The equivalence of Mann iteration and Ishikawa iteration for non-Lipschitzian operators

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B. E. Rhoades
Department of Mathematics, Indiana University, Bloomington, IN47405-7106, USA

Stefan M. Soltuz:
“T. Popoviciu” Institute of Numerical Analysis, P.O. Box 68-1, 3400Cluj-Napoca, Romania

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B.E. Rhoades and Ş.M. Şoltuz, The equivalence of Mann iteration and Ishikawa iteration for non-Lipschitzian operators, Internat. J. Math. Math. Sci. 2003 (42), 2645-2652.

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International Journal of Mathematics and Mathematical Sciences – 2003 – Rhoades – The equivalence of Mann iteration and

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[11] X. Weng, Fixed point iteration for local strictly pseudocontractive mapping, Proc.Amer. Math. Soc. 113 (1991), no. 3, 727–731

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International Journal of Mathematics and Mathematical Sciences - 2003 - Rhoades - The equivalence of

THE EQUIVALENCE OF MANN ITERATION AND ISHIKAWA ITERATION FOR NON-LIPSCHITZIAN OPERATORS

B. E. RHOADES and ŞTEFAN M. ŞOLTUZ

Received 25 November 2002
We show that the convergence of Mann iteration is equivalent to the convergence of Ishikawa iteration for various classes of non-Lipschitzian operators.
2000 Mathematics Subject Classification: 47H10.
  1. Introduction. Let X X XXX be a real Banach space, B B BBB a nonempty, convex subset of X X XXX, and T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B an operator. Let u 1 , x 1 ∈ B u 1 , x 1 ∈ B u_(1),x_(1)in Bu_{1}, x_{1} \in Bu1,x1∈B. The following iteration is known as Mann iteration (see [6]):
(1.1) u n + 1 = ( 1 − α n ) u n + α n T u n (1.1) u n + 1 = 1 − α n u n + α n T u n {:(1.1)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{1.1} \end{equation*}(1.1)un+1=(1−αn)un+αnTun
The sequence ( 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) is convergent such that
(1.2) lim n → ∞ α n = 0 , ∑ n = 1 ∞ α n = ∞ . (1.2) lim n → ∞   α n = 0 , ∑ n = 1 ∞   α n = ∞ . {:(1.2)lim_(n rarr oo)alpha_(n)=0","quadsum_(n=1)^(oo)alpha_(n)=oo.:}\begin{equation*} \lim _{n \rightarrow \infty} \alpha_{n}=0, \quad \sum_{n=1}^{\infty} \alpha_{n}=\infty . \tag{1.2} \end{equation*}(1.2)limn→∞αn=0,∑n=1∞αn=∞.
Ishikawa iteration is given by (see [5])
x n + 1 = ( 1 − α n ) x n + α n T y n (1.3) y n = ( 1 − β n ) x n + β n T x n , n = 1 , 2 , … . x n + 1 = 1 − α n x n + α n T y n (1.3) y n = 1 − β n x n + β n T x n , n = 1 , 2 , … . {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Ty_(n)],[(1.3)y_(n)=(1-beta_(n))x_(n)+beta_(n)Tx_(n)","quad n=1","2","dots.]:}\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}, \quad n=1,2, \ldots . \tag{1.3} \end{align*}xn+1=(1−αn)xn+αnTyn(1.3)yn=(1−βn)xn+βnTxn,n=1,2,….
The sequences ( α n ) n , ( β n ) n ⊂ ( 0 , 1 ) α n n , β n n ⊂ ( 0 , 1 ) (alpha_(n))_(n),(beta_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n} \subset(0,1)(αn)n,(βn)n⊂(0,1) are convergent such that
(1.4) lim n → ∞ α n = 0 , lim n → ∞ β n = 0 , ∑ n = 1 ∞ α n = ∞ . (1.4) lim n → ∞   α n = 0 , lim n → ∞   β n = 0 , ∑ n = 1 ∞   α n = ∞ . {:(1.4)lim_(n rarr oo)alpha_(n)=0","quadlim_(n rarr oo)beta_(n)=0","quadsum_(n=1)^(oo)alpha_(n)=oo.:}\begin{equation*} \lim _{n \rightarrow \infty} \alpha_{n}=0, \quad \lim _{n \rightarrow \infty} \beta_{n}=0, \quad \sum_{n=1}^{\infty} \alpha_{n}=\infty . \tag{1.4} \end{equation*}(1.4)limn→∞αn=0,limn→∞βn=0,∑n=1∞αn=∞.
Moreover, the sequence ( α n ) n α n n (alpha_(n))_(n)\left(\alpha_{n}\right)_{n}(αn)n from (1.3) is the same as in (1.1).
The map J : X → 2 X ∗ J : X → 2 X ∗ J:X rarr2^(X^(**))J: X \rightarrow 2^{X^{*}}J:X→2X∗ given by J x := { f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ } J x := f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ Jx:={f inX^(**):(:x,f:)=||x||^(2),||f||=||x||}J x:=\left\{f \in X^{*}:\langle x, f\rangle=\|x\|^{2},\|f\|=\|x\|\right\}Jx:={f∈X∗:⟨x,f⟩=‖x‖2,‖f‖=‖x‖}, for all x ∈ X x ∈ X x in Xx \in Xx∈X, is called the normalized duality mapping. The Hahn-Banach theorem assures that J x ≠ ∅ J x ≠ ∅ Jx!=O/J x \neq \varnothingJx≠∅, for all x ∈ X x ∈ X x in Xx \in Xx∈X. It is easy to see that we have ⟨ j ( x ) , y ⟩ ≤ ‖ x ‖ ‖ y ‖ ⟨ j ( x ) , y ⟩ ≤ ‖ x ‖ ‖ y ‖ (:j(x),y:) <= ||x||||y||\langle j(x), y\rangle \leq \|x\|\|y\|⟨j(x),y⟩≤‖x‖‖y‖, for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X and for all j ( x ) ∈ J ( x ) j ( x ) ∈ J ( x ) j(x)in J(x)j(x) \in J(x)j(x)∈J(x).
