METRIC SPACES AND COMMON FIXED POINT THEOREMS

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METRIC SPACES AND COMMON FIXED POINT THEOREMS

LJILJANA GAJI ´C AND VLADIMIR RAKO ˇCEVI ´C

Received 29 September 2004 and in revised form 24 January 2005

We consider quasicontraction nonself-mappings on Takahashi convex metric spaces and common fixed point theorems for a pair of maps. Results generalizing and unifying fixed point theorems of Ivanov, Jungck, Das and Naik, and ´Ciri´c are established.

1. Introduction and preliminaries

LetXbe a complete metric space. A mapT:X→Xsuch that for some constantλ∈(0, 1) and for everyx,y∈X

d(Tx,T y)≤λ·maxd(x,y),d(x,Tx),d(y,T y),d(x,T y),d(y,Tx) (1.1) is calledquasicontraction. Let us remark that Ćirić [1] introduced and studied quasicontraction as one of the most general contractive type map. The well known Ćirić’s result (see, e.g., [1,6,11]) is that quasicontractionTpossesses a unique fixed point.

For the convenience of the reader we recall the following recent ´Ciri´c’s result.

Theorem1.1 [2, Theorem 2.1]. LetXbe a Banach space,Ca nonempty closed subset ofX, and∂Cthe boundary ofC. LetT:C→Xbe a nonself mapping such that for some constant λ∈(0, 1)and for everyx,y∈C

d(Tx,T y)≤λ·maxd(x,y),d(x,Tx),d(y,T y),d(x,T y),d(y,Tx). (1.2) Suppose that

T(∂C)⊂C. (1.3)

ThenThas a unique fixed point inC.

Following ´Ciri´c [3], let us remark thatproblem to extend the known fixed point theorem for self mappingsT:C→C, defined by (1.1), to corresponding nonself mappingsT:C→X, C=X, was open more than20years.

In 1970, Takahashi [15] introduced the definition of convexity in metric space and generalized same important fixed point theorems previously proved for Banach spaces. In

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this paper we consider quasicontraction nonself-mappings on Takahashi convex metric spaces and common fixed point theorems for a pair of maps. Results generalizing and unifying fixed point theorems of Ivanov [7], Jungck [8], Das and Naik [3], Cirić [2], Gajić [5] and Rakoˇcević [12] are established.

Let us recall that (see Jungck [9]) the self maps f andg on a metric space (X,d) are said to be acompatible pairif

nlim→∞dg f xn,f gxn

=0 (1.4)

whenever{x_n}is a sequence inXsuch that

nlim→∞gx_n₌lim

n→∞f x_n₌x (1.5)

for somexinX.

Following Sessa [14] we will say that f,g:X→Xareweakly commutingif

d(f gx,g f x)≤d(f x,gx) for everyx∈X. (1.6) Clearly weak commutativity of f andg is a generalization of the conventional commutativity of f andg, and the concept of compatibility of two mappings includes weakly commuting mappings as a proper subclass.

We recall the following definition of a convex metric space (see [15]).

Definition 1.2. LetXbe a metric space andI=[0, 1] the closed unit interval. A Takahashi convex structure onXis a functionW:X×X×I→Xwhich has the property that for everyx,y∈Xandλ∈I

dz,W(x,y,λ)≤λd(z,x) + (1−λ)d(z,y) (1.7) for everyz∈X. If (X,d) is equipped with a Takahashi convex structure, thenXis called a Takahashi convex metric space.

If (X,d) is a Takahashi convex metric space, then forx,y∈Xwe set seg[x,y]=

W(x,y,λ) :λ∈[0, 1]. (1.8) Let us remark that any convex subset of normed space is a convex metric space with W(x,y,λ)=λx+ (1−λ)y.

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2. Main results

The next theorem is our main result.

Theorem2.1. Let(X,d)be a complete Takahashi convex metric space with convex struc- tureWwhich is continuous in the third variable,Ca nonempty closed subset ofXand∂C the boundary ofC. Letg:C→X, f :X→X and f :C→C. Suppose that∂C= ∅, f is continuous, and let us assume that f andgsatisfy the following conditions.

