Theorem 2.1 Let (X,\u2291) be a partially ordered set, and let (X,\mu ,\ast ) be a complete FNS such that the tnorm ∗ is of Htype. Let F:X\to X be a mapping such that F is nondecreasing and
\mu (F(x)F(u),kt)\ge {\ast}^{2}\mu (xu,t),
(2.1)
for which x\u2291u and t>0, where 0<k<1. Suppose either

(a)
F is continuous, or

(b)
if \{{x}_{n}\} is a nondecreasing sequence and {lim}_{n\to \mathrm{\infty}}{x}_{n}=x, then {x}_{n}\u2291x for all n\in \mathbb{N}.
If there exists
{x}_{0}\in X
such that
then F has a fixed point in X.
Proof Let {x}_{0}\in X such that {x}_{0}\u2291F({x}_{0}), and let {x}_{n}=F({x}_{n1}), n=1,2,\dots , then we have that
{x}_{0}\u2291{x}_{1}\u2291{x}_{2}\u2291\cdots \u2291{x}_{n}\u2291{x}_{n+1}\u2291\cdots .
Now, put
{\delta}_{n}(t):=\mu ({x}_{n}{x}_{n+1},t).
Then, by using (2.1), we have
\mu ({x}_{n}{x}_{n+1},kt)=\mu (F({x}_{n1})F({x}_{n}),kt)\ge {\ast}^{2}\mu ({x}_{n1}{x}_{n},t)={\ast}^{2}{\delta}_{n1}(t).
Thus, it follows that {\delta}_{n}(kt)\ge {\ast}^{2}{\delta}_{n1}(t), and so
{\delta}_{n}(t)\ge {\ast}^{2}{\delta}_{n1}\left(\frac{t}{k}\right)\ge {\ast}^{{2}^{n}}{\delta}_{0}\left(\frac{t}{{k}^{n}}\right).
(2.2)
On the other hand, we have
t(1k)(1+k+\cdots +{k}^{mn1})<t,\phantom{\rule{1em}{0ex}}\mathrm{\forall}m>n,0<k<1.
By Definition 1.4, we get that
\begin{array}{rcl}\mu ({x}_{n}{x}_{m},t)& \ge & \mu ({x}_{n}{x}_{m},t(1k)(1+k+\cdots +{k}^{mn1}))\\ \ge & \mu ({x}_{n}{x}_{n+1},t(1k))\ast \mu ({x}_{n+1}{x}_{m},t(1k)(k+\cdots +{k}^{mn1}))\\ \ge & \mu ({x}_{n}{x}_{n+1},t(1k))\ast \mu ({x}_{n+1}{x}_{n+2},t(1k)k)\\ \ast \cdots \ast \mu ({x}_{m1}{x}_{m},t(1k){k}^{mn1}).\end{array}
(2.3)
It follows from (2.2) and (2.3) that
\begin{array}{rcl}\mu ({x}_{n}{x}_{m},t)& \ge & \mu ({x}_{n}{x}_{n+1},t(1k))\ast \mu ({x}_{n+1}{x}_{n+2},t(1k)k)\\ \ast \cdots \ast \mu ({x}_{m1}{x}_{m},t(1k){k}^{mn1})\\ \ge & \left[{\ast}^{{2}^{n}}{\delta}_{0}\left(\frac{t(1k)}{{k}^{n}}\right)\right]\ast \cdots \ast \left[{\ast}^{{2}^{m1}}{\delta}_{0}\left(\frac{t(1k)}{{k}^{n}}\right)\right]\\ =& {\ast}^{{2}^{m}{2}^{n}}{\delta}_{0}\left(\frac{t(1k)}{{k}^{n}}\right).\end{array}
(2.4)
By the hypothesis, the tnorm ∗ is of Htype; for all \epsilon \in (0,1), there exists \eta >0 such that
{\ast}^{p}(s)>1\epsilon ,
for all s\in (1\eta ,1] and for all p. Note that
\underset{n\to \mathrm{\infty}}{lim}{\delta}_{0}\left(\frac{t(1k)}{{k}^{n}}\right)=1
for all t>0 and 0<k<1, we have that there exists {n}_{0} such that
\mu ({x}_{n}{x}_{m},t)>1\epsilon ,
for all m>n>{n}_{0}. Thus, \{{x}_{n}\} is a Cauchy sequence. Since X is complete, there exists x\in X such that
\underset{n\to \mathrm{\infty}}{lim}{x}_{n}=x.
If the assumption (a) holds, then by the continuity of F, we get that
x=\underset{n\to \mathrm{\infty}}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty}}{lim}F({x}_{n})=F(x).
If the assumption (b) holds, then we have that {x}_{n}\u2291x for all n\in \mathbb{N}. It follows from (2.1) that
\underset{n\to \mathrm{\infty}}{lim}\mu ({x}_{n+1}F(x),kt)=\underset{n\to \mathrm{\infty}}{lim}\mu (F({x}_{n})F(x),kt)\ge \underset{n\to \mathrm{\infty}}{lim}{\ast}^{2}\mu ({x}_{n}x,t)=1.
Thus, \mu (xF(x),kt)=1, that is, x=F(x). Therefore, x is a fixed point of F. The proof is completed. □
Theorem 2.2 Let (X,\u2291) be a partially ordered set, and let (X,\mu ,\ast ) be a complete FNS such that the tnorm ∗ is of Htype and a\ast b>0 for any a,b\in (0,1]. Let F:X\times X\to X be a mapping such that F has the mixed monotone property and
\mu (F(x,y)F(u,v),kt)\ge \mu (xu,t)\ast \mu (yv,t),
(2.5)
for which x\u2291u,y\u2292v and t>0, where 0<k<1. Suppose either

