On best simultaneous approximation in operator and function spaces

On best simultaneous approximation in operator and function spaces

Sharifa Al-Sharif

Abstract. Let X be a Banach space, (I,P

, µ) a finite measure space andL¹(µ, X) the Banach space of all X-valuedµ-integrable functions on the unit interval I equipped with the usual 1-norm. In this paper we prove that for a closed subspaceGofX, L¹(µ, G) is simultaneously Chebyshev inL¹(µ, X) if and only ifGis simultaneously Chebyshev in X.Further results are obtained in the space of bounded linear operators L(l¹, X) and in the space of continuous functionsC¹(I, l^p) with respect to theL¹ norm.

Mathematics Subject Classification (2010):41A65, 41A50.

Keywords:Best approximation, simultaneous approximation, spaces of vector functions.

1. Introduction

Let X be a Banach space and (I,P

, µ) be a finite measure space. Let us denote byL¹(µ, X),the Banach space of allX-valuedµ-integrable functions on the unit intervalI equipped with the usual 1-norm.

For a closed subspace G of X, let us recall that G is simultaneously proximinal inX if for allm-tuples (x1, x2, ..., xm)∈X^m,there existsg∈G such that

m

X

i=1

kxi−gk= dist(x₁, x₂, ..., x_m, G) = inf (_m

X

i=1

kxi−zk:z∈G )

. In this case,g is called a best simultaneous approximation of (x₁, x₂, ..., x_m) inG.If this best approximation is unique for all (x1, x2, ..., xm)∈X^m,then Gis called simultaneously Chebyshev.

Of course form= 1 the preceding concepts are just best approximation and proximinality.

The problem of best simultaneous approximation can be viewed as a special case of vector valued approximation. Recent results in this area are

due to Pinkus [10], where he considered the problem when a finite dimensional subspace is a uniqueness space. Results on best simultaneous approximation in general Banach spaces may be found in [9] and [11]. Related results on L^p(µ, X), 1 ≤ p < ∞, are given in [12]. In [12], it is shown that if G is a reflexive subspace of a Banach space X, then L^p(µ, G) is simultaneously proximinal inL^p(µ, X).Ifp= 1,Abu Sarhan and Khalil [1],proved that ifG is a reflexive subspace of the Banach spaceX orGis a 1-summand subspace ofX,thenL¹(µ, G) is simultaneously proximinal inL¹(µ, X).

It is the aim of this paper to give some sufficient conditions forL¹(µ, G) to be a Chebyshev subspace ofL¹(µ, X).Further results are obtained in the space of bounded linear operators L(l¹, X) and in the space of continuous functionsC¹(I, l^p) with respect to theL¹norm.

Throughout this paper,X is a Banach space andGis a closed subspace ofX.

2. Main results

In [1] it is shown that ifm= 1 andGis a finite dimensional subspace of a Ba- nach spaceX,thenGis Chebyshev inX if and only ifL¹(µ, G) is Chebyshev in L¹(µ, X). The main result in this section is: IfG is a reflexive subspace of X, then G is simultaneously Chebyshev in X if and only if L¹(µ, G) is simultaneously Chebyshev inL¹(µ, X).

Theorem 2.1. Let G be a reflexive subspace of X. Then G is simultane- ously Chebyshev inX if and only if L¹(µ, G)is simultaneously Chebyshev in L¹(µ, X).

Proof. Let f1, f2, ..., fm ∈ L¹(µ, X). Since G is reflexive, it follows that [Th.4,12],there existsg∈L¹(µ, G) such that

m

X

i=1

kfi−gk₁= dist(f₁, f₂, ..., f_m, L¹(µ, G)).

Thus by [Th.2.2, 2],we have:

m

X

i=1

kfi(t)−g(t)k= dist(f1(t), f2(t), ..., fm(t), G),

for almost allt∈I. ButGis simultaneously Chebyshev. Sog(t) is unique.

Thusg is determined uniquely, andL¹(µ, G) is simultaneously Chebyshev in L¹(µ, X).

