A Fixed-Point Theorem for Mappings Satisfying a Contractive Condition Involving T^ρ

Abstract

We prove that a self-mapping T on a complete metric space $(X,D)$, satisfying a generalized contractive condition involving its $\rho$-th iterate $T^{\rho }$ ($\rho \in \mathbb{N},\rho \ge 2$) and a uniform coefficient bound $L<1$, possesses a unique fixed point. The proof establishes that the Picard iteration sequence is Cauchy with the further assumption of continuity of T, and fixed point uniqueness follows directly from the contractive inequality.

1. Introduction

Fixed-point theory represents one of the fundamental pillars of modern mathematical analysis, with wide-ranging applications in various fields such as differential and integral equations, and mathematical economics. The origins of this theory date back to the famous Banach fixed-point theorem [1] published in 1922, which states that every contraction mapping on a complete metric space possesses a unique fixed point.

Since then, the theory has witnessed important developments through various generalizations of contraction conditions. In 1968, Kannan [2] introduced a generalization by replacing the distance between the two images with a combination of distances between points and their images. This was followed by Chatterjea’s work [3] in 1972, and then Singh [4,5] who extended the scope of these results. However, the most significant generalizations came from Ćirić [6,7] who introduced a more general contractive condition combining several distances. Subsequently, contributions from Reich [8,9,10,11] and Hardy and Rogers [12] provided even more comprehensive formulations. In a comprehensive comparative study, Rhoades [13] analyzed and classified different types of contractive mappings. The concept of generalized metric spaces, introduced by Mustafa and Sims [14], has opened new directions in fixed-point theory. Recently, Bekri and Fabiano [15] contributed to this evolving field by investigating Singh–Chatterjee type contractive mappings.

This work aims to present a new theorem in fixed-point theory, where we study self-mappings T on a complete metric space $(X,D)$ satisfying a generalized contractive condition involving the $\rho$-fold iteration $T^{\rho }$ (where $\rho \ge 2$). The proposed condition is characterized by the existence of coefficient functions $a,b,c,d:X\times X\to [0,\infty )$ satisfying the uniform bound $L<1$ such that:

$$\begin{array}{cc}\hfill D(Tx,Ty)& \le a(x,y)D(x,y)+b(x,y)D(x,T^{\rho }x)+c(x,y)D(y,T^{\rho }y)\hfill \\ \hfill & +d(x,y)[D(x,T^{\rho }y)+D(y,T^{\rho }x)].\hfill \end{array}$$

(1)

The main significance of this work lies in demonstrating that despite using $T^{\rho }$ instead of T in the contractive conditions, one can prove the existence of a unique fixed point for T. This constitutes an important extension of classical theory, as we have shown through examples that the class of mappings covered by our theorem (with $\rho \ge 2$) is strictly larger than the class covered by traditional generalized contractions such as Ćirić’s condition (which corresponds to $\rho =1$).

This paper is organized as follows: after the introduction, Section 2 presents the preliminary concepts and auxiliary results. Section 3 is devoted to proving the main theorem in three stages: constructing a Cauchy sequence via Picard iteration, proving that its limit is a fixed point of T, and then proving the uniqueness of this point. Section 4 provides illustrative examples, including one showing that $T\left(x\right)=-x/(1+x^2)$ satisfies the theorem’s conditions with $\rho =2$ despite failing any Ćirić-type contractive condition with $\rho =1$. Section 5 is dedicated to applying the results to a nonlinear Fredholm integral equation of the second kind and to a Second-Order Boundary Value Problem. The paper concludes with a final Section 6 presenting conclusions and a discussion of the results.

2. Preliminaries

Let $(X,D)$ be a complete metric space, and let $T:X\to X$ be a continuous mapping. Fix an integer $\rho \ge 2$, and suppose there exist functions

$$a,b,c,d:X\times X\to [0,\infty )$$

such that for all $x,y\in X$,

$$\begin{array}{cc}\hfill D(Tx,Ty)& \le a(x,y)D(x,y)+b(x,y)D(x,T^{\rho }x)+c(x,y)D(y,T^{\rho }y)\hfill \\ \hfill & +d(x,y)\left[D(x,T^{\rho }y)+D(y,T^{\rho }x)\right],\hfill \end{array}$$

(2)

and the coefficients satisfy the uniform bound

$$\underset{x,y\in X}{sup}\left[a(x,y)+b(x,y)+c(x,y)+2d(x,y)\right]=:L<1.$$

(3)

Here, $T^{\rho }$ denotes the $\rho$-fold composition of T with itself: $T^{\rho }=\underset{\rho \mathrm{times}}{\underbrace{T\circ T\circ \dots \circ T}}$.

Theorem 1. 

Under assumptions (2) and (3), the mapping T has a unique fixed point in X; that is, there exists a unique $x^{*}\in X$ such that

$$T{x}^{*}=x^{*}.$$

The proof proceeds in three stages: (i) construction of a Cauchy sequence via Picard iteration, (ii) verification that its limit is a fixed point of T, and (iii) proof of uniqueness.

3. Proof of Theorem 1

Fix an arbitrary $x_0\in X$, and define the Picard iteration sequence

$$x_n:=T^n{x}_0,n=0,1,2,\dots .$$

(4)

For convenience, denote

$$s_n:=D(x_n,x_{n+1}),t_n:=D(x_n,x_{n+\rho })=D(x_n,T^{\rho }{x}_n),n\ge 0.$$

(5)

Note that $t_n$ measures the $\rho$-step displacement from $x_n$.

We first establish a fundamental estimate for the sequence $\left\{t_n\right\}$.

Lemma 1. 

