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On the intermixed method for mixed variational inequality problems: another look and some corrections
Journal of Inequalities and Applications volume 2024, Article number: 42 (2024)
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
where ∂f is the subdifferential operator of f, that is,
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 \(i=1,2\),
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:
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
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
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:
where the sequences \(\{\alpha _{n}\}_{n=1}^{\infty},\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]\) satisfy the following conditions:
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
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
We refine the preceding estimates by considering \(\|\cdot \|^{2}\) as follows:
and
It follows that
Since \(\sum_{n}\alpha _{n}\beta _{n}=\infty \), we have
where
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
In particular, the sequences
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
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
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:
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
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:
It follows from \(\sum_{n}\alpha _{n}\beta _{n}=\infty \) that
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:
where the sequences \(\{\alpha _{n}\}_{n=1}^{\infty},\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]\) satisfy the following conditions:
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
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
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
Note that \(f(x)\le r\). This implies that \(\langle Ax,y-x\rangle +f(y)-f(x)\ge 0\). Moreover, we have
This implies that \(f(x)\ge r\), and hence \(r=f(x)\). In particular, we have
(⇐) 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
(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
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
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
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
To deduce and correct the conclusion in Theorem KK, let us fix the following notation.
Let
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
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
Define \(\boldsymbol{A}:\boldsymbol{C}\to \boldsymbol{\mathcal{H}}\) and \(\boldsymbol{S}:\boldsymbol{C}\to \boldsymbol{C}\) by
and
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
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
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
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
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
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
where the sequences \(\{\alpha _{n}\}_{n=1}^{\infty},\{\beta _{n}\}_{n=1}^{\infty}\subset [0,1]\) satisfy the following conditions:
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:
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 \(y^{*}=P_{\operatorname{VI}(C,A_{2},f_{2})\cap \operatorname{Fix}(T_{2})}g_{1}(x^{*})\).
Data Availability
No datasets were generated or analysed during the current study.
References
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The research was supported by the Fundamental Fund of Khon Kaen University and the National Science, Research and Innovation Fund (NSRF), Thailand.
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Saejung, S. On the intermixed method for mixed variational inequality problems: another look and some corrections. J Inequal Appl 2024, 42 (2024). https://doi.org/10.1186/s13660-024-03123-3
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DOI: https://doi.org/10.1186/s13660-024-03123-3