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# Discussion of several contractions by Jachymski’s approach

Fixed Point Theory and Applications20162016:91

https://doi.org/10.1186/s13663-016-0581-9

• Accepted: 8 September 2016
• Published:

## Abstract

We discuss several contractions of integral type by using Jachymski’s approach. We give alternative proofs of recent generalizations of the Banach contraction principle due to Ri (Indag. Math. 27:85-93, 2016) and Wardowski (Fixed Point Theory Appl. 2012:94, 2012).

## Keywords

• the Banach contraction principle
• Boyd-Wong contraction
• Meir-Keeler contraction
• Matkowski contraction
• contraction of integral type
• fixed point

• 47H09
• 54H25

## 1 Introduction

The Banach contraction principle [3, 4] is an elegant, forceful tool in nonlinear analysis and has many generalizations. See, e.g., . For example, Boyd and Wong in  proved the following.

### Theorem 1

(Boyd and Wong )

Let $$(X,d)$$ be a complete metric space and let T be a mapping on X. Assume that T is a Boyd-Wong contraction, that is, there exists a function φ from $$[0, \infty)$$ into itself satisfying the following:
1. (i)

φ is upper semicontinuous from the right.

2. (ii)

$$\varphi(t) < t$$ holds for any $$t \in(0, \infty)$$.

3. (iii)

$$d(Tx,Ty) \leq\varphi\circ d(x,y)$$ for any $$x,y \in X$$.

Then T has a unique fixed point.
Branciari in  introduced contractions of integral type as follows: A mapping T on a metric space $$(X,d)$$ is a Branciari contraction if there exist $$r \in[0, 1)$$ and a locally integrable function f from $$[0, \infty)$$ into itself such that
$$\int_{ 0}^{ s} f(t)\,dt > 0 \quad\mbox{and}\quad \int_{ 0}^{ d(Tx, Ty)} f(t)\,dt \leq r \int_{ 0}^{ d(x,y)} f(t)\,dt$$
for all $$s >0$$ and $$x, y \in X$$. We have studied contractions of integral type in .

In this paper, we discuss several contractions of integral type by using Jachymski’s approach. As applications, we give alternative proofs of recent generalizations of the Banach contraction principle due to Ri  and Wardowski .

## 2 Preliminaries

Throughout this paper we denote by $$\mathbb {N}$$ the set of all positive integers and by $$\mathbb {R}$$ the set of all real numbers.

Let f be a function from a subset Q of $$\mathbb {R}$$ into $$\mathbb {R}$$. Then f is said to satisfy (UR) f if the following holds:
(UR) f

For any $$t \in Q$$, there exist $$\delta> 0$$ and $$\varepsilon> 0$$ such that $$f(s) \leq t - \varepsilon$$ holds for any $$s \in[t,t+\delta) \cap Q$$.

We give some lemmas concerning (UR).

### Lemma 2

Let f be a function from a subset Q of $$\mathbb {R}$$ into $$\mathbb {R}$$. Then the following are equivalent:
1. (i)

f satisfies (UR) f .

2. (ii)

$$\limsup[ f(u) : u \to t, u \in Q, t \leq u ] < t$$ holds for any $$t \in Q$$.

3. (iii)

$$\limsup[ f(u) : u \to t, u \in Q, t < u ] < t$$ and $$f(t) < t$$ hold for any $$t \in Q$$.

Obvious. □

### Lemma 3

Let f be a function from a subset Q of $$\mathbb {R}$$ into $$\mathbb {R}$$ such that $$f(t) < t$$ for any $$t \in Q$$. Assume that f is upper semicontinuous from the right. Then f satisfies (UR) f .

