Kurdyka–Łojasiewicz Property of Zero-Norm Composite Functions

Abstract

This paper focuses on a class of zero-norm composite optimization problems. For this class of nonconvex nonsmooth problems, we establish the Kurdyka–Łojasiewicz property of exponent being a half for its objective function under a suitable assumption and provide some examples to illustrate that such an assumption is not very restricted which, in particular, involve the zero-norm regularized or constrained piecewise linear–quadratic function, the zero-norm regularized or constrained logistic regression function, the zero-norm regularized or constrained quadratic function over a sphere.

Appendices

Appendices

1.1 A: KL Property Relative to a Manifold

Let ${\mathcal {M}}\subset {\mathbb {R}}^p$ be a ${\mathcal {C}}^2$-smooth manifold and $f\!:{\mathcal {M}}\rightarrow {\mathbb {R}}$ be a ${\mathcal {C}}^2$-smooth function. The set of critical points of the problem $\min _{x\in {\mathcal {M}}}f(x)$ is ${\mathcal {X}}:=\big \{x\in {\mathcal {M}}:\ \nabla _{\!{\mathcal {M}}}f(x)=0\big \}$, where $\nabla _{\!{\mathcal {M}}}f(z)$ is the projection of $\nabla \!f(z)$ onto the tangent space ${\mathcal {T}}_{{\mathcal {M}}}(z)$ of ${\mathcal {M}}$ at z. We say that f is a KL function of exponent 1/2 relative to ${\mathcal {M}}$ if f has the KL property of exponent 1/2 at each ${\overline{x}}\in {\mathcal {X}}$, i.e., there exist $\delta >0$ and $\gamma >0$ such that

$$\begin{aligned} \Vert \nabla _{\!{\mathcal {M}}}f(z)\Vert \ge \gamma \sqrt{|f(z)-f({\overline{x}})|} \quad \ \forall z\in {\mathbb {B}}({\overline{x}},\delta )\cap {\mathcal {M}}. \end{aligned}$$

(13)

This part states the relation between the KL property of exponent 1/2 of f relative to ${\mathcal {M}}$ and the KL property of exponent 1/2 for its extended ${\widetilde{f}}(x):=f(x)+\delta _{{\mathcal {M}}}(x)$ for $x\in {\mathbb {R}}^p$.

Lemma A.1

Let ${\mathcal {M}}\!\subset \!{\mathbb {R}}^p$ be a ${\mathcal {C}}^2$-smooth manifold and $f\!:{\mathcal {M}}\rightarrow {\mathbb {R}}$ be a ${\mathcal {C}}^2$-smooth function. If f is a KL function of exponent 1/2 relative to ${\mathcal {M}}$, then ${\widetilde{f}}$ is a KL function of exponent 1/2. Conversely, if ${\widetilde{f}}$ is a KL function of exponent 1/2 and each critical point is a local minimizer, then f is a KL function of exponent 1/2 relative to ${\mathcal {M}}$.

Proof

Notice that $\partial \!{\widetilde{f}}(x)=\nabla \!f(x) +{\mathcal {N}}_{{\mathcal {M}}}(x)$ for any $x\in {\mathcal {M}}$. Clearly, ${\mathcal {X}}=\mathrm{crit}{\widetilde{f}}$. Fix an arbitrary ${\overline{x}}\in {\mathcal {X}}$. Since f has the KL property of exponent 1/2 relative to ${\mathcal {M}}$ at ${\overline{x}}$, there exist $\delta >0$ and $\gamma >0$ such that (13) holds for all $z\in {\mathbb {B}}({\overline{x}},\delta )\cap {\mathcal {M}}$. Fix an arbitrary $\eta >0$ and an arbitrary $x\in {\mathbb {B}}({\overline{x}},\delta ) \cap [{\widetilde{f}}({\overline{x}})<{\widetilde{f}} <{\widetilde{f}}({\overline{x}})+\eta ]$. Clearly, $x\in {\mathcal {M}}$. Moreover,

$$\begin{aligned}&\mathrm{dist}(0,\partial \!{\widetilde{f}}(x)) =\Vert \nabla f(x)-\varPi _{{\mathcal {N}}_{{\mathcal {M}}}(x)}(\nabla \!f(x))\Vert \nonumber \\&\quad =\Vert \varPi _{{\mathcal {T}}_{{\mathcal {M}}}(x)}(\nabla \!f(x))\Vert =\Vert \nabla \!f_{{\mathcal {M}}}(x)\Vert . \end{aligned}$$

(14)

Along with (13), $\mathrm{dist}(0,\partial \!{\widetilde{f}}(x))\ge \gamma \sqrt{f(x)-f({\overline{x}})}$. So, the first part of the results follows.

