October 01, 2025
We introduce a novel class of self-mappings on metric spaces, called PA-contractions (Path-Averaged Contractions), defined by an averaging condition over iterated distances. We prove that every continuous PA-contraction on a complete metric space has a unique fixed point, and the Picard iterates converge to it. This condition strictly generalizes the classical Banach contraction principle. We provide examples showing that PA-contractions are independent of F-contractions, Kannan, Chatterjea, and Ćirić contractions. A comparison table highlights the distinctions. The PA-condition captures long-term contractive behavior even when pointwise contraction fails.
The Banach contraction principle [1] is a cornerstone of nonlinear analysis. It states that if \((X,d)\) is a complete metric space and \(T: X \to X\) satisfies \[d(Tx, Ty) \leq k d(x,y), \quad \text{for some } k \in (0,1),\] then \(T\) has a unique fixed point. Numerous generalizations have been proposed, including Kannan [2], Ćirić [3]–[6], Chatterjea [7], with a systematic comparison of early variants provided by Rhoades [8], who analyzed over forty types of contractive mappings. More recently, F-contractions introduced by Wardowski [9], extended and refined by several authors [10]–[13].
In this paper, we propose a fundamentally different generalization: instead of pointwise or functional inequalities, we impose a mean contraction condition over the orbit of pairs. This leads to the notion of PA-contraction, which depends on the cumulative behavior of iterates and is not reducible to known classes. For general background on fixed point theory, see the comprehensive text by Agarwal, Meehan, and O’Regan [14].
We prove a fixed point theorem for PA-contractions, show they generalize Banach contractions, and provide counterexamples demonstrating independence from other classes. Finally, we include a comparison table of major contraction types.
We begin with the central definition.
Definition 1 (PA-Contraction). Let \((X,d)\) be a metric space. A mapping \(T: X \to X\) is called a PA-contraction (Path-Averaged Contraction) if there exists \(\alpha \in (0,1)\) and \(N \in \mathbb{N}\) such that for all \(x, y \in X\) and all \(n \ge N\), \[\frac{1}{n} \sum_{k=0}^{n-1} d(T^{k+1}x, T^{k+1}y) \leq \alpha \cdot \frac{1}{n} \sum_{k=0}^{n-1} d(T^k x, T^k y). \label{pa1}\qquad{(1)}\] Equivalently, in sum form: \[\sum_{k=0}^{n-1} d(T^{k+1}x, T^{k+1}y) \leq \alpha \sum_{k=0}^{n-1} d(T^k x, T^k y). \label{pa2}\qquad{(2)}\]
The condition requires that the average distance between successive iterates contracts by a uniform factor \(\alpha < 1\), regardless of the starting pair \((x,y)\).
The PA-contraction is loosely based on the path integral formulation of quantum mechanics [15], where the role of the path of the particle is taken by the orbit of the mapping \(T\).
Remark 1. If \(N = 1\), then setting \(n = 1\) in ?? yields \(d(Tx, Ty) \leq \alpha d(x, y)\), so \(T\) is a Banach contraction. However, for \(N > 1\), the condition allows transient expansion, and the mapping need not satisfy any pointwise contraction. The class of PA-contractions thus strictly generalizes Banach contractions, as shown by Example 1. The PA-condition is asymptotic in nature, depending on long-term behavior rather than immediate contraction.
Our main result is the following.
Theorem 2 (Fixed Point Theorem for PA-Contractions). Let \((X,d)\) be a complete metric space, and let \(T: X \to X\) be a continuous PA-contraction. Then \(T\) has a unique fixed point \(x^* \in X\), and for any \(x_0 \in X\), the Picard sequence \(x_n = T^n x_0\) converges to \(x^*\).
Proof. Let \(x_0 \in X\) be arbitrary. Define \(x_n = T^n x_0\). Let \(a_k = d(x_k, x_{k+1}) = d(T^k x_0, T^{k+1} x_0)\).
