2021 1]Faculty of Mathematics, TU Chemnitz, Reichenhainer Straße 41, Chemnitz,
09126, Germany

# Extended convergence analysis of the Scholtes-type regularization for
cardinality-constrained optimization problems

Sebastian Lämmel<EMAIL_ADDRESS>Vladimir
Shikhman<EMAIL_ADDRESS>[

###### Abstract

We extend the convergence analysis of the Scholtes-type regularization method
for cardinality-constrained optimization problems. Its behavior is clarified
in the vicinity of saddle points, and not just of minimizers as it has been
done in the literature before. This becomes possible by using as an
intermediate step the recently introduced regularized continuous reformulation
of a cardinality-constrained optimization problem. We show that the Scholtes-
type regularization method is well-defined locally around a nondegenerate
T-stationary point of this regularized continuous reformulation. Moreover, the
nondegenerate Karush-Kuhn-Tucker points of the corresponding Scholtes-type
regularization converge to a T-stationary point having the same index, i.e.
its topological type persists. Overall, we conclude that the global structure
of the regularized continuous reformulation and its Scholtes-type
regularization essentially coincide.

###### keywords:

cardinality-constrained optimization problem, Scholtes-type regularization
method, nondegenerate T-stationarity, index, genericity

##  1 Introduction

In nonconvex optimization Scholtes-type regularization methods became popular
since the seminal paper scholtes:2001 . Typically, hard nonsmooth constraints
are relaxed by means of a parameter. Then, Karush-Kuhn-Tucker points of the
induced nonlinear programs need to be computed. They are finally shown to
converge towards some suitably defined stationary points of the original
optimization problem as the regularization parameter tends to zero. Scholtes-
type regularization methods for mathematical programs with complementarity
(MPCC), vanishing (MPVC), switching (MPSC), and orthogonality type constrains
(MPOC) were examined along these lines in the literature so far, see e.g.
scholtes:2001 , izmailov:2009 , kanzow:2021 , and laemmel:mpoc  for further
details, respectively.

In this paper, we study the Scholtes-type regularization method for the class
of cardinality-constrained optimization problems:

$\mbox{CCOP}:\quad\min_{x}\,\,f(x)\quad\mbox{s.\,t.}\quad h(x)=0,\quad
g(x)\geq 0,\quad\left\|x\right\|_{0}\leq s$

with the feasible set given by equality, inequality, and cardinality
constraints, where the so-called zero ”norm”
$\left\|x\right\|_{0}=\left|\left\\{i\in\\{1,\ldots,n\\}\;|\;x_{i}\neq
0\right\\}\right|$ is counting non-zero entries of $x$. Here, we assume that
the objective function $f$, as well as the equality and inequality constraints
$h=\left(h_{p},p\in P\right)$, $g=\left(g_{q},q\in Q\right)$ are twice
continuously differentiable, and $s\in\\{0,1,\ldots,n-1\\}$ is an integer. In
order to arrive at the Scholtes-type regularization, the so-called continuous
reformulation of CCOP from burdakov:2016  is helpful:

$\begin{array}[]{rl}\displaystyle\min_{x,y}\,\,f(x)\quad\mbox{s.\,t.}&h(x)=0,\quad
g(x)\geq 0,\quad\displaystyle\sum_{i=1}^{n}y_{i}\geq n-s,\\\
&x_{i}y_{i}=0,\quad 0\leq y_{i}\leq 1,\quad i=1,\ldots,n.\end{array}$ (1)

As pointed out there, $\bar{x}$ solves CCOP if and only if there exists a
vector $\bar{y}$ such that $\left(\bar{x},\bar{y}\right)$ solves (1). In order
to tackle (1) numerically, bucher:2018  suggests to regularize the
orthogonality type constraints by using the Scholte’s idea, cf. scholtes:2001
:

$\begin{array}[]{rl}\displaystyle\min_{x,y}\,\,f(x)\quad\mbox{s.\,t.}&h(x)=0,\quad
g(x)\geq 0,\quad\displaystyle\sum_{i=1}^{n}y_{i}\geq n-s,\\\ &-t\leq
x_{i}y_{i}\leq t,\quad 0\leq y_{i}\leq 1,\quad i=1,\ldots,n,\end{array}$ (2)

where $t>0$. Further in branda:2018 , the authors prove that – under some
suitable constraint qualification and second-order sufficient condition – the
Scholtes-type regularization method is well-defined locally around a minimizer
of (1). Moreover, the Karush-Kuhn-Tucker points of (2) converge to an
S-stationary point of (1) whenever $t\rightarrow 0$.

Our goal is to extend the convergence analysis of the Scholtes-type
regularization method beyond the case of minimizers of (1), but also for all
kinds of its saddle points. By doing so, we intend to relate the indices of
nondegenerate Karush-Kuhn-Tucker points of the Scholtes-type regularization
with those of T-stationary points of the regularized continuous reformulation.
Here, nondegeneracy refers to some tailored versions of linear independence
constraint qualification, strict complementarity and second-order regularity.
Assuming nondegeneracy, Karush-Kuhn-Tucker points and T-stationary points can
be classified according to their quadratic and T-index, respectively. The
index encodes the local structure of the optimization problem under
consideration in algebraic terms and its global structure in the sense of
Morse theory, see Jongen:2000  and laemmel:reform . We note that for our
purpose we need to preliminarily regularize the continuous reformulation (1).
The reason is that all T-stationary points of (1) – considered as an MPOC
instance – turn out to be degenerate, cf. laemmel:mpoc . To overcome this
obstacle, it has been suggested in laemmel:reform  not only to linearly
perturb the objective function in (1) with respect to $y$-variables, but also
to additionally relax the upper bounds on them. As for our main results, the
Scholtes-type regularization method proves to be well-defined locally around a
nondegenerate T-stationary point of the regularized continuous reformulation.
Moreover, the nondegenerate Karush-Kuhn-Tucker points of its Scholtes-type
regularization converge to a T-stationary point having the same index.

The article is organized as follows. In Section 2 we discuss some preliminary
results on the regularized continuous reformulation of CCOP. Section 3 is
devoted to the extended convergence analysis of its Scholtes-type
regularization.

Our notation is standard. The cardinality of a finite set $A$ is denoted by
$|A|$. The $n$-dimensional Euclidean space is denoted by $\mathbb{R}^{n}$ with
the coordinate vectors $e_{i},i=1,\ldots,n$. The vector consisting of ones is
denoted by $e$. Given a twice continuously differentiable function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}$, $\nabla f$ denotes its gradient, and
$D^{2}f$ stands for its Hessian.

##  2 Regularized continuous reformulation

We associate with CCOP the regularized continuous reformulation from
laemmel:reform :

$\begin{array}[]{rl}\displaystyle\mathcal{R}(c,\varepsilon):\quad\min_{x,y}\,\,f(x)+c^{T}y\quad\mbox{s.\,t.}&h(x)=0,\quad
g(x)\geq 0,\quad\displaystyle\sum_{i=1}^{n}y_{i}\geq n-s,\\\
&x_{i}y_{i}=0,\quad 0\leq y_{i}\leq 1+\varepsilon,\quad
i=1,\ldots,n,\end{array}$

where the components of $c\in\mathbb{R}^{n}$ $c$ are positive and pairwise
different, and $0<\varepsilon\leq\frac{1}{n-s}$. Given a feasible point
$(\bar{x},\bar{y})$ of $\mathcal{R}$, we define the index sets which
correspond to the orthogonality type constraints $x_{i}y_{i}=0$, $y_{i}\geq
0$:

$\begin{array}[]{c}a_{01}\left(\bar{x},\bar{y}\right)=\left\\{i\,\left|\,\bar{x}_{i}=0,\bar{y}_{i}>0\right.\right\\},\quad
a_{10}\left(\bar{x},\bar{y}\right)=\left\\{i\,\left|\,\bar{x}_{i}\neq
0,\bar{y}_{i}=0\right.\right\\},\\\
a_{00}\left(\bar{x},\bar{y}\right)=\left\\{i\,\left|\,\bar{x}_{i}=0,\bar{y}_{i}=0\right.\right\\}.\end{array}$

The index sets of the active inequality constraints will be denoted by

$Q_{0}(\bar{x})=\left\\{q\in
Q\,\left|\,g_{q}(\bar{x})=0\right.\right\\},\quad\mathcal{E}(\bar{y})=\left\\{i\,\left|\,\bar{y}_{i}=1+\varepsilon\right.\right\\}.$

The regularized continuous reformulation $\mathcal{R}$ is a special case of
MPOC. The latter class was examined in laemmel:mpoc , where the MPOC-tailored
linear independence constraint qualification and the notion of (nondegenerate)
T-stationary points with the corresponding T-index were introduced. In
laemmel:reform , these concepts were applied to the regularization
$\mathcal{R}$.

######  Definition 1 (MPOC-LICQ, laemmel:reform ).

We say that a feasible point $(\bar{x},\bar{y})$ of $\mathcal{R}$ satisfies
the MPOC-tailored linear independence constraint qualification (MPOC-LICQ) if
the following vectors are linearly independent:

$\begin{array}[]{c}\begin{pmatrix}\nabla h_{p}(\bar{x})\\\ 0\end{pmatrix},p\in
P,\quad\begin{pmatrix}\nabla g_{q}(\bar{x})\\\ 0\end{pmatrix},q\in
Q_{0}(\bar{x}),\quad\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in\mathcal{E}(\bar{y}),\quad\begin{pmatrix}0\\\
e\end{pmatrix}\mbox{ if }\sum\limits_{i=1}^{n}\bar{y}_{i}=n-s,\\\
\begin{pmatrix}e_{i}\\\ 0\end{pmatrix},i\in
a_{01}\left(\bar{x},\bar{y}\right)\cup
a_{00}\left(\bar{x},\bar{y}\right),\quad\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in a_{10}\left(\bar{x},\bar{y}\right)\cup
a_{00}\left(\bar{x},\bar{y}\right).\end{array}$

######  Definition 2 (T-stationary point, laemmel:reform ).

A feasible point $(\bar{x},\bar{y})$ of $\mathcal{R}$ is called T-stationary
if there exists multipliers

$\begin{array}[]{l}\bar{\lambda}_{p},p\in P,\bar{\mu}_{1,q},q\in
Q_{0}(\bar{x}),\bar{\mu}_{2,i},i\in\mathcal{E}(\bar{y}),\bar{\mu}_{3},\\\
\bar{\sigma}_{1,i},i\in
a_{01}\left(\bar{x},\bar{y}\right),\bar{\sigma}_{2,i},i\in
a_{10}\left(\bar{x},\bar{y}\right),\bar{\varrho}_{1,i},\bar{\varrho}_{2,i},i\in
a_{00}\left(\bar{x},\bar{y}\right),\end{array}$

such that the following conditions hold:

$\begin{array}[]{rcl}\begin{pmatrix}\nabla f(\bar{x})\\\
c\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\bar{\lambda}_{p}\begin{pmatrix}\nabla h_{p}(\bar{x})\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(\bar{x})}\bar{\mu}_{1,q}\begin{pmatrix}\nabla g_{q}(\bar{x})\\\
0\end{pmatrix}-\sum\limits_{i\in\mathcal{E}(\bar{y})}\bar{\mu}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&+\bar{\mu}_{3}\begin{pmatrix}0\\\
e\end{pmatrix}\displaystyle+\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\bar{\sigma}_{1,i}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}+\sum\limits_{i\in
a_{10}\left(\bar{x},\bar{y}\right)}\bar{\sigma}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle+\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)}\left(\bar{\varrho}_{1,i}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}+\bar{\varrho}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\right),\end{array}$ (3) $\bar{\mu}_{1,q}\geq 0,q\in
Q_{0}\left(\bar{x}\right),\bar{\mu}_{2,i}\geq
0,i\in\mathcal{E}(\bar{y}),\bar{\mu}_{3}\geq
0,\bar{\mu}_{3}\left(\sum\limits_{i=1}^{n}\bar{y}_{i}-(n-s)\right)=0,$ (4)
$\bar{\varrho}_{1,i}=0\mbox{ or }\bar{\varrho}_{2,i}\leq 0,i\in
a_{00}\left(\bar{x},\bar{y}\right).$ (5)

We define the appropriate Lagrange function:

$\begin{array}[]{rcl}L^{\mathcal{R}}(x,y)&=&\displaystyle
f(x)+c^{T}y-\sum\limits_{p\in P}\bar{\lambda}_{p}h_{p}(x)-\sum\limits_{q\in
Q_{0}(\bar{x})}\bar{\mu}_{1,q}g_{q}(x)\\\
&&\displaystyle+\sum\limits_{i\in\mathcal{E}(\bar{y})}\bar{\mu}_{2,i}\left(y_{i}-(1+\varepsilon)\right)-\bar{\mu}_{3}\left(\sum\limits_{i=1}^{n}y_{i}-(n-s)\right)\\\
&&\displaystyle-\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\bar{\sigma}_{1,i}x_{i}-\sum\limits_{i\in
a_{10}\left(\bar{x},\bar{y}\right)}\bar{\sigma}_{2,i}y_{i}-\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)}\left(\bar{\varrho}_{1,i}x_{i}+\bar{\varrho}_{2,i}y_{i}\right).\end{array}$

Moreover, we set for the corresponding tangent space:

$\mathcal{T}^{\mathcal{R}}_{(\bar{x},\bar{y})}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(\bar{x}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(\bar{x}),0\end{pmatrix}\xi=0,q\in Q_{0}(\bar{x}),\\\
\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(\bar{y}),\begin{pmatrix}0,e\end{pmatrix}\xi=0\mbox{
if }\displaystyle\sum_{i=1}^{n}\bar{y}_{i}=n-s,\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in a_{00}(\bar{x},\bar{y})\cup
a_{01}(\bar{x},\bar{y}),\\\ \begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\cup
a_{10}(\bar{x},\bar{y})\end{array}\right.\right\\}.$

######  Definition 3 (Nondegenerate T-stationary point, laemmel:reform ).

A T-stationary point $(\bar{x},\bar{y})$ of $\mathcal{R}$ with multipliers
$(\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho})$ is called nondegenerate
if

NDT1: MPOC-LICQ holds at $(\bar{x},\bar{y})$,

NDT2: $\bar{\mu}_{1,q}>0$, $q\in Q_{0}\left(\bar{x}\right)$,
$\bar{\mu}_{2,i}>0$, $i\in\mathcal{E}\left(\bar{y}\right)$, and
$\bar{\mu}_{3}>0$ if $\sum\limits_{i=1}^{n}\bar{y}_{i}=n-s$,

NDT3: $\bar{\varrho}_{1,i}\neq 0$ and $\bar{\varrho}_{2,i}<0$, $i\in
a_{00}\left(\bar{x},\bar{y}\right)$,

NDT4: the matrix
$D^{2}L^{\mathcal{R}}(\bar{x},\bar{y})\restriction_{\mathcal{T}^{\mathcal{R}}_{(\bar{x},\bar{y})}}$
is nonsingular.

For a nondegenerate T-stationary point we eventually use an additional
condition:

NDT6: $\bar{\sigma}_{1,i}\neq 0$, $i\in a_{01}(\bar{x},\bar{y})$.

######  Definition 4 (T-index, laemmel:reform ).

Let $(\bar{x},\bar{y})$ be a nondegenerate T-stationary point of $\mathcal{R}$
with unique multipliers
$\left(\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho}\right)$. The number
of negative eigenvalues of the matrix
$D^{2}L^{\mathcal{R}}(\bar{x},\bar{y})\restriction_{\mathcal{T}^{\mathcal{R}}_{(\bar{x},\bar{y})}}$
is called its quadratic index ($QI$). The cardinality of
$a_{00}\left(\bar{x},\bar{y}\right)$ is called the biactive index ($BI$) of
$(\bar{x},\bar{y})$. We define the T-index ($TI$) as the sum of both, i. e.
$TI=QI+BI$.

