A Nonmonotone Smoothing Newton Algorithm for Weighted Complementarity Problem

Abstract

The weighted complementarity problem (denoted by WCP) significantly extends the general complementarity problem and can be used for modeling a larger class of problems from science and engineering. In this paper, by introducing a one-parametric class of smoothing functions which includes the weight vector, we propose a smoothing Newton algorithm with nonmonotone line search to solve WCP. We show that any accumulation point of the iterates generated by this algorithm, if exists, is a solution of the considered WCP. Moreover, when the solution set of WCP is nonempty, under assumptions weaker than the Jacobian nonsingularity assumption, we prove that the iteration sequence generated by our algorithm is bounded and converges to one solution of WCP with local superlinear or quadratic convergence rate. Promising numerical results are also reported.

Appendix: The proof of the example given in remark (ii) for Assumption 5.1

Since M is positive definite, all of its diagonal elements are greater than zero. Without loss of generality, here we assume that $M-I_n$ is positive definite where $I_n$ represents the $n\times n$ identity matrix. Consider the smoothing function

$$\begin{aligned}\psi _c(\mu ,a,b)=a+b-\sqrt{a^2+b^2+(\tau -2)a b+(4-\tau )c+4\mu ^t },~\forall ~(\mu ,a,b)\in \mathcal {R}_+\times \mathcal {R}^2,\end{aligned}$$

where $t\in [1,2]$ and $\tau \in [0,4)$. By (3.6), we have

$$\begin{aligned}\psi _c(0,a,b)=0\Longleftrightarrow a\ge 0,~ b\ge 0,~ ab=c.\end{aligned}$$

Then we can reformulate the problem (5.7) as the following nonlinear smooth equations:

$$\begin{aligned}\mathcal {H}(z):=\mathcal {H}(\mu ,x,s)=\left( \begin{array}{c} {\mu }\\ s-Mx-a\\ \psi _{w_1}(\mu ,x_1,s_1)\\ \vdots \\ \psi _{w_n}(\mu ,x_n,s_n) \end{array} \right) =0\end{aligned}$$

and apply Algorithm 4.1 to solve it. Let

$$\begin{aligned}g(\mu ,a,b):=\sqrt{a^2+b^2+(\tau -2)ab+(4-\tau )c+4\mu ^t},~\forall ~(\mu ,a,b)\in \mathcal {R}_+\times \mathcal {R}^2.\end{aligned}$$

Then, $\mathcal {H}(z)$ is continuously differentiable at any $z\in \mathcal {R}_{++}\times \mathcal {R}^{2n}$ with its Jacobian

$$\begin{aligned}\mathcal {H}'(z)=\left[ \begin{array}{cccc} 1&{}0&{}0\\ 0&{}-M&{}I\\ d_\mu &{}\mathbf{diag }(d_x)&{}\mathbf{diag }(d_s) \end{array} \right] , \end{aligned}$$

where

$$\begin{aligned}&d_\mu :=\bigg (-\frac{2t}{\mu ^{1-t}g(\mu ,x_1,s_1)},...,-\frac{2t}{\mu ^{1-t}g(\mu ,x_n,s_n)}\bigg )^T,\\&d_x:=\bigg (1-\frac{x_1+(\tau /2-1)s_1}{g(\mu ,x_1,s_1)},...,1-\frac{x_n+(\tau /2-1)s_n}{g(\mu ,x_n,s_n)}\bigg )^T,\\&d_s:=\bigg (1-\frac{s_1+(\tau /2-1)x_1}{g(\mu ,x_1,s_1)},...,1-\frac{s_n+(\tau /2-1)x_n}{g(\mu ,x_n,s_n)}\bigg )^T. \end{aligned}$$

Since M is positive definite, $\mathcal {H}'(z)$ is nonsingular at any $z\in \mathcal {R}_{++}\times \mathcal {R}^{2n}$ and we can find its inverse

$$\begin{aligned} \mathcal {H}'(z)^{-1}=\left[ \begin{array}{cccc} 1&{}0&{}0\\ z_{21}&{}z_{22}&{}z_{23}\\ z_{31}&{}z_{32}&{}z_{33} \end{array} \right] , \end{aligned}$$

where

$$\begin{aligned}&z_{21}=-[\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}d_\mu ,\\&z_{22}=-[\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}\mathbf{diag }(d_s),\\&z_{23}=[\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}, \\&z_{31}=-M[\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}d_\mu ,\\&z_{32}=I-M[\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}\mathbf{diag }(d_s),\\&z_{33}=M[\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}. \end{aligned}$$

In what follows, we divide the analysis into three parts.

