Published: 19 October 2019

# Fixed point problems of nonexpansive mappings for nononvex set in Hilbert spaces

Xianbing Wu1
1Department of Mathematics Yangtze Normal University, Chongqing, China
Views 85

#### Abstract

In this paper, we introduce a new concept of W-nonexpansive mappings and obtain fixed point theorems for nonexpansive mappings for non-convex set. Our results resolve fixed pointed problem that nonexpansive mappings be not on closed convex set, and it extends fixed point theorems for nonexpansive mappings.

## 1. Introduction and preliminaries

Fixed point theory is widely applied in engineering. Browder (1965) , Kirk (1965)  obtained fixed points theorem for nonexpansive mapping. Non-expansion fixed point theory has made great progress, large number of results are obtained by authors (e.g. See [3-11]). let’s come up with some definitions.

Definition 1.1 Let $X$ be a nonempty set, the function $W:X×X\to \left[0,\infty \right)$ is called triangular if for all $x\text{,}$$y\in X\text{,}$ if $W\left(x,y\right)\ge 1\text{,}$$W\left(y,z\right)\ge 1$ or $W\left(y,x\right)\ge 1\text{,}$$W\left(y,z\right)\ge 1\text{,}$ then $W\left(x,z\right)\ge 1.$

Definition 1.2 Let $\left(X,d\right)$ be a metric space and $T:X\to X$ be a given mapping, if there exists a function $W:X×X\to \left[0,\infty \right)$ such that $W\left(x,y\right)d\left(Tx,Ty\right)\le d\left(x,y\right)$, $\forall x,y\in X$, then we say that $T$ is a $W$-nonexpansive mapping.

Clearly, any nonexpansive mapping is a $W$-nonexpansive mapping with $W\left(x,y\right)=1$ for all $x,y\in X$.

Definition 1.3 Let $T:X\to X$ be a mapping and $W:X×X\to \left[0,\infty \right)$ be a function. We say that $T$ is a $W$-admissible if $W\left(x,y\right)\ge 1⇒W\left(Tx,Ty\right)\ge 1$, $\forall x,y\in X.$

Definition 1.4  Let $H$ be a Hilbert space, $T:H\to H$ is called demicompact if whenever $\left\{{x}_{n}\right\}\subset H$ is bounded and $\left\{T{x}_{n}-T{x}_{n}\right\}$ strongly convergent, then there exists a subsequence $\left\{{x}_{nk}\right\}$ of $\left\{{x}_{n}\right\}$ which is strongly convergent.

Next our main results are presented.

## 2. Main results

Theorem 2.1 Let $E$ be a bounded closed convex subset of a Hilbert space $H$, $W:E×E\to \left[0,\infty \right)$ is triangular function, $T:E\to E$ is a $W$-nonexpansive mapping and it is $W$-admissible. If the following conditions are satisfied:

(w1) there exists ${x}_{0}\in E$ such that $W\left({x}_{0},T{x}_{0}\right)\ge 1$;

(w2) there exists a sequence $\left\{{s}_{j}\right\}\subseteq \left[0,1\right)$ with ${lim}_{n\to \infty }{s}_{j}=1$ such that for all $x,y\in E$, if $W\left(x,y\right)\ge 1$, then $W\left(x,\left(1-{s}_{j}\right)x+{s}_{j}y\right)\ge 1$, $\forall {s}_{j}\in \left\{{s}_{j}\right\}$;

(w3) if $\left\{{x}_{n}\right\}\subseteq E$ is satisfied $W\left({x}_{0},{x}_{n}\right)\ge 1\text{,}$ moreover ${x}_{n}\to {x}^{*}$ or ${x}_{n}\to {x}^{*}\in E\text{,}$ then $W\left({x}_{n},x*\right)\ge 1.$

Then $T$ has a fixed point.

