let X be a Banach lattice, U be a free ultrafilter on N, r ∈ (0, 1) and n ∈ N, N ≥ 2. If X is r-N-OUNC, then (X )U is r ′ -N-OUNC for every r ′ ∈ (r , 1). Proof. Assume that (X )U is not r ′ -N-OUNC. There exist x1 , . . . , xN ∈
N −1 B(X )U N
such that ∥ xi − xj ∥ ≤ 1 for all i, j,
(| xi | ∨ | xj |) > r ′ i,j=1,...,N i̸=j and for any i, j = 1, . . . , N , i ̸= j, there exists yij ∈ (X )U such that | yij | ≤ | xi − xj |, ∥ yij ∥ ≥ r ′ and ∥ | xi − xj | − | yij | ∥ > r ′ . ′ i Let ε ∈ (0, 1) be such that r /r > 1 + ε . For every i = 1, . . . , N we choose a sequence (xn ) ∈ l∞ (X ) in xi . We have j
limn→U ∥xin − xn ∥ ≤ 1, limn→U ∥xin ∥ ≤
N −1 N
and
i j lim (|xn | ∨ |xn |) > r ′ > (1 + ε)r . n→U N i,j=i1̸=,..., j Fix i ̸= j. There exist sequences (yn ) ∈ l∞ (X ) and (zn ) ∈ l∞ (X ) in the equivalence classes yij and xi − xj respectively such j that |yn | ≤ |zn | for every n ∈ N. We put un = xin − xn − zn . Then limn→U ∥un ∥ = 0 and
|yn | ≤ |xin − xjn − un | ≤ |xin − xjn | + |un | j
ij
j
ij
for every n. Hence |yn | − |un | ≤ |xin − xn |, thus vn ≤ |xin − xn |, where vn = (|yn | − |un |)+ ≥ 0. Applying inequality (i) from Lemma 2.1, we obtain
∥vnij ∥ ≥ ∥yn ∥ − ∥ |yn | − (|yn | − |un |)+ ∥ ≥ ∥yn ∥ − ∥un ∥, ij so limn→U ∥vn ∥ ≥ limn→U ∥yn ∥ ≥ r ′ > (1 + ε)r. Applying again the same inequality, we get
∥ |xin − xjn | − vnij ∥ = ∥ |xin − xjn | − |yn | + |yn | − (|yn | − |un |)+ ∥ ≥ ∥ |xin − xjn | − |yn | ∥ − ∥ |yn | − (|yn | − |un |)+ ∥ ≥ ∥ |xin − xjn | − |yn | ∥ − ∥un ∥. j ij j Therefore limn→U ∥ |xin − xn | − vn ∥ ≥ limn→U ∥ |xin − xn | − |yn | ∥ > r ′ > (1 + ε)r.
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Since U is a nontrivial ultrafilter, there exists n such that N −1 ∥xin − xjn ∥ < 1 + ε, ∥xin ∥ < (1 + ε), N i j (|xn | ∨ |xn |) > r (1 + ε), i,j=1,...,N i̸=j
∥vnij ∥ > r (1 + ε),
∥ |xin − xjn | − vnij ∥ > r (1 + ε) ij
1 1 xin and yn = 1+ε vn for i, j = 1, . . . , N , i ̸= j. Then for all i, j = 1, . . . , N , i ̸= j. We put xn = 1+ε i
ij
N −1 ∥xin − xjn ∥ ≤ 1, ∥xin ∥ ≤ , N i j (|xn | ∨ |xn |) > r , i,j=1,...,N i̸=j
|yijn | ≤ |xin − xjn |,
∥yijn ∥ > r ,
which shows that X is not r-N-OUNC.
∥ |xin − xjn | − |yijn | ∥ > r ,
Corollary 2.7. Let X be a Banach lattice and U be a free ultrafilter on N. If X is N-OUNC, then (X )U is N-OUNC. 3. Applications to fixed point theory A self-mapping T of a subset E of a Banach space X is said to satisfy condition (C) if 1 2
∥x − Tx∥ ≤ ∥x − y∥ implies ∥Tx − Ty∥ ≤ ∥x − y∥
for all x, y ∈ E. Clearly, if a mapping T is nonexpansive, i.e.,
∥Tx − Ty∥ ≤ ∥x − y∥ for all x, y in E, then T satisfies condition (C). However, there are mappings satisfying condition (C) which are not even continuous (see [12]). We say that X has the weak fixed point property for continuous mappings satisfying condition (C) if every such mapping defined on a nonempty weakly compact convex subset of X has a fixed point. Suzuki [12] proved the existence of approximate fixed point sequences for mappings satisfying condition (C). Recall that (xn ) is an approximate fixed point sequence for a mapping T if limn→∞ ∥xn − Txn ∥ = 0. We will generalize Theorem 4.16 in [9]. For this purpose we need the following results. Theorem 3.1 Let (X , d) be a complete metric space and T be a mapping on X . Define a nonincreasing function θ from ([13]). [0, 1) onto 21 , 1 by
1 θ (r ) = (1 − r )r −2 ( 1 + r ) −1
√
if 0√ ≤ r ≤ ( 5 − 1)/2,
if ( 5 − 1)/2 ≤ r ≤ 2−1/2 , if 2−1/2 ≤ r < 1.
Assume that there exists r ∈ [0, 1) such that
θ (r )d(x, Tx) ≤ d(x, y) implies d(Tx, Ty) ≤ rd(x, y) for all x, y ∈ X . Then T has a fixed point. Remark 3.2 ([9]). If we assume in addition that T is continuous and there exist r , r1 ∈ [0, 1) such that
θ (r )d(x, Tx) ≤ d(x, y) implies d(Tx, Ty) ≤ r1 d(x, y) for all x, y ∈ X , then T has a fixed point.
A. Betiuk-Pilarska, S. Prus / J. Math. Anal. Appl. 399 (2013) 459–471
467
We will also need the concepts of asymptotic center and asymptotic radius. Let E be a nonempty bounded closed convex subset of X and (xn ) be a bounded sequence in X . We define the asymptotic radius of (xn ) in E as follows
r (E , (xn )) = inf lim sup ∥xn − x∥ : x ∈ E
n→∞
and the asymptotic center of (xn ) in E as follows A(E , (xn )) =
x ∈ E : lim sup ∥xn − x∥ = r (E , (xn )) . n→∞
A sequence (xn ) is called regular relative to E if r (E , (xn )) = r (E , (xni )) for all subsequences (xni ) of (xn ) and (xn ) is called asymptotically uniform relative to E if A(E , (xn )) = A(E , (xni )) for all subsequences (xni ) of (xn ). Theorem 3.3 ([9]). Let T be a mapping on a nonempty weakly compact convex subset E of a Banach space X . Assume that T satisfies condition (C) and K is a minimal invariant set for T . Then there exists an approximate fixed point sequence (xn ) for T in K such that r (K , (xn )) = inf{r (K , (yn ))}, where the infimum is taken over all approximate fixed point sequences (yn ) in K . Lemma 3.4 ([14]). Let X be a Banach space and let (xn ) be a bounded sequence in X . Then (xn ) has a subsequence (xnk ) such that the following double limit exists lim
k,i→∞,k̸=i
∥xnk − xni ∥.
Remark 3.5 ([9]). It is very easy to verify that every approximate point sequence (xn ) for T obtained by Theorem 3.3 is regular and asymptotically uniform. Moreover, for all x ∈ K , limn→∞ ∥xn − x∥ = r0 , where r0 = inf{r (K , (yn ))} and the infimum is taken over all approximate fixed point sequences (yn ) for T in K . Furthermore, if limn,m→∞,n̸=m ∥xn − xm ∥ exists, then the limit must be r0 . Since there always exists a subsequence (x′n ) of (xn ) such that the limit limm′ ,n′ →∞,n′ ̸=m′ ∥x′n − x′m ∥ exists, we can assume that the sequence itself has this property. Now we are in the position to prove the aforementioned generalization of Theorem 4.16 in [9]. Theorem 3.6. Let K be nonempty weakly compact convex subset of a Banach space X which is minimal invariant for a continuous mapping T satisfying condition (C). Assume that T does not have a fixed point and let r0 > 0 be given by Remark 3.5. If (x1n ), (x2n ), . . . , (xNn ) are approximate fixed point sequences for T in K such that lim ∥xin − xjn ∥ = r0
n→∞
for all i ̸= j, then there exist an increasing sequence (nk ) of positive integers and an approximate fixed point sequence (zk ) for T in K such that N −1
lim sup ∥zk − xink ∥ ≤
N
k→∞
lim sup ∥zk − x∥ ≥ r0
r0
for every i = 1, . . . , N
and
for every x ∈ K .
k→∞
Proof. For each n ∈ N let
δn = max{∥xin − Txin ∥ : i = 1, . . . , N }, dn = max{∥xin − xjn ∥ : i, j = 1, . . . , N , i ̸= j}. n Choose (εn ) such that εn ∈ (0, 1) and limn→∞ εn = limn→∞ εδn = 0. Let ηn = 1−ε δn . For each n ∈ N we put ε n
Kn =
z ∈ K : max{∥xin − z ∥ : i = 1, . . . , N } ≤
N −1 N
n
dn + ηn .
