Let
\(H=\mathbb{R}^{n}\) and
\(C= \{x\in\mathbb{R}^{n}:-5 \leq x_{i}\leq 5,\forall i\in\{1,2,\ldots,n\}\}\). Let the bifunction
\(g: \mathbb {R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}\) be defined by
$$g(x,y)= \langle Px+Qy, y-x\rangle\quad \text{for all } x,y\in \mathbb{R}^{n}, $$
where
P and
Q are randomly symmetric positive semidefinite matrices such that
\(P-Q\) is positive semidefinite. Then
g is pseudomonotone on
\(\mathbb{R}^{n}\). Indeed, let
\(g(x,y)\geq0\) for every
\(x,y\in\mathbb{R}^{n}\), we have
$$\begin{aligned} g(y,x)\leq g(x,y)+g(y,x) =& \langle P x+ Q y, y-x\rangle+\langle P y+ Q x, x-y\rangle \\ =& -\bigl\langle (P-Q) (x-y), x-y\bigr\rangle \leq0. \end{aligned}$$
Next, we obtain that
g is Lipschitz-type continuous with
\(L_{1}=L_{2}=\frac{1}{2}\|P-Q\|\). Indeed, for each
\(x,y,z \in\mathbb{R}^{n}\),
$$\begin{aligned} g(x,y)+g(y,z)-g(x,z) =&\langle P x + Q y, y-x\rangle+\langle P y+ Q z, z-y\rangle-\langle P x + Q z , z-x\rangle \\ =& \bigl\langle (P-Q) (x-y), y-z\bigr\rangle \\ \geq& -2\frac{ \Vert P-Q \Vert }{2} \Vert x-y \Vert \Vert y-z \Vert \\ \geq& -\frac{ \Vert P-Q \Vert }{2} \Vert x-y \Vert ^{2}-\frac{ \Vert P-Q \Vert }{2} \Vert y-z \Vert ^{2}, \end{aligned}$$
where
\(\|P-Q\|\) is the spectral norm of the matrix
\(\|P-Q\|\), that is, the square root of the largest eigenvalue of the positive semidefinite matrix
\((P-Q)^{T}(P-Q)\). It is easy to check that
\(\varOmega\neq\emptyset\). Furthermore, we define the bifunction
\(f:\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}\) as
$$f(x,y)= \langle Ax +By, y-x\rangle\quad \text{for all } x,y\in \mathbb{R}^{n}, $$
with
A and
B being positive definite matrices defined by
$$B = N^{T}N + n I_{n}\quad \text{and}\quad A = B + M^{T}M+ n I_{n}, $$
where
\(M,N\) are randomly
\(n\times n\) matrices and
\(I_{n}\) is the identity matrix. Then we have
f is
n-strongly monotone on
\(\mathbb{R}^{n}\). Indeed, let
\(x,y\in\mathbb{R}^{n}\), we get
$$\begin{aligned} f(x,y)+f(y,x) =& \langle A x + B y, y-x\rangle+ \langle A y + B x, x-y\rangle \\ =& -\bigl\langle (A-B) (x-y), x-y\bigr\rangle \\ =& -\bigl\langle M^{T}M+ n I_{n}(x-y), x-y\bigr\rangle \\ =& -\bigl\langle M^{T}M (x-y), x-y\bigr\rangle -\bigl\langle n I_{n}(x-y), x-y\bigr\rangle \\ =& - \bigl\Vert M(x-y) \bigr\Vert ^{2}-n \Vert x-y \Vert ^{2} \\ \leq& -n \Vert x-y \Vert ^{2}. \end{aligned}$$
Moreover,
\(\partial_{2}f(x,x)=\{(A+B) x\}\) and
\(\|(A+B)x - (A+B)y\|\leq\|A+B\|\|x-y\|\) for all
\(x,y\in\mathbb{R}^{n}\). Thus the mapping
\(x \rightarrow\partial_{2}g(x,x)\) is bounded and
\(\| A+B\|\)-Lipschitz continuous on every bounded subset of
H. In this example, we consider the quadratic optimization
$$ \min_{x\in C} \biggl\{ \frac{1}{2}x^{T}Hx+f^{T}x \biggr\} , $$
(76)
where
H is a matrix,
f and
x are vectors. From the subproblem of solving
\(y_{k}\) and
\(z_{k}\) in Algorithm
1, we can consider problem (
76).
We have tested for this example where
\(n=5, 10, 50,100\), and 500. Starting point
\(x_{0}\) is a randomly initial point. Take the parameters
$$\alpha_{k}=\frac{1}{k+4},\qquad \eta_{k}= \frac{k+1}{3(k+4)},\qquad \lambda_{k}=\frac{1}{2 \Vert P-Q \Vert },\qquad \mu= \frac{2}{ \Vert A+B \Vert ^{2}}. $$
We have implemented Algorithm
1 for this problem in Matlab R2015 running on a Desktop with Intel(R) Core(TM) i5-7200u CPU 2.50 GHz, and 4 GB RAM, and we used the stopping criteria
\(\|x_{k+1} - x_{k}\|<\varepsilon \) with
\(\varepsilon= 0.001\) is a tolerance to cease the algorithm. Denote that
-
N.P: the number of the tested problems.
-
Average iteration: the average number of iterations.
-
Average times: the average CPU-computation times (in s).
The computation results are reported in the following tables.