СИММЕТРИЧНАЯ ФИНИТНАЯ ПРЕДСТАВИМОСТЬ ℓp В ПРОСТРАНСТВАХ ОРЛИЧА

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Introduction

While a Banach space X need not contain any subspace isomorphic to a space ℓp (1 � p < ∞) or c0 (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)

p

to ℓn

0

for some 1 � p < ∞ or cn. This fact is the content of the famous result proved by Krivine in [2]

(see also [3]). To state it we need some definitions.

i=1

Suppose X is a Banach space, 1 � p � ∞, and {zi}∞

is a bounded sequence in X. The space ℓp is said

i=1

to be block finitely representable in {zi}∞

if for every n ∈ N and ε > 0 there exist 0 = m0 < m1 < . . . < mn

i=mk−1 +1

and αi ∈ R such that the vectors uk = ∑mk

αizi, k = 1, 2, . . . , n, satisfy the inequality

1 n 1

(1 + ε)−1∥a∥p � 1 ∑ ak uk 1

� (1 + ε)∥a∥p

k=1

for arbitrary a = (ak )n

∈ Rn. In what follows,

1 1

1 k=1 1X

∥a∥p :=

( n

∑

|ak|p

k=1

)1/p

if p < ∞, and ∥a∥∞

:= max

k=1,2,...,n

|ak|

The space ℓp, 1 � p � ∞, is said to be finitely representable in X if for every n ∈ N and ε > 0 there exist

k=1

x1, x2, . . . , xn ∈ X such that for any a = (ak )n

∈ Rn

1 n 1

(1 + ε)−1∥a∥p � 1 ∑ ak xk 1

� (1 + ε)∥a∥p

1 1

1 k=1 1X

1The 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.

Astashkin S.V. Symmetric finite representability of ℓp in Orlicz spaces

16 Асташкин С.В. Симметричная финитная представимость ℓp в пространствах Орлича

(alternatively, in the case p = ∞, one might say that c0 is finitely representable in X).

i=1

Clearly, if ℓp is block finitely representable in some sequence {zi}∞

⊂ X, then ℓp 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 ℓp for some 1 � p � ∞ in any Banach space.

Theorem (Krivine)

i=1

Let {zi}∞

be an arbitrary normalized sequence in a Banach space X such that the vectors zi do not

form a relatively compact set. Then ℓp

i=1

is block finitely representable in {zi}∞ for some p ∈ [1, ∞].

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 ℓp is symmetrically finitely representable in such a space. To introduce the notion of symmetric finite representability, we need some more definitions.

k=1

A sequence y = (yk )∞

k=1

will be called a copy of a sequence x = (xk )∞

if x and y have the same entries,

that is, there is a permutation π of the set of positive integers such that yπ(k) = xk for all k = 1, 2, . . ..

Given a measurable function x(t) on [0, 1], we set

nx(τ ) := m({t ∈ [0, α) : |x(t)| > τ }), τ > 0.

Here and in the sequel, m denotes the Lebesgue measure. Functions x(t) and y(t) are called equimeasurable

if nx(τ ) = ny (τ ) for each τ > 0.

Let X be a symmetric sequence space (see e.g. [5]), 1 � p � ∞. We say that ℓp is symmetrically finitely representable in X if for every n ∈ N and each ε > 0 there exists an element x0 ∈ X such that for its

k=1

disjoint copies xk , k = 1, 2, . . . , n, and for every (ak )n

∈ Rn we have

1 n 1

(1 + ε)−1∥a∥p � 1 ∑ ak xk 1

� (1 + ε)∥a∥p

1 1

1 k=1 1X

Similar notion will be defined also in the function case. Let X be a symmetric function space on [0, 1]

[5]. The space ℓp is symmetrically finitely representable in X if for every n ∈ N and ε > 0 there exist

k=1

equimeasurable and disjointly supported on [0, 1] functions ui(t), i = 1, 2, . . . , n, such that for all (ak )n

∈ Rn

n

1

(1 − ε)∥a∥p � ∑

1

1

i=1

1

aiui1

1X

� (1 + ε)∥a∥p

The set of all p ∈ [1, ∞] such that ℓp is symmetrically finitely representable in X (in both sequence and function cases) we will denote by F (X).

