1 Introduction
We have introduced many types of contractions. We give the definitions of three of them. Define Φ as follows: \(\varphi\in\varPhi\) iff φ is a nondecreasing, right continuous function from \([0, \infty)\) into itself satisfying \(\varphi(t) < t\) for any \(t > 0\). It is obvious that \(\varphi(0) = 0\) holds.
Definition 1
Let T be a mapping on a metric space \((X,d)\).
-
T is said to be a Browder contraction (BroC, for short) [3] if there exists \(\varphi\in\varPhi\) such that \(d(Tx,Ty) \leq\varphi\circ d(x,y)\) for any \(x,y \in X\).
Anzeige
We know the following implications: There are some conditions equivalent to BroC; see [7]. Lemma 5 in [8] gives seven equivalent conditions connected with BroC. See [9] and the references therein for further information as regards contractions.
$$\text{C} \Longrightarrow \text{BroC} \Longrightarrow \text{CJMC}. $$
Branciari [10] 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 \(x, y \in X\).
$$\int_{ 0}^{ s} f(t)\,dt > 0 \quad \text{and}\quad \int_{ 0}^{ d(Tx, Ty)} f(t)\,dt \leq r \int_{ 0}^{ d(x,y)} f(t)\,dt $$
We have studied contractions of integral type in [11, 12]. Very recently, Jleli and Samet [13] introduced the following new type of contractions.
Theorem 2
(Jleli and Samet [13])
Anzeige
Let
\((X,d)\)
be a complete generalized metric space and let
θ
be a function from
\((0,\infty)\)
into
\((1,\infty)\)
satisfying the following:
Let
T
be a mapping on
X. Assume that there exists
\(k \in(0,1)\)
such that
for any
\(x, y \in X\). Then
T
has a unique fixed point.
(θ1)
θ
is nondecreasing.
(θ2)
For any sequence
\(\{ t_{n} \}\)
in
\((0,\infty)\), \(\lim_{n} \theta(t_{n}) = 1\)
iff
\(\lim_{n} t_{n} = 0\).
(θ3)
There exist
\(r \in(0,1)\)
and
\(\ell\in(0,\infty]\)
such that
\(\lim_{t \to+0} (\theta(t) - 1 )/t^{r} = \ell\).
$$Tx \neq Ty \quad \textit{implies}\quad \theta\circ d(Tx,Ty) \leq \bigl( \theta \circ d(x,y) \bigr)^{k} $$
Remark
-
Considering the domain and range of θ and (θ1), it is obvious that (θ2) is equivalent to \(\inf \{ \theta(t) : t \in(0,\infty) \} = 1\).
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. For a function f, we denote by \(\operatorname {Dom}(f)\) the domain of f.
Let f be a function from a subset of \(\mathbb {R}\) into \(\mathbb {R}\). Then f is said to satisfy (U)
f
if the following holds:
(U)
f
For any \(t \in \operatorname {Dom}(f)\), there exist \(\delta> 0\) and \(\varepsilon> 0\) such that \(f(s) \leq t - \varepsilon\) holds for any \(s \in(t-\delta,t+\delta) \cap \operatorname {Dom}(f)\).
We list some further notation in order to simplify the statement of the results of this paper:
(A1)
Let Y be an arbitrary set and let h be a function from Y into \([0,\infty)\). Let S be a mapping on Y satisfying that \(h(x) = 0\) implies \(h(Sx) = 0\) for any \(x \in Y\).
(A2)
Let θ be a function from \((0,\infty)\) into \(\mathbb {R}\). Put \(\Theta= \theta ( (0,\infty) )\) and
$$\Theta_{\leq}= \bigcup \bigl\{ [t,\infty) : t \in\Theta \bigr\} . $$
The purpose of this study is to obtain the mathematical structure of contractions mentioned in Section 1. So, we simplify our setting as follows:
Definition 3
Assume (A1).
