The convergence of mean value iteration for a family of maps

Stefan Soltuz
(original), paper

Abstract


We consider a mean value iteration for a family of functions, which corresponds to the Mann iteration with \(\lim_{n\rightarrow \infty}an\neq0\). We prove convergence results for this iteration when applied to strongly pseudocontractive or strongly accretive maps.

Authors

Stefan M. Soltuz
Tiberiu Popoviciu Institute of Numerical Analysis

B.E. Rhoades
Department of Mathematics, Indiana University, Bloomington, IN 47405-7106,USA

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B.E. Rhoades, Ş.M. Şoltuz , The convergence of mean value iteration for a family of maps, Int. J. Math. Math. Sci. 2005: 21, 3479-3485. (published in 28 dec. 2005)

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International Journal of Mathematics and Mathematical Sciences

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Wiley

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0161-1712

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1687-0425

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[1] C. E. Chidume and S. A. Mutangadura, An example of the Mann iteration method for Lipschitz pseudocontractions, Proc. Amer. Math. Soc. 129 (2001), no. 8, 2359–2363.
[2] S. Ishikawa, Fixed points by a new iteration method, Proc. Amer. Math. Soc. 44 (1974), no. 1,147–150.
[3] W. R. Mann, Mean value methods in iteration, Proc. Amer. Math. Soc. 4 (1953), 506–510.
[4] C. H. Morales and J. S. Jung, Convergence of paths for pseudocontractive mappings in Banachspaces, Proc. Amer. Math. Soc. 128 (2000), no. 11, 3411–3419.
[5] B. E. Rhoades and S. M. Soltuz, The equivalence of Mann iteration and Ishikawa iteration fornon-Lipschitzian operators, Int. J. Math. Math. Sci. 2003 (2003), no. 42, 2645–2651.
[6] , Mean value iteration for a family of functions, to appear in Nonlinear Funct. Anal.Appl.
[7] S. M. Soltuz, Some sequences supplied by inequalities and their applications, Rev. Anal. Numer.Theor. Approx. 29 (2000), no. 2, 207–212.

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2005-Soltuz-IJMMS-The-convergence-of-mean-value

THE CONVERGENCE OF MEAN VALUE ITERATION FOR A FAMILY OF MAPS

B. E. RHOADES AND ŞTEFAN M. ŞOLTUZ

Received 3 January 2005 and in revised form 6 September 2005
We consider a mean value iteration for a family of functions, which corresponds to the Mann iteration with lim n → ∞ α n ≠ 0 lim n → ∞   α n ≠ 0 lim_(n rarr oo)alpha_(n)!=0\lim _{n \rightarrow \infty} \alpha_{n} \neq 0limn→∞αn≠0. We prove convergence results for this iteration when applied to strongly pseudocontractive or strongly accretive maps.

1. Introduction

Let X X XXX be a real Banach space. The map J : X → 2 X ∗ J : X → 2 X ∗ J:X rarr2^(X^(**))J: X \rightarrow 2^{X^{*}}J:X→2X∗ given by
(1.1) J x := { f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ } , ∀ x ∈ X (1.1) J x := f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ , ∀ x ∈ X {:(1.1)Jx:={f inX^(**):(:x,f:)=||x||^(2),||f||=||x||}","quad AA x in X:}\begin{equation*} J x:=\left\{f \in X^{*}:\langle x, f\rangle=\|x\|^{2},\|f\|=\|x\|\right\}, \quad \forall x \in X \tag{1.1} \end{equation*}(1.1)Jx:={f∈X∗:⟨x,f⟩=‖x‖2,‖f‖=‖x‖},∀x∈X
is called the normalized duality mapping. Let y ∈ X y ∈ X y in Xy \in Xy∈X and j ( y ) ∈ J ( y ) j ( y ) ∈ J ( y ) j(y)in J(y)j(y) \in J(y)j(y)∈J(y); note that ⟨ ⋅ , j ( y ) ⟩ ⟨ ⋅ , j ( y ) ⟩ (:*,j(y):)\langle\cdot, j(y)\rangle⟨⋅,j(y)⟩ is a Lipschitzian map.
Remark 1.1. The above J J JJJ satisfies
(1.2) ⟨ x , j ( y ) ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x ∈ X , ∀ j ( y ) ∈ J ( y ) . (1.2) ⟨ x , j ( y ) ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x ∈ X , ∀ j ( y ) ∈ J ( y ) . {:(1.2)(:x","j(y):) <= ||x||||y||","quad AA x in X","AA j(y)in J(y).:}\begin{equation*} \langle x, j(y)\rangle \leq\|x\|\|y\|, \quad \forall x \in X, \forall j(y) \in J(y) . \tag{1.2} \end{equation*}(1.2)⟨x,j(y)⟩≤‖x‖‖y‖,∀x∈X,∀j(y)∈J(y).
Definition 1.2. Let B B BBB be a nonempty subset of X X XXX. The map T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is strongly pseudocontractive if there exist k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) and j ( x − y ) ∈ J ( x − y ) j ( x − y ) ∈ J ( x − y ) j(x-y)in J(x-y)j(x-y) \in J(x-y)j(x−y)∈J(x−y) such that
(1.3) ⟨ T x − T y , j ( x − y ) ⟩ ≤ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . (1.3) ⟨ T x − T y , j ( x − y ) ⟩ ≤ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . {:(1.3)(:Tx-Ty","j(x-y):) <= k||x-y||^(2)","quad AA x","y in B.:}\begin{equation*} \langle T x-T y, j(x-y)\rangle \leq k\|x-y\|^{2}, \quad \forall x, y \in B . \tag{1.3} \end{equation*}(1.3)⟨Tx−Ty,j(x−y)⟩≤k‖x−y‖2,∀x,y∈B.
A map S : B → B S : B → B S:B rarr BS: B \rightarrow BS:B→B is called strongly accretive if there exist k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) and j ( x − y ) ∈ J ( x − y ) j ( x − y ) ∈ J ( x − y ) j(x-y)in J(x-y)j(x-y) \in J(x-y)j(x−y)∈J(x−y) such that
(1.4) ⟨ S x − S y , j ( x − y ) ⟩ ≥ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . (1.4) ⟨ S x − S y , j ( x − y ) ⟩ ≥ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . {:(1.4)(:Sx-Sy","j(x-y):) >= k||x-y||^(2)","quad AA x","y in B.:}\begin{equation*} \langle S x-S y, j(x-y)\rangle \geq k\|x-y\|^{2}, \quad \forall x, y \in B . \tag{1.4} \end{equation*}(1.4)⟨Sx−Sy,j(x−y)⟩≥k‖x−y‖2,∀x,y∈B.
In (1.3), take k = 1 k = 1 k=1k=1k=1 to obtain a pseudocontractive map. In (1.4), take k = 0 k = 0 k=0k=0k=0 to obtain an accretive map.
Let B B BBB be a nonempty and convex subset of X , T : B → B X , T : B → B X,T:B rarr BX, T: B \rightarrow BX,T:B→B and x 0 , u 0 ∈ B x 0 , u 0 ∈ B x_(0),u_(0)in Bx_{0}, u_{0} \in Bx0,u0∈B. The Mann iteration (see [3]) is defined by
(1.5) u n + 1 = ( 1 − α n ) u n + α n T u n . (1.5) u n + 1 = 1 − α n u n + α n T u n . {:(1.5)u_(n+1)=(1-alpha_(n))u_(n)+alpha_(n)Tu_(n).:}\begin{equation*} u_{n+1}=\left(1-\alpha_{n}\right) u_{n}+\alpha_{n} T u_{n} . \tag{1.5} \end{equation*}(1.5)un+1=(1−αn)un+αnTun.
The Ishikawa iteration is defined (see [2]) by
x n + 1 = ( 1 − α n ) x n + α n T y n , (1.6) y n = ( 1 − β n ) x n + β n T x n , x n + 1 = 1 − α n x n + α n T y n , (1.6) y n = 1 − β n x n + β n T x n , {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Ty_(n)","],[(1.6)y_(n)=(1-beta_(n))x_(n)+beta_(n)Tx_(n)","]:}\begin{align*} x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T y_{n}, \\ y_{n} & =\left(1-\beta_{n}\right) x_{n}+\beta_{n} T x_{n}, \tag{1.6} \end{align*}xn+1=(1−αn)xn+αnTyn,(1.6)yn=(1−βn)xn+βnTxn,
where { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1) and { β n } ⊂ [ 0 , 1 ) β n ⊂ [ 0 , 1 ) {beta_(n)}sub[0,1)\left\{\beta_{n}\right\} \subset[0,1){βn}⊂[0,1).
Let s ≥ 2 s ≥ 2 s >= 2s \geq 2s≥2 be fixed. Let T i : B → B , 1 ≤ i ≤ s T i : B → B , 1 ≤ i ≤ s T_(i):B rarr B,1 <= i <= sT_{i}: B \rightarrow B, 1 \leq i \leq sTi:B→B,1≤i≤s, be a family of functions. We consider the following multistep procedure:
x n + 1 = ( 1 − α n ) x n + α n T 1 y n 1 (1.7) y n i = ( 1 − β n i ) x n + β n i T i + 1 y n i + 1 , i = 1 , … , s − 2 , y n s − 1 = ( 1 − β n s − 1 ) x n + β n s − 1 T s x n x n + 1 = 1 − α n x n + α n T 1 y n 1 (1.7) y n i = 1 − β n i x n + β n i T i + 1 y n i + 1 , i = 1 , … , s − 2 , y n s − 1 = 1 − β n s − 1 x n + β n s − 1 T s x n {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)T_(1)y_(n)^(1)],[(1.7)y_(n)^(i)=(1-beta_(n)^(i))x_(n)+beta_(n)^(i)T_(i+1)y_(n)^(i+1)","quad i=1","dots","s-2","],[y_(n)^(s-1)=(1-beta_(n)^(s-1))x_(n)+beta_(n)^(s-1)T_(s)x_(n)]:}\begin{gather*} x_{n+1}=\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T_{1} y_{n}^{1} \\ y_{n}^{i}=\left(1-\beta_{n}^{i}\right) x_{n}+\beta_{n}^{i} T_{i+1} y_{n}^{i+1}, \quad i=1, \ldots, s-2, \tag{1.7}\\ y_{n}^{s-1}=\left(1-\beta_{n}^{s-1}\right) x_{n}+\beta_{n}^{s-1} T_{s} x_{n} \end{gather*}xn+1=(1−αn)xn+αnT1yn1(1.7)yni=(1−βni)xn+βniTi+1yni+1,i=1,…,s−2,yns−1=(1−βns−1)xn+βns−1Tsxn
Let A , b ∈ ( 0 , 1 ) A , b ∈ ( 0 , 1 ) A,b in(0,1)A, b \in(0,1)A,b∈(0,1) be fixed. The sequence { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1) satisfies
(1.8) 0 < A ≤ α n ≤ b < 2 ( 1 − k ) , ∀ n ∈ N , (1.9) { β n i } ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 . (1.8) 0 < A ≤ α n ≤ b < 2 ( 1 − k ) , ∀ n ∈ N , (1.9) β n i ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 . {:[(1.8)0 < A <= alpha_(n) <= b < 2(1-k)","quad AA n inN","],[(1.9){beta_(n)^(i)}sub[0","1)","quad i=1","dots","s-1.]:}\begin{gather*} 0<A \leq \alpha_{n} \leq b<2(1-k), \quad \forall n \in \mathbb{N}, \tag{1.8}\\ \left\{\beta_{n}^{i}\right\} \subset[0,1), \quad i=1, \ldots, s-1 . \tag{1.9} \end{gather*}(1.8)0<A≤αn≤b<2(1−k),∀n∈N,(1.9){βni}⊂[0,1),i=1,…,s−1.
Let F ( T 1 , … , T s ) F T 1 , … , T s F(T_(1),dots,T_(s))F\left(T_{1}, \ldots, T_{s}\right)F(T1,…,Ts) denote the common fixed points set with respect to B B BBB for the family T 1 , … , T s T 1 , … , T s T_(1),dots,T_(s)T_{1}, \ldots, T_{s}T1,…,Ts. In this paper, we will prove convergence results for iteration (1.7), for strongly pseudocontractive (accretive) maps when { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} satisfies (1.8). These results improve the recently obtained results from [6], in which { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} and { β n } β n {beta_(n)}\left\{\beta_{n}\right\}{βn} converge to zero. We give two numerical examples in which iteration (1.7), when { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} satisfies (1.8), converges faster as in [6]. Note that, in both cases, iteration (1.7) converges faster than Ishikawa iteration.
Lemma 1.3 [4]. Let X X XXX be a real Banach space, and let J : X → 2 X ∗ J : X → 2 X ∗ J:X rarr2^(X^(**))J: X \rightarrow 2^{X^{*}}J:X→2X∗ be the duality mapping. Then for any given x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X,
(1.10) ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , j ( x + y ) ⟩ , ∀ x , y ∈ X , ∀ j ( x + y ) ∈ J ( x + y ) . (1.10) ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , j ( x + y ) ⟩ , ∀ x , y ∈ X , ∀ j ( x + y ) ∈ J ( x + y ) . {:(1.10)||x+y||^(2) <= ||x||^(2)+2(:y","j(x+y):)","quad AA x","y in X","AA j(x+y)in J(x+y).:}\begin{equation*} \|x+y\|^{2} \leq\|x\|^{2}+2\langle y, j(x+y)\rangle, \quad \forall x, y \in X, \forall j(x+y) \in J(x+y) . \tag{1.10} \end{equation*}(1.10)‖x+y‖2≤‖x‖2+2⟨y,j(x+y)⟩,∀x,y∈X,∀j(x+y)∈J(x+y).
Lemma 1.4 [7]. Let { a n } a n {a_(n)}\left\{a_{n}\right\}{an} be a nonnegative sequence which satisfies the inequality
(1.11) a n + 1 ≤ ( 1 − t ) a n + σ n , (1.11) a n + 1 ≤ ( 1 − t ) a n + σ n , {:(1.11)a_(n+1) <= (1-t)a_(n)+sigma_(n)",":}\begin{equation*} a_{n+1} \leq(1-t) a_{n}+\sigma_{n}, \tag{1.11} \end{equation*}(1.11)an+1≤(1−t)an+σn,
where t ∈ ( 0 , 1 ) t ∈ ( 0 , 1 ) t in(0,1)t \in(0,1)t∈(0,1) is fixed, lim n → ∞ σ n = 0 lim n → ∞   σ n = 0 lim_(n rarr oo)sigma_(n)=0\lim _{n \rightarrow \infty} \sigma_{n}=0limn→∞σn=0. Then lim n → ∞ a n = 0 lim n → ∞   a n = 0 lim_(n rarr oo)a_(n)=0\lim _{n \rightarrow \infty} a_{n}=0limn→∞an=0.

