On Mappings with Contractive Iterate at a Point in Generalized Metric Spaces

Volume 2010, Article ID 458086,16pages doi:10.1155/2010/458086

Research Article

On Mappings with Contractive Iterate at a Point in

Generalized Metric Spaces

Ljiljana Gaji ´c and Zagorka Lozanov-Crvenkovi ´c

Department of Mathematics and Informatics, University of Novi Sad, Trg Dositeja Obradov´ca 4, 21000 Novi Sad, Serbia

Correspondence should be addressed to Ljiljana Gaji´c,[email protected]

Received 7 September 2010; Revised 1 December 2010; Accepted 29 December 2010 Academic Editor: B. Rhoades

Copyrightq2010 L. Gaji´c and Z. Lozanov-Crvenkovi´c. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Using the setting of generalized metric space, the so-calledG-metric space, fixed point theorems for mappings with a contractive and a generalized contractive iterate at a point are proved. These results generalize some comparable results in the literature. A common fixed point result is also proved.

1. Introduction

Sehgal in1proved fixed point theorem for mappings with a contractive iterate at a point and therefore generalized a well-known Banach theorem.

Theorem 1.1. LetX, dbe a complete metric space and letT :X → Xbe a continuous mapping with property that for everyx∈Xthere existsnx∈Nso that for everyy∈X

dTnxx, Tnxy≤q·dx, y, whereq∈0,1. 1.1

ThenT has a unique fixed pointuinXandlimkTkx0 u, for eachx0∈X.

Guseman2extended Sehgal’s result by removing the condition of continuity ofT

and weakening 1.1to hold on some subset Bof X such thatTB ⊆ B, where, for some

x0 ∈B,Bcontains the closure of the iterates ofx0. Further extensions appear in3,4. Our

results obtained in the two settings. Because of that, Dhage5introduced a new concept of the measure of nearness between three or more object. But topological structure of so called

D-metric spaces was incorrect. Finally, Mustafa and Sims6introduced correct definition of generalized metric space as follows.

Definition 1.2see6. LetXbe a nonempty set, and letG :X×X×X → Rbe a function

satisfying the following properties:

G1Gx, y, z 0 ifxyz;

G20< Gx, x, y; for allx, y∈X, with_{x /}y;

G3Gx, x, y≤Gx, y, z, for allx, y, z∈X, with_{z /}y;

G4Gx, y, z Gx, z, y Gy, z, x · · ·,symmetry in all three variables;

G5Gx, y, z≤Gx, a, a Ga, y, z, for allx, y, z, a∈X.

Then the functionGis called a generalized metric, or, more specifically, aG-metric onX, and the pairX, Gis called aG-metric space.

Clearly these properties are satisfied whenGx, y, zis the perimeter of triangle with vertices atx,y, andz∈R2_{, moreover takingain the interior of the triangle shows thatG5}

is the best possible.

Example 1.3. LetX, dbe an ordinary metric apace, thenX, dcan defineG-metrics onXby

EsGsx, y, z dx, y dy, z dx, z,

EmGmx, y, z max{dx, y, dy, z, dx, z}.

Example 1.4see6. Letx{a, b}. DefineGonX×X×Xby

Ga, a, a Gb, b, b 0, Ga, a, b 1, Ga, b, b 2, 1.2

and extendGtoX×X×Xby using the symmetry in the variables. Then it is clear theX, G

is aG-metric space.

Definition 1.5see6. LetX, Gbe aG-metric space, and let{xn}be sequence of points of

X, a pointx∈Xis said to be the limit of the sequence{xn}, if limn,m→ ∞Gx, xn, xm 0, and

one says that the sequence{xn}isG-convergent tox

Thus, ifxn → xin aG-metric spaceX, G, then for any >0, there existsN∈Nsuch

thatGx, xn, xm< , for alln, m≥N.

