The convergence of Mann iteration with delay

Stefan Soltuz
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Stefan M. Soltuz
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy

Diana Otrocol
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy

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Ş.M. Şoltuz, D. Otrocol, The convergence of Mann iteration with delay, Mathematical Sciences Research, v. 11 (2007) no. 3, pp. 390-393

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Mathematical Sciences Research

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2007-SoltuzOtrocol-MathSciRes-The convergence of Mann iteration with delay

The convergence of Mann iteration with delay ŞTEFAN M. ŞOLTUZ and DIANA OTROCOL

Abstract

We show the convergence of Mann iteration with delay for various classes of non-Lipschitzian operators.

AMS Subject classification: 47 H 10 .
Key words: Mann iteration with delay, strongly pseudocontractive, strongly accretive map.

1 Introduction

Let X X XXX be a real Banach space, B B BBB be a nonempty, convex subset of X X XXX, and T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B be an operator. Let u 1 ∈ B u 1 ∈ B u_(1)in Bu_{1} \in Bu1∈B be given and s > 0 s > 0 s > 0s>0s>0 a fixed number. We consider the following iteration, to which we further refer as Mann iteration with delay, see [5]:
(1) u n + 1 = ( 1 − α n ) u n − s + α n T u n − s , (1) u n + 1 = 1 − α n u n − s + α n T u n − s , {:(1)u_(n+1)=(1-alpha_(n))u_(n-s)+alpha_(n)Tu_(n-s)",":}\begin{equation*} u_{n+1}=\left(1-\alpha_{n}\right) u_{n-s}+\alpha_{n} T u_{n-s}, \tag{1} \end{equation*}(1)un+1=(1−αn)un−s+αnTun−s,
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
(2) lim n → ∞ α n = 0 , ∑ n = 1 ∞ α n = ∞ . (2) lim n → ∞   α n = 0 , ∑ n = 1 ∞   α n = ∞ . {:(2)lim_(n rarr oo)alpha_(n)=0","sum_(n=1)^(oo)alpha_(n)=oo.:}\begin{equation*} \lim _{n \rightarrow \infty} \alpha_{n}=0, \sum_{n=1}^{\infty} \alpha_{n}=\infty . \tag{2} \end{equation*}(2)limn→∞αn=0,∑n=1∞αn=∞.
We are inspired for such delays from economics and biology problems in which fixed point are required. Usually, T T TTT is a contraction. It is wellknown that Mann iteration is desirable when T T TTT is not a contraction.