2. Main result

Theorem 2.1. Let s ≥ 2 s ≥ 2 s >= 2s \geq 2s≥2 be fixed, X X XXX a real Banach space, and B B BBB a nonempty, closed, convex subset of X X XXX. Let T 1 : B → B T 1 : B → B T_(1):B rarr BT_{1}: B \rightarrow BT1:B→B be a strongly pseudocontractive operator and T 2 , … , T s : B → B T 2 , … , T s : B → B T_(2),dots,T_(s):B rarr BT_{2}, \ldots, T_{s}: B \rightarrow BT2,…,Ts:B→B,
with T i ( B ) T i ( B ) T_(i)(B)T_{i}(B)Ti(B) bounded for all 1 ≤ i ≤ s 1 ≤ i ≤ s 1 <= i <= s1 \leq i \leq s1≤i≤s, such that F ( T 1 , … , T s ) ≠ ∅ F T 1 , … , T s ≠ ∅ F(T_(1),dots,T_(s))!=O/F\left(T_{1}, \ldots, T_{s}\right) \neq \varnothingF(T1,…,Ts)≠∅. If A , b ∈ ( 0 , 1 ) , { α n } ⊂ ( 0 , 1 ) A , b ∈ ( 0 , 1 ) , α n ⊂ ( 0 , 1 ) A,b in(0,1),{alpha_(n)}sub(0,1)A, b \in(0,1),\left\{\alpha_{n}\right\} \subset (0,1)A,b∈(0,1),{αn}⊂(0,1) satisfies (1.8), x 0 ∈ B x 0 ∈ B x_(0)in Bx_{0} \in Bx0∈B, and the following condition is satisfied:
(2.1) lim n → ∞ ‖ T 1 x n + 1 − T 1 y n 1 ‖ = 0 (2.1) lim n → ∞   T 1 x n + 1 − T 1 y n 1 = 0 {:(2.1)lim_(n rarr oo)||T_(1)x_(n+1)-T_(1)y_(n)^(1)||=0:}\begin{equation*} \lim _{n \rightarrow \infty}\left\|T_{1} x_{n+1}-T_{1} y_{n}^{1}\right\|=0 \tag{2.1} \end{equation*}(2.1)limn→∞‖T1xn+1−T1yn1‖=0
then iteration (1.7) converges to the unique common fixed point of T 1 , … , T s T 1 , … , T s T_(1),dots,T_(s)T_{1}, \ldots, T_{s}T1,…,Ts, which is the unique fixed point of T 1 T 1 T_(1)T_{1}T1.
Proof. Any common fixed point of T 1 , … , T s T 1 , … , T s T_(1),dots,T_(s)T_{1}, \ldots, T_{s}T1,…,Ts, in particular, is a fixed point of T 1 T 1 T_(1)T_{1}T1. However, T 1 T 1 T_(1)T_{1}T1 can have at most one fixed point since it is strongly pseudocontractive. Let x ∗ = F ( T 1 , … , T s ) x ∗ = F T 1 , … , T s x^(**)=F(T_(1),dots,T_(s))x^{*}=F\left(T_{1}, \ldots, T_{s}\right)x∗=F(T1,…,Ts). Denote
(2.2) M = sup n ∈ N { ‖ T 1 y n 1 ‖ , ‖ x 0 ‖ , ‖ x ∗ ‖ } (2.2) M = sup n ∈ N   T 1 y n 1 , x 0 , x ∗ {:(2.2)M=s u p_(n inN){||T_(1)y_(n)^(1)||,||x_(0)||,||x^(**)||}:}\begin{equation*} M=\sup _{n \in \mathbb{N}}\left\{\left\|T_{1} y_{n}^{1}\right\|,\left\|x_{0}\right\|,\left\|x^{*}\right\|\right\} \tag{2.2} \end{equation*}(2.2)M=supn∈N{‖T1yn1‖,‖x0‖,‖x∗‖}
Then if we assume ‖ x n ‖ ≤ M x n ≤ M ||x_(n)|| <= M\left\|x_{n}\right\| \leq M‖xn‖≤M, by
(2.3) ‖ x n + 1 ‖ ≤ ( 1 − α n ) ‖ x n ‖ + α n ‖ T 1 y n 1 ‖ ≤ M , (2.3) x n + 1 ≤ 1 − α n x n + α n T 1 y n 1 ≤ M , {:(2.3)||x_(n+1)|| <= (1-alpha_(n))||x_(n)||+alpha_(n)||T_(1)y_(n)^(1)|| <= M",":}\begin{equation*} \left\|x_{n+1}\right\| \leq\left(1-\alpha_{n}\right)\left\|x_{n}\right\|+\alpha_{n}\left\|T_{1} y_{n}^{1}\right\| \leq M, \tag{2.3} \end{equation*}(2.3)‖xn+1‖≤(1−αn)‖xn‖+αn‖T1yn1‖≤M,
we get ‖ x n + 1 ‖ ≤ M x n + 1 ≤ M ||x_(n+1)|| <= M\left\|x_{n+1}\right\| \leq M‖xn+1‖≤M.
From (1.2) and (1.10), with
x : = ( 1 − α n ) ( x n − x ∗ ) , (2.4) y := α n ( T 1 y n 1 − T 1 x ∗ ) , x + y = x n + 1 − x ∗ , x : = 1 − α n x n − x ∗ , (2.4) y := α n T 1 y n 1 − T 1 x ∗ , x + y = x n + 1 − x ∗ , {:[x:=(1-alpha_(n))(x_(n)-x^(**))","],[(2.4)y:=alpha_(n)(T_(1)y_(n)^(1)-T_(1)x^(**))","],[x+y=x_(n+1)-x^(**)","]:}\begin{align*} x: & =\left(1-\alpha_{n}\right)\left(x_{n}-x^{*}\right), \\ y:= & \alpha_{n}\left(T_{1} y_{n}^{1}-T_{1} x^{*}\right), \tag{2.4}\\ & x+y=x_{n+1}-x^{*}, \end{align*}x:=(1−αn)(xn−x∗),(2.4)y:=αn(T1yn1−T1x∗),x+y=xn+1−x∗,
we get
‖ x n + 1 − x ∗ ‖ 2 = ‖ ( 1 − α n ) ( x n − x ∗ ) + α n ( T 1 y n 1 − T 1 x ∗ ) ‖ 2 ≤ ( 1 − α n ) 2 ‖ x n − x ∗ ‖ 2 + 2 α n ⟨ T 1 y n 1 − T 1 x ∗ , j ( x n + 1 − x ∗ ) ⟩ = ( 1 − α n ) 2 ‖ x n − x ∗ ‖ 2 + 2 α n ⟨ T 1 x n + 1 − T 1 x ∗ , j ( x n + 1 − x ∗ ) ⟩ + 2 α n ⟨ T 1 y n 1 − T 1 x n + 1 , j ( x n + 1 − x ∗ ) ⟩ (2.5) ≤ ( 1 − α n ) 2 ‖ x n − x ∗ ‖ 2 + 2 α n k ‖ x n + 1 − x ∗ ‖ 2 + 2 α n ⟨ T 1 y n 1 − T 1 x n + 1 , j ( x n + 1 − x ∗ ) ⟩ ≤ ( 1 − α n ) 2 ‖ x n − x ∗ ‖ 2 + 2 α n k ‖ x n + 1 − x ∗ ‖ 2 + 2 α n ‖ T 1 y n 1 − T 1 x n + 1 ‖ ‖ x n + 1 − x ∗ ‖ ≤ ( 1 − α n ) 2 ‖ x n − x ∗ ‖ 2 + 2 α n k ‖ x n + 1 − x ∗ ‖ 2 + 4 α n ‖ T 1 y n 1 − T 1 x n + 1 ‖ M . x n + 1 − x ∗ 2 = 1 − α n x n − x ∗ + α n T 1 y n 1 − T 1 x ∗ 2 ≤ 1 − α n 2 x n − x ∗ 2 + 2 α n T 1 y n 1 − T 1 x ∗ , j x n + 1 − x ∗ = 1 − α n 2 x n − x ∗ 2 + 2 α n T 1 x n + 1 − T 1 x ∗ , j x n + 1 − x ∗ + 2 α n T 1 y n 1 − T 1 x n + 1 , j x n + 1 − x ∗ (2.5) ≤ 1 − α n 2 x n − x ∗ 2 + 2 α n k x n + 1 − x ∗ 2 + 2 α n T 1 y n 1 − T 1 x n + 1 , j x n + 1 − x ∗ ≤ 1 − α n 2 x n − x ∗ 2 + 2 α n k x n + 1 − x ∗ 2 + 2 α n T 1 y n 1 − T 1 x n + 1 x n + 1 − x ∗ ≤ 1 − α n 2 x n − x ∗ 2 + 2 α n k x n + 1 − x ∗ 2 + 4 α n T 1 y n 1 − T 1 x n + 1 M . {:[||x_(n+1)-x^(**)||^(2)=||(1-alpha_(n))(x_(n)-x^(**))+alpha_(n)(T_(1)y_(n)^(1)-T_(1)x^(**))||^(2)],[ <= (1-alpha_(n))^(2)||x_(n)-x^(**)||^(2)+2alpha_(n)(:T_(1)y_(n)^(1)-T_(1)x^(**),j(x_(n+1)-x^(**)):)],[=(1-alpha_(n))^(2)||x_(n)-x^(**)||^(2)+2alpha_(n)(:T_(1)x_(n+1)-T_(1)x^(**),j(x_(n+1)-x^(**)):)],[+2alpha_(n)(:T_(1)y_(n)^(1)-T_(1)x_(n+1),j(x_(n+1)-x^(**)):)],[(2.5) <= (1-alpha_(n))^(2)||x_(n)-x^(**)||^(2)+2alpha_(n)k||x_(n+1)-x^(**)||^(2)],[+2alpha_(n)(:T_(1)y_(n)^(1)-T_(1)x_(n+1),j(x_(n+1)-x^(**)):)],[ <= (1-alpha_(n))^(2)||x_(n)-x^(**)||^(2)+2alpha_(n)k||x_(n+1)-x^(**)||^(2)],[+2alpha_(n)||T_(1)y_(n)^(1)-T_(1)x_(n+1)||||x_(n+1)-x^(**)||],[ <= (1-alpha_(n))^(2)||x_(n)-x^(**)||^(2)+2alpha_(n)k||x_(n+1)-x^(**)||^(2)],[+4alpha_(n)||T_(1)y_(n)^(1)-T_(1)x_(n+1)||M.]:}\begin{align*} \left\|x_{n+1}-x^{*}\right\|^{2}= & \left\|\left(1-\alpha_{n}\right)\left(x_{n}-x^{*}\right)+\alpha_{n}\left(T_{1} y_{n}^{1}-T_{1} x^{*}\right)\right\|^{2} \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-x^{*}\right\|^{2}+2 \alpha_{n}\left\langle T_{1} y_{n}^{1}-T_{1} x^{*}, j\left(x_{n+1}-x^{*}\right)\right\rangle \\ = & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-x^{*}\right\|^{2}+2 \alpha_{n}\left\langle T_{1} x_{n+1}-T_{1} x^{*}, j\left(x_{n+1}-x^{*}\right)\right\rangle \\ & +2 \alpha_{n}\left\langle T_{1} y_{n}^{1}-T_{1} x_{n+1}, j\left(x_{n+1}-x^{*}\right)\right\rangle \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-x^{*}\right\|^{2}+2 \alpha_{n} k\left\|x_{n+1}-x^{*}\right\|^{2} \tag{2.5}\\ & +2 \alpha_{n}\left\langle T_{1} y_{n}^{1}-T_{1} x_{n+1}, j\left(x_{n+1}-x^{*}\right)\right\rangle \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-x^{*}\right\|^{2}+2 \alpha_{n} k\left\|x_{n+1}-x^{*}\right\|^{2} \\ & +2 \alpha_{n}\left\|T_{1} y_{n}^{1}-T_{1} x_{n+1}\right\|\left\|x_{n+1}-x^{*}\right\| \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-x^{*}\right\|^{2}+2 \alpha_{n} k\left\|x_{n+1}-x^{*}\right\|^{2} \\ & +4 \alpha_{n}\left\|T_{1} y_{n}^{1}-T_{1} x_{n+1}\right\| M . \end{align*}‖xn+1−x∗‖2=‖(1−αn)(xn−x∗)+αn(T1yn1−T1x∗)‖2≤(1−αn)2‖xn−x∗‖2+2αn⟨T1yn1−T1x∗,j(xn+1−x∗)⟩=(1−αn)2‖xn−x∗‖2+2αn⟨T1xn+1−T1x∗,j(xn+1−x∗)⟩+2αn⟨T1yn1−T1xn+1,j(xn+1−x∗)⟩(2.5)≤(1−αn)2‖xn−x∗‖2+2αnk‖xn+1−x∗‖2+2αn⟨T1yn1−T1xn+1,j(xn+1−x∗)⟩≤(1−αn)2‖xn−x∗‖2+2αnk‖xn+1−x∗‖2+2αn‖T1yn1−T1xn+1‖‖xn+1−x∗‖≤(1−αn)2‖xn−x∗‖2+2αnk‖xn+1−x∗‖2+4αn‖T1yn1−T1xn+1‖M.
Using (1.8), we obtain
(2.6) ( 1 − α n ) 2 ≤ 1 − 2 α n + α n b < 1 − 2 α n + α n 2 ( 1 − k ) = 1 − 2 α n k (2.6) 1 − α n 2 ≤ 1 − 2 α n + α n b < 1 − 2 α n + α n 2 ( 1 − k ) = 1 − 2 α n k {:(2.6)(1-alpha_(n))^(2) <= 1-2alpha_(n)+alpha_(n)b < 1-2alpha_(n)+alpha_(n)2(1-k)=1-2alpha_(n)k:}\begin{equation*} \left(1-\alpha_{n}\right)^{2} \leq 1-2 \alpha_{n}+\alpha_{n} b<1-2 \alpha_{n}+\alpha_{n} 2(1-k)=1-2 \alpha_{n} k \tag{2.6} \end{equation*}(2.6)(1−αn)2≤1−2αn+αnb<1−2αn+αn2(1−k)=1−2αnk
thus,
(2.7) ‖ x n + 1 − x ∗ ‖ 2 ≤ ( 1 − α n ) 2 1 − 2 α n k ‖ x n − x ∗ ‖ 2 + 4 α n M 1 − 2 α n k ‖ T 1 y n 1 − T 1 x n + 1 ‖ (2.7) x n + 1 − x ∗ 2 ≤ 1 − α n 2 1 − 2 α n k x n − x ∗ 2 + 4 α n M 1 − 2 α n k T 1 y n 1 − T 1 x n + 1 {:(2.7)||x_(n+1)-x^(**)||^(2) <= ((1-alpha_(n))^(2))/(1-2alpha_(n)k)||x_(n)-x^(**)||^(2)+(4alpha_(n)M)/(1-2alpha_(n)k)||T_(1)y_(n)^(1)-T_(1)x_(n+1)||:}\begin{equation*} \left\|x_{n+1}-x^{*}\right\|^{2} \leq \frac{\left(1-\alpha_{n}\right)^{2}}{1-2 \alpha_{n} k}\left\|x_{n}-x^{*}\right\|^{2}+\frac{4 \alpha_{n} M}{1-2 \alpha_{n} k}\left\|T_{1} y_{n}^{1}-T_{1} x_{n+1}\right\| \tag{2.7} \end{equation*}(2.7)‖xn+1−x∗‖2≤(1−αn)21−2αnk‖xn−x∗‖2+4αnM1−2αnk‖T1yn1−T1xn+1‖
The following inequality is satisfied:
( 1 − α n ) 2 1 − 2 α n k = ( 1 − α n ) 2 ( 1 − 2 α n k + 2 α n k ) 1 − 2 α n k = ( 1 − α n ) 2 ( 1 + 2 α n k 1 − 2 α n k ) = ( 1 − α n ) 2 + 2 α n k ( 1 − α n ) 2 1 − 2 α n k ≤ ( 1 − α n ) 2 + 2 α n k ≤ 1 − 2 α n + α n b + 2 α n k (2.8) = 1 − ( 2 ( 1 − k ) − b ) α n ≤ 1 − ( 2 ( 1 − k ) − b ) A 1 − α n 2 1 − 2 α n k = 1 − α n 2 1 − 2 α n k + 2 α n k 1 − 2 α n k = 1 − α n 2 1 + 2 α n k 1 − 2 α n k = 1 − α n 2 + 2 α n k 1 − α n 2 1 − 2 α n k ≤ 1 − α n 2 + 2 α n k ≤ 1 − 2 α n + α n b + 2 α n k (2.8) = 1 − ( 2 ( 1 − k ) − b ) α n ≤ 1 − ( 2 ( 1 − k ) − b ) A {:[((1-alpha_(n))^(2))/(1-2alpha_(n)k)=((1-alpha_(n))^(2)(1-2alpha_(n)k+2alpha_(n)k))/(1-2alpha_(n)k)=(1-alpha_(n))^(2)(1+(2alpha_(n)k)/(1-2alpha_(n)k))],[=(1-alpha_(n))^(2)+(2alpha_(n)k(1-alpha_(n))^(2))/(1-2alpha_(n)k) <= (1-alpha_(n))^(2)+2alpha_(n)k <= 1-2alpha_(n)+alpha_(n)b+2alpha_(n)k],[(2.8)=1-(2(1-k)-b)alpha_(n) <= 1-(2(1-k)-b)A]:}\begin{align*} \frac{\left(1-\alpha_{n}\right)^{2}}{1-2 \alpha_{n} k} & =\frac{\left(1-\alpha_{n}\right)^{2}\left(1-2 \alpha_{n} k+2 \alpha_{n} k\right)}{1-2 \alpha_{n} k}=\left(1-\alpha_{n}\right)^{2}\left(1+\frac{2 \alpha_{n} k}{1-2 \alpha_{n} k}\right) \\ & =\left(1-\alpha_{n}\right)^{2}+\frac{2 \alpha_{n} k\left(1-\alpha_{n}\right)^{2}}{1-2 \alpha_{n} k} \leq\left(1-\alpha_{n}\right)^{2}+2 \alpha_{n} k \leq 1-2 \alpha_{n}+\alpha_{n} b+2 \alpha_{n} k \\ & =1-(2(1-k)-b) \alpha_{n} \leq 1-(2(1-k)-b) A \tag{2.8} \end{align*}(1−αn)21−2αnk=(1−αn)2(1−2αnk+2αnk)1−2αnk=(1−αn)2(1+2αnk1−2αnk)=(1−αn)2+2αnk(1−αn)21−2αnk≤(1−αn)2+2αnk≤1−2αn+αnb+2αnk(2.8)=1−(2(1−k)−b)αn≤1−(2(1−k)−b)A
Substituting (2.6) and (2.8) into (2.7), we obtain
(2.9) ‖ x n + 1 − x ∗ ‖ 2 ≤ ( 1 − ( 2 ( 1 − k ) − b ) A ) ‖ x n − x ∗ ‖ 2 + 4 b M 1 − 2 b k ‖ T 1 y n 1 − T 1 x n + 1 ‖ (2.9) x n + 1 − x ∗ 2 ≤ ( 1 − ( 2 ( 1 − k ) − b ) A ) x n − x ∗ 2 + 4 b M 1 − 2 b k T 1 y n 1 − T 1 x n + 1 {:(2.9)||x_(n+1)-x^(**)||^(2) <= (1-(2(1-k)-b)A)||x_(n)-x^(**)||^(2)+(4bM)/(1-2bk)||T_(1)y_(n)^(1)-T_(1)x_(n+1)||:}\begin{equation*} \left\|x_{n+1}-x^{*}\right\|^{2} \leq(1-(2(1-k)-b) A)\left\|x_{n}-x^{*}\right\|^{2}+\frac{4 b M}{1-2 b k}\left\|T_{1} y_{n}^{1}-T_{1} x_{n+1}\right\| \tag{2.9} \end{equation*}(2.9)‖xn+1−x∗‖2≤(1−(2(1−k)−b)A)‖xn−x∗‖2+4bM1−2bk‖T1yn1−T1xn+1‖
Set
a n := ‖ x n − x ∗ ‖ 2 (2.10) t := ( 2 ( 1 − k ) − b ) A ∈ ( 0 , 1 ) σ n := 4 b M 1 − 2 b k ‖ T 1 y n 1 − T 1 x n + 1 ‖ a n := x n − x ∗ 2 (2.10) t := ( 2 ( 1 − k ) − b ) A ∈ ( 0 , 1 ) σ n := 4 b M 1 − 2 b k T 1 y n 1 − T 1 x n + 1 {:[a_(n):=||x_(n)-x^(**)||^(2)],[(2.10)t:=(2(1-k)-b)A in(0","1)],[sigma_(n):=(4bM)/(1-2bk)||T_(1)y_(n)^(1)-T_(1)x_(n+1)||]:}\begin{gather*} a_{n}:=\left\|x_{n}-x^{*}\right\|^{2} \\ t:=(2(1-k)-b) A \in(0,1) \tag{2.10}\\ \sigma_{n}:=\frac{4 b M}{1-2 b k}\left\|T_{1} y_{n}^{1}-T_{1} x_{n+1}\right\| \end{gather*}an:=‖xn−x∗‖2(2.10)t:=(2(1−k)−b)A∈(0,1)σn:=4bM1−2bk‖T1yn1−T1xn+1‖
From (2.1), we know that lim n → ∞ σ n = 0 lim n → ∞   σ n = 0 lim_(n rarr oo)sigma_(n)=0\lim _{n \rightarrow \infty} \sigma_{n}=0limn→∞σn=0; all the assumptions of Lemma 1.4 are fulfilled and consequently we have lim n → ∞ ‖ x n − x ∗ ‖ = 0 lim n → ∞   x n − x ∗ = 0 lim_(n rarr oo)||x_(n)-x^(**)||=0\lim _{n \rightarrow \infty}\left\|x_{n}-x^{*}\right\|=0limn→∞‖xn−x∗‖=0.
In Theorem 2.1, { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} does not converge to zero, while in [6], { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} converges to zero.
Theorem 2.2 [6]. Let s ≥ 2 s ≥ 2 s >= 2s \geq 2s≥2 be fixed, X X XXX a real Banach space with a uniformly convex dual, and B B BBB a nonempty, closed, convex subset of X X XXX. Let T 1 : B → B T 1 : B → B T_(1):B rarr BT_{1}: B \rightarrow BT1:B→B be a strongly pseudocontractive operator and T 2 , … , T s : B → B T 2 , … , T s : B → B T_(2),dots,T_(s):B rarr BT_{2}, \ldots, T_{s}: B \rightarrow BT2,…,Ts:B→B, with T i ( B ) T i ( B ) T_(i)(B)T_{i}(B)Ti(B) bounded for all 1 ≤ i ≤ s 1 ≤ i ≤ s 1 <= i <= s1 \leq i \leq s1≤i≤s, such that F ( T 1 , … , T s ) ≠ ∅ F T 1 , … , T s ≠ ∅ F(T_(1),dots,T_(s))!=O/F\left(T_{1}, \ldots, T_{s}\right) \neq \varnothingF(T1,…,Ts)≠∅. If { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1) satisfies lim n → ∞ α n = 0 , ∑ n = 1 ∞ α n = + ∞ lim n → ∞   α n = 0 , ∑ n = 1 ∞   α n = + ∞ lim_(n rarr oo)alpha_(n)=0,sum_(n=1)^(oo)alpha_(n)=+oo\lim _{n \rightarrow \infty} \alpha_{n}=0, \sum_{n=1}^{\infty} \alpha_{n}=+\inftylimn→∞αn=0,∑n=1∞αn=+∞, and { β n i } ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 β n i ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 {beta_(n)^(i)}sub[0,1),i=1,dots,s-1\left\{\beta_{n}^{i}\right\} \subset [0,1), i=1, \ldots, s-1{βni}⊂[0,1),i=1,…,s−1, satisfy lim n → ∞ β n 1 = 0 lim n → ∞   β n 1 = 0 lim_(n rarr oo)beta_(n)^(1)=0\lim _{n \rightarrow \infty} \beta_{n}^{1}=0limn→∞βn1=0, then iteration (1.7) converges to a fixed point of T 1 , … , T s T 1 , … , T s T_(1),dots,T_(s)T_{1}, \ldots, T_{s}T1,…,Ts.
The Banach space in Theorem 2.1 contains no restrictions.

