Open Access

Triple fixed point theorems on FLM algebras

Fixed Point Theory and Applications20132013:16

https://doi.org/10.1186/1687-1812-2013-16

Received: 4 July 2012

Accepted: 9 January 2013

Published: 25 January 2013

Abstract

This paper considers tripled fixed point theorems on unital without of order semi-simple fundamental locally multiplicative topological algebras (abbreviated by FLM algebras).

MSC:46H.

Keywords

tripled fixed point fundamental topological algebras FLM algebras holomorphic function semi-simple algebras without of order

1 Introduction

Ansari in [1] introduced the notion of fundamental topological spaces and algebras and proved Cohen’s factorization theorem for these algebras. A topological linear space A is said to be fundamental if there exists b > 1 such that for every sequence ( x n ) of A , the convergence of b n ( x n x n 1 ) to zero in A implies that ( x n ) is Cauchy. A fundamental topological algebra is an algebra whose underlying topological linear space is fundamental.

A fundamental topological algebra is called locally multiplicative if there exists a neighborhood U 0 of zero such that for every neighborhood V of zero, the sufficiently large powers of U 0 lie in V. The fundamental locally multiplicative topological algebras (FLM) were introduced by Ansari in [2]. Some celebrated theorems in Banach algebras were generalized to FLM algebras in [3], and authors investigated some fixed points theorems for holomorphic functions on these algebras (see Theorems 3.5, 3.6 and 3.7 of [3]).

An algebra A is called without of order if for every a , b A , a b = 0 , then a = 0 or b = 0 .

In [4], Bhaskar and Lakshmikantham introduced the notions of a mixed monotone mapping and a coupled fixed point, proved some coupled fixed point theorems for the mixed monotone mapping and discussed the existence and uniqueness of a solution for a periodic boundary value problem. Also, Samet and Vetro studied a coupled fixed point of N-order in [5]. There are many works on a coupled fixed point of contraction, weak contraction and generalized contraction mappings on various metric spaces such as [69].

Let A be a metric space and let F : A × A × A A be a function. An element ( x , y , z ) A × A × A is said to be a tripled fixed point of the mapping F if F ( x , y , z ) = F ( x , z , y ) = x , F ( y , x , z ) = F ( y , z , x ) = y and F ( z , x , y ) = F ( z , y , x ) = z . Tripled fixed point theorems in partially ordered metric spaces were studied by Berinde and Borcut in [10], and this concept was considered by Aydi et al. for weak compatible mappings in abstract metric spaces [11].

In this paper, at first (Section 2) we obtain some basic results for FLM algebras, and next we consider tripled fixed point theorems on FLM algebras.

2 Some results on FLM algebras

By Ω A we mean the set of all elements a A such that ρ A ( a ) < 1 , where ρ A ( a ) is the spectral radius of a A . We denote the center of topological algebra A by Z ( A ) such that
Z ( A ) = { a A : a x = x a  for all  x A } .
Definition 2.1 Let ( A , d ) be a metrizable topological algebra. We say A is a submultiplicatively metrizable topological algebra if
d ( 0 , x y z ) d ( 0 , x ) d ( 0 , y ) d ( 0 , z ) and d ( 0 , λ x ) < | λ | d ( 0 , x )

for each x , y , z A and λ C . For abbreviation, we denote d A ( 0 , x ) by D A ( x ) for any x A .

Let A , and C be metric spaces with meters d A , d B and d C , respectively. Then A × B × C becomes a metric space with the following meter:
d ( ( a 1 , b 1 , c 1 ) , ( a 2 , b 2 , c 2 ) ) = d A ( a 1 , a 2 ) + d B ( b 1 , b 2 ) + d C ( c 1 , c 2 )
(2.1)

for every a 1 , a 2 A , b 1 , b 2 B and c 1 , c 2 C . When A , and C are algebras, then by the usual point-wise definitions for addition, scalar multiplication and product, A × B × C becomes an algebra.

Proposition 2.2 Let A , and C be complete metrizable FLM algebras with submultiplicative meters d A , d B and d C , respectively. Then A × B × C is a complete metrizable FLM algebra with a submultiplicative meter d.

