TRANSACTIONS O F T H E AlIERICAN lIATHElIATICAL SOCIETY Volume 3 4 8 . Number 7 Jul? 1996
FINITE-DIMENSIONAL LATTICE-SUBSPACES
OF C(R) AND CURVES OF Rn
IOANNIS A. P O L Y R A K I S
.
ABSTRACT.L et X I . . . . xn b e linearly independent positive functions i n C ( Q ) . let X b e t h e vector subspace generated b y t h e x , and let 3 denote t h e curve ( x l ( t ) x2 ( t ) , z n ( t ) ) where o f Rn determined b y t h e functlon 3 ( t ) = ~, ~ ( t=) x ~ ( t+)x 2 ( t )+. . . + x n ( t ) . W e establish t h a t X is a vector lattice under t h e induced ordering f r o m C ( R ) i f and only i f there exists a convex polygon o f Rn w i t h n vertices containing t h e curve 3 and having its vertices i n t h e closure o f t h e range o f 3 . W e also present a n algorithm which determines whether or not X is a vector lattice and in case X is a vector lattice it constructs a positive basis o f X . T h e results are also shown t o be valid for general normed vector lattices.
It is well known that each separable Banach space is isometric to a closed subspace of C[O,11. By a minor modification of the existing proof of the universality of C[O, 11, we can show (see [12]) that C[O,11 is also a universal Banach lattice. More precisely, we can show that each separable Banach lattice is order isomorphic to a closed lattice-subspace of C[O,11,i.e., it is order isomorphic to a closed subspace of C[O, 11 which with the induced ordering is a vector lattice in its own right. Since the class of vector sublattices is not enough for this representation, the class of latticesubspaces seems to be the proper class of subspaces for studying Banach lattices. The structure of lattice-subspaces has not been yet systematically studied. In [8] it is shown that a subspace X of a vector lattice is a lattice-subspace if and only if there exists a positive projection from the vector sublattice generated by X onto X. In [ll]and [12]the existence of positive bases in lattice-subspaces has been studied. A recent survey of lattice-subspaces and positive projections as well as some new results on lattice-subspaces can be found in [I]. Recently, lattice-subspaces have been employed in economics where they appear naturally in incomplete markets and the theory of finance [4, 51. Now let E be an ordered Banach space with posit'ive cone E+. A sequence { e n ) of E is a positive basis if { e n } is a (Shauder) basis and
(
E+ = r
=
hiei E E : hi E B+ for each i .
Received b y t h e editors April 24. 1993.
1991 Mathematzcs Subject Classzficatzon. Primary 46B42. 52.421. 15A48. 53A04.
T h i s research was supported i n part b y t h e NATO Collaborative Research Grant #941059
0 1 9 9 6 A m e n c a n hlathematical Society
A positive basis {en) of E is unique in the sense that if {b,) is another positive basis of E, then each element of {b,) is a positive multiple of an element of {en). If {e,) is a positive basis of E, then the following statements are equivalent. (i) The basis {en) is unconditional. (ii) The cone E+ is generating and normal. (iii) E is a locally solid vector lattice (i.e., E is a Banach lattice with respect to an equivalent norm). For a proof see [13, Theorem 16.3, p. 4731 and [6, Theorems 3.5.2 and 4.1.51. The cone E+ is generating if E = E+ - E+ and E+ is normal (or self-allied) if there exists some c E R+ such that: 0 5 x 5 y implies jjxlj 5 cjjylj. If {en) is a positive unconditional basis of E. then the lattice operations in E are given by x V y = C ( A , V p Z ) e i and
xAy=C(A,Ap,)t.i,
for each x = CEl Ate,: y = CEl p,e, E E. If E is an n-dimensional space and the cone E+ is closed and generating, then, by the Choquet-Kendall Theorem [7] (see also [9: Theorem 3.11: p. 30]), E is a vector lattice if and only if a base B for E+ is an ( n - 1)-dimensional simplex (i.e., B is a convex polygon with n vertices). In such a case, if b l , b2,. . . , bn are the vertices of B , then the set {bl, bz, . . . , b,) is a positive basis for E . Therefore: for finite-dimensional spaces the Choquet-Kendall Theorem can be stated as follows. Proposition 1.1 (Choquet-Kendall). A finite-dimensional ordered vector space E with a closed and generating cone E+ is a vector lattice if and only if E has a positive basis. If E is 2-dimensional and E+ is closed and generating, then E is a vector lattice. This is true because each base B for E+ is a closed line segment. Therefore, B