The following theorem is the main result of the paper and is a generalization of [[14], Theorem 2.1] for fuzzy mappings in ordered metric spaces.

**Theorem 4** *Let*(X,d,\le )*be a complete ordered metric space*. *If an ordered fuzzy mapping*F:X\to {W}_{\alpha}(X)*satisfies*

\sigma (r){p}_{\alpha}(x,Fx)\le d(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}{D}_{\alpha}(Fx,Fy)\le r{M}_{\alpha}(F)

(2)

*for all*(x,y)\in \mathrm{\nabla}, *where*

{M}_{\alpha}(F)=max\{d(x,y),{p}_{\alpha}(x,Fx),{p}_{\alpha}(y,Fy),\frac{{p}_{\alpha}(x,Fy)+{p}_{\alpha}(y,Fx)}{2}\}.

*Then there exists a point*x\in X*such that*{x}_{\alpha}\subset Fx*provided that* *X* *satisfies the order sequential limit property*.

*Proof* Let {r}_{1} be a real number such that 0\le r<{r}_{1}<1 and {u}_{1}\in X. Since {(F{u}_{1})}_{\alpha} is nonempty and compact, there exists {u}_{2}\in {(F{u}_{1})}_{\alpha} such that

d({u}_{1},{u}_{2})={p}_{\alpha}({u}_{1},F{u}_{1}).

By the given assumption, we have ({u}_{1},{u}_{2})\in \mathrm{\nabla}. Since {(F{u}_{2})}_{\alpha} is nonempty and compact, there exists {u}_{3}\in {(F{u}_{2})}_{\alpha} such that

d({u}_{2},{u}_{3})={p}_{\alpha}({u}_{2},F{u}_{2})\le {D}_{\alpha}(F{u}_{1},F{u}_{2}).

Also, ({u}_{2},{u}_{3})\in \mathrm{\nabla}. Since \sigma (r)<1, we obtain

\sigma (r){\rho}_{\alpha}({u}_{1},F{u}_{1})\le {p}_{\alpha}({u}_{1},F{u}_{1})=d({u}_{1},{u}_{2}).

That is,

\sigma (r){p}_{\alpha}({u}_{1},F{u}_{1})\le d({u}_{1},{u}_{2}).

So, we have

\begin{array}{rcl}d({u}_{2},{u}_{3})& \le & {D}_{\alpha}(F{u}_{1},F{u}_{2})\\ \le & rmax\{d({u}_{1},{u}_{2}),{p}_{\alpha}({u}_{1},F{u}_{1}),{p}_{\alpha}({u}_{2},F{u}_{2}),\frac{{p}_{\alpha}({u}_{1},F{u}_{2})+{p}_{\alpha}({u}_{2},F{u}_{1})}{2}\}\\ \le & {r}_{1}max\{d({u}_{1},{u}_{2}),d({u}_{1},{u}_{2}),d({u}_{2},{u}_{3}),\frac{d({u}_{1},{u}_{2})+{p}_{\alpha}({u}_{2},F{u}_{2})}{2}\}\\ \le & {r}_{1}max\{d({u}_{1},{u}_{2}),d({u}_{2},{u}_{3}),\frac{d({u}_{1},{u}_{2})+d({u}_{2},{u}_{3})}{2}\}.\end{array}

Note that d({u}_{2},{u}_{3})\le d({u}_{1},{u}_{2}). If not, then the above inequality gives

\begin{array}{rcl}d({u}_{2},{u}_{3})& \le & {r}_{1}max\{d({u}_{2},{u}_{3}),d({u}_{2},{u}_{3}),\frac{d({u}_{2},{u}_{3})+d({u}_{2},{u}_{3})}{2}\}\\ =& {r}_{1}d({u}_{2},{u}_{3})<d({u}_{2},{u}_{3})\phantom{\rule{1em}{0ex}}\text{as}{r}_{1}1,\end{array}

a contradiction. Hence, d({u}_{2},{u}_{3})\le {r}_{1}d({u}_{1},{u}_{2}). Continuing this process, we construct a sequence \{{u}_{n}\} in *X* such that {u}_{n+1}\in {(F{u}_{n})}_{\alpha} and {u}_{n+2}\in {(F{u}_{n+1})}_{\alpha} with

d({u}_{n+1},{u}_{n+2})={p}_{\alpha}({u}_{n+1},F{u}_{n+1})\le {D}_{\alpha}(F{u}_{n},F{u}_{n+1}).

By the given assumption, we have ({u}_{n},{u}_{n+1})\in \mathrm{\nabla} and ({u}_{n+1},{u}_{n+2})\in \mathrm{\nabla}. As \sigma (r)<1, so

\sigma (r){p}_{\alpha}({u}_{n},F{u}_{n})\le {p}_{\alpha}({u}_{n},F{u}_{n})=d({u}_{n},{u}_{n+1}).

