1 Introduction and preliminaries Report issue for preceding element
In [10], Takahashi introduced a new concept of convexity in metric spaces and proved that all normed convex spaces and their convex subsets are convex metric spaces. Moreover, he gave some examples of convex metric spaces. We recall the basic definitions and properties of convex metric spaces. For details, we let the reader see [ 2 , 4 , 9 ] [2,4,9] and [10].
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Definition 1 Let ( X , d X,d ) be a metric space.
A function W : X × X × [ 0 , 1 ] → X W:X\times X\times[0,1]\rightarrow X is called a convexity structure, if
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d ​ ( u , W ​ ( x , y , λ ) ) ≤ λ ​ d ​ ( u , x ) + ( 1 − λ ) ​ d ​ ( u , y ) , d(u,W(x,y,\lambda))\leq\lambda d(u,x)+(1-\lambda)d(u,y),
for each x , y ∈ X x,y\in X and λ ∈ [ 0 , 1 ] \lambda\in[0,1] .
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A metric space ( X , d X,d ) endowed with a convexity structure W W is called a convex metric space. A nonempty subset K K of X X is convex, if W ​ ( x , y , λ ) ∈ K W(x,y,\lambda)\in K , for each x , y ∈ K x,y\in K and λ ∈ [ 0 , 1 ] \lambda\in[0,1] .
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Let X X be a convex metric space. Takahashi showed that the open balls and the closed balls are convex subsets of X X .
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If { K α } α ∈ J \left\{K_{\alpha}\right\}_{\alpha\in J} is a family of nonempty, convex subsets of X X , then the intersection ⋂ α ∈ J K α \bigcap_{\alpha\in J}K_{\alpha} is a convex subset of X X .
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Also, in [4], Berinde has mentioned an important property of convex metric spaces, i.e.:
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d ​ ( u , W ​ ( x , y , λ ) ) ≥ ( 1 − λ ) ​ d ​ ( u , y ) − λ ​ d ​ ( u , x ) , d(u,W(x,y,\lambda))\geq(1-\lambda)d(u,y)-\lambda d(u,x),
for each x , y ∈ X x,y\in X and λ ∈ [ 0 , 1 ] \lambda\in[0,1] .
Also, from [6-8] and [11], we recall some important lemmas regarding convex metric spaces:
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Lemma 1 Let ( X , d , W X,d,W ) be a convex metric space. Then the following statements hold:
(i) d ​ ( x , y ) = d ​ ( x , W ​ ( x , y , λ ) ) + d ​ ( y , W ​ ( x , y , λ ) ) d(x,y)=d(x,W(x,y,\lambda))+d(y,W(x,y,\lambda)) , for each ( x , y ) ∈ X × X (x,y)\in X\times X and λ ∈ [ 0 , 1 ] \lambda\in[0,1] .
(ii) d ​ ( x , W ​ ( x , y , λ ) ) = ( 1 − λ ) ​ d ​ ( x , y ) d(x,W(x,y,\lambda))=(1-\lambda)d(x,y) , for each x , y ∈ X x,y\in X .
(iii) d ​ ( y , W ​ ( x , y , λ ) ) = λ ​ d ​ ( x , y ) d(y,W(x,y,\lambda))=\lambda d(x,y) , for each x , y ∈ X x,y\in X .
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Lemma 2 Let ( X , d , W ) (X,d,W) be a convex metric space. Then:
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d ​ ( x , W ​ ( x , y , 1 2 ) ) = d ​ ( y , W ​ ( x , y , 1 2 ) ) = 1 2 ​ d ​ ( x , y ) , d\left(x,W\left(x,y,\frac{1}{2}\right)\right)=d\left(y,W\left(x,y,\frac{1}{2}\right)\right)=\frac{1}{2}d(x,y),
for each x , y ∈ X x,y\in X .
Also, for the sake of completeness we remind the classical iterative processes, such as Krasnoselskii, Mann and Ishikawa in convex metric spaces.
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In [3], Asadi used Krasnoselskij iteration for a contractive-type nonlinear mapping, i.e.:
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x n = W ​ ( x n − 1 ⋅ T ​ x n − 1 , λ ) , x_{n}=W\left(x_{n-1}\cdot Tx_{n-1},\lambda\right),
with λ ∈ [ 0 , 1 ] \lambda\in[0,1] . Also, the Mann iteration is defined as:
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x n = W ​ ( x n − 1 ⋅ T ​ x n − 1 , α n ) , x_{n}=W\left(x_{n-1}\cdot Tx_{n-1},\alpha_{n}\right),
with α n ∈ [ 0 , 1 ] \alpha_{n}\in[0,1] , for each n ∈ ℕ n\in\mathbb{N} .
Moreover, the classical Ishikawa iteration is as follows
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{ x n + 1 = W ​ ( x n , T ​ y n , α n ) y n = W ​ ( x n , T ​ x n , β n ) , \left\{\begin{array}[]{l}x_{n+1}=W\left(x_{n},Ty_{n},\alpha_{n}\right)\\
y_{n}=W\left(x_{n},Tx_{n},\beta_{n}\right)\end{array},\right.
with α n , β n ∈ [ 0 , 1 ] \alpha_{n},\beta_{n}\in[0,1] , for each n ∈ ℕ n\in\mathbb{N} .
We make the following remark: that although Berinde, Assadi and Moosaei used the same definition of convex metric spaces, in [4], Berinde defined Mann iteration as
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x n = W ​ ( T ​ x n − 1 , x n − 1 , α n ) , x_{n}=W\left(Tx_{n-1},x_{n-1},\alpha_{n}\right),
with α n ∈ [ 0 , 1 ] \alpha_{n}\in[0,1] , for each n ∈ ℕ n\in\mathbb{N} .
Since many authors use the property that W ​ ( x , y , λ ) = W ​ ( y , x , 1 − λ ) W(x,y,\lambda)=W(y,x,1-\lambda) , the Mann iterations can be transformed one to another, so it lies no confusion.
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Now, since we work with generalized type of contractype mappings, we recall some generalization of Banach contraction principle in convex metric spaces.
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In [5] Karapinar developed existence and uniqueness theorems for mappings satisfying certain contractive conditions in cone Banach spaces, such as
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d ​ ( T ​ x , T ​ y ) ≥ a ​ d ​ ( x , y ) \displaystyle d(Tx,Ty)\geq ad(x,y)
d ​ ( x , T ​ x ) + d ​ ( y , T ​ y ) ≤ p ​ d ​ ( x , y ) \displaystyle d(x,Tx)+d(y,Ty)\leq pd(x,y)
a ​ d ​ ( T ​ x , T ​ y ) + b ​ [ d ​ ( x , T ​ x ) + d ​ ( y , T ​ y ) ] ≤ s ​ d ​ ( x , y ) \displaystyle ad(Tx,Ty)+b[d(x,Tx)+d(y,Ty)]\leq sd(x,y)
Karapinar used the following conditions on this contractive type-operators, such as:
a > 1 , p ∈ [ 0 , 2 ) a>1,p\in[0,2) , respectively 0 ≤ s + | a | − 2 ​ b < 2 ​ ( a + b ) 0\leq s+|a|-2b<2(a+b) .
In [3], Asadi generalized the above conditions for mappings endowed with more coefficients, i.e.
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a ​ d ​ ( x , T ​ x ) + b ​ d ​ ( y , T ​ y ) + c ​ d ​ ( T ​ x , T ​ y ) + e ​ d ​ ( x , T ​ y ) + f ​ d ​ ( y , T ​ x ) ≤ k ​ d ​ ( x , y ) . ad(x,Tx)+bd(y,Ty)+cd(Tx,Ty)+ed(x,Ty)+fd(y,Tx)\leq kd(x,y).
Asadi, developed existence and uniqueness for above contractions, using the condition:
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b + e − | f | ​ ( 1 − λ ) − | c | ​ λ 1 − λ ≤ k < a + b + c + e + f − | c | ​ λ − | f | ​ ( 1 − λ ) 1 − λ \frac{b+e-|f|(1-\lambda)-|c|\lambda}{1-\lambda}\leq k<\frac{a+b+c+e+f-|c|\lambda-|f|(1-\lambda)}{1-\lambda}
Independently, in [11] Wang and Zhang developed a theorem for the existence of fixed points of a pair of contractive mappings, such as
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k ​ d ​ ( T ​ x , S ​ y ) ≤ a ​ d ​ ( x , y ) + b ​ [ d ​ ( x , T ​ x ) + d ​ ( y , S ​ y ) ] + c ​ [ d ​ ( x , S ​ y ) + d ​ ( y , T ​ x ) ] , kd(Tx,Sy)\leq ad(x,y)+b[d(x,Tx)+d(y,Sy)]+c[d(x,Sy)+d(y,Tx)],
using a Jungk-Krasnoselskii iteration, as follows
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{ x 2 ​ n + 1 = W ​ ( x 2 ​ n , T ​ x 2 ​ n , λ ) x 2 ​ n + 2 = W ​ ( x 2 ​ n + 1 , T ​ x 2 ​ n + 1 , λ ) , \left\{\begin{array}[]{l}x_{2n+1}=W\left(x_{2n},Tx_{2n},\lambda\right)\\
x_{2n+2}=W\left(x_{2n+1},Tx_{2n+1},\lambda\right)\end{array},\right.
and give an exemple of a pointwise mapping satysfing the above condition.
Also, Wang and Zhang imposed the conditions a + 2 ​ c < k a+2c<k and ( k , a , b , c ) ∈ Γ ρ (k,a,b,c)\in\Gamma_{\rho} , with ρ ∈ { 1 , 2 , 3 , 4 } \rho\in\{1,2,3,4\} , where Γ ρ \Gamma_{\rho} were defined in terms of the coefficients a , b , c a,b,c and k k .
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In [6] Moosaei developed a theorem for a Banach operator pair ( f , g f,g ) satysfing
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a ​ d ​ ( g ​ x , f ​ x ) + b ​ d ​ ( g ​ y , f ​ y ) + c ​ d ​ ( f ​ x , f ​ y ) ≤ k ​ d ​ ( g ​ x , g ​ y ) , ad(gx,fx)+bd(gy,fy)+cd(fx,fy)\leq kd(gx,gy),
with 2 ​ b − | c | ≤ k < 2 ​ ( a + b + c ) − | c | 2b-|c|\leq k<2(a+b+c)-|c| and in [8] the same author presented a theorem for existence and uniqueness of a common fixed point of a pair of mappings ( S , T S,T ), where ( S , T ) (S,T) weakly compatible pair, i.e.:
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a ​ d ​ ( S ​ x , T ​ x ) + b ​ d ​ ( S ​ y , T ​ y ) + c ​ d ​ ( T ​ x , T ​ y ) ≤ e ​ d ​ ( S ​ x , S ​ y ) . ad(Sx,Tx)+bd(Sy,Ty)+cd(Tx,Ty)\leq ed(Sx,Sy).
Likewise, in [7], Moosaei presented sufficient conditions for the existence of a coincidence point of a generalized contraction pair ( S , T S,T ), as follows
α ​ d ​ ( T ​ x , T ​ Y ) + β ​ [ d ​ ( S ​ x , T ​ y ) + d ​ ( S ​ y , T ​ x ) ] + γ ​ [ d ​ ( S ​ x , T ​ y ) + d ​ ( S ​ y , T ​ x ) ] ≤ η ​ d ​ ( S ​ x , S ​ y ) \alpha d(Tx,TY)+\beta[d(Sx,Ty)+d(Sy,Tx)]+\gamma[d(Sx,Ty)+d(Sy,Tx)]\leq\eta d(Sx,Sy) ,
using as Wang and Zhang a Jungk-Krasnoselskii iteration.
In [1], Abbas et al. discussed the simplicity of the modified Krasnoselskij iteration
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x n + 1 = ( 1 − λ ) ​ x n + λ ​ T n ​ x n , x_{n+1}=(1-\lambda)x_{n}+\lambda T^{n}x_{n},
where λ ∈ ( 0 , 1 ) \lambda\in(0,1) and the operator T T is defined on a nonempty, closed convex subset of a uniformly convex Banach space. Furthermore, one can easily extend this iteration to the case of convex metric spaces in the sense of Takahashi. Abbas, Khan and Rhoades suggested that a number of fixed point iteration problems can be solved using the above iteration, which is
much simpler than Ishikawa iteration. However, the simplification principle does not directly apply to the setting of our research since we are dealing with a generalized type of nonlinear contractions in contrast with the case when the operator T T is defined by it’s iterates, for example when T T is an asymptotically nonexpansive mapping. In this case it satisfies the following condition: there exists a sequence ( k n ) n ∈ ℕ ⊂ [ 1 , ∞ ) \left(k_{n}\right)_{n\in\mathbb{N}}\subset[1,\infty) , with ∑ n ∈ ℕ ( k n − 1 ) < ∞ \sum_{n\in\mathbb{N}}\left(k_{n}-1\right)<\infty , such that
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‖ T n ​ x − T n ​ y ‖ ≤ k n ​ ‖ x − y ‖ . \left\|T^{n}x-T^{n}y\right\|\leq k_{n}\|x-y\|.
So, in the above case, the condition would act on the iterates of the operator T T , contrary to the case when T T is a nonlinear generalized contraction-type mapping. In our case, as we do not assume an asymptotic condition for the nonlinear operator of interest, especially from a computational point of view, the modified Krasnoselskij iteration process would be less cost-efficient than the well-known Ishikawa iteration. The cause of this relies on the fact that the modified Krasnoselskij iteration requires the computation of T n T^{n} in each step.
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2 Main results Report issue for preceding element
In this section the existence and uniqueness of fixed points of a generalized contraction mapping are established.
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Because of the heavy computations, we work with contraction type mappings developed by Karapinar, i.e
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c ​ d ​ ( T ​ x , T ​ y ) + a ​ [ d ​ ( x , T ​ x ) + d ​ ( y , T ​ y ) ] ≤ k ​ d ​ ( x , y ) , cd(Tx,Ty)+a[d(x,Tx)+d(y,Ty)]\leq kd(x,y),
although the same method can be applied to all the above nonlinear mappings.
Since Karapinar, Moosaei and the already mentioned authors have used Krasnoselskii iteration (when λ ∈ [ 0 , 1 ] \lambda\in[0,1] and on the particular case when λ = 1 2 \lambda=\frac{1}{2} ), we give a new method of proof through Ishikawa iteration.
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We also give an example to illustrate the theorem we develop.
Theorem 1 Let ( X , d X,d ) be a complete convex metric space.
Let K K a nonempty, closed and convex subset of X X and T : K → K T:K\rightarrow K be a map satisfying the following contractive condition
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c ​ d ​ ( T ​ x , T ​ y ) + a ​ [ d ​ ( x , T ​ x ) + d ​ ( y , T ​ y ) ] ≤ k ​ d ​ ( x , y ) , cd(Tx,Ty)+a[d(x,Tx)+d(y,Ty)]\leq kd(x,y),
for each x , y ∈ K x,y\in K .
Let ( α n ) \left(\alpha_{n}\right) and ( β n ) \left(\beta_{n}\right) be two sequences in ( 0 , 1 ) (0,1) and let a 1 , a 2 , b 1 , b 2 ∈ [ 0 , 1 ] a_{1},a_{2},b_{1},b_{2}\in[0,1] , such that α n ∈ [ a 1 , a 2 ] ⊂ [ 0 , 1 ) \alpha_{n}\in\left[a_{1},a_{2}\right]\subset[0,1) and β n ∈ [ b 1 , b 2 ] ⊂ [ 0 , 1 ) \beta_{n}\in\left[b_{1},b_{2}\right]\subset[0,1) , for each n ∈ ℕ n\in\mathbb{N} .
