A UNIQUE COMMON TRIPLE FIXED POINT THEOREM IN PARTIALLY ORDERED CONE METRIC SPACES (COMMUNICATED BY SIMEON RICH) K.P.R.RAO AND G.N.V.KISHORE Abstract

ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 3 Issue 4(2011), Pages 213 - 222.

A UNIQUE COMMON TRIPLE FIXED POINT THEOREM IN PARTIALLY ORDERED CONE METRIC SPACES

(COMMUNICATED BY SIMEON RICH)

K.P.R.RAO AND G.N.V.KISHORE

Abstract. The notion of triple fixed point was introduced by V.Berinde and M.Borcut [8] and obtained some fixed point theorems in partially ordeded metric spaces. In this paper, we obtain a unique common triple fixed point theorem in partially ordered cone metric spaces in which cone is not necessarilly normal and also mention one supported example.

1. Introduction

In 2007, Huang and Zhang [10] introduced the concept of cone metric spaces by using ordered Banach space instead of the set of real numbers as a codomain and established Banach contraction principle. Later several authors proved fixed and common fixed point theorems in cone and partially ordered metric spaces. Some interesting references are [1, 3, 4, 5, 6, 11, 12, 15, 16, 19, 21, 22, 23].

The notion of coupled fixed point is introduced by Bhaskar and Lakshmikantham [9] and studied some fixed point theorems in partially ordered metric spaces.

Recently some of authors proved coupled and common coupled fixed point theorems in partially ordered cone metric spaces see [2, 7, 13, 14, 17, 18, 20].

In 2011, V.Berinde and M.Borcut [8] introduced triple fixed point and obtained some fixed point theorems for contractive type maps in partially ordeded metric spaces. The aim of this paper is to study unique common triple fixed point theorem for two maps by using w - compatible maps over partially ordered cone metric spaces,in which the underlying cone is not necessarily normal.

Throughout this paper, letZ⁺ denote the set of all positive integers.

Definition 1.1. [10] Let E be a real Banach space and P be a subset of E. P is called a cone if and only if :

(i)P is closed, non - empty andP 6={0};

(ii) a, b∈R, a, b≥0, x, y∈P ⇒ax+by∈P;

(iii)x∈P and−x∈P⇒x= 0.

2000Mathematics Subject Classification. 54H25, 47H10, 54E50.

Key words and phrases. cone metric,w- compatible maps, triple fixed point, complete space.

c

2011 Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e.

Submitted September 28, 2011. Published November 26, 2011.

213

Given a coneP⊂E, we define a partial ordering≤with respect toP byx≤y if and only if x−y ∈P. We shal writex < y to indicate that x≤y but x6=y, whilex << y will stand fory−x∈intP, intP denotes the interior ofP.

The coneP is callednormalif there is a numberK >0 such that for allx, y∈E, 0≤x≤y⇒ ||x|| ≤K||y||.

The least positive number satisfying above is called the normal constant ofP. There are no normal cones with normal constantK <1 (see [19]).

Example 1.2. [19] Let E = C_R²[0,1] with the norm ||f|| = ||f||_∞+||f⁰||_∞ and consider the cone P = {f ∈ E : f ≥ 0}. For each K > 1, put f(x) = x and g(x) =x^2K. Then0≤g≤f, ||f||= 2and||g||= 2K+ 1. Since K||f||<||g||, K is not the normal constant of P. Therefore, P is a non - normal cone.

Definition 1.3. [10]LetX be a nonempty set. Suppose the mappingd:X×X →E satisfies

(i) 0< d(x, y)for all x, y∈X andd(x, y) = 0if and only if x=y;

(ii) d(x, y) =d(y, x)for all x, y∈X;

(iii) d(x, y)≤d(x, z) +d(z, y)for allx, y, z∈X.

Thendis called a cone metric onX and(X, d)is called a cone metric space.

Definition 1.4. [10] Let (X, d) be a cone metric space. Let {xn} be a sequence in X and x ∈ X. If for every c ∈ E with 0 << c, there is n0 ∈ Z⁺ such that d(xn, x)<< cfor alln≥n0, then{xn} is said to be convergent toxandxis called the limit of{xn}. We denote this by lim

n→∞xn =xorxn→xasn→ ∞.

