Iterative Procedure for Finite Family of Total Asymptotically Nonexpansive Maps (TAN)

In this paper, CQ Algorithms for iterative approximation of a common fixed point of a finite family of nonlinear maps were introduced and sufficient conditions for the strong convergence of this process to a common fixed point of the family of Total asymptotically Nonexpansive maps (TAN) were proved.

Preliminaries

Let H be a normed space, K be a nonempty closed convex subset of H and $T:K\to K$ be a map. The mapping T is called asymptotically nonexpansive mapping if and only if there exists a sequence $\{{\mu }_n{\}}_{n\ge 1}\subset [0,+\mathrm{\infty })$, with ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\mu }_n=0}$ such that for all $x.y\in K,$

(1) $$\begin{array}{r}\Vert T^{n}x-T^{n}y\Vert \le (1+{\mu }_n)\Vert x-y\Vert \mathrm{\forall }n\in \mathbb{N}\end{array}$$

The class of asymptotically nonexpansive mappings was introduced by Goebel and Kirk [1] as a generalisation of nonexpansive mappings. As further generalisation of class of nonexpansive mappings, Alber et al. [2] introduced the class of total asymptotically nonexpansive mappings, where a mapping $T:K⟶K$ is called total asymptotically nonexpansive (TAN) if and only if there exist two sequences $\{{\mu }_n{\}}_{n\ge 1},\{{\eta }_n{\}}_{n\ge 1}\subset [0,+\mathrm{\infty })$, with ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\mu }_n=0={\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\eta }_n}}$ and nondecreasing continuous function $\varphi :[0,+\mathrm{\infty })⟶[0,+\mathrm{\infty })$ with $\varphi (0)=0$ such that for all $x,y\in K,$

(2) $$\begin{array}{r}\Vert T^{n}x-T^{n}y\Vert \le \Vert x-y\Vert +{\mu }_n\varphi (\Vert x-y\Vert )+{\eta }_{n}n\ge 1\end{array}$$

In Ofoedu and Nnubia [3], an example to show that the class of asymptotically nonexpansive mappings is properly contained in the class of total asymptotically nonexpansive mappings was given. The class of asymptotically nonexpansive type mappings includes the class of mappings which are asymptotically nonexpansive in the intermediate sense. These classes of mappings had been studied extensively by several authors (see e.g., [4]–[9]).

A map T is said to satisfies condition B if there exists $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ strictly increasing, continuous, $f(0)=0,f(r)>0\mathrm{\forall }r>0$ such that for all $x\in D(T),\Vert x-Tx\Vert \ge f(d(x,F))$ where $F=F(T)=\{x\in D(T):x=Tx\}$ and $d(x,F)=inf\{\Vert x-y\Vert :y\in F\}$.

Lemma 1.1 Takahashi [10]

Let $\{{\mu }_n\},\{{\beta }_n\},\{{\gamma }_n\}$ be sequences of nonnegative numbers satisfying the conditons: ${\displaystyle \sum _{n\ge 0}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty },{\beta }_n⟶0}$ as $n⟶\mathrm{\infty }$ and ${\gamma }_n=0({\beta }_n)$. Suppose that where $\psi :[0,1)\to [0,1)$ is a strictly increasing function with $\psi (0)=0.$ Then ${\mu }_n\to 0$ as $n\to \mathrm{\infty }$.

$$\begin{array}{rl}& {\mu }_{n+1}^2\le {\mu }_n^2-{\beta }_n\psi ({\mu }_{n+1})+{\gamma }_n;n=1,2,...\end{array}$$

Lemma 1.2 Moore and Nnoli [11]

