Correction to: Journal of Optimization Theory and Applications (2020) 186:86–101 https://doi.org/10.1007/s10957-020-01691-0

1 Introduction

In [1], some parts of the proofs of Theorems 3.1 and 3.2(ii) are needed to be revised. The conclusion of [1, Theorem 3.1] holds under an additional assumption, thus it is restated and proved. Also, the proof of [1, Theorem 3.2(ii)] is modified. With this method of proof, the USRC of the function \({\tilde{d}}(u)\) at \(u=0\) becomes a smaller set in comparison with its counterpart in [1, Theorem 3.2(ii)]. Thus we can say, the statement of [1, Theorem 3.2(ii)] is improved. Furthermore, [1, Lemmas 3.2 and 3.3] are not required and hence are omitted. Accordingly, the paragraph before these Lemmas is also removed. Further, [1, Theorem 3.3] gives a better form of the optimality condition and is updated. No other changes are required regarding the preliminaries, definitions, main conclusions and examples.

2 Modified Results

First, we update [1, Theorem 3.1] by adding the following additional assumption from [2]:

We say that the function F is calm at \({\bar{x}}\) with some modulus \(l>0\), if there exists a positive scalar \(\delta \) satisfying \(\Vert F(x)-F({\bar{x}})\Vert \leqslant l\Vert x-{\bar{x}}\Vert ,\) for each \(x\in {\bar{x}}+\delta {\mathbb {B}}_n.\)

Theorem 2.1

([1, Theorem 3.1] updated) Assume that the function F is calm at \({\bar{x}}\) with some modulus \(l>0\) and \(d_{\varLambda }\) is directionally differentiable at \(F({\bar{x}})\). If EBCQ holds at \({\bar{x}}\) with a constant \(\sigma \), then ACQ is satisfied at \({\bar{x}}\) with the same constant.

Proof

(modified) Let \(u\notin T({\bar{x}};F^{-1}(\varLambda ))\) (otherwise there is nothing to prove) and EBCQ be satisfied at \({\bar{x}}\) with \(\sigma =1\). Assume also that \(0\leqslant {\tilde{d}}(u)<\infty \) (if \({\tilde{d}}(u)=+\infty \), the ACQ obviously holds). Thus, there is a sequence \(t_k\downarrow 0\) such that

$$\begin{aligned} {\tilde{d}}(u)=\lim _{k\rightarrow \infty }d_{T(F({\bar{x}});\varLambda )}\left( \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}\right) . \end{aligned}$$

The closedness of \(T(F({\bar{x}});\varLambda )\), gives us a sequence \(\{w_k\}\) such that for each k, 

$$\begin{aligned} d_{T(F({\bar{x}});\varLambda )}\left( \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}\right) =\left\| \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}-w_k\right\| . \end{aligned}$$
(1)

We assert that the sequence \(\{w_k\}\) is bounded. Fixing \(\varepsilon >0\) and observing (1), we obtain the following inequalities for all k sufficiently large:

$$\begin{aligned} \Vert w_k\Vert \leqslant \left\| \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}-w_k\right\| +\left\| \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}\right\| <{\tilde{d}}(u)+\varepsilon +l\Vert u\Vert , \end{aligned}$$

which shows the boundedness of \(\{w_k\}\) and the assertion is proved. Thus by passing to a subsequence, without relabelling, \(\{w_k\}\) converges to some vector \(w\in T(F({\bar{x}});\varLambda )\). Now, By EBCQ one has

$$\begin{aligned} \begin{aligned} \frac{d_{F^{-1}(\varLambda )}({\bar{x}}+t_ku)}{t_k}&\leqslant \frac{d_{\varLambda }(F({\bar{x}}+t_ku))}{t_k}\\&\leqslant \frac{d_{\varLambda }(F({\bar{x}})+t_kw_k)}{t_k}+\left\| \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}-w_k\right\| . \end{aligned} \end{aligned}$$
(2)

Next, we claim that \(\limsup _{k\rightarrow {\infty }}\frac{d_{\varLambda }(F({\bar{x}})+t_kw_k)}{t_k}=0.\) From [1, Lemma 3.1] and the fact that \(d_{\varLambda }\) is directionally differentiable at \(F({\bar{x}}),\) we get

$$\begin{aligned} \begin{aligned} 0\leqslant \limsup _{k\rightarrow \infty }\frac{d_{\varLambda }(F({\bar{x}})+t_kw_k)}{t_k}&\leqslant \lim _{k\rightarrow \infty }\left\{ \frac{d_{\varLambda }(F({\bar{x}})+t_kw )}{t_k} +\Vert w_k-w\Vert \right\} \\&=d_{\varLambda }'(F({\bar{x}});w)=0, \end{aligned} \end{aligned}$$
(3)

which proves the claim. Now, it follows from (2), (3) and (1) that

$$\begin{aligned}\begin{aligned} \liminf _{k\rightarrow \infty }\frac{d_{F^{-1}(\varLambda )}({\bar{x}}+t_ku)}{t_k}&\leqslant \limsup _{k\rightarrow \infty } \bigg \{\frac{d_{\varLambda }(F({\bar{x}})+t_kw_k)}{t_k}\\&\quad +\left\| \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}-w_k\right\| \bigg \}\\&\leqslant \limsup _{k\rightarrow \infty }\frac{d_{\varLambda }(F({\bar{x}})+t_kw_k)}{t_k}\\&\quad +\lim _{k\rightarrow \infty }\left\| \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}-w_k\right\| \\&={\tilde{d}}(u). \end{aligned}\end{aligned}$$

