On The Fixed Points Of Large-Kannan Contraction Mappings And Applications

On The Fixed Points Of Large-Kannan Contraction Mappings And Applications

Abdelkader Dehici

, Mouataz Billah Mesmouli

, Erdal Karapinar

Received 10 August 2018

Abstract

In this paper, we establish some …xed point results for large-Kannan mappings in complete metric spaces. Our results are applied to solve some implicit integral equations.

1 Introduction and Preliminaries

It is well known that Banach’s contraction principle (1922) is a powerful tool in analy- sis, that most mathematicians applying about to solve many of their problems. It is remarkable by its simplicity; this comes from the fact that the contractive condition on the mapping is easy to check and it requires only a complete metric space for its framework. Banach’s contraction principle appeared in explicit form in Banach’s thesis [2], where it was used to obtain the solution of an integral equation in the functional space C([0;1]). One of the other application of the Banach contraction principle is for example, the study of existence and uniqueness of solutions of the Itô stochastic equation with deviating argument. Indeed, it was proved (see [5]) that if the stochas- tic parameters satisfy Lipschitz condition in the second variable, then the associated integral operator turns out to be a contraction and the solution is obtained by using the method of successive approximations. In Banach spaces setting, recall that Kras- noselskii’s …xed point result is a combination of Banach and Schauder’s …xed point theorems, this result can be seen as a consequence of the fact that the measure of noncompactness is invariant by compact perturbations. The sum of two operators is a powerful tool used to solve delay integral equations, neutral functional equations, Cauchy problems for ordinary di¤erential equations and partial di¤erential equations modeled by Hammerstein integral operators in Lp-spaces. For more details, see [1, 3, 4].

In this work, we establish existence and uniqueness results for large-Kannan map- pings extending those in [8, 9, 12]. In particular, we distinguish the continuous and

Mathematics Sub ject Classi…cations: 47H10.

yLaboratory of Informatics and Mathematics, University of Souk-Ahras, P.O.Box 1553, Souk-Ahras 41000, Algeria

zMathematics Department, Faculty of Science, University of Hail, Kingdom of Saudi Arabia

xDepartment of Medical Research, China Medical University Hospital, China Medical University, 40402, Taichung, Taiwan

535

non-continuous cases. Finally, our results are explored to solve an implicit functional equation.

The classical statement of Banach …xed point theorem is the following:

THEOREM 1. Let (X; d) be a complete metric space and T : X ! X be a contraction mapping, i.e.,

d(T x; T y) d(x; y);

for all x; y2X, where0< <1. ThenT has a unique …xed pointy₀ inX. Moreover for each x₀2X, the sequence of iteratesfTⁿx₀gn converges toy₀.

In 1968, R. Kannan [8, 9] obtained the following …xed point result.

THEOREM 2. Let (X; d) be a complete metric space and T : X ! X be a selfmapping on X. Assume that there exists 2 0;¹₂ such that

d(T x; T y) [d(x; T x) +d(y; T y)];

for allx; y2X. ThenT has a unique …xed pointz0in X. Moreover for eachx02X, the sequence of iteratesfTⁿx0gn converges toz0.

It has be seen that contraction mappings are continuous which is not in general the case of Kannan mappings as shown in the following examples.

EXAMPLE 1. Let(X; d) = (R;j j)andT :R!Rbe de…ned by T x= 0 ifx 2;

1

4 ifx >2:

For allx; y2R, we prove easily that jT x T yj 1

4(jx T xj+jy T yj):

EXAMPLE 2. Let(X; d) = ([0;1];j j)andT : [0;1]![0;1]be given by T x=

x

16 ifx2[0;1[;

1

18 ifx= 1:

Letx; y2[0;1[. Thus

jT x T yj= x 16

y 16 = 1

16jx yj; and

jx T xj= 15x

16 ; jy T yj= 15y 16;

which implies that

jT x T yj= 1

16jx yj 1

14(jx T xj+jy T yj): Now, ifx2[0;1[andy= 1, we get

jT x T yj= x 16

1 18 ; and

jx T xj=15x

16 ; jT1 1j=17 18: Consequently, for all x; y2[0;1], we have

jT x T yj x 16+ 1

18 1

14(jx T xj+jy T yj):

EXAMPLE 3. Let(X; d) = ([0;1];j j)and T : [0;1]![0;1]be de…ned by T x=

x

6 if0 x < ¹₂;

x

4 if ¹₂ x 1:

Ifx; y2 0;¹₂ , we get

jT x T yj 1

6(x+y) =1

5(jx T xj+jy T yj): On the other hand, ifx; y2 ¹₂;1 , we get

jT x T yj 1

4(x+y) =1

3(jx T xj+jy T yj): Now, if0 x < ¹₂ y, then

jT x T yj= 1 6x 1

4y 1

4(x+y) 1

3(jx T xj+jy T yj): Consequently, for all x; y2[0;1], we obtain that

jT x T yj 1

3(jx T xj+jy T yj):

There is a large literature dealing with Kannan mappings and their generalizations, we can quote for examples [6, 7, 8, 9, 10] and [12]. In [3], Burton observed that Theorem 1 is more interesting in applications through some modi…cation and formulated it as follows:

DEFINITION 1. Let(X; d)be a metric space and letT :X!X be a selfmapping on X. T is said to be a large contraction, if for x; y 2 X, with x 6= y, we have d(T x; T y)< d(x; y), and if for all >0, there exists <1 such that

[x; y2X; d(x; y) ] =)d(T x; T y) d(x; y):

REMARK 1. We observe that every contraction mapping is a large contraction.

The converse does not hold in general, as the following example shows [3].

EXAMPLE 4. Let(X; d) = (R;j:j)and letT :R !Rbe de…ned byT x=x x³. Then forx; y2R, by applying the Mean Value Theorem, we get

jT x T yj= x x³ y+y³ 1 3c² jx yj; where c2]minfx; yg;maxfx; yg[.

Afterwards, from the inequality given above, it is easy to observe that there exists su¢ ciently small such that for allx; y2[ ; ](x6=y), we havejT x T yj<jx yj. Additionally, it was proved in [3] that for a given >0, ifjx yj , then

jT x T yj 1

2

4 jx yj;

moreover, since T0 = 0andlim_x!0 ^{x x}_x³ = 1, we deduce thatT is not a contraction selfmapping on [ ; ].

EXAMPLE 5. Letf : [0;1]![0;1]be given byf(x) =x ^x₄⁴. Then jf(x) f(y)j = x x⁴

4 y y⁴

4 = (x y) 1

4 x^{2 2} y^{2 2}

= (x y) 1

4 x² y² x²+y²

= (x y) 1

4(x y) (x+y) x²+y²

= (x y) 1 1

4(x+y) x²+y² :

Sincejx yj jx+yjandjx yj²=x²+y² 2xy 2 x²+y² , it follows that jf(x) f(y)j = (x y) 1 1

4(x+y) x²+y² jx yj 1 jx yj³

8

! :

Next, if jx yj , we infer that

jf(x) f(y)j jx yj 1

3

8 :

Hence, to deduce thatf is a large contraction, it su¢ ces to take ( ) = 1 ₈³ . Now, to show thatf is not a contraction mapping, it su¢ ces to see that

xlim!0

x ^x₄⁴

x = 1:

Thus, there is nok2(0;1)such that x x⁴

4 y y⁴

4 kjx yj; 8x; y2(0;1):

THEOREM 3 ([3]). Let(X; d)be a complete metric space and letT :X!X be a large contraction selfmapping. Assume that there exist x₀ 2X andL >0, such that d(x₀; Tⁿx₀) Lfor alln 1. ThenT has a unique …xed point in X.

REMARK 2. Notice that, if(X; d)is a compact metric space, then the assumption that there exist x0 2X and L > 0, such that d(x0; Tⁿx0) L for alln 1 can be dropped. Indeed, in this case, the existence and uniqueness of the …xed point is ensured by Edelstein’s theorem.

REMARK 3. If(X; d)is a bounded complete metric space, then for allx02X and for all integern 1, we haved(x0; Tⁿx0) (X), where (X)is the diameter ofX. So the boundedness assumption given above is trivially satis…ed in this setting.

2 Main Results

We start this section by the following lemma which asserts that the set of contraction mappings has an in…nite subset of Kannan mappings.

LEMMA 1. Let(X; d)be a metric space. Assume thatT :X !Xbe a selfmapping onX satisfying that

d(T x; T y) d(x; y)for allx; y2X;

where 2 0;¹₃ . ThenT is a Kannan mapping with a constant of contraction equals to ₁ .

PROOF. Letx; y2X. Then by assumption, we have d(T x; T y) d(x; y);

where 2 0;¹₃ . On the other hand, by using the triangle inequality, we get d(x; y) d(x; T x) +d(T x; T y) +d(T y; y):

Multiplying the previous inequality by , it follows that

d(T x; T y) d(x; y) (d(x; T x) +d(T x; T y) +d(T y; y));

therefore,

d(T x; T y)

1 (d(x; T x) +d(y; T y)):

Since 2 0;¹₃ , then ₁ 2 0;¹₂ . Consequently,T is a Kannan mapping.

