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Fixed point theorems for Ćirić type mapping and application to integral equation
Fixed Point Theory and Applications volume 2015, Article number: 117 (2015)
Abstract
In this paper, we introduce a new class of Ćirić type single-valued mapping with respect to u-distance and prove some fixed point theorems for this mapping. An example is given to show that our results are a proper extension of many well-known results. As an application, we establish the existence of a solution for an integral equation.
1 Introduction
The Banach contraction principle is a remarkable result in metric fixed point theory. Over the years, it has been generalized in different directions and spaces by several mathematicians, see [1–17] and the references therein. In 1974, Ćirić [5] proved the following fixed point theorem on a complete metric space, which generalizes the Banach contraction principle: Let X be a complete metric space and let \(T:X\rightarrow X\) be a quasi-contractive mapping; i.e., there exists a constant \(q \in[0,1)\) such that, for all \(x, y \in X\),
Then:
-
(1)
T has a unique fixed point w in X.
-
(2)
\(\lim_{n \rightarrow\infty} T^{n} x= w\) for every \(x \in X\).
-
(3)
\(d(T^{n} x, w) \leq[ \frac{q^{n}}{1-q} ] d(x, Tx)\) for every x in X.
Recently, Ume [15] generalized the notion of τ-distance [18] by introducing u-distance as follows.
Let X be metric space with metric d. Then a function \(p:X \times X \rightarrow R_{+} \) is called a u-distance on X if there exists a function \(\theta: X \times X \times R_{+} \times R_{+} \rightarrow R_{+} \) such that the following hold for \(x, y,z \in X \):
- (u1):
-
\(p(x, z) \leq p(x, y) + p(y, z)\);
- (u2):
-
\(\theta(x, y, 0, 0) = 0\) and \(\theta(x, y, s, t) \geq\min\{s,t\}\) for each \(s, t \in R_{+} \), and for any \(x \in X\) and for every \(\varepsilon> 0\), there exists \(\delta > 0\) such that \(| s- s_{0} |< \delta\), \(|t - t_{0} |< \delta\), \(s, s_{0}, t, t_{0} \in R_{+}\) and \(y \in X\) imply \(|\theta(x,y,s,t) -\theta(x,y,s_{0} , t_{0} ) |< \varepsilon\);
- (u3):
-
\(\lim_{n \rightarrow\infty} x_{n} =x\) and \(\lim_{n \rightarrow \infty} \sup\{ \theta(w_{n} , z_{n} , p(w_{n} , x_{m}), p(z_{n} ,x_{m} )) : m\geq n \} = 0\) imply \(p(y,x) \leq\lim_{n \rightarrow\infty} \inf p(y, x_{n} )\) for all \(y\in X\);
- (u4):
-
\(\lim_{n \rightarrow\infty} \sup\{p(x_{n} , w_{m}):m \geq n \} =0\), \(\lim_{n \rightarrow\infty} \sup\{p(y_{n} , z_{m}):m \geq n \} =0\), \(\lim_{n \rightarrow\infty} \theta(x_{n} , w_{n} , s_{n} ,t_{n} )=0\), \(\lim_{n \rightarrow\infty} \theta(y_{n} , z_{n} , s_{n} ,t_{n} )=0\) imply \(\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , s_{n} ,t_{n} )=0\) or \(\lim_{n \rightarrow\infty} \sup\{p(w_{m} , x_{n} ): m \geq n \} =0\), \(\lim_{n \rightarrow\infty} \sup\{p(z_{m} , y_{n}): m \geq n \} =0\), \(\lim_{n \rightarrow\infty} \theta(x_{n} , w_{n} , s_{n} ,t_{n} )=0\), \(\lim_{n \rightarrow\infty} \theta(y_{n} , z_{n} , s_{n} ,t_{n} )=0\) imply \(\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , s_{n} ,t_{n} )=0\);
- (u5):
-
\(\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , p(w_{n} , x_{n}), p(z_{n} ,x_{n} ))=0\), \(\lim_{n \rightarrow\infty} \theta(w_{n} , z_{n} , p(w_{n} , y_{n}), p(z_{n} ,y_{n} ))=0\) imply \(\lim_{n \rightarrow\infty} d(x_{n}, y_{n})=0\) or \(\lim_{n \rightarrow\infty} \theta(a_{n} , b_{n} , p(x_{n} , a_{n}), p(x_{n} ,b_{n} ))=0\), \(\lim_{n \rightarrow\infty} \theta(a_{n} , b_{n} , p(y_{n} , a_{n}), p(y_{n} ,b_{n} ))=0\) imply \(\lim_{n \rightarrow\infty} d(x_{n}, y_{n})=0\).
