# Generalized Köthe-Toeplitz Duals of Some Vector-Valued Sequence Spaces

Authors: Yılmaz Yılmaz
Journal: International Journal of Analysis
 (2013)

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## Abstract

We know from the classical sequence spaces theory that there is a useful relationship between continuous andβ-duals of a scalar-valued FK-space E originated by the AK-property. Our main interest in this work is to expose relationships between the operator space ℒ(E,Y) and Eβ and the generalized β-duals of some X-valued AK-space E where X and Y are Banach spaces and Eβ={(Ak),Ak∈ℒ(X,Y):∑k=1∞AkxkconvergesinY,forallx∈E}. Further, by these results, we obtain the generalized β-duals of some vector-valued Orlicz sequence spaces.


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## Body

## 1. Introduction

The idea of dual sequence space was introduced by Köthe and Toeplitz [1]. Then, Maddox, [2], generalized this notion to X-valued sequence classes where X is a Banach space. This brings an important contribution to the operator matrix transformation of Banach space-valued sequence spaces. Remember that β- and α-duals of a (complex-valued) sequence space E, denoted by Eβ and Eα, respectively, are defined to be
(1)Eβ={(ak)∈w:∑k=1∞akxkconvergentforallx=(xk)∈E},Eα={(ak)∈w:∑k=1∞|akxk|<∞forallx∈E},
where w is the space of all complex-valued sequences. The (ak) in the classical definitions of Köthe-Toeplitz duals is replaced by a sequence (Ak)k=1∞ of linear operators, not necessarily continuous, from X into another Banach space Y. Thus, if E is a nonempty set of sequences x=(xk) with xk∈X, then generalized β- and α-duals of E are defined to be
(2)Eβ={(Ak):∑k=1∞AkxkconvergentintheY-normforallx∈E∑},(3)Eα={(Ak):∑k=1∞∥Akxk∥<∞forallx∈E}
respectively. It is clear that this notion depends on the space Y and if E⊂s(X), then
(4)Eα(orβ)⊂s(L(X,Y)),
where L(X,Y) is the space of all linear operators from X into Y. Without the loss of generality we can restrict ourselves in this work to continuous operators and ℒ(X,Y) being the space of all continuous linear operators from X into Y and s(X) being the space of all X-valued sequences which is a natural generalization of w=s(ℂ).We know from the classical sequence spaces theory that there is a useful relationship between continuous andβ-duals of a sequence space whenever it has the AK-property. Related results are also expressed in [3, page 176]. Here, we are going to show that there is an analogue relationship for X-valued sequence spaces in the context of generalized β-duals with respect to another fixed Banach space Y. Further, by applying this result, we obtain generalized β-duals of some vector-valued Orlicz sequence spaces. We think that our results give a fruitful way to find generalized duals of this kind of vector-valued sequence spaces.


