Three Existence Results in the Fixed Point Theory

Abstract

In the present paper, we obtain three results on the existence of a fixed point for nonexpansive mappings. Two of them are generalizations of the result for F-contraction, while third one is a generalization of a recent result for set-valued contractions.

1. Introduction

The work referenced here [1] was the starting point of the fixed point theory of nonexpansive maps, which is a growing field of research; see [2,3,4,5,6,7]. In particular, the convergence of iterated Bregman projections and of the alternating algorithm were studied in [2], fixed point theorems under Mizoguchi–Takahashi-type conditions were obtained in [3], Ciric-type results were proved in [4], fixed point theory in modular function spaces was discussed in [5] and cyclic contractions were analyzed in [6]. Note that the research on nonexpansive maps in spaces with graphs is of great importance; see [8,9,10,11,12] and the references mentioned therein. In particular, fixed point results on metric spaces with a graph are obtained in [9,11], Reich-type contractions were studied in [8], hybrid methods were studied in [10] and the convergence of fixed points in graphical spaces was considered in [12]. In [13], D. Wardowski introduced an interesting class of mappings which contains Banach contractions and showed the existence of fixed points for these mappings. More precisely, we can assume that $(X,\rho )$ is a complete metric space, $\tau >0$, $F:(0,\infty )\to R^1$ is a strictly increasing function and that $T:X\to X$ is a mapping such that for every pair of points $u,v\in X$ satisfying $u\ne v$,

$$F\left(\rho \right(T\left(u\right),T\left(v\right)\left)\right)+\tau \le F\left(\rho \right(u,v\left)\right).$$

Assuming two additional assumptions on F, D. Wardowski showed that the mapping of T has a unique fixed point. In the subsequent research it was shown that these two additional assumptions are not necessary. One of the examples of F is the function $ln(·)$, and in this case the Wardowski contraction is a strict contraction. Wardowski-type contractions were studied in [14,15,16]. In this work, we obtain three results on the existence of a fixed point for nonexpansive mappings in a complete metric space. Two of them are generalizations of the result by D. Wardowski for F-contraction, while the third one is a generalization of a recent result by S.-H. Cho for set-valued contractions [17].

Assume that $(X,\rho )$ is a complete metric space. Let N be the set of all natural numbers. We assume that the sum over empty set is zero. For every element $x\in X$ and every real number $r>0$, set

$$B(x,r)=\{y\in X:\rho (x,y)\le r\}.$$

For every element $\xi \in X$ and each nonempty set $A\subset X$ set

$$\rho (\xi ,A)=inf\left\{\rho \right(\xi ,\eta ):\eta \in A\}.$$

2. The First Result

Assume that $(X,\rho )$ is a complete metric space where $\rho$ is a metric, $K\subset X$ is a nonempty closed set, $\tau >0$, and $F:(0,\infty )\to R^1$ is an increasing function satisfying

$$F\left(t\right)\le F\left(s\right)$$

for each $s>t>0$ and that $T:K\to X$ is an operator such that for every pair of points $x,y\in K$ satisfying $x\ne y$,

$$F\left(\rho \right(T\left(x\right),T\left(y\right)\left)\right)+\tau \le F\left(\rho \right(x,y\left)\right).$$

(1)

Set $T^0\left(x\right)=x$, $x\in K$.

D. Wardowski in [13] proved the existence of a fixed point of T in the case when $K=X$₀, assuming that F is strictly increasing. Here, we assume that F is merely increasing in general.

Since the function F is increasing, the following proposition holds [18].

Proposition 1.

There is a countable set $E\subset (0,\infty )$ such that the function F is continuous at every element $z\in (0,\infty )\setminus E.$

Equation (1) implies the following proposition.

Proposition 2.

For every pair of elements $x,y\in K$, the relation $\rho \left(T\right(x),T(y\left)\right)\le \rho (x,y)$ is true.

Theorem 1.

Assume that $K_0\subset X$ is a nonempty bounded set and for each integer $n\ge 1$ there is an element $x_n\in K_0$ for which $T^n\left(x_n\right)\in X$ is defined. Then, $x_{*}\in K$ can be found, satisfying $T\left(x_{*}\right)=x_{*}$.

Proof. 

Fix $\theta \in K_0$. $M_0>0$ is found, for which

$$\rho (z,\theta )\le M_0,z\in K_0.$$

(2)

Proposition 2 and (2) imply that for each $n,m\in \mathbf{N}$,

$$\rho (x_n,x_m)\le 2M_0,$$

$$\rho (T\left(x_n\right),T\left(\theta \right))\le \rho (x_n,\theta )\le M_0,$$

$$\rho (T\left(x_n\right),\theta )\le M_0+\rho (\theta ,T\left(\theta \right)).$$

(3)

Let $ϵ\in (0,1)$. We show that the following property holds:

(P1) There exists a natural number $n_0$ such that for each integer $n>n_0$ and each integer $i\in [n_0,n)$, we have

$$\rho (T^i\left(x_n\right),T^{i+1}\left(x_n\right))\le ϵ.$$

Choose an integer:

$$n_0>1+{\tau }^{-1}(F(2M_0+\rho (\theta ,T\left(\theta \right)))-F\left(ϵ\right)).$$