Definition 1.6see6. LetX, Gbe aG-metric space, a sequence{xn}is calledG-Cauchy

if for every > 0, there is N ∈ Nsuch thatGxn, xm, xl < , for alln, m, l ≥ N; that is, if

Gxn, xm, xl → 0 asn, m, l → ∞.

AG-metric spaceX, Gis said to beG-completeor completeG-metricif everyG -Cauchy sequence inX, GisG-convergent inX, G.

Definition 1.8 see 6. A G-metric space X, G is called symmetric G-metric space if

Gx, y, y Gy, x, x, for allx, y∈X.

Proposition 1.9see6. EveryG-metric spaceX, Gwill define a metric spaceX, dGby

dG

x, yGx, y, yGy, x, x, ∀x, y∈X. 1.3

Note that ifX, Gis a symmetricG-metric space, then

dG

x, y2Gx, y, y, ∀x, y∈X. 1.4

However, ifX, Gis nonsymmetric, then by theG-metric properties it follows that

3 2G

x, y, y≤dG

x, y≤3Gx, y, y, ∀x, y∈X, 1.5

and that in general these inequalities cannot be improved.

Proposition 1.10 see 6. A G-metric space X, G is G-complete if and only ifX, dG is a

complete metric space.

In recent years a lot of interesting papers were published with fixed point results inG -metric spaces, see7–18. This paper is our contribution to the fixed point theory inG-metric spaces.

2. Fixed Point Results

LetX, Gbe aG-metric space,f : X → Xa mapping,B⊆ Xsuch that for someq∈ 0,1 and each forx∈Bthere exists a positive integernnxsuch that

Gfnxz, fnxx, fnxx

≤q·maxGz, x, x, Gz, fnxx, fnxx, Gfnxz, x, x

2.1

for allz∈B. Then we writef∈ 1. If

Gfnxz, fnxx, fnxx

≤q·max

Gz, x, x,1

2 G

z, fnxz, fnxzGx, fnxx, fnxx,

1 2 G

z, fnxx, fnxxGfnxz, x, x

2.2

Theorem 2.1. Letf ∈ 1orf∈ 2. LetB⊆X, withfB⊆B. If there existsu∈Bsuch that for

nnu,fn_{u u, thenuis the unique fixed point offinB. Moreover,f}k_y

0 → u,k → ∞, for

anyy0∈Bforf ∈ 1, and forf ∈ 2ifq <2/3.

Proof. IfX, Gis a symmetric space thandGz, x 2Gz, x, xand2.1becomes

dG

fnxz, fnxx≤qmaxdGz, x, dG

z, fnxx, dG

x, fnxz, 2.3

and2.2becomes

dG

fnxz, fnxx≤qmax

dGz, x,

1 2 dG

z, fnxzdG

x, fnxx,

1 2 dG

z, fnxxdG

x, fnxz,

2.4

thus the result follows from Theorem 12 in3and it is valid for anyq <1. Suppose now that

X, Gis nonsymmetric space. Then by inequality1.5we have that2.1becomes

Gfnxz, fnxx, fnxx

≤2q·maxGz, x, x, Gz, fnxx, fnxx, Gfnxz, x, x,

2.5

and2.2becomes

Gfnxz, fnxx, fnxx

≤2q·max

Gz, x, x,1

2 G

z, fnxz, fnxzGx, fnxx, fnxx,

1 2 G

z, fnxx, fnxxGfnxz, x, x.

2.6

Since 2qneed not be less then 1 we can use metric fixed point results only forq <1/2. On the other side, using the concept ofG-metric space, we are going to prove the result, if the first case for any 0< q <1, and in the second one for 0< q <2/3. This means that our results are real generalization in the case of nonsymmetricG-metric spaces.

