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

that is,

d ( x n + 1 , x n , a ) ρ s 2 d ( x n , x n 1 , a ) , since ρ s 2 [ 0 , 1 ) , by Lemma 2.6, we get { x n } is a Cauchy sequence.

Since X is complete, there exists z in X, such that lim n x n = z , that is lim n g x 2 n = lim n x 2 n + 1 = z , and lim n f x 2 n + 1 = lim n x 2 n + 2 = z .

Now let us give that

d ( f x , z , a ) ρ d ( x , z , a ) , for every x z . For { d ( x 2 n , g x 2 n , a ) } is convergent to 0, and by Lemma 2.5, we get

1 s d ( x , z , a ) lim n sup d ( x 2 n , x , a ) , thus we have lim n sup d ( x 2 n , x , a ) > 0 , thus from the above relation, there exists a point x 2 n k in X such that

1 s ϕ ( ρ ) min { d ( x 2 n k , g x 2 n k , a ) , d ( x 2 n k , f x 2 n k , a ) } d ( x 2 n k , x , a ) .

For such x 2 n k , (3.2) implies that

d ( g x 2 n k , f x , a ) max { d ( g x 2 n k , g x , a ) , d ( g x 2 n k , f x , a ) , d ( f x 2 n k , f x , a ) , d ( g x , f x 2 n k , a ) } ρ s 2 d ( x 2 n k , x , a ) ,

therefore by Lemma 3.5,

1 s d ( f x , z , a ) lim sup n d ( g x 2 n k , f x , a ) ρ s 2 lim sup n d ( x 2 n k , x , a ) ρ s d ( x , z , a ) ,

therefore we get

d ( f x , z , a ) ρ d ( x , z , a ) , for each x z . (3.9)

Now we show that for each n N ,

d ( f n z , z , a ) d ( f z , z , a ) , (3.10)

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

d ( z , f m + 1 z , a ) ρ d ( z , f m z , a ) ρ 2 d ( z , f m 1 z , a ) ρ m + 1 d ( z , f z , a ) ρ d ( f z , z , a ) d ( f z , z , a ) ,

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 f z = z .

Case 1 0 ρ < 1 2 , therefore, ϕ ( ρ ) 1 ρ ρ 2 . First, we prove the following relation

d ( f n z , f z , a ) ρ s d ( f z , z , a ) , for n N . (3.11)

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

d ( f n z , z , f z ) d ( f z , z , f z ) = 0 , then we get d ( f n z , f z , z ) = 0 .

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

d ( f z , z , a ) s ( d ( z , f n z , a ) + d ( f n z , f z , a ) + d ( f n z , f z , z ) ) s d ( z , f n z , a ) + s d ( z , f z , a ) ,

Therefore, we get

( 1 ρ ) d ( z , f z , a ) s d ( z , f n z , a ) , that is d ( z , f z , a ) s 1 ρ s d ( z , f n z , a ) , (3.11.1)

then by taking x = f n 1 z in (3.6)

d ( f n z , f n + 1 z , a ) ρ s 2 d ( f n 1 z , f n z , a ) ρ n s 2 n d ( z , f z , a ) , (3.11.2)

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

1 s ϕ ( ρ ) min { d ( g f n z , f n z , a ) , d ( f n z , f n + 1 z , a ) } 1 ρ s ρ 2 d ( f n z , f n + 1 z , a ) 1 ρ s ρ n d ( f n z , f n + 1 z , a ) 1 ρ s ρ n ρ n s 2 n d ( z , f z , a ) = 1 ρ s 2 n + 1 d ( z , f z , a ) 1 ρ s 2 n + 1 s 1 ρ d ( z , f n z , a ) 1 s 2 n d ( z , f n z , a ) d ( z , f n z , a ) .

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

max { d ( g f n z , g z , a ) , d ( g f n z , f z , a ) , d ( f n + 1 z , f z , a ) , d ( g z , f n + 1 z , a ) } ρ s 2 d ( z , f n z , a ) ρ s 2 d ( z , f z , a ) ρ s d ( z , f z , a ) .

Therefore,

d ( f n + 1 z , f z , a ) ρ s d ( f z , z , a ) . (3.12)

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

Now (3.11) and f z z show that for every n N f n z z , thus, (3.9) shows that

d ( z , f n + 1 z , a ) ρ d ( z , f n z , a ) ρ 2 d ( z , f n 1 z , a ) ρ n d ( z , f z , a ) .

Therefore lim n d ( z , f n + 1 z , a ) = 0 . Furthermore by using Lemma 2.5, we get

1 s d ( z , lim inf n f n + 1 z , a ) lim inf n d ( z , f n + 1 z , a ) = 0 ,

so

d ( z , lim inf n f n + 1 z , a ) = 0.

In the same way,

d ( z , lim sup n f n + 1 z , a ) = 0 , thus we have d ( z , lim n f n + 1 z , a ) = 0 , that is f n + 1 z z , and by using Lemma 2.5 in (3.12), we get

1 s d ( z , f z , a ) lim sup n d ( f n + 1 z , f z , a ) ρ s d ( z , f z , a ) , which claims that d ( z , f z , a ) = 0 , and that is a contraction.