(4)

Let $n>n_0$ be an integer. In order to prove that (P1) holds in view of Proposition 2, it is enough to prove that

$$\rho (T^{n_0}\left(x_n\right),T^{n_0+1}\left(x_n\right))\le ϵ.$$

Assume the contrary. Then, according to Proposition 2,

$$\rho (T^{n_0}\left(x_n\right),T^{n_0+1}\left(x_n\right))\ge ϵ,i=0,\dots ,n_0.$$

(5)

Equations (1) and (5) imply that for every $i\in \{0,\dots ,n_0-1\}\setminus \{n_0-1\}$,

$$F\left(\rho (T^{i+1}\left(x_n\right),T^{i+2}\left(x_n\right))\right)+\tau \le F\left(\rho (T^i\left(x_n\right),T^{i+1}\left(x_n\right))\right),$$

$$F\left(\rho (T^{n_0}\left(x_n\right),T^{n_0+1}\left(x_n\right))\right)\le \tau (1-n_0)+F\left(\rho (T\left(x_n\right),x_n)\right).$$

(6)

It follows from (3), (5) and (6) that

$$F\left(ϵ\right)\le F\left(\rho (T^{n_0}\left(x_n\right),T^{n_0+1}\left(x_n\right))\right)\le \tau (1-n_0)+F(2M_0+\rho (\theta ,T\left(\theta \right))),$$

$$\tau (n_0-1)\le F(2M_0+\rho (\theta ,T\left(\theta \right)))-F\left(ϵ\right),$$

$$n_0\le 1+{\tau }^{-1}(F(2M_0+\rho (\theta ,T\left(\theta \right)))-F\left(ϵ\right)).$$

This contradicts inequality (4) and proves property (P1).

We show that the following property holds:

(P2) $n_0\in \mathbf{N}$ is found, such that for each triplet of integers $i, n_1, n_2$ satisfying $n_1, n_2, i\ge n_0,$$i\le n_1, n_2,$

$$\rho (T^i\left(x_{n_1}\right),T^i\left(x_{n_2}\right))\le ϵ.$$

Choose an integer:

$$n_0>1+{\tau }^{-1}(F\left(2M_0\right)-F\left(ϵ\right)).$$

(7)

Let $n_1, n_2\ge n_0$ be integers. In view of Proposition 2, in order to show that property (P2) holds, it is enough to show that

$$\rho (T^{n_0}\left(x_{n_1}\right),T^{n_0}\left(x_{n_2}\right))\le ϵ.$$

Assume the contrary. Then,

$$\rho (T^{n_0}\left(x_{n_1}\right),T^{n_0}\left(x_{n_2}\right))>ϵ.$$

(8)

Proposition 2 and (1), (2) and (8) imply that for each $i\in \{0,\dots ,n_0-1\}$,

$$\rho (T^i\left(x_{n_1}\right),T^i\left(x_{n_2}\right))>ϵ,$$

$$F\left(\rho (T^{i+1}\left(x_{n_1}\right),T^{i+1}\left(x_{n_2}\right))\right)+\tau \le F\left(\rho (T^i\left(x_{n_1}\right),T^i\left(x_{n_2}\right))\right),$$

$$F\left(\rho (T^{n_0}\left(x_{n_1}\right),T^{n_0}\left(x_{n_2}\right))\right)\le (1-n_0)\tau +F\left(\rho (x_{n_1},x_{n_2})\right)$$

$$\le (1-n_0)\tau +F\left(2M_0\right).$$

(9)

According to (8) and (9),

$$F\left(ϵ\right)\le \tau (1-n_0)+F\left(2M_0\right),n_0\le 1+{\tau }^{-1}(F\left(2M_0\right)-F\left(ϵ\right)).$$

This contradicts (7). Therefore, (P2) holds.

We will prove that the following property is fulfilled:

(P3) $n_{*}\in \mathbf{N}$ is found, such that for each triplet $i_1,i_2,n\in \mathbf{N}$ satisfying $n_{*}\le i_1,i_2\le n$,

$$\rho (T^{i_1}\left(x_n\right),T^{i_2}\left(x_n\right))\le ϵ.$$

According to Proposition 1, we may assume that F is continuous at $ϵ$. Assume that property (P3) does not hold. Then, for any $k\in \mathbf{N}$, there are integers $i_{k,1}, i_{k_2}, n_k$ for which

$$k\le i_{k,1}<i_{k,2}\le n_k,$$

(10)

$$\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))>ϵ.$$

(11)

In view of property (P1), we may assume that for every integer $k\ge 1$,

$$\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,1}+1}\left(x_{n_k}\right))\le ϵ.$$

(12)

Let $k\ge 1$. According to (11) and (12),

$$i_{k,1}+1<i_{k,2}$$

and we may assume that

$$\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}-1}\left(x_{n_k}\right))\le ϵ.$$

