SYMMETRIC FINITE REPRESENTABILITY OF ℓ p IN ORLICZ SPACES 1

It is well known that a Banach space need not contain any subspace isomorphic to a space ℓ p (1 6 p < ∞ ) or c 0 (it was shown by Tsirel’son in 1974). At the same time, by the famous Krivine’s theorem, every Banach space X always contains at least one of these spaces locally , i.e., there exist ﬁnite-dimensional subspaces of X of arbitrarily large dimension n which are isomorphic (uniformly) to ℓ np for some 1 6 p < ∞ or c n 0 . In this case one says that ℓ p (resp. c 0 ) is ﬁnitely representable in X . The main purpose of this paper is to give a characterization (with a complete proof) of the set of p such that ℓ p is symmetrically ﬁnitely representable in a separable Orlicz space. Information about the conﬂict of interests: author and reviewers declare no conﬂict of interests.


Introduction
While a Banach space X need not contain any subspace isomorphic to a space ℓ p (1 p < ∞) or c 0 (as was shown by Tsirel'son in [1]), it will always contain at least one of these spaces locally. This means that there exist finite-dimensional subsets of X of arbitrarily large dimension n which are isomorphic (uniformly) to ℓ n p for some 1 p < ∞ or c n 0 . This fact is the content of the famous result proved by Krivine in [2] (see also [3]). To state it we need some definitions.
Suppose X is a Banach space, 1 p ∞, and {z i } ∞ i=1 is a bounded sequence in X. The space ℓ p is said to be block finitely representable in {z i } ∞ i=1 if for every n ∈ N and ε > 0 there exist 0 = m 0 < m 1 < . . . < m n and α i ∈ R such that the vectors u k = ∑ m k i=m k−1 +1 α i z i , k = 1, 2, . . . , n, satisfy the inequality The space ℓ p , 1 p ∞, is said to be finitely representable in X if for every n ∈ N and ε > 0 there exist x 1 , x 2 , . . . , x n ∈ X such that for any a = (a k ) n k=1 ∈ R n (1 + ε) −1 ∥a∥ p n ∑ k=1 a k x k X (1 + ε)∥a∥ p 1 The work was completed as a part of the implementation of the development program of the Scientific and Educational Mathematical Center the Volga Federal District, agreement no. 075-02-2021-1393.
(alternatively, in the case p = ∞, one might say that c 0 is finitely representable in X).
Clearly, if ℓ p is block finitely representable in some sequence {z i } ∞ i=1 ⊂ X, then ℓ p is finitely representable in X. Therefore, the following famous result proved by Krivine in [2] (see also [3] and [4,Theorem 11.3.9]) implies the finite representability of ℓ p for some 1 p ∞ in any Banach space.
be an arbitrary normalized sequence in a Banach space X such that the vectors z i do not form a relatively compact set. Then ℓ p is block finitely representable in Here, we consider both Orlicz sequence and function spaces (see the next section for the definition) and in the separable case we give a characterization of the set of p such that ℓ p is symmetrically finitely representable in such a space. To introduce the notion of symmetric finite representability, we need some more definitions.
A sequence y = (y k ) ∞ k=1 will be called a copy of a sequence x = (x k ) ∞ k=1 if x and y have the same entries, that is, there is a permutation π of the set of positive integers such that y π(k) = x k for all k = 1, 2, . . ..
Given a measurable function x(t) on [0, 1], we set Here and in the sequel, m denotes the Lebesgue measure. Functions x(t) and y(t) are called equimeasurable if n x (τ ) = n y (τ ) for each τ > 0. Let X be a symmetric sequence space (see e.g. [5]), 1 p ∞. We say that ℓ p is symmetrically finitely representable in X if for every n ∈ N and each ε > 0 there exists an element x 0 ∈ X such that for its disjoint copies x k , k = 1, 2, . . . , n, and for every (a k ) n k=1 ∈ R n we have Similar notion will be defined also in the function case. Let X be a symmetric function space on [0, 1] [5]. The space ℓ p is symmetrically finitely representable in X if for every n ∈ N and ε > 0 there exist equimeasurable and disjointly supported on The set of all p ∈ [1, ∞] such that ℓ p is symmetrically finitely representable in X (in both sequence and function cases) we will denote by F(X).
From the definition 2 of the Matuszewska-Orlicz indices α 0 N and β 0 N (resp. α ∞ N and β ∞ N ) of an Orlicz sequence space ℓ N (resp. an Orlicz function space The main purpose of this paper is to give a detailed proof of the opposite embedding for both Orlicz sequence and function spaces. To this end, following the idea mentioned in [6, p. 140-141] we will make use of the proof of Theorem 4.a.9 from [7]. Similar problems for Orlicz function spaces (and more generally symmetric spaces) on (0, ∞) were considered in the paper [8].

