Correction to: Kurdyka–Łojasiewicz Property of Zero-Norm Composite Functions

1 Correction to: Journal of Optimization Theory and Applications (2021) 188:94–112 https://doi.org/10.1007/s10957-020-01779-7

2 Introduction

In our paper [6], there is a gap for the statement of Proposition 3.2 and Remark 3.2. In addition, on line 14 of page 110, the set $[x'\in C,\ \mathrm{supp}(x')=J,\ x'\rightarrow \overline{x}, \ x'\ne \overline{x}]$ may be empty. In this erratum, we provide the correct statements for Proposition 3.2 and Remark 3.2 and update the proof of Proposition 3.2.

3 Corrected Result

First, we give the correct statement of [6, Proposition 3.2 & Remark 3.2].

Proposition 2.1

([6, Proposition 3.2] corrected) (i) When $h(\cdot )=\nu \Vert \cdot \Vert _0$ for a constant $\nu >0$, if $\psi \!:\mathbb {R}^p\rightarrow ]-\infty ,+\infty ]$ is a proper closed piecewise linear regular function, then for any $\overline{x}\in \mathrm{dom}\,\psi$,

$$\begin{aligned} \widehat{\partial }(\psi \!+\!h)(\overline{x})&=\partial \psi (\overline{x})\!+\!\partial h(\overline{x}) =\partial (\psi \!+\!h)(\overline{x}) \end{aligned}$$

(1)

$$\begin{aligned} \partial ^{\infty }(\psi \!+\!h)(\overline{x})&=[\partial \psi (\overline{x})\!+\!\partial h(\overline{x})]^{\infty }; \end{aligned}$$

(2)

if $\psi$ is an indicator function of some closed convex set $C\subseteq \mathbb {R}^p$, then for any $\overline{x}\in C$ such that $\{x\in \mathbb {R}^p\,|\, x_i=0\ \mathrm{for}\ i\notin \mathrm{supp}(\overline{x})\}\cap \mathrm{ri}(C)\ne \emptyset$, it holds that

$$\begin{aligned} \widehat{\partial }(\psi \!+\!h)(\overline{x}) =\partial \psi (\overline{x})\!+\!\partial h(\overline{x}) =\!\partial (\psi \!+\!h)(\overline{x}) =\!\partial ^{\infty }(\psi \!+\!h)(\overline{x}) =[\widehat{\partial }(\psi \!+\!h)(\overline{x})]^{\infty }. \end{aligned}$$

(ii) When $h=\delta _{\varOmega }$, the indicator function of $\varOmega :=\{x\in \mathbb {R}^p\!:\,\Vert x\Vert _0\le \kappa \}$ for an integer $\kappa >0$, the results of part (i) hold at any $\overline{x}\in \mathrm{dom}\psi$ with $\Vert \overline{x}\Vert _0=\kappa$, and at any $\overline{x}\in \mathrm{dom}\psi$ with $\Vert \overline{x}\Vert _0<\kappa$ it holds that $\partial (\psi +h)(\overline{x}) \subseteq \partial \psi (x)+\partial h(\overline{x})$.

Remark 2.1

([6, Remark 3.2] corrected) When $\psi$ is a locally Lipschitz regular function, the first part of Proposition 2.1 still holds by [5, Theorem 9.13(b) & Corollary 10.9], and now equality (2) is also given in [2, Proposition 1.107(iii)] and [3, Prop. 1.29].

In what follows, we provide the proof of Proposition 2.1.

The proof of Proposition 2.1: First, we consider that $\psi$ is a proper closed piecewise linear regular function. Fix any $\overline{x}\in \mathrm{dom}\,\psi$. Notice that $\mathrm{epi}\psi$ and $\mathrm{epi}h$ are the union of finitely many polyhedral sets. By combining [4, Proposition 1] and [1, Section 3.2], it then follows that

$$\begin{aligned} \partial (\psi +h)(\overline{x})\subseteq \partial \psi (\overline{x})+\partial h(\overline{x}) \ \ \mathrm{and}\ \ \partial ^{\infty }(\psi +h)(\overline{x})\subseteq \partial ^{\infty }\psi (\overline{x})+\partial ^{\infty } h(\overline{x}). \end{aligned}$$

(3)