Definition 1.1. Let X X XXX be a real Banach space and let B B BBB be a nonempty subset. A map T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is called strongly pseudocontractive if there exist k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) and a 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
(1.5) ⟨ T x − T y , j ( x − y ) ⟩ ≤ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . (1.5) ⟨ T x − T y , j ( x − y ) ⟩ ≤ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . {:(1.5)(: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{1.5} \end{equation*}(1.5)⟨Tx−Ty,j(x−y)⟩≤k‖x−y‖2,∀x,y∈B.
A map S : D ( S ) → X S : D ( S ) → X S:D(S)rarr XS: D(S) \rightarrow XS:D(S)→X 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 a 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
(1.6) ⟨ S x − S y , j ( x − y ) ⟩ ≥ k ‖ x − y ‖ 2 , ∀ x , y ∈ D ( S ) . (1.6) ⟨ S x − S y , j ( x − y ) ⟩ ≥ k ‖ x − y ‖ 2 , ∀ x , y ∈ D ( S ) . {:(1.6)(:Sx-Sy","j(x-y):) >= k||x-y||^(2)","quad AA x","y in D(S).:}\begin{equation*} \langle S x-S y, j(x-y)\rangle \geq k\|x-y\|^{2}, \quad \forall x, y \in D(S) . \tag{1.6} \end{equation*}(1.6)⟨Sx−Sy,j(x−y)⟩≥k‖x−y‖2,∀x,y∈D(S).
In (1.5), when k = 1 , T k = 1 , T k=1,Tk=1, Tk=1,T is called pseudocontractive. In (1.6), when k = 0 , S k = 0 , S k=0,Sk=0, Sk=0,S is called accretive. We denote by I I III the identity map.
REMARK 1.2. (i) The operator T T TTT is (strongly) pseudocontractive map if and only if ( I − T I − T I-TI-TI−T ) is (strongly) accretive.
(ii) If S S SSS is accretive map, then T = f − S T = f − S T=f-ST=f-ST=f−S is strongly pseudocontractive map.
Remark 1.2(i) is obvious from (1.5) and (1.6). For Remark 1.2(ii), supposing that x , y ∈ B x , y ∈ B x,y in Bx, y \in Bx,y∈B 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), one obtains
⟨ S x − S y , j ( x − y ) ⟩ ≥ 0 ⟺ ⟨ ( f − T ) x − ( f − T ) y , j ( x − y ) ⟩ ≥ 0 (1.7) ⟺ ⟨ T x − T y , j ( x − y ) ⟩ ≤ 0 ≤ k ‖ x − y ‖ 2 ⟨ S x − S y , j ( x − y ) ⟩ ≥ 0 ⟺ ⟨ ( f − T ) x − ( f − T ) y , j ( x − y ) ⟩ ≥ 0 (1.7) ⟺ ⟨ T x − T y , j ( x − y ) ⟩ ≤ 0 ≤ k ‖ x − y ‖ 2 {:[(:Sx-Sy","j(x-y):) >= 0 Longleftrightarrow(:(f-T)x-(f-T)y","j(x-y):) >= 0],[(1.7) Longleftrightarrow(:Tx-Ty","j(x-y):) <= 0 <= k||x-y||^(2)]:}\begin{align*} \langle S x-S y, j(x-y)\rangle \geq 0 & \Longleftrightarrow\langle(f-T) x-(f-T) y, j(x-y)\rangle \geq 0 \\ & \Longleftrightarrow\langle T x-T y, j(x-y)\rangle \leq 0 \leq k\|x-y\|^{2} \tag{1.7} \end{align*}⟨Sx−Sy,j(x−y)⟩≥0⟺⟨(f−T)x−(f−T)y,j(x−y)⟩≥0(1.7)⟺⟨Tx−Ty,j(x−y)⟩≤0≤k‖x−y‖2
for all k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1).
In [10], it was shown that the Mann and Ishikawa iterations are equivalent for various classes of Lipschitzian operators. We prove here the equivalence for non-Lipschitzian operators. For this purpose, we need several lemmas.
Lemma 1.3 [11]. Let ( a n ) n a n n (a_(n))_(n)\left(a_{n}\right)_{n}(an)n be a nonnegative sequence which satisfies the following inequality:
(1.8) a n + 1 ≤ ( 1 − λ n ) a n + σ n , (1.8) a n + 1 ≤ 1 − λ n a n + σ n , {:(1.8)a_(n+1) <= (1-lambda_(n))a_(n)+sigma_(n)",":}\begin{equation*} a_{n+1} \leq\left(1-\lambda_{n}\right) a_{n}+\sigma_{n}, \tag{1.8} \end{equation*}(1.8)an+1≤(1−λn)an+σn,
where λ n ∈ ( 0 , 1 ) λ n ∈ ( 0 , 1 ) lambda_(n)in(0,1)\lambda_{n} \in(0,1)λn∈(0,1), for all n ∈ N , ∑ n = 1 ∞ λ n = ∞ n ∈ N , ∑ n = 1 ∞   λ n = ∞ n inN,sum_(n=1)^(oo)lambda_(n)=oon \in \mathbb{N}, \sum_{n=1}^{\infty} \lambda_{n}=\inftyn∈N,∑n=1∞λn=∞, and σ n = o ( λ n ) σ n = o λ n sigma_(n)=o(lambda_(n))\sigma_{n}=o\left(\lambda_{n}\right)σn=o(λn). 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.
It is known that J ( x ) = ∂ ϕ ( x ) J ( x ) = ∂ ϕ ( x ) J(x)=del phi(x)J(x)=\partial \phi(x)J(x)=∂ϕ(x), where ϕ ( x ) = ( 1 / 2 ) ‖ x ‖ 2 ϕ ( x ) = ( 1 / 2 ) ‖ x ‖ 2 phi(x)=(1//2)||x||^(2)\phi(x)=(1 / 2)\|x\|^{2}ϕ(x)=(1/2)‖x‖2 and ∂ ϕ ( x ) ∂ ϕ ( x ) del phi(x)\partial \phi(x)∂ϕ(x) denotes the subdifferential of ϕ ( x ) ϕ ( x ) phi(x)\phi(x)ϕ(x) at x x xxx, so the following inequality is satisfied, see also [1, Lemma 2.1] or [9, Lemma 1].