The sequence $\left\{t_n\right\}$ satisfies

$$t_{n+1}\le L·max\{t_n,t_{n+\rho }\},\forall n\ge 0.$$

Proof. 

Apply inequality (2) to the pair $(x_n,x_{n+\rho })$. Using $T^{\rho }{x}_n=x_{n+\rho }$ and $T^{\rho }{x}_{n+\rho }=x_{n+2\rho }$, we compute

$$\begin{array}{cc}\hfill t_{n+1}& =D(x_{n+1},x_{n+\rho +1})=D(T{x}_n,T{x}_{n+\rho })\hfill \\ \hfill & \le a_n^{\prime }D(x_n,x_{n+\rho })+b_n^{\prime }D(x_n,T^{\rho }{x}_n)+c_n^{\prime }D(x_{n+\rho },T^{\rho }{x}_{n+\rho })\hfill \\ \hfill & +d_n^{\prime }\left[D(x_n,T^{\rho }{x}_{n+\rho })+D(x_{n+\rho },T^{\rho }{x}_n)\right],\hfill \end{array}$$

where $a_n^{\prime }=a(x_n,x_{n+\rho })$, etc. Now observe

• $D(x_n,x_{n+\rho })=t_n$,
• $D(x_n,T^{\rho }{x}_n)=D(x_n,x_{n+\rho })=t_n$,
• $D(x_{n+\rho },T^{\rho }{x}_{n+\rho })=D(x_{n+\rho },x_{n+2\rho })=t_{n+\rho }$,
• $D(x_n,T^{\rho }{x}_{n+\rho })=D(x_n,x_{n+2\rho })\le D(x_n,x_{n+\rho })+D(x_{n+\rho },x_{n+2\rho })=t_n+t_{n+\rho }$,
• $D(x_{n+\rho },T^{\rho }{x}_n)=D(x_{n+\rho },x_{n+\rho })=0$.

Substituting these bounds yields

$$\begin{array}{cc}\hfill t_{n+1}& \le a_n^{\prime }{t}_n+b_n^{\prime }{t}_n+c_n^{\prime }{t}_{n+\rho }+d_n^{\prime }(t_n+t_{n+\rho })\hfill \\ \hfill & =(a_n^{\prime }+b_n^{\prime }+d_n^{\prime })t_n+(c_n^{\prime }+d_n^{\prime })t_{n+\rho }\hfill \\ \hfill & \le (a_n^{\prime }+b_n^{\prime }+c_n^{\prime }+2d_n^{\prime })·max\{t_n,t_{n+\rho }\}\hfill \\ \hfill & \le L·max\{t_n,t_{n+\rho }\},\hfill \end{array}$$

where the last inequality follows from (3). This completes the proof. □

Let $u_n:={sup}_{k\ge n}{t}_k$. We first show that the sequence $\left\{t_n\right\}$ is bounded. Indeed, $t_0=D(x_0,T^{\rho }{x}_0)$ is a finite non-negative real number. We claim that for all $n\ge 0$,

$$t_n\le t_0{L}^{\lfloor n/\rho \rfloor }.$$

We prove this by induction. For $n=0$, the inequality holds trivially. Assume that it holds for all indices up to n. Then, using the estimate $t_{n+1}\le Lmax\{t_n,t_{n+\rho }\}$ from Lemma 1 and the induction hypothesis, we obtain

$$t_{n+1}\le L·t_0{L}^{\lfloor n/\rho \rfloor }=t_0{L}^{\lfloor n/\rho \rfloor +1}\le t_0{L}^{\lfloor (n+1)/\rho \rfloor },$$

where the last inequality follows from the fact that $\lfloor (n+1)/\rho \rfloor \le \lfloor n/\rho \rfloor +1$. Consequently, $t_n\le t_0$ for all n, and, thus,

$$u_0=\underset{n\ge 0}{sup}{t}_n\le t_0<\infty .$$

Hence, $\left(u_n\right)$ is a well-defined non-increasing sequence bounded below by zero, and, therefore, it converges.

From Lemma 1,

$$t_{n+1}\le L{u}_n,$$

so taking supremum over all $k\ge n+1$ gives

$$u_{n+1}=\underset{k\ge n+1}{sup}{t}_k\le L{u}_n.$$

Iterating yields

$$u_n\le L^n{u}_0,\mathrm{hence}{t}_n\le u_n\le u_0{L}^n\forall n\ge 0.$$

(6)

In particular,

$$\underset{n\to \infty }{lim}{t}_n=0.$$

(7)

Next, we show that the full Picard sequence $\left\{x_n\right\}$ is Cauchy.

Lemma 2. 

The Picard sequence $\left\{x_n\right\}$ defined by $x_n=T^n{x}_0$ is a Cauchy sequence in $(X,D)$.

Proof. 

Let $s_n=D(x_n,x_{n+1})$ and $t_n=D(x_n,x_{n+\rho })=D(x_n,T^{\rho }{x}_n)$. From Lemma 2 and the subsequent argument, we have the exponential decay

$$t_n\le t_0{L}^n\mathrm{for}\mathrm{all}n\ge 0.$$

(8)

Now, apply the contraction condition (2) to the pair $(x_n,x_{n-1})$

$$\begin{array}{cc}\hfill s_n& =D(T{x}_n,T{x}_{n-1})\hfill \\ & \le a(x_n,x_{n-1})D(x_n,x_{n-1})+b(x_n,x_{n-1})D(x_n,T^{\rho }{x}_n)\hfill \\ & +c(x_n,x_{n-1})D(x_{n-1},T^{\rho }{x}_{n-1})\hfill \\ & +d(x_n,x_{n-1})\left[D(x_n,T^{\rho }{x}_{n-1})+D(x_{n-1},T^{\rho }{x}_n)\right].\hfill \end{array}$$

Estimate each term using the triangle inequality

• $D(x_n,x_{n-1})=s_{n-1}$,
• $D(x_n,T^{\rho }{x}_n)=t_n$,
• $D(x_{n-1},T^{\rho }{x}_{n-1})=t_{n-1}$,
• $D(x_n,T^{\rho }{x}_{n-1})=D(x_n,x_{n-1+\rho })\le D(x_n,x_{n-1})+D(x_{n-1},x_{n-1+\rho })=s_{n-1}+t_{n-1}$,
• $D(x_{n-1},T^{\rho }{x}_n)=D(x_{n-1},x_{n+\rho })\le D(x_{n-1},x_n)+D(x_n,x_{n+\rho })=s_{n-1}+t_n$.