Proof. Let ${x}_{0}\in X$ such that $\alpha \left({x}_{0},T{x}_{0}\right)\ge 1$. Take ${x}_{n+1j}=\left(1-{s}_{j}\right){x}_{0}+{s}_{j}T{x}_{nj}$ for all $j$, $n\in N$, there ${x}_{0}={x}_{0j}$. Now we fix $j$, for each $j\in N$, from (w2), we may obtain $W\left({x}_{0j},{x}_{1j}\right)\ge 1.$

Also, for $T$ is $W$-admissible, then $W\left(T{x}_{0j},T{x}_{1j}\right)\ge 1$ is obtained. According to $W$ is a triangular function and (w1), then $W\left({x}_{0j},T{x}_{1j}\right)\ge 1.$

Once again use (w2), then $W\left({x}_{0j},{x}_{2j}\right)\ge 1$ is also obtained. Continuously, we easily obtain:

1
$W\left({x}_{0j},{x}_{nj}\right)\ge 1,\mathrm{}\mathrm{}\mathrm{}\mathrm{}\forall n\in N.$

Based on that $W$ is triangular, we may get:

2
$W\left({x}_{nj},{x}_{mj}\right)\ge 1,\mathrm{}\mathrm{}\mathrm{}\forall n,m\in N,n

So from Eq. (2) and for $T$ is $W$-nonexpansive, we have:

3
$‖{x}_{nj}-{x}_{mj}‖={s}_{j}‖T{x}_{n-1j}-T{x}_{m-1j}‖\le {s}_{j}W\left({x}_{n-1j},{x}_{m-1j}\right)‖T{x}_{n-1j}-T{x}_{m-1j}‖$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le {s}_{j}‖{x}_{n-1j}-{x}_{m-1j}‖...\le {s}_{j}^{n}‖{x}_{0j}-{x}_{m-nj}‖.$

Let $n\to \mathrm{\infty }$, for $E$ is bounded we may get $‖{x}_{nj}-{x}_{mj}‖\to 0$, hence $\left\{{x}_{nj}\right\}$ is Cauchy sequence, it means there exists ${x}_{j}^{\mathrm{*}}\in E$ such that $\left\{{x}_{nj}\right\}$ convergent to ${x}_{j}^{\mathrm{*}}$, that is:

4
$\mathrm{l}\mathrm{i}{\mathrm{m}}_{n\to \mathrm{\infty }}‖{x}_{nj}-{x}_{j}^{\mathrm{*}}‖=0.$

Also, from Eq. (1) and (w3), we have:

5
$W\left({x}_{nj},{x}_{j}^{\mathrm{*}}\right)\ge 1.$

Once again by Eq. (1), for $W$ is triangular, so we have:

6
$W\left({x}_{0j},{x}_{j}^{\mathrm{*}}\right)\ge 1.$

Since $E$ is bounded, closed and convex in Hilbert $H$, then it is weakly compact. Hence there exists a ${x}^{\mathrm{*}}\in E$ such that:

7
${x}_{j}^{\mathrm{*}}\to {x}^{\mathrm{*}},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\left(j\to \mathrm{\infty }\right).$

From Eqs. (6, 7), applying (w3) we have:

8
$W\left({x}_{j}^{\mathrm{*}},{x}^{\mathrm{*}}\right)\ge 1.$

Next, we show that ${x}_{j}^{\mathrm{*}}=\left(1-{s}_{j}\right){x}_{0j}+{s}_{j}T{x}_{j}^{\mathrm{*}}$.

Indeed, according to ${x}_{nj}=\left(1-{s}_{j}\right){x}_{0}+{s}_{j}T{x}_{n-1j}$, $T$ is $W$-nonexpiansive and Eq. (5), we have:

9
$‖{x}_{j}^{\mathrm{*}}-\left(\left(1-{s}_{j}\right){x}_{0j}+{s}_{j}T{x}_{j}^{\mathrm{*}}\right)‖=‖{x}_{j}^{\mathrm{*}}-{x}_{nj}+{x}_{nj}-\left(\left(1-{s}_{j}\right){x}_{0j}+{s}_{j}T{x}_{j}^{\mathrm{*}}\right)‖$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}=‖{x}_{nj}-{x}_{j}^{\mathrm{*}}‖+{s}_{j}‖T{x}_{n-1j}-T{x}_{j}^{\mathrm{*}}‖\le ‖{x}_{nj}-{x}_{j}^{\mathrm{*}}‖+{s}_{j}W\left({x}_{nj},{x}_{j}^{\mathrm{*}}\right)‖T{x}_{n-1j}-T{x}_{j}^{\mathrm{*}}‖$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le ‖{x}_{nj}-{x}_{j}^{\mathrm{*}}‖+{s}_{j}‖{x}_{n-1j}-{x}_{j}^{\mathrm{*}}‖.$