If for every c∈E with 0<< c, there isn0∈Z⁺ such thatd(xn, xm)<< c for alln, m≥n0, then{xn}is called Cauchy sequence inX. If every Cauchy sequence is convergent inX thenX is called a complete cone metric space.

Remark. Let E be an ordered Banach space with coneP. Then (1)if u≤v andv << wthen u << w,

(2)if u << v andv << w thenu << w, (3)if 0≤u << cfor each c∈intP, thenu= 0,

(4)c∈intP if and only if[−c, c]is a neighborhood of0,

(5)ifP is a solid cone and if a sequence{xn}is convergent in a cone metric space (X, d), then the limit of{xn} is unique.

Definition 1.5. [9]Let (X, ≺) be a partially ordered set and F :X ×X →X.

Then the map F is said to have mixed monotone property if F(x, y) is monotone non - decreasing in x and is monotone non - increasing in y; that is, for any x, y∈X,

x₁ ≺x₂ impliesF(x₁, y)≺F(x₂, y)for ally∈X and

y1 ≺y2 implies F(x, y2)≺F(x, y1)for all x∈X.

Inspired by Definition 1.5, Lakshmikantham and ´Ciri´c in [17] introduced the concept of ag - mixed monotone mapping.

Definition 1.6. [17] Let (X, ≺ )be a partially ordered set andF :X×X →X.

Then the mapF is said to have mixedg - monotone property ifF(x, y)is monotone

g - non - decreasing in xand is monotone g - non - increasing in y; that is, for any x, y∈X,

gx1 ≺gx2 impliesF(x1, y)≺F(x2, y)for ally∈X and

gy1 ≺gy2 impliesF(x, y2)≺F(x, y1)for allx∈X.

Recently V.Berinde and M.Borcut [8] introduced the notion of triple fixed point of a mapping as folloows.

Definition 1.7. [8]An element(x, y, z)∈X×X×X is called a triple fixed point of mappingF:X×X×X→X ifx=F(x, y, z), y=F(y, x, y)andz=F(z, y, x) .

In [8] the authors obtained the following theorem.

Theorem 1.8. (Theorem.7.[8]): Let (X,≤) be a partially order set and suppose there is a metric d on X such that (X, d) is a complete metric space. Let F : X×X×X →X be a continuous mapping having the mixed monotone property on X. Assume that there exist constantsj, k, l∈[0,1) withj+k+l <1for which

d(F(x, y, z), F(u, v, w))≤jd(x, u) +kd(y, v) +ld(z, w),

∀ x≥u, y≤v, z≥w. If there existx0, y0, z0∈X such that

x0≤F(x0, y0, z0), y0≥F(y0, x0, y0) and z0≤F(z0, y0, x0), then there existx, y, z∈X such that

x=F(x, y, z), y=F(y, x, y)and z=F(z, y, x).

Now we give the following definitions.

Definition 1.9. An element (x, y)∈X×X×X is called

(i) a triple coincident point of mapping F : X×X ×X →X andf :X →X if f x=F(x, y, z),f y=F(y, x, y)andf z=F(z, y, x);

(ii)a common triple fixed point of mapping F :X×X×X →X andf :X →X if x=f x=F(x, y, z),y=f y=F(y, x, y) andz=f z=F(z, y, x).

Definition 1.10. The mappings F : X ×X ×X → X and f : X → X are calledw- compatible iff(F(x, y, z)) =F(f x, f y, f z),f(F(y, x, y)) =F(f y, f x, f y) and f(F(z, y, x)) = F(f z, f y, f x) whenever f x =F(x, y, z), f y =F(y, x, y) and f z=F(z, y, x).

Now we prove our main result.

2. Main Result

Theorem 2.1. Let (X, ≺, d)be a partially ordered cone metric space and let T :X×X×X →X andf :X →X be mappings satisfying

(i)d(T(x, y, z), T(u, v, w))≤jd(f x, f u) +kd(f y, f v) +ld(f z, f w)

∀x, y, z, u, v, w ∈ X with f x f u, f y ≺ f v, f z f w and j, k, l ∈ [0,1) with j+k+l <1,

(ii)T(X×X×X)⊆f(X)andf(X)is a complete subspace ofX, (iii)T has the mixedf - monotone property,

(iv) (a)If a non - decreasing sequence{xn} →x, thenx_n ≺xfor all n, (b)If a non - increasing sequence {xn} →x, thenx≺x_n for alln.

