Abstract
In this paper, first we introduce certain new classes of Suzuki type contractions in triangular and non-Archimedean fuzzy metric spaces. Further we establish fixed point theorems for such kind of mappings in non-Archimedean and triangular fuzzy metric spaces. We also prove Suzuki type fixed point results in non-Archimedean and triangular ordered fuzzy metric spaces. The results presented here improve and generalize certain recent results from the literature. Two illustrative examples and an application to integral equations are given to support the usability of our results.
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1 Introduction and preliminaries
The concept of fuzzy metric space was introduced in different ways by some authors (see, i.e., [1, 2]) and further to this, the fixed point theory in this kind of spaces has been intensively studied (see [3–11]). Here, we consider the notion of fuzzy metric space introduced by Kramosil and Michálek [2] and modified by George and Veeramani [12, 13] who obtained a Hausdorff topology for the class of fuzzy metric spaces. Recently, Miheţ [14] enlarged the class of fuzzy contractive mappings of Gregori and Sapena [7] and proved a fuzzy Banach contraction result for complete non-Archimedean fuzzy metric spaces [7] (see also Vetro [15]).
The applications of fixed point theorems are remarkable in different disciplines of mathematics, engineering and economics in dealing with problems arising in approximation theory, game theory and many others (see [16] and the references therein). Consequently, many researchers, following the Banach contraction principle, investigated the existence of weaker contractive conditions or extended previous results under relatively weak hypotheses on the metric space. On the other hand, Samet et al. [17] introduced the concepts of α-ψ-contractive and α-admissible mappings and established various fixed point theorems for such mappings defined on complete metric spaces. Afterwards Salimi et al. [18] and Hussain et al. [19, 20] modified the notions of α-ψ-contractive and α-admissible mappings and established certain fixed point theorems (see also [21–25]). In this paper, we introduce certain new classes of contraction mappings and establish fixed point theorems for such kind of mappings in non-Archimedean fuzzy metric spaces. The results presented in this paper generalize and extend some recent results in non-Archimedean fuzzy metric spaces. Some examples are given to support the usability of our results. For the sake of completeness, we now briefly recall some basic concepts.
Definition 1.1
A binary operation \(\star: [0,1]\times[0,1]\to[0,1]\) is called a continuous t-norm if it satisfies the following assertions:
-
(TN1)
⋆ is commutative and associative;
-
(TN2)
⋆ is continuous;
-
(TN3)
\(a\star1=a\) for all \(a\in[0,1]\);
-
(TN4)
\(a\star b\leq c\star d\) when \(a\leq c\) and \(b\leq d\) and \(a,b,c,d \in[0,1]\).
Definition 1.2
(George and Veeramani [12])
A fuzzy metric space is an ordered triple \((X,M,\star)\) such that X is a nonempty set, ⋆ is a continuous t-norm and M is a fuzzy set on \(X\times X \times(0,+\infty)\) satisfying the following conditions, for all \(x,y,z\in X\) and \(t,s>0\):
-
(FM1)
\(M(x,y,t)>0\) for all \(t>0\);
-
(FM2)
\(M(x,y,t)=1\) if and only if \(x=y\);
-
(FM3)
\(M(x,y,t)=M(y,x,t)\);
-
(FM4)
\(M(x,y,t)\star M(y,z,s)\leq M(x,z,t+s)\);
-
(FM5)
\(M(x,y,\cdot): (0,+\infty)\to(0,1]\) is left continuous.
Then the triple \((X,M,\star)\) is called a fuzzy metric space. If we replace (FM4) by
-
(FM6)
\(M(x,y,t)\star M(y,z,s)\leq M(x,z,\max\{t,s\})\),
then the triple \((X,M,\star)\) is called a non-Archimedean fuzzy metric space. Since (FM6) implies (FM4), then each non-Archimedean fuzzy metric space is a fuzzy metric space.
Definition 1.3
Let \((X,M,\star)\) be a fuzzy metric space (or non-Archimedean fuzzy metric space). Then
-
(i)
a sequence \(\{x_{n}\}\) converges to \(x\in{X}\) if and only if \(\lim_{n\to+\infty}M(x_{n},x,t)=1\) for all \(t>0\);
-
(ii)
a sequence \(\{x_{n}\}\) in X is a Cauchy sequence if and only if for all \(\epsilon\in(0,1)\) and \(t>0\), there exists \(n_{0}\) such that \(M(x_{n},x_{m},t)>1 -\epsilon\) for all \(m, n \geq n_{0}\);
-
(iii)
the fuzzy metric space (or the non-Archimedean fuzzy metric space) is called complete if every Cauchy sequence converges to some \(x\in{X}\).
