A generalised fixed point theorem of presic type in cone metric spaces and application to markov process

Abstract

A generalised common fixed point theorem of Presic type for two mappings f: X → X and T: X^k → X in a cone metric space is proved. Our result generalises many well-known results.

2000 Mathematics Subject Classification

47H10

1. Introduction

Considering the convergence of certain sequences, Presic [1] proved the following:

Theorem 1.1. Let (X, d) be a metric space, k a positive integer, T: X^k → X be a mapping satisfying the following condition:

$$d\left(T\left(x_1,x_2,\dots ,x_k\right),T\left(x_2,x_3,\dots ,x_{k+1}\right)\right)\le q_1\cdot d\left(x_1,x_2\right)+q_2\cdot d\left(x_2,x_3\right)+\cdots +q_k\cdot d\left(x_k,x_{k+1}\right)$$

(1.1)

where x₁, x₂, ..., x_{k+ 1}are arbitrary elements in X and q₁, q₂, ..., q _k are non-negative constants such that q₁ + q₂ + · · · + q _k < 1. Then, there exists some x ∈ X such that x = T(x, x, ..., x). Moreover if x₁, x₂, ..., x _k are arbitrary points in X and for n ∈ N x _n+k = T(x _n , x_n+1, ..., x_n+k-1), then the sequence < x _n > is convergent and lim x _n = T(lim x _n , lim x _n , ..., lim x _n ).

Note that for k = 1 the above theorem reduces to the well-known Banach Contraction Principle. Ciric and Presic [2] generalising the above theorem proved the following:

Theorem 1.2. Let (X, d) be a metric space, k a positive integer, T: X^k → X be a mapping satisfying the following condition:

$$d\left(T\left(x_1,x_2,\dots ,x_k\right),T\left(x_2,x_3,\dots ,x_{k+1}\right)\right)\le \lambda .max\{d\left(x_1,x_2\right),d\left(x_2,x_3\right),\dots d\left(x_k,x_{k+1}\right)$$

(1.2)

where x₁, x₂, ..., x_k+1are arbitrary elements in X and λ ∈ (0,1). Then, there exists some x ∈ X such that x = T(x, x, ..., x). Moreover if x₁, x₂, ..., x _k are arbitrary points in X and for n ∈ Nx _n+k = T(x _n , x_n+1, ..., x_n+k-1), then the sequence < x _n > is convergent and lim x _n = T(lim x _n , lim x _n , ..., lim x _n ). If in addition T satisfies D(T(u, u, ... u), T(v, v, ... v)) < d(u, v), for all u, v ∈ X then x is the unique point satisfying x = T(x, x, ..., x).

Huang and Zang [3] generalising the notion of metric space by replacing the set of real numbers by ordered normed spaces, defined a cone metric space and proved some fixed point theorems of contractive mappings defined on these spaces. Rezapour and Hamlbarani [4], omitting the assumption of normality, obtained generalisations of results of [3]. In [5], Di Bari and Vetro obtained results on points of coincidence and common fixed points in non-normal cone metric spaces. Further results on fixed point theorems in such spaces were obtained by several authors, see [5–15].

The purpose of the present paper is to extend and generalise the above Theorems 1.1 and 1.2 for two mappings in non-normal cone metric spaces and by removing the requirement of D(T(u, u, ... u), T(v, v, ... v)) < d(u, v), for all u, v ∈ X for uniqueness of the fixed point, which in turn will extend and generalise the results of [3, 4].

2. Preliminaries

Let E be a real Banach space and P a subset of E. Then, P is called a cone if

• (i) P is closed, non-empty, and satisfies P ≠ {0},

• (ii) ax + by ∈ P for all x, y ∈ P and non-negative real numbers a, b

• (iii) x ∈ P and - x ∈ P ⇒ x = 0, i.e. P ∩ (-P) = 0

Given a cone P ⊂ E, we define a partial ordering ≤ with respect to P by x ≤ y if and only if y - x ∈ P. We shall write x < y if x ≤ y and x ≠ y, and x ≪ y if y - x ∈ intP, where intP denote the interior of P. The cone P is called normal if there is a number K > 0 such that for all x, y ∈ E, 0 ≤ x ≤ y implies || x || ≤ K || y || .