Therefore,

\begin{array}{rcl}d({u}_{n+1},{u}_{n+2})& \le & {D}_{\alpha}(F{u}_{n},F{u}_{n+1})\\ \le & rmax\{d({u}_{n},{u}_{n+1}),{p}_{\alpha}({u}_{n},F{u}_{n}),{p}_{\alpha}({u}_{n+1},F{u}_{n+1}),\\ \frac{{p}_{\alpha}({u}_{n},F{u}_{n+1})+{p}_{\alpha}({u}_{n+1},F{u}_{n})}{2}\}\\ \le & {r}_{1}max\{d({u}_{n},{u}_{n+1}),d({u}_{n},{u}_{n+1}),d({u}_{n+1},{u}_{n+2}),\\ \frac{d({u}_{n},{u}_{n+1})+{p}_{\alpha}({u}_{n+1},F{u}_{n+1})}{2}\}\\ \le & {r}_{1}max\{d({u}_{n},{u}_{n+1}),d({u}_{n+1},{u}_{n+2}),\frac{d({u}_{n},{u}_{n+1})+d({u}_{n+1},{u}_{n+2})}{2}\}.\end{array}

We claim that d({u}_{n+1},{u}_{n+2})\le d({u}_{n},{u}_{n+1}). If not, then by the above inequality, we obtain

\begin{array}{rcl}d({u}_{n+1},{u}_{n+2})& \le & {r}_{1}max\{d({u}_{n+1},{u}_{n+2}),d({u}_{n+1},{u}_{n+2}),\frac{d({u}_{n+1},{u}_{n+2})+d({u}_{n+1},{u}_{n+2})}{2}\}\\ \le & {r}_{1}d({u}_{n+1},{u}_{n+2})<d({u}_{n+1},{u}_{n+2}),\end{array}

a contradiction as {r}_{1}<1. So, we have

d({u}_{n+1},{u}_{n+2})\le {r}_{1}d({u}_{n},{u}_{n+1})\le \cdots \le {({r}_{1})}^{n}d({u}_{1},{u}_{2})

and

\sum _{n=1}^{\mathrm{\infty}}d({u}_{n+1},{u}_{n+2})\le \sum _{n=1}^{\mathrm{\infty}}{({r}_{1})}^{n}d({u}_{1},{u}_{2})<\mathrm{\infty}.

(3)

Hence, \{{u}_{n}\} is a Cauchy sequence in *X*. Since *X* is complete, there is some point z\in X such that {lim}_{n\to \mathrm{\infty}}{u}_{n}=z. As ({u}_{n},{u}_{n+1})\in \mathrm{\nabla} for all *n*, then by the assumption, ({u}_{n},z)\in \mathrm{\nabla}. Now, we show that for every pair (x,z)\in \mathrm{\nabla} with x\ne z, the following inequality holds:

{p}_{\alpha}(z,Tx)\le rmax\{d(z,x),{p}_{\alpha}(x,Fx)\}.

As {lim}_{n\to \mathrm{\infty}}{u}_{n}=z, there exists a positive integer {n}_{0}\in N such that for all n\ge {n}_{0}, we have

d(z,{u}_{n})\le \frac{1}{3}d(z,x).

(4)

Now, for all n\ge {n}_{0},

\begin{array}{rcl}\sigma (r){p}_{\alpha}({u}_{n},F{u}_{n})& \le & {p}_{\alpha}({u}_{n},F{u}_{n})\\ \le & d({u}_{n},z)+{p}_{\alpha}(z,F{u}_{n})\le d({u}_{n},z)+d(z,{u}_{n+1})\\ \le & \frac{1}{3}d(z,x)+\frac{1}{3}d(z,x)\le \frac{2}{3}d(z,x)\\ =& d(z,x)-\frac{1}{3}d(z,x)\\ \le & d(z,x)-d({u}_{n},z)\le d({u}_{n},x)\end{array}

implies that

\begin{array}{rcl}{p}_{\alpha}({u}_{n+1},Fx)& \le & {D}_{\alpha}(F{u}_{n},Fx)\\ \le & rmax\{d({u}_{n},x),{p}_{\alpha}({u}_{n},F{u}_{n}),{p}_{\alpha}(x,Fx),\frac{{p}_{\alpha}({u}_{n},Fx)+{p}_{\alpha}(x,F{u}_{n})}{2}\}\\ \le & rmax\{d({u}_{n},x),d({u}_{n},{u}_{n+1}),{p}_{\alpha}(x,Fx),\frac{{p}_{\alpha}({u}_{n},Fx)+d(x,{u}_{n+1})}{2}\},\end{array}

which on taking limit as n\to \mathrm{\infty} gives

{p}_{\alpha}(z,Fx)\le rmax\{d(z,x),{p}_{\alpha}(x,Fx),\frac{{p}_{\alpha}(z,Fx)+d(x,z)}{2}\}.

