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A generalization of HegedüsSzilágyi’s fixed point theorem in complete metric spaces
Fixed Point Theory and Applications volume 2018, Article number: 1 (2018)
Abstract
In 1980, Hegedüs and Szilágyi proved some fixed point theorem in complete metric spaces. Introducing a new contractive condition, we generalize HegedüsSzilágyi’s fixed point theorem. We discuss the relationship between the new contractive condition and other contractive conditions. We also show that we cannot extend HegedüsSzilágyi’s fixed point theorem to MeirKeeler type.
1 Introduction and 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 T be a mapping on a metric space \((X,d)\). Throughout this paper, we define \(D_{T}(x)\) and \(D_{T}(x,y)\) by
for any \(x, y \in X\). That is, \(D_{T}(x)\) is the diameter of the orbit \(\{ x, Tx, T^{2} x,\ldots\}\) of x.
Hegedüs and Szilágyi in [1] proved the following fixed point theorem. The author thinks that the proof in [1] is splendid.
Theorem 1
(Theorem 5 in [1])
Let \((X,d)\) be a complete metric space, and let T be a mapping on X. Assume \(D_{T}(x) < \infty\) for all \(x \in X\). Assume also that there exists a function φ from \([0, \infty)\) into itself satisfying the following:

(i)
\(\varphi(t) < t\) holds for all \(t \in(0, \infty)\);

(ii)
φ is upper semicontinuous from the right;

(iii)
\(d(Tx,Ty) \leq\varphi\circ D_{T}(x,y) \) holds for all \(x, y \in X\).
Then T has a unique fixed point z. Moreover, \(\{ T^{n} x \}\) converges to z for any \(x \in X\).
Remark 1
See also [2–4]. Note that in the proof of Theorem 1 in [3], we need an additional assumption such as the nondecreasingness of φ.
We state BoydWong’s [5], MeirKeeler’s [6] and Matkowski’s [7] fixed point theorems.
Theorem 2
(Theorem 1 in [5])
Let \((X,d)\) be a complete metric space, and let T be a mapping on X. Assume that there exists a function φ from \([0, \infty)\) into itself satisfying (i) and (ii) of Theorem 1 and the following:

(iii)
\(d(Tx,Ty) \leq\varphi\circ d(x,y) \) holds for all \(x, y \in X\).
Then T has a unique fixed point z. Moreover, \(\{ T^{n} x \}\) converges to z for any \(x \in X\).
Theorem 3
([6])
Let \((X,d)\) be a complete metric space, and let T be a mapping on X. Assume that, for any \(\varepsilon> 0\), there exists \(\delta> 0\) such that
for all \(x, y \in X\). Then T has a unique fixed point z. Moreover, \(\{ T^{n} x \}\) converges to z for any \(x \in X\).
Theorem 4
(Theorem 1.2 in [7])
Let \((X,d)\) be a complete metric space, and let T be a mapping on X. Assume that there exists a function φ from \([0, \infty)\) into itself satisfying the following:

(i)
φ is nondecreasing;

(ii)
\(\lim_{n} \varphi^{n}(t) = 0\) holds for all \(t \in(0, \infty)\);

(iii)
\(d(Tx,Ty) \leq\varphi\circ d(x,y) \) holds for all \(x, y \in X\).
Then T has a unique fixed point z. Moreover, \(\{ T^{n} x \}\) converges to z for any \(x \in X\).
From the above, we can tell that Theorem 1 is of BoydWong [5] type (see Definition 8). So it is a very natural question of whether we can extend Theorem 1 to MeirKeeler [6] type. It is also a natural question of whether we can prove a Matkowski [7] type fixed point theorem.
In this paper, we answer the above two questions; one is negative and the other is affirmative. Indeed, we generalize Theorem 1. The assumption of the new theorem (Theorem 5) is weaker than a Matkowski type condition (see Corollary 7). We also give a counterexample for a MeirKeeler type condition (Example 16). We further discuss the relationship between the assumption of Theorem 5 and other contractive conditions.
2 Main results
In this section, we generalize Theorem 1.
Theorem 5
Let \((X,d)\) be a complete metric space, and let T be a mapping on X. Assume \(D_{T}(x) < \infty\) for all \(x \in X\). Assume also that there exists a function φ from \([0, \infty)\) into itself satisfying the following:

(i)
\(\varphi(t) < t\) holds for all \(t \in(0, \infty)\);

(ii)
For any \(\varepsilon> 0\), there exists \(\delta> 0\) such that, for any \(t \in(0,\infty)\),
$$\varepsilon< t < \varepsilon+\delta \quad\textit{implies}\quad \varphi(t) \leq \varepsilon. $$ 
(iii)
For any \(x, y \in X\),
$$d(Tx,Ty) \leq\varphi\circ D_{T}(x,y) $$holds.
Then T has a unique fixed point z. Moreover, \(\{ T^{n} x \}\) converges to z for any \(x \in X\).
Remark 2

\(D_{T}(x,y) < \infty\) obviously holds for any \(x, y \in X\).

