# Characterization of the equality
of weak efficiency and efficiency
on convex free disposal hulls

Naoki Hamada  KLab Inc., Roppongi Hills Mori Tower, 6-10-1 Roppongi, Minato-
ku, Tokyo, 106-6122, Japan; RIKEN AIP-FUJITSU Collaboration Center, The RIKEN
Center for Advanced Intelligence Project, Nihonbashi 1-chome Mitsui Building,
15th floor, 1-4-1 Nihonbashi, Chuo-ku, Tokyo 103-0027, Japan
<EMAIL_ADDRESS>and  Shunsuke Ichiki Department of Mathematical and
Computing Science, School of Computing, Tokyo Institute of Technology, Tokyo
152-8552, Japan<EMAIL_ADDRESS>
###### Abstract.

In solving a multi-objective optimization problem by scalarization techniques,
solutions to a scalarized problem are, in general, weakly efficient rather
than efficient to the original problem. Thus, it is crucial to understand what
problem ensures that all weakly efficient solutions are efficient. In this
paper, we give a characterization of the equality of the weakly efficient set
and the efficient set, provided that the free disposal hull of the domain is
convex. By using this characterization, we obtain various mathematical
applications. As a practical application, we show that all weakly efficient
solutions to a multi-objective LASSO with mild modification are efficient.

##  1\. Introduction

The aim of multi-objective optimization is to find efficient solutions to a
given problem. In order to do so, various scalarization techniques have been
developed so far (see for example [17, 4, 15, 14, 26, 6, 21]). Nevertheless,
there is no scalarization method that ensures for a wide variety of problems
that all solutions optimal to scalarized problems are efficient to the
original problem. In general, scalarization methods only ensure that their
solutions are _weakly_ efficient to the original problem, which means users
may waste computation resources for finding inefficient, undesirable
solutions. Thus, it is crucial to understand conditions that the weak
efficiency coincides with the efficiency.

In the literature, the relationship between the weak efficiency and the
efficiency has been investigated. In some cases, the set of weakly efficient
solutions to a given problem can be described as the union of the sets of
efficient solutions to its subproblems [10, 24, 12, 2]. This property was
named the _Pareto reducibility_ [18] and further investigated [19, 20, 9].
Some relationships of the weak efficiency and the efficiency on quasi-convex
problems are collected in [11, 5]. However, the equality between the weak
efficiency and the efficiency, both of which are of the _original_ problem
(rather than subproblems), is still unclear.

In this paper, we give a characterization of the equality of the set of weakly
efficient solutions and the set of efficient solutions, provided that the free
disposal hull [1] of the image of an objective mapping is convex (see
Proposition 2.3 in Section 2). This claim is derived from our main theorem
(Theorem 2.1 in Section 2), which gives a similar characterization of the
equality of the weakly efficient set and the efficient set on a partially
ordered Euclidean space without objective functions. Furthermore, Proposition
2.3 yields various mathematical applications (see Corollaries 6.1, 6.2, 6.3
and 6.4 in Section 6). Moreover, as a practical application of Proposition
2.3, we show that all weakly efficient solutions to a multi-objective LASSO
with mild modification are efficient.

This paper is organized as follows. First, in Section 2, we present the main
results (Theorems 2.1 and 2.3). Implications of Theorem 2.1 are discussed with
illustrative examples in Section 3. Section 4 is devoted to the proof of
Theorem 2.1. In order to state and prove Corollaries 6.1, 6.2, 6.3 and 6.4 in
Section 6, we prepare some definitions and lemmas in Section 5. In Section 7,
we investigate a multi-objective version of the LASSO with mild modification
as a practical application of our result. Section 8 provides concluding
remarks.

##  2\. Preliminaries and the statements of the main results

Unless otherwise stated, it is not necessary to assume that mappings are
continuous. Throughout this paper, we set

$\displaystyle M=\set{1,\ldots,m},$

where $m$ is a positive integer. We denote a nonempty subset of $M$ by $I$.
Let $y=(y_{1},\ldots,y_{m})$ and
$y^{\prime}=(y_{1}^{\prime},\ldots,y_{m}^{\prime})$ be two elements of
$\mathbb{R}^{m}$. The inequality $y\leq_{I}y^{\prime}$ (resp.,
$y<_{I}y^{\prime}$) means that $y_{i}\leq y_{i}^{\prime}$ (resp.,
$y_{i}<y_{i}^{\prime}$) for all $i\in I$. The inequality
$y\lneq_{I}y^{\prime}$ means that $y_{i}\leq y_{i}^{\prime}$ for all $i\in I$
and there exists $j\in I$ such that $y_{j}<y^{\prime}_{j}$.

Let $Y$ be a subset of $\mathbb{R}^{m}$. Let $\operatorname{Min}_{I}Y$ (resp.,
$\operatorname{WMin}_{I}Y$) be the set consisting of all elements
$y^{\prime}\in Y$ such that there does not exist any element $y\in Y$
satisfying $y\lneq_{I}y^{\prime}$ (resp., $y<_{I}y^{\prime}$). For simplicity,
set $\operatorname{Min}Y=\operatorname{Min}_{M}Y$ and
$\operatorname{WMin}Y=\operatorname{WMin}_{M}Y$. Then, the set
$\operatorname{Min}Y$ (resp., $\operatorname{WMin}Y$) is called the _efficient
set_ (resp., the _weakly efficient set_) of $Y$.

