# Monotonicity and the Dominated Farthest Points Problem in Banach Lattice

Authors: H. R. Khademzadeh; H. Mazaheri
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We introduce the dominated farthest points problem in Banach lattices. We prove that for two equivalent norms such thatX becomes an STM and LLUM space the dominated farthest points problem has the same solution. We give some conditions such that under these conditions
the Fréchet differentiability of the farthest point map is equivalent to the continuity of metric antiprojection in the dominated farthest points problem. Also
we prove that these conditions are equivalent to strong solvability of the dominated farthest points problem. We prove these results in STM, reflexive STM, and UM spaces. Moreover, we give some applications of the stated results
in Musielak-Orlicz spaces L​ϕ(μ) and E​ϕ(μ) over nonatomic measure spaces in terms of the function ϕ. We will prove that the Fréchet differentiability of the farthest point map and the conditions ϕ∈Δ2 and ϕ>0 in reflexive Musielak-Orlicz function spaces are equivalent.


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## Body

## 1. Introduction

The problem of farthest points in Banach spaces is studied with many authors (see [1–3]). An interesting question in this field is, under what conditions on the set A does the point x0∈A farthest from point x in the spaces exist and is this unique? We recall that the mapping
(1)fA(x)=sup⁡{∥x-y∥:y∈A}
is called farthest point map and the mapping
(2)FA(x)={x0∈A:∥x-x0∥=fA(x)}
is called a metric antiprojection. Fitzpatrick in [2] gives some conditions such that farthest point map is Fréchet differentiable and the metric antiprojection is continuous; in fact he showed that these conditions are equivalent. Balashov and Ivanov in [1] proved that in Hilbert spaces, the set of conditions for the existence, uniqueness, and Lipschitz dependence (on x) of the metric antiprojection of x on the set A for points x that are sufficiently far from the set A is equivalent to the strong convexity of the set A. Ivanov in [3] showed that the results of [1] generalized to uniformly convex Banach spaces with Fréchet differentiable norm.Kurc in [4] introduces the dominated best approximation problem and examines the relations between monotonicity properties and the existence and uniqueness of the dominated best approximation problem. Hudzik and Kurc proved that strictly monotone and order continuity of the norm on X is equivalent to unique solvability of the dominated best approximation problem (e.g., [5]).In this paper we introduce the dominated farthest point problem in a Banach lattice and try to examine the relation between the dominated farthest point problem and monotonicity, Fréchet differentiability of farthest point map, and the continuity of antiprojection map in Banach lattices.In preliminaries section we recall main definitions and some lemmas that will be used in this context. In Section3, we introduce the dominated farthest points problem and state some conditions such that guaranteed, existence and uniqueness of the dominated farthest points problem. We give some criteria for strict monotonicity, lower locally uniformly monotone, upper locally uniformly monotone and uniformly monotone. Also we prove that for two equivalent norms such that X becomes an STM space and LLUM space the dominated farthest point has the same solution. This note will prove that the Fréchet differentiability of farthest point map is equivalent to continuity of antiprojection map under some conditions. In fact these conditions are equivalent to strong solvability of the set A. We give some conditions such that it is proved that if FA(x) is a singleton set then A is a singleton set.Finally we will say some application of the stated results in Musielak-Orlicz function spaceLϕ(μ) and Eϕ(μ) over nonatomic measure spaces in terms of the function ϕ. Equivalency of the Fréchet differentiability of the farthest point map and the conditions ϕ∈Δ2 and ϕ>0 in reflexive Musielak-Orlicz function spaces is the final result which will be proved.


## 2. Preliminaries

LetX be a Banach lattice and A a bounded sublattice in X. Suppose that x∈X such that x≥A (i.e., x≥y for each y∈A); we define FA(x) as 1.1; we always refer to such problems as to the dominated Farthest points problem.The dominated farthest points problem is called solvable ifFA(x)≠∅. The problem is said to be uniquely solvable if card⁡(FA(x))=1 and is to be stable if for every maximizing sequence {xn} in A, that is, a sequence in A such that lim⁡n→∞∥x-xn∥=fA(x), there holds d(xn,FA(x))→0 as n→∞. Finally, the problem is said to be strongly solvable if it is uniquely solvable and stable. A sequence {xn} in A is a maximizing sequence for x0 if lim⁡n→∞∥xn-x0∥=fA(x0).In this section, we recall some definitions and lemmas which we need in main results.Definition 1 (see [4]).
