A unified framework for three accelerated extragradient methods and further acceleration for variational inequality problems

Abstract

The main strategy of this paper is intended to speed up the convergence of the inertial Mann iterative method and further speed up it through the normal S-iterative method for a certain class of nonexpansive-type operators that are linked with variational inequality problems. Our new convergence theory permits us to settle down the difficulty of unification of Korpelevich’s extragradient method, Tseng’s extragardient method, and subgardient extragardient method for solving variational inequality problems through an auxiliary algorithmic operator, which is associated with the seed operator. The paper establishes an interesting the fact that the relaxed inertial normal S-iterative extragradient methods do influence much more on convergence behaviour. Finally, the numerical experiments are carried out to illustrate that the relaxed inertial iterative methods; in particular, the relaxed inertial normal S-iterative extragradient methods may have a number of advantages over other methods in computing solutions to variational inequality problems in many cases.

Appendices

Appendix-I: Proof of Lemma 2.5

Proof

(a) Set $K:=\varepsilon -\theta (1+\varepsilon +\max \{1,\varepsilon \})$. Define $K_{n}=\theta _{n}(1+\theta _{n}+\varepsilon (1-\theta _{n}))$, $n=1,2,\cdots$ and $\phi _{n}=\Vert x_{n}-v\Vert ^{2}$, $n=0,1,2,\cdots .$ From (11), we have

$$\begin{aligned}{} & {} \phi _{n+1} \le (1+\theta _{n})\phi _{n}-\theta _{n}\phi _{n-1} -\varepsilon (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}\nonumber \\{} & {} \quad +K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2} \text { for all } n\in \mathbb {N} \end{aligned}$$

(50)

and hence

$$\begin{aligned}{} & {} \phi _{n+1}-(1+\theta _{n})\phi _{n}+\theta _{n}\phi _{n-1} \le -\varepsilon (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}\nonumber \\{} & {} \quad +K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2}\text { for all } n\in \mathbb {N}. \end{aligned}$$

Since $\theta _{n}\le \theta _{n+1}$ for all $n\in \mathbb {N}$, it follows that

$$\begin{aligned} \varphi _{n+1}-\varphi _{n}= & {} \phi _{n+1}-\theta _{n+1}\phi _{n}+K_{n+1}\Vert x_{n+1}-x_{n}\Vert ^{2} \nonumber \\{} & {} \quad -(\phi _{n}-\theta _{n}\phi _{n-1}+K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2}) \\= & {} \phi _{n+1}-(1+\theta _{n+1})\phi _{n}+\theta _{n}\phi _{n-1}\\{} & {} \quad +K_{n+1}\Vert x_{n+1}-x_{n}\Vert ^{2}-K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2} \\\le & {} -\varepsilon (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}+K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2}\\{} & {} \quad +K_{n+1}\Vert x_{n+1}-x_{n}\Vert ^{2}-K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2} \\= & {} -[\varepsilon (1-\theta _{n})-K_{n+1}]\Vert x_{n+1}-x_{n}\Vert ^{2}. \end{aligned}$$

Note $\theta _{n}\le \theta _{n+1}\le \theta$ for all $n\in \mathbb {N}$ and $K_{n}=\theta _{n}\left( 1+\theta _{n}+\varepsilon (1-\theta _{n})\right) \le \theta _{n}\left( 1+\max \{1,\varepsilon \}\right) \text { for all } n\in \mathbb {N}.$ Hence

$$\begin{aligned}{} & {} -[\varepsilon (1-\theta _{n})-K_{n+1}] \\{} & {} \quad = -\varepsilon (1-\theta _{n})+\theta _{n+1}\left( 1+\theta _{n+1}+\varepsilon (1-\theta _{n+1})\right) \\{} & {} \quad \le -\varepsilon (1-\theta _{n})+\theta _{n+1}\left( 1+\max \{1,\varepsilon \}\right) \\{} & {} \quad \le -\varepsilon (1-\theta )+\theta \left( 1+\max \{1,\varepsilon \}\right) \\{} & {} \quad =-[\varepsilon -\theta (1+\varepsilon +\max \{1,\varepsilon \})] =-K. \end{aligned}$$

Note $K>0$ by the inequality (12). Hence

$$\begin{aligned} \varphi _{n+1}-\varphi _{n}\le -K\Vert x_{n+1}-x_{n}\Vert ^{2}\text { for all }n\in \mathbb {N}, \end{aligned}$$

(51)

which implies that $\varphi _{n+1}\le \varphi _{n}$ for all $n\in \mathbb {N }.$ Thus, the sequence $\{\varphi _{n}\}$ is non-increasing.

(b) For $n\in \mathbb {N},$ we have

$$\begin{aligned} \varphi _{n}=\phi _{n}-\theta _{n}\phi _{n-1}+K_{n}\Vert x_{n}-x_{n-1}\Vert ^{2}\ge \phi _{n}-\theta _{n}\phi _{n-1} \end{aligned}$$

(52)

and

$$\begin{aligned} \varphi _{n+1}=\phi _{n+1}-\theta _{n+1}\phi _{n}+K_{n+1}\Vert x_{n+1}-x_{n}\Vert ^{2}\ge -\theta _{n+1}\phi _{n}. \end{aligned}$$

(53)

Since $\varphi _{1}>0,$ from (52), we have

$$\begin{aligned} \phi _{n}\le & {} \theta _{n}\phi _{n-1}+\varphi _{n} \nonumber \\\le & {} \theta \phi _{n-1}+\varphi _{1} \nonumber \\&\cdot \cdot \cdot \nonumber \\\le & {} \theta ^{n}\phi _{0}+(1+\theta +\theta ^{2}+\cdots +\theta ^{n-1})\varphi _{1} \nonumber \\\le & {} \theta ^{n}\phi _{0}+\frac{\varphi _{1}}{1-\theta }\text { for all } n\in \mathbb {N}. \end{aligned}$$

(54)

