Fixed point theorems for discontinous map- [625568]

Fixed point theorems for discontinous map-
pings of Kannan and Bianchini type in dis-
tance spaces
Paula Homorodan
Abstract. In this paper we establish xed point theorems for two classes
of discontinous mappings (Kannan type mappings, Bianchini type map-
pings) in the very general setting of a space with a distance, thus extend-
ing some results in the paper [Choban, M. M., Fixed point of mappings
dened on spaces with distance , Carpathian J. Math., 32(2016), no. 2,
173{188]. We also indicate some particular cases of our main results and
present some examples to illustrate the theoretical results and show that
our generalizations are eective.
Keywords. Space with distance, H-distance space, Kannan mapping,
Bianchini mapping, xed point.
1. Introduction
It is well known that the xed point theorem of Banach, for contraction
mappings, is one of the pivotal results in analysis. It has been used in many
dierent elds of mathematics but has one major drawback due to the fact
that any Banach contraction Tis continuous. So, Banach contraction map-
ping principle can be applied only to operator equations involving continuous
operators.[8]
A natural question arises: could we nd contractive conditions which
will imply the existence of a xed point in a complete metric space but will
not imply continuity?
Kannan [6] proved the following result giving an armative answer to
the above question.
Theorem 1.1. [1]If(X;d)is a complete metric space and the mapping
T:X!Xsatises
d(Tx;Ty )a[d(x;Tx ) +d(y;Ty ) ];8x;y2X (1.1)
where 0<a<1
2, thenThas a unique xed point.

2 Paula Homorodan
Bianchini [4] obtained an extension of Theorem 1.1, as follows.
Theorem 1.2. Let(X;d)be a metric space, and T:X!Xbe a Bianchini
mapping, i.e., there exists h2[0;1)such that
d(Tx;Ty )hmaxfd(x;Tx );d(y;Ty )g;8x;y2X: (1.2)
ThenThas a unique xed point.
It is easy to see that every Kannan mapping is a Bianchini mapping but
the converse is not more true, as shown by the next example.
Example. [7]LetX= [0;1]be endowed with the usual metric. Then the
functionf(x) =x=3;0x < 1, satises Bianchini condition (1.2) , but it
does not satisfy Kannan condition (1.1) .
In a recent paper [5], Choban stated and proved a xed point theorem
in a space with distance, which is a signicant extension of the well known
Banach's contraction mapping principle.
He proved that in a H-distance space with a mapping T, exists a com-
parison function such thatd(Tx;Ty )(d(x;y)), for allx;y2X, thenT
has a unique xed point.
So we try to answer to the following questions:
1. Is it possible to establish a xed point theorem for Kannan mappings
in a distance space?
2. Is it possible to establish a xed point theorem for Bianchini mappings
in a distance space?
The main aim of this paper is to give positive but partial answers to both
these questions.
2. Preliminaries
For a mapping T:X!Xwe denote
Fix(T) =fx2X:T(x) =xg; (2.1)
the set of xed points of T. We recall in this section the basic notions and
results on spaces with a distance, mainly taken from [5].
LetXbe a non-empty set and d:XX!Rbe a mapping such that
for allx;y2Xwe have:
(im)d(x;y)0;
(iim)d(x;y) +d(y;x) = 0 if and only if x=y.
Then, according to [5], ( X;d) is called a distance space anddis called
adistance onX.
Let (X;d) be a distance space, fxn:n2N=f1;2;:::ggbe a sequence
inXand letx2X. We say that the sequence fxn:n2Ngis:
1)convergent toxif and only if lim
n!1d(x;xn) = 0. We denote this by
xn!xorx= lim
n!1xn;
2)convergent if it converges to some point in X;

