Mann iteration for generalized pseudocontractive maps in Hilbert spaces

Abstract

If \(X\) is a real Hilbert space, \(B\) is a nonempty, bounded, convex and closed subset, \(T:B\rightarrow B\) is a generalized pseudocontraction; then the iteration%

\begin{align*}
x_{1} & \in B,\\
x_{n+1} & =(1-\U{3b1} _{n})x_{n}+\U{3b1} _{n}Tx_{n},\\
(\U{3b1} _{n})_{n} & \subset(0,1),\sum \limits_{n-1}^{\infty}\alpha_{n}%
=\infty,\\
\sum \limits_{n-1}^{\infty}\left \vert \alpha_{n+1}-\alpha_{n}\right \vert &
<\infty,\lim_{n\rightarrow \infty}\alpha_{n}=0,
\end{align*}

strongly converges to the fixed point of \(T\).

Authors

Stefan M. Soltuz
Tiberiu Popoviciu Institute of Numerical Analysis

Keywords

Mann iteration; fixed points

Paper coordinates

Ş.M. Şoltuz, Mann iteration for generalized pseudocontractive maps in Hilbert spaces, Math. Commun. 6 (2001) no. 1, 97-100.

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Journal

Mathematical Communications

Publisher Name

 Osijek: Department of Mathematics J. J. Strossmayer University of Osijek; Osijek Mathematical Society

DOI
Print ISSN
Online ISSN

ISSN 1848-8013

google scholar link

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[6] J. A. Park, Mann-iteration for strictly pseudocontractive maps, J. Korean Math. Soc. 31(1994), 333-337.
[7] N. Shioji, W. Takahashi, Strong convergence of approximated sequences for nonexpansive mappings in Banach spaces, Proc. Amer. Math. Soc. 125(1997), 3641-3645.
[8] R. U. Verma, A fixed point theorem involving Lipschitzian generalised pseudocontractions, Proc. Royal Irish Acad. 97A(1997), 83-86.
[9] X.Weng, Fixed point iteration for local strictly pseudocontractive mapping, Proc. Amer. Math. Soc. 113(1991), 727-731.

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Mann iteration for generalized pseudocontractive maps in Hilbert spaces

Ştefan M. Şoltuz*

Abstract

If X X XXX is a real Hilbert space, B B BBB is a nonempty, bounded, convex and closed subset, T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is a generalized pseudocontraction; then the iteration

(1) x 1 ∈ B , x n + 1 = ( 1 − α n ) x n + α n T x n , ( α n ) n ⊂ ( 0 , 1 ) , ∑ n = 1 ∞ α n = ∞ , ∑ n = 1 ∞ | α n + 1 − α n | < ∞ , lim n → ∞ α n = 0 , (1) x 1 ∈ B , x n + 1 = 1 − α n x n + α n T x n , α n n ⊂ ( 0 , 1 ) , ∑ n = 1 ∞ &hairsp; α n = ∞ , ∑ n = 1 ∞ &hairsp; α n + 1 − α n < ∞ , lim n → ∞ &hairsp; α n = 0 , {:[(1)x_(1) in B","],[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Tx_(n)","],[(alpha_(n))_(n) sub(0","1)","sum_(n=1)^(oo)alpha_(n)=oo","],[sum_(n=1)^(oo)|alpha_(n+1)-alpha_(n)| < oo","lim_(n rarr oo)alpha_(n)=0","]:}\begin{align*} x_{1} & \in B, \tag{1}\\ x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T x_{n}, \\ \left(\alpha_{n}\right)_{n} & \subset(0,1), \sum_{n=1}^{\infty} \alpha_{n}=\infty, \\ \sum_{n=1}^{\infty}\left|\alpha_{n+1}-\alpha_{n}\right| & <\infty, \lim _{n \rightarrow \infty} \alpha_{n}=0, \end{align*}(1)x1∈B,xn+1=(1−αn)xn+αnTxn,(αn)n⊂(0,1),∑n=1∞αn=∞,∑n=1∞|αn+1−αn|<∞,limn→∞αn=0,
strongly converges to the fixed point of T T TTT.
Key words: Mann iteration, fixed points
AMS subject classifications: 47 H 10 , 47 H 06 47 H 10 , 47 H 06 47H10,47H0647 \mathrm{H} 10,47 \mathrm{H} 0647H10,47H06
Received June 10, 2000
Accepted May 18, 2001

