Regularity of bounded tri-linear and the fourth adjiont of tri-derivation

TL;DR

This paper establishes a criterion for the regularity of bounded tri-linear mappings and explores conditions under which the fourth adjoint of a tri-derivation remains a tri-derivation.

Contribution

It provides a simple criterion for regularity of tri-linear mappings and characterizes when the fourth adjoint of a tri-derivation is also a tri-derivation.

Findings

A necessary and sufficient condition for the regularity of a tri-linear mapping.

Characterization of when the fourth adjoint of a tri-derivation is a tri-derivation.

A criterion linking the regularity of a tri-linear map to the regularity of its composition with a bounded linear map.

Abstract

In this Article, we give a simple criterion for the regularity of a tri-linear mapping. We provide if $f:X\times Y\times Z\longrightarrow W $ is a bounded tri-linear mapping and $h:W\longrightarrow S$ is a bounded linear mapping, then $f$ is regular if and only if $hof$ is regular. We also shall give some necessary and sufficient conditions such the fourth adjoint $D^{***}$ of a tri-derivation $D$ is again tri-derivation.

Equations

$$a^{\ast \ast }\square b^{\ast \ast }={\pi }^{\ast \ast \ast }(a^{\ast \ast },b^{\ast \ast }),a^{\ast \ast }◊b^{\ast \ast }={\pi }^{r\ast \ast \ast r}(a^{\ast \ast },b^{\ast \ast }).$$

$$w^{\ast }-\underset{\alpha }{\lim }{w}^{\ast }-\underset{\beta }{\lim }\pi (a_{\alpha },b_{\beta })=w^{\ast }-\underset{\beta }{\lim }{w}^{\ast }-\underset{\alpha }{\lim }\pi (a_{\alpha },b_{\beta }).$$

$$D_1(\pi (a,b))={\pi }_1^{\ast }(D_1(a),b)+{\pi }_2^{r\ast r}(a,D_1(b)).$$

$$D_2(\pi (a,b),c)={\pi }_1(a,D_2(b,c))+{\pi }_2(D_2(a,c),b),D_2(a,\pi (b,c))={\pi }_1(b,D_2(a,c))+{\pi }_2(D_2(a,b),c).$$

$$w^{\ast }-\underset{\alpha }{\lim }{w}^{\ast }-\underset{\beta }{\lim }{w}^{\ast }-\underset{\gamma }{\lim }f(x_{\alpha },y_{\beta },z_{\gamma })=w^{\ast }-\underset{\gamma }{\lim }{w}^{\ast }-\underset{\beta }{\lim }{w}^{\ast }-\underset{\alpha }{\lim }f(x_{\alpha },y_{\beta },z_{\gamma }),$$

$$f^{\ast \ast \ast r\ast }(X^{\ast \ast },W^{\ast },Z^{\ast \ast })\subseteq Y^{\ast \ast \ast }=Y^{\ast }(2-1)$$

$$f^{\ast \ast \ast \ast \ast }(W^{\ast \ast \ast },X^{\ast \ast },Y^{\ast \ast })\subseteq Z^{\ast \ast \ast }=Z^{\ast }(2-2)$$

$$f:L^1(G)\times L^p(G)\times L^q(G)⟶L^r(G)$$

$$f:L^1(G)\times L^1(G)\times L^1(G)⟶L^1(G)$$

$$\begin{gathered} \int \psi d({\mu }_1\ast {\mu }_2)=\int \int \psi (xy)d{\mu }_1(x)d{\mu }_2(y), \\ (\psi \in C_0(G)). \end{gathered}$$

$$\int \psi d({\mu }_1\ast ({\mu }_2\ast {\mu }_3))=\int \int \int \psi (xyz)d{\mu }_1(x)d{\mu }_2(y)d{\mu }_3(z)=\int \psi d(({\mu }_1\ast {\mu }_2)\ast {\mu }_3)$$

$$f:M(G)\times M(G)\times M(G)⟶M(G)$$

$$(hof{)}^{r\ast \ast \ast \ast r}(x^{\ast \ast },y^{\ast \ast },z^{\ast \ast })=(hof{)}^{\ast \ast \ast \ast }(x^{\ast \ast },y^{\ast \ast },z^{\ast \ast }).$$

$$⟨f^{\ast }(w^{\ast },x,y),z⟩=⟨w^{\ast },f(x,y,z)⟩=⟨w^{\ast },g_1(h_1(x),y,z)⟩=⟨g_1^{\ast }(w^{\ast },h_1(x),y),z⟩.$$

$$⟨f^{\ast \ast }(z^{\ast \ast },w^{\ast },x),y⟩=⟨z^{\ast \ast },f^{\ast }(w^{\ast },x,y)⟩=⟨z^{\ast \ast },g_1^{\ast }(w^{\ast },h_1(x),y)⟩=⟨g_1^{\ast \ast }(z^{\ast \ast },w^{\ast },h_1(x)),y⟩.$$

$$f^{\ast \ast \ast r\ast }(X^{\ast \ast },W^{\ast },Z^{\ast \ast })\subseteq Y^{\ast },f^{\ast \ast \ast \ast \ast }(W^{\ast \ast \ast },X^{\ast \ast },Y^{\ast \ast })\subseteq Z^{\ast }.$$

$$f^{\ast \ast \ast r\ast }(X^{\ast \ast },W^{\ast },Z)\subseteq Y^{\ast },f^{\ast \ast \ast \ast \ast }(W^{\ast },X^{\ast \ast },Y^{\ast \ast })\subseteq Z^{\ast }.$$

$$\parallel a\parallel =({\Sigma }_{i,j}|{\alpha }_{ij}{|}^2{)}^{\frac{1}{2}},(a=({\alpha }_{ij})\in A).$$

$$D(a,b,c)=[\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),abc],(a,b,c\in A).$$

$$(case1)D^{\ast \ast \ast \ast }:(A^{\ast \ast },\square )\times (A^{\ast \ast },\square )\times (A^{\ast \ast },\square )⟶X^{\ast \ast }$$

$$(case2)D^{\ast \ast \ast \ast }:(A^{\ast \ast },◊)\times (A^{\ast \ast },\square )\times (A^{\ast \ast },\square )⟶X^{\ast \ast }$$

$$(case3)D^{\ast \ast \ast \ast }:(A^{\ast \ast },\square )\times (A^{\ast \ast },◊)\times (A^{\ast \ast },\square )⟶X^{\ast \ast }$$

$$(case4)D^{\ast \ast \ast \ast }:(A^{\ast \ast },\square )\times (A^{\ast \ast },\square )\times (A^{\ast \ast },◊)⟶X^{\ast \ast }$$

$$(case5)D^{\ast \ast \ast \ast }:(A^{\ast \ast },◊)\times (A^{\ast \ast },◊)\times (A^{\ast \ast },\square )⟶X^{\ast \ast }$$

$$(case6)D^{\ast \ast \ast \ast }:(A^{\ast \ast },◊)\times (A^{\ast \ast },\square )\times (A^{\ast \ast },◊)⟶X^{\ast \ast }$$

$$(case7)D^{\ast \ast \ast \ast }:(A^{\ast \ast },\square )\times (A^{\ast \ast },◊)\times (A^{\ast \ast },◊)⟶X^{\ast \ast }$$

$$(case8)D^{\ast \ast \ast \ast }:(A^{\ast \ast },◊)\times (A^{\ast \ast },◊)\times (A^{\ast \ast },◊)⟶X^{\ast \ast }$$

$$w^{\ast }-\underset{\alpha }{\lim }{w}^{\ast }-\underset{\tau }{\lim }{w}^{\ast }-\underset{\beta }{\lim }{w}^{\ast }-\underset{\gamma }{\lim }{\pi }_2(D(a_{\alpha },b_{\beta },c_{\gamma }),d_{\tau })={\pi }_2^{\ast \ast \ast }(D^{\ast \ast \ast \ast }(a^{\ast \ast },b^{\ast \ast },c^{\ast \ast }),d^{\ast \ast }).$$

$$D^{\ast \ast \ast \ast }({\pi }^{\ast \ast \ast }(a^{\ast \ast },d^{\ast \ast }),b^{\ast \ast },c^{\ast \ast })={\pi }_2^{\ast \ast \ast }(D^{\ast \ast \ast \ast }(a^{\ast \ast },b^{\ast \ast },c^{\ast \ast }),d^{\ast \ast })+{\pi }_1^{\ast \ast \ast }(a^{\ast \ast },D^{\ast \ast \ast \ast }(d^{\ast \ast },b^{\ast \ast },c^{\ast \ast })).(4-1)$$

$$w^{\ast }-\underset{\alpha }{\lim }{w}^{\ast }-\underset{\beta }{\lim }{w}^{\ast }-\underset{\tau }{\lim }{w}^{\ast }-\underset{\gamma }{\lim }{\pi }_2(D(a_{\alpha },b_{\beta },c_{\gamma }),d_{\tau })={\pi }_2^{\ast \ast \ast }(D^{\ast \ast \ast \ast }(a^{\ast \ast },b^{\ast \ast },c^{\ast \ast }),d^{\ast \ast }),$$

Full text

1. Introduction and Preliminaries

Richard Arens showed in [3] that a bounded bilinear map $m:X\times Y\longrightarrow Z$ on normed spaces, has two natural different extensions $m^{***}$, $m^{r***r}$ from $X^{**}\times Y^{**}$ into $Z^{**}$. When these extensions are equal, $m$ is said to be Arens regular. A Banach algebra $A$ is said to be Arens regular, if its product $\pi(a,b)=ab$ considered as a bilinear mapping $\pi:A\times A\longrightarrow A$ isَ Arens regular. The first and second Arens products of $A^{**}$ by symbols $\square$ and $\lozenge$ respectively defined by

[TABLE]

Arens regularity of $\pi$ is equivalent to the following

[TABLE]

given by $\{a_{\alpha}\}$ and $\{b_{\beta}\}$ are bounded nets in $A$ and $a^{**}=w^{*}-\lim\limits_{\alpha}a_{\alpha}$ and $b^{**}=w^{*}-\lim\limits_{\beta}b_{\beta}$ are elements of $A^{**}$. Some characterizations for the Arens regularity of bounded bilinear map $m$ and Banach algebra $A$ are proved in [1, 2, 3, 4, 5, 11, 14, 16] and [17]. Suppose $X,Y,Z,W$ and $S$ be normed spaces and $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping. In this paper we first define regularity of $f$ map and showing that $f$ is regular if and only if $f^{***r*}(X^{**},W^{*},Z)\subseteq Y^{*}$ and $f^{*****}(W^{*},X^{**},Y^{**})\subseteq Z^{*}$. Also we show that be a bounded tri-linear maps $f:X\times Y\times Z\longrightarrow W$ and $h:W\longrightarrow S$, $f$ is regular if and only if $hof$ is regular.