Let K be closed convex nonempty subset of a real Hilbert Space H. Let $\{x_n\}$ be a sequence in H, $y\in H$ and $z=P_{K}y$ be such that ${\omega }_w(x_n)\subseteq K$ and $\Vert x_n-y\Vert \le \Vert y-z\Vert \mathrm{\forall }n\ge 1,$ then $\{x_n\}$ converges strongly to $z.$

$${\omega }_w(x_n)=\{z\in H:\mathrm{\exists }\{x_{nj(z)}\}\subset \{x_n\}\ni {x_n}_{j(z)}{\to }^{w}z\text{as}j\to \mathrm{\infty }\}$$

Lemma 1.3 Ofoedu and Nnubia [9, p. 703]

Let E be a reflexive Banach space with weakly continuous normalized duality mapping. Let K be a closed convex subset of E and $T:K\to K$ a uniformly continuous total asymptotically nonexpansive mapping with bounded orbits. Then I − T is demiclosed at zero.

Proposition 1.1 Ofoedu and Nnubia [9, p. 704]

Let H be a real Hilbert space, let K be a nonempty closed convex subset of H and let $T_i:K⟶K,\text{where}i\in I=\{1,2,...,m\},$ be m uniformly continuous total asymptotically nonexpansive mappings from K into itself with sequences $\{{\mu }_{n,i}{\}}_{n\ge 1},\{{\eta }_{n,i}{\}}_{n\ge 1}\subset [0,+\mathrm{\infty })$ such that ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\mu }_{n,i}=0={\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\eta }_{n,i}}}$ and with function ${\varphi }_i:[0,+\mathrm{\infty })⟶[0,+\mathrm{\infty })$ satisfying ${\varphi }_i(t)\le M_{0}t\mathrm{\forall }t>M_1$ for some constants$M_0,M_1>0.$ Let ${\mu }_n={\displaystyle \underset{i\in I}{max}\{{\mu }_{n,i}\}}$ and ${\eta }_n={\displaystyle \underset{i\in I}{max}\{{\eta }_{n,i}\}}$ and, $\varphi (t)={\displaystyle \underset{i\in I}{max}\{{\varphi }_i(t)\}\mathrm{\forall }t\in [0,\mathrm{\infty })}$. Suppose that $F(T)=\bigcap _{i=1}^{m}F(T_i),$ then $F(T)$ is closed and convex.

Proposition 1.2 Nnubia and Bishop [6, p. 74]

Let K be a nonempty subset of a real normed space E and $T_i:K⟶K,\text{where}i\in I=\{1,2,...,m\},$ be m total asymptotically nonexpansive mappings, then there exist sequences $\{{\mu }_n{\}}_{n\ge 1},\{{\eta }_n{\}}_{n\ge 1}\subset [0,+\mathrm{\infty })$, with ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\mu }_n=0={\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\eta }_n}}$ and nondecreasing continuous function $\varphi :[0,+\mathrm{\infty })⟶[0,+\mathrm{\infty })$ with $\varphi (0)=0$ such that for all $x,y\in K,$

(3) $$\begin{array}{r}\Vert T_i^{n}x-T_i^{n}y\Vert \le \Vert x-y\Vert +{\mu }_n\varphi (\Vert x-y\Vert )+{\eta }_n;n\ge 1,\mathrm{\forall }i\in I.\end{array}$$

Main Result

Proposition 2.1 Result

Suppose that there exist $c>0,k>0$ constants such that $\varphi (t)\le ct\mathrm{\forall }t\ge k,$ then T is total asymptotically nonexpansive if $\mathrm{\exists }{\nu }_n={\mu }_{n}c$ and ${\gamma }_n={\mu }_n{c}_0+{\eta }_n$ such that

$$\begin{array}{rcl}\Vert T^{n}x-T^{n}y\Vert & \le & (1+{\nu }_n)\Vert x-y\Vert +{\gamma }_n\end{array}$$

Proof

Suppose T is total asymptotically nonexpansive, that is, let T be such that

(4) $$\begin{array}{r}\Vert T^{n}x-T^{n}y\Vert \le \Vert x-y\Vert +{\mu }_n\varphi (\Vert x-y\Vert )+{\eta }_{n}n\ge 1\end{array}$$