A Banach latticeX is said to be strictly monotone (X∈ STM) if, for all x,y∈X+, the conditions x≥y, y≠0, and ∥x∥=∥y∥ imply x=y.Definition 2 (see [4]).
A Banach latticeX is said to be uniformly monotone (X∈UM) if, for all yn≥xn≥0, such that lim⁡n→∞∥xn∥=lim⁡n→∞∥yn∥, then lim⁡n→∞⁡∥xn-yn∥=0.Definition 3 (see [5]).
A Banach latticeX is said to be upper (lower) locally uniformly monotone, X∈ ULUM (X∈ LLUM), if, for each x,yn∈X, such that yn≥x≥0 (yn≤x≤0) and lim⁡n→∞∥xn∥→∥x∥,  then lim⁡n→∞∥xn-x∥=0.Definition 4 (see [6]).
A Banach latticeX is said to be decreasing (increasing) uniformly monotone, X∈ DUM (X∈IUM), if, for each yn,xn∈X+, such that yn≥xn↓(xn≤yn↑) and lim⁡n→∞∥xn∥=lim⁡n→∞∥yn∥,  then lim⁡n→∞∥xn-yn∥=0.Definition 5 (see [5]).
A Banach latticeX is said to be CWLLUM if for any nonnegative x′∈X′ with ∥x′∥=1 and any x∈X, with x≥0, ∥x∥=1 and any sequence {xn} in X satisfying 0≤xn≤x for all n the condition x′(x-xn)→∥x∥ implies ∥xn∥→0.Definition 6 (see [7]).
A lattice seminormρ on a Riesz space is said to be order continuous whenever xα↓0 implies ρ(xα)↓0. If the above condition holds for sequences, that is, xn↓0 implies ρ(xn)↓0, then ρ is said to be σ-order continuous. If ρ is a lattice norm then the norm is order continuous.Definition 7 (see [7]).
A Banach latticeX is said to be a Kantorovich-Banach space (or briefly a KB-space) whenever every increasing norm bounded sequence of X+ is norm convergent.Definition 8 (see [8]).
We say that the norm of the Banach spaceX is Fréchet differentiable at x0∈S(X) whenever
(3)lim⁡λ→0∥x0+λy∥-∥x0∥λ
exists uniformly for y∈S(X). If the norm of X is Fréchet differentiable at x∈S(X), then we say that X has a Fréchet differentiable norm, where S(X)={x∈X:∥x∥=1}.Definition 9 (see [9]).
For a functionf from a Banach space X into a Banach space Y the Gâteaux derivative at a point x0∈X is by definition a bounded linear operator T:X→Y such that, for every u∈X,
(4)lim⁡t→0f(x0+tu)-f(x0)t=Tu.
The operator T is called the Fréchet derivative of f at x0 if it is a Gâteaux derivative of f at x0 and the limit in (4) holds uniformly in u in the unit ball (or unit sphere) in X.Definition 10 (e.g., [10–12]).
Let(T,Σ,μ) be a σ-finite, complete (nontrivial), positive measure space and ϕ(t,r):T×ℝ+→ℝ¯+ a function such that for μ-a.e. t∈T, ϕ(t,0)=0, and ϕ(t,·) is nontrivial (continuous at zero with nonzero values), convex, and lsc.
Moreover, ifϕ(·,r) is measurable, for all r>0, then we call ϕ the Musielak-Orlicz function.Definition 11.