Remark 1.1
([15])
-
(a)
Suppose that θ from \(X \times X \times R_{+} \times R_{+} \) into \(R_{+} \) is a mapping satisfying \((\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})\). Then there exists a mapping η from \(X \times X \times R_{+} \times R_{+} \) into \(R_{+} \) such that η is nondecreasing in its third and fourth variable, satisfying \((\mathrm{u}_{2})_{\eta}{\sim}(\mathrm{u}_{5})_{\eta}\), where \((\mathrm{u}_{2})_{\eta}{\sim}(\mathrm{u}_{5})_{\eta}\) stand for substituting η for θ in \((\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})\), respectively.
-
(b)
On account of (a), we may assume that θ is nondecreasing in its third and fourth variables, respectively, for a function θ from \(X \times X \times R_{+} \times R_{+} \) into \(R_{+} \) satisfying \((\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})\).
-
(c)
Each τ-distance p on a metric space \((X,d)\) is also a u-distance on X. We present some example of u-distance which are not τ-distance. (For details, see [15].)
Example 1.2
Let \(X=R_{+}\) with the usual metric. Define \(p: X \times X \rightarrow R_{+} \) by \(p(x,y) = (\frac{1}{4})x^{2}\). Then p is a u-distance on X but not a τ-distance on X.
Example 1.3
Let X be a normed space with norm \(\| \cdot\|\). Then a function \(p:X \times X \rightarrow R_{+} \) defined by \(p(x,y)=\| x \|\) for every \(x, y \in X \) is a u-distance on X but not a τ-distance.
It follows from the above example and Remark 1.1(c) that u-distance is a proper extension of τ-distance. Other useful examples on u-distance are given in [15].
2 Preliminaries
Throughout this paper we denote by N the set of all positive integers, by R the set of all real numbers and by \(R_{+}\) the set of all nonnegative real numbers.
Definition 2.1
([15])
Let X be a metric space with a metric d and let p be a u-distance on X. Then a sequence \(\{x_{n}\}\) in X is called p-Cauchy if there exists a function θ from \(X \times X \times R_{+} \times R_{+} \) into \(R_{+} \) satisfying \((\mathrm{u}_{2}){\sim}(\mathrm{u}_{5})\) and a sequence \(\{z_{n}\}\) of X such that
Lemma 2.2
([15])
Let X be a metric space with a metric d and let p a u-distance on X. If \(\{x_{n}\}\) is a p-Cauchy sequence, then \(\{x_{n} \}\) is a Cauchy sequence.
Lemma 2.3
([15])
Let X be a metric space with a metric d and let p be a u-distance on X.
-
(1)
If sequences \(\{x_{n}\}\) and \(\{y_{n}\}\) of X satisfy \(\lim_{n \rightarrow\infty} p(z,x_{n}) =0 \) and \(\lim_{n \rightarrow\infty} p(z,y_{n}) =0\) for some \(z \in X\), then \(\lim_{n\rightarrow\infty} d(x_{n}, y_{n}) =0\).
-
(2)
If \(p(z,x)=0\) and \(p(z,y)=0\), then \(x=y\).
-
(3)
Suppose that sequences \(\{x_{n}\}\) and \(\{y_{n}\}\) of X satisfy \(\lim_{n \rightarrow\infty} p(x_{n},z) =0 \) and \(\lim_{n \rightarrow\infty} p(y_{n} ,z) =0\) for some \(z \in X\), then \(\lim_{n\rightarrow\infty} d(x_{n}, y_{n}) =0\).
-
(4)
If \(p(x,z)=0\) and \(p(y,z)=0\), then \(x=y\).