Conversely. Letx1, x2, ..., xm∈X.Fori= 1,2, ..., m,consider the func- tions:fi :I →X, fi(t) = xi, for allt∈I. SinceL¹(µ, G) is simultaneously Chebyshev inL¹(µ, X), there exists g∈L¹(µ, G) such that

dist(f₁, f₂, ..., f_m, L¹(µ, G)) =

m

X

i=1

kfi−gk₁≤

m

X

i=1

kfi−hk₁

for allh∈L¹(µ, G).Thus by [Th.2.2, 2],we have:

m

X

i=1

kfi(t)−g(t)k ≤

m

X

i=1

kfi(t)−h(t)k (2.1) for almost allt∈I.But sinceGis reflexive, there existsw∈Gsuch that

m

X

i=1

kxi−wk ≤

m

X

i=1

kxi−zk

for all z ∈G,[Lemma 1.12]. Hence the functionb(t) =w for all t∈ I is a best simultaneous approximation off₁, f₂, ..., f_min L¹(µ, G).Equation (2.1) and since L¹(µ, G) is simultaneously Chebyshev in L¹(µ, X) it follows that g(t) =b(t) = w and w is unique. Hence Gis simultaneously Chebyshev in

X.

For 0< p <∞, let us denote byl^p(X),the space of all sequences (xn) inX such that

∞

P

n=1

kx_nk^p <∞.Forx= (x_n)∈l^p(X), let

kxk_p =









 _∞

P

k=1

kxnk^{p 1}_p

1≤p <∞

∞

P

k=1

kx_nk^p 0< p <1 In the spacel¹(X),we have the following result:

Theorem 2.2. G is simultaneously Chebyshev in X if and only if l¹(G) is simultaneously Chebyshev inl¹(X).

Proof. For 1≤i≤m, letx_i= (x_in)∈l¹(X).Ifg_n∈Gis such that

m

X

i=1

kxin−gnk ≤

m

X

i=1

kxin−zk (2.2)

for allz∈G.Using triangle inequality and takingz= 0 in (2.2) we get

m

X

i=1

kgnk − kxink ≤

m

X

i=1

kxin−gnk ≤

m

X

i=1

kxink

and this implies

mkg_nk=

m

X

i=1

kg_nk ≤2

m

X

i=1

kx_ink. (2.3)

Thus ∞

X

n=1

kgnk ≤ 2 m

m

X

i=1

∞

X

n=1

kxink<∞.

Hence the element g = (g_n) ∈ l¹(G) andg is a best simultaneous approxi- mation of the m-tuple ((x_in))^m_i=1 inl¹(G).The uniqueness ofg_nimplies that g= (gn) is unique and l¹(G) is simultaneously Chebyshev inl¹(X).

Conversely. Let x1, x2, ...., xm ∈ X. For each i = 1,2, ..., m, consider the sequence (xi,0, ....)∈l¹(X).Sincel¹(G) is simultaneously Chebyshev in l¹(X), it follows that there exists a sequence of the form (g,0, ....) inl¹(G) such that

m

X

i=1

k(x_i,0, ...)−(g,0, ...)k<

m

X

i=1

k(x_i,0, ....)−(z₁, z₂, ...)k

for all (z_n)∈l¹(G)\ {(g,0, ...)}.This implies that

m

X

i=1

kxi−gk<

m

X

i=1

kxi−zk

for allz∈G\ {g}.

For the space of bounded linear operators, L(l¹, X), from l¹ into X, wherel¹ is the space of all summable real sequences it has been proved in [1]

that G is proximinal inX if and only ifL(l¹, G) is proximinal in L(l¹, X).

For the case of simultaneous approximation we have the following result:

Theorem 2.3. Gis simultaneously proximinal inX if and only ifL(l¹, G)is simultaneously proximinal in L(l¹, X).

Proof. Let T1, T2, ..., Tm ∈ L(l¹, X). If (δn) is the natural basis of l¹, then Tiδn∈X, i= 1,2, ..., m.