N

From the definition2 of the Matuszewska-Orlicz indices α0

N

and β0

N

(resp. α∞

N

and β∞) of an Orlicz

sequence space ℓN (resp. an Orlicz function space LN ) it follows that F (X) ⊂ [α0 , β0 ] (resp. F (X) ⊂

N N

[αN , βN ]). 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. Preliminaries

   1. Orlicz sequence spaces

      A detailed information related to Orlicz sequence and function spaces see in monographs [9–11].

      The Orlicz sequence spaces are a natural generalization of the ℓp-spaces, 1 � p � ∞, which equipped

      with the usual norms

      

       

       (∑∞

       

      p 1/p

      k=1|ak| )

      , 1 � p < ∞

      ∥a∥ℓp :=

      sup

       k=1,2,...

      |ak| , p = ∞. .

      Let N be an Orlicz function, that is, an increasing convex continuous function on [0, ∞) such that N (0) =

      k=1

       

      = 0 and limt→∞ N (t) = ∞. The Orlicz sequence space ℓN consists of all sequences a = (ak )∞

      , for which

      the following (Luxemburg) norm

       

      

      2See the next section.

       

      ∥a∥ℓN := inf

      {

      u > 0 :

       

      ∞

      ∑

       

      k=1

      }

       

      N ( |ak| ) � 1

      u

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      Vestnik of Samara University. Natural Science Series. 2020, vol. 26, no. 4, pp. 15–24 17

       

      is finite. Without loss of generality, we will assume that N (1) = 1. In particular, if N (t) = tp, we get the

      ℓp-space, 1 � p < ∞.

      N

       

      Recall that the Matuszewska-Orlicz indices (at zero) α0

      p

      N

       

      and β0

      of an Orlicz function N are defined by

      p

      α0

       

      N := sup {p : sup

      x,y:s1

      N (x)y

      N (xy)

      N

       

      < ∞}, β0

      := inf {p : inf

      x,y:s1

      N (x)y

      N (xy)

      > 0}.

      N

       

      It can be easily checked that 1 � α0

      N

       

      � β0

      � ∞. It is well known also that an Orlicz sequence space ℓN is

      N

       

      separable if and only if β0

      < ∞, or equivalently, if the function N satisfies the ∆2-condition at zero, i.e.,

      N (2u)

      lim sup

      u→0

       

      

      N (u)

      < ∞.

      k=1

       

      The subset hN of an Orlicz sequence space ℓN consists of all (ak )∞

      ∈ ℓN such that

      ∞

      

      ( a ) ∞

       

      ∑ N | k| < for each u > 0.

      u

      k=1

      One can easily check (see also [7, Proposition 4.a.2]) that hN is a separable closed subspace of ℓN and

      the canonical unit vectors en = (ei ) such that en = 1 and ei

      = 0 if i ̸= n, n = 1, 2, . . ., form a symmetric

      n n n

      basis of the space

      n=1

       

      hN . Recall that a basis {xn}∞

      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 an ∈ R we have

      ∞ ∞

      C−11 ∑

       

      1

       

      1

       

      1 1 1 ∑

      1

       

      ∞

       

      1

       

      1

       

      1 1 ∑ 1

      1

      n=1

      anxn1X � 1

       

      n=1

      anxπ(n)1X � C1

       

      n=1

      anxn1X .