The natural extensions of $f$ are as follows:

1. (1)

$f^{*}:W^{*}\times X\times Y\longrightarrow Z^{*}$, given by $\langle f^{*}(w^{*},x,y),z\rangle=\langle w^{*},f(x,y,z)\rangle$ where $x\in X,y\in Y,z\in Z,w^{*}\in W^{*}$ ($f^{*}$ is said the adjoint of $f$ and is a bounded tri-linear map). 2. (2)

$f^{**}=(f^{*})^{*}:Z^{**}\times W^{*}\times X\longrightarrow Y^{*}$, given by $\langle f^{**}(z^{**},w^{*},x),y\rangle=\langle z^{**},f^{*}(w^{*},x,y)$ where $x\in X,y\in Y,z^{**}\in Z^{**},w^{*}\in W^{*}$. 3. (3)

$f^{***}=(f^{**})^{*}:Y^{**}\times Z^{**}\times W^{*}\longrightarrow X^{*}$, given by $\langle f^{***}(y^{**},z^{**},w^{*}),x\rangle=\langle y^{**},f^{**}(z^{**},w^{*},x)\rangle$ where $x\in X,y^{**}\in Y^{**},z^{**}\in Z^{**},w^{*}\in W^{*}$. 4. (4)

$f^{****}=(f^{***})^{*}:X^{**}\times Y^{**}\times Z^{**}\longrightarrow W^{**}$, given by $\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle x^{**},f^{***}(y^{**},z^{**},w^{*})\rangle$ where $x^{**}\in X^{**},y^{**}\in Y^{**},z^{**}\in Z^{**},w^{*}\in W^{*}$.

Now let $f^{r}:Z\times Y\times X\longrightarrow W$ be the flip of $f$ defined by $f^{r}(z,y,x)=f(x,y,z)$, for every $x\in X,y\in Y$ and $z\in Z$. Then $f^{r}$ is a bounded tri-linear map and it may extends as above to $f^{r****}:Z^{**}\times Y^{**}\times X^{**}\longrightarrow W^{**}$. When $f^{****}$ and $f^{r****r}$ are equal, then $f$ is called regular.

Suppose $A$ be a Banach algebra and $\pi_{1}:A\times X\longrightarrow X$ is a bounded bilinear map. The pair $(\pi_{1},X)$ is said to be a left Banach $A-$module when $\pi_{1}(\pi_{1}(a,b),x)=\pi_{1}(a,\pi_{1}(b,x))$, for each $a,b\in A$ and $x\in X$. A right Banach $A-$module may is defined similarly. Let $\pi_{2}:X\times A\longrightarrow X$ be a bounded bilinear map. The pair $(X,\pi_{2})$ is said to be a right Banach $A-$module if $\pi_{2}(x,\pi_{2}(a,b))=\pi_{2}(\pi_{2}(x,a),b)$. A triple $(\pi_{1},X,\pi_{2})$ is said to be a Banach $A-$module if $(X,\pi_{1})$ and $(X,\pi_{2})$ are left and right Banach $A-$modules, respectively, and $\pi_{1}(a,\pi_{2}(x,b))=\pi_{2}(\pi_{1}(a,x),b)$. Let $(\pi_{1},X,\pi_{2})$ be a Banach $A-$module. Then $(\pi_{2}^{r*r},X^{*},\pi_{1}^{*})$ is the dual Banach $A-$module of $(\pi_{1},X,\pi_{2})$.

A bounded linear mapping $D_{1}:A\longrightarrow X^{*}$ is said to be a derivation if for each $a,b\in A$

[TABLE]

A bounded bilinear map $D_{2}:A\times A\longrightarrow X$(or $X^{*}$) is called a bi-derivation, if for each $a,b,c$ and $d\in A$

[TABLE]

Let $D_{1}:A\longrightarrow A^{*}$ be a derivation. Dales, Rodriguez and Velasco, in [8] showed that $D_{1}^{**}:(A^{**},\square)\longrightarrow A^{***}$ is a derivation if and only if $\pi^{r****}(D_{1}^{**}(A^{**}),A^{**})\subseteq A^{*}$. In [15], S. Mohamadzadeh and H.R.E Vishki extends this and showed that second adjont $D_{1}^{**}:(A^{**},\square)\longrightarrow A^{***}$ is a derivation if and only if $\pi_{2}^{****}(D_{1}^{**}(A^{**}),X^{**})\subseteq A^{*}$ and which $D_{1}^{**}:(A^{**},\lozenge)\longrightarrow A^{***}$ is a derivation if and only if $\pi_{1}^{r****}(D_{1}^{**}(A^{**}),X^{**})\subseteq A^{*}$. A. Erfanian Attar et al, provide condition such that the third adjoint $D_{2}^{***}$ of a bi-derivation $D_{2}:A\times A\longrightarrow X$ (or $X^{*}$) is again a bi-derivation, see [10]. For a Banach $A-$module $(\pi_{1},X,\pi_{2})$, the fourth adjoint $D^{****}$ of a tri-derivation $D:A\times A\times A\longrightarrow X^{*}$ is trivially a tri-linear extension of $D$. A problem which is of interest is under what conditions we need that $D^{****}$ is again a tri-derivation. The section 4 we will extend above mentioned result. A bounded trilinear mapping $f:X\times Y\times Z\longrightarrow W$ is said to factor if it is surjective, that is $f(X\times Y\times Z)=W$.

Throughout the article, we usually identify a normed space with its canonical image in its second dual.

2. regularity of bounded tri-linear maps

Theorem 2.1**.**

Let $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear map. Then $f$ is regular if and only if

[TABLE]

where $\{x_{\alpha}\},\{y_{\beta}\}$ and $\{z_{\gamma}\}$ are nets in $X,Y$ and $Z$ which converge to $x^{**}\in X^{**},y^{**}\in Y^{**}$ and $z^{**}\in Z^{**}$ in the $w^{*}-$topologies, respectively.

Proof.

For every $w^{*}\in W^{*}$ we have

$\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle x^{**},f^{***}(y^{**},z^{**},w^{*})\rangle=\lim\limits_{\alpha}\langle f^{***}(y^{**},z^{**},w^{*}),x_{\alpha}\rangle$

$=\lim\limits_{\alpha}\langle y^{**},f^{**}(z^{**},w^{*},x_{\alpha})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{**}(z^{**},w^{*},x_{\alpha}),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle z^{**},f^{*}(w^{*},x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle f^{*}(w^{*},x_{\alpha},y_{\beta}),z_{\gamma}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle f(x_{\alpha},y_{\beta},z_{\gamma}),w^{*}\rangle$.

Therefore $f^{****}(x^{**},y^{**},z^{**})=w^{*}-\lim\limits_{\alpha}w^{*}-\lim\limits_{\beta}w^{*}-\lim\limits_{\gamma}f(x_{\alpha},y_{\beta},z_{\gamma})$. In the other hands

$\langle f^{r****r}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle f^{r****}(z^{**},y^{**},x^{**}),w^{*}\rangle=\langle z^{**},f^{r***}(y^{**},x^{**},w^{*})\rangle$

$=\lim\limits_{\gamma}\langle f^{r***}(y^{**},x^{**},w^{*}),z_{\gamma}\rangle=\lim\limits_{\gamma}\langle y^{**},f^{r**}(x^{**},w^{*},z_{\gamma})$

$=\lim\limits_{\gamma}\lim\limits_{\beta}\langle f^{r**}(x^{**},w^{*},z_{\gamma}),y_{\beta}\rangle=\lim\limits_{\gamma}\lim\limits_{\beta}\langle x^{**},f^{r*}(w^{*},z_{\gamma},y_{\beta})\rangle$

$=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f^{r*}(w^{*},z_{\gamma},y_{\beta}),x_{\alpha}\rangle=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},f^{r}(z_{\gamma},y_{\beta},x_{\alpha})\rangle$

$=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f(x_{\alpha},y_{\beta},z_{\gamma}),w^{*}\rangle$.

