On the intermixed method for mixed variational inequality problems: another look and some corrections

Abstract

We explore the intermixed method for finding a common element of the intersection of the solution set of a mixed variational inequality and the fixed point set of a nonexpansive mapping. We point out that Khuangsatung and Kangtunyakarn’s statement [J. Inequal. Appl. 2023:1, 2023] regarding the resolvent utilized in their paper is not correct. To resolve this gap, we employ the epigraphical projection and the product space approach. In particular, we obtain a strong convergence theorem with a weaker assumption.

1 Introduction

Let $\mathcal{H}$ be a real Hilbert space with an inner product $\langle \cdot ,\cdot \rangle$ and the induced norm $\|\cdot \|$. Let $C\subset \mathcal{H}$, $S:C\to \mathcal{H}$, and $\alpha >0$. We say that

• S is α-Lipschitzian if $\|Sx-Sy\|\le \alpha \|x-y\|$ for all $x,y\in C$;

• S is α-inverse strongly monotone if $\langle Sx-Sy,x-y\rangle \ge \alpha \|Sx-Sy\|^{2}$ for all $x,y\in C$.

An α-Lipschitzian mapping with $\alpha \in (0,1)$ ($\alpha =1$, resp.) is called a contraction (a nonexpansive mapping, resp.). The following two classical nonlinear problems have been widely studied:

Fixed Point Problem::

Find $x\in C$ such that $x=Sx$ (see [2]).

Variational Inequality Problem::

Find $x\in C$ such that $\langle Sx,y-x\rangle \ge 0$ for all $y\in C$ (see [3]).

The solution sets of the preceding two problems are denoted by $\operatorname{Fix}(S)$ and $\operatorname{VI}(C,S)$, respectively. The following two observations are well known.

• If $S:C\to C$ is any mapping and $\operatorname{Id}:C\to C$ is the identity mapping, then $\operatorname{Fix}(S)=\operatorname{VI}(C,\operatorname{Id}-S)$. In fact, if $x=Sx$, then $\langle (\operatorname{Id}-S)x,y-x\rangle =0$ for all $y\in C$. Hence $\operatorname{Fix}(S)\subset \operatorname{VI}(C,\operatorname{Id}-S)$. On the other hand, let $x\in C$ be such that $\langle (\operatorname{Id}-S)x,y-x\rangle \ge 0$ for all $y\in C$. Let $y:=Sx\in C$. It follows that $-\|x-Sx\|^{2}=\langle x-Sx,Sx-x\rangle \ge 0$, and hence $x=Sx$. This implies that reverse inclusion, and the statement is proved.

• If C is a closed convex subset of $\mathcal{H}$ and $S:C\to \mathcal{H}$ is any mapping, then $\operatorname{VI}(C,S)=\operatorname{Fix}(P_{C}\circ ( \operatorname{Id}-S))$, where $P_{C}$ is the metric projection onto C. Note that for $x\in \mathcal{H}$ and $z\in C$, $z=P_{C}x$ if and only if $\langle z-x,y-z\rangle \ge 0$ for all $y\in C$ (for example, see [4]). To see this, let $x\in C$. It follows that

  $$\langle Sx,y-x\rangle \ge 0 \text{ for all }y\in C \iff \langle x-(\operatorname{Id}-S)x,y-x\rangle \ge 0 \text{ for all }y\in C.$$

  Hence $x\in \operatorname{VI}(C,S)\iff x=P_{C}(\operatorname{Id}-S)x\iff x \in \operatorname{Fix}(P_{C}\circ (\operatorname{Id}-S))$, and the statement is proved.

Recently, Khuangsatung and Kangtunyakarn [1] studied the following problem:

Let $f:\mathcal{H}\to (-\infty ,\infty ]$ be a proper, convex, and lower semicontinuous function. Let $C\subset \mathcal{H}$ be a closed convex set, and let $S:C\to C$. The mixed variational inequality problem is to find an element $x\in C$ such that

$$ \langle Sx,y-x\rangle +f(y)-f(x)\ge 0 \quad \text{for all }y\in C. $$

The solution of this problem is denoted by $\operatorname{VI}(C,S,f)$. If $f\equiv 0$, then the mixed variational inequality problem becomes the (classical) variational inequality problem. They claimed in their Lemma 2.6 that

$$ \operatorname{VI}(C,S,f)=\operatorname{Fix} \bigl((\operatorname{Id}+ \gamma \partial f)^{-1}\circ (\operatorname{Id}-\gamma S) \bigr)\quad (\text{for all }\gamma >0), $$

where ∂f is the subdifferential operator of f, that is,

$$ \partial f(x):=\bigl\{ z\in \mathcal{H}:\langle z,y-x\rangle +f(x)\le f(y) \text{ for all }y\in \mathcal{H}\bigr\} . $$

Unfortunately, their claim is not correct. To see this, let $C:=[1,2]\subset \mathbb{R}$, $Sx:=2x$ for all $x\in C$, and $f(x):=0$ for all $x\in \mathbb{R}$. It follows that $\operatorname{VI}(C,S,f)=\{1\}$ and $\operatorname{Fix}((\operatorname{Id}+\gamma \partial f)^{-1}\circ ( \operatorname{Id}-\gamma S))=\operatorname{Fix}(\operatorname{Id}- \gamma S)=\varnothing$ for all $\gamma >0$. In this paper, we propose an alternative way to address this gap. Moreover, we use the product space approach to deduce the intermixed method [5] and show that the convergence result can be established under a weaker assumption.

Let us recall their main result.

Theorem KK

Let C be a closed convex subset of a real Hilbert space $\mathcal{H}$. Suppose that $A_{1},A_{2},B_{1},B_{2}:C\to \mathcal{H}$ are α-inverse strongly monotone operators and $T_{1},T_{2}:C\to C$ are nonexpansive mappings. Suppose that $f_{1},f_{2}:\mathcal{H}\to (-\infty ,\infty ]$ are proper, convex, and lower semicontinuous functions. Assume that for $i=1,2$,

$$ \Omega _{i}:=\operatorname{Fix}(T_{i})\cap \operatorname{VI}(C,A_{i},f_{i}) \cap \operatorname{VI}(C,B_{i},f_{i}) \neq \varnothing . $$