-
S is said to be a Browder contraction if there exists \(\varphi\in\varPhi\) such thatfor any \(x \in Y\).$$ h(Sx) \leq\varphi\circ h(x) $$
-
S is said to be a CJM contraction if the following hold:(j)For any \(\varepsilon> 0\), there exists \(\delta> 0\) such that \(h(x) < \varepsilon+ \delta\) implies \(h(Sx) \leq\varepsilon\).(jj)\(h(x) > 0 \) implies \(h(Sx) < h(x) \).
Remark
Let \((X,d)\) and T be as in Definition 1. Put \(Y = X \times X\) and define h and S by \(h ( (x, y) ) = d(x, y)\) and \(S(x, y) = (Tx, Ty)\). Then the concept of Browder contraction in Definition 3 becomes that in Definition 1 and the concept of CJM contraction in Definition 3 becomes that in Definition 1.
We give some lemmas concerning (U)
f
.
Lemma 4
Let
f
be a function from a subset of
\(\mathbb {R}\)
into
\(\mathbb {R}\). Then
f
satisfies (U)
f
iff
holds for any
\(t \in \operatorname {Dom}(f)\).
$$\limsup \bigl[ f(u) : u \to t, u \in \operatorname {Dom}(f) \bigr] < t $$
Remark
We define \(\limsup [ f(u) : u \to t, u \in \operatorname {Dom}(f) ] = \gamma\) iff the following hold: Note that we do not exclude the case of \(u_{n} = t\).
-
There exists a sequence \(\{ u_{n} \}\) in \(\operatorname {Dom}(f)\) such that \(\{ u_{n} \}\) converges to t and \(\{ f(u_{n}) \}\) converges to γ.
-
\(\limsup_{n} f(u_{n}) \leq\gamma\) holds for any sequence \(\{ u_{n} \}\) in \(\operatorname {Dom}(f)\) converging to t.
Proof of Lemma 4
Obvious. □
Lemma 5
Let
f
be an upper semicontinuous function from a subset of
\(\mathbb {R}\)
into
\(\mathbb {R}\)
such that
\(f(t) < t\)
for any
\(t \in \operatorname {Dom}(f)\). Then
f
satisfies (U)
f
.
Proof
Obvious. □
Lemma 6
Let
ψ
be a function from a subset
D
of
\(\mathbb {R}\)
into
\(\mathbb {R}\)
satisfying (U)
ψ
. Let
M
be a real number with
\(M>1\). Define a function
φ
from
D
into
\(\mathbb {R}\)
by
for
\(t \in D\). Then the following hold:
$$\begin{aligned} \varphi(t) ={}& \frac{t}{ 2 } + \frac{1}{ 2 } \max \Bigl\{ \sup \bigl\{ \psi(u) : u \in D, u \leq t \bigr\} , \\ &{} \sup \bigl\{ \psi(u) + M (t-u) : u \in D, t \leq u \bigr\} \Bigr\} \end{aligned}$$
(i)
φ
satisfies (U)
φ
.
(ii)
φ
is strictly increasing and Lipschitz continuous.
(iii)
\(\psi(t) < \varphi(t)\)
holds for any
\(t \in D\).