All sets Kn are closed, convex and nonempty. Next define a mapping Tn : Kn → K by Tn z = (1 − εn )Tz +
N εn
N j =1
xjn
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for z ∈ Kn . Now we will show that Tn maps Kn into Kn for n large enough. First observe that 12 ∥xin − Txin ∥ ≤ ∥xin − z ∥ for all
i = 1, . . . , N. Indeed, since δn tends to 0, for every ε > 0 there exists n0 such that 12 ∥xin − Txin ∥ < r0 − ε for all n ≥ n0 . Also by Remark 3.5 there exists n1 such that ∥xin − z ∥ > r0 − ε for all n ≥ n1 . Hence 1 2
∥xin − Txin ∥ < r0 − ϵ < ∥xin − z ∥
for n ≥ n2 = max{n0 , n1 }. Since T satisfies condition (C), we have ∥Txin − Tz ∥ ≤ ∥xin − z ∥ for all i = 1, . . . , N and n ≥ n2 . Using the equality (1 − εn )δn = εn ηn , we obtain
N εn i i j i ∥xn − Tn z ∥ = (1 − εn )(xn − Tz ) + (xn − xn ) N j=1 j̸=i εn i ∥xn − xjn ∥ ≤ (1 − εn ) ∥xin − Txin ∥ + ∥Txin − Tz ∥ + N
≤ (1 − εn ) δn + =
N −1 N
N −1 N
dn + η n
+ εn
j=1 j̸=i
N −1 N
dn
dn + ηn .
This means that Tn (Kn ) ⊂ Kn for every n ≥ n2 . Let
αn = inf ∥x − Tn x∥. x∈Kn
We will show that lim infn→∞ αn = 0. Assuming the contrary, we can find α0 > 0 such that αn > α0 for every sufficiently large n. Since limn→∞ εn = 0, there exists n ≥ n2 for which αn > α0 ,
√
εn diam(K ) <
1 1 = 2 −√ Then θ (r ) = 1+ , where θ is the function defined in Theorem 3.1. r εn Now let x, y ∈ Kn and θ (r )∥x − Tn x∥ ≤ ∥x − y∥. We will show that
α0 2
and r := 1 −
√
1
εn ∈ [2− 2 , 1).
∥Tn x − Tn y∥ ≤ (1 − εn )∥x − y∥. Since
∥x − Tx∥ ≤ ∥x − Tn x∥ + ∥Tn x − Tx∥ ≤ (2 −
√ εn )∥x − y∥ + εn diam(K )
and
εn diam(K ) ≤
√ α0 √ ∥x − Tn x∥ ≤ εn ≤ εn 2
2
√
εn
2
(2 −
√ εn )∥x − y∥,
we have
∥x − Tx∥ ≤ (2 −
√ εn εn √ εn ) 1 + ∥x − y∥ = 2 − ∥x − y∥ < 2∥x − y∥. 2
2
The mapping T satisfies condition (C), so ∥Tx − Ty∥ ≤ ∥x − y∥. Therefore
(1 − εn )∥Tx − Ty∥ ≤ (1 − εn )∥x − y∥. Consequently we have ∥Tn x − Tn y∥ ≤ (1 − εn )∥x − y∥. By Remark 3.2, Tn has a fixed point, which contradicts the condition αn > α0 > 0. We therefore see that lim infn→∞ αn = 0 and hence limk→∞ αnk = 0 for some subsequence (αnk ). Choose zk ∈ Knk so that ∥zk − Tnk zk ∥ ≤ αnk + 1k . Thus
∥zk − Tzk ∥ ≤ ∥zk − Tnk zk ∥ + ∥Tnk zk − Tzk ∥ N 1 j = ∥zk − Tnk zk ∥ + εnk xnk − Tzk N j=1 ≤ ∥zk − Tnk zk ∥ + εnk diam(K ) ≤ αnk +
1 k
+ εnk diam(K ).
A. Betiuk-Pilarska, S. Prus / J. Math. Anal. Appl. 399 (2013) 459–471
469
This implies that limk→∞ ∥zk − Tzk ∥ = 0. Clearly we have lim sup ∥xink − zk ∥ ≤
N −1
k→∞
N
lim dk =
N −1
k→∞
N
r0
for all i = 1, . . . , N and by Remark 3.5 we obtain lim supk→∞ ∥zk − x∥ ≥ r0 for every x ∈ K .
Definition. A Banach lattice X is said to be weakly orthogonal if lim inf lim inf ∥ |xn | ∧ |xm |∥ = 0 n→∞
m→∞
whenever (xn ) is a sequence in X which converges weakly to 0. This definition was given by Borwein and Sims in [8], but the reader should be aware that also a different property is called weak orthogonality in the literature (see [15]). It is easy to show that c0 , c , lp (1 ≤ p < ∞) are weakly orthogonal while l∞ and Lp ([0, 1]) do not have this property. A slight change in the proof of Lemma 3.4 enables us to establish the next lemma. Lemma 3.7. Let X be a Banach lattice and let (xn ) be a bounded sequence in X . Then (xn ) has a subsequence (xni ) such that the following double limit exists lim
i,j→∞,i̸=j
∥ |xni | ∧ |xnj | ∥.
We will also need the following auxiliary lemma. Lemma 3.8. Let X be a Banach lattice and N ≥ 2, N ∈ N. Then for all z , x1 , . . . , xN ∈ X the following inequality holds
N ∥ |xi | ∧ |xj | ∥. |z − xi | ∨ |z − xj | + ∥z ∥ ≤ i,j=1 i,j=1,...,N i̸=j i̸=j Proof. Applying Lemma 2.1 (ii), we get |z | ≤ |z − xi | ∨ |z − xj | + |xi | ∧ |xj | for all i, j = 1, . . . , N , i ̸= j. Hence
|z − xi | ∨ |z − xj | + |xi | ∧ |xj |
|z | ≤
i,j=1,...,N i̸=j
N |xi | ∧ |xj |. |z − xi | ∨ |z − xj | +
≤
i,j=1,...,N i̸=j
i,j=1 i̸=j
This clearly gives us the conclusion of our lemma.
Observe that from the conclusion of Lemma 3.8 it follows that
∥z ∥ ≤ αN (X )
∥z − xi ∥ +
i=1,...,N
N
∥ |xi | ∧ |xj | ∥.
i,j=1 i̸=j
Our next result generalize a fixed point theorem proved by S. Dhompongsa and A. Kaewcharoen in [9]. Theorem 3.9. A Banach space X has the weak fixed point property for continuous mappings satisfying condition (C) if there exists a weakly orthogonal r-N-OUNC Banach lattice Y such that d(X , Y )r < 1. Proof. Assume that X fails to have the weak fixed point property for continuous mappings satisfying condition (C). Then there exists a nonempty weakly compact convex set E ⊂ X and a continuous mapping T : E → E satisfying condition (C) which does not have a fixed point. Moreover, from the Zorn Lemma there exists a nonempty weakly compact convex minimal and T -invariant subset K ⊂ E. By Theorem 3.3, there exists an approximate fixed point sequence (xn ) for T in K such that r (K , (xn )) = inf{r (K , (yn ))}, where the infimum is taken over all approximate fixed point sequences (yn ) in K . We can assume that r (K , (xn )) = 1. From Remark 3.5 we see that (xn ) is regular and asymptotically uniform. Therefore limn→∞ ∥xn − x∥ = 1 for every x ∈ K . Since K is weakly compact, we can assume that (xn ) converges weakly to 0 (by translation if necessary) and hence 0 ∈ K . Let Y be a weakly orthogonal Banach lattice such that d(X , Y )r < 1. There exist r ′ > r such that d(X , Y )r ′ < 1 and an isomorphism S : X → Y satisfying ∥S ∥ ∥S −1 ∥r ′ < 1. From Lemmas 3.4 and 3.7, by passing to a subsequence, we can assume that the following double limits exist lim
i,j→∞,i̸=j
∥xi − xj ∥ and
lim
i,j→∞,i̸=j
∥ |Sxi | ∧ |Sxj | ∥.