Since \(D_{T}(x,y) = 0\) implies \(d(Tx,Ty) = 0\), without loss of generality, we may assume \(\varphi(0) = 0\).

We do not assume that φ is nondecreasing. So, in general, \(D_{T}(Tx,Ty) \leq\varphi\circ D_{T}(x,y)\) does not hold.
Before proving Theorem 5, we need one lemma.
Lemma 6
Let \(x, y \in X\). Assume that either of the following holds:

(a)
\(x = y\);

(b)
\(\lim_{n} D_{T}(T^{n} x) = \lim_{n} D_{T}(T^{n} y) = 0\).
Then \(\lim_{n} D_{T}(T^{n} x, T^{n} y) = 0\) holds.
Proof
Since
for \(n \in \mathbb {N}\), \(\{ D_{T}(T^{n} x,T^{n} y) \}\) is nonincreasing. So \(\{ D_{T}(T^{n} x,T^{n} y) \}\) converges to some \(\varepsilon\in[0,\infty)\). Arguing by contradiction, we assume \(\varepsilon> 0\). We consider the following two cases:

\(\varepsilon< D_{T}(T^{n} x,T^{n} y) \) holds for any \(n \in \mathbb {N}\);

\(\varepsilon= D_{T}(T^{n} x,T^{n} y) \) holds for some \(n \in \mathbb {N}\).
In the first case, we choose \(\delta\in(0,\infty)\) such that
We choose \(\nu\in \mathbb {N}\) satisfying
In the case of (b), without loss of generality, we may assume
Fix \(m \geq\nu\) and \(n \geq\nu\). Then since
we have
Since m, n are arbitrary, considering (1), we obtain
which implies a contradiction. In the second case, we choose \(\nu\in \mathbb {N}\) satisfying
In the case of (b), without loss of generality, we may assume
Fix \(m \geq\nu\) and \(n \geq\nu\). Then since
we have
Since m, n are arbitrary, considering (2), we obtain
which implies a contradiction. Therefore we have shown \(\lim_{n} D_{T}(T^{n} x,T^{n} y) = 0\). □
Proof of Theorem 1
Fix \(x \in X\). By Lemma 6(a), \(\{ D_{T}(T^{n} x) \}\) converges to 0. Thus \(\{ T^{n} x\}\) is a Cauchy sequence in X. Since X is complete, \(\{ T^{n} x \}\) converges to some \(z \in X\). By Lemma 6(a) again, \(\{ D_{T}(T^{n} z) \}\) also converges to 0. So, by Lemma 6(b), we obtain
So \(\{ T^{n} z \}\) also converges to z. Hence
holds. Arguing by contradiction, we assume \(\varepsilon:= D_{T}(z) > 0\). Since \(\lim_{n} D_{T}(T^{n} z) = 0\) holds, there exists \(\nu\in \mathbb {N}\) satisfying
where \(T^{0} z = z\). This implies
For \(n > \nu\), we have
Since n is arbitrary, we obtain
which implies a contradiction. Therefore we have shown \(D_{T}(z) = 0\). Hence z is a fixed point of T. Since (3) holds for any \(x \in X\), we obtain the uniqueness of the fixed point. □
By Theorem 5, we obtain a Matkowski type fixed point theorem.
Corollary 7
Let \((X,d)\) be a complete metric space, and let T be a mapping on X. Assume \(D_{T}(x) < \infty\) for all \(x \in X\). Assume also that there exists a function φ from \([0, \infty)\) into itself satisfying the following:

(i)
φ is nondecreasing;

(ii)
\(\lim_{n} \varphi^{n}(t) = 0\) holds for all \(t \in(0, \infty)\);

(iii)
\(d(Tx,Ty) \leq\varphi\circ D_{T}(x,y) \) holds for all \(x, y \in X\).
Then T has a unique fixed point z. Moreover, \(\{ T^{n} x \}\) converges to z for any \(x \in X\).
3 Comparison
In this section, using subsets of \((0,\infty)^{2}\), we discuss the relationship between the new contractive condition in Theorem 5 and other contractive conditions. See [1, 8–11] and the references therein.
Definition 8
Let Q be a subset of \((0,\infty)^{2}\).

