Definition 8 Let (X,d) be a metric space and \psi \in \mathrm{\Psi}. Suppose that f:X\to X is a triangular αadmissible mapping satisfying the following condition: for each \epsilon >0 there exists \delta >0 such that
\epsilon \le \psi (d(x,y))<\epsilon +\delta \phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}\alpha (x,y)\psi (d(fx,fy))<\epsilon
(2.1)
for all x,y\in X. Then f is called an αψMeirKeeler contractive mapping.
Remark 9 Let f be an αψMeirKeeler contractive mapping. Then
\alpha (x,y)\psi (d(fx,fy))<\psi (d(x,y))
for all x,y\in X when x\ne y. Also, if x=y then d(fx,fy)=0. i.e.,
\alpha (x,y)\psi (d(fx,fy))\le \psi (d(x,y))
for all x,y\in X.
Theorem 10 Let (X,d) be a complete metric space. Suppose that f is a continuous αψMeirKeeler contractive mapping and that there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1, then f has a fixed point.
Proof Let {x}_{0}\in X and define a sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0} for all n\in \mathbb{N}. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N}\cup \{0\}, then obviously f has a fixed point. Hence, we suppose that
{x}_{n}\ne {x}_{n+1}
(2.2)
for all n\in \mathbb{N}\cup \{0\}. We have d({x}_{n},{x}_{n+1})>0 for all n\in \mathbb{N}\cup \{0\}. Now define {s}_{n}=\psi (d({x}_{n},{x}_{n+1})). By Remark 9, we deduce that for all n\in \mathbb{N}\cup \{0\},
\alpha ({x}_{n},{x}_{n+1})\psi (d({x}_{n+1},{x}_{n+2}))=\alpha ({x}_{n},{x}_{n+1})\psi (d(f{x}_{n},f{x}_{n+1}))<\psi (d({x}_{n},{x}_{n+1})).
Then by applying Lemma 7
\psi (d({x}_{n+1},{x}_{n+2}))<\psi (d({x}_{n},{x}_{n+1})).
Hence, the sequence \{{s}_{n}\} is decreasing in {\mathbb{R}}_{+} and so it is convergent to s\in {\mathbb{R}}_{+}. We will show that s=0. Suppose, to the contrary, that s>0. Hence, we have
0<s<\psi (d({x}_{n},{x}_{n+1}))\phantom{\rule{1em}{0ex}}\text{for all}n\in \mathbb{N}\cup \{0\}.
(2.3)
Let \epsilon =s>0. Then by hypothesis, there exists a \delta (\epsilon )>0 such that (2.1) holds. On the other hand, by the definition of ε, there exists {n}_{0}\in \mathbb{N} such that
\epsilon <{s}_{{n}_{0}}=\psi (d({x}_{{n}_{0}},{x}_{{n}_{0}+1}))<\epsilon +\delta .
Now by (2.1), we have
\begin{array}{rcl}{s}_{{n}_{0}+1}& =& \psi (d({x}_{{n}_{0}+1},{x}_{{n}_{0}+2}))\le \alpha ({x}_{{n}_{0}},{x}_{{n}_{0}+1})\psi (d({x}_{{n}_{0}+1},{x}_{{n}_{0}+2}))\\ =& \alpha ({x}_{{n}_{0}},{x}_{{n}_{0}+1})\psi (d(f{x}_{{n}_{0}},f{x}_{{n}_{0}+1}))<\epsilon =s\end{array}
which is a contradiction. Hence, s=0, that is, {lim}_{n\to +\mathrm{\infty}}{s}_{n}=0. Now, by the continuity of ψ in t=0, we have {lim}_{n\to +\mathrm{\infty}}d({x}_{n},{x}_{n+1})=0. For given \epsilon >0, by the hypothesis, there exists a \delta =\delta (\epsilon )>0 such that (2.1) holds. Without loss of generality, we assume \delta <\epsilon. Since s=0, then there exists N\in \mathbb{N} such that
{s}_{n1}=\psi (d({x}_{n1},{x}_{n}))<\delta ,\phantom{\rule{1em}{0ex}}\text{for all}n\ge N.