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Let:
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ε ∈ ( 0 , 1 2 ) , w 1 := { a 1 − a 2 , a ≥ 0 2 ​ a 3 ​ ( 1 − a 1 ) , a < 0 , w 2 := { a ​ ( 1 + b 2 ) 1 − b 2 , a ≥ 0 a ​ ( 1 + b 1 ) 1 − b 1 , a < 0 \displaystyle\varepsilon\in\left(0,\frac{1}{2}\right),w_{1}=\left\{\begin{array}[]{ll}\frac{a}{1-a_{2}},&a\geq 0\\
\frac{2a}{3\left(1-a_{1}\right)},&a<0\end{array},w_{2}:=\begin{cases}\frac{a\left(1+b_{2}\right)}{1-b_{2}},&a\geq 0\\
\frac{a\left(1+b_{1}\right)}{1-b_{1}},&a<0\end{cases}\right.
w 3 := { 1 2 ​ ( a + c ) − | a | ​ ( 1 + b 2 ) 1 − b 1 , k > 0 1 2 ​ ( a + c ) − | a | ​ ( 1 + b 2 ) 1 − b 2 , k < 0 , τ := { 1 3 , a 2 < 1 2 1 3 ⋅ a 2 1 − a 2 , a 1 > 1 2 , \displaystyle w_{3}=\left\{\begin{array}[]{ll}\frac{\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right)}{1-b_{1}},&k>0\\
\frac{\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right)}{1-b_{2}},&k<0\end{array},\tau:=\left\{\begin{array}[]{ll}\frac{1}{3},&a_{2}<\frac{1}{2}\\
\frac{1}{3}\cdot\frac{a_{2}}{1-a_{2}},&a_{1}>\frac{1}{2}\end{array},\right.\right.
s := { 3 ​ ( 1 − a 1 ) 2 ​ ( a + c ) , a ≥ 0 ​ and ​ c ≥ 0 , a 2 ​ ( 1 − a 1 ) + c ε ​ ( 1 + b 2 ) ( 1 − a 2 ) ​ ( 1 − b 2 ) , a ≥ 0 ​ and ​ c < 0 , a ε ​ ( 1 + b 2 ) ( 1 − a 2 ) ​ ( 1 − b 2 ) + c 2 ​ ( 1 − a 1 ) , a < 0 ​ and ​ c ≥ 0 s:=\begin{cases}\frac{3\left(1-a_{1}\right)}{2(a+c)},&a\geq 0\text{ and }c\geq 0,\\
\frac{a}{2\left(1-a_{1}\right)}+\frac{c}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)},&a\geq 0\text{ and }c<0,\\
\frac{a}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)},&a<0\text{ and }c\geq 0\end{cases}
Suppose the following conditions hold:
(1) c ≠ 0 c\neq 0 ,
(2) a + c > 0 a+c>0 ,
(3) k ≥ w 1 k\geq w_{1} ,
(4) k ≥ w 2 k\geq w_{2} ,
(5) k < w 3 k<w_{3} ,
(6) s > 0 s>0 ,
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| c | ​ τ < s − k − | a | ε ​ ( 1 + b 2 ) ( 1 − a 2 ) ​ ( 1 − b 2 ) . |c|\tau<s-k-\frac{|a|}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}.
(7)
Then, the operator T T admits a fixed point.
Moreover, if k < c k<c , then the fixed point is unique.
Proof For each n ∈ ℕ n\in\mathbb{N} , define the Ishikawa iteration
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{ x n + 1 = W ​ ( x n , T ​ y n , α n ) y n = W ​ ( x n , T ​ x n , β n ) \left\{\begin{array}[]{l}x_{n+1}=W\left(x_{n},Ty_{n},\alpha_{n}\right)\\
y_{n}=W\left(x_{n},Tx_{n},\beta_{n}\right)\end{array}\right.
for α n \alpha_{n} and β n ∈ [ 0 , 1 ) \beta_{n}\in[0,1) , for each n ∈ ℕ n\in\mathbb{N} .
From the contractive condition for the operator T T applied on the pair ( x n , x n − 1 x_{n},x_{n-1} ), we obtain that
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c ​ d ​ ( T ​ x n − 1 , T ​ x n ) + a ​ d ​ ( x n , T ​ x n ) + a ​ d ​ ( x n − 1 , T ​ x n − 1 ) ≤ k ​ d ​ ( x n − 1 , x n ) cd\left(Tx_{n-1},Tx_{n}\right)+ad\left(x_{n},Tx_{n}\right)+ad\left(x_{n-1},Tx_{n-1}\right)\leq kd\left(x_{n-1},x_{n}\right)
Moreover, from now on, let’s denote by D D the distance d ​ ( x n , T ​ x n ) d\left(x_{n},Tx_{n}\right) , for each n ∈ ℕ n\in\mathbb{N} fixed, unless we specify it explicitly.
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We divide our proof in the following steps, as follows:
(I) We make the following estimations:
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d ​ ( x n , x n + 1 ) = d ​ ( x n , W ​ ( x n , T ​ y n , α n ) ) = ( 1 − α n ) ​ d ​ ( x n , T ​ y n ) \displaystyle d\left(x_{n},x_{n+1}\right)=d\left(x_{n},W\left(x_{n},Ty_{n},\alpha_{n}\right)\right)=\left(1-\alpha_{n}\right)d\left(x_{n},Ty_{n}\right)
d ​ ( x n , y n ) = d ​ ( x n , W ​ ( x n , T ​ x n , β n ) ) = ( 1 − β n ) ​ d ​ ( x n , T ​ x n ) = ( 1 − β n ) ​ D \displaystyle d\left(x_{n},y_{n}\right)=d\left(x_{n},W\left(x_{n},Tx_{n},\beta_{n}\right)\right)=\left(1-\beta_{n}\right)d\left(x_{n},Tx_{n}\right)=\left(1-\beta_{n}\right)D
d ​ ( T ​ x n , y n ) = d ​ ( T ​ x n , W ​ ( x n , T ​ x n , β n ) ) = β n ​ d ​ ( x n , T ​ x n ) = β n ​ D \displaystyle d\left(Tx_{n},y_{n}\right)=d\left(Tx_{n},W\left(x_{n},Tx_{n},\beta_{n}\right)\right)=\beta_{n}d\left(x_{n},Tx_{n}\right)=\beta_{n}D
Also, we have that
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d ​ ( T ​ y n , x n + 1 ) = d ​ ( T ​ y n , W ​ ( x n , T ​ y n , α n ) ) = α n ​ d ​ ( x n , T ​ y n ) . d\left(Ty_{n},x_{n+1}\right)=d\left(Ty_{n},W\left(x_{n},Ty_{n},\alpha_{n}\right)\right)=\alpha_{n}d\left(x_{n},Ty_{n}\right).
So
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d ​ ( x n , x n + 1 ) = 1 − α n α n ​ d ​ ( T ​ y n , x n + 1 ) d\left(x_{n},x_{n+1}\right)=\frac{1-\alpha_{n}}{\alpha_{n}}d\left(Ty_{n},x_{n+1}\right)
(II) Now, we estimate d ​ ( T ​ x n , T ​ y n ) d\left(Tx_{n},Ty_{n}\right) :
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Case (A): a ≥ 0 , c ≥ 0 a\geq 0,c\geq 0 :
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d ​ ( x n , x n + 1 ) \displaystyle d\left(x_{n},x_{n+1}\right)
= ( 1 − α n ) ​ d ​ ( x n , T ​ y n ) ≤ ( 1 − α n ) ​ [ d ​ ( x n , y n ) + d ​ ( y n , T ​ y n ) ] \displaystyle=\left(1-\alpha_{n}\right)d\left(x_{n},Ty_{n}\right)\leq\left(1-\alpha_{n}\right)\left[d\left(x_{n},y_{n}\right)+d\left(y_{n},Ty_{n}\right)\right]
= ( 1 − α n ) ​ d ​ ( x n , y n ) + ( 1 − α n ) ​ d ​ ( y n , T ​ y n ) \displaystyle=\left(1-\alpha_{n}\right)d\left(x_{n},y_{n}\right)+\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right)
= ( 1 − α n ) ​ ( 1 − β n ) ​ D + ( 1 − α n ) ​ d ​ ( y n , T ​ y n ) \displaystyle=\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)D+\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right)
By the definition of a convex metric space, we have that
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d ​ ( y n , T ​ y n ) \displaystyle d\left(y_{n},Ty_{n}\right)
= d ​ ( W ​ ( x n , T ​ x n , β n ) , T ​ y n ) ≤ β n ​ d ​ ( x n , T ​ y n ) + ( 1 − β n ) ​ d ​ ( T ​ x n , T ​ y n ) \displaystyle=d\left(W\left(x_{n},Tx_{n},\beta_{n}\right),Ty_{n}\right)\leq\beta_{n}d\left(x_{n},Ty_{n}\right)+\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
= β n 1 − α n ​ d ​ ( x n , x n + 1 ) + ( 1 − β n ) ​ d ​ ( T ​ x n , T ​ y n ) \displaystyle=\frac{\beta_{n}}{1-\alpha_{n}}d\left(x_{n},x_{n+1}\right)+\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
So
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d ​ ( x n , x n + 1 ) ≤ ( 1 − α n ) ​ ( 1 − β n ) ⋅ D + β n ​ d ​ ( x n , x n + 1 ) + ( 1 − α n ) ​ ( 1 − β n ) ​ d ​ ( T ​ x n , T ​ y n ) \displaystyle d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)\cdot D+\beta_{n}d\left(x_{n},x_{n+1}\right)+\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
⟹ d ​ ( x n , x n + 1 ) ≤ ( 1 − α n ) ​ D + ( 1 − α n ) ​ d ​ ( T ​ x n , T ​ y n ) . \displaystyle\Longrightarrow d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right).
From the contraction-type condition for the pair ( x n , y n ) \left(x_{n},y_{n}\right) , we obtain that
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c ​ d ​ ( T ​ x n , T ​ y n ) + a ​ d ​ ( x n , T ​ x n ) + a ​ d ​ ( y n , T ​ y n ) ≤ k ​ d ​ ( x n , y n ) \displaystyle cd\left(Tx_{n},Ty_{n}\right)+ad\left(x_{n},Tx_{n}\right)+ad\left(y_{n},Ty_{n}\right)\leq kd\left(x_{n},y_{n}\right)
c ​ d ​ ( T ​ x n , T ​ y n ) ≤ k ​ d ​ ( x n , y n ) − a ​ d ​ ( x n , T ​ x n ) − a ​ d ​ ( y n , T ​ y n ) \displaystyle cd\left(Tx_{n},Ty_{n}\right)\leq kd\left(x_{n},y_{n}\right)-ad\left(x_{n},Tx_{n}\right)-ad\left(y_{n},Ty_{n}\right)
c ​ d ​ ( T ​ x n , T ​ y n ) ≤ k ​ ( 1 − β n ) ⋅ D − a ⋅ D − a ​ d ​ ( y n , T ​ y n ) \displaystyle cd\left(Tx_{n},Ty_{n}\right)\leq k\left(1-\beta_{n}\right)\cdot D-a\cdot D-ad\left(y_{n},Ty_{n}\right)
c ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ] ⋅ D − a ​ d ​ ( y n , T ​ y n ) . \displaystyle cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]\cdot D-ad\left(y_{n},Ty_{n}\right).
Since c > 0 c>0 , we obtain that
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d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a c ] ⋅ D + ( − a c ) ​ d ​ ( y n , T ​ y n ) d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]\cdot D+\left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)
Since a ≥ 0 a\geq 0 and c > 0 c>0 , we have that
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( − a c ) ≤ 0 ​ and ​ k ​ ( 1 − β n ) − a c ≥ 0 \left(-\frac{a}{c}\right)\leq 0\text{ and }\frac{k\left(1-\beta_{n}\right)-a}{c}\geq 0
because k ≥ a 1 − β n k\geq\frac{a}{1-\beta_{n}} from the hypothesis.
We have that ( − a c ) ​ d ​ ( y n , T ​ y n ) ≤ 0 \left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)\leq 0 , so, by triangle inequality, it follows that
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d ​ ( y n , T ​ y n ) ≥ d ​ ( T ​ x n , T ​ y n ) − d ​ ( T ​ x n , y n ) ≥ d ​ ( T ​ x n , T ​ y n ) − β n ⋅ D . d\left(y_{n},Ty_{n}\right)\geq d\left(Tx_{n},Ty_{n}\right)-d\left(Tx_{n},y_{n}\right)\geq d\left(Tx_{n},Ty_{n}\right)-\beta_{n}\cdot D.
So
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( − a c ) ​ d ​ ( y n , T ​ y n ) ≤ ( − a c ) ​ d ​ ( T ​ x n , T ​ y n ) + a c ​ β n ⋅ D . \left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)\leq\left(-\frac{a}{c}\right)d\left(Tx_{n},Ty_{n}\right)+\frac{a}{c}\beta_{n}\cdot D.
From the above relations, we get
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d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a c ] ⋅ D − a c ​ d ​ ( T ​ x n , T ​ y n ) + a c ​ β n ⋅ D . \displaystyle d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]\cdot D-\frac{a}{c}d\left(Tx_{n},Ty_{n}\right)+\frac{a}{c}\beta_{n}\cdot D.
( 1 + a c ) ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a + a ​ β n c ] ​ D \displaystyle\left(1+\frac{a}{c}\right)d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a+a\beta_{n}}{c}\right]D
⟹ a + c c ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ​ ( 1 − β n ) c ] ​ D . \displaystyle\Longrightarrow\frac{a+c}{c}d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a\left(1-\beta_{n}\right)}{c}\right]D.
Since c > 0 c>0 , we have ( a + c ) ​ d ​ ( T ​ x n , T ​ y n ) ≤ ( 1 − β n ) ​ ( k − a ) ⋅ D (a+c)d\left(Tx_{n},Ty_{n}\right)\leq\left(1-\beta_{n}\right)(k-a)\cdot D . Moreover, since a + c > 0 a+c>0 from the hypothesis, we obtain:
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d ​ ( T ​ x n , T ​ y n ) ≤ ( 1 − β n ) ​ ( k − a ) a + c ⋅ D . d\left(Tx_{n},Ty_{n}\right)\leq\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\cdot D.
We observe that we have used the following conditions:
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c > 0 , a + c > 0 , k > a c>0,a+c>0,k>a
(3)
Now
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d ​ ( x n , x n + 1 ) ≤ ( 1 − α n ) ⋅ D + ( 1 − α n ) ​ d ​ ( T ​ x n , T ​ y n ) ⟹ \displaystyle d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\cdot D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right)\Longrightarrow
d ​ ( x n , x n + 1 ) ≤ ( 1 − α n ) ​ [ 1 + ( 1 − β n ) ​ ( k − a ) a + c ] ​ D . \displaystyle d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\left[1+\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.
Here, we observe, that we need the same condition, i.e. k > a k>a . In a similar way, we estimate a lower bound for d ​ ( x n , x n + 1 ) d\left(x_{n},x_{n+1}\right) :
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We know that
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d ​ ( x n , x n + 1 ) \displaystyle d\left(x_{n},x_{n+1}\right)
= ( 1 − α n ) ​ d ​ ( x n , T ​ y n ) ≥ ( 1 − α n ) ​ [ d ​ ( x n , y n ) − d ​ ( y n , T ​ y n ) ] \displaystyle=\left(1-\alpha_{n}\right)d\left(x_{n},Ty_{n}\right)\geq\left(1-\alpha_{n}\right)\left[d\left(x_{n},y_{n}\right)-d\left(y_{n},Ty_{n}\right)\right]
= ( 1 − α n ) ​ ( 1 − β n ) ⋅ D − ( 1 − α n ) ​ d ​ ( y n , T ​ y n ) . \displaystyle=\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)\cdot D-\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right).
But we have shown that
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d ​ ( y n , T ​ y n ) ≤ β n 1 − α n ​ d ​ ( x n , x n + 1 ) + ( 1 − β n ) ​ d ​ ( T ​ x n , T ​ y n ) ⟹ \displaystyle d\left(y_{n},Ty_{n}\right)\leq\frac{\beta_{n}}{1-\alpha_{n}}d\left(x_{n},x_{n+1}\right)+\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)\Longrightarrow
− ( 1 − α n ) ​ d ​ ( y n , T ​ y n ) ≥ − β n ​ d ​ ( x n , x n + 1 ) − ( 1 − α n ) ​ ( 1 − β n ) ​ d ​ ( T ​ x n , T ​ y n ) \displaystyle-\left(1-\alpha_{n}\right)d\left(y_{n},Ty_{n}\right)\geq-\beta_{n}d\left(x_{n},x_{n+1}\right)-\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
So
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d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) ​ D − β n ​ d ​ ( x n , x n + 1 ) − ( 1 − α n ) ​ ( 1 − β n ) ​ d ​ ( T ​ x n , T ​ y n ) d\left(x_{n},x_{n+1}\right)\geq\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)D-\beta_{n}d\left(x_{n},x_{n+1}\right)-\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)d\left(Tx_{n},Ty_{n}\right)
Then
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d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ⋅ D − ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ⋅ d ​ ( T ​ x n , T ​ y n ) . d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\cdot D-\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\cdot d\left(Tx_{n},Ty_{n}\right).