Case 2. 1 2 ρ < 1 , and that is when ϕ ( ρ ) = 1 1 + ρ . We now prove that we can find a subsequence { x n k } of { x n } such that

1 s ( 1 + ρ ) min { d ( x n k , g x n k , a ) , d ( x n k , f x n k , a ) } d ( x n k , z , a ) , for k N . (3.13)

The contraries of the above relation are as follows

1 s ( 1 + ρ ) d ( x n , f x n , a ) 1 s ( 1 + ρ ) min { d ( x n , g x n , a ) , d ( x n , f x n , a ) } > d ( x n , z , a ) ,

and

1 s ( 1 + ρ ) d ( x n , f x n , a ) 1 s ( 1 + ρ ) min { d ( x n , g x n , a ) , d ( x n , f x n , a ) } > d ( x n , z , a ) ,

for n N . If n is even we have

1 s ( 1 + ρ ) d ( x 2 n , g x 2 n , a ) 1 s ( 1 + ρ ) min { d ( x 2 n , g x 2 n , a ) , d ( x 2 n , f x 2 n , a ) } > d ( x 2 n , z , a ) ,

if n is odd then we get

1 s ( 1 + ρ ) d ( x 2 n + 1 , f x 2 n + 1 , a ) 1 s ( 1 + ρ ) min { d ( x 2 n + 1 , g x 2 n + 1 , a ) , d ( x 2 n + 1 , f x 2 n + 1 , a ) } > d ( x 2 n + 1 , z , a ) ,

for n N . By (3.8) we have

d ( x 2 n , x 2 n + 1 , a ) s ( d ( x 2 n , z , a ) + d ( x 2 n + 1 , z , a ) + d ( x 2 n , x 2 n + 1 , z ) ) < s s ( 1 + ρ ) d ( x 2 n , g x 2 n , a ) + s s ( 1 + ρ ) d ( x 2 n + 1 , f x 2 n + 1 , a ) + s s ( 1 + ρ ) d ( x 2 n , g x 2 n , x 2 n + 1 ) = 1 1 + ρ ( d ( x 2 n , x 2 n + 1 , a ) + d ( x 2 n + 1 , x 2 n + 2 , a ) + d ( x 2 n , x 2 n + 1 , x 2 n + 1 ) )

1 1 + ρ d ( x 2 n , x 2 n + 1 , a ) + ρ s 2 ( 1 + ρ ) d ( x 2 n + 1 , x 2 n , a ) 1 1 + ρ d ( x 2 n , x 2 n + 1 , a ) + ρ 1 + ρ d ( x 2 n + 1 , x 2 n , a ) = d ( x 2 n , x 2 n + 1 , a ) ,

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

1 s ϕ ( ρ ) min { d ( x 2 n , g x 2 n , a ) , d ( x 2 n , f x 2 n , a ) } d ( x 2 n , z , a ) ,

or

1 s ϕ ( ρ ) min { d ( x 2 n + 1 , g x 2 n + 1 , a ) , d ( x 2 n + 1 , f x 2 n + 1 , a ) } d ( x 2 n + 1 , z , a ) .

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

d ( g x 2 n , f z , a ) max { d ( f x 2 n , g z , a ) , d ( f z , g x 2 n , a ) , d ( f x 2 n , f z , a ) , d ( g z , f x 2 n , a ) } ρ s 2 d ( x 2 n , z , a ) .

or

d ( f x 2 n + 1 , f z , a ) max { d ( g x 2 n + 1 , g z , a ) , d ( f z , g x 2 n + 1 , a ) , d ( f x 2 n + 1 , f z , a ) , d ( g z , f x 2 n + 1 , a ) } ρ s 2 d ( x 2 n + ! , z , a ) .

From Lemma 2.5, we have

1 s d ( z , f z , a ) lim sup n d ( g x 2 n , f z , a ) ρ s 2 lim sup n d ( x 2 n , z , a ) ρ s d ( z , z , a ) = 0 ,

or

1 s d ( z , f z , a ) lim sup n d ( f x 2 n + 1 , f z , a ) ρ s 2 lim sup n d ( x 2 n + 1 , z , a ) ρ s d ( z , z , a ) = 0 ,

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

0 = 1 s ϕ ( ρ ) min { d ( z , f z , a ) , d ( z , g z , a ) } d ( z , f z , a ) ,

it implies

d ( g z , z , a ) max { d ( g z , g f z , a ) , d ( g z , f 2 z , a ) , d ( f z , f 2 z , a ) , d ( g f z , f z , a ) } ρ s 2 d ( f z , z , a ) = 0 ,

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

1 s ϕ ( ρ ) min { d ( z , f z , a ) , d ( z , g z , a ) } d ( z , z , a ) , so (3.2) shows that

d ( z , z , a ) = max { d ( g z , g z , a ) , d ( f z , f z , a ) , d ( g z , f z , a ) , d ( g z , f z , a ) } ρ s 2 d ( z , z , a ) ,

which is impossible unless z = z . □

NOTES

*Corresponding author.

Conflicts of Interest

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

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