# 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  introduced the notion of fundamental topological spaces and algebras and proved Cohen’s factorization theorem for these algebras. A topological linear space $\mathcal{A}$ is said to be fundamental if there exists $b>1$ such that for every sequence $\left({x}_{n}\right)$ of $\mathcal{A}$, the convergence of ${b}^{n}\left({x}_{n}-{x}_{n-1}\right)$ to zero in $\mathcal{A}$ implies that $\left({x}_{n}\right)$ 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 . Some celebrated theorems in Banach algebras were generalized to FLM algebras in , and authors investigated some fixed points theorems for holomorphic functions on these algebras (see Theorems 3.5, 3.6 and 3.7 of ).

An algebra $\mathcal{A}$ is called without of order if for every $a,b\in \mathcal{A}$, $ab=0$, then $a=0$ or $b=0$.

In , 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 . There are many works on a coupled fixed point of contraction, weak contraction and generalized contraction mappings on various metric spaces such as .

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

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 ${\mathrm{\Omega }}_{\mathcal{A}}$ we mean the set of all elements $a\in \mathcal{A}$ such that ${\rho }_{\mathcal{A}}\left(a\right)<1$, where ${\rho }_{\mathcal{A}}\left(a\right)$ is the spectral radius of $a\in \mathcal{A}$. We denote the center of topological algebra $\mathcal{A}$ by $Z\left(\mathcal{A}\right)$ such that

Definition 2.1 Let $\left(\mathcal{A},d\right)$ be a metrizable topological algebra. We say $\mathcal{A}$ is a submultiplicatively metrizable topological algebra if

$d\left(0,xyz\right)\le d\left(0,x\right)d\left(0,y\right)d\left(0,z\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left(0,\lambda x\right)<|\lambda |d\left(0,x\right)$

for each $x,y,z\in \mathcal{A}$ and $\lambda \in \mathbb{C}$. For abbreviation, we denote ${d}_{\mathcal{A}}\left(0,x\right)$ by ${D}_{\mathcal{A}}\left(x\right)$ for any $x\in \mathcal{A}$.

Let $\mathcal{A}$, and $\mathcal{C}$ be metric spaces with meters ${d}_{\mathcal{A}}$, ${d}_{\mathcal{B}}$ and ${d}_{\mathcal{C}}$, respectively. Then $\mathcal{A}×\mathcal{B}×\mathcal{C}$ becomes a metric space with the following meter:

$d\left(\left({a}_{1},{b}_{1},{c}_{1}\right),\left({a}_{2},{b}_{2},{c}_{2}\right)\right)={d}_{\mathcal{A}}\left({a}_{1},{a}_{2}\right)+{d}_{\mathcal{B}}\left({b}_{1},{b}_{2}\right)+{d}_{\mathcal{C}}\left({c}_{1},{c}_{2}\right)$
(2.1)

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

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

Proof Let $\mathcal{A}$, and $\mathcal{C}$ be FLM algebras with meters ${d}_{\mathcal{A}}$, ${d}_{\mathcal{B}}$ and ${d}_{\mathcal{C}}$, respectively. By the definition of FLM algebras, obviously, $\mathcal{A}×\mathcal{B}×\mathcal{C}$ is a complete metrizable FLM algebra with a meter d (the meter defined in (2.1)). For submultiplicativity, we have (2.2)

for every ${a}_{1},{a}_{2}\in \mathcal{A}$, ${b}_{1},{b}_{2}\in \mathcal{B}$ and ${c}_{1},{c}_{2}\in \mathcal{C}$. Also,

$\begin{array}{rcl}d\left(\left(0,0,0\right),\left(\lambda a,\lambda b,\lambda c\right)\right)& =& {d}_{\mathcal{A}}\left(0,\lambda a\right)+{d}_{\mathcal{B}}\left(0,\lambda b\right)+{d}_{\mathcal{C}}\left(0,\lambda c\right)\\ <& |\lambda |{d}_{\mathcal{A}}\left(0,a\right)+|\lambda |{d}_{\mathcal{B}}\left(0,b\right)+|\lambda |{d}_{\mathcal{C}}\left(0,c\right)\\ =& |\lambda |\left({d}_{\mathcal{A}}\left(0,a\right)+{d}_{\mathcal{B}}\left(0,b\right)+{d}_{\mathcal{C}}\left(0,c\right)\right)\\ =& |\lambda |\left(d\left(\left(0,0,0\right),\left(a,b,c\right)\right)\right).\end{array}$
(2.3)

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

Similar to Definition 2.1, we write ${D}_{\mathcal{A}×\mathcal{B}×\mathcal{C}}\left(a,b,c\right)$ as an abbreviation for $d\left(\left(0,0,0\right),\left(a,b,c\right)\right)$. We recall the following theorem from .

Theorem 2.3 [, Theorem 3.3]

Let $\mathcal{A}$ be a complete metrizable FLM algebra with a submultiplicative meter ${d}_{\mathcal{A}}$. Then $\rho \left(x\right)={lim}_{n\to \mathrm{\infty }}{D}_{\mathcal{A}}{\left({x}^{n}\right)}^{1/n}$.