Combining (53) and (54), we have

$$\begin{aligned} -\varphi _{n+1} \le \theta _{n+1}\phi _{n} \le \theta \phi _{n} \le \theta ^{n+1}\phi _{0}+\frac{\theta \varphi _{1}}{1-\theta }. \end{aligned}$$

From (51), we get

$$\begin{aligned}{} & {} K\sum _{i=1}^{n}\Vert x_{i+1}-x_{i}\Vert ^{2} \le \varphi _{1}-\varphi _{n+1} \\{} & {} \quad \le \varphi _{1}+\theta ^{n+1}\phi _{0}+\frac{\theta \varphi _{1}}{1-\theta } \text { for all } n\in \mathbb {N}. \end{aligned}$$

Taking limit as $n\rightarrow \infty ,$ we have $K\sum _{i=1}^{\infty }\Vert x_{i+1}-x_{i}\Vert ^{2}\le \frac{ \varphi _{1}}{1-\theta }<\infty .$

(c) From (50), we have

$$\begin{aligned} \phi _{n+1}\le & {} (1+\theta _{n})\phi _{n}-\theta _{n}\phi _{n-1}\\{} & {} \quad +\theta \left( 1+\max \{1,\varepsilon \}\right) \Vert x_{n}-x_{n-1}\Vert ^{2}. \end{aligned}$$

From Lemma 2.4, we obtain that $\lim _{n\rightarrow \infty }\Vert x_{n}-v\Vert$ exists. $\square$

Appendix-II: Proof of Proposition 3.1

Proof

(a)–(b). Let $v\in \textrm{Fix}(S)$. Note $\displaystyle Tw_{n}-w_{n}=\frac{ 1}{\alpha _{n}}(x_{n+1}-w_{n})$ and $\displaystyle \varepsilon \le \frac{ 1+\kappa -\alpha _{n}}{\alpha _{n}}$ for all $n\in \mathbb {N}.$ Note T is $\kappa$-strongly quasi-nonexpansive operator. From (13) and Lemma 2.6, we have

$$\begin{aligned} \Vert x_{n+1}-v\Vert ^{2}&=\Vert (1-\alpha _{n})w_{n}+\alpha _{n}Tw_{n}-v\Vert ^{2} \nonumber \\&\le \Vert w_{n}-v\Vert ^{2}-\alpha _{n}(1+\kappa -\alpha _{n})\Vert w_{n}-Tw_{n}\Vert ^{2} \end{aligned}$$

(55)

$$\begin{aligned}&=\Vert w_{n}-v\Vert ^{2}-\frac{1+\kappa -\alpha _{n}}{\alpha _{n}}\Vert x_{n+1}-w_{n}\Vert ^{2}. \end{aligned}$$

(56)

Again, from (13) and Lemma 2.2(ii), we have

$$\begin{aligned} \Vert w_{n}-v\Vert ^{2}= & {} \Vert (1+\theta _{n})(x_{n}-v)-\theta _{n}(x_{n-1}-v)\Vert ^{2} \nonumber \\= & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \nonumber \\{} & {} +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}. \end{aligned}$$

(57)

Combining (56) and (57), we get

$$\begin{aligned} \Vert x_{n+1}-v\Vert ^{2}\le & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \nonumber \\{} & {} +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}\nonumber \\{} & {} -\varepsilon \Vert x_{n+1}-w_{n}\Vert ^{2}. \end{aligned}$$

(58)

From (10), we have

$$\begin{aligned}{} & {} \Vert x_{n+1}-w_{n}\Vert ^{2} = \Vert x_{n+1}-x_{n}-\theta _{n}(x_{n}-x_{n-1})\Vert ^{2} \nonumber \\{} & {} \quad \ge (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}-\theta _{n}(1-\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}. \end{aligned}$$

From (58), we obtain

$$\begin{aligned} \Vert x_{n+1}-v\Vert ^{2}\le & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \nonumber \\{} & {} +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}\nonumber \\{} & {} -\varepsilon (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}\nonumber \\{} & {} +\varepsilon \theta _{n}(1-\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2} \nonumber \\= & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \nonumber \\{} & {} -\varepsilon (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2} \nonumber \\{} & {} +\theta _{n}\left( 1+\theta _{n}+\varepsilon (1-\theta _{n})\right) \Vert x_{n}-x_{n-1}\Vert ^{2}.\nonumber \\ \end{aligned}$$

(59)

Noticing that $\{\theta _{n}\}$ is an increasing sequence in [0, 1) by the assumption (C2) and that the condition (12) holds by the assumption (C3). Apply Lemma 2.5 on (59), we conclude that $\sum _{n=1}^{\infty }\Vert x_{n+1}-x_{n}\Vert ^{2}<\infty \text { and }\lim _{n\rightarrow \infty }\Vert x_{n}-v\Vert ~\text { exists.}$

Clearly, $\Vert x_{n+1}-x_{n}\Vert \rightarrow 0$ as $n\rightarrow \infty$. Hence, from (13), we have

$$\begin{aligned} \Vert w_{n}-x_{n+1}\Vert \le \Vert x_{n}-x_{n+1}\Vert +\theta \Vert x_{n}-x_{n-1}\Vert \rightarrow 0. \end{aligned}$$

It follows that $\lim _{n\rightarrow \infty }\Vert w_{n}-x_{n}\Vert =0$. We may assume that $\lim _{n\rightarrow \infty }\Vert x_{n}-v\Vert =\ell >0$. It follows, from $\lim _{n\rightarrow \infty }\Vert x_{n}-w_{n}\Vert =0$, that $\lim _{n\rightarrow \infty }\Vert w_{n}-v\Vert =\ell$. In view of condition (C1), we obtain, from (55), that $\lim _{n\rightarrow \infty }\Vert w_{n}-Tw_{n}\Vert =0.$

