In this paper, we present a new viscosity technique of nonexpansive mappings in the framework of CAT(0) spaces. The strong convergence theorems of the proposed technique is proved under certain assumptions imposed on the sequence of parameters. The results presented in this paper extend and improve some recent announced in the current literature.
Theorem 1.1 Let \(C\) be a nonempty closed convex subset of the real CAT(0) space \(X\). Let \(T\) be a nonexpansive mapping of \(C\) into itself such that \(Fix(T)\) is nonempty. Let \(f\) be a contraction of \(C\) into itself with coefficient \(\theta\in [0,1)\). Pick any \(x_{0}\in [0,1)\), let \(\{x_{n}\}\) be a sequence generated by $$x_{n+1}=\frac{\gamma_{n}}{1+\gamma_{n}}f(x_{n})+\frac{1}{1+\gamma_{n}}T(x_{n}),\;\;\;n\geq 0.$$ Where \(\{\gamma_{n}\}\) is a sequence in \((0,1)\) satisfying the following conditions:
Lemma 2.1. Let \(X\) be a CAT(0) spaces.
Definition 2.2 Let \(X\) be a CAT(0) space and \(T: X\rightarrow X\) be a mapping. Then \(T\) is called nonexpensive if $$d(T(x), T(y))\leq d(x,y), \;\;\; x,y\in C$$
Definition 2.3 Let \(X\) be a CAT(0) space and \(T: X\rightarrow X\) be a mapping. Then \(T\) is called contraction if $$d(T(x), T(y))\leq \theta d(x,y), \;\;\; x,y\in C\;\; \theta \in [0,1)$$
Berg and Nikolaev [4] introduce the concept of quasilinearization as follow. Let us denote the pair \((a,b)\in X\times X\) by the \(\overrightarrow{ab}\) and call it a vector. Then, quasilinearization is defined as a map $$\langle .,.\rangle: (x\times X)\times (X\times X) \longrightarrow\mathbb{R}$$ defined asDefinition 2.4. Let \(P_{c}: X\rightarrow C\) is called the metric projection if for every \(x\in X\) there exist a unique nearest point in \(C\), denoted by \(P_{c}x\), such that $$d(x, P_{c}x)\leq d(x,y), \;\;\; y\in C.$$
The following theorem gives you the conditions for a projection mapping to be non-expensive.Theorem 2.5 Let \(C\) be a non-empty closed convex subset of a real CAT(0) space \(X\) and \(P_{c}: X\rightarrow X\) a metric projection. Then
Lemma 2.6 (The demiclosedness principle) Let \(C\) be a nonempty closed convex subset of the real CAT(0) space \(X\) and \(T:C\rightarrow C\) such that $$x_n\rightharpoonup x^\ast \in C\,\, \mbox{and}\,\, (I-T)x_n \rightarrow 0.$$ Then \(x^\ast=Tx^\ast\). (Here \(\rightarrow\) (respectively ⇀) denotes strong (respectively weak) convergence.) Moreover, the following result gives the conditions for the convergence of a nonnegative real sequences.
Lemma 2.7. Assume that \(\{a_n\}\) is a sequence of nonnegative real numbers such that \(a_{n+1}\leq(1-\beta_n)a_n+\delta_n, \forall n\geq0\), where \(\{\beta_n\}\) is a sequence in \((0,1)\) and \(\{\delta_n\}\) is a sequence with
