Let (X,d) be a metric space, and let \alpha :X\times X\to \mathbb{R} be a function. A map T:X\to X is called a generalized αGeraghty contraction type map if there exists \beta \in \mathcal{F} such that for all x,y\in X,
\alpha (x,y)d(Tx,Ty)\le \beta (M(x,y))M(x,y),
(3)
where M(x,y)=max\{d(x,y),d(x,Tx),d(y,Ty)\}.
Remark 2.1 Since the functions belonging to ℱ are strictly smaller than one, condition (3) implies that
d(Tx,Ty)<M(x,y)\phantom{\rule{1em}{0ex}}\text{for any}x,y\in X\text{with}x\ne y.
Theorem 2.1 Let (X,d) be a complete metric space, \alpha :X\times X\to \mathbb{R} be a function, and let T:X\to X be a map. Suppose that the following conditions are satisfied:

(1)
T is a generalized αGeraghty contraction type map;

(2)
T is triangular αadmissible;

(3)
there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1;

(4)
T is continuous.
Then T has a fixed point {x}_{\ast}\in X, and T is a Picard operator, that is, \{{T}^{n}{x}_{1}\} converges to {x}_{\ast}.
Proof Let {x}_{1}\in X be such that \alpha ({x}_{1},T{x}_{1})\ge 1. Define a sequence \{{x}_{n}\}\subset X by {x}_{n+1}=T{x}_{n} for n\in \mathbb{N}. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N}, then {x}_{{n}_{0}} is a fixed point of T, and hence the proof is completed. Thus, throughout the proof, we assume that {x}_{n}\ne {x}_{n+1} for all n\in \mathbb{N}.
By Lemma 1.1, we have
\alpha ({x}_{n},{x}_{n+1})\ge 1
(4)
for all n\in \mathbb{N}. Then we have
\begin{array}{rl}d({x}_{n+1},{x}_{n+2})& =d(T{x}_{n},T{x}_{n+1})\\ \le \alpha ({x}_{n},{x}_{n+1})d(T{x}_{n},T{x}_{n+1})\\ \le \beta (M({x}_{n},{x}_{n+1}))M({x}_{n},{x}_{n+1}),\end{array}
(5)
for all n\in \mathbb{N}, where
\begin{array}{rl}M({x}_{n},{x}_{n+1})& =max\{d({x}_{n},{x}_{n+1}),d({x}_{n},T{x}_{n}),d({x}_{n+1},T{x}_{n+1})\}\\ =max\{d({x}_{n},{x}_{n+1}),d({x}_{n},{x}_{n+1}),d({x}_{n+1},{x}_{n+2})\}.\end{array}
Regarding the definition of β, the case M({x}_{n},{x}_{n+1})=d({x}_{n+1},{x}_{n+2}) is impossible. Indeed,
\begin{array}{rl}d({x}_{n+1},{x}_{n+2})& \le \beta (M({x}_{n},{x}_{n+1}))M({x}_{n},{x}_{n+1})\\ \le \beta (d({x}_{n+1},{x}_{n+2}))d({x}_{n+1},{x}_{n+2})<d({x}_{n+1},{x}_{n+2}),\end{array}
a contradiction.
Thus, we conclude that d({x}_{n+1},{x}_{n+2})<d({x}_{n},{x}_{n+1}) for all n\in \mathbb{N}. So, the sequence \{d({x}_{n},{x}_{n+1})\} is nonnegative and nonincreasing. Hence, there exists r\ge 0 such that {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=r. We claim that r=0. Suppose, on the contrary, that r>0. Then, due to (5), we have
\frac{d({x}_{n+1},{x}_{n+2})}{d({x}_{n},{x}_{n+1})}\le \beta (d({x}_{n},{x}_{n+1}))<1,
which yields that {lim}_{n\to \mathrm{\infty}}\beta (d({x}_{n},{x}_{n+1}))=1. Since \beta \in \mathcal{F}, we derive that
