1 Introduction

In this paper we investigate the fixed point property of a new class of multivalued mappings defined by putting together the ideas of contraction satisfying rational types inequalities and Pata type inequalities, respectively. The former class was first introduced in the work of Dass et al. [13] in which the inequality to be satisfied by the contraction contains rational terms. These contractions are known as rational contractions. Their considerations occupy a large area of fixed point theory. Some of the fixed point results of rational type contractions are in [1, 4, 6,7,8, 10, 11, 14, 15, 21, 22]. Pata type inequalities are recent introductions in fixed point theory. It was initiated in a paper of Pata [24] in which a generalization of the Banach contraction mapping principle was established. The inequality in this case is not a single one, instead it is a group of inequalities which is obtained by varying a parameter over a certain range. A new methodology was introduced in [24] to ensure the existence of the fixed points of such mappings. Several works have appeared by following the idea of Pata-type contraction. Some works along this line of research are noted in [5, 12, 16, 17]. The idea was extended to the domain of setvalued analysis through the work of Kolagar et al. [20] where a multivalued version of the result due to Pata was established in metric spaces.

In this paper, by combining a rational term containing Kannan type expressions with Pata type terms, we define a hybrid multivalued contraction and establish that such mappings have fixed points. It may be mentioned that the class of Kannan type mappings [18, 19, 25, 26] is considered to be an important class of contractive mappings with discontinuities and has been considered quite extensively in fixed point theory. Our function satisfies two types of inequalities of the above type with Hausdorff and \(\delta\)-distances, respectively, in two different situations. The main result has a corollary and is illustrated with examples. One of which shows that the corollary is properly contained in main theorem. We establish our theorem without assuming the continuity of the function.

2 Mathematical preliminaries

The following are the concepts from setvalued analysis which we use in this paper. Let \((X,\ d)\) be a metric space. Then

\(N(X) = \{ A : A\ \text {is a non-empty subset of}\ X\}\),

\(CB(X) = \{A : A\ \text {is a non-empty closed and bounded subset of}\ X\}\) and

\(C(X) = \{ A : A\ \text {is a non-empty compact subset of}\ X\}\).

For \(x\in X\) and \(A,\ B\in CB(X)\), the functions \(D(x,\ B)\), \(\delta (A,\ B)\) and \(H(A,\ B)\) are defined as follows:

$$\begin{aligned} D(x,\ B) = \text {inf}\ \{d(x,\ y) : y \in B\}, \ \delta (A,\ B)= \text {sup}\ \{d(x,\ y) : x \in A,\ y \in B\} \end{aligned}$$

and

$$\begin{aligned} H(A,B) = \text {max}\ \{\underset{x\in A }{ \text {sup} }\ D(x,\ B),\underset{y \in B }{\text {sup} }\ D(y,\ A)\}. \end{aligned}$$

The \(\delta\)-distance has all the properties of a metric except one. It has been used in works like [2, 3, 9]. H is known as the Hausdorff metric induced by the metric d on CB(X) [23]. Further, if \((X,\ d)\) is complete then \((CB(X),\ H)\) is also complete.

We use the following result in our subsequent discussion.

Lemma 2.1

Let\((X,\ d)\)be a metric space and\(B \in C(X)\). Then for every\(x\in X\)there exists\(y\in B\)such that\(d(x,\ y)= D(x,\ B)\).

Definition 2.1

Let \(T: X \longrightarrow CB(Y)\) be a multivalued mapping, where \((X,\ \rho )\), \((Y,\ d)\) are two metric spaces and H is the Hausdorff metric on CB(Y). The mapping T is said to be continuous at \(x\in X\) if for any sequence \(\{x_{n}\}\) in X, \(H(Tx_n,\ Tx)\longrightarrow 0\) whenever \(\rho (x_n,\ x)\longrightarrow 0\) as \(n \longrightarrow \infty\).

Definition 2.2

Let X be a non-empty set, \(f : X\longrightarrow X\) be a single valued mapping and \(T : X\longrightarrow N(X)\) be a multivalued mapping. A point \(x\in X\) is a fixed point of f (resp. T ) if and only if \(x = fx\) (resp. \(x \in Tx\)).

We will require the following class of functions in our results. Let \(\Phi\) denote the family of all functions \(\varphi : [0,1] \longrightarrow [0, \infty )\) such that \(\varphi\) is continuous at zero with \(\varphi (0)= 0.\)

3 Main results

Theorem 3.1

Let\((X,\ d)\)be a complete metric space and\(T: X \longrightarrow C(X)\)be a multivalued mapping. Suppose there exist\(\Lambda \ge 0\), \(L \ge 0\), \(\eta > 0\), \(\alpha \ge 1\), \(\beta \in [0,\alpha ]\)and\(\varphi \in \Phi\)such that for every\(\varepsilon \in [0,1]\)and for all\(x,\ y \in X\),

$$\begin{aligned} H(Tx,\ Ty)\le (1-\varepsilon ) \Big [M(x,\ y)+ LN(x,\ y)\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x||+||y||+||Tx||+||Ty||\Big ]^\beta \end{aligned}$$
(1)

whenever\(d(x,\ y)\ge \eta\)and

$$\begin{aligned} \delta (Tx,\ Ty)\le (1-\varepsilon ) \Big [M(x,\ y)+ LN(x,\ y)\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x||+||y||+||Tx||+||Ty||\Big ]^\beta \end{aligned}$$
(2)

whenever\(d(x,\ y)< \eta\),

where

\(M(x,y)= \max \ \Big \{d(x,\ y), \dfrac{D(x,\ Tx)D(y,\ Ty)}{1+d(x,\ y)}\Big \}\),