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By Remark 3.5 we have limi,j→∞,i̸=j ∥xi − xj ∥ = 1 and by the weak orthogonality of Y , limi,j→∞,i̸=j ∥ |Sxi | ∧ |Sxj | ∥ = 0. We choose N subsequences of (xn ) in the following way: x1k = xNk , x2k = xNk+1 , . . . , xNk = xNk+N −1 . j
j
For i, j = 1, . . . , N , i ̸= j, we get limk→∞ ∥xik ∥ = limk→∞ ∥xik − xk ∥ = 1 and limk→∞ ∥ |Sxik | ∧ |Sxk | ∥ = 0. From Theorem 3.6 there exists an approximate fixed point sequence (zk ) in K such that lim sup ∥zk − xik ∥ ≤ k→∞
N −1 N
for i = 1, . . . , N and lim supk→∞ ∥zk ∥ ≥ 1. Passing to appropriate subsequences, we can assume that the limit limk→∞ ∥zk ∥ exists. Let U be a free ultrafilter on N and
i
u =
Sxin
∥S ∥
,
u=
U
Szn
∥S ∥
,
v =u −u= i
i
U
S (xin − zn )
∥S ∥
U
j − zn ∥ ≤ and ∥v − v ∥ ≤ limk→∞ ∥xin − xn ∥ ≤ 1. Since i Sxk = ∥S −1 ∥ ∥S ∥ ∥ui ∥, 1 = lim ∥xik ∥ = lim ∥xik ∥ ≤ ∥S −1 ∥ ∥S ∥ lim k→∞ k→U k→U ∥S ∥
for i = 1, . . . , N. Then ∥v ∥ ≤ lim supk→∞ ∥ i
xin
N −1 N
i
j
j
we have ∥ui ∥ ≥ ∥S −11∥ ∥S ∥ > r ′ . Moreover, ∥ |ui |∧|uj | ∥ = limk→∞ ∥ |Sxik |∧|Sxk | ∥ = 0, if i ̸= j. This means that |ui |∧|uj | = 0, so using Lemma 2.1 (iii) we conclude that |ui − uj | = |ui | + |uj | ≥ |ui | and hence ∥ |ui − uj | − |ui | ∥ > r ′ . Finally, from Lemma 3.8 we see that 1 = lim ∥zk ∥ ≤ ∥S −1 ∥ lim ∥Szk ∥ = ∥S −1 ∥ ∥S ∥ ∥u∥ k→∞
k→U
i j −1 (|u − u | ∨ |u − u |) ≤ ∥S ∥ ∥S ∥ i,j=1,...,N i̸=j and hence
1 i j (|v | ∨ |v |) ≥ > r ′. ∥S ∥ ∥S −1 ∥ i,j=1,...,N i̸=j It follows that (Y )U is not r ′ -N-OUNC, so by Theorem 2.6, Y is not r-N-OUNC.
Corollary 3.10. A Banach space X has the weak fixed point property for continuous mappings satisfying condition (C) if there exists N ≥ 2 and weakly orthogonal Banach lattice Y such that d(X , Y )αN (Y ) < NN−1 . Corollary 3.11. Let X be a weakly orthogonal N-OUNC Banach lattice. Then X has the weak fixed point property for continuous mappings satisfying condition (C). Acknowledgment The authors were partially supported by the Polish MNiSW grant N N201 393737. References [1] B. Maurey, Points fixes des contractions de certains faiblement compacts de L1 , in: Seminaire d’Analyse Fonctionelle 1980–1981, Vol. Exp. No. VIII, École Polytech., Palaiseau, 1981. [2] J. Elton, P.-K. Lin, E. Odell, S. Szarek, Remarks on the fixed point problem for nonexpansive maps, Contemp. Math. 18 (1983) 87–120. [3] M.A. Khamsi, W.A. Kirk, An Introduction to Metric Spaces and Fixed Point Theory, Wiley-Interscience, New York, 2001. [4] W.A. Kirk, B. Sims (Eds.), Handbook of Metric Fixed Point Theory, Kluwer Academic Publishers, Dordrecht, 2001. [5] S. Prus, Banach spaces which are uniformly noncreasy, Nonlinear Anal. 30 (1997) 2317–2324. [6] A. Betiuk-Pilarska, S. Prus, Banach lattices which are order uniformly noncreasy, J. Math. Anal. Appl. 342 (2008) 1271–1279. [7] W. Kurc, A dual property to uniform monotonicity in Banach lattices, Collect. Math. 44 (1993) 155–165. [8] J.M. Borwein, B. Sims, Non-expansive mappings on Banach lattices and related topics, Houston J. Math. 10 (1984) 339–356. [9] S. Dhompongsa, A. Kaewcharoen, Fixed point theorems for nonexpansive mappings and Suzuki-generalized nonexpansive mappings on a Banach lattice, Nonlinear Anal. 71 (2009) 5344–5353. [10] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces II, Springer-Verlag, New York, 1979. [11] A.G. Aksoy, M.A. Khamsi, Nonstandard Methods in Fixed Point Theory, Springer-Verlag, New York, 1990.
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T. Suzuki, Fixed point theorems and convergence theorems for some generalized nonexpansive mappings, J. Math. Anal. Appl. 340 (2008) 1088–1095. T. Suzuki, A generalized Banach contraction principle that characterizes metric completeness, Proc. Amer. Math. Soc. 136 (2008) 1861–1869. D. van Dulst, Reflexive and Superreflexive Banach Spaces, in: Mathematical Centre Tracts, vol. 102, Mathematisch Centrum, Amsterdam, 1978. B. Sims, Orthogonality and fixed points of nonexpansive maps, Proc. Centre Math. Appl. Austral. Nat. Univ. 20 (1988) 178–186.
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Journal of Mathematical Analysis and Applications journal homepage: www.elsevier.com/locate/jmaa
Banach lattices which are N-order uniformly noncreasy Anna Betiuk-Pilarska, Stanisław Prus ∗ Institute of Mathematics, M. Curie-Skłodowska University, 20-031 Lublin, Poland
article
info
Article history: Received 30 June 2012 Available online 22 October 2012 Submitted by T.D. Benavides
abstract An N-dimensional Riesz angle for a Banach lattice and the class of N-order uniformly noncreasy Banach lattices are introduced and studied. A fixed point theorem for these lattices is proved. © 2012 Elsevier Inc. All rights reserved.
Keywords: Uniform monotonicity Order uniform smoothness N-dimensional Riesz angle N-order uniform noncreasiness Mapping with condition (C) Weak fixed point property
1. Introduction One of the main problems of metric fixed point theory concerns existence of fixed points of nonexpansive mappings. In [1], Maurey applied an ultrapower technique to this problem (see also [2]). His approach turned out to be very fruitful. In particular it enabled many authors to prove fixed point theorems for nonexpansive mappings in Banach spaces satisfying various geometric conditions (see [3,4]). One of them was introduced in [5] as a combination of two classical geometric properties: uniform convexity and uniform smoothness. Banach spaces with this property are called uniformly noncreasy. In [6] a new property which is related to order was introduced. Banach lattices with this property are said to be order uniformly noncreasy. This definition was inspired by Kurc [7] who considered uniform monotonicity and order uniform smoothness of Banach lattices and showed a duality theorem for these properties. Order uniform noncreasiness can be seen as a combination of uniform monotonicity and order uniform smoothness and it is essentially weaker than each of them. Modulus of order uniform smoothness is strictly related to the Riesz angle which was introduced by Borwein and Sims [8] to prove a fixed point theorem for nonexpansive mappings. In [6] their result was extended to the class of order uniformly noncreasy lattices. Later on a further extension was given by S. Dhompongsa and A. Kaewcharoen in [9]. Their result covers the class of continuous mappings satisfying condition (C), which is larger than the class of nonexpansive mappings. In this paper we introduce a new coefficient αN (X ) for a Banach lattice X which we call the N-dimensional Riesz angle. For N = 2 it coincides with the Riesz angle considered by Borwein and Sims. We give estimates for αN (X ) and find the exact value of this coefficient for some lattices, including Lp spaces. We show relations between αN (X ) and some geometric lattice properties. Following an idea from [6], we introduce a new class of lattices: N-order uniformly noncreasy Banach lattices. Then we extend the fixed point result from [9] to this class. 2. Geometric properties of Banach lattices Let X be a Banach space. By BX we denote the unit ball of X . For notation and terminology concerning Banach lattices we refer the reader to [10]. Let us recall that if an inequality involves lattice and algebraic operations, then it is enough to
∗
Corresponding author. E-mail addresses: [email protected] (A. Betiuk-Pilarska), [email protected] (S. Prus).