(13)

Equations (1) and (13) imply that

$$ϵ<\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))$$

$$\le \rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}-1}\left(x_{n_k}\right))+\rho (T^{i_{k,2}-1}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))$$

$$\le ϵ+\rho (T^{i_{k,2}-1}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right)).$$

(14)

Property (P1) and (10), (14) imply that

$$\underset{k\to \infty }{lim}\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))=ϵ.$$

(15)

Clearly, for each $k\in \mathbf{N}$,

$$\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))\le \rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,1}+1}\left(x_{n_k}\right))$$

$$+\rho (T^{i_{k,1}+1}\left(x_{n_k}\right),T^{i_{k,2}+1}\left(x_{n_k}\right))+\rho (T^{i_{k,2}}\left(x_{n_k}\right),T^{i_{k,2}+1}\left(x_{n_k}\right)).$$

(16)

According to (1) and (11), for each integer $k\ge 1$,

$$F\left(\rho (T^{i_{k,1}+1}\left(x_{n_k}\right),T^{i_{k,2}+1}\left(x_{n_k}\right))\right)\le F\left(\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))\right)-\tau .$$

(17)

In view of Proposition 2 and (15), we may assume that there exists

$$\Delta =\underset{k\to \infty }{lim}\rho (T^{i_{k,1}+1}\left(x_{n_k}\right),T^{i_{k,2}+1}\left(x_{n_k}\right)).$$

(18)

It follows from (15), (17) and (18) that

$$0\le \Delta \le ϵ.$$

(19)

Property (P1) and (15), (18) and (19) imply that

$$\Delta =ϵ$$

(20)

and

$$ϵ=\underset{k\to \infty }{lim}\rho (T^{i_{k,1}+1}\left(x_{n_k}\right),T^{i_{k,2}+1}\left(x_{n_k}\right)).$$

(21)

Since the function F is continuous at $ϵ$, Equations (15) and (21) imply that

$$F\left(ϵ\right)=\underset{k\to \infty }{lim}F\left(\rho (T^{i_{k,1}+1}\left(x_{n_k}\right),T^{i_{k,2}+1}\left(x_{n_k}\right))\right)$$

$$=\underset{k\to \infty }{lim}F\left(\rho (T^{i_{k,1}}\left(x_{n_k}\right),T^{i_{k,2}}\left(x_{n_k}\right))\right).$$

This contradicts (17) and proves property (P3).

Let $ϵ>0$. Property (P3) implies that there exists $n_0\in \mathbf{N}$ such that for any triplet of integers $i_1, i_2, n$ satisfying $n_0\le i_1,i_2\le n$,

$$\rho (T^{i_1}\left(x_n\right),T^{i_2}\left(x_n\right))\le ϵ/4.$$

(22)

Property (P2) implies that there exists a natural number $m_0\ge n_0$ such that for any triplet of integers $i, n_1, n_2$ satisfying $n_1,n_2,i\ge m_0,i\le n_j,j=1,2$ we have

$$\rho (T^i\left(x_{n_1}\right),T^i\left(x_{n_2}\right))\le ϵ/4.$$

(23)

Assume that integers $n_1, n_2, i_1, i_2$ satisfy

$$m_0\le i_1\le n_1,m_0\le i_2\le n_2.$$

(24)

According to (22)–(24),

$$\rho (T^{i_1}\left(x_{n_1}\right),T^{i_2}\left(x_{n_1}\right))\le ϵ/4,$$

$$\rho (T^{i_1}\left(x_{n_1}\right),T^{i_2}\left(x_{n_2}\right))\le ϵ/4,$$

$$\rho (T^{i_1}\left(x_{n_1}\right),T^{i_1}\left(x_{n_2}\right))\le ϵ/4,\rho (T^{i_2}\left(x_{n_1}\right),T^{i_2}\left(x_{n_2}\right))\le ϵ/4.$$

These relations imply that

$$\rho (T^{i_1}\left(x_{n_1}\right),T^{i_2}\left(x_{n_2}\right))\le \rho (T^{i_1}\left(x_{n_1}\right),T^{i_2}\left(x_{n_1}\right))+\rho (T^{i_2}\left(x_{n_1}\right),T^{i_2}\left(x_{n_2}\right))\le ϵ/2.$$

(25)

Thus, we have shown that the following property holds:

For (P4), if integers $i_1, i_2, n_1, n_2$ satisfy (24), then (25) holds.