\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},{x}_{n+1})=0.
(6)
We shall show that \{{x}_{n}\} is a Cauchy sequence. Suppose, on the contrary, that \{{x}_{n}\} is not a Cauchy sequence. Thus, there exists \u03f5>0 such that, for all k>0, there exist m(k)>n(k)>k with (the smallest number satisfying the condition below)
d({x}_{m(k)},{x}_{n(k)})\ge \u03f5\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d({x}_{m(k)1},{x}_{n(k)})<\u03f5.
Then we have
\begin{array}{rl}\u03f5& \le d({x}_{m(k)},{x}_{n(k)})\\ \le d({x}_{m(k)},{x}_{m(k)1})+d({x}_{m(k)1},{x}_{n(k)})\\ <d({x}_{m(k)},{x}_{m(k)1})+\u03f5.\end{array}
Letting k\to \mathrm{\infty} in the above inequality, we have
\underset{k\to \mathrm{\infty}}{lim}d({x}_{m(k)},{x}_{n(k)})=\u03f5.
(7)
By using (6) and (7), we obtain {lim}_{k\to \mathrm{\infty}}d({x}_{m(k)1},{x}_{n(k)1})=\u03f5.
By Lemma 1.1, \alpha ({x}_{n(k)1},{x}_{m(k)1})\ge 1. Thus we have
\begin{array}{r}d({x}_{m(k)},{x}_{n(k)})\\ \phantom{\rule{1em}{0ex}}=d(T{x}_{m(k)1},T{x}_{n(k)1})\\ \phantom{\rule{1em}{0ex}}\le \alpha ({x}_{n(k)1},{x}_{m(k)1})d(T{x}_{n(k)1},T{x}_{m(k)1})\\ \phantom{\rule{1em}{0ex}}\le \beta (M({x}_{n(k)1},{x}_{m(k)1}))M({x}_{n(k)1},{x}_{m(k)1}),\end{array}
where
\begin{array}{rl}M({x}_{n(k)1},{x}_{m(k)1})& =max\{d({x}_{n(k)1},{x}_{m(k)1}),d({x}_{n(k)1},T{x}_{n(k)1}),d({x}_{m(k)1},T{x}_{m(k)1})\}\\ =max\{d({x}_{n(k)1},{x}_{m(k)1}),d({x}_{n(k)1},{x}_{n(k)}),d({x}_{m(k)1},{x}_{m(k)})\}.\end{array}
Hence, we conclude that
\frac{d({x}_{m(k)},{x}_{n(k)})}{M({x}_{n(k)1},{x}_{m(k)1})}\le \beta (M({x}_{n(k)1},{x}_{m(k)1})).
Keeping (6) in mind and letting n\to \mathrm{\infty} in the above inequality, we derive that {lim}_{k\to \mathrm{\infty}}\beta (d({x}_{n(k)1},{x}_{m(k)1}))=1, and so {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)1},{x}_{m(k)1})=0. Hence, \u03f5=0, which is a contradiction. So, we conclude that \{{x}_{n}\} is a Cauchy sequence. It follows from the completeness of X that there exists
{x}_{\ast}=\underset{n\to \mathrm{\infty}}{lim}{x}_{n}\in X.
Since T is continuous, we get {lim}_{n\to \mathrm{\infty}}{x}_{n}=T{x}_{\ast}, and so {x}_{\ast}=T{x}_{\ast}, which completes the proof. □
The continuity of the mapping T can be dropped. In the following theorem, we replace the continuity of the operator T by a suitable condition.
Theorem 2.2 Let (X,d) be a complete metric space, \alpha :X\times X\to \mathbb{R} be a function, and let T:X\to X be a map. Suppose that the following conditions are satisfied:

(1)
T is a generalized αGeraghty contraction type map;

(2)
T is triangular αadmissible;

(3)
there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1;

