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Generalized multi-valued mappings satisfy some inequalities conditions on metric spaces
Journal of Inequalities and Applications volume 2015, Article number: 343 (2015)
Abstract
In this paper, we prove a condition of the existence for generalized multi-valued mappings satisfying some inequalities in metric spaces. These results are improved versions of results of Boško Damjanović and Dragan Dorić (Filomat 25:125-131, 2011).
1 Introduction and preliminaries
Let \((X,d)\) be a metric space. We denote by \(\operatorname {CB}(X)\) the family of all non-empty closed bounded subsets of X. Let \(H(\cdot,\cdot)\) be the Hausdorff metric, i.e.,
for \(A, B\in \operatorname {CB}(X)\), where
-
(i)
Let T be a self-mapping on X. Then T is called a Banach contraction mapping if there exists \(r\in[0, 1)\) such that
$$d(Tx,Ty)\leq r d(x,y) $$for all \(x, y \in X\).
-
(ii)
T is called a Kannan mapping if there exists \(a\in[0, \frac{1}{2})\) such that
$$d(Tx, Ty) \leq a d(x,Tx) + a d(y, Ty) $$for all \(x, y \in X\).
-
(iii)
If T is a mapping such that
$$d(Tx, Ty) \leq r\max\bigl\{ d(x,Tx), d(y, Ty)\bigr\} , $$such that \(r\in[0, 1)\) and all \(x, y \in X\), then T is called a generalized Kannan mapping.
In 1973, Hardy and Rogers [1] introduced a condition as follows:
-
(iv)
Let \(x, y \in X\). Then there exists \(a_{i}\geq 0\) such that
$$d(Tx, Ty) \leq a_{1}d(x,y)+a_{2}d(x,Tx)+a_{3}d(y, Ty)+a_{4}d(x,Ty)+a_{5}d(y, Tx), $$where \(\sum_{i=1}^{5}a_{i}<1\).
-
(v)
Ciric [2] defined the following condition which generalizes the Banach contraction and Kannan mapping, that is,
$$d(Tx, Ty) \leq r\max\bigl\{ d(x,y),d(x,Tx),d(y, Ty),d(x,Ty),d(y, Tx)\bigr\} , $$such that \(r\in[0, 1)\) and all \(x, y \in X\).
If X is complete and at least one of (i), (ii), (iii), (iv), and (v) holds, then T has a unique fixed point (see [1–5]).
In 2008, Suzuki [6] introduced the condition C as follows. T is said to satisfy condition C if
for all \(x, y \in C\).
In the same year, Kikkawa and Suzuki [7] generalized the Kannan mapping resulting in the following theorem.
Theorem 1.1
(Kikkawa and Suzuki [7])
Let T be a mapping on complete metric space \((X, d)\) and let φ be a non-increasing function from \([0,1)\) into \((\frac{1}{2},1]\) defined by
Let \(\alpha\in[0,\frac{1}{2})\) and put \(r = \frac{\alpha}{1-\alpha} \in [0,1)\). Suppose that
for all \(x, y\in X\). Then T has a unique fixed point z and \(\lim_{n\to\infty} T^{n}x = z\) holds for every \(x \in X\).
Theorem 1.2
(Kikkawa and Suzuki [7])
Let T be a mapping on a complete metric space \((X, d)\) and let θ be a non-increasing function from \([0,1)\) into \((\frac{1}{2},1]\) defined by
Suppose that \(r\in[0,1)\) such that
for all \(x, y\in X\). Then T has a unique fixed point z and \(\lim_{n\to\infty} T^{n}x = z\) holds for every \(x \in X\).
In 2011, Karapinar and Tas [8] stated some new conditions which are modifications of Suzuki’s condition C, as follows. T is said to satisfy condition \(SCC\) if
for all \(x, y \in K\), where
In 1969, Nadler [9] proved a multi-valued extension of the Banach contraction theorem as follows.
Theorem 1.3
(Nadler [10])
Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Assume that there exists \(r \in[0, 1)\) such that
for all \(x, y \in X\). Then there exists \(z \in X\) such that \(z \in Tz\).
Next, the result of Kikkawa and Suzuki [9] is a generalization of Nadler.
Theorem 1.4
(Kikkawa and Suzuki [9])
Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Define a strictly decreasing function η from \([0, 1)\) onto \((\frac{1}{2},1]\) by
and assume that there exists \(r \in[0, 1)\) such that
for all \(x, y \in X\). Then there exists \(z \in X\) such that \(z \in Tz\).