is a simplex. Now let X be a subspace of a partially ordered vector space E. The cone X n E+ is called the induced cone of X and the ordering defined in X by this cone is called the induced ordering. An ordered subspace of E is a subspace of E ordered by the induced cone. A lattice-subspace of E is an ordered subspace of E which is also a vector lattice (Riesz space).' If X is a lattice-subspace, then for each x, y E X we denote by xVy (resp. xAy) the supremum (resp. infimum) of {x, y) in X. It is clear that xAy 5 x A y and x V y 5 xVy: whenever x A y and x V y exist. Now let E be a vector lattice. If X is a lattice-subspace and xVy = x V y, xAy = x A y for all x: y E X , then X is a sublattice (Riesz subspace) of E . If X is the range of a positive projection P : E -+ E: then X is a lattice-subspace with xVy = P ( x V Y), xAy = P ( x A y) for each x , y E X . For notation and terminology not explained here, we refer the reader to 13, 6: 91. 2. LATTICE-SUBSPACES O F C ( 0 ) WITH
POSITIVE BASES
In this paper, we shall denote by R a compact Hausdorff topological space and by C(R) the Banach lattice of continuous real valued functions defined on R. he term "lattice-subspace" has been introduced in [lo]. In [8] a lattice-subspace is called a "cluasi-sublattice."
LATTICE-SUBSPACES
2796
Let Y be a closed subspace of C(R) with a basis {b,). Fix t E R and m E N. If bm(t) # 0 and b,(t) = 0 for each n # m, then we shall say that the point t is an m-node (or simply a node) of the basis {b,). If for each n there exists an n-node t, of the basis {b,), then we shall say that {b,) is a basis of Y with nodes and that {t,) is a sequence of nodes of {b,). If dimY = n and for each m E {1,2,.. . , n) there exists an m-node t, of the basis of Y, then we shall say that {bl, b2,.. . , b,) is a basis of Y with nodes and that the points t l , t2, . . . , t, are nodes of the basis {bl,bz,...,bn). Recall that the support of a function x E C(R), in symbols supp x, is the closure of the set {t E R : x(t) > 0).
Theorem 2.1. For a closed ordered subspace Y of C(R) having a basis {b,) consisting of positive functions we have the following. (i) If {b,) is a positive basis of Y, then (a) for each m there ensts a sequence {w,) of R such that lim,+, -= 0 for each i # m, and (b) there exists a sequence {t,) of R with t, E supp b, and bm(tn) = 0 for m # n. (ii) If {t,) is a sequence of nodes of {b,), then {b,) is a positive basis of Y and X for each x = Aibi E Y we have hi = - for each i.
+
Proof. (i) For each k let zr,= - i b m C ik= l , i f b i Since {b,) is a positive basis, zk @ Y+ and so there exists some wk E R (depending on m) such that zk(wk) < 0 or
for each k. Also, let tm be a limit point of the sequence {w,). Now (a) and (b) follow by letting k -+ m. (ii) Let {t,) be a sequence of nodes of {b,). It is easy to see that if x = X I l Aibi E X , then A, = - In particular, x E X+ implies hi E W+ for each i which shows that {b,) is also a positive basis.
Proposition 2.2. For a closed lattice-subspace Y of C(R) with a positive basis {b,) the following statements are equivalent. 1. Y is a sublattice of C(R). 2. If bm(t) > 0 for some m and t, then t is an m-node of the basis {b,). Proof. (1) + (2) Suppose that Y is a sublattice of C(R) and that bm(t) > 0 and let n # m. Then em A en = emAen = 0. This implies b,(t) = 0 and hence t is an m-node of the basis {b,). (2) + (1) Let x = Czl hibi, y = X I l pibi and let bm(t) > 0. Then t is an m-node of {b,) which implies b,(t) = 0 for each n # m. Therefore,
I . A. P O L Y R A K I S
2796
I f b,(t) = 0 for all i , t h e n ( x V y ) ( t ) = ( x V y ) ( t )= 0 is also true. T h u s . x V y = x V y. and so Y is a sublattice o f C ( R ) . Proposition 2.3. Let Y be a n n-dimensional subspace of C ( R ) and let bl, b2, . . . , b, i n Y+. T h e n { b l , . . . , b,} is a positive basis of Y if and only if for each 1 5 m 5 n there exists a sequence {u,} of 0 satisfying limy-, = 0 for each i # rn.
a
Proof. TVe show t h e -only if" part. So, assume t h a t t h e vectors bl , b2, . . . . b, satisfy t h e stated property. W e must show t h a t { b l , . . . , b,} is a positive basis. T o this end, assume x = x:=l A2b2E Y + . T h e n f r o m n
0 5 ---
bi ( u u )
>-
y,, Am, 2=1 we have A, 2 0 for each m. Also, i f x = 0 , t h e n as before we see t h a t A, = 0 for each m. Hence { b l , b2:. . . : b,} is a positive basis o f Y .