Substituting these bounds

$$\begin{array}{cc}\hfill s_n& \le a_n{s}_{n-1}+b_n{t}_n+c_n{t}_{n-1}+d_n\left[(s_{n-1}+t_{n-1})+(s_{n-1}+t_n)\right]\hfill \\ & =(a_n+2d_n)s_{n-1}+(b_n+d_n)t_n+(c_n+d_n)t_{n-1},\hfill \end{array}$$

where $a_n=a(x_n,x_{n-1})$, etc.

By the coefficient bound (3), we have

$$a_n+2d_n\le L,b_n+d_n\le L,c_n+d_n\le L.$$

Therefore,

$$s_n\le L{s}_{n-1}+L{t}_n+L{t}_{n-1}.$$

Using the exponential decay (6) $t_n\le t_0{L}^n$ and $t_{n-1}\le t_0{L}^{n-1}$, we obtain

$$s_n\le L{s}_{n-1}+L\left(t_0{L}^n\right)+L\left(t_0{L}^{n-1}\right)=L{s}_{n-1}+t_0(L^{n+1}+L^n).$$

Let $K=t_0(L+1)$. Then

$$s_n\le L{s}_{n-1}+K{L}^n.$$

Divide both sides by $L^n$

$${\displaystyle \frac{s_n}{L^n}}\le {\displaystyle \frac{s_{n-1}}{L^{n-1}}}+K.$$

Iterating this inequality

$${\displaystyle \frac{s_n}{L^n}}\le {\displaystyle \frac{s_0}{L^0}}+nK=s_0+nK.$$

Hence,

$$s_n\le (s_0+nK)L^n.$$

Since $L<1$, the series ${\sum }_{n=0}^{\infty }{s}_n$ converges by the root test (or comparison to a convergent series). Therefore, for any $m>n$,

$$D(x_n,x_m)\le \sum _{k=n}^{m-1}{s}_k\le \sum _{k=n}^{\infty }{s}_k\to 0\mathrm{as}n\to \infty ,$$

which proves that $\left\{x_n\right\}$ is Cauchy. □

Since $(X,D)$ is complete, there exists $x^{*}\in X$ such that

$$\underset{n\to \infty }{lim}{x}_n=x^{*}.$$

(9)

We now prove that $x^{*}$ is a fixed point of T.

Lemma 3. 

$T{x}^{*}=x^{*}$.

Proof. 

By construction, $x_{n+1}=T{x}_n$ for all $n\ge 0$. Since $x_n\to x^{*}$ and T is continuous, we have

$$T{x}^{*}=T\left(\underset{n\to \infty }{lim}{x}_n\right)=\underset{n\to \infty }{lim}T{x}_n=\underset{n\to \infty }{lim}{x}_{n+1}=x^{*}.$$

Hence, $x^{*}$ is a fixed point of T. □

Finally, we verify uniqueness.

Lemma 4. 

The fixed point of T is unique.

Proof. 

Suppose $x^{*},y^{*}\in X$ satisfy $T{x}^{*}=x^{*}$ and $T{y}^{*}=y^{*}$. Then $T^{\rho }{x}^{*}=x^{*}$ and $T^{\rho }{y}^{*}=y^{*}$. Substituting into (2) gives

$$\begin{array}{cc}\hfill D(x^{*},y^{*})& =D(T{x}^{*},T{y}^{*})\hfill \\ \hfill & \le a(x^{*},y^{*})D(x^{*},y^{*})+b(x^{*},y^{*})·0+c(x^{*},y^{*})·0\hfill \\ \hfill & +d(x^{*},y^{*})\left[D(x^{*},y^{*})+D(y^{*},x^{*})\right]\hfill \\ \hfill & =\left[a(x^{*},y^{*})+2d(x^{*},y^{*})\right]D(x^{*},y^{*})\hfill \\ \hfill & \le LD(x^{*},y^{*}),\hfill \end{array}$$

where the last inequality uses (3). Since $L<1$, this implies $(1-L)D(x^{*},y^{*})\le 0$, hence $D(x^{*},y^{*})=0$, i.e., $x^{*}=y^{*}$. □

4. Illustrative Examples

In this section, we present two explicit examples that clarify the scope of Theorem 1. The first demonstrates a nontrivial mapping satisfying the contractive condition (2) with $\rho =2$. The second shows that the class of mappings covered by our theorem with $\rho \ge 2$ is strictly larger than the class covered by classical generalized contractions such as those of Ćirić (which correspond to $\rho =1$).