3. Further results

Denote by I I III the identity map.
Remark 3.1. Let T , S : X → X T , S : X → X T,S:X rarr XT, S: X \rightarrow XT,S:X→X and let f ∈ X f ∈ X f in Xf \in Xf∈X be given. Then,
(i) a fixed point for the map T x = f + ( I − S ) x T x = f + ( I − S ) x Tx=f+(I-S)xT x=f+(I-S) xTx=f+(I−S)x, for all x ∈ X x ∈ X x in Xx \in Xx∈X, is a solution for S x = f S x = f Sx=fS x=fSx=f;
(ii) a fixed point for T x = f − S x T x = f − S x Tx=f-SxT x=f-S xTx=f−Sx is a solution for x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f.
Remark 3.2 [5]. The following are true.
(i) The operator T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X is a (strongly) pseudocontractive map if and only if ( I − T ) : X → X I − T ) : X → X I-T):X rarr XI- T): X \rightarrow XI−T):X→X is (strongly) accretive.
(ii) If S : X → X S : X → X S:X rarr XS: X \rightarrow XS:X→X is an accretive map, then T = f − S : X → X T = f − S : X → X T=f-S:X rarr XT=f-S: X \rightarrow XT=f−S:X→X is a strongly pseudocontractive map.
We consider iteration (1.7), with T i x = f i + ( I − S i ) x , 1 ≤ i ≤ s T i x = f i + I − S i x , 1 ≤ i ≤ s T_(i)x=f_(i)+(I-S_(i))x,1 <= i <= sT_{i} x=f_{i}+\left(I-S_{i}\right) x, 1 \leq i \leq sTix=fi+(I−Si)x,1≤i≤s and s ≥ 2 , { α n } ⊂ ( 0 , 1 ) s ≥ 2 , α n ⊂ ( 0 , 1 ) s >= 2,{alpha_(n)}sub(0,1)s \geq 2,\left\{\alpha_{n}\right\} \subset(0,1)s≥2,{αn}⊂(0,1), { β n i } ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 β n i ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 {beta_(n)^(i)}sub[0,1),i=1,dots,s-1\left\{\beta_{n}^{i}\right\} \subset[0,1), i=1, \ldots, s-1{βni}⊂[0,1),i=1,…,s−1 satisfying (1.8):
x n + 1 = ( 1 − α n ) x n + α n ( f 1 + ( I − S 1 ) y n 1 ) (3.1) y n i = ( 1 − β n i ) x n + β n i ( f i + 1 + ( I − S i + 1 ) y n i + 1 ) , i = 1 , … , s − 2 y n s − 1 = ( 1 − β n s − 1 ) x n + β n s − 1 ( f s − 1 + ( I − S s ) x n ) x n + 1 = 1 − α n x n + α n f 1 + I − S 1 y n 1 (3.1) y n i = 1 − β n i x n + β n i f i + 1 + I − S i + 1 y n i + 1 , i = 1 , … , s − 2 y n s − 1 = 1 − β n s − 1 x n + β n s − 1 f s − 1 + I − S s x n {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)(f_(1)+(I-S_(1))y_(n)^(1))],[(3.1)y_(n)^(i)=(1-beta_(n)^(i))x_(n)+beta_(n)^(i)(f_(i+1)+(I-S_(i+1))y_(n)^(i+1))","quad i=1","dots","s-2],[y_(n)^(s-1)=(1-beta_(n)^(s-1))x_(n)+beta_(n)^(s-1)(f_(s-1)+(I-S_(s))x_(n))]:}\begin{gather*} x_{n+1}=\left(1-\alpha_{n}\right) x_{n}+\alpha_{n}\left(f_{1}+\left(I-S_{1}\right) y_{n}^{1}\right) \\ y_{n}^{i}=\left(1-\beta_{n}^{i}\right) x_{n}+\beta_{n}^{i}\left(f_{i+1}+\left(I-S_{i+1}\right) y_{n}^{i+1}\right), \quad i=1, \ldots, s-2 \tag{3.1}\\ y_{n}^{s-1}=\left(1-\beta_{n}^{s-1}\right) x_{n}+\beta_{n}^{s-1}\left(f_{s-1}+\left(I-S_{s}\right) x_{n}\right) \end{gather*}xn+1=(1−αn)xn+αn(f1+(I−S1)yn1)(3.1)yni=(1−βni)xn+βni(fi+1+(I−Si+1)yni+1),i=1,…,s−2yns−1=(1−βns−1)xn+βns−1(fs−1+(I−Ss)xn)
Theorem 2.1, Remark 3.1(i), and Remark 3.2(i) lead to the following result.
Corollary 3.3. Let s ≥ 2 s ≥ 2 s >= 2s \geq 2s≥2 be fixed, X X XXX a real Banach space, and S 1 : X → X S 1 : X → X S_(1):X rarr XS_{1}: X \rightarrow XS1:X→X a strongly accretive operator, S 2 , … , S s : X → X S 2 , … , S s : X → X S_(2),dots,S_(s):X rarr XS_{2}, \ldots, S_{s}: X \rightarrow XS2,…,Ss:X→X, such that the equations S i x = f i , 1 ≤ i ≤ s S i x = f i , 1 ≤ i ≤ s S_(i)x=f_(i),1 <= i <= sS_{i} x=f_{i}, 1 \leq i \leq sSix=fi,1≤i≤s, have a common solution and T i ( X ) , 1 ≤ i ≤ s T i ( X ) , 1 ≤ i ≤ s T_(i)(X),1 <= i <= sT_{i}(X), 1 \leq i \leq sTi(X),1≤i≤s, are bounded. If A , b ∈ ( 0 , 1 ) , { α n } ⊂ ( 0 , 1 ) A , b ∈ ( 0 , 1 ) , α n ⊂ ( 0 , 1 ) A,b in(0,1),{alpha_(n)}sub(0,1)A, b \in(0,1),\left\{\alpha_{n}\right\} \subset(0,1)A,b∈(0,1),{αn}⊂(0,1) satisfies (1.8), and condition (2.1) is satisfied, then iteration (3.1) converges to a common solution of S i x = f i S i x = f i S_(i)x=f_(i)S_{i} x=f_{i}Six=fi, 1 ≤ i ≤ s 1 ≤ i ≤ s 1 <= i <= s1 \leq i \leq s1≤i≤s.
We consider iteration (1.7), with T i x = f i − S i x , 1 ≤ i ≤ s T i x = f i − S i x , 1 ≤ i ≤ s T_(i)x=f_(i)-S_(i)x,1 <= i <= sT_{i} x=f_{i}-S_{i} x, 1 \leq i \leq sTix=fi−Six,1≤i≤s, and s ≥ 2 , { α n } ⊂ ( 0 , 1 ) s ≥ 2 , α n ⊂ ( 0 , 1 ) s >= 2,{alpha_(n)}sub(0,1)s \geq 2,\left\{\alpha_{n}\right\} \subset(0,1)s≥2,{αn}⊂(0,1), { β n i } ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 β n i ⊂ [ 0 , 1 ) , i = 1 , … , s − 1 {beta_(n)^(i)}sub[0,1),i=1,dots,s-1\left\{\beta_{n}^{i}\right\} \subset[0,1), i=1, \ldots, s-1{βni}⊂[0,1),i=1,…,s−1, satisfying (1.8):
x n + 1 = ( 1 − α n ) x n + α n ( f 1 − S 1 y n 1 ) (3.2) y n i = ( 1 − β n i ) x n + β n i ( f i + 1 − S i + 1 y n i + 1 ) , i = 1 , … , s − 2 y n s − 1 = ( 1 − β n s − 1 ) x n + β n s − 1 ( f s − 1 − S s x n ) x n + 1 = 1 − α n x n + α n f 1 − S 1 y n 1 (3.2) y n i = 1 − β n i x n + β n i f i + 1 − S i + 1 y n i + 1 , i = 1 , … , s − 2 y n s − 1 = 1 − β n s − 1 x n + β n s − 1 f s − 1 − S s x n {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)(f_(1)-S_(1)y_(n)^(1))],[(3.2)y_(n)^(i)=(1-beta_(n)^(i))x_(n)+beta_(n)^(i)(f_(i+1)-S_(i+1)y_(n)^(i+1))","quad i=1","dots","s-2],[y_(n)^(s-1)=(1-beta_(n)^(s-1))x_(n)+beta_(n)^(s-1)(f_(s-1)-S_(s)x_(n))]:}\begin{gather*} x_{n+1}=\left(1-\alpha_{n}\right) x_{n}+\alpha_{n}\left(f_{1}-S_{1} y_{n}^{1}\right) \\ y_{n}^{i}=\left(1-\beta_{n}^{i}\right) x_{n}+\beta_{n}^{i}\left(f_{i+1}-S_{i+1} y_{n}^{i+1}\right), \quad i=1, \ldots, s-2 \tag{3.2}\\ y_{n}^{s-1}=\left(1-\beta_{n}^{s-1}\right) x_{n}+\beta_{n}^{s-1}\left(f_{s-1}-S_{s} x_{n}\right) \end{gather*}xn+1=(1−αn)xn+αn(f1−S1yn1)(3.2)yni=(1−βni)xn+βni(fi+1−Si+1yni+1),i=1,…,s−2yns−1=(1−βns−1)xn+βns−1(fs−1−Ssxn)
Theorem 2.1, Remark 3.1(ii), and Remark 3.2(ii) lead to the following result.
Corollary 3.4. Let s ≥ 2 s ≥ 2 s >= 2s \geq 2s≥2 be fixed, X X XXX a real Banach space, and S 1 : X → X S 1 : X → X S_(1):X rarr XS_{1}: X \rightarrow XS1:X→X an accretive operator, S 2 , … , S s : X → X S 2 , … , S s : X → X S_(2),dots,S_(s):X rarr XS_{2}, \ldots, S_{s}: X \rightarrow XS2,…,Ss:X→X, such that the equations x + S i x = f i , 1 ≤ i ≤ s x + S i x = f i , 1 ≤ i ≤ s x+S_(i)x=f_(i),1 <= i <= sx+S_{i} x=f_{i}, 1 \leq i \leq sx+Six=fi,1≤i≤s, have a common solution and S i ( X ) , 1 ≤ i ≤ s S i ( X ) , 1 ≤ i ≤ s S_(i)(X),1 <= i <= sS_{i}(X), 1 \leq i \leq sSi(X),1≤i≤s, are bounded. If A , b ∈ ( 0 , 1 ) , { α n } ⊂ ( 0 , 1 ) A , b ∈ ( 0 , 1 ) , α n ⊂ ( 0 , 1 ) A,b in(0,1),{alpha_(n)}sub(0,1)A, b \in(0,1),\left\{\alpha_{n}\right\} \subset(0,1)A,b∈(0,1),{αn}⊂(0,1) satisfies (1.8), and condition (2.1) is satisfied, then iteration (3.2) converges to a common solution of x + S i x = f i , 1 ≤ i ≤ s x + S i x = f i , 1 ≤ i ≤ s x+S_(i)x=f_(i),1 <= i <= sx+S_{i} x= f_{i}, 1 \leq i \leq sx+Six=fi,1≤i≤s.

