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Since nonlinear schur theorem was proposed, it broke the limitation of linear operator matrices. And in this paper we study the summability theory for a class of matrices of nonlinear mapping, and the characterizations of a class of infinite matrix transformations are obtained. These results enrich the results on infinite matrices transformations, and have important meaning for the study of Banach space.

A decisive break in the theory of matrix transformations was in 1950, when Robinson considered the action of infinite matrices of linear operators from a Banach space on sequences of elements of that space [

Let X and Y be topological vector spaces, and F 0 ( X , Y ) = { f ∈ Y X : f ( 0 ) = 0 } . For sequence families λ ( X ) ⊆ X ℕ and μ ( Y ) ⊆ Y ℕ , the matrix ( T i , j ) ∈ ( λ ( X ) , μ ( Y ) ) means that ∑ j = 1 ∞ T i j ( x j ) converges when ( x j ) ∈ λ ( X ) , i ∈ ℕ and { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ ∈ μ ( Y ) for each ( x j ) ∈ λ ( X ) .

As usual,

c 0 = { ( x j ) ⊆ ℂ : x j → 0 } , c 0 ( X ) = { ( x j ) ∈ X ℕ : x j → 0 } ,

c ( X ) = { ( x j ) ∈ X ℕ : lim x j exists } and

l ∞ ( X ) = { ( x j ) ∈ X ℕ : x j isbounded } .

In 2001, Li Ronglu depicted the nonlinear operator matrices transformation with some restrictive condition on topological vector spaces [

( l ∞ ( X ) , c ( Y ) ) , ( l ∞ ( X ) , l ∞ ( Y ) ) , ( c 0 ( X ) , c 0 ( Y ) ) ( c 0 ( X ) , c 0 ( Y ) ) .

All of the researches enrich the results on infinite matrices transformations, and have important meaning for the study of Banach space.

In 1993, nonlinear Schur Theorem was given by Li Ronglu and C. Swartz, and broke the limitations of linear operator matrices.

Theorem A. [

As a special case, the following theorem is a nice result for the matrix family ( l ∞ ( X ) , c ( Y ) ) .

Theorem B. [

Note that theorem B exceeded the restriction of linear operators, and a characterization of ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c ( Y ) ) was given. For Banach spaces X , Y , it is useful to discuss the characterization of a variety of matrix families, where the mapping need not be linear.

As preparation of the proves of the main results, we also need following lemma.

Lemma [

Unless otherwise noted X , Y below are Banach spaces, and the mapping we studied in this section need not be linear.

Theorem 1. Let T i j ∈ F 0 ( X , Y ) for all i , j ∈ ℕ , then ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c ( Y ) ) if and only if

(i) lim i T i j ( x ) exists for all j ∈ ℕ and x ∈ X ;

(ii) For any ε > 0 , M > 0 there exists m 0 ∈ N such that ‖ ∑ j = m ∞ T i j ( x j ) ‖ < ε for all natural number m > m 0 , i ∈ ℕ , and { x j } ⊆ X with sup j ‖ x j ‖ ≤ M .

Proof. Necessity of condition (i) and (ii) is easy to prove by the theorem B in Introduction.

Now suppose that (i) and (ii) are hold, and ( x j ) ∈ l ∞ ( X ) , then for any ε > 0 , there exists m 0 ∈ ℕ such that ‖ ∑ j = m 0 + 1 ∞ T i j ( x j ) ‖ < ε 3 for all i ∈ ℕ by the condition (ii). And because of condition (i) there is i 0 ∈ ℕ , such that ‖ T k j ( x j ) − T i j ( x j ) ‖ < ε 3 m 0 for all k , i > i 0 . Hence we have

‖ ∑ j = 1 m 0 T k j ( x j ) − ∑ j = 1 m 0 T i j ( x j ) ‖ ≤ ∑ j = 1 m 0 ‖ T k j ( x j ) − T i j ( x j ) ‖ < ε 3 (1)

for all k , i > i 0 . Therefore

‖ ∑ j = 1 ∞ T k j ( x j ) − ∑ j = 1 ∞ T i j ( x j ) ‖ = ‖ ∑ j = 1 m 0 T k j ( x j ) − ∑ j = 1 m 0 T i j ( x j ) + ∑ j = m 0 + 1 ∞ T k j ( x j ) − ∑ j = m 0 + 1 ∞ T i j ( x j ) ‖ ≤ ‖ ∑ j = 1 m 0 T k j ( x j ) − ∑ j = 1 m 0 T i j ( x j ) ‖ + ‖ ∑ j = m 0 + 1 ∞ T k j ( x j ) ‖ + ‖ ∑ j = m 0 + 1 ∞ T i j ( x j ) ‖ < ε 3 + ε 3 + ε 3 = ε

So { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ is a Cauchy sequence in Y. Therefore { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ converges by the completeness of Y, and then ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c ( Y ) ) . The sufficiency is proved. Q.E.D.

