Monotonicity properties and solvability of dominated best approximation problem in Orlicz spaces equipped with s-norms

In the paper, Wisła (J Math Anal Appl 483(2):123659, 2020, 10.1016/j.jmaa.2019.123659), it was proved that the classical Orlicz norm, Luxemburg norm and (introduced in 2009) p-Amemiya norm are, in fact, special cases of the s-norms defined by the formula xΦ,s=infk>01ks∫TΦ(kx)dμ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left\| x\right\| _{\Phi ,s}=\inf _{k>0}\frac{1}{k}s\left( \int _T \Phi (kx)d\mu \right) $$\end{document}, where s and Φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Phi $$\end{document} are an outer and Orlicz function respectively and x is a measurable real-valued function over a σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document}-finite measure space (T,Σ,μ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(T,\Sigma ,\mu )$$\end{document}. In this paper the strict monotonicity, lower and upper uniform monotonicity and uniform monotonicity of Orlicz spaces equipped with the s-norm are studied. Criteria for these properties are given. In particular, it is proved that all of these monotonicity properties (except strict monotonicity) are equivalent, provided the outer function s is strictly increasing or the measure μ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu $$\end{document} is atomless. Finally, some applications of the obtained results to the best dominated approximation problems are presented.


Introduction
Many geometric properties of Banach spaces are directly related to the property of its norm. One such important property is the property of the existence of an element that realizes the distance of a set from a point (the so-called solvability of best approximation problem). More precisely, for a given nonempty proper subset A of a Banach space X and x / ∈ A the question is stated whether there exists z ∈ A such that In this paper, we will give results on solvability (even more, on strong solvability) of the dominated best approximation problem, i.e., for the case when A is a closed sublattice of an Orlicz space X and z is an upper or lower boundary of A.
The dominated best approximation problem is strictly connected with the monotonicity properties of the norm. Thus in the first part of the paper we will investigate the strict monotonicity, uniform monotonicity, local lower uniform monotonicity and upper local uniform monotonicity of Orlicz spaces equipped with the so-called s-norms [36].
Orlicz spaces L , introduced by W. Orlicz in 1932 (see [33]) form a wide class of Banach spaces of measurable functions or sequences. Originally W. Orlicz defined the norm as follows where is the complementary function to in the sense of Young defined by In 1955, Luxemburg [25] investigated topologically equivalent norm to the Orlicz one which was defined as follows where I (x) = T (x(t))dμ. In the fifties Ichiro Amemiya (see [30, p. 218]) considered the norm defined by the following formula Krasnoselskii and Rutickii [18], Nakano [30], Luxemburg and Zaanen [26] proved, under additional assumptions on the function , that the Orlicz norm can be expressed exactly by the Amemiya formula, i.e. · o = · A . In the most general case of Orlicz function , the similar result was obtained by Hudzik and Maligranda [12]. Moreover, it is not difficult to verify that the Luxemburg norm can also be expressed by an Amemiya-like formula (see [4,32]), namely In the paper, Hudzik and Maligranda [12] proposed to investigate another class of norms given by the so-called p-Amemiya formula where 1 ≤ p ≤ ∞ (if p = ∞ then we use the formula (1)).
However it is easy to observe that both formulae (1) and (2) can be expressed by the norm (called s-norm) defined by where s is the so-called outer function (see [36]). Let us underline that this concept covers all the previously investigated cases of norms. Moreover, each outer function corresponds to a random nonnegative variable X (over a probability space ( , M, P)) with expected value 0 ≤ E(X ) ≤ 1 (see the Preliminaries chapter below). So the results presented in this paper may find applications in probability theory or stochastic modelling. In recent years several papers have been published devoted to the study of the problem of monotonicity of Orlicz space equipped with a certain norm (and often under additional assumptions about the function or the measure μ) see, e.g., [2,6,9,11,[19][20][21]24]. In this paper we consider the most general case of s-norms and Orlicz functions. Moreover, we will investigate the general case of measure space-in particular, there may be atoms among measurable sets.
In the section 3 it was proved that strict monotonicity of the function on (0, ∞) is necessary for L ,s (the Orlicz space L equipped with the s-norm · ,s ) to be strictly monotone, for all possible choices of the outer function s (Lemma 3.7). The strict monotonicity of is also sufficient for L ,s to be strictly monotone as long the outer function s is strictly increasing on (0, ∞) (Theorem 3.9). If the outer function s is constant on some nontrivial interval (0, ε) and the Orlicz function takes finite values only, then the Orlicz space L ,s is strictly monotone if and only if is strictly increasing on (0, ∞), satisfies the 2 -condition (Theorem 3.9).
