R E S E A R C H
Open Access
On mappings with
ϕ
-contractive iterate at a
point on generalized metric spaces
Ljiljana Gaji´c
1and Mila Stojakovi´c
2**Correspondence: [email protected] 2Department of Mathematics, Faculty of Technical Sciences, University of Novi Sad, Novi Sad, Serbia
Full list of author information is available at the end of the article
Abstract
We prove a set of fixed point theorems for mappings with
ϕ
-contractive iterate at a point in a class of generalized metric spaces. This is a generalization of some well-known results for Guseman and Matkowski types of fixed point results in metric and generalized metric spaces.MSC: 47H10; 54H25
Keywords:
ϕ
-contraction; fixed point; mapping with contractive iterate at a point; generalized metric space1 Introduction and preliminaries
In Matkowski introduced the following class of mappings.
Definition .[] LetTbe a mapping on a metric space (X,d). ThenT is called a weak contraction if there exists a functionϕfrom [,∞) to itself satisfying the following:
(i) ϕis nondecreasing, (ii) limnϕn(t) = for allt> ,
(iii) d(Tx,Ty)≤ϕ(d(x,y))for allx,y∈X.
In the same paper he proved the existence and uniqueness of a fixed point for such type of mappings. This result is significant because the concept of weak contraction of Matkowski type is independent of the Meir-Keeler contraction [], and it was generalized in different directions [–]. Matkowski generalized his own result proving a theorem of Segal-Guseman type [].
Theorem .[] Let (X,d) be a complete metric space, T :X→X,and ϕ: [,∞)→ [,∞).Ifϕ is nondecreasing,limt→∞(t–ϕ(t)) =∞,limk→∞ϕk(t) = for t> ,and for
each x∈X there is a positive integer n=n(x)such that for all y∈X,
dTn(x)x,Tn(x)y≤ϕd(x,y),
then T has a unique fixed point a∈X.Moreover,for each x∈X,limk→∞Tk(x) =a.
The aim of this paper is to show that this result is valid in a more general class of spaces and wide class of functionsϕ.
In , Gähler introduced -metric spaces, but other authors proved that there is no relation between the two distance functions and there is no easy relationship between
sults obtained in the two settings. Dhage introduced a new concept of the measure of near-ness between three or more objects. But the topological structure of so-calledD-metric spaces was incorrect. Finally, Mustafa and Sims [] introduced the correct definition of a generalized metric space as follows.
Definition .[] LetXbe a nonempty set, and letG:X×X×X→R+be a function
satisfying the following properties:
(G) G(x,y,z) = ifx=y=z;
(G) <G(x,x,y), for allx,y∈X, withx=y;
(G) G(x,x,y)≤G(x,y,z), for allx,y,z∈X, withz=y;
(G) G(x,y,z) =G(x,z,y) =G(y,z,x) =· · · (symmetry in all three variables); (G) G(x,y,z)≤G(x,a,a) +G(a,y,z), for allx,y,z,a∈X.
Then the functionGis called a generalized metric, abbreviatedG-metric, onX, and the pair (X,G) is called aG-metric space.
Clearly these properties are satisfied whenG(x,y,z) is the perimeter of the triangle with vertices atx,y, andz∈R. Moreover, takingain the interior of the triangle shows that
(G) is the best possible.
Example .[] Let (X,d) be an ordinary metric space, then (X,d) definesG-metrics on Xby
Gs(x,y,z) =d(x,y) +d(y,z) +d(x,z),
Gm(x,y,z) =max
d(x,y),d(y,z),d(x,z).
Example .[] LetX={a,b}. DefineGonX×X×Xby
G(a,a,a) =G(b,b,b) = , G(a,a,b) = , G(a,b,b) = ,
and extendGtoX×X×Xby using the symmetry in the variables. Then it is clear that (X,G) is aG-metric space.
The following useful properties of aG-metric are readily derived from the axioms.
