Existence of common xed points on modular spaces with [620032]

Existence of common xed points on modular spaces with
respect toC-class functions
Arslan Hojat Ansari1, Liliana Guran2
Abstract. In this paper, we prove the existence of common xed points for a generalized weak
contractive mapping in modular spaces concerning a new type of functions- Cclass functions. Then,
our results generalize various comparable results in the existing literature.
1. Introduction and mathematical preliminaries
Over the years the well-know Banach's contraction principle has been generalized in many
ways, see([ 8], [9], [11], [26], [33], [32] [34]).
For the case of weakly contractions are studied some extensions of Banach's contraction prin-
ciple. Let us recall the notion of weakly contraction.
A mapping T:X!Xwhere (X;d) is a metric space, is said to be weakly contraction if
(1) d(T(x);T(y))d(x;y)'(d(x;y));
where': [0;1)![0;1) is a continuous and nondecreasing function such that '(t) = 0 if
and only if t= 0.
In 2008 Dutta and Choudhury ([ 14]) introduced a new generalization of contractions in metric
spaces and proved the following theorem.
Theorem 1.1.Let(X;d)be a complete metric space, T:X!Xbe a self-mapping satisfying
the inequality
(2) (d(Tx;Ty )) (d(x;y))'(d(x;y));
where ;' : [0;1)![0;1)are both continuous and monotone nondecreasing functions with
(t) ='(t) = 0 if and only if t= 0.
Then T has a unique xed point.
We note that, if one takes (t) =t, then (2) reduces to (1).
In 2009 Zhang and Song [ 39] used generalized '-weak contractions which is dened for two
mappings and gave conditions for existence of a common xed point.
Theorem 1.2.Let (X, d) be a complete metric space and T;S :X!Xtwo mapping such
that for all x;y2X
(3) d(Tx;Ty )M(x;y)'(M(x;y));
2010 Mathematics Subject Classication. 46T99; 47H10; 54H25.
Key words and phrases. C-class function, common xed point, -modular space.
1

2 A.H. ANSARI, L. GURAN
where': [0;1)![0;1)is lower semi-continuous function with '(t)>0fort2(0;1),
'(0) = 0 , andM(x;y) = maxfd(x;y);d(Tx;x );d(Sy;y );d(y;Tx )+d(x;Sy)
2g.
Then there exists the unique point u2Xsuch thatu=Tu=Su.
Later D. Doric [ 13] has generalized Theorem 1.2 and proved the following theorem.
Theorem 1.3.Let(X;d)be a complete metric space, T:X!Xbe a self-mapping where
(a) : [0;1)![0;1)is a continuous monotone nondecreasing function with (t) = 0 if and
only ift= 0,
(b)': [0;1)![0;1)is a lower semi-continuous function with '(t) = 0 if and only if t= 0,
(c) M is dened in Theorem 1.2.
Then there exists the unique point u2Xsuch thatu=Tu=Su.
Next, let us present a brief recollection of basic concepts and facts in modular space.
Definition 1.1.LetXbe a vector space over R(orC). A functional :X![0;1]is called
a modular if for arbitrary xandy, elements of X, it satises the following conditions:
(m1)(x) = 0 if and only if x= 0;
(m2)(x) =(x)for all scalar ; withjj= 1;
(m3)(x+y)(x) +(y), whenever ;0and+= 1.
If we replace (m3)by
(m4)(x+y)s(x) +s(y), for;0;s+s= 1 with ans2(0;1], then the
modularis calleds-convex modular, and if s= 1,is called convex modular.
Proposition 1.1.(1)(x)is a nondecreasing function of 0;
(2) Ifis s – convex then s(x)is a nondecreasing function of 0.
Definition 1.2.The vector space Xgiven by
(4) X=fx2X;(x)!0as!0g
is called a modular space. Xis a vector subspace of X.
Followings are some consequences of condition ( m3):
Remark 1.1.([10])
(r1)Fora,b2Rwithjaj<jbjwe have(ax)<(bx)for allx2X;
(r2)Fora1;:::;an2R+withnP
i=1ai= 1, we have
 nX
i=1aixi!
nX
i=1(xi);for anyx1;:::;xn2X:
Proposition 1.2.([29]) LetXbe a modular space. If a;b2R+withba;then(ax)
(bx).
Remark 1.2.A function modular is said to satisfy:
a)
2type condition if there exists K > 0such that for any x2X, we have(2x)K(x)
b)
2-condition if (2xn)!0 asn!1 , whenever (xn)!0 asn!1 .
Definition 1.3.A sequencefxngin modular space Xis said to be:
(t1)convergent to x2Xif(xnx)!0asn!1
(t2)Cauchy if(xnxm)!0asn; m!1:
Xis calledcomplete if any -Cauchy sequence is -convergent. Note that, convergence
does not imply -Cauchy since does not satisfy the triangle inequality. In fact, one can show that
this will happen if and only if satises the
2condition.