2.1 Large-Kannan Contractions in the Continuous Sense

Now, we give the following de…nition of a large-Kannan contraction (in the continuous sense) as an extension of the classical ones.

DEFINITION 2. Let(X; d)be a metric space and letT :X!X be a selfmapping on X. T is said to be a large-Kannan contraction (in the continuous sense), if for x; y 2X, withx6=y, we haved(T x; T y)< d(x; y), and if for all >0, there exists

<¹₂ such that

[x; y2X; d(x; y) ] =)d(T x; T y) [d(x; T x) +d(y; T y)]:

REMARK 4. Large-Kannan contractions (in the continuous sense) are continuous.

This is an immediate consequence of the inequality d(T x; T y)< d(x; y)forx6=y.

REMARK 5. Let(X; d)be a metric space and letT :X !Xbe a large contraction on X. Assume that 2 [0;¹₃[, then by Lemma 1, it is easy to conclude that T is a large-Kannan contraction mapping.

Now, we give the following …xed point result for large-Kannan contractions.

THEOREM 4. Let(X; d)be a complete metric space and T :X !X be a large- Kannan contraction mapping (in the continuous sense). Then T has a unique …xed point in X.

PROOF. Let x0 2 X, if there exists an integer m 1 such that T^m(x0) = T^m+1(x0), thenT(T^mx0) =T^mx0 andT^mx0is a …xed point ofT.

Now, assume thatTⁿx₀6=Tⁿ⁺¹x₀for every integern 1. SinceT is large-Kannan contraction (in the continuous sense), then

d Tⁿ⁺¹x0; Tⁿx0 < d Tⁿx0; T^{n 1}x0 < ::: < d(T x0; x0):

This proves that the sequence _n = d Tⁿ⁺¹x0; Tⁿx0 is strictly decreasing, hence

n lim!+1 n= 0. If >0, then for alln 1, we get

d Tⁿ⁺¹x0; Tⁿx0 :

Consequently, there exists <¹₂ such that

d Tⁿ⁺¹x₀; Tⁿ⁺²x₀ = d T(Tⁿx₀); T Tⁿ⁺¹x₀

d Tⁿx₀; Tⁿ⁺¹x₀ +d Tⁿ⁺¹x₀; Tⁿ⁺²x₀ : This implies that

(1 )d Tⁿ⁺¹x0; Tⁿ⁺²x0 d Tⁿx0; Tⁿ⁺¹x0 : Thus, we have

d Tⁿ⁺¹x0; Tⁿ⁺²x0

1 d Tⁿx0; Tⁿ⁺¹x0

1

2

d T^{n 1}x0; Tⁿx0

...

1

n

d T x₀; T²x₀

1

n+1

d(x₀; T x₀): (1) Since <¹₂, we see thatk= ₁ <1. So, by using (1), it follows that

n!1lim d Tⁿx₀; Tⁿ⁺¹x₀ = 0; (2) which is a contradiction. Hence = 0 and achieves the proof of this step.

Now, we shall prove that fxngⁿ given by xn = Tⁿx0 is a Cauchy sequence in X. Suppose, to the contrary thatfxngⁿis not a Cauchy sequence. Thus, there exist >0 and subsequences of integers(Nk),(nk),(mk)such that

N_k! 1,m_k> n_k> N_k; and

d(xmk; xnk): (3)

SinceT is large-Kannan mapping, by using (3), there exists <¹₂ such that d(x_mk; x_nk) =d(T x_{mk 1}; T x_{nk 1}) [d(x_{mk 1}; x_mk) +d(x_{nk 1}; x_nk)]: Lettingk! 1, from (2), follows that

klim!1d(x_{mk 1}; x_mk) = lim

k!1d(x_{nk 1}; x_nk) = 0:

Hence lim

k!1d(xmk; xnk) = 0, which is a contradiction. Thus fxngⁿ is a Cauchy sequence in X. Finally, since X is complete, then there exists l 2 X such that

nlim!1x_n= lim

n!1Tⁿx₀=l. The continuity ofT implies thatT(l) =l, which proves that l is a …xed point ofT.