Let $n\to \mathrm{\infty }$ in Eq. (9), utilize Eq. (4) we obtain $‖{x}_{j}^{\mathrm{*}}-\left[\left(1-{s}_{j}\right){x}_{0j}+{s}_{j}T{x}_{j}^{\mathrm{*}}\right]‖\to 0$, it implies that ${x}_{j}^{\mathrm{*}}=\left(1-{s}_{j}\right){x}_{0j}+{s}_{j}T{x}_{j}^{\mathrm{*}}$.

Finally, we show that ${x}^{\mathrm{*}}$ is a fixed point of $T$. If $y$ is any arbitrary point in $H$, we have:

10
${‖{x}_{j}^{\mathrm{*}}-y‖}^{2}={‖\left({x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}\right)+\left({x}^{\mathrm{*}}-y\right)‖}^{2}={‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖}^{2}+{‖{x}^{\mathrm{*}}-y‖}^{2}+2〈{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}},{x}^{\mathrm{*}}-y〉.$

Since ${x}_{j}^{\mathrm{*}}\to {x}^{\mathrm{*}}$, then $2〈{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}},\mathrm{}{x}^{\mathrm{*}}-y〉\to 0$, $\left(j\to \mathrm{\infty }\right).$

So, based on the above inequality and Eq. (10), we get:

11
$\underset{j\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left({‖{x}_{j}^{\mathrm{*}}-y‖}^{2}-{‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖}^{2}\right)={‖{x}^{\mathrm{*}}-y‖}^{2}.$

Setting $y=T{x}^{\mathrm{*}}$ in Eq. (11), we have:

12
$\underset{j\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left({‖{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖}^{2}-{‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖}^{2}\right)={‖{x}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖}^{2}.$

Moreover, since ${x}_{j}^{\mathrm{*}}=\left(1-{s}_{j}\right){x}_{0j}+{s}_{j}T{x}_{j}^{\mathrm{*}}$, then:

13
$‖T{x}_{j}^{\mathrm{*}}-{x}_{j}^{\mathrm{*}}‖=‖T{x}_{j}^{\mathrm{*}}-\left(1-{s}_{j}\right){x}_{0j}-{s}_{j}T{x}_{j}^{\mathrm{*}}‖=\left(1-{s}_{j}\right)‖T{x}_{j}^{\mathrm{*}}-{x}_{0j}‖.$

So, in Eq. (13) as $j\to \mathrm{\infty }$, for ${\mathrm{l}\mathrm{i}\mathrm{m}}_{j\to \mathrm{\infty }}{s}_{j}=1$ we have:

14
$‖T{x}_{j}^{\mathrm{*}}-{x}_{j}^{\mathrm{*}}‖\to 0.$

On the other hand, from Eq. (8) and since $T$ is $W$-nonexpansive mapping, we have:

$‖T{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖\le W\left({x}_{j}^{\mathrm{*}},{x}^{\mathrm{*}}\right)‖T{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖\le ‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖.$

Thus:

15
$‖{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖\le ‖{x}_{j}^{\mathrm{*}}-T{x}_{j}^{\mathrm{*}}‖+‖T{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖\le ‖{x}_{j}^{\mathrm{*}}-T{x}_{j}^{\mathrm{*}}‖+‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖,$

in turn:

16
$‖{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖-‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖\le ‖{x}_{j}^{\mathrm{*}}-T{x}_{j}^{\mathrm{*}}‖.$

Hence by Eq. (14), we have:

17
$\underset{j\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(‖{x}_{j}^{\mathrm{*}}-T{x}^{\mathrm{*}}‖-‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖\right)\le \underset{j\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}‖{x}_{j}^{\mathrm{*}}-T{x}_{j}^{\mathrm{*}}‖=0.$

And, due to $E$ is bounded, we have also:

18
$\underset{j\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left({‖{x}_{j}^{\mathrm{*}}-T{x}_{\mathrm{*}}‖}^{2}-{‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖}^{2}\right)$
$=\underset{j\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\left(‖{x}_{j}^{\mathrm{*}}-T{x}_{\mathrm{*}}‖-‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖\right)\left(‖{x}_{j}^{\mathrm{*}}-T{x}_{\mathrm{*}}‖+‖{x}_{j}^{\mathrm{*}}-{x}^{\mathrm{*}}‖\right)\le 0.$

So, by Eq. (12), we get ${‖{x}^{\mathrm{*}}-T{x}_{\mathrm{*}}‖}^{2}=0$, that is, ${x}^{\mathrm{*}}$ is fixed point of $T$.

