Fixed points of a new type of contractive mappings in complete metric spaces

Abstract

In the article, we introduce a new concept of contraction and prove a fixed point theorem which generalizes Banach contraction principle in a different way than in the known results from the literature. The article includes an example which shows the validity of our results, additionally there is delivered numerical data which illustrates the provided example.

MSC: 47H10; 54E50

1 Introduction

Throughout the article denoted by ℝ is the set of all real numbers, by ℝ₊ is the set of all positive real numbers and by ℕ is the set of all natural numbers. (X, d), (X for short), is a metric space with a metric d.

In the literature, there are plenty of extensions of the famous Banach contraction principle [1], which states that every self-mapping T defined on a complete metric space (X, d) satisfying

$${\forall }_{x,y\in X}\mathsf{\text{d}}\left(Tx,Ty\right)\le \lambda \mathsf{\text{d}}\left(x,y\right),\mathsf{\text{ where }}\lambda \in \left(0,1\right),$$

(1)

has a unique fixed point and for every x₀ ∈ X a sequence {Tⁿx₀}_n∈ℕis convergent to the fixed point. Some of the extensions weaken right side of inequality in the condition (1) by replacing λ with a mapping, see e.g. [2, 3]. In other results, the underlying space is more general, see e.g [4–7]. The Nadler's paper [8] started the invatigations concerning fixed point theory for set-valued contractions, see e.g. [9–20]. There are many theorems regarding asymptotic contractions, see e.g. [21–23], contractions of Meir-Keeler type [24], see e.g [19, 23, 25] and weak contractions, see e.g. [26–28]. There are also lots of different types of fixed point theorems not mentioned above extending the Banach's result.

In the present article, using a mapping F: ℝ⁺ → ℝ we introduce a new type of contraction called F-contraction and prove a new fixed point theorem concerning F-contraction. For the concrete mappings F, we obtain the contractions of the type known from the literature, Banach contraction as well. The article includes the examples of F-contractions and an example showing that the obtained extension is significant. Theoretical considerations that we support by computational data illustrate the nature of F-contractions.

2 The result

Definition 2.1 Let F: ℝ₊ → ℝ be a mapping satisfying:

(F1) F is strictly increasing, i.e. for all α, β ∈ ℝ₊ such that α < β, F (α) < F (β);

(F2) For each sequence {α _n }_n∈ℕof positive numbers lim_n→∞α _n = 0 if and only if lim_n→∞F (α _n ) = -∞;

(F3) There exists k ∈ (0, 1) such that lim_α→0+ α^kF(α) = 0.

A mapping T: X → X is said to be an F-contraction if there exists τ > 0 such that

$${\forall }_{x,y\in X}\left(\mathsf{\text{d}}\left(Tx,Ty\right)>0\Rightarrow \tau +F\left(\mathsf{\text{d}}\left(Tx,Ty\right)\right)\le F\left(\mathsf{\text{d}}\left(x,y\right)\right)\right).$$

(2)

When we consider in (2) the different types of the mapping F then we obtain the variety of contractions, some of them are of a type known in the literature. See the following examples:

Example 2.1 Let F : ℝ₊ → ℝ be given by the formula F (α) = ln α. It is clear that F satisfies (F1)-(F3) ((F3) for any k ∈ (0, 1)). Each mapping T : X → X satisfying (2) is an F-contraction such that

$$\mathsf{\text{d}}\left(Tx,Ty\right)\le e^{-\tau }\mathsf{\text{d}}\left(x,y\right),\mathsf{\text{for all }}x,y\in X,Tx\ne Ty.$$

(3)

It is clear that for x, y ∈ X such that Tx = Ty the inequality d(Tx, Ty) ≤ e^-τd(x, y) also holds, i.e. T is a Banach contraction [1].

Example 2.2 If F(α) = ln α + α, α > 0 then F satisfies (F1)-(F3) and the condition (2) is of the form

$$\frac{\mathsf{\text{d}}\left(Tx,Ty\right)}{\mathsf{\text{d}}\left(x,y\right)}{e}^{\mathsf{\text{d}}\left(Tx,Ty\right)-\mathsf{\text{d}}\left(x,y\right)}\le e^{-\tau },\mathsf{\text{for all}}x,y\in X,Tx\ne Ty.$$

(4)

Example 2.3 Consider $F\left(\alpha \right)=-1/\sqrt{\alpha }$, α > 0. F satisfies (F1)-(F3) ((F3) for any k ∈ (1/2, 1)). In this case, each F-contraction T satisfies

$$\mathsf{\text{d}}\left(Tx,Ty\right)\le \frac{1}{{\left(1+\tau \sqrt{\mathsf{\text{d}}\left(x,y\right)}\right)}^2}\mathsf{\text{d}}\left(x,y\right),\mathsf{\text{for all }}x,y\in X,Tx\ne Ty.$$

Here, we obtained a special case of nonlinear contraction of the type d(Tx, Ty) ≤ α(d(x, y))d(x, y). For details see [2, 3].