1. Preliminaries

In this note we study the convergence of the Mann iteration process (1) for generalized pseudocontractions. According to [8] the generalized pseudocontractions are more general than the pseudocontractions introduced by Browder.
Definition 1. [8]. Let X X XXX be a Hilbert 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 said to be a generalized pseudocontraction if for x , y ∈ B x , y ∈ B x,y in Bx, y \in Bx,y∈B there exists r > 0 r > 0 r > 0r>0r>0 such that
(2) ⟨ T x − T y , x − y ⟩ ≤ r ‖ x − y ‖ 2 . (2) ⟨ T x − T y , x − y ⟩ ≤ r ‖ x − y ‖ 2 . {:(2)(:Tx-Ty","x-y:) <= r||x-y||^(2).:}\begin{equation*} \langle T x-T y, x-y\rangle \leq r\|x-y\|^{2} . \tag{2} \end{equation*}(2)⟨Tx−Ty,x−y⟩≤r‖x−y‖2.
Clearly, (2) is equivalent to
⟨ ( I − T ) x − ( I − T ) y , x − y ⟩ ≥ ( 1 − r ) ‖ x − y ‖ 2 . ⟨ ( I − T ) x − ( I − T ) y , x − y ⟩ ≥ ( 1 − r ) ‖ x − y ‖ 2 . (:(I-T)x-(I-T)y,x-y:) >= (1-r)||x-y||^(2).\langle(I-T) x-(I-T) y, x-y\rangle \geq(1-r)\|x-y\|^{2} .⟨(I−T)x−(I−T)y,x−y⟩≥(1−r)‖x−y‖2.
The map T T TTT is a strong pseudocontraction if there exists k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) such that for all x , y ∈ B x , y ∈ B x,y in Bx, y \in Bx,y∈B,
⟨ ( I − T ) x − ( I − T ) y , x − y ⟩ ≥ k ‖ x − y ‖ 2 , ⟨ ( I − T ) x − ( I − T ) y , x − y ⟩ ≥ k ‖ x − y ‖ 2 , (:(I-T)x-(I-T)y,x-y:) >= k||x-y||^(2),\langle(I-T) x-(I-T) y, x-y\rangle \geq k\|x-y\|^{2},⟨(I−T)x−(I−T)y,x−y⟩≥k‖x−y‖2,
see, for example [6]. Remark that both generalized pseudocontractivity and strong pseudocontractivity generalize the pseudocontractivity, but in a different manner. Iteration (1), where T T TTT is a strong pseudocontraction in Banach spaces, was studied in [1], [2], [3], [4], [6], [9].
The following lemma can be found in [9] as Lemma4. Also, it can be found in [4] as Lemma 1.2, with another proof. A more general case is in Lemma 2 from [5]. The proof from [5] is similar to the proof of Lemma 4 from [9].
Lemma 1. [9], [4]. Let ( ρ n ) n ρ n n (rho_(n))_(n)\left(\rho_{n}\right)_{n}(ρn)n be a nonnegative real sequence satisfying
ρ n + 1 ≤ ( 1 − λ n ) ρ n + σ n ρ n + 1 ≤ 1 − λ n ρ n + σ n rho_(n+1) <= (1-lambda_(n))rho_(n)+sigma_(n)\rho_{n+1} \leq\left(1-\lambda_{n}\right) \rho_{n}+\sigma_{n}ρn+1≤(1−λn)ρn+σn
where λ n ∈ ( 0 , 1 ) , ∀ n ∈ N , ∑ n = 1 ∞ λ n = ∞ λ n ∈ ( 0 , 1 ) , ∀ n ∈ N , ∑ n = 1 ∞ &hairsp; λ n = ∞ lambda_(n)in(0,1),AA n in N,sum_(n=1)^(oo)lambda_(n)=oo\lambda_{n} \in(0,1), \forall n \in N, \sum_{n=1}^{\infty} \lambda_{n}=\inftyλn∈(0,1),∀n∈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 → ∞ ρ n = 0 lim n → ∞ &hairsp; ρ n = 0 lim_(n rarr oo)rho_(n)=0\lim _{n \rightarrow \infty} \rho_{n}=0limn→∞ρn=0.
The normalized duality mapping J J JJJ is the identity, when X X XXX is a Hilbert space, see [4]. Thus Lemma 1.1 from [4] becomes:
Lemma 2. [4]. If X X XXX is a Hilbert space, then
‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , ( x + y ) ⟩ , ‖ x + y ‖ 2 ≤ ‖ x ‖ 2 + 2 ⟨ y , ( x + y ) ⟩ , ||x+y||^(2) <= ||x||^(2)+2(:y,(x+y):),\|x+y\|^{2} \leq\|x\|^{2}+2\langle y,(x+y)\rangle,‖x+y‖2≤‖x‖2+2⟨y,(x+y)⟩,
for all x , y ∈ X x , y ∈ X x,y in Xx, y \in Xx,y∈X.
The following result is a corollary of Lemma 1 from [7]:
Lemma 3. [7]. If X X XXX is a real Hilbert space, B B BBB is a nonempty, bounded, convex and closed subset, and T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is a generalized pseudocontraction, then the sequence given by (1) satisfies
lim n → ∞ ‖ x n + 1 − x n ‖ = 0 lim n → ∞ &hairsp; x n + 1 − x n = 0 lim_(n rarr oo)||x_(n+1)-x_(n)||=0\lim _{n \rightarrow \infty}\left\|x_{n+1}-x_{n}\right\|=0limn→∞‖xn+1−xn‖=0
In [7], the map T T TTT is nonexpansive. If we consider the proof of Lemma 1 from [7], we see that the result is true, when our assumptions are fulfilled.