Proof
In this proof, we always assume \(s, t, u \in D\). Define a function ω from D into \(\mathbb {R}\) by For each u, we define a function \(\psi_{u}\) from D into \(\mathbb {R}\) by Then from the definition of ω, we note So it is obvious that \(\psi(t) \leq\omega(t)\). We can easily check that \(\psi_{u}\) is nondecreasing and \(\vert { \psi_{u}(s) - \psi_{u}(t)} \vert \leq M \vert s - t \vert \) holds. Hence ω is nondecreasing and holds. Hence ω is continuous. Fix t; and choose δ and ε as in the definition of (U)
f
. We have Hence by Lemma 5, ω satisfies (U)
ω
. It is obvious that hold. We can easily prove the remainders. □
$$\omega(t) = \max \bigl\{ \sup \bigl\{ \psi(u) : u \leq t \bigr\} , \sup \bigl\{ \psi(u) + M (t-u) : t \leq u \bigr\} \bigr\} . $$
$$\psi_{u}(t) = \textstyle\begin{cases} \psi(u) + M (t - u) & \text{if }t \leq u, \\ \psi(u) & \text{if }t \geq u. \end{cases} $$
$$\omega(t) = \sup \bigl\{ \psi_{u}(t) : u \bigr\} . $$
$$\bigl\vert \omega(s) - \omega(t) \bigr\vert \leq M \vert s - t \vert $$
$$\begin{aligned} \omega(t) ={}& \max \Bigl\{ \sup_{u \leq t-\delta} \psi_{u}(t), \sup_{t-\delta< u \leq t} \psi_{u}(t), \sup_{t \leq u < t+\delta} \psi_{u}(t), \sup_{t+\delta\leq u} \psi_{u}(t) \Bigr\} \\ ={}& \max \Bigl\{ \sup_{u \leq t-\delta} \psi(u), \sup _{t-\delta< u \leq t} \psi(u), \\ &{} \sup_{t \leq u < t+\delta} \bigl( \psi(u) + M (t-u) \bigr),\sup_{t+\delta\leq u} \bigl( \psi(u) + M (t-u) \bigr) \Bigr\} \\ \leq{}&\max \Bigl\{ \sup_{u \leq t-\delta} u, \sup_{t-\delta< u < t+\delta} \psi(u), \sup_{t+\delta\leq u} \bigl( u + M (t-u) \bigr) \Bigr\} \\ \leq{}&\max\{ t - \delta, t - \varepsilon, t + \delta- M \delta\} \\ < {}& t. \end{aligned}$$
$$\bigl\vert \varphi(s) - \varphi(t) \bigr\vert \leq\frac{M+1}{2} \vert s - t \vert \quad \text{and}\quad \psi(t) \leq\omega(t) < \varphi(t) < t $$
3 Browder contraction
In this section, we discuss Browder contractions.
Lemma 7
Assume (A1). Let
ψ
be a function from
\((0,\infty)\)
into
\([0,\infty)\)
such that (U)
ψ
holds and
for any
\(x \in Y\)
with
\(h(x)>0\)
and
\(h(Sx)>0\). Then
S
is a Browder contraction.
$$h(Sx) \leq\psi\circ h(x) $$
Proof
By Lemma 6, there exists a nondecreasing continuous function η from \((0,\infty)\) into \([0,\infty)\) such that (U)
η
is satisfied and \(\psi(t) \leq\eta(t)\) holds for any \(t \in(0,\infty)\). Define a function φ from \([0,\infty)\) into itself by It is obvious that φ is nondecreasing continuous function such that \(\varphi(t) < t\) for any \(t \in(0,\infty)\). Thus, \(\varphi\in\varPhi\). Fix \(x \in Y\). We consider the following two cases: In the first case, \(h(Sx) \leq\varphi\circ h(x)\) obviously holds. In the second case, we note that \(h(Sx) > 0\) implies \(h(x) > 0\). So \(h(x) > 0\) holds. We have Therefore S is a Browder contraction. □
$$\varphi(t) = \textstyle\begin{cases} \eta(t) & \text{if } t > 0 ,\\ 0 & \text{if } t = 0 . \end{cases} $$
-
\(h(Sx) = 0\),
-
\(h(Sx) > 0\).
$$h(Sx) \leq\psi\circ h(x) \leq\eta\circ h(x) = \varphi\circ h(x). $$
Proposition 8
Assume (A1), (A2) and the following:
Then
S
is a Browder contraction.
(i)
θ
is nondecreasing and continuous.