The uniform monotonicity of L ,s does not depend on the outer function s in the case of atomless measure μ. Namely, if the measure μ is atomless (and for any choice of s) or s is strictly increasing on (0, ∞) (and for any choice of measure μ), the Orlicz space L ,s is uniformly monotone if and only if is strictly increasing on (0, ∞) and satisfies the 2 -condition (Theorem 4.10). Finally, using the results in [19] we proved that, if s is an outer function strictly increasing on (0, ∞) or μ is an atomless measure then, for every closed sublattice A of L ,s , the dominated best approximation problem is strongly solvable for any (lower or upper) boundary of A as long as is strictly increasing on (0, ∞) and satisfies the 2 -condition (Corollary 5.5).

Preliminaries
For any map : If a = 0 then is positive and strictly increasing on (0, ∞). We will also write > 0 in that case. Analogously, we will write < ∞, provided b = ∞. A map : R → [0, ∞] is said to be an Orlicz function if (0) = 0, is not identically equal to zero, it is even, convex on the interval (−b , b ) and left-continuous at b , i.e., lim u→b − (u) = (b ).
Throughout the paper we will assume that (T , , μ) is a non-trivial measure space with a σ -finite and complete measure μ. By T a we will denote the (countable) set of all μ-atoms. We will say that the atomless part of T is not empty, if T contains a measurable set of positive measure that does not contain any atoms. Moreover we will always assume that either the atomless part of T is not empty or T a does not reduce to a finite set of atoms. Thus the σ -algebra is non-trivial, i.e., it contains a nonempty and proper subset of T of positive and finite measure (indeed, otherwise T would be an atom or μ would not be σ -finite).
By L 0 = L 0 (μ) we will denote the set of all μ-equivalence classes of -measurable real functions defined on T equipped with the topology of convergence in measure on μ-finite sets. For a given Orlicz function , on the space L 0 (μ) we define a convex functional (called a pseudomodular [29]) by The Orlicz space L generated by an Orlicz function is a linear lattice of measurable functions defined by the formula with the μ-a.e. partial order, i.e., x ≤ y if and only if y(t) − x(t) ≥ 0 for μ-a.e. t ∈ T . By the space E we will mean a linear subspace of L defined as follows In the case of purely atomic measure (i.e., if T \T a = ∅), the spaces L and E are usually denoted by and h respectively. We will use this notation only if it is more convenient or necessary. Let us note that the space E may degenerate into one element set {0}. For instance, this is what happens when the values jump to infinity (i.e., b < ∞).
We say that an Orlicz function satisfies the condition 2 if (a) there exists a constant K > 0 such that (2u) ≤ K (u) for every u ∈ R (resp., for every |u| ≥ u 0 , where (u 0 ) < ∞) provided the measure μ is atomless and μ(T ) = ∞ (resp. μ(T ) < ∞), (b) there are constants a > 0, K > 0 and a nonnegative sequence (c n ) ∈ 1 such that (2u)μ(e n ) ≤ K (u)μ(e n ) + c n for every u ≥ 0 with (u)μ(e n ) ≤ a and every n ∈ N , provided the measure μ is purely atomic and {e n : n ∈ N } is the set of atoms of T (this condition is known as the δ 2 -condition), (c) both conditions (a) and (b) are satisfied in the case when T contains an atomless part and the set of atoms is not empty.
It is well known that L = E if and only if ∈ 2 and < ∞ (see, e.g., [3,34] for the atomless measure case and [15] for the purely atomic case). Note that ∈ 2 implies that < ∞ provided the measure μ is atomless. But this implication is not true for purely atomic measures.