Suppose that $g_{1},g_{2}:\mathcal{H}\to \mathcal{H}$ are contractions and $\{x_{n}\}_{n=1}^{\infty}$ and $\{y_{n}\}_{n=1}^{\infty}$ are iterative sequences generated by the following scheme:

$$\begin{aligned} &x_{1},y_{1}\in C\textit{ are arbitrarily chosen}, \\ &x'_{n}:=b_{1}x_{n}+(1-b_{1})T_{1}x_{n}, \\ &y'_{n}:=b_{2}y_{n}+(1-b_{2})T_{2}y_{n}, \\ &x''_{n}:=J_{\gamma _{1}f_{1}} \bigl(x_{n}-\gamma _{1}\bigl(a_{1}A_{1}+(1-a_{1})B_{1} \bigr)x_{n}\bigr), \\ &y''_{n}:=J_{\gamma _{2}f_{2}} \bigl(y_{n}-\gamma _{2}\bigl(a_{2}A_{2}+(1-a_{2})B_{2} \bigr)y_{n}\bigr), \\ &x_{n+1}:=(1-\beta _{n})x'_{n}+\beta _{n} P_{C}\bigl(\alpha _{n}g_{2}(y_{n})+(1- \alpha _{n})x''_{n}\bigr), \\ &y_{n+1}:=(1-\beta _{n})y'_{n}+\beta _{n} P_{C}\bigl(\alpha _{n}g_{1}(x_{n})+(1- \alpha _{n})y''_{n}\bigr), \end{aligned}$$

where $\gamma _{1},\gamma _{2}\in (0,2\alpha )$, $a_{1},a_{2},b_{1},b_{2}\in (0,1)$, and the sequences $\{\alpha _{n}\}_{n=1}^{\infty}$ and $\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]$ satify the following conditions:

• (C1) $\lim_{n}\alpha _{n}=0$ and $\sum_{n}\alpha _{n}=\infty$,

• (C2) $\beta _{n}\in [k,l]\subset (0,1)$ for all $n\ge 1$,

• (C3) $\sum_{n}|\alpha _{n}-\alpha _{n+1}|<\infty$ and $\sum_{n}|\beta _{n}-\beta _{n+1}|<\infty$.

Then there are two elements $x^{*}$ and $y^{*}$ such that $x^{*}=P_{\Omega _{1}}g_{2}(y^{*})$, $y^{*}=P_{\Omega _{2}}g_{1}(x^{*})$, and the iterative sequences $\{x_{n}\}_{n=1}^{\infty}$ and $\{y_{n}\}_{n=1}^{\infty}$ converge strongly to $x^{*}$ and $y^{*}$, respectively.

We need the following lemma.

Lemma 1

([6])

Let $\{s_{n}\}_{n=1}^{\infty}$ be a sequence of nonnegative real numbers, let $\{t_{n}\}_{n=1}^{\infty}$ be a sequence of real numbers, and let $\{\alpha _{n}\}_{n=1}^{\infty}$ be a sequence in $[0,1]$ such that

$$ s_{n+1}\le (1-\alpha _{n})s_{n}+\alpha _{n}t_{n}\quad \textit{for all }n\ge 1. $$

If $\sum_{n}\alpha _{n}=\infty$ and $\limsup_{n}t_{n}\le 0$, then $\lim_{n}s_{n}=0$.

Lemma 2

Let $C\subset \mathcal{H}$ and $S:C\to \mathcal{H}$. Then:

• (a) If C is closed and convex and S is nonexpansive, then $\operatorname{Fix}(S)$ is closed and convex.

• (b) If S is α-inverse strongly monotone, then $\operatorname{Id}-\lambda S$ is nonexpansive for all $\lambda \in [0,2\alpha ]$.

2 Main results

2.1 A Halpern-type method

Recall that a nonexpansive mapping $S:C\to C$ is r-strongly quasi-nonexpansive ($r>0$) if $\operatorname{Fix}(S)\neq \varnothing$ and

$$ \Vert Sx-p \Vert ^{2}\le \Vert x-p \Vert ^{2}-r \Vert x-Sx \Vert ^{2} \quad \text{for all }x\in C\text{ and }p\in \operatorname{Fix}(S). $$

It is well known that every nonexpansive mapping $S:C\to C$ satisfies the Browder demiclosedness principle: $p\in \operatorname{Fix}(S)$ whenever $\{x_{n}\}_{n=1}^{\infty}$ is a sequence in C such that $\lim_{n}\|x_{n}-Sx_{n}\|=0$ and $\{x_{n}\}_{n=1}^{\infty}$ converges weakly to $p\in C$ (see [7]). The technique we used in the following result is taken from Wang et al. [8].

Theorem 3

Let $C\subset \mathcal{H}$ be closed and convex, and let $S,U:C\to C$ be nonexpansive mappings such that $F:=\operatorname{Fix}(S)\cap \operatorname{Fix}(U)\neq \varnothing$. Suppose that S is r-strongly quasinonexpansive, where $r>0$. Suppose that $u\in \mathcal{H}$ and $\{x_{n}\}_{n=1}^{\infty}$ is an iterative sequence generated by the following scheme:

$$\begin{aligned} &x_{1}\in C\textit{ is arbitrarily chosen}, \\ &x_{n+1}:=(1-\beta _{n})Sx_{n}+\beta _{n}P_{C}\bigl(\alpha _{n} u+(1- \alpha _{n})Ux_{n}\bigr)\quad (n\ge 1), \end{aligned}$$

where the sequences $\{\alpha _{n}\}_{n=1}^{\infty},\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]$ satisfy the following conditions:

$$ \lim_{n}\frac{\alpha _{n}}{1-\beta _{n}}=0\quad \textit{and}\quad \sum _{n} \alpha _{n}\beta _{n}=\infty . $$

Then the iterative sequence $\{x_{n}\}_{n=1}^{\infty}$ converges strongly to $P_{F}u$.

