Research article

On some new vector valued sequence spaces E(X,λ,p)

  • Received: 15 December 2022 Revised: 24 March 2023 Accepted: 27 March 2023 Published: 04 April 2023
  • MSC : 46A45, 46B45, 40C05

  • To define a new sequence space and determine the Köthe-Toeplitz duals of this sequence space, characterizing the matrix transformation classes between the defined sequence spaces and classical sequence spaces has been an important area of work for researchers. Defining and examining a new vector-valued sequence space is also a considerable field of study since it generalizes classical sequence spaces. In this study, new vector-valued sequence spaces E(X,λ,p) are introduced. The Köthe-Toeplitz duals of E(X,λ,p) spaces are identified. Also, necessary and sufficient conditions are determined for A=(Ank) to belong to the matrix classes (E(X,λ,p),c(q)); where AnkB(X,Y), X{c,} and Y is any Banach spaces.

    Citation: Osman Duyar. On some new vector valued sequence spaces E(X,λ,p)[J]. AIMS Mathematics, 2023, 8(6): 13306-13316. doi: 10.3934/math.2023673

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  • To define a new sequence space and determine the Köthe-Toeplitz duals of this sequence space, characterizing the matrix transformation classes between the defined sequence spaces and classical sequence spaces has been an important area of work for researchers. Defining and examining a new vector-valued sequence space is also a considerable field of study since it generalizes classical sequence spaces. In this study, new vector-valued sequence spaces E(X,λ,p) are introduced. The Köthe-Toeplitz duals of E(X,λ,p) spaces are identified. Also, necessary and sufficient conditions are determined for A=(Ank) to belong to the matrix classes (E(X,λ,p),c(q)); where AnkB(X,Y), X{c,} and Y is any Banach spaces.



    Let (X,) be a Banach space and p=(pk) be any bounded sequence of strictly positive real numbers. Let w(X) be the set of all sequences x=xk with xkX. Bounded, convergent and null vector-valued sequence spaces in X are denoted by (X), c(X), and c0(X), respectively. These sequence spaces are linear subspace of w(X). When X=R or C (the real or complex numbers) it is used the familiar notations , c and c0. Here, (X), c(X), and c0(X) are Banach spaces with the norm

    x=supkxk,

    where xkX for k=1,2,.... The set

    S={xX:x1}

    is the unit ball. B(X,Y) denotes the set of all bounded operators from a Banach space X into the Banach space Y. If TB(X,Y), the operator norm of T is

    T=supxSTx.

    In this paper, to indicate the continuous dual of X we use the notation X instead of B(X,C). Also, e and e(x) denote the sequences (1,1,1,...) and (x,x,x,...), respectively.

    In the classical theory of matrix transformations, one of the basic problems is the characterization of matrices that map a sequence space E into a sequence space F. The first step of this characterization is the determination of the Köthe-Toeplitz duals of E. Also, the Köthe-Toeplitz duals of E are called β-dual and α-dual of E.

    Let X and Y be Banach spaces and (Ak) be the sequence of operators Ak defined from X to Y, which are linear but not necessarily bounded. Then, the β-dual and α-dual of E are defined as follows:

    Eα={A=(Ak):k=1Akxk<forallxE},Eβ={A=(Ak):k=1AkxkconvergesintheYnormforallxE}.

    Generalized Köthe-Toeplitz duals of X-valued sequence spaces (X), c(X) and c0(X) were defined by Maddox [6].

    Let Ax=(Ank) be an infinite matrix of linear, but not necessarily bounded, operators Ank from X to Y. Suppose E is a nonempty subset of w(X) and suppose F is a nonempty subset of w(Y). We define the matrix classes (E,F) by saying that A(E,F) if and only if, for every x=(xk)E,

    Anx=k=1Ankxk,

    converges for each n in the norm topology of Y and the sequence

    Ax=(Ankxk)n belongstoF.

    X-valued paranormed sequence spaces c0(X,p), c(X,p), (X,p) and (X,p) that are the generalization of Maddox spaces c0(p), c(p), (p) and (p) were first defined by Rath [1]. Rath also determined the generalized Köthe-Toeplitz duals of c0(X,p), c(X,p), (X,p) and (X,p). Suantai [8] and Suantai and Sudsukh [7] gave the matrix characterizations (ξ(X,p),c(q)) where

    ξ{c0,c0_,c,,Er,Fr}.

