数理解析研究所講究録
第 829 巻 1993 年 25-41

25

無限次元空間上の順序を保存する作用素のresolvent 方程式について

お茶の水女子大理

荻原俊子 (Toshiko Ogiwara)

1. Introduction
The Perron-Frobenius theorem, concerning the properties of eigenvalues
and eigenvectors of square matrices whose components are nonnegative,
has been extended and applied in various ways. Recently, in the interestof mathematical economics, nonlinear extensions have been obtained by
H. Nikaido [5], M. Morishima [6], T. Fujimoto [7], Y. Oshime [8] [9] [10].
Many of them have been studied in a n-dimensional Euclidean space andthe resolvent problem has been discussed in the nonlinear Leontief model.
In this paper, we extend their results to a mapping on an infinite dimensional space. In doing so, we introduced the notion of indecomposability
for a nonlinear mapping on an infinite dimensional space. We consider
the resolvent problem of an order-preserving mapping on a positive cone(aclosed convex cone with vertex at ) of a strongly ordered Banach space(an ordered Banach space with positive cone having nonempty interior).
We define the resolvent mapping and study the solutions of the resolevent
equation applying the results on eigenvalue problem which is obtained by
the author. Though we deal with a class of positively homogeneous mappings in this paper, we obtain the same results on a more general classof subhomogeneous mappings along the argument of Y. Oshime [9] for a
finite dimensional space.
$0$

2. Notations and assumptions
Let $E$ be an ordered Banach space, that is, a real Banach space provided(a closed convex cone with vertex at such thatwith an order cone
$E_{+}$$0$26$E+\cap(-E_{+})=\{0\})$. We assume that $E+has$ nonempty interior (wecall such a space a strongly ordered Banach space) and $\dim E\geq 2$ . Let
Note that
denote the interior of $E+and(E_{+})^{b}$ the boundary of$(E_{+})^{i}$$E+\cdot$is not empty, which follows from $\dim E\geq 2$ .
For $x,$ $y\in E$ , we write $x\gg y$ if $x-y\in(E_{+})^{i},$ $x>y$ if $x-y\in E+\backslash \{0\}$ ,and $x\geq y$ if $x-y\in E_{+}$ . For $x\in E$ , we say is strongly positive, positive,nonnegative if and only if $x\gg O,$ $x>0,$ $x\geq 0$ , respectively.
We assume order-preserving norm on $E$ , namely, $0\leq x\leq y$ implies$(E_{+})^{b}\backslash \{0\}$$x$$\Vert x||\leq||y||$.

Fix $e\gg O$ , e-norm can be defined on $E$ as$||x||_{e}= \inf\{m\geq 0| -me\leq x\leq me\}$ ,because of closedness of $E+$$||x||_{e}= \min\{m\geq 0| -me\leq x\leq me\}$.

For each $e\gg O$ , e-norm is well-defined and equivalent to the original norm.
Letbe a mapping from $E+into$ itself. We consider the following
assumptions:
Al:(compact) is continuous and the image of a bounded set by isrelatively compact,
$T$

$T$

$T$

A2:(positive1y homogeneous)$T(\lambda x)=\lambda Tx$$A3$

for any$\lambda>0,$ $x\geq 0$,: (order-preserving)$x\leq y$

An order-preserving map
For $x\geq 0$ , we denote$T$

implies$Tx\leq Ty$.

is also called monotone or isotone.

$E_{x}=\{y\geq 0|\exists\lambda>0, y\leq\lambda x\}$.

27
We can see that $E_{x}=\{0\}$ if and only if $x=0$ , and $E_{x}=E_{+}$ if and only if$x\gg 0$.

A4:(indecomposab1e)$\{0\}\subsetneq E_{x-y}\subsetneq E_{+}$

implies$Tx-Ty\not\in E_{x-y}$.

This is an infinite dimensional extension of indecomposability defined by
M. Morishima [6], where $E=R^{n},$ $E+=R_{+}^{n}=\{(x_{1}, \cdot\cdot x_{n})\in R^{n}|$$x_{i}\geq 0,0\leq i\leq n\}$ . It is also an nonlinear extension of indecomposability found in M. A. $Krasnose1’ ski_{1}\cdot[4]$ and order irreducibility for a linear
operator introduced by H. Nikaido [5]. Furthermore, under the assumption A3, when is a bounded linear operator, is indecomposable if andonly if is irreducible [2], semi-non-support [3]. Strongly order-preserving
mappings(x $<y$ implies $Tx\ll Ty$ ) are indecomposable.
We denote
$T$

$T$

$T$

$VP(T)=$$=${{$\lambda|\lambda$

is an eigenvalue of$\lambda|Tx=\lambda x$$T$}for $\exists x>0$ },$\lambda_{0}(T)=\{\begin{array}{l}\sup\{\lambda|\lambda\in VP(T)\},(VP(T)\neq\emptyset)-\infty,(VP(T)=\emptyset)\end{array}$$||T||= \sup\{||Tx|||x\in B_{1}\}$$\lambda_{0}(T)$.

is called the spectral radius, as in the case of linear operators.

