1 Introduction and Preliminaries

The concept of a multiplier first appears in harmonic analysis in connection with the theory of summability for Fourier series. Subsequently, the notion has been employed in other areas of harmonic analysis, such as the investigation of homomorphisms of group algebras, in the general theory of Banach algebras, and so on; see [5]. Many authors generalized the notion of a multiplier in different ways. See [1, 6], for one of this generalizations.

In this paper, our main concern will not be with these applications of the theory of multipliers and its generalizations. We only develop the theory of multipliers differently from the previous ways, by introducing a new class of operators from a Banach algebra A into a Banach A-bimodule X.

Let A be a Banach algebra and a,b∈A. Define a bounded bilinear functional on A ∗×A ∗ as

$$(a\otimes b)(f,g)=f(a)g(b)\qquad (f, g\in A^{*}). $$

The projective tensor product space \(A\widehat {\otimes }A\) is a Banach algebra and a Banach A-bimodule that is characterized as follows

$$\left\{\sum\limits_{n=1}^{\infty}a_{n}\otimes b_{n}: n\in \mathbb{N}, a_{n},b_{n}\in A, \sum\limits_{n=1}^{\infty}\|a_{n}\| \|b_{n}\|<\infty\right\}, $$

and its module actions are defined by

$$a\cdot(b\otimes c)=ab\otimes c,\quad (b\otimes c)\cdot a=b\otimes ca\qquad (a,b,c\in A). $$

A Banach algebra A is called nilpotent if there exists an integer n≥2 such that

$$A^{n}=\{a_{1}a_{2}a_{3}{\dots} a_{n}: a_{1}, a_{2}, a_{3},\dots,a_{n}\in A\}=\{0\}. $$

The minimum of numbers n that A n={0} is called the index of A which we denote by I(A), i.e., if I(A) = n, then there exists a 1,a 2,…,a n−1∈A such that a 1 a 2…a n−1≠0.

To see an example of a nilpotent Banach algebra, suppose that B is a Banach algebra and let A be defined as follows

$$A=\left[\begin{array}{lllll} 0 & B & B & B & B\\ 0 & 0 & B & B & B\\ 0 & 0 & 0 & B & B\\ 0 & 0 & 0 & 0 & B\\ 0 & 0 & 0 & 0 & 0 \end{array} \right]. $$

Then, A is a Banach algebra equipped with the usual matrix-like operations and l ∞ -norm such that A is nilpotent with I(A)=5.

For undefined concepts and notations appearing in the sequel, one can consult [3].

2 n-Multipliers

We start this section with the main object of the paper.

Definition 1

Let A be a Banach algebra, X be a Banach A-bimodule and T:A→X be a bounded linear map. We say that T is an n-multiplier (n≥2) if

$$T(a_{1}a_{2}{\dots} a_{n})=a_{1}\cdot T(a_{2}{\dots} a_{n})\quad(a_{1}, a_{2}, a_{3},\dots,a_{n}\in A). $$

We will denote by Mul n (A,X) the set of all n-multipliers of Banach algebra A into X. Now, we study in more details the space Mul n (A,X) when n≥3 (in the case n=2 this is the space of all multipliers in the classical sense).

Let A be a Banach algebra and X be a Banach A-bimodule. The set Mul n (A,X) is a vector subspace of B(A,X); the space of all bounded linear maps from A into X.

As the first result, we show that Mul n (A,X) is a closed vector subspace of B(A,X).

Theorem 1

Let A be a Banach algebra and X be a Banach A-bimodule. Then for all integers n≥3, the space Mul n (A,X) is a closed vector subspace of B(A,X).

Proof

We claim that Mul n (A,X) is closed in B(A,X). Suppose that {T m } is a sequence in Mul n (A,X) converging to T∈B(A,X).