Lemma 2.4 Let $\mathcal{A}$, and $\mathcal{C}$ be complete metrizable FLM algebras with submultiplicative meters ${d}_{\mathcal{A}}$, ${d}_{\mathcal{B}}$ and ${d}_{\mathcal{C}}$, respectively. Then

$\rho \left(x,y,z\right)\le {\rho }_{\mathcal{A}}\left(x\right)+{\rho }_{\mathcal{B}}\left(y\right)+{\rho }_{\mathcal{C}}\left(z\right)$

for any element $\left(x,y,z\right)\in \mathcal{A}×\mathcal{B}×\mathcal{C}$.

Proof For given $a\in \mathcal{A}$, $b\in \mathcal{B}$ and $c\in \mathcal{C}$, we have ${\rho }_{\mathcal{A}}\left(a\right)={lim}_{n\to \mathrm{\infty }}{D}_{\mathcal{A}}{\left({a}^{n}\right)}^{1/n}$, ${\rho }_{\mathcal{B}}\left(b\right)={lim}_{n\to \mathrm{\infty }}{D}_{\mathcal{B}}{\left({b}^{n}\right)}^{1/n}$ and ${\rho }_{\mathcal{C}}\left(c\right)={lim}_{n\to \mathrm{\infty }}{D}_{\mathcal{C}}{\left({c}^{n}\right)}^{1/n}$ (Theorem 2.3). From Proposition 2.2, it follows that $\mathcal{A}×\mathcal{B}×\mathcal{C}$ is a complete metrizable FLM algebra with a submultiplicative meter d. Then again, Theorem 2.3 implies that

$\begin{array}{rcl}\rho \left(x,y,z\right)& =& \underset{n\to \mathrm{\infty }}{lim}{D}_{\mathcal{A}×\mathcal{B}×\mathcal{C}}{\left({\left(x,y,z\right)}^{n}\right)}^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}{D}_{\mathcal{A}×\mathcal{B}×\mathcal{C}}{\left(\left({x}^{n},{y}^{n},{z}^{n}\right)\right)}^{\frac{1}{n}}\\ =& \underset{n\to \mathrm{\infty }}{lim}{\left({D}_{\mathcal{A}}\left({x}^{n}\right)+{D}_{\mathcal{B}}\left({y}^{n}\right)+{D}_{\mathcal{C}}\left({z}^{n}\right)\right)}^{\frac{1}{n}}\\ \le & \underset{n\to \mathrm{\infty }}{lim}{D}_{\mathcal{A}}{\left({x}^{n}\right)}^{\frac{1}{n}}+\underset{n\to \mathrm{\infty }}{lim}{D}_{\mathcal{B}}{\left({y}^{n}\right)}^{\frac{1}{n}}+\underset{n\to \mathrm{\infty }}{lim}{D}_{\mathcal{C}}{\left({z}^{n}\right)}^{\frac{1}{n}}\\ =& {\rho }_{\mathcal{A}}\left(x\right)+{\rho }_{\mathcal{B}}\left(y\right)+{\rho }_{\mathcal{C}}\left(z\right)\end{array}$
(2.4)

for every $x\in \mathcal{A}$, $y\in \mathcal{B}$ and $z\in \mathcal{C}$. □

Similar to ${\mathrm{\Omega }}_{\mathcal{A}}$ and $Z\left(\mathcal{A}\right)$, we define these sets for $\mathcal{A}×\mathcal{A}×\mathcal{A}$ as follows:

${\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}=\left\{\left(x,y,z\right)\in \mathcal{A}×\mathcal{A}×\mathcal{A}:\rho \left(x,y,z\right)<1\right\},$

and

Clearly, if $\left(x,y,z\right)\in Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$, then $x,y,z\in Z\left(\mathcal{A}\right)$ and $Z\left(\mathcal{A}\right)\subseteq Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$. Also, if $\left(x,y,z\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}$, then $\left(x,0,0\right)$, $\left(0,y,0\right)$ and $\left(0,0,z\right)$ are in ${\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}$, and by Lemma 2.4 and its proof, we have $x,y,z\in {\mathrm{\Omega }}_{\mathcal{A}}$.

Let $E\left(\mathcal{A}\right)$ be the set of all elements $x\in \mathcal{A}$ for which $E\left(x\right)={\sum }_{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n!}$ can be defined. If $\mathcal{A}$ is a complete metrizable FLM algebra, then $E\left(\mathcal{A}\right)=\mathcal{A}$ ([, Theorem 5.4]). Therefore, in the light of Theorem 5.4 of  and Proposition 2.2, we have the following theorem.

Theorem 2.5 Let $\mathcal{A}$ be a complete metrizable FLM algebra, then $E\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)=\mathcal{A}×\mathcal{A}×\mathcal{A}$.

## 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 ${\mathrm{id}}_{\mathcal{A}}$, we mean the identity map on $\mathcal{A}$.