Part 1. We show that $\mathbf{diag }(d_x)$ and $\mathbf{diag }(d_s)$ are positive semidefinite and bounded when $\mu \rightarrow 0^+$. This result holds by noticing that $g(\mu ,a,b)$ can be written as

$$\begin{aligned} g(\mu ,a,b)= & {} \sqrt{[a+(\tau /2-1)b]^2+\tau (1-\tau /4)b^2+(4-\tau )c+4\mu ^t}\\= & {} \sqrt{[b+(\tau /2-1)a]^2+\tau (1-\tau /4)a^2+(4-\tau )c+4\mu ^t}, \end{aligned}$$

and hence for any $(\mu ,a,b)\in \mathcal {R}_+\times \mathcal {R}^2$,

$$\begin{aligned}0\le 1-\frac{a+(\tau /2-1)b}{g(\mu ,a,b)}\le 2,~0\le 1-\frac{b+(\tau /2-1)a}{g(\mu ,a,b)}\le 2.\end{aligned}$$

Part 2. We show that $\Vert [\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}\Vert \le \frac{\sqrt{n}\mathrm {cond}(M-I_n)}{2-\sqrt{\tau }}$ when $\mu \rightarrow 0^+$, where $\mathrm {cond}(M-I_n)$ is the conditional number of $M-I_n$. First, we have

$$\begin{aligned}\mathbf{diag }(d_x)+\mathbf{diag }(d_s)=\mathbf {diag}(d_{xs}),\end{aligned}$$

where

$$\begin{aligned}d_{xs}:=\bigg (2-\frac{\tau /2(x_1+s_1)}{g(\mu ,x_1,s_1)},...,2-\frac{\tau /2(x_n+s_n)}{g(\mu ,x_n,s_n)}\bigg )^T.\end{aligned}$$

Since $\big |\frac{\tau /2(a+b)}{g(\mu ,a,b)}\big |\le \big |\frac{\tau /2(a+b)}{\sqrt{a^2+b^2+(\tau -2)ab}}\big |\le \sqrt{\tau }$ holds for any $\tau \in [0,4)$ and $(\mu ,a,b)\in \mathcal {R}_+\times \mathcal {R}^2$, we have

$$\begin{aligned}0<2-\sqrt{\tau }<2-\frac{\tau /2(x_i+s_i)}{g(\mu ,x_i,s_i)}<2+\sqrt{\tau },~~\forall ~i=1,...,n.\end{aligned}$$

Hence, $\mathbf {diag}(d_{xs})$ is positive definite and so is $\mathbf {diag}(d_{xs})^{-1}$. Since $M-I_n$ is positive definite, it is invertible and $(M-I_n)^{-1}$ is also positive definite. So, we have

$$\begin{aligned} \mathbf{diag }(d_x)+\mathbf{diag }(d_s)M= & {} \mathbf {diag}(d_{xs})-\mathbf{diag }(d_s)+\mathbf{diag }(d_s)M\\= & {} \mathbf {diag}(d_{xs})+\mathbf{diag }(d_s)(M-I_n)\\= & {} \mathbf {diag}(d_{xs}) [(M-I_n)^{-1}+\mathbf {diag}(d_{xs})^{-1}\mathbf{diag }(d_s)](M-I_n). \end{aligned}$$