The nondegeneracy conditions NDT1-NDT4 are tailored for $\mathcal{R}$. Note
that NDT2 corresponds to the strict complementarity and NDT4 to the second-
order regularity as they are typically defined in the context of nonlinear
programming. NDT1 substitutes the usual linear independence constraint
qualification. NDT3 is new and says that the multipliers corresponding to
biactive orthogonality type constraints must not vanish. With a nondegenerate
T-stationary point $(\bar{x},\bar{y})$ a T-index can be associated. The
T-index captures the structure of $\mathcal{R}$ locally around
$(\bar{x},\bar{y})$ and defines the type of a T-stationary point, see
laemmel:reform  for details. In particular, nondegenerate minimizers of
$\mathcal{R}$ are characterized by a vanishing T-index. If the T-index does
not vanish, we get all kinds of saddle points. We note that nondegenerate
M-stationary points of CCOP, see e.g. laemmel:ccop , naturally correspond to
nondegenerate T-stationary points of $\mathcal{R}$ and vice versa. As shown in
laemmel:reform , also their M- and T-indices coincide. Thus, the regularized
continuous reformulation $\mathcal{R}$ can be likewise studied instead of (1).

Next Lemma 1 provides insights into the structure of auxiliary $y$-variables
corresponding to a T-stationary point of $\mathcal{R}$. Its proof will be
useful in what follows as well.

######  Lemma 1 (Auxiliary $y$-variables in $\mathcal{R}$, laemmel:reform ).

Let $(\bar{x},\bar{y})$ be a T-stationary point of $\mathcal{R}$, then it
holds:

a) the summation inequality constraint is active, i. e.
$\sum\limits_{i=1}^{n}\bar{y}_{i}=n-s$,

b) the index set $a_{01}(\bar{x},\bar{y})$ consists of exactly $n-s$ elements,

c) $n-s-1$ components of $\bar{y}$ are equal to $1+\varepsilon$, one component
is equal to $1-(n-s-1)\varepsilon$, and $s$ remaining components vanish.

Proof: a) Let $(\bar{x},\bar{y})$ be a T-stationary point of $\mathcal{R}$ and
$\displaystyle\sum_{i=1}^{n}\bar{y}_{i}>n-s$. Then, there exist multipliers
$(\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho})$ such that (3)–(5) are
fulfilled. Since $\bar{\mu}_{3}=0$, we have that the $(n+i)$-th row of (3)
reads as

$c_{i}=\left\\{\begin{array}[]{ll}-\bar{\mu}_{2,i},&\mbox{for
}i\in\mathcal{E}(\bar{y}),\\\ \bar{\sigma}_{2,i},&\mbox{for }i\in
a_{10}\left(\bar{x},\bar{y}\right),\\\ \bar{\varrho}_{2,i},&\mbox{for }i\in
a_{00}\left(\bar{x},\bar{y}\right),\\\ 0,&\mbox{else}.\end{array}\right.$

Due to $c>0$, the sets $a_{01}\left(\bar{x},\bar{y}\right)$ and
$\mathcal{E}(\bar{y})$ have to be equal and, moreover, due to
$\bar{\mu}_{2,i}\geq 0$, they have to be empty. But then, clearly,
$\sum_{i=1}^{n}\bar{y}_{i}=0<n-s$, a contradiction.

b) Since $(\bar{x},\bar{y})$ is a T-stationary point, there exist
$(\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho})$ such that (3)–(5). By
the proof of statement a) we can conclude that $\bar{\mu}_{3}>0$. Hence, the
$(n+i)$-th row reads as

$c_{i}=\left\\{\begin{array}[]{ll}-\bar{\mu}_{2,i}+\bar{\mu}_{3},&\mbox{for
}i\in\mathcal{E}(\bar{y}),\\\ \bar{\sigma}_{2,i}+\bar{\mu}_{3},&\mbox{for
}i\in a_{10}\left(\bar{x},\bar{y}\right),\\\
\bar{\varrho}_{2,i}+\bar{\mu}_{3},&\mbox{for }i\in
a_{00}\left(\bar{x},\bar{y}\right),\\\
\bar{\mu}_{3},&\mbox{else}.\end{array}\right.$ (6)

Let us assume that the index set $a_{01}(\bar{x},\bar{y})$ consists of fewer
than $n-s$ elements. Then, we have by using the assumption on $\varepsilon$:

$\sum_{i=1}^{n}\bar{y}_{i}\leq(n-s-1)\cdot(1+\varepsilon)\leq
n-s-1+\frac{n-s-1}{n-s}<n-s,$

a contradiction to feasibility. Let us assume that the index set
$a_{01}(\bar{x},\bar{y})$ consists of more than $n-s$ elements instead. Since
the components of $c$ are assumed to be pairwise different, we see from (6)
that there exists at most one element in
$a_{01}\left(\bar{x},\bar{y}\right)\backslash\mathcal{E}(\bar{y})$ and,
consequently, there are at least $n-s$ elements in $\mathcal{E}(\bar{y})$.
Therefore, we have:

$\sum_{i=1}^{n}\bar{y}_{i}\geq(n-s)\cdot(1+\varepsilon)>n-s,$

which contradicts a).

c) Due to b), $a_{01}(\bar{x},\bar{y})$ consists of exactly $n-s$ elements. We
conclude as in b) that there is at most one element in
$a_{01}\left(\bar{x},\bar{y}\right)\backslash\mathcal{E}(\bar{y})$. In view of
statement a), $\mathcal{E}(\bar{y})$ cannot consist of $n-s$ elements and,
thus, must consist of $n-s-1$ elements. Hence, the statement follows
immediately.

##  3 Scholtes-type regularization

Let us now regularize the orthogonality type constraints in $\mathcal{R}$ by
using the Scholte’s idea, cf. scholtes:2001 :

$\begin{array}[]{rl}\displaystyle\mathcal{S}(t):\quad\min_{x,y}\,\,f(x)+c^{T}y\quad\mbox{s.\,t.}&h(x)=0,\quad
g(x)\geq 0,\quad\displaystyle\sum_{i=1}^{n}y_{i}\geq n-s,\\\ &-t\leq
x_{i}y_{i}\leq t,\quad 0\leq y_{i}\leq 1+\varepsilon,\quad
i=1,\ldots,n,\end{array}$

where $t>0$. Note that $\mathcal{S}$ from above falls into the scope of
nonlinear programming, i.e. just smooth equality and inequality constraints
are present. The notation for the sets $Q_{0}(x)$ and $\mathcal{E}(y)$, which
were used for $\mathcal{R}$, will be used here again. Furthermore, we define
for a feasible point $\left(x,y\right)$ of $\mathcal{S}$ the index set of
vanishing $y$-components as well as the index sets of active relaxed
orthogonality type constraints:

$\mathcal{N}(y)=\left\\{i\,\left|\,y_{i}=0\right.\right\\},\quad\mathcal{H}^{\geq}\left(x,y\right)=\left\\{i\,\left|\,x_{i}y_{i}=-t\right.\right\\},\mathcal{H}^{\leq}\left(x,y\right)=\left\\{i\,\left|\,x_{i}y_{i}=t\right.\right\\}.$

We also eventually use the following index sets:

$\mathcal{H}\left(x,y\right)=\mathcal{H}^{\geq}\left(x,y\right)\cup\mathcal{H}^{\leq}\left(x,y\right),\quad\mathcal{O}\left(x,y\right)=\left(\mathcal{E}\left(y\right)\cup\mathcal{N}\left(y\right)\cup\mathcal{H}\left(x,y\right)\right)^{c}.$

For the sake of completeness we state the linear independence constraint
qualification for the nonlinear programming problem $\mathcal{S}$.

######  Definition 5 (LICQ).

We say that a feasible point $(x,y)$ of $\mathcal{S}$ satisfies the linear
independence constraint qualification (LICQ) if the following vectors are
linearly independent:

$\begin{array}[]{c}\begin{pmatrix}\nabla h_{p}(x)\\\ 0\end{pmatrix},p\in
P,\quad\begin{pmatrix}\nabla g_{q}(x)\\\ 0\end{pmatrix},q\in
Q_{0}(x),\quad\begin{pmatrix}0\\\ e_{i}\end{pmatrix},i\in\mathcal{E}(y),\\\
\begin{pmatrix}0\\\ e\end{pmatrix}\mbox{ if
}\sum\limits_{i=1}^{n}y_{i}=n-s,\quad\begin{pmatrix}y_{i}e_{i}\\\
x_{i}e_{i}\end{pmatrix},i\in\mathcal{H}\left(x,y\right),\quad\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in\mathcal{N}(y).\end{array}$

Let us relate MPOC-LICQ for $\mathcal{R}$ with LICQ for $\mathcal{S}$.

######  Theorem 1 (MPOC-LICQ vs. LICQ).

Let a feasible point $(\bar{x},\bar{y})$ of $\mathcal{R}$ fulfill MPOC-LICQ.
Then, LICQ holds at all feasible points $(x,y)$ of $\mathcal{S}$ for all
sufficiently small $t$, whenever they are sufficiently close to
$(\bar{x},\bar{y})$.

Proof: Let us contrarily assume that there exists a sequence of feasible
points $\left(x^{t},y^{t}\right)$ of $\mathcal{S}$ violating LICQ, which
converges to $(\bar{x},\bar{y})$ for $t\rightarrow 0$. Additionally, suppose
that along some subsequence, which we index by $t$ again, it holds
$\sum\limits_{i=1}^{n}y^{t}_{i}=n-s$. Then, we have
$\sum\limits_{i=1}^{n}\bar{y}_{i}=n-s$. Due to MPOC-LICQ at
$\left(\bar{x},\bar{y}\right)$ as well as continuity of $\nabla h$ and $\nabla
g$, we have that for $t$ sufficiently small all multipliers
$\bar{\lambda}^{t},\bar{\mu}^{t},\bar{\sigma}^{t},\bar{\varrho}^{t}$ in the
following equation vanish:

$\begin{array}[]{rcl}\begin{pmatrix}0\\\
0\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\bar{\lambda}^{t}_{p}\begin{pmatrix}\nabla h_{p}\left(x^{t}\right)\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(\bar{x})}\bar{\mu}^{t}_{1,q}\begin{pmatrix}\nabla
g_{q}\left(x^{t}\right)\\\
0\end{pmatrix}-\sum\limits_{i\in\mathcal{E}(\bar{y})}\bar{\mu}^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle+\bar{\mu}^{t}_{3}\begin{pmatrix}0\\\
e\end{pmatrix}+\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\bar{\sigma}^{t}_{1,i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{i\in
a_{10}\left(\bar{x},\bar{y}\right)}\bar{\sigma}^{t}_{2,i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\ &&\displaystyle+\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)}\left(\bar{\varrho}^{t}_{1,i}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}+\bar{\varrho}^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\right).\end{array}$ (7)

Moreover, due to the violation of LICQ at $\left(x^{t},y^{t}\right)$, there
exist multipliers $\lambda^{t},\mu^{t},\eta^{t},\nu^{t}$, not all vanishing,
with

$\begin{array}[]{rcl}\begin{pmatrix}0\\\
0\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\lambda^{t}_{p}\begin{pmatrix}\nabla h_{p}(x^{t})\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(x^{t})}\mu^{t}_{1,q}\begin{pmatrix}\nabla g_{q}(x^{t})\\\
0\end{pmatrix}+\sum\limits_{i\in\mathcal{E}(y^{t})}\mu^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle+\mu^{t}_{3}\begin{pmatrix}0\\\
e\end{pmatrix}+\sum\limits_{i\in\mathcal{H}(x^{t},y^{t})}\eta^{t}_{i}\begin{pmatrix}y^{t}_{i}e_{i}\\\
x^{t}_{i}e_{i}\end{pmatrix}+\sum\limits_{i\in\mathcal{N}(y^{t})}\nu^{t}_{i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}.\end{array}$

For $t$ sufficiently small we have $Q_{0}\left(x^{t}\right)\subset
Q_{0}(\bar{x})$ and
$\mathcal{E}\left(y^{t}\right)\subset\mathcal{E}\left(\bar{y}\right)$. In
addition, it holds $\mathcal{H}\left(x^{t},y^{t}\right)\subset
a_{01}(\bar{x},\bar{y})\cup a_{10}(\bar{x},\bar{y})\cup
a_{00}(\bar{x},\bar{y})$ and $\mathcal{N}\left(y^{t}\right)\subset
a_{10}(\bar{x},\bar{y})\cup a_{00}(\bar{x},\bar{y})$. By setting some
$\mu$-multipliers to be zero if needed, we equivalently obtain:

$\begin{array}[]{rcl}\begin{pmatrix}0\\\
0\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\lambda^{t}_{p}\begin{pmatrix}\nabla h_{p}(x^{t})\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(\bar{x})}\mu^{t}_{1,q}\begin{pmatrix}\nabla g_{q}(x^{t})\\\
0\end{pmatrix}+\sum\limits_{i\in\mathcal{E}(\bar{y})}\mu^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle+\mu^{t}_{3}\begin{pmatrix}0\\\
e\end{pmatrix}+\sum\limits_{i\in\mathcal{H}(x^{t},y^{t})\cap
a_{01}\left(\bar{x},\bar{y}\right)}\eta^{t}_{i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{{i}\in\mathcal{H}(x^{t},y^{t})\cap
a_{10}\left(\bar{x},\bar{y}\right)}\eta^{t}_{i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\
&&+\displaystyle\sum\limits_{i\in\mathcal{H}(x^{t},y^{t})\cap
a_{00}\left(\bar{x},\bar{y}\right)}\eta^{t}_{i}\begin{pmatrix}y^{t}_{i}e_{i}\\\
x^{t}_{i}e_{i}\end{pmatrix}\\\
&&+\displaystyle\sum\limits_{i\in\mathcal{N}(y^{t})\cap
a_{10}\left(\bar{x},\bar{y}\right)}\nu^{t}_{i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}+\sum\limits_{i\in\mathcal{N}(y^{t})\cap
a_{00}\left(\bar{x},\bar{y}\right)}\nu^{t}_{i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}.\end{array}$

This, however, implies that not all multipliers in the following equation
vanish:

$\begin{array}[]{rcl}\begin{pmatrix}0\\\
0\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\hat{\lambda}^{t}_{p}\begin{pmatrix}\nabla h_{p}(x^{t})\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(\bar{x})}\hat{\mu}^{t}_{1,q}\begin{pmatrix}\nabla g_{q}(x^{t})\\\
0\end{pmatrix}+\sum\limits_{i\in\mathcal{E}(\bar{y})}\hat{\mu}^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle+\hat{\mu}^{t}_{3}\begin{pmatrix}0\\\
e\end{pmatrix}+\sum\limits_{i\in\mathcal{H}(x^{t},y^{t})\cap
a_{01}\left(\bar{x},\bar{y}\right)}\hat{\eta}^{t}_{i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{{i}\in\mathcal{H}(x^{t},y^{t})\cap
a_{10}\left(\bar{x},\bar{y}\right)}\hat{\eta}^{t}_{i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\
&&+\displaystyle\sum\limits_{i\in\mathcal{H}(x^{t},y^{t})\cap
a_{00}\left(\bar{x},\bar{y}\right)}\hat{\eta}^{t}_{1,i}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}+\sum\limits_{i\in\mathcal{H}(x^{t},y^{t})\cap
a_{00}\left(\bar{x},\bar{y}\right)}\hat{\eta}^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&+\displaystyle\sum\limits_{i\in\mathcal{N}(y^{t})\cap
a_{10}\left(\bar{x},\bar{y}\right)}\hat{\nu}^{t}_{i}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{i\in\mathcal{N}(y^{t})\cap
a_{00}\left(\bar{x},\bar{y}\right)}\hat{\nu}^{t}_{i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}.\end{array}$

A contradiction to (7) follows by taking into account that
$\mathcal{H}(x^{t},y^{t})\cap\mathcal{N}(y^{t})=\emptyset$. If instead we
suppose that there is no subsequence with
$\sum\limits_{i=1}^{n}y^{t}_{i}=n-s$, then we can consider a subsequence with
$\sum\limits_{i=1}^{n}y^{t}_{i}>n-s$. By following a similar argumentation, we
produce a contradiction to (7) again.