Proposition 2.4. Let { b l , . . . , b,} be a positive basis of a n n-dimensional latticesubspace Y of C ( 0 ) . For a function x = x y = l Aib2 € Y , we have the following. ( i ) If a point t , is a n i-node of the basis, then A, (ii) If {u.}
is a sequence o f 0 such that limy-,
we have X i = limy,,
a. #
=
=
0 for each j
# i , then
m. .(mu)
Proof. I f t h e point t , is a n 2-node, t h e n x ( t , ) = A,b,(t,) and t h e validity o f ( i ) follows. For (ii) notice t h a t lim,,, = limy+, x ; = , A, b ( 1 = A'. Theorem 2.5 ( [ 1 2 ] ) .Let E be a Banach lattice with a positive basis. T h e n there exists a closed lattice-subspace Z of C[O:11 with positive basis {b,) having nodeintervals (i.e., besides {b,} being a positive basis of Z there exists a sequence {J,) of intervals of [ O , l ] satisfying b,(t) > 0 f o r t E J , and b,(t) = 0 for all t E J , with m # n ) and a n onto order-isomorphism T : E -+ Z such that for all x E E Now let Y b e a closed lattice-subspace o f C ( R ) w i t h a positive basis. A s we shall see i n Example 3.l(iii) below, i n general Y need not have a positive basis w i t h nodes. However, according t o T h e o r e m 2 . 5 . t h e space Y is order-isomorphic t o a closed lattice-subspace Z o f C[O,11 which has a positive basis w i t h node-intervals (and therefore Z also has a positive basis w i t h nodes).
For our discussion here, we shall fix n linearly independent positive functions . . . , x , o f C ( R ) . T h e ordered subspace o f C ( R )generated b y these functions will b e denoted b y X or b y [ X I , 2 2 , . . . : x,], i.e.,
21,22,
X = [ x l ,x 2 , . . . , x,]. W e now state t h e main problem o f our study. W h e n is X a lattice-subspace of C ( R ) ? Or equivalently, when does X have a positive basis?
LATTICE-SUBSPACES
We shall answer this question by completely characterizing the lattice-subspaces. As a matter of fact, we shall not only characterize the lattice-subspaces X but we shall also present an algorithm of determining the positive basis of X. This is important since once the positive basis has been found, we can determine the coordinates of the elements of X and the lattice operations in X; see Examples 4.3 and 4.5 below. :"=)i, and In the sequel, we shall denote by z the sum of X I , .. . , x, (i.e., z = C= by ,8 the function p : R -+ Rn such that
IWT
for each t E R with z(t) > 0. So, p defines a curve on the base B = {y E : C:=l yi = 1) for the cone We shall refer to ,8 as the basic curve of the vectors 21, 22,. . . , x,. The reader should keep in mind that the cone X+ of X is always generating and normal. We start our study with several examples of ordered subspaces of C(R).
IWT.
Example 3.1. (i) Let R = {(u, v) E C(R) defined by
E
R2 : u2
+ v2 5 9) and consider the functions
21,22
See Figure 1. Let X be the subspace of C(R) generated by 21, x2. Then X as a 2-dimensional subspace, is a lattice-subspace of C(R). The points (0,O) and (0,2) are nodes of the basis {xl, x2), and so {xl, 2 2 ) is a positive basis of X. The space X is a sublattice of C(R) because each (u, v) E R with ( X I x2) (u, v) > 0 is a node of the basis { X I , 22). (ii) Let R = [O, 11, x1 (t) = 1, 2 2 (t) = t and X = [ X I , x2]. Then X is a latticesubspace and the set {bl(t) = 1- t , b2(t) = t) is a basis of X with nodes the points t l = 0 and t2 = 1. Therefore, {bl, b2) is a positive basis. The space X is not a sublattice of C(R) because the point t = is not a node of the basis {bl, b2). Also, it should be clear that b1Vb2 = 1 and b l a b 2 = 0, where 1 and 0 are the constant functions one and zero; see Figure 2.