2. Main result

We are now able to give the following result:
Theorem 1. If X X XXX is a real Hilbert space, B B BBB is a nonempty, bounded, convex and closed subset, and T : B → B T : B → B T:B rarr BT: B \rightarrow BT:B→B is a generalized pseudocontraction, then the iteration (1) :
x 1 ∈ B , x n + 1 = ( 1 − α n ) x n + α n T x n ( α n ) n ⊂ ( 0 , 1 ) , ∑ n = 1 ∞ α n = ∞ , ∑ n = 1 ∞ | α n + 1 − α n | < ∞ , lim n → ∞ α n = 0 . x 1 ∈ B , x n + 1 = 1 − α n x n + α n T x n α n n ⊂ ( 0 , 1 ) , ∑ n = 1 ∞ &hairsp; α n = ∞ , ∑ n = 1 ∞ &hairsp; α n + 1 − α n < ∞ , lim n → ∞ &hairsp; α n = 0 . {:[x_(1) in B","],[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Tx_(n)],[(alpha_(n))_(n) sub(0","1)","sum_(n=1)^(oo)alpha_(n)=oo","sum_(n=1)^(oo)|alpha_(n+1)-alpha_(n)| < oo","],[lim_(n rarr oo)alpha_(n)=0.]:}\begin{aligned} x_{1} & \in B, \\ x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T x_{n} \\ \left(\alpha_{n}\right)_{n} & \subset(0,1), \sum_{n=1}^{\infty} \alpha_{n}=\infty, \sum_{n=1}^{\infty}\left|\alpha_{n+1}-\alpha_{n}\right|<\infty, \\ \lim _{n \rightarrow \infty} \alpha_{n} & =0 . \end{aligned}x1∈B,xn+1=(1−αn)xn+αnTxn(αn)n⊂(0,1),∑n=1∞αn=∞,∑n=1∞|αn+1−αn|<∞,limn→∞αn=0.
strongly converges to the fixed point of T T TTT.
Proof. Theorem 2.1 from [8] gives us the existence and the uniqueness of the fixed point of T T TTT. Let us denote this fixed point by q q qqq. Using Lemma 3 and (2), we have
‖ x n + 1 − q ‖ 2 = ‖ ( 1 − α n ) ( x n − q ) + α n ( T x n − q ) ‖ 2 ≤ ( 1 − α n ) 2 ‖ x n − q ‖ 2 + 2 α n ⟨ T x n − q , x n + 1 − q ⟩ = ( 1 − α n ) 2 ‖ x n − q ‖ 2 + 2 α n ⟨ T x n − q , x n − q ⟩ + + 2 α n ⟨ T x n − q , x n + 1 − x n ⟩ ≤ ( 1 − α n ) 2 ‖ x n − q ‖ 2 + 2 α n r ‖ x n − q ‖ 2 + 2 α n ⟨ T x n − q , x n + 1 − x n ⟩ ≤ [ 1 − α n ( 2 ( 1 − r ) − α n ) ] ‖ x n − q ‖ 2 + 2 α n ⟨ T x n − q , x n + 1 − x n ⟩ . x n + 1 − q 2 = 1 − α n x n − q + α n T x n − q 2 ≤ 1 − α n 2 x n − q 2 + 2 α n T x n − q , x n + 1 − q = 1 − α n 2 x n − q 2 + 2 α n T x n − q , x n − q + + 2 α n T x n − q , x n + 1 − x n ≤ 1 − α n 2 x n − q 2 + 2 α n r x n − q 2 + 2 α n T x n − q , x n + 1 − x n ≤ 1 − α n 2 ( 1 − r ) − α n x n − q 2 + 2 α n T x n − q , x n + 1 − x n . {:[||x_(n+1)-q||^(2)=||(1-alpha_(n))(x_(n)-q)+alpha_(n)(Tx_(n)-q)||^(2)],[ <= (1-alpha_(n))^(2)||x_(n)-q||^(2)+2alpha_(n)(:Tx_(n)-q,x_(n+1)-q:)],[=(1-alpha_(n))^(2)||x_(n)-q||^(2)+2alpha_(n)(:Tx_(n)-q,x_(n)-q:)+],[+2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):)],[ <= (1-alpha_(n))^(2)||x_(n)-q||^(2)+2alpha_(n)r||x_(n)-q||^(2)],[+2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):)],[ <= [1-alpha_(n)(2(1-r)-alpha_(n))]||x_(n)-q||^(2)],[+2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):).]:}\begin{aligned} \left\|x_{n+1}-q\right\|^{2}= & \left\|\left(1-\alpha_{n}\right)\left(x_{n}-q\right)+\alpha_{n}\left(T x_{n}-q\right)\right\|^{2} \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-q\right\|^{2}+2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-q\right\rangle \\ = & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-q\right\|^{2}+2 \alpha_{n}\left\langle T x_{n}-q, x_{n}-q\right\rangle+ \\ & +2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle \\ \leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-q\right\|^{2}+2 \alpha_{n} r\left\|x_{n}-q\right\|^{2} \\ & +2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle \\ \leq & {\left[1-\alpha_{n}\left(2(1-r)-\alpha_{n}\right)\right]\left\|x_{n}-q\right\|^{2} } \\ & +2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle . \end{aligned}‖xn+1−q‖2=‖(1−αn)(xn−q)+αn(Txn−q)‖2≤(1−αn)2‖xn−q‖2+2αn⟨Txn−q,xn+1−q⟩=(1−αn)2‖xn−q‖2+2αn⟨Txn−q,xn−q⟩++2αn⟨Txn−q,xn+1−xn⟩≤(1−αn)2‖xn−q‖2+2αnr‖xn−q‖2+2αn⟨Txn−q,xn+1−xn⟩≤[1−αn(2(1−r)−αn)]‖xn−q‖2+2αn⟨Txn−q,xn+1−xn⟩.
Let us denote
A n : = ⟨ T x n − q , x n + 1 − x n ⟩ λ n : = α n ( 2 ( 1 − r ) − α n ) ρ n : = ‖ x n − q ‖ 2 σ n : = 2 α n A n . A n : = T x n − q , x n + 1 − x n λ n : = α n 2 ( 1 − r ) − α n ρ n : = x n − q 2 σ n : = 2 α n A n . {:[A_(n):=(:Tx_(n)-q,x_(n+1)-x_(n):)],[lambda_(n):=alpha_(n)(2(1-r)-alpha_(n))],[rho_(n):=||x_(n)-q||^(2)],[sigma_(n):=2alpha_(n)A_(n).]:}\begin{aligned} A_{n}: & =\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle \\ \lambda_{n}: & =\alpha_{n}\left(2(1-r)-\alpha_{n}\right) \\ \rho_{n}: & =\left\|x_{n}-q\right\|^{2} \\ \sigma_{n}: & =2 \alpha_{n} A_{n} . \end{aligned}An:=⟨Txn−q,xn+1−xn⟩λn:=αn(2(1−r)−αn)ρn:=‖xn−q‖2σn:=2αnAn.
Thus, we have
ρ n + 1 ≤ ( 1 − λ n ) ρ n + σ n ρ n + 1 ≤ 1 − λ n ρ n + σ n rho_(n+1) <= (1-lambda_(n))rho_(n)+sigma_(n)\rho_{n+1} \leq\left(1-\lambda_{n}\right) \rho_{n}+\sigma_{n}ρn+1≤(1−λn)ρn+σn
We observe that