(2.4)
We will prove that for any fixed k\ge {N}_{0},
\psi (d({x}_{k},{x}_{k+l}))\le \epsilon ,\phantom{\rule{1em}{0ex}}\text{for all}l\in \mathbb{N},
(2.5)
holds. Note that (2.5), by (2.4), holds for l=1. Suppose the condition (2.1) is satisfied for some m\in \mathbb{N}. For l=m+1, by (2.4), we get
\begin{array}{rcl}\psi (d({x}_{k1},{x}_{k+m}))& \le & \psi (d({x}_{k1},{x}_{k})+d({x}_{k},{x}_{k+m}))\\ \le & \psi (d({x}_{k1},{x}_{k}))+\psi (d({x}_{k},{x}_{k+m}))\\ <& \epsilon +\delta .\end{array}
(2.6)
If \psi (d({x}_{k1},{x}_{k+m}))\ge \epsilon, then by (2.1) we get
\begin{array}{rcl}\psi (d({x}_{k},{x}_{k+m+1}))& \le & \alpha ({x}_{k1},{x}_{k+m})\psi (d({x}_{k},{x}_{k+m+1}))\\ =& \alpha ({x}_{k},{x}_{k+m+1})\psi (d(f{x}_{k1},f{x}_{k+m}))<\epsilon \end{array}
and hence (2.5) holds.
If \psi (d({x}_{k1},{x}_{k+m}))<\epsilon, by Remark 9, we get
\psi (d({x}_{k},{x}_{k+m+1}))\le \alpha ({x}_{k1},{x}_{k+m})\psi (d({x}_{k},{x}_{k+m+1}))\le \psi (d({x}_{k1},{x}_{k+m}))<\epsilon .
Consequently, (2.5) holds for l=m+1. Hence, \psi (d({x}_{k},{x}_{k+l}))\le \epsilon for all k\ge {N}_{0} and l\ge 1, which means
d({x}_{n},{x}_{m})<\epsilon ,\phantom{\rule{1em}{0ex}}\text{for all}m\ge n\ge {N}_{0}.
(2.7)
Hence \{{x}_{n}\} is a Cauchy sequence. Since (X,d) is complete, there exists z\in X such that {x}_{n}\to z as n\to \mathrm{\infty}. Now, since, f is continuous then
fz=f\left(\underset{n\to \mathrm{\infty}}{lim}{x}_{n}\right)=\underset{n\to \mathrm{\infty}}{lim}{x}_{n+1}=z,
that is, f has a fixed point. □
Theorem 11 Let (X,d) be a complete metric space and let f be a αψMeirKeeler contractive mapping. If the following conditions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1,

(ii)
if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x as n\to +\mathrm{\infty}, then \alpha ({x}_{n},x)\ge 1 for all n.
Then f has a fixed point.
Proof Following the proof of the Theorem 10, we say that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and that there exist z\in X such that {x}_{n}\to z as n\to +\mathrm{\infty}. Hence, from (ii) \alpha ({x}_{n},z)\ge 1. By Remark 9, we have
\begin{array}{rcl}\psi (d(fz,z))& \le & \psi (d(fz,f{x}_{n})+d(f{x}_{n},z))\\ \le & \psi (d(fz,f{x}_{n}))+\psi (d(f{x}_{n},z))\\ \le & \alpha ({x}_{n},z)\psi (d(fz,f{x}_{n}))+\psi (d(f{x}_{n},z))\\ \le & \psi (d(z,{x}_{n}))+\psi (d({x}_{n+1},z)).\end{array}
By taking limit as n\to +\mathrm{\infty}, in the above inequality, we get \psi (d(fz,z))\le 0, that is, d(fz,z)=0. Hence, fz=z. □
Next, we give some examples to validate our main result.
Example 12 Let X=[0,\mathrm{\infty}) and d(x,y)=xy be a metric on X. Define f:X\to X by
fx=\{\begin{array}{cc}\frac{{x}^{2}}{4},\hfill & \text{if}x\in [0,1],\hfill \\ 2x,\hfill & \text{if}x\in (1,\mathrm{\infty}),\hfill \end{array}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\alpha (x,y)=\{\begin{array}{cc}1,\hfill & \text{if}x,y\in [0,1],\hfill \\ 1,\hfill & \text{otherwise}\hfill \end{array}
and \psi (t)=\frac{1}{4}t. Clearly, (X,d) is a complete metric space. We show that f is a triangular αadmissible mapping. Let x,y\in X, if \alpha (x,y)\ge 1 then x,y\in [0,1]. On the other hand, for all x,y\in [0,1] we have fx\le 1 and fy\le 1. It follows that \alpha (fx,fy)\ge 1. Also, if \alpha (x,z)\ge 1 and \alpha (z,y)\ge 1 then x,y,z\in [0,1] and hence, \alpha (x,y)\ge 1. Thus, the assertion holds by the same arguments. Notice that \alpha (0,f0)\ge 1.
