Algorithm 4.1 (ACPM) can be well implemented if the structure of the set C is very simple and the projection operator \(P_{C}\) is easy to calculate. However, the calculation of a projection onto a closed convex subset is generally difficult. To overcome this difficulty, Fukushima [13] suggested a relaxation projection method to calculate the projection onto a level set of a convex function by computing a sequence of projections onto half-spaces containing the original level set. Since its inception, the relaxation technique has received much attention and has been used by lots of authors to construct iterative algorithms for solving nonlinear problems, see [25] and the references therein.
We now consider the inverse variational inequality problem \(\operatorname{IVI}(C,f)\), where \(f: \mathcal{H}\rightarrow \mathcal{H}\) is a Lipschitz continuous and strongly monotone operator. Let the closed convex subset C be the level set of a convex function, i.e.,
$$ C=\bigl\{ z\in \mathcal{H}\mid c(x)\leq 0\bigr\} , $$
(22)
where \(c: \mathcal{H}\rightarrow \mathbb{R}\) is a convex function. We always assume that c is weakly lower semi-continuous, subdifferentiable on \(\mathcal{H}\), and ∂c is a bounded operator (i.e., bounded on bounded sets). It is worth noting that the subdifferential operator is bounded for a convex function defined on a finite dimensional Hilbert space (see [3, Corollary 7.9]).
Take the constants \(\tilde{L}=\frac{1}{\eta }\), \(\tilde{\eta }=\frac{\eta }{L^{2}}\), μ, and α and the sequence of positive numbers \(\{\varepsilon _{n}\}_{n=0}^{\infty }\) as in the last section.
Adopting the relaxation technique of Fukushima [13], we introduce a relaxed projection algorithm for computing the unique solution \(\xi ^{*}\) of \(\operatorname{IVI}(C,f)\), where C is given as in (22).
Algorithm 5.1
(The alternating contraction relaxation projection method)
-
Step 1:
Take \(u_{0}\in C\) and \(\xi _{0}^{(0)}\in \mathcal{H}\) arbitrarily and set \(n:=0\).
-
Step 2:
For the current \(u_{n}\) and \(\xi _{n}^{(0)}\) (\(n\geq 0\)), calculate
$$ \xi ^{(m+1)}_{n}=\xi ^{(m)}_{n}- \alpha f\bigl(\xi ^{(m)}_{n} \bigr)+\alpha u_{n}, \quad m=0,1,\ldots,m _{n}, $$
(23)
where \(m_{n}\) is the smallest positive integer such that
$$ \frac{(1-\tau )^{m_{n}+1}}{\tau } \bigl\Vert \xi ^{(1)}_{n}- \xi ^{(0)}_{n} \bigr\Vert \leq \varepsilon _{n}, $$
(24)
where \(\tau =\frac{1}{2}\alpha (2\eta -\alpha L^{2})\).
Set
$$ \xi _{n}=\xi ^{(m_{n}+1)}_{n}. $$
(25)
-
Step 3:
Calculate
$$ u_{n+1}=P_{C_{n}}(u_{n}-\lambda _{n}\mu \xi _{n}), $$
(26)
where
$$\begin{aligned}& \begin{aligned}[b] &C_{n}=\bigl\{ z\in \mathcal{H}\mid c(u_{n})+ \langle \nu _{n}, z-u_{n}\rangle \leq 0\bigr\} , \\&\quad \nu _{n}\in \partial c(u_{n}) \quad \text{and} \quad \lambda _{n}\in (0, 1).\end{aligned} \end{aligned}$$
(27)
Set
$$ \xi ^{(0)}_{n+1}=\xi _{n}, $$
\(n:=n+1\) and return to Step 2.
Next theorem establishes the strong convergence of Algorithm 4.1.