For a subset $Z$ of $\mathbb{R}^{m}$, the set $Z+\mathbb{R}^{m}_{\geq 0}$ is
called the _free disposal hull_ of $Z$ (denoted by $\operatorname{FDH}Z$),
where

$\mathbb{R}^{m}_{\geq 0}=\Set{(y_{1},\ldots,y_{m})\in\mathbb{R}^{m}}{y_{1}\geq
0,\ldots,y_{m}\geq 0}.$

For details on free disposal hulls, see [1]. A subset $Z$ of $\mathbb{R}^{m}$
is said to be convex if $tx+(1-t)y\in Z$ for all $x,y\in Z$ and all
$t\in[0,1]$.

The main theorem of this paper is the following.

######  Theorem 2.1.

Let $Y$ be a subset of $\mathbb{R}^{m}$. If the free disposal hull of $Y$ is
convex, then the following $(\alpha)$ and $(\beta)$ are equivalent:

  1. $(\alpha)$

$\operatorname{WMin}Y=\operatorname{Min}Y$.

  2. $(\beta)$

$\displaystyle\bigcup_{\emptyset\neq I\subseteq
M}\operatorname{Min}_{I}Y\subseteq\operatorname{Min}Y$.

######  Remark 2.2.

As in the proof of Theorem 2.1, the hypothesis that the free disposal hull of
$Y$ is convex is used only in the proof of $(\beta)\Rightarrow(\alpha)$ (see
Section 4.2). In the proof of $(\alpha)\Rightarrow(\beta)$ of Theorem 2.1, it
is not necessary to assume that the free disposal hull of $Y$ is convex (see
Section 4.1).

Now, in order to state Proposition 2.3, we will prepare some definitions. Let
$f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ be a mapping, and
$I=\Set{i_{1},\ldots,i_{k}}$ $(i_{1}<\cdots<i_{k})$ be a nonempty subset of
$M$, where $X$ is a given set and $k$ is the number of the elements of $I$.
Let $f_{I}:X\to\mathbb{R}^{k}$ be the mapping defined by
$f_{I}=(f_{i_{1}},\ldots,f_{i_{k}})$. A point $x^{*}\in X$ is called an
_efficient solution_ (resp., a _weakly efficient solution_) to the following
multi-objective optimization problem:

$\displaystyle\operatorname*{minimize}_{x\in
X}f_{I}(x)=(f_{i_{1}}(x),\ldots,f_{i_{k}}(x)),$

if $f(x^{*})\in\operatorname{Min}_{I}f(X)$ (resp.,
$f(x^{*})\in\operatorname{WMin}_{I}f(X)$). By $\operatorname{S}(f_{I},X)$
(resp., $\operatorname{WS}(f_{I},X)$), we denote the set consisting of all
efficient solutions (resp., all weakly efficient solutions). Namely,

$\displaystyle\operatorname{S}(f_{I},X)$
$\displaystyle=f^{-1}(\operatorname{Min}_{I}f(X)),$
$\displaystyle\operatorname{WS}(f_{I},X)$
$\displaystyle=f^{-1}(\operatorname{WMin}_{I}f(X)).$

It is well known that a solution to a weighting problem is a weekly efficient
solution (for example, see [15, Theorem 3.1.1 (p. 78)]). On the other hand, a
solution to a weighting problem is not necessarily an efficient solution. For
a given mapping $f:X\to\mathbb{R}^{m}$, if
$\operatorname{WS}(f,X)=\operatorname{S}(f,X)$, then a solution to weighting
problem is always an efficient solution. Therefore, characterizations of
$\operatorname{WS}(f,X)=\operatorname{S}(f,X)$ are useful and significant.

As an application of Theorem 2.1 to multi-objective optimization problems, we
have the following, which can be easily shown by setting $Y=f(X)$ in Theorem
2.1.

######  Proposition 2.3.

Let $f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ be a mapping, where $X$ is a
given set. If the free disposal hull of $f(X)$ is convex, then the following
$(\alpha)$ and $(\beta)$ are equivalent:

  1. $(\alpha)$

$\operatorname{WS}(f,X)=\operatorname{S}(f,X)$.

  2. $(\beta)$

${\displaystyle\bigcup_{\emptyset\neq I\subseteq
M}\operatorname{S}(f_{I},X)\subseteq\operatorname{S}(f,X)}$.

##  3\. Illustration of Theorem 2.1

In this section, we denote by $\overline{xy}$ the line segment with end points
$x,y\in\mathbb{R}^{2}$. First, we see an example that $(\beta)$ implies
$(\alpha)$ in Theorem 2.1.

######  Example 3.1.

Let us consider the situation shown in Figure 1. The domain $Y$ in this case
is defined by the convex hull of four points $p_{1}=(0,1)$, $p_{2}=(1,0)$,
$p_{3}=(2,1)$, $p_{4}=(1,2)$, as shown in dark gray in the figure.

Figure 1. The condition $(\beta)$ implies $(\alpha)$ on convex
$\operatorname{FDH}Y$.

It is easy to check that

$\displaystyle\operatorname{Min}_{\Set{1}}Y$ $\displaystyle=\Set{p_{1}},$
$\displaystyle\operatorname{Min}_{\Set{2}}Y$ $\displaystyle=\Set{p_{2}},$
$\displaystyle\operatorname{WMin}Y=\operatorname{Min}Y$
$\displaystyle=\overline{p_{1}p_{2}}.$

Since $\operatorname{Min}_{\Set{1}}Y\subseteq\operatorname{Min}Y$ and
$\operatorname{Min}_{\Set{2}}Y\subseteq\operatorname{Min}Y$, we can see the
condition $(\beta)$ in Theorem 2.1 holds. The free disposal hull of $Y$ is the
region shown in light gray in the figure, which is a convex set. Thus, we can
apply Theorem 2.1 and obtain $(\alpha)$. Actually, the condition
$\operatorname{WMin}Y=\operatorname{Min}Y$ holds in this example.

On the other hand, Example 3.2 shows that $\operatorname{FDH}Y$ is convex, but
both $(\alpha)$ and $(\beta)$ do not hold.