In view of (P4), the sequences ${\left\{T^{n-1}\left(x_n\right)\right\}}_{n=1}^{\infty }$ and ${\left\{T^{n-2}\left(x_n\right)\right\}}_{n=2}^{\infty }$ converge and there exists

$$x_{*}=\underset{n\to \infty }{lim}{T}^{n-1}\left(x_n\right)=\underset{n\to \infty }{lim}{T}^{n-2}\left(x_n\right).$$

In view of Proposition 2, $x_{*}=T\left(x_{*}\right)$. Theorem 1 is proved. □

The following example illustrates Theorem 1. Assume that X is the collection of all continuous functions on $[0,1]$,

$$\rho (f_1,f_2)=sup\left\{\right|f_1\left(t\right)-f_2\left(t\right)|:t\in [0,1]\},f_1,f_2\in X,$$

$$K=\{f\in X:f([0,1])\subset [0,1\left]\right\}$$

and

$$T\left(f\right)=2^{-1}f+2/3,f\in K.$$

Clearly, T is a Wardowski contraction with $F\left(t\right)=ln\left(t\right),t>0$ and $\tau =ln\left(2\right).$ The formula above defines T for all $x\in X,$ but we consider T to be a mapping from K to X. Evidently, T does not have a fixed point in T. In view of Theorem 1, $n\in \mathbf{N}$ can be found, such that $T^n\left(x\right)\notin K$ for any $x\in K$. A direct calculation shows that $n=3$.

Let us consider the same mapping T with the same F and $\tau$ and

$$K=\{f\in X:f([0,1])\subset [0,2\left]\right\}\cup \{f\in X:f([0,1])\subset [10,11\left]\right\}.$$

It is not difficult to see that

$$T\left(K\right)\not\subset K$$

but T has a fixed point in K.

3. The Second Result

Assume that $(X,\rho )$ is a complete metric space, $T:X\to 2^X\setminus \{Ø\}$; for each $x\in X$, the set $T\left(x\right)$ is closed and a function

$$\varphi :[0,\infty )\to [0,\infty )$$

satisfies

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

(26)

In [17], it was shown that the set-valued T has a fixed point if for each $(x,y)\in X^2$ and each $u\in T\left(x\right)$ there is $v\in T\left(y\right)$, such that

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

Examples of such mappings are considered in [17]. Here, we show that T possesses a fixed point under a weaker assumption. Namely, we assume that the following assumption holds:

(A) For each $x,y\in X$ and each $u\in T\left(x\right)$,

$$inf\left\{\rho \right(u,v)+\varphi (\rho (u,v)):v\in T(y\left)\right\}\le \varphi (x,y).$$

(27)

Theorem 2.

1. Assume that ${\left\{{ϵ}_i\right\}}_{i=0}^{\infty }\subset (0,\infty )$,

$$\sum _{i=0}^{\infty }{ϵ}_i<\infty ,$$

(28)

${\left\{x_i\right\}}_{i=0}^{\infty }\subset X$, for each $i\in \mathbf{N}\cup \left\{0\right\}$,

$$x_{i+1}\in T\left(x_i\right),$$

(29)

$$\rho (x_{i+1},x_{i+2})+\varphi \left(\rho (x_{i+1},x_{i+2})\right)\le \varphi \left(\rho (x_i,x_{i+1})\right)+{ϵ}_i.$$

(30)

(This sequence exists according to assumption (A)). Then, the sequence ${\left\{x_i\right\}}_{i=0}^{\infty }$ converges to a fixed point of T.

2. Let $ϵ>0$. Then, $\delta \in (0,ϵ)$ can be found such that for each $x_0\in X$ satisfying $\rho (x_0,T\left(x_0\right))<\delta$, there exists $x_{*}\in B(x_0,ϵ)$ such that $x_{*}\in T\left(x_{*}\right)$.

Proof. 

Let us prove Assertion 1. According to (30), for each integer $i\ge 0$,

$$\rho (x_{i+1},x_{i+2})\le \varphi \left(\rho (x_i,x_{i+1})\right)-\varphi \left(\rho (x_{i+1},x_{i+2})\right)+{ϵ}_i,$$

(31)

$$\varphi \left(\rho (x_{i+1},x_{i+2})\right)\le \varphi \left(\rho (x_i,x_{i+1})\right)+{ϵ}_i,$$

(32)

$$\varphi \left(\rho (x_{i+1},x_{i+2})\right)\le \varphi \left(\rho (x_0,x_1)\right)+\sum _{j=0}^{\infty }{ϵ}_j.$$

(33)

Let $m>n\ge 1$ be integers. In view of (30),

$$\rho (x_n,x_m)\le \sum _{i=n-1}^{m-1}\rho (x_{i+1},x_{i+2})$$

$$\le \sum _{i=n-1}^{m-1}(\varphi \left(\rho (x_i,x_{i+1})\right)-\varphi \left(\rho (x_{i+1},x_{i+2})\right)+{ϵ}_i)$$

$$=\varphi \left(\rho (x_{n-1},x_n)\right)-\varphi \left(\rho (x_{m-1},x_m)\right)+\sum _{i=n-1}^{m-1}{ϵ}_i.$$

(34)

We claim that there exists

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

According to (33), the sequence ${\left\{\varphi \left(\rho (x_n,x_{n+1})\right)\right\}}_{n=0}^{\infty }$ is bounded. Let

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

(35)

We claim that

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

Let $ϵ>0$. According to (35), there exists a natural number $n_0$ such that for each integer $n\ge n_0$,

$$\varphi \left(\rho (x_n,x_{n+1})\right)\ge r-ϵ/8.$$

(36)

