In this section, we state and prove our convergence result.
Theorem 4.1
Let K be a nonempty, closed, and convex subset of a Hadamard manifold M. Let \(A: K \to TM\) be a monotone vector field and \(F: K \times K \to \mathbb{R}\) such that \(F(x,x)=0\) for all \(x \in K\) be a bifunction satisfying conditions (A1)–(A3). Let \(f: M \to M\) be a ψcontraction and \(S: K \to K\) be a nonexpansive mapping. Assume \(\mathrm{Fix}(S) \cap \mathrm{GEP}(F,A) \neq \emptyset \). For arbitrary \(x_{1} \in K\), sequences \(\{r_{n}\} \in (0,\infty )\), \(\beta _{n}, \alpha _{n} \in (0,1)\), let the sequence \(\{x_{n}\}\) be defined iteratively by
$$\begin{aligned} \textstyle\begin{cases} y_{n}=\exp _{x_{n}}(1\beta _{n})\exp _{x_{n}}^{1}Sx_{n}, \\ x_{n+1}=\exp _{f(x_{n})}(1\alpha _{n})\exp _{f(x_{n})}^{1}T_{r_{n}}^{F,A}y_{n}. \end{cases}\displaystyle \end{aligned}$$
(4.1)
Suppose the following conditions hold:

(i)
\(\lim_{n \to \infty}\alpha _{n}=0\) and \(\sum_{n =1}^{\infty}\alpha _{n}=0\);

(ii)
\(0< a\leq \beta _{n} \leq b < 1\) for some \(a,b > 0\) for all \(n \geq 1\);

(iii)
\(0 < r \leq r_{n}\).
If \(0 < \kappa =\sup \{\frac{\psi (d(x_{n},p))}{d(x_{n},p)}: x_{n} \ne p, n \ge 0\}< 1\) for all \(p \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)\), then the sequence \(\{x_{n}\}\) converges to a point \(p \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)\)
Proof
Let \(p \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)\). We can rewrite \(y_{n}\) as \(y_{n}=\gamma _{n}^{1}(1\beta _{n})\), where \(\gamma _{n}^{1}: [0,1] \to M\) is a sequence of geodesics joining \(x_{n}\) to \(Sx_{n}\). Then by the nonexpansivity of S, we have
$$\begin{aligned} d(y_{n},p) &= d\bigl(\gamma _{n}^{1}(1 \beta _{n}),p\bigr) \\ &\leq \beta _{n} d\bigl(\gamma _{n}^{1}(0),p \bigr)+(1\beta _{n})d\bigl(\gamma _{n}^{1}(1),p \bigr) \\ &= \beta _{n} d(x_{n},p)+(1\beta _{n})d(Sx_{n},p) \\ &\leq \beta _{n} d(x_{n},p)+(1\beta _{n})d(x_{n},p) \\ &=d(x_{n},p). \end{aligned}$$
(4.2)
Note also that \(x_{n+1}\) can be written in the form \(x_{n+1}=\gamma _{n}^{2}(1\alpha _{n})\), where \(\gamma _{n}^{2}: [0,1] \to M\) is a sequence of geodesics joining \(f(x_{n})\) to \(T_{r_{n}}^{F,A}y_{n}\), i.e \(\gamma _{n}^{2}(0)=f(x_{n})\) and \(\gamma _{n}^{2}(1)=T_{r_{n}}^{F,A}y_{n}\). By the convexity of Riemannian distance and the fact that \(0 < \kappa =\sup \{\frac{\psi (d(x_{n},p))}{d(x_{n},p)}: x_{n} \ne p, n \ge 0\}< 1\), we obtain