Proposition .[] Let(X,G)be a G-metric space,then for any x,y,z and a from X it follows that:
. ifG(x,y,z) = ,thenx=y=z, . G(x,y,z)≤G(x,x,y) +G(x,x,z), . G(x,y,y)≤G(y,x,x),
. G(x,y,z)≤G(x,a,z) +G(a,y,z),
. G(x,y,z)≤(G(x,y,a) +G(x,a,z) +G(a,y,z)), . G(x,y,z)≤G(x,a,a) +G(y,a,a) +G(z,a,a).
Definition .[] Let (X,G) be aG-metric space, and let{xn}be a sequence of points ofX. A pointx∈Xis said to be thelimitof the sequence{xn}iflimn,m→∞G(x,xn,xm) = ,
Proposition .[] Let(X,G)be a G-metric space,then for a sequence{xn} ⊆X and a point x∈X the following are equivalent:
. {xn}isG-convergent tox, . G(xn,xn,x)→asn→ ∞,
. G(xn,x,x)→asn→ ∞.
Definition .[] Let (X,G) be aG-metric space, a sequence{xn}is calledG-Cauchy if for everyε> , there isN∈Nsuch thatG(xn,xm,xl) <ε, for alln,m,l≥N, that is, if
G(xn,xm,xl)→ asn,m,l→ ∞.
Proposition .[] In a G-metric space(X,G),the following are equivalent:
. the sequence{xn}isG-Cauchy,
. for everyε> ,there exists ann∈Nsuch thatG(xn,xm,xm) <ε,for alln,m≥n.
AG-metric space (X,G) isG-complete (or completeG-metric), if everyG-Cauchy se-quence in (X,G) isG-convergent in (X,G).
Proposition .[] Let(X,G)be a G-metric space,then the function G(x,y,z)is jointly continuous in all three of its variables.
Definition . (X,G) is symmetricG-metric space ifG(x,y,y) =G(y,x,x) for allx,y∈X.
Fixed point theorems in symmetricG-metric space are mostly consequences of the re-lated fixed point results in metric spaces. In this paper we discuss the non-symmetric case. In [] it was shown that if (X,G) is aG-metric space, puttingδ(x,y) =G(x,y,y), (X,δ) is a quasi-metric space (δis not symmetric). It is well known that any quasi-metric induces different metrics and mostly used are
(μ) μ(x,y) =δ(x,y) +δ(y,x), (ρ) ρ(x,y) =max{δ(x,y),δ(y,x)}.
The following result is an immediate consequence of above definitions and relations.
Theorem . Let(X,G)be a G-metric space and let D∈ {δ,ρ}.Then
. {xn} ⊂XisG-convergent tox∈Xif and only if{xn}is convergent toxin(X,D); . {xn} ⊂XisG-Cauchy if and only if{xn}is Cauchy in(X,D);
. (X,G)isG-complete if and only if(X,D)is complete.
For more fixed point results for mappings defined inG-metric spaces, we refer the reader to [, , , , –].
2 Main result
A generalization of the contraction principle can be obtained using a different type of a nondecreasing function ϕ: [,∞)→[,∞). The most usual additional properties im-posed onϕare given using a combination of the next seven conditions:
(ϕ) ϕ() = ,
(ϕ) ϕ(t) <t, for allt> ,
(ϕ) limi→∞ϕi(t) = , for allt> ,
(ϕ) if{ti} ⊂[,∞)is a sequence such thatti+≤ϕ(ti), thenlimi→∞ti= ,
(ϕ) for anyy≥there exists at(y)≥,t(y) =supt≥{t≤y+ϕ(t)},
(ϕ) limt→∞(t–ϕ(t)) =∞,
(ϕ)
∞
i=ϕi(t) <∞, for allt> .
Some of the noted properties ofϕare equivalent, some of them imply others, some of them are incompatible. The next lemma discusses some of the relations between proper-ties (ϕ)-(ϕ), especially those which are used in this paper to define a generalized
con-traction.