Equations (28) and (35) imply that there exists a natural number $n_1>n_0$ such that

$$\sum _{i=n_1}^{\infty }{ϵ}_i<ϵ/8,$$

(37)

$$\varphi \left(\rho (x_{n_1},x_{n_1+1})\right)<r+ϵ/8.$$

(38)

According to (32), (37) and (38), for any integer $n>n_1$,

$$\varphi \left(\rho (x_n,x_{n+1})\right)\le \varphi \left(\rho (x_{n_1},x_{n_1+1})\right)+\sum _{i=n_1}^{\infty }{ϵ}_i<r+ϵ/8+ϵ/8.$$

Thus, for any integer $n>n_1$,

$$|\varphi \left(\rho (x_n,x_{n+1})\right)-r|<ϵ/4$$

(39)

and

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

(40)

According to (34), (37) and (39), for any pair of integers $m>n>n_1$,

$$\rho (x_n,x_m)\le \varphi \left(\rho (x_{n-1},x_n)\right)-\varphi \left(\rho (x_{m-1},x_m)\right)+\sum _{i=n_1}^{\infty }{ϵ}_i<ϵ.$$

Thus, ${\left\{x_n\right\}}_{n=0}^{\infty }$ is a Cauchy sequence and there exists

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

Assumption (A) and (29) imply that for any $n\in \mathbf{N}$, there exists

$$v_n\in T\left(x_{*}\right)$$

such that

$$\rho (x_{n+1},v_n)+\varphi \left(\rho (x_{n+1},v_n)\right)\le \varphi \left(\rho (x_n,x_{*})\right)+{ϵ}_n\to 0 \mathrm{as} n\to \infty .$$

According to (26), (28) and (40),

$$\underset{n\to \infty }{lim}\rho (x_{n+1},v_n)=0,\underset{n\to \infty }{lim}{v}_n=x_{*}$$

and $x_{*}\in T\left(x_{*}\right)$. Assertion 1 is proved.

Let us proved Assertion 2. According to (26), there exists $\delta \in (0,ϵ)$ such that for each $t\in [0,\delta ]$,

$$\varphi \left(t\right)<ϵ/4.$$

(41)

Assume that $x_0\in X$ satisfies

$$\rho (x_0,T\left(x_0\right))<\delta .$$

(42)

Choose ${\left\{{ϵ}_i\right\}}_{i=0}^{\infty }\subset (0,1)$ such that

$$\sum _{i=0}^{\infty }{ϵ}_i<ϵ/4.$$

(43)

In view of (42), there is

$$x_1\in T\left(x_0\right)$$

for which

$$\rho (x_0,x_1)<\delta .$$

Assume that a sequence ${\left\{x_i\right\}}_{i=2}^{\infty }$ is as in Assertion 1. Then, there exists

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

such that

$$x_{*}\in T\left(x_{*}\right)$$

and in view of (34) and (41)–(43),

$$\sum _{i=1}^{\infty }\rho (x_i,x_{i+1})\le \varphi \left(\rho (x_0,x_1)\right)+\sum _{i=0}^{\infty }{ϵ}_i$$

$$\le \varphi \left(\rho (x_0,x_1)\right)+ϵ/4<ϵ/2,$$

$$\sum _{i{=}_1}^{\infty }\rho (x_i,x_{i+1})<ϵ,\rho (x_{*},x_0)<ϵ.$$

Assertion 2 is proved. □

Theorem 3.

Assume that the function ϕ is bounded, $ϵ\in (0,1)$,

$$M_1>\varphi \left(t\right),t\in [0,\infty ),$$

(44)

$$n_0>2+2{ϵ}^{-1}{M}_1$$

(45)

is an integer and that ${\left\{x_i\right\}}_{i=0}^{n_0}\subset X$ satisfies for each integer $i\in \{0,\dots ,n_0-1\}$,

$$x_{i+1}\in T\left(x_i\right)$$

(46)

and for any $i\in \{0,\dots ,n_0-2\}$,

$$\rho (x_{i+1},x_{i+2})+\varphi \left(\rho (x_{i+1},x_{i+2})\right)\le \varphi \left(\rho (x_i,x_{i+1})\right)+ϵ/2.$$

(47)

Then, there exists $j\in \{1,\dots ,n_0-1\}$, for which

$$\rho (x_j,x_{j+1})\le ϵ,B(x_j,ϵ)\cap T\left(x_j\right)\ne Ø.$$

Proof. 