Theorem 3.1 Let $\mathcal{A}$ be a unital without of order complete semi-simple metrizable FLM algebra with a submultiplicative meter ${d}_{\mathcal{A}}$. If $F:{\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\subseteq \mathcal{A}×\mathcal{A}×\mathcal{A}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ is a holomorphic map that satisfies the conditions $F\left(0,0,0\right)=0$, $\frac{\partial F}{\partial x}\left(0,0,0\right)={\mathrm{id}}_{\mathcal{A}}$, $\frac{\partial F}{\partial y}\left(0,0,0\right)=0$, $\frac{\partial F}{\partial z}\left(0,0,0\right)=0$, $\frac{{\partial }^{2}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=2$, $i,j,k=0,1,2$, and $\frac{{\partial }^{3}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=3$, $i,j,k=0,1,2,3$, then every $\left(a,b,c\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\cap Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$ is a tripled fixed point for F.

Proof Fix $\left(a,b,c\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\cap Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$ and consider the map $f:\mathbb{C}×\mathbb{C}×\mathbb{C}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ with $f\left(\alpha ,\beta ,\gamma \right)=F\left(\alpha a,\beta b,\gamma c\right)$. Clearly, f is a holomorphic function on Since $F\left(0,0,0\right)=0$, $\frac{\partial F}{\partial x}\left(0,0,0\right)={\mathrm{id}}_{\mathcal{A}}$, $\frac{\partial F}{\partial y}\left(0,0,0\right)=0$, $\frac{\partial F}{\partial z}\left(0,0,0\right)=0$, $\frac{{\partial }^{2}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=2$, $i,j,k=0,1,2$, and $\frac{{\partial }^{3}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=3$, $i,j,k=0,1,2,3$, then F has a Taylor expansion about $\left(0,0,0\right)$:

$\begin{array}{rcl}F\left(x,y,z\right)& =& \sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}\frac{{x}^{i}{y}^{j}{z}^{k}}{i!j!k!}\left(\frac{{\partial }^{i+j+k}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\right)\left(0,0,0\right)\\ =& x+\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}\sum _{j=0}^{k}\left(\genfrac{}{}{0}{}{k}{j}\right)\sum _{i=0}^{j}\left(\genfrac{}{}{0}{}{k-j}{i}\right){x}^{i}{y}^{j}{z}^{k-i-j}\left(\frac{{\partial }^{k}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k-i-j}}\right)\left(0,0,0\right)\end{array}$

for every $\left(x,y,z\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\cap Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$. Therefore,

$\begin{array}{rcl}F\left(\alpha a,\beta b,\gamma c\right)& =& \alpha a+\sum _{k=4}^{\mathrm{\infty }}\frac{1}{k!}\sum _{j=0}^{k}\left(\genfrac{}{}{0}{}{k}{j}\right)\sum _{i=0}^{j}\left(\genfrac{}{}{0}{}{k-j}{i}\right){\alpha }^{i}{a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{k-i-j}{c}^{k-i-j}\\ ×\left(\frac{{\partial }^{k}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k-i-j}}\right)\left(0,0,0\right).\end{array}$
(3.1)

We claim that

$\sum _{j=0}^{k}\left(\genfrac{}{}{0}{}{k}{j}\right)\sum _{i=0}^{j}\left(\genfrac{}{}{0}{}{k-j}{i}\right){\alpha }^{i}{a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{k-i-j}{c}^{k-i-j}\left(\frac{{\partial }^{k}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k-i-j}}\right)\left(0,0,0\right),$
(3.2)

is zero for every $k\ge 4$. Assume towards a contradiction that there exists $k\ge 4$ such that (3.2) is non-zero. Let $l\ge 4$ be an integer such that

$\sum _{j=0}^{k}\left(\genfrac{}{}{0}{}{l}{j}\right)\sum _{i=0}^{j}\left(\genfrac{}{}{0}{}{l-j}{i}\right){\alpha }^{i}{a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{l-i-j}{c}^{l-i-j}\left(\frac{{\partial }^{k}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\right)\left(0,0,0\right)\ne 0.$
(3.3)

Suppose that q is an element of $\mathcal{A}$ such that ${\rho }_{\mathcal{A}}\left(q\right)=0$. 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\le i+j,

5. (5)

$i=j=0$.

Case (1). In this case, we have ${\alpha }^{l}{a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\ne 0$. Let $n\ge 1$, by (3.1) and (3.3), we have

$\begin{array}{rcl}F\left({n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q,\beta b,\gamma c\right)& =& {n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q+\frac{1}{l!}{\left({n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q\right)}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\\ =& {n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q+\frac{1}{l!}\left({n}^{l}{\alpha }^{{l}^{2}}{q}^{l}+l{n}^{\frac{1}{l}}\alpha a{n}^{l-1}{\alpha }^{l\left(l-1\right)}{q}^{l-1}\\ +\cdots +n{\alpha }^{l}{a}^{l}\right)\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\\ =& {n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}\left(q+\frac{1}{l!}{a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\right)+P\left(\alpha \right)\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right).\end{array}$
(3.4)

In (3.4), by $P\left(\alpha \right)$, we mean the remaining part of ${\left({n}^{\frac{1}{l}}\alpha a+n{\alpha }^{k}q\right)}^{k}$. Since $a\in Z\left(\mathcal{A}\right)$, therefore $aq=qa$. Then Lemma 2.4 and Lemma 3.6 of  imply