Lemma 1.4 [1, 9]. If X X XXX is a real Banach space, then the following relation is true:
(1.9) ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , j ( x + y ) ⟩ , ∀ x , y ∈ X , ∀ j ( x + y ) ∈ J ( x + y ) . (1.9) ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , j ( x + y ) ⟩ , ∀ x , y ∈ X , ∀ j ( x + y ) ∈ J ( x + y ) . {:(1.9)||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{1.9} \end{equation*}(1.9)‖x+y‖2≤‖x‖2+2⟨y,j(x+y)⟩,∀x,y∈X,∀j(x+y)∈J(x+y).
  1. Main result. We are now able to prove the following result.
Theorem 2.1. Let X X XXX be a real Banach space with a uniformly convex dual and B B BBB a nonempty, closed, convex, and bounded subset of X X XXX. Let T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B be a continuous and strongly pseudocontractive operator. Then for u 1 = x 1 ∈ B u 1 = x 1 ∈ B u_(1)=x_(1)in Bu_{1}=x_{1} \in Bu1=x1∈B, the following assertions are equivalent:
(a) Mann iteration (1.1) converges to the fixed point of T T TTT;
(b) Ishikawa iteration (1.3) converges to the fixed point of T T TTT.
Proof. Deimling [3, Corollary 1] assures the existence of a fixed point. The uniqueness of the fixed point comes from (1.5). Because X ∗ X ∗ X^(**)X^{*}X∗ is uniformly convex, the duality map is single valued (see, e.g., [4]). Using (1.1), (1.3), and Lemma 1.4, we get
‖ x n + 1 − u n + 1 ‖ 2 = ‖ ( 1 − α n ) ( x n − u n ) + α n ( T y n − T u n ) ‖ 2 ≤ ( 1 − α n ) 2 ‖ x n − u n ‖ 2 + 2 α n ⟨ T y n − T u n , J ( x n + 1 − u n + 1 ) ⟩ = ( 1 − α n ) 2 ‖ x n − u n ‖ 2 + 2 α n ⟨ T y n − T u n , J ( x n + 1 − u n + 1 ) − J ( y n − u n ) ⟩ + 2 α n ⟨ T y n − T u n , J ( y n − u n ) ⟩ ≤ ( 1 − α n ) 2 ‖ x n − u n ‖ 2 + 2 α n k ‖ y n − u n ‖ 2 + 2 α n ⟨ T y n − T u n , J ( x n + 1 − u n + 1 ) − J ( y n − u n ) ⟩ ≤ ( 1 − α n ) 2 ‖ x n − u n ‖ 2 + 2 α n k ‖ y n − u n ‖ 2 + 2 α n ‖ T y n − T u n ‖ ‖ J ( x n + 1 − u n + 1 ) − J ( y n − u n ) ‖ ≤ ( 1 − α n ) 2 ‖ x n − u n ‖ 2 + 2 α n k ‖ y n − u n ‖ 2 (2.1) + 2 α n M 1 ‖ J ( x n + 1 − u n + 1 ) − J ( y n − u n ) ‖ x n + 1 − u n + 1 2 = 1 − α n x n − u n + α n T y n − T u n 2 ≤ 1 − α n 2 x n − u n 2 + 2 α n T y n − T u n , J x n + 1 − u n + 1 = 1 − α n 2 x n − u n 2 + 2 α n T y n − T u n , J x n + 1 − u n + 1 − J y n − u n + 2 α n T y n − T u n , J y n − u n ≤ 1 − α n 2 x n − u n 2 + 2 α n k y n − u n 2 + 2 α n T y n − T u n , J x n + 1 − u n + 1 − J y n − u n ≤ 1 − α n 2 x n − u n 2 + 2 α n k y n − u n 2 + 2 α n T y n − T u n J x n + 1 − u n + 1 − J y n − u n ≤ 1 − α n 2 x n − u n 2 + 2 α n k y n − u n 2 (2.1) + 2 α n M 1 J x n + 1 − u n + 1 − J y n − u n {:[||x_(n+1)-u_(n+1)||^(2)=||(1-alpha_(n))(x_(n)-u_(n))+alpha_(n)(Ty_(n)-Tu_(n))||^(2)],[ <= (1-alpha_(n))^(2)||x_(n)-u_(n)||^(2)+2alpha_(n)(:Ty_(n)-Tu_(n),J(x_(n+1)-u_(n+1)):)],[=(1-alpha_(n))^(2)||x_(n)-u_(n)||^(2)],[+2alpha_(n)(:Ty_(n)-Tu_(n),J(x_(n+1)-u_(n+1))-J(y_(n)-u_(n)):)],[+2alpha_(n)(:Ty_(n)-Tu_(n),J(y_(n)-u_(n)):)],[ <= (1-alpha_(n))^(2)||x_(n)-u_(n)||^(2)+2alpha_(n)k||y_(n)-u_(n)||^(2)],[+2alpha_(n)(:Ty_(n)-Tu_(n),J(x_(n+1)-u_(n+1))-J(y_(n)-u_(n)):)],[ <= (1-alpha_(n))^(2)||x_(n)-u_(n)||^(2)+2alpha_(n)k||y_(n)-u_(n)||^(2)],[+2alpha_(n)||Ty_(n)-Tu_(n)||||J(x_(n+1)-u_(n+1))-J(y_(n)-u_(n))||],[ <= (1-alpha_(n))^(2)||x_(n)-u_(n)||^(2)+2alpha_(n)k||y_(n)-u_(n)||^(2)],[(2.1)+2alpha_(n)M_(1)||J(x_(n+1)-u_(n+1))-J(y_(n)-u_(n))||]:}\begin{align*} \left\|x_{n+1}-u_{n+1}\right\|^{2}= & \left\|\left(1-\alpha_{n}\right)\left(x_{n}-u_{n}\right)+\alpha_{n}\left(T y_{n}-T u_{n}\right)\right\|^{2} \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2 \alpha_{n}\left\langle T y_{n}-T u_{n}, J\left(x_{n+1}-u_{n+1}\right)\right\rangle \\ = & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2} \\ & +2 \alpha_{n}\left\langle T y_{n}-T u_{n}, J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}-u_{n}\right)\right\rangle \\ & +2 \alpha_{n}\left\langle T y_{n}-T u_{n}, J\left(y_{n}-u_{n}\right)\right\rangle \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2 \alpha_{n} k\left\|y_{n}-u_{n}\right\|^{2} \\ & +2 \alpha_{n}\left\langle T y_{n}-T u_{n}, J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}-u_{n}\right)\right\rangle \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2 \alpha_{n} k\left\|y_{n}-u_{n}\right\|^{2} \\ & +2 \alpha_{n}\left\|T y_{n}-T u_{n}\right\|\left\|J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}-u_{n}\right)\right\| \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2 \alpha_{n} k\left\|y_{n}-u_{n}\right\|^{2} \\ & +2 \alpha_{n} M_{1}\left\|J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}-u_{n}\right)\right\| \tag{2.1} \end{align*}‖xn+1−un+1‖2=‖(1−αn)(xn−un)+αn(Tyn−Tun)‖2≤(1−αn)2‖xn−un‖2+2αn⟨Tyn−Tun,J(xn+1−un+1)⟩=(1−αn)2‖xn−un‖2+2αn⟨Tyn−Tun,J(xn+1−un+1)−J(yn−un)⟩+2αn⟨Tyn−Tun,J(yn−un)⟩≤(1−αn)2‖xn−un‖2+2αnk‖yn−un‖2+2αn⟨Tyn−Tun,J(xn+1−un+1)−J(yn−un)⟩≤(1−αn)2‖xn−un‖2+2αnk‖yn−un‖2+2αn‖Tyn−Tun‖‖J(xn+1−un+1)−J(yn−un)‖≤(1−αn)2‖xn−un‖2+2αnk‖yn−un‖2(2.1)+2αnM1‖J(xn+1−un+1)−J(yn−un)‖
for some positive constant M 1 M 1 M_(1)M_{1}M1. Observe that ( ‖ T y n − T u n ‖ ) n T y n − T u n n (||Ty_(n)-Tu_(n)||)_(n)\left(\left\|T y_{n}-T u_{n}\right\|\right)_{n}(‖Tyn−Tun‖)n is bounded. We now prove that
(2.2) J ( x n + 1 − u n + 1 ) − J ( y n − u n ) ⟶ 0 ( n → ∞ ) . (2.2) J x n + 1 − u n + 1 − J y n − u n ⟶ 0 ( n → ∞ ) . {:(2.2)J(x_(n+1)-u_(n+1))-J(y_(n)-u_(n))longrightarrow0quad(n rarr oo).:}\begin{equation*} J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}-u_{n}\right) \longrightarrow 0 \quad(n \rightarrow \infty) . \tag{2.2} \end{equation*}(2.2)J(xn+1−un+1)−J(yn−un)⟶0(n→∞).
Deimling [4, Proposition 12.3, page 115] assures that when X ∗ X ∗ X^(**)X^{*}X∗ is uniformly convex, J J JJJ is uniformly continuous on every bounded set of X X XXX. To prove (2.2), it is sufficient to see that
‖ ( x n + 1 − u n + 1 ) − ( y n − u n ) ‖ = ‖ ( x n + 1 − y n ) − ( u n + 1 − u n ) ‖ (2.3) = ‖ − α n x n + α n T y n + β n x n − β n T x n + α n u n − α n T u n ‖ ≤ α n ( ‖ x n ‖ + ‖ T y n ‖ + ‖ u n ‖ + ‖ T u n ‖ ) + β n ( ‖ x n ‖ + ‖ T x n ‖ ) ≤ ( α n + β n ) M → 0 ( n → ∞ ) , x n + 1 − u n + 1 − y n − u n = x n + 1 − y n − u n + 1 − u n (2.3) = − α n x n + α n T y n + β n x n − β n T x n + α n u n − α n T u n ≤ α n x n + T y n + u n + T u n + β n x n + T x n ≤ α n + β n M → 0 ( n → ∞ ) , {:[||(x_(n+1)-u_(n+1))-(y_(n)-u_(n))||],[quad=||(x_(n+1)-y_(n))-(u_(n+1)-u_(n))||],[(2.3)quad=||-alpha_(n)x_(n)+alpha_(n)Ty_(n)+beta_(n)x_(n)-beta_(n)Tx_(n)+alpha_(n)u_(n)-alpha_(n)Tu_(n)||],[quad <= alpha_(n)(||x_(n)||+||Ty_(n)||+||u_(n)||+||Tu_(n)||)+beta_(n)(||x_(n)||+||Tx_(n)||)],[quad <= (alpha_(n)+beta_(n))M rarr0quad(n rarr oo)","]:}\begin{align*} & \left\|\left(x_{n+1}-u_{n+1}\right)-\left(y_{n}-u_{n}\right)\right\| \\ & \quad=\left\|\left(x_{n+1}-y_{n}\right)-\left(u_{n+1}-u_{n}\right)\right\| \\ & \quad=\left\|-\alpha_{n} x_{n}+\alpha_{n} T y_{n}+\beta_{n} x_{n}-\beta_{n} T x_{n}+\alpha_{n} u_{n}-\alpha_{n} T u_{n}\right\| \tag{2.3}\\ & \quad \leq \alpha_{n}\left(\left\|x_{n}\right\|+\left\|T y_{n}\right\|+\left\|u_{n}\right\|+\left\|T u_{n}\right\|\right)+\beta_{n}\left(\left\|x_{n}\right\|+\left\|T x_{n}\right\|\right) \\ & \quad \leq\left(\alpha_{n}+\beta_{n}\right) M \rightarrow 0 \quad(n \rightarrow \infty), \end{align*}‖(xn+1−un+1)−(yn−un)‖=‖(xn+1−yn)−(un+1−un)‖(2.3)=‖−αnxn+αnTyn+βnxn−βnTxn+αnun−αnTun‖≤αn(‖xn‖+‖Tyn‖+‖un‖+‖Tun‖)+βn(‖xn‖+‖Txn‖)≤(αn+βn)M→0(n→∞),
where
(2.4) M = sup n ( ( ‖ x n ‖ + ‖ T y n ‖ + ‖ u n ‖ + ‖ T u n ‖ ) , ( ‖ x n ‖ + ‖ T x n ‖ ) ) < ∞ . (2.4) M = sup n   x n + T y n + u n + T u n , x n + T x n < ∞ . {:(2.4)M=s u p_(n)((||x_(n)||+||Ty_(n)||+||u_(n)||+||Tu_(n)||),(||x_(n)||+||Tx_(n)||)) < oo.:}\begin{equation*} M=\sup _{n}\left(\left(\left\|x_{n}\right\|+\left\|T y_{n}\right\|+\left\|u_{n}\right\|+\left\|T u_{n}\right\|\right),\left(\left\|x_{n}\right\|+\left\|T x_{n}\right\|\right)\right)<\infty . \tag{2.4} \end{equation*}(2.4)M=supn((‖xn‖+‖Tyn‖+‖un‖+‖Tun‖),(‖xn‖+‖Txn‖))<∞.