Since $\varphi$ is continuous, it follows that $\varphi$ attains its maximum (say $c_0$) on the interval $[0,k];$ moreover, $\varphi (t)\le ct$ whenever $t>k$. Thus,

(5) $$\begin{array}{r}\varphi (t)\le c_0+ct\mathrm{\forall }t\in [0,+\mathrm{\infty }).\end{array}$$

So, we have, where ${\nu }_n={\mu }_{n}c$ and ${\gamma }_n={\mu }_n{c}_0+{\eta }_n$ Thus completing the proof.

$$\begin{array}{rcl}\Vert T^{n}x-T^{n}y\Vert & \le & \Vert x-y\Vert +{\mu }_n(c_0+c\Vert x-y\Vert )+{\eta }_{n}n\ge 1\\ & =& (1+{\mu }_{n}c)\Vert x-y\Vert +{\mu }_n{c}_0+{\eta }_n\\ & =& (1+{\nu }_n)\Vert x-y\Vert +{\gamma }_n\end{array}$$

Theorem 2.1 Let H, K, $T_i$, and F be as in Propostion 1.1, then $\{x_n{\}}_{n\ge 1}$ generated iteratively by: converges to $P_F{x}_0$ where ${\sigma }_n={\alpha }_n(k_{m(n),i(n)}^2-1)(diam.K{)}^2+{\alpha }_n(2k_{m(n),i(n)}diam.K+v_{m(n),i(n)})v_{m(n),(n)}$ and $\{{\alpha }_n\}\subset [a,b]\subset (0,1).$

(6) $$\begin{array}{rcl}{y}_n& =& (1-{\alpha }_n)x_n+{\alpha }_n{T}_{i(n)}^{m(n)}{x}_n;i(n)\equiv n\mathrm{mod}m\mathrm{\forall }n\in \mathbb{Z};m(n)=1+\left[\frac{n}{m}\right]\\ K_n& =& \{z\in K:\Vert y_n-z{\Vert }^2\le \Vert x_n-z{\Vert }^2-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2+{\sigma }_n\}\\ Q_n& =& \{z\in K:⟨x_n-z,x_o-x_n⟩\ge 0\}\\ x_{n+1}& =& P_{K_n\cap Q_n}{x}_0.\end{array}$$

Proof

Let $x^{\ast }\in F$.

$$\begin{array}{rcl}\Vert y_n-x^{\ast }{\Vert }^2& =& \Vert (1-{\alpha }_n)(x_n-x\ast )+{\alpha }_n(T_{i(n)}^{m(n)}{x}_n-x^{\ast }){\Vert }^2\\ & \le & (1-{\alpha }_n)\Vert x_n-x\ast {\Vert }^2+{\alpha }_n\Vert T_{i(n)}^{m(n)}{x}_n-x^{\ast }{\Vert }^2-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2\\ & \le & (1-{\alpha }_n)\Vert x_n-x\ast {\Vert }^2+{\alpha }_n(1+k_{i(n)m(n)}{)}^2\Vert x_n-x^{\ast }{\Vert }^2+(2(1+k_{i(n)m(n)})\Vert x_n-x^{\ast }\Vert \\ & & +v_{i(n)m(n)})v_{i(n)m(n)}-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2\\ & =& (1+{\alpha }_n[1+k_{i(n)m(n)}-1])\Vert x_n-x^{\ast }{\Vert }^2+(2(1+k_{i(n)m(n)})\Vert x_n-x^{\ast }\Vert \\ & & +v_{i(n)m(n)})v_{i(n)m(n)}-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2\\ & \le & \Vert x_n-x\ast {\Vert }^2-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2\\ & & +{\alpha }_n[(1+k_{i(n)m(n)}{)}^2-1](diam.K{)}^2+{\alpha }_n[2(1+k_{i(n)m(n)})(diam.K)\\ & & +v_{i(n)m(n)}]v_{i(n)m(n)}\\ & =& \Vert x_n-x^{\ast }{\Vert }^2-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2+{\sigma }_n.\end{array}$$

So that $x^{\ast }\in K_n\mathrm{\forall }n.$ Hence, $F\subset K_n\mathrm{\forall }n$. For $n=0,Q_o=K.$ $F\subset Q_o$.

