= d ( g x 2 n , f g x 2 n , a ) ρ s 2 d ( x 2 n , g x 2 n , a ) = ρ s 2 d ( x 2 n , x 2 n + 1 , a ) . (3.8)

From (3.7) and (3.8) we get

$d\left({x}_{2n+1},{x}_{2n},a\right)=d\left(gf{x}_{2n-1},f{x}_{2n-1},a\right)\le \frac{\rho }{{s}^{2}}d\left({x}_{2n},f{x}_{2n-1},a\right)=\frac{\rho }{{s}^{2}}d\left({x}_{2n-1},{x}_{2n},a\right)$,

that is,

$d\left({x}_{n+1},{x}_{n},a\right)\le \frac{\rho }{{s}^{2}}d\left({x}_{n},{x}_{n-1},a\right)$, since $\frac{\rho }{{s}^{2}}\in \left[0,1\right)$, by Lemma 2.6, we get $\left\{{x}_{n}\right\}$ is a Cauchy sequence.

Since X is complete, there exists z in X, such that $\underset{n\to \infty }{\mathrm{lim}}{x}_{n}=z$, that is $\underset{n\to \infty }{\mathrm{lim}}g{x}_{2n}=\underset{n\to \infty }{\mathrm{lim}}{x}_{2n+1}=z$, and $\underset{n\to \infty }{\mathrm{lim}}f{x}_{2n+1}=\underset{n\to \infty }{\mathrm{lim}}{x}_{2n+2}=z$.

Now let us give that

$d\left(fx,z,a\right)\le \rho d\left(x,z,a\right)$, for every $x\ne z$. For $\left\{d\left({x}_{2n},g{x}_{2n},a\right)\right\}$ is convergent to 0, and by Lemma 2.5, we get

$\frac{1}{s}d\left(x,z,a\right)\le \underset{n\to \infty }{\mathrm{lim}}\mathrm{sup}d\left({x}_{2n},x,a\right)$, thus we have $\underset{n\to \infty }{\mathrm{lim}}\mathrm{sup}d\left({x}_{2n},x,a\right)>0$, thus from the above relation, there exists a point ${x}_{2{n}_{k}}$ in X such that

$\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left({x}_{2{n}_{k}},g{x}_{2{n}_{k}},a\right),d\left({x}_{2{n}_{k}},f{x}_{2{n}_{k}},a\right)\right\}\le d\left({x}_{2{n}_{k}},x,a\right).$

For such ${x}_{2{n}_{k}}$, (3.2) implies that

$\begin{array}{l}d\left(g{x}_{2{n}_{k}},fx,a\right)\\ \le \mathrm{max}\left\{d\left(g{x}_{2{n}_{k}},gx,a\right),d\left(g{x}_{2{n}_{k}},fx,a\right),d\left(f{x}_{2{n}_{k}},fx,a\right),d\left(gx,f{x}_{2{n}_{k}},a\right)\right\}\\ \le \frac{\rho }{{s}_{2}}d\left({x}_{2{n}_{k}},x,a\right),\end{array}$

therefore by Lemma 3.5,

$\frac{1}{s}d\left(fx,z,a\right)\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left(g{x}_{2{n}_{k}},fx,a\right)\le \frac{\rho }{{s}^{2}}\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left({x}_{2{n}_{k}},x,a\right)\le \frac{\rho }{s}d\left(x,z,a\right),$

therefore we get

$d\left(fx,z,a\right)\le \rho d\left(x,z,a\right)$, for each $x\ne z$. (3.9)

Now we show that for each $n\in N$,

$d\left({f}^{n}z,z,a\right)\le d\left(fz,z,a\right),$ (3.10)

It is obvious that the above inequality is true for $n=1$, assume that the relation holds for some $m\in N$. We get (3.10) is true when we have ${f}^{m}z=fz$ if ${f}^{m}z=z$, then if ${f}^{m}z\ne z$, we get the following relation from (3.9) and induction hypothesis, and that is

$\begin{array}{c}d\left(z,{f}^{m+1}z,a\right)\le \rho d\left(z,{f}^{m}z,a\right)\le {\rho }^{2}d\left(z,{f}^{m-1}z,a\right)\le \cdots \le {\rho }^{m+1}d\left(z,fz,a\right)\\ \le \rho d\left(fz,z,a\right)\le d\left(fz,z,a\right),\end{array}$

then (3.10) is proved.

