Caristi’s fixed point theorem in semimetric spaces

We give a negative answer to one of the questions raised by Kirk and Shahzad (J Fixed Point Theory Appl 17:541–555, 2015). Motivated by this question, we prove Caristi’s fixed point theorem in (∑,≠)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\sum ,\ne )$$\end{document}-complete semimetric spaces.


Introduction
We begin by recalling the concept of a semimetric space. Kirk and Shahzad in [10] introduced the following concept.

Theorem 3 (Caristi and Kirk
In [10], Kirk and Shahzad extended Theorem 3 to complete twogeneralized metric spaces satisfying the (CS) criterion. Also, in [10], they raised two open questions. The following is one of them: • Does Caristi's fixed point theorem still hold in a complete semimetric space which is regular with a strong triangle function?
Very recently Dung and Hang in [5] gave a negative answer to the above question. In this paper, we give another answer (see Example 10), which makes it easy to understand the essence of the above question. On the other hand, it is a natural question what condition is sufficient to deduce Caristi's fixed point theorem in semimetric spaces. We give an answer to this question, too. Indeed under the assumption of ( , =)-completeness, we prove Caristi's fixed point theorem (Corollary 19).

Preliminaries, part 1
Throughout this paper we denote by N the set of all positive integers and by R the set of all real numbers. For an arbitrary set A, we also denote by #A the cardinal number of A.
In this section, we give some preliminaries.

Definition 4.
Let (X, d) be a semimetric space, let {x n } be a sequence in X and let x, y ∈ X. Let g be a function from X into R and let h be a function from X into (−∞, +∞]. • X is said to be Hausdorff if lim n d(x n , x) = 0 and lim n d(x n , y) = 0 imply x = y. • X is said to be complete if every Cauchy sequence converges. • X is said to be -complete if every -Cauchy sequence converges. • X is said to be ( , =)-complete if every ( , =)-Cauchy sequence converges. • g is said to be sequentially continuous if lim n g(x n ) = g(x) holds provided {x n } converges to x.
Remark. We note that X does not necessarily have a topology.
We will discuss strong triangle functions.

Lemma 5.
Let (X, d) be a semimetric space. Then the following are equivalent: (i) X is regular with a strong triangle function Φ.

Lemma 7.
Let X be a nonempty set and let e be a function from X × X into [0, ∞] satisfying the following: Assume that for any x, y ∈ X, there exists a finite sequence Then the following hold: Proof. By Lemma 6 (i), we first note that ϕ is strictly increasing. From the definition of d, obviously holds. To prove (iii), we fix x, y, z ∈ X. Let ε > 0 be arbitrary. Then since ϕ is continuous and strictly increasing, there exists δ > 0 satisfying Then we have by Lemma 6 (i) and (iii) Since ε > 0 is arbitrary, we obtain (iii).

Lemma 8. Let ϕ be a continuous function from
.
Proof. Using the mean value theorem, we obtain the desired result.
The following function ϕ plays a very important role in this paper.

Example
In this section, we give a negative answer to the question raised by Kirk and Shahzad, which is stated in Sect. 1.
Define a mapping T on X and a function h from X into [0, ∞) by Then the following assertions hold: (xi) h is sequentially continuous.
(xii) T does not have a fixed point.
Proof. By Lemma 9 (ii) and (iv), we first note that ϕ is continuous and (2) holds for any s, t ∈ [0, ∞) with s ≤ t.
To show (vii) and (xi), we let {x n } be a sequence in X converging to x ∈ X. Then by (5), x n = x holds for sufficiently large n ∈ N. Hence, (vii) and (xi) hold.
Let us prove (viii). We will show for any m ∈ N. Indeed, fix ν ∈ N with r ν < e −2 . Then we have holds. Using this, we can obtain (6). Considering (5) and (6), for any Cauchy sequence {x n } in X, we have x m = x n for sufficiently large m, n ∈ N. Hence, X is complete.
Since the sequence {n} in X is ( , =)-Cauchy and it does not converge, X is not ( , =)-complete. We have shown (ix).

Preliminaries, part 2
To prove fixed point theorems, we give some preliminaries.  Proof. The conclusion follows from Lemma 11.

