On Saddle Points in Semidefinite Optimization via Separation Scheme

Abstract

This paper aims at investigating saddle point conditions for augmented Lagrangian functions for semidefinite optimization problems. By means of the image space analysis, the existence of a saddle point is shown to be equivalent to a regular weak nonlinear separation of two suitable subsets in the image space (IS) associated with the given problem. Especially, three classes of augmented Lagrangians based on smooth spectral penalty functions can be derived, as particular cases, from a nonlinear separation scheme in the IS. Without requiring the strict complementarity, it is proved that, under strong second-order sufficiency conditions, all these augmented Lagrangian functions admit a local saddle point, and their Hessians become positive definite in a neighborhood of a local optimal point of the original problem. The existence of global saddle points is then obtained under additional assumptions that do not require the compactness of the feasible set.

Appendix: Proof of Proposition 4.2

We first recall a well-known result regarding the differentiability properties of the composite matrix function. We use the following definition from [41].

Definition 6.1

Let $G: {{\mathrm{I}\!\mathrm{R}}}^n\rightarrow {{\mathrm{I}\!\mathrm{R}}}^{m\times m}$. Let $\lambda _1(x),\ldots ,\lambda _{\mu (x)}(x)$ be the $\mu (x)$ increasingly ordered eigenvalues of $G(x)$. Let $G(x)=U(x)\varLambda (x)U(x)^T$ be the spectral decomposition of $G(x)$, with $\varLambda (x)=\mathrm{diag}\left( \lambda _1(x), \ldots , \lambda _1(x),\right.$ $\left. \ldots ,\lambda _{\mu (x)}(x),\ldots ,\right.$ $\left. \lambda _{\mu (x)}(x)\right)$. Then, the Frobenius covariance matrices is defined by

$$\begin{aligned} P_i(x):= U(x)\mathrm{diag} (0,\ldots ,0,1,\ldots ,1,0,\ldots ,0)U(x)^T,~i=1,\ldots ,\mu (x), \end{aligned}$$

(90)

where the non-zeros in the diagonal matrix occur exactly in the positions of $\lambda _i(x)$ in $\varLambda (x)$.

Note that $P_i(x)P_j(x)=0$ if $i\ne j$, $P_i(x)^2=P_i(x)$, $i,j=1,\ldots ,\mu (x)$, and $\sum _{i=1}^{\mu (x)}P_i(x)=I_m$ (see [41], p. 403).

Definition 6.2

Let $t_1,\ldots ,t_m$ be real values and let $f_0 : {{\mathrm{I}\!\mathrm{R}}}\rightarrow {{\mathrm{I}\!\mathrm{R}}}$ be twice continuously differentiable. Then, we define

$$\begin{aligned}&\Delta f_0(t_k,t_l):=\left\{ \begin{array}{ll} \frac{f_0(t_k)-f_0(t_l)}{t_k-t_l},&{}k\ne l,\\ f_0^\prime (t_k),&{}k=l, \end{array}\right. \\&\Delta ^2 f_0(t_k,t_l,t_q):=\left\{ \begin{array}{ll} \frac{\Delta f_0(t_k,t_q)-\Delta f_0(t_l,t_q)}{t_k-t_l},&{}k\ne l,\\ \frac{\Delta f_0(t_k,t_l)-\Delta f_0(t_q,t_l)}{t_k-t_q},&{}k=l\ne q,\\ f_0^{\prime \prime }(t_k),&{}k=l=q. \end{array}\right. \end{aligned}$$

The following result is from [41] (Theorem 6.6.30), which will be used to prove Proposition 4.2.

Lemma 6.1

Let $f_0 : {{\mathrm{I}\!\mathrm{R}}}\rightarrow {{\mathrm{I}\!\mathrm{R}}}$ and $G: {{\mathrm{I}\!\mathrm{R}}}^n\rightarrow {\mathcal {S}}^m$ be twice continuously differentiable. Define $F : {\mathcal {S}}^m\rightarrow {\mathcal {S}}^m$ by $F(Z):= U f_1(D)U^T$, where $Z= UD U^T$ is the spectral decomposition of $Z$, $D=\mathrm{diag}\left( \lambda _1, \ldots , \lambda _m\right)$ and $f_1(D) = \mathrm{diag}\left( f_0(\lambda _1),\ldots , f_0(\lambda _m)\right)$, where $\lambda _i$, $i=1,\ldots ,m$ are eigenvalues of $Z$ listed in the decreasing order. Let $\lambda _1(x),\ldots ,\lambda _{\mu (x)}(x)$ be the $\mu (x)$ increasingly ordered eigenvalues of the matrix $G(x)$. Then, $F(G(x))$ is twice continuously differentiable with