Lemma 2.4
([15])
Let X be a metric space with a metric d and let p be a u-distance on X. Suppose that a sequence \(\{x_{n}\}\) of X satisfies
Then:
-
(i)
\(\{ x_{n} \}\) is a p-Cauchy sequence.
-
(ii)
If \(\{x_{n}\}\) is a p-Cauchy sequence, then \(\{x_{n}\}\) is a Cauchy sequence.
3 Fixed point theorems
The following lemma plays an important role in proving our theorems.
Lemma 3.1
Let \((X,d)\) be a metric space with a u-distance p on X and \(\{a_{n}\}\) and \(\{b_{n}\}\) be sequences of X such that
Then there exist a subsequence \(\{a_{k_{n}}\}\) of \(\{a_{n}\}\) and a subsequence \(\{b_{k_{n}}\}\) of \(\{b_{n}\}\) such that \(\lim_{n \rightarrow\infty} d(a_{k_{n}}, b_{k_{n}})=0\).
Proof
Since p is a u-distance on X,
For each \(n\in N\), let
By the hypotheses and (3.2), we have
Let \(k_{1} \in N\) be an arbitrary and fixed element. Then, by (u2), for this \(a_{k_{1}} \in X\) and \(\varepsilon=1\), there exists \(\delta_{1} >0 \) such that
By virtue of (3.3) and (3.4), for this \(\delta_{1}>0\), there exists \(M_{1} \in N\) such that
Let \(k_{2} \in N\) be such that
Due to (3.6), we have
From (3.4), (3.5), (3.6), and (3.7) we get
In terms of (u2) and (3.6), for this \(a_{k_{2}} \in X\) and \(\varepsilon=\frac{1}{2} \), there exists \(\delta_{2} >0 \) such that \(| s |=s < \delta_{2}\), \(| t |=t < \delta_{2}\), \(y \in X \) imply
In view of (3.3) and (3.9), for this \(\delta_{2}>0\), there exists \(M_{2}\in N\) such that
Let \(k_{3} \in N\) be such that
On account of (3.9), (3.10), (3.11), we obtain
Continuing this process, there exist a subsequence \(\{a_{k_{n}}\}\) of \(\{a_{n}\}\), and a subsequence \(\{b_{k_{n}}\}\) of \(\{b_{n}\}\) such that, for all \(n \in N\),
Using (3.2), (3.3), and (3.13), we know that
Using (3.1), (3.2), (3.14) and putting \(x_{n} =y_{n} = a_{k_{n}}\), \(w_{m} =z_{m} = a_{k_{m+1}}\) and \(s_{n} =t_{n} = \alpha_{k_{n+1}} + \beta_{k_{n+1}}\) in (u4) we deduce
Using (3.15) and putting \(w_{n} =z_{n} = a_{k_{n+1}}\), \(x_{n}= a_{k_{n+2}}\), and \(y_{n}=b_{k_{n+2}}\) in (u5), we have
Due to (3.13) and (3.16), there exist a subsequence \(\{a_{k_{n}}\}\) of \(\{a_{n}\}\) and a subsequence \(\{b_{k_{n}}\}\) of \(\{b_{n}\}\) such that
□
Definition 3.2
Let \((X,d)\) be a metric space with a u-distance p on X and let T be a selfmapping on X. For \(A\subseteq X\), let \(\delta(A)= \sup\{ p(x,y) : x,y \in A \}\) and for each \(x, y \in X\), \(n \in N\), let
where \(T^{0} x =x\) and \(T^{i}\) is the i times repeated composition of T with itself. Let
for each \(x, y \in X\).
Lemma 3.3
Let \((X,d)\) be a metric space with a u-distance p on X. Let \(T : X \rightarrow X \) and \(\varphi: R_{+} \rightarrow R_{+}\) be mappings that satisfy the following conditions:
for all \(x,y \in X\);
where \(\varphi^{n}\) is n-times repeated composition of φ with itself.