Therefore $f^{r****r}(x^{**},y^{**},z^{**})=w^{*}-\lim\limits_{\gamma}w^{*}-\lim\limits_{\beta}w^{*}-\lim\limits_{\alpha}f(x_{\alpha},y_{\beta},z_{\gamma})$, and proof follows. ∎

In the following theorem, we provide a criterion concerning to the regularity of a bounded tri-linear map.

Theorem 2.2**.**

For a bounded tri-linear map $f:X\times Y\times Z\longrightarrow W$ the following statements are equivalent:

1. (1)

$f$* is regular.* 2. (2)

$f^{*****}=f^{r*******r}.$** 3. (3)

$f^{***r*}(X^{**},W^{*},Z)\subseteq Y^{*}$* and $f^{*****}(W^{*},X^{**},Y^{**})\subseteq Z^{*}.$*

Proof.

(1) $\Rightarrow$ (2), if $f$ is regular, then $f^{****}=f^{r****r}$. For every $x^{**}\in X^{**},y^{**}\in Y^{**},z^{**}\in Z^{**}$ and $w^{***}\in W^{***}$ we have

$\langle f^{*****}(w^{***},x^{**},y^{**}),z^{**}\rangle=\langle w^{***},f^{****}(x^{**},y^{**},z^{**})\rangle=\langle w^{***},f^{r****r}(x^{**},y^{**},z^{**})\rangle$

$=\langle w^{***},f^{r****}(z^{**},y^{**},x^{**})\rangle=\langle f^{r*****}(w^{***},z^{**},y^{**}),x^{**}\rangle$

$=\langle f^{r******}(x^{**},w^{***},z^{**}),y^{**}\rangle=\langle f^{r*******}(y^{**},x^{**},w^{***}),z^{**}\rangle=\langle f^{r*******r}(w^{***},x^{**},y^{**}),z^{**}\rangle$

as claimed.

(2) $\Rightarrow$ (1), let $f^{*****}=f^{r*******r}$, then for every $w^{*}\in W^{*}$,

$\langle f^{r****r}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle f^{r****}(z^{**},y^{**},x^{**}),w^{*}\rangle=\langle f^{r*****}(w^{*},z^{**},y^{**}),x^{**}\rangle$

$=\langle f^{r******}(x^{**},w^{*},z^{**}),y^{**}\rangle=\langle f^{r*******}(y^{**},x^{**},w^{*}),z^{**}\rangle$

$=\langle f^{r*******r}(w^{*},x^{**},y^{**}),z^{**}\rangle=\langle f^{*****}(w^{*},x^{**},y^{**}),z^{**}\rangle=\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle$.

It follows that $f$ is regular.

(1) $\Rightarrow$ (3), let $f$ is regular and $x^{**}\in X^{**},y^{**}\in Y^{**},z\in Z,w^{*}\in W^{*}$. Then we have

$\langle f^{***r*}(x^{**},w^{*},z),y^{**}\rangle=\langle x^{**},f^{***r}(w^{*},z,y^{**})\rangle=\langle x^{**},f^{***}(y^{**},z,w^{*})\rangle$

$=\langle f^{****}(x^{**},y^{**},z),w^{*}\rangle=\langle f^{r****r}(x^{**},y^{**},z),w^{*}\rangle$

$=\langle f^{r****}(z,y^{**},x^{**}),w^{*}\rangle=\langle f^{r***}(y^{**},x^{**},w^{*}),z\rangle=\langle f^{r**}(x^{**},w^{*},z),y^{**}\rangle$.

Therefore $f^{***r*}(x^{**},w^{*},z)=f^{r**}(x^{**},w^{*},z)\in Y^{*}$. So $f^{***r*}(X^{**},W^{*},Z)\subseteq Y^{*}$. A similar argument shows that $f^{*****}(w^{*},x^{**},y^{**})=f^{r***r}(w^{*},x^{**},y^{**})\in Z^{*}$. Thus $f^{*****}(W^{*},X^{**},Y^{**})\subseteq Z^{*}$, as claimed.

(3) $\Rightarrow$ (1), let $\{x_{\alpha}\},\{y_{\beta}\}$ and $\{z_{\gamma}\}$ are nets in $X,Y$ and $Z$ which converge to $x^{**},y^{**}$ and $z^{**}$ in the $w^{*}-$topologies, respectively. For every $w^{*}\in W^{*}$ we have

$\langle f^{r****r}(x^{**},y^{**},z^{**}),w^{*}\rangle=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f(x_{\alpha},y_{\beta},z_{\gamma}),w^{*}\rangle=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f^{*}(w^{*},x_{\alpha},y_{\beta}),z_{\gamma}\rangle$

$=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f^{**}(z_{\gamma},w^{*},x_{\alpha}),y_{\beta}\rangle=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f^{***}(y_{\beta},z_{\gamma},w^{*}),x_{\alpha}\rangle$

$=\lim\limits_{\gamma}\lim\limits_{\beta}\langle x^{**},f^{***}(y_{\beta},z_{\gamma},w^{*})=\lim\limits_{\gamma}\lim\limits_{\beta}\langle x^{**},f^{***r}(w^{*},z_{\gamma},y_{\beta})\rangle$

$=\lim\limits_{\gamma}\lim\limits_{\beta}\langle f^{***r*}(x^{**},w^{*},z_{\gamma}),y_{\beta}\rangle=\lim\limits_{\gamma}\langle f^{***r*}(x^{**},w^{*},z_{\gamma}),y^{**}\rangle$

$=\lim\limits_{\gamma}\langle x^{**},f^{***r}(w^{*},z_{\gamma},y^{**})\rangle=\lim\limits_{\gamma}\langle x^{**},f^{***}(y^{**},z_{\gamma},w^{*})\rangle$

$=\lim\limits_{\gamma}\langle f^{****}(x^{**},y^{**},z_{\gamma}),w^{*}\rangle=\lim\limits_{\gamma}\langle f^{*****}(w^{*},x^{**},y^{**}),z_{\gamma}\rangle$

$=f^{*****}(w^{*},x^{**},y^{**}),z^{**}\rangle=\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle$.

It follows that $f$ is regular and this completes the proof. ∎

Corollary 2.3**.**

For a bounded tri-linear map $f:X\times Y\times Z\longrightarrow W$ the following statements are equivalent:

1. (1)

$f$* is regular.* 2. (2)

$f^{r*****r}=f^{*******}$. 3. (3)

$f^{r***r*}(Z^{**},W^{*},X)\subseteq Y^{*}$* and $f^{*****}(W^{*},Z^{**},Y^{**})\subseteq X^{*}.$*

Proof.

The mapping $f$ is regular if and only if $f^{r}$ is regular. Therefore by Theorem 2.2, the desired result is obtained. ∎

Corollary 2.4**.**

For a bounded tri-linear map $f:X\times Y\times Z\longrightarrow W$, if from $X,Y$ or $Z$ at least two reflexive then f is regular.

Proof.

Without having to enter the whole argument, let $Y$ and $Z$ are reflexive. Since $Y$ is reflexive, $Y^{*}=Y^{***}$. Therefore

[TABLE]

In the other hands, because $Z$ is the reflexive space, thus

[TABLE]

Now Using (2-1), (2-2) and Theorem 2.2, the result holds. ∎

Corollary 2.5**.**

Let bounded tri-linear map $f:X\times Y\times Z\longrightarrow W$ is regular. Then

1. (1)

If $f^{***r*}(X^{**},W^{*},Z)$ factors, then $Y$ is reflexive space. 2. (2)

If $f^{*****}(W^{*},X^{**},Y^{**})$ factors, then $Z$ is reflexive space. 3. (3)

If $f^{****r*}(W^{*},Z,Y)$ factors, then $X$ is reflexive space.

Proof.

(1) Let $f$ be regular. It follows that $f^{***r*}(X^{**},W^{*},Z)\subseteq Y^{*}$. In the other hands, $f^{***r*}(X^{**},W^{*},Z)$ is factor. So for each $y^{***}\in Y^{***}$ there exist $x^{**}\in X^{**},w^{*}\in W^{*}$ and $z\in Z$ such that $f^{***r*}(x^{**},w^{*},z)=y^{***}$. Therefore $Y^{***}\subseteq Y^{*}$.

(2) The proof similar to (1).

(3) Enough show that, if $f$ is regular, then $f^{****r*}(W^{*},Z,Y)\subseteq X^{*}$. For every $x^{**}\in X^{**},y\in Y,z\in Z$ and $w^{*}\in W^{*}$ we have

$\langle f^{****r*}(w^{*},z,y),x^{**}\rangle=\langle w^{*},f^{****r}(z,y,x^{**})\rangle=\langle w^{*},f^{****}(x^{**},y,z)\rangle$

$=\langle f^{r****r}(x^{**},y,z),w^{*}\rangle=\langle f^{r****}(z,y,x^{**}),w^{*}\rangle$

$=\langle z,f^{r***}(y,x^{**},w^{*})\rangle=\langle f^{r**}(x^{**},w^{*},z),y\rangle$

$=\langle f^{r*}(w^{*},z,y),x^{**}\rangle$.

Therefore $f^{****r*}(w^{*},z,y)=f^{r*}(w^{*},z,y)\in X^{*}$. The rest of proof has similar argument such as (1). ∎

The next corollary, let $X,Y$ and $Z$ be Banach spaces, $I_{X}:X\longrightarrow X$ is identity map on $X$ and $I_{Y},I_{Z}$ are identity on $Y,Z$ respectively.

Corollary 2.6**.**

If $I_{X}$, $I_{Y}$ and $I_{Z}$ are weakly compact identity mapping, then all of them and all of their adjoints are regular.