\(N(x,y)=\min \ \{D(x,\ Tx),\ D(y,\ Ty),\ D(x,\ Ty),\ D(y,\ Tx)\}\) and

\(||x|| = d(x,z)\)and\(||Tx|| = D(z, Tx)\)for an arbitrary but fixed\(z \in X\). Also, suppose that there exists\(x_0\in X\)such that\(Tx_{0}\)is singleton. ThenThas a fixed point.

Proof

Suppose that \(x_{0}\in X\) be such that \(Tx_{0}\) is singleton. Let \(Tx_{0} = \{x_{1}\}\). Then clearly \(d(x_{0},\ x_{1})= D(x_{0},\ Tx_{0})\). By Lemma 2.1, there exists \(x_{2}\in Tx_{1}\) such that \(d(x_{1},\ x_{2})= D(x_{1},\ Tx_{1})\). Again, by Lemma 2.1, we can find \(x_{3}\in Tx_{2}\) such that \(d(x_{2},\ x_{3})= D(x_{2},\ Tx_{2})\). Continuing this process we construct a sequence \(\{x_{n}\}\) such that

$$\begin{aligned} x_{n+1}\in Tx_{n}\ \ \text {and}\ \ d(x_{n},\ x_{n+1})= D(x_{n},\ Tx_{n})\ \ \text {for all}\ \ n\ge 0. \end{aligned}$$
(3)

Let

$$\begin{aligned} c_{n}=||x_{n}|| = d(x_{n},\ x_{0}),\ \ \ \text {for all}\ \ \ n\ge 0. \end{aligned}$$
(4)

Since \(Tx_{0}\) is singleton and \(Tx_{0} = \{x_{1}\}\), we have

$$\begin{aligned} d(x_{n+1},\ x_{1})=D(x_{n+1},\ Tx_{0}) \le H(Tx_{n},\ Tx_{0})=\delta (Tx_{n},\ Tx_{0})\ \text {for all}\ n\ge 0. \end{aligned}$$
(5)

If \(d(x_{n},\ x_{n+1})\ge \eta\), applying the contraction condition (1) for \(0 <\varepsilon \le 1\), we have

$$\begin{aligned} d(x_{n+1},\ x_{n+2})&=D(x_{n+1},\ Tx_{n+1})\le H(Tx_{n},\ Tx_{n+1})\nonumber \\&\le (1-\varepsilon ) \Big [M(x_{n},\ x_{n+1})+ L N(x_{n},\ x_{n+1})\Big ] \nonumber \\&\ \ + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+||x_{n+1}||+||Tx_{n}|| +||Tx_{n+1}||\Big ]^\beta \nonumber \\&\le (1-\varepsilon ) \Big [M(x_{n},\ x_{n+1})+ LN(x_{n},\ x_{n+1})\Big ]\nonumber \\&\ \ + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+||x_{n+1}||+||x_{n+1}|| +||x_{n+2}||\Big ]^\beta . \end{aligned}$$
(6)

If \(d(x_{n},\ x_{n+1})< \eta\), applying the contraction condition (2) for \(0<\varepsilon \le 1\), we have

$$\begin{aligned} d(x_{n+1},\ x_{n+2})&=D(x_{n+1},\ Tx_{n+1})\le \delta (Tx_{n},\ Tx_{n+1})\nonumber \\&\le (1-\varepsilon ) \Big [M(x_{n},\ x_{n+1})+ LN(x_{n},\ x_{n+1})\Big ]\nonumber \\&\ \ + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+||x_{n+1}||+||Tx_{n}|| +||Tx_{n+1}||\Big ]^\beta \nonumber \\&\le (1-\varepsilon ) \Big [M(x_{n},\ x_{n+1})+ LN(x_{n},\ x_{n+1})\Big ]\nonumber \\&\ \ + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+||x_{n+1}||+||x_{n+1}|| +||x_{n+2}||\Big ]^\beta . \end{aligned}$$
(7)

Combining (6) and (7), we have

$$\begin{aligned} d(x_{n+1},\ x_{n+2})&\le (1-\varepsilon ) \Big [M(x_{n},\ x_{n+1})+ LN(x_{n},\ x_{n+1})\Big ]\nonumber \\&\ \ + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+2||x_{n+1}|| +||x_{n+2}||\Big ]^\beta . \end{aligned}$$
(8)

