Fixed Point Theorems for Generalized Set Valued F-contractions in Metric Spaces

1. Introduction

Wardowski [21] introduced the notion of F-contraction mappings and generalized Banach contraction principle by proving that every F-contractions on complete metric spaces have only one fixed point, where $F:(0,\infty )\to (-\infty ,\infty )$ is a function such that

(F1)

F is stlictly increasing;

(F2)

$\forall \left\{s_n\right\}\subset (0,\infty )$,

$$\underset{n\to \infty }{lim}{s}_n=0⟺\underset{n\to \infty }{lim}F\left(s_n\right)=-\infty ;$$

(F3)

$\exists q\in (0,1):{lim}_{t\to 0^{+}}{t}^{q}F\left(t\right)=0$.

Among several results ([1,2,3,4,7,8,10,11,12,13,15,16,18,19,20,22,23]) generalizing Wardowski’s result, Piri and Kumam [17] introduced the concept of Suzuki type F-contractions and obtained related fixed point results in complete metric spaces, where $F:(0,\infty )\to (-\infty ,\infty )$ is a stlictly increasing function such that

(F4)

$infF=-\infty$;

(F5)

F is continuous on $(0,\infty )$.

Note that for a function $F:(0,\infty )\to (-\infty ,\infty )$, the following are equivalent.

(1)

(F2) is satisfied;

(2)

(F4) is satisfied;

(3)

${lim}_{t\to 0^{+}}F\left(t\right)=-\infty$.

Hence, we have that

$$\underset{n\to \infty }{lim}{s}_n=0\Rightarrow \underset{n\to \infty }{lim}F\left(s_n\right)=-\infty$$

whenever (F4) holds.

Very recently, Fabiano et al. [9] gave a generalization of Wardowski’s result [21] by reducing the condition on function $F:(0,\infty )\to (-\infty ,\infty )$ and by using the right limit of function $F:(0,\infty )\to (-\infty ,\infty )$. They proved the following theorem.

Theorem 1.(Theorem 2.3 [9]) Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to E$ is a map such that for all $x,y\in E\mathrm{with}\rho (Tx,Ty)>0$

$$\tau +F\left(\rho \right(Tx,Ty\left)\right)\le F\left(\rho \right(x,y\left)\right)$$

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function If $\left(F1\right)$ is satisfied, then T possesses only one fixed point.

In [9], Fabiano et al. asked the following question:

Question.(Question 4.3 [9]) Can conditions for the function F be reduced to ($F1$) and ($F2$), and can the proof be made simpler in some results for multivalued mappings in the same way as it was presented in [9] for single-valued mappings?

In this paper, we give a positive answer to the above question by extending above theorem to set-valued maps and obtain a fixed point result for Işik type set valued contractions. We give examples to interpret main results and an application to integral inclusion.

Let $(E,\rho )$ be a metric space. We denote by $CL\left(E\right)$ the family of all nonempty closed subsets of E.

Let $H(·,·)$ be the generalized Pompeiu-Hausdorff distance[5] on $CL\left(E\right)$, i.e., for all $A,B\in CL\left(E\right)$,

$$H(A,B)=\left\{\begin{array}{cc}max\{\underset{a\in A}{sup}\rho (a,B),\underset{b\in B}{sup}\rho (b,A)\},\hfill & \mathrm{if}\mathrm{the}\mathrm{maximum}\mathrm{exists},\hfill \\ \infty ,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where $\rho (a,B)=inf\left\{\rho \right(a,b):b\in B\}$ is the distance from the point a to the subset B.

Let $\delta (A,B)=sup\left\{\rho \right(a,b):a\in A,b\in B\}$. When $A=\left\{x\right\}$, we denote $\delta (A,B)$ by $\delta (x,B)$.

For $A,B\in CL\left(E\right)$, let $D(A,B)={sup}_{x\in A}d(x,B)={sup}_{x\in A}{inf}_{y\in B}d(x,y).$

Then, we have that for all $A,B\in CL\left(E\right)$

$$D(A,B)\le H(A,B)\le \delta (A,B).$$

Lemma 1. 

Let $l>0$, and let $\left\{t_n\right\},\left\{s_n\right\}\subset (l,\infty )$ be non-increasing sequences such that

$$t_n<s_n,\forall n=1,2,3,\cdots \mathrm{and}\underset{n\to \infty }{lim}{t}_n=\underset{n\to \infty }{lim}{s}_n=l.$$

If $F:(0,\infty )\to (-\infty ,\infty )$ is non-decreasing, then we have

$$\underset{n\to \infty }{lim}F\left(t_n\right)=\underset{n\to \infty }{lim}F\left(s_n\right)=F\left(l^{+}\right)>0$$

where $F\left(l^{+}\right)$ denotes ${lim}_{t\to l^{+}}F\left(t\right)$.

Proof. 

Since $\left\{t_n\right\}$ and $\left\{s_n\right\}$ are non-increasing, it follows from the nondecreasingness of F that

$$\underset{t\to l^{+}}{lim}F\left(t\right)=\underset{n\to \infty }{lim}F\left(t_{n+1}\right)\le \underset{n\to \infty }{lim}F\left(t_n\right)\le \underset{n\to \infty }{lim}F\left(s_n\right)\le \underset{n\to \infty }{lim}F\left(s_{n-1}\right)\le \underset{t\to l^{+}}{lim}F\left(t\right).$$

Thus we have

$$\underset{n\to \infty }{lim}F\left(t_n\right)=\underset{n\to \infty }{lim}F\left(s_n\right)=\underset{t\to l^{+}}{lim}F\left(t\right)=F\left(l^{+}\right)>0.$$

□

Lemma 2. 