(ii)
There exists a function
ψ
from Θ into
\(\mathbb {R}\)
satisfying (U)
ψ
and
for any
\(x \in Y\)
with
\(h(x) > 0 \)
and
\(h(Sx) > 0\).
$$ \theta\circ h(Sx) \leq\psi\circ\theta\circ h(x) $$
(1)
Proof
Define a function \(\theta_{+}^{-1}\) from \(\mathbb {R}\) into \([0, \infty]\) by Put \(\varphi= \theta_{+}^{-1} \circ\psi\circ\theta\). We note Since \(\psi\circ\theta(t) < \theta(t) \leq\theta(s)\) for any \(t, s \in(0,\infty)\) with \(t \leq s\), we have \(\varphi(t) \leq t\) for any \(t \in(0,\infty)\). Thus, φ is a function from \((0,\infty)\) into \([0,\infty)\). Arguing by contradiction, we assume that (U)
φ
does not hold. Then there exist \(t \in(0,\infty)\) and a sequence \(\{ t_{n} \}\) in \((0,\infty)\) such that \(\{ t_{n} \}\) converges to t and holds for \(n \in \mathbb {N}\). Since \(\varphi(t_{n}) > 0\), 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 because θ is continuous. This contradicts (U)
ψ
. Therefore (U)
φ
holds. For any \(x \in Y\) with \(h(x) > 0\) and \(h(Sx) > 0\), we have by (1) Therefore by Lemma 7, S is a Browder contraction. □
$$\theta_{+}^{-1}(\tau) = \textstyle\begin{cases} \sup \{ s \in(0,\infty) : \theta(s) \leq\tau \} & \text{if } \{ s \in(0,\infty) : \theta(s) \leq\tau \} \neq\varnothing, \\ 0 & \text{otherwise.} \end{cases} $$
$$\varphi(t) = \sup \bigl\{ s \in(0,\infty) : \theta(s) \leq\psi\circ\theta(t) \bigr\} \quad \text{provided }\varphi(t) > 0 . $$
$$\varphi(t_{n}) > (1 - 1/n) t $$
$$\sup \bigl\{ s \in(0,\infty) : \theta(s) \leq\psi\circ\theta(t_{n}) \bigr\} > (1-1/n) t $$
$$\theta(u_{n}) \leq\psi\circ\theta(t_{n}) < \theta(t_{n}) \quad \text{and}\quad u_{n} > (1-2/n) t $$
$$\theta(t) \leq\limsup_{n \to\infty} \psi\circ\theta(t_{n}) \leq\limsup \bigl[ \psi(\tau) : \tau\to\theta(t), \tau\in\Theta \bigr] , $$
$$h(Sx) \leq\sup \bigl\{ s \in(0,\infty) : \theta(s) \leq\psi\circ\theta\circ h(x) \bigr\} = \varphi\circ h(x). $$
By Lemma 5, we obtain the following.
Corollary 9
The following examples tell that the continuity of θ in Proposition 8 is needed. In Example 10, θ is right continuous, however, it is not left continuous. On the other hand, in Example 11, θ is left continuous, however, it is not right continuous.
Example 10
(Example 2.3 in [12])
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 \(\psi(t) = t / 2\). Put \(Y = X \times X\) and define h and S by \(h ( (x, y) ) = d(x, y)\) and \(S(x, y) = (Tx, Ty)\) for \((x, y) \in Y\). Then all the assumptions of Proposition 8 except the left continuity of θ are satisfied. However, T is not a Browder contraction.
$$X = [0, 1] \cup[2, \infty) \quad \text{and}\quad d(x,y) = \textstyle\begin{cases} \min\{ x + y, 2 \} & \text{if } x \neq y,\\ 0 & \text{if } x = y. \end{cases} $$
$$Tx = \textstyle\begin{cases} 0 & \text{if } x \leq1 ,\\ 1 - 1/x & \text{if } x \geq2, \end{cases}\displaystyle \qquad \theta(t) = \textstyle\begin{cases} 1 & \text{if } t < 2 ,\\ 2 & \text{if } t \geq2, \end{cases} $$
Example 11
(Example 2.6 in [11])
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 \(\psi(t) = t / 2\). Let Y, h, and S be as in Example 10. Then all the assumptions of Proposition 8 except the right continuity of θ are satisfied. However, T is not a Browder contraction.
$$Tx = \textstyle\begin{cases} 0 & \text{if } x \leq1 ,\\ 1 & \text{if } x > 1, \end{cases}\displaystyle \qquad \theta(t) = \textstyle\begin{cases} t & \text{if } t \leq1 ,\\ 1 + t & \text{if } t > 1, \end{cases} $$
4 CJM contraction
In this section, we discuss CJM contractions.