EXISTENCE OF COMMON FIXED POINTS ON MODULAR SPACES WITH RESPECT TO C-CLASS FUNCTIONS 3
Definition 1.4.LetXbe a modular space, where satises the
2-condition. Two self-
mappings T and f of Xare called- compatible if (fTxnTfxn)!0asn!1 , whenever
fxngn2Nis a sequence in Xsuch thatfxn!zandTxn!zfor some point z2X.
Mongkolkeha and Kumam (see [ 27], [28]) proved the existence of common xed points as
follows.
Theorem 1.4.([27]) LetXbe a-complete modular space, where satises the
2-condition.
Letc;l2R+;c>l andT;f:X!Xare two-compatible mappings such that T(X)f(X)
and
(5)Z(c(TxTy))
0'(t) dtZ(l(fxfy))
0'(t) dt Z(l(fxfy))
0'(t) dt!
);
for anyx;y2X;where': [0;1)![0;1)is a Lebesgue integrable which is summable, nonneg-
ative, and for all " >0;R"
0'(t) dt >0;and: [0;1)![0;1)is lower semicontinuous function
with(t)>0for allt>0and(t) = 0 if and only if t= 0. If one of Torfis continuous, then
there exists a unique common xed point of Tandf.
Theorem 1.5.([28]) LetXbe a-complete modular space, where satises the
2-condition.
Letc;l2R+;c>l andT;f:X!Xare two-compatible mappings such that T(X)f(X)
and satisfying the inequality
(6) ((c(TxTy))) ((l(fxfy)))((l(fxfy)));
for anyx;y2X;where ;: [0;1)![0;1)are both continuous and monotone nondecreasing
functions with (t) =(t) = 0 if and only if t= 0. If one of Torfis continuous, then there
exists a unique common xed point of Tandf.
In [29] Mongkolkeha and Kumam proved the existence of xed points as follows.
Theorem 1.6.([29]) LetXbe a-complete modular space, where satises the
2-condition.
Letc;l2R+;c>l andT:X!Xbe a mapping satisfying the inequality
(7) ((c(TxTy))) ((l(xy)))((l(xy)));
for anyx;y2X;where ;: [0;1)![0;1)are both continuous and monotone nondecreasing
functions with (t) =(t) = 0 if and only if t= 0. Then,Thas a unique xed point.
With note to ((l(TxTy))) ((c(TxTy))) andlT(X)lf(X);they can revise to
following forms.
Theorem 1.7.LetXbe a-complete modular space, where satises the
2-condition. Let
T;f:X!Xare two-compatible mappings such that T(X)f(X)and
Z(TxTy)
0'(t) dtZ(fxfy)
0'(t) dt Z(fxfy)
0'(t) dt!
);
for anyx;y2X;where': [0;1)![0;1)is a Lebesgue integrable which is summable, nonneg-
ative, and for all " >0;R"
0'(t) dt >0;and: [0;1)![0;1)is lower semicontinuous function
with(t)>0for allt>0and(t) = 0 if and only if t= 0. If one of Torfis continuous, then
there exists a unique common xed point of Tandf.
Theorem 1.8.LetXbe a-complete modular space, where satises the
2-condition. Let
T;f :X!Xare two-compatible mappings such that T(X)f(X)and satisfying the
inequality
((TxTy)) ((fxfy))((fxfy));