Next we focus on the second part. Fix an arbitrary ${\overline{x}}\in {\mathcal {X}}$. By the given assumption, clearly, ${\overline{x}}$ is a local optimal solution of $\min _{x\in {\mathcal {M}}}f(x).$ Hence, there exists $\varepsilon '>0$ such that

$$\begin{aligned} f(z)\ge f({\overline{x}})\quad \ \forall z\in {\mathbb {B}}({\overline{x}},\varepsilon ')\cap {\mathcal {M}}. \end{aligned}$$

By the KL property of exponent 1/2 of ${\widetilde{f}}$ at ${\overline{x}}$, there exist $\varepsilon ,c>0$ and $\eta >0$ such that

$$\begin{aligned} \mathrm{dist}(0,\partial \!{\widetilde{f}}(x))\ge c\sqrt{f(x)-{\widetilde{f}}(x)} \quad \forall x\in {\mathbb {B}}({\overline{x}},\varepsilon ) \cap [{\widetilde{f}}({\overline{x}})<{\widetilde{f}} <{\widetilde{f}}({\overline{x}})+\eta ]. \end{aligned}$$

(15)

Since f is ${\mathcal {C}}^2$-smooth around ${\overline{x}}$, there exists $\varepsilon ''>0$ such that for all $z\in {\mathbb {B}}({\overline{x}},\varepsilon '')\cap {\mathcal {M}}$, $f(z)<f({\overline{x}})+\eta .$ Take $\delta =\min (\varepsilon ,\varepsilon ',\varepsilon '')$. Fix an arbitrary $x\in {\mathbb {B}}({\overline{x}},\delta )\cap {\mathcal {M}}$. Clearly, $f({\overline{x}})\le f(x)\le f({\overline{x}})+\eta .$ If $f(x)>f({\overline{x}})$, then $x\in {\mathbb {B}}({\overline{x}},\varepsilon ) \cap [{\widetilde{f}}({\overline{x}})<{\widetilde{f}} <{\widetilde{f}}({\overline{x}})+\eta ]$, and from (15) and (14), $\Vert \nabla \!f_{{\mathcal {M}}}(x)\Vert \ge c\sqrt{|f(x)-f({\overline{x}})|}.$ If $f(x)=f({\overline{x}})$, this inequality holds automatically. $\square$

1.2 B: KL Property of the Quadratic Function over a Sphere

For any integer $m\ge 1$ and any given $m\times m$ real symmetric H, define $g(z):=z^{{\mathbb {T}}}Hz+\delta _{{\mathcal {S}}}(z)$ for $z\in {\mathbb {R}}^m$. Lemma 1 in Appendix A and [19, Theorem 1] imply that g is a KL function of exponent 1/2. This part gives a different proof, which needs the following lemmas.

Lemma B.1

The critical point set of g takes the form of $\mathrm{crit}g=\big \{z\in {\mathcal {S}}:\ Hz=\langle z,Hz\rangle z\big \}.$ So, by letting H have the eigenvalue decomposition $P\varLambda P^{{\mathbb {T}}}$ with $\varLambda =\mathrm{diag}(\lambda _1,\ldots ,\lambda _m)$ for $\lambda _1\ge \cdots \ge \lambda _m$ and $P\in {\mathbb {O}}^m$, $\mathrm{crit}g =PW$ with $W=\big \{y\in {\mathcal {S}}:\ \varLambda y=\langle y,\varLambda y\rangle y\big \}.$

Proof

By [27, Exercise 8.8] and Lemma 3.1, it immediately follows that for any $z\in {\mathbb {R}}^m$,

$$\begin{aligned} \partial g(z)=2Hz+\partial \delta _{{\mathcal {S}}}(z)=2Hz+[\![z]\!]. \end{aligned}$$