Obvious. □

### Lemma 4

Let f be a function from a subset Q of $$\mathbb {R}$$ into $$\mathbb {R}$$ satisfying (UR) f . Define a function g from Q into $$\mathbb {R}$$ by
$$g(t) = \limsup\bigl[ f(u) : u \to t, u \in Q, t \leq u \bigr]$$
for $$t \in Q$$. Define a mapping L from Q into the power set of $$\mathbb {R}$$, a function from Q into $$[-\infty,\infty)$$ and a function h from Q into $$\mathbb {R}$$ by
\begin{aligned} & L(t) = \bigl\{ s \in Q : s \leq t , \limsup\bigl[ g(u) : u \to s, u \in Q, u \leq s \bigr] = s \bigr\} , \\ & \ell(t) = \textstyle\begin{cases} \sup L(t) & \textit{if }L(t) \neq\varnothing, \\ - \infty& \textit{if }L(t) = \varnothing, \end{cases}\displaystyle \quad\textit{and} \\ & h(t) = \sup\bigl\{ g(s) : s \in Q, \ell(t) \leq s \leq t \bigr\} \end{aligned}
for $$t \in Q$$. Define a function φ from Q into $$\mathbb {R}$$ by
$$\varphi(t) = \frac{h(t)+t}{2}$$
for $$t \in Q$$. Then the following hold:
1. (i)

g is upper semicontinuous from the right.

2. (ii)

h and φ are right continuous.

3. (iii)

$$f(t) \leq g(t) \leq h(t) < \varphi(t) < t$$ holds for any $$t \in Q$$.

### Proof

Since f satisfies (UR) f , we have $$f(t) \leq g(t) < t$$ for any $$t \in Q$$. In order to show (i), we fix $$t \in Q$$ and let $$\{ t_{n} \}$$ be a strictly decreasing sequence in Q converging to t. Fix $$\varepsilon> 0$$. Then for every $$n \in \mathbb {N}$$, there exists $$s_{n} \in Q$$ satisfying $$t_{n} \leq s_{n} \leq t_{n} + 1/n$$ and $$g(t_{n}) \leq f(s_{n}) + \varepsilon$$. Since $$\{ s_{n} \}$$ converges to t, we have
$$\limsup_{n \to\infty} g(t_{n}) \leq\limsup _{n \to\infty} f(s_{n}) + \varepsilon \leq g(t) + \varepsilon.$$
Since $$\varepsilon> 0$$ is arbitrary, we obtain $$\limsup_{n} g(t_{n}) \leq g(t)$$. Therefore we have shown (i). We shall show $$h(t) < t$$ for any $$t \in Q$$. Arguing by contradiction, we assume $$h(t) \geq t$$ for some $$t \in Q$$. Then since $$g(t) < t$$, there exists a strictly increasing sequence $$\{ s_{n} \}$$ such that $$\lim_{n} s_{n} = t$$ and $$\lim_{n} g(s_{n}) = h(t)$$. Since $$g(s_{n}) < s_{n}$$ for $$n \in \mathbb {N}$$, we have $$h(t) = t$$. Therefore $$t \in L(t)$$, which implies $$h(t) = g(t) < t$$. This is a contradiction. So $$h(t) < t$$ holds. It is obvious that $$h(t) < \varphi(t) < t$$ for any $$t \in Q$$. Therefore we have shown (iii). In order to show (ii), we fix $$t \in Q$$ and $$\varepsilon> 0$$ with $$h(t) + \varepsilon< t$$. From (i), there exists $$\delta> 0$$ such that
$$g(s) \leq g(t) + \varepsilon \leq h(t) + \varepsilon< t$$
for $$s \in(t,t+\delta) \cap Q$$. Let $$\{ t_{n} \}$$ be a strictly decreasing sequence $$\{ t_{n} \}$$ in Q such that $$t_{1} < t + \delta$$ and $$\{ t_{n} \}$$ converges to t. Then we note $$\ell(t) = \ell(t_{n})$$ for $$n \in \mathbb {N}$$. So we have
\begin{aligned} h(t) &\leq h(t_{n}) \\ &= \max \bigl\{ h(t), \sup\bigl\{ g(s) : s \in Q, t < s \leq t_{n} \bigr\} \bigr\} \\ &\leq\max\bigl\{ h(t), g(t) + \varepsilon\bigr\} \\ &\leq h(t) + \varepsilon \end{aligned}
for $$n \in \mathbb {N}$$. Hence
$$h(t) \leq\liminf_{n \to\infty} h(t_{n}) \leq\limsup _{n \to\infty} h(t_{n}) \leq h(t) + \varepsilon.$$
Since $$\varepsilon> 0$$ is arbitrary, we obtain $$\lim_{n} h(t_{n}) = h(t)$$. Thus, h is right continuous. It is obvious that φ is also right continuous. We have shown (ii). □