Now, if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, then \{{x}_{n}\}\subset [0,1] and hence x\in [0,1]. This implies that \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}. Let x,y\in [0,1]. Without loss of generality, take x\le y. Then
Clearly, by taking \delta =\epsilon the condition (2.1) holds. Otherwise, \alpha (x,y)=1. Hence, for given \epsilon >0 we have \alpha (x,y)\psi (d(fx,fy))\le 0<\epsilon. Hence, conditions of Theorem 11 holds and f has a fixed point. But, if x,y\in (1,\mathrm{\infty}) and
\epsilon \le d(x,y)<\epsilon +\delta ,
where \epsilon >0 and \delta >0. Then
d(fx,fy)=2xy=2d(x,y)\ge d(x,y)\ge \epsilon .
That is, the MeirKeeler theorem cannot applied for this example.
Example 13 Let X=[0,\mathrm{\infty}) and d(x,y)=xy be a metric on X. Define f:X\to X by
fx=\{\begin{array}{cc}\frac{1}{2}(1x),\hfill & \text{if}x\in [0,1],\hfill \\ \frac{\sqrt{4{x}^{2}}}{2+sin({x}^{2})},\hfill & \text{if}x\in (1,2],\hfill \\ \frac{\sqrt[3]{x2}}{{x}^{3}+1},\hfill & \text{if}x\in (2,\mathrm{\infty})\hfill \end{array}
and \psi (t)=\frac{1}{2}t,
\alpha (x,y)=\{\begin{array}{cc}1,\hfill & \text{if}x,y,z\in [0,1],\hfill \\ 1,\hfill & \text{otherwise.}\hfill \end{array}
Let x,y\in [0,1]. Without loss of generality, take x\le y. Then
Clearly, by taking \delta =\epsilon the condition (2.1) holds. Otherwise, \alpha (x,y)=1. Hence, for given \epsilon >0 we have \alpha (x,y)\psi (d(fx,fy))\le 0<\epsilon. Hence, conditions of Theorem 11 holds and f has a fixed point.
Denote with {\mathrm{\Psi}}_{st} the family of strictly nondecreasing functions {\psi}_{st}:[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) continuous in t=0 such that
Definition 14 Let (X,d) be a metric space and {\psi}_{st}\in {\mathrm{\Psi}}_{st}. Suppose that f:X\to X is a triangular αadmissible mapping satisfying the following condition: for each \epsilon >0 there exists \delta >0 such that
\epsilon \le {\psi}_{st}(M(x,y))<\epsilon +\delta \phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}\alpha (x,y){\psi}_{st}(d(fx,fy))<\epsilon
(2.8)
for all x,y\in X where
M(x,y)=max\{d(x,y),d(fx,x),d(fy,y),\frac{1}{2}[d(fx,y)+d(x,fy)]\}.
Then f is called generalized an α{\psi}_{st}MeirKeeler contractive mapping.
Remark 15 Let f be a generalized α{\psi}_{st}MeirKeeler contractive mapping. Then
\alpha (x,y){\psi}_{st}(d(fx,fy))<{\psi}_{st}(M(x,y))
for all x,y\in X when M(x,y)>0. Also, if M(x,y)=0 then x=y which implies \psi (d(fx,fy))=0, i.e.,
\alpha (x,y){\psi}_{st}(d(fx,fy))\le {\psi}_{st}(M(x,y))
for all x,y\in X.
Proposition 16 Let (X,d) be a metric space and f:X\to X a generalized α{\psi}_{st}MeirKeeler contractive mapping, if there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1 . Then {lim}_{n\to \mathrm{\infty}}d({f}^{n+1}{x}_{0},{f}^{n}{x}_{0})=0.