Now we consider the following two possible cases in order to prove that f has a fixed point z in X, and that is $fz=z$.

Case 1 $0\le \rho <\frac{1}{\sqrt{2}}$, therefore, $\varphi \left(\rho \right)\le \frac{1-\rho }{{\rho }^{2}}$. First, we prove the following relation

$d\left({f}^{n}z,fz,a\right)\le \frac{\rho }{s}d\left(fz,z,a\right)$, for $n\in N$. (3.11)

When $n=1$ it is obvious, and it follows from (3.6) when $n=2$, from (3.10) and take $a=fz$ we have

$d\left({f}^{n}z,z,fz\right)\le d\left(fz,z,fz\right)=0$, then we get $d\left({f}^{n}z,fz,z\right)=0$.

Now suppose that (3.11) holds for some $n>2$,

$\begin{array}{c}d\left(fz,z,a\right)\le s\left(d\left(z,{f}^{n}z,a\right)+d\left({f}^{n}z,fz,a\right)+d\left({f}^{n}z,fz,z\right)\right)\\ \le sd\left(z,{f}^{n}z,a\right)+sd\left(z,fz,a\right),\end{array}$

Therefore, we get

$\left(1-\rho \right)d\left(z,fz,a\right)\le sd\left(z,{f}^{n}z,a\right)$, that is $d\left(z,fz,a\right)\le \frac{s}{1-\rho }sd\left(z,{f}^{n}z,a\right)$, (3.11.1)

then by taking $x={f}^{n-1}z$ in (3.6)

$d\left({f}^{n}z,{f}^{n+1}z,a\right)\le \frac{\rho }{{s}^{2}}d\left({f}^{n-1}z,{f}^{n}z,a\right)\le \cdots \le \frac{{\rho }^{n}}{{s}^{2n}}d\left(z,fz,a\right)$, (3.11.2)

using the above two relations, (3.11.1) and (3.11.2) we have

$\begin{array}{l}\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left(g{f}^{n}z,{f}^{n}z,a\right),d\left({f}^{n}z,{f}^{n+1}z,a\right)\right\}\\ \le \frac{1-\rho }{s{\rho }^{2}}d\left({f}^{n}z,{f}^{n+1}z,a\right)\le \frac{1-\rho }{s{\rho }^{n}}d\left({f}^{n}z,{f}^{n+1}z,a\right)\\ \le \frac{1-\rho }{s{\rho }^{n}}\cdot \frac{{\rho }^{n}}{{s}^{2n}}d\left(z,fz,a\right)=\frac{1-\rho }{{s}^{2n+1}}d\left(z,fz,a\right)\\ \le \frac{1-\rho }{{s}^{2n+1}}\cdot \frac{s}{1-\rho }d\left(z,{f}^{n}z,a\right)\le \frac{1}{{s}^{2n}}d\left(z,{f}^{n}z,a\right)\le d\left(z,{f}^{n}z,a\right).\end{array}$

From (3.2) and (3.10) with $x={f}^{n}z$ and $y=z$, we have

$\begin{array}{l}\mathrm{max}\left\{d\left(g{f}^{n}z,gz,a\right),d\left(g{f}^{n}z,fz,a\right),d\left({f}^{n+1}z,fz,a\right),d\left(gz,{f}^{n+1}z,a\right)\right\}\\ \le \frac{\rho }{{s}^{2}}d\left(z,{f}^{n}z,a\right)\le \frac{\rho }{{s}^{2}}d\left(z,fz,a\right)\le \frac{\rho }{s}d\left(z,fz,a\right).\end{array}$

Therefore,

$d\left({f}^{n+1}z,fz,a\right)\le \frac{\rho }{s}d\left(fz,z,a\right).$ (3.12)

So by induction we prove the relation of (3.11).