If \((X,M,\star)\) is a fuzzy metric space and \((X,\preceq)\) is partially ordered, then \((X,M,\star)\) is called a partially ordered fuzzy metric space. Then \(x,y\in X\) are called comparable if \(x\preceq y\) or \(y\preceq x\) holds. Let \(f:X\rightarrow X\) be a mapping, f is said to be non-decreasing if \(fx\preceq fy\) whenever \(x,y\in X\) and \(x\preceq y\).
Definition 1.4
[3]
Let \((X,M,\ast)\) be a triangular fuzzy metric space. The fuzzy metric M is called triangular whenever
Definition 1.5
[17]
Let \(f:X\to X\) and \(\alpha: X \times X \to[0,+\infty)\). f is an α-admissible mapping if
Definition 1.6
[26]
Let \((X,M,\ast)\) be a fuzzy metric space, \(T:X\to X\) and \(\alpha: X\times X \times(0,\infty)\to[0,\infty)\). We say that T is an α-admissible mapping if
for all \(t>0\).
2 Fixed point results in triangular fuzzy metric spaces
Let \((X,M,\ast)\) be a fuzzy metric space, \(f:X\to X\) be a self-mapping on X. We define \(P^{f}(x,y,t)\), \(Q^{f}(x,y,t)\) and \(R^{f}(x,y,t)\) as follows:
and
We now state and prove our first result of this section.
Theorem 2.1
Let \((X,M,\ast)\) be a complete triangular fuzzy metric space and f be a self-mapping on X. Also suppose that \(\alpha:X\times X\times[0,\infty)\to[0,\infty)\) is a mapping. Assume that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\);
-
(ii)
f is an α-admissible mapping;
-
(iii)
if \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\) and all \(t>0\) with \(x_{n}\to x\) as \(n\to\infty\), then \(\alpha(x_{n},x,t)\geq t\) for all \(n\in\mathbb{N}\cup\{0\}\);
-
(iv)
for all \(x,y\in X\) and all \(t>0\) with \(\frac{1}{1+\lambda}(\frac{1}{M(x,fx,t)}-1)\leq \frac{1}{M(x,y,t)}-1\), we have
$$ \frac{\alpha(x,y,t)}{tM(fx,fy,t)}\leq\lambda P^{f}(x,y,t)+\bigl\vert Q^{f}(x,y,t)-\lambda\bigr\vert +LR^{f}(x,y,t), $$(2.1)where \(\lambda\in (0,1)\) and \(L\geq0\).
Then f has a fixed point.
Proof
Let \(x_{0} \in X\) be such that \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\). Define a sequence \(\{x_{n}\}\) in X by \(x_{n}=f^{n}x_{0}=fx_{n-1}\) for all \(n\in{\mathbb{N}}\). If \(x_{n+1}= x_{n}\) for some \(n\in{\mathbb{N}}\), then \(x=x_{n}\) is a fixed point for f and the result is proved. Hence, we suppose that \(x_{n+1}\neq x_{n}\) for all \(n\in{\mathbb{N}}\). Since f is an α-admissible mapping and \(\alpha(x_{0},fx_{0},t)\geq t\), we deduce that \(\alpha(x_{1},x_{2},t)=\alpha(fx_{0},f^{2}x_{0},t)\geq t\). Continuing this process, we get
for all \(n\in\mathbb{N}\cup\{0\}\) and all \(t>0\). Now since
then from (iv) with \(x=x_{n-1}\) and \(y=x_{n}\) we get
That is,
for all \(t>0\) and all \(n\in\mathbb{N}\), where
which implies
and
Also,
From (2.3), (2.4), (2.5) and (2.6) we obtain
which implies
Now if \(P^{f}(x_{n-1},x_{n},t)= \frac{1}{M(x_{n},x_{n+1},t)}\), then we get
which is a contradiction. Hence,
for all \(n\in\mathbb{N}\) and all \(t>0\). This implies
Thus, for all \(n>m\), we have
That is, \(\{x_{n}\}\) is a Cauchy sequence. Since, X is a complete fuzzy metric space, then there exists \(x^{*}\in X\) such that \(x_{n}\to x^{*}\) as \(n\to\infty\). By (iii), \(\alpha(x_{n},x^{*},t)\geq t\) holds for all \(n\in\mathbb{N}\) and all \(t>0\). Suppose that there exists \(n_{0}\in \mathbb{N}\) such that
