Isometric copies of $l^\infty$ in Ces\`aro-Orlicz function spaces

0.2 MB·

by  Tomasz Kiwerski

Isometric copies of l∞ in Ces`aro-Orlicz
function spaces
Tomasz Kiwerski
Faculty of Mathematics, Computer Science and Econometrics
University of Zielona Go´ra,
prof. Z. Szafrana 4a, 65-516 Zielona Go´ra, Poland
e-mail address: [email protected]
7
1
Pawe l Kolwicz
0
2 Institute of Mathematics,
Faculty of Electrical Engineering,
n
a Poznan´ University of Technology,
J Piotrowo 3A, 60-965 Poznan´, Poland

0 e-mail address: [email protected]
3

]
A ABSTRACT. We characterize Cesa`ro-Orlicz function spaces Ces containing order isomorphi-
ϕ
F cally isometric copy of l∞. We discuss also some useful applicable conditions sufficient for the
.
h existence of such a copy.
t
a
m
2010 Mathematics Subject Classification. 46A80, 46B04, 46B20, 46B42, 46E30.
[
Key words and phrases. Cesa`ro-Orlicz function space; isometric copy of l∞; order continuous

1
norm.
v
4
9
7
8
0
1. INTRODUCTION
.
1
0 The structure of different types of Cesa`ro spaces has been widely investigated during the last
7
decades from the isomorphic as well as isometric point of view. The spaces generated by the
1
Cesa`ro operator (including abstract Cesa`ro spaces) have been considered by Curbera, Ricker
:
v
and Le´snik, Maligranda in several papers (see [12], [13], [14], [15], [26], [27], [28]). The classical
i
X Cesa`ro sequence ces and function Ces spaces have been studied by many authors (see [2],
p p
r [3] - also for further references, [1], [4], [10], [11]). It has been proved among others that some
a
properties are fulfilled in the sequence case and are not in function case. Moreover, sometimes
the cases Ces [0,1] andCes [0,∞) are essentially different (see an isomorphicdescription of the
p p
Ko¨the dual of Cesa`ro spaces in [2] and [26]).
The Cesa`ro-Orlicz sequence spaces denoted by ces are generalization of the Cesa`ro sequence
ϕ
spaces ces . Of course, the structure of the spaces ces is richer than of the space ces . The
p ϕ p
spaces ces have beenstudiedintensively (see[9], [19], [22]and[34]). Wewant toinvestigate the
ϕ
Cesa`ro-Orlicz functionspace Ces (I). Thespaces Ces whichcontain anorderisomorphiccopy
ϕ ϕ
of l∞ (equivalently, are not order continuous) have been characterized in [21]. The monotonicity
properties have been also considered in [21]. In this paper we want to describe Cesa`ro-Orlicz
function spaces Ces containing order isomorphically isometric copy of l∞. We discuss also
ϕ
someusefulapplicableconditions sufficientfortheexistence ofsuchacopy.Weadmitthelargest
possible class of Orlicz functions giving the maximal generality of spaces under consideration.
1
2. PRELIMINARIES
Let R, R and N be the sets of real, nonnegative real and natural numbers, respectively.
+
Denote by µ the Lebesgue measureon I and by L0 = L0(I) the space of all classes of real-valued
Lebesgue measurable functions defined on I, where I = [0,1] or I = [0,∞).
A Banach space E = (E,k·k) is said to be a Banach ideal space on I if E is a linear subspace
of L0(I) and satisfies two conditions:
(i) if g ∈ E, f ∈ L0 and |f|≤ |g| a.e. on I then f ∈ E and kfk ≤ kgk,
(ii) there is an element f ∈E that is positive on whole I.
Sometimes we write k·k to be sure in which space the norm has been taken.
E
For two Banach ideal spaces E and F on I the symbol E ֒→ F means that the embedding
E ⊂ F is continuous, i.e., there exists constant a C > 0 such that kxk ≤ Ckxk for all x ∈ E.
F E
Moreover, E = F means that the spaces are the same as the sets and the norms are equivalent.
A Banach ideal space is called order continuous (E ∈ (OC) shortly) if every element of E is
order continuous, that is, for each f ∈ E and for each sequence (f ) ⊂ E satisfying 0≤ f ≤ |f|
n n
and f → 0 a.e. on I, we have kf k → 0. By E we denote the subspace of all order continuous
n n a
elements of E. It is worth to notice that in case of Banach ideal spaces on I, x ∈ E if and only
a
if xχ → 0 for any decreasing sequence of Lebesgue measurable sets A ⊂ I with empty
An n
intersection (see [5, Proposition 3.5, p. 15]).