Now, suppose thatl⁰is another …xed point forT such thatl6=l⁰. Thusd(l; l⁰) 0

for some ₀>0. SinceT is Kannan-large mapping, there exists ₀<¹₂ such that d(l; l⁰) =d(T(l); T(l⁰)) 0[d(l; T(l)) +d(l⁰; T(l⁰))]:

Hence, we get d(l; l⁰) = 0, which is a contradiction. Thus, we must havel=l⁰. REMARK 6. For the case of large-Kannan mappings (in the continuous sense), the existence of …xed points is proved without any assumption on the boundedness of the set fd(x₀; Tⁿx₀)g for somex₀2X.

COROLLARY 1. Let (X; d) be a complete metric space and T : X ! X be a selfmapping onX such thatT^m0 is a large-Kannan mapping (in the continuous sense) for some integerm0 1. ThenT has a unique …xed point inX.

PROOF. From Theorem 4, there existsz₀2X such thatT^m0z₀=z₀, then T(T^m0z0) =T^m0+1z0=T z0:

This gives T^m0(T z0) = T z0 and implies that T z0 is a …xed point for T^m0z0. The uniqueness of the …xed point for the mapping T^m0 (given by Theorem 4) shows that T z0 =z0. Now, if z1 is another …xed point for T, thenz1 is a …xed point for T^m0. Hence z0=z1, which achieves the proof.

EXAMPLE 6. Let f : R! R, de…ned by f(x) = x³: If x 0, we have jxj = x x+x³ = x x³ = jx f(x)j. If x 0, we have jxj = x x x³ = x+ x³ = jx f(x)j. Let us prove that f is not a Kannan mapping. For all x; y2R, we have

jf(x) f(y)j= x³ y³ : Thus

jx f(x)j+jy f(y)j= x+x³ + y+y³ : By takingy= 0and letting x!+1, we get

x!lim+1

jf(x) f0j

jx f(x)j+j0 f0j = lim

x!+1

jf(x)j

jx+x³j = lim

x!+1

jx³j

jx+x³j = 1> 1 2: Consequently, f need not be a Kannan mapping.

Now, set

= (x; y)2[ 1;1]²: x²+y²+xy +1

2jx yj² 1 2 :

Ifx; y2 , then

jf(x) f(y)j = x³ y³

= jx yj x²+y²+xy (jxj+jyj) x²+y²+xy

(jx f(x)j+jy f(y)j) x²+y²+xy (jx f(x)j+jy f(y)j) 1 jx yj²

2

! :

So, for a given (su¢ ciently small) >0, ifx; y2 satisfying thatjx yj , we have jf(x) f(y)j (jx f(x)j+jy f(y)j) 1 ²

2 :

Finally, to conclude that f is a large-Kannan mapping, it su¢ ces to take ( ) = ¹₂² which achieves the proof.

By Rakotch [11], let denote the class of real-valued control functions (not neces- sarily continuous) which satisfy the condition

= f : (0;1)! 0;1

2 ; f(t_n)7! 1

2 )t_n!0 (n! 1) :

Now, we are in position to prove a general version of Theorem 4 given as follows:

THEOREM 5. Let (X; d) be a complete metric space and T : X ! X be a selfmapping such that, for x; y2X, withx6=y, we haved(T x; T y)< d(x; y)and for all >0, there existsf 2 such that

[x; y2X; d(x; y) ] =)d(T x; T y) f (d(x; y)) [d(x; T x) +d(y; T y)]: ThenT has a unique …xed pointz₀ inX.

PROOF. Let x₀ 2 X, if there exists an integer m₀ 1 such that T^m0(x₀) = T^m0+1(x0), thenT(T^m0x0) =T^m0x0andT^m0x0 is a …xed point ofT.

Now, assume that Tⁿx0 6=Tⁿ⁺¹x0 for every integer n 1. De…ne the sequence fxngn byxn =Tⁿx0. Hence

d(x_n; x_n+1) =d Tⁿx₀; Tⁿ⁺¹x₀ < d Tⁿx₀; T^{n 1}x₀ =d(x_{n 1}; x_n);

this implies that the sequence _n = d(x_n; x_n+1) is strictly decreasing, consequently

n lim!+1 n= 0. If >0, by assumption, there existsf 2 such that

d(xn; xn+1) =d Tⁿx0; Tⁿ⁺¹x0 f (d(xn 1; xn)) [d(xn; xn+1) +d(xn 1; xn)]: Therefore

d(x_n; x_n+1)

d(xn; xn+1) +d(xn 1; xn) f (d(xn 1; xn))<1 2:

Letting n! 1, it follows that

nlim!1

d(xn; xn+1)

d(x_n; x_n+1) +d(x_{n 1}; x_n)=

2 = 1 2 lim

n!1f (d(x_{n 1}; x_n))<1 2; (4) which is a contradiction, then we must have = 0.