Next, we give the definitions of a (nondegenerate) Karush-Kuhn-Tucker point of
$\mathcal{S}$ and of its quadratic index as it is meanwhile standard in
nonlinear programming, see e.g. Jongen:2000 .

######  Definition 6 (Karush-Kuhn-Tucker point).

A feasible point $(x,y)$ of $\mathcal{S}$ is called Kurush-Kuhn-Tucker if
there exist multipliers

$\lambda_{p},p\in P,\mu_{1,q},q\in
Q_{0}(x),\mu_{2,i},i\in\mathcal{E}(y),\mu_{3},\eta^{\geq}_{i},\eta^{\leq}_{i},\nu_{i},i\in\left\\{1,\ldots,n\right\\},$

such that the following conditions hold:

$\begin{array}[]{rcl}\begin{pmatrix}\nabla f(x)\\\
c\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\lambda_{p}\begin{pmatrix}\nabla h_{p}(x)\\\
0\end{pmatrix}+\sum\limits_{q\in Q_{0}(x)}\mu_{1,q}\begin{pmatrix}\nabla
g_{q}(x)\\\
0\end{pmatrix}-\sum\limits_{i\in\mathcal{E}(y)}\mu_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}+\mu_{3}\begin{pmatrix}0\\\ e\end{pmatrix}\\\
&&\displaystyle+\sum\limits_{i\in\mathcal{H}^{\geq}(x,y)}\eta^{\geq}_{i}\begin{pmatrix}y_{i}e_{i}\\\
x_{i}e_{i}\end{pmatrix}-\sum\limits_{i\in\mathcal{H}^{\leq}(x,y)}\eta^{\leq}_{i}\begin{pmatrix}y_{i}e_{i}\\\
x_{i}e_{i}\end{pmatrix}+\sum\limits_{i\in\mathcal{N}(y)}\nu_{i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix},\end{array}$ (8) $\mu_{1,q}\geq 0,q\in
Q_{0}\left(x\right),\mu_{2,i}\geq 0,i\in\mathcal{E}(y),\mu_{3}\geq
0,\mu_{3}\left(\sum\limits_{i=1}^{n}y_{i}-(n-s)\right)=0,$ (9)
$\eta^{\geq}_{i}\geq 0,i\in\mathcal{H}^{\geq}(x,y),\eta^{\leq}_{i}\geq
0,i\in\mathcal{H}^{\leq}(x,y),\nu_{i}\geq 0,i\in\mathcal{N}(y).$ (10)

We again define the Lagrange function as

$\begin{array}[]{rcl}L^{\mathcal{S}}(x,y)&=&\displaystyle
f(x)+c^{T}y-\sum\limits_{p\in P}\lambda_{p}h_{p}(x)-\sum\limits_{q\in
Q_{0}(x)}\mu_{1,q}g_{q}(x)\\\
&&\displaystyle+\sum\limits_{i\in\mathcal{E}(y)}\mu_{2,i}\left(y_{i}-(1+\varepsilon)\right)-\mu_{3}\left(\sum\limits_{i=1}^{n}y_{i}-(n-s)\right)\\\
&&\displaystyle-\sum\limits_{i\in\mathcal{H}^{\geq}(x,y)}\eta^{\geq}_{i}\left(x_{i}y_{i}+t\right)+\sum\limits_{i\in\mathcal{H}^{\leq}(x,y)}\eta^{\leq}_{i}\left(x_{i}y_{i}-t\right)-\sum\limits_{i\in\mathcal{N}(y)}\nu_{i}y_{i}.\end{array}$

The tangent space is given by

$\mathcal{T}^{\mathcal{S}}_{(x,y)}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(x),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(x),0\end{pmatrix}\xi=0,q\in Q_{0}(x),\\\
\displaystyle\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(y),\begin{pmatrix}0,e\end{pmatrix}\xi=0\mbox{
if }\sum_{i=1}^{n}y_{i}=n-s,\\\
\begin{pmatrix}y_{i}e_{i},x_{i}e_{i}\end{pmatrix}\xi=0,i\in\mathcal{H}\left(x,y\right),\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{N}(y)\end{array}\right.\right\\}.$

######  Definition 7 (Nondegenerate Karush-Kuhn-Tucker point).

A Karush-Kuhn-Tucker point $(x,y)$ of $\mathcal{S}$ with multipliers
$(\lambda,\mu,\eta,\nu)$ is called nondegenerate if

ND1: LICQ holds at $(x,y)$,

ND2: $\mu_{1,q}>0$, $q\in Q_{0}\left(x\right)$, $\mu_{2,i}>0$,
$i\in\mathcal{E}\left(y\right)$, $\eta^{\geq}_{i}>0$,
$i\in\mathcal{H}^{\geq}(x,y)$, $\eta^{\leq}_{i}>0$,
$i\in\mathcal{H}^{\leq}(x,y)$, $\nu_{i}>0$, $i\in\mathcal{N}(y)$, and
$\mu_{3}>0$ if $\sum\limits_{i=1}^{n}y_{i}=n-s$,

ND3: the matrix
$D^{2}L^{\mathcal{S}}(x,y)\restriction_{\mathcal{T}^{\mathcal{S}}_{(x,y)}}$ is
nonsingular.

######  Definition 8 (Quadratic index).

Let $(x,y)$ be a Karush-Kuhn-Tucker point of $\mathcal{S}$ with unique
multipliers $(\lambda,\mu,\eta,\nu)$. The number of negative eigenvalues of
the matrix
$D^{2}L^{\mathcal{S}}(x,y)\restriction_{\mathcal{T}^{\mathcal{S}}_{(x,y)}}$ is
called its quadratic index ($QI$).

Note that ND1-ND3 are usual assumptions in nonlinear programming. ND1 refers
to the linear independence constraint qualification, ND2 means the strict
complementarity, and ND3 describes the second-order regularity. For the index
of a nondegenerate Karush-Kuhn-Tucker point just the quadratic part is
essential.

Next Lemma 2 examines the structure of $y$-components of a Karush-Kuhn-Tucker
point of $\mathcal{S}$.

######  Lemma 2 (Auxiliary $y$-variables in $\mathcal{S}$).

Let $(x,y)$ be a Karush-Kuhn-Tucker point of $\mathcal{S}$. Then, it holds:

a) the summation inequality constraint is active, i. e.
$\sum\limits_{i=1}^{n}y_{i}=n-s$,

b) the index set $\mathcal{E}(y)\cup\mathcal{H}(x,y)$ consists of at least
$n-s-1$ elements, and the index set $\mathcal{N}(y)$ consists of at most $s$
elements. Additionally, there is at most one index, that does not belong to
any of these sets, i. e. $\left|\mathcal{O}\left(x,y\right)\right|\leq 1$.

Proof: a) Let $(x,y)$ be a Karush-Kuhn-Tucker point of $\mathcal{S}$ and
$\sum\limits_{i=1}^{n}y_{i}>n-s$. Then, there exist multipliers
$(\lambda,\mu,\eta,\nu)$, such that (8)–(10) are fulfilled. Since $\mu_{3}=0$,
we have that the $(n+i)$-th row of (8) reads as

$c_{i}=\left\\{\begin{array}[]{ll}-\mu_{2,i},&\mbox{for
}i\in\mathcal{E}(y)\backslash\mathcal{H}(x,y),\\\
-\mu_{2,i}+\eta^{\geq}_{i}x_{i},&\mbox{for
}i\in\mathcal{E}(y)\cap\mathcal{H}^{\geq}(x,y),\\\
-\mu_{2,i}-\eta^{\leq}_{i}x_{i},&\mbox{for
}i\in\mathcal{E}(y)\cap\mathcal{H}^{\leq}(x,y),\\\
\eta^{\geq}_{i}x_{i},&\mbox{for
}i\in\mathcal{H}^{\geq}(x,y)\backslash\mathcal{E}(y),\\\
-\eta^{\leq}_{i}x_{i},&\mbox{for
}i\in\mathcal{H}^{\leq}(x,y)\backslash\mathcal{E}(y),\\\ \nu_{i},&\mbox{for
}i\in\mathcal{N}(y),\\\ 0,&\mbox{else}.\end{array}\right.$

Due to (9), (10), and $c>0$, it must hold that $i\in\mathcal{N}(y)$ for all
$i\in\left\\{1,\ldots,n\right\\}$. This, however, contradicts
$\sum\limits_{i=1}^{n}y_{i}>n-s$.

b) As in the proof of statement a), we conclude that $\mu_{3}>0$ for a Karush-
Kuhn-Tucker point $(x,y)$ of $\mathcal{S}$. Hence, the $(n+i)$-th row now
reads as

$c_{i}=\left\\{\begin{array}[]{ll}-\mu_{2,i}+\mu_{3},&\mbox{for
}i\in\mathcal{E}(y)\backslash\mathcal{H}(x,y),\\\
-\mu_{2,i}+\mu_{3}+\eta^{\geq}_{i}x_{i},&\mbox{for
}i\in\mathcal{E}(y)\cap\mathcal{H}^{\geq}(x,y),\\\
-\mu_{2,i}+\mu_{3}-\eta^{\leq}_{i}x_{i},&\mbox{for
}i\in\mathcal{E}(y)\cap\mathcal{H}^{\leq}(x,y),\\\
\mu_{3}+\eta^{\geq}_{i}x_{i},&\mbox{for
}i\in\mathcal{H}^{\geq}(x,y)\backslash\mathcal{E}(y),\\\
\mu_{3}-\eta^{\leq}_{i}x_{i},&\mbox{for
}i\in\mathcal{H}^{\leq}(x,y)\backslash\mathcal{E}(y),\\\
\mu_{3}+\nu_{i},&\mbox{for }i\in\mathcal{N}(y),\\\
\mu_{3},&\mbox{else}.\end{array}\right.$ (11)

It follows from (11) and the components of $c$ being pairwise different that
there can be at most one element ${\bar{i}}\in\mathcal{O}\left(x,y\right)$. If
$\mathcal{E}(y)\cup\mathcal{H}(x,y)$ consists of fewer than $n-s-1$ elements,
we get:

$\sum\limits_{i=1}^{n}y_{i}\leq(n-s-2)\cdot(1+\varepsilon)+y_{\bar{i}}<(n-s-1)\cdot(1+\varepsilon)<n-s,$

a contradiction. Finally, we assume that $\mathcal{N}(y)$ consists of more
than $s$ elements. In this case, there are at most $n-s-1$ nonvanishing
components of $y$. Consequently,

$\sum\limits_{i=1}^{n}y_{i}\leq(n-s-1)\cdot(1+\varepsilon)<n-s$

provides a contradiction.

We apply the general result on the Scholtes-type regularization of MPOC in our
context for the regularized continuous reformulation $\mathcal{R}$, see
laemmel:mpoc .

######  Theorem 2 (Convergence from $\mathcal{S}$ to $\mathcal{R}$, cf.
laemmel:mpoc ).

Suppose that a sequence of Karush-Kuhn-Tucker points $(x^{t},y^{t})$ of
$\mathcal{S}$ converges to $\left(\bar{x},\bar{y}\right)$ for $t\rightarrow
0$. If MPOC-LICQ holds at $\left(\bar{x},\bar{y}\right)$, then it is a
T-stationary point of $\mathcal{R}$.

From the proof of Theorem 2 in laemmel:mpoc  also the convergence of the
corresponding multipliers can be deduced.

######  Remark 1 (Convergence of multipliers).

Let $\left(\lambda^{t},\mu^{t},\eta^{t},\nu^{t}\right)$ be the multipliers of
the Karush-Kuhn-Tucker points $\left(x^{t},y^{t}\right)$ of $\mathcal{S}$ and
$\left(\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho}\right)$ of the
T-stationary point $\left(\bar{x},\bar{y}\right)$ of $\mathcal{R}$ as in
Theorem 2. Due to MPOC-LICQ at $\left(\bar{x},\bar{y}\right)$, we have:

a) $\lim\limits_{t\to 0}\lambda^{t}=\bar{\lambda}$, $\lim\limits_{t\to
0}\mu^{t}=\bar{\mu}$,

b) $\lim\limits_{t\to
0}\left(\eta_{i}^{\geq,t}-\eta_{i}^{\leq,t}\right)y_{i}^{t}=\bar{\sigma}_{1,i}$,
$i\in a_{01}\left(\bar{x},\bar{y}\right)$,

c) $\lim\limits_{t\to
0}\nu_{i}^{t}+\left(\eta_{i}^{\geq,t}-\eta_{i}^{\leq,t}\right)x_{i}^{t}=\bar{\sigma}_{2,i}$,
$i\in a_{10}\left(\bar{x},\bar{y}\right)$,

d) $\lim\limits_{t\to
0}\left(\eta_{i}^{\geq,t}-\eta_{i}^{\leq,t}\right)y_{i}^{t}=\bar{\varrho}_{1,i}$,
$\lim\limits_{t\to
0}\nu_{i}^{t}+\left(\eta_{i}^{\geq,t}-\eta_{i}^{\leq,t}\right)x_{i}^{t}=\bar{\varrho}_{2,i}$,
$i\in a_{00}\left(\bar{x},\bar{y}\right)$.

The convergence of nondegenerate Karush-Kuhn-Tucker points of $\mathcal{S}$
does not prevent the limiting T-stationary point of $\mathcal{S}$ from being
degenerate. Let us present in Example 1 the failure of NDT2. Examples with the
failure of NDT1, NDT3, or NDT4 are not difficult to construct analogously.