and
By (2.7) we deduce
which is a contradiction. Hence, either
or
holds for all \(n\in\mathbb{N}\). Let
Then from (iv) we have
for all \(t>0\), where
and so
Also,
and so
Similarly,
By taking limit as \(n\to\infty\) in (2.10) we get
for all \(t>0\). Now, assume that there exists \(t_{0}\) such that \(M(x^{*},fx^{*},t_{0})<1\). Then by the above inequality we have \(1\leq\lambda\), which is a contradiction. Hence, \(M(x^{*},fx^{*},t)=1\) for all \(t>0\); i.e., \(x^{*}=fx^{*}\). Similarly we can deduce that \(x^{*}\) is a fixed point of f when
□
Example 2.1
Let \(X = \mathbb{R}^{2}\). We define \(\alpha: X\times X\times[0, \infty) \to[0,\infty)\) by
Define M on \(X\times X\times(0,\infty)\) by \(M((x_{1},x_{2}),(y_{1},y_{2}),t) = \frac{1}{1+ |x_{1} - y_{1}|+|x_{2} - y_{2}|}\) and \(a\star b=\min\{a,b\}\). Clearly, \((M,X,\star)\) is a complete triangular fuzzy metric space. Also, define \(f: X \to X\) and \(\psi: [0,\infty)\to[0,\infty)\) by
First we assume
and \(\alpha(x,y,t)\geq t\) (or \(x,y\in U\)). Then
Since \(M(fx,fy,t)=M(fy,fx,t)\), \(P^{f}(x,y,t)=P^{f}(y,x,t)\) and \(Q^{f}(x,y,t)=Q^{f}(y,x,t)\), hence without any loss of generality we can reduce the above set to the following:
Now, we consider the following cases:
• Let \((x,y)= ((0,0), (4,0) )\), then
and
and so
• Let \((x,y)= ((0,0), (0,4) )\), then
and
and so
• Let \((x,y)= ((0,0), (4,5) )\), then
and
and so
• Let \((x,y)= ((0,0), (5,4) )\), then
and
and so
• Let \((x,y)= ((4,0), (0,4) )\), then
and
and so
• Let \((x,y)= ((4,0), (5,4) )\), then
and
and so
• Let \((x,y)= ((4,0), (4,5) )\), then
and
and so
• Let \((x,y)= ((0,4), (5,4) )\), then
and
and so
• Let \((x,y)= ((0,4), (4,5) )\), then
and
and so
Otherwise, \(\alpha(x,y,t)=0\), and so
That is,
implies
where \(L\geq0\). Let \(\alpha(x,y,t)\geq t\), then \(x,y\in U\). On the other hand, \(fw\in U\) for all \(w\in U\). Then \(\alpha(fx,fy,t)\geq t\). That is, f is an α-admissible mapping. If \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1},t)\geq t\) with \(x_{n}\to x\) as \(n\to\infty\), then \(x_{n}\in U\) for all \(n\in\mathbb{N}\). Also, U is a closed set, so \(x\in U\). That is, \(\alpha(x_{n},x,t)\geq t\) for all \(n\in \mathbb{N}\cup\{0\}\). Clearly, \(\alpha((0,0),f(0,0),t)\geq t\).
Therefore all conditions of Theorem 2.1 hold and f has a fixed point. Here, \(x=(0,0)\) is a fixed point of f.
Corollary 2.1
Let \((X,M,\ast)\) be a complete triangular fuzzy metric space and f be a self-mapping on X. Also suppose that \(\alpha:X\times X\times[0,\infty)\to[0,\infty)\) is a mapping. Assume that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\);
-
(ii)
f is an α-admissible mapping;
-
(iii)
if \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\) and all \(t>0\) with \(x_{n}\to x\) as \(n\to\infty\), then \(\alpha(x,x_{n},t)\geq t\) for all \(n\in\mathbb{N}\cup\{0\}\);
-
(iv)
for all \(x,y\in X\) and all \(t>0\), we have
$$ \frac{\alpha(x,y,t)}{tM(fx,fy,t)}\leq\lambda P^{f}(x,y,t)+\bigl\vert Q^{f}(x,y,t)-\lambda\bigr\vert +LR^{f}(x,y,t) $$holds where \(\lambda\in (0,1)\) and \(L\geq0\).
Then f has a fixed point.
By taking \(L=0\) in Corollary 2.1, we obtain the following corollary.