0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.10.025
460
A. Betiuk-Pilarska, S. Prus / J. Math. Anal. Appl. 399 (2013) 459–471
check its validity for real numbers to be sure that it holds for vectors in an arbitrary Banach lattice (see [10, p. 1]). In the next lemma we collect some inequalities which will be used in the sequel. Lemma 2.1. Let X be a Banach lattice. Then (i) 0 ≤ |x| − (|x| − |y|)+ ≤ |y|, (ii) |z | ≤ |x − z | ∨ |y − z | + |x| ∧ |y|, (iii) |x| + |y| − 2(|x| ∧ |y|) ≤ |x − y| ≤ |x| + |y| for all x, y, z ∈ X . Given a Banach lattice X and ε ∈ [0, 1], we put
δm,X (ε) = inf{1 − ∥x − y∥ : 0 ≤ y ≤ x, ∥x∥ ≤ 1, ∥y∥ ≥ ε}. We say that X is uniformly monotone if δm,X (ε) > 0 for all ε ∈ (0, 1]. The coefficient
ε0,m (X ) = sup{ε ∈ [0, 1) : δm,X (ε) = 0} is called the characteristic of monotonicity of the lattice X . In [7], the modulus of order smoothness of a Banach lattice X was introduced as follows
ρm,X (t ) = sup{∥x ∨ ty∥ − 1 : x, y ∈ BX , x, y ≥ 0} where t ∈ [0, 1]. A Banach lattice X is called order uniformly smooth if lim
ρm,X (t )
t →0
t
= 0.
The coefficient
ρm,X (1) + 1 = sup{∥x ∨ y∥ : x, y ∈ BX , x, y ≥ 0} is called the Riesz angle of X and denoted by α(X ) (see [8]). Recall that 1 ≤ α(X ) ≤ 2 and α(X ) = 1 if and only if X is an abstract M space. Moreover, if X is an abstract L1 space, then α(X ) = 2. In particular, α(c0 ) = α(c ) = 1 and α(l1 ) = 2. Kurc [7] proved the duality formulas
ρm,X ∗ (t ) = sup (ε t − δm,X (ε)) 0≤ε≤1
δm,X (ε) = sup (εt − ρm,X ∗ (t )) 0≤t ≤1
for all ε, t ∈ [0, 1]. They show that a Banach lattice X is uniformly monotone (resp. order uniformly smooth) if and only if the dual lattice X ∗ is order uniformly smooth (resp. uniformly monotone). Using the above formulas it is also easy to see that ρm,X ∗ (1) < 1 if and only if ε0,m (X ) < 1. In [6] it was proved that if ρm,X (1) < 1 and ε0,m (X ) < 1, then X is superreflexive. In particular, if a Banach lattice X is order uniformly smooth and uniformly monotone, then X is superreflexive. Now we introduce a generalized Riesz angle. Definition. Let X be a Banach lattice and N ∈ N, N ≥ 2. The N-dimensional Riesz angle of X is defined as
αN (X ) = sup (xi ∨ xj ) : x1 , . . . , xN ∈ BX , x1 , . . . , xN ≥ 0 . N i,j=i1̸=,..., j Of course αN (X ) ≥ 1. Moreover, α2 (X ) = α(X ) and αN (X ) ≤ αN −1 (X ) for every N ≥ 3. Hence αN (X ) ≤ α(X ). This estimate is not sharp. To get a better upper bound for αN (X ) we need the following lemma. Lemma 2.2. For N ∈ N, N ≥ 2 and for all nonnegative scalars x1 , . . . , xN ∈ R the following inequality holds
(xi ∨ xj ) ≤
i,j=1,...,N i̸=j
1
N − 1 i=1
Proof. We have
i,j=1,...,N i̸=j
N
(xi ∨ xj ) = xk
xi .
A. Betiuk-Pilarska, S. Prus / J. Math. Anal. Appl. 399 (2013) 459–471
461
for some k. Let xm = min{xi : i = 1, . . . , N }. Then xk ≤ xi ∨ xm = xi for every i ̸= m. Hence xk ≤
N
1 N −1
xi ≤
i=1 i̸=m
N
1
N − 1 i=1
xi .
Proposition 2.3. Let X be a Banach lattice and N ∈ N, N ≥ 2. Then
αN (X ) ≤
N N −1
.
Proof. Let x1 , . . . , xN ∈ BX , x1 , . . . , xN ≥ 0. Lemma 2.2 leads to the inequality
z :=
(xi ∨ xj ) ≤
i,j=1,...,N i̸=j
Hence ∥z ∥ ≤
1 N −1
N
i =1
N
1
N − 1 i=1
∥x i ∥ ≤
N , N −1
xi .
which proves that αN (X ) ≤
N . N −1
We will show relations between αN (X ) and some geometric lattice properties. Let us recall that a Banach lattice X is said to satisfy an upper p-estimate, where 1 ≤ p < +∞, if there exists a constant M > 0 such that
1p n n p ∥x i ∥ x≤M i=1 i i =1 for all pairwise disjoint elements x1 , . . . , xn in X . Our next result generalizes Theorem 2.3(i) in [6]. Theorem 2.4. Let X be a Banach lattice and N ∈ N, N ≥ 2. If αN (X ) < p ∈ (1, +∞).
N , N −1
then X satisfies an upper p-estimate for some
Proof. Suppose that X does not satisfy an upper p-estimate for any p ∈ (1, +∞). By [10, Theorem 1.f.12], for every ε > 0 there exist disjoint elements x1 , . . . , xN ∈ X such that
(1 − ε)
N i=1
N N ai x i ≤ |a | |ai | ≤ i=1 i=1 i
(1)
for all scalars a1 , . . . , aN . Put yk =
N
1 N −1
|xi |
i=1 i̸=k
for k = 1, . . . , N. By (1) we have ∥xk ∥ ≤ 1 and consequently, ∥yk ∥ ≤ 1. Moreover, since x1 , . . . , xN are pairwise disjoint, yk ∨ yl =
1
N
N − 1 i =1
| xi |
whenever k ̸= l. Substituting ai =
1 N −1
into (1), we obtain
N N N 1 1 (1 − ε) ≤ xi ≤ |xi | = (yk ∨ yl ) . N − 1 i=1 N − 1 i=1 k,l=1,...,N N −1 k̸=l This and Proposition 2.3 show that αN (X ) =
N . N −1
Let us now recall that a Banach lattice X is said to be p-convex, where 1 ≤ p < +∞, if there exists a constant M > 0 such that
1p 1p n n p p ≤M |xi | ∥x i ∥ i=1 i=1 for all vectors x1 , . . . , xn ∈ X . If a Banach lattice X is p-convex, then X satisfies the upper p-estimate and if X satisfies an upper r-estimate for some 1 < r < +∞, then X is p-convex for every p ∈ [1, r ) (see [10, p. 85]). We give an estimate for αN (X ) for a p-convex Köthe function space X (see [10, p. 28] for the definition).
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Theorem 2.5. Let N ∈ N, N ≥ 2. If a Köthe function space X is p-convex for some p ≥ 1 with the constant M = 1, then
αN (X ) ≤
1p
N N −1
.