Now, we shall prove thatfx_ngn given byx_n=Tⁿx₀ is a Cauchy sequence inX. Suppose, to the contrary, thatfx_ngn is not a Cauchy sequence inX. Thus, there exist ₀>0, subsequences of positive integers(N_k),(n_k)and(m_k)such that

N_k! 1,m_k> n_k> N_k; and

0 d(xmk; xnk) =d(T xnk 1; T xmk 1): (5) This last inequality shows that x_{mk 1} 6=x_{nk 1}. Now, by assumptions and using the relation 5, there existsf 2 such that

0 d(x_nk; x_mk) =d(T x_{nk 1}; T x_{mk 1})

f (d(xnk 1; xmk 1)) [d(xnk 1; xnk) +d(xmk 1; xmk)]

< 1

2[d(x_{nk 1}; x_nk) +d(x_{mk 1}; x_mk)]: Lettingk! 1, it follows that

0 lim

k!1d(x_nk; x_mk) = 0;

which is a contradiction. Thusfx_ngnis a Cauchy sequence inX. SinceX is a complete metric space, there existsz₀2X such that lim

n !+1x_n=z₀. To prove thatz₀is a …xed point forT, we argue as follows:

Select an arbitrary integern 1. By using the triangle inequality, we get d(z₀; T z₀) d(z₀; x_n+1) +d(T z₀; x_n+1)

=d(z0; xn+1) +d(T z0; T xn):

Without loss of generality, we assume that x_n 6=z₀. Hence d(z₀; T z₀) d(z₀; x_n+1) +d(T z₀; x_n+1) d(z0; xn+1) +¹₂(d(xn; xn+1) +d(z0; T z0)):

Similarly

0 ¹₂d(z₀; T z₀) d(z₀; x_n+1) +¹₂d(x_n; x_n+1):

Lettingn !+1, it follows that 0 ¹₂d(z0; T z0) lim

n !+1d(z0; xn+1) +1 2 lim

n !+1d(xn; xn+1) = 0:

Consequently,z₀ is a …xed point forT.

If T has two …xed points z0; z1 2 X; z0 6= z1, thus, by our assumption and putting

=d(z₀; z₁), there existsf

2 (taking its values in the interval[0;¹₂[) such that 0 <

2 < d(z0; z1) =d(T z0; T z1) f₂(d(z0; z1)) (d(z0; T z0) +d(z1; T z1))

< 1

2(d(z0; T z0) +d(z1; T z1)) = 0:

This is a contradiction. Consequently, we must havez₀=z₁which achieves the proof.

REMARK 7. It is worth noting that the condition d(T x; T y) < d(x; y); (x6=y) does not imply the existence of the …xed point. To see this, it su¢ ces to take(X; d) = (R;j j)andT x=p

x²+ 1.

DEFINITION 3. Let(X; d)be a metric space andT be a selfmapping onX. T is said to be asymptotically regular if for eachx2X we have lim

n!1d Tⁿx; Tⁿ⁺¹x = 0.

De…ne now the class of functions ⁰ by

0 =ff : (0;1)![0;1[; f(tn)7!1)tn!0 (n! 1)g

In the following …xed point theorem, we drop the condition d(T x; T y)< d(x; y), (x6=y)and we replace it by T continuous and asymptotically regular.

THEOREM 6. Let(X; d)be a metric space and letT be a continuous selfmapping onX which is asymptotically regular. Assume that for every >0, there existf 2 ⁰ such that

[x; y2X; d(x; y) ] =)d(T x; T y) f (d(x; y)) [d(x; T x) +d(y; T y)]: Then T has a unique …xed pointz₀ inX. Moreover for eachx₀2X, the sequence of iterates fTⁿx₀gn converges toz₀.

PROOF. Letx02X and de…ne the sequence fxngn byxn =Tⁿx0 for all integer n 1. If there exists m₀ 1 such that T^m0(x₀) = T^m0+1(x₀), thenT(T^m0x₀) = T^m0x₀ andT^m0x₀ is a …xed point ofT.

Now, suppose that Tⁿx₀ 6= Tⁿ⁺¹x₀ for all n 1. We shall prove that fx_ngn is a Cauchy sequence inX.

