Lemma 4.1
Let {xn} be a sequence generated by Algorithm 3.2 and {Γn} be defined by \({\Gamma }_{n}=\frac {1}{2}\|x_{n}-z\|^{2}\) for any z ∈Γ. Then, under assumption (8) and Assumption 3.1(c),(d), we have that
$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{i=1}^{n-1}\left[t_{i+1, n}\left( (1-3\theta_{i})-(1-\theta_{i})\right)+t_{i, n}(1-\theta_{i-1})\right] \|x_{i}-x_{i-1}\|^{2}\\ &&{\kern22pt}\leq 2t_{1}|{\Gamma}_{1}-{\Gamma}_{0}|+2{\Gamma}_{0}+t_{1}(1-\theta_{0})\|x_{1}-x_{0}\|^{2}, \end{array} $$
where {ti, n} is defined in Eq. 11.
Proof
First observe that
$$ \begin{array}{@{}rcl@{}} \|x_{n+1}-x_{n}\|^{2}&=&\|x_{n+1}-2x_{n}+x_{n-1}-(x_{n-1}-x_{n})\|^{2}\\ &=& \|x_{n+1}-2x_{n}+x_{n-1}\|^{2}+\|x_{n-1}-x_{n}\|^{2}\\ &&+2\langle x_{n+1}-2x_{n}+x_{n-1}, x_{n}-x_{n-1}\rangle, \end{array} $$
which implies that
$$2\langle x_{n+1}-2x_{n}+x_{n-1}, x_{n}-x_{n-1}\rangle =\|x_{n+1}-x_{n}\|^{2}- \|x_{n+1}-2x_{n}+x_{n-1}\|^{2}-\|x_{n-1}-x_{n}\|^{2}.$$
Thus, we obtain that
$$ \begin{array}{@{}rcl@{}} {}\|x_{n+1}-v_{n}\|^{2}&=&\|x_{n+1}-x_{n}-(x_{n}-x_{n-1})+(1-\theta_{n})(x_{n}-x_{n-1})\|^{2}\\ &=& \|x_{n+1}-2x_{n}+x_{n-1}\|^{2}+(1-\theta_{n})^{2}\|x_{n}-x_{n-1}\|^{2}\\ &&+2(1-\theta_{n})\langle x_{n+1}-2x_{n}+x_{n-1}, x_{n}-x_{n-1}\rangle \\ &=& \|x_{n+1}-2x_{n}+x_{n-1}\|^{2}+(1-\theta_{n})^{2}\|x_{n}-x_{n-1}\|^{2}\\ &&+(1-\theta_{n})\left[\|x_{n+1}-x_{n}\|^{2}- \|x_{n}-x_{n-1}\|^{2}-\|x_{n+1}-2x_{n}+x_{n-1}\|^{2}\right]\\ &=& \theta_{n} \|x_{n+1}-2x_{n}+x_{n-1}\|^{2}+(1-\theta_{n})^{2}\|x_{n}-x_{n-1}\|^{2}\\ &&+(1-\theta_{n})\left[\|x_{n+1}-x_{n}\|^{2}- \|x_{n}-x_{n-1}\|^{2}\right]\\ &\geq & (1-\theta_{n})^{2}\|x_{n}-x_{n-1}\|^{2}+(1-\theta_{n})\left[\|x_{n+1}-x_{n}\|^{2}- \|x_{n}-x_{n-1}\|^{2}\right]. \end{array} $$
(29)
Let z ∈Γ, then by Remark 3.5, we have that \(z\in C_{n}^{*}\). Thus, we obtain from Lemma 2.11 and Eq. 29 that
$$ \begin{array}{@{}rcl@{}} {}{\Gamma}_{n+1}-{\Gamma}_{n}-\theta_{n}({\Gamma}_{n}-{\Gamma}_{n-1})&=& \frac{1}{2}(\theta_{n}+{\theta_{n}^{2}})\|x_{n}-x_{n-1}\|^{2}+\langle x_{n+1}-v_{n}, x_{n+1}-z\rangle \\ &&-\frac{1}{2}\|x_{n+1}-v_{n}\|^{2}\\ &\leq & \frac{1}{2}(\theta_{n}+{\theta_{n}^{2}})\|x_{n}-x_{n-1}\|^{2}-\frac{1}{2}\|x_{n+1}-v_{n}\|^{2}\\ &\leq & \frac{1}{2}(\theta_{n}+{\theta_{n}^{2}})\|x_{n}-x_{n-1}\|^{2}-\frac{1}{2}(1-\theta_{n})^{2}\|x_{n}-x_{n-1}\|^{2}\\ &&-\frac{1}{2}(1-\theta_{n})\left[\|x_{n+1}-x_{n}\|^{2}- \|x_{n}-x_{n-1}\|^{2}\right]\\ &=& \frac{1}{2}(3\theta_{n}-1)\|x_{n}-x_{n-1}\|^{2} -\frac{1}{2}(1-\theta_{n})\\ &&\times\left[\|x_{n+1}-x_{n}\|^{2}- \|x_{n}-x_{n-1}\|^{2}\right], \end{array} $$