Fixed point theorems for discontinous mappings 3
3)Cauchy orfundamental if lim
n;m!1d(xn;xm) = 0.
A distance space ( X;d) iscomplete if every Cauchy sequence in Xcon-
verges to some point in X.
LetXbe a non-empty set and dbe a distance on X. Then:
(X;d) is called a symmetric space anddis called a symmetric onXif,
for allx;y2X, we have (iiim)d(x;y) =d(y;x);
(X;d) is called a quasimetric space anddis called a quasimetric onX
if, for allx;y;z2X, we have (ivm)d(x;z)d(x;y) +d(y;z);
(X;d) is called a metric space anddis called a metric ifdis asymmetric
and quasimetric simultaneously.
We say that a distance don a space ( X;d) isbalanced if for every Cauchy
sequencefxn:n2Ngconvergent to xin X and any point y2Xwe have
d(y;x) = lim
n!1d(y;xn).
LetXbe a non-empty set and d(x;y) be a distance on Xwith the
following property:
(N) for each point x2Xand any >0 there exists =(x;)>0
such that from d(x;y)andd(y;z)it followsd(x;z).
Then (X;d) is called an N-distance space anddis called an N-distance
onX.
Ifdis a symmetric, then we say that dis an N-symmetric .
Ifdsatises the condition:
(F) for any >0 there exists =()>0 such that from d(x;y)
andd(y;z)it followsd(x;z),
thendis called an F-distance or a Frechet distance and (X;d) is called an
F-distance space .
Ifdis a symmetric and an F-distance on a spaceX, then we say that d
is an F-symmetric .
Remark 2.1. If(X;d)is an F-symmetric space, then any convergent sequence
is a Cauchy sequence. For N-symmetric spaces and for quasimetric spaces this
assertion is not more true.
A distance space ( X;d) is called an H-distance space if for any two
distinct points x;y2Xthere exists =(x;y)>0 such that B(x;d; )\
B(y;d; ) =;.
Remark 2.2. [2][3][5] Let(X;d)be a distance space. Then (X;d)is anH-
distance space if and only if any convergent sequence has a unique limit point.
Lemma 2.1. [5]Let(X;d)be a distance space and the space (X;(d))is
Hausdor. Then dis anH-distance.
Proposition 2.1. [5]Let(X;d)be anH-distance space and T:X!Xbe a
continuous mapping. Then
1.Fix(T)is closed.
2. If for some point x2X, the Picard iteration O(T;x)is convergent,
then the set of xed points Fix(T)of the mapping Tis non-empty.

4 Paula Homorodan
Fix a distance space ( X;d) and a mapping T:X!X. We say that the
space (X;d) isT-bounded if, for each x2X, there exists a positive number
(x) such that d(Tn(x);x) +d(x;Tn(x))(x), for alln2N.
The space ( X;d) is called weakly T-bounded if for each x2Xthere
exist a positive number (x) andp=p(x)2Nsuch that
d(Tn(x);Tp1(x)) +d(Tp1(x);Tn(x))(x), for eachnp.
Proposition 2.2. [5]Let(X;d)be a distance space and the mapping
T:X!Xbe a contraction. If the space (X;d)is weakly T-bounded, then:
1) For each point x2Xthe Picard sequence O(T;x)is a Cauchy;
2) The mapping Thas a unique xed point provided (X;d)is a complete
H-distance space;
3) The mapping Thas a unique xed point provided (X;d)is a complete
balanced distance space.
A function : [0;1)![0;1) is called a comparison function if it
satises the following conditions:
(i)is increasing;
(ii) lim
n!1n(t) = 0 for each t2[0;1).
Remark 2.3. [5]If: [0;1)![0;1)is a comparison function then it
satises the following conditions: (0) = 0 and(t)<t, for eacht2[0;1).
Proposition 2.3. [5]Let(X;d)be a distance space and T:X!Xbe a map-
ping such that the space X;d is weaklyT-bounded. If there exists a comparison
functionsuch thatd(Tx;Ty )(d(x;y)), for allx;y2X, then:
1) For each point x2Xthe Picard sequence O(T;x)is Cauchy;
2) The mapping Thas a unique xed point provided (X;d)is a complete
H-distance space.;
3) The mapping Thas a unique xed point provided (X;d)is a complete
balanced distance space.
Corollary 2.1. [5]Let(X;d)be a bounded complete H-distance space or a
bounded balanced distance space, T:X!Xbe a mapping and suppose there
exists a comparison function such that:
d(Tx;Ty )(d(x;y));for allx;y2X:
ThenThas a unique xed point. Moreover, for each ">0there exists n02N
such thatd(Tn(x);Tm(x)<", for allX2Xandn;mn0.
3. Main results
Theorem 3.1. Let(X;d)be a complete H-distance space and let T:X!X
be a mapping for wich there exists 0<a<1
2such that:
d(Tx;Ty )a[d(x;Tx ) +d(y;Ty ) ];8x;y2X: (3.1)
Then the Picard iteration at the any point x2Xis convergent. If, addition-
ally, the limit xof the Picard sequence is a xed point of T, then xis the
unique xed point of T.

