Fixed-Point Theorem with a Novel Contraction Approach in Banach Algebras

Abstract

In this paper, we establish a fixed-point theorem for mixed monotone operators in ordered Banach algebras by introducing a novel contraction condition formulated in terms of the product law, which represents a significant departure from the traditional additive approach. By exploiting the underlying algebraic structure, our method ensures both the existence and uniqueness of fixed points under broader conditions. To illustrate the effectiveness of the proposed theorem, we also provide a concrete example that demonstrates its applicability.

1. Introduction

The study of fixed points for mixed monotone operators constitutes a central theme in nonlinear analysis, owing to its wide-ranging applications in disciplines such as engineering, physics, and biology. This class of operators was systematically investigated for the first time by Guo and Lakshmikantham [1], who developed their theory within the framework of partially ordered metric spaces and established fundamental existence and uniqueness results. Their pioneering work laid the foundation for subsequent research, where mixed monotone operators have been used as a key tool in addressing nonlinear integral and differential equations, boundary value problems, and dynamical systems. Over time, this line of study has evolved into a well-developed branch of fixed-point theory, attracting continuous attention due to its strong theoretical significance and practical relevance.

It is well known that Banach’s contraction principle [2] remains one of the most influential results in fixed-point theory. Its extension to the framework of mixed monotone operators has significantly enriched the field of nonlinear analysis, particularly through applications to nonlinear integral and differential equations. In this context, Bhaskar and Lakshmikantham [3] introduced a systematic study of mixed monotone mappings in ordered metric spaces. Subsequently, several new developments have emerged, notably in the setting of partially ordered hyperbolic metric spaces, where fixed-point results for mixed monotone operators have been established. For instance, in [4], the authors proved a fixed point theorem for a new class of nearly asymptotically nonexpansive mixed monotone operators. More recently, in [5], the authors employed a new iterative technique to establish several fixed-point theorems for mixed monotone operators, ensuring both the existence and uniqueness of fixed points. In addition, they derived a number of new results under certain concavity and convexity assumptions, without relying on the existence of coupled upper or lower solutions.

Recently, considerable attention has been devoted to generalizing and extending Banach’s contraction principle to more sophisticated settings, notably ordered Banach algebras, where the interplay between the algebraic structure, the underlying topology, and the partial order provides powerful tools for establishing the existence and uniqueness of fixed points; see, for example, [6,7,8,9]. One of the central motivations behind this line of research lies in the search for more flexible frameworks that capture both the algebraic and ordered nature of these spaces, thus allowing a finer analysis of contractive mappings beyond the classical setting.

In this paper, we contribute to this line of research by establishing a new fixed-point theorem for mixed monotone operators in ordered Banach algebras. Unlike the classical additive approach commonly employed in contractive conditions, our theorem is based on a novel contractive condition that relies on the product law in the algebras. This shift in perspective provides a deeper understanding of how algebraic operations can be used to guarantee the existence and uniqueness of fixed points. Furthermore, the result presented here not only complement earlier findings but also open new avenues and may serve as a foundation for further advancements in fixed-point theory.

2. Preliminaries

This section contains some useful definitions and results which are required for our main results.

A Banach algebra A is a Banach space together with a norm compatible algebra structure, namely: for all $x,y\in A$, $\parallel xy\parallel \le \parallel x\parallel \parallel y\parallel$. If A has a multiplicative identity (denoted by 1) whose norm is 1, we say that the algebra A is unital.

Let A be a Banach algebra with unit 1 and 0 be the zero element of A. We call a nonempty subset C of A an algebra cone of A if C satisfies the following:

(i)

C is closed and $\{0,1\}\subset C$;

(ii)

$\lambda C+C\subset C$ for all $\lambda \ge 0$;

(iii)

$C\cap (-C)=\left\{0\right\}$;

(iv)

$CC\subset C$.

A cone C induces a partial ordering ⪯ on A, defined by $x⪯y$ (or $y⪰x$) if $y-x\in C$. We write $x\prec y$ to mean $x-y\in \dot{C}$, where $\dot{C}=C-\left\{0\right\}$, and $x\ll y$ will stand for $x-y\in \mathring{C}$, where $\mathring{C}$ denotes the interior of C. If $\mathring{C}\ne \varnothing$, then C is called a solid cone. C is called normal if there exists a constant $N>0$, such that, for all $x,y\in A$, $0⪯x⪯y$ implies $\parallel x\parallel \le N\parallel y\parallel$, in this case N is called the normality constant of C. A Banach algebra A with unit 1 is called an ordered Banach algebra (OBA) if A is ordered by a relation ⪯ in such a manner that for every $x,y\in A$ we have,

(i)

$x,y⪰0\Rightarrow x+y⪰0$;

(ii)

$x⪰0,\lambda \ge 0\Rightarrow \lambda x⪰0$;

(iii)

$x,y⪰0\Rightarrow xy⪰0$;

(iv)

$1⪰0$.