[6] Let $(E,\rho )$ be a metric space. If $\left\{x_n\right\}$ is not a Cauchy sequence, then there exists $ϵ>0$ for which we can find subsequences $\left\{x_{m\left(k\right)}\right\}$ and $\left\{x_{n\left(k\right)}\right\}$ of $\left\{x_n\right\}$ such that $m\left(k\right)$ is the smallest index for which

$$m\left(k\right)>n\left(k\right)>k,\rho (x_{m\left(k\right)},x_{n\left(k\right)})\ge ϵ\mathrm{and}\rho (x_{m\left(k\right)-1},x_{n\left(k\right)})<ϵ.\left(1.1\right)$$

Further if

$$\underset{n\to \infty }{lim}\rho (x_n,x_{n+1})=0,$$

then we have that

$$\underset{k\to \infty }{lim}\rho (x_{n\left(k\right)},x_{m\left(k\right)})=\underset{k\to \infty }{lim}\rho (x_{n\left(k\right)+1},x_{m\left(k\right)})$$

$$=\underset{k\to \infty }{lim}\rho (x_{n\left(k\right)},x_{m\left(k\right)+1})=\underset{k\to \infty }{lim}\rho (x_{n\left(k\right)+1},x_{m\left(k\right)+1})=ϵ.\left(1.2\right)$$

Lemma 3. 

Let $(E,\rho )$ be a metric space, and let $A,B\in CL\left(E\right)$.

If $a\in A$ and $\rho (a,B)<c$, then there exists $b\in B$ such that $\rho (a,b)<c$.

Proof. 

Let $ϵ=c-\rho (a,B)$.

It follows from definition of infimum that there exists $b\in B$ such that $\rho (a,b)<\rho (a,B)+ϵ$. Hence, $\rho (a,b)<c$. □

Lemma 4. 

Let $(E,\rho )$ be a metric space, and let $A,B\in CL\left(E\right)$ and $\varphi :[0,\infty )\to [0,\infty )$ be a strictly increasing function. If $a\in A$ and $\rho (a,B)+\varphi \left(\rho \right(a,B\left)\right)<c$, then there exists $b\in B$ such that $\rho (a,b)+\varphi \left(\rho \right(a,b\left)\right)<c$.

Proof. 

Since $\varphi$ is strictly increasing,

$$\rho (a,B)<{\varphi }^{-1}(c-\rho (a,B)).$$

By Lemma 3, there exists $b^{\prime }\in B$ such that

$$\rho (a,b^{\prime })<{\varphi }^{-1}(c-\rho (a,B))$$

which yields

$$\rho (a,B)<c-\varphi \left(\rho (a,b^{\prime })\right).$$

Again, by applying Lemma 3, there exists $b^{\prime \prime }\in B$ such that

$$\rho (a,b^{\prime \prime })<c-\varphi \left(\rho (a,b^{\prime })\right).$$

Let $min\{\rho (a,b^{\prime }),\rho (a,b^{\prime \prime })\}=\rho (a,b).$

Then, we have that

$$\rho (a,b)+\varphi \left(\rho \right(a,b\left)\right)<c.$$

□

Lemma 5. 

If $(E,\rho )$ is a metric space, then $K\left(E\right)\subset CL\left(E\right)$, where $K\left(E\right)$ is the family of nonempty compact subsets of E.

Proof. 

Let $A\in K\left(E\right)$, and let $\left\{x_n\right\}\subset A$ be a sequence such that

$$\underset{n\to \infty }{lim}\rho (x_n,x)=0,\mathrm{where}x\in E.$$

It follows from compactness of A that there exists a convergent subsequence $\left\{x_{n\left(k\right)}\right\}$ of $\left\{x_n\right\}$.

Let

$$\underset{n\to \infty }{lim}\rho (x_{n\left(k\right)},a)=0,\mathrm{for}a\in A.$$

Since

$$\underset{n\to \infty }{lim}\rho (x_{n\left(k\right)},x)=0,$$

$x=a\in A.$ Hence $A\in CL\left(E\right)$. □

2. Fixed point results

Let $(E,\rho )$ be a metric space.

A set valued map $T:E\to CL\left(E\right)$ is called generalized F-contraction if the following condition holds:

for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\tau +F\left(H\right(Tx,Ty\left)\right)\le F\left(m\right(x,y\left)\right)\left(2.1\right)$$

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function, and

$$\begin{array}{cc}\hfill m(x,y)=& max\{\rho (x,y),\rho (x,Tx),\rho (y,Ty),\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)]\}.\hfill \end{array}$$

We now prove our main result.

Theorem 2. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a generalized F-contraction. If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Proof. 

Let $x_0\in E$ be a point, and let $x_1\in T{x}_0$.

If $x_1\in T{x}_1$, then the proof is completed.

Assume that $x_1\notin T{x}_1$.