Since
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d ​ ( T ​ x n , T ​ y n ) ≤ ( 1 − β n ) ​ ( k − a ) a + c ⋅ D , we get \displaystyle d\left(Tx_{n},Ty_{n}\right)\leq\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\cdot D,\text{ we get }
d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ⋅ [ 1 − ( 1 − β n ) ​ ( k − a ) a + c ] ⋅ D . \displaystyle d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\cdot\left[1-\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]\cdot D.
We observe that we have used the following condition, such that the right hand side is positive:
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( 1 − β n ) ​ ( k − a ) a + c < 1 \frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1
(4)
Case (D): a < 0 , c < 0 a<0,c<0 :
This case is not possible, since from the hypothesis we have that a + c > 0 a+c>0 and this leads to a contradiction.
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Case (B): a < 0 , c > 0 a<0,c>0 :
We have that
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d ​ ( y n , T ​ y n ) ≤ d ​ ( T ​ x n , T ​ y n ) + d ​ ( y n , T ​ x n ) = d ​ ( T ​ x n , T ​ y n ) + β n ⋅ D . d\left(y_{n},Ty_{n}\right)\leq d\left(Tx_{n},Ty_{n}\right)+d\left(y_{n},Tx_{n}\right)=d\left(Tx_{n},Ty_{n}\right)+\beta_{n}\cdot D.
We have shown that
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c ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ] ⋅ D − a ​ d ​ ( y n , T ​ y n ) . cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]\cdot D-ad\left(y_{n},Ty_{n}\right).
Since a < 0 a<0 , this means that − a > 0 -a>0 , we have
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− a ​ d ​ ( y n , T ​ y n ) ≤ − a ​ d ​ ( T ​ x n , T ​ y n ) − a ​ β n ​ D . -ad\left(y_{n},Ty_{n}\right)\leq-ad\left(Tx_{n},Ty_{n}\right)-a\beta_{n}D.
So
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c ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ] ​ D − a ​ d ​ ( T ​ x n , T ​ y n ) − a ​ β n ​ D ⟹ cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]D-ad\left(Tx_{n},Ty_{n}\right)-a\beta_{n}D\Longrightarrow
( a + c ) ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ​ ( 1 + β n ) ] ​ D (a+c)d\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)\right]D
Since c > 0 c>0 and a + c > 0 a+c>0 , it follows that
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d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c ] ​ D d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D
Now, we have used the following condition:
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k > a ​ ( 1 + β n 1 − β n ) k>a\left(\frac{1+\beta_{n}}{1-\beta_{n}}\right)
(5)
Also, we know that
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d ​ ( x n , x n + 1 ) ≤ ( 1 − α n ) ​ D + ( 1 − α n ) ​ d ​ ( T ​ x n , T ​ y n ) d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right)
So
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d ​ ( x n , x n + 1 ) ≤ ( 1 − α n ) ⋅ [ 1 + k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c ] ​ D . d\left(x_{n},x_{n+1}\right)\leq\left(1-\alpha_{n}\right)\cdot\left[1+\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D.
We recall that this estimation does not depend on the sign of the coefficients a a and c c . Also, we notice that we have k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c > 0 \frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}>0 , which is the condition imposed above. Now, for a lower bound of d ​ ( x n , x n + 1 ) d\left(x_{n},x_{n+1}\right) , we proceed in a similar way:
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We have shown that
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d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c ] ​ D . d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D.
Moreover, we know that
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d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ D − ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ d ​ ( T ​ x n , T ​ y n ) d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}D-\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}d\left(Tx_{n},Ty_{n}\right)
We mention that this estimation does not depend on the coefficients from the contractivecondition. From these two relations, we obtain that
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d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ 1 − k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c ] ​ D . d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]D.
We observe that we have used the condition:
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k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c < 1 \frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}<1
(6)
Case (C): a ≥ 0 , c < 0 a\geq 0,c<0 :
From the contraction condition, we have shown that
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c ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a ] ​ D − a ​ d ​ ( y n , T ​ y n ) cd\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a\right]D-ad\left(y_{n},Ty_{n}\right)
Since c < 0 c<0 , we get
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d ​ ( T ​ x n , T ​ y n ) ≥ [ k ​ ( 1 − β n ) − a c ] ​ D + ( − a c ) ​ d ​ ( y n , T ​ y n ) d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]D+\left(-\frac{a}{c}\right)d\left(y_{n},Ty_{n}\right)
Now, we estimate the following:
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d ​ ( y n , T ​ y n ) ≥ d ​ ( T ​ x n , T ​ y n ) − d ​ ( T ​ x n , y n ) = d ​ ( T ​ x n , T ​ y n ) − β n ​ D ⟹ d\left(y_{n},Ty_{n}\right)\geq d\left(Tx_{n},Ty_{n}\right)-d\left(Tx_{n},y_{n}\right)=d\left(Tx_{n},Ty_{n}\right)-\beta_{n}D\Longrightarrow
Because − a c ≥ 0 -\frac{a}{c}\geq 0 , we have that
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d ​ ( T ​ x n , T ​ y n ) ≥ [ k ​ ( 1 − β n ) − a c ] ​ D − a c ​ d ​ ( T ​ x n , T ​ y n ) + a c ​ β n ​ D ⟹ \displaystyle d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}\right]D-\frac{a}{c}d\left(Tx_{n},Ty_{n}\right)+\frac{a}{c}\beta_{n}D\Longrightarrow
( 1 + a c ) ​ d ​ ( T ​ x n , T ​ y n ) ≥ [ k ​ ( 1 − β n ) − a c + a c ​ β n ] ​ D . \displaystyle\left(1+\frac{a}{c}\right)d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}+\frac{a}{c}\beta_{n}\right]D.
This means that
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( a + c c ) ​ d ​ ( T ​ x n , T ​ y n ) ≥ [ k ​ ( 1 − β n ) − a c + a c ​ β n ] ​ D . \left(\frac{a+c}{c}\right)d\left(Tx_{n},Ty_{n}\right)\geq\left[\frac{k\left(1-\beta_{n}\right)-a}{c}+\frac{a}{c}\beta_{n}\right]D.
Whereas c < 0 c<0 the right hand side is negative and the left hand side is also negative, because c < 0 c<0 and a + c > 0 a+c>0 . Moreover, we notice that k ​ ( 1 − β n ) − a ≥ 0 k\left(1-\beta_{n}\right)-a\geq 0 , which means that
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k ​ ( 1 − β n ) ≥ a . k\left(1-\beta_{n}\right)\geq a.
What with c c cannot be 0 , we divide by c c and obtain:
( a + c ) ​ d ​ ( T ​ x n , T ​ y n ) ≤ [ k ​ ( 1 − β n ) − a + a ​ β n ] ​ D ⟹ d ​ ( T ​ x n , T ​ y n ) ≤ [ ( 1 − β n ) ​ ( k − a ) a + c ] ​ D (a+c)d\left(Tx_{n},Ty_{n}\right)\leq\left[k\left(1-\beta_{n}\right)-a+a\beta_{n}\right]D\Longrightarrow d\left(Tx_{n},Ty_{n}\right)\leq\left[\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D .
Furthermore, we notice that we used the following condition:
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such that the right hand side is strict positive.
The above inequalities imply that
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d ​ ( x n , x n + 1 ) \displaystyle d\left(x_{n},x_{n+1}\right)
≤ ( 1 − α n ) ​ D + ( 1 − α n ) ​ d ​ ( T ​ x n , T ​ y n ) ⟹ \displaystyle\leq\left(1-\alpha_{n}\right)D+\left(1-\alpha_{n}\right)d\left(Tx_{n},Ty_{n}\right)\Longrightarrow
d ​ ( x n , x n + 1 ) \displaystyle d\left(x_{n},x_{n+1}\right)
≤ ( 1 − α n ) ​ [ 1 + ( 1 − β n ) ​ ( k − a ) a + c ] ​ D . \displaystyle\leq\left(1-\alpha_{n}\right)\left[1+\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.
Also, we observe that the condition k > a k>a is satisfied by hypothesis assumptions. In a similar way, for a lower bound of d ​ ( x n , x n + 1 ) d\left(x_{n},x_{n+1}\right) , we have shown that
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d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ D − d ​ ( T ​ x n , T ​ y n ) ] d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[D-d\left(Tx_{n},Ty_{n}\right)\right]
This means that
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d ​ ( x n , x n + 1 ) ≥ ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ 1 − ( 1 − β n ) ​ ( k − a ) a + c ] ​ D . d\left(x_{n},x_{n+1}\right)\geq\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]D.
Given that the right hand side must pe positive, we use the following condition
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( 1 − β n ) ​ ( k − a ) a + c < 1 \frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1
(8)
(III) Now we find θ n \theta_{n} , such that d ​ ( x n , x n + 1 ) ⋅ θ n ≤ d ​ ( x n , T ​ x n ) d\left(x_{n},x_{n+1}\right)\cdot\theta_{n}\leq d\left(x_{n},Tx_{n}\right) :
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We have a major bound for d ​ ( x n , x n + 1 ) : d ​ ( x n , x n + 1 ) ≤ η ′ ​ d ​ ( x n , T ​ x n ) d\left(x_{n},x_{n+1}\right):d\left(x_{n},x_{n+1}\right)\leq\eta^{\prime}d\left(x_{n},Tx_{n}\right) . This means that
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d ​ ( x n , T ​ x n ) ≥ 1 η ′ ​ d ​ ( x n , x n + 1 ) ≥ θ n ​ d ​ ( x n , x n + 1 ) d\left(x_{n},Tx_{n}\right)\geq\frac{1}{\eta^{\prime}}d\left(x_{n},x_{n+1}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)
Let us define
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η ′ = { η 1 , in the case ​ ( A ) η 2 , in the case ​ ( B ) η 3 , in the case ​ ( C ) \eta^{\prime}=\left\{\begin{array}[]{l}\eta_{1},\text{ in the case }(A)\\
\eta_{2},\text{ in the case }(B)\\
\eta_{3},\text{ in the case }(C)\end{array}\right.
where
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η 1 = ( 1 − α n ) ​ [ 1 + ( 1 − β n ) ​ ( k − a ) a + c ] = ( 1 − α n ) ​ [ 1 + k ​ ( 1 − β n ) a + c + − a ​ ( 1 − β n ) a + c ] , \displaystyle\eta_{1}=\left(1-\alpha_{n}\right)\left[1+\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right]=\left(1-\alpha_{n}\right)\left[1+\frac{k\left(1-\beta_{n}\right)}{a+c}+\frac{-a\left(1-\beta_{n}\right)}{a+c}\right],
η 2 = ( 1 − α n ) ​ [ 1 + k ​ ( 1 − β n ) a + c + − a ​ ( 1 + β n ) a + c ] ​ and ​ η 3 ​ is equal to ​ η 1 . \displaystyle\eta_{2}=\left(1-\alpha_{n}\right)\left[1+\frac{k\left(1-\beta_{n}\right)}{a+c}+\frac{-a\left(1+\beta_{n}\right)}{a+c}\right]\text{ and }\eta_{3}\text{ is equal to }\eta_{1}.
We observe that η 1 , η 2 , η 3 \eta_{1},\eta_{2},\eta_{3} and η ′ \eta^{\prime} depend on the index n ∈ ℕ n\in\mathbb{N} . Moreover, we observe that :
For the cases ( A ) (A) and ( C ) (C) , since a + c > 0 a+c>0 and a ≥ 0 a\geq 0 , we have that
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− a ​ ( 1 − β n ) a + c ≤ 0 ≤ a ​ ( 1 + β n ) a + c = | a | ​ 1 + β n a + c . \frac{-a\left(1-\beta_{n}\right)}{a+c}\leq 0\leq\frac{a\left(1+\beta_{n}\right)}{a+c}=|a|\frac{1+\beta_{n}}{a+c}.
Also, for the case ( B B ), since a + c > 0 a+c>0 and a < 0 a<0 , we get
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− a ​ ( 1 + β n ) a + c = | a | ​ 1 + β n a + c . \frac{-a\left(1+\beta_{n}\right)}{a+c}=|a|\frac{1+\beta_{n}}{a+c}.
So 0 < η ′ ≤ η 0<\eta^{\prime}\leq\eta , where
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η := ( 1 − α n ) ​ [ 1 + k ​ ( 1 − β n ) a + c + | a | ​ 1 + β n a + c ] . \eta:=\left(1-\alpha_{n}\right)\left[1+\frac{k\left(1-\beta_{n}\right)}{a+c}+|a|\frac{1+\beta_{n}}{a+c}\right].
Simplifying, it follows that
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η = 1 − α n a + c ​ [ k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) + ( a + c ) ] . \eta=\frac{1-\alpha_{n}}{a+c}\left[k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)\right].
Moreover
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1 η ≤ 1 η ′ \frac{1}{\eta}\leq\frac{1}{\eta^{\prime}}
So
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d ​ ( x n , x n + 1 ) ≤ η ′ ​ d ​ ( x n , T ​ x n ) ≤ η ​ d ​ ( x n , T ​ x n ) ⟹ \displaystyle d\left(x_{n},x_{n+1}\right)\leq\eta^{\prime}d\left(x_{n},Tx_{n}\right)\leq\eta d\left(x_{n},Tx_{n}\right)\Longrightarrow
d ​ ( x n , T ​ x n ) ≥ θ n ​ d ​ ( x n , x n + 1 ) \displaystyle d\left(x_{n},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)
where ​ θ n = a + c 1 − α n ⋅ 1 k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) + ( a + c ) = 1 η \displaystyle\text{ where }\theta_{n}=\frac{a+c}{1-\alpha_{n}}\cdot\frac{1}{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)}=\frac{1}{\eta}
and ​ η ​ depends on the index ​ n ∈ ℕ ​ . \displaystyle\text{ and }\eta\text{ depends on the index }n\in\mathbb{N}\text{. }
(IV) Now we find θ n ′ \theta_{n}^{\prime} , such that d ​ ( x n , T ​ x n ) ≤ θ n ′ ​ d ​ ( x n , x n + 1 ) d\left(x_{n},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right) :
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We have that
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d ​ ( x n , x n + 1 ) ≥ μ ′ ​ d ​ ( x n , T ​ x n ) ⟹ d ​ ( x n , T ​ x n ) ≤ 1 μ ′ ​ d ​ ( x n , x n + 1 ) ≤ θ n ′ ​ d ​ ( x n , x n + 1 ) d\left(x_{n},x_{n+1}\right)\geq\mu^{\prime}d\left(x_{n},Tx_{n}\right)\Longrightarrow d\left(x_{n},Tx_{n}\right)\leq\frac{1}{\mu^{\prime}}d\left(x_{n},x_{n+1}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)
We denote by:
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μ ′ = { μ 1 , in the case ​ ( A ) , μ 2 , in the case ​ ( B ) , μ 3 , in the case ​ ( C ) \mu^{\prime}=\left\{\begin{array}[]{l}\mu_{1},\text{ in the case }(A),\\
\mu_{2},\text{ in the case }(B),\\
\mu_{3},\text{ in the case }(C)\end{array}\right.
where
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μ 1 = ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ 1 − ( 1 − β n ) ​ ( k − a ) a + c ] , \mu_{1}=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}\right],
μ 2 = ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ 1 − k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c ] ​ and \displaystyle\mu_{2}=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}\right]\text{ and }
μ 3 ​ is equal to ​ μ 1 \displaystyle\mu_{3}\text{ is equal to }\mu_{1}
Moreover, we observe that
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μ 3 = μ 1 = ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ 1 + − k ​ ( 1 − β n ) a + c + a ​ ( 1 − β n ) a + c ] . \mu_{3}=\mu_{1}=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1+\frac{-k\left(1-\beta_{n}\right)}{a+c}+\frac{a\left(1-\beta_{n}\right)}{a+c}\right].