Now, we provide a method for computation of that fixed point ${x}^{\mathrm{*}}$.

Theorem 2.2 Suppose all conditions of the Theorem 2.1 are satisfied. Then the Krasnoselskij iteration $\left\{{x}_{n}{\right\}}_{0}^{\infty }$ given by:

19
${x}_{n+1}=\left(1-s\right){x}_{n}+sT{x}_{n},s\in \left\{{s}_{j}{\right\}}_{j\in N},n=0,1,2,\dots ,$

converges to a fixed point of $T$.

Proof. Take the same ${x}_{0}\in E$ as Theorem 2.1, and such that $W\left({x}_{0},T{x}_{0}\right)\ge 1$. From (w2) we get:

20
$W\left({x}_{0},\left(1-s\right){x}_{0}+sT{x}_{0}\right)=W\left({x}_{0},{x}_{1}\right)\ge 1.$

For $W$ is triangular, so:

21
$W\left(T{x}_{0},{x}_{1}\right)\ge 1.$

Since $T$ is a $W$-admissible, from Eq. (20) we have:

22
$W\left(T{x}_{0},T{x}_{1}\right)\ge 1.$

Once again for $W$ is triangular, by Eqs. (21) and (22) we have $W\left({x}_{1},T{x}_{1}\right)\ge 1.$

Also, from (w2) we have $W\left({x}_{1},\left(1-s\right){x}_{1}+sT{x}_{1}\right)=W\left({x}_{1},{x}_{2}\right)\ge 1.$

Continuously, we can obtain:

23
$W\left({x}_{n},{x}_{m}\right)\ge 1,\mathrm{}\mathrm{}\mathrm{}\forall n,m\in N,\mathrm{}\mathrm{}\mathrm{}\mathrm{}n

Hence:

24
$W\left({x}_{0},{x}_{n}\right)\ge 1.$

Also form Theorem 2.1, we know that ${x}_{\mathrm{*}}$ is fixed point of $T$, and Based on all conditions of Theorem 2.1 are satisfied in Theorem 2.2, similarly we have:

25
$W\left({x}_{0},{x}_{j}^{\mathrm{*}}\right)\ge 1,$
26
$W\left({x}_{j}^{\mathrm{*}},{x}^{\mathrm{*}}\right)\ge 1.$

From Eqs. (25) and (26), for $W$ is triangular, then:

27
$W\left({x}_{0},{x}^{\mathrm{*}}\right)\ge 1.$

Also, by Eqs. (24) and (27), use $W$ is triangular, we get:

28
$W\left({x}_{n},{x}^{\mathrm{*}}\right)\ge 1.$

Based on Eq. (28), since $T$ is $W$-nonexpansive mapping, then we have:

29
$‖T{x}_{n}-T{x}^{\mathrm{*}}‖\le W\left({x}_{n},{x}^{\mathrm{*}}\right)‖T{x}_{n}-T{x}^{\mathrm{*}}‖\le ‖{x}_{n}-{x}^{\mathrm{*}}‖.$

So:

30
$‖{x}_{n+1}-{x}^{\mathrm{*}}‖=‖\left(1-s\right)\left({x}_{n}-{x}^{\mathrm{*}}\right)+s\left(T{x}_{n}-T{x}^{\mathrm{*}}\right)‖$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le \left(1-s\right)‖\left({x}_{n}-{x}^{\mathrm{*}}\right)‖+s‖T{x}_{n}-T{x}^{\mathrm{*}}‖$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le \left(1-s\right)‖\left({x}_{n}-{x}^{\mathrm{*}}\right)‖+s‖{x}_{n}-{x}^{\mathrm{*}}‖=‖\left({x}_{n}-{x}^{\mathrm{*}}\right)‖.$

Continuously, we have $‖{x}_{n+1}-{x}^{\mathrm{*}}‖\le ‖{x}_{0}-{x}^{\mathrm{*}}‖$, which implies that $\left\{‖{x}_{n+1}-{x}^{\mathrm{*}}‖\right\}$ is monotone decrease bounded sequence. So ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \mathrm{\infty }}‖{x}_{n+1}-{x}^{\mathrm{*}}‖$ exists.