$\begin{array}{rcl}\rho \left({n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q,\beta b,\gamma c\right)& \le & {\rho }_{\mathcal{A}}\left({n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q\right)+{\rho }_{\mathcal{A}}\left(\beta b\right)+{\rho }_{\mathcal{A}}\left(\gamma c\right)\\ <& {n}^{\frac{1}{l}}|\alpha |{\rho }_{\mathcal{A}}\left(a\right)+|\beta |{\rho }_{\mathcal{A}}\left(b\right)+|\gamma |{\rho }_{\mathcal{A}}\left(c\right)\\ <& \mu \left({\rho }_{\mathcal{A}}\left(a\right)+{\rho }_{\mathcal{A}}\left(b\right)+{\rho }_{\mathcal{A}}\left(c\right)\right),\end{array}$

where $\mu =max\left\{{n}^{\frac{1}{l}}|\alpha |,|\beta |,|\gamma |\right\}$. Now, we define a holomorphic function H from $\left\{\alpha \in \mathbb{C}:0<|\alpha |<\frac{1}{\rho \left(a,b,c\right)}\right\}$ into $\mathcal{A}$ as follows:

$H\left(\alpha \right)=\frac{F\left({n}^{\frac{1}{l}}\alpha a+n{\alpha }^{l}q,\beta b,\gamma c\right)-{n}^{\frac{1}{l}}\alpha a}{n{\alpha }^{l}}.$

By (3.4) we conclude that $H\left(0\right)=q+\frac{1}{l!}{a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)$. Vesentini’s theorem ([, Theorem 3.4.7]) implies that ${\rho }_{\mathcal{A}}\circ H$ is a subharmonic function on $\left\{\alpha \in \mathbb{C}:0<|\alpha |<\frac{1}{\rho \left(a,b,c\right)}\right\}$. Moreover, by the maximum principle, we can write ${\rho }_{\mathcal{A}}\left(H\left(0\right)\right)\le {max}_{|\alpha |=1}{\rho }_{\mathcal{A}}\left(H\left(\alpha \right)\right)$. Then Lemma 3.6 of  implies that

$\begin{array}{rcl}{\rho }_{\mathcal{A}}\left(q+\frac{1}{l!}{a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\right)& \le & \underset{|\alpha |=1}{max}\rho \left(H\left(\alpha \right)\right)<\frac{1}{nl!}{\rho }_{\mathcal{A}}\left({a}^{l}\right){\rho }_{\mathcal{A}}\left(\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\right)\\ <& \frac{1}{nl!{|\alpha |}^{l}}{\rho }_{\mathcal{A}}\left(\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\right).\end{array}$
(3.5)

The above inequality holds for every $n\ge 1$. Therefore, if $n⟶\mathrm{\infty }$, then

${\rho }_{\mathcal{A}}\left(q+\frac{1}{l!}{a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)\right)=0$

for every $q\in \mathcal{A}$ with ${\rho }_{\mathcal{A}}\left(q\right)=0$. Hence, Theorem 3.4 of  implies that ${a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)$ is in radical of $\mathcal{A}$. Since $\mathcal{A}$ is semi-simple, therefore ${a}^{l}\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)=0$. Since $a\in {\mathrm{\Omega }}_{\mathcal{A}}\cap Z\left(\mathcal{A}\right)$, so ${a}^{l}\ne 0$, and since $\mathcal{A}$ is without of order, therefore $\frac{{\partial }^{l}F}{\partial {x}^{l}}\left(0,0,0\right)=0$, a contradiction. Thus, our claim is true, and from (3.1), we conclude that $F\left(a,b,c\right)=a$. Similarly, we have $F\left(a,c,b\right)=a$, $F\left(b,a,c\right)=F\left(b,c,a\right)=b$ and $F\left(c,a,b\right)=F\left(c,b,a\right)=c$.

Case (2). In this case, we have ${\beta }^{l}{b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\ne 0$. Again, by (3.1) and (3.3), we have

$\begin{array}{rcl}F\left(\alpha a+n{\beta }^{l}q,{n}^{\frac{1}{l}}\beta b,\gamma c\right)& =& \alpha a+n{\beta }^{l}q+\frac{1}{l!}n{\beta }^{l}{b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\\ =& \alpha a+n{\beta }^{l}\left(q+\frac{1}{l!}{b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\right).\end{array}$

