View of Simultaneous proximinality in \(L^{\infty}(\mu,X)\)

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Rev. Anal. Num´er. Th´eor. Approx., vol. 42 (2013) no. 2, pp. 85–93 ictp.acad.ro/jnaat

SIMULTANEOUS PROXIMINALITY IN L^∞(µ, X)

EYAD ABU-SIRHAN^∗

Abstract. Let X be a Banach space and G be a closed subspace of X. Let us denote byL^∞(µ, X) the Banach space of all X-valued essentially bounded functions on aσ-finite complete measure space (Ω,Σ, µ).In this paper we show that ifGis separable, thenL^∞(µ, G) is simultaneously proximinal inL^∞(µ, X) if and only ifGis simultaneously proximinal inX.

MSC 2000. 41A28.

Keywords. Simultaneous approximation, Banach spaces.

1. INTRODUCTION

LetZ be a Banach space andY be a closed subspace of Z. For a subset B ofZ, define

d(B, Y) = inf

y∈Y sup

b∈B

kb−yk.

An element y^∗ ∈ Y is said to be a best simultaneous approximant to the subsetB if

sup

b∈B

kb−y^∗k=d(B, Y).

Definition 1.1. If every finite subset of Z admits a best simultaneous approximation in Y, then Y is said to be simultaneously proximinal inZ.

The theory of best simultaneous approximation has been studied by many authors. Most of these works have dealt with the space of continuous functions with values in a Banach space e.g. [18, 10, 4]. Some recent results for best simultaneous approximation in the Banach space ofP-Bochner integrable (essentially bounded) functions have been obtained in [5, 6, 9, 16, 19]. We consider here a problem of simultaneous approximation completing the work done in [6,19]. In this paper (Ω,Σ, µ) stands for a complete σ-finite measure space andL^∞(µ, X) the Banach space of all essentially bounded functions on (Ω,Σ, µ) with values in a Banach space X, endowed with the usual norm

kfk_∞= ess supkf(t)k.

∗Department of Mathematics, Tafila Technical University, Tafila, Jordan, e-mail:

[email protected].

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In [19], it is shown that if G is a reflexive subspace of a Banach space X, then L^∞(µ, G) is simultaneously proximinal inL^∞(µ, X).In [6], it is shown that ifGis a separablew^∗-closed subspace of a dual spaceX, thenL^∞(µ, G) is simultaneously proximinal in L^∞(µ, X). The aim of this paper is to show that ifG is a closed separable subspace of a Banach spaceX, thenL^∞(µ, G) is simultaneously proximinal in L^∞(µ, X) if and only if G is simultaneously proximinal in X.

2. PRELIMINARY RESULTS

The following Theorem is a generalization of the distance formula given in [6]. Here χ_A is denoted to the characteristic function of A.

Theorem2.1. LetX be a Banach space,G be a closed subspace of X, and f1, f2, ..., fm be any finite number of elements inL^∞(µ, X).Then the function

s→d({f_i(s) : 1≤i≤m}, G),

which we denote by d({f_i(·) : 1≤i≤m}, G), is measurable, and d({f_i: 1≤i≤m}, L^∞(µ, G)) =kd({f_i(s) : 1≤i≤m}, G)k_∞ Proof. Letf1, f2, ..., fm∈L^∞(µ, X). Being strongly measurable functions, there exist sequences of simple functions (f_(i,n))^∞_n=1, i= 1,2, ..., m,such that

lim

f_(i,n)(t)−f_i(t) = 0,

fori= 1,2, ..., m, and for almost allt’s. We may write, [9],

f_(i,n) =

k(n)

X

j=1

χ_A(n,j)(·)x_(i,n,j) , i= 1,2, ..., m,

whereA(n, j) are disjoint and ^k(n)∪

j=1A(n, j) = Ω.Define dn(·) : Ω→Rby dn(s) =d

f_(i,n)(s) : 1≤i≤m , G .

Then

d_n(s) =

k(n)

X

j=1

χ_A(n,j)d

x_(i,n,j): 1≤i≤m , G and

limd_n(s) =d({f_i(s) : 1≤i≤m}, G), for almost all s. Thus d({f_i(·) : 1≤i≤m}, G) is measurable.