Musielak-Orlicz spacesLϕ(μ) consist of all μ-measurable functions f:T→ℝ¯ such that
(5)Iϕ(αf)=∫T‍ϕ(α|f(t)|,t)dμ<+∞
for some α>0 (depending on f).Musielak-Orlicz spaces under the natural ordering, when endowed with each of the following norms, become a Banach lattice (e.g., [12, 13]). Luxemburg norm is
(6)∥f∥ϕ=inf⁡{λ>0:Iϕ(fλ)≤1},
and Orlicz norm is
(7)∥f∥ϕ0=sup⁡⁡{|〈f,g〉|:Iϕ*(g)≤1},
where 〈f,g〉=∫T‍f(t)g(t)dμ and ϕ* denote the young conjugate of ϕ. The amemiya norm (see [14–16] for the Orlicz spaces and [17] for the general case) is
(8)∥f∥ϕA=inf⁡k>01k(1+Iϕ(kf)).
All norms defined above are lattice monotone norms and they are equivalent:
(9)∥f∥ϕ≤∥f∥ϕ0≤2∥f∥ϕ,∥f∥ϕ0≤∥f∥ϕA.
In the following we will write, for short, ϕ>0 or ϕ<+∞, if for μ-a.e. t∈T the function ϕ(t,·) is strictly positive (except zero) or assumes finite values only, respectively. In the case that ϕ is finitely valued and the ∞-condition is satisfied then ∥f∥ϕ0=∥f∥ϕA for all f∈Lϕ(μ). We recall that ϕ is satisfied ∞-condition if ϕ(t,u)/u→∞ as u→∞ for μ-a.e.  t∈T.The functionϕ is said to satisfy a Δ2, condition (ϕ∈Δ2), if there exist a set T0 of zero measure, a constant K>0, and an integrable (nonnegative) function h, such that, for all t∈T∖T0 and r>0, there holds
(10)ϕ(2r,t)≤Kϕ(r,t)+h(t).
Suppose that Laϕ(μ) is a subspace of functions with order continuous norm
(11)Laϕ(μ)={f∈Lϕ(μ):|f|≥fn↓0⟹∥fn∥ϕ↓0},Eϕ(μ)={f∈Lϕ(μ):Iϕ(αf)<∞∀α>0}.
Then Eϕ(μ)⊂Laϕ(μ)⊂Lϕ(μ) as closed ideals (see [18], [13, p. 17], and [19]). If ϕ<+∞ then Eϕ(μ) is super order dense in Lϕ(μ) and Laϕ(μ)=Eϕ(μ) [13, p. 19], and Lϕ(μ) has an order continuous norm precisely when Laϕ(μ)=Lϕ(μ). Clearly the norm in Eϕ(μ) is order continuous.Lemma 12 (see [20]).
The following assertions are equivalent:(i)
the norm onX is order continuous,(ii)
X is Dedekind complete (σ-Dedekind complete) satisfying ∥xn∥→0 as n→∞ for any decreasing sequence {xn}⊂X+ with inf⁡{xn:n∈ℕ}=0,(iii)
every monotone order bounded sequence ofX is convergent,(iv)
every disjoint order bounded sequence ofX+ is convergent to zero,(v)
X is an ideal in X′′,(vi)
every order interval ofX is weakly compact,(vii)
X′=Xn′.Lemma 13 (see [20]).
The following assertions are equivalent:(i)
X is reflexive,(ii)
X and X′ are KB-spaces,(iii)
X does not contain any subspace isomorphic to l1 or to c0,(iv)
X does not contain any sublattice isomorphic to l1 or to c0.Lemma 14 (see [21]).
A Banach latticeX has an order-continuous norm if and only if it has an equivalent locally uniformly convex lattice norm.Lemma 15 (see [22]).
IfX is a Banach lattice, the following properties are equivalent: (i)
X and X′ have order continuous norms,(ii)
there exists an equivalent lattice norm onX which is locally uniformly convexand Fréchet differentiable, such that its dual norm is also locally uniformly convex on X′.Lemma 16 (see [22]).
A Banach lattice is reflexive if and only if it can be given an equivalent lattice norm such that both the space and its dual are simultaneously locally uniformly convex and Fréchet differentiable.Lemma 17 (see [2]).