SinceGis simultaneously proximinal, so for eachnthere existsxn∈G such that

m

X

i=1

kTi(δ_n)−x_nk= dist (T₁(δ_n), T₂(δ_n), ..., T_m(δ_n), G).

DefineS:l¹→G, S(δn) =xn.ThenS is a bounded linear operator froml¹ intoG.It is clear thatSis linear. To prove thatSis bounded, lety= (αn)∈ l¹,kyk₁=P∞

n=1|αn| ≤1.Then kS(y)k=

∞

X

n=1

αnS(δn)

≤

∞

X

n=1

|αn| kS(δn)k=

∞

X

n=1

|αn| kxnk.

Using (2.3) in Theorem 2.2 we get kS(y)k ≤

∞

X

n=1

|αn| 2 m

m

X

i=1

kTi(δ_n)k ≤

∞

X

n=1

|αn| 2 m

m

X

i=1

kTik= 2 m

m

X

i=1

kTik

∞

X

n=1

|αn|

HenceSis a bounded linear operator withkSk ≤ _m² Pm

i=1kTik.Now for any x= (βn)∈l¹ we have

m

X

i=1

kTi(x)−S(x)k =

m

X

i=1

Ti

∞

X

n=1

βnδn

!

−S

∞

X

n=1

βnδn

!

=

m

X

i=1

∞

X

n=1

βnTi(δn)−

∞

X

n=1

βnS(δn)

=

m

X

i=1

∞

X

n=1

β_n(T_i(δ_n)−S(δ_n))

≤

m

X

i=1

∞

X

n=1

|βn| kTi(δn)−S(δn)k

=

∞

X

n=1

|βn|dist (T1(δn), T2(δn), ..., Tm(δn), G).

≤

∞

X

n=1

|β_n|

m

X

i=1

kT_i(δ_n)−gk for everyg∈G.In particular for everyA∈L(l¹, G)

m

X

i=1

kT_i(x)−S(x)k ≤

∞

X

n=1

|β_n|

m

X

i=1

kT_i(δ_n)−A(δ_n)k

≤

∞

X

n=1

|βn|

m

X

i=1

kTi−Ak

=

m

X

i=1

kTi−Ak

∞

X

n=1

|βn|=

m

X

i=1

kTi−Ak kxk.

Taking supremum over allx∈l¹,kxk= 1 we get

m

X

i=1

kTi−Sk ≤

m

X

i=1

kTi−Ak.

HenceL(l¹, G) is simultaneously proximinal in L(l¹, X).

Conversely.Let x₁, x₂, ..., x_m ∈X. For each i = 1,2, ..., m, define T_i : l¹→X,

Tiδn =

xi n= 1 0 n6= 1 .

ThenTi∈L(l¹, X) andkTik=kxik.By assumption there existsA∈L(l¹, G) such that

m

X

i=1

kTi−Ak ≤

m

X

i=1

kTi−Bk

for allB ∈L(l¹, G). Hence

m

X

i=1

kxi−Aδ1k=

m

X

i=1

k(Ti−A)δ1k ≤

m

X

i=1

kTi−Ak ≤

m

X

i=1

kTi−Bk. IfB runs over all functions of the form

Bδ_n=

w n= 1 0 n6= 1 for allw ∈G, we obtain Pm

i=1kxi−Aδ1k ≤ Pm

i=1kxi−wk for allw ∈G.

HenceGis simultaneously proximinal inX.

Theorem 2.4. IfL(l¹, G)is simultaneously Chebyshev in L(l¹, X),thenG is simultaneously Chebyshev inX.

Proof. SupposeGis not Chebyshev inX. Then there existg1, g2∈G and x1, x2, ..., xm∈X such that

m

X

i=1

kxi−g1k=

m

X

i=1

kxi−g2k= dist(x1, x2, ..., xm, G).