Now, for any h∈L^∞(µ, G),

ess supd({f_i(s) : 1≤i≤m}, G)≤ess sup sup

1≤i≤m

kf_i(s)−h(s)k

= sup

1≤i≤m

kf_i−hk_∞

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Hence

ess supd({f_i(s) : 1≤i≤m}, G)≤d({f_i: 1≤i≤m}, L^∞(µ, G)). To prove the reverse inequality, let > 0 be given and wi, i = 1,2, ..., m, be countably valued functions in L^∞(µ, X) such that

kf_i−w_ik_∞< ₃. We may write w_i =P∞

k=1χ_A_k(·) x_(i,k), whereA_k are disjoint, ^∞∪

k=1A_k = Ω, and µ(A_k)>0, for allk.Leth_k∈G be such that

sup

1≤i≤m

x_(i,k)−h_k < d

x_(i,k): 1≤i≤m , G +₃, for all k.Letg=P∞

k=1χ_A_k(·)h_k.It is clear thatg∈L^∞(µ, G).Further d({f_i : 1≤i≤m}, L^∞(µ, G))≤

≤ sup

1≤i≤m

kf_i−wik_∞+d({w_i: 1≤i≤m}, L^∞(µ, G))

≤ ₃ + sup

1≤i≤m

kg−w_ik_∞

= ₃ + ess sup sup

1≤i≤m

kg(t)−w_i(t)k

= ₃ + ess sup sup

1≤i≤m

∞

X

k=1

χ_A_k(t)

x_(i,k)−h_k

= ₃ + ess sup

∞

X

k=1

χAk(t) sup

1≤i≤m

x_(i,k)−hk

< ²₃ + ess sup

∞

X

k=1

χA_k(t)d

x_(i,k): 1≤i≤m , G

= ²₃ + ess supd({w_i(t) : 1≤i≤m}, G)

≤ ²₃ + ess sup

d({f_i(t) : 1≤i≤m}, G) + sup

1≤i≤m

kf_i(t)−wi(t)k

≤ ²₃ + ess supd({f_i(t) : 1≤i≤m}, G) + sup

1≤i≤m

kf_i−wik_∞

< + ess supd({f_i(t) : 1≤i≤m}, G).

Corollary2.2. LetXbe a Banach space,Gbe a closed subspace ofX, and f₁, f₂, ..., f_m be any finite number of elements in L^∞(µ, X). Let g : Ω → G be a measurable function such that g(s) is a best simultaneous approximation of f1(s), f2(s), ..., fn(s) for almost all s. Then g is a best simultaneous approximation of f₁, f₂, ..., f_n in L^∞(µ, G) (and therefore g∈L^∞(µ, G)).

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Proof. Assume that g(s) is a best simultaneous approximation of f₁(s), f₂(s), ..., f_m(s) for almost all s.Then

sup

1≤i≤m

kf_i(s)−g(s)k ≤ sup

1≤i≤m

kf_i(s)−zk, for almost all s, and for all z∈G. Then

kg(s)k ≤2 sup

1≤i≤m

kf_i(s)k ≤2 sup

1≤i≤m

kf_ik_∞, for almost all s,therefore g∈L^∞(µ, G).By Theorem 2.1,

d({f_i: 1≤i≤m}, L^∞(µ, G)) = ess supd({f_i(s) : 1≤i≤m}, G)

= ess sup sup

1≤i≤m

kf_i(s)−g(s)k

= sup

1≤i≤m

kf_i−gk_∞.

Therefore g is a best simultaneous approximation for f1, f2, .., fm in

L^∞(µ, G).

The condition in Corollary 2.2 is sufficient;g(s) is a best simultaneous approximation off1(s), f2(s), ..., fm(s) for almost allsinG,impliesgis a best simultaneous approximation of f₁, f₂, ..., f_m in L^∞(µ, G). For the converse, we need the following easy lemma.

Lemma2.3. LetX be a Banach space,Gbe a closed subspace of X, A⊂Ω be such that µ(A)>0, and f1, f2, ..., fm ∈ L^∞(µ, X) be such that

d({f_i(s) : 1≤i≤m}, G) =

(1, if s∈A 0, if s∈ΩrA.

Then d({f_i: 1≤i≤m}, L^∞(µ, G)) = 1.

Proof. Letg∈L^∞(µ, G),then sup

1≤i≤m

kf_i(s)−g(s)k ≥d({f_i(s) : 1≤i≤m}, G), for all s∈Ω.

ess sup sup

1≤i≤m

kf_i(s)−g(s)k ≥ess supd({f_i(s) : 1≤i≤m}, G)

= 1.