If there exist x0, y0, z0 ∈ X such that f x0 T(x0, y0, z0), f y0 ≺ T(y0, x0, y0) andf z0T(z0, y0, x0), then T andf have triple coincidence point inX×X×X.

Proof. Letx₀, y₀, z₀∈X such thatf x₀ T(x₀, y₀, z₀), f y₀ ≺T(y₀, x₀, y₀) andf z₀ T(z₀, y₀, x₀).

SinceT(X×X×X)⊆f(X), we choosex₁, y₁, z₁∈X such that f x1 =T(x0, y0, z0)≺f x0,

f y1 =T(y0, x0, y0)f y0and f z1 =T(z0, y0, x0)≺f z0. Now choosex2, y2, z2∈X such that

f x2 =T(x1, y1, z1), f y2 =T(y1, x1, y1) and f z2 =T(z1, y1, x1).

SinceT has the mixedf - monotone property we have f x0 f x1 f x2, f y0 ≺f y1 ≺f y2 and f z0 f z1 f z2.

Continuing this process, we can construct three sequences {xn},{yn} and{zn} in X such that

f x_n+1 =T(x_n, y_n, z_n), f y_n+1 =T(y_n, x_n, y_n) and

f z_n+1 =T(z_n, y_n, x_n), n= 0,1,2,· · · with

f x0 f x1 f x2 · · ·, f y0 ≺f y1 ≺f y2 ≺ · · · and f z0 f z1 f z2 · · · To simplify we denote

d^x_n=d(f x_n−1, f xn), d^y_n=d(f y_n−1, f yn) andd^z_n=d(f z_n−1, f zn).

Then by (i) we obtain

d^x₂ =d(f x1, f x2)

=d(f x2, f x1)

=d(T(x2, y2, z2), T(x1, y1, z1))

≤jd(f x₂, f x₁) +kd(f y₂, f y₁) +ld(f z₂, f z₁)

=jd^x₁+kd^y₁+ld^z₁. Similarly, we obtain

d^y₂ ≤kd^x₁+ (j+l)d^y₁ d^z₂ ≤ld^x₁+kd^y₁+jd^z₁. Also

d^x₃ ≤(j²+k²+l²)d^x₁+ (2jk+ 2lk)d^y₁+ (2jl)d^z₁ d^y₃ ≤(2jk+lk)d^x₁+ [(j+l)²+k²]d^y₁+kld^z₁ d^z₃ ≤(2jl+k²)d^x₁+ 2[jk+lk]d^y₁+ (j²+l²)d^z₁. In order to simplify we consider the matrix

A=





j k l

k j+l 0

l k j



 denoted by





a1 b1 c1

d1 e1 f1

g1 b1 h1





and further denote

A²=





j²+k²+l² 2jk+ 2lk 2jl 2jk+lk (j+l)²+k² kl 2jl+k² 2jk+ 2lk j²+l²



 by





a₂ b₂ c₂ d₂ e₂ f₂ g2 b2 h2





where

a₂+b₂+c₂=d₂+e₂+f₂=g₂+b₂+h₂= (j+k+l)²<(j+k+l)<1.

Now we prove by the induction that

Aⁿ =





an bn cn

dn en fn

gn bn hn



 (2.1)

where

an+bn+cn=dn+en+fn=gn+bn+hn= (j+k+l)ⁿ<(j+k+l)<1.

Clearly (2.1) is true forn= 1 and n= 2.

Assume that (2.1) is true for somen.

Consider

Aⁿ⁺¹ =Aⁿ.A

=





an bn cn

dn en fn

gn bn hn









j k l

k j+l 0

l k j





=





jan+kbn+lcn kan+ (j+l)bn+kcn lan+jcn

jdn+ken+lfn kdn+ (j+l)en+kfn ldn+jfn

jg_n+kb_n+lh_n kg_n+ (j+l)b_n+kh_n lg_n+jh_n





we have

an+1+bn+1+cn+1 =jan+kbn+lcn+kan+ (j+l)bn+kcn+lan+jcn

= (j+k+l)an+ (j+k+l)bn+ (j+k+l)cn

= (j+k+l)(an+bn+cn)

= (j+k+l)(j+k+l)ⁿ

= (j+k+l)ⁿ⁺¹

<(j+k+l)

<1.