4 A.H. ANSARI, L. GURAN
for anyx;y2X;where where ;: [0;1)![0;1)are both continuous and monotone nonde-
creasing functions with (t) =(t) = 0 if and only if t= 0. If one of Torfis continuous, then
there exists a unique common xed point of Tandf.
Another useful xed point result was proved by Mongkolkeha and Kumam (see [ 29]) as follows.
Theorem 1.9.LetXbe a-complete modular space, where satises the
2-condition. Let
T:X!Xbe a mapping satisfying the inequality
((TxTy)) ((xy))((l(xy)));
for anyx;y2X;where ;: [0;1)![0;1)are both continuous and monotone nondecreasing
functions with (t) =(t) = 0 if and only if t= 0. Then,Thas a unique xed point.
Beygmohammadi and Razani [ 10] proved the existence of xed points as follows.
Theorem 1.10.([10]) LetXbe a-complete modular space, where satises the
2-condition.
Assume that ':R+![0;1)is an increasing and upper semicontinuous function satisfying
(t)t;8t>0
Let': [0;1)![0;1)is a Lebesgue integrable which is summable, nonnegative, and for all
">0;R"
0'(t) dt>0;and letf:X!Xbe a mapping such that there are c;l2Rwherel<c ,
(8)Z(c(fxfy))
0'(t) dt Z(l(xy))
0'(t) dt!
;
for anyx;y2X:Thenfhas a unique xed point in X:
As a particular case we have the following result.
Theorem 1.11.LetXbe a-complete modular space, where satises the
2-condition.
Assume that ':R+![0;1)is an increasing and upper semicontinuous function satisfying
(t)t;8t>0
Let': [0;1)![0;1)is a Lebesgue integrable which is summable, nonnegative, and for all
">0;R"
0'(t) dt>0;and letf:X!Xbe a mapping such that ,
(9)Z(fxfy)
0'(t) dt Z(xy)
0'(t) dt!
;
for anyx;y2X:Thenfhas a unique xed point in X:
T.L. Shateri (see [ 37]) proved the existence of common xed points as follows.
Theorem 1.12.LetXbe a complete modular space, where satises the
2-condition. Suppose
that':R+![0;1)is an increasing and upper semicontinuous function satisfying '(t)t, with
t>0.
LetCbe a-closed subset of Xand letT;S :C!Cbe mappings such that there exist
;2R+with> , and((TxSy))'(((xy))), for allx;y2C. ThenTandShave
a unique common xed point in C.
As a particular case we have the following result.
Theorem 1.13.LetXbe a complete modular space, where satises the
2-condition. Suppose
that':R+![0;1)is an increasing and upper semicontinuous function satisfying '(t)t, with
t>0.
LetCbe a-closed subset of Xand letT;S :C!Cbe mappings such that (TxSy)
'((xy));for allx;y2C. ThenTandShave a unique common xed point in C.

