# Some Properties oflp(A,X) Spaces

Authors: Xiaohong Fu; Songxiao Li
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We provide a representation of elements of the spacelp(A,X) for a locally convex space X and 1≤p<∞ and determine
its continuous dual for normed space X and 1<p<∞. In particular, we study the extension and characterization of isometries on lp(N,X) space, when X is a normed space with an unconditional basis and with a
symmetric norm. In addition, we give a simple proof of the main result
of G. Ding (2002).


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## Body

## 1. Introduction

LetX be a Hausdorff locally convex space, let R be a family of seminorms on X determining its topology and, let A be a set. We say that x belongs to lp(A,X) if and only if(1.1)∑a∈A‍[(r∘x)(a)]p<∞
for each r in R, where 1≤p<+∞. Obviously, lp(A,X) is a Hausdorff locally convex space with the seminorms (∑a∈A‍[(r∘x)(a)]p)1/p, for each r in R. When p=1, Yilmaz in [1] investigated some structural properties of the function space l1(A,X) for a Hausdorff locally convex space X and obtained the continuous duals of l1(A,X) and c0(A,X) for a normed space X. It should be mentioned that [2] is a powerful tool in the detailed investigation of mentioned function spaces.LetX be a real F space with the F-norm ∥x∥ and with an unconditional basis {en}. The norm ∥x∥ is called symmetric if, for any permutation {pn} and for an arbitrary sequence {εn} of numbers equal either to 1 or to −1, the following equality holds (see [3]):(1.2)∥t1e1+⋯+tnen+⋯∥=∥ε1t1ep1+⋯+εntnepn+⋯∥.As follows from the definition of symmetric norms, the operatorV defined by the formula(1.3)V(t1e1+⋯+tnen+⋯)=ε1t1ep1+⋯+εntnepn+⋯
is an isometry of the X onto itself.LetE and F be normed spaces. A mapping V:E→F is called an isometry if ∥Vx-Vy∥=∥x-y∥ for all x,y∈E (see, e.g., [4]). The classical Mazur-Ulam theorem in [5] describes the relation between isometry and linearity and states that every onto isometry V between two normed spaces with V(0)=0 is linear. So far, this has been generalized in several directions (see, e.g., [6]). One of them is the study of the isometric extension problem.Mankiewicz in [7] showed that an isometry which maps a connected subset of a normed space X onto an open subset of another normed space Y can be extended to an affine isometry from X to Y. In 1987, Tingley [8] posed the problem of extending an isometry between unit spheres as follows.LetE and F be two real Banach spaces. Suppose that V0 is a surjective isometry between the two unit spheres S1(E) and S1(F). Is V0 necessarily a restriction of a linear or affine transformation to S1(E)?It is very difficult to answer this question, even in two dimensional cases. In the same paper, Tingley proved that ifE and F are finite-dimensional Banach spaces and V0:S1(E)→S1(F) is a surjective isometry, then V0(x)=-V0(-x) for all x∈S1(E). In [9], Ding gave an affirmative answer to Tingley problem, when E and F are Hilbert spaces. In the case E and F are metric vector spaces, the corresponding extension problem was investigated in [10] and [11]. See [12] for some related results.In this paper we obtain some structural properties oflp(A,X) for 1<p<∞. We mainly provide a representation of the elements of lp(A,X) space and obtain continuous duals of lp(A,X) for a normed space X, where 1<p<∞. We also study the extension and characterization of isometries on lp(N,X) space, when X is a normed space with an unconditional basis and with a symmetric norm. Finally, we give a simple proof of an isometric extension theorem of [9].


