#  On Lagrange multipliers of the KKT system in Hilbert spaces

Zhiyu Tan ⋆

⋆ Center for Computation and Technology, Louisiana State University, Baton
Rouge, LA 70803, USA. Email<EMAIL_ADDRESS>or<EMAIL_ADDRESS>
Abstract: In this paper we develop a new decomposition framework to deal with
Lagrange multipliers of the Karush-Kuhn-Tucker (KKT) system of constrained
optimization problems and variational inequalities in Hilbert spaces. It is
different from existing frameworks based on separation theorems. We introduce
the essential Lagrange multiplier and establish the basic theory of this new
multiplier. The essential Lagrange multiplier poses essentially different
existence results in finite and infinite-dimensional spaces. It can also be
used to give an essential characterization of the convergence of multipliers
generated by the classical augmented Lagrangian method. Our analysis reveals
that the essential Lagrange multiplier is at the core of both theories and
applications of Lagrange multipliers.

Keywords: Hilbert spaces, Constrained optimization, KKT system, Weak form
asymptotic KKT system, Essential Lagrange multiplier, Classical augmented
Lagrangian method.

Mathematics Subject Classification: 46N10, 49J27, 90C25, 90C46, 90C48.

##  1\. Introduction

The KKT (Karush-Kuhn-Tucker) system and the related Lagrange multipliers are
of great significance to the theories and algorithms of constrained
optimization problems (cf. [4, 5, 18, 22, 23, 24, 26, 29, 32, 33]). In this
paper we attempt to move away from the classical approach based on separation
theorems to develop a new framework to investigate Lagrange multipliers of the
KKT system of constrained optimization problems, which can also cover some
cases of variational inequalities related to constrained optimization
problems. In our approach we first construct a surrogate model that should
share the same KKT system with the optimization problem at a given local
minimizer by making use of the linearization of the problem, and then discuss
the KKT system of the surrogate model at the minimizer. The approach here is
based on a key observation that the KKT system only involves the linearized
information of the optimization problem at a minimizer and is somehow a
reformulation of the first order necessary condition with respect to the
linearizing cone at the minimizer.

For the surrogate model, we will prove the following existence theorem (cf.
Section 2).

######  Theorem 1.1.

For a given minimizer $u^{*}$ of the constrained optimization problem (2.1),
the surrogate model exists for all $f\in\mathcal{F}(u^{*})$ at the minimizer
if and only if Guignard’s condition (2.17) holds, where $\mathcal{F}(u^{*})$
is the set of all Fréchet differentiable objective functions which have a
local constrained minimizer at $u^{*}$.

In general the surrogate model is a special case of the following model
problem

(1.1) $\min\theta(u)\quad\mbox{s.t.}\ \ Su\in K,$

where $\mathcal{U}$ and $\mathcal{X}$ are real Hilbert spaces, $\emptyset\neq
K\subseteq\mathcal{X}$ is closed and convex, and $S$ is a bounded linear
operator from $\mathcal{U}$ to $\mathcal{X}$. According to the Riesz
representation theorem (cf. [9, Theorem 3.4, Chapter I]), we set
$\mathcal{U}^{\prime}=\mathcal{U}$ and $\mathcal{X}^{\prime}=\mathcal{X}$. We
assume that the feasible set $\mathrm{R}(S)\cap K\neq\emptyset$, where
$\mathrm{R}(S)$ is the range of $S$ and $\theta(u)$ is continuously Fréchet
differentiable and strongly convex on $\mathcal{U}$, i.e., there exists
$c_{0}>0$ such that

(1.2) $\langle u-v,D_{u}\theta(u)-D_{u}\theta(v)\rangle_{\mathcal{U}}\geq
c_{0}\|u-v\|^{2}_{\mathcal{U}},$

where $D_{u}\theta(v)$ is the first order Fréchet derivative of
$\theta(\cdot)$ at $v$, $\langle\cdot,\cdot\rangle_{\mathcal{U}}$ is the inner
product of $\mathcal{U}$ and $\|\cdot\|_{\mathcal{U}}$ is the induced norm.
The inner product and the induced norm on $\mathcal{X}$ are denoted by
$\langle\cdot,\cdot\rangle$ and $\|\cdot\|$ respectively.

It follows from the assumptions on the model problem (1.1) and the classical
convex optimization theory that there exists a unique global minimizer $u^{*}$
of the model problem (1.1). We will investigate the KKT system of the model
problem at $u^{*}$, and the results can be applied to the surrogate model. To
avoid separation theorems, we use an optimization procedure regularization
approach to derive the KKT system at $u^{*}$, which will be realized by the
classical augmented Lagrangian method (ALM, for short)(cf. [16, 30]) in this
paper. By carrying out the convergence analysis of the classical ALM without
using any information of Lagrange multipliers of the model problem (1.1) at
$u^{*}$, we will prove the following theorem (cf. Appendix A).

######  Theorem 1.2.

A feasible point $u^{*}$ is a global minimizer of the model problem (1.1) if
and only if there exists $\\{\lambda^{k}\\}_{k=1}^{+\infty}\subset\mathcal{X}$
such that the following weak form asymptotic KKT system (W-AKKT, for short)
holds

(1.3) $\left\\{\begin{aligned} &\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=0\quad\forall\
v\in\mathcal{U};\\\ &-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta-Su^{*}\rangle\geq
0\quad\forall\ \zeta\in K.\\\ \end{aligned}\right.$

Motivated by the results in Theorem 1.2, we introduce the essential Lagrange
multiplier.

######  Definition 1.1.

An element $\lambda^{*}\in\overline{R(S)}$ is called an essential Lagrange
multiplier of the model problem (1.1) at $u^{*}$ if it satisfies

(1.4) $\left\\{\begin{aligned} &\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\lambda^{*},Sv\rangle=0\quad\forall\
v\in\mathcal{U};\\\ &-\langle\lambda^{*},\zeta-Su^{*}\rangle\geq
0\quad\forall\ \zeta\in\overline{K\cap R(S)},\\\ \end{aligned}\right.$

where $\overline{R(S)}$ and $\overline{K\cap R(S)}$ are the closure of $R(S)$
and $K\cap R(S)$, respectively.

We also recall the definition of the proper Lagrange multiplier and the
classical KKT system as follows (cf. [2, p. 160]).

######  Definition 1.2.

An element $\bar{\lambda}\in\mathcal{X}$ is called a proper Lagrange
multiplier of the model problem (1.1) at $u^{*}$ if it satisfies the classical
KKT system

(1.5) $\left\\{\begin{aligned} &\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\bar{\lambda},Sv\rangle=0\quad\forall\
v\in\mathcal{U};\\\ &-\langle\bar{\lambda},\zeta-Su^{*}\rangle\geq
0\quad\forall\ \zeta\in K.\\\ \end{aligned}\right.$

Note that the essential Lagrange multiplier of the model problem (1.1) is
actually the proper Lagrange multiplier of the optimization problem

$\min\theta(u)\quad\mbox{s.t.}\ \ Su\in\overline{K\cap
R(S)}\subset\overline{R(S)},$

which implies that the essential Lagrange multiplier is only related to the
feasible set of the model problem (1.1). This observation inspires us to
investigate the proper Lagrange multiplier with the help of the essential
Lagrange multiplier. More details on the essential Lagrange multiplier will be
given in Section 3. These results indicate that the essential Lagrange
multiplier is a fundamental concept in the theory of Lagrange multipliers of
constrained optimization.

As an application of the essential Lagrange multiplier, we will consider the
convergence of the multipliers generated by the classical ALM for the model
problem (1.1). Our results show some equivalence between the convergence of
the multipliers and the existence of the essential Lagrange multiplier (see
Theorem 3.9). This indicates that the essential Lagrange multiplier is of
fundamental importance in the application of Lagrange multipliers.

The rest of the paper is organized as follows. In Section 2 a general
optimization problem and the related variational inequalities are given, and
some results of the surrogate model are also presented there, especially the
proof of Theorem 1.1. A thorough discussion of the essential Lagrange
multiplier and the proper Lagrange multiplier of the model problem (1.1) is
included in Section 3. The results in this section also theoretically confirm
the necessity of using asymptotic or approximate KKT systems (cf [1, 6, 20,
38]) to give the optimality conditions of constrained optimization problems in
infinite-dimensional spaces. Section 4 is devoted to some further applications
of the W-AKKT system. We give an elementary proof of the existence of the
proper Lagrange multiplier under Robinson’s condition that is widely used in
the theories and applications of optimization problems in both finite and
infinite-dimensional spaces (cf. [18, 27, 28, 31, 42]). An application of the
theory to optimal control problems with pointwise constraints is given in
Section 5. The paper ends up with some concluding remarks in Section 6.
Details for the proof of Theorem 1.2 are provided in Appendix A.

In this paper we use the standard notations from functional analysis, convex
analysis and partial differential equations, see for example in [2, 3, 9, 11,
14, 33, 41].

##  2\. A general optimization problem and the surrogate model

Let us consider the following general optimization problem

(2.1) $\min f(u)\quad\mbox{s.t.}\ \ G(u)\in\mathcal{K},$

where $f:\ \mathcal{U}\rightarrow\mathbb{R}$, $G:\
\mathcal{U}\rightarrow\mathcal{X}$, $\mathcal{K}$ is a closed and convex set
in $\mathcal{X}$, and $\mathcal{U}$, $\mathcal{X}$ are two real Hilbert
spaces. Assume that $G(u)\cap\mathcal{K}\neq\emptyset$ and $u^{*}$ is a
minimizer of the optimization problem (2.1). We will investigate the KKT
system of the optimization problem (2.1) at $u^{*}$. At present we assume that
$f$ and $G$ are Fréchet differentiable, and we will extend the results to some
nonsmooth cases later. As aforementioned, we will denote by
$\mathcal{F}(u^{*})$ the set of all Fréchet differentiable objective functions
which have a local constrained minimizer at $u^{*}$.

###  2.1. Some preliminary results

Let $M$ be the feasible set, i.e.,

(2.2) $M=\\{u\in\mathcal{U}:\ G(u)\in\mathcal{K}\\}.$

We denote by $T(M,\bar{u})$, $T_{w}(M,\bar{u})$ and $L(\mathcal{K},\bar{u})$
the sequential tangent cone, the weak sequential tangent cone and the
linearizing cone at $\bar{u}\in M$ respectively, which are defined by

(2.3) $T(M,\bar{u})=\\{v\in\mathcal{U}:\ \exists\\{u_{n}\\}\subset M,\
\\{t_{n}\\}\subset\mathbb{R}^{+},\ u_{n}\rightarrow\bar{u},\ t_{n}\rightarrow
0^{+},\frac{1}{t_{n}}(u_{n}-\bar{u})\rightarrow v\\},$ (2.4)
$T_{w}(M,\bar{u})=\\{v\in\mathcal{U}:\ \exists\\{u_{n}\\}\subset M,\
\\{t_{n}\\}\subset\mathbb{R}^{+},\ u_{n}\rightarrow\bar{u},\ t_{n}\rightarrow
0^{+},\frac{1}{t_{n}}(u_{n}-\bar{u})\rightharpoonup v\\}$

and

(2.5) $L(\mathcal{K},\bar{u})=\\{tv\in\mathcal{U}:\
G(\bar{u})+G^{\prime}(\bar{u})v\in\mathcal{K},\ \forall\ t>0\\}.$

Let $C\subset\mathcal{U}$. The polar cone of $C$ is defined by

(2.6) $C^{\circ}=\\{v\in\mathcal{U}:\ \langle v,w\rangle_{\mathcal{U}}\leq
0\quad\forall\ w\in C\\}.$

The following property of the sequential tangent cone, the weak sequential
tangent cone and the linearizing cone is crucial to establish our theory.

######  Lemma 2.1.

If $G$ is a bounded linear operator, for any $\bar{u}\in M$, it holds

(2.7) $\overline{L(\mathcal{K},\bar{u})}=T(M,\bar{u})=T_{w}(M,\bar{u}).$

###### Proof.

Let $v\in T(M,\bar{u})$, i.e., there exist $\\{u_{n}\\}\subset\mathcal{U}$ and
$\\{t_{n}\\}\subset\mathbb{R}^{+}$ such that

$v=\lim\limits_{n\rightarrow+\infty}\frac{1}{t_{n}}(u_{n}-\bar{u}),\
t_{n}\rightarrow 0^{+},\ u_{n}\in M,$

and set $v_{n}=u_{n}-\bar{u}$. Since $G$ is linear and $u_{n}\in M$, we have

$G(\bar{u})+G^{\prime}(\bar{u})v_{n}=G(\bar{u}+v_{n})=G(u_{n})\in\mathcal{K},$

which implies $\frac{1}{t_{n}}v_{n}\in L(\mathcal{K},\bar{u})$ and
$v\in\overline{L(\mathcal{K},\bar{u})}$ by
$v=\lim\limits_{n\rightarrow+\infty}\frac{1}{t_{n}}v_{n}$. This leads to

$T(M,\bar{u})\subset\overline{L(\mathcal{K},\bar{u})}.$

For any $0\neq v\in L(\mathcal{K},\bar{u})$, there exist $v_{0}\in\mathcal{U}$
and $t_{0}>0$, such that $v=t_{0}v_{0}$ and

$G(\bar{u}+v_{0})=G(\bar{u})+G(v_{0})=G(\bar{u})+G^{\prime}(\bar{u})v_{0}\in\mathcal{K}.$

Let $u_{n}=\bar{u}+\frac{1}{n}v_{0}.$ Since $\mathcal{K}$ is convex, we have

$G(u_{n})=G(\bar{u}+\frac{1}{n}v_{0})=\frac{n-1}{n}G(\bar{u})+\frac{1}{n}G(\bar{u}+v_{0})\in\mathcal{K},$

which implies $u_{n}\in M$ for any $n\in\mathbb{N}^{+}$. Taking
$t_{n}=1/(nt_{0})>0$, it follows

$v=\lim\limits_{n\rightarrow+\infty}\frac{1}{t_{n}}(u_{n}-\bar{u}),$

which gives $v\in T(M,\bar{u})$ and

$L(\mathcal{K},\bar{u})\subset T(M,\bar{u}).$

Note that $T(M,\bar{u})$ is closed (cf. [19, Theorem 4.10]). Therefore,

$\overline{L(\mathcal{K},\bar{u})}=T(M,\bar{u}).$

Since $G$ is a bounded linear operator, $M$ is closed and convex. This further
implies

$T(M,\bar{u})=T_{w}(M,\bar{u})$

by Proposition 6.1 of [13]. This completes the proof. ∎

###  2.2. A first order necessary condition

According to the classical optimization theory, at the minimizer $u^{*}$, the
following first order necessary condition holds (cf. [15, Theorem 1], [18,
Proposition 1.2] or [19, Theorem 4.14])

(2.8) $\langle D_{u}f(u^{*}),v\rangle_{\mathcal{U}}\geq 0\quad\forall\ v\in
T_{w}(M,u^{*}),$

which is equivalent to

$-D_{u}f(u^{*})\in T_{w}^{\circ}(M,u^{*}).$

We can also consider the variational inequality: Find $u^{*}\in M$ such that

(2.9) $\langle F(u),v\rangle_{\mathcal{U}}\geq 0\quad\forall\ v\in
T_{w}(M,u),$

where $F:\ \mathcal{U}\rightarrow\mathcal{U}^{\prime}(=\mathcal{U})$ is a
given mapping. It is obvious that at a solution point $u^{*}$, there holds

$\langle F(u^{*}),v\rangle_{\mathcal{U}}\geq 0\quad\forall\ v\in
T_{w}(M,u^{*}),$

which is in the same form of (2.8). Therefore, we can deal with these two
problems in the same framework, and we will only give the arguments for (2.8).