The sequences ( u n ) n , ( x n ) n , ( T x n ) n , ( T u n ) n u n n , x n n , T x n n , T u n n (u_(n))_(n),(x_(n))_(n),(Tx_(n))_(n),(Tu_(n))_(n)\left(u_{n}\right)_{n},\left(x_{n}\right)_{n},\left(T x_{n}\right)_{n},\left(T u_{n}\right)_{n}(un)n,(xn)n,(Txn)n,(Tun)n, and ( T y n ) n T y n n (Ty_(n))_(n)\left(T y_{n}\right)_{n}(Tyn)n are bounded being in the bounded set B B BBB. Hence one can see that the M M MMM above is finite and (2.2) holds. We define
(2.5) σ n := 2 α n M 1 ‖ J ( x n + 1 − u n + 1 ) − J ( y n − u n ) ‖ . (2.5) σ n := 2 α n M 1 J x n + 1 − u n + 1 − J y n − u n . {:(2.5)sigma_(n):=2alpha_(n)M_(1)||J(x_(n+1)-u_(n+1))-J(y_(n)-u_(n))||.:}\begin{equation*} \sigma_{n}:=2 \alpha_{n} M_{1}\left\|J\left(x_{n+1}-u_{n+1}\right)-J\left(y_{n}-u_{n}\right)\right\| . \tag{2.5} \end{equation*}(2.5)σn:=2αnM1‖J(xn+1−un+1)−J(yn−un)‖.
Again, using (1.1) and (1.3), we get
‖ y n − u n ‖ 2 = ‖ ( 1 − β n ) ( x n − u n ) + β n ( T x n − u n ) ‖ 2 (2.6) ≤ ( 1 − β n ) 2 ‖ x n − u n ‖ 2 + 2 β n ⟨ T x n − u n , J ( y n − u n ) ⟩ ≤ ‖ x n − u n ‖ 2 + β n M 2 y n − u n 2 = 1 − β n x n − u n + β n T x n − u n 2 (2.6) ≤ 1 − β n 2 x n − u n 2 + 2 β n T x n − u n , J y n − u n ≤ x n − u n 2 + β n M 2 {:[||y_(n)-u_(n)||^(2)=||(1-beta_(n))(x_(n)-u_(n))+beta_(n)(Tx_(n)-u_(n))||^(2)],[(2.6) <= (1-beta_(n))^(2)||x_(n)-u_(n)||^(2)+2beta_(n)(:Tx_(n)-u_(n),J(y_(n)-u_(n)):)],[ <= ||x_(n)-u_(n)||^(2)+beta_(n)M_(2)]:}\begin{align*} \left\|y_{n}-u_{n}\right\|^{2} & =\left\|\left(1-\beta_{n}\right)\left(x_{n}-u_{n}\right)+\beta_{n}\left(T x_{n}-u_{n}\right)\right\|^{2} \\ & \leq\left(1-\beta_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2 \beta_{n}\left\langle T x_{n}-u_{n}, J\left(y_{n}-u_{n}\right)\right\rangle \tag{2.6}\\ & \leq\left\|x_{n}-u_{n}\right\|^{2}+\beta_{n} M_{2} \end{align*}‖yn−un‖2=‖(1−βn)(xn−un)+βn(Txn−un)‖2(2.6)≤(1−βn)2‖xn−un‖2+2βn⟨Txn−un,J(yn−un)⟩≤‖xn−un‖2+βnM2
The last inequality is true because ( ⟨ T x n − u n , J ( y n − u n ) ⟩ ) n T x n − u n , J y n − u n n ((:Tx_(n)-u_(n),J(y_(n)-u_(n)):))_(n)\left(\left\langle T x_{n}-u_{n}, J\left(y_{n}-u_{n}\right)\right\rangle\right)_{n}(⟨Txn−un,J(yn−un)⟩)n is bounded, with a constant M 2 > 0 M 2 > 0 M_(2) > 0M_{2}>0M2>0. Replacing (2.5) and (2.6) in (2.1), we obtain
‖ x n + 1 − u n + 1 ‖ 2 ≤ ( 1 − α n ) 2 ‖ x n − u n ‖ 2 + 2 α n k ‖ x n − u n ‖ 2 (2.7) + σ n + α n ( 2 k ) β n M 2 = ( 1 − 2 ( 1 − k ) α n + α n 2 ) ‖ x n − u n ‖ 2 + o ( α n ) x n + 1 − u n + 1 2 ≤ 1 − α n 2 x n − u n 2 + 2 α n k x n − u n 2 (2.7) + σ n + α n ( 2 k ) β n M 2 = 1 − 2 ( 1 − k ) α n + α n 2 x n − u n 2 + o α n {:[||x_(n+1)-u_(n+1)||^(2) <= (1-alpha_(n))^(2)||x_(n)-u_(n)||^(2)+2alpha_(n)k||x_(n)-u_(n)||^(2)],[(2.7)+sigma_(n)+alpha_(n)(2k)beta_(n)M_(2)],[=(1-2(1-k)alpha_(n)+alpha_(n)^(2))||x_(n)-u_(n)||^(2)+o(alpha_(n))]:}\begin{align*} \left\|x_{n+1}-u_{n+1}\right\|^{2} \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-u_{n}\right\|^{2}+2 \alpha_{n} k\left\|x_{n}-u_{n}\right\|^{2} \\ & +\sigma_{n}+\alpha_{n}(2 k) \beta_{n} M_{2} \tag{2.7}\\ = & \left(1-2(1-k) \alpha_{n}+\alpha_{n}^{2}\right)\left\|x_{n}-u_{n}\right\|^{2}+o\left(\alpha_{n}\right) \end{align*}‖xn+1−un+1‖2≤(1−αn)2‖xn−un‖2+2αnk‖xn−un‖2(2.7)+σn+αn(2k)βnM2=(1−2(1−k)αn+αn2)‖xn−un‖2+o(αn)