######  Example 3.2.

Let us consider the situation shown in Figure 2. The domain $Y$ is the convex
hull of four points $p_{1}=(0,2)$, $p_{2}=(0,1)$, $p_{3}=(1,0)$,
$p_{4}=(2,0)$, as shown in dark gray in the figure. We have the same free
disposal hull as in Example 3.1, which is a convex set, and thus we can apply
Theorem 2.1 to this case.

Figure 2. The condition $\lnot(\beta)$ implies $\lnot(\alpha)$ on convex
$\operatorname{FDH}Y$.

We can easily check:

$\displaystyle\operatorname{Min}_{\Set{1}}Y$
$\displaystyle=\overline{p_{1}p_{2}},$
$\displaystyle\operatorname{Min}_{\Set{2}}Y$
$\displaystyle=\overline{p_{3}p_{4}},$ $\displaystyle\operatorname{Min}Y$
$\displaystyle=\overline{p_{2}p_{3}},$ $\displaystyle\operatorname{WMin}Y$
$\displaystyle=\overline{p_{1}p_{2}}\cup\overline{p_{2}p_{3}}\cup\overline{p_{3}p_{4}}.$

Since $\operatorname{Min}_{\Set{1}}Y\not\subseteq\operatorname{Min}Y$, the
condition $(\beta)$ in Theorem 2.1 does not hold. By Theorem 2.1, the
condition $(\alpha)$ $\operatorname{WMin}Y=\operatorname{Min}Y$ does not hold,
as shown in the above equations.

The following example shows why the assumption of Theorem 2.1 is required.

######  Example 3.3.

Let us consider the situation shown in Figure 3 where the domain $Y$ is the
nonconvex polygon with five vertices $p_{1}=(0,3)$, $p_{2}=(1,2)$,
$p_{3}=(1,1)$, $p_{4}=(2,0)$, $p_{5}=(2,3)$, shown in dark gray.

Figure 3. The condition $(\beta)$ holds, but $(\alpha)$ does not on nonconvex
$\operatorname{FDH}Y$ .

We can easily check:

$\displaystyle\operatorname{Min}_{\Set{1}}Y$ $\displaystyle=\Set{p_{1}},$
$\displaystyle\operatorname{Min}_{\Set{2}}Y$ $\displaystyle=\Set{p_{4}},$
$\displaystyle\operatorname{Min}Y$
$\displaystyle=\overline{p_{1}p_{2}}\cup\overline{p_{3}p_{4}}\setminus\Set{p_{2}},$
$\displaystyle\operatorname{WMin}Y$
$\displaystyle=\overline{p_{1}p_{2}}\cup\overline{p_{2}p_{3}}\cup\overline{p_{3}p_{4}}.$

Since $\operatorname{Min}_{\Set{1}}Y\subseteq\operatorname{Min}Y$ and
$\operatorname{Min}_{\Set{2}}Y\subseteq\operatorname{Min}Y$, the condition
$(\beta)$ holds. However, the free disposal hull of $Y$ is a nonconvex set, as
shown in light gray in the figure. Hence, we cannot apply Theorem 2.1 to this
case. In such a case, $(\alpha)$ can be false even if $(\beta)$ is true.
Actually, in this example, the condition $(\alpha)$ does not hold as seen in
the above equations.

In Theorem 2.1, the assumption (the free disposal hull of $Y$ is convex) is
not a necessary condition. In the following example, we will give a case where
the free disposal hull is nonconvex but the condition $(\alpha)$ holds (thus,
$(\beta)$ also holds).

######  Example 3.4.

Let us consider the situation shown in Figure 4 where the domain $Y$ is a
nonconvex polygon with four vertices $p_{1}=(0,3)$, $p_{2}=(2,2)$,
$p_{3}=(3,0)$, $p_{4}=(3,3)$, as shown in dark gray in the figure.

Figure 4. The condition $(\beta)$ holds, and $(\alpha)$ does on nonconvex
$\operatorname{FDH}Y$.

We can easily check:

$\displaystyle\operatorname{Min}_{\Set{1}}Y$ $\displaystyle=\Set{p_{1}},$
$\displaystyle\operatorname{Min}_{\Set{2}}Y$ $\displaystyle=\Set{p_{3}},$
$\displaystyle\operatorname{WMin}Y=\operatorname{Min}Y$
$\displaystyle=\overline{p_{1}p_{2}}\cup\overline{p_{2}p_{3}}.$

Since $\operatorname{Min}_{\Set{1}}Y\subseteq\operatorname{Min}Y$ and
$\operatorname{Min}_{\Set{2}}Y\subseteq\operatorname{Min}Y$, the condition
$(\beta)$ holds. The free disposal hull of $Y$ is a nonconvex set, as shown in
light gray in the figure. Hence we cannot apply Theorem 2.1 to this case.
Nevertheless, the condition $(\alpha)$ actually holds as seen in the above
equations.

##  4\. Proof of Theorem 2.1

In the case $Y=\emptyset$, it is trivially seen that both $(\alpha)$ and
$(\beta)$ hold. Hence, in what follows, we will consider the case
$Y\not=\emptyset$.