(c) Let $v\in \textrm{Fix}(S)$. Note $R_{T}(n):=R_{T,\{w_{n}\}}(n)\le \Vert w_{n}-Tw_{n}\Vert$ for all $n\in \mathbb {N}$. From (55) and (57), we have

$$\begin{aligned} \Vert x_{n+1}-v\Vert ^{2}\le & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \\{} & {} +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2} \\{} & {} -\alpha _{n}(1+\kappa -\alpha _{n})\Vert w_{n}-Tw_{n}\Vert ^{2} \\\le & {} \Vert x_{n}-v\Vert ^{2}+( \theta _{n+1}\Vert x_{n}-v\Vert ^{2} \\{} & {} -\theta _{n}||x_{n-1}-v\Vert ^{2}) +\theta (1+\theta )\Vert x_{n}-x_{n-1}\Vert ^{2}\\{} & {} -\alpha _{n}(1+\kappa -\alpha _{n})\Vert w_{n}-Tw_{n}\Vert ^{2}. \end{aligned}$$

Hence, from the condition (C1), we have

$$\begin{aligned}{} & {} a(1+\kappa -b)\sum \limits _{i=1}^{n}\Vert w_{i}-Tw_{i}\Vert ^{2} \\{} & {} \quad \le \sum \limits _{i=1}^{n}(\Vert x_{i}-v\Vert ^{2}-\Vert x_{i+1}-v\Vert ^{2})\\{} & {} \quad +\sum \limits _{i=1}^{n}(\theta _{i+1}\Vert x_{i}-v\Vert ^{2}-\theta _{i}\Vert x_{i-1}-v\Vert ^{2}) \\{} & {} \quad +\theta (1+\theta )\sum \limits _{i=1}^{n}\Vert x_{i}-x_{i-1}\Vert ^{2} \\{} & {} \quad =\Vert x_{1}-v\Vert ^{2}-\Vert x_{n+1}-v\Vert ^{2}+\theta _{n+1}\Vert x_{n}-v\Vert ^{2}\\{} & {} \qquad \quad -\theta _{1}\Vert x_{0}-v\Vert ^{2}+\theta (1+\theta )\sum \limits _{i=1}^{n}\Vert x_{i}-x_{i-1}\Vert ^{2} \\{} & {} \quad \le \ \Vert x_{1}-v\Vert ^{2}+\theta _{n+1}\Vert x_{n}-v\Vert ^{2}\\{} & {} \qquad \quad +\theta (1+\theta )\sum \limits _{i=1}^{\infty }\Vert x_{i}-x_{i-1}\Vert ^{2} \\{} & {} \quad \le M \end{aligned}$$

for all $n\in \mathbb {N}$ and for some $M>0.$ Thus, $\sum \limits _{n=1}^{ \infty }(R_{T}(n))^{2}<\infty .$ Note $\{R_{T}(n)\}$ is decreasing. Hence, from Lemma 2.3, we see that $R_{T,\{w_{n}\}}(n)=o\left( \frac{1}{\sqrt{n}}\right) .$ $\square$

Appendix-III: Proof of Proposition 3.2

Proof

(a)–(b). Let $v\in \textrm{Fix}(S)$. Set $y_{n}:=(1-\alpha _{n})w_{n}+\alpha _{n}Tw_{n}$. Hence $\displaystyle Tw_{n}-w_{n}=\frac{1}{ \alpha _{n}}(y_{n}-w_{n})$. From (19), we have $\displaystyle \mathcal {E}\le \frac{\alpha _{n}(1+\kappa -\alpha _{n})}{2(1+{\alpha _{n}^{2}\beta }^{2})}$ for all $n\in \mathbb {N}.$ For $\rho =\frac{1}{2}$, from (9), we obtain

$$\begin{aligned}{} & {} \Vert x_{n+1}-Tw_{n}\Vert ^{2} = \Vert x_{n+1}-w_{n}+w_{n}-Tw_{n}\Vert ^{2} \\{} & {} \quad \ge \frac{1}{2}\Vert x_{n+1}-w_{n}\Vert ^{2}-\Vert w_{n}-Tw_{n}\Vert ^{2}. \end{aligned}$$

Since T is $\beta$-Lipschitz continuous, we have

$$\begin{aligned}{} & {} \Vert w_{n}-Tw_{n}\Vert ^{2} = \frac{1}{{\alpha _{n}^{2}}}\Vert y_{n}-w_{n}\Vert ^{2} \\{} & {} \quad \ge \frac{1}{{\alpha _{n}^{2}\beta }^{2}}\Vert Ty_{n}-Tw_{n}\Vert ^{2}= \frac{1}{{\alpha _{n}^{2}\beta }^{2}}\Vert x_{n+1}-Tw_{n}\Vert ^{2} \\{} & {} \quad \ge \frac{1}{{\alpha _{n}^{2}\beta }^{2}}\left( \frac{1}{2}\Vert x_{n+1}-w_{n}\Vert ^{2}-\Vert w_{n}-Tw_{n}\Vert ^{2}\right) . \end{aligned}$$

Thus,

$$\begin{aligned} \Vert w_{n}-Tw_{n}\Vert ^{2}\ge \frac{1}{2(1+{\alpha _{n}^{2}\beta }^{2}){\ }}\Vert x_{n+1}-w_{n}\Vert ^{2}. \end{aligned}$$

(60)

From Lemma 2.6, we get

$$\begin{aligned}{} & {} \Vert y_{n}-v\Vert ^{2}\nonumber \\{} & {} \quad = \Vert (1-\alpha _{n})w_{n}+\alpha _{n}Tw_{n}-v\Vert ^{2} \nonumber \\{} & {} \quad = \Vert w_{n}-v\Vert ^{2}-\alpha _{n}(1+\kappa -\alpha _{n})\Vert w_{n}-Tw_{n}\Vert ^{2}. \end{aligned}$$

(61)

Using (61) and then (60), we obtain

$$\begin{aligned}{} & {} \Vert x_{n+1}-v\Vert ^{2} = \Vert Ty_{n}-v\Vert ^{2}\nonumber \\{} & {} \quad \le \Vert y_{n}-v\Vert ^{2}-\kappa \Vert y_{n}-Ty_{n}\Vert ^{2} \nonumber \\{} & {} \quad \le \Vert w_{n}-v\Vert ^{2}-\alpha _{n}(1+\kappa -\alpha _{n})\Vert w_{n}-Tw_{n}\Vert ^{2}\nonumber \\{} & {} \qquad -\kappa \Vert y_{n}-Ty_{n}\Vert ^{2} \end{aligned}$$