If

max\{d(z,x),{p}_{\alpha}(x,Fx),\frac{{p}_{\alpha}(z,Fx)+d(x,z)}{2}\}=\frac{{p}_{\alpha}(z,Fx)+d(x,z)}{2},

then

Hence,

{p}_{\alpha}(z,Fx)\le rmax\{d(z,x),{p}_{\alpha}(x,Fx)\}.

(5)

Now, we show that {z}_{\alpha}\subset Fz for each \alpha \in [0,1]. First, consider the case 0\le r<1/2. Assume on the contrary that {z}_{\alpha}\u2288Fz, that is, z\notin {(Fz)}_{\alpha}. Let a\in {(Fz)}_{\alpha}, as {(Fz)}_{\alpha} is nonempty and compact, so for each \alpha \in [0,1], we have

2rd(a,z)<{p}_{\alpha}(z,Fz).

(6)

Now, a\in {(Fz)}_{\alpha} implies (a,z)\in \mathrm{\nabla} and a\ne z. From (5) we have

{p}_{\alpha}(z,Fa)\le rmax\{d(z,a),{p}_{\alpha}(a,Fa)\}.

(7)

Now,

\sigma (r){p}_{\alpha}(z,Fz)\le {p}_{\alpha}(z,Fz)=d(z,a)

implies that

\begin{array}{rcl}{D}_{\alpha}(Fz,Fa)& \le & rmax\{d(z,a),{p}_{\alpha}(z,Fz),{p}_{\alpha}(a,Fa),\frac{{p}_{\alpha}(z,Fa)+{p}_{\alpha}(a,Fz)}{2}\}\\ \le & rmax\{d(z,a),{p}_{\alpha}(z,Fz),{p}_{\alpha}(a,Fa),\frac{{p}_{\alpha}(z,Fa)}{2}\}\\ \le & rmax\{d(z,a),{p}_{\alpha}(z,Fz),{p}_{\alpha}(a,Fa),\frac{d(z,a)+{p}_{\alpha}(a,Fa)}{2}\}\\ \le & rmax\{d(z,a),{p}_{\alpha}(z,Fz),{p}_{\alpha}(a,Fa)\}\\ \le & rmax\{d(z,a),{p}_{\alpha}(a,Fa)\}.\end{array}

Hence,

{D}_{\alpha}(Fz,Fa)\le rmax\{d(z,a),{p}_{\alpha}(a,Fa)\},

which further implies that

{p}_{\alpha}(a,Fa)\le rmax\{d(z,a),{p}_{\alpha}(a,Fa)\}\phantom{\rule{1em}{0ex}}\text{as}{p}_{\alpha}(a,Fa)\le {D}_{\alpha}(Fz,Fa).

We claim that {p}_{\alpha}(a,Fa)\le d(z,a). If not, then the above inequality becomes

{p}_{\alpha}(a,Fa)\le r{p}_{\alpha}(a,Fa)<{p}_{\alpha}(a,Fa)\phantom{\rule{1em}{0ex}}\text{as}r1,

a contradiction, so we deduce that {p}_{\alpha}(a,Fa)\le rd(z,a). From inequality (7), we have

{p}_{\alpha}(z,Fa)\le rd(z,a).

Therefore,

\begin{array}{rcl}{p}_{\alpha}(z,Fz)& \le & {p}_{\alpha}(z,Fa)+{D}_{\alpha}(Fz,Fa)\\ \le & rd(z,a)+rmax\{d(z,a),{p}_{\alpha}(a,Fa)\}\\ \le & rd(z,a)+rd(z,a)\le 2rd(z,a)<{p}_{\alpha}(z,Fz),\end{array}

a contradiction. Hence, {z}_{\alpha}\subset Fz.

Now, when 1/2\le r<1, we first prove that

{D}_{\alpha}(Fx,Fz)\le rmax\{d(x,z),{p}_{\alpha}(x,Fx),{p}_{\alpha}(z,Fz),\frac{{p}_{\alpha}(x,Fz)+{p}_{\alpha}(z,Fx)}{2}\}

(8)

for all (x,z)\in \mathrm{\nabla}. If x=z, then (8) holds trivially. So, assume that x\ne z. For every n\in N, one may find a sequence {y}_{n}\in {(Fx)}_{\alpha} such that

d(z,{y}_{n})\le {p}_{\alpha}(z,Fx)+\frac{1}{n}d(x,z).