Lemma 2.1 Under the assumptions $A2$ and $A3_{f}\Vert T\Vert<+\infty,$.
exists and satisfies$\lim_{narrow\infty}||T^{n}||^{\frac{1}{n}}\leq\Vert T||$

Denote$r(T)= \lim_{narrow\infty}||T^{n}||^{\frac{1}{n}}$.

3. Fundamental results
We summarize the results obtained before.

$||T^{n}||^{\frac{1}{n}}$$narrow\infty$科

28
that $T:E+arrow E_{+}$ satisfies the assumptions $A1,A2,A3$ ,and $r(T)>0$ . Then $r(T)$ is an eigenvalue of and $r(T)=\lambda_{0}(T)$ .

Theprem 3.1$\prime s_{uppose}$$T$satisfy the assumptions $A1_{f}A2_{f}A3$ and $A4$ . Then
Theorem 3.2 Let$r(T)$ is a positive eigenvalue having strongly positive eigenvectors uniquely
without positive scalar multiplication. Moreover there is no other eigen$T$

value.

Under some conditions, $r(T)$ coincides with the maximal eigenvalue.
Now we give a characterization of $r(T)$ , which is a key to solve the resolvent
problem.
Theorem 3.3 Let satisfy the assumptions $A1,A2,A3$ , then(:) $\lambda>r(T)$ if and only if there exists some $x\gg O$ such that $Tx\ll\lambda x$ .
(ii).
if and only if there exists some $x>0$ such that
Further let satisfy $A4$ . Then(iii).
if and only if there exists some $x>0$ such that(iv) $\lambda<r(T)$ if and only if there exists some $x\gg O$ such that $Tx\gg\lambda x$ .
$T$

$Tx\geq\lambda x$$\lambda\leq r(T)$$T$$Tx\leq\lambda x$$\lambda\geq r(T)$

4. Resolvent problem

In this subsection, we treat the resolvent problem. We denote by
the set of the nonnegative solutions of the resolvent equation$\lambda x=Tx+c$

where$\lambda$

is a nonnegative number and

$c$

,

(1)

is an element of $E+$ ’ that is,$R_{\lambda}(c)=\{x\geq 0|\lambda x=Tx+c\}$

This defines a multivalued mapping$R_{\lambda}$$R_{\lambda}(c)$.

can be defined$R_{\lambda}:E_{+}arrow E_{+}$$E_{+}$

into itself:,

which we call the resolvent mapping of corresponding to . Throughtout
this paper the $symbolarrow$ is used to indicate the domain and the range of$T$$\lambda$

29
a multivalued mapping. When the equation (1) has no solution for some$c\geq 0$ , we define.
In Theorem 4.1 , it is shown that under the condition $A1,A2,A3,$
for all $c\geq 0,$ $\lambda>r(T)$ , on the other hand, the resolvent equation (1) has$c\gg O$ .
no nonnegative solution for any, it is obvious that when
By definition of
for all $c\geq 0$ thecomposed mapping $(\lambda I-T)\circ R_{\lambda}=I$ and, where is the identical mapping on $E+,$
means therestriction of on a subset $K$ of.
$\lambda\geq 0,$$R_{\lambda}(c)=\emptyset$$R_{\lambda}(c)\neq$$\emptyset$$\lambda\leq r(T),$$R_{\lambda}(c)\neq\emptyset$$R_{\lambda}$$R_{\lambda}\circ(\lambda I-T)_{|(\lambda I-T)^{-1}E_{+}\cap E+}\supseteq$$I$$I_{|(\lambda I-T)^{-1}E_{+}\cap E+}$$T$

