The convergence of mann iteration for an asymptotic hemicontractive map

Dedeculed to Costica MUSTATA on his 60 th annwersary

Abstract. We prove that the Mam iteration converges to a fixed point of an asymptotic hemicontractive map.

MSC: 47H10
Keywords: Mann iteration, fixed point, asymptotic hemicontractive map

Let $X$$X$XX$X$ be a real Hilbert space, let $B\subset X$$B\subset X$B sub XB \subset X$B\subset X$ be a nonempty, convex set. Lot $T:B\to B$$T:B\to B$T:B rarr BT: B \rightarrow B$T:B\to B$ be a map. Let $\mathbb{N}=\{1,2,\dots \}$$\mathbb{N}=\{1,2,\dots \}$N={1,2,dots}\mathbb{N}=\{1,2, \ldots\}$\mathbb{N}=\{1,2,\dots \}$. Let $x_1\in B$$x_1\in B$x_(1)in Bx_{1} \in B$x_1\in B$, be an arbitrary fixed point. We consider the iteration

The sequence ${\left({\alpha }_n\right)}_{n\ge 1}$${\left({\alpha }_n\right)}_{n\ge 1}$(alpha_(n))_(n >= 1)\left(\alpha_{n}\right)_{n \geq 1}${\left({\alpha }_n\right)}_{n\ge 1}$ satisfies:

A prototype for ${\left({\kappa }_n\right)}_{n\ge 1}$${\left({\kappa }_n\right)}_{n\ge 1}$(kappa_(n))_(n >= 1)\left(\kappa_{n}\right)_{n \geq 1}${\left({\kappa }_n\right)}_{n\ge 1}$ is $(1/n{)}_{n⩾1}$$(1/n{)}_{n⩾1}$(1//n)_(n >= 1)(1 / n)_{n \geqslant 1}$(1/n{)}_{n⩾1}$. Iteration (1) is known as Mam iteration, sox $[8]$$[8]$[8][8]$[8]$. We consider the following iteration, known as Ishikawa iteration, soe [6]:

where ${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n\subset (0,1)$${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n\subset (0,1)$(alpha_(n))_(n),(beta_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n} \subset(0,1)${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n\subset (0,1)$. Choosing ${\beta }_n=0,\mathrm{\forall }n\in \mathbb{N}$${\beta }_n=0,\mathrm{\forall }n\in \mathbb{N}$beta_(n)=0,AA n inN\beta_{n}=0, \forall n \in \mathbb{N}${\beta }_n=0,\mathrm{\forall }n\in \mathbb{N}$, from ( $Tsh$$Tsh$TshT s h$Tsh$ ) we get (1).
Let us denote by $F(T)$$F(T)$F(T)F(T)$F(T)$ the set of fixed points of the operator $T$$T$TT$T$. We need the following . definition, see for example [10]:

Definition 1. The map $T:B\to B$$T:B\to B$T:B rarr BT: B \rightarrow B$T:B\to B$ is culled asymptotically bemicumtractme with sequence ${\left(k_n\right)}_n$${\left(k_n\right)}_n$(k_(n))_(n)\left(k_{n}\right)_{n}${\left(k_n\right)}_n$, if and only if $\underset{n\to \mathrm{\infty }}{lim},k_n,-1,F(T)\ne \mathrm{\varnothing }$$\underset{n\to \mathrm{\infty }}{lim} ,k_n,-1,F(T)\ne \mathrm{\varnothing }$lim_(n rarr oo),k_(n),-1,F(T)!=O/\lim _{n \rightarrow \infty}, k_{n},-1, F(T) \neq \emptyset$\underset{n\to \mathrm{\infty }}{lim},k_n,-1,F(T)\ne \mathrm{\varnothing }$ such that