A function s : [0, ∞) → [1, ∞) is said to be an outer function, if it is convex and To simplify notations, we extend the domain and range of s to the interval [0, ∞] by setting s(∞) = ∞. Let us note that each outer function corresponds to a random nonnegative variable X (over a probability space ( , M, P)) with the expected value 0 ≤ E(X ) ≤ 1 in the sense that for u ≥ 0, where F (2) X : R → [0, ∞) denotes the the second performance function of X , which is defined as the integral of the cumulative distribution function F X : R → [0, 1], F X (u) = P(X ≤ u), over the half-line (−∞, u), i.e., [31]). Indeed, the second performance function F (2) X is nonnegative, finite, convex on R and admits an asymptote at infinity with the slope 1 and the y-intercept −E(X ) [31]. Moreover, if the random variable X is nonnegative and E(X ) ∈ [0, 1] then F (2) If s is an outer function and is an Orlicz function then the functional is a norm on the Orlicz space L [36]. In the following by L ,s (resp. E ,s ) we will denote the Orlicz space L (resp. E ) equipped with the s-norm · ,s . Put s L (u) = max {1, u} and s o (u) = 1 + u. Then the norms · ,s L and · ,s o are the Luxemburg norm · and the Orlicz norm · o respectively [25,32]. Since, for any outer function s, s L (u) ≤ s(u) ≤ s o (u) for all u ≥ 0, we have the following inequality for all x ∈ L , so all s-norms are (topologically) equivalent to each other. In general, for a fixed Orlicz function two different outer functions s 1 = s 2 create two different norms · ,s 1 ≡ · ,s 2 . But in the case of power functions (u) = c |u| p , 1 ≤ p < ∞, c > 0, all Orlicz spaces L ,s , no matter how the outer function s is chosen, coincide with the Lebesgue spaces L p and · ,s ≡ d · p up to a some constant d > 0 depending on p and s. Analogously, if (u) = 0 for |u| ≤ 1 and (u) = ∞ for |u| > 1 then, for all outer functions s, the Orlicz space L ,s coincides with the Lebesgue space L ∞ and · ,s = · ∞ (see [36]).
Note that, for any outer function s, the s-norm convergence to 0 of a sequence (x n ) of functions of L implies its modular convergence to 0. Moreover, these two types of convergence are equivalent, i.e., as n → ∞ if and only if ∈ 2 and, in the case of purely atomic measure, > 0 (see, e.g., [3,15]).
Recall that an element x of a Banach lattice (X , · ) is said to be order continuous if x n → 0 for any sequence (x n ) in X such that 0 ≤ x n ≤ |x| and x n → 0 μ-a.e. By X a we will denote the subspace of all order continuous elements in X . If X = X a then the Banach lattice X is called order continuous (OC).
If μ is purely atomic, then (L ) a = ( ) a = {0} for any Orlicz function (consider finite sequences of numbers). In general E ⊂ (L ) a ⊂ L .
Another nice characterization of 2 -condition is as follows: the modular unit sphere and the Luxemburg-norm unit sphere coincide, i.e., for all x ∈ L if and only if ∈ 2 and, if the measure μ is purely atomic, on each atom e ∈ T a the function u → (u)μ(e) achieves the value 1, i.e., (b )μ(e) ≥ 1 for all e ∈ T a (see [3,6,16]). More details about Orlicz spaces equipped with the Luxemburg or the Orlicz norm, can be found in [3,18,22,23,[27][28][29]34]. In the paper [12] Hudzik and Maligranda proposed to investigate the family of so-called p-Amemiya norms generated by the outer functions defined by s p (u) = (1 + u p ) 1/ p . Then several geometric properties of p-Amemiya norms were investigated in a number of papers, e.g., [4,6,7,13,14,35]. Basic properties of Orlicz spaces equipped with the s-norm (and, among others, the theorem on duality) are presented in [36].

Strict monotonicity properties of Orlicz spaces L 8,s
Let X be a Banach lattice with a norm · . By X + we will denote the positive cone of X , i.e. X + = {x ∈ X , x ≥ 0} and by S(A) the interception of the unit sphere of X with the set In order to calculate the value of s-norm it is important to know whether the infimum in the formula (3) is achieved at some k > 0. Denote then we say that the s-norm is k-unique. In [36], it was proved, that Moreover, the criteria for s-norm to be k *finite and k * * -finite were also established in that paper. We will start with a characterization of the set of those function x for which k * * (x) = ∞.

Lemma 3.1 For any outer function s and any Orlicz function , if the space L ,s is not k * * -finite then takes only finite values. Moreover
In consequence, the s-norm · ,s may be not k * * -finite only if the set E \ {0} is not empty.
In order to prove the inclusion of sets, take any x ∈ L \E . Then there exists k 0 > 0 such that I (k 0 x) = ∞. Then To make this paper clearer, we present the following lemma here, although some of its statements can also be deduced form the results given in [36].