Example 2.7**.**

1. (1)

Let $G$ be a compact group. Let $1<p,q<\infty$ and $\frac{1}{p}+\frac{1}{q}=1+\frac{1}{r}$. Then by [12, Sections 2.4 and 2.5], we conclude that $L^{1}(G)\star L^{p}(G)\subset L^{p}(G)$ and $L^{p}(G)\star L^{q}(G)\subset L^{r}(G)$ where $(g\star h)(x)=\int_{G}g(y)h(y^{-1}x)dy$ for $x\in G$. The Banach spaces $L^{p}(G)$ and $L^{q}(G)$ are reflexive, thus by corollary 2.4 we conclude that the bounded tri-linear mapping

[TABLE]

defined by $f(k,g,h)=(k\star g)\star h$, is regular for every $k\in L^{1}(G),g\in L^{p}(G)$ and $h\in L^{q}(G)$. 2. (2)

Let $G$ be a locally compact group. We know from [18] that $L^{1}(G)$ is regular if and only if it is reflexive or equivalent to $G$ being finite. It follows that for every finite locally compact group $G$, by corollary 2.4, the bounded tri-linear mapping

[TABLE]

defined by $f(k,g,h)=k\star g\star h$, is regular for every $k,g$ and $h\in L^{1}(G)$. 3. (3)

$C^{*}-$algebras are standard examples of Banach algebras that are Arens regular (see[6]). We know that a $C^{*}-$algebra is reflexive if and only if it is of finite dimension. Since if $A$ be a finite dimension $C^{*}$-algebra, then we conclude that the bounded tri-linear mapping $f:A\times A\times A\longrightarrow A$ is regular by corollary 2.4. 4. (4)

Let $G$ be a locally compact group and let $M(G)$ be measure algebra of $G$, see [12, Section 2.5]. The convolution for $\mu_{1},\mu_{2}\in M(G)$ defined by

[TABLE]

We have

[TABLE]

for $\mu_{1},\mu_{2}$ and $\mu_{3}\in M(G)$. Therefore convolution is associative. Now we define the bounded tri-linear mapping

[TABLE]

by $f(\mu_{1},\mu_{2},\mu_{3})=\int\psi d(\mu_{1}*\mu_{2}*\mu_{3})$. If $G$ is finite, then $f$ is regular.

3. Some results for regularity

Dales, Rodriguez-Palacios and Velasco, in the theorem of their paper, [[8], Theorem 4.1], for a bonded bilinear map $m:X\times Y\longrightarrow Z$ have shown that, $m^{r*r***}=m^{***r*r}$ if and only if both $m$ and $m^{r*}$ are Arens regular. The We study some theorems from this paper for general case as following.

Remark 3.1*.*

In the next theorem, $f^{n}$ is $n-$th adjoint of $f$ for each $n\in N$.

Theorem 3.2**.**

If $f$ and $f^{rn}$ are reular, then $f^{4rnr}=f^{rnr4}$.

Proof.

Since $f$ is regular, so $f^{4r}=f^{r4}$. Therefore $f^{4rn}=f^{r(n+4)}$. In the other hands, regularity of $f^{rn}$ follows that $f^{r(n+4)}=f^{rnr4r}$. Thus $f^{rnr4r}=f^{4rn}$ and this completes the proof. ∎

In general the converse of preceding theorem not holds as following theorem.

Theorem 3.3**.**

Let $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping. Then

1. (1)

$f^{****r**r}=f^{r**r****}$* if and only if both $f$ and $f^{r**}$ are regular.* 2. (2)

$f^{****r***r}=f^{r***r****}$* if and only if both $f$ and $f^{r***}$ are regular.*

Proof.

We prove only (1), the other part has the same argument. If both $f$ and $f^{r**}$ are regular, then by applying Theorem 3.2, for $n=2$, $f^{****r**r}=f^{r**r****}$.

Conversely, suppose that $f^{****r**r}=f^{r**r****}$. First we show that $f$ is regular. Let $\{z_{\gamma}\}$ is net in $Z$ which converge to $z^{**}\in Z^{**}$ in the $w^{*}-$topologies. Then for every $x^{**}\in X^{**},y^{**}\in Y^{**}$ and $w^{*}\in W^{*}$ we have

$\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle f^{****r}(z^{**},y^{**},x^{**}),w^{*}\rangle=\langle f^{****r*}(w^{*},z^{**},y^{**}),x^{**}\rangle$

$=\langle f^{****r**}(x^{**},w^{*},z^{**}),y^{**}\rangle=\langle f^{****r**r}(z^{**},w^{*},x^{**}),y^{**}\rangle$

$=\langle f^{r**r****}(z^{**},w^{*},x^{**}),y^{**}\rangle=\langle z^{**},f^{r**r***}(w^{*},x^{**},y^{**})\rangle$

$=\lim\limits_{\gamma}\langle f^{r**r***}(w^{*},x^{**},y^{**}),z_{\gamma}\rangle=\lim\limits_{\gamma}\langle w^{*},f^{r**r**}(x^{**},y^{**},z_{\gamma})\rangle$

$=\lim\limits_{\gamma}\langle f^{r**r*}(y^{**},z_{\gamma},w^{*}),x^{**}\rangle=\lim\limits_{\gamma}\langle y^{**},f^{r**r}(z_{\gamma},w^{*},x^{**})\rangle$

$=\lim\limits_{\gamma}\langle y^{**},f^{r**}(x^{**},w^{*},z_{\gamma})\rangle=\lim\limits_{\gamma}\langle f^{r***}(y^{**},x^{**},w^{*}),z_{\gamma}\rangle$

$=\langle z^{**},f^{r***}(y^{**},x^{**},w^{*})\rangle=\langle f^{r****}(z^{**},y^{**},x^{**}),w^{*}\rangle$

$=\langle f^{r****r}(x^{**},y^{**},z^{**}),w^{*}\rangle$.

Therefore $f$ is regular. Now we show that $f^{r**}$ is regular. Let $\{x_{\alpha}^{**}\}$ is net in $X^{**}$ which converge to $x^{****}\in X^{****}$ in the $w^{*}-$topologies. Then for every $y^{**}\in Y^{**},z^{**}\in Z^{**}$ and $w^{***}\in W^{***}$ we have

$\langle f^{r**r****r}(x^{****},w^{***},z^{**}),y^{**}\rangle=\langle f^{r**r****}(z^{**},w^{***},x^{****}),y^{**}\rangle$

$=\langle f^{****r**r}(z^{**},w^{***},x^{****}),y^{**}\rangle=\langle f^{****r**}(x^{****},w^{***},z^{**}),y^{**}\rangle\rangle$

$=\langle x^{****},f^{****r*}(w^{***},z^{**},y^{**})=\lim\limits_{\alpha}\langle f^{****r*}(w^{***},z^{**},y^{**}),x_{\alpha}^{**}\rangle$

$=\lim\limits_{\alpha}\langle w^{***},f^{****r}(z^{**},y^{**},x_{\alpha}^{**})\rangle=\lim\limits_{\alpha}\langle w^{***},f^{****}(x_{\alpha}^{**},y^{**},z^{**})\rangle$

$=\lim\limits_{\alpha}\langle w^{***},f^{r****r}(x_{\alpha}^{**},y^{**},z^{**})\rangle=\lim\limits_{\alpha}\langle w^{***},f^{r****}(z^{**},y^{**},x_{\alpha}^{**})\rangle$

$=\lim\limits_{\alpha}\langle f^{r*****}(w^{***},z^{**},y^{**}),x_{\alpha}^{**}\rangle=\langle x^{****},f^{r*****}(w^{***},z^{**},y^{**})\rangle$

$=\langle f^{r******}(x^{****},w^{***},z^{**}),y^{**}\rangle$.

It follows that $f^{r**}$ is regular and this completes the proof. ∎

Arens has shown [3] that bounded bilinear map $m$ is regular if and only if for each $z^{*}\in Z^{*}$, the bilinear form $z^{*}om$ is regular. In the next theorem we give an important characterization of regularity bounded tri-linear mappings.

Lemma 3.4**.**

Suppose $X,Y,Z,W$ and $S$ be normed spaces and $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping and $h:W\longrightarrow S$ be a bounded linear mapping. Then we have

1. (1)

$h^{**}of^{****}=(hof)^{****}$. 2. (2)

$h^{**}of^{r****r}=(hof)^{r****r}$.

Proof.

(1) Let $\{x_{\alpha}\},\{y_{\beta}\}$ and $\{z_{\gamma}\}$ be nets in $X,Y$ and $Z$ which converge to $x^{**}\in X^{**},y^{**}\in Y^{**}$ and $z^{**}\in Z^{**}$ in the $w^{*}-$topologies, respectively. For each $s^{*}\in S^{*}$ we have

$\langle h^{**}of^{****}(x^{**},y^{**},z^{**}),s^{*}\rangle=\langle h^{**}(f^{****}(x^{**},y^{**},z^{**})),s^{*}\rangle=\langle f^{****}(x^{**},y^{**},z^{**}),h^{*}(s^{*})\rangle$

$=\langle x^{**},f^{***}(y^{**},z^{**},h^{*}(s^{*}))\rangle=\lim\limits_{\alpha}\langle f^{***}(y^{**},z^{**},h^{*}(s^{*})),x_{\alpha}\rangle$

$=\lim\limits_{\alpha}\langle y^{**},f^{**}(z^{**},h^{*}(s^{*}),x_{\alpha})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{**}(z^{**},h^{*}(s^{*}),x_{\alpha}),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle z^{**},f^{*}(h^{*}(s^{*}),x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle f^{*}(h^{*}(s^{*}),x_{\alpha},y_{\beta}),z_{\gamma}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle h^{*}(s^{*}),f(x_{\alpha},y_{\beta},z_{\gamma})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle s^{*},h(f(x_{\alpha},y_{\beta},z_{\gamma}))\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle s^{*},hof(x_{\alpha},y_{\beta},z_{\gamma})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle(hof)^{*}(s^{*},x_{\alpha},y_{\beta}),z_{\gamma}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle z^{**},(hof)^{*}(s^{*},x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle(hof)^{**}(z^{**},s^{*},x_{\alpha}),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\langle y^{**},(hof)^{**}(z^{**},s^{*},x_{\alpha})\rangle=\lim\limits_{\alpha}\langle(hof)^{***}(y^{**},z^{**},s^{*}),x_{\alpha})\rangle$

$=\langle x^{**},(hof)^{***}(y^{**},z^{**},s^{*})\rangle=\langle(hof)^{****}(x^{**},y^{**},z^{**}),s^{*}\rangle$.