(30)
which implies from Lemma 2.9 (a) that
$$ \begin{array}{@{}rcl@{}} &&{\Gamma}_{n}-{\Gamma}_{0}= \sum\limits_{i=1}^{n}\left( {\Gamma}_{i}-{\Gamma}_{i-1}\right)\\ &\leq & t_{1, n}\left( {\Gamma}_{1}-{\Gamma}_{0}\right)+ \sum\limits_{i=1}^{n-1}t_{i+1, n}\left[\frac{1}{2}(3\theta_{i}-1)\|x_{i}-x_{i-1}\|^{2}-\frac{1}{2}(1-\theta_{i})\right.\\ &&\times\left. \left( \|x_{i+1}-x_{i}\|^{2}-\|x_{i}-x_{i-1}\|^{2}\right)\right]. \end{array} $$
Notice that t1,n ≤ t1. Thus, we obtain that
$$ \begin{array}{@{}rcl@{}} && \sum\limits_{i=1}^{n-1}t_{i+1, n}\left[(1-3\theta_{i})\|x_{i}-x_{i-1}\|^{2}+(1-\theta_{i})\left( \|x_{i+1}-x_{i}\|^{2}-\|x_{i}-x_{i-1}\|^{2}\right) \right]\\ &\leq & 2t_{1, n}({\Gamma}_{1}-{\Gamma}_{0})+2({\Gamma}_{0}-{\Gamma}_{n})\\ &\leq & 2t_{1}|{\Gamma}_{1}-{\Gamma}_{0}|+2{\Gamma}_{0}. \end{array} $$
(31)
□
Now, observe that
$$ \begin{array}{@{}rcl@{}} && \sum\limits_{i=1}^{n-1}t_{i+1, n}(1-\theta_{i})\left( \|x_{i+1}-x_{i}\|^{2}-\|x_{i}-x_{i-1}\|^{2}\right)\\ &= & \sum\limits_{i=1}^{n-1}\left( t_{i, n}(1-\theta_{i-1})-t_{i+1, n}(1-\theta_{i})\right)\|x_{i}-x_{i-1}\|^{2}\\ &&+t_{n, n}(1-\theta_{n-1})\|x_{n}-x_{n-1}\|^{2}-t_{1, n}(1-\theta_{0})\|x_{1}-x_{0}\|^{2}\\ &\geq & \sum\limits_{i=1}^{n-1}\left( t_{i, n}(1-\theta_{i-1})-t_{i+1, n}(1-\theta_{i})\right)\|x_{i}-x_{i-1}\|^{2}-t_{1}(1-\theta_{0})\|x_{1}-x_{0}\|^{2}. \end{array} $$
(32)
Combining (31) and (32), we get that
$$ \begin{array}{@{}rcl@{}} && \sum\limits_{i=1}^{n-1}t_{i+1, n}(1-3\theta_{i})\|x_{i}-x_{i-1}\|^{2}+\sum\limits_{i=1}^{n-1}\left( t_{i, n}(1-\theta_{i-1})-t_{i+1, n}(1-\theta_{i})\right)\|x_{i}-x_{i-1}\|^{2}\\ &\leq & 2t_{1}|{\Gamma}_{1}-{\Gamma}_{0}|+2 {\Gamma}_{0}+t_{1}(1-\theta_{0})\|x_{0}-x_{1}\|^{2}. \end{array} $$
That is,
$$ \begin{array}{@{}rcl@{}} && \sum\limits_{i=1}^{n-1}\left[t_{i+1, n}\left( (1-3\theta_{i})-(1-\theta_{i})\right)+t_{i, n}(1-\theta_{i-1})\right]\|x_{i}-x_{i-1}\|^{2}\\ &\leq & 2t_{1}|{\Gamma}_{1}-{\Gamma}_{0}|+2 {\Gamma}_{0}+t_{1}(1-\theta_{0})\|x_{0}-x_{1}\|^{2}. \end{array} $$
(33)
Lemma 4.2
Let {xn} be a sequence generated by Algorithm 3.2. Then, under assumption (8) and Assumption 3.1(c),(d), we have that \(\sum \limits _{n=1}^{\infty }(1-\theta _{n-1}) \|x_{n}-x_{n-1}\|^{2} < \infty \) and \(\sum \limits _{n=1}^{\infty }\theta _{n}t_{n+1} \|x_{n}-x_{n-1}\|^{2}<\infty \).