(cid:13) (cid:13)
(cid:13)A func(cid:13)tion ϕ : [0,∞) → [0,∞] is called an Orlicz function if:
(i) ϕ is convex,
(ii) ϕ(0) = 0,
(iii) ϕ is neither identically equal to zero nor infinity on (0,∞),
(iv) ϕ is left continuous on (0,∞), i.e., limu→b−ϕ ϕ(u) = ϕ(bϕ) if bϕ < ∞, where
b = sup{u > 0 :ϕ(u) < ∞}.
ϕ
For more information about Orlicz functions see [8] and [24].
If we denote
a = sup{u ≥ 0 :ϕ(u) = 0},
ϕ
then 0 ≤ a ≤ b ≤ ∞. Moreover, a < ∞ and b > 0, since an Orlicz function is neither
ϕ ϕ ϕ ϕ
identically equal to zero nor infinity on (0,∞). The function ϕ is continuous and non-decreasing
on [0,b ) and is strictly increasing on [a ,b ). We use notations ϕ > 0, ϕ < ∞ when a = 0,
ϕ ϕ ϕ ϕ
b = ∞, respectively.
ϕ
We say an Orlicz function ϕ satisfies the condition ∆ for large arguments (ϕ ∈ ∆ (∞) for
2 2
short) if there exists K > 0 and u > 0 such that ϕ(u )< ∞ and
0 0
ϕ(2u) ≤ Kϕ(u)
for all u ∈ [u ,∞). Similarly, we can define the condition ∆ for small, with ϕ(u ) > 0
0 2 0
(ϕ ∈ ∆ (0)) or for all arguments (ϕ ∈ ∆ (R )). These conditions play a crucial role in the
2 2 +
theoryof Orliczspaces, see[8], [24], [33]and[36]. Wewillwriteϕ∈ ∆ intwocases: ϕ ∈ ∆ (∞)
2 2
if I = [0,1] and ϕ ∈∆ (R ) if I = [0,∞).
2 +
The Orlicz function space Lϕ = Lϕ(I) generated by an Orlicz function ϕ is defined by
Lϕ ={f ∈ L0(I) :I (f/λ) < ∞ for some λ = λ(f) > 0},
ϕ
where I (f) = ϕ(|f(t)|)dt is a convex modular (for the theory of Orlicz spaces and modular
ϕ I
spaces see [33] and [36]). The space Lϕ is a Banach ideal space with the Luxemburg-Nakano
R
norm
kfk = inf{λ > 0 : I (f/λ)≤ 1}.
ϕ ϕ
It is well known that kfk ≤ 1 if and only if I (f)≤ 1. Moreover, the set
ϕ ϕ
(2.1) KLϕ = KLϕ(I) = {f ∈ L0(I) : I (f)< ∞},
ϕ
2
will be called the Orlicz class.
The Cesa`ro operator C :L0(I) → L0(I) is defined by
1 x
Cf(x)= f(t)dt,
x
Z0
for0 < x ∈ I. ForaBanach idealspaceX onI wedefineanabstractCesa`rospaceCX = CX(I)
by
CX = {f ∈ L0(I) : C|f|∈ X}
with the norm kfk = kC|f|k (see [26], [27], [28]).
CX X
The Cesa`ro-Orlicz function space Ces = Ces (I) is defined by Ces (I) = CLϕ(I). Conse-
ϕ ϕ ϕ
quently, the norm in the space Ces is given by the formula
ϕ
kfk = inf{λ >0 : ρ (f/λ) ≤ 1},
Ces(ϕ) ϕ
where ρ (f) = I (C|f|) is a convex modular. We always assume that Ces 6= {0}. If I = [0,∞)
ϕ ϕ ϕ
then Ces [0,∞) 6= {0} if and only if the function x → 1χ (x),x > 0 belongs to Lϕ[0,∞)
ϕ x [a,∞)
for some a > 0 (see Theorem 1 (a) in [26] and Proposition 3 in [21]). However, Ces [0,1] 6= {0}
ϕ
for any Orlicz function ϕ. Indeed, Lϕ[0,1] is symmetric and Ces [0,1] 6= {0} if and only if
ϕ
χ ∈ Lϕ[0,1] for some 0 < a < 1 (see Theorem 1 (b) in [26]).
[a,1]
In this paper we accept the convention that k x = 0 if k < m.
n=m n
Note that if 0 < a = b then Lϕ = L∞ and kxk = 1 kxk , see Example 1 in [33, p.
ϕ ϕ P ϕ bϕ ∞
98]. Consequently, Ces = Ces in that case (see [2], [26]). Therefore we can assume that if
ϕ ∞
b < ∞, then a < b .
ϕ ϕ ϕ
3. Isometric copies of l∞ in Ces
ϕ
Define a set
C = C (I) = {x ∈ Ces (I) : ρ (kx) < ∞ for all k > 0}.