###  2.3. The surrogate model

In this part we will give the definition of the surrogate model at a minimizer
of the optimization problem (2.1) and prove a fundamental theorem in our
theory, i.e., Theorem 1.1.

Note that the linearization problem of (2.1) at $u^{*}$ is

$\min f(u^{*})+\langle
D_{u}f(u^{*}),u-u^{*}\rangle_{\mathcal{U}}\quad\mbox{s.t.}\ \
G(u^{*})+G^{\prime}(u^{*})(u-u^{*})\in\mathcal{K},$

which is equivalent to

$\min f(u^{*})+\langle
D_{u}f(u^{*}),u-u^{*}\rangle_{\mathcal{U}}\quad\mbox{s.t.}\ \
G^{\prime}(u^{*})u\in\mathcal{K}-G(u^{*})+G^{\prime}(u^{*})u^{*}.$

Let us consider the following optimization problem

(2.10) $\min f(u^{*})+\langle
D_{u}f(u^{*}),u-u^{*}\rangle_{\mathcal{U}}+\frac{c}{2}\|u-u^{*}\|_{\mathcal{U}}^{2}\quad\mbox{s.t.}\
\ G^{\prime}(u^{*})u\in K,$

where $c>0$ and $K=\mathcal{K}-G(u^{*})+G^{\prime}(u^{*})u^{*}$. Note that the
optimization problem (2.10) is a special case of the model problem (1.1).

The feasible set of this problem is

$\tilde{M}=\\{u\in\mathcal{U}:\ G^{\prime}(u^{*})u\in K\\},$

and we have some key observations

(2.11) $\left\\{\begin{aligned} K-G^{\prime}(u^{*})u^{*}&\ \ =\ \
\mathcal{K}-G(u^{*});\\\ G^{\prime}(u^{*})u^{*}\in K&\Longleftrightarrow
G(u^{*})\in\mathcal{K}\end{aligned}\right.$

and

(2.12) $\displaystyle L(K,u^{*})$ $\displaystyle=\\{tv\in\mathcal{U}:\
G^{\prime}(u^{*})u^{*}+G^{\prime}(u^{*})v\in K,\ \forall\ t>0\\}$
$\displaystyle=\\{tv\in\mathcal{U}:\
G^{\prime}(u^{*})u^{*}+G^{\prime}(u^{*})v\in\mathcal{K}-G(u^{*})+G^{\prime}(u^{*})u^{*},\
\forall\ t>0\\}$ $\displaystyle=\\{tv\in\mathcal{U}:\
G(u^{*})+G^{\prime}(u^{*})v\in\mathcal{K},\ \forall\ t>0\\}$
$\displaystyle=L(\mathcal{K},u^{*}).$

######  Lemma 2.2.

The classical KKT system (1.5) of the optimization problem (2.10) holds at
$u^{*}$ with a proper Lagrange multiplier $\bar{\lambda}\in\mathcal{X}$, i.e.,

(2.13) $\left\\{\begin{aligned} &\langle
D_{u}f(u^{*}),v\rangle_{\mathcal{U}}+\langle\bar{\lambda},G^{\prime}(u^{*})v\rangle=0\quad\forall\
v\in\mathcal{U};\\\
&-\langle\bar{\lambda},\zeta-G^{\prime}(u^{*})u^{*}\rangle\geq 0\quad\forall\
\zeta\in K,\\\ \end{aligned}\right.$

if and only if the classical KKT system of the optimization problem (2.1)
holds at $u^{*}$ with the same proper Lagrange multiplier $\bar{\lambda}$,
i.e.,

(2.14) $\left\\{\begin{aligned} &\langle
D_{u}f(u^{*}),v\rangle_{\mathcal{U}}+\langle\bar{\lambda},G^{\prime}(u^{*})v\rangle=0\quad\forall\
v\in\mathcal{U};\\\ &-\langle\bar{\lambda},\zeta-G(u^{*})\rangle\geq
0\quad\forall\ \zeta\in\mathcal{K}.\\\ \end{aligned}\right.$

###### Proof.

This follows from (2.11) directly. ∎

Note that if $u^{*}$ satisfies (2.13), $u^{*}$ is the global minimizer of the
optimization problem (2.10). Meanwhile, according to the convex optimization
theory, $u^{*}$ is the global minimizer of the optimization problem (2.10) if
and only if (cf. [15, Theorem 1] or [19, Theorem 4.14 and Theorem 4.19])

$\langle D_{u}f(u^{*}),v\rangle_{\mathcal{U}}\geq 0\quad\forall\ v\in
T(\tilde{M},u^{*}),$

which is equivalent to

$\langle D_{u}f(u^{*}),v\rangle_{\mathcal{U}}\geq 0\quad\forall\
v\in\overline{L(\mathcal{K},u^{*})}$

by Lemma 2.1 and (2.12). That is

(2.15)
$-D_{u}f(u^{*})\in\overline{L(\mathcal{K},u^{*})}^{\circ}=L^{\circ}(\mathcal{K},u^{*}).$

On the other hand, Lemma 4.2 of [13] states that

(2.16) $L^{\circ}(\mathcal{K},u^{*})\subset T_{w}^{\circ}(M,u^{*}).$

Therefore, the first order necessary condition (2.8) holds automatically if
$u^{*}$ is the global minimizer of the optimization problem (2.10).

This inspires us to use the optimization problem (2.10) to investigate the KKT
system of the optimization problem (2.1) at $u^{*}$.

######  Definition 2.3.

For $f\in\mathcal{F}(u^{*})$, the optimization problem (2.10) is called a
surrogate model of the optimization problem (2.1) at $u^{*}$ if $u^{*}$ is the
global minimizer of the optimization problem (2.10).

Theorem 1.1 establishes the existence results of the surrogate model of the
optimization problem (2.1) at $u^{*}$. We recall here Guignard’s condition
(cf. [13, 15]), which is

(2.17) $T_{w}^{\circ}(M,u^{*})=L^{\circ}(\mathcal{K},u^{*}).$

###  2.4. Proof of Theorem 1.1

For any $f\in\mathcal{F}(u^{*})$, since $u^{*}$ is a global minimizer of the
optimization problem (2.10), the condition (2.15) should be held, i.e.,
$-D_{u}f(u^{*})\in L^{\circ}(\mathcal{K},u^{*})$, which yields

$D\mathcal{F}(u^{*})\subset L^{\circ}(\mathcal{K},u^{*}).$

Here $D\mathcal{F}(u^{*})=\\{-D_{u}f(u^{*})\in\mathcal{U}:\
f\in\mathcal{F}(u^{*})\\}$. Note that Theorem 3.2 of [13] gives

$D\mathcal{F}(u^{*})=T_{w}^{\circ}(M,u^{*}).$

Hence, $T_{w}^{\circ}(M,u^{*})=D\mathcal{F}(u^{*})\subset
L^{\circ}(\mathcal{K},u^{*})$. Combining with (2.16), we arrive at (2.17).

For the other direction, (2.8) holds, i.e.,

$-D_{u}f(u^{*})\in T_{w}^{\circ}(M,u^{*})$

for any $f\in\mathcal{F}(u^{*})$. If (2.17) holds, we have (2.15) holds, which
implies $u^{*}$ is a global minimizer of the optimization problem (2.10).
Hence, the optimization problem (2.10) is a surrogate model of the
optimization problem (2.1) at $u^{*}$. This completes the proof.

###  2.5. Nonsmooth cases

In the previous arguments, we assume that $f$ and $G$ are Fréchet
differentiable. It is worth noting that the theory in this paper can also be
applied to some nonsmooth cases, for example the case where $f$ and $G$ are
only semismooth. In this case, we can choose $p_{u^{*}}\in\partial f(u^{*})$
and $S_{u^{*}}\in\partial G(u^{*})$ to do the analysis. If there exist
$p_{u^{*}}\in\partial f(u^{*})$ and $S_{u^{*}}\in\partial G(u^{*})$ such that

  1. (1)

$\langle p_{u^{*}},v\rangle_{\mathcal{U}}\geq 0\quad\forall\ v\in
T_{w}(M,u^{*})$;

  2. (2)

$T_{w}^{\circ}(M,u^{*})=L^{\circ}(\mathcal{K},u^{*},S_{u^{*}})$, where
$L(\mathcal{K},u^{*},S_{u^{*}})$ is $L(\mathcal{K},u^{*})$ with
$G^{\prime}(u^{*})=S_{u}^{*}$,

then the surrogate model can be defined as

$\min f(u^{*})+\langle
p_{u^{*}},u-u^{*}\rangle_{\mathcal{U}}+\frac{c}{2}\|u-u^{*}\|_{\mathcal{U}}^{2}\quad\mbox{s.t.}\
\ S_{u^{*}}u\in K=\mathcal{K}-G(u^{*})+S_{u^{*}}u^{*}$

for any $c>0$.

##  3\. The essential Lagrange multiplier

In this section we will establish the basic theory of the essential Lagrange
multiplier and the proper Lagrange multiplier of the model problem (1.1). As
an application of the essential Lagrange multiplier, we will use it to
characterize the convergence of the multipliers generated by the classical ALM
(see (A.3)).

Without further explanation, we always assume that $u^{*}$ is the global
minimizer of the model problem (1.1) in this section.

###  3.1. The essential Lagrange multiplier

We will establish the existence and uniqueness theory of the essential
Lagrange multiplier (see Definition 1.1) here. Our results indicate the
existence theory of the essential Lagrange multiplier is different in finite
and infinite-dimensional cases. More precisely, the essential Lagrange
multiplier always exists in the finite-dimensional case, while in the
infinite-dimensional case, this is no longer true.

######  Theorem 3.1.

The essential Lagrange multiplier is unique.

###### Proof.

This can be derived from Definition 1.1 directly. If there exist two essential
Lagrange multipliers $\lambda^{*}_{1}\in\overline{R(S)}$ and
$\lambda^{*}_{2}\in\overline{R(S)}$ at $u^{*}$, it holds

$\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\lambda^{*}_{1},Sv\rangle=0\quad\forall\
v\in\mathcal{U}$

and

$\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\lambda^{*}_{2},Sv\rangle=0\quad\forall\
v\in\mathcal{U},$

which follows

$\langle\lambda^{*}_{1}-\lambda_{2}^{*},Sv\rangle=0\quad\forall\
v\in\mathcal{U}.$

This gives $\lambda^{*}_{1}=\lambda_{2}^{*}$, and it completes the proof. ∎

######  Theorem 3.2.

The essential Lagrange multiplier exists at the global minimizer $u^{*}$ of
the model problem (1.1) if and only if

$-D_{u}\theta(u^{*})\in R(S^{*}),$

where $S^{*}$ is the adjoint operator of $S$.

###### Proof.

If the essential Lagrange multiplier $\lambda^{*}$ exists, according to its
definition, we have

$-D_{u}\theta(u^{*})=S^{*}\lambda^{*},$

which means $-D_{u}\theta(u^{*})\in R(S^{*})$.

If $-D_{u}\theta(u^{*})\in R(S^{*})$, there exists
$\bar{\lambda}^{*}\in\mathcal{X}$ such that

$-D_{u}\theta(u^{*})=S^{*}\bar{\lambda}^{*},$

which is equivalent to

(3.1) $\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\bar{\lambda}^{*},Sv\rangle=0\quad\forall\
v\in\mathcal{U}.$

Since $u^{*}$ is the global minimizer, by Theorem 1.2, there exists
$\\{\lambda^{k}\\}\subset\mathcal{X}$ such that the W-AKKT system (1.3) holds.
Therefore,

$\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=\langle\bar{\lambda}^{*},Sv\rangle\quad\forall\
v\in\mathcal{U},$

by (3.1), and then

$-\langle\bar{\lambda}^{*},Sv-
Su^{*}\rangle=-\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv-
Su^{*}\rangle\geq-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv-Su^{*}\rangle\geq
0\quad\forall\ v\in\mathcal{U},$

by (1.3). Finally, by the boundedness of $\bar{\lambda}^{*}$, we arrive at

$-\langle\bar{\lambda}^{*},\zeta-Su^{*}\rangle\geq 0\quad\forall\
\zeta\in\overline{K\cap R(S)},$

which, together with (3.1), implies that the restriction of
$\bar{\lambda}^{*}$ to $\overline{R(S)}$ is the essential Lagrange multiplier.
This gives the existence of the essential Lagrange multiplier. ∎

######  Theorem 3.3.

If $R(S)$ is closed in $\mathcal{X}$, then the essential Lagrange multiplier
exists. Conversely, if the essential Lagrange multiplier always exists at
$u^{*}$ for any $K$ and $\theta(\cdot)$ satisfying the assumptions of the
model problem (1.1), then $R(S)$ is closed in $\mathcal{X}$.

###### Proof.

According to Theorem 1.2, there exists
$\\{\lambda^{k}\\}_{k=1}^{+\infty}\subset\mathcal{X}$ such that

$\lim\limits_{k\rightarrow+\infty}\langle
S^{*}\lambda^{k}+D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}=0\quad\forall\
v\in\mathcal{U},$

i.e., $\\{S^{*}\lambda^{k}\\}_{k=1}^{+\infty}$ weakly converges to
$-D_{u}\theta(u^{*})$ in $\mathcal{U}$.

If $R(S)$ is closed, by the closed range theorem (cf. [41, p. 205]),
$R(S^{*})$ is closed, which further is weakly closed. Therefore,
$-D_{u}\theta(u^{*})\in R(S^{*})$ and then by Theorem 3.2, we have the
existence of the essential Lagrange multiplier.

Let $u_{0}\in\mathrm{Ker}(S)^{\perp}$ be arbitrary,
$\theta(u)=\frac{1}{2}\|u\|_{\mathcal{U}}^{2}$ and $K=\\{Su_{0}\\}$. In this
case the global minimizer $u^{*}=u_{0}$ and
$D_{u}\theta(u^{*})=u_{0}\in\mathrm{Ker}(S)^{\perp}$. Note that
$\mathrm{Ker}(S)^{\perp}=\overline{R(S^{*})}$. Since $u_{0}$ is arbitrary, by
the assumption on the existence of the essential Lagrange multiplier and
Theorem 3.2, we have $\mathrm{Ker}(S)^{\perp}\subset R(S^{*})$, which gives
$\overline{R(S^{*})}\subset R(S^{*})$. Therefore,
$R(S^{*})=\overline{R(S^{*})}$, which is equivalent to $R(S)=\overline{R(S)}$
by the closed range theorem, i.e., $R(S)$ is closed. ∎

######  Remark 3.1.

Theorem 3.3 implies that the condition that $R(S)$ is closed in $\mathcal{X}$
is sufficient and almost necessary for the existence of the essential Lagrange
multiplier.