4.1. Example Satisfying the $\rho =2$ Condition

Let $X=\mathbb{R}$ equipped with the standard metric $D(x,y)=|x-y|$, which is complete. Define the mapping $T:X\to X$ by

$$Tx={\displaystyle \frac{1}{8}}x.$$

Then T is continuous, and its second iterate is

$$T^{2}x=T\left(Tx\right)={\displaystyle \frac{1}{64}}x.$$

We verify that T satisfies inequality (2) with $\rho =2$ and constant coefficients

$$a(x,y)={\displaystyle \frac{1}{4}},b(x,y)={\displaystyle \frac{1}{8}},c(x,y)={\displaystyle \frac{1}{8}},d(x,y)=0,$$

so that the uniform bound is

$$L=a+b+c+2d={\displaystyle \frac{1}{2}}<1.$$

Indeed, for any $x,y\in \mathbb{R}$,

$$|Tx-Ty|={\displaystyle \frac{1}{8}}|x-y|.$$

On the other hand,

$$\begin{array}{cc}\hfill |x-T^{2}x|& =\left|x-{\displaystyle \frac{1}{64}}x\right|={\displaystyle \frac{63}{64}}\left|x\right|,\hfill \\ \hfill |y-T^{2}y|& ={\displaystyle \frac{63}{64}}\left|y\right|.\hfill \end{array}$$

Therefore, the right-hand side of (2) becomes

$${\displaystyle \frac{1}{4}}|x-y|+{\displaystyle \frac{1}{8}}|x-T^{2}x|+{\displaystyle \frac{1}{8}}|y-T^{2}y|={\displaystyle \frac{1}{4}}|x-y|+{\displaystyle \frac{1}{8}}·{\displaystyle \frac{63}{64}}\left|x\right|+{\displaystyle \frac{1}{8}}·{\displaystyle \frac{63}{64}}\left|y\right|$$

$$={\displaystyle \frac{1}{4}}|x-y|+{\displaystyle \frac{63}{512}}\left(\right|x|+\left|y\right|).$$

Since $|x-y|\le \left|x\right|+\left|y\right|$, we have

$${\displaystyle \frac{1}{8}}|x-y|\le {\displaystyle \frac{1}{8}}\left(\right|x|+\left|y\right|)={\displaystyle \frac{64}{512}}\left(\right|x|+\left|y\right|)<{\displaystyle \frac{1}{4}}|x-y|+{\displaystyle \frac{63}{512}}\left(\right|x|+\left|y\right|),$$

because

$${\displaystyle \frac{1}{4}}|x-y|+{\displaystyle \frac{63}{512}}\left(\right|x|+\left|y\right|)\ge \left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{63}{512}}\right)|x-y|={\displaystyle \frac{191}{512}}|x-y|>{\displaystyle \frac{64}{512}}|x-y|={\displaystyle \frac{1}{8}}|x-y|.$$

Thus, inequality (2) holds for all $x,y\in \mathbb{R}$ with the stated coefficients. Since $L=1/2<1$, all assumptions of Theorem 1 are satisfied. The unique fixed point is $x^{*}=0$, as $Tx=x$ implies $x/8=x$, so $x=0$.

This example illustrates that even simple linear contractions can be embedded in the framework of (2) with $\rho =2$, where the terms involving $D(x,T^{2}x)$ and $D(y,T^{2}y)$ provide additional flexibility, essential in nonlinear contexts such as integral equations.

4.2. Comparison with Ćirić’s Condition ($\rho =1$)

We now construct a mapping that satisfies the hypotheses of Theorem 1 for $\rho =2$, but does not satisfy any Ćirić-type contractive condition with $\rho =1$.

Let $X=\mathbb{R}$ with the usual metric, and define $T:X\to X$ by

$$Tx=-{\displaystyle \frac{x}{1+x^2}},x\in \mathbb{R}.$$

This map is continuous, odd, and satisfies $\left|Tx\right|<\left|x\right|$ for all $x\ne 0$, with $T_0=0$. Its second iterate is

$$T^{2}x=T\left(Tx\right)=-{\displaystyle \frac{Tx}{1+{\left(Tx\right)}^2}}={\displaystyle \frac{x}{(1+x^2)\left(1+\frac{x^2}{{(1+x^2)}^2}\right)}}.$$

A key observation is that $T^{2}x$ and x have the same sign, whereas $Tx$ has the opposite sign.

Now consider the pair $(x,-x)$ for $x>0$. Then:

• $Tx=-{\displaystyle \frac{x}{1+x^2}}$, $T(-x)={\displaystyle \frac{x}{1+x^2}}$,
• $|Tx-T(-x)|={\displaystyle \frac{2x}{1+x^2}}$,
• $|x-(-x\left)\right|=2x$,
• $|x-Tx|=x+{\displaystyle \frac{x}{1+x^2}}=x\left(1+{\displaystyle \frac{1}{1+x^2}}\right)$,
• $|-x-T(-x)|=x+{\displaystyle \frac{x}{1+x^2}}$ (same as above).

Suppose, for contradiction, that T satisfies a Ćirić-type inequality with $\rho =1$, i.e., there exist constants $a,b,c,d\ge 0$ with $a+b+c+2d=L<1$ such that for all $u,v\in \mathbb{R}$,

$$|Tu-Tv|\le a|u-v|+b|u-Tu|+c|v-Tv|+d\left(|u-Tv|+|v-Tu|\right).$$

Apply this to $u=x$, $v=-x$. Using the estimates above and the triangle inequality $|x-T(-x)|\le \left|x\right|+\left|T(-x)\right|=x+\frac{x}{1+x^2}$ (and similarly for $|-x-Tx|$), we obtain

$${\displaystyle \frac{2x}{1+x^2}}\le a\left(2x\right)+(b+c)x\left(1+{\displaystyle \frac{1}{1+x^2}}\right)+2dx\left(1+{\displaystyle \frac{1}{1+x^2}}\right).$$

Dividing both sides by $x>0$ and rearranging the inequality gives

$${\displaystyle \frac{2}{1+x^2}}\le 2a+(b+c+2d)\left(1+{\displaystyle \frac{1}{1+x^2}}\right).$$

Taking $x\to 0^{+}$, the left-hand side tends to 2, while the right-hand side tends to $2a+2(b+c+2d)=2(a+b+c+2d)=2L$. Thus we require $2\le 2L$, i.e., $L\ge 1$, contradicting the assumption $L<1$.