The operator 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 ‖ } , ∀ x ∈ X J x := f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ } , ∀ x ∈ X Jx:={f inX^(**):(:x,f:)=||x||^(2),||f||=:}||x||},AA x in XJ x:=\left\{f \in X^{*}:\langle x, f\rangle=\|x\|^{2},\|f\|=\right. \|x\|\}, \forall x \in XJx:={f∈X∗:⟨x,f⟩=‖x‖2,‖f‖=‖x‖},∀x∈X, is called the normalized duality mapping. The Hahn-Banach theorem assures that J x ≠ ∅ , ∀ x ∈ X J x ≠ ∅ , ∀ x ∈ X Jx!=O/,AA x in XJ x \neq \emptyset, \forall x \in XJx≠∅,∀x∈X. It is easy to see that we have ⟨ j ( x ) , y ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x , y ∈ X , ∀ j ( x ) ∈ J ( x ) ⟨ j ( x ) , y ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x , y ∈ X , ∀ j ( x ) ∈ J ( x ) (:j(x),y:) <= ||x||||y||,AA x,y in X,AA j(x)in J(x)\langle j(x), y\rangle \leq \|x\|\|y\|, \forall x, y \in X, \forall j(x) \in J(x)⟨j(x),y⟩≤‖x‖‖y‖,∀x,y∈X,∀j(x)∈J(x).
Definition 1 Let X X XXX be a real Banach space. Let B B BBB be a nonempty subset. A A AAA map T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is called strongly pseudocontractive if there exists 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
(3) ⟨ T x − T y , j ( x − y ) ⟩ ≤ ( 1 − k ) ‖ x − y ‖ 2 , ∀ x , y ∈ B . (3) ⟨ T x − T y , j ( x − y ) ⟩ ≤ ( 1 − k ) ‖ x − y ‖ 2 , ∀ x , y ∈ B . {:(3)(:Tx-Ty","j(x-y):) <= (1-k)||x-y||^(2)","AA x","y in B.:}\begin{equation*} \langle T x-T y, j(x-y)\rangle \leq(1-k)\|x-y\|^{2}, \forall x, y \in B . \tag{3} \end{equation*}(3)⟨Tx−Ty,j(x−y)⟩≤(1−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 exists 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
(4) ⟨ S x − S y , j ( x − y ) ⟩ ≥ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . (4) ⟨ S x − S y , j ( x − y ) ⟩ ≥ k ‖ x − y ‖ 2 , ∀ x , y ∈ B . {:(4)(:Sx-Sy","j(x-y):) >= k||x-y||^(2)","AA x","y in B.:}\begin{equation*} \langle S x-S y, j(x-y)\rangle \geq k\|x-y\|^{2}, \forall x, y \in B . \tag{4} \end{equation*}(4)⟨Sx−Sy,j(x−y)⟩≥k‖x−y‖2,∀x,y∈B.
In (3) when k = 1 k = 1 k=1k=1k=1, then T T TTT is called pseudocontractive. In (4) when k = 0 k = 0 k=0k=0k=0, then S S SSS is called accretive. Let us denote by I I III the identity map.
Lemma 2 If X X XXX is a real Banach space, then the following relation is true
(5) ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , j ( x + y ) ⟩ , ∀ x , y ∈ X , ∀ j ( x + y ) ∈ J ( x + y ) . (5) ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , j ( x + y ) ⟩ , ∀ x , y ∈ X , ∀ j ( x + y ) ∈ J ( x + y ) . {:(5)||x+y||^(2) <= ||x||^(2)+2(:y","j(x+y):)","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, \forall x, y \in X, \forall j(x+y) \in J(x+y) . \tag{5} \end{equation*}(5)‖x+y‖2≤‖x‖2+2⟨y,j(x+y)⟩,∀x,y∈X,∀j(x+y)∈J(x+y).
Lemma 3 [6] Let ( ρ n ) n ρ n n (rho_(n))_(n)\left(\rho_{n}\right)_{n}(ρn)n be a nonnegative sequence which satisfies the following inequality
