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Coupled coincidence point theorems in (intuitionistic) fuzzy normed spaces
Journal of Inequalities and Applications volume 2013, Article number: 104 (2013)
Abstract
In this paper, we prove some coupled coincidence point theorems in fuzzy normed spaces. Our results improve and restate the proof lines of the main results given in the papers (Eshaghi Gordji et al. in Math. Comput. Model. 54:18971906, 2011) and (Sintunavarat et al. in Fixed Point Theory Appl. 2011:81, 2011).
MSC:47H10, 54H25, 34B15.
1 Introduction
Recently, many authors have shown the existence of coupled fixed points and common fixed points for some contractions in cone metric spaces, partially ordered metric spaces, fuzzy metric spaces, fuzzy normed spaces, intuitionistic fuzzy normed spaces and others ([1–11]).
Especially in [11], Sintunavarat et al. proved some coupled fixed point theorems for contractive mappings in partially complete intuitionistic fuzzy normed spaces, which extended and improved coupled coincidence point theorems in Gordji et al. [5]. But the authors found some mistakes in the proof lines of the main result (Theorem 2.5) of [5] and the same mistakes in [11].
In Section 2 of this paper, we restate some definitions and the main results in [11]. In Section 3, we give some comments about the incorrect proof lines of the main results given in [5] and [11] and explain why the lines of the proofs are wrong. Finally, in Section 4, we extend and improve some coupled fixed point theorems.
2 Preliminaries
A tnorm (resp., a tconorm) is a mapping \ast :{[0,1]}^{2}\to [0,1] (resp., \diamond :{[0,1]}^{2}\to [0,1]) that is associative, commutative and nondecreasing in both arguments and has 1 (resp., 0) as identity.
Definition 2.1 [9]
A fuzzy normed space (in brief, FNS) is a triple (X,\mu ,\ast ), where X is a vector space, ∗ is a continuous tnorm and \mu :X\times (0,\mathrm{\infty})\to [0,1] is a fuzzy set such that, for all x,y\in X and t,s>0,
(F1) \mu (x,t)>0;
(F2) \mu (x,t)=1 for all t>0 if and only if x=0;
(F3) \mu (\alpha x,t)=\mu (x,\frac{t}{\alpha }) for all \alpha \ne 0;
(F4) \mu (x,t)\ast \mu (y,s)\le \mu (x+y,t+s);
(F5) \mu (x,\cdot ):(0,\mathrm{\infty})\to [0,1] is continuous;
(F6) {lim}_{t\to \mathrm{\infty}}\mu (x,t)=1 and {lim}_{t\to 0}\mu (x,t)=0.
Using the continuous tnorms and tconorms, Saadati and Park [12] introduced the concept of an intuitionistic fuzzy normed space.
Definition 2.2 An intuitionistic fuzzy normed space (in brief, IFNS) is a 5tuple (X,\mu ,\nu ,\ast ,\diamond ) where X is a vector space, ∗ is a continuous tnorm, ⋄ is a continuous tconorm and \mu ,\nu :X\times (0,\mathrm{\infty})\to [0,1] are fuzzy sets such that, for all x,y\in X and t,s>0,
(IF1) \mu (x,t)+\nu (x,t)\le 1;
(IF2) \mu (x,t)>0 and \nu (x,t)<1;
(IF3) \mu (x,t)=1 for all t>0 if and only if x=0 if and only if \nu (x,t)=0 for all t>0;
(IF4) \mu (\alpha x,t)=\mu (x,\frac{t}{\alpha }) and \nu (\alpha x,t)=\nu (x,\frac{t}{\alpha }) for all \alpha \ne 0;
(IF5) \mu (x,t)\ast \mu (y,s)\le \mu (x+y,t+s) and \nu (x,t)\diamond \nu (y,s)\ge \nu (x+y,t+s);
(IF6) \mu (x,\cdot ),\nu (x,\cdot ):(0,\mathrm{\infty})\to [0,1] are continuous;
(IF7) {lim}_{t\to \mathrm{\infty}}\mu (x,t)=1={lim}_{t\to 0}\nu (x,t) and {lim}_{t\to 0}\mu (x,t)=0={lim}_{t\to \mathrm{\infty}}\nu (x,t).
Obviously, if (X,\mu ,\nu ,\ast ,\diamond ) is an IFNS, then (X,\mu ,\ast ) is an FNS. We refer to this space as its support.
Lemma 2.1 \mu (x,\cdot ) is a nondecreasing function on (0,\mathrm{\infty}) and \nu (x,\cdot ) is a nonincreasing function on (0,\mathrm{\infty}).
Some properties and examples of IFNS and the concepts of convergence and a Cauchy sequence in IFNS are given in [12].
Definition 2.3 [12]
Let (X,\mu ,\nu ,\ast ,\diamond ) be an IFNS.