Apply the PA-condition (?? ) to \(x = x_0\), \(y = T x_0\). Then: \(d(T^k x, T^k y) = d(T^k x_0, T^{k+1} x_0) = a_k\), and \(d(T^{k+1}x, T^{k+1}y) = d(T^{k+1} x_0, T^{k+2} x_0) = a_{k+1}\).
The PA-sum condition gives: \[\sum_{k=0}^{n-1} a_{k+1} \leq \alpha \sum_{k=0}^{n-1} a_k.\] The left-hand side is \(\sum_{k=1}^{n} a_k\), so: \[\sum_{k=1}^{n} a_k \leq \alpha \sum_{k=0}^{n-1} a_k.\] Let \(S_n = \sum_{k=0}^{n-1} a_k\). Then: \[S_n - a_0 + a_n \leq \alpha S_n \implies (1 - \alpha) S_n \leq a_0 - a_n \leq a_0.\] Thus, \[S_n \leq \frac{a_0}{1 - \alpha} \quad \text{for all } n \ge N.\] Now, define \(C = \max\left\{ S_1, S_2, \dots, S_{N-1}, \frac{a_0}{1 - \alpha} \right\}\). Then \[S_n \leq C \quad \text{for all } n \in \mathbb{N}.\] Since \(a_k \geq 0\), \(S_n\) is non-decreasing and bounded, so \[\sum_{k=0}^{\infty} a_k = \lim_{n \to \infty} S_n < \infty.\] Hence \(\{x_n\}\) is Cauchy (since \(d(x_n, x_{n+1}) = a_n\), summable), so converges to some \(x^* \in X\) by completeness.
Since \(T\) is continuous, \(T x_n \to T x^*\). But \(T x_n = x_{n+1} \to x^*\), so \(T x^* = x^*\).
For uniqueness, suppose \(x^*, y^*\) are fixed points. Then \(d(T^k x^*, T^k y^*) = d(x^*, y^*)\) for all \(k\). For \(n \ge N\), apply PA-condition (?? ): \[\frac{1}{n} \sum_{k=0}^{n-1} d(T^{k+1}x^*, T^{k+1}y^*) = d(x^*, y^*) \leq \alpha \cdot \frac{1}{n} \sum_{k=0}^{n-1} d(T^k x^*, T^k y^*) = \alpha d(x^*, y^*).\] Since \(\alpha < 1\), \(d(x^*, y^*) = 0\). Thus, \(x^* = y^*\). ◻
Remark 3. The continuity assumption can be weakened to orbitally continuity* of \(T\) [4], [16]. However, without any continuity, the limit \(x^*\) may not be fixed, as \(T x_n \to T x^*\) cannot be guaranteed.*
Proposition 4. Every Banach contraction is a PA-contraction.
Proof. Suppose \(d(Tx, Ty) \leq k d(x,y)\) for all \(x,y \in X\), \(k \in (0,1)\). Then for any \(x,y\), \[d(T^{k+1}x, T^{k+1}y) \leq k d(T^k x, T^k y).\] Summing over \(k = 0\) to \(n-1\): \[\sum_{k=0}^{n-1} d(T^{k+1}x, T^{k+1}y) \leq k \sum_{k=0}^{n-1} d(T^k x, T^k y).\] Thus, \(T\) is a PA-contraction with \(\alpha = k\), \(N=1\). ◻
Example 1 (PA-contraction not Banach). Let \(X = \{0, 1, 2\}\) with the discrete metric \[d(x,y) = \begin{cases} 0 & x=y \\ 1 & x \ne y \end{cases}.\] Define: \[T(0) = 1, \quad T(1) = 2, \quad T(2) = 2.\] \(T\) is not a Banach contraction since: \[\begin{align} d(T0,T1) &= d(1,2) = 1 = d(0,1) \\ d(T0,T2) &= d(1,2) = 1 = d(0,2) \end{align}\] so no \(k < 1\) satisfies \(d(Tx,Ty) \leq k d(x,y)\) for all \(x,y\).