In 2011, Damjanović and Dorić [11] generalized the result of Kannan (iii) and Nadler.
Theorem 1.5
(Damjanović and Dorić [11])
Define a non-increasing function φ from \([0,1)\) into \((0,1]\) by
Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Assume that
for all \(x, y\in X\). Then there exists \(z\in X\) such that \(z \in T z\).
Corollary 1.6
(Damjanović and Dorić [11]) Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Let \(\alpha\in[0, \frac{1}{2})\) and put \(r = 2\alpha \). Suppose that
for all \(x, y\in X\), where the function φ is defined as in Theorem 1.5. Then there exists \(z\in X\) such that \(z \in T z\).
In this paper, we prove a condition of the existence for generalized multi-valued mappings under SCC conditions in metric spaces. These results are improved versions of results of Boško Damjanović and Dragan Dorić [11].
2 Main results
Theorem 2.1
Define a non-increasing function φ from \([0,\frac{1}{2})\) into \((0,1]\) by
Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Assume that
where \(M(x,y)=\max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}\), for all \(x, y\in X\). Then there exists \(z\in X\) such that \(z \in T z\).
Proof
Let \(r_{1}\) be a real number such that \(0 \leq r < r_{1} < \frac {1}{2}\). Let \(u_{1} \in X\) and \(u_{2} \in T u_{1}\) be arbitrary. Since \(u_{2} \in T u_{1}\), we have \(d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})\) and
Thus from the assumption (2.1),
where \(M(u_{1},u_{2})=\max\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),d(u_{1},Tu_{2}),d(u_{2},Tu_{1})\}\). Consider
If \(\max\{d(u_{1},u_{2}),d(u_{1},Tu_{2})\}=d(u_{1},Tu_{2})\), then
and then
If \(\max\{d(u_{1},u_{2}),d(u_{1},Tu_{2})\}=d(u_{1},u_{2})\), then
So
So there exists \(u_{3}\in Tu_{2}\) such that \(d(u_{2},u_{3})\leq (\frac {r_{1}}{1-r_{1}})d(u_{1},u_{2})\). Thus, we can construct a sequence \(\{x_{n}\}\) in X such that \(u_{n+1} \in Tu_{n}\) and
Hence, by induction,
Then by the triangle inequality, we have
Hence we conclude that \(\{u_{n}\}\) is a Cauchy sequence. Since X is complete, there is some point \(z \in X\) such that
Now, we will show that \(d(z,Tx)\leq rd(x,Tx)\) for all \(x\in X\setminus \{z\}\).
Let \(x\in X\setminus\{z\}\). Since \(u_{n}\to z\), there exists \(n_{0} \in N\) such that \(d(z, u_{n})\leq (\frac{1}{3})d(z, x)\) for all \(n\geq n_{0}\). Then we have
Then from (2.1) we have
Since \(u_{n+1} \in Tu_{n}\), \(d(u_{n+1}, Tx) \leq H(Tu_{n}, Tx)\), so that
for all \(n\geq n_{0}\). Letting \(n \to\infty\), we obtain
It follows that
Next, we show that \(z \in Tz\). Suppose that z is not an element in Tz.
Case (i): \(0\leq r<\frac{\sqrt{5}-1}{\sqrt{5}+1}\). Let \(a \in Tz\). Then \(a\neq z\) and so by (2.3), we have
On the other hand, since \(\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)\), from (2.1) we have
So
It implies that
Since \(d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)\), we have
Using (2.3), (2.4), and (2.5), we have
where \(k=\frac{r}{1-r}\). Since \(r<\frac{\sqrt{5}-1}{\sqrt{5}+1}\), we have \(k^{2}+k<1\) and so \(d(z, Tz)< d(z, Tz)\), a contradiction. Thus \(z\in Tz\).
