数理解析研究所講究録
1187 巻 2001 年 88-94

88

SOME PROPERTIES OF GENERALIZED SUPREMUM
IN PARTIALLY ORDERED LINEAR SPACES

NAOTO KOMUROMathematics Laboratory, AsahikawaCampus, Hokkaido University of Education\S 1 INTRODUCTION AND BASIC RESULTS
Let $E$ be a linear space over , and be a convex cone in $E$ satisfying(P1) $E=P-P$ ,(P2) $P\cap(-P)=\{0\}$ .
An order relation in $E$ can be defined by $x\leq y\Leftrightarrow y-x\in P$ . We call alinear space $E$ equipped with such a positive cone a partially orderedlinear space, and denote it by $(E, P)$ .
For a subsetof $E$ , the generalized supremum
is defined tobe the set of all minimal elements of $U(A)$ , where $U(A)$ is the set of allupper bound of . In other words, $U(A)=\{x\in E|y\leq x, \forall y\in A\}$ ,$\{a \in E|b\leq a, b\in U(A)\Rightarrow a=b\}$ . The generalized
andinfimum Inf can be defined similarly. In order to distinguish this notion
from the least upper bound and the greatest lower bound, we denotethe latter ones by $\sup$ and inf respectively. If $E$ is order complete,thenholds whenever the subset
is upper bounded(i.e.,). Whenandis closed and not a lattice cone,
becomes an infinite set in most cases. However, it is possibly empty,
even whenis upper bounded. For the preparation, we recall somebasic results of the generalized supremum. The proofs of the following
propositions can be found in previous papers $([4],[5],[6])$ .
$P$$\mathbb{R}$

$P$

$A$

$\mathrm{S}\mathrm{u}\mathrm{p}$$A$$\mathrm{S}\mathrm{u}\mathrm{p}A=$

$A$

$A$

$A$

$A$

$\mathrm{S}\mathrm{u}\mathrm{p}A=\{\sup A\}$

$U(A)\neq\emptyset$$\mathrm{S}\mathrm{u}\mathrm{p}$$P$$E=\mathbb{R}^{n}$

$A$

$A$

Proposition 1. For $a\in E$ and(1)
,
(2)
,
(3)
.

$\lambda>0$$\mathrm{S}\mathrm{u}\mathrm{p}(A+a)=\mathrm{S}\mathrm{u}\mathrm{p}A+a$$\mathrm{S}\mathrm{u}\mathrm{p}\lambda A=\lambda \mathrm{S}\mathrm{u}\mathrm{p}A$$\mathrm{S}\mathrm{u}\mathrm{p}A=-\mathrm{I}\mathrm{n}\mathrm{f}(-A)$, we have$A$89
Proposition 2. For an arbitrary set $A\subset E$ with

$U(A)\neq\emptyset$,$\mathrm{S}\mathrm{u}\mathrm{p}A=\mathrm{S}\mathrm{u}\mathrm{p}(coA)$

holds where $coA$ is the convex hull of$A$.

Prposition 3. For a,the converse is also true. Moreover,$b\in E_{f}\mathrm{S}\mathrm{u}\mathrm{p}\{a, b\}\neq\emptyset$

implies$\mathrm{I}\mathrm{n}\mathrm{f}\{a, b\}\neq\emptyset$

and$a_{+}=\mathrm{S}\mathrm{u}\mathrm{p}\{a, 0\}$and

$a+b-\mathrm{S}\mathrm{u}\mathrm{p}\{a, b\}=\mathrm{I}\mathrm{n}\mathrm{f}\{a, b\}$holds and in particular we have $a\in a_{+}+a_{-}$ where.