Using again [1, Lemma 3.1], the above especially implies that

$$\begin{aligned} d_{T({\bar{x}};F^{-1}(\varLambda ))}(u)=d_{F^{-1}(\varLambda )}^-({\bar{x}};u)\leqslant {\tilde{d}}(u), \end{aligned}$$

and completes the proof of the theorem. \(\square \)

In what follows, the proof of [1, Theorem 3.2(ii)] is modified. By this modification, USRC of the function \({\tilde{d}}(.)\) at \(u=0\) becomes a smaller set and gives a better result; hence its statement is also improved.

Theorem 2.2

([1, Theorem 3.2(ii)] updated) Assume that \(\partial F({\bar{x}})\) is an u.s.c. PJ of \(F:{\mathbb {R}}^n\rightarrow {\mathbb {R}}^m\) at \({\bar{x}}\). Suppose also that \(F({\bar{x}})\in \varLambda \subseteq {\mathbb {R}}^m\) and \(\partial d_{\varLambda }(F({\bar{x}}))\) is a bounded USRC of \( d_{\varLambda }\) at \(F({\bar{x}})\). Then the closure of the set

$$\begin{aligned} \partial d_{\varLambda }(F({\bar{x}}))\circ \{\textrm{conv}~\partial F({\bar{x}})\cup [(\partial F({\bar{x}}))_{\infty }\setminus \{0\}]\} \end{aligned}$$

is an USRC of the function \( {\tilde{d}}\) at \(u=0\).

Proof

(revised) Put \(A:=\partial d_{\varLambda }(F({\bar{x}}))\circ \{\textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }{\setminus } \{0\}]\}\) and fix \(u\in {\mathbb {R}}^n\). First, let us show that \(\sup _{\eta \in A}\left<\eta ,u\right>\geqslant 0.\) For given \(M\in \textrm{conv}\partial F({\bar{x}})\cup [(\partial F({\bar{x}}))_{\infty }{\setminus } \{0\}]\), we have

$$\begin{aligned} \sup _{\xi \in \partial d_{\varLambda }(F({\bar{x}}))}\langle \xi , Mu\rangle \geqslant d_{\varLambda }^+(F({\bar{x}});Mu)\geqslant 0. \end{aligned}$$

Thus, using the definition of A, we get

$$\begin{aligned}\begin{aligned} \sup _{\eta \in A}\left<\eta ,u\right>&=\sup _{{\begin{array}{c}\xi \in \partial d_{\varLambda }(F({\bar{x}}))\\ M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }\setminus \{0\}]\end{array}}}\left<\xi M,u\right>\\&=\sup _{{\begin{array}{c}\xi \in \partial d_{\varLambda }(F({\bar{x}}))\\ M\in \textrm{conv}\partial F({\bar{x}})\cup [(\partial F({\bar{x}}))_{\infty }\setminus \{0\}]\end{array}}}\left<\xi ,Mu\right>\geqslant 0.\end{aligned}\end{aligned}$$

There are two possible cases: If \({\tilde{d}}^+(0; u)=0\), then trivially we obtain

$$\begin{aligned} {\tilde{d}}^+(0; u)\leqslant \sup _{\eta \in A}\left<\eta ,u\right>. \end{aligned}$$
(4)

Hence, let \({\tilde{d}}^+(0; u)>0.\) If the following inequality holds:

$$\begin{aligned} \sup _{{\begin{array}{c}\xi \in \partial d_{\varLambda }(F({\bar{x}}))\\ M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }\setminus \{0\}]\end{array}}}\left<\xi , Mu\right>>0, \end{aligned}$$

due to the cone property of \(( \partial F({\bar{x}}))_{\infty }{\setminus } \{0\}\), we get

$$\begin{aligned} \sup _{{\begin{array}{c}\xi \in \partial d_{\varLambda }(F({\bar{x}}))\\ M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }\setminus \{0\}]\end{array}}}\left<\xi , Mu\right>=+\infty , \end{aligned}$$

and the inequality in (4) holds trivially. Finally, the following case remains

$$\begin{aligned} \sup _{{\begin{array}{c}\xi \in \partial d_{\varLambda }(F({\bar{x}}))\\ M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }\setminus \{0\}]\end{array}}}\left<\xi , Mu\right>=0. \end{aligned}$$
(5)