## 2. Prerequisites

We use the notationsℕ,ℂ, and ℝ for the sets of all positive integers, complex numbers, and real numbers, respectively. For some locally convex (lc, for short) space X,X* denotes the continuous dual of X and we denote by BX and SX the closed unit ball and the sphere of some normed space X, respectively.An FH-space is an lc Fréchet spaceE such that E is a vector subspace of a Hausdorff topological vector space H and the topology of E is larger than the restricted topology of H to E; that is, the inclusion map: E→H is continuous. If H=w then an FH-space is called an FK-space. With a little extension, an X-valued sequence space E is called an FK-space whenever H=s(X) where X is a Banach space. In fact, the theory of FK-spaces can be developed without the local convexity. However, we are interested only in locally convex FK-spaces. Note that, s(X)=Xℕ and so its topology is the weakest topology such that the projections
(5)Pn:s(X)⟶X,Pn(x)=xn,n=1,2,…
are continuous.An Orlicz function is a functionM:[0,∞)→[0,∞) which is continuous, nondecreasing, and convex with M(0)=0,M(u)>0 for all u>0 and M(u)→∞ as u→∞. An Orlicz function M can always be represented in the following integral form:
(6)M(u)=∫0u‍p(t)dt,
where p, known as the kernel of M, is right differentiable for t⩾0,p(0)=0,p(t)>0 for t>0, and p is nondecreasing and p(t)→∞ as t→∞.Consider the kernelp(t) associated with Orlicz function M(u), and let
(7)q(s)=sup{t:p(t)⩽s}.
Then q possesses the same properties as the function p. Suppose now
(8)N(v)=∫0v‍q(s)ds.
Then N is an Orlicz function. The functions M and N are called mutually complementary Orlicz functions, and they satisfy the Young inequality,
(9)uv⩽M(u)+N(v)foru,v⩾0.An Orlicz functionM is said to satisfy the Δ2-condition for small u at 0 if for each k>0 there exist Rk>0 and uk>0 such that M(ku)⩽RkM(u), for all u∈(0,uk] [4].The spaceℓM consists of all sequences (xk) of scalars such that
(10)∑k=1∞M(|xk|ρ)<∞forsomeρ>0,
and it becomes a Banach space which is called an Orlicz sequence space with the Luxemburg norm
(11)∥x∥(M)=inf{ρ>0:∑k=1∞M(|xk|ρ)⩽1}.
The space ℓM is closely related to the space ℓp which is an Orlicz sequence space with M(u)=up,(1⩽p<∞).Another definition ofℓM, [4], is given by the complementary function to M as follows:
(12)ℓM={x=(xk)∈w:∑k=1∞xkykconverges,forally∈ℓ~N},
where N is the complementary function to M and  ℓ~N is the collection of all x in w with ∑k=1∞‍N(|xk|)<∞. Clearly,  ℓ~N⊆ℓN and ℓM are normed by the Orlicz norm
(13)∥x∥M=sup{|∑k=1∞xkyk|:∑k=1∞N(|yk|)⩽1}.
It was shown that these two norms on ℓM are equivalent.An important closed subspace ofℓM, introduced by Y. Garibanov, is hM which is defined by
(14)hM={x=(xk)∈w:∑k=1∞M(|xk|ρ)<∞,forallρ>0}.Immediately, we can introduce the vector-valued extension of the spacesℓM and hM for any Banach space X. Therefore,
(15)ℓM(X)={x∈s(X):∑k=1∞M(∥xk∥ρ)<∞,forsomeρ>0},
where s(X) is the space of all X-valued sequences and ∥·∥ is the norm of X. ℓM(X) is a Banach space with the Luxemburg norm
(16)∥x∥(M)=inf{ρ>0:∑k=1∞M(∥xk∥ρ)⩽1},
and it coincides with ℓM whenever X=ℂ. Further, define the closed subspace hM(X) of ℓM(X) by x=(xk)∈hM(X) if and only if
(17)∑k=1∞M(∥xk∥ρ)<∞∀ρ>0.
If M satisfies the Δ2-condition then hM(X)=ℓM(X).


