In this section, based on Censor, Gibali and Reich’s recent work and combining it with Byrne’s CQ algorithm and Korpelevich’s extragradient method, we propose a new iterative method for solving a class of the SVIP under weaker conditions in a Hilbert space.
First, we consider a class of the generalized split feasibility problems (GSFP): finding an element that solves a variational inequality problem such that its image under a given bounded linear operator is in a fixed point set of a nonexpansive mapping, i.e., find \(x^{*}\) satisfying
$$ x^{*}\in \operatorname {VI}(C,f) \quad \text{and} \quad Ax^{*}\in \operatorname {Fix}(T). $$
(3.1)
Theorem 3.1
Let
\(H_{1}\)
and
\(H_{2}\)
be real Hilbert spaces. Let
C
be a nonempty closed convex subset of
\(H_{1}\). Let
\(A:H_{1}\rightarrow H_{2}\)
be a bounded linear operator such that
\(A\neq0\), \(f:C\rightarrow H_{1}\)
be a monotone and
kLipschitz continuous mapping and
\(T:H_{2}\rightarrow H_{2}\)
be a nonexpansive mapping. Setting
\(\Gamma=\{z\in \operatorname {VI}(C,f): Az\in \operatorname {Fix}(T)\}\), assume that
\(\Gamma\neq\emptyset\). Let the sequences
\(\{x_{n}\}\), \(\{y_{n}\}\)
and
\(\{t_{n}\}\)
be generated by
\(x_{1}=x\in C\)
and
$$ \textstyle\begin{cases} y_{n}=P_{C}(x_{n}\gamma_{n}A^{*}(IT)Ax_{n}),\\ t_{n}=P_{C}(y_{n}\lambda_{n}f(y_{n})),\\ x_{n+1}=P_{C}(y_{n}\lambda_{n}f(t_{n})), \end{cases} $$
(3.2)
for each
\(n\in\mathbb{N}\), where
\(\{\gamma_{n}\}\subset[a,b]\)
for some
\(a,b\in(0,\frac{1}{ \Vert A \Vert ^{2}})\)
and
\(\{\lambda_{n}\} \subset[c,d]\)
for some
\(c,d\in(0,\frac{1}{k})\). Then the sequence
\(\{x_{n}\}\)
converges weakly to a point
\(z\in\Gamma\), where
\(z=\lim_{n\rightarrow \infty} P_{\Gamma}x_{n}\).
Proof
From Lemma 2.2(ii), (iii), (iv) and Lemma 2.3, we can easily know that \(P_{C}(I\gamma_{n}A^{*}(IT)A)\) is \(\frac{1+\gamma_{n} \Vert A \Vert ^{2}}{2}\)averaged. So, \(y_{n}\) can be rewritten as
$$ y_{n}=(1\alpha_{n})x_{n}+ \alpha_{n}V_{n}x_{n}, $$
(3.3)
where \(\alpha_{n}=\frac{1+\gamma_{n} \Vert A \Vert ^{2}}{2}\) and \(V_{n}\) is a nonexpansive mapping for each \(n\in\mathbb{N}\).
Let \(u\in\Gamma\), we have
$$\begin{aligned}& \Vert y_{n}u \Vert ^{2} \\& \quad = \bigl\Vert (1\alpha_{n}) (x_{n}u)+ \alpha_{n}(V_{n}x_{n}u) \bigr\Vert ^{2} \\& \quad = (1\alpha_{n}) \Vert x_{n}u \Vert ^{2}+ \alpha_{n} \Vert V_{n}x_{n}u \Vert ^{2} \alpha_{n}(1\alpha_{n}) \Vert x_{n}V_{n}x_{n} \Vert ^{2} \\& \quad \leq \Vert x_{n}u \Vert ^{2} \alpha_{n}(1 \alpha_{n}) \Vert x_{n}V_{n}x_{n} \Vert ^{2} \\& \quad \leq \Vert x_{n}u \Vert ^{2}. \end{aligned}$$
(3.4)
So, we obtain that
$$ \alpha_{n}(1\alpha_{n}) \Vert x_{n}V_{n}x_{n} \Vert ^{2}\leq \Vert x_{n}u \Vert ^{2} \Vert y_{n}u \Vert ^{2}. $$
(3.5)
On the other hand, from Lemma 2.5, we have