###  4.1. Proof of $(\alpha)\Rightarrow(\beta)$

Let $I$ be a nonempty subset of $M$. Then, it is clearly seen that

$\operatorname{Min}_{I}Y\subseteq\operatorname{WMin}_{I}Y\subseteq\operatorname{WMin}Y.$

By $(\alpha)$, we get $\operatorname{Min}_{I}Y\subseteq\operatorname{Min}Y$.
Thus, we have $(\beta)$. $\Box$

###  4.2. Proof of $(\beta)\Rightarrow(\alpha)$

It is sufficient to show that
$\operatorname{WMin}Y\subseteq\operatorname{Min}Y$. Let
$y^{*}=(y^{*}_{1},\ldots,y^{*}_{m})\in\operatorname{WMin}Y$ be an arbitrary
element. Set

$\displaystyle A$
$\displaystyle=\Set{(y_{1}-y^{*}_{1},\ldots,y_{m}-y^{*}_{m})\in\mathbb{R}^{m}}{(y_{1},\ldots,y_{m})\in\operatorname{FDH}Y},$
$\displaystyle B$ $\displaystyle=\Set{y\in\mathbb{R}^{m}}{y<_{M}0}.$

Here, note that $0=(0,\ldots,0)\in\mathbb{R}^{m}$ in the above description of
$B$. Then, we will have $A\cap B=\emptyset$ by contradiction. Suppose that
$A\cap B\neq\emptyset$. Then, there exist $y^{\prime}\in Y$ and
$z\in\mathbb{R}^{m}_{\geq 0}$ satisfying $y^{\prime}+z-y^{*}<_{M}0$. Since
$z\in\mathbb{R}^{m}_{\geq 0}$, we get $y^{\prime}-y^{*}<_{M}0$. This
contradicts $y^{*}\in\operatorname{WMin}Y$. Hence, we have $A\cap
B=\emptyset$.

In the following lemma, $\left\langle,\right\rangle$ stands for the inner
product in $\mathbb{R}^{m}$.

######  Lemma 4.1 (Separation theorem [13]).

Let $D_{1}$ and $D_{2}$ be nonempty convex subsets of $\mathbb{R}^{m}$
satisfying $D_{1}\cap D_{2}=\emptyset$. Then, there exist
$a=(a_{1},\ldots,a_{m})\in\mathbb{R}^{m}$ $(a\neq 0)$ and $b\in\mathbb{R}$
such that the following both assertions hold.

  1. $(1)$

For any $y\in D_{1}$, we have $\left\langle a,y\right\rangle\geq b$.

  2. $(2)$

For any $y\in D_{2}$, we have $\left\langle a,y\right\rangle\leq b$.

Note that $A$ and $B$ are nonempty convex subsets of $\mathbb{R}^{m}$. Hence,
by Lemma 4.1, there exist $a=(a_{1},\ldots,a_{m})\in\mathbb{R}^{m}$ $(a\neq
0)$ and $b\in\mathbb{R}$ such that the following both assertions hold.

  1. (1’)

For any $y\in A$, we have $\left\langle a,y\right\rangle\geq b$.

  2. (2’)

For any $y\in B$, we have $\left\langle a,y\right\rangle\leq b$.

Then, we will show that $b=0$. Since $0=(0,\ldots,0)\in A$, we have
$\left\langle a,0\right\rangle\geq b$ by (1’). Namely, we get $b\leq 0$. Since
$(-\varepsilon,\ldots,-\varepsilon)\in B$ for any sufficiently small
$\varepsilon>0$, it is clearly seen that $b=0$ by (2’).

We will show that $a_{i}\geq 0$ for any $i\in M$ by contradiction. Suppose
that there exists an element $i^{\prime}\in M$ satisfying $a_{i^{\prime}}<0$.
Let $y=(y_{1},\ldots,y_{m})\in\mathbb{R}^{m}$ be the element given by

$y_{i}=\begin{cases}-1&\text{if $i\neq i^{\prime}$},\\\
\dfrac{\left(\sum_{j=1,j\neq
i^{\prime}}^{m}\left|a_{j}\right|\right)+1}{a_{i^{\prime}}}&\text{if
$i=i^{\prime}$}.\end{cases}$

Then, we get $y\in B$ and $\left\langle a,y\right\rangle>0$. This contradicts
(2’). Hence, it follows that $a_{i}\geq 0$ for any $i\in M$.

Now, set

$I=\Set{i\in M}{a_{i}>0}.$

Notice that $I\neq\emptyset$. Set $I=\Set{i_{1},\ldots,i_{k}}$, where $k$ is
an integer $(1\leq k\leq m)$ and $i_{1}<\cdots<i_{k}$.

We will show that $y^{*}\in\operatorname{Min}_{I}Y$. Let
$y=(y_{1},\ldots,y_{m})\in Y$ be any element. Since
$(y_{1}-y^{*}_{1},\ldots,y_{m}-y^{*}_{m})\in A$ and $b=0$, by (1’), we have

$\displaystyle
a_{i_{1}}(y_{i_{1}}-y^{*}_{i_{1}})+\cdots+a_{i_{k}}(y_{i_{k}}-y^{*}_{i_{k}})\geq
0.$

Since $a_{i_{1}}>0,\ldots,a_{i_{k}}>0$, the element $y\in Y$ does not satisfy
$y\lneq_{I}y^{*}$. Therefore, we obtain $y^{*}\in\operatorname{Min}_{I}Y$. By
the assumption $(\beta)$, it follows that $y^{*}\in\operatorname{Min}Y$.
$\Box$

##  5\. Preliminaries for applications of Proposition 2.3

Let $X$ be a convex subset of $\mathbb{R}^{n}$. A function $f:X\to\mathbb{R}$
is said to be _convex_ if

$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$

for all $x,y\in X$ and all $t\in[0,1]$. A function $f:X\to\mathbb{R}$ is said
to be _strongly convex_ if there exists $\alpha>0$ satisfying

$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)-\frac{1}{2}\alpha
t(1-t)\left\|x-y\right\|^{2}.$

for all $x,y\in X$ and all $t\in[0,1]$, where $\left\|x-y\right\|$ denotes the
Euclidean norm of $x-y$. For details on convex functions and strongly convex
functions, see [16]. A mapping $f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ is
said to be _convex_ (resp., _strongly convex_) if every $f_{i}$ is convex
(resp., strongly convex).