Definition 2.1. [3] Let X be a non empty set. Suppose that the mapping d: X × X → E satisfies:

(d₁) 0 ≤ d (x, y) for all x, y ∈ X and d (x, y) = 0 if and only if x = y

(d₂)d (x, y) = d (y, x) for all x, y ∈ X

(d₃)d (x, y) ≤ d (x, z) + d (z, y) for all x, y, z ∈ X

Then, d is called a conemetric on X and (X, d) is called a conemetricspace.

Definition 2.2. [3] Let (X, d) be a cone metric space. The sequence {x _n } in X is said to be:

• (a) A convergent sequence if for every c ∈ E with 0 ≪ c, there is n₀ ∈ N such that for all n ≥ n₀, d (x _n , x) ≪ c for some x ∈ X. We denote this by lim_n→∞x _n = x.

• (b) A Cauchy sequence if for all c ∈ E with 0 ≪ c, there is no ∈ N such that d (x _m , x _n ) ≪ c, for all m, n ≥ n₀.

• (c) A cone metric space (X, d) is said to be complete if every Cauchy sequence in X is convergent in X.

• (d) A self-map T on X is said to be continuous if lim _n→∞ x _n = x implies that lim _n→∞ T(x _n ) = T(x), for every sequence {x _n }in X.

Definition 2.3. Let (X, d) be a metric space, k a positive integer, T: X^k → X and f : X → X be mappings.

• (a) An element x ∈ X said to be a coincidence point of f and T if and only if f(x) = T(x, x, ..., x). If x = f(x) = T(x, x, ..., x), then we say that x; is a common fixed point of f and T. If w = f(x) = T(x, x, ..., x), then w is called a point of coincidence of f and T.

• (b) Mappings f and T are said to be commuting if and only if f(T(x, x, ... x)) = T(fx, fx, ... fx) for all x ∈ X.

• (c) Mappings f and T are said to be weakly compatible if and only if they commute at their coincidence points.

Remark 2.4. For k = 1, the above definitions reduce to the usual definition of commuting and weakly compatible mappings in a metric space.

The set of coincidence points of f and T is denoted by C(f, T).

3. Main results

Consider a function ϕ: E^k → E such that

• (a) ϕ is an increasing function, i.e x₁ < y₁, x₂ < y₂, ..., x _k < y _k implies ϕ(x₁, x₂, ..., x _k ) < ϕ(y₁, y₂, ..., y _k ).

• (b) ϕ(t, t, t, ...) ≤ t, for all t ∈ X

• (c) ϕ is continuous in all variables.

Now, we present our main results as follows:

Theorem 3.1. Let (X, d) be a cone metric space with solid cone P contained in a real Banach space E. For any positive integer k, let T: X^k → X and f: X → X be mappings satisfying the following conditions:

$$T\left(X^k\right)\subseteq f\left(X\right)$$

(3.1)

$$d\left(T\left(x_1,x_2,\dots ,x_k\right),T\left(x_2,x_3,\dots ,{x_{k+}}_1\right)\right)\le \lambda \varphi \left(d\left(f{x}_1,f{x}_2\right),d\left(f{x}_2,f{x}_3\right),\dots ,\left(f{x}_k,f{x_{k+}}_1\right)\right)$$

(3.2)

where x₁, x₂, ..., x_k+1are arbitrary elements in X and $\lambda \in \left(0,\frac{1}{k}\right)$ and

$$f\left(X\right)iscomplete$$

(3.3)

there exist elements x₁, x₂, ..., x _k in X such that

$$R=max\left(\frac{d\left(f{x}_1,f{x}_2\right)}{\theta },\frac{d\left(f{x}_2,f{x}_3\right)}{{\theta }^2},\dots ,\frac{d\left(f{x}_k,T\left(x_1,x_2,\dots ,x_k\right)\right)}{{\theta }^k}\right)existinE$$

(3.4)

where $\theta ={\lambda }^{\frac{1}{k}}$. Then, f and T have a coincidence point, i.e. C(f, T) ≠ ∅.