(16)

Choose an arbitrary ${\overline{z}}\in \mathrm{crit}g$. From (16), there exists ${\overline{t}}\in {\mathbb {R}}$ such that $0=2H{\overline{z}}+{\overline{t}}{\overline{z}}$. Along with $\Vert {\overline{z}}\Vert =1$, we have ${\overline{t}}=-2\langle {\overline{z}},H{\overline{z}}\rangle$, and hence ${\overline{z}}\in \!\big \{z\in {\mathcal {S}}:\ Hz=\!\langle z,Hz\rangle z\big \}$. Consequently, $\mathrm{crit}g\subseteq \big \{z\in {\mathcal {S}}:\ Hz=\langle z,Hz\rangle z\big \}.$ The converse inclusion is immediate to check by Lemma 3.1. Thus, the first part follows. The second part is immediate. $\square$

Lemma B.2

Let $D=\mathrm{diag}(d_1,d_2,\ldots ,d_p)$ with $d_1\ge d_2\ge \cdots \ge d_p$. Define the function $\psi (x):=x^{{\mathbb {T}}}Dx+\delta _S(x)$ for $x\in {\mathbb {R}}^p$. Then, $\psi$ is a KL function of exponent 1/2.

Proof

By Lemma B.1, it is immediate to obtain the following characterization for $\mathrm{crit}\psi$:

$$\begin{aligned} \mathrm{crit}\,\psi =\big \{x\in {\mathcal {S}}:\ Dx=\langle x,Dx\rangle x\big \}. \end{aligned}$$

(17)

Clearly, for each $x\in \mathrm{crit}\,\psi$, $d_i=\langle x,Dx\rangle$ with $i\in \mathrm{supp}(x)$. For any $z\in \mathrm{dom}\,\partial \psi$, we have

$$\begin{aligned} \begin{aligned} \text{ dist }(0,\partial \psi (z))^2&= \min _{u\in \partial \psi (z)} \Vert u\Vert ^2= \min _{w\in {\mathbb {R}}} \Vert 2Dz + wz\Vert ^2\\&=\min _{w\in {\mathbb {R}}}\Big \{4\langle z,D^{{\mathbb {T}}}Dz\rangle +w^2+4w\langle z,Dz\rangle \Big \}\\&= 4\langle z,D^{{\mathbb {T}}}Dz\rangle -4(\langle z,Dz\rangle )^2 =4\Vert Dz-\langle z,Dz\rangle z\Vert ^2. \end{aligned} \end{aligned}$$

(18)

Now fix an arbitrary ${\overline{x}}\in \mathrm{crit}\,\psi$. From (17), it immediately follows that $-D{\overline{x}}+\langle {\overline{x}},D{\overline{x}}\rangle {\overline{x}}=0.$ We next proceed the arguments by two cases as will be shown below.

Case 1: $d_1=\cdots =d_p=\gamma$ for some $\gamma \in {\mathbb {R}}$. Choose an arbitrary $\eta >0$ and an arbitrary $\delta >0$. Fix an arbitrary $x \in {\mathbb {B}}({\overline{x}},\delta )\cap [\psi ({\overline{x}})<\psi (x) < \psi ({\overline{x}})+\eta ]$. Clearly, $x\in {\mathcal {S}}$ and $\langle x,Dx\rangle =\gamma$. Combining $\langle {\overline{x}},D{\overline{x}}\rangle {\overline{x}}=D{\overline{x}}$ and Eq. (18) yields that

$$\begin{aligned} \text{ dist }(0, \partial \psi (x)) =4\Vert Dx-\langle x,Dx\rangle x-(D{\overline{x}}-\langle {\overline{x}},D{\overline{x}}\rangle {\overline{x}})\Vert =0. \end{aligned}$$

In addition, $\psi (x)=\psi ({\overline{x}})=\gamma$. This means that $\text{ dist }(0, \partial \psi (x))=\sqrt{\psi (x)-\psi ({\overline{x}})}.$