## 2. Some Results oflp(A,X) Spaces

In this section we obtain some structural properties of the function spacelp(A,X)(1≤p<∞). For this purpose, we need a lemma that will be used in the proofs of our main results. We begin with the following well-known result (see [3]).Lemma 2.1.
LetXbe a real infinite-dimensionalF-spacewith a basis{en}and with a symmetric norm∥x∥. Then eitherXis a Hilbert space or each isometry is of type(1.3).Now we are in position to state and prove the main results in this section.Theorem 2.2.
LetXbe a Hausdorff locally convex space, let Rbe a family of seminorms onXdetermining its topology, and letAbe a set. Then eachx∈lp(A,X)(1≤p<∞)is represented by(2.1)x=Σa∈A(Ia∘x)(a),
where Ia:X→lp(A,X) is defined by
(2.2)Ia(t)(b)={t,b=a,0,b≠a,b∈A.Proof.
We denote byℱ the family of all finite subsets of the index set A. We write x=Σa∈A(Ia∘x)(a) if the net (∑a∈F‍(Ia∘x)(a):F∈ℱ) converges to x. Define
(2.3)SF(x)=∑a∈F‍(Ia∘x)(a)
for a finite subset F of A. We must prove that the net (SF(x):ℱ) converges to x in lp(A,X). By the definition of SF(x), we have
(2.4)SF(x)(a)={x(a),a∈F,0,a∈A∖F.
For U∈𝒩0(lp(A,X))  (where 𝒩0(lp(A,X)) denotes a base of neighborhoods of the origin of lp(A,X)), there exist ε>0 and r1,r2,…,rn∈R such that
(2.5)U⊇⋂i=1n‍{z:∑a∈A‍[(ri∘z)(a)]p<ε}.
Since ∑a∈A‍[(r∘x)(a)]p<∞ for each r∈R, then for i(1≤i≤n), we can find Fi∈ℱ such that
(2.6)∑a∈A∖Fi‍[(ri∘x)(a)]p<ε.
Hence, setting F0:=⋃i=1n‍Fi, we have
(2.7)∑a∈A‍[(ri∘[x-SF(x)])(a)]p=∑a∈A∖F‍[(ri∘x)(a)]p<ε
for each F⊇F0. This implies x-SF(x)∈U. That is x=Σa∈A(Ia∘x)(a).Remark 2.3.
IfX is a normed space and ∥∥p denotes the norm of lp(A,X), it holds that ∥Ia(t)∥p=∥t∥ and ∥Ia∥=1.Theorem 2.4.
LetXbe a normed space and letAbe a set. Then for eachf∈lp(A,X)', there existsψ∈lq(A,X')such that(2.8)f(x)=∑a∈A‍ψ(a)[x(a)],
and lp(A,X)'=lq(A,X'), where 1/p+1/q=1 and 1<p<∞.Proof.
By Theorem2.2, x∈lp(A,X) is represented by
(2.9)x=∑a∈A‍Ia[x(a)].
If f∈lp(A,X)', then
(2.10)f(x)=∑a∈Af∘Ia[x(a)].
Define ψ:A→X' by ψ(a)=f∘Ia. Next, we prove that ψ∈lq(A,X').
LetF be an arbitrary finite subset of A. Since Bishop and Phelps showed that the norm-attainers are dense in B(X,Y) for every Banach space X when Y=𝔽 (the symbol 𝔽 denotes a field that can be either ℝ and ℂ), there exists ξ(a) in the closed unit ball of X such that
(2.11)∥ψ(a)∥=|ψ(a)[ξ(a)]|
for each a∈F. Let us write ψ(a)[ξ(a)] in the polar form, that is,
(2.12)ψ(a)[ξ(a)]=eiθa|ψ(a)[ξ(a)]|,
and define the function x from A to X by
(2.13)x(a)={∥ψ(a)∥q-1e-iθaξ(a),ifa∈Fandψ(a)[ξ(a)]≠0,0,ifa∉Forψ(a)[ξ(a)]=0.
Obviously, x∈lp(A,X). Therefore, for this x, we have
(2.14)|f(x)|=|∑a∈A‍ψ(a)[x(a)]|=|∑a∈F‍∥ψ(a)∥q-1e-iθaeiθa|ψ(a)[ξ(a)]||=∑a∈F‍∥ψ(a)∥q≤∥f∥∥x∥≤∥f∥(∑a∈F‍(∥ψ(a)∥q-1)p)1/p=∥f∥(∑a∈F‍∥ψ(a)∥q)1/p.
Thus
(2.15)(∑a∈F‍∥ψ(a)∥q)1/q≤∥f∥<∞.
Since F is an arbitrary finite subset of A, we have
(2.16)∥ψ∥=(∑a∈A‍∥ψ(a)∥q)1/q≤∥f∥<∞,
and so ψ∈lq(A,X'). Moreover, by Hölder inequality, we have
(2.17)|f(x)|≤∑a∈A‍∥ψ(a)∥∥x(a)∥≤(∑a∈A‍∥ψ(a)∥q)1/q(∑a∈A‍∥x(a)∥p)1/p=∥ψ∥∥x∥,
from which we get
(2.18)∥f∥≤∥ψ∥.
Combining (2.15) and (2.18) yields ∥f∥=∥ψ∥. Thus we define a linear isometry T:lp(A,X)'→lq(A,X') with Tf=ψ. To prove that T is surjective. Indeed, for ψ∈lq(A,X'), there exists f defined on lp(A,X) such that
(2.19)f(x)=∑a∈A‍ψ(a)[x(a)],
that is, Tf=ψ. By Mazur-Ulam theorem (see [5]), T is a linear isometry from lp(A,X)' onto lq(A,X'), thus
(2.20)lp(A,X)'=lq(A,X').
The proof of this Theorem is finished.Theorem 2.5.
LetXbe a normed space with an unconditional basis and with a symmetric norm. Thenlp(N,X)is also a normed space with an unconditional basis and with a symmetric norm. Moreover, eitherlp(N,X)is a Hilbert space or each isometry is of type (1.3).Proof.
Suppose that{ek} is an unconditional basis for X with ||ek||=1. Let
(2.21)eik=(o,…,ek,o,…)︸ithplace.
By Theorem 2.2, if x(i)=Σk=1∞aikek then x∈lp(N,X) is represented by
(2.22)x=∑i∈Nk∈N‍aikeik,
that is {eik}i∈N,k∈N is a basis for lp(N,X). Note that x=∑i∈N,k∈Naikeik is an unconditionally convergent series in lp(N,X) and that {ek} is an unconditional basis for X. Thus {eik}i∈N,k∈N is an unconditional basis for lp(N,X). by the definition of norm on lp(N,X) and symmetry of norm on X it follows that
(2.23)∥∑‍aikeik∥=(∑‍∥aikeik∥p)1/p=(∑‍|aik|p)1/p.
For any permutation of positive integers {pik}, we have
(2.24)∥∑‍εikaikepik∥=(∑‍|aik|p)1/p,
thus lp(N,X) has symmetric norm. By Lemma 2.1, either lp(N,X) is a Hilbert space or each isometry is of type (1.3).