Hence $h^{**}of^{****}(x^{**},y^{**},z^{**})=(hof)^{****}(x^{**},y^{**},z^{**})$. A similar argument applies for (2). ∎

Theorem 3.5**.**

Let $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping and $h:W\longrightarrow S$ be a bounded linear mapping. Then $f$ is regular if and only if $hof$ is regular.

Proof.

Let $f$ is regular. Then for every $x^{**}\in X^{**},y^{**}\in Y^{**},z^{**}\in Z^{**}$ and $s^{*}\in S^{*}$ we have

$\langle h^{**}(f^{r****r}(x^{**},y^{**},z^{**})),s^{*}\rangle=\langle f^{r****r}(x^{**},y^{**},z^{**}),h^{*}(s^{*})\rangle$

$=\langle f^{****}(x^{**},y^{**},z^{**}),h^{*}(s^{*})\rangle=\langle h^{**}(f^{****}(x^{**},y^{**},z^{**})),s^{*}\rangle$.

Therefore $h^{**}of^{r****r}(x^{**},y^{**},z^{**})=h^{**}of^{****}(x^{**},y^{**},z^{**})$ and by applying Lemma 3.4, we implies that

[TABLE]

Thus $hof$ is regular.

For the converse, suppose that $hof$ is regular. By contradiction, let $f$ be not regular. There exist $x^{**}\in X^{**},y^{**}\in Y^{**}$ and $z^{**}\in Z^{**}$ such that $f^{****}(x^{**},y^{**},z^{**})\neq f^{r****r}(x^{**},y^{**},z^{**})$. Therefore we have

$(hof)^{****}(x^{**},y^{**},z^{**})=w^{*}-\lim\limits_{\alpha}w^{*}-\lim\limits_{\beta}w^{*}-\lim\limits_{\gamma}(hof)(x_{\alpha},y_{\beta},z_{\gamma})$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\gamma}\langle f(x_{\alpha},y_{\beta},z_{\gamma}),h\rangle=\langle f^{****}(x^{**},y^{**},z^{**}),h\rangle$

$\neq\langle f^{r****r}(x^{**},y^{**},z^{**}),h\rangle=\lim\limits_{\gamma}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f(x_{\alpha},y_{\beta},z_{\gamma}),h\rangle$

$=w^{*}-\lim\limits_{\gamma}w^{*}-\lim\limits_{\beta}w^{*}-\lim\limits_{\alpha}(hof)(x_{\alpha},y_{\beta},z_{\gamma})=(hof)^{r****r}(x^{**},y^{**},z^{**})$.

It follows that $(hof)^{****}(x^{**},y^{**},z^{**})\neq(hof)^{r****r}(x^{**},y^{**},z^{**})$. ∎

Another interesting case of regularity is in the following.

Theorem 3.6**.**

Let $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping, $m:X\times Y\longrightarrow Z$ be a bounded bilinear mapping, $T:X\times Y\longrightarrow W$ defined by $T(x,y)=f(x,y,m(x,y))$ for each $x\in X,y\in Y$ and $m^{***}$ is factors. Then $T$ is regular if and only if $f$ is regular.

Proof.

Let $T$ is regular. The mapping $m^{***}:X^{**}\times Y^{**}\longrightarrow Z^{**}$ is onto, so for each $z^{**}\in Z^{**}$ There must then exist $x^{**}\in X^{**}$ and $y^{**}\in Y^{**}$ such that $m^{***}(x^{**},y^{**})=z^{**}$. Let $\{x_{\alpha}\}$ and $\{y_{\beta}\}$ are nets in $X$ and $Y$ which converge to $x^{**}$ and $y^{**}$ in the $w^{*}-$topologies, respectively. For every $w^{*}\in W^{*}$ we have

$\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle f^{****}(x^{**},y^{**},m^{***}(x^{**},y^{**})),w^{*}\rangle$

$=\langle x^{**},f^{***}(y^{**},m^{***}(x^{**},y^{**}),w^{*})\rangle=\lim\limits_{\alpha}\langle f^{***}(y^{**},m^{***}(x^{**},y^{**}),w^{*}),x_{\alpha}\rangle$

$=\lim\limits_{\alpha}\langle y^{**},f^{**}(m^{***}(x^{**},y^{**}),w^{*},x_{\alpha})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{**}(m^{***}(x^{**},y^{**}),w^{*},x_{\alpha}),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle(m^{***}(x^{**},y^{**}),f^{*}(w^{*},x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle x^{**},m^{**}(y^{**},f^{*}(w^{*},x_{\alpha},y_{\beta}))\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\langle m^{**}(y^{**},f^{*}(w^{*},x_{\alpha},y_{\beta})),x_{\alpha}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\langle y^{**},m^{*}(f^{*}(w^{*},x_{\alpha},y_{\beta}),x_{\alpha})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\lim\limits_{\beta}\langle m^{*}(f^{*}(w^{*},x_{\alpha},y_{\beta}),x_{\alpha}),y_{\beta}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{*}(w^{*},x_{\alpha},y_{\beta}),m(x_{\alpha},y_{\beta})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\lim\limits_{\beta}\langle w^{*},f(x_{\alpha},y_{\beta},m(x_{\alpha},y_{\beta}))\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\lim\limits_{\beta}\langle w^{*},T(x_{\alpha},y_{\beta})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\lim\limits_{\beta}\langle T^{*}(w^{*},x_{\alpha}),y_{\beta}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\langle y^{**},T^{*}(w^{*},x_{\alpha})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\alpha}\langle T^{**}(y^{**},w^{*}),x_{\alpha}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle x^{**},T^{**}(y^{**},w^{*})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle T^{***}(x^{**},y^{**}),w^{*}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle T^{r***r}(x^{**},y^{**}),w^{*}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle T^{r***}(y^{**},x^{**}),w^{*}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle y^{**},T^{r**}(x^{**},w^{*})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\langle T^{r**}(x^{**},w^{*}),y_{\beta}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\langle x^{**},T^{r*}(w^{*},y_{\beta})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\lim\limits_{\alpha}\langle T^{r*}(w^{*},y_{\beta}),x_{\alpha}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},T^{r}(y_{\beta},x_{\alpha})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},T(x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},f(x_{\alpha},y_{\beta},m(x_{\alpha},y_{\beta}))\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},f^{r}(m(x_{\alpha},y_{\beta}),y_{\beta},x_{\alpha})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\lim\limits_{\alpha}\langle f^{r*}(w^{*},m(x_{\alpha},y_{\beta}),y_{\beta}),x_{\alpha}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\langle x^{**},f^{r*}(w^{*},m(x_{\alpha},y_{\beta}),y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\lim\limits_{\beta}\langle f^{r**}(x^{**},w^{*},m(x_{\alpha},y_{\beta})),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle y^{**},f^{r**}(x^{**},w^{*},m(x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{r***}(y^{**},x^{**},w^{*}),m(x_{\alpha},y_{\beta})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle m^{*}(f^{r***}(y^{**},x^{**},w^{*}),x_{\alpha}),y_{\beta}\rangle=\lim\limits_{\alpha}\langle y^{**},m^{*}(f^{r***}(y^{**},x^{**},w^{*}),x_{\alpha})\rangle$

$=\lim\limits_{\alpha}\langle m^{**}(y^{**},f^{r***}(y^{**},x^{**},w^{*})),x_{\alpha}\rangle=\langle x^{**},m^{**}(y^{**},f^{r***}(y^{**},x^{**},w^{*}))\rangle$

$=\langle m^{***}(x^{**},y^{**}),f^{r***}(y^{**},x^{**},w^{*})\rangle=\langle f^{r****}(m^{***}(x^{**},y^{**}),y^{**},x^{**}),w^{*}\rangle$

$=\langle f^{r****r}(x^{**},y^{**},m^{***}(x^{**},y^{**})),w^{*}\rangle=\langle f^{r****r}(x^{**},y^{**},z^{**}),w^{*}\rangle$.

Therefore $f$ is regular.