Similarly we have

d_n+1+e_n+1+f_n+1=g_n+1+b_n+1+h_n+1= (j+k+l)ⁿ⁺¹<(j+k+l)<1.

Thus (2.1) is true forn+ 1.

Hence by induction, (2.1) is true for alln.

Therefore,



 d^x_n+1 d^y_n+1 d^z_n+1



≤





a_n b_n c_n d_n e_n f_n gn bn hn







 d^x₁ d^y₁ d^z₁



 f or all n= 1,2,3,· · ·

That is

d^x_n+1≤a_nd^x₁+b_nd^y₁+c_nd^z₁ d^y_n+1≤dnd^x₁+end^y₁+fnd^z₁ d^z_n+1≤gnd^x₁+bnd^y₁+hnd^z₁







(2.2)

for alln= 1,2,3,· · · Letm, n∈N withm > n.

d(f xm, f xn) ≤d(f xm, f x_m−1) +d(f x_m−1, f x_m−2) +· · ·+d(f xn+2, f xn+1) +d(f xn+1, f xn)

≤a_m−1d^x₁+b_m−1d^y₁+c_m−1d^z₁+a_m−2d^x₁+b_m−2d^y₁+c_m−2d^z₁ +· · ·+an+1d^x₁+bn+1d^y₁+cn+1d^z₁+and^x₁+bnd^y₁+cnd^z₁

= [a_m−1+a_m−2+a_m−3+· · ·+an]d^x₁ +[b_m−1+b_m−2+b_m−3+· · ·+bn]d^y₁ +[c_m−1+c_m−2+c_m−3+· · ·+cn]d^z₁

≤(µ^m−1+µ^m−2+· · ·+µⁿ⁺¹+µⁿ)d^x₁ +(µ^m−1+µ^m−2+· · ·+µⁿ⁺¹+µⁿ)d^y₁ +(µ^m−1+µ^m−2+· · ·+µⁿ⁺¹+µⁿ)d^z₁

= (µ^m−1+µ^m−2+· · ·+µⁿ⁺¹+µⁿ)(d^x₁+d^y₁+d^z₁)

≤ µⁿ

1−µ(d^x₁+d^y₁+d^z₁) (2.3)

whereµ=j+k+l <1.

It follows from (2.3) that forc∈E, 0<< cand largen, we have _1−µ^µn (d^x₁+d^y₁+d^z₁)<< c.

Thus

d(f xm, f xn)<< c.

Hence{f xn} is a Cauchy sequence in the metric space (X, d).

Similarly {f yn} and {f zn} are also Cauchy sequences in the cone metric space (X, d).

Supposef(X) is complete.

Since{f xn} ⊆f(X),{f yn} ⊆f(X) and{f zn} ⊆f(X) are Cauchy sequences in the complete cone metric space (f(X), d), it follows that the sequences {f xn},{f yn} and{f zn}are converge to someα, βandγ in (f(X), d) respectively.

There existx, y, z∈X such thatα=f x, β=f yandγ=f z.

Since{f xn},{f yn}and{f zn}are Cauchy sequences inXand{f xn} →α,{f yn} → β and{f zn} →γ , it follows that{f xn+1} →α,{f yn+1} →β and{f zn+1} →γ.

Since{f x_n} is a non - increasing sequence and{f x_n} →f xwe have f x≺f x_n, {f y_n}is a non - decreasing sequence and{f y_n} →f y we havef y_n ≺f y and {f z_n} is a non - increasing sequence and{f z_n} →f z we havef z ≺f z_n for alln.

Now,

d(T(x, y, z), α) ≤d(T(x, y, z), f xn+1) +d(f xn+1, α)

=d(T(x, y, z), T(xn, yn, zn)) +d(f xn+1, α)

≤jd(f x, f xn) +kd(f y, f yn) +ld(f z, f zn) +d(f xn+1, α)

=jd(α, f x_n) +kd(β, f y_n) +ld(γ, f z_n) +d(f x_n+1, α)

<< j_4j^c +k_4k^c +l_4l^c +^c₄ =c.

It follows thatα=T(x, y, z).