When $\mu \rightarrow 0^+$, since $\mathbf{diag }(d_s)$ is positive semidefinite by Part 1, $\mathbf {diag}(d_{xs})^{-1}\mathbf{diag }(d_s)$ is positive semidefinite. So, $(M-I_n)^{-1}+\mathbf {diag}(d_{xs})^{-1}\mathbf{diag }(d_s)$ is positive definite. Hence, when $\mu \rightarrow 0^+,$ $\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M$ is nonsingular and

$$\begin{aligned}{}[\mathbf{diag }(d_x)+ & {} \mathbf{diag }(d_s)M]^{-1}=(M-I_n)^{-1}[(M-I_n)^{-1}\\+ & {} \mathbf {diag}(d_{xs})^{-1}\mathbf{diag }(d_s)]^{-1}\mathbf {diag}(d_{xs})^{-1},\end{aligned}$$

which implies that

$$\begin{aligned}\Vert [\mathbf{diag }(d_x)+\mathbf{diag }(d_s)M]^{-1}\Vert \le \frac{\sqrt{n}\mathrm {cond}(M-I_n)}{2-\sqrt{\tau }},\end{aligned}$$

where $\mathrm {cond}(M-I_n):=\Vert M-I_n\Vert \Vert (M-I_n)^{-1}\Vert$.

Part 3. We show that when $\mu \rightarrow 0^+$, if $w>0$, then $\Vert d_\mu \Vert \le \varrho$ where $\varrho >0$ is a constant, and if $w\ge 0$, then $\Vert d_\mu \Vert \le \frac{\sqrt{n}t}{\mu ^{1-\frac{t}{2}}}$. This result holds since for any $(\mu ,a,b)\in \mathcal {R}_+\times \mathcal {R}^2$,

$$\begin{aligned} \frac{2t}{\mu ^{1-t}g(\mu ,a,b)}= & {} \frac{2t}{\mu ^{1-t}\sqrt{[a+(\tau /2-1)b]^2+\tau (1-\tau /4)b^2+(4-\tau )c+4\mu ^t}}\\\le & {} \left\{ \begin{array}{c} \frac{2t\mu ^{t-1}}{\sqrt{(4-\tau )c}},~\mathrm {if}~c>0,~~~~~~~~~~~~~ \\ \frac{t}{\mu ^{1-\frac{t}{2}}}, ~~~\mathrm {if}~c=0.~~~~~~~~~~~~\\ \end{array} \right. \end{aligned}$$

Therefore, when $\mu \rightarrow 0^+$, from Parts 1,2 and 3, we have that $z_{22},z_{23},z_{32}$ and $z_{33}$ are bounded, and $z_{21}, z_{31}$ are bounded or

$$\begin{aligned}\Vert z_{21}\Vert \le \frac{n\mathrm {cond}(M-I_n)t}{(2-\sqrt{\tau })\mu ^{1-\frac{t}{2}}}~~\mathrm {and}~~\Vert z_{31}\Vert \le \frac{n\Vert M\Vert \mathrm {cond}(M-I_n)t}{(2-\sqrt{\tau })\mu ^{1-\frac{t}{2}}}.\end{aligned}$$

Hence, for any $t\in [1,2]$, we can conclude that for any $z=(\mu ,x,s)\in \mathcal {R}_{++}\times \mathcal {R}^{2n}$, when $\mu \rightarrow 0^+$, $\Vert \mathcal {H}'(z)^{-1}\Vert$ is bounded or there exists a constant ${C}_t>0$ such that

$$\begin{aligned}\Vert \mathcal {H}'(z)^{-1}\Vert \le \frac{{C}_t}{\mu ^{1-\frac{t}{2}}}.\end{aligned}$$

Since $1-\frac{t}{2}\in [0,1/2],$ there exist constants $C>0$ and $d\in [0,1/2)$ such that Assumption 5.1 holds. This completes the proof.$\square$

The proof of Lemma 4

For any $\xi =(\mu ,a,b)^T, \xi '=(\mu ',a',b')^T\in \mathcal {R}^3$, we have

$$\begin{aligned} |\psi _c(\xi )-\psi _c(\xi ')|= & {} |\psi _c(\mu ,a,b)-\psi _c(\mu ',a',b')|\\= & {} |a+b-\Vert (a,b,\sqrt{2c},2\mu )^T\Vert -(a'+b')+\Vert (a',b',\sqrt{2c},2\mu ')^T\Vert |\\\le & {} |a-a'|+|b-b'|+|\Vert (a,b,\sqrt{2c},2\mu )^T\Vert -\Vert (a',b',\sqrt{2c},2\mu ')^T\Vert |\\\le & {} |a-a'|+|b-b'|+|\Vert (a,b,\sqrt{2c},2\mu )^T-(a',b',\sqrt{2c},2\mu ')^T\Vert |\\= & {} |a-a'|+|b-b'|+\Vert (a-a',b-b',2(\mu -\mu '))^T\Vert \\\le & {} \sqrt{2[(a-a')^2+(b-b')^2]}+\sqrt{(a-a')^2+(b-b')^2+4(\mu -\mu ')^2}\\\le & {} (\sqrt{2}+2)\Vert \xi -\xi '\Vert . \end{aligned}$$