Proof

Note that F is closed and convex. Let $z:=P_{F}u$. It follows that $z=P_{F}z=Sz=Uz$ and

$$\begin{aligned} \Vert x_{n+1}-z \Vert &\le (1-\beta _{n}) \Vert Sx_{n}-z \Vert +\beta _{n} \bigl\Vert P_{C} \bigl( \alpha _{n} u+(1-\alpha _{n})Ux_{n} \bigr)-P_{F}z \bigr\Vert \\ &\le (1-\beta _{n}) \Vert x_{n}-z \Vert +\beta _{n}\alpha _{n} \Vert u-z \Vert +\beta _{n}(1- \alpha _{n}) \Vert Ux_{n}-z \Vert \\ &\le (1-\beta _{n}\alpha _{n}) \Vert x_{n}-z \Vert +\beta _{n}\alpha _{n} \Vert u-z \Vert \\ &\le \max \bigl\{ \Vert x_{n}-z \Vert , \Vert u-z \Vert \bigr\} . \end{aligned}$$

It follows by induction that $\{x_{n}\}_{n=1}^{\infty}$ is a bounded sequence. In particular, the sequences $\{Sx_{n}\}_{n=1}^{\infty}$, $\{Ux_{n}\}_{n=1}^{\infty}$, and $\{P_{C}(\alpha _{n} u+(1-\alpha _{n})Ux_{n})\}_{n=1}^{\infty}$ are all bounded. For convenience, we denote

$$ u_{n}:=\alpha _{n} u+(1-\alpha _{n})Ux_{n}. $$

We refine the preceding estimates by considering $\|\cdot \|^{2}$ as follows:

$$\begin{aligned} \Vert Sx_{n}-z \Vert ^{2} &\le \Vert x_{n}-z \Vert ^{2}-r \Vert x_{n}-Sx_{n} \Vert ^{2}, \end{aligned}$$

and

$$\begin{aligned} \Vert u_{n}-z \Vert ^{2}&= \bigl\Vert \alpha _{n}(u-z)+(1-\alpha _{n}) (Ux_{n}-z) \bigr\Vert ^{2} \\ &\le \bigl\Vert (1-\alpha _{n}) (Ux_{n}-z) \bigr\Vert ^{2}+2\bigl\langle \alpha _{n}(u-z),u_{n}-z \bigr\rangle \\ &\le (1-\alpha _{n}) \Vert x_{n}-z \Vert ^{2}+2\alpha _{n}\langle u-z,u_{n}-z \rangle . \end{aligned}$$

It follows that

$$\begin{aligned} & \Vert x_{n+1}-z \Vert ^{2} \\ &\quad = \bigl\Vert (1-\beta _{n}) (Sx_{n}-z)+\beta _{n}(P_{C}u_{n}-z) \bigr\Vert ^{2} \\ &\quad =(1-\beta _{n}) \Vert Sx_{n}-z \Vert ^{2} + \beta _{n} \Vert P_{C}u_{n}-z \Vert ^{2}- \beta _{n}(1-\beta _{n}) \Vert Sx_{n}-P_{C}u_{n} \Vert ^{2} \\ &\quad \le (1-\beta _{n}) \bigl( \Vert x_{n}-z \Vert ^{2}- r \Vert x_{n}-Sx_{n} \Vert ^{2} \bigr)+\beta _{n} \bigl((1-\alpha _{n}) \Vert x_{n}-z \Vert ^{2}+2\alpha _{n} \langle u-z,u_{n}-z\rangle \bigr) \\ &\qquad {} -\beta _{n}(1-\beta _{n}) \Vert Sx_{n}-P_{C}u_{n} \Vert ^{2} \\ &\quad =(1-\alpha _{n}\beta _{n}) \Vert x_{n}-z \Vert ^{2}+2\alpha _{n}\beta _{n} \langle u-z,u_{n}-z\rangle \\ &\qquad {} -r(1-\beta _{n}) \Vert x_{n}-Sx_{n} \Vert ^{2}-\beta _{n}(1-\beta _{n}) \Vert Sx_{n}-P_{C}u_{n} \Vert ^{2}. \end{aligned}$$

Since $\sum_{n}\alpha _{n}\beta _{n}=\infty$, we have

$$ \limsup_{n} \Vert x_{n}-z \Vert ^{2}\le L, $$

where

$$ L:=\limsup_{n} \biggl(2\langle u-z,u_{n}-z\rangle - \frac{r(1-\beta _{n})}{\alpha _{n}\beta _{n}} \Vert x_{n}-Sx_{n} \Vert ^{2}- \frac{1-\beta _{n}}{\alpha _{n}} \Vert Sx_{n}-P_{C}u_{n} \Vert ^{2} \biggr). $$

Note that $L\le 2\limsup_{n}\langle u-z,u_{n}-z\rangle <\infty$ because $\{u_{n}\}_{n=1}^{\infty}$ is bounded. If $L=-\infty$, then it follows that $\limsup_{n}\|x_{n}-z\|^{2}\le 0$, and we are done. We now assume that L is finite. Let $\{n_{k}\}_{k=1}^{\infty}$ be a strictly increasing sequence such that $\{u_{n_{k}}\}_{k=1}^{\infty}$ converges weakly to some element $q\in C$ and

$$ \lim_{k} \biggl(2\langle u-z,u_{n_{k}}-z\rangle - \frac{r(1-\beta _{n_{k}})}{\alpha _{n_{k}}\beta _{n_{k}}} \Vert x_{n_{k}}-Sx_{n_{k}} \Vert ^{2}-\frac{1-\beta _{n_{k}}}{\alpha _{n_{k}}} \Vert Sx_{n_{k}}-P_{C}u_{n_{k}} \Vert ^{2} \biggr)=L. $$

In particular, the sequences

$$ \biggl\{ \frac{1-\beta _{n_{k}}}{\alpha _{n_{k}}\beta _{n_{k}}} \Vert x_{n_{k}}-Sx_{n_{k}} \Vert ^{2} \biggr\} _{k=1}^{\infty}\quad \text{and} \quad \biggl\{ \frac{1-\beta _{n_{k}}}{\alpha _{n_{k}}} \Vert Sx_{n_{k}}-P_{C}u_{n_{k}} \Vert ^{2} \biggr\} _{k=1}^{\infty} $$

are both bounded. Note that $\lim_{n}\frac{\alpha _{n}\beta _{n}}{1-\beta _{n}}=\lim_{n} \frac{\alpha _{n}}{1-\beta _{n}}=0$. It follows that

$$ \lim_{k} \Vert x_{n_{k}}-Sx_{n_{k}} \Vert ^{2}=\lim_{k} \Vert Sx_{n_{k}}-P_{C}u_{n_{k}} \Vert ^{2}=0. $$

Moreover, we have $\lim_{k}\|u_{n_{k}}-Ux_{n_{k}}\|=0$. In particular, $\lim_{k}\|P_{C}u_{n_{k}}-Ux_{n_{k}}\|=0$ and $x_{n_{k}}\rightharpoonup q$. Then it follows that

$$ \lim_{k} \Vert x_{n_{k}}-Ux_{n_{k}} \Vert \le \lim_{k} \bigl( \Vert x_{n_{k}}-Sx_{n_{k}} \Vert + \Vert Sx_{n_{k}}-P_{C}u_{n_{k}} \Vert + \Vert P_{C}u_{n_{k}}-Ux_{n_{k}} \Vert \bigr)=0. $$