Proof
From Eq. 12 and since ti+ 1,n ≤ ti+ 1, we obtain
$$ \begin{array}{@{}rcl@{}} &&t_{i+1, n}\left[(1-3\theta_{i})-(1-\theta_{i})\right]+t_{i, n}(1-\theta_{i-1})\\ &= & t_{i+1, n}\left[(1-3\theta_{i})-(1-\theta_{i})\right]+(1-\theta_{i-1})+\theta_{i} t_{i+1, n}(1-\theta_{i-1})\\ &= & t_{i+1, n}\left[(1-3\theta_{i})-(1-\theta_{i})+\theta_{i} (1-\theta_{i-1})\right]+(1-\theta_{i-1})\\ &= & (1-\theta_{i-1})-\theta_{i} t_{i+1, n}\left( 1-\theta_{i-1}\right)\\ &\geq & (1-\theta_{i-1})-\theta_{i} t_{i+1}\left( 1-\theta_{i-1}\right)\\ &\geq & (1-\theta_{i-1})-\theta_{i} t_{i+1}\left( 1+\theta_{i}+\left[\theta_{i-1}-\theta_{i}\right]_{+}\right). \end{array} $$
(34)
Using Eq. 34 in Lemma 4.1, we obtain that
$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{i=1}^{n-1}(1-\theta_{i-1})-\theta_{i} t_{i+1}\left( 1+\theta_{i}+\left[\theta_{i-1}-\theta_{i}\right]_{+}\right)\|x_{i}-x_{i-1}\|^{2}\\ &\leq & 2t_{1}|{\Gamma}_{1}-{\Gamma}_{0}|+2 {\Gamma}_{0}+t_{1}(1-\theta_{0})\|x_{0}-x_{1}\|^{2}. \end{array} $$
We may assume without loss of generality that assumption (14) holds for every n ≥ 1. Then, we obtain that
$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n-1}\varepsilon (1-\theta_{i-1})\|x_{i}-x_{i-1}\|^{2} &\leq & 2t_{1}|{\Gamma}_{1}-{\Gamma}_{0}|+2 {\Gamma}_{0}+t_{1}(1-\theta_{0})\|x_{0}-x_{1}\|^{2}. \end{array} $$
Now, taking limit as \(n\to \infty \), we get that
$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{\infty}(1-\theta_{i-1}) \|x_{i}-x_{i-1}\|^{2} < \infty. \end{array} $$
(35)
Thus, the first conclusion of the lemma is established. To establish the second conclusion of the lemma, we employ assumption (14) again in Eq. 35 and obtain
$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{\infty}\theta_{i} t_{i+1} \|x_{i}-x_{i-1}\|^{2} < \infty. \end{array} $$
□
Lemma 4.3
Let {xn} be a sequence generated by Algorithm 3.2. Then, under assumption (8) and Assumption 3.1(c),(d), we have that
-
(a)
\(\lim \limits _{n\to \infty } \|x_{n}-z\|\) exists for all z ∈Γ.
-
(b)
\(\lim \limits _{n\to \infty } \|v_{n}-x_{n+1}\|=0\).
Proof
-
(a)
From Eq. 30, we obtain that
$$ \begin{array}{@{}rcl@{}} {\Gamma}_{n+1}-{\Gamma}_{n} &\leq & \theta_{n}({\Gamma}_{n}-{\Gamma}_{n-1})+ \frac{1}{2}(\theta_{n}+{\theta_{n}^{2}})\|x_{n}-x_{n-1}\|^{2}-\frac{1}{2}\|x_{n+1}-v_{n}\|^{2}\\ &\leq & \theta_{n}({\Gamma}_{n}-{\Gamma}_{n-1})+\theta_{n}\|x_{n}-x_{n-1}\|^{2}-\frac{1}{2}\|x_{n+1}-v_{n}\|^{2}\\ &\leq & \theta_{n}({\Gamma}_{n}-{\Gamma}_{n-1})+\theta_{n}\|x_{n}-x_{n-1}\|^{2}. \end{array} $$
(36)
Thus, from Lemma 4.2 and Lemma 2.9 (b), we obtain that \(\sum \limits _{n=1}^{\infty } \left [{\Gamma }_{n}-{\Gamma }_{n-1}\right ]_{+}<\infty \). This implies that \(\lim \limits _{n\to \infty }{\Gamma }_{n}=\lim \limits _{n\to \infty }\frac {1}{2}\|x_{n}-z\|^{2}\) exists, which further gives that \(\lim \limits _{n\to \infty }\|x_{n}-z\|\) exists for all z ∈Γ.