If it’s not the case, there exist ₀>0and subsequences of positive integers(N_k),(n_k) and (m_k) such that lim

k !+1N_k = +1, m_k > n_k > N_k and d(x_mk; x_nk) ₀. Thus, by using our assumptions and the triangle inequality, we get

d(xmk; xnk) d(xmk; xmk+1) +d(xmk+1; xnk+1) +d(xnk+1; xnk)

d(x_mk; x_mk+1) +f₀(d(x_nk; x_mk)) [d(x_nk; x_nk+1) +d(x_mk; x_mk+1)]

+d(xnk+1; xnk);

for some f₀ 2 ⁰. Then,

[1 f ₀(d(x_nk; x_mk))]d(x_nk; x_mk) d(x_nk; x_mk)

[1 +f ₀(d(xnk; xmk))]

(d(xnk; xnk+1) +d(xmk; xmk+1)): (6) Dividing each right side in (6) by

[1 f₀(d(x_nk; x_mk))] (d(x_nk; x_nk+1) +d(x_mk; x_mk+1)); and using the fact thatd(x_mk; x_nk) ₀, we conclude

0

d(x_nk; x_nk+1) +d(x_mk; x_mk+1)

d(xnk; xmk)

d(x_nk; x_nk+1) +d(x_mk; x_mk+1) 1 +f ₀(d(x_nk; x_mk))

1 f ₀(d(xnk; xmk)): Lettingk! 1, thus

k!lim+1

1 +f ₀(d(x_nk; x_mk))

1 f ₀(d(xnk; xmk))= +1: Consequently,

lim sup

n;m!1

f ₀(d(xn; xm)) = 1: (7)

Sincef ₀2 ⁰, we obtain

n;mlim!1d(xn; xm) = 0;

which is a contradiction. Hence fxngⁿ must be a Cauchy sequence.

Therefore, sinceXis a complete metric space, there existsz⁰₀2Xsuch thatlimxn= z₀⁰. The continuity ofT implies thatT z⁰₀=z₀⁰.

Ifz₁⁰ is another …xed point forT such that z⁰₀6=z⁰₁. Setd(z⁰₀; z₁⁰) = ₁> ₂¹. Then by assumption there exists f 1

2 2 ⁰ such that h

x; y2X; d(x; y) ¹ 2 i

=)d(T x; T y) f ¹

2 (d(x; y)) [d(x; T x) +d(y; T y)]: It follows that

d(z₀⁰; z₁⁰) f₁(d(z₀⁰; z₁⁰)) [d(z₀⁰; T z₀⁰) +d(z₁⁰; T z₁⁰)] = 0;

which is a contradiction. Hence z⁰₀=z₁⁰ which achieves the proof.

2.2 Large-Kannan Contractions in the Non (Necessarily) Con- tinuous Sense

In this section, we focus our study on …xed point results for Kannan-large mappings which are not necessarily continuous. First of all, we start by following de…nition

DEFINITION 4. Let(X; d)be a metric space and letT :X !X be a selfmapping.

T is said to be a large-Kannan contraction (in the non-necessarily-continuous sense), if for x; y2X, withx6=y, we haved(T x; T y)< ¹₂(d(x; T x) +d(y; T y)), and if for all

>0, there exists <¹₂ such that

[x; y2X; d(x; y) ] =)d(T x; T y) [d(x; T x) +d(y; T y)]:

REMARK 8. The following example given in [7] shows that mappings satisfying that d(T x; T y)< ¹₂(d(x; T x) +d(y; T y))may fail to have …xed points.

EXAMPLE 7. LetX =f1 +¹_n; n= 1;2; :::gandd0:X X ![0;+1[de…ned by

d0(x; y) = 0 ifx=y;

x+y ifx6=y:

Thus(X; d₀)is a complete metric space. On the other hand, letT : (X; d₀) !(X; d₀) de…ned by T(1 + ¹_n) = 1 + _n+1¹ . In [7], it was proved thatT satis…es the inequality d0(T x; T y)<¹₂(d0(x; T x) +d0(y; T y))forx6=y butT has no …xed points.

THEOREM 7. Let(X; d)be a complete metric space and let T : (X; d) !(X; d) be a large-Kannan mapping (in the non-necessarily continuous sense). Then T has a unique …xed point.

PROOF.The uniqueness: IfT has two …xed pointsx0; x12X; x06=x1. Thus 0 d(x0; x1) =d(T x0; T x1)<1

2(d(x0; T x0) +d(x1; T x1)) = 0;

which is a contradiction.

The existence: Step 1: Letx02X and de…ne the Picard sequence fxngⁿ by xn = Tⁿx₀ for all integer n 1. If there exists an integer m₀ 1 such that T^m0x₀ = T^m0+1x₀, thus T^m0x₀ is a …xed point point for T and the proof is achieved. Now, assume that x_n = Tⁿx₀ 6= Tⁿ⁺¹x₀ = x_n+1 for all n 1. We shall prove that the sequence _n =d(x_n; x_n+1)is strictly decreasing.