Now (3.11) and $fz\ne z$ show that for every $n\in N$ ${f}^{n}z\ne z$, thus, (3.9) shows that

$d\left(z,{f}^{n+1}z,a\right)\le \rho d\left(z,{f}^{n}z,a\right)\le {\rho }^{2}d\left(z,{f}^{n-1}z,a\right)\le \cdots \le {\rho }^{n}d\left(z,fz,a\right).$

Therefore $\underset{n\to \infty }{\mathrm{lim}}d\left(z,{f}^{n+1}z,a\right)=0$. Furthermore by using Lemma 2.5, we get

$\frac{1}{s}d\left(z,\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}{f}^{n+1}z,a\right)\le \underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}d\left(z,{f}^{n+1}z,a\right)=0,$

so

$d\left(z,\underset{n\to \infty }{\mathrm{lim}\mathrm{inf}}{f}^{n+1}z,a\right)=0.$

In the same way,

$d\left(z,\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}{f}^{n+1}z,a\right)=0$, thus we have $d\left(z,\underset{n\to \infty }{\mathrm{lim}}{f}^{n+1}z,a\right)=0$, that is ${f}^{n+1}z\to z$, and by using Lemma 2.5 in (3.12), we get

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left({f}^{n+1}z,fz,a\right)\le \frac{\rho }{s}d\left(z,fz,a\right)$, which claims that $d\left(z,fz,a\right)=0$, and that is a contraction.

Case 2. $\frac{1}{\sqrt{2}}\le \rho <1$, and that is when $\varphi \left(\rho \right)=\frac{1}{1+\rho }$. We now prove that we can find a subsequence $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$ such that

$\frac{1}{s\left(1+\rho \right)}\mathrm{min}\left\{d\left({x}_{{n}_{k}},g{x}_{{n}_{k}},a\right),d\left({x}_{{n}_{k}},f{x}_{{n}_{k}},a\right)\right\}\le d\left({x}_{{n}_{k}},z,a\right)$, for $k\in N$. (3.13)

The contraries of the above relation are as follows

$\frac{1}{s\left(1+\rho \right)}d\left({x}_{n},f{x}_{n},a\right)\ge \frac{1}{s\left(1+\rho \right)}\mathrm{min}\left\{d\left({x}_{n},g{x}_{n},a\right),d\left({x}_{n},f{x}_{n},a\right)\right\}>d\left({x}_{n},z,a\right),$

and

$\frac{1}{s\left(1+\rho \right)}d\left({x}_{n},f{x}_{n},a\right)\ge \frac{1}{s\left(1+\rho \right)}\mathrm{min}\left\{d\left({x}_{n},g{x}_{n},a\right),d\left({x}_{n},f{x}_{n},a\right)\right\}>d\left({x}_{n},z,a\right),$

for $n\in N$. If n is even we have

$\begin{array}{l}\frac{1}{s\left(1+\rho \right)}d\left({x}_{2n},g{x}_{2n},a\right)\\ \ge \frac{1}{s\left(1+\rho \right)}\mathrm{min}\left\{d\left({x}_{2n},g{x}_{2n},a\right),d\left({x}_{2n},f{x}_{2n},a\right)\right\}\\ >d\left({x}_{2n},z,a\right),\end{array}$

if n is odd then we get

$\begin{array}{l}\frac{1}{s\left(1+\rho \right)}d\left({x}_{2n+1},f{x}_{2n+1},a\right)\\ \ge \frac{1}{s\left(1+\rho \right)}\mathrm{min}\left\{d\left({x}_{2n+1},g{x}_{2n+1},a\right),d\left({x}_{2n+1},f{x}_{2n+1},a\right)\right\}\\ >d\left({x}_{2n+1},z,a\right),\end{array}$