### Remark

See Theorem 2 in . Note that the domain of h is Q. We cannot extend the domain of h to $$\bigcup [ [t,\infty) : t \in Q ]$$, considering the function f from $$(-\infty,0) \cup(0,\infty)$$ into $$\mathbb {R}$$ defined by
$$f(t) = \textstyle\begin{cases} - 2 t & \mbox{if } t < 0, \\ t/2 & \mbox{if } t > 0 . \end{cases}$$

## 3 Definitions

We list the following notation in order to simplify the statement of the results of this paper:
1. (A1)

Let D be a subset of $$(0,\infty)^{2}$$.

2. (A2)
Let θ be a function from $$(0,\infty)$$ into $$\mathbb {R}$$. Put $$\Theta= \theta ( (0,\infty) )$$ and
$$\Theta_{\leq}= \bigcup \bigl[ [t,\infty) : t \in\Theta \bigr] .$$

Jachymski in  discussed several contractions by using subsets of $$[0,\infty)^{2}$$. Since this approach seems to be very reasonable for considering future studies, we use an approach similar to Jachymski’s.

### Definition 5

Assume (A1).
1. (1)

D is said to be contractive (Cont for short) [3, 4] if there exists $$r \in(0,1)$$ such that $$u \leq r t$$ holds for any $$(t,u) \in D$$.

2. (2)
D is said to be a Browder (Bro, for short)  if there exists a function φ from $$(0, \infty)$$ into itself satisfying the following:
1. (2-i)

φ is nondecreasing and right continuous.

2. (2-ii)

$$\varphi(t) < t$$ holds for any $$t \in(0, \infty)$$.

3. (2-iii)

$$u \leq\varphi(t)$$ holds for any $$(t,u) \in D$$.

3. (3)
D is said to be Boyd-Wong (BW for short)  if there exists a function φ from $$(0, \infty)$$ into itself satisfying the following:
1. (3-i)

φ is upper semicontinuous from the right.

2. (3-ii)

$$\varphi(t) < t$$ holds for any $$t \in(0, \infty)$$.

3. (3-iii)

$$u \leq\varphi(t)$$ holds for any $$(t,u) \in D$$.

4. (4)

D is said to be Meir-Keeler (MK for short)  if for any $$\varepsilon> 0$$, there exists $$\delta> 0$$ such that $$u < \varepsilon$$ holds for any $$(t,u) \in D$$ with $$t < \varepsilon+ \delta$$; see also .

5. (5)
D is said to be Matkowski (Mat for short)  if there exists a function φ from $$(0, \infty)$$ into itself satisfying the following:
1. (5-i)

φ is nondecreasing.

2. (5-ii)

$$\lim_{n} \varphi^{n}(t) = 0$$ for every $$t \in(0, \infty)$$.

3. (5-iii)

$$u \leq\varphi(t)$$ holds for any $$(t,u) \in D$$.

6. (6)
D is said to be CJM [6, 2224] if the following hold:
1. (6-i)

For any $$\varepsilon> 0$$, there exists $$\delta> 0$$ satisfying $$u \leq\varepsilon$$ holds for any $$(t,u) \in D$$ with $$t < \varepsilon+ \delta$$.

2. (6-ii)

$$u < t$$ holds for any $$(t,u) \in D$$.

### Remark

We know the following implications; see, e.g., [5, 7, 10].
• Cont Bro BW MK CJM;

• Cont Bro Mat CJM.

We give one proposition on the concept of Boyd-Wong. Note that we can easily obtain similar results on the other concepts.