    Srivastava and Srivastava [2] introduced the set c0(X,λ,p) and determined the generalized Köthe-Toeplitz duals of these spaces where λ=(λk) is any sequence of non-zero complex numbers.

    Now, we define the new X-valued sequence spaces as follows:

    c(X,λ,p)={x=(xk):lX,limkλkxklpk=0},(X,λ,p)={x=(xk):supkλkxkpk<}.

    The spaces c(X,λ,p) and (X,λ,p) are BK spaces with the norm

    xλ,p=supλkxkpk/M,

    where

    p=(pk),

    and

    M=max{1,supkpk}.

    c(X,λ,p) and (X,λ,p) are generalization of several sequence spaces. Some of these generalizations are as follows:

    If λk=1 for all k and X=C, then we obtain spaces c(p) and (p) introduced by Maddox in [5].

    If λk=1 for all k, then we obtain spaces c(X,p) and (X,p) introduced by Rath in [1] and also introduced by Suantai and Sudsukh [7].

    If λk=1 and pk=1 for all k, then we obtain spaces c(X) and (X) introduced by Maddox in [6].

    This section is concerned with the generalized Köthe-Toeplitz duals of the spaces c(X,λ,p), (X,λ,p).

    Now we define the group norm and consider an inequality to be used to calculate the β- duals of our spaces. If (Fk) is a sequence in B(X,Y), the group norm of (Tk) is

    (Fk)=supnk=1Fkxk, (2.1)

    where the supremum is taken over all nN and all choices of xkS. Inequality

    n+pk=nFkxkRn.max{xk:nkn+p} (2.2)

    holds for any xk and all n1, and all non-negative integers p, where

    Rn=(Fn,Fn+1,Fn+2,...)

    is called nth tail of (Fn), see [5].

    Some results related to Köthe-Toeplitz duals of the space c0(X,λ,p) were obtained by Srivastava and Srivastava [2] as follows:

    Lemma 1. A=(Ak)cα0(X,λ,p) if and only if

    (i) There exist an integer m1 such that AkB(X,Y) for each km.

    (ii) There exist an integer N>1 such that k=m|λk|1AkNrk<.

    Lemma 2. Let (Tk) is a sequence in B(X,Y). Then, exactly one of the following is true:

    (i) Rk= for all k1.

    (ii) Rk< for all k1.

    Lemma 3. A=(Ak)cβ0(X,λ,p) if and only if

    (i) There exist an integer m1 such that AkB(X,Y) for each km.

    (ii) Rm(λ,N)< for some N>1, where

    Rm(λ,N)=(λ1mNrmAm,λ1m+1Nrm+1Am+1,λ1m+2Nrm+2Am+2,...).

    Theorem 1. A=(Ak)cβ(X,λ,p) if and only if

    (i) A=(Ak)cβ0(X,λ,p).

    (ii) k=1λ1kAkx converges for each xX.

    Proof. Suppose xc(X,λ,p) and suppose (i) and (ii) hold. We can write

    c(X,λ,p)=c0(X,λ,p)+E,

    where

    E={e(z):zX}.

    See Theorem 3.9 [9]. Then, we can write

    λkxk=λkyk+z

    for all kN where the sequence z in X and the sequence (yk) in c0(X,λ,p).

    Therefore, k=1λ1kAkz is convergent by condition (ii) and k=1Akyk is convergent by Lemma 3. Thus we have that k=1Akxk converges.

    Since cβ(X,λ,p)cβ0(X,λ,p) condition (i) follows from Lemma 3. If we take any xc(X,λ,p) and yc0(X,λ,p), then k=1Akxk and k=1Akyk are converge. Morever, we consider

    λkxk=λkyk+z

    for all kN where the sequence z in X and the sequence (yk) in c0(X,λ,p). Hence, k=1λ1kAkz converges for each zX.