For the converse, suppose that $f$ is regular. For every $w^{*}\in W^{*}$ we have

$\langle T^{***}(x^{**},y^{**}),w^{*}\rangle=\langle x^{**},T^{**}(y^{**},w^{*})\rangle=\lim\limits_{\alpha}\langle T^{**}(y^{**},w^{*}),x_{\alpha}\rangle$

$=\lim\limits_{\alpha}\langle y^{**},T^{*}(w^{*},x_{\alpha})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle T^{*}(w^{*},x_{\alpha}),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle w^{*},T(x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle w^{*},f(x_{\alpha},y_{\beta},m(x_{\alpha},y_{\beta}))\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{*}(w^{*},x_{\alpha},y_{\beta}),m(x_{\alpha},y_{\beta})\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle m^{*}(f^{*}(w^{*},x_{\alpha},y_{\beta}),x_{\alpha}),y_{\beta}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle m^{**}(y_{\beta},f^{*}(w^{*},x_{\alpha},y_{\beta})),x_{\alpha}\rangle=\lim\limits_{\alpha}\lim\limits_{\beta}\langle m^{***}(x_{\alpha},y_{\beta}),f^{*}(w^{*},x_{\alpha},y_{\beta})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\beta}\langle f^{**}(m^{***}(x_{\alpha},y_{\beta}),w^{*},x_{\alpha}),y_{\beta}\rangle=\lim\limits_{\alpha}\langle y^{**},f^{**}(m^{***}(x_{\alpha},y_{\beta}),w^{*},x_{\alpha})\rangle$

$=\lim\limits_{\alpha}\langle f^{***}(y^{**},m^{***}(x_{\alpha},y_{\beta}),w^{*}),x_{\alpha}\rangle=\langle x^{**},f^{***}(y^{**},m^{***}(x_{\alpha},y_{\beta}),w^{*})\rangle$

$=\langle f^{****}(x^{**},y^{**},m^{***}(x_{\alpha},y_{\beta})),w^{*}\rangle=\langle f^{r****r}(x^{**},y^{**},m^{***}(x_{\alpha},y_{\beta})),w^{*}\rangle$

$=\langle f^{r****}(m^{***}(x_{\alpha},y_{\beta}),y^{**},x^{**}),w^{*}\rangle=\langle m^{***}(x_{\alpha},y_{\beta}),f^{r***}(y^{**},x^{**},w^{*})\rangle$

$=\langle x_{\alpha},m^{**}(y_{\beta},f^{r***}(y^{**},x^{**},w^{*}))\rangle=\langle y_{\beta},m^{*}(f^{r***}(y^{**},x^{**},w^{*}),x_{\alpha})\rangle$

$=\langle f^{r***}(y^{**},x^{**},w^{*}),m(x_{\alpha},y_{\beta})\rangle=\langle y^{**},f^{r**}(x^{**},w^{*},m(x_{\alpha},y_{\beta}))\rangle$

$=\lim\limits_{\beta}\langle f^{r**}(x^{**},w^{*},m(x_{\alpha},y_{\beta})),y_{\beta}\rangle=\lim\limits_{\beta}\langle x^{**},f^{r*}(w^{*},m(x_{\alpha},y_{\beta}),y_{\beta})\rangle$

$=\lim\limits_{\beta}\lim\limits_{\alpha}\langle f^{r*}(w^{*},m(x_{\alpha},y_{\beta}),y_{\beta}),x_{\alpha}\rangle=\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},f^{r}(m(x_{\alpha},y_{\beta}),y_{\beta},x_{\alpha})\rangle$

$=\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},f(x_{\alpha},y_{\beta},m(x_{\alpha},y_{\beta}))\rangle=\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},T(x_{\alpha},y_{\beta})\rangle$

$=\lim\limits_{\beta}\lim\limits_{\alpha}\langle w^{*},T^{r}(y_{\beta},x_{\alpha})\rangle=\lim\limits_{\beta}\lim\limits_{\alpha}\langle T^{r*}(w^{*},y_{\beta}),x_{\alpha}\rangle$

$=\lim\limits_{\beta}\langle x^{**},T^{r*}(w^{*},y_{\beta})\rangle=\lim\limits_{\beta}\langle T^{r**}(x^{**},w^{*}),y_{\beta}\rangle$

$=\langle y^{**},T^{r**}(x^{**},w^{*})\rangle=\langle T^{r***}(y^{**},x^{**}),w^{*}\rangle=\langle T^{r***r}(x^{**},y^{**}),w^{*}\rangle$.

Therefore $T^{***}=T^{r***r}$ , as claimed. ∎

Theorem 3.7**.**

Let $X,Y,Z,W$ and $S$ be Banach spaces, $f:X\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping and $x\in X,y\in Y,z\in Z$ arbitrary. Then

1. (1)

Let $g_{1}:S\times Y\times Z\longrightarrow W$ be a bounded tri-linear mapping and let $h_{1}:X\longrightarrow S$ be a bounded linear mapping such that $f(x,y,z)=g_{1}(h_{1}(x),y,z)$. If $h_{1}$ is weakly compact, then $f^{****r*}(W^{***},Z^{**},Y^{**})\subseteq X^{*}$. 2. (2)

Let $g_{2}:X\times S\times Z\longrightarrow W$ be a bounded tri-linear mapping and let $h_{2}:Y\longrightarrow S$ be a bounded linear mapping such that $f(x,y,z)=g_{2}(x,h_{2}(y),z)$. If $h_{2}$ is weakly compact, then $f^{***r*}(X^{**},W^{*},Z^{**})\subseteq Y^{*}$. 3. (3)

Let $g_{3}:X\times Y\times S\longrightarrow W$ be a bounded tri-linear mapping and let $h_{3}:Z\longrightarrow S$ be a bounded linear mapping such that $f(x,y,z)=g_{3}(x,y,h_{3}(z))$. If $h_{3}$ is weakly compact, then $f^{*****}(W^{***},X^{**},Y^{**})\subseteq Z^{*}$.

Proof.

We prove only (1), the other parts have the same argument. For every $x\in X,y\in Y,z\in Z$ and $w^{*}\in W^{*}$ we have

[TABLE]

Therefore $f^{*}(w^{*},x,y)=g_{1}^{*}(w^{*},h_{1}(x),y)$, and implies that for every $z^{**}\in Z^{**}$,

[TABLE]

So $f^{**}(z^{**},w^{*},x)=g_{1}^{**}(z^{**},w^{*},h_{1}(x))$ and implies that for every $y^{**}\in Y^{**}$,

$\langle f^{***}(y^{**},z^{**},w^{*}),x\rangle=\langle y^{**},f^{**}(z^{**},w^{*},x)\rangle=\langle y^{**},g_{1}^{**}(z^{**},w^{*},h_{1}(x))\rangle$

$=\langle g_{1}^{***}(y^{**},z^{**},w^{*}),h_{1}(x)\rangle=\langle h_{1}^{*}(g_{1}^{***}(y^{**},z^{**},w^{*})),x\rangle$.

Thus $f^{***}(y^{**},z^{**},w^{*})=h_{1}^{*}(g_{1}^{***}(y^{**},z^{**},w^{*}))$ and implies that for every $x^{**}\in X^{**}$,

$\langle f^{****}(x^{**},y^{**},z^{**}),w^{*}\rangle=\langle x^{**},f^{***}(y^{**},z^{**},w^{*})\rangle=\langle x^{**},h_{1}^{*}(g_{1}^{***}(y^{**},z^{**},w^{*}))\rangle$

$=\langle h_{1}^{**}(x^{**}),(g_{1}^{***}(y^{**},z^{**},w^{*})\rangle=\langle g_{1}^{****}(h_{1}^{**}(x^{**}),y^{**},z^{**}),w^{*}\rangle$.

Therefore for every $w^{***}\in W^{***}$ we have

$\langle f^{****r*}(w^{***},z^{**},y^{**}),x^{**}\rangle=\langle w^{***},f^{****r}(z^{**},y^{**},x^{**})\rangle=\langle w^{***},f^{****}(x^{**},y^{**},z^{**})\rangle$

$=\langle w^{***},g_{1}^{****}(h_{1}^{**}(x^{**}),y^{**},z^{**})\rangle=\langle w^{***},g_{1}^{****r}(z^{**},y^{**},h_{1}^{**}(x^{**}))\rangle$

$=\langle g_{1}^{****r*}(w^{***},z^{**},y^{**}),h_{1}^{**}(x^{**})\rangle=\langle h_{1}^{***}(g_{1}^{****r*}(w^{***},z^{**},y^{**})),x^{**}\rangle$.

Therefore $f^{****r*}(w^{***},z^{**},y^{**})=h_{1}^{***}(g_{1}^{****r*}(w^{***},z^{**},y^{**}))$. The weak compactness of $h_{1}$ implies that of $h_{1}^{*}$, from which we have $h_{1}^{***}(S^{***})\subseteq X^{*}$. Thus $h_{1}^{***}(g_{1}^{****r*}(w^{***},z^{**},y^{**}))\in X^{*}$ and this completes the proof. ∎

This theorem, combined with Theorem 2.2, yields the next result.

Corollary 3.8**.**

With the assumptions Theorem $3.7$, if $h_{2}$ and $h_{3}$ are weakly compact, then $f$ is regular.

Proof.

Both $h_{2}$ and $h_{3}$ are weakly compact, so by Theorem 3.7 we have

[TABLE]

In particular

[TABLE]

Now by Theorem 2.2, $f$ is regular. ∎

The converse of previous result is not true in general sense as following corollary.

Corollary 3.9**.**

With the assumptions Theorem $3.7$, if $f$ is regular and both $g_{2}^{***r*}$ and $g_{3}^{*****}$ are factors, then $h_{2}$ and $h_{3}$ are weakly compact.

Proof.