SupposeA is a closed subset of a Banach space X such that the norm of X′ is Fréchet differentiable. If A is bounded and fA is Fréchet differentiable at some x∈X, then every maximizing sequence for x converges; also FA is continuous at x.Lemma 18 (see [2]).
Suppose thatX is a Banach space such that the norms of X and X′ are both Fréchet differentiable. If A is a closed bounded subset of X and x is a point of X, then the following are equivalent: (i)
the metric antiprojection is continuous at x,(ii)
every maximizing sequence inA for x converges,(iii)
the functionfA is Fréchet differentiable at x.Lemma 19 (see [23]).
Given a Banach latticeX the following hold true: (i)
ifX+ is rotund, then X is strictly monotone;(ii)
ifX+ is locally uniformly rotund, then X is upper and lower locally uniformly monotone;(iii)
ifX+ is uniformly rotund then X is uniformly monotone,(iv)
in the order intervals in the positive coneX+ the inverse statement of each of the above is also true.Lemma 20 (see [24]).
The following statements are equivalent:(i)
ϕ∈Δ2,(ii)
∥f∥ϕ=1 implies Iϕ(f)=1,(iii)
∥fn∥ϕ↑1 implies Iϕ(fn)→1,(iv)
Lϕ(μ) dose not contain an isometric copy of l∞,(v)
Lϕ(μ) dose not contain a lattice isometric copy of l∞,(vi)
Lϕ(μ)=Laϕ(μ); that is, the norm ∥·∥ϕ is order continuous on Lϕ(μ).


## 3.  Dominated Farthest Points Problem in Banach Lattices

In this section we introduce the dominated farthest points problem in Banach lattices and give some criteria for strict monotonicity, lower locally uniformly monotone, upper locally uniformly monotone, and uniformly monotone. We prove that if(X,∥·∥1) and (X,∥·∥2) are two Banach lattices with the same order such that ∥·∥1 and ∥·∥2 are equivalent norms and (X,∥·∥1) is an STM space and (X,∥·∥2) is an LLUM space, then the problem of the dominated farthest points has the same solution in two spaces. We give some condition such that the Fréchet differentiability of farthest point map is equivalent to continuity of antiprojection map.Lemma 21.
LetX be a Banach lattice, A a sublattice, and x∈X such that x≥A. If a1∈FA(x) and a2∈A such that a1≥a2, then a2∈FA(x).Proof.
Sincea1≥a2 and x≥A, so 0≤x-a1≤x-a2 and thus ∥x-a1∥≤∥x-a2∥. Since a1∈FA(x) we have ∥x-a1∥=∥x-a2∥. Therefore, a2∈FA(x).Theorem 22.
LetX be a Banach lattice. Then X is an STM space if and only if for every sublattice A in X and x∈X such that x≥A, card(FA(x))≤1.Proof.
Suppose thatX is an STM space and A a sublattice in X. Suppose that x∈X such that x≥A and s,t∈FA(x). Since A is a sublattice s∧t∈A and s∧t≤s, from Lemma 21, s∧t∈FA(x), so ∥x-s∧t∥=∥x-s∥; since X is an STM space s∧t=s; similarly s∧t=t; therefore s=t.
Conversely ifX is not an STM space, then there exist x,y∈X such that x≥y≥0 and ∥x-y∥=∥x∥. Define A=[0,y]; then A is a sublattice and x≥A; since x≥x-t for each t∈A, so ∥x∥≥∥x-t∥ for each t∈A, and thus 0∈FA(x), by the assumption ∥x∥=∥x-y∥; so y∈FA(x); therefore card(FA(x))>1.Theorem 23.
A Banach latticeX is an STM space and has order continuous norm if and only if the dominated farthest points problem with respect to closed order bounded sublattices is uniquely solvable.Proof.
Suppose thatX is an STM space and has order continuous norm and A is a closed sublattice. We assume that x≥A and {yn} is a maximizing sequence; that is, fA(x)=sup⁡y∈A∥x-y∥=lim⁡n→∞∥x-yn∥. Put zn=∧k=1nyk. Since A is a sublattice and 0≤x-yn≤x-zn, {zn} is a decreasing maximizing sequence. Since there exists z=∧nyn≤x, we have 0≤zn-z↓0, by the order continuity of the norm ∥zn-z∥→0. On the other hand A is a closed sublattice so z∈A and, since ∥x-z∥=fA(x), therefore z∈FA(x).