Proof. By (3.1) and (3.4) we define sequence < y _n > in f(X) as y _n = fx _n for n = 1, 2, ..., k and y _n+k = f(x _n+k ) = T(x _n , x_n+1, ..., x_n+k-1), n = 1, 2, ... Let α _n = d(y _n , y_n+1). By the method of mathematical induction, we will now prove that

$${\alpha }_n\le R.{\theta }^n$$

(3.5)

for all n. Clearly by the definition of R, (3.5) is true for n = 1, 2, ..., k. Let the k inequalities α _n ≤ Rθⁿ, α_n+1≤ Rθⁿ⁺¹, ..., α_n+k-1≤ Rθ^n+k-1be the induction hypothesis. Then, we have

$$\begin{array}{c}{\alpha }_{n+k}=d\left(y_{n+k},y_{n+k+1}\right)\\ =d\left(T\left(x_n,x_{n+1},\dots ,x_{n+k-1}\right),T\left(x_{n+1},x_{n+2},\dots ,x_{n+k}\right)\right)\\ \le \lambda \varphi \left(d\left(f{x}_n,f{x}_{n+1}\right),d\left(f{x}_{n+1},f{x}_{n+2}\right),\dots ,d\left(f{x}_{n+k-1},f{x}_{n+k}\right)\right)\\ =\lambda \varphi \left({\alpha }_n,{\alpha }_{n+1},\dots ,{\alpha }_{n+k-1}\right)\\ \le \lambda \varphi \left(R{\theta }^n,R{\theta }^{n+1},\dots ,R.{\theta }^{n+k-1}\right)\\ \le \lambda \varphi \left(R{\theta }^n,R{\theta }^n,\dots ,R{\theta }^n\right)\le \lambda R{\theta }^n=R.{\theta }^{n+k}.\end{array}$$

Thus inductive proof of (3.5) is complete. Now for n, p ∈ N, we have

$$\begin{array}{c}d\left(y_n,y_{n+p}\right)\le d\left(y_n,y_{n+1}\right)+d\left(y_{n+1},y_{n+2}\right)+\cdots +d\left(y_{n+p-1},y_{n+p}\right),\\ \le R{\theta }^n+R{\theta }^{n+1}+\cdots +R{\theta }^{n+p-1}\\ \le R{\theta }^n\left(1+\theta +{\theta }^2+\cdots \right)\\ =\frac{R{\theta }^n}{1-\theta }\end{array}$$

Let 0 ≪ c be given. Choose δ > 0 such that c + N _δ (0) ⊆ P where N _δ (0) = {y ∈ E; || y || < δ}. Also choose a natural number N₁ such that $\frac{R{\theta }^n}{1-\theta }\in N_{\delta }\left(0\right)$, for all n > N₁. Then, $\frac{R{\theta }^n}{1-\theta }\ll c$ for all n ≥ N₁. Thus, $d\left(y_n,y_{n+p}\right)\le \frac{R{\theta }^n}{1-\theta }\ll c$ for all n ≥ N₁. Hence, sequence < y _n > is a Cauchy sequence in f(X), and since f(X) is complete, there exists v, u ∈ X such that lim_n→∞y _n = v = f(u). Choose a natural number N₂ such that $d\left(y_n,y_{n+1}\right)\ll \frac{c}{\lambda \left(k+1\right)}$ and $d\left(x,y_{n+1}\right)\ll \frac{c}{k+1}$ for all n ≥ N ₂ .