### Proposition 6

Let T be a mapping on a metric space $$(X,d)$$ and define a subset D of $$(0,\infty)^{2}$$ by
$$D = \bigl\{ \bigl( d(x,y), d(Tx,Ty) \bigr) : x, y \in X \bigr\} \cap(0,\infty)^{2}.$$
(1)
Then T is a Boyd-Wong contraction iff D is Boyd-Wong.

### Proof

We first note
\begin{aligned} D &= \bigl\{ \bigl( d(x,y), d(Tx,Ty) \bigr) : x, y \in X, x \neq y, Tx \neq Ty \bigr\} \\ &= \bigl\{ \bigl( d(x,y), d(Tx,Ty) \bigr) : x, y \in X, Tx \neq Ty \bigr\} \end{aligned}
because $$Tx \neq Ty$$ implies $$x \neq y$$. We assume that D is Boyd-Wong. Then there exists φ satisfying (3-i)-(3-iii) in Definition 5. Define a function η from $$[0,\infty)$$ into itself by $$\eta(0) = 0$$ and $$\eta(t) = \varphi(t)$$ for $$t \in(0,\infty)$$. Then we have $$(\mathrm{i})_{\eta}$$ and $$(\mathrm{ii})_{\eta}$$ in Theorem 1. If either $$x=y$$ or $$Tx=Ty$$ holds, then $$d(Tx,Ty) \leq\eta\circ d(x,y)$$ obviously holds. Considering this fact, we have $$(\mathrm{iii})_{\eta}$$ in Theorem 1. Therefore T is a Boyd-Wong contraction. Conversely, we next assume that T is a Boyd-Wong contraction. Then there exists η satisfying $$(\mathrm{i})_{\eta}$$-$$(\mathrm{iii})_{\eta}$$ in Theorem 1. Define a function φ from $$(0,\infty)$$ into itself by
$$\varphi(t) = \max\bigl\{ \eta(t), t/2 \bigr\}$$
for any $$t \in(0,\infty)$$. Then φ satisfies (3-i) and (3-ii) in Definition 5. We also have
$$d(Tx,Ty) \leq\eta\circ d(x,y) \leq\varphi\circ d(x,y)$$
for any $$x,y \in X$$ with $$Tx \neq Ty$$. So (3-iii) holds. Therefore D is Boyd-Wong. □

The following are variants of Corollaries 9 and 14 in .

### Proposition 7

()

Assume (A1), (A2) and the following:
1. (i)

θ is nondecreasing and continuous.

2. (ii)

There exists an upper semicontinuous function ψ from Θ into $$\mathbb {R}$$ satisfying $$\psi(\tau) < \tau$$ for any $$\tau\in\Theta$$ and $$\theta(u) \leq\psi\circ\theta(t)$$ for any $$(t,u) \in D$$.

Then D is Browder.

### Proposition 8

()

Assume (A1), (A2), and the following:
1. (i)

θ is nondecreasing.

2. (ii)

There exists an upper semicontinuous function ψ from $$\Theta_{\leq}$$ into $$\mathbb {R}$$ satisfying $$\psi(\tau) < \tau$$ for any $$\tau\in\Theta_{\leq}$$ and $$\theta(u) \leq\psi\circ\theta(t)$$ for any $$(t,u) \in D$$.

Then D is CJM.

### Remark

From the proof in , we can weaken (ii) of Proposition 8 to the following:
(ii)′:

There exists a function ψ from $$\Theta_{\leq}$$ into $$\mathbb {R}$$ such that ψ is upper semicontinuous from the right, $$\psi(\tau) < \tau$$ for any $$\tau\in\Theta_{\leq}$$ and $$\theta(u) \leq\psi\circ\theta(t)$$ for any $$(t,u) \in D$$.

## 4 Main results

In this section, we prove our main results. We begin with Boyd-Wong.

### Proposition 9

Assume (A1), (A2), and the following:
1. (i)

θ is nondecreasing and continuous.