Proof Let A , and C be FLM algebras with meters d A , d B and d C , respectively. By the definition of FLM algebras, obviously, A × B × C is a complete metrizable FLM algebra with a meter d (the meter defined in (2.1)). For submultiplicativity, we have
(2.2)
for every a 1 , a 2 A , b 1 , b 2 B and c 1 , c 2 C . Also,
d ( ( 0 , 0 , 0 ) , ( λ a , λ b , λ c ) ) = d A ( 0 , λ a ) + d B ( 0 , λ b ) + d C ( 0 , λ c ) < | λ | d A ( 0 , a ) + | λ | d B ( 0 , b ) + | λ | d C ( 0 , c ) = | λ | ( d A ( 0 , a ) + d B ( 0 , b ) + d C ( 0 , c ) ) = | λ | ( d ( ( 0 , 0 , 0 ) , ( a , b , c ) ) ) .
(2.3)

Therefore, (2.2) and (2.3) show that d is submultiplicative. □

Similar to Definition 2.1, we write D A × B × C ( a , b , c ) as an abbreviation for d ( ( 0 , 0 , 0 ) , ( a , b , c ) ) . We recall the following theorem from [3].

Theorem 2.3 [[3], Theorem 3.3]

Let A be a complete metrizable FLM algebra with a submultiplicative meter d A . Then ρ ( x ) = lim n D A ( x n ) 1 / n .

Lemma 2.4 Let A , and C be complete metrizable FLM algebras with submultiplicative meters d A , d B and d C , respectively. Then
ρ ( x , y , z ) ρ A ( x ) + ρ B ( y ) + ρ C ( z )

for any element ( x , y , z ) A × B × C .

Proof For given a A , b B and c C , we have ρ A ( a ) = lim n D A ( a n ) 1 / n , ρ B ( b ) = lim n D B ( b n ) 1 / n and ρ C ( c ) = lim n D C ( c n ) 1 / n (Theorem 2.3). From Proposition 2.2, it follows that A × B × C is a complete metrizable FLM algebra with a submultiplicative meter d. Then again, Theorem 2.3 implies that
ρ ( x , y , z ) = lim n D A × B × C ( ( x , y , z ) n ) 1 n = lim n D A × B × C ( ( x n , y n , z n ) ) 1 n = lim n ( D A ( x n ) + D B ( y n ) + D C ( z n ) ) 1 n lim n D A ( x n ) 1 n + lim n D B ( y n ) 1 n + lim n D C ( z n ) 1 n = ρ A ( x ) + ρ B ( y ) + ρ C ( z )
(2.4)

for every x A , y B and z C . □

Similar to Ω A and Z ( A ) , we define these sets for A × A × A as follows:
Ω A × A × A = { ( x , y , z ) A × A × A : ρ ( x , y , z ) < 1 } ,
and
Z ( A × A × A ) = { ( x , y , z ) A × A × A : ( x , y , z ) ( a , b , c ) = ( a , b , c ) ( x , y , z ) , for every  a , b , c A } = { ( x , y , z ) A × A × A : ( x a , y b , z c ) = ( a x , b y , c z ) , for every  a , b , c A } .

Clearly, if ( x , y , z ) Z ( A × A × A ) , then x , y , z Z ( A ) and Z ( A ) Z ( A × A × A ) . Also, if ( x , y , z ) Ω A × A × A , then ( x , 0 , 0 ) , ( 0 , y , 0 ) and ( 0 , 0 , z ) are in Ω A × A × A , and by Lemma 2.4 and its proof, we have x , y , z Ω A .

Let E ( A ) be the set of all elements x A for which E ( x ) = n = 1 x n n ! can be defined. If A is a complete metrizable FLM algebra, then E ( A ) = A ([[12], Theorem 5.4]). Therefore, in the light of Theorem 5.4 of [12] and Proposition 2.2, we have the following theorem.

Theorem 2.5 Let A be a complete metrizable FLM algebra, then E ( A × A × A ) = A × A × A .