Theorem 5.1
Let
C
be given by (22), and let
\(f: \mathcal{H}\rightarrow \mathcal{H}\)
be an
L-Lipschitz continuous and
η-strongly monotone operator. Assume that
\(c: \mathcal{H}\rightarrow \mathbb{R}\)
is weakly lower semi-continuous and subdifferentiable on
\(\mathcal{H}\)
and
∂c
is a bounded operator. Suppose that the sequence
\(\{\lambda _{n}\}_{n=0}^{\infty }\subset (0, 1)\)
satisfies (i) \(\lambda _{n}\rightarrow 0\)
as
\(n\rightarrow \infty \)
and (ii) \(\sum_{n=0}^{\infty }\lambda _{n}=\infty \). Then the two sequences
\(\{\xi _{n}\}_{n=0}^{\infty }\)
and
\(\{u_{n}\}_{n=0}^{ \infty }\)
generated by Algorithm 5.1
converge strongly to the unique solution
\(\xi ^{*} \)
of
\(\operatorname{IVI}(C,f)\)
and the unique solution
\(u^{*}\)
of
\(\operatorname{VI}(C,f^{-1})\), respectively.
Proof
For convenience, we denote by M a positive constant, which represents different values in different places. Firstly, we verify that \(\{u_{n}\}_{n=0}^{\infty }\) is bounded. Indeed, from the subdifferential inequality (7) and the definition of \(C_{n}\), it is easy to verify that \(C_{n}\supset C\) for all \(n\geq 0\). Similar to (15), we also have
$$ \bigl\Vert \xi _{n}-f^{-1}(u_{n}) \bigr\Vert \leq \varepsilon _{n}, \quad n\geq 0. $$
(28)
Noting that the projection operator \(P_{C_{n}}\) is nonexpansive, we obtain from (26), (28), and Lemma 2.4 that
$$ \begin{aligned} \bigl\Vert u_{n+1}-u^{*} \bigr\Vert &= \bigl\Vert P_{C_{n}}(u_{n}-\lambda _{n} \mu \xi _{n})-P _{C_{n}}u^{*} \bigr\Vert \\ &= \bigl\Vert P_{C_{n}}(u_{n}-\lambda _{n}\mu \xi _{n})-P_{C _{n}}\bigl(u_{n}-\lambda _{n} \mu f^{-1}(u_{n})\bigr) \bigr\Vert \\ &\quad {} + \bigl\Vert P_{C_{n}}\bigl(u_{n}- \lambda _{n} \mu f^{-1}(u_{n})\bigr)-P_{C_{n}}u^{*} \bigr\Vert \\ &\leq \bigl\Vert P_{C_{n}}\bigl(u _{n}-\lambda _{n}\mu f^{-1}(u_{n})\bigr)-P_{C_{n}} \bigl(u^{*}-\lambda _{n}\mu f ^{-1} \bigl(u^{*}\bigr)\bigr) \bigr\Vert \\ & \quad {} + \bigl\Vert P_{C_{n}}\bigl(u^{*}-\lambda _{n}\mu f^{-1}\bigl(u^{*} \bigr)\bigr)-P_{C_{n}}u^{*} \bigr\Vert +\lambda _{n}\mu \varepsilon _{n} \\ &\leq (1- \tilde{\tau }\lambda _{n}) \bigl\Vert u_{n}-u^{*} \bigr\Vert +\tilde{\tau }\lambda _{n}\frac{ \mu }{\tilde{\tau }}\bigl(\varepsilon _{n}+ \bigl\Vert f^{-1}\bigl(u^{*}\bigr) \bigr\Vert \bigr) \\ &\leq (1- \tilde{\tau }\lambda _{n}) \bigl\Vert u_{n}-u^{*} \bigr\Vert +\tilde{\tau }\lambda _{n}\frac{ \mu }{\tilde{\tau }}M, \end{aligned} $$