ϕ ϕ ϕ ϕ
Note that C = {0} whenever b < ∞. If b = ∞, then (Ces ) = C by Theorem 5 from [21]
ϕ ϕ ϕ ϕ a ϕ
(recall that (Ces ) is the subspace of all order continuous elements in Ces ).
ϕ a ϕ
We say that a measurable set Ω is a support of a Banach ideal space E, we write suppE = Ω
whenever
(i) for each x ∈ E there is a measurable set A with µ(A) = 0 and suppx ⊂ A∪Ω.
(ii) there is x ∈ E such that µ(Ω\suppx)= 0.
Lemma 1. Suppose b < ∞. Then (Lϕ) = {0} and supp(Ces ) = I. Moreover, if 0 ≤ x ∈
ϕ a ϕ a
(Ces ) , then lim Cx(t)= 0.
ϕ a t→0+
Proof. The equality (Lϕ) = {0} is well known, it is enough to consider the element x = αχ
a A
for α > 0 and 0 < µ(A) < ∞. Taking a sequence (A ) of measurable subsets of A with
n
µ(A ) → 0 we conclude that xχ 9 0 because I λxχ = ∞ for λ > b /α and
n An ϕ ϕ An ϕ
each n. We will prove that supp(Ces ) = I. Let x = χ for any 0 < a < b < m(I).
(cid:13) (cid:13)ϕ a (cid:0)(a,b) (cid:1)
Take a decreasing sequence of Le(cid:13)besgue(cid:13)measurable sets A ⊂ I with empty intersection. Then
n
µ(A ∩(a,b)) → 0.SetB = A ∩(a,b)andx = xχ ThenCx → 0uniformly. Consequently,
n n n n Bn n
ρ (λx )= I (λCx ) → 0 for each λ > 0, which for I = [0,1] follows directly and for I = [0,∞)
ϕ n ϕ n
we need Proposition 3 from [21]. Consequently, kx k → 0. Thus x ∈ (Ces ) which gives
n Ces(ϕ) ϕ a
supp(Ces ) = I.
ϕ a
Finally, suppose limsup Cx(t) = δ > 0. Then there is a number γ > 0 and a sequence
t→0+
(t )∞ such that t → 0+ and Cx(t )≥ δ/2 for each k. Since the function Cx is continuous on
k k=1 k k
the interval (0,m(I)), for each k there is a measurable set B of positive measure such that
k
Cx(t)≥ δ/4
3
for each t ∈ B . Let A = 0, 1 for n ∈ N. Consequently, for each n we find a number k with
k n n n
µ(B ∩A ) > 0. Thus
kn n (cid:0) (cid:1)
ρ λxχ = I λC xχ ≥ I λ(Cx)χ ≥ I λ(Cx)χ = ∞
ϕ An ϕ An ϕ An ϕ Bkn∩An
(cid:16) (cid:17)
for each λ >(cid:0) 4b /δ,(cid:1)whenc(cid:0)e x(cid:0)χ (cid:1)(cid:1) 9(cid:0)0. Thus x ∈/(cid:1)(Ces ) . (cid:3)
ϕ An Ces(ϕ) ϕ a
Remark 2. Let I = [0,1] or(cid:13)I = [0(cid:13),∞).
(cid:13) (cid:13)
(i) Suppose there is a sequence (z )∞ in Ces (I) of pairwise disjoint supports satisfying
n n=1 ϕ
conditions:
(a) kz k = 1, for each n.
n Ces(ϕ)
(b) ksup z k = 1.
n n Ces(ϕ)
Then Ces (I) contains an order isomorphically isometric copy of l∞.
ϕ
(ii) Assume that ϕ< ∞ and there is an element z ∈ Ces (I) such that kzk = 1 and
ϕ Ces(ϕ)
z
δ(z) := inf λ > 0 :ρ < ∞ = 1.
ϕ λ
n (cid:16) (cid:17) o
Then Ces (I) contains an order isomorphically isometric copy of l∞.
ϕ
(iii) Let b <∞. Suppose there is an element z ∈ Ces (I) such that kzk = 1 and
ϕ ϕ Ces(ϕ)
z−u
(3.1) ρ = ∞
ϕ λ
(cid:18) (cid:19)
for all λ < 1 and u ∈ (Ces ) . Then Ces (I) contains an order isomorphically isometric copy
ϕ a ϕ
of l∞.
Proof. (i) It is enough to apply Theorem 1 in [18]. Note that each Banach ideal space (in
particular Ces ) satisfies the assumption of Theorem 1 in [18].