We have

d(x_n; x_n+1) =d(T T^{n 1}x₀; T Tⁿx₀))<¹₂(d(T^{n 1}x₀; Tⁿx₀) +d(Tⁿx₀; Tⁿ⁺¹x₀))

=¹₂(d(xn 1; xn) +d(xn; xn+1)):

So, we conclude that d(x_n; x_n+1)< d(x_{n 1}; x_n), which proves that _n =d((x_{n 1}; x_n) is strictly decreasing. Hence, there exists ₀ 0 such that lim

n !+1d(x_n; x_n+1) = ₀. Step 2: ₀= 0 :Suppose that ₀>0, since the sequence _n=d(x_n; x_n+1)is decreasing, we get ₀ < d(x_n; x_n+1) for all integer n 1. Thus, by assumption there exists 0< 0<¹₂ such that

d(x_n; x_n+1) =d(T T^{n 1}x₀; T Tⁿx₀)) ₀(d(T^{n 1}x₀; Tⁿx₀) +d(Tⁿx₀; Tⁿ⁺¹x₀))

= 0(d(xn 1; xn) +d(xn; xn+1));

which gives

d(x_n; x_n+1) ₁ ⁰

0d(x_{n 1}; x_n):

By induction, it follows that

d(xn; xn+1) (₁ ⁰₀)ⁿd(x0; x1):

Afterwards, since 0< 0 < ¹₂, then ₁ ⁰

0 <1. This proves that lim

n !+1( ⁰

1 0

)ⁿ = 0 and implies lim

n !+1d(xn; xn+1) = 0which is a contradiction. Consequently 0= 0.

Step 3: fxngⁿ is a Cauchy sequence inX:

If it is not the case, then there exists ₀ and subsequences of integersfN_kg;fn_kg and fm_kg with m_k > n_k > N_k such that ₀ d(x_nk; x_mk) =d(T x_{nk 1}; T x_{mk 1}), which leads to deduce thatxmk 16=xnk 1.

Thus, by assumption and using the fact that the sequence n=d(xn; xn+1)is decreas- ing, we get

0 d(x_mk; x_nk) =d(T x_{nk 1}; T x_{mk 1})

1

2(d(xnk 1; xnk) +d(xmk 1; xmk)) d(xnk 1; xnk):

Lettingk !+1, it follows that

0 lim

k!+1d(xmk; xnk) lim

k!+1d(xnk; xnk 1) = 0;

which is a contradiction. Hence fxngⁿ is a Cauchy sequence.

SinceX is a complete metric space, there existsz₀2X such that lim

n !+1x_n=z₀. Step 4: z₀ is a …xed point forT:

Select an arbitrary integern 1 and using the triangle inequality, we obtain d(z0; T z0) d(z0; xn+1) +d(T z0; xn+1):

=d(z₀; x_n+1) +d(T z₀; T x_n):

Without loss of generality, we assume that xn 6=z0. Thus d(z0; T z0) d(z0; xn+1) +d(T z0; xn+1):

d(z₀; x_n+1) +¹₂(d(x_n; x_n+1) +d(z₀; T z₀)):

Similarly

0 ¹₂d(z₀; T z₀) d(z₀; x_n+1) +¹₂d(x_n; x_n+1):

Lettingn !+1, we deduce 0 ¹₂d(z0; T z0) lim

n !+1d(z0; xn+1) +1 2 lim

n !+1d(xn; xn+1) = 0:

Henced(z₀; T z₀) = 0. Consequently,z₀is a …xed point forT which achieves the proof.

REMARK 9. Following the same way given in Theorem 7, we can establish a variants of Corollary 1 and Theorem 6 for the case of large-Kannan mappings (in the non-necessarily continuous sense).

3 Applications

In this last section, we prove that our results established in the previous section enable us to solve some implicit functional integral equations.

Let (8) be the integral equation formulated as a …xed point problem of the following nonlinear mapping

T x(t) = x(t) + Z t

1

(t; s; x(s); T x(s))dswhere0< <1

3; (8)

in Banach spaceE=C([ 1;1];R)of scalar continuous functions where the investiga- tion is essentially based on the properties of the kernel (; ; ; ).

Under the following assumptions:

1. (t; s; x(s); T x(s)) 0fort; s2[ 1;1]such that (; ;0; )6= 0andT(M) M whereM =fz2E: 1 z(t) 1g.

