Fixed Point Theorems for a System of Mappings in Generalized b-Metric Spaces

Abstract

In this paper we present some results of both global and local type on the existence of fixed points for a system of mappings in generalized b-metric spaces. In particular, we obtain a strict generalization of the Banach contraction principle for mappings in ordinary complete metric spaces.

1 Introduction

We shall utilize the idea of a generalized b-metric space (gbms shortly). For details see [1]. Assume that X is a nonempty set. A function d : X × X → [0, ∞] is said to be a generalized b-metric on X, iff for x, y, z ∈ X it holds:

• 1. d(x, y) = 0 iff x = y,

• 2. d(x, y) = d(y, x),

• 3. $d(x,y) \leqslant s[d(x,z) + d(z,y)]$, where $s \geqslant 1$ is a fixed real constant.

A pair (X, d) is called a generalized b-metric space with a generalized b-metric d.

For any f : X → X, by f ⁿ we denote the n-th iterate of f, defined by

$$\displaystyle \begin{aligned} f^0(x)=x \text{ for } x \in X, \quad f^{n+1} = f(f^n), \quad n \in \mathbb{N}_0. \end{aligned}$$

By $\mathbb {N}_0, \mathbb {N}, \mathbb {R}, \mathbb {R}_{+}$ as usual, we denote the set of all nonnegative integers, the set of all natural numbers, the set of all real numbers, or the set of all nonnegative real numbers, respectively.

One can find more about b-metric spaces, for example, in [4, 5] (see also [13, 14] for related topics).

Let’s note also the following (see [1, 6]) theorem, which will be used later on.

Theorem 1

Let (X, d) be a complete b-metric space and T : X → X satisfy

$$\displaystyle \begin{aligned} d(T(x),T(y)) \leqslant \varphi [d(x,y)], \quad x,y \in X, \end{aligned} $$

where φ : R ₊ → R ₊ is a nondecreasing function such that

$$\displaystyle \begin{aligned} \lim_{n \to \infty} \varphi^n(t) = 0 \end{aligned} $$

for each t > 0. Then T has exactly one fixed point u ∈ X, and

$$\displaystyle \begin{aligned} \lim_{n \to \infty} d[T^n(x), u] = 0 \end{aligned}$$

for each x ∈ X.

2 Main Results

2.1 Basic Theorem

In this section we prove the result for a system of mappings in generalized b-metric spaces (we apply the ideas of [7]).

Theorem 2

Let (X _i, d _i), i = 1, …, n be complete generalized b-metric spaces. Assume that there exist nonnegative real numbers a _i,k, i, k = 1, …, n such that the mappings T _i : X ₁ ×… × X _n → X _i, i = 1, …, n satisfy

$$\displaystyle \begin{aligned} d_i(T_i(x_1, \ldots, x_n), T_i(z_1,\ldots,z_n)) \leqslant \displaystyle\sum_{k=1}^{n} a_{i,k} d_k (x_k, z_k), \end{aligned} $$

(1)

for x _k, z _k ∈ X _k, d _k(x _k, z _k) < ∞, k = 1, …, n.

Moreover, there exists a system of positive real numbers r _i, i = 1, …, n satisfying the inequalities

$$\displaystyle \begin{aligned} \displaystyle\sum_{i=1}^{n} r_i a_{i,k} < r_k, \quad k=1,\ldots,n. \end{aligned} $$

(2)

For any fixed x ⁰ ∈ X = X ₁ ×… × X _n consider the sequence of successive approximations

$$\displaystyle \begin{aligned} x_i^{m+1} = T_i (x_1^m, \ldots, x_n^m), \quad m=0,1,\ldots, \quad i=1,\ldots,n. \end{aligned} $$

(3)

Then either

• (A) for any non-negative integer v there exists an i ∈{1, …, n} such that

  $$\displaystyle \begin{aligned} d_i(x_i^v, T_i(x_1^v, \ldots, x_n^v)) = \infty, \end{aligned} $$

  or

• (B) there exists a non-negative integer v such that for every i = 1, …, n,

  $$\displaystyle \begin{aligned} d_i (x_i^v, x_i^{v+1}) < \infty. \end{aligned} $$