Lemma 3.2 Let s be an outer function and an Orlicz function. If the space L
admits an oblique asymptote at infinity if and only if (c) < ∞, where is the Orlicz function complementary to in the sense of Young, in fact, the slope and the y-intercept of the asymptote are equal to c > 0 and − (c) respectively, (iii) the Lebesgue space L 1 is continuously imbedded into the Orlicz space L ,s and y ,s ≤ c y 1 for every y ∈ L 1 , Proof Let x ∈ L \ {0} be such that k * * (x) = ∞. Then, by Lemma 3.1, has to take finite values only and x ∈ E \ {0}. Moreover, Since is convex and (0) = 0, the function u → (u)/u is nondecreasing on the half-line (0, ∞). By the Beppo Levy theorem, If admits an oblique asymptote then, by (i), its slope is equal to c = lim u→∞ (u)/u and 0 < c < ∞. Since is convex, the slope of each secant of is less or equal to c. Thus Thus admits the oblique asymptote with the slope c if and only if (c) < ∞ and its y-intercept is equal to − (c).
Observe that (1) = ∞ in this case as well, where is the Orlicz function complementary to in the sense of Young. Indeed, the right-hand derivative p + of is equal to p + (u) = u/(1 + u) for all u ≥ 0. Hence the inverse function q + = ( p + ) −1 , which is the right-hand derivative of , is given by the formula q Evidently In order to illustrate the relationship between k * * -finiteness of the s-norm and the space E consider the following example.

Example 3.3 Let μ be an atomless measure with μ(T ) = ∞ and let
(Note: the set K * * depends on and s, but the sets L , E depend on only.) i.e., x ∈ L . Conversely, let x ∈ L 0 and k > 0 be such that I (kx) < ∞. Define Let (a n ) be any sequence of positive numbers converging to 0. Without loss of generality we can assume that (a n ) decreases to 0. Take a sequence (A n ) of pairwise disjoint measurable sets such that μ(A n ) = 1 a n for all n ∈ N and define x = ∞ n=1 a n χ A n . Evidently x / ∈ L 1 . For any k > 0 put n k = min {n ∈ N : ka n ≤ 1}. If n k = 1 then ka n ≤ 1 for all n ∈ N , whence I (kx) = 0 < ∞. So, assume that n k > 1. Then Let A be any measurable subset of T (with finite or infinite measure). Take k = 1. Since (1) = 0, for any outer function s we have whence χ A ,s ≤ 1. Further, for every 0 < k < 1, so the norm χ A ,s cannot be achieved at any 0 < k < 1. From now on we will assume that s(u) = 1 + u. Finally, for every k > 1 put The inclusion E \K * * ⊂ L 1 follows directly from Lemma 3.2. Moreover, χ A ∈ L 1 ∩ K * * for all A ∈ with 1 < μ(A) < ∞, so E \K * * L 1 .

Example 3.4
If the space L 1 + L ∞ is equipped with the Luxemburg norm, i.e., in the Example 3.3 we consider the outer function s(u) = max {1, u}, then the Orlicz space L ,s is k * * -finite and, for all measurable sets A ∈ , Proof The k * * -finitness follows from the definition of the outer function s (see [36]). Let A ∈ . For any 0 < λ < ∞ define By (5) and (6) As it was pointed out in the previous section, the 2 -condition plays an importatnt role in Orlicz spaces. In the sequel, we will need the following lemma.

Lemma 3.5 Let s be an outer function. If
/ ∈ 2 then, for every 0 < ε < 1, we can find x ,s . Hence, in the following, we will assume that d = b − a > 0.
In the case of Luxemburg norm it is well known that if / ∈ 2 then there exists x ∈ L \ {0} such that I (x) < x = 1 (see [3,16]). This implies that I (kx) = ∞ for all k > 1 (otherwise, applying the Lebesgue dominated convergence theorem we would get that x < 1).
and the thesis is proved. Thus we can assume that I k 0 xχ A 0 < ∞ for some k 0 > 1. For each n ≥ 1 define Evidently, the sets A 0 , A 1 , ... are pairwise disjoint, {t ∈ T : |x(t)| > a } ⊂ ∞ n=0 A n and Thus we can find m > 1 such that I (y) < ε, where y = ∞ n=m xχ A n . Suppose that I (ky) < ∞ for some k > 1. Without loss of generality we can assume Thus and this contradicts to the assumption that I (kx) = ∞ for all k > 1. Thus whence, putting z = y · y −1 ,s , and the proof is completed.
As the immediate consequence of Lemma 3.5 we get the following corollary.