(62)

$$\begin{aligned}{} & {} \quad =\Vert w_{n}-v\Vert ^{2}-\frac{\alpha _{n}(1+\kappa -\alpha _{n})}{2(1+{ \alpha _{n}^{2}\beta }^{2})}\Vert x_{n+1}-w_{n}\Vert ^{2}\nonumber \\{} & {} \qquad -\kappa \Vert y_{n}-Ty_{n}\Vert ^{2} \nonumber \\{} & {} \quad \le \Vert w_{n}-v\Vert ^{2}-\mathcal {E}\Vert x_{n+1}-w_{n}\Vert ^{2} -\kappa \Vert y_{n}-Ty_{n}\Vert ^{2}. \end{aligned}$$

(63)

From (17) and Lemma 2.2(ii), we have

$$\begin{aligned} \Vert w_{n}-v\Vert ^{2}= & {} \Vert x_{n}+\theta _{n}(x_{n}-x_{n-1})-v\Vert ^{2} \nonumber \\= & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}||x_{n-1}-v\Vert ^{2} \nonumber \\{} & {} \quad +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}. \end{aligned}$$

(64)

From (10), we have

$$\begin{aligned}{} & {} \Vert x_{n+1}-w_{n}\Vert ^{2} = \Vert x_{n+1}-x_{n}-\theta _{n}(x_{n}-x_{n-1})\Vert ^{2} \\{} & {} \quad \ge (1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2} -\theta _{n}(1-\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}. \end{aligned}$$

From (63), we obtain

$$\begin{aligned} \Vert x_{n+1}-v\Vert ^{2}\le & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \\{} & {} \quad +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2} \\{} & {} -\mathcal {E}\Vert x_{n+1}-w_{n}\Vert ^{2}-\kappa \Vert y_{n}-Ty_{n}\Vert ^{2} \\\le & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \\{} & {} \quad +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}\\{} & {} \qquad -\mathcal {E}(1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}\\{} & {} +\mathcal {E}\theta _{n}(1-\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2}\\{} & {} \qquad -\kappa \Vert y_{n}-Ty_{n}\Vert ^{2} \\= & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2}\\{} & {} \quad -\mathcal {E}(1-\theta _{n})\Vert x_{n+1}-x_{n}\Vert ^{2}\\{} & {} +\theta _{n}\left( 1+\theta _{n}+\mathcal {E}(1-\theta _{n})\right) \Vert x_{n}-x_{n-1}\Vert ^{2}\\{} & {} \quad -\kappa \Vert y_{n}-Ty_{n}\Vert ^{2}. \end{aligned}$$

Note $\{\theta _{n}\}$ is an increasing sequence in [0, 1) by the assumption (C2) and the condition (12) holds by the assumption (C5) with $\varepsilon =\mathcal {E}$. Thus, Lemma 2.5 infers us that $\sum _{n=1}^{\infty }\Vert x_{n+1}-x_{n}\Vert ^{2}<\infty$ and $\lim _{n\rightarrow \infty }\Vert x_{n}-v\Vert$ exists. It immediately follows, from (17), that $\lim _{n\rightarrow \infty }\Vert w_{n}-x_{n}\Vert =0$.

Since $\ell :=\lim _{n\rightarrow \infty }\Vert x_{n}-v\Vert$ exists, it follows, from $\lim _{n\rightarrow \infty }\Vert x_{n}-w_{n}\Vert =0$, that $\lim _{n\rightarrow \infty }\Vert w_{n}-v\Vert =\ell$. Hence, from (62), we obtain that $\lim _{n\rightarrow \infty }\Vert w_{n}-Tw_{n}\Vert$ $=$ 0 $=$ $\lim _{n\rightarrow \infty }\Vert y_{n}-Ty_{n}\Vert .$

(c) Let $v\in \textrm{Fix}(S)$. Note $\{\theta _{n}\}$ is an increasing sequence in [0, 1). From (62) and (64), we get

$$\begin{aligned} \Vert x_{n+1}-v\Vert ^{2}\le & {} (1+\theta _{n})\Vert x_{n}-v\Vert ^{2}-\theta _{n}\Vert x_{n-1}-v\Vert ^{2} \\{} & {} \quad +\theta _{n}(1+\theta _{n})\Vert x_{n}-x_{n-1}\Vert ^{2} \\{} & {} \quad -\alpha _{n}(1+\kappa -\alpha _{n})\Vert w_{n}\\{} & {} \quad -Tw_{n}\Vert ^{2}-\kappa \Vert y_{n}-Ty_{n}\Vert ^{2}. \end{aligned}$$

Hence, from the condition (C1), we have

$$\begin{aligned}{} & {} a(1+\kappa -b)\sum \limits _{i=1}^{n}\Vert w_{i}-Tw_{i}\Vert ^{2}+\kappa \sum \limits _{i=1}^{n}\Vert y_{i}-Ty_{i}\Vert ^{2} \\{} & {} \quad \le \sum \limits _{i=1}^{n}(\Vert x_{i}-v\Vert ^{2}-\Vert x_{i+1}-v\Vert ^{2}) +\sum \limits _{i=1}^{n}(\theta _{i+1}\Vert x_{i}\\{} & {} \qquad \quad -v\Vert ^{2}-\theta _{i}\Vert x_{i-1}-v\Vert ^{2}) +\theta (1+\theta )\sum \limits _{i=1}^{n}\Vert x_{i}-x_{i-1}\Vert ^{2} \\{} & {} \quad \le \Vert x_{1}-v\Vert ^{2}-\Vert x_{n+1}-v\Vert ^{2}+\theta _{n+1}\Vert x_{n}-v\Vert ^{2}\\{} & {} \qquad \quad -\theta _{1}\Vert x_{0}-v\Vert ^{2}+\theta (1+\theta )\sum \limits _{i=1}^{n}\Vert x_{i}-x_{i-1}\Vert ^{2} \\{} & {} \quad \le \Vert x_{1}-v\Vert ^{2}+\theta _{n+1}\Vert x_{n}-v\Vert ^{2} +\theta (1+\theta )\\{} & {} \qquad \sum \limits _{i=1}^{\infty }\Vert x_{i}-x_{i-1}\Vert ^{2} \\{} & {} \quad \le M\text { for all } n\in \mathbb {N}\text { and for some } M>0. \end{aligned}$$