Let a 1,a 2,…,a n be arbitrary elements of A. So, we have

$$\begin{array}{@{}rcl@{}} \|T(a_{1}{\dots} a_{n})-a_{1}\cdot T(a_{2}{\dots} a_{n})\|&\leq& \|T(a_{1}{\dots} a_{n})-T_{m}(a_{1}{\dots} a_{n})\|\\ &&+\|T_{m}(a_{1}{\dots} a_{n})-a_{1}\cdot T(a_{2}{\dots} a_{n})\|\\ &\leq& \|T-T_{m}\| \|a_{1}{\dots} a_{n}\|\\ &&+ \|a_{1}\cdot T_{m}(a_{2}{\dots} a_{n})-a_{1}\cdot T(a_{2}{\dots} a_{n})\|\\ &\leq& \|T-T_{m}\| \|a_{1}{\dots} a_{n}\| + \|T-T_{m}\| \|a_{1}\| \|a_{2}{\dots} a_{n}\|. \end{array} $$

If m→∞, we conclude that T(a 1…a n ) = a 1⋅T(a 2…a n ). The rest of the proof is easy. □

Similarly, one can see that Mul n (A,X) is complete in the strong operator topology (SOT), i.e., in the topology on B(A,X) for which a net {T α } converges to T if and only if for each a∈A, ∥T a−T α a∥→0.

In the next two theorems, we give some relations between the spaces of n-multipliers.

Theorem 2

There exist a Banach algebra A and a Banach A-bimodule X such that for all positive integers n≥3

$$\text{Mul}_{2}(A,X)\subsetneq \text{Mul}_{3}(A,X)\subsetneq{\cdots} \subsetneq \text{Mul}_{n}(A,X). $$

Proof

For every positive integer n≥3, take A being a nilpotent Banach algebra with I(A) = n and \(X=A\widehat {\otimes }A\). So, there exist non-zero elements a 1,a 2,…,a n−1∈A such that a 1 a 2…a n−1≠0.

The verification of the above chain of inclusions is easy. We only show each of the strict relations. For every integer number i such that 2≤i<n, define a linear map \(T_{i}:A\rightarrow A\widehat {\otimes }A\) by

$$T_{i}(a)=a_{1}a_{2}{\dots} a_{n-1}\otimes a_{1}a_{2}{\dots} a_{n-(i+1)}a\qquad (a\in A). $$

Thus, T i is an element of \(\text {Mul}_{i+1}(A, A\widehat {\otimes }A)\), but it does not belong to \(\text {Mul}_{i}(A, A\widehat {\otimes }A)\). To see this, let f∈A ∗ be a functional such that f(a 1 a 2…a n−1)≠0. So

$$\begin{array}{@{}rcl@{}} T_{i}(a_{n-i}{\ldots} a_{n-1})(f,f)&=&(a_{1}a_{2}{\ldots} a_{n-1}\otimes a_{1}a_{2}{\ldots} a_{n-1})(f,f)\\ &=&f(a_{1}a_{2}{\ldots} a_{n-1})^{2}\neq 0. \end{array} $$

Therefore, T i (a n−i …a n−1)≠0, but a n−i ⋅T i (a (n−i)+1…a n−1)=0 and this completes the proof. □

For a Banach algebra A, let A 2=span{a b:a,b∈A}. The Banach algebra A is essential if \(\overline {A^{2}}=A\).

Theorem 3

Let A be an essential Banach algebra and X be a Banach A-bimodule. Then for all integers n≥3, we have

$$\text{Mul}_{n-1}(A,X)=\text{Mul}_{n}(A,X). $$

Specially, Mul 2 (A,X)=Mul n (A,X) for all n≥2.

Proof

Let T∈Mul n (A,X), A be essential and a 1,…,a n−1 be arbitrary elements of A. We show that T∈Mul n−1(A,X). Since \(a_{2}\in A=\overline {A^{2}}\), there exists a net {a 2,α }∈A 2 with \(a_{2,\alpha }={\sum }_{i_{\alpha }} \beta _{i_{\alpha }}b_{i_{\alpha }}c_{i_{\alpha }}\) for \(b_{i_{\alpha }}, c_{i_{\alpha }}\in A\) and \(\beta _{i_{\alpha }}\in \mathbb {C}\), such that \(a_{2}=\lim _{\alpha }a_{2,\alpha }\). So, we have

$$\begin{array}{@{}rcl@{}} T(a_{1}a_{2}{\dots} a_{n-1})&=&\lim\limits_{\alpha}T\left( a_{1}\left( \sum\limits_{i_{\alpha}} \beta_{i_{\alpha}}b_{i_{\alpha}}c_{i_{\alpha}}\right)a_{3}{\dots} a_{n-1}\right)\\ &=&\lim\limits_{\alpha}\sum\limits_{i_{\alpha}}\beta_{i_{\alpha}}T(\overbrace{a_{1}b_{i_{\alpha}}c_{i_{\alpha}}a_{3}{\dots} a_{n-1}}^{n})\\ &=&\lim\limits_{\alpha}\sum\limits_{i_{\alpha}}\beta_{i_{\alpha}}a_{1}\cdot T(b_{i_{\alpha}}c_{i_{\alpha}}a_{3}{\dots} a_{n-1})\\ &=&a_{1}\cdot T(a_{2}{\dots} a_{n-1}), \end{array} $$

which completes the proof. □

3 Relations with n-Homomorphisms

Let A be a Banach algebra and n≥3 be a positive integer. Here, we denote the space Mul n (A,A) briefly by Mul n (A).