-
(b)
Now, using Eq. 36 and Lemma 2.9 (a), we obtain that
$$ \begin{array}{@{}rcl@{}} {\Gamma}_{n}-{\Gamma}_{0}&=&\sum\limits_{i=1}^{n}\left( {\Gamma}_{i}-{\Gamma}_{i-1}\right)\\ &\leq & t_{1, n}\left( {\Gamma}_{1}-{\Gamma}_{0}\right)+\sum\limits_{i=1}^{n-1}t_{i+1, n}\left[\theta_{i} \|x_{i}-x_{i-1}\|^{2}-\frac{1}{2}\|x_{i+1}-v_{i}\|^{2}\right]. \end{array} $$
(37)
Since ti+ 1,n ≤ ti+ 1, we obtain from Eq. 37 and Lemma 4.2 that
$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{n-1}t_{i+1, n}\|x_{i+1}-v_{i}\|^{2}{}&\leq &{} 2 {\Gamma}_{0}+2t_{1, n}({\Gamma}_{1}-{\Gamma}_{0}) +2\sum\limits_{i=1}^{n-1}t_{i+1, n}\theta_{i} \|x_{i}-x_{i-1}\|^{2}\\ {}&\leq &{}2 {\Gamma}_{0}{}+{}2t_{1}|{\Gamma}_{1}{}-{}{\Gamma}_{0}| +2\sum\limits_{i=1}^{\infty}t_{i+1}\theta_{i} \|x_{i}-x_{i-1}\|^{2}<\infty. \end{array} $$
Since ti+ 1,n = 0 for i ≥ n, letting n tend to \(\infty \), we obtain that
$$ \begin{array}{@{}rcl@{}} \sum\limits_{i=1}^{\infty} t_{i+1}\|x_{i+1}-v_{i}\|^{2}<\infty. \end{array} $$
(38)
Replacing i with n in Eq. 38 and since tn ≥ 1 for every n ≥ 1, we obtain from Eq. 38 that \(\sum \limits _{n=1}^{\infty } \|x_{n+1}-v_{n}\|^{2}<\infty \). This implies that \(\lim \limits _{n\to \infty } \|v_{n}-x_{n+1}\|=0\).
□
Remark 4.4
The main role of assumption (14) is to guarantee the condition
$$ \begin{array}{@{}rcl@{}} \sum\limits_{n=1}^{\infty}t_{n+1}\theta_{n}\|x_{n}-x_{n-1}\|^{2}<\infty, \end{array} $$
(39)
obtained in Lemma 4.2 above. Note that Lemma 4.3 holds true if we assume condition (39) directly. Moreover, if 𝜃n ∈ [0,𝜃] for every n ≥ 1, where 𝜃 ∈ [0,1), then \(t_{n}\leq \frac {1}{(1-\theta )}~~~\forall n\geq 1\). Under this setting, we have that condition (39) is guaranteed by the condition
$$ \begin{array}{@{}rcl@{}} \sum\limits_{n=1}^{\infty}\theta_{n}\|x_{n}-x_{n-1}\|^{2} <\infty. \end{array} $$
(40)
In other words, if we assume that condition (40) holds for 𝜃n ∈ [0,𝜃] ∀n ≥ 1, with 𝜃 ∈ [0,1), then Lemma 4.3 holds. This assumption has been used by numerous authors to ensure convergence of inertial methods (see, for example, Alvarez and Attouch 2001; Chuang 2017; Lorenz and Pock 2015; Mainge 2008; Moudafi and Oliny 2003 and the references therein).
Furthermore, under the assumptions of Proposition 3.1, we obtain the following as corollaries of Lemma 4.2 and Lemma 4.3 respectively.
Corollary 4.5
Let {xn} be a sequence generated by Algorithm 3.2 such that Assumption 3.1(c) holds. Suppose that {𝜃n} is a nondecreasing sequence that satisfies 𝜃n ∈ [0,1[ ∀n ≥ 1 with \(\lim \limits _{n \rightarrow \infty }\theta _{n}=\theta \) such that 1 − 3𝜃 > 0. Then, we have that \(\sum \limits _{n=1}^{\infty }(1-\theta _{n-1}) \|x_{n}-x_{n-1}\|^{2} < \infty \) and \(\sum \limits _{n=1}^{\infty }\theta _{n}t_{n+1} \|x_{n}-x_{n-1}\|^{2}<\infty \).