Corollary 2.2
Let \((X,M,\ast)\) be a complete triangular fuzzy metric space and f be a self-mapping on X. Also suppose that \(\alpha:X\times X\times[0,\infty)\to[0,\infty)\) is a mapping. Assume that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\);
-
(ii)
f is an α-admissible mapping;
-
(iii)
if \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\) and all \(t>0\) with \(x_{n}\to x\) as \(n\to\infty\), then \(\alpha(x,x_{n},t)\geq t\) for all \(n\in\mathbb{N}\cup\{0\}\);
-
(iv)
for all \(x,y\in X\) and all \(t>0\), we have
$$ \frac{\alpha(x,y,t)}{tM(fx,fy,t)}\leq\lambda P^{f}(x,y,t)+\bigl\vert Q^{f}(x,y,t)-\lambda\bigr\vert $$holds where \(\lambda\in (0,1)\).
Then f has a fixed point.
By taking \(\alpha(x,y,t)=t\) for all \(x,y\in X\) and all \(t>0\) in Corollary 2.1, we obtain the following result.
Corollary 2.3
Let \((X,M,\ast)\) be a complete triangular fuzzy metric space and f be a self-mapping on X. Assume that for all \(x,y\in X\) and all \(t>0\),
holds where \(\lambda\in (0,1)\) and \(L\geq0\). Then f has a fixed point.
By taking \(L=0\) in Corollary 2.3, we obtain the following corollary.
Corollary 2.4
Let \((X,M,\ast)\) be a complete triangular fuzzy metric space and f be a self-mapping on X. Assume that for all \(x,y\in X\) and all \(t>0\),
holds where \(\lambda\in (0,1)\). Then f has a fixed point.
Theorem 2.2
Let \((X,M,\ast, \preceq)\) be a complete triangular partially ordered fuzzy metric space and f be a self-mapping on X. Assume that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(x_{0}\preceq fx_{0}\);
-
(ii)
f is an increasing mapping;
-
(iii)
if \(\{x_{n}\}\) is an increasing sequence in X such that with \(x_{n}\to x\) as \(n\to\infty\), then \(x\preceq x_{n}\) for all \(n\in\mathbb{N}\cup \{0\}\);
-
(iv)
for all \(x,y\in X\) and all \(t>0\) with \(\frac{1}{1+\lambda }(\frac{1}{M(x,fx,t)}-1)\leq \frac{1}{M(x,y,t)}-1\) and \(x\preceq y\), we have
$$ \frac{1}{M(fx,fy,t)}\leq\lambda P^{f}(x,y,t)+\bigl\vert Q^{f}(x,y,t)-\lambda(t)\bigr\vert +LR^{f}(x,y,t), $$where \(\lambda\in (0,1)\) and \(L\geq0\).
Then f has a fixed point.
Proof
Define \(\alpha:X\times X\times(0,\infty)\to[0,+\infty)\) by \(\alpha(x,y,t)= \bigl\{\scriptsize{ \begin{array}{l@{\quad}l} t, & \mbox{if } x \preceq y, \\ 0, & \mbox{otherwise}. \end{array}} \bigr. \) At first we prove that f is an α-admissible mapping. Let \(\alpha(x,y,t)\geq t\), then \(x\preceq y\). Now, since f is increasing, then we have \(fx\preceq fy\). That is, \(\alpha(fx,fy,t)\geq t\). Therefore f is an α-admissible mapping. From (i) there exists \(x_{0}\in X\) such that \(x_{0}\preceq fx_{0}\). That is, \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\). If \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\) and all \(t>0\) with \(x_{n}\to x\) as \(n\to\infty\), then \(x_{n}\preceq x_{n+1}\) for all \(n\in\mathbb{N}\) with \(x_{n}\to x\) as \(n\to\infty\). Hence from (iii) we get \(x_{n}\preceq x\) for all \(n\in\mathbb{N}\cup\{0\}\). That is, \(\alpha(x_{n}, x,t)\geq t\) for all \(n\in\mathbb{N}\cup\{0\}\).