For the cases ( A ) (A) and ( C ) (C) , we have that
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a ​ ( 1 − β n ) = | a | ​ ( 1 − β n ) ≥ − | a ​ | ( 1 − β n ) > − | ​ a | ( 1 + β n ) a\left(1-\beta_{n}\right)=|a|\left(1-\beta_{n}\right)\geq-|a|\left(1-\beta_{n}\right)>-|a|\left(1+\beta_{n}\right)
, since 1 − β n < 1 + β n 1-\beta_{n}<1+\beta_{n} .
Also, for the case ( B B ), we have
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a ​ ( 1 + β n ) = − | a | ​ ( 1 + β n ) a\left(1+\beta_{n}\right)=-|a|\left(1+\beta_{n}\right)
From all the inequalities presented above, we derive
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where
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μ := ( 1 − α n ) ​ ( 1 − β n ) 1 + β n ​ [ 1 − k ​ ( 1 − β n ) a + c − | a | ​ ( 1 + β n ) a + c ] . \mu:=\frac{\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}{1+\beta_{n}}\left[1-\frac{k\left(1-\beta_{n}\right)}{a+c}-\frac{|a|\left(1+\beta_{n}\right)}{a+c}\right].
So
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d ​ ( x n , x n + 1 ) \displaystyle d\left(x_{n},x_{n+1}\right)
≥ μ ′ ​ d ​ ( x n , T ​ x n ) ≥ μ ​ d ​ ( x n , T ​ x n ) ⟹ \displaystyle\geq\mu^{\prime}d\left(x_{n},Tx_{n}\right)\geq\mu d\left(x_{n},Tx_{n}\right)\Longrightarrow
d ​ ( x n , T ​ x n ) \displaystyle d\left(x_{n},Tx_{n}\right)
≤ 1 μ ​ d ​ ( x n , x n + 1 ) = θ n ′ ​ d ​ ( x n , x n + 1 ) \displaystyle\leq\frac{1}{\mu}d\left(x_{n},x_{n+1}\right)=\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)
where θ n ′ = 1 μ \theta_{n}^{\prime}=\frac{1}{\mu} , where μ \mu depends on the index n ∈ ℕ n\in\mathbb{N} .
If μ > 0 \mu>0 , then μ ′ > 0 \mu^{\prime}>0 , where μ ′ \mu^{\prime} is μ 1 , μ 2 , μ 3 > 0 \mu_{1},\mu_{2},\mu_{3}>0 , depending on the cases ( A ) , ( B ) (A),(B) and (C). For μ > 0 \mu>0 , we have that
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) a + c < 1 . \frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1.
So, we impose the following condition:
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) < ( a + c ) k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)<(a+c)
(9)
Now, we have that
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θ n ​ d ​ ( x n , x n + 1 ) ≤ d ​ ( x n , T ​ x n ) ≤ θ n ′ ​ d ​ ( x n , x n + 1 ) \theta_{n}d\left(x_{n},x_{n+1}\right)\leq d\left(x_{n},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)
If there exists n ∈ ℕ n\in\mathbb{N} such that x n = x n + 1 := x ∗ x_{n}=x_{n+1}:=x^{*} , then, by the inequality above d ​ ( x ∗ , T ​ x ∗ ) = d\left(x^{*},Tx^{*}\right)= 0 , so the sequence ( x n ) \left(x_{n}\right) is convergent to the fixed point x ∗ x^{*} .
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So, let’s suppose that x n ≠ x n + 1 x_{n}\neq x_{n+1} . On the other hand, because μ > 0 \mu>0 and η ′ > 0 ⟹ η > 0 \eta^{\prime}>0\Longrightarrow\eta>0 , we have that
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θ n , θ n ′ > 0 , which implies ​ θ n ≤ θ n ′ . \theta_{n},\theta_{n}^{\prime}>0,\text{ which implies }\theta_{n}\leq\theta_{n}^{\prime}.
(V) At this step, we can show that θ n < 1 1 − α n \theta_{n}<\frac{1}{1-\alpha_{n}} , for each n ∈ ℕ n\in\mathbb{N} :
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We recall the following inequalities
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For the case (A):
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d ​ ( T ​ x n − 1 , T ​ y n − 1 ) ≤ ( 1 − β n − 1 ) ​ ( k − a ) a + c ​ d ​ ( x n − 1 , T ​ x n − 1 ) d\left(Tx_{n-1},Ty_{n-1}\right)\leq\frac{\left(1-\beta_{n-1}\right)(k-a)}{a+c}d\left(x_{n-1},Tx_{n-1}\right)
For the case ( B B ):
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d ​ ( T ​ x n − 1 , T ​ y n − 1 ) ≤ k ​ ( 1 − β n − 1 ) − a ​ ( 1 + β n − 1 ) a + c ​ d ​ ( x n − 1 , T ​ x n − 1 ) ​ and d\left(Tx_{n-1},Ty_{n-1}\right)\leq\frac{k\left(1-\beta_{n-1}\right)-a\left(1+\beta_{n-1}\right)}{a+c}d\left(x_{n-1},Tx_{n-1}\right)\text{ and }
For the case ( C ) (C) :
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d ​ ( T ​ x n − 1 , T ​ y n − 1 ) ≤ ( 1 − β n − 1 ) ​ ( k − a ) a + c ​ d ​ ( x n − 1 , T ​ x n − 1 ) d\left(Tx_{n-1},Ty_{n-1}\right)\leq\frac{\left(1-\beta_{n-1}\right)(k-a)}{a+c}d\left(x_{n-1},Tx_{n-1}\right)
Moreover, for all three cases we have the same conditions for positivity. Also, for ( A ) , ( B ) (A),(B) and (C), we get
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d ​ ( T ​ x n − 1 , T ​ y n − 1 ) ≤ [ k ​ ( 1 − β n − 1 ) a + c + | a | ​ ( 1 + β n − 1 ) a + c ] ​ D , because d\left(Tx_{n-1},Ty_{n-1}\right)\leq\left[\frac{k\left(1-\beta_{n-1}\right)}{a+c}+\frac{|a|\left(1+\beta_{n-1}\right)}{a+c}\right]D,\text{ because }
In the cases ( A ) (A) and ( C ) (C) , when a ≥ 0 a\geq 0 , we have
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− a ​ ( 1 − β n − 1 ) ≤ a ​ ( 1 − β n − 1 ) ≤ a ​ ( 1 + β n − 1 ) = | a | ​ ( 1 + β n − 1 ) -a\left(1-\beta_{n-1}\right)\leq a\left(1-\beta_{n-1}\right)\leq a\left(1+\beta_{n-1}\right)=|a|\left(1+\beta_{n-1}\right)
and for the case ( B ) (B) , when a < 0 a<0 , we have
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− a ​ ( 1 + β n − 1 ) = | a | ​ ( 1 + β n − 1 ) -a\left(1+\beta_{n-1}\right)=|a|\left(1+\beta_{n-1}\right)
At the same time
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η = 1 − α n a + c ​ [ k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) + ( a + c ) ] \eta=\frac{1-\alpha_{n}}{a+c}\left[k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)\right]
We can denote by η n \eta_{n} the right hand side since η \eta depends on the index n ∈ ℕ n\in\mathbb{N} . So
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) a + c + 1 = η n 1 − α n = η 1 − α n , for each ​ n ∈ ℕ \frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}+1=\frac{\eta_{n}}{1-\alpha_{n}}=\frac{\eta}{1-\alpha_{n}},\text{ for each }n\in\mathbb{N}
This means that
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) a + c = η 1 − α n − 1 . \frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}=\frac{\eta}{1-\alpha_{n}}-1.
It implies that
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d ​ ( T ​ x n − 1 , T ​ y n − 1 ) ≤ [ η 1 − α n − 1 − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) = [ 1 θ n − 1 ​ ( 1 − α n − 1 ) − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) d\left(Tx_{n-1},Ty_{n-1}\right)\leq\left[\frac{\eta}{1-\alpha_{n-1}}-1\right]d\left(x_{n-1},Tx_{n-1}\right)=\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]d\left(x_{n-1},Tx_{n-1}\right)
From the condition of positivity from above, the right hand side is positive, since the upper bound for d ​ ( T ​ x n , T ​ y n ) d\left(Tx_{n},Ty_{n}\right) is positive, we deduce that
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1 θ n − 1 ​ ( 1 − α n − 1 ) > 1 ​ , i.e. ​ θ n < 1 1 − α n ​ , for each ​ n ∈ ℕ ​ . \frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}>1\text{, i.e. }\theta_{n}<\frac{1}{1-\alpha_{n}}\text{, for each }n\in\mathbb{N}\text{. }
(VI) Now we evaluate d ​ ( T ​ x n − 1 , T ​ x n ) d\left(Tx_{n-1},Tx_{n}\right) :
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•
The case when c ≥ 0 c\geq 0 :
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We have that
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d ​ ( x n , T ​ x n ) ≤ d ​ ( x n , T ​ x n − 1 ) + d ​ ( T ​ x n − 1 , T ​ x n ) d\left(x_{n},Tx_{n}\right)\leq d\left(x_{n},Tx_{n-1}\right)+d\left(Tx_{n-1},Tx_{n}\right)
So
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d ​ ( T ​ x n − 1 , T ​ x n ) \displaystyle d\left(Tx_{n-1},Tx_{n}\right)
≥ d ​ ( x n , T ​ x n ) − d ​ ( x n , T ​ x n − 1 ) ⟹ d ​ ( T ​ x n − 1 , T ​ x n ) \displaystyle\geq d\left(x_{n},Tx_{n}\right)-d\left(x_{n},Tx_{n-1}\right)\Longrightarrow d\left(Tx_{n-1},Tx_{n}\right)
≥ θ n ​ d ​ ( x n , x n + 1 ) − d ​ ( x n , T ​ x n − 1 ) \displaystyle\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-d\left(x_{n},Tx_{n-1}\right)
Moreover, we know that
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d ​ ( x n , T ​ x n − 1 ) = \displaystyle d\left(x_{n},Tx_{n-1}\right)=
d ​ ( W ​ ( x n − 1 , T ​ y n − 1 , α n − 1 ) , T ​ x n − 1 ) \displaystyle d\left(W\left(x_{n-1},Ty_{n-1},\alpha_{n-1}\right),Tx_{n-1}\right)
≤ α n − 1 ​ d ​ ( x n − 1 , T ​ x n − 1 ) + ( 1 − α n − 1 ) ​ d ​ ( T ​ y n − 1 , T ​ x n − 1 ) \displaystyle\leq\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)d\left(Ty_{n-1},Tx_{n-1}\right)
≤ α n − 1 ​ d ​ ( x n − 1 , T ​ x n − 1 ) + ( 1 − α n − 1 ) ​ [ 1 θ n − 1 ​ ( 1 − α n − 1 ) − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) \displaystyle\leq\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]d\left(x_{n-1},Tx_{n-1}\right)
= [ α n − 1 + 1 θ n − 1 − 1 + α n − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) \displaystyle=\left[\alpha_{n-1}+\frac{1}{\theta_{n-1}}-1+\alpha_{n-1}\right]d\left(x_{n-1},Tx_{n-1}\right)
So
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d ​ ( x n , T ​ x n − 1 ) ≤ [ 2 ​ α n − 1 − 1 + 1 θ n − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) . d\left(x_{n},Tx_{n-1}\right)\leq\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right).
We show that the condition 2 ​ α n − 1 − 1 + 1 θ n − 1 ≥ 0 2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\geq 0 is true for each n ∈ ℕ n\in\mathbb{N} , so the right hand side is positive:
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If
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α n < 1 2 , i.e. ​ 1 − 2 ​ α n > 0 , \alpha_{n}<\frac{1}{2},\text{ i.e. }1-2\alpha_{n}>0,
then
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θ n < 1 1 − 2 ​ α n , since ​ θ n < 1 1 − α n < 1 1 − 2 ​ α n . \theta_{n}<\frac{1}{1-2\alpha_{n}},\text{ since }\theta_{n}<\frac{1}{1-\alpha_{n}}<\frac{1}{1-2\alpha_{n}}.
Moreover, if
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α n > 1 2 , i.e. ​ 1 − 2 ​ α n < 0 , \alpha_{n}>\frac{1}{2},\text{ i.e. }1-2\alpha_{n}<0,
then
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1 − 2 ​ α n < 0 < 1 θ n . 1-2\alpha_{n}<0<\frac{1}{\theta_{n}}.
So
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d ​ ( T ​ x n − 1 , T ​ x n ) ≥ θ n ​ d ​ ( x n , x n + 1 ) − [ 2 ​ α n − 1 − 1 + 1 θ n − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) . d\left(Tx_{n-1},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right).
But
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d ​ ( x n − 1 , T ​ x n − 1 ) ≤ d ​ ( x n − 1 , x n ) + d ​ ( x n , T ​ x n − 1 ) = d ​ ( x n − 1 , x n ) + d ​ ( W ​ ( x n − 1 , T ​ y n − 1 , α n − 1 ) , T ​ x n − 1 ) \displaystyle d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right)+d\left(x_{n},Tx_{n-1}\right)=d\left(x_{n-1},x_{n}\right)+d\left(W\left(x_{n-1},Ty_{n-1},\alpha_{n-1}\right),Tx_{n-1}\right)
≤ d ​ ( x n − 1 , x n ) + α n − 1 ​ d ​ ( x n − 1 , T ​ x n − 1 ) + ( 1 − α n − 1 ) ​ d ​ ( T ​ y n − 1 , T ​ x n − 1 ) \displaystyle\quad\leq d\left(x_{n-1},x_{n}\right)+\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)d\left(Ty_{n-1},Tx_{n-1}\right)
≤ d ​ ( x n − 1 , x n ) + α n − 1 ​ d ​ ( x n − 1 , T ​ x n − 1 ) + ( 1 − α n − 1 ) ​ [ 1 θ n − 1 ​ ( 1 − α n − 1 ) − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) \displaystyle\quad\leq d\left(x_{n-1},x_{n}\right)+\alpha_{n-1}d\left(x_{n-1},Tx_{n-1}\right)+\left(1-\alpha_{n-1}\right)\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]d\left(x_{n-1},Tx_{n-1}\right)
This implies that
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( 1 − α n − 1 ) ​ d ​ ( x n − 1 , T ​ x n − 1 ) ≤ d ​ ( x n − 1 , x n ) + ( 1 − α n − 1 ) ​ [ 1 θ n − 1 ​ ( 1 − α n − 1 ) − 1 ] \displaystyle\left(1-\alpha_{n-1}\right)d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right)+\left(1-\alpha_{n-1}\right)\left[\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}-1\right]
d ​ ( x n − 1 , T ​ x n − 1 ) ⟹ \displaystyle d\left(x_{n-1},Tx_{n-1}\right)\Longrightarrow
( 1 − α n − 1 ) ​ [ 1 − 1 θ n − 1 ​ ( 1 − α n − 1 ) + 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) ≤ d ​ ( x n − 1 , x n ) ⟹ \displaystyle\left(1-\alpha_{n-1}\right)\left[1-\frac{1}{\theta_{n-1}\left(1-\alpha_{n-1}\right)}+1\right]d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right)\Longrightarrow
[ 2 ​ ( 1 − α n − 1 ) − 1 θ n − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) ≤ d ​ ( x n − 1 , x n ) , with ​ θ n > 1 2 ​ ( 1 − α n ) \displaystyle{\left[2\left(1-\alpha_{n-1}\right)-\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right)\leq d\left(x_{n-1},x_{n}\right),\text{ with }\theta_{n}>\frac{1}{2\left(1-\alpha_{n}\right)}}
This means that
d ​ ( x n − 1 , T ​ x n − 1 ) ≤ 1 2 ​ ( 1 − α n − 1 ) − 1 θ n − 1 ​ d ​ ( x n − 1 , x n ) = θ n − 1 2 ​ θ n − 1 ​ ( 1 − α n − 1 ) − 1 ​ d ​ ( x n − 1 , x n ) d\left(x_{n-1},Tx_{n-1}\right)\leq\frac{1}{2\left(1-\alpha_{n-1}\right)-\frac{1}{\theta_{n-1}}}d\left(x_{n-1},x_{n}\right)=\frac{\theta_{n-1}}{2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1}d\left(x_{n-1},x_{n}\right) .