Case (ii): \(\frac{\sqrt{5}-1}{\sqrt{5}+1}\leq r <\frac{1}{2}\). Let \(x\in X\). If \(x = z\), then \(H(T x, T z) \leq r \max\{ d(x,z),d(x,Tx),d(z,Tz), d(x,Tz),d(z,Tx)\}\) holds. If \(x \neq z\), then for all \(n\in\mathbb{N}\), there exists \(y_{n} \in Tx\) such that
We consider
Thus, \(( \frac{1-2r}{1-r})d(x, Tx)\leq(1+\frac{1}{n})d(x,z)\). Take \(n\to\infty\), we obtain
by using (2.1), implies \(H(T x, T z) \leq r \max\{ d(x,z),d(x,Tx),d(z,Tz),d(x,Tz),d(z,Tx)\}\). Hence, as \(u_{n+1}\in Tu_{n}\), it follows that with \(x=u_{n}\)
Therefore, \((1-r)d(z, Tz)\leq0\), which implies \(d(z, Tz)=0\). Since Tz is closed, we have \(z\in Tz\). This completes the proof. □
Example 2.2
Let \(X=[0,\infty)\) be endowed with the usual metric d. Define \(T: X\to \operatorname {CB}(X)\) by
Proof
We show that T satisfies (2.1). Let \(x,y\in X\). We prove by cases.
Case (i): Suppose that \(x,y \in[0,\frac{1}{2}]\). Thus, if \(x^{2}\leq y\), then
But if \(x^{2}> y\), then
and
where \(r=\frac{1}{4}\). Hence T satisfies (2.1).
Case (ii): Suppose that \(x,y \in(\frac{1}{2},1)\). Thus, if \(\frac {x}{3}\leq y\), then
But if \(\frac{x}{3}> y\), then
and
where \(r=\frac{1}{3}\). Hence T satisfies (2.1).
Case (iii): Suppose that \(x,y \in[1,\infty]\). Thus, if \(\log(x) \leq y\), then
But if \(\log(x)> y\), then
and
where \(r=\frac{1}{3}\). Hence T satisfies (2.1).
Case (iv): Suppose that \(x \in[0,\frac{1}{2}]\) and \(y \in(\frac {1}{2},1)\). Then \(x^{2}< x< y\). Thus, \(\varphi(\frac{1}{3})d(x, T x)=\vert x-x^{2}\vert \geq \vert x-y\vert =d(x,y)\). Hence T satisfies (2.1).
Case (v): Suppose that \(x \in(\frac{1}{2},1)\) and \(y \in[0,\frac {1}{2}]\). So \(x>y\). Thus, if \(\frac{x}{3} \leq y\), then
But if \(\frac{x}{3}> y\), then
and
where \(r=\frac{1}{3}\). Hence T satisfies (2.1).
Case (vi): Suppose that \(x \in[0,\frac{1}{2}]\) and \(y \in[1,\infty]\).
and
where \(r=\frac{1}{3}\). Hence T satisfies (2.1).
Case (vii): Suppose that \(x \in[1,\infty]\) and \(y \in[0,\frac {1}{2}]\). Thus, if \(\log(x) \leq y\), then
But if \(\log(x) > y\), then
and
where \(r=\frac{1}{4}\). Hence T satisfies (2.1).
Case (viii): Suppose that \(x \in(\frac{1}{2},1)\) and \(y \in[1,\infty]\).
and
where \(r=\frac{1}{3}\). Hence T satisfies (2.1).
Case (ix): Suppose that \(x \in[1,\infty]\) and \(y \in(\frac{1}{2},1)\). Thus, if \(\log(x) \leq y\), then
But if \(\log(x) > y\), then
and
where \(r=\frac{1}{3}\). Hence T satisfies (2.1).
Thus we see that T satisfies condition (2.1) and satisfies Theorem 2.1. So there exists \(z\in X\) such that \(z \in T z\). Moreover, \(0\in T(0)\). □
Theorem 2.3
Define a non-increasing function φ from \([0,\frac{1}{5})\) into \((0,1]\) by
Let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Assume that
where \(S(x,y)=r d(x,y)+r d(x,Tx)+r d(y,Ty)+r d(x,Ty)+r d(y,Tx)\) for all \(x, y\in X\). Then there exists \(z\in X\) such that \(z \in T z\).