$a_{-}=\mathrm{I}\mathrm{n}\mathrm{f}\{a, 0\}$A partially ordered linear space $(E, P)$ is said to be monotone ordercomplete (m.o.c. for short) if every upper bounded totally ordered subset$(E, P)$ is m.o.c.
of $E$ has the least upper bound in $E$ . In the case
if and only if $P$ is closed. In the case when $E$ is a Banach space with aclosed positive cone $P$ satisfying $P^{*}-P^{*}=E^{*},$ $(E^{*}, P^{*})$ is m.o.c. whereis the topological dual of $E$ and $P^{*}=\{x^{*}\in E^{*}|x^{*}(x)\geq 0, x\in P\}$ .
The proofs of these facts can be seen in a previous paper [6].
$E=\mathbb{R}^{d},$$E^{*}$

Proposition 4.
Suppose that a partially ordered linear space $(E, P)$
is monotone order complete. Then for every subset of $E$ ,$A$$U(A)=(\mathrm{S}\mathrm{u}\mathrm{p}A)+P$

holds. In particular,$\mathrm{S}\mathrm{u}\mathrm{p}\{a, b\}\neq\emptyset,$$U(a, b)=(\mathrm{S}\mathrm{u}\mathrm{p}\{a, b\})+P$$\mathrm{I}\mathrm{n}\mathrm{f}\{a, b\}\neq\emptyset$.

for every a,$b\in E_{f}$andLet $(E, P)$ be a partially ordered linear space, and suppose that $P$ isalgebraically closed, that is, every straight line of $E$ meets $P$ by a closedinterval. A pointof a convex subset $A\subset E$ is called an algebraic
interior point of
if for every $z\in E$ , there exists $\lambda>0$ such that$x+\lambda z\in A$ .
Algebraic exterior points are defined similarly, and wedenote the algebraic interior (exterior) of by int (ext ) respectively.
Moreover,int
is called the algebraic boundary of .
A convex subset
of $P$ is called an exposed face of $P$ if there exists asupporting hyperplane $H$ of $P$ such that $C=P\cap H$ . By $S(P)$ , we denote$\dim C$ is defined as thethe set of all exposed faces of $P$ . For
dimension of affC where affC denotes the affine hull of .
$x$

$A$

$A$

$\partial A=$$($$A$

$A$

$A$

$A\cup \mathrm{e}\mathrm{x}\mathrm{t}A)^{c}$$C$$C\in \mathfrak{F}(P),$$C$

Propositon 5.

Suppose that $P$ is algebraically closed and int
If $\dim C<\infty$ for every $C\in ff(P)$ , then$U(A)=(\mathrm{S}\mathrm{u}\mathrm{p}A)+P$

holds for every subset $A\subset E$ .

$P\neq\emptyset$.

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Corollary 1. Suppose that $(E, P)$ satisfies the hypotheses in Proposition
consists of a4 or Proposition 5, and let be a subset of E. If
single element a, then is the least upper bound of .
$A$$\mathrm{S}\mathrm{u}\mathrm{p}$

$A$

$A$

$a$

Corollary 2. For every subsetof $E,$ $U(L(U(A)))=U(A)$ holds
where $L(U(A))$ denotes the lower bound of $U(A)$ . Moreover, if $(E, P)$satisfies the hypotheses in Proposition 4 or Propositon 5, then we have.
Inf$A$$\mathrm{S}\mathrm{u}\mathrm{p}A=\mathrm{S}\mathrm{u}\mathrm{p}A$$\mathrm{S}\mathrm{u}\mathrm{p}$The proofs of these results can be seen in $[4],[5],[6]$ , and [7].

\S 2 PROPERTIES OF THE SET OF UPPER BOUNDS AND LOWER BOUNDS
the finiteThrough this section, we consider only the case whendimensional Euclidean space and the positive cone $P$ is a closed convex. Under this assumptions, it is easy to observe
cone satisfying. Moreoverthat $U(A)$ and $L(A)$ are closed convex sets for every
is monotone order complete, and by Proposition 4, the formula$E=\mathbb{R}^{d}$$(\mathrm{P}1),(\mathrm{P}2)$$A\subset \mathbb{R}^{d}$$(\mathbb{R}^{d}, P)$(2.1)$U(A)=(\mathrm{S}\mathrm{u}\mathrm{p}A)+P$

be the family of all upper bounded subset
andalways holds. Let
respectively, i.e.
and lower bounded subset in$\mathfrak{B}’$$\mathfrak{B}$$\mathbb{R}^{d}$$\mathfrak{B}=\{A\subset \mathbb{R}^{d}|A\neq\emptyset, U(A)\neq\emptyset\}$,$\mathfrak{B}’=\{B\subset \mathbb{R}^{d}|B\neq\emptyset, L(B)\neq\emptyset\}$We define an equivalence$\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\sim \mathrm{i}\mathrm{n}\mathfrak{B}$.

by$A\sim B\Leftrightarrow U(A)=U(B)$$(A, B\in \mathfrak{B})$.

