# Extension of The Best Approximation  Operator in  Orlicz Spaces

Authors: Ivana Carrizo; Sergio Favier; Felipe Zó
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

Let(Ω,𝒜,μ) be a probability space and ℒ⊂𝒜 a sub-σ-lattice of the σ-algebra 𝒜. We study an extension of the best ϕ-approximation operator from an Orlicz space Lϕ to the space Lϕ′, where ϕ′ denotes the derivative of the convex, but not necessarily a  strictly convex function ϕ. We obtain convergence results when a sequence of σ-algebras ℬn converges to ℬ∞ in a suitable way.


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## Body

## 1. Introduction and Notations

Let(Ω,𝒜,μ) be a
probability space and let ℳ=ℳ(Ω,𝒜,μ) be the set of
all 𝒜-measurable
real-valued functions defined on Ω. Given a C1 convex function ϕ:[0,∞)→[0,∞) such that ϕ(0)=0,ϕ(t)>0,  when t>0, let Lϕ=Lϕ(Ω,𝒜,μ) be the space of
all functions f∈ℳ such
that (1.1)∫Ωϕ(λ|f|)dμ<∞, for some λ>0. Since we only
deal with a Δ2 function ϕ, that is, there
exists a constant K>0 such that ϕ(2x)≤Kϕ(x) for all x≥0, the space Lϕ can be
defined as the space of all
function f∈ℳ, where (1.1) holds for every positive number λ. The space Lϕ′ is analogously
defined, where ϕ′ is the
derivative of the function ϕ. Besides, observe that for a Δ2 function ϕ it holds the
next inequality (1.2)ϕ(x)≤xϕ′(x)≤ϕ(2x)≤Kϕ(x) for all x≥0, and therefore Lϕ⊂Lϕ′. Moreover φ is a Δ2 function if and
only if φ′ is a Δ2 function.We say, according to [1], that a collection ℒ of sets in 𝒜 is a σ-lattice if it
is closed under countable unions and intersections and contains ∅ and Ω. Given a σ-lattice ℒ, we denote by ℒ¯ the σ-lattice of all
the complementary sets of ℒ, that is, ℒ¯={Ac:A∈ℒ}. Denote by Lϕ(ℒ) all ℒ-measurable
functions in Lϕ.A setC⊂Lϕ is called ϕ-closed if and
only if fn∈C and fn↗f∈Lϕ or (fn↘f∈Lϕ) then f∈C. Then Lϕ(ℒ) is a ϕ-closed convex
set and a lattice, that is closed for the maximum and minimum of functions.We will use the notationf∧g=min⁡(f,g) and f∨g=max⁡(f,g). A set C⊂Lϕ is called a σ-complete
lattice if and only if ⋀n∈ℕfn=inf⁡n∈ℕfn∈C and ⋁n∈ℕfn=sup⁡n∈ℕfn∈C for all
sequence {fn}n∈ℕ⊂C.It is well known, see [1], that for every f∈Lϕ there exists an
element g∈Lϕ(ℒ) such
that (1.3)∫Ωϕ(|f−g|)dμ=inf⁡h∈Lϕ(ℒ)∫Ωϕ(|f−h|)dμ. Denote by μϕℒ(f) the set of all g∈Lϕ(ℒ) satisfying
(1.3). Each element of μϕℒ(f) will be called
a best ϕ-approximation
of f given Lϕ(ℒ), and we will
refer to the mapping f→μϕℒ(f) defined on Lφ as the best
approximation operator.It is showed in [1] that for f∈Lϕ the set μϕℒ(f) is a nonvoid σ-complete
lattice. Also it was proved that if f≤g, both in Lϕ,⁢f1∈μϕℒ(f) and g1∈μϕℒ(g) then we have f1∧f2∈μϕℒ(f) and f1∨f2∈μϕℒ(g). In this case,
we say that the multivalued operator μϕℒ(f) is a monotone
operator.The main purpose of this paper is to extend the best
approximation operator to the setLφ′. The case φ(t)=tp,⁢p>1, was extensively treated in [2] and the best L1 approximation
operator is extended to all measurable functions in [3]. The extension from Lφ to Lφ′ is considered in
[4] for a C1 function φ which is
strictly convex and φ′(0)=0.Now in this paper we consider aC1 convex function ϕ,ϕ′(0)=0, 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 isμϕℒ(f); for f∈Lϕ′, we will be denoted by μ˜ϕℒ(f). In Theorem
2.12, we prove that μ˜ϕℒ(f)≠∅ for every f∈Lϕ′, while in
Theorem 2.16 it is proved that it is indeed an extension, that is, μ˜ϕℒ(f)=μϕℒ(f) for f∈Lϕ. Additional
properties are obtained for the set μ˜ϕℒ(f) when the σ-lattice ℒ is a σ-algebra (see
Theorem 2.17) and similar results hold when ℒ is the class of
monotone functions in Lϕ′ (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 Bestϕ-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
functionϕ to satisfy the Δ2 condition is
that there exists a constant α>1 such that (2.1)uϕ′(u)≤αϕ(u)∀u≥0.Lemma 2.2.