Case 2: there exist $i\ne j\in \{1,2,\ldots ,p\}$ such that $d_i\ne d_j$. Write $J=\mathrm{supp}({\overline{x}})$ and ${\overline{J}}=\{1,\ldots ,p\}\backslash J$. By (17), we know that $d_i=\langle {\overline{x}},D{\overline{x}}\rangle$ for all $i\in J$. This means that there must exist an index $\kappa \in {\overline{J}}$ such that $d_{\kappa }\ne \langle {\overline{x}},D{\overline{x}}\rangle$. Write ${\overline{J}}_1:=\big \{i\in {\overline{J}}:\ d_i\ne \langle {\overline{x}},D{\overline{x}}\rangle \big \}.$ By the continuity of the function $\langle \cdot ,D\cdot \rangle$, there exists $\delta >0$ such that for all $z\in {\mathbb {B}}({\overline{x}},\delta )\cap {\mathcal {S}}$,

$$\begin{aligned} \frac{1}{2}|d_j-\langle {\overline{x}},D{\overline{x}} \rangle |\le |d_j-\langle z,Dz\rangle |\le \frac{3}{2}|d_j-\langle {\overline{x}},D{\overline{x}}\rangle |\quad \forall j\in {\overline{J}}_1. \end{aligned}$$

(19)

Choose an arbitrary $\eta >0$. Fix an arbitrary $x\in {\mathbb {B}}({\overline{x}},\delta ) \cap [\psi ({\overline{x}})<\psi (x)<\psi ({\overline{x}})+\eta ]$. Clearly, $x \in {\mathcal {S}}$. From Eq. (18), it follows that

$$\begin{aligned} \frac{1}{4}\text{ dist }(0,\partial \psi (x))^2&=\sum _{j\in {\overline{J}}}\big (d_j-\langle x,Dx\rangle \big )^2x_j^2 +\sum _{j\in J}\big (d_j-\langle x,Dx\rangle \big )^2x_j^2\nonumber \\&=\sum _{j\in {\overline{J}}}\big (d_j-\langle x,Dx\rangle \big )^2x_j^2 +\sum _{j\in J}\big (\langle {\overline{x}},D{\overline{x}}\rangle -\langle x,Dx\rangle \big )^2x_j^2\nonumber \\&\ge \sum _{j\in {\overline{J}}_1}\big (d_j-\langle x,Dx\rangle \big )^2x_j^2 \ge \frac{1}{4}\sum _{j\in {\overline{J}}_1}\big (d_j-\langle {\overline{x}},D{\overline{x}}\rangle \big )^2x_j^2 \end{aligned}$$

(20)

where the third equality is due to (17), the first inequality is by the definition of ${\overline{J}}_1$, and the last inequality is due to (19). On the other hand, by the definition of $\psi$,

$$\begin{aligned} \psi (x)-\psi ({\overline{x}})&=\langle x,Dx\rangle -\langle {\overline{x}},D{\overline{x}}\rangle =\sum _{j\in {\overline{J}}}d_jx_j^2+\sum _{j\in J}d_jx_j^2-\langle {\overline{x}},D{\overline{x}}\rangle \Vert x\Vert ^2\nonumber \\&=\sum _{j\in {\overline{J}}}\big (d_j-\langle {\overline{x}},D{\overline{x}}\rangle \big )x_j^2 +\sum _{j\in J}\big (d_j-\langle {\overline{x}},D{\overline{x}}\rangle \big )x_j^2\nonumber \\&=\sum _{j\in {\overline{J}}}\big (d_j-\langle {\overline{x}},D{\overline{x}}\rangle \big )x_j^2 =\sum _{j\in {\overline{J}}_1}\big (d_j-\langle {\overline{x}},D{\overline{x}}\rangle \big )x_j^2\nonumber \\&\le \sum _{j\in {\overline{J}}_1}|\langle {\overline{x}},D{\overline{x}}\rangle -d_j| x_j^2 \le \max _{j\in {\overline{J}}_1}|d_j-\langle {\overline{x}},D{\overline{x}}\rangle |\Vert x_{{\overline{J}}_1}\Vert ^2 \end{aligned}$$