Let $F\subset Q_v$, we show that $F\subset Q_{v+1}$. Now, $x_{v+1}$ is the projection of $x_o$ onto $K_v\cap Q_v$. then (i) $⟨x_{v+1}-z,x_o-x_{v+1}⟩\ge 0.\mathrm{\forall }z\in K_v\cap Q_v.$

Since, $F\subset K_v\cap Q_v,$ then $⟨x_{v+1}-x^{\ast },x_o-x_{v+1}⟩\ge 0\mathrm{\forall }{x}^{\ast }\in F.$ So, $F\subset Q_{v+1}$ and hence $F\subset Q_n\mathrm{\forall }n\ge 0.$

Now,

$$\begin{array}{rl}& \Vert x_o-P_{Q_n}{x}_o\Vert \le \Vert x_o-y\Vert \mathrm{\forall }y\in Q_n\end{array}$$

Hence, $\mathrm{\forall }n$ and since $F\subset Q_n$, then $\Vert x_n-x_o\Vert \le \Vert x_o-x^{\ast }\Vert \mathrm{\forall }{x}^{\ast }\in F.$

$$\begin{array}{rl}& \Vert x_n-x_o\Vert =\Vert x_o-P_{Q_n}{x}_o\Vert \le \Vert x_o-y\Vert \mathrm{\forall }y\in Q_n.\end{array}$$

In particular, $\Vert x_n-x_o\Vert \le \Vert x_o-x^{\ast }\Vert ;x^{\ast }=P_F{x}_o.$

Since, $x_{n+1}\in Q_n,⟨x_{n+1}-x_n,x_n-x_o⟩\ge 0.$

So,

$$\begin{array}{rcl}\Vert x_{n+1}-x_n{\Vert }^2& =& \Vert x_{n+1}-x_o-(x_n-x_o){\Vert }^2\\ & =& \Vert x_{n+1}-x_o^{\ast }{\Vert }^2-\Vert x_n-x_o{\Vert }^2-2⟨x_{n+1}-x_n,x_n-x_o⟩\\ & \le & \Vert x_{n+1}-x_o{\Vert }^2-\Vert x_n-x_o{\Vert }^2\end{array}$$

Then, $\Vert x_n-x_o\Vert \le \Vert x_{n+1}-x_o\Vert$.

Since $\{\Vert x_n-x_o\Vert \}$ is bounded, then ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_n-x_o\Vert }$ exists.

Thus,

(7) $$\begin{array}{r}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_{n+1}-x_n\Vert =0.}\end{array}$$

Observe that by (7)

(8) $$\begin{array}{r}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_{n+i}-x_n\Vert =0={\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_{n-i}-x_n\Vert \mathrm{\forall }i\in \{1,...,m\}.}}\end{array}$$

Now,

$$\begin{array}{rcl}{\alpha }_n^2\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2& =& \Vert y_n-x_n{\Vert }^2\\ & \le & (\Vert y_n-x_{n+1}\Vert +\Vert x_{n+1}-x_n{)}^2\\ & =& \Vert y_n-x_{n+1}{\Vert }^2+2\Vert y_n-x_{n+1}\Vert .\Vert x_{n+1}-x_n\Vert +\Vert x_{n+1}-x_n{\Vert }^2\end{array}$$

$x_{n+1}\in K_n,$ so that

$$\begin{array}{rl}& \Vert y_n-x_{n+1}{\Vert }^2\le \Vert x_n-x_{n+1}{\Vert }^2-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2+{\sigma }_n.\end{array}$$