Then:
-
(1)
For each \(x,y \in X\) and \(n\in N\),
$$\begin{aligned}& \max\bigl\{ p\bigl(T^{i} x, T^{j} x\bigr), p \bigl(T^{i} x, T^{j} y\bigr), p\bigl(T^{i} y, T^{j} x\bigr), p\bigl(T^{i} y, T^{j} y\bigr)\mid i,j \in N, i,j \leq n \bigr\} \\& \quad \leq\varphi\bigl(\delta\bigl(O(x,y,n)\bigr)\bigr). \end{aligned}$$ -
(2)
For each \(x,y \in X\), \(n \in N\) and for each \(i\in N\) with \(i \leq8\), there exists \(l_{i} \in N\) with \(l_{i} \leq n\) such that
$$\begin{aligned} \begin{aligned} \delta\bigl(O(x,y,n)\bigr) ={}& \max\bigl\{ p(x,x),p(x,y),p(y,x),p(y,y),p \bigl(x,T^{l_{1}} x\bigr), p\bigl(x,T^{l_{2}} y\bigr), \\ &p\bigl(y, T^{l_{3}} x\bigr),p\bigl(y, T^{l_{4}} y\bigr), p\bigl(T^{l_{5}} x, x\bigr), p \bigl(T^{l_{6}} x ,y \bigr),p\bigl(T^{l_{7}} y , x\bigr), p \bigl(T^{l_{8}} y , y\bigr) \bigr\} . \end{aligned} \end{aligned}$$ -
(3)
For each \(x,y \in X\),
$$\delta\bigl(O(x,y,\infty)\bigr) \leq(I-\varphi)^{-1} \bigl(b(x,y) \bigr), $$where \(b(x,y)=p(x,x)+p(y,y)+p(x,y)+p(y,x)+p(x,Tx)+p(Tx,x) +p(y,Ty)+ p(Ty,y)\).
-
(4)
For each \(x \in X\), \(\{T^{n} x \}\) is a Cauchy sequence.
-
(5)
For each \(x,y \in X\) and \(n \in N\),
$$p\bigl(T^{n} x, T^{n} y\bigr) \leq\varphi^{n-1} \bigl((I-\varphi)^{-1} \bigl(b(x,y)\bigr)\bigr). $$ -
(6)
For each \(x,y \in X\), \(\lim_{n\rightarrow\infty} p(T^{n} x, T^{n} y)=0\).
Proof
Let \(x,y \in X\) and \(n \in N\), and let i and j be natural numbers with \(i,j \leq n\). Then \(T^{i-1} x, T^{i} x, T^{j-1} x, T^{j} x, T^{i-1} y, T^{i} y, T^{j-1} y, T^{j} y \in O(x,y,n)\).
From (3.18) and hypothesis (ii), we have
which proves (1).
From (1), it follows that for each \(x,y \in X\), \(n\in N\) and for each \(i \in N\) with \(i \leq8 \), there exists \(l_{i} \in N \) with \(l_{i} \le n \) such that
which proves (2).
Applying the triangle inequality, hypothesis (iii), (1), and (2), we have
Therefore \(\delta(O(x,y,n)) \leq(I- \varphi)^{-1} (b(x,y))\).
Since n is arbitrary, the proof of (3) is complete.
To prove (4), let x be an arbitrary point of X and define \(x_{n} = T^{n} x\) for every \(n \in N\). On account of (3.18) and hypothesis (ii), we have
Substituting (3.21)∼(3.26) into (3.20), proceeding in this manner and by hypotheses, (1), (2), and (3) of Lemma 3.3, we have
where \(a(x)=4[p(x,x)+p(x, Tx) +p(Tx,x)]\).
If \(n< m\), then, by (3.27),
Combining (3.19) and (3.28), we get
By means of Lemma 2.4 and (3.29),
for each \(x \in X\). This is the proof of (4).
To prove (5), let \(x,y \in X\) and define \(x_{n} = T^{n} x\) and \(y_{n}=T^{n} y\) for every \(n \in N\). By the same method as in (3.18)∼(3.27), we get
which proves (5).