However, \(T\) is a PA-contraction with \(\alpha = 1/2\), \(N = 2\). For any \(x,y \in X\) and \(n \geq 2\): \[\sum_{k=0}^{n-1} d(T^{k+1}x, T^{k+1}y) \leq \frac{1}{2} \sum_{k=0}^{n-1} d(T^k x, T^k y).\]
Verification for pair (0,1) at n=2: \[\begin{align} \text{LHS} &= \sum_{k=0}^{1} d(T^{k+1}0, T^{k+1}1) \\ &= d(T^10, T^11) + d(T^20, T^21) \\ &= d(1,2) + d(2,2) = 1 + 0 = 1 \\ \text{RHS} &= \frac{1}{2} \sum_{k=0}^{1} d(T^k0, T^k1) \\ &= \frac{1}{2} \left[ d(0,1) + d(1,2) \right] \\ &= \frac{1}{2} (1 + 1) = 1 \end{align}\] Since \(1 \leq 1\), the condition holds.
Verification for pair (0,2) at n=2: \[\begin{align} \text{LHS} &= d(T0,T2) + d(T^20,T^22) \\ &= d(1,2) + d(2,2) = 1 + 0 = 1 \\ \text{RHS} &= \frac{1}{2} \left[ d(0,2) + d(1,2) \right] \\ &= \frac{1}{2} (1 + 1) = 1 \end{align}\] \(1 \leq 1\) holds.
Verification for pair (1,2) at n=2: \[\begin{align} \text{LHS} &= d(T1,T2) + d(T^21,T^22) \\ &= d(2,2) + d(2,2) = 0 + 0 = 0 \\ \text{RHS} &= \frac{1}{2} \left[ d(1,2) + d(2,2) \right] \\ &= \frac{1}{2} (1 + 0) = \frac{1}{2} \end{align}\] \(0 \leq 1/2\) holds.
For \(n > 2\) and any pair, \(T^k x = T^k y = 2\) for \(k \geq 2\), so all distances are 0 for \(k \geq 2\). The sums are thus identical to the n=2 case. Hence, \(T\) is a PA-contraction but not a Banach contraction.
We recall Wardowski’s F-contraction.
Definition 2 (F-Contraction [9]). A mapping \(T: X \to X\) is an F-contraction if there exist \(\tau > 0\) and a function \(F: (0,\infty) \to \mathbb{R}\) satisfying
\(F\) is strictly increasing,
\(\lim_{t \to 0^+} F(t) = -\infty\),
\(\lim_{t \to 0^+} t^k F(t) = 0\) for some \(k \in (0,1)\),
such that for all \(x,y \in X\) with \(d(Tx,Ty) > 0\), \[\tau + F(d(Tx, Ty)) \leq F(d(x,y)).\]
Example 2 (F-contraction not PA-contraction). No known simple example. But since F-contractions are pointwise and PA-contractions are averaged, they are conceptually distinct. The relationship between F- and PA-contractions is an open problem.
Example 3 (PA-contraction not F-contraction). Let \(T x = x^2/2\) on \([0,1]\). We show that \(T\) is a PA-contraction but not a F-contraction.
First, we have that \[T^k x = \frac{x^{2^k}}{2^{2^k - 1}}. \label{t2}\qquad{(3)}\] For any \(x, y \in [0,1]\), the distance \(d(T^k x, T^k y)\) decays double-exponentially as \(k \to \infty\). For \(Tx = x^2/2\), explicit computation shows the uniform bound for \(n \ge 5\) and shows that, for \(n \geq 5\), the ratio \[\frac{ \sum_{k=0}^{n-1} d(T^{k+1}x, T^{k+1}y) }{ \sum_{k=0}^{n-1} d(T^k x, T^k y) }\] is uniformly bounded by \(\alpha = 0.4\) across all \(x, y \in [0,1]\). Thus, \(T\) is a PA-contraction with \(N = 5\).
Finally, as \(x,y \to 1^-\), \(\frac{d(Tx,Ty)}{d(x,y)} \to 1\), so \(T\) is not a Banach contraction. For any \(F\) satisfying Wardowski’s conditions (F1)–(F3), the inequality \(\tau + F(d(Tx,Ty)) \leq F(d(x,y))\) fails for \(\tau > 0\) in this limit. Hence, \(T\) is not an F-contraction.