3 Tripled fixed point theorems

In this section, we consider some results about tripled fixed point theorems on unital complete semi-simple metrizable FLM algebras, and we extend these results to Banach algebras. By id A , we mean the identity map on A .

Theorem 3.1 Let A be a unital without of order complete semi-simple metrizable FLM algebra with a submultiplicative meter d A . If F : Ω A × A × A A × A × A Ω A is a holomorphic map that satisfies the conditions F ( 0 , 0 , 0 ) = 0 , F x ( 0 , 0 , 0 ) = id A , F y ( 0 , 0 , 0 ) = 0 , F z ( 0 , 0 , 0 ) = 0 , 2 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 2 , i , j , k = 0 , 1 , 2 , and 3 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 3 , i , j , k = 0 , 1 , 2 , 3 , then every ( a , b , c ) Ω A × A × A Z ( A × A × A ) is a tripled fixed point for F.

Proof Fix ( a , b , c ) Ω A × A × A Z ( A × A × A ) and consider the map f : C × C × C Ω A with f ( α , β , γ ) = F ( α a , β b , γ c ) . Clearly, f is a holomorphic function on
Since F ( 0 , 0 , 0 ) = 0 , F x ( 0 , 0 , 0 ) = id A , F y ( 0 , 0 , 0 ) = 0 , F z ( 0 , 0 , 0 ) = 0 , 2 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 2 , i , j , k = 0 , 1 , 2 , and 3 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 3 , i , j , k = 0 , 1 , 2 , 3 , then F has a Taylor expansion about ( 0 , 0 , 0 ) :
F ( x , y , z ) = i = 0 j = 0 k = 0 x i y j z k i ! j ! k ! ( i + j + k F x i y j z k ) ( 0 , 0 , 0 ) = x + k = 0 1 k ! j = 0 k ( k j ) i = 0 j ( k j i ) x i y j z k i j ( k F x i y j z k i j ) ( 0 , 0 , 0 )
for every ( x , y , z ) Ω A × A × A Z ( A × A × A ) . Therefore,
F ( α a , β b , γ c ) = α a + k = 4 1 k ! j = 0 k ( k j ) i = 0 j ( k j i ) α i a i β j b j γ k i j c k i j × ( k F x i y j z k i j ) ( 0 , 0 , 0 ) .
(3.1)
We claim that
j = 0 k ( k j ) i = 0 j ( k j i ) α i a i β j b j γ k i j c k i j ( k F x i y j z k i j ) ( 0 , 0 , 0 ) ,
(3.2)
is zero for every k 4 . Assume towards a contradiction that there exists k 4 such that (3.2) is non-zero. Let l 4 be an integer such that
j = 0 k ( l j ) i = 0 j ( l j i ) α i a i β j b j γ l i j c l i j ( k F x i y j z l i j ) ( 0 , 0 , 0 ) 0 .
(3.3)
Suppose that q is an element of A such that ρ A ( q ) = 0 . Now, we consider the following five cases:
  1. (1)

    i = l , j = 0 ,

     
  2. (2)

    i = 0 , j = l ,

     
  3. (3)

    i + j = l ,

     
  4. (4)

    1 i + j < l ,

     
  5. (5)

    i = j = 0 .

