On the viscosity rule for common fixed points of two nonexpansive mappings in Hilbert spaces

1. Introduction

In this paper, we shall take \(H\) as a real Hilbert space, \(\langle, \rangle\) as inner product, \(\|.\|\) as the induced norm, and \(C\) as a nonempty closed subset of \(H\).

Definition 1.1. Let \(T : H \to H\) be a mapping. \(T\) is called non-expansive if $$\|T(x) - T(y)\|\leq\|x-y\|, \quad \forall x,y \in H$$

Definition 1.2. A mapping \(f:H\rightarrow H\) is called a contraction if for all \(x,y\in H\) and \(\theta\in[0,1)\) $$\|f(x)-f(y)\|\leq \theta\|x-y\|.$$

Definition 1.3. \(P_c: H\rightarrow C\) is called a metric projection if for every \(x\in H\) there exists a unique nearest point in \(C\), denoted by \(P_cx\), such that $$\|x-P_cx\|\leq \|x-y\|, \quad \forall\,\, y\in C.$$

In order to verify the weak convergence of an algorithm to a fixed point of a non-expansive mapping we need the demiclosedness principle:

Theorem 1.4. [1] (The demiclosedness principle) Let \(C\) be a nonempty closed convex subset of the real Hilbert space \(H\) 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, \(\rightarrow\) and \(\rightharpoonup\) denotes strong and weak convergence respectively.

Moreover, the following result gives the conditions for the convergence of a nonnegative real sequence.

Theorem 1.5. [2] Assume that \(\{a_n\}\) is a sequence of nonnegative real numbers such that \(a_{n+1}\leq(1-\gamma_n)a_n+\delta_n, \forall n\geq0\), where \(\{\gamma_n\}\) is a sequence in \((0,1)\) and \(\{\delta_n\}\) is a sequence with

1. \(\sum_{n=0}^\infty\gamma_n=\infty\),
2. \( \lim_{n\rightarrow\infty}\sup\frac{\delta_n}{\gamma_n}\leq0\) or \(\sum_{n=0}^{\infty}|\delta_n|< \infty\).

Then \(a_n\rightarrow0\).

The following strong convergence theorem, which is also called the viscosity approximation method, for non-expansive mappings in real Hilbert spaces is given by Moudafi, [3], in (2000).

Theorem 1.6. Let \(C\) be a non-empty closed convex subset of the real Hilbert space \(H\). Let \(T\) be a non-expansive mapping of \(C\) into itself such that \(F(T) = \{x \in C : T(x) = x \}\) is nonempty. Let \(f\) be a contraction of \(C\) into itself. Consider the sequence $$x_{n+1} = \frac{\epsilon_n}{1+\epsilon_n}f(x_n)+\frac{1}{1+\epsilon_n}T(x_n), \quad n\geq 0,$$ where the sequence \(\{\epsilon_n\}\in (0,1)\) satisfies

1. \(\lim_{n\rightarrow\infty} \epsilon_n= 0\),
2. \(\sum_{n=0}^{\infty}\epsilon_n=\infty\), and
3. \(\lim_{n\rightarrow\infty}|\frac{1}{\epsilon_{n+1}}- \frac{1}{\epsilon_n}|=0\).

Then \({x_n}\) converges strongly to a fixed point \(x^\ast\) of the non-expansive mapping \(T\), which is also the unique solution of the variational inequality \begin{equation*} \langle(I-f)x, y-x\rangle \geq 0, \quad \forall\in F(T). \end{equation*}

In (2015), Xu et al. [2] applied viscosity method on the midpoint rule for non-expansive mappings and give the generalized viscosity implicit rule (GVIR): $$x_{n+1}=\alpha_nf(x_n)+(1-\alpha_n)T\left(\frac{x_n+x_{n+1}}{2}\right), \quad \forall n \geq 0$$ This, using contraction, regularizes the implicit midpoint rule for nonexpansive mappings. They also proved that the sequence generated by GMIR converges strongly to a fixed point of \(T\). Ke et al. [4], motivated and inspired by the idea of Xu et al. [2], proposed two generalized viscosity implicit rules: \begin{equation*} x_{n+1}=\alpha_nf(x_n)+(1-\alpha_n)T\left(s_nx_n+(1-s_n)x_{n+1}\right), \end{equation*} \begin{equation*} x_{n+1}=\alpha_nx_n+\beta f(x_n)+\gamma_nT(s_nx_n+(1-s_n)x_{n+1}). \end{equation*} Our contribution in this direction is the following viscosity rule for common fixed points of two nonexpansive mappings in Hilbert spaces: \begin{equation*} x_{n+1} = \alpha_n f(x_n) + \beta_n S\left(\frac{x_{n+1} + x_n}{2}\right) + \gamma_n T\left(\frac{x_{n+1}+x_n}{2}\right) \end{equation*}

2. Main Result

Theorem 2.1. Let \(C\) be a nonempty closed convex subset of the real Hilbert space \(H\). Let \(S:C\rightarrow C\) and \(T:C\rightarrow C\) be two nonexpansive mappings with \(U:=F(T)\cap F(S)\neq \phi\) and \(f:C\rightarrow C\) be a contraction with coefficient \(\theta \in [0,1)\). Let \(\{x_n\}\) be a sequence in \(C\) generated by \begin{equation}\label{eq3.1} x_{n+1}=\alpha_nf(x_n)+\beta_nS\left(\frac{x_{n+1}+x_n}{2}\right)+\gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right) , \end{equation} where \(\{\alpha_n\},\{\beta_n\},\{\gamma_n\}\subset(0,1)\), satisfying the following conditions:

1. \(\alpha_n + \beta_n + \gamma_n = 1\) and \(\lim\limits_{n\rightarrow \infty} \gamma_n = 1\),
2. \(\sum\limits_{n=0}^\infty |\alpha_{n+1}-\alpha_n|< \infty\) and \(\sum\limits_{n=0}^\infty |\beta_{n+1}-\beta_n|< \infty\),
3. \(\sum\limits_{n=0}^\infty\alpha_n=\infty\),
4. \(\lim\limits_{n\rightarrow\infty}||{T\left(\frac{x_{n+1}+x_n}{2}\right)-S\left(\frac{x_{n+1}+x_n}{2}\right)}=0\) and \(\lim\limits_{n\rightarrow \infty}\alpha_n=\lim\limits_{n\rightarrow \infty}\beta_n=0\).

Then \(\{x_n\}\) converges strongly to a common fixed point \(x^\ast\) of the nonexpansive mappings \(T\) and \(S\) which is also the unique solution of the variational inequality $$\langle(I-f)x,y-x\rangle, \quad \forall \quad y\in U$$ In other words, \(x^\ast\) is the unique fixed point of the contraction \(P_Uf\).

Proof. We will prove this theorem into the following five steps:
Step 1. Firstly, we want to show that the sequence \(\{x_n\}\) is bounded. Indeed, take \(p\in U\) arbitrarily, we have \begin{eqnarray*} \|x_{n+1}-p\|&=&\left\|- p + \alpha_nf(x_n)+\beta_nS\left(\frac{x_{n+1}+x_n}{2}\right)+\gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right)\right\|\\ &=&\left\| -(\alpha_n+\beta_n+\gamma_n)p + \alpha_nf(x_n)+\beta_nS\left(\frac{x_{n+1}+x_n}{2}\right) \right. \\ && \left. + \gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right) \right\| \\ &\leq&\alpha_n\|f(x_n)-p\|+\beta_n\left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-p\right\| \\ && + \gamma_n \left\| T\left(\frac{x_{n+1}+x_n}{2}\right) - p \right\| \\ &\leq&\alpha_n\|f(x_n)-f(p)\|+\alpha_n\|f(p)-p\|+\beta_n\left\|\frac{x_{n+1}+x_n}{2}-p\right\| \\ && + \gamma_n\left\|\frac{x_{n+1}+x_n}{2}-p\right\|\\ &\leq&\theta\alpha_n\|x_n-p\|+\alpha_n\|f(p)-p\|+(\beta_n+\gamma_n)\left\|\frac{x_{n+1}+x_n}{2}-p\right\|\\ &=&\theta\alpha_n\|x_n-p\|+\alpha_n\|f(p)-p\|+(1-\alpha_n)\left\|\frac{x_{n+1}+x_n}{2}-p\right\|\\ &\leq&\theta\alpha_n\|x_n-p\|+\alpha_n\|f(p)-p\|+\frac{1-\alpha_n}{2}\|x_{n+1}-p\| \\ && + \frac{1-\alpha_n}{2}\|x_n-p\|. \end{eqnarray*} This is equivalent to \begin{equation*} \left(1-\frac{1-\alpha_n}{2}\right)\|x_{n+1}-p\| \leq \left(\frac{1-\alpha_n}{2}+\alpha_n\theta\right)\|x_n-p\|+\alpha_n\|f(p)-p\| \end{equation*} \(\Rightarrow\) \begin{equation*} (1+\alpha_n)\|x_{n+1}-p\| \leq (1-\alpha_n+2\alpha_n\theta)\|x_n-p\|+2\alpha_n\|f(p)-p\| \end{equation*} \(\Rightarrow\) \begin{eqnarray*} \|x_{n+1}-p\| &\leq& \frac{1+\alpha_n-2\alpha_n+2\alpha_n\theta}{1+\alpha_n}\|x_n-p\| + \frac{2\alpha_n}{1+\alpha_n}\|f(p)-p\| \\ &=&\left(1-\frac{2\alpha_n(1-\theta)}{1+\alpha_n}\right)\|x_n-p\| \\ && + \frac{2\alpha_n(1-\theta)}{1+\alpha_n} \left(\frac{1}{1-\theta} \|f(p) - p\| \right). \end{eqnarray*} Thus, $$\|x_{n+1}-p\|\leq\max\left\{\|x_n-p\|,\frac{1}{1-\theta}\|f(p)-p\|\right\}.$$ Similarly, $$\|x_n-p\|\leq\max\left\{\|x_{n-1}-p\|,\left(\frac{1}{1-\theta}\|f(p)-p\|\right)\right\}.$$ From this, we obtain, \begin{eqnarray*} \|x_{n+1}-p\|&\leq&\max\left\{\|x_n-p\|,\frac{1}{1-\theta}\|f(p)-p\| \right\} \\ &\leq& \max\left\{\|x_{n-1}-p\|, \frac{1}{1-\theta}\|f(p)-p\| \right\} \\ &.&\\ &.&\\ &.&\\ &.&\\ &\leq&\max\left\{\|x_0-p\|, \frac{1}{1-\theta}\|f(p)-p\| \right\}. \end{eqnarray*} Hence, we concluded that \(\{x_n\}\) is a bounded sequence. Consequently, \(\{f(x_n)\}\), \(\big\{S\big(\frac{x_{n+1}+x_n}{2}\big)\big\}\) and \(\big\{T\big(\frac{x_{n+1}+x_n}{2}\big)\big\}\) are bounded.