$$\begin{aligned} d(x_{n+1},p)&=d\bigl(\gamma _{n}^{2}(1\alpha _{n}),p\bigr) \\ &\leq \alpha _{n}d\bigl(\gamma _{n}^{2}(0),p \bigr)+(1\alpha _{n})d\bigl(\gamma _{n}^{2}(1),p \bigr) \\ &= \alpha _{n} d\bigl(f(x_{n}),p\bigr)+(1\alpha _{n})d\bigl(T_{r_{n}}^{F,A}y_{n},p\bigr) \\ &\leq \alpha _{n} \bigl[d\bigl(f(x_{n}),f(p)\bigr)+d \bigl(f(p),p\bigr)\bigr]+(1\alpha _{n})d\bigl(T_{r_{n}}^{F,A}y_{n},T_{r_{n}}^{F,A}p \bigr) \\ &\leq \alpha _{n}\bigl[\psi \bigl( d(x_{n},p)\bigr)+d \bigl(f(p),p\bigr)\bigr]+(1\alpha _{n})d(y_{n},p) \\ &\leq \alpha _{n}\bigl[\kappa d(x_{n},p)+d\bigl(f(p),p \bigr)\bigr]+(1\alpha _{n})d(x_{n},p) \\ &\leq \bigl[1\alpha _{n}(1\kappa )\bigr]d(x_{n},p)+ \alpha _{n} d\bigl(f(p),p\bigr) \\ &\leq \max \biggl\{ d(x_{n},p),\frac {1}{(1\kappa )}d\bigl(f(p),p\bigr) \biggr\} \\ &\vdots \\ &\leq \max \biggl\{ d(x_{1},p),\frac {1}{(1\kappa )}d\bigl(f(p),p\bigr) \biggr\} . \end{aligned}$$
It implies that the sequence \(\{x_{n}\}\) is bounded. It follows from (4.2) that \(\{y_{n}\}\) is bounded, and thus \(\{Sx_{n}\}\) and \(T_{r_{n}}^{F,A}y_{n}\) are bounded.
Fix \(n \geq 1\), let \(u_{n}=f(x_{n})\), \(v_{n}=T_{r_{n}}^{F,A}y_{n}\), \(w=f(p)\), \(v=Sp\), \(p_{n}=x_{n}\) and \(q_{n}=Sx_{n}\). Consider the geodesic triangles \(\Delta (u_{n},v_{n},p)\), \(\Delta (w,v_{n},p)\), \(\Delta (u_{n},v_{n},w)\) and \(\Delta (p_{n},q_{n},p)\). Then by Lemma 2.5, there exist comparison triangles \(\Delta (u_{n}^{\prime},v_{n}^{\prime},p)\), \(\Delta (w^{\prime},v_{n}^{\prime},p^{\prime})\), \(\Delta (u_{n}^{\prime},v_{n}^{\prime},w^{\prime})\) and \(\Delta (p_{n}^{\prime},q_{n}^{\prime},p^{\prime})\) such that
$$\begin{aligned} &d(p_{n},q_{n})= \bigl\Vert p_{n}^{\prime}q_{n}^{\prime} \bigr\Vert , d(p_{n},p)= \bigl\Vert p_{n}^{ \prime}p^{\prime} \bigr\Vert \quad\text{and}\quad d(q_{n},p)= \bigl\Vert q_{n}^{\prime}p^{ \prime} \bigr\Vert ,\\ &d(u_{n},v_{n})= \bigl\Vert u_{n}^{\prime}v_{n}^{\prime} \bigr\Vert , d(u_{n},p)= \bigl\Vert u_{n}^{ \prime}p^{\prime} \bigr\Vert \quad \text{and}\quad d(v_{n},p)= \bigl\Vert v_{n}^{\prime}p^{ \prime} \bigr\Vert \end{aligned}$$
and
$$\begin{aligned} d(w,p)= \bigl\Vert w^{\prime}p^{\prime} \bigr\Vert , d(u_{n},w)= \bigl\Vert u_{n}^{\prime}w^{ \prime} \bigr\Vert \quad\text{and}\quad d(q_{n},v)= \bigl\Vert q_{n}^{\prime}v^{\prime} \bigr\Vert . \end{aligned}$$