  (4)

  In (B) the sequence $x^m = (x_1^m, \ldots , x_n^m)$ given by (3) converges to a fixed point u = (u ₁, …, u _n) ∈ X of T = (T ₁, …, T _n), i.e.

  $$\displaystyle \begin{aligned} T_i(u_1, \ldots, u_n) = u_i, i=1,\ldots, n. \end{aligned}$$

  In the space K = K ₁ ×… × K _n where

  $$\displaystyle \begin{aligned} K_i = \{x_i \in X_i : d_i(x_i^v, x_i) < \infty \}, \quad i=1,\ldots,n, \end{aligned} $$

  (5)

  the point u is the unique fixed point of T.

Proof

In (B) by (4) we get for i = 1, …, n,

$$\displaystyle \begin{aligned}d_i(T_i(x^v), T_i(x^{v+1})) \leqslant \displaystyle\sum_{k=1}^{n} a_{i,k} d_k (x_k^v, x_k^{v+1}) < \infty, \end{aligned} $$

and consequently by induction $d_i(x_i^{v+l}, x_i^{v+l+1}) < \infty$ for all

$l \in \mathbb {N}_0, \quad i=1,\ldots ,n.$

Consider the number

$$\displaystyle \begin{aligned} \alpha = \underset{k}{\max} \Big\{ \frac{1}{r_k} \displaystyle\sum_{i=1}^{n} r_i a_{i,k} \Big\}. \end{aligned} $$

Clearly

$$\displaystyle \begin{aligned} 0 \leqslant \alpha < 1\end{aligned} $$

(6)

and

$$\displaystyle \begin{aligned} \displaystyle\sum_{i=1}^{n} r_i a_{i,k} \leqslant \alpha r_k, \quad k=1,\ldots,n. \end{aligned} $$

(7)

Now we verify that T : K → K. For if x ∈ K,

$$\displaystyle \begin{aligned} \begin{aligned} d_k(x_k^v, T_k(x)) \leqslant s_k[d_k(x_k^v, T_k(x^v)) + d_k(T_k(x^v), T_k(x))] \\ \leqslant s_k [d_k(x_k^v, x_k^{v+1}) + \displaystyle\sum_{l=1}^{n} a_{k,l} d_l(x_l^v, x_l)] < \infty. \end{aligned} \end{aligned}$$

Define

$$\displaystyle \begin{aligned} D(x,y) := \displaystyle\sum_{i=1}^{n} r_i d_i (x_i, y_i), \quad x,y \in K. \end{aligned} $$

(8)

We can show that D is a b-metric in K with $s = \underset {i}{\max } (s_i) > 0$. It is also easy to prove that (K, D) is a complete b-metric space (see also [7]).

Now we prove that T is a contraction mapping in K. In fact, for x, z ∈ K we obtain

$$\displaystyle \begin{aligned} \begin{aligned} D(T(x), T(z)) = \displaystyle\sum_{i=1}^{n} r_i d_i (T_i(x), T_i(z)) \\ \leqslant \displaystyle\sum_{i=1}^{n} r_i \Big[ \sum_{k=1}^{n} a_{i,k} d_k (x_k, z_k) \Big] \\ \leqslant \displaystyle\sum_{k=1}^{n} \Big( \sum_{i=1}^{n} r_i a_{i,k} \Big) d_k(x_k, z_k) \\ \leqslant \displaystyle\sum_{k=1}^{n} \alpha r_k d_k (x_k, z_k) = \alpha D(x,z), \end{aligned} \end{aligned}$$

which means that

$$\displaystyle \begin{aligned} D(T(x), T(z)) \leqslant \alpha D(r, z), \quad x,z \in K. \end{aligned}$$

Since by (6), $0 \leqslant \alpha <1$, T is a strict contraction in K.

Eventually, in view of Theorem 1 for $\varphi (t) = \alpha t, \quad t \geqslant 0$, T has in K exactly one fixed point u which is the limit of successive approximations with any initial element from K (and hence T has a fixed point in X). This concludes the proof.