Thus sup

1≤i≤m

kf_i−gk_∞≥1.Since g∈L^∞(µ, G) was arbitrary, then d({f_i: 1≤i≤m}, L^∞(µ, G))≥1.

To prove the converse inequality. Let > 0 be given. Let f₁⁰, f₂⁰, ..., f_m⁰ ∈ L^∞(µ, X) be countably valued functions such that f_i⁰(Ω) ⊂ f_i(Ω), i= 1,2, . . . , m, and

f_i⁰ −fi

< ₃.

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We may write f_i⁰ = P∞

k=1χ_A_k(·) x_(i,k), with the subsets A_k disjoint and measurable, and x_(i,k)∈ f_i(Ω).For eachktake h_k ∈Gsuch that

sup

1≤i≤m

x_(i,k)−hk

< d

x_(i,k): 1≤i≤m , G +₃

≤1 +₃. It is clear that g defined by

g=

∞

X

k=1

χA_k(·)hk

belongs to L^∞(µ, G) and

d {f_i: 1≤i≤m}, L^∞(µ, G)

≤

≤ sup

1≤i≤m

fi−f_i⁰

∞+d

n

f_i⁰ : 1≤i≤m o

, L^∞(µ, G)

≤ ₃ + sup

1≤i≤m

g−f_i⁰

_∞

= ₃ + ess sup sup

1≤i≤m

g(t)−f_i⁰(t)

= ₃ + ess sup sup

1≤i≤m

∞

X

k=1

χ_A_k(t)

x_(i,k)−h_k

< ²₃ + ess sup

∞

X

k=1

χAk(t)d

x_(i,k): 1≤i≤m , G

≤+ ess supd({f_i(t) : 1≤i≤m}, G)

=+ 1.

Therefore,

d({f_i : 1≤i≤m}, L^∞(µ, G))≤+ 1.

Theorem 2.4. Let X be a Banach space and G be a closed subspace of X. Then L^∞(µ, G) is simultaneously proximinal in L^∞(µ, X) if and only if for any finite number of elements f1, f2, ..., fm in L^∞(µ, X), there exists g ∈ L^∞(µ, G) such that g(s) is a best simultaneous approximation of f1(s), f2(s), ..., fn(s) for almost all s.

Proof. Sufficiency of the condition is an immediate consequence of Corollary 2.2. We will show the necessity. Assume that L^∞(µ, G) is simultaneously proximinal in L^∞(µ, X) and take f₁, f₂, ..., f_m in L^∞(µ, X). Consider the non-negative measurable function

h: Ω→[0,∞)

s→d({f_i(s) : 1≤i≤m}, G).

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Take Ω0 = {s∈Ω :h(s) = 0}, and for each n = 1,2, ..., take Ω_n = {s∈Ω :n−1< h(s)≤n}. Of course, we may forget those Ω_n which areµ−null sets, so, without loss of generality, we will assume thatµ(Ωn)>0 for all n. Now for eachn= 1,2, ..., we definef_in : Ω→X by

fin(s) = ( ₁

h(s)f_i(s), if s∈Ω_n 0, if s∈ΩrΩn. It is clear that fin ∈L^∞(µ, X), i= 1,2, ..., m, and also that

d({f_in(s) : 1≤i≤m}, G) =dn

1

h(s)f_i(s) : 1≤i≤mo , G

= _h(s)¹ d({f_i(s) : 1≤i≤m}, G)

= 1,

for all s∈Ω_n.So, it follows from proceeding lemma that d({f_in: 1≤i≤m}, L^∞(µ, G)) = 1.

On the other hand, using simultaneous proximinality of L^∞(µ, G),we de- duce that there exists g_n∈L^∞(µ, G) such that

sup

1≤i≤m

kf_in−gnk_∞=d({f_in: 1≤i≤m}, L^∞(µ, G))

= 1.