Now,

$$\begin{aligned} M(x_{n},\ x_{n+1})= & {} \max \Big \{d(x_{n},\ x_{n+1}),\dfrac{D(x_{n},\ T x_{n})\ D(x_{n+1},\ T x_{n+1})}{1+d(x_{n},\ x_{n+1})}\Big \}\nonumber \\\le & {} \max \Big \{d(x_{n},\ x_{n+1}),\ \dfrac{d(x_{n},\ x_{n+1}) d(x_{n+1},\ x_{n+2})}{1+d(x_{n},\ x_{n+1})}\Big \}\nonumber \\\le & {} \max \{d(x_{n},\ x_{n+1}),\ d(x_{n+1},\ x_{n+2})\} \end{aligned}$$
(9)

and

$$\begin{aligned} N(x_{n},\ x_{n+1})& = \min \{D(x_{n},\ Tx_{n}),\ D(x_{n+1},\ Tx_{n+1}),\ D(x_{n},\ T x_{n+1}),\ D(x_{n+1},\ T x_{n})\}\nonumber \\& \le \min \{d(x_{n},\ x_{n+1}),\ d(x_{n+1},\ x_{n+2}),\ d(x_{n},\ x_{n+2}),\ d(x_{n+1},\ x_{n+1})\}\nonumber \\& = 0. \end{aligned}$$
(10)

Suppose that \(d(x_{n+1},\ x_{n+2}) > d(x_{n},\ x_{n+1})\). Then \(d(x_{n+1},\ x_{n+2}) > 0\). From (8)–(10), we have

$$\begin{aligned} d(x_{n+1},\ x_{n+2})\le (1-\varepsilon ) d(x_{n+1},\ x_{n+2})\ \ + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+2||x_{n+1}|| +||x_{n+2}||\Big ]^\beta , \end{aligned}$$

which implies that

$$\begin{aligned} \varepsilon \ d(x_{n+1},\ x_{n+2})\le \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+2||x_{n+1}|| +||x_{n+2}||\Big ]^\beta , \end{aligned}$$

that is,

$$\begin{aligned} d(x_{n+1},\ x_{n+2}) \le \Lambda \ \varepsilon ^{\alpha -1} \varphi (\varepsilon )\Big [1+||x_{n}||+2||x_{n+1}||+||x_{n+2}||\Big ]^\beta , \end{aligned}$$

Since \(\alpha \ge 1\), taking \(\varepsilon \longrightarrow 0\) in the above inequality and using the property of \(\varphi\), we have \(d(x_{n+1},\ x_{n+2})\le 0\), which contradicts the assumption that \(d(x_{n+1},\ x_{n+2}) > 0\). So we have

$$\begin{aligned} d(x_{n+1},\ x_{n+2})\le d(x_{n},\ x_{n+1})\ \ \text {for all}\ n\ge 0, \end{aligned}$$
(11)

that is, \(\{d(x_{n},\ x_{n+1})\}\) is a decreasing sequence of nonnegative real numbers. So

$$\begin{aligned} d(x_{n}, \ x_{n+1}) \le d(x_{0},\ x_{1}) = c_{1} = ||x_{1}||,\ \ \text {for all}\ n\ge 0, \end{aligned}$$
(12)

and also there exists a nonnegative real number l such that

$$\begin{aligned} d(x_{n},\ x_{n+1}) \longrightarrow l\ \ \text {as}\ \ n \longrightarrow \infty . \end{aligned}$$
(13)

Also, from (8)–(11), we have

$$\begin{aligned} d(x_{n+1},\ x_{n+2})\le (1-\varepsilon ) d(x_{n},\ x_{n+1}) + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+2||x_{n+1}|| +||x_{n+2}||\Big ]^\beta . \end{aligned}$$
(14)

Now, \(c_{n}=d(x_{n},\ x_{0})\le d(x_{n},\ x_{n+1})+d(x_{n+1},\ x_{0})\le d(x_{n},\ x_{n+1})+d(x_{n+1},\ x_{1})\,+\,d(x_{1},\ x_{0})\).

By (4), (5) and (12), we have

$$\begin{aligned} c_{n}&\le H(Tx_{n},\ Tx_{0})+c_{1}+c_{1} = \delta (Tx_{n},\ Tx_{0})+c_{1}+c_{1}\\&= H(Tx_{n},\ Tx_{0})+2c_{1} = \delta (Tx_{n},\ Tx_{0})+2c_{1}. \end{aligned}$$

For both the cases where \(d(x_{n},\ x_{0})\ge \eta\) and \(d(x_{n},\ x_{0})< \eta\), we have from (1), (2), (12) and the above inequality that

$$\begin{aligned} c_{n}&\le (1-\varepsilon ) \Big [M(x_{n}, x_{0})+ LN(x_{n}, x_{0})\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x_{n}||+||x_{0}||+||Tx_{n}||+||Tx_{0}||\Big ]^\beta +2c_{1}\\&\le (1-\varepsilon ) \Big [M(x_{n}, x_{0})+ LN(x_{n}, x_{0})\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x_{n}||+||x_{n+1}||+||x_{1}||\Big ]^\beta +2c_{1}\\&= (1-\varepsilon ) \Big [M(x_{n}, x_{0})+ LN(x_{n}, x_{0})\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x_{n}||+d(x_{n+1},\ x_{0})+||x_{1}||\Big ]^\beta +2c_{1}\\&\le (1-\varepsilon ) \Big [M(x_{n}, x_{0})+ LN(x_{n}, x_{0})\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+c_{n}+d(x_{n+1}, x_{n})+ d(x_{n}, x_{0})+c_{1}\Big ]^\beta +2c_{1}\\&\le (1-\varepsilon ) \Big [M(x_{n}, x_{0})+ LN(x_{n}, x_{0})\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+c_{n}+c_{1}+ c_{n}+c_{1}\Big ]^\beta +2c_{1}\\&\le (1-\varepsilon ) \Big [M(x_{n}, x_{0})+ LN(x_{n}, x_{0})\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+2c_{n}+2c_{1}\Big ]^\alpha +2c_{1},\ (\text {since}\ \beta \le \alpha ) \end{aligned}$$