Since $f^{***r*}(X^{**},W^{*},Z^{**})=h_{2}^{***}(g_{2}^{***r*}(X^{**},W^{*},Z^{**}))$, thus $h_{2}^{***}(g_{2}^{***r*}(X^{**},W^{*},Z^{**}))\subseteq Y^{*}$. In the other hands $g_{2}^{***r*}$ is factors, so implies that $h_{2}^{***}(S^{***})\subseteq Y^{*}$. Therefore $h_{2}^{*}$ is weakly compact and implies that $h_{2}$ is weakly compact. The other part has the same argument for $h_{3}$. ∎

4. The fourth adjoint of a tri-derivation

Definition 4.1**.**

Let $(\pi_{1},X,\pi_{2})$ be a Banach $A-$module. A bounded tri-linear mapping $D:A\times A\times A\longrightarrow X$ is said to be a tri-derivation when

1. (1)

$D(\pi(a,d),b,c)=\pi_{2}(D(a,b,c),d)+\pi_{1}(a,D(d,b,c))$, 2. (2)

$D(a,\pi(b,d),c)=\pi_{2}(D(a,b,c),d)+\pi_{1}(b,D(a,d,c))$, 3. (3)

$D(a,b,\pi(c,d))=\pi_{2}(D(a,b,c),d)+\pi_{1}(c,D(a,b,d))$,

for each $a,b,c,d\in A$. If $(\pi_{1},X,\pi_{2})$ is a Banach $A-$module, then $(\pi_{2}^{r*r},X^{*},\pi_{1}^{*})$ is the dual Banach $A-$module of $(\pi_{1},X,\pi_{2})$. Therefore a bounded tri-linear mapping $D:A\times A\times A\longrightarrow X^{*}$ is a tri-derivation when

1. (1)

$D(\pi(a,d),b,c)=\pi_{1}^{*}(D(a,b,c),d)+\pi_{2}^{r*r}(a,D(d,b,c))$, 2. (2)

$D(a,\pi(b,d),c)=\pi_{1}^{*}(D(a,b,c),d)+\pi_{2}^{r*r}(b,D(a,d,c))$, 3. (3)

$D(a,b,\pi(c,d))=\pi_{1}^{*}(D(a,b,c),d)+\pi_{2}^{r*r}(c,D(a,b,d))$.

It can also be written, a bounded tri-linear mapping $D:A\times A\times A\longrightarrow A$ is said to be a tri-derivation when

1. (1)

$D(\pi(a,d),b,c)=\pi(D(a,b,c),d)+\pi(a,D(d,b,c))$, 2. (2)

$D(a,\pi(b,d),c)=\pi(D(a,b,c),d)+\pi(b,D(a,d,c))$, 3. (3)

$D(a,b,\pi(c,d))=\pi(D(a,b,c),d)+\pi(c,D(a,b,d))$.

Example 4.2**.**

Let $A$ be a Banach algebra, for any $a,b\in A$ the symbol $[a,b]=ab-ba$ stands for multiplicative commutator of $a$ and $b$. Let $M_{n\times n}(C)$ be the Banach algebra of all $n\times n$ matrix and $A=\{\begin{pmatrix}x&y\\ 0&0\end{pmatrix}\in M_{n\times n}(C)|x,y\in C\}$. Then $A$ is Banach algebra with the norm

[TABLE]

We define $D:A\times A\times A\longrightarrow A$ to be the bounded tri-linear map given by

[TABLE]

Then for $a=\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix},b=\begin{pmatrix}x_{2}&y_{2}\\ 0&0\end{pmatrix},c=\begin{pmatrix}x_{3}&y_{3}\\ 0&0\end{pmatrix}$ and $d=\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}\in A$ we have

$D(\pi(a,d),b,c)=D(\begin{pmatrix}x_{1}x_{4}&x_{1}y_{4}\\ 0&0\end{pmatrix},\begin{pmatrix}x_{2}&y_{2}\\ 0&0\end{pmatrix},\begin{pmatrix}x_{3}&y_{3}\\ 0&0\end{pmatrix})=[\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\begin{pmatrix}x_{1}x_{2}x_{3}x_{4}&x_{1}x_{2}x_{4}y_{3}\\ 0&0\end{pmatrix}]$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\begin{pmatrix}0&-x_{1}x_{2}x_{3}x_{4}\\ 0&0\end{pmatrix}=\begin{pmatrix}0&-x_{1}x_{2}x_{3}\\ 0&0\end{pmatrix}\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}+\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}\begin{pmatrix}0&-x_{2}x_{3}x_{4}\\ 0&0\end{pmatrix}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(\begin{pmatrix}0&0\\ 0&0\end{pmatrix}-\begin{pmatrix}0&x_{1}x_{2}x_{3}\\ 0&0\end{pmatrix})\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}+\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}(\begin{pmatrix}0&0\\ 0&0\end{pmatrix}-\begin{pmatrix}0&x_{2}x_{3}x_{4}\\ 0&0\end{pmatrix})$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}x_{1}x_{2}x_{3}&x_{1}x_{2}y_{3}\\ 0&0\end{pmatrix}-\begin{pmatrix}x_{1}x_{2}x_{3}&x_{1}x_{2}y_{3}\\ 0&0\end{pmatrix}\begin{pmatrix}0&1\\ 0&0\end{pmatrix})\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}(\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\begin{pmatrix}x_{2}x_{3}x_{4}&x_{2}x_{4}y_{3}\\ 0&0\end{pmatrix}-\begin{pmatrix}x_{2}x_{3}x_{4}&x_{2}x_{4}y_{3}\\ 0&0\end{pmatrix}\begin{pmatrix}0&1\\ 0&0\end{pmatrix})$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =[\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\begin{pmatrix}x_{1}x_{2}x_{3}&x_{1}x_{2}y_{3}\\ 0&0\end{pmatrix}]\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}+\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}[\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\begin{pmatrix}x_{2}x_{3}x_{4}&x_{2}x_{4}y_{3}\\ 0&0\end{pmatrix}]$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =[\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}\begin{pmatrix}x_{2}&y_{2}\\ 0&0\end{pmatrix}\begin{pmatrix}x_{3}&y_{3}\\ 0&0\end{pmatrix}]\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}[\begin{pmatrix}0&1\\ 0&0\end{pmatrix},\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}\begin{pmatrix}x_{2}&y_{2}\\ 0&0\end{pmatrix}\begin{pmatrix}x_{3}&y_{3}\\ 0&0\end{pmatrix}]$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =D(\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix},\begin{pmatrix}x_{2}&y_{2}\\ 0&0\end{pmatrix},\begin{pmatrix}x_{3}&y_{3}\\ 0&0\end{pmatrix})\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\begin{pmatrix}x_{1}&y_{1}\\ 0&0\end{pmatrix}D(\begin{pmatrix}x_{4}&y_{4}\\ 0&0\end{pmatrix},\begin{pmatrix}x_{2}&y_{2}\\ 0&0\end{pmatrix},\begin{pmatrix}x_{3}&y_{3}\\ 0&0\end{pmatrix})$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\pi(D(a,b,c),d)+\pi(a,D(d,b,c)).$

Similarly, we have $D(a,\pi(b,d),c)=\pi(D(a,b,c),d)+\pi(b,D(a,d,c))$ and $D(a,b,\pi(c,d))=\pi(D(a,b,c),d)+\pi(c,D(a,b,d))$. Thus $D$ is tri-derivation.

In the section, we provide a necessary and sufficient condition such that the fourth adjoint $D^{****}$ of a tri-derivation $D:A\times A\times A\longrightarrow X$ is again a tri-derivation. For the fourth adjoint $D^{****}$ of a tri-derivation $D:A\times A\times A\longrightarrow X$, we are faced with the case eight:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

In the follwing, we prove the state of case 1. The remaining state are proved in the same way.

Theorem 4.3**.**

Let $(\pi_{1},X,\pi_{2})$ be a Banach $A-$module and $D:A\times A\times A\longrightarrow X$ be a tri-derivation. Then $D^{****}:(A^{**},\square)\times(A^{**},\square)\times(A^{**},\square)\longrightarrow X^{**}$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{**r*}(D^{****}(A,A,A^{**}),X^{*})\subseteq A^{*}$, 2. (2)

$\pi_{2}^{****}(X^{*},D^{****}(A,A^{**},A^{**}))\subseteq A^{*}$, 3. (3)

$D^{****r*}(\pi_{1}^{****}(X^{*},A^{**}),A^{**},A^{**})\subseteq A^{*}$, 4. (4)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 5. (5)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

Proof.