This proves (i). Now we prove (ii) by considering the following three cases.

Case 1. If $c>0$, then $\psi _c$ is differentiable at any $\xi \in \mathcal {R}^3$ with

$$\begin{aligned} {\psi }_c'(\xi )= & {} \bigg (-\frac{4\mu }{\sqrt{a^2+b^2+2c+4\mu ^2}}, 1-\frac{a}{\sqrt{a^2+b^2+2c+4\mu ^2}},\\&1-\frac{b}{\sqrt{a^2+b^2+2c+4\mu ^2}}\bigg )^T\end{aligned}$$

and

$$\begin{aligned}{\psi }_c''(\xi )=\frac{1}{(\sqrt{a^2+b^2+2c+4\mu ^2})^3}\times M\end{aligned}$$

where

$$\begin{aligned}M:=\left[ \begin{array}{cccc} -4(a^2+b^2+2c)&{}4a\mu &{}4b\mu \\ 4a\mu &{}-(b^2+2c+4\mu ^2)&{}ab\\ 4b\mu &{}ab&{}-(a^2+2c+4\mu ^2) \end{array} \right] , \end{aligned}$$

which yields

$$\begin{aligned} \Vert {\psi }_c''(\xi )\Vert\le & {} \frac{\Vert M\Vert }{(\sqrt{a^2+b^2+2c+4\mu ^2})^3}\nonumber \\\le & {} \frac{ \sqrt{18(a^2+b^2+2c+4\mu ^2)^2}}{(\sqrt{a^2+b^2+2c+4\mu ^2})^3}\nonumber \\= & {} \frac{3\sqrt{2}}{\sqrt{a^2+b^2+2c+4\mu ^2}}\nonumber \\\le & {} \frac{3}{\sqrt{c}}. \end{aligned}$$

(7.1)

By (7.1), for any $u,v\in \mathcal {R}^3$ we have

$$\begin{aligned} \Vert {\psi }_c'(u)-{\psi }_c'(v)\Vert \le \frac{3}{\sqrt{c}}\Vert u-v\Vert .\end{aligned}$$

(7.2)

For any $\xi \in \mathcal {R}^3$ and $h\in \mathcal {R}^3$, $\psi _c$ is differentiable at $\xi +h$ and hence $\partial {\psi }_c(\xi +h)=\{{\psi }_c'(\xi +h)^T\}$. So, from (7.2) we have for any $V\in \partial {\psi }_c(\xi +h)$ and $h\rightarrow 0$,

$$\begin{aligned} |{\psi }_c(\xi +h)-{\psi }_c(\xi )-Vh|= & {} |{\psi }_c(\xi +h)-{\psi }_c(\xi )-{\psi }_c'(\xi +h)^Th|\\\le & {} |{\psi }_c(\xi +h)-{\psi }_c(\xi )-{\psi }_c'(\xi )^Th|+|[{\psi }_c'(\xi +h)-{\psi }_c'(\xi )]^Th|\\\le & {} |{\psi }_c(\xi +h)-{\psi }_c(\xi )-{\psi }_c'(\xi )^Th|+\frac{3}{\sqrt{c}}\Vert h\Vert ^2\\= & {} \bigg |\int _0^1 [{\psi }_c'(\xi +th)-{\psi }_c'(\xi )]^T hdt\bigg |+\frac{3}{\sqrt{c}}\Vert h\Vert ^2\\\le & {} \frac{3}{\sqrt{c}}\Vert h\Vert ^2\int _0^1 tdt+\frac{3}{\sqrt{c}}\Vert h\Vert ^2\\= & {} \frac{9}{2\sqrt{c}} \Vert h\Vert ^2. \end{aligned}$$