+
I . A. POLYRAKIS
(iii) Let R
=
[-I, 11, x1 (t) = It1 and
Then X is a lattice-subspace of C(R) and x l(t) = 0 and lim x2(t)
t-0-
{XI, 2 2 )
is a basis of X since
lim x1(t)
t-o+
The space X does not have a positive basis with nodes; see Figure 3. (iv) Let R = [0,1], x l ( t ) = 1, xz(t) = t , x3(t) = t 2 a n d X = [x1,x2,x3]. We claim that X is not a lattice-subspace. To see this, assume by way of contradiction that X is a lattice-subspace of C[O,11. Let {bl , b2, b3} be a positive basis of X . Then for each a E [O,1] the function x,(t) = (t - a ) 2 , as a positive element of X , is a positive linear combination of b l , bz, bg. This implies bi (a) = 0 for at least one i and so bi = 0 for at least one i, a contradiction. Therefore, X is not a lattice-subspace of C[O,11. (v) Let X be the subspace of C[0,1] generated by the functions x l ( t ) = 11 - tj(2 - t), x2(t) = t(2 - t) and x3(t) = tll - tj. The points tl = 0, t2 = 1 ). X is a lattice-subspace with and tg = 2 are nodes of the basis 1x1, ~ 2 ~ x 3 Hence, a positive basis with nodes; Figure 4. (vi) Let {t,), {w,) and {r,} be strictly increasing sequences of [O,1] all convergent to satisfying t, < w, < r, < t , + ~ for each v. Also, let X I , 2 2 and x3 be elements of C[O, 11 with z i ( t ) > 0 for each t # and each i = 1 , 2 , 3 such that:
i
i
LATTICE-SCBSPACES
Then x , ( i ) = 0 for each i . The graph of might be as in Figure 5. Now note that
x z (t " ) = 0 for i = 2.3; lim 2 1 (t")
vice
XI, 22
and x3 over the interval [t,, t,+l]
x ' ( W y ) = 0 for i = 1,3; lim -
u-lcc 2 2 (w")
and
x,(ru) = O f o r i = 1 , 2 . lim x3(ru) By Proposition 2.3, the set { x l ,x2, 23) is a positive basis of X and so X is a latticesubspace of C[O,11. Since x , ( t ) > 0 for each t # and each z, it follows that X does not have a positive basis with nodes. u'm
:
Recall that the Wronskian of the functions pi E c("-') ( a .b), i = 1.2, . . . , m, is the determinant function: 'P2(t)
W(y1
~
2
. .,. , ~
cp: ( t )
cpk(t)
... . ..
cpm(t)
cp:,(t)
( t ))= det
m
Proposition 3.2. Assume [a,b] G ( c ,d ) and fo, f l , . . . , fm-1 E C ( c ,d ) . If m > 2, then the vector space of all solutions of the linear differential equation
is not a lattice-subspace of C [ a ,b ].
I . .4 P O L Y R A K I S
2800
Proof. Suppose that the vector space L of all solutions of (*) is a lattice-subspace having a positive basis { b l . bz.. . . , b,). Choose t, E [a,b],j = 1 , . . . , m, such that b,(t,) = 0 for each i # j . Since W ( b 1 ,b2, . . . , b,)(t,) # 0 . we infer that b, (t,) > 0 . This implies that t, # t, for z # j and so (in view of m > 2 ) there exists some t k E ( a ,b). But then for each i # k , the function b, attains a local minimum at t k . which implies that b i ( t k ) = 0 for each z # k . In turn. the latter implies W ( b 1 .b2, . . . , b m ) ( t k )= 0 , a contradiction. Corollary 3.3. A s s u m e [a,b] C ( c ,d ) and cpl, y 2 , . . . , y , E ~ ( ~ ) d() ,cwhere , m> 2. I f W(cp1,y2, . . . , y,) ( t ) # 0 for each t E ( c ,d ) , t h e n the subspace generated by the functions y l , . . . , cp, is n o t a lattice-subspace of C [ a ,b]. Theorem 3.4. T h e following statements are equzvalent. 1. T h e subspace X has a positive basis with nodes. 2. T h e subspace X is the range of a positive projection P : C ( R ) + C ( R ) . I n particular, i f { b l , b z , . . . , b,} is a positive basis with nodes the points t l ,t 2 ,. . . , t,, then the operator P : C ( R ) + C ( R ) , defined by
is a positive projection with range X Proof. ( 1 ) + ( 2 ) Clearly, if { b l , b2.. . . . b,} is a positive basis with nodes the points
t l ,t 2 , . . . , t,, then the operator P : C ( R ) -+ C ( R ) , defined by