ϕ
(ii) We apply Theorem 2 from [18]. Note that if ϕ < ∞ then (Ces ) = C (by Theorem 5 from
ϕ a ϕ
[21]). Next, if ϕ < ∞ and Ces 6= {0} then supp(Ces ) = suppCes = I (for I = [0,∞) it
ϕ ϕ a ϕ
is enough to apply Proposition 3 from [21]). Moreover, since Lϕ ∈ (FP), so Ces ∈ (FP) (see
ϕ
Theorem 1 in [26]), in particular Ces is monotone complete. Consequently, the space Ces
ϕ ϕ
satisfies the assumptions of Theorem 2 from [18]. On the other hand, Ces is a modular space
ϕ
generated by a convex modular ρ . By Theorem 2.1 in [17] we get
ϕ
x
dist z,(Ces ) = inf λ > 0 :ρ < ∞ .
ϕ a ϕ λ
n (cid:16) (cid:17) o
Thus the conclusion follows f(cid:0)rom Theor(cid:1)em 2 from [18].
(iii) Note that supp(Ces ) = I by Lemma 1. Similarly as in (ii) above we conclude that the
ϕ a
space Ces satisfies the assumptions of Theorem 2 from [18]. We prove that dist z,(Ces ) =
ϕ ϕ a
1. Clearly, dist z,(Ces ) ≤ 1. We claim that kz−uk ≥ 1 for each u ∈ (Ces ) . Let
ϕ a Ces(ϕ) (cid:0) ϕ a (cid:1)
u ∈(Ces ) . The set λ > 0 :ρ z−u ≤ 1 is nonempty. By the assumption (3.1) we have
ϕ a (cid:0) (cid:1) ϕ λ
(cid:8) (cid:0) (cid:1) (cid:9) z−u
kz−uk = inf λ > 0 :ρ ≤ 1 ≥ 1,
Ces(ϕ) ϕ λ
(cid:26) (cid:18) (cid:19) (cid:27)
which proves the claim. Thus dist z,(Ces ) = 1 and, by Theorem 2 from [18], Ces (I)
ϕ a ϕ
contains an order isomorphically isometric copy of l∞. (cid:3)
(cid:0) (cid:1)
The following characterization of order continuity is well known.
Theorem A. (G. Ya. Lozanovski˘ı, see [31]) A Banach ideal space E is order continuous if and
only if E contains no isomorphic copy of l∞.
4
Theorem 3. Let ϕ be an Orlicz function.
(i) In the case when ϕ(b ) = ∞ we assume that for each ǫ > 0 there exists a constant D =
ϕ
D(ε) > 0 such that
(3.2) ρ (f) ≤ DI (f)+ǫ,
ϕ ϕ
forall f ∈ KLϕ[0,1] satisfying|f (t)| ≥ ϕ−1(ε)fora.e. t ∈ supp(f)(see (2.1)forthedefinition).
If ϕ ∈/ ∆ (∞) then Ces [0,1] contains an order isomorphically isometric copy of l∞.
2 ϕ
(ii) If ϕ ∈∆ (∞) then Ces [0,1] does not contain an order isomorphic copy of l∞.
2 ϕ
Proof. (i). We divide the proof into two parts.
(A) Suppose bϕ = ∞. Since ϕ ∈/ ∆2(∞), for each i ∈ N there is a sequence (uin)n∈N ⊂ R+ such
that ui ր ∞ as n → ∞,
n
1
ϕ 1+ ui > 2n+iϕ(ui)
i n n
(cid:18)(cid:18) (cid:19) (cid:19)
and ϕ(ui) ≥ 1 for all n ∈ N. Moreover, there exists a sequence of pairwise disjoint intervals Ai
n
such that ∞ Ai = [0,1) and µ(Ai) = 2−i for i ∈ N. Consequently, we can choose a sequence
i=1
of intervals (Ai)∞ with the following properties:
(a) (Ai)⊂SAinfonr=a1ll n ∈ N,
n
(b) Ai ∩Ai = ∅ for all n6= m, n,m ∈ N,
n m
(c) µ(Ai) = 2−n−i for all n ∈ N,
n ϕ(ui)
n
for all i ∈ N. This is possible, since ∞ 2−n−i < 2−i for all i ∈ N. Finally, for i = 1,2,3,...
n=1 ϕ(ui)
n
define the elements
P ∞
x = uiχ ,
i n Ai
n
n=1
X
and x = ∞ x . From the construction it follows that supp(x )∩supp(x ) = ∅ for all i 6= j.
i=1 i i j
Moreover,
P 1 1 ∞
I (x )= ϕ(x (t))dt = ϕ(ui)χ (t)dt
ϕ i i n Ai
n
Z0 Z0 n=1
X
∞ ∞ 2−n−i
= ϕ(ui)µ(Ai) = ϕ(ui) = 2−i,
n n n ϕ(ui)
n=1 n=1 n
X X
and
1 ∞ ∞ 1
I (x) = ϕ(x (t))dt = ϕ(x (t))dt = 1.