(1)
A sequence \{{x}_{n}\}\subset X is called a Cauchy sequence if, for any \u03f5>0 and t>0, there exists {n}_{0}\in \mathbb{N} such that \mu ({x}_{n}{x}_{m},t)>1\u03f5 and \nu ({x}_{n}{x}_{m},t)<\u03f5 for all n,m\ge {n}_{0}.

(2)
A sequence \{{x}_{n}\}\subset X is said to be convergent to a point x\in X denoted by {x}_{n}\to x or by {lim}_{n\to \mathrm{\infty}}{x}_{n}=x if, for any \u03f5>0 and t>0, there exists {n}_{0}\in \mathbb{N} such that \mu ({x}_{n}x,t)>1\u03f5 and \nu ({x}_{n}x,t)<\u03f5 for all n\ge {n}_{0}.

(3)
An IFNS in which every Cauchy sequence is convergent is said to be complete.
Most of the following definitions were introduced in [8].
Definition 2.4 Let F:X\times X\to X and g:X\to X be two mappings.

(1)
F and g are said to be commuting if gF(x,y)=F(gx,gy) for all x,y\in X.

(2)
A point (x,y)\in X\times X is called a coupled coincidence point of the mappings F and g if F(x,y)=gx and F(y,x)=gy. If g is the identity, (x,y) is called a coupled fixed point of F.

(3)
If (X,\u2291) is a partially ordered set, then F is said to have the mixed gmonotone property if it verifies the following properties:
If g is the identity mapping, then F is said to have the mixed monotone property.

(4)
If (X,\u2291) is a partially ordered set, then X is said to have the sequential gmonotone property if it verifies the following properties:
(B1) If \{{x}_{n}\} is a nondecreasing sequence and {lim}_{n\to \mathrm{\infty}}{x}_{n}=x, then g{x}_{n}\u2291gx for all n\in \mathbb{N}.
(B2) If \{{x}_{n}\} is a nonincreasing sequence and {lim}_{n\to \mathrm{\infty}}{y}_{n}=y, then g{y}_{n}\u2292gy for all n\in \mathbb{N}.
If g is the identity mapping, then X is said to have the sequential monotone property.
Definition 2.5 Let X and Y be two IFNS. A function f:X\to Y is said to be continuous at a point {x}_{0}\in X if, for any sequence \{{x}_{n}\} in X converging to {x}_{0}, the sequence \{f({x}_{n})\} in Y converges to f({x}_{0}). If f is continuous at each x\in X, then f is said to be continuous on X.
Definition 2.6 [5]
Let (X,\mu ,\nu ,\ast ,\diamond ) be an IFNS. The pair (\mu ,\nu ) is said to satisfy the nproperty on X\times (0,\mathrm{\infty}) if {lim}_{n\to \mathrm{\infty}}{[\mu (x,{k}^{n}t)]}^{{n}^{p}}=1 and {lim}_{n\to \mathrm{\infty}}{[\nu (x,{k}^{n}t)]}^{{n}^{p}}=0 whenever x\in X, k>1 and p>0.
The following lemma proved by Haghi et al. [7] is useful for our main results.
Lemma 2.2 Let X be a nonempty set and g:X\to X be a mapping. Then there exists a subset E\subset X such that g(E)=g(X) and g:E\to X is onetoone.
In order to state our results, we give the main results given in [5] and [11].
Lemma 2.3 [[5], Lemma 2.1]
Let (X,\mu ,\nu ,\ast ,\diamond ) be an IFNS. Let a\ast b\ge ab, a\diamond b\le ab for all a,b\in [0,1] and (\mu ,\nu ) satisfy the nproperty. Suppose that \{{x}_{n}\} is a sequence in X such that
for all t>0 and n\in \mathbb{N}, where 0<k<1. Then the sequence \{{x}_{n}\} is a Cauchy sequence in X.
Theorem 2.4 ([[5], Theorem 2.5], [[11], Theorem 3.1])
Let (X,\u2291) be a partially ordered set and suppose that a\diamond b\le ab\le a\ast b for all a,b\in [0,1]. Let (X,\mu ,\nu ,\ast ,\diamond ) be a complete IFNS such that (\mu ,\nu ) has the nproperty. Let F:X\times X\to X and g:X\to X be two mappings such that F has the mixed gmonotone property and
for which gx\u2291gu and gy\u2292gv, where 0<k<1, F(X\times X)\subseteq g(X) and g is continuous. Suppose either