By virtue of (3.19) and (3.30), we deduce that
for each \(x,y\in X\). This is the proof of (6). □
Definition 3.4
Let \((X,d)\) be a metric space, a mapping \(T:X\rightarrow X\) is called Ćirić type φ-generalized single-valued p-contractive if it satisfies the following:
-
(c1)
There exist a u-distance p on X and \(\varphi: [0, \infty) \rightarrow[0, \infty) \) such that
$$\begin{aligned} p(Tx,Ty) \leq&\varphi\bigl(\max\bigl[p(x,y),p(x,Tx),p(y,Ty),p(x,Ty),p(y,Tx), \\ &p(y,x),p(Tx,x),p(Ty,y),p(Ty,x),p(Tx,y)\bigr]\bigr) \end{aligned}$$for all \(x,y \in X\).
-
(c2)
For each \(x \in X\) with \(\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X\), there exists \(y \in X\) such that \(\lim_{n \rightarrow \infty} T^{n}y=Tc_{x}\).
Theorem 3.5
Let \((X,d)\) be a complete metric space with a u-distance p. Let \(T:X\rightarrow X\) be Ćirić type φ-generalized single-valued p-contractive satisfying (ii)∼(iv) of Lemma 3.3. Then:
-
(1)
\(\lim_{n \rightarrow\infty} T^{n} x=z\) for each \(x \in X\).
-
(2)
\(p(T^{n} x, z) \leq\sum_{k=n-1}^{\infty} \varphi^{k} ((I-\varphi)^{-1} (a(x))) \) for each \(x \in X\), where \(a(x) = [p(x,x)+p(x,Tx)+p(Tx,x)] \times4\).
-
(3)
T has a unique fixed point z in X and \(p(z,z)=0\).
Proof
Let \(x,y \in X\) and let \(x_{n} = T^{n} x\) and \(y_{n}=T^{n} y\) for every \(n \in N\). Then, by (4) of Lemma 3.3, \(\{ x_{n} \}\) is a Cauchy sequence.
Since X is complete, \(\{ x_{n} \}\) converges to some \(z \in X\). This is the proof of (1). Due to (3.28), (iv) of Lemma 3.3, Lemma 2.4, Definition 2.1, and (u3), we have
which proves (2).
By (1) and (c2) of Definition 3.4, there exists \(y \in X\) such that
In view of (3.29) and (3.31), we get
Due to (3.33), we obtain
In terms of (3.29), (3.34), and Lemma 3.1, there exist a subsequence \(\{x_{k_{n}}\}\) of \(\{x_{n}\}\) and a subsequence \(\{y_{k_{n}}\}\) of \(\{y_{n}\}\) such that
From (1), (3.32), and (3.35), we have
Thus z is a fixed point of T.
To prove the unique fixed point of T, let \(z=Tz\) and \(w=Tw\).
Then, by hypothesis, we obtain
By (3.36) and the hypothesis
From Lemma 2.3 and (3.37), we have
□
From Theorem 3.5, we have the following corollary.
Corollary 3.6
Let \((X,d)\) be a complete metric space with a u-distance p on X. Let \(T:X \rightarrow X\) be a mapping that satisfies the following conditions:
for all \(x, y \in X\) and for some \(k \in(0,1)\);
Then T has a unique fixed point z in X and \(p(z,z)=0\).
Proof
Let \(\varphi: R_{+} \rightarrow R_{+} \) be defined by
Then, by (3.39), all the conditions of Theorem 3.5 are satisfied.
Thus T has a unique fixed point z in X and \(p(z,z)=0\). □
Lemma 3.7
Let \((X,d)\) be a complete metric space with a u-distance p on X and let \(T:X \rightarrow X\) be a mapping satisfying (3.38) and
for every \(y \in X\) with \(y \neq Ty\).
Then, for each \(x \in X\) with \(\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X\), there exists \(y \in X\) such that \(\lim_{n \rightarrow\infty} T^{n}y=Tc_{x}\).
Proof
Suppose that there exists some \(x \in X\) with \(\lim_{n \rightarrow\infty} T^{n}x=c_{x} \in X\) such that
From (3.41) we get
Then, by (3.42), the same method as in Theorem 3.5 and simple calculations, we have
On account of (3.43) and the hypotheses of Lemma 3.7, we obtain
This is a contradiction. □
From Corollary 3.6 and Lemma 3.7 we have the following corollary.