where \(\tilde{\tau }=\frac{1}{2}\mu (2\tilde{\eta }-\mu \tilde{L}^{2})\). Inductively, it turns out that
$$ \Vert u_{n}-u \Vert \leq \max \biggl\{ \Vert u_{0}-u \Vert ,\frac{\mu }{\tilde{\tau }} M\biggr\} , \quad \forall n\geq 1, $$
which implies that \(\{u_{n}\}_{n=0}^{\infty }\) is bounded and so is \(\{f^{-1}(u_{n})\}_{n=0}^{\infty }\). Similar to (17), we have
$$ \begin{aligned}[b] \bigl\Vert \xi _{n}- \xi ^{*} \bigr\Vert &\leq \bigl\Vert \xi _{n}-f^{-1}(u_{n}) \bigr\Vert + \bigl\Vert f^{-1}(u_{n})-\xi ^{*} \bigr\Vert \\ &\leq \varepsilon _{n}+ \bigl\Vert f^{-1}(u_{n})-f^{-1} \bigl(u^{*}\bigr) \bigr\Vert \\ & \leq \varepsilon _{n}+\tilde{L} \bigl\Vert u_{n}-u^{*} \bigr\Vert , \end{aligned} $$
(29)
which implies that \(\{\xi _{n}\}_{n=0}^{\infty }\) is bounded. Using (26), (28), Lemma 2.1, and Lemma 2.4, we have
$$\begin{aligned} & \bigl\Vert u_{n+1}-u^{*} \bigr\Vert ^{2} \\ &\quad = \bigl\Vert P_{C_{n}}(u_{n}-\lambda _{n}\mu \xi _{n})-P _{C_{n}}u^{*} \bigr\Vert ^{2} \\ &\quad\leq \bigl\Vert u_{n}-\lambda _{n}\mu \xi _{n}-u^{*} \bigr\Vert ^{2} \\ &\quad= \bigl\Vert (u_{n}-\lambda _{n}\mu \xi _{n})-\bigl(u_{n}-\lambda _{n}\mu f^{-1}(u _{n})\bigr)+\bigl(u_{n}-\lambda _{n}\mu f^{-1}(u_{n})\bigr)-u^{*} \bigr\Vert ^{2} \\ &\quad\leq \bigl\Vert \bigl(u _{n}-\lambda _{n}\mu f^{-1}(u_{n})\bigr)-u^{*} \bigr\Vert ^{2}+2\lambda _{n}\mu \varepsilon _{n} \bigl\Vert u_{n}-\lambda _{n}\mu f^{-1}(u_{n})-u^{*} \bigr\Vert \\ &\quad\quad {} +\lambda ^{2}_{n}\mu ^{2}\varepsilon ^{2}_{n} \\ &\quad\leq \bigl\Vert \bigl(u_{n}-\lambda _{n}\mu f ^{-1}(u_{n})\bigr)-\bigl(u^{*}-\lambda _{n}\mu f^{-1}\bigl(u^{*}\bigr)\bigr)-\lambda _{n}\mu f ^{-1}\bigl(u^{*}\bigr) \bigr\Vert ^{2}+\lambda _{n}\varepsilon _{n} M \\ &\quad\leq (1- \tilde{\tau }\lambda _{n}) \bigl\Vert u_{n}-u^{*} \bigr\Vert ^{2}-2\lambda _{n}\mu \bigl\langle f^{-1}\bigl(u^{*} \bigr),u_{n}-u^{*}-\lambda _{n}\mu f^{-1}(u_{n})\bigr\rangle \\ &\quad\quad {} + \lambda _{n}\varepsilon _{n} M \\ &\quad=(1-\tilde{\tau }\lambda _{n}) \bigl\Vert u_{n}-u ^{*} \bigr\Vert ^{2}+\tilde{\tau }\lambda _{n} \biggl[\frac{1}{\tilde{\tau }}\bigl(-2\mu \bigl\langle f^{-1} \bigl(u^{*}\bigr),u_{n} -u^{*}\bigr\rangle \\ &\quad\quad {} +2\lambda _{n}\mu ^{2} \bigl\Vert f ^{-1} \bigl(u^{*}\bigr) \bigr\Vert \bigl\Vert f^{-1}(u_{n}) \bigr\Vert +\varepsilon _{n}M\bigr)\biggr]. \end{aligned}$$
(30)
Since the projection operator \(P_{C_{n}}\) is firmly nonexpansive, we get
$$ \bigl\Vert P_{C_{n}}u_{n}-P_{C_{n}}u^{*} \bigr\Vert ^{2}\leq \bigl\Vert u_{n}-u^{*} \bigr\Vert ^{2}- \Vert u_{n}-P _{C_{n}}u_{n} \Vert ^{2}. $$
(31)
Using (28) and (31), we have
$$ \begin{aligned}[b] & \bigl\Vert u_{n+1}-u^{*} \bigr\Vert ^{2} \\ &\quad = \bigl\Vert P_{C_{n}}(u_{n}-\lambda _{n}\mu \xi _{n})-P _{C_{n}}u^{*} \bigr\Vert ^{2} \\ &\quad= \bigl\Vert P_{C_{n}}(u_{n}-\lambda _{n}\mu \xi _{n})-P _{C_{n}}\bigl(u_{n}-\lambda _{n} \mu f^{-1}(u_{n})\bigr) \\ &\quad\quad {} +P_{C_{n}}\bigl(u_{n}- \lambda _{n}\mu f^{-1}(u_{n})\bigr)-P_{C_{n}}u^{*} \bigr\Vert ^{2} \\ &\quad\leq \bigl\Vert P_{C_{n}}\bigl(u _{n}-\lambda _{n}\mu f^{-1}(u_{n})\bigr)-P_{C_{n}}u^{*} \bigr\Vert ^{2}+2\lambda _{n} \mu \varepsilon _{n} \bigl\Vert u_{n}-u^{*}-\lambda _{n}\mu f^{-1}(u_{n}) \bigr\Vert \\ &\quad\quad {} + \lambda ^{2}_{n}\mu ^{2}\varepsilon ^{2}_{n} \\ &\quad\leq \bigl\Vert P_{C_{n}}\bigl(u_{n}- \lambda _{n}\mu f^{-1}(u_{n})\bigr)-P_{C_{n}}u_{n}+P_{C_{n}}u_{n}-P_{C_{n}}u ^{*} \bigr\Vert ^{2}+\lambda _{n} \varepsilon _{n}M \\ &\quad\leq \bigl\Vert P_{C_{n}}u _{n}-P_{C_{n}}u^{*} \bigr\Vert ^{2}+2\lambda _{n}\mu \bigl\Vert f^{-1}(u_{n}) \bigr\Vert \bigl\Vert u_{n}-u ^{*} \bigr\Vert + \lambda ^{2}_{n}\mu ^{2} \bigl\Vert f^{-1}(u_{n}) \bigr\Vert ^{2} \\ & \quad\quad {} +\lambda _{n} \varepsilon _{n}M \\ &\quad\leq \bigl\Vert P_{C_{n}}u_{n}-P_{C_{n}}u^{*} \bigr\Vert ^{2}+\lambda _{n}M \\ &\quad\leq \bigl\Vert u_{n}-u^{*} \bigr\Vert ^{2}- \Vert u_{n}-P_{C_{n}}u_{n} \Vert ^{2}+\lambda _{n}M. \end{aligned} $$
