Isaac V. Shragin
On some properties of Musielak-Orlicz sequence spaces
Abstract. We consider a nontrivial vector space X and a semimodular M : X→[0, ∞]
with property: (∀x ∈ X)(∃α > 0) M(αx) < ∞ (in other words, M is normal (i.e.
(∀x ∈ X \ {0}) (∃α > 0) M(αx) > 0) pregenfunction). The function M generates in X a metric d with
d(x, y) := inf{a > 0 : M(a−1(x− y)) ≤ a}.
At the same time M generates a metric ρ in Musielak-Orlicz sequence space lM, namely
ρ(ϕ, ψ) := inf{a > 0 : I(a−1(ϕ− ψ)) ≤ a}.
with I(ϕ) =P
n≥1M (ϕ(n)).
It is proved that the space (lM, ρ) is complete if and only if the space (X, d) is complete.
We consider also the closed subspace GM ⊂ lM of sequences ϕ ={ϕ(n)} such that (∀α > 0) (∃m ∈ N) P
n≥mM (αϕ(n)) <∞ and prove that (GM, ρ) is separable if and only if (X, d) is the same. Several examples are considered.
2000 Mathematics Subject Classification: 46A45.
Key words and phrases: normal pregenfunction, Musielak-Orlicz sequence space, completeness, separability.
1. Preliminaries. In what follows α, γ, δ, ε denote positive numbers, and j, k, m, n - positive integers. Let X be a vector space (X 6= {0}) over scalar field K with K = R or C.
Definition 1.1 A function M : X→[0, ∞] is called pregenfunction if the following conditions are fulfilled:
(a) M(0) = 0;
(b) if λ ∈ K with |λ| = 1, then(∀x ∈ X) M(λx) = M(x);
(c) for any x, y ∈ X and a ∈ (0, 1)
M (ax + (1− a)y) ≤ M(x) + M(y);
(d) (∀x ∈ X) (∃α) M(αx) < ∞.
In other words, the pregenfunction is a pseudomodular [1] that satisfies to condition (d). If moreover the condition
(e) (∀x ∈ X \ {0}) (∃α > 0) M(αx) > 0
is fulfiled, we shall call such function M by normal pregenfunction (in this case M is semimodular [1] with property (d)).
A normal pregenfunction M generates in X a metric. Really, if p(x) := inf{a > 0 : M(a−1(x)) ≤ a}, ∀x ∈ X,
then [1] 0 ≤ p(x) < ∞; p(x) = 0 if and only if x = 0; p(−x) = p(x); p(x + y) ≤ p(x) + p(y) (it is evidently also that (∀b > p(x)) M(b−1x) < b). Hence the formula
d(x, y) := p(x− y) = inf{a > 0 : M(a−1(x − y)) ≤ a}
defines a metric in X.
Remark 1.2 Suppose taht condition (d) is not fulfilled, i.e. ∃x06= 0 (∀α) M(αx0) =
∞. Then {a > 0 : M(a−1x0) ≤ a} = ∅, and p(x0) is not defined (or one may set that p(x0) = ∞). In this situation it is possible to define p (and hence the metric d) in the subspace
X0:= {x ∈ X : ∃α M(αx) < ∞}.
In particular, if (∀x 6= 0) M(x) = ∞, then X0= {0}.
Note yet that if in place of (d) to accept more strong condition: (∀x ∈ X) limα→0M (αx) = 0, then p is a F -norm in X [1]. However, in this paper it is enough to have a metric.
Lemma 1.3 Let M be a normal pregenfunction, and xk ∈ X, k = 0, 1, 2 . . . . Then (1) (limk→∞d(xk, x0) = 0)⇔ ((∀α) limk→∞M (α(xk− x0)) = 0);
(2) (limj,k→∞d(xj, xk) = 0)⇔ ((∀α) limj,k→∞M (α(xj− xk)) = 0).