Proof Define a sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0} for all n\in \mathbb{N}. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N}\cup \{0\}, then obviously f has a fixed point. Hence, we suppose that
{x}_{n}\ne {x}_{n+1}
(2.9)
for all n\in \mathbb{N}\cup \{0\}. Then we have M({x}_{n+1},{x}_{n})>0 for every n\ge 0. Then by Lemma 7 and Remark 15, we have
\begin{array}{rcl}{\psi}_{st}(d({x}_{n+1},{x}_{n+2}))& \le & \alpha ({x}_{n},{x}_{n+1}){\psi}_{st}(d({x}_{n+1},{x}_{n+2}))\\ =& \alpha ({x}_{n},{x}_{n+1}){\psi}_{st}(d(f{x}_{n},f{x}_{n+1}))<{\psi}_{st}(M({x}_{n},{x}_{n+1}))\\ =& {\psi}_{st}(max\{d({x}_{n},{x}_{n+1}),d(f{x}_{n},{x}_{n}),d(f{x}_{n+1},{x}_{n+1}),\\ \frac{1}{2}[d(f{x}_{n},{x}_{n+1})+d({x}_{n},f{x}_{n+1})]\left\}\right)\\ \le & {\psi}_{st}(max\{d({x}_{n},{x}_{n+1}),d({x}_{n+1},{x}_{n+2})\}).\end{array}
Now, since {\psi}_{st} is strictly nondecreasing then, we get
d({x}_{n+2},{x}_{n+1})<max\{d({x}_{n+1},{x}_{n}),d({x}_{n+2},{x}_{n+1})\}.
Hence, the case where
max\{d({x}_{n+1},{x}_{n}),d({x}_{n+2},{x}_{n+1})\}=d({x}_{n+2},{x}_{n+1})
is not possible. Therefore, we deduce that
d({x}_{n+2},{x}_{n+1})<d({x}_{n+1},{x}_{n})
(2.10)
for all n. That is, {\{d({x}_{n+1},{x}_{n})\}}_{n=0}^{\mathrm{\infty}} is a decreasing sequence in {\mathbb{R}}_{+} and it converges to \epsilon \in {\mathbb{R}}_{+}, that is,
\underset{n\to \mathrm{\infty}}{lim}{\psi}_{st}(d({x}_{n+1},{x}_{n}))=\underset{n\to \mathrm{\infty}}{lim}{\psi}_{st}(M({x}_{n+1},{x}_{n}))={\psi}_{st}(\epsilon ).
(2.11)
Notice that \epsilon =inf\{p({x}_{n},{x}_{n+1}):n\in \mathbb{N}\}. Let us prove that \epsilon =0. Suppose, to the contrary, that \epsilon >0. Then \psi (\epsilon )>0. Regarding (2.11) together with the assumption that f is a generalized α{\psi}_{st}MeirKeeler contractive mapping, for {\psi}_{st}(\epsilon ), there exists \delta >0 and a natural number m such that
{\psi}_{st}(\epsilon )\le {\psi}_{st}(M({x}_{m},{x}_{m+1}))<{\psi}_{st}(\epsilon )+\delta
implies that
\begin{array}{rcl}{\psi}_{st}(d({x}_{m+1},{x}_{m+2}))& \le & \alpha ({x}_{m},{x}_{m+1}){\psi}_{st}(d({x}_{m+1},{x}_{m+2}))\\ =& \alpha ({x}_{m},{x}_{m+1}){\psi}_{st}(d(f{x}_{m},f{x}_{m+1}))<{\psi}_{st}(\epsilon ).\end{array}
Now, since {\psi}_{st} is strictly nondecreasing then we get
d({x}_{m+2},{x}_{m+1})<\epsilon
which is a contradiction since \epsilon =inf\{p({x}_{n},{x}_{n+1}):n\in \mathbb{N}\}. Then \epsilon =0 and so {lim}_{n\to \mathrm{\infty}}d({x}_{n+1},{x}_{n})=0. □
Theorem 17 Let (X,d) be a complete metric space and f:X\to X a orbitally continuous generalized α{\psi}_{st}MeirKeeler contractive mapping, if there exist {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Then, f has a fixed point.
Proof Define {x}_{n+1}={f}^{n+1}{x}_{0} for all n\ge 0. We want to prove that {lim}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0. If this is not, then there exist \epsilon >0 and a subsequence \{{x}_{n(i)}\} of \{{x}_{n}\} such that
d({x}_{n(i)},{x}_{n(i+1)})>2\epsilon .