## 3. A Simple Proof of an Isometric Extension Result in Hilbert Space

Lemma 3.1.
LetEandFbe normed spaces and letV0be an isometric operator mappingS1(E)intoS1(F). If for anyλ∈Rand anyx,y∈S1(E),(3.1)∥V0x-|λ|V0y∥≤∥x-|λ|y∥,
then V0 can be isometrically extended to the whole space. Furthermore, when V0 is surjective, V0 can be linearly and isometrically extended to the whole space.Proof.
Set(3.2)Vx={∥x∥V0(x∥x∥),ifx≠0,0,ifx=0.
It is easy to see that ∥Vx-Vy∥≤∥x-y∥ for all x,y∈E. In particular, when ∥x∥=∥y∥ either x or y is zero element, we have
(3.3)∥Vx-Vy∥=∥x-y∥.
Thus, it suffices to prove (3.3) whenever ∥x∥>∥y∥>0.
Suppose, on the contrary, there existx0,y0∈E such that ∥x0∥>∥y0∥>0 and ∥Vx0-Vy0∥<∥x0-y0∥. Define a function on R by
(3.4)φ(λ)=∥x0+λ(y0-x0)∥.
The facts that φ(λ) is a continuous function, φ(1)=∥y0∥<∥x0∥ and limλ→+∞φ(λ)=+∞ assure that there exists λ0∈(1,+∞) such that φ(λ0)=∥x0∥ (by the intermediate value theorem). Let z0=x0+λ0(y0-x0). We see that x0,y0, and z0 lie on a straight line and ∥z0∥=∥x0∥. Hence
(3.5)∥z0-x0∥=∥z0-y0∥+∥y0-x0∥>∥Vz0-Vy0∥+∥Vx0-Vy0∥≥∥Vz0-Vx0∥=∥z0-x0∥,
a contradiction. Thus V0 can be isometrically extended to the whole space, and V is an extension of V0.
IfV0 is surjective, then the conclusion follows easily from the Mazur-Ulam Theorem.Theorem 3.2.
Suppose thatEandFare Hilbert spaces andV0is a surjective isometric operator mappingS1(E)ontoS1(F). ThenV0can be linearly and isometrically extended to the whole space.Proof.
SinceV0 is an isometry, we have for all x,y in S1(E) that
(3.6)〈V0(x)-V0(y),V0(x)-V0(y)〉=〈x-y,x-y〉,
that is,
(3.7)2-〈V0(x),V0(y)〉-〈V0(y),V0(x)〉=2-〈x,y〉-〈y,x〉,
and thus we have
(3.8)〈V0(x),V0(y)〉+〈V0(y),V0(x)〉=〈x,y〉+〈y,x〉.
The last equality gives that
(3.9)〈V0(x),V0(x)〉-λ〈V0(x),V0(y)〉-λ〈V0(y),V0(x)〉+λ2〈V0(y),V0(y)〉=1+λ2-λ〈V0(x),V0(y)〉-λ〈V0(y),V0(x)〉=1+λ2-λ〈x,y〉-λ〈y,x〉=〈x,x〉-λ〈x,y〉-λ〈y,x〉+λ2〈y,y〉.
Thus
(3.10)∥V0(x)-λV0(y)∥=∥x-λy∥
holds for all λ in R. Now we can apply Lemma 3.1 to obtain the desired result.



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