$$\begin{aligned} \frac{\partial F(G(x))}{\partial x_i}&= \sum _{k,l=1}^{\mu (x)}\Delta f_0(\lambda _k(x),\lambda _l(x)) P_k(x) G_i^\prime (x) P_l(x),\\ \frac{\partial ^2 F(G(x))}{\partial x_i\partial x_j}&= \sum _{k,l=1}^{\mu (x)}\Delta f_0(\lambda _k(x),\lambda _l(x)) P_k(x) G_{ij}^{\prime \prime }(x) P_l(x) \\&\quad +\sum _{k,l,q=1}^{\mu (x)}\Delta ^2 f_0(\lambda _k(x),\lambda _l(x), \lambda _q(x))(M_{klq} +M^T_{klq} ), \end{aligned}$$

where $P_k(x)$, $k=1,\ldots ,\mu (x)$ are Frobenius covariance matrices of $G(x)$, and the matrix $M_{klq}$ is given by $M_{klq}:=P_k(x) G_i^{\prime }(x) P_l(x) G_j^{\prime }(x)P_q(x)$.

Proof of Proposition 4.2. We first show conclusion (i). Let $\lambda _i^*\ge 0$, $i=1,\ldots ,m$ be eigenvalues of $G(x^*)$ of rank $r$ listed in the decreasing order. Since $\varLambda ^*\bullet G(x^*)=0$ is equivalent to $\lambda _i(\varLambda ^*) \lambda _i^*=0$ for $i=1,\ldots ,m$, we obtain $\lambda _i(\varLambda ^*)=0$ for $i=1,\ldots ,r$. Let us define $\psi _c(t):=c^{-1}\psi (ct)$ for any $t\in {{\mathrm{I}\!\mathrm{R}}}$. By the property $\psi (0) = 0$, we have $\psi _c(\lambda _i^*)=0$ for $i=r+1,\ldots ,m$. Thus, $\lambda _i(\varLambda ^*) \psi _c(\lambda _i^*)=0$ for $i=1,\ldots ,m$, which, by the definition of $\varPsi _c$, means that $\varLambda ^*\bullet \varPsi _c(G(x^*))=0$. Similarly, $\varLambda ^*\bullet \varPhi _c(G(x^*))=0$. Observe from condition (D2) that $G(x^*)\succeq 0$ implies $\mathrm{Tr}\left( \varXi _c(G(x^*))\right) =0$. Hence, $L_i(x^*,\varLambda ^*,\mu ^*,c)= f(x^*)$ $(i=1,2,3)$ for any $c > 0$.

Next, we prove conclusion (ii) for $L_1$. The proof for $L_2$ can be constructed by using similar arguments and the properties (C1)–(C3).

Let $\lambda _1^*,\ldots ,\lambda ^*_{\mu (x^*)}$ be the $\mu (x^*)$ distinct eigenvalues of $G(x^*)$ and let $P_i(x^*)$ be defined by (90) for $i=1,\ldots ,\mu (x^*)$. Note that $P_i(x^*)P_j(x^*)=0$ for all $i\ne j$ and $P_i(x^*)^2=P_i(x^*)$ for all $i,j=1,\ldots ,\mu (x^*)$. Let $P_0:=EE^T$ with $E=[e_{r+1},\ldots ,e_m]$. We see that $P_0=P_{\mu (x^*)}(x^*)$. Since $\varLambda ^*\bullet G(x^*)=0$ and $\varLambda ^*,G(x^*) \in {\mathcal {S}}^m_+$ imply $\varLambda ^* G(x^*)=G(x^*)\varLambda ^*$, by Theorem 1.3.12 of [41], $\varLambda ^*$ and $G(x^*)$ are simultaneously diagonalizable. So, $\varLambda ^*={\bar{ E}}D_{\varLambda ^*}{\bar{E}}^T$, where $D_{\varLambda ^*}:=\mathrm{diag}\left( 0,\ldots ,0,\lambda _{r+1}(\varLambda ^*),\ldots ,\lambda _m(\varLambda ^*)\right)$, and ${\bar{E}}:=[e_1,\ldots ,e_m]$ with ${\bar{E}}^T\bar{E}=I_m$. Let $D_0:=\mathrm{diag}{({\underbrace{0,\ldots ,0}_{r}},\underbrace{1\ldots ,1}_{m-r})}$. Then, $P_0={\bar{E}}D_0\bar{E}^T$ and

