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Abstract and Applied Analysis
Volume 2008 (2008), Article ID 374742, 15 pages
http://dx.doi.org/10.1155/2008/374742
Research Article

Extension of The Best Approximation Operator in Orlicz Spaces

Instituto de Matemática Aplicada San Luis, UNSL-CONICET, Avda, Ejército de los Andes 950, 5700 San Luis, Argentina

Received 17 December 2007; Accepted 26 February 2008

Academic Editor: Jean-Pierre Gossez

Copyright © 2008 Ivana Carrizo et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let be a probability space and a sub--lattice of the -algebra . We study an extension of the best -approximation operator from an Orlicz space to the space , where denotes the derivative of the convex, but not necessarily a strictly convex function . We obtain convergence results when a sequence of -algebras converges to in a suitable way.

1. Introduction and Notations

Let be a probability space and let be the set of all -measurable real-valued functions defined on . Given a convex function such that when , let be the space of all functions such that for some Since we only deal with a function that is, there exists a constant such that for all the space can be defined as the space of all function , where (1.1) holds for every positive number The space is analogously defined, where is the derivative of the function Besides, observe that for a function it holds the next inequality for all , and therefore Moreover is a function if and only if is a function.

We say, according to [1], that a collection of sets in is a -lattice if it is closed under countable unions and intersections and contains and Given a -lattice , we denote by the -lattice of all the complementary sets of that is, Denote by all -measurable functions in .

A set is called -closed if and only if and or () then Then is a -closed convex set and a lattice, that is closed for the maximum and minimum of functions.

We will use the notation and A set is called a -complete lattice if and only if and for all sequence

It is well known, see [1], that for every there exists an element such that Denote by the set of all satisfying (1.3). Each element of will be called a best -approximation of given and we will refer to the mapping defined on as the best approximation operator.

It is showed in [1] that for the set is a nonvoid -complete lattice. Also it was proved that if both in and then we have and In this case, we say that the multivalued operator is a monotone operator.

The main purpose of this paper is to extend the best approximation operator to the set The case , was extensively treated in [2] and the best approximation operator is extended to all measurable functions in [3]. The extension from to is considered in [4] for a function which is strictly convex and

Now in this paper we consider a convex function but not necessarily a strictly convex function. Extension of best approximation operator when the approximation classes are the constants is treated in [57].

The extension of the best approximation operator is ; for , we will be denoted by In Theorem 2.12, we prove that for every while in Theorem 2.16 it is proved that it is indeed an extension, that is, for Additional properties are obtained for the set when the -lattice is a -algebra (see Theorem 2.17) and similar results hold when is the class of monotone functions in (see Theorem 3.2). A martingale-type result is given in Theorem 4.1 which generalizes [8, Theorem 2.8] for the particular family of measures considered in this paper.

2. Extension of the Best -Approximation Operator

We begin with some definitions and auxiliary results. The proof of the next two lemmas can be found in [4].

Lemma 2.1. A necessary and sufficient condition for a continuously differentiable and convex function to satisfy the condition is that there exists a constant such that

Lemma 2.2. Let be in Then is an integrable function.

According to Brunk and Johansen [8], we set the following definitions.

Definition 2.3. Let be a signed measure on and let be a -lattice contained in Say that is a -positive set, if for all , then A set is called -negative if for all one has

Definition 2.4. Let be a family of measures on and let be a -lattice contained in An -measurable function is called a Lebesgue-Radon-Nikodym function (LRN function) for given if and only if the set is -positive for all and the set is -negative for all

Remark 2.5. We note that in Definition 2.4 it is sufficient to impose the conditions for all in a dense set in , see [8, page 588].

For and , we define the following measures on : where Note that when and , the measure and are well defined.

The next theorem is a characterization of , see [9, Theorem 3.2].

Theorem 2.6. Let be a -lattice and . Then the following statement are equivalent.
(1)(2)(a) a set is -positive for all and(b) a set is -negative for all (3) is an LRN function for the family given

Now we extend the operator to the space

Definition 2.7. Let be a -lattice and let Then is an extended best -approximation if and only if and
(i)the set is -positive for all ;(ii)the set is -negative for all . For we denote by the set of all extended best -approximation functions.