where

$$\begin{aligned} M(x_{n}, x_{0})&= \max \Big \{d(x_{n}, x_{0}),\ \dfrac{D(x_{n}, Tx_{n})D(x_{0}, Tx_{0})}{1+d(x_{n}, x_{0})}\Big \} \le \max \Big \{d(x_{n}, x_{0}),\ \dfrac{d(x_{n}, x_{n+1})d(x_{0}, x_{1})}{1+d(x_{n}, x_{0})}\Big \}\\&\le \max \Big \{c_{n},\ \dfrac{c_{1}^2}{1+c_{n}}\Big \} \le \max \{c_{n},\ c_{1}^2\} \end{aligned}$$

and

$$\begin{aligned} N(x_{n},\ x_{0})&=\min \{D(x_{n},\ T x_{n}),\ D(x_{0},\ T x_{0}),\ D(x_{n},\ T x_{0}),\ D(x_{0},\ T x_{n})\}\\&\le \min \{d(x_{n},\ x_{n+1}),\ d(x_{0},\ x_{1}),\ d(x_{n},\ x_{1}),\ d(x_{0},\ x_{n+1})\}\\&\le \min \{d(x_{n},\ x_{n+1}),\ d(x_{0},\ x_{1}),\ d(x_{n},\ x_{0})+d(x_{1},\ x_{0}),\ d(x_{n},\ x_{0})+d(x_{n},\ x_{n+1})\}\\&\le \min \{c_{1},\ c_{1},\ c_{n}+c_{1},\ c_{n}+c_{1}\}= c_{1}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} c_{n}\le (1-\varepsilon )\Big [\max \{c_{n},\ c_{1}^2\}+L\ c_{1}\Big ] +\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+2c_{n}+2c_{1}\Big ]^\alpha +2c_{1}. \end{aligned}$$
(15)

If possible, suppose that the sequence \(\{c_{n}\}\) is unbounded. Then there exists a subsequence \(\{c_{n_{k}}\}\) with \(c_{n_{k}}\longrightarrow \infty\) as \(k \longrightarrow \infty\). So there exists a natural number N such that

$$\begin{aligned} c_{n_{k}}\ge \max \{c_{1}^2,\ 1+(L+2)c_{1}\}\ \text {for all}\ k \ge N. \end{aligned}$$
(16)

So, for all \(k \ge N\), we have

$$\begin{aligned} c_{n_{k}}\le (1-\varepsilon )(c_{n_{k}}+Lc_{1}) + \Lambda \ \varepsilon ^\alpha \ \varphi (\varepsilon )\ \Big [1+2c_{n_{k}}+2c_{1}\Big ]^\alpha + 2c_{1}. \end{aligned}$$
(17)

Now

$$\begin{aligned} \Big (1+2c_{n_{k}}+2c_{1}\Big )^\alpha&= (1+2c_{n_{k}})^\alpha \Big ( 1 + \dfrac{2c_{1}}{1+2c_{n_{k}}}\Big )^\alpha \le (1+2c_{n_{k}})^\alpha (1+2c_{1})^\alpha \nonumber \\&=2^\alpha \ c_{n_{k}}^\alpha \ \Big (1 + \dfrac{1}{2c_{n_{k}}}\Big )^\alpha \ (1+2c_{1})^\alpha \nonumber \\&\le 2^\alpha c_{n_{k}}^\alpha ( 1+ 1)^\alpha (1+2c_{1})^\alpha = 2^{2\alpha }\ c_{n_{k}}^\alpha \ (1+2c_{1})^\alpha . \end{aligned}$$
(18)

Then, for all \(k \ge N\), we have from (17) and (18) that

$$\begin{aligned} c_{n_{k}}\le (1-\varepsilon )(c_{n_{k}} + Lc_{1}) + \Lambda \ \varepsilon ^\alpha \ \varphi (\varepsilon )\ 2^{2\alpha }\ c_{n_{k}}^\alpha \ (1+2c_{1})^\alpha + 2c_{1}, \end{aligned}$$

that is,

$$\begin{aligned} \varepsilon \ c_{n_{k}}&\le (1-\varepsilon )L\ c_{1} + \Lambda \ \varepsilon ^\alpha \ \varphi (\varepsilon )\ 2^{2\alpha }\ c_{n_{k}}^\alpha \ (1+2c_{1})^\alpha + 2c_{1}\\&\le L\ c_{1} + \Lambda \ \varepsilon ^\alpha \ \varphi (\varepsilon )\ 2^{2\alpha }\ c_{n_{k}}^\alpha \ (1+2c_{1})^\alpha + 2c_{1},\ \ \Big ( \text {since}\ 1-\varepsilon \le 1 \Big )\\&= \Big [\Lambda \ 2^{2\alpha }\ (1+2c_{1})^\alpha \Big ]\ \varepsilon ^\alpha \ \varphi (\varepsilon )\ c_{n_{k}}^\alpha + (L+2)c_{1}. \end{aligned}$$

