Metric subregularity and/or calmness of the normal cone mapping to the p-order conic constraint system

Abstract

For a finite convex function, we show that its subdifferential mapping is metrically subregular if and only if the normal cone mapping to its epigraph is metrically subregular. Then, for the nonconvex composition $\psi =\theta \circ G$ where G is a continuously differentiable mapping and $\theta$ is an extended real-valued function, we develop a criterion to identify the metric subregularity and calmness of the subdifferential mapping of $\psi$ in terms of that of the subdifferential mapping of $\theta$. Together with the existing results, we obtain the metric subregularity of the normal cone mapping to the vector and matrix p-order cone $K_p$ and the conic constraint system $g^{-1}(K_p)$ with $p\in [1,2]\cup \{+\infty \}$, where g is a continuously differentiable mapping. We also establish the calmness of the normal cone mapping to $K_p$ and $g^{-1}(K_p)$ with $p\in [2,+\infty ]\cup \{1\}$.

Appendix

In this part, we shall establish the calmness of the subdifferential mappings of the vector and matrix $\ell _p$ norms with $p\in \{1\}\cup [2,+\infty ]$. The following lemma is required which follows directly by the convexity of $\phi (t)=t^{\tau } (\tau \ge 1)$ for $t\ge 0$.

Lemma 3

For any given $a,b\in \mathbb {R}_{+}$ and $\tau \ge 1$, the following inequality holds:

$$\begin{aligned} |a^\tau -b^\tau |\le \tau \max (a^{\tau -1},b^{\tau -1})|a-b|. \end{aligned}$$

If in addition $a,b\in [0,1]$, the inequality can be simplified as $|a^\tau -b^\tau |\le \tau |a-b|$.

Proposition 1

Suppose that $p\in \{1\}\cup [2,+\infty ]$. The following statements hold.

• (i) $\partial f_p$ is calm at any $\widehat{x}\in \mathbb {R}^n$ for any $\widehat{y}\in \partial f_p(\widehat{x})$;

• (ii) $\partial F_p$ is calm at any $\widehat{X}\in \mathbb {R}^{n\times m}$ for any $\widehat{W}\in \partial f_p(\widehat{X})$.

Proof

(i) By the remarks after Definition 2, it suffices to consider $p\in [2,+\infty )$. Fix an arbitrary $\widehat{x}\in \mathbb {R}^n$ and an arbitrary $\widehat{y}\in \partial f_p(\widehat{x})$. The calmness of $\partial f_p$ at $\widehat{x}$ for $\widehat{y}$ requires the existence of $\kappa >0$ along with $\varepsilon >0$ and $\delta >0$ such that

$$\begin{aligned} \partial f_p(x)\cap \mathbb {B}(\widehat{y},\delta ) \subset \partial f_p(\widehat{x})+\kappa \Vert x-\widehat{x}\Vert \mathbb {B}_{\mathbb {R}^n} \quad \mathrm{for\ all}\ x\in \mathbb {B}(\widehat{x},\varepsilon ). \end{aligned}$$

(25)

Recall that the subdifferential mapping of $f_p$ for $p\in (1,+\infty )$ has the form of

$$\begin{aligned} \partial f_p(x)=\left\{ \begin{array}{ll} \big \{{\varXi _p(x)}/{\Vert x\Vert _p^{p-1}}\big \} &{}\quad \mathrm{if}\ x\ne 0;\\ \{w\in \mathbb {R}^n: \Vert w\Vert _q\le 1\} &{}\quad \mathrm{if}\ x=0 \end{array}\right. \end{aligned}$$

(26)

For any given $x\in \mathbb {R}^n$, write $J(x):=\{i\in \{1,\ldots ,n\}: x_i\ne 0\}$ and define $J_0(x):=\{i\in J(x): \widehat{y}_i=0\}$, $J_{+}(x):=\{i\in J(x)\backslash J_0(x): \mathrm{sign}(x_i)=\mathrm{sign}(\widehat{y}_i)\}$ and $J_{-}(x):=J(x)\backslash (J_0(x)\cup J_{+}(x))$. We proceed the arguments by two cases.

Case 1$\widehat{x}=0$. Fix an arbitrary $\varepsilon >0$ and an arbitrary $\delta >0$. Let x be an arbitrary point from $\mathbb {B}(\widehat{x},\varepsilon )$. By Eq. (26), it is immediate to check that

$$\begin{aligned} \partial f_p(x)\subseteq \{w\in \mathbb {R}^n: \Vert w\Vert _q\le 1\} =\partial f_p(\widehat{x}), \end{aligned}$$

which implies the inclusion in (25) holds for any $\kappa >0$. This means that any $\kappa >0$ along with any $\varepsilon >0$ and $\delta >0$ satisfies the requirement of (25).