Since T is $\beta$ -Lipschitz continuous, we have

$$\begin{aligned} \Vert x_{n+1}-Tx_{n+1}\Vert= & {} \Vert Ty_{n}-Tx_{n+1}\Vert \le \beta \Vert y_{n}-x_{n+1}\Vert \nonumber \\{} & {} \quad =\beta \Vert y_{n}-Ty_{n}\Vert \text { for all } n\in \mathbb {N}. \end{aligned}$$

(65)

Since $\sum \limits _{n=1}^{\infty }\Vert y_{n}-Ty_{n}\Vert ^{2}<\infty ,$ it follows that $\sum \limits _{n=1}^{\infty }\Vert x_{n}-Tx_{n}\Vert ^{2}<\infty$. Noticing that $R_{T}(n):=R_{T,\{x_{n}\}}(n)\le \Vert x_{n}-Tx_{n}\Vert$ for all $n\in \mathbb {N}$ and that $\{R_{T}(n)\}$ is decreasing. Hence $\sum \limits _{n=1}^{\infty }(R_{T}(n))^{2}<\infty .$ Therefore, from Lemma 2.3, we conclude that (20) holds. $\square$

Appendix-IV: Proof of Proposition 4.2

Proof

Let $\{x_{n}\}$ be a sequence in X such that $x_{n}$ $\rightharpoonup z$ and $(I-S_{\lambda })x_{n}\rightarrow 0$ as $n\rightarrow \infty$. We now show that $(I-S_{\lambda })z=0.$ If $Fz=0,$ then $z\in \varOmega [ \textrm{VI}(C,F)],$ i.e, $(I-S_{\lambda })z=0.$ Assume that $Fz\ne 0.$ Let $x\in C.$ Set $y_{n}=S_{\lambda }x_{n}.$ From Lemma 2.1(a), we have

$$\begin{aligned}{} & {} \left\langle x_{n}-\lambda Fx_{n}-y_{n},x-y_{n}\right\rangle \\{} & {} \quad =\left\langle x_{n}-\lambda Fx_{n}-P_{C}(I-\lambda F)x_{n}, x-P_{C}(I-\lambda F)x_{n}\right\rangle \\{} & {} \quad \le 0 \text { for all } n\in \mathbb {N}. \end{aligned}$$

Hence $\left\langle x_{n}-y_{n},x-y_{n}\right\rangle \le \lambda \left\langle Fx_{n},x-y_{n}\right\rangle ,$ which implies that

$$\begin{aligned} \left\langle x_{n}-y_{n},x-y_{n}\right\rangle -\lambda \left\langle Fx_{n},x_{n}-y_{n}\right\rangle \le \lambda \left\langle Fx_{n},x-x_{n}\right\rangle . \nonumber \\ \end{aligned}$$

(66)

Since $x_{n}\rightharpoonup z,\Vert x_{n}-y_{n}\Vert \rightarrow 0$ and F is sequentially weak-to-weak continuous, we have $y_{n}\rightharpoonup z$ and $Fx_{n}\rightarrow Fz.$ Hence, from (66), we have

$$\begin{aligned} 0\le \liminf _{n\rightarrow \infty }\left\langle Fy_{n},x-y_{n}\right\rangle . \end{aligned}$$

(67)

Since the norm is sequentially weakly lower semicontinuous, we get

$$\begin{aligned} 0<\Vert Fz\Vert \le \liminf _{n\rightarrow \infty }\Vert Fy_{n}\Vert . \end{aligned}$$

(68)

It follows that there exists $n_{0}\in \mathbb {N}$ such that $Fy_{n}\ne 0$ for all $n\ge n_{0}.\$Now we choose a sequence $\{\varepsilon _{k}\}$ in $(0,\infty )$ such that $\{\varepsilon _{k}\}_{_{k\ge 0}}$ is strictly decreasing and $\varepsilon _{k}\rightarrow 0.\$For $\{\varepsilon _{k}\}_{k\ge 0}$, from (67) and (68), there exists a strictly increasing sequence $\{n_{k}\}_{k\ge 0}$ of positive integers such that

$$\begin{aligned} \left\langle Fy_{n_{k}},x-y_{n_{k}}\right\rangle +\varepsilon _{k}\ge 0 \text { and }Fy_{n_{k}}\ne 0\text { for all }k\ge 0. \end{aligned}$$

(69)

Set $p_{k}=\frac{1}{\Vert Fy_{n_{k}}\Vert ^{2}}Fy_{n_{k}}.$ Then $\left\langle Fy_{n_{k}},p_{k}\right\rangle =1.$ From (69), we obtain

$$\begin{aligned}{} & {} \left\langle Fy_{n_{k}},x-y_{n_{k}}+\varepsilon _{k}p_{k}\right\rangle = \left\langle Fy_{n_{k}},x-y_{n_{k}}\right\rangle \\{} & {} \quad + \varepsilon _{k}\left\langle Fy_{n_{k}},p_{k}\right\rangle \ge 0\text { for all }k\ge 0. \end{aligned}$$

By the pseudo-monotonicity of F on X, we have

$$\begin{aligned} \left\langle F(x+\varepsilon _{k}p_{k}),x-y_{n_{k}}+\varepsilon _{k}p_{k}\right\rangle \ge 0\text { for all }k\ge 0. \end{aligned}$$

(70)