######  Example 1 (Failure of NDT2).

We consider the following Scholtes-type regularization $\mathcal{S}$ with
$n=2$ and $s=1$:

$\begin{array}[]{rl}\mathcal{S}:\quad\min\limits_{x,y}&(x_{1}-1)^{2}+(x_{2}-1)^{2}+c_{1}y_{1}+(c_{1}+\frac{5}{36})y_{2}\\\
\mbox{s.t.}&1+x_{1}-x_{2}\geq 0,\\\ &y_{1}+y_{2}\geq 1,\quad-t\leq
x_{i}y_{i}\leq t,\quad 0\leq y_{i}\leq 1+\varepsilon,\quad i=1,2,\end{array}$

as well as the point $(x^{t},y^{t})=(t,1,1,0)$.We claim that this point is a
nondegenerate Karush-Kuhn-Tucker point for
$t<\frac{1}{2}-\sqrt{\frac{13}{72}}$. Indeed, it holds:

$\begin{pmatrix}2t-2\\\ 0\\\ c_{1}\\\
c_{1}+\frac{5}{36}\end{pmatrix}=\mu_{3}^{t}\begin{pmatrix}0\\\ 0\\\ 1\\\
1\end{pmatrix}-\eta_{1}^{\leq,t}\begin{pmatrix}1\\\ 0\\\ t\\\
0\end{pmatrix}+\nu_{2}^{t}\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix}$

with the positive multipliers $\mu_{3}^{t}=c_{1}+2t-2t^{2}$,
$\eta_{1}^{\leq,t}=2-2t$, $\nu_{2}^{t}=\frac{5}{36}-2t+2t^{2}$. The tangent
space is
$\mathcal{T}^{\mathcal{S}}_{(x^{t},y^{t})}=\left\\{\xi\in\mathbb{R}^{4}\,\left|\,\xi_{1}=\xi_{3}=\xi_{4}=0\right.\right\\}$.
The Hessian of the corresponding Lagrange function is

$D^{2}L^{\mathcal{S}}(x^{t},y^{t})=\begin{pmatrix}2&0&2-2t&0\\\ 0&2&0&0\\\
2-2t&0&0&0\\\ 0&0&0&0\end{pmatrix}.$

Therefore, it is straightforward to see that
$D^{2}L^{\mathcal{S}}(x^{t},y^{t})\restriction_{\mathcal{T}^{\mathcal{S}}_{(x^{t},y^{t})}}$
is nonsingular. We conclude that ND1-ND3 are fulfilled at the $(x^{t},y^{t})$.
Moreover, $(x^{t},y^{t})$ converges to $(\bar{x},\bar{y})=(0,1,1,0)$ if $t\to
0$. This point is T-stationary for the corresponding regularized continuous
reformulation $\mathcal{R}$ according to Theorem 2, since MPOC-LICQ is
fulfilled. Indeed, we obtain the T-stationarity condition

$\begin{pmatrix}-2\\\ 0\\\ c_{1}\\\
c_{1}+\frac{5}{36}\end{pmatrix}=\bar{\mu}_{1}\begin{pmatrix}1\\\ -1\\\ 0\\\
0\end{pmatrix}+\bar{\mu}_{3}\begin{pmatrix}0\\\ 0\\\ 1\\\
1\end{pmatrix}+\bar{\sigma}_{1,1}\begin{pmatrix}1\\\ 0\\\ 0\\\
0\end{pmatrix}+\bar{\sigma}_{2,2}\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix}$

with the unique multipliers
$\bar{\mu}_{1}=0,\bar{\mu}_{3}=c_{1},\bar{\sigma}_{1,1}=-2,\bar{\sigma}_{2,2}=\frac{5}{36}$.
However, NDT2 is violated at $(\bar{x},\bar{y})$.  $\blacksquare$

Due to Example 1, we cannot expect that a T-stationary point of $\mathcal{R}$,
which is the limit of a sequence of nondegenerate Karush-Kuhn-Tucker points of
$\mathcal{S}$, is also nondegenerate. Instead, we intend to examine its type
if assuming nondegeneracy. Next Lemma 3 provides some valuable insights into
the relations between active index sets while doing so.

######  Lemma 3 (Active index sets).

Suppose a sequence of Karush-Kuhn-Tucker points $\left(x^{t},y^{t}\right)$ of
$\mathcal{S}(t)$ converges to $\left(\bar{x},\bar{y}\right)$ for $t\to 0$.
Moreover, let $(\bar{x},\bar{y})$ be a nondegenerate T-stationary point of
$\mathcal{R}$. Then, for all sufficiently small $t$ it holds:

a) $Q_{0}\left(\bar{x}\right)=Q_{0}\left(x^{t}\right)$,

b) $\mathcal{E}\left(\bar{y}\right)=\mathcal{E}\left(y^{t}\right)$,

c)
$a_{00}\left(\bar{x},\bar{y}\right)\subset\mathcal{H}\left(x^{t},y^{t}\right)$,

d) $\mathcal{N}\left(y^{t}\right)\subset
a_{10}\left(\bar{x},\bar{y}\right)\subset\mathcal{N}\left(y^{t}\right)\cup\mathcal{H}\left(x^{t},y^{t}\right)$.

Proof: a) We start by proving
$Q_{0}\left(\bar{x}\right)=Q_{0}\left(x^{t}\right)$. Due to continuity
arguments, we have $Q_{0}\left(x^{t}\right)\subset Q_{0}\left(\bar{x}\right)$
for all sufficiently small $t$. Let us now assume that there exists
$\bar{i}\in Q_{0}\left(\bar{x}\right)\backslash Q_{0}\left(x^{t}\right)$ along
a subsequence. Hence, for the corresponding multipliers it holds
$\mu_{\bar{i}}^{t}=0$. NDT1 allows us to apply Remark 1, and we thus have
$\bar{\mu}_{\bar{i}}=\lim\limits_{t\rightarrow\infty}\mu_{\bar{i}}^{t}=0$, a
contradiction to NDT2. Consequently,
$Q_{0}\left(\bar{x}\right)=Q_{0}\left(x^{t}\right)$ holds for all sufficiently
small $t$.

b) Next, we prove
$\mathcal{E}\left(\bar{y}\right)=\mathcal{E}\left(y^{t}\right)$. Again,
continuity arguments provide
$\mathcal{E}\left(y^{t}\right)\subset\mathcal{E}\left(\bar{y}\right)$ for all
sufficiently small $t$. Similar to the first part of the proof, we now assume
there exists
$\bar{i}\in\mathcal{E}\left(\bar{y}\right)\backslash\mathcal{E}\left(y^{t}\right)$
along a subsequence. As we have seen in Lemma 1, T-stationarity of
$(\bar{x},\bar{y})$ implies in particular
$c_{\bar{i}}=-\bar{\mu}_{2,\bar{i}}+\bar{\mu}_{3,\bar{i}}$. Moreover, NDT1 and
Remark 1 provide $\lim\limits_{t\to 0}\mu_{3}^{t}=\bar{\mu}_{3}$. Since
$\bar{i}\notin\mathcal{N}\left(y^{t}\right)$, we distinguish the following
cases:

$(i)$
$\bar{i}\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)\backslash\mathcal{E}\left(y^{t}\right)$.
Karush-Kuhn-Tucker conditions for $\left(x^{t},y^{t}\right)$ imply
$c_{\bar{i}}=\mu^{t}_{3,\bar{i}}+\eta^{\geq,t}_{\bar{i}}x_{\bar{i}}$, cf.
(11). It follows
$-\bar{\mu}_{2,\bar{i}}+\bar{\mu}_{3,\bar{i}}=\mu^{t}_{3,\bar{i}}+\eta^{\geq,t}_{\bar{i}}x_{\bar{i}}$.
By taking the limit, we can cancel out $\bar{\mu}_{3,\bar{i}}$ and
$\mu^{t}_{3,\bar{i}}$. This leads to a contradiction because the left-hand
side of the equation is strictly negative due to NDT2 and the right-hand side
is nonnegative since $\eta^{\geq,t}_{\bar{i}}$ is nonnegative and
$x_{\bar{i}}$ is positive.

$(ii)$
$\bar{i}\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)\backslash\mathcal{E}\left(y^{t}\right)$.
By using (11), we get
$c_{\bar{i}}=\mu^{t}_{3,\bar{i}}-\eta^{\leq,t}_{\bar{i}}x_{\bar{i}}$. This
leads to a contradiction just as in the previous case.

$(iii)$ $\bar{i}\in\mathcal{O}\left(x^{t},y^{t}\right)$. Analogously, we
obtain $c_{\bar{i}}=\mu^{t}_{3,\bar{i}}$ from (11). It follows
$-\bar{\mu}_{2,\bar{i}}+\bar{\mu}_{3,\bar{i}}=\mu^{t}_{3,\bar{i}}$. Taking the
limits leads to $\bar{\mu}_{2,\bar{i}}=0$, a contradiction with NDT2.

Altogether,
$\mathcal{E}\left(\bar{y}\right)\backslash\mathcal{E}\left(y^{t}\right)=\emptyset$
for all sufficiently small $t$, and the assertion follows.

c) Clearly,
$a_{00}\left(\bar{x},\bar{y}\right)\cap\mathcal{E}(y^{t})=\emptyset$ for
sufficiently small $t$. Let us assume there exists an $\bar{i}\in
a_{00}(\bar{x},\bar{y})\cap\mathcal{N}\left(y^{t}\right)$. Due to (11), we
then have $c_{\bar{i}}=\mu_{3}^{t}+\nu_{\bar{i}}^{t}$, whereas (6) provides
$c_{\bar{i}}=\bar{\mu}_{3}+\bar{\varrho}_{2,\bar{i}}$. According to Remark 1,
we have $\lim\limits_{t\to 0}\mu_{3}^{t}=\bar{\mu}_{3}$. Consequently, it must
hold $\lim\limits_{t\to 0}\nu_{\bar{i}}^{t}=\bar{\varrho}_{2,\bar{i}}$. This,
however, cannot be true since $\nu_{\bar{i}}^{t}\geq 0$, while
$\bar{\varrho}_{2,\bar{i}}<0$ due to NDT3 from the nondegeneracy of
$(\bar{x},\bar{y})$, a contradiction. Let us assume now that there exists an
$\bar{i}\in a_{00}(\bar{x},\bar{y})\cap\mathcal{O}\left(x^{t},y^{t}\right)$.
Analogously, we get $\bar{\varrho}_{2,\bar{i}}=0$, again a contradiction to
NDT3. Overall, we get the assertion.

d) Clearly,
$a_{01}\left(\bar{x},\bar{y}\right)\cap\mathcal{N}(y^{t})=\emptyset$ for
sufficiently small $t$. From c) we also know that
$a_{00}\left(\bar{x},\bar{y}\right)\cap\mathcal{N}(y^{t})=\emptyset$.
Altogether, the first inclusion of the assertion follows immediately. Further,
it also holds
$a_{10}\left(\bar{x},\bar{y}\right)\cap\mathcal{E}(y^{t})=\emptyset$ for
sufficiently small $t$. Let us assume there exists an $\bar{i}\in
a_{10}(\bar{x},\bar{y})\cap\mathcal{O}\left(x^{t},y^{t}\right)$. Due to (11),
we have $c_{\bar{i}}=\mu_{3}^{t}$. In view of Lemma 1c), there exists an index
$\tilde{i}\in
a_{01}(\bar{x},\bar{y})\backslash\mathcal{E}\left(\bar{y}\right)$. Thus,
T-stationarity of $(\bar{x},\bar{y})$ implies via (6) that
$c_{\tilde{i}}=\bar{\mu}_{3}$. By taking the limit and Remark 1, we obtain
$c_{\bar{i}}=c_{\tilde{i}}$, but $\bar{i}\not=\tilde{i}$, a contradiction to
the choice of $c$.

Next Theorem 3 highlights the convergence properties of the Scholtes-type
regularization method.

######  Theorem 3 (Convergence from $\mathcal{S}$ to $\mathcal{R}$ again).

Suppose that a sequence of nondegenerate Karush-Kuhn-Tucker points
$(x^{t},y^{t})$ of $\mathcal{S}$ with quadratic index $m$ converges to
$\left(\bar{x},\bar{y}\right)$ for $t\to 0$. If $(\bar{x},\bar{y})$ is a
nondegenerate T-stationary point of $\mathcal{R}$, then we have for its
T-Index:

$\max\left\\{m-\left|\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}=0\right.\right\\}\right|,0\right\\}\leq
TI\leq m.$

If additionally NDT6 holds at $(\bar{x},\bar{y})$, then the indices coincide,
i.e. $TI=m$.

Proof: The proof will be divided into 4 major steps.

Step 1a. We rewrite the tangent space corresponding to the Karush-Kuhn-Tucker
point $\left(x^{t},y^{t}\right)$. For that, we use Lemma 2a) which provides
that the summation constraint is active:

$\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(x^{t}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(x^{t}),0\end{pmatrix}\xi=0,q\in Q_{0}(x^{t}),\\\
\begin{pmatrix}0,e\end{pmatrix}\xi=0,\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(y^{t}),\\\
\begin{pmatrix}y_{i}^{t}e_{i},x_{i}^{t}e_{i}\end{pmatrix}\xi=0,i\in\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{00}\left(\bar{x},\bar{y}\right),\\\
\begin{pmatrix}y_{i}^{t}e_{i},x_{i}^{t}e_{i}\end{pmatrix}\xi=0,i\in\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right),\\\
\begin{pmatrix}y_{i}^{t}e_{i},x_{i}^{t}e_{i}\end{pmatrix}\xi=0,i\in\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{10}\left(\bar{x},\bar{y}\right),\\\
\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{N}\left(y^{t}\right)\end{array}\right.\right\\}.$

In total there are, due to LICQ,

$\begin{array}[]{rcl}\alpha^{\mathcal{S}}_{t}&=&\left|P\right|+\left|Q_{0}\left(x^{t}\right)\right|+1+\left|\mathcal{E}\left(y^{t}\right)\right|+\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{00}\left(\bar{x},\bar{y}\right)\right|\\\
&&+\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|+\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{10}\left(\bar{x},\bar{y}\right)\right|+\left|\mathcal{N}\left(y^{t}\right)\right|\end{array}$

linearly independent vectors involved. We use Lemma 3a) and 3b) to substitute
$\left|Q_{0}\left(x^{t}\right)\right|$ with
$\left|Q_{0}\left(\bar{x}\right)\right|$ and
$\left|\mathcal{E}\left(y^{t}\right)\right|$ with
$\left|\mathcal{E}\left(\bar{y}\right)\right|$, respectively. The latter set
has cardinality of $n-s-1$ due to Lemma 1c). Additionally, we use Lemma 3c)
and 3d) to conclude:

$\begin{array}[]{rcl}\alpha^{\mathcal{S}}_{t}&=&\left|P\right|+\left|Q_{0}\left(\bar{x}\right)\right|+1+n-s-1+\left|a_{00}\left(\bar{x},\bar{y}\right)\right|\\\
&&+\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|+\left|a_{10}\left(\bar{x},\bar{y}\right)\right|.\end{array}$

Finally,
$\left|a_{00}\left(\bar{x},\bar{y}\right)\right|+\left|a_{10}\left(\bar{x},\bar{y}\right)\right|=s$,
cf. Lemma 1b). Thus, we have:

$\begin{array}[]{rcl}\alpha^{\mathcal{S}}_{t}&=&\left|P\right|+\left|Q_{0}\left(\bar{x}\right)\right|+\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|+n.\end{array}$

Step 1b. We examine the tangent space corresponding to the T-stationary point
$(\bar{x},\bar{y})$. For this purpose, we consider the following vectors from
its definition:

$\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in\mathcal{E}(\bar{y}),\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in a_{10}\left(\bar{x},\bar{y}\right)\cup
a_{00}\left(\bar{x},\bar{y}\right),\begin{pmatrix}0\\\ e\end{pmatrix}.$

The latter vector is involved due to Lemma 1a). The number of these vectors is
due to Lemma 1c) equal to $(n-s-1)+s+1=n$. Moreover, they are linearly
independent due to MPOC-LICQ. Hence, we can write the respective tangent space
as follows:

$\mathcal{T}^{\mathcal{R}}_{(\bar{x},\bar{y})}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(\bar{x}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(\bar{x}),0\end{pmatrix}\xi=0,q\in Q_{0}(\bar{x}),\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in a_{00}(\bar{x},\bar{y})\cup
a_{01}(\bar{x},\bar{y}),\xi_{n+1}=\ldots=\xi_{2n}=0.\end{array}\right.\right\\}$

In total there are, due to MPOC-LICQ,

$\alpha^{\mathcal{R}}=\left|P\right|+\left|Q_{0}\left(\bar{x}\right)\right|+\left|a_{00}(\bar{x},\bar{y})\right|+\left|a_{01}(\bar{x},\bar{y})\right|+n$

linearly independent vectors involved.