$T_{|K}$$E_{+}$

Theorem 4.1 Suppose that $T:E+arrow E+satisfiesA1,$ $A2,$ $A3$ .
(1) Let $\lambda>r(T)$ be fixed. Then the equation (1) is solvable for all $c\geq 0$ ,is not empty.
that is,is a singleton if $c\gg O$ or $c=0$ .
Moreover(1.1).
is positively homogeneous(1.2)
when $c\gg O$ .
(1.3) $O\ll R_{\lambda}(c)<R_{\lambda}(c’)$
when $O\ll c<c’$ .
(1.4)
continuously maps $(E_{+})^{i}\cup\{0\}$ into itself.
.
(1.5)
is not necessarily a singleton when
which
(1.6)
contains the limit of a converging sequence.
satisfies $c_{n}\gg c$ and $c_{n}arrow c$ as $narrow\infty$ , that is,$R_{\lambda}(c)$$R_{\lambda}(c)$$(R_{\lambda}(\alpha c)=\alpha R_{\lambda}(c) \alpha\geq 0, c\geq 0)$$R_{\lambda}$$R_{\lambda}(c)\gg O$$R_{\lambda}$$c\in(E_{+})^{b}\backslash \{0\}$$R_{\lambda}(c)$$\{R_{\lambda}(c_{n})\}$$R_{\lambda}(c)$$\lim_{c_{n}\downarrow c}R_{\lambda}(c_{n})\in R_{\lambda}(c)$Moreover, this limit is independent of the choice of the sequence$\{c_{n}\}$.

is compact-valued and upper semicontinueous.
is not empty for some $c\gg O$ , then $\lambda>r(T)$ .

(1.7) $R_{\lambda}:E_{+}arrow E_{+}$

(2)

If

$R_{\lambda}(c)$Remark 4.2 Under the condition of Theorem 4.1, for each $\lambda>r(T)$$(1.6)$ shows that.
. We denote this limit bycontains$R_{\lambda}(c)$$\overline{R_{\lambda}}(c)$$\lim_{c_{n}\downarrow c}R_{\lambda}(c_{n})$the following properties:
Then a single-valued mapping(1)
when $c\gg O$ or $c=0$ ,(2)when $0\leq c<c’$ ,$\overline{R_{\lambda}}:E+arrow E+has$$R_{\lambda}(c)=\overline{R_{\lambda}}(c)$$0\leq\overline{R_{\lambda}}(c)<\overline{R_{\lambda}}(c’)$

30

(3)

$0\leq\overline{R_{\lambda}}(c)\ll\overline{R_{\lambda}}(c’)$

when$0\leq c\ll c’$.

Proof of Theorem 4.1 (i). Let $\lambda>r(T)$ be fixed. We begin with thesolvability of the resolvent equation (1).
Since $\lambda>r(T)$ , from Theorem 3.3 we can choose $x_{0}\gg 0$ such that, that is,$Tx_{0}\ll\lambda x_{0}$$\Vert\frac{Tx_{0}}{\lambda}||_{x_{0}}<1$.

Then, for each $c\geq 0$ , there exists a sufficiently large $M>0$ such that$|| \frac{c}{\lambda M}||_{x_{0}}<1-||\frac{Tx_{0}}{\lambda}||_{x_{0}}$.

Thus$\frac{1}{\lambda}(Tx_{0}+\frac{c}{M})\leq x_{0}$.

(2)Therefore the compact mapping $F$ defined as$Fx= \frac{1}{\lambda}(Tx+\frac{c}{M})$leaves the interval$Tx\leq Tx_{0}$ , then,$[0, x_{0}]$invariant. In fact, $0\leq x\leq x_{0}$ implies $0=TO\leq$$0 \leq\frac{c}{\lambda M}\leq\frac{1}{\lambda}(Tx+\frac{c}{M})\leq\frac{1}{\lambda}(Tx_{0}+\frac{c}{M})$.

Hence, from (2)$0\leq Fx\leq x_{0}$.

Since $F([0, x_{0}])$ is relatively compact, continuously maps a compact set$K$ into itself where. By virtue of Shauder’s fixed
points theorem,
has fixed points in $K$ . Then there exists
such that$F$$K=\overline{co}(F([0, x_{0}])\subseteq[0, x_{0}]$$F$$x_{0}\in K\subseteq E_{+}$$x_{0}=Fx_{0}= \frac{1}{\lambda}(Tx_{0}+\frac{c}{M})$,that is,$\lambda(Mx_{0})=T(Mx_{0})+c$.

This shows that $Mx_{0}$ is a solution of the equation (1).

31
Next we prove thatis a singleton if $c\gg O$ or $c=0$ . Let $c\gg 0$, where we can see $x\gg O,$ $y\gg O$ from the inequalities
and$R_{\lambda}(c)$$x,\cdot y\in R_{\lambda}(c)$$\lambda x=Tx+c\geq c\gg O,$ $\lambda y=Ty+c\geq c\gg O$.

(3)

If $||x||_{y}>1$ , then $x\leq||x||_{y}y$ implies$Tx\leq||x||_{y}Ty$,$\lambda x-c\leq||x||_{y}(\lambda y-c)$$x\leq$,$||x||_{y}y+ \frac{1-||x||_{y}}{\lambda}c$$\leq(||x||_{y}+\frac{1-||x||_{y}}{\lambda}\cdot\frac{1}{||y||_{c}})y$.