      Observe that the definition of an Orlicz sequence space ℓN is determined (up to equivalence of norms) by the behaviour of the function N near zero. More precisely, the following conditions are equivalent: 1) ℓN = ℓM (with equivalence of norms); 2) the canonical vector bases of the spaces hN и hM are equivalent; 3) there are constants C > 0, c > 0 and t0 > 0 such that for all 0 � t � t0 it holds

      cN (C−1t) � M (t) � c−1N (Ct)

      (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 t0 > 0 we have N (t) = 0 if 0 � t � t0, then ℓN = l∞ (with equivalence of norms).

      

      2

       

      Given Orlicz function N , we define the following subsets of the space C[0, 1 ]:

       

      

      E0 { N (xy) 0 ∩ 0

       

      and

      N,a :=

      : 0 < y < a}, EN :=

      N (y)

       

      0<a<1

      EN,a

      

      N,a := convEN,a, CN := ∩

      CN,a,

      C0 0

      0 0

      0<a<1

      

      

      2

       

      2

       

      where 0 < a < 1 and the closure is taken in the norm topology of C[0, 1 ]. All these sets are non-void norm compact subsets of the space C[0, 1 ] [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 < ∞,

      then tp ∈ C0

      if and only if p ∈ [α0 , β0 ] [7, Theorem 4.a.9].

      N N N

      N satisfies the ∆2-condition at zero, the sets E0

      , E0 , C0

      and

      In the case when an Orlicz function

      C0

      N,a N

      N,a

      N can be considered as subsets of the space C[0, 1] (see the remark after Lemma 4.a.6 in [7]).

       

   2. Orlicz function spaces

      Let N be an Orlicz function such that N (1) = 1. Denote by LN the Orlicz space on [0, 1] endowed with the Luxemburg norm

      ∥x∥LN := inf{u > 0 :

      1

      ∫ N ( |x(t)| )

      u

      0

      dt � 1}.

      In particular, if N (t) = tp, 1 � p < ∞, we obtain the space Lp = Lp[0, 1] with the usual norm.

      N

       

      The Matuszewska-Orlicz indices α∞

      N

       

      and β∞

      (at infinity) of an Orlicz function N are defined by the

      formulae

      α∞

       

      N = sup {p : sup

      

      N (x)yp

      <

       

      ∞

       

      ∞

       

      }, βN = inf

      {

       

      N (x)yp

      p : inf

      > 0}.

      x,y?:1

      N (xy)

      x,y?:1

      N (xy)

      Astashkin S.V. Symmetric finite representability of ℓp in Orlicz spaces

      

      18 Асташкин С.В. Симметричная финитная представимость ℓp в пространствах Орлича

       

      Again 1 � α∞ � β∞ � ∞. As in the case of sequence spaces, an Orlicz space LN is separable if and only

      N N

      β∞

      if N < ∞, or equivalently, if the function N satisfies the ∆2-condition at infinity, i.e.,

      N (2u)

      lim sup

      u→∞

       

      

      N (u)

      < ∞.

      In contrast to the sequence case, the definition of an Orlicz function space LN on [0, 1] is determined (up to equivalence of norms) by the behaviour of the function N (t) for large values of t.

      

      2

       

      For every Orlicz function N we define the following subsets of the space C[0, 1 ]:

       

      

      E∞

       

      N,A :=

      { N (xy) : y > A}, E∞ =

      N (y)

      ∩ E∞

      

      , C∞ := convE∞,

      N

      A>0

      N,A N N

      

      

      2

       

      2

       

      where the closure is taken in the norm topology of C[0, 1 ]. Again all these sets are non-void norm compact subsets of the space C[0, 1 ] and they determine largely the structure of disjoint sequences in Orlicz function

      spaces (see [12, Propositions 3 and 4]). Moreover, if 1 � p < ∞, then tp ∈ C∞

      if and only if p ∈ [α∞, β∞]

      [12].

      N N N

      N,A

       

      Finally, if an Orlicz function N satisfies the ∆2-condition at infinity, the sets E∞

      N

       

      , E∞

      N

       

      and C∞

      can

      be considered as subsets of the space C[0, 1].