Let \(\frac{1}{1+\lambda}(\frac{1}{M(x,fx,t)}-1)\leq \frac{1}{M(x,y,t)}-1\) and \(x\preceq y\) (or \(\alpha(x,y,t)=t\)), then from (iv) we have
Also if \(\alpha(x,y,t)=0\), then
That is, for all \(x,y\in X\) and all \(t>0\) with \(\frac{1}{1+\lambda}(\frac{1}{M(x,fx,t)}-1)\leq \frac{1}{M(x,y,t)}-1\), we have
Hence, all conditions of Theorem 2.1 are satisfied and f has a fixed point. □
Corollary 2.5
Let \((X,M,\ast, \preceq)\) be a complete triangular partially ordered fuzzy metric space and f be a self-mapping on X. Assume that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(x_{0}\preceq fx_{0}\);
-
(ii)
f is an increasing mapping;
-
(iii)
if \(\{x_{n}\}\) is an increasing sequence in X such that with \(x_{n}\to x\) as \(n\to\infty\), then \(x\preceq x_{n}\) for all \(n\in\mathbb{N}\cup \{0\}\);
-
(iv)
for all \(x,y\in X\) and all \(t>0\) with \(x\preceq y\), we have
$$ \frac{1}{M(fx,fy,t)}\leq\lambda P^{f}(x,y,t)+\bigl\vert Q^{f}(x,y,t)-\lambda\bigr\vert +LR^{f}(x,y,t), $$where \(\lambda\in (0,1)\) and \(L\geq0\).
Then f has a fixed point.
Corollary 2.6
Let \((X,M,\ast, \preceq)\) be a complete triangular partially ordered fuzzy metric space and f be a self-mapping on X. Assume that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(x_{0}\preceq fx_{0}\);
-
(ii)
f is an increasing mapping;
-
(iii)
if \(\{x_{n}\}\) is an increasing sequence in X such that with \(x_{n}\to x\) as \(n\to\infty\), then \(x\preceq x_{n}\) for all \(n\in\mathbb{N}\cup \{0\}\);
-
(iv)
for all \(x,y\in X\) and all \(t>0\) with \(x\preceq y\), we have
$$ \frac{1}{M(fx,fy,t)}\leq\lambda P^{f}(x,y,t)+\bigl\vert Q^{f}(x,y,t)-\lambda\bigr\vert , $$where \(\lambda\in (0,1)\).
Then f has a fixed point.
3 Some results in non-Archimedean fuzzy metric spaces
In this section we state and prove certain fixed point results in the setting of non-Archimedean fuzzy metric space to generalize the work of Miheţ [14].
Theorem 3.1
Let \((X,M,\star, \preceq)\) be a partially ordered complete non-Archimedean fuzzy metric space and f be an α-admissible and non-increasing mapping. Also suppose that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\) and \(x_{0}\preceq fx_{0}\);
-
(ii)
if \(\{x_{n}\}\) is an increasing sequence such that \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\) and all \(t>0\) with \(x_{n}\to x\) as \(n\to+\infty\), then \(x_{n}\preceq x\) for all \(n\in\mathbb{N}\) where \(\alpha(x,fx,t)\geq t\);
-
(iii)
for all comparable \(x,y\in X\) and all \(t>0\), we have
$$\begin{aligned}& \bigl[\alpha(x,fx,t)\alpha(y,fy,t)+1 \bigr]^{ \psi(M(fx,fy,t))} \biggl[t^{2}+ \frac{\alpha(x,fx,t)+\alpha(y,fy,t)}{2t} \biggr]^{ \phi(M(x,y,t))} \\& \quad \leq \bigl[t^{2}+1 \bigr]^{ \psi(M(x,y,t))}, \end{aligned}$$where \(\psi, \phi: [0,1]\to[0,1]\) are two continuous functions such that ψ is decreasing, \(\psi(t)=0\) iff \(t=1\) and \(\phi(t)>0\) for all \(t\in(0,1)\).
Then f has a fixed point.