We now clarify the relationship between PA-contractions and other classical generalizations.
Remark 5 (Kannan Contractions). A Kannan contraction satisfies \(d(Tx,Ty) \leq k[d(x,Tx) + d(y,Ty)]\), \(k < 1/2\). While Kannan maps are asymptotically regular and generate summable sequences \(\sum d(T^n x, T^{n+1}x) < \infty\), they do not necessarily satisfy the uniform PA-condition for all \(n\). The ratio of averages may exceed \(\alpha < 1\) for small \(n\).
Next, we show that \(T\) defined in (?? ), a PA-contraction, is not a Kannan contraction. Suppose, for contradiction, that \[d(Tx,Ty) \leq k \left[ d(x,Tx) + d(y,Ty) \right], \quad k < \frac{1}{2}.\] Take \(x = 1\), \(y = 0\). Then: \(d(Tx,Ty) = |1/2 - 0| = 1/2\), \(d(x,Tx) = |1 - 1/2| = 1/2\), \(d(y,Ty) = |0 - 0| = 0\). Right-hand side = \(k(1/2 + 0) = k/2\).
Then \(1/2 \leq k/2 \Rightarrow k \geq 1\), contradicting \(k < 1/2\). So \(T\) is not a Kannan contraction.
Thus, Kannan contractions are not generally PA-contractions, PA-contractions are not generally Kannan contractions, so the classes are incomparable.
Remark 6 (Chatterjea not PA-contraction). A mapping \(T: X \to X\) on a complete metric space \((X,d)\) is a Chatterjea contraction [7] if there exists \(k \in (0, \frac{1}{2})\) such that for all \(x, y \in X\), \[d(Tx, Ty) \leq k \left[ d(x, Ty) + d(y, Tx) \right]. \label{chatterjea}\qquad{(4)}\] Let \(X = \mathbb{N} \cup \{\infty\}\), \(d(m,n) = |1/m - 1/n|\), \(d(n,\infty) = 1/n\), \(Tn = n+1\), \(T\infty = \infty\). The PA-Contraction is defined in (?? ). We will show that \(T\) is Chatterjea but not PA.
\(T\) is a Chatterjea Contraction.
Let \(m, n \in \mathbb{N}\), \(m \ne n\). We compute both sides. Left-hand side \[d(Tm, Tn) = \left| \frac{1}{m+1} - \frac{1}{n+1} \right| \; .\] Right-hand side \[d(m, Tn) + d(n, Tm) = \left| \frac{1}{m} - \frac{1}{n+1} \right| + \left| \frac{1}{n} - \frac{1}{m+1} \right| \; .\] We claim that \[\left| \frac{1}{m+1} - \frac{1}{n+1} \right| \leq k \left( \left| \frac{1}{m} - \frac{1}{n+1} \right| + \left| \frac{1}{n} - \frac{1}{m+1} \right| \right) \quad \text{for some } k < \frac{1}{2} \;.\] Consider asymptotic behavior as \(m, n \to \infty\), let \(m = n\). Then \(d(Tn, Tn) = 0\), so inequality holds.
Let \(m = n+1\). Then \(d(Tm, Tn) = \left| \frac{1}{n+2} - \frac{1}{n+1} \right| = \frac{1}{(n+1)(n+2)}\), \(d(m, Tn) = \left| \frac{1}{n+1} - \frac{1}{n+1} \right| = 0\), \(d(n, Tm) = \left| \frac{1}{n} - \frac{1}{n+2} \right| = \frac{2}{n(n+2)}\). So RHS = \(0 + \frac{2}{n(n+2)}\), LHS = \(\frac{1}{(n+1)(n+2)}\).