Proof (1) Let d(xk, x0)−→0. Take α > 1 nad ε < 1. Then (∃m) (∀k > m)k d(xk, x0) < ε/α, hence M(α(xk − x0)) ≤ M(ε−1α(xk − x0)) < ε/α < ε, i.e.
M (α(xk− x0))−→0.k
Conversely, let (∀α) M(α(xk−x0))−→0. Then (∀ε) (∃m) (∀k > m) M(εk −1(xk− x0)) < ε. Hence d(xk, x0) ≤ ε.
(2) It is proved similarly to (1).
Remark 1.4 Lemma 1.3 coincides in essence with Theorem 1.6 from [1], which is proved for pseudomodular ρ and subspace {x ∈ X : limλ→0ρ(λx) = 0}.
2. Sequence spaces generated by pregenfunctions. . Consider the vector space XN of all sequences ϕ = (ϕ(1), ϕ(2), . . . , ), where (∀n) ϕ(n) ∈ X. Denote the zero-element of XN by θ, i.e. (∀ n) θ(n) = 0 ∈ X.
Let M : X→[0, ∞] be a pregenfunction. Put I(ϕ) :=X
n≥1
M (ϕ(n)), ϕ∈ XN,
lM := {ϕ ∈ XN : (∃α)I(αϕ) < ∞}.
The conditions (a), (b), (c) imply that lM is a vector subspace (in other words, lineal) in XN. By condition (d) lM 6= {θ}. Indeed, let x0 6= 0 and M(αx0) < ∞.
Put ϕ(1) = x0 and (∀n > 1) ϕ(n) = 0. Then ϕ 6= θ and I(αϕ) < ∞, i.e. ϕ ∈ lM.
Lemma 2.1 The functional I : lM→[0, ∞] is a pregenfunction. Moreover, it is normal if and only if M is the same.
Proof The properties (a), (b), (c) for I follow directly from this properties for M . The property (d) is contained in the definition of the lineal lM.
Let M be a normal pregenfunction. If ϕ ∈ lM \ {θ}, then (∃n0) ϕ(n0) 6= 0.
Hence, by (e), (∃α) M(αϕ(n0)) > 0, so that I(αϕ) > 0, i.e. I is normal. Evidently
also, if M is not normal, then I is the same.
The Lemma 2.1 implies that if the pregenfunction M is normal, then the formula ρ(ϕ, ψ) := inf{a > 0 : I(a−1(ϕ − ψ)) ≤ a}
defines a metric in lM. Moreover, according to Lemma 1.3 the following statement is valid. Let ϕk∈ lM, k = 0, 1, . . . . Then
(1) (limk→∞ρ(ϕk, ϕ0) = 0)⇔ ((∀α) limk→∞I(α(ϕk− ϕ0)) = 0);
(2) (limj,k→∞ρ(ϕj, ϕk) = 0)⇔ ((∀α) limj,k→∞I(α(ϕj− ϕk)) = 0).
Consider now some other lineals in XN. Denote by F the collection of all sequences from XN, that have only finite set of nonzero members. Further, by the given pregenfunction M put
GM = {ϕ ∈ XN : (∀α) (∃m) Im(αϕ) < ∞}, with Im(ϕ) = Pn≥mM (ϕ(n)). It is easy to check that GM is a lineal.
Lemma 2.2 F ⊂ GM ⊂ lM.
Proof Let ϕ∈ F \ {θ}, and m = max{n : ϕ(n) 6= 0} + 1. Then (∀α) Im(αϕ) = 0, so that ϕ ∈ GM. Thus F ⊂ GM.
Let now ϕ ∈ GM. Then (∃m) Im(ϕ) < ∞. If m = 1, i.e. I(ϕ) < ∞, then ϕ ∈ lM. Let m > 1. Then, by (d), (∀n < m) (∃αn) M(αnϕ(n)) < ∞. Put α = min(1, α1, . . . , αm−1). Then I(αϕ) < ∞, i.e. ϕ ∈ lM. Thus GM ⊂ lM. The following statement is valid: GM = lM if and only if there exist γ and δ such that if M(x) < δ, then M(2x) ≤ γM(x) (this is generalization of the known in the theory of Orlicz spaces ”∆2-condition” by small x). We publish the proof of this statement in [2].