Let \(a = \Lambda \ 2^{2\alpha }\ (1+2c_{1})^\alpha\) and \(b = (L+2)c_{1}\). So, we have

$$\begin{aligned} \varepsilon \ c_{n_{k}}\le a\ \varepsilon ^\alpha \ \varphi (\varepsilon )\ c_{n_{k}}^\alpha + b,\ \text {where}\ a\ \text {and}\ b\ \text {are constants}. \end{aligned}$$

Choose \(\varepsilon = \varepsilon _{k} = \dfrac{1+b}{c_{n_{k}}}=\dfrac{1+(L+2)c_{1}}{c_{n_{k}}}\), where \(k \ge N\). Then by (16), we have \(0 < \varepsilon \le 1\). Now we have

$$\begin{aligned} 1\le a\ (1+b)^\alpha \ \varphi (\varepsilon _{k}) \longrightarrow 0\ \ \text {as}\ \ k\longrightarrow \infty , \end{aligned}$$

which is a contradiction. Hence \(\{c_{n}\}\) is bounded. Therefore, there exists a real number \(P>0\) such that

$$\begin{aligned} c_{n}=||x_{n}||\le P\ \ \text {for all}\ \ n\ge 0. \end{aligned}$$
(19)

Now from (14) and (19), we get

$$\begin{aligned} d(x_{n+1},\ x_{n+2})\le (1-\varepsilon )d(x_{n},\ x_{n+1}) +\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )[1+4P]^\beta . \end{aligned}$$

Taking limit \(n \longrightarrow \infty\) in the above inequality, using (13), we get

$$\begin{aligned} l\le (1-\varepsilon )l +\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )[1+4P]^\beta , \end{aligned}$$

that is,

$$\begin{aligned} \varepsilon \ l\le \Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )[1+4P]^\beta , \end{aligned}$$

that is,

$$\begin{aligned} l\le \Lambda \varepsilon ^{\alpha -1}\ \varphi (\varepsilon )[1+4P]^\beta . \end{aligned}$$

Since \(\alpha \ge 1\), taking \(\varepsilon \longrightarrow 0\) in the above inequality and using the property of \(\varphi\), we have \(l \le 0\), which is a contradiction unless \(l = 0\). So, we have

$$\begin{aligned} \displaystyle \lim _{n \longrightarrow \infty }\,d(x_{n},\ x_{n+1})= 0. \end{aligned}$$
(20)

Now we prove that \(\{x_{n}\}\) is a Cauchy sequence. Suppose \(\{x_{n}\}\) is not a Cauchy sequence. Then there exists a \(\xi\) satisfying \(0<\xi <\eta\) for which we can find two sequences of positive integers \(\{m(k)\}\) and \(\{n(k)\}\) such that for all positive integers k, \(n(k)> m(k) > k\) and \(d(x_{m(k)},\ x_{n(k)})\ge \xi\). Assuming that n(k) is the smallest such positive integer, we get

$$\begin{aligned} d(x_{m(k)},\ x_{n(k)})\ge \xi \ \ \text {and}\ \ d(x_{m(k)},\ x_{n(k)-1})< \xi . \end{aligned}$$

Now,

$$\begin{aligned} \xi \le d(x_{m(k)},\ x_{n(k)})&\le d(x_{m(k)},\ x_{n(k)-1})+ d(x_{n(k)-1},\ x_{n(k)})\\&< \xi + d(x_{n(k)-1},\ x_{n(k)}). \end{aligned}$$

Taking the limit as \(k\longrightarrow \infty\) in the above inequality and using (20), we have

$$\begin{aligned} \displaystyle \lim _{k \longrightarrow \infty }\,d(x_{m(k)},\ x_{n(k)})= \xi . \end{aligned}$$
(21)

Again,

$$\begin{aligned} d(x_{m(k)},\ x_{n(k)})\le d(x_{m(k)},\ x_{m(k)+1}) + d(x_{m(k)+1},\ x_{n(k)+1})+ d(x_{n(k)+1},\ x_{n(k)}) \end{aligned}$$

and

$$\begin{aligned} d(x_{m(k)+1},\ x_{n(k)+1})\le d(x_{m(k)+1},\ x_{m(k)})+ d(x_{m(k)},\ x_{n(k)}) + d(x_{n(k)},\ x_{n(k)+1}). \end{aligned}$$

Taking the limit as \(k\longrightarrow \infty\) in the above inequalities, using (20) and (21), we have

$$\begin{aligned} \displaystyle \lim _{k\longrightarrow \infty }\,d(x_{m(k)+1},\ x_{n(k)+1})= \xi . \end{aligned}$$
(22)

Again,

$$\begin{aligned} d(x_{m(k)},\ x_{n(k)})\le d(x_{m(k)},\ x_{m(k)+1})+ d(x_{m(k)+1},\ x_{n(k)}) \end{aligned}$$

and

$$\begin{aligned} d(x_{m(k)+1},\ x_{n(k)})\le d(x_{m(k)+1},\ x_{m(k)})+d(x_{m(k)},\ x_{n(k)}). \end{aligned}$$

Taking the limit as \(k\longrightarrow \infty\) in the above inequalities, using (20) and (21), we have

$$\begin{aligned} \displaystyle \lim _{k\longrightarrow \infty }\,d(x_{m(k)+1},\ x_{n(k)})= \xi . \end{aligned}$$
(23)