2. (ii)

There exists a function ψ from Θ into $$\mathbb {R}$$ satisfying $$(\mathrm{UR})_{\psi}$$ and $$\theta(u) \leq\psi\circ\theta(t)$$ for any $$(t,u) \in D$$.

Then D is Boyd-Wong.

### Proof

Define a function $$\theta_{+}^{-1}$$ from $$\mathbb {R}$$ into $$[0, \infty]$$ by
$$\theta_{+}^{-1}(\tau) = \textstyle\begin{cases} \sup\{ s \in(0,\infty) : \theta(s) \leq\tau\} & \mbox{if } \{ s \in(0,\infty) : \theta(s) \leq\tau\} \neq\varnothing, \\ 0 & \mbox{otherwise}. \end{cases}$$
We also define a function η from $$(0,\infty)$$ into $$[0,\infty)$$ by $$\eta= \theta_{+}^{-1} \circ\psi\circ\theta$$. We note
$$\eta(t) = \sup\bigl\{ s \in(0,\infty) : \theta(s) \leq\psi\circ\theta(t) \bigr\} \quad\mbox{provided } \eta(t) > 0 .$$
Since $$\psi(\tau) < \tau$$ for any $$\tau\in\Theta$$, we have $$\psi\circ\theta(t) < \theta(t) \leq\theta(s)$$ for any $$t, s \in(0,\infty)$$ with $$t \leq s$$. Hence $$\eta(t) \leq t$$ holds for any $$t \in(0,\infty)$$. Arguing by contradiction, we assume that $$(\mathrm{UR})_{\eta}$$ does not hold. Then there exist $$t \in(0,\infty)$$ and a sequence $$\{ t_{n} \}$$ in $$[t,\infty)$$ such that $$\{ t_{n} \}$$ converges to t and
$$\eta(t_{n}) > (1 - 1/n) t$$
holds for $$n \in \mathbb {N}$$. Since $$\eta(t_{n}) > 0$$,
$$\sup\bigl\{ s \in(0,\infty) : \theta(s) \leq\psi\circ\theta(t_{n}) \bigr\} = \eta(t_{n}) > (1-1/n) t$$
holds. Hence there exists a sequence $$\{ u_{n} \}$$ in $$(0,\infty)$$ satisfying
$$\theta(u_{n}) \leq\psi\circ\theta(t_{n}) < \theta(t_{n}) \quad\mbox{and}\quad u_{n} > (1-2/n) t$$
for $$n \in \mathbb {N}$$. Since θ is nondecreasing, $$u_{n} < t_{n}$$ holds for any $$n \in \mathbb {N}$$. Thus $$\{ u_{n} \}$$ also converges to t. Hence by the continuity of θ,
$$\theta(t) \leq\limsup_{n \to\infty} \psi\circ\theta(t_{n}) \leq\limsup\bigl[ \psi(\tau) : \tau\to\theta(t), \tau\geq\theta(t), \tau\in\Theta \bigr] .$$
This contradicts $$(\mathrm{UR})_{\psi}$$. Therefore $$(\mathrm{UR})_{\eta}$$ holds. For any $$(t,u) \in D$$, since $$\theta(u) \leq\psi\circ\theta(t)$$, we have
$$u \leq\theta_{+}^{-1} \circ\theta(u) \leq\theta_{+}^{-1} \circ\psi \circ\theta(t) = \eta(t) .$$
By Lemma 4, there exists a right continuous function φ from $$(0,\infty)$$ into itself satisfying $$\eta(t) < \varphi(t) < t$$. It is obvious that $$u \leq\eta(t) < \varphi(t)$$ for any $$(t,u) \in D$$. Therefore D is Boyd-Wong. □

### Remark

There appears $$\theta_{+}^{-1}$$ in Proposition 2.1 in .

We next discuss Meir-Keeler.

### Proposition 10

Assume (A1), (A2), and the following:
1. (i)

θ is nondecreasing and right continuous.

