Posts by Stefan Soltuz

For $T$$T$TT$T$ a direct pseudocontractive map, we prove the convergence of Mann iteration to the fixed point of $T$$T$TT$T$.

MSC. 47H10
Keywords: preudocontractive maps, fix points, iterative method, convergence

The sequence ${\left({\alpha }_n\right)}_{n\ge 1}$${\left({\alpha }_n\right)}_{n\ge 1}$(alpha_(n))_(n >= 1)\left(\alpha_{n}\right)_{n \geq 1}${\left({\alpha }_n\right)}_{n\ge 1}$ satisfics: ${\left({\alpha }_n\right)}_{n\ge 1}\subset (0,1),\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$${\left({\alpha }_n\right)}_{n\ge 1}\subset (0,1),\sum _{n=1}^{\mathrm{\infty }} {\alpha }_n=\mathrm{\infty }$(alpha_(n))_(n >= 1)sub(0,1),sum_(n=1)^(oo)alpha_(n)=oo\left(\alpha_{n}\right)_{n \geq 1} \subset(0,1), \sum_{n=1}^{\infty} \alpha_{n}=\infty${\left({\alpha }_n\right)}_{n\ge 1}\subset (0,1),\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, and $\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n^2<\mathrm{\infty }$$\sum _{n=1}^{\mathrm{\infty }} {\alpha }_n^2<\mathrm{\infty }$sum_(n=1)^(oo)alpha_(n)^(2) < oo\sum_{n=1}^{\infty} \alpha_{n}^{2}< \infty$\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n^2<\mathrm{\infty }$. The last relation implies that $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0$$\underset{n\to \mathrm{\infty }}{lim} {\alpha }_n=0$lim_(n rarr oo)alpha_(n)=0\lim _{n \rightarrow \infty} \alpha_{n}=0$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0$. A prototype for ${\left({\alpha }_n\right)}_{n\ge 1}$${\left({\alpha }_n\right)}_{n\ge 1}$(alpha_(n))_(n >= 1)\left(\alpha_{n}\right)_{n \geq 1}${\left({\alpha }_n\right)}_{n\ge 1}$ is $(1/n{)}_{n\ge 1}$$(1/n{)}_{n\ge 1}$(1//n)_(n >= 1)(1 / n)_{n \geq 1}$(1/n{)}_{n\ge 1}$.

Definition 1 The map $T$$T$TT$T$ is called pseudocontractive if

In [7] we can see an example of a Lipechitz pseudocontractive map with a unique fixed point for which every non trivial Mann sequence fails to converge. The set $B$$B$BB$B$ is nonempty, convex and compact.

Definition 2 The map $T$$T$TT$T$ is called strongly pseudocontractive if there exists $q\in (0,1)$$q\in (0,1)$q in(0,1)q \in(0,1)$q\in (0,1)$ such that

In [1], [2], [3], [5], [8], [11] the map $T$$T$TT$T$ is considered strongly pseudocontractive. The sequence ${\left(x_n\right)}_{n\ge 1}$${\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) strongly converges to a fixed point of $T$$T$TT$T$.

We introduce the following class of maps:
Definition 3 The map $T$$T$TT$T$ is called direct pseudocontractive if there exists $k\in (0,1)$$k\in (0,1)$k in(0,1)k \in(0,1)$k\in (0,1)$ such that

The class of direct pseudocontractive maps is nonempty. If $T$$T$TT$T$ is a contraction, then $T$$T$TT$T$ is a direct pseudocontractive map.Picard -Banach Theorem can't be used to find the fixed point of a direct psendocontractive map. Instead, Mann iteration (1) can be successfully used. Our aim is to give a convergence result for (1). We denote by $F(T):=\{x\in B:Tx=x\}$$F(T):=\{x\in B:Tx=x\}$F(T):={x in B:Tx=x}F(T):=\{x \in B: T x=x\}$F(T):=\{x\in B:Tx=x\}$.
Remark 1 If $T$$T$TT$T$ is a direct pseudocontractive map and has $F(T)\ne \mathrm{\varnothing }$$F(T)\ne \mathrm{\varnothing }$F(T)!=O/F(T) \neq \emptyset$F(T)\ne \mathrm{\varnothing }$, then $T$$T$TT$T$ hass a unique fixed point.