In this note, we will consider

In context of Hilbert spaces, the convergence of Ishikawa iteration (I ah) to the fixed point of $T$$T$TT$T$, when we deal with s asymptotically hemicontractive map (with $k_n>1,\mathrm{V}n\in$$k_n>1,\mathrm{V}n\in$k_(n) > 1,Vn ink_{n}>1, \mathrm{~V} n \in$k_n>1,\mathrm{V}n\in$ N), could be found in [10]. A convergence result in normed spaces for (Ish) (with ${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n\subset (0,1)$${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n\subset (0,1)$(alpha_(n))_(n),(beta_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n} \subset(0,1)${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n\subset (0,1)$ being not convergent to zero), could be found in [0]. Anyway, in [5], we deal with an asymptotic hemicontractive like map, which is not the same as in Definition 2 with condition (2). Let us remark that:
(i) In $[10]$$[10]$[10][10]$[10]$ and $(5)$$(5)$(5)(5)$(5)$, the sequentes ${\left({\alpha }_n\right)}_n{\left({\beta }_n\right)}_n\subset (0,1)$${\left({\alpha }_n\right)}_n{\left({\beta }_n\right)}_n\subset (0,1)$(alpha_(n))_(n)(beta_(n))_(n)sub(0,1)\left(\alpha_{n}\right)_{n}\left(\beta_{n}\right)_{n} \subset(0,1)${\left({\alpha }_n\right)}_n{\left({\beta }_n\right)}_n\subset (0,1)$ are not convergent to zero, because of the cxistonce of an $c>0$$c>0$c > 0c>0$c>0$ such that $c\le {\alpha }_n\le {\beta }_{n,}\mathrm{\forall }n\in \mathbb{N}$$c\le {\alpha }_n\le {\beta }_{n,}\mathrm{\forall }n\in \mathbb{N}$c <= alpha_(n) <= beta_(n,)AA n inNc \leq \alpha_{n} \leq \beta_{n,} \forall n \in \mathbb{N}$c\le {\alpha }_n\le {\beta }_{n,}\mathrm{\forall }n\in \mathbb{N}$. From the convergence of (Ish) (which is proved in [10] for Hilbert spaces and $T$$T$TT$T$ asymptotic hemicontractive), with that sequences ${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n$${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n$(alpha_(n))_(n),(beta_(n))_(n)\left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n}${\left({\alpha }_n\right)}_n,{\left({\beta }_n\right)}_n$, we can not dednces convergence of Mann iteration. It is impossible to bave ${\beta }_e=0,\mathrm{\forall }n\in \mathbb{N}$${\beta }_e=0,\mathrm{\forall }n\in \mathbb{N}$beta_(e)=0,AA n inN\beta_{e}=0, \forall n \in \mathbb{N}${\beta }_e=0,\mathrm{\forall }n\in \mathbb{N}$.
(ii) According to [5], convergence results of (Ish) for asymptotic hemicontractive maps, exist only in [10], (and of course in [5]). So far as we know, no other papers are dealing with asymptotic hemicontractive maps and Mann iteration (1).

This two reasons lead us to remark a lack of a convergenoe result which deal with Mann iteration (1) for an bsymptotic hemicontractive map. In particular for a map asymptotically hemicontractive with sequence ${\left(k_n\right)}_n,k_n<1$${\left(k_n\right)}_n,k_n<1$(k_(n))_(n),k_(n) < 1\left(k_{n}\right)_{n}, k_{n}<1${\left(k_n\right)}_n,k_n<1$, ''n $\in N$$\in N$in N\in N$\in N$. Our aim is to prove a result which deal with the convergence of Mamn-iteration (1) for an asymptotic hemicontractive map as in Definition 1.

The following lemma can be found in [11] as Lemma 4. Also, it can be found in [12] as Lemma 1.2, with an other proof. In $|1|$$|1|$|1||1|$|1|$ can be found as Lemma 2.