(i)For everyx,y∈C

d(gx,g y)≤Mω(x,y), (2.1)

where

Mω(x,y)=maxωd(f x,f y),ωd(f x,gx),ωd(f y,g y),

ωd(f x,g y),ωd(f y,gx), (2.2) ω: [0, +∞)→[0, +∞)is a nondecreasing semicontinuous function from the right, such that ω(r)< r, forr >0, andlimr→∞[r−ω(r)]=+∞.

(ii) f andgare a compatible pair onC, that is,

nlim→∞dg f xn,f gxn

=0 (2.3)

whenever{xn}is a sequence inCsuch that

nlim→∞gx_n=lim

n→∞f x_n=x (2.4)

for somexinX.

(iii)

g(C)C⊂ f(C). (2.5)

(iv)

g(∂C)_⊂C. (2.6)

(v)

f(∂C)⊃∂C. (2.7)

Then f andghave a unique common fixed pointzinC.

Proof. Starting with an arbitraryx0∈∂C, we construct a sequence{xn} of points in C as follows. By (2.6) g(x0)∈C. Hence, (2.5) implies that there is x1∈C such that f(x1)=g(x0). Let us consider g(x1). If g(x1)∈C, again by (2.5) there is x2∈C such that f(x2)=g(x1). Suppose thatg(x1)∈C. Now, becauseWis continuous in the third

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variable, there existsλ11∈[0, 1] such that Wfx1

,gx1

,λ11

∈∂Csegfx1

,gx1

. (2.8)

By (2.7) there isx2∈∂Csuch thatf(x2)=W(f(x1),g(x1),λ11).

Hence, by induction we construct a sequence{xn}of points inCas follows. Ifg(xn)∈ C, than by (2.5) f(x_n+1)=g(x_n) for somex_n+1∈C; ifg(x_n)∈C, then there existsλ_nn∈ [0, 1] such that

Wfxn

,gxn

,λnn

∈∂Csegfxn

,gxn

. (2.9)

Now, by (2.7) pickx_n+1∈∂Csuch that fxn+1

=Wfxn ,gxn

,λnn

. (2.10)

Let us remark (see [6]) that for everyx,y∈Xand everyλ∈[0, 1]

d(x,y)=dx,W(x,y,λ)+dW(x,y,λ),y. (2.11) Furthermore, ifu∈Xandz=W(x,y,λ)∈seg[x,y] then

d(u,z)=du,W(x,y,λ)≤maxd(u,x),d(u,y). (2.12) First let us prove that

fxn+1

=gxn

=⇒ fxn

=gxn−1

. (2.13)

Suppose the contrary that f(x_n)=g(x_n₋1). Thenx_n_∈∂C. Now, by (2.5)g(x_n)∈C, hence f(x_n+1)=g(x_n), a contradiction. Thus we prove (2.13).

We will prove thatg(xn) andf(xn) are Cauchy sequences. First we will prove that these sequences are bounded, that is that the set

A= _∞

i=0

fxi _∞

i=0

g(xi)

(2.14) is bounded.

For eachn≥1 set

A_n= _n₋₁

i=0

fx_i

_n₋₁

i=0

gx_i

, an=diamAn

.

(2.15)

We will prove that

an=maxdfx0

,gxi

: 0≤i≤n−1. (2.16)

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Ifa_n=0, then f(x0)=g(x0). We will prove thatg(x0) is a common fixed point for f and g. By (2.3) it follows that

f gx0

=g fx0

=ggx0

. (2.17)

Now we obtain

dggx0

,gx0

≤Mω

gx0,x0

=ωdgg(x0

,gx0

, (2.18)

and hencegg(x0)=g(x0). From (2.17), we conclude thatg(x0)=zis also a fixed point of f. To prove the uniqueness of the common fixed point, let us suppose that f u=gu=u for someu∈C. Now, by (2.1) we have

d(z,u)=d(gz,gu)≤Mω(z,u)=ωd(z,u), (2.19) and so,z=u.

Suppose thata_n>0. To prove (2.16) we have to consider three cases.

Case 1. Suppose thatan=d(f xi,gxj) for some 0≤i,j≤n−1.