Proof
Let \(x_{0}\preceq fx_{0}\). If \(x_{0}= fx_{0}\), then the result is proved. Hence we suppose that \(x_{0}\prec fx_{0}\). Define a sequence \(\{x_{n}\}\) by \(x_{n}=f^{n}x_{0}=fx_{n-1}\) for all \(n\in{\mathbb{N}}\). Since f is non-decreasing and \(x_{0}\prec fx_{0}\), then
and hence \(\{x_{n}\}\) is a non-decreasing sequence. If \(x_{n}=x_{n+1}= fx_{n}\) for some \(n \in\mathbb{N}\), then the result is proved as \(x_{n}\) is a fixed point of f. In what follows we suppose that \(0< M(x_{n},x_{n+1}, t)<1\). Since f is an α-admissible mapping and
we deduce that
Continuing this process, we get \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\cup\{0\}\) and all \(t>0\). From (iii) with \(x=x_{n-1}\) and \(y=x_{n}\), we obtain
and so
Since ψ is decreasing, so \(M(x_{n-1},x_{n},t)\leq M(x_{n},x_{n+1},t)\). Hence, \(\{M(x_{n},x_{n+1},t)\}\) is an increasing sequence in \((0,1]\). Then there exists \(l(t)\in(0,1]\) such that
for all \(t>0\). Let us prove that \(l(t)=1\) for all \(t>0\). Suppose that there exists \(t_{0}>0\) such that \(0< l(t_{0})<1\). By taking the limit as \(n\to+\infty\) in (3.2), we have
Then \(\phi(l(t_{0}))=0\), which is a contradiction, and so \(l(t)=1\) for all \(t>0\). Now, we want to show that \(\{x_{n}\}\) is a Cauchy sequence. Assume that it is not true. Then there exist \(\epsilon\in(0,1)\) and \(t_{0}>0\) such that for all \(k\in\mathbb{N}\) there exist \(n(k),m(k)\in \mathbb{N}\) with \(m(k)>n(k)\geq k\) and
Assume that \(m(k)\) is the least integer exceeding \(n(k)\) satisfying the above inequality. Equivalently,
and so for all k we get
By taking limit as \(n\to+\infty\) in (3.5), we deduce that
for \(t>0\).
From
and
we get
From (iii) with \(x=x_{m(k)}\) and \(y=x_{n(k)}\), we deduce
which implies
Applying the continuity of the functions ϕ and ψ, by taking the limit as \(k\to+\infty\) in the above inequality, we get
and so \(\phi(1-\epsilon)=0\), which is a contradiction. Then \(\{x_{n}\}\) is a Cauchy sequence. Since \((X,M,\star)\) is a complete non-Archimedean fuzzy metric space, then the sequence \(\{x_{n}\}\) converges to some \(z\in{X}\), that is, for all \(t>0\),
Assume that there exists \(t_{0}>0\) such that \(0< M(z,fz,t_{0})<1\). Then by (2.11) and (ii) we get
and hence
By taking the limit as \(n\to+\infty\) in the above inequality, we have
Then \(\psi(M(z,fz,t_{0}))=0\), i.e., \(M(z,fz,t_{0})=1\), which is a contradiction. Hence, \(M(z,fz,t)=1\) for all \(t>0\), that is, \(z=fz\). □
If in Theorem 3.1 we take \(\alpha(x,y,t)=t\) for all \(x,y\in X\) and all \(t>0\), then we deduce the following corollary.
Corollary 3.1
Let \((X,M,\star,\preceq)\) be a partially ordered complete non-Archimedean fuzzy metric space, \(\psi, \phi: [0,1]\to[0,1]\) as in Theorem 3.1 and \(f:X\to X\) be an increasing mapping such that
holds for all comparable \(x,y\in X\). If the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(x_{0}\preceq fx_{0}\);
-
(ii)
if \(\{x_{n}\}\) is an increasing sequence such that \(x_{n}\to x\) as \(n\to +\infty\), then \(x_{n}\preceq x\) for all \(n\in\mathbb{N}\).
Then f has a fixed point.
Theorem 3.2
Let \((X,M,\star, \preceq)\) be a partially ordered complete non-Archimedean fuzzy metric space and f be an α-admissible and non-increasing mapping such that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},fx_{0},t)\geq t\) for all \(t>0\) and \(x_{0}\preceq fx_{0}\);
-
(ii)
if \(\{x_{n}\}\) is an increasing sequence such that \(\alpha(x_{n},x_{n+1},t)\geq t\) for all \(n\in\mathbb{N}\cup\{0\}\), all \(t>0\) and \(x_{n}\to x\) as \(n\to+\infty\), then \(x_{n}\preceq x\) for all \(n\in\mathbb{N}\) where \(\alpha(x,fx,t)\geq t\);
-
(iii)
assume that there exists a function \(\beta:[0,1] \to [1,+\infty)\) such that for any sequence \(\{t_{n}\}\subseteq[0,1]\) of positive reals, \(\beta(t_{n})\to1\) implies \(t_{n} \to1\) such that
$$\begin{aligned}& \bigl[t^{2}+1 \bigr]^{ M(fx,fy,t)} \\& \quad \geq \biggl[ \frac{\alpha(x,fx,t)\alpha(y,fy,t)}{t^{2}} \biggr] \bigl[\alpha(x,fx,t)\alpha(y,fy,t)+1 \bigr]^{ \beta(M(x,y,t))M(x,y,t)} \end{aligned}$$(3.6)holds for all comparable \(x,y\in{X}\) and all \(t>0\).
Then f has a fixed point.