We notice that the right hand side is positive, so
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2 ​ θ n − 1 ​ ( 1 − α n − 1 ) − 1 > 0 , i.e. ​ θ n ≥ 1 2 ​ ( 1 − α n ) . 2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1>0,\text{ i.e. }\theta_{n}\geq\frac{1}{2\left(1-\alpha_{n}\right)}.
We imposed the condition that:
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θ n > 1 2 ​ ( 1 − α n ) , for each ​ n ∈ ℕ \theta_{n}>\frac{1}{2\left(1-\alpha_{n}\right)},\text{ for each }\mathrm{n}\in\mathbb{N}
(10)
This does not contradict the fact that
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θ n ≤ 1 1 − α n , since ​ α n < 1 , for each ​ n ∈ ℕ . \theta_{n}\leq\frac{1}{1-\alpha_{n}},\text{ since }\alpha_{n}<1,\text{ for each }n\in\mathbb{N}.
It implies that
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d ​ ( T ​ x n − 1 , T ​ x n ) ≥ θ n ​ d ​ ( x n , x n + 1 ) − [ 2 ​ α n − 1 − 1 + 1 θ n − 1 ] ⋅ [ θ n − 1 2 ​ θ n − 1 ​ ( 1 − α n − 1 ) − 1 ] ​ d ​ ( x n − 1 , x n ) d\left(Tx_{n-1},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]\cdot\left[\frac{\theta_{n-1}}{2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1}\right]d\left(x_{n-1},x_{n}\right)
Now, let’s denote by
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τ n := [ 2 ​ α n − 1 + 1 θ n ] ⋅ [ θ n 2 ​ θ n ​ ( 1 − α n ) − 1 ] = θ n ​ ( 2 ​ α n − 1 ) + 1 2 ​ θ n ​ ( 1 − α n ) − 1 . \tau_{n}:=\left[2\alpha_{n}-1+\frac{1}{\theta_{n}}\right]\cdot\left[\frac{\theta_{n}}{2\theta_{n}\left(1-\alpha_{n}\right)-1}\right]=\frac{\theta_{n}\left(2\alpha_{n}-1\right)+1}{2\theta_{n}\left(1-\alpha_{n}\right)-1}.
Then
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d ​ ( T ​ x n − 1 , T ​ x n ) ≥ θ n ​ d ​ ( x n , x n + 1 ) − τ n − 1 ​ d ​ ( x n − 1 , x n ) ⟹ \displaystyle d\left(Tx_{n-1},Tx_{n}\right)\geq\theta_{n}d\left(x_{n},x_{n+1}\right)-\tau_{n-1}d\left(x_{n-1},x_{n}\right)\Longrightarrow
c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ​ d ​ ( x n , x n + 1 ) − c ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) . \displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-c\tau_{n-1}d\left(x_{n-1},x_{n}\right).
From the definition of τ n \tau_{n} , we have that τ n > 0 \tau_{n}>0 , for each n ∈ ℕ n\in\mathbb{N} . Also, we have seen that the inequality θ n < 1 1 − α n \theta_{n}<\frac{1}{1-\alpha_{n}} holds. The condition that 1 2 ​ ( 1 − α n ) < θ n \frac{1}{2\left(1-\alpha_{n}\right)}<\theta_{n} can be written as follows
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1 2 < a + c k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) + ( a + c ) ⟹ \displaystyle\frac{1}{2}<\frac{a+c}{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)}\Longrightarrow
k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) + ( a + c ) < 2 ​ ( a + c ) , i.e. \displaystyle k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)<2(a+c),\text{ i.e. }
k ​ ( 1 − β n ) ​ < ( a + c ) − | ​ a | ( 1 + β n ) . \displaystyle k\left(1-\beta_{n}\right)<(a+c)-|a|\left(1+\beta_{n}\right).
So, we impose the condition that:
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) a + c < 1 \frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1
(11)
•
The case when c < 0 c<0 :
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We have
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d ​ ( T ​ x n − 1 , T ​ x n ) ≤ d ​ ( x n , T ​ x n ) + d ​ ( x n , T ​ x n − 1 ) . d\left(Tx_{n-1},Tx_{n}\right)\leq d\left(x_{n},Tx_{n}\right)+d\left(x_{n},Tx_{n-1}\right).
Moreover, we have shown that
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d ​ ( x n , T ​ x n − 1 ) ≤ [ 2 ​ α n − 1 − 1 + 1 θ n − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) d\left(x_{n},Tx_{n-1}\right)\leq\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right)
Also, we mention that this inequality is universal, i.e. it does not depend on the cases defined by the condition of positivity of a a and c c .
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So
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d ​ ( T ​ x n − 1 , T ​ x n ) ≤ d ​ ( x n , T ​ x n ) + [ 2 ​ α n − 1 − 1 + 1 θ n − 1 ] ​ d ​ ( x n − 1 , T ​ x n − 1 ) d\left(Tx_{n-1},Tx_{n}\right)\leq d\left(x_{n},Tx_{n}\right)+\left[2\alpha_{n-1}-1+\frac{1}{\theta_{n-1}}\right]d\left(x_{n-1},Tx_{n-1}\right)
But
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d ​ ( x n − 1 , T ​ x n − 1 ) ≤ θ n − 1 2 ​ θ n − 1 ​ ( 1 − α n − 1 ) − 1 ​ d ​ ( x n − 1 , x n ) ⟹ \displaystyle d\left(x_{n-1},Tx_{n-1}\right)\leq\frac{\theta_{n-1}}{2\theta_{n-1}\left(1-\alpha_{n-1}\right)-1}d\left(x_{n-1},x_{n}\right)\Longrightarrow
d ​ ( T ​ x n − 1 , T ​ x n ) ≤ θ n ′ ​ d ​ ( x n , x n + 1 ) + τ n − 1 ​ d ​ ( x n − 1 , x n ) ⟹ \displaystyle d\left(Tx_{n-1},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)+\tau_{n-1}d\left(x_{n-1},x_{n}\right)\Longrightarrow
c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ′ ​ d ​ ( x n , x n + 1 ) + c ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) \displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)+c\tau_{n-1}d\left(x_{n-1},x_{n}\right)
•
Now, for these two cases we have that:
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If c ≥ 0 c\geq 0 , then
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c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)
If c < 0 c<0 , then
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c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ′ ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)
(VII) Now we analyze cases ( A ) , ( B ) (A),(B) and ( C ) (C) on the contraction-type condition:
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Case ( A ) (A) , when a ≥ 0 a\geq 0 and c > 0 c>0 :
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We get that
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a ​ d ​ ( x n , T ​ x n ) ≥ a ​ θ n ​ d ​ ( x n , x n + 1 ) \displaystyle ad\left(x_{n},Tx_{n}\right)\geq a\theta_{n}d\left(x_{n},x_{n+1}\right)
a ​ d ​ ( x n − 1 , T ​ x n − 1 ) ≥ a ​ θ n − 1 ​ d ​ ( x n − 1 , x n ) ​ and \displaystyle ad\left(x_{n-1},Tx_{n-1}\right)\geq a\theta_{n-1}d\left(x_{n-1},x_{n}\right)\text{ and }
c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) \displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)
By the contraction-type condition on the pair ( x n − 1 , x n x_{n-1},x_{n} ), it follows that
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a ​ θ n ​ d ​ ( x n , x n + 1 ) + a ​ θ n − 1 ​ d ​ ( x n − 1 , x n ) + c ​ θ n ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) \displaystyle a\theta_{n}d\left(x_{n},x_{n+1}\right)+a\theta_{n-1}d\left(x_{n-1},x_{n}\right)+c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)
≤ a ​ d ​ ( x n , T ​ x n ) + a ​ d ​ ( x n − 1 , T ​ x n − 1 ) + c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≤ k ​ d ​ ( x n − 1 , x n ) \displaystyle\leq ad\left(x_{n},Tx_{n}\right)+ad\left(x_{n-1},Tx_{n-1}\right)+cd\left(Tx_{n-1},Tx_{n}\right)\leq kd\left(x_{n-1},x_{n}\right)
So
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d ​ ( x n , x n + 1 ) ≤ k − a ​ θ n − 1 + | c | ​ τ n − 1 a ​ θ n + c ​ θ n ⏟ δ 1 ​ d ​ ( x n − 1 , x n ) . d\left(x_{n},x_{n+1}\right)\leq\underbrace{\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{a\theta_{n}+c\theta_{n}}}_{\delta_{1}}d\left(x_{n-1},x_{n}\right).
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Case ( C ) (C) , when a ≥ 0 a\geq 0 and c < 0 c<0 :
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We get that
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a ​ d ​ ( x n , T ​ x n ) ≥ a ​ θ n ​ d ​ ( x n , x n + 1 ) \displaystyle ad\left(x_{n},Tx_{n}\right)\geq a\theta_{n}d\left(x_{n},x_{n+1}\right)
a ​ d ​ ( x n − 1 , T ​ x n − 1 ) ≥ a ​ θ n − 1 ​ d ​ ( x n − 1 , x n ) ​ and \displaystyle ad\left(x_{n-1},Tx_{n-1}\right)\geq a\theta_{n-1}d\left(x_{n-1},x_{n}\right)\text{ and }
c ​ d ​ ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ′ ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) \displaystyle cd\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)
By the contraction-type condition on the pair ( x n − 1 , x n x_{n-1},x_{n} ), it follows that
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a ​ θ n ​ d ​ ( x n , x n + 1 ) + a ​ θ n − 1 ​ d ​ ( x n − 1 , x n ) + c ​ θ n ′ ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) ≤ k ​ d ​ ( x n − 1 , x n ) a\theta_{n}d\left(x_{n},x_{n+1}\right)+a\theta_{n-1}d\left(x_{n-1},x_{n}\right)+c\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)\leq kd\left(x_{n-1},x_{n}\right)
So
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d ​ ( x n , x n + 1 ) ≤ k − a ​ θ n − 1 + | c | ​ τ n − 1 a ​ θ n + c ​ θ n ′ ⏟ δ 2 ​ d ​ ( x n , x n − 1 ) . d\left(x_{n},x_{n+1}\right)\leq\underbrace{\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{a\theta_{n}+c\theta_{n}^{\prime}}}_{\delta_{2}}d\left(x_{n},x_{n-1}\right).
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Case (B), when a < 0 a<0 and c ≥ 0 c\geq 0 :
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We have that
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ad ( x n , T ​ x n ) ≥ a ​ θ n ′ ​ d ​ ( x n , x n + 1 ) , \displaystyle\operatorname{ad}\left(x_{n},Tx_{n}\right)\geq a\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right),
ad ( x n − 1 , T ​ x n − 1 ) ≥ a ​ θ n − 1 ′ ​ d ​ ( x n − 1 , x n ) ​ and \displaystyle\operatorname{ad}\left(x_{n-1},Tx_{n-1}\right)\geq a\theta_{n-1}^{\prime}d\left(x_{n-1},x_{n}\right)\text{ and }
cd ( T ​ x n − 1 , T ​ x n ) ≥ c ​ θ n ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) . \displaystyle\operatorname{cd}\left(Tx_{n-1},Tx_{n}\right)\geq c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right).
By the contraction-type condition on the pair ( x n − 1 , x n x_{n-1},x_{n} ), it follows that
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a ​ θ n ′ ​ d ​ ( x n , x n + 1 ) + a ​ θ n − 1 ′ ​ d ​ ( x n − 1 , x n ) + c ​ θ n ​ d ​ ( x n , x n + 1 ) − | c | ​ τ n − 1 ​ d ​ ( x n − 1 , x n ) ≤ k ​ d ​ ( x n − 1 , x n ) . a\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)+a\theta_{n-1}^{\prime}d\left(x_{n-1},x_{n}\right)+c\theta_{n}d\left(x_{n},x_{n+1}\right)-|c|\tau_{n-1}d\left(x_{n-1},x_{n}\right)\leq kd\left(x_{n-1},x_{n}\right).
So
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d ​ ( x n , x n + 1 ) ≤ k − a ​ θ n − 1 ′ + | c | ​ τ n − 1 a ​ θ n ′ + c ​ θ n ⏟ δ 3 ​ d ​ ( x n − 1 , x n ) . d\left(x_{n},x_{n+1}\right)\leq\underbrace{\frac{k-a\theta_{n-1}^{\prime}+|c|\tau_{n-1}}{a\theta_{n}^{\prime}+c\theta_{n}}}_{\delta_{3}}d\left(x_{n-1},x_{n}\right).
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Now we impose conditions on the already defined δ 1 , δ 2 \delta_{1},\delta_{2} and δ 3 \delta_{3} , which depend on the index n ∈ ℕ n\in\mathbb{N} and give some bounds that do not depend on n n for these numbers such that the sequence ( x n x_{n} ) will be Cauchy:
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We have shown that
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θ n ∈ [ 1 2 ​ ( 1 − α n ) , 1 1 − α n ] ⊂ [ 1 2 ​ ( 1 − a 1 ) , 1 1 − a 2 ] . \theta_{n}\in\left[\frac{1}{2\left(1-\alpha_{n}\right)},\frac{1}{1-\alpha_{n}}\right]\subset\left[\frac{1}{2\left(1-a_{1}\right)},\frac{1}{1-a_{2}}\right].
Moreover, we know that
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τ n = ( 2 ​ α n − 1 + 1 θ n ) ⋅ ( θ n 2 ​ θ n ​ ( 1 − α n ) − 1 ) . \tau_{n}=\left(2\alpha_{n}-1+\frac{1}{\theta_{n}}\right)\cdot\left(\frac{\theta_{n}}{2\theta_{n}\left(1-\alpha_{n}\right)-1}\right).
Now, we impose a little more restrictive condition on θ n \theta_{n} :
Let
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1 2 ​ ( 1 − α n ) < 1 q ​ ( 1 − α n ) < θ n < 1 1 − α n , so ​ q < 2 ​ . \frac{1}{2\left(1-\alpha_{n}\right)}<\frac{1}{q\left(1-\alpha_{n}\right)}<\theta_{n}<\frac{1}{1-\alpha_{n}},\text{ so }q<2\text{. }
Also
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1 q < θ n ​ ( 1 − α n ) < 1 , so ​ q > 1 . \frac{1}{q}<\theta_{n}\left(1-\alpha_{n}\right)<1,\text{ so }q>1.
Since q ∈ ( 1 , 2 ) q\in(1,2) , taking q = 3 2 q=\frac{3}{2} , it implies that
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2 ​ θ n ​ ( 1 − α n ) − 1 > 2 q − 1 = 1 3 , so ​ θ n > 2 3 ​ ( 1 − α n ) . 2\theta_{n}\left(1-\alpha_{n}\right)-1>\frac{2}{q}-1=\frac{1}{3},\text{ so }\theta_{n}>\frac{2}{3\left(1-\alpha_{n}\right)}.
This means that
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θ n ∈ [ 2 3 ​ ( 1 − α n ) , 1 1 − α n ] . \theta_{n}\in\left[\frac{2}{3\left(1-\alpha_{n}\right)},\frac{1}{1-\alpha_{n}}\right].
Now, the condition becomes
k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) + ( a + c ) ​ < 3 ​ ( a + c ) 2 ⟹ k ​ ( 1 − β n ) ​ < ( a + c ) 2 − | ​ a | ​ ( 1 + β n ) k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)+(a+c)<\frac{3(a+c)}{2}\Longrightarrow k\left(1-\beta_{n}\right)<\frac{(a+c)}{2}-|a|\left(1+\beta_{n}\right) .