Note that if $R(S)$ is a finite-dimensional space, $R(S)$ is closed. We have
the following corollaries.

######  Corollary 3.2.

If $R(S)$ is a finite-dimensional space, then the essential Lagrange
multiplier exists.

######  Corollary 3.3.

If $\mathcal{X}$ is a finite-dimensional space, then the essential Lagrange
multiplier exists.

The essential Lagrange multiplier can also be used to characterize the
optimality of a feasible point.

######  Theorem 3.4.

If the essential Lagrange multiplier exits at a feasible point $u^{*}$, then
$u^{*}$ is the global minimizer of the model problem (1.1).

###### Proof.

According to the convex optimization theory, it suffices to show that (cf. [3,
Proposition 26.5], [10, Proposition 2.1, Chapter II] or [19, Corollary 4.20])

$\langle D_{u}\theta(u^{*}),v-u^{*}\rangle_{\mathcal{U}}\geq 0\quad\ \forall\
Sv\in K.$

Let $\lambda^{*}$ be the essential Lagrange multiplier. The definition of
$\lambda^{*}$ gives

$\langle
D_{u}\theta(u^{*}),v-u^{*}\rangle_{\mathcal{U}}=-\langle\lambda^{*},S(v-u^{*})\rangle=-\langle\lambda^{*},Sv-
Su^{*}\rangle\geq 0\quad\forall\ Sv\in K,$

which completes the proof. ∎

Furthermore, Theorem 3.3 and Theorem 3.4 lead to the following corollary.

######  Corollary 3.4.

Assume that $R(S)$ is closed in $\mathcal{X}$, a feasible point $u^{*}$ is the
global minimizer of the model problem (1.1) if and only if the essential
Lagrange multiplier exists at $u^{*}$.

###  3.2. The proper Lagrange multiplier

According to the definition of the proper Lagrange multiplier (see Definition
1.2) and the definition of the essential Lagrange multiplier (see Definition
1.1), the existence of the proper Lagrange multiplier always implies the
existence of the essential Lagrange multiplier and

(3.2) $\lambda^{*}=\bar{\lambda}|_{\overline{R(S)}},$

where $\bar{\lambda}|_{\overline{R(S)}}$ is the restriction of $\bar{\lambda}$
to $\overline{R(S)}$. In other words, if the essential Lagrange multiplier
does not exist, neither does the proper Lagrange multiplier. Therefore, we can
consider the theory of the proper Lagrange multiplier under the assumption
that the essential Lagrange multiplier exists, and which can be verified by
the results in the previous subsection. We can also assume that
$\lambda^{*}\neq 0$. Otherwise $\bar{\lambda}=0$ is a proper Lagrange
multiplier. We will establish the existence and uniqueness theory of the
proper Lagrange multiplier under Assumption 3.6.

######  Remark 3.5.

If the essential Lagrange multiplier does not exist, neither does the
classical KKT system. This happens in infinite-dimensional cases as shown in
the previous subsection. Therefore, it is necessary to use the asymptotic or
approximate KKT system to characterize the optimality in some infinite-
dimensional cases, which has been used in the literature (cf. [1, 6, 20, 37,
38]) as a technique, but did not confirm its necessity theoretically.

######  Assumption 3.6.

The essential Lagrange multiplier $\lambda^{*}$ exists at $u^{*}$ and
$\lambda^{*}\neq 0$.

Let $\zeta^{*}=Su^{*}$ and $\mathcal{N}(\zeta^{*},K)$ be the normal cone to
$K$ at $\zeta^{*}$, i.e.,

$\mathcal{N}(\zeta^{*},K)=\\{\lambda\in\mathcal{X}:\
-\langle\lambda,\zeta-\zeta^{*}\rangle\geq 0\quad\forall\ \zeta\in K\\}.$

######  Theorem 3.5.

Suppose that Assumption 3.6 holds. The proper Lagrange multiplier exists at
$u^{*}$ if and only if there exist
$\tilde{\lambda}\in\mathcal{N}(\zeta^{*},K)$ and
$\bar{\zeta}_{0}\in\overline{R(S)}$ such that

(3.3)
$\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}\quad\quad(\mbox{Compatibility\
Condition})$

or equivalently

$\mathrm{Span}\\{\tilde{\lambda}\\}+\mathrm{Ker}(S^{*})=\mathrm{Span}\\{\lambda^{*}\\}+\mathrm{Ker}(S^{*})$

and

(3.4) $\left\\{\begin{aligned}
&\langle\tilde{\lambda},\bar{\zeta}_{0}\rangle>0,\\\
&\langle\lambda^{*},\bar{\zeta}_{0}\rangle>0.\end{aligned}\right.\quad\quad(\mbox{Consistency\
Condition})$

###### Proof.

A direct calculation gives

$\displaystyle\quad\quad\quad\quad\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}$
$\displaystyle\ \ \Longleftrightarrow\ \
[\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}]^{\perp}=[\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}]^{\perp}$
$\displaystyle\ \ \Longleftrightarrow\ \
\mathrm{cl}\\{[\mathrm{Ker}(\tilde{\lambda})]^{\perp}+\mathrm{Ker}(S^{*})\\}=\mathrm{cl}\\{[\mathrm{Ker}(\lambda^{*})]^{\perp}+\mathrm{Ker}(S^{*})\\}$
$\displaystyle\ \ \Longleftrightarrow\ \
[\mathrm{Ker}(\tilde{\lambda})]^{\perp}+\mathrm{Ker}(S^{*})=[\mathrm{Ker}(\lambda^{*})]^{\perp}+\mathrm{Ker}(S^{*})$
$\displaystyle\ \ \Longleftrightarrow\ \
\mathrm{Span}\\{\tilde{\lambda}\\}+\mathrm{Ker}(S^{*})=\mathrm{Span}\\{\lambda^{*}\\}+\mathrm{Ker}(S^{*}).$

If the proper Lagrange multiplier $\bar{\lambda}$ exists, we have
$\bar{\lambda}\in\mathcal{N}(\zeta^{*},K)$ and
$\lambda^{*}=\bar{\lambda}|_{\overline{R(S)}}$, which follows

$\mathrm{Ker}(\bar{\lambda})\cap\overline{R(S)}=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}.$

Since $\lambda^{*}\neq 0$ on $\overline{R(S)}$, there exists
$\zeta_{0}\in\overline{R(S)}$ such that

$\langle\lambda^{*},\zeta_{0}\rangle\neq 0.$

We assume that

$\langle\lambda^{*},\zeta_{0}\rangle>0.$

Otherwise, we can choose $-\zeta_{0}$. Taking $\bar{\zeta}_{0}=\zeta_{0}$, the
condition (3.4) holds for $\bar{\lambda}$. Hence, we can take
$\tilde{\lambda}=\bar{\lambda}$.

Now we prove the other direction. Let $\tilde{\lambda}$ be an element in
$\mathcal{N}(\zeta^{*},K)$ such that (3.3) and (3.4) hold.

Let

$\bar{\lambda}=t_{0}\tilde{\lambda},$

where
$t_{0}=\langle\lambda^{*},\bar{\zeta}_{0}\rangle/\langle\tilde{\lambda},\bar{\zeta}_{0}\rangle$
and $\bar{\zeta}_{0}$ satisfies (3.4).

We will prove that $\bar{\lambda}$ is a proper Lagrange multiplier. Note that
the condition (3.4) implies $t_{0}>0$, which further gives
$\bar{\lambda}\in\mathcal{N}(\zeta^{*},K)$.

Now, by the definition of the proper Lagrange multiplier, we only need to show
that

$\lambda^{*}=\bar{\lambda}|_{\overline{R(S)}}.$

By (3.4) and (3.3), we have
$\overline{R(S)}=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}+\mathrm{Span}\\{\bar{\zeta}_{0}\\}=\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}+\mathrm{Span}\\{\bar{\zeta}_{0}\\}$.
Therefore, for any $\zeta\in\overline{R(S)}$, there exist $s\in\mathbb{R}$ and
$\zeta_{0}\in\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}$ such that
$\zeta=\zeta_{0}+s\bar{\zeta}_{0}$. It follows

$\langle\bar{\lambda},\zeta\rangle=t_{0}\langle\tilde{\lambda},\zeta\rangle=t_{0}\langle\tilde{\lambda},\zeta_{0}+s\bar{\zeta}_{0}\rangle=t_{0}s\langle\tilde{\lambda},\bar{\zeta}_{0}\rangle=s\langle\lambda^{*},\bar{\zeta}_{0}\rangle=\langle\lambda^{*},\zeta_{0}+s\bar{\zeta}_{0}\rangle=\langle\lambda^{*},\zeta\rangle,$

which implies $\lambda^{*}=\bar{\lambda}|_{\overline{R(S)}}.$

∎

######  Theorem 3.6.

Suppose that Assumption 3.6 holds. There exists a unique proper Lagrange
multiplier at $u^{*}$ if and only if (1) there exists
$\tilde{\lambda}\in\mathcal{N}(\zeta^{*},K)$ which satisfies (3.3) and (3.4),
and (2) for any $\hat{\lambda}\in\mathcal{N}(\zeta^{*},K)$ which satisfies
(3.3) and (3.4), it holds
$\mathrm{Ker}(\hat{\lambda})=\mathrm{Ker}(\tilde{\lambda})$.

###### Proof.

Since all the proper Lagrange multipliers belong to $\mathcal{N}(\zeta^{*},K)$
and their restrictions to $\overline{R(S)}\ (\neq\\{0\\})$ are the same,
according to the proof of Theorem 3.5, we have
$\bar{\lambda}=\langle\lambda^{*},\bar{\zeta}_{0}\rangle/\langle\tilde{\lambda},\bar{\zeta}_{0}\rangle\tilde{\lambda}$
is the unique proper Lagrange multiplier.

Let $\bar{\lambda}$ be the unique proper Lagrange multiplier. According to the
condition (3.4) of Theorem 3.5, $\bar{\lambda}\in\mathcal{N}(\zeta^{*},K)$,
and (3.3) and (3.4) hold for $\bar{\lambda}$.

Suppose that there exists $\tilde{\lambda}\in\mathcal{N}(\zeta^{*},K)$ which
satisfies (3.3) and (3.4), it holds
$\mathrm{Ker}(\tilde{\lambda})\neq\mathrm{Ker}(\bar{\lambda})$. By Theorem
3.5, there exists a proper Lagrange multiplier $\bar{\lambda}_{0}$ with
$\mathrm{Ker}(\bar{\lambda}_{0})\neq\mathrm{Ker}(\bar{\lambda})$. This is a
contradiction to the uniqueness of the Lagrange multiplier. ∎

######  Remark 3.7.

If $\mathcal{X}=\overline{R(S)}$ or equivalently
$\mathrm{Ker}(S^{*})=\\{0\\}$, the proper Lagrange multiplier is the essential
Lagrange multiplier, which implies that the proper Lagrange multiplier (if
exists) is unique.

######  Remark 3.8.

In the finite-dimensional case, the condition $\mathrm{Ker}(S^{*})=\\{0\\}$ is
the LICQ condition. The connections of the LICQ condition and the uniqueness
of the proper Lagrange multiplier have been investigated in [40]. It has also
been proved that the proper Lagrange multiplier is unique if and only if it
satisfies SMFC in [25] for an optimization problem with both equality and
inequality constraints under the assumption that the proper Lagrange
multiplier exists. The uniqueness results for the case of general cone
constraints can be found in [34, 35].

In the following part we will use the results in quotient spaces to simplify
the presentation of the results in Theorem 3.5 and Theorem 3.6.

Let $Y=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}$ and
$[\mathcal{X}]=\mathcal{X}/Y$ be the quotient space with respect to $Y$. Since
$Y=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}$ is a closed subspace of
$\mathcal{X}$, $[\mathcal{X}]$ is a Banach space under the conventional norm

(3.5) $\|[\zeta]\|_{q}=\inf\\{\|\zeta-\zeta^{\prime}\|:\
\zeta^{\prime}\in\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}\\}.$

Let $K$ be a set in $\mathcal{X}$ and $[K]=\\{[\zeta]\in[\mathcal{X}]:\
\zeta\in K\\}$.

######  Lemma 3.9.

For any $\zeta\in\mathcal{X}$, there exists a unique $\zeta_{0}\in Y^{\perp}$
such that

(3.6) $[\zeta]=[\zeta_{0}],\ \ \|[\zeta]\|_{q}=\|\zeta_{0}\|\ \ \mbox{and}\ \
\|\zeta_{0}\|\leq\|\zeta\|.$

###### Proof.

This follows directly from the orthogonal decomposition $\mathcal{X}=Y\oplus
Y^{\perp}$. ∎

Let

$\mathcal{X}_{Y}=\\{f\in\mathcal{X}:\ Y\subset\mathrm{Ker}(f)\\}.$

It can be checked that $\mathcal{X}_{Y}$ is a closed subspace of
$\mathcal{X}$.

######  Lemma 3.10.

The space $\mathcal{X}_{Y}$ is isometrically isomorphic to the dual space
$[\mathcal{X}]^{\prime}$ of $([\mathcal{X}],\|\cdot\|_{q})$.

###### Proof.

For any $f\in\mathcal{X}_{Y}$, we define $F$ by

(3.7) $F([\zeta])=\langle f,\zeta\rangle\quad\forall\
[\zeta]\in[\mathcal{X}].$

If $[\zeta_{1}]=[\zeta_{2}]$, we have $\zeta_{1}-\zeta_{2}\in
Y\subset\mathrm{Ker}(f)$, which gives $F([\zeta_{1}])=F([\zeta_{2}])$ and $F$
is well defined. According to Lemma 3.9, $F\in[\mathcal{X}]^{\prime}$ and

(3.8) $\|F\|_{[\mathcal{X}]^{\prime}}=\|f\|.$

On the other hand, for any $F\in[\mathcal{X}]^{\prime}$, we can define $f$ by

(3.9) $\langle f,\zeta\rangle=F([\zeta])\quad\forall\ \zeta\in\mathcal{X}.$

It is obvious that $Y\subset\mathrm{Ker}(f)$. Again by Lemma 3.9, we have
$f\in\mathcal{X}_{Y}$ and (3.8) holds.

Define $T:\mathcal{X}_{Y}\rightarrow[\mathcal{X}]^{\prime}$ by

(3.10) $T:f\rightarrow F,$

where $F$ is defined by (3.7). By (3.7), (3.9) and (3.8), $T$ is an
isometrical isomorphism. This completes the proof. ∎

######  Lemma 3.11.

For any $f\in\mathcal{X}_{Y}$, let $F$ be defined as in (3.7). It holds
$[\mathrm{Ker}(f)]=\mathrm{Ker}(F)$.