Proof
Let \(x_{0}\preceq fx_{0}\). If \(x_{0}= fx_{0}\), then the result is proved. Hence we suppose that \(x_{0}\prec fx_{0}\). Define a sequence \(\{x_{n}\}\) by \(x_{n}=f^{n}x_{0}=fx_{n-1}\) for all \(n\in{\mathbb{N}}\). Since f is non-decreasing and \(x_{0}\prec fx_{0}\), then
and hence \(\{x_{n}\}\) is a non-decreasing sequence. If \(x_{n}=x_{n+1}= fx_{n}\) for some \(n \in\mathbb{N}\), then the result is proved as \(x_{n}\) is a fixed point of f. In what follows we suppose that \(0< M(x_{n},x_{n+1}, t)<1\). Since f is an α-admissible mapping with respect to η and \(\alpha(x_{0},fx_{0},t)= \alpha(x_{0},x_{1},t)\geq t\), we deduce that \(\alpha(x_{1},x_{2},t)=\alpha(fx_{0},fx_{1},t)\geq t\). By continuing this process, we get \(\alpha(x_{n}, x_{n+1}, t)\geq t\) for all \(n\in \mathbb{N}\cup\{0\}\) and all \(t>0\). From (3.6) we get
Thus
Hence,
That is, \(\{S_{n}=M(x_{n},x_{n+1},t)\}\) is an increasing sequence in \((0,1]\). Then there exists \(l(t)\in(0,1]\) such that \(\lim_{n\to +\infty}M(x_{n},x_{n+1},t)=l(t)\) for all \(t>0\). We shall prove that \(l(t)=1\) for all \(t>0\). By (3.8) we deduce
which implies \(\lim_{n\to+\infty}\beta(M(x_{n-1},x_{n},t))=1\). Regarding the property of the function β, we conclude that
Next, we shall prove that \(\{x_{n}\}\) is a Cauchy sequence. Suppose, to the contrary, that \(\{x_{n}\}\) is not a Cauchy sequence. Proceeding as in the proof of Theorem 3.1, there exist \(\epsilon\in(0,1)\) and \(t_{0}>0\) such that for all \(k\in\mathbb{N}\) there exist \(n(k),m(k)\in\mathbb{N}\) with \(m(k)>n(k)\geq k\) such that
and
From (3.6) with \(x=x_{m(k)}\) and \(y=x_{n(k)}\) we deduce
which implies
and so
Taking the limit as \(k\to+\infty\) in the above inequality, we get
which implies
and so \(\epsilon=0\), which is a contradiction. Then \(\{x_{n}\}\) is a Cauchy sequence. Since \((X,M,\star)\) is a complete space, then the sequence \(\{x_{n}\}\) converges to some \(z\in{X}\) such that for all \(t>0\),
By (3.6) and (ii) we get
and hence
Taking the limit as \(n\to+\infty\) in the above inequality, we have \(\lim_{n\to+\infty}M(fx_{n},fz,t)=1\) for all \(t>0\) and then
that is, \(z=fz\). □
Example 3.1
Let \((X,M,\star)\) be the non-Archimedean fuzzy metric space where \(M(x,y,t)=\frac{\min\{x,y\}}{\max\{x,y\}}\) for all \(t>0\) and \(a\star b=\min\{a,b\}\). Define \(f:X\to X\) with
Also define \(\alpha(x,y,t)= \bigl\{\scriptsize{ \begin{array}{l@{\quad}l} t, & \mbox{if } x,y\in[1,3], \\ 0, & \mbox{otherwise} \end{array}} \bigr.\) and \(\beta(t)=1\). Also, \(x\preceq y\) iff \(x\leq y\).
Let, \(x,y\in[1,3]\) and \(x\leq y\). Then
Otherwise, \(\alpha(x,fx,t)\alpha(y,fy,t)=0\), and so
Clearly, \(\alpha(1, f1, t)\geq t\) and \(1\leq f1\). Now, if \(\{x_{n}\}\) is an increasing sequence in X such that \(\alpha(x_{n}, x_{n+1}, t)\geq1\) for all \(n\in\mathbb{N}\cup\{0\}\) and \(x_{n} \to x\) as \(n\to\infty\), then \(\{x_{n}\}\subset[0, 1]\) and hence \(x\in[0, 1]\). This implies that \(x_{n} \to x\) for all \(n\in\mathbb{N}\) and \(\alpha(x, fx, t)\geq t\). Hence, all the conditions of Theorem 3.2 hold and f has a fixed point. Then, by Theorem 3.2, f has a fixed point.