       

2. Symmetric finite representability of ℓp in Orlicz sequence spaces

   Theorem 1

    

   Let M be an Orlicz function satisfying the ∆2-condition at zero. Then ℓp is symmetrically finitely representable in the Orlicz sequence space ℓM if and only if p ∈ [α0 , β0 ], i.e., F (ℓM ) = [α0 , β0 ].

   Proof.

   M M M M

   As was observed in § 1, we always have F (ℓM ) ⊂ [α0 , β0 ]. Therefore, it suffices to prove only the opposite

   M M

   M

    

   embedding. In other words, we need to show that for every p ∈ [α0 ,

   M

    

   β0 ], m ∈ N and each ε > 0 there exists

   k=1

    

   an element x0 ∈ ℓM such that for its disjoint copies xk , k = 1, 2, . . . , m, and for every c = (ck )m

   ∈ Rn we

   have

   1 m 1

   (1 + ε)−1∥c∥p � 1 ∑ ck xk 1

   � (1 + ε)∥c∥p. (1)

   1

   1 k=1

   1

   1ℓM

   M

    

   According to the proof of Theorem 4.a.9 in [7] and a comment followed this proof on p. 144, tp ∈ C0

   (see also § 2.1). Since

   M

    

   M satisfies the ∆2-condition at zero, the set C0

   may be considered as a subset of the

   M

    

   space C[0, 1] (see the remark after Lemma 4.a.6 in [7] or again § 2.1). Therefore, since C0

   := ∩

   0<a<1

   C

    

   0

    

   M,a,

   M,2−n

    

   we conclude that tp ∈ C0

   Note that the mapping

   for each n ∈ N.

   λ 1→ Mλ(t) := M (λt)/M (λ) (2)

   M,2−n

    

   is continuous from In := (0, 2−n] into the subset E0

   of C[0, 1]. Indeed, as it is well known (see e.g. [9,

   Theorem 1.1]),

   M (t) =

   ∫ t

   ρ(s) ds, (3)

   0

   where ρ is a nondecreasing right-continuous function.

   Therefore, for arbitrary λ2 > λ1 > 0 and all 0 � t � 1 we have

   M (λ1)M (λ2t) − M (λ2)M (λ1t)|

   |Mλ2 (t) − Mλ1 (t)| = |

   1

   M (λ1)M (λ2)

   (M (λ2t) − M (λ1t) + M (λ2) − M (λ1))

   

   � M (λ2)

   1

   

   � M (λ2)

   ( ∫ λ2 t λ1 t

    

   ρ(s) ds +

   ∫ λ2 λ1

    

   ρ(s) ds)

   

   � 2ρ(λ2) (λ

   M (λ2) 2

   − λ1).

   Thus, mapping (2) may be extended uniquely to a map ω 1→ Mω from the Stone-C˘ ech compactification βIn

   M,2−n−1

    

   of In onto the set E0

   M,2−n

    

   . Since tp ∈ C0

   M,2−n

    

   and the extreme points of C0

   are contained in the

   Вестник Самарского университета. Естественнонаучная серия. 2020. Том 26, № 4. С. 15–24

   Vestnik of Samara University. Natural Science Series. 2020, vol. 26, no. 4, pp. 15–24 19

    

   M,2−n

    

   compact set E0

   , by the Krein-Milman theorem (see e.g. [13, Theorem 3.28]), there exists a probability

   measure µn on the set βIn such that

    

   tp =

    

   Let us show that

    

   ∫

    

   βIn

    

   Mω (t) dµn(ω), 0 � t � 1. (4)

   for some probability measure νn on In we have

   2−n

   ∫

   p

    

   n

    

    

    

    

   t −

   0

   Mλ(t) dνn(λ) < 2−

   , 0 � t � 1. (5)

