In this article we present non-convex hybrid iteration algorithm corollaryresponding to Karakaya iterative scheme [1] as done by Guan et al. in [2] corollaryresponding to Mann iterative scheme [3]. We also prove some strong convergence results about common fixed points for a uniformly closed asymptotic family of countable quasi-Lipschitz mappings in Hilbert spaces.
Definition 2.1. \(\{T_n\}\) is said to be asymptotic, if \(\lim_{n\rightarrow \infty} L_n=1\)
Proposition 2.2. For \(x\in H\) and \(z\in C\), \(z=P_Cx\) iff we have $$\langle x-z,z-y\rangle\geq 0$$ for all \(y\in C\).
Proposition 2.3. The common fixed point set \(F\) of above said \({T_n}\) is closed and convex.
Proposition 2.4. For any given \(x_0\in H\), we have \(p=P_Cx_0 \) \(\Longleftrightarrow\) \(\langle p-z,x_0-p\rangle\geq 0\), \(\forall z\in C\).
Theorem 2.5. Assume that \(\alpha_n\), \(\beta_n\), \(\gamma_n\), \(a_n\) and \(b_n\in [0,1]\), \(\alpha_n+\beta_n\in[0,1]\) and \(a_n+b_n\in[0,1]\) for all \(n\in N\) and \(\sum_{n=0}^\infty(\alpha_n+\beta_n)=\infty\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n-\beta_n)z_n+\alpha_nT_nz_n+\beta_nT_nt_n, & n\geq 0,\\ z_n=(1-a_n-b_n)t_n+a_nT_nt_n+b_nT_nx_n , & n\geq 0,\\ t_n=(1-\gamma_n)x_n+\gamma_nT_nx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq[1+(L_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2b_n\gamma_n)+L_{n}^2(-3a_n\gamma_n+\gamma_n-b_n\gamma_n+a_n+b_n)+a_n\gamma_nL_{n}^3\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n+(L_n(1-a_n-b_n-2\gamma_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2a_n\gamma_n-b_n\gamma_n)+L_{n}^2(-a_n\gamma_n+\gamma_n)-b_n\gamma_n-a_n\gamma_n-b_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+a_n+\gamma_n-1)\beta_n+(L_n(1-2a_n-2b_n)+a_nL_{n}^2+b_n-a_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1)\gamma_n-a_n-b_n+L_n(a_n+b_n)]\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_Fx_0\).
Proof. We partition our proof in following seven steps.
Step 2. We know that \(\overline{co}C_n\) and \(Q_n\) are closed and convex for all \(n\geq 0\). Next, we show that \(F\cap A\subset\overline{co}C_n\) for all \(n\geq 0\). Indeed, for each \(p\in F\cap A\), we have \begin{align*}\\ \nonumber \|y_n-p\|&=\|(1-\alpha_n-\beta_n)z_n+\alpha_nT_nz_n+\beta_nT_nt_n-p\|\\ &=\|(1-\alpha_n-\beta_n)[(1-a_n-b_n)t_n+a_nT_nt_n+b_nT_nx_n]\\&+\alpha_nT_n[(1-a_n-b_n)t_n+a_nT_nt_n+b_nT_nx_n]+\beta_nT_nt_n-p\|\\ &=\|(1-\alpha_n-\beta_n)[(1-a_n-b_n)((1-\gamma_n)x_n+\gamma_nT_nx_n)\\& +a_nT_n((1-\gamma_n)x_n+\gamma_nT_nx_n)+b_nT_nx_n ]\\&+\alpha_nT_n[(1-a_n-b_n)((1-\gamma_n)x_n+\gamma_nT_nx_n)\\&+a_nT_n((1-\gamma_n)x_n+\gamma_nT_nx_n)+b_nT_nx_n ]+\beta_nT_n[(1-\gamma_n)x_n+\gamma_nT_nx_n]-p\|\\ \nonumber &=\|(1-\gamma_n-a_n-b_n-\alpha_n-\beta_n-a_n\gamma_n+b_n\gamma_n+\alpha_n\gamma_n+a_n\alpha_n\\& +b_n\alpha_n+\beta_n\gamma_n+a_n\beta_n-b_n\beta_n-a_n\alpha_n\gamma_n-b_n\alpha_n\gamma_n- a_n\beta_n\gamma_n-b_n\beta_n\gamma_n)\\&\times (x_n-p)+(\gamma_n+a_n+b_n+\beta_n+\alpha_n-b_n\alpha_n-a_n\beta_n-b_n\beta_n-2a_n\gamma_n\\& -2b_n\gamma_n-2\alpha_n\gamma_n-2a_n\alpha_n-2\beta_n\gamma_n+3a_n\alpha_n\gamma_n+2b_n\alpha_n\gamma_n+2a_n\beta_n\gamma_n\\& -b_n\beta_n\gamma_n)(T_nx_n-p)+(a_n\gamma_n-3a_n\alpha_n\gamma_n-a_n\beta_n\gamma_n+\alpha_n\gamma_n-b_n\alpha_n\gamma_n\\& +a_n\alpha_n+b_n\alpha_n+\beta_n\gamma_n)(T_{n}^2x_n-p)+(a_n\alpha_n\gamma_n)(T_{n}^3x_n-p)\|\\ \nonumber &\leq (1-\gamma_n-a_n-b_n-\alpha_n-\beta_n-a_n\gamma_n+b_n\gamma_n+\alpha_n\gamma_n+a_n\alpha_n\\& +b_n\alpha_n+\beta_n\gamma_n+a_n\beta_n-b_n\beta_n-a_n\alpha_n\gamma_n-b_n\alpha_n\gamma_n-a_n\beta_n\gamma_n\\& -b_n\beta_n\gamma_n)\|x_n-p\|+(\gamma_n+a_n+b_n+\beta_n+\alpha_n-b_n\alpha_n-a_n\beta_n-b_n\beta_n\\& -2a_n\gamma_n-2b_n\gamma_n-2\alpha_n\gamma_n-2a_n\alpha_n-2\beta_n\gamma_n+3a_n\alpha_n\gamma_n+2b_n\alpha_n\gamma_n\\& +2a_n\beta_n\gamma_n-b_n\beta_n\gamma_n)L_n\|x_n-p\|+(a_n\gamma_n-3a_n\alpha_n\gamma_n-a_n\beta_n\gamma_n\\& +\alpha_n\gamma_n-b_n\alpha_n\gamma_n+a_n\alpha_n+b_n\alpha_n+\beta_n\gamma_n)L_{n}^2\|x_n-p\|+(a_n\alpha_n\gamma_n)L_{n}^3\|x_n-p\|\\ \nonumber &=[1+(L_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n+2b_n\gamma_n)+L_{n}^2(-3a_n\gamma_n\\& +\gamma_n-b_n\gamma_n+a_n+b_n)+a_n\gamma_nL_{n}^3+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n\\& +(L_n(1-a_n-b_n-2\gamma_n+2a_n\gamma_n-b_n\gamma_n)+L_{n}^2(-a_n\gamma_n+\gamma_n)-b_n\gamma_n\\&-a_n\gamma_n-b_n +a_n+\gamma_n-1)\beta_n+(L_n(1-2a_n-2b_n)+a_nL_{n}^2+b_n\\& -a_n-1)\gamma_n-a_n-b_n+L_n(a_n+b_n)]\|x_n-p\| \end{align*} and \(p\in A\), so \(p\in C_n\) which implies that \(F\cap A\subset C_n\) for all \(n\geq 0\). Therefore, \(F\cap A\subset\overline{co}C_n\) for all \(n\geq 0\).
Step 2. We show that \(F\cap A\subset\overline{co}C_n\cap Q_n\) for all \(n\geq 0\). It suffices to show that \(F\cap A\subset Q_n\), for all \(n\geq 0\). We prove this by mathematical induction. For \(n=0\) we have \(F\cap A\subset C=Q_0\). Assume that \(F\cap A\subset Q_n\). Since \(x_{n+1}\) is the projection of \(x_0\) onto \(\overline{co}C_n\cap Q_n\), from Proposition 2.2, we have \(\langle x_{n+1}-z,x_{n+1}-x_0\rangle\leq 0\), \(\forall z\in \overline{co}C_n\cap Q_n\) as \(F\cap A\subset\overline{co}C_n\cap Q_n\), the last inequality holds, in particular, for all \(z\in F\cap A\). This together with the definition of \(Q_{n+1}\) implies that \(F\cap A\subset Q_{n+1}\). Hence the \(F\cap A\subset\overline{co}C_n\cap Q_n\) holds for all \(n\geq 0\).