Letf,⁢g be in Lϕ. Then ϕ′(|f|)g is an
integrable function.According to Brunk and Johansen [8], we set the following definitions.Definition 2.3.
Letν be a signed
measure on 𝒜 and let ℒ be a σ-lattice
contained in 𝒜.
Say that
P∈ℒ is a ν-positive set,
if for all D∈ℒ¯, then
ν(P∩D)≥0. A set N∈ℒ¯ is called ν-negative if
for all C∈ℒ
one has
ν(N∩C)≤0.Definition 2.4.
Let{νa}a∈ℝ be a family of
measures on 𝒜, and let ℒ be a σ-lattice
contained in 𝒜. An ℒ-measurable
function g is called a
Lebesgue-Radon-Nikodym function (LRN function) for {νa} given ℒ if and only if
the set {g>a} is νa-positive for
all a∈ℝ and the set {g<a} is νa-negative for
all a∈ℝ.Remark 2.5.
We note that in Definition2.4 it is sufficient to
impose the conditions for all a in a dense set
in  ℝ, see [8, page 588].Forf∈Lϕ,⁢g∈Lϕ(ℒ), and a∈ℝ, we define the following measures on 𝒜: (2.2)μg(A)=∫Aϕ̲′(f−g)dμ,μa(A)=∫Aϕ̲′(f−a)dμ, where ϕ̲′(x)=ϕ′(|x|)sign(x). Note that when f∈Lϕ′ and g∈Lϕ′(ℒ), the measure μg and μa are well
defined.The next theorem is a characterization ofμϕℒ(f), see [9, Theorem 3.2].Theorem 2.6.
Letf∈Lϕ,ℒ⊂𝒜 be a σ-lattice and g∈Lϕ(ℒ). Then the following statement are equivalent.
(1)
g∈μϕℒ(f).(2)
(a) a set{g>a} is μg-positive for
all a∈ℝ; and(b) a set{g<a} is μg-negative for
all a∈ℝ.(3)
g is an LRN
function for the family {μa}a∈ℝ given ℒ.Now we extend the operatorμϕℒ(⋅) to the space Lϕ′.Definition 2.7.
Letℒ be a σ-lattice
and let f∈Lϕ′. Then g is an extended
best ϕ-approximation
if and only if g∈Lϕ′(ℒ) and
(i)
the set{g>a} is μg-positive
for all a∈ℝ;(ii)
the set{g<a} is μg-negative
for all a∈ℝ.
For f∈Lϕ′ we denote by μ˜ϕℒ(f) the set of all
extended best ϕ-approximation
functions.Remark 2.8.
Letf∈Lϕ′ and let g be a function
in Lϕ′(ℒ) such that the
set {g>a} is μg-positive
and the set {g<a} is μg-negative
for all a∈ℝ. Then we have the following.
(i)
For allh∈L∞(ℒ¯) and h≥0, (2.3)∫{g>a}ϕ̲′(f−g)h⁢dμ≥0. For all h∈L∞(ℒ) and h≥0, (2.4)∫{g<a}ϕ̲′(f−g)h⁢dμ≤0.(ii)
(2.5)∫ϕ̲′(f−g)dμ=0.Proof.
We prove inequality (2.3). Since the
set {g>a} is μg-positive,
that is, for each D∈ℒ¯ and a∈ℝ, we have (2.6)∫{g>a}∩Dϕ̲′(f−g)dμ≥0. For h=∑k=1Nck𝒳Dk, where Dk∈ℒ¯ and ck≥0, k=1,…,N, then by (2.6), we have (2.7)∫{g>a}ϕ̲′(f−g)h⁢dμ=∑k=1Nck∫{g>a}∩Dkϕ̲′(f−g)dμ≥0. All nonnegative h∈L∞(ℒ¯) 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 a→∞ with h=1 in (2.4) and if a→−∞ consider in
(2.3) also h=1.As a reference, we note that (2) is
equivalent to (3) in Theorem2.6 for f∈Lϕ′,⁢g∈Lϕ′. We have the
next remark.Remark 2.9.
Forf∈Lϕ′ and g∈Lϕ′(ℒ), the following
statements are equivalent:
(1)
g∈μ˜ϕℒ(f);(2)
g is an LRN
function for the family {μa}a∈ℝ given ℒ.The next lemma is a particular case  of [8, Theorem 1.8].Lemma 2.10.
Letf∈Lϕ′ and g∈Lϕ′(ℒ). Then the following statements are equivalent.
(1)
g is an LRN
function for the family {μa}a∈ℝ given ℒ.(2)
There exists a
countable setD such that {g≤a} is μa-negative
for all a∈Dc and the set {g≥a} is μa-positive
for all a∈Dc.We need the following auxiliary result.Lemma 2.11.