$$\begin{aligned} P_0\varLambda ^*P_0={\bar{E}} D_0{\bar{E}}^T\bar{E}D_{\varLambda ^*}{\bar{E}}^T\bar{E}D_0{\bar{E}}^T=\bar{E}D_0 D_{\varLambda ^*} D_0{\bar{E}}^T ={\bar{E}} D_{\varLambda ^*} {\bar{E}}^T=\varLambda ^*.\nonumber \\ \end{aligned}$$

(91)

By Lemma 6.1 and (91), we can obtain

$$\begin{aligned}&\left[ \varLambda ^*\bullet \frac{\partial }{\partial x_i}\varPsi _c(G(x^*))\right] _{i =1}^n = \left[ (P_0\varLambda ^*P_0)\bullet \sum _{k,l=1}^{\mu (x^*)} \Delta \psi _c(\lambda ^*_k,\lambda ^*_l)P_k(x^*)G_i^\prime (x^*)P_l(x^*)\right] _{i =1}^n\\&= \left[ \varLambda ^* \bullet \sum _{k,l=1}^{\mu (x^*)} \Delta \psi _c(\lambda ^*_k,\lambda ^*_l)P_0P_k(x^*)G_i^\prime (x^*)P_l(x^*)P_0\right] _{i =1}^n\\&= \left[ \varLambda ^* \bullet \sum _{k,l=1}^{\mu (x^*)} \Delta \psi _c(\lambda ^*_k,\lambda ^*_l)P_{\mu (x^*)}(x^*)P_k(x^*)G_i^\prime (x^*)P_l(x^*)P_{\mu (x^*)}(x^*)\right] _{i =1}^n\\&= \left[ \varLambda ^* \bullet \left( \Delta \psi _c(\lambda ^*_{\mu (x^*)},\lambda ^*_{\mu (x^*)})P_{\mu (x^*)}(x^*) G_i^\prime (x^*) P_{\mu (x^*)}(x^*)\right) \right] _{i =1}^n\\&= \left[ \varLambda ^* \bullet \left( \psi _c^\prime (0)P_0 G_i^\prime (x^*) P_0\right) \right] _{i =1}^n = \psi ^\prime (0)\left[ \varLambda ^*\bullet G_i^\prime (x^*)\right] _{i =1}^n\!, \end{aligned}$$

from which and by $\psi ^\prime (0)=1$ and $h(x^*)=0$, we draw immediately that

$$\begin{aligned} \nabla _x L_1(x^*,\varLambda ^*,\mu ^*,c)&= \nabla f(x^*) -\psi ^\prime (0)\left[ \varLambda ^*\bullet G_i^\prime (x^*)\right] _{i =1}^n+\nabla h(x^*)(\mu ^*+ch(x^*))\\&= \nabla _x L_0(x^*,\varLambda ^*,\mu ^*)=0. \end{aligned}$$