EXISTENCE OF COMMON FIXED POINTS ON MODULAR SPACES WITH RESPECT TO C-CLASS FUNCTIONS 5
Now, we recall the notion of C-class function introduced by Ansari in [ 5]. For examples and
more applications see also [ 15, 16, 24 ].
Definition 1.5.[5]A mapping F: [0;+1)2!Ris calledC-class function if it is continuous
and the following conditions hold:
(1)F(s;t)sfor alls;t2[0;+1);
(2)F(s;t) =simplies that either s= 0 ort= 0.
Denote Cthe family ofC-class functions.
Following examples show that the class Cis nonempty.
Example 1.1.[5]LetF: [0;+1)2!R, withs;t2[0;+1). Then we have:
(1)F(s;t) =st,F(s;t) =s)t= 0;
(2)F(s;t) =ms,0<m< 1,F(s;t) =s)s= 0;
(3)F(s;t) =s
(1+t)r,r2(0;+1),F(s;t) =s)s= 0 ort= 0;
(4)F(s;t) = loga[(t+as)=(1 +t)],a>1,F(s;t) =s)s= 0 ort= 0;
(5)F(s;t) = ln[(1 + as)=2],e>a> 1,F(s;t) =s)s= 0;
(6)F(s;t) = (s+l)(1=(1+t)r)l,l>1;r2(0;+1),F(s;t) =s)t= 0;
(7)F(s;t) =slogt+aa,a>1,F(s;t) =s)s= 0 ort= 0;
(8)F(s;t) =s(1+s
2+s)(t
1+t),F(s;t) =s)t= 0;
(9)F(s;t) =s(s),: [0;+1)![0;1)a continuous function, F(s;t) =s)s= 0;
(10)F(s;t) =st
k+t;F(s;t) =s)t= 0;
(11)F(s;t) =s'(t),F(s;t) =s)t= 0;here': [0;+1)![0;+1)is a continuous
function such that '(t) = 0,t= 0;
(12)F(s;t) =sh(s;t),F(s;t) =s)s= 0;hereh: [0;+1)[0;+1)![0;+1)is a
continuous function such that h(t;s)<1for allt;s> 0;
(13)F(s;t) =s(2+t
1+t)t,F(s;t) =s)t= 0;
(14)F(s;t) =np
ln(1 +sn),F(s;t) =s)s= 0;
(15)F(s;t) =(s),F(s;t) =s)s= 0;here: [0;+1)![0;+1)is a continuous function
such that(0) = 0 and(t)<tfort>0;
(16)F(s;t) =s
(1+s)r;r2(0;+1),F(s;t) =s)s= 0
2. Common Fixed Point of Almost Generalized ( ;')-Contraction
In this section, we obtain common xed point results for a pair of mappings satisfying gener-
alized ( ;')-contractive condition in the framework of a modular space.
Let us dene the following sets:
 =f : [0;1)![0;1) : is continuous nondecreasing function and (t) = 0 if and
only ift= 0g:
 =f': [ 0;1)![0;1) :'is lower-semi continuous function and '(t) = 0 if and only
ift= 0g:
u=f': [0;1)![0;1) :'is lower-semi continuous function and '(0)0g:
Theorem 2.1.LetXbe a-complete modular space, where satises the
2-condition. Let
T;f:X!Xare two-compatible mappings such that T(X)f(X)and
(10) Z(TxTy)
0'(t) dt!
F( Z(fxfy)
0'(t) dt!
; Z(fxfy)
0'(t) dt!
)

6 A.H. ANSARI, L. GURAN
for anyx;y2X;whereF2C, 2 ; 2u': [0;1)![0;1)is a Lebesgue integrable which
is summable, nonnegative, and for all " >0;R"
0'(t) dt >0. If one of Torfis continuous, then
there exists a unique common xed point of Tandf.
Proof. Letx2Xand generate inductively the sequence fTxngn2Nas follow:Txn=fxn+1:
First, we prove that the sequence f(TxnTxn1)gconverges to 0. Then we have:
(11) R(TxnTxn1)
0'(t) dt
F( R(fxnfxn1)
0'(t) dt
;R(fxnfxn1)
0'(t) dt
)
=F( R(Txn1Txn2)
0'(t) dt
;R(Txn1Txn2)
0'(t) dt
)
 R(Txn1Txn2)
0'(t) dt
=)R(TxnTxn1)
0'(t) dtR(Txn1Txn2)
0'(t) dt:
This means that the sequence fR(TxnTxn1)
0'(t) dtgis decreasing and bounded below. Hence,
there exists r0 such that
lim
n!1Z(TxnTxn1)
0'(t) dt=r:
On taking limit as n!1 on both sides of the inequality (11), we have
(r)F( (r);'(r));
which implies that (r) = 0 or'(r) = 0, that is r= 0. Hence
(12) lim
n!1(TxnTxn1) = 0:
Now we show that fTxngis a-Cauchy sequence.
Assume on contrary, there exists ">0 such that we can nd two subsequences fmkgandfnkg
of positive integers satisfying nk>mkksuch the following inequalities hold:
(13) (TxnkTxmk)";  (2(Txnk1Txmk))<":
From (13) and Remark1.1, it follows that
"(TxnkTxmk)
=(TxnkTxnk1+Txnk1Txmk)
(2(TxnkTxnk1)) +(2(Txnk1Txmk))
<"+(2(TxnkTxnk1)):
On taking limit as k!1 , we obtain that
(14) lim
k!1(TxnkTxmk) ="
Usingx=xnkandy=xmk1in (19), we have
(15) R(TxnkTxmk)
0'(t) dt
F( R(fxnkfxmk)
0'(t) dt
;R(fxnkfxmk)
0'(t) dt
)
F( R(Txnk1Txmk1)
0'(t) dt
;R(Txnk1Txmk1)
0'(t) dt
)