Assume that the theorem does not hold. Then,

$$\rho (x_j,x_{j+1})>ϵ,j\in \{1,\dots ,n_0-1\}.$$

(48)

According to (44), (47) and (48), for each $i\in \{0,\dots ,n_0-2\}$,

$$ϵ<\rho (x_{i+1},x_{i+2})\le \varphi \left(\rho (x_i,x_{i+1})\right)-\varphi \left(\rho (x_{i+1},x_{i+2})\right)+ϵ/2$$

and

$$(n_0-1)ϵ<\sum _{i=0}^{n_0-2}\rho (x_{i+1},x_{i+2})$$

$$\le 2^{-1}ϵ(n_0-1)+\sum _{i=0}^{n_0-2}\varphi \left(\rho (x_i,x_{i+1})\right)-\varphi \left(\rho (x_{i+1},x_{i+2})\right)$$

$$\le 2^{-1}ϵ(n_0-1)+\varphi \left(\rho (x_0,x_1)\right),$$

$$2^{-1}ϵ(n_0-1)\le M_1$$

and

$$n_0\le 1+2{ϵ}^{-1}{M}_1.$$

This contradicts (45) and proves Theorem 3. □

Applying Theorem 3 by induction, we can obtain the following result.

Corollary 1.

Assume that the function ϕ is bounded, $ϵ\in (0,1)$,

$$M_1>\varphi \left(t\right),t\in [0,\infty ),$$

$$n_0>2+2{ϵ}^{-1}{M}_1$$

is an integer and that a sequence ${\left\{x_i\right\}}_{i=0}^{\infty }\subset X$ satisfies for any $i\in \mathbf{N}\cup \left\{0\right\}$,

$$x_{i+1}\in T\left(x_i\right)$$

and

$$\rho (x_{i+1},x_{i+2})+\varphi \left(\rho (x_{i+1},x_{i+2})\right)\le \varphi \left(\rho (x_i,x_{i+1})\right)+ϵ/2.$$

Then, there exists ${\left\{n_k\right\}}_{k=1}^{\infty }\subset \mathbf{N}$ such that $n_1\le n_0$ and that for any $k\in \mathbf{N}$,

$$1\le n_{k+1}-n_k,\rho (x_{n_k},x_{n_k+1})\le ϵ.$$

Theorem 4.

Assume that the function ϕ is bounded, $ϵ\in (0,1)$, ${ϵ}_0\in (0,ϵ)$,

$$M_1>\varphi \left(t\right),t\in [0,\infty ),$$

$$\varphi \left(t\right)\le ϵ/2,t\in [0,{ϵ}_0),$$

(49)

$$n_0>2+2{ϵ}_0^{-1}{M}_1$$

(50)

is an integer,

$$\delta ={\left(2n_0\right)}^{-1}{ϵ}_0$$

(51)

and that a sequence ${\left\{x_i\right\}}_{i=0}^{\infty }\subset X$ satisfies for any $i\in \mathbf{N}\cup \left\{0\right\}$,

$$x_{i+1}\in T\left(x_i\right)$$

(52)

and

$$\rho (x_{i+1},x_{i+2})+\varphi \left(\rho (x_{i+1},x_{i+2})\right)\le \varphi \left(\rho (x_i,x_{i+1})\right)+\delta .$$

(53)

Then, $\rho (x_i,x_{i+1})\le ϵ$ for each integer $i\ge n_0$.

Proof. 

Corollary 1 implies that there exists ${\left\{n_k\right\}}_{k=1}^{\infty }\subset \mathbf{N}$ such that $n_1\le n_0$ and that for any $k\in \mathbf{N}$,

$$1\le n_{k+1}-n_k\le n_0,$$

(54)

$$\rho (x_{n_k},x_{n_k+1})\le {ϵ}_0.$$

(55)

Let $k\ge 1$ be an integer. According to (49), (51), (53) and (55),

$$\sum _{i=n_k}^{n_{k+1}}\rho (x_i,x_{i+1})\le \sum _{i=n_k}^{n_{k+1}}(\varphi \left(\rho (x_i,x_{i+1})\right)-\varphi \left(\rho (x_{i+1},x_{i+2})\right))+n_0\delta$$

and

$$ϵ>ϵ/2+{ϵ}_0/2\ge \varphi \left(\rho (x_{n_k},x_{n_k+1})\right)+n_0\delta \ge \sum _{i=n_k+1}^{n_{k+1}}\rho (x_i,x_{i+1}).$$

This completes the proof of Theorem 4. □

Our results from this section can be applied to the following problem, considered in [17]. Assume that $a<b$ are real numbers, X is the space $C\left(\right[a,b\left]\right)$ of all real-valued continuous functions and that

$$\rho (f_1,f_2)=sup\left\{\right|f_1\left(t\right)-f_2\left(t\right)|:t\in [a,b]\},f_1,f_2\in X.$$

We consider a Fredholm-type integral inclusion:

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

It was shown in [17] that the study of this problem is reduced to the analysis of a fixed point problem

$$T\left(z\right)=\{y\in X:y\left(\tau \right)\in {\int }_a^{b}K(\tau ,s,z\left(s\right))ds+f\left(\tau \right),\tau \in [a,b]\},z\in X$$

and that for mapping T, all the assumptions made in this section hold. Therefore, all the results can be applied for T.