Suppose that the norm onX is not order continuous; then there exists a sequence {yn} such that 0≤yn↓0 but inf⁡∥yn∥>0. We can assume that ∥yn∥>∥yn+1∥ for n∈ℕ, otherwise replacing yn by (1+1/n)yn, for n∈ℕ. Suppose that x∈X such that x>y1 and put A={yn}; then A is a sublattice and FA(x)=∅. On the other hand by Dini’s theorem A is norm closed. Indeed if ∥ynk-z∥→0, ynk∈A and z∉A, then we use the fact that if ynk is downward directed sequence which is weakly convergent to z then z=inf⁡k(ynk), and thus z=0; hence ∥ynk∥→0; this is a contradiction and so z∈A. By the assumption FA(x)≠∅ this contradiction completes the proof.Remark 24.
In Theorem23, if X is an LLUM, CWLLUM, IUM, or DUM space then the theorem is also true.Theorem 25.
LetX be an order continuous Banach lattice with the ULUM property; then the dominated farthest points problem with respect to closed order bounded sublattices is strongly solvable.Proof.
From Theorem23 the dominated farthest points problem with respect to closed sublattices is uniquely solvable. The proof of stability is the same as the proof of Theorem  4.4 in [5].Theorem 26.
LetX be an STM space and A a sublattice in X. If x,y∈X such that x,y≥A (or x,y≤A) and FA(x),FA(y)≠∅, then FA(x)=FA(y).Proof.
Suppose thatx,y∈X such that x,y≥A and FA(x)={x0},FA(y)={y0}. Since A is a sublattice so x0∧y0∈A; from Lemma 21, x0∧y0∈FA(x). Similarly x0∧y0∈FA(y); therefore x0∧y0=x0=y0.Corollary 27.
LetX be an STM space with order continuous norm. If A is a sublattice in X, x∈X such that x≥A or x≤A; then the metric antiprojection is in this form:
(12)FA(x)={{x0},x≥A,{y0},x≤A,
for some x0,y0∈A.Proof.
It is a consequence of Theorems23 and 26.Theorem 28.
LetX be an STM space and A a sublattice in X. If x,y∈X such that x≥A and y≤A and FA(x)=FA(y)≠∅, then A is singleton.Proof.
Suppose thatFA(x)=FA(y)={x0} and y0∈A. Since A is a sublattice so x0∧y0∈A, from Lemma 21, x-x0∧y0≥x-x0≥0, thus x0∧y0∈FA(x)={x0}, thus x0≤y0, and again, from Lemma 21, we have y0∈FA(y)={x0}. Therefore A={x0}.Lemma 29.
Let(X,∥·∥1) and (X,∥·∥2) be two Banach lattices with the same order such that ∥·∥1 and ∥·∥2 are equivalent norm. If (X,∥·∥1) is an order continuous Banach lattice, then (X,∥·∥2) is also a Banach lattice with order continuous norm.Proof.
It is a consequence of equivalency of two norms.Example 30.
(a) Every equivalent norm onLp(μ) with ∥·∥p for 1<p<∞ is order continuous.
(b) Every equivalent norm onC(X) (X is not a finite set) with sup-norm is not order continuous.Theorem 31.
Let(X,∥·∥1) and (X,∥·∥2) be two Banach lattices with the same order such that ∥·∥1 and ∥·∥2 are equivalent norms. Suppose that (X,∥·∥1) is an STM space and (X,∥·∥2) is an LLUM space. If A is a sublattice of X and x∈X such that x≥A (or x≤A), then FA1(x)=FA2(x), where FAi(x) is the set of farthest points with respect to ∥·∥i for i=1,2.Proof.