Step 3. We prove \(\{x_n\}\) is bounded. Since \(F\) is a nonepmty, closed, and convex subset of \(C\), there exists a unique element \(z_0\in F\) such that \(z_0=P_Fx_0\). From \(x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0\), we have \(\|x_{n+1}-x_0\|\leq \|z-x_0\|\) for every \(z\in \overline{co}C_n\cap Q_n\). As \(z_0\in F\cap A\subset\overline{co}C_n\cap Q_n\), we get \(\|x_{n+1}-x_0\|\leq \|z_0-x_0\|\) for each \(n\geq 0\). This implies that \(\{x_n\}\) is bounded.
Step 4. We show that \(\{x_n\}\) converges strongly to a point of \(C\) by showing that \(\{x_n\}\) is a cauchy sequence. As \(x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0\subset Q_n\) and \(x_n=P_{Q_n}x_0\) (Proposition 2.4), we have \(\|x_{n+1}-x_0\|\geq \|x_n-x_0\|\) for every \(n\geq 0\), which together with the boundedness of \(\|x_n-x_0\|\) implies that there exsists the limit of \(\|x_n-x_0\|\). On the other hand, from \(x_{n+m}\in Q_n\), we have \(\langle x_n-x_{n+m},x_n-x_0\rangle\leq 0\) and hence \begin{align*} \|x_{n+m}-x_n\|^2&=\|(x_{n+m}-x_0)-(x_n-x_0)\|^2\\ \nonumber &\leq\|x_{n+m}-x_0\|^2-\|x_n-x_0\|^2-2\langle x_{n+m}-x_n,x_n-x_0\rangle\\ \nonumber &\leq\|x_{n+m}-x_0\|^2-\|x_n-x_0\|^2\rightarrow0,\ n\rightarrow\infty \end{align*} for any \(m\geq 1\). Therefore \(\{x_n\}\) is a cauchy sequence in \(C\), then there exists a point \(q\in C\) such that \(\lim_{n\rightarrow \infty} x_n=q\).
Step 5. We show that \(y_n\rightarrow q\), as \(n\rightarrow\infty\). Let \(D_n=\{z\in C:\|y_n-z\|^2\leq\|x_n-z\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\}\). From the definition of \(D_n\), we have \begin{align*} D_n&=\{z\in C:\langle y_n-z,y_n-z\rangle\leq\langle x_n-z,x_n-z\rangle+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\}\\ \nonumber &=\{z\in C:\|y_n\|^2-2\langle y_n,z\rangle+\|z\|^2\leq\|x_n\|^2-2\langle x_n,z\rangle\\&+\|z\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\}\\ \nonumber &=\{z\in C:2\langle x_n-y_n,z\rangle\leq\|x_n\|^2-\|y_n\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\} \end{align*} This shows that \(D_n\) is convex and closed, \(n \in \mathbb{Z^{+}}\cup \{0\}\). Next, we want to prove that \(C_n\subset D_n\),\(n\geq 0\). In fact, for any \(z\in C_n\), we have \begin{align*} \|y_n-z\|^2&\leq[1+(L_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n+2b_n\gamma_n)+L_{n}^2(-3a_n\gamma_n+\gamma_n\\&-b_n\gamma_n+a_n+b_n) +a_n\gamma_nL_{n}^3+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n\\&+(L_n(1-a_n-b_n-2\gamma_n+2a_n\gamma_n-b_n\gamma_n) +L_{n}^2(-a_n\gamma_n+\gamma_n)\\&-b_n\gamma_n-a_n\gamma_n-b_n+a_n+\gamma_n-1)\beta_n+(L_n(1-2a_n-2b_n)\\&+a_nL_{n}^2 +b_n-a_n-1)\gamma_n-a_n-b_n+L_n(a_n+b_n)]^2\|x_n-z\|^2\\ &=\|x_n-z\|^2+2[(L_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n+2b_n\gamma_n)\\&+L_{n}^2(-3a_n\gamma_n+\gamma_n-b_n\gamma_n+a_n+b_n) +a_n\gamma_nL_{n}^3+a_n+b_n-a_n\gamma_n\\&-b_n\gamma_n-1)\alpha_n+(L_n(1-a_n-b_n-2\gamma_n+2a_n\gamma_n -b_n\gamma_n)\\& +L_{n}^2(-a_n\gamma_n+\gamma_n)-b_n\gamma_n-a_n\gamma_n-b_n+a_n+\gamma_n-1)\beta_n\\& +(L_n(1-2a_n-2b_n)+a_nL_{n}^2+b_n-a_n-1)\gamma_n-a_n-b_n+L_n(a_n+b_n)]\\& +[(L_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n+2b_n\gamma_n) +L_{n}^2(-3a_n\gamma_n+\gamma_n\\&-b_n\gamma_n+a_n+b_n)+a_n\gamma_nL_{n}^3+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n \\&+(L_n(1-a_n-b_n-2\gamma_n+2a_n\gamma_n-b_n\gamma_n)+L_{n}^2(-a_n\gamma_n+\gamma_n)\\&-b_n\gamma_n-a_n\gamma_n-b_n+a_n +\gamma_n-1)\beta_n+(L_n(1-2a_n-2b_n)\\&+a_nL_{n}^2+b_n-a_n-1)\gamma_n-a_n-b_n+L_n(a_n+b_n)]^2\alpha_n^2]\|x_n-z\|^2\\ &\leq\|x_n-z\|^2+[2(L_{n}^3-2L_n-6)+(L_{n}^3-2L_n-6)^2]\|x_n-z\|^2\\ &=\|x_n-z\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\|x_n-z\|^2. \end{align*} From \(C_n=\{z\in C:\|y_n-z\|\leq[1+(L_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n+2b_n\gamma_n)+L_{n}^2(-3a_n\gamma_n+\gamma_n-b_n\gamma_n+a_n+b_n)+a_n\gamma_nL_{n}^3+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n+(L_n(1-a_n-b_n-2\gamma_n+2a_n\gamma_n-b_n\gamma_n)+L_{n}^2(-a_n\gamma_n+\gamma_n)-b_n\gamma_n-a_n\gamma_n-b_n+a_n+\gamma_n-1)\beta_n+(L_n(1-2a_n-2b_n)+a_nL_{n}^2+b_n-a_n-1)\gamma_n-a_n-b_n+L_n(a_n+b_n)]\|x_n-z\|\}\cap A,\ n\geq 0\), we have \(C_n\subset A\), \(n\geq 0\). Since \(A\) is convex, we also have \(\overline{co}C_n\subset A\), \(n\geq 0\). Consider \(x_n\in\overline{co}C_{n-1}\), we know that \begin{align*} \|y_n-z\|&\leq\|x_n-z\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\|x_n-z\|^2\\ &\leq\|x_n-z\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4). \end{align*} This implies that \(z\in D_n\) and hence \(C_n\subset D_n\), \(n\geq 0\). Sinnce \(D_n\) is convex, we have \(\overline{co}(C_n)\subset D_n\), \(n\geq 0\). Therefore \(\|y_n-x_{n+1}\|^2\leq\|x_n-x_{n+1}\|^2+(L_{n}^3-2L_n-6)(L_{n}^3-2L_n-4)\rightarrow 0,\) as \(n\rightarrow\infty\). That is, \(y_n\rightarrow q\) as \(n\rightarrow\infty\).
Step 6. To prove that \(q\in F\), we use definition of \(y_n\). So we have \((\alpha_n+\beta_n+\gamma_n+a_n+b_n-a_n\gamma_n-b_n\gamma_n-\alpha_n\gamma_n-a_n\alpha_n+a_n\alpha_n\gamma_n-b_n\alpha_n+b_n\alpha_n\gamma_n-\beta_n\gamma_n-a_n\beta_n+a_n\beta_n\gamma_n-b_n\beta_n+b_n\beta_n\gamma_n-a_n\beta_n\gamma_nT_n-a_n\alpha_n\gamma_nT_n+\alpha_n\gamma_nT_n+a_n\alpha_nT_n-a_n\alpha_n\gamma_nT_n+b_n\alpha_nT_n-b_n\alpha_n\gamma_nT_n+\beta_n\gamma_nT_n+a_n\alpha_n\gamma_nT_{n}^2+a_n\gamma_nT_n)\|T_nx_n-x_n\|=\|y_n-x_n\|\rightarrow 0,\) as \(n\rightarrow \infty\). Since \(\alpha_n\in(a,1]\subset[0,1]\), from the above limit we have \(\lim_n\rightarrow\infty\|T_nx_n-x_n\|=0.\) Since \(\{T_n\}\) is uniformly closed and \(x_n\rightarrow q\), we have \(q\in F\).