The following is a modification of Proposition 2.6 in [12].
Proposition 12
Assume (A1), (A2), and the following:
Then
S
is a CJM contraction.
(i)
θ
is nondecreasing.
(ii)
For any
\(\varepsilon\in\Theta_{\leq}\), there exists
\(\delta> 0\)
such that
\(h(x) > 0 \), \(h(Sx) > 0\), and
\(\theta\circ h(x) < \varepsilon+ \delta\)
imply
\(\theta\circ h(Sx) \leq\varepsilon\)
for all
\(x \in Y\).
(iii)
\(h(x) > 0 \)
and
\(h(Sx) > 0 \)
imply
\(\theta\circ h(Sx) < \theta\circ h(x) \).
Proof
In order to show (jj), we fix \(x \in Y\) with \(h(x) > 0\). Arguing by contradiction, we assume \(h(Sx) \geq h(x)\). Then \(h(Sx) > 0\) obviously holds. We have from (iii), which contradicts (i). Therefore we obtain \(h(Sx) < h(x)\). We have shown (jj). We shall prove (j). Fix \(\varepsilon> 0\) and put \(\beta= \lim [ \theta(t) : t \to\varepsilon+ 0 ]\). We consider the following two cases: In the first case, since \(\theta(\varepsilon) \leq\beta\), \(\beta\in\Theta_{\leq}\) holds. From (ii), there exists \(\alpha> 0\) such that We can choose \(\delta_{1} > 0\) satisfying \(\theta(\varepsilon+ \delta_{1}) < \beta+ \alpha\). Fix \(x \in Y\) with \(h(x) < \varepsilon+ \delta_{1}\). Arguing by contradiction, we assume \(h(Sx) > \varepsilon\). Then we have \(h(Sx) > 0\) and hence \(h(x) > 0\). We have and hence which is a contradiction. Therefore we obtain \(h(Sx) \leq\varepsilon\). In the second case, we also fix \(x \in Y\) with \(h(x) < \varepsilon+ \delta_{2}\). Arguing by contradiction, we assume \(h(Sx) > \varepsilon\). Then we have \(h(Sx) > 0\) and hence \(h(x) > 0\). By (iii), we have which is a contradiction. Therefore we obtain \(h(Sx) \leq\varepsilon\). We have proven (j). □
$$\theta\circ h(Sx) < \theta\circ h(x), $$
-
\(\beta< \theta(\varepsilon+ \gamma) \) holds for any \(\gamma> 0\).
-
There exists \(\delta_{2} > 0\) such that \(\beta= \theta(\varepsilon+ \delta_{2}) \).
$$h(x) > 0, \qquad h(Sx) > 0 \quad \text{and}\quad \theta\circ h(x) < \beta+ \alpha \quad \text{imply}\quad \theta\circ h(Sx) \leq\beta. $$
$$\begin{aligned} \theta\circ h(x) \leq\theta( \varepsilon+ \delta_{1} ) < \beta+ \alpha \end{aligned}$$
$$\begin{aligned} \beta< \theta\circ h(Sx) \leq\beta, \end{aligned}$$
$$\beta \leq\theta\circ h(Sx) < \theta\circ h(x) \leq\theta( \varepsilon+ \delta_{2} ) = \beta, $$
The proof of the following is obvious, however, we give a proof in order to show how differently Θ and \(\Theta_{\leq}\) work.