Again, by Lemma 2.4 and Lemma 3.6 of , we have

$\begin{array}{rcl}\rho \left(\alpha a+n{\beta }^{l}q,{n}^{\frac{1}{l}}\beta b,\gamma c\right)& \le & {\rho }_{\mathcal{A}}\left(\alpha a+n{\beta }^{l}q\right)+{\rho }_{\mathcal{A}}\left({n}^{\frac{1}{l}}\beta b\right)+{\rho }_{\mathcal{A}}\left(\gamma c\right)\\ <& |\alpha |{\rho }_{\mathcal{A}}\left(a\right)+{n}^{\frac{1}{l}}|\beta |{\rho }_{\mathcal{A}}\left(b\right)+|\gamma |{\rho }_{\mathcal{A}}\left(c\right)\\ <& \mu \left({\rho }_{\mathcal{A}}\left(a\right)+{\rho }_{\mathcal{A}}\left(b\right)+{\rho }_{\mathcal{A}}\left(c\right)\right),\end{array}$

where $\mu =max\left\{|\alpha |,{n}^{\frac{1}{l}}|\beta |,|\gamma |\right\}$. Now, we define a holomorphic function H from $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,{n}^{\frac{1}{l}}|\beta |,|\gamma |\right\}\right\}$ into $\mathcal{A}$ as follows:

$H\left(\alpha \right)=\frac{F\left(\alpha a+n{\beta }^{l}q,{n}^{\frac{1}{l}}\beta b,\gamma c\right)-\alpha a}{n{\beta }^{l}}.$

Then from (3.7) it follows that $H\left(0\right)=q+\frac{1}{l!}{b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)$. Then ${\rho }_{\mathcal{A}}\circ H$ is a subharmonic function on $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,{n}^{\frac{1}{l}}|\beta |,|\gamma |\right\}\right\}$. Moreover, Lemma 3.6 of  implies that

$\begin{array}{rcl}{\rho }_{\mathcal{A}}\left(q+\frac{1}{l!}{b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\right)& \le & \underset{|\alpha |=1}{max}\rho \left(H\left(\alpha \right)\right)\\ <& \frac{1}{nl!}{\rho }_{\mathcal{A}}\left({b}^{l}\right){\rho }_{\mathcal{A}}\left(\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\right)\\ <& \frac{1}{nl!{|\beta |}^{l}}{\rho }_{\mathcal{A}}\left(\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\right).\end{array}$
(3.6)

The above inequality holds for every $n\ge 1$. Therefore, if $n⟶\mathrm{\infty }$, then

${\rho }_{\mathcal{A}}\left(q+\frac{1}{l!}{b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)\right)=0$

for every $q\in \mathcal{A}$ with ${\rho }_{\mathcal{A}}\left(q\right)=0$. Hence, Theorem 3.4 of  implies that ${b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)$ is in radical of $\mathcal{A}$. Since $\mathcal{A}$ is semi-simple, therefore ${b}^{l}\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)=0$. Since $b\in {\mathrm{\Omega }}_{\mathcal{A}}\cap Z\left(\mathcal{A}\right)$, so ${b}^{l}\ne 0$. By using that $\mathcal{A}$ is without of order, we conclude that $\frac{{\partial }^{l}F}{\partial {y}^{l}}\left(0,0,0\right)=0$, a contradiction. Thus, our claim is true, and from (3.1), we conclude that $F\left(a,b,c\right)=a$. Similarly, we have $F\left(a,c,b\right)=a$, $F\left(b,a,c\right)=F\left(b,c,a\right)=b$ and $F\left(c,a,b\right)=F\left(c,b,a\right)=c$.

Case (3). In this case, we suppose that $i+j=l$, $i,j\in \left\{0,1,2,3,\dots \right\}$ (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 , we have

$\begin{array}{rcl}\rho \left(\alpha a+n{\alpha }^{i}q,{n}^{\frac{1}{l-i}}\beta b,\gamma c\right)& \le & {\rho }_{\mathcal{A}}\left(\alpha a+n{\alpha }^{i}q\right)+{\rho }_{\mathcal{A}}\left({n}^{\frac{1}{l-i}}\beta b\right)+{\rho }_{\mathcal{A}}\left(\gamma c\right)\\ <& |\alpha |{\rho }_{\mathcal{A}}\left(a\right)+{n}^{\frac{1}{l-i}}|\beta |{\rho }_{\mathcal{A}}\left(b\right)+|\gamma |{\rho }_{\mathcal{A}}\left(c\right)\\ <& \mu \left({\rho }_{\mathcal{A}}\left(a\right)+{\rho }_{\mathcal{A}}\left(b\right)+{\rho }_{\mathcal{A}}\left(c\right)\right),\end{array}$

where $\mu =max\left\{|\alpha |,{n}^{\frac{1}{l-i}}|\beta |,|\gamma |\right\}$. Now, we define a holomorphic function H from $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,{n}^{\frac{1}{l-i}}|\beta |,|\gamma |\right\}\right\}$ into $\mathcal{A}$ as follows:

$H\left(\lambda \right)=\frac{F\left(\alpha a+n{\alpha }^{i}q,{n}^{\frac{1}{l-i}}\beta b,\gamma c\right)-\alpha a}{n{\alpha }^{i}}.$

Then from (3.7) it follows that $H\left(0\right)=q+\frac{1}{\left(l-i\right)!i!}{a}^{i}{\beta }^{l-i}{b}^{l-i}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{l-i}}\left(0,0,0\right)$. Then ${\rho }_{\mathcal{A}}\circ H$ is a subharmonic function on $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,{n}^{\frac{1}{l-i}}|\beta |,|\gamma |\right\}\right\}$. Moreover, Lemma 3.6 of  implies that (3.8)