2. (ii)

For any $$\varepsilon\in\Theta$$, there exists $$\delta> 0$$ such that $$\theta(t) < \varepsilon+ \delta$$ implies $$\theta(u) < \varepsilon$$ for any $$(t,u) \in D$$.

Then D is Meir-Keeler.

### Proof

Fix $$\varepsilon> 0$$. Then from (ii), there exists $$\alpha> 0$$ such that
$$\theta(t) < \theta(\varepsilon) + \alpha \quad\mbox{implies}\quad \theta(u) < \theta(\varepsilon)$$
for any $$(t,u) \in D$$. From the right continuity of θ, there exists $$\delta> 0$$ such that $$\theta(\varepsilon+ \delta) < \theta(\varepsilon) + \alpha$$. Fix $$(t,u) \in D$$ with $$t < \varepsilon+ \delta$$. Then we have
$$\theta(t) \leq\theta( \varepsilon+ \delta) < \theta(\varepsilon) + \alpha$$
and hence $$\theta(u) < \theta(\varepsilon)$$. Therefore $$u < \varepsilon$$ holds. So D is Meir-Keeler. □

We obtain the following, which is a generalization of Corollary 17 in .

### Corollary 11

Assume (A1), (A2), (i) of Proposition  10, and (ii) of Proposition  9. Then D is Meir-Keeler.

Let us discuss Matkowski.

### Proposition 12

Assume (A1), (A2), and the following:
1. (i)

θ is nondecreasing and left continuous.

2. (ii)

minΘ does not exist.

3. (iii)
There exist a subset Q of $$\mathbb {R}$$ and a nondecreasing function ψ from Q into Q satisfying $$\Theta\subset Q \subset\Theta_{\leq}$$,
$$\lim_{n \to\infty} \psi^{n} (\tau) = \inf\Theta$$
for any $$\tau\in Q$$ and $$\theta(u) \leq\psi\circ\theta(t)$$ for any $$(t,u) \in D$$.

Then D is Matkowski.

### Proof

We first note that $$\inf\Theta= \inf Q = \inf\Theta_{\leq}$$ holds and neither minΘ, minQ nor $$\min\Theta_{\leq}$$ does exist. So, from (ii) and (iii), $$\psi(\tau) < \tau$$ holds for any $$\tau\in Q$$. Define a function $$\theta_{+}^{-1}$$ from Q into $$(0,\infty]$$ by
$$\theta_{+}^{-1}(\tau) = \sup\bigl\{ s \in(0,\infty) : \theta(s) \leq\tau \bigr\} .$$
Since θ is left continuous, we have $$\tau< \theta(t)$$ implies $$\theta_{+}^{-1}(\tau) < t$$. We also have
$$\theta_{+}^{-1}(\tau) = \max\bigl\{ s \in(0,\infty) : \theta(s) \leq\tau \bigr\}$$
provided $$\tau< \sup\Theta$$. Hence $$\theta\circ\theta_{+}^{-1}(\tau) \leq\tau$$ provided $$\tau< \sup\Theta$$. It is obvious that $$\theta_{+}^{-1}$$ is nondecreasing. Define a function φ from $$(0,\infty)$$ into itself by $$\varphi= \theta_{+}^{-1} \circ\psi\circ\theta$$. Then for any $$t \in(0, \infty)$$, since $$\psi\circ\theta(t) < \theta(t)$$, we have $$\varphi(t) < t$$. Since θ, ψ, and $$\theta_{+}^{-1}$$ are nondecreasing, φ is also nondecreasing. Noting $$\psi\circ\theta(t) < \theta(t) \leq\sup\Theta$$, we have
$$\varphi^{2} (t) = \theta_{+}^{-1} \circ\psi\circ\theta \circ \theta_{+}^{-1} \circ\psi\circ\theta(t) \leq\theta_{+}^{-1} \circ \psi^{2} \circ\theta(t) .$$
Continuing this argument, we can prove $$\varphi^{n} (t) \leq\theta_{+}^{-1} \circ\psi^{n} \circ\theta(t)$$ by induction. Since $$\lim_{n} \psi^{n} \circ\theta(t) = \inf\Theta$$, we have $$\lim_{n} \theta_{+}^{-1} \circ\psi^{n} \circ\theta(t) = 0$$ from (ii). Therefore we obtain
$$\lim_{n \to\infty} \varphi^{n} (t) \leq\lim _{n \to\infty} \theta_{+}^{-1} \circ\psi^{n} \circ \theta(t) = 0$$
for any $$t \in(0, \infty)$$. Since $$u \leq\theta_{+}^{-1} \circ\theta(u) \leq\theta_{+}^{-1} \circ\psi\circ\theta(t)$$, we obtain $$u \leq\varphi(t)$$ for any $$(t,u) \in D$$. Therefore D is Matkowski. □