Let now the pregenfunction M be normal and consider the metric space (lM, ρ).
Lemma 2.3 The lineal GM is closed in (lM, ρ). Moreover, GM = F .
Proof Let ϕk ∈ GM, k = 1, 2, . . . , and ρ(ϕk, ϕ)−→0 with ϕ ∈ lk M. Take an arbitrary α. Since I(2α(ϕk− ϕ))−→0, there exists j such that I(2α(ϕk j− ϕ)) < ∞.
As ϕj ∈ GM (∃m) Im(2αϕj) < ∞. Then, by condition (c), Im(αϕ) ≤ Im(2α(ϕ − ϕj)) + Im(2αϕj) < ∞,
so that ϕ ∈ GM. Thus the lineal GM is closed. From here and Lemma 2.2 it follows that F ⊂ GM. Prove that GM ⊂ F . Indeed, let ϕ ∈ GM. Put (∀k) ϕk(n) = ϕ(n) by n ≤ k, and ϕk(n) = 0 by n > k. Then (∀k) ϕk ∈ F and (∀α) I(α(ϕ− ϕk)) = Pn>kM (αϕ(n))−→0, i.e. ρ(ϕk k, ϕ)−→0. Hence ϕ ∈ F .k
3. On completeness of lM and separability of GM. let M be a normal pregenfunction, so that (X, d), (lM, ρ) and (GM, ρ) are metric spaces.
Theorem 3.1 The space (lM, ρ) is complete if and only if (X, d) is the same.
Proof Let (lM, ρ) be complete. Take a Cauchy sequence {xk} ∈ (X, d) and put (∀k) ϕk(1) = xk and (∀n > 1) ϕk(n) = 0. Since
(∀α) lim
j,k→∞I(α(ϕj− ϕk)) = lim
j,k→∞M (α(xj− xk)) = 0,
then {ϕk} is a Cauchy sequence in (lM, ρ). Hence, (∃ϕ ∈ lM) ρ(ϕk, ϕ)−→0, i.e.k (∀α) I(α(ϕk− ϕ))−→0, whence (∀α) M(α(xk k− ϕ(1)))−→0, i.e. d(xk k, ϕ(1))−→0.k Thus (X, d) is complete.
Let now (X, d) be complete, and let {ϕk} be a Cauchy sequence in (lM, ρ), i.e.
(∀α) limj,k→∞I(α(ϕj− ϕk)) = 0. Hence
∀(n, α) lim
j,k→∞M (α(ϕj(n) − ϕk(n))) = 0,
i.e. (∀n) {ϕk(n)} is a Cauchy sequence in (X, d). Since (X, d) is complete, then (∀n) (∃xn ∈ X) d(ϕk(n), xn)−→0. Put (∀n) ϕ(n) = xk n and show taht (∀α) I(α(ϕk− ϕ))−→0. To this end fix α and ε and find kk 0 such that
(1) (∀j, k > k0) I(2α(ϕk− ϕj)) < ε.
Further fix k > k0. Then (∀n, j)
M (α(ϕk(n) − ϕ(n))) ≤ M(2α(ϕk(n) − ϕj(n))) + M(2α(ϕj(n) − ϕ(n))).
Since M(2α(ϕj(n) − ϕ(n)))−→0, then (∀n)j M (α(ϕk(n) − ϕ(n))) ≤ lim inf
j→∞ M (2α(ϕk(n) − ϕj(n))).
Denote here the right side by ak(n). Then
(2) I(α(ϕk− ϕ)) ≤X
n≥1
ak(n).