Letfn∈Lϕ′ be a sequence
of functions such that fn↗f(fn↘f), where f∈Lϕ′. Let gn∈μ˜ϕℒ(fn) be such that gn↗g(gn↘g). Then g∈μ˜ϕℒ(f).Proof.
We will prove the result just for
the increasing case, the proof for the decreasing case follows the same
pattern. The functiong=lim⁡n→∞gn is obviously ℒ-measurable
function. Now, we prove that g∈Lϕ′, and it
satisfies (i) and (ii) of Definition 2.7. We have that (2.8)fn−gn≤f−gn≤f−g1. Using (2.5), (2.9)∫Ωϕ̲′(fn−gn)dμ=0foreachn∈ℕ. According to (2.8) and (2.9), we have, since ϕ̲′(x) is an
increasing function, (2.10)∫Ωϕ̲′(f−gn)dμ≥0. Since ϕ′ is a continuous
function and ϕ′(0)=0, we have by
(2.8) and Lebesgue's theorem (2.11)∫Ω(ϕ̲′(f−g)∨0)dμ≤∫Ω(ϕ̲′(f−g1)∨0)dμ. Now by (2.10) it
holds (2.12)∫Ω(−ϕ̲′(f−gn)∨0)dμ≤∫Ω(ϕ̲′(f−gn)∨0)dμ. Using Fatou in (2.12), we
have (2.13)∫Ω(−ϕ̲′(f−g)∨0)dμ≤∫Ω(ϕ̲′(f−g)∨0)dμ. Therefore, using (2.11) and
(2.12), we get g∈Lϕ′(Ω) and ∫Ωϕ̲′(f−g)dμ≥0.
Leta∈ℝ,⁢D∈ℒ¯, and C∈ℒ we know for
each n(2.14)∫{gn>a}∩Dϕ̲′(fn−gn)dμ≥0,∫{gn<a}∩Cϕ̲′(fn−gn)dμ≤0. Since {gn}n∈ℕ is an
increasing sequence, we get ⋃n∈ℕ{gn>a}={g>a}, and by (2.14),
we have (2.15)∫{g>a}∩Dϕ̲′(f−g)dμ≥0.Hence, the set {g>a} is μg-positive
for all a∈ℝ.
Now, fora∈ℝ and k,n∈ℕ, we define Bn={gn<a} and Bn,k={gn<a−1/k}. We have that,
for n→∞,Bn,k↘Ak for some 𝒜-measurable
set such that (2.16){g<a−1k}⊂Ak⊂{g≤a−1k}. We observe that 𝒳Ak→𝒳{g<a},a.e. Then taking
limit as k→∞,n→∞ and using
Lebesgue's theorem, we obtain (2.17)∫{g<a}∩Cϕ̲′(f−g)dμ=lim⁡k→∞∫ϕ̲′(f−g)𝒳Ak∩C⁢dμ=lim⁡k→∞⁢lim⁡n→∞∫Bn,k∩Cϕ̲′(f−g)dμ≤0.Theorem 2.12.
Letℒ be a σ-lattice
and f∈Lϕ′, then μ˜ϕℒ(f)≠∅.Proof.
Forf∈Lϕ′ we can define
the following sequences. For each m∈ℕ, let fm=f∨(−m) and when m→∞ we have fm↘f. Set fmn=(f∨(−m))∧n for all n,m in  ℕ, then we have fmn↗fm, when n→∞. Since for each m,n∈ℕ we have fmn∈Lϕ, there exist gmn∈μϕℒ(fmn). As μϕℒ(⋅) is a monotone
operator over Lϕ we can take a
new sequence that we call again gmn∈μϕℒ(fmn), such that gmn≤gmn+1 for all n∈ℕ.
Sincefmn≥fm+1n and using again
that μϕℒ(⋅) is a monotone
operator, we have g˜m+1n≤g˜mn, where g˜mn∈μϕℒ(fmn) is the sequence
defined by g˜1n=g1n and g˜m+1n=g˜mn∧gm+1n. Furthermore, it
is easy to check that g˜mn≤g˜mn+1.
Then, we have that for eachm∈ℕ that fmn↗fm, when n→∞, and since μϕℒ(fmn)⊂μ˜ϕℒ(fmn) we have g˜mn∈μ˜ϕℒ(fmn) and if we
define gm=lim⁡n→∞g˜mn by Lemma 2.11
we obtain gm∈μ˜ϕℒ(fm) and gm≥gm+1 for all m∈ℕ. If we take m→∞, we have fm↘f and by Lemma
2.11 we get g∈μ˜ϕℒ(f), where g=lim⁡m→∞gm.To see that the extended bestϕ-approximation
is an extension of the best ϕ-approximation
operator, we must prove μ˜ϕℒ(f)=μϕℒ(f) for every f∈Lϕ. First, we need
to prove the following lemmas.Lemma 2.13.
Letϕ be a C1 convex function
and assume that it satisfies the Δ2-condition.
Then (2.18)ϕ(a)+K2aϕ̲′(x−a)≤K+22ϕ(x), for a,x≥0, where K is the constant
for the Δ2 condition.Proof.