The condition lim n → ∞ α n = 0 lim n → ∞   α n = 0 lim_(n rarr oo)alpha_(n)=0\lim _{n \rightarrow \infty} \alpha_{n}=0limn→∞αn=0 implies the existence of an n 0 n 0 n_(0)n_{0}n0 such that, for all n ≥ n 0 n ≥ n 0 n >= n_(0)n \geq n_{0}n≥n0, we have
(2.8) α n ≤ ( 1 − k ) (2.8) α n ≤ ( 1 − k ) {:(2.8)alpha_(n) <= (1-k):}\begin{equation*} \alpha_{n} \leq(1-k) \tag{2.8} \end{equation*}(2.8)αn≤(1−k)
Substituting (2.8) into (2.7), we get
(2.9) 1 − 2 ( 1 − k ) α n + α n 2 ≤ 1 − 2 ( 1 − k ) α n + ( 1 − k ) α n = 1 − ( 1 − k ) α n (2.9) 1 − 2 ( 1 − k ) α n + α n 2 ≤ 1 − 2 ( 1 − k ) α n + ( 1 − k ) α n = 1 − ( 1 − k ) α n {:(2.9)1-2(1-k)alpha_(n)+alpha_(n)^(2) <= 1-2(1-k)alpha_(n)+(1-k)alpha_(n)=1-(1-k)alpha_(n):}\begin{equation*} 1-2(1-k) \alpha_{n}+\alpha_{n}^{2} \leq 1-2(1-k) \alpha_{n}+(1-k) \alpha_{n}=1-(1-k) \alpha_{n} \tag{2.9} \end{equation*}(2.9)1−2(1−k)αn+αn2≤1−2(1−k)αn+(1−k)αn=1−(1−k)αn
Finally,
(2.10) ‖ x n + 1 − u n + 1 ‖ 2 ≤ ( 1 − ( 1 − k ) α n ) ‖ x n − u n ‖ 2 + o ( α n ) . (2.10) x n + 1 − u n + 1 2 ≤ 1 − ( 1 − k ) α n x n − u n 2 + o α n . {:(2.10)||x_(n+1)-u_(n+1)||^(2) <= (1-(1-k)alpha_(n))||x_(n)-u_(n)||^(2)+o(alpha_(n)).:}\begin{equation*} \left\|x_{n+1}-u_{n+1}\right\|^{2} \leq\left(1-(1-k) \alpha_{n}\right)\left\|x_{n}-u_{n}\right\|^{2}+o\left(\alpha_{n}\right) . \tag{2.10} \end{equation*}(2.10)‖xn+1−un+1‖2≤(1−(1−k)αn)‖xn−un‖2+o(αn).
With a n := ‖ x n − u n ‖ 2 , λ n := ( 1 − k ) α n ∈ ( 0 , 1 ) a n := x n − u n 2 , λ n := ( 1 − k ) α n ∈ ( 0 , 1 ) a_(n):=||x_(n)-u_(n)||^(2),lambda_(n):=(1-k)alpha_(n)in(0,1)a_{n}:=\left\|x_{n}-u_{n}\right\|^{2}, \lambda_{n}:=(1-k) \alpha_{n} \in(0,1)an:=‖xn−un‖2,λn:=(1−k)αn∈(0,1), and using Lemma 1.3, we obtain lim n → ∞ a n = lim n → ∞ ‖ x n − u n ‖ 2 = 0 lim n → ∞   a n = lim n → ∞   x n − u n 2 = 0 lim_(n rarr oo)a_(n)=lim_(n rarr oo)||x_(n)-u_(n)||^(2)=0\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left\|x_{n}-u_{n}\right\|^{2}=0limn→∞an=limn→∞‖xn−un‖2=0, that is,
(2.11) lim n → ∞ ‖ x n − u n ‖ = 0 (2.11) lim n → ∞   x n − u n = 0 {:(2.11)lim_(n rarr oo)||x_(n)-u_(n)||=0:}\begin{equation*} \lim _{n \rightarrow \infty}\left\|x_{n}-u_{n}\right\|=0 \tag{2.11} \end{equation*}(2.11)limn→∞‖xn−un‖=0
Let x ∗ x ∗ x^(**)x^{*}x∗ be the fixed point of T T TTT. Suppose that lim n → ∞ u n = x ∗ lim n → ∞   u n = x ∗ lim_(n rarr oo)u_(n)=x^(**)\lim _{n \rightarrow \infty} u_{n}=x^{*}limn→∞un=x∗. The inequality
(2.12) 0 ≤ ‖ x ∗ − x n ‖ ≤ ‖ u n − x ∗ ‖ + ‖ x n − u n ‖ (2.12) 0 ≤ x ∗ − x n ≤ u n − x ∗ + x n − u n {:(2.12)0 <= ||x^(**)-x_(n)|| <= ||u_(n)-x^(**)||+||x_(n)-u_(n)||:}\begin{equation*} 0 \leq\left\|x^{*}-x_{n}\right\| \leq\left\|u_{n}-x^{*}\right\|+\left\|x_{n}-u_{n}\right\| \tag{2.12} \end{equation*}(2.12)0≤‖x∗−xn‖≤‖un−x∗‖+‖xn−un‖
and (2.11) imply that lim n → ∞ x n = x ∗ lim n → ∞   x n = x ∗ lim_(n rarr oo)x_(n)=x^(**)\lim _{n \rightarrow \infty} x_{n}=x^{*}limn→∞xn=x∗. Analogously, lim n → ∞ x n = x ∗ lim n → ∞   x n = x ∗ lim_(n rarr oo)x_(n)=x**\lim _{n \rightarrow \infty} x_{n}=x *limn→∞xn=x∗ implies lim n → ∞ u n = x ∗ lim n → ∞   u n = x ∗ lim_(n rarr oo)u_(n)=x^(**)\lim _{n \rightarrow \infty} u_{n}=x^{*}limn→∞un=x∗.