Then, $\rho (x_1,T{x}_1)>0$, because $T{x}_1\in CL\left(X\right)$. Hence, $H(T{x}_0,T{x}_1)\ge d(x_1,T{x}_1)>0.$ From (2.1) we have that

$$\tau +F\left(H(T{x}_0,T{x}_1)\right)\le F\left(m(x_0,x_1)\right).\left(2.2\right)$$

We infer that

$$\begin{array}{cc}\hfill & m(x_0,x_1)=max\{\rho (x_0,x_1),\rho (x_0,T{x}_0),\rho (x_1,T{x}_1),\frac{1}{2}[\rho (x_0,T{x}_1)+\rho (x_1,T{x}_0)]\}\hfill \\ \hfill =& max\{\rho (x_0,x_1),\rho (x_1,T{x}_1)\},\mathrm{because}\mathrm{that}\rho (x_0,T{x}_0)\le \rho (x_0,x_1)\mathrm{and}\hfill \\ \hfill & \frac{1}{2}[\rho (x_0,T{x}_1)+\rho (x_1,T{x}_0)]\le \frac{1}{2}[\rho (x_0,x_1)+\rho (x_1,T{x}_1)].\hfill \end{array}$$

If $m(x_0,x_1)=\rho (x_1,T{x}_1)$, then from (2.2) we obtain that

$$F\left(\rho (x_1,T{x}_1)\right)<\tau +F\left(H(T{x}_0,T{x}_1)\right)\le F\left(\rho (x_1,T{x}_1)\right)$$

which is a contradiction.

Thus, $m(x_0,x_1)=\rho (x_0,x_1)$. It follows from (2.2) that

$$\begin{array}{c}\hfill \frac{1}{2}\tau +F\left(\rho (x_1,T{x}_1)\right)<\tau +F\left(H(T{x}_0,T{x}_1)\right)\le F\left(\rho (x_0,x_1)\right).\end{array}$$

(2.3)

Since (F1) is satisfied, we obtain that

$$\rho (x_1,T{x}_1)<F^{-1}(\frac{1}{2}\tau +F\left(H(T{x}_0,T{x}_1)\right)).$$

Applying Lemma 3, there exists $x_2\in T{x}_1$ such that

$$\rho (x_1,x_2)<F^{-1}(\frac{1}{2}\tau +F\left(H(T{x}_0,T{x}_1)\right))$$

which implies

$$F\left(\rho (x_1,x_2)\right)<\frac{1}{2}\tau +F\left(H(T{x}_0,T{x}_1)\right)\le F\left(\rho (x_0,x_1)\right)-\frac{1}{2}\tau .\left(2.4\right)$$

Again from (2.1) we have that

$$\frac{1}{2}\tau +F\left(\rho (x_2,T{x}_2)\right)<\tau +F\left(H(T{x}_1,T{x}_2)\right)\le F\left(\rho (x_1,x_2)\right)\left(2.5\right)$$

which implies

$$\rho (x_2,T{x}_2)<F^{-1}(\frac{1}{2}\tau +F\left(H(T{x}_1,T{x}_2)\right)).$$

By Lemma 3, there exists $x_3\in T{x}_2$ such that

$$\rho (x_2,x_3)<F^{-1}(\frac{1}{2}\tau +F\left(H(T{x}_1,T{x}_2)\right)).$$

Hence, we obtain that

$$F\left(\rho (x_2,x_3)\right)<\frac{1}{2}\tau +F\left(H(T{x}_1,T{x}_2)\right)\le F\left(\rho (x_1,x_2)\right)-\frac{1}{2}\tau .\left(2.6\right)$$

Inductively, we have that for all $n=1,2,3,\cdots ,$

$$x_n\in T{x}_{n-1}$$

and

$$F\left(\rho (x_n,x_{n+1})\right)<\frac{1}{2}\tau +F\left(H(T{x}_{n-1},x_n)\right)\le F\left(\rho (x_{n-1},x_n)\right)-\frac{1}{2}\tau .\left(2.7\right)$$

Because (F1) is satisfied,

$$\rho (x_n,x_{n+1})<\rho (x_{n-1},x_n),\forall n=1,2,3,\cdots .$$

Hence, there exists $r\ge 0$ such that

$$\underset{n\to \infty }{lim}\rho (x_n,x_{n+1})=r.$$

Assume that $r>0$.

By Lemma 1, we have that

$$\underset{n\to \infty }{lim}F\left(\rho (x_n,x_{n+1})\right)=\underset{n\to \infty }{lim}F\left(\rho (x_{n-1},x_n)\right)=\underset{t\to r^{+}}{lim}F\left(t\right)=F\left(r^{+}\right).\left(2.8\right)$$

Taking limit $n\to \infty$ in (2.7) and using (2.8), we obtain that

$$F\left(r^{+}\right)\le F\left(r^{+}\right)-\frac{1}{2}\tau$$

which is a contradiction, because $\tau >0.$

Thus, we obtain that

$$\underset{n\to \infty }{lim}\rho (x_n,x_{n+1})=0.$$

(2.9)

Now, we show that $\left\{x_n\right\}$ is a Cauchy sequence.

Assume that $\left\{x_n\right\}$ is not a Cauchy sequence.

Then, there exists $ϵ>0$ for which we can find subsequences $\left\{x_{m\left(k\right)}\right\}$ and $\left\{x_{n\left(k\right)}\right\}$ of $\left\{x_n\right\}$ such that $m\left(k\right)$ is the smallest index for which (1.1) holds.