(1)
Q is said to be contractive (Cont for short) [12, 13] if there exists \(r \in(0,1)\) such that \(u \leq r t\) holds for any \((t,u) \in Q\).

(2)
Q is said to be Browder (Bro for short) [14] if there exists a function φ from \((0, \infty)\) into itself satisfying the following:

(2i)
φ is nondecreasing and rightcontinuous;

(2ii)
\(\varphi(t) < t\) holds for any \(t \in(0, \infty)\);

(2iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in Q\).

(2i)

(3)
Q is said to be BoydWong (BW for short) [5] if there exists a function φ from \((0, \infty)\) into itself satisfying the following:

(3i)
φ is upper semicontinuous from the right;

(3ii)
\(\varphi(t) < t\) holds for any \(t \in(0, \infty)\);

(3iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in Q\).

(3i)

(4)
Q is said to be MeirKeeler (MK for short) [6] if, for any \(\varepsilon> 0\), there exists \(\delta> 0\) such that \(u < \varepsilon\) holds for any \((t,u) \in Q\) with \(t < \varepsilon+ \delta\).

(5)
Q is said to be Matkowski (Mat for short) [7] if there exists a function φ from \((0, \infty)\) into itself satisfying the following:

(5i)
φ is nondecreasing;

(5ii)
\(\lim_{n} \varphi^{n}(t) = 0\) for any \(t \in(0, \infty)\);

(5iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in Q\).

(5i)

(6)
Q is said to be of Newtype (NT for short) if there exists a function φ from \((0, \infty)\) into itself satisfying the following:

(6i)
\(\varphi(t) < t\) for any \(t \in(0,\infty)\);

(6ii)
For any \(\varepsilon> 0\), there exists \(\delta> 0\) such that \(\varepsilon< t < \varepsilon+ \delta\) implies \(\varphi(t) \leq\varepsilon\);

(6iii)
\(u \leq\varphi(t)\) holds for any \((t,u) \in Q\).

(6i)

(7)
Q is said to be CJM [15–18] if the following hold:

(7i)
For any \(\varepsilon> 0\), there exists \(\delta> 0\) satisfying \(u \leq\varepsilon\) holds for any \((t,u) \in Q\) with \(t < \varepsilon+ \delta\);

(7ii)
\(u < t\) holds for any \((t,u) \in Q\).

(7i)
It is obvious that the following implications hold:
It is well known that the converse implication of (Cont → Bro) does not hold. The following three examples tell us that for each implication except (Cont → Bro), there exists a counterexample for its converse implication. In particular, MK and NT are independent.
Example 9
Let \(u \in(0,\infty)\) and define Q by
Then Q is Mat. However, Q is not MK.
Remark 3
We note that the converse implication of (BW → NT) does not hold.
Example 10
Let \(t, u \in(0,\infty)\) with \(t < u\). Define Q by
Then Q is BW. However, Q is not Mat.
Remark 4
We note that the converse implication of (Mat → NT) does not hold.
Example 11
Let \(t \in(0,\infty)\) and define Q by
Then Q is MK. However, Q is not NT.
Remark 5
We note that the converse implication of (NT → CJM) does not hold.
In the remainder of this section, we let \((X,d)\) be a complete metric space, and let T be a mapping on X satisfying \(D_{T}(x) < \infty\) for all \(x \in X\). Define subsets \(P_{T}\) and \(Q_{T}\) of \((0,\infty)^{2}\) by
We will give three mappings such that \(Q_{T}\) for each mapping matches one of Examples 911, respectively.
Lemma 12
Let X be a nonempty set. Let f be a function from X into \([0, \infty)\) such that \(\{ x \in X : f(x) = 0 \}\) consists of at most one element. Define a function d from \(X \times X\) into \([0, \infty)\) by
Let T be a mapping on X satisfying the following:

\(f(x) > 0\) implies \(Tx \neq x\) and \(f(Tx) \leq f(x)\);

\(f(x) = 0\) implies \(Tx = x\).
Then the following hold:

(i)
\((X, d)\) is a metric space;

(ii)
if either \(\{ x \in X : f(x) = 0 \} \neq\varnothing\) or \(\inf f(X) > 0\) holds, then X is complete;

(iii)
\(P_{T} = Q_{T}\).
Proof
We have essentially proved (i) and (ii); see Lemma 7 in [19]. Let us prove (iii). Fix \(x, y \in X\) with \(x \neq y\) and \(f(x) \leq f(y)\). Then we have \(f(y) > 0\) and hence \(Ty \neq y\). We have
Hence
holds. Therefore \(P_{T}=Q_{T}\) holds. □
Example 13
Let \(X = [0,\infty)\) and define a function d from \(X \times X\) into \([0,\infty)\) by (5), where \(f(x) = x\). That is,
holds. Define a mapping T on X by
Then the following hold:

(i)
\((X,d)\) is a complete metric space;

(ii)
\(f(x) > 0\) implies \(f(Tx) < f(x)\);

(iii)
\(f(x) = 0\) implies \(Tx = x\);

(iv)
\(P_{T} = Q_{T} = \{ (t,1) : 1 < t \}\);

(v)
\(P_{T}\) and \(Q_{T}\) are Mat;

(vi)
neither \(P_{T}\) nor \(Q_{T}\) are MK.
Proof
We can prove (i)(iii) easily. Using Lemma 12, we can prove (iv). (v) and (vi) follow from Example 9. □
Example 14
Put \(X = [0,2)\) and define f and d as in Example 13. Define a mapping T on X by
Then (i)(iii) of Example 13 and the following hold:

(iv)
\(P_{T} = Q_{T} = \{ (1+\lambda,2 \lambda) : \lambda\in(0,1) \}\);

(v)
\(P_{T}\) and \(Q_{T}\) are BW;

(vi)
neither \(P_{T}\) nor \(Q_{T}\) are Mat.
Proof
We can prove (i)(iii) easily. Using Lemma 12, we can prove (iv). (v) and (vi) follow from Example 10. □
Example 15
Let \(X = [0,1) \cup(1,\infty)\) and define a function d from \(X \times X\) into \([0,\infty)\) by (5), where \(f(x) = \min\{ x, 1 \}\). Define a mapping T on X by
Then (i)(iii) of Example 13 and the following hold:

(iv)
\(P_{T} = Q_{T} = \{ (1,u) : 0 < u < 1 \}\);

(v)
\(P_{T}\) and \(Q_{T}\) are MK;

(vi)
neither \(P_{T}\) nor \(Q_{T}\) are NT.
Proof
We can prove (i)(iii) easily. Using Lemma 12, we can prove (iv). (v) and (vi) follow from Example 11. □
We finally give the following example, which tells us that we cannot extend Theorem 1 to a MeirKeeler type contractive condition.
Example 16
Let \(X = [0,1)\) and define a function d from \(X \times X\) into \([0,\infty)\) by (6). Define a mapping T on X by
Then the following hold:

(i)
\((X,d)\) is a complete metric space;

(ii)
\(d(x,y) < 1\) holds for any \(x, y \in X\);

(iii)
for any \(x \in X\), \(\{ T^{n} x \}\) converges to 1 in the Euclidean space \(\mathbb {R}^{1}\);

(iv)
\(D_{T}(x) = 1\) holds for any \(x \in X\);

(v)
\(TX = (0,1)\);

(vi)
\(Q_{T} = \{ (1,u) : 0 < u < 1 \}\);

(vii)
\(Q_{T}\) is MK;

(viii)
\(Q_{T}\) is not NT;

(ix)
T does not have a fixed point.
Proof
We can easily prove (i)(vi) and (ix). (vii) and (viii) follow from Example 11. □
4 Conclusions
In this paper, introducing a new contractive condition (see Definition 8(6)), we generalize HegedüsSzilágyi’s fixed point theorem (Theorem 1) in complete metric spaces proved in 1980. In Section 3, we discuss the relationship between the new contractive condition and other contractive conditions. We also show that we cannot extend Theorem 1 to MeirKeeler type (see Example 16).
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Acknowledgements
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. A generalization of HegedüsSzilágyi’s fixed point theorem in complete metric spaces. Fixed Point Theory Appl 2018, 1 (2018). https://doi.org/10.1186/s1366301706259
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DOI: https://doi.org/10.1186/s1366301706259