   First, the fact that the set Qn := Q ∩ In (Q is the set of rationals) is dense in βIn implies that the set

   k=1

    

   {Mr , r ∈ Qn} is dense in the subset {Mω , ω ∈ βIn} of C[0, 1]. Consequently, putting Qn = {rk}∞

   and

   n

    

   Ek := {ω ∈ βIn : |Mω (t) − Mrk (t)| < 2−

   for all 0 � t � 1}, (6)

   k=1

    

   we have βIn = ∪∞

   k=1

    

   Ek . Now, if Fm := Em \ (∪m−1 Ek ), m = 1, 2, . . ., then Fm are pairwise disjoint and

   ∞

    

   βIn = ∪m=1Fm. Define the measure νn on σ-algebra of Borel subsets U of the interval In by

   νn(U ) := ∑

   {k: rk ∈U}

   where µn is the probability measure from (4). Since

   µn(Fk ),

   ∞

   νn(In) = ∑ µn(Fk ) = µn(βIn) = 1,

   k=1

   then νn is a probability measure on In. Moreover, by (4) and (6), for all 0 � t � 1

    

   p

    

    

   t −

   ∫ 2−n

   0

    

   Mλ(t) dνn(λ) =

    

   ∫

    

   βIn

   Mω (t) dµn(ω) −

   n

    

   ∫ 2−

    

   0

    

   Mλ(t) dνn(λ)

    

   ∞

    

   � ∑ ∫ M ∫

    

   k=1 Fk

   ω (t) dµn(ω) −

   {rk }

   Mλ(t) dνn(λ)

    

   ∞

    

   � ∑ ∫ n ∫

    

   k=1

   {rk }

   ∞

   (Mλ(t) + 2−

   ) dνn(λ) −

   {rk }

   Mλ(t) dνn(λ)

    

    

   and inequality (5) is proved.

   � 2−n ∑ νn({rk}) = 2−nνn(In) = 2−n−1,

   k=1

   Next, for any s ∈ (0, 1) and n, j ∈ N we set

    

   aj,n :=

    

   ∫ sj−1 2−n

    

   dνn(λ)

    

   . (7)

    

   Then, by inequality (5), we have

   ∞

   sj 2−n

    

   ∞

   M (λ)

   ∑[aj,n]M (sj 2−nt) − 2−n < tp < ∑[aj,n]M (sj−12−nt) + M (t)2−n/(1 − s) + 2−n,

   j=1

   j=1

   where by [z] we denote the integer part of a real number z. Choosing now kn such that

    

   as M (t) � M (1) = 1, we get

   ∞

   ∑

    

   j=kn +1

   [aj,n]M (sj−12−n) < 2−n,

    

   where

   Fn(st) − 2−n+1 < tp < Fn(t) + 2−n/(1 − s) + 2−n+1, 0 � t � 1, (8)

    

   kn

   Fn(t) := ∑[aj,n]M (sj−12−nt). (9)

   j=1

   Astashkin S.V. Symmetric finite representability of ℓp in Orlicz spaces

   

   20 Асташкин С.В. Симметричная финитная представимость ℓp в пространствах Орлича

    

   Since the right derivative ρ of M (see (3)) is a nondecreasing function and 0 < s < 1, from (7) it follows that

   Fn(t) − Fn(st) �

   kn

   ∑ aj,n(M (sj−12−nt) − M (sj 2−nt))

   j=1

   k

   

   n

    

   � ∑ 2−

   nsj−1(1 − s)ρ(sj−12−

   n) ∫ s

   j−1 2−n

    

   dνn

    

   (λ).