We know that the space of all multipliers on A is a Banach subalgebra of B(A) = B(A,A) with composition of operators as product and the operator norm. But in general, the space of n-multipliers on A is not an algebra with composition of operators. So, we should define another product on this space to make Mul n (A) into a Banach algebra.

Now, let a 0∈A and consider \(\bullet _{a_{0}}:\text {Mul}_{n}(A)\times \text {Mul}_{n}(A)\rightarrow \text {Mul}_{n}(A)\) which is defined by

$$ S\bullet_{a_{0}} T(a):=S(T(a)a_{0}^{n-2})\qquad(a\in A). $$
(1)

Without losing the generality we assume that ∥a 0∥≤1.

Theorem 4

Let A be a Banach algebra. Then for all positive integers n≥3, Mul n (A) is a Banach algebra, with the product “ \(\bullet _{a_{0}}\) ” and the operator norm.

Proof

Clearly, Mul n (A) is a vector space with operations that inherit from B(A). Let S,T∈Mul n (A). First, we show that \(S\bullet _{a_{0}}T\) is well-defined, i.e., \(S\bullet _{a_{0}}T\in \text {Mul}_{n}(A)\). Let a 1,a 2,…,a n ∈A, we have

$$\begin{array}{@{}rcl@{}} S\bullet_{a_{0}}T(a_{1}{\dots} a_{n})=S(T(a_{1}{\dots} a_{n})a_{0}^{n-2})&=&S(a_{1}T(a_{2}{\dots} a_{n})a_{0}^{n-2})\\ &=&a_{1} S(T(a_{2}{\dots} a_{n})a_{0}^{n-2})\\ &=&a_{1}(S\bullet_{a_{0}}T)(a_{2}{\dots} a_{n}). \end{array} $$

Therefore, \(S\bullet _{a_{0}}T\) is an n-multiplier on A.

Let T 1,T 2, and T 3 be elements of Mul n (A). We have

$$\begin{array}{@{}rcl@{}} (T_{1}\bullet_{a_{0}} T_{2})\bullet_{a_{0}} T_{3}(a)&=&(T_{1}\bullet_{a_{0}} T_{2})(T_{3}(a)a_{0}^{n-2})=T_{1}(T_{2}(T_{3}(a)a_{0}^{n-2})a_{0}^{n-2}),\\ T_{1}\bullet_{a_{0}} (T_{2}\bullet_{a_{0}} T_{3})(a)&=&T_{1}((T_{2}\bullet_{a_{0}}T_{3})(a)a_{0}^{n-2})=T_{1}(T_{2}(T_{3}(a)a_{0}^{n-2})a_{0}^{n-2}). \end{array} $$

Hence, the product “\(\bullet _{a_{0}}\)” is associative.

On the other hand, Theorem 1 shows that Mul n (A) is a closed vector subspace of B(A) with the operator norm. The investigation of the other properties is easy. □

Recall that for Banach algebras A and B a linear map ϕ:A→B is called an n-homomorphism if, ϕ(a 1 a 2…a n ) = ϕ(a 1)ϕ(a 2)…ϕ(a n ) for all a 1,a 2,…,a n ∈A [4].

For each integer n≥2, suppose that Δ n (A) denotes the n-character space of A, i.e., the space consisting of all non-zero n-homomorphisms from A into \(\mathbb {C}\). It is clear that for every integer n≥3, Δ2(A)⊆Δ n (A). The last inclusion may be strict. As an example for n=3, if ϕ∈Δ2(A), then φ:=−ϕ is in Δ3(A), but φ is not a 2-character from A into \(\mathbb {C}\).