Let $X$ be the quotient set $\mathfrak{B}/\sim=\{[A]|A\in \mathfrak{B}\}$ where $[A]$ denotes theequivalence class of .
$A$

Proposition 6. $[A]=[L(U(A))]=[L(\mathrm{S}\mathrm{u}\mathrm{p}A)]$ holds for every. Moreover if $[L(B)]=[A]$ for
and $[L(B)]=$ [Inf $B$ ] for every
then $A\subset L(B)$ .
someand$A\in \mathfrak{B}$$B\in \mathfrak{B}’$$A\in \mathfrak{B}$$B\in \mathfrak{B}_{f}’$proof. By (2.1) we can easily see that
$U(A)=U(L(U(A)))$$=U(L(\mathrm{S}\mathrm{u}\mathrm{p}A+P))$$=U(L(\mathrm{S}\mathrm{u}\mathrm{p}A))$.

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This directly shows the first formula. Since we also have
$P$by (2.1), the second formula follows similarly. Indeed,$U$(Inf $B$ ) $=U((\mathrm{I}\mathrm{n}\mathrm{f}B)-P)=U(L(B))$ . The latter statement follows
from Corollary 2. Indeed,$L(B)=(\mathrm{I}\mathrm{n}\mathrm{f}B)-$$(B\in \mathfrak{B}’)$$A\subset L(U(A))$$=L(U(L(B)))$$=L(B)$ .

For every $[A]\in X$ , two operations $u([A])=U(A)$ and $l([A])=$$L(U(A))$ are well defined. By virtue of (2.1), $X$ can be identified with. We now define an
the set $\{U(A)|A\in \mathfrak{B}\}$ or the set$X$
byorder relation in$\{\mathrm{S}\mathrm{u}\mathrm{p}A|A\in \mathfrak{B}\}$$[A],$ $[B]\in X$$[A]\leq[B]\Leftrightarrow u([B])\subset u([A])$.

By this definition $X$ becomes a partially ordered set. Moreover, we shallshow that $X$ is an order complete lattice and that $X$ has a subset which. Let
be the set of all $[A]\in X$ such thatis order isomorphic to. Note that the correspondence which$u([A])=a+P$ for some
to $[A]\in X_{1}$ such that $u([A])=a+P$ is one to one.
assigns$(\mathbb{R}^{d}, P)$$X_{1}$$a\in \mathbb{R}^{d}$$a\in \mathbb{R}^{d}$

Theorem 1. $X$ is an order complete lattice with respect to the order$‘\leq’$by the correspondence
Moreover,is order isomorphic towhere $u([A])=a+P$ .
$(\mathbb{R}^{d}, P)$$X_{1}$$\mathbb{R}^{d}\ni arightarrow[A]\in X_{1}$Lemma 1. Let
such that(1)(2)$\{A_{\sigma}\}_{\sigma\in\Sigma}\subset \mathfrak{B}_{f}$$\bigcup_{\sigma\in\Sigma}A_{\sigma}\in \mathfrak{B}$

and

and

$\bigcup_{\lambda\in\Lambda}B_{\lambda}\in \mathfrak{B}’$$\bigcap_{\sigma\in\Sigma}u([A_{\sigma}])=u([\bigcup_{\sigma\in\Sigma}A_{\sigma}])$,$\bigcap_{\lambda\in\Lambda}l([L(B_{\lambda})])=l([L(\bigcup_{\lambda\in\Lambda}B_{\lambda})])$$U(L( \bigcap_{\sigma\in\Sigma}u([A_{\sigma}])))=\bigcap_{\sigma\in\Sigma}u([A_{\sigma}])$$\bigcap_{\lambda\in\Lambda}l([L(B_{\lambda})])$, be arbitrary families
Then$\{B_{\lambda}\}_{\lambda\in\Lambda}\subset \mathfrak{B}’$,$L(U( \bigcap_{\lambda\in\Lambda}l([L(B_{\lambda})])))=$.