    If we take λk=1 and pk=1 for all k in the Theorem 1, then we have Maddox [6] Proposition 3.2 as follows:

    Corollary 1. (Ak)cβ(X) if and only if

    (i) There exists mN such that AkB(X,Y) for all km.

    (ii) k=1Ak converges in the strong operator topology.

    Theorem 2. A=(Ak)β(X,λ,p) if and only if

    (i) There exist an integer m1 such that AkB(X,Y) for each km.

    (ii) Rm(λ,N)0 for all N>1, where

    Rm(λ,N)=(λ1mNrmAm,λ1m+1Nrm+1Am+1,λ1m+2Nrm+2Am+2,...).

    Proof. Let (i) and (ii) hold and x(X,λ,p). Then, there exists an integer N0>1 such that xk<|λk|1Nrk0 for all kN. If we use inequality (2.2), then we obtain

    n+ik=nAkxx=n+ik=nλ1kNrkAkλkNrkxkRm(λ,N).maxnkn+i{|λk|Nrkxk}Rm(λ,N).

    Hence, k=1Akxk is convergent since Y is a Banach space.

    Since β(X,λ,p)cβ0(X,λ,p) Lemma 3.2 yields condition (i). Now, suppose that (ii) fails, say

    lim supnRm(λ,N)=3κ>0.

    Then, there exist integer 0<mm1n1 and zm1,zm1+1,...,zn1 in S such that

    n1m1λ1kNrkAkzk>κ.

    Choose n1<m2 such that Rm2(λ)>2κ. Then, there exist m2n2 and zm2,zm2+1,...,zn2 in S such that

    n2m2λ1kNrkAkzk>κ.

    Now, by proceeding in this way, we can define the sequence x=(xk) as follows

    xk={λ1kNrkzk,mikni,θ,otherwise,

    for all natural numbers i. Then, we have that x(X,λ,p), since

    x=supλkxkpkN.

    On the other hand k=0Akxk is divergent, that is a contradiction. Hence the proof is completed.

    The Köthe-Toeplitz duals of c0(X,λ,p), c(X,λ,p) and (X,λ,p) can be rewritten in the case of AkB(X,Y) for all integers k. For these concept let define the following sets, which are the subspace of w(B(X,Y)):

    M0(B(X,Y),λ,p)=N>1{A=(Ak):k=1|λk|1AkNrk<}, (2.3)
    M0(X,λ,p)=N>1{f=(fk):k=1|λk|1fkNrk<}, (2.4)
    M(B(X,Y),λ,p)=N>1{A=(Ak):k=1|λk|1AkNrk<}, (2.5)
    M(X,λ,p)=N>1{f=(fk):k=1|λk|1fkNrk<}. (2.6)

    Srivastava and Srivastava [2] showed that in the Theorem 3.1

    cα0(X,λ,p)=M0(B(X,Y),λ,p).

    Moreover, they showed that if Y=C and fkX for k1, then

    cα0(X,λ,p)=cβ0(X,λ,p)=M0(X,λ,p)

    in the Theorem 3.5.

    The following result can now be proved in the same way as Theorem 3.1 in [2].

    Theorem 3. α(X,λ,p)=M(B(X,Y),λ,p).

    Theorem 4. Let Y be complex field C and f=(fk) be a sequence of continuous linear functionals on X. Then,

    α(X,λ,p)=β(X,λ,p)=M(X,λ,p).

    Proof.

    α(X,λ,p)=M(X,λ,p)

    from Theorem 3. Now, we show that

    β(X,λ,p)M(X,λ,p).

    Following Srivastava and Srivastava [2] as in Theorem 3.5; suppose

    f=(fk)M(X,λ,p),

    then we obtain

    k=1|λk|1fkNrk>2,

    for some N>1. Also, there exists zkS such that fk<2|fk(zk)| for each k1. If define the sequence x=(xk) as

    xk=sgn(fk(zk))|λk|1Nrkzk,

    for kS(N), then the sequence x=(xk) in (X,λ,p) but

    k=1fk(xk)=.

    These show that f=(fk)β(X,λ,p), so β(X,λ,p)M(X,λ,p).

    On the other hand α(X,λ,p)β(X,λ,p) since C is complete.