Lemma . Letϕ: [,∞)→[,∞)be a nondecreasing function.Then
(i) (ϕ)⇔(ϕ)⇒(ϕ),
(ii) ifϕis right continuous,then(ϕ)⇔(ϕ)⇔(ϕ),
(iii) (ϕ)⇒(ϕk)⇒(ϕ)⇒(ϕ),wherek∈ {, },
(iv) (ϕ)⇔(ϕ),
(v) (ϕ)(ϕ), (ϕ)(ϕ),
(vi) (ϕ)+(ϕ)(ϕ)and(ϕ)+(ϕ)(ϕ),
(vii) (ϕ)(ϕ)and(ϕ)(ϕ).
Proof (i) (ϕ)⇒(ϕ): If for somet> ,ϕ(t)≥t, then, knowing thatϕis nondecreasing, ϕi(t)≥ϕi–(t)≥ · · · ≥ϕ(t)≥t> . It means thatlimi→∞ϕi(t)= , which contradicts (ϕ).
(ϕ)⇒ (ϕ): Let {ti} ⊂[,∞) be any sequence such thatti+ ≤ϕ(ti). Using the
im-plication (ϕ) ⇒ (ϕ), we get ti ≤ ϕ(ti–)≤ ϕ(ti–)≤ · · · ≤ ϕi(t) and limi→∞ti ≤
limi→∞ϕi(t) = .
(ϕ)⇒(ϕ): We assume that for somet> ,limi→∞ϕi(t) =a> . Since (ϕ)⇒(ϕ), the
sequenceti=ϕi(t) satisfies conditionti+=ϕi+(t) =ϕ(ϕi(t))≤ϕi(t) =ti, but it converges
toa> . That contradicts (ϕ).
(ii) It is enough to prove that (ϕ)⇒(ϕ): We assume that for somet> ,limi→∞ϕi(t) =
a> . Since {ϕi(t)} is a nonincreasing sequence, by the right continuity ofϕ, ϕ(a) = ϕlimi→∞ϕi(t) =limi→∞ϕi+(t) =a> ,i.e. <a=ϕ(a), which contradicts (ϕ).
(iii), (iv) are obvious, so the proof is omitted. (v) The function
ϕ(t) =
t
, ≤t≤,
t
+ , <t
(vi) The function
ϕ(t) =
(n+ )–, (n+ )–≤t<n–,n∈N,
–, ≤t
satisfies (ϕ), (ϕ), and (ϕ), but not (ϕ).
(vii) The function
ϕ(t) =
(t), ≤t< ,
t– , ≤t
satisfies (ϕ), but not (ϕ), nor (ϕ).
Theorem . Let(X,G)be a complete G-metric space,T:X→X,where the nondecreas-ing functionϕsatisfies(ϕ)or(ϕ)together with(ϕ)or(ϕ)and for each x∈X there exists
a positive integer n=n(x)such that
GTn(x)x,Tn(x)x,Tn(x)y≤ϕG(x,x,y) ()
for all y∈X.Then T has a unique fixed point a∈X.Moreover,for each x∈X,limkTkx=a
and Tn(a)is continuous at a.
Proof Let the nondecreasing functionϕsatisfy (ϕ) together with (ϕ) (weak contraction
in the sense of Matkowski). Then by Proposition .(), in a non-symmetricG-metric space we have
GTn(x)x,Tn(x)y,Tn(x)y≤GTn(x)x,Tn(x)x,Tn(x)y≤ϕG(x,x,y).
The last inequality together with () implies
ρTn(x)x,Tn(x)y≤ϕρ(x,y)
for allx,y∈X. So, one can apply the Matkowski fixed point theorem if the function ϕ=ϕ˜
satisfies the conditions (ϕ) and (ϕ). Since there exist functionsϕwhich satisfy (ϕ) and
(ϕ), but ϕdoes not (for exampleϕ(t) =+tt ,t≥), the Jleli-Samet technique [] is not
applicable. We are going to prove our theorem using theG-metricG.