is a positive projection with range X. ( 2 ) + ( 1 ) Since X is the range of a positive projection, we know that X is a lattice-subspace of C ( R ) . Let { b l , b2.. . . . b,) be a positive basis of X. Then there are t l ,t 2 , . . . , t , in R such that b,(t,) = 0 for each i # j . Let q, = cp, o P , where cp, denotes the ith coefficient functional of the basis { b l , . . . , b,}. By the Riesz Representation Theorem, for each i there exists a unique Bore1 regular measure p, such that
for each I(: E C ( R ) . Let b = z : = , b,, u, = b - b,, A = b-I ( ( 0 ,oo)),A, = b', and B, = u ~ ~ ( m)). ( 0 , Since
yi(bi) =
b, d p , = 1 and
p , ( ~ i )=
S,
U,
((0,oo))
dpi = 0 ,
it follows that the measure p, restricted to A is supported by a subset S, of A\B, such that S , n A, # ia. Now notice that if t , E S , n A,, then the points t l , . . . , t , are nodes of the basis { b l ,b 2 , . . . , b,). In Banach spaces each finite-dimensional subspace is complemented. The above result shows that an order-analogue result is not valid in Banach lattices. For instance, the lattice-subspace of Example S.l(iii) is not positively complemented
LATTICE-SUBSPACES
2801
because it does not have a positive basis w i t h nodes. O n t h e other hand, t h e lattice-subspace X o f Example 3 . l ( v ) has a positive basis w i t h nodes and therefore is positively complemented. As a matter o f fact, a positive projection P : C[O,21 + C[O,21 w i t h range X is given b y P ( x ) = q x l + x ( 1 ) x 2 + y x 3 . Proposition 3.5. Let X be a lattice-subspace of C ( R ) . If z ( t ) > 0 for each t E R , then X has a positive basis with nodes (and therefore X is the range of a positive projection). Proof. Let { b l , b 2 , .. . , b,) b e a positive basis o f X . T h e n there exist t l ,t 2 , .. . , t , E such tha,t b,(t,) = 0 for each j f i . I f bi(t,) = 0 , t h e n z ( t i ) = 0 , a contradiction. Hence b,(ti) > 0 for each i and so t h e points t l ,t 2 , .. . , tn are nodes o f • {b1,b2,. . .
R
W e shall denote b y D ( P ) t h e domain and b y R(P) t h e range o f t h e basic curve o f x1, 2 2 , . . . , x,. AS usual, i f K is a subset o f a topological space F , we shall denote b y I n t ( K ) t h e interior o f K , b y t h e closure o f K and b y 8 K t h e boundary o f K . Also (whenever F is a linear topological space) we shall denote b y co K t h e convex hull o f K , b y m K t h e closed convex hull o f K (i.e., t h e closure o f co K ) . I f A is a matrix, t h e n we shall denote b y AT t h e transpose o f A.
p
Theorem 3.6. The following statements are equivalent. ( i ) X is a lattice-subspace of C ( 0 ) . (ii) There exist n linearly independent vectors P I , P 2 , . . . , P, of Rn,belonging to the closure of the range of p such that for each t E D ( P ) the vector P ( t ) is a convex combination of the vectors of P I , . . . , P,, i.e., R(P) C c o { P l , . . . , P,}. If (ii) is true, P, = lim,,, P(w,,)for each i , A is the n x n matrix whose ith column is the vector P, and b l , b2, . . . , b, are the functions defined by the formula
then X has the following properties: ( a ) The set { b l , b2, . . . , b,) is a positive basis of X . I n addition, if t, is a limit point of the sequence {a,, : v = 1 , 2 , . . .}, then ti E supp b, and bk(t,) = 0 for each k f i . ( b ) The closed convex hull of R(P) and the convex polygon with vertices the points P I , P 2 , . . . , P, coincide. ( c ) If Pk = P ( t k ) , then ticis a k-node of the basis { b l , . . . , b,}. ( d ) If R C Rm, Pic = P(tk) for some interior point t k of 0 and the functions xi are c2-functions i n a neighborhood of t k , then
where D, denotes the operator of the j t h partial derivative. Proof. W e assume that (ii) and t h e other assumptions are true. W e shall establish that ( a ) , ( b ) ,( c ) and ( d ) (and therefore ( i ) ) are true. Since 1x1, 2 2 , . . . , x,} is a basis o f X , it follows t h a t { b l , b2, . . . , b,} is likewise a basis o f X . Let
I. A. POLYRAKIS
2602
Since each Pi is a vector of the base B = {y E IW? : Therefore 23
yr = I), it follows that
Oi=Cjn,ia , . - 1 for each i.