Step 7. We claim that \(q=z_0=P_Fx_0\), if not, we have that \(\|x_0-p\|>\|x_0-z_0\|\). There must exist a positive integer \(N\), if \(n>N\) then \(\|x_0-x_n\|>\|x_0-z_0\|\), which leads to \(\|z_0-x_n\|^2=\|z_0-x_n+x_n-x_0\|^2=\|z_0-x_n\|^2+\|x_n-x_0\|^2+2\langle z_0-x_n,x_n-x_0\rangle\). It follows that \(\langle z_0-x_n,x_n-x_0\rangle<0\) which implies that \(z_0\overline{\in} Q_n\), so that \(z_0\overline{\in} F\), this is a contradiction. This completes the proof.
Now, we present an example of \(C_n\) which does not involve a convex subset.Example 2.6 Take \(H=R^2\), and a sequence of mappings \(T_n:R^2\rightarrow R^2\) given by \(T_n:(t_1,t_2)\mapsto(\frac{1}{8}t_1,t_2)\),\ \(\forall(t_1,t_2)\in R^2\), \(\forall n\geq 0\). It is clear that \(\{T_n\}\) satisfies the desired definition of with \(F=\{(t_1,0):t_1\in(-\infty,+\infty)\}\) common fixed point set. Take \(x_0=(4,0)\), \(a_0=\frac{6}{7}\), we have \(y_0=\frac{1}{7}x_0+\frac{6}{7}T_0x_0=(4\times\frac{1}{7}+\frac{4}{8}\times\frac{6}{7},0)=(1,0)\). Take \(1+(L_0-1) a_0=\sqrt{\frac{5}{2}}\), we have \(C_0=\{z\in R^2:\|y_0-z\|\leq\sqrt{\frac{5}{2}}\|x_0-z\|\}\). It is easy to show that \(z_1=(1,3)\), \(z_2=(-1,3)\in C_0\). But \(z^{‘}=\frac{1}{2}z_1+\frac{1}{2}z_2=(0,3)\overline{\in} C_0\), since \(\|y_0-z\|=2\), \(\|x_0-z\|=1\). Therefore \(C_0\) is not convex.
Corollary 2.7 Assume that \(\alpha_n\), \(\beta_n\), \(\gamma_n\), \(a_n\) and \(b_n\in [0,1]\), \(\alpha_n+\beta_n\in[0,1]\) and \(a_n+b_n\in[0,1]\) for all \(n\in N\) and \(\sum_{n=0}^\infty(\alpha_n+\beta_n)=\infty\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n-\beta_n)z_n+\alpha_nTz_n+\beta_nTt_n, & n\geq 0,\\ z_n=(1-a_n-b_n)t_n+a_nTt_n+b_nTx_n , & n\geq 0,\\ t_n=(1-\gamma_n)x_n+\gamma_nTx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq[1-(\gamma_n-b_n)\alpha_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(2a_n+b_n)\gamma_n-2b_n\beta_n(1+\gamma_n)]\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_{F(T)}x_0\).
Proof. Take \(T_n\equiv T\), \(L_n\equiv 1\) in Theorem 2.5, in this case, \(C_n\) is convex and closed and , for all \(n\geq 0\), by using Theorem 2.5, we obtain Corollary 2.7. Take \(T_n\equiv T\), \(L_n\equiv 1\) in Theorem 2.5, in this case, \(C_n\) is closed and convex, for all \(n\geq 0\), by using Theorem 2.5, we obtain Corollary 2.7.
Corollary 2.8 Assume that \(\alpha_n\), \(\beta_n\), \(\gamma_n\), \(a_n\) and \(b_n\in [0,1]\), \(\alpha_n+\beta_n\in[0,1]\) and \(a_n+b_n\in[0,1]\) for all \(n\in N\) and \(\sum_{n=0}^\infty(\alpha_n+\beta_n)=\infty\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n-\beta_n)z_n+\alpha_nTz_n+\beta_nTt_n, & n\geq 0,\\ z_n=(1-a_n-b_n)t_n+a_nTt_n+b_nTx_n , & n\geq 0,\\ t_n=(1-\gamma_n)x_n+\gamma_nTx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq[1-(\gamma_n-b_n)\alpha_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-(2a_n+b_n)\gamma_n-2b_n\beta_n(1+\gamma_n)]\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_{F(T)}x_0\).