We first prove by mathematical induction that, for every x∈X, the orbit {Tkx}k is
bounded.
Fixx∈X, fix the integers, ≤s<n=n(x) and put
uk=G
x,x,Tkn(x)+sx, k= , , , . . . ,
h=maxGx,x,Tn(x)x,Gx,x,Txs.
By (ϕ) there existc,c>h, such that
The last inequalities imply thatu<c. Suppose that there exists a positive integerjsuch
thatuj≥c, butui<cfori<j.
Using (), we get
uj =G
x,x,Tjn(x)+sx≤Gx,x,Tn(x)x+GTn(x)x,Tn(x)x,Tjn(x)+sx
≤h+ϕ(uj–)≤h+ϕ(uj),
i.e. uj–ϕ(uj)≤h, which contradicts the choice ofc. Thereforeuj<cforj= , , . . . , and
consequently the orbit{Tkx}kis bounded, sosup
kG(x,x,Tkx, ) =M<∞.
For anyx∈X, we define sequence{xk}kas follows:
xk+=Tnkxk, nk=n(xk),k= , , . . . . ()
We shall prove that{xk}kis a Cauchy sequence. Letk,j∈N. From () we obtain
xk+j=Tnk+j–+···+nkxk.
With the notations=nk+j–+· · ·+nk, we have
G(xk,xk,xk+j) =G
xk,xk,Tsxk
=GTnk–(x
k–),Tnk–(xk–),Tnk–Tsxk–
≤ϕGxk–,xk–,Tsxk–
≤ · · · ≤ϕkGx,x,Tsx
≤ϕk(M).
Sincelimkϕk(t) = ,{xk}kis a Cauchy sequence in a completeG-metric space,limkxk=a,
a∈X.
In order to prove thatTn(a)a=a, we assume thatG(a,Tn(a)a,Tn(a)a) =ε> . Using the
same arguments as in the previous part of the proof, we see that
lim
k G
xk,xk,Tn(a)xk
= ,
meaning that there existsk∈Nsuch that
lim
k G
xk,xk,Tn(a)xk
<
ε–ϕ(ε), G(a,a,xk) <
ε–ϕ(ε).
Hence,
ε=GTn(a)a,Tn(a)a,a
≤GTn(a)a,Tn(a)a,Tn(a)xk
+GTn(a)xk,Tn(a)xk,xk
+G(xk,xk,a)
≤ϕG(a,a,xk)
+
ε–ϕ(ε)<
ε+ϕ(ε)<ε.
Suppose that there is a pointb∈X,b=a, such thatTn(a)b=b. Then, by (),
G(a,a,b) =GTn(a)a,Tn(a)a,Tn(a)b≤ϕG(a,a,b)<G(a,a,b).
This contradiction proves thatais a unique fixed point ofTn(a). According toTa=Tn(a)Ta
and from the uniqueness which has been proved already, we deduce thatTa=a.
Next, we claim thatlimkTkx=a, for eachx∈X. To prove this, fixx∈X,s∈N, ≤s<
n(a) and put
ak=G
a,a,Tkn(a)+sx, k= , , , . . . .
Then
ak=G
Tn(a)a,Tn(a)a,Tn(a)T(k–)n(a)+sx≤ϕGa,a,T(k–)n(a)+sx
=ϕ(ak–) <· · ·<ϕk(a).
By (ϕ),limkak= , which implies thatlimkTkx=a.
To prove continuity ofTn(a)ata, we consider any sequence{y
m}m⊂Xconverging toa.
For anym∈N
Ga,a,Tn(a)ym
=GTn(a)a,Tn(a)a,Tn(a)ym
≤ϕG(a,a,ym)
<G(a,a,ym)
forym=a. Lettingm→ ∞, we getlimmG(a,a,Tn(a)ym) = . Hence,{Tn(a)ym}mconverges
toa=Tn(a)a, meaning thatTn(a)is continuous ata.