REMARK 2.2. (i) If T T TTT has a fixed point, then Theorem 2.1 holds without the continuity of T T TTT.
(ii) If B B BBB is not bounded, then Theorem 2.1 holds, supposing that ( x n ) n x n n (x_(n))_(n)\left(x_{n}\right)_{n}(xn)n is bounded. The point was to prove that if Mann iteration is convergent (thus bounded), then Ishikawa iteration is convergent too. We remark that having the convergence of Ishikawa iteration, one can immediately deduce the convergence of Mann iteration by setting β n = 0 β n = 0 beta_(n)=0\beta_{n}=0βn=0 for all n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N in (1.3).
Theorem 2.1 does not completely generalize the Lipschitzian case from [10] because the operator there is not necessarily bounded.
Theorem 2.3 [10]. Let K K KKK be a closed convex (not necessary bounded) subset of an arbitrary Banach space X X XXX and let T T TTT be a Lipschitzian pseudocontractive selfmap of K K KKK. We consider Mann iteration and Ishikawa iteration with the same initial point and with the conditions lim n → ∞ α n = 0 , lim n → ∞ β n = 0 lim n → ∞   α n = 0 , lim n → ∞   β n = 0 lim_(n rarr oo)alpha_(n)=0,lim_(n rarr oo)beta_(n)=0\lim _{n \rightarrow \infty} \alpha_{n}=0, \lim _{n \rightarrow \infty} \beta_{n}=0limn→∞αn=0,limn→∞βn=0, and ∑ n = 1 ∞ α n = ∞ ∑ n = 1 ∞   α n = ∞ sum_(n=1)^(oo)alpha_(n)=oo\sum_{n=1}^{\infty} \alpha_{n}=\infty∑n=1∞αn=∞. Let x ∗ ∈ F ( T ) x ∗ ∈ F ( T ) x^(**)in F(T)x^{*} \in F(T)x∗∈F(T). Then the following conditions are equivalent:
(i) Mann iteration (1.1) converges to x ∗ ∈ F ( T ) x ∗ ∈ F ( T ) x^(**)in F(T)x^{*} \in F(T)x∗∈F(T);
(ii) Ishikawa iteration (1.3) converges to x ∗ ∈ F ( T ) x ∗ ∈ F ( T ) x^(**)in F(T)x^{*} \in F(T)x∗∈F(T).
3. Further equivalences. Let S S SSS be a strongly accretive operator. We consider when the equation S x = f S x = f Sx=fS x=fSx=f has a solution for a given f ∈ X f ∈ X f in Xf \in Xf∈X. It easy to see that
(3.1) T x = x + f − S x , ∀ x ∈ X , (3.1) T x = x + f − S x , ∀ x ∈ X , {:(3.1)Tx=x+f-Sx","quad AA x in X",":}\begin{equation*} T x=x+f-S x, \quad \forall x \in X, \tag{3.1} \end{equation*}(3.1)Tx=x+f−Sx,∀x∈X,
is a strongly pseudocontractive operator. A fixed point for T T TTT is the solution of S x = f S x = f Sx=fS x=fSx=f, and conversely. Theorem 2.1 assures that the convergence of Mann and Ishikawa iterations to the fixed point of T T TTT are equivalent for bounded strongly pseudocontractive maps. A similar result holds for the convergence of Mann and Ishikawa iterations to the solution of S x = f S x = f Sx=fS x=fSx=f. Suppose that the operator S S SSS is strongly accretive. It is well known that if S S SSS is bounded, ( I − S I − S I-SI-SI−S ) could be unbounded. Take, for example, S : R → B = [ − 1 , 1 ] S : R → B = [ − 1 , 1 ] S:R rarr B=[-1,1]S: R \rightarrow B=[-1,1]S:R→B=[−1,1] with S ( x ) = ( 1 / 2 ) cos ⁡ x S ( x ) = ( 1 / 2 ) cos ⁡ x S(x)=(1//2)cos xS(x)=(1 / 2) \cos xS(x)=(1/2)cos⁡x. According to [2], the map ( I − S ) ( x ) = x − ( 1 / 2 ) cos ⁡ x ( I − S ) ( x ) = x − ( 1 / 2 ) cos ⁡ x (I-S)(x)=x-(1//2)cos x(I-S)(x)=x-(1 / 2) \cos x(I−S)(x)=x−(1/2)cos⁡x is strongly accretive and ( I − S ) ( R ) = R ( I − S ) ( R ) = R (I-S)(R)=R(I-S)(R)=R(I−S)(R)=R. Thus, if B B BBB is bounded and x ∈ B x ∈ B x in Bx \in Bx∈B does not mean that T x = x − S x + f ∈ B T x = x − S x + f ∈ B Tx=x-Sx+f in BT x=x-S x+f \in BTx=x−Sx+f∈B. For the same ( α n ) n , ( β n ) n ⊂ ( 0 , 1 ) α n n , β n n ⊂ ( 0 , 1 ) (alpha_(n))_(n),(beta_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n} \subset(0,1)(αn)n,(βn)n⊂(0,1) as in (1.4), iterations (1.1) and (1.3) become
x n + 1 = ( 1 − α n ) x n + α n ( f + ( I − S ) y n ) (3.2) y n = ( 1 − β n ) x n + β n ( f + ( I − S ) x n ) , (3.3) u n + 1 = ( 1 − α n ) u n + α n ( f + ( I − S ) u n ) , n = 1 , 2 , … x n + 1 = 1 − α n x n + α n f + ( I − S ) y n (3.2) y n = 1 − β n x n + β n f + ( I − S ) x n , (3.3) u n + 1 = 1 − α n u n + α n f + ( I − S ) u n , n = 1 , 2 , … {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)(f+(I-S)y_(n))],[(3.2)y_(n)=(1-beta_(n))x_(n)+beta_(n)(f+(I-S)x_(n))","],[(3.3)u_(n+1)=(1-alpha_(n))u_(n)+alpha_(n)(f+(I-S)u_(n))","quad n=1","2","dots]:}\begin{align*} x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n}\left(f+(I-S) y_{n}\right) \\ y_{n} & =\left(1-\beta_{n}\right) x_{n}+\beta_{n}\left(f+(I-S) x_{n}\right), \tag{3.2}\\ u_{n+1} & =\left(1-\alpha_{n}\right) u_{n}+\alpha_{n}\left(f+(I-S) u_{n}\right), \quad n=1,2, \ldots \tag{3.3} \end{align*}xn+1=(1−αn)xn+αn(f+(I−S)yn)(3.2)yn=(1−βn)xn+βn(f+(I−S)xn),(3.3)un+1=(1−αn)un+αn(f+(I−S)un),n=1,2,…
The existence of the solution for S x = f S x = f Sx=fS x=fSx=f when S S SSS is a continuous and strongly accretive operator results from [8]. This argument and Remark 2.2(ii) lead us to the following corollary.