Let P(t) = C y = d i ( t ) P i be the expansion of P(t) relative to the basis {PI, P2,. . . , Pn). Then
and in view of (1) we get
Since P(t) is a convex combination of the vectors Pi, we get Ei(t) E R+ and so bi(t) E R+ for each i. Thus, bi E X+ for each i. From ( I ) , we have
Replacing t by w,, and taking limits, we get bj where qij = lim,,, z(w,") Since the solution of the system (2) is unique, we have ~ i =i 1 and qzj = 0 for j f i. Since ~(wi,) > 0 for each v and ~ i = i I , we have bi(wi,) > 0 for each v. Therefore, (w"W)
'
Hence lim
(3)
U-30
) (;
(wi,) = 0 for each j
# i,
and so, from Proposition 2.3, the set {bl, . . . , b,) is a positive basis of X . In other words, X is a lattice-subspace. Now let ti be an accumulation point of the sequence {wi, : v = 1 , 2 , .. . ). We have shown that bi(wi,) > 0 for each v and therefore ti E supp bi. Also, from (3), we see that b, (ti) = 0 for each j # i. Thus, the validity of (a) has been established. By our assumptions, R(P) C_ co{P~,Pz, . . . , P,) and so .., WR(/3) CG{Pl,... ,Pn)= C O { P ~ , . Pn). Since Pi E R(P) 2 W R(/3), we get co{Pl, . . . , P,) C W R(/3), and the validity of (b) follows. To establish (c) let Pk = @(tk). Without loss of generality, we can assume that wk, = t k for each v. Then bk(tk) > 0. Also, by (a), we have bj(tk) = 0 for each j # k. Hence tk is a k-node of the basis {bl, . . . , b,), and therefore statement (c) is true. Finally, assume that the hypotheses of claim (d) are valid. Then tk is a k-node = 0 for each p # k. Since t k is an interior point of R, for each p # k and so bCL(tk) -
LATTICE-SUBSPACES
2803
the function b, attains a local minimum at the point tk. This implies D, b,(tk) = 0 for each j and all p # k. Now let x, = C:=, c,,b,. Then x,(tk) = c,kbk(tk) and D3x, (tk) = c,k D, b k (tk). Hence ~ ( t k ) D j x ~ ( t-k ~) ~ ( t k ) D j ~ ( t k )
and the validity of (d) also follows. (ii) Suppose that X is a lattice-subspace of C(R) and that {bl, bz, . . . , b,) (i) is a positive basis of X. Then, by Proposition 2.3, for each i there exists a sequence = 0 for each j # 1. Let {w,,) such that lirnu+,
*
Since {bl, . . . , b,} is a positive basis, we see that A,, E R+ for all i and j . Moreover, we have z = C,"=l x, = C:=l azb,, where ai = C;=, A,,. Hence,
from which it follows that
Now let A be the n x n matrix with columns the vectors PI,Pz, . . . , P,. Then, from (4), we see that So, the vectors P I , . . . , P, are linearly independent since {xl, . . . , x,} and {bl, . . . , b,) are both bases of X. Let P(t) = x:=l
and E3(t)= ; t ( t - l ) ' ,
from which it follows that E,(t) 0 for each i and each t E [O, 21. Since P(t) belongs to the base B = {y E IW; : y1 y2 y3 = 1) of IW; we observe that tl (t) Ez (t) &(t) = 1, which shows that {PI, P2,P3) is an extreme subset of p. A positive basis {bl, b2, b3) of X is the following:
+
+
+ +
2(x1 - 53)
hence bl (t) = 2(t - 1)' (2 - t), bz (t) = 4t(2 - t ) and b3 (t) = 2t(t - I ) ~ The . points tl = 0, t2 = 1 and tg = 2 are nodes for the basis {bl, b2, b3). See Figure 8. Now, we can determine the coefficients of the points of X and the lattice operations of X. For example,
and similarly xz = $ bl
+ $b2.