(21)

where the fourth equality is due to (17), the fifth one is by the definition of ${\overline{J}}_1$, and the inequality is since $\psi (x)-\psi ({\overline{x}})>0$. From the above inequalities (20) and (21),

$$\begin{aligned} \text{ dist }(0, \partial \psi (x))&\ge \sqrt{\sum _{j\in {\overline{J}}_1} \big (d_j-\langle {\overline{x}},D{\overline{x}}\rangle \big )^2x_j^2} \ge \min _{j\in {\overline{J}}_1}|d_j-\langle {\overline{x}}, D{\overline{x}}\rangle |\Vert x_{{\overline{J}}_1}\Vert \\&\ge \frac{\min _{j\in {\overline{J}}_1}|d_j- \langle {\overline{x}},D{\overline{x}}\rangle |}{\sqrt{\max _{j\in {\overline{J}}_1}|d_j- \langle {\overline{x}},D{\overline{x}}\rangle |}} \sqrt{\psi (x)-\psi ({\overline{x}})}. \end{aligned}$$

By the arbitrariness of x, Cases 1 and 2 show that $\psi$ has the KL property with exponent 1/2 at ${\overline{x}}$. From the arbitrariness of ${\overline{x}}$ in $\mathrm{crit}\psi$, $\psi$ is a KL function of exponent 1/2. $\square$

Now we prove that g is a KL function of exponent 1/2. Fix an arbitrary ${\overline{z}}\in \mathrm{crit}g$. Let H have the eigenvalue decomposition as in Lemma B.1. Then, ${\overline{y}}=P^{{\mathbb {T}}}{\overline{z}}\in \mathrm{crit}\psi$ where $\psi$ is defined in Lemma B.2 with $D=\varLambda$. By Lemma B.2, there exist $\eta>0,\delta >0$ and $c>0$ such that

$$\begin{aligned} \mathrm{dist}(0,\partial \psi (y))\ge c\sqrt{\psi (y)-\psi ({\overline{y}})} \quad \ \forall y\in {\mathbb {B}}({\overline{y}},\delta ) \cap [\psi ({\overline{y}})<\psi <\psi ({\overline{y}})+\eta ]. \end{aligned}$$

Fix an arbitrary $z\in {\mathbb {B}}({\overline{z}},\delta )\cap [g({\overline{z}})<g<g({\overline{z}})+\eta ]$. Clearly, $z\in {\mathcal {S}}$. Write $y=P^{{\mathbb {T}}}z$. Then, $y\in {\mathcal {S}}$ and $g(z)=\psi (y)$. Since $g({\overline{z}})=g({\overline{y}})$, $y\in {\mathbb {B}}({\overline{y}},\delta )\cap [\psi ({\overline{y}})<\psi (y) < \psi ({\overline{y}})+\eta ].$ In addition, from (16) and the eigenvalue decomposition of H, $\partial g(z)=P\partial \psi (y)$. Thus,

$$\begin{aligned} \mathrm{dist}(0,\partial g(z))=\mathrm{dist}(0,P\partial \psi (y)) =\mathrm{dist}(0,\partial \psi (y))\ge c\sqrt{\psi (y)-\psi ({\overline{y}})}. \end{aligned}$$

Together with $\psi (y)-\psi ({\overline{y}})=g(z)-g({\overline{z}})$, it follows that g has the KL property with exponent of 1/2 at ${\overline{z}}$. By the arbitrariness of ${\overline{z}}$ in $\mathrm{crit}g$, g is a KL function of exponent 1/2.

1.3 C: Supplementary Lemma and Proofs

The following lemma extends the result of [34, Section 2.3] for the differentiable strongly convex function to the setting of closed proper strongly convex functions. In particular, it implies that the composite g is a KL function of exponent 1/2 without surjectivity of ${\mathcal {A}}$.

Lemma C.1

Consider $g(x)\!:=\vartheta ({\mathcal {A}}x)$ for $x\in {\mathbb {X}}$ where ${\mathcal {A}}\!:{\mathbb {X}}\rightarrow {\mathbb {Z}}$ is a linear mapping, and $\vartheta \!:{\mathbb {Z}}\rightarrow ]-\infty ,\infty ]$ is a proper closed strongly convex function with modulus $\mu$. Here, ${\mathbb {X}}$ and ${\mathbb {Z}}$ are two finite dimensional vector spaces equipped with the inner product $\langle \cdot ,\cdot \rangle$ and its induced norm $\Vert \cdot \Vert$. If $\mathrm{ri}(\mathrm{dom}\vartheta )\cap \mathrm{range}{\mathcal {A}}\ne \emptyset$, then there exists a constant ${\overline{c}}>0$ such that

$$\begin{aligned} \mathrm{dist}(0, \partial g(x))\ge \frac{\sqrt{2\mu }}{{\overline{c}}} \sqrt{g(x)-g^*}\quad \ \forall x\in {\mathbb {X}} \end{aligned}$$

(22)

where $g^*$ denotes the minimum value of the function g.