Note F satisfies the condition $(\mathscr {B})$. Then, by the boundedness of $\{Fy_{n_{k}}\},$ there exist constants $m,M>0$ such that $m\le \Vert Fy_{n_{k}}\Vert \le M$ for all $k\ge 0.$ Hence $\Vert \varepsilon _{k}p_{k}\Vert =\varepsilon _{k}\Vert p_{k}\Vert =\frac{\varepsilon _{k}}{ \Vert Fy_{n_{k}}\Vert }\rightarrow 0$ as $k\rightarrow \infty .$ Thus, from ( 70), we get $\left\langle Fx,x-z\right\rangle \ge 0.$ Since x is an arbitrary element in C, we infer that $z\in \varOmega [\textrm{DVI}(C,F)]$. Therefore, from Lemma 4.2, we conclude that $z\in \varOmega [\textrm{VI}(C,F)],~$i.e., $(I-S_{\lambda })z=0.$ $\square$

Appendix-V: Proof of Proposition 4.3

Proof

(a) Let $u,v\in X.$ Then, from Remark 4.1, we have

$$\begin{aligned} \Vert E_{\lambda }(u)-E_{\lambda }(v)\Vert\le & {} \Vert u-v-\lambda (FS_{\lambda }(u)-FS_{\lambda }(v)\Vert \\\le & {} \Vert u-v\Vert +\lambda L\Vert S_{\lambda }(u)-S_{\lambda }(v)\Vert \\\le & {} (1+\lambda L+\lambda ^{2}L^{2})\Vert u-v\Vert . \end{aligned}$$

(b) Let $\lambda \in (0,1/L)$, $x\in X$ and $v\in \varOmega [\textrm{VI}(C,F)]$. Set $y=P_{C}(x-\lambda Fx)$ and $z=P_{C}(x-\lambda Fy)$. Then $\left\langle Fv,y-v\right\rangle \ge 0.$ By the pseudo-monotonicity of F, we have $\left\langle Fy,y-v\right\rangle \ge 0.$ From Lemma 2.1(d), we have

$$\begin{aligned}{} & {} \Vert P_{C}(x-\lambda Fy)-v\Vert ^{2}\nonumber \\{} & {} \quad \le \Vert x-\lambda Fy-v\Vert ^{2}-\Vert x-\lambda Fy-P_{C}(x-\lambda Fy)\Vert ^{2} \nonumber \\{} & {} \quad =\Vert x-v-\lambda Fy\Vert ^{2}-\Vert x-z-\lambda Fy\Vert ^{2} \nonumber \\{} & {} \quad =\Vert x-v\Vert ^{2}-2\lambda \left\langle Fy,x-v\right\rangle +\lambda ^{2}\Vert Fy\Vert ^{2} -\Vert x-z\Vert ^{2}\nonumber \\{} & {} \qquad +2\lambda \left\langle Fy,x-z\right\rangle -\lambda ^{2}\Vert Fy\Vert ^{2} \nonumber \\{} & {} \quad =\Vert x-v\Vert ^{2}-\Vert x-z\Vert ^{2}+2\lambda \left\langle Fy,v-z\right\rangle \nonumber \\{} & {} \quad =\Vert x-v\Vert ^{2}-\Vert x-z\Vert ^{2}+2\lambda [ \left\langle Fy,v-y\right\rangle +\left\langle Fy,y-z\right\rangle ] \nonumber \\{} & {} \quad \le \Vert x-v\Vert ^{2}-\Vert x-z\Vert ^{2}+2\lambda \left\langle Fy,y-z\right\rangle . \end{aligned}$$

(71)

Substituting x by $x-\lambda Fx$ and y by z in Lemma 2.1(a), we get

$$\begin{aligned} \left\langle x-\lambda Fx-P_{C}(x-\lambda Fx),z-P_{C}(x-\lambda Fx)\right\rangle \le 0, \end{aligned}$$

which gives us that

$$\begin{aligned} \left\langle x-\lambda Fx-y,z-y\right\rangle \le 0. \end{aligned}$$

(72)

Thus, from (72), we have

$$\begin{aligned}{} & {} \left\langle x-\lambda Fy-y,z-y\right\rangle \nonumber \\{} & {} \quad =\left\langle x-\lambda Fx-y,z-y\right\rangle +\left\langle \lambda Fx-\lambda Fy,z-y\right\rangle \nonumber \\{} & {} \quad \le \lambda \left\langle Fx-Fy,z-y\right\rangle \nonumber \\{} & {} \quad \le \lambda L\left\| x-y\right\| \left\| z-y\right\| \nonumber \\{} & {} \quad \le \frac{\lambda L}{2}(\Vert x-y\Vert ^{2}+\Vert y-z\Vert ^{2}). \end{aligned}$$

(73)

Hence, from (71) and (73), we get

$$\begin{aligned} \Vert z-v\Vert ^{2}\le & {} \Vert x-v\Vert ^{2}-\Vert x-y+(y-z)\Vert ^{2} \\{} & {} +2\lambda \left\langle Fy,y-z\right\rangle \\= & {} \Vert x-v\Vert ^{2}-\Vert x-y\Vert ^{2}-\Vert y-z\Vert ^{2}\\{} & {} -2\left\langle x-y,y-z\right\rangle +2\lambda \left\langle Fy,y-z\right\rangle \\= & {} \Vert x-v\Vert ^{2}-\Vert x-y\Vert ^{2}-\Vert y-z\Vert ^{2}\\{} & {} -2\left\langle x-\lambda Fy-y,y-z\right\rangle \\\le & {} \Vert x-v\Vert ^{2}-\Vert x-y\Vert ^{2}-\Vert y-z\Vert ^{2}\\{} & {} +\lambda L(\Vert x-y\Vert ^{2}+\Vert y-z\Vert ^{2}) \\= & {} \Vert x-v\Vert ^{2}-(1-\lambda L)\Vert x-y\Vert ^{2}\\{} & {} -(1-\lambda L)\Vert y-z\Vert ^{2}. \end{aligned}$$