(a)
F is continuous or

(b)
X has the sequential gmonotone property.
If there exist {x}_{0},{y}_{0}\in X such that g{x}_{0}\u2291F({x}_{0},{y}_{0}) and g{y}_{0}\u2292F({y}_{0},{x}_{0}), then there exist x,y\in X such that gx=F(x,y) and gy=F(y,x).
3 Comments and suggestions
In this section, we show that the conditions of the above Lemma 2.3 and Theorem 2.4 in [5] are inadequate and, furthermore, the proof lines of Theorem 2.4 are not correct. We also would like to point out that the results in [5] can be corrected under the appropriate conditions on the tnorm and the FNS.
First of all, in the conditions of Lemma 2.3 and Theorem 2.4, we have a tconorm ⋄ such that a\diamond b\le ab for all a,b\in [0,1]. If we take b=0, then a=a\diamond 0\le 0 for all a\in [0,1]. It is obviously impossible. Moreover, it is well known and easy to see that if ∗ is a tnorm and ⋄ is a tconorm, then a\ast b\le a\diamond b for all a,b\in [0,1]. In this sense, Lemma 2.3 and Theorem 2.4 have to be corrected.
Secondly, from the property (IF1), it follows that a sequence \{{x}_{n}\}\subset X is a Cauchy sequence if, for any \u03f5>0 and t>0, there exists {n}_{0}\in \mathbb{N} such that \mu ({x}_{n}{x}_{m},t)>1\u03f5 for all n,m\ge {n}_{0}. That is, the sequence \{{x}_{n}\}\subset X is a Cauchy sequence on the IFNS (X,\mu ,\nu ,\ast ,\diamond ) if it also is on the FNS (X,\mu ,\ast ). A similar comment is valid for the convergence.
Furthermore, the completeness of an IFNS is equivalent to the completeness of its support FNS and so we can deduce any fixed point theorem for IFNS (when the conditions on μ and ν are splitting) as an immediate consequence of its associated fixed point theorem for FNS. In particular, it is sufficient to prove Theorem 2.4 just for FNS. Therefore, we only develop Theorem 4.2 for FNS.
Also, some proof lines of Lemma 2.3 are not correct (see p.1900, lines 923):
where q>0 such that m<{n}^{q}, and
where p>0 such that m<{n}^{p}. Hence the sequence \{{x}_{n}\} is a Cauchy sequence. This is not correct since the same q (or p) would not be valid for all positive integers m>n\ge {n}_{0}. For instance, let (X,\parallel \cdot \parallel ) be an ordinary normed space, define \mu (x,t)=\frac{t}{t+\parallel x\parallel} for any x\in X and t>0 and a\ast b=ab for all a,b\in [0,1]. Then (X,\mu ,1\mu ,\ast ,{\ast}^{\mathrm{\prime}}) is an IFNS. If k=1/2 and m={2}^{n}, we have
Also, the proof lines of Theorem 2.4 that are not correct are the following ones (see p.1902, lines 2232):
where q>0 such that m<{n}^{q}. In general, we cannot obtain \mu ({x}_{n}{x}_{m},t)\to 1 as n,m\to \mathrm{\infty}. It is not shown that \{{x}_{n}\} is a Cauchy sequence. Moreover, a similar conclusion can be obtained for \nu ({x}_{n}{x}_{m},t)\to 0. Thus, from the hypothesis of Theorem 2.4, the conclusion cannot be guaranteed.
4 The modification in FNS
In this section, by replacing the hypothesis that μ satisfies the nproperty with the one that the tnorm is of Htype, we state and prove a coupled fixed point theorem as a modification.
Definition 4.1 [13]
For any a\in [0,1], let the sequence {\{{\ast}^{n}a\}}_{n=1}^{\mathrm{\infty}} be defined by {\ast}^{1}a=a and {\ast}^{n}a=({\ast}^{n1}a)\ast a. Then a tnorm ∗ is said to be of Htype if the sequence {\{{\ast}^{n}a\}}_{n=1}^{\mathrm{\infty}} is equicontinuos at a=1.
Theorem 4.1 Let (X,\u2291) be a partially ordered set and (X,\mu ,\ast ) be a complete FNS such that ∗ is of Htype and a\ast b\ge ab for all a,b\in [0,1]. Let k\in (0,1) be a number and F:X\times X\to X be a mapping such that F has the mixed monotone property and
for which x\u2291u and y\u2292v. Suppose that either