1.1.
Orlicz sequence spaces A detailed information related to Orlicz sequence and function spaces see in monographs [9][10][11]. The Orlicz sequence spaces are a natural generalization of the ℓ p -spaces, 1 p ∞, which equipped with the usual norms Let N be an Orlicz function, that is, an increasing convex continuous function on [0, ∞) such that N (0) = = 0 and lim t→∞ N (t) = ∞. The Orlicz sequence space ℓ N consists of all sequences a = (a k ) ∞ k=1 , for which the following (Luxemburg) norm is finite. Without loss of generality, we will assume that N (1) = 1. In particular, if N (t) = t p , we get the Recall that the Matuszewska-Orlicz indices (at zero) α 0 N and β 0 N of an Orlicz function N are defined by It can be easily checked that 1 α 0 It is well known also that an Orlicz sequence space ℓ N is separable if and only if β 0 N < ∞, or equivalently, if the function N satisfies the ∆ 2 -condition at zero, i.e., One can easily check (see also [7, Proposition 4.a.2]) that h N is a separable closed subspace of ℓ N and the canonical unit vectors e n = (e i n ) such that e n n = 1 and e i n = 0 if i ̸ = n, n = 1, 2, . . ., form a symmetric basis of the space h N . Recall that a basis {x n } ∞ n=1 of a Banach space X is said to be symmetric if there exists C > 0 such that for any permutation π of the set of positive integers and all a n ∈ R we have Observe that the definition of an Orlicz sequence space ℓ N is determined (up to equivalence of norms) by the behaviour of the function N near zero. More precisely, the following conditions are equivalent: 1) ℓ N = ℓ M (with equivalence of norms); 2) the canonical vector bases of the spaces h N и h M are equivalent; 3) there are constants C > 0, c > 0 and t 0 > 0 such that for all 0 t t 0 it holds (see e.g. [7, Proposition 4.a.5] or [11,Theorem 3.4]). In particular, if N is a degenerate Orlicz function, i. e., for some t 0 > 0 we have N (t) = 0 if 0 t t 0 , then ℓ N = l ∞ (with equivalence of norms).
Given Orlicz function N , we define the following subsets of the space C[0, 1 2 ]: where 0 < a < 1 and the closure is taken in the norm topology of C[0, 1 2 ]. All these sets are non-void norm compact subsets of the space C[0, 1 2 ] [7, Lemma 4.a.6]. It is well known that they determine to a large extent the structure of disjoint sequences of Orlicz sequence spaces (see [7, § 4.a] and [12]). Moreover, if 1 p < ∞, ]. In the case when an Orlicz function N satisfies the ∆ 2 -condition at zero, the sets

Orlicz function spaces
Let N be an Orlicz function such that N (1) = 1. Denote by L N the Orlicz space on [0, 1] endowed with the Luxemburg norm ∥x∥ LN := inf{u > 0 : In particular, if N (t) = t p , 1 p < ∞, we obtain the space L p = L p [0, 1] with the usual norm. The Matuszewska-Orlicz indices α ∞ N and β ∞ N (at infinity) of an Orlicz function N are defined by the formulae In contrast to the sequence case, the definition of an Orlicz function space L N on [0, 1] is determined (up to equivalence of norms) by the behaviour of the function N (t) for large values of t.
For every Orlicz function N we define the following subsets of the space C[0, 1 2 ]: where the closure is taken in the norm topology of C[0, 1 2 ]. Again all these sets are non-void norm compact subsets of the space C[0, 1 2 ] and they determine largely the structure of disjoint sequences in Orlicz function spaces (see [12,Propositions 3 and 4]). Moreover, if Finally, if an Orlicz function N satisfies the ∆ 2 -condition at infinity, the sets E ∞ N,A , E ∞ N and C ∞ N can be considered as subsets of the space C[0, 1].