Hence, and hence, ${\alpha }_n^2\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2\le 2\Vert x_{n+1}-x_n{\Vert }^2+2\Vert y_n-x_{n+1}\Vert .\Vert x_{n+1}-x_n\Vert +{\sigma }_n$

$$\begin{array}{r}{\alpha }_n^2\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2\le \Vert x_n-x_{n+1}{\Vert }^2-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_{i(n)}^{m(n)}{x}_n{\Vert }^2+{\sigma }_n\\ +2\Vert y_n-x_{n+1}\Vert .\Vert x_{n+1}-x_n\Vert +\Vert x_{n+1}-x_n{\Vert }^2\end{array}$$

Thus,

(9) $$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_{i(n)}^{m(n)}{x}_n\Vert =0\end{array}$$

Now, $\mathrm{\forall }n>m$, we have $n=(n-m)(modm)$ and since $n=(m(n)-1)m+i(n)$, we obtain $n-m=(m(n)-1)N+i(n)-m=(m(n-m)-1)+i(n-m)$, so that $n-m=[(m(n)-1)-1]m+i(n)=(m(n-m)-1)m+i(n-m)$. Hence, $m(n)-1=m(n-m)$ and $i(n)=i(n-m)$. Similarly, $m(n+1-m)=m(n+1)-1$ and i(n + 1 − N) = i(n + 1). Using this we obtain but, so that by hypothesis, (9), (8) and the uniform countinuity of $T_{i}i\in I$ we have:

$$\begin{array}{rcl}\Vert x_n-T_{i(n+1)}{x}_{n+1}\Vert & \le & \Vert x_n-T_{i(n+1)}^{m(n+1)}{x}_{n+1}\Vert +\Vert T_{i(n+1)}^{m(n+1)}{x}_{n+1}-T_{i(n+1)}{x}_{n+1}\Vert \\ & \le & \Vert x_n-x_{n+1}\Vert +\Vert x_{n+1}-T_{i(n+1)}^{m(n+1)}{x}_{n+1}\Vert +\Vert T_{i(n+1)}^{m(n+1)}{x}_{n+1}-T_{i(n+1)}{x}_{n+1}\Vert \end{array}$$

(10) $$\begin{array}{rcl}\Vert T_{i(n+1)}^{m(n+1)-1}{x}_{n+1}-x_{n+1}\Vert & \le & \Vert T_{i(n+1)}^{m(n+1)-1}{x}_{n+1}-T_{i(n+1)}^{m(n+1)-1}{m(n+1)-1}_{n+1-m}\Vert \\ & & +\Vert T_{i(n+1)}^{m(n+1)-1}{x}_{n+1-m}-x_{n+1-m}\Vert +\Vert x_{n+1-m}-x_{n+1}\Vert \\ & \le & (2+k_{m(n+1)-1})\Vert x_{n+1}-x_{n+1-m}\Vert +\Vert T_{i(n+1-m)}^{m(n+1-m}{x}_{n+1-m}-x_{n+1-m}\Vert \\ & & +v_{m(n+1)-1}\end{array}$$

(11) $$\begin{array}{r}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_{i(n+1)}{x}_{n+1}\Vert =0}\end{array}$$

Furthermore,

(12) $$\begin{array}{r}\Vert x_{n+1}-T_{i(n+1)}{x}_{n+1}\Vert \le \Vert x_{n+1}-x_n\Vert +\Vert x_n-T_{i(n+1)}{x}_{n+1}\Vert .\end{array}$$

So that using (7) and (11) we have,

(13) $$\begin{array}{r}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_{n+1}-T_{i(n+1)}{x}_{n+1}\Vert =0.}\end{array}$$