Therefore, if A is ordered by an algebra cone C, then A, or more specifically $(A,C)$, is an OBA. Conversely, if A is an OBA the set $C=\{x\in A:x⪰0\}$ is an algebra cone that induces the ordering on A.

An element $x\in A$ is said to be invertible if there exists $x^{-1}\in A$, such that $x{x}^{-1}=x^{-1}x=1$. Moreover, if every non-zero element in A has an inverse in A, then A is called a divisible Banach algebra.

Let $x,y\in A$ be such that $x⪯y$. Then, the set

$$\left[x,y\right]=\left\{z\in A;x⪯z⪯y\right\}$$

is called an order interval in A. Since C is closed, then every order interval is closed in A.

Definition 1

([10]). An algebra cone C is said to be inverse-closed, if C has the property that if $x\in C$ and x is invertible, then $x^{-1}\in C$.

Lemma 1

(Zorn’s lemma [11]). If X is a partially ordered set such that every chain in X has an upper bound, then X contains a maximal element.

Lemma 2

([10]). Let $(A,C)$ be an ordered Banach algebra, and let $x\in C$. If $\lambda >r\left(x\right)$, then ${(\lambda 1-x)}^{-1}\in C$, where $r\left(x\right)={\lim }_{n\to \infty }{\parallel x^n\parallel }^{\frac{1}{n}}$ is the spectral radius of x.

Definition 2

([1]). Let $\left(X,⪯\right)$ be a partially ordered set, and let $T:X\times X⟶X$ be an operator. T is said to be mixed monotone if $T(x,y)$ is nondecreasing in x and is nonincreasing in y, that is,

$$\begin{array}{cc}\hfill & x_1,x_2,y\in X,x_1⪯x_2\Rightarrow T(x_1,y)⪯T(x_2,y),\hfill \\ \hfill & y_1,y_2,x\in X,y_1⪯y_2\Rightarrow T(x,y_2)⪯T(x,y_1).\hfill \end{array}$$

Definition 3.

Let $\left(X,⪯\right)$ be a partially ordered set, and let $T:X\times X\to X$ be a mixed monotone operator. An element $(x,y)\in X\times X$ is called a coupled lower and upper fixed point of T if

$$x⪯T(x,y)\mathit{and}T(y,x)⪯y.$$

If $T(x,y)=x$ and $T(y,x)=y$, then $(x,y)$ is called a coupled fixed point of T. Moreover, if $x=y$, then x is called a fixed point of T. Consequently, every fixed point is a coupled fixed point, and every coupled fixed point is, in turn, a coupled lower and upper fixed point. However, the converse implications do not necessarily hold. This hierarchy can be summarized as

$$\mathit{Fixed}\mathit{point}\Rightarrow \mathit{Coupled}\mathit{fixed}\mathit{point}\Rightarrow \mathit{Coupled}\mathit{lower}\mathit{and}\mathit{upper}\mathit{fixed}\mathit{point}.$$

3. Fixed-Point Theorem

In this section, we present the main result of our work, in which we establish sufficient conditions for a mixed monotone operator to have a unique fixed point in a Banach algebra. The proof is carried out using Zorn’s lemma.

Theorem 1.

Let C be a normal, solid, and inverse-closed algebra cone in a unital Banach algebra A, where $\mathring{C}$ consists of invertible elements. Let $T:\mathring{C}\times \mathring{C}⟶\mathring{C}$ be a mixed monotone operator. Suppose that

$\left(H_1\right)$

T has a coupled lower and upper fixed point $(x_0,x^0)\in \mathring{C}\times \mathring{C}$, with $x_0⪯x^0$;

$\left(H_2\right)$

For all $x,y\in \mathring{C}$, with $x_0⪯x\prec y⪯x^0$, there exists a positive integer n satisfying

$$∥T^{n+1}(y,x){\left(T^{n+1}(x,y)\right)}^{-1}∥<{∥T^n(x,y){\left(T^n(y,x)\right)}^{-1}∥}^{-1}.$$

Then, T has a unique fixed point $x^{*}\in \left[x_0,x^0\right]$, that is, $T(x^{*},x^{*})=x^{*}$ and $x_0⪯x^{*}⪯x^0$.

Proof. 

The proof of the theorem will be carried out in four steps.