###### Proof.

For any $[\zeta]\in[\mathrm{Ker}(f)]$, we can assume
$\zeta\in\mathrm{Ker}(f)$. It gives

$F([\zeta])=\langle f,\zeta\rangle=0,$

i.e., $[\zeta]\in\mathrm{Ker}(F)$, which yields
$[\mathrm{Ker}(f)]\subset\mathrm{Ker}(F)$.

On the other hand, for any $[\zeta]\in\mathrm{Ker}(F)$, we have
$F([\zeta])=0$, which gives $\langle f,\zeta\rangle=F([\zeta])=0.$ This means
$\zeta\in\mathrm{Ker}(f)$, which further implies $[\zeta]\in[\mathrm{Ker}(f)]$
and $\mathrm{Ker}(F)\subset[\mathrm{Ker}(f)]$. This completes the proof. ∎

######  Lemma 3.12.

Let $K$ be a convex set and $\zeta^{*}\in K$. Then
$f\in\mathcal{X}_{Y}\cap\mathcal{N}(\zeta^{*},K)$ if and only if
$T(f)\in\mathcal{N}([\zeta^{*}],[K])$, where $T$ is defined by (3.10) and

(3.11) $\mathcal{N}([\zeta^{*}],[K])=\\{\tilde{F}\in[\mathcal{X}]^{\prime}:\
-\tilde{F}([\zeta]-[\zeta^{*}])\geq 0\quad\forall\ [\zeta]\in[K]\\}.$

###### Proof.

This follows from the fact

$-T(f)([\zeta]-[\zeta^{*}])=-T(f)([\zeta-\zeta^{*}])=-\langle
f,\zeta-\zeta^{*}\rangle\quad\forall\ \zeta\in K.$

∎

######  Theorem 3.7.

Suppose that Assumption 3.6 holds. The proper Lagrange multiplier exists at
$u^{*}$ if and only if there exist
$\bar{\Lambda}\in\mathcal{N}([\zeta^{*}],[K])$ and
$\bar{\zeta}_{0}\in\overline{R(S)}$ such that

(3.12) $\left\\{\begin{aligned} &\bar{\Lambda}([\bar{\zeta}_{0}])>0,\\\
&\langle\lambda^{*},\bar{\zeta}_{0}\rangle>0.\end{aligned}\right.$

###### Proof.

If the proper Lagrange multiplier exists, by Theorem 3.5, there exists
$\tilde{\lambda}\in\mathcal{N}(\zeta^{*},K)$ satisfying (3.3) and (3.4). This
gives $\tilde{\lambda}\in\mathcal{X}_{Y}$ and we can define
$\bar{\Lambda}\in[\mathcal{X}]^{\prime}$ by $\tilde{\lambda}$ and (3.7). Since
$\tilde{\lambda}\in\mathcal{N}(\zeta^{*},K)$, by Lemma 3.12, we have
$\bar{\Lambda}\in\mathcal{N}([\zeta^{*}],[K])$. According to the consistency
condition (3.4) and the definition of $\bar{\Lambda}$, it holds

$\bar{\Lambda}([\bar{\zeta}_{0}])=\langle\tilde{\lambda},\
\bar{\zeta}_{0}\rangle>0.$

For the other direction, assume that such
$\bar{\Lambda}\in\mathcal{N}([\zeta^{*}],[K])$ exists. We can define
$\tilde{\lambda}$ by $\bar{\Lambda}$ and (3.9). By Lemma 3.12, (3.9) and
(3.12), we have

$\tilde{\lambda}\in\mathcal{X}_{Y},\ \
\tilde{\lambda}\in\mathcal{N}(\zeta^{*},K)\ \ \mbox{and}\ \
\langle\tilde{\lambda},\ \bar{\zeta}_{0}\rangle>0.$

Now by Theorem 3.5, we only need to prove that the compatibility condition
(3.3) holds.

Since $\tilde{\lambda}\in\mathcal{X}_{Y}$, we have

(3.13)
$\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}=Y\subset\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}.$

By (3.12), $[\overline{R(S)}]$ is not a subspace of
$\mathrm{Ker}(\bar{\Lambda})$, which implies

$\mathrm{Ker}(\bar{\Lambda})\cap[\overline{R(S)}]=[0],$

by $\dim([\overline{R(S)}])\leq 1$. Note that Lemma 3.11 gives

$[\mathrm{Ker}(\tilde{\lambda})]=\mathrm{Ker}(\bar{\Lambda}).$

Then we have

$[\mathrm{Ker}(\tilde{\lambda})]\cap[\overline{R(S)}]=\mathrm{Ker}(\bar{\Lambda})\cap[\overline{R(S)}]=[0].$

Since

$[\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}]\subset[\mathrm{Ker}(\tilde{\lambda})]\cap[\overline{R(S)}],$

it follows

$[\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}]=[0].$

This gives

$\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}\subset\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}.$

Together with (3.13), we have

$\mathrm{Ker}(\tilde{\lambda})\cap\overline{R(S)}=\mathrm{Ker}(\lambda^{*})\cap\overline{R(S)}.$

This completes the proof. ∎

For the uniqueness of the proper Lagrange multiplier, it follows from the
proof of Theorem 3.7 directly.

######  Theorem 3.8.

Suppose that Assumption 3.6 holds. The proper Lagrange multiplier is unique if
and only if there exists a unique (up to multiplication by a positive
constant) $\bar{\Lambda}\in\mathcal{N}([\zeta^{*}],[K])$ satisfying (3.12) for
some $\bar{\zeta}_{0}\in\overline{R(S)}$.

######  Remark 3.13.

Since the canonical map from $\mathcal{X}$ to $[X]$ is continuous, the open
mapping theorem indicates that if $K$ has an interior point, so does $[K]$.
But the other direction is not true. This is useful in exploring the existence
of the proper Lagrange multiplier.

The following examples are helpful in understanding the previous theoretical
results.

######  Example 3.14.

We consider the optimization problem

$\min\frac{1}{2}[(x_{1}-\alpha)^{2}+x_{2}^{2}]\ \ \mbox{s.t.}\ \ Sx\in K,$

where $\alpha\in\mathbb{R}$, $x\in\mathbb{R}^{2}$, $S=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}$ and $K\subset\mathbb{R}^{2}$ is a closed convex set.

  1. $\bullet$

Let $K=K_{1}$, where

$K_{1}=\\{(\zeta_{1},\zeta_{2})^{T}\in\mathbb{R}^{2}:\
\zeta_{1}^{2}+(\zeta_{2}-1)^{2}\leq 1\\}.$

In this case $K\cap R(S)=(0,0)^{T}$ and the feasible set is

$M=\\{(0,x_{2})^{T}\in\mathbb{R}^{2}:\ x_{2}\in\mathbb{R}\\}.$

The global minimizer of this problem is $x^{*}=(0,0)^{T}$. The gradient of the
objective function at this point is $(-\alpha,0)^{T}$ and the essential
Lagrange multiplier is $\lambda^{*}=\alpha$. Note that the solution of

$\begin{pmatrix}-\alpha\\\ 0\end{pmatrix}+S^{*}\lambda=0$

is $\lambda=(\alpha,\lambda_{2})^{T}$, for any $\lambda_{2}\in\mathbb{R}$.
Since the proper Lagrange multiplier must satisfy the above equation, we
assume that $\bar{\lambda}=(\alpha,\bar{\lambda}_{2})^{T}$ for some
$\bar{\lambda}_{2}\in\mathbb{R}$. Now we consider the condition

$-\langle\bar{\lambda},\zeta-\zeta^{*}\rangle\geq 0\quad\forall\ \zeta\in K,$

where $\zeta^{*}=Sx^{*}=(0,0)^{T}$. It is equivalent to

$-\alpha\zeta_{1}-\bar{\lambda}_{2}\zeta_{2}\geq 0\quad\forall\
\zeta=(\zeta_{1},\zeta_{2})^{T}\in K.$

If $\alpha=0$, we can choose $\bar{\lambda}_{2}=0$. If $\alpha\neq 0$, there
is no $\bar{\lambda}_{2}\in\mathbb{R}$ to satisfy the inequality. This means
that unless $\alpha=0$, the proper Lagrange multiplier does not exist for this
problem at the global minimizer. Note that if $\alpha\neq 0$, we have

$\mathcal{N}(\zeta^{*},K):=\\{(0,\lambda_{2})^{T}\in\mathbb{R}^{2}:\
\lambda_{2}\leq
0\\}\subset\mathrm{Ker}(S^{*}):=\\{(0,\lambda_{2})^{T}\in\mathbb{R}^{2}:\
\lambda_{2}\in\mathbb{R}\\}.$

Both the compatibility condition (3.3) and the consistency condition (3.4) are
not satisfied.

  2. $\bullet$

Let $K=K_{2}$, where

$K_{2}=\\{(\zeta_{1},\zeta_{2})^{T}\in\mathbb{R}^{2}:\
\zeta_{1}^{2}+(\zeta_{2}-1)^{2}\leq
1\\}\setminus\\{(\zeta_{1},\zeta_{2})^{T}\in\mathbb{R}^{2}:\
\zeta_{1}-\zeta_{2}\leq 0\\}.$

The feasible set $M$, the global minimizer and the essential Lagrange
multiplier are the same as those in the case $K=K_{1}$. As before, for the
proper Lagrange multiplier $\bar{\lambda}=(\alpha,\bar{\lambda}_{2})^{T}$ we
have

$-\alpha\zeta_{1}-\bar{\lambda}_{2}\zeta_{2}\geq 0\quad\forall\
\zeta=(\zeta_{1},\zeta_{2})^{T}\in K.$

On the other hand, the normal cone $\mathcal{N}(\zeta^{*},K)$ is given by

$\mathcal{N}(\zeta^{*},K)=\\{(\lambda_{1},\lambda_{2})^{T}\in\mathbb{R}^{2}:\
\lambda_{2}\leq 0,\ \lambda_{1}+\lambda_{2}\geq 0\\}.$

    1. (i)

If $\alpha<0$, $(\alpha,\bar{\lambda}_{2})^{T}\not\in\mathcal{N}(\zeta^{*},K)$
for any $\bar{\lambda}_{2}\in\mathbb{R}$. Therefore, the proper Lagrange
multiplier does not exist. Note that the condition (3.4) can not be satisfied,
while the condition (3.3) is always true for any
$0\neq\lambda\in\mathcal{N}(\zeta^{*},K)$.

    2. (ii)

If $\alpha>0$, $\bar{\lambda}$ is a proper Lagrange multiplier for any
$\bar{\lambda}_{2}\in[-\alpha,0]$. In this case, both the compatibility
condition (3.3) and the consistency condition (3.4) can be fulfilled.

Figure 1 gives an illustration of the results above.

$\zeta_{1}$$\zeta_{2}$No $\bar{\lambda}$ for $\alpha\neq
0$$K_{1}$$R(S)$$\zeta_{1}$$\zeta_{2}$No $\bar{\lambda}$ for
$\alpha<0$$R(S)$$K_{2}$$\zeta_{1}$$\zeta_{2}$$\bar{\lambda}=(\alpha,\bar{\lambda}_{2})^{T}$
for $\alpha\geq 0$$K_{2}$$R(S)$$\mathrm{Figure\ 1.}$

######  Example 3.15.

We consider the optimization problem

$\min\frac{1}{2}[(x_{1}-\alpha)^{2}+x_{2}^{2}]\ \ \mbox{s.t.}\ \ Sx\in K,$

where $\alpha>0$, $x\in\mathbb{R}^{2}$, $S=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}$ and $K\subset\mathbb{R}^{2}$ is a closed convex set. Let
$0\leq 2r<\alpha$.

  1. $\bullet$

Let $K=K_{3}$, where

$K_{3}=\\{(\zeta_{1},\zeta_{2})^{T}\in\mathbb{R}^{2}:\
(\zeta_{1}-r)^{2}+\zeta_{2}^{2}\leq r^{2}\\}.$

The feasible set of this example is

$M=\\{(x_{1},x_{2})^{T}\in\mathbb{R}^{2}:\ 0\leq x_{1}\leq 2r\\}$

and the global minimizer of this problem is $x^{*}=(2r,0)^{T}$. The gradient
of the objective function at this point is $(2r-\alpha,0)^{T}$. The essential
Lagrange multiplier is $\lambda^{*}=\alpha-2r>0$ and
$\zeta^{*}=Sx^{*}=(2r,0)^{T}$. The proper Lagrange multiplier should be in the
form $\bar{\lambda}=(\alpha-2r,\lambda_{2})^{T}$ for some
$\lambda_{2}\in\mathbb{R}$. By the condition

$-\langle\bar{\lambda},\zeta-\zeta^{*}\rangle\geq 0\quad\forall\ \zeta\in K,$

or equivalently

$-(\alpha-2r)(\zeta_{1}-2r)-\lambda_{2}\zeta_{2}\geq 0\quad\forall\
\zeta=(\zeta_{1},\zeta_{2})^{T}\in K,$

we have

$\lambda_{2}=0.$

Hence the proper Lagrange multiplier for this example is
$\bar{\lambda}=(\alpha-2r,0)^{T}$ and it is unique. Note that Robinson’s
condition (4.2) holds in this case.

  2. $\bullet$

Let $K=K_{4}$, where

$K_{4}=\\{(\zeta_{1},\zeta_{2})^{T}\in\mathbb{R}^{2}:\
(\zeta_{1}-r)^{2}+\zeta_{2}^{2}\leq r^{2},\ \ \zeta_{2}\geq 0\\}.$

The global minimizer and the essential Lagrange multiplier are the same as in
the previous case. In this case, the proper Lagrange multiplier
$\bar{\lambda}=(\alpha-2r,\bar{\lambda}_{2})^{T}$ for any
$\bar{\lambda}_{2}\leq 0$, which is not unique. Note that Robinson’s condition
(4.2) is not true in this case.

Figure 2 gives an illustration of the results above.