## 3. Relative Weak Topologies

An operatorT∈ℒ(V,Y), for Banach spaces V and Y, is called a Hahn-Banach operator if for every Banach space X containing V as a subspace there exists an operator T^∈ℒ(V,Y) such that ∥T^∥=∥T∥ and T^x=Tx, for every x∈V. Thus, the classical Hahn-Banach theorem can be restated in the following way for operators (see also [5]).Theorem 1.
LetX, Y be Banach spaces and let T:X→Y be a continuous linear operator of rank 1. Then T is a Hahn-Banach operator.By some modification on the assertion: norm preserving, the result remains true ifX is taken as a locally convex space. The evidence of this assertion can be found in [6, Section 7.2.].Hence, by using Theorem1 we derived some tools for later sections as in the way that is similar to classical treatments. The proof of the following result is also given in [7]. Nevertheless, it will be convenient to restate it here.Corollary 2.
LetX,Y be Banach spaces and x∈X. Then, for some a∈SY, there exists a corresponding operator Ta(∈ℒ(X,Y)) such that
(18)∥Ta∥=1,Ta(x)=∥x∥a.Proof.
Ifx is not null, take V={λx}. Then V is a closed subspace of X hence is a Banach space with the same norm. Define Ta′(λx)=λ∥x∥a,  for some a∈SY, from V into Y. Then Ta′ satisfies the required condition of Theorem 1 on V. Hence it is a Hahn-Banach operator such that ∥Ta′∥=1. The norm preserving extension Ta of Ta′ to X has the desired properties. The result is obvious for x=0.Corollary 3.
LetX be an lc space, Y be a Banach space, S be a vector subspace of X, and x∈X∖S¯. Then there exists an a∈SY and a corresponding operator Ta(∈L(X,Y)) such that
(19)Ta(x)=a,Ta=0onS.Proof.
LetV={λx} and consider Z=S⊕V. Fix some a∈SY and define the operator
(20)Ta′:Z⟶Y,Ta′(z)=Ta′(s+λx)=λa,
for z∈Z (equivalently, for some s∈S and λ∈𝕂 (=ℂ or ℝ) such that z=s+λx). Clearly, the hyphothesis x∈X∖S¯ says that S is not dense in Z. Thus, S must be closed in Z since it is a maximal subspace of Z (see [6, Prob. 4.2.5]). But S=KerTa′={z∈Z:T(z)=0}, hence Ta′ is continuous. Further, Ta′ is an operator of rank 1 such that Ta′(x)=a and Ta′=0 on S. Thus, the extension Ta of Ta′ is the desired operator.Let us establish an lc topology on a Banach spaceX with respect to another Banach space Y. Let wY be a topology on X such that, for each net x=(xδ) in X,xδ→0(wY) if and only if ∥T(xδ)∥Y→0 for each T∈ℒ(X,Y). It is an lc topology generated by the family PY={∥·∥Y∘T:T∈ℒ(X,Y)} of the seminorms ∥·∥Y∘T on X. Obviously, the norm topology of X is stronger than wY, in general. If Y is a scalar field of X then wY coincide with the usual weak topology. It is clear that, a net which is wY-convergent to 0 is also weak convergent to 0. The converse of this assertion is not true.Example 4.