Corollary 13
Proof
We first note that from (U)
ψ
, \(\psi(t) < t\) for any \(t \in \operatorname {Dom}(\psi) = \Theta_{\leq}\). For \(x \in Y\) with \(h(x) > 0 \) and \(h(Sx) > 0 \), we have We have shown (iii) in Proposition 12. Let us prove (ii) in Proposition 12. Fix \(\varepsilon\in\Theta_{\leq}\). Then from (U)
ψ
, there exists \(\delta> 0\) such that \(\psi(s) < \varepsilon\) holds for any \(s \in[\varepsilon,\varepsilon+\delta)\). Fix \(x \in Y\) with \(h(x) > 0 \), \(h(Sx) > 0\), and \(\theta\circ h(x) < \varepsilon+ \delta\). Then if \(\theta\circ h(x) < \varepsilon\), then we have If \(\theta\circ h(x) \geq\varepsilon\), then we have Therefore by Proposition 12, S is a CJM contraction. □
$$\theta\circ h(Sx) \leq\psi\circ\theta\circ h(x) < \theta\circ h(x). $$
$$\theta\circ h(Sx) < \theta\circ h(x) < \varepsilon. $$
$$\theta\circ h(Sx) \leq\psi\circ\theta\circ h(x) < \varepsilon. $$
By Lemma 5, we obtain the following.
Corollary 14
Assume (A1), (A2), (i) in Proposition
12
and the following:
Then
S
is a CJM contraction.
(ii)
There exists an upper semicontinuous function
ψ
from
\(\Theta_{\leq}\)
into
\(\mathbb {R}\)
satisfying
\(\psi(t) < t\)
for any
\(t \in\Theta_{\leq}\)
and (1) for any
\(x \in Y\)
with
\(h(x) > 0 \)
and
\(h(Sx) > 0\).
There is a counterexample if we replace \(\Theta_{\leq}\) with Θ in Corollary 14.
Example 15
Define a complete metric space \((X, d)\) by Define a mapping T on X by \(Tx = 2 x\) for \(x \in X\). Define functions θ from \((0,\infty)\) into \(\mathbb {R}\) and ψ from \((0, 1] \cup(2,\infty)\) into \(\mathbb {R}\) by Let Y, h, and S be as in Example 10. Then all the assumptions of Corollary 14 except \(\operatorname {Dom}(\psi) = \Theta_{\leq}\) are satisfied. However, T is not a CJM contraction.
$$X = [1, \infty) \quad \text{and}\quad d(x,y) = \textstyle\begin{cases} 1 + 1/x + 1/y & \text{if } x \neq y,\\ 0 & \text{if } x = y. \end{cases} $$
$$\theta(t) = \textstyle\begin{cases} t & \text{if } t \leq1 ,\\ 1 + t & \text{if } t > 1, \end{cases}\displaystyle \quad \text{and}\quad \psi(t) = \textstyle\begin{cases} t/2 & \text{if } t \leq1 ,\\ 1 + t/2 & \text{if } t > 2 . \end{cases} $$
Remark
We note that θ is left continuous, however, θ is not right continuous.
Proof
For any \(x,y \in X\) with \(x \neq y\) and \(Tx \neq Ty\), we have and Thus, (1) holds. Since T has no fixed point, T is not a CJM contraction. □
$$\begin{aligned} \psi\circ\theta\circ d(x,y) = \psi\circ\theta \biggl( 1 + \frac{ 1 }{x} + \frac{ 1 }{y} \biggr) = \psi \biggl( 2 + \frac{ 1 }{x} + \frac{ 1 }{y} \biggr) = 2 + \frac{1}{2x} + \frac{1}{2y} \end{aligned}$$
$$\begin{aligned} \theta\circ d(Tx,Ty) = \theta\circ d(2x,2y) = \theta \biggl( 1 + \frac{1}{2x} + \frac{1}{2y} \biggr) = 2 + \frac{1}{2x} + \frac{1}{2y}. \end{aligned}$$
We assume additionally that θ is right continuous. Then we can prove the following.
Proposition 16
Assume (A1), (A2), (iii) in Proposition
12
and the following:
Then
S
is a CJM contraction.
(i)
θ
is nondecreasing and right continuous.