EXISTENCE OF COMMON FIXED POINTS ON MODULAR SPACES WITH RESPECT TO C-CLASS FUNCTIONS 7
Also, from (13) we have
(16)(Txnk1Txmk1) =(Txnk1Txnk+TxnkTxmk1)
(2(Txmk1Txmk)) +(2(TxmkTxnk1))
<(2(Txmk1Txmk)) +":
Taking limit as k!1 on both sides of (15) and by using (12), Proposition 1.2, we get
Z"
0'(t) dt
F( Z"
0'(t) dt
;'Z"
0'(t) dt
)
which implies that R"
0'(t) dt

= 0;or; 'R"
0'(t) dt

= 0, that is,R"
0'(t) dt= 0, a contradic-
tion.
HencefTxngis a-Cauchy sequence.
Next, let us prove the existence of a xed point of one mapping.
AsXis a-complete, there exists a u2Xsuch that(Txnu)!0 asn!1 .
That isTxn!uand implies that fxn!uasn!1 . IfTis continuous, then T2xn!Tu
andTfxn!Tuasn!1 .
By-compatible,( fTxnTfxn)!0 asn!1 , thus,fTxn!Tuasn!1 .
Next, we prove that uis a unique xed point of T. Indeed,
(17) R(T2xnTxn)
0 '(t) dt
= R(TTx nTxn)
0'(t) dt
F( R(fTx nfxn)
0'(t) dt
;R(fTx nfxn)
0'(t) dt
)
Takingn!1 in the inequality (32), we have
Z(uTu)
0'(t) dt!
F( Z(uTu)
0'(t) dt!
;' Z(uTu)
0'(t) dt!
)
which implies that R(uTu)
0'(t) dt
= 0;or;'R(uTu)
0'(t) dt
= 0 and soR(uTu)
0'(t) dt=
0;therefore(uTu) = 0.
Then, we obtain that Tu=u.
Thuszis a xed point of T.
SinceT(X)f(X), then there exists a point u1such thatu=Tu=fu1. The inequality,
Z(T2xnTu1)
0'(t) dt!
F( Z(fTx nfu1)
0'(t) dt!
;' Z(fTx nfu1)
0'(t) dt!
)
Asn!1 , yields
Z(TuTu1)
0'(t) dt!
F( Z(Tufu1)
0'(t) dt!
;' Z(Tufu1)
0'(t) dt!
)
Thus,
Z(uTu1)
0'(t) dt!
F( Z(uu)
0'(t) dt!
;' Z(uu)
0'(t) dt!
)
F( Z0
0'(t) dt
;'Z0
0'(t) dt
)0
Which implies that, u=Tu1=fu1and alsofu=fTu 1=Tfu 1=Tu=u.
Hence,fu=Tu=u.