Let θ and \(\theta ^{\prime}\) be the angles at p and \(p^{\prime}\) in the triangles \(\Delta (w,x_{n+1},p)\) and \(\Delta (w^{\prime},x_{n+1}^{\prime},p^{\prime})\), respectively. Hence \(\theta \leq \theta ^{\prime}\) and \(\cos \theta ^{\prime} \leq \cos \theta \). Let \(y_{n}^{\prime}\) and \(x_{n+1}^{\prime}\) be the comparison point of \(y_{n}\) and \(x_{n+1}\), respectively, then
$$\begin{aligned} y_{n}^{\prime}=\beta _{n} p_{n}^{\prime} +(1\beta _{n})q_{n}^{\prime} \quad\text{and}\quad x_{n+1}^{\prime}=\alpha _{n}u_{n}^{\prime}+(1 \alpha _{n})v_{n}^{ \prime}. \end{aligned}$$
Then by Lemma 2.10, we have
$$\begin{aligned} d^{2}(x_{n+1},p) &\leq \bigl\Vert x_{n+1}^{\prime}p^{\prime} \bigr\Vert ^{2} \\ &= \bigl\Vert \alpha _{n}u_{n}^{\prime}+(1\alpha _{n})v_{n}^{\prime}p^{\prime} \bigr\Vert ^{2} \\ &\leq \bigl\Vert \alpha _{n}\bigl(u_{n}^{\prime}w^{\prime} \bigr)+(1\alpha _{n}) \bigl(v_{n}^{ \prime}p^{\prime} \bigr) \bigr\Vert +2\alpha _{n} \bigl\langle x_{n+1}^{\prime}p^{\prime},w^{ \prime}p^{\prime} \bigr\rangle \\ &\leq (1\alpha _{n}) \bigl\Vert v_{n}^{\prime}p^{\prime} \bigr\Vert ^{2} +\alpha _{n} \bigl\Vert u_{n}^{ \prime}w^{\prime} \bigr\Vert ^{2} +2 \beta _{n} \bigl\Vert x_{n+1}^{\prime}p^{\prime} \bigr\Vert \bigl\Vert w^{\prime}p^{\prime} \bigr\Vert \cos \theta ^{\prime} \\ &\leq (1\alpha _{n})d(v_{n},p)+\alpha _{n} d(u_{n},w)+2\alpha _{n}d(x_{n+1},p)d(w,p) \cos \theta \\ &=(1\alpha _{n})d\bigl(T_{r_{n}}^{F,A}y_{n},p \bigr)+\alpha _{n}d\bigl(f(x_{n}),f(p)\bigr)+2 \alpha _{n} d(x_{n+1},p)d\bigl(f(p),p\bigr) \cos \theta \\ &\le (1\alpha _{n})d\bigl(T_{r_{n}}^{F,A}y_{n},p \bigr)+\alpha _{n}\psi \bigl(d(x_{n},p)\bigr)+2 \alpha _{n} d(x_{n+1},p)d\bigl(f(p),p\bigr) \cos \theta . \end{aligned}$$
Using Lemma 3.3, \(0 < \kappa =\sup \{\frac{\psi (d(x_{n},p))}{d(x_{n},p)}: x_{n} \ne p, n \ge 0\}< 1\) and the fact that \(\langle \exp _{p}^{1}x_{n+1}, \exp _{p}^{1}f(p) \rangle =d(x_{n+1},p)d(f(p),p) \cos \theta \), we obtain
$$\begin{aligned} &d^{2}(x_{n+1},p) \\ &\quad \leq (1\alpha _{n})d^{2} \bigl(T_{r_{n}}^{F,A}y_{n},p\bigr)+ \kappa d(x_{n},p)+2\alpha _{n} \bigl\langle \exp _{p}^{1}x_{n+1}, \exp _{p}^{1}f(p) \bigr\rangle \\ &\quad\leq (1\alpha _{n})d^{2}(y_{n},p)(1\alpha _{n})d^{2}\bigl(y_{n},T_{r_{n}}^{F,A}y_{n} \bigr)+ \kappa d(x_{n},p)+2\alpha _{n} \bigl\langle \exp _{p}^{1}x_{n+1}, \exp _{p}^{1}f(p) \bigr\rangle \\ &\quad\leq (1\alpha _{n})d^{2}(x_{n},p)+\kappa d(x_{n},p)+2\alpha _{n} \bigl\langle \exp _{p}^{1}x_{n+1}, \exp _{p}^{1}f(p) \bigr\rangle (1\alpha _{n})d^{2}\bigl(y_{n},T_{r_{n}}^{F,A}y_{n} \bigr) \\ &\quad\leq \bigl[1\alpha _{n}(1\kappa )\bigr]d^{2}(x_{n},p)+ \alpha _{n}(1\kappa )b_{n}(1 \alpha _{n})d^{2} \bigl(y_{n},T_{r_{n}}^{F,A}y_{n}\bigr), \end{aligned}$$
(4.3)
where \(b_{n} =\frac{2}{(1\kappa )}\langle \exp _{p}^{1}x_{n+1},\exp _{p}^{1}f(p) \rangle \). It follows from (4.3) that
$$\begin{aligned} (1\alpha _{n})d^{2}\bigl(y_{n},T_{r_{n}}^{F,A}y_{n} \bigr) \leq d^{2}(x_{n},p)d^{2}(x_{n+1},p)+ \alpha _{n}(1\kappa )M^{\prime}, \end{aligned}$$
(4.4)
where \(M^{\prime}=\sup_{n \in \mathbb{N}}b_{n}\).
We proceed to show that \(\{x_{n}\}\) converges strongly to \(p=P_{\mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)}f(p)\). Let \(a_{n}=d^{2}(x_{n},p)\) and \(\delta _{n}=\alpha _{n}(1\kappa )\), then we have that
$$\begin{aligned} a_{n+1}\leq (1\alpha _{n})a_{n}+ \delta _{n} b_{n} \end{aligned}$$
holds from (4.3). Next, we show that \(\limsup_{k \to \infty} b_{n_{k}} \leq 0\) whenever a subsequence \(\{a_{n_{k}}\}\) of \(\{a_{n}\}\) satisfies
$$\begin{aligned} \liminf_{k \to \infty}(a_{n_{k} +1}a_{n_{k}}) \geq 0. \end{aligned}$$
Suppose such a subsequence exists, then by (4.4) and condition (i), we obtain
$$\begin{aligned} \limsup_{k \to \infty}(1\alpha _{n_{k}})d^{2} \bigl(y_{n_{k}},T_{r_{n_{k}}}^{F,A}y_{n_{k}}\bigr)& \leq \limsup_{k \to \infty}(a_{n_{k}}a_{n_{k} +1})+(1 \kappa )M^{\prime}\lim_{k \to \infty}\alpha _{n_{k}} \\ &=\liminf_{k \to \infty}(a_{n_{k} +1}a_{n_{k}}) \\ &\leq 0, \end{aligned}$$
(4.5)
thus
$$\begin{aligned} \lim_{k \to \infty}d\bigl(y_{n_{k}},T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr)=0. \end{aligned}$$
(4.6)
It implies by Lemma 3.3, that
$$\begin{aligned} \lim_{k \to \infty}d\bigl(y_{n_{k}},T_{r}^{F,A}y_{n_{k}} \bigr)=0. \end{aligned}$$
(4.7)
Observe also that
$$\begin{aligned} d^{2}(y_{n},p) &\leq \bigl\Vert y_{n}^{\prime}p^{\prime} \bigr\Vert ^{2} \\ & = \bigl\Vert \beta _{n}\bigl(p_{n}^{\prime}p^{\prime} \bigr)+(1\beta _{n}) \bigl(q_{n}^{ \prime}p^{\prime} \bigr) \bigr\Vert ^{2} \\ &=\beta _{n} \bigl\Vert p_{n}^{\prime}p^{\prime} \bigr\Vert ^{2}+(1\beta _{n}) \bigl\Vert q_{n}^{ \prime}v^{\prime} \bigr\Vert ^{2}\beta _{n}(1\beta _{n}) \bigl\Vert p_{n}^{\prime}q_{n}^{ \prime} \bigr\Vert ^{2} \\ &\leq \beta _{n} d^{2}(p_{n},p)+(1\beta _{n})d^{2}(q_{n},v)\beta _{n}(1 \beta _{n})d^{2}(p_{n},q_{n}) \\ &\leq \beta _{n} d^{2}(x_{n},p)+(1\beta _{n})d^{2}(Sx_{n},Sp)\beta _{n}(1 \beta _{n})d^{2}(x_{n},Sx_{n}) \\ &\leq d^{2}(x_{n},p)\beta _{n}(1\beta _{n})d^{2}(x_{n},Sx_{n}). \end{aligned}$$
(4.8)
Using this in (4.3), we have
$$\begin{aligned} d^{2}(x_{n+1},p) \leq {}&(1\alpha _{n})d^{2}(y_{n},p) +\kappa d(x_{n},p)+2 \alpha _{n} \bigl\langle \exp _{p}^{1}x_{n+1},\exp _{p}^{1}f(p) \bigr\rangle \\ \leq{}& (1\alpha _{n})\bigl[d(x_{n},p)\beta _{n}(1\beta _{n})d^{2}(x_{n},Sx_{n}) \bigr] \\ &{}+\kappa d(x_{n},p)+2\alpha _{n} \bigl\langle \exp _{p}^{1}x_{n+1},\exp _{p}^{1}f(p) \bigr\rangle \\ \leq{}& \bigl[1\alpha _{n}(1\kappa )\bigr]d^{2}(x_{n},p)+2 \alpha _{n}\bigl\langle \exp _{p}^{1}x_{n+1}, \exp _{p}^{1}f(p) \bigr\rangle \\ &{}\beta _{n}(1\beta _{n}) (1 \alpha _{n})d^{2}(x_{n},Sx_{n}). \end{aligned}$$
Proceeding as before, we obtain
$$\begin{aligned} \lim_{k \to \infty}\beta _{n_{k}}(1\beta _{n_{k}}) (1 \alpha _{n_{k}})d^{2}(x_{n_{k}},Sx_{n_{k}})&\leq \limsup_{k \to \infty}(a_{n_{k} +1}a_{n_{k}})+(1\kappa )M^{\prime}\lim_{k \to \infty}\alpha _{n_{k}} \\ &= \liminf_{k \to \infty}(a_{n_{k} +1}a_{n_{k}}) \\ &\leq 0, \end{aligned}$$
which implies using condition (i) and (ii) that
$$\begin{aligned} \lim_{k \to \infty}d(x_{n_{k}},Sx_{n_{k}})=0. \end{aligned}$$
(4.9)
Now, from convexity of the Riemannian manifold, we have
$$\begin{aligned} d\bigl(x_{n_{k} +1},T_{r_{n_{k}}}^{F,A}y_{n_{k}}\bigr)&=d \bigl(\gamma _{n_{k}}^{2}(1 \alpha _{n_{k}}),T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr) \\ &\leq \alpha _{n_{k}}d\bigl(\gamma _{n_{k}}^{2}(0),T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr)+(1 \alpha _{n_{k}})d\bigl(\gamma _{n_{k}}^{2}(1),T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr) \\ &=\alpha _{n_{k}}d\bigl(f(x_{n_{k}}),T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr)+(1 \alpha _{n_{k}})d\bigl(T_{r_{n_{k}}}^{F,A}y_{n_{k}},T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr) \\ &\leq \alpha _{n_{k}}d\bigl(f(x_{n_{k}}),T_{r_{n_{k}}}^{F,A}y_{n_{k}} \bigr), \end{aligned}$$
which by condition (i), implies
$$\begin{aligned} d\bigl(x_{n_{k} +1},T_{r_{n_{k}}}^{F,A}y_{n_{k}}\bigr) \to 0 \quad\text{as } k \to \infty. \end{aligned}$$