Example 2.4 Let F(α) = ln(α² + α), α > 0. Obviously F satisfies (F1)-(F3) and for F-contraction T, the following condition holds:

$$\frac{\mathsf{\text{d}}\left(Tx,Ty\right)\left(\mathsf{\text{d}}\left(Tx,Ty\right)+1\right)}{\mathsf{\text{d}}\left(x,y\right)\left(\mathsf{\text{d}}\left(x,y\right)+1\right)}\le e^{-\tau },\mathsf{\text{for all}}x,y\in X,Tx\ne Ty.$$

Let us observe that in Examples 2.1-2.4 the contractive conditions are satisfied for x, y ∈ X, such that Tx = Ty.

Remark 2.1 From (F1) and (2) it is easy to conclude that every F-contraction T is a contractive mapping, i.e.

$$\mathsf{\text{d}}\left(Tx,Ty\right)<\mathsf{\text{d(}}x\mathsf{\text{, }}y\mathsf{\text{),}}\mathsf{\text{for all }}x\mathsf{\text{, }}y\in X,Tx\ne Ty.$$

Thus every F-contraction is a continuous mapping.

Remark 2.2 Let F₁, F₂ be the mappings satisfying (F1)-(F3). If F₁(α) ≤ F₂(α) for all α > 0 and a mapping G = F₂ - F₁ is nondecreasing then every F₁-contraction T is F₂-contraction.

Indeed, from Remark 2.1 we have G(d(Tx, Ty)) ≤ G(d(x, y)) for all x, y ∈ X, Tx ≠ Ty. Thus, for all x, y ∈ X, Tx ≠ Ty we obtain

$$\begin{array}{ll}\hfill \tau +F_2\left(\mathsf{\text{d}}\right(Tx,Ty\left)\right)& =\tau +F_1\left(\mathsf{\text{d}}\right(Tx,Ty\left)\right)+G\left(\mathsf{\text{d}}\right(Tx,Ty\left)\right)\\ \le F_1\left(\mathsf{\text{d}}\right(x,y\left)\right)+G\left(\mathsf{\text{d}}\right(x,y\left)\right)=F_2\left(\mathsf{\text{d}}\right(x,y\left)\right).\end{array}$$

Now we state the main result of the article.

Theorem 2.1 Let (X, d) be a complete metric space and let T : X → X be an F-contraction. Then T has a unique fixed point x* ∈ X and for every x₀ ∈ X a sequence {Tⁿx₀}_n∈ℕis convergent to x*.

Proof. First, let us observe that T has at most one fixed point. Indeed, if $x_1^{*}$, $x_2^{*}\in X$, $T{x}_1^{*}=x_1^{*}\ne x_2^{*}=T{x}_2^{*}$, then we get

$$\tau \le F\left(\mathsf{\text{d}}\left(x_1^{*},x_2^{*}\right)\right)-F\left(\mathsf{\text{d}}\left(T{x}_1^{*},T{x}_2^{*}\right)\right)=0,$$

which is a contradiction.

In order to show that T has a fixed point let x₀ ∈ X be arbitrary and fixed. We define a sequence {x _n }_n∈ℕ⊂ X, x_n+1= Tx _n , n = 0, 1, .... Denote γ _n = d(x_n+1, x _n ), n = 0, 1, ....

If there exists n₀ ∈ ℕ for which $x_{n_0+1}=x_{n_0}$, then $T{x}_{n_0}=x_{n_0}$ and the proof is finished.

Suppose now that x_n+1≠ x _n , for every n ∈ ℕ. Then γ _n > 0 for all n ∈ ℕ and, using (2), the following holds for every n ∈ ℕ:

$$F\left({\gamma }_n\right)\le F\left({\gamma }_{n-1}\right)-\tau \le F\left({\gamma }_{n-2}\right)-2\tau \le \dots \le F\left({\gamma }_0\right)-n\tau .$$

(5)

From (5), we obtain lim_n→∞F(γ _n ) = -∞ that together with (F2) gives

$$\underset{n\to \infty }{\mathsf{\text{lim}}}{\gamma }_n=0.$$

(6)

From (F3) there exists k ∈ (0, 1) such that

$$\underset{n\to \infty }{\mathsf{\text{lim}}}{{\gamma }_n}^{k}F\left({\gamma }_n\right)=0.$$

(7)

By (5), the following holds for all n ∈ ℕ:

$${{\gamma }_n}^{k}F\left({\gamma }_n\right)-{{\gamma }_n}^{k}F\left({\gamma }_0\right)\le {{\gamma }_n}^k\left(F\left({\gamma }_0\right)-n\tau \right)-{{\gamma }_n}^{k}F\left({\gamma }_0\right)=-{{\gamma }_n}^{k}n\tau \le 0.$$

(8)

Letting n → ∞ in (8), and using (6) and (7), we obtain

$$\underset{n\to \infty }{\mathsf{\text{lim}}}n{{\gamma }_n}^k=0.$$

(9)