First, we give the following well-known result. For the sake of the readers’
convenience, we also give the proof.

######  Lemma 5.1.

Let $X$ be a convex subset of $\mathbb{R}^{n}$, and $f:X\to\mathbb{R}^{m}$ be
a convex mapping. Then, the free disposal hull of $f(X)$ is convex.

###### Proof of Lemma 5.1.

Let $y=(y_{1},\ldots,y_{m})$,
$\widetilde{y}=(\widetilde{y}_{1},\ldots,\widetilde{y}_{m})\in\operatorname{FDH}f(X)$
be arbitrary points and $t\in[0,1]$ be an arbitrary element. Then, there exist
$x\in X$ (resp., $\widetilde{x}\in X$) and
$z=(z_{1},\ldots,z_{m})\in\mathbb{R}^{m}_{\geq 0}$ (resp.,
$\widetilde{z}=(\widetilde{z}_{1},\ldots,\widetilde{z}_{m})\in\mathbb{R}^{m}_{\geq
0})$ such that $y=f(x)+z$ (resp.,
$\widetilde{y}=f(\widetilde{x})+\widetilde{z}$). Let $i$ be an arbitrary
integer satisfying $1\leq i\leq m$. Since $f_{i}$ is convex, we have

$\displaystyle f_{i}(tx+(1-t)\widetilde{x})\leq
tf_{i}(x)+(1-t)f_{i}(\widetilde{x}),$

where $f=(f_{1},\ldots,f_{m})$. Since $z_{i}\geq 0$ and $\widetilde{z}_{i}\geq
0$ for any $i=1,\ldots,m$, we also get

$\displaystyle tf_{i}(x)+(1-t)f_{i}(\widetilde{x})\leq
t(f_{i}(x)+z_{i})+(1-t)(f_{i}(\widetilde{x})+\widetilde{z}_{i})=ty_{i}+(1-t)\widetilde{y}_{i}.$

Hence, we obtain

(5.1) $\displaystyle f_{i}(tx+(1-t)\widetilde{x})\leq
ty_{i}+(1-t)\widetilde{y}_{i}.$

Since we have 5.1 for any integer $i$ satisfying $1\leq i\leq m$, it follows
that $ty+(1-t)\widetilde{y}\in\operatorname{FDH}f(X)$. ∎

In the following, for two sets $U,V$, and a subset $W$ of $U$, the restriction
of a given mapping $g:U\to V$ to $W$ is denoted by $g|_{W}:W\to V$.

######  Lemma 5.2.

Let $f:X\to\mathbb{R}^{m}$ be a mapping, where $X$ is a given set. Let
$I=\Set{i_{1},\ldots,i_{k}}$ $(i_{1}<\cdots<i_{k})$ be a nonempty subset of
$M$, where $k$ is the number of the elements of $I$. If
$f_{I}|_{\operatorname{S}(f_{I},X)}:\operatorname{S}(f_{I},X)\to\mathbb{R}^{k}$
is injective, then we have
$\operatorname{S}(f_{I},X)\subseteq\operatorname{S}(f,X)$.

###### Proof of Lemma 5.2.

Suppose that there exists an element $x\in\operatorname{S}(f_{I},X)$ such that
$x\not\in\operatorname{S}(f,X)$. Then, there exists an element $y\in X$
($y\neq x$) satisfying $f_{i}(y)\leq f_{i}(x)$ for any $i\in M$. Since
$I\subseteq M$, it follows that $f_{i}(y)\leq f_{i}(x)$ for any $i\in I$.
Since $x\in\operatorname{S}(f_{I},X)$, we get $f_{I}(x)=f_{I}(y)$. Therefore,
we have $y\in\operatorname{S}(f_{I},X)$. This contradicts the assumption that
$f_{I}|_{\operatorname{S}(f_{I},X)}:\operatorname{S}(f_{I},X)\to\mathbb{R}^{k}$
is injective. ∎

######  Lemma 5.3.

Let $X$ be a convex subset of $\mathbb{R}^{n}$, and $f:X\to\mathbb{R}^{m}$ be
a strongly convex mapping. Then,
$f|_{\operatorname{S}(f,X)}:\operatorname{S}(f,X)\to\mathbb{R}^{m}$ is
injective.

###### Proof of Lemma 5.3.

Suppose that
$f|_{\operatorname{S}(f,X)}:\operatorname{S}(f,X)\to\mathbb{R}^{m}$ is not
injective. Then, there exist $x,y\in\operatorname{S}(f,X)$ such that $x\neq y$
and $f|_{\operatorname{S}(f,X)}(x)=f|_{\operatorname{S}(f,X)}(y)$. Let $i$ be
an arbitrary integer satisfying $1\leq i\leq m$. Since
$f=(f_{1},\ldots,f_{m})$ is strongly convex, there exists $\alpha_{i}>0$
satisfying

$f_{i}(tx+(1-t)y)\leq
tf_{i}(x)+(1-t)f_{i}(y)-\frac{1}{2}\alpha_{i}t(1-t)\left\|x-y\right\|^{2}$

for the points $x,y\in\operatorname{S}(f,X)$ and all $t\in[0,1]$. Set
$t=\frac{1}{2}$. Then, we get

$f_{i}\left(\frac{x+y}{2}\right)\leq\frac{f_{i}(x)+f_{i}(y)}{2}-\frac{\alpha_{i}}{8}\left\|x-y\right\|^{2}.$

Since $f|_{\operatorname{S}(f,X)}(x)=f|_{\operatorname{S}(f,X)}(y)$, we have

$f_{i}\left(\frac{x+y}{2}\right)\leq
f_{i}(x)-\frac{\alpha_{i}}{8}\left\|x-y\right\|^{2}.$

Since $x\neq y$ and $\alpha_{i}>0$, it follows that

$f_{i}\left(\frac{x+y}{2}\right)<f_{i}(x).$

This contradicts $x\in\operatorname{S}(f,X)$. ∎

##  6\. Mathematical applications of Proposition 2.3

In this section, as mathematical applications of Proposition 2.3, we give
Corollaries 6.1, 6.2, 6.3 and 6.4.