Corollary 3.6
If / ∈ 2 then, for any outer function s, there exists a sequence (x n ) of functions in the unit sphere S(L ,s ) such that the modular I (x n ) converges to 0, i.e., x n ,s = 1 for all n ∈ N and I (x n ) → 0 as n → ∞.
Now we turn back to the monotonicity properties of Orlicz spaces. We will start with the lemma that will allow us to restrict the number of cases of outer and Orlicz functions that we have to dealt with.

Lemma 3.7 Let s and
be an outer and Orlicz function respectively and denote a s = sup {u ≥ 0 : s(u) = 1}. The space L ,s is not strictly monotone if one of the following conditions is satisfied

If a > 0 and E = {0} then the space E ,s is not strictly monotone.
Proof If T contains a nonempty atomless subset T 0 then take a set A ⊂ T 0 such that μ(A) > 0 and μ(T 0 \A) > 0. In the other case, the set of atoms T a is infinite, so take A ⊂ T a such that the set A is infinite and T a \A = ∅. Moreover, let B ⊂ T a \A be such that 0 < μ(B) < ∞.
(i) Assume that a > 0. Put z = χ B . Then z ∈ L \ {0}. Further, let ε > 0 be a fixed arbitrary positive number and take k z > 0 such that 1 k z s (I (k z z)) ≤ z ,s + ε. Define y = a k −1 z χ A and x = y + z. Evidently x, y ∈ L , 0 ≤ y ≤ x and y = 0. Moreover, whence, by arbitrariness of ε > 0, x − y ,s ≥ x ,s . Since y = 0, the space L ,s is not strictly monotone.
(iii) If the set of atoms T a is infinite and (b )μ(e) < a s for an atom e ∈ T a then, putting z = b χ {e} , we have I (z) < a s ≤ 1 and z ,s = 1. Without loss of generality we can assume that e ∈ A. Thus, we can repeat the part (ii) of the proof to conclude that L ,s is not strictly monotone.
Assume that E = {0}. This implies that takes only finite values, whence the functions z = χ B and y = a k −1 z χ A in the part (i) of the proof belong to E \ {0}. Repeating the arguments of (i) we conclude that E ,s is not strictly monotone.
In the proof of the theorem on strict monotonicity of the Orlicz space L ,s we will need the following lemma. Proof Let x ∈ L be such that I (x) < c. We claim that there exists k > 1 such that I (kx) < ∞. If T \T a = ∅ then, by 2 -condition, I 2xχ T \T a < ∞. Now, assume that the set of atoms T a is infinite. For simplicity, put T a = n e n , e n = {n} and a n = x(n) for all n ∈ N . Let a, K > 0 be constants and (c n ) ∈ 1 a sequence that appear in 2 there exists m ∈ N such that (a n )μ(e n ) ≤ a for all n ≥ m. Thus Moreover, for each n ∈ N , (a n )μ(e n ) ≤ I (x) < c ≤ (b )μ(e n ).
Thus |a n | < b for all n ∈ N . Hence, for each 1 ≤ n < m we can find k n > 1 such that k n |a n | < b . Put k 0 = min {k 1 , . . . , k m−1 , 2}. Then k 0 > 1 and and the claim is proved.
Therefore k → I (kx) is a nondecreasing continuous function on the interval (1, k 0 ) and I (x) < c. Hence we can find k ∈ (1, k 0 ) such that I (kx) < c as well.
The next theorem provides criteria for the Orlicz space equipped with the s-norm to be strictly monotone. Proof The necessity part of the proof follows directly from Lemma 3.7. Sufficiency. Assume that vanishes only at 0. Take any x, y ∈ L (respectively, x, y ∈ E , provided E = {0}) with 0 ≤ y ≤ x and y = 0. Since vanishes only at 0, we have 0 < I (k(x − y)) < I (kx) for all k > 0. Without loss of generality we can also assume that x ,s = 1.
If K (x) = ∅ then, by Lemma 3.2, admits an asymptote at infinity with the slope c > 0 and x ∈ L 1 . By strict monotonicity of the L 1 -norm, x − y 1 < x 1 < ∞, so x − y ∈ L 1 and, again by Lemma 3.2, Thus, in the following part of the proof we can assume that If I (k x (x − y)) ≥ a s then I (k x x) > a s and, applying the fact that s is strictly increasing on (a s , ∞), we get If a s = 0, the above inequality completes the sufficiency part of the proof. So, let 0 < a s ≤ 1 and assume that I (k x (x − y)) < a s . If x, y ∈ E then we can find k 0 > k x such that I (k 0 (x − y)) < a s . Therefore whence E ,s is strictly monotone and the statement (b) is proved. If x, y ∈ L then, by (i)(a), (i)(b) and Lemma 3.8, we can also find k 0 > k x such that I (k 0 (x − y)) < a s . Thus (7) holds true and the proof is completed.