Therefore,
xlVxz = i b l + i b 2 + i b g
and xlAxz =
ibl f i b 2 .
LATTICE-SUBSPACES
A positive projection P with range X is given by
x(t1) bl P(x) = bi (ti)
+
x(t2) b2
x(t3) b3
+
b2 (t2)
b3 (t3)
-
4 0 ) b1 4
f
~ ( 1b2) 4
f
4 2 ) b3,
4
for each x E C[O,21. Example 4.4. Let R = [-I, 21, xl (t) = t2(t - I ) ~x2(t) , = t4(t - 1)2 and x3(t) = see Figure 9. Also let X be the subspace of C[-1,2] generated by t4(t 21, x2,x3. A computation shows that the Wronskian of these functions is W(t) = 4t8(t - 7t + 3) which has roots in the interval [-I, 21. Notice that z(t) = xl(t) x2(t) x3(t) = xl(t)a(t),where a(t) = 1 t2 t2(t - 1)2 and
+
+
+ +
is the basic curve of
XI,
x2,x3. Also,
p(as2) = {P(-1) = p1
(i,:,
4 ;) , ~ ( 2=) p2( 51 , ,4 ,))
and limt,o P(t) = P3(1,0,O) and limt+l P(t) = ~ 4 ( + 4,O). , Therefore, L(P) = {P3,P4) NOWit is easy to show that the equation P'(t) = 0 does not have any root in the set (-1,2) n D(P). Therefore, I ( @ )= 0 and hence
The possible extreme subsets of
P are the following:
S1 = {PI, p 2 , p3), S 2 = {PI, p 2 , p 4 ) , S3 = {PI,P3, P 4 ) , and S4 = {P2,P3, P4).
I. A. POLYRAKIS
2808
We can show that no set from the above candidates is an extreme subset of P. For instance if the set S1 is an extreme subset of P, then let P(t) = [ l ( t ) P l + [2(t)P2 2t3(t-2) c3(t)P3and note that [l(t) = 7 < 0 for each t E (0,2), a contradiction. The conclusion here is that X is not a lattice-subspace of C[-l,2].
+
+
+
Example 4.5. Let R = { ( s , t ) E R2 : s2 t2 5 I ) , x l ( s , t ) = s 2 t2, x z ( s , t ) = 1 - e-(s2+t2)and let X be the lattice-subspace of C(R) generated by xl and 2 2 . We shall determine a positive basis of X. Notice that z ( s , t) = ( X I x2)(s,t ) = s2 t2 1 - e-(s2+t2)and that the basic curve of xl and x2 is P = (%, For each ( s , t ) E dR, we have P ( s , t ) = PI(&, and so P(dR) = {PI). Also, P(s, t) = P ~ ( $ $),, we see that L(P) = {Pz}. dD(P) = ((0,O)). Since lim(s,tj+(o.o) Now the system of equations
+
+ +
a)
5).
does not have solutions in the set Int(R) n D(P) and so I(P) = 0.This implies that extreme subset of p is E(P) = {PI,P2). A positive basis {bl, b2) of X is the following:
Therefore b l (s,t ) = (2e - 1)(s2
+ t2 - 1 + e-(s2+t2))and + + e - e1-(s2+t2)1.
b2(s, t) = 2[(l - e)(s2 t2)
The restriction of bl and b2 in the s-axis is as in Figure 10.
Next, we shall determine the coefficients of xl relative to the basis {bl, bn). Let xl = Albl A2b2. Then A 1 = -= & because the point ( 0 , l ) is a 1-node of
+
the basis of X. Also, lims-o A 2 = lims+o -=
$.
= 0, which (in view of Proposition 2.4) implies
Therefore, x1 = &bl
+ ib2.
Suppose now that R = {1,2,.. . , m). Then C(R) = Rm and X is the subspace of Rm generated by the linearly independent positive vectors
xi
=
(xI(I),xi( 2 ) ,. . . , xi ( m ) ,)
i = 1,2, . . . , n,
LATTICE-SUBSPACES
+ +
of Rm. Then z = xl x2 . . . function p : R + Rm defined by
+ x,
2809
and the basic curve of x l , x2,. . . , x, is the
)
for each k with z(k)
> 0.