That is, the following are satisfied:

$$m\left(k\right)>n\left(k\right)>k,\rho (x_{m\left(k\right)},x_{n\left(k\right)})\ge ϵ\mathrm{and}\rho (x_{m\left(k\right)-1},x_{n\left(k\right)})<ϵ.$$

It follows from (2.1) that

$$\begin{array}{cc}\hfill & F(\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})<\tau +F(\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})\hfill \\ \hfill \le & \tau +F(H(T{x}_{n\left(k\right)},T{x}_{m\left(k\right)})\le F\left(m(x_{n\left(k\right)},x_{m\left(k\right)})\right).\hfill \end{array}$$

(2.10)

We infer that

$$\begin{array}{cc}\hfill & ϵ\le \rho (x_{n\left(k\right)},x_{m\left(k\right)})\le m(x_{n\left(k\right)},x_{m\left(k\right)})\hfill \\ \hfill =& max\{\rho (x_{n\left(k\right)},x_{m\left(k\right)}),\rho (x_{n\left(k\right)},T{x}_{n\left(k\right)}),\rho (x_{m\left(k\right)},T{x}_{m\left(k\right)}),\hfill \\ \hfill & \frac{1}{2}[\rho (x_{n\left(k\right)},T{x}_{m\left(k\right)})+\rho (x_{m\left(k\right)},T{x}_{n\left(k\right)})]\}\hfill \\ \hfill \le & max\{\rho (x_{n\left(k\right)},x_{m\left(k\right)}),\rho (x_{n\left(k\right)},x_{n\left(k\right)+1}),\rho (x_{m\left(k\right)},x_{m\left(k\right)+1}),\hfill \\ \hfill & \frac{1}{2}[\rho (x_{n\left(k\right)},x_{m\left(k\right)+1})+\rho (x_{m\left(k\right)},x_{n\left(k\right)+1})]\}\hfill \end{array}$$

(2.11)

Taking limit as $k\to \infty$ on both sides of (2.11) and using (1.2), we obtain that

$$\underset{k\to \infty }{lim}m(x_{n\left(k\right)},x_{m\left(k\right)})=ϵ.\left(2.12\right)$$

Since (F1) is satisfied, from (2.10) we have that

$$\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})<F^{-1}(\tau +F\left(\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})\right).$$

By applying Lemma 3, there exists $y_{m\left(k\right)}\in T{x}_{m\left(k\right)}$ such that

$$\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})<F^{-1}(\tau +F\left(\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})\right).$$

Hence,

$$F\left(\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})\right)<\tau +F(\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)}).$$

Thus, it follows from (2.10) that

$$\begin{array}{cc}\hfill & F\left(\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})\right)\hfill \\ \hfill <& \tau +F\left(\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})\right)<\tau +F(\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})\hfill \\ \hfill \le & \tau +F(H(T{x}_{n\left(k\right)},T{x}_{m\left(k\right)})\hfill \\ \hfill \le & F\left(m(x_{n\left(k\right)},x_{m\left(k\right)})\right)\hfill \end{array}$$

(2.13)

which leads to

$$\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})<m(x_{n\left(k\right)},x_{m\left(k\right)}),\forall k=1,2,3,\cdots .\left(2.14\right)$$

By taking $limsup$ as $k\to \infty$ in (2.14) and using (2.12), we have that

$$\underset{k\to \infty }{lim}sup\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})\le ϵ.\left(2.15\right)$$

Since

$$\rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})\le \rho (x_{n\left(k\right)+1},y_{m\left(k\right)}),$$

$$\begin{array}{cc}\hfill & \rho (x_{n\left(k\right)+1},x_{m\left(k\right)})\hfill \\ \hfill \le & \rho (x_{n\left(k\right)+1},T{x}_{m\left(k\right)})+\rho (T{x}_{m\left(k\right)},x_{m\left(k\right)})\hfill \\ \hfill \le & \rho (x_{n\left(k\right)+1},y_{m\left(k\right)})+\rho (x_{m\left(k\right)+1},x_{m\left(k\right)})\hfill \end{array}$$

(2.16)

Taking $liminf$ as $k\to \infty$ in (2.16) and using (1.2), we obtain that

$$ϵ\le \underset{k\to \infty }{lim}inf\rho (x_{n\left(k\right)+1},y_{m\left(k\right)}).\left(2.17\right)$$

It follows from (2.15) and (2.17) that

$$\underset{k\to \infty }{lim}\rho (x_{n\left(k\right)+1},y_{m\left(k\right)})=ϵ.\left(2.18\right)$$

By applying Lemma 1 to (2.13) with (2.12), (2.14) and (2.18), we obtain that

$$F\left({ϵ}^{+}\right)\le \tau +F\left({ϵ}^{+}\right)\le F\left({ϵ}^{+}\right)$$

which leads to a contradiction.

Hence, $\left\{x_n\right\}$ is a Cauchy sequence.

From the completeness of E, there exists

$$x_{*}=\underset{n\to \infty }{lim}{x}_n\in E.$$

It follows from (2.1) that

$$\begin{array}{cc}\hfill & F\left(\rho (x_{n+1},T{x}_{*})\right)<\tau +F\left(\rho (x_{n+1},T{x}_{*})\right)\hfill \\ \hfill \le & \tau +F\left(H(T{x}_n,T{x}_{*})\right)\le F\left(m(x_n,x_{*})\right),\hfill \end{array}$$

(2.19)

where $m(x_n,x_{*})=max\{\rho (x_n,x_{*}),\rho (x_n,x_{n+1}),\rho (x_{*},T{x}_{*}),$

$\frac{1}{2}[\rho (x_{*},x_{n+1})+\rho (x_n,T{x}_{*})]\}$.