So, we imposed the following condition:
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k ​ ( 1 − β n ) ​ < ( a + c ) 2 − | ​ a | ( 1 + β n ) k\left(1-\beta_{n}\right)<\frac{(a+c)}{2}-|a|\left(1+\beta_{n}\right)
(12)
Also θ n ​ ( 2 ​ α n − 1 ) + 1 ≥ 0 \theta_{n}\left(2\alpha_{n}-1\right)+1\geq 0 is equivalent to θ n ​ ( 1 − 2 ​ α n ) ≤ 1 \theta_{n}\left(1-2\alpha_{n}\right)\leq 1 . In a simiar manner, for θ n ​ ( 2 ​ α n − 1 ) + 1 \theta_{n}\left(2\alpha_{n}-1\right)+1 , we have two cases:
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When
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a 2 < 1 2 ⟹ α n − 1 < 1 2 , a_{2}<\frac{1}{2}\Longrightarrow\alpha_{n-1}<\frac{1}{2},
it implies that
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2 ​ α n − 1 − 1 < 0 ⟹ ( 2 ​ α n − 1 − 1 ) ​ θ n − 1 + 1 < 1 , i.e. \displaystyle 2\alpha_{n-1}-1<0\Longrightarrow\left(2\alpha_{n-1}-1\right)\theta_{n-1}+1<1,\text{ i.e. }
τ n − 1 < 1 1 3 = 3 , for each ​ n ∈ ℕ , \displaystyle\tau_{n-1}<\frac{1}{\frac{1}{3}}=3,\text{ for each }n\in\mathbb{N},
since ​ θ n > 2 3 ​ ( 1 − α n ) ⟹ 1 2 ​ θ n ​ ( 1 − α n ) − 1 < 3 . \displaystyle\text{ since }\theta_{n}>\frac{2}{3\left(1-\alpha_{n}\right)}\Longrightarrow\frac{1}{2\theta_{n}\left(1-\alpha_{n}\right)-1}<3.
When
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a 1 > 1 2 ⟹ α n − 1 > 1 2 , a_{1}>\frac{1}{2}\Longrightarrow\alpha_{n-1}>\frac{1}{2},
it implies that
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2 ​ α n − 1 − 1 > 0 ⟹ θ n − 1 ​ ( 2 ​ α n − 1 − 1 ) + 1 < 1 + 2 ​ a 2 − 1 1 − a 2 = a 2 1 − a 2 . 2\alpha_{n-1}-1>0\Longrightarrow\theta_{n-1}\left(2\alpha_{n-1}-1\right)+1<1+\frac{2a_{2}-1}{1-a_{2}}=\frac{a_{2}}{1-a_{2}}.
So
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τ n − 1 < a 2 1 − a 2 ⋅ 1 3 . \tau_{n-1}<\frac{a_{2}}{1-a_{2}}\cdot\frac{1}{3}.
Taking these two cases into account, we have that τ n − 1 < τ \tau_{n-1}<\tau , where
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τ := { 1 3 a 2 < 1 2 1 3 ⋅ a 2 1 − a 2 a 1 > 1 2 \tau:=\begin{cases}\frac{1}{3}&a_{2}<\frac{1}{2}\\
\frac{1}{3}\cdot\frac{a_{2}}{1-a_{2}}&a_{1}>\frac{1}{2}\end{cases}
Now, let ε ∈ ( 0 , 1 ) \varepsilon\in(0,1) . Furthermore, we impose the condition that
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θ n ′ ≤ 1 + β n ε ​ ( 1 − α n ) ​ ( 1 − β n ) \theta_{n}^{\prime}\leq\frac{1+\beta_{n}}{\varepsilon\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}
So
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( a + c ) ( a + c ) − k ​ ( 1 − β n ) − | a | ​ ( 1 + β n ) < 1 ε ⟹ \displaystyle\frac{(a+c)}{(a+c)-k\left(1-\beta_{n}\right)-|a|\left(1+\beta_{n}\right)}<\frac{1}{\varepsilon}\Longrightarrow
ε ​ ( a + c ) ( a + c ) − k ​ ( 1 − β n ) − | a | ​ ( 1 + β n ) < 1 ⟹ \displaystyle\frac{\varepsilon(a+c)}{(a+c)-k\left(1-\beta_{n}\right)-|a|\left(1+\beta_{n}\right)}<1\Longrightarrow
( ε − 1 ) ​ ( a + c ) ​ < − k ​ ( 1 − β n ) − | ​ a | ( 1 + β n ) ⟹ ( ε − 1 ) ​ ( a + c ) \displaystyle(\varepsilon-1)(a+c)<-k\left(1-\beta_{n}\right)-|a|\left(1+\beta_{n}\right)\Longrightarrow(\varepsilon-1)(a+c)
+ | a | ​ ( 1 + β n ) < − k ​ ( 1 − β n ) , i.e. \displaystyle\quad+|a|\left(1+\beta_{n}\right)<-k\left(1-\beta_{n}\right),\text{ i.e. }
( 1 − ε ) ​ ( a + c ) − | a | ​ ( 1 + β n ) > k ​ ( 1 − β n ) . (1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right)>k\left(1-\beta_{n}\right).
We observe that we have imposed the following condition:
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k ​ ( 1 − β n ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + β n ) k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right)
(13)
Now, for δ 1 , δ 2 \delta_{1},\delta_{2} and δ 3 \delta_{3} , we analyze all three cases:
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Case ( A ) (A) , when a ≥ 0 a\geq 0 and c > 0 c>0 :
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We have that
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δ 1 = k − a ​ θ n − 1 + | c | ​ τ n − 1 ( a + c ) ​ θ n ≤ k + | c | ​ τ n − 1 ( a + c ) ​ θ n ≤ k + | c | ​ τ ( a + c ) ​ θ n . \delta_{1}=\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{(a+c)\theta_{n}}\leq\frac{k+|c|\tau_{n-1}}{(a+c)\theta_{n}}\leq\frac{k+|c|\tau}{(a+c)\theta_{n}}.
At the same time
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( a + c ) ​ θ n > 2 ​ ( a + c ) 3 ​ ( 1 − α n ) ≥ 2 ​ ( a + c ) 3 ​ ( 1 − a 1 ) ⟹ 1 ( a + c ) ​ θ n < 3 ​ ( 1 − a 1 ) 2 ​ ( a + c ) ⟹ \displaystyle(a+c)\theta_{n}>\frac{2(a+c)}{3\left(1-\alpha_{n}\right)}\geq\frac{2(a+c)}{3\left(1-a_{1}\right)}\Longrightarrow\frac{1}{(a+c)\theta_{n}}<\frac{3\left(1-a_{1}\right)}{2(a+c)}\Longrightarrow
δ 1 ≤ 3 ​ ( k + | c | ​ τ ) ​ ( 1 − a 1 ) 2 ​ ( a + c ) , i.e. ​ δ 1 ≤ 3 ​ ( 1 − a 1 ) 2 ​ ( a + c ) ⋅ [ k + | c | ​ τ ] . \displaystyle\delta_{1}\leq\frac{3(k+|c|\tau)\left(1-a_{1}\right)}{2(a+c)},\text{ i.e. }\delta_{1}\leq\frac{3\left(1-a_{1}\right)}{2(a+c)}\cdot[k+|c|\tau].
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Case ( C ) (C) , when a ≥ 0 a\geq 0 and c < 0 c<0 :
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We know that
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δ 2 = k − a ​ θ n − 1 + | c | ​ τ n − 1 a ​ θ n + c ​ θ n ′ . \delta_{2}=\frac{k-a\theta_{n-1}+|c|\tau_{n-1}}{a\theta_{n}+c\theta_{n}^{\prime}}.
But
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k − a ​ θ n − 1 + | c | ​ τ n − 1 ≤ k − a 1 − a 2 + | c | ​ τ ≤ k + | c | ​ τ k-a\theta_{n-1}+|c|\tau_{n-1}\leq k-\frac{a}{1-a_{2}}+|c|\tau\leq k+|c|\tau
Now
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θ n ′ ≤ 1 + β n ε ​ ( 1 − α n ) ​ ( 1 − β n ) ≤ 1 + b 2 ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) \theta_{n}^{\prime}\leq\frac{1+\beta_{n}}{\varepsilon\left(1-\alpha_{n}\right)\left(1-\beta_{n}\right)}\leq\frac{1+b_{2}}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}
where a 2 a_{2} and b 2 b_{2} are upper bounds for the sequences ( α n \alpha_{n} ), respectively ( β n \beta_{n} ).
Furthermore
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a ​ θ n + c ​ θ n ′ ≥ a 2 ​ ( 1 − a 1 ) + c ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) . a\theta_{n}+c\theta_{n}^{\prime}\geq\frac{a}{2\left(1-a_{1}\right)}+\frac{c\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.
Since
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θ n ≥ 2 3 ​ ( 1 − a 1 ) > 1 2 ​ ( 1 − a 1 ) , \theta_{n}\geq\frac{2}{3\left(1-a_{1}\right)}>\frac{1}{2\left(1-a_{1}\right)},
in this case it is appropriate to use the above inequality with 1 2 ​ ( 1 − a 1 ) \frac{1}{2\left(1-a_{1}\right)} . Also, we notice that we have the condition :
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a 2 ​ ( 1 − a 1 ) + c ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) > 0 \frac{a}{2\left(1-a_{1}\right)}+\frac{c\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}>0
(14)
Then
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δ 2 ≤ k + | c | ​ τ a 2 ​ ( 1 − a 1 ) + c ​ ( 1 + b 2 ) ( 1 − a 2 ) ​ ( 1 − b 2 ) \delta_{2}\leq\frac{k+|c|\tau}{\frac{a}{2\left(1-a_{1}\right)}+\frac{c\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}}
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Case ( B ) (B) , when a < 0 a<0 and c ≥ 0 c\geq 0 :
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We know that
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δ 3 = k − a ​ θ n − 1 ′ + | c | ​ τ n − 1 a ​ θ n ′ + c ​ θ n \delta_{3}=\frac{k-a\theta_{n-1}^{\prime}+|c|\tau_{n-1}}{a\theta_{n}^{\prime}+c\theta_{n}}
But
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k − a ​ θ n − 1 ′ + | c | ​ τ n − 1 ≤ k − a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + | c | ​ τ ​ and \displaystyle k-a\theta_{n-1}^{\prime}+|c|\tau_{n-1}\leq k-a\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+|c|\tau\text{ and }
c ​ θ n ≥ c 2 ​ ( 1 − a 1 ) \displaystyle c\theta_{n}\geq\frac{c}{2\left(1-a_{1}\right)}
As in the previous case it is more convenient to use the inequality for θ n \theta_{n} with 1 2 ​ ( 1 − a 1 ) \frac{1}{2\left(1-a_{1}\right)} , than with 2 3 ​ ( 1 − a 1 ) \frac{2}{3\left(1-a_{1}\right)} . Moreover
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a ​ θ n ′ + c ​ θ n ≥ a ​ θ n ′ + c 2 ​ ( 1 − a 1 ) a\theta_{n}^{\prime}+c\theta_{n}\geq a\theta_{n}^{\prime}+\frac{c}{2\left(1-a_{1}\right)}
It follows that
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a ​ θ n ′ + c ​ θ n ≥ a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + c 2 ​ ( 1 − a 1 ) a\theta_{n}^{\prime}+c\theta_{n}\geq\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)}
Also, we imposed the condition:
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a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + c 2 ​ ( 1 − a 1 ) > 0 \frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)}>0
(15)
Finally, we get the upper bound for δ 3 \delta_{3} , i.e.
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δ 3 ≤ k − a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + c ​ | τ | a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + c 2 ​ ( 1 − a 1 ) . \delta_{3}\leq\frac{k-\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+c|\tau|}{\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)}}.
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Now, we can get an upper bound for all δ i \delta_{i} ’s:
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k + | a | ​ θ n − 1 ′ + | c | ​ τ n − 1 ≤ k + | a | ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + | c | ​ τ . k+|a|\theta_{n-1}^{\prime}+|c|\tau_{n-1}\leq k+|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+|c|\tau.
Now, we denote the denominator by
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s := { 3 ​ ( 1 − a 1 ) 2 ​ ( a + c ) , in the case (A), a 2 ​ ( 1 − a 1 ) + c ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) , in the case (C), a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + c 2 ​ ( 1 − a 1 ) , in the case (B) s:=\begin{cases}\frac{3\left(1-a_{1}\right)}{2(a+c)},&\text{ in the case (A), }\\
\frac{a}{2\left(1-a_{1}\right)}+c\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)},&\text{ in the case (C), }\\
\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+\frac{c}{2\left(1-a_{1}\right)},&\text{ in the case (B) }\end{cases}
Also, let’s denote
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t := k + | a | ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + | c | ​ τ s . t:=\frac{k+|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+|c|\tau}{s}.
In the step ( I ​ X IX ), we shall show that the numerators of all δ i \delta_{i} are positive, so that δ i ≥ 0 \delta_{i}\geq 0 . For the moment, since δ i ≥ 0 \delta_{i}\geq 0 , for each i ∈ { 1 , 2 , 3 } i\in\{1,2,3\} , then t ≥ 0 t\geq 0 . Moreover, we want that t < 1 t<1 , that is
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| c | ​ τ ​ < s − k − | ​ a | ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) . |c|\tau<s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.
We now have the condition:
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τ < 1 | c | ⋅ [ s − k − | a | ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) ] \tau<\frac{1}{|c|}\cdot\left[s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}\right]
(16)
(VIII) Now we show that the sequence ( x n x_{n} ) is Cauchy:
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Let’s take t ∈ [ 0 , 1 ) t\in[0,1) as before. For all the three cases, ( A ) , ( B ) (A),(B) and ( C ) (C) , we get
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d ​ ( x n , x n + 1 ) ≤ t ​ d ​ ( x n − 1 , x n ) ⟹ d ​ ( x n , x n + 1 ) ≤ t n ​ d ​ ( x 1 , x 0 ) , \displaystyle\quad d\left(x_{n},x_{n+1}\right)\leq td\left(x_{n-1},x_{n}\right)\Longrightarrow d\left(x_{n},x_{n+1}\right)\leq t^{n}d\left(x_{1},x_{0}\right),
where ​ { x 1 = W ​ ( x 0 , T ​ y 0 , α 0 ) y 0 = W ​ ( x 0 , T ​ x 0 , β 0 ) , ​ with ​ x 0 ∈ K ​ is a fixed arbitrary element \displaystyle\text{ where }\left\{\begin{array}[]{l}x_{1}=W\left(x_{0},Ty_{0},\alpha_{0}\right)\\
y_{0}=W\left(x_{0},Tx_{0},\beta_{0}\right),\end{array}\text{ with }x_{0}\in\mathrm{~K}\right.\text{ is a fixed arbitrary element }
Let m > n m>n . We now have that
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d ​ ( x m , x n ) ≤ [ t n + t n + 1 + ⋯ + t m − 1 ] ​ d ​ ( x 1 , x 0 ) ≤ t n 1 − t ​ d ​ ( x 1 , x 0 ) d\left(x_{m},x_{n}\right)\leq\left[t^{n}+t^{n+1}+\cdots+t^{m-1}\right]d\left(x_{1},x_{0}\right)\leq\frac{t^{n}}{1-t}d\left(x_{1},x_{0}\right)
Letting m m and n n go to ∞ \infty , it follows that ( x n x_{n} ) is a Cauchy sequence, because ( x n x_{n} ) ⊂ K ⊂ X \subset K\subset X and K K is a convex subset of the convex metric space ( X , d , W X,d,W ). Since ( x n x_{n} ) is Cauchy and K K is closed in the complete metric space ( X , d , W X,d,W ), then ( x n x_{n} ) is convergent, i.e. there exists p ∈ K p\in K , such that lim n → ∞ x n = p \lim_{n\rightarrow\infty}x_{n}=p . Taking in the contraction-type conditions p instead of x and x n x_{n} instead of y , it follows
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c ​ d ​ ( T ​ x n , T ​ p ) + a ​ d ​ ( p , T ​ p ) + a ​ d ​ ( x n , T ​ x n ) ≤ k ​ d ​ ( x n , p ) cd\left(Tx_{n},Tp\right)+ad(p,Tp)+ad\left(x_{n},Tx_{n}\right)\leq kd\left(x_{n},p\right)
(*)
But
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d ​ ( T ​ x n , p ) ≤ d ​ ( p , x n ) + d ​ ( x n , T ​ x n ) ≤ d ​ ( x n , p ) + θ n ′ ​ d ​ ( x n , x n + 1 ) d\left(Tx_{n},p\right)\leq d\left(p,x_{n}\right)+d\left(x_{n},Tx_{n}\right)\leq d\left(x_{n},p\right)+\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)
Also, we now that
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d ​ ( x n , p ) → 0 , d ​ ( x n , x n + 1 ) → 0 ​ and that ​ lim n → ∞ θ n ′ < ∞ d\left(x_{n},p\right)\rightarrow 0,d\left(x_{n},x_{n+1}\right)\rightarrow 0\text{ and that }\lim_{n\rightarrow\infty}\theta_{n}^{\prime}<\infty
because
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θ n ′ ≤ ( a + c ) ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) . \theta_{n}^{\prime}\leq\frac{(a+c)\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}.