We can also prove the following.

Corollary 1

Let the assumptions of Theorem 2be satisfied. If, moreover,

$$\displaystyle \begin{aligned} s \alpha < 1, \end{aligned} $$

(9)

then

$$\displaystyle \begin{aligned} D(y,u) \leqslant \frac{s}{1-s\alpha} D(y, T(y)), \quad y \in B. \end{aligned} $$

(10)

Proof

For (B) and y ∈ K, (u = T(u)) one has

$$\displaystyle \begin{aligned} \begin{aligned} D(y, u) \leqslant s [D(y,T(y)) + D(T(y),T(u))] \\ \leqslant s [D(y,T(y)) + \alpha D(y,u)] \end{aligned} \end{aligned}$$

whence

$$\displaystyle \begin{aligned} D(y,u) \leqslant \frac{s}{1-s\alpha} D(y, T(y)), \quad y \in K. \end{aligned}$$

Corollary 2

Let the assumptions of Corollary 1 be satisfied. If, moreover, d _i, i = 1, …, n, are continuous (with respect to one variable), then

$$\displaystyle \begin{aligned} D(T^m(y),u) \leqslant \frac{s \alpha^m}{1-s\alpha} D(y, T(y)), \quad y \in B. \end{aligned} $$

(11)

Proof

From Theorem 1, for $\varphi (t) = \alpha t, \quad t \geqslant 0, \quad z \in K$ and continuity of D, one gets

$$\displaystyle \begin{aligned} \begin{aligned} D(T^m(z), u) \leqslant \sum_{k=0}^{\infty} s^{k+1} \varphi^{m+k} [D(z,T(z))] \\ = \sum_{k=0}^{\infty} s^{k+1} \alpha^{m+k} D(z, T(z)) = \frac{s \alpha^{m}}{1-s\alpha} D(z, T(z)), \end{aligned} \end{aligned}$$

i.e., the inequality (11), which completes the proof.

Remark 1

A function D (b-metric) may not be continuous (cf. [15]).

Remark 2

From Theorem 2 we get theorem of Diaz, Margolis [8], Luxemburg [10], Banach [2], Matkowski [11], Czerwik [4, 5].

2.2 Local Theorems

First of all we present the local result for a system of mappings in generalized b-metric spaces. Namely, we have

Theorem 3

Suppose that (X _i, d _i), i = 1, …, m are complete generalized b-metric spaces. Assume that there exist non-negative real numbers a _i,k, i, k = 1, …, n and c > 0 such that the mappings T _i : X ₁ ×… × X _n → X _i, i = 1, …, n fulfill the inequalities

$$\displaystyle \begin{aligned} d_i(T_i(x_1,\ldots,x_n), T_i(z_1,\ldots,z_n) \leqslant \sum_{k=1}^{n} a_{i,k} d_k (x_k, z_k) \end{aligned} $$

(12)

for d _k(x _k, z _k) < c, and x _k, z _k ∈ X _k, k = 1, …, n. Additionally, let the characteristic roots d _i, i = 1, …, n of the matrix [a _i,k] satisfy

$$\displaystyle \begin{aligned} \alpha = \max \{ |d_i| : i=1,\ldots,n \} < 1. \end{aligned} $$

(13)

Let x ⁰ ∈ X = X ₁ ×…X _n be arbitrarily fixed. Consider the sequence of successive approximations (3). Then the following alternative holds: either

• (C) for any $v \in \mathbb {N}_0$ there exists an i ∈{1, …, n} = A such that

  $$\displaystyle \begin{aligned} d_i(x_i^v, T_i(x_i^v, \ldots, x_n^v)) \geqslant c, \end{aligned} $$

  (14)

or

• (D) there exists a non-negative integer v such that for every i = 1, …, n,

  $$\displaystyle \begin{aligned} d_i(x_i^v, x_i^{v+1}) < c. \end{aligned} $$

  (15)

In the case (D), if, moreover, the numbers in (15) are sufficiently small and (9) holds true, then T = (T ₁, …, T _n) has a fixed point u ∈ X.

