Let
\(x_{0}\in H\) be a solution of SVMQVIP (
1.1), then
$$\bigl\langle A(u_{0},u_{0}), y-x_{0}\bigr\rangle +f (x_{0},y)-f (x_{0},x_{0})\geq0,\quad \forall y\in K(x_{0}). $$
Substituting
\(y=P_{K(x_{0})}^{f}[x-\theta A(u,u)]\) in the above inequality, we have
$$ \bigl\langle A(u_{0},u_{0}), P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr]-x_{0} \bigr\rangle +f \bigl(x_{0},P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr] \bigr)-f (x_{0},x_{0}) \geq0. $$
(2.6)
For any fixed
\(x\in H\), and
\(\theta> 0\), we observe that
$$x-\theta A(u,u)\in(I+\theta\partial f) (I+\theta\partial f)^{-1} \bigl(x-\theta A(u,u) \bigr)=(I+\theta\partial f)P_{K}^{f} \bigl[x-\theta A(u,u)\bigr], $$
which is equivalent to
$$-A(u,u)+{1\over \theta} \bigl[x-P_{K}^{f}\bigl[x- \theta A(u,u)\bigr] \bigr]\in \partial f \bigl(P_{K}^{f} \bigl[x-\theta A(u,u)\bigr] \bigr). $$
By the definition of a sub-differential, we have
$$\begin{aligned}& \biggl\langle A(u,u)-{1\over \theta} \bigl(x-P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr] \bigr), y-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \biggr\rangle \\& \quad {}+f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr],y \bigr)-f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr],P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \bigr)\geq0. \end{aligned}$$
Taking
\(y=x_{0}\) in the above we get
$$\begin{aligned}& \biggl\langle A(u,u)-{1\over \theta} \bigl(x-P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr] \bigr), x_{0}-P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr] \biggr\rangle \\& \quad {}+f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr],x_{0} \bigr)-f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr],P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \bigr) \geq0. \end{aligned}$$
This implies that
$$\begin{aligned}& \biggl\langle -A(u,u)+{1\over \theta} \bigl(x-P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr] \bigr), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x_{0} \biggr\rangle \\& \quad {}+f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr],x_{0} \bigr) \\& \quad {}-f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr],P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \bigr)\geq 0. \end{aligned}$$
(2.7)
Adding (
2.6) and (
2.7), we get
$$\begin{aligned}& \biggl\langle A(u_{0},u_{0})-A(u,u)+{1\over \theta} \bigl(x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \bigr), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x_{0} \biggr\rangle \\& \quad {}+f \bigl(x_{0}, P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr] \bigr)-f (x_{0}, x_{0} )+f \bigl(P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr],x_{0} \bigr) \\& \quad {}-f \bigl(P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr], P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \bigr)\geq0. \end{aligned}$$
Since
f is skew-symmetric,
$$\biggl\langle A(u_{0},u_{0})-A(u,u)+{1\over \theta} \bigl(x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \bigr), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x_{0} \biggr\rangle \geq0. $$
This also can be written as
$$\begin{aligned}& \theta \bigl\langle A(u_{0},u_{0})-A(u,u), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x \bigr\rangle +\theta \bigl\langle A(u_{0},u_{0})-A(u,u), x-x_{0} \bigr\rangle \\& \quad {}+ \bigl\langle x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr], P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x \bigr\rangle \\& \quad {}+ \bigl\langle x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr], x-x_{0} \bigr\rangle \geq0, \end{aligned}$$