By definition of y-norm to the contrary,$||x||_{y} \leq||x\Vert_{y}+\frac{1-\Vert x\Vert_{y}}{\lambda}\cdot\frac{1}{||y||_{c}}<||x\Vert_{y}$.

, which means $x\leq y$ . Similarly $x\geq y$ can be shown, hence$x=y$ . Therefore it is proved that the equation (1) has a unique solution.
When $c=0,$ $x=0$ is a trivial solution. If there exists a positive solution$x’>0$ ,Thus$||x||_{y}\leq 1$$\lambda x’=Tx’$, which contrais an eigenvalue of . This implies
dicts that $\lambda>r(T)$ . Therefore $x=0$ is a unique solution.
When $c\gg O$ or $c=0$ we use the convenient notationmeans that$T$$\lambda$$\lambda\leq r(T)$$R_{\lambda}(c)=\{x_{c}\}$.

Now we prove several properties of.
Positively homogenuity ofis evident from the fact that
$R_{\lambda}(1.1)-(1.7)$$R_{\lambda}$$\lambda x=Tx+c$

32
implies$\lambda\alpha x=T(\alpha x)+\alpha c$$(\forall\alpha>0)$.

The statement of (1.2) have been noted by (3).
We show the statement (1.3). Let $O\ll c<c’$ , then, as we have shown,are singletons and strongly positive. Further, it is clear that
$R_{\lambda}(c),$ $R_{\lambda}(c’)$$R_{\lambda}(c)\neq R_{\lambda}(c’)$. Assume that $x_{c}<x_{c’}$ is not true. Then, 1$x_{c}||_{x_{c}},$$>1$

implies$\lambda x_{c}=Tx_{c}+c\leq$$||x_{c}||_{x_{c}},Tx_{c’}+c$$<$$||x_{c}||_{x_{c’}}Tx_{c’}+c’$$||x_{c}||_{x_{c’}}Tx_{c’}+||x_{c}||_{x_{c’}}c’$$=$$||x_{c}||_{x_{c’}}(\lambda x_{c’})$.

Thus$x_{c}\ll||x_{c}||_{x_{c’}}x_{c’}$which contradicts to definition of$x_{d}$,-norm. Hence we obtain$0\ll x_{c}<x_{c’}$.

We prove the continuity ofon $(E_{+})^{i}\cup\{0\}$ . Let $c_{0}\in(E_{+})^{i}\cup\{0\}$ andsuch that $c_{n}arrow c_{0}(narrow\infty)$ . Fix $e\gg O$ . Since
is bounded,is also bounded in e-norm. Then, for a sufficiently large$M>0$ ,
$R_{\lambda}$$\{c_{n}\}\subseteq(E_{+})^{i}\cup\{0\}$$\{c_{n}\}$$\{c_{n}\}$$c_{n}<Me$$(\forall n\in N)$.

From (1.3), which shows thatis also bounded. Since
is a relatively compact subset in $E+$ , there exists a subsequence, the limit of which being denoted by , then$x_{c_{n}}\leq R_{\lambda}(Me)$$\{x_{c_{n}}\}$$\{Tx_{c_{n}}\}$$\{Tx_{c_{n_{j}}}\}\subseteq\{Tx_{c_{n}}\}$$z_{0}$$Tx_{c_{n_{j}}}arrow z_{0}$Hence, as$n_{j}arrow\infty$$(n_{j}arrow\infty)$

.

(4)

,
$\lambda x_{c_{n_{j}}}=Tx_{c_{n_{j}}}+c_{n_{j}}arrow z_{0}+c$,33which means$x_{c_{n_{j}}} arrow\frac{1}{\lambda}(z_{0}+c)$.

Therefore$Tx_{c_{n_{j}}} arrow T(\frac{1}{\lambda}(z_{0}+c))$

From this and (4) it must be that$(n_{j}arrow\infty)$$z_{0}=T( \frac{1}{\lambda}(z_{0}+c))$.

, that is,$\lambda(\frac{1}{\lambda}(z_{0}+c))=z_{0}+c=T(\frac{1}{\lambda}(z_{0}+c))+c$.

Therefore$x_{c_{n_{j}}} arrow\frac{1}{\lambda}(z_{0}+c)=R_{\lambda}(c)$.

This shows.
As
is a singleton, we have$\lim_{n_{j}arrow\infty}R_{\lambda}(c_{n_{j}})=R_{\lambda}(c)$$R_{\lambda}(c)$$\lim_{narrow\infty}R_{\lambda}(c_{n})=R_{\lambda}(c)$.