(1i) Now, ifi≥1 and f x_i=gx_i₋1, we have an=df xi,gxj

=dgxi−1,gxj

≤Mω xi−1,xj

≤ωan

< an. (2.20) and we get a contradiction. Hencei=0.

(1ii) Ifi≥1 and f x_i=gx_i₋1, we havei≥2, andf x_i₋1=gx_i₋2. Hence f xi∈seggxi−2

,gxi−1

, (2.21)

we have

an=df xi,gxj

≤maxdgxi−2,gxj

,dgxi−1,gxj

≤maxM_ωx_i₋2,x_j,M_ωx_i₋1,x_j≤ωa_n)< a_n (2.22) and we get a contradiction.

Case 2. Suppose thata_n=d(f x_i,f x_j) for some 0≤i,j≤n−1.

(2i) Iff xj=gxj−1, then Case (2i) reduces to Case (1i).

(2ii) Iff xj=gxj−1, then as in the Case (1ii) we havej≥2, f xj−1=gxj−2, and f xj∈∂Cseggxj−2,gxj−1

. (2.23)

Hence

an=df xi,f xj

≤maxdf xi,gxj−2

,df xi,gxj−1

(2.24)

and Case (2ii) reduces to Case (1i).

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Case 3. The remaining casea_n=d(gx_i,gx_j) for some 0≤i,j≤n−1, is not possible (see Case (1i)). Hence we proved (2.16).

Now

an=df x0,gxi

≤df x0,gx0

+dgx0,gxi

≤df x0,gx0

+ω(an), (2.25) a_n−ωa_n≤df x0,gx0

. (2.26)

By (i) there isr0∈[0, +∞) such that

r−ω(r)> df x0,g y0

, forr > r0. (2.27) Thus, by (2.26)

an≤r0, n=1, 2,. . ., (2.28)

and clearly

a=lim

n→∞an=diam(A)≤r0. (2.29)

Hence we proved thatgxnand f xnare bounded sequences.

To prove thatgx_nand f x_nare Cauchy sequences, let us consider the set Bn=

_∞

i=n

f xi

_∞

i=n

gxi

, n=2, 3,. . . . (2.30) By (2.16) we have

bn≡diamBn

=sup

j≥n

df xn,gxj

, n=1, 2,. . . . (2.31)

Iff xn=gxn−1, then as in Case (1i) for each j≥n bn=df xn,gxj

=dgxn−1,gxj

≤ωbn−1

, n=1, 2,. . . . (2.32) Iff x_n=gx_n₋1, then as in Case (1ii) for eachn≥1 andj≥n

b_n=df x_n,gx_j≤maxdgx_n₋2,gx_j,dgx_n₋1,gx_j≤ωb_n₋2

. (2.33)

By (2.32) and (2.33) we get

b_n≤ωb_n₋2

, n=2, 3,. . . . (2.34)

Clearly,bn≥bn+1for eachn, and set limnbn=b. We will prove thatb=0. Ifb >0, then (2.34) and (i) implyb≤ω(b)< b, and we get a contradiction. It follows that both f x_nand gxnare Cauchy sequences. Since f xn∈CandCis a closed subset of a complete metric spaceXwe conclude that limnf xn=y∈C. Furthermore,

dfxn

,gxn

−→0, n−→ ∞, (2.35)

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implies limg(x_n)=y. Hence, limgxn

=limfxn

=y∈C. (2.36)

By continuity of f

limfgxn

=limffxn

=f(y)∈C. (2.37)

Now, by (2.3), we have

dg fx_n),f(y)≤dg fx_n,f gx_n+df gx_n,f(y)−→0, n−→ ∞, (2.38) that is

lim(g f)xn

=f(y). (2.39)

Now,

M_ωf x_n,y−→ωd(f y,g y) n−→ ∞, dg f xn,g y≤Mω

f xn,y n−→ ∞, (2.40)

implies

d(f y,g y)≤ωd(f y,g y). (2.41) Hence, f(y)=g(y), andg yis a common fixed point of f andg(see (2.17)).

In the special case, whenω(r)=λ·rwhere 0< λ <1, we obtain the following result.