4. Numerical examples

Let X = R 2 X = R 2 X=R^(2)X=\mathbb{R}^{2}X=R2 be the euclidean plane, consider x = ( a , b ) ∈ R 2 x = ( a , b ) ∈ R 2 x=(a,b)inR^(2)x=(a, b) \in \mathbb{R}^{2}x=(a,b)∈R2, with x ⊥ = ( b , − a ) ∈ R 2 x ⊥ = ( b , − a ) ∈ R 2 x^(_|_)=(b,-a)inR^(2)x^{\perp}=(b,-a) \in \mathbb{R}^{2}x⊥=(b,−a)∈R2. We know that ⟨ x , x ⊥ ⟩ = 0 , ‖ x ‖ = ‖ x ⊥ ‖ , ⟨ x ⊥ , y ⊥ ⟩ = ⟨ x , y ⟩ , ‖ x ⊥ − y ⊥ ‖ = ‖ x − y ‖ x , x ⊥ = 0 , ‖ x ‖ = x ⊥ , x ⊥ , y ⊥ = ⟨ x , y ⟩ , x ⊥ − y ⊥ = ‖ x − y ‖ (:x,x^(_|_):)=0,||x||=||x^(_|_)||,(:x^(_|_),y^(_|_):)=(:x,y:),||x^(_|_)-y^(_|_)||=||x-y||\left\langle x, x^{\perp}\right\rangle=0,\|x\|=\left\|x^{\perp}\right\|,\left\langle x^{\perp}, y^{\perp}\right\rangle=\langle x, y\rangle,\left\|x^{\perp}-y^{\perp}\right\|=\|x-y\|⟨x,x⊥⟩=0,‖x‖=‖x⊥‖,⟨x⊥,y⊥⟩=⟨x,y⟩,‖x⊥−y⊥‖=‖x−y‖, and ⟨ x ⊥ , y ⟩ + ⟨ x , y ⊥ ⟩ = 0 x ⊥ , y + x , y ⊥ = 0 (:x^(_|_),y:)+(:x,y^(_|_):)=0\left\langle x^{\perp}, y\right\rangle+\left\langle x, y^{\perp}\right\rangle=0⟨x⊥,y⟩+⟨x,y⊥⟩=0, for all x , y ∈ R 2 x , y ∈ R 2 x,y inR^(2)x, y \in \mathbb{R}^{2}x,y∈R2. Denote by B B BBB the closed unit ball. In [1], we can get the following example in which Ishikawa iteration (1.6) converges and (1.5) is not convergent.
Table 4.1
\Iteration (1.7) Case 1 Case 2
Step 10 (0.2217,0.1480) (0.0151, -0.0023)
Step 15 (0.1837, 0.1184) (0.0017,-0.0006)
Step 20 (0.1603, 0.1015) (0.0002, -0.0001)
Step 21 (0.1566, 0.0989) 10 − 3 ⋅ ( 0.1156 , − 0.0686 ) 10 − 3 ⋅ ( 0.1156 , − 0.0686 ) 10^(-3)*(0.1156,-0.0686)10^{-3} \cdot(0.1156,-0.0686)10−3⋅(0.1156,−0.0686)
Step 22 (0.1531, 0.0965) 10 − 4 ⋅ ( 0.7406 , − 0.4641 ) 10 − 4 ⋅ ( 0.7406 , − 0.4641 ) 10^(-4)*(0.7406,-0.4641)10^{-4} \cdot(0.7406,-0.4641)10−4⋅(0.7406,−0.4641)
Step 23 (0.1499, 0.0942) 10 − 4 ⋅ ( 0.4743 , − 0.3129 ) 10 − 4 ⋅ ( 0.4743 , − 0.3129 ) 10^(-4)*(0.4743,-0.3129)10^{-4} \cdot(0.4743,-0.3129)10−4⋅(0.4743,−0.3129)
Step 24 (0.1468, 0.0921) 10 − 4 ⋅ ( 0.3037 , − 0.2103 ) 10 − 4 ⋅ ( 0.3037 , − 0.2103 ) 10^(-4)*(0.3037,-0.2103)10^{-4} \cdot(0.3037,-0.2103)10−4⋅(0.3037,−0.2103)
Step 25 (0.1440, 0.0902) 10 − 4 ⋅ ( 0.1945 , − 0.1409 ) 10 − 4 ⋅ ( 0.1945 , − 0.1409 ) 10^(-4)*(0.1945,-0.1409)10^{-4} \cdot(0.1945,-0.1409)10−4⋅(0.1945,−0.1409)
\Iteration (1.7) Case 1 Case 2 Step 10 (0.2217,0.1480) (0.0151, -0.0023) Step 15 (0.1837, 0.1184) (0.0017,-0.0006) Step 20 (0.1603, 0.1015) (0.0002, -0.0001) Step 21 (0.1566, 0.0989) 10^(-3)*(0.1156,-0.0686) Step 22 (0.1531, 0.0965) 10^(-4)*(0.7406,-0.4641) Step 23 (0.1499, 0.0942) 10^(-4)*(0.4743,-0.3129) Step 24 (0.1468, 0.0921) 10^(-4)*(0.3037,-0.2103) Step 25 (0.1440, 0.0902) 10^(-4)*(0.1945,-0.1409)| \Iteration (1.7) | Case 1 | Case 2 | | :--- | :--- | :--- | | Step 10 | (0.2217,0.1480) | (0.0151, -0.0023) | | Step 15 | (0.1837, 0.1184) | (0.0017,-0.0006) | | Step 20 | (0.1603, 0.1015) | (0.0002, -0.0001) | | Step 21 | (0.1566, 0.0989) | $10^{-3} \cdot(0.1156,-0.0686)$ | | Step 22 | (0.1531, 0.0965) | $10^{-4} \cdot(0.7406,-0.4641)$ | | Step 23 | (0.1499, 0.0942) | $10^{-4} \cdot(0.4743,-0.3129)$ | | Step 24 | (0.1468, 0.0921) | $10^{-4} \cdot(0.3037,-0.2103)$ | | Step 25 | (0.1440, 0.0902) | $10^{-4} \cdot(0.1945,-0.1409)$ |
Example 4.1 [1]. Let H = R 2 H = R 2 H=R^(2)H=\mathbb{R}^{2}H=R2 and let
(4.1) B 1 = { x ∈ R 2 : ‖ x ‖ ≤ 1 2 } , B 2 = { x ∈ R 2 : 1 2 ≤ ‖ x ‖ ≤ 1 } . (4.1) B 1 = x ∈ R 2 : ‖ x ‖ ≤ 1 2 , B 2 = x ∈ R 2 : 1 2 ≤ ‖ x ‖ ≤ 1 . {:(4.1)B_(1)={x inR^(2):||x|| <= (1)/(2)}","quadB_(2)={x inR^(2):(1)/(2) <= ||x|| <= 1}.:}\begin{equation*} B_{1}=\left\{x \in \mathbb{R}^{2}:\|x\| \leq \frac{1}{2}\right\}, \quad B_{2}=\left\{x \in \mathbb{R}^{2}: \frac{1}{2} \leq\|x\| \leq 1\right\} . \tag{4.1} \end{equation*}(4.1)B1={x∈R2:‖x‖≤12},B2={x∈R2:12≤‖x‖≤1}.
The map T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is given by
(4.2) T x = { x + x ⊥ , x ∈ B 1 x ‖ x ‖ − x + x ⊥ , x ∈ B 2 . (4.2) T x = x + x ⊥ , x ∈ B 1 x ‖ x ‖ − x + x ⊥ , x ∈ B 2 . {:(4.2)Tx={[x+x^(_|_)",",x inB_(1)],[(x)/(||x||)-x+x^(_|_)",",x inB_(2).]:}:}T x= \begin{cases}x+x^{\perp}, & x \in B_{1} \tag{4.2}\\ \frac{x}{\|x\|}-x+x^{\perp}, & x \in B_{2} .\end{cases}(4.2)Tx={x+x⊥,x∈B1x‖x‖−x+x⊥,x∈B2.
Then the following are true:
(i) T T TTT is Lipschitz and pseudocontractive;
(ii) for all ( α n ) n ⊂ ( 0 , 1 ) α n n ⊂ ( 0 , 1 ) (alpha_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n} \subset(0,1)(αn)n⊂(0,1), the Mann iteration does not converge to the fixed point of T T TTT (which is the point ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) and it is unique).
The main result from [2] assures the convergence of the Ishikawa iteration (1.6) applied to the map T T TTT given by (4.2). The convergence is very slow. In [6], for the same T T TTT, it was shown that iteration (1.7) converges faster. Here, we give an example for which (1.7) with { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} satisfying (1.8) converges even faster as in [6].
Case 1 [6]. Consider now T 1 ( x , y ) = 0.5 ⋅ ( x , y ) T 1 ( x , y ) = 0.5 ⋅ ( x , y ) T_(1)(x,y)=0.5*(x,y)T_{1}(x, y)=0.5 \cdot(x, y)T1(x,y)=0.5⋅(x,y), for all ( x , y ) ∈ B , T 2 = T ( x , y ) ∈ B , T 2 = T (x,y)in B,T_(2)=T(x, y) \in B, T_{2}=T(x,y)∈B,T2=T, and s = 2 s = 2 s=2s=2s=2, where T T TTT is given by (4.2), the initial point x 0 = ( 0.5 , 0.7 ) x 0 = ( 0.5 , 0.7 ) x_(0)=(0.5,0.7)x_{0}=(0.5,0.7)x0=(0.5,0.7), and α n = β n = 1 / ( n + 1 ) α n = β n = 1 / ( n + 1 ) alpha_(n)=beta_(n)=1//(n+1)\alpha_{n}=\beta_{n}=1 /(n+1)αn=βn=1/(n+1) in (1.7). The main result from [6] assures the convergence of (1.7).