Similarlyβ=T(y, x, y) andγ=T(z, y, x).

Thus

α=f x=T(x, y, z), β=f y=T(y, x, y) andγ=f z=T(z, y, x).

Hence (x, y, z) is a triple coincidence point ofT andf. Theorem 2.2. In addition to the hypothesis of Theorem 2.1. Suppose that for every(x, y, z),(x^∗, y^∗, z^∗)∈X×X×X there exists(u, v, w)∈X×X×X such that

(T(u, v, w), T(v, u, v), T(w, v, u)) is comparable to (T(x, y, z), T(y, x, y), T(z, y, x)) and(T(x^∗, y^∗, z^∗), T(y^∗, x^∗, y^∗), T(z^∗, y^∗, x^∗)). If(x, y, z) and(x^∗, y^∗, z^∗)are triple coincidence points ofT andf, then

T(x, y, z) =f x=f x^∗=T(x^∗, y^∗, z^∗), T(y, x, y) =f y=f y^∗=T(y^∗, x^∗, y^∗)and T(z, y, x) =f z=f z^∗=T(z^∗, y^∗, x^∗).

Moreover if(T, f)isw- compatible, then T andf have a unique common triple fixed point in X×X×X.

Proof. From Theorem 2.1, there exists (x, y, z)∈X×X×such that T(x, y, z) =f x=α, T(y, x, y) =f y=β andT(z, y, x) =f z=γ.

Thus the existence of triple coincidence point ofT andf is conformed.

Now let (x^∗, y^∗, z^∗) be another triple coincidence point ofT andf. That is

T(x^∗, y^∗, z^∗) =f x^∗, T(y^∗, x^∗, y^∗) =f y^∗ andT(z^∗, y^∗, x^∗) =f z^∗. By additional assumption, there is (u, v, w)∈X×X×X such that

(T(u, v, w), T(v, u, v), T(w, v, u)) is comparable to (T(x, y, z), T(y, x, y), T(z, y, x)) and (T(x^∗, y^∗, z^∗), T(y^∗, x^∗, y^∗), T(z^∗, y^∗, x^∗)).

Letu0 =u, v0=v, w0 =w, x0=x, y0=y, z0 =z, x^∗₀=x^∗, y₀^∗ =y^∗ andz₀^∗ =z^∗. SinceT(X×X×X)⊆f(X), we can construct the sequences

{f un},{f vn},{f wn},{f xn},{f yn},{f zn},{f x^∗_n},{f y_n^∗} and{f z_n^∗}.

f un+1 =T(un, vn, wn), f vn+1=T(vn, un, vn), f wn+1=T(wn, vn, un), f xn+1 =T(xn, yn, zn), f yn+1=T(yn, xn, yn), f zn+1=T(zn, yn, xn),

f x^∗_n+1 =T(x^∗_n, y_n^∗, z_n^∗), f y^∗_n+1=T(y^∗_n, x^∗_n, y_n^∗) andf z_n+1^∗ =T(z_n^∗, y_n^∗, x^∗_n), n= 0,1,2,· · · Since (f x, f y, f z) = (T(x, y, z), T(y, x, y), T(z, y, x)) = (f x1, f y1, f z1) and

(T(u, v, w), T(v, u, v), T(w, v, u)) = (f u₁, f v₁, f w₁) are comparable, thenf xf u₁, f y ≺f v₁ andf z f w₁ .

One can show thatf x f u_n, f y≺f v_n andf z f w_n for alln.

As in Theorem 2.1 we conclude that





d(f x, f un+1) d(f y, f vn+1) d(f z, f wn+1)



 ≤





an bn cn

dn en fn

gn bn hn









d(f x, f u1) d(f y, f v1) d(f z, f w1)





wherean+bn+cn=dn+en+fn=gn+bn+hn = (j+k+l)ⁿ<(j+k+l)<1.

Thus

d(f x, f u_n+1) ≤a_nd(f x, f u₁) +b_nd(f y, f v₁) +c_nd(f z, f w₁)

≤[an+bn+cn][d(f x, f u1) +d(f y, f v1) +d(f z, f w1)]

≤µⁿ[d(f x, f u1) +d(f y, f v1) +d(f z, f w1)] (2.4) whereµ=j+k+l <1.