For each fixed \(M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }{\setminus } \{0\}],\) one has

$$\begin{aligned}\begin{aligned} 0&\leqslant d_{\varLambda }^+(F({\bar{x}});Mu) \leqslant \sup _{\xi \in \partial d_{\varLambda }(F({\bar{x}}))}\langle \xi , Mu\rangle \\&\leqslant \sup _{{\begin{array}{c}\xi \in \partial d_{\varLambda }(F({\bar{x}}))\\ M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }\setminus \{0\}]\end{array}}}\left<\xi , Mu\right>=0 \end{aligned}\end{aligned}$$

Utilizing [1, Lemma 3.1], we have

$$\begin{aligned} 0\leqslant d_{T(F({\bar{x}});\varLambda )}(Mu)= d_{\varLambda }^-(F({\bar{x}});Mu)\leqslant d_{\varLambda }^+(F({\bar{x}});Mu)=0, \end{aligned}$$

which means that \(Mu\in T(F({\bar{x}});\varLambda )\), for all \(M\in \textrm{conv}\partial F({\bar{x}})\cup [( \partial F({\bar{x}}))_{\infty }{\setminus } \{0\}].\) Now, since \({\tilde{d}}^+(0; u)>0,\) there exits some positive number c such that \(c<{\tilde{d}}^+(0; u)={\tilde{d}}(u)\). Thus for some sequence \(t_k\downarrow 0\) and for all k sufficiently large, one has

$$\begin{aligned} c<d_{T(F({\bar{x}});\varLambda )}\left( \frac{F({\bar{x}}+t_ku)-F({\bar{x}})}{t_k}\right) . \end{aligned}$$
(6)

Applying now the mean value Theorem in [1, Propostion 2.3], we have for each k, 

$$\begin{aligned} F({\bar{x}}+t_ku)-F({\bar{x}})\in \textrm{cl conv}\{\partial F[{\bar{x}}+t_ku, {\bar{x}}] t_ku\}. \end{aligned}$$

Using the upper semicontinuity of \(\partial F(.)\) at \({\bar{x}}\), for given sequence \(r_s\downarrow 0\), there exits \(k_s>k_{s-1}\) satisfying

$$\begin{aligned}\begin{aligned} F({\bar{x}}+t_{k_s}u)-F({\bar{x}})&\in \textrm{cl conv}\{\partial F[{\bar{x}}+t_{k_s}u, {\bar{x}}] t_{k_s}u\}\\&\subseteq \textrm{cl conv}\{\{\partial F({\bar{x}})+\frac{r_s}{2}{\mathbb {B}}_{m\times n}\} t_{k_s}u\}\\&\subseteq \textrm{cl}\{\{\textrm{conv}\partial F({\bar{x}})+\frac{r_s}{2}{\mathbb {B}}_{m\times n}\} t_{k_s}u\}. \end{aligned}\end{aligned}$$

Thus, there exists \(M_{k_s}\in \textrm{conv}\partial F({\bar{x}}) \) such that

$$\begin{aligned} \left\| \frac{F({\bar{x}}+t_{k_s}u)-F({\bar{x}})}{t_{k_s}}-M_{k_s}u\right\| < r_s\Vert u\Vert . \end{aligned}$$

Choosing now subsequences \( M_s:=M_{k_s}\) and \( t_s:=t_{k_s}\), and using the inequality in (6), we deduce that

$$\begin{aligned} c<d_{T (F({\bar{x}});\varLambda )}\left( \frac{F({\bar{x}}+t_su)-F({\bar{x}})}{t_s}\right) <d_{T (F({\bar{x}});\varLambda )}(M_su)+r_s\Vert u\Vert . \end{aligned}$$

Observing that \(d_{T (F({\bar{x}});\varLambda )}(M_su)=0\) and taking limit as \(s\rightarrow \infty \) in the latter inequality, we arrive at the contradiction \( c\leqslant 0\), which shows the case \({\tilde{d}}^+(0; u)>0 \) and the equality (5) do not occur together and the proof is completed. \(\square \)

Since the USRC of the function \({\tilde{d}}\) is changed, the optimality condition in [1, Theorem 3.3] is improved and updated, accordingly.

Theorem 2.3

([1, Theorem 3.3] updated) Suppose that ACQ is satisfied at the local optimal point \({\bar{x}}\) of GOP. Let \(\partial f({\bar{x}})\) and \(\partial F({\bar{x}})\) are USRC and u.s.c. PJ of f and F at \({\bar{x}},\) respectively and \(\partial d_{\varLambda }( F({\bar{x}}))\) is a bounded USRC of \(d_{\varLambda }\) at \(F({\bar{x}})\). Then

$$\begin{aligned} 0\in \mathrm{cl~conv}\{\partial f({\bar{x}})+ l\sigma \partial d_{\varLambda }( F({\bar{x}}))\circ \{\textrm{conv}\partial F({\bar{x}})\cup [(\partial F({\bar{x}}))_{\infty }\setminus \{0\}]\}\}, \end{aligned}$$

where \(\sigma \) is the positive constant of ACQ and l is the Lipschitz constant of the function f in a neighborhood of \({\bar{x}}\).

3 Conclusion

The proofs of [1, Theorems 3.1 and 3.2(ii)] are rectified and their statements are updated. Also, [1, Theorem 3.3] gives a better form of the optimality condition which is improved.