Fixed point theorems for discontinous mappings 5
Proof. Letx02Xand consider the Picard sequence fxng,xn+1=Txn;
n0. We prove that
d(xn;xn+1)a
1an

d(x0;x1); n= 0;1;::: (3.2)
First, we note that a2
0;1
2

=)a
1a2(0;1). Inequality (3.2) is
obviously true for n= 0.
Now, we take x:=x0; y:=x1in (3.1) and obtain
d(Tx0;Tx 1)a[d(x0;Tx 0) +d(x1;Tx 1) ];
that is,
d(x1;x2)a[d(x0;x1) +d(x1;x2) ]
which yields
d(x1;x2)a
1a1

d(x0;x1);
and so (3.2) holds for n= 1.
Next, we take x:=xn1; y:=xnin (3:1) and obtain
d(xn;xn+1)a[d(xn1;xn) +d(xn;xn+1) ]
and we have
d(xn;xn+1)a
1a

d(xn1;xn): (3.3)
Suppose (3.2) holds for n=k, that is,
d(xk;xk+1)a
1ak

d(x0;x1)
and prove that (3.2) also holds for n=k+ 1. Indeed, in view of (3.3)
d(xk+1;xk+2)a
1a
d(xk;xk+1)a
1a
a
1ak

d(x0;x1)
i.e.,
d(xk+1;xk+2)a
1ak+1

d(x0;x1):
Then, by mathematical induction we have that (3.2) holds for all n0.
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1and
y:=xn1in (3.1) and so we have
d(Txn+p1;Txn1)a[d(xn+p1;xn+p) +d(xn1;xn) ]()
d(xn+p;xn)a[d(xn+p1;xn+p) +d(xn1;xn) ]
and by using (3.2) we have
d(xn+p;xn)a"a
1an+p1
d(x0;x1) +a
1an
d(x0;x1)#
=

6 Paula Homorodan
=a"a
1an+p1
+a
1an#
d(x0;x1)
We obtain
d(xn+p;xn)a
a
1an

"a
1ap1
+ 1#
d(x0;x1); (3.4)
by which we immediately conclude that fxngis a Cauchy sequence. As ( X;d)
is complete, it follows that fxngis convergent, which proves the rst part of
the theorem.
Now, if we denote  x= lim
n!1xn2Fix(T), then the uniqueness is imme-
diate. Indeed, suppose that Twould have two xed points
x;y2Fix(T), x6= y. Then
d(x;y) =d(Tx;Ty)a[d(x;Tx) +d(y;Ty) ] = 0;
a contradiction. So  x6= y, and hence  xis the unique xed point of T.
Remark 3.1.
1.IfTis continous then the limit xof the Picard sequence is always a
xed point of T. Indeed, we have:
x= lim
n!1xn+1= lim
n!1T(xn) =T( lim
n!1xn) =T(x);
i.e., xis a xed point of T;
2.In general, the limit xof the Picard sequence is not a xed point of T;
3.Ifdis actually a quasimetric, then Theorem 3.1reduces to the well
known Kannan xed point theorem in metric spaces [6].
Example. LetX=f0;1g[f 2n:n2Ng. Consider on Xthe
F-symmetric d, dened as: d(x;x) = 0 ,d(0;x) =d(x;0) = 0 ;d(1;21) =
d(21;1) = 1 ,d(1;2n) =d(2n;1) = 5 ,n6= 0;1;d(21;2n) =d(2n;21) =
32,n6= 0;1;d(2n;2n1) =d(2n1;2n) = 32,n6= 0;1andd(2m;2n) =j
2m2nj;m+ 16=n, for allm;n2N.
Note thatdis not a metric, because the triangle inequality is not satis-
ed: 1 = (1;21)>d(1;0) +d(0;21) = 0 .
Now we consider the mapping T:X!X, whereT(0) = 0 ,T(1) = 21
andT(2n) = 2n1. Condition (3.1) is obviously satised for the following
cases: 1)x= 0;y= 1and 2)x= 0;y= 2n, since the left hand side term of
the inequality is zero. We now check the remaining cases:
3)x= 1;y= 2n, when condition (3.1) witha=1
4, reduces to
d(T(1);T(2n))1
4
[d(1;T(1)) +d(2n;T(2n))]()1
95
2:
4)x= 2m;y= 2n, when condition (3.1) witha=1
4, reduces to
d(T(2m);T(2n))1
4
[d(2m;T(2m) +d(2n;T(2n)]()