Fori= 1,2, ..., m,let

T_iδ_n =

x_i n= 1 0 n6= 1 , and

A1δn =

g1 n= 1

0 n6= 1 , A2δn=

g2 n= 1 0 n6= 1 . Then

m

X

i=1

kTi−A₁k=

m

X

i=1

kTi−A₂k=dist(T₁, T₂, ..., T_m, L(l¹, G)).

This contradict the fact thatL(l¹, G) is simultaneously Chebyshev.

We remark that the converse of Theorem 2.4 is not true. To see this, let Gbe a Chebyshev subspace of X and x₁, x₂, ..., x_m ∈ X. For eachi = 1,2, ..., m,define T_i:l¹→X

Tiδn =

xi n= 1 0 n6= 1 , then ifz∈Gis such thatPm

i=1kxi−zk=dist(x1, x2, ..., xm, G),the operator A:l¹→X,

Aδ_n=

z n= 1 0 n6= 1 ,

is a best simultaneous approximation ofT1, T2, ..., Tm inL(l¹, G) that is

m

X

i=1

kTi−Ak ≤

m

X

i=1

kTi−Bk

for allB ∈L(l¹, G).Letr= min

1≤i≤mkxi−zk.Consider the map S :l¹→G, Sδn=

z n= 1 zn n6= 1 where|zn|< r.ThenPm

i=1kTi−Sk=Pm

i=1kTi−Ak.HenceL(l¹, G) is not a Chebyshev subspace ofL(l¹, X).

As a corollary from Theorem 2.2 for the Banach spacec₀ we have:

Corollary 2.5. Gis simultaneously Chebyshev inX if and only if L(c₀, G)is simultaneously Chebyshev inL(c₀, X).

Proof. By the result of Grothendieck [6],page 86, we haveL(c0, G) =l¹(G).

the result follows from Theorem 2.2.

3. Further results

Ann-dimensional subspaceVn ofC(I),the space of continuous functions on a compact setI,is called a Haar subspace if anyf ∈Vn\ {0}, f has at most n−1 zero’s on I. Haar subspaces on intervals of real numbers are called T-Systems. For each natural number n, let Mn be an n-dimensional Haar subspace. Set

U =

g∈L¹(µ, l^p) :g= (g_i), g_i∈M_i . We remark thatU is a closed subspace ofL¹(µ, l^p),[1].

On the space of continuous functions C¹(I, l^p), we have the following result

Theorem 3.1. For 1 ≤p <∞, U is proximinal in C¹(I, l^p) with respect to the L¹ norm.

Proof. Let p = 1 and let S1, S2, ...Sm ∈ C¹(I, l¹). Then for each i = 1,2, ..., m, S_i = (f_i,k)^∞_k=1 and kS_ik = R

I

∞

P

k=1

|f_i,k(t)|dt. Hence Pm

i=1kS_ik = Pm

i=1

R

I

∞

P

k=1

|fi,k(t)|dt.Using the Monotone Convergence Theorem, we get:

m

X

i=1

kSik=

∞

X

k=1 m

X

i=1

Z

I

|fi,k(t)|dt=

∞

X

k=1 m

X

i=1

kfi,kk₁.

Since for eachk, M_k is finite dimensional, there existsg_k ∈M_k such that

m

X

i=1

kf_i,k−g_kk₁≤

m

X

i=1

kf_i,k−h_kk₁ for allhk ∈Mk. Note that

m

X

i=1

kfi,k−h_kk₁≥

m

X

i=1

kfi,k−g_kk₁≥

m

X

i=1

kfi,kk₁− kgkk₁ .

for allhk ∈Mk. Since 0∈Mk, we get:

mkgkk₁≤2

m

X

i=1

kfi,kk₁ and so

∞

X

k=1

kgkk₁≤ 2 m

m

X

i=1

∞

X

k=1

kfi,kk₁= 2 m

m

X

i=1

kfik Henceg= (g_k)∈U and

m

X

i=1

kSi−gk=

∞

X

k=1 m

X

i=1

Z

I

|fi,k(t)−gk(t)|dt≤

∞

X

k=1 m

X

i=1

Z

I

|fi,k(t)−hk(t)|dt for allh_k ∈M_k. In particular we getPm

i=1kSi−gk ≤Pm

i=1kSi−hkfor all h∈U.HenceU is proximinal inC¹(I, l¹) with respect to theL¹norm.