(6) ρ n + 1 ≤ ( 1 − λ n ) ρ n + ε n (6) ρ n + 1 ≤ 1 − λ n ρ n + ε n {:(6)rho_(n+1) <= (1-lambda_(n))rho_(n)+epsi_(n):}\begin{equation*} \rho_{n+1} \leq\left(1-\lambda_{n}\right) \rho_{n}+\varepsilon_{n} \tag{6} \end{equation*}(6)ρn+1≤(1−λn)ρn+εn
where λ n ∈ ( 0 , 1 ) , ∀ n ∈ N , ∑ n = 1 ∞ λ n = ∞ λ n ∈ ( 0 , 1 ) , ∀ n ∈ N , ∑ n = 1 ∞   λ n = ∞ lambda_(n)in(0,1),AA n inN,sum_(n=1)^(oo)lambda_(n)=oo\lambda_{n} \in(0,1), \forall n \in \mathbb{N}, \sum_{n=1}^{\infty} \lambda_{n}=\inftyλn∈(0,1),∀n∈N,∑n=1∞λn=∞, and ε n = o ( λ n ) ε n = o λ n epsi_(n)=o(lambda_(n))\varepsilon_{n}=o\left(\lambda_{n}\right)εn=o(λn). Then lim n → ∞ ρ n = 0 lim n → ∞   ρ n = 0 lim_(n rarr oo)rho_(n)=0\lim _{n \rightarrow \infty} \rho_{n}=0limn→∞ρn=0.
Lemma 4 Let s ≥ 0 s ≥ 0 s >= 0s \geq 0s≥0 be a fixed number, { a n } a n {a_(n)}\left\{a_{n}\right\}{an} a nonnegative sequence which satisfies the following inequality
(6) a n + 1 ≤ ( 1 − α n ) a n − s + σ n (6) a n + 1 ≤ 1 − α n a n − s + σ n {:(6)a_(n+1) <= (1-alpha_(n))a_(n-s)+sigma_(n):}\begin{equation*} a_{n+1} \leq\left(1-\alpha_{n}\right) a_{n-s}+\sigma_{n} \tag{6} \end{equation*}(6)an+1≤(1−αn)an−s+σn
where α n ∈ ( 0 , 1 ) , ∀ n ∈ N , ∑ n = 1 ∞ α n = ∞ α n ∈ ( 0 , 1 ) , ∀ n ∈ N , ∑ n = 1 ∞   α n = ∞ alpha_(n)in(0,1),AA n inN,sum_(n=1)^(oo)alpha_(n)=oo\alpha_{n} \in(0,1), \forall n \in \mathbb{N}, \sum_{n=1}^{\infty} \alpha_{n}=\inftyαn∈(0,1),∀n∈N,∑n=1∞αn=∞, and σ n = o ( α n ) σ n = o α n sigma_(n)=o(alpha_(n))\sigma_{n}=o\left(\alpha_{n}\right)σn=o(αn). 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.
Proof. Note that sequence { a n } a n {a_(n)}\left\{a_{n}\right\}{an} is the reunion of s + 1 s + 1 s+1s+1s+1 independent subsequences. If all such subsequences converges to zero then { a n } a n {a_(n)}\left\{a_{n}\right\}{an} shall converge. The generic subsequence satisfies (6). Set ρ n := a n − s , λ n := α n , ε n := σ n ρ n := a n − s , λ n := α n , ε n := σ n rho_(n):=a_(n-s),lambda_(n):=alpha_(n),epsi_(n):=sigma_(n)\rho_{n}:=a_{n-s}, \lambda_{n}:=\alpha_{n}, \varepsilon_{n}:=\sigma_{n}ρn:=an−s,λn:=αn,εn:=σn, and use Lemma 3 to obtain 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 5 Let s ≥ 0 s ≥ 0 s >= 0s \geq 0s≥0 be a fixed number, X X XXX a real Banach space with a uniformly convex dual X ∗ , B X ∗ , B X^(**),BX^{*}, BX∗,B a nonempty closed convex bounded subset of X X XXX, and T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B be a continuous strongly pseudocontractive mapping. Then the Mann iteration with delay { x n } n = 1 ∞ x n n = 1 ∞ {x_(n)}_(n=1)^(oo)\left\{x_{n}\right\}_{n=1}^{\infty}{xn}n=1∞ defined by (1) converges strongly to the unique fixed point of T T TTT.