Case 2. Consider T 1 ( x , y ) = 0.5 ⋅ ( x , y ) T 1 ( x , y ) = 0.5 ⋅ ( x , y ) T_(1)(x,y)=0.5*(x,y)T_{1}(x, y)=0.5 \cdot(x, y)T1(x,y)=0.5⋅(x,y), for all ( x , y ) ∈ B , T 2 = T ( x , y ) ∈ B , T 2 = T (x,y)in B,T_(2)=T(x, y) \in B, T_{2}=T(x,y)∈B,T2=T, and s = 2 s = 2 s=2s=2s=2, where T T TTT is given by (4.2), the initial point x 0 = ( 0.5 , 0.7 ) , α n = 0.7 x 0 = ( 0.5 , 0.7 ) , α n = 0.7 x_(0)=(0.5,0.7),alpha_(n)=0.7x_{0}=(0.5,0.7), \alpha_{n}=0.7x0=(0.5,0.7),αn=0.7, for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N, and β n = 1 / ( n + 1 ) β n = 1 / ( n + 1 ) beta_(n)=1//(n+1)\beta_{n}=1 /(n+1)βn=1/(n+1) in (1.7). The fixed point for both functions is ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0). Observe that k = 0.5 k = 0.5 k=0.5k=0.5k=0.5, and { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} satisfies (1.8):
(4.3) A = 0.7 = α n = b ≤ 2 ( 1 − k ) = 1 , ∀ n ∈ N . (4.3) A = 0.7 = α n = b ≤ 2 ( 1 − k ) = 1 , ∀ n ∈ N . {:(4.3)A=0.7=alpha_(n)=b <= 2(1-k)=1","quad AA n inN.:}\begin{equation*} A=0.7=\alpha_{n}=b \leq 2(1-k)=1, \quad \forall n \in \mathbb{N} . \tag{4.3} \end{equation*}(4.3)A=0.7=αn=b≤2(1−k)=1,∀n∈N.
Note that Mann iteration does not converge for any { α n } ⊂ ( 0 , 1 ) α n ⊂ ( 0 , 1 ) {alpha_(n)}sub(0,1)\left\{\alpha_{n}\right\} \subset(0,1){αn}⊂(0,1). Using a Matlab program, we obtain Table 4.1.
Case 3. Consider in (1.7) the same T 1 , T 2 , s = 2 T 1 , T 2 , s = 2 T_(1),T_(2),s=2T_{1}, T_{2}, s=2T1,T2,s=2, and x 0 x 0 x_(0)x_{0}x0 as in Case 1 and α n = β n = 1 / n + 1 α n = β n = 1 / n + 1 alpha_(n)=beta_(n)=1//sqrt(n+1)\alpha_{n}=\beta_{n}= 1 / \sqrt{n+1}αn=βn=1/n+1.
Table 4.2
\Iteration Case 3 (1.7) Case 4 (1.7) Ishikawa iteration
Step 10 (0.0631, -0.0333) (0.0044, -0.0164) (0.4545, 0.2689)
Step 15 (0.0256, -0.0221) (-0.0010, -0.0018) (0.1289, -0.4827)
Step 20 (0.0117,-0.0139) 10 − 5 ⋅ ( − 22.6516 , − 11.0267 ) 10 − 5 ⋅ ( − 22.6516 , − 11.0267 ) 10^(-5)*(-22.6516,-11.0267)10^{-5} \cdot(-22.6516,-11.0267)10−5⋅(−22.6516,−11.0267) (-0.4456, -0.1532)
Step 11 (0.0101, -0.0126) 10 − 5 ⋅ ( − 15.5657 , − 5.4373 ) 10 − 5 ⋅ ( − 15.5657 , − 5.4373 ) 10^(-5)*(-15.5657,-5.4373)10^{-5} \cdot(-15.5657,-5.4373)10−5⋅(−15.5657,−5.4373) (-0.4651,-0.0274)
Step 22 (0.0087,-0.0115) 10 − 5 ⋅ ( − 10.5234 , − 2.3727 ) 10 − 5 ⋅ ( − 10.5234 , − 2.3727 ) 10^(-5)*(-10.5234,-2.3727)10^{-5} \cdot(-10.5234,-2.3727)10−5⋅(−10.5234,−2.3727) (-0.4511, 0.0941)
Step 23 (0.0075,-0.0105) 10 − 5 ⋅ ( − 7.0134 , − 0.7743 ) 10 − 5 ⋅ ( − 7.0134 , − 0.7743 ) 10^(-5)*(-7.0134,-0.7743)10^{-5} \cdot(-7.0134,-0.7743)10−5⋅(−7.0134,−0.7743) (-0.4077, 0.2037)
Step 24 (0.0066, -0.0096) 10 − 5 ⋅ ( − 4.6140 , − 0.0022 ) 10 − 5 ⋅ ( − 4.6140 , − 0.0022 ) 10^(-5)*(-4.6140,-0.0022)10^{-5} \cdot(-4.6140,-0.0022)10−5⋅(−4.6140,−0.0022) (-0.3407, 0.2954)
Step 25 (0.0057,-0.0088) 10 − 5 ⋅ ( − 2.9993 , 0.3215 ) 10 − 5 ⋅ ( − 2.9993 , 0.3215 ) 10^(-5)*(-2.9993,0.3215)10^{-5} \cdot(-2.9993,0.3215)10−5⋅(−2.9993,0.3215) (-0.2562, 0.3654)
Step 1500 - - (0.0790, -0.0311)
\Iteration Case 3 (1.7) Case 4 (1.7) Ishikawa iteration Step 10 (0.0631, -0.0333) (0.0044, -0.0164) (0.4545, 0.2689) Step 15 (0.0256, -0.0221) (-0.0010, -0.0018) (0.1289, -0.4827) Step 20 (0.0117,-0.0139) 10^(-5)*(-22.6516,-11.0267) (-0.4456, -0.1532) Step 11 (0.0101, -0.0126) 10^(-5)*(-15.5657,-5.4373) (-0.4651,-0.0274) Step 22 (0.0087,-0.0115) 10^(-5)*(-10.5234,-2.3727) (-0.4511, 0.0941) Step 23 (0.0075,-0.0105) 10^(-5)*(-7.0134,-0.7743) (-0.4077, 0.2037) Step 24 (0.0066, -0.0096) 10^(-5)*(-4.6140,-0.0022) (-0.3407, 0.2954) Step 25 (0.0057,-0.0088) 10^(-5)*(-2.9993,0.3215) (-0.2562, 0.3654) Step 1500 - - (0.0790, -0.0311)| \Iteration | Case 3 (1.7) | Case 4 (1.7) | Ishikawa iteration | | :--- | :--- | :--- | :--- | | Step 10 | (0.0631, -0.0333) | (0.0044, -0.0164) | (0.4545, 0.2689) | | Step 15 | (0.0256, -0.0221) | (-0.0010, -0.0018) | (0.1289, -0.4827) | | Step 20 | (0.0117,-0.0139) | $10^{-5} \cdot(-22.6516,-11.0267)$ | (-0.4456, -0.1532) | | Step 11 | (0.0101, -0.0126) | $10^{-5} \cdot(-15.5657,-5.4373)$ | (-0.4651,-0.0274) | | Step 22 | (0.0087,-0.0115) | $10^{-5} \cdot(-10.5234,-2.3727)$ | (-0.4511, 0.0941) | | Step 23 | (0.0075,-0.0105) | $10^{-5} \cdot(-7.0134,-0.7743)$ | (-0.4077, 0.2037) | | Step 24 | (0.0066, -0.0096) | $10^{-5} \cdot(-4.6140,-0.0022)$ | (-0.3407, 0.2954) | | Step 25 | (0.0057,-0.0088) | $10^{-5} \cdot(-2.9993,0.3215)$ | (-0.2562, 0.3654) | | Step 1500 | - | - | (0.0790, -0.0311) |
Case 4. Consider in (1.7) T 1 , T 2 , s = 2 T 1 , T 2 , s = 2 T_(1),T_(2),s=2T_{1}, T_{2}, s=2T1,T2,s=2, and x 0 x 0 x_(0)x_{0}x0 as above and α n = 0.7 α n = 0.7 alpha_(n)=0.7\alpha_{n}=0.7αn=0.7, for all n ∈ N , β n = 1 / n + 1 n ∈ N , β n = 1 / n + 1 n inN,beta_(n)=1//sqrt(n+1)n \in \mathbb{N}, \beta_{n}= 1 / \sqrt{n+1}n∈N,βn=1/n+1.
Also, consider the Ishikawa iteration with the same T T TTT as in (4.2), x 0 = ( 0.5 , 0.7 ) , α n = β n = 1 / n + 1 x 0 = ( 0.5 , 0.7 ) , α n = β n = 1 / n + 1 x_(0)=(0.5,0.7),alpha_(n)=beta_(n)=1//sqrt(n+1)x_{0}=(0.5,0.7), \alpha_{n}= \beta_{n}=1 / \sqrt{n+1}x0=(0.5,0.7),αn=βn=1/n+1, for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N. The main result from [2] assures the convergence of Ishikawa iteration. Note that in this case the convergence is very slow. Eventually, Example 4.1 assures that for the same map, Mann iteration does not converge. A Matlab program leads to the evaluations illustrated in Table 4.2.

