Abstract and Applied Analysis

Volume 2008 (2008), Article ID 374742, 15 pages

http://dx.doi.org/10.1155/2008/374742

## 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 [5–7].

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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