8 A.H. ANSARI, L. GURAN
Suppose that wis an another common xed point, that is, w=Sw,w=Tw, andw6=z.
(18) R(zw)
0'(t) dt
= R(TzTw)
0'(t) dt
F( R(fufw)
0'(t) dt
;'R(fufw)
0'(t) dt
)
F( R(zw)
0'(t) dt
;'R(zw)
0'(t) dt
)
which implies that R(zw)
0'(t) dt
= 0;or;'R(zw)
0'(t) dt
= 0 and so (zw) = 0 a
contradiction. Hence z=w. 
As a consequence of the theorem above let us present the following result.
Corollary 2.1.LetXbe a-complete modular space, where satises the
2-condition.
LetT;f:X!Xare two-compatible mappings such that T(X)f(X)and
(19)  Z(TxTy)
0'(t) dt!
F( Z(fxfy)
0'(t) dt!
; Z(fxfy)
0'(t) dt!
)
for anyx;y2X;where2(0;1),F2C,2u': [0;1)![0;1)is a Lebesgue integrable
which is summable, nonnegative, and for all ">0;R"
0'(t) dt>0. If one of Torfis continuous,
then there exists a unique common xed point of Tandf.
Proof. By taking (t) =twhere2(0;1) follow the conclusion. 
Next, let us present another xed point result.
Theorem 2.2.LetXbe a-complete modular space, where satises the
2-condition. Let
T;f:X!Xare two-compatible mappings such that T(X)f(X)and
(20) Z(TxTy)
0'(t) dt!
F( Zm(x;y)
0'(t) dt!
; Zm(x;y)
0'(t) dt!
)
for anyx;y2X;wherem(x;y) = maxf(xy);(xfx);(yfy);(1
2(xfy))+(1
2(yfx))
2g,F2C,
2 ; 2u': [0;1)![0;1)is a Lebesgue integrable which is summable, nonnegative, all
">0;R"
0'(t) dt>0. If one ofTorfis continuous, then there exists a unique common xed point
ofTandf.
Proof. Letx2Xand generate inductively the sequence fTxngn2Nas follow:Txn=fxn+1:
First, we prove that the sequence f(TxnTxn1)gconverges to 0. Then we have:
(21) R(TxnTxn1)
0'(t) dt
F( Rm(fxn;fxn1)
0'(t) dt
;Rm(fxn;fxn1)
0'(t) dt
)
 Rm(fxn;fxn1)
0'(t) dt
:
Then we have:
(22)Z(TxnTxn1)
0'(t) dtZm(fxn;fxn1)
0'(t) dt:
By the denition of m(x;y) we obtain:
(23)m(fxn;fxn1) = maxf(fxn+1fxn);(fxnfxn1);1
2(fxn+1fxn1)
2g;

EXISTENCE OF COMMON FIXED POINTS ON MODULAR SPACES WITH RESPECT TO C-CLASS FUNCTIONS 9
(1
2(fxn+1fxn1))
2(fxn+1fxn) +(fxnfxn1)
2maxf(fxn+1fxn);(fxnfxn1)g:
Hence:
(24) m(fxn;fxn1) = maxf(fxn+1fxn);(fxnfxn1)g:
Replacing in (22) we obtain:
Z(TxnTxn1)
0'(t) dtZm(fxn;fxn1)
0'(t) dt=Zmaxf(fxn+1fxn);(fxnfxn1)g
0'(t) dt=
(25) = maxfZ(fxn+1fxn)
0'(t) dt;Z(fxnfxn1)
0'(t) dtg=Z(fxnfxn1)
0'(t) dt=
=Z(Txn1Txn2)
0'(t) dt:
This means that the sequence fR(TxnTxn1)
0'(t) dtgis decreasing and bounded below. Hence,
there exists r0 such that:
lim
n!1Z(TxnTxn1)
0'(t) dt=r:
On taking limit as n!1 on both sides of the inequality (21), we have
(r)F( (r);'(r));
which implies that (r) = 0 or'(r) = 0, that is r= 0. Hence:
(26) lim
n!1(TxnTxn1) = 0:
Now we show that fTxngis a-Cauchy sequence.
As in the proof of Theorem 2.1, we assume on contrary, there exists ">0 such that we can nd
two subsequences fmkgandfnkgof positive integers satisfying nk> mkksuch the following
inequalities hold:
(27) (TxnkTxmk)";  (2(Txnk1Txmk))<":
Then, it is easy to prove that:
(28) lim
k!1(TxnkTxmk) ="
Usingx=xnkandy=xmk1in (20), we have:
(29)
R(TxnkTxmk1)
0 '(t) dt
F( Rm(fxnk;fxmk1)
0 '(t) dt
;Rm(fxnk;fxmk1)
0 '(t) dt
)
(Rm(fxnk;fxmk1)
0 '(t) dt):
By the denition of m(x;y) we get:
(30)m(fxnk;fxmk1) = maxf(fxnkfxmk1);(fxnkfxnk+1);(fxmk1fxmk);
(1
2(fxnkfxmk)) +(1
2(fxmk1fxnk+1))
2g:

10 A.H. ANSARI, L. GURAN
Also, from (27) we have
(31)(Txnk1Txmk1) =(Txnk1Txnk+TxnkTxmk1)
(2(Txmk1Txmk)) +(2(TxmkTxnk1))
<(2(Txmk1Txmk)) +":
Taking limit as k!1 on both sides of (29) and by using (26), Proposition 1.2, we get
Z"
0'(t) dt
F( Z"
0'(t) dt
;'Z"
0'(t) dt
)
which implies that R"
0'(t) dt

= 0;or; 'R"
0'(t) dt

= 0, that is,R"
0'(t) dt= 0, a contradic-
tion.
HencefTxngis a-Cauchy sequence.
Next, let us prove the existence of a xed point of one mapping.
AsXis a-complete, there exists a u2Xsuch that(Txnu)!0 asn!1 .
That isTxn!uand implies that fxn!uasn!1 . IfTis continuous, then T2xn!Tu
andTfxn!Tuasn!1 .
By-compatible,( fTxnTfxn)!0 asn!1 , thus,fTxn!Tuasn!1 .
Next, we prove that uis a unique xed point of T. Indeed,
(32) R(T2xnTxn)
0 '(t) dt
= R(TTx nTxn)
0'(t) dt
F( R(fTx nfxn)
0'(t) dt
;R(fTx nfxn)
0'(t) dt
)
Takingn!1 in the inequality (32), we have
Z(uTu)
0'(t) dt!
F( Z(uTu)
0'(t) dt!
;' Z(uTu)
0'(t) dt!
)
which implies that R(uTu)
0'(t) dt
= 0;or;'R(uTu)
0'(t) dt
= 0 and soR(uTu)
0'(t) dt=
0;therefore(uTu) = 0.
Then, we obtain that Tu=u.
Thuszis a xed point of T.
SinceT(X)f(X), then there exists a point u1such thatu=Tu=fu1. The inequality,
Z(T2xnTu1)
0'(t) dt!
F( Z(fTx nfu1)
0'(t) dt!
;' Z(fTx nfu1)
0'(t) dt!
)
Asn!1 , yields
Z(TuTu1)
0'(t) dt!
F( Z(Tufu1)
0'(t) dt!
;' Z(Tufu1)
0'(t) dt!
)
Thus,
Z(uTu1)
0'(t) dt!
F( Z(uu)
0'(t) dt!
;' Z(uu)
0'(t) dt!
)
F( Z0
0'(t) dt
;'Z0
0'(t) dt
)0
Which implies that, u=Tu1=fu1and alsofu=fTu 1=Tfu 1=Tu=u.
Hence,fu=Tu=u.

EXISTENCE OF COMMON FIXED POINTS ON MODULAR SPACES WITH RESPECT TO C-CLASS FUNCTIONS 11
Suppose that wis an another common xed point, that is, w=Sw,w=Tw, andw6=z.
(33) R(zw)
0'(t) dt
= R(TzTw)
0'(t) dt
F( R(fufw)
0'(t) dt
;'R(fufw)
0'(t) dt
)
F( R(zw)
0'(t) dt
;'R(zw)
0'(t) dt
)
which implies that R(zw)
0'(t) dt
= 0;or;'R(zw)
0'(t) dt
= 0 and so (zw) = 0 a
contradiction. Hence z=w. 
Con
ict of Interests
The authors declare that there is no con
ict of interests regarding the publication of this article.
Acknowledgement
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1Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran.
Email address :analsisamirmath2@gmail.com
2Department of Pharmaceutical Sciences, "Vasile Goldis " Western University of Arad,
Revolut iei Avenue, no. 94-96, 310414 Arad, Roumania.
Email address :lguran@uvvg.ro, gliliana.math@gmail.com