Proof
Let \(r_{1}\) be a real number such that \(0 \leq r < r_{1} < 1\). Let \(u_{1} \in X\) and \(u_{2} \in T u_{1}\) be arbitrary. Since \(u_{2} \in T u_{1}\), we have \(d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})\) and
Thus, from the assumption (2.15),
where \(S(u_{1},u_{2})=r d(u_{1},u_{2})+r d(u_{1},Tu_{1})+r d(u_{2},Tu_{2})+r d(u_{1},Tu_{2})+r d(u_{2},Tu_{1})\). Consider
So
So there exists \(u_{3}\in Tu_{2}\) such that \(d(u_{2},u_{3})\leq( \frac {3r_{1}}{1-2r_{1}})d(u_{1},u_{2})\). Thus, we can construct a sequence \(\{x_{n}\}\) in X such that \(u_{n+1} \in Tu_{n}\) and
Hence, by induction,
Then by the triangle inequality, we have
Hence we conclude that \(\{u_{n}\}\) is a Cauchy sequence. Since X is complete, there is some point \(z \in X\) such that
Now, we will show that \(d(z,Tx)\leq( \frac{3r}{1-2r})d(x,Tx)\) for all \(x\in X\setminus\{z\}\).
Let \(x\in X\setminus\{z\}\). Since \(u_{n}\to z\), there exists \(n_{0} \in N\) such that \(d(z, u_{n})\leq (\frac{1}{3})d(z, x)\) for all \(n\geq n_{0}\). By using (2.2), we get
Then from (2.15) we have
Since \(u_{n+1} \in Tu_{n}\), \(d(u_{n+1}, Tx) \leq H(Tu_{n}, Tx)\), so that
for all \(n\geq n_{0}\). Letting \(n \to\infty\), we obtain
It follows that
Next, we show that \(z \in Tz\). Suppose that z is not an element in Tz.
Case (i): \(0\leq r<\frac{\sqrt{5}-1}{4+2\sqrt{5}}\). Let \(a \in Tz\). Then \(a\neq z\) and so by (2.16), we have
On the other hand, since \(\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)\), from (2.15) we have
So
Since \(d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)\), we have
Using (2.15), (2.16), and (2.17), we have
where \(k=\frac{3r}{1-2r}\).
Since \(0\leq r<\frac{\sqrt{5}-1}{4+2\sqrt{5}}\), we have \(0\leq k^{2}+k<1\) and so, \(d(z, Tz)< d(z, Tz)\), a contradiction. Thus \(z\in Tz\).
Case (ii): \(\frac{\sqrt{5}-1}{4+2\sqrt{5}}\leq r <\frac{1}{5}\). Let \(x\in X\).
If \(x = z\), then \(H(T x, T z) \leq r [d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]\) holds. If \(x \neq z\), then for all \(n\in\mathbb{N}\), there exists \(y_{n} \in Tx\) such that
We consider
Thus, \((\frac{1-5r}{1-2r})d(x, Tx)\leq(1+\frac{1}{n})d(x,z)\). Take \(n\to\infty\), we obtain
by using (2.15), implies \(H(T x, T z) \leq S(x,z)\), where \(S(x,z)=r [d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]\).
Hence, as \(u_{n+1}\in Tu_{n}\), it follows that with \(x=u_{n}\)
Using (2.18), we have \((1-2r)d(z, Tz)\leq0\), which implies \(d(z, Tz)=0\). Since Tz is closed, we have \(z\in Tz\). This completes the proof. □
Example 2.4
Let \(X=[0,\frac{1}{2}]\) with the metric \(d(x,y)=\frac {\vert x-y\vert }{\vert x-y\vert +1}\) for all \(x,y \in X\). Define \(T: X\to \operatorname {CB}(X)\) by
Proof
We show that T satisfies (2.15). Let \(x,y\in X\). Thus, if \(x^{2}\leq y\), then
But if \(x^{2}> y\), then
and
where \(r= \frac{1}{6}\).
Thus we see that T satisfies condition (2.15) and satisfies Theorem 2.3. So there exists \(z\in X\) such that \(z \in T z\). Moreover, \(0\in T(0)\). □
Theorem 2.5
Define a non-increasing function φ from \([0,1)\) into \((0,1]\) by
Let \(\alpha\in[0,\frac{1}{2})\) and \(r=\frac{\alpha}{1-\alpha}\), and let \((X, d)\) be a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\).
Assume that
where \(M(x,y)=\max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}\), for all \(x, y\in X\). Then there exists \(z\in X\) such that \(z \in T z\).
Proof
Let \(r_{1}\) be a real number such that \(0 \leq r < r_{1} < \frac {1}{2}\). Let \(u_{1} \in X\) and \(u_{2} \in T u_{1}\) be arbitrary. Since \(u_{2} \in T u_{1}\), we have \(d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})\) and
Thus, from the assumption (2.20),
where \(M(u_{1},u_{2})=\max\{ d(u_{1},u_{2}),d(u_{1},Tu_{1}),d(u_{2},Tu_{2}),d(u_{1},Tu_{2}),d(u_{2},Tu_{1})\}\).