Let f ∈ 1. Uniqueness follows from 2.1, since for fn_{z z, it follows that}

Lety0∈B, and assumefmy0/ufor eachm. Formsufficiently large writemknr,

k≥1, and 1≤r < n. Then

Gfmy0

, u, uGfknry0

, fnu, fnu

≤qmaxGfk−1nry0

, u, u, Gfk−1nry0

, fnu, fnu,

Gfm_y 0

, u, u,

qGfk−1nry0

, u, u≤ · · · ≤qkGfry0

, u, u

≤qkmaxGfpy0

, u, u: 1≤p < n,

2.7

soGfm_y

0, u, u → 0,m → ∞.

Iff ∈ 2, uniqueness follows from2.2since forfn_{z z, it follows thatGz, u, u}

Gfn_{z, f}n_{u, f}n_{u≤qmax{Gz, u, u,0}and furtherfu u. Now for anyy} 0∈B

Gfmy0

, u, uGfknry0

, fnu, fnu≤qMy0, m, u

, 2.8

where

My0, m, u

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

Gfk−1nr_y 0

, u, u,

1 2

Gfk−1nr_y 0

, fm_y 0

, fm_y 0

0,

1 2

Gfk−1nr_{y0, u, uGf}m_y 0

, u, u.

2.9

ForMy0, m, u 1/2Gfk−1nry0, u, u Gfmy0, u, uwe have

1 2G

fk−1nry0

, u, u< 1

2G

fmy0

, u, u,

1 2G

fmy0

, u, u< 1

2G

fk−1nry0

, u, u

2.10

which is a contradiction, and therefore

My0, m, u

max

Gfk−1nry0

, u, u,1

2G

fk−1nry0

, fmy0

, fmy0

IfMy0, m, u 1/2Gfk−1nry0, fmy0, fmy0then

Gfmy0

, u, u≤ q

2G

fk−1nry0

, u, uqGu, u, fmy0

. 2.12

So

21−qGfmy0

, u, u≤qGfk−1nry0

, u, u. 2.13

Therefore,Gfm_y

0, u, u≤hGfk−1nry0, u, u, wherehmax{q, q/21−q}. Forh <1,

Gfm_y

0, u, u → ∞,m → ∞.

Forf:X → Xthe setOf;x0 {fnx0:n∈N}is called the orbit forx0∈X. Theorem 2.2. LetX, Gbe a completeG-metric space and letf :X → X be a mapping. Suppose that for somex0∈Xthe orbitOf;x0is complete, and that: for someq∈0,1and eachx∈ Of;x0

there is an integernx≥1such that

Gfnxz, fnxx, fnxx≤q·Gz, x, x 2.14

for allz∈ Of;x0.

Thenxkfnxk−1xk−1,k∈N, converges to someu∈Xand for allm, k∈N,m > k

Gxk, xk, xm≤ q k

1−q2max

Gfp_x

0, x0, x0

: 1≤p≤nx0

2.15

If inequality in2.14holds for allx∈ Of;x0, thenfnuu uandfkx0 → u,k → ∞.

Moreover, iffOf;x0⊆ Of;x0, thenuis the fixed point off.

Proof. IfX, G is a symmetric G-metric space the statement easily follows from Guseman

fixed point result2. LetX, Gbe nonsymmetricG-metric space. Then by inequality1.5

dG

fnxz, fnxx≤2qdGz, x. 2.16

Thus, one can use the fixed point result in metric space only forq <1/2. But here, using the concept ofG-metric, we prove the result for any 0< q <1. At first let us show that

sup

m G

fmx0, x0, x0

For anym ∈ N, sufficiently large, there existk, r ∈ N, 1 ≤ r ≤ nx0−1 such that

mk·nx0 r. Then

Gfmx0, x0, x0

≤Gfknx0r_x

0, fnx0x0, fnx0x0

Gfnx0_x

0, x0, x0

≤qGfk−1nx0r_x

0, x0, x0

Gfnx0_x

0, x0, x0

≤qGfk−1nx0r_x

0, fnx0x0, fnx0x0

1qGfnx0_x

0, x0, x0

≤q2Gfk−2nx0r_x

0, x0, x0

1qGfnx0_x

0, x0, x0

≤ · · ·

≤qkGfrx0, x0, x0

1q· · ·qk−1Gfnx0_x

0, x0, x0

≤ 1

1−qmax

Gfpx0, x0, x0

: 1≤p≤nx0

M <∞.