Remark 2.8. Let and let be a function in such that the set is -positive and the set is -negative for all . Then we have the following.
(i)For all and , For all and , (ii)

Proof. We prove inequality (2.3). Since the set is -positive, that is, for each and , we have For , where and , , then by (2.6), we have All nonnegative can be obtained as a limit of functions of the above type. The proof of inequality (2.4) is similar.
The equality (2.5) is obtained using Lebesgue's theorem when with in (2.4) and if consider in (2.3) also

As a reference, we note that (2) is equivalent to (3) in Theorem 2.6 for We have the next remark.

Remark 2.9. For and the following statements are equivalent:
(1);(2) is an LRN function for the family given

The next lemma is a particular case of [8, Theorem 1.8].

Lemma 2.10. Let and . Then the following statements are equivalent.
(1) is an LRN function for the family given (2)There exists a countable set such that is -negative for all and the set is -positive for all

We need the following auxiliary result.

Lemma 2.11. Let be a sequence of functions such that , where Let be such that . Then

Proof. We will prove the result just for the increasing case, the proof for the decreasing case follows the same pattern. The function is obviously -measurable function. Now, we prove that and it satisfies (i) and (ii) of Definition 2.7. We have that Using (2.5), According to (2.8) and (2.9), we have, since is an increasing function, Since is a continuous function and we have by (2.8) and Lebesgue's theorem Now by (2.10) it holds Using Fatou in (2.12), we have Therefore, using (2.11) and (2.12), we get and
Let , and we know for each Since is an increasing sequence, we get and by (2.14), we have Hence, the set is -positive for all
Now, for and , we define and We have that, for for some -measurable set such that We observe that Then taking limit as and using Lebesgue's theorem, we obtain

Theorem 2.12. Let be a -lattice and , then .

Proof. For we can define the following sequences. For each , let and when we have Set for all in then we have , when Since for each we have there exist As is a monotone operator over we can take a new sequence that we call again such that for all
Since and using again that is a monotone operator, we have , where is the sequence defined by and Furthermore, it is easy to check that
Then, we have that for each that , when , and since we have and if we define by Lemma 2.11 we obtain and for all If we take , we have and by Lemma 2.11 we get , where

To see that the extended best -approximation is an extension of the best -approximation operator, we must prove for every First, we need to prove the following lemmas.

Lemma 2.13. Let be a convex function and assume that it satisfies the -condition. Then for , where is the constant for the condition.

Proof. We consider two cases. First, we assume Since is -convex function, we have that Using for all we get Then we obtain For , we have

Lemma 2.14. Let and , then

Proof. Since is -positive for all , then for all , we have that In particular, it holds that for all , that is, Now, we have By the Fubini's theorem, we get Thus To see that inequality (2.22) is equivalent to (2.28) we will prove that . In fact Since by Lemma 2.2, the last integral is finite.

The following properties of the set can be easily proved.

Proposition 2.15. Let , then
(1),(2) for all

Now we prove that the operator is in fact an extension of the operator

Theorem 2.16. Let , then

Proof. For we will prove only that The other inclusion follows from Theorem 2.6. Let and again using Theorem 2.6 it remains to prove that Recall that then By Lemma 2.13 we obtain the following inequality: Applying Lemma 2.14 and taking into account that is an increasing function, we get Thus using (2.32) in (2.31), we have For the set again by Lemma 2.13, we obtain Since , we have Thus By (1) in Proposition 2.15, and by Lemma 2.14, we have that Therefore, . By (2.33), we have and therefore

Now, if consider a -subalgebra instead of a -lattice , the extended best -approximation operator has the following properties.