Step 1. Construct the sequences

$$\begin{array}{c}\hfill x_{n+1}=T(x_n,x^n)\mathrm{and}{x}^{n+1}=T(x^n,x_n),n=0,1,2,\dots .\end{array}$$

Since T is mixed monotone, and by invoking assumption $\left(H_1\right)$, it follows by induction that

$$x_0⪯x_1⪯\dots ⪯x_n⪯\dots ⪯x^n⪯\dots ⪯x^1⪯x^0.$$

(1)

If there is $i\ge 1$, such that $x_i=x^i$, then $x_{i+1}=x_i=x^i$, which implies that $T(x_i,x_i)=x_i$, i.e., $x_i$ is a fixed point of T in $\left[x_0,x^0\right]$. Let y be another fixed point of T in $\left[x_0,x^0\right]$. Then, by $\left(H_1\right)$ and the mixed monotonicity of T, we have, $x_i⪯y⪯x^i=x_i$, hence $x_i=y$. Thus, T has a unique fixed point in $\left[x_0,x^0\right]$.

Suppose that for any $n\ge 0$, $x_n\prec x^n$. From $\left(H_2\right)$, for every $n\ge 0$, there exist $\gamma \left(n\right)\ge n$ and ${\alpha }_n>0$, where $\left\{\gamma \right(n\left)\right\}$ is an increasing sequence, such that

$$∥x^{\gamma \left(n\right)+1}{\left(x_{\gamma \left(n\right)+1}\right)}^{-1}∥<{\alpha }_n<{∥x_{\gamma \left(n\right)}{\left(x^{\gamma \left(n\right)}\right)}^{-1}∥}^{-1}.$$

By Lemma 2, we have ${\left({\alpha }_n-x^{\gamma \left(n\right)+1}{\left(x_{\gamma \left(n\right)+1}\right)}^{-1}\right)}^{-1}\in C$ (because $\parallel a\parallel \ge r\left(a\right)$). Since C is inverse-closed, then $\left({\alpha }_n-x^{\gamma \left(n\right)+1}{\left(x_{\gamma \left(n\right)+1}\right)}^{-1}\right)\in C$, it follows that ${\alpha }_n⪰x^{\gamma \left(n\right)+1}{\left(x_{\gamma \left(n\right)+1}\right)}^{-1}$. Which implies that,

$${\alpha }_n{x}_{\gamma \left(n\right)+1}⪰x^{\gamma \left(n\right)+1},\forall n=0,1,2,\dots$$

(2)

Similarly, since ${\alpha }_n<{∥x_{\gamma \left(n\right)}{\left(x^{\gamma \left(n\right)}\right)}^{-1}∥}^{-1}$, then $∥x_{\gamma \left(n\right)}{\left(x^{\gamma \left(n\right)}\right)}^{-1}∥<{\displaystyle \frac{1}{{\alpha }_n}}$; this implies

$${\alpha }_n{x}_{\gamma \left(n\right)}⪯x^{\gamma \left(n\right)},\forall n=0,1,2,\dots$$

(3)

$\left\{\gamma \right(n\left)\right\}$ is an increasing sequence, so for any $n\ge 1$, we have, $\gamma \left(n\right)\ge \gamma (n-1)+1$, then $x_{\gamma \left(n\right)}⪰x_{\gamma (n-1)+1}$ and $x^{\gamma (n-1)+1}⪰x^{\gamma \left(n\right)}$; therefore, by (2), we obtain,

$${\alpha }_{n-1}{x}_{\gamma \left(n\right)}⪰{\alpha }_{n-1}{x}_{\gamma (n-1)+1}⪰x^{\gamma (n-1)+1}⪰x^{\gamma \left(n\right)}.$$

(4)

From (3) and (4), we obtain

$${\alpha }_n{x}_{\gamma \left(n\right)}⪯x^{\gamma \left(n\right)}⪯{\alpha }_{n-1}{x}_{\gamma \left(n\right)},\forall n=1,2,\dots$$

(5)

Notice that $\left\{{\alpha }_n\right\}$ is a nonincreasing sequence. Moreover, we have ${\alpha }_n{x}_{\gamma \left(n\right)+1}⪰x^{\gamma \left(n\right)+1}⪰x_{\gamma \left(n\right)+1}$, which implies that ${\alpha }_n\ge 1$, for any $n\in \mathbb{N}$. It follows that there exists $\alpha \ge 1$, such that ${\lim }_{n\to \infty }{\alpha }_n=\alpha$.