Proof. Our space X is a space of locally integrable functions on some set Ω with a σ -finite measure µ. Let f1 , . . . , fN ∈ BX be nonnegative. We put
z ( x) =
(fi (x) ∨ fj (x))
i,j=1,...,N i̸=j
for every x ∈ Ω . Let Ak = {x ∈ Ω : fk (x) < z (x)} , k = 1, . . . , N. The sets A1 , . . . , AN are pairwise disjoint. Indeed, if x ∈ Ai ∩ Aj for some i ̸= j, then fi (x) < z (x) and fj (x) < z (x). But z (x) ≤ fi (x) ∨ fj (x), which is a contradiction. For a nonempty set D ⊂ Ω by 1D we denote the characteristic function of D. In addition, we put 1∅ = 0. We have z p − z p 1Ak = z p 1Ac for every k = 1, . . . , N, where Ack = Ω \ Ak . Adding these equalities, we obtain k
N
Nz p −
z p 1Ak =
k=1
N
z p 1Ac . k
k=1
But N
z p 1Ak = z p 1A ≤ z p ,
k=1
where A =
N
k=1
Ak , so we have
(N − 1)z p ≤
N
z p 1Ac . k
k=1
By our assumption of p-convexity we therefore get
1p 1p 1p N N N 1 p p p ∥z1Ack ∥ (z1Ack ) ∥f k ∥ ≤ ≤ Np. (N − 1) ∥z ∥ ≤ ≤ k=1 k=1 k=1 1 p
This clearly gives us the estimate αN (X ) ≤
N N −1
1p
.
Consider some special cases. Let us start with the trivial one, when Ω = {1, . . . , N − 1} with the counting measure, i.e. X = RN −1 endowed with a lattice norm. Following the first part of the proof of Theorem 2.5, we obtain pairwise disjoint subsets A1 , . . . , AN of the set Ω of N − 1 elements. Consequently, Ak = ∅ for some k, which means that z ≤ fk and consequently, ∥z ∥ ≤ 1. This shows that αN (X ) = 1. Consider now the case when X = Lp (Ω ) and Ω has N pairwise disjoint subsets A1 , . . . , AN with finite positive measure. We put gk = ((N − 1)µ(Ak )) fi =
N
− 1p
− 1p
1Ak for k = 1, . . . , N. Then ∥gk ∥ = (N − 1)
. Let
gk
k=1 k̸=i
for i = 1, . . . , N. We have ∥fi ∥ = 1 and z :=
(fi ∨ fj ) =
i,j=1,...,N i̸=j
N
gk .
k=1
Hence
∥z ∥p =
N k=1
∥gk ∥p =
N N −1
.
In view of Theorem 2.5, this shows that αN (Lp (Ω )) = and lp .
N N −1
1p
. In particular, this formula is true for the spaces Lp ([0, 1])
A. Betiuk-Pilarska, S. Prus / J. Math. Anal. Appl. 399 (2013) 459–471
463
Definition. Let r ∈ (0, 1]. A Banach lattice X is said to be r-N-order uniformly noncreasy (r-N-OUNC) if for all u1 , . . . , uN ∈ N −1 BX such that ∥ui − uj ∥ ≤ 1 we have either N
(|ui | ∨ |uj |) ≤ r i,j=1,...,N i̸=j or there exist i ̸= j such that for every y ∈ X the conditions |y| ≤ |ui − uj |, ∥y∥ ≥ r imply ∥ |ui − uj | − |y| ∥ ≤ r. A Banach lattice X is N-order uniformly noncreasy (N-OUNC) if it is r-N-OUNC for some r ∈ (0, 1). For N = 2 this definition coincides with the definition of an order uniformly noncreasy Banach lattice given in [6]. Of course, each Banach lattice X is r-N-OUNC with r = NN−1 αN (X ). Therefore, if αN (X ) < NN−1 , then X is N-OUNC. It is also easy to see that X is r-N-OUNC where r = max ε, 1 − δm,X (ε) for any ε ∈ (0, 1). It follows that if ε0,m (X ) < 1, then
X is N-OUNC. The class of all N-OUNC Banach lattices contains therefore all Banach lattices X with αN (X ) < NN−1 and all uniformly monotone lattices. The following example shows that for different values of N the classes of N-OUNC Banach lattices are essentially different. N −1 Let N ∈ N, N ≥ 3 and X = R ⊕ i=1 c0 . Elements of X are of the form x = (ξ , x1 , . . . , xN −1 ), where ξ ∈ R and x1 , . . . , xN −1 ∈ c0 . We define the norm of such vector x by the formula
∥x∥ = |ξ | +
N −1
∥x i ∥0 ,
i=1
where ∥·∥0 is the standard norm in c0 . In the sequel, given a vector y ∈ c0 we write y = (y(n)). We will show that X is N-OUNC, but it is not M-OUNC for any M < N. We begin by proving that X is r-N-OUNC where r = 1 − N (N1+2) . Let u1 , . . . , uN ∈ X be such that ∥ui ∥ ≤ NN−1 and ∥ui − uj ∥ ≤ 1 for any i, j = 1, . . . , N. We put
z=
(|ui | ∨ |uj |).
i,j=1,...,N i̸=j
Then z = (ζ , z1 , . . . , zN −1 ) for some ζ ∈ R and some z1 , . . . , zN −1 ∈ c0 . For every k = 1, . . . , N − 1 we choose nk ∈ N such that ∥zk ∥0 = zk (nk ). Given x = (ξ , x1 , . . . , xN −1 ), where ξ ∈ R and x1 , . . . , xN −1 ∈ c0 , we put P 1 ( x) = ξ ,
Pk+1 (x) = xk (nk )
for k = 1, . . . , N − 1. Consider the sets Ai = {k : |Pk (ui )| < Pk (z )}, i = 1, . . . , N. It easy to show that they are pairwise disjoint. If Ai = ∅ for some i, then Pk (z ) ≤ |Pk (ui )| for every k = 1, . . . , N and hence
∥z ∥ =
N
Pk (z ) ≤
k=1
N
|Pk (ui )| ≤ ∥ui ∥ ≤
k=1
N −1 N
< r.
Now suppose that sets A1 , . . . , AN are nonempty. Then we can assume that Ai = {i} for every i = 1, . . . , N. Therefore |Pi (ui )| < Pi (z ) and Pk (z ) ≤ |Pk (ui )| for all k ̸= i. Thus
(N − 1)Pk (z ) ≤
N
|Pk (ui )|
i=1 i̸=k
for k = 1, . . . , N, and from this we obtain
∥z ∥ =
N
Pk (z ) ≤
k=1
=
1 N −1
1
N N
N − 1 k=1
N N
|Pk (ui )| −
i=1 k=1
|Pk (ui )| =
i=1 i̸=k
N k=1
1
N N
N − 1 k=1
i=1
|Pk (uk )| ≤
1 N −1
|Pk (ui )| − |Pk (uk )|
N
∥ ui ∥ − s ≤ 1 −
i =1
s N −1
,
|Pk (uk )|. Consequently, if s ≥ N (NN−+12) , then ∥z ∥ ≤ r. Now let us consider the case when s < N (NN−+12) . Of course it suffices to examine the case ∥z ∥ > r. We have
where s =
N
k=1
r < ∥z ∥ =
N k=1
Pk ( z ) ≤ P1 ( z ) +
N k=2
|Pk (u1 )| ≤ P1 (z ) + ∥u1 ∥ ≤ P1 (z ) +
N −1 N
,
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hence N −1
P1 ( z ) ≥ r −
N
.
(2)
For x = (ξ , x1 , . . . , xN −1 ), where ξ ∈ R and x1 , . . . , xN −1 ∈ c0 we put Q (x) = (0, x1 , . . . , xN −1 ). Applying inequality (2), we obtain 1 ≥ ∥u2 − u1 ∥ = |P1 (u2 − u1 )| + ∥Q (u2 − u1 )∥
≥ |P1 (u2 )| − |P1 (u1 )| + ∥Q (u2 − u1 )∥ ≥ P1 (z ) − s + ∥Q (u2 − u1 )∥ N −1 ≥r− − s + ∥Q (u2 − u1 )∥. N
Therefore, ∥Q (u2 − u1 )∥ ≤ 2 + s − N1 − r. Let y ∈ X be such that 0 ≤ y ≤ |u2 − u1 | and ∥y∥ ≥ r. Then Q (y) ≤ Q (|u2 − u1 |), so r ≤ ∥y∥ = P1 (y) + ∥Q (y)∥ ≤ P1 (y) + ∥Q (u2 − u1 )∥ ≤ P1 (y) + 2 + s − Hence P1 (y) ≥ 2r − 2 − s +
1 N
1 N
− r.
and we obtain
∥ |u2 − u1 | − y∥ = |P1 (u2 − u1 )| − P1 (y) + ∥Q (|u2 − u1 | − y)∥ ≤ |P1 (u2 − u1 )| − P1 (y) + ∥Q (u2 − u1 )∥ = ∥u2 − u1 ∥ − P1 (y) ≤ 3 − 2r + s − But s <
N −1 , N (N +2)
1 N
.
which gives us the inequality
∥ |u2 − u1 | − y∥ < 3 − 2r +
N −1 N (N + 2)
−
1 N
= r.