Case 2$\widehat{x}\ne 0$. Since $\lim _{z\rightarrow \widehat{x}}\Vert z\Vert _p=\Vert \widehat{x}\Vert _p>0$, there is $\widehat{\varepsilon }\in (0,1)$ such that

$$\begin{aligned} 0<\frac{1}{2}\Vert \widehat{x}\Vert _p\le \Vert z\Vert _p\quad \ \mathrm{for\ all}\ z\in \mathbb {B}(\widehat{x},\widehat{\varepsilon }). \end{aligned}$$

(27)

Write $\gamma :={\displaystyle \min \nolimits _{z\in \mathbb {B}(\widehat{x},\widehat{\varepsilon })}}\Vert z\Vert _p$. Clearly, $\gamma >0$. Let $\delta :={\displaystyle \max _{z\in \mathbb {B}(\widehat{x},\widehat{\varepsilon })}\max _{y\in \partial f_p(z)}}\Vert y-\widehat{y}\Vert +1$. Fix an arbitrary $x\in \mathbb {B}(\widehat{x},\widehat{\varepsilon })$. Take an arbitrary $y\in \partial f_p(x)\cap \mathbb {B}(\widehat{y},\delta )$. From (26),

$$\begin{aligned} \Vert y-\widehat{y}\Vert ^2&=\sum _{i:\,x_i=0}\frac{|\widehat{x}_i|^{2p-2}}{\Vert \widehat{x}\Vert _p^{2p-2}} +\sum _{i\in J_{+}(x)}\bigg |\frac{|x_i|^{p-1}}{\Vert x\Vert _p^{p-1}}-\frac{|\widehat{x}_i|^{p-1}}{\Vert \widehat{x}\Vert _p^{p-1}}\bigg |^2\nonumber \\&\quad +\sum _{i\in J_{-}(x)}\bigg [\frac{|\widehat{x}_i|^{p-1}}{\Vert \widehat{x}\Vert _p^{p-1}}+\frac{|x_i|^{p-1}}{\Vert x\Vert _p^{p-1}}\bigg ]^2 +\sum _{i\in J_0(x)}\frac{|x_i|^{2p-2}}{\Vert x\Vert _p^{2p-2}}. \end{aligned}$$

(28)

For the second summand on the right hand side of (28), by Lemma 3 we have

$$\begin{aligned} \sum _{i\in J_{+}(x)} \bigg |\frac{|x_i|^{p-1}}{\Vert x\Vert _p^{p-1}}-\frac{|\widehat{x}_i|^{p-1}}{\Vert \widehat{x}\Vert _p^{p-1}}\bigg |^2 \le (p-1)^2\sum _{i\in J_{+}(x)}\bigg |\frac{|x_i|}{\Vert x\Vert _p}-\frac{|\widehat{x}_i|}{\Vert \widehat{x}\Vert _p}\bigg |^2. \end{aligned}$$

(29)

For each $i\in J_{+}(x)$, let $h_i(z):=|z_i|/\Vert z\Vert _p$ for $z\in \mathbb {B}(\widehat{x},\widehat{\varepsilon })$. Since $\widehat{x}\ne 0$ and $\widehat{y}\in \partial f_p(\widehat{x})$, clearly, $\widehat{x}_i\ne 0$ for $i\in J_{+}(x)$. This means that each $h_i$ is continuously differentiable at $\widehat{x}$, and hence is continuously differentiable over $\mathbb {B}(\widehat{x},\widehat{\varepsilon })$ (making reduction on $\widehat{\varepsilon }$ if necessary). From the mean-valued theorem, there exists $\widetilde{x}$ in the segment between x and $\widehat{x}$ such that

$$\begin{aligned} |h_i(x)-h_i(\widehat{x})|&=\left| \frac{\mathrm{sign}(\widetilde{x}_i)}{\Vert \widetilde{x}\Vert _p}(x_i-\widehat{x}_i) -\frac{|\widetilde{x}_i|\langle \varXi _p(\widetilde{x}),x-\widehat{x}\rangle }{\Vert \widetilde{x}\Vert _p^{p+1}}\right| \\&\le \frac{2}{\Vert \widehat{x}\Vert _p}|x_i-\widehat{x}_i| +\frac{2}{\Vert \widehat{x}\Vert _p}\Vert x-\widehat{x}\Vert \le \frac{4}{\Vert \widehat{x}\Vert _p}\Vert x-\widehat{x}\Vert \end{aligned}$$