Hereafter we consider the case $c\geq 0$ .
See Remark 1 of Y. Osime [10] for the detail of the statement (1.5).
such that $c_{n}\gg c$
We prove the statement (1.6). Take a sequencewhich converges to . Since
is bounded, from (1.3) we can see thatis a relatively compact set inis also bounded and hencewith the limit. Then there exists a subsequence$z_{0}\in E+$ , namely,$\{c_{n}\}$$c$$\{x_{c_{n}}\}$$E_{+}$$\{c_{n}\}$$\{T(x_{c_{n}})\}$$\{T(x_{c_{n_{j}}})\}\subseteq\{T(x_{c_{n}})\}$$Tx_{c_{n_{j}}}arrow z_{0}\in E+$$(n_{j}arrow\infty)$.

In a similar fashion to that in the proof of the continuity of$\lambda(\frac{z_{0}+c}{\lambda})=z_{0}+c=T(\frac{z_{0}+c}{\lambda})+c$

holds, that is,$\lim_{c_{n_{j}}\downarrow c}R_{\lambda}(c_{n_{j}})=\frac{1}{\lambda}(z_{0}+c)\in R_{\lambda}(c)$.

,$R_{\lambda}$, the equality

34This limit does not depend on the choice ofsubsequenceswhere$\{c_{n_{j}}\}$. In fact, if we have two$\{c_{n_{j}}\},$ $\{c_{n_{j}},\}$$\lim_{c_{n_{j}}\downarrow c}R_{\lambda}(c_{n_{j}})=x_{0},\lim_{c_{n_{j}},\downarrow c}R_{\lambda}(c_{n_{j}},)=x_{0}’\in R_{\lambda}(c)$.

(5)

Let us make a new subsequence
combining them such thatofwhen
is odd,whenis even and$c_{n_{j}}>c_{n_{j}}\gg c$ . Then, using (1.3), we can note$\{c_{n}\}$$\{c_{n_{j}},,\}$$c_{n_{j}},,$$\in\{c_{n_{j}}\}$$n_{j’’}$$c_{n_{j}},,$$\in\{c_{n_{j}},\}$$n_{j’’}$$c_{n_{j}},,$$>$$x_{c_{n_{j’’}}}>x_{c_{n_{j’’}+1}}>x_{c_{n_{j’’}+2}}>\cdots$

On the other hand, from (5)$x_{n_{j}},,$$arrow x_{0}$($n_{j’’}$

is odd),$x_{n_{j}},,$$arrow x_{0}’$($n_{j’’}$

is even).

Therefore, $x_{0}=x_{0}’$ follows from the closedness of $E+andE+\cap(-E_{+})=$. We have proved.
Moreover, in the same way as above, we can show thatis$\{0\}$$x_{c_{n}} arrow\frac{1}{\lambda}(z_{0}+c)\in R_{\lambda}(c)$$\lim_{c_{n}\downarrow c}R_{\lambda}(c)$independent of the choice of$narrow\infty$

which satisfies $c_{n}\gg c$ and $c_{n}arrow c$ as$\{c_{n}\}$.

Thus we have completed the proof of (1.6).
Finally we show (1.7). It is clear thatis compact-valued and uppersemicontinueous on $(E_{+})^{i}\cup\{0\}$ from (1.4). Fix $e\gg O$ . Then, in the same
way as in the proof of (1.3),$R_{\lambda}$$x_{c}\leq R_{\lambda}(c+e)$

can be obtained. This shows that$(\forall x_{c}\in R_{\lambda}(c))$$R_{\lambda}(c)$is bounded, and$R_{\lambda}(c)= \frac{1}{\lambda}TR_{\lambda}(c)+\frac{c}{\lambda}$

is a relatively compact set. Moreoveris closed,
which follows from the fact thatis continuous. Thereforeiscompact. Next we prove that
: $E+arrow E+is$ upper semicontinuous.
implies that

$R_{\lambda}(c)$$R_{\lambda}(c)$$T$$R_{\lambda}$$R_{\lambda}(c)$35
is compact-valued, it is sufficient to show that if $c_{n}arrow c$ and, there exists some converging subsequenceofwith the limit in. Fix a sufficiently. Let $c_{n}arrow c$ andlarge $e\gg O$ such that $c_{n}\leq c+e(\forall n)$ . Repeating the above argument leads

Since$R_{\lambda}$$x_{n}\in R_{\lambda}(c_{n})(\forall n)$$\{x_{n_{j}}\}$$x_{n}\in R_{\lambda}(c_{n})$$R_{\lambda}(c)$$x_{n}= \frac{Tx_{n}}{\lambda}+\frac{c_{n}}{\lambda}\leq R_{\lambda}(c+e)$

which implies thatverging subsequence
limit by , then, as$\{x_{n}\}$$x_{0}$$\{x_{n}\}$$(\forall n)$,

is relatively compact. Thencontains a conwith the limit in. In fact, let denote the,$\{x_{n}\}$$R_{\lambda}(c)$$\{x_{n_{j}}\}$$n_{j}arrow\infty$$c_{n_{j}}arrow c$,$x_{n_{j}}arrow x_{0}$,$Tx_{n_{j}}arrow Tx_{0}$,which implies$\lambda x_{0}=Tx_{0}+c$.