     
Case (1). In this case, we have α l a l l F x l ( 0 , 0 , 0 ) 0 . Let n 1 , by (3.1) and (3.3), we have
F ( n 1 l α a + n α l q , β b , γ c ) = n 1 l α a + n α l q + 1 l ! ( n 1 l α a + n α l q ) l l F x l ( 0 , 0 , 0 ) = n 1 l α a + n α l q + 1 l ! ( n l α l 2 q l + l n 1 l α a n l 1 α l ( l 1 ) q l 1 + + n α l a l ) l F x l ( 0 , 0 , 0 ) = n 1 l α a + n α l ( q + 1 l ! a l l F x l ( 0 , 0 , 0 ) ) + P ( α ) l F x l ( 0 , 0 , 0 ) .
(3.4)
In (3.4), by P ( α ) , we mean the remaining part of ( n 1 l α a + n α k q ) k . Since a Z ( A ) , therefore a q = q a . Then Lemma 2.4 and Lemma 3.6 of [3] imply
ρ ( n 1 l α a + n α l q , β b , γ c ) ρ A ( n 1 l α a + n α l q ) + ρ A ( β b ) + ρ A ( γ c ) < n 1 l | α | ρ A ( a ) + | β | ρ A ( b ) + | γ | ρ A ( c ) < μ ( ρ A ( a ) + ρ A ( b ) + ρ A ( c ) ) ,
where μ = max { n 1 l | α | , | β | , | γ | } . Now, we define a holomorphic function H from { α C : 0 < | α | < 1 ρ ( a , b , c ) } into A as follows:
H ( α ) = F ( n 1 l α a + n α l q , β b , γ c ) n 1 l α a n α l .
By (3.4) we conclude that H ( 0 ) = q + 1 l ! a l l F x l ( 0 , 0 , 0 ) . Vesentini’s theorem ([[13], Theorem 3.4.7]) implies that ρ A H is a subharmonic function on { α C : 0 < | α | < 1 ρ ( a , b , c ) } . Moreover, by the maximum principle, we can write ρ A ( H ( 0 ) ) max | α | = 1 ρ A ( H ( α ) ) . Then Lemma 3.6 of [3] implies that
ρ A ( q + 1 l ! a l l F x l ( 0 , 0 , 0 ) ) max | α | = 1 ρ ( H ( α ) ) < 1 n l ! ρ A ( a l ) ρ A ( l F x l ( 0 , 0 , 0 ) ) < 1 n l ! | α | l ρ A ( l F x l ( 0 , 0 , 0 ) ) .
(3.5)
The above inequality holds for every n 1 . Therefore, if n , then
ρ A ( q + 1 l ! a l l F x l ( 0 , 0 , 0 ) ) = 0

for every q A with ρ A ( q ) = 0 . Hence, Theorem 3.4 of [3] implies that a l l F x l ( 0 , 0 , 0 ) is in radical of A . Since A is semi-simple, therefore a l l F x l ( 0 , 0 , 0 ) = 0 . Since a Ω A Z ( A ) , so a l 0 , and since A is without of order, therefore l F x l ( 0 , 0 , 0 ) = 0 , a contradiction. Thus, our claim is true, and from (3.1), we conclude that F ( a , b , c ) = a . Similarly, we have F ( a , c , b ) = a , F ( b , a , c ) = F ( b , c , a ) = b and F ( c , a , b ) = F ( c , b , a ) = c .

Case (2). In this case, we have β l b l l F y l ( 0 , 0 , 0 ) 0 . Again, by (3.1) and (3.3), we have
F ( α a + n β l q , n 1 l β b , γ c ) = α a + n β l q + 1 l ! n β l b l l F y l ( 0 , 0 , 0 ) = α a + n β l ( q + 1 l ! b l l F y l ( 0 , 0 , 0 ) ) .
Again, by Lemma 2.4 and Lemma 3.6 of [3], we have
ρ ( α a + n β l q , n 1 l β b , γ c ) ρ A ( α a + n β l q ) + ρ A ( n 1 l β b ) + ρ A ( γ c ) < | α | ρ A ( a ) + n 1 l | β | ρ A ( b ) + | γ | ρ A ( c ) < μ ( ρ A ( a ) + ρ A ( b ) + ρ A ( c ) ) ,
where μ = max { | α | , n 1 l | β | , | γ | } . Now, we define a holomorphic function H from { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , n 1 l | β | , | γ | } } into A as follows:
H ( α ) = F ( α a + n β l q , n 1 l β b , γ c ) α a n β l .
Then from (3.7) it follows that H ( 0 ) = q + 1 l ! b l l F y l ( 0 , 0 , 0 ) . Then ρ A H is a subharmonic function on { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , n 1 l | β | , | γ | } } . Moreover, Lemma 3.6 of [3] implies that
ρ A ( q + 1 l ! b l l F y l ( 0 , 0 , 0 ) ) max | α | = 1 ρ ( H ( α ) ) < 1 n l ! ρ A ( b l ) ρ A ( l F y l ( 0 , 0 , 0 ) ) < 1 n l ! | β | l ρ A ( l F y l ( 0 , 0 , 0 ) ) .
(3.6)
The above inequality holds for every n 1 . Therefore, if n , then
ρ A ( q + 1 l ! b l l F y l ( 0 , 0 , 0 ) ) = 0