Step 2. Let $\mathcal{T}\subset\mathbb{R}^{2n}$ be a linear subspace. We
denote the number of negative eigenvalues of
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}}$ by
$QI^{\mathcal{S}}_{t,\mathcal{T}}$. Analogously,
$QI^{\mathcal{R}}_{t,\mathcal{T}}$ stands for the number of negative
eigenvalues of
$D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}}$ and
$\overline{QI}^{\mathcal{R}}_{\mathcal{T}}$ stands for the number of negative
eigenvalues of
$D^{2}L^{\mathcal{R}}\left(\bar{x},\bar{y}\right)\restriction_{\mathcal{T}}$.
We have the following relation between the involved Hessians of the Lagrange
functions by denoting $E(i)=e_{i}e_{n+i}^{T}+e_{n+i}e_{i}^{T}$,
$i=1,\ldots,n$:

$\begin{array}[]{rcl}\displaystyle
D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)&=&\displaystyle
D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)-\sum\limits_{i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)}\eta_{i}^{\geq,t}E(i)\displaystyle+\sum\limits_{i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)}\eta_{i}^{\leq,t}E(i).\end{array}$
(12)

Step 2a. It holds for $t$ sufficiently small:

$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}=\overline{QI}^{\mathcal{R}}_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}.$

Indeed, by using (12), we derive for any
$\xi\in\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$:

$\begin{array}[]{rcl}\xi^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi&=&\xi^{T}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi-\displaystyle\sum\limits_{i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)}2\eta_{i}^{\geq,t}\xi_{i}\xi_{n+i}\\\
&&\displaystyle+\sum\limits_{i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)}2\eta_{i}^{\leq,t}\xi_{i}\xi_{n+i}=\xi^{T}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi,\end{array}$
(13)

since $\xi_{n+1}=\ldots=\xi_{2n}=0$ as seen in Step 1b. Hence, we get
$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}=QI^{\mathcal{R}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$.
Due to NDT4, continuity arguments provide
$QI^{\mathcal{R}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}=\overline{QI}^{\mathcal{R}}_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$.

Step 2b. We claim that the numbers of positive and negative eigenvalues of
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$
and of
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\prime}}$,
respectively, coincide, where

$\mathcal{T}^{\prime}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(x^{t}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(x^{t}),0\end{pmatrix}\xi=0,q\in Q_{0}(\bar{x}),\\\
\begin{pmatrix}0,e\end{pmatrix}\xi=0,\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(\bar{y}),\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y}),\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y}),\\\
\begin{pmatrix}y^{t}_{i}e_{i},x^{t}_{i}e_{i}\end{pmatrix}\xi=0,i\in
a_{01}(\bar{x},\bar{y})\cup
a_{10}(\bar{x},\bar{y})\end{array}\right.\right\\}.$

Let $\left\\{\lambda^{+}_{1},\ldots,\lambda^{+}_{k^{+}}\right\\}$ be the
positive eigenvalues of
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$
with corresponding eigenvectors
$\left\\{\xi^{+}_{1},\ldots,\xi^{+}_{k^{+}}\right\\}$. Hence, for all
$k=1,\ldots,k^{+}$:

${\xi^{+}_{k}}^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi^{+}_{k}>0.$

We rewrite the tangent space as follows:

$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(\bar{x}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(\bar{x}),0\end{pmatrix}\xi=0,q\in Q_{0}(\bar{x}),\\\
\begin{pmatrix}0,e\end{pmatrix}\xi=0,\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(\bar{y}),\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y}),\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y}),\\\
\begin{pmatrix}\bar{y}_{i}e_{i},\bar{x}_{i}e_{i}\end{pmatrix}\xi=0,i\in
a_{01}(\bar{x},\bar{y})\cup
a_{10}(\bar{x},\bar{y})\end{array}\right.\right\\}.$

Due to MPOC-LICQ, the application of the implicit function theorem provides
the existence of $\delta_{2},\delta_{3}>0$ such that for all
$k=1,\ldots,k^{+}$ and $t<\delta_{2}$ there exists $\xi_{k,t}$ with
$\left\|\xi_{k,t}-\xi^{+}_{k}\right\|<\delta_{3}$ and
$\xi_{k,t}\in\mathcal{T}^{\prime}$. We can choose $t$ even smaller, such that
$\xi_{1,t},\ldots,\xi_{k^{+},t}$ remain linearly independent and for all
$k=1,\ldots,k^{+}$ it holds:

${\xi_{k,t}}^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi_{k,t}>0.$

Hence,
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\prime}}$
has at least $k^{+}$ positive eigenvalues. If we repeat the above reasoning
for negative eigenvalues, the matrix
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\prime}}$
has at least as many negative eigenvalues as
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$.
Additionally, we show that the dimensions of
$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$ and
$\mathcal{T}^{\prime}$ coincide. By Step 1b, we have $2n-\alpha^{\mathcal{R}}$
for the dimension of
$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$. Since MPOC-LICQ
remains valid in the neighborhood of $(\bar{x},\bar{y})$, we get again
$2n-\alpha^{\mathcal{R}}$ for the dimension of $\mathcal{T}^{\prime}$. By
continuity arguments, NDT4 and (13) provide that
$D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\restriction_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$
is nonsingular. Altogether, the assertion follows.

Step 2c. We claim that

$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}}\leq
QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}+\alpha^{\mathcal{R}}-\alpha^{\mathcal{S}}_{t}.$

For that, we focus on the dimension of
$\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$. As a consequence of
Step 1a it is $2n-\alpha^{\mathcal{S}}_{t}$. Due to continuity arguments, we
can choose $t$ small enough to ensure $x^{t}_{i}\neq 0$, $i\in
a_{10}(\bar{x},\bar{y})$ and $y^{t}_{i}\neq 0$, $i\in
a_{01}(\bar{x},\bar{y})$. Using this and Lemma 3a), 3b), and 3d), it follows
that
$\mathcal{T}^{\prime}\subset\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$.
Therefore, using Step 2b,
$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}}\leq
2n-\alpha^{\mathcal{S}}_{t}-k^{+}$. We observe in view of NDT4 and Step 1b
that
$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}=2n-\alpha^{\mathcal{R}}-k^{+}$.
The assertion follows immediately.

Step 3. Let us show that

$\max\left\\{m-\left|\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}=0\right.\right\\}\right|,0\right\\}\leq
TI.$

In view of Step 2a, Step 2c and due to continuity, we have for $t$
sufficiently small:

$\begin{array}[]{rcl}m=QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}}&\overset{Step\,2c}{\leq}&QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}+\alpha^{\mathcal{R}}-\alpha^{\mathcal{S}}_{t}\overset{Step\,2a}{=}\overline{QI}^{\mathcal{R}}_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}+\alpha^{\mathcal{R}}-\alpha^{\mathcal{S}}_{t}\\\
&\overset{Step\,1}{=}&\overline{QI}^{\mathcal{R}}_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}+\left|a_{00}(\bar{x},\bar{y})\right|+\left|a_{01}(\bar{x},\bar{y})\right|-\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|\\\
&=&TI+\left|a_{01}(\bar{x},\bar{y})\right|-\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|.\end{array}$

We show for $t$ sufficiently small:

$\left|a_{01}(\bar{x},\bar{y})\right|-\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|\leq\left|\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}=0\right.\right\\}\right|,$

and the assertion will follow immediately since $TI\geq 0$. Clearly,

$\left|a_{01}(\bar{x},\bar{y})\right|-\left|\mathcal{H}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right)\right|=\left|a_{01}(\bar{x},\bar{y})\backslash\mathcal{H}\left(x^{t},y^{t}\right)\right|.$

Suppose $\bar{i}\in a_{01}\left(\bar{x},\bar{y}\right)$ with
$\bar{\sigma}_{1,\bar{i}}\neq 0$. In view of Remark 1, the difference
$\eta^{\geq,t}_{\bar{i}}-\eta^{\leq,t}_{\bar{i}}$ cannot vanish for all $t$
sufficiently small. In particular, one of the multipliers
$\eta^{\geq,t}_{\bar{i}}$ or $\eta^{\leq,t}_{\bar{i}}$ has to be not vanishing
for all $t$ sufficiently small. Hence,
$\bar{i}\in\mathcal{H}\left(x^{t},y^{t}\right)$. We therefore have:

$a_{01}(\bar{x},\bar{y})\backslash\mathcal{H}\left(x^{t},y^{t}\right)\subset\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}=0\right.\right\\}.$

Step 4. Without loss of generality – considering subsequences if needed – we
can assume that for any $i\in a_{00}(\bar{x},\bar{y})$ at least one of the
sequences $\frac{x_{i}^{t}}{y_{i}^{t}}$ or $\frac{y_{i}^{t}}{x_{i}^{t}}$ is
convergent. First, we note that the quotients are well defined due to Lemma
3c). Moreover, if the former sequence does not contain a convergent
subsequence, we find a subsequence that tends to plus or minus infinity.
Consequently, the corresponding subsequence of the latter reciprocal sequence
has to converge to zero. We define the following auxiliary sets:

$a_{00}^{x}(\bar{x},\bar{y})=\left\\{i\in
a_{00}(\bar{x},\bar{y})\,\left|\,\frac{x^{t}_{i}}{y^{t}_{i}}\mbox{ converges
for }t\to
0\right.\right\\},a_{00}^{y}(\bar{x},\bar{y})=a_{00}(\bar{x},\bar{y})\backslash
a_{00}^{x}(\bar{x},\bar{y}).$

For $\bar{i}\in a_{00}^{x}\left(\bar{x},\bar{y}\right)$ we consider
$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$ and replace two of
the involved equations, namely
$\begin{pmatrix}e_{\bar{i}},0\end{pmatrix}\xi=0$ and
$\begin{pmatrix}0,e_{\bar{i}}\end{pmatrix}\xi=0$ by one equation
$\begin{pmatrix}e_{\bar{i}},\lim\limits_{t\to
0}\frac{x_{\bar{i}}^{t}}{y_{\bar{i}}^{t}}e_{\bar{i}}\end{pmatrix}\xi=0$.
Clearly, the vectors involved in the definition of the newly generated linear
space, i. e.

$\mathcal{T}^{{\bar{i}}}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(\bar{x}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(\bar{x}),0\end{pmatrix}\xi=0,q\in Q_{0}(\bar{x}),\\\
\begin{pmatrix}0,e\end{pmatrix}\xi=0,\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(\bar{y}),\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\backslash\\{\bar{i}\\},\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\backslash\\{\bar{i}\\},\\\
\begin{pmatrix}e_{\bar{i}},\lim\limits_{t\to
0}\frac{x_{\bar{i}}^{t}}{y_{\bar{i}}^{t}}e_{\bar{i}}\end{pmatrix}\xi=0,\begin{pmatrix}\bar{y}_{i}e_{i},\bar{x}_{i}e_{i}\end{pmatrix}\xi=0,i\in
a_{01}(\bar{x},\bar{y})\cup
a_{10}(\bar{x},\bar{y})\\}\end{array}\right.\right\\},$

remain linearly independent. The dimension of $\mathcal{T}^{{\bar{i}}}$ is
greater than the dimension of
$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$ by one. Moreover,
there exists $\xi^{\bar{i}}\in\mathcal{T}^{{\bar{i}}}$ with
$\xi^{\bar{i}}_{n+\bar{i}}\not=0$. Indeed, assume that no such $\xi^{\bar{i}}$
exists, then we can add the equation
$\begin{pmatrix}0,e_{\bar{i}}\end{pmatrix}\xi=0$ to the defining equations of
$\mathcal{T}^{{\bar{i}}}$ wihout changing it. The resulting space, however, is
identical to $\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$, a
contradiction. Without loss of generality, we assume
$\xi^{\bar{i}}_{n+\bar{i}}=1$. Further, by straightforward application of the
implicit function theorem and due to Lemma 3a) and 3b), we find a sequence of
vectors $\xi^{\bar{i}}_{t}\in\mathcal{T}^{\bar{i}}_{t}$ that converges to
$\xi^{\bar{i}}$ for $t\to 0$. For this, we define

$\mathcal{T}^{\bar{i}}_{t}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(x^{t}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(x^{t}),0\end{pmatrix}\xi=0,q\in Q_{0}(x^{t}),\\\
\begin{pmatrix}0,e\end{pmatrix}\xi=0,\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(y^{t}),\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\backslash\\{\bar{i}\\},\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\backslash\\{\bar{i}\\},\\\
\begin{pmatrix}e_{\bar{i}},\frac{x_{\bar{i}}^{t}}{y_{\bar{i}}^{t}}e_{\bar{i}}\end{pmatrix}\xi=0,\begin{pmatrix}y^{t}_{i}e_{i},x^{t}_{i}e_{i}\end{pmatrix}\xi=0,i\in
a_{01}(\bar{x},\bar{y})\cup
a_{10}(\bar{x},\bar{y})\end{array}\right.\right\\}.$

We again have, due to continuity arguments, that
$\xi^{\bar{i}}_{t,n+\bar{i}}\not=0$. For $\bar{i}\in
a_{00}^{y}\left(\bar{x},\bar{y}\right)$ we proceed analogously by considering
$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$ again and replace
two of the involved equations $\begin{pmatrix}e_{\bar{i}},0\end{pmatrix}\xi=0$
and $\begin{pmatrix}0,e_{\bar{i}}\end{pmatrix}\xi=0$ by the equation
$\begin{pmatrix}\lim\limits_{t\to
0}\frac{y_{\bar{i}}^{t}}{x_{\bar{i}}^{t}}e_{\bar{i}},e_{\bar{i}}\end{pmatrix}\xi=0$.
By the same arguments as before, we find
$\xi^{\bar{i}}\in\mathcal{T}^{\bar{i}}$ with $\xi^{\bar{i}}_{\bar{i}}\neq 0$.
Again we will assume $\xi^{\bar{i}}_{\bar{i}}=1$ and find a sequence of
vectors $\xi^{\bar{i}}_{t}\in\mathcal{T}^{\bar{i}}_{t}$ that converges to
$\xi^{\bar{i}}$ for $t\to 0$. Due to continuity, it holds then
$\xi^{\bar{i}}_{t,\bar{i}}\neq 0$.