口We haVe Completed the proof
This shows
Proof of Theorem 4.1 (2). Let the condition be satisfied, that is, wehave $c\gg O$ and $x\geq 0$ such that$x_{0}\in R_{\lambda}(c)$$\lambda x=Tx+c$.

Then $\lambda x\geq c\gg 0$ implies $x\gg 0$ and $\lambda x=Tx+c\gg Tx$ . Applying
口Theorem 3.3, we obtain $\lambda>r(T)$ .

is a
Proof of Remark 4.2. The statement (1) is clear because
singleton if $c\gg O$ or $c=0$ . We prove the statements (2),(3). It is clear thateach $x\geq 0$ cannot be a solution of the equation (1) for different , $c’\geq 0$ .
This means
when $c\neq c’$ . Let $0\leq c<c’$ . Let $e\gg 0$ be,
fixed, then, for each
$R_{\lambda}(c)$$c$$\overline{R_{\lambda}}(c)\neq\overline{R_{\lambda}}(c’)$$\epsilon>0$$0\ll c+\epsilon e<c’+\epsilon e$.

From (1.3) we have
$O\ll R_{\lambda}(c+\epsilon e)<R_{\lambda}(c’+\epsilon e)$$(\forall\epsilon>0)$.

36Because$c\neq c^{l}$implies$\overline{R_{\lambda}}(c)\neq\overline{R_{\lambda}}(c’)$, as$\epsilon\downarrow 0$$0\leq\overline{R_{\lambda}}(c)<\overline{R_{\lambda}}(c’)$

Let now $0\leq c\ll c’$ . For a sufficiently small

we obtain.

$\epsilon>0$$0\leq c<(1-\epsilon)c’\ll c’$

From (1.1) and the fact

$R_{\lambda}(c’)\gg O$,.

, we can prove$0\leq\overline{R_{\lambda}}(c)<(1-\epsilon)R_{\lambda}(c’)\ll R_{\lambda}(c’)$.

口Besides we inquire into the problem when is indecomposable. For each$\lambda>r(T)$ , the resolvent mapping
which is defined as a single-valued$T$$R_{\lambda}$

mapping is continuous, positively homogeneous, strongly order-preserving) and hence indecomposable. The resolvent( $x<y$ implies$c>0$ .
equation (1) has no nonnegative solution for any$R_{\lambda}(x)\ll R_{\lambda}(y)$$\lambda\leq r(T),$

Theorem 4.3 In addition to the assumptions in Theorem 4.1, assume
further that $A4$ holds.
(1) Let $\lambda>r(T)$ be fixed. Then
is a singleton for all $c\geq 0$ , andis also an indecomposable and continuous mapping.
(1.1)
when $c>0$ .
(1.2)
when $0\leq c<c’$ .
(2) Ifis not empty for some $c>0$ , then $\lambda>r(T)$ .
$R_{\lambda}(c)$$R_{\lambda}:E_{+}arrow E_{+}$$R_{\lambda}(c)\gg O$$R_{\lambda}(c)\ll R_{\lambda}(c’)$$R_{\lambda}(c)$Proof of Theorem 4.3 (1). Let $\lambda>r(T)$ be fixed. It is clear thatis a singleton if $c\gg O$ or $c=0$ from the preceding theorem. We consider, then it is. Assume that there exist $x,$
the caseclear that $x,$ are linearly independent. Since$R_{\lambda}(c)$$c\in(E_{+})^{b}\backslash \{0\}$$y\in R_{\lambda}(c)$$y$$\lambda x=Tx+c$,$\lambda y=Ty+c$

implies $\lambda x\geq c>0,$ $\lambda y\geq c>0$ , we have $x,$ $y>0$ . Further,$Ty$ means $E_{x}=E_{y}=E_{+}$ by the definition of indecomposability. Then we$\lambda x\geq Tx,$$\lambda y\geq$37
can note $x,$ $y\gg O$ .
Hence, we have $x\leq||x||_{y}y$ which satisfies
assumption $A4$
$E$

国 yy-x$\not\supset$$\{0\}\subsetneq E_{||x||_{y}y-x}\subset\wedge E+\cdot$

From the$||x||_{y}Ty-Tx$$=||x||_{y}(\lambda y-c)-(\lambda x-c)$$=\lambda(||x||_{y}y-x)+(1-||x||_{y})c$.