Proof

Pick an arbitrary $x^*\in \mathrm{crit}g$ (if $\mathrm{crit}g=\emptyset$, the conclusion holds automatically). We first prove that $\mathrm{crit}g =\{x\in {\mathbb {X}}\!:\,{\mathcal {A}}x={\mathcal {A}}x^*\}$. To this end, pick any $x'\in {\mathbb {X}}$ with ${\mathcal {A}}x'={\mathcal {A}}x^*$. Since $\mathrm{ri}(\mathrm{dom}\vartheta )\cap \mathrm{range}{\mathcal {A}}\ne \emptyset$, by [30, Theorem 23.9], we have $\partial g(x')={\mathcal {A}}^*\partial \vartheta ({\mathcal {A}}x') = {\mathcal {A}}^*\partial \vartheta ({\mathcal {A}}x^*) = \partial g(x^*).$ Notice that $0\in \partial g(x^*)$, we obtain $0\in \partial g(x')$, which implies that $x'\in \mathrm{crit}g$. This means that $\{x\in {\mathbb {X}}\!:\,{\mathcal {A}}x={\mathcal {A}}x^*\}\subseteq \mathrm{crit}g$. Suppose that there exist ${\overline{x}}\in \mathrm{crit} g$ such that ${\mathcal {A}}{\overline{x}}\ne {\mathcal {A}}x^*$. Then, by the strong convexity of $\vartheta$, we have

$$\begin{aligned} g(({\overline{x}} + x^*)/2) = \vartheta (({\mathcal {A}}{\overline{x}}+{\mathcal {A}}x^*)/2) <(g({\overline{x}})+ g(x^*))/2. \end{aligned}$$

This contradicts the fact that $x', x^* \in \mathrm{crit}g$. Thus, the equality $\mathrm{crit} g= \{x \in {\mathbb {X}}\!:\,{\mathcal {A}}x = {\mathcal {A}}x^*\}$ holds. By Hoffman inequality [35], there exists a constant ${\overline{c}}>0$ such that for any $z \in {\mathbb {X}}$,

$$\begin{aligned} \Vert \varPi _{\mathrm{crit} g} (z) -z\Vert \le {\overline{c}} \Vert {\mathcal {A}}(\varPi _{\mathrm{crit} g} (z) - z)\Vert , \end{aligned}$$

(23)

where $\varPi _{\mathrm{crit} g}$ is the projection mapping onto $\mathrm{crit} g$. Fix an arbitrary $x\in {\mathbb {X}}$. If $x\notin \mathrm{dom}\partial g$, the inequality (22) holds trivially. So, it suffices to consider the case $x\in \mathrm{dom}\partial g$. By [30, Theorem 23.9], $\partial g(x)={\mathcal {A}}^* \partial \vartheta ({\mathcal {A}}x)$. Obviously, $\partial \vartheta ({\mathcal {A}}x) \ne \emptyset$. Pick any $\xi \in \partial \vartheta ({\mathcal {A}}x)$. By the strong convexity of $\vartheta$ and [36, Theorem 6.1.2], it follows that

$$\begin{aligned} g(z) \ge g(x)+\langle \xi ,{\mathcal {A}}(z-x)\rangle + \frac{\mu }{2}\Vert {\mathcal {A}}(z-x)\Vert ^2 \quad \ \forall z\in {\mathbb {X}}. \end{aligned}$$