for every q A with ρ A ( q ) = 0 . Hence, Theorem 3.4 of [3] implies that b l l F y l ( 0 , 0 , 0 ) is in radical of A . Since A is semi-simple, therefore b l l F y l ( 0 , 0 , 0 ) = 0 . Since b Ω A Z ( A ) , so b l 0 . By using that A is without of order, we conclude that l F y l ( 0 , 0 , 0 ) = 0 , a contradiction. Thus, our claim is true, and from (3.1), we conclude that F ( a , b , c ) = a . Similarly, we have F ( a , c , b ) = a , F ( b , a , c ) = F ( b , c , a ) = b and F ( c , a , b ) = F ( c , b , a ) = c .

Case (3). In this case, we suppose that i + j = l , i , j { 0 , 1 , 2 , 3 , } (without loss of generality, we prove this case for only one i and one j such that i + j = l ). Again by (3.1) and (3.3), we have
(3.7)
By Lemma 2.4 and Lemma 3.6 of [3], we have
ρ ( α a + n α i q , n 1 l i β b , γ c ) ρ A ( α a + n α i q ) + ρ A ( n 1 l i β b ) + ρ A ( γ c ) < | α | ρ A ( a ) + n 1 l i | β | ρ A ( b ) + | γ | ρ A ( c ) < μ ( ρ A ( a ) + ρ A ( b ) + ρ A ( c ) ) ,
where μ = max { | α | , n 1 l i | β | , | γ | } . Now, we define a holomorphic function H from { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , n 1 l i | β | , | γ | } } into A as follows:
H ( λ ) = F ( α a + n α i q , n 1 l i β b , γ c ) α a n α i .
Then from (3.7) it follows that H ( 0 ) = q + 1 ( l i ) ! i ! a i β l i b l i l F x i y l i ( 0 , 0 , 0 ) . Then ρ A H is a subharmonic function on { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , n 1 l i | β | , | γ | } } . Moreover, Lemma 3.6 of [3] implies that
(3.8)
The above inequality holds for every n 1 . Therefore, if n , then
ρ A ( q + 1 ( l i ) ! i ! a i β l i b l i l F x i y l i ( 0 , 0 , 0 ) ) = 0

for every q A with ρ A ( q ) = 0 . Hence, a i β l i b l i l F x i y l i ( 0 , 0 , 0 ) is in radical of A , therefore a i β l i b l i l F x i y l i ( 0 , 0 , 0 ) = 0 . Since β l i 0 and a , b Ω A Z ( A ) , so a i 0 and b l i 0 . Again, by using that A is without of order, we conclude that l F x i y l i ( 0 , 0 , 0 ) = 0 , a contradiction. Thus, our claim is true, and from (3.1), we conclude that F ( a , b , c ) = a . Similarly, we have F ( a , c , b ) = a , F ( b , a , c ) = F ( b , c , a ) = b and F ( c , a , b ) = F ( c , b , a ) = c .

Case (4). Let 1 i + j l . Then we have α i a i β j b j γ l i j c l i j l F x i y j z l i j ( 0 , 0 , 0 ) 0 . Again, by (3.1) and (3.3), we have
(3.9)
Then
ρ ( α a + n α i q , β b , n 1 l i j γ c ) ρ A ( α a + n α i q ) + ρ A ( β b ) + ρ A ( n 1 l i j γ c ) < | α | ρ A ( a ) + | β | ρ A ( b ) + n 1 l i j | γ | ρ A ( c ) < μ ( ρ A ( a ) + ρ A ( b ) + ρ A ( c ) ) ,
where μ = max { | α | , | β | , n 1 l i j | γ | } . Now, we define a holomorphic function H from { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , | β | , n 1 l i j | γ | } } into A as follows:
H ( α ) = F ( α a + n α i q , β b , n 1 l i j γ c ) α a n α i .
Then from (3.9) it follows that H ( 0 ) = q + 1 i ! j ! ( l i j ) ! ( a i β j b j γ l i j c l i j l F x i y j z l i j ( 0 , 0 , 0 ) ) . Then ρ A H is a subharmonic function on { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , | β | , n 1 l i j | γ | } } . Therefore,
(3.10)
Therefore, if n , then
ρ A ( q + 1 i ! j ! ( l i j ) ! a i β j b j γ l i j c l i j l F x i y j z l i j ( 0 , 0 , 0 ) ) = 0