We consider two cases. First, we
assume0<x≤a. Since ϕ is Δ2-convex
function, we have that ϕ(a)=ϕ(a−x+x)≤(K/2)(ϕ(a−x)+ϕ(x)). Using xϕ′(x)≥ϕ(x) for all x, we
get (2.19)ϕ(x)+aϕ′(a−x)=ϕ(x)+(a−x+x)ϕ′(a−x)≥ϕ(x)+ϕ(a−x)+xϕ′(a−x)≥2Kϕ(a)+xϕ′(a−x)≥2Kϕ(a). Then we obtain (2.20)ϕ(a)−K2aϕ′(a−x)≤K2ϕ(x). For 0≤a<x, we have (2.21)ϕ(a)+aK2ϕ′(x−a)≤ϕ(a)+K2∫x−axϕ′(t)dt≤ϕ(x)+K2ϕ(x)=K+22ϕ(x).Lemma 2.14.
Letf∈Lϕ and g∈μ˜ϕℒ(f), then (2.22)∫{g>0}ϕ̲′(f−g)gdμ≥0.Proof.
Since{g>a} is μg-positive
for all a∈ℝ, then for all D∈ℒ¯, we have that (2.23)∫{g>a}∩Dϕ̲′(f−g)dμ≥0. In particular, it holds that for
all a∈ℝ, (2.24)∫{g>a}ϕ̲′(f−g)dμ≥0, that is, (2.25)∫{g>a}∩{f>g}ϕ′(|f−g|)dμ≥∫{g>a}∩{f≤g}ϕ′(|f−g|)dμ. Now, we have (2.26)∫0∞∫{g>a}∩{f>g}ϕ′(|f−g|)dμda≥∫0∞∫{g>a}∩{f≤g}ϕ′(|f−g|)dμ⁢da. By the Fubini's theorem, we get (2.27)∫{g>0}∫0gϕ′(|f−g|)𝒳{f>g}dadμ≥∫{g>0}∫0gϕ′(|f−g|)𝒳{f≤g}da⁢dμ. Thus (2.28)∫{g>0}ϕ′(|f−g|)𝒳{f>g}g⁢dμ≥∫{g>0}ϕ′(|f−g|)𝒳{f≤g}gdμ. To see that inequality (2.22) is
equivalent to (2.28) we will prove that ϕ′(f−g)𝒳{f>g}g∈L1({g>0}). In fact (2.29)∫{g>0}ϕ′(f−g)𝒳{f>g}g⁢dμ≤∫{g>0}ϕ′(f−g)𝒳{f>g}fdμ≤∫{g>0}ϕ′(f)f.Since f∈Lϕ by Lemma 2.2,
the last integral is finite.The following properties of the setμ˜ϕℒ(f) can be easily
proved.Proposition 2.15.
Letf∈Lϕ′, then
(1)
−μ˜ϕℒ¯(−f)=μ˜ϕℒ(f),(2)
μ˜ϕℒ(f+t)=μ˜ϕℒ(f)+t for all h∈ℝ.Now we prove that the operatorμ˜ϕℒ(f) is in fact an
extension of the operator μϕℒ(f).Theorem 2.16.
Letf∈Lϕ, then μ˜ϕℒ(f)=μϕℒ(f).Proof.
Forf∈Lϕ, we will prove
only that μ˜ϕℒ(f)⊂μϕℒ(f). The other
inclusion follows from Theorem 2.6. Let g∈μ˜ϕℒ(f) and again using
Theorem 2.6 it remains to prove that g∈Lϕ. Recall that ϕ(0)=0, then (2.30)∫Ωϕ(|g|)dμ=∫{g>0}ϕ(g)dμ+∫{g<0}ϕ(−g)dμ. By Lemma 2.13 we obtain the
following inequality: (2.31)∫{g>0}ϕ(g)dμ+K2∫{g>0}ϕ̲′(|f|−g)gdμ≤K+22∫{g>0}ϕ(|f|)dμ. Applying Lemma 2.14 and taking
into account that ϕ̲′(x) is an
increasing function, we get (2.32)0≤∫{g>0}ϕ̲′(f−g)gdμ≤∫{g>0}ϕ̲′(|f|−g)gdμ. Thus using (2.32) in (2.31), we
have (2.33)∫{g>0}ϕ(g)dμ≤K+22∫Ωϕ(|f|)dμ. For the set {g<0}, again by Lemma
2.13, we obtain (2.34)∫{g<0}ϕ(−g)dμ+K2∫{g<0}ϕ̲′(|f|+g)(−g)dμ≤K+22∫{g<0}ϕ(|f|)dμ. Since −f+g≤|f|+g, we have ϕ̲′(−f+g)≤ϕ̲′(|f|+g). Thus (2.35)∫{g<0}ϕ̲′(−f+g)(−g)dμ≤∫{g<0}ϕ̲′(|f|+g)(−g)dμ. By (1) in Proposition 2.15, −g∈μ˜ϕℒ¯(−f), and by Lemma
2.14, we have that (2.36)∫{g<0}ϕ̲′(−f+g)(−g)dμ≥0. Therefore, ∫{g<0}ϕ(−g)dμ≤((K+2)/2⁢)∫ϕ(|f|)dμ. By (2.33), we have (2.37)∫Ωϕ(|g|)dμ≤K+22∫Ωϕ(|f|)dμ, and therefore g∈Lϕ.Now, if consider aσ-subalgebra ℬ instead of a σ-lattice ℒ, the extended best ϕ-approximation
operator has the following properties.Theorem 2.17.