Considering Pn≥1ak(n) as an integral over N, we have by Fatou lemma X
n≥1
ak(n) ≤ lim inf
j→∞ I(2α(ϕk− ϕj)).
Then, by (1) and (2), (∀k > k0) I(α(ϕk− ϕ)) ≤ ε. From here it follows that ϕ ∈ lM and ρ(ϕk, ϕ)−→0, i.e. the space (lk M, ρ) is complete.
Corollary 3.2 If X is a Banach space (with norm | · |), and for pregenfunction M the following conditions are fulfilled:
|x|lim→0M (x) = 0, (3)
lim inf
|x|→∞M (x) > 0, (4)
(note that (4) implies condition (e)), then the space (lM, ρ) is complete.
Proof Show that the space (X, d) is complete. To this end take in (X, d) a Cauchy sequence {xk}, i.e.
(5) (∀α) lim
j,k→∞M (α(xj− xk)) = 0, and show that {xk} is a Cauchy sequence in (X, | · |).
Suppose the contrary. Then there exist γ and indexes k(1) < k(2) < . . . such that (∀m) |xk(m)− xk(m+1)| > γ. On the other hand, by (4),
(∃δ, ξ > 0) inf{M(x) : |x| > ξ} > δ.
Therefore, if α > ξ/γ, then (∀m) M(α(xk(m)− xk(m+1))) > δ, that contradicts to (5). So {xk} is a Cauchy sequence in Banach space (X, | · |). Hence, (∃x ∈ X)
|xk− x|−→0. Then, by (3), (∃α) M(α(xk k− x))−→0, so that (X, d) is complete. Itk
remains to applay Theorem 3.1.
Example 3.3 Let X be the space of continuous functions x : [0, 1]→R with the norm |x| =R1
0 |x(t)|dt, and M(x) := maxt|x(t)|. Evidently M is a pregenfunction with property (4), but condition (3) is not fulfilled. In addition, the space (X, | · |) is not complete. Therefore, Corollary 3.2 is not applicable here. Nevertheless, the space (lM, ρ) is complete by Theorem 3.1 since the space (X, d) is complete (it is easy to check that d(x, y) = (maxt|x(t) − y(t)|)1/2).
Example 3.4 Let X = L∞[0, 1], so that X is a Banach space with the norm |x| = ess supt∈[0,1]|x(t)|. Let futher M(x) =R1
0 |x(t)|dt (Lebesgue integral). Evidently M is a normal pregenfunction, for which (3) is fulfilled, but (4) is not. So that Corollary 3.2 is not applicable. But Theorem 3.1 does not help here too since (X, d) is not complete. Indeed, consider a sequence {xk} ⊂ L∞[0, 1] that converges by the norm of the space L[0, 1] to a function x ∈ L[0, 1] \ L∞[0, 1]. Hence, {xk} is a Cauchy sequence in (X, d) that has not a limit in this space.
Thus in Example 3.4 the space (lM, ρ) is not complete. We have a similar situation if X = C[0, 1] (|x| = maxt|x(t)|) and M(x) =R1
0 |x(t)|dt.
Pass to the question on separability of the space (GM, ρ).
Theorem 3.5 The space (GM, ρ) is separable if and only if (X, d) is the same.
Proof Since, by Lemma 2.3 GM = F , it is possible to consider F in place of GM. So, let F be separable in metric ρ, i.e. there exists a countable set A ⊂ F , such that F ⊂ A. Show that the countable set A1 := {ϕ(1) : ϕ ∈ A} is dense in (X, d).
To this end fix x ∈ X and put ϕ(1) = x and (∀n > 1) ϕ(n) = 0. Then ϕ ∈ F , so that (∃ϕk ∈ A, k = 1, 2, . . . ) (∀α) I(α(ϕk−ϕ))−→0, whence (∀α) M(α(ϕk k(1)−x))−→0.k Hence d(ϕk(1), x)−→0 with ϕk k(1) ∈ A, k = 1, 2, . . . . Thus (X, d) is separable.