Next, we prove that $‖{x}_{n}-T{x}_{n}‖\to 0$:

31
${‖{x}_{n+1}-{x}^{\mathrm{*}}‖}^{2}={‖\left(1-s\right)\left({x}_{n}-{x}^{\mathrm{*}}\right)-s\left(T{x}_{n}-T{x}^{\mathrm{*}}\right)‖}^{2}$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}=\left(1-s{\right)}^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}+{s}^{2}{‖T{x}_{n}-T{x}^{\mathrm{*}}‖}^{2}+2\left(1-s\right)s〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le \left(1-s{\right)}^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}+{s}^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}+2\left(1-s\right)s〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}=\left(\left(1-s{\right)}^{2}+{s}^{2}\right){‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}+2\left(1-s\right)s〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉.$

Also, on the other hand for any constant $\lambda$:

32
${\lambda }^{2}{‖{x}_{n}-T{x}_{n}‖}^{2}={‖\left({x}_{n}-{x}^{\mathrm{*}}\right)-\left(T{x}_{n}-T{x}^{\mathrm{*}}\right)‖}^{2}$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}={\lambda }^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}+{\lambda }^{2}{‖T{x}_{n}-T{x}^{\mathrm{*}}‖}^{2}-2{\lambda }^{2}〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le {\lambda }^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}+{\lambda }^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}-2{\lambda }^{2}〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}=2{\lambda }^{2}{‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}-2{\lambda }^{2}〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉$

Adding Eq. (31) to Eq. (32) and let ${\lambda }^{2}\le \left(1-s\right)s$ , we may obtain:

33
${‖{x}_{n+1}-{x}^{\mathrm{*}}‖}^{2}+{\lambda }^{2}{‖{x}_{n}-T{x}_{n}‖}^{2}\le \left(\left(1-s{\right)}^{2}+{s}^{2}+2{\lambda }^{2}\right){‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}+\left(2\left(1-s\right)s-2{\lambda }^{2}\right)〈{x}_{n}-{x}^{\mathrm{*}},T{x}_{n}-T{x}^{\mathrm{*}}〉$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\le \left(\left(1-s{\right)}^{2}+{s}^{2}+2{\lambda }^{2}+2\left(1-s\right)s-2{\lambda }^{2}\right){‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}={‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}.$

It implies ${\lambda }^{2}{‖{x}_{n}-T{x}_{n}‖}^{2}\le {‖{x}_{n}-{x}^{\mathrm{*}}‖}^{2}-{‖{x}_{n+1}-{x}^{\mathrm{*}}‖}^{2}.$

Since ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \mathrm{\infty }}‖{x}_{n+1}-{x}^{\mathrm{*}}‖$ exists, in the above inequality let $n\to \mathrm{\infty }\text{,}$ it results ${\lambda }^{2}{‖{x}_{n}-T{x}_{n}‖}^{2}\to 0.$

It means $‖{x}_{n}-T{x}_{n}‖\to 0.$

For $T$ is demicompact, it results that there exists a strongly convergent subsequence $\left\{{x}_{{n}_{i}}\right\}\subseteq \left\{{x}_{n}\right\}$ such that ${x}_{{n}_{i}}\to {x}^{\mathrm{*}}\in F\left(T\right)$, that is, $‖{x}_{{n}_{i}}\to {x}^{\mathrm{*}}‖\to 0$. Also $\left\{‖{x}_{n}\to {x}^{\mathrm{*}}‖\right\}$ is convergent, it implies that $‖{x}_{n}\to {x}^{\mathrm{*}}‖\to 0$. Hence that $\left\{{x}_{n}\right\}$ is convergent to ${x}^{\mathrm{*}}\in F\left(T\right)$.

#### Acknowledgements

This work was supported by the Educational Science Foundation of Chongqing.