LetX=Y=ℓ2. Then the sequence (en)n=1∞ of unit vectors is weak convergent to 0 in ℓ2 [8, page 99]. But, it is not wℓ2-convergent to 0. Therefore, for the identity operator on ℓ2, we have ∥Ien∥=∥en∥=1↛0.However, we cannot work this example inℓ1 (in fact, in a Banach space which has the Schur property) since weak convergence implies the norm convergence in this case. Hence the following result is obvious from the definition of the Schur property.Theorem 5.
LetX be a Banach space having the Schur property. Then weak convergence implies wY-convergence in X for every Banach space Y.Now consider the canonical embeddingX→ℒ2(X,Y), where ℒ2(X,Y) is the space of all continuous operators from ℒ(X,Y) into Y and X and Y are Banach spaces, which assigns each x∈X to the operator Fx on ℒ(X,Y) defined by
(21)Fx(T)=Tx,foreachT∈ℒ(X,Y).
Clearly,
(22)∥Fx(T)∥Y=∥Tx∥Y≤∥T∥∥x∥X
so that Fx∈ℒ2(X,Y) and
(23)∥Fx∥≤∥x∥X.
Theorem 1 and the succeeding corollary assert that the canonical embedding is a linear isometry from X into ℒ2(X,Y) as is in the classical case.Now, let us investigate how do the bounded subsets of theX in the wY-topology behave. Note that a subset A of X is called wY-bounded if T(A) is bounded in Y for each T∈ℒ(X,Y). It is clear that, for every pair of the Banach spaces X and Y,A⊆X is wY-bounded if it is norm bounded. The converse of this assertion is the following theorem.Theorem 6.
LetX and Y be Banach spaces. Then wY-bounded sets are norm bounded.Proof.
LetV⊂X be wY-bounded and V^ be canonical embedding of V into ℒ2(X,Y). A hypothesis says that V^ is pointwise bounded so it is uniformly (norm) bounded by the uniform boundedness principle. Hence there exists a K>0 such that ∥Fx∥≤K for each x∈V. So
(24)∥x∥=∥Fx∥≤K
for each x∈V.Theorem 7.
LetX be an lc space and Y be a Banach space. Then wY-bounded sets are also bounded in the lc topology of X.Proof.
LetV⊂X be wY-bounded. We are going to show that p[V] is bounded for each seminorm p in P where P is the family of all seminorms generating the lc topology of X. For an arbitrary p,Z=(X,p) is a seminormed space. Thus we can show as in Theorem 6 that V^ is bounded in ℒ2(Z,Y), whence, V is bounded in Z, that is, p[V] is a bounded subset of ℝ.We conclude this section with a brief discussion of equicontinuity. A setΩ of linear maps from one topological vector space X into another one Y is called equicontinuous if, for each neighborhood N of 0 in Y,
(25)⋂{T-1[N]:T∈Ω}
is a neighborhood of 0 in X. Equicontinuity is a generalization of the uniform boundedness of the family of linear maps between seminormed spaces.Theorem 8 (see [9]).
LetΩ be a collection of continuous linear mappings T from the Fréchet space X into the topological vector space Y. Then Ω is equicontinuous if and only if the set
(26)Ω(x)={Tx:T∈Ω}
is bounded in Y, for each x∈X.Lemma 9 (see [6]).
Let(Tδ) be a net of continuous operators Tδ from a Fréchet space X into the topological vector space Y. Then the set {x:Tδx→0} is a closed subspace of X.