Proof
By Proposition 3.1, we have that assumptions (8) and (14) hold. Hence, the proof follows from Lemma 4.2. □
Corollary 4.6
Let {xn} be a sequence generated by Algorithm 3.2 such that Assumption 3.1(c) holds. Suppose that {𝜃n} is a nondecreasing sequence that satisfies 𝜃n ∈ [0,1[ ∀n ≥ 1 with \(\lim \limits _{n \rightarrow \infty }\theta _{n}=\theta \) such that 1 − 3𝜃 > 0. Then,
-
(a)
\(\lim \limits _{n\to \infty } \|x_{n}-z\|\) exists for all z ∈Γ.
-
(b)
\(\lim \limits _{n\to \infty } \|v_{n}-x_{n+1}\|=0\).
Proof
It is similar to the proof of Corollary 4.5. □
Remark 4.7
Observe that Eq. 18 and Proposition 2.7 imply that condition (8) also holds in Proposition 3.2. Hence, by replacing assumptions (8) and (14) with the assumptions of Proposition 3.2 in Lemma 4.2 and Lemma 4.3, we also obtain corollaries of Lemma 4.2 and Lemma 4.3 in the same manner as Corollaries 4.5 and 4.6 respectively.
Remark 4.8
If we take the inertial factor 𝜃n to be a constant (that is 𝜃n = 𝜃 ∀n ≥ 1), then we obtain the following corollaries of Lemma 4.2 and Lemma 4.3.
Corollary 4.9
Let {xn} be a sequence generated by Algorithm 3.2 such that Assumption 3.1(c) holds. Suppose that 𝜃n = 𝜃 ∀n ≥ 1 with 𝜃 ∈ [0,1) such that
$$ \begin{array}{@{}rcl@{}} (1-\theta)^{2}>\theta(1+\theta). \end{array} $$
(41)
Then, we have that \(\sum \limits _{n=1}^{\infty }(1-\theta ) \|x_{n}-x_{n-1}\|^{2} < \infty \) and \(\sum \limits _{n=1}^{\infty }\frac {\theta }{1-\theta } \|x_{n}-x_{n-1}\|^{2}<\infty \). Consequently, we have \(\sum \limits _{n=1}^{\infty }\|x_{n}-x_{n-1}\|^{2}<\infty \).
Proof
Since 𝜃n = 𝜃 ∈ [0,1), we obtain for i ≥ 1 that \(t_{i}=\sum \limits _{l=i-1}^{\infty }\theta ^{l-i+1}=\frac {1}{1-\theta }<\infty \). Thus, we get that assumption (8) holds. Note also from Eq. 41 that there exists 𝜖 ∈ (0,1) such that
$$ \begin{array}{@{}rcl@{}} (1-\epsilon)(1-\theta)\geq \frac{\theta(1+\theta)}{1-\theta}, \end{array} $$
which is equivalent to condition (14) since 𝜃n = 𝜃 ∀n ≥ 1. Hence, all the assumptions of Lemma 4.2 are satisfied. Thus, the rest of the proof follows from Lemma 4.2. □
Corollary 4.10
Let {xn} be a sequence generated by Algorithm 3.2 such that Assumption 3.1(c) holds. Suppose that 𝜃n = 𝜃 ∀n ≥ 1 with 𝜃 ∈ [0,1) such that (1 − 𝜃)2 > 𝜃(1 + 𝜃). Then,
-
(a)
\(\lim \limits _{n\to \infty } \|x_{n}-z\|\) exists for all z ∈Γ.
-
(b)
\(\lim \limits _{n\to \infty } \|v_{n}-x_{n+1}\|=0\).
Proof
The proof is similar to the proof of Corollary 4.9. □
We now return to a very important result for our convergence analysis, whose proof rely on the linesearch given in Algorithm 3.2.
Lemma 4.11
Let assumption (8) and Assumption 3.1 hold, and let the sequence {xn} be generated by Algorithm 3.2. Then, \(\lim \limits _{n\to \infty }\alpha _{n} \|y_{n}-v_{n}\|^{2}=0\). Moreover, if there exists a subsequence \(\{x_{n_{k}}\}\) of {xn} such that \(\{x_{n_{k}}\}\) converges to x∗ and x∗∉Γ, then
-
(a)
\(\liminf \limits _{k\to \infty } \alpha _{n_{k}}>0\);
-
(b)
\(\lim \limits _{k\to \infty }\|v_{n_{k}}-y_{n_{k}}\|=0\).
