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1 Introduction
The Banach contraction principle [3, 4] is an elegant, forceful tool in nonlinear analysis and has many generalizations. See, e.g., [5–10]. For example, Boyd and Wong in [11] proved the following.
Theorem 1
(Boyd and Wong [11])
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:
-
(i)
φ is upper semicontinuous from the right.
-
(ii)
\(\varphi(t) < t\) holds for any \(t \in(0, \infty)\).
-
(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 [12] 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
for all \(s >0\) and \(x, y \in X\). We have studied contractions of integral type in [13–15].
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 [1] and Wardowski [2].
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:
-
(i)
f satisfies (UR) f .
-
(ii)
\(\limsup[ f(u) : u \to t, u \in Q, t \leq u ] < t \) holds for any \(t \in Q\).
-
(iii)
\(\limsup[ f(u) : u \to t, u \in Q, t < u ] < t \) and \(f(t) < t \) hold for any \(t \in Q\).
Proof
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 .
Proof
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
for \(t \in Q\). Define a mapping L from Q into the power set of \(\mathbb {R}\), a function ℓ from Q into \([-\infty,\infty)\) and a function h from Q into \(\mathbb {R}\) by
for \(t \in Q\). Define a function φ from Q into \(\mathbb {R}\) by
for \(t \in Q\). Then the following hold:
-
(i)
g is upper semicontinuous from the right.
-
(ii)
h and φ are right continuous.
-
(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
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
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
for \(n \in \mathbb {N}\). Hence
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 [7]. 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
3 Definitions
We list the following notation in order to simplify the statement of the results of this paper:
-
(A1)
Let D be a subset of \((0,\infty)^{2}\).
-
(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 [8] 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)
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)
D is said to be a Browder (Bro, for short) [16] if there exists a function φ from \((0, \infty)\) into itself satisfying the following:
-
(2-i)
φ is nondecreasing and right continuous.
-
(2-ii)
\(\varphi(t) < t\) holds for any \(t \in(0, \infty)\).
-
(2-iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in D\).
-
(2-i)
-
(3)
D is said to be Boyd-Wong (BW for short) [11] if there exists a function φ from \((0, \infty)\) into itself satisfying the following:
-
(3-i)
φ is upper semicontinuous from the right.
-
(3-ii)
\(\varphi(t) < t\) holds for any \(t \in(0, \infty)\).
-
(3-iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in D\).
-
(3-i)
-
(4)
D is said to be Meir-Keeler (MK for short) [17] 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 [18–20].
-
(5)
D is said to be Matkowski (Mat for short) [21] if there exists a function φ from \((0, \infty)\) into itself satisfying the following:
-
(5-i)
φ is nondecreasing.
-
(5-ii)
\(\lim_{n} \varphi^{n}(t) = 0\) for every \(t \in(0, \infty)\).
-
(5-iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in D\).
-
(5-i)
-
(6)
D is said to be CJM [6, 22–24] if the following hold:
-
(6-i)
For any \(\varepsilon> 0\), there exists \(\delta> 0\) satisfying \(u \leq\varepsilon\) holds for any \((t,u) \in D\) with \(t < \varepsilon+ \delta\).
-
(6-ii)
\(u < t\) holds for any \((t,u) \in D\).
-
(6-i)
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
Then T is a Boyd-Wong contraction iff D is Boyd-Wong.
Proof
We first note
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
for any \(t \in(0,\infty)\). Then φ satisfies (3-i) and (3-ii) in Definition 5. We also have
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 [14].
Proposition 7
([14])
Assume (A1), (A2) and the following:
-
(i)
θ is nondecreasing and continuous.
-
(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
([14])
Assume (A1), (A2), and the following:
-
(i)
θ is nondecreasing.
-
(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 [14], 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:
-
(i)
θ is nondecreasing and continuous.
-
(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
We also define a function η from \((0,\infty)\) into \([0,\infty)\) by \(\eta= \theta_{+}^{-1} \circ\psi\circ\theta\). We note
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
holds for \(n \in \mathbb {N}\). Since \(\eta(t_{n}) > 0\),
holds. Hence there exists a sequence \(\{ u_{n} \}\) in \((0,\infty)\) satisfying
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,
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
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 [15].
We next discuss Meir-Keeler.
Proposition 10
Assume (A1), (A2), and the following:
-
(i)
θ is nondecreasing and right continuous.
-
(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
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
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 [14].
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:
-
(i)
θ is nondecreasing and left continuous.
-
(ii)
minΘ does not exist.
-
(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
Since θ is left continuous, we have \(\tau< \theta(t)\) implies \(\theta_{+}^{-1}(\tau) < t\). We also have
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
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
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 [15], Example 10 in [14])
Define a complete metric space \((X, d)\) by
Define a mapping T on X and functions θ and ψ from \((0, \infty)\) into itself by
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
Then \(E \subset \{ 2 \} \times (1/4,1)\) holds. Hence E is contractive.
Proof
We have
Hence D is neither Boyd-Wong nor Matkowski. □
Example 14
(Example 2.6 in [13], Example 11 in [14])
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
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
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
Define D by (1). Then all the assumptions of Proposition 12 except (ii) are satisfied. However, D is not Matkowski.
Proof
Obvious. □
6 Applications
In this section, as applications, we give alternative proofs of some recent generalizations of the Banach contraction principle. Ri in [1] proved the following fixed point theorem.
Theorem 16
(Ri [1])
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:
-
(R1)
\(\psi(t) < t\) for any \(t \in(0,\infty)\).
-
(R2)
\(\limsup_{s \to t+0} \psi(s) < t\) for any \(t \in(0,\infty)\).
-
(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 [2] proved a fixed point theorem on F-contraction.
Theorem 17
(Wardowski [2])
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:
-
(F1)
F is strictly increasing.
-
(F2)
For any sequence \(\{ \alpha_{n} \}\) of positive numbers, \(\lim_{n} \alpha_{n} = 0\) iff \(\lim_{n} F(\alpha_{n})=-\infty\).
-
(F3)
\(\lim_{t \to+0} t^{k} F(t) = 0\) holds.
-
(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:
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The author is supported in part by JSPS KAKENHI Grant Number 16K05207 from Japan Society for the Promotion of Science.
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Suzuki, T. Discussion of several contractions by Jachymski’s approach. Fixed Point Theory Appl 2016, 91 (2016). https://doi.org/10.1186/s13663-016-0581-9
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DOI: https://doi.org/10.1186/s13663-016-0581-9