Moreover, we have

$$\begin{aligned}&\left[ \varLambda ^*\bullet \frac{\partial ^2}{\partial x_i\partial x_j}\varPsi _c(G(x^*))\right] _{i,j =1}^n=\left[ (P_0\varLambda ^*P_0)\bullet \frac{\partial ^2}{\partial x_i\partial x_j}\varPsi _c(G(x^*))\right] _{i,j =1}^n\nonumber \\&= \left[ \varLambda ^*\bullet \left( \sum _{k,l=1}^{\mu (x^*)} \Delta \psi _c(\lambda ^*_k,\lambda ^*_l)P_0P_k(x^*)G_{ij}^{\prime \prime }(x^*)P_l(x^*)P_0\right) \right] _{i,j =1}^n\nonumber \\&+\left[ \varLambda ^*\bullet \left( \sum _{k,l,q=1}^{\mu (x^*)} \Delta ^2 \psi _c(\lambda ^*_k,\lambda ^*_l,\lambda ^*_q)P_0P_k(x^*)G_i^{\prime }(x^*)P_l(x^*)G_j^{\prime }(x^*)P_q(x^*)P_0\right) \right] _{i,j =1}^n\nonumber \\&+\left[ \varLambda ^*\bullet \left( \sum _{k,l,q=1}^{\mu (x^*)} \Delta ^2 \psi _c(\lambda ^*_k,\lambda ^*_l,\lambda ^*_q)P_0P_q(x^*)G_j^{\prime }(x^*)P_l(x^*)G_i^{\prime }(x^*)P_k(x^*)P_0\right) \right] _{i,j =1}^n\nonumber \\&= \left[ \varLambda ^* \bullet \left( \Delta \psi _c(\lambda ^*_{\mu (x^*)},\lambda ^*_{\mu (x^*)})P_{\mu (x^*)}(x^*) G_{ij}^{\prime \prime }(x^*) P_{\mu (x^*)}(x^*)\right) \right] _{i,j =1}^n\nonumber \\&+ \left[ \varLambda ^* \bullet \left( P_{\mu (x^*)}(x^*) N_{ij}(x^*) P_{\mu (x^*)}(x^*) \right) \right] _{i,j =1}^n\nonumber \\&+\left[ \varLambda ^* \bullet \left( P_{\mu (x^*)}(x^*) N_{ji}(x^*) P_{\mu (x^*)}(x^*)\right) \right] _{i,j =1}^n\nonumber \\&= \left[ \varLambda ^* \bullet \left( \psi _c^\prime (0)P_0 G_{ij}^{\prime \prime }(x^*) P_0\right) \right] _{i,j =1}^n+2\left[ \varLambda ^* \bullet (P_0 N_{ij}(x^*)P_0)\right] _{i,j =1}^n\nonumber \\&= \psi ^\prime (0)\left[ \varLambda ^*\bullet G_{ij}^{\prime \prime }(x^*)\right] _{i,j =1}^n+2\left[ \varLambda ^* \bullet N_{ij}(x^*) \right] _{i,j =1}^n\!, \end{aligned}$$

(92)

where

$$\begin{aligned} N_{ij}(x^*) =G_i^\prime (x^*)\left( \sum _{k=1}^{\mu (x^*)} \Delta ^2 \psi _c(\lambda ^*_k,0,0)P_k(x^*)\right) G_j^{\prime }(x^*),\quad i,j=1,\ldots ,n. \end{aligned}$$

By Definition 6.2, via a straightforward computation, we obtain

$$\begin{aligned} \Delta ^2 \psi _c(\lambda ^*_k,0,0)=\left\{ \begin{array}{ll} c \psi ^{\prime \prime }(0), &{}\quad k=\mu (x^*)\\ \frac{c^{-1}\psi (c\lambda ^*_k)}{(\lambda ^*_k)^2}-\frac{\psi ^\prime (0)}{\lambda ^*_k},&{}\quad k<\mu (x^*) \end{array}\right. . \end{aligned}$$

Note that $\psi ^\prime (0)=1$. It then follows from (92) and the definition of $L_1$ that

$$\begin{aligned} \nabla _{xx}^2 L_1(x^*,\varLambda ^*,\mu ^*,c)&= \underbrace{{\nabla ^2f(x^*) -\psi ^{\prime }(0)\left[ \varLambda ^*\bullet G_{ij}^{\prime \prime }(x^*)\right] _{i,j=1}^n+\sum _{i=1}^p\mu ^*_i\nabla ^2 h_i(x^*)}}_{{\nabla _{xx}^2 L_0(x^*,\varLambda ^*,\mu ^*)}}\nonumber \\&+\,H_c(x^*,\varLambda ^*)+c M(x^*,\varLambda ^*)+c\nabla h(x^*)\nabla h(x^*)^T , \end{aligned}$$

(93)

which is just (38), where

$$\begin{aligned}&H_c(x^*,\varLambda ^*)=-2\nonumber \\&\quad \times \left[ \varLambda ^*\bullet \left( G_i^\prime (x^*) \sum _{k=1}^{\mu (x^*)-1}\left( \frac{c^{-1}\psi (c\lambda _k^* )}{(\lambda ^*_k)^2}-\frac{1}{\lambda _k^* }\right) P_k(x^*) G_j^\prime (x^*)\right) \right] _{i,j=1}^n \end{aligned}$$

(94)

$$\begin{aligned}&M(x^*,\varLambda ^*)=-2\psi ^{\prime \prime }(0)\left[ \varLambda ^*\bullet \left( G_i^\prime (x^*)P_{\mu (x^*)}G_j^\prime (x^*)\right) \right] _{i,j=1}^n. \end{aligned}$$