(2.12)
For this \epsilon >0, there exists \delta >0 such that \epsilon \le {\psi}_{st}(M(x,y))<\epsilon +\delta implies that \alpha (x,y){\psi}_{st}(d(fx,fy))<\epsilon. Put r=min\{\epsilon ,\delta \} and {s}_{n}=d({x}_{n},{x}_{n+1}) for all n\ge 1. From Proposition 16, there exists {n}_{0} such that
{s}_{n}=d({x}_{n},{x}_{n+1})<\frac{r}{4}
(2.13)
for all n\ge {n}_{0}. Let n(i)>{n}_{0}. We get n(i)\le n(i+1)1. If d({x}_{n(i)},{x}_{n(i+1)1})\le \epsilon +\frac{r}{2}, then
\begin{array}{rcl}d({x}_{n(i)},{x}_{n(i+1)})& \le & d({x}_{n(i)},{x}_{n(i+1)1})+d({x}_{n(i+1)1},{x}_{n(i+1)})\\ \le & d({x}_{n(i)},{x}_{n(i+1)1})+d({x}_{n(i+1)1},{x}_{n(i+1)})\\ <& \epsilon +\frac{r}{2}+{s}_{n(i+1)1}<\epsilon +\frac{3r}{4}<2\epsilon \end{array}
which contradicts the assumption (2.12). Therefore, there are values of k such that n(i)\le k\le n(i+1) and d({x}_{n(i)},{x}_{k})>\epsilon +\frac{r}{2}. Now if d({x}_{n(i)},{x}_{n(i)+1})\ge \epsilon +\frac{r}{2}, then
{s}_{n(i)}=d({x}_{n(i)},{x}_{n(i)+1})\ge \epsilon +\frac{r}{2}>r+\frac{r}{2}>\frac{r}{4}
which is a contradiction with (2.13). Hence, there are values of k with n(i)\le k\le n(i+1) such that d({x}_{n(i)},{x}_{k})<\epsilon +\frac{r}{2}. Choose smallest integer k with k\ge n(i) such that d({x}_{n(i)},{x}_{k})\ge \epsilon +\frac{r}{2}. Thus, d({x}_{n(i)},{x}_{k1})<\epsilon +\frac{r}{2} and so
\begin{array}{rcl}d({x}_{n(i)},{x}_{k})& \le & d({x}_{n(i)},{x}_{k1})+d({x}_{k1},{x}_{k})\\ \le & d({x}_{n(i)},{x}_{k1})+d({x}_{k1},{x}_{k})<\epsilon +\frac{r}{2}+\frac{r}{4}=\epsilon +\frac{3r}{4}.\end{array}
Now, we can choose a natural number k satisfying n(i)\le k\le n(i+1) such that
\epsilon +\frac{r}{2}\le d({x}_{n(i)},{x}_{k})<\epsilon +\frac{3r}{4}.
(2.14)
Therefore, we obtain
and
d({x}_{k},{x}_{k+1})={d}_{k}<\frac{r}{4}<\epsilon +r.
(2.17)
Thus, we have
Now, the inequalities (2.15)(2.18) imply that M({x}_{n(i)},{x}_{k})<\epsilon +r\le \epsilon +\delta and so {\psi}_{st}(M({x}_{n(i)},{x}_{k}))<{\psi}_{st}(\epsilon +\delta )\le {\psi}_{st}(\epsilon )+{\psi}_{st}(\delta ) the fact that f is a generalized α{\psi}_{st}MeirKeeler contractive mapping yields that,
d({x}_{n(i)+1},{x}_{k+1})\le \alpha ({x}_{n(i)},{x}_{k}){\psi}_{st}(d({x}_{n(i)+1},{x}_{k+1}))<{\psi}_{st}(\epsilon ).