Then, we obtain that lim n → ∞ T ​ x n = p \lim_{n\rightarrow\infty}Tx_{n}=p . Also, we know that d ​ ( x n , T ​ x n ) ≤ θ n ′ ​ d ​ ( x n , x n + 1 ) → 0 d\left(x_{n},Tx_{n}\right)\leq\theta_{n}^{\prime}d\left(x_{n},x_{n+1}\right)\rightarrow 0 . From (*), we get c ​ d ​ ( p , T ​ p ) + a ​ d ​ ( p , T ​ p ) ≤ 0 cd(p,Tp)+ad(p,Tp)\leq 0 , i.e. ( a + c ) ​ d ​ ( p , T ​ p ) ≤ 0 (a+c)d(p,Tp)\leq 0 . Since a + c > 0 a+c>0 , then d ​ ( p , T ​ p ) = 0 ⟺ p = T ​ p d(p,Tp)=0\Longleftrightarrow p=Tp , i.e. the limit of the sequence ( x n ) \left(x_{n}\right) is a fixed point for the operator T T .
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•
The uniqueness of the fixed point: Let’s consider p , q p,q two fixed points for T T , i.e. p = T ​ p p=Tp and q = T ​ q q=Tq . Applying the contraction-type condition for the pair ( p , q p,q ), it implies that:
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c ​ d ​ ( T ​ p , T ​ q ) + a ​ d ​ ( p , T ​ p ) + a ​ d ​ ( q , T ​ q ) ≤ k ​ d ​ ( p , q ) cd(Tp,Tq)+ad(p,Tp)+ad(q,Tq)\leq kd(p,q)
Since d ​ ( T ​ p , T ​ q ) = d ​ ( p , q ) , d ​ ( p , T ​ p ) = 0 d(Tp,Tq)=d(p,q),d(p,Tp)=0 and d ​ ( q , T ​ q ) = 0 d(q,Tq)=0 , we get ( c − k ) ​ d ​ ( p , q ) ≤ 0 (c-k)d(p,q)\leq 0 . Also, because k < c k<c , then we have that d ​ ( p , q ) = 0 d(p,q)=0 , i.e. p = q p=q .
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We have imposed the condition:
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k < c ​ , for the uniqueness of the fixed point for ​ T k<c\text{, for the uniqueness of the fixed point for }\mathrm{T}
(IX) Now we verify that all the conditions we imposed are valid because of the hypothesis assumptions:
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We recall the conditions:
For the case ( A A ):
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c ≠ 0 , a + c > 0 , k > a ​ and ​ ( 1 − β n ) ​ ( k − a ) a + c < 1 c\neq 0,a+c>0,k>a\text{ and }\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1
For the case ( B B ):
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k > a ​ 1 + β n 1 − β n ​ and ​ k ​ ( 1 − β n ) − a ​ ( 1 + β n ) a + c < 1 k>a\frac{1+\beta_{n}}{1-\beta_{n}}\text{ and }\frac{k\left(1-\beta_{n}\right)-a\left(1+\beta_{n}\right)}{a+c}<1
For the case ( C C ):
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c ≠ 0 , k > a ​ and ​ ( 1 − β n ) ​ ( k − a ) a + c < 1 . c\neq 0,k>a\text{ and }\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1.
For θ n ′ \theta_{n}^{\prime} , we have
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) a + c < 1 . \frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1.
The conditions (10) and (11):
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k ​ ( 1 − β n ) + | a | ​ ( 1 + β n ) a + c < 1 . \frac{k\left(1-\beta_{n}\right)+|a|\left(1+\beta_{n}\right)}{a+c}<1.
The condition (12):
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k ​ ( 1 − β n ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + β n ) k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right)
The condition (13):
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k ​ ( 1 − β n ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + β n ) , with ​ ε ∈ ( 0 , 1 ) k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right),\text{ with }\varepsilon\in(0,1)
The conditions (14) and (15):
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The condition (16):
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| c | ​ τ ​ < s − k − | ​ a | ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) |c|\tau<s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}
We analyze these conditions by cases:
Casew ( A A ) and ( C C ) imply that
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⟹ k > a , ( 1 − β n ) ​ ( k − a ) a + c < 1 , i.e. ​ ( 1 − β n ) ​ k < ( a + c ) + a ​ ( 1 − β n ) . \Longrightarrow k>a,\frac{\left(1-\beta_{n}\right)(k-a)}{a+c}<1,\text{ i.e. }\left(1-\beta_{n}\right)k<(a+c)+a\left(1-\beta_{n}\right).
Case (B) implies that
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⟹ k > a ​ 1 + β n 1 − β n ​ and ​ k ​ ( 1 − β n ) < ( a + c ) + a ​ ( 1 + β n ) ​ . \Longrightarrow k>a\frac{1+\beta_{n}}{1-\beta_{n}}\text{ and }k\left(1-\beta_{n}\right)<(a+c)+a\left(1+\beta_{n}\right)\text{. }
Also, for a ≥ 0 a\geq 0 , for k > a k>a , we take k > a ​ ( 1 + β n ) 1 − β n > a k>\frac{a\left(1+\beta_{n}\right)}{1-\beta_{n}}>a . In a similar way, for a < 0 a<0 , we have the same inequality. So, for the both cases involving a a , we can take
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k > a ​ ( 1 + β n ) ( 1 − β n ) . k>a\frac{\left(1+\beta_{n}\right)}{\left(1-\beta_{n}\right)}.
Now, the second conditions for a a can be written as:
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{ a ≥ 0 ⟹ ( 1 − β n ) ​ k ​ < ( a + c ) + | ​ a | ( 1 − β n ) , a ​ < 0 ⟹ ( 1 − β n ) ​ k ​ < ( a + c ) − | ​ a | ​ ( 1 + β n ) \left\{\begin{array}[]{l}a\geq 0\Longrightarrow\left(1-\beta_{n}\right)k<(a+c)+|a|\left(1-\beta_{n}\right),\\
a<0\Longrightarrow\left(1-\beta_{n}\right)k<(a+c)-|a|\left(1+\beta_{n}\right)\end{array}\right.
Also, the condition for θ n ′ \theta_{n}^{\prime} is
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k ​ ( 1 − β n ) ​ < ( a + c ) − | ​ a | ( 1 + β n ) . k\left(1-\beta_{n}\right)<(a+c)-|a|\left(1+\beta_{n}\right).
In a similar manner, the condition (12) is
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k ​ ( 1 − β n ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + β n ) . k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right).
From the above two relations, we impose the restrictive one, i.e.
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k ​ ( 1 − β n ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + β n ) . k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right).
Moreover, at condition (13), we have
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k ​ ( 1 − β n ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + β n ) . k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right).
For the cases involving a a , condition (**) can be restrained as follows:
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( 1 − β n ) ​ k ​ < ( a + c ) − | ​ a | ( 1 + β n ) . \left(1-\beta_{n}\right)k<(a+c)-|a|\left(1+\beta_{n}\right).
So, the valid condition remains only condition (12), that is
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k ​ ( 1 − β n ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + β n ) . k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right).
•
Now, we impose condition such that numerators of δ i \delta_{i} ’s are positive:
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We recall the cases reminded above:
Cases ( A ) (A) and ( C ) (C) :
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k ≥ a ​ θ n − 1 − | c | ​ τ n − 1 , k\geq a\theta_{n-1}-|c|\tau_{n-1},
Case (B):
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k ≥ a ​ θ n − 1 ′ − | c | ​ τ n − 1 . k\geq a\theta_{n-1}^{\prime}-|c|\tau_{n-1}.
At case ( B ) (B) , since a < 0 a<0 and θ n − 1 ≤ θ n − 1 ′ \theta_{n-1}\leq\theta_{n-1}^{\prime} , we have that
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a ​ θ n − 1 ≥ a ​ θ n − 1 ′ . a\theta_{n-1}\geq a\theta_{n-1}^{\prime}.
So, for all the cases we need to impose
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k ≥ a ​ θ n − 1 − | c | ​ τ n − 1 . k\geq a\theta_{n-1}-|c|\tau_{n-1}.
•
Finally, the conditions that are valid are:
(a) k ≥ a ​ θ n − 1 − | c | ​ τ n − 1 k\geq a\theta_{n-1}-|c|\tau_{n-1} ,
(b) k ​ ( 1 − β n ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + β n ) k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right) ,
(c) k ​ ( 1 − β n ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + β n ) k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right) ,
(d) k > a ​ ( 1 + β n ) ( 1 − β n ) k>\frac{a\left(1+\beta_{n}\right)}{\left(1-\beta_{n}\right)} ,
(e) s ≥ 0 s\geq 0 and | c | ​ τ ​ < s − k − | ​ a | ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) |c|\tau<s-k-|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)} .
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Also, now we analyze in detail conditions (a),(c) and (d):
For condition (c), we have that
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k < 1 2 ​ ( a + c ) − | a | ​ ( 1 + β n ) 1 − β n . k<\frac{\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right)}{1-\beta_{n}}.
So, for k > 0 k>0 , we impose k ​ ( 1 − b 1 ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + b 2 ) k\left(1-b_{1}\right)<\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right) and for k < 0 k<0 we impose k ​ ( 1 − b 2 ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + b 2 ) k\left(1-b_{2}\right)<\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right) . Furthermore, for condition (b), we have that
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k ​ ( 1 − β n ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + β n ) . k\left(1-\beta_{n}\right)<(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right).
Now, in a similar manner, using the boundaries for the sequence ( β n ) \left(\beta_{n}\right) , we have two cases:
If k > 0 k>0 , then
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k ​ ( 1 − b 1 ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + b 2 ) k\left(1-b_{1}\right)<(1-\varepsilon)(a+c)-|a|\left(1+b_{2}\right)
and for k < 0 k<0 we impose
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k ​ ( 1 − b 2 ) ​ < ( 1 − ε ) ​ ( a + c ) − | ​ a | ( 1 + b 2 ) . k\left(1-b_{2}\right)<(1-\varepsilon)(a+c)-|a|\left(1+b_{2}\right).
Taking ε ∈ ( 0 , 1 2 ] \varepsilon\in\left(0,\frac{1}{2}\right] , the condition (c) implies that:
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k ​ ( 1 − β n ) ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + β n ) ≤ ( 1 − ε ) ​ ( a + c ) − | a | ​ ( 1 + β n ) , k\left(1-\beta_{n}\right)<\frac{1}{2}(a+c)-|a|\left(1+\beta_{n}\right)\leq(1-\varepsilon)(a+c)-|a|\left(1+\beta_{n}\right),
for each n ∈ ℕ n\in\mathbb{N} , so condition (c) implies condition (b).
For condition (d), we have two cases:
If a ≥ 0 a\geq 0 , then
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k > a ​ ( 1 + b 2 ) 1 − b 2 , k>a\frac{\left(1+b_{2}\right)}{1-b_{2}},
and if a < 0 a<0 , we have
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k ≥ a ​ 1 + b 1 1 − b 1 k\geq a\frac{1+b_{1}}{1-b_{1}}
For condition (a), it follows that:
Since
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a ​ θ n − 1 − | c | ​ τ n − 1 ≤ a ​ θ n − 1 , a\theta_{n-1}-|c|\tau_{n-1}\leq a\theta_{n-1},
then we can take
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k ≥ a ​ θ n − 1 . k\geq a\theta_{n-1}.
If a ≥ 0 a\geq 0 , since a ​ θ n − 1 ≤ a 1 − a 2 a\theta_{n-1}\leq\frac{a}{1-a_{2}} , then we need to impose
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k ≥ a 1 − a 2 . k\geq\frac{a}{1-a_{2}}.
If a < 0 a<0 , since θ n − 1 ≥ 2 3 ​ ( 1 − α n − 1 ) ≥ 2 3 ​ ( 1 − a 1 ) \theta_{n-1}\geq\frac{2}{3\left(1-\alpha_{n-1}\right)}\geq\frac{2}{3\left(1-a_{1}\right)} , then
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a ​ θ n − 1 ≤ 2 ​ a 3 ​ ( 1 − a 1 ) a\theta_{n-1}\leq\frac{2a}{3\left(1-a_{1}\right)}
We impose that
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k ≥ 2 ​ a 3 ​ ( 1 − a 1 ) k\geq\frac{2a}{3\left(1-a_{1}\right)}
Remark 1 For the sake of completeness involving the proof from (Theorem 1), we make the following crucial remark: if a ≥ 0 a\geq 0 , then k ≥ 0 k\geq 0 .
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As a consequence of the above result, we can obtain a fixed point theorem for a certain iterate of a self-operator in a complete convex metric space.
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Corollary 1 Let ( X , d X,d ) be a complete convex metric space. Let K K a nonempty, closed and convex subset of X X and T : K → K T:K\rightarrow K be a self-mapping. Suppose, there exists p > 1 p>1 , such that T T satisfies:
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c ​ d ​ ( T p ​ x , T p ​ y ) + a ​ [ d ​ ( x , T p ​ x ) + d ​ ( y , T p ​ y ) ] ≤ k ​ d ​ ( x , y ) cd\left(T^{p}x,T^{p}y\right)+a\left[d\left(x,T^{p}x\right)+d\left(y,T^{p}y\right)\right]\leq kd(x,y)
for each x , y ∈ K x,y\in K . Also, suppose the operator T p : K → K T^{p}:K\rightarrow K satisfies conditions (1)-(8) from (Theorem 1). Then, T T has a fixed point. Moreover, if k < c k<c then the fixed point is unique.
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Proof Let T ′ := T p T^{\prime}:=T^{p} . Applying (Theorem 1) for the operator T ′ T^{\prime} , we find that T ′ T^{\prime} has a fixed point u ∈ K u\in K , i.e. T ′ ​ u = u T^{\prime}u=u . Also u u is the unique fixed point of T ′ T^{\prime} if k < c k<c . But T p ​ ( T ​ u ) = T ​ ( T p ​ u ) = T ​ ( T ′ ​ u ) = T ​ ( u ) T^{p}(Tu)=T\left(T^{p}u\right)=T\left(T^{\prime}u\right)=T(u) . Also, since T p ​ ( T ​ u ) = T ′ ​ ( T ​ u ) T^{p}(Tu)=T^{\prime}(Tu) , from these two equalities, we find that T ​ u = u Tu=u , so T T has a fixed point. Moreover, if k < c k<c , then u u is the unique element in K K such that T ′ ​ u = u T^{\prime}u=u . Finally, with the same reasoning we find that u u is the unique fixed point of T T .
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Example 1 Let X := [ 0 , ∞ ) X:=[0,\infty) and T : X → X T:X\rightarrow X , such that :
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T ​ ( x ) = { x 194 , x ∈ [ 0 , x 0 ) x 250 , x ∈ [ x 0 , ∞ ) T(x)=\begin{cases}\frac{x}{194},&x\in\left[0,x_{0}\right)\\
\frac{x}{250},&x\in\left[x_{0},\infty\right)\end{cases}
where x 0 x_{0} can be any point in [ 0 , ∞ ) [0,\infty) . We remark that because x 0 x_{0} is a discontinuity point for T T , then T can’t be a contraction type mapping. Then T T satisfies the inequality:
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1800 ​ d ​ ( T ​ x , T ​ y ) + 1 2 ​ d ​ ( x , y ) ≤ 10 ​ [ d ​ ( x , T ​ x ) + d ​ ( y , T ​ y ) ] 1800d(Tx,Ty)+\frac{1}{2}d(x,y)\leq 10[d(x,Tx)+d(y,Ty)]
Also, T T has a unique fixed point, i.e. 0 ∈ X 0\in X .