(a)
F is continuous, or

(b)
X has the following property:

(i)
if \{{x}_{n}\} is a nondecreasing sequence and {lim}_{n\to \mathrm{\infty}}{x}_{n}=x, then {x}_{n}\u2291x for all n\in \mathbb{N},

(ii)
if \{{y}_{n}\} is a nonincreasing sequence and {lim}_{n\to \mathrm{\infty}}{y}_{n}=y, then {y}_{n}\u2292y for all n\in \mathbb{N}.
If there exist
{x}_{0},{y}_{0}\in X
such that
{x}_{0}\u2291F({x}_{0},{y}_{0}),\phantom{\rule{2em}{0ex}}{y}_{0}\u2292F({y}_{0},{x}_{0}),
then F has a coupled fixed point x,y\in X, that is,
x=F(x,y),\phantom{\rule{2em}{0ex}}y=F(y,x).
Proof First, we define a partial order ≼ on X\times X as follows: (x,y)\preccurlyeq (u,v) if and only if x\u2291u and y\u2292v. Second, we define a fuzzy set on X\times X as follows: \tilde{\mu}((x,y),t)=\mu (x,t)\ast \mu (y,t) for any (x,y)\in X\times X and any t>0. Since (X,\mu ,\ast ) is a complete FNS, we can easily prove that (X\times X,\tilde{\mu},\ast ) is a complete FNS. Lastly, we define a mapping \tilde{F}:X\times X\to X\times X by
\tilde{F}(x,y)=(F(x,y),F(y,x)),\phantom{\rule{1em}{0ex}}\mathrm{\forall}(x,y)\in X\times X.
Since F has the mixed monotone property, if (x,y)\preccurlyeq (u,v), we have that
that is, \tilde{F}(x,y)\preccurlyeq \tilde{F}(u,v). Therefore, \tilde{F}:X\times X\to X\times X is a nondecreasing mapping. Since {x}_{0},{y}_{0}\in X and
{x}_{0}\u2291F({x}_{0},{y}_{0}),\phantom{\rule{2em}{0ex}}{y}_{0}\u2292F({y}_{0},{x}_{0}),
we have that ({x}_{0},{y}_{0})\preccurlyeq (F({x}_{0},{y}_{0}),F({y}_{0},{x}_{0}))=\tilde{F}({x}_{0},{y}_{0}). If (x,y)\preccurlyeq (u,v), by (2.5) we have that
\begin{array}{rcl}\tilde{\mu}(\tilde{F}(x,y)\tilde{F}(u,v),kt)& =& \tilde{\mu}((F(x,y)F(u,v),F(v,u)F(y,x)),kt)\\ =& \mu (F(x,y)F(u,v),kt)\ast \mu (F(v,u)F(y,x),kt)\\ \ge & (\mu (xu,t)\ast \mu (yv,t))\ast (\mu (xu,t)\ast \mu (yv,t))\\ =& {\ast}^{2}\tilde{\mu}((x,y)(u,v),t).\end{array}
Thus, all the assumptions of Theorem 2.1 hold for \tilde{F} and (X\times X,\tilde{\mu},\ast ). By Theorem 2.1 we get that \tilde{F} has a fixed point (x,y)\in X\times X, that is, (x,y)=(F(x,y),F(y,x)). This implies that x=F(x,y), y=F(y,x), that is, (x,y) is a coupled fixed point of F. The proof is completed. □
By using Theorem 2.2, we can prove the following coupled fixed point theorem in intuitionistic fuzzy normed spaces.
Theorem 2.3 Let (X,\u2291) be a partially ordered set, and let (X,\mu ,\nu ,\ast ,\star ) be a complete IFNS such that the tnorm ∗ is of Htype and a\ast b>0 for any a,b\in (0,1]. Let F:X\times X\to X be a mapping such that F has the mixed monotone property and
\mu (F(x,y)F(u,v),kt)\ge \mu (xu,t)\ast \mu (yv,t),
(2.6)
for which x\u2291u,y\u2292v and t>0, where 0<k<1. Suppose either