Let $D:A\times A\times A\longrightarrow X$ be a tri-derivation and (1),(2),(3),(4),(5) holds. If $\{a_{\alpha}\},\{b_{\beta}\},\{c_{\gamma}\}$ and $\{d_{\tau}\}$ be bounded nets in $A$ , converging in $w^{*}-$topology to $a^{**},b^{**},c^{**}$ and $d^{**}\in A^{**}$ respectively, in this case using (2), we conclude that

[TABLE]

Thus for every $x^{*}\in X^{*}$ we have

$\langle D^{****}(\pi^{***}(a^{**},d^{**}),b^{**},c^{**}),x^{*}\rangle=\langle\pi^{***}(a^{**},d^{**}),D^{***}(b^{**},c^{**},x^{*})\rangle$

$=\langle a^{**},\pi^{**}(d^{**},D^{***}(b^{**},c^{**},x^{*}))\rangle=\lim\limits_{\alpha}\langle\pi^{**}(d^{**},D^{***}(b^{**},c^{**},x^{*})),a_{\alpha}\rangle$

$=\lim\limits_{\alpha}\langle d^{**},\pi^{*}(D^{***}(b^{**},c^{**},x^{*}),a_{\alpha})\rangle=\lim\limits_{\alpha}\lim\limits_{\tau}\langle\pi^{*}(D^{***}(b^{**},c^{**},x^{*}),a_{\alpha}),d_{\tau}\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\tau}\langle D^{***}(b^{**},c^{**},x^{*}),\pi(a_{\alpha},d_{\tau})\rangle=\lim\limits_{\alpha}\lim\limits_{\tau}\langle b^{**},D^{**}(c^{**},x^{*},\pi(a_{\alpha},d_{\tau}))\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\langle D^{**}(c^{**},x^{*},\pi(a_{\alpha},d_{\tau})),b_{\beta}\rangle=\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\langle c^{**},D^{*}(x^{*},\pi(a_{\alpha},d_{\tau}),b_{\beta})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle D^{*}(x^{*},\pi(a_{\alpha},d_{\tau}),b_{\beta}),c_{\gamma}\rangle=\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle x^{*},D(\pi(a_{\alpha},d_{\tau}),b_{\beta},c_{\gamma})\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle x^{*},\pi_{2}(D(a_{\alpha},b_{\beta},c_{\gamma}),d_{\tau})+\pi_{1}(a_{\alpha},D(d_{\tau},b_{\beta},c_{\gamma}))\rangle$

$=\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle x^{*},\pi_{2}(D(a_{\alpha},b_{\beta},c_{\gamma}),d_{\tau})\rangle+\lim\limits_{\alpha}\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle x^{*},\pi_{1}(a_{\alpha},D(d_{\tau},b_{\beta},c_{\gamma}))\rangle$

$=\langle x^{*},\pi_{2}^{***}(D^{****}(a^{**},b^{**},c^{**}),d^{**})\rangle+\langle x^{*},\pi_{1}^{***}(a^{**},D^{****}(d^{**},b^{**},c^{**}))\rangle$

$=\langle\pi_{2}^{***}(D^{****}(a^{**},b^{**},c^{**}),d^{**})+\pi_{1}^{***}(a^{**},D^{****}(d^{**},b^{**},c^{**})),x^{*}\rangle$.

Therefore

[TABLE]

Applying (1) and (3) respectively, we can deduce that

[TABLE]

[TABLE]

So, by proof very similar we can deduce that

[TABLE]

Applying (4) and (5), we can deduce that

[TABLE]

Thus we deduce that

[TABLE]

By comparing equations (4-1),(4-2) and (4-3) follows that $D^{****}:(A^{**},\square)\times(A^{**},\square)\times(A^{**},\square)\longrightarrow X^{**}$ is a tri-derivation.

For the converse, let $D$ and $D^{****}:(A^{**},\square)\times(A^{**},\square)\times(A^{**},\square)\longrightarrow X^{**}$ are tri-derivation. We show that (1),(2),(3),(4),(5) holds. We shall only prove (2) the others have similar argument. Fourth adjoint $D^{****}$ is tri-derivation, thus we have

[TABLE]

In the other hands, the mapping $D$ is tri-derivation, thus we have

[TABLE]

Therefore follows that $\pi_{2}^{***}(D^{****}(a,b^{**},c^{**}),d^{**})=w^{*}-\lim\limits_{\tau}w^{*}-\lim\limits_{\beta}w^{*}-\lim\limits_{\gamma}\pi_{2}(D(a,b_{\beta},c_{\gamma}),d_{\tau})$. So, for every $d^{**}\in A^{**}$ we have

$\langle\pi_{2}^{****}(x^{*},D^{****}(a,b^{**},c^{**})),d^{**}\rangle=\langle x^{*},\pi_{2}^{***}(D^{****}(a,b^{**},c^{**}),d^{**})\rangle$

$=\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle x^{*},\pi_{2}(D(a,b_{\beta},c_{\gamma}),d_{\tau})\rangle=\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle x^{*},\pi_{2}^{r}(d_{\tau},D(a,b_{\beta},c_{\gamma}))\rangle$

$=\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle\pi_{2}^{r*}(x^{*},d_{\tau}),D(a,b_{\beta},c_{\gamma})\rangle=\lim\limits_{\tau}\lim\limits_{\beta}\lim\limits_{\gamma}\langle D^{*}(\pi_{2}^{r*}(x^{*},d_{\tau}),a,b_{\beta}),c_{\gamma}\rangle$

$=\lim\limits_{\tau}\lim\limits_{\beta}\langle c^{**},D^{*}(\pi_{2}^{r*}(x^{*},d_{\tau}),a,b_{\beta})\rangle=\lim\limits_{\tau}\lim\limits_{\beta}\langle D^{**}(c^{**},\pi_{2}^{r*}(x^{*},d_{\tau}),a),b_{\beta}\rangle$

$=\lim\limits_{\tau}\langle b^{**},D^{**}(c^{**},\pi_{2}^{r*}(x^{*},d_{\tau}),a)\rangle=\lim\limits_{\tau}\langle D^{***}(b^{**},c^{**},\pi_{2}^{r*}(x^{*},d_{\tau})),a\rangle$

$=\lim\limits_{\tau}\langle D^{****}(a,b^{**},c^{**}),\pi_{2}^{r*}(x^{*},d_{\tau})\rangle=\lim\limits_{\tau}\langle D^{****}(a,b^{**},c^{**}),\pi_{2}^{r*r}(d_{\tau},x^{*})\rangle$

$=\lim\limits_{\tau}\langle\pi_{2}^{r*r*}(D^{****}(a,b^{**},c^{**}),d_{\tau}),x^{*}\rangle=\lim\limits_{\tau}\langle\pi_{2}^{r*r**}(x^{*},D^{****}(a,b^{**},c^{**})),d_{\tau}\rangle$

$=\langle\pi_{2}^{r*r**}(x^{*},D^{****}(a,b^{**},c^{**})),d^{**}\rangle$.

As $\pi_{2}^{r*r**}(x^{*},D^{****}(a,b^{**},c^{**}))$ always lies in $A^{*}$, we have reached (2). ∎

For case 2, forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{**r*}(D^{****}(A^{**},A^{**},A^{**}),X^{*})\subseteq A^{*}$, 2. (2)

$D^{****r*}(\pi_{1}^{****}(X^{*},A^{**}),A^{**},A^{**})\subseteq A^{*}$, 3. (3)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 4. (4)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

For case 3, Forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{****}(X^{*},D^{****}(A,A^{**},A^{**}))\subseteq A^{*}$, 2. (2)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 3. (3)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

For case 4, Forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{**r*}(D^{****}(A,A,A^{**}),X^{*})\subseteq A^{*}$, 2. (2)

$\pi_{2}^{****}(X^{*},D^{****}(A,A^{**},A^{**}))\subseteq A^{*}$, 3. (3)

$D^{****r*}(\pi_{1}^{****}(X^{*},A^{**}),A^{**},A^{**})\subseteq A^{*}$, 4. (4)

$D^{*****}(\pi_{1}^{****}(X^{*},A^{**}),A,A)\subseteq A^{*}$, 5. (5)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 6. (6)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

For case 5, Forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{**r*}(D^{****}(A^{**},A^{**},A^{**}),X^{*})\subseteq A^{*}$, 2. (2)

$\pi_{2}^{****}(X^{*},D^{****}(A,A^{**},A^{**}))\subseteq A^{*}$, 3. (3)

$D^{****r*}(\pi_{1}^{****}(X^{*},A^{**}),A^{**},A^{**})\subseteq A^{*}$, 4. (4)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 5. (5)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

For case 6, Forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{**r*}(D^{****}(A^{**},A^{**},A^{**}),X^{*})\subseteq A^{*}$, 2. (2)

$D^{****r*}(\pi_{1}^{****}(X^{*},A^{**}),A^{**},A^{**})\subseteq A^{*}$, 3. (3)

$D^{*****}(\pi_{1}^{****}(X^{*},A^{**}),A,A)\subseteq A^{*}$, 4. (4)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 5. (5)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

For case 7, Forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{****}(X^{*},D^{****}(A,A^{**},A^{**}))\subseteq A^{*}$, 2. (2)

$\pi_{2}^{**r*}(D^{****}(A,A,A^{**}),X^{*})\subseteq A^{*}$, 3. (3)

$D^{*****}(\pi_{1}^{****}(X^{*},A^{**}),A,A)\subseteq A^{*}$, 4. (4)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 5. (5)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

For case 8, Forth adjoint $D^{****}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ is a tri-derivation if and only if

1. (1)

$\pi_{2}^{****}(X^{*},D^{****}(A,A^{**},A^{**}))\subseteq A^{*}$, 2. (2)

$\pi_{2}^{**r*}(D^{****}(A^{**},A^{**},A^{**}),X^{*})\subseteq A^{*}$, 3. (3)

$D^{****r*}(\pi_{1}^{****}(X^{*},A^{**}),A^{**},A^{**})\subseteq A^{*}$, 4. (4)

$D^{*****}(\pi_{1}^{****}(X^{*},A^{**}),A,A)\subseteq A^{*}$, 5. (5)

$D^{******}(A^{**},\pi_{1}^{****}(X^{*},A^{**}),A)\subseteq A^{*}$, 6. (6)

$D^{*******}(A^{**},A^{**},\pi_{1}^{****}(X^{*},A^{**}))\subseteq A^{*}$.

Remark 4.4*.*

For adjoint $D^{r****r}$ of tri-derivation $D:A\times A\times A\longrightarrow X$ we can use the same argument.