For 1< p <∞, letS1, S2, ...Sm∈C¹(I, l^p). Consider the operator Pk : L¹(µ, l^p)→L¹(µ, l^p_k)

Pkf = (f1, f2, ..., fk)

where f = (fi)^∞_i=1. Then Pk is continuous. For 1 ≤ k < ∞, set Uk =

g= (gi)∈

k

Q

i=1

Mi

.SinceUkis finite dimensional, there exists somebg∈Uk

such that

m

X

i=1

kPkSi−bgk₁≤

m

X

i=1

kPkSi−hk₁ (3.1) for allh∈U_k, .Let us write g^k forbg.We shall prove that the sequence g^k must have a subsequence that converges to someg∈U.

Since P_kS_i → S_i, then the setsE_i ={P₁S_i, P₂S_i, P₃S_i, ..., S_i}, i = 1,2, ..., m are weakly compact inL¹(µ, l^p).SetEb=

g¹, g², g³, ..., gⁿ, ... . We want to prove thatEb is weakly relatively compact. Since l^p is reflexive, then by the Dunford Theorem [4, p.101], it is enough to prove that Eb is bounded and uniformly integrable. Note that

m

X

i=1

kPkSi−hk₁≥

m

X

i=1

PkSi−g^k ₁≥

m

X

i=1

kPkSik₁− g^k

₁ . for allh∈Uk.Since 0∈Uk, we get

m g^k

₁≤2

m

X

i=1

kPkSik₁

HenceEb is bounded.

To see thatEb is uniformly integrable, first note that for eachk kPkSik₁≤ kSik₁

i = 1,2, ...m. Thus lim

µ(Ω)→0

R

Ω

kh(t)kdµ(t) = 0 uniformly for h in Ei, i = 1,2, ..., m.

Now let >0 be given.By the uniform integrability ofEi there exists δi >0 such thatR

Ω

kh(t)kdµ(t)<₂ wheneverµ(Ω)< δi for allh∈Ei.Hence forµ(Ω)< δ= min

1≤i≤m(δ_i) Z

Ω

g^k(t)

dµ(t)< 2 m

m

X

i=1

Z

Ω

kP_kS_ikdµ(t)< .

Sinceδdepends only onE₁, E₂, ..., E_mandit follows thatEbis uniformly in- tegrable and hence weakly relatively compact. Thus there existsg∈L¹(µ, l^p) such thatg^k→g weakly.

Since the sequence g^k

in U converges weakly to some g ∈ L¹(µ, l^p) andU is a closed subspace ofL¹(µ, l^p), hence weakly closed, it follows that g∈U.

For h ∈ U, we have kPkh−hk₁ → 0. Hence for each i = 1,2, ..., m, kPkSi−Pkhk₁ →

k kSi−hk₁. Now let ϕ ∈ L^∞(µ, l^p∗) = L¹(µ, l^p)^∗ , the dual ofL¹(µ, l^p).Then

m

X

i=1

|hSi−g, ϕi| = lim

k→∞

m

X

i=1

|hPkS_i−g, ϕi|

≤ lim

m

X

i=1

P_kS_i−g^k

≤ lim

m

X

i=1

kPkSi−Pkhk for allh∈Uk,sinceUk is proximinal. Hence

m

X

i=1

|hSi−g, ϕi| ≤

m

X

i=1

kSi−hk. ConsequentlyPm

i=1kS_i−gk ≤Pm

i=1kS_i−hkfor allh∈U.

ThusU is proximinal in C¹(I, l^p),with theL¹-norm, 1< p <∞.

Acknowledgements. The author would like to thank the referee for his valu- able comments that improved the presentation of the paper.

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Sharifa Al-Sharif Yarmouk University Faculty of Science Mathematics Department Irded, Jordan

e-mail:sharifa@yu.edu.jo