In particular, it follows from the Browder demiclosedness principle that $q\in F$, and hence $\langle z-u,q-z\rangle \ge 0$. This implies that $\limsup_{n}\|x_{n}-z\|^{2}\le L\le 2\lim_{k}\langle u-z,u_{n_{k}}-z \rangle =2\langle u-z,q-z\rangle \le 0$. □

Corollary 4

Suppose that C, S, U, F, r, $\{\alpha _{n}\}_{n=1}^{\infty}$ and $\{\beta _{n}\}_{n=1}^{\infty}$ are as in the preceding theorem. Suppose that $h:\mathcal{H}\to \mathcal{H}$ is a contraction and $\{x_{n}\}_{n=1}^{\infty}$ is an iterative sequence generated by the following scheme:

$$\begin{aligned} &x_{1}\in C\textit{ is arbitrarily chosen}, \\ &x_{n+1}:=(1-\beta _{n})Sx_{n}+\beta _{n}P_{C}\bigl(\alpha _{n} h(x_{n})+(1- \alpha _{n})Ux_{n}\bigr)\quad (n\ge 1). \end{aligned}$$

Proof

Note that $P_{F}\circ h:\mathcal{H}\to \mathcal{H}$ is a contraction, and thus it follows that there exists a unique element $z\in \mathcal{H}$ such that $z= (P_{F}\circ h)(z)$. It is clear that $z\in F$. We let $u:=h(z)$ and define

$$\begin{aligned} &y_{1}:=x_{1}, \\ &y_{n+1}:=(1-\beta _{n})Sy_{n}+\beta _{n}P_{C}\bigl(\alpha _{n} u+(1- \alpha _{n})Uy_{n}\bigr)\quad (n\ge 1). \end{aligned}$$

It follows from the preceding theorem that $\lim_{n}\|y_{n}-z\|=0$. Suppose that h is γ-Lipschitzian with $\gamma \in (0,1)$. We have the following estimate:

$$\begin{aligned} \Vert x_{n+1}-y_{n+1} \Vert \le{}& (1-\beta _{n}) \Vert Sx_{n}-Sy_{n} \Vert +\beta _{n} \bigl\Vert P_{C}\bigl( \alpha _{n} h(x_{n})+(1-\alpha _{n})Ux_{n}\bigr) \\ &{}-P_{C}\bigl(\alpha _{n} u+(1- \alpha _{n})Uy_{n}\bigr) \bigr\Vert \\ \le{}& (1-\beta _{n}) \Vert x_{n}-y_{n} \Vert +\beta _{n} \bigl\Vert \alpha _{n} h(x_{n})+(1- \alpha _{n})Ux_{n} \\ &{}-\bigl(\alpha _{n} h(z)+(1- \alpha _{n})Uy_{n}\bigr) \bigr\Vert \\ \le{}& (1-\beta _{n}) \Vert x_{n}-y_{n} \Vert +\alpha _{n}\beta _{n} \bigl\Vert h(x_{n})-h(z) \bigr\Vert +(1-\alpha _{n})\beta _{n} \Vert Ux_{n}-Uy_{n} \Vert \\ \le{}& (1-\alpha _{n}\beta _{n}) \Vert x_{n}-y_{n} \Vert +\alpha _{n}\beta _{n} \bigl\Vert h(x_{n})-h(z) \bigr\Vert . \end{aligned}$$

It follows from $\sum_{n}\alpha _{n}\beta _{n}=\infty$ that

$$\begin{aligned} \limsup_{n} \Vert x_{n}-y_{n} \Vert &\le \limsup_{n} \bigl\Vert h(x_{n})-h(z) \bigr\Vert \\ &\le \limsup_{n}\gamma \Vert x_{n}-z \Vert \\ &\le \gamma \limsup_{n}\bigl( \Vert x_{n}-y_{n} \Vert + \Vert y_{n}-z \Vert \bigr) \\ &=\gamma \limsup_{n} \Vert x_{n}-y_{n} \Vert . \end{aligned}$$

In particular, since $\gamma <1$, we have $\lim_{n}\|x_{n}-y_{n}\|=0$, and hence $\lim_{n}\|x_{n}-z\|=0$. The proof is complete. □

Let $S:=\operatorname{Id}$ and $u\in C$. We immediately obtain the following result.

Corollary 5

Let $C\subset \mathcal{H}$ be closed and convex, and let $U:C\to C$ be a nonexpansive mapping such that $\operatorname{Fix}(U)\neq \varnothing$. Suppose that $u\in C$ and $\{x_{n}\}_{n=1}^{\infty}$ is an iterative sequence generated by the following scheme:

$$\begin{aligned} &x_{1}\in C\textit{ is arbitrarily chosen}, \\ &x_{n+1}:=(1-\beta _{n})x_{n}+\beta _{n}\bigl(\alpha _{n} u+(1-\alpha _{n})Ux_{n} \bigr) \quad (n\ge 1), \end{aligned}$$

where the sequences $\{\alpha _{n}\}_{n=1}^{\infty},\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]$ satisfy the following conditions:

$$ \lim_{n}\frac{\alpha _{n}}{1-\beta _{n}}=0 \quad \textit{and}\quad \sum _{n} \alpha _{n}\beta _{n}=\infty . $$

Then the iterative sequence $\{x_{n}\}_{n=1}^{\infty}$ converges strongly to $P_{\operatorname{Fix}(U)}u$.

2.2 Comments and remarks on the mixed variational inequality problem

Let $C\subset \mathcal{H}$ be closed and convex, let $A:C\to \mathcal{H}$, and let $f:\mathcal{H}\to (-\infty ,\infty ]$ be a proper convex and lower semicontinuous function. The mixed variational inequality problem is to find $x\in C$ such that

$$ \langle Ax,y-x\rangle +f(y)- f(x)\ge 0\quad \text{for all }y\in C. $$

(⋆)

As pointed out in the introduction of the paper, the resolvent proposed by Khuangsatung and Kangtunyakarn [1] is not correct. Moreover, without any further assumption on C and domf, it is possible to encounter the experession $\infty -\infty$ in (⋆). For example, let $Ax:=0$ for all $x\in C:=[1,2]\subset \mathbb{R}\mathbbm{.}$ and let $f(x):=0$ if $x\in [3,4]$ and $f(x):=\infty$ if $x\notin [3,4]$. To be on the right track, we discuss the problem with an additional assumption.