Then d({x}_{n(i)+1},{x}_{k+1})<\epsilon. We deduce,
\begin{array}{rcl}d({f}^{n(i)}{x}_{0},{f}^{k}{x}_{0})& \le & d({f}^{n(i)}{x}_{0},{f}^{n(i)+1}{x}_{0})+d({f}^{n(i)+1}{x}_{0},{f}^{k}{x}_{0})\\ \le & d({f}^{n(i)}{x}_{0},{f}^{n(i)+1}{x}_{0})+d({f}^{n(i)+1}{x}_{0},{f}^{k}{x}_{0})\\ \le & d({f}^{n(i)}{x}_{0},{f}^{n(i)+1}{x}_{0})+d({f}^{n(i)+1}{x}_{0},{f}^{k+1}{x}_{0})+d({f}^{k+1}{x}_{0},{f}^{k}{x}_{0}).\end{array}
Hence, from this with (2.14), (2.16) and (2.17), we obtain
\begin{array}{rcl}d({x}_{n(i)+1},{x}_{k+1})& \ge & d({x}_{n(i)},{x}_{k})d({x}_{n(i)},{x}_{n(i)+1})d({x}_{k},{x}_{k+1})\\ >& \epsilon +\frac{r}{2}\frac{r}{4}\frac{r}{4}=\epsilon ,\end{array}
which is a contradiction. We obtained that {lim}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0 and so \{{x}_{n}={f}^{n}{x}_{0}\} is a Cauchy sequence. Since, X is complete, then there exists z\in X such that {f}^{n}{x}_{0}\to z as n\to \mathrm{\infty}. Now, since f is orbitally continuous, then z=fz. □
Example 18 Let X=[0,\mathrm{\infty}) and d(x,y)=xy be a metric on X. Define f:X\to X by
fx=\{\begin{array}{cc}\frac{{x}^{2}}{8},\hfill & \text{if}x\in [0,1],\hfill \\ \frac{1}{2}+lnx,\hfill & \text{if}x\in (1,\mathrm{\infty})\hfill \end{array}
and {\psi}_{st}(t)=\frac{1}{2}t,
\alpha (x,y)=\{\begin{array}{cc}1,\hfill & \text{if}x,y\in [0,1],\hfill \\ 0,\hfill & \text{otherwise.}\hfill \end{array}
Clearly, f is a triangular αadmissible mapping and orbitally continuous. Let x,y\in [0,1]. Without loss of generality, take x\le y. Then
Clearly, by taking \delta =3\epsilon the condition (2.8) holds. Otherwise, \alpha (x,y)=0. Hence, for given \epsilon >0, we have 0=\alpha (x,y){\psi}_{st}(d(fx,fy))<\epsilon. Hence, condition of Theorem 17 is held and f has a fixed point.
Theorem 19 Assume that all the hypotheses of Theorem 10 (11 and 17) hold. Adding the following conditions:

(iii)
for all x\ne y\in X there exists v\in X such that \alpha (x,v)\ge 1 and \alpha (v,y)\ge 1,
we obtain the uniqueness of the fixed point of f.
Proof Suppose that z and {z}^{\ast} are two fixed points of f such that z\ne {z}^{\ast}. Then \alpha (z,v)\ge 1 and \alpha (v,{z}^{\ast})\ge 1. Hence, from (T2), we have \alpha (z,{z}^{\ast})\ge 1. Now, by Remark 9, we get
d(z,{z}^{\ast})=d(fz,f{z}^{\ast})\le \alpha (z,{z}^{\ast})d(fz,f{z}^{\ast})<d(z,{z}^{\ast})
which is a contradiction and so z={z}^{\ast}. Similarly, for Theorem 17, we can observe that f has a unique fixed point. □
We can obtain the following corollaries intermediately.
Corollary 20 Let (X,d) be a complete metric space and f:X\to X is selfmapping. Suppose that for each \epsilon >0, there exists \delta >0 such that
\epsilon \le \psi (d(x,y))<\epsilon +\delta \phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}\psi (d(fx,fy))<\frac{\epsilon}{L},
where \psi \in \mathrm{\Psi} and L\ge 1. Then f has a unique fixed points.
Corollary 21 Let (X,d) be a complete metric space and f:X\to X a orbitally continuous selfmapping. Suppose that for each \epsilon >0 there exists \delta >0 such that
\epsilon \le {\psi}_{st}(M(x,y))<\epsilon +\delta \phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}{\psi}_{st}(d(fx,fy))<\frac{\epsilon}{L},
where {\psi}_{st}\in {\mathrm{\Psi}}_{st}, L\ge 1 and
M(x,y)=max\{d(x,y),d(fx,x),d(fy,y),\frac{1}{2}[d(fx,y)+d(x,fy)]\}.
Then f has unique fixed points.