Theorem 2.17. Let be in if is a sub--algebra of the -algebra then the following hold.
(1)The set-valued function is a monotone operator.(2)The set is a -complete lattice, and there exist such that a.e. for every

Proof. To prove (1), recall that this set-valued operator is monotone if ; then if and , we have that and Since is a lattice, we know and We will prove first that Set where and We will see that is an LRN function for the family of measures given First, we will see that for each and for all , we have Since and and using that is an LRN function of the family , we obtain that for all Now, we see that is -negative for all For , we have Using and that is a nondecreasing function, we obtain Thus By (2.39) and (2.42), we have
Now we show Since and is a -positive for all and for all we have Since the inequalities (2.43) and (2.44) prove that
As the statement (1) proves in particular that is a lattice, we will see that the set is a -complete lattice. Given a sequence in we have that then, from Lemma 2.11 we obtain that The proof is similar.
By [10, Proposition II.4.1] there exists a sequence such that for every Set and then and are in since this set is a -complete lattice.

3. Extended Best -Approximation with Nondecreasing Functions

When the approximation class is the monotone functions defined on we can obtain similar results as those of Theorem 2.17. Now is Lebesgue measure on the measurable sets, and Therefore, is the set of nondecreasing functions in

Remark 3.1. Let be a nondecreasing function on Given the set is one of the intervals or and similarly the set is or with Then is a dense set in . In fact, the complement set of is a countable set.

Note that each is of the form or and is or Thus

Theorem 3.2. Let be the class of the -integrable nondecreasing functions in Then the following hold.
(1)The set mapping is a monotone operator.(2)For every , the set is a -complete lattice.

Proof. First we prove (1), that is given , in with for each we will see that and Let be where is the set given in Remark 3.1. Recall that if and only if for each and Also if and only if for each and , First we prove (3.1). Now we see that with Since to prove (3.3), we have to see that where Indeed by (), we get Since is a nondecreasing function, we have As (), we have that is (3.4).
Now we will prove that In fact The last two integrals in (3.8) are less or equal than zero, so (3.7) holds. A similar argument shows that Therefore, the extended best -approximation operator is a monotone operator. By (1), we have that is a lattice, just setting Now by Lemma 2.11 we obtain that is a -complete lattice.

4. A Limit Theorem for Extended Best -Approximations

Given a sequence of -algebras contained in the -algebra we consider two cases, for all and we set for the -algebra generated by and if for all , we set

The next result is a particular case of [8, Theorem 2.8] when is a strictly convex function. This assumption on the function assures that the family of measures decreases at zero as required by Brunk and Johansen in that theorem.

Theorem 4.1. Let be an increasing or decreasing sequence of -algebras in and let be the limit of the sequence. If , then we have for all , that are in

Proof. Define and then we only prove that , when is an increasing sequence of -algebras, the proof for the decreasing case is similar.
First, we prove for each that the set is -positive for all Let and for where decreases to zero, we have that and for all Now for each , we define the following disjoint sets. For , set Thus, and then As and for , we have that As then and for Thus and by Lebesgue's theorem, we get for all Now we have (4.2) for all In fact, the set is a monotone class, that is, the set is closed for increasing and decreasing sequences of sets. As and this union is an algebra of sets, the monotone class generated by it is that is,
Now let us prove that the set is -negative for all As and then we have to prove that for all the set is -negative. Since where and we have for all that Then for a fixed set
Set and note that Then for we have Now, we prove the following inequality: We can see that where are the following disjoint sets Then Since , where we have (4.6). Therefore by (4.5), we have for all Thus by (4.10), (4.4), and (4.3), we get for all Therefore, the result is satisfied for all Thus
We have , where and then for all Since we have for all that For define the following disjoint sets and Then for we have Now if where Thus Then by (4.12) and (4.14), we have for all that Therefore, we have (4.15) for all
Let us see now that for all
As we have to prove that for all the set is -positive. We have that where is the increasing sequence and Then we have Set and note that Then for we have Now, we prove We can see that where are the following disjoint sets: Then Since , where we have (4.19). Therefore by (4.18), we have for all Thus by (4.23) and (4.17), we get for all Therefore, the result is satisfied for all Thus

Acknowledgment

This work was supported by CONICET and UNSL grants.

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