Proof. Corollary 1 of [2] assures the existence of a fixed point. The uniqueness of the fixed point comes from (3). Because X ∗ X ∗ X^(**)X^{*}X∗ is uniformly convex the duality map is singled valued (see, e.g., [1]). Let x ∗ x ∗ x^(**)x^{*}x∗ be the fixed point of T T TTT. Using (1), (3) and Lemma 2 we get
‖ u n + 1 − x ∗ ‖ 2 = ‖ ( 1 − α n ) ( u n − s − x ∗ ) + α n ( T u n − s − T x ∗ ) ‖ 2 ≤ ( 1 − α n ) 2 ‖ u n − s − x ∗ ‖ 2 + 2 α n ⟨ T u n − s − T x ∗ , J ( u n + 1 − x ∗ ) ⟩ = ( 1 − α n ) 2 ‖ u n − s − x ∗ ‖ 2 + 2 α n ⟨ T u n − s − T x ∗ , J ( u n − s − x ∗ ) ⟩ + + 2 α n ⟨ T u n − s − T x ∗ , J ( u n + 1 − x ∗ ) − J ( u n − s − x ∗ ) ⟩ ≤ ( 1 − α n ) 2 ‖ u n − s − x ∗ ‖ 2 + 2 α n ( 1 − k ) ‖ u n − s − x ∗ ‖ 2 + 2 α n σ n u n + 1 − x ∗ 2 = 1 − α n u n − s − x ∗ + α n T u n − s − T x ∗ 2 ≤ 1 − α n 2 u n − s − x ∗ 2 + 2 α n T u n − s − T x ∗ , J u n + 1 − x ∗ = 1 − α n 2 u n − s − x ∗ 2 + 2 α n T u n − s − T x ∗ , J u n − s − x ∗ + + 2 α n T u n − s − T x ∗ , J u n + 1 − x ∗ − J u n − s − x ∗ ≤ 1 − α n 2 u n − s − x ∗ 2 + 2 α n ( 1 − k ) u n − s − x ∗ 2 + 2 α n σ n {:[||u_(n+1)-x^(**)||^(2)=||(1-alpha_(n))(u_(n-s)-x^(**))+alpha_(n)(Tu_(n-s)-Tx^(**))||^(2)],[ <= (1-alpha_(n))^(2)||u_(n-s)-x^(**)||^(2)+2alpha_(n)(:Tu_(n-s)-Tx^(**),J(u_(n+1)-x^(**)):)],[=(1-alpha_(n))^(2)||u_(n-s)-x^(**)||^(2)+2alpha_(n)(:Tu_(n-s)-Tx^(**),J(u_(n-s)-x^(**)):)+],[+2alpha_(n)(:Tu_(n-s)-Tx^(**),J(u_(n+1)-x^(**))-J(u_(n-s)-x^(**)):)],[ <= (1-alpha_(n))^(2)||u_(n-s)-x^(**)||^(2)+2alpha_(n)(1-k)||u_(n-s)-x^(**)||^(2)+2alpha_(n)sigma_(n)]:}\begin{aligned} \left\|u_{n+1}-x^{*}\right\|^{2} & =\left\|\left(1-\alpha_{n}\right)\left(u_{n-s}-x^{*}\right)+\alpha_{n}\left(T u_{n-s}-T x^{*}\right)\right\|^{2} \\ & \leq\left(1-\alpha_{n}\right)^{2}\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n}\left\langle T u_{n-s}-T x^{*}, J\left(u_{n+1}-x^{*}\right)\right\rangle \\ & =\left(1-\alpha_{n}\right)^{2}\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n}\left\langle T u_{n-s}-T x^{*}, J\left(u_{n-s}-x^{*}\right)\right\rangle+ \\ & +2 \alpha_{n}\left\langle T u_{n-s}-T x^{*}, J\left(u_{n+1}-x^{*}\right)-J\left(u_{n-s}-x^{*}\right)\right\rangle \\ & \leq\left(1-\alpha_{n}\right)^{2}\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n}(1-k)\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n} \sigma_{n} \end{aligned}‖un+1−x∗‖2=‖(1−αn)(un−s−x∗)+αn(Tun−s−Tx∗)‖2≤(1−αn)2‖un−s−x∗‖2+2αn⟨Tun−s−Tx∗,J(un+1−x∗)⟩=(1−αn)2‖un−s−x∗‖2+2αn⟨Tun−s−Tx∗,J(un−s−x∗)⟩++2αn⟨Tun−s−Tx∗,J(un+1−x∗)−J(un−s−x∗)⟩≤(1−αn)2‖un−s−x∗‖2+2αn(1−k)‖un−s−x∗‖2+2αnσn