Consider
If \(\max\{d(u_{1},u_{2}),d(u_{1},Tu_{2})\}=d(u_{1},Tu_{2})\), then
and then
where \(r=\frac{\alpha}{1-\alpha}\).
So there exists \(u_{3}\in Tu_{2}\) such that \(d(u_{2},u_{3})\leq r_{1}d(u_{1},u_{2})\). Thus, we can construct a sequence \(\{x_{n}\}\) in X such that \(u_{n+1} \in Tu_{n}\) and
Hence, by induction
Then by the triangle inequality, we have
Hence we conclude that \(\{u_{n}\}\) is a Cauchy sequence. Since X is complete, there is some point \(z \in X\) such that
Now, we will show that \(d(z,Tx)\leq rd(x,Tx)\) for all \(x\in X\setminus \{z\}\).
Let \(x\in X\setminus\{z\}\). Since \(u_{n}\to z\), there exists \(n_{0} \in N\) such that \(d(z, u_{n})\leq (\frac{1}{3})d(z, x)\) for all \(n\geq n_{0}\). By using (2.2), we get
Then from (2.20), we have
Since \(u_{n+1} \in Tu_{n}\), we have \(d(u_{n+1}, T_{x}) \leq H(Tu_{n}, T_{x})\), so that
for all \(n\geq n_{0}\). Letting \(n \to\infty\), we obtain
We obtain
Next, we show that \(z \in Tz\). Suppose that z is not an element in Tz.
Case (i): \(0\leq r<\frac{\sqrt{5}-1}{2}\). Let \(a \in Tz\). Then \(a\neq z\) and so by (2.21), we have
On the other hand, since \(\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)\), from (2.20) we have
So
It implies that
Since \(d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)\), we have
Using (2.20), (2.21), (2.22), and (2.23), we have
where \(r=\frac{\alpha}{1-\alpha}\).
Since \(r<\frac{\sqrt{5}-1}{2}\), we have \(r^{2}+r<1\) and so \(d(z, Tz)< d(z, Tz)\), a contradiction. Thus \(z\in Tz\).
Case (ii) \(\frac{\sqrt{5}-1}{2}\leq r <1\). Let \(x\in X\). If \(x = z\), then \(H(T x, T z) \leq\alpha\max\{ d(x,z),d(x,Tx),d(z,Tz), d(x,Tz),d(z,Tx)\}\) holds. If \(x \neq z\), then for all \(n\in\mathbb{N}\), there exists \(y_{n} \in Tx\) such that
We consider
Thus, \(( 1-r)d(x, Tx)\leq(1+\frac{1}{n})d(x,z)\). Take \(n\to\infty\), we obtain
by using (2.20), this implies \(H(T x, T z) \leq\alpha\max\{ d(x,z),d(x,Tx),d(z,Tz),d(x,Tz),d(z,Tx)\}\). Hence, as \(u_{n+1}\in Tu_{n}\), it follows that with \(x=u_{n}\)
Therefore, \((1-\alpha)d(z, Tz)\leq0\), which implies \(d(z, Tz)=0\). Since Tz is closed, we have \(z\in Tz\). This completes the proof. □
Corollary 2.6
Let be \((X, d)\) a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\). Let \(\alpha\in[0,\frac{1}{5})\) and \(r=5\alpha\). Assume that
where \(S(x,y)=\alpha d(x,y)+\alpha d(x,Tx)+\alpha d(y,Ty)+\alpha d(x,Ty)+\alpha d(y,Tx)\) for all \(x, y\in X\), where the function φ is defined as Theorem 2.5. Then there exists \(z\in X\) such that \(z \in T z\).
Remark 2.7
We see that Theorem 2.5 is a multi-valued mapping generalization of Theorem 2.3 of Kikkawa and Suzuki [7] and therefore the Kannan fixed point theorem [6] for generalized Kannan mappings.
Theorem 2.8
Define a non-increasing function φ from \([0,1)\) into \((0,1]\) by
Let \(\alpha\in[0,\frac{1}{5})\) and \(r=\frac{3\alpha}{1-2\alpha}\), and let be \((X, d)\) a complete metric space and let T be a mapping from X into \(\operatorname {CB}(X)\).