Similarly, we have

$$\begin{aligned} \displaystyle \lim _{k\longrightarrow \infty }\,d(x_{m(k)},\ x_{n(k)+1})= \xi . \end{aligned}$$
(24)

Since \(\xi < \eta\), (21) implies that there exists a positive integer \(k_0\) such that \(d(x_{m(k)},\ x_{n(k)})<\eta\) for all \(k>k_0\). When \(k>k_0\), applying (2) for \(0 < \varepsilon \le 1\), we have

$$\begin{aligned} d(x_{m(k)+1},\ x_{n(k)+1}) & \le \delta (Tx_{m(k)},\ Tx_{n(k)})\nonumber \\ & \le (1-\varepsilon )\Big [M(x_{m(k)},\ x_{n(k)})+ LN(x_{m(k)},\ x_{n(k)})\Big ]\nonumber \\& + \Lambda \ \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x_{m(k)}||+||x_{n(k)}||+||Tx_{m(k)}|| +||Tx_{n(k)}||\Big ]^\beta \nonumber \\ & \le (1-\varepsilon )\Big [M(x_{m(k)},\ x_{ n(k)})+ LN(x_{m(k)},\ x_{n(k)})\Big ]\nonumber \\& +\Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{m(k)}||+||x_{n(k)}||+||x_{m(k)+1}|| +||x_{n(k)+1}||\Big ]^\beta \nonumber \\ & \le (1-\varepsilon )\Big [M(x_{m(k)},\ x_{n (k)})+ LN(x_{m(k)},\ x_{n(k)})\Big ] \nonumber \\& +\Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+4P\Big ]^\beta ,\ (\text {by}\ (19)). \end{aligned}$$
(25)

Now,

$$\begin{aligned} M(x_{m(k)},\ x_{n(k)})&=\max \Big \{d(x_{m(k)},\ x_{n(k)}),\ \dfrac{D(x_{m(k)},\ Tx_{m(k)})D(x_{n(k)},\ Tx_{n(k)})}{1+d(x_{m(k)},\ x_{n(k)})}\Big \}\\&\le \max \Big \{d(x_{m(k)},\ x_{n(k)}),\ \dfrac{d(x_{m(k)},x_{m(k)+1})\ d(x_{n(k)},\ x_{n(k)+1})}{1+d(x_{m(k)},\ x_{n(k)})}\Big \}. \end{aligned}$$

So, we have

$$\begin{aligned} d(x_{m(k)},\ x_{n(k)}) \le M(x_{m(k)},\ x_{n(k)}) \le \max \Big \{d(x_{m(k)},\ x_{n(k)}), \dfrac{d(x_{m(k)},x_{m(k)+1})\ d(x_{n(k)}, x_{n(k)+1})}{1+d(x_{m(k)},\ x_{n(k)})}\Big \}. \end{aligned}$$

Taking limit \(k\longrightarrow \infty\) in the above inequality, using (20) and (21), we have

$$\begin{aligned} \lim _{k\longrightarrow \infty }M(x_{m(k)},\ x_{n(k)})=\xi . \end{aligned}$$
(26)
$$\begin{aligned} N(x_{m(k)},\ x_{n(k)})& =\min \Big \{D(x_{m(k)},\ T x_{m(k)}), D(x_{n(k)},\ T x_{n(k)}),\\&\quad D(x_{m(k)},\ Tx_{n(k)}),\ D(x_{n(k)},\ T x_{m(k)})\Big \}\\&\le \min \Big \{d(x_{m(k)},\ x_{m(k)+1}),\ d(x_{n(k)},\ x_{n(k)+1}),\\& \quad d(x_{m(k)},\ x_{n(k)+1}),\ d(x_{n(k)},\ x_{m(k)+1})\Big \}. \end{aligned}$$

Taking limit \(k\longrightarrow \infty\) in the above inequality, using (20), (23) and (24), we have

$$\begin{aligned} \lim _{k\longrightarrow \infty }N(x_{m(k)},\ x_{n(k)})=0. \end{aligned}$$
(27)

Taking limit as \(k\longrightarrow \infty\) in (25), using (22), (26) and (27), we have

$$\begin{aligned} \xi \le (1-\varepsilon )\xi + \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+4P\Big ]^\beta , \end{aligned}$$

that is,

$$\begin{aligned} \varepsilon \ \xi \le \Lambda \ \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+4P\Big ]^\beta , \end{aligned}$$

that is,

$$\begin{aligned} \xi \le \Lambda \ \varepsilon ^{\alpha -1} \varphi (\varepsilon )\Big [1+4P\Big ]^\beta . \end{aligned}$$

Since \(\alpha \ge 1\), taking limit as \(\varepsilon \longrightarrow 0\) in the above inequality and using the property of \(\varphi\), we have \(\xi \le 0\), which is a contradiction. Therefore, \(\{x_{n}\}\) is a Cauchy sequence in X.