Since c 0 ( Y ) ⊆ c ( Y ) , we can get the next corollary by the theorem.

Corollary 1. Suppose that i , j ∈ ℕ , T i j ∈ F 0 ( X , Y ) , then ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) if and only if lim i T i j ( x ) = 0 for all j ∈ ℕ , x ∈ X , and for any ε > 0 , M > 0 , there exists m 0 ∈ ℕ such that ‖ ∑ j = m ∞ T i j ( x j ) ‖ ≤ ε for all m > m 0 , i ∈ ℕ , and { x j } ⊆ X with sup j ‖ x j ‖ ≤ M .

Proof. Necessity is clear by above theorem and the definition of c 0 ( Y ) .

Conversely, let ( x j ) ∈ l ∞ ( X ) , then for any ε > 0 , there exists m 0 ∈ ℕ such that ‖ ∑ j = m 0 + 1 ∞ T i j ( x j ) ‖ < ε 2 for all i ∈ ℕ . Since lim i T i j ( x ) = 0 for all j ∈ ℕ and x ∈ X , there is i 0 ∈ ℕ , such that ‖ T i j ( x j ) ‖ < ε 2 m 0 for all i > i 0 and j ∈ ℕ .

Hence we have

‖ ∑ j = 1 ∞ T i j ( x j ) − 0 ‖ ≤ ‖ ∑ j = 1 m 0 T i j ( x j ) ‖ + ‖ ∑ j = m 0 + 1 ∞ T i j ( x j ) ‖ ≤ ε 2 + ε 2 = ε . (2)

So column { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ converges to 0, and then ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) . The sufficiency is proved. Q.E.D.

Theorem 2. Let T i j ∈ F 0 ( X , Y ) with respect to i , j ∈ ℕ , then ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , l ∞ ( Y ) ) if and only if

(i) sup i ‖ T i j ( x ) ‖ < + ∞ for all j ∈ ℕ , x ∈ X ;

(ii) For any ε > 0 , M > 0 and t i ∈ c 0 , there exists m 0 ∈ ℕ such that | t i | ‖ ∑ j = m ∞ T i j ( x j ) ‖ < ε for all m > m 0 , i ∈ ℕ and { x j } ⊆ X with sup j ‖ x j ‖ ≤ M .

Proof. ⇒) Suppose that T i j ∈ F 0 ( X , Y ) , the condition (i) is clear.

Since { ∑ j = 1 ∞ T i j ( x j ) } ∈ l ∞ ( Y ) for every ( x j ) ∈ l ∞ ( X ) , { t i ∑ j = 1 ∞ T i j ( x j ) } ∈ c 0 ( Y ) for every ( t i ) ∈ c 0 by lemma 1, that is ∑ j = 1 ∞ ( t i T i j ) ( x j ) ∈ c 0 ( Y ) . Hence, for every ( t i ) ∈ c 0 , we have ( t i T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) . Therefore, by above corollary, for every ε > 0 , M > 0 and ( t i ) ∈ c 0 there is m 0 ∈ ℕ such that for all m > m 0 , i ∈ ℕ , and sup j ‖ x j ‖ ≤ M , we have

| t i | ‖ ∑ j = m ∞ T i j ( x j ) ‖ = ‖ ∑ j = m ∞ t i T i j ( x j ) ‖ < ε . (3)

condition (ii) is proved.

⇐) For every j ∈ ℕ , x ∈ X , and ( t i ) ∈ c 0 , we have lim i t i T i j ( x ) = 0 by the condition (i). Because of the condition (ii), we have ( t i T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) by the Corollary 1, and then for ( x j ) ∈ l ∞ ( X ) , we have { t i ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ ∈ c 0 ( Y ) is hold for every ( t i ) ∈ c 0 . Therefore { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ ∈ l ∞ ( Y ) by lemma 1, and then ( T i j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , l ∞ ( Y ) ) . Q.E.D.

Theorem 3. Let T i j ∈ F 0 ( X , Y ) with respect to i , j ∈ ℕ , then ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , c 0 ( Y ) ) if and only if

(i) lim i T i j ( x ) = 0 for all j ∈ ℕ and x ∈ X ;

(ii) For any ε > 0 , M > 0 and ( t j ) ∈ c 0 , there exists m 0 ∈ ℕ such that

sup i ∈ ℕ , m > m 0 , ‖ x j ‖ ≤ M ‖ ∑ j = m ∞ T i j ( t j x j ) ‖ < ε (4)

Proof. For T i j ∈ F 0 ( X , Y ) , since T i j ∘ t j ( x ) = T i j ( t j x ) , lim i T i j ∘ t j ( x ) = lim i T i j ( t j x ) = 0 for any j ∈ ℕ , x ∈ X and t j ∈ c 0 , by condition (i). So condition (i) and (ii) is equivalent to ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) by corollary 1.