2. The mappingGde…ned byGx(t) = Z t

1

(t; s; x(s); T x(s))dssatis…esGx2M for allx2M and

kGx Gyk<(1 )kx yk; 8x; y2M; (x6=y):

3. For a given >0, there exists < ^{1 3}₂ such that ifx; y2M and kx yk , we have for allt2[ 1;1]

jGx(t) Gy(t)j (jx(t) T x(t)j+jy(t) T y(t)j):

ThenT has unique …xed point inM. PROOF. We have

x(t) T x(t) = (1 )x(t) Z t

1

(t; s; x(s); T x(s))ds:

Letx; y2M withkx yk . Then, by using our assumptions, we get jT x(t) T y(t)j

= (x(t) y(t)) + Z t

1

(t; s; x(s); T x(s))ds Z t

1

(t; s; y(s); T y(s))ds jx(t) +T x(t) T x(t) +T y(t) T y(t) y(t)j

+ Z t

1

[ (t; s; x(s); T x(s)) (t; s; y(s); T y(s))]ds (kx T xk+ky T yk+kT x T yk)

+ (kx T xk+ky T yk); which gives that

kT x T yk kT x T yk+ ( + ) (kx T xk+ky T yk): Hence

kT x T yk +

1 (kx T xk+ky T yk):

Now, since0< < ^{1 3}₂ , then ₁⁺ < ¹₂ and the result is an immediate consequence of Theorem 4.

REMARK 10. In equation (8), if T is a large-Kannan mapping (in the non- necessarily continuous sense) which is satis…ed under assumptions (1) and (3) and if(2)is replaced by the following;

(2⁰)The mappingGde…ned byGx(t) = Z t

1

(t; s; x(s); T x(s))dssatis…es thatGx2 M for allx2M and for allx; y2M withx6=y, we have

kGx Gyk<^{1 3}₂ (kx Gxk+ky Gyk):

Thus, by the same reasoning given above and using Theorem 7, we prove that the equation (8) has a unique solution.

REMARK 11. IfGis continuous andG(M)is a compact set inM, then the …xed point of T can be deduced as an immediate consequence of Krasnoselskii’s theorem since the mappingA:M !M de…ned byA = is a contraction and satis…es that A(M) M.

4 Conclusion

In applications, the implicit functional equation (8) is related to a large class of inter- esting problems. Several authors have studied important properties of their solutions (stability, controlability, ....). A major problem appears when the inversion of a per- turbed di¤erential operator does not yield a contraction and a compact mapping or

when it is hard to check this fact, since in this situation the classical Krasnoselskii’s

…xed point result or analog does not apply. To overcome this constrainst, we solve this equation by treating the continuous and non-continuous cases provided that the kernel or its associated mapping satis…es large-Kannan assumptions which is the main motivation of this work.

Acknowledgment. The authors would like to thank the anonymous referees for their valuable remarks and suggestions which help us to improve the paper.

References

[1] R. R. Akhmerov, M. I. Kamenskii, A. S. Potapov, A. E. Rodkina and B. N.

Sadovskii, Measure of Noncompactness and Condensing Operators, Birkhäuser Basel, 1992.

[2] S. Banach, Sur les opérations dans les ensembles abstraits et leur applications aux équations intégrales, Fundam. Math., 3(1922), 133–181.

[3] T. A. Burton, Integral equations, implicit relations and …xed points, Proc. Amer.

Math. Soc., 124(1996), 2383–2390.

[4] T. A. Burton, Krasnoselskii’s inversion principle and …xed points, Nonlinear Anal., 30(1997), 3975–3986.

[5] I. I. Gikhman and A. V. Skorohod, The theory of stochastic processes, Vols I and III, "Nauka", Moscou, 1971 and 1975.

[6] J. Górnicki, Fixed point theorems for Kannan maps type mappings, J. Fixed.

Point. Theor. Appl., 19(2017), 2145–2152.

[7] J. Górnicki, Various extensions of Kannan’s …xed point theorem, J. Fixed. Point.

Theor. Appl., 20(2018), 1–12.

[8] R. Kannan, Some results on …xed points, Bull. Calcutta Math. Soc., 60(1968), 71–76.

[9] R. Kannan, Some results on …xed points, II. Amer. Math. Mon., 76(1969), 405–

408.

[10] L. Nova, Fixed point theorems for some discontinuous operators, Paci…c. J. Math., 123(1986), 189–196.

[11] E. Racotch, A note of contractive mappings, Proc. Amer. Math. Soc., 13(1962), 459–465.

[12] S. Reich, Kannan’s …xed point theorem, Boll. Un. Mat. Ital., 4(1971), 1–11.