If in Theorem 3.2 we take \(\alpha(x,y,t)=t\) for all \(x,y\in X\), then we deduce the following result.
Corollary 3.2
Let \((X,M,\star,\preceq)\) be a complete non-Archimedean fuzzy metric space, and f be an increasing mapping on X. Assume that there exists a function \(\beta:[0,1] \to [1,+\infty)\) such that for any sequence \(\{t_{n}\}\subseteq[0,1]\) of positive reals, \(\beta(t_{n})\to 1\) implies \(t_{n} \to1\) and
for all comparable \(x,y\in{X}\) and all \(t>0\). Also suppose that the following assertions hold:
-
(i)
there exists \(x_{0}\in X\) such that \(x_{0}\preceq fx_{0}\);
-
(ii)
if \(\{x_{n}\}\) is an increasing sequence such that \(x_{n}\to x\) as \(n\to +\infty\), then \(x_{n}\preceq x\) for all \(n\in\mathbb{N}\).
Then f has a fixed point.
4 Application to integral equations
Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various applications in differential and integral equations (see [27–31] and the references therein). Let \(X = C([0, T],\mathbb{R})\) be the set of real continuous functions defined on \([0, T]\) and \(M: X\times X \times(0,+\infty) \to[0,1]\) be defined by
for all \(x,y \in X\) and all \(r>0\). Also define \(a\star b=\min\{a,b\}\). Then \((M,X,\star)\) is a complete triangular fuzzy metric space.
Consider the integral equation
and the mapping \(F: X \to X\) defined by
where
-
(A)
\(f: [0, T] \times\mathbb{R} \to\mathbb{R}\) is continuous;
-
(B)
\(p : [0, T] \to\mathbb{R}\) is continuous;
-
(C)
\(S : [0, T] \times[0, T] \to[0,+\infty)\) is continuous;
-
(D)
there exist \(\theta: X\times X \to \mathbb{R}\) and \(\lambda\in(0,1)\) such that if \(\theta(x,y)\geq 0\) for \(x,y\in X\), then for all \(s \in[0,T]\) and all \(r>0\) we have
$$\begin{aligned} \bigl\vert f\bigl(s,x(s)\bigr)-f\bigl(s,y(s)\bigr)\bigr\vert \leq& \lambda \max\bigl\{ |x-y|, \bigl\vert x(s)-Fx(s)\bigr\vert ,\bigl\vert y(s)-Fy(s) \bigr\vert \bigr\} \\ &{}+r\bigl\vert Q^{F}(x,y,r)-\lambda\bigr\vert +\lambda r-r; \end{aligned}$$ -
(F)
$$\int_{0}^{T}S(t, s)\, ds\leq1\quad \mbox{for all } t\in[0,T]; $$
-
(G)
there exists \(x_{0} \in X \) such that \(\theta (x_{0},Fx_{0})\geq0\);
-
(H)
$$\theta(x,y)\geq0 \quad \mbox{for some } x\in X \quad \mbox{implies}\quad \theta (Fx,Fy)\geq0; $$
-
(I)
if \(\{x_{n}\}\) is a sequence in X such that \(\theta(x_{n},x_{n+1})\geq0\) for all \(n\in\mathbb{N}\cup\{0\}\) and \(x_{n}\to x\) as \(n\to+\infty\), then \(\theta(x_{n},x)\geq0\).
Theorem 4.1
Under the assumptions (A)-(I), the integral equation (4.1) has a solution in \(X= C([0, T],\mathbb{R})\).
Proof
Let \(F:X\to X\) be defined by (4.2) and let \(x,y \in X\) be such that \(\theta(x,y)\geq0\). By the condition (D), we deduce that
and so
Now we define \(\alpha:X\times(0,\infty) \to[0,+\infty)\) by
Since \(\theta(x,y)\geq0\), so, \(\alpha(x,y,r)=r\). Therefore we can write
Thus all of the conditions of Corollary 2.2 are satisfied and hence the mapping F has a fixed point which is a solution of the integral equation (4.1) in \(X = C([0, T],\mathbb{R})\). □
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This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the first author acknowledges with thanks DSR, KAU for financial support.
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Hussain, N., Hezarjaribi, M. & Salimi, P. Suzuki type theorems in triangular and non-Archimedean fuzzy metric spaces with application. Fixed Point Theory Appl 2015, 134 (2015). https://doi.org/10.1186/s13663-015-0386-2
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DOI: https://doi.org/10.1186/s13663-015-0386-2