For $\rho =1/2$, from (9), we have

$$\begin{aligned} \Vert y-z\Vert ^{2}=\Vert y-x+x-z\Vert ^{2}\ge \frac{1}{2}\Vert x-z\Vert ^{2}-\Vert y-x\Vert ^{2}. \nonumber \\ \end{aligned}$$

(74)

Hence, from (74), we have

$$\begin{aligned} \Vert z-v\Vert ^{2}\le & {} \Vert x-v\Vert ^{2}-(1-\lambda L)\Vert x-y\Vert ^{2}\\{} & {} -\frac{1-\lambda L}{2}\Vert x-z\Vert ^{2}+(1-\lambda L)\Vert y-x\Vert ^{2}\\= & {} \Vert x-v\Vert ^{2}-\frac{1-\lambda L}{2}\Vert x-z\Vert ^{2}. \end{aligned}$$

(c) Let $\lambda \in (0,1/L)$. Lemma 4.1 shows that $\textrm{Fix}(S_{\lambda })=\varOmega [\textrm{VI}(C,F)].$ Note $\emptyset \ne \textrm{Fix} (S_{\lambda })\subseteq \textrm{Fix}(E_{\lambda }).$ We now show that $\textrm{Fix}(E_{\lambda })\subseteq \textrm{Fix}(S_{\lambda }).$ Let $u\in \textrm{Fix}(E_{\lambda })$ and $v\in \textrm{Fix}(S_{\lambda }).$ From Part b(i), we get

$$\begin{aligned}{} & {} \Vert u-v\Vert ^{2} = \Vert E_{\lambda }(u)-v\Vert ^{2} \le \Vert u-v\Vert ^{2}\\{} & {} \quad -(1-\lambda L)\Vert u-S_{\lambda }(u)\Vert ^{2} -(1-\lambda L)\Vert S_{\lambda }(u)-E_{\lambda }(u)\Vert ^{2} \\{} & {} \quad =\Vert u-v\Vert ^{2}-2(1-\lambda L)\Vert u-S_{\lambda }(u)\Vert ^{2}, \end{aligned}$$

which implies that $u\in \textrm{Fix}(S_{\lambda }).$

(d) Let $\lambda \in (0,1/L)$. Let $\{x_{n}\}$ be a bounded sequence in X such that $\lim _{n\rightarrow \infty }\Vert x_{n}-E_{\lambda }(x_{n})\Vert =0.$ Let $v\in \varOmega [\textrm{VI}(C,F)].$ From Part (b)(i), we have

$$\begin{aligned} \Vert E_{\lambda }(x_{n})-v\Vert ^{2}\le & {} \Vert x_{n}-v\Vert ^{2}\\{} & {} \quad -(1-\lambda L)\Vert x_{n}-S_{\lambda }(x_{n})\Vert ^{2}\text { for all } n\in \mathbb {N}, \end{aligned}$$

it follows that

$$\begin{aligned}{} & {} (1-\lambda L)\Vert x_{n}-S_{\lambda }(x_{n})\Vert ^{2}\\{} & {} \quad \le \Vert x_{n}-v\Vert ^{2}-\Vert E_{\lambda }(x_{n})-v\Vert ^{2} \\{} & {} \quad =(\Vert x_{n}-v\Vert -\Vert E_{\lambda }(x_{n})-v\Vert ) (\Vert x_{n}-v\Vert +\Vert E_{\lambda }(x_{n})-v\Vert ) \\{} & {} \quad \le 2\Vert x_{n}-v\Vert \Vert x_{n}-E_{\lambda }(x_{n})\Vert \text { for all }n\in \mathbb {N}. \end{aligned}$$

By the boundedness of $\{x_{n}\}$, we have $\lim _{n\rightarrow \infty }\Vert x_{n}-S_{\lambda }(x_{n})\Vert =0.$ $\square$

Appendix-VI: Proof of Proposition 4.4

Proof

(a) Let $u,v\in X.$ Then

$$\begin{aligned} \Vert T_{\lambda }(u)-T_{\lambda }(v)\Vert\le & {} \Vert S_{\lambda }(u)-S_{\lambda }(v)\Vert +\lambda \Vert FS_{\lambda }(u)\\{} & {} -FS_{\lambda }(v)\Vert +\lambda \Vert Fu-Fv\Vert \\\le & {} (1+3\lambda L+\lambda ^{2}L^{2})\Vert u-v\Vert . \end{aligned}$$

(b) Let $\lambda \in (0,1/L),$ $x\in X$ and $z\in \varOmega [\textrm{VI} (C,F)].$ Set $y:=P_{C}(x-\lambda Fx).~$Hence

$$\begin{aligned}{} & {} \Vert T_{\lambda }(x)-x\Vert =\Vert y-x-\lambda (Fy-Fx)\Vert \\{} & {} \quad \le \Vert y-x\Vert +\lambda \Vert Fy-Fx\Vert \le (1+\lambda L)\Vert y-x\Vert \end{aligned}$$

and

$$\begin{aligned} \Vert S_{\lambda }(x)-x\Vert&=\Vert T_{\lambda }x-x+\lambda (Fy-Fx)\Vert \le \Vert T_{\lambda }x-x\Vert \\&+\lambda L\Vert y-x\Vert , \end{aligned}$$

which implies that $(1-\lambda L)\Vert x-S_{\lambda }(x)\Vert \le \Vert x-T_{\lambda }x\Vert .$ From Lemma 2.1(a), we have

$$\begin{aligned}{} & {} y=P_{C}(x-\lambda Fx) \Leftrightarrow \langle y-x+\lambda Fx,y-z\rangle \\{} & {} \quad \le 0 \Leftrightarrow \langle y-x,y-z\rangle \le -\lambda \langle Fx,y-z\rangle . \end{aligned}$$

