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Triple fixed point theorems on FLM algebras

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.

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,bA, ab=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×AA 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 aA such that ρ A (a)<1, where ρ A (a) is the spectral radius of aA. We denote the center of topological algebra A by Z(A) such that

Z(A)={aA:ax=xa for all xA}.

Definition 2.1 Let (A,d) be a metrizable topological algebra. We say A is a submultiplicatively metrizable topological algebra if

d(0,xyz)d(0,x)d(0,y)d(0,z)andd(0,λx)<|λ|d(0,x)

for each x,y,zA and λC. For abbreviation, we denote d A (0,x) by D A (x) for any xA.

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 aA, bB and cC, 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 xA, yB and zC. □

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,zZ(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 xA 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(αa,βb,γc). 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 k4. Assume towards a contradiction that there exists k4 such that (3.2) is non-zero. Let l4 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)

    1i+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 n1, 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 aZ(A), therefore aq=qa. 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 {αC: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 {αC: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 n1. Therefore, if n, then

ρ A ( q + 1 l ! a l l F x l ( 0 , 0 , 0 ) ) =0

for every qA 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 {ηC: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 {ηC: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 n1. Therefore, if n, then

ρ A ( q + 1 l ! b l l F y l ( 0 , 0 , 0 ) ) =0

for every qA 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 {ηC: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 {ηC: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 n1. 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 qA 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 1i+jl. 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 {ηC: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 {ηC: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 qA 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 {ηC: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 {ηC: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 qA 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

(ua,ub,uc)(au,bu,cu).

Let D A × A × A (u,u,u)<1, then D A (u)<1. Define U:=log(eu), then

e U a e U a, e U b e U band 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×XX 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×XX 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.

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Correspondence to Abdolrahman Razani.

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All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Keywords

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