(a)
F is continuous or

(b)
X has the sequential monotone property.
If there exist {x}_{0},{y}_{0}\in X such that {x}_{0}\u2291F({x}_{0},{y}_{0}) and {y}_{0}\u2292F({y}_{0},{x}_{0}), then F has a coupled fixed point. Furthermore, if {x}_{0} and {y}_{0} are comparable, then x=y, that is, x=F(x,x).
Proof Let {x}_{0},{y}_{0}\in X be such that {x}_{0}\u2291F({x}_{0},{y}_{0}) and {y}_{0}\u2292F({y}_{0},{x}_{0}). Since F(X\times X)\subseteq X, we can choose {x}_{1},{y}_{1}\in X such that {x}_{1}=F({x}_{0},{y}_{0}) and {y}_{1}=F({y}_{0},{x}_{0}). Again, from F(X\times X)\subseteq X, we can choose {x}_{2},{y}_{2}\in X such that {x}_{2}=F({x}_{1},{y}_{1}) and {y}_{2}=F({y}_{1},{x}_{1}). Continuing this process, we can construct two sequences \{{x}_{n}\} and \{{y}_{n}\} in X such that, for each n\ge 0,
The proof is divided into two steps.
Step 1. Prove that \{{x}_{n}\} and \{{y}_{n}\} are Cauchy sequences. Firstly, we show by induction that, for each n\ge 0,
For n=0, (4.3) holds trivially. Suppose that, for some fixed n\ge 0, (4.3) holds. Since {x}_{n}\u2291{x}_{n+1} and {x}_{n}\u2292{x}_{n+1} and F has the mixed monotone property, it follows from (4.2) that
Similarly, we have
Thus, combining (4.4) and (4.5), (4.3) holds.
Let {\delta}_{n}(t)={[\mu ({x}_{n}{x}_{n+1},t)]}^{1/2}\ast {[\mu ({y}_{n}{y}_{n+1},t)]}^{1/2} for all n\ge 0. Then it follows from (4.1), (4.3) and (F3) that
and
Then it follows from the tnorm and a\ast b\ge ab that {\delta}_{n}(kt)\ge {\delta}_{n1}(t) for all n\ge 1. This implies that
Since {lim}_{n\to \mathrm{\infty}}{\delta}_{0}(\frac{t}{{k}^{n}})=1 for all t>0, we have {lim}_{n\to \mathrm{\infty}}{\delta}_{n}(t)=1 for all t>0.
Now, we claim that, for any p\ge 1,
In fact, it is obvious for p=1 by (4.6), (4.7) and Lemma 2.1 since t/k\ge tkt and {\delta}_{n1}^{1} is nondecreasing. Assume that (4.9) holds for some p\ge 1. By (4.7), we have
and so
Thus, from (4.1), (4.9) and a\ast b\ge ab, we have
Hence, by the monotonicity of the tnorm ∗, we have
Similarly, we have
Therefore, by induction, (4.9) holds for all p\ge 1. Suppose that t>0 and \u03f5\in (0,1] are given. By hypothesis, since ∗ is a tnorm of Htype, there exists 0<\eta <1 such that {\ast}^{p}(a)>1\u03f5 for all a\in (1\eta ,1] and p\ge 1. Since {lim}_{n\to \mathrm{\infty}}{\delta}_{n}(t)=1, there exists {n}_{0} such that {\delta}_{n}(tkt)>1\eta for all n\ge {n}_{0}. Hence, from (4.9), we get
Therefore, \{{x}_{n}\} and \{{y}_{n}\} are Cauchy sequences.
Step 2. We prove that F has a coupled fixed point. Since X is complete, there exist x,y\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=x and {lim}_{n\to \mathrm{\infty}}{y}_{n}=y. Suppose that the assumption (a) holds. By the continuity of F, we get
Similarly, we can show that F(y,x)=y.
Suppose now that (b) holds. Since \{{x}_{n}\} is a nondecreasing sequence with {x}_{n}\to x and \{{y}_{n}\} is a nonincreasing sequence with {x}_{y}\to y, from (B1) and (B2), we have {x}_{n}\u2291x and {y}_{n}\u2292y for all n\ge 1. Then, by (4.1), we obtain
Letting n\to \mathrm{\infty}, we have {lim}_{n\to \mathrm{\infty}}{x}_{n}=F(x,y). Hence F(x,y)=x.
Similarly, we can show that F(y,x)=y.
Suppose that {x}_{0}\u2291{y}_{0}. By induction and the mixed monotone property of F, it follows that {x}_{n}=F({x}_{n1},{y}_{n1})\u2291F({y}_{n1},{x}_{n1})={y}_{n}. From (4.1), it follows that
By the iterative procedure, we have
Taking n\to \mathrm{\infty}, since {lim}_{n\to \mathrm{\infty}}\mu ({x}_{0}{y}_{0},\frac{t}{{k}^{n1}})=1 for all t>0, we conclude that \mu (xy,t)\ge 1 for all t>0, i.e., x=y. This completes the proof. □
Next, we prove the existence of a coupled coincidence point theorem, where we do not require that F and g are commuting.
Theorem 4.2 Let (X,\u2291) be a partially ordered set and let (X,\mu ,\ast ) be a complete FNS such that ∗ is of Htype and a\ast b\ge ab for all a,b\in [0,1]. Let k\in (0,1) be a number and F:X\times X\to X and g:X\to X be two mappings such that F has the mixed gmonotone property and
for which gx\u2291gu and gy\u2292gv. Suppose that F(X\times X)\subseteq g(X), g is continuous and either