Let 0<< cbe given choose a natural numbern₀ such that µⁿ[d(f x, f u₁) +d(f y, f v₁) +d(f z, f w₁)]<< cfor alln > n₀. Thus

d(f x, f u_n+1)<< c.

Therefore{f un+1}converges to f xin (f(X), d).

Similarly we may show that {f vn+1} converges to f y and {f wn+1} converges to f z.

Analogously we can show that {f un+1} converges to f x^∗, {f vn+1} converges to f y^∗ and{f wn+1} converges tof z^∗ in (f(X), d).

Since the coneP is closed and{f un+1}converges tof xandf x^∗, we getf x=f x^∗. Similarlyf y=f y^∗ andf z=f z^∗.

Thus we have that if (x, y, z) and (x^∗, y^∗, z^∗) are triple coincidence points ofT and f, then

α =T(x, y, z) =f x=f x^∗=T(x^∗, y^∗, z^∗), β =T(y, x, y) =f y=f y^∗=T(y^∗, x^∗, y^∗) and γ =T(z, y, x) =f z=f z^∗=T(z^∗, y^∗, x^∗).

Since (T, f) isw- compatible, then

f α =f(f x) =f(T(x, y, z)) =T(f x, f y, f z) =T(α, β, γ), f β =f(f y) =f(T(y, x, y)) =T(f y, f x, f y) =T(β, α, β) and f γ =f(f z) =f(T(z, y, x)) =T(f z, f y, f x) =T(γ, β, α).

Hence the triple (α, β, γ) is also triple coincidence point ofT andf. Thus we have f α=f x, f β=f y andf γ=f z.

Therfore

α=f α=T(α, β, γ), β=f β=T(β, α, β) andγ=f γ=T(γ, β, α).

Thus (α, β, γ) is common triple fixed point ofT andf.

To prove uniqueness, let (p, s, t) be any common triple fixed point of T andf. Thenp=f p=T(p, s, t), s=f s=T(s, p, s) andt=f t=T(t, s, p).

Since the triple (p, s, t) is a triple coincidence point ofT andf. We have

f p=f x, f s=f y andf t=f z.

Thus

p=f p=f x=α, s=f s=f y=β andt=f t=f z=γ.

Hence the common triple fixed point is unique.

Example 2.3. Let X =R⁺= [0,+∞)and the order relation ≺, be defined by x≺ y ⇔ {(x=y) or(x, y∈[0,1]withx≤y)}.Let E =C_R¹[0,1]with the norm

||x|| = ||x||_∞+||x⁰||_∞ and consider the cone P = {x ∈ E : x(t) ≥ 0 on [0,1]}

(this cone is not normal). Define d : X×X → E by d(x, y) =|x−y|ϕ, where ϕ: [0,1]→R such that ϕ(t) =e^t. It is easy to see thatdis a cone metric onX.

Consider the mappings T : X ×X ×X → X and f : X → X are defined as F(x, y, z) =^x+y+z₁₂ and f(x) =^x₂ respectively.

Clearly

d(T(x, y, z), T(u, v, w)) =|^x+y+z₁₂ −^u+v+w₁₂ |ϕ

=|₁₂^x +₁₂^y +₁₂^z −₁₂^u −₁₂^v −₁₂^w|ϕ

≤ |₁₂^x −₁₂^u|+|₁₂^y −₁₂^v|+|₁₂^z −₁₂^w| ϕ

=¹₆|^x₂ −^u₂|ϕ+¹₆|^y₂ −^v₂|ϕ+¹₆|^z₂ −^w₂|ϕ

=¹₆d(f x, f u) +¹₆d(f y, f v) +¹₆d(f z, f w).

Also all conditions of Theorem 2.1 and Theorem 2.2 are hold.

Clearly(0,0,0) is the unique common triple fixed point ofT andf.

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K. P. R. Rao,

Department of Applied Mathematics, A.N.U.-Dr.M.R.Appa Row Campus, Nuzvid-521 201, Andhra Pradesh, India

E-mail address:[email protected]

G.N.V.Kishore,

Department of Mathematics, Swarnandhra Institute of Engineering and Technology, Seetharampuram, Narspur- 534 280 , West Godavari District, Andhra Pradesh, India

E-mail address:[email protected]