for every q A with ρ A ( q ) = 0 . Hence, a i β j b j γ l i j c l i j l F x i y j z l i j ( 0 , 0 , 0 ) is in radical of A . Since A is semi-simple, therefore a i β j b j γ l i j c l i j l F x i y j z l i j ( 0 , 0 , 0 ) = 0 . Since β j 0 , γ l i j 0 and a , b Ω A Z ( A ) , so a i 0 , b j and c l i j 0 , we conclude that l F x i y j z l i j ( 0 , 0 , 0 ) = 0 , a contradiction. Thus, our claim is true, and from (3.1), we conclude that F ( a , b , c ) = a . Similarly, we have F ( a , c , b ) = a , F ( b , a , c ) = F ( b , c , a ) = b and F ( c , a , b ) = F ( c , b , a ) = c .

Case (5). Now, let i = j = 0 . Then we have γ l c l l F z l ( 0 , 0 , 0 ) 0 . Similar to the previous cases, we have
F ( α a + n γ l q , β b , n 1 l γ c ) = α a + n γ l q + 1 l ! n γ l c l l F z l ( 0 , 0 , 0 ) = α a + n γ l ( q + 1 l ! c l l F z l ( 0 , 0 , 0 ) ) .
(3.11)
Then
ρ ( α a + n γ l q , β b , n 1 l γ c ) ρ A ( α a + n γ l q ) + ρ A ( β b ) + ρ A ( n 1 l γ c ) < | α | ρ A ( a ) + | β | ρ A ( b ) + n 1 l | γ | ρ A ( c ) < μ ( ρ A ( a ) + ρ A ( b ) + ρ A ( c ) ) ,
where μ = max { | α | , | β | , n 1 l | γ | } . Now, we define a holomorphic function H from { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , | β | , n 1 l | γ | } } into A as follows:
H ( α ) = F ( α a + n γ l q , β b , n 1 l γ c ) α a n γ l .
Then from (3.11) it follows that H ( 0 ) = q + 1 l ! c l l F z l ( 0 , 0 , 0 ) . Then ρ A H is a subharmonic function on { η C : μ < 1 ρ ( a , b , c ) , μ = | η | = max { | α | , | β | , n 1 l | γ | } } , and
ρ A ( q + 1 l ! c l l F z l ( 0 , 0 , 0 ) ) max | α | = 1 ρ ( H ( α ) ) < 1 n l ! ρ A ( c l ) ρ A ( l F z l ( 0 , 0 , 0 ) ) < 1 n l ! | γ | l ρ A ( l F z l ( 0 , 0 , 0 ) ) .
(3.12)
Therefore, if n , then
ρ A ( q + 1 l ! c l l F z l ( 0 , 0 , 0 ) ) = 0

for every q A with ρ A ( q ) = 0 . Hence, c l l F z l ( 0 , 0 , 0 ) is in radical of A . Therefore, c l l F z l ( 0 , 0 , 0 ) = 0 . Since c Ω A Z ( A ) , so c l 0 , then l F z l ( 0 , 0 , 0 ) = 0 , a contradiction. Thus, (3.1) implies that our claim is true, and from (3.1), we conclude that F ( a , b , c ) = a . Similarly, we have F ( a , c , b ) = a , F ( b , a , c ) = F ( b , c , a ) = b and F ( c , a , b ) = F ( c , b , a ) = c .