Now, let $k\in I$ be arbitrarily choosen, then,

$$\begin{array}{rcl}\Vert x_n-T_{i(n+k)}{x}_n\Vert & \le & \Vert x_n-x_{n+k}\Vert +\Vert x_{n+k}-T_{i(n+k)}{x}_{n+k}\Vert +\Vert T_{i(n+k)}{x}_{n+k}-T_{i(n+k)}{x}_n\Vert \end{array}$$

By uniform continuity of $T_i\mathrm{\forall }i\in I$ and from (8) and (13) we have that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_{i(n+k)}{x}_n\Vert =0\mathrm{\forall }k\in I=\{1,...,m\}\end{array}$$

Observe that $\mathrm{\forall }k\in I=\{1,..,m\}\mathrm{\exists }{\eta }_k\in I\text{such that}i(n)+{\eta }_k\equiv kmodm$, put in another way, $\mathrm{\forall }k\in I\mathrm{\exists }{i}_k\in I\text{such that}i(n+k)\equiv imodN$ Hence,

$$\begin{array}{rl}& {\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_i{x}_n\Vert =0;\mathrm{\forall }i\in \{1,...,m\}.}\end{array}$$

Since $(1-T_i)$ is demiclosed at 0 $\in H\mathrm{\forall }i.\{x_n\}$ is bounded and H is reflexive,

so, $\mathrm{\exists }z\in K$ and $\{{x_n}_j\}\subset \{x_n\}$ such that, ${x_n}_j{\to }^{w}z$ as $j\to \mathrm{\infty }$. Since, ${x_n}_j-T_i{x_n}_j\to 0$ as $j\to \mathrm{\infty }\mathrm{\forall }i$ then $z\in F(T_i)\mathrm{\forall }i$ and so $z\in F=\bigcap _{i=1}^{m}F(T_i)$

Let $q\in {\omega }_w(x_n)$ arbitrary. Then $\mathrm{\exists }\{{x_n}_r\}\subset \{x_n\}\ni {x_n}_r{\to }^{w}q$ and ${x_n}_r-T_i{x_n}_r\to 0$ as $r\to \mathrm{\infty }\mathrm{\forall }i.$ So that since $1-T_i$ is demiclosed at $0\mathrm{\forall }i,q\in F.$ Hence, ${\omega }_w(x_n)\subseteq F.$ Moreover, $\Vert x_n-x_o\Vert \le \Vert x_o-x^{\ast }\Vert \mathrm{\forall }n\ge 0$ where $x^{\ast }=P_F{x}_o.$ Then by the lemma 1.2 $\{x_n\}$ converges strongly to $x^{\ast }=P_F{x}_o$ (that is the common fixed point nearest to $x_0$).

Theorem 2.2 Let H, K, $T_i$, and F be as in Theorem 2.1, then $\{x_n{\}}_{n\ge 1}$ generated iteratively by converges to $P_F{x}_0$ where ${\sigma }_{n,i}=d_o{\displaystyle \sum _{j=1}^i[(k_{n,j}^2-1)+v_{n,j}],i\in I,0<a\le {\alpha }_n\le b<1.}$

(14) $$\begin{array}{rcl}{y}_{n,0}& =& x_n;y_{n,i}=(1-{\alpha }_n)x_n+{\alpha }_n{T}_i^n{y}_{n,i-1};i=1,...,m\\ K_{n,i}& =& \{z\in K:\Vert y_{n,i}-z{\Vert }^2\le \Vert x_n-z{\Vert }^2-{\alpha }_n(1-{\alpha }_n)a^i{\displaystyle \sum _{j=0}^{i-1}\Vert x_n-T_{i-j}^n{y}_{n,i-j-1}{\Vert }^2+{\sigma }_{n,i}\}}\\ K_n& =& \bigcap _{i=1}^m{K}_{n,i}\end{array}$$

(15) $$\begin{array}{rcl}{Q}_n& =& \{z\in K:⟨x_n-z,x_o-x_n⟩\ge 0\}\\ x_{n+1}& =& P_{K_n\cap Q_n}{x}_0.\end{array}$$