(4)
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\in X as n\to \mathrm{\infty}, then there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},x)\ge 1 for all k.
Then T has a fixed point {x}_{\ast}\in X, and T is a Picard operator, that is, \{{T}^{n}{x}_{1}\} converges to {x}_{\ast}.
Proof Following the lines in the proof of Theorem 2.1, we conclude that the sequence \{{x}_{n}\}, defined by {x}_{n+1}=T{x}_{n} for all n\ge 0, converges to {x}_{\ast}\in X. Regarding (4) together with condition (4), we deduce that there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},{x}_{\ast})\ge 1 for all k. Applying (3), for all k, we get that
\begin{array}{rl}d({x}_{n(k)+1},T{x}_{\ast})& =d(T{x}_{n(k)},T{x}_{\ast})\\ \le \alpha ({x}_{n(k)},{x}_{\ast})d(T{x}_{n(k)},T{x}_{\ast})\\ \le \beta (M({x}_{n(k)},{x}_{\ast}))M({x}_{n(k)},{x}_{\ast}).\end{array}
(8)
On the other hand, we have
\begin{array}{rl}M({x}_{n},{x}_{\ast})& =max\{d({x}_{n},{x}_{\ast}),d({x}_{n},T{x}_{n}),d({x}_{\ast},T{x}_{\ast})\}\\ =max\{d({x}_{n},{x}_{\ast}),d({x}_{n},{x}_{n+1}),d({x}_{\ast},T{x}_{\ast})\}.\end{array}
Letting k\to \mathrm{\infty} in the above equality, we get that
\underset{k\to \mathrm{\infty}}{lim}M({x}_{n(k)},{x}_{\ast})=d({x}_{\ast},T{x}_{\ast}).
(9)
Suppose that d({x}_{\ast},T{x}_{\ast})>0. From (9), for k large enough, we have M({x}_{n(k)},{x}_{\ast})>0, which implies that \beta (M({x}_{n(k)},{x}_{\ast}))<M({x}_{n(k)},{x}_{\ast}). By Remark 2.1, we have
d({x}_{n(k)+1},T{x}_{\ast})<M({x}_{n(k)},{x}_{\ast}).
Letting k\to \mathrm{\infty} in the above inequality, using (9), we obtain that
d({x}_{\ast},T{x}_{\ast})<d({x}_{\ast},T{x}_{\ast}),
a contradiction. Thus, we find d({x}_{\ast},T{x}_{\ast})=0, that is, {x}_{\ast}=T{x}_{\ast}. □
For the uniqueness of a fixed point of a generalized αGeraghty contractive mapping, we consider the following hypothesis.

(H)
For all x,y\in Fix(T), there exists z\in X such that \alpha (x,z)\ge 1 and \alpha (y,z)\ge 1.
Here, Fix(T) denotes the set of fixed points of T.
Theorem 2.3 Adding condition (H) to the hypotheses of Theorem 2.1 (resp. Theorem 2.2), we obtain that {x}_{\ast} is the unique fixed point of T.
Proof Due to Theorem 2.1 (resp. Theorem 2.2), we have a fixed point, namely {x}_{\ast}\in X. Let {y}_{\ast}\in X be another fixed point of T. Then, by assumption, there exists z\in X such that
\alpha ({x}_{\ast},z)\ge 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\alpha ({y}_{\ast},z)\ge 1.
(10)
Since T is αadmissible, from (10), we derive that
\alpha ({x}_{\ast},{T}^{n}z)\ge 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\alpha ({y}_{\ast},{T}^{n}z)\ge 1\phantom{\rule{1em}{0ex}}\text{for all}n.
(11)
Hence we have
\begin{array}{rl}d({x}_{\ast},{T}^{n}z)& \le \alpha ({x}_{\ast},{T}^{n1}z)d(T{x}_{\ast},T{T}^{n1}z)\\ \le \beta \left(d({x}_{\ast},{T}^{n1}z)\right)d({x}_{\ast},{T}^{n1}z)\\ <d({x}_{\ast},{T}^{n1}z)\end{array}
(12)
for all n\in \mathbb{N}. Thus, the sequence \{d({x}_{\ast},{T}^{n}z)\} is nonincreasing, and so there exists u\ge 0 such that {lim}_{n\to \mathrm{\infty}}d({x}_{\ast},{T}^{n}z)=u. From (12) we have
\frac{d({x}_{\ast},{T}^{n}z)}{d({x}_{\ast},{T}^{n1}z)}\le \beta \left(d({x}_{\ast},{T}^{n1}z)\right),
and so {lim}_{n\to \mathrm{\infty}}\beta (d({x}_{\ast},{T}^{n}z))=1. Consequently, we have {lim}_{n\to \mathrm{\infty}}d({x}_{\ast},{T}^{n}z)=0, and hence {lim}_{n\to \mathrm{\infty}}{T}^{n}z={x}_{\ast}.
Similarly, we find that {lim}_{n\to \mathrm{\infty}}{T}^{n}z={y}_{\ast}. Thus, we get {x}_{\ast}={y}_{\ast}. □