Thus,

$$\begin{aligned} \Vert T_{\lambda }(x)-z\Vert ^{2}= & {} \Vert y-\lambda (Fy-Fx)-z\Vert ^{2}\\= & {} \Vert y-x+x-z\Vert ^{2}+\lambda ^{2}\Vert Fy-Fx\Vert ^{2}\\{} & {} -2\lambda \langle Fy-Fx,y-z\rangle \\= & {} \Vert y-x\Vert ^{2}+\Vert x-z\Vert ^{2}+2\langle y-x,x-z\rangle \\{} & {} +\lambda ^{2}\Vert Fy-Fx\Vert ^{2}-2\lambda \langle Fy-Fx,y-z\rangle \\= & {} \Vert y-x\Vert ^{2}+\Vert x-z\Vert ^{2}+2\langle y-x,x-y\rangle \\{} & {} +2\langle y-x,y-z\rangle \\{} & {} +\lambda ^{2}\Vert Fy-Fx\Vert ^{2}-2\lambda \langle Fy-Fx,y-z\rangle \\= & {} \Vert x-z\Vert ^{2}-\Vert y-x\Vert ^{2}+2\langle y-x,y-z\rangle \\{} & {} +\lambda ^{2}\Vert Fy-Fx\Vert ^{2}-2\lambda \langle Fy-Fx,y-z\rangle \\\le & {} \Vert x-z\Vert ^{2}-\Vert y-x\Vert ^{2}+\lambda ^{2}\Vert Fy-Fx\Vert ^{2}\\{} & {} -2\lambda \langle Fx,y-z\rangle -2\lambda \langle Fy-Fx,y-z\rangle \\\le & {} \Vert x-z\Vert ^{2}-\Vert y-x\Vert ^{2}+\lambda ^{2}L^{2}\Vert y-x\Vert ^{2}\\{} & {} -2\lambda \langle Fy,y-z\rangle . \end{aligned}$$

Since $z\in \varOmega [\text {VI}(C,F)],$ we have $\langle Fz,y-z\rangle \ge 0.$ By pseudo-monotonicity of F,  we have $\langle Fy,y-z\rangle \ge 0.$ From (75), we have

$$\begin{aligned} \Vert T_{\lambda }(x)-z\Vert ^{2}\le & {} \Vert x-z\Vert ^{2}-(1-\lambda ^{2}L^{2})\Vert x-S_{\lambda }(x)\Vert ^{2}\\{} & {} \quad =\Vert x-z\Vert ^{2}-\kappa \Vert x-T_{\lambda }(x)\Vert ^{2}. \end{aligned}$$

(c) Let $\lambda \in (0,1/L)$. Note $\textrm{Fix}(S_{\lambda })=\varOmega [\textrm{VI}(C,F)].$ Observe that $\textrm{Fix}(S_{\lambda })\subseteq \textrm{Fix}(T_{\lambda }).\$We now show that $\textrm{Fix} (T_{\lambda })\subseteq \textrm{Fix}(S_{\lambda }).$ Let $u\in \textrm{Fix} (T_{\lambda })$ and $v\in \textrm{Fix}(S_{\lambda }).$ From Part b(iii), we get

$$\begin{aligned} \Vert u-v\Vert ^{2}\!=\!\Vert T_{\lambda }(u)\!-\!v\Vert ^{2}\!\le \! \Vert u-v\Vert ^{2}\!-\!(1\!-\!\lambda ^{2}L^{2})\Vert u\!-\!S_{\lambda }(u)\Vert ^{2}, \end{aligned}$$

which implies that $u\in \textrm{Fix}(S_{\lambda }).$

(d) Let $\lambda \in (0,1/L)$. Let $\{x_{n}\}$ be a bounded sequence in X such that $\lim _{n\rightarrow \infty }\Vert x_{n}-T_{\lambda }(x_{n})\Vert =0.$ Using Part (b)(ii), we obtain $(1-\lambda L)\Vert x_{n}-S_{\lambda }(x_{n})\Vert \le \Vert x_{n}-T_{\lambda }(x_{n})\Vert \text { for all }n\in \mathbb {N}.$ Thus, we conclude that the operator $S_{\lambda }$ has property $(\mathscr {A} )$ with respect to the operator $T_{\lambda }$. $\square$

Appendix-VII: Proof of Example 5.2

To prove the pseudo-monotonicity and Lipschitz continuity of operator F defined by (46), we define $g:X\rightarrow \mathbb {R}$ such that $g(x)=e^{-\langle x,Ux\rangle }+\alpha \ \text {for all}\ x\in X.$ Let $x,y\in X$ such that $\langle Fx,y-x\rangle \ge 0.$ Since $g(x)>0,$ it follows that $\langle Vx+q,y-x\rangle \ge 0.$ Hence

$$\begin{aligned} \langle Fy,y-x\rangle= & {} g(y)\langle Vy+q,y-x\rangle \\\ge & {} g(y)(\langle Vy+q,y-x\rangle -\langle Vx+q,y-x\rangle )\\= & {} g(y)\langle V(y-x),y-x\rangle \\\ge & {} 0. \end{aligned}$$

Thus, F is pseudo-monotonous.

Now, let $x,h\in X.$ Then $\nabla F(x)(h) =2e^{-\langle x,Ux\rangle }\langle Ux,h\rangle (Vx+p)+\left( e^{-\langle x,Ux\rangle }+\alpha \right) Vh$ and

$$\begin{aligned}{} & {} \Vert \nabla F(x)(h)\Vert \le 2e^{-\eta \Vert x\Vert ^{2}}\Vert U\Vert \Vert x\Vert (\Vert V\Vert \Vert x\Vert +\Vert p\Vert )\Vert h\Vert \\{} & {} \quad +\left( e^{-\eta \Vert x\Vert ^{2}} +\alpha \right) \Vert V\Vert \Vert h\Vert . \end{aligned}$$

Hence

$$\begin{aligned} \Vert \nabla F(x)\Vert\le & {} 2e^{-\eta \Vert x\Vert ^{2}}\Vert U\Vert \Vert x\Vert (\Vert V\Vert \Vert x\Vert +\Vert p\Vert )\\{} & {} +\left( e^{-\eta \Vert x\Vert ^{2}}+\alpha \right) \Vert V\Vert . \end{aligned}$$