It is straightforward to verify the following observations for $t$
sufficiently small:

a) Let $\left\\{\xi^{\prime,1},\ldots,\xi^{\prime,\ell}\right\\}$ be a base of
$\mathcal{T}^{\prime}$, cf. Step 2b, then
$\left\\{\xi^{\prime,1},\ldots,\xi^{\prime,\ell}\right\\}\cup\left\\{\xi^{\bar{i}}_{t}\,\left|\,\bar{i}\in
a_{00}(\bar{x},\bar{y})\right.\right\\}$ is a set of linear independent
vectors. In fact, suppose for some coefficients $b_{i},\in\mathbb{R}$, $i\in
a_{00}\left(\bar{x},\bar{y}\right),\beta_{i}\in\mathbb{R},i=1,\ldots,\ell$ it
holds:

$\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)}b_{i}\xi^{i}_{t}+\sum\limits_{i=1}^{\ell}\beta_{i}\xi^{\prime,i}=0.$

For $\bar{i}\in a_{00}^{x}\left(\bar{x},\bar{y}\right)$ we consider the
$(n+\bar{i})$-th row of this sum

$b_{\bar{i}}\underbrace{\xi^{\bar{i}}_{t,n+\bar{i}}}_{\neq
0}+\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)\backslash\left\\{\bar{i}\right\\}}b_{i}\underbrace{\xi^{i}_{t,n+\bar{i}}}_{=0}+\sum\limits_{i=1}^{\ell}\beta_{i}\underbrace{\xi^{\prime,i}_{n+\bar{i}}}_{=0}=0.$

If instead $\bar{i}\in a_{00}^{y}\left(\bar{x},\bar{y}\right)$ we consider the
$\bar{i}$-th row of the sum

$b_{\bar{i}}\underbrace{\xi^{\bar{i}}_{t,\bar{i}}}_{\neq 0}+\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)\backslash\left\\{\bar{i}\right\\}}b_{i}\underbrace{\xi^{i}_{t,\bar{i}}}_{=0}+\sum\limits_{i=1}^{\ell}\beta_{i}\underbrace{\xi^{\prime,i}_{\bar{i}}}_{=0}=0.$

Altogether, it must hold $b_{i}=0$, $i\in a_{00}\left(\bar{x},\bar{y}\right)$.
However, this implies

$\sum\limits_{i=1}^{\ell}\beta_{i}\xi^{\prime,i}=0.$

Hence, $\beta_{i}=0$ for $i=1,\ldots,\ell$.

b) It holds
$\xi^{\bar{i}}_{t}\in\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$,
cf. Step 1a, for any $\bar{i}\in a_{00}(\bar{x},\bar{y})$.

c) It holds
$\left(\eta_{i}^{\leq,t}-\eta_{i}^{\geq,t}\right)\xi^{\bar{i}}_{t,i}\xi^{\bar{i}}_{t,n+i}\leq
0,i\in\mathcal{H}\left(x^{t},y^{t}\right)$. Since
$\xi^{\bar{i}}_{t}\in\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$, we
obtain:

$\begin{array}[]{rcl}\left(\eta_{i}^{\leq,t}-\eta_{i}^{\geq,t}\right)\xi^{\bar{i}}_{t,i}\xi^{\bar{i}}_{t,n+i}&=&\left(\eta_{i}^{\geq,t}-\eta_{i}^{\leq,t}\right)\left(\xi^{\bar{i}}_{t,n+i}\right)^{2}\frac{x^{t}_{i}}{y^{t}_{i}}.\end{array}$

If $i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)$, we have
$x_{i}^{t}<0,y_{i}^{t}>0$ and $\eta_{i}^{\leq,t}=0$. Moreover, due to ND2, we
have $\eta_{i}^{\geq,t}>0$. The assertion follows immediately. The other case
$i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)$ is completely analogous.

d) It holds $\lim\limits_{t\to
0}\left(\eta_{\bar{i}}^{\leq,t}-\eta_{\bar{i}}^{\geq,t}\right)\xi^{\bar{i}}_{t,\bar{i}}\xi^{\bar{i}}_{t,n+\bar{i}}=-\infty$.
We calculate:

$\begin{array}[]{rcl}\left(\eta_{\bar{i}}^{\leq,t}-\eta_{\bar{i}}^{\geq,t}\right)\xi^{\bar{i}}_{t,{\bar{i}}}\xi^{\bar{i}}_{t,n+{\bar{i}}}&=&\left\\{\begin{array}[]{ll}\left(\eta_{\bar{i}}^{\geq,t}-\eta_{\bar{i}}^{\leq,t}\right)\left(\xi^{\bar{i}}_{t,n+\bar{i}}\right)^{2}\frac{x^{t}_{\bar{i}}}{y^{t}_{\bar{i}}}&\mbox{for
}\bar{i}\in a_{00}^{x}(\bar{x},\bar{y}),\\\
\left(\eta_{\bar{i}}^{\geq,t}-\eta_{\bar{i}}^{\leq,t}\right)\left(\xi^{\bar{i}}_{t,\bar{i}}\right)^{2}\frac{y^{t}_{\bar{i}}}{x^{t}_{\bar{i}}}&\mbox{for
}\bar{i}\in a_{00}^{y}(\bar{x},\bar{y}).\end{array}\right.\end{array}$

Let us suppose $\bar{i}\in a_{00}^{x}(\bar{x},\bar{y})$. We have
$y^{t}_{\bar{i}}\neq 0$ and, thus, $\nu^{t}_{\bar{i}}=0$. We use Remark 1 and
NDT3 to conclude that the sequence
$\left(\eta_{\bar{i}}^{\geq,t}-\eta_{\bar{i}}^{\leq,t}\right)x^{t}_{\bar{i}}$
converges to $\varrho_{2,\bar{i}}<0$ for $t\to 0$. Further,
$\frac{1}{y^{t}_{\bar{i}}}>0$ tends to infinity for $t\to 0$. Finally,
$\left(\xi^{\bar{i}}_{t,n+\bar{i}}\right)^{2}$ converges to $1$ for $t\to 0$,
due to the construction of $\xi^{\bar{i}}_{t}$. Thus, the assertion follows.
Instead, let us suppose $\bar{i}\in a_{00}^{y}(\bar{x},\bar{y})$. This time,
we have that $\left(\xi^{\bar{i}}_{t,\bar{i}}\right)^{2}$ converges to $1$ for
$t\to 0$. Due to Remark 1 and NDT3,
$\left(\eta_{\bar{i}}^{\geq,t}-\eta_{\bar{i}}^{\leq,t}\right)y^{t}_{\bar{i}}$
converges to $\varrho_{1,\bar{i}}\neq 0$ for $t\to 0$. If
$\bar{i}\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)$, then
$\varrho_{1,\bar{i}}$ is positive from here. Also,
$\frac{1}{x^{t}_{\bar{i}}}<0$ tends to minus infinity for $t\to 0$. The other
case $\bar{i}\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)$ is completely
analogous.

e) We notice that
$\displaystyle\xi^{\bar{i}^{T}}_{t}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi^{\bar{i}}_{t}$
converges for $t\to 0$ due to the construction above.

Finally, for $\bar{i}\in a_{00}(\bar{x},\bar{y})$ we estimate:

$\begin{array}[]{rcl}\xi_{t}^{\bar{i}^{T}}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi_{t}^{\bar{i}}&\overset{(\ref{eq:Hessetandbar})}{=}&\displaystyle\xi^{\bar{i}^{T}}_{t}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi^{\bar{i}}_{t}\\\
&&\displaystyle-\sum\limits_{i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)}2\eta_{i}^{\geq,t}\xi_{t,i}^{\bar{i}}\xi_{t,n+i}^{\bar{i}}+\sum\limits_{i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)}2\eta_{i}^{\leq,t}\xi_{t,i}^{\bar{i}}\xi_{t,n+i}^{\bar{i}}\\\
&=&\displaystyle\xi^{\bar{i}^{T}}_{t}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi^{\bar{i}}_{t}+2\sum\limits_{i\in\mathcal{H}\left(x^{t},y^{t}\right)}\left(\eta_{i}^{\leq,t}-\eta_{i}^{\geq,t}\right)\xi_{t,i}^{\bar{i}}\xi_{t,n+i}^{\bar{i}}\\\
&\overset{c)}{\leq}&\displaystyle\xi^{\bar{i}^{T}}_{t}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi^{\bar{i}}_{t}+2\left(\eta_{\bar{i}}^{\leq,t}-\eta_{\bar{i}}^{\geq,t}\right)\xi^{\bar{i}}_{t,\bar{i}}\xi^{\bar{i}}_{t,n+\bar{i}}.\end{array}$

Thus, due to d) and e),
$\xi^{\bar{i}^{T}}_{t}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi^{\bar{i}}_{t}$
has to be negative for $t$ small enough. As we have seen in Step 2c, it holds
$\mathcal{T}^{\prime}\subset\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$.
Then, due a) and b), we have therefore:

$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}}\geq
QI^{\mathcal{S}}_{t,\mathcal{T}^{\prime}}+\left|a_{00}(\bar{x},\bar{y})\right|.$

By Step 2a, we have
$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}=\overline{QI}^{\mathcal{R}}_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}$,
and by Step 2b,
$QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}=QI^{\mathcal{S}}_{t,\mathcal{T}^{\prime}}$.
Overall, we obtain:

$m=QI^{\mathcal{S}}_{t,\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}}\geq\overline{QI}^{\mathcal{R}}_{\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}}+\left|a_{00}(\bar{x},\bar{y})\right|=TI.$

Let us illustrate the necessity of NDT6 for the validity of Theorem 3.

######  Example 2 (Necessity of NDT6).

We consider the following Scholtes-type regularization $\mathcal{S}$ with
$n=2$, $s=1$ and $0<c_{1}<c_{2}$:

$\begin{array}[]{rl}\mathcal{S}:\quad\min\limits_{x,y}&(1+x_{1})^{2}+(3-2x_{2})^{2}+c_{1}y_{1}+c_{2}y_{2}\\\
\mbox{s.t.}&x_{1}+x_{2}-1\geq 0,\\\ &y_{1}+y_{2}\geq 1,\quad-t\leq
x_{i}y_{i}\leq t,\quad 0\leq y_{i}\leq 1+\varepsilon,\quad i=1,2,\end{array}$

as well as the point $\left(x^{t},y^{t}\right)=(0,1,1,0)$. We claim that this
point is a nondegenerate Karush-Kuhn-Tucker point. Indeed, it holds:

$\begin{pmatrix}2\\\ 2\\\ c_{1}\\\
c_{2}\end{pmatrix}=\mu_{1,1}^{t}\begin{pmatrix}1\\\ 1\\\ 0\\\
0\end{pmatrix}+\mu_{3}^{t}\begin{pmatrix}0\\\ 0\\\ 1\\\
1\end{pmatrix}+\nu_{2}^{t}\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix}$

with the positive multipliers
$\mu_{1,1}^{t}=2,\mu_{3}^{t}=c_{1},\nu_{2}^{t}=c_{2}-c_{1}$. Obviously, LICQ
and strict complementarity, i.e. ND1 and ND2, respectively, are fulfilled. We
show that
$D^{2}L^{\mathcal{S}}(x^{t},y^{t})\restriction_{\mathcal{T}^{\mathcal{S}}_{(x^{t},y^{t})}}$
is nonsingular and calculate the number of its negative eigenvalues. The
tangent space is
$\mathcal{T}^{\mathcal{S}}_{(x^{t},y^{t})}=\left\\{\xi\in\mathbb{R}^{4}\,\left|\,\xi_{1}+\xi_{2}=0,\xi_{3}=\xi_{4}=0\right.\right\\}$.
For the Hessian of the corresponding Lagrange function we have:

$D^{2}L^{\mathcal{S}}(x^{t},y^{t})=\begin{pmatrix}2&0&0&0\\\ 0&-4&0&0\\\
0&0&0&0\\\ 0&0&0&0\end{pmatrix}.$

Thus, for $\xi\in\mathcal{T}^{\mathcal{S}}_{(x^{t},y^{t})}$ it holds:

$\xi^{T}D^{2}L^{\mathcal{S}}(x^{t},y^{t})\xi=2\xi_{1}^{2}-4\xi_{2}^{2}=-2\xi_{1}^{2}.$

Hence, ND3 is also fulfilled, the Karush-Kuhn-Tucker point
$\left(x^{t},y^{t}\right)$ is nondegenerate and its quadratic index equals
one, i.e. $m=1$ in Theorem 3. The limiting point is
$(\bar{x},\bar{y})=(0,1,1,0)$. This point is T-stationary for the
corresponding regularized continuous reformulation $\mathcal{R}$ according to
Theorem 2, since MPOC-LICQ is fulfilled. Indeed, we have:

$\begin{pmatrix}2\\\ 2\\\ c_{1}\\\
c_{2}\end{pmatrix}=\bar{\mu}_{1,1}\begin{pmatrix}1\\\ 1\\\ 0\\\
0\end{pmatrix}+\bar{\mu}_{3}\begin{pmatrix}0\\\ 0\\\ 1\\\
1\end{pmatrix}+\bar{\sigma}_{1,1}\begin{pmatrix}1\\\ 0\\\ 0\\\
0\end{pmatrix}+\bar{\sigma}_{2,2}\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix}$

with the unique multipliers
$\bar{\mu}_{1,1}=2,\bar{\mu}_{3}=c_{1},\bar{\sigma}_{1,1}=0,\bar{\sigma}_{2,2}=c_{2}-c_{1}.$
It is easy to see that this point is nondegenerate with vanishing T-index,
i.e. $TI=0$, since $a_{00}(\bar{x},\bar{y})=\emptyset$ and
$\mathcal{T}^{\mathcal{R}}_{(\bar{x},\bar{y})}=\\{0\\}$. Note that
additionally $\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}=0\right.\right\\}=\\{1\\}$.
Although all assumptions of Theorem 3 are fulfilled, we have here:

$TI=\max\left\\{m-\left|\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}=0\right.\right\\}\right|,0\right\\}.$

With other words, the saddle points of the Scholtes-type regularization
$\mathcal{S}$ approximate a minimizer of the regularized continuous
reformulation $\mathcal{R}$. The reason is that the $\sigma$-multipliers
corresponding to zero $x$\- and nonzero $y$-variables vanish. The lower bound
given in Theorem 3 is attained.  $\blacksquare$

Next, we point out that the assumption NDT6 is not restrictive at all.

######  Remark 2 (Genericity for NDT6).

Let us briefly sketch why condition NDT6 must be generically fulfilled at the
T-stationary points of $\mathcal{R}$. First, we note that all T-stationary
points of $\mathcal{R}$ are generically nondegenerate, see laemmel:reform .
Now, let us count the losses of freedom induced by the definition of a
T-stationary point. For feasibility we have $\left|P\right|$ equality
constraints, $\left|Q_{0}\right|$ active inequality constraints,
$\left|\mathcal{E}\right|$ bounding constraints on the $y$-variables,
eventually one summation constraint, and
$\left|a_{01}\right|+\left|a_{10}\right|+2\left|a_{00}\right|$ orthogonality
type constraints. Additional losses of freedom come from the T-stationarity
condition. They amount to
$2n-\left|P\right|-\left|Q_{0}\right|-\left|\mathcal{E}\right|-1-\left|a_{01}\right|-\left|a_{10}\right|-2\left|a_{00}\right|$
if the summation constraint is active, and to
$2n-\left|P\right|-\left|Q_{0}\right|-\left|\mathcal{E}\right|-\left|a_{01}\right|-\left|a_{10}\right|-2\left|a_{00}\right|$
otherwise. In both cases, the losses of freedom are equal to the number of
variables $2n$. The violation of NDT6 would produce an additional loss of
freedom, which would imply that the total available degrees of freedom $2n$
are exceeded. By virtue of the structured jet transversality theorem from
guenzel:2008 , this cannot happen generically.  $\blacksquare$

Now, we prove that the Scholtes-type regularization method is well-defined.

######  Theorem 4 (Well-posedness of $\mathcal{S}$ from $\mathcal{R}$).

Let $(\bar{x},\bar{y})$ be a nondegenerate T-stationary point of $\mathcal{R}$
with T-index $m$, additionally, fulfilling NDT6. Then, for all sufficiently
small $t$ there exists a nondegenerate Karush-Kuhn-Tucker point
$(x^{t},y^{t})$ of $S$ within a neighborhood of $(\bar{x},\bar{y})$, which has
the same quadratic index $m$.