This shows $1-||x||_{y}>0$ , that is, $x\ll y$ . we can similarly obtain the
opposite inequality $y\ll x$ . This is a contradiction. Thus
is singlevalued for each $c\geq 0$ .
provided $c>0$ as we have shown as above, the stateSInce
ment (1.1) is true.
$R_{\lambda}(c)$$R_{\lambda}(c)\gg O$

Next we prove (1.2). Let $0\leq c<c’$ . From Theorem 4.1 it is clear that$0\leq R_{\lambda}(c)<R_{\lambda}(c’)$.

, where ,. Assume that
We proveare convinient notations ofdefined in the proof of thepreceding theorem, respectively. Then, by the assumption $A4$ , we have
$R_{\lambda}(c)\ll R_{\lambda}(c’)$$\{0\}\subsetneq E_{x_{c}’-x_{C}}\subsetneq E_{+}$$R_{\lambda}(c),$$x_{c’}$$E_{x_{c’}-x_{C}}$$x_{c}$$R_{\lambda}(c’)$$\not\supset Tx_{c’}-Tx_{c}$$=(\lambda x_{c};-c’)-(\lambda x_{c}-c)$$=\lambda(x_{c’}-x_{c})-(c’-c)$$\leq\lambda(x_{c’}-x_{c})$.

This is a contradiction.
is indecomposable by the property (1.2). In fact, let
We can see that, which shows. Then from (1.2) we have
$R_{\lambda}$$R_{\lambda}(x)\ll R_{\lambda}(y)$$\{0\}\subsetneq E_{y-x}\subsetneq E_{+}$$0\ll R_{\lambda}(y)-R_{\lambda}(x)\not\in E_{y-x}$.

is single-valued and upper semicontinuous in the sense of a multiSince
is clearly continuous regarded
valued mapping from Theorem 4.1 (1.7),$R_{\lambda}$$R_{\lambda}$

38
as a Single-Valued mapping口Proof of Theorem 4.3 (2). Let the condition be satisfied, that is, wehave $c>0$ and $x\geq 0$ such that$\lambda x=Tx+c$.

Then, since $\lambda x\geq c>0$ , it must be that $x>0$ , and since $\lambda x>Tx$ , it must

be that $x\gg O$ by the indecomposability of .
From Theorem 3.2 it is shown that $r(T)$ is an eigenvalue ofstrongly positive eigenvector. Let $x_{0}\gg 0$ be the eigenvector. If
$T$

$T$

with a$x$and$x_{0}$

are linearly dependent, we have$\lambda x>Tx=r(T)x$ ,which implies $\lambda>r(T)$ . If

$x$

and

$x_{0}$

are linearly independent, we have$x_{0}\leq||x_{0}||_{x}x$where$\{0\}\subsetneq E_{||x_{0}||_{x}x-x_{0}}\subsetneq E_{+}$$E_{||x_{0}||_{x}x-x_{0}}$,. From the indecomposability of ,$T$$\geq$$||x_{0}||_{x}Tx-Tx_{0}$$\leq$$||x_{0}||_{x}\lambda x-r(T)x_{0}$$= \lambda(||x_{0}||_{x}x-\frac{r(T)}{\lambda}x_{0})$Therefore, we obtain$\frac{r(T)}{\lambda}<1$.

, namely, $r(T)<\lambda$

Next we consider the relation between$\lambda$and$R_{\lambda}(c)$口, where $c\geq 0$ is given.

Theorem 4.4 Let satisfy the assumptions $A1,A2,A3$ , andtrarily fixed element in
Then the multivalued mapping
is compact-valued and upper semicontinuous on $(r(T), \infty)$ ,In case $c\gg O$ , the following statements hold:,(i)
implies$T$$E+\cdot$$\lambda\leq r(T)$$R_{\lambda}(c)=\emptyset$$c$be an arbi-$R_{\lambda}(c)$on$[0, \infty$)

39(ii) $r(T)<\lambda’<\lambda$ implies $O\ll R_{\lambda}(c)<R_{\lambda’}(c)$ ,(iii)$\cdot\lim_{\lambda\downarrow r(T)}||R_{\lambda}(c)$Il$=\infty$.

satisfy the assumptions $A1,A2,A3,A4$ , and be an
Theorem 4.5 Let
arbitrarily fixed element inis single-valued and continuousThen
on $(r(T), \infty)$ .
In case $c>0$ , the following statements hold:(i)
implies,,(ii) $r(T)<\lambda’<\lambda$ implies
$T$

$c$

$E+\cdot$$\lambda\leq r(T)$$R_{\lambda}(c)$$R_{\lambda}(c)=\emptyset$$O\ll R_{\lambda}(c)\ll R_{\lambda’}(c)$(iii)$\lim||R_{\lambda}(c)||=\infty$.