Proof

Let $x^{\ast }\in F$.

$$\begin{array}{rcl}\Vert y_{n,i}-x^{\ast }{\Vert }^2& \le & (1-{\alpha }_n)\Vert x_n-x^{\ast }{\Vert }^2+{\alpha }_n{k}_{n,i}^2\Vert y_{n,i-1}-x^{\ast }{\Vert }^2\\ & & +{\alpha }_n(2k_{n,i}\Vert y_{n,i-1}-x^{\ast }\Vert +v_{n,i}){\eta }_{n,i}-{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_i^n{y}_{n,i-1}{\Vert }^2\end{array}$$

So,

(16) $$\begin{array}{rcl}\Vert y_{n,1}-x^{\ast }{\Vert }^2& \le & (1+{\alpha }_n(k_{n,1}^2-1))\Vert x_n-x^{\ast }{\Vert }^2+{\alpha }_n(2k_{n,1}\Vert x_n-x^{\ast }\Vert +v_{n,1})v_{n,1}\\ & & -{\alpha }_n(1-{\alpha }_n)\Vert x_n-T_1^n{x}_n\Vert \end{array}$$

So, where $d_o=b{q}^{i}max\{d_1,(diamK{)}^2\}$ so, $x^{\ast }\in K_{n,i}\mathrm{\forall }i$ and hence $x^{\ast }\in \bigcap _{i=1}^m{K}_{n,i}\mathrm{\forall }n.$ So, $F\subset K_{n,i}\mathrm{\forall }n\ge 0,\mathrm{\forall }i$ and hence, $F\subset K_n$. Following the argument as in Theorem 2.1, we have that $F\subset Q_n\mathrm{\forall }n\ge 0$ and

(17) $$\begin{array}{rcl}\Vert y_{n,i}-x^{\ast }{\Vert }^2& \le & (1+{\displaystyle \sum _{j=0}^{i-1}{\alpha }_n^{j+1}{\displaystyle {\mathrm{\Pi }}_{t=0}^{j-1}{k}_{n,i-t}^2(k_{n,i-j}^2-1))\Vert x_n-x^{\ast }{\Vert }^2}}\\ & & +{\displaystyle \sum _{j=0}^{i-1}{\alpha }_n^{j+1}(2k_{n,i-j}\Vert y_{n,i-j-1}-x^{\ast }{\Vert }^2+v_{n,i-j})v_{n,i-j}{\displaystyle {\mathrm{\Pi }}_{t=0}^{j-1}{k}_{n,i-t}^2}}\\ & & -{\displaystyle \sum _{j=0}^{j-1}{\alpha }_n^{j+1}(1-{\alpha }_n)\Vert x_n-T_{i-j}^n{y}_{n,i-j-1}{\Vert }^2{\displaystyle {\mathrm{\Pi }}_{t=0}^{j-1}{k}_{n,i-t}^2}}\\ & \le & (1+b{q}^i{\displaystyle \sum _{j=1}^i(k_{n,j}^2-1))\Vert x_n-x^{\ast }{\Vert }^2}\\ & & +b{q}^i{d}_1{\displaystyle \sum _{j=1}^i{v}_{n,j}-a^i(1-b){\displaystyle \sum _{j=1}^i\Vert x_n-T_j^n{y}_{n,j-1}{\Vert }^2}}\\ & \le & \Vert x_n-x^{\ast }{\Vert }^2-a^i(1-b){\displaystyle \sum _{j=1}^i\Vert x_n-T_j^n{y}_{n,j-1}{\Vert }^2+{\sigma }_{n,i}}\end{array}$$

$$\begin{array}{rl}& {\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_{n+1}-x_n\Vert =0.}\end{array}$$