## 4. Sectional Properties and Operator Spaces

For anx∈s(X),
(27)x(n)=(x1,x2,…,xn,0,…)
is called nth  section  of  x. Further, ϕ(X) denotes the space of all finite sequences in s(X).Definition 10.
LetE⊃ϕ(X) be an FK-space. If, for each x∈E,
(28)x(n)⟶xinE,
then E is called an AK-space. Further, E is called an AD-space whenever ϕ(X) is dense in E. If, for each x∈E, the sequence (x(n)) is bounded in E then E is called an AB-space.An AK-(AD-, AB-) space is also called to have AK-(AD-, AB-) property.LetE⊃ϕ(X) be an FK-space and define the set
(29)B={x∈E:(x(n))isboundedinE}.
Clearly, B⊂E and B=E whenever E is an AB-space.To define another important classes we consider the mappings(30)Ik:X⟶EIk(a)=(0,…,0,akthposition,0,…)
and define the set WY, for some Banach space Y, by
(31)WY={x∈E:Tx=∑‍(T∘Ik)(xk)∀T∈ℒ(E,Y)}.Proposition 11.
For each Banach spaceY,WY⊂B.Proof.
Letx∈WY and T∈ℒ(E,Y) then we can write Tx=∑‍(T∘Ik)(xk), that is,
(32)∑k=1n(T∘Ik)(xk)⟶Tx(n→∞)inE.
Since
(33)∑k=1n(T∘Ik)(xk)=Tx(n),Tx(n)→Tx; that is, the sequence {x(n)} is wY-convergent hence it is wY-bounded. Thus, it is also bounded in the lc Fréchet topology of E by Theorem 7.Proposition 12.
For each Banach spaceY,
(34)WY⊂ϕ(X)¯,
where ϕ(X)¯ is the closure of ϕ(X) in E.Proof.
Letx∈WY. Then, for every T∈ℒ(E,Y) such that T=0 on ϕ(X),
(35)Tx=∑‍(T∘Ik)(xk)=∑‍T(0,…,0,xkkth,0,…)=∑‍0=0.
This implies x∈ϕ(X)¯. If this is not so, then there exists an a∈SY and a corresponding operator Ta(∈ℒ(E,Y)) such that
(36)Ta(x)=a≠0,Ta=0onϕ(X)
by Corollary 3. This is a contradiction.Proposition 13.
LetE⊃ϕ(X) be an FK-space with AD- and AB-property. Then E also has the AK-property.Proof.
Define(37)An:E⟶E,An(x)=x(n)-x,forn=1,2,….
Then the sequence {An(x)} is bounded by the AB-property. Therefore {An} is equicontinuous by Theorem 8. On the other hand An(x)→0 for each x∈ϕ(X), That is,
(38)ϕ(X)⊂{x∈E;An(x)⟶0}=Λ.
Since Λ is a closed subspace of E from Lemma 9, we obtain that ϕ(X)¯⊆Λ. Thus
(39)An(x)=x(n)-x⟶0
for each x∈ϕ(X)¯ (=E by the AD-property), whence, E has the AK-property.Theorem 14.
LetE⊃ϕ(X) be an FK-space. Then E is an AK-space if and only if Eβ and ℒ(E,Y) are isomorphic for every Banach space Y.Proof.
Let(Tk)∈Eβ where each Tk∈ℒ(X,Y) and define A[(Tk)]=T by
(40)Tx=∑k=1∞‍Tkxk
for each x∈E. Write
(41)ηn=∑k=1n‍Tk∘Pk,
where Pk:E→X,k=1,2,… is the kth (continuous) projection defined by Pk(x)=xk. Then each ηn is continuous and the sequence (ηn) is pointwise convergent since the series ∑‍Tkxk is convergent. So, the operator T, which is also defined by
(42)Tx=limηn(x),
is continuous by the Banach-Steinhauss closure theorem, whence, T∈ℒ(E,Y). That A is injective comes from the following discussion. Let A[(Tk)]=T=0. Then, for each a∈X,
(43)(T∘Ik)(a)=Tka=0.
This implies each Tk=0, that is, (Tk)=0. Further, for each T∈ℒ(E,Y), let us consider
(44)Tk=T∘Ik,fork=1,2,…,
from X to Y. For each x∈E,
(45)∑k=1n‍Tkxk=∑k=1n‍(T∘Ik)(xk)⟶Txr(n⟶∞)
since E is an AK-space, whence, (Tk)∈Eβ. This means that A is surjective.
Conversely, letEβ and ℒ(E,Y) be isomorphic. Then each T∈ℒ(E,Y) has the representation (Tk) such that each
(46)Tk=T∘Ik∈ℒ(E,Y)
and also, for each x∈E,
(47)Tx=∑‍Tkxk.
This shows that x∈WY, that is,
(48)E⊆WY.
Thus, we obtain E=ϕ(X)¯ by the Proposition 12, whence, E has the AD-property. Also, E has the AB-property by Proposition 11. Hence, E is an AK-space by Proposition 13.