Since(X,∥·∥2) is order continuous and ∥·∥1 and ∥·∥2 are equivalent, from Lemma 29, (X,∥·∥1) is order continuous. From Theorem 23 and Remark 24, the dominated farthest points problem is uniquely solvable for two norms. Suppose that x0∈FA2(x); if x0∉FA1(x), there exists x0≠y0∈A such that ∥x-y0∥1>∥x-x0∥1. Since A is a sublattice x0∧y0∈A; since x0≠y0, we have either x0∧y0<x0 or x0∧y0<y0.
Ifx0∧y0<x0, then x-x0∧y0>x-x0≥0 and since (X,∥·∥2) is an LLUM space ∥x-x0∧y0∥2>∥x-x0∥2 that is a contradiction.
Ifx0∧y0<y0, then x-x0∧y0>x-y0 and so ∥x-x0∧y0∥1>∥x-y0∥1; also x-x0∧y0≥x-x0 and so ∥x-x0∧y0∥2≥∥x-x0∥2; since {x0}=FA2(x), we have x0∧y0=x0 and so ∥x-x0∥1=∥x-x0∧y0∥1>∥x-y0∥1 that is a contradiction; therefore x0=y0; thus FA1(x)=FA2(x).Remark 32.
If we assume that(X,∥·∥1) is an STM space with order continuous norm, then Lemmas 14 and 19 guaranteed the existence of ∥·∥2, such that (X,∥·∥2) is a LLUM space.Theorem 33.
LetX be a Banach lattice and A a closed order bounded subset of X. If x∈X, x≥A, the following statements are equivalent: (i)
any sequence{an}⊂A such that lim⁡n→∞∥x-an∥=fA(x) is convergent;(ii)
the metric antiprojection is uniquely defined onx≥A and, for any vector x≥A, any maximizing sequences have a convergent subsequence;(iii)
dominated farthest point problem with respect to closed bounded sublattices is strongly solvable.Proof.
(i)→ (ii). Suppose that there exists an x≥A, such that card(FA(x))>1. We assume that x0,y0∈FA(x), x0≠y0, and {an},{bn} are two maximizing sequences in A convergent to x0 and y0, respectively.
We define(13)cn={an,nisodd,bn,niseven;
then cn is not convergent, that is, a contradiction, so card(FA(x))=1. Since every maximizing sequence is convergent the proof is complete.
(ii)→ (i). Suppose that FA(x)={x0} and {xn} is a maximizing sequence. If the condition (i) is not true, then {xn} has a subsequent {xnj} such that ∥x0-xnj∥≥ɛ>0, for any j∈ℕ. Since lim⁡j→∞∥x-xnj∥=fA(x), the sequence {xnj} has a limit point of  x0′, we have x0′∈FA(x) and x0≠x0′ that is a contradiction; therefore (i) is true.
(ii)→ (iii). By the assumption FA(x)={x0}, suppose that {an} is a maximizing sequence in A for x. From (ii), {an} has a convergent subsequence to x0, so d({an},FA(x))=d({an},x0)=0.
(iii)→ (ii). From the definition of strongly solvable  card(FA(x))=1, put FA(x)={x0}. Suppose that {an} is a maximizing sequence in A for x. So d({an},x0)=0 so x0 is a limit point of {an}; thus there exists a subsequence {anj} convergent to x0; this completes the proof.Corollary 34.
LetX be a Banach lattice with ULUM property and order continuous norm and A a closed order bounded sublattice of X; then (i)
any sequence{an}⊂A such that lim⁡n→∞∥x-an∥=fA(x) is convergent;(ii)
the metric antiprojection is uniquely defined onx≥A and, for any vector x≥A, any maximizing sequences have a convergence subsequence,(iii)
dominated farthest point problem with respect to closed order bounded sublattices is strongly solvable.Proof.
It is a compound of Theorems25 and 33.Theorem 35.
Let(X,∥·∥) be an STM space such that X′ and X′′ have order continuous norm, A a closed order bounded sublattice in X, and x0∈X, such that x0≥A. If fA is Fréchet differentiable at x0 then (i)
card(FA(x0))=1,(ii)
any sequence{an}⊂A such that lim⁡n→∞∥x0-an∥=fA(x0) is convergent,(iii)
ifxn→x0 and {ak}=FA(xk) for k∈ℕ∪{0} then an→a0,(iv)
the dominated farthest points problem is strongly solvable.Proof.