proof. (1) can be shown directly by the definitions. Indeed,$\bigcap_{\sigma\in\Sigma}u([A_{\sigma}])=\bigcap_{\sigma\in\Sigma}U(A_{\sigma})$$=U( \bigcup_{\sigma\in\Sigma}A_{\sigma})$$=u([ \bigcup_{\sigma\in\Sigma}A_{\sigma}])$,and$\bigcap_{\lambda\in\Lambda}l([L(B_{\lambda})])=\bigcap_{\lambda\in\Lambda}L(U(L(B_{\lambda})))$$= \bigcap_{\lambda\in\Lambda}L(B_{\lambda})$$=L( \bigcup_{\lambda\in\Lambda}B_{\lambda})$$=L(U(L( \bigcup_{\lambda\in\Lambda}B_{\lambda})))$$=l([L( \bigcup_{\lambda\in\Lambda}B_{\lambda})])$.

.

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Moreover, we can see by (1) and Corollary 2 that$U(L( \bigcap_{\sigma\in\Sigma}u([A_{\sigma}])))=U(L(u([\bigcup_{\sigma\in\Sigma}A_{\sigma}])))$$=u([ \bigcup_{\sigma\in\Sigma}A_{\sigma}])$.

The latter formula can be shown similarly.
proof of Theorem 1. Let be an upper bounded subset of $X$ . Then there
such that $U(B)\subset u([A])$ for all $[A]\in Y$. Let
exists a subset$\mathrm{Y}$$B\in \mathfrak{B}$$C=L(\cap u([A]))$$[A]\in Y$

Then$C\in \mathfrak{B}$and by Lemma 1,$U(C)=$$u([A])$$\cap$$[A]\in Y$$\supset U(B)$.

This means that $[C]$ is the least upper bound of Y. Next we suppose thatis a lower bounded subset of $X$ . We put$\mathrm{Y}’$$C’=$$\cap$$L(u([A]))$$[A]\in Y’$.
Thenand $U(C’)\supset U(L(u([A])))=u([A])$ for every$Y’$. Let $[B’]$ be an arbitrary lower
Hence $[C’]$ is a lower bound ofbound of $Y’$ then $u([A])\subset U(B’)$ for every $[A]\in Y’$ , and we have$\bigcap_{[A]\in Y’}L(u([A]))\supset L(U(B’))$ . Thus
$C’\in \mathfrak{B}$$[A]\in \mathrm{Y}’$$U(C’)=U(\cap L(u([A])))$$[A]\in Y’$$\subset U(L(U(B’)))$$=u([B’])$ .

This means that $[C’]$ is the greatest lower bound of $Y’$ . Thus we have
proved that $X$ is order complete. To prove that $X$ forms a lattice it issufficient to show that $\{[A], [B]\}$ is bounded for every pair $[A],$ $[B]\in X$ .
For $a\in u([A])$ and $b\in u([B])$ we can choose $q\in P$ such that $a-b=p-q$
by the condition (P1). Hence $a+q=b+p\in u([A])\cap u([B])$ . Thus$u([A])\cap u([B])$ and $L(u([A]))\cap L(u([B]))$ are both nonempty, and we put$C_{1}=L(u([A])\cap u([B]))$ , and $C_{2}=L(u([A]))\cap L(u([B]))$ . It is easy to
see that $[C_{1}]\geq[A],$ $[B]$ and $[C_{2}]\leq[A],$ $[B]$ , and this is what we wantedto show. The second statement of this theorem is obvious.
$p,$

By $[A]\vee[B]$ , and $[A]$ A $[B]$ we denote the least upper bound and thegreatest lower bound of $\{[A], [B]\}$ in $X$ respectively. Repeating the samearguement of the proof of Theorem 1, we obtain

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Proposition 7. For $[A],$ $[B]\in X_{f}$$[A]\vee[B]=[L(u([A])\cap u([B]))]$ ,(1)$[A]$ A $[B]=[L(u([A]))\cap L(u([B]))]$ .
(2)