Lemma 1 [1]. , [11]. [12] Let ${\left({\alpha }_n\right)}_{n\ge 1}$${\left({\alpha }_n\right)}_{n\ge 1}$(alpha_(n))_(n >= 1)\left(\alpha_{n}\right)_{n \geq 1}${\left({\alpha }_n\right)}_{n\ge 1}$ be a nonnegotive sequence which verzfies

where: ${\left({\lambda }_n\right)}_{n>1}\subset (0.1),\sum _{\pi =1}^{\mathrm{\infty }}{\lambda }_n=\cos$${\left({\lambda }_n\right)}_{n>1}\subset (0.1),\sum _{\pi =1}^{\mathrm{\infty }} {\lambda }_n=\cos$(lambda_(n))_(n > 1)sub(0.1),sum_(pi=1)^(oo)lambda_(n)=cos\left(\lambda_{n}\right)_{n>1} \subset(0.1), \sum_{\pi=1}^{\infty} \lambda_{n}=\cos${\left({\lambda }_n\right)}_{n>1}\subset (0.1),\sum _{\pi =1}^{\mathrm{\infty }}{\lambda }_n=\cos$ and ${\sigma }_n=o\left({\lambda }_n\right)$${\sigma }_n=o\left({\lambda }_n\right)$sigma_(n)=o(lambda_(n))\sigma_{n}=o\left(\lambda_{n}\right)${\sigma }_n=o\left({\lambda }_n\right)$. Then lim $n\to \mathrm{\infty },a_n\ne 0$$n\to \mathrm{\infty },a_n\ne 0$n rarr oo,a_(n)!=0n \rightarrow \infty, a_{n} \neq 0$n\to \mathrm{\infty },a_n\ne 0$.
The following result is from 61 :
Lemma 2 [6]. Let $X$$X$XX$X$ be a real Hilbert space, the following relation is frue for all $x,y\in X$$x,y\in X$x,y in Xx, y \in X$x,y\in X$, and for all $\lambda \in (0,1)$$\lambda \in (0,1)$lambda in(0,1)\lambda \in(0,1)$\lambda \in (0,1)$ :

We are able now to give the following result:
Theorem 1. Let $X$$X$XX$X$ he a real Hilbert syme and let $B\subset X$$B\subset X$B sub XB \subset X$B\subset X$ be a nonempty convex bounded set, and $T:B\to B$$T:B\to B$T:B rarr BT: B \rightarrow B$T:B\to B$ be an asymptotic homicontractive map with ${\left({\alpha }_n\right)}_{n\ge 1}{k}_n<1$${\left({\alpha }_n\right)}_{n\ge 1}{k}_n<1$(alpha_(n))_(n >= 1)k_(n) < 1\left(\alpha_{n}\right)_{n \geq 1} k_{n}<1${\left({\alpha }_n\right)}_{n\ge 1}{k}_n<1$. Let $\left\{x^{\ast }\right\}=F(T)$$\left\{x^{\ast }\right\}=F(T)${x^(**)}=F(T)\left\{x^{*}\right\}=F(T)$\left\{x^{\ast }\right\}=F(T)$. Ir ${\left({\alpha }_n\right)}_n$${\left({\alpha }_n\right)}_n$(alpha_(n))_(n)\left(\alpha_{n}\right)_{n}${\left({\alpha }_n\right)}_n$ given by (1) verifies ${\alpha }_n=o\left((1-k_n)\right)$${\alpha }_n=o\left(\left(1-k_n\right)\right)$alpha_(n)=o((1-k_(n)))\alpha_{n}=o\left(\left(1-k_{n}\right)\right)${\alpha }_n=o\left((1-k_n)\right)$, then the Mann iteration ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$, given by $(1)$$(1)$(1)(1)$(1)$, is comprgent to $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$, for all $x_1\in B$$x_1\in B$x_(1)in Bx_{1} \in B$x_1\in B$.