First, Proposition 2.3 gives a characterization of the equality of the weak
efficiency and the efficiency for possibly nonconvex problems having a convex
image $f(X)$ as follows:

######  Corollary 6.1.

Let $f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ be a mapping, where $X$ is a
given set. If $f(X)$ is convex, then the following $(\alpha)$ and $(\beta)$
are equivalent:

  1. $(\alpha)$

$\operatorname{WS}(f,X)=\operatorname{S}(f,X)$.

  2. $(\beta)$

${\displaystyle\bigcup_{\emptyset\neq I\subseteq
M}\operatorname{S}(f_{I},X)\subseteq\operatorname{S}(f,X)}$.

###### Proof of Corollary 6.1.

Since $f(X)$ is convex, it is clearly seen that the free disposal hull of
$f(X)$ is also convex. Thus, by Proposition 2.3, we have Corollary 6.1. ∎

Unfortunately, it is not easy to check the convexity of the image of a given
mapping which is possibly nonconvex. A more workable condition ensuring this
characterization is the convexity of a given mapping itself.

######  Corollary 6.2.

Let $X$ be a convex subset of $\mathbb{R}^{n}$ and
$f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ be a convex mapping. Then, the
following $(\alpha)$ and $(\beta)$ are equivalent:

  1. $(\alpha)$

$\operatorname{WS}(f,X)=\operatorname{S}(f,X)$.

  2. $(\beta)$

${\displaystyle\bigcup_{\emptyset\neq I\subseteq
M}\operatorname{S}(f_{I},X)\subseteq\operatorname{S}(f,X)}$.

###### Proof of Corollary 6.2.

Since $f$ is convex, by Lemma 5.1, the free disposal full of $f(X)$ is convex.
Therefore, by Proposition 2.3, we get Corollary 6.2. ∎

As a direct consequence, the characterization is valid for convex programming
problems.

######  Corollary 6.3.

Let $X$ be a convex subset of $\mathbb{R}^{n}$ and
$f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ be a convex mapping. Let
$g_{1},\ldots,g_{\ell}$ be convex functions of $X$ into $\mathbb{R}$, where
$\ell$ is a positive integer. Set

$\Omega=\Set{x\in X}{g_{1}(x)\leq 0,\ldots,g_{\ell}(x)\leq 0}.$

Then, the following $(\alpha)$ and $(\beta)$ are equivalent:

  1. $(\alpha)$

$\operatorname{WS}(f|_{\Omega},\Omega)=\operatorname{S}(f|_{\Omega},\Omega)$.

  2. $(\beta)$

${\displaystyle\bigcup_{\emptyset\neq I\subseteq
M}\operatorname{S}((f|_{\Omega})_{I},\Omega)\subseteq\operatorname{S}(f|_{\Omega},\Omega)}$.

###### Proof of Corollary 6.3.

Since $g_{1},\ldots,g_{\ell}$ are convex functions, it is clearly seen that
$\Omega$ is convex. Since the mapping $f|_{\Omega}:\Omega\to\mathbb{R}^{m}$ is
convex, by Corollary 6.2, we get Corollary 6.3. ∎

Let $f=(f_{1},\ldots,f_{m}):X\to\mathbb{R}^{m}$ be a mapping, where $X$ is a
set. Then, $x^{*}\in X$ is called a _strictly efficient solution_ if there
does not exist $x\in X$ $(x\neq x^{*})$ such that $f_{i}(x)\leq f_{i}(x^{*})$
for all $i=1,\ldots,m$. We denote the set of all strictly efficient solutions
to the problem minimizing $f$ by $\operatorname{SS}(f,X)$.

######  Corollary 6.4.

Let $X$ be a convex subset of $\mathbb{R}^{n}$, and $f:X\to\mathbb{R}^{m}$ be
a strongly convex mapping. Then, we have

$\displaystyle\operatorname{WS}(f,X)=\operatorname{S}(f,X)=\operatorname{SS}(f,X).$

###### Proof of Corollary 6.4.

Since $f|_{\operatorname{S}(f,X)}$ is injective by Lemma 5.3, it is not hard
to see that $\operatorname{S}(f,X)=\operatorname{SS}(f,X)$.

Now, we will show that $\operatorname{WS}(f,X)=\operatorname{S}(f,X)$. Since
$f$ is strongly convex, by Lemma 5.1, the set $\operatorname{FDH}f(X)$ is
convex. Thus, by Proposition 2.3, in order to show that
$\operatorname{WS}(f,X)=\operatorname{S}(f,X)$, it is sufficient to show

(6.1) $\bigcup_{\emptyset\neq I\subseteq
M}\operatorname{S}(f_{I},X)\subseteq\operatorname{S}(f,X).$

Let $I$ be any nonempty subset of $M$. Since
$f_{I}|_{\operatorname{S}(f_{I},X)}:\operatorname{S}(f_{I},X)\to\mathbb{R}^{k}$
is strongly convex, by Lemma 5.3, the mapping
$f_{I}|_{\operatorname{S}(f_{I},X)}$ is injective, where $k$ is the number of
the elements of $I$. By Lemma 5.2, we have
$\operatorname{S}(f_{I},X)\subseteq\operatorname{S}(f,X)$. Thus, we obtain
6.1. ∎