Theorem 3.1. Let \(C\) be a non-empty closed convex subset of a complete CAT(0) space \(X\) and \(T:C\longrightarrow C\) be a non-expensive mapping with \(\textrm{Fix}(T)\neq\emptyset\). Let \(f:C\longrightarrow C\) be a contraction with coefficient \(\theta\in [0,1)\) and for arbitrary initial point \(x_{0}\in C\). Let \(\{x_{n}\}\) be a sequence generated by $$\left\{ \begin{array}{ll} x_{n+1}=T(y_{n}),\\ y_{n}=\alpha_{n}(w_{n})\oplus\beta_{n}f(w_{n})\oplus\gamma_{n}T(w_{n}), \\ w_{n}=\frac{x_{n}\oplus x_{n+1}}{2}. \end{array} \right.$$ Where \(\{\alpha_{n}\},\{\beta_{n}\}\) and \(\{\gamma_{n}\}\) are the sequence in \((0,1)\) satisfying the following conditions:
Proof. We divide the proof into four steps:
Step 1: Firstly we show that the sequence \(\{x_{n}\}\) is bounded. Indeed take \(p\in \textrm{Fix}(T)\) arbitrary, we have \begin{eqnarray*} d(x_{n+1},p)&=& d(T(y_{n}),p)\\ &=& d(T(\alpha_{n}(w_{n})\oplus\beta_{n}(w_{n})\oplus\gamma_{n}(w_{n})), p)\\ &\leq& d(\alpha_{n}(w_{n})\oplus\beta_{n}(w_{n})\oplus\gamma_{n}T(w_{n}), p)\\ &=& d(\alpha_{n}(w_{n})-\alpha_{n}p+\beta_{n}(w_{n})-\beta_{n}p+\gamma_{n}T(w_{n})+\alpha_{n}p+\beta_{n}p, p)\\ &\leq & \alpha_{n}d((w_{n}),p)+\beta_{n}d((w_{n}),p)+\gamma_{n}d(T(w_{n}), p)\\ &\leq &\frac{\alpha_{n}}{2}d((x_{n}),p)+\frac{\alpha_{n}}{2}d((x_{n+1}),p)+\beta_{n}d((w_{n}),f(p))+\beta_{n}d(f(p),p)+\gamma_{n}d(T(w_{n}), p)\\ &=& \frac{\alpha_{n}}{2}d((x_{n}),p)+\frac{\alpha_{n}}{2}d((x_{n+1}),p)+\theta\beta d((w_{n}),p)+\beta d(f(p),p)\\ &+&\gamma_{n}\left(\frac{1}{2}d(x_{n}, p)+\frac{1}{2}d(x_{n+1}, p)\right)\\ &=& \left(\frac{\alpha_{n}+\gamma_{n}+\theta\beta_{n}}{2}\right)d(x_{n},p)+\left(\frac{\alpha_{n}+\gamma_{n}+\theta\beta_{n}}{2}\right)d(x_{n+1},p)\\ &+& \frac{\gamma_{n}}{2}d(x_{n+1}),p)+\beta_{n}d(f(p),p)\\ &=& \left(\frac{1-\beta_{n}+\theta\beta_{n}}{2}\right)d(x_{n},p)+\left(\frac{1-\beta_{n}+\theta\beta_{n}}{2}\right)d(x_{n+1},p)\\ &+& \frac{\gamma_{n}}{2}d(x_{n+1}),p)+\beta_{n}d(f(p),p). \end{eqnarray*} It follows that \begin{eqnarray*} \left(1-\frac{1-\beta_{n}+\theta\beta_{n}}{2}\right)d(x_{n+1},p)&=&\left(\frac{1-\beta_{n}+\theta\beta_{n}}{2}\right)d(x_{n},p)\\ &&+\beta_{n}d(f(p),p). \end{eqnarray*} implies that
Step 2: Now, we prove that \(\lim\limits_{n\rightarrow \infty}d(x_{n+1},x_{n})=0\) \begin{eqnarray*} d(x_{n+1},x_{n})&=& d(T(y_{n}),T(y_{n-1}))\\ &=& d(T(\alpha_{n}(w_{n})\oplus\beta_{n}(w_{n})\oplus\gamma_{n}(w_{n})), T(\alpha_{n-1}(w_{n-1})\oplus\beta_{n-1}(w_{n-1})\oplus\gamma_{n-1}(w_{n-1})))\\ &\leq& d(\alpha_{n}(w_{n})\oplus\beta_{n}(w_{n})\oplus\gamma_{n}T(w_{n}), [\alpha_{n-1}(w_{n-1})\oplus\beta_{n-1}(w_{n-1})\oplus\gamma_{n-1}T(w_{n-1})])\\ &\leq &\frac{\alpha_{n}}{2}d(x_{n+1},x_{n})+\frac{\alpha_{n}}{2}d(x_{n},x_{n-1})+\frac{1}{2}|\alpha_{n}-\alpha_{n-1}|d\left((x_{n-1}+x_{n}),2T(w_{n-1})\right)\\ &+&\beta_{n}d(f(w_{n}),f(w_{n-1}))+|\beta_{n}-\beta_{n-1}|d(f(w_{n-1}),T(w_{n-1}))+\gamma_{n}d(T(w_{n}),T(w_{n-1}))\\ &=& \frac{\alpha_{n}}{2}d(x_{n+1},x_{n})+\frac{\alpha_{n}}{2}d(x_{n},x_{n-1})+\left(\frac{1}{2}|\alpha_{n}-\alpha_{n-1}|+|\beta_{n}-\beta_{n-1}|\right)M\\ &+&\theta\beta_{n}d(w_{n},w_{n-1})+\gamma_{n}(w_{n},w_{n-1})\\ &=& \frac{\alpha_{n}}{2}d(x_{n+1},x_{n})+\frac{\alpha_{n}}{2}d(x_{n},x_{n-1})+\left(\frac{1}{2}|\alpha_{n}-\alpha_{n-1}|+|\beta_{n}-\beta_{n-1}|\right)M\\ &+&\frac{\theta\beta_{n}}{2}d(x_{n+1},x_{n})+\frac{\theta\beta_{n}}{2}d(x_{n},x_{n-1})+ \frac{\gamma_{n}}{2}d(x_{n+1},x_{n})+\frac{\gamma_{n}}{2}d(x_{n},x_{n-1})\\ &=& \left(\frac{\alpha_{n}+\gamma_{n}+\theta\beta_{n}}{2}\right)d(x_{n+1},x_{n})+\left(\frac{\alpha_{n}+\gamma_{n}+\theta\beta_{n}}{2}\right)d(x_{n},x_{n-1})\\ &+& \left(\frac{1}{2}|\alpha_{n}-\alpha_{n-1}|+|\beta_{n}-\beta_{n-1}|\right)M \end{eqnarray*} Where \(M>0\) is constant such that $$M\geq \max\left\{\sup_{n\geq 0}d((x_{n}+x_{n+1},2T(w_{n-1})),\sup_{n\geq 0}d(f(w_{n-1}),T(w_{n-1}))\right\}$$ It gives \begin{eqnarray*} \left(1-\frac{\alpha_{n}+\theta\beta_{n}+\gamma_{n}}{2}\right)d(x_{n+1},x_{n})&=&\left(\frac{\alpha_{n}+\theta\beta_{n}+\gamma_{n}}{2}\right)d(x_{n},x_{n-1})\\ &+& \left(\frac{1}{2}|\alpha_{n}-\alpha_{n-1}|+|\beta_{n}-\beta_{n-1}|\right)M \end{eqnarray*} implies that \begin{eqnarray*} \left(1-\frac{1-\beta_{n}+\theta\beta_{n}}{2}\right)d(x_{n+1},x_{n})&=&\left(\frac{1-\beta_{n}+\theta\beta_{n}}{2}\right)d(x_{n},x_{n-1})\\ &+& \left(\frac{1}{2}|\alpha_{n}-\alpha_{n-1}|+|\beta_{n}-\beta_{n-1}|\right)M \end{eqnarray*} implies \begin{eqnarray*} (1+\beta_{n}(1-\theta))d(x_{n+1},x_{n})&\leq &(1-\beta_{n}(1-\theta))d(x_{n},x_{n-11})\\ &+& \left(|\alpha_{n}-\alpha_{n-1}|+2|\beta_{n}-\beta_{n-1}|\right)M. \end{eqnarray*} Thus, we have \begin{eqnarray*} d(x_{n+1},x_{n})&\leq&\left(\frac{1-\beta_{n}(1-\theta)}{1+\beta_{n}(1-\theta)}\right)d(x_{n},x_{n-1})\\ &+&\frac{M}{(1+\beta_{n}(1-\theta))} \left(|\alpha_{n}-\alpha_{n-1}|+2|\beta_{n}-\beta_{n-1}|\right). \end{eqnarray*} Since \(\beta_{n}, \theta\in (0,1), 1+\beta_{n}(1-\theta)\geq 1\) and \(\left(\frac{1-\beta_{n}(1-\theta)}{1+\beta_{n}(1-\theta)}\right)\leq 1-\beta_{n}(1-\theta)\) Thus \begin{eqnarray*} d(x_{n+1},x_{n})&\leq&[1-\beta_{n}(1-\theta)]d(x_{n},x_{n-1})\\ &+&\frac{M}{(1+\beta_{n}(1-\theta))} \left(|\alpha_{n}-\alpha_{n-1}|+2|\beta_{n}-\beta_{n-1}|\right). \end{eqnarray*} Since \(\sum_{n=0}^{\infty}\beta_{n}=\infty\), \(\sum_{n=0}^{\infty}|\alpha_{n+1}-\alpha_{n}|<\infty\),and \(\sum_{n=0}^{\infty}|\beta_{n+1}-\beta_{n}|<\infty\), by the Lemma (2.7) we have \(\lim\limits_{n\rightarrow \infty}d(x_{n+1},x_{n})=0\).