Corollary 3.8
([15])
Let \((X,d) \) be a complete metric space with a u-distance p on X. Let \(T:X\rightarrow X\) be a mapping satisfying (3.38) and (3.40). Then T has a unique fixed point z in X and \(p(z,z)=0\).
Proof
Since all the conditions of Corollary 3.8 satisfy all the conditions of Corollary 3.6, we obtain result of Corollary 3.8. □
In the next example we shall show that all the conditions of Theorem 3.5 are satisfied, but condition (3.38) in Corollary 3.6 and condition (3.40) in Lemma 3.7 are not satisfied.
Example 3.9
Let \(k \in(0,1)\) and let \(X=[0,1]\) be closed interval with the usual metric, and \(p:X \times X \rightarrow R_{+} \), \(T:X \rightarrow X\) and \(\varphi: R_{+} \rightarrow R_{+}\) be mappings defined as follows:
Define \(\theta: X \times X \times R_{+} \times R_{+} \rightarrow R_{+} \) by
for all \(x,y \in X\) and \(s,t \in R_{+}\).
Then, by (3.44)∼(3.47) and simple calculations, we know that p is a u-distance on X and φ satisfies (ii) and (iii) in Lemma 3.3. We now show that φ satisfies (iv) in Lemma 3.3.
On account of (3.46), if \(0 \leq t \leq\frac{1-k}{1+k}\), then \(\varphi^{n} (t) = (\frac{1+k}{2})^{n} t \) for all \(n \in N\) and so (iv) holds for all \(t \in [0,\frac{1-k}{1+k}]\).
If \(t > \frac{1-k}{1+k}\), then there exists \(M \in N\) such that
Suppose that \(\varphi^{n} (t)> \frac{1-k}{1+k}\) for all \(n \in N\) and \(t \in(\frac{1-k}{1+k}, \infty)\).
Then, by (3.46), \(\frac{1-k}{1+k} < \varphi^{n}(t) = \frac{t}{1+nt}\) for all \(n \in N\) and \(t \in(\frac{1-k}{1+k}, \infty)\). Thus \(0< \frac{1-k}{1+k} \leq \lim_{n \rightarrow \infty} \varphi^{n}(t) = \lim_{n \rightarrow\infty} \frac{t}{1+nt} =0 \), a contradiction.
Hence (3.48) holds.
By virtue of (3.46) and (3.48), we get
for all \(n \in N\) with \(n >M\). Thus φ satisfies (iv) in Lemma 3.3 for \(t \in(\frac{1-k}{1+k}, \infty)\). Therefore (iv) in Lemma 3.3 holds. Using (3.44)∼(3.46), we have
for all \(x, y \in X\).
By (3.49), (c1) of Definition 3.4 is satisfied.
Due to (3.45), since \(\lim_{n \rightarrow\infty} T^{n} x = \lim_{n \rightarrow\infty} (\frac{1+k}{2})^{n} x=0\) for each \(x \in X\), there exists \(y = \frac{x}{2} \in X\) such that \(\lim_{n \rightarrow\infty} T^{n} y =\lim_{n \rightarrow\infty} (\frac{1+k}{2})^{n} \cdot\frac{x}{2}=0=T0\).
This implies (c2) of Definition 3.4.
Therefore all the conditions of Theorem 3.5 are satisfied.
By means of (3.44) and (3.45), there exist \(a=1 \in X\) and \(b= 0 \in X\) such that
On account of (3.50), (3.38) in Corollary 3.6 is not satisfied.
In terms of (3.44) and (3.45) we obtain
for all \(y \in X \) with \(y \neq Ty \). This means that (3.40) in Lemma 3.7 is not satisfied.
Remark 3.10
It follows from Lemma 3.7 and Example 3.9 that Theorem 3.5 is a proper extension of Corollary 3.6 and Corollary 3.8, the results of Ćirić [5], Kannan [12] and Ume [15].
The following theorem is a generalization of Suzuki’s fixed point theorem [18].