Proof
From Eq. 4, Step 1, Step 2 and the fact that \(x_{n+1}\in C_{n}^{*},\) we obtain that
$$ \begin{array}{@{}rcl@{}} \alpha_{n}\|v_{n}-y_{n}\|^{2}&=&\alpha_{n}\langle v_{n}-y_{n}, v_{n}-y_{n}\rangle\\ &\leq & \alpha_{n} \langle v_{n}-y_{n}, v_{n}-y_{n}\rangle +\alpha_{n} \langle y_{n}-v_{n}+\rho_{n} u_{n}, v_{n}-y_{n}\rangle\\ &=& \alpha_{n} \rho_{n} \langle u_{n}, v_{n}-y_{n}\rangle\\ &\leq & \frac{\alpha_{n} \rho_{n}}{\sigma} \langle w_{n}, v_{n}-y_{n}\rangle\\ &=& \frac{\rho_{n}}{\sigma} \langle w_{n}, v_{n}-z_{n}\rangle\\ &\leq & \frac{\rho_{n}}{\sigma} \left( \langle w_{n}, v_{n}-x_{n+1}\rangle +\langle w_{n}, x_{n+1}-z_{n}\rangle\right) \\ &\leq &\frac{\rho_{n}}{\sigma} \|w_{n}\| \|v_{n}-x_{n+1}\|. \end{array} $$
(42)
Since by Lemma 4.3, {xn} is bounded, we have that {zn} is also bounded. Moreover, since F is locally bounded, we obtain from Proposition 2.3 that {wn} is bounded. Using this and the boundedness of {ρn}, we obtain from Eq. 42 and Lemma 4.3 that
$$ \begin{array}{@{}rcl@{}} \lim\limits_{n\to\infty}\alpha_{n} \|y_{n}-v_{n}\|^{2}=0. \end{array} $$
(43)
-
(a)
By Step 2, we have that {αn}⊂ (0,1) is bounded. Thus, there exists a subsequence \(\{\alpha _{n_{k}}\}\) of {αn} such that \(\liminf \limits _{k\to \infty } \alpha _{n_{k}}\geq 0\). In fact, we claim that \(\liminf \limits _{k\to \infty } \alpha _{n_{k}}> 0\). Suppose on the contrary that \(\liminf \limits _{k\to \infty } \alpha _{n_{k}}= 0\). Then, without loss of generality, we can choose a subsequence of \(\{\alpha _{n_{k}}\}\) still denoted by \(\{\alpha _{n_{k}}\}\) such that \(\lim \limits _{k\to \infty } \alpha _{n_{k}}=0\).
Now, define \(\bar {\alpha }_{n_{k}}:=\frac {\alpha _{n_{k}}}{\gamma },~~\bar {z}_{n_{k}}:= \bar {\alpha }_{n_{k}}y_{n_{k}}+(1-\bar {\alpha }_{n_{k}}) v_{n_{k}}\). Then, by the boundedness of \(\{y_{n_{k}}-v_{n_{k}}\}\) and since \(\alpha _{n_{k}}\to 0\) as \(k\to \infty \), we obtain that
$$ \begin{array}{@{}rcl@{}} \lim_{k\to\infty}\| \bar{z}_{n_{k}}-v_{n_{k}}\|=0. \end{array} $$
(44)
Also, by Lemma 4.2, we obtain that \(\lim \limits _{k\to \infty }\| x_{n_{k}}-v_{n_{k}}\|=\lim \limits _{k\to \infty } \theta _{n_{k}}\|x_{n_{k}}-x_{n_{k}-1}\|=0\). Thus, since \(x_{n_{k}}\to x^{*}\), we have that \(v_{n_{k}}\to x^{*}\). Using Assumption 3.1 (b), the boundedness of \(\{v_{n_{k}}\}\) and Proposition 2.3, we obtain that \(\{u_{n_{k}}\}\) is also bounded. Thus, we can choose a subsequence of \(\{u_{n_{k}}\}\) still denoted by \(\{u_{n_{k}}\}\) such that \(u_{n_{k}}\to \bar {u}\). Since F is continuous, it is outer-semicontinuous. Hence, \(\bar {u}\in F(x^{*})\). We also assume without loss of generality that \(\rho _{n_{k}}\to \rho \in [\rho _{0}, \rho _{1}]\). Therefore, we obtain from the continuity of PC that \(y_{n_{k}}\to y^{*}\) as \(k\to \infty \), where \(y^{*}=P_{C}(x^{*}-\rho \bar {u})\).
Again, from Eq. 44, we obtain that \(\bar {z}_{n_{k}}\to x^{*}\). Since F is inner-semicontinuous and \(\bar {u}\in F(x^{*})\), we can choose a subsequence \(w_{n_{k}}\in F(\bar {z}_{n_{k}})\) such that \(\bar {w}_{n_{k}}\to \bar {u}\).