## 5. Applications on Vector-Valued Orlicz Sequence Spaces

It is not hard to see as in the classical case, [4], that another definition of ℓM(X) by the complementary function N to M is
(49)ℓM(X)={x∈s(X):∑k=1∞‍fk(xk)converges,21forallf=(fk)∈ℓ~N(X*)∑k=1∞‍fk(xk)},
where ℓ~N(X*) is the class of all sequences f=(fk) such that ∑k=1∞‍N(∥fk∥)<∞ and each fk∈X*. Further, for each x∈ℓM(X),
(50)∥x∥M=sup{|∑k=1∞‍fk(xk)|:∑k=1∞‍N(∥fk∥)⩽1}<∞
defines a norm on ℓM(X). This norm is said to be Orlicz norm on ℓM(X).Lemma 15.
OnℓM(X), the norms ∥·∥M and ∥·∥(M) are equivalent, and ∥x∥(M)⩽∥x∥M⩽2∥x∥(M).Proofs of this lemma and the above assertion can be given in a similar way followed in [4, Theorem 8.9], by using the inequality
(51)|∑k=1∞‍fk(xk)|⩽∑k=1∞‍∥fk∥∥xk∥,
and by using the fact that x=(xk)∈ℓM(X) if and only if (∥xk∥)k=1∞∈ℓM.Lemma 16.
LetM be an Orlicz function. The sets
(52)Λ1={x∈s(X):∑k=1∞‍M(∥xk∥)⩽1},Λ2={x∈s(X):∥x∥(M)⩽1}
are identical.Proof.
Letx∈Λ1, this means ∑k=1∞‍M(∥xk∥/ρ)⩽1 for ρ=1. Hence, ∥x∥(M)⩽1, that is, x∈Λ2. Conversely, let x∈Λ2, that is
(53)inf{ρ>0:∑k=1∞‍M(∥xk∥ρ)⩽1}⩽1.
This means ∑k=1∞‍M(∥xk∥/ρ)⩽1 for some ρ⩽1. Therefore ∑k=1∞‍M(∥xk∥)⩽1 since M is nondecreasing.In generalhM(X) has no Schauder basis in classical manner. In [10] we introduce a new kind basis notion. Let us give this definition and prove that hM(X) has a basis in this manner.Definition 17 (see [10]).
LetX and Y be Banach spaces and 𝔸 be a set. A family {ηa:a∈𝔸} of continuous linear functions ηa:Y→X is called  Y-basis for X if  the following condition is satisfied. There exists a directed subset 𝒟 (by some relation ≪) of ℱ satisfying the property; for each a∈𝔸 there is some F∈𝒟 such that a∈F, and there exists a unique family {Ra:a∈𝔸} of linear functions Ra from X onto Y such that, for each x∈X, the net (πF(x):𝒟,≪) converges to x in X where
(54)πF(x)=∑a∈F(ηa∘Ra)(x),
for each F∈𝒟 and ℱ is the family of all finite subsets of the index set 𝔸 which is directed by the inclusion relation ⊆. Furthermore, {ηa} is called a Y-Schauder basis forX whenever each Ra is continuous.Thus we say that eachx∈X has the representation
(55)x=∑a∈𝔸(ηa∘Ra)(x),
in this case.Definition 18.
TheY-basis {ηa:a∈𝔸} in the above definition is called unconditional whenever 𝒟=ℱ with the inclusion relation ⊆.By taking𝔸=ℕ in the Definition [10] we now prove that hM(X) has an unconditional X-Schauder basis.Theorem 19.
Fork∈ℕ consider again the operators Ik:X→hM(X) such that
(56)Ik(u)=(0,0,…,0,ukthposition,0,…).
Then, the sequence {Ik} is an unconditional X-Schauder basis for hM(X).Proof.
Let us takeRk=Pk:hM(X)→X,Pk(x)=xk as a coordinate projection in the Definition [10]. We should prove that the net (πF(x):ℱ,⊆) converges to x in hM(X). This means, for each ϵ>0, we should find an F0∈ℱ such that ∥x-πF(x)∥(M)<ϵ for F0⊆F. Now, let ϵ>0 be given. Since ∑k=1∞‍M(∥xk∥/ρ)<∞ for every ρ>0, especially for ϵ>0, the series ∑k=1∞‍M(∥xk∥/ϵ) is absolutely convergent and hence it is unconditional convergent in real numbers. Hence we can find an n0(ϵ) such that ∑k=n0+1∞‍M(∥xk∥/ϵ)≤1. Now let F0={1,2,…,n0}. Obviously, F0 is dependent on ϵ and the set