Now, but $x_{n+1}\in K_n,$ so, $\mathrm{\forall }i\in I$ so that $\mathrm{\forall }i\in I$ and so $\mathrm{\forall }i\in I$

$$\begin{array}{rcl}{\alpha }_n^2\Vert x_n-T_i^n{y}_{n,i-1}{\Vert }^2& \le & \Vert y_{n,i}-x_{n+1}{\Vert }^2+2\Vert y_{n,i}-x_{n+1}\Vert .\Vert x_{n+1}-x_n\Vert +\Vert x_{n+1}-x_n{\Vert }^2\end{array}$$

$$\begin{array}{rl}& \Vert y_{n,i}-x_{n+1}{\Vert }^2\le \Vert x_n-x_{n+1}{\Vert }^2-{\alpha }_n(1-{\alpha }_n)a^i{\displaystyle \sum _{j=1}^i\Vert x_n-T_j^n{y}_{n,j-1}{\Vert }^2+{\sigma }_{n,i}.}\end{array}$$

$$\begin{array}{rcl}{\alpha }_n^2\Vert x_n-T_i^n{y}_{n,i-1}{\Vert }^2& \le & \Vert x_{n+1}-x_n{\Vert }^2+{\sigma }_{n,i}-{\alpha }_n(1-{\alpha }_n)a^i{\displaystyle \sum _{j=1}^i\Vert x_n-T_j^n{y}_{n,j-1}{\Vert }^2}\\ & & +2\Vert y_{n,i}-x_{n+1}\Vert .\Vert x_{n+1}-x_n\Vert +\Vert x_{n+1}-x_n{\Vert }^2\end{array}$$

$$\begin{array}{rcl}{a}^2\Vert x_n-T_i^n{y}_{n,i-1}{\Vert }^2& \le & {\alpha }_n^2\Vert x_n-T_i^n{y}_{n,i-1}{\Vert }^2\\ & \le & 2\Vert x_{n+1}-x_n{\Vert }^2+{\sigma }_{n,i}+2\Vert y_{n,i}-x_{n+1}\Vert .\Vert x_{n+1}-x_n\Vert \end{array}$$

Hence, $\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_i^n{y}_{n,i-1}\Vert =0\mathrm{\forall }i\in I$ and hence, $\underset{n\to \mathrm{\infty }}{lim}\Vert y_{n,i}-x_n\Vert =0\mathrm{\forall }i\in I$

$$\begin{array}{rcl}\Vert x_n-T_i^n{x}_n\Vert & \le & \Vert x_n-T_i^n{y}_{n,i-1}\Vert +k_{n,i}\Vert y_{n,i-1}-x_n\Vert +v_{n,i}\end{array}$$

Thus, $\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_i^n{x}_n\Vert =0\mathrm{\forall }i\in I$.

$$\begin{array}{rcl}\Vert x_n-T_i{x}_n\Vert & \le & \Vert x_n-T_i^n{x}_n\Vert +\Vert T_i^n{x}_n-T_i{x}_n\Vert \end{array}$$

$$\begin{array}{rcl}\Vert x_n-T_i^{n-1}{x}_n\Vert & \le & (1+k_{n-1,i})\Vert x_n-x_{n-1}\Vert +\Vert x_{n-1}-T_i^{n-1}{x}_{n-1}\Vert \end{array}$$

Thus, $\underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_i^{n-1}{x}_n\Vert =0\mathrm{\forall }i\in I$, so that by uniform continuity of $T_i\mathrm{\forall }i\in I$, ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_n-T_i{x}_n\Vert =0}$. Now, since $(1-T_i)$ is demiclosed at 0, the same argument in Theorem 2.1 completes the Proof.

Conclusion

CQ Algorithms for iterative approximation of a common fixed point of a finite family of nonlinear maps were introduced and sufficient conditions for the strong convergence of this process to a common fixed point of the family of Total asymptotically Nonexpansive maps were proved. Our iterative processes generalise some of the existing ones, our theorems improve, generalise and extend several known results and our method of proof is of independent interest.