Proof: First, we show that for all $i\in a_{10}\left(\bar{x},\bar{y}\right)$
it holds $\bar{\sigma}_{2,i}\neq 0$. Assume contrarily that
$\bar{\sigma}_{2,\bar{i}}=0$ for some $\bar{i}\in
a_{10}\left(\bar{x},\bar{y}\right)$. We then have due to T-stationarity, cf.
(6):

$c_{\bar{i}}=\bar{\sigma}_{2,\bar{i}}+\bar{\mu}_{3}=\bar{\mu}_{3}.$

Moreover, we have in view of Lemma 1c) an index $\tilde{i}\in
a_{01}\left(\bar{x},\bar{y}\right)\backslash\mathcal{E}\left(\bar{y}\right).$
Thus it holds, cf. (6), $c_{\tilde{i}}=\bar{\mu}_{3}$. Due to the assumption
on $c$, we have $\bar{i}=\tilde{i}$, a contradiction. Hence, we may write:

$a_{10}\left(\bar{x},\bar{y}\right)=\left\\{i\in
a_{10}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{2,i}<0\right.\right\\}\cup\left\\{i\in
a_{10}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{2,i}>0\right.\right\\}=a_{10}^{<}\left(\bar{x},\bar{y}\right)\cup
a_{10}^{>}\left(\bar{x},\bar{y}\right).$

Due to NDT6 and NDT3, we may split the other index sets as follows:

$a_{01}\left(\bar{x},\bar{y}\right)=\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}<0\right.\right\\}\cup\left\\{i\in
a_{01}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\sigma}_{1,i}>0\right.\right\\}=a_{01}^{<}\left(\bar{x},\bar{y}\right)\cup
a_{01}^{>}\left(\bar{x},\bar{y}\right),$
$a_{00}\left(\bar{x},\bar{y}\right)=\left\\{i\in
a_{00}\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\varrho}_{1,i}<0\right.\right\\}\cup\left\\{i\in
a\left(\bar{x},\bar{y}\right)\,\left|\,\bar{\varrho}_{1,i}>0\right.\right\\}=a_{00}^{<}\left(\bar{x},\bar{y}\right)\cup
a_{00}^{>}\left(\bar{x},\bar{y}\right).$

We consider the auxiliary system of equations
$F(t,x,y,\lambda,\mu,\sigma,\varrho)=0$ given by (14)-(20), which mimics
stationarity and feasibility. For stationarity we use:

$-L\left(t,x,y,\lambda,\mu,\sigma,\varrho\right)=0,$ (14)

where

$\begin{array}[]{rcl}L&=&\begin{pmatrix}\nabla f(x)\\\
c\end{pmatrix}-\displaystyle\sum\limits_{p\in
P}\lambda_{p}\begin{pmatrix}\nabla h_{p}(x)\\\
0\end{pmatrix}-\sum\limits_{q\in Q_{0}(\bar{x})}\mu_{1,q}\begin{pmatrix}\nabla
g_{q}(x)\\\
0\end{pmatrix}+\sum\limits_{i\in\mathcal{E}(\bar{y})}\mu_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle-\mu_{3}\begin{pmatrix}0\\\
e\end{pmatrix}-\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\frac{\sigma_{1,i}}{\bar{y}_{i}}\begin{pmatrix}y_{i}e_{i}\\\
x_{i}e_{i}\end{pmatrix}-\sum\limits_{i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right)}\frac{\sigma_{2,i}}{\bar{x}_{i}}\begin{pmatrix}y_{i}e_{i}\\\
x_{i}e_{i}\end{pmatrix}\\\ &&\displaystyle-\sum\limits_{i\in
a_{10}^{>}\left(\bar{x},\bar{y}\right)}\sigma_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}-\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)}\left(\varrho_{1,i}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}+\varrho_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\right).\end{array}$

For feasibility we use:

$h_{p}(x)=0,p\in P,\quad g_{q}(x)=0,q\in Q_{0}\left(\bar{x}\right),$ (15)
$1+\varepsilon-
y_{i}=0,i\in\mathcal{E}\left(\bar{y}\right),\quad\displaystyle\sum_{i=1}^{n}y_{i}-(n-s)=0,$
(16) $\frac{1}{\bar{y}_{i}}\left(x_{i}y_{i}-t\right)=0,i\in
a_{01}^{<}\left(\bar{x},\bar{y}\right),\quad\frac{1}{\bar{y}_{i}}\left(x_{i}y_{i}+t\right)=0,i\in
a_{01}^{>}\left(\bar{x},\bar{y}\right),$ (17)
$\frac{1}{|\bar{x}_{i}|}\left(\mbox{sgn}\left(\bar{x}_{i}\right)x_{i}y_{i}-t\right)=0,i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right),\quad y_{i}=0,i\in
a_{10}^{>}\left(\bar{x},\bar{y}\right),$ (18)
$x_{i}+\displaystyle\frac{\varrho_{2,i}\sqrt{t}}{\sqrt{\varrho_{1,i}\varrho_{2,i}}}=0,i\in
a_{00}^{<}\left(\bar{x},\bar{y}\right),\quad
x_{i}-\displaystyle\frac{\varrho_{2,i}\sqrt{t}}{\sqrt{-\varrho_{1,i}\varrho_{2,i}}}=0,i\in
a_{00}^{>}\left(\bar{x},\bar{y}\right),$ (19)
$y_{i}+\displaystyle\frac{\varrho_{1,i}\sqrt{t}}{\sqrt{\varrho_{1,i}\varrho_{2,i}}}=0,i\in
a_{00}^{<}\left(\bar{x},\bar{y}\right),\quad
y_{i}-\displaystyle\frac{\varrho_{1,i}\sqrt{t}}{\sqrt{-\varrho_{1,i}\varrho_{2,i}}}=0,i\in
a_{00}^{>}\left(\bar{x},\bar{y}\right).$ (20)

In view of feasibility and T-stationarity of $\left(\bar{x},\bar{y}\right)$
for $\mathcal{R}$, the vector
$(0,\bar{x},\bar{y},\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho})$
solves (14)-(20). We consider the Jacobian matrix
$DF(t,x,y,\lambda,\mu,\sigma,\rho)=\begin{bmatrix}A&B\\\
B^{T}&D\end{bmatrix}$, where

$\begin{array}[]{rcl}A&=&\displaystyle-D^{2}f(x)+\sum\limits_{p\in
P}\lambda_{p}D^{2}h_{p}(x)+\sum\limits_{q\in
Q_{0}\left(\bar{x}\right)}\mu_{1,q}D^{2}g_{q}(x)\\\
&&\displaystyle+\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\frac{1}{\bar{y}_{i}}E(i)+\sum\limits_{i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right)}\frac{1}{\bar{x}_{i}}E(i),\end{array}$

the columns of $B$ are give by the vectors:

$\begin{pmatrix}\nabla h_{p}(x)\\\ 0\end{pmatrix},p\in
P,\quad\begin{pmatrix}\nabla g_{q}(x)\\\ 0\end{pmatrix},q\in
Q_{0}\left(\bar{x}\right),\quad\begin{pmatrix}0\\\
-e_{i}\end{pmatrix},i\in\mathcal{E}\left(\bar{y}\right),\quad\begin{pmatrix}0\\\
e\end{pmatrix},$ $\begin{pmatrix}\frac{y_{i}}{\bar{y}_{i}}e_{i}\\\
\frac{x_{i}}{\bar{y}_{i}}e_{i}\end{pmatrix},i\in
a_{01}\left(\bar{x},\bar{y}\right),\quad\begin{pmatrix}\frac{y_{i}}{\bar{x}_{i}}e_{i}\\\
\frac{x_{i}}{\bar{x}_{i}}e_{i}\end{pmatrix},i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right),\quad\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in a_{10}^{>}\left(\bar{x},\bar{y}\right),$
$\begin{pmatrix}e_{i}\\\ 0\end{pmatrix},\begin{pmatrix}0\\\
e_{i}\end{pmatrix},i\in a_{00}\left(\bar{x},\bar{y}\right),$

and $D$ consists of
$|P|+\left|Q_{0}\left(\bar{x}\right)\right|+\left|\mathcal{E}\left(\bar{y}\right)\right|+1+\left|a_{01}\left(\bar{x},\bar{y}\right)\right|+\left|a_{10}\left(\bar{x},\bar{y}\right)\right|$
vanishing rows. The remaining rows of $D$ are given by the vectors:

$\begin{pmatrix}0\\\
-\frac{\sqrt{-\varrho_{2,i}t}}{2\sqrt{-\varrho_{1,i}^{3}}}e_{i}\\\
\frac{\sqrt{t}}{2\sqrt{\varrho_{1,i}\varrho_{2,i}}}e_{i}\end{pmatrix},i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right),\quad\begin{pmatrix}0\\\
\frac{\sqrt{-\varrho_{2,i}t}}{2\sqrt{\varrho_{1,i}^{3}}}e_{i}\\\
-\frac{\sqrt{t}}{2\sqrt{-\varrho_{1,i}\varrho_{2,i}}}e_{i}\end{pmatrix},i\in
a_{10}^{>}\left(\bar{x},\bar{y}\right),$ $\begin{pmatrix}0\\\
\frac{\sqrt{t}}{2\sqrt{\varrho_{1,i}\varrho_{2,i}}}e_{i}\\\
-\frac{\sqrt{-\varrho_{1,i}t}}{2\sqrt{-\varrho_{2,i}^{3}}}e_{i}\end{pmatrix},i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right),\quad\begin{pmatrix}0\\\
-\frac{\sqrt{t}}{2\sqrt{-\varrho_{1,i}\varrho_{2,i}}}e_{i}\\\
-\frac{\sqrt{\varrho_{1,i}t}}{2\sqrt{-\varrho_{2,i}^{3}}}e_{i}\end{pmatrix},i\in
a_{10}^{>}\left(\bar{x},\bar{y}\right).$

Additionally we have $D=0$ at
$(0,\bar{x},\bar{y},\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\varrho})$.
Hence, we can apply Theorem 2.3.2 from Jongen:2004 , which says that
$DF(0,\bar{x},\bar{y},\bar{\lambda},\bar{\mu},\bar{\sigma},\bar{\rho})=\begin{bmatrix}A&B\\\
B^{T}&0\end{bmatrix}$ is nonsingular if and only if $\xi^{T}A\xi\neq 0$ for
all $\xi\in B^{\perp}$, the orthogonal complement of the subspace spanned by
the columns of $B$. In view of
$B^{\perp}=\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$, we
check:

$\begin{array}[]{rcl}\xi^{T}A\xi&=&\xi^{T}\left(-D^{2}L^{\mathcal{R}}\left(\bar{x},\bar{y}\right)+\displaystyle\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\frac{1}{\bar{y}_{i}}E(i)+\sum\limits_{i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right)}\frac{1}{\bar{x}_{i}}E(i)\right)\xi\\\
&=&-\underbrace{\xi^{T}D^{2}L^{\mathcal{R}}\left(\bar{x},\bar{y}\right)\xi}_{\neq
0\mbox{\footnotesize\ due to NDT4}}+\displaystyle\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\frac{1}{\bar{y}_{i}}2\xi_{i}\underbrace{\xi_{n+i}}_{=0}+\sum\limits_{i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right)}\frac{1}{\bar{x}_{i}}2\xi_{i}\underbrace{\xi_{n+i}}_{=0}.\end{array}$

Hence, by means of the implicit function theorem we obtain for any $t>0$
sufficiently small a solution
$\left(t,x^{t},y^{t},\lambda^{t},\mu^{t},\sigma^{t},\varrho^{t}\right)$ of the
system of equations (14)-(20).

By choosing $t$ even smaller, if necessary, we can ensure due to continuity
reasons as well as NDT2, NDT3, and NDT6 that the following holds:

(i) $g_{q}\left(x^{t}\right)>0$, $q\in Q\backslash Q_{0}\left(\bar{x}\right)$
and $\mu_{1,q}^{t}>0$, $q\in Q_{0}\left(\bar{x}\right)$,

(ii)
$\mbox{sgn}\left(\sigma^{t}_{1,i}\right)=\mbox{sgn}\left(\bar{\sigma}_{1,i}\right)$,
$i\in a_{01}\left(\bar{x},\bar{y}\right)$,

(iii)
$\mbox{sgn}\left(\sigma^{t}_{2,i}\right)=\mbox{sgn}\left(\bar{\sigma}_{2,i}\right)$,
$\mbox{sgn}\left(x^{t}_{i}\right)=\mbox{sgn}\left(\bar{x}_{i}\right)$, $i\in
a_{10}\left(\bar{x},\bar{y}\right)$,

(iv)
$\mbox{sgn}\left(\varrho^{t}_{1,i}\right)=\mbox{sgn}\left(\bar{\varrho}_{1,i}\right)$,
$\varrho_{2,i}^{t}<0$, $i\in a_{00}\left(\bar{x},\bar{y}\right)$,

(v) $y_{i}^{t}\geq 0$, $i\in\left\\{1,\ldots,n\right\\}$ and
$y^{t}_{i}<1+\varepsilon$,
$i\in\left\\{1,\ldots,n\right\\}\backslash\mathcal{E}\left(\bar{y}\right)$.

From here it is straightforward to see that $\left(x^{t},y^{t}\right)$ is
feasible for $\mathcal{S}$ and we have:

(i) $Q_{0}\left(x^{t}\right)=Q_{0}\left(\bar{x}\right)$,
$\mathcal{E}\left(y^{t}\right)=\mathcal{E}\left(\bar{y}\right)$,
$\mathcal{N}\left(y^{t}\right)=a^{>}_{10}\left(\bar{x},\bar{y}\right)$,
$\mathcal{O}\left(x^{t},y^{t}\right)=\emptyset$,

(ii)
$\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)=a^{>}_{01}\left(\bar{x},\bar{y}\right)\cup\left\\{i\in
a^{<}_{10}\left(\bar{x},\bar{y}\right)\,\left|\,x^{t}_{i}<0\right.\right\\}\cup
a^{>}_{00}\left(\bar{x},\bar{y}\right)$,

(iii)
$\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)=a^{<}_{01}\left(\bar{x},\bar{y}\right)\cup\left\\{i\in
a^{<}_{10}\left(\bar{x},\bar{y}\right)\,\left|\,x^{t}_{i}>0\right.\right\\}\cup
a^{<}_{00}\left(\bar{x},\bar{y}\right)$.