The above inequality holds for every $n\ge 1$. Therefore, if $n⟶\mathrm{\infty }$, then

${\rho }_{\mathcal{A}}\left(q+\frac{1}{\left(l-i\right)!i!}{a}^{i}{\beta }^{l-i}{b}^{l-i}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{l-i}}\left(0,0,0\right)\right)=0$

for every $q\in \mathcal{A}$ with ${\rho }_{\mathcal{A}}\left(q\right)=0$. Hence, ${a}^{i}{\beta }^{l-i}{b}^{l-i}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{l-i}}\left(0,0,0\right)$ is in radical of $\mathcal{A}$, therefore ${a}^{i}{\beta }^{l-i}{b}^{l-i}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{l-i}}\left(0,0,0\right)=0$. Since ${\beta }^{l-i}\ne 0$ and $a,b\in {\mathrm{\Omega }}_{\mathcal{A}}\cap Z\left(\mathcal{A}\right)$, so ${a}^{i}\ne 0$ and ${b}^{l-i}\ne 0$. Again, by using that $\mathcal{A}$ is without of order, we conclude that $\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{l-i}}\left(0,0,0\right)=0$, a contradiction. Thus, our claim is true, and from (3.1), we conclude that $F\left(a,b,c\right)=a$. Similarly, we have $F\left(a,c,b\right)=a$, $F\left(b,a,c\right)=F\left(b,c,a\right)=b$ and $F\left(c,a,b\right)=F\left(c,b,a\right)=c$.

Case (4). Let $1\le i+j\le l$. Then we have ${\alpha }^{i}{a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{l-i-j}{c}^{l-i-j}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\left(0,0,0\right)\ne 0$. Again, by (3.1) and (3.3), we have (3.9)

Then

$\begin{array}{rcl}\rho \left(\alpha a+n{\alpha }^{i}q,\beta b,{n}^{\frac{1}{l-i-j}}\gamma c\right)& \le & {\rho }_{\mathcal{A}}\left(\alpha a+n{\alpha }^{i}q\right)+{\rho }_{\mathcal{A}}\left(\beta b\right)+{\rho }_{\mathcal{A}}\left({n}^{\frac{1}{l-i-j}}\gamma c\right)\\ <& |\alpha |{\rho }_{\mathcal{A}}\left(a\right)+|\beta |{\rho }_{\mathcal{A}}\left(b\right)+{n}^{\frac{1}{l-i-j}}|\gamma |{\rho }_{\mathcal{A}}\left(c\right)\\ <& \mu \left({\rho }_{\mathcal{A}}\left(a\right)+{\rho }_{\mathcal{A}}\left(b\right)+{\rho }_{\mathcal{A}}\left(c\right)\right),\end{array}$

where $\mu =max\left\{|\alpha |,|\beta |,{n}^{\frac{1}{l-i-j}}|\gamma |\right\}$. Now, we define a holomorphic function H from $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,|\beta |,{n}^{\frac{1}{l-i-j}}|\gamma |\right\}\right\}$ into $\mathcal{A}$ as follows:

$H\left(\alpha \right)=\frac{F\left(\alpha a+n{\alpha }^{i}q,\beta b,{n}^{\frac{1}{l-i-j}}\gamma c\right)-\alpha a}{n{\alpha }^{i}}.$

Then from (3.9) it follows that $H\left(0\right)=q+\frac{1}{i!j!\left(l-i-j\right)!}\left({a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{l-i-j}{c}^{l-i-j}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\left(0,0,0\right)\right)$. Then ${\rho }_{\mathcal{A}}\circ H$ is a subharmonic function on $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,|\beta |,{n}^{\frac{1}{l-i-j}}|\gamma |\right\}\right\}$. Therefore, (3.10)

Therefore, if $n⟶\mathrm{\infty }$, then

${\rho }_{\mathcal{A}}\left(q+\frac{1}{i!j!\left(l-i-j\right)!}{a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{l-i-j}{c}^{l-i-j}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\left(0,0,0\right)\right)=0$

for every $q\in \mathcal{A}$ with ${\rho }_{\mathcal{A}}\left(q\right)=0$. Hence, ${a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{l-i-j}{c}^{l-i-j}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\left(0,0,0\right)$ is in radical of $\mathcal{A}$. Since $\mathcal{A}$ is semi-simple, therefore ${a}^{i}{\beta }^{j}{b}^{j}{\gamma }^{l-i-j}{c}^{l-i-j}\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\left(0,0,0\right)=0$. Since ${\beta }^{j}\ne 0$, ${\gamma }^{l-i-j}\ne 0$ and $a,b\in {\mathrm{\Omega }}_{\mathcal{A}}\cap Z\left(\mathcal{A}\right)$, so ${a}^{i}\ne 0$, ${b}^{j}$ and ${c}^{l-i-j}\ne 0$, we conclude that $\frac{{\partial }^{l}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{l-i-j}}\left(0,0,0\right)=0$, a contradiction. Thus, our claim is true, and from (3.1), we conclude that $F\left(a,b,c\right)=a$. Similarly, we have $F\left(a,c,b\right)=a$, $F\left(b,a,c\right)=F\left(b,c,a\right)=b$ and $F\left(c,a,b\right)=F\left(c,b,a\right)=c$.