(4.10)
In a similar vein, we have
$$\begin{aligned} d(y_{n_{k}},x_{n_{k}}) &=d\bigl(\gamma _{n_{k}}^{1}(1 \beta _{n_{k}}),x_{n_{k}}\bigr) \\ &\leq \beta _{n_{k}}d\bigl(\gamma _{n_{k}}^{1}(0),x_{n_{k}} \bigr)+(1\beta _{n_{k}})d\bigl( \gamma _{n_{k}}^{1}(1),Sx_{n_{k}} \bigr) \\ &= \beta _{n_{k}}d(x_{n_{k}},x_{n_{k}})+(1\beta _{n_{k}})d(x_{n_{k}},Sx_{n_{k}}) \\ &\leq (1\beta _{n_{k}})d(x_{n_{k}},Sx_{n_{k}}), \end{aligned}$$
it implies using (4.9), that
$$\begin{aligned} \lim_{k \to \infty}d(y_{n_{k}},x_{n_{k}})=0. \end{aligned}$$
(4.11)
It is easy to see from (4.6) and (4.10) that
$$\begin{aligned} d(x_{n_{k} +1},y_{n_{k}}) \to 0 \quad\text{as } k \to \infty. \end{aligned}$$
Using this and (4.11), we get
$$\begin{aligned} \lim_{k \to \infty}d(x_{n_{k} +1},x_{n_{k}})=0. \end{aligned}$$
(4.12)
To conclude this process, we now show that \(\lim_{k \to \infty} b_{n_{k}} \leq 0\). Indeed, since \(\{x_{n_{k}}\}\) is bounded, there exists a subsequence \(\{x_{n_{k_{j}}}\}\) of \(\{x_{n_{k}}\}\), which converges weakly to \(q \in M\). Thus, we obtain by (4.12) that
$$\begin{aligned} \limsup_{k \to \infty} \bigl\langle \exp _{p}^{1}f(p),\exp _{p}^{1}x_{n_{k} +1} \bigr\rangle &= \lim_{j \to \infty}\bigl\langle \exp _{p}^{1}f(p), \exp _{p}^{1}x_{n_{k_{j}}+1} \bigr\rangle \\ & =\bigl\langle \exp _{p}^{1}f(p), \exp _{p}^{1}q \bigr\rangle , \end{aligned}$$
(4.13)
It also follows from \(x_{n_{k_{j}}}\rightharpoonup q\) and (4.11) that \(y_{n_{k_{j}}} \rightharpoonup q\). Therefore, by (4.9) and (4.11), we obtain that \(q \in \mathrm{Fix}(S)\) and \(q\in \mathrm{Fix}(T_{r}^{F,A})=\mathrm{GEP}(F,A)\), respectively. Thus, \(q \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)\). From Lemma 2.3, (4.13), and \(p=P_{\mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)}f(p)\), we get
$$\begin{aligned} \limsup_{k \to \infty} \bigl\langle \exp _{p}^{1}f(p), \exp _{p}^{1}x_{n_{k} +1} \bigr\rangle &=\lim _{j \to \infty}\bigl\langle \exp _{p}^{1}f(p), \exp _{p}^{1}x_{n_{k_{j}}+1} \bigr\rangle \\ &=\bigl\langle \exp _{p}^{1}f(p),\exp _{p}^{1}q \bigr\rangle \\ &\leq 0. \end{aligned}$$
(4.14)
Therefore, we conclude by Lemma 2.11 on (4.3) that \(x_{n} \to p\). □
The following is a corollary of our main result. For \(A=0\), we obtain a result for approximating a common solution of an equilibrium problem and a fixed point of a nonexpansive mapping.