Proof. Let $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$ and $y^{\ast }$$y^{\ast }$y^(**)y^{*}$y^{\ast }$ be two distinct fixed points. From (2) we have

Hence $x^{\ast }=y^{\ast }$$x^{\ast }=y^{\ast }$x^(**)=y^(**)x^{*}=y^{*}$x^{\ast }=y^{\ast }$. Thus $F(T)=\left\{x^{\ast }\right\}$$F(T)=\left\{x^{\ast }\right\}$F(T)={x^(**)}F(T)=\left\{x^{*}\right\}$F(T)=\left\{x^{\ast }\right\}$.
The following lemma can be found in [10] as Lemma 4. Also, it can be found in [12] as Lemma 1.2, with an other proof. In [1] can be found as Lemma 2, the proof is similar to the proof of Lemma 1 from [8].
Lemma 4 [1], [10], [12] Let ${\left(a_n\right)}_{n\ge 1}$${\left(a_n\right)}_{n\ge 1}$(a_(n))_(n >= 1)\left(a_{n}\right)_{n \geq 1}${\left(a_n\right)}_{n\ge 1}$ be a nonnegative sequence which verifies where $a_{n+1}\le (1-{\lambda }_n)a_n+{\sigma }_n,{\left({\lambda }_n\right)}_{n\ge 1}\subset (0,1),\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$$a_{n+1}\le \left(1-{\lambda }_n\right)a_n+{\sigma }_n,{\left({\lambda }_n\right)}_{n\ge 1}\subset (0,1),\sum _{n=1}^{\mathrm{\infty }} {\lambda }_n=\mathrm{\infty }$a_(n+1) <= (1-lambda_(n))a_(n)+sigma_(n),(lambda_(n))_(n >= 1)sub(0,1),sum_(n=1)^(oo)lambda_(n)=ooa_{n+1} \leq\left(1-\lambda_{n}\right) a_{n}+\sigma_{n},\left(\lambda_{n}\right)_{n \geq 1} \subset(0,1), \sum_{n=1}^{\infty} \lambda_{n}=\infty$a_{n+1}\le (1-{\lambda }_n)a_n+{\sigma }_n,{\left({\lambda }_n\right)}_{n\ge 1}\subset (0,1),\sum _{n=1}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$ and ${\sigma }_n=o\left({\lambda }_n\right)$${\sigma }_n=o\left({\lambda }_n\right)$sigma_(n)=o(lambda_(n))\sigma_{n}=o\left(\lambda_{n}\right)${\sigma }_n=o\left({\lambda }_n\right)$. Then $\underset{n-\mathrm{\infty }}{lim}{a}_n=0$$\underset{n-\mathrm{\infty }}{lim} a_n=0$lim_(n-oo)a_(n)=0\lim _{n-\infty} a_{n}=0$\underset{n-\mathrm{\infty }}{lim}{a}_n=0$.

The following result is proved in [4].
Lemma 5 (4) Let $H$$H$HH$H$ be a Hilbert space, the following relation is true for all $x,y\in H$$x,y\in H$x,y in Hx, y \in H$x,y\in H$, and for all $\lambda \in (0,1)$$\lambda \in (0,1)$lambda in(0,1)\lambda \in(0,1)$\lambda \in (0,1)$ :

2.The main result.

We are able now to give the main result:

Theorem 6 Let $H$$H$HH$H$ be a real Hilbert space, let $B\subset H$$B\subset H$B sub HB \subset H$B\subset H$ be a nonempty, convex, bounded and closed set and let $T:B\to B$$T:B\to B$T:B rarr BT: B \rightarrow B$T:B\to B$ be a continuous, direct pseudocontructive map, with $F(T)\ne \mathrm{\varnothing }$$F(T)\ne \mathrm{\varnothing }$F(T)!=O/F(T) \neq \emptyset$F(T)\ne \mathrm{\varnothing }$. Then for each $x_1$$x_1$x_(1)x_{1}$x_1$ a fixed point in $B$$B$BB$B$, the sequence ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ given by (1) converges strongly to the unique fixed point of $T$$T$TT$T$.