## 5 Counterexamples

In this section, we give counterexamples connected with the results in Section 4.

### Example 13

(Example 2.3 in , Example 10 in )

Define a complete metric space $$(X, d)$$ by
$$X = [0, 1] \cup[2, \infty) \quad\mbox{and}\quad d(x,y) = \textstyle\begin{cases} \min\{ x + y, 2 \} & \mbox{if } x \neq y, \\ 0 & \mbox{if } x = y . \end{cases}$$
Define a mapping T on X and functions θ and ψ from $$(0, \infty)$$ into itself by
$$Tx = \textstyle\begin{cases} 0 & \mbox{if } x \leq1, \\ 1 - 1/x & \mbox{if } x \geq2, \end{cases}\displaystyle \qquad \theta(t) = \textstyle\begin{cases} t/2 & \mbox{if } t < 2, \\ 2 & \mbox{if } t \geq2, \end{cases}$$
and $$\psi(t) = t / 2$$. Define D by (1). Then all the assumptions of Propositions 9 and 12 except the left continuity of θ are satisfied. However, D is neither Boyd-Wong nor Matkowski.

### Remark

By Corollary 11, D is Meir-Keeler. We define E by
$$E = \bigl\{ \bigl( \theta\circ d(x,y), \theta\circ d(Tx,Ty) \bigr) : x, y \in X \bigr\} \cap(0,\infty)^{2} .$$
(2)
Then $$E \subset \{ 2 \} \times (1/4,1)$$ holds. Hence E is contractive.

### Proof

We have
\begin{aligned} D &\supset \bigl\{ \bigl( d(x,y), d(Tx,Ty) \bigr) : x, y \geq2, x \neq y \bigr\} \\ &= \bigl\{ ( 2, 2-1/x-1/y ) : x, y \geq2, x \neq y \bigr\} \\ &= \{ 2 \} \times(1,2) . \end{aligned}
Hence D is neither Boyd-Wong nor Matkowski. □

### Example 14

(Example 2.6 in , Example 11 in )

Define a complete metric space $$(X, d)$$ by $$X = [0, \infty)$$ and $$d(x,y) = x + y$$ for $$x, y \in X$$ with $$x \neq y$$. Define a mapping T on X and functions θ and ψ from $$(0, \infty)$$ into itself by
$$Tx = \textstyle\begin{cases} 0 & \mbox{if } x \leq1, \\ 1 & \mbox{if } x > 1, \end{cases}\displaystyle \qquad \theta(t) = \textstyle\begin{cases} t & \mbox{if } t \leq1, \\ 2 & \mbox{if } t > 1, \end{cases}$$
and $$\psi(t) = t / 2$$. Define D by (1). Then all the assumptions of Proposition 10 except the right continuity of θ are satisfied. However, D is not Meir-Keeler. Therefore D is not Boyd-Wong.

### Remark

By Proposition 12, D is Matkowski. We define E by (2). Then $$E = \{ (2,1) \}$$ holds. Hence E is contractive.