Conversely, let some countable set B be dense in (X, d) with 0 ∈ B. Put A ={ϕ ∈ F : (∀n)ϕ(n) ∈ B}. Evidently, A is countable. It remains to show that F ⊂ A.
So, let ϕ ∈ F , i.e. (∃m) (∀n > m) ϕ(n) = 0. Then (∀n ≤ m) (∃xn,k ∈ B, k = 1, 2, . . . ) d(xn,k, ϕ(n))−→0, i.e. (∀α) M(α(xk n,k− ϕ(n)))−→0. Put (∀k) (∀n ≤ m)k ϕk(n) = xn,k and (∀n > m) ϕk(n) = 0. Then (∀k) ϕk ∈ A and
(∀α) I(α(ϕk− ϕ)) = X
n≤m
M (α(xn,k− ϕ(n)))−→0,k
i.e. ρ(ϕk, ϕ)−→0, so that ϕ ∈ A.k
Corollary 3.6 Let X be a separable normed space (with a norm | · |), and for M the condition (3) is valid. Then (GM, ρ) is separable.
Proof Let a countable set A be dense in (X,| · |). Then (∀x ∈ x) (∃ak ∈ A, k = 1, 2, . . . ) |ak− x|−→0, whence, by (3), (∀α) M(α(ak k− x))−→0, i.e. d(ak k, x)−→0.k Thus the space (X, d) is separable. It remains to applay Theorem 3.5.
Example 3.7 Let a normed space (X,| · |) be separable, M(x) = 0 by |x| ≤ 1 and M (x) =∞ by |x| > 1. Then
lM = l∞(X) := {ϕ ∈ XN : sup
n |ϕ(n)| < ∞}, GM = c0(X) := {ϕ ∈ XN : |ϕ(n)|−→0},n
and, by Corollary 3.6, (GM, ρ) is separable. By the way, in this example ρ(ϕ, ψ) = supn|ϕ(n) − ψ(n)|, so that the space (lM, ρ) is not separable.
Example 3.8 Let X be the space of bounded functions x : [0, 1]→R, M(x) = supt|x(t)|. Since d(x, y) = (supt|x(t) − y(t)|)1/2, then (X, d) is not separable.
Hence, (GM, ρ)(=(lM, ρ)) is the same.
In Example 3.3 the space (X, | · |) is separable, but (3) is not valid, so that Corol- lary 3.6 is not applicable. But since M represents a norm in separable space C[0, 1], then (X, d) is separable (note that d(x, y) = (maxt|x(t) − y(t)|)1/2). So that (GM, ρ)(=(lM, ρ)) is separable.
In Example 3.4 the space (X, |·|) is not separable, so that Corollary 3.6 does not work. At the same time, since M represents a norm in a subspace of the separable space L[0, 1], then (X, d) is separable (note that d(x, y) = (R1
0 |x(t) − y(t)|dt)1/2).
So that (GM, ρ)(=(lM, ρ)) is separable.
Remark 3.9 The stated results it is possible to extend to situation when on X normal pregenfunction Mn, n = 1, 2, . . . are given, so that every Mn generates in X the metric dn. In this case lineals lM and GM and metric ρ are defined as before, but now
I(ϕ) :=X
n≥1
Mn(ϕ(n)), Im(ϕ) := X
n≥m
Mn(ϕ(n)).
In this situation, in Theorems 3.1 and 3.5 it ought to replace (X, d) by (X, dn), n = 1, 2, . . . . Similar modifications it is necessary to bring in Corollaries 3.2 and 3.6.
References
[1] J. Musielak,Orlicz spaces and modular spaces, Springer-Verlag, 1983.
[2] I.V. Xragin, Prostranstva vektornyh posledovatel~noste$i i operator superpozicii v tih prostranstvah, Odessa, Astroprint 2008.
Isaac V. Shragin K¨oln, Germany
E-mail: marina spektor@gmx.de
(Received: 22.07.2008)