Thus, it holds:

$\begin{array}[]{rcl}\begin{pmatrix}\nabla f\left(x^{t}\right)\\\
c\end{pmatrix}&=&\displaystyle\sum\limits_{p\in
P}\lambda_{p}^{t}\begin{pmatrix}\nabla h_{p}\left(x^{t}\right)\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(\bar{x})}\mu_{1,q}^{t}\begin{pmatrix}\nabla g_{q}\left(x^{t}\right)\\\
0\end{pmatrix}-\sum\limits_{i\in\mathcal{E}(\bar{y})}\mu_{2,i}^{t}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&+\mu_{3}^{t}\begin{pmatrix}0\\\
e\end{pmatrix}\displaystyle+\sum\limits_{i\in
a_{01}\left(\bar{x},\bar{y}\right)}\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{i\in
a_{10}^{<}\left(\bar{x},\bar{y}\right)}\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\ &&\displaystyle+\sum\limits_{i\in
a^{>}_{10}\left(\bar{x},\bar{y}\right)}\sigma_{2,i}^{t}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\displaystyle+\sum\limits_{i\in
a_{00}\left(\bar{x},\bar{y}\right)}\left(\varrho^{t}_{1,i}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}+\varrho^{t}_{2,i}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\right)\\\ &=&\displaystyle\sum\limits_{p\in
P}\lambda_{p}^{t}\begin{pmatrix}\nabla h_{p}\left(x^{t}\right)\\\
0\end{pmatrix}+\sum\limits_{q\in
Q_{0}(x^{t})}\mu_{1,q}^{t}\begin{pmatrix}\nabla g_{q}\left(x^{t}\right)\\\
0\end{pmatrix}-\sum\limits_{i\in\mathcal{E}(y^{t})}\mu_{2,i}^{t}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}\\\ &&\displaystyle+\mu_{3}^{t}\begin{pmatrix}0\\\
e\end{pmatrix}+\sum\limits_{\begin{subarray}{c}i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right),\\\
i\in
a_{01}\left(\bar{x},\bar{y}\right)\end{subarray}}\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{\begin{subarray}{c}i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right),\\\
i\in
a_{10}\left(\bar{x},\bar{y}\right)\end{subarray}}\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\
&&\displaystyle+\sum\limits_{\begin{subarray}{c}i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right),\\\
i\in
a_{00}\left(\bar{x},\bar{y}\right)\end{subarray}}\displaystyle\frac{\sqrt{-\varrho^{t}_{1,i}\varrho^{t}_{2,i}}}{\sqrt{t}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{\begin{subarray}{c}i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right),\\\
i\in
a_{01}\left(\bar{x},\bar{y}\right)\end{subarray}}\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\
&&\displaystyle\par+\sum\limits_{\begin{subarray}{c}i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right),\\\
i\in
a_{10}\left(\bar{x},\bar{y}\right)\end{subarray}}\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}+\sum\limits_{\begin{subarray}{c}i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right),\\\
i\in
a_{00}\left(\bar{x},\bar{y}\right)\end{subarray}}\displaystyle-\frac{\sqrt{\varrho^{t}_{1,i}\varrho^{t}_{2,i}}}{\sqrt{t}}\begin{pmatrix}y_{i}^{t}e_{i}\\\
x_{i}^{t}e_{i}\end{pmatrix}\\\
&&+\displaystyle\sum\limits_{i\in\mathcal{N}\left(y^{t}\right)}\sigma_{2,i}^{t}\begin{pmatrix}0\\\
e_{i}\end{pmatrix}.\end{array}$

We rename the multipliers as follows:

$\eta^{\geq,t}_{i}=\left\\{\begin{array}[]{ll}\displaystyle\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}},&\mbox{for
}i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right),\\\
\displaystyle\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}},&\mbox{for
}i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)\cap
a_{10}\left(\bar{x},\bar{y}\right),\\\
\displaystyle\frac{\sqrt{-\varrho^{t}_{1,i}\varrho^{t}_{2,i}}}{\sqrt{t}},&\mbox{for
}i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)\cap
a_{00}\left(\bar{x},\bar{y}\right),\\\ 0,&\mbox{else,}\end{array}\right.$
$\eta^{\leq,t}_{i}=\left\\{\begin{array}[]{ll}\displaystyle-\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}},&\mbox{for
}i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)\cap
a_{01}\left(\bar{x},\bar{y}\right),\\\
\displaystyle-\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}},&\mbox{for
}i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)\cap
a_{10}\left(\bar{x},\bar{y}\right),\\\
\displaystyle\frac{\sqrt{\varrho^{t}_{1,i}\varrho^{t}_{2,i}}}{\sqrt{t}},&\mbox{for
}i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)\cap
a_{00}\left(\bar{x},\bar{y}\right),\\\ 0,&\mbox{else,}\end{array}\right.$
$\nu^{t}_{i}=\left\\{\begin{array}[]{ll}\displaystyle\sigma_{2,i}^{t},&\mbox{for
}i\in\mathcal{N}\left(y^{t}\right)\cap a_{10}\left(\bar{x},\bar{y}\right),\\\
0,&\mbox{else.}\end{array}\right.\\\ $

Hence, $\left(x^{t},y^{t}\right)$ fulfills (8). Also it is straightforward to
check that (9) and (10) are fulfilled. Thus, $\left(x^{t},y^{t}\right)$ is a
Karush-Kuhn-Tucker point of $\mathcal{S}$. Moreover, ND1 is satisfied as well
in view of Theorem 1. Similar to (9) and (10), ND2 holds at
$\left(x^{t},y^{t}\right)$. It remains to show ND3, i.e.

$\xi_{k}^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi_{k}\neq
0\quad\mbox{for }k=1,\ldots,2n-\alpha_{t}^{\mathcal{S}},$

where $\xi_{1},\ldots\xi_{2n-\alpha_{t}^{\mathcal{S}}}$ form a basis of
$\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$, cf. Step 1a from the
proof of Theorem 2. Note, that by construction $\alpha_{t}^{\mathcal{S}}$ is
constant for $t$ sufficiently small. Thus, we refer to it as
$\alpha^{\mathcal{S}}$. Next, we construct for $t>0$ sufficiently small such a
basis as follows. First, we choose eigenvectors
$\bar{\xi}_{1},\ldots,\bar{\xi}_{2n-\alpha^{\mathcal{R}}}$ of
$D^{2}L^{\mathcal{R}}\left(\bar{x},\bar{y}\right)$ forming a basis of
$\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$, cf. Step 1b from
the proof of Theorem 2. With similar arguments as in Step 2b of the proof of
Theorem 2 and by using the implicit function theorem, we find
$\xi_{j}^{t}\in\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$,$j=1,\ldots,2n-\alpha^{\mathcal{R}}$,
still linearly independent. The remaining
$2n-\alpha^{\mathcal{S}}-\left(2n-\alpha^{\mathcal{R}}\right)=\left|a_{00}\left(\bar{x},\bar{y}\right)\right|$
vectors are chosen as follows. Namely, for $\bar{i}\in
a_{00}(\bar{x},\bar{y})$ we consider

$\mathcal{T}^{\bar{i}}_{t}=\left\\{\xi\in\mathbb{R}^{2n}\,\left|\,\begin{array}[]{l}\begin{pmatrix}Dh_{p}(x^{t}),0\end{pmatrix}\xi=0,p\in
P,\begin{pmatrix}Dg_{q}(x^{t}),0\end{pmatrix}\xi=0,q\in Q_{0}(\bar{x}),\\\
\begin{pmatrix}0,e\end{pmatrix}\xi=0,\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in\mathcal{E}(\bar{y}),\\\
\begin{pmatrix}e_{i},0\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\backslash\\{\bar{i}\\},\begin{pmatrix}0,e_{i}\end{pmatrix}\xi=0,i\in
a_{00}(\bar{x},\bar{y})\backslash\\{\bar{i}\\},\\\
\begin{pmatrix}e_{\bar{i}},\frac{\varrho^{t}_{2,\bar{i}}}{\varrho^{t}_{1,\bar{i}}}{e_{\bar{i}}}\end{pmatrix}\xi=0,\begin{pmatrix}y^{t}_{i}e_{i},x^{t}_{i}e_{i}\end{pmatrix}\xi=0,i\in
a_{01}(\bar{x},\bar{y})\cup
a_{10}(\bar{x},\bar{y})\end{array}\right.\right\\}.$

As in Step 4 of the proof of Theorem 2, we can find
$\bar{\xi}_{\bar{i}}\in\mathcal{T}^{\bar{i}}_{0}\backslash\mathcal{T}^{\mathcal{R}}_{\left(\bar{x},\bar{y}\right)}$.
Especially, $\left(e_{\bar{i}},0\right)\bar{\xi}_{\bar{i}}\neq 0$. We note
that $\lim\limits_{t\to
0}\frac{x^{t}_{\bar{i}}}{y^{t}_{\bar{i}}}=\frac{\bar{\varrho}_{2,\bar{i}}}{\bar{\varrho}_{1,\bar{i}}}$.
Using this and the implicit function theorem, we find for $t$ sufficiently
small a vector
$\xi_{\bar{i}}^{t}\in\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$. It
is then straightforward to check, that
$\xi_{j}^{t},j\in\left\\{1,\ldots,2n-\alpha^{\mathcal{R}}\right\\}$ and
$\xi_{\bar{i}}^{t}$, $\bar{i}\in a_{00}(\bar{x},\bar{y})$, indeed form a basis
of $\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$.

We continue by considering the following limits with respect to the subsets of
$\mathcal{H}\left(x^{t},y^{t}\right)$ for any sequence of vectors
$\xi^{t}\in\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$ from the
constructed base, cf. the definition of $\eta$-multipliers:

$\begin{array}[]{lclcl}\displaystyle\lim\limits_{t\to 0}\sum\limits_{i\in
a^{<}_{01}\left(\bar{x},\bar{y}\right)}\eta^{\leq,t}_{i}\xi^{t}_{i}\xi^{t}_{n+i}&=&\displaystyle\lim\limits_{t\to
0}\sum\limits_{i\in
a^{<}_{01}\left(\bar{x},\bar{y}\right)}\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}}\frac{x^{t}_{i}}{y^{t}_{i}}\left(\xi^{t}_{n+i}\right)^{2}&=&0,\\\
\displaystyle\lim\limits_{t\to 0}\sum\limits_{i\in
a^{>}_{01}\left(\bar{x},\bar{y}\right)}\eta^{\geq,t}_{i}\xi_{i}^{t}\xi^{t}_{n+i}&=&\displaystyle\lim\limits_{t\to
0}\sum\limits_{i\in
a^{>}_{01}\left(\bar{x},\bar{y}\right)}-\sigma_{1,i}^{t}\frac{1}{\bar{y}_{i}}\frac{x^{t}_{i}}{y^{t}_{i}}\left(\xi^{t}_{n+i}\right)^{2}&=&0,\\\
\displaystyle\lim\limits_{t\to 0}\sum\limits_{i\in
a^{<}_{10}\left(\bar{x},\bar{y}\right),x^{t}_{i}>0}\eta^{\leq,t}_{i}\xi^{t}_{i}\xi^{t}_{n+i}&=&\displaystyle\lim\limits_{t\to
0}\sum\limits_{i\in
a^{<}_{10}\left(\bar{x},\bar{y}\right),x^{t}_{i}>0}\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}}\frac{y^{t}_{i}}{x^{t}_{i}}\left(\xi^{t}_{i}\right)^{2}&=&0,\\\
\displaystyle\lim\limits_{t\to 0}\sum\limits_{i\in
a^{<}_{10}\left(\bar{x},\bar{y}\right),x^{t}_{i}<0}\eta^{\geq,t}_{i}\xi^{t}_{i}\xi^{t}_{n+i}&=&\displaystyle\lim\limits_{t\to
0}\sum\limits_{i\in
a^{<}_{10}\left(\bar{x},\bar{y}\right),x^{t}_{i}<0}-\sigma_{2,i}^{t}\frac{1}{\bar{x}_{i}}\frac{y^{t}_{i}}{x^{t}_{i}}\left(\xi^{t}_{i}\right)^{2}&=&0.\end{array}$

If for some $i\in a^{<}_{00}(\bar{x},\bar{y})$ or $i\in
a^{>}_{00}(\bar{x},\bar{y})$ it holds $\lim\limits_{t\to
0}\left(e_{i},0\right)\xi^{t}\neq 0$, we observe:

$\begin{array}[]{lclcl}\displaystyle\lim\limits_{t\to 0}\sum\limits_{i\in
a^{<}_{00}\left(\bar{x},\bar{y}\right)}\eta^{\leq,t}_{i}\xi^{t}_{i}\xi^{t}_{n+i}&=&\displaystyle\lim\limits_{t\to
0}\sum\limits_{i\in
a^{<}_{00}\left(\bar{x},\bar{y}\right)}-\frac{\sqrt{\varrho_{1,i}^{t}\varrho_{2,i}^{t}}}{\sqrt{t}}\frac{\varrho_{1,i}^{t}}{\varrho_{2,i}^{t}}\left(\xi^{t}_{i}\right)^{2}&=&-\infty,\\\
\displaystyle\lim\limits_{t\to 0}\sum\limits_{i\in
a^{>}_{00}\left(\bar{x},\bar{y}\right)}\eta^{\geq,t}_{i}\xi^{t}_{i}\xi^{t}_{n+i}&=&\displaystyle\lim\limits_{t\to
0}\sum\limits_{i\in
a^{>}_{00}\left(\bar{x},\bar{y}\right)}-\frac{\sqrt{-\varrho_{1,i}^{t}\varrho_{2,i}^{t}}}{\sqrt{t}}\frac{\varrho_{1,i}^{t}}{\varrho_{2,i}^{t}}\left(\xi^{t}_{i}\right)^{2}&=&\infty.\end{array}$

Finally, we calculate as in (13):

$\begin{array}[]{rcl}\left(\xi^{t}\right)^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi^{t}&=&\displaystyle\left(\xi^{t}\right)^{T}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi^{t}-\sum\limits_{i\in\mathcal{H}^{\geq}\left(x^{t},y^{t}\right)}2\eta_{i}^{\geq,t}\xi^{t}_{i}\xi^{t}_{n+i}\\\
&&\displaystyle+\sum\limits_{i\in\mathcal{H}^{\leq}\left(x^{t},y^{t}\right)}2\eta_{i}^{\leq,t}\xi^{t}_{i}\xi^{t}_{n+i}.\end{array}$

Altogether, we obtain for any basis vector
$\xi_{j}^{t},j\in\left\\{1,\ldots,2n-\alpha^{\mathcal{R}}\right\\}$ of
$\mathcal{T}^{\mathcal{S}}_{\left(x^{t},y^{t}\right)}$:

$\lim\limits_{t\to
0}\left(\xi_{j}^{t}\right)^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi_{j}^{t}=\bar{a}_{j}\left\lVert\bar{\xi}_{j}\right\rVert^{2},$
(21)

where $\bar{a}_{j}$ is a nonzero eigenvalue of
$D^{2}L^{\mathcal{R}}\left(\bar{x},\bar{y}\right)$ due to the choice of
$\bar{\xi}_{j}$ and NDT4. Let us now focus on the basis vectors
$\xi^{t}_{\bar{i}}$, $\bar{i}\in a_{00}(\bar{x},\bar{y})$. We then have:

$\lim\limits_{t\to
0}\left(\xi_{\bar{i}}^{t}\right)^{T}D^{2}L^{\mathcal{S}}\left(x^{t},y^{t}\right)\xi_{\bar{i}}^{t}=-\infty.$
(22)

This is due to the following reasoning. First,
$\left(\xi_{\bar{i}}^{t}\right)^{T}D^{2}L^{\mathcal{R}}\left(x^{t},y^{t}\right)\xi_{\bar{i}}^{t}$
is bounded due to the construction of $\xi_{\bar{i}}^{t}$, and, moreover,
$\lim\limits_{t\to 0}\left(e_{\bar{i}},0\right)\xi^{t}_{{\bar{i}}}\neq 0$. We
conclude that ND3 is fulfilled. Additionally, the T-index of
$(\bar{x},\bar{y})$ is equal to the sum of its quadratic index and its
biactive index, i.e.
$m=\overline{QI}^{\mathcal{R}}+\left|a_{00}(\bar{x},\bar{y})\right|$. In view
of (21) and (22), the quadratic index ${QI}^{\mathcal{S}}_{t}$ of
$(x^{t},y^{t})$ is then exactly $m$ for $t$ sufficiently small.

## Conclusions

In laemmel:reform , the number of saddle points for the regularized continuous
reformulation of CCOP has been estmated. Namely, each saddle point of CCOP
generates exponentially many saddle points of $\mathcal{R}$, all of them
having the same index. It has been concluded there that the introduction of
auxiliary $y$-variables shifts the complexity of dealing with the cardinality
constraint in CCOP into the appearance of multiple saddle points for its
continuous reformulation. From our extended convergence analysis of the
Scholtes-type regularization it follows that the number of its saddle points
also grows exponentially as compared to that of CCOP. We emphasize that this
issue is at the core of numerical difficulties if solving CCOP up to global
optimality by means of the Scholtes-type regularization method. To the best of
our knowledge this is the first paper studying convergence properties of the
Scholtes-type regularization method in the vicinity of saddle points, rather
than of minimizers. The ideas from our analysis can be potentially applied not
only for other classes of nonsmooth optimization problems, such as MPCC, MPVC,
MPSC, and MPOC, but also for other regularization schemes known from the
literature.

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