Hence, no Ćirić-type contractive condition with $\rho =1$ and $L<1$ can hold for this T.

However, with $\rho =2$, the mapping T does satisfy the hypotheses of Theorem 1. To see this, note that a Taylor expansion of $T^{2}x$ around $x=0$ yields

$$T^{2}x=x-2x^3+O\left(x^5\right),\mathrm{as}x\to 0,$$

and, therefore,

$$|x-T^2{x|=2|x|}^3+{O\left(\right|x|}^5).$$

In particular, $|x-T^2{x|=O(\left|x\right|}^3)$ as $x\to 0$, which decays faster than $\left|x\right|$.

Now consider the contractive inequality (2) with $\rho =2$. For $x,y$ in a bounded neighborhood of 0, the term $|Tx-Ty|=\left|{\displaystyle \frac{x}{1+x^2}}-{\displaystyle \frac{y}{1+y^2}}\right|$ is Lipschitz with constant strictly less than 1, while the quantities $|x-T^{2}x|$ and $|y-T^{2}y|$ are of higher order. Consequently, one can choose constants $a,b,c,d\ge 0$ with $a+b+c+2d<1$ such that

$$|Tx-Ty|\le a|x-y|+b|x-T^{2}x|+c|y-T^{2}y|+d\left(|x-T^{2}y|+|y-T^{2}x|\right)$$

holds for all $x,y\in \mathbb{R}$. For instance, on any compact interval $[-M,M]$, the function

$${\displaystyle \frac{|Tx-Ty|}{|x-y|+|x-T^{2}x|+|y-T^{2}y|+|x-T^{2}y|+|y-T^{2}x|}}$$

is continuous (with the convention $0/0=0$) and attains a maximum strictly less than 1, due to the cubic decay of $|x-T^{2}x|$ near zero and the global boundedness of T and $T^2$. Extending this bound globally (e.g., by noting that $\left|Tx\right|\le 1/2$ for all x) ensures the existence of a uniform constant $L<1$ satisfying (3).

Thus, while T fails all $\rho =1$ contractive conditions (as shown above), it satisfies the $\rho =2$ condition of Theorem 1, confirming that the latter framework strictly extends classical single-step contraction theories.

5. Application to a Nonlinear Fredholm Integral Equation

Consider the nonlinear Fredholm integral equation of the second kind

$$x\left(t\right)=f\left(t\right)+\lambda {\int }_0^{1}K\left(t,s,x\left(s\right),x\left({\varphi }_1\left(s\right)\right),\dots ,x\left({\varphi }_{\rho -1}\left(s\right)\right)\right)ds,t\in [0,1],$$

(10)

where

• $f\in C[0,1]$ is a given continuous function;
• $\lambda \in \mathbb{R}$ is a real parameter;
• ${\varphi }_k:[0,1]\to [0,1]$ ($k=1,\dots ,\rho -1$) are continuous self-maps;
• the kernel $K:{[0,1]}^2\times {\mathbb{R}}^{\rho }\to \mathbb{R}$ is continuous and satisfies the following generalized Lipschitz condition: there exist non-negative integrable functions $\alpha ,\beta ,\gamma ,\delta \in L^1[0,1]$ such that for all $t,s\in [0,1]$ and all $u_0,\dots ,u_{\rho -1},v_0,\dots ,v_{\rho -1}\in \mathbb{R}$,

  $$\begin{array}{cc}\hfill & \left|K(t,s,u_0,\dots ,u_{\rho -1})-K(t,s,v_0,\dots ,v_{\rho -1})\right|\hfill \\ \hfill & \le \alpha \left(s\right)|u_0-v_0|+\beta \left(s\right)\left|u_0-K\left(s,r,u_0,u_1,\dots ,u_{\rho -2}\right)(\mathrm{integrated}\mathrm{over}r)\right|\hfill \\ \hfill & +\gamma \left(s\right)\left|v_0-K\left(s,r,v_0,v_1,\dots ,v_{\rho -2}\right)(\mathrm{integrated}\mathrm{over}r)\right|\hfill \\ \hfill & +\delta \left(s\right)\left(|u_0-v_{\rho -1}|+|v_0-u_{\rho -1}|\right).\hfill \end{array}$$

Consider the explicit iterative scheme

$$\left(Tx\right)\left(t\right)=f\left(t\right)+\lambda {\int }_0^{1}K\left(t,s,x\left(s\right),x\left({\psi }_1\left(s\right)\right),\dots ,x\left({\psi }_{\rho -1}\left(s\right)\right)\right)ds,$$

where $T:C[0,1]\to C[0,1]$, and each ${\psi }_k:[0,1]\to [0,1]$ is a fixed continuous map (e.g., ${\psi }_k\left(s\right)=s$, or delay maps like ${\psi }_k\left(s\right)=max\{0,s-{\tau }_k\}$). In this case, the $\rho$-th iterate satisfies

$$\left(T^{\rho }x\right)\left(t\right)=f\left(t\right)+\lambda {\int }_0^{1}K\left(t,s,\left(T^{\rho -1}x\right)\left(s\right),\left(T^{\rho -1}x\right)\left({\psi }_1\left(s\right)\right),\dots ,\left(T^{\rho -1}x\right)\left({\psi }_{\rho -1}\left(s\right)\right)\right)ds,$$

and for any $x,y\in C[0,1]$, one can estimate (using uniform continuity of K on compact sets and Fubini’s theorem) that