Step 2. Now, we prove that \(\lim\limits_{n\rightarrow\infty}\|x_{n+1}-x_n\| = 0\). \begin{eqnarray*} &&\|x_{n+1}-x_n\|\\ &=&\left\|\alpha_nf(x_n)+\beta_nS\left(\frac{x_{n+1}+x_n}{2}\right)+\gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right) \right.\\ && - \left. \alpha_{n-1}f(x_{n-1})+\beta_{n-1}S\left(\frac{x_n+x_{n-1}}{2}\right)+\gamma_{n-1}T\left(\frac{x_n+x_{n-1}}{2}\right) \right\|\\ &=& \left\|\alpha_n(f(x_n)-f(x_{n-1}))+(\alpha_n-\alpha_{n-1})f(x_{n-1}) + \beta_n\left(S\left(\frac{x_{n+1}+x_n}{2}\right) \right. \right.\\ && - \left. S\left(\frac{x_n+x_{n-1}}{2}\right)\right) + (\beta_n-\beta_{n-1})S\left(\frac{x_n+x_{n-1}}{2}\right) \\ && + \gamma_n\left(T\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_n+x_{n-1}}{2}\right)\right) \\ && + \left. (\gamma_n-\gamma_{n-1})T\left(\frac{x_n+x_{n-1}}{2}\right)\right\| \\ &=& \left\|\alpha_n(f(x_n)-f(x_{n-1}))+(\alpha_n-\alpha_{n-1})f(x_{n-1}) + \beta_n\left(S\left(\frac{x_{n+1}+x_n}{2}\right) \right. \right. \\ && - \left. S\left(\frac{x_n+x_{n-1}}{2}\right) \right) + (\beta_n-\beta_{n-1})S\left(\frac{x_n+x_{n-1}}{2}\right) \\ && + \gamma_n\left(T\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_n+x_{n-1}}{2}\right)\right)\\ && + \left.(\alpha_n-\alpha_{n-1}+\beta_n-\beta_{n-1})T\left(\frac{x_n+x_{n-1}}{2}\right)\right\| \\ &=& \left\|\alpha_n(f(x_n)-f(x_{n-1}))+(\alpha_n-\alpha_{n-1})\left(f(x_{n-1})-T\left(\frac{x_n+x_{n-1}}{2}\right)\right) \right. \\ && + \beta_n\left(S\left(\frac{x_{n+1}+x_n}{2}\right)-S\left(\frac{x_n+x_{n-1}}{2}\right)\right) \\ && + (\beta_n-\beta_{n-1})\left(S\left(\frac{x_n+x_{n-1}}{2}\right)-T\left(\frac{x_n+x_{n-1}}{2}\right)\right)\\ && + \left. \gamma_n\left(T\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_n+x_{n-1}}{2}\right)\right)\right\| \\ &\leq& \alpha_n \big\|f(x_n)-f(x_{n-1})\big\|+|\alpha_n-\alpha_{n-1}|\left\|f(x_{n-1})-T\left(\frac{x_n+x_{n-1}}{2}\right)\right\| \\ && + \beta_n\left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-S\left(\frac{x_n+x_{n-1}}{2}\right)\right\| \\ && + |\beta_n-\beta_{n-1}| \left\|S\left(\frac{x_n+x_{n-1}}{2}\right)-T\left(\frac{x_n+x_{n-1}}{2}\right)\right\| \\ && + \gamma_n \left\| T\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_n+x_{n-1}}{2}\right)\right\|. \end{eqnarray*} Let \(M_2\) be a number such that \(M_2 \geq \max \left\{\sup\limits_{n\geq0} \left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\|, \right.\) \(\left.\sup\limits_{n\geq0} \left\|f(x_n) - T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \right\}\). Thus, the above is equivalent to \begin{eqnarray*} && \|x_{n+1}-x_n\| \\ &\leq& \alpha_n \theta \left\|x_n - x_{n-1}\right\| + \beta_n \left\|\frac{x_{n+1} + x_n}{2} - \frac{x_n + x_{n-1}}{2}\right\| \\ && + \gamma_n \left\|\frac{x_{n+1} + x_n}{2} - \frac{x_n + x_{n-1}}{2}\right\| + \{|\alpha_n - \alpha_{n-1}| + |\beta_n - \beta_{n-1}|\} M_2 \\ &\leq& \alpha_n \theta \left\|x_n - x_{n-1}\right\| \\ &&+ \frac{\beta_n}{2} \left\|x_{n+1} - x_n \right\|+ \frac{\beta_n}{2} \left\|x_n - x_{n-1}\right\| + \frac{\gamma_n}{2} \left\|x_{n+1} - x_n\right\| \\ && + \frac{\gamma_n}{2} \left\|x_n - x_{n-1}\right\| + \{|\alpha_n - \alpha_{n-1}| + |\beta_n - \beta_{n-1}|\}M_2 \\ &=&\left(\alpha_n\theta+\frac{\beta_n}{2}+\frac{\gamma_n}{2}\right)\|x_n-x_{n-1}\|+\left(\frac{\beta_n}{2}+\frac{\gamma_n}{2}\right)\|x_{n+1}-x_n\| \\ && + (|\alpha_n-\alpha_{n-1}|+|\beta_n-\beta_{n-1}|)M_2\\ &=&\left(\alpha_n\theta+\frac{1-\alpha_n}{2}\right)\|x_n-x_{n-1}\|+\frac{1-\alpha_n}{2}\|x_{n+1}-x_n\| \\ && + (|\alpha_n-\alpha_{n-1}|+|\beta_n-\beta_{n-1}|)M_2. \end{eqnarray*} Combining the common terms from left and right hand sides, we get, \begin{eqnarray*} \left(1-\frac{1-\alpha_n}{2}\right)\|x_{n+1}-x_n\| &\leq& \left(\alpha_n\theta+\frac{1-\alpha_n}{2}\right)\|x_n-x_{n-1}\| \\ && + (|\alpha_n-\alpha_{n-1}|+|\beta_n-\beta_{n-1}|)M_2. \end{eqnarray*} This implies that \begin{eqnarray*} &&\|x_{n+1}-x_n\| \\ &\leq& \frac{1+\alpha_n-2\alpha_n+2\alpha_n\theta}{1+\alpha_n}\|x_n-x_{n-1}\|\\&& + \{\frac{2(|\alpha_n-\alpha_{n-1}|+|\beta_n-\beta_{n-1}|)}{1+\alpha_n}\} M_2 \\ &=&\left(1-\frac{2\alpha_n(1-\theta)}{1+\alpha_n}\right)\|x_n-x_{n-1}\|\\&& + \{\frac{2(|\alpha_n-\alpha_{n-1}|+|\beta_n-\beta_{n-1}|)}{1+\alpha_n}\} M_2. \end{eqnarray*} Note that \(\sum\limits_{n=0}^\infty \alpha_n = \infty\), \(\sum\limits_{n=0}^\infty |\alpha_{n+1}-\alpha_n| < \infty\) and \(\sum\limits_{n=0}^\infty|\beta_{n+1}-\beta_n| < \infty \). Using Theorem 1.5, we have \(\|x_{n+1}-x_n\|\rightarrow0\) as \(n\rightarrow\infty\).
Step 3. Now, we will show that \(\lim\limits_{n\rightarrow\infty} \|x_n-Sx_n\|=0\) and \(\lim\limits_{n\rightarrow\infty} \|x_n-Tx_n\|=0\). Consider \begin{eqnarray*} && \|x_n-S(x_n)\| \\ &=& \left\|x_n-x_{n+1}+x_{n+1}-T\left(\frac{x_{n+1}+x_n}{2}\right)+T\left(\frac{x_{n+1}+x_n}{2}\right) \right. \\ && - \left. S \left(\frac{x_{n+1}+x_n}{2}\right)+S\left(\frac{x_{n+1}+x_n}{2}\right)-S(x_n) \right\| \\ &\leq& \|x_n-x_{n+1}\|+ \left\|x_{n+1} - T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| + \left\| T\left(\frac{x_{n+1} + x_n}{2}\right) \right. \\ && \left. - S\left(\frac{x_{n+1}+x_n}{2}\right) \right\| + \left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-S(x_n)\right\| \\ &\leq& \|x_n-x_{n+1}\| + \left\| \alpha_n f(x_n) + \beta_n S\left(\frac{x_{n+1} + x_n}{2} \right) + \gamma_n T\left(\frac{x_{n+1} + x_n}{2} \right) \right. \\ && \left. - T\left(\frac{x_{n+1} + x_n}{2}\right) \right\| + \left\|\frac{x_{n+1}+x_n}{2} - x_n\right\| \\ && + \left\|T\left(\frac{x_{n+1}+x_n}{2}\right)-S\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \\ &\leq& \|x_n-x_{n+1}\| + \alpha_n \left\| f(x_n) - T\left(\frac{x_{n+1}+x_n}{2}\right) \right\| \\ && + \beta_n \left\| S\left(\frac{x_{n+1}+x_n}{2}\right) - T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| + \frac{1}{2}\|x_{n+1}-x_n\|\\ && + \left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \\ &\leq& \frac{3}{2}\|x_n-x_{n+1}\| + \alpha_n \left\|f(x_n) - T\left(\frac{x_{n+1} + x_n}{2}\right)\right\| \\ && + (1+\beta_n)\left\|S\left(\frac{x_{n+1}+x_n}{2}\right) - T\left(\frac{x_{n+1}+x_n}{2}\right)\right\|. \end{eqnarray*} Since, \(\lim\limits_{n\rightarrow\infty} \alpha_n = \lim\limits_{n\rightarrow\infty} \left\|T\left(\frac{x_{n+1} + x_n}{2}\right) - S\left(\frac{x_{n+1} + x_n}{2}\right)\right\| = 0\) and \(\lim\limits_{n\rightarrow\infty} \|x_{n+1} - x_n\| \rightarrow 0\), we get \(\|x_n-S(x_n)\|\rightarrow0\) as \(n\rightarrow\infty\) Moreover, we have \begin{eqnarray*} \left\|S \left(\frac{x_{n+1} + x_n}{2}\right) - x_n\right\| &=& \left\|S\left(\frac{x_{n+1} + x_n}{2} \right) - S(X_n) + S(x_n) - x_n\right\| \\ &\leq& \left\|S \left(\frac{x_{n+1} + x_n}{2}\right) - S(x_n)\right\| + \left\|S(x_n) - x_n\right\| \\ &\leq& \left\|\frac{x_{n+1} + x_n}{2} - x_n\right\| + \|S(x_n)-x_n\| \\ &=& \frac{1}{2}\|x_{n+1} - x_n\| + \|S(x_n) - x_n\| \\ &\to& 0, \quad \quad \text{ as } (n \to \infty). \end{eqnarray*} Now, consider \begin{eqnarray*} &&\|x_n-T(x_n)\|\\ &=& \left\|x_n-x_{n+1}+x_{n+1}-S\left(\frac{x_{n+1}+x_n}{2}\right)+S\left(\frac{x_{n+1}+x_n}{2}\right) \right.