Letf,⁢f1,andf2 be in Lϕ′, if ℬ is a sub-σ-algebra
of the σ-algebra 𝒜, then
the following hold.
(1)
The set-valued
functionμ˜ϕℬ(f) is a monotone
operator.(2)
The setμ˜ϕℬ(f) is a σ-complete
lattice, and there exist Uℬ,Vℬ∈μ˜ϕℬ(f) such that Uℬ≤g≤Vℬ a.e. for every g∈μ˜ϕℬ(f).Proof.
To prove (1), recall that this
set-valued operator is monotone iff1≤f2; then if g1∈μ˜ϕℬ(f1) and g2∈μ˜ϕℬ(f2), we have that g1∧g2∈μ˜ϕℬ(f1) and g1∨g2∈μ˜ϕℬ(f2). Since Lϕ(ℬ) is a lattice,
we know g1∧g2∈Lϕ(ℬ) and g1∨g2∈Lϕ(ℬ). We will prove
first that g1∧g2∈μ˜ϕℬ(f1). Set μafi(A)=∫Aϕ̲′(fi−a)dμ, where a∈ℝ and i=1,2. We will see
that g1∧g2 is an LRN function for the family of measures {μaf1}a∈ℝ given ℬ. First, we will
see that for each a∈ℝ and for all B∈ℬ, we have (2.38)μaf1({g1∧g2>a}∩B)≥0. Since {g1∧g2>a}∩B={g1>a}∩{g2>a}∩B and {g2>a}∩B∈ℬ and using that g1 is an LRN
function of the family {μaf1}a∈ℝ, we obtain that for all B∈ℬ(2.39)∫{g1∧g2>a}∩Bϕ̲′(f1−a)dμ=∫{g1>a}∩{g2>a}∩Bϕ̲′(f1−a)dμ≥0. Now, we see that {g1∧g2<a} is μaf1-negative
for all a∈ℝ. For B∈ℬ, we have (2.40){g1∧g2<a}∩B=({g1<a}∩B)∪({g1≥a}∩{g2<a}∩B). Using f1≤f2 and that ϕ̲′(⋅) is a
nondecreasing function, we obtain (2.41)∫{g1∧g2<a}∩Bϕ̲′(f1−a)dμ=∫{g1<a}∩Bϕ̲′(f1−a)dμ+∫{g1≥a}∩{g2<a}∩Bϕ̲′(f1−a)dμ≤∫{g1<a}∩Bϕ̲′(f1−a)dμ+∫{g1≥a}∩{g2<a}∩Bϕ̲′(f2−a)dμ≤0. Thus (2.42)∫{g1∧g2<a}∩Bϕ̲′(f1−a)dμ≤0. By (2.39) and (2.42), we have g1∧g2∈μ˜ϕℬ(f1).
Now we showg1∨g2∈μ˜ϕℬ(f2). Since f1≤f2 and {g1>a} is a μaf1-positive
for all a∈ℝ and for all B∈ℬ, we
have (2.43)∫{g1∨g2>a}∩Bϕ̲′(f2−a)dμ=∫{g2>a}∩Bϕ̲′(f2−a)dμ+∫{g1>a}∩{g2≤a}∩Bϕ̲′(f2−a)dμ≥∫{g2>a}∩Bϕ̲′(f2−a)dμ+∫{g1>a}∩{g2≤a}∩Bϕ̲′(f1−a)dμ≥0. Since (2.44)∫{g1∨g2<a}∩Bϕ̲′(f2−a)dμ=∫{g1<a}∩{g2<a}∩Bϕ̲′(f2−a)dμ≤0, the inequalities (2.43) and (2.44) prove that g1∨g2∈μ˜ϕℬ(f2).
As the statement (1) proves in particular thatμ˜ϕℬ(f) is a lattice,
we will see that the set is a σ-complete
lattice. Given a sequence {gn}n∈ℕ in μ˜ϕℬ(f), we have that ⋁1ngk∈μ˜ϕℬ(f); then, from
Lemma 2.11 we obtain that ⋁n∈ℕgn=lim⁡n→∞⋁1ngn∈μ˜ϕℬ(f). The proof ⋀n∈ℕgn∈μ˜ϕℬ(f) is similar.