Now, from the definition of \(\bar {z}_{n_{k}}\) and Step 2, we obtain that
$$ \begin{array}{@{}rcl@{}} \langle \bar{w}_{n_{k}}, v_{n_{k}}-y_{n_{k}}\rangle <\sigma \langle u_{n_{k}}, v_{n_{k}}-y_{n_{k}}\rangle. \end{array} $$
(45)
Thus, taking limit as \(k\to \infty \), we obtain that
$$ \begin{array}{@{}rcl@{}} \langle \bar{u}, x^{*}-y^{*}\rangle \leq 0. \end{array} $$
(46)
On the hand, since x∗∉Γ, we have from Lemma 2.10 that x∗≠y∗. Hence, we get
$$ \begin{array}{@{}rcl@{}} \langle \bar{u}, x^{*}-y^{*}\rangle=\frac{1}{\rho} \langle y^{*}-(x^{*}-\rho \bar{u})+(x^{*}-y^{*}), x^{*}-y^{*}\rangle>\frac{1}{\rho} \langle x^{*}-y^{*}, x^{*}-y^{*}\rangle>0, \end{array} $$
(47)
which is a contradiction to Eq. 46. Therefore, \(\liminf \limits _{k\to \infty } \alpha _{n_{k}}>0\).
-
(b)
From (a), we have that \(\liminf \limits _{k\to \infty } \alpha _{n_{k}}>0\). Thus, we obtain from Eq. 43 that
$$ \begin{array}{@{}rcl@{}} 0\leq \limsup\limits_{k \rightarrow\infty}\|v_{n_{k}}-y_{n_{k}}\|^{2}&\leq \limsup\limits_{k\rightarrow \infty}\left( \alpha_{n_{k}}\|v_{n_{k}}-y_{n_{k}}\|^{2}\right)\left( \limsup\limits_{k\rightarrow\infty}\frac{1}{\alpha_{n_{k}}}\right)\\ &= \left( \limsup\limits_{k\rightarrow \infty}\alpha_{n_{k}}\|v_{n_{k}}-y_{n_{k}}\|^{2}\right)\left( \frac{1}{\liminf\limits_{k\rightarrow\infty}\alpha_{n_{k}}}\right)\\ &=0. \end{array} $$
Therefore, we obtain that
$$\lim\limits_{k\rightarrow\infty}\|v_{n_{k}}-y_{n_{k}}\|=0.$$
□
Now, we are in position to give the main theorem of this section.
Theorem 4.12
Let {xn} be a sequence generated by Algorithm 3.2. Then, under assumption (8) and Assumption 3.1, we have that {xn} converges to an element of Γ.
Proof
By Lemma 4.3, {xn} is bounded. Thus, there exists a subsequence \(\{x_{n_{k}}\}\) of {xn} such that \(\{x_{n_{k}}\}\) converges to some point x∗. Also, we have that
$$ \begin{array}{@{}rcl@{}} \|v_{n_{k}}-x_{n_{k}}\|=\theta_{n_{k}}\|x_{n_{k}}-x_{{n_{k}}-1}\|\to 0,~\text{as}~k\to \infty. \end{array} $$
(48)
We now claim that x∗∈Γ.
Suppose on the contrary that x∗∉Γ. Then, it follows from Lemma 4.11 (b) and Eq. 48 that
$$ \begin{array}{@{}rcl@{}} \lim_{k\to \infty} y_{n_{k}}=\lim_{k\to \infty} P_{C}(v_{n_{k}}-\rho_{n_{k}}u_{n_{k}})=\lim_{k\to \infty} x_{n_{k}}=x^{*}. \end{array} $$
(49)
Now, without loss of generality, we may assume that \(\rho _{n_{k}}\to {\rho ^{*}}\) and \(u_{n_{k}}\to {u^{*}}\). Since F is continuous, it is outer-semicontinuous. Thus, we obtain that u∗∈ F(x∗). Therefore, we obtain from Eq. 49 that
$$ P_{C}(x^{*}-\rho^{*}u^{*})=x^{*}.$$
It then follows from Lemma 2.10 that x∗∈Γ, which is a contraction. Hence, our claim holds.
We now show that {xn} converges to x∗.
Replacing z by x∗ in Lemma 4.3, we obtain that \(\lim \limits _{n\rightarrow \infty }\|x_{n}-x^{*}\|^{2}\) exists. Since x∗ is an accumulation point of {xn}, we obtain that {xn} converges to x∗. □
Remark 4.13
In view of Corollaries 4.5-4.10, we can obtain various corollaries of Theorem 4.12. Furthermore, in the case that 𝜃n = 0 for all n ≥ 1, assumptions (8) and (14) are automatically satisfied. Moreover, we have in this case that tn = 1 for all n ≥ 1. Hence, we can employ Procedure A (see page 1) to obtain similar result as in He et al. (2019, Theorem 3.1).