From the first inclusion, $\partial (\psi +h)(\overline{x})\supseteq \widehat{\partial }(\psi +h)(\overline{x})\supseteq \widehat{\partial }\psi (\overline{x})+\widehat{\partial } h(\overline{x})$, and the regularity of $\psi$ and h, we obtain the equalities in (1). When $\partial \psi (\overline{x})=\emptyset$, obviously, the equalities in (2) hold. So, it suffices to consider the case where $\partial \psi (\overline{x})\ne \emptyset$. From the second inclusion in (3), it follows that

$$\begin{aligned}{}[\partial ^{\infty }(\psi +h)(\overline{x})]^{\circ } \supseteq [\partial ^{\infty }\psi (\overline{x})+\partial ^{\infty } h(\overline{x})]^{\circ } =[\partial ^{\infty }\psi (\overline{x})]^{\circ }\cap [\partial ^{\infty } h(\overline{x})]^{\circ } \end{aligned}$$

where $K^{\circ }$ denotes the negative polar of a cone K. By combining this inclusion with [5, Exercise 8.23] and the second equality of (1), for any $w\in \mathbb {R}^p$ we have

$$\begin{aligned} \widehat{d}(\psi +h)(\overline{x})(w)&\le \widehat{d}\psi (\overline{x})(w)+\widehat{d}h(\overline{x})(w) =d\psi (\overline{x})(w)+dh(\overline{x})(w)\\&\le d(\psi +h)(\overline{x})(w)\le \widehat{d}(\psi +h)(\overline{x})(w) \end{aligned}$$

where $\widehat{d}h(\overline{x})$ and $dh(\overline{x})$, respectively, denote the regular subderivative and the subderivative of $\psi +h$ at $\overline{x}$, the equality is due to the regularity of $\psi$ and h, and the second inequality is using [5, Corollary 10.9]. By [5, Corollary 8.19], this shows that $\psi +h$ is regular, and hence $\partial ^{\infty }(\psi +h)(\overline{x}) =[\widehat{\partial }(\psi +h)(\overline{x})]^{\infty } =[\partial \psi (\overline{x})+\partial h(\overline{x})]^{\infty }.$ Thus, we obtain the first part.

Next we consider the case $\psi =\delta _C$. Fix any $\overline{x}\in C$ with $\mathrm{ri}(C)\cap L_{\overline{x}}\ne \emptyset$, where $L_{\overline{x}}\!:=\{x\in \mathbb {R}^p\,|\, x_i=0\ \mathrm{for}\ i\notin \mathrm{supp}(\overline{x})\}$. Let $J=\mathrm{supp}(\overline{x})$. We first argue

$$\begin{aligned} \widehat{\partial }(\delta _C\!+h)(\overline{x}) \subseteq \partial \delta _{C\cap L_{\overline{x}}}(\overline{x}). \end{aligned}$$

(4)

If there exists $\varepsilon >0$ such that $[\mathbb {B}(\overline{x},\varepsilon )\backslash \{\overline{x}\}]\cap [C\cap L_{\overline{x}}]=\emptyset$, then $\partial \delta _{C\cap L_{\overline{x}}}(\overline{x}) =\mathcal {N}_{C\cap L_{\overline{x}}}(\overline{x})=\mathbb {R}^p$, and the inclusion in (4) clearly holds. So, it suffices to consider that for any $\varepsilon >0$, $[\mathbb {B}(\overline{x},\varepsilon )\backslash \{\overline{x}\}]\cap [C\cap L_{\overline{x}}]\ne \emptyset$. Pick any $v\in \widehat{\partial }(\delta _C\!+\!h)(\overline{x})$. By the definition of regular subgradient, it follows that

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

which implies that $v\in \widehat{\partial } \delta _{C\cap L_{\overline{x}}}(\overline{x}) =\partial \delta _{C\cap L_{\overline{x}}}(\overline{x})$. The inclusion in (4) holds. By combining (4) with [5, Corollary 10.9] and $\partial h(\overline{x})=\mathcal {N}_{L_{\overline{x}}}(\overline{x})$, we have

$$\begin{aligned} \begin{aligned} \partial \delta _C(\overline{x})+\partial h(\overline{x})= \widehat{\partial }\delta _C(\overline{x})+\widehat{\partial }h(\overline{x}) \subseteq \widehat{\partial }(\delta _C+h)(\overline{x})\subseteq \partial (\delta _C+\delta _{L_{\overline{x}}})(\overline{x})\\ =\partial \delta _C(\overline{x})+\partial \delta _{L_{\overline{x}}}(\overline{x}) =\partial \delta _C(\overline{x})+\partial h(\overline{x}).\qquad \qquad \quad \ \end{aligned} \end{aligned}$$

(5)

where the second equality is due to $\mathrm{ri}C \cap L_{\overline{x}}\ne \emptyset$. In fact, from the above arguments, we conclude that for all $x\in C\ \mathrm{with}\ \mathrm{ri}(C)\cap L_{x}\ne \emptyset$,

$$\begin{aligned} \widehat{\partial }(\delta _{C}\!+\!h)(x) =\partial \delta _C(x)+\partial h(x)=\partial \delta _{C\cap L_{x}}(x). \end{aligned}$$

(6)