By [10, Proposition II.4.1] there exists a sequence gn∈μ˜ϕℬ(f) such that infgn≤g≤sup⁡gn, for every g∈μ˜ϕℬ(f). Set Uℬ=inf⁡gn and Vℬ=sup⁡gn, then Uℬ and Vℬ are in μ˜ϕℬ(f) since this set
is a σ-complete
lattice.


## 3. Extended Bestϕ-Approximation with Nondecreasing Functions

When the
approximation class is the monotone functions defined on[0,1] we can obtain
similar results as those of Theorem 2.17. Now Ω=[0,1],μ is Lebesgue
measure on the measurable sets, and ℒ={(a,1),[a,1),∅,ℝ}a∈ℝ. Therefore, Lϕ′(ℒ) is the set of
nondecreasing functions in Lϕ′[0,1].Remark 3.1.
Letg be a
nondecreasing function on [0,1]. Given a∈ℝ, the set {g<a} is one of the
intervals [0,αa) or [0,αa] and similarly
the set {g>a} is (βa,1] or [βa,1] with 0≤αa≤βa≤1. Then Hg={a∈ℝ:αa=βa} is a dense set
in  ℝ. In fact, the
complement set of Hg is a countable set.Note that eachC∈ℒ is of the form (c,1] or [c,1] and D∈ℒ¯ is D=[0,d) or [0,d]. Thus (★)∫{g>a}∩Dϕ̲′(f−a)dμ=∫βadϕ̲′(f−a)dμ,∫{g<a}∩Cϕ̲′(f−a)dμ=∫cαaϕ̲′(f−a)dμ.Theorem 3.2.
LetLϕ(ℒ) be the class of
the ϕ′-integrable
nondecreasing functions in [0,1]. Then
the following hold.
(1)
The set mappingμ˜ϕℒ(f) is a monotone
operator.(2)
For everyf∈Lϕ′, the set μ˜ϕℒ(f) is a σ-complete
lattice.Proof.
First we prove (1), that is givenf1, f2 in Lϕ′ with f1≤f2, for each gi∈μ˜ϕℒ(fi),i=1,2 we will see
that g1∧g2∈μ˜ϕℒ(f1) and g1∨g2∈μ˜ϕℒ(f2). Let H be Hg1∩Hg2, where Hgi is the set
given in Remark 3.1. Recall that g1∧g2∈μ˜ϕℒ(f1) if and only if
for each a∈H and c,d∈ℝ(3.1)∫{g1∧g2>a}∩(0,d)ϕ̲′(f1−a)dx≥0,∫{g1∧g2<a}∩(c,1)ϕ̲′(f1−a)dx≤0. Also g1∨g2∈μ˜ϕℒ(f2) if and only if
for each a∈H, and c,d∈ℝ, (3.2)∫{g1∨g2>a}∩(0,d)ϕ̲′(f2−a)dx≥0,∫{g1∨g2<a}∩(c,1)ϕ̲′(f2−a)dx≤0. First we prove (3.1). Now we see
that (3.3)∫{g1∧g2>a}∩(0,d)ϕ̲′(f1−a)dx≥0, with {g1∧g2>a}=(β1a,1]∩(β2a,1]. Since ∫(β1a,d)ϕ̲′(f1−a)dx≥0 to prove (3.3),
we have to see that (3.4)∫(β2a,d)ϕ̲′(f1−a)dx≥0, where β1a≤β2a. Indeed by (★), we
get (3.5)0≤∫β1aadϕ̲′(f1−a)dx=∫β1aβ2aϕ̲′(f1−a)dx+∫β2adϕ̲′(f1−a)dx. Since ϕ̲′(⋅) is a
nondecreasing function, we have (3.6)0≤∫β1aβ2aϕ̲′(f2−a)dx+∫β2adϕ̲′(f1−a)dx. As ∫β1aβ2aϕ̲′(f2−a)dx≤0 (β2a=α2a), we have ∫β2adϕ̲′(f1−a)dx≥0, that is (3.4).
Now we will prove that(3.7)∫{g1∧g2<a}∩(c,1)ϕ̲′(f1−a)dx≤0. In fact (3.8)∫{g1∧g2<a}∩(c,1)ϕ̲′(f1−a)dx=∫{g1<a}∩(c,1)ϕ̲′(f1−a)dx+∫{g1≥a}∩{g2<a}∩(c,1)ϕ̲′(f1−a)dx≤∫{g1<a}∩(c,1)ϕ̲′(f1−a)dx+∫{g2<a}∩[{g1≥a}∩(c,1)]ϕ̲′(f2−a)dx. The last two integrals in (3.8)
are less or equal than zero, so (3.7) holds. A similar argument shows that g1∨g2∈μ˜ϕℒ(f2). Therefore, the
extended best ϕ-approximation
operator is a monotone operator. By (1), we have that μ˜ϕℒ(f) is a lattice,
just setting f=f1=f2. Now by Lemma
2.11 we obtain that μ˜ϕℒ(f) is a σ-complete lattice.