Corollary 3.1. Let X X XXX be a real Banach space with a uniformly convex dual and B B BBB a nonempty, convex, and closed subset of X X XXX. Let S : B → B S : B → B S:B rarr BS: B \rightarrow BS:B→B be a continuous and strongly accretive operator and let ( x n ) n x n n (x_(n))_(n)\left(x_{n}\right)_{n}(xn)n, given by (3.2), be bounded. Then, for u 1 = x 1 ∈ B u 1 = x 1 ∈ B u_(1)=x_(1)in Bu_{1}=x_{1} \in Bu1=x1∈B, the following assertions are equivalent:
(a) Mann iteration (3.3) converges to the solution of S x = f S x = f Sx=fS x=fSx=f;
(b) Ishikawa iteration (3.2) converges to the solution of S x = f S x = f Sx=fS x=fSx=f.
Let S S SSS be an accretive operator. From Remark 1.2(ii), the operator T x = f − S x T x = f − S x Tx=f-SxT x=f-S xTx=f−Sx is strongly pseudocontractive for a given f ∈ X f ∈ X f in Xf \in Xf∈X. A solution for T x = x T x = x Tx=xT x=xTx=x becomes a solution for x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f. For ( α n ) n , ( β n ) n ⊂ ( 0 , 1 ) α n n , β n n ⊂ ( 0 , 1 ) (alpha_(n))_(n),(beta_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n} \subset(0,1)(αn)n,(βn)n⊂(0,1), as in (1.4), iterations (1.1) and (1.3) become
x n + 1 = ( 1 − α n ) x n + α n ( f − S y n ) (3.4) y n = ( 1 − β n ) x n + β n ( f − S x n ) , n = 1 , 2 , … (3.5) u n + 1 = ( 1 − α n ) u n + α n ( f − S u n ) , n = 1 , 2 , … x n + 1 = 1 − α n x n + α n f − S y n (3.4) y n = 1 − β n x n + β n f − S x n , n = 1 , 2 , … (3.5) u n + 1 = 1 − α n u n + α n f − S u n , n = 1 , 2 , … {:[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)(f-Sy_(n))],[(3.4)y_(n)=(1-beta_(n))x_(n)+beta_(n)(f-Sx_(n))","quad n=1","2","dots],[(3.5)u_(n+1)=(1-alpha_(n))u_(n)+alpha_(n)(f-Su_(n))","quad n=1","2","dots]:}\begin{align*} x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n}\left(f-S y_{n}\right) \\ y_{n} & =\left(1-\beta_{n}\right) x_{n}+\beta_{n}\left(f-S x_{n}\right), \quad n=1,2, \ldots \tag{3.4}\\ u_{n+1} & =\left(1-\alpha_{n}\right) u_{n}+\alpha_{n}\left(f-S u_{n}\right), \quad n=1,2, \ldots \tag{3.5} \end{align*}xn+1=(1−αn)xn+αn(f−Syn)(3.4)yn=(1−βn)xn+βn(f−Sxn),n=1,2,…(3.5)un+1=(1−αn)un+αn(f−Sun),n=1,2,…
The existence of a solution for this equation follows from [7]. We are now able to give the following result.
Corollary 3.2. Let X X XXX be a real Banach space with a uniformly convex dual and B B BBB a nonempty, convex, and closed subset of X X XXX. Let S : B → B S : B → B S:B rarr BS: B \rightarrow BS:B→B be a continuous and accretive operator and let ( x n ) n x n n (x_(n))_(n)\left(x_{n}\right)_{n}(xn)n, given by (3.4), be bounded. Then, for u 1 = x 1 ∈ B u 1 = x 1 ∈ B u_(1)=x_(1)in Bu_{1}=x_{1} \in Bu1=x1∈B, the following assertions are equivalent:
(a) Mann iteration (3.5) converges to the solution of x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f;
(b) Ishikawa iteration (3.4) converges to the solution of x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f.
Observe that if S S SSS is not continuous, and the equations S x = f S x = f Sx=fS x=fSx=f, respectively, x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f, have solutions, then Corollary 3.1, respectively, Corollary 3.2 hold.
We remark that all the results from this paper hold in the multivalued case, provided that these multivalued maps admit appropriate single-valued selections.

References

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[6] W. R. Mann, Mean value methods in iteration, Proc. Amer. Math. Soc. 4 (1953), 506-510.
[7] R. H. Martin Jr., A global existence theorem for autonomous differential equations in a Banach space, Proc. Amer. Math. Soc. 26 (1970), 307-314.
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[10] B. E. Rhoades and S. M. Soltuz, On the equivalence of Mann and Ishikawa iteration methods, Int. J. Math. Math. Sci. 2003 (2003), no. 7, 451-459.
[11] X. Weng, Fixed point iteration for local strictly pseudocontractive mapping, Proc. Amer. Math. Soc. 113 (1991), no. 3, 727-731.
B. E. Rhoades: Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
E-mail address: rhoades@indiana.edu
Ştefan M. Şoltuz: "T. Popoviciu" Institute of Numerical Analysis, P.O. Box 68-1, 3400 Cluj-Napoca, Romania
E-mail address: stefansoltuz@personal.ro
2003

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