Theorem 3.11
Let \((X,d)\) be a complete metric space with a u-distance p on X. Let \(\varphi: R_{+}\rightarrow R_{+} \) be a mapping satisfying conditions (ii)∼(iv) of Lemma 3.3.
Let \(T: X \rightarrow X\) be a mapping that satisfies the following conditions:
Then there exists \(z \in X\) such that \(Tz=z\) and \(p(z,z)=0\).
Proof
By (3.51), the same methods in Theorem 3.5 and simple calculations, we deduce that
By means of (3.52) and hypotheses (i), (ii), we obtain
□
Corollary 3.12
([18])
Let \((X,d)\) be a complete metric space with a τ-distance p on X. Let \(T: X \rightarrow X\) be a mapping satisfying (ii) of Theorem 3.11 and
for all \(x \in X\) and some \(k \in(0,1)\).
Then T has a unique fixed point z in X and \(p(z,z)=0\).
Proof
Let \(\varphi: R_{+}\rightarrow R_{+} \) be defined by
Since p is a τ-distance, p is a u-distance. Thus all the conditions of Theorem 3.11 are satisfied.
Therefore T has a unique fixed point z in X and \(p(z,z)=0\). □
4 Existence of a solution for an integral equation
In what follows, we assume that \(X=C([0,1])\) is the set of all continuous functions defined on \([0,1]\) and \(\varphi: R_{+} \rightarrow R_{+}\) satisfy conditions (ii), (iii), and (iv) of Lemma 3.3. Let \(d,p:X \times X \rightarrow R_{+}\) and \(\theta:X \times X \times R_{+} \times R_{+} \rightarrow R_{+}\) be mappings defined as follows:
and
for all \(x,y \in X\) and \(s,t \in R_{+} \). Then clearly \((X,d)\) is a complete metric space and p is a u-distance on X. Now we prove the existence theorem for a solution of the following integral equation by using Theorem 3.5:
where \(x \in X\), \(r: X \times R \rightarrow R\), \(G: [0,1] \times[0,1] \rightarrow R\) and \(f : [0,1] \times R \rightarrow R\) are given mappings.
Theorem 4.1
Suppose that the following hypotheses hold:
- (I1):
-
\(r: X \times R \rightarrow R\) is a continuous mapping such that
$$\bigl\vert r(x,t)\bigr\vert \leq{{1}\over {2}} \varphi\bigl(\bigl\vert x(t)\bigr\vert \bigr) \quad \textit{for all } x \in X \textit{ and } t \in R . $$ - (I2):
-
\(G: [0,1] \times[0,1] \rightarrow R\) is a continuous mapping such that
$$\bigl\vert G(t,s)\bigr\vert \leq{{1}\over {2}} \quad \textit{for all } t,s \in [0,1]. $$ - (I3):
-
\(f: [0,1] \times R \rightarrow R\) is a continuous mapping such that
$$\bigl\vert f\bigl(s,x(s)\bigr)\bigr\vert \leq \varphi\bigl(\bigl\vert x(s) \bigr\vert \bigr) \quad \textit{for all } x \in X \textit{ and } s \in [0,1] . $$ - (I4):
-
For each \(x \in X\) with \(\lim_{n \rightarrow\infty} {T^{n} x=c_{x} \in X} \), there exists \(y \in X\) such that \(\lim_{n \rightarrow\infty} T^{n} y=Tc_{x} \).
Then the integral equation (4.1) has a solution \(x \in X\).
Proof
Let \(T: X\rightarrow X\) be a mapping defined by
for all \(x \in X\) and \(t \in[0,1]\). By conditions (I1), (I2), and (I3), we have
for all \(x \in X\) and \(t \in[0,1]\). Then
for all \(x,y \in X\).
Thus all of the hypotheses of Theorem 3.5 are satisfied. Hence the mapping T has a fixed point that is a solution in \(X=C([0,1])\) of the integral equation (4.1). □
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Acknowledgement
This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2013R1A1A2057665).
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Ume, J.S. Fixed point theorems for Ćirić type mapping and application to integral equation. Fixed Point Theory Appl 2015, 117 (2015). https://doi.org/10.1186/s13663-015-0370-x
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DOI: https://doi.org/10.1186/s13663-015-0370-x