(a)
F is continuous or

(b)
X has the sequential gmonotone property.
If there exist {x}_{0},{y}_{0}\in X such that g{x}_{0}\u2291F({x}_{0},{y}_{0}) and g{y}_{0}\u2292F({y}_{0},{x}_{0}), then there exist x,y\in X such that gx=F(x,y) and gy=F(y,x), that is, F and g have a coupled coincidence point.
Proof Using Lemma 2.2, there exists E\subset X such that g(E)=g(X) and g:E\to X is onetoone. We define a mapping \mathcal{A}:g(E)\times g(E)\to X by \mathcal{A}(gx,gy)=F(x,y). Since g is onetoone on g(E), \mathcal{A} is well defined. Thus it follows from (4.10) that
for which gx\u2291gu and gy\u2292gv. Since F has the mixed gmonotone property, we have
which implies that \mathcal{A} has the mixed monotone property.
Suppose that the assumption (a) or (b) holds. Using Theorem 4.1 with the mapping \mathcal{A}, it follows that \mathcal{A} has a coupled fixed point (u,v)\in g(X)\times g(X), i.e., u=\mathcal{A}(u,v) and v=\mathcal{A}(v,u). Since (u,v)\in g(X)\times g(X), there exists (\tilde{u},\tilde{v})\in X\times X such that g\tilde{u}=u and g\tilde{v}=v. Thus g\tilde{u}=u=\mathcal{A}(u,v)=\mathcal{A}(g\tilde{u},g\tilde{v}). Similarly, g\tilde{v}=\mathcal{A}(g\tilde{v},g\tilde{u}). This completes the proof. □
Now, we show the existence and uniqueness of coupled coincidence points. Note that if (S,\u2291) is a partially ordered set, then we endow the product S\times S with the following partial order:
We say that (x,y) and (u,v) are comparable if (x,y)\u2291(u,v) or (x,y)\u2292(u,v).
Theorem 4.3 In addition to the hypotheses of Theorem 4.2, suppose that, for any pair of coupled coincidence points (x,y),({x}^{\ast},{y}^{\ast})\in X\times X, there exists a point (u,v)\in X\times X such that (gu,gv) is comparable to (gx,gy) and (g{x}^{\ast},g{y}^{\ast}). Then F and g have a unique coupled coincidence point, that is, there exists a unique (x,y)\in X\times X such that x=gx=F(x,y) and y=gy=F(y,x).
Proof From Theorem 4.2, the set of coupled coincidences is nonempty. Now, we show that if (x,y) and ({x}^{\ast},{y}^{\ast}) are coupled coincidence points, that is, gx=F(x,y), gy=F(y,x) and g{x}^{\ast}=F({x}^{\ast},{y}^{\ast}), g{y}^{\ast}=F({y}^{\ast},{x}^{\ast}), then
Put {u}_{0}=u and {v}_{0}=v and choose {u}_{1},{v}_{1}\in X such that g{u}_{1}=F({u}_{0},{v}_{0}) and g{v}_{1}=F({v}_{0},{u}_{0}). Then, as in the proof of Theorem 4.2, we can inductively define the sequences \{{u}_{n}\} and \{{v}_{n}\} such that
Since (F(x,y),F(y,x))=(gx,gy) and (F(u,v),F(v,u))=(g{u}_{1},g{v}_{1}) are comparable, we can suppose that gx\u2291g{u}_{1} and gy\u2292g{v}_{1}. It is easy to show, by induction and gmonotonicity, that gx\u2291g{u}_{n} and gy\u2292g{v}_{n} for all n\ge 1. From (4.1), we obtain