$\zeta_{1}$$\zeta_{2}$$K_{3}$$\bar{\lambda}=(\alpha-2r,0)^{T}$$R(S)$$\zeta_{1}$$\zeta_{2}$$R(S)$$K_{4}$$\bar{\lambda}=(\alpha-2r,\bar{\lambda}_{2})^{T}$$\mathrm{Figure\
2.}$

######  Example 3.16.

We consider the problem (cf. [11, Section 8.4])

$\min\limits_{u\in H_{0}^{1}(\Omega)}\int_{\Omega}\nabla u\cdot\nabla u\
dx\quad\mbox{s.t.}\ \ \|u\|_{L^{2}(\Omega)}^{2}=1.$

Let

$G(u)=\|u\|_{L^{2}(\Omega)}^{2}\quad\mbox{and}\quad\mathcal{K}=\\{1\\}\subset\mathbb{R}.$

For this problem, we have

$T_{w}(M,u)=T(M,u)=\overline{L(\mathcal{K},u)}$

for any $u$. Note that $G^{\prime}(u)$ is bounded from $H_{0}^{1}(\Omega)$ to
$\mathbb{R}$ and $R(G^{\prime}(u))=\mathbb{R}$. According to our theory, for a
global minimizer $u^{*}\in H_{0}^{1}(\Omega)$, the essential (or proper)
Lagrange multiplier $\lambda^{*}\in\mathbb{R}$ exists and the following KKT
system holds

$\int_{\Omega}\nabla u^{*}\cdot\nabla v\ dx-\lambda^{*}\int_{\Omega}u^{*}v\
dx\quad\forall\ v\in
H_{0}^{1}(\Omega)\quad\mbox{and}\quad\|u^{*}\|_{L^{2}(\Omega)}=1.$

This is actually the weak form of the following eigenvalue problem

$\left\\{\begin{aligned} -\Delta u^{*}&=\lambda^{*}u^{*}\quad\mbox{in}\ \
\Omega;\\\ u^{*}&=0\quad\ \ \ \mbox{on}\ \ \partial\Omega;\\\
\|u^{*}\|_{L^{2}(\Omega)}&=1.\end{aligned}\right.$

###  3.3. The convergence of multipliers of the classical ALM

The classical ALM for the model problem is given in (A.3). We will give an
essential characterization of the convergence of the multipliers generated by
the algorithm.

######  Theorem 3.9.

Let $\\{\lambda^{k}\\}_{k=1}^{+\infty}$ be the multipliers generated in the
classical ALM (A.3).

  1. (1)

If the essential Lagrange multiplier $\lambda^{*}$ exists at $u^{*}$, then

(3.14)
$\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=\langle\lambda^{*},Sv\rangle\quad\forall\
v\in\mathcal{U}.$

  2. (2)

If the restriction of $\\{\lambda^{k}\\}_{k=1}^{+\infty}$ to $\overline{R(S)}$
weakly converges in $\overline{R(S)}$ to some element
$\lambda^{*}\in\overline{R(S)}$, then $\lambda^{*}$ is the essential Lagrange
multiplier at $u^{*}$.

###### Proof.

Note that $\\{\lambda^{k}\\}_{k=1}^{+\infty}$ satisfies (1.3). The first part
follows from (1.3) and Definition 1.1 directly.

For the second part, if the restriction of $\\{\lambda^{k}\\}_{k=1}^{+\infty}$
to $\overline{R(S)}$ weakly converges in $\overline{R(S)}$ to some element
$\lambda^{*}\in\overline{R(S)}$, i.e.,

$\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=\langle\lambda^{*},Sv\rangle\quad\forall\
v\in\mathcal{U},$

then we have

$\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\lambda^{*},Sv\rangle=\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=0\quad\forall\
v\in\mathcal{U}$

and

$-\langle\lambda^{*},Sv-
Su^{*}\rangle=-\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv-
Su^{*}\rangle\geq-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv-Su^{*}\rangle\geq
0\quad\forall\ Sv\in K,$

by (1.3). Therefore, $\lambda^{*}$ is the essential Lagrange multiplier by the
boundedness of $\lambda^{*}$ and we finish the proof.

∎

The following corollary follows from Theorem 3.3 and Theorem 3.9 directly.

######  Corollary 3.17.

If $R(S)$ is closed in $\mathcal{X}$, then the restriction of
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ to $R(S)$ always weakly converges in
$R(S)$ to the essential Lagrange multiplier $\lambda^{*}$ at $u^{*}$.

######  Example 3.18.

Let us consider the optimization problem

$\min\frac{1}{2}x^{2}\quad\mbox{s.t.}\ \left\\{\begin{aligned} x&=1;\\\
2x&=2.\end{aligned}\right.$

The global minimizer of this problem is $x^{*}=1$ and the essential Lagrange
multiplier at $x^{*}$ is

$\lambda^{*}=-\frac{1}{5}\begin{pmatrix}1\\\ 2\end{pmatrix}.$

All the proper Lagrange multipliers of this example define the set

$\Lambda=\\{\bar{\lambda}=(\lambda_{1},\lambda_{2})^{T}\in\mathbb{R}^{2}:\
\lambda_{1}+2\lambda_{2}+1=0\\}.$

The iterators of the classical ALM satisfy

$\left\\{\begin{aligned} x^{k+1}-1&=\frac{1}{1+5\beta}(x^{k}-1);\\\
\lambda^{k+1}+\frac{1}{5}\begin{pmatrix}1\\\
2\end{pmatrix}&=J\left[\lambda^{k}+\frac{1}{5}\begin{pmatrix}1\\\
2\end{pmatrix}\right];\\\
\lambda^{k+1}_{1}+2\lambda^{k+1}_{2}+1&=\frac{1}{1+5\beta}(\lambda^{k}+2\lambda^{k}+1),\end{aligned}\right.$

where $k=1,2,\dots$,

$J=\frac{1}{1+5\beta}\begin{pmatrix}1+4\beta&-2\beta\\\
-2\beta&1+\beta\end{pmatrix}$

and $\beta>0$.

Note that $J$ is symmetric, positive definite and $\det J=1$, which implies
$\rho(J)>1$. We have the following convergence results.

  * •

$\\{x^{k}\\}_{k=1}^{+\infty}$ converges to $x^{*}=1$.

  * •

$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ is unbounded.

  * •

$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ converges to the essential Lagrange
multiplier in $R(S)$, i.e.,

$\lambda^{k}_{1}+2\lambda^{k}_{2}\rightarrow-1=-\frac{1}{5}(1\times 1+2\times
2)\ \ \mbox{as}\ \ k\rightarrow+\infty.$

  * •

The distance between $\lambda^{k}\ (k=1,2,\dots)$ and $\Lambda$ converges to
$0$.

##  4\. Revisiting the weak form asymptotic KKT system

In the previous arguments we notice that in the infinite-dimensional case, the
essential Lagrange multiplier may not exist. Here we derive a sufficient
condition to guarantee the existence of the essential Lagrange multiplier by
the W-AKKT system (1.3). We will also use the W-AKKT system (1.3) to give a
proof of the existence of the proper Lagrange multiplier under Robinson’s
condition (cf. [18, p. 5], [28, 31, 42] or (4.2)). A sufficient condition,
which includes Robinson’s condition as a special case, to guarantee the
existence of the proper Lagrange multiplier is also given. It is worth noting
that all these results also give the convergence results of the multipliers
generated by the classical ALM for the model problem.

According to the W-AKKT system (1.3), we have

$\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=0\quad\forall\
v\in\mathcal{U}.$

A linear functional $\tilde{\lambda}$ on $\mathrm{R}(S)\subset\mathcal{X}$ can
be defined as

$\tilde{\lambda}(\zeta)=\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta\rangle\quad\forall\
\zeta\in\mathrm{R}(S).$

The second condition leads to

$\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta\rangle\leq\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta^{*}\rangle\
\ \Longleftrightarrow\ \
\tilde{\lambda}(\zeta)\leq\tilde{\lambda}(\zeta^{*})\quad\forall\ \zeta\in
K\cap R(S).$

If $\tilde{\lambda}$ can be extended to a bounded linear functional in
$\overline{R(S)}^{\prime}$, then we have the essential Lagrange multiplier.
Therefore, the existence of the essential Lagrange multiplier is equivalent to
the boundedness of the linear functional $\tilde{\lambda}$ in
$\overline{\mathrm{R}(S)}^{\prime}$.

If the topology of $\mathcal{X}$ is too weak to guarantee the boundedness of
$\tilde{\lambda}$, we can set the problem on a small subspace of
$\mathcal{X}$, where a stronger topology can be equipped on the subspace. More
precisely, we can set the problem on a subspace $\tilde{\mathcal{X}}$ where
$\mathrm{R}(S)\subseteq\tilde{\mathcal{X}}\subseteq\mathcal{X}$. In this case,
we can define a strong topology on $\tilde{\mathcal{X}}$, which is stronger
than that of $\mathcal{X}$, and the stronger topology will make it easier to
guarantee the boundedness of $\tilde{\lambda}$. Motivated by these arguments,
we give the following results.

######  Theorem 4.1.

Let $\|\cdot\|_{*}$ be a norm on $R(S)$ such that $(R(S),\|\cdot\|_{*})$ is a
Banach space which is continuously embedded into $(\mathcal{X},\|\cdot\|)$.
There exists a unique essential Lagrange multiplier
$\tilde{\lambda}^{*}\in(R(S),\|\cdot\|_{*})^{\prime}$ which is the dual space
of $(R(S),\|\cdot\|_{*})$ such that the following KKT system holds, i.e.,

(4.1) $\left\\{\begin{aligned} &\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\tilde{\lambda}^{*},Sv\rangle_{*}=0\quad\forall\
v\in\mathcal{U};\\\ &-\langle\tilde{\lambda}^{*},\zeta-Su^{*}\rangle_{*}\geq
0\quad\forall\ \zeta\in\overline{K\cap R(S)}^{\|\cdot\|_{*}},\\\
\end{aligned}\right.$

where $\langle\cdot,\cdot\rangle_{*}$ is the duality pair of
$(R(S),\|\cdot\|_{*})^{\prime}$ and $(R(S),\|\cdot\|_{*})$.

###### Proof.

Since $(R(S),\|\cdot\|_{*})$ is continuously embedded into
$(\mathcal{X},\|\cdot\|)$ and $\\{\lambda^{k}\\}\subseteq\mathcal{X}$,
$\\{\lambda^{k}\\}\subseteq(R(S),\|\cdot\|_{*})^{\prime}$. By the first
condition in the W-AKKT system (1.3), we know

$\sup\\{|\langle\lambda^{k},\zeta\rangle_{*}|:\
k=1,\dots,+\infty\\}<+\infty\quad\forall\ \zeta\in R(S).$

According to the uniform boundedness principle (cf. [36]), we know that
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ is bounded in
$(R(S),\|\cdot\|_{*})^{\prime}$. Note that the unit ball in
$(R(S),\|\cdot\|_{*})^{\prime}$ is weak-* compact. Again, by the first
condition in the W-AKKT system (1.3), we know that there exists
$\tilde{\lambda}^{*}\in(R(S),\|\cdot\|_{*})^{\prime}$ such that (4.1) holds.
The uniqueness follows from (4.1) directly. ∎

The W-AKKT system (1.3) can also be used to derive the existence of the proper
Lagrange multiplier under Robinson’s condition (cf. [18, p. 5] and [28, 31,
42]), which is

(4.2) $\mathcal{X}=S(\mathcal{U})-\mathcal{K}_{K,\zeta^{*}},$

where

$\mathcal{K}_{K,\zeta^{*}}=\\{t(\zeta-\zeta^{*}):\ \forall\ \zeta\in K\ \
\mbox{and}\ \ \forall\ t\geq 0\\}.$

The following results cover the results under Robinson’s condition and the
proof below is based on the generalized open mapping theorem (cf. [18, Theorem
1.4], [28, Lemma 2.3] or [42, Theorem 2.1]).

######  Theorem 4.2.

Let $(\tilde{\mathcal{X}},\|\cdot\|_{*})$ be a Banach space which is
continuously embedded into $(\mathcal{X},\|\cdot\|)$. Assume that the
following condition holds for $(\tilde{\mathcal{X}},\|\cdot\|_{*})$ at
$(u^{*},\zeta^{*})=(u^{*},Su^{*})$:

(4.3) $\tilde{\mathcal{X}}=S(\mathcal{U})-\mathcal{K}_{K,\zeta^{*}}.$

Then there exists a proper Lagrange multiplier
$\tilde{\lambda}\in(\tilde{\mathcal{X}},\|\cdot\|_{*})^{\prime}$ such that

(4.4) $\left\\{\begin{aligned} &\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\tilde{\lambda},Sv\rangle_{*}=0\quad\forall\
v\in\mathcal{U};\\\ &-\langle\tilde{\lambda},\zeta-Su^{*}\rangle_{*}\geq
0\quad\forall\ \zeta\in K,\\\ \end{aligned}\right.$

where $\langle\cdot,\cdot\rangle_{*}$ is the duality pair of
$(\tilde{X},\|\cdot\|_{*})^{\prime}$ and $(\tilde{X},\|\cdot\|_{*})$.

###### Proof.

Since $(\tilde{\mathcal{X}},\|\cdot\|_{*})$ is continuously embedded into
$(\mathcal{X},\|\cdot\|)$, we have
$\\{\lambda^{k}\\}\subseteq(\tilde{\mathcal{X}},\|\cdot\|_{*})^{\prime}$. By
(4.3) and the generalized open mapping theorem (cf. [18, Theorem 1.4], [28,
Lemma 2.3] or [42, Theorem 2.1]), there exists $r>0$ such that

$\overline{B_{r,\tilde{\mathcal{X}}}}\subseteq
S(B_{1,\mathcal{U}})-\mathcal{K}_{K,\zeta^{*}}\cap B_{1,\tilde{\mathcal{X}}},$

where $B_{\rho,V}$ is the ball in $V$ with center at the original and radius
$\rho>0$, $V=\tilde{\mathcal{X}}$ or $\mathcal{U}$, $\rho=r$ or $1$ and
$\overline{B_{\rho,V}}$ is its closure in $V$.

Let $\\{x^{k}\\}\subseteq\tilde{\mathcal{X}}$ be a sequence of unit vectors
such that

$\langle\lambda^{k},x^{k}\rangle_{*}\geq\frac{1}{2}\|\lambda^{k}\|_{(\tilde{X},\|\cdot\|_{*})^{\prime}}.$

Then

$-rx^{k}=Sv^{k}-t_{k}(\zeta^{k}-\zeta^{*}),$

where $\\{v^{k}\\}\subseteq B_{1,\mathcal{U}}$,
$\\{t_{k}(\zeta^{k}-\zeta^{*})\\}\subseteq\mathcal{K}_{K,\zeta^{*}}\cap
B_{1,\tilde{\mathcal{X}}}$ and $t_{k}\geq 0,\ \ \forall\ k=1,\dots,\infty$.

Note that

$\langle
D_{u}\theta(u^{*}),v^{k}\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv^{k}\rangle_{*}=0\quad\forall\
k=1,\dots,\infty$

and

$-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},t_{k}(\zeta^{k}-\zeta^{*})\rangle_{*}\geq
0\quad\forall\ k=1,\dots,\infty.$

Hence, it holds at least for a subsequence that for any $\epsilon>0$ there
exists $N_{\epsilon}>0$ such that for any $k>N_{\epsilon}$,

$-\langle\lambda^{k},Sv^{k}\rangle_{*}+\langle\lambda^{k},t_{k}(\zeta^{k}-\zeta^{*})\rangle_{*}\leq\epsilon+|\langle
D_{u}\theta(u^{*}),v^{k}\rangle_{\mathcal{U}}|\leq\epsilon+\|D_{u}\theta(u^{*})\|_{\mathcal{U}}<+\infty.$

Therefore, for a fixed $\epsilon>0$ and any $k>N_{\epsilon}$, we have

$\displaystyle\frac{r}{2}\|\lambda^{k}\|_{(\tilde{X},\|\cdot\|_{*})^{\prime}}$
$\displaystyle\leq\langle\lambda^{k},rx^{k}\rangle_{*}=-\langle\lambda^{k},Sv^{k}-t_{k}(\zeta^{k}-\zeta^{*})\rangle_{*}<+\infty.$

It follows that $\\{\lambda^{k}\\}_{k=1}^{+\infty}$ is bounded in
$(\tilde{\mathcal{X}},\|\cdot\|_{*})^{\prime}$. There exists a subsequence of
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ which weak-* converges to some element
$\tilde{\lambda}$ in $(\tilde{\mathcal{X}},\|\cdot\|_{*})^{\prime}$. According
to the W-AKKT (1.3), (4.4) is valid and we complete the proof. ∎

Now we give a proof of the existence of the proper Lagrange multiplier under
the condition (4.2) based on the W-AKKT system (1.3) and the uniform
boundedness principle that has an elementary proof in [36].