As X is a complete, there exists \(z \in X\) such that

$$\begin{aligned} x_{n} \longrightarrow z\ \ \ \text {as}\ \ n \longrightarrow \infty . \end{aligned}$$
(28)

As \(\eta > 0\), there exists a positive integer \(n_{0}\) such that \(d(x_{n},\ z) < \eta\) for all \(n\ge n_{0}\). When \(n\ge n_{0}\), applying (2) for \(0 < \varepsilon \le 1\), we have

$$\begin{aligned} D(x_{n+1},\ Tz) & \le \delta (Tx_{n},\ Tz)\nonumber \\ & \le (1-\varepsilon ) \Big [M(x_{n}, z)+ LN(x_{n}, z)\Big ]+ \Lambda \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+||z||+||Tx_{n}||+||Tz||\Big ]^\beta \nonumber \\ & \le (1-\varepsilon ) \Big [M(x_{n}, z)+ LN(x_{n}, z)\Big ]+\Lambda \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x_{n}||+||z||+||x_{n+1}||+||Tz||\Big ]^\beta \nonumber \\ & \le (1-\varepsilon ) \Big [M(x_{n}, z)+ LN(x_{n}, z)\Big ]+\Lambda \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+2P+||z||+||Tz||\Big ]^\beta ,\ (\text {by}\ ({19})). \end{aligned}$$
(29)

Now

$$\begin{aligned} M(x_{n},\ z)&=\max \ \Big \{d(x_{n},\ z), \dfrac{D(x_{n},\ Tx_{n})\ D(z,\ Tz)}{1+d(x_{n},\ z)}\Big \}\\&\le \max \ \Big \{d(x_{n},\ z), \dfrac{d(x_{n},\ x_{n+1})\ D(z,\ Tz)}{1+d(x_{n},\ z)}\Big \}. \end{aligned}$$

Taking limit as \(n\longrightarrow \infty\) in the above inequality and using (28), we get

$$\begin{aligned} \lim _{n\longrightarrow \infty }M(x_{n},\ z)=0. \end{aligned}$$
$$\begin{aligned} N(x_{n},\ z)&=\min \Big \{D(x_{n},\ T x_{n}),\ D(z,\ Tz),\ D(x_{n},\ Tz),\ D(z,\ T x_{n})\Big \}\\&\le \min \Big \{d(x_{n},\ x_{n+1}),\ D(z,\ Tz),\ D(x_{n},\ Tz),\ d(z,\ x_{n+1})\Big \}. \end{aligned}$$

Taking limit as \(n\longrightarrow \infty\) in the above inequality and using (28), we get

$$\begin{aligned} \lim _{n\longrightarrow \infty }N(x_{n},\ z)=0. \end{aligned}$$

Letting \(n\longrightarrow \infty\) in (29), we obtain

$$\begin{aligned} D(z,\ Tz)\le \Lambda \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+2P+||z||+||Tz||\Big ]^\beta . \end{aligned}$$

Since \(\alpha \ge 1\), taking limit as \(\varepsilon \longrightarrow 0\) in the above inequality and using the property of \(\varphi\), we have \(D(z,\ Tz)\le 0\), which implies that \(D(z,\ Tz) = 0\). Now \(D(z,\ Tz) = 0\) implies \(z \in \overline{Tz}\), where \(\overline{Tz}\) is the closure of Tz. Since \(Tz \in C(X)\), \(\overline{Tz} = Tz\). So \(z \in Tz\), that is, z is a fixed point of T.

Corollary 3.1

Let\((X,\ d)\)be a complete metric space and\(T: X \longrightarrow C(X)\)be a multivalued mapping. Suppose there exist\(\Lambda \ge 0\), \(L \ge 0\), \(\alpha \ge 1\), \(\beta \in [0,\ \alpha ]\)and\(\varphi \in \Phi\)such that the inequality (2) is satisfied, that is, for every\(\varepsilon \in [0,\ 1]\)and for all\(x,\ y \in X\),

$$\begin{aligned} \delta (Tx,\ Ty)\le (1-\varepsilon ) \Big [M(x,\ y)+ LN(x,\ y)\Big ]+\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x||+||y||+||Tx||+||Ty||\Big ]^\beta , \end{aligned}$$

where\(M(x,\ y)\), \(N(x,\ y)\), ||x|| and ||Tx|| are same as defined in Theorem3.1. Also, suppose that there exists\(x_0\in X\) such that \(Tx_{0}\)is singleton. ThenThas a fixed point.

Proof

Since \(H(Tx,\ Ty)\le \delta (Tx,\ Ty)\) for all \(x,\ y\in X\), for any \(\eta > 0\) the inequality of the corollary can be trivially decomposed into (1) and (2). Then all the conditions of Theorem 3.1 are satisfied and hence by an application of Theorem 3.1 we conclude that T has a fixed point.

Example 3.1

Let \(X= [0,\ 1]\bigcup \{3,\ 6\}\) and \(`` d ''\) be the usual metric on X. Let \(T : X\longrightarrow C(X)\) be defined as follows:

$$Tx = \left\{ {\begin{array}{*{20}l} {\left\{ {x - \frac{{x^{2} }}{2}} \right\}} \hfill & {{\text{if }}\quad 0 \le x \le 1}, \hfill \\ {\{ 0,\;1\} } \hfill & {{\text{if }}\quad x = 3}, \hfill \\ {\{ 1,\;3\} } \hfill & {{\text{if }}\quad x = 6.} \hfill \\ \end{array} } \right.$$

Let \(\varphi :[0,\ 1] \longrightarrow [0,\ \infty )\) be defined as follows:

$$\varphi (t) = \left\{ {\begin{array}{*{20}l} t \hfill & {{\text{if}}\;\;0 \le t \le \frac{1}{{100}}}, \hfill \\ {4t} \hfill & {{\text{otherwise}}}. \hfill \\ \end{array} } \right.$$

Let \(\eta = \frac{1}{1000}\), \(\Lambda = 100\), \(\alpha = 2\), \(\beta = 1\) and \(L\ge 0\) be arbitrary. Then all the conditions of the Theorem 3.1 are satisfied and 0 is a fixed point of T.