Fixed point theorems for discontinous mappings 7
1<9
2, ifm6=n. Ifm=n, thend(T(2m);T(2n)) = 0 and(3.1) it is also
true.
Therefore,Tis a Kannan contraction, i.e.,
d(T(x);T(y))1
4
[d(x;T(x) +d(y;T(y)];x;y2X;
the Picard iteration is a convergent Cauchy sequence and Fix(T) =f0g.
The next Corollary is a generalization of Kannan xed point theorem
in metric spaces [6].
Corollary 3.1. Let(X;d)be a complete symetric H-distance space and let
T:X!Xbe a mapping for which there exists 0< a <1
2such that (3.1)
holds.
Then:
(i) The Picard iteration at the point xis convergent.
(ii) If, additionally, the limit xof the Picard sequence is a xed point
ofT, then xis the unique xed point of T.
Proof. If (X;d) is a complete symetric H-distance space, then it is a complete
H-distance and conclusion follows by Theorem 3.1. 
Theorem 3.2. Let(X;d)be a complete H-distance space and let T:X!X
be a mapping. Assume that there exists a number h,0<h< 1such that
d(Tx;Ty )hmaxfd(x;Tx );d(y;Ty )g;8x;y2X: (3.5)
Then the Picard iteration of the point xis convergent. If, additionally the
limit xof the Picard sequence is a xed point of T, then xis the unique xed
point ofT.
Proof. Letx02Xand denefxngbyxn+1=Txn,n0.
We prove that
d(xn;xn+1)hmaxfd(xn1;xn);d(xn;xn+1)g; n0: (3.6)
We have two cases:
Case 1.
If maxfd(xn1;xn);d(xn;xn+1)g=d(xn;xn+1) then by (3.6) it follows
d(xn;xn+1)h d(xn;xn+1)<d(xn;xn+1), a contradiction.
Case 2.
If maxfd(xn1;xn);d(xn;xn+1)g=d(xn1;xn) then we obtain
d(xn;xn+1)h d(xn1;xn): (3.7)
By (3.7), we have d(xn;xn+1)hn
d(x0;x1); n0
To prove thatfxngis a Cauchy sequence, we take x:=xn+p1and
y:=xn1in (3.5) and so we have

8 Paula Homorodan
d(Txn+p1;Txn1)hmaxfd(xn+p1;xn+p);d(xn1;xn)g ()
d(xn+p;xn)hmaxfd(xn+p1;xn+p);d(xn1;xn)g;
and by using (3.7) we have
d(xn+p;xn)h
maxfhn+p1
d(x0;x1);hn1
d(x0;x1)g=
=h
d(x0;x1) maxfhn+p1;hn1g=
=h
hn1
d(x0;x1) =hn
d(x0;x1); (3.8)
by which we immediately conclude that fxngis a Cauchy sequence. As ( X;d)
is complete, it follows that fxngis convergent, which proves the rst part of
the theorem.
Now, if we denote  x= lim
n!1xn2Fix(T), then the uniqueness is imme-
diate. Indeed, suppose that Twould have two xed points
x;y2Fix(T), x6= y.
d(x;y) =d(Tx;Ty)h
maxfd(x;Tx);d(y;Ty)g= 0;
a contradiction. So  x6= y, and hence  xis the unique xed point of T.