Proof. The sequence ${\left(\Vert x_n-x^{\ast }\Vert \right)}_n$${\left(‖x_n-x^{\ast }‖\right)}_n$(||x_(n)-x^(**)||)_(n)\left(\left\|x_{n}-x^{*}\right\|\right)_{n}${\left(\Vert x_n-x^{\ast }\Vert \right)}_n$ is woll-defined, bexause $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$ is unique. Using (3) and (2) we have, w'v $\in \mathrm{M}$$\in \mathrm{M}$inM\in \mathrm{M}$\in \mathrm{M}$,

The last inequality is true, since the sequence ${\left({\Vert T^n{x}_n-x_n\Vert }^2\right)}_{n\ge 1}$${\left({‖T^n{x}_n-x_n‖}^2\right)}_{n\ge 1}$(||T^(n)x_(n)-x_(n)||^(2))_(n >= 1)\left(\left\|T^{n} x_{n}-x_{n}\right\|^{2}\right)_{n \geq 1}${\left({\Vert T^n{x}_n-x_n\Vert }^2\right)}_{n\ge 1}$ is bounded, because $B$$B$BB$B$ is bounded. There exists $M>0$$M>0$M > 0M>0$M>0$ such that ${\Vert T^n{x}_n-x_n\Vert }^2<M$${‖T^n{x}_n-x_n‖}^2<M$||T^(n)x_(n)-x_(n)||^(2) < M\left\|T^{n} x_{n}-x_{n}\right\|^{2}<M${\Vert T^n{x}_n-x_n\Vert }^2<M$, for all $n\in \mathbb{N}$$n\in \mathbb{N}$n inNn \in \mathbb{N}$n\in \mathbb{N}$. We denote by $a_n:=\Vert x_n-x^{\ast }\Vert ,{\lambda }_n={\alpha }_n(1-k_n),{\sigma }_n={\alpha }_n^{2}M,\mathrm{\forall }n\in \mathbb{N}$$a_n:=‖x_n-x^{\ast }‖,{\lambda }_n={\alpha }_n\left(1-k_n\right),{\sigma }_n={\alpha }_n^{2}M,\mathrm{\forall }n\in \mathbb{N}$a_(n):=||x_(n)-x^(**)||,lambda_(n)=alpha_(n)(1-k_(n)),sigma_(n)=alpha_(n)^(2)M,AA n inNa_{n}:=\left\|x_{n}-x^{*}\right\|, \lambda_{n}=\alpha_{n}\left(1-k_{n}\right), \sigma_{n}=\alpha_{n}^{2} M, \forall n \in \mathbb{N}$a_n:=\Vert x_n-x^{\ast }\Vert ,{\lambda }_n={\alpha }_n(1-k_n),{\sigma }_n={\alpha }_n^{2}M,\mathrm{\forall }n\in \mathbb{N}$. Because ${\alpha }_n=o\left((1-k_n)\right)$${\alpha }_n=o\left(\left(1-k_n\right)\right)$alpha_(n)=o((1-k_(n)))\alpha_{n}= o\left(\left(1-k_{n}\right)\right)${\alpha }_n=o\left((1-k_n)\right)$, we have $\underset{n\to \mathrm{\infty }}{lim}\frac{{\kappa }_n}{1-k_n}M=0$$\underset{n\to \mathrm{\infty }}{lim} \frac{{\kappa }_n}{1-k_n}M=0$lim_(n rarr oo)(kappa_(n))/(1-k_(n))M=0\lim _{n \rightarrow \infty} \frac{\kappa_{n}}{1-k_{n}} M=0$\underset{n\to \mathrm{\infty }}{lim}\frac{{\kappa }_n}{1-k_n}M=0$. All the conditions from Lemma 2 are fulfilled. Thus $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$$\underset{n\to \mathrm{\infty }}{lim} a_n=0$lim_(n rarr oo)a_(n)=0\lim _{n \rightarrow \infty} a_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$. The proof is complete.

In Theorem 4 we don't need any Lipschitx condition for $T$$T$TT$T$ as in [10]

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Receivext: 21.04.2002