Part (i). From Theorem23, the dominated farthest points problem is uniquely solvable so card(FA(x0))=1.
Part (ii). From (i)FA(x0)={a0}, suppose that {an} is a maximizing sequence in A, lim⁡n→∞∥x0-an∥=∥x0-a0∥. a0 is a limit point of {an} and so there is a subsequence {ank} such that ank→a0 from Lemma 15, there exists an equivalent norm ∥·∥1 such that X′ and X′′ are locally uniformly convex space and X′ has Fréchet differentiable norm, from Theorem 31, and  a0 is unique farthest point in A from x0 with ∥·∥1. Since {an} has a subsequence convergent to a0, {an} is also a maximizing sequence with ∥·∥1 and hence lim⁡n→∞∥x0-an∥1=∥x0-a0∥1. From Lemma 17, {an} converges with ∥·∥1, so it is convergent with ∥·∥ and this completes the proof.
Part (iii). From Lemma17, FA is continuous so if xn→x0; then FA(xn)→FA(x0) or equivalently an→a0.
Part (iv). By part (i),card(FA(x))=1. Suppose that {xn} is a maximizing sequence in A; that is, lim⁡n→∞∥x0-xn∥=∥x0-a0∥. Since xn→x0 we have d(xn,FA(x0))→0; therefore the dominated farthest points problem is strongly solvable.Theorem 36.
Let(X,∥·∥) be a reflexive, STM space A, a closed order bounded sublattice in X, and x0∈X, such that x0≥A. The following statements are equivalent: (i)
the dominated farthest points problem is strongly solvable;(ii)
any sequence{an}⊂A such that lim⁡n→∞∥x0-an∥=fA(x0) is convergent;(iii)
ifxn→x0 and {ak}=FA(xk) for k∈ℕ∪{0}, then an→a0;(iv)
fA is Fréchet differentiable at x0.Proof.
(i)↔ (ii). From Theorem 33 thus is true.
(iv)→ (ii), (iv)→ (iii). From Lemma 13, X and X′ are KB-spaces and so from Lemma 12 they have order continuous norm. Therefore it is a part of Theorem 35.
(ii)→ (iv), (iii)→ (iv). From Lemma 16, there exists an equivalent norm ∥·∥1 such that X′ and X′′ are locally uniformly convex space and X′, X′′ has Fréchet differentiable norm, and from Lemma 18, (ii), (iii), and (iv) are equivalent to norm ∥·∥1, so fA is Fréchet differentiable with norm ∥·∥1 since ∥·∥ is equivalent to ∥·∥1 from [25, p. 3]; fA is Fréchet differentiable with norm ∥·∥; this completes the proof.Theorem 37.
LetX be a uniformly convex Banach lattice space, A a closed order bounded sublattice in X, and x0∈X, such that x0≥A. The following statements are equivalent: (i)
the dominated farthest points problem is strongly solvable;(ii)
any sequence{an}⊂A such that lim⁡n→∞∥x0-an∥=fA(x0) is convergent;(iii)
ifxn→x0 and {ak}=FA(xk) for k∈ℕ∪{0}, then an→a0;(iv)
fA is Fréchet differentiable at x0.Proof.
SinceX is uniformly convex, so it is a reflexive Banach lattice and from Lemma 19, it is a UM space; from Theorem 36 the proof is complete.Theorem 38.
LetX be a UM space with an order unit 1, A a closed order bounded sublattice in X, and x0∈X, such that x0≥A. The following statements are equivalent: (i)
the dominated farthest points problem is strongly solvable;(ii)
any sequence{an}⊂A such that lim⁡n→∞∥x-an∥=fA(x) is convergent;(iii)
ifxn→x0 and {ak}=FA(xk) for k∈ℕ∪{0}, then an→a0;(iv)
fA is Fréchet differentiable at x0.Proof.