which implies that
$$\begin{aligned}& \theta \bigl\langle A(u_{0},u_{0})-A(u,u), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x \bigr\rangle + \bigl\langle x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr], x-x_{0} \bigr\rangle \\& \quad \geq \theta \bigl\langle A(u_{0},u_{0})-A(u,u), x_{0}-x \bigr\rangle \\& \qquad {}+ \bigl\langle x-P_{K(x_{0})}^{f} \bigl[x-\theta A(u,u)\bigr], x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr] \bigr\rangle . \end{aligned}$$
By using the strong monotonicity of
A, we get
$$\begin{aligned}& \theta \bigl\langle A(u_{0},u_{0})-A(u,u), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]-x \bigr\rangle + \bigl\langle x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr], x-x_{0} \bigr\rangle \\& \quad \geq\alpha\theta\|x_{0}-x\|^{2}+\bigl\Vert R(x, \theta)\bigr\Vert ^{2}. \end{aligned}$$
Also the above inequality can be written as
$$\begin{aligned}& \theta \bigl\langle A(u_{0},u_{0})-A(u,u), P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr] \\& \qquad {}-x-P_{K(x)}^{f} \bigl[x-\theta A(u,u)\bigr]+P_{K(x)}^{f}\bigl[x-\theta A(u,u) \bigr] \bigr\rangle \\& \qquad {}+ \bigl\langle x-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr]-P_{K(x)}^{f}\bigl[x-\theta A(u,u)\bigr]+P_{K(x)}^{f} \bigl[x-\theta A(u,u)\bigr], x-x_{0} \bigr\rangle \\& \quad \geq\alpha \theta\| x_{0}-x\|^{2}+\bigl\Vert R(x, \theta)\bigr\Vert ^{2}. \end{aligned}$$
By using the Cauchy-Schwarz inequality along with the triangular inequality, we have
$$\begin{aligned}& \theta\bigl\Vert A(u_{0},u_{0})-A(u,u)\bigr\Vert \cdot \bigl\Vert P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u) \bigr]-P_{K(x)}^{f}\bigl[x-\theta A(u,u)\bigr]\bigr\Vert \\& \qquad {}+\theta\bigl\Vert A(u_{0},u_{0})-A(u,u)\bigr\Vert \cdot\bigl\Vert P_{K(x)}^{f}\bigl[x-\theta A(u,u) \bigr]-x\bigr\Vert \\& \qquad {}+\bigl\Vert x- P_{K(x)}^{f}\bigl[x-\theta A(u,u)\bigr]\bigr\Vert \cdot \Vert x-x_{0}\Vert \\& \qquad {}+ \bigl\Vert P_{K(x)}^{f}\bigl[x-\theta A(u,u) \bigr]-P_{K(x_{0})}^{f}\bigl[x-\theta A(u,u)\bigr]\bigr\Vert \cdot \Vert x-x_{0}\Vert \\& \quad \geq\alpha\theta \Vert x_{0}-x\Vert ^{2}+\bigl\Vert R(x,\theta)\bigr\Vert ^{2}. \end{aligned}$$
Now using the Lipschitz continuity of the operator
A and assumption on
\(P_{K(x)}^{f}(\cdot)\), we have
$$\begin{aligned}& \theta\beta \Vert u_{0}-u\Vert \cdot k\Vert x_{0}-x \Vert +\theta\beta \Vert u_{0}-u\Vert \cdot\bigl\Vert R(x,\theta) \bigr\Vert +\bigl\Vert R(x,\theta)\bigr\Vert \cdot \Vert x-x_{0} \Vert + k\Vert x-x_{0}\Vert ^{2} \\& \quad \geq\alpha\theta \Vert x_{0}-x\Vert ^{2}+\bigl\Vert R(x,\theta)\bigr\Vert ^{2}. \end{aligned}$$
Now using the
M-Lipschitz continuity of
T, we have
$$\begin{aligned}& k\theta\beta\mu \Vert x_{0}-x\Vert ^{2}+\theta\beta\mu \Vert x_{0}-x\Vert \cdot\bigl\Vert R(x,\theta)\bigr\Vert +\bigl\Vert R(x,\theta)\bigr\Vert \cdot \Vert x-x_{0}\Vert + k\Vert x-x_{0}\Vert ^{2} \\& \quad \geq\alpha\theta \Vert x_{0}-x \Vert ^{2}+\bigl\Vert R(x,\theta)\bigr\Vert ^{2}. \end{aligned}$$
Therefore, we have
$$\|x-x_{0}\|\leq{1+\theta\beta\mu\over \alpha\theta-(1+\theta\beta \mu)k}\bigl\Vert R(x,\theta)\bigr\Vert ,\quad \forall x\in H, $$
where
\(\theta>{k\over \alpha-\beta\mu k}\). This completes the proof. □