4. The Third Result

Assume that $(X,\rho )$ is endowed with a graph G. Let $V\left(G\right)$ be the set of its vertices, $E\left(G\right)$ be the set of its edges and let

$$(x,x)\in E\left(G\right),x\in X.$$

Assume that $\tau >0$, $F:(0,\infty )\to R^1$ is an increasing function and that $T:X\to X$ is a mapping such that for any $(x,y)\in E\left(G\right)$ for which $x\ne y$,

$$\left(T\right(x),T(y\left)\right)\in E\left(G\right)\mathrm{and}F\left(\rho \right(T\left(x\right),T\left(y\right)\left)\right)+\tau \le F\left(\rho \right(x,y\left)\right).$$

(56)

Equation (56) implies the following proposition.

Proposition 3.

For any $(x,y)\in E\left(G\right)$, $\rho \left(T\right(x),T(y\left)\right)\le \rho (x,y)$.

Proposition 4.

Let $(x,y)\in E\left(G\right)$. Then

$$\underset{n\to \infty }{lim}\rho (T^n\left(x\right),T^n\left(y\right))=0.$$

Proof. 

We may assume that

$$T^n\left(x\right)\ne T^n\left(y\right)$$

for each integer $n\ge 0$. According to (56), for any integer $n\in \mathbf{N}\cup \left\{0\right\}$,

$$(T^n\left(x\right),T^n\left(y\right))\in E\left(G\right),$$

$$F\left(\rho (T^{n+1}\left(x\right),T^{n+1}\left(y\right))\right)+\tau \le F\left(\rho (T^n\left(x\right),T^n\left(y\right))\right),$$

$$F\left(\rho (T^n\left(x\right),T^n\left(y\right))\right)\le F\left(\rho (x,y)\right)-n\tau .$$

(57)

Proposition 3 implies that for any $n\in \mathbf{N}\cup \left\{0\right\}$,

$$\rho (T^{n+1}\left(x\right),T^{n+1}\left(y\right))\le \rho (T^n\left(x\right),T^n\left(y\right)).$$

Assume that the proposition does not hold. Then, $ϵ>0$ can be found such that for any integer $n\ge 0$,

$$ϵ\le \rho (T^n\left(x\right),T^n\left(y\right)).$$

(58)

In view of (57) and (58), for each $n\in \mathbf{N}\cup \left\{0\right\}$,

$$F\left(ϵ\right)\le F\left(\rho (T^n\left(x\right),T^n\left(y\right))\right)\le F\left(\rho (x,y)\right)-n\tau \to -\infty \mathrm{as} n\to \infty .$$

This contradiction proves Proposition 4. □

Proposition 4 implies the following result.

Proposition 5.

Let $x,y\in X$, $q\in \mathbf{N}$, ${\left\{x_i\right\}}_{i=0}^q\subset X$,

$$x_0=x,x_q=y,$$

$$(x_i,x_{i+1})\in E\left(G\right),i=0,\dots ,q-1.$$

Then,

$$\underset{n\to \infty }{lim}\rho (T^n\left(x\right),T^n\left(y\right))=0.$$

Proposition 5 implies the following result.

Theorem 5.

Assume that $x\in X$ and there are $q\in \mathbf{N}$ and points ${\left\{y_i\right\}}_{i=0}^q\subset X$ such that

$$y_0=x,y_q=T\left(x\right),$$

$$(y_i,y_{i+1})\in E\left(G\right),i=0,\dots ,q-1.$$

Then,

$$\underset{i\to \infty }{lim}\rho (T^i\left(x\right),T^{i+1}\left(x\right))=0.$$

Theorem 6.

Assume that $x\in X$ and there exist $q\in \mathbf{N}$ and ${\left\{y_i\right\}}_{i=0}^q$ such that

$$y_0=x,y_q=T\left(x\right),$$

$$(y_i,y_{i+1})\in E\left(G\right),i=0,\dots ,q-1.$$

Assume that there exists $m_0\in \mathbf{N}$ such that the following property holds:

(P) for any pair of nonnegative integers $i<j$, there is

$$p\in \{j,\dots ,j+m_0\}$$

for which

$$(T^i\left(x\right),T^p\left(x\right))\in E\left(G\right).$$

Then, the sequence ${\left\{T^n\left(x\right)\right\}}_{n=0}^{\infty }$ converges and its limit is a fixed point of T if the graph of T is closed.

Proof. 

According to Theorem 5,

$$\underset{n\to \infty }{lim}\rho (T^n\left(x\right),T^{n+1}\left(x\right))=0.$$

(59)

Let $ϵ\in (0,1)$. We show that for all sufficiently large $i,j\in \mathbf{N}$,

$$\rho (T^i\left(x\right),T^j\left(x\right))\le ϵ.$$

Since the collection of all points at which F is not continuous is countable, we may assume that the function F is continuous at $ϵ$. Choose

$$\delta \in (0,ϵ/4)$$

such that

$$\left|F\right(\xi )-F(ϵ\left)\right|\le \tau /2 \mathrm{for} \mathrm{any} \xi \in [ϵ-4\delta ,ϵ+4\delta ]$$

(60)

and set

$${\delta }_0=\delta {\left(4m_0\right)}^{-1}.$$

(61)