$$\begin{aligned}& \Vert x_{n+1}u \Vert ^{2} \\& \quad \leq \bigl\Vert y_{n}\lambda_{n}f(t_{n})u \bigr\Vert ^{2} \bigl\Vert y_{n}\lambda_{n}f(t_{n})x_{n+1} \bigr\Vert ^{2} \\& \quad = \Vert y_{n}u \Vert ^{2} \Vert y_{n}x_{n+1} \Vert ^{2}+2\lambda_{n} \bigl\langle f(t_{n}), ux_{n+1} \bigr\rangle \\& \quad = \Vert y_{n}u \Vert ^{2} \Vert y_{n}x_{n+1} \Vert ^{2}+2\lambda_{n} \bigl( \bigl\langle f(t_{n}), ut_{n} \bigr\rangle \\& \quad \quad{} + \bigl\langle f(t_{n}), t_{n}x_{n+1} \bigr\rangle \bigr) \\& \quad = \Vert y_{n}u \Vert ^{2} \Vert y_{n}x_{n+1} \Vert ^{2}+2\lambda_{n} \bigl( \bigl\langle f(t_{n})f(u), ut_{n} \bigr\rangle \\& \quad \quad{} + \bigl\langle f(u), ut_{n} \bigr\rangle + \bigl\langle f(t_{n}), t_{n}x_{n+1} \bigr\rangle \bigr) \\& \quad \leq \Vert y_{n}u \Vert ^{2} \Vert y_{n}x_{n+1} \Vert ^{2}+2\lambda_{n} \bigl\langle f(t_{n}), t_{n}x_{n+1} \bigr\rangle \\& \quad = \Vert y_{n}u \Vert ^{2} \Vert y_{n}t_{n} \Vert ^{2}2\langle y_{n}t_{n}, t_{n}x_{n+1}\rangle \\& \quad\quad{}  \Vert t_{n}x_{n+1} \Vert ^{2}+2 \lambda_{n} \bigl\langle f(t_{n}), t_{n}x_{n+1} \bigr\rangle \\& \quad = \Vert y_{n}u \Vert ^{2} \Vert y_{n}t_{n} \Vert ^{2} \Vert t_{n}x_{n+1} \Vert ^{2} \\& \quad\quad{} +2 \bigl\langle y_{n}\lambda_{n}f(t_{n})t_{n}, x_{n+1}t_{n} \bigr\rangle . \end{aligned}$$
(3.6)
Then, from Lemma 2.4, we obtain that
$$\begin{aligned}& \bigl\langle y_{n}\lambda_{n}f(t_{n})t_{n}, x_{n+1}t_{n} \bigr\rangle \\& \quad = \bigl\langle y_{n}\lambda_{n}f(y_{n})t_{n}, x_{n+1}t_{n} \bigr\rangle + \bigl\langle \lambda_{n}f(y_{n}) \lambda_{n}f(t_{n}), x_{n+1}t_{n} \bigr\rangle \\& \quad \leq \bigl\langle \lambda_{n}f(y_{n}) \lambda_{n}f(t_{n}), x_{n+1}t_{n} \bigr\rangle \\& \quad = \lambda_{n} \bigl\langle f(y_{n})f(t_{n}), x_{n+1}t_{n} \bigr\rangle \\& \quad \leq \lambda_{n} \bigl\Vert f(y_{n})f(t_{n}) \bigr\Vert \Vert x_{n+1}t_{n} \Vert \\& \quad \leq \lambda_{n}k \Vert y_{n}t_{n} \Vert \Vert x_{n+1}t_{n} \Vert . \end{aligned}$$
(3.7)
So, we have
$$\begin{aligned}& \Vert x_{n+1}u \Vert ^{2} \\& \quad \leq \Vert y_{n}u \Vert ^{2} \Vert y_{n}t_{n} \Vert ^{2} \Vert t_{n}x_{n+1} \Vert ^{2} \\& \quad\quad{} +2\lambda_{n}k \Vert y_{n}t_{n} \Vert \Vert x_{n+1}t_{n} \Vert \\& \quad \leq \Vert y_{n}u \Vert ^{2} \Vert y_{n}t_{n} \Vert ^{2} \Vert t_{n}x_{n+1} \Vert ^{2} \\& \quad \quad{} +\lambda_{n}^{2}k^{2} \Vert y_{n}t_{n} \Vert ^{2}+ \Vert x_{n+1}t_{n} \Vert ^{2} \\& \quad \leq \Vert y_{n}u \Vert ^{2}+ \bigl( \lambda_{n}^{2}k^{2}1 \bigr) \Vert y_{n}t_{n} \Vert ^{2} \\& \quad \leq \Vert y_{n}u \Vert ^{2} \\& \quad \leq \Vert x_{n}u \Vert ^{2}. \end{aligned}$$
(3.8)
Therefore, there exists
$$ c=\lim_{n\rightarrow\infty} \Vert x_{n}u \Vert , $$
(3.9)
and the sequence \(\{x_{n}\}\) is bounded. From (3.5), we also get