## 4. A Limit Theorem for Extended Bestϕ-Approximations

Given a sequence{ℬn}n∈ℕ of σ-algebras
contained in the σ-algebra 𝒜, we consider two
cases, ℬn⊂ℬn+1 for all n∈ℕ and we set ℬ∞ for the σ-algebra
generated by ⋃nℬn, and if ℬn⊇ℬn+1 for all n∈ℕ, we set ℬ∞=⋂n∈ℕℬn.The next result is a particular case of [8, Theorem
2.8] when ϕ is a strictly
convex function. This assumption on the function ϕ assures that
the family of measures {μa}a∈ℝ decreases at
zero as required by Brunk and Johansen in that theorem.Theorem 4.1.
Let{ℬn}n∈ℕ be an
increasing or decreasing sequence of σ-algebras in 𝒜, and let ℬ∞ be the limit of
the sequence. If f∈Lϕ′, then we have for all gn∈μ˜ϕℬn(f),n∈ℕ, that lim inf⁢n→∞gn,andlim sup⁡n→∞gn are in μ˜ϕℬ∞(f).Proof.
Defineg¯=lim sup⁡n→∞gn and g¯=lim inf⁡n→∞gn, then we only
prove that g¯,g¯∈μ˜ϕℬ∞(f), when {ℬn}n∈ℕ is an
increasing sequence of σ-algebras, the
proof for the decreasing case is similar.
First, we prove for eachf∈Lϕ′ that the set {g¯≥a} is μa-positive for
all a∈ℝ. Let B∈ℬm, and for Hn=⋃k≥n{gk>a−ϵn}, where ϵn decreases to
zero, we have that {g¯≥a}=⋂n≥1Hn, and for all n∈ℕ,Hn+1⊂Hn. Now for each n∈ℕ, we define the following disjoint sets. For p≥n, set Hn,n={gn>a−ϵn},…,Hn,p={gp>a−ϵn}∩{gp−1≤a−ϵn}∩⋯∩{gn≤a−ϵn}. Thus, Hn=⋃p≥nHn,p, and then (4.1)∫Hn∩Bϕ̲′(f−(a−ϵn))dμ=∑p=n∞∫Hn,p∩Bϕ̲′(f−(a−ϵn))dμ. As ℬn⊂ℬn+1, and for p≥n, we have that {gp−1≤a−ϵn}∩⋯∩{gn≤a−ϵn}∈ℬp. As B∈ℬm then B∈ℬp and ∫Hn,p∩Bϕ̲′(f−(a−ϵn))dμ≥0 for m≤n≤p. Thus ∫Hn∩Bϕ̲′(f−(a−ϵn))dμ≥0, and by
Lebesgue's theorem, we get (4.2)∫{g¯≥a}∩Bϕ̲′(f−a)dμ=lim⁡n→∞∫Hn∩Bϕ̲′(f−(a−ϵn))dμ≥0 for all B∈⋃n∈ℕℬn. Now we have
(4.2) for all B∈ℬ∞. In fact, the
set D={B∈𝒜:∫{g¯≥a}∩Bϕ̲′(f−a)dμ≥0} is a monotone
class, that is, the set D is closed for
increasing and decreasing sequences of sets. As ⋃n∈ℕℬn⊂D and this union
is an algebra of sets, the monotone class generated by it is ℬ∞, that is, ℬ∞⊂D.
Now let us prove that the set{g¯≤a} is μa-negative for
all a∈ℝ. As {g¯≤a}=⋂n∈ℕ{g¯<a+1/n} and (4.3)∫{g¯≤a}∩Bϕ̲′(f−a)dμ=lim⁡n→∞∫{g¯<a+1/n}∩Bϕ̲′(f−a)dμ, then we have to prove that for all a∈ℝ, the set {g¯<a} is μa-negative.
Since {g¯<a}=⋃n≥1⋂k≥n{gk<a−ϵn}=⋃n≥1Hn, where Hn=⋂k≥n{gk<a−ϵn} and ϵn↘0, we have for all n∈ℕ that Hn⊂Hn+1. Then (4.4)∫{g¯<a}∩Bϕ̲′(f−a)dμ=lim⁡n→∞∫Hn∩Bϕ̲′(f−(a−ϵn))dμ, for a fixed set B∈ℬm.
SetGn=⋂k≥n+1{gk<a−ϵn} and note that Hn=Gn∩{gn<a−ϵn}. Then for m≤n, we
have (4.5)∫{gn<a−ϵn}∩B∩Gnϕ̲′(f−(a−ϵn))dμ+∫{gn<a−ϵn}∩B∩Gncϕ̲′(f−(a−ϵn))dμ=∫{gn<a−ϵn}∩Bϕ̲′(f−(a−ϵn))dμ≤0. Now, we prove the following
inequality: (4.6)∫{gn<a−ϵn}∩B∩Gncϕ̲′(f−(a−ϵn))dμ≥0. We can see that (4.7){gn<a−ϵn}∩B∩Gnc=⋃k≥n+1Ak,where Ak are the
following disjoint sets (4.8)An+1={gn+1≥a−ϵn}∩B∩{gn<a−ϵn},⋮Ak={gk≥a−ϵn}∩⋂i=n+1k−1{gi<a−ϵn}∩{gn<a−ϵn}∩B. Then (4.9)∫{gn<a−ϵn}∩B∩Gncϕ̲′(f−(a−ϵn))dμ=∑k≥n+1∫Akϕ̲′(f−(a−ϵn))dμ. Since Ak={gk≥a−ϵn}∩Bk, where Bk∈ℬk, we have (4.6).