Now we argue that $\partial (\delta _C\!+h)(\overline{x})\subseteq \partial \delta _C(\overline{x})+\partial h(\overline{x})$. To this end, pick any $v\in \partial (\delta _C+h)(\overline{x})$. Then, there exist sequences $x^k\xrightarrow [\delta _{C}+h]{}\overline{x}$ and $v^k\in \widehat{\partial }(\delta _{C}\!+\!h)(x^k)$ with $v^k\rightarrow v$ as $k\rightarrow \infty$. 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 k large enough. The latter, along with $\mathrm{supp}(x^k)\supseteq J$, implies that $\mathrm{supp}(x^k)=J$ for all sufficiently large k. By invoking (6), for all sufficiently large k, $v^k\in \partial \delta _{C}(x^k)+\partial h(x^k).$ By passing to the limit $k\rightarrow \infty$ and using $h(x^k)\rightarrow h(\overline{x})$, we obtain $v\in \partial \delta _{C}(\overline{x})+\partial h(\overline{x})$. By the arbitrariness of v in $\partial (\delta _C+h)(\overline{x})$, the stated inclusion follows. In particular, together with $\partial (\delta _C\!+h)(\overline{x})\supseteq \widehat{\partial } (\delta _C\!+h)(\overline{x}) =\partial \delta _C(\overline{x})+\partial h(\overline{x})$ and (5),

$$\begin{aligned} \widehat{\partial }(\delta _C\!+h)(\overline{x}) =\partial (\delta _C\!+h)(\overline{x}) =\mathcal {N}_{C}(\overline{x})+\partial h(\overline{x}) =\partial \delta _{C\cap L_{\overline{x}}}(\overline{x}). \end{aligned}$$

(7)

Next we argue $\partial ^{\infty }(\delta _C\!+h)(\overline{x})= \partial ^{\infty }\delta _{C\cap L_{\overline{x}}}(\overline{x})$. Pick any $u\in \partial ^{\infty }(\delta _C\!+h)(\overline{x})$. Then, there exist sequences $x^k\xrightarrow [\delta _{C}+h]{}\overline{x}$ and $u^k\in \widehat{\partial }(\delta _{C}\!+\!h)(x^k)$ with $\lambda _ku^k\rightarrow u$ for some $\lambda _k\downarrow 0$ as $k\rightarrow \infty$. By following the same arguments as above, $\mathrm{supp}(x^k)=J$ for all k large enough. Together with (6) and $u^k\in \widehat{\partial }(\delta _{C}\!+\!h)(x^k)$, we have $u^k\in \widehat{\partial }\delta _{C\cap L_{\overline{x}}}(x^k)$ for all k large enough. Notice that $x^k\xrightarrow [C\cap L_{\overline{x}}]{}\overline{x}$. So, $u\in \partial ^{\infty }(\delta _C\!+\!\delta _{L_{\overline{x}}})(\overline{x})$ and $\partial ^{\infty }(\delta _C\!+h)(\overline{x})\subseteq \partial ^{\infty }\delta _{C\cap L_{\overline{x}}}(\overline{x})$. Conversely, pick any $u\in \partial ^{\infty }\delta _{C\cap L_{\overline{x}}}(\overline{x})$. There exist $x^k\xrightarrow [C\cap L_{\overline{x}}]{}\overline{x}$ and $u^k\in \widehat{\partial }\delta _{C\cap L_{\overline{x}}}(x^k)$ with $\lambda _ku^k\rightarrow u$ for some $\lambda _k\downarrow 0$ as $k\rightarrow \infty$. Clearly, $(\delta _C+h)(x^k)\rightarrow (\delta _C+h)(\overline{x})$. Also, from (6) and $u^k\in \widehat{\partial }\delta _{C\cap L_{\overline{x}}}(x^k)$, we have $u^k\in \widehat{\partial }(\delta _{C}+h)(x^k)$. So, $u\in \partial ^{\infty }(\delta _C+h)(\overline{x})$, and $\partial ^{\infty }(\delta _C\!+h)(\overline{x})\supseteq \partial ^{\infty }\delta _{C\cap L_{\overline{x}}}(\overline{x})$. The stated equality follows. From [5, Exercise 8.14 & Proposition 8.12], $\partial \delta _{C\cap L_{\overline{x}}}(\overline{x}) =\partial ^{\infty }\delta _{C\cap L_{\overline{x}}}(\overline{x}) =[\widehat{\partial }\delta _{C\cap L_{\overline{x}}}(\overline{x})]^{\infty }$. Thus,

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

Together with (7), we obtain the conclusion for $h(\cdot )=\nu \Vert \cdot \Vert _0$. By following the same arguments as above, one may obtain the second part. $\square$

4 Conclusion

Since [6, Proposition 3.2] is only used to check [6, Assumption 4.1(iii)], the results in [6, Section 4] are all correct by Proposition 2.1.