Proof

According to the result of Perron-Frobenius ([9, pp. 354–355]) the number given in (13) is the characteristic root of the matrix [a _i,k] with the eigenvector (r ₁, …, r _n), r _i > 0, i = 1, …, n, i.e.

$$\displaystyle \begin{aligned} \displaystyle\sum_{k=1}^{n} a_{i,k} r_k = \alpha r_i, \quad i=1,\ldots,n. \end{aligned} $$

(16)

Since the equations (16) are homogeneous, also ar ₁, …, ar _n, a > 0 is a solution of (16). So we can assume that

$$\displaystyle \begin{aligned} r_i < \frac{c}{2s}, \quad i=1,\ldots,n. \end{aligned} $$

(17)

By the assumptions, suppose that

$$\displaystyle \begin{aligned} d_k(T_k(x^v), x_k^v) \leqslant (1 - \alpha s_k) \frac{r_k}{s_k}, \quad k=1,\ldots,n. \end{aligned} $$

(18)

Define

$$\displaystyle \begin{aligned} B_k := \{ z_k \in X_k: d_k(z_k, x_k^v) \leqslant r_k \}, \quad k=1,\ldots,n. \end{aligned} $$

(19)

Let B = B ₁ ×… × B _k. For z ∈ B, by (12), (17), (19), (16) and (18), we get (k = 1, …, n):

$$\displaystyle \begin{aligned} \begin{aligned} d_k (T_k(z), x_k^v) \leqslant s_k [ d_k(T_k(z), T_k(x^v)) + d_k( T_k(x^v), x_k^v) ] \\ \leqslant s_k \Big[ \sum_{l=1}^n a_{k,l} d_l (z_l, x_l^v) + d_k(T_k(x^v), x_k^v) \Big] \\ \leqslant s_k \Big[ \alpha r_k + \frac{r_k}{s_k} - \alpha r_k \Big] = r_k. \end{aligned} \end{aligned}$$

Thus T _k(z) ∈ B _k, k = 1, …, n which means that T(z) ∈ B for z ∈ B, i.e. T(B) ⊂ B.

Note that also the matrix [a _i,k]^T has exactly the same characteristic roots as [a _i,k], so by the Perron-Frobenius theorem there exists a system of positive numbers ξ _i, i = 1, …, n, which is the solution of the system of equations (inequalities)

$$\displaystyle \begin{aligned} \sum_{i=1}^{n} \xi_i a_{i,k} \leqslant \alpha \xi_k, \quad k=1,\ldots,n. \end{aligned} $$

(20)

Define

$$\displaystyle \begin{aligned} D(x,y) := \sum_{i=1}^{n} \xi_i d_i (x_i,y_i), \quad x,y \in B \end{aligned} $$

(21)

and consider the space (B, D). Then (B, D) is a b-metric space with $s = \underset {i}{\max } (s_i)$. For any x, y ∈ B, we have for k = 1, …, n (see (17))

$$\displaystyle \begin{aligned} \begin{aligned} d_k(x_k,y_k) \leqslant s_k [ d_k(x_k, x_k^v) + d_k(x_k^v, y_k)] \\ \leqslant s_k [r_k + r_k] < 2s_k \frac{c}{2s} \leqslant c. \end{aligned} \end{aligned}$$

Therefore, by (12), (21), and (20), for x, y ∈ B, we can verify that (see the proof of Theorem 2):

$$\displaystyle \begin{aligned} D(T(x),T(y)) \leqslant \alpha D(x,y) \mathrm{ for} x,y \in B. \end{aligned} $$

(22)

Consequently for any x ∈ B, {T ⁿ(x)}⊂ B that is a Cauchy sequence. Indeed, we have

$$\displaystyle \begin{aligned} D(T(x),T^2(x)) \leqslant \alpha D(x, T(x)), \end{aligned}$$

and by the induction principle

$$\displaystyle \begin{aligned} D(T^m(x),T^{m+1}(x)) \leqslant \alpha^m D(x,T(x)), \quad m \geqslant 1. \end{aligned} $$