From (5) we obtain, for all positive integers n and p,

$$\begin{array}{cc}\hfill & 0⪯x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}⪯{\alpha }_{n-1}{x}_{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}⪯({\alpha }_{n-1}-\alpha )x^0,\hfill \\ \hfill & 0⪯x^{\gamma \left(n\right)}-x^{\gamma (n+p)}⪯x^{\gamma \left(n\right)}-{\alpha }_{(n+p)}{x}_{\gamma (n+p)}⪯x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)},\hfill \\ \hfill & 0⪯\alpha x_{\gamma (n+p)}-\alpha x_{\gamma \left(n\right)}⪯{\alpha }_{(n+p)}{x}_{\gamma (n+p)}-\alpha x_{\gamma \left(n\right)}⪯x^{\gamma (n+p)}-\alpha x_{\gamma \left(n\right)}\hfill \\ \hfill & ⪯x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}.\hfill \end{array}$$

Since C is a normal cone and ${\lim }_{n\to \infty }{\alpha }_{n-1}=\alpha$, then ${\lim }_{n\to \infty }\parallel x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}\parallel =0$. Moreover,

$$\parallel x^{\gamma \left(n\right)}-x^{\gamma (n+p)}\parallel \le N\parallel x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}\parallel \mathrm{and}\parallel \alpha x_{\gamma (n+p)}-\alpha x_{\gamma \left(n\right)}\parallel \le N\parallel x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}\parallel .$$

So, $\left\{x^{\gamma \left(n\right)}\right\}$ and $\left\{\alpha x_{\gamma \left(n\right)}\right\}$ are Cauchy sequences in the Banach algebra A. Thus, there exists $x^{*}$ and $x_{*}$, such that $x^{\gamma \left(n\right)}⟶x^{*}$ and $\alpha x_{\gamma \left(n\right)}⟶x_{*}$ (as $n⟶\infty$).

From (1), we have $0⪯x^{*}-x_{*}⪯x^{\gamma \left(n\right)}-\alpha x_{\gamma \left(n\right)}$, which implies that $x^{*}=x_{*}$. Hence, $x^{\gamma \left(n\right)}⟶x^{*}$ and $x_{\gamma \left(n\right)}⟶{\displaystyle \frac{1}{\alpha }}{x}^{*}$ (as $n⟶\infty$). Moreover, $x^{*}⪯x^{\gamma \left(n\right)}$ and $x_{\gamma \left(n\right)}⪯{\displaystyle \frac{1}{\alpha }}{x}^{*}$. T is a mixed monotone operator, then

$$\begin{array}{c}{x}_{\gamma \left(n\right)+1}=T(x_{\gamma \left(n\right)},x^{\gamma \left(n\right)})⪯T({\displaystyle \frac{1}{\alpha }}{x}^{*},x^{*}),\\ T(x^{*},{\displaystyle \frac{1}{\alpha }}{x}^{*})⪯T(x^{\gamma \left(n\right)},x_{\gamma \left(n\right)})=x^{\gamma \left(n\right)+1}.\end{array}n=0,1,2,\dots$$

(6)

By (1) and since $\left\{\gamma \right(n\left)\right\}$ is increasing, we obtain $x_{\gamma \left(n\right)}⪯x_{\gamma \left(n\right)+1}⪯x_{\gamma (n+1)}$ and $x^{\gamma (n+1)}⪯x^{\gamma \left(n\right)+1}⪯x^{\gamma \left(n\right)}$. Since C is a normal cone, then $x_{\gamma \left(n\right)+1}⟶{\displaystyle \frac{1}{\alpha }}{x}^{*}$ and $x^{\gamma \left(n\right)+1}⟶x^{*}$ (as $n⟶\infty$). Consequently,

$${\displaystyle \frac{1}{\alpha }}{x}^{*}⪯T({\displaystyle \frac{1}{\alpha }}{x}^{*},x^{*})\mathrm{and}T(x^{*},{\displaystyle \frac{1}{\alpha }}{x}^{*})⪯x^{*},$$

It is easy to see, by assumption $\left(H_1\right)$, that $x^0⪯{\displaystyle \frac{1}{\alpha }}{x}^{*}\mathrm{and}{x}^{*}⪯x^0$.

Step 2. If $\alpha =1$, then $T(x^{*},x^{*})=x^{*}$, that is, $x^{*}$ is a fixed point of T. In this case, if $y^{*}\in [x_0,x^0]$ is another fixed point of T. Then, for any integer n, $x_n⪯y^{*}⪯x^n$. Since C is a normal cone, it follows that $x^{*}=y^{*}$.

Suppose that $\alpha >1$. Let K be the set of coupled lower and upper fixed points of T, which can be written in the form $({\displaystyle \frac{1}{t}}x,x)$, with $t\ge 1 \mathrm{and} x_0⪯x⪯x^0$.