This proves that X is r-N-OUNC. Now we show that X is not M-OUNC for any M < N. Let (en ) be the standard basis of c0 and for k = 1, . . . , M let ekn = (0, x1 , . . . , xN −1 ) be the element of X such that xk = en and xi = (0, 0, . . .) for i ̸= k. Put
vk =
M 2k 1
M
eij
j =1
i=1 i̸=k
for k = 1, . . . , M. Of course
M 2k M −1 1 i ej = . ∥vk ∥ = M i=1 j=1 M 0
i̸=k
Assume that 1 ≤ l < k ≤ M. Then
2k M 2k 2l 1 elj + |vk − vl | = eij + ekj , M
j =1
i=1 j=2l+1 i̸=k,l
j =1
so ∥vk − vl ∥ = 1. Moreover, for
M
1 i y= e2k + ek2l M i=1 i̸=k
we have 0 ≤ y ≤ |vk − vl |, ∥y∥ = 1 and
M 2k −1 2l−1 1 |vk − vl | − y = elj + eij + ekj , 2k−1
M
j =1
i=1 j=2l+1 i̸=k,l
which shows that ∥ |vk − vl | − y∥ = 1.
j=1
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465
Finally we see that
M 2k 2l M 2 1 1 i i k vk ∨ vl = ej + ej ≥ ej M
i=1 i̸=k
j=1
M
j =1
i=1 j=1
and hence
M 2 1 i ej = 1. (vk ∨ vl ) ≥ k,l=1,...,M M i=1 j=1 k̸=l Note also that X contains a copy of c0 , so X is not uniformly monotone. Moreover, X contains a copy of the lattice RN with the l1 -norm and consequently, αN (X ) = NN−1 . To prove our fixed point theorem we will use the ultrapower method (see [4]). Let us therefore briefly recall the construction of the ultrapower. Given a Banach space X , by l∞ (X ) we denote the space of all bounded sequences in X with the supremum norm. Let U be a free ultrafilter on N. The ultrapower (X )U is the quotient space l∞ (X )/N where
N = (xn ) ∈ l∞ (X ) : lim ∥xn ∥ = 0 . n→U
Here limn→U ∥xn ∥ stands for the limit over the ultrafilter U (see [11, p. 13]). The equivalence class of the sequence (xn ) ∈ l∞ (X ) will be denoted by (xn )U . It is easy to see that the quotient norm is given by the formula (see [11, p. 23])
∥(xn )U ∥ = lim ∥xn ∥. n→U
If X is a Banach lattice, then (X )U is also a Banach lattice with the nonnegative cone consisting of all equivalent classes of sequences (xn ) such that xn ≥ 0 for every n. Our next result generalizes Theorem 3.3 in [6]. Theorem 2.6. Let X be a Banach lattice, U be a free ultrafilter on N, r ∈ (0, 1) and n ∈ N, N ≥ 2. If X is r-N-OUNC, then (X )U is r ′ -N-OUNC for every r ′ ∈ (r , 1). Proof. Assume that (X )U is not r ′ -N-OUNC. There exist x1 , . . . , xN ∈
N −1 B(X )U N
such that ∥ xi − xj ∥ ≤ 1 for all i, j,
(| xi | ∨ | xj |) > r ′ i,j=1,...,N i̸=j and for any i, j = 1, . . . , N , i ̸= j, there exists yij ∈ (X )U such that | yij | ≤ | xi − xj |, ∥ yij ∥ ≥ r ′ and ∥ | xi − xj | − | yij | ∥ > r ′ . ′ i Let ε ∈ (0, 1) be such that r /r > 1 + ε . For every i = 1, . . . , N we choose a sequence (xn ) ∈ l∞ (X ) in xi . We have j
limn→U ∥xin − xn ∥ ≤ 1, limn→U ∥xin ∥ ≤
N −1 N
and
i j lim (|xn | ∨ |xn |) > r ′ > (1 + ε)r . n→U N i,j=i1̸=,..., j Fix i ̸= j. There exist sequences (yn ) ∈ l∞ (X ) and (zn ) ∈ l∞ (X ) in the equivalence classes yij and xi − xj respectively such j that |yn | ≤ |zn | for every n ∈ N. We put un = xin − xn − zn . Then limn→U ∥un ∥ = 0 and
|yn | ≤ |xin − xjn − un | ≤ |xin − xjn | + |un | j
ij
j
ij
for every n. Hence |yn | − |un | ≤ |xin − xn |, thus vn ≤ |xin − xn |, where vn = (|yn | − |un |)+ ≥ 0. Applying inequality (i) from Lemma 2.1, we obtain
∥vnij ∥ ≥ ∥yn ∥ − ∥ |yn | − (|yn | − |un |)+ ∥ ≥ ∥yn ∥ − ∥un ∥, ij so limn→U ∥vn ∥ ≥ limn→U ∥yn ∥ ≥ r ′ > (1 + ε)r. Applying again the same inequality, we get
∥ |xin − xjn | − vnij ∥ = ∥ |xin − xjn | − |yn | + |yn | − (|yn | − |un |)+ ∥ ≥ ∥ |xin − xjn | − |yn | ∥ − ∥ |yn | − (|yn | − |un |)+ ∥ ≥ ∥ |xin − xjn | − |yn | ∥ − ∥un ∥. j ij j Therefore limn→U ∥ |xin − xn | − vn ∥ ≥ limn→U ∥ |xin − xn | − |yn | ∥ > r ′ > (1 + ε)r.
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Since U is a nontrivial ultrafilter, there exists n such that N −1 ∥xin − xjn ∥ < 1 + ε, ∥xin ∥ < (1 + ε), N i j (|xn | ∨ |xn |) > r (1 + ε), i,j=1,...,N i̸=j
∥vnij ∥ > r (1 + ε),
∥ |xin − xjn | − vnij ∥ > r (1 + ε) ij
1 1 xin and yn = 1+ε vn for i, j = 1, . . . , N , i ̸= j. Then for all i, j = 1, . . . , N , i ̸= j. We put xn = 1+ε i
ij
N −1 ∥xin − xjn ∥ ≤ 1, ∥xin ∥ ≤ , N i j (|xn | ∨ |xn |) > r , i,j=1,...,N i̸=j
|yijn | ≤ |xin − xjn |,
∥yijn ∥ > r ,
which shows that X is not r-N-OUNC.
∥ |xin − xjn | − |yijn | ∥ > r ,
Corollary 2.7. Let X be a Banach lattice and U be a free ultrafilter on N. If X is N-OUNC, then (X )U is N-OUNC. 3. Applications to fixed point theory A self-mapping T of a subset E of a Banach space X is said to satisfy condition (C) if 1 2
∥x − Tx∥ ≤ ∥x − y∥ implies ∥Tx − Ty∥ ≤ ∥x − y∥
for all x, y ∈ E. Clearly, if a mapping T is nonexpansive, i.e.,
∥Tx − Ty∥ ≤ ∥x − y∥ for all x, y in E, then T satisfies condition (C). However, there are mappings satisfying condition (C) which are not even continuous (see [12]). We say that X has the weak fixed point property for continuous mappings satisfying condition (C) if every such mapping defined on a nonempty weakly compact convex subset of X has a fixed point. Suzuki [12] proved the existence of approximate fixed point sequences for mappings satisfying condition (C). Recall that (xn ) is an approximate fixed point sequence for a mapping T if limn→∞ ∥xn − Txn ∥ = 0. We will generalize Theorem 4.16 in [9]. For this purpose we need the following results. Theorem 3.1 Let (X , d) be a complete metric space and T be a mapping on X . Define a nonincreasing function θ from ([13]). [0, 1) onto 21 , 1 by
1 θ (r ) = (1 − r )r −2 ( 1 + r ) −1
√
if 0√ ≤ r ≤ ( 5 − 1)/2,
if ( 5 − 1)/2 ≤ r ≤ 2−1/2 , if 2−1/2 ≤ r < 1.