Let ϕ n ∈Δ n (A). Define \(\widetilde {\phi }_{n}: (\text {Mul}_{n}(A), \bullet _{a_{0}}) \to \mathbb {C}\) by

$$\widetilde{\phi}_{n}(T)=\phi_{n}(T(a_{0}^{n-1}))\quad (T\in \text{Mul}_{n}(A)). $$

Clearly, \(\widetilde {\phi }_{n}\) is a linear operator. We say that \(\widetilde {\phi }_{n}\) extends ϕ n if, \(\widetilde {\phi }_{n}(L_{a})=\phi _{n}(a)\) for all a∈A.

In the next theorem, under some mild conditions, we show that Δ n (Mul n (A))≠∅. Recall that Z(A) denotes the center of A, i.e.,

$$Z(A)=\{a\in A : ab=ba\quad (b\in A)\}. $$

Theorem 5

Let A be a Banach algebra and let a 0 ∈Z(A)∖{0}. Then \(\widetilde {\phi }_{n}\in {\Delta }_{n}(\text {Mul}_{n}(A))\) is an extension of ϕ n ∈Δ n (A) if ϕ n (a 0 )=1.

Proof

Suppose that there exists ϕ n ∈Δ n (A) with ϕ n (a 0)=1. We must show that \(\widetilde {\phi }_{n}\) is a non-zero n-homomorphism. For each a∈A, we have

$$\widetilde{\phi}_{n}(L_{a})=\phi_{n}(L_{a}(a_{0}^{n-1}))=\phi_{n}(aa_{0}^{n-1})=\phi_{n}(a)\phi_{n}(a_{0})^{n-1}= \phi_{n}(a). $$

Therefore, in the special case when a = a 0, we have \(\widetilde {\phi }_{n}(L_{a_{0}})=1\). So, \(\widetilde {\phi }_{n}\) is a non-zero extension of ϕ n .

On the other hand, for T 1,T 2,…,T n ∈Mul n (A), we have

$$\begin{array}{@{}rcl@{}} \widetilde{\phi}_{n}(T_{1}\bullet_{a_{0}}\ldots\bullet_{a_{0}}T_{n})&=&\phi_{n}(T_{1}\bullet_{a_{0}}{\ldots} \bullet_{a_{0}}T_{n}(a_{0}^{n-1}))\\ &=&\phi_{n}(T_{1}(T_{2}\overbrace{(\ldots(}^{n-2} T_{n}(a_{0}^{n-1})a_{0}^{n-2}\overbrace{)\ldots)}^{n-2}a_{0}^{n-2}))\\ &=&\phi_{n}(a_{0})^{n-1}\phi_{n}(T_{1}(T_{2}\overbrace{(\ldots(}^{n-2} T_{n}(a_{0}^{n-1})a_{0}^{n-2}\overbrace{)\ldots)}^{n-2}a_{0}^{n-2}))\\ &=&\phi_{n}(a_{0}^{n-1}T_{1}(T_{2}\overbrace{(\ldots(}^{n-2} T_{n}(a_{0}^{n-1})a_{0}^{n-2}\overbrace{)\ldots)}^{n-2}a_{0}^{n-2}))\\ &=&\phi_{n}(T_{1}(T_{2}\overbrace{(\ldots(}^{n-2} T_{n}(a_{0}^{n-1})a_{0}^{n-1}\overbrace{)\ldots)}^{n-2}a_{0}^{n-1}))\\ &=&\cdots\\ &=&\phi_{n}(T_{1}(a_{0}^{n-1})T_{2}(a_{0}^{n-1}){\ldots} T_{n}(a_{0}^{n-1}))\\ &=&\phi_{n}(T_{1}(a_{0}^{n-1}))\phi_{n}(T_{2}(a_{0}^{n-1})){\ldots} \phi_{n}(T_{n}(a_{0}^{n-1}))\\ &=&\widetilde{\phi}_{n}(T_{1})\widetilde{\phi}_{n}(T_{2}){\ldots} \widetilde{\phi}_{n}(T_{n}). \end{array} $$

Therefore, \(\widetilde {\phi }_{n}\) is an n-homomorphism and this completes the proof. □

4 Approximate Local n-Multipliers

In [7], Samei investigated the approximately local left 2-multipliers and study some of its relations with left 2-multipliers on a Banach algebra A. In this section, we give two theorems similar to Theorem 2.2 and Proposition 2.3 of [7] for n-multipliers. Indeed, we are interested in determining when an approximately local n-multiplier (Definition 2) is an n-multiplier. First, we give the following definition.

