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

1. Introduction and preliminaries

Fixed point theory is widely applied in engineering. Browder (1965) [1], Kirk (1965) [2] 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 $W:X\times X\to [0,\infty )$ is called triangular if for all $x\text{,}$$y\in X\text{,}$ if $W(x,y)\ge 1\text{,}$$W(y,z)\ge 1$ or $W(y,x)\ge 1\text{,}$$W(y,z)\ge 1\text{,}$ then $W(x,z)\ge 1.$

Definition 1.2 Let $(X,d)$ be a metric space and $T:X\to X$ be a given mapping, if there exists a function $W:X\times X\to [0,\infty )$ 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 $W(x,y)=1$ for all $x,y\in X$.

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

Definition 1.4 [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 $\{T{x}_n-T{x}_n\}$ 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 $H$, $W:E\times E\to [0,\infty )$ 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 $W(x_0,T{x}_0)\ge 1$;

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

(w3) if $\left\{x_n\right\}\subseteq E$ is satisfied $W(x_0,x_n)\ge 1\text{,}$ moreover $x_n\to x^{*}$ or $x_n\to x^{*}\in E\text{,}$ then $W(x_n,x*)\ge 1.$

Then $T$ has a fixed point.

Proof. Let $x_0\in X$ such that $\alpha (x_0,T{x}_0)\ge 1$. Take $x_{n+1j}=(1-s_j)x_0+s_{j}T{x}_{nj}$ 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 $W(x_{0j},x_{1j})\ge 1.$

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

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

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

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

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:

Also, from Eq. (1) and (w3), we have:

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

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:

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

Next, we show that $x_j^{\mathrm{*}}=(1-s_j)x_{0j}+s_{j}T{x}_j^{\mathrm{*}}$.

Indeed, according to $x_{nj}=(1-s_j)x_0+s_{j}T{x}_{n-1j}$, $T$ is $W$-nonexpiansive and Eq. (5), we have:

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 $x_j^{\mathrm{*}}=(1-s_j)x_{0j}+s_{j}T{x}_j^{\mathrm{*}}$.

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

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

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

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

Moreover, since $x_j^{\mathrm{*}}=(1-s_j)x_{0j}+s_{j}T{x}_j^{\mathrm{*}}$, then:

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:

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

Thus:

in turn:

Hence by Eq. (14), we have:

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

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 $\{x_n{\}}_0^{\infty }$ given by:

converges to a fixed point of $T$.

Proof. Take the same $x_0\in E$ as Theorem 2.1, and such that $W(x_0,T{x}_0)\ge 1$. From (w2) we get:

For $W$ is triangular, so:

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

Once again for $W$ is triangular, by Eqs. (21) and (22) we have $W(x_1,T{x}_1)\ge 1.$

Also, from (w2) we have $W(x_1,(1-s)x_1+sT{x}_1)=W(x_1,x_2)\ge 1.$

Continuously, we can obtain:

Hence:

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:

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

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

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

So:

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 $‖x_n-T{x}_n‖\to 0$:

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

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

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 $x^{\mathrm{*}}\in F\left(T\right)$.