Proof. Let $x^{\ast }\in F(T)$$x^{\ast }\in F(T)$x^(**)in F(T)x^{*} \in F(T)$x^{\ast }\in F(T)$. From remark 2 we know that $F(T)=\left\{x^{\ast }\right\}$$F(T)=\left\{x^{\ast }\right\}$F(T)={x^(**)}F(T)=\left\{x^{*}\right\}$F(T)=\left\{x^{\ast }\right\}$. Using (2) and (3) we get

The sequence ${\left({\Vert T{x}_n-x_n\Vert }^2\right)}_{n\ge 1}$${\left({‖T{x}_n-x_n‖}^2\right)}_{n\ge 1}$(||Tx_(n)-x_(n)||^(2))_(n >= 1)\left(\left\|T x_{n}-x_{n}\right\|^{2}\right)_{n \geq 1}${\left({\Vert T{x}_n-x_n\Vert }^2\right)}_{n\ge 1}$ is bounded, because $B$$B$BB$B$ is bounded. There exists $M>0$$M>0$M > 0M>0$M>0$ such that ${\Vert T{x}_n-x_n\Vert }^2<M$${‖T{x}_n-x_n‖}^2<M$||Tx_(n)-x_(n)||^(2) < M\left\|T x_{n}-x_{n}\right\|^{2}<M${\Vert T{x}_n-x_n\Vert }^2<M$, for all $n\ge 1$$n\ge 1$n >= 1n \geq 1$n\ge 1$. We denote $a_n:={\Vert x_n-x^{\ast }\Vert }^2$$a_n:={‖x_n-x^{\ast }‖}^2$a_(n):=||x_(n)-x^(**)||^(2)a_{n}:= \left\|x_{n}-x^{*}\right\|^{2}$a_n:={\Vert x_n-x^{\ast }\Vert }^2$, and we get:

Let us denote by

Observe that ${\lambda }_n=(1-k){\alpha }_n\subset (0,1)$${\lambda }_n=(1-k){\alpha }_n\subset (0,1)$lambda_(n)=(1-k)alpha_(n)sub(0,1)\lambda_{n}=(1-k) \alpha_{n} \subset(0,1)${\lambda }_n=(1-k){\alpha }_n\subset (0,1)$, for all $n\ge 1$$n\ge 1$n >= 1n \geq 1$n\ge 1$. We have $\sum _{n-1}^{\mathrm{\infty }}{\lambda }_n=(1-k)\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$$\sum _{n-1}^{\mathrm{\infty }} {\lambda }_n=(1-k)\sum _{n=1}^{\mathrm{\infty }} {\alpha }_n=\mathrm{\infty }$sum_(n-1)^(oo)lambda_(n)=(1-k)sum_(n=1)^(oo)alpha_(n)=oo\sum_{n-1}^{\infty} \lambda_{n}= (1-k) \sum_{n=1}^{\infty} \alpha_{n}=\infty$\sum _{n-1}^{\mathrm{\infty }}{\lambda }_n=(1-k)\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$. The following relation is true

Thus, we have ${\sigma }_n=o\left({\lambda }_n\right)$${\sigma }_n=o\left({\lambda }_n\right)$sigma_(n)=o(lambda_(n))\sigma_{n}=o\left(\lambda_{n}\right)${\sigma }_n=o\left({\lambda }_n\right)$. From Lemma 1 we get $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$$\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$. Hence $\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-x^{\ast }\Vert =0$$\underset{n\to \mathrm{\infty }}{lim} ‖x_n-x^{\ast }‖=0$lim_(n rarr oo)||x_(n)-x^(**)||=0\lim _{n \rightarrow \infty}\left\|x_{n}-x^{*}\right\|=0$\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-x^{\ast }\Vert =0$. The proof is complete.

Using the Schauder fixed point theorem we give the following corollary:
Corollary 7 Let $H$$H$HH$H$ be a real Hilbert space, let $B\subset H$$B\subset H$B sub HB \subset H$B\subset H$ be a nonempty, convex, compact set and let $T:B\to B$$T:B\to B$T:B rarr BT: B \rightarrow B$T:B\to B$ be a continuous, direct pseudocontractive map. Then for exach $x_1$$x_1$x_(1)x_{1}$x_1$ a fixed point in $B$$B$BB$B$, the sequence ${\left(x_n\right)}_{n\ge 1}$${\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) converges strongly to the unique fixed poinl of $T$$T$TT$T$.

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125 (1997), 1705-1709.
Received: 12.03.2001
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