    Taking f=(fk)B(X,C) instead of A=(Ak)B(X,Y) in Theorem 1, the following result can be obtained.

    Theorem 5. Let f=(fk) be a sequence such that fkX for all kN. Then, f=(fk)cβ(X,λ,p) if and only if

    (i) k=1|λk|1fkN1/pk< for some NN.

    (ii) k=1λ1kfk(x) converges for each xX.

    In this section, some matrix classes will be characterized using similar methods developed by Wu and Liu [11] and Suantai and Sudsukh [7]. Throughout these section, F=(fnk) be an infinite matrix and fnkX for all k,nN. Moreover, let u=(uk) and v=(vk) be scalar sequences and uk0, vk0 and we can define two sets as follow:

    Eu={x=(xk)w(X):(ukxk)E},

    and

    u(E,F)v={F=(fnk):(unvkfnk)n,k(E,F)}.

    An X-valued sequence space E is solid when x=(xk)E and ykxk for all kN together imply that y=(yk)E.

    Now we give two lemmas, their proof can be found in [7].

    Lemma 4. Let E be a solid X-valued sequence space which is an FK-space and contain Φ(X). Then

    (E,c(q))=(E,c0(q))(E,e).

    Lemma 5. Let E and En be X-valued sequence spaces, and F and Fn be scalar-valued sequence spaces for all nN. Then,

    (i) (n=1En,F)=n=1(En,F). (ii) (E,n=1Fn)=n=1(E,Fn). (iii) (E1+E2,F)=(E1,F)(E2,F). (iv) (Eu,Fv)=v(E,F)u1.

    Theorem 6. F=(fnk)(c0(X,λ,p),c(q)) if and only if there is a sequence (fk)X such that

    (i) k=1|λk|1fkN1/pk< for some NN.

    (ii) m1/qn(fnkfk)0 as n for every k,mN.

    (iii) m1/qnk=1fnjr1/pk0 as n,r for each mN.

    Proof. For the sufficiency assume that

    F=(fnk)(c0(X,λ,p),c(q)).

    Since

    cq=c0(q)e

    from [10] (page-301), then we obtain

    F=(fnk)(c0(X,λ,p),c0(q)e).

    Now, we can choose

    D(c0(X,λ,p),c0(q))

    and

    E=(hnk)(c0(X,λ,p),e)

    such that F=D+E by Lemma 4. In addition, (hnk(x))n=1e for all xX and kN since Φ(X)c0(X,λ,p). These yields that hnk=hn+1k for all k,nN. Let choose fk=h1k, so we obtain

    D=(fnkfk)n,k(c0(X,λ,p),c0(q)).

    We know that

    c0(q)=m=1c0(m1/pk)

    from Theorem 0 in [10], which yields that

    (m1/qn(fnkfk))n,k(c0(X,λ,p),c0)

    for all mN by Lemma 5(ii) and (iv). If we use Theorem 2.4 in [11], then conditions (ii) and (iii) hold. Further k=1fk(xk) converges for all

    x=(xk)c0(X,λ,p),

    since

    E=(fk)n,k(c0(X,λ,p),e).

    So (i) holds by Theorem 3.5 in [2].

    For the necessity suppose that (fk) be a sequence such that fkX for all kN and conditions (i)(iii) hold. Also, let F=D+E where

    D=(fnkfk)n,k

    and E=(fk)n,k. Hence

    D(c0(X,λ,p),c0(q)),

    by conditions (i), (ii), Proposition 3.1 (ⅱ) and Theorem 2.4 in [11]. Moreover, k=1fk(xk) converges for all

    x=(xk)c0(X,λ,p),

    by proposition 3.3 and the condition (i). Hence E(c0(X,λ,p),e). So Lemma 4 yields

    F=(fnk)(c0(X,λ,p),c(q)).

    As an easy consequence of Theorem 4.3 in [7], we have

    Lemma 6. F=(fnk)((X,λ),c(q)) if and only if there is a sequence (fk)X such that

    (i) k=1|λk|1fk<.

    (ii) m1/qn(fnkfk)0 as n for every k,mN.

    (iii) j=k+1m1/qnfnjfjr1/pk0 as k uniformly on nN, for each m,rN.