$$ \alpha_{n}(1\alpha_{n}) \Vert x_{n}V_{n}x_{n} \Vert ^{2}\leq \Vert x_{n}u \Vert ^{2} \Vert x_{n+1}u \Vert ^{2}. $$
(3.10)
Hence
$$ x_{n}V_{n}x_{n}\rightarrow0, \quad n\rightarrow\infty. $$
(3.11)
From (3.3), we can get
$$ x_{n}y_{n}\rightarrow0, \quad n\rightarrow \infty. $$
(3.12)
From (3.8), we also get
$$ \Vert y_{n}t_{n} \Vert ^{2} \leq\frac{1}{1\lambda_{n}^{2}k^{2}} \bigl( \Vert x_{n}u \Vert ^{2} \Vert x_{n+1}u \Vert ^{2} \bigr). $$
(3.13)
So,
$$ y_{n}t_{n}\rightarrow0, \quad n\rightarrow \infty. $$
(3.14)
From (3.12) and (3.14), we have
$$ x_{n}t_{n}\rightarrow0, \quad n\rightarrow \infty. $$
(3.15)
Notice that \(P_{C}\) is firmly nonexpansive, so
$$\begin{aligned}& \Vert x_{n+1}t_{n} \Vert \\& \quad = \bigl\Vert P_{C} \bigl(y_{n} \lambda_{n}f(t_{n}) \bigr)P_{C} \bigl(y_{n}\lambda _{n}f(y_{n}) \bigr) \bigr\Vert \\& \quad \leq \bigl\Vert y_{n}\lambda_{n}f(t_{n})y_{n}+ \lambda _{n}f(y_{n}) \bigr\Vert \\& \quad = \bigl\Vert \lambda_{n}f(t_{n}) \lambda_{n}f(y_{n}) \bigr\Vert \\& \quad = \lambda_{n} \bigl\Vert f(t_{n})f(y_{n}) \bigr\Vert \\& \quad \leq \lambda_{n}k \Vert t_{n}y_{n} \Vert . \end{aligned}$$
(3.16)
We obtain
$$ x_{n+1}t_{n}\rightarrow0, \quad n\rightarrow \infty. $$
(3.17)
From (3.15) and (3.17), we have
$$ x_{n+1}x_{n}\rightarrow0, \quad n\rightarrow \infty. $$
(3.18)
Since \(\{x_{n}\}\) is bounded, for each \(z\in\omega_{w}(x_{n})\), there exists a subsequence \(\{x_{n_{i}}\}\) of \(\{x_{n}\}\) converging weakly to z. Without loss of generality, the subsequence \(\{\gamma_{n_{i}}\}\) of \(\{\gamma_{n}\}\) converges to a point \(\hat{\gamma}\in(0,\frac{1}{ \Vert A \Vert ^{2}})\). Since \(A^{*}(IT)A\) is inverse strongly monotone, we know that \(\{A^{*}(IT)Ax_{n_{i}}\}\) is bounded. Since \(P_{C}\) is firmly nonexpansive,
$$ \bigl\Vert P_{C} \bigl(I\gamma_{n_{i}}A^{*}(IT)A \bigr)x_{n_{i}}P_{C} \bigl(I\hat {\gamma}A^{*}(IT)A \bigr)x_{n_{i}} \bigr\Vert \leq \vert \gamma_{n_{i}}\hat{ \gamma} \vert \bigl\Vert A^{*}(IT)Ax_{n_{i}} \bigr\Vert . $$
From \(\gamma_{n_{i}}\rightarrow\hat{\gamma}\), we have
$$ P_{C} \bigl(I\gamma_{n_{i}}A^{*}(IT)A \bigr)x_{n_{i}}P_{C} \bigl(I\hat{\gamma }A^{*}(IT)A \bigr)x_{n_{i}} \rightarrow0, \quad i\rightarrow\infty. $$
(3.19)
From (3.12), we have
$$ x_{n_{i}}P_{C} \bigl(I\gamma_{n_{i}}A^{*}(IT)A \bigr)x_{n_{i}}\rightarrow0, \quad i\rightarrow\infty. $$
(3.20)
Since
$$\begin{aligned}& \bigl\Vert x_{n_{i}}P_{C} \bigl(I\hat{ \gamma}A^{*}(IT)A \bigr)x_{n_{i}} \bigr\Vert \\& \quad \leq \bigl\Vert x_{n_{i}}P_{C} \bigl(I\gamma _{n_{i}}A^{*}(IT)A \bigr)x_{n_{i}} \bigr\Vert \\& \quad\quad{} + \bigl\Vert P_{C} \bigl(I\gamma _{n_{i}}A^{*}(IT)A \bigr)x_{n_{i}}P_{C} \bigl(I\hat{\gamma }A^{*}(IT)A \bigr)x_{n_{i}} \bigr\Vert , \end{aligned}$$