Acknowledgment

The authors are indebted to the referee for carefully reading the paper and for making useful suggestions.

References

[1] C. E. Chidume and S. A. Mutangadura, An example of the Mann iteration method for Lipschitz pseudocontractions, Proc. Amer. Math. Soc. 129 (2001), no. 8, 2359-2363.
[2] S. Ishikawa, Fixed points by a new iteration method, Proc. Amer. Math. Soc. 44 (1974), no. 1, 147-150.
[3] W. R. Mann, Mean value methods in iteration, Proc. Amer. Math. Soc. 4 (1953), 506-510.
[4] C. H. Morales and J. S. Jung, Convergence of paths for pseudocontractive mappings in Banach spaces, Proc. Amer. Math. Soc. 128 (2000), no. 11, 3411-3419.
[5] B. E. Rhoades and Ş. M. Şoltuz, The equivalence of Mann iteration and Ishikawa iteration for non-Lipschitzian operators, Int. J. Math. Math. Sci. 2003 (2003), no. 42, 2645-2651.
[6] -, Mean value iteration for a family of functions, to appear in Nonlinear Funct. Anal. Appl.
[7] Ş. M. Şoltuz, Some sequences supplied by inequalities and their applications, Rev. Anal. Numér. Théor. Approx. 29 (2000), no. 2, 207-212.
  1. B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
    E-mail address: rhoades@indiana.edu
    Ştefan M. Şoltuz: Institute of Numerical Analysis, P.O. Box 68-1, 400110 Cluj-Napoca, Romania
    E-mail address: smsoltuz@gmail.com
2005

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