Assume that
where \(S(x,y)=\alpha d(x,y)+\alpha d(x,Tx)+\alpha d(y,Ty)+\alpha d(x,Ty)+\alpha d(y,Tx)\) for all \(x, y\in X\). Then there exists \(z\in X\) such that \(z \in T z\).
Proof
Let \(r_{1}\) be a real number such that \(0 \leq r < r_{1} < 1\). Let \(u_{1} \in X\) and \(u_{2} \in T u_{1}\) be arbitrary. Since \(u_{2} \in T u_{1}\), we have \(d(u_{2},T u_{2}) \leq H(T u_{1},T u_{2})\) and
Thus, from the assumption (2.24),
where \(S(u_{1},u_{2})=\alpha d(u_{1},u_{2})+\alpha d(u_{1},Tu_{1})+\alpha d(u_{2},Tu_{2})+\alpha d(u_{1},Tu_{2})+\alpha d(u_{2},Tu_{1})\). Consider
Then
where \(r=\frac{3\alpha}{1-2\alpha}\).
So there exists \(u_{3}\in Tu_{2}\) such that \(d(u_{2},u_{3})\leq r_{1}d(u_{1},u_{2})\). Thus, we can construct a sequence \(\{x_{n}\}\) in X such that \(u_{n+1} \in Tu_{n}\) and
Hence, by induction
Then by the triangle inequality, we have
Hence we conclude that \(\{u_{n}\}\) is a Cauchy sequence. Since X is complete, there is some point \(z \in X\) such that
Now, we will show that \(d(z,Tx)\leq rd(x,Tx)\) for all \(x\in X\setminus \{z\}\).
Let \(x\in X\setminus\{z\}\). Since \(u_{n}\to z\), there exists \(n_{0} \in N\) such that \(d(z, u_{n})\leq (\frac{1}{3})d(z, x)\) for all \(n\geq n_{0}\). By using (2.2), we get
Then from (2.1), we have
Since \(u_{n+1} \in Tu_{n}\), \(d(u_{n+1}, T_{x}) \leq H(Tu_{n}, T_{x})\), so that
for all \(n\geq n_{0}\). Letting \(n \to\infty\), we obtain
It follows that
Next, we show that \(z \in Tz\). Suppose that z is not element in Tz.
Case (i): \(0\leq r<\frac{1}{2}\). Let \(a \in Tz\). Then \(a\neq z\) and so by (2.25), we have
On the other hand, since \(\varphi(r) d(z, Tz) = d(z, Tz)\leq d(z, a)\), from (2.24) we have
So
Since \(d(z,a)\leq d(z, Tz)+d( Tz,a)=d(z, Tz)\), we have
Using (2.24), (2.26), and (2.27), we have
Since \(0\leq r<\frac{1}{2}\), we have \(0\leq2r^{2}+r<1\) and so, \(d(z, Tz)< d(z, Tz)\), a contradiction. Thus \(z\in Tz\).
Case (ii): \(\frac{1}{2}\leq r <1\). Let \(x\in X\). If \(x = z\), then \(H(T x, T z) \leq\alpha [d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]\) holds. If \(x \neq z\), then for all \(n\in\mathbb{N}\), there exists \(y_{n} \in Tx\) such that
We consider
Thus, \((1-r)d(x, Tx)\leq(1+\frac{1}{n})d(x,z)\). Take \(n\to\infty\), we obtain
by using (2.24), this implies \(H(T x, T z) \leq S(x,z)\), where \(S(x,z)=\alpha[d(x,z)+d(x,Tx)+d(z,Tz)+d(x,Tz)+d(z,Tx)]\). Hence, as \(u_{n+1}\in Tu_{n}\), it follows that with \(x=u_{n}\)
Therefore, \((1-2\alpha)d(z, Tz)\leq0\), which implies \(d(z, Tz)=0\). Since Tz is closed, we have \(z\in Tz\). This completes the proof. □
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Acknowledgements
The authors would like to thank the Thailand Research Fund under the project RTA5780007 and Science Achievement Scholarship of Thailand, which provides funding for our research.
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Klin-eam, C., Suanoom, C. & Suantai, S. Generalized multi-valued mappings satisfy some inequalities conditions on metric spaces. J Inequal Appl 2015, 343 (2015). https://doi.org/10.1186/s13660-015-0864-4
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DOI: https://doi.org/10.1186/s13660-015-0864-4