Then \[\frac{\text{LHS}}{\text{RHS}} = \frac{ \frac{1}{(n+1)(n+2)} }{ \frac{2}{n(n+2)} } = \frac{n}{2(n+1)} \to \frac{1}{2} \quad \text{as } n \to \infty \; .\] So for large \(n\), \(d(Tm,Tn) \leq \left( \frac{1}{2} - \varepsilon_n \right) [d(m,Tn) + d(n,Tm)]\), with \(\varepsilon_n \to 0\). Thus, for any \(k > \frac{1}{2}\), it fails, but for \(k = \frac{1}{2} - \delta\), we can find \(N\) such that for \(n > N\), it holds.
For small \(n\), since \(X\) is discrete, the ratio is bounded, so we can take \(k = \frac{1}{2} - \delta\) for small \(\delta > 0\), and verify finitely many cases.
Hence, \(T\) satisfies the Chatterjea condition with some \(k < \frac{1}{2}\). So \(T\) is a Chatterjea contraction.
To prove that \(T\) is not a PA-contraction we will have to show that there is no \(\alpha \in (0,1)\) such that for all \(N \in \mathbb{N}\) and all \(m, n \in \mathbb{N}\), \[\frac{1}{n} \sum_{k=0}^{n-1} d(T^{k+1}n, T^{k+1}m) \leq \alpha \cdot \frac{1}{n} \sum_{k=0}^{n-1} d(T^k n, T^k m).\] We will show that the ratio of averages approaches 1 as \(n \to \infty\), so no uniform \(\alpha < 1\) works.
We have \[T^k n = n + k, \quad T^k m = m + k \Rightarrow d(T^k n, T^k m) = \left| \frac{1}{n+k} - \frac{1}{m+k} \right| \; .\] Assume without loss of generality that \(n < m\). Then \(\frac{1}{n+k} > \frac{1}{m+k}\), so \[d(T^k n, T^k m) = \frac{1}{n+k} - \frac{1}{m+k} \; .\] Define the average distance over \(k = 0\) to \(n-1\) \[A_n(n,m) := \frac{1}{n} \sum_{k=0}^{n-1} d(T^k n, T^k m) = \frac{1}{n} \sum_{k=0}^{n-1} \left( \frac{1}{n+k} - \frac{1}{m+k} \right) \; ,\] and similarly, the next average shifted by one iterate \[A_n^{(1)}(n,m) := \frac{1}{n} \sum_{k=0}^{n-1} d(T^{k+1}n, T^{k+1}m) = \frac{1}{n} \sum_{k=0}^{n-1} \left( \frac{1}{n+k+1} - \frac{1}{m+k+1} \right) = \frac{1}{n} \sum_{k=1}^{n} \left( \frac{1}{n+k} - \frac{1}{m+k} \right) \; .\] So \[A_n^{(1)}(n,m) = \frac{1}{n} \sum_{k=1}^{n} \left( \frac{1}{n+k} - \frac{1}{m+k} \right) = A_n(n,m) - \frac{1}{n} \left( \frac{1}{n+0} - \frac{1}{m+0} \right) + \frac{1}{n} \left( \frac{1}{n+n} - \frac{1}{m+n} \right) \;\] that is \[A_n^{(1)} = A_n - \frac{1}{n} \left( \frac{1}{n} - \frac{1}{m} \right) + \frac{1}{n} \left( \frac{1}{2n} - \frac{1}{m+n} \right) \; .\] We are interested in whether \[\frac{A_n^{(1)}}{A_n} \leq \alpha < 1 \quad \text{uniformly in } n\] for some \(\alpha < 1\) and all \(n, m\).