ϕ i i
Z0 i=1 i=1Z0
X X
Withoutloss ofgenerality wemayassumethatx isincreasingforeach i ∈ N. Considerelements
i
ui in the following way
n
u1 u1 u1 ··· u1 ···
1 2 3 n
u2 u2 u2 ··· u2 ···
1 2 3 n
(3.3) u3 u3 u3 ··· u3 ···
1 2 3 n
.. .. .. .. .. ..
. . . . . .
Ordertheset{u1,u2,u1}fromthesmallesttothelargestinthesenseoforder≤. Denoteordered
1 1 2
elements by v ,v and v . That is, v ≤ v ≤ v . Since for each i ∈ N, ui → ∞ as n → ∞,
1 2 3 1 2 3 n
so there exist indices i2,i2,i2 ∈ N, i2 > 2 and i2 > 1 such that u1 ,u2 ,u3 ≥ v . Order the
1 2 3 1 2 i2 i2 i2 3
1 2 3
set {u1 ,u2 ,u3 } as before and denote ordered elements by v ,v and v . Clearly, the sequence
i2 i2 i2 4 5 6
1 2 3
(v )6 is non-decreasing. Generally, in the m-th step we find indexes im,im,...,im ∈ N,
n n=1 1 2 m+1
im ≥ im−1, im ≥ im−1,...,im ≥ im−1 such that u1 ,u2 ,...,um+1 ≥ v and, ordering the set
1 1 2 2 m m im1 im2 imm+1 lm
5
{u1 ,u2 ,...,um+1}, we obtain a sequence (v )lm , where l = 3 and l = l +m+1 for
im im im i i=1 1 m m−1
1 2 m+1
m ≥ 2, which is non-decreasing. By the induction, we construct a sequence (v )∞ which is
n n=1
non-decreasing. Corresponding to each value v the respective interval Ai we can denote by A
n n n
for simplicity. Define an element
∞
y = y ,
n
n=1
X
wherey = v χ , B = a− n µ(A ),a− n−1 µ(A ) anda = ∞ µ(A ). We have
n n Bn n m=1 m m=1 m n=1 n
Iϕ(y) ≤ Iϕ(x) ≤ 1.Applyi(cid:16)ngcoPndition(3.2)forǫ =P1/2wefind(cid:17)t ∈(0,1]suPchthatIϕ(z) ≤ 1/2D
and |z(t)| ≥ ϕ−1(1/2) for a.e. t ∈ supp(z), where z = yχ . Therefore, we have
[0,t)
ρ (z) ≤ DI (z)+1/2 = 1.
ϕ ϕ
Moreover, there exists n ∈ N such that ∞ µ(B ) ≤ t. Set λ > 0. There is i with λ ≥ 1/i .
0 n=n0 n 0 0
We have
P
1
ρ ((1+λ)z) = I (C((1+λ)z)) ≥ I ((1+λ)z) ≥ I 1+ z .
ϕ ϕ ϕ ϕ i
(cid:18)(cid:18) 0(cid:19) (cid:19)
ConsiderthemodularI ((1+1/i )z). Bytheconstructionofelementz,wehavechoseninfinitely
ϕ 0
many terms from the i -th row of matrix (3.3). Denote them by ui0 for k = 1,2,.... We have
0 nk
1
ϕ 1+ ui0 > 2nk+i0ϕ(ui0 ),
i nk nk
(cid:18)(cid:18) 0(cid:19) (cid:19)
for each k and µ(Ai0 )= 2−nk−i0/ϕ(ui0 ). Therefore,
nk nk
∞ ∞
1 1
I 1+ z ≥ ϕ 1+ ui0 µ(Ai0 ) ≥ 1 = ∞.
ϕ i i nk nk
(cid:18)(cid:18) 0(cid:19) (cid:19) kX=k0 (cid:18)(cid:18) 0(cid:19) (cid:19) kX=k0
Consequently, kzk = 1and,byRemark2(ii),weconcludethatCes [0,1]contains anorder
Ces(ϕ) ϕ
isomorphically isometric copy of l∞.
(B) Suppose b < ∞. We apply Remark 2(i).