-lower locally uniformly monotone (LLUM) if for every ε ∈ (0, 1] and x ∈ S(X + ) there exists δ(x, ε) > 0 such that for every 0 ≤ y ≤ x, -upper locally uniformly monotone (ULUM) if for every ε > 0 and x ∈ S(X + ) there exists δ(x, ε) > 0 such that It is known that a Banach lattice X is LLUM (resp. ULUM) if and only if for every x ∈ S(X + ) and every sequence (x n ) with 0 ≤ x n ≤ x (resp. 0 ≤ x ≤ x n ) the implication holds true. Evidently, each of the properties UM, LLUM and ULUM of X implies strict monotonicity (SM) of X .
We will say that an Orlicz function satisfies the Kamińska condition for a constant for all u > 0 and each atom e ∈ T a (this condition, for d = 1, was introduced by Kamińska in the paper [17] and it was denoted by (*) there).
An important consequence of Kamińska condition is a significant simplification of 2condition in the purely atomic measure case.
for each n ∈ N and all u > 0 with (u)μ(e n ) < d − ε.
Proof By 2 -condition, there exist constants a > 0, K 0 > 1 and a sequence (c n ) ∈ 1 such that (2u)μ(e n ) ≤ K 0 (u)μ(e n ) + c n for all u > 0 with (u)μ(e n ) ≤ a and n ∈ N . Let ε > 0 and let d > 0, δ > 0 be taken from Kamińska condition. Without loss of generality we can assume that δ < 1. If d − ε ≤ a then (10) is evident. So, let d − ε > a. Then, for all u > 0 with a < (u)μ(e n ) < d − ε and each n ∈ N we have whence where K = max K 0 , 1 a . Conversely, if (10) holds then, using the fact that 2 ≤ (1 + δ) i for some i ∈ N and modifying (decreasing) the number a = d − ε > 0 if necessary, we get 2 -condition on T a (for detailed proof see Lemma 4 in [16]).
Before we prove the criteria for the Orlicz space L ,s to be uniformly monotone, we need three more result on s-norms generated by outer functions that are constant nearby 0. (i) For all x ∈ L , x ,s ≥ 1 ⇒ I (x) ≥ a s , (ii) If T a = ∅ then, for any 0 < ε < a s we can find η > 0 such that for all x ∈ L . (iii) If T a = ∅ then (11) holds true if and only if also satisfies Kamińska condition with the constant a s .
Proof (i) Let x ,s ≥ 1 and suppose that I (x) < a s . By Lemma 3.8, we can find k > 1 such that I (kx) < a s , whence (ii) (also proof of the sufficiency part of (iii)). Suppose that we can find ε ∈ (0, a s ) and a sequence (x n ) of elements of L \ {0} such that I (x n ) ≤ a s − ε and x n ,s ≥ (1 − 1 n+2 ) for all n ∈ N . By (i) we infer that x n ,s < 1 for all n ∈ N .