In other cases (when we use (ϕ) instead of (ϕ) or (ϕ) instead of (ϕ)), by Lemma .,
the same conclusion can be drawn.
Corollary . Let(X,G)be a complete G-metric space,T:X→X and for each x∈X there exists a positive integer n=n(x)such that
GTn(x)x,Tn(x)x,Tn(x)y≤qG(x,x,y)
for all y∈X and some q∈(, ).Then T has a unique fixed point a∈X.Moreover,for each x∈X,limkTkx=a and Tn(a)is continuous at a.
Proof The functionϕ(t) =q·t,t∈[,∞), satisfies (ϕ) and (ϕ), so the corollary is a
con-sequence of Theorem ..
From the proof of Theorem . we can see that it would be enough to impose certain assumptions not for all elements fromX but only over some subsetBofX, just as was done by Guseman []. The next theorem is a Guseman type of fixed point theorem in a G-metric space.
B,then there exists a unique u∈B such that Tu=u andlimk→∞Tky=u for each y∈B.
Furthermore,if T satisfies()over X,then u is unique fixed point in X andlimk→∞Tky=u
for each y∈X.
Remark . Takingϕ(t) =q·t, <q< , by Theorem . we obtain the fixed point result from [] or [], so Theorem . is also a generalization of the Guseman fixed point result from [].
Corollary . Let(X,G)be a complete G-metric space,T :X→X,and for each x∈X there exists a positive integer n=n(x)such that
GTn(x)x,Tn(x)x,Tn(x)y≤ G(x,x,y) +G(x,x,y)
for all y∈X.Then T has a unique fixed point a∈X.Moreover,for each x∈X,limkTkx=a
and Tn(a)is continuous at a.
Proof Since the functionϕ(t) = +tt ,t∈[,∞), satisfies (ϕ), (ϕ), (ϕ), and (ϕ), we can
apply Theorem .. Also for thatϕ, the appropriate version of Theorem . can be formu-lated in a similar way as was done in this corollary.
Ifn(x) =m∈N, for eachx∈X, it is easy to see that condition (ϕ) or (ϕ) in Theorem .
can be omitted. This version of Theorem . is an improvement and another proof of Theorem . (Corollary .) from []. But in that case it would be more appropriate to use the metricρ, which reduces () toρ(Tmx,Tmy)≤ϕ(ρ(x,y)),x,y∈Xand enables the use of well-known results in metric spaces.
Proposition . Let(X,G)be a complete G-metric space,whereϕis a nondecreasing func-tion satisfying(ϕ).If T:X→X satisfies
GTmx,Tmx,Tmy≤ϕG(x,x,y) ()
for all x,y∈X and some m∈N,then T has a unique fixed point a∈X.Moreover,for each x∈X,limkTkx=a and Tn(a)is continuous at a.
The next theorem is also a Guseman type of fixed point theorem in aG-metric space. The assumptions about the contractorϕis different with respect to Theorem .. Similarly as in previous analysis, the next theorem can be applied in a metric space and in cases where some special form of functionϕis used.
Theorem . Let f :X→X,where(X,G)is a G-metric space and letϕ: [,∞)→[,∞) be a subadditive mapping satisfying(ϕ).If for some x∈X the closure of orbitO(f;x)is
complete and for each x∈O(f;x)there exists an n(x)∈Nsuch that
Gfn(x)y,fn(x)x,fn(x)x≤ϕG(y,x,x) ()
for all y∈O(f;x),then the sequence xi+=fn(xi)xi,i∈N,converges to some x∗∈X.