ϕ
(B1) Let ϕ(b ) = ∞. Let (u ) ⊂ R be the sequence with u ր b−. Since ϕ(u ) → ∞,
ϕ n + n ϕ n
therefore we can assume that ϕ(u ) ≥ 1 for all n ∈ N. Put a = 1/2nϕ(u ) for
n n n
n ∈ N and denote a = ∞ a . Define a sequence of pairwise disjoint open intervals
n=1 n
(An)n∈N ⊂ [0,1] as follows
P
n n−1
1 1
A = a− ,a− ,
n 2kϕ(u ) 2kϕ(u )
k k !
k=1 k=1
X X
for all n ∈ N. Let
∞
x = u χ .
n An
n=1
X
We have
∞ ∞ ∞
I (x) = ϕ(x)dµ = ϕ(u )χ dµ
ϕ n An
Z0 Z0 n=1
X
∞ ∞ ∞
1
= ϕ(u )dµ = ϕ(u )µ(A )= = 1.
n n n 2n
n=1ZAn n=1 n=1
X X X
6
Now, like in (A), we can find t ∈ (0,1] with ρ (z) ≤ 1, where z = xχ . Hence,
ϕ [0,t)
relabeling if necessary, we may assume that
∞
z = u χ .
n An
n=1
X
Becauseforallλ > 0thereexistsN =N(λ)∈ Nsuchthatforallk ≥ N,(1+λ)u > b ,
k ϕ
hence
ρ ((1+λ)z) = I (C((1+λ)z)) = I ((1+λ)Cz)
ϕ ϕ ϕ
∞
≥ I ((1+λ)z) = ϕ((1+λ)u )µ(A ) = ∞.
ϕ n n
n=1
X
Therefore kzk = 1. We will show that
Ces(ϕ)
ρ (λ(z−u)) = ∞
ϕ
for all λ > 1 and u ∈ (Ces ) which would imply that Ces [0,1] contains an order
ϕ a ϕ
isomorphically isometric copy of l∞ (see Remark 2(iii)). Let λ > 1 and u ∈ (Ces ) .
ϕ a
We have
ρ (λ(z−u))= I (C(|λ(z−u)|)) ≥I (λC||z|−|u||).
ϕ ϕ ϕ
By Lemma 1, lim C|u|(t) = 0. Clearly, Cz ≥ z, because z is nonincreasing. Then,
t→0+
since the function C|u| is continuous, there is γ = γ(λ) > 0 such that
λCz(t)≥ λz(t)> b +λC|u|(t)
ϕ
for t ∈ (0,γ). Consequently,
γ
I (λC(||z|−|u||)) ≥ ϕ(λ(Cz(t)−C|u|(t)))dt = ∞,
ϕ
Z0
which finishes the proof.
(B2) Suppose ϕ(b ) < ∞. Let us define an element x = b χ . Then x ∈ Ces [0,1].
ϕ ϕ [0,1] ϕ
Indeed,
1
ρ (x)= I (Cx)= I (x) = ϕ(b )dµ = ϕ(b ) < ∞.
ϕ ϕ ϕ ϕ ϕ
Z0
There exists t > 0 with
t 1 b
ϕ
ρ xχ = ϕ(b )ds+ ϕ t ds ≤ 1.
ϕ [0,t) ϕ s
(cid:16) (cid:17) Z0 Zt (cid:18) (cid:19)
Moreover, for each λ > 0
ρ ((1+λ)xχ ) ≥ tϕ((1+λ)b ) = ∞.
ϕ [0,t) ϕ
For z = xχ we have kzk = 1. Then, like in (B1) above, we conclude that
[0,t) Ces(ϕ)
Ces [0,1] contains an order isomorphically isometric copy of l∞.
ϕ
(ii). Ifϕ∈ ∆ (∞)thenLϕ[0,1] ∈ (OC)(seeforexample[33,p. 21]). Consequently,Ces [0,1] ∈
2 ϕ
(OC) (see [28, Lemma 1 (a)]). By Theorem A, Ces [0,1] cannot contain isomorphic copy of
ϕ
l∞. (cid:3)
Theorem 4. Let ϕ be an Orlicz function.
(i) In the case when ϕ(b )= ∞ we assume that there exists constant D > 0 such that
ϕ
(3.4) ρ (f) ≤ DI (f),
ϕ ϕ
for all f ∈ KLϕ[0,∞) (see (2.1)). If ϕ ∈/ ∆ (R ) then the space Ces [0,∞) contains an order
2 + ϕ
isomorphically isometric copy of l∞.
(ii) If ϕ ∈∆ (R ) then Ces [0,1] does not contain an order isomorphic copy of l∞.
2 + ϕ
7
Proof. (i). We divide the proof into three parts.
(A) Assume that ϕ > 0 and ϕ < ∞. Suppose ϕ ∈/ ∆ (R ). This means that ϕ ∈/ ∆ (∞) or
2 + 2
ϕ ∈/ ∆ (0). In the first case we can proceed as in the proof of Theorem 3. The proof in the
2
second case is similar to the proof of Theorem 1 in [19]. We present the details for the reader’s
convenience.