If μ(T \T a ) > 0 then, by 2 -condition, there exist u 0 ≥ 0 and K 1 > 0 such that for all n ∈ N , where A n = {t ∈ T \T a : |x n (t)| < u 0 } and (u 0 )μ(T \T a ) < ∞ (with the convention 0 · ∞ = ∞). Further, if T a = ∅ then, by Kamińska condition with d = a s and Lemma 4.1, < ∞ and there exist δ > 0, K 2 > 0 and a sequence (c n ) ∈ 1 such that ,s − 1 for n ∈ N . Then 0 < α n < 1 and α n → 0, so α n < δ for all n large enough. Applying (i) we obtain for all n large enough, a contradiction. Thus the condition (11) holds true. Proof of the necessity part of (iii). Suppose that (b )μ(e) < a s for an atom e ∈ T a . Denote x = b χ e and ε = 1 2 (a s − I (x)). Then, evidently, I (x) < a s − ε. Moreover, s(I (x)) = s(a s − ε) = 1, 1 k s(I (kx)) ≥ 1 k > 1 for all 0 < k < 1 and 1 k s(I (kx)) = s(∞) = ∞ for all k > 1. Therefore x ,s = 1 and this equality contradicts to the condition (11). Let 0 < ε < a s . Take any u > 0 and n ∈ N such that (u)μ(e n ) ≤ a s − ε. For x = uχ e n we have I (x) ≤ a s − ε, whence, by (11), x ≤ x ,s < 1 − η for some 0 < η < 1. Thus, putting δ = η 1−η , and applying the fact, that (for the Luxemburg norm · ) if z ≤ 1 then I (z) ≤ z ≤ 1, we obtain so Kamińska condition with the constant d = a s holds true. Proof Suppose that there are v > a s , ε ∈ (0, v − a s ) and a sequence (u n ) of numbers from Since the sequence (u n ) is bounded, passing to a subsequence if necessary, we can assume that u n → u 0 ∈ [ε, v] as n → ∞. Letting n → ∞ and applying the continuity of s, we get s(v − u 0 ) = s(v). Since u 0 > 0, the last equality can hold true only if v − u 0 , v ∈ [0, a s ], whence v ≤ a s , a contradiction. Now we will prove the main theorem of this section. Proof Necessity. Since UM implies SM, by Theorem 3.9 we infer that > 0. Moreover, UM implies LLUM and this property implies order continuity of the norm (see [8]), so the s-norm norm · ,s is order continuous. In consequence, the Luxemburg norm · is order continuous as well, whence ∈ 2 (see, e.g., [3,5]). Finally, if a s > 0 and T a = ∅, Kamińska condition with the constant a s follows from Lemma 4.3.
If K (x) = ∅ then x ∈ L 1 . Hence x − y ∈ L 1 as well because 0 ≤ x − y ≤ x. Thus applying Lemma 3.2 and the fact that the Lebesgue space L 1 is UM, we can find 0 < η 1 < 1 such that where c > 0 is the slope of the oblique asymptote of at infinity. Hence, in the following, we can assume that K (x) = ∅.
Let k ≥ 1 be such that k ∈ K (x) and let us assume that I (kx) − I (ky) ≥ a s . Since ∈ 2 and > 0, the modular convergence of a sequence of functions to 0 implies its norm convergence to 0. Hence we can find δ > 0 such that I (y) > δ for all y ,s ≥ ε. Thus I (ky) > δ as well, whence I (kx) > a s . Moreover, δ < I (ky) ≤ I (kx) − a s . By superaddtivity of and by Lemma 4.4, we can find 0 < η 2 < 1 such that That completes the proof in the case when the outer function s is strictly increasing (i.e, a s = 0). Now, assume that a s > 0 and I (kx) − I (ky) < a s . If I (kx) ≤ a s + δ/2 then By Lemma 4.2, we can find η 3 > 0 such that Finally, assume that a s > 0, I (kx) − I (ky) < a s and I (kx) > a s + δ/2. Since x ,s = 1 and k ∈ K (x), we have where η 4 = 1 − 1/s(a s + δ/2) ∈ (0, 1). Thus and all 0 ≤ y ≤ x with y ,s ≥ ε, i.e., the space L ,s is uniformly monotone.
We say that a sequence (x n ) of nonnegative elements of a Banach lattice X is nearly order convergent to x ∈ X + from below, if 0 ≤ x n ≤ x and x n → x . Proof Let (x n ) be a sequence of nonnegative elements of (L ) + , x ∈ (L ) + , 0 ≤ x n ≤ x and x n ,s → x ,s . Without loss of generality we can assume that x ,s = 1.
Suppose that (x n ) does not converge to x in measure μ as n → ∞, i.e., we can find ε, δ > 0 and a strictly increasing sequence (n m ) of natural numbers such that inf m μ(A m ) ≥ 2δ, where Then 0 ≤ εχ A m ≤ x − x n m ∈ L , whence εχ A m ∈ L and μ(A n m ) < ∞ for all m ∈ N .
If K (x) = ∅ then, by Lemma 3.2, x ∈ L 1 . Thus x n ∈ L 1 as well and whence x n m ,s → 1, a contradiction. Now, assume that K (x) = ∅, i.e., x ,s = 1 k x s(I (k x x)) = 1 for some k x ≥ 1. Define η = I (k x εδ), v = I (k x x) and u m = I k x εχ A m for m ∈ N . Since > 0 we have 0 < η < u m ≤ v for all m ∈ N . Since a s = 0, by Lemma 4.4 we can find γ > 0 such that Thus, by superadditivity of , for all m ∈ N . Therefore x n m ,s → 1, a contradiction that ends the proof. Proof Evidently we can assume that E = {0}. Let x ∈ S + (E ) and let (x n ) be a sequence of E such that 0 ≤ x n ≤ x and x n ,s → 1. Let λ > 0 be an arbitrary positive number.