Since (F1) is satisfied, from (2.19) we have that

$$\rho (x_{n+1},T{x}_{*})<m(x_n,x_{*}).\left(2.20\right)$$

We have that

$$\underset{n\to \infty }{lim}\rho (x_{n+1},T{x}_{*})=\underset{n\to \infty }{lim}m(x_n,x_{*})=\rho (x_{*},T{x}_{*}).\left(2.21\right)$$

Assume that $\rho (x_{*},T{x}_{*})>0$.

By Lemma 1, we have that

$$\begin{array}{cc}\hfill & \underset{n\to \infty }{lim}F\left(\rho (x_{n+1},T{x}_{*})\right)=\underset{n\to \infty }{lim}F\left(m(x_n,x_{*})\right)\hfill \\ \hfill =& \underset{t\to \rho {(x_{*},T{x}_{*})}^{+}}{lim}F\left(t\right)=F\left(\rho {(x_{*},T{x}_{*})}^{+}\right).\hfill \end{array}$$

(2.22)

Applying (2.22) to (2.19), we obtain that

$$F\left(\rho {(x_{*},T{x}_{*})}^{+}\right)\le \tau +F\left(\rho {(x_{*},T{x}_{*})}^{+}\right)\le F\left(\rho {(x_{*},T{x}_{*})}^{+}\right)$$

which leads to a contradiction.

Hence, $\rho (x_{*},T{x}_{*})=0$, and hence $x_{*}\in T{x}_{*}.$ □

The following example interprets Theorem 2.

Example 1. 

Let $E=[0,1]$ and $\rho (x,y)=|x-y|,\forall x,y\in E$.

Then $(E,\rho )$ is a complete metric space.

Define a set-valued map $T:E\to CL\left(E\right)$ by

$$Tx=\left\{\begin{array}{cc}\left\{1\right\},\hfill & (x=0)\hfill \\ \{\frac{2}{5},\frac{1}{2}\},\hfill & (0<x\le 1).\hfill \end{array}\right.$$

Let $\tau =ln\frac{3}{2}$ and $F\left(t\right)=lnt,\forall t>0$.

We show that T is a generalized F-contraction.

We have that for all $x,y\in E$,

$$H(Tx,Ty)>0⟺(x=0\mathrm{and}0<y\le 1),(0<x\le 1\mathrm{and}y=0).$$

$1^{\circ }$: Let $x=0\mathrm{and}0<y\le 1$.

Then, we obtain that

$$\begin{array}{cc}\hfill & \tau +F\left(H\right(Tx,Ty\left)\right)-F\left(\rho \right(x,Tx\left)\right)\hfill \\ \hfill =& \tau +F\left(\frac{3}{5}\right)-F\left(1\right)\hfill \\ \hfill =& ln\frac{3}{2}+ln\frac{3}{5}-ln1\hfill \\ \hfill =& ln9-ln10<0.\hfill \end{array}$$

Thus,

$$\tau +F\left(H\right(Tx,Ty\left)\right)<F\left(\rho \right(x,Tx\left)\right)$$

which implies

$$\tau +F\left(H\right(Tx,Ty\left)\right)<F\left(m\right(x,y\left)\right).$$

$2^{\circ }$: Let $0\le x<1\mathrm{and}y=1$. Then, we have that

$$\begin{array}{cc}\hfill & \tau +F\left(H\right(Tx,Ty\left)\right)-F\left(\rho \right(y,Ty\left)\right)\hfill \\ \hfill =& \tau +F\left(\frac{3}{5}\right)-F\left(1\right)\hfill \\ \hfill =& ln\frac{3}{2}+ln\frac{3}{5}-ln1\hfill \\ \hfill =& ln9-ln10<0.\hfill \end{array}$$

Thus,

$$\tau +F\left(H\right(Tx,Ty\left)\right)<F\left(\rho \right(y,Ty\left)\right)$$

which leads to

$$\tau +F\left(H\right(Tx,Ty\left)\right)<F\left(m\right(x,y\left)\right).$$

Hence, F is generalized F-contraction. The assumptions of Theorem 2 are satisfied. By Theorem 2, T possesses two fixed points, $\frac{2}{5}$ and $\frac{1}{2}.$

Remark 1. 

Theorem 2 is a positive answer to Question 4.3 of [9].

By Theorem 2, we have the following result.

Corollary 1. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a set-valued map such that for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\tau +F\left(H\right(Tx,Ty\left)\right)\le F\left(l\right(x,y\left)\right)$$

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function, and

$$\begin{array}{cc}\hfill l(x,y)=& max\{\rho (x,y),\frac{1}{2}[\rho (x,Tx)+\rho (y,Ty)],\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)]\}.\hfill \end{array}$$

If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Corollary 2. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CB\left(E\right)$ is a set-valued map such that for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\tau +F\left(H\right(Tx,Ty\left)\right)\le F\left(\rho \right(x,y\left)\right)$$

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function. If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Corollary 3. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a set-valued map such that for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\begin{array}{cc}\hfill & \tau +F\left(H\right(Tx,Ty\left)\right)\hfill \\ \hfill \le & F\left(a\rho \right(x,y)+b\rho (x,Tx)+c\rho (y,Ty)+e[\rho (x,Ty)+\rho (y,Tx)\left]\right)\hfill \end{array}$$

(2.23)

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function, and $a,b,c,e\ge 0$ and $a+b+c+2e=1$. If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Proof. 