Now, let us observe that from (9) there exists n₁ ∈ ℕ such that $n{{\gamma }_n}^k\le 1$ for all n ≥ n₁. Consequently we have

$${\gamma }_n\le \frac{1}{n^{1/k}},\mathsf{\text{for all}}n\ge n_1.$$

(10)

In order to show that {x _n }_n∈ℕis a Cauchy sequence consider m, n ∈ ℕ such that m > n ≥ n₁. From the definition of the metric and from (10) we get

$\mathsf{\text{d}}\left(x_m,x_n\right)\le {\gamma }_{m-1}+{\gamma }_{m-2}+\cdots +{\gamma }_n<\sum _{i=n}^{\infty }{\gamma }_i\le \sum _{i=n}^{\infty }\frac{1}{i^{1/k}}.$

From the above and from the convergence of the series ${\sum }_{i=1}^{\infty }1/i^{\frac{1}{k}}$ we receive that {x _n }_n∈ℕis a Cauchy sequence.

From the completeness of X there exists x* ∈ X such that lim_n→∞x _n = x*. Finally, the continuity of T yields

$$\mathsf{\text{d}}\left(T{x}^{*},x^{*}\right)=\underset{n\to \infty }{\mathsf{\text{lim}}}\mathsf{\text{d}}\left(T{x}_n,x_n\right)=\underset{n\to \infty }{\mathsf{\text{lim}}}\mathsf{\text{d}}\left(x_{n+1},x_n\right)=0,$$

which completes the proof. □

Note that for the mappings F₁(α) = ln(α), α > 0, F₂(α) = ln(α) + α, α > 0, F₁< F₂ and a mapping F₂ - F₁ is strictly increasing. Hence, by Remark 2.2, we obtain that every Banach contraction (3) satisfies the contraction condition (4). On the other side in Example 2.5, we present a metric space and a mapping T which is not F₁-contraction (Banach contraction), but still is an F₂-contraction. Consequently, Theorem 2.1 gives the family of contractions which in general are not equivalent.

Example 2.5 Consider the sequence {S _n }_n∈ℕas follows:

$$\begin{array}{l}{S}_1=1,\\ S_2=1+2,\\ \dots \\ S_n=1+2+\cdots +n=\frac{n\left(n+1\right)}{2},n\in \mathbb{N},\\ \dots \end{array}$$

Let X = {S _n : n ∈ ℕ} and d(x, y) = |x - y|, x, y ∈ X. Then (X, d) is a complete metric space. Define the mapping T : X → X by the formulae:

$$\begin{array}{l}T\left(S_n\right)=S_{n-1}\mathsf{\text{for}}n>1,\\ T\left(S_1\right)=S_1.\end{array}$$

First, let us consider the mapping F₁ defined in Example 2.1. The mapping T is not the F₁-contraction in this case (which actually means that T is not the Banach contraction). Indeed, we get

$$\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{\mathsf{\text{d}}\left(T\right(S_n),T(S_1\left)\right)}{\mathsf{\text{d}}\left(S_n,S_1\right)}=\underset{n\to \infty }{\mathsf{\text{lim}}}\frac{S_{n-1}-1}{S_n-1}=1.$$

On the other side taking F₂ as in Example 2.2, we obtain that T is F₂-contraction with τ = 1. To see this, let us consider the following calculations:

First, observe that

$${\forall }_{m,n\in \mathbb{N}}\left[T\left(S_m\right)\ne T\left(S_n\right)\iff \left(\left(m>2\wedge n=1\right)\vee \left(m>n>1\right)\right)\right].$$

For every m ∈ ℕ, m > 2 we have

$$\begin{array}{ll}\hfill \frac{\mathsf{\text{d}}\left(T\right(S_m),T(S_1\left)\right)}{\mathsf{\text{d}}(S_m,S_1)}{e}^{\mathsf{\text{d}}\left(T\left(S_m\right),T\left(S_1\right)\right)-\mathsf{\text{d}}\left(S_m,S_1\right)}& =\frac{S_{m-1}-1}{S_m-1}{e}^{S_{m-1}-S_m}\\ =\frac{m^2-m-2}{m^2+m-2}{e}^{-m}<e^{-m}<e^{-1}.\end{array}$$

For every m, n ∈ ℕ, m > n > 1 the following holds

$$\begin{array}{ll}\hfill \frac{\mathsf{\text{d}}\left(T\right(S_m),T(S_n\left)\right)}{\mathsf{\text{d}}(S_m,S_n)}{e}^{\mathsf{\text{d}}\left(T\left(S_m\right),T\left(S_n\right)\right)-\mathsf{\text{d}}\left(S_m,S_n\right)}& =\frac{S_{m-1}-S_{n-1}}{S_m-S_n}{e}^{S_n-S_{n-1}+S_{m-1}-S_m}\\ =\frac{m+n-1}{m+n+1}{e}^{n-m}<e^{n-m}\le e^{-1}.\end{array}$$

Clearly S₁ is a fixed point of T. To see the computational data confirming the above calculations the reader is referred to Table 1.

Table 1 The comparison of Banach contraction condition with F-contraction condition