It is well known that ϕ ∈ ∆ (0) if and only if there exist L > 1 and K,u > 0 such that
2 0
ϕ(u ) > 0, and ϕ(Lu) ≤ Kϕ(u) for all u ∈ [0,u ] (see [8, p. 9]). Assume that ϕ ∈/ ∆ (0). For
0 0 2
each L > 1 and every sequence (K )∞ we find a sequence u ց 0+ with
n n=1 n
ϕ(Lu ) > K ϕ(u ).
n n n
Take a sequence (K )∞ satisfying
n n=1
∞
1
< ∞.
K
n
n=1
X
Let (ǫ )∞ be any positive, decreasing sequence converging to zero. For any m ∈ N take a
m m=1
sequence (Km)∞ of positive reals numbers such that
n n=1
∞
1 1
≤ .
Km 2m
n=1 n
X
Now by the first part, for any m ∈ N we can find a decreasing sequence (um)∞ with um → 0
n n=1 n
as n → ∞ and ϕ((1+ǫ )um) > Kmϕ(um) for all m ∈ N. Thus
m n n n
∞ ϕ(um) ∞ 1 1
n < ≤ ,
ϕ((1+ǫ )um) Km 2m
n=1 m n n=1 n
X X
for all m ∈ N. In view of um → 0 as n → ∞, we can find a subsequence (n )⊂ N such that
n k
u1 > u2 >u3 > ... .
n1 n2 n3
Hence, without loss of generality, we can assume that for m > 1,
um < um−1.
m m−1
Let
1
c = ,
n ϕ((1+ǫ )un)
n n
for n ∈ N. Note that 1 ≤ c < ∞ for all n ∈ N. Define
n
∞
f(t)= unχ (t),
n Ωn
n=1
X
where Ω = [c +c +···+c ,c +c +···+c +c )⊂ R . It is clear that the function f
n 1 2 n−1 1 2 n−1 n +
is decreasing. We have
∞ ∞
I (f)= ϕ(|f(t)|)dt = c ϕ(un)
ϕ n n
Z0 n=1
X
∞ ϕ(un) ∞ 1
≤ n ≤ = 1.
ϕ((1+ǫ )un) 2n
n=1 n n n=1
X X
For any ǫ > 0 and for any positive integer M, we get
∞ ∞
ϕ((1+ǫ)|f(t)|)dt ≥ ϕ((1+ǫ)|f(t)|)dt
ZM ZM1
∞ ∞ ϕ((1+ǫ)un) ∞
= c ϕ((1+ǫ)un)= n ≥ 1 = ∞,
n n ϕ((1+ǫ )un)
nX=n1 nX=n1 n n nX=n2
8
for some M ≥ M and n ≥ n . Since I (f)≤ 1, there is N ∈ (0,∞) such that
1 2 1 ϕ
1
I (fχ )≤ .
ϕ [N,∞) D
Consequently, by the assumption, ρ (fχ ) ≤ 1. Put g = fχ . Since f is a decreasing
ϕ [N,∞) [N,∞)
function, for any ǫ > 0,
1 t t−N
(1+ǫ) |f(x)|dx ≥(1+ǫ) |f(t)|
t t
ZN (cid:18) (cid:19)
N ǫ
= (1+ǫ) 1− |f(t)| ≥ 1+ |f(t)|,
t 2
(cid:18) (cid:19)
(cid:16) (cid:17)
for t large enough. Now, for any ǫ > 0 and N large enough we get
1
∞ 1 t
ρ ((1+ǫ)g) = ϕ (1+ǫ) |f(x)|dx dt
ϕ t
ZN (cid:18) ZN (cid:19)
∞ ǫ
≥ ϕ 1+ |f(t)| dt = ∞.
2
ZN1 (cid:16)(cid:16) (cid:17) (cid:17)
We have constructed an element g ∈ Ces [0,∞) such that ρ (g) ≤ 1 and for every ǫ > 0,
ϕ ϕ
ρ ((1+ǫ)g) = ∞. By Remark 2 we conclude that Ces [0,∞) contains an order isomorphically
ϕ ϕ
isometric copy of l∞.
(B) Suppose b < ∞ and
ϕ
(B1) ϕ(b )= ∞. The proof is similar to the proof of case (B1) of Theorem 3.
ϕ
(B2) ϕ(b ) < ∞. We just proceed as in the proof of Theorem 3 case (B2). We additionally
ϕ
use the assumption Ces 6= {0} and Proposition 3 from [21].
ϕ
(C) Assume that ϕ <∞ and a > 0. Put x = a χ . Then kxk = 1 and
ϕ ϕ [0,∞) Ces(ϕ)
δ(x) = inf{λ > 0: ρ (x/λ)< ∞} = 1.
ϕ
Indeed, for any λ > 0
ρ ((1+λ)x) = I ((1+λ)Cx) =I ((1+λ)x) = ϕ((1+λ)a )m([0,∞)) = ∞.