Theorem2.2. Let(X,d)be a complete Takahashi convex metric space with convex struc- tureWwhich is continuous in the third variable,Ca nonempty closed subset ofXand∂C the boundary ofC. Letg:C→X, f :X→X and f :C→C. Suppose that∂C= ∅, f is continuous, and let us assume that f andgsatisfy the following conditions.

(i)There exists a constantλ∈(0, 1)such that for everyx,y∈C

d(gx,g y)≤λ·M(x,y), (2.42)

where

M(x,y)=maxd(f x,f y),d(f x,gx),d(f y,g y),d(f x,g y),d(f y,gx). (2.43) Suppose that the conditions (ii)–(v) inTheorem 2.1are satisfied. Thenf andghave a unique common fixed pointzinCandgis continuous atz. Moreover, ifz_n∈C,n=1, 2,. . .,then

limdf zn,gzn

=0 iﬀ lim

n zn=z. (2.44)

Proof. ByTheorem 2.1we know that f andghave a unique common fixed pointzinC.

Now, we show thatg is continuous atz. Let{y_n}be a sequence inCsuch thaty_n→z.

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Now we have

dg y_n,gz≤λ·My_n,z

=λ·maxdf yn,f z,df yn,g yn

,df z,g yn

=λ·maxdf y_n,f z,df y_n,g y_n

≤λ·

df yn,f z+df z,g yn

,

(2.45)

that is

dg yn,gz≤(1−λ)⁻¹λ·df yn,f z. (2.46) Therefore, we haveg y_n→gzand sogis continuous atz. To prove (2.44), let us suppose thatw∈C. Now, since f z=gz=z, we have

d(f w,gw)≤d(f w,f z) +d(gw,gz)≤d(f w,f z) +λ·M(w,z)

≤d(f w,f z) +λ·maxd(f w,f z),d(f w,gw),d(f z,gw)

≤d(f w,f z) +λ·

d(f w,f z) +d(f w,gw),

(2.47)

that is

(1−λ)d(f w,gw)≤(1 +λ)d(f w,f z). (2.48) Let us remark that

d(f w,f z)≤d(f w,gw) +d(gw,gz)≤d(f w,gw) +λ·M(w,z)

≤d(f w,gw) +λ·maxd(f w,f z),d(f w,gw),d(f z,gw)

≤d(f w,gw) +λ·

d(f w,f z) +d(f w,gw),

(2.49)

that is

(1−λ)d(f w,f z)≤(1 +λ)d(f w,gw). (2.50) By (2.48) and (2.50) we obtain

(1−λ)d(f w,gw)≤(1 +λ)d(f w,f z)

≤(1−λ)⁻¹(1 +λ)²d(f w,gw). (2.51)

Clearly (2.51) implies (2.44).

Remark 2.3. Let (K,ρ) be a bounded metric space. It is said that the fixed point problem for a mappingA:K→K iswell posed if there exists a uniquexA∈K such that Ax_A=x_Aand the following property holds: If {x_n} ⊂K andρ(x_n,Ax_n)→0 asn→ ∞, thenρ(x_n,x_A)→0 asn→ ∞. Let us remark that condition (2.44) is related to the notion

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of well posed fixed point problem, and the notion of well-posedness is of central impor- tance in many areas of Mathematics and its applications ([4,10,13]).

Remark 2.4. If inTheorem 2.1we let f be the identity map onXandω(r)=λ·rwhere 0< λ <1, we get Ćirić’sTheorem 1.1(Gajić’s theorem [5]) stated for a Banach (convex complete metric) spaceX.

Remark 2.5. If in Theorem 2.1we let f be the identity map onX andC=X, we get Ivanov’s result [6,7] stated for a Banach spaceX.

Remark 2.6. Let us recall that the first part ofTheorem 2.2, that is the existence of the unique common fixed point of f andg was proved by Rakoˇcevi´c [12].

By the proof ofTheorem 2.1we can recover some results of Das and Naik [3] and Jungck [8].