Case 2. If $c=0$ and $\xi =0$, then for any nonzero direction vector $h=(\tilde{\mu },\tilde{a},\tilde{b})^T\in \mathcal {R}^3$, ${\psi }_c$ is smooth at the point $0+h=h$. Thus, $V\in \partial {\psi }_c(0+h)=\{{\psi }_c'(h)^T\}$ is uniquely given by

$$\begin{aligned}V=\bigg (-\frac{4\tilde{\mu }}{\sqrt{\tilde{a}^2+\tilde{b}^2+4\tilde{\mu }^2}}, 1-\frac{\tilde{a}}{\sqrt{\tilde{a}^2+\tilde{b}^2+4\tilde{\mu }^2}}, 1-\frac{\tilde{b}}{\sqrt{\tilde{a}^2+\tilde{b}^2+4\tilde{\mu }^2}}\bigg ).\end{aligned}$$

Then, for any $V\in \partial {\psi }_c(0+h)$ and $h\rightarrow 0$,

$$\begin{aligned} |{\psi }_c(0+h)-{\psi }_c(0)-Vh|= & {} \bigg |-\sqrt{\tilde{a}^2+\tilde{b}^2+4\tilde{\mu }^2}+\frac{\tilde{a}^2+\tilde{b}^2+4\tilde{\mu }^2}{\sqrt{\tilde{a}^2+\tilde{b}^2+4\tilde{\mu }^2}}\bigg |\\= & {} 0.\end{aligned}$$

Case 3. If $c=0$ and $\xi \ne 0$, then $\psi _c$ is differentiable at $\xi$ and from (7.1) we have

$$\begin{aligned} \Vert {\psi }_c''(\xi )\Vert\le & {} \frac{3\sqrt{2}}{\sqrt{a^2+b^2+4\mu ^2}}\nonumber \\\le & {} \frac{3\sqrt{2}}{\sqrt{a^2+b^2+\mu ^2}}=\frac{3\sqrt{2}}{\Vert \xi \Vert }. \end{aligned}$$

(7.3)

Next we show that ${\psi }_c'(\xi )$ is locally Lipschitz continuous at $\xi$ with the constant $\frac{6\sqrt{2}}{\Vert \xi \Vert }$. In fact, let $u,v\in N(\xi ,\frac{\Vert \xi \Vert }{2})$, we have

$$\begin{aligned} {\psi }_c'(u)={\psi }_c'(v)+\int _0^1{\psi }_c''(v+t(u-v))(u-v)dt.\end{aligned}$$

(7.4)

By (7.3), we have

$$\begin{aligned} \Vert {\psi }_c''(v+t(u-v))\Vert \le \frac{3\sqrt{2}}{\Vert v+t(u-v)\Vert }.\end{aligned}$$

(7.5)

Since $u,v\in N(\xi ,\frac{\Vert \xi \Vert }{2})$, we have

$$\begin{aligned}\Vert v+t(u-v)\Vert =\Vert tu+(1-t)v\Vert \in N(\xi ,\frac{\Vert \xi \Vert }{2}).\end{aligned}$$

It follows that

$$\begin{aligned} \Vert v+t(u-v)\Vert \ge \Vert \xi \Vert -\Vert v+t(u-v)-\xi \Vert \ge \Vert \xi \Vert -\frac{\Vert \xi \Vert }{2}=\frac{\Vert \xi \Vert }{2}.\end{aligned}$$

This together with (7.5) yields

$$\begin{aligned} \Vert {\psi }_c''(v+t(u-v))\Vert \le \frac{6\sqrt{2}}{\Vert \xi \Vert }.\end{aligned}$$

(7.6)

Thus, from (7.4) and (7.6), we have for any $u,v\in N(\xi ,\frac{\Vert \xi \Vert }{2})$,

$$\begin{aligned} \Vert {\psi }_c'(u)-{\psi }_c'(v)\Vert \le \frac{6\sqrt{2}}{\Vert \xi \Vert }\Vert u-v\Vert .\end{aligned}$$

(7.7)