(95)

By condition (B1), $\psi (0)=0$ and $\psi ^\prime (0)=1$, we know that $\psi (t)< t$ for any $t\ne 0$, and hence $\frac{c^{-1}\psi (c\lambda _k^* )}{(\lambda ^*_k)^2}-\frac{1}{\lambda _k^* }<0$ for $k=1,\ldots ,\mu (x^*)-1$. Together with $P_k(x^*)\succeq 0$ for all $k$ and $\varLambda ^*\succeq 0$, we then infer from (94) that $H_c(x^*,\varLambda ^*)$ is positive semidefinite. By the condition (B3) for $\psi$, we see that $\lim _{c\rightarrow \infty }\frac{\psi (c\tau )}{c}=0$ for any $\tau >0$. Note also that $\lambda _k^*>0$ for $k=1,\ldots ,r$ and that, by the definition of $P_k(x)$,

$$\begin{aligned} \sum _{k=1}^{\mu (x^*)-1}\frac{1}{\lambda _k^* } P_k(x^*)=\sum _{k=1}^r \frac{1}{\lambda _k^* } e_ke_k^T. \end{aligned}$$

We then deduce from (94) that

$$\begin{aligned} \lim _{c\rightarrow \infty }H_{c}(x^{*},\varLambda ^*)&= 2\left[ \varLambda ^{*}\bullet \left( G_i^{\prime }(x^{*})\left( \sum _{k=1}^{r} \frac{1}{\lambda _{k}^{*}} e_ke_k^T\right) G_j^\prime (x^*)\right) \right] _{i,j=1}^n\nonumber \\&= 2\left[ \varLambda ^{*}\bullet \left( G_{i}^{\prime }(x^{*})[G(x^{*})]^{\dag } G_{j}^{\prime }(x^{*})\right) \right] _{i,j=1}^{n}\nonumber \\&= H(x^{*},\varLambda ^{*}), \end{aligned}$$

(96)

proving (39). Note that $s=m-\mathrm{rank}(\varLambda ^*)\ge r$. Then, $\lambda _i(\varLambda ^*)=0$ for $i=1,\ldots ,s$ and

$$\begin{aligned} D_{\varLambda ^*}=\mathrm{diag}\left( 0,\ldots ,0,\lambda _{s+1}(\varLambda ^*),\ldots ,\lambda _m(\varLambda ^*)\right) \!. \end{aligned}$$

Note that $\varLambda ^*={\bar{E}}D_{\varLambda ^*}{\bar{E}}^T$ with ${\bar{E}}=[e_1,\ldots ,e_m]$ and $P_{\mu (x^*)}=P_0=EE^T$ with $E=[e_{r+1},\ldots ,e_m]$. It then follows from (95) that

$$\begin{aligned} M(x^*,\varLambda ^*)&= -2\psi ^{\prime \prime }(0)\left[ ({\bar{E}}D_{\varLambda ^*}{\bar{E}}^T)\bullet \left( G_i^\prime (x^*)EE^TG_j^\prime (x^*)\right) \right] _{i,j=1}^n\nonumber \\&= -2\psi ^{\prime \prime }(0)\left[ \sum _{k=s+1}^m \lambda _k(\varLambda ^*)e_k^T G_i^\prime (x^*)\left( \sum _{l=r+1}^me_le^T_l\right) G_j^\prime (x^*) e_k\right] _{i,j=1}^n\nonumber \\&= -2\psi ^{\prime \prime }(0)\left[ \sum _{k=s+1}^m \sum _{l=r+1}^m\lambda _k(\varLambda ^*)\left( e_k^T G_i^\prime (x^*)e_le^T_lG_j^\prime (x^*) e_k\right) \right] _{i,j=1}^n\nonumber \\&= -2\psi ^{\prime \prime }(0)BQ^*_sB^T, \end{aligned}$$