Case (5). Now, let $i=j=0$. Then we have ${\gamma }^{l}{c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\ne 0$. Similar to the previous cases, we have

$\begin{array}{rcl}F\left(\alpha a+n{\gamma }^{l}q,\beta b,{n}^{\frac{1}{l}}\gamma c\right)& =& \alpha a+n{\gamma }^{l}q+\frac{1}{l!}n{\gamma }^{l}{c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\\ =& \alpha a+n{\gamma }^{l}\left(q+\frac{1}{l!}{c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\right).\end{array}$
(3.11)

Then

$\begin{array}{rcl}\rho \left(\alpha a+n{\gamma }^{l}q,\beta b,{n}^{\frac{1}{l}}\gamma c\right)& \le & {\rho }_{\mathcal{A}}\left(\alpha a+n{\gamma }^{l}q\right)+{\rho }_{\mathcal{A}}\left(\beta b\right)+{\rho }_{\mathcal{A}}\left({n}^{\frac{1}{l}}\gamma c\right)\\ <& |\alpha |{\rho }_{\mathcal{A}}\left(a\right)+|\beta |{\rho }_{\mathcal{A}}\left(b\right)+{n}^{\frac{1}{l}}|\gamma |{\rho }_{\mathcal{A}}\left(c\right)\\ <& \mu \left({\rho }_{\mathcal{A}}\left(a\right)+{\rho }_{\mathcal{A}}\left(b\right)+{\rho }_{\mathcal{A}}\left(c\right)\right),\end{array}$

where $\mu =max\left\{|\alpha |,|\beta |,{n}^{\frac{1}{l}}|\gamma |\right\}$. Now, we define a holomorphic function H from $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,|\beta |,{n}^{\frac{1}{l}}|\gamma |\right\}\right\}$ into $\mathcal{A}$ as follows:

$H\left(\alpha \right)=\frac{F\left(\alpha a+n{\gamma }^{l}q,\beta b,{n}^{\frac{1}{l}}\gamma c\right)-\alpha a}{n{\gamma }^{l}}.$

Then from (3.11) it follows that $H\left(0\right)=q+\frac{1}{l!}{c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)$. Then ${\rho }_{\mathcal{A}}\circ H$ is a subharmonic function on $\left\{\eta \in \mathbb{C}:\mu <\frac{1}{\rho \left(a,b,c\right)},\mu =|\eta |=max\left\{|\alpha |,|\beta |,{n}^{\frac{1}{l}}|\gamma |\right\}\right\}$, and

$\begin{array}{rcl}{\rho }_{\mathcal{A}}\left(q+\frac{1}{l!}{c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\right)& \le & \underset{|\alpha |=1}{max}\rho \left(H\left(\alpha \right)\right)\\ <& \frac{1}{nl!}{\rho }_{\mathcal{A}}\left({c}^{l}\right){\rho }_{\mathcal{A}}\left(\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\right)\\ <& \frac{1}{nl!{|\gamma |}^{l}}{\rho }_{\mathcal{A}}\left(\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\right).\end{array}$
(3.12)

Therefore, if $n⟶\mathrm{\infty }$, then

${\rho }_{\mathcal{A}}\left(q+\frac{1}{l!}{c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)\right)=0$

for every $q\in \mathcal{A}$ with ${\rho }_{\mathcal{A}}\left(q\right)=0$. Hence, ${c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)$ is in radical of $\mathcal{A}$. Therefore, ${c}^{l}\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)=0$. Since $c\in {\mathrm{\Omega }}_{\mathcal{A}}\cap Z\left(\mathcal{A}\right)$, so ${c}^{l}\ne 0$, then $\frac{{\partial }^{l}F}{\partial {z}^{l}}\left(0,0,0\right)=0$, a contradiction. Thus, (3.1) implies that our claim is true, and from (3.1), we conclude that $F\left(a,b,c\right)=a$. Similarly, we have $F\left(a,c,b\right)=a$, $F\left(b,a,c\right)=F\left(b,c,a\right)=b$ and $F\left(c,a,b\right)=F\left(c,b,a\right)=c$.

By gathering the above five cases, we conclude $\left(a,b,c\right)$ is a tripled fixed point for F, and since $\left(a,b,c\right)$ was arbitrary, so every point of ${\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\cap Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$ is a tripled fixed point for F. □