where
σ n = ⟨ T u n − s − T x ∗ , J ( u n + 1 − x ∗ ) − J ( u n − s − x ∗ ) ⟩ . σ n = T u n − s − T x ∗ , J u n + 1 − x ∗ − J u n − s − x ∗ . sigma_(n)=(:Tu_(n-s)-Tx^(**),J(u_(n+1)-x^(**))-J(u_(n-s)-x^(**)):).\sigma_{n}=\left\langle T u_{n-s}-T x^{*}, J\left(u_{n+1}-x^{*}\right)-J\left(u_{n-s}-x^{*}\right)\right\rangle .σn=⟨Tun−s−Tx∗,J(un+1−x∗)−J(un−s−x∗)⟩.
Now we shall show σ n → 0 σ n → 0 sigma_(n)rarr0\sigma_{n} \rightarrow 0σn→0 as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞. Observe that ( ‖ T u n − s − T x ∗ ‖ ) n T u n − s − T x ∗ n (||Tu_(n-s)-Tx^(**)||)_(n)\left(\left\|T u_{n-s}-T x^{*}\right\|\right)_{n}(‖Tun−s−Tx∗‖)n is bounded. We prove now that
(7) J ( u n + 1 − x ∗ ) − J ( u n − s − x ∗ ) → 0 as n → ∞ (7) J u n + 1 − x ∗ − J u n − s − x ∗ → 0  as  n → ∞ {:(7)J(u_(n+1)-x^(**))-J(u_(n-s)-x^(**))rarr0" as "n rarr oo:}\begin{equation*} J\left(u_{n+1}-x^{*}\right)-J\left(u_{n-s}-x^{*}\right) \rightarrow 0 \text { as } n \rightarrow \infty \tag{7} \end{equation*}(7)J(un+1−x∗)−J(un−s−x∗)→0 as n→∞
Proposition 12.3 on page 115, of [2] assures that, when X ∗ X ∗ X^(**)X^{*}X∗ is uniformly convex, then J J JJJ is uniformly continuous on every bounded set of X X XXX. To prove (7) it is sufficient to see that
‖ u n + 1 − u n − s ‖ = α n ‖ u n − s − T u n − s ‖ ≤ α n ( ‖ u n − s ‖ + ‖ T u n − s ‖ ) ≤ 2 α n M → 0 u n + 1 − u n − s = α n u n − s − T u n − s ≤ α n u n − s + T u n − s ≤ 2 α n M → 0 ||u_(n+1)-u_(n-s)||=alpha_(n)||u_(n-s)-Tu_(n-s)|| <= alpha_(n)(||u_(n-s)||+||Tu_(n-s)||) <= 2alpha_(n)M rarr0\left\|u_{n+1}-u_{n-s}\right\|=\alpha_{n}\left\|u_{n-s}-T u_{n-s}\right\| \leq \alpha_{n}\left(\left\|u_{n-s}\right\|+\left\|T u_{n-s}\right\|\right) \leq 2 \alpha_{n} M \rightarrow 0‖un+1−un−s‖=αn‖un−s−Tun−s‖≤αn(‖un−s‖+‖Tun−s‖)≤2αnM→0
where M = sup { ‖ u n − s ‖ , ‖ T u n − s ‖ } M = sup u n − s , T u n − s M=s u p{||u_(n-s)||,||Tu_(n-s)||}M=\sup \left\{\left\|u_{n-s}\right\|,\left\|T u_{n-s}\right\|\right\}M=sup{‖un−s‖,‖Tun−s‖}. The sequences ( u n − s ) n , ( T u n − s ) n u n − s n , T u n − s n (u_(n-s))_(n),(Tu_(n-s))_(n)\left(u_{n-s}\right)_{n},\left(T u_{n-s}\right)_{n}(un−s)n,(Tun−s)n are bounded being in the bounded set B B BBB. Hence (7) holds. Then