for $n\in N$. By (3.8) we have

$\begin{array}{l}d\left({x}_{2n},{x}_{2n+1},a\right)\\ \le s\left(d\left({x}_{2n},z,a\right)+d\left({x}_{2n+1},z,a\right)+d\left({x}_{2n},{x}_{2n+1},z\right)\right)\\ <\frac{s}{s\left(1+\rho \right)}d\left({x}_{2n},g{x}_{2n},a\right)+\frac{s}{s\left(1+\rho \right)}d\left({x}_{2n+1},f{x}_{2n+1},a\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\frac{s}{s\left(1+\rho \right)}d\left({x}_{2n},g{x}_{2n},{x}_{2n+1}\right)\\ =\frac{1}{1+\rho }\left(d\left({x}_{2n},{x}_{2n+1},a\right)+d\left({x}_{2n+1},{x}_{2n+2},a\right)+d\left({x}_{2n},{x}_{2n+1},{x}_{2n+1}\right)\right)\end{array}$

$\begin{array}{l}\le \frac{1}{1+\rho }d\left({x}_{2n},{x}_{2n+1},a\right)+\frac{\rho }{{s}^{2}\left(1+\rho \right)}d\left({x}_{2n+1},{x}_{2n},a\right)\\ \le \frac{1}{1+\rho }d\left({x}_{2n},{x}_{2n+1},a\right)+\frac{\rho }{1+\rho }d\left({x}_{2n+1},{x}_{2n},a\right)\\ =d\left({x}_{2n},{x}_{2n+1},a\right),\end{array}$

this is impossible. Therefore, one of the following relations is true for every $n\in N$,

$\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left({x}_{2n},g{x}_{2n},a\right),d\left({x}_{2n},f{x}_{2n},a\right)\right\}\le d\left({x}_{2n},z,a\right),$

or

$\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left({x}_{2n+1},g{x}_{2n+1},a\right),d\left({x}_{2n+1},f{x}_{2n+1},a\right)\right\}\le d\left({x}_{2n+1},z,a\right).$

That means there exists a subsequence $\left\{{x}_{{n}_{k}}\right\}$ of $\left\{{x}_{n}\right\}$ such that (3.13) is true for every $k\in N$. Thus (3.2) shows that

$\begin{array}{l}d\left(g{x}_{2n},fz,a\right)\\ \le \mathrm{max}\left\{d\left(f{x}_{2n},gz,a\right),d\left(fz,g{x}_{2n},a\right),d\left(f{x}_{2n},fz,a\right),d\left(gz,f{x}_{2n},a\right)\right\}\\ \le \frac{\rho }{{s}^{2}}d\left({x}_{2n},z,a\right).\end{array}$

or

$\begin{array}{l}d\left(f{x}_{2n+1},fz,a\right)\\ \le \mathrm{max}\left\{d\left(g{x}_{2n+1},gz,a\right),d\left(fz,g{x}_{2n+1},a\right),d\left(f{x}_{2n+1},fz,a\right),d\left(gz,f{x}_{2n+1},a\right)\right\}\\ \le \frac{\rho }{{s}^{2}}d\left({x}_{2n+!},z,a\right).\end{array}$

From Lemma 2.5, we have

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left(g{x}_{2n},fz,a\right)\le \frac{\rho }{{s}^{2}}\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left({x}_{2n},z,a\right)\le \frac{\rho }{s}d\left(z,z,a\right)=0,$

or

$\frac{1}{s}d\left(z,fz,a\right)\le \underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left(f{x}_{2n+1},fz,a\right)\le \frac{\rho }{{s}^{2}}\underset{n\to \infty }{\mathrm{lim}\mathrm{sup}}d\left({x}_{2n+1},z,a\right)\le \frac{\rho }{s}d\left(z,z,a\right)=0,$