2.18

Now, for eachk∈N

Gxk, xk, xk1 G

fnxk−1_x

k−1, fnxk−1xk−1, fnxkfnxk−1xk−1

≤qGxk−1, xk−1, fnxkxk−1

≤ · · · ≤qkGx0, x0, fnxkx0

≤qkM.

2.19

For allm, k∈N,m > k, it follows that

Gxk, xk, xm≤Gxk, xk, xk1 Gxk1, xk1, xk2 · · ·Gxm−1, xm−1, xm≤ q k

1−qM,

2.20

so{xk}is Cauchy sequence and there existsulimkxk, for someu∈X, and inequality2.15

is proved.

If we suppose that inequality in2.14is satisfied for all x ∈ Of;x0, then, for all

k∈N,

Gfnuu, fnuu, fnuxk

≤qGu, u, xk 2.21

so limkfnuxk fnuu.

On the other hand,

Gfnuxk, xk, xk

Gfnufnxk−1_x

k−1, fnxk−1xk−1, fnxk−1xk−1

≤qGfnuxk−1, xk−1, xk−1

≤ · · · ≤qkGfnux0, x0, x0

2.22

SinceGis continuous it means that

Gfnuu, u, u0. 2.23

Hencefnuu u.

Now, let us suppose thatfOf;x0⊆ Of;x0. Sincef ∈ 1byTheorem 2.1uis the

fixed point offinXand limkfkx0 u.

Fornx 1, in inequality2.14independently onx, we are going to simplify the proof and to relax the condition in2.14.

Corollary 2.3. LetX, Gbe a completeG-metric apace and letf:X → X. Suppose that there exist a pointx0∈Xandq∈0,1withOf;x0complete and

Gfz, fx, fx≤qGz, x, x 2.24

for eachx, zfx∈ Of;x0. Then{fkx0}converges to some pointu∈Xand for allk, m∈N,

m > k,

Gxk, xk, xm≤ q k

1−qG

x0, x0, fx0

. 2.25

If 2.24holds, for allx∈ Of;x0orfis orbitally continuous atu, thenuis a fixed point off.

Proof. IfX, Gis a symmetric space thandGx, z 2Gz, x, xso2.24becomes

dG

fz, fx≤qdGz, x, 2.26

and result follows from Theorem 2 in19.

Now, letX, Gbe a nonsymmetricG-metric space. Then sincexkfxk−1,k∈N,

Gxk, xk, xk1≤qGxk−1, xk−1, xk ≤ · · · ≤qkG

x0, x0, fx0

, 2.27

so for allm, k∈N,m > k,

Gxk, xk, xm≤

qk

1−qG

x0, x0, fx0

, 2.28

and there existsu limkxk. If2.24holds for allx ∈ Of;x0, then byTheorem 2.2, since

nu 1, it follows thatfu u.

The fact thatfis orbitally continuous atx u, and that limkfkx0 u, implies that

Remark 2.4. Let us note that this result is very close to Theorem 2.1 in8.

Remark 2.5. In the statements abovefdoes not have to be continuous.

The next theorems are generalizations of ´Ciri´c fixed point results in4.

Theorem 2.6. LetX, Gbe a complete metric space andT :X → X a mapping. Suppose that for eachx∈Xthere exists a positive integernnxsuch that

GTnx, Tnx, Tny

≤qmax

Gx, x, y, Gx, x, Ty, . . . , Gx, x, Tny,1

2Gx, x, T

n_{x Gx, T}n_{x, T}n_x

2.29

holds for someq < 2/3and ally∈ X. ThenT has a unique fixed pointu∈X. Moreover, for every

x∈X,limmTmx u.