Example 3.2

Let \(X= [0,\ \frac{1}{4}]\bigcup \{\frac{1}{2},\ 1\}\bigcup \{n\in \mathbb {N} : n\ge 2\}\) and \(`` d ''\) be the usual metric on X. Let \(T : X\longrightarrow C(X)\) be defined as follows:

$$Tx = \left\{ {\begin{array}{*{20}l} {\left\{ {\frac{1}{{20}}} \right\}} \hfill & {{\text{if}}\;0 \le x < \frac{1}{4}}, \hfill \\ {\left[ {\frac{1}{{20}},\frac{x}{5}} \right]} \hfill & {{\text{if}}\;x = \frac{1}{4},\frac{1}{2},1}, \hfill \\ {\{ 1\} } \hfill & {{\text{if}}\;x = 2,3,4 \ldots } \hfill \\ \end{array} } \right..$$

Let \(\varphi :[0,\ 1] \longrightarrow [0,\ \infty )\) be defined as follows:

$$\begin{aligned} \varphi (t)=\left\{ \begin{array}{ll} t^{2}\ &{}\mathrm{if}\ 0 \le t \le \dfrac{1}{3},\\ 4t &{}\mathrm{otherwise}. \end{array} \right. \end{aligned}$$

Let \(\Lambda = 2000\), \(\alpha = 2\), \(\beta = 1\) and \(L\ge 0\) be arbitrary. Then all the conditions of the Corollary 3.1 are satisfied and \(\frac{1}{20}\) is a fixed point of T.

Remark

In Example 3.1, when \(x=6\), \(y=3\) and \(\varepsilon = \dfrac{1}{10^{6}}\), the inequality of the Corollary 3.1, that is, (2) is not satisfied. So the Example 3.1 is not applicable to Corollary 3.1 and hence Theorem 3.1 properly contains Corollary 3.1. Further, it is observed that for any \(\varepsilon > 0\), the example does not satisfy the inequality \(H(Tx,\ Ty)\le (1-\varepsilon ) \Big [M(x,\ y)+ LN(x,\ y)\Big ]\) or \(\delta (Tx,\ Ty)\le (1-\varepsilon ) \Big [M(x,\ y)+ LN(x,\ y)\Big ]\) for all \(x,\ y \in X\). This indicates the second term \(\Lambda \varepsilon ^\alpha \ \varphi (\varepsilon )\Big [1+||x||+||y||+||Tx||+||Ty||\Big ]^\beta\) is essential for the theorem which is the spirit of Pata type results. The theorem is proved without any continuity assumption on the function.

The following theorem is the special case of Theorem 3.1 when we treat \(T : X \longrightarrow X\) as a multivalued mapping in which case Tx can be treated as a singleton set for every \(x \in X\).

Theorem 3.2

Let\((X,\ d)\)be a complete metric space and\(T: X \longrightarrow X\). Suppose there exist\(\Lambda \ge 0\), \(L \ge 0\), \(\alpha \ge 1\), \(\beta \in [0,\ \alpha ]\)and\(\varphi \in \Phi\)such that for every\(\varepsilon \in [0,\ 1]\)and for all\(x,\ y \in X\),

$$\begin{aligned} d(Tx,\ Ty)\le (1-\varepsilon ) \Big [M^*(x,\ y)+ LN^*(x,\ y)\Big ]+\Lambda \varepsilon ^\alpha \varphi (\varepsilon )\Big [1+||x||+||y||+||Tx||+||Ty||\Big ]^\beta , \end{aligned}$$
(30)

where

\(M^*(x,\ y)= \max \ \Big \{d(x,\ y), \dfrac{d(x,\ Tx)\ d(y,\ Ty)}{1+d(x,\ y)}\Big \}\),

\(N^*(x,\ y)=\min \ \{d(x,\ Tx),\ d(y,\ Ty),\ d(x,\ Ty),\ d(y,\ Tx)\}\) and

\(||x|| = d(x,\ z)\)for an arbitrary but fixed\(z \in X\).

ThenThas a fixed point.

Proof

We define \(S: X \longrightarrow C(X)\) by \(Sx=\{Tx\}\). Then, for all \(x,\ y\in X\), we have

$$\begin{aligned} D(x,\ Sx)=d(x,\ Tx)\ \ H(Sx,\ Sy)=\delta (Sx,\ Sy)=d(Tx,\ Ty)\ \ \text {and}\ \ ||Sx||=||Tx||. \end{aligned}$$

Then for any \(\eta > 0\), (30) can be expressed in the form of (1) and (2). Then all the conditions of Theorem 3.1 are satisfied and hence by an application of Theorem 3.1, there exists \(z \in X\) such that \(z \in Sz = \{Tz\}\), which implies that \(z= Tz\), that is, z is a fixed point of T.