Definition 2

Let X be a Banach A-module and T:A→X be a bounded linear operator. We say that T is an approximately local n-multiplier if, for each a∈A, there exists a sequence {T a,m } of n-multipliers such that \(T(a)=\lim _{m}T_{a,m}(a)\).

We recall the algebraic reflexivity from [2]. Let X and Y be Banach spaces and S be a subset of B(X,Y). Put

$$\text{ref}(S)=\{T\in B(X,Y): T(x)\in \overline{\{s(x):s\in S\}} (x\in X)\}. $$

Then S is algebraically reflexive if, S=ref(S) or just ref(S)⊆S.

Theorem 6

Let A be a Banach algebra and X be a Banach A-module. Then the following statements are equivalent.

  1. 1.

    Every approximately local n-multiplier from A into X is an n-multiplier (n≥3).

  2. 2.

    Mul n (A,X) is algebraically reflexive.

Proof

(1)⇒(2) Let T∈ref(Mul n (A,X)). So, for all a∈A, there exists a sequence {T m } in Mul n (A,X) such that, \(T(a)=\lim _{m}T_{a,m}(a)\). Hence, T is an approximately local n-multiplier. Therefore, T is an n-multiplier by assumption and this shows that Mul n (A,X) is algebraically reflexive.

(2)⇒(1) Let T:A→X be an approximately local n-multiplier. So, for all a∈A, there exists a sequence {T a,m } such that, \(T(a)=\lim _{m}T_{a,m}(a)\). Hence, T∈ref(Mul n (A,X)) and reflexivity of Mul n (A,X) implies that T is an n-multiplier. □

Let A be a Banach algebra and X be a Banach A-module. Then for each x∈X, the left annihilator of x in A is defined by x ⊥={a∈A:a⋅x=0}.

Theorem 7

Suppose that A is a Banach algebra such that Mul n (A,A ∗ ) is algebraically reflexive and X is a left Banach A-module with {x∈X:x ⊥ =A}=0. Then every approximately local n-multiplier from A into X is an n-multiplier.

Proof

Let T:A→X be an approximately local n-multiplier and f∈X ∗. Define a map \(\mathfrak {M}_{f}:X\rightarrow A^{*}\) as follows

$$\mathfrak{M}_{f}(x)=x\bullet f\quad(x\in X), $$

where x∙f∈A ∗ is defined by x∙f(a) = f(a⋅x) for all a∈A. Therefore, \(\mathfrak {M}_{f}\) is a bounded right A-module morphism. Because, for a∈A and x∈X, we have

$$\mathfrak{M}_{f}(a\cdot x)=(a\cdot x)\bullet f=a\cdot (x\bullet f)=a\cdot \mathfrak{M}_{f}(x), $$

so

$$\mathfrak{M}_{f}\circ T\in \text{ref}(\text{Mul}_{n}(A,A^{*}))=\text{Mul}_{n}(A,A^{*}). $$

Thus, \(\mathfrak {M}_{f}\circ T\in \text {Mul}_{n}(A,A^{*})\). Now, for a 1,a 2,a 3,…,a n ∈A, we have

$$\begin{array}{@{}rcl@{}} \mathfrak{M}_{f}(T(a_{1}a_{2}a_{3}{\dots} a_{n}))=\mathfrak{M}_{f}\circ T(a_{1}a_{2}a_{3}{\dots} a_{n})&=&a_{1}\cdot\mathfrak{M}_{f}\circ T(a_{2}a_{3}{\dots} a_{n})\\ &=&a_{1}\cdot\mathfrak{M}_{f}(T(a_{2}a_{3}{\dots} a_{n}))\\ &=&\mathfrak{M}_{f}(a_{1}\cdot T(a_{2}a_{3}{\dots} a_{n})). \end{array} $$

Therefore, \(\mathfrak {M}_{f}(T(a_{1}a_{2}a_{3}{\dots } a_{n})-a_{1}\cdot T(a_{2}a_{3}{\dots } a_{n}))=0\). If we put

$$u=T(a_{1}a_{2}a_{3}{\dots} a_{n})-a_{1}\cdot T(a_{2}a_{3}{\dots} a_{n}), $$

then f(a⋅u)=0 for all a∈A. So, by Hahn-Banach’s theorem, we have a⋅u=0 for all a∈A. So, u ⊥ = A and this implies that u=0. Hence, T is an n-multiplier. □