Firstly, note that $({\displaystyle \frac{1}{\alpha }}{x}^{*},x^{*})\in K$, hence $K\ne \varnothing$. Fix $({\displaystyle \frac{1}{t}}x,x)\in K$ with $t>1$ and construct the sequences

$$\begin{array}{cc}\hfill & x_{n+1}=T(x_n,x^n),x_1=T({\displaystyle \frac{1}{t}}x,x),\hfill \\ \hfill & x^{n+1}=T(x^n,x_n),x^1=T(x,{\displaystyle \frac{1}{t}}x).\hfill \end{array}$$

Following the same procedure as in the first step (used to find $(x^{*},\alpha )$), we show that there exists $(x^{\prime },t^{\prime })$ satisfying

$${\displaystyle \frac{1}{t}}x⪯T({\displaystyle \frac{1}{t}}x,x)⪯{\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime }⪯T({\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime },x^{\prime })\mathrm{and}T(x^{\prime },{\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime })⪯x^{\prime }⪯T(x,{\displaystyle \frac{1}{t}}x)⪯x.$$

(7)

Assume that for every $({\displaystyle \frac{1}{t}}x,x)$, we have $t>1$. Then, we can define the operator $\Gamma :K\to K$ as follows

$$\Gamma ({\displaystyle \frac{1}{t}}x,x)=({\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime },x^{\prime }).$$

Next, we prove that the relation ⊑, defined for any $({\displaystyle \frac{x}{\lambda }},x)$ and $({\displaystyle \frac{y}{\beta }},y)$ in K, by

$$({\displaystyle \frac{x}{\lambda }},x)⊑({\displaystyle \frac{y}{\beta }},y)\iff \beta \le \lambda ,{\displaystyle \frac{1}{\lambda }}x⪯{\displaystyle \frac{1}{\beta }}yandy⪯x$$

is a partial order in K.

(i)

Clearly, the relation ⊑ is reflexive.

(ii)

Let $({\displaystyle \frac{x}{\lambda }},x)⊑({\displaystyle \frac{y}{\beta }},y)$ and $({\displaystyle \frac{y}{\beta }},y)⊑({\displaystyle \frac{x}{\lambda }},x)$, then we have $\left(\beta \le \lambda ,{\displaystyle \frac{x}{\lambda }}⪯{\displaystyle \frac{y}{\beta }}\mathrm{and}y⪯x\right)\mathrm{and}\left(\lambda \le \beta ,{\displaystyle \frac{y}{\beta }}⪯{\displaystyle \frac{x}{\lambda }}\mathrm{and}x⪯y\right)$; therefore, $\lambda =\beta$ and $x=y$. Then, ⊑ is antisymmetric.

(iii)

Let $({\displaystyle \frac{x}{\lambda }},x)⊑({\displaystyle \frac{y}{\beta }},y)$ and $({\displaystyle \frac{y}{\beta }},y)⊑({\displaystyle \frac{z}{\gamma }},z)$ then, we have $\left(\beta \le \lambda ,{\displaystyle \frac{x}{\lambda }}⪯{\displaystyle \frac{y}{\beta }}\mathrm{and}y⪯x\right)\mathrm{and}\left(\gamma \le \beta ,{\displaystyle \frac{y}{\beta }}⪯{\displaystyle \frac{z}{\gamma }}\mathrm{and}z⪯y\right)$ which implie that $\gamma \le \lambda ,{\displaystyle \frac{x}{\lambda }}⪯{\displaystyle \frac{z}{\gamma }}\mathrm{and}z⪯x$ i.e., $({\displaystyle \frac{x}{\lambda }},x)⊑({\displaystyle \frac{z}{\gamma }},z)$. Then, ⊑ is transitive.

Let $({\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime },x^{\prime })=\Gamma ({\displaystyle \frac{1}{t}}x,x)$, it is obviously $({\displaystyle \frac{1}{t}}x,x)⊑({\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime },x^{\prime })$. Moreover, if $({\displaystyle \frac{1}{t}}x,x)=({\displaystyle \frac{1}{t^{\prime }}}{x}^{\prime },x^{\prime })$, then by (7), we have $T({\displaystyle \frac{1}{t}}x,x)={\displaystyle \frac{1}{t}}x$ and $T(x,{\displaystyle \frac{1}{t}}x)=x$, assume that $t>1$, then ${\displaystyle \frac{1}{t}}x\prec x$, by $\left(H_2\right)$, we obtain

$$\parallel x{\left({\displaystyle \frac{1}{t}}x\right)}^{-1}\parallel <{\parallel \left({\displaystyle \frac{1}{t}}x\right)x^{-1}\parallel }^{-1}⟹|t|<|{\displaystyle \frac{1}{t}}{|}^{-1}.$$

Therefore, we conclude this step by observing that for all $\left({\displaystyle \frac{1}{t}}x,x\right)\in K$

$$({\displaystyle \frac{1}{t}}x,x)⊑\Gamma ({\displaystyle \frac{1}{t}}x,x)\mathrm{and}\left[\left({\displaystyle \frac{1}{t}}x,x\right)=\Gamma \left({\displaystyle \frac{1}{t}}x,x\right)⟹t=1\right].$$

(8)

Step 3. In what follows, we use the following definitions.