ϕ ϕ ϕ ϕ
Therefore, by Remark 2(ii), Ces [0,∞) contains an order isomorphically isometric copy of l∞.
ϕ
(ii). We can use analogue arguments as in the proof of Theorem 3. (cid:3)
Now we will discuss the technical assumption from Theorem 3 and Theorem 4.
Let us recall the standard form of the integral Hardy inequality (see [23]). Supposep > 1 and
f is a nonnegative p-integrable function on I. Then f is integrable over the interval (0,x)∩I
for each positive x and
1 x p p p
f(t)dt dx ≤ f(x)pdx.
x p−1
ZI(cid:18) Z0 (cid:19) (cid:18) (cid:19) ZI
The dilation operator D , s > 0, defined on L0(I) by
s
D x(t)= x(t/s)χ (t/s) =x(t/s)χ (t),
s I [0,min{1,s})
for t ∈ I, is bounded in any symmetric space E on I and kD k ≤ max{1,s} (see [5, p.
s E→E
148]). For the theory of symmetric (rearrangement invariant) spaces the reader is referred to [5]
and [25]. Moreover, the lower and upper Boyd indices of E are defined by
lnkD k
p(E) = lim s E→E,
s→0+ lns
lnkD k
q(E) = lim s E→E.
s→∞ lns
9
In particular, they satisfy the inequalities
1≤ p(E) ≤ q(E) ≤ ∞.
In the case when E is the Orlicz space Lϕ, these indices equal to the so-called lower and upper
Orlicz-Matuszewska indices α and β of Orlicz functions generating the Orlicz spaces, i.e.,
ϕ ϕ
α = p(Lϕ) and β = q(Lϕ) (see [30, Proposition 2.b.5 and Remark 2 on page 140]). For more
ϕ ϕ
details see [6], [7], [32], [33] and [36].
Let us mention the important result about boundedness of the operator C.
Theorem B. [23, p. 127] For any symmetric space E on I the operator C : E → E is bounded
if and only if the lower Boyd index satisfies p(E) > 1.
Note that if (X,k·k ) is a Banach ideal space then C : X → X implies C is bounded. In fact, if
X
C : X → X,thenX ֒→ CX. ThismeansthatthereisM > 0withkxk = kC|x|k ≤ M kxk
CX X X
for all x ∈ X, i.e. C is bounded. However, it may happen that X 6֒→ CX (see [16, Proposition
2.1]). Moreover, C : CX → X is always bounded (from the definition of CX) and CX is so-
called optimal domain of C for X (cf. [16] and [28]). The immediate consequence of Theorem
B and the above discussion about indices of the Orlicz space Lϕ is a next corollary.
Corollary 5. The embedding Lϕ ֒→ Ces holds if and only if α > 1.
ϕ ϕ
Remark 6. If f ∈ KLϕ[0,1] then
t
f(s)ds < ∞
Z0
for each t ∈ [0,1], because KLϕ[0,1] ֒→ Lϕ[0,1] ֒→ L1[0,1]. However, we present also the
direct proof, since we need this fact also in the proof of Proposition 8 in the case of I = [0,∞).
Conversely, suppose t0|f(s)|ds = ∞ for some t ∈ (0,1]. Consequently, f is unbounded on
0 0
(0,t ). Since ϕ(u)/u ր a, a ≤ ∞, then there is a > 0 such that ϕ(u) ≥ a u/2 for u ≥ u .
0 0 0 3
R
Setting I = {t ∈[0,1] : |f(t)| ≥ u } we have
0 3
a
0
I (f)≥ ϕ(|f(s)|)ds ≥ |f(s)|ds = ∞,
ϕ
2
ZI0 ZI0∩[0,t0]
whence f ∈/ KLϕ[0,1] and the claim is proved.
Proposition 7. Let ϕ be an Orlicz function with ϕ < ∞. Consider the following conditions:
(i) There exists p >1, a convex function γ and constants A,B,u > 0 such that ϕ(u )> 0
0 0
and for all u∈ [u ,∞),
0
Aγ(u) ≤ ϕ(u)1/p ≤ Bγ(u),
i.e. ϕ1/p is equivalent to a convex function for ”large arguments”.
(ii) For each ǫ > 0 there exists a constant D = D(ε) > 0 such that
ρ (f) ≤ DI (f)+ǫ,
ϕ ϕ
for all f ∈ KLϕ[0,1] satisfying |f (t)| ≥ ϕ−1(ε) for a.e. t ∈ supp(f). Moreover, the
Orlicz class KLϕ[0,1] is closed under the operator C.
(iii) C :Lϕ[0,1] → Lϕ[0,1].
Then (i)⇒(ii)⇒(iii).
10