Corollary2.7 [3, Theorem 2.1]. LetXbe a complete metric space. Let f be a continuous self-map onXandgbe any self-map onXthat commutes with f. Further let f andgsatisfy

g(X)⊂ f(X) (2.52)

and there exists a constantλ∈(0, 1)such that for everyx,y∈X

d(gx,g y)≤λ·M(x,y), (2.53)

where

M(x,y)=maxd(f x,f y),d(f x,gx),d(f y,g y),d(f x,g y),d(f y,gx). (2.54) Then f andghave a unique fixed point.

Proof. We follow the proof ofTheorem 2.1. Let us remark that the condition (2.52) implies that starting with an arbitraryx0∈X, we construct a sequence {x_n}of points in Xsuch that f(xn+1)=g(xn),n=0, 1, 2,. . . .The rest of the proof follows by the proof of

Theorem 2.1.

Corollary2.8 [3, Theorem 3.1]. LetXbe a complete metric space. Letf²be a continuous self-map onXandgbe any self-map onXthat commutes with f. Further let f andgsatisfy

g f(X)⊂ f²(X) (2.55)

and f(g(x))=g(f(x))whenever both sides are defined. Further, let there exist a constant λ∈(0, 1)such that for everyx,y∈f(X)

d(gx,g y)≤λ·M(x,y), (2.56)

where

M(x,y)=maxd(f x,f y),d(f x,gx),d(f y,g y),d(f x,g y),d(f y,gx). (2.57) Then f andghave a unique common fixed point.

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Proof. Again, we follow the proof ofTheorem 2.1. By (2.55) starting with an arbitrary x0∈ f(X), we construct a sequence{x_n}of points in f(X) such that f(x_n+1)=g(x_n)= yn,n=0, 1, 2,. . . .Nowf(yn)= f(g(xn))=g(f(xn))=g(yn−1)=zn,n=1, 2,. . ., and from the proof ofTheorem 2.1we conclude that{zn}is a Cauchy sequence in Xand hence convergent to somez∈X. Now, for eachn≥1

df²gx_n,g f(z)

=dg f²xn

,g f(z)≤λ·Mf²xn ,f(z)

=λ·maxdf²fx_n,f²(z),df²fx_n,f²gx_n,

df²(z),g f(z),df²fx_n,g f(z),df²(z),f²gx_n.

(2.58)

Now, by continuity of f²

df²(z),g f(z)≤λ·df²(z),g f(z). (2.59) Whence, f²(z)=g f(z), andg f zis a unique common fixed of f andg. Let us remark that fromTheorem 2.1and the proof ofCorollary 2.7, we get the following.

Corollary2.9. LetXbe a complete metric space. Letf be a continuous self-map onXand gbe any self-map onXthat weakly commutes with f. Further letf andgsatisfy (2.52) and (2.53). Then f andghave a unique common fixed point.

Now as a corollary we get the following result of Jungck [8].

Corollary2.10. LetXbe a complete metric space. Let f be a continuous self-map onX andg be any self-map onX that commutes with f. Further let f andg satisfy (2.52) and there exists a constantλ∈(0, 1)such that for everyx,y∈X

d(gx,g y)≤λ·d(f x,f y). (2.60) Then f andghave a unique common fixed point.

Corollary2.11. LetXbe a convex complete metric space,Ca nonempty compact subset of X, and∂Cthe boundary ofC. Letg:C→X,f :X→Xand f :C→C. Suppose thatgand f are continuous,f andgsatisfy the conditions(ii)–(v)inTheorem 2.1, and for allx,y∈C, x=y

d(gx,g y)< M(x,y), (2.61)

where

M(x,y)=maxd(f x,f y),d(f x,gx),d(f y,g y),d(f x,g y),d(f y,gx). (2.62) Then f andghave a unique common fixed point inC.

Proof. ByTheorem 2.2and the proof of [12, Theorem 4].

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Acknowledgment

The authors are grateful to the referees for some helpful comments and suggestions.

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Ljiljana Gaji´c: Institute of Mathematics, Faculty of Science, University of Novi Sad, Trg D.

Obradovi´ca 4, 21000 Novi Sad, Serbia and Montenegro E-mail address:[email protected]

Vladimir Rakoˇcevi´c: Department of Mathematics, Faculty of Sciences and Mathematics, University of Niˇs, Viˇsegradska 33, 18000 Niˇs, Serbia and Montenegro

E-mail address:[email protected]

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