Algorithm 4.1.
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Step 0: Let \(x_{1}\in \mathbb {R}^{N}\) be given arbitrary and fix \(\gamma , \sigma \in (0, 1), 0<\rho _{0}\leq \rho _{1}<\infty \). Set \(C_{1}=\mathbb {R}^{N}\), \(\bar {x}_{1}=x_{1}\) and n = 1.
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Step 1. Apply Procedure A to obtain \(x_{n}=R(\bar {x}_{n})\).
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Step 2. Choose un ∈ F(xn) and ρn ∈ [ρ0, ρ1]. Then, compute
yn = PC(xn − ρnun). If xn = yn: STOP. Otherwise, go to Step 2.
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Step 3. Compute
$$z_{n}=\alpha_{n}y_{n}+(1-\alpha_{n}) x_{n}$$
and choose the largest \(\alpha \in \{\gamma , \gamma ^{2}, \gamma ^{3}, \dots \}\) such that there exists a point wn ∈ F(zn) satisfying
$$ \begin{array}{@{}rcl@{}} \langle w_{n}, x_{n}-y_{n}\rangle \geq \sigma\langle u_{n}, x_{n}-y_{n}\rangle. \end{array} $$
(50)
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Step 4. Set \(C_{n}=\{y\in \mathbb {R}^{N} : \langle w_{n}, y-z_{n}\rangle \leq 0\}\) for n ≥ 2 and \(C_{n}^{*}=\cap _{i=1}^{n} C_{i}\). Then, compute
$$\bar{x}_{n+1}=P_{C_{n}^{*}}(x_{n}).$$
If \(\bar {x}_{n+1}=x_{n}\), then stop. Otherwise, let n = n + 1 and return to Step 1.
Corollary 4.14 (see for example, He et al. (2019, Theorem 3.1))
Let {xn} be a sequence generated by Algorithm 4.1 such that the following assumptions hold:
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(a)
The set C is described as in procedure A (see page 1).
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(b)
\(F:C\rightrightarrows \mathbb {R}^{N}\) is locally bounded and continuous.
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(c)
Γ is nonempty and satisfies condition (7).
Then, we have that {xn} converges to an element of Γ.
Proof
It follows carefully from Lemma 2.13 and Theorem 4.12. □
Remark 4.15
Under the settings of Remark 4.13, we can obtain in general, similar result as in He et al. (2019, Theorem 3.2) without Procedure A.
Algorithm 4.2.
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Step 0: Let x1 ∈ C be given arbitrary and fix \(\gamma , \sigma \in (0, 1), 0<\rho _{0}\leq \rho _{1}<\infty \). Set \(C_{1}=\mathbb {R}^{N}\) and n = 1.
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Step 1. Choose un ∈ F(xn) and ρn ∈ [ρ0, ρ1]. Then, compute
yn = PC(xn − ρnun). If xn = yn: STOP. Otherwise, go to Step 2.
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Step 2. Compute
$$z_{n}=\alpha_{n}y_{n}+(1-\alpha_{n}) x_{n}$$
and choose the largest \(\alpha \in \{\gamma , \gamma ^{2}, \gamma ^{3}, \dots \}\) such that there exists a point wn ∈ F(zn) satisfying
$$ \begin{array}{@{}rcl@{}} \langle w_{n}, x_{n}-y_{n}\rangle \geq \sigma\langle u_{n}, x_{n}-y_{n}\rangle. \end{array} $$
(51)
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Step 3. Set \(C_{n}=\{y\in \mathbb {R}^{N} : \langle w_{n}, y-z_{n}\rangle \leq 0\}\) for n ≥ 2 and \(C_{n}^{*}=\cap _{i=1}^{n} C_{i}\). Then, compute
$${x}_{n+1}=P_{C}\cap {C_{n}^{*}}(x_{n}).$$
If xn+ 1 = xn, then stop. Otherwise, let n = n + 1 and return to Step 1.
Corollary 4.16 (see, for example, He et al. (2019, Theorem 3.2))
Let {xn} be a sequence generated by Algorithm 4.2 such that the following assumptions hold:
-
(a)
The feasible set C is a nonempty closed and convex subset of \(\mathbb {R}^{N}\).
-
(b)
\(F:C\rightrightarrows \mathbb {R}^{N}\) is locally bounded and continuous.
-
(c)
Γ is nonempty and satisfies condition (7).
Then, we have that {xn} converges to an element of Γ.
Proof
It follows directly from Corollary 4.14. □