    Lemma 7. (X,λ,p)=m=1(X,λ)(m1/pk).

    Proof. Let x(X,λ,p), then there is some mN such that supkλkxkpk<m. Therefore,

    λkxkm1/pk1

    for all kN. Hence,

    xm=1(X,λ)(m1/pk).

    Conversely, suppose that

    xm=1(X,λ)(m1/pk).

    Then, there are some mN and K>1 such that

    λkxkm1/pkK

    for all kN. Hence, this yields

    λkxkpkmKpkmKt

    for all kN, where t=supkpk, so that x(X,λ,p).

    Theorem 7. F=(fnk)((X,λ,p),c(q)) if and only if there is a sequence (fk)X such that

    (i) k=1|λk|1fkN1/pk< for all NN.

    (ii) r1/qn(m1/pkfnkfk)0 as n for every k,mN.

    (iii) r1/qnj=k+1m1/pjfnjfjs1/pj0 as k uniformly on nN, for each m,r,sN.

    Proof. Let consider the Lemmas 5–7. Then, we derive from two-sided implication Fx is in c(q) whenever x(X,λ,p) if and only if

    (m1/pkfnk)n,k((X,λ),c(q)).

    So we have for the necessary and sufficient (i)(iii) hold.

    Theorem 8. F=(fnk)(c(X,λ,p),c(q)) if and only if there is a sequence (fk)X such that

    (i) k=1|λk|1fkN1/pk< for some NN.

    (ii) m1/qn(fnkfk)$w$0 as n for every k,mN.

    (iii) m1/qnk=1fnjr1/pk0 as n,r for each mN.

    (iv) (k=1fnk(x))nc(q) for all xX.

    Proof. As in the proof of Theorem 1

    c(X,λ,p)=c0(X,λ,p)+{e(x):xX}.

    Hence, Lemma 5 show that

    F(c(X,λ,p),c(q))

    if and only if

    F(c0(X,λ,p),c(q))

    and

    F({e(x):xX},c(q)).

    Then,

    F(c0(X,λ,p),c(q))

    if and only if the conditions (i)(iii) hold, follows easily from Theorem 6. Also, one can show that

    F({e(x):xX},c(q))

    if and only if (iv) holds.

    The computation of Köthe-Toeplitz duals of sequence spaces is an important field of study in order to determine the matrix transformation classes between two spaces. Many researchers try to derive new sequence spaces using well-known sequence spaces. They also investigate for matrix transformation classes between the proposed new sequence spaces and classical sequence spaces by identifying the duals of the spaces they have obtained.

    Many different methods have been used to obtain new sequence spaces. One of these is paranormed sequence spaces defined by Maddox. Maddox has constructed new sequence spaces from classical sequence spaces , c0 and c. These sequence spaces are (p), c0(p), c0(p) where

    p=(pk)

    and

    M=max{1,supkpk}.

    With these sequence spaces defined by Maddox a wide working area has been opened. An other method to obtaining new sequence spaces is "matrix domain" method. For an arbitrary sequence space X, the matrix domain of an infinite matrix A in X is defined by

    XA={xw:AxX}

    as in [3,4]. The other method is to generate a new sequence space from classical sequence spaces using any scalar sequence λ=(λk). For example, Srivastava and Srivastava[2] introduced the sequence space

    c0(X,λ,p)={x=(xk):limkλkxkpk=0}.

    The researchers can obtain new sequence spaces by using above methods.

    In this study, we introduced generalized vector valued sequence spaces c(X,λ,p) and (X,λ,p). We determined the Köthe-Toeplitz duals of these spaces and characterized some matrix classes (E(X,λ,p),c(q)) where E{,c}. These spaces are the generalization of some sequence spaces given in the references.

    I would like to thank the editor and anonymous reviewers for their insightful and constructive comments and suggestions that have been helpful in providing a better version of the present work.

    Author declares no conflict of interest in this paper.



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    [9] S. Suantai, W. Sanhan, On β-dual of vector-valued sequence spaces of Maddox, Int. J. Math. Math. Sci., 30 (2001), 385–392. http://doi.org/10.1155/S0161171202012772 doi: 10.1155/S0161171202012772
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