##  7\. A practical application

In this section, as a practical application of Proposition 2.3, we show that
all weakly efficient solutions to a multi-objective LASSO with mild
modification are efficient. The LASSO is a sparse modeling method that is
originally proposed as a single-objective optimization problem [22] and
sometimes treated as a multi-objective one [3]. Let us consider a linear
regression model:

$\displaystyle y=\theta_{1}x_{1}+\theta_{2}x_{2}+\dots+\theta_{n}x_{n}+\xi,$

where $x_{i}$ and $\theta_{i}$ $(i=1,\ldots,n)$ are a predictor and its
coefficient, $y$ is a response to be predicted, and $\xi$ is a Gaussian noise.
Given a matrix $X$ with $m$ rows of observations and $n$ columns of predictors
and a row vector $y$ of $m$ responses, the (original) LASSO regressor is the
solution to the following problem:

(7.1)
$\operatorname*{minimize}_{\theta\in\mathbb{R}^{n}}g_{\lambda}(\theta):=\frac{1}{2m}\left\|X\theta-y\right\|^{2}+\lambda\left|\theta\right|,$

where $\left|\;\cdot\;\right|$ is the $\ell_{1}$-norm and $\lambda$ is a user-
specified positive number to force the solution to be sparse (i.e., the
optimal $\theta$ contains many zeros). Note that with $\lambda=0$, the problem
7.1 reduces to the ordinary least squares (OLS) regression. Choosing an
appropriate value for $\lambda$ requires repeated solution of 7.1 with varying
$\lambda$, which is the most time-consuming part of this method.

To find a good solution without such a costful hyper-parameter search, the
problem 7.1 is sometimes reformulated as a multi-objective one whose efficient
solutions are optimal solutions to the original problem 7.1 with different
$\lambda$’s (for example, see [3]). We treat the OLS term and the
regularization term as individual objective functions:

$f_{1}(\theta)=\frac{1}{2m}\left\|X\theta-y\right\|^{2},\quad
f_{2}(\theta)=\left|\theta\right|.$

In the multi-objective problem of minimizing $f=(f_{1},f_{2})$, the equality
$\operatorname{WS}(f,\mathbb{R}^{n})=\operatorname{S}(f,\mathbb{R}^{n})$ does
not necessarily hold (for example, if $X$ is a zero matrix, then
$\operatorname{WS}(f,\mathbb{R}^{n})=\mathbb{R}^{n}$ and
$\operatorname{S}(f,\mathbb{R}^{n})=\set{(0,\ldots,0)}$).

To avoid such a corner case, we consider a modified version of multi-objective
LASSO:

(7.2) $\begin{split}\operatorname*{minimize}_{\theta\in\mathbb{R}^{n}}\
&\widetilde{f}(\theta):=(\widetilde{f}_{1}(\theta),\widetilde{f}_{2}(\theta))\\\
\text{where }&\widetilde{f}_{i}(\theta)=f_{i}(\theta)+\varepsilon
f_{2}(\theta)\quad(i=1,2).\end{split}$

In 7.2, we assume that $\varepsilon$ is a positive real number. Note that
$\widetilde{f}$ in 7.2 is a non-differentiable mapping that is convex but
never strongly convex. As a practical application of Proposition 2.3, we can
show the following.

######  Theorem 7.1.

In 7.2, we have
$\operatorname{S}(\widetilde{f},\mathbb{R}^{n})=\operatorname{WS}(\widetilde{f},\mathbb{R}^{n})$.

###### Proof of Theorem 7.1.

Since $\widetilde{f}$ is convex, it is sufficient to show that
$\operatorname{S}(\widetilde{f_{i}},\mathbb{R}^{n})\subseteq\operatorname{S}(\widetilde{f},\mathbb{R}^{n})$
for $i=1,2$ by Corollary 6.2, which is one of the applications of Proposition
2.3. Since $\widetilde{f_{2}}$ has the unique minimizer
$(0,\ldots,0)\in\mathbb{R}^{n}$, we have
$\operatorname{S}(\widetilde{f_{2}},\mathbb{R}^{n})\subseteq\operatorname{S}(\widetilde{f},\mathbb{R}^{n})$.
In order to show
$\operatorname{S}(\widetilde{f_{1}},\mathbb{R}^{n})\subseteq\operatorname{S}(\widetilde{f},\mathbb{R}^{n})$,
we prepare the following.

######  Lemma 7.2.

Let $\theta^{\ast},\widetilde{\theta}^{\ast}$ be elements of
$\operatorname{S}(\widetilde{f_{1}},\mathbb{R}^{n})$. Then, we have

$\displaystyle\left\|X\theta^{\ast}-y\right\|^{2}=\|X\widetilde{\theta}^{\ast}-y\|^{2}\mbox{
and }\left|\theta^{\ast}\right|=|\widetilde{\theta}^{\ast}|.$

###### Proof of Lemma 7.2.

It is sufficient to consider the case
$\theta^{\ast}\not=\widetilde{\theta}^{\ast}$. Since $\widetilde{f_{1}}$ is
convex, there exists a mapping $c=(c_{1},\ldots,c_{n}):[0,1]\to\mathbb{R}^{n}$
given by

$\displaystyle c(t)=(a_{1}t+b_{1},\ldots,a_{n}t+b_{n}),\ \ c(0)=\theta^{*},\ \
c(1)=\widetilde{\theta}^{*}$

such that $c(t)$ is the minimizer of $\widetilde{f}_{1}$ for all $t\in[0,1]$,
where $a_{1},\ldots,a_{n},b_{1},\ldots,b_{n}\in\mathbb{R}$ and $a_{i}\not=0$
for at least one index $i$. Set

$\displaystyle N$ $\displaystyle=\set{1,\ldots,n},$
$\displaystyle\widetilde{N}$ $\displaystyle=\set{i\in N}{\mbox{$a_{i}=0$ and
$b_{i}=0$}}.$

Since $\widetilde{N}\subsetneq N$, we set
$N\setminus\widetilde{N}=\set{i_{1},\ldots,i_{\rho}}$. Since $c$ is
continuous, it is not hard to see that there exists an open interval $I$
$(\subseteq[0,1])$ such that for any $k=1,\ldots,\rho$, either one of the
following two holds:

  1. (1)

We have $c_{i_{k}}(t)>0$ for all $t\in I$.