(23)

Consequently, for $m,l \in \mathbb {N}_0$, by (23) and (9) one gets

$$\displaystyle \begin{aligned} \begin{aligned} D(T^m(x),T^{m+l}(x)) \leqslant sD(T^m(x),T^{m+1}(x)) + \ldots + s^l D(T^{m+l-1}(x),T^{m+l}(x)) \\ \leqslant s \alpha^m D(x,T(x)) + \ldots + s^l \alpha^{m+l-1} D(x, T(x)) \\ \leqslant s \alpha^m [1 + (s \alpha) + \ldots + (s \alpha)^{l-1}] D(x, T(x)) \\ \leqslant \frac{s \alpha^m}{1 - s \alpha} D(x, T(x)). \end{aligned} \end{aligned}$$

Eventually, for x ∈ B and $m,l \in \mathbb {N}_0$,

$$\displaystyle \begin{aligned} D(T^m(x),T^{m+l}(x)) \leqslant \frac{s \alpha^m}{1 - s \alpha} D(x,T(x)). \end{aligned} $$

(24)

Hence, it follows that {T ⁿ(x)} is a Cauchy sequence for x ∈ B.

Since X _k, k = 1, …, n are complete, there exist u _k ∈ X _k, k = 1, …, n such that

$$\displaystyle \begin{aligned} T_k^m(x) \to u_k, \quad k=1,\ldots,n \mathrm{ as} m \to \infty. \end{aligned}$$

Therefore for i = 1, …, n, m sufficiently large and 𝜖 > 0

$$\displaystyle \begin{aligned} \begin{aligned} d_i(T_i(u), u_i) \leqslant s_i [d_i(T_i(u), T_i(x^m)) + d_i(T_i(x^m), u_i)] \\ \leqslant s_i \Big[ \sum_{k=1}^{n} a_{i,k} d_k (u_k, x_k^m) + d_i(x_i^{m+1}, u_i) \Big] \\ \leqslant s_i \Big[ \sum_{k=1}^{n} a_{i,k} \epsilon + \epsilon \Big] \\ \leqslant s \epsilon \Big[ \sum_{k=1}^{n} a_{i,k} + 1 \Big] \to 0 \mathrm{ as} \epsilon \to 0. \end{aligned} \end{aligned}$$

Therefore we get for i = 1, …, n the following:

$$\displaystyle \begin{aligned} d_i(T_i(u), u_i) = 0 \Rightarrow T_i(u) = u_i \Rightarrow T(u) = u, \end{aligned}$$

which means that u ∈ X is a fixed point of T, and this concludes the proof of the theorem.

Remark 3

The point u = (u ₁, …, u _n) satisfies the condition

$$\displaystyle \begin{aligned} d_k(u_k, x_k^v) \leqslant s_k r_k, \quad k=1,\ldots,n. \end{aligned} $$

(25)

In fact, we have for $m \in \mathbb {N}$ sufficiently large, 𝜖 > 0 and x ∈ B:

$$\displaystyle \begin{aligned} \begin{aligned} d_k(u_k, x_k^v) \leqslant s_k [ d_k(u_k, T_k^m(\alpha) + d_k(T_k^m(\alpha), x_k^v)] \\ \leqslant s_k [ \epsilon + r_k] \to s_k r_k \mathrm{ as} \epsilon \to 0, \end{aligned} \end{aligned}$$

whence we get (25).

Remark 4

It is an open question whether this result is true without the assumption that T does not displace the center of the ball B too far.

