In this paper, we present a new non-convex hybrid iteration algorithm for common fixed points of a uniformly closed asymptotically family of countable quasi-Lipschitz mappings in the domains of Hilbert spaces.
Definition 1. \(\{T_n\}\) is said to be asymptotic, if \(\lim_{n\rightarrow \infty} L_n=1\)
Proposition 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 3. The common fixed point set \(F\) of above said \({T_n}\) is closed and convex.
Proposition 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 5. Suppose that \(\alpha_n\in (0,1]\), and \(\beta_n\in [0,1]\) for all \(n\in N\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n)T_nx_n+\alpha_nT_nz_n, & n\geq 0,\\ z_n=(1-\beta_n)+\beta_nT_nx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq L_n(1+(L_n-1)\alpha_n\beta_n)\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_Fx_0\).
Proof.
We give our proof in following steps.
Step 1. 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*}
\|y_n-p\|&=\|(1-\alpha_n)T_nx_n+\alpha_nT_nz_n-p\|\\
&=\|(1-\alpha_n)T_nx_n+\alpha_nT_n((1-\beta_n)+\beta_nT_nx_n)-p\|\\
\nonumber
&=\|(1-\alpha_n\beta_n)(T_nx_n-p)+(\alpha_n\beta_n)(T_{n}^2x_n)\|\\
\nonumber
&\leq (1-\alpha_n\beta_n)\|T_nx_n-p\|+(\alpha_n\beta_n)\|T_{n}^2x_n\|\\
\nonumber
&=L_n(1+(L_n-1)\alpha_n\beta_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 3, 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 nonempty, 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 \(C\) (we show 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 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 exists 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
\(D_n=\{z\in C:\|y_n-z\|^2\leq\|x_n-z\|^2+L_{n}^2(L_n-1)(L_n+1)\}\).
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}^2(L_n-1)(L_n+1)\}\\
\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}^2(L_n-1)(L_n+1)\}\\
\nonumber
&=\{z\in C:2\langle x_n-y_n,z\rangle\leq\|x_n\|^2-\|y_n\|^2+L_{n}^2(L_n-1)(L_n+1)\}
\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[L_n(1+(L_n-1)\alpha_n\beta_n)]^2\|x_n-z\|^2\\
&=\|x_n-z\|^2L_{n}^2+L_{n}^2[2(L_n-1)\alpha_n\beta_n+(L_n-1)^2\alpha_{n}^2\beta_{n}^2]\|x_n-z\|^2\\
&\leq\|x_n-z\|^2L_{n}^2+L_{n}^2[2(L_n-1)+(L_n-1)^2]\|x_n-z\|^2\\
&=\|x_n-z\|^2L_{n}^2+L_{n}^2(L_n-1)(L_n+1)\|x_n-z\|^2.
\end{align*}\
From
\(C_n=\{z\in C:\|y_n-z\|\leq[L_n(1+(L_n-1)\alpha_n\beta_n)]\|x_n-z\|\}\cap A, n\geq 0\), 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\|^2L_{n}^2+L_{n}^2(L_n-1)(L_n+1)\|x_n-z\|^2\\
&\leq\|x_n-z\|^2+L_{n}^2(l_n-1)(L_n+1).
\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
\(\|y_n-x_{n+1}\|^2\leq\|x_n-x_{n+1}\|^2+L_{n}^2(L_n-1)(L_n-1)\rightarrow 0\)
as \(n\rightarrow\infty\). That is, \(y_n\rightarrow q\) as \(n\rightarrow\infty\).
Step 6. We show that \(q\in F\). From the definition of \(y_n\), we have
\((1+\alpha_n\beta_nT_n)\|T_nx_n-x_n\|=\|y_n-x_n\|\rightarrow 0\)
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.
Example 1.
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 6. Assume that \(\alpha_n\in (0,1]\), and \(\beta_n\in [0,1]\) for all \(n\in N\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n)Tx_n+\alpha_nTz_n, & n\geq 0,\\ z_n=(1-\beta_n)+\beta_nTx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_{F(T)}x_0\).
Proof. Take \(T_n\equiv T\), \(L_n\equiv 1\) in Theorem 5, in this case, \(C_n\) is convex and closed and , for all \(n\geq 0\), by using Theorem 1.9, we obtain our desired result.
Corollary 7. Assume that \(\alpha_n\in (0,1]\), and \(\beta_n\in [0,1]\) for all \(n\in N\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n)Tx_n+\alpha_nTz_n, & n\geq 0,\\ z_n=(1-\beta_n)+\beta_nTx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_{F(T)}x_0\).
Theorem 8. Let \(\{T_n\}_{n=0}^{N-1}: C\rightarrow C\) be a finite uniformly L-Lipschitz family of asymptotically quasi-nonexpansive mappings with nonempty common fixed point set \(F\). Assume that \(\alpha_n\in (0,1]\), and \(\beta_n\in [0,1]\) for all \(n\in N\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n)T_{i(n)}^{j(n)}x_n+\alpha_nT_{i(n)}^{j(n)}z_n, & n\geq 0,\\ z_n=(1-\beta_n)+\beta_nT_{i(n)}^{j(n)}x_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq k_{i(n),j(n)}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1+(k_{i(n),j(n)}-1)\alpha_n\beta)\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_Fx_0\).
We can drive the prove from the following two conclusions.Conclusion 9. \(\{T_{n=0}^{N-1}\}_{n=0}^{\infty}\) is a uniformly closed asymptotically family of countable quasi-\(L_n\)-Lipschitz mappings from \(C\) into itself.
Conclusion 10. \(F=\bigcap_{n=0}^{N}F(T_n)=\bigcap_{n=0}^{\infty}F(T_{i(n)}^{j(n)})\), where \(F(T)\) denotes the fixed point set of the mappings \(T\).
Corollary 11. Let \(T: C\rightarrow C\) be a L-Lipschitz asymptotically quasi-nonexpansive mappings with nonempty common fixed point set \(F\). Assume that \(\alpha_n\in (0,1]\), and \(\beta_n\in [0,1]\) for all \(n\in N\). Then \(\{x_n\}\) generated by $$\left\{ \begin{array}{ll} x_0\in C=Q_0, & \text{choosen arbitrarily,}\\ y_n=(1-\alpha_n)T^nx_n+\alpha_nT^nz_n, & n\geq 0,\\ z_n=(1-\beta_n)+\beta_nT^nx_n, & n\geq 0,\\ C_n=\{z\in C:\|y_n-z\|\leq k_n(1+(k_n-1)\alpha_n\beta)\|x_n-z\|\}\cap A, & n\geq 0,\\ Q_n=\{z\in Q_{n-1}:\langle x_n-z,x_0-x_n\rangle\geq 0\},& n\geq 1,\\ x_{n+1}=P_{\overline{co}C_n\cap Q_n}x_0, \end{array} \right.$$ converges strongly to \(P_Fx_0\).
Proof. Take \(T_n\equiv T\) in Theorem 8, we get the desired result.