Therefore $d\left(z,fz,a\right)\le 0$, which is impossible unless $fz=z$. hence z in X is a fixed point of f. From the process of the above proof, we know $fz=z$, then by

$0=\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left(z,fz,a\right),d\left(z,gz,a\right)\right\}\le d\left(z,fz,a\right)$,

it implies

$\begin{array}{c}d\left(gz,z,a\right)\le \mathrm{max}\left\{d\left(gz,gfz,a\right),d\left(gz,{f}^{2}z,a\right),d\left(fz,{f}^{2}z,a\right),d\left(gfz,fz,a\right)\right\}\\ \le \frac{\rho }{{s}^{2}}d\left(fz,z,a\right)=0,\end{array}$

this proves that $gz=z$. By (3.2) we can prove the uniqueness of the common fixed point z,

$\frac{1}{s}\varphi \left(\rho \right)\mathrm{min}\left\{d\left(z,fz,a\right),d\left(z,gz,a\right)\right\}\le d\left(z,{z}^{\prime },a\right)$, so (3.2) shows that

$\begin{array}{c}d\left(z,{z}^{\prime },a\right)=\mathrm{max}\left\{d\left(gz,g{z}^{\prime },a\right),d\left(fz,f{z}^{\prime },a\right),d\left(gz,f{z}^{\prime },a\right),d\left(g{z}^{\prime },fz,a\right)\right\}\\ \le \frac{\rho }{{s}^{2}}d\left(z,{z}^{\prime },a\right),\end{array}$

which is impossible unless $z={z}^{\prime }$. □

NOTES

*Corresponding author.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

  Banach, S. (1922) Sur les opérations dans les ensembles abtraits et leur applications aux équations intégrales. Fundamenta Mathematicae, 3, 133-181. https://doi.org/10.4064/fm-3-1-133-181  Ekeland, I. (1974) On the Variational Principle. Journal of Mathematical Analysis and Applications, 47, 324-353. https://doi.org/10.1016/0022-247X(74)90025-0  Meir, A. and Keeler, E. (1969) A Theorem on Contraction Mappings. Journal of Mathematical Analysis and Applications, 28, 326-329. https://doi.org/10.1016/0022-247X(69)90031-6  Nadler Jr., S.B. (1969) Multi-Valued Contraction Mappings. Pacific Journal of Mathematics, 30, 475-488. https://doi.org/10.2140/pjm.1969.30.475  Caristi, J. (1976) Fixed Point Theorems for Mappings Satisfying Inwardness Conditions. Transactions of the American Mathematical Society, 215, 241-251. https://doi.org/10.1090/S0002-9947-1976-0394329-4  Caristi, J. and Kirk, W.A. (1975) Geometric Fixed Point Theory and Inwardness Conditions. Lecture Notes in Mathematics, 499, 74-83. https://doi.org/10.1007/BFb0081133  Subrahmanyam, P.V. (1974) Remarks on Some Fixed Point Theorems Related to Banach’s Contraction Principle. Electronic Journal of Mathematical and Physical Sciences, 8, 445-457.  Suzuki, T. (2004) Generalized Distance and Existence Theorems in Complete Metric Spaces. Journal of Mathematical Analysis and Applications, 253, 440-458. https://doi.org/10.1006/jmaa.2000.7151  Mustafa, Z., Parvaech, V., Roshan, J.R. and Kadelburg, Z. (2014) b2-Metric Spaces and Some Fixed Point Theorems. Fixed Point Theory and Applications, 2014, Article Number: 144. https://doi.org/10.1186/1687-1812-2014-144  Fadail, Z.M., Ahmad, A.G.B., Ozturk, V. and Radenovi?, S. (2015) Some Remarks on Fixed Point Results of b2-Metric Spaces. Far East Journal of Mathematical Sciences, 97, 533-548. https://doi.org/10.17654/FJMSJul2015_533_548 