Let F be a nonempty subset of K

(i)

We say that F is stable under $\Gamma$ if, for every $({\displaystyle \frac{1}{t}}x,x)\in F$, we have $\Gamma ({\displaystyle \frac{1}{t}}x,x)\in F$.

(ii)

We say that F is stable under mixed monotonic convergence if, for any sequence $\left\{(y_n,x_n)\right\}$ in F, such that $\left\{y_n\right\}$ is nondecreasing, $\left\{x_n\right\}$ is nonincreasing and $(y_n,x_n)\to (y,x)$ (as $n⟶\infty$), we have $(y,x)\in F$.

Now, we can easily observe that K is stable under $\Gamma$. We prove that K is stable under mixed monotonic convergence. Let $\left\{({\displaystyle \frac{1}{t_n}}{x}_n,x_n)\right\}$ be a sequence in K, such that $\left\{{\displaystyle \frac{1}{t_n}}{x}_n\right\}$ is nondecreasing and $\left\{x_n\right\}$ is nonincreasing, with $({\displaystyle \frac{1}{t_n}}{x}_n,x_n)\to (y,x)$ (as $n⟶\infty$). By the monotonicity of $\left\{{\displaystyle \frac{1}{t_n}}{x}_n\right\}$ and $\left\{x_n\right\}$, it is easy to show that $\left\{t_n\right\}$ is nonincreasing sequence. Since $t_n\ge 1$ for all n, there exist $t\ge 1$, such that ${\lim }_{n\to \infty }{t}_n=t$. Moreover,

$$0⪯{\displaystyle \frac{x_n}{t}}-{\displaystyle \frac{x_n}{t_n}}⪯\left({\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{t_n}}\right)x_n⪯\left({\displaystyle \frac{1}{t_n}}-{\displaystyle \frac{1}{t}}\right)x^0.$$

Then,

$$\begin{array}{c}\hfill \parallel y-{\displaystyle \frac{x}{t}}\parallel \le \parallel y-{\displaystyle \frac{x_n}{t_n}}\parallel +\parallel {\displaystyle \frac{x_n}{t_n}}-{\displaystyle \frac{x_n}{t}}\parallel +\parallel {\displaystyle \frac{x_n}{t}}-{\displaystyle \frac{x}{t}}\parallel ⟶0(\mathrm{as}n⟶\infty ).\end{array}$$

Therefore, $y={\displaystyle \frac{x}{t}}$. Using the monotonicity of $\left\{{\displaystyle \frac{1}{t_n}}{x}_n\right\}$ and $\left\{x_n\right\}$, we obtain for any $n\in \mathbb{N}$, ${\displaystyle \frac{1}{t_n}}{x}_n⪯{\displaystyle \frac{1}{t}}x$ and $x⪯x_n$, which implies that $T({\displaystyle \frac{1}{t_n}}{x}_n,x_n)⪯T({\displaystyle \frac{1}{t}}x,x)$ and $T(x,{\displaystyle \frac{1}{t}}x)⪯T(x_n,{\displaystyle \frac{1}{t_n}}{x}_n)$. Since $({\displaystyle \frac{1}{t_n}}{x}_n,x_n)\in K$, ${\displaystyle \frac{x_n}{t_n}}⪯T({\displaystyle \frac{x_n}{t_n}},x_n)$ and $T(x_n,{\displaystyle \frac{x_n}{t_n}})⪯x_n$, for any $n\in \mathbb{N}$, we have

$${\displaystyle \frac{1}{t}}x⪯T\left({\displaystyle \frac{1}{t}}x,x\right)\mathrm{and}T\left(x,{\displaystyle \frac{1}{t}}x\right)⪯x.$$

Therefore, $(y,x)\in K$. Consequently, K is stable under mixed monotonic convergence.

Let $\mathcal{F}$ be the collection of all subsets of $C\times C$ which are stable under $\Gamma$, stable under mixed monotonic convergence and contain $({\displaystyle \frac{x^{*}}{\alpha }},x^{*})$. We set $\sigma (x^{*},\alpha )=\underset{L\in \mathcal{F}}{\cap }L.$ Since $K\in \mathcal{F}$, it follows that $\sigma (x^{*},\alpha )\subset K$.