(32)
Setting
$$\begin{aligned}& s_{n}= \bigl\Vert u_{n}-u^{*} \bigr\Vert ^{2}, \quad\quad \gamma _{n}=\tilde{\tau } \lambda _{n}, \\& \delta _{n}=-\frac{1}{\tilde{\tau }}\bigl(2\mu \bigl\langle f^{-1}\bigl(u^{*}\bigr),u_{n}-u ^{*}\bigr\rangle \bigr)+\frac{1}{\tilde{\tau }}\bigl(2\lambda _{n}\mu ^{2} \bigl\Vert f^{-1}\bigl(u ^{*}\bigr) \bigr\Vert \bigl\Vert f^{-1}(u_{n}) \bigr\Vert + \varepsilon _{n}M \bigr), \\& \eta _{n}= \Vert u_{n}-P_{C_{n}}u_{n} \Vert ^{2}, \quad\quad \alpha _{n}= M\lambda _{n}, \end{aligned}$$
then (30) and (32) can be rewritten as the following forms, respectively:
$$ s_{n+1}\leq (1-\gamma _{n})s_{n}+\gamma _{n}\delta _{n}, \quad n\geq 0, $$
(33)
and
$$ s_{n+1}\leq s_{n}-\eta _{n}+\alpha _{n}, \quad n\geq 0. $$
(34)
From the conditions \(\lambda _{n}\rightarrow 0\) and \(\sum^{+\infty }_{n=1} \lambda _{n}=\infty \), it follows \(\alpha _{n} \rightarrow 0\) and \(\sum^{\infty }_{n=1} \gamma _{n}=\infty \). So, in order to use Lemma 2.6 to complete the proof, it suffices to verify that
$$ \lim_{k\rightarrow \infty }\eta _{n_{k}}=0 $$
implies
$$ \limsup_{k\rightarrow \infty }\delta _{n_{k}}\leq 0 $$
for any subsequence \(\{n_{k}\}_{k=0}^{\infty }\subset \{n\}_{n=0}^{ \infty }\). In fact, from \(\Vert u_{n_{k}}-P_{C_{n_{k}}}u _{n_{k}} \Vert \rightarrow 0\) and the fact that ∂c is bounded on bounded sets, it follows that there exists a constant \(\delta > 0 \) such that \(\Vert \nu _{n_{k}} \Vert \leq \delta \) for all \(k\geq 0\). Using (27) and the trivial fact that \(P_{C_{n_{k}}}u_{n_{k}} \in C_{n_{k}}\), we have
$$ c(u_{n_{k}})\leq \langle \nu _{n_{k}},u_{n_{k}}-P_{C_{n_{k}}}u_{n_{k}} \rangle \leq \delta \Vert u_{n_{k}}-P_{C_{n_{k}}}u_{n_{k}} \Vert . $$
(35)
For any \(u^{\prime }\in \omega _{\omega }(u_{n_{k}})\), without loss of generality, we assume that \(u_{n_{k}} \rightharpoonup u^{\prime }\). Using w-lsc of c and (35), we have
$$ c\bigl(u^{\prime }\bigr)\leq \liminf_{k\rightarrow \infty }c(u_{n_{k}}) \leq 0, $$
which implies that \(u^{\prime }\in C\). Hence \(\omega _{\omega }(u_{n_{k}})\subset C\).