we have
$$ x_{n_{i}}P_{C} \bigl(I\hat{ \gamma}A^{*}(IT)A \bigr)x_{n_{i}}\rightarrow0, \quad i \rightarrow \infty. $$
(3.21)
From Lemma 2.10, we obtain that
$$ z\in \operatorname {Fix}\bigl(P_{C} \bigl(I\hat{\gamma}A^{*}(IT)A \bigr) \bigr). $$
From Corollary 2.9, we obtain
$$ z\in C\cap A^{1}\operatorname {Fix}(T). $$
(3.22)
Now, we show that \(z\in \operatorname {VI}(C,f)\). From (3.12), (3.15) and (3.18), we have \(y_{n_{i}}\rightharpoonup z\), \(t_{n_{i}}\rightharpoonup z\) and \(x_{n_{i}+1}\rightharpoonup z\).
Let
$$ Bv= \textstyle\begin{cases} f(v)+N_{C}v, &\forall v\in C,\\ \emptyset,&\forall v\notin C. \end{cases} $$
From Lemma 2.7, we know that B is maximal monotone and \(0\in Bv\) if and only if \(v\in \operatorname {VI}(C,f)\).
For each \((v,w)\in G(B)\), we have
$$ w\in Bv=f(v)+N_{C}v. $$
Hence
So, we obtain that
$$ \bigl\langle vp, wf(v) \bigr\rangle \geq0, \quad\forall p\in C. $$
(3.23)
On the other hand, from \(v\in C\) and
$$ x_{n+1}=P_{C} \bigl(y_{n}\lambda_{n}f(t_{n}) \bigr), $$
we get
$$ \bigl\langle y_{n}\lambda_{n}f(t_{n})x_{n+1}, x_{n+1}v \bigr\rangle \geq0 $$
and hence
$$ \biggl\langle vx_{n+1}, \frac{x_{n+1}y_{n}}{\lambda_{n}}+f(t_{n}) \biggr\rangle \geq0. $$
(3.24)
Therefore, from (3.23) and (3.24), we obtain that
$$\begin{aligned}& \langle vx_{n_{i}+1}, w\rangle \\& \quad \geq \bigl\langle vx_{n_{i}+1}, f(v) \bigr\rangle \\& \quad \geq \bigl\langle vx_{n_{i}+1}, f(v) \bigr\rangle  \biggl\langle vx_{n_{i}+1}, \frac{x_{n_{i}+1}y_{n_{i}}}{\lambda _{n_{i}}}+f(t_{n_{i}}) \biggr\rangle \\& \quad = \bigl\langle vx_{n_{i}+1}, f(v)f(t_{n_{i}}) \bigr\rangle  \biggl\langle vx_{n_{i}+1}, \frac{x_{n_{i}+1}y_{n_{i}}}{\lambda _{n_{i}}} \biggr\rangle \\& \quad = \bigl\langle vx_{n_{i}+1}, f(v)f(x_{n_{i}+1}) \bigr\rangle + \bigl\langle vx_{n_{i}+1}, f(x_{n_{i}+1})f(t_{n_{i}}) \bigr\rangle \\& \quad\quad{}  \biggl\langle vx_{n_{i}+1}, \frac{x_{n_{i}+1}y_{n_{i}}}{\lambda _{n_{i}}} \biggr\rangle \\& \quad \geq \bigl\langle vx_{n_{i}+1}, f(x_{n_{i}+1})f(t_{n_{i}}) \bigr\rangle  \biggl\langle vx_{n_{i}+1}, \frac{x_{n_{i}+1}y_{n_{i}}}{\lambda_{n_{i}}} \biggr\rangle . \end{aligned}$$
(3.25)
As \(i\rightarrow\infty\), we have
$$ \langle vz, w\rangle\geq0. $$
(3.26)
Since B is maximal monotone, we have \(0\in Bz\) and hence \(z\in \operatorname {VI}(C,f)\). So, we obtain that
$$ \omega_{w}(x_{n})\subset\Gamma. $$
(3.27)
From Lemma 2.11, we get
$$ x_{n}\rightharpoonup z\in\Gamma, $$
(3.28)
and from Lemma 2.12, we obtain
$$ z=\lim_{n\rightarrow\infty}P_{\Gamma}x_{n}. $$
(3.29)
□
Remark 3.2

(i)
If \(f=0\) and \(T=P_{Q}\), problem (3.1) reduces to the SFP introduced by Censor and Elfving, and the algorithm in Theorem 3.1 reduces to Byrne’s CQ algorithm for solving the SFP.