Let H be a totally ordered subset of $\sigma (x^{*},\alpha )$, and let $p:H⟶[1,\alpha ]$ be the operator defined by $p({\displaystyle \frac{x}{t}},x)=t$, for each $({\displaystyle \frac{x}{t}},x)\in H$. Then, it is clear that $p\left(H\right)$ has a lower bound $\underline{t}$ in $[1,\alpha ]$. Therefore, there exists a nonincreasing sequence $\left\{t_n\right\}$ of $p\left(H\right)$ converging to $\underline{t}$. It follows that, there exists a sequence $\left\{({\displaystyle \frac{x_n}{t_n}},x_n)\right\}$ of H, which is a totally ordered set, and since $\left\{t_n\right\}$ is nonincreasing, we obtain

$$\left({\displaystyle \frac{x_n}{t_n}},x_n\right)⊑\left({\displaystyle \frac{x_{n+1}}{t_{n+1}}},x_{n+1}\right),n\in \mathbb{N}.$$

Then,

$${\displaystyle \frac{x_n}{t_n}}⪯{\displaystyle \frac{x_{n+1}}{t_{n+1}}}\mathrm{and}{x}_{n+1}⪯x_n,n\in \mathbb{N}.$$

So, for any $n,p$ in $\mathbb{N}$, we have

$$0⪯{\displaystyle \frac{x_n}{\underline{t}}}-{\displaystyle \frac{x_{n+p}}{\underline{t}}}⪯{\displaystyle \frac{x_n}{\underline{t}}}-{\displaystyle \frac{x_{n+p}}{t_{n+p}}}⪯\left({\displaystyle \frac{1}{\underline{t}}}-{\displaystyle \frac{1}{t_{n+p}}}\right)x^0.$$

Since C is a normal cone and ${\lim }_{n\to \infty }\left({\displaystyle \frac{1}{\underline{t}}}-{\displaystyle \frac{1}{t_{n+p}}}\right)=0$, $\left\{{\displaystyle \frac{x_n}{\underline{t}}}\right\}$ is a Cauchy sequence in the Banach algebra A. Thus, there exists $\underline{x}\in A$, such that ${\lim }_{n\to \infty }{\displaystyle \frac{x_n}{\underline{t}}}=\underline{x}$. It follows that $x_n⟶\underline{t}\underline{x}$ and ${\displaystyle \frac{x_n}{t_n}}⟶\underline{x}$ (as $n⟶\infty$). Notice that $\underline{t}\underline{x}⪯x_n$ and ${\displaystyle \frac{x_n}{t_n}}⪯\underline{x}$. Since $\sigma (x^{*},\alpha )$ is stable under mixed monotonic convergence and $({\displaystyle \frac{x_n}{t_n}},x_n)⟶(\underline{x},\underline{t}\underline{x})$ (as $n⟶\infty$), then $({\displaystyle \frac{1}{\underline{t}}}X,X)=(\underline{x},\underline{t}\underline{x})\in \sigma (x^{*},\alpha )$. Let $({\displaystyle \frac{1}{t}}x,x)\in H$, thus $t\ge \underline{t}$. Since H is a totally ordered subset, then $({\displaystyle \frac{1}{t}}x,x)⊑({\displaystyle \frac{1}{\underline{t}}}X,X)$, which means that $({\displaystyle \frac{1}{\underline{t}}}X,X)$ is an upper bound of H. Applying Lemma 1, we conclude that $\sigma (x^{*},\alpha )$ has a maximal element $({\displaystyle \frac{1}{\delta }}\overline{x},\overline{x})$. Since $\sigma (x^{*},\alpha )$ is stable under $\Gamma$, then $\Gamma ({\displaystyle \frac{1}{\delta }}\overline{x},\overline{x})\in \sigma (x^{*},\alpha )$. From (8) we have $\delta =1$. Consequently $(\overline{x},\overline{x})$ is a fixed point of T.

Step 4. To prove the uniqueness of the fixed point of T in $[x_0,x^0]$. Let $y\in [x_0,x^0]$ be another fixed point of T. We define

$$I=\left\{\left({\displaystyle \frac{1}{t}}x,x\right)\in K,such that{\displaystyle \frac{x}{t}}⪯y⪯x\right\}.$$

Considering the sequences $\left\{x_n\right\}$ and $\left\{x^n\right\}$ defined in the first step of the proof, it is easy to show by the mixed monotonicity of T that $x_n⪯y⪯x^n$, which gives ${\displaystyle \frac{x^{*}}{\alpha }}⪯y⪯x^{*}$, and hence $({\displaystyle \frac{x^{*}}{\alpha }},x^{*})\in I$.