$\lambda\downarrow r(T)$Proof of Theorem 45 The proof is similar to that of Theorem 44 口
is a nonempty compact set
Proof of Theorem 4.4. It is clear thatfor each $\lambda>r(T),$ $c\geq 0$ , and a singleton for each $\lambda>r(T),$ $c\gg O$ from
Theorem 4.1 (1).
Fix $c\gg 0$ . Then the statement (i) is true by Theorem 4.1 (2). Let$R_{\lambda}(c)$where $r(T)<\lambda’<\lambda$ . Suppose that the$>1$ , since. On
statement (ii) is not true. Then we have
impliesthe other hand,$\{x_{\lambda}\}=R_{\lambda}(c),$$\{x_{\lambda’}\}=R_{\lambda’}(c)$$||x_{\lambda}\Vert_{x_{\lambda}},$$x_{\lambda},$$x_{\lambda}\leq\Vert x_{\lambda}\Vert_{x_{\lambda}},x_{\lambda’}$$Tx_{\lambda}\leq||x_{\lambda}\Vert_{x_{\lambda’}}Tx_{\lambda’}$$\lambda x_{\lambda}-c\leq\Vert x_{\lambda}||_{x_{\lambda’}}(\lambda’x_{\lambda’}-c)$.

Thus$x_{\lambda}$$\leq$$\leq$$||x_{\lambda}||_{x_{\lambda’}} \frac{\lambda’}{\lambda}x_{\lambda}+\frac{(1-||x_{\lambda}||_{x_{\lambda’}})}{\lambda}c$$||x_{\lambda}||_{x_{\lambda’}}x_{\lambda}+ \frac{(1-||x_{\lambda}||_{x_{\lambda’}})}{\lambda}c$$\Vert x_{\lambda}||_{x_{\lambda’}}x_{\lambda}$,

which is a contradiction. We proved the statement (ii).

$x_{\lambda’}\gg 0$

40
is upperNow we sh\’ow that the compact-valued mapping
semicontinuous on $(r(T), \infty)$ for any $c\geq 0$ . Let $c\geq 0$ be fixed and. As we mentioned in the proof of Theorem 4.1
where(1), for any fixed $e\gg O$ and $\lambda>r(T)$ ,$\lambda\vdash\succ+R_{\lambda}(c)$$\lambda_{n}arrow\lambda_{0}$$\lambda_{n},$$\lambda_{0}\in(r(T), \infty)$$x<R_{\lambda}(c+e)$$(\forall x\in R_{\lambda}(c))$.

By the statement (ii), for a sufficiently small $M>r(T)$ such that
$M(\forall n)$$\lambda_{n}>$,$x<R_{\lambda_{n}}(c+e)<R_{M}(c+e)$$(\forall x\in R_{\lambda_{n}}(c))$.

Then, in the same way as in the proof of Theorem 4.1 (1), we can choose, the limit ofa converging subsequenceofwhere. Thus we can show thatwhich is contained in
is upper semicontinuous on $(r(T), \infty)$ .
Last we prove that the statement (iii). Fix $c\gg 0$ and assume that
there exists a sequencesuch that 1
is bounded and$r(T)$ . Since
is compact,
contains a converging subsequence, the limit of which we denote by . Then,
implies, as$x_{n}\in R_{\lambda_{n}}(c)$$\{x_{n}\}$$\{x_{n_{j}}\}$$R_{\lambda}(c)$$R_{\lambda_{0}}(c)$$\{R_{\lambda_{n}}(c)\}$$T$$\lambda_{n}\downarrow$$\{TR_{\lambda_{n}}(c)\}$$\{TR_{\lambda_{n_{j}}}(c)\}$$c$$R_{\lambda_{n}}(c)\Vert$$\lambda_{n}R_{\lambda_{n}}(c)=TR_{\lambda_{n}}(c)+$$x_{0}$$n_{j}arrow\infty$$\lambda_{n_{j}}R_{\lambda_{n_{j}}}(c)arrow x_{0}+c\gg 0$.

Thus we can see that $r(T)>0$ and$R_{\lambda_{n_{j}}}(c) arrow\frac{x_{0}+c}{r(T)}$Hence$T( \frac{x_{0}+c}{r(T)})=x_{0}$$\frac{x_{0}+c}{r(T)}\in R_{r(T)}(c)$pleted the proof, that is,$(n_{j}arrow\infty)$.

$T( \frac{x_{0}+c}{r(T)})+c=r(T)\cdot\frac{x_{0}+c}{r(T)}$. This means, which contradicts to Theorem 4.1 (2). We have com口41
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