Proof. IfX, Gis a symmetric space thendGx, y 2Gx, x, yand inequality2.29becomes

dG

Tnx, Tny≤qmaxdG

x, y, dG

x, Ty, . . . , dGx, Tnx

, 2.30

for ally∈X. Then the result follows from Theorem 2.1 in4and it is true for allq <1. Now suppose thatX, Gis nonsymmetric space. Then, by definition of the metricdG

and inequality1.5we have

dG

Tnx, Tny≤2qmaxdG

x, y, dG

x, Ty, . . . , dG

x, Tny, dGx, Tnx

. 2.31

But 2qneed not to be less than 1, so we will prove the statement by usingG-metric. First, let us prove prove that

Gx, x, Tmx≤ 1

1−qbx, m1,2, . . . , 2.32

where

bx max

Gx, x, Tx, Gx, x, T2_{x, . . . , Gx, x, T}n_x,1

2Gx, x, T

n_{x Gx, T}n_{x, T}n_x

.

2.33

Clearly2.32is true form1,2, . . . , n. Suppose thatm > n, and that2.32is true fori≤m

and let us prove it forim1. Letm1−nr. Now

Gx, x, Tm1x,≤Gx, x, Tnx GTnx, Tnx, Tm1x,

GTn_{x, T}n_{x, T}m1_x≤qbx,

where

bx max

Gx, x, Trx, Gx, x, Tr1x, . . . , Gx, x, Trnx,

1

2Gx, x, T

n_{x Gx, T}n_{x, T}n_x

.

2.35

Ifbx Gx, x, Tnr, then2.34imply

Gx, x, Tm1x≤ 1

1−qGx, x, T

n_x≤ 1

1−qbx. 2.36

Ifbx/Gx, x, Tnr, then2.34imply

Gx, x, Tm1x≤Gx, x, Tnx q

1−qbx≤

1

1−qbx. 2.37

Thus by induction we obtain2.32. Let us prove that{Tm_x}

mis a Cauchy sequence. Letx0x,n0 nx0,x1Tn0x0, and

we define inductively a sequence of integers and a sequence of points{xk}kinXas follows:

nk nxk, andxk1 Tnkxk,k 0,1, . . .. Evidently, {xk}k is a subsequence of the orbit {Tm_x

0}m. Using this sequence we will prove that{Tmx0}mis a Cauchy sequence.

Letxk be any fixed member of {xk}k and letxp Tpx0 and xq Tqx0 be any two

members of the orbit which follow afterxk. Thenxp Trxkandxq Tsxkfor somerands,

respectively. Now, using2.29we get

Gxk, xk, xp

GTnk−1_x

k−1, Tnk−1xk−1, Tnk−1Trxk−1

≤ 3

2qGxk−1, xk−1, T

r1_x

k−1, 2.38

where

Gxk−1, xk−1, Tr1xk−1

maxGxk−1, xk−1, Trxk−1, G

xk−1, xk−1, Tr1xk−1

, . . . ,

Gxk−1, xk−1, Trnk−1xk−1, Gxk−1, xk−1, Tnk−1xk−1

.

2.39

Similarly,Gxk−1, xk−1, Tr1xk−1≤3/2qGxk−2, xk−2, Tr2xk−2, where

Gxk−2, xk−2, Tr2xk−2 max{Gxk−2, xk−2, Tr1xk−2, . . . , Gxk−2, xk−2, Tnk−2xk−2} 2.40

Repeating this argumentktimes we get

Gxk, xk, xp

≤

3 2q

k

HenceGxk, xk, xp≤3/2qkbx0. SimilarlyGxk, xk, xq≤3/2qkbx0, so

Gxp, xp, xq

≤2Gxk, xk, xp

Gxk, xk, xq

≤

3 2q

k

·3bx0. 2.42

Sinceq <2/3, it follows that{Tm_x

0}mis a Cauchy sequence. Let limmTmx0 u∈X.