If inequality()holds for all x∈O(f;x),then fn(x
∗)
x∗=x∗andlimifi(x) =x∗ for every
Proof First, we show that{xi}i∈N⊂Xis a Cauchy sequence. For sufficiently largem∈N,
there existk,r∈N, ≤r<n(x) such thatm=k·n(x) +r. Using (), we get
Gfmx,x,x
≤Gfkn(x)+rx
,fn(x)x,fn(x)x
+Gfn(x)x ,x,x
≤ϕGf(k–)n(x)+rx ,x,x
+Gfn(x)x ,x,x
≤ϕGf(k–)n(x)+rx
,fn(x)x,fnx
+Gfn(x)x ,x,x
+Gfn(x)x ,x,x
≤ϕGf(k–)n(x)+rx ,x,x
+ϕGfn(x)x ,x,x
+Gfn(x)x ,x,x
≤ · · ·
≤ϕkGfrx,x,x
+
k–
i=
ϕiGfn(x)x ,x,x
.
PuttingA=max{G(fpx
,x,x) : ≤p≤n(x)}, for allm∈N, the next inequality holds:
Gfmx,x,x
≤ k
s=
ϕs(A)≤
∞
s=
ϕs(A) =B<∞, ()
and consequently,
G(xm,xm,xm+) =G
fn(xm–)x
m–,fn(xm–)xm–,fn(xm)fn(xm–)xm–
≤ϕGxm–,xm–,fn(xm)xm–
≤ · · · ≤ϕmGx,x,fn(xm)x
≤ϕm(B)
for allm∈N. Using the last inequality, for everyi,j∈N,i<j, we have
G(xi,xi,xj)≤G(xi,xi,xi+) +· · ·+G(xk–,xk–,xk)≤ j
s=i ϕs(B)
implying that{xi}i∈Nis a Cauchy sequence. SinceO(f;x) is complete, and there exists an
x∗∈O(f;x) such thatlimi→∞xi=x∗.
In the second part of the theorem, inequality () holds for all x∈O(f;x). Then the
elementsxiof the sequence{xi}i∈Nfrom the previous part of the proof satisfy the next two
relations:
Gfn(x∗)x∗,fn(x∗)x∗,fn(x∗)xi
≤ϕGx∗,x∗,xi
<Gx∗,x∗,xi
()
and
Gfn(x∗)xi,xi,xi
=Gfn(x∗)fn(xi–)x
i–,fn(xi–)xi–,fn(xi–)xi–
≤ϕGfn(x∗)xi–,xi–,xi–
≤ϕiGfn(x∗)x,x,x
By ()
lim
i→∞f n(x∗)x
i=fn(x
∗) x∗
and by ()
lim
i→∞G
fn(x∗)xi,xi,xi
=Gfn(x∗)x∗,x∗,x∗= .
Hence,fn(x∗)x∗=x∗.
Next, we claim thatlimkfix=x∗, for eachx∈O(f;x
). Puttingi=kn(x∗) +s,s∈N, ≤
s<n(x∗), we get
Gfkn(mz)+sx,x∗,x∗≤ϕGf(k–)n(mz)+sx,x∗,x∗≤ · · ·
≤ϕkGfsx,x∗,x∗=ϕk(M),
whereM=max{G(fsx,x∗,x∗) : ≤s<n(x∗)}. Since (ϕ
)⇒(ϕ),limkfix=x∗.
To show thatx∗is a unique fixed point offn(x∗)inO(f;x
), we assume that there exists
another pointx∗∗∈O(f;x) with the same property. Then
Gx∗∗,x∗,x∗=Gfn(x∗)x∗∗,fn(x∗)x∗,fn(x∗)x∗≤ϕGx∗∗,x∗,x∗,
that is,x∗∗=x∗. Further, iff(O(f;x))⊆O(f;x), thenfx∗=f(fn(x
∗)
x∗) =fn(x∗)(fx∗),
imply-ingfx∗=x∗.
In the last theorem in this paper we consider a common fixed point for a family of self-mappings with the property of a contractive iterate at a point. The generalized contractive condition is imposed over a subset of aG-metric space.