######  Lemma 4.1.

The W-AKKT system (1.3) is equivalent to

(4.5) $\langle D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-
Su^{*})\rangle\leq 0\quad\forall\ v\in\mathcal{U},\ \forall\ \zeta\in K,\ \
t\geq 0.$

###### Proof.

It is obvious that the W-AKKT system (1.3) implies (4.5).

Now we prove that if (4.5) holds, so does the W-AKKT system (1.3). Taking
$t=0$, we have

$\langle D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle\leq
0\quad\forall\ v\in\mathcal{U}.$

If we change $v$ to $-v$, we have

$\langle D_{u}\theta(u^{*}),-v\rangle_{\mathcal{U}}-\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=\langle
D_{u}\theta(u^{*}),-v\rangle_{\mathcal{U}}+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},S(-v)\rangle\leq 0.$

These two inequalities give

$\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle-\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle\leq 0.$

On the other hand, by the definitions of $\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}$ and $\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}$, we have

$\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle\geq\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle.$

Therefore,

$\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle$

and

$\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=0\quad\forall\
v\in\mathcal{U}.$

By taking $v=0$ and $t=1$, we have

$-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta-Su^{*}\rangle\geq
0\quad\forall\ \zeta\in K.$

This completes the proof. ∎

Let $\zeta^{*}=Su^{*}$ and

$\mathcal{C}(S,\zeta^{*},K)=\\{Sv+t(\zeta-\zeta^{*})\in\mathcal{X}:\ \forall\
v\in\mathcal{U},\ \forall\ \zeta\in K\ \ \mbox{and}\ \ t\geq 0\\}.$

Note that $\mathcal{C}(S,\zeta^{*},K)$ is a convex cone in $\mathcal{X}$.

######  Theorem 4.3.

Assume that there exists a norm $\|\cdot\|_{*}$ such that
$(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})$ is a Banach space and it is
continuously embedded into $(\mathcal{X},\|\cdot\|)$. Then there exists
$\tilde{\lambda}\in(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})^{\prime}$ such
that

(4.6) $\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\tilde{\lambda},Sv+t(\zeta-\zeta^{*})\rangle_{*}\leq
0\quad\forall\ v\in\mathcal{U},\ \forall\ \zeta\in K\ \ \mbox{and}\ \ t\geq
0,$

or equivalently,

(4.7) $\left\\{\begin{aligned} &\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\tilde{\lambda},Sv\rangle_{*}=0\quad\forall\
v\in\mathcal{U};\\\ &-\langle\tilde{\lambda},\zeta-Su^{*}\rangle_{*}\geq
0\quad\forall\ \zeta\in K,\end{aligned}\right.$

where $\langle\cdot,\cdot\rangle_{*}$ is the duality pair of
$(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})^{\prime}$ and
$(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})$.

###### Proof.

Let $\\{\lambda^{k}\\}_{k=1}^{+\infty}$ be the same as that in the W-AKKT
system (1.3). Since $(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})$ is a Banach
space and it is continuously embedded into $(\mathcal{X},\|\cdot\|)$,
$\\{\lambda^{k}\\}\subset(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})^{\prime}$.

Since $\mathcal{C}(S,\zeta^{*},K)$ is a linear space, for any
$v\in\mathcal{U}$, $\zeta\in K$ and $t\geq 0$, we have

$-[Sv+t(\zeta-\zeta^{*})]\in\mathcal{C}(S,\zeta^{*},K),$

i.e., there exist $w\in\mathcal{U}$, $\bar{\zeta}\in K$ and $s\geq 0$ such
that

$-[Sv+t(\zeta-\zeta^{*})]=Sw+s(\zeta-\zeta^{*}).$

Therefore, by (4.5), we have

$\langle D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-\zeta^{*})\rangle\leq
0$

and

$\displaystyle\quad\ \langle
D_{u}\theta(u^{*}),w\rangle_{\mathcal{U}}-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-\zeta^{*})\rangle$
$\displaystyle\leq\langle
D_{u}\theta(u^{*}),w\rangle_{\mathcal{U}}-\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-\zeta^{*})\rangle$
$\displaystyle=\langle
D_{u}\theta(u^{*}),w\rangle_{\mathcal{U}}+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},-[Sv+t(\zeta-\zeta^{*})]\rangle$
$\displaystyle=\langle
D_{u}\theta(u^{*}),w\rangle_{\mathcal{U}}+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sw+s(\bar{\zeta}-\zeta^{*})\rangle$
$\displaystyle\leq 0,$

which follows

$\langle D_{u}\theta(u^{*}),w\rangle_{\mathcal{U}}\leq\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-\zeta^{*})\rangle\leq-\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}.$

Furthermore, we have

$|\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-\zeta^{*})\rangle|\leq\|D_{u}\theta(u^{*})\|_{\mathcal{U}}(\|v\|_{\mathcal{U}}+\|w\|_{\mathcal{U}})<+\infty\quad\forall\
v\in\mathcal{U},\ \forall\zeta\in K.$

It follows that there exists a subsequence of
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ (which will still be denoted by
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$), such that

$|\langle\lambda^{k},\eta\rangle_{*}|=|\langle\lambda^{k},\eta\rangle|<+\infty\quad\forall\
\eta\in\mathcal{C}(S,\zeta^{*},K),\ \forall\ k=1,2,\dots.$

According to the uniform boundedness principle,
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ is bounded in
$(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})^{\prime}$. Then there exists a
subsequence of $\\{\lambda^{k}\\}_{k=1}^{+\infty}$ (which will still be
denoted by $\\{\lambda^{k}\\}_{k=1}^{+\infty}$) converges to
$\tilde{\lambda}\in(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})^{\prime}$ in the
weak-* topology of $(\mathcal{C}(S,\zeta^{*},K),\|\cdot\|_{*})^{\prime}$.
Therefore,

$\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},\eta\rangle=\langle\tilde{\lambda},\eta\rangle_{*}\quad\forall\
\eta\in\mathcal{C}(S,\zeta^{*},K).$

By (4.5), we have (4.6) holds. By taking $t=0$ and $v=0$, we can get the
equivalence of (4.6) and (4.7). This completes the proof.

∎

######  Remark 4.2.

Note that $\mathcal{X}=\mathcal{C}(S,\zeta^{*},K)$, which means Robinson’s
condition (4.2) holds, is a special case of the results in Theorem 4.3.

######  Remark 4.3.

If there exists $\lambda_{0}$ such that

$\langle\lambda_{0},Sv+t(\zeta-\zeta^{*})\rangle\leq\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv+t(\zeta-\zeta^{*})\rangle\quad\forall\
v\in\mathcal{U},\ \forall\ \zeta\in K\ \ \mbox{and}\ \ t\geq 0,$

then we have

$\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\langle\lambda_{0},Sv+t(\zeta-
Su^{*})\rangle\leq 0,\quad\forall\ v\in\mathcal{U},\ \forall\ \zeta\in K\ \
\mbox{and}\ \ t\geq 0.$

This implies $\lambda_{0}$ is a proper Lagrange multiplier.

Therefore, every weak accumulation point (if exists) of
$\\{\lambda^{k}\\}_{k=1}^{+\infty}$ is a proper Lagrange multiplier, and
furthermore, the weak convergence of the multipliers generated by the
classical ALM in $\mathcal{X}$ implies the existence of a proper Lagrange
multiplier. However, the existence of proper Lagrange Lagrange multipliers can
not guarantee the convergence of the multipliers generated by the classical
ALM in $\mathcal{X}$, see e.g., Example 3.18 in this paper.

This will be also useful for developing new constrained qualification
conditions to guarantee the existence of the proper Lagrange multiplier.

##  5\. Optimal control problems with pointwise constraints

In this section we apply our theory to optimal control problems with pointwise
state or control constraints. We will discuss the existence of Lagrange
multipliers for optimal control problems with pointwise state or control
constraints.

###  5.1. Optimal control problems with pointwise state constraints

We consider the following optimal control problem with pointwise state
constraints

$\displaystyle\min\limits_{u\in L^{2}(\Omega),y\in Y_{ad}}\
J(y,u)=\frac{1}{2}\|y-y_{d}\|_{L^{2}(\Omega)}^{2}+\frac{\alpha}{2}\|u\|_{L^{2}(\Omega)}^{2}$

subject to

$\left\\{\begin{aligned} -\Delta y&=u\quad\ \mbox{in}\ \Omega,\\\ y&=0\
\quad\mbox{on}\ \partial\Omega,\\\ \end{aligned}\right.$

where $u\in L^{2}(\Omega)$ is the control variable, $y_{d}\in L^{2}(\Omega)$
is the desired state or observation, $\alpha>0$ is the regularization
parameter and $Y_{ad}$ is the set of the state constraints. For simplicity, we
assume that $\Omega\subset\mathbb{R}^{2}$ is a convex domain in the following
arguments. The extension to general cases is straightforward.

We have two different state constraint sets $Y_{ad,1}$ and $Y_{ad,2}$, where

$Y_{ad,1}=\\{y\in C(\bar{\Omega}):\ a(x)\leq y(x)\leq b(x)\ \ \mbox{a.e. in}\
\Omega\\},$ $Y_{ad,2}=\\{y\in L^{2}(\Omega):\ a(x)\leq y(x)\leq b(x)\ \
\mbox{a.e. in}\ \Omega\\}$

and $a(x),b(x)\in L^{\infty}(\Omega)$. We refer the case $Y_{ad}=Y_{ad,1}$ as
the continuous setting and the case $Y_{ad}=Y_{ad,2}$ as the $L^{2}$ setting.
In both cases we can prove the existence and uniqueness of the solution to the
optimal control problem.

Let $(\bar{u},\bar{y})\in L^{2}(\Omega)\times Y_{ad,2}$ be the solution of

$\min\limits_{u\in L^{2}(\Omega),y\in Y_{ad,2}}\ J(y,u).$

By the regularity results of elliptic partial differential equations (cf.
[14]), we have

$\bar{y}\in C(\bar{\Omega}),$

which implies

$\bar{y}\in Y_{ad,1}.$

Since $Y_{ad,1}\subset Y_{ad,2}$, we have

$J(\bar{y},\bar{u})=\min\limits_{u\in L^{2}(\Omega),y\in Y_{ad,2}}\
J(y,u)\leq\min\limits_{u\in L^{2}(\Omega),y\in Y_{ad,1}}\ J(y,u)\leq
J(\bar{y},\bar{u}).$

It gives

$\min\limits_{u\in L^{2}(\Omega),y\in Y_{ad,2}}\ J(y,u)=\min\limits_{u\in
L^{2}(\Omega),y\in Y_{ad,1}}\ J(y,u),$

which yields that $(\bar{u},\bar{y})$ is also the unique solution of the
problem in the continuous setting.

In general, one can not expect the existence of a proper Lagrange multiplier
in $L^{2}(\Omega)$, see e.g., Example 7.1 in [7].

In the continuous setting, we can obtain a multiplier in the measure space if
Slater’s condition is satisfied (cf. [8]), which implies Robinson’s condition
(cf. [17, Theorem 1.55]) in this case.

In the $L^{2}$ setting, we can not get the existence results of the
multipliers from the classical optimization theory. Note that there are not
interior points in $Y_{ad,2}$ and Slater’s condition can not be imposed.

###  5.2. A general problem in the form of the model problem

Let us consider the following general optimal control problem with state
and/or control constraints

$\displaystyle\min\limits_{(u,y)\in\mathcal{U}_{ad}\times\mathcal{Y}_{ad}}\
J(y,u)=\frac{1}{2}\|y-y_{d}\|_{\mathcal{H}}^{2}+\frac{\alpha}{2}\|u\|_{\mathcal{U}}^{2}$

subject to

$y=\mathcal{L}u,$

where $\mathcal{H}$, $\mathcal{U}$ are two real Hilbert spaces, $\mathcal{L}$
is a bounded operator from $\mathcal{U}$ to $\mathcal{W}\
(\subset\mathcal{H})$ and $\mathcal{U}_{ad}$, $\mathcal{Y}_{ad}$ are two
closed and convex sets in $\mathcal{U}$ and $\mathcal{H}$, respectively.

Let

$\theta(u)=J(\mathcal{L}u,u)$

and set

$\mathcal{L}u=z,\quad u=w,\quad Su=\begin{pmatrix}\mathcal{L}u\\\
u\end{pmatrix}\quad\mbox{and}\quad\zeta=\begin{pmatrix}z\\\ w\end{pmatrix}.$

We also denote

$K=\mathcal{Y}_{ad}\times\mathcal{U}_{ad}.$

With the help of these notations, we can reformulate the general optimal
control problem in the desired form

$\min\limits_{u\in\mathcal{U}}\theta(u)\quad\mbox{s.t.}\ \ Su\in K,$

where $K\subset\mathcal{X}=\mathcal{W}\times\mathcal{U}$. Note that in this
case $\theta(u)$ is strongly convex on $\mathcal{U}$ and $c_{0}$ can be chosen
as $\alpha$.

If we take $\mathcal{W}=\mathcal{H}=\mathcal{U}=L^{2}(\Omega)$,
$\mathcal{L}=\Delta^{-1}:L^{2}(\Omega)\mapsto H_{0}^{1}(\Omega)\hookrightarrow
L^{2}(\Omega)$, $\mathcal{Y}_{ad}=Y_{ad,2}$ and
$\mathcal{U}_{ad}=\mathcal{U}$, this is the problem with pointwise state
constraints in the $L^{2}$ setting. In this case, $S^{*}=(\Delta^{-1},I_{d})$
where $I_{d}$ is the identity operator on $L^{2}(\Omega)$. Since
$\mathrm{R}(\Delta^{-1})$ is dense in $L^{2}(\Omega)$, by Theorem 3.2, in
general we can not obtain an essential Lagrange multiplier (which is also a
proper Lagrange multiplier) in $L^{2}(\Omega)$ for the state constraints in
this case. This theoretically clarifies the existence results of Lagrange
multipliers in $L^{2}(\Omega)$ for optimal control problems with pointwise
state constraints.