To analyze asymptotic behavior, fix \(m = n + 1\). Then \[d(T^k n, T^k m) = \left| \frac{1}{n+k} - \frac{1}{n+1+k} \right| = \frac{1}{n+k} - \frac{1}{n+k+1} = \frac{1}{(n+k)(n+k+1)} \; .\]
So \[A_n = \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{(n+k)(n+k+1)} \; ,\] observe that \(n+k\) runs from \(n\) to \(2n-1\). Thus, \[A_n = \frac{1}{n} \sum_{j=n}^{2n-1} \frac{1}{j(j+1)} \; .\] Now use \[\frac{1}{j(j+1)} = \frac{1}{j} - \frac{1}{j+1} \Rightarrow \sum_{j=n}^{2n-1} \left( \frac{1}{j} - \frac{1}{j+1} \right) = \frac{1}{n} - \frac{1}{2n} = \frac{1}{2n} \;,\] so \[A_n = \frac{1}{n} \cdot \frac{1}{2n} = \frac{1}{2n^2} \; .\] Compute \(A_n^{(1)} = \frac{1}{n} \sum_{k=0}^{n-1} d(T^{k+1}n, T^{k+1}m) = \frac{1}{n} \sum_{k=1}^{n} \frac{1}{(n+k)(n+k+1)}\), where \(j = n+k\) runs from \(n+1\) to \(2n\), \[\sum_{j=n+1}^{2n} \frac{1}{j(j+1)} = \sum_{j=n+1}^{2n} \left( \frac{1}{j} - \frac{1}{j+1} \right) = \frac{1}{n+1} - \frac{1}{2n+1} \; .\] Thus \[A_n^{(1)} = \frac{1}{n} \left( \frac{1}{n+1} - \frac{1}{2n+1} \right) = \frac{1}{n} \left( \frac{(2n+1) - (n+1)}{(n+1)(2n+1)} \right) = \frac{1}{n} \cdot \frac{n}{(n+1)(2n+1)} = \frac{1}{(n+1)(2n+1)} \; .\] Now calculate the ratio \[\frac{A_n^{(1)}}{A_n} = \frac{ \frac{1}{(n+1)(2n+1)} }{ \frac{1}{2n^2} } = \frac{2n^2}{(n+1)(2n+1)} \; ,\] and simplify \[= \frac{2n^2}{2n^2 + n + 2n + 1} = \frac{2n^2}{2n^2 + 3n + 1} = \frac{2}{2 + \frac{3}{n} + \frac{1}{n^2}} \to 1 \quad \text{as } n \to \infty \; ,\] so we conclude that \[\lim_{n \to \infty} \frac{A_n^{(1)}}{A_n} = 1 \; .\] Therefore, for any \(\alpha < 1\), there exists \(n\) large enough such that: \[A_n^{(1)} > \alpha A_n \; ,\] hence, the PA-condition (?? ) \[\frac{1}{n} \sum_{k=0}^{n-1} d(T^{k+1}n, T^{k+1}m) \leq \alpha \cdot \frac{1}{n} \sum_{k=0}^{n-1} d(T^k n, T^k m)\] cannot hold uniformly for all \(n \in \mathbb{N}\) with a fixed \(\alpha < 1\). Thus, \(T\) is not a PA-contraction.
Remark 7 (Ćirić not PA-contraction). A mapping \(T: X \to X\) on a complete metric space \((X,d)\) is a Ćirić contraction if there exists \(k \in (0,1)\) such that for all \(x, y \in X\), \[d(Tx, Ty) \leq k \cdot \max\left\{ d(x,y),\; d(x,Tx),\; d(y,Ty),\; \frac{d(x,Ty) + d(y,Tx)}{2} \right\}. \label{ciric}\qquad{(5)}\] The PA-Contraction is defined in (?? ). We will use the same definitions of Example 6 and show that \(T\) is Ćirić but not PA.
\(T\) is a Ćirić contraction.
Let \(m, n \in \mathbb{N}\), \(m \ne n\). Without loss of generality, assume \(m < n\).
We compute \(d(Tm, Tn) = \left| \frac{1}{m+1} - \frac{1}{n+1} \right|\), \(d(m,n) = \left| \frac{1}{m} - \frac{1}{n} \right|\), \(d(m, Tm) = \left| \frac{1}{m} - \frac{1}{m+1} \right| = \frac{1}{m(m+1)}\), \(d(n, Tn) = \frac{1}{n(n+1)}\), \(d(m, Tn) = \left| \frac{1}{m} - \frac{1}{n+1} \right|\), \(d(n, Tm) = \left| \frac{1}{n} - \frac{1}{m+1} \right|\), \(\frac{d(m,Tn) + d(n,Tm)}{2}\).