(97)

where $Q^*_s: =\mathrm{diag}\left( \lambda _{s+1}(\varLambda ^*),\ldots ,\lambda _m(\varLambda ^*),\ldots , \lambda _{s+1}(\varLambda ^*),\ldots ,\lambda _m(\varLambda ^*)\right) \in {{\mathrm{I}\!\mathrm{R}}}^{{\bar{m}}\times {\bar{m}}}$ with ${\bar{m}}:=(m-s)(m-r)$, and the $i$-th row of matrix $B\in {{\mathrm{I}\!\mathrm{R}}}^{n\times {\bar{m}}}$ is defined by

$$\begin{aligned} b_i\!:=\!\left( e_{s+1}^TG_i^\prime (x^*)e_{r+1},\ldots ,e_{m}^TG_i^\prime (x^*)e_{r+1},\ldots ,e_{s+1}^TG_i^\prime (x^*)e_m,\ldots ,e_{m}^TG_i^\prime (x^*)e_m\right) ^T\!\!\!. \end{aligned}$$

Note that, since $\mathrm{rank}(\varLambda ^*)=m-s$ implies $\lambda _i(\varLambda ^*)>0$ for $i=s+1,\ldots ,m$, $Q^*_s$ is positive definite and hence, by $\psi ^{\prime \prime }(0)<0$, we see from (97) that $M(x^*,\varLambda ^*)$ is positive semidefinite. Now assume that for some ${\bar{d}}\ne 0\in {{\mathrm{I}\!\mathrm{R}}}^n$, it holds

$$\begin{aligned} {\bar{d}}^T M(x^*,\varLambda ^*){\bar{d}} =0. \end{aligned}$$

Since $Q^*_s$ is positive definite, it then follows from (97) that $B^T{\bar{d}}=\sum _{i=1}^n{\bar{d}}_ib_i=0$. So,

$$\begin{aligned} e_{k}^T\left( \sum _{i=1}^n{\bar{d}}_iG_i^\prime (x^*)\right) e_j=0, ~k=s+1,\ldots ,m,~j=r+1,\ldots ,m. \end{aligned}$$

This proves that (40) holds.

Finally, we prove conclusion (ii) for $L_3$. Note that $\mathrm{Tr}\left( \varXi _c(G(x))\right) =I_m\bullet \varXi _c(G(x))$. By Lemma 6.1, similar to the computations of (92) and (92) above, we have

$$\begin{aligned} \left[ I_m\bullet \frac{\partial }{\partial x_i}\varXi _c(G(x^*))\right] _{i =1}^n&= \left[ I_m\bullet \sum _{k,l=1}^{\mu (x^*)} \Delta \xi _c(\lambda ^*_k,\lambda ^*_l)P_k(x^*)G_i^\prime (x^*)P_l(x^*)\right] _{i =1}^n \nonumber \\&= \left[ \sum _{k,l=1}^{\mu (x^*)} \Delta \xi _c(\lambda ^*_k,\lambda ^*_l)\mathrm{Tr}\left( P_k(x^*)G_i^\prime (x^*)P_l(x^*)\right) \right] _{i =1}^n\nonumber \\&= \left[ \sum _{k,l=1}^{\mu (x^*)} \Delta \xi _c(\lambda ^*_k,\lambda ^*_l)\mathrm{Tr}\left( P_k(x^*)P_l(x^*)G_i^\prime (x^*)\right) \right] _{i =1}^n\nonumber \\&= \left[ \Delta \xi _c(\lambda ^*_{\mu (x^*)},\lambda ^*_{\mu (x^*)})\mathrm{Tr}\left( P_{\mu (x^*)}(x^*)G_i^\prime (x^*)\right) \right] _{i =1}^n\nonumber \\&= \xi ^\prime (0)\left[ P_{\mu (x^*)}(x^*)\bullet G_i^\prime (x^*)\right] _{i =1}^n\!. \end{aligned}$$

(98)