Our theory also indicates that in the case with only pointwise control
constraints, i.e., $\mathcal{W}=\mathcal{H}=\mathcal{U}=L^{2}(\Omega)$,
$\mathcal{L}=\Delta^{-1}$, $\mathcal{Y}_{ad}=\mathcal{W}$ and
$\mathcal{U}_{ad}=\\{u\in L^{2}(\Omega):\ u_{a}(x)\leq u(x)\leq u_{b}(x)\ \
\mbox{a.e.}\ x\in\Omega\\}$, a proper Lagrange multiplier in $L^{2}(\Omega)$
always exists for the control constraints, since the identity operator related
to the control variable is closed. The same results can be obtained by a
direct construction (cf. [17, 39]) or the results under Robinson’s condition
by the classical optimization theory (cf. [39]). Our theory can guarantee this
naturally.

######  Remark 5.1.

Let

$W=\\{y\in H_{0}^{1}(\Omega):\ \Delta y\in L^{2}(\Omega)\\}.$

$W$ is a Hilbert space under the following norm (cf. [14, p. 59])

$\|\varphi\|_{W}^{2}=\|\varphi\|_{H^{1}(\Omega)}^{2}+\|\Delta\varphi\|_{L^{2}(\Omega)}^{2}.$

Note that $W\hookrightarrow C(\bar{\Omega})$ and

$(\Delta)^{-1}\ :\ L^{2}(\Omega)\rightarrow W$

is onto, i.e., $R((\Delta)^{-1})=W$.

If we take

$Y_{ad,3}=\\{y\in W:\ a(x)\leq y(x)\leq b(x)\ \mbox{a.e.\ in}\ \Omega\\},$

then we have the same solution $(\bar{u},\bar{y})$ of the optimal control
problem with pointwise state constraints as in the continuous and $L^{2}$
settings. In this case, by Theorem 3.3 or Theorem 4.1, we have the existence
of the essential Lagrange multiplier $\lambda^{*}$ in $W^{\prime}$.

##  6\. Concluding remarks

In this paper, we systematically developed a new decomposition framework to
investigate Lagrange multipliers of the KKT system of constrained optimization
problems and variational inequalities in Hilbert spaces. Our new framework is
totally different from existing frameworks based on separation theorems. We
derived the weak form asymptotic KKT system and introduced the essential
Lagrange multiplier. The basic theory of the essential Lagrange multiplier has
been established in this paper. The existence theory of the essential Lagrange
multiplier shows essential differences in finite and infinite-dimensional
cases. Based on it, we also gave necessary and sufficient conditions for the
existence and uniqueness of the proper Lagrange multiplier. The results
theoretically confirm the necessity of using the asymptotic or approximate KKT
system in the infinite-dimensional case as well. We proved the convergence of
the classical augmented Lagrangian method without using the information of
Lagrange multipliers of the problem, and essentially characterized the
convergence properties of the multipliers generated by the classical augmented
Lagrangian method. Some sufficient conditions to guarantee the existence of
the essential Lagrange multiplier and the proper Lagrange multiplier were also
derived.

## Acknowledgement

##  Appendix A The proof of Theorem 1.2

We first prove that if a feasible point $u^{*}$ satisfies the W-AKKT system
(1.3), then $u^{*}$ is the global minimizer of the model problem (1.1).
According to the convex optimization theory (cf. [3, Proposition 26.5], [10,
Proposition 2.1, Chapter II] or [19, Corollary 4.20]), it suffices to show
that

$\langle D_{u}\theta(u^{*}),v-u^{*}\rangle_{\mathcal{U}}\geq 0\quad\ \forall\
Sv\in K.$

Since $u^{*}$ satisfies the W-AKKT system (1.3), there exists
$\\{\lambda^{k}\\}\subset\mathcal{X}$ such that

$\langle
D_{u}\theta(u^{*}),v-u^{*}\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle
S^{*}\lambda^{k},v-u^{*}\rangle=0\quad\forall\ v\in\mathcal{U}$

and

$-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta-Su^{*}\rangle\geq
0\quad\forall\ \zeta\in K.$

Therefore,

$-\operatorname*{lim\ sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv-
Su^{*}\rangle\geq 0\quad\forall\ Sv\in K,$

i.e.,

$-\operatorname*{lim\ sup}\limits_{k\rightarrow+\infty}\langle
S^{*}\lambda^{k},v-u^{*}\rangle_{\mathcal{U}}\geq 0\quad\forall\ Sv\in K.$

It follows

(A.1) $\displaystyle\langle D_{u}\theta(u^{*}),v-u^{*}\rangle_{\mathcal{U}}$
$\displaystyle=-\lim\limits_{k\rightarrow+\infty}\langle
S^{*}\lambda^{k},v-u^{*}\rangle_{\mathcal{U}}$
$\displaystyle\geq-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle
S^{*}\lambda^{k},v-u^{*}\rangle_{\mathcal{U}}\geq 0\quad\forall\ Sv\in K,$

which implies that $u^{*}$ is the global minimizer of the model problem (1.1).

For the other direction, we will use the classical ALM for the model problem
as an optimization procedure regularization to prove it. The algorithm will be
given and the convergence analysis will be carried out without using any
assumptions on Lagrange multipliers of the model problem. The convergence
analysis borrows some ideas of that in [12] and [21] for ADMM.

######  Remark A.1.

A comprehensive discussion of the classical ALM in Banach spaces based on a
different approach can be found in [37, Chapter 4], and a further application
of these results to develop new constraint qualification conditions in Banach
spaces can be found in [6].

We first rewrite the model problem (1.1) into the following equivalent form

(A.2)
$\min\limits_{u\in\mathcal{U},\zeta\in\mathcal{X}}\theta(u)+I_{K}(\zeta)\ \
\mbox{s.t.}\ \ Su=\zeta,$

where $I_{K}(\cdot)$ is the indicator function of $K$, which is defined by

$I_{K}(\zeta)=\left\\{\begin{aligned} &0,&\quad&\mbox{if}\ \zeta\in K,\\\
&+\infty,&\quad&\mbox{if}\ \zeta\not\in K.\end{aligned}\right.$

Note that the global minimizer of the problem (A.2) is given by
$(u^{*},\zeta^{*})$, where $u^{*}$ is the global minimizer of the model
problem (1.1) and $\zeta^{*}=Su^{*}$.

###  A.1. The classical augmented Lagrangian method

The augmented Lagrangian functional
$L_{\beta}(u,\zeta;\lambda):(\mathcal{U}\times\mathcal{X})\times\mathcal{X}\rightarrow\mathbb{R}\cup\\{+\infty\\}$
of the model problem (1.1) based on (A.2) is given by

$L_{\beta}(u,\zeta;\lambda)=\theta(u)+I_{K}(\zeta)+\langle\lambda,Su-\zeta\rangle+\frac{\beta}{2}\|Su-\zeta\|^{2},$

where $\beta>0$.

The classical augmented Lagrangian method for the model problem (1.1) based on
this augmented Lagrangian functional is given by

(A.3) $\left\\{\begin{aligned}
(u^{k+1},\zeta^{k+1})&=\arg\min\limits_{(u,\zeta)\in\mathcal{U}\times\mathcal{X}}L_{\beta}(u,\zeta;\lambda^{k}),\\\
\lambda^{k+1}&=\lambda^{k}+\beta(Su^{k+1}-\zeta^{k+1}),\end{aligned}\right.$

where $\lambda^{1}\in\mathcal{X}$ is given.

###  A.2. Convergence analysis

In this part we will give an elementary proof of the convergence of the
algorithm. The proof is composed of several steps.

We first give a characterization of the iterators
$\\{(u^{k},\zeta^{k},\lambda^{k})\\}_{k=1}^{+\infty}$ by the first order
optimality system of the subproblem in the first step. Since the subproblem in
the first step is a convex problem, solving the subproblem is equivalent to
solve

$\left\\{\begin{aligned}
&D_{u}\theta(u^{k+1})+S^{*}[\lambda^{k}+\beta(Su^{k+1}-\zeta^{k+1})]=0;\\\
&I_{K}(\zeta)-I_{K}(\zeta^{k+1})-\langle\lambda^{k}+\beta(Su^{k+1}-\zeta^{k+1}),\zeta-\zeta^{k+1}\rangle\geq
0\quad\forall\ \zeta\in\mathcal{X}.\end{aligned}\right.$

Hence, the iterators of the classical ALM satisfy

(A.4) $\left\\{\begin{aligned} &D_{u}\theta(u^{k+1})+S^{*}\lambda^{k+1}=0;\\\
&I_{K}(\zeta)-I_{K}(\zeta^{k+1})-\langle\lambda^{k+1},\zeta-\zeta^{k+1}\rangle\geq
0\quad\forall\ \zeta\in\mathcal{X};\\\ &\beta
r^{k+1}=\lambda^{k+1}-\lambda^{k},\end{aligned}\right.$

where $r^{k+1}=Su^{k+1}-\zeta^{k+1}$. According to (A.4), it holds

(A.5) $\lambda^{k}\in\partial I_{K}(\zeta^{k})\quad\forall\ k=2,3,\dots.$

Without loss of generality, we also assume that (A.5) holds for $k=1$.

####  A.2.1. Convergence of $\\{(u^{k},\zeta^{k})\\}_{k=1}^{+\infty}$

We first prove that $\\{\|r^{k}\|\\}_{k=1}^{+\infty}$ is Fejér monotone. It
follows from (A.4) that

$\beta
S^{*}r^{k+1}=S^{*}(\lambda^{k+1}-\lambda^{k})=S^{*}\lambda^{k+1}-S^{*}\lambda^{k}=D_{u}\theta(u^{k})-D_{u}\theta(u^{k+1})$

and

$\displaystyle\beta\langle r^{k+1},\zeta^{k+1}-\zeta^{k}\rangle$
$\displaystyle=\langle\lambda^{k+1}-\lambda^{k},\zeta^{k+1}-\zeta^{k}\rangle$
$\displaystyle=I_{K}(\zeta^{k})-I_{K}(\zeta^{k+1})-\langle\lambda^{k+1},\zeta^{k}-\zeta^{k+1}\rangle\quad(\geq
0)$
$\displaystyle\quad+I_{K}(\zeta^{k+1})-I_{K}(\zeta^{k})-\langle\lambda^{k},\zeta^{k+1}-\zeta^{k}\rangle\quad(\geq
0)$ $\displaystyle\geq 0,$

which yield

$\displaystyle\beta\langle r^{k+1}-r^{k},r^{k+1}\rangle$
$\displaystyle=\beta\langle(Su^{k+1}-\zeta^{k+1})-(Su^{k}-\zeta^{k}),r^{k+1}\rangle$
$\displaystyle=\beta\langle
S(u^{k+1}-u^{k}),r^{k+1}\rangle-\beta\langle\zeta^{k+1}-\zeta^{k},r^{k+1}\rangle$
$\displaystyle=\langle u^{k+1}-u^{k},\beta
S^{*}r^{k+1}\rangle_{\mathcal{U}}-\beta\langle
r^{k+1},\zeta^{k+1}-\zeta^{k}\rangle$ $\displaystyle\leq-\langle
u^{k+1}-u^{k},D_{u}\theta(u^{k+1})-D_{u}\theta(u^{k})\rangle_{\mathcal{U}}$
$\displaystyle\leq-c_{0}\|u^{k+1}-u^{k}\|^{2}_{\mathcal{U}},$

where we used (1.2) in the last inequality. Furthermore, by applying the
identity

$\beta\langle
r^{k+1}-r^{k},r^{k+1}\rangle=\frac{\beta}{2}\left(\|r^{k+1}\|^{2}-\|r^{k}\|^{2}+\|r^{k+1}-r^{k}\|^{2}\right),$

we have

$\frac{\beta}{2}\left(\|r^{k+1}\|^{2}-\|r^{k}\|^{2}+\|r^{k+1}-r^{k}\|^{2}\right)\leq-
c_{0}\|u^{k+1}-u^{k}\|^{2}_{\mathcal{U}}$

or equivalently

(A.6)
$\|r^{k+1}\|^{2}\leq\|r^{k}\|^{2}-\|r^{k+1}-r^{k}\|^{2}-2c_{0}\beta^{-1}\|u^{k+1}-u^{k}\|^{2}_{\mathcal{U}}\leq\|r^{k}\|^{2},$

which shows that $\\{\|r^{k}\|\\}_{k=1}^{+\infty}$ is Fejér monotone.

Secondly, we prove that

$\sum\limits_{k=1}^{+\infty}\|r^{k}\|^{2}<+\infty.$

We denote by $\eta=(u,\zeta)\in\mathcal{U}\times\mathcal{X}$ and recall the
Bregman distance induced by the convex functional $\theta(u)+I_{K}(\zeta)$ at
$(D_{u}\theta(u),\lambda)$ with $\lambda\in\partial I_{K}(\zeta)$, which is
given by

(A.7)
$\mathcal{D}_{\eta}(\hat{\eta},\eta;\lambda):=\theta(\hat{u})-\theta(u)-\langle
D_{u}\theta(u),\hat{u}-u\rangle_{\mathcal{U}}+I_{K}(\hat{\zeta})-I_{K}(\zeta)-\langle\lambda,\hat{\zeta}-\zeta\rangle\geq
0,$

for any $\hat{\eta}\in\mathcal{U}\times\mathcal{X}$.