By taking $z = \varPi _{\mathrm{crit} g}(x)$, from the last inequality we obtain that

$$\begin{aligned} \begin{aligned} g(\varPi _{\mathrm{crit} g}(x))&\ge g(x) + \langle \xi , {\mathcal {A}}(\varPi _{\mathrm{crit} g}(x)-x) \rangle + \frac{\mu }{2} \Vert {\mathcal {A}}(\varPi _{\mathrm{crit} g}(x)-x)\Vert ^2 \\&\ge g(x) + \langle \xi , {\mathcal {A}}(\varPi _{\mathrm{crit} g}(x)-x)\rangle + \frac{\mu }{2{\overline{c}}^2} \Vert \varPi _{\mathrm{crit} g}(x)-x\Vert ^2 \\&\ge g(x) + \min _{y\in {\mathbb {X}}}\left[ \langle \xi , {\mathcal {A}}(y-x)\rangle + \frac{\mu }{2{\overline{c}}^2} \Vert y-x\Vert ^2 \right] \\&\ge g(x) - 0.5 ({\overline{c}}^2/\mu ) \Vert {\mathcal {A}}^*\xi \Vert ^2, \end{aligned} \end{aligned}$$

where the second inequality follows from (23). Note that $g(\varPi _{\mathrm{crit} g}(x))=g(x^*)$. The last inequality implies that $\Vert {\mathcal {A}}^*\xi \Vert ^2\ge (2\mu /{\overline{c}}^2)[g(x)-g(x^*)]$. Together with $\partial g(x)={\mathcal {A}}^*\partial \vartheta ({\mathcal {A}}x)$,

$$\begin{aligned} \mathrm{dist}(0, \partial g(x))^2\ge \min _{\xi \in \partial \vartheta ({\mathcal {A}}x)}\Vert {\mathcal {A}}^*\xi \Vert ^2 \ge (2\mu /{\overline{c}}^2)[g(x)-g(x^*)]. \end{aligned}$$

This implies that the desired inequality (22) holds. $\square$

The proof of Proposition 3.2: First, we assume that $\psi$ is a proper closed piecewise linear–quadratic convex function. From Lemma 3.2 and 3.3, we observe that the multifunction $\partial h$ is piecewise, i.e., its graph is the union of finitely many polyhedral sets. So, $\partial h$ is locally upper Lipschitzian at each point $x\in {\mathbb {R}}^p$ by [33, Proposition 1], which implies that $\partial h$ is metrically subregular at each point of its graph. In addition, Sun [37] showed that a proper closed convex function $\psi$ is piecewise linear–quadratic if and only if $\partial \psi$ is piecewise polyhedral. Thus, by combining [33, Proposition 1] and [38, Section 3.2], we obtain the conclusion.

Now assume $\psi =\delta _C$. Fix an arbitrary ${\overline{x}}\in C$. Write $J=\mathrm{supp}({\overline{x}})$ and ${\overline{J}}=\{1,\ldots ,p\}\backslash J$. Define the subspace $L\!:=\{x\in {\mathbb {R}}^p:\ x_i=0\ \mathrm{for}\ i\in {\overline{J}}\}$. By Lemma 3.2, $\partial h({\overline{x}})={\mathcal {N}}_{L}({\overline{x}})$. Take an arbitrary $v\in \widehat{\partial }(\delta _C+h)({\overline{x}})$. From Definition 3.1, it follows that

$$\begin{aligned} 0&\le \liminf _{x'\rightarrow {\overline{x}}, x'\ne {\overline{x}}}\frac{h(x')+\delta _C(x') -h({\overline{x}})-\delta _C({\overline{x}})-\langle v,x'-{\overline{x}}\rangle }{\Vert x'-{\overline{x}}\Vert }\\&\le \liminf _{x'\in C,\mathrm{supp}(x')=J,x'\rightarrow {\overline{x}}, x'\ne {\overline{x}}} \frac{h(x')-h({\overline{x}})-\langle v,x'-{\overline{x}}\rangle }{\Vert x'-{\overline{x}}\Vert }\\&=\liminf _{x'\in C,\mathrm{supp}(x')=J,x'\rightarrow {\overline{x}}, x'\ne {\overline{x}}} \frac{-\langle v,x'-{\overline{x}}\rangle }{\Vert x'-{\overline{x}}\Vert }\\&=\liminf _{x'\in C\cap L,x'\rightarrow {\overline{x}}, x'\ne {\overline{x}}} \frac{\delta _{C\cap L}(x')-\delta _{C\cap L}({\overline{x}}) -\langle v,x'-{\overline{x}}\rangle }{\Vert x'-{\overline{x}}\Vert } \end{aligned}$$