This mixed type problem was also considered by Mosco [9] in 1969. From now on, we also assume that $\operatorname{dom}f\subset C$ is as in Mosco’s setting. In particular, we also have $\operatorname{VI}(C,A,f)\subset \operatorname{dom}f$.

Mosco proved that the mixed and the classical variational inequality problems are equivalent. To see this, let $\widehat{\mathcal{H}}:=\mathcal{H}\times \mathbb{R}$ with $\bigl\langle\bigl\langle \widehat{x},\widehat{y}\bigr\rangle\bigr\rangle := \langle x,y\rangle +rs$ for all $\widehat{x}:=(x,r)$ and $\widehat{y}:=(y,s)\in \widehat{\mathcal{H}}$, and let $\widehat{C}:=C\times \mathbb{R}$. Note that ${|\!|\!| \widehat{x} |\!|\!|}^{2}=\bigl\langle\bigl\langle \widehat{x},\widehat{x} \bigr\rangle\bigr\rangle =\|x\|^{2}+r^{2}$. Define $\widehat{A}:\widehat{C}\to \widehat{\mathcal{H}}$ by

$$ \widehat{A}(x,r):=(Ax,1) \quad \text{for all }(x,r)\in \widehat{C}. $$

Here $\operatorname{epi}f:=\{(x,r)\in \widehat{\mathcal{H}}:f(x)\le r\}$ is the epigraph of f, which is closed and convex because of the lower semicontinuity and convexity of f.

Theorem 6

Suppose that $\operatorname{dom}f\subset C$. The following statements are true:

• (1) $\operatorname{VI}(C,A,f)=\{x\in C:\langle Ax,y-x\rangle +f(y)- f(x) \ge 0\textit{ for all }y\in \operatorname{dom}f\}$;

• (2) $(x,r)\in \operatorname{VI}(\operatorname{epi}f,\widehat{A}) \iff x \in \operatorname{VI}(C,A,f) \textit{ and }r=f(x)$;

• (3) If A is α-inverse strongly monotone, then so is $\widehat{A,}$ and hence $\operatorname{Id}-\lambda \widehat{A}$ is nonexpansive for all $\lambda \in (0,2\alpha ]$.

Proof

(1) is straight forward. (2) was proved by Mosco. For completeness, we give a proof of (2).

(⇒) Let $(x,r)\in \operatorname{VI}(\operatorname{epi}f,\widehat{A})$, and let $y\in \operatorname{dom}f$. This implies that $(y,f(y))\in \operatorname{epi}f$ and

$$ \langle Ax,y-x\rangle +f(y)-r=\bigl\langle \bigl\langle \widehat{A}(x,r), \bigl(y,f(y)\bigr)-(x,r) \bigr\rangle \bigr\rangle \ge 0. $$

Note that $f(x)\le r$. This implies that $\langle Ax,y-x\rangle +f(y)-f(x)\ge 0$. Moreover, we have

$$ f(x)-r=\bigl\langle \bigl\langle \widehat{A}(x,r),\bigl(x,f(x)\bigr)-(x,r)\bigr\rangle \bigr\rangle \ge 0. $$

This implies that $f(x)\ge r$, and hence $r=f(x)$. In particular, we have

$$ \langle Ax,y-x\rangle +f(y)-f(x)\ge 0. $$

(⇐) Suppose that $x\in \operatorname{VI}(C,A,f)$. We prove that $(x,f(x))\in \operatorname{VI}(\operatorname{epi}f,\widehat{A})$. To see this, let $(y,s)\in \operatorname{epi}f$. It follows that $f(y)\le s$ and

$$ \bigl\langle \bigl\langle \widehat{A}\bigl(x,f(x)\bigr),(y,s)-\bigl(x,f(x)\bigr) \bigr\rangle \bigr\rangle =\langle Ax,y-x \rangle +s-f(x)\ge \langle Ax,y-x \rangle +f(y)-f(x) \ge 0. $$

(3) Suppose that A is α-inverse strongly monotone. We show that $\widehat{A}:\widehat{C}\to \widehat{\mathcal{H}}$ is also α-inverse strongly monotone. To see this, let $\widehat{x}:=(x,r)$, $\widehat{y}:=(y,s)\in \widehat{C}$. It follows that

$$\begin{aligned} \bigl\langle \bigl\langle \widehat{A}\widehat{x}-\widehat{A}\widehat{y}, \widehat{x}- \widehat{y} \bigr\rangle \bigr\rangle =\langle Ax-Ay,x-y\rangle \ge \alpha \Vert Ax-Ay \Vert ^{2}= \alpha {\big|\!\big|\!\big| \widehat{A}\widehat{x}-\widehat{A} \widehat{y} \big|\!\big|\!\big|}^{2}. \end{aligned}$$

In particular, $\operatorname{Id}-\lambda \widehat{A}$ is nonexpansive for $\lambda \in (0,2\alpha ]$. □

Because of the error of the resolvent proposed by the authors of [1], we cannot infer the closedness and the convexity of $\operatorname{VI}(C,A,f)$. However, the conclusion remains true as follows.

Corollary 7

Let $A:C\to \mathcal{H}$ be α-inverse strongly monotone, and let $f:\mathcal{H}\to (-\infty ,\infty ]$ be a proper convex and lower semicontinuous function. Suppose that $\operatorname{dom}f\subset C$. Then $\operatorname{VI}(C,A,f)$ is closed and convex.

Proof

We assume that $\operatorname{VI}(C,A,f)$ is nonempty. Note that $\operatorname{VI}(\operatorname{epi}f,\widehat{A})= \operatorname{Fix}(P_{\operatorname{epi}f}\circ (\operatorname{Id}- \alpha \widehat{A}))$ is closed and convex. To prove the closedness of $\operatorname{VI}(C,A,f)$, let $\{x_{n}\}_{n=1}^{\infty}$ be a sequence in $\operatorname{VI}(C,A,f)$ and assume that $\{x_{n}\}_{n=1}^{\infty}$ is strongly convergent to a point $x\in C$. It suffices to show that $(x,f(x))\in \operatorname{VI}(\operatorname{epi}f,\widehat{A})$. Put $r:=f(x)$ and $r_{n}:=f(x_{n})$. From the lower semicontinuity of f it follows that $r\le \liminf_{n}r_{n}$. Note that for $(y,s)\in \widehat{C}:=C\times \mathbb{R}\mathbbm{,}$ we have

$$ \langle Ax_{n},y-x_{n}\rangle +s-r_{n}=\bigl\langle \bigl\langle \widehat{A}(x_{n},r_{n}),(y,s)-(x_{n}-r_{n}) \bigr\rangle \bigr\rangle \ge 0. $$

Since A is $(1/\alpha )$-Lipschitzian and hence continuous, we obtain that $\lim_{n}\langle Ax_{n},y-x_{n}\rangle =\langle Ax,y-x\rangle$. In particular, $\langle Ax,y-x\rangle +s\ge \limsup_{n} r_{n}\ge r$. Hence $\langle \langle \widehat{A}(x,r),(y,s)-(x,r)\rangle \rangle = \langle Ax,y-x\rangle +s-r\ge 0$, that is, $(x,f(x))=(x,r)\in \operatorname{VI}(\operatorname{epi}f,\widehat{A})$.