We have 0 ≤ λ(x − x n ) ≤ λx and, by Lemma 4.6, λx n μ −→ λx. Since x ∈ E , we have I (λx) < ∞ so, by the Lebesgue dominated convergence theorem, I (λ(x − x n )) → 0, whence, by arbitrariness of λ, x − x n ,s → 0. Thus the space E ,s is lower locally uniformly monotone. Proof Denote by X the space L in the case when L is ULUM or the space E whenever E = {0} and E is ULUM. Since ULUM property implies SM, by Lemma 3.7 we infer that > 0. Suppose that / ∈ 2 . Then, since X = {0}, we can find a sequence (z n ) of elements of X + such that (z n ) is modular convergent to 0 but it is not norm convergent to 0 with respect to the norm · ,s . Without loss of generality we can assume that μ(T \ n suppz n ) > 0, I (z n ) ≤ 2 −n and z n ,s ≥ η for all n ∈ N and some η > 0. Take x ∈ S + (X ) such that supp(x) ⊂ T \ n suppz n and let ε > 0 be an arbitrary number. We can find k x ≥ 1 such that 1 k x s (I (k x x)) ≤ x ,s + ε. Put y n = k −1 x z n . We have I (k x y n ) = I (z n ) ≤ 2 −n and y n ,s = k −1 x z n ,s ≥ k −1

Corollary 4.8 If the outer function s is strictly increasing then the the following conditions are equivalent
x η > 0. Thus By arbitrariness of ε > 0 we infer that lim sup n→∞ x + y n , p ≤ 1, whence the space X is not ULUM, a contradiction.
The following theorem summarizes the relationship between uniform monotonicity, lower locally uniform monotonicity and upper local uniform monotonicity properties.  (b) If all of the assumptions of (a) are satisfied and, moreover, < ∞ then in each of the conditions (i), (ii) and (iii) we can put the space E ,s instead of L ,s .

The dominated best approximation problem
Recall that if X = (X , · ) is a Banach space and ∅ = A ⊂ X , then for any x ∈ X the number is called the distance of x from A and the sequence (y n ) of elements of A is said to be an x-minimizing sequence whenever lim n→∞ x − y n = d(x, A). It is obvious that for any nonempty set A in X and any x ∈ X the distance d(x, A) is finite and d(x, A) = 0 for any x ∈ A. Further, the function P A (x) = X → 2 X defined by is called the projection from X onto A and, for any x ∈ X , the set P A (x) is called the projection of x onto A.
The best approximation problem deals with the description of the elements of the set P A (x). If P A (x) = ∅ (resp. cardP A (x) = 1) then we say that the best approximation problem is solvable (resp. uniquely solvable) for x ∈ X . Further, the best approximation problem is said to be stable for x ∈ A, if for every x-minimizing sequence (y n ) in A we have d(y n , P A (x)) → 0 as n → ∞. Finally, the best approximation problem is called strongly solvable if it is uniquely solvable and stable.
A set A is called a sublattice of the Banach lattice X , if A ⊂ X and for any x, y ∈ A there exist x ∧ y ∈ A and x ∨ y ∈ A. The best approximation problem restricted to A being a sublattice and x being a boundary (lower or upper) of A is called the dominated best approximation problem. Let us recall theorems that present conditions under which the dominated best approximation problem is solvable (see [9,11,19]).  [11,Theorem 4.3]) Let X be a σ -Dedekind complete Banach lattice and let A be a closed sublattice of X . If X is lower locally uniformly monotone then the dominated best approximation property is uniquely solvable for any (lower or upper) boundary of A. [11,Theorem 4.4]) Let X be a σ -Dedekind complete Banach lattice and let A be a closed sublattice of X . If X is order continuous and upper locally uniformly monotone then the dominated best approximation property is strongly solvable for every (lower or upper) boundary of A. Now, applying Theorem 3.9 and Theorem 4.10 we get the following corollaries on solvability of the dominated best approximation problem in Orlicz spaces equipped with s-norms. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.