### Proof

We have
\begin{aligned} D &\supset \bigl\{ \bigl( d(0,y), d(T0,Ty) \bigr) : y > 1 \bigr\} \\ &= \bigl\{ ( y, 1 ) : y > 1 \bigr\} = (1,\infty) \times\{ 1 \} . \end{aligned}
Hence D is not Meir-Keeler. □

### Example 15

Define a complete metric space $$(X, d)$$ by $$X = \{ 0, 1 \}$$ and $$d(0,1) = 1$$. Define a mapping T on X and functions θ and ψ from $$(0, \infty)$$ into itself by
$$Tx = 1 - x \quad\mbox{and}\quad \theta(t) = \psi(t) = 1 .$$
Define D by (1). Then all the assumptions of Proposition 12 except (ii) are satisfied. However, D is not Matkowski.

Obvious. □

## 6 Applications

In this section, as applications, we give alternative proofs of some recent generalizations of the Banach contraction principle. Ri in  proved the following fixed point theorem.

### Theorem 16

(Ri )

Let $$(X,d)$$ be a complete metric space and let T be a mapping on X. Assume there exists a function ψ from $$[0,\infty)$$ into itself satisfying the following:
1. (R1)

$$\psi(t) < t$$ for any $$t \in(0,\infty)$$.

2. (R2)

$$\limsup_{s \to t+0} \psi(s) < t$$ for any $$t \in(0,\infty)$$.

3. (R3)

$$d(Tx, Ty) \leq\psi ( d(x, y) )$$ for any $$x, y \in X$$.

Then T has a unique fixed point.

We give an alternative proof of Theorem 16 by showing that a mapping T in Theorem 16 is a Boyd-Wong contraction.

### Proof of Theorem 16

By Lemma 2, the restriction ψ to $$(0,\infty)$$ satisfies $$(\mathrm{UR})_{\psi}$$. Then by Lemma 4, there exists a right continuous function φ from $$(0,\infty)$$ into itself satisfying $$\psi(t) < \varphi(t) < t$$ for $$t \in(0,\infty)$$. Thus T is a Boyd-Wong contraction. So T has a unique fixed point. □

Wardowski in  proved a fixed point theorem on F-contraction.

### Theorem 17

(Wardowski )

Let $$(X,d)$$ be a complete metric space and let T be a F-contraction on X, that is, there exist a function F from $$(0,\infty)$$ into $$\mathbb {R}$$ and real numbers $$\eta\in(0,\infty)$$ and $$k \in(0,1)$$ satisfying the following:
1. (F1)

F is strictly increasing.

2. (F2)

For any sequence $$\{ \alpha_{n} \}$$ of positive numbers, $$\lim_{n} \alpha_{n} = 0$$ iff $$\lim_{n} F(\alpha_{n})=-\infty$$.

3. (F3)

$$\lim_{t \to+0} t^{k} F(t) = 0$$ holds.

4. (F4)
If $$Tx \neq Ty$$, then
$$F \bigl( d(Tx,Ty) \bigr) \leq F \bigl( d(x,y) \bigr) - \eta$$
holds.

Then T has a unique fixed point.

### Remark

By (F1), we note that (F2) is equivalent to the following:
(F2)′:

$$\lim_{t \to+0} F(t) = - \infty$$ holds.

We give an alternative proof of Theorem 17 by showing that mappings satisfying (F1) and (F4) are CJM contractions.

### Proof of Theorem 17

Define a subset D of $$(0,\infty)^{2}$$ by (1). Define θ and ψ by $$\theta= F$$ and $$\psi(\tau) = \tau- \eta$$. Then all the assumptions of Proposition 8 hold. So, by Proposition 8, D is CJM. Therefore T has a unique fixed point. □

### Remark

We assume (F4) and that F is nondecreasing instead of (F1)-(F4). Then D defined by (1) is CJM. Moreover, the following hold:
• If we assume additionally that F is right continuous, then D is Meir-Keeler by Corollary 11.

• If we assume additionally that F is left continuous, then D is Matkowski by Proposition 12.

• If we assume additionally that F is continuous, then D is Browder by Proposition 7.

## Declarations

### Acknowledgements

The author is supported in part by JSPS KAKENHI Grant Number 16K05207 from Japan Society for the Promotion of Science. 