Assume that there exists r ∈ [0, 1) such that
θ (r )d(x, Tx) ≤ d(x, y) implies d(Tx, Ty) ≤ rd(x, y) for all x, y ∈ X . Then T has a fixed point. Remark 3.2 ([9]). If we assume in addition that T is continuous and there exist r , r1 ∈ [0, 1) such that
θ (r )d(x, Tx) ≤ d(x, y) implies d(Tx, Ty) ≤ r1 d(x, y) for all x, y ∈ X , then T has a fixed point.
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467
We will also need the concepts of asymptotic center and asymptotic radius. Let E be a nonempty bounded closed convex subset of X and (xn ) be a bounded sequence in X . We define the asymptotic radius of (xn ) in E as follows
r (E , (xn )) = inf lim sup ∥xn − x∥ : x ∈ E
n→∞
and the asymptotic center of (xn ) in E as follows A(E , (xn )) =
x ∈ E : lim sup ∥xn − x∥ = r (E , (xn )) . n→∞
A sequence (xn ) is called regular relative to E if r (E , (xn )) = r (E , (xni )) for all subsequences (xni ) of (xn ) and (xn ) is called asymptotically uniform relative to E if A(E , (xn )) = A(E , (xni )) for all subsequences (xni ) of (xn ). Theorem 3.3 ([9]). Let T be a mapping on a nonempty weakly compact convex subset E of a Banach space X . Assume that T satisfies condition (C) and K is a minimal invariant set for T . Then there exists an approximate fixed point sequence (xn ) for T in K such that r (K , (xn )) = inf{r (K , (yn ))}, where the infimum is taken over all approximate fixed point sequences (yn ) in K . Lemma 3.4 ([14]). Let X be a Banach space and let (xn ) be a bounded sequence in X . Then (xn ) has a subsequence (xnk ) such that the following double limit exists lim
k,i→∞,k̸=i
∥xnk − xni ∥.
Remark 3.5 ([9]). It is very easy to verify that every approximate point sequence (xn ) for T obtained by Theorem 3.3 is regular and asymptotically uniform. Moreover, for all x ∈ K , limn→∞ ∥xn − x∥ = r0 , where r0 = inf{r (K , (yn ))} and the infimum is taken over all approximate fixed point sequences (yn ) for T in K . Furthermore, if limn,m→∞,n̸=m ∥xn − xm ∥ exists, then the limit must be r0 . Since there always exists a subsequence (x′n ) of (xn ) such that the limit limm′ ,n′ →∞,n′ ̸=m′ ∥x′n − x′m ∥ exists, we can assume that the sequence itself has this property. Now we are in the position to prove the aforementioned generalization of Theorem 4.16 in [9]. Theorem 3.6. Let K be nonempty weakly compact convex subset of a Banach space X which is minimal invariant for a continuous mapping T satisfying condition (C). Assume that T does not have a fixed point and let r0 > 0 be given by Remark 3.5. If (x1n ), (x2n ), . . . , (xNn ) are approximate fixed point sequences for T in K such that lim ∥xin − xjn ∥ = r0
n→∞
for all i ̸= j, then there exist an increasing sequence (nk ) of positive integers and an approximate fixed point sequence (zk ) for T in K such that N −1
lim sup ∥zk − xink ∥ ≤
N
k→∞
lim sup ∥zk − x∥ ≥ r0
r0
for every i = 1, . . . , N
and
for every x ∈ K .
k→∞
Proof. For each n ∈ N let
δn = max{∥xin − Txin ∥ : i = 1, . . . , N }, dn = max{∥xin − xjn ∥ : i, j = 1, . . . , N , i ̸= j}. n Choose (εn ) such that εn ∈ (0, 1) and limn→∞ εn = limn→∞ εδn = 0. Let ηn = 1−ε δn . For each n ∈ N we put ε n
Kn =
z ∈ K : max{∥xin − z ∥ : i = 1, . . . , N } ≤
N −1 N
n
dn + ηn .
All sets Kn are closed, convex and nonempty. Next define a mapping Tn : Kn → K by Tn z = (1 − εn )Tz +
N εn
N j =1
xjn
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for z ∈ Kn . Now we will show that Tn maps Kn into Kn for n large enough. First observe that 12 ∥xin − Txin ∥ ≤ ∥xin − z ∥ for all
i = 1, . . . , N. Indeed, since δn tends to 0, for every ε > 0 there exists n0 such that 12 ∥xin − Txin ∥ < r0 − ε for all n ≥ n0 . Also by Remark 3.5 there exists n1 such that ∥xin − z ∥ > r0 − ε for all n ≥ n1 . Hence 1 2
∥xin − Txin ∥ < r0 − ϵ < ∥xin − z ∥
for n ≥ n2 = max{n0 , n1 }. Since T satisfies condition (C), we have ∥Txin − Tz ∥ ≤ ∥xin − z ∥ for all i = 1, . . . , N and n ≥ n2 . Using the equality (1 − εn )δn = εn ηn , we obtain
N εn i i j i ∥xn − Tn z ∥ = (1 − εn )(xn − Tz ) + (xn − xn ) N j=1 j̸=i εn i ∥xn − xjn ∥ ≤ (1 − εn ) ∥xin − Txin ∥ + ∥Txin − Tz ∥ + N
≤ (1 − εn ) δn + =
N −1 N
N −1 N
dn + η n
+ εn
j=1 j̸=i
N −1 N
dn
dn + ηn .
This means that Tn (Kn ) ⊂ Kn for every n ≥ n2 . Let
αn = inf ∥x − Tn x∥. x∈Kn
We will show that lim infn→∞ αn = 0. Assuming the contrary, we can find α0 > 0 such that αn > α0 for every sufficiently large n. Since limn→∞ εn = 0, there exists n ≥ n2 for which αn > α0 ,
√
εn diam(K ) <
1 1 = 2 −√ Then θ (r ) = 1+ , where θ is the function defined in Theorem 3.1. r εn Now let x, y ∈ Kn and θ (r )∥x − Tn x∥ ≤ ∥x − y∥. We will show that
α0 2
and r := 1 −
√
1
εn ∈ [2− 2 , 1).
∥Tn x − Tn y∥ ≤ (1 − εn )∥x − y∥. Since
∥x − Tx∥ ≤ ∥x − Tn x∥ + ∥Tn x − Tx∥ ≤ (2 −
√ εn )∥x − y∥ + εn diam(K )
and
εn diam(K ) ≤
√ α0 √ ∥x − Tn x∥ ≤ εn ≤ εn 2
2
√
εn
2
(2 −
√ εn )∥x − y∥,
we have
∥x − Tx∥ ≤ (2 −
√ εn εn √ εn ) 1 + ∥x − y∥ = 2 − ∥x − y∥ < 2∥x − y∥. 2
2
The mapping T satisfies condition (C), so ∥Tx − Ty∥ ≤ ∥x − y∥. Therefore
(1 − εn )∥Tx − Ty∥ ≤ (1 − εn )∥x − y∥. Consequently we have ∥Tn x − Tn y∥ ≤ (1 − εn )∥x − y∥. By Remark 3.2, Tn has a fixed point, which contradicts the condition αn > α0 > 0. We therefore see that lim infn→∞ αn = 0 and hence limk→∞ αnk = 0 for some subsequence (αnk ). Choose zk ∈ Knk so that ∥zk − Tnk zk ∥ ≤ αnk + 1k . Thus
∥zk − Tzk ∥ ≤ ∥zk − Tnk zk ∥ + ∥Tnk zk − Tzk ∥ N 1 j = ∥zk − Tnk zk ∥ + εnk xnk − Tzk N j=1 ≤ ∥zk − Tnk zk ∥ + εnk diam(K ) ≤ αnk +
1 k
+ εnk diam(K ).
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This implies that limk→∞ ∥zk − Tzk ∥ = 0. Clearly we have lim sup ∥xink − zk ∥ ≤
N −1
k→∞
N
lim dk =
N −1
k→∞
N
r0
for all i = 1, . . . , N and by Remark 3.5 we obtain lim supk→∞ ∥zk − x∥ ≥ r0 for every x ∈ K .