where the second inequality is by (27) and the expression of $\varXi _p(\cdot )$. By (29),

$$\begin{aligned} \sum _{i\in J_{+}(x)} \left| \frac{|x_i|^{p-1}}{\Vert x\Vert _p^{p-1}}-\frac{|\widehat{x}_i|^{p-1}}{\Vert \widehat{x}\Vert _p^{p-1}}\right| ^2\le & {} (p-1)^2\sum _{i\in J_{+}(x)}\big |h_i(x)-h_i(\widehat{x})\big |^2 \\\le & {} \frac{16n(p-1)^2}{\Vert \widehat{x}\Vert _p^2}\Vert x-\widehat{x}\Vert ^2. \end{aligned}$$

For the rest three summands on the right hand side of (28), by virtue of (27),

$$\begin{aligned}&\sum _{i:\,x_i=0}\frac{|\widehat{x}_i|^{2p-2}}{\Vert \widehat{x}\Vert _p^{2p-2}} \le \frac{1}{\Vert \widehat{x}\Vert _p^2}\sum _{i:\,x_i=0}|\widehat{x}_i|^2; \sum _{i\in J_{0}(x)}\frac{|x_i|^{2p-2}}{\Vert x\Vert _p^{2p-2}} \le \sum _{i\in J_{0}(x)}\frac{|x_i|^{2}}{\Vert x\Vert _p^{2}} \le \frac{1}{\gamma ^2}\sum _{i\in J_{0}(x)}|x_i|^2;\\&\sum _{i\in J_{-}(x)}\left[ \frac{|\widehat{x}_i|^{p-1}}{\Vert \widehat{x}\Vert _p^{p-1}}+\frac{|x_i|^{p-1}}{\Vert x\Vert _p^{p-1}}\right] ^2 \le \sum _{i\in J_{-}(x)}\left[ \frac{|\widehat{x}_i|}{\Vert \widehat{x}\Vert _p}+\frac{|x_i|}{\Vert x\Vert _p}\right] ^2\\&\qquad \le \frac{1}{\min (\gamma ^2,\Vert \widehat{x}\Vert _p^2)}\sum _{i\in J_{-}(x)}\big ||x_i|+|\widehat{x}_i|\big |^2. \end{aligned}$$

By combining the last three equations with (28), it is not difficult to obtain

$$\begin{aligned} \Vert y-\widehat{y}\Vert ^2\le \frac{32n(p-1)^2}{\min (\gamma ^2,\Vert \widehat{x}\Vert _p^2)}\Vert x-\widehat{x}\Vert ^2 \le \frac{32n(p-1)^2}{\gamma ^2}\Vert x-\widehat{x}\Vert ^2, \end{aligned}$$

which implies that $\mathrm{dist}(y,\partial f_p(\widehat{x}))\le \Vert y-\widehat{y}\Vert \le \frac{4\sqrt{2n}(p-1)}{\gamma }\Vert x-\widehat{x}\Vert$. By the arbitrariness of y in $\partial f_p(x)\cap \mathbb {B}(\widehat{y},\delta )$, (25) holds for $\kappa =\frac{4\sqrt{2n}(p-1)}{\gamma }$ and $\widehat{\varepsilon }$.

(ii) From $(\widehat{X},\widehat{W})\in \mathrm{gph}\,\partial F_{p}$ and [22, Corollary 2.5], $\sigma (\widehat{W})\in \partial f_p(\sigma (\widehat{X}))$. By [22, Lemma 2.3(b)], the conjugate $f_p^*$ of $f_p$ is absolutely symmetric, closed and convex. By the von Neumann’s trace inequality, it is easy to verify that $F_p^*(X)=f_p^*(\sigma (X))$ for $X\in \mathbb {R}^{n\times m}$, and hence $F_p^*$ is closed and convex by [22, Corollary 2.5]. Recall that $\partial f_p^*=(\partial f_p)^{-1}$. By the first part, $\partial f_p^*$ is metrically subregular at $\sigma (\widehat{W})$ for $\sigma (\widehat{X})$. Using [5, Proposition 14] again, we conclude that $\partial F_p^*=(\partial F_p)^{-1}$ is metrically subregular at $\widehat{W}$ for $\widehat{X}$. Consequently, the mapping $\partial F_p$ is calm at $\widehat{X}$ for $\widehat{W}$ when $p\in \{1\}\cup [2,+\infty ]$. $\square$