\\ && - \left. T\left(\frac{x_{n+1}+x_n}{2}\right)+T\left(\frac{x_{n+1}+x_n}{2}\right)-T(x_n) \right\| \\ &\leq& \|x_n-x_{n+1}\| + \left\|x_{n+1} - S\left(\frac{x_{n+1}+x_n}{2}\right)\right\| + \left\|T\left(\frac{x_{n+1}+x_n}{2}\right)-T(x_n)\right\| \\ && + \left\|S\left(\frac{x_{n+1}+x_n}{2}\right) - T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \\ &\leq& \|x_n-x_{n+1}\| + \left\|T\left(\frac{x_{n+1}+x_n}{2}\right) - S\left(\frac{x_{n+1}+x_n}{2}\right)\right\| + \\ && \left\|\alpha_nf(x_n) + \beta_nS\left(\frac{x_{n+1}+x_n}{2}\right) + \gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right) - S\left(\frac{x_{n+1}+x_n}{2}\right) \right\| \\ && + \left\|\frac{x_{n+1}+x_n}{2}-x_n\right\| \\ &\leq& \left\|x_n-x_{n+1}\right\| + \alpha_n\left\|f(x_n)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| +\frac{1}{2}\|x_{n+1}-x_n\| \\ && + \gamma_n\left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \\ && + \left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \\ &\leq& \frac{3}{2}\|x_n-x_{n+1}\| + \alpha_n \left\|f(x_n)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| \\ && + (1+\gamma_n)\left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-T\left(\frac{x_{n+1}+x_n}{2}\right)\right\| . \end{eqnarray*} Since, \(\lim\limits_{n \to \infty} \alpha_n = \lim\limits_{n \to \infty} \left\|T\left(\frac{x_{n+1} + x_n}{2}\right) - S\left(\frac{x_{n+1} + x_n}{2}\right)\right\| = 0\) and \(\lim\limits_{n \to \infty} \|x_{n+1}-x_n\| \to 0\), we get \(\|x_n-Tx_n\| \to 0\) as \(n \to \infty\). Also, \begin{eqnarray*} \left\|T\left(\frac{x_{n+1}+x_n}{2}\right)-x_n\right\| &=& \left\|T\left(\frac{x_{n+1}+x_n}{2}\right)-T(X_n)+T(x_n)-x_n\right\| \\ &\leq& \left\|T \left(\frac{x_{n+1} + x_n}{2}\right) - T(x_n)\right\| + \left\|T(x_n) - x_n\right\| \\ &\leq& \left\|\frac{x_{n+1}+x_n}{2}-x_n\right\| + \left\|T(x_n) - x_n\right\| \\ &=& \frac{1}{2}\left\|x_{n+1} - x_n\right\| + \left\|T(x_n) - x_n\right\| \\ &\to& 0 \quad \quad \quad (\text{ as } n \to \infty) \end{eqnarray*} Step 4. In this step, we will show that \(\limsup\limits_{n \to \infty} \langle x^\ast - f(x^\ast), x^\ast - x_n \rangle \leq 0\), where, \(x^\ast = P_Uf(x^\ast)\).
Indeed, we take a subsequence, \(\{x_{n_i}\}\) of \(\{x_n\}\), which converges weakly to a fixed point \(p \in U = F(T) \cap F(S)\). Without loss of generality, we may assume that \(\{x_{n_i}\} \rightharpoonup p\). From \(\lim\limits_{n \to \infty} ||{x_n - S(x_n)} = 0\), \(\lim\limits_{n \to \infty} ||{x_n - T(x_n)} = 0\) and Theorem 1.4, we have \(p = S(p)\) and \(p = T(p)\). This together with the property of the metric projection implies that \begin{equation*} \limsup\limits_{n \to \infty} \langle x^\ast - f(x^\ast), x^\ast - x_n \rangle = \langle x^\ast - f(x^\ast), x^\ast - p \rangle \leq 0. \end{equation*} Step 5. Finally, we show that \(x_n \rightarrow x^\ast\) as \(n\rightarrow\infty\). Again, take \)x^\ast \in U\) to be the unique fixed point of the contraction \(P_Uf\). Consider \begin{eqnarray*} &&\|x_{n+1}-x_n\|^2 \nonumber \\ &=& \left\|\alpha_nf(x_n)+\beta_nS\left(\frac{x_{n+1}+x_n}{2}\right)+\gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast\right\| ^2\nonumber\\ &=& \left\|\alpha_nf(x_n)+\beta_nS\left(\frac{x_{n+1}+x_n}{2}\right)+\gamma_nT\left(\frac{x_{n+1}+x_n}{2}\right)-(\alpha_n+\beta_n+\gamma_n)x^\ast\right\|^2\nonumber\\ &=& \left\|\alpha_n(f(x_n)-x^\ast)+\beta_n \{S\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast\} + \gamma_n \{T\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast\}\right\|^2 \nonumber\\ &=& \alpha^2_n \left\|f(x_n) - x^\ast\right\|^2 + \beta^2_n \left\|S\left(\frac{x_{n+1}+x_n}{2}\right) - x^\ast\right\|^2 \\ &&+ \gamma^2_n \left\|T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast\right\|^2\nonumber\\ && + 2\alpha_n \beta_n \left\langle f(x_n) - x^\ast, S\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast \right\rangle \nonumber \\ && + 2\alpha_n \gamma_n \left\langle f(x_n)-x^\ast,T\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast \right\rangle \nonumber \\ && + 2\beta_n \gamma_n \left\langle S\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast, T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \nonumber \\ &\leq& \alpha^2_n \|f(x_n) - x^\ast\|^2 + \beta^2_n \left\|\frac{x_{n+1} + x_n}{2} - x^\ast\right\|^2 + \gamma^2_n \left\|\frac{x_{n+1} + x_n}{2} - x^\ast\right\|^2 \nonumber \\ && + 2\alpha_n \beta_n \left\langle f(x_n) - f(x^\ast), S\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \nonumber \\ && + 2\alpha_n \beta_n \left\langle f(x^\ast) - x^\ast, S\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \nonumber \\ && + 2 \alpha_n \gamma_n \left\langle f(x_n) - f(x^\ast), T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \nonumber \\ && + 2 \alpha_n \gamma_n \left\langle f(x^\ast) - x^\ast, T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \nonumber \\ && + 2\beta_n \gamma_n \left\langle S\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast, T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \nonumber \\ &\leq& \{\beta^2_n + \gamma^2_n\} \left\|\frac{x_{n+1} + x_n}{2} - x^\ast\right\|^2 + 2\alpha_n\beta_n\| f(x_n) - f(x^\ast)\|.\nonumber\\ &&\left\|S\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast\right\|\nonumber\\ && + 2\alpha_n \gamma_n \|f(x_n) - f(x^\ast)\|.\left\|T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast\right\| \nonumber \\ && + 2\beta_n \gamma_n \left\|S\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast\right\|.\left\|T\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast\right\| + K_n, \nonumber \end{eqnarray*} where, \begin{eqnarray*} K_n &=& \alpha^2_n \|f(x_n)-x^\ast\|^2 + 2\alpha_n\beta_n \left\langle f(x^\ast) - x^\ast, S\left(\frac{x_{n+1} + x_n}{2}\right) - x^\ast \right\rangle \\ && + 2\alpha_n\gamma_n \left\langle f(x^\ast)-x^\ast,T\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast \right\rangle \end{eqnarray*} This implies that \begin{eqnarray*} &&\|x_{n+1}-x_n\|^2 \nonumber \\ &\leq& (\beta^2_n + \gamma^2_n) \left\|\frac{x_{n+1} + x_n}{2} - x^\ast\right\|^2 + 2\alpha_n\beta_n \theta \left\|x_n - x^\ast\right\|.\left\|\frac{x_{n+1} + x_n}{2} - x^\ast\right\| \nonumber \\ && + 2\alpha_n\gamma_n\theta\left\|x_n - x^\ast\right\|.\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\| \nonumber \\ && + 2\beta_n\gamma_n\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\|.\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\| + K_n \nonumber \\ &=&(\beta^2_n+\gamma^2_n+2\beta_n\gamma_n)\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\|^2 \\ && + 2 \alpha_n\theta(\beta_n+\gamma_n)\left\|x_n-x^\ast\right\|.\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\| + K_n \nonumber\\ &=&(\beta_n+\gamma_n)^2 \left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\|^2 \\ && + 2\alpha_n\theta(\beta_n+\gamma_n)\left\|x_n-x^\ast\right\|.\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\| + K_n\nonumber\\ &=& (1-\alpha_n)^2\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\|^2 \\ && + 2\alpha_n\theta(1-\alpha_n)\left\|x_n-x^\ast\right\|.\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\|+K_n. \end{eqnarray*} The above calculation shows that \begin{eqnarray*} 0 &\leq& 2\alpha_n\theta(1-\alpha_n)\left\|x_n-x^\ast\right\|.