$$\begin{array}{cc}\hfill {\parallel Tx-Ty\parallel }_{\infty }& \le \left|\lambda \right|{\int }_0^1{[\alpha \left(s\right)\parallel x-y\parallel }_{\infty }+\beta \left(s\right)\parallel x-T^{\rho }{x\parallel }_{\infty }+\gamma \left(s\right){\parallel y-T^{\rho }y\parallel }_{\infty }\hfill \\ \hfill & +\delta \left(s\right)\left(\parallel x-T^{\rho }{y\parallel }_{\infty }+{\parallel y-T^{\rho }x\parallel }_{\infty }\right)]ds\hfill \\ \hfill & ={a\parallel x-y\parallel }_{\infty }+b\parallel x-T^{\rho }{x\parallel }_{\infty }+c{\parallel y-T^{\rho }y\parallel }_{\infty }\hfill \\ \hfill & +d\left(\parallel x-T^{\rho }{y\parallel }_{\infty }+{\parallel y-T^{\rho }x\parallel }_{\infty }\right),\hfill \end{array}$$

where we set

$$a=\left|\lambda \right|{\int }_0^1\alpha \left(s\right)ds,b=\left|\lambda \right|{\int }_0^1\beta \left(s\right)ds,c=\left|\lambda \right|{\int }_0^1\gamma \left(s\right)ds,d=\left|\lambda \right|{\int }_0^1\delta \left(s\right)ds.$$

Assume the uniform bound

$$L:=a+b+c+2d=\left|\lambda \right|{\int }_0^1\left[\alpha \left(s\right)+\beta \left(s\right)+\gamma \left(s\right)+2\delta \left(s\right)\right]ds<1.$$

(11)

Then the operator T satisfies condition (2) on the complete metric space $(C[0,1],\parallel ·{\parallel }_{\infty })$; continuity of T follows from the continuity of K and dominated convergence. Hence, by Theorem 1, T admits a unique fixed point $x^{*}\in C[0,1]$, which is precisely the unique continuous solution of (10).

For a concrete instance with nontrivial coefficients, let $\rho =2$, $X=C[0,1]$ equipped with the sup-norm metric $D(x,y)={\parallel x-y\parallel }_{\infty }$, and define $T:X\to X$ by

$$\left(Tx\right)\left(t\right)={\displaystyle \frac{1}{4}}{\int }_0^1\left({\displaystyle \frac{1}{8}}x\left(s\right)+{\displaystyle \frac{1}{8}}\left(Tx\right)\left(s\right)\right)ds,t\in [0,1].$$

As $\left(Tx\right)\left(t\right)$ is constant in t, solving for this constant yields $\left(Tx\right)\left(t\right)\equiv {\displaystyle \frac{1}{31}}{\int }_0^{1}x\left(s\right)ds$. A direct computation shows that for all $x,y\in X$,

$$D(Tx,Ty)=\left|{\displaystyle \frac{1}{31}}{\int }_0^1(x\left(s\right)-y\left(s\right))ds\right|\le {\displaystyle \frac{1}{32}}D(x,y)+{\displaystyle \frac{1}{32}}D(x,T^{2}x)+{\displaystyle \frac{1}{32}}D(y,T^{2}y),$$

since the right-hand side dominates the left for all extremal choices (e.g., constant or sign-aligned functions), and the deficit from the factor $\frac{1}{32}<\frac{1}{31}$ is compensated by the non-negative terms involving $D(x,T^{2}x)$ and $D(y,T^{2}y)$. Thus, inequality (2) holds with

$$a(x,y)=b(x,y)=c(x,y)={\displaystyle \frac{1}{32}},d(x,y)=0,$$

and the uniform bound is

$$L={\displaystyle \frac{1}{32}}+{\displaystyle \frac{1}{32}}+{\displaystyle \frac{1}{32}}+0={\displaystyle \frac{3}{32}}<1.$$

Hence, all hypotheses of Theorem 1 are satisfied. The unique fixed point is the zero function, which indeed solves $x=Tx$.

Example 1. 

Nonzero fixed point – failure of the Banach condition

Consider the space $X=C[0,1]$ equipped with the uniform metric $D(x,y)={\parallel x-y\parallel }_{\infty }={max}_{t\in [0,1]}|x\left(t\right)-y\left(t\right)|$, which is a complete metric space. Let $\rho =2$, and define the operator $T:X\to X$ as follows

$$\left(Tx\right)\left(t\right)={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}{\int }_0^1{\left(x\left(s\right)\right)}^{2}ds+{\displaystyle \frac{1}{8}}{\int }_0^{1}x\left(s\right)ds,\forall t\in [0,1].$$

(12)

Note that $\left(Tx\right)\left(t\right)$ does not depend on t; that is, T maps any function to a constant function. This simplification allows us to compute the fixed point explicitly.

• Step 1: Finding the fixed point.

Suppose $x^{*}\left(t\right)\equiv c$ (constant). Then

$$c={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}{\int }_0^1{c}^{2}ds+{\displaystyle \frac{1}{8}}{\int }_0^{1}cds={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}{c}^2+{\displaystyle \frac{1}{8}}c.$$

Multiplying both sides by 24

$$24c=8+6c^2+3c\Rightarrow 0=6c^2-21c+8.$$

The discriminant is $\Delta =441-192=249>0$, so there are two real solutions. The solution lying in the domain to be determined later is

$$c^{*}={\displaystyle \frac{21-\sqrt{249}}{12}}\approx 0.437.$$

Thus $x^{*}\left(t\right)\equiv c^{*}$ is a nonzero fixed point.

• Step 2: Verifying the conditions of Theorem 1.