Corollary 4.2
Let K be a nonempty, closed, and convex subset of a Hadamard manifold M. Let \(F: K \times K \to \mathbb{R}\) such that \(F(x,x)=0\) for all \(x \in K\) be a bifunction satisfying conditions (A1)–(A3). Let \(f: M \to M\) be a ψcontraction and \(S: K \to K\) be a nonexpansive mapping. Assume \(\mathrm{Fix}(S) \cap \mathrm{GEP}(F,0) \neq \emptyset \). For arbitrary \(x_{1} \in K\), sequences \(\{r_{n}\} \in (0,\infty )\), \(\beta _{n}, \alpha _{n} \in (0,1)\), let the sequence \(\{x_{n}\}\) be defined iteratively by
$$\begin{aligned} \textstyle\begin{cases} y_{n}=\exp _{x_{n}}(1\beta _{n})\exp _{x_{n}}^{1}Sx_{n}, \\ x_{n+1}=\exp _{f(x_{n})}(1\alpha _{n})\exp _{f(x_{n})}^{1}T_{r_{n}}^{F,0}y_{n}. \end{cases}\displaystyle \end{aligned}$$
(4.15)
Suppose the following conditions hold:

(i)
\(\lim_{n \to \infty}\alpha _{n}=0\) and \(\sum_{n =1}^{\infty}\alpha _{n}=0\);

(ii)
\(0< a\leq \beta _{n} \leq b < 1\) for some \(a,b > 0\) for all \(n \geq 1\);

(iii)
\(0 < r \leq r_{n}\).
If \(0 < \kappa =\sup \{\frac{\psi (d(x_{n},p))}{d(x_{n},p)}: x_{n} \ne p, n \ge 0\}< 1\) for all \(p \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,0)\), then the sequence \(\{x_{n}\}\) converges to a point \(p \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,0)\).
Suppose \(M=H\) is a real Hilbert space, then we have the following as a consequence of Theorem 4.1:
Corollary 4.3
Let K be a nonempty, closed, and convex subset of a real Hilbert space H. Let \(A: K \to K\) be a monotone vector field and \(F: K \times K \to \mathbb{R}\) such that \(F(x,x)=0\) for all \(x \in K\) be a bifunction satisfying conditions (A1)–(A3). Let \(f: M \to M\) be a contraction and \(S: K \to K\) be a nonexpansive mapping. Assume \(\mathrm{Fix}(S) \cap \mathrm{GEP}(F,A) \neq \emptyset \). For arbitrary \(x_{1} \in K\), sequences \(\{r_{n}\} \in (0,\infty )\), \(\beta _{n}, \alpha _{n} \in (0,1)\), let the sequence \(\{x_{n}\}\) be defined iteratively by
$$\begin{aligned} \textstyle\begin{cases} y_{n}=\beta _{n} x_{n} +(1\beta _{n})Sx_{n}, \\ x_{n+1}=\alpha _{n}f(x_{n})+(1\alpha _{n})T_{r_{n}}^{F,A}y_{n}. \end{cases}\displaystyle \end{aligned}$$
(4.16)
Suppose the following conditions hold:

(i)
\(\lim_{n \to \infty}\alpha _{n}=0\) and \(\sum_{n =1}^{\infty}\alpha _{n}=0\);

(ii)
\(0< a\leq \beta _{n} \leq b < 1\) for some \(a,b > 0\) for all \(n \geq 1\);

(iii)
\(0 < r \leq r_{n}\).
Then, the sequence \(\{x_{n}\}\) converges to a point \(p \in \mathrm{Fix}(S) \cap \mathrm{GEP}(F,A)\).