Moreover, we have

$$\begin{aligned}&\left[ I_m\bullet \frac{\partial ^2}{\partial x_i\partial x_j}\varXi _c(G(x^*))\right] _{i,j =1}^n \nonumber \\&= \left[ \sum _{k,l=1}^{\mu (x^*)} \Delta \xi _c(\lambda ^*_k,\lambda ^*_l)\mathrm{Tr}\left( P_k(x^*)G_{ij}^{\prime \prime }(x^*)P_l(x^*) \right) \right] _{i,j =1}^n\nonumber \\&+\left[ \sum _{k,l,q=1}^{\mu (x^*)} \Delta ^2 \xi _c(\lambda ^*_k,\lambda ^*_l,\lambda ^*_q)\mathrm{Tr}\left( P_k(x^*)G_i^{\prime }(x^*)P_l(x^*)G_j^{\prime }(x^*)P_q(x^*) \right) \right] _{i,j =1}^n \nonumber \\&+\left[ \sum _{k,l,q=1}^{\mu (x^*)} \Delta ^2 \xi _c(\lambda ^*_k,\lambda ^*_l,\lambda ^*_q)\mathrm{Tr}\left( P_q(x^*)G_j^{\prime }(x^*)P_l(x^*)G_i^{\prime }(x^*)P_k(x^*) \right) \right] _{i,j =1}^n\nonumber \\&= \left[ \Delta \xi _c(\lambda ^*_{\mu (x^*)},\lambda ^*_{\mu (x^*)})\mathrm{Tr}\left( P_{\mu (x^*)}(x^*) G_{ij}^{\prime \prime }(x^*) \right) \right] _{i,j =1}^n\nonumber \\&+ \left[ \mathrm{Tr}\left( P_{\mu (x^*)}(x^*) N_{ij}(x^*) \right) \right] _{i,j =1}^n+\left[ \mathrm{Tr}\left( P_{\mu (x^*)}(x^*) N_{ji}(x^*) \right) \right] _{i,j =1}^n\nonumber \\&= \xi ^\prime (0)\left[ P_{\mu (x^*)}(x^*)\bullet G_{ij}^{\prime \prime }(x^*)\right] _{i,j =1}^n+2\left[ P_{\mu (x^*)}(x^*) \bullet N_{ij}(x^*) \right] _{i,j =1}^n\!, \end{aligned}$$

(99)

where

$$\begin{aligned}&N_{ij}(x^*) =G_i^\prime (x^*)\left( \sum _{k=1}^{\mu (x^*)} \Delta ^2 \xi _c(\lambda ^*_k,0,0)P_k(x^*)\right) G_j^{\prime }(x^*),\quad i,j=1,\ldots ,n.\\&\Delta ^2 \xi _c(\lambda ^*_k,0,0)=\left\{ \begin{array}{ll} c \xi ^{\prime \prime }(0), &{}\quad k=\mu (x^*)\\ \frac{c^{-1}\xi (c\lambda ^*_k)}{(\lambda ^*_k)^2}-\frac{\xi ^\prime (0)}{\lambda ^*_k},&{}\quad k<\mu (x^*) \end{array}\right. . \end{aligned}$$

We note from conditions (D1) and (D2) that $\xi (t) = 0$ and $\xi ^\prime (t) = 0$ for $t \ge 0$ and $\xi ^{\prime \prime }(0)=0$. Note also that $\lambda ^*_k>0$ for all $k=1,\ldots ,r$. Therefore, from (98) and (99), we can obtain

$$\begin{aligned} \left[ I_m\bullet \frac{\partial }{\partial x_i}\varXi _c(G(x^*))\right] _{i =1}^n=0,\quad \left[ I_m\bullet \frac{\partial ^2}{\partial x_i\partial x_j}\varXi _c(G(x^*))\right] _{i,j =1}^n=0. \end{aligned}$$

(100)

By the definition of $L_3$ and $\psi ^\prime (0)=1$, it follows from (92), (92), and (100) that

$$\begin{aligned} \nabla _{x} L_{3}(x^{*},\varLambda ^{*},\mu ^*,c)&= \nabla f(x^*) - \psi ^\prime (0) [\varLambda ^*\bullet G_i^\prime (x^*)]_{i =1}^n+ \nabla h (x^*)\mu ^* \\&= \nabla _x L_0(x^*,\varLambda ^*,\mu ^*),\\ \nabla _{xx}^2 L_3(x^*,\varLambda ^*,\mu ^*,c)&= {\underbrace{\nabla ^2f(x^*) -\psi ^\prime (0) [\varLambda ^*\bullet G_{ij}^{\prime \prime }(x^*)]_{i,j=1}^n+\sum _{i=1}^p\mu ^*_i\nabla ^2 h_i(x^*)}_{\nabla _{xx}^2 L_0(x^*,\varLambda ^*,\mu ^*)}}\\&+\,H_c(x^*,\varLambda ^*)+cM(x^*,\varLambda ^*)+c\nabla h(x^*)\nabla h(x^*)^T , \end{aligned}$$

where $H_c(x^*,\varLambda ^*)$ and $M(x^*,\varLambda ^*)$ are given in (94) and (95), and satisfy (39) and (40). The proof of the proposition is completed. $\square$