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Remark 2 We will construct a self-mapping T T defined on [ 0 , ∞ ) [0,\infty) , which satisfies the above generalized contraction condition with a ≤ 0 , k ≤ 0 a\leq 0,k\leq 0 , and c > 0 c>0 . as well as the assumptions from (Theorem 1). For this purpose, let b 1 = 1 10 , b 2 = 9 10 , a 1 = 1 10 b_{1}=\frac{1}{10},b_{2}=\frac{9}{10},a_{1}=\frac{1}{10} and a 2 = 2 5 a_{2}=\frac{2}{5} . Following the notations from (Theorem 1), we have that
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a 2 < 1 2 ​ and ​ τ = 1 3 a_{2}<\frac{1}{2}\text{ and }\tau=\frac{1}{3}
Also,
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s = a ​ ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) + c ​ 1 2 ​ ( 1 − a 1 ) s=\frac{a\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)}+c\frac{1}{2\left(1-a_{1}\right)}
Since τ | c | < s − k − − | a | ( 1 + b 2 ) ε ​ ( 1 − a 2 ) ​ ( 1 − b 2 ) \tau|c|<s-k--|a|\frac{\left(1+b_{2}\right)}{\varepsilon\left(1-a_{2}\right)\left(1-b_{2}\right)} , that is the condition (e) in the proof of (Theorem 1), then we get that
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c 3 ​ < s − k − | ​ a | ( 1 + b 2 ) ( 1 − a 2 ) ​ ( 1 − b 2 ) , i.e. \displaystyle\frac{c}{3}<s-k-|a|\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)},\text{ i.e. }
c 3 < 2 ​ a ε ​ ( 1 + b 2 ) ( 1 − a 2 ) ​ ( 1 − b 2 ) − k \displaystyle\frac{c}{3}<\frac{2a}{\varepsilon}\frac{\left(1+b_{2}\right)}{\left(1-a_{2}\right)\left(1-b_{2}\right)}-k
Furthermore,
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c 3 < 2 ​ a ε ​ 190 6 − k + 5 ​ c 9 \frac{c}{3}<\frac{2a}{\varepsilon}\frac{190}{6}-k+\frac{5c}{9}
Also a + c > 0 a+c>0 , i.e. c > − a c>-a and k ≥ 2 ​ a 3 ​ ( 1 − a 1 ) k\geq\frac{2a}{3\left(1-a_{1}\right)} , so k ≥ 20 27 ​ a k\geq\frac{20}{27}a , which is the condition k ≥ a ​ θ n − 1 k\geq a\theta_{n-1} , i.e. which is the condition (a) in the proof of (Theorem 1). Moreover,
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( 1 − β n ) ​ k < [ 1 2 ​ ( a + c ) − | a | ​ ( 1 + b 2 ) ] \left(1-\beta_{n}\right)k<\left[\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right)\right]
for each n ∈ ℕ n\in\mathbb{N} . Since k ≤ 0 k\leq 0 , then we impose the condition that ( 1 − b 2 ) ​ k ​ < 1 2 ​ ( a + c ) − | ​ a | ( 1 + b 2 ) \left(1-b_{2}\right)k<\frac{1}{2}(a+c)-|a|\left(1+b_{2}\right) , which is the condition (c) in the proof of (Theorem 1):
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k < 10 ⋅ [ a + c 2 + 19 ​ a 10 ] k<10\cdot\left[\frac{a+c}{2}+\frac{19a}{10}\right]
Since k ≥ a ​ ( 1 + b 1 ) ( 1 − b 1 ) k\geq a\frac{\left(1+b_{1}\right)}{\left(1-b_{1}\right)} , then
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k ≥ 11 9 ​ a k\geq\frac{11}{9}a
which is the condition (d) in the proof of (Theorem 1). Furthermore, because s > 0 s>0 , then
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19 6 ​ ε ​ a + 5 ​ c 9 > 0 \frac{19}{6\varepsilon}a+\frac{5c}{9}>0
which is the condition (e) in the proof of (Theorem 1). Now, let’s take a = − 10 a=-10 . Then c > 10 c>10 . The other inequalities become :
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k ≥ 20 27 ​ ( − 10 ) ⟹ k ≥ − 200 27 = − 7.4 k\geq\frac{20}{27}(-10)\Longrightarrow k\geq-\frac{200}{27}=-7.4
k ≥ 11 9 ​ a ⟹ k ≥ − 110 9 \displaystyle k\geq\frac{11}{9}a\Longrightarrow k\geq-\frac{110}{9}
k ≤ 10 ⋅ [ a 2 + c 2 + 19 ​ a 10 ] = 10 ⋅ [ c 2 − 24 ] . \displaystyle k\leq 0\cdot\left[\frac{a}{2}+\frac{c}{2}+\frac{19a}{10}\right]=0\cdot\left[\frac{c}{2}-24\right].
Since 5 ​ c 9 ≥ 1900 6 ​ ε \frac{5c}{9}\geq\frac{1900}{6\varepsilon} , then
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c ≥ 570 ε . c\geq\frac{570}{\varepsilon}.
On the other hand, we have that
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− 2 ​ c 9 < − 1900 ε − k . -\frac{2c}{9}<\frac{-1900}{\varepsilon}-k.
So, it follows that
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c ≥ 570 ε , k ≥ − 200 27 c\geq\frac{570}{\varepsilon},k\geq-\frac{200}{27}
and the following conditions hold:
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(1) ​ k < 10 ⋅ ( c 2 − 24 ) ​ and \displaystyle\text{ (1) }k<0\cdot\left(\frac{c}{2}-24\right)\text{ and }
(2) − 2 ​ c 9 < − 1900 ε − k . \displaystyle\text{ (2) }-\frac{2c}{9}<\frac{-1900}{\varepsilon}-k.
Also, taking ε = 1 3 ∈ ( 0 , 1 2 ) \varepsilon=\frac{1}{3}\in\left(0,\frac{1}{2}\right) , it follows that c ≥ 1710 c\geq 1710 . Now, condition (2) becomes k < 5700 − 2 ​ c 9 k<5700-\frac{2c}{9} . Since k < 0 k<0 , we put the condition that 5700 − 2 ​ c 9 ≥ 0 5700-\frac{2c}{9}\geq 0 , i.e. c ≥ 25650 c\geq 25650 , so that condition (2) is satisfied. Furthermore, condition (1) is k ≤ 5 ​ c − 240 k\leq 5c-240 . We put 5 ​ c − 240 > 0 5c-240>0 , so c > 48 c>48 . Taking c ≥ 570 c\geq 570 , condition (1) is satisfied, since k < 0 k<0 . Now, we can take c = 1800 c=1800 . So, let’s construct T T as a linear piecewise function, such as
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T ​ ( x ) = { x p 1 , x ∈ [ 0 , x 0 ) x p 2 , x ∈ [ x 0 , ∞ ) , T(x)=\left\{\begin{array}[]{ll}\frac{x}{p_{1}},&x\in\left[0,x_{0}\right)\\
\frac{x}{p_{2}},&x\in\left[x_{0},\infty\right)\end{array},\right.
where x 0 x_{0} is a discontinuity point such that T T is no a contraction mapping. Furthermore p 1 , p 2 > 0 p_{1},p_{2}>0 with p 1 ≠ p 2 p_{1}\neq p_{2} and p 1 , p 2 > 0 p_{1},p_{2}>0 . Now, denote
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I 1 := [ 0 , x 0 ) ​ and ​ I 2 := [ x 0 , ∞ ) . I_{1}:=\left[0,x_{0}\right)\text{ and }I_{2}:=\left[x_{0},\infty\right).
We have four cases two analyze the generalized contraction condition:
(I) When x , y ∈ I 1 x,y\in I_{1} , we get, by the contraction-type condition that:
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( c p 1 − k ) ​ | x − y | ≤ − p 1 − 1 p 1 ​ ( x + y ) ​ a \displaystyle\left(\frac{c}{p_{1}}-k\right)|x-y|\leq-\frac{p_{1}-1}{p_{1}}(x+y)a
( c p 1 − k ) ≤ ( − p 1 + 1 p 1 ) ​ a ⟹ c − p 1 ​ k ≤ ( 1 − p 1 ) ​ a \displaystyle\left(\frac{c}{p_{1}}-k\right)\leq\left(\frac{-p_{1}+1}{p_{1}}\right)a\Longrightarrow c-p_{1}k\leq\left(1-p_{1}\right)a
(II) When x ∈ I 1 x\in I_{1} and y ∈ I 2 y\in I_{2} , by contraction-type condition we get:
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c ​ | x p 1 − y p 2 | − k ​ | x − y | = c p 1 ​ p 2 ​ | p 2 ​ x − p 1 ​ y | − k ​ | x − y | c\left|\frac{x}{p_{1}}-\frac{y}{p_{2}}\right|-k|x-y|=\frac{c}{p_{1}p_{2}}\left|p_{2}x-p_{1}y\right|-k|x-y|
≤ ( c p 1 ​ x + c p 2 ​ y ) − k ​ x − k ​ y ⟹ \displaystyle\leq\left(\frac{c}{p_{1}}x+\frac{c}{p_{2}}y\right)-kx-ky\Longrightarrow
( c p 1 − k ) ​ x + ( c p 2 − k ) ​ y ≤ − ( p 1 − 1 ) ​ x p 1 − ( p 2 − 1 ) ​ y p 2 \displaystyle\left(\frac{c}{p_{1}}-k\right)x+\left(\frac{c}{p_{2}}-k\right)y\leq-\frac{\left(p_{1}-1\right)x}{p_{1}}-\frac{\left(p_{2}-1\right)y}{p_{2}}
⟹ { c p 1 − k + p 1 − 1 p 1 ≤ 0 c p 2 − k + p 2 − 1 p 2 ≤ 0 \displaystyle\Longrightarrow\left\{\begin{array}[]{l}\frac{c}{p_{1}}-k+\frac{p_{1}-1}{p_{1}}\leq 0\\
\frac{c}{p_{2}}-k+\frac{p_{2}-1}{p_{2}}\leq 0\end{array}\right.
(III) When x ∈ I 2 x\in I_{2} and y ∈ I 1 y\in I_{1} : it’s analogous to the case above.
(IV) When x , y ∈ I 2 x,y\in I_{2} :
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Like in case (I), we get that c − p 2 ​ k ≤ 1 − p 2 c-p_{2}k\leq 1-p_{2} . Now, we have shown that k ∈ [ − 200 27 , 0 ) = [ − 7.4 , 0 ) k\in\left[-\frac{200}{27},0\right)=[-7.4,0) . By the conditions above, we have that
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1800 − p 1 ​ k ≤ ( 1 − p 1 ) ​ ( − 10 ) ⟹ 1810 ≤ p 1 ​ ( 10 + k ) ⟹ k ≥ − 10 1800-p_{1}k\leq\left(1-p_{1}\right)(-10)\Longrightarrow 1810\leq p_{1}(10+k)\Longrightarrow k\geq-10
Let’s take k = − 1 2 k=-\frac{1}{2} . Then 1810 ≤ p 1 ​ ( 10 − 1 2 ) 1810\leq p_{1}\left(10-\frac{1}{2}\right) , so we can take p 1 ≥ 3620 19 = 190.52 ​ In p_{1}\geq\frac{3620}{19}=190.52\mathrm{In} a similar way, we obtain that p 2 ≥ 190.52 p_{2}\geq 190.52 . So, we can take, for example p 1 = 194 p_{1}=194 and p 2 = 205 p_{2}=205 .
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•
Now, let’s construct the sequences ( α n \alpha_{n} ) and ( β n \beta_{n} ):
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We know that
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α n ∈ [ a 1 , a 2 ] = [ 1 10 , 2 5 ] \alpha_{n}\in\left[a_{1},a_{2}\right]=\left[\frac{1}{10},\frac{2}{5}\right]
and that
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β n ∈ [ b 1 , b 2 ] = [ 1 10 , 9 10 ] . \beta_{n}\in\left[b_{1},b_{2}\right]=\left[\frac{1}{10},\frac{9}{10}\right].
Let’s take α n = 1 n + w + a 1 + a 2 2 ⟹ α n > 0 \alpha_{n}=\frac{1}{n+w}+\frac{a_{1}+a_{2}}{2}\Longrightarrow\alpha_{n}>0 , for each n ∈ ℕ n\in\mathbb{N} . Then α n > a 1 \alpha_{n}>a_{1} , which means that 1 n + w + a 1 + a 2 2 > 1 n + w + a 1 > a 1 \frac{1}{n+w}+\frac{a_{1}+a_{2}}{2}>\frac{1}{n+w}+a_{1}>a_{1} . Also, α n < a 2 \alpha_{n}<a_{2} means that 1 n + w + a 1 + a 2 2 < a 2 \frac{1}{n+w}+\frac{a_{1}+a_{2}}{2}<a_{2} , i.e.
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1 n + w < a 2 − a 1 2 = 3 20 ⟹ n + w > 20 3 , for each ​ n ∈ ℕ . \frac{1}{n+w}<\frac{a_{2}-a_{1}}{2}=\frac{3}{20}\Longrightarrow n+w>\frac{20}{3},\text{ for each }n\in\mathbb{N}.
Taking n = 0 n=0 and observing that the sequence ( n + w ) n ∈ ℕ (n+w)_{n\in\mathbb{N}} is increasing, we get that w > 20 3 w>\frac{20}{3} . So, we can take
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w = 21 3 = 7 ⟹ α n = 1 n + 7 + 1 4 w=\frac{21}{3}=7\Longrightarrow\alpha_{n}=\frac{1}{n+7}+\frac{1}{4}
In a similar manner, let β n = 1 n + w ′ + b 1 + b 2 2 \beta_{n}=\frac{1}{n+w^{\prime}}+\frac{b_{1}+b_{2}}{2} . Analogous as in the previous case: β n > b 1 \beta_{n}>b_{1} . Furthermore: β n < b 2 \beta_{n}<b_{2} means that
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1 n + w ′ < b 2 − b 1 2 ⟹ n + w ′ > 5 2 \frac{1}{n+w^{\prime}}<\frac{b_{2}-b_{1}}{2}\Longrightarrow n+w^{\prime}>\frac{5}{2}
As before, the sequence ( n + w ′ ) n ∈ ℕ \left(n+w^{\prime}\right)_{n\in\mathbb{N}} is increasing. Taking n = 0 n=0 , it follows that w ′ > 5 2 w^{\prime}>\frac{5}{2} and taking w ′ = 6 2 = 3 w^{\prime}=\frac{6}{2}=3 , we get that β n = 1 n + 3 + 1 2 \beta_{n}=\frac{1}{n+3}+\frac{1}{2} . In conclusion, we have constructed the operator T T as:
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T ​ ( x ) = { x 194 , x ∈ [ 0 , x 0 ) x 250 , x ∈ [ x 0 , ∞ ) , T(x)=\left\{\begin{array}[]{ll}\frac{x}{194},&x\in\left[0,x_{0}\right)\\
\frac{x}{250},&x\in\left[x_{0},\infty\right)\end{array},\right.
which verifies that
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1800 ​ d ​ ( T ​ x , T ​ y ) + 1 2 ​ d ​ ( x , y ) ≤ 10 ​ [ d ​ ( x , T ​ x ) + d ​ ( y , T ​ y ) ] , for each ​ x , y ∈ [ 0 , ∞ ) . 1800d(Tx,Ty)+\frac{1}{2}d(x,y)\leq 10[d(x,Tx)+d(y,Ty)],\text{ for each }x,y\in[0,\infty).
Finally, the sequences ( α n \alpha_{n} ) and ( β n \beta_{n} ) from the Ishikawa iteration are:
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α n = 1 n + 7 + 1 4 ​ and ​ β n = 1 n + 3 + 1 2 . \alpha_{n}=\frac{1}{n+7}+\frac{1}{4}\text{ and }\beta_{n}=\frac{1}{n+3}+\frac{1}{2}.