Theorem 3.1 Let \(\{T_n\}_{n=0}^{N-1}: C\rightarrow C\) be a uniformly L-Lipschitz finit family of asymptotically quasi-nonexpansive mappings with nonempty common fixed point set \(F\). Assume that \(\alpha_n\), \(\beta_n\), \(\gamma_n\), \(a_n\) and \(b_n\in [0,1]\), \(\alpha_n+\beta_n\in[0,1]\) and \(a_n+b_n\in[0,1]\) for all \(n\in N\) and \(\sum_{n=0}^\infty(\alpha_n+\beta_n)=\infty\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n-\beta_n)z_n+\alpha_nT_{i(n)}^{j(n)}z_n+\beta_nT_{i(n)}^{j(n)}t_n, & n\geq 0,\\ z_n=(1-a_n-b_n)t_n+a_nT_{i(n)}^{j(n)}t_n+b_nT_{i(n)}^{j(n)}x_n , & n\geq 0,\\ t_n=(1-\gamma_n)x_n+\gamma_nT_{i(n)}^{j(n)}x_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq[1+(k_{i(n),j(n)}(1-b_n-2\gamma_n-2a_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3a_n\gamma_n+2b_n\gamma_n)+k_{i(n),j(n)}^2(-3a_n\gamma_n+\gamma_n-b_n\gamma_n+a_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+b_n)+a_n\gamma_nk_{i(n),j(n)}^3+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(k_{i(n),j(n)}(1-a_n-b_n-2\gamma_n+2a_n\gamma_n-b_n\gamma_n)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+k_{i(n),j(n)}^2(-a_n\gamma_n+\gamma_n)-b_n\gamma_n-a_n\gamma_n-b_n+a_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\gamma_n-1)\beta_n+(k_{i(n),j(n)}(1-2a_n-2b_n)+a_nk_{i(n),j(n)}^2\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+b_n-a_n-1)\gamma_n-a_n-b_n+k_{i(n),j(n)}(a_n+b_n)]\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_Fx_0\), where \(n=(j(n)-1)N+i(n)\) for all \(n\geq0\).
Proof
We can drive the prove from the following two conclusions:
conclusion1
\(\{T_{n=0}^{N-1}\}_{n=0}^{\infty}\) is a uniformly closed asymptotically family of countable quasi-\(L_n\)-Lipschitz mappings from \(C\) into itself.
conclusion1
\(F=\bigcap_{n=0}^{N}F(T_n)=\bigcap_{n=0}^{\infty}F(T_{i(n)}^{j(n)})\), where \(F(T)\) denotes the fixed point set of the mappings \(T\).
Corollary 3.2 Let \(T: C\rightarrow C\) be a L-Lipschitz asymptotically quasi-nonexpansive mappings with nonempty common fixed point set \(F\). Assume that \(\alpha_n\), \(\beta_n\), \(\gamma_n\), \(a_n\) and \(b_n\in [0,1]\), \(\alpha_n+\beta_n\in[0,1]\) and \(a_n+b_n\in[0,1]\) for all \(n\in N\) and \(\sum_{n=0}^\infty(\alpha_n+\beta_n)=\infty\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n-\beta_n)z_n+\alpha_nT^nz_n+\beta_nT^nt_n, & n\geq 0,\\ z_n=(1-a_n-b_n)t_n+a_nT^nt_n+b_nT^nx_n , & n\geq 0,\\ t_n=(1-\gamma_n)x_n+\gamma_nT^nx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq[1+(k_n(1-b_n-2\gamma_n-2a_n+3a_n\gamma_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2b_n\gamma_n)+k_{n}^2(-3a_n\gamma_n+\gamma_n-b_n\gamma_n+a_n+b_n)+a_n\gamma_nk_{n}^3\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+a_n+b_n-a_n\gamma_n-b_n\gamma_n-1)\alpha_n+(k_n(1-a_n-b_n-2\gamma_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+2a_n\gamma_n-b_n\gamma_n)+k_{n}^2(-a_n\gamma_n+\gamma_n)-b_n\gamma_n-a_n\gamma_n-b_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+a_n+\gamma_n-1)\beta_n+(k_n(1-2a_n-2b_n)+a_nk_{n}^2+b_n-a_n\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1)\gamma_n-a_n-b_n+k_n(a_n+b_n)]\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_Fx_0\), where \(\overline{co}C_n\) denotes the closed convex closure of \(C_n\) for all \(n\geq 1\), \(A=\{z\in H:\|z-P_Fx_0\|\leq 1\}\).
Proof. Take \(T_n\equiv T\) in Theorem 3.1, we proved.