    

   Furthermore, the estimate

   j=1

    

   ∫ 2x

   M (sj 2−n)

   sj 2−n

   F (2x) ;;:

   x

   ρ(s) ds ;;: xρ(x), 0 � x � 1,

   combined with the hypothesis that M satisfies the ∆2-condition at zero, shows that

   xρ(x)

    

   Hence,

   

   K := sup

   0<x:s1 M (x)

   < ∞.

   j−1 −n

    

   kn M (sj−12−n) ∫ s 2

    

   n − n −

    

   F (t) F (st) � K(1 s) ∑

   M (sj 2−n)

    

   sj 2−n

   dνn(λ).

   j=1

   M

    

   Moreover, one can readily check that the upper Matuszewska-Orlicz index β0

    

   is finite (see also § 2.1) and,

   by its definition, for each q > βM there is a constant c0 > 0 such that

   M (sj 2−n) ;;: c0M (sj−12−n)sq .

   As a result, since νn is a probability measure, we conclude

   kn ∫ sj−1 2−n

   0

    

   Fn(t) − Fn(st) � K(1 − s)s−q c−1 ∑

   j=1

    

   sj 2−n

   0

    

   dνn(λ) � K(1 − s)s−q c−1. (10)

   Let m ∈ N and ε > 0 be arbitrary. Choose and fix s ∈ (0, 1) so that

   0

    

   K(1 − s)s−q c−1 < ε/(2m). (11)

   Then, from (8) and (10) it follows

   ε

    

   n+1

    

   −n+1 p

   

   Fn(t) − 2m − 2−

   Now, taking n ∈ N satisfying the inequality

   < Fn(st) − 2

    

   2−n

   < t , 0 � t � 1. (12)

    

   from (8) and (12), we obtain

   1 − s ε p

   

   

   + 2−n+1 < ε

   2m

    

   ε

   , (13)

   Fn(t) − m < t

   Therefore, for any ci ∈ [0, 1], i = 1, 2, . . . , m,

   < Fn(t) + m, 0 � t � 1. (14)

    

   k=1

    

   whence for all c = (ck )n

   m

   ∑ cp

    

   i − ε <

   i=1

   ∈ Rn, ck ;;: 0,

   m

   ∑

    

   i=1

    

   Fn(ci) <

   m

   ∑

    

   i=1

   i

    

   cp + ε,

   m

   1 − ε < ∑ Fn

   ( ci

   )

   

   < 1 + ε.

   i=1

   ∥c∥p

   Moreover, since Fn is a convex function, from the latter inequality it follows that

   m

   ∑ Fn

   ( ci )

   

   � 1

    

   and

   i=1

    

   m

   ∑ Fn

   i=1

   (1 + ε)∥c∥p

   ( ci )

   

   (1 − ε)∥c∥p

    

   > 1.

   Вестник Самарского университета. Естественнонаучная серия. 2020. Том 26, № 4. С. 15–24

   Vestnik of Samara University. Natural Science Series. 2020, vol. 26, no. 4, pp. 15–24 21

    

   k=1

    

   Therefore, by the definition of the norm in an Orlicz sequence space, for every m ∈ N and all c = (ck )n

   ∈ Rn

   we have

   m

   1

   (1 − ε)∥c∥p � ∑

    

   1

   1

   i=1

   1

    

   ciei1

   1ℓFn

   � (1 + ε)∥c∥p, (15)

   where ei, i = 1, 2, . . ., are the canonical unit vectors in ℓFn .

   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, . . . , kn

   j,n

    

   denote by Ai

   j,n

    

   pairwise disjoint subsets of positive integers such that card Ai

   kn

   = [aj,n]. Then, the vectors

   ui := 2−n ∑ sj−1 ∑

   ek , i = 1, 2, . . . , m,

   j=1

   k∈A

    

   i j,n

   are copies of an element from lm. Moreover, by formula (9), we have

   m

   1

   ∑

    

   1

   1

   i=1

   1

   ciui1

   1ℓM

   m

   1

   ∑

    

   = 1

   1

   i=1

   1

   ciei1

   1ℓFn

   for all ci ∈ R. Combining this with (15), we get (1), which completes the proof.