\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\| \\ && + (1-\alpha_n)^2\left\|\frac{x_{n+1}+x_n}{2}-x^\ast\right\|^2 - \left\|x_{n+1}-x^\ast\right\|^2 + K_n, \end{eqnarray*} which is a quadratic inequality in \(||{\frac{x_{n+1}+x_n}{2}-x^\ast}\). Solving the above inequality for \(||{\frac{x_{n+1}+x_n}{2}-x^\ast}\), we have, \begin{eqnarray*} ||{\frac{x_{n+1}+x_n}{2}-x^\ast} &\geq& \frac{-2\theta\alpha_n(1-\alpha_n)\|x_n-x^\ast\|}{2(1-\alpha_n)^2} \end{eqnarray*} \begin{eqnarray*} && + \frac{\sqrt{4\theta^2\alpha^2_n(1-\alpha_n)^2\|x_n-x^\ast\|^2-4(1-\alpha_n)^2(K_n-\|x_{n+1}-x^\ast\|^2 )}}{2(1-\alpha_n)^2} \\ &=& \frac{-\theta\alpha_n\|x_n-x^\ast\|+\sqrt{\theta^2\alpha^2_n\|x_n-x^\ast\|^2-K_n+\|x_{n+1}-x^\ast\|^2 }}{1-\alpha_n}. \end{eqnarray*} This will give \begin{eqnarray*} && \frac{1}{2}\left(\|x_{n+1}-x^\ast\|+\|x_n-x^\ast\|\right) \\ &\geq& \frac{-\theta\alpha_n\|x_n-x^\ast\|+\sqrt{\theta^2\alpha^2_n\|x_n-x^\ast\|^2-K_n+\|x_{n+1}-x^\ast\|^2 }}{1-\alpha_n} \end{eqnarray*} \(\Rightarrow\) \begin{eqnarray*} && \frac{1}{2}\left((1-\alpha_n)||{x_{n+1}-x^\ast} + (1 + (2\theta - 1) \alpha_n) ||{x_n-x^\ast} \right) \\ &\geq&\sqrt{\theta^2\alpha^2_n\|x_n-x^\ast\|^2-K_n+\|x_{n+1}-x^\ast\|^2} \end{eqnarray*} \(\Rightarrow\) \begin{eqnarray*} && \frac{1}{4}\left((1-\alpha_n)\|x_{n+1}-x^\ast\|+(1+(2\theta-1)\alpha_n)\|x_n-x^\ast\|\right)^2 \\ &\geq&\theta^2\alpha^2_n\|x_n-x^\ast\|^2-K_n+\|x_{n+1}-x^\ast\|^2, \end{eqnarray*} which is reduced to \begin{eqnarray*} &&\frac{1}{4}(1-\alpha_n)^2\|x_{n+1}-x^\ast\|^2+\frac{1}{4}(1+(2\theta-1)\alpha_n)^2\|x_n-x^\ast\|^2\\ &+&\frac{1}{2}(1-\alpha_n)(1+(2\theta-1)\alpha_n)\|x_{n+1}-x^\ast\|\|x_n-x^\ast\|\\ &\geq&\theta^2\alpha^2_n\|x_n-x^\ast\|^2-K_n+\|x_{n+1}-x^\ast\|^2. \end{eqnarray*} This inequality is further reduced by using the elementary inequality \begin{eqnarray*} 2\|x_{n+1}-x^\ast\|\|x_n-x^\ast\|&\leq&\|x_{n+1}-x^\ast\|^2+\|x_n-x^\ast\|^2, \end{eqnarray*} to the following inequality \begin{eqnarray*} &&\frac{1}{4}(1-\alpha_n)^2\|x_{n+1}-x^\ast\|^2+\frac{1}{4}(1+(2\theta-1)\alpha_n)^2\|x_n-x^\ast\|^2\\ &+&\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)(\|x_{n+1}-x^\ast\|^2+\|x_n-x^\ast\|^2)\\ &\geq&\theta^2\alpha^2_n\|x_n-x^\ast\|^2-K_n+\|x_{n+1}-x^\ast\|^2. \end{eqnarray*} This implies that \begin{eqnarray*} &&\left(1-\frac{1}{4}(1-\alpha_n)^2-\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)\right)\|x_{n+1}-x^\ast\|^2\\ &\leq&\left(\frac{1}{4}(1+(2\theta-1)\alpha_n)^2+\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)-\theta^2\alpha^2_n\right)\|x_n-x^\ast\|^2 \\ && + K_n, \end{eqnarray*} or \begin{eqnarray} \|x_{n+1}-x^\ast\|^2 &\leq& \frac{\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)-\theta^2\alpha^2_n}{1-\frac{1}{4}(1-\alpha_n)^2-\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)}\|x_n-x^\ast\|^2 \nonumber\\ && + \frac{\frac{1}{4}(1+(2\theta-1)\alpha_n)^2}{1-\frac{1}{4}(1-\alpha_n)^2-\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)} + K'_n,\label{a} \end{eqnarray} where, \begin{eqnarray*} K'_n &=& \frac{K_n}{1-\frac{1}{4}(1-\alpha_n)^2-\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)}. \end{eqnarray*} Note that, \begin{eqnarray*} && 1-\frac{1}{4}(1-\alpha_n)^2-\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n) \\ &=&1-\frac{1}{4}(1-\alpha_n)(1-\alpha_n+1+(2\theta-1)\alpha_n)\\ &=&1-\frac{1}{4}(1-\alpha_n)(1-\alpha_n+1+2\theta\alpha_n-\alpha_n)\\ &=&1-\frac{1}{4}(1-\alpha_n)(2-2\alpha_n+2\theta\alpha_n)\\ &=&1-\frac{1}{2}(1-\alpha_n)(1-\alpha_n+\theta\alpha_n)\\ &=&1-\frac{1}{2}(1-\alpha_n)(1-\alpha_n(1-\theta)), \end{eqnarray*} and \begin{eqnarray*} &&\frac{1}{4}(1+(2\theta-1)\alpha_n)^2+\frac{1}{4}(1-\alpha_n)(1+(2\theta-1)\alpha_n)-\theta^2\alpha^2_n\\ &=&\frac{1}{4}(1+(2\theta-1)\alpha_n)(1+(2\theta-1)\alpha_n+1-\alpha_n)-\theta^2\alpha^2_n\\ &=&\frac{1}{4}(1+(2\theta-1)\alpha_n)(2+2\theta\alpha_n-2\alpha_n)-\theta^2\alpha^2_n\\ &=&\frac{1}{2}(1+(2\theta-1)\alpha_n)(1+\theta\alpha_n-\alpha_n)-\theta^2\alpha^2_n\\ &=&\frac{1}{2}(1+(2\theta-1)\alpha_n)(1-(1-\theta)\alpha_n)-\theta^2\alpha^2_n. \end{eqnarray*} Now from (2), \begin{eqnarray} && \|x_{n+1}-x^\ast\|^2 \nonumber\\ &\leq& \frac{\frac{1}{2}(1 + (2\theta - 1)\alpha_n)(1 - (1 - \theta)\alpha_n) - \theta^2 \alpha^2_n}{1 - \frac{1}{2}(1 - \alpha_n)(1 - \alpha_n(1 - \theta))}\|x_n - x^\ast\|^2 + K'_n.\label{b} \end{eqnarray} Consider the following function, for \(t > 0\). $$g(t):=\frac{1}{t}\left\{1-\frac{\frac{1}{2}(1+(2\theta-1)t)(1-(1-\theta)t)-\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}$$ \begin{eqnarray*} g(t)&=&\frac{1}{t}\left\{\frac{1-\frac{1}{2}(1-t)(1-t(1-\theta))-\frac{1}{2}(1+(2\theta-1)t)(1-(1-\theta)t)+\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}\\ &=&\frac{1}{t}\left\{\frac{1-\frac{1}{2}(1-t(1-\theta))(1-t+1+2\theta t-t)+\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}\\ &=&\frac{1}{t}\left\{\frac{1-\frac{1}{2}(1-t(1-\theta))(2-2t+2\theta t)+\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}\\ &=&\frac{1}{t}\left\{\frac{1-(1-t+\theta t))(1-t+\theta t ))+\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}\\ &=&\frac{1}{t}\left\{\frac{1-(1+t^2+\theta^2 t^2-2t-2\theta t^2+2\theta t)+\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}\\ &=&\frac{1}{t}\left\{\frac{1-1-t^2-\theta^2 t^2+2t+2\theta t^2-2\theta t+\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\}\\ &=&\frac{-t+2+2\theta t-2\theta }{1-\frac{1}{2}(1-t)(1-t(1-\theta))}. \end{eqnarray*} By applying limit \(t\rightarrow0\), we have \begin{eqnarray*} \lim_{t\rightarrow0}g(t)=4(1-\theta)>0. \end{eqnarray*} Let \(\delta>0\) be such that for all \(0< t < \delta \), \(g(t) > \epsilon := 4(1-\theta) > 0\). This is equivalent to \begin{eqnarray*} \frac{1}{t}\left\{1-\frac{\frac{1}{2}(1+(2\theta-1)t)(1-(1-\theta)t)-\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}\right\} &>& \epsilon \end{eqnarray*} This implies, \begin{eqnarray*} 1-t\epsilon>\frac{\frac{1}{2}(1+(2\theta-1)t)(1-(1-\theta)t)-\theta^2t^2}{1-\frac{1}{2}(1-t)(1-t(1-\theta))}. \end{eqnarray*} Since \(\alpha_n\rightarrow0\) as \(n\rightarrow\infty\), there exist some integer \(N\), such that \(\alpha_n< \delta\), \(\forall\) \(n\geq N\). From (3), we have \begin{equation*} \|x_{n+1}-x^\ast\|^2\leq(1-\alpha_n\epsilon)\|x_n-x^\ast\|^2+K'_n \end{equation*} On the other hand, we have \begin{eqnarray*} \limsup\limits_{n \to \infty} \frac{K_n}{\alpha_n} &=& \limsup\limits_{n \to \infty} \left\{ 2\beta_n \left\langle f(x^\ast)-x^\ast,S\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast \right\rangle \right. \\ && \hspace{11.5mm} + 2\gamma_n \left\langle f(x^\ast)-x^\ast,T\left(\frac{x_{n+1}+x_n}{2}\right)-x^\ast \right\rangle \\ && \left. \hspace{11.5mm} + \alpha_n\|f(x_n)-x^\ast\|^2\right\}\leq 0. \end{eqnarray*} The above inequality implies that \begin{eqnarray*} \limsup\limits_{n\rightarrow\infty}\frac{K'_n}{\alpha_n} &\leq& 0. \end{eqnarray*} From the above two inequalities and Theorem 1.4 we have $$\lim_{n\rightarrow\infty}\|x_{n+1}-x^\ast\|^2=0,$$ which implies that \(x_n\rightarrow x^\ast\) as \(n\rightarrow\infty\). This completes the proof.

Competing Interests

The author(s) do not have any competing interests in the manuscript.