Theorem . Let(X,G)be G-metric space and B⊆X.Further,let{fi} be the sequence of selfmappings of X such that for all i∈N,fi(B)⊆B and for each x∈X there exists an
n(x)∈Nsuch that
Gfin(x)y,fjn(x)x,fjn(x)x≤ϕmaxG(y,x,x),Gy,fjn(x)x,fjn(x)x,Gx,fjn(x)x,fjn(x)x,
– Gy,fin(x)x,fin(x)x+Gfin(x)y,x,x ()
for all i,j∈N,i=j,and all y∈B,whereϕ: [,∞)→[,∞)is a nondecreasing right con-tinuous function satisfying(ϕ).If there exists x∗∈B such that fn(x
∗)
i (x∗) =x∗for all i∈N,
then x∗ is a unique common fixed point for{fi} in B and for every x∈B,the sequence
xi+=fn(x
∗)
i (xi),i∈N,converges to x∗.
Proof First we prove thatx∗ is a unique point inBwith the property thatfin(x∗)x∗=x∗, i∈N. Ifx∗∗∈B,x∗∗=x∗,fin(x∗)x∗∗=x∗∗,i∈N, then
Gx∗∗,x∗,x∗≤Gfin(x∗)x∗∗,fjn(x∗)x∗,fjn(x∗)x∗≤ϕGx∗∗,x∗,x∗.
Further, since
fi
x∗=fi
fin(x∗)x∗=fin(x∗)+x∗=fin(x∗)fix∗
,
it follows thatfix∗=x∗for alli∈N.
Now, for somex∈B, we form the sequencexi+=fn(x
∗)
i xi.
If x=x∗, thenxi=fn(x
∗)
i– (f n(x∗) i– · · ·(f
n(x∗)
x∗)· · ·) =x∗ and the sequence{xi} converges
tox∗.
Ifxi=x∗, in order to prove that the sequence{xi}converges tox∗, we consider the
se-quenceG(xi+,x∗,x∗),i∈N,
Gxi+,x∗,x∗
=Gfin(x∗)xi,fn(x
∗)
j x∗,f n(x∗) j x∗
≤ϕmaxGxi,x∗,x∗
,Gxi,x∗,x∗
,Gx∗,x∗,x∗,
– Gxi,x∗,x∗
+Gxi+,x∗,x∗
=ϕmaxGxi,x∗,x∗
, – Gxi,x∗,x∗
+Gxi+,x∗,x∗
.
If we choose the option that
maxGxi,x∗,x∗
, – Gxi,x∗,x∗
+Gxi+,x∗,x∗
= – Gxi,x∗,x∗
+Gxi+,x∗,x∗
,
it implies that
Gxi,x∗,x∗
≤Gxi+,x∗,x∗
. ()
On the other hand, in that case
Gxi+,x∗,x∗
≤ϕ– Gxi,x∗,x∗
+Gxi+,x∗,x∗
< –Gxi,x∗,x∗
+ –Gxi+,x∗,x∗
,
that is,
Gxi+,x∗,x∗
<Gxi,x∗,x∗
. ()
It is obvious that () contradicts (). So,
Gxi+,x∗,x∗
≤ϕGxi,x∗,x∗
.
Now, applying that procedureitimes and lettingi→ ∞, we get
Gxi+,x∗,x∗
≤ϕGxi,x∗,x∗
≤ · · · ≤ϕiGx,x∗,x∗
.
Sincex=x∗,G(xi,x∗,x∗) > andlimi→∞G(xi+,x∗,x∗) = . The last relation proves that
the sequence{xi}converges tox∗.
Competing interests
Authors’ contributions
Both authors have equal contribution in the paper and they read and approved the final manuscript.
Author details
1Department of Mathematics and Informatics, Faculty of Science, University of Novi Sad, Novi Sad, Serbia.2Department of Mathematics, Faculty of Technical Sciences, University of Novi Sad, Novi Sad, Serbia.
Acknowledgements
The authors are very grateful to the anonymous referees for their careful reading of the paper and suggestions, which have contributed to the improvement of the paper. This work is supported by Ministry of Science and Technological Development, Republic of Serbia.
Received: 18 July 2013 Accepted: 6 February 2014 Published:21 Feb 2014
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