Noting that \(u^{*}\) is the unique solution of \(\operatorname{VI}(C,f^{-1})\), it turns out that
$$ \begin{aligned} \limsup_{k\rightarrow \infty }\biggl\{ - \frac{2\mu }{\tau }\bigl\langle f^{-1}\bigl(u ^{*} \bigr),u_{n_{k}}-u^{*}\bigr\rangle \biggr\} = {}&-\frac{2\mu }{\tau } \liminf_{k\rightarrow \infty }\bigl\langle f^{-1}\bigl(u^{*} \bigr),u_{n_{k}}-u^{*} \bigr\rangle \\ ={}&-\frac{2\mu }{\tau } \inf_{\omega \in \omega _{\omega }(u_{n_{k}})}\bigl\langle f^{-1} \bigl(u^{*}\bigr), \omega -u^{*}\bigr\rangle \leq 0. \end{aligned} $$
Since \(\lambda _{n}\rightarrow 0\), \(\varepsilon _{n}\rightarrow 0\), and \(\{f^{-1}(u_{n})\}_{n=0}^{\infty }\) is bounded, it is easy to verify that \(\limsup_{k\rightarrow 0} \delta _{n_{k}}\leq 0\). Therefore, by using Lemma 2.6 we get that \(u_{n}\rightarrow u^{*}\) as \(n\rightarrow \infty \). Consequently, this together with (29) leads to \(\xi _{n}\rightarrow \xi ^{*}\) and the proof is completed. □
Next we estimate the convergence rate of Algorithm 5.1. Note that the conditions \(\lambda _{n}\rightarrow 0\) and \(\sum_{n=0}^{\infty }\lambda _{n}=\infty \) guarantee the strong convergence, but slow down the convergence rate. Since it is difficult to estimate the asymptotic convergence rate of Algorithm 5.1, we will focus on the convergence rate of Algorithm 5.1 in the non-asymptotic sense. Based on Lemma 1.1 and Theorem 5.1, estimating the convergence rate of Algorithm 5.1 for \(\operatorname{IVI}(C,f)\) is equivalent to estimating the convergence rate of Algorithm 5.1 for \(\operatorname{VI}(C,f^{-1})\), so we will analyze the convergence rate of Algorithm 5.1 for \(\operatorname{VI}(C,f^{-1})\).
The analysis of the convergence rate is based on the fundamental equivalence: a point \(u\in C\) is a solution of \(\operatorname{VI}(C,f^{-1})\) if and only if \(\langle f^{-1}(v),v-u\rangle \geq 0\) holds for all \(v\in C\cap S(u,1)\), where \(S(u,1)\) is the closed sphere with the center u and the radius one (see [4] and [10] for details).
A useful inequality for estimating the convergence rate of Algorithm 5.1 is given as follows.
Lemma 5.1
Let
\(\{u_{n}\}_{n=1}^{\infty }\)
be the sequence generated by Algorithm 5.1. Assume that the conditions in Theorem 5.1
hold. Suppose
\(\sum_{n=0}^{\infty }\lambda _{n}^{2}<\infty \)
and
\(\sum_{n=0}^{\infty }\lambda _{n}\varepsilon _{n}<\infty \). Then, for any integer
\(n\geq 1\), we have a sequence
\(\{w_{n}\}_{n=1}^{ \infty }\)
which converges strongly to the unique solution
\(u^{*}\)
of
\(\operatorname{VI}(C,f^{-1})\)
and
$$ \begin{aligned}[b] & \bigl\langle f^{-1}(v),w_{n}-v \bigr\rangle \\ &\quad \leq \frac{ \Vert u _{0}-v \Vert ^{2}+\mu ^{2}(\sigma _{1}+\sigma _{2})+2\mu (\sigma _{3}+\sigma _{4} \Vert v \Vert +\mu \sigma _{5})}{\varUpsilon _{n}}, \quad \forall v\in C, \end{aligned} $$
(36)
where