We show thatuis a fixed point ofT. First, let us prove thatTn_{u u, wheren nu. For}

m≥nnu, we now have

GTnu, Tnu, TnTmx0

≤qmax

Gu, u, Tm_x 0, G

u, u, Tm1_x 0

, . . . , Gu, u, Tmnx0,

1

2Gu, u, T

n_{u Gu, T}n_{u, T}n_u

,

2.43

and on lettingmtend to infinity it follows that

GTnu, Tnu, u≤qmax

0,1

2Gu, u, T

n_{u Gu, T}n_{u, T}n_u

. 2.44

Forq <2/3 we haveTn_uu.

To show thatuis a fixed point ofT, let us suppose that_{Tu /}uand letGu, u, Tk_u

max{Gu, u, Tr_{u: 1≤r≤nnu}. Then}

Gu, u, TkuGTnu, Tnu, TnTku

≤qmax

Gu, u, Tku, Gu, u, Tk1u, . . . , Gu, u, Tknu, . . . ,

1

2Gu, u, T

n_{u Gu, T}n_{u, T}n_u

≤ 3

2qG

u, u, Tk_u.

2.45

Sinceq <2/3, it follows thatGu, u, Tk_{u 0, which implies thatuis a fixed point ofT.}

Let us suppose that for somez∈X,Tzz. Then,

Gu, u, z GTnu, Tnu, Tnz≤qmax{Gu, u, z,0} 2.46

If we suppose thatT is continuous, then we may prove the following theorem.

Theorem 2.7. LetX, Gbe a completeG-metric space and letT :X → Xbe a continuous mapping which satisfies the condition: for eachx∈Xthere is a positive integernnxsuch that

GTn_{x, T}n_{x, T}n_y

≤maxGx, x, y, Gx, x, Ty, . . . , Gx, x, Tny, Gx, x, Tx, . . . , Gx, x, Tnx 2.47

for someq <1and ally∈X. ThenThas a unique fixed pointu∈XandlimmTmx u∈X, for

everyx∈X.

Proof. Letxbe an arbitrary point inX. Then, as in the proof ofTheorem 2.6, the orbit{Tm_x} m

is bounded and is a Cauchy sequence in the completeG-metric spaceXand so it has a limit

uinX. Since by the hypothesisTis continuous,

TnuuTnulim

m T

m_xlim m T

mnu_{u. 2.48}

Therefore,uis a fixed point ofTnu. By the same argument as in the proof ofTheorem 2.6, it follows thatuis a unique fixed point ofT.

Remark 2.8. The condition thatT is a continuous mapping can be relaxed by the following

condition:Tnxis continuous at a pointx∈X.

3. A Common Fixed Point Result

Now, we are going to prove Hadˇzi´c20fixed point theorem in 2-metric space, in a manner ofG-metric spaces.

Theorem 3.1. LetX, Gbe a completeG-metric space,SandT : X → X one to one continuous

mappings,A:X → SX∩TXcontinuous mapping commutative withSandT. Suppose that there

exists a pointx0∈Xsuch thatOA;x0is complete and that the following conditions are satisfied:

iFor everyx∈ OA;x0there existsnx∈Nso that for allz∈Xand someq∈0,1

GAnxz, Anxx, Anxx

≤qmin{GTx, Tx, Sz, GTx, Sx, Tz, GTx, Sx, Sz, GSx, Sx, Tz}.

3.1

iiThere existsM >0such that for allz∈ OA;x0

Then there exists one and only one elementu∈Xsuch that

AuSuTuu. 3.3

e.g., there exists a unique common fixed point forA, S, andT

Proof. SinceAX⊆SX∩TXstarting withx0we can define the sequence{xn} ⊆Xsuch that

Tx2k−1Anx2k−2x2k−2,

Sx2kAnx2k−1x2k−1.