Corollary 3.2 Let $\mathcal{A}$ be a unital without of order semi-simple Banach algebra. If $F:{\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\subseteq \mathcal{A}×\mathcal{A}×\mathcal{A}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ is a holomorphic map that satisfies the conditions $F\left(0,0,0\right)=0$, $\frac{\partial F}{\partial x}\left(0,0,0\right)={\mathrm{id}}_{\mathcal{A}}$, $\frac{\partial F}{\partial y}\left(0,0,0\right)=0$, $\frac{\partial F}{\partial z}\left(0,0,0\right)=0$, $\frac{{\partial }^{2}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=2$, $i,j,k=0,1,2$, and $\frac{{\partial }^{3}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=3$, $i,j,k=0,1,2,3$, then every $\left(a,b,c\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\cap Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$ 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 $\mathcal{A}$ be a unital without of order complete semi-simple metrizable FLM algebra. For given $\left(a,b,c\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\mathrm{\setminus }Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$, there is a holomorphic map $F:{\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ satisfying the conditions $F\left(0,0,0\right)=0$, $\frac{\partial F}{\partial x}\left(0,0,0\right)={\mathrm{id}}_{\mathcal{A}}$, $\frac{\partial F}{\partial y}\left(0,0,0\right)=0$, $\frac{\partial F}{\partial z}\left(0,0,0\right)=0$, $\frac{{\partial }^{2}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=2$, $i,j,k=0,1,2$, and $\frac{{\partial }^{3}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=3$, $i,j,k=0,1,2,3$, such that $F\left(a,b,c\right)\ne a$, $F\left(b,a,c\right)\ne b$ and $F\left(c,a,b\right)\ne c$.

Proof Let $\left(a,b,c\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\mathrm{\setminus }Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$. Then there exist $\left(u,u,u\right)\in \mathcal{A}×\mathcal{A}×\mathcal{A}$ such that

$\left(ua,ub,uc\right)\ne \left(au,bu,cu\right).$

Let ${D}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\left(u,u,u\right)<1$, then ${D}_{\mathcal{A}}\left(u\right)<1$. Define $U:=log\left(e-u\right)$, then

${e}^{-U}a{e}^{U}\ne a,\phantom{\rule{2em}{0ex}}{e}^{-U}b{e}^{U}\ne b\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{-U}c{e}^{U}\ne c.$

Now, define $F:{\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ as follows:

$F\left(x,y,z\right)={e}^{-\frac{{x}^{2}{z}^{2}U}{{a}^{2}{c}^{2}}}x{e}^{\frac{{y}^{2}{z}^{2}U}{{b}^{2}{c}^{2}}}$
(3.13)

for every $\left(x,y,z\right)$ in ${\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}$. Clearly, F is a holomorphic function, $F\left(0,0,0\right)=0$, $\frac{\partial F}{\partial x}\left(0,0,0\right)={\mathrm{id}}_{\mathcal{A}}$, $\frac{\partial F}{\partial y}\left(0,0,0\right)=0$, $\frac{\partial F}{\partial z}\left(0,0,0\right)=0$, $\frac{{\partial }^{2}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=2$, $i,j,k=0,1,2$, and $\frac{{\partial }^{3}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=3$, $i,j,k=0,1,2,3$, but $F\left(a,b,c\right)\ne a$, and similarly, we can show that there is a holomorphic map $F:{\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ with the required conditions such that $F\left(b,a,c\right)\ne b$ and $F\left(c,a,b\right)\ne c$. □

Example 3.4 Let $X=\mathbb{R}$ be the space of real numbers and let $F:X×X⟶X$ be a function defined by $F\left(x,y,z\right)=x$ 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 $F\left(x,y,z\right)={e}^{{y}^{2}{z}^{2}}x{e}^{-{y}^{2}{z}^{2}}$ that satisfies the conditions of Theorem 3.1. For example, let $X=M\left(G\right)$ be the measure space on a locally compact Hausdorff space G. Another algebra that we can choose is ${\ell }^{1}\left(G\right)$, where G is a locally compact discrete group.

Corollary 3.6 Let $\mathcal{A}$ be a unital without of order semi-simple Banach algebra. For given $\left(a,b,c\right)\in {\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}\mathrm{\setminus }Z\left(\mathcal{A}×\mathcal{A}×\mathcal{A}\right)$, there is a holomorphic map $F:{\mathrm{\Omega }}_{\mathcal{A}×\mathcal{A}×\mathcal{A}}⟶{\mathrm{\Omega }}_{\mathcal{A}}$ satisfying the conditions $F\left(0,0,0\right)=0$, $\frac{\partial F}{\partial x}\left(0,0,0\right)={\mathrm{id}}_{\mathcal{A}}$, $\frac{\partial F}{\partial y}\left(0,0,0\right)=0$, $\frac{\partial F}{\partial z}\left(0,0,0\right)=0$, $\frac{{\partial }^{2}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=2$, $i,j,k=0,1,2$, and $\frac{{\partial }^{3}F}{\partial {x}^{i}\phantom{\rule{0.2em}{0ex}}\partial {y}^{j}\phantom{\rule{0.2em}{0ex}}\partial {z}^{k}}\left(0,0,0\right)=0$, where $i+j+k=3$, $i,j,k=0,1,2,3$, such that $F\left(a,b,c\right)\ne a$, $F\left(b,a,c\right)\ne b$ and $F\left(c,a,b\right)\ne c$.

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Correspondence to Abdolrahman Razani.

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Razani, A., Hosseinzadeh, H. Triple fixed point theorems on FLM algebras. Fixed Point Theory Appl 2013, 16 (2013). https://doi.org/10.1186/1687-1812-2013-16

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### Keywords

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