Now we show that ${\psi }_c(\xi )$ is strongly semismooth at $\xi \ne 0$. Since $\xi \ne 0$, when $h\rightarrow 0$, $\xi +h\ne 0$ and hence $\partial {\psi }_c(\xi +h)=\{{\psi }_c'(\xi +h)^T\}$. So, from (7.7), similarly as the proof of Case 1, we have for any $V\in \partial {\psi }_c(\xi +h)$ and $h\rightarrow 0$,

$$\begin{aligned} |{\psi }_c(\xi +h)-{\psi }_c(\xi )-Vh| \le \frac{6\sqrt{2}}{\Vert \xi \Vert }\Vert h\Vert ^2\int _0^1 tdt+\frac{6\sqrt{2}}{\Vert \xi \Vert }\Vert h\Vert ^2 =\frac{9\sqrt{2}}{\Vert \xi \Vert } \Vert h\Vert ^2.\end{aligned}$$

Thus, the proof is completed.$\square$

The proof of Lemma 5

Let $\mathcal {{H}}(z)$ be defined by (5.37). Then, $\mathcal {{H}}(z)$ is continuously differentiable at any $z=(\mu ,x,s,y)\in \mathcal {R}_{++}\times \mathcal {R}^{2n+m}$ and its Jacobian is

$$\begin{aligned}\mathcal {{H}}'(z)=\left[ \begin{array}{cccc} 1&{}0&{}0&{}0\\ 0&{}P&{}Q&{}R\\ D_\mu &{}\mathbf{diag }(D_x) &{}\mathbf{diag }(D_s)&{}0 \end{array} \right] ,\end{aligned}$$

where

$$\begin{aligned}&D_\mu :=\Bigg (-\frac{4\mu }{\sqrt{x_1^2+s_1^2+2 w_1+4\mu ^2}},....,-\frac{4\mu }{\sqrt{x_n^2+s_n^2+2 w_n+4\mu ^2}}\Bigg )^T, \\&D_x:=\Bigg (1-\frac{x_1}{\sqrt{x_1^2+s_1^2+2 w_1+4\mu ^2}},...,1-\frac{x_n}{\sqrt{x_n^2+s_n^2+2 w_n+4\mu ^2}}\Bigg )^T,\\&D_s:=\Bigg (1-\frac{s_1}{\sqrt{x_1^2+s_1^2+2 w_1+4\mu ^2}},....,1-\frac{s_n}{\sqrt{x_n^2+s_n^2+2 w_n+4\mu ^2}}\Bigg )^T. \end{aligned}$$

We now divide the proof by the following three parts.

Part (i) We show that $\mathcal {{{H}}}(z)$ is Lipschitz continuous on $\mathcal {R}^{1+2n+m}$. In fact, since $(Px+Qs+Ry-a)'=[P, Q, R]$, we have that $Px+Qs+Ry-a$ is Lipschitz continuous on $\mathcal {R}^{2n+m}$. Moreover, from (i) of Lemma 4, ${\psi }_c$ is Lipschitz continuous on $\mathcal {R}^{3}$. Hence, $\mathcal {{{H}}}(z)$ is Lipschitz continuous on $\mathcal {R}^{1+2n+m}$.

Part (ii) For any $\theta >0$, we show that $\mathcal {{{H}}}'(z)$ is bounded and Lipschitz continuous on the set

$$\begin{aligned}\varOmega :=\{z=(\mu ,x,s,y)\in \mathcal {R}_{++}\times \mathcal {R}^{2n+m}|~\mu \ge \theta \}.\end{aligned}$$

In fact, for any $i=1,...,n,$ since

$$\begin{aligned}&0\le \frac{4\mu }{\sqrt{x_i^2+s_i^2+2 w_i+4\mu ^2}}\le 2,\\&0\le 1-\frac{x_i}{\sqrt{x_i^2+s_i^2+2 w_i+4\mu ^2}}\le 2,\\&0\le 1-\frac{s_i}{\sqrt{x_i^2+s_i^2+2 w_i+4\mu ^2}}\le 2, \end{aligned}$$

$D_\mu$, $D_x$ and $D_s$ are bounded and hence $\mathcal {{H}}'(z)$ is bounded. Let

$$\begin{aligned}\varGamma :=\{\upsilon =(\mu ,a,b)\in \mathcal {R}_{++}\times \mathcal {R}^{2}|~\mu \ge \theta \}.\end{aligned}$$