It follows from (2.23) that

$$\begin{array}{cc}\hfill & \tau +F\left(H\right(Tx,Ty\left)\right)\hfill \\ \hfill \le & F\left(a\rho \right(x,y)+b\rho (x,Tx)+c\rho (y,Ty)+e[\rho (x,Ty)+\rho (y,Tx)\left]\right)\hfill \\ \hfill =& F(a\rho (x,y)+b\rho (x,Tx)+c\rho (y,Ty)]+2e\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)])\hfill \\ \hfill \le & F((a+b+c+2e)max\{\rho (x,y),\rho (x,Tx),\rho (y,Ty),\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)]\})\hfill \\ \hfill =& F\left(m\right(x,y\left)\right).\hfill \end{array}$$

By Theorem 2, T possesses a fixed point. □

Corollary 4. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a set-valued map such that for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\begin{array}{cc}\hfill & \tau +F\left(H\right(Tx,Ty\left)\right)\hfill \\ \hfill \le & F\left(a\rho \right(x,y)+b[\rho (x,Tx)+\rho (y,Ty)]+c[\rho (x,Ty)+\rho (y,Tx)\left]\right)\hfill \end{array}$$

(2.24)

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function, and $a,b,c\ge 0$ and $a+2b+2c=1$. If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Proof. 

It follows from (2.24) that

$$\begin{array}{cc}\hfill & \tau +F\left(H\right(Tx,Ty\left)\right)\hfill \\ \hfill \le & F\left(a\rho \right(x,y)+b[\rho (x,Tx)+\rho (y,Ty)]+c[\rho (x,Ty)+\rho (y,Tx)\left]\right)\hfill \\ \hfill =& F(a\rho (x,y)+2b\frac{1}{2}[\rho (x,Tx)+\rho (y,Ty)]+2c\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)])\hfill \\ \hfill \le & F((a+2b+2c)max\{\rho (x,y),\frac{1}{2}[\rho (x,Tx)+\rho (y,Ty)],\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)]\})\hfill \\ \hfill =& F\left(l\right(x,y\left)\right).\hfill \end{array}$$

By Corollary 2.2, T possesses a fixed point. □

Corollary 5. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a set-valued map such that for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\tau +F\left(H(Tx,Ty)\right)\le F\left(\frac{1}{2}[\rho (x,Tx)+\rho (y,Ty)]\right)$$

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function. If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Corollary 6. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a set-valued map such that for all $x,y\in E\mathrm{with}H(Tx,Ty)>0$

$$\tau +F\left(H(Tx,Ty)\right)\le F\left(\frac{1}{2}[\rho (x,Ty)+\rho (y,Tx)]\right)$$

where $\tau >0$ and $F:(0,\infty )\to (-\infty ,\infty )$ is a function. If $\left(F1\right)$ is satisfied, then T possesses a fixed point.

Theorem 3. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is an Işik type set-valued contraction, i.e. for each $x,y\in E$ and each $u\in Tx$, there exists $v\in Ty$ such that

$$\rho (u,v)\le \varphi \left(\rho \right(x,y\left)\right)-\varphi \left(\rho \right(u,v\left)\right)$$

(2.25)

where $\varphi :[0,\infty )\to [0,\infty )$ is a function such that

$$\underset{t\to 0^{+}}{lim}\varphi \left(t\right)=0.$$

(2.26)

Then, T possesses a fixed point.

Proof. 

Let $x_0\in E$, and let $x_1\in T{x}_0$. Then there exits $x_2\in T{x}_1$ such that

$$\rho (x_1,x_2)\le \varphi \left(\rho (x_0,x_1)\right)-\varphi \left(\rho (x_1,x_2)\right).$$

Again, there exists $x_3\in T{x}_2$ such that

$$\rho (x_2,x_3)\le \varphi \left(\rho (x_1,x_2)\right)-\varphi \left(\rho (x_2,x_3)\right).$$

Inductively, we have a sequence $\left\{x_n\right\}\subset E$ such that for all $n=1,2,3,\cdots$,

$$x_n\in T{x}_{n-1}\mathrm{and}\rho (x_n,x_{n+1})\le \varphi \left(\rho (x_{n-1},x_n)\right)-\varphi \left(\rho (x_n,x_{n+1})\right).\left(2.27\right)$$

From (2.27) $\left\{\varphi \left(\rho (x_{n-1},x_n)\right)\right\}$ is a non-increasing sequence and bounded below by 0. Hence, there exists $r\ge 0$ such that

$$\underset{n\to \infty }{lim}\varphi \left(\rho (x_{n-1},x_n)\right)=r.$$

We show that $\left\{x_n\right\}$ is a Cauchy sequence.

Let $m,n$ be any positive integers such that $m>n$.

Then, we have that

$$\begin{array}{cc}\hfill & \rho (x_n,x_m)\hfill \\ \hfill \le & \rho (x_n,x_{n+1})+\rho (x_{n+1},x_{n+2})+\cdots +\rho (x_{m-1},x_m)\hfill \\ \hfill \le & \varphi \left(\rho (x_{n-1},x_n)\right)-\varphi \left(\rho (x_{m-1},x_m)\right)\hfill \\ \hfill \le & \varphi \left(\rho (x_{n-1},x_n)\right)-r.\hfill \end{array}$$

(2.28)

Letting $m,n\to \infty$ in (2.28), we obtain that

$$\underset{n,m\to \infty }{lim}\rho (x_n,x_m)=0.$$

Thus, $\left\{x_n\right\}$ is a Cauchy sequence. It follows from the completeness of E that

$$x_{*}=\underset{n\to \infty }{lim}{x}_n\mathrm{exists}.$$

(2.29)

Now, we show that $x_{*}$ is a fixed point for T.