$$ \begin{gathered} \sigma _{1}= \sum_{k=0}^{\infty }\lambda _{k}^{2} \bigl\Vert f^{-1}(u_{k}) \bigr\Vert ^{2}, \quad \quad \sigma _{2}=\sum_{k=0}^{\infty } \lambda _{k}^{2}\varepsilon _{k}^{2}, \\ \sigma _{3}=\sum_{k=0}^{\infty } \lambda _{k}\varepsilon _{k} \Vert u_{k} \Vert , \quad\quad \sigma _{4}=\sum_{k=0}^{\infty } \lambda _{k}\varepsilon _{k}, \\ \sigma _{5}=\sum_{k=0}^{\infty } \lambda _{k}^{2}\varepsilon _{k} \bigl\Vert f^{-1}(u_{k}) \bigr\Vert , \quad\quad w_{n}= \frac{\sum_{k=0}^{n}2\mu \lambda _{k}u_{k}}{\varUpsilon _{n}},\quad \textit{and} \quad \varUpsilon _{n}=\sum _{k=0}^{n}2\mu \lambda _{k}. \end{gathered} $$
(37)
Proof
For each \(k\geq 0\) and any \(v\in C\), using (26) and (28), we have
$$ \begin{aligned}[b] \Vert u_{k+1}-v \Vert & = \bigl\Vert P_{C_{k}}(u_{k}-\lambda _{k}\mu \xi _{k})-P _{C_{k}}v \bigr\Vert \\ &\leq \Vert u_{k}-\lambda _{k}\mu \xi _{k}-v \Vert \\ &\leq \bigl\Vert u _{k}-v-\lambda _{k}\mu f^{-1}(u_{k})+\lambda _{k}\mu f^{-1}(u_{k})- \lambda _{k}\mu \xi _{k} \bigr\Vert \\ &\leq \bigl\Vert u_{k}-v-\lambda _{k}\mu f^{-1}(u _{k}) \bigr\Vert +\lambda _{k}\mu \varepsilon _{k}. \end{aligned} $$
Consequently, we obtain
$$ \begin{aligned}[b] & \Vert u_{k+1}-v \Vert ^{2} \\ &\quad \leq \Vert u_{k}-v \Vert ^{2}-2\lambda _{k}\mu \bigl\langle f^{-1}(u_{k}), u_{k}-v\bigr\rangle +\mu ^{2}\lambda _{k}^{2} \bigl\Vert f^{-1}(u_{k}) \bigr\Vert ^{2}+\mu ^{2}\lambda _{k}^{2}\varepsilon _{k}^{2} \\ &\quad\quad{} +2 \mu \lambda _{k}\varepsilon _{k} \Vert u_{k} \Vert +2\mu \lambda _{k}\varepsilon _{k} \Vert v \Vert +2\mu ^{2} \lambda _{k}^{2} \varepsilon _{k} \bigl\Vert f^{-1}(u_{k}) \bigr\Vert , \end{aligned} $$
(38)
which together with the monotonicity of \(f^{-1}\) yields
$$ \begin{aligned}[b] &2\lambda _{k}\mu \bigl\langle f^{-1}(v), u_{k}-v\bigr\rangle \\ &\quad \leq \Vert u _{k}-v \Vert ^{2}- \Vert u_{k+1}-v \Vert ^{2} +\mu ^{2}\lambda _{k} ^{2} \bigl\Vert f^{-1}(u_{k}) \bigr\Vert ^{2}+\mu ^{2}\lambda _{k}^{2} \varepsilon _{k}^{2} \\ &\quad\quad {} +2\mu \lambda _{k}\varepsilon _{k} \Vert u_{k} \Vert +2\mu \lambda _{k}\varepsilon _{k} \Vert v \Vert +2\mu ^{2}\lambda _{k}^{2} \varepsilon _{k} \bigl\Vert f^{-1}(u_{k}) \bigr\Vert . \end{aligned} $$
(39)
Note the fact that \(\{u_{k}\}_{k=0}^{\infty }\) and \(\{ \Vert f^{-1}(u_{k}) \Vert \}_{k=0}^{\infty }\) are all bounded. So, from the conditions \(\sum_{k=0}^{\infty }\lambda _{k}^{2}<\infty \) and \(\sum_{k=0}^{\infty }\lambda _{k}\varepsilon _{k}<\infty \), it follows that \(\sigma _{k}<\infty \), \(k=1,2,3,4\). Summing inequality (39) over \(k=0,\ldots,n\), we get
$$ \begin{aligned}[b] & \Biggl\langle f^{-1}(v), \sum_{k=0}^{n}2\mu \lambda _{k} u_{k}- \sum_{k=0}^{n}2\mu \lambda _{k} v \Biggr\rangle \\ &\quad \leq \Vert u_{0}-v \Vert ^{2}+\mu ^{2}( \sigma _{1}+\sigma _{2})+2\mu \bigl(\sigma _{3}+ \sigma _{4} \Vert v \Vert +\mu \sigma _{5}\bigr) \quad \forall v\in C. \end{aligned} $$
(40)
Thus (36) follows from (37) and (40).