Remark 5

In B, the mapping T may have at most one fixed point. Indeed, assume, on the contrary, that u, w ∈ B, u ≠ w, and T(u) = u, T(w) = w. Note that (see (17))

$$\displaystyle \begin{aligned} d_k(u_k,w_k) < c, \quad k=1,\ldots,n, \end{aligned}$$

and hence by (12) and (16), we obtain for k = 1, …, n and

$$\displaystyle \begin{aligned} \begin{aligned} \xi = \underset{k}{\max} \Big(\frac{1}{r_k} d_k(u_k,w_k)\Big), \\ d_k(u_k,w_k) = d_k(T_k(u), T_k(w)) \leqslant \sum_{l=1}^{n} a_{k,l} d_l (u_l, w_l) \\ \leqslant \sum_{l=1}^{n} a_{k,l} r_l \cdot \frac{1}{r_l} d_l(u_l,w_l) \leqslant \xi \sum_{l=1}^{n} a_{k,l} r_l \leqslant \xi \alpha r_k. \end{aligned} \end{aligned}$$

Similarly,

$$\displaystyle \begin{aligned} d_k(u_k,w_k) = \sum_{l=1}^{n} a_{k,l} d_l (u_l, w_l) \leqslant \sum_{l=1}^{n} a_{k,l} \xi \alpha r_l \leqslant \xi \alpha^2 r_k.\end{aligned}$$

By induction one gets

$$\displaystyle \begin{aligned} d_k(u_k,w_k) \leqslant \xi \alpha^m r_k, \quad m \in \mathbb{N}, \quad k=1,\ldots,n.\end{aligned}$$

Thus, in view of (13), one has

$$\displaystyle \begin{aligned} d_k(u_k,w_k) = 0, \quad k=1,\ldots,n \Rightarrow u=w,\end{aligned}$$

and the proof is completed.

Remark 6

For some conditions equivalent to the condition (13) cf. [3, 12].

In the sequel we prove a similar result for systems of mappings in b-metric spaces.

Theorem 4

Let (X _i, d _i), i = 1, …, n be complete b-metric spaces with $s_i \geqslant 1,$ i = 1, …, n. Suppose that $a_{i,k} \geqslant 0$, i, k = 1, …, n and the characteristic roots λ _i, i = 1, …, n of the matrix [a _i,k] satisfy (13). Consider any

$$\displaystyle \begin{aligned}x^0 \in X_1 \times \ldots \times X_n = X\end{aligned}$$

and set

$$\displaystyle \begin{aligned} B_k := \{ z_k \in X_k: d_k (z_k, x_k^0) \leqslant r_k \}, \quad k=1,\ldots,n, \end{aligned}$$

and B = B ₁ ×… × B _n, where r _k > 0, k = 1, …, n satisfy (16).

Assume that T _k : B → X _k, k = 1, …, n satisfy

$$\displaystyle \begin{aligned} d_k(T_k(x), T_k(y)) \leqslant \sum_{l=1}^{n} a_{k,l} d_l (x_l, y_l), \quad x,y \in B. \end{aligned} $$

(26)

If, moreover,

$$\displaystyle \begin{aligned} d_k(T_k(x^0), x_k^0) \leqslant (1 - \alpha s_k) \frac{r_k}{s_k}, \quad k=1,\ldots,n, \end{aligned} $$

(27)

and

$$\displaystyle \begin{aligned} \alpha s < 1, \quad s = \underset{i}{\max} (s_i), \end{aligned} $$

(28)

then T = (T ₁, …, T _n) has a fixed point in X.

Proof

We verify that T(B) ⊂ B. By (26), (16) and (27), it can be carried through exactly as in the proof of Theorem 3.

Next, consider numbers ξ _k > 0, k = 1, …, n, fulfilling inequalities (20) and the number α satisfying inequality (13).

Consider the b-metric defined by

$$\displaystyle \begin{aligned} D(x,y) := \sum_{k=1}^n \xi_k d_k (x_k, y_k), \quad x,y \in B. \end{aligned} $$

(29)

Then (B, D) is a b-metric space with $s = \underset {i}{\max } (s_i)$. Also T : B → B is a contraction: for x, y ∈ B, we have by (29), (26), and (20),

$$\displaystyle \begin{aligned} D(T(x), T(y)) \leqslant \alpha D(x,y), \quad x,y \in B. \end{aligned}$$

Therefore, for any x ∈ B, {T ⁿ(x)} is a Cauchy sequence (see (28)). The rest of the proof follows the same arguments as in the proof of Theorem 3.