It follows from (2.25) that for $x_n\in T{x}_{n-1}$, there exists $v\in T{x}_{*}$ such that

$$\rho (x_n,v)\le \varphi \left(\rho (x_{n-1},x_{*})\right)-\varphi \left(\rho (x_n,v)\right)\le \varphi \left(\rho (x_{n-1},x_{*})\right).$$

(2.30)

Taking limit $n\to \infty$ in equation (2.30) and using (2.26), we infer that

$$\underset{n\to \infty }{lim}\rho (x_n,v)=0$$

which implies

$$x_{*}=v\in T{x}_{*}.$$

□

Example 2. 

Let $E=\{x_n:x_n=1+2+3+\cdots +n,n=1,2,3,\cdots \}$ and $\rho (x,y)=|x-y|,\forall x,y\in E$.

Then, $(E,\rho )$ is a complete metric space.

Define a map $T:E\to CL\left(E\right)$ by

$$Tx=\left\{\begin{array}{cc}\left\{x_1\right\},\hfill & (x=x_1)\hfill \\ \{x_1,x_2,x_3,\cdots x_{n-1}\},\hfill & (x=x_n).\hfill \end{array}\right.$$

Let $\varphi \left(t\right)=\frac{1}{2}t,\forall t\ge 0$.

We show that condition (2.25) is satisfied.

Consider the following two cases.

$1^{\circ }$: Let $x=x_1\mathrm{and}y=x_n,n=2,3,4,\cdots$.

Then, for $u=x_1\in Tx$, there exists $v=x_1\in Ty$ such that

$$\rho (u,v)=0<\frac{1}{2}\rho (x_1,x_n)=\varphi \left(\rho (x_1,x_n)\right)=\varphi \left(\rho (x_1,x_n)\right)-\varphi \left(\rho (u,v)\right).$$

$2^{\circ }$: Let $x=x_n\mathrm{and}y=x_m,m>n,n=2,3,4,\cdots$.

For $u=x_k\in Tx(k=1,2,3,\cdots ,n-1)$ , there exists $v=x_k\in Ty$ such that

$$\rho (u,v)=0<\frac{1}{2}\rho (x_n,x_m)=\varphi \left(\rho (x_n,x_m)\right)=\varphi \left(\rho (x_n,x_m)\right)-\varphi \left(\rho (u,v)\right).$$

This show that T satisfies condition (2.25). Thus, all conditions of Theorem 3 hold. From Theorem 3T possesses a fixed point, $x_{*}=x_1$.

Corollary 7. 

Let $(E,\rho )$ be a complete metric space. Suppose that $T:E\to CL\left(E\right)$ is a set-valued map such that for each $x,y\in E$,

$$H(Tx,Ty)<\varphi \left(\rho \right(x,y\left)\right)-\varphi \left(H\right(Tx,Ty\left)\right),$$

where $\varphi :[0,\infty )\to [0,\infty )$ is a strictly increasing function such that

$$\underset{t\to 0^{+}}{lim}\varphi \left(t\right)=0.$$

Then, T possesses a fixed point.

Proof. 

Let $x,y\in E$ and let $u\in Tx$.

As $\varphi$ is strictly increasing,

$$\rho (u,Ty)+\varphi \left(\rho \right(u,Ty\left)\right)<\varphi \left(\rho \right(x,y\left)\right).$$

Applying Lemma 4, there exists $v\in Ty$ such that

$$\rho (u,v)+\varphi \left(\rho \right(u,v\left)\right)<\varphi \left(\rho \right(x,y\left)\right).$$

By Theorem 3, T possesses a fixed point. □

From Theorem 3 we have the following result.

Corollary 8. 

[14] Let $(E,\rho )$ be a complete metric space. Suppose that $f:E\to E$ is a map such that for each $x,y\in E$,

$$\rho (fx,fy)\le \varphi \left(\rho \right(x,y\left)\right)-\varphi \left(\rho \right(fx,fy\left)\right)$$

where $\varphi :[0,\infty )\to [0,\infty )$ is a function such that

$$\underset{t\to 0^{+}}{lim}\varphi \left(t\right)=0.$$

Then, f possesses a fixed point.

3. Application

In this section, we give an application of our result to integral inclusion.

Let $[a,b]\subset R$ be a closed interval, and let $C\left(\right[a,b],\mathbb{R})$ be the family of continuous mapping from $[a,b]\mathrm{into} \mathbb{R}$.

Let $E=C\left(\right[a,b],R)$ and $\rho (x,y)={sup}_{t\in [a,b]}|x\left(t\right)-y\left(t\right)|$ for all $x,y\in E$.

Then, $(E,\rho )$ is a complete metric space.

Consider the Fredholm type integral inclusion:

$$x\left(t\right)\in {\int }_a^{b}K(t,s,x\left(s\right))ds+f\left(t\right),t\in [a,b]$$

(3.1)

where $f\in E$, $K:[a,b]\times [a,b]\times R\to CB\left(R\right)$, and $x\in E$ is the unknown function.