Then for all n ≥ N₂

$$\begin{array}{c}d\left(fu,T\left(u,u,\dots u\right)\right)\le d\left(fu,y_{n+k}\right)+d\left(y_{n+k},T\left(u,u,\dots u\right)\right)\\ =d\left(fu,y_{n+k}\right)+d\left(T\left(x_n,x_{n+1},\dots x_{n+k-1}\right),T\left(u,u,\dots u\right)\right)\\ \le d\left(fu,y_{n+k}\right)+d\left(T\left(u,u,\dots u\right),T\left(u,u,\dots x_n\right)\right)+d\left(T\left(u,u,\dots x_n\right),T\left(u,u,\dots x_n,x_{n+1}\right)\right)\\ +\cdots d(T\left(u,x_n,\dots x_{n+k-2}\right),T\left(x_n,x_{n+1}\dots x_{n+k-1}\right)\\ \le d\left(fu,y_{n+k}\right)+\lambda \varphi \left\{d\left(fu,fu\right),d\left(fu,fu\right),\dots ,d\left(fu,f{x}_n\right)\right\}\\ +\lambda \varphi \left\{d\left(fu,fu\right),d\left(fu,fu\right),\dots ,d\left(fu,f{x}_n\right),d\left(f{x}_n,f{x}_{n+1}\right)\right\}+\cdots \\ +\lambda \varphi \left\{d\left(fu,f{x}_n\right),d\left(f{x}_n,f{x}_{n+1}\right),\dots d\left(f{x}_{n+k-2},f{x}_{n+k-1}\right)\right\}.\\ =d\left(fu,y_{n+k}\right)+\lambda \varphi \left(0,0,\dots ,d\left(fu,f{x}_n\right)\right)\\ +\lambda \varphi \left(0,0,\dots ,d\left(fu,f{x}_n\right),d\left(f{x}_n,f{x}_{n+1}\right)\right)+\cdots \\ +\lambda \varphi \left(d\left(fu,f{x}_n\right),d\left(f{x}_n,f{x}_{n+1}\right),\dots d\left(f{x}_{n+k-2},f{x}_{n+k-1}\right)\right).\\ \ll \frac{c}{k+1}+\lambda \varphi \left(\frac{c}{\lambda \left(k+1\right)},\frac{c}{\lambda \left(k+1\right)},\dots ,\frac{c}{\lambda \left(k+1\right)}\right)+\lambda \varphi \left(\frac{c}{\lambda \left(k+1\right)},\frac{c}{\lambda \left(k+1\right)},\dots ,\frac{c}{\lambda \left(k+1\right)}\right)\\ +\cdots +\lambda \varphi \left(\frac{c}{\lambda \left(k+1\right)},\frac{c}{\lambda \left(k+1\right)},\dots ,\frac{c}{\lambda \left(k+1\right)}\right)\\ \ll \frac{c}{k+1}+\lambda \frac{c}{\lambda \left(k+1\right)}\dots +\lambda \frac{c}{\lambda \left(k+1\right)}=c.\end{array}$$

Thus, $d\left(fu,T\left(u,u,\dots u\right)\right)\ll \frac{c}{m}$ for all m ≥ 1.

So, $\frac{c}{m}-d\left(fu,T\left(u,u,\dots u\right)\right)\in P$ for all m ≥ 1. Since $\frac{c}{m}\to 0$ as m → ∞ and P is closed, -d(fu, T(u, u, ... u)) ∈ P, but P ∩(-P) = /0/. Therefore, d(fu, T(u, u, ... u)) = 0. Thus, fu = T(u, u, u, ..., u), i.e. C(f, T) ≠ ∅. □

Theorem 3.2. Let (X, d) be a cone metric space with solid cone P contained in a real Banach space E. For any positive integer k, let T: X^k → X and f: X → X be mappings satisfying (3.1), (3.2), (3.3) and let there exist elements x₁, x₂, ... x _k in X satisfying (3.4). If f and T are weakly compatible, then f and T have a unique common fixed point. Moreover if x₁, x₂, ...,x _k are arbitrary points in X and for n ∈ N, y _n+k = f(x _n+k ) = T(x _n , x_n+1, ... x_n+k-1), n = 1, 2, ..., then the sequence < y _n > is convergent and lim y _n = f(lim y _n ) = T(lim y _n , lim y _n , ..., lim y _n ).

Proof. As proved in Theorem 3.1, there exists v, u ∈ X such that lim_n→∞y _n = v = f(u) = T(u, u, u ... u). Also since f and T are weakly compatible f(T(u, u, ... u) = T(fu, fu, fu ... fu). By (3.2) we have,