Definition. A Banach lattice X is said to be weakly orthogonal if lim inf lim inf ∥ |xn | ∧ |xm |∥ = 0 n→∞
m→∞
whenever (xn ) is a sequence in X which converges weakly to 0. This definition was given by Borwein and Sims in [8], but the reader should be aware that also a different property is called weak orthogonality in the literature (see [15]). It is easy to show that c0 , c , lp (1 ≤ p < ∞) are weakly orthogonal while l∞ and Lp ([0, 1]) do not have this property. A slight change in the proof of Lemma 3.4 enables us to establish the next lemma. Lemma 3.7. Let X be a Banach lattice and let (xn ) be a bounded sequence in X . Then (xn ) has a subsequence (xni ) such that the following double limit exists lim
i,j→∞,i̸=j
∥ |xni | ∧ |xnj | ∥.
We will also need the following auxiliary lemma. Lemma 3.8. Let X be a Banach lattice and N ≥ 2, N ∈ N. Then for all z , x1 , . . . , xN ∈ X the following inequality holds
N ∥ |xi | ∧ |xj | ∥. |z − xi | ∨ |z − xj | + ∥z ∥ ≤ i,j=1 i,j=1,...,N i̸=j i̸=j Proof. Applying Lemma 2.1 (ii), we get |z | ≤ |z − xi | ∨ |z − xj | + |xi | ∧ |xj | for all i, j = 1, . . . , N , i ̸= j. Hence
|z − xi | ∨ |z − xj | + |xi | ∧ |xj |
|z | ≤
i,j=1,...,N i̸=j
N |xi | ∧ |xj |. |z − xi | ∨ |z − xj | +
≤
i,j=1,...,N i̸=j
i,j=1 i̸=j
This clearly gives us the conclusion of our lemma.
Observe that from the conclusion of Lemma 3.8 it follows that
∥z ∥ ≤ αN (X )
∥z − xi ∥ +
i=1,...,N
N
∥ |xi | ∧ |xj | ∥.
i,j=1 i̸=j
Our next result generalize a fixed point theorem proved by S. Dhompongsa and A. Kaewcharoen in [9]. Theorem 3.9. A Banach space X has the weak fixed point property for continuous mappings satisfying condition (C) if there exists a weakly orthogonal r-N-OUNC Banach lattice Y such that d(X , Y )r < 1. Proof. Assume that X fails to have the weak fixed point property for continuous mappings satisfying condition (C). Then there exists a nonempty weakly compact convex set E ⊂ X and a continuous mapping T : E → E satisfying condition (C) which does not have a fixed point. Moreover, from the Zorn Lemma there exists a nonempty weakly compact convex minimal and T -invariant subset K ⊂ E. By Theorem 3.3, there exists an approximate fixed point sequence (xn ) for T in K such that r (K , (xn )) = inf{r (K , (yn ))}, where the infimum is taken over all approximate fixed point sequences (yn ) in K . We can assume that r (K , (xn )) = 1. From Remark 3.5 we see that (xn ) is regular and asymptotically uniform. Therefore limn→∞ ∥xn − x∥ = 1 for every x ∈ K . Since K is weakly compact, we can assume that (xn ) converges weakly to 0 (by translation if necessary) and hence 0 ∈ K . Let Y be a weakly orthogonal Banach lattice such that d(X , Y )r < 1. There exist r ′ > r such that d(X , Y )r ′ < 1 and an isomorphism S : X → Y satisfying ∥S ∥ ∥S −1 ∥r ′ < 1. From Lemmas 3.4 and 3.7, by passing to a subsequence, we can assume that the following double limits exist lim
i,j→∞,i̸=j
∥xi − xj ∥ and
lim
i,j→∞,i̸=j
∥ |Sxi | ∧ |Sxj | ∥.
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By Remark 3.5 we have limi,j→∞,i̸=j ∥xi − xj ∥ = 1 and by the weak orthogonality of Y , limi,j→∞,i̸=j ∥ |Sxi | ∧ |Sxj | ∥ = 0. We choose N subsequences of (xn ) in the following way: x1k = xNk , x2k = xNk+1 , . . . , xNk = xNk+N −1 . j
j
For i, j = 1, . . . , N , i ̸= j, we get limk→∞ ∥xik ∥ = limk→∞ ∥xik − xk ∥ = 1 and limk→∞ ∥ |Sxik | ∧ |Sxk | ∥ = 0. From Theorem 3.6 there exists an approximate fixed point sequence (zk ) in K such that lim sup ∥zk − xik ∥ ≤ k→∞
N −1 N
for i = 1, . . . , N and lim supk→∞ ∥zk ∥ ≥ 1. Passing to appropriate subsequences, we can assume that the limit limk→∞ ∥zk ∥ exists. Let U be a free ultrafilter on N and
i
u =
Sxin
∥S ∥
,
u=
U
Szn
∥S ∥
,
v =u −u= i
i
U
S (xin − zn )
∥S ∥
U
j − zn ∥ ≤ and ∥v − v ∥ ≤ limk→∞ ∥xin − xn ∥ ≤ 1. Since i Sxk = ∥S −1 ∥ ∥S ∥ ∥ui ∥, 1 = lim ∥xik ∥ = lim ∥xik ∥ ≤ ∥S −1 ∥ ∥S ∥ lim k→∞ k→U k→U ∥S ∥
for i = 1, . . . , N. Then ∥v ∥ ≤ lim supk→∞ ∥ i
xin
N −1 N
i
j
j
we have ∥ui ∥ ≥ ∥S −11∥ ∥S ∥ > r ′ . Moreover, ∥ |ui |∧|uj | ∥ = limk→∞ ∥ |Sxik |∧|Sxk | ∥ = 0, if i ̸= j. This means that |ui |∧|uj | = 0, so using Lemma 2.1 (iii) we conclude that |ui − uj | = |ui | + |uj | ≥ |ui | and hence ∥ |ui − uj | − |ui | ∥ > r ′ . Finally, from Lemma 3.8 we see that 1 = lim ∥zk ∥ ≤ ∥S −1 ∥ lim ∥Szk ∥ = ∥S −1 ∥ ∥S ∥ ∥u∥ k→∞
k→U
i j −1 (|u − u | ∨ |u − u |) ≤ ∥S ∥ ∥S ∥ i,j=1,...,N i̸=j and hence
1 i j (|v | ∨ |v |) ≥ > r ′. ∥S ∥ ∥S −1 ∥ i,j=1,...,N i̸=j It follows that (Y )U is not r ′ -N-OUNC, so by Theorem 2.6, Y is not r-N-OUNC.
Corollary 3.10. A Banach space X has the weak fixed point property for continuous mappings satisfying condition (C) if there exists N ≥ 2 and weakly orthogonal Banach lattice Y such that d(X , Y )αN (Y ) < NN−1 . Corollary 3.11. Let X be a weakly orthogonal N-OUNC Banach lattice. Then X has the weak fixed point property for continuous mappings satisfying condition (C). Acknowledgment The authors were partially supported by the Polish MNiSW grant N N201 393737. References [1] B. Maurey, Points fixes des contractions de certains faiblement compacts de L1 , in: Seminaire d’Analyse Fonctionelle 1980–1981, Vol. Exp. No. VIII, École Polytech., Palaiseau, 1981. [2] J. Elton, P.-K. Lin, E. Odell, S. Szarek, Remarks on the fixed point problem for nonexpansive maps, Contemp. Math. 18 (1983) 87–120. [3] M.A. Khamsi, W.A. Kirk, An Introduction to Metric Spaces and Fixed Point Theory, Wiley-Interscience, New York, 2001. [4] W.A. Kirk, B. Sims (Eds.), Handbook of Metric Fixed Point Theory, Kluwer Academic Publishers, Dordrecht, 2001. [5] S. Prus, Banach spaces which are uniformly noncreasy, Nonlinear Anal. 30 (1997) 2317–2324. [6] A. Betiuk-Pilarska, S. Prus, Banach lattices which are order uniformly noncreasy, J. Math. Anal. Appl. 342 (2008) 1271–1279. [7] W. Kurc, A dual property to uniform monotonicity in Banach lattices, Collect. Math. 44 (1993) 155–165. [8] J.M. Borwein, B. Sims, Non-expansive mappings on Banach lattices and related topics, Houston J. Math. 10 (1984) 339–356. [9] S. Dhompongsa, A. Kaewcharoen, Fixed point theorems for nonexpansive mappings and Suzuki-generalized nonexpansive mappings on a Banach lattice, Nonlinear Anal. 71 (2009) 5344–5353. [10] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces II, Springer-Verlag, New York, 1979. [11] A.G. Aksoy, M.A. Khamsi, Nonstandard Methods in Fixed Point Theory, Springer-Verlag, New York, 1990.
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