(a)
F is continuous, or

(b)
X has the following property:

(i)
if \{{x}_{n}\} is a nondecreasing sequence and {lim}_{n\to \mathrm{\infty}}{x}_{n}=x, then {x}_{n}\u2291x for all n\in \mathbb{N},

(ii)
if \{{y}_{n}\} is a nonincreasing sequence and {lim}_{n\to \mathrm{\infty}}{y}_{n}=y, then {y}_{n}\u2292y for all n\in \mathbb{N}.
If there exist
{x}_{0},{y}_{0}\in X
such that
{x}_{0}\u2291F({x}_{0},{y}_{0}),\phantom{\rule{2em}{0ex}}{y}_{0}\u2292F({y}_{0},{x}_{0}),
then F has a coupled fixed point x,y\in X, that is,
x=F(x,y),\phantom{\rule{2em}{0ex}}y=F(y,x).
Proof Assume that \{{x}_{n}\} is a sequence in (X,\mu ,\nu ,\ast ,\star ). Let t>0, 0<\epsilon <1. If \mu ({x}_{m}{x}_{n},t)>1\epsilon, then by Definition 1.5(i) we can deduce that \nu ({x}_{m}{x}_{n},t)<\epsilon. Thus, a sequence \{{x}_{n}\} in (X,\mu ,\nu ,\ast ,\star ) is Cauchy if and only if \{{x}_{n}\} is Cauchy in (X,\mu ,\ast ). By Lemma 1.1, we know that the topology of (X,\mu ,\nu ,\ast ,\star ) is the same as the topology of (X,\mu ,\ast ). This implies that (X,\mu ,\nu ,\ast ,\star ) is a complete IFNS if and only if (X,\mu ,\ast ) is a complete FNS. Therefore, by using Theorem 2.2 to (X,\mu ,\ast ) and F, we get that F has a coupled fixed point x,y\in X. The proof is completed. □
Theorem 2.4 Let (X,\u2291) be a partially ordered set, and let (X,\mu ,\nu ,\ast ,\star ) be a complete IFNS such that the tnorm ∗ is of Htype and a\ast b>0 for any a,b\in (0,1]. Let F:X\times X\to X, g:X\to X be two mappings such that F has the mixed g monotone property and
\mu (F(x,y)F(u,v),kt)\ge \mu (gxgu,t)\ast \mu (gygv,t),
(2.7)
for which g(x)\u2291g(u) , g(y)\u2292g(v) and t>0, 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 following property:

(i)
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},

(ii)
if \{{y}_{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 there exist
{x}_{0},{y}_{0}\in X
such that
g({x}_{0})\u2291F({x}_{0},{y}_{0}),\phantom{\rule{2em}{0ex}}g({y}_{0})\u2292F({y}_{0},{x}_{0}),
then there exist
x,y\in X
such that
g(x)=F(x,y),\phantom{\rule{2em}{0ex}}g(y)=F(y,x),
that is, F and g have a coupled coincidence point in X.
Proof The conclusion of Theorem 2.4 can be proved by using Lemma 1.2 and Theorem 2.3. Since the proof is similar to the proof of Theorem 3.2 in [17], we delete the details of the proof. The proof is completed. □
Remark 2.1 It follows from the proof of the above theorems that the following implications hold: Theorem 2.1 ⟹ Theorem 2.2 ⟹Theorem 2.3 ⟹ Theorem 2.4. Conversely, it is clear that the following implications hold: Theorem 2.4 ⟹ Theorem 2.3 ⟹ Theorem 2.2. Thus, we have the following conclusion.
Theorem 2.5 Theorem 2.2Theorem 2.4 are equivalent.
Remark 2.2 In [13] and [17], the condition a\star b\le ab for all a,b\in [0,1] is used. But this condition cannot hold in intuitionistic fuzzy normed spaces. In fact, if this condition holds, by using (iii) and (iv) in the definition of IFNS, we can get 1=1\star 0\le 1\cdot 0=0, which yields a contradiction. Furthermore, the proofs of the results in [13] and [17] have the same errors as noted in [18]. Therefore, our results improve and correct results in [13] and [17].
In the following, we give an example to show that our contractive conditions are a real improvement over the contractive conditions used in [13] and [17].
Example 2.1 Let X=\mathbb{R}, \mu (x,t)=\frac{t}{t+x}, \nu (x,t)=1\frac{t}{t+x} for every x\in X, and let t>0, a\ast b=min\{a,b\}, a\star b=max\{a,b\} for all a,b\in [0,1]. Then (X,\mu ,\nu ,\ast ,\star ) is a complete intuitionistic fuzzy normed linear space, and the tnorm ∗ and tconorm ⋆ are of Htype. If X is endowed with the usual order x\u2291y\iff xy\le 0, then (X,\u2291) is a partially ordered set. Let 0<k<1, and define F(x,y)=\frac{xy}{4}, gx=\frac{x}{k} for any x,y\in X. Then F:X\times X\to X is a mixed gmonotone mapping, F(X,X)\subseteq g(X), and g is continuous. Let {x}_{0}=1 and {y}_{0}=1, then
\frac{1}{k}=g{x}_{0}\u2291F({x}_{0},{y}_{0})=\frac{1}{2},\phantom{\rule{2em}{0ex}}\frac{1}{2}=F({y}_{0},{x}_{0})\u2291\frac{1}{k}=g{y}_{0}.
For any x,y,u,v\in X, with gx\u2291gu,gv\u2291gy, we have
\begin{array}{rcl}\mu (F(x,y)F(u,v),kt)& =& \frac{kt}{kt+\frac{ux+yv}{4}}\\ \ge & min\{\frac{kt}{kt+ux},\frac{kt}{kt+yv}\}\\ =& min\{\frac{t}{t+ux/k},\frac{t}{t+yv/k}\}\\ =& min\{\mu (gxgu,t),\mu (gygv,t)\}\\ =& \mu (gxgu,t)\ast \mu (gygv,t).\end{array}
Thus, all the conditions of Theorem 2.4 are satisfied. By Theorem 2.4, there is ({x}^{\ast},{y}^{\ast})\in X\times X such that F({x}^{\ast},{y}^{\ast})=g({x}^{\ast}) and F({y}^{\ast},{x}^{\ast})=g({y}^{\ast}). But υ does not satisfy the contractive conditions in [13] and [17]. In fact, for u=x+1, y=v+1,
\begin{array}{rcl}\nu (F(x,y)F(u,v),kt)& =& 1\frac{kt}{kt+\frac{ux+yv}{4}}=1\frac{kt}{kt+\frac{1}{2}}\\ >& max\{1\frac{kt}{kt+ux},1\frac{kt}{kt+yv}\}\\ =& max\{1\frac{t}{t+ux/k},1\frac{t}{t+yv/k}\}\\ =& max\{\nu (gxgu,t),\nu (gygv,t)\}\\ =& \nu (gxgu,t)\star \nu (gygv,t).\end{array}
This shows that υ does not satisfy the contractive conditions in [13] and [17].