We compute the second iterate $T^2$

$$\left(T^{2}x\right)\left(t\right)={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}{\int }_0^1{\left(\left(Tx\right)\left(s\right)\right)}^{2}ds+{\displaystyle \frac{1}{8}}{\int }_0^1\left(Tx\right)\left(s\right)ds.$$

Since $Tx$ is a constant function, $\left(Tx\right)\left(s\right)\equiv {\alpha }_x$ where

$${\alpha }_x={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}{\int }_0^1{\left(x\left(s\right)\right)}^{2}ds+{\displaystyle \frac{1}{8}}{\int }_0^{1}x\left(s\right)ds.$$

Consequently

$$\left(T^{2}x\right)\left(t\right)={\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}{\alpha }_x^2+{\displaystyle \frac{1}{8}}{\alpha }_x\left(a\mathit{constant}\mathit{value}\right).$$

Now we verify inequality 2 from Theorem 1. For all $x,y\in X$

$${\parallel Tx-Ty\parallel }_{\infty }=\left|{\displaystyle \frac{1}{4}}{\int }_0^1\left(x^2\left(s\right)-y^2\left(s\right)\right)ds+{\displaystyle \frac{1}{8}}{\int }_0^1\left(x\left(s\right)-y\left(s\right)\right)ds\right|.$$

Using the following estimates

$$\begin{array}{cc}\hfill \left|{\int }_0^1(x^2-y^2)\right|& \le {\int }_0^1|x-y|\left(\right|x|+|y\left|\right)\le {\parallel x-y\parallel }_{\infty }{(\parallel x\parallel }_{\infty }+{\parallel y\parallel }_{\infty }),\hfill \\ \hfill \left|{\int }_0^1(x-y)\right|& \le {\parallel x-y\parallel }_{\infty },\hfill \end{array}$$

we obtain

$${\parallel Tx-Ty\parallel }_{\infty }\le {\displaystyle \frac{1}{4}}{\parallel x-y\parallel }_{\infty }{(\parallel x\parallel }_{\infty }+{\parallel y\parallel }_{\infty })+{\displaystyle \frac{1}{8}}{\parallel x-y\parallel }_{\infty }.$$

Since we aim to obtain the form 2 containing the terms $D(x,T^{2}x)$ and $D(y,T^{2}y)$, we note that:

$$\parallel x-T^2{x\parallel }_{\infty }=max\left\{|x\left(t\right)-c_x|\right\},$$

where $c_x$ is the constant resulting from $T^{2}x$. In a neighborhood of the fixed point $x^{*}$, the quantity $\parallel x-T^2{x\parallel }_{\infty }$ is of higher order than ${\parallel x-y\parallel }_{\infty }$ (due to the squaring in the definition of T).

By choosing a closed ball containing $x^{*}$, for instance $B=\{x\in X:\parallel x{\parallel }_{\infty }\le 1\}$, one can find constants $a,b,c,d\ge 0$ such that $L=a+b+c+2d<1$ and the inequality holds for all $x,y\in B$

$${\parallel Tx-Ty\parallel }_{\infty }\le {a\parallel x-y\parallel }_{\infty }+b\parallel x-T^2{x\parallel }_{\infty }+c{\parallel y-T^{2}y\parallel }_{\infty }+d\left(\parallel x-T^2{y\parallel }_{\infty }+{\parallel y-T^{2}x\parallel }_{\infty }\right).$$

For example, one may take

$$a={\displaystyle \frac{1}{2}},b=c={\displaystyle \frac{1}{8}},d=0,L={\displaystyle \frac{3}{4}}<1.$$

• Step 3: Failure of the Banach contraction condition.

Consider the functions $x_n\left(t\right)\equiv n$ (constant and large). Then

$$\parallel T{x}_n-T_0{\parallel }_{\infty }=\left|{\displaystyle \frac{1}{4}}{n}^2+{\displaystyle \frac{1}{8}}n\right|,$$

while $\parallel x_n{-0\parallel }_{\infty }=n$. The ratio

$${\displaystyle \frac{\parallel T{x}_n-T_0{\parallel }_{\infty }}{\parallel x_n{-0\parallel }_{\infty }}}={\displaystyle \frac{1}{4}}n+{\displaystyle \frac{1}{8}}\to \infty \mathrm{as}n\to \infty .$$

Thus there is no constant $k<1$ satisfying ${\parallel Tx-Ty\parallel }_{\infty }\le k{\parallel x-y\parallel }_{\infty }$ for all $x,y$ in the space. Hence T is not a contraction in the sense of Banach.

By Conclusion. The operator T satisfies the conditions of Theorem 1 with $\rho =2$ and has a unique nonzero fixed point $x^{*}\left(t\right)\equiv c^{*}$, yet it does not satisfy the standard Banach contraction principle. This confirms that the proposed theorem genuinely generalizes Banach’s theorem and extends the scope of applications to integral equations that cannot be studied by classical methods.

Example 2. 

Application to a Second-Order Boundary Value Problem 

Consider the following nonlinear second-order ordinary differential equation with Dirichlet boundary conditions

$$\left\{\begin{array}{c}{x}^{\prime \prime }\left(t\right)=F(t,x\left(t\right)),t\in [0,1],\hfill \\ x\left(0\right)=0,x\left(1\right)=0,\hfill \end{array}\right.$$

(13)

where $F:[0,1]\times \mathbb{R}\to \mathbb{R}$ is a continuous function. This type of problem arises in various physical contexts, such as steady-state heat conduction with a nonlinear source term.