$$\begin{array}{c}d\left(ffu,fu\right)=d\left(fT\left(u,u,\dots u\right),T\left(u,u,\dots u\right)\right)=d\left(T\left(fu,fu,fu,\dots fu\right),T\left(u,u,\dots u\right)\right)\\ \le d\left(T\left(fu,fu,fu,\dots fu\right),T\left(fu,fu,\dots fu,u\right)\right)+d(T\left(fu,fu,\dots fu,u\right),\\ T\left(fu,fu,\dots ,u,u\right))+\cdots +d\left(T\left(fu,u,\dots u,u\right),T\left(u,u,\dots u\right)\right)\\ \le \lambda \varphi \left(d\left(ffu,ffu\right),\dots d\left(ffu,ffu\right),d\left(ffu,fu\right)\right)+\lambda \varphi (d\left(ffu,ffu\right),\dots d\left(ffu,fu\right),\\ d\left(fu,fu\right))+\cdots \lambda \varphi \left(d\left(ffu,fu\right),\dots d\left(fu,fu\right),d\left(fu,fu\right)\right)\\ =\lambda \varphi \left(0,0,0,\dots d\left(ffu,fu\right)\right)+\lambda \varphi \left(0,0\dots 0,d\left(ffu,fu\right),0\right)+\cdots .\lambda \varphi \left(d\left(ffu,fu\right),0,0\dots 0\right)=k\lambda d\left(ffu,fu\right).\end{array}$$

Repeating this process n times we get, d(f fu, fu) < kⁿ λⁿ d(f fu, fu). So kⁿ λⁿ d(f fu, fu) - d(f fu, fu) ∈ P for all n ≥ 1. Since kⁿ λⁿ → 0 as n → ∞ and P is closed, --d(f fu, fu) ∈ P, but P ∩ (-P) = {0}. Therefore, d(f fu, fu) = 0 and so f fu = fu. Hence, we have, fu = f fu = f(T(u, u, ... u)) = T(fu, fu, fu ... fu), i.e. fu is a common fixed point of f and T, and lim y _n = f(lim y _n ) = T(lim y _n , lim y _n , ... lim y _n ). Now suppose x, y be two fixed points of f and T. Then,

$$\begin{array}{c}d\left(x,y\right)=d\left(T\left(x,x,x\dots x\right),T\left(y,y,y\dots y\right)\right)\\ \le d\left(T\left(x,x,\dots x\right),T\left(x,x,\dots x,y\right)\right)+d\left(T\left(x,x,\dots x,y\right),T\left(x,x,x\dots x,y,y\right)\right)\\ +\cdots +d\left(T\left(x,y,y,\dots y\right),T\left(y,y,\dots y\right)\right)\\ \le \lambda \varphi \left\{d\left(fx,fx\right),d\left(fx,fx\right),\dots ,d\left(fx,fy\right)\right\}\\ +\lambda \varphi \left\{d\left(fx,fx\right),d\left(fx,fx\right),\dots d\left(fx,fy\right),d\left(fy,fy\right)\right\}\\ +\cdots +\lambda \varphi \left\{d\left(fx,fy\right),d\left(fy,fy\right),\dots d\left(fy,fy\right)\right\}.\\ =\lambda \varphi \left(0,0,\dots ,d\left(fx,fy\right)\right)+\lambda \varphi \left(0,0,\dots d\left(fx,fy\right),0\right)+\cdots +\lambda \varphi \left(d\left(fx,fy\right),0,0,0,\dots 0\right)).\\ =k\lambda d\left(fx,fy\right)=k\lambda d\left(x,y\right).\end{array}$$

Repeating this process n times we get as above, d(x, y) ≤ kⁿ λⁿ d(x, y) and so as n → ∞d(x, y) = 0, which implies x = y. Hence, the common fixed point is unique. □

Remark 3.3. Theorem 3.2 is a proper extension and generalisation of Theorems 1.1 and 1.2.

Remark 3.4. If we take k = 1 in Theorem, 3.2, we get the extended and generalised versions of the result of [3] and [4].