(ii)
For any
\(\varepsilon\in\Theta\), there exists
\(\delta> 0\)
such that
\(h(x) > 0 \), \(h(Sx) > 0\), and
\(\theta\circ h(x) < \varepsilon+ \delta\)
imply
\(\theta\circ h(Sx) \leq\varepsilon\)
for all
\(x \in Y\).
Proof
We will show (ii) in Proposition 12. Fix \(\varepsilon\in\Theta_{\leq}\setminus\Theta\). Then there exists \(\delta> 0\) such that If not so, then there exists a sequence \(\{ t_{n} \}\) in \((0,\infty)\) such that \(\{ \theta(t_{n}) \}\) is strictly decreasing and converges to ε. Since θ is nondecreasing, \(\{ t_{n} \}\) is strictly decreasing. Since \(\{ t_{n} \}\) is bounded from below, \(\{ t_{n} \}\) converges to some \(t \in[0,\infty)\). Since \(\varepsilon\in\Theta_{\leq}\), \(t > 0 \) holds. Since θ is right continuous, we obtain \(\theta(t) = \varepsilon\), which implies \(\varepsilon\in\Theta\). This is a contradiction. Fix \(x \in Y\) with \(h(x) > 0 \), \(h(Sx) > 0\), and \(\theta\circ h(x) < \varepsilon+ \delta\). Then we have \(\theta\circ h(x) < \varepsilon\). Hence we obtain from (iii) (in Proposition 12) So all the assumptions in Proposition 12 are satisfied. Therefore S is a CJM contraction. □
$$[\varepsilon, \varepsilon+\delta] \subset\Theta_{\leq}\setminus\Theta . $$
$$\theta\circ h(Sx) < \theta\circ h(x) < \varepsilon. $$
Corollary 17
Proof
By Lemma 5, we also obtain the following.
Corollary 18
Proof of Theorem 2
Let Y, h, and S be as in Example 10. Define a function ψ from \((1,\infty)\) into \((1,\infty)\) by \(\psi(t) = t^{k}\). Since ψ is continuous and \(\psi(t) < t\) for any \(t \in(1,\infty)\), all the assumption of Corollary 14 are satisfied. So by Corollary 14, S is a CJM contraction. By Theorem 13 in [16], T has a unique fixed point z. Moreover, \(\lim_{n} d(T^{n} x, z) = 0\) for any \(x \in X\). □
Remark
(ii)
If we assume additionally that θ is continuous, then T is a Browder contraction.
5 Some comments
As stated in Section 1, Lemma 5 in Jachymski [8] gives seven equivalent conditions connected with BroC. Finally, by Proposition 8 and Corollary 9 we can obtain the following, which is similar to Lemma 5 in [8].
Lemma 19
Let
D
be a subset of
\((0,\infty)^{2}\). Then the following statements are equivalent:
(i)
There exists
\(\varphi\in\varPhi\)
such that for any
\((t,u) \in D\), \(u \leq\varphi(t)\).
(ii)
There exist a nondecreasing, continuous function
θ
from
\((0,\infty)\)
into
\(\mathbb {R}\)
and a function
ψ
from
\(\theta ( (0,\infty) )\)
into
\(\mathbb {R}\)
satisfying (U)
ψ
such that for any
\((t,u) \in D\), \(\theta(u) \leq\psi\circ\theta(t)\).
(iii)
There exist a nondecreasing, continuous function
θ
from
\((0,\infty)\)
into
\(\mathbb {R}\)
and an upper semicontinuous function
ψ
from
\(\theta ( (0,\infty) )\)
into
\(\mathbb {R}\)
satisfying
\(\psi(t) < t\)
such that for any
\((t,u) \in D\), \(\theta(u) \leq\psi\circ\theta(t)\).
Acknowledgements
The author is grateful to the referees for many suggestions to improve the exposition of the paper. The author is supported in part by JSPS KAKENHI Grant Number 25400141 from Japan Society for the Promotion of Science.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The author declares that he has no competing interests.