It is well known that the boundary value problem can be reformulated as a Fredholm integral equation of the second kind. Indeed, using the Green’s function

$$G(t,s)=\left\{\begin{array}{cc}t(1-s),\hfill & 0\le t\le s\le 1,\hfill \\ s(1-t),\hfill & 0\le s\le t\le 1,\hfill \end{array}\right.$$

(14)

the solution x satisfies the equivalent integral equation

$$x\left(t\right)={\int }_0^{1}G(t,s)F(s,x\left(s\right))ds,t\in [0,1].$$

(15)

Define the operator $T:C[0,1]\to C[0,1]$ by

$$\left(Tu\right)\left(t\right)={\int }_0^{1}G(t,s)F(s,x\left(s\right))ds.$$

(16)

The continuity of F and the kernel G ensures that T is well-defined and continuous. Moreover, the space $C[0,1]$ equipped with the supremum norm ${\parallel u\parallel }_{\infty }={max}_{t\in [0,1]}\left|u\left(t\right)\right|$ is a complete metric space.

• Lipschitz Condition:

Assume that F satisfies the following generalized Lipschitz condition: there exist non-negative constants $a,b,c,d\ge 0$ such that for all $t\in [0,1]$ and all $x,y\in \mathbb{R}$,

$$|F(t,x)-F(t,y)|\le a|x-y|+b|x-T^{2}x|+c|y-T^{2}y|+d\left(|x-T^{2}y|+|y-T^{2}x|\right),$$

(17)

where $T^2$ denotes the second iterate of the operator T. This condition reflects the structure of our contractive inequality.

• Verification of Theorem 1 with $\rho =2$:

For any $x,y\in C[0,1]$, we estimate

$$\begin{array}{cc}\hfill \left|\right(Tx\left)\right(t)-(Ty\left)\right(t\left)\right|& \le {\int }_0^{1}G(t,s)|F(s,x\left(s\right))-F(s,y\left(s\right))|ds\hfill \\ \hfill & \le {\int }_0^{1}G(t,s)\left[a\right|x\left(s\right)-y\left(s\right)|+b|x\left(s\right)-T^{2}x\left(s\right)|\hfill \\ \hfill & +c|y\left(s\right)-T^{2}y\left(s\right)|+d\left(|x\left(s\right)-T^{2}y\left(s\right)|+|y\left(s\right)-T^{2}x\left(s\right)|\right)]ds.\hfill \end{array}$$

Taking the supremum over $t\in [0,1]$ and noting that ${max}_{t\in [0,1]}{\int }_0^{1}G(t,s)ds={\displaystyle \frac{1}{8}}$ (a standard property of the Green’s function), we obtain

$$\begin{array}{cc}\hfill {\parallel Tx-Ty\parallel }_{\infty }& \le {\displaystyle \frac{1}{8}}{[a\parallel x-y\parallel }_{\infty }+b\parallel x-T^2{x\parallel }_{\infty }+c{\parallel y-T^{2}y\parallel }_{\infty }\hfill \\ \hfill & +d\left(\parallel x-T^2{y\parallel }_{\infty }+{\parallel y-T^{2}x\parallel }_{\infty }\right)].\hfill \end{array}$$

Thus, inequality (2) holds with

$$a(x,y)={\displaystyle \frac{a}{8}},b(x,y)={\displaystyle \frac{b}{8}},c(x,y)={\displaystyle \frac{c}{8}},d(x,y)={\displaystyle \frac{d}{8}}.$$

(18)

The uniform bound condition (3) becomes

$$L={\displaystyle \frac{a+b+c+2d}{8}}<1.$$

(19)

Hence, all hypotheses of Theorem 1 are satisfied for $\rho =2$.

• Existence and Uniqueness Result:

By Theorem 1, the operator T has a unique fixed point $x^{*}\in C[0,1]$, which satisfies

$$x^{*}\left(t\right)={\int }_0^{1}G(t,s)F(s,x^{*}\left(s\right))ds,$$

(20)

and consequently solves the original boundary value problem (13). This demonstrates the applicability of our fixed-point theorem to differential equations, expanding its scope beyond integral equations.

• Illustrative Example:

Consider the specific case where $F(t,x)={\displaystyle \frac{1}{16}}sin\left(x\left(t\right)\right)$. Then the Green’s function bound gives

$${\parallel Tx-Ty\parallel }_{\infty }\le {\displaystyle \frac{1}{8}}·{\displaystyle \frac{1}{16}}{\parallel x-y\parallel }_{\infty }={\displaystyle \frac{1}{128}}{\parallel x-y\parallel }_{\infty }.$$

(21)

Here, we may take $a={\displaystyle \frac{1}{16}}$, $b=c=d=0$, so that $L={\displaystyle \frac{1}{128}}<1$. Theorem 1 with $\rho =2$ (or even $\rho =1$ in this simple linear case) guarantees a unique solution. However, the flexibility of our theorem allows for more complex nonlinearities where the terms involving $T^2$ become essential for verifying the contractive condition.

Remark 1. 

This application illustrates that our fixed-point theorem is not limited to integral equations but extends naturally to boundary value problems for ordinary differential equations, thereby broadening its practical relevance. The use of $\rho =2$ is particularly significant when the nonlinearity F depends on the second iterate in a nontrivial way, as the contractive condition (17) explicitly involves $T^2$.

6. Conclusions

By Lemmas 2–4, the mapping T possesses a unique fixed point in X. This completes the proof of Theorem 1.

Remark 2. 

This result generalizes several classical fixed point theorems:

• When $\rho =1$ and $b=c=d=0$, we recover the Banach Contraction Principle.
• Other choices of coefficients recover theorems of Ćirić, Hardy–Rogers, and Reich.

The key innovation is that the contractive control involves $D(x,T^{\rho }x)$ rather than $D(x,Tx)$, yet the uniform bound $L<1$ ensures sufficient decay in the ρ-step distances to guarantee convergence.