(8) ‖ u n + 1 − x ∗ ‖ 2 ≤ ( ( 1 − α n ) 2 + 2 α n ( 1 − k ) ) ‖ u n − s − x ∗ ‖ 2 + 2 α n σ n = ( 1 − 2 α n + α n 2 + 2 α n − 2 α n k ) ‖ u n − s − x ∗ ‖ 2 + 2 α n σ n = ( 1 + α n 2 − 2 α n k ) ‖ u n − s − x ∗ ‖ 2 + 2 α n σ n (8) u n + 1 − x ∗ 2 ≤ 1 − α n 2 + 2 α n ( 1 − k ) u n − s − x ∗ 2 + 2 α n σ n = 1 − 2 α n + α n 2 + 2 α n − 2 α n k u n − s − x ∗ 2 + 2 α n σ n = 1 + α n 2 − 2 α n k u n − s − x ∗ 2 + 2 α n σ n {:[(8)||u_(n+1)-x^(**)||^(2) <= ((1-alpha_(n))^(2)+2alpha_(n)(1-k))||u_(n-s)-x^(**)||^(2)+2alpha_(n)sigma_(n)],[=(1-2alpha_(n)+alpha_(n)^(2)+2alpha_(n)-2alpha_(n)k)||u_(n-s)-x^(**)||^(2)+2alpha_(n)sigma_(n)],[=(1+alpha_(n)^(2)-2alpha_(n)k)||u_(n-s)-x^(**)||^(2)+2alpha_(n)sigma_(n)]:}\begin{align*} \left\|u_{n+1}-x^{*}\right\|^{2} & \leq\left(\left(1-\alpha_{n}\right)^{2}+2 \alpha_{n}(1-k)\right)\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n} \sigma_{n} \tag{8}\\ & =\left(1-2 \alpha_{n}+\alpha_{n}^{2}+2 \alpha_{n}-2 \alpha_{n} k\right)\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n} \sigma_{n} \\ & =\left(1+\alpha_{n}^{2}-2 \alpha_{n} k\right)\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n} \sigma_{n} \end{align*}(8)‖un+1−x∗‖2≤((1−αn)2+2αn(1−k))‖un−s−x∗‖2+2αnσn=(1−2αn+αn2+2αn−2αnk)‖un−s−x∗‖2+2αnσn=(1+αn2−2αnk)‖un−s−x∗‖2+2αnσn
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
(9) α n < k (9) α n < k {:(9)alpha_(n) < k:}\begin{equation*} \alpha_{n}<k \tag{9} \end{equation*}(9)αn<k
Substituting (9) into (8) we get 1 + α n 2 − 2 α n k < 1 − α n k 1 + α n 2 − 2 α n k < 1 − α n k 1+alpha_(n)^(2)-2alpha_(n)k < 1-alpha_(n)k1+\alpha_{n}^{2}-2 \alpha_{n} k<1-\alpha_{n} k1+αn2−2αnk<1−αnk. Finally, the above inequality yields
‖ u n + 1 − x ∗ ‖ 2 ≤ ( 1 − α n k ) ‖ u n − s − x ∗ ‖ 2 + 2 α n σ n u n + 1 − x ∗ 2 ≤ 1 − α n k u n − s − x ∗ 2 + 2 α n σ n ||u_(n+1)-x^(**)||^(2) <= (1-alpha_(n)k)||u_(n-s)-x^(**)||^(2)+2alpha_(n)sigma_(n)\left\|u_{n+1}-x^{*}\right\|^{2} \leq\left(1-\alpha_{n} k\right)\left\|u_{n-s}-x^{*}\right\|^{2}+2 \alpha_{n} \sigma_{n}‖un+1−x∗‖2≤(1−αnk)‖un−s−x∗‖2+2αnσn
Setting a n = ‖ u n − s − x ∗ ‖ 2 , λ n = α n k ∈ ( 0 , 1 ) a n = u n − s − x ∗ 2 , λ n = α n k ∈ ( 0 , 1 ) a_(n)=||u_(n-s)-x^(**)||^(2),lambda_(n)=alpha_(n)k in(0,1)a_{n}=\left\|u_{n-s}-x^{*}\right\|^{2}, \lambda_{n}=\alpha_{n} k \in(0,1)an=‖un−s−x∗‖2,λn=αnk∈(0,1), and using Lemma 4, we obtain lim n → ∞ a n = lim n → ∞ ‖ u n − s − x ∗ ‖ 2 = 0 lim n → ∞   a n = lim n → ∞   u n − s − x ∗ 2 = 0 lim_(n rarr oo)a_(n)=lim_(n rarr oo)||u_(n-s)-x^(**)||^(2)=0\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left\|u_{n-s}-x^{*}\right\|^{2}=0limn→∞an=limn→∞‖un−s−x∗‖2=0 i.e.
lim n → ∞ ‖ u n − s − x ∗ ‖ = 0 lim n → ∞   u n − s − x ∗ = 0 lim_(n rarr oo)||u_(n-s)-x^(**)||=0\lim _{n \rightarrow \infty}\left\|u_{n-s}-x^{*}\right\|=0limn→∞‖un−s−x∗‖=0