SinceX is a UM space with order unit 1, B1(0)=[-1,1]. So from Lemma 19, rotundity properties are equivalent to monotonicity properties on B1(0), so X is a uniformly convex Banach lattice from Theorem 37; the proof is complete.


## 4.  Some Applications of the Dominated Farthest Points Problem in Musielak-Orlicz Spaces

In this section we give some applications of the theorems that were proved in Section3 in Musielak-Orlicz function space. The most important result that will be proved in this section is the equivalency of the Fréchet differentiability of farthest point map and the conditions ϕ∈Δ2 and ϕ>0 in reflexive Musielak-Orlicz function spaces.Theorem 39.
For the Musielak-Orlicz spaceLϕ(μ) the following statements are equivalent: (i)
ϕ∈Δ2;(ii)
for each closed order bounded sublatticeA in Lϕ(μ) and x∈Lϕ(μ) such that x≥A (or x≤A), card(FA(x))≥1.
Moreover (ii) is true forEϕ(μ) with ϕ<∞.Proof.
It is a consequence of Theorem23.Remark 40.
In Musielak-Orlicz spaceLϕ with μ(T)<∞ and nonatomic measure μ, if ϕ∈Δ2 and ϕ>0 with any norm the sets of farthest points in dominated farthest points problem are coinciding; this means that FA1(x)=FA2(x)=FA3(x), for every x∈X and sublattice A⊂X with x≥A, and that FAi(x) for i=1,2,3 denote the set of farthest points with respect to Luxemburg norm, Orlicz norm, and Ammemyia norm, respectively.Corollary 41.
The dominated farthest points problem inLϕ(μ) (Eϕ(μ) with ϕ<∞) with respect to closed order bounded sublattices is unique if and only if ϕ>0 and ϕ∈Δ2 (resp., ϕ>0).Proof.
From [4, Theorem 2.7] (resp., [4, Theorem 2.8]) Lϕ(μ) has order continuous norm and it is an STM space (resp., Eϕ(μ)); so from Theorem 22 the proof is complete.Theorem 42.
In the Musielak-Orlicz spaces the following statements are equivalent:(i)
ϕ∈Δ2;(ii)
for all closed order bounded linear sublatticesA⊂Lϕ(μ), FA(x)≠∅ for each x∈Lϕ(μ).
Moreover (ii) is true forEϕ(μ) with ϕ<∞.Proof.
(ii)→ (i). It is an immediate consequence of (ii)→ (i), in Theorem 39.
(i)→ (ii). Suppose that {xn} is a maximizing sequence in A; for example,
(14)lim⁡n→∞∥x-xn∥ϕ=fA(x).
SinceA is a bounded subset of Lϕ(μ), so sup⁡n{∥xn∥ϕ}<∞. Lϕ(μ) is a KB-space [4, Theorem 3.5] so {xn} is convergent to some x0∈X. Since A is a closed linear sublattice, so x0∈A; this complete the proof.Remark 43.
LetX be a KB-space; with the same argument as Theorem 44, we can show that for all closed bounded linear sublattices A⊂X, FA(x)≠∅ for each x∈X.Theorem 44.
LetA be a closed order bounded sublattice in the reflexive Musielak-Orlicz spaces with the Luxemburg norm; the following statements are equivalent: (i)
the dominated farthest points problem is strongly solvable;(ii)
any sequence{an}⊂A such that lim⁡n→∞∥x-an∥=fA(x) is convergent;(iii)
ifxn→x0 and {ak}=FA(xk) for k∈ℕ∪{0}, then an→a0;(iv)
fA is Fréchet differentiable at x0;(v)
ϕ>0 and ϕ∈Δ2;(vi)
Lϕ(μ) (resp. Eϕ(μ) with ϕ<∞) is a UM (resp. STM) space.Proof.
From Theorem36 (i)–(iv) are equivalent. From [4, Theorem 2.7, 2.8] (v) and (vi) are equivalent. From Theorems 38 and 25 (i) and (v) are equivalent.



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*Source: 616989-2014-03-27.xml*