and
Now, let {\beta}_{n}(t)={[\mu (gxg{u}_{n},t)]}^{1/2}\ast {[\mu (gyg{v}_{n},t)]}^{1/2}. By (4.14) and (4.15), we have
and
Since {lim}_{n\to \mathrm{\infty}}{\beta}_{0}(\frac{t}{{k}^{n}})=1, we conclude that {lim}_{n\to \mathrm{\infty}}g{u}_{n}=gx and {lim}_{n\to \mathrm{\infty}}g{v}_{n}=gy.
Similarly, {lim}_{n\to \mathrm{\infty}}g{u}_{n}=g{x}^{\ast} and {lim}_{n\to \mathrm{\infty}}g{v}_{n}=g{y}^{\ast}. Hence gx=g{x}^{\ast} and gy=g{y}^{\ast} and so (4.12) is proved.
Since gx=F(x,y) and gy=F(y,x), if z=gx and w=gy, by the commutativity of F and g, we have
and
i.e., (z,w) is a coupled coincidence point. In particular, from (4.12), we have z=gx=gz and w=gy=gw. Therefore, (z,w) is a coupled common fixed point of F and g.
To prove the uniqueness of the coupled common fixed point of F and g, assume that (p,q) is another coupled common fixed point. Then, by (4.12), we have p=gp=gz=z and q=gq=gw=w. This completes the proof. □
Finally, we present an intuitionistic version of Theorem 4.2 with the dual conditions on tconorms. The proof is just reduced to apply Theorem 4.2 to the support FNS.
Corollary 4.4 Let (X,\u2291) be a partially ordered set and suppose that a\ast b\ge ab and (1a)\diamond (1b)\le 1ab for all a,b\in [0,1]. Let (X,\mu ,\nu ,\ast ,\diamond ) be a complete IFNS such that ∗ and ⋄ are of Htype. Let F:X\times X\to X and g:X\to X be two mappings such that F has the mixed gmonotone property,
and
for which g(x)\u2291g(u) and g(y)\u2292g(v), where 0<k<1, F(X\times X)\subseteq g(X) and g is continuous. Suppose that either

(a)
F is continuous or

(b)
X has the sequential gmonotone property.
If there exist {x}_{0},{y}_{0}\in X such that g({x}_{0})\u2291F({x}_{0},{y}_{0}) and g({y}_{0})\u2292F({y}_{0},{x}_{0}), then there exist x,y\in X such that g(x)=F(x,y) and g(y)=F(y,x), that is, F and g have a coupled coincidence point.
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Acknowledgements
This work was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 20120008170) and by the Junta de Andalucía and projects FQM268, FQM178 of the Andalusian CICYE, Spain.
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Cho, Y., MartínezMoreno, J., Roldán, A. et al. Coupled coincidence point theorems in (intuitionistic) fuzzy normed spaces. J Inequal Appl 2013, 104 (2013). https://doi.org/10.1186/1029242X2013104
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DOI: https://doi.org/10.1186/1029242X2013104
Keywords
 fuzzy normed space
 intuitionistic fuzzy normed space
 coupled fixed point theorem
 tnorm
 Htype