Suppose that the following conditions are satisfied:

($1^{\circ }$)

for each $x\in E,K(·,·,x\left(s\right))=K_x(·,·)$ is continuous;

($2^{\circ }$)

there exists a continuous function $Z:[a,b]\times [a,b]\to [0,\infty )$ such that for all $t,s\in [a,b]$ and all $u,v\in E$,

$$|k_u(t,s)-k_v(t,s)|\le Z(t,s)\rho (u\left(s\right),v\left(s\right))$$

where $k_u(t,s)\in K_u(t,s),k_v(t,s)\in K_v(t,s);$

($3^{\circ }$)

there exists $\alpha >1$ such that

$$\underset{t\in [a,b]}{sup}{\int }_a^{b}Z(t,s)ds\le \frac{1}{2+\alpha }.$$

Theorem 4. 

Let $(E,\rho )$ be a complete metric space. If conditions $\left(1^{\circ }\right)$,$\left(2^{\circ }\right)$ and $\left(3^{\circ }\right)$ are satisfied, then the integral inclusion (3.1) has a solution.

Proof. 

Define a set-valued map $T:E\to CB\left(E\right)$ by

$$Tx=\{y\in E:y\left(t\right)\in {\int }_a^{b}K(t,s,x\left(s\right))ds+f\left(t\right),t\in [a,b]\}.$$

Let $x\in E$ be given.

For the set-valued map $K_x(t,s):[a,b]\times [a,b]\to CB\left(\mathbb{R}\right)$, by applying Michael’s selection theorem, there exists a continuous map $k_x(t,s):[a,b]\times [a,b]\to \mathbb{R}$ such that

$$k_x(t,s)\in K_x(t,s),\forall t,s\in [a,b].$$

Thus,

$${\int }_a^b{k}_x(t,s)ds+f\left(t\right)\in Tx,$$

and so $Tx\ne \varnothing .$

Since f and $k_x$ are continuous, $Tx\in CB\left(E\right)$ for each $x\in E$.

Let $y_1\in T{x}_1$.

Then,

$$y_1\left(t\right)\in {\int }_a^{b}K(t,s,x_1\left(s\right))ds+f\left(t\right),t\in [a,b].$$

Hence, there exists $k_{x_1}(t,s)\in K_{x_1}(t,s),\forall t,s\in [a,b]$ such that

$$y_1\left(t\right)={\int }_a^b{k}_{x_1}(t,s)ds+f\left(t\right),\forall t,s\in [a,b].$$

It follows from ($2^{\circ }$) that there exists $z(t,s)\in K_{x_2}(t,s)$ such that

$$|k_{x_1}(t,s)-z(t,s)|\le Z(t,s)\rho (x_1\left(s\right),x_2\left(s\right)),\forall t,s\in [a,b].$$

Let $U:[a,b]\times [a,b]\to CB\left(\mathbb{R}\right)$ be defined by

$$U(t,s)=K_{x_2}(t,s)\cap \{u\in \mathbb{R}:\rho (k_{x_1}(t,s),u)\le \rho (x_1\left(s\right),x_2\left(s\right))\}.$$

From ($1^{\circ }$) U is continuous. Hence, it follows that there exists a continuous map $k_{x_2}:[a,b]\times [a,b]\to \mathbb{R}$ such that

$$k_{x_2}(t,s)\in U(t,s),\forall t,s\in [a,b].$$

Let

$$y_2\left(t\right)={\int }_a^b{k}_{x_2}(t,s)ds+f\left(t\right),\forall t,s\in [a,b].$$

Then,

$$y_2\left(t\right)\in {\int }_a^b{K}_{x_2}(t,s)ds+f\left(t\right)={\int }_a^{b}K(t,s,x_2\left(s\right))ds+f\left(t\right),\forall t,s\in [a,b],$$

and so $y_2\in T{x}_2.$

Thus, we obtain that

$$\begin{array}{cc}\hfill & \rho (y_1,y_2)=|{\int }_a^b{k}_{x_1}(t,s)-k_{x_2}(t,s)ds|\hfill \\ \hfill \le & \underset{t\in [a,b]}{sup}{\int }_a^b|k_{x_1}(t,s)-k_{x_2}(t,s)|ds\hfill \\ \hfill \le & \underset{t\in [a,b]}{sup}{\int }_a^{b}Z(t,s)ds\rho (x_1\left(s\right),x_2\left(s\right))\hfill \\ \hfill \le & \frac{1}{2+\alpha }\rho (x_1\left(s\right),x_2\left(s\right)).\hfill \end{array}$$

Thus, we have that

$$(1+\frac{1}{2}\alpha )\delta (T{x}_1,T{x}_2)\le \frac{1}{2}\rho (x_1,x_2)$$

which implies

$$(1+\frac{1}{2}\alpha )H(T{x}_1,T{x}_2)\le \frac{1}{2}\rho (x_1,x_2).$$

Hence, we obtain that

$$\begin{array}{cc}\hfill & H(T{x}_1,T{x}_2))\le \varphi \left(\rho (x_1,x_2)\right)-\varphi \left(\alpha H(T{x}_1,T{x}_2)\right)\hfill \\ \hfill <& \varphi \left(\rho (x_1,x_2)\right)-\varphi \left(H(T{x}_1,T{x}_2)\right)\mathrm{where}\varphi \left(t\right)=\frac{1}{2}t,\forall t\ge 0.\hfill \end{array}$$

By Corollary 7, T possesses a fixed point, and hence the integral inclusion (3.1) has a solution. □

4. Conclusions

Our results are generalizations and extensions of F-contractions and Işik contractions to set-valued maps on metric spaces. We give a positive answer to the Question 4.3 of [9] and an application to integral inclusion.