Finally, we prove that $\operatorname{VI}(C,A,f)$ is convex. To this end, let $x,x'\in \operatorname{VI}(C,A,f)$ and $t\in (0,1)$. It follows that $(x,r),(x',r')\in \operatorname{VI}(\operatorname{epi}f,\widehat{A})$, where $r:=f(x)$ and $r':=f(x')$. Put $x'':=(1-t)x+tx'$. Since $\operatorname{VI}(C,A,f)$ is convex, it follows that $(x'',(1-t)r+tr')\in \operatorname{VI}(\operatorname{epi}f, \widehat{A})$. In particular, for $(y,s)\in \widehat{C}:=C\times \mathbb{R}$ and $r'':=f(x'')$, we have $r''\le (1-t)r+tr'$ and

$$\begin{aligned} \bigl\langle \bigl\langle \widehat{A}\bigl(x'',r'' \bigr),(y,s)-\bigl(x'',r'' \bigr)\bigr\rangle \bigr\rangle &=\bigl\langle Ax'',y-x'' \bigr\rangle +s-r'' \\ &\ge \bigl\langle Ax'',y-x'' \bigr\rangle +s-\bigl((1-t)r+tr'\bigr) \\ &=\bigl\langle \bigl\langle \widehat{A}\bigl(x'',(1-t)r+tr' \bigr),(y,s)-\bigl(x'',(1-t)r+tr'\bigr) \bigr\rangle \bigr\rangle \ge 0. \end{aligned}$$

It follows that $(x'',r'')\in \operatorname{VI}(\operatorname{epi}f,\widehat{A})$, and hence $x''\in \operatorname{VI}(C,A,f)$. □

2.3 Another look at the intermixed method via a product space approach

Suppose that C, $\mathcal{H}$, $A_{i}$, $B_{i}$, $T_{i}$, $f_{i}$, $g_{i}$ ($i=1,2$) are as in Theorem KK. Note that we can show that $\operatorname{VI}(C,A_{1},f_{1})\cap \operatorname{VI}(C,B_{1},f_{1})= \operatorname{VI}(C,a_{1}A_{1}+(1-a_{1})B_{1},f_{1})$ for $0< a_{1}<1$ if $\operatorname{VI}(C,A_{1},f_{1})\cap \operatorname{VI}(C,B_{1},f_{1}) \neq \varnothing$ and if $A_{1}$ and $B_{1}$ are α-inverse strongly monotone. Corresponding to this note, we assume for simplicity that $A_{1}=B_{1}$ and $A_{2}=B_{2}$. We also assume that

$$ \Omega _{i}:=\operatorname{Fix}(T_{i})\cap \operatorname{VI}(C,A_{i},f_{i}) \neq \varnothing \quad \text{for }i=1,2. $$

To deduce and correct the conclusion in Theorem KK, let us fix the following notation.

Let

$$ \boldsymbol{\mathcal{H}}:=\widehat{\mathcal{H}}\times \widehat{\mathcal{H}} \quad \text{and} \quad \boldsymbol{C}:=\widehat{C} \times \widehat{C}, $$

where $\widehat{\mathcal{H}}:=\mathcal{H}\times \mathbb{R}$ and $\widehat{C}:=C\times \mathbb{R}$. Note that $\boldsymbol{\mathcal{H}}$ is a Hilbert space endowed with the inner product $[\cdot ,\cdot ]$ defined by

$$\begin{aligned} \bigl[\boldsymbol{x},\boldsymbol{x'}\bigr]:=\bigl\langle x,x'\bigr\rangle +rr'+\bigl\langle y,y' \bigr\rangle +ss' \end{aligned}$$

for all $\boldsymbol{x}:=((x,r),(y,s))$ and $\boldsymbol{x'}:=((x',r'),(y',s'))\in \boldsymbol{\mathcal{H}}$. Moreover, the induced norm of each element $\boldsymbol{x}:=((x,r),(y,s))\in \boldsymbol{\mathcal{H}}$ is given by

$$ \Vert \boldsymbol{x} \Vert := \bigl( \Vert x \Vert ^{2}+r^{2}+ \Vert y \Vert ^{2}+s^{2} \bigr)^{1/2}. $$

Define $\boldsymbol{A}:\boldsymbol{C}\to \boldsymbol{\mathcal{H}}$ and $\boldsymbol{S}:\boldsymbol{C}\to \boldsymbol{C}$ by

$$\begin{aligned} \boldsymbol{A}\boldsymbol{x} &:= \bigl((A_{1}x,1),(A_{2}y,1) \bigr) \end{aligned}$$

and

$$\begin{aligned} \boldsymbol{S}\boldsymbol{x} &:= \bigl( \bigl(b_{1}x+(1-b_{1})T_{1}x,r\bigr), \bigl(b_{2}y+(1-b_{2})T_{2}y,s\bigr) \bigr) \end{aligned}$$

for $\boldsymbol{x}:=((x,r),(y,s))\in \boldsymbol{C}$.

Using the preceding setting, we obtain the following results.

Proposition 8

(Properties of A)

Let $\boldsymbol{x}:= ((x,r),(y,s) )\in \boldsymbol{\mathcal{H}}$ and $\boldsymbol{E}:=\operatorname{epi}f_{1}\times \operatorname{epi}f_{2}$. Then the following two statements are equivalent:

• (a) $\boldsymbol{x}\in \operatorname{VI}(\boldsymbol{E},\boldsymbol{A})$;

• (b) $x\in \operatorname{VI}(C,A_{1},f_{1})$, $y\in \operatorname{VI}(C,A_{2},f_{2})$, $r=f_{1}(x)$, and $s=f_{2}(x)$.