(ii)
If \(T=I\), problem (3.1) reduces to the VIP, and the algorithm in Theorem 3.1 reduces to Korpelevich’s extragradient method for solving the VIP with a monotone mapping.

(iii)
If \(A=I\), \(f=0\) and \(C=H_{1}\), problem (3.1) reduces to the fixed point problem of a nonexpansive mapping.
Now, we consider SVIP (1.6).
Theorem 3.3
Let
\(H_{1}\)
and
\(H_{2}\)
be real Hilbert spaces. Let
C
be a nonempty closed convex subset of
\(H_{1}\)
and
Q
be a nonempty closed convex subset of
\(H_{2}\). Let
\(A:H_{1}\rightarrow H_{2}\)
be a bounded linear operator such that
\(A\neq0\), let
\(f:C\rightarrow H_{1}\)
be a monotone and
kLipschitz continuous mapping and
\(g:H_{2}\rightarrow H_{2}\)
be an
αinverse strongly monotone mapping. Setting
\(\Gamma=\{z\in \operatorname {VI}(C,f): Az\in \operatorname {VI}(Q,g)\}\), assume that
\(\Gamma\neq\emptyset\). Let the sequences
\(\{x_{n}\}\), \(\{y_{n}\}\)
and
\(\{t_{n}\}\)
be generated by
\(x_{1}=x\in C\)
and
$$ \textstyle\begin{cases} y_{n}=P_{C}(x_{n}\gamma_{n}A^{*}(IP_{Q}(I\mu g))Ax_{n}),\\ t_{n}=P_{C}(y_{n}\lambda_{n}f(y_{n})),\\ x_{n+1}=P_{C}(y_{n}\lambda_{n}f(t_{n})), \end{cases} $$
(3.30)
for each
\(n\in\mathbb{N}\), where
\(\{\gamma_{n}\}\subset[a,b]\)
for some
\(a,b\in(0,\frac{1}{ \Vert A \Vert ^{2}})\), \(\{\lambda_{n}\} \subset[c,d]\)
for some
\(c,d\in(0,\frac{1}{k})\), \(\mu\in(0,2\alpha)\). Then the sequence
\(\{x_{n}\}\)
converges weakly to a point
\(z\in\Gamma\), where
\(z=\lim_{n\rightarrow \infty} P_{\Gamma}x_{n}\).
Proof
From Lemma 2.4, we can easily know that \(z\in \operatorname {VI}(Q,g)\) if and only if \(z=P_{Q}(I\mu g)z\) for \(\mu>0\), and for \(\mu\in(0,2\alpha)\), \(P_{Q}(I\mu g)\) is nonexpansive. Putting \(T=P_{Q}(I\mu g)\) in Theorem 3.1, we get the desired result. □
Remark 3.4

(i)
If \(f=0\) and \(g=0\), SVIP (1.6) reduces to the SFP introduced by Censor and Elfving, and the algorithm in Theorem 3.3 reduces to Byrne’s CQ algorithm for solving the SFP.

(ii)
If \(f=0\), \(C=H_{1}\) and \(A=I\), SVIP (1.6) reduces to the VIP and the algorithm in Theorem 3.3 reduces to (1.2) for solving the VIP with an inverse strongly monotone mapping.

(iii)
If \(g=0\) and \(Q=H_{2}\), SVIP (1.6) reduces to the VIP and the algorithm in Theorem 3.1 reduces to Korpelevich’s extragradient method for solving the VIP with a monotone mapping.