By Theorem 5.1, \(\{u_{n}\}_{n=0}^{\infty }\) converges strongly to the unique solution \(u^{*}\) of \(\operatorname{VI}(C,f^{-1})\). Since \(w_{n}\) is a convex combination of \(u_{0}, u_{1},\ldots, u_{n}\), it is easy to see that \(\{w_{n}\}_{n=1}^{\infty }\) also converges strongly to \(u^{*}\). □
Finally we are in a position to estimate the convergence rate of Algorithm 5.1.
Theorem 5.2
Assume that the conditions in Theorem 5.1
hold and the condition
\(\sum_{n=0}^{\infty }\lambda _{n}\varepsilon _{n}<\infty \)
is satisfied. Then, in the ergodic sense, Algorithm 5.1
has the
\(O (\frac{1}{n^{1-\alpha }} )\)
convergence rate if
\(\{\lambda _{n}\}_{n=1}^{\infty }=\{\frac{1}{n^{\alpha }}\}_{n=1}^{ \infty }\)
with
\(\frac{1}{2}<\alpha <1\)
and
\(\lambda _{0}=\frac{1}{1- \alpha }\), and has the
\(O (\frac{1}{\ln n} )\)
convergence rate if
\(\{\lambda _{n}\}_{n=1}^{\infty }=\{\frac{1}{n}\}_{n=1}^{ \infty }\).
Proof
For \(\{\lambda _{n}\}_{n=1}^{\infty }=\{\frac{1}{n^{ \alpha }}\}_{n=1}^{\infty }\) with \(\frac{1}{2}<\alpha <1\), one has that \(\sum_{n=1}^{\infty }\lambda _{n}=\infty \) and \(\sum_{n=1}^{\infty } \lambda _{n}^{2}< \infty \). For any integer \(k\geq 1\), it is easy to verify that
$$ \frac{1}{1-\alpha }\bigl\{ (k+1)^{1-\alpha }-k^{1-\alpha }\bigr\} \leq \frac{1}{k ^{\alpha }}. $$
Consequently, for all \(n\geq 1\), we have
$$ \frac{1}{1-\alpha }\bigl\{ (n+1)^{1-\alpha }-1\bigr\} \leq \sum _{k=1}^{n}\frac{1}{k ^{\alpha }}. $$
(41)
It concludes from (37) and (41) that
$$ \varUpsilon _{n}\geq \frac{2\mu }{(1-\alpha )}(n+1)^{1-\alpha } \geq \frac{2 \mu }{(1-\alpha )}n^{1-\alpha }, $$
(42)
which implies that Algorithm 5.1 has the \(O (\frac{1}{n^{1-\alpha }} )\) convergence rate. In fact, for any bounded subset \(D\subset C\), put \(\gamma =\sup \{ \Vert v \Vert \mid v\in D\}\), then from (36) and (42), we obtain
$$ \begin{aligned}[b] & \bigl\langle f^{-1}(v),w_{n}-v \bigr\rangle \\ &\quad \leq \frac{(1- \alpha )\{( \Vert u_{0} \Vert +\gamma ) ^{2}+\mu ^{2}(\sigma _{1}+\sigma _{2})+2 \mu (\sigma _{3}+\sigma _{4}\gamma +\mu \sigma _{5})\}}{2\mu n^{1-\alpha }}, \quad \forall v\in D. \end{aligned} $$
(43)
The conclusion can be similarly proved for \(\{\lambda _{n}\}_{n=1}^{\infty }=\{\frac{1}{n}\}_{n=1}^{\infty }\). □