If, in addition, $A_{1},A_{2}:C\to \mathcal{H}$ are α-inverse strongly monotone, then $\boldsymbol{A}:\boldsymbol{C}\to \boldsymbol{\mathcal{H}}$ is α-inverse strongly monotone.

Proof

(a) ⇒ (b) Let $\boldsymbol{x}:= ((x,r),(y,s) )\in \operatorname{VI}( \boldsymbol{E},\boldsymbol{A})$. Let $\boldsymbol{x'}:= ((x',r'),(y,s) )$, where $(x',r')\in \operatorname{epi}f_{1}$. It follows that $\boldsymbol{x'}\in \boldsymbol{E}$, and hence $\bigl\langle\bigl\langle \widehat{A_{1}}(x,r),(x',r')-(x,r)\bigr\rangle\bigr\rangle = [\boldsymbol{Ax},\boldsymbol{x'-x}]\ge 0$. This means that $(x,r)\in \operatorname{VI}(\operatorname{epi}f_{1},\widehat{A_{1}})$. It follows from Theorem 6 that $x\in \operatorname{VI}(C,A_{1},f_{1})$ and $r=f_{1}(x)$. Using a similar technique, we obtain the remaining conclusion.

(b) ⇒ (a) is trivial.

Suppose that $A_{1},A_{2}:C\to \mathcal{H}$ are α-inverse strongly monotone. To see that $\boldsymbol{A}:\boldsymbol{C}\to \boldsymbol{\mathcal{H}}$ is α-inverse strongly monotone, let $\boldsymbol{x}:= ((x,r),(y,s) )$ and $\boldsymbol{x'}:= ((x',r'),(y',s') )\in \boldsymbol{C}$. It follows that

$$\begin{aligned} &\bigl[\boldsymbol{Ax}-\boldsymbol{Ax'},\boldsymbol{x-x'} \bigr] \\ &\quad = \bigl[ \bigl(\bigl(A_{1}x-A_{1}x',0\bigr), \bigl(A_{2}y-A_{2}y',0\bigr) \bigr), \bigl( \bigl(x-x',r-r'\bigr),\bigl(y-y',s-s' \bigr) \bigr) \bigr] \\ &\quad =\bigl\langle A_{1}x-A_{1}x',x-x' \bigr\rangle +\bigl\langle A_{2}y-A_{2}y',y-y' \bigr\rangle \\ &\quad \ge \alpha \bigl( \bigl\Vert A_{1}x-A_{1}x' \bigr\Vert ^{2}+ \bigl\Vert A_{2}y-A_{2}y' \bigr\Vert ^{2}\bigr)=\alpha \bigl\Vert \boldsymbol{Ax}- \boldsymbol{Ax'} \bigr\Vert ^{2}. \end{aligned}$$

This completes the proof. □

Proposition 9

(Properties of S)

Let $\boldsymbol{x}:= ((x,r),(y,s) )\in \boldsymbol{C}$. Then the following two statements are equivalent:

• (a) $\boldsymbol{x}\in \operatorname{Fix}(\boldsymbol{S})$;

• (b) $x\in \operatorname{Fix}(T_{1})$ and $y\in \operatorname{Fix}(T_{2})$.

If, in addition, $T_{1}$, $T_{2}$ are nonexpansive and $\operatorname{Fix}(T_{1})\times \operatorname{Fix}(T_{2})\neq \varnothing$, then S is nonexpansive and r-strongly quasinonexpansive where $r:=\min \{b_{1}(1-b_{1}),b_{2}(1-b_{2})\}$.

Proof

(a) ⇔ (b) is trivial. Now we suppose that $T_{1}$ and $T_{2}$ are nonexpansive and $\operatorname{Fix}(T_{1})\times \operatorname{Fix}(T_{2})\neq \varnothing$. It is clear that S is nonexpansive. Let $r:=\min \{b_{1}(1-b_{1}),b_{2}(1-b_{2})\}$. We show that S is r-strongly quasinonexpansive. To see this, let $\boldsymbol{x}:= ((x,r),(y,s) )\in \boldsymbol{C}$ and $\boldsymbol{p}:= ((p,r'),(q,s') )\in \operatorname{Fix}( \boldsymbol{S})$. It follows that

$$\begin{aligned} \bigl\Vert \bigl(b_{1}x+(1-b)T_{1}x\bigr)-p \bigr\Vert ^{2} &\le b_{1} \Vert x-p \Vert ^{2}+(1-b_{1}) \Vert T_{1}x-p \Vert ^{2}-b_{1}(1-b_{1}) \Vert x-T_{1}x \Vert ^{2} \\ &\le \Vert x-p \Vert ^{2}-r \Vert x-T_{1}x \Vert ^{2}. \end{aligned}$$

Similarly, $\|(b_{2}y+(1-b_{2})T_{2}y)-q\|^{2}\le \|y-q\|^{2}-r\|y-T_{2}y\|^{2}$. This implies that

$$\begin{aligned} \Vert \boldsymbol{Sx-p} \Vert ^{2}={}& \bigl\Vert \bigl(b_{1}x+(1-b)T_{1}x\bigr)-p \bigr\Vert ^{2}+ \bigl(r-r'\bigr)^{2} \\ &{}+ \bigl\Vert \bigl(b_{2}y+(1-b_{2})T_{2}y \bigr)-q \bigr\Vert ^{2}+\bigl(s-s'\bigr)^{2} \\ \le{}& \Vert x-p \Vert ^{2}+\bigl(r-r' \bigr)^{2}+ \Vert y-q \Vert ^{2}+\bigl(s-s' \bigr)^{2}-r\bigl( \Vert x-T_{1}x \Vert ^{2}+ \Vert y-T_{2}y \Vert ^{2}\bigr) \\ ={}& \Vert \boldsymbol{x-p} \Vert ^{2}-r \Vert \boldsymbol{x-Sx} \Vert ^{2}. \end{aligned}$$

The proof is complete. □

Proposition 10

If $g_{1},g_{2}:\mathcal{H}\to \mathcal{H}$ are α-Lipschitzian, then $\boldsymbol{h}:\boldsymbol{\mathcal{H}}\to \boldsymbol{\mathcal{H}}$ defined by

$$ \boldsymbol{h}(\boldsymbol{x}) := \bigl(\bigl(g_{2}(y),\alpha s\bigr), \bigl(g_{1}(x), \alpha r\bigr) \bigr) \quad \textit{for all }\boldsymbol{x}:= \bigl((x,r),(y,s)\bigr)\in \boldsymbol{\mathcal{H}} $$

is also α-Lipschitzian.