Step 3: In this step, we claim that $$\limsup\limits_{x\rightarrow\infty}\langle\overrightarrow{x^{\ast}f(x^\ast)},\overrightarrow{x^{\ast}x_{n}}\rangle\leq 0,$$ where \(x^{\ast}=P_{Fix(T)}f(x^{\ast})\). Indeed, we take a subsequence \(\{x_{n_{i}}\}\) of \(\{x_{n}\}\) which converges weakly to a fixed point \(p\) of \(T\). Without loss of generality, we may assume that \(\{x_{n_{i}}\}\rightharpoonup p\). From \(\lim\limits_{n\rightarrow \infty}d(x_{n},T(x_{n})=0\) and the Lemma (2.6) we have \(p=Tp\). This together, with the properity of metric projection implies that \begin{eqnarray*} \limsup\limits_{x\rightarrow\infty}\langle\overrightarrow{x^{\ast}f(x^\ast)},\overrightarrow{x^{\ast}x_{n}}\rangle &=&\limsup\limits_{x\rightarrow\infty}\langle\overrightarrow{x^{\ast}f(x^\ast)},\overrightarrow{x^{\ast}x_{n_{i}}}\rangle\\ &=&\limsup\limits_{x\rightarrow\infty}\langle\overrightarrow{x^{\ast}f(x^\ast)},\overrightarrow{x^{\ast}p}\rangle\\ &\leq & 0. \end{eqnarray*}
Step 4: Finally, we show that \(x_{n}\rightarrow x^{\ast}\) as \(n\rightarrow \infty\). Now, we prove that \(\lim\limits_{n\rightarrow \infty}d(x_{n+1},x_{n})=0\). Now, we again take \(x^{\ast}\in \textrm{Fix}(T)\) is the unique fixed point of the contraction \(P_{\textrm{Fix}(T)}f\). Consider \begin{eqnarray*} d^2(x_{n+1},x_{n})&=& d^2(T(y_{n}),x^{\ast})\\ &=& d^2(T(\alpha_{n}(w_{n})\oplus\beta_{n}(w_{n})\oplus\gamma_{n}(w_{n})), x^{\ast})\\ &\leq& d^{2}(\alpha_{n}(w_{n})\oplus\beta_{n}(w_{n})\oplus\gamma_{n}T(w_{n}),x^{\ast} )\\ &=& d^{2}(\alpha_{n}(w_{n})-\alpha_{n}x^{\ast}+\beta_{n}(w_{n})-\beta_{n}x^{\ast}+\gamma_{n}T(w_{n})+\alpha_{n}x^{\ast}+\beta_{n}x^{\ast}, x^{\ast})\\ &=& \alpha^{2}_{n}d^2((w_{n}),x^{\ast})+\beta^{2}_{n}d^2((w_{n}),x^{\ast})+\gamma^{2}_{n}d^2((w_{n}),x^{\ast})\\ &+&2\alpha_{n}\beta_{n}\langle\overrightarrow{x^{\ast}w_{n}},\overrightarrow{x^{\ast}f(w_{n})}\rangle+2\alpha_{n}\gamma_{n}\langle\overrightarrow{x^{\ast}w_{n}},\overrightarrow{x^{\ast}T(w_{n})}\rangle\\ &+& 2\beta_{n}\gamma_{n}\langle\overrightarrow{x^{\ast}f(w_{n})},\overrightarrow{x^{\ast}T(w_{n})}\rangle\\ &=& \alpha^{2}_{n}d^2((w_{n}),x^{\ast})+\beta^{2}_{n}d^2((w_{n}),x^{\ast})+\gamma^{2}_{n}d^2((w_{n}),x^{\ast})\\ &+&2\alpha_{n}\beta_{n}\langle\overrightarrow{x^{\ast}w_{n}},\overrightarrow{x^{\ast}f(w_{n})}\rangle+2\alpha_{n}\gamma_{n}d(x_{n},x^{\ast})d(T(w_{n}),x^{\ast})\\ &+& 2\beta_{n}\gamma_{n}\langle\overrightarrow{x^{\ast}f(w_{n})},\overrightarrow{x^{\ast}T(w_{n})}\rangle\\ &\leq& (\alpha^{2}_{n}+\gamma^{2}_{n})d^{2}(w_{n},x^{\ast})+2\alpha_{n}\gamma_{n}d^{2}(w_{n},x^{\ast})+2\beta_{n}\gamma_{n}d^{2}(f(w_{n}),f(x^{\ast}))d^{2}(w_{n},x^{\ast})+K_{n}\\ &\leq& (\alpha^{2}_{n}+\gamma^{2}_{n})d^{2}(w_{n},x^{\ast})+2\theta\beta_{n}\gamma_{n}d^{2}(w_{n},x^{\ast})+K_{n}\\ &\leq&(\alpha^{2}_{n}+\gamma^{2}_{n}+2\theta\beta_{n}\gamma_{n})d^{2}(w_{n},x^{\ast})+K_{n}\\ &\leq&((1-\beta^{2}_{n})^2+2\theta\beta_{n}\gamma_{n})d^{2}(w_{n},x^{\ast})+K_{n} \end{eqnarray*} where \begin{eqnarray*} K_{n}&=&\beta^{2}_{n}d^2(f(w_{n}),x^{\ast})+2\alpha_{n}\beta_{n}\langle\overrightarrow{x^{\ast}w_{n}},\overrightarrow{x^{\ast}f(w_{n})}\rangle\\ &+&2\beta_{n}\gamma_{n}\langle\overrightarrow{f(w_{n})}x^{\ast},\overrightarrow{T(w_{n})x^{\ast}}\rangle \end{eqnarray*} it become $$[(1-\beta)^2+2\theta\beta_{n}\gamma_{n})]d^{2}(w_{n},x^{\ast})\geq d^{2}(x_{n+1},x_{n})-K_{n}$$ implies $$\sqrt{(1-\beta)^2+2\theta\beta_{n}\gamma_{n}}d(w_{n},x^{\ast})\geq \sqrt{d^{2}(x_{n+1},x_{n})-K_{n}}$$ implies \begin{align*} \frac{1}{2}\sqrt{(1-\beta)^2+2\theta\beta_{n}\gamma_{n}}d(w_{n},x^{\ast})(d(x_{n+1},x^{\ast})+d(x_{n},x^{\ast}))\\ \geq \sqrt{d^{2}(x_{n+1},x_{n})-K_{n}} \end{align*} implies \begin{eqnarray*} \frac{1}{4}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})(d^{2}(x_{n+1},x^{\ast})+d^{2}(x_{n},x^{\ast}))&+&2(d(x_{n+1},x^{\ast})+d(x_{n},x^{\ast}))\\ &\geq& d^{2}(x_{n+1},x_{n})-K_{n} \end{eqnarray*} implies \begin{eqnarray*} \frac{1}{4}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})(d^{2}(x_{n+1},x^{\ast})+d^{2}(x_{n},x^{\ast}))&+&(d^{2}(x_{n+1},x^{\ast})+d^{2}(x_{n},x^{\ast}))\\ &\geq& d^{2}(x_{n+1},x_{n})-K_{n} \end{eqnarray*} implies \begin{align*} \left[1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})\right]d^{2}(x_{n+1},x^{\ast})\\ \leq \left[\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})\right]d^{2}(x_{n+1},x^{\ast})+K_{n}. \end{align*} Thus, we have \begin{eqnarray*} d^2(x_{n+1},x_{n})&\leq& \frac{\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}d^{2}(x_{n+1},x^{\ast})\\ &+&\frac{K_{n}}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}\\ &=&\frac{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})-1+((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}d^{2}(x_{n+1},x^{\ast})\\ &+&\frac{K_{n}}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}\\ &=&\left[1-\frac{1-(1-\beta)^2+2\theta\beta_{n}\gamma_{n})}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}\right]d^{2}(x_{n+1},x^{\ast})\\ &+&\frac{K_{n}}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}. \end{eqnarray*} Note that $$0<1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})<1$$ implies $$\frac{1-(1-\beta)^2+2\theta\beta_{n}\gamma_{n})}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}\geq 1-((1-\beta)^2+2\theta\beta_{n}\gamma_{n}). $$ Thus, we have \begin{eqnarray*} d^2(x_{n+1},x_{n})&\leq&1-((1-\beta)^2+2\theta\beta_{n}\gamma_{n})d^{2}(x_{n+1},x^{\ast})\\ &+&\frac{K_{n}}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}\\ &=&[((1-\beta)^2+2\theta\beta_{n}\gamma_{n})]d^{2}(x_{n+1},x^{\ast})\\ &+&\frac{K_{n}}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}\\ &=&(1-\beta)^2d^{2}(x_{n+1},x^{\ast})\\ &+&\frac{K_{n}}{1-\frac{1}{2}((1-\beta)^2+2\theta\beta_{n}\gamma_{n})}. \end{eqnarray*} Since \(0< 1-\beta_{n}<1\), this gives \((1-\beta_{n})^2<(1-\beta_{n})\) and