By the assumption (1.2) of $\theta(u)$, we have

(A.8)
$\mathcal{D}_{\eta}(\hat{\eta},\eta;\lambda)\geq\theta(\hat{u})-\theta(u)-\langle
D_{u}\theta(u),\hat{u}-u\rangle_{\mathcal{U}}\geq\frac{c_{0}}{2}\|\hat{u}-u\|_{\mathcal{U}}^{2}.$

Let $\hat{\eta}=(\hat{u},\hat{\zeta})$ satisfy $\hat{\zeta}=S\hat{u}\in K$. We
will always assume this in the following analysis. It follows from (A.4),
(A.5) and (A.7) that

(A.9) $\displaystyle\quad\
\mathcal{D}_{\eta^{k}}(\hat{\eta},\eta^{k};\lambda^{k})-\mathcal{D}_{\eta^{n}}(\hat{\eta},\eta^{n};\lambda^{n})+\mathcal{D}_{\eta^{n}}(\eta^{k},\eta^{n};\lambda^{n})$
$\displaystyle=\langle
D_{u}\theta(u^{n})-D_{u}\theta(u^{k}),\hat{u}-u^{k}\rangle_{\mathcal{U}}+\langle\lambda^{n}-\lambda^{k},\hat{\zeta}-\zeta^{k}\rangle$
$\displaystyle=\langle\lambda^{n}-\lambda^{k},Su^{k}-S\hat{u}\rangle+\langle\lambda^{n}-\lambda^{k},S\hat{u}-\zeta^{k}\rangle$
$\displaystyle=\langle\lambda^{n}-\lambda^{k},Su^{k}-\zeta^{k}\rangle=-\sum\limits_{i=n}^{k-1}\langle\lambda^{i+1}-\lambda^{i},Su^{k}-\zeta^{k}\rangle$
$\displaystyle=-\beta\sum\limits_{i=n}^{k-1}\langle r^{i+1},r^{k}\rangle,$

for any $k>n$. This gives

$\mathcal{D}_{\eta^{k}}(\eta^{k+1},\eta^{k};\lambda^{k})+\beta\|r^{k+1}\|^{2}=\mathcal{D}_{\eta^{k}}(\hat{\eta},\eta^{k};\lambda^{k})-\mathcal{D}_{\eta^{k+1}}(\hat{\eta},\eta^{k+1};\lambda^{k+1})\quad\forall\
k=1,2,\dots.$

Summing over $k$ from $n$ to $m\ (>n)$ on both sides, we have

(A.10)
$\sum\limits_{k=n}^{m}\left(\mathcal{D}_{\eta^{k}}(\eta^{k+1},\eta^{k};\lambda^{k})+\beta\|r^{k+1}\|^{2}\right)=\mathcal{D}_{\eta^{n}}(\hat{\eta},\eta^{n};\lambda^{n})-\mathcal{D}_{\eta^{m}}(\hat{\eta},\eta^{m};\lambda^{m}).$

Then by taking $n=1$, using (A.8) and letting $m\rightarrow+\infty$, we have

$\sum\limits_{k=1}^{+\infty}\left(\mathcal{D}_{\eta^{k}}(\eta^{k+1},\eta^{k};\lambda^{k})+\beta\|r^{k+1}\|^{2}\right)\leq\mathcal{D}_{\eta^{1}}(\hat{\eta},\eta^{1};\lambda^{1})<+\infty,$

which yields

(A.11) $\sum\limits_{k=1}^{+\infty}\|r^{k}\|^{2}<\infty$

and

(A.12)
$\\{\mathcal{D}_{\eta^{k}}(\hat{\eta},\eta^{k};\lambda^{k})\\}_{k=1}^{+\infty}\
\ \mbox{is\ a\ Cauchy\ sequence},$

by (A.10). Moreover, we have

(A.13) $\|Su^{k}-\zeta^{k}\|=\|r^{k}\|\ \rightarrow 0\ \ \mbox{as}\ \
k\rightarrow+\infty.$

Now, we prove that $\\{u^{k}\\}_{k=1}^{+\infty}$ and
$\\{\zeta^{k}\\}_{k=1}^{+\infty}$ are Cauchy sequences. For any $k>n$, it
follows from (A.6) that

(A.14) $\displaystyle|\beta\sum\limits_{i=n}^{k-1}\langle
r^{i+1},r^{k}\rangle|\leq\frac{\beta}{2}\sum\limits_{i=n}^{k-1}(\|r^{i+1}\|^{2}+\|r^{k}\|^{2})\leq\beta\sum\limits_{i=n}^{k}\|r^{i}\|^{2}.$

For any $k>n$, by (A.8), (A.9), (A.14), (A.11) and (A.12), we have

$\displaystyle\frac{c_{0}}{2}\|u^{k}-u^{n}\|_{\mathcal{U}}^{2}$
$\displaystyle\leq\mathcal{D}_{\eta^{n}}(\eta^{k},\eta^{n};\lambda^{n})$
$\displaystyle=-\beta\sum\limits_{i=n}^{k-1}\langle
r^{i+1},r^{k}\rangle-[\mathcal{D}_{\eta^{k}}(\hat{\eta},\eta^{k};\lambda^{k})-\mathcal{D}_{\eta^{n}}(\hat{\eta},\eta^{n};\lambda^{n})]$
$\displaystyle\leq\beta\sum\limits_{i=n}^{k}\|r^{i}\|^{2}+|\mathcal{D}_{\eta^{k}}(\hat{\eta},\eta^{k};\lambda^{k})-\mathcal{D}_{\eta^{n}}(\hat{\eta},\eta^{n};\lambda^{n})|\quad\rightarrow\
\ 0\ \ \mbox{as}\ n,k\rightarrow+\infty.$

Therefore, $\\{u^{k}\\}_{k=1}^{+\infty}$ is a Cauchy sequence and
$\\{\zeta^{k}\\}_{k=1}^{+\infty}$ is a Cauchy sequence by (A.13).

Let $\bar{u}\in\mathcal{U}$ and $\bar{\zeta}\in\mathcal{X}$ be the limits of
$\\{u^{k}\\}_{k=1}^{+\infty}$ and $\\{\zeta^{k}\\}_{k=1}^{+\infty}$
respectively, i.e.,

(A.15)
$\lim\limits_{k\rightarrow+\infty}u^{k}=\bar{u}\quad\mbox{and}\quad\lim\limits_{k\rightarrow+\infty}\zeta^{k}=\bar{\zeta}.$

Since $\\{\zeta^{k}\\}\subset K$ and $K$ is closed, we have $\bar{\zeta}\in
K$. It follows from (A.13) that

(A.16) $S\bar{u}=\bar{\zeta}.$

####  A.2.2. Global convergence of the algorithm

Now, we prove that $(\bar{u},\bar{\zeta})$ is the global minimizer of the
problem (A.2). Since

$\theta(u)+I_{K}(\zeta)$

is lower semicontinuous, we have

(A.17) $\theta(\bar{u})+I_{K}(\bar{\zeta})\leq\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}[\theta(u^{k})+I_{K}(\zeta^{k})].$

Note that $\lambda^{k}\in\partial I_{K}(\zeta^{k})$. The convexity of the
objective functional gives

$\theta(\bar{u})+I_{K}(\bar{\zeta})-[\theta(u^{k})+I_{K}(\zeta^{k})]-[\langle\lambda^{k},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k}),\bar{u}-u^{k}\rangle_{\mathcal{U}}]\geq 0,$

i.e.,

(A.18)
$\theta(u^{k})+I_{K}(\zeta^{k})\leq\theta(\bar{u})+I_{K}(\bar{\zeta})-[\langle\lambda^{k},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k}),\bar{u}-u^{k}\rangle_{\mathcal{U}}].$

For any fixed $n$ and $k>n$, it follows from (A.16) and (A.9) with
$\hat{\eta}=(\bar{u},\bar{\zeta})$ that

(A.19) $\displaystyle\quad\
|\langle\lambda^{k},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k}),\bar{u}-u^{k}\rangle_{\mathcal{U}}|$
$\displaystyle=|\langle\lambda^{k}-\lambda^{n},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k})-D_{u}\theta(u^{n}),\bar{u}-u^{k}\rangle_{\mathcal{U}}$
$\displaystyle\quad+\langle\lambda^{n},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{n}),\bar{u}-u^{k}\rangle_{\mathcal{U}}|$
$\displaystyle=|\beta\sum\limits_{i=n}^{k-1}\langle
r^{i+1},r^{k}\rangle+[\langle\lambda^{n},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{n}),\bar{u}-u^{k}\rangle_{\mathcal{U}}]|$
$\displaystyle\leq\beta\sum\limits_{i=n}^{k}\|r^{i}\|^{2}+|\langle\lambda^{n},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{n}),\bar{u}-u^{k}\rangle_{\mathcal{U}}|.$

Hence, by (A.18), (A.19), (A.11) and (A.15), we get

$\displaystyle\quad\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}[\theta(u^{k})+I_{K}(\zeta^{k})]$
$\displaystyle\leq\theta(\bar{u})+I_{K}(\bar{\zeta})+\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\left\\{\beta\sum\limits_{i=n}^{k}\|r^{i}\|^{2}+|\langle\lambda^{n},\bar{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{n}),\bar{u}-u^{k}\rangle_{\mathcal{U}}|\right\\}$
$\displaystyle=\theta(\bar{u})+I_{K}(\bar{\zeta})+\beta\sum\limits_{i=n}^{+\infty}\|r^{i}\|^{2}+|\langle\lambda^{n},\bar{\zeta}-\bar{\zeta}\rangle+\langle
D_{u}\theta(u^{n}),\bar{u}-\bar{u}\rangle_{\mathcal{U}}|$
$\displaystyle=\theta(\bar{u})+I_{K}(\bar{\zeta})+\beta\sum\limits_{i=n}^{+\infty}\|r^{i}\|^{2}.$

If we further take $n\rightarrow+\infty$ and use (A.11), we obtain

(A.20)
$\lim\limits_{k\rightarrow+\infty}\sup[\theta(u^{k})+I_{K}(\zeta^{k})]\leq\theta(\bar{u})+I_{K}(\bar{\zeta}).$

This together with (A.17) yields

(A.21)
$\theta(\bar{u})+I_{K}(\bar{\zeta})=\lim\limits_{k\rightarrow+\infty}[\theta(u^{k})+I_{K}(\zeta^{k})].$

For any $(\hat{u},\hat{\zeta})$ satisfing $S\hat{u}=\hat{\zeta}$, we will
prove that

$\theta(\bar{u})+I_{K}(\bar{\zeta})\leq\theta(\hat{u})+I_{K}(\hat{\zeta}).$

The proof is similar to that of (A.20). Since

$\theta(u^{k})+I_{K}(\zeta^{k})\leq\theta(\hat{u})+I_{K}(\hat{\zeta})-[\langle\lambda^{k},\hat{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k}),\hat{u}-u^{k}\rangle_{\mathcal{U}}]$

and

$\displaystyle\quad|\langle\lambda^{k},\hat{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k}),\hat{u}-u^{k}\rangle_{\mathcal{U}}|$
$\displaystyle=|\beta\sum\limits_{i=n}^{k-1}\langle
r^{i+1},r^{k}\rangle|+|\langle\lambda^{n},\hat{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{n}),\hat{u}-u^{k}\rangle_{\mathcal{U}}|$
$\displaystyle\leq\beta\sum\limits_{i=n}^{k}\|r^{i}\|^{2}+|\langle\lambda^{n},r^{k}\rangle|$

where $n$ is fixed and $k>n$, we have

$\displaystyle\theta(u^{k})+I_{K}(\zeta^{k})$
$\displaystyle\leq\theta(\hat{u})+I_{K}(\hat{\zeta})-[\langle\lambda^{k},\hat{\zeta}-\zeta^{k}\rangle+\langle
D_{u}\theta(u^{k}),\hat{u}-u^{k}\rangle_{\mathcal{U}}]$
$\displaystyle\leq\theta(\hat{u})+I_{K}(\hat{\zeta})+\beta\sum\limits_{i=n}^{k}\|r^{i}\|^{2}+|\langle\lambda^{n},r^{k}\rangle|.$

By (A.11) and (A.21), we get

$\displaystyle\theta(\bar{u})+I_{K}(\bar{\zeta})=\lim\limits_{k\rightarrow+\infty}\theta(u^{k})+I_{K}(\zeta^{k})\leq\theta(\hat{u})+I_{K}(\hat{\zeta})+\beta\sum\limits_{i=n}^{+\infty}\|r^{i}\|^{2}.$

By taking $n\rightarrow+\infty$ and using (A.11) again, it leads to

$\theta(\bar{u})+I_{K}(\bar{\zeta})\leq\theta(\hat{u})+I_{K}(\hat{\zeta}),$

which implies $(\bar{u},\bar{\zeta})$ is the global minimizer of the model
problem.

###  A.3. Deriving the weak form asymptotic KKT system

Now we derive the weak form asymptotic KKT system (1.3). This can be achieved
by exploring the limiting case of (A.4).

We have proved that $\\{u^{k}\\}_{k=1}^{+\infty}$ strongly converges to the
global minimizer $u^{*}$ of the model problem (1.1). Since $\theta(u)$ is
continuously differentiable, it gives

$\lim\limits_{k\rightarrow+\infty}D_{u}\theta(u^{k})=D_{u}\theta(u^{*})$

and

(A.22) $\langle
D_{u}\theta(u^{*}),v\rangle_{\mathcal{U}}+\lim\limits_{k\rightarrow+\infty}\langle\lambda^{k},Sv\rangle=0\quad\forall\
v\in\mathcal{U},$

by (A.4).

In order to get (1.3), it suffices to show that

$-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta-Su^{*}\rangle\geq
0\quad\forall\ \zeta\in K.$

Denote by $\zeta^{*}=Su^{*}\in K$. According to (A.4), for any $\zeta\in K$,
we have

$-\langle\lambda^{k+1},\zeta-\zeta^{k+1}\rangle\geq 0$

and

(A.23)
$\displaystyle\quad-\langle\lambda^{k+1},\zeta-\zeta^{*}\rangle=-\langle\lambda^{k+1},\zeta-\zeta^{k+1}\rangle-\langle\lambda^{k+1},\zeta^{k+1}-\zeta^{*}\rangle$
$\displaystyle\geq\langle\lambda^{k+1},\zeta^{*}-\zeta^{k+1}\rangle$
$\displaystyle=\langle\lambda^{k+1},\zeta^{*}-\zeta^{k+1}\rangle+\langle
D_{u}\theta(u^{k+1}),u^{*}-u^{k+1}\rangle_{\mathcal{U}}-\langle
D_{u}\theta(u^{k+1}),u^{*}-u^{k+1}\rangle_{\mathcal{U}}.$

For any fixed $n$ and $k+1>n$, applying (A.19) to $(u^{*},\zeta^{*})$, we have

$\displaystyle\quad|\langle\lambda^{k+1},\zeta^{*}-\zeta^{k+1}\rangle+\langle
D_{u}\theta(u^{k+1}),u^{*}-u^{k+1}\rangle_{\mathcal{U}}|$
$\displaystyle\leq\beta\sum\limits_{i=n}^{k+1}\|r^{i}\|^{2}+|\langle\lambda^{n},\zeta^{*}-\zeta^{k+1}\rangle+\langle
D_{u}\theta(u^{n}),u^{*}-u^{k+1}\rangle_{\mathcal{U}}|.$

Following the same arguments as before (see (A.19)-(A.20)), we get

(A.24) $|\langle\lambda^{k+1},\zeta^{*}-\zeta^{k+1}\rangle+\langle
D_{u}\theta(u^{k+1}),u^{*}-u^{k+1}\rangle_{\mathcal{U}}|\rightarrow
0\quad\mbox{as}\quad k\rightarrow+\infty.$

On the other hand, the strong convergence of $\\{u^{k}\\}_{k=1}^{+\infty}$ and
the boundedness of $D_{u}\theta(u)$ around $u^{*}$ give

(A.25) $\lim\limits_{k\rightarrow+\infty}\langle
D_{u}\theta(u^{k+1}),u^{*}-u^{k+1}\rangle_{\mathcal{U}}=0.$

Together with (A.23), (A.24) and (A.25), we arrive at

(A.26) $-\operatorname*{lim\
sup}\limits_{k\rightarrow+\infty}\langle\lambda^{k},\zeta-\zeta^{*}\rangle=\operatorname*{lim\
inf}\limits_{k\rightarrow+\infty}[-\langle\lambda^{k},\zeta-\zeta^{*}\rangle]\geq
0\quad\forall\ \zeta\in K.$

Therefore, by (A.22), (A.26) and $\zeta^{*}=Su^{*}$, we have the weak form
asymptotic KKT system (1.3) at $u^{*}$.

This completes the proof of Theorem 1.2.

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