Proof

To see this, let $\boldsymbol{x}:=((x,r),(y,s))$, $\boldsymbol{x'}:=((x',r'),(y',s'))\in \boldsymbol{\mathcal{H}}$. It follows that

$$\begin{aligned} \bigl\Vert \boldsymbol{h}(\boldsymbol{x})-\boldsymbol{h}\bigl( \boldsymbol{x'}\bigr) \bigr\Vert ^{2} &= \bigl\Vert \bigl(\bigl(g_{2}(y)-g_{2}\bigl(y'\bigr),\alpha \bigl(s-s'\bigr)\bigr),\bigl(g_{1}(x)-g_{1} \bigl(x'\bigr), \alpha \bigl(r-r'\bigr)\bigr) \bigr) \bigr\Vert ^{2} \\ &= \bigl\Vert g_{2}(y)-g_{2}\bigl(y'\bigr) \bigr\Vert ^{2}+\alpha ^{2}\bigl(s-s' \bigr)^{2})+ \bigl\Vert g_{1}(x)-g_{1} \bigl(x'\bigr) \bigr\Vert ^{2}+\alpha ^{2} \bigl(r-r'\bigr)^{2} \\ &\le \alpha ^{2} \bigl( \bigl\Vert y-y' \bigr\Vert ^{2}+\bigl(s-s'\bigr)^{2}+ \bigl\Vert x-x' \bigr\Vert ^{2}+\bigl(r-r' \bigr)^{2} \bigr) \\ &=\alpha ^{2} \bigl\Vert \boldsymbol{x}-\boldsymbol{x'} \bigr\Vert ^{2}. \end{aligned}$$

This completes the proof. □

The intermixed algorithm can be regarded as a classical algorithm of Theorem 3, and we obtain the following convergence theorem.

Theorem 11

Let $\boldsymbol{U}:=\boldsymbol{P_{E}}(\boldsymbol{\operatorname{Id}}- \lambda \boldsymbol{A})$ and $\boldsymbol{F}:=\operatorname{VI}(\boldsymbol{E,A})\cap \operatorname{Fix}(\boldsymbol{S})$. Suppose that $\boldsymbol{x}_{1}\in \boldsymbol{C}$ is arbitrarily chosen and

$$ \boldsymbol{x_{n+1}}:=(1-\beta _{n})\boldsymbol{Sx_{n}}+ \beta _{n} \boldsymbol{P_{C}}\bigl(\alpha _{n} \boldsymbol{h}(\boldsymbol{x_{n}})+(1- \alpha _{n}) \boldsymbol{Ux_{n}}\bigr), $$

where the sequences $\{\alpha _{n}\}_{n=1}^{\infty},\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]$ satisfy the following conditions:

$$ \lim_{n}\frac{\alpha _{n}}{1-\beta _{n}}=0 \quad \textit{and}\quad \sum _{n} \alpha _{n}\beta _{n}=\infty . $$

Then the iterative sequence $\{\boldsymbol{x_{n}}\}_{n=1}^{\infty}$ converges to $\boldsymbol{z}=\boldsymbol{P_{F}\circ h}(\boldsymbol{z})$.

Remark 12

Our result is simultaneously a correction and an improvement of Theorem KK in the following ways.

• (1) We use a product space approach to consider the mixed variational inequality problem and the intermixed algorithm.

• (2) The resolvent proposed for the mixed variational inequality problem in the original work is not correct, and we propose a correction.

• (3) The assumptions on the parameters $\{\alpha _{n}\}_{n=1}^{\infty}$ and $\{\beta _{n}\}_{n=1}^{\infty}$ are more general than those in Theorem KK. Moreover, Condition (C3) is superfluous. The choice $\alpha _{n}=\beta _{n}:=1/\sqrt{n}$ is applicable in our result, but it is not in Theorem KK.

Finally, we express the iterative sequence in our Theorem 11 as follows:

$$\begin{aligned} &(x_{1},r_{1}),(y_{1},s_{1})\in C\times \mathbb{R} \text{are arbitrarily chosen}, \\ &\bigl(x'_{n},r'_{n}\bigr):= \bigl(b_{1}x_{n}+(1-b_{1})T_{1}x_{n},r_{n} \bigr), \\ &\bigl(y'_{n},s'_{n}\bigr):= \bigl(b_{2}y_{n}+(1-b_{2})T_{2}y_{n},s_{n} \bigr), \\ &\bigl(x''_{n},r''_{n} \bigr):=P_{\operatorname{epi}f_{1}}(x_{n}-\lambda _{n}A_{1}x_{n},r_{n}- \lambda _{n}), \\ &\bigl(y''_{n},s''_{n} \bigr):=P_{\operatorname{epi}f_{2}}(x_{n}-\lambda _{n}A_{2}x_{n},s_{n}- \lambda _{n}), \\ &(x_{n+1},r_{n+1}):= \bigl((1-\beta _{n})x'_{n}+ \beta _{n}P_{C} \bigl( \alpha _{n}h(x_{n})+(1- \alpha _{n})x''_{n} \bigr), \\ &\hphantom{(x_{n+1},r_{n+1}):=}{} (1- \beta _{n})r'_{n}+ \beta _{n}\bigl( \alpha _{n}\alpha r_{n}+(1-\alpha _{n})r''_{n} \bigr) \bigr), \\ &(y_{n+1},s_{n+1}):= \bigl((1-\beta _{n})y'_{n}+ \beta _{n}P_{C} \bigl( \alpha _{n}h(y_{n})+(1- \alpha _{n})y''_{n} \bigr), \\ &\hphantom{(y_{n+1},s_{n+1}):=} (1- \beta _{n})s'_{n}+ \beta _{n}\bigl( \alpha _{n}\alpha s_{n}+(1-\alpha _{n})s''_{n} \bigr) \bigr). \end{aligned}$$

For more detail on epigraphical projection, we refer to the book of Bauschke and Combettes [4]. It follows from our Theorem 11 that $\{x_{n}\}_{n=1}^{\infty}$ and $\{y_{n}\}_{n=1}^{\infty}$ converge strongly to $x^{*}$ and $y^{*}$, respectively, where $x^{*}=P_{\operatorname{VI}(C,A_{1},f_{1})\cap \operatorname{Fix}(T_{1})}g_{2}(y^{*})$ and $y^{*}=P_{\operatorname{VI}(C,A_{2},f_{2})\cap \operatorname{Fix}(T_{2})}g_{1}(x^{*})$.