For any $\upsilon \in \varGamma$, define

$$\begin{aligned}f_c(\upsilon ):=1-\frac{a}{\sqrt{a^2+b^2+2c+4\mu ^2}}.\end{aligned}$$

Since $\mu \ge \theta >0,$ $f_c(\upsilon )$ is continuously differentiable and

$$\begin{aligned}f_c'(\upsilon )=\frac{1}{(\sqrt{a^2+b^2+2c+4\mu ^2})^3}\bigg [4a\mu ,-(b^2+2c+4\mu ^2),~ab\bigg ],\end{aligned}$$

which yields

$$\begin{aligned}\Vert f_c'(\upsilon )\Vert =\frac{\sqrt{16a^2\mu ^2 + (b^2+2c+4\mu ^2)^2+ a^2b^2}}{(\sqrt{a^2+b^2+2c+4\mu ^2})^3}.\end{aligned}$$

By noticing that

$$\begin{aligned} 16 a^2\mu ^2\le 2(a^2+4\mu ^2)^2\le & {} 2(a^2+b^2+2c+4\mu ^2)^2,\\ (b^2+2c+4\mu ^2)^2\le & {} (a^2+b^2+2c+4\mu ^2)^2,\\ a^2b^2\le (a^2+b^2)^2\le & {} (a^2+b^2+2c+4\mu ^2)^2, \end{aligned}$$

we have

$$\begin{aligned}\Vert f_c'(\upsilon )\Vert \le \frac{2}{\sqrt{a^2+b^2+2c+4\mu ^2}}\le \frac{1}{\mu }\le \frac{1}{\theta }.\end{aligned}$$

Thus, for any $\tilde{\upsilon }, \upsilon \in \varGamma$, we have

$$\begin{aligned}\Vert f_c(\tilde{\upsilon })-f_c(\upsilon )\Vert |\le \frac{1}{\theta }\Vert \tilde{\upsilon }-\upsilon \Vert .\end{aligned}$$

This implies that $f_c$ is Lipschitz continuous on $\varGamma$ and hence $D_x$ is Lipschitz continuous on $\varOmega$. By a similar way, we can show that $D_s$ and $D_\mu$ are also Lipschitz continuous on $\varOmega$ and so is $\mathcal {{{H}}}'(z)$.

Part (iii) We show that $\mathcal {{{M}}}'(z)$ is Lipschitz continuous on the set $\varTheta$ defined by (5.9). In fact, $\mathcal {{{M}}}(z)$ is continuous differentiable at any $z\in \mathcal {R}_{++}\times \mathcal {R}^{2n+m}$ and

$$\begin{aligned}\mathcal {M}'(z)=2\mathcal {H}(z)^T\mathcal {H}'(z).\end{aligned}$$

So, for any $\tilde{z},z\in \varTheta$, by Parts (i) and (ii) and $\Vert \mathcal {H}(\tilde{z})\Vert \le 2 \Vert \mathcal {H}(z^0)\Vert$, there exists a constant $M>0$ such that

$$\begin{aligned} \Vert \mathcal {M}'(\tilde{z})-\mathcal {M}'(z)\Vert= & {} 2\Vert \mathcal {H}(\tilde{z})^T\mathcal {H}'(\tilde{z})-\mathcal {H}(z)^T\mathcal {H}'(z)\Vert \\= & {} 2\Vert \mathcal {H}(\tilde{z})^T[\mathcal {H}'(\tilde{z})-\mathcal {H}'({z})]-[\mathcal {H}(z)-\mathcal {H}(\tilde{z})]^T\mathcal {H}'(z)\Vert \\\le & {} 2[\Vert \mathcal {H}(\tilde{z})\Vert \Vert \mathcal {H}'(\tilde{z})-\mathcal {H}'(z)\Vert +\Vert \mathcal {H}(z)-\mathcal {H}(\tilde{z})\Vert \Vert \mathcal {H}'({z})\Vert ]\\= & {} M\Vert z-\tilde{z}\Vert . \end{aligned}$$

This completes the proof.$\square$