-
Then R(P) = R(P). Therefore, if X is a lattice-subspace, then there exists {jl,j 2 , . . . ,j,} & D ( p ) such that E(P) = {p(j,) : i = 1,.. . , n ) . So, in order to determine whether X is a lattice-subspace of Rm we find the subsets of R(P) consisting of n linearly independent vectors (i.e., we find the possible extreme subsets of p ) and examine if one of them is an extreme subset of p . As a consequence of Theorem 3.6 we have the following result-which was also proven in [l,Theorem 2.61.
Theorem 5.1. The space X is a lattice-subspace of Rm if and only if there exist indices jl, j2,. . . ,j, of D(P) such that for each k E D(P) the vector P(k) is a convex combination of the vectors P ( j l ) ,P ( j z ) ,. . . ,P(j,). In this case, a positive basis {bl, bz, . . . , b,} of X , with nodes the points jl, j 2 , . . . , j, is given by the formula
where A is the n x n matrix whose columns are the vectors P(j,), i
=
1 , . . . , n.
6. LATTICE-SUBSPACES OF NORMED LATTICES Now let E be a normed vector lattice (normed Riesz space) and let U i denote the positive part of the closed unit ball of the norm dual E* of E, i.e.,
Suppose that U i is equipped with the weak* topology and let R be a weak* closed subset of UT defining the positive cone of E, i.e.,
E+ = {x E E : f ( x )
>0
for each f E R).
For each x E E denote by li: the function li:(f) = f (x), f E R, and for any subspace X of E denote by 2 the subspace j? = ( 2 E C(R) : x E X} of C(R). Then, x H T ( x ) = li: is a linear, one-to-one operator from E into C(R) such that T and T-' are both positive.
Theorem 6.1. A subspace X of E is a lattice-subspace if and only if 2 is a latticesubspace of C(R). This study of lattice-subspaces is more interesting if there exists a sequence
{ f,} of U: defining the positive cone of E and R is the weak* closure of { f,}. For instance, if E is a Banach lattice with positive basis, then we can take {f,} to be the sequence of the coefficient functionals of the basis and R = {f, : n E W} U (0). On the other hand, if E = l,, then we can take {f,) to be the sequence of functionals of the standard biorthogonal system of 1, and R to be the weak* closure of the sequence {f,). For details on determining R see [2, p. 4481.
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I. A . POLYRAKIS
1. Y. A. Abramovich, C. D. Aliprantis and I. A. Polyrakis, Lattice-subspaces and positive projections, Proc. Roy. Irish Acad., 9 4 A (1994), 237-253. 2. C. D. Aliprantis and K. C. Border, Infinite dimensional analysis: A hitchhickers guzde, Studies in Economic Theory. #4, Springer-Verlag, New York and Heidelberg, 1994. 3. C. D . Aliprantis and 0. Burkinshaw, Positive operators, Academic Press, New York and London, 1985. hIR 87h:47086 4. P. Henrotte, Existence and optimality of equilibria i n m a r k e t s w i t h tradable derivative securities, Stanford Institute for Theoretical Economics, Technical Report, No. 48, 1992. 5 . , Three essays i n financial economics. Ph.D. Dissertation, Department of Economics, Standford University, 1993. 6. G. J. 0.Jameson, Ordered linear spaces, Lecture Notes in hIath.. vol. 141, Springer-Verlag, Heidelberg and New York, 1970. MR 55:10996 7. D. G. Kendall, Simplexes and vector lattices, J . London h'lath. Soc. 3 7 (1962), 365-371. hIR 25:2423 8. S. hIiyajima, Structure of B a n a c h quasisublattices. Hokkaido hlath. J . 1 2 (1983), 83-91. hlR 84g:46033 9. A. Peressini, Ordered topological vector spaces, Harper & Row, New York, 1967. MR 37:3315 10. I. A. Polyrakis. Lattice B a n a c h spaces order-isomorphic t o 1 1 , Math. Proc. Cambridge Philos. Soc. 9 4 (1983), 519-522. htIR 85f:46042 11. , S c h a u d e r bases in locally solid lattice B a n a c h spaces. Math. Proc. Cambridge Philos. Soc. 1 0 1 (1987), 91-105. MR 89b:46020 12. , Lattice-subspaces of C[O. 11 and positive bases, J . Math. Anal. Appl. 1 8 4 (1994), 1-18. hIR 95g:46040 13. I. Singer, Bases in B a n a c h spaces. I, Springer-Verlag, Berlin and New York. 1970. hIR 45:7451
E-mail address: ypolyQmath .ntua.gr