Suppose that ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , c 0 ( Y ) ) , then { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ ∈ c 0 ( Y ) for all ( x j ) ∈ c 0 ( X ) . By lemma 1, { ∑ j = 1 ∞ T i j ( t j x j ) } i = 1 ∞ ∈ c 0 ( Y ) , for all ( x j ) ∈ l ∞ ( X ) and ( t j ) ∈ c 0 . Hence ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) .

On the other hand, suppose that ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , c 0 ( Y ) ) . For every ( x j ) ∈ c 0 ( X ) , there exists ( t j ) ∈ c 0 and ( z j ) ∈ c 0 ( X ) ⊂ l ∞ ( X ) , such that ( x j ) = ( t j z j ) , and so { ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ = { ∑ j = 1 ∞ ( T i j ∘ t j ) ( z j ) } i = 1 ∞ ∈ c 0 ( Y ) . Hence ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , c 0 ( Y ) ) . Q.E.D.

Theorem 4. Let T i j ∈ F 0 ( X , Y ) with respect to i , j ∈ ℕ , then ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , l ∞ ( Y ) ) if and only if

(i) sup i ‖ T i j ( x ) ‖ < + ∞ for all j ∈ ℕ and x ∈ X ;

(ii) For any ε > 0 , M > 0 and ( s j ) , ( t j ) ∈ c 0 , there exists m 0 ∈ ℕ , such that ‖ s j ∑ j = m ∞ T i j ( t j x j ) ‖ < ε for all m > m 0 , i ∈ ℕ and sup j ‖ x j ‖ ≤ M .

Proof. By condition (i), for all ( t j ) ∈ c 0 and x ∈ X ,

sup i ‖ T i j ∘ t j ( x ) ‖ = sup i ‖ T i j ( t j ( x ) ) ‖ < + ∞ . (5)

By theorem 2, condition (i) and (ii) are equivalent to ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , l ∞ ( Y ) ) . Next, we prove that ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , l ∞ ( Y ) ) for all ( t j ) ∈ c 0 is equivalent to ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , l ∞ ( Y ) ) .

In fact, If ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , l ∞ ( Y ) ) , and let ( x j ) ∈ l ∞ ( X ) , then ∑ j = 1 ∞ ( T i j ∘ t j ) ( x j ) = ∑ j = 1 ∞ T i j ( t j x j ) . Since ( t j x j ) ∈ c 0 ( X ) for all ( t j ) ∈ c 0 by lemma, we have { ∑ j = 1 ∞ T i j ( t j x j ) } i = 1 ∞ ∈ l ∞ ( Y ) . So ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , l ∞ ( Y ) ) . On the other hand, suppose that ( T i j ∘ t j ) i , j ∈ ℕ ∈ ( l ∞ ( X ) , l ∞ ( Y ) ) . Since for any ( x j ) ∈ c 0 ( X ) , there must be ( t j ) ∈ c 0 , ( z j ) ∈ c 0 ( X ) ⊂ l ∞ ( X ) , such that ( x j ) = ( t j z j ) , we have

{ ∑ j = 1 ∞ T i j ( x j ) } i = 1 ∞ = { ∑ j = 1 ∞ T i j ( t j z j ) } i = 1 ∞ = { ∑ j = 1 ∞ T i j ∘ t j ( z j ) } i = 1 ∞ , (6)

and then ( T i j ) i , j ∈ ℕ ∈ ( c 0 ( X ) , l ∞ ( Y ) ) . Q.E.D.

In this paper, we first review the research history of infinite matrix transformation, and then we mainly study the summability of a class of nonlinear mapping matrices in Banach space.

And some new results about, matrix transformation theorems are obtained: we characterize the matrix classes such as ( l ∞ ( X ) , c ( Y ) ) , ( l ∞ ( X ) , l ∞ ( Y ) ) , ( c 0 ( X ) , c 0 ( Y ) ) , ( c 0 ( X ) , l ∞ ( Y ) ) .

This research was supported by the Science and Technology Project of Jilin Provincial Department of Education (JJKH20180891KJ).

The authors declare no conflicts of interest regarding the publication of this paper.

Hua, N., Kang, N. and Liao, H. (2020) Operator Matrices on Banach Spaces. Open Access Library Journal, 7: e6813. https://doi.org/10.4236/oalib.1106813