As {y}_{n}\in {(Fx)}_{\alpha}, this implies ({y}_{n},x)\in \mathrm{\nabla}. Using (7) we have

\begin{array}{rcl}{p}_{\alpha}(x,Fx)& \le & d(x,{y}_{n})\\ \le & d(x,z)+d(z,{y}_{n})\\ \le & d(x,z)+{p}_{\alpha}(z,Fx)+\frac{1}{n}d(x,z)\\ \le & d(x,z)+rmax\{d(z,x),{p}_{\alpha}(x,Fx)\}+\frac{1}{n}d(x,z)\end{array}

for all n\in N. If d(x,z)\ge {p}_{\alpha}(x,Fx), then

\begin{array}{rcl}{p}_{\alpha}(x,Fx)& \le & d(x,z)+rd(z,x)+\frac{1}{n}d(x,z)\\ \le & d(x,z)+rd(z,x)+\frac{1}{n}d(x,z)\le (1+r+\frac{1}{n})d(x,z).\end{array}

This implies that

\frac{1}{(1+r)}{p}_{\alpha}(x,Fx)\le (1+\frac{1}{(1+r)n})d(x,z).

Hence, for \frac{1}{2}\le r<1, we obtain

\begin{array}{rcl}\sigma (r){p}_{\alpha}(x,Fx)& =& (1-r){p}_{\alpha}(x,Fx)\\ \le & \frac{1}{(1+r)}{p}_{\alpha}(x,Fx)\le (1+\frac{1}{(1+r)n})d(x,z).\end{array}

On taking the limit as n\to \mathrm{\infty}, we have

\sigma (r){p}_{\alpha}(x,Fx)\le d(x,z).

If d(x,z)\le {p}_{\alpha}(x,Fx), then

On taking the limit as n\to \mathrm{\infty}, we have

\sigma (r){p}_{\alpha}(x,Fx)\le d(x,z).

By the given assumption, we have

{D}_{\alpha}(Fx,Fz)\le rmax\{d(x,z),{p}_{\alpha}(x,Fx),{p}_{\alpha}(z,Fz),\frac{{p}_{\alpha}(x,Fz)+{p}_{\alpha}(z,Fx)}{2}\}.

Thus, for any x\ne z, (8) holds true. Put x={u}_{n} in the above inequality to obtain

\begin{array}{rcl}{p}_{\alpha}(z,Fz)& \le & \underset{n\to \mathrm{\infty}}{lim}{p}_{\alpha}({u}_{n+1},Fz)\le \underset{n\to \mathrm{\infty}}{lim}{D}_{\alpha}(F{u}_{n},Fz)\\ \le & \underset{n\to \mathrm{\infty}}{lim}rmax\{d({u}_{n},z),d({u}_{n},{u}_{n+1}),{p}_{\alpha}(z,Fz),\frac{{p}_{\alpha}({u}_{n},Fz)+{p}_{\alpha}(z,F{u}_{n})}{2}\}\\ =& r{p}_{\alpha}(z,Fz)\end{array}

as r<1, we get {p}_{\alpha}(z,Fz)=0. Hence by Lemma 2, {z}_{\alpha}\subset Fz. □

**Corollary 5** *Let*(X,d,\le )*be a complete ordered metric space*. *If an ordered fuzzy mapping*F:X\to {W}_{\alpha}(X)*satisfies*

\sigma (r){p}_{\alpha}(x,Fx)\le d(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}{D}_{\alpha}(Fx,Fy)\le r{M}_{\alpha}(F)

*for all*(x,y)\in \mathrm{\nabla}, *where*

{M}_{\alpha}(F)=max\{d(x,y),{p}_{\alpha}(x,Fx),{p}_{\alpha}(y,Fy)\}.

*Then there exists a point*x\in X*such that*{x}_{\alpha}\subset Fx*provided that* *X* *satisfies the order sequential limit property*.

**Corollary 6** *Let*(X,d,\le )*be a complete ordered metric space*. *If an ordered fuzzy mapping*F:X\to {W}_{\alpha}(X)*satisfies*

\sigma (r){p}_{\alpha}(x,Fx)\le d(x,y)\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}{D}_{\alpha}(Fx,Fy)\le \lambda {M}_{\alpha}(F)

*for all*(x,y)\in \mathrm{\nabla}, *where*

{M}_{\alpha}(F)=d(x,y)+{p}_{\alpha}(x,Fx)+{p}_{\alpha}(y,Fy)\}

*and*\lambda \in [0,\frac{1}{3}), r=3\lambda. *Then there exists a point*x\in X*such that*{x}_{\alpha}\subset Fx*provided that* *X* *satisfies the order sequential limit property*.