Symmetric finite representability of ℓ p in Orlicz sequence spaces
According to the proof of Theorem 4.a.9 in [7] and a comment followed this proof on p. 144, t p ∈ C 0 M (see also § 2.1). Since M satisfies the ∆ 2 -condition at zero, the set C 0 M may be considered as a subset of the space C[0, 1] (see the remark after Lemma 4.a.6 in [7] or again § 2.1). Therefore, since C 0 M := ∩ 0<a<1 C 0 M,a , we conclude that t p ∈ C 0 M,2 −n for each n ∈ N. Note that the mapping where ρ is a nondecreasing right-continuous function. Therefore, for arbitrary λ 2 > λ 1 > 0 and all 0 t 1 we have Thus, mapping (2) may be extended uniquely to a map ω → M ω from the Stone-Cech compactification βI n of I n onto the set E 0 M,2 −n−1 . Since t p ∈ C 0 M,2 −n and the extreme points of C 0 M,2 −n are contained in the compact set E 0 M,2 −n , by the Krein-Milman theorem (see e.g. [13,Theorem 3.28]), there exists a probability measure µ n on the set βI n such that Let us show that for some probability measure ν n on I n we have First, the fact that the set Q n := Q ∩ I n (Q is the set of rationals) is dense in βI n implies that the set . ., then F m are pairwise disjoint and βI n = ∪ ∞ m=1 F m . Define the measure ν n on σ-algebra of Borel subsets U of the interval I n by where µ n is the probability measure from (4). Since then ν n is a probability measure on I n . Moreover, by (4) and (6), for all 0 t 1 and inequality (5) is proved. Next, for any s ∈ (0, 1) and n, j ∈ N we set a j,n := Then, by inequality (5), we have where by [z] we denote the integer part of a real number z. Choosing now k n such that as M (t) M (1) = 1, we get where Since the right derivative ρ of M (see (3)) is a nondecreasing function and 0 < s < 1, from (7) it follows that Furthermore, the estimate combined with the hypothesis that M satisfies the ∆ 2 -condition at zero, shows that Hence, Moreover, one can readily check that the upper Matuszewska-Orlicz index β 0 M is finite (see also § 2.1) and, by its definition, for each q > β M there is a constant c 0 > 0 such that As a result, since ν n is a probability measure, we conclude Let m ∈ N and ε > 0 be arbitrary. Choose and fix s ∈ (0, 1) so that Then, from (8) and (10) it follows Now, taking n ∈ N satisfying the inequality from (8) and (12), we obtain Therefore, for any whence for all c = (c k ) n k=1 ∈ R n , c k 0, Moreover, since F n is a convex function, from the latter inequality it follows that Therefore, by the definition of the norm in an Orlicz sequence space, for every m ∈ N and all c = (c k ) n k=1 ∈ R n we have where e i , i = 1, 2, . . ., are the canonical unit vectors in ℓ Fn . Given m ∈ N and ε > 0, select s and n to satisfy (11) and (13). For any i = 1, 2, . . . , m and j = 1, 2, . . . , k n denote by A i j,n pairwise disjoint subsets of positive integers such that card A i j,n = [a j,n ]. Then, the vectors are copies of an element from l m . Moreover, by formula (9), we have for all c i ∈ R. Combining this with (15), we get (1), which completes the proof.

Theorem 2
Let M be an Orlicz function satisfying ∆ 2 -condition at infinity. Then ℓ p is symmetrically finitely representable in the Orlicz function space As in the sequence case, we need only to prove the embedding , m ∈ N and each ε > 0 there exist equimeasurable and disjointly supported functions u k , k = 1, 2, . . . , m, satisfying for all c = (c k ) m k=1 ∈ R m the inequality: First, t p ∈ C ∞ M ⊂ C[0, 1] and then the same reasoning as in the proof of Theorem 1 shows that and that for every n ∈ N there is a probabilistic measure ν n on [2 n , ∞) such that for all t ∈ [0, 1] For any s > 1 and n, j ∈ N we define a j,n := Then, by the preceding inequality, a j,n M (s j 2 n t) + 2 −n .
Combining this inequality with the estimate 2 −n ∞ ∑ j=1 a j,n M (s j−1 2 n t) < 2 −2n + 2 −n t p < 2 −n+1 , 0 t 1, On the other hand, since M (u) u for all u 1, we have a j,n 2 M (2 n s j−1 2 −n s −j+1 , which implies that Let m ∈ N and ε > 0 be arbitrary. Fix n so that The first of the inequalities (18) M (2 n s j−1 |c i |)a j,n .
Therefore, by (17) and the second inequality in (18), we get Repeating further the arguments from the end of the proof of Theorem 1, we come to (16) and so complete the proof.