(57){ρ>0:∑k∈ℕ∖F0M(∥xk∥ρ)⩽1}
includes the ϵ. This means
(58)inf{ρ>0∑k∈ℕ∖FM(∥xk∥ρ)⩽1}⩽ϵ.
Now, remember that
(59)πF(x)=∑k∈F(Ik∘Pk)(x).
Hence, for some F∈ℱ such that F0⊆F, we have
(60)∥x-πF(x)∥(M)=∥(x1,…,xn1-1,0,xn1+1…xn2-1,0,xn2+1,…xnm-1,0,xnm+1…)∥(M)=inf{ρ>0:∑k∈ℕ∖FM(∥xk∥ρ)⩽1}⩽ϵ.
The continuity of each Pk and uniqueness of the sequence {Pk} in the representation can be done similarly in the classical case. This completes the proof.One of our main results is the following theorem which states the generalizedβ-dual of hM(X) with respect to the Banach space Y. The above theorem brings that hM(X) is an AK-space and we can use Theorem 14 to find the generalized β-dual of hM(X).Theorem 20.
LetX, Y be Banach spaces and M,N be mutually complementary Orlicz functions. Then, hM(X)β is isomorphic by the mapping T→(T∘Ik) to the Banach space
(61)VN={supf∈BY*∥(Ak*f)k=1∞∥NA=(Ak)∈s(ℬ(X,Y)),2∥A∥=supf∈BY*∥(Ak*f)k=1∞∥N<∞},
where each Ik is defined as in Theorem 19.Proof.
We prove thatℒ(hM(X),Y) is isometrically isomorphic to VN.
A routine calculation shows that∥A∥=supf∈BY*∥(Ak*f)k=1∞∥N really defines a norm on VN and it is a Banach space with this norm. Let T∈ℒ(hM(X),Y) and say Ak=T∘Ik for each k. This implies ∥(Ak∘Pk)(x)∥=0 so that Ak(xk)=0 for each k. Since each x∈hM(X) has the unconditional representation x=∑k=1∞‍(Ik∘Pk)(x), we can write
(62)Tx=∑k=1∞‍(T∘Ik)(xk)=∑k=1∞‍Akxk.
Immediately each Ak∈ℒ(X,Y) since ∥Ak∥⩽∥T∥∥Ik∥=∥T∥. Now, let us define the mapping
(63)Ψ:ℒ(hM(X),Y)→VN,byΨ(T)=A=(Ak)k=1∞;Ak=T∘Ik.Ψ(T)=0 if and only if each T∘Ik=0 so T=0 by the definition of each Ik, that is, Ψ is one to one. Also, for an arbitrary A∈VN, if we define the operator T by
(64)Tx=∑k=1∞‍Akxk
on hM(X) then, by using the Young inequality, we have
(65)∥∑k=mn‍Akxk∥=supf∈BY*|f(∑k=mn‍Akxk)|⩽supf∈BY*∑k=mn‍∥Ak*f∥∥xk∥⩽supf∈BY*∑k=mn‍N(∥Ak*f∥∥(Ak*f)k=1∞∥N)+supf∈BY*∑k=mn‍M(∥(Ak*f)k=1∞∥N∥xk∥)⩽supf∈BY*∑k=mn‍N(∥Ak*f∥∥(Ak*f)k=1∞∥N)+∑k=mn‍M(∥xk∥1/∥A∥).
Since (Ak*f)k=1∞∈ℓN(ℒ(X,Y)) for each f∈Y* and
(66)∑k‍N(∥Ak*f∥∥(Ak*f)k=1∞∥N)⩽1
from [4, Prop. 8.12], we have
(67)supf∈BY*∑k=mn‍N(∥Ak*f∥∥(Ak*f)k=1∞∥N)⟶0asm,n⟶∞.
Also
(68)∑k=mn‍M(∥xk∥1/∥A∥)⟶0asm,n⟶∞
since x∈hM(X). This means the series ∑‍Ak(xk) is convergent, that is, T is well defined. Further, that the mapping Ψ is onto, that is, T∈ℬ(hM(X),Y) comes from the following equalities:
(69)∥T∥=supx∈BM∥Tx∥=supx∈BM∥∑k=1∞‍Ak(xk)∥=supx∈BMsupf∈BY*|f(∑k=1∞‍Ak(xk))|=supf∈BY*supx∈BM|∑k=1∞‍(Ak*f)(xk)|=supf∈BY*sup{|∑k=1∞‍(Ak*f)(xk)|:∑k=1∞‍M(∥xk∥)⩽1},(byLemma16)=supf∈BY*∥(Ak*f)k=1∞∥N=∥A∥.
This shows at the same time that Ψ is an isometry.



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*Source: 862949-2013-01-03.xml*