By gathering the above five cases, we conclude ( a , b , c ) is a tripled fixed point for F, and since ( a , b , c ) was arbitrary, so every point of Ω A × A × A Z ( A × A × A ) is a tripled fixed point for F. □

Corollary 3.2 Let A be a unital without of order semi-simple Banach algebra. If F : Ω A × A × A A × A × A Ω A is a holomorphic map that satisfies the conditions F ( 0 , 0 , 0 ) = 0 , F x ( 0 , 0 , 0 ) = id A , F y ( 0 , 0 , 0 ) = 0 , F z ( 0 , 0 , 0 ) = 0 , 2 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 2 , i , j , k = 0 , 1 , 2 , and 3 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 3 , i , j , k = 0 , 1 , 2 , 3 , then every ( a , b , c ) Ω A × A × A Z ( A × A × A ) is a tripled fixed point for F.

In the following theorem, we characterize tripled fixed points of holomorphic functions on FLM algebras.

Theorem 3.3 Let A be a unital without of order complete semi-simple metrizable FLM algebra. For given ( a , b , c ) Ω A × A × A Z ( A × A × A ) , there is a holomorphic map F : Ω A × A × A Ω A satisfying the conditions F ( 0 , 0 , 0 ) = 0 , F x ( 0 , 0 , 0 ) = id A , F y ( 0 , 0 , 0 ) = 0 , F z ( 0 , 0 , 0 ) = 0 , 2 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 2 , i , j , k = 0 , 1 , 2 , and 3 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 3 , i , j , k = 0 , 1 , 2 , 3 , such that F ( a , b , c ) a , F ( b , a , c ) b and F ( c , a , b ) c .

Proof Let ( a , b , c ) Ω A × A × A Z ( A × A × A ) . Then there exist ( u , u , u ) A × A × A such that
( u a , u b , u c ) ( a u , b u , c u ) .
Let D A × A × A ( u , u , u ) < 1 , then D A ( u ) < 1 . Define U : = log ( e u ) , then
e U a e U a , e U b e U b and e U c e U c .
Now, define F : Ω A × A × A Ω A as follows:
F ( x , y , z ) = e x 2 z 2 U a 2 c 2 x e y 2 z 2 U b 2 c 2
(3.13)

for every ( x , y , z ) in Ω A × A × A . Clearly, F is a holomorphic function, F ( 0 , 0 , 0 ) = 0 , F x ( 0 , 0 , 0 ) = id A , F y ( 0 , 0 , 0 ) = 0 , F z ( 0 , 0 , 0 ) = 0 , 2 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 2 , i , j , k = 0 , 1 , 2 , and 3 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 3 , i , j , k = 0 , 1 , 2 , 3 , but F ( a , b , c ) a , and similarly, we can show that there is a holomorphic map F : Ω A × A × A Ω A with the required conditions such that F ( b , a , c ) b and F ( c , a , b ) c . □

Example 3.4 Let X = R be the space of real numbers and let F : X × X X be a function defined by F ( x , y , z ) = x that satisfies the conditions of Theorem 3.1.

Example 3.5 Let X be a unital without of order complete semi-simple Banach algebra and let F : X × X X be a function defined by F ( x , y , z ) = e y 2 z 2 x e y 2 z 2 that satisfies the conditions of Theorem 3.1. For example, let X = M ( G ) be the measure space on a locally compact Hausdorff space G. Another algebra that we can choose is 1 ( G ) , where G is a locally compact discrete group.

Corollary 3.6 Let A be a unital without of order semi-simple Banach algebra. For given ( a , b , c ) Ω A × A × A Z ( A × A × A ) , there is a holomorphic map F : Ω A × A × A Ω A satisfying the conditions F ( 0 , 0 , 0 ) = 0 , F x ( 0 , 0 , 0 ) = id A , F y ( 0 , 0 , 0 ) = 0 , F z ( 0 , 0 , 0 ) = 0 , 2 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 2 , i , j , k = 0 , 1 , 2 , and 3 F x i y j z k ( 0 , 0 , 0 ) = 0 , where i + j + k = 3 , i , j , k = 0 , 1 , 2 , 3 , such that F ( a , b , c ) a , F ( b , a , c ) b and F ( c , a , b ) c .

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Karaj Branch, Islamic Azad University

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© Razani and Hosseinzadeh; licensee Springer. 2013

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