Remark 3.2.
1.IfTis continous then the limit xof the Picard sequence is always a
xed point of T. Indeed, we have:
x= lim
n!1xn+1= lim
n!1T(xn) =T( lim
n!1xn) =T(x),
i.e., xis a xed point of T.
2.In general the limit xof the Picard sequence is not a xed point of T.
3.Ifdis a quasimetric, Theorem 3.2reduces to the well known Bianchini
xed point theorem in metric spaces [4].
Example. LetX=f0;1g[f 2n:n2Ng. Consider on Xthe
F-symmetric d, dened as: d(x;x) = 0 ,d(0;x) =d(x;0) = 0 ;d(1;22) =
d(22;1) = 21;d(1;2n) =d(2n;1) = 5 ,n6= 0;1;2;d(21;2n1) =
d(2n1;21) = 21,n6= 0;1;d(22;2n1) =d(2n1;22) = 21,n6=
0;1;2;d(2n;2n1) = 1 andd(2m;2n) =j2m2nj; m+ 16=n, for
allm;n2N.
Note thatdis not a metric, because the triangle inequality is not satis-
ed: 21=d(1;22)>d(1;0) +d(0;22) = 0 .
Now we consider the mapping T:X!X, whereT(0) = 0 ,T(1) = 22,
T(2n) = 2n1. Condition (3.5) is obviously satised for the following cases:
1)x= 0;y= 1 and 2)x= 0;y= 2n, since the left hand side term of the
inequality is zero. We now check the remaining cases:
3)x= 1;y= 2n, when condition (3.5) withh=1
2, reduces to
d(T(1);T(2n))1
2
maxfd(1;T(1)); d(2n;T(2n))g ()1
21
2:

Fixed point theorems for discontinous mappings 9
4)x= 2m;y= 2n, when condition (3.5) withh=1
2, reduces to
d(T(2m);T(2n))1
2
maxfd(2m;T(2m)); d(2n;T(2n))g ()
1
21
2,m6=n:Ifm=n, thend(T(2m);T(2n)) = 0 and (3.5) it is also
true.
Therefore,Tis a Bianchini contraction, i.e.,
d(T(x);T(y))1
2
maxfd(x;T(x); d(y;T(y)g;x;y2X;
the Picard iteration is a convergent Cauchy sequence and Fix(T) =f0g.
The next Corollary is a generalization of Bianchini xed point theorem
in metric spaces [4].
Corollary 3.2. Let(X;d)be a complete symetric H-distance space and let
T:X!Xbe a mapping. Assume that there exists a number h,0<h< 1
such that (3.5) holds.
Then:
(i) The Picard iteration of the point xis convergent.
(ii) If, additionally the limit xof the Picard sequence is a xed point of
T, then xis the unique xed point of T.
Proof. If (X;d) is a complete symmetric H-distance space, then it is a com-
pleteH-distance and conclusion follows by Theorem 3.2. 
Conclusion
By working in the general setting of a complete H-distance space, we obtained
signicant generalizations of Kannan and Bianchini xed point theorems in
usual metric spaces.
We note that, so far, we were not able to obtain a Banach type xed point
theorem in distance spaces which are not quasimetric spaces.
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[1] V. Berinde, \Approximating xed points of weak contractions using the Picard
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[4] R. Bianchini, \Su un problema di S.Reich riguardante la teori dei punti ssi."
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10 Paula Homorodan
[5] M. Choban, \Fixed point of mappings dened on spaces with distance."
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[6] R. Kannan, \Some results on xed points II." Am.Math.Mon. , vol. 76, pp. 405{
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Paula Homorodan
Department of Mathematics and Computer Science
North University Center at Baia Mare
Technical University of Cluj-Napoca
Victoriei 76, 430122, Baia Mare, Romania
e-mail: paula homorodan@yahoo.com