Therefore by (4.5), we have (4.10)∫Hn∩Bϕ̲′(f−(a+ϵn))dμ≤0, for all B∈⋃n∈ℕℬn. Thus by (4.10),
(4.4), and (4.3), we get (4.11)∫{g¯≤a}∩Bϕ̲′(f−a)dμ≤0 for all B∈⋃n∈ℕℬn. Therefore, the
result is satisfied for all B∈ℬ∞. Thus g¯∈μ˜ϕℬ∞(f).
We have{g¯≤a}=⋂n≥1Hn, where Hn=⋃k≥n{gk<a+ϵn} and ϵn↘0, then Hn+1⊂Hn for all n∈ℕ. Since f∈Lϕ′ we have for all B∈ℬ∞ that (4.12)∫{g¯≤a}∩Bϕ̲′(f−a)dμ=lim⁡n→∞∫Hn∩Bϕ̲′(f−(a+ϵn))dμ. For p>n define the
following disjoint sets Hn,n={gn<a+ϵn} and Hn,p={gp<a+ϵn}∩{gp−1≥a+ϵn}∩⋯∩{gn≥a+ϵn}. Then for B∈ℬm we
have (4.13)∫Hn∩Bϕ̲′(f−(a+ϵn))dμ=∑p≥n∫Hn,p∩Bϕ̲′(f−(a+ϵn))dμ. Now if m≤n≤p,Hn,p∩B={gk<a+ϵn}∩B⋆, where B⋆∈ℬp. Thus (4.14)∫Hn,p∩Bϕ̲′(f−(a+ϵn))dμ≤0. Then by (4.12) and (4.14), we have
for all B∈⋃n∈ℕℬn that (4.15)∫{g¯≤a}∩Bϕ̲′(f−a)dμ≤0. Therefore, we have (4.15) for
all B∈ℬ∞.
Let us see now that(4.16)∫{g¯≥a}∩Bϕ̲′(f−a)dμ≥0 for all B∈⋃n∈ℕℬn.
As{g¯≥a}=⋂n∈ℕ{g¯>a−1/n}, we have to
prove that for all a∈ℝ, the set {g¯>a} is μa-positive. We
have that {g¯>a}=⋃n≥1⋂k≥n{gk≥a+ϵn}=⋃n≥1Hn, where Hn is the
increasing sequence Hn=⋂k≥n{gk≥a+ϵn} and ϵn↘0. Then we
have (4.17)∫{g¯>a}∩Bϕ̲′(f−a)dμ=lim⁡n→∞⁢∫Hn∩Bϕ̲′(f−(a+ϵn))dμ. Set Gn=⋂k≥n+1{gk≥a+ϵn} and note that Hn=Gn∩{gn≥a+ϵn}. Then for B∈ℬm,⁢m≤n, we
have (4.18)0≤∫{gn≥a+ϵn}∩Bϕ̲′(f−(a+ϵn))dμ=∫{gn≥a+ϵn}∩B∩Gnϕ̲′(f−(a+ϵn))dμ+∫{gn≥a+ϵn}∩B∩Gncϕ̲′(f−(a+ϵn))dμ. Now, we prove (4.19)∫{gn≥a+ϵn}∩B∩Gncϕ̲′(f−(a+ϵn))dμ≤0. We can see that (4.20){gn≥a+ϵn}∩B∩Gnc=⋃k≥n+1Ak, where Ak are the
following disjoint sets: (4.21)An+1={gn+1<a+ϵn}∩B∩{gn≥a+ϵn},⋮Ak={gk<a+ϵn}∩⋂i=n+1k−1{gi≥a+ϵn}∩{gn≥a+ϵn}∩B.Then (4.22)∫{gn≥a+ϵn}∩B∩Gncϕ̲′(f−(a+ϵn))dμ=∑k≥n+1∫Akϕ̲′(f−(a+ϵn))dμ. Since Ak={gk<a+ϵn}∩Bk, where Bk∈ℬk, we have (4.19).
Therefore by (4.18), we have (4.23)∫Hn∩Bϕ̲′(f−(a+ϵn))dμ≥0 for all B∈⋃n∈ℕℬn. Thus by
(4.23) and (4.17), we get (4.24)∫{g¯≥a}∩Bϕ̲′(f−a)dμ≥0 for all B∈⋃n∈ℕℬn. Therefore, the
result is satisfied for all B∈ℬ∞. Thus g¯∈μ˜ϕℬ∞(f).



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*Source: 374742-2008-03-24.xml*