For
we can characterize $U(A)$ by using the support function ofand the dual cone $P^{*}=\{x^{*}\in \mathbb{R}^{d}| <x^{*}, x>\geq 0 x\in P\}$ . In theconditions we have assumed, the relation$A\in \mathfrak{B}$$A$(2.2)$P=P^{**}=\{x\in \mathbb{R}^{d}|<x^{*}, x>\geq 0 x^{*}\in P^{*}\}$

holds. If$A\in \mathfrak{B}$finite on
all $x\in A$ .

$P^{*}$

then the support function $f_{A}(x^{*})= \sup_{x\in A}<x^{*},$ $x>$ isIndeed if $x_{0}\in U(A)$ , then $<x^{*},$ $x>\leq<x^{*},$ $x_{0}>$ holds forTheorem 2. For every$A\in \mathfrak{B}$,$U(A)= \bigcap_{x^{*}\in\partial P^{*}}\{x|<x^{*}, x>\geq f_{A}(x^{*})\}$

where$\partial P^{*}$.

denotes the boundary of,$P^{*}$

satisfies (P1) and (P2), if $P$ is closed
It is known that the dual conein. For the proof of Theorem 2, we prepare a basic lemma.
$P^{*}$$\mathbb{R}^{d}$be a closed positive cone satisfying (P1) andLemma 2. Let(P2). Then,such that(1)
if $0\leq b\leq a$ and $b\neq 0$ , there exists, then there exists $t>0$(2)
if is an interior point of andsuch that $a+t(a-b)\in\partial P$ .
$P\subset \mathbb{R}^{d}$$nb\not\leq a$$n\in \mathbb{N}$

$P$

$a$

$b\not\leq a$. Then the closedfor every $n=1,2,3,$
proof. Suppose thatness ofwhich contradicts (P1). Hence there exists
such that $a-nb\not\geq 0$ and (1) follows immediately. Next we suppose that$a+t(a-b)\geq 0$ for every $t>0$ . Then $\frac{t+1}{t}a-b\geq 0$ $(t>0)$ and theclosedness of $P$ yields $a-b\geq 0$ which contradicts the assumption. Hence
we can choose $t_{0}= \sup\{t>0|a+t(a-b)\in P\}$ , and $a+t_{0}(a-b)\in\partial P$ .
$\cdots$$\frac{a}{n}-b\geq 0$$n\in \mathrm{N}$$P\mathrm{y}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}\mathrm{s}-b\geq 0$’
is obvious we will prove only the converse.
proof of Theorem 2. Since
By (1) in Lemma 2, we can takeLet
be an arbirary element ofMoreover, by (2) in Lemma 2, there exists
such thatfor some $0<\lambda<1$ . Suppose
such thatthat $x \in\bigcap_{x^{*}\in\partial P^{*}}\{x|<x^{*}, x>\geq f_{A}(x^{*})\}$ and $y\in A$ , then$‘\subset$$P^{*}$$x^{*}$$x_{1}^{*}\in\partial P^{*}$$x_{1}^{*}\not\leq x^{*}$$x_{2}^{*}\in\partial P^{*}$$x^{*}=\lambda x_{1}^{*}+(1-\lambda)x_{2}^{*}$$<x^{*},$ $x-y>=\lambda<x_{1}^{*},$ $x-y>+(1-\lambda)<x_{2}^{*},$$\geq 0$.

$x-y>$

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Since $x^{*}\in P^{*}$ and $y\in A$ are arbirary, we can conclude by (2.2) that$x\in U(A)$ .
The following is an immediate consequence of this theorem.

Corollary 3. Let $A,$
then $[A]=[B]$ .

$B\in \mathfrak{B}$and suppose that $f_{A}(x^{*})=f_{B}(x^{*})$ on$\partial P^{*}$,REFERENCES
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$properties\backslash$

N. Komuro
Hokkaido University of Education at Asahikawa
Hokumoncho 9 chome Asahikawa
070 Japan-mail: komuro@ atson. asa. hokkyodai. ac.jp$e$