Example 3.5. Let E = R², P = {(x, y) ∈ E\x, y ≥ 0}, X = [0, 2] and d: X × X → E such that d(x, y) = (|x - y |, |x - y |). Then, d is a cone metric on X. Let T: X² → X and f: X → X be defined as follows:

$$\begin{array}{c}T\left(x,y\right)=\frac{\left(x^2+y^2\right)}{4}+\frac{1}{2}if\left(x,y\right)\in \left[0,1\right]\times \left[0,1\right]\\ T\left(x,y\right)=\frac{\left(x+y\right)}{4}+\frac{1}{2}if\left(x,y\right)\in \left[1,2\right]\times \left[1,2\right]\\ T\left(x,y\right)=\frac{\left(x^2+y\right)}{4}+\frac{1}{2}if\left(x,y\right)\in \left[0,1\right]\times \left[1,2\right]\\ T\left(x,y\right)=\frac{\left(x+y^2\right)}{4}+\frac{1}{2}if\left(x,y\right)\in \left[1,2\right]\times \left[0,1\right]\\ f\left(x\right)=x^{2}ifx\in \left[0,1\right]\\ f\left(x\right)=xifx\in \left[1,2\right]\end{array}$$

T and f satisfies condition (3.2) as follows:

Case 1. x, y, z ∈ [0, 1]

$$\begin{array}{c}d\left(T\left(x,y\right),T\left(y,z\right)\right)=\left(\mid T\left(x,y\right)-T\left(y,z\right)\mid ,\mid T\left(x,y\right)-T\left(y,z\right)\mid \right)\\ =\left(\mid \frac{x^2-z^2}{4}\mid ,\mid \frac{x^2-z^2}{4}\mid \right)\\ \le \left(\mid \frac{x^2-z^2}{4}\mid +\mid \frac{y^2-z^2}{4}\mid ,\mid \frac{x^2-y^2}{4}\mid +\mid \frac{y^2-z^2}{4}\mid \right)\\ \le \frac{1}{2}.max\left\{d\left(fx,fy\right),d\left(fy,fz\right)\right\}\end{array}$$

Case 2. x, y ∈ [0, 1] and z ∈ [1, 2]

$$\begin{array}{c}d\left(T\left(x,y\right),T\left(y,z\right)\right)=\left(\mid \frac{x^2+y^2}{4}-\frac{y^2+z}{4}\mid ,\mid \frac{x^2+y^2}{4}-\frac{y^2+z}{4}\mid \right)\\ \le \left(\mid \frac{x^2-y^2}{4}\mid +\mid \frac{y^2-z}{4}\mid ,\mid \frac{x^2-y^2}{4}\mid +\mid \frac{y^2-z}{4}\mid \right)\\ \le \frac{1}{2}.max\left\{\left(fx,fy\right),d\left(fy,fz\right)\right\}\end{array}$$

Case 3. x ∈ [0, 1] and y; z ∈ [1, 2]

$$\begin{array}{c}d\left(T\left(x,y\right),T\left(y,z\right)\right)=\left(\mid \frac{x^2+y}{4}-\frac{y+z}{4}\mid ,\mid \frac{x^2+y}{4}-\frac{y+z}{4}\mid \right)\\ =\left(\mid \frac{x^2-z}{4}\mid ,\mid \frac{x^2-z}{4}\mid \right)\\ \le \left(\mid \frac{x^2-y}{4}1\mid +\mid \frac{y-z}{4}\mid ,\mid \frac{x^2-y}{4}\mid +\mid \frac{y-z}{4}\mid \right)\\ \le \frac{1}{.∕2}.max\left\{d\left(fx,fy\right),d\left(fy,fz\right)\right\}\end{array}$$

Case 4. x, y, z ∈ [1, 2]

$$\begin{array}{c}d\left(T\left(x,y\right),T\left(y,z\right)\right)=\left(\mid \frac{x+y}{4}-\frac{y+z}{4}\mid ,\mid \frac{x+y}{4}-\frac{y+z}{4}\mid \right)\\ \le \left(\mid \frac{x-y}{4}\mid +\mid \frac{y-z}{4}\mid ,\mid \frac{x-y}{4}\mid +\mid \frac{y-z}{4}\mid \right)\\ \le \frac{1}{2}.max\left\{\left(fx,fy\right),d\left(fy,fz\right)\right\}.\end{array}$$

Similarly in all other cases $d\left(T\left(x,y\right),T\left(y,z\right)\right)\le \frac{1}{2}.max\left\{\left(fx,fy\right),d\left(fy,fz\right)\right\}$. Thus, f and T satisfy condition (3.2) with ϕ(x₁, x₂) = max{x₁, x₂}. We see that C(f, T) = 1, f and T commute at 1. Finally, 1 is the unique common fixed point of f and T.