  2. (2)

We have $c_{i_{k}}(t)<0$ for all $t\in I$.

Set $X=(x_{ij})_{1\leq i\leq m,1\leq j\leq n}$ and $y=(y_{1},\ldots,y_{m})$.
Then, the composition $\widetilde{f}_{1}\circ c:I\to\mathbb{R}$ is expressed
by

$\displaystyle(\widetilde{f}_{1}\circ
c)(t)=\frac{1}{2m}\sum_{j=1}^{m}\left(\sum_{k=1}^{\rho}x_{j,i_{k}}(a_{i_{k}}t+b_{i_{k}})-y_{j}\right)^{2}+\varepsilon\sum_{k=1}^{\rho}q_{i_{k}}(a_{i_{k}}t+b_{i_{k}}),$

where

$\displaystyle q_{i_{k}}$ $\displaystyle=\begin{cases}1&\mbox{ if
$c_{i_{k}}(t)>0$ for all $t\in I$},\\\ -1&\mbox{ if $c_{i_{k}}(t)<0$ for all
$t\in I$}.\end{cases}$

By calculation, we obtain

$\displaystyle\frac{d^{2}(\widetilde{f}_{1}\circ
c)}{dt^{2}}(t)=\frac{1}{m}\sum_{j=1}^{m}\left(\sum_{k=1}^{\rho}x_{j,i_{k}}a_{i_{k}}\right)^{2},$

where $t\in I$. Since $\widetilde{f}_{1}\circ c:[0,1]\to\mathbb{R}$ is a
constant function, we have

(7.3) $\displaystyle\sum_{k=1}^{\rho}x_{j,i_{k}}a_{i_{k}}=0$

for all $j=1,\ldots,m$.

For simplicity, set $h(\theta)=\left\|X\theta-y\right\|^{2}$. Then, we have

$\displaystyle(h\circ c)(t)$
$\displaystyle=\sum_{j=1}^{m}\left(\sum_{k=1}^{\rho}x_{j,i_{k}}\left(a_{i_{k}}t+b_{i_{k}}\right)-y_{j}\right)^{2}$
$\displaystyle=\sum_{j=1}^{m}\left(\left(\sum_{k=1}^{\rho}x_{j,i_{k}}a_{i_{k}}\right)t+\sum_{k=1}^{\rho}x_{j,i_{k}}b_{i_{k}}-y_{j}\right)^{2}$
$\displaystyle=\sum_{j=1}^{m}\left(\sum_{k=1}^{\rho}x_{j,i_{k}}b_{i_{k}}-y_{j}\right)^{2}$

for all $t\in[0,1]$. The last equality above is obtained by 7.3. Hence, since
$h\circ c:[0,1]\to\mathbb{R}$ is a constant function, we obtain

$\displaystyle h(\theta^{\ast})=(h\circ c)(0)=(h\circ
c)(1)=h(\widetilde{\theta}^{\ast}).$

Since
$\widetilde{f}_{1}(\theta^{\ast})=\widetilde{f}_{1}(\widetilde{\theta}^{\ast})$,
we also have $|\theta^{\ast}|=|\widetilde{\theta}^{\ast}|$. ∎

Now, let $\theta\in\operatorname{S}(\widetilde{f}_{1},\mathbb{R}^{n})$ be an
arbitrary element. Suppose that
$\theta\not\in\operatorname{S}(\widetilde{f},\mathbb{R}^{n})$. Then, there
exists $\widetilde{\theta}\in\mathbb{R}^{n}$ such that
$\widetilde{f}_{1}(\widetilde{\theta})=\widetilde{f}_{1}(\theta)$ and
$\widetilde{f}_{2}(\widetilde{\theta})<\widetilde{f}_{2}(\theta)$. Thus, we
have
$\theta,\widetilde{\theta}\in\operatorname{S}(\widetilde{f}_{1},\mathbb{R}^{n})$
and $|\widetilde{\theta}|<|\theta|$. This contradicts Lemma 7.2. Hence, we
obtain $\theta\in\operatorname{S}(\widetilde{f},\mathbb{R}^{n})$. ∎

##  8\. Conclusion

In this paper, we have given a characterization of the equality of weak
efficiency and efficiency when the free disposal hull of the domain is convex.
By this fact, we have presented four classes of optimization problems where
this characterization holds, including convex problems. As a practical
application, we have also shown that all the weakly efficient solutions to a
multi-objective LASSO defined by 7.2 are efficient. We expect that the scope
of the multi-objective reformulation discussed in Section 7 is not limited to
the LASSO. The same idea may be applied to a wide range of sparse modeling
methods, including the group lasso [25], the fused lasso [23], the graphical
lasso [7], the smooth lasso [8], the elastic net [27], etc.

## Acknowledgements

The authors are grateful to Kenta Hayano, Yutaro Kabata, and Hiroshi Teramoto
for their kind comments. Shunsuke Ichiki was supported by JSPS KAKENHI Grant
Numbers JP19J00650 and JP17H06128. This work is based on the discussions at
2018 IMI Joint Use Research Program, Short-term Joint Research “multi-
objective optimization and singularity theory: Classification of Pareto point
singularities” in Kyushu University.

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