lim n → ∞ σ n λ n = lim n → ∞ 2 α n ⟨ T x n − q , x n + 1 − x n ⟩ α n ( 2 ( 1 − r ) − α n ) = 2 lim n → ∞ ⟨ T x n − q , x n + 1 − x n ⟩ ( 2 ( 1 − r ) − α n ) = 0 lim n → ∞ &hairsp; σ n λ n = lim n → ∞ &hairsp; 2 α n T x n − q , x n + 1 − x n α n 2 ( 1 − r ) − α n = 2 lim n → ∞ &hairsp; T x n − q , x n + 1 − x n 2 ( 1 − r ) − α n = 0 {:[lim_(n rarr oo)(sigma_(n))/(lambda_(n))=lim_(n rarr oo)(2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):))/(alpha_(n)(2(1-r)-alpha_(n)))],[=2lim_(n rarr oo)((:Tx_(n)-q,x_(n+1)-x_(n):))/((2(1-r)-alpha_(n)))=0]:}\begin{aligned} \lim _{n \rightarrow \infty} \frac{\sigma_{n}}{\lambda_{n}} & =\lim _{n \rightarrow \infty} \frac{2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle}{\alpha_{n}\left(2(1-r)-\alpha_{n}\right)} \\ & =2 \lim _{n \rightarrow \infty} \frac{\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle}{\left(2(1-r)-\alpha_{n}\right)}=0 \end{aligned}limn→∞σnλn=limn→∞2αn⟨Txn−q,xn+1−xn⟩αn(2(1−r)−αn)=2limn→∞⟨Txn−q,xn+1−xn⟩(2(1−r)−αn)=0
the last equality is true. From Lemma 4, we have lim n → ∞ ‖ x n + 1 − x n ‖ = 0 lim n → ∞ &hairsp; x n + 1 − x n = 0 lim_(n rarr oo)||x_(n+1)-x_(n)||=0\lim _{n \rightarrow \infty}\left\|x_{n+1}-x_{n}\right\|=0limn→∞‖xn+1−xn‖=0. The sequence ( ‖ T x n − q ‖ ) n T x n − q n (||Tx_(n)-q||)_(n)\left(\left\|T x_{n}-q\right\|\right)_{n}(‖Txn−q‖)n is bounded, being in the bounded set B B BBB. Hence we have lim n → ∞ ⟨ T x n − q , x n + 1 − x n ⟩ = 0 lim n → ∞ &hairsp; T x n − q , x n + 1 − x n = 0 lim_(n rarr oo)(:Tx_(n)-q,x_(n+1)-x_(n):)=0\lim _{n \rightarrow \infty}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle=0limn→∞⟨Txn−q,xn+1−xn⟩=0. The assumptions from Lemma 2 are fulfilled. Hence ρ n → 0 ρ n → 0 rho_(n)rarr0\rho_{n} \rightarrow 0ρn→0 as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞. Thus x n → q x n → q x_(n)rarr qx_{n} \rightarrow qxn→q as n → ∞ n → ∞ n rarr oon \rightarrow \inftyn→∞.
A prototype for ( α n ) n α n n (alpha_(n))_(n)\left(\alpha_{n}\right)_{n}(αn)n is ( 1 / n ) n ≥ 1 ( 1 / n ) n ≥ 1 (1//sqrtn)_(n >= 1)(1 / \sqrt{n})_{n \geq 1}(1/n)n≥1.

References

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  1. *str. Avram Iancu 13, ap. 1, 3400 Cluj-Napoca, Romania, e-mail: ssoltuz@ictp-acad. math.ubbcluj.ro
2001

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