4. An application to markov process

Let ${\Delta }_{n-1}=\left\{x\in R_{+}^n:{\Sigma }_{i=1}^n{x}_i=1\right\}$ denote the n - 1 dimensional unit simplex. Note that any x ∈ Δ_n-1may be regarded as a probability over the n possible states. A random process in which one of the n states is realised in each period t = 1, 2, ... with the probability conditioned on the current realised state is called Markov Process. Let a _ij denote the conditional probability that state i is reached in succeeding period starting in state j. Then, given the prior probability vector x^t in period t, the posterior probability in period t + 1 is given by $x_i^{t+1}=\sum a_{ij}{x}_j^t$ for each i = 1, 2, .... To express this in matrix notation, we let x^t denote a column vector. Then, x^t+1 = Ax^t. Observe that the properties of conditional probability require each a _ij ≥ 0 and ${\sum }_{i=1}^n{a}_{ij}=1$ for each j. If for any period t, x^{t+ 1}= x^t then x^t is a stationary distribution of the Markov Process. Thus, the problem of finding a stationary distribution is equivalent to the fixed point problem Ax^t = x^t.

For each i, let ε _i = min _j a _ij and define $\epsilon ={\sum }_{i=1}^n{\epsilon }_i$.

Theorem 4.1. Under the assumption a _i,j > 0, a unique stationary distribution exist for the Markov process.

Proof. Let d: Δ_n-1× Δ_n-1→ R² be given by $d\left(x,y\right)=\left({\sum }_{i=1}^n\mid x_i-y_i\mid ,\alpha {\sum }_{i=1}^n\mid x_i-y_i\mid \right)$ for all x, y ∈ Δ_n-1and some α ≥ 0.

Clearly d(x, y) ≥ (0, 0) for all x, y ∈ Δ_n-1and $d\left(x,y\right)=\left(0,0\right)\Rightarrow \left({\sum }_{i=1}^n\mid x_i-y_i\mid ,\alpha {\sum }_{i=1}^n\mid x_i-y_i\mid \right)=\left(0,0\right)\Rightarrow \mid x_i-y_i\mid =0$ for all i ⇒ x = y. Also x = y ⇒ x _i = y _i for all $i\Rightarrow \mid x_i-y_i\mid =0\Rightarrow {\sum }_{i=1}^n\mid x_i-y_i\mid =0\Rightarrow d\left(x,y\right)=\left(0,0\right)$

$$\begin{array}{c}d\left(x,y\right)=\left({\sum }_{i=1}^n\mid x_i-y_i\mid ,\alpha {\sum }_{i=1}^n\mid x_i-y_i\mid \right)\\ =\left({\sum }_{i=1}^n\mid y_i-x_i\mid ,\alpha {\sum }_{i=1}^n\mid y_i-x_i\mid \right)=d\left(y,x\right)\\ d\left(x,y\right)=\left({\sum }_{i=1}^n\mid x_i-y_i\mid ,\alpha {\sum }_{i=1}^n\mid x_i-y_i\mid \right)\\ =\left({\sum }_{i=1}^n\mid \left(x_i-z_i\right)+\left(z_i-y_i\right)\mid ,\alpha {\sum }_{i=1}^n\mid \left(x_i-z_i\right)\mid +\mid \left(z_i-y_i\right)\mid \right)\\ \le \left({\sum }_{i=1}^n\mid \left(x_i-z_i\right)\mid +\mid \left(z_i-y_i\right)\mid ,\alpha {\sum }_{i=1}^n\mid \left(x_i-z_i\right)\mid +\mid \left(z_i-y_i\right)\mid \right)\\ =\left({\sum }_{i=1}^n\mid \left(x_i-z_i\right)\mid ,\alpha {\sum }_{i=1}^n\mid \left(x_i-z_i\right)\mid \right)+\left({\sum }_{i=1}^n\mid \left(z_i-y_i\right)\mid ,\alpha {\sum }_{i=1}^n\mid \left(z_i-y_i\right)\mid \right)\\ =d\left(x,z\right)+d\left(z,x\right).\end{array}$$