In view of (59), there exists a natural number $n_0$ for which

$$\rho (T^i\left(x\right),T^{i+1}\left(x\right))\le {\delta }_0\mathrm{for} \mathrm{each} \mathrm{integer} i\ge n_0.$$

(62)

Assume that

$$j>i\ge n_0$$

are integers. We show that

$$\rho (T^i\left(x\right),T^j\left(x\right))\le ϵ.$$

Assume the contrary. Then,

$$\rho (T^i\left(x\right),T^j\left(x\right))>ϵ.$$

(63)

According to (61) and (62), we may assume that for any $s\in \{i,\dots ,j-1\}$,

$$\rho (T^i\left(x\right),T^s\left(x\right))\le ϵ.$$

(64)

According to Equation (62), for any $s\in \{i+1,\dots ,i+m_0\}$,

$$\rho (T^i\left(x\right),T^s\left(x\right))\le {\delta }_0{m}_0<\delta <ϵ.$$

(65)

According to (63) and (65),

$$j>i+m_0,j-1\ge i+m_0.$$

(66)

In view of (66),

$$j-m_0>i$$

and (P) implies that there is

$$p\in \{j-m_0-1,\dots ,j-1\}$$

(67)

for which

$$(T^i\left(x\right),T^p\left(x\right))\in E\left(G\right).$$

(68)

According to (67) and (68),

$$p\ge i.$$

If

$$T^p\left(x\right)=T^i\left(x\right),$$

then according to (62) and (67),

$$\rho (T^i\left(x\right),T^j\left(x\right))\le \sum \{\rho (T^s\left(x\right),T^{s+1}\left(x\right)):s\in \{p,\dots ,j-1\}\}\le m_0{\delta }_0\le ϵ$$

and this contradicts (63). Therefore,

$$T^p\left(x\right)\ne T^i\left(x\right).$$

(69)

According to (56), (64), (67) and (69),

$$F\left(\rho (T^{i+1}\left(x\right),T^{p+1}\left(x\right))\right)+\tau \le F\left(\rho (T^i\left(x\right),T^p\left(x\right))\right)\le F\left(ϵ\right).$$

(70)

It follows from (60) and (70) that

$$\rho (T^{i+1}\left(x\right),T^{p+1}\left(x\right))\le ϵ-4\delta .$$

(71)

According to (61), (62), (67) and (71),

$$\rho (T^i\left(x\right),T^j\left(x\right))\le \rho (T^i\left(x\right),T^{i+1}\left(x\right))+\rho (T^{i+1}\left(x\right),T^{p+1}\left(x\right))$$

$$+\rho (T^{p+1}\left(x\right),T^j\left(x\right))\le {\delta }_0+ϵ-4\delta$$

$$+\sum \{\rho (T^s\left(x\right),T^{s+1}\left(x\right)):s\in \{p,\dots ,j-1\}\setminus \left\{p\right\}\}\le {\delta }_0+ϵ-4\delta +m_0{\delta }_0<ϵ.$$

This contradicts (63) and proves Theorem 6. □

Theorem 6 was proved under assumption (P). It holds in the following two cases. In the first case, X is equipped with an order ≤ and $(x,y)\in E\left(G\right)$ if and only if $x\le y$. In the second case, m is a natural number $A_i\subset X$, $i=1,\dots ,m$ are nonempty closed sets, $A_{m+1}=A_1$, ${\cup }_{i=1}^m{A}_i=X$,

$$T:{\cup }_{i=1}^m{A}_i\to {\cup }_{i=1}^m{A}_i$$

and for any $i\in \{1,\dots ,m\}$, any $a\in A_i$, any $b\in A_{i+1}$,

$$T\left(A_i\right)\subset A_{i+1},$$

$$F\left(\rho \right(T\left(a\right),T\left(b\right)\left)\right)+\tau \le F\left(\rho \right(a,b\left)\right)$$

and

$$E\left(G\right)={\cup }_{i=1}^m(A_i\times A_{i+1}).$$

In this case, T is called a cyclical operator [6].

Let us consider the following example [19]. Assume that $X=[0,\infty )$, $\rho (u,v)=|u-v|$, $u,v\in [0,\infty )$ and that $(u,v)\in E\left(G\right)$ if and only if $u\le v$ and

$$(u,v)\in [0,1]\times [0,1]\cup {\cup }_{n=1}^{\infty }(n,n+1]\times (n,n+1],$$

$$T\left(0\right)=0,T\left(u\right)=2^{-1}u+n/2,u\in (n,n+1],n=0,1,\dots .$$

It was shown in [19] that for each $(u,v)\in E\left(G\right)$,

$$\rho (T\left(u\right),T\left(v\right))\le 2^{-1}\rho (u,v).$$

It is not difficult to see that T is a Wardowski contraction with $F\left(t\right)=ln\left(t\right),t>0$ and $\tau =ln\left(2\right).$

5. Conclusions

We consider three fixed point problems, and for each of them establish the existence of a fixed point. In the first and the third cases, we consider single-valued Wardowski type contraction, while in the second case we study Cho-type set-valued contractions. In the second case, we also study approximate fixed points.