Proposition 13. Let (X, d) be a -complete semimetric space. Then X is Hausdorff.
Proof. Let {x n } be a sequence converging to z and w. We can choose a subsequence {f (n)} of the sequence {n} in N satisfying for n ∈ N. Define a sequence {y n } in X by Since X is -complete, {y n } converges to some v. From the definition of {y n }, we have z = v = w. Thus, X is Hausdorff.
( , =)-completeness is stronger than completeness under Hausdorffness. See also [16,17]. Proof. Let {x n } be a Cauchy sequence in X. We consider the following two cases: • #{n ∈ N : x n = z} = ∞ holds for some z ∈ X.
In the first case, we have .
thus, {x f (n) } is ( , =)-Cauchy. Since X is ( , =)-complete, there exists z ∈ X satisfying lim n d(x f (n) , z) = 0. Arguing by contradiction, we assume lim sup n d(x n , z) > 0. Then there exist ε > 0 and a subsequence {g(n)} of {n} in N satisfying the following: Vol. 20 (2018) Caristi's fixed point theorem Page 9 of 15 30 Define a subsequence {h(n)} of {n} in N as follows: In the case where n is odd, we put h(n) = f (k) for some k ∈ N with k ≥ n and f (k) > h(n − 1), where we put h(0) = 0 temporarily. In the other case, where n is even, we put h(n) = g(k) for some k ∈ N with g(k) ≥ f (n) and g(k) > h(n − 1). Then since h(n) ≥ f (n) holds for any n ∈ N, Since X is Hausdorff, we obtain z = w. So we have which implies a contradiction. Therefore, we have shown lim n d(x n , z) = 0.

Proposition 15.
Let (X, d) be a -complete semimetric space. Then X is complete.
Proof. The conclusion follows from Propositions 12,13 and 14 In the remainder of this paper, we let Ω be the first uncountable ordinal number. Let α be an ordinal number. We denote by α + and α − the successor and the predecessor of α, respectively. We recall that α is isolated if α − exists. α is limit ordinal if α = 0 holds and α − does not exist. For κ ∈ N, we define α + κ by It is well known that for any limit α ∈ Ω, there exists a strictly increasing sequence {β n } in Ω converging to α, that is, the following hold: • β n < α.
Since X is Hausdorff, we obtain x = z. Thus, we obtain the desired result.

Fixed point theorems
In this section, we prove fixed point theorems. We first extend Kirk and Saliga's fixed point theorem (Theorem 2.3 in [9]) to ( , =)-complete semimetric spaces. and where T 0 is the identity mapping on X. We note that H is well defined because h(x) = −∞ holds for x ∈ X. We also note that H is bounded from below. We have by (8) for any x ∈ X. Also, we have by (7) H Arguing by contradiction, we assume T x = x for any x ∈ X. Using transfinite induction, we will define a net {u α : α ∈ Ω} satisfying the following (P : α): (P 3 : α) For any ε > 0 and for any β ∈ Ω with β < α, there exists a finite sequence (γ 0 , . . . , γ n ) ∈ Ω n+1 satisfying β = γ 0 < γ 1 < · · · < γ n = α and Fix u ∈ X with h(u) < ∞. Then we note H(u) < ∞. Put u 0 = u. Then it is obvious that (P : 0) is satisfied.
In the second case, let {β n } be a strictly increasing sequence in Ω converging to α. For any n ∈ N, from (P 3 : β n+1 ), we can choose a finite sequence (γ Since H is bounded from below, we have Since X is ( , =)-complete, the sequence 1 , . . . converges to some u α ∈ X. We note that {u βn } converge to u α ∈ X because {u βn } is a subsequence of the above sequence. Fix ε > 0 and β ∈ Ω with β < α. We can choose ν ∈ N satisfying β < β ν and d(u βν , u α ) < ε.
If we assume that X is Hausdorff, then we can obtain a finer result.
Remark. lim[u β : β < α] is defined as follows: Proof. Arguing by contradiction, we assume T u α = u α for any α ∈ Ω. We have to confirm that we can use the same proof of Theorem 17.
Let α ∈ Ω be isolated. Then it is obvious that we can use the same proof.
Let α ∈ Ω be limit. Then from the proof of Theorem 17, for any strictly increasing sequence {β n } in Ω converging to α, {u βn } converges. By Lemma 16, there exists a unique element u α ∈ X satisfying lim[d(u β , u α ) : β < α] = 0. It is obvious that {u βn } converges to u α for any strictly increasing sequence {β n } in Ω converging to α. So we can use the same proof. Also we have shown that we can define the net {u α : α ∈ Ω}.