We claim that \(d(Tm, Tn) \leq k \cdot \max\{ \dots \}\) for some \(k < 1\).
Consider the full max of (?? ).
For large \(m, n\), \(d(m,Tm) = \frac{1}{m(m+1)} \approx \frac{1}{m^2}\), very small, and \(d(m,n) \approx \frac{|m-n|}{m^2}\) if \(m \approx n\), also small. And \(d(Tm, Tn) \approx \frac{|m-n|}{(m+1)^2}\).
Now consider \(\frac{d(m,Tn) + d(n,Tm)}{2}\). If \(m < n\), \(Tn = n+1 < n\), so \(\frac{1}{m} > \frac{1}{n+1}\), so \[d(m,Tn) = \frac{1}{m} - \frac{1}{n+1}, \quad d(n,Tm) = \frac{1}{n} - \frac{1}{m+1} \quad \text{(if } m+1 < n\text{)} \; .\] So \[\begin{gather} \frac{d(m,Tn) + d(n,Tm)}{2} = \frac{1}{2} \left( \frac{1}{m} - \frac{1}{n+1} + \frac{1}{n} - \frac{1}{m+1} \right) = \frac{1}{2} \left( \left( \frac{1}{m} - \frac{1}{m+1} \right) + \left( \frac{1}{n} - \frac{1}{n+1} \right) \right) = \nonumber \\ \frac{1}{2} \left( \frac{1}{m(m+1)} + \frac{1}{n(n+1)} \right) \; , \end{gather}\] and this is small.
\(d(m,n) = \frac{1}{m} - \frac{1}{n} = \frac{n - m}{mn}\), and \(d(Tm,Tn) = \frac{n - m}{(m+1)(n+1)}\). So \[\frac{d(Tm,Tn)}{d(m,n)} = \frac{n - m}{(m+1)(n+1)} \cdot \frac{mn}{n - m} = \frac{mn}{(m+1)(n+1)} < 1\]
Let \(k = \sup \frac{mn}{(m+1)(n+1)}\). Since this \(< 1\) for all \(m,n\), and \(\to 1\) as \(m,n \to \infty\), we can pick any \(k \in (0,1)\), say \(k = 0.99\), and for all \(m,n\) such that \(\frac{mn}{(m+1)(n+1)} \leq k\), it holds.
For small \(m,n\), verify finitely many cases: since all values are bounded away from equality, we can always find a \(k < 1\) such that: \[d(Tm,Tn) \leq k \cdot d(m,n) \leq k \cdot M(m,n) \; .\]
Thus, \(T\) is a Ćirić contraction with \(k < 1\).
To prove that \(T\) is not a PA-Contraction, fix \(m, n \in \mathbb{N}\), \(m = n - 1\). The proof goes exactly as in Remark 6.
Whether PA-contractions imply Chatterjea or Ćirić contractions remains open. Conversely, Remarks 6 and 7 show they do not.
The following table summarizes key properties of major contraction types.
Remark 8. While we have shown that PA-contractions are not implied by classical types, the converse — whether PA-contractions imply other classes — is only partially resolved. The mapping \(T x = x^2/2\) shows that PA-contractions do not imply Kannan or F-contractions. Whether every F-contraction is a PA-contraction, and if every PA-contraction is a Chatterjea or Ćirić contraction remains open.
We introduced PA-contractions, a new class based on averaged contraction over orbits. This condition strictly generalizes Banach contractions, is independent of F-contractions, Kannan, Chatterjea, and Ćirić types, ensures unique fixed points under continuity. Future investigation could go in the directions of solving open problems, multivalued mappings and common fixed points.
PA-contractions offer a new lens for analyzing asymptotic behavior in fixed point theory. Our work continues the long tradition of generalizing the Banach principle, as studied by Rhoades [8] and extended by Wardowski [9], Proinov [17], Ran and Reurings [18], Alam and Imdad [19], and others.
MSC2020: 47H10, 54E50
Keywords: Path Averaged Contractions, contractive mapping, fixed point, complete metric space.↩︎