Therefore, we have

1 =d({f_in(s) : 1≤i≤m}, G)

≤ sup

1≤i≤m

kf_in(s)−g_n(s)k

≤ sup

1≤i≤m

kf_in−g_nk_∞

= 1, for almost all s∈Ω_n.Then,

sup

1≤i≤m

kf_in(s)−g_n(s)k= 1, for almost all s∈Ωn.Thus

d({f_i(s) : 1≤i≤m}, G) =h(s)

=h(s) sup

1≤i≤m

kf_in(s)−g_n(s)k

= sup

1≤i≤m

kh(s)fin(s)−h(s)gn(s)k

= sup

1≤i≤m

kf_i(s)−h(s)gn(s)k,

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for almost all s ∈ Ωn. We do notice that, if s ∈Ω0, then fi(s) = fj(s), for 1≤i≤j≤m. Hence, it is clear thatg defined by

g(s) =χ_Ω₀(s)f₁(s) +

∞

X

n=1

χ_Ω_n(s)h(s)g_n(s), for all s∈Ω enjoys the required property.

3. MAIN RESULT

The following lemmas will be used to prove our main result.

Lemma 3.1. [21] Let (Ω,Σ, µ) be a complete measure space, X a Banach space, andf a strongly measurable function fromΩ toX. Then f is measurable in the classical sense; f⁻¹(O) is measurable for every open setO ⊂X.

Lemma 3.2. [21] Let (Ω,Σ, µ) be a complete measure space, X a Banach space. If f : Ω → X is measurable in the classical sense and has essentially separable range, then f is strongly measurable.

Let Φ be a set-valued mapping, taking each point of a measurable space Ω into a subset of a metric space X. We say that Φ is weakly measurable if Φ⁻¹(O) is measurable in Ω whenever O is open in X. Here we have put, for any A ⊂X,

Φ⁻¹(A) ={s∈Ω :φ(s)∩A6=φ}.

The following theorem is due to Kuratowski [17], it is known as Measurable Selection Theorem.

Theorem3.3. [17]Let Φbe a weakly measurable set-valued map which car- ries each point of measurable spaceΩ to a closed nonvoid subset of a complete separable metric spaceX. ThenΦhas a measurable selection; i.e., there exists a function f : Ω → X such that f(s) ∈ φ(s) for each s ∈ Ω and f⁻¹(O) is measurable in Ωwhenever O is open in X.

Theorem 3.4. Let X be a Banach space and G be a closed separable subspace of X. Then the following are equivalent:

(1) Gis simultaneously proximinal in X.

(2) L^∞(µ, G) is simultaneously proximinal in L^∞(µ, X).

Proof. (2)⇒(1) : Let x₁, x₂, ..., x_m be any finite number of elements inX.

Define fi: Ω→X, i= 1,2, ..., m,by

fi(s) =xi .

Using simultaneous proximinality ofL^∞(µ, G) and Theorem 2.4, we getg∈ L^∞(µ, G) such that g(s) is a best simultaneous approximation of f1(s), f2(s), ..., fn(s) for almost all s. Choose s0 ∈ Ω so that g(s0) is a best simultaneous approximation of x₁, x₂, ..., x_m inG.

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(1)⇒(2) : Letf1, f2, ..., fmbe any finite number of elements in L^∞(µ, X). For eachs∈Ω define

Φ (s) =n

g∈G: sup

1≤i≤m

kf_i(s)−gk=d({f_i(s) : 1≤i≤m}, G)o . For eachs∈Ω,Φ (s) is closed, bounded, and nonvoid subset ofG.We shall show that Φ is weakly measurable. LetO be an open set inX, the set

Φ⁻¹(O) ={s∈Ω : Φ(s)∩O 6=φ}

can be also be described as Φ⁻¹(O) =n

s∈Ω : inf

g∈G sup

1≤i≤m

kf_i(s)−gk= inf

g∈O sup

1≤i≤m

kf_i(s)−gko . Since (Ω,Σ, µ) is complete, fi is measurable in the classical sense for i = 1,2, ..., m,by Lemma 3.1. Since subtraction inX,sup,and the norm inX are continuous, then the map

s→ inf

g∈A sup

1≤i≤m

kf_i(s)−gk

is measurable for any setA.It follows that Φ⁻¹(O) is measurable. By Theorem 3.3, Φ has a measurable selection; i.e., there exists a functionf : Ω→G such that f(s) ∈ φ(s) for each sΩ and f is measurable in the classical sense.

By Lemma 3.2, f is strongly measurable. Hence f is a best simultaneous approximation forf1, f2, ..., fm inL^∞(µ, G) by Theorem 2.4.

Acknowledgement. I would like to thank J. Mendoza for providing me a reprint of his paper.

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Received by the editors: October 21, 2013.