which implies that $v\in \widehat{\partial }\delta _{C\cap L}({\overline{x}})$. Consequently, $\widehat{\partial }(\delta _C+h)({\overline{x}}) \subseteq \widehat{\partial }\delta _{C\cap L}({\overline{x}})$. Together with [27, Corollary 10.9], Lemma 3.2, $\partial h({\overline{x}})={\mathcal {N}}_{L}({\overline{x}})$ and the convexity of C, we have

$$\begin{aligned} \begin{aligned} \partial \delta _C({\overline{x}})+\partial h({\overline{x}})&= \widehat{\partial }\delta _C({\overline{x}})+\widehat{\partial }h({\overline{x}}) \subseteq \widehat{\partial }(\delta _C+h)({\overline{x}})\subseteq \partial (\delta _C+\delta _L)({\overline{x}})\\&=\partial \delta _C({\overline{x}})+\partial \delta _L({\overline{x}}) =\partial \delta _C({\overline{x}})+\partial h({\overline{x}}). \end{aligned} \end{aligned}$$

(24)

By the arbitrariness of ${\overline{x}}$, this implies that $\partial \delta _C(x)+\partial h(x)=\widehat{\partial }(\delta _C+h)(x)$ for any $x\in C$. Next we argue that $\partial (\delta _C+h)({\overline{x}})\subseteq \partial \delta _C({\overline{x}})+\partial h({\overline{x}})$. Take an arbitrary $v\in \partial (\delta _C+h)({\overline{x}})$. There exist $x^k\xrightarrow [\delta _{C}+h]{}{\overline{x}}$ and $v^k\in \widehat{\partial }(\delta _{C}\!+\!h)(x^k)$ with $v^k\rightarrow v$ as $k\rightarrow \infty$. From the previous arguments, $v^k\in \partial \delta _{C}(x^k)+\partial h(x^k)$ for each k. Since $x^k\rightarrow {\overline{x}}$, we have $x^k\ne 0$ and $\mathrm{supp}(x^k)\supseteq J$ for all sufficiently large k. Since $\delta _{C}(x^k)+h(x^k)\rightarrow \delta _{C}({\overline{x}})+h({\overline{x}})$, we must have $x^k\in C$ and $h(x^k)\rightarrow h({\overline{x}})$ for all sufficiently large k. The latter, along with $\mathrm{supp}(x^k)\supseteq J$, implies that $\mathrm{supp}(x^k)=J$ for all sufficiently large k. So, $\partial h(x^k)=\partial \delta _{L}(x^k)$ for large enough k. Combing with (24) and $v^k\in \partial \delta _{C}(x^k)+\partial h(x^k)$, we have $v^k\in \partial \delta _{C\cap L}(x^k)$. Then, $v\in \partial (\delta _C+\delta _{L})({\overline{x}}) =\partial \delta _C({\overline{x}})+\partial \delta _{L}({\overline{x}}) =\partial \delta _C({\overline{x}})+\partial h({\overline{x}})$. The stated inclusion holds. The previous arguments imply that $\widehat{\partial }(\delta _C+h)({\overline{x}}) =\partial (\delta _C+h)({\overline{x}}) ={\mathcal {N}}_{C}({\overline{x}})+\partial h({\overline{x}}) =\partial \delta _{C\cap L}({\overline{x}}).$ Suppose that $\partial \delta _{C\cap L}({\overline{x}})\ne \emptyset$ (if not, the last equation implies the result). Then, we have

$$\begin{aligned} \partial (\delta _{C}+h)({\overline{x}}) =\partial \delta _{C\cap L}({\overline{x}}) =\partial ^{\infty }\delta _{C\cap L}({\overline{x}}) =[\partial \delta _{C\cap L}({\overline{x}})]^{\infty } =[\widehat{\partial }(\delta _{C}\!+h)({\overline{x}})]^{\infty } \end{aligned}$$

where the second equality is by [27, Exercise 8.14] and the third one is due to [27, Proposition 8.12]. Thus, the first part of the desired results follows. Using the same arguments as above, we can obtain the second part. The proof is completed. $\square$