3 The accretive and strongly accretive cases

Let I I III denote the identity map.
Remark 6 The operator T T TTT is a (strongly) pseudocontractive map if and only if ( I − T ) ( I − T ) (I-T)(I-T)(I−T) is a (strongly) accretive map.

Remark 7

  1. Let T , S : X → X T , S : X → X T,S:X rarr XT, S: X \rightarrow XT,S:X→X, and f ∈ X f ∈ X f in Xf \in Xf∈X be given. A fixed point for the map T x = f + ( I − S ) x , ∀ x ∈ X T x = f + ( I − S ) x , ∀ x ∈ X Tx=f+(I-S)x,AA x in XT x=f+(I-S) x, \forall x \in XTx=f+(I−S)x,∀x∈X is a solution for S x = f S x = f Sx=fS x=fSx=f.
  2. Let f ∈ X f ∈ X f in Xf \in Xf∈X be a given point. If S S SSS is an accretive map then T = f − S T = f − S T=f-ST=f-ST=f−S is a strongly pseudocontractive map.
Consider Mann iteration with delay and set 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 to obtain
(10) u n + 1 = ( 1 − α n ) u n − s + α n ( f + ( I − S ) u n − s ) (10) u n + 1 = 1 − α n u n − s + α n f + ( I − S ) u n − s {:(10)u_(n+1)=(1-alpha_(n))u_(n-s)+alpha_(n)(f+(I-S)u_(n-s)):}\begin{equation*} u_{n+1}=\left(1-\alpha_{n}\right) u_{n-s}+\alpha_{n}\left(f+(I-S) u_{n-s}\right) \tag{10} \end{equation*}(10)un+1=(1−αn)un−s+αn(f+(I−S)un−s)
Remarks 6 and 7 and Theorem 5 lead to the following results.
Corollary 8 Let X X XXX be a real Banach space with a uniformly convex dual X ∗ X ∗ X^(**)X^{*}X∗, and S : X → X S : X → X S:X rarr XS: X \rightarrow XS:X→X a continuous and strongly accretive map with ( I − S ) ( X ) ( I − S ) ( X ) (I-S)(X)(I-S)(X)(I−S)(X) bounded, { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} satisfies (2), and u 0 = x 0 ∈ X u 0 = x 0 ∈ X u_(0)=x_(0)in Xu_{0}=x_{0} \in Xu0=x0∈X, then, the Mann iteration with delay (10) converges to the solution of S x = f S x = f Sx=fS x=fSx=f.
Let S S SSS be an accretive operator. 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. Consider Mann iteration with delay, set T x := f − S x T x := f − S x Tx:=f-SxT x:=f-S xTx:=f−Sx such that
(11) u n + 1 = ( 1 − α n ) u n − s + α n ( f − S u n − s ) (11) u n + 1 = 1 − α n u n − s + α n f − S u n − s {:(11)u_(n+1)=(1-alpha_(n))u_(n-s)+alpha_(n)(f-Su_(n-s)):}\begin{equation*} u_{n+1}=\left(1-\alpha_{n}\right) u_{n-s}+\alpha_{n}\left(f-S u_{n-s}\right) \tag{11} \end{equation*}(11)un+1=(1−αn)un−s+αn(f−Sun−s)
Again, using the Remarks 6 and 7 and Theorem 5 we obtain the following result.
Corollary 9 Let X X XXX be a real Banach space with a uniformly convex dual X ∗ X ∗ X^(**)X^{*}X∗, and B B BBB a nonempty, convex, 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 with ( I − S ) ( X ) ( I − S ) ( X ) (I-S)(X)(I-S)(X)(I−S)(X) bounded, { α n } α n {alpha_(n)}\left\{\alpha_{n}\right\}{αn} satisfies (2). Then, the Mann iteration with delay (11) converges to the solution of x + S x = f x + S x = f x+Sx=fx+S x=fx+Sx=f.

References

[1] C.E. Chidume, Approximation of fixed points of strongly pseudocontractive mappings, Proc. Amer. Math. Soc., 120, 2, 1994, 545-551.
[2] K. Deimling, Zeros of accretive operators, Manucripta Math. 13 (1974), 283288.
[3] Z. Haiyun, J. Yuting, Approximation on fixed points on strongly psedocontractive maps without Lipschitz assumption, Proc. Amer. Math. Soc., 125, 6, 1997, 1705-1709.
[4] S. Ishikawa, Fixed Points by a New Iteration Method, Proc. Amer. Math. Soc. 44 (1974), 147-150.
[5] W. R. Mann, Mean Value in Iteration, Proc. Amer. Math. Soc. 4 (1953), 506-510.
[6] X. Weng, Fixed Point Iteration for Local Strictly Pseudocontractive Mapping, Proc. Amer. Math. Soc. 113 (1991), 727-731.
2007

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