So Δ_n-1is a cone metric space. For x ∈ Δ_n-1, let y = Ax. Then each $y_i={\sum }_{j=1}^n{a}_{ij}{x}_j\ge 0$. Further more, since each ${\sum }_{i=1}^n{a}_{ij}=1$, we have ${\sum }_{i=1}^n{y}_i={\sum }_{i=1}^n{\sum }_{j=1}^n{a}_{ij}{x}_j={\sum }_{j=1}^n{x}_j{\sum }_{i=1}^n{a}_{ij}={\sum }_{j=1}^n{x}_j=1$, so y ∈ Δ_n-1. Thus, we see that A: Δ_n-1→ Δ_n-1. We will show that A is a contraction. Let A _i denote the i th row of A. Then for any x, y ∈ Δ_n-1, we have

$$\begin{array}{l}d(Ax,Ay)=({\displaystyle {\sum }_{i=1}^n|}{(Ax)}_i-{(Ay)}_i|,\alpha {\displaystyle {\sum }_{i=1}^n|}{(Ax)}_i-{(Ay)}_i|)\\ =({\displaystyle {\sum }_{i=1}^n|}{\displaystyle {\sum }_{j=1}^n{a}_{ij}{x}_j}-a_{ij}{y}_j|,\alpha {\displaystyle {\sum }_{i=1}^n|}{\displaystyle {\sum }_{j=1}^n{a}_{ij}}{x}_j-a_{ij}{y}_j|)\\ =({\displaystyle {\sum }_{i=1}^n|}{\displaystyle {\sum }_{j=1}^n(}{a}_{ij}-{\epsilon }_i)(x_j-y_j)+{\epsilon }_i(x_j-y_j)|,\\ \alpha {\displaystyle {\sum }_{i=1}^n|}{\displaystyle {\sum }_{j=1}^n(}{a}_{ij}-{\epsilon }_i)(x_j-y_j)+{\epsilon }_i(x_j-y_j)|)\\ \le ({\displaystyle {\sum }_{i=1}^n(}|{\displaystyle {\sum }_{j=1}^n(}{a}_{ij}-{\epsilon }_i)(x_j-y_j)|+{\epsilon }_i|{\displaystyle {\sum }_{j=1}^n(}{x}_j-y_j)|),\\ \alpha ({\displaystyle {\sum }_{i=1}^n(}|{\displaystyle {\sum }_{j=1}^n(}{a}_{ij}-{\epsilon }_i)(x_j-y_j)|+{\epsilon }_i|{\displaystyle {\sum }_{j=1}^n(}{x}_j-y_j)|)\\ \le ({\displaystyle {\sum }_{i=1}^n{\displaystyle {\sum }_{j=1}^n(}}{a}_{ij}-{\epsilon }_i)|x_j-y_j|,\alpha {\displaystyle {\sum }_{i=1}^n{\displaystyle {\sum }_{j=1}^n(}}{a}_{ij}-{\epsilon }_i)|x_j-y_j|)\\ (Since{\displaystyle {\sum }_{j=1}^n(}{x}_j-y_j)=0)\\ =({\displaystyle {\sum }_{j=1}^n|}{x}_j-y_j|{\displaystyle {\sum }_{i=1}^n(}{a}_{ij}-{\epsilon }_i),\\ \alpha {\displaystyle {\sum }_{j=1}^n|}{x}_j-y_j|{\displaystyle {\sum }_{i=1}^n(}{a}_{ij}-{\epsilon }_i)\\ =({\displaystyle {\sum }_{j=1}^n|}{x}_j-y_j|(1-\epsilon ),\alpha {\displaystyle {\sum }_{j=1}^n|}{x}_j-y_j|(1-\epsilon ))\\ (Since{\displaystyle {\sum }_{i=1}^n{a}_{ij}}=1\text{and}{\displaystyle {\sum }_{i=1}^n{\epsilon }_i}=\epsilon )\\ =d(x,y)(1-\epsilon )\end{array}$$

which establishes that A is a contraction mapping. Thus, Theorem 3.2 with k = 1 and f as identity mapping ensures a unique stationary distribution for the Markov Process. Moreover for any x⁰ ∈ Δ_n-1, the sequence < Aⁿx⁰ > converges to the unique stationary distribution. □