    

3. Symmetric finite representability of ℓp in Orlicz function spaces

Theorem 2

Let M be an Orlicz function satisfying ∆2-condition at infinity. Then ℓp is symmetrically finitely representable in the Orlicz function space LM if and only if p ∈ [α∞ , β∞], i.e., F (LM ) = [α∞ , β∞].

Proof.

M M M M

As in the sequence case, we need only to prove the embedding [α∞ , β∞] ⊂ F (LM ). More precisely, we

M M

M

M

have to check that for every p ∈ [α∞ , β∞

], m ∈ N and each ε > 0 there exist equimeasurable and disjointly

supported functions

k=1

uk , k = 1, 2, . . . , m, satisfying for all c = (ck )m

∈ Rm the inequality:

1 m 1

(1 + ε)−1∥c∥p � 1 ∑ ck uk 1

� (1 + ε)∥c∥p (16)

1

1 k=1

1

1LM

M

First, tp ∈ C∞ ⊂ 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 νn on [2n, ∞) such that for all t ∈ [0, 1]

p

t −

∫ ∞ M (λt)

dνn(λ) < 2−n.

For any s > 1 and n, j ∈ N we define

2n M (λ)

j n

∫ s 2

dµn(λ)

Then, by the preceding inequality,

∞

aj,n :=

sj−1 2n

.

M (λ)

∞

∑ aj,nM (sj−12nt) − 2−n < tp < ∑ aj,nM (sj 2nt) + 2−n.

j=1

Next, as M satisfies the ∆2-condition at infinity, we have

j=1

M (sj 2nt) � (1 + 2−n)M (sj−12nt)

for all j ∈ N and t ∈ [0, 1] whenever s is sufficiently close to 1. Fixing such a s, we get

∞ ∞

∑ aj,nM (sj−12nt) − 2−n < tp < ∑(1 + 2−n)aj,nM (sj−12nt) + 2−n.

j=1

Combining this inequality with the estimate

∞

j=1

we deduce

2−n ∑ aj,nM (sj−12nt) < 2−2n + 2−ntp < 2−n+1, 0 � t � 1,

j=1

∞ ∞

∑ aj,nM (sj−12nt) − 2−n < tp < ∑ aj,nM (sj−12nt) + 2−n+2. (17)

j=1

j=1

Astashkin S.V. Symmetric finite representability of ℓp in Orlicz spaces

22 Асташкин С.В. Симметричная финитная представимость ℓp в пространствах Орлича

On the other hand, since M (u) ;;: u for all u ;;: 1, we have

2 −n

−j+1

which implies that

aj,n � M (2nsj−1 � 2 s ,

∞ ∞ s

∑ aj,n � 2−n ∑ s−j+1 = 2−n+1 ·

.

s − 1

j=1

j=1

Let m ∈ N and ε > 0 be arbitrary. Fix n so that

2−n+1s 1

<

s − 1 m

and 2−n+2m < ε. (18)

j

The first of the inequalities (18) allows us to take pairwise disjoint sets Ei ⊂ [0, 1], j ∈ N, i = 1, 2, . . . , m,

j

with m(Ei ) = aj,n. Then, the functions

∞

E

ui := ∑ 2nsj−1χ i

j

j=1

are equimeasurable and disjointly supported on [0, 1]. Moreover, for all ci ∈ R

∫ 1 ( m

) m ∞

M ∑ ciui(t)

dt = ∑ ∑ M (2nsj−1|ci|)aj,n.

0

i=1

i=1 j=1

Therefore, by (17) and the second inequality in (18), we get

|ci|p + ε.

0

Repeating further the arguments from the end of the proof of Theorem 1, we come to (16) and so complete the proof.

Об авторах

С. В. Асташкин

Автор, ответственный за переписку.
Email: astash56@mail.ru
ORCID iD: 0000-0002-8239-5661

доктор физико-математических наук, профессор, заведующий кафедрой функционального анализа и теории функций

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