Let $({\displaystyle \frac{1}{t}}x,x)\in I$ construct the sequences:

$$\begin{array}{cc}\hfill & x_{n+1}=T(x_n,x^n),x_1=T\left({\displaystyle \frac{1}{t}}x,x\right),\hfill \\ \hfill & x^{n+1}=T(x^n,x_n),x^1=T\left(x,{\displaystyle \frac{1}{t}}x\right).\hfill \end{array}$$

One can see that $x_n⪯y⪯x^n$, so the definition of the operator $\Gamma$ implies $\Gamma ({\displaystyle \frac{1}{t}}x,x)\in I$. Thus, I is stable under $\Gamma$. By reasoning analogous to the proof that K is stable under mixed monotonic convergence, we deduce that I is likewise stable under mixed monotonic convergence.

Since $({\displaystyle \frac{x^{*}}{\alpha }},x^{*})\in I$, then $\sigma (x^{*},\alpha )\subset I$. Consequently, $(\overline{x},\overline{x})\in I$. Thus, $\overline{x}⪯y⪯\overline{x}$, which implies that $\overline{x}=y$. □

Example 1.

Take the ordered Banach algebra $A=\mathbb{R}$. $C={\mathbb{R}}_{+}=[0,+\infty )$ and $\mathring{C}=(0,+\infty )$. It is clear that ${\mathbb{R}}_{+}$ is an inverse-closed and normal cone algebra. Define the operator $T:(0,+\infty )\times (0,+\infty )⟶(0,+\infty )$ by

$$\begin{array}{c}\hfill T(x,y)=\sqrt{x}+{\displaystyle \frac{1}{\sqrt{y}}},\mathit{for}\mathit{all}x,y\in (0,+\infty ).\end{array}$$

(9)

Then, T is mixed monotone in $(0,+\infty )\times (0,+\infty )=\mathring{C}\times \mathring{C}$. Furthermore, one has $T(1,4)={\displaystyle \frac{3}{2}}>1$ and $T(4,1)=3<4$.

Let $1\le x<y\le 4$, then

$$\left|T(y,x){\left(T(x,y)\right)}^{-1}\right|=\left|{\displaystyle \frac{\sqrt{y}+\frac{1}{\sqrt{x}}}{\sqrt{x}+\frac{1}{\sqrt{y}}}}\right|=\sqrt{{\displaystyle \frac{y}{x}}}<{\displaystyle \frac{y}{x}}=|x{y}^{-1}{|}^{-1}.$$

Thus, all hypotheses of Theorem 1 are satisfied. Therefore, the operator T defined by (9) has a unique fixed point in $(1,4)$.

Corollary 1.

Let C be a normal and inverse-closed algebra cone in a unital Banach algebra A, where $\mathring{C}$ consists of invertible elements. Let $T:\mathring{C}⟶\mathring{C}$ be a nondecreasing operator. Suppose that

$\left(H_1\right)$

There exists $x_0⪯x^0$ in $\mathring{C}$, such that $x_0⪯T\left(x_0\right)⪯T\left(x^0\right)⪯x^0$;

$\left(H_2\right)$

For all $x,y\in \mathring{C}$, with $x_0⪯x\prec y⪯x^0$, there exists a positive integer n satisfying

$$∥T^{n+1}\left(y\right){\left(T^{n+1}\left(x\right)\right)}^{-1}∥<{∥T^n\left(x\right){\left(T^n\left(y\right)\right)}^{-1}∥}^{-1}.$$

Then, T has a unique fixed point $x^{*}\in \left[x_0,x^0\right]$.

Proof. 

Let $\tilde{T}:\mathring{C}\times \mathring{C}⟶\mathring{C}$ be the operator defined by $\tilde{T}(x,y)=T\left(x\right)$, for every $x,y\in \mathring{C}$. We show that the operator $\tilde{T}$ satisfies all the hypotheses of Theorem 1.

□

4. Conclusions and Perspectives

In this paper, we established a fixed-point theorem for mixed monotone operators in ordered Banach algebras by introducing a novel contractive condition within a Banach algebra, employing the multiplication operation instead of the classical addition-based approach commonly used in traditional contraction theory. Our results leverage the underlying algebraic structure to ensure the existence and uniqueness of fixed point, providing a framework more suitable for certain nonlinear problems.

The presented example illustrates the applicability of our theorem, demonstrating its effectiveness in concrete settings. Future research may explore generalizations to partial metric spaces or applications in functional problems, particularly in solving differential and integral equations.