3.4

Let

yn

⎧ ⎨ ⎩

Tx2k−1, n2k−1,

Sx2k, n2k,

k∈N. 3.5

We are going to prove that{yn}is Cauchy sequence

Gy2k−1, y2k−1, y2k

GAnx2k−2_x

2k−2, Anx2k−2x2k−2, Anx2k−1x2k−1

GAnx2k−2_x

2k−2, Anx2k−2x2k−2, Anx2k−1T−1Anx2k−2x2k−2

≤qGSx2k−2, Sx2k−2, Anx2k−1x2k−2

qGAnx2k−3_x

2k−3, Anx2k−3x2k−3, Anx2k−1S−1Anx2k−3x2k−3

≤ · · · ≤q2k−2_GTx

1, Tx1, Anx2k−1x1

q2k−2GAnx0_x

0, Anx0x0, Anx2k−1T−1Anx0x0

≤q2k−1GSx0, Sx0, Anx2k−1x0

≤q2k−1M.

3.6

Similarly one can prove thatGy2k, y2k, y2k1≤q2kM,k∈N, for allm, k∈N,m > k,

Gyk, yk, ym

≤Gyk, yk, yk1

Gyk1, yk1, ym

≤ · · ·

≤m−1 jk

Gyj, yj, yj1

≤ qk

1−qM.

Thus we proved that{yn}is a Cauchy sequence, so there existsu∈Xsuch that

lim

n ynu. 3.8

It obvious that limkTx2k−1limkSx2ku.

At first we will prove thatAuu

Gy2k, y2k, Ay2k1

GSx2k, Sx2k, ATx2k1

GAnx2k−1_x

2k−1, Anx2k−1x2k−1, AAnx2kx2k

GAnx2k−1_x

2k−1, Anx2k−1x2k−1, AAnx2kS−1Anx2k−1x2k−1

≤qGTx2k−1, Tx2k−1, AAnx2kx2k−1

qGAnx2k−2_x

2k−2, Anx2k−2x2k−2, AAnx2kT−1Anx2k−2x2k−2

≤q2GSx2k−2, Sx2k−2, AAnx2kx2k−2

≤ · · ·

≤q2kGSx0, Sx0, AAnx2kx0

≤q2k·M,

3.9

so limkGy2k, y2k, Ay2k1 0.

Now, since thatGandAare continuous we have thatGu, u, Au 0 soAuu. Further, let us prove thatTuu.

Gy2k, y2k, Ty2k

GAnx2k−1_x

2k−1, Anx2k−1x2k−1, TAnx2k−1x2k−1

≤qGTx2k−1, Tx2k−1, STx2k−1

qGAnx2k−2_x

2k−2, Anx2k−2x2k−2, SAnx2k−2x2k−2

≤q2GSx2k−2, Sx2k−2, TSx2k−2

q2GAnx2k−3_x

2k−3, Anx2k−3x2k−3, TAnx2k−3x2k−3

≤ · · ·

≤q2kGSx0, Sx0, TSx0

3.10

implies that

lim

k G

y2k, y2k, Ty2k

Gu, u, Tu 0_{, 3.11}

andTuu. Similarly one can see thatSuu, so we prove that

If we suppose thatωis some other common fixed point forA, S, andT then we have that

Gu, u, ω GAnuu, Anuu, Anuω

≤qGSu, Su, Tω qGu, u, ω< Gu, u, ω

3.13

which is contradiction!

So, the common fixed point forA, S, andTis unique, and proof is completed.

Remark 3.2. ForS T IdX condition2.14is satisfied but theTheorem 2.2is not just a

consequence ofTheorem 3.1since inTheorem 2.2we do not suppose thatfis continuous.

Acknowledgments

The authors are thankful to professor B. E. Rhoades, for his advice which helped in improving the results. This work was supported by grants approved by the Ministry of Science and Technological Development, Republic of Serbia, for the first author by Grant no. 144016, and for the second author by Grant no. 144025.

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