Chapter 2 Elementary Properties of Ordered Vector Spaces and Riesz Spaces

CHAPTER 2

Elementary Properties of Ordered Vector Spaces and Riesz Spaces

The contents of the present chapter, except for the last section (on order convergence and relatively uniform convergence), are mostly of an elementary nature. Many proofs are so obvious that brief indications will be sufficient. The only notable exceptions are in Theorem 11.9 and in the discussion of Example 11.2(x). The notions introduced here appeared already in the earlier development of the subject in the years between 1935 and 1942, a development mainly due to F. Riesz, L. V. Kantorovitch, H. Freudenthal, G. Birkhoff, K. Yosida, H. Nakano and T. Ogasawara. 11. Ordered vector spaces and Riesz spaces

We begin immediately with a definition.

..

Definition 11.1. The real linear space L, with elementsf, g, ., is called an ordered vector space if L is partially ordered in such a manner that the partial ordering is compatible with the algebraic structure of L, i.e., (i) f 5 g implies f i - h 5 g+h for every h E L, (ii) f 2 0 implies af 2 0 for every real number a 1 0. The ordered vector space L is called a Riesz space iffor every pair f and g in L, the supremum sup (f,9 ) with respect to the partial ordering exists in L. The terminology in the above definition is taken from Bourbaki (espace de Riesz). A Riesz space is also often called a vector lattice or, in the Soviet literature, a K-lineal. In the terminology of H. Nakano and his school, a Riesz space is called a semi-ordered linear space. I t will be proved in Theorem 11.5(v) that a Riesz space is indeed a lattice. We list some examples of Riesz spaces.

Example 11.2. (i) Let R" (n 2 1) be the real linear space of all real ntuples f = ( f l , . . .,f.)with coordinatewise addition and multiplication by real numbers. If we define thatf 5 g means that& 5 gk holds for 1 S k 5 n, then R" is a Riesz space with respect to the thus introduced partial ordering. 48

CH.2, 5 111

ORDERED VECTOR SPACES AND RIESZ SPACES

49

(ii) The partial ordering as defined in (i) above is not the only manner to make R" (n 2 2 ) into a Riesz space. There exists also the lexicographical ordering, as follows. For R2,letf S g forf = (fl ,fz)and g = (gl,g2)if either fl < g1 or f l = g1,f2 5 g 2 . The ordering is now a linear ordering, and so RZ is a Riesz space with respect to the lexicographical ordering. Similarly for n > 2. (iii) If L is the real linear space of all real finitevalued functionsf(x) on the arbitrary non-empty point set X with pointwise addition and multiplication by real constants, then L is a Riesz space with respect to the partial ordering introduced by defining thatf S g means thatf(x) 5 g(x) holds for every x E X. The linear subspace of all real bounded functions on X , with the same partial ordering, is a Riesz space by itself. If Xconsists of a finite number of points, say n points, we obtain essentially the example in part (i), i.e., the space R" with coordinatewise ordering. If X consists of a countably infinite number of points, we obtain the sequence space (s) of all real sequences and the subspace I , of all bounded real sequences. (iv) If X is a non-empty topological space and C ( X ) is the real linear space of all real continuous functions on X , then C ( X ) is a Riesz space with respect to the partial ordering introduced by defining thatf 5 g means that f ( x ) 6 g(x) holds for every x E X . The linear subspace of all real bounded continuous functions on X , with the same partial ordering, is a Riesz space by itself. If Xis a locally compact space, then the linear subspace of all real continuous functions on X with a compact carrier (also called compact support) is a Riesz space by itself. (v) As in section 3 and section 9, let ( X , A, p) be a measure space, i.e., p is a countably additive non-negative measure on the a-field A of subsets of the non-empty point set X . We assume that sets of measure zero are neglected; more precisely, A is identified with the Boolean measure algebra A/&, where A , is the ideal of sets of measure zero. By M(') = M'"(X, p) we denote the set of all real p-almost everywhere finitevalued p-measurable functions on X , with identification of p-almost equal functions, and pointwise addition and multiplication by real numbers. Then M(') is a real linear space, partially ordered by defining that f 5 g means that f ( x ) 5 g ( x ) holds for p-almost every x E X . The space M'"(X, p) is a Riesz space with respect to the thus defined partial ordering. (vi) Let M") = M")(X, p) be the same as in the preceding part (v). Many linear subspaces of M") are Riesz spaces by themselves. By way of example,

50

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH. 2 , s 1 1

-= -=

let p be a real number satisfying 0 p 00, and let L, = L,(X, p) be the well-known linear space of all (rea1)fE M"' satisfying n

Then L, is a Riesz space with respect to the partial ordering inherited from M"'. The same holds for the space L , consisting of all essentially bounded f i n M"'. (vii) The spaces Lp (1 5 p 5 co) are normed linear spaces with respect to the norm

less sup If(x)l

for p = co.

These spaces are special examples of Orlicz spaces, and Orlicz spaces are special examples of normed Kothe spaces. All these spaces are Riesz spaces with respect to the partial ordering inherited from M"'(X, p). (viii) Let p be a finitely additive signed measure (also sometimes called a signed charge) on the field (also called algebra) r of subsets of the nonempty point set X such that the number

llpll = SUP (IP(4I :A E r )

is finite. Under the natural definitions of addition and multiplication by real numbers the set L of all p of this kind is a real linear space, partially ordered by defining that p1 p z means that pl(A) _S p z ( A ) holds for every A E r. The space L is a Riesz space. Indeed, the ordering is compatible with the algebraic structure and furthermore, given pl and p z in L, the element v = sup (pl ,p z ) exists in L and is given for any A E r by

v(A) = sup { p l ( B ) + p z ( A - B ) :A

=,

B E r}.

(1)

For the proof, note that it is not difficult to see that v, as given by (l), is finitely additive and llvll 5 llplll + llpzll, so v EL. Furthermore, it is evident that v ( A ) 2 p l ( A ) as well as v(A) 2 p z ( A ) for every A E I', so v is an upper bound of p1 and p z Finally, if z is also an upper bound of pl and p z and if A E r, then z ( A ) = z ( B ) + z ( A - B ) 2 pi(B)+pZ(A-B)

.

for every B E I' satisfying B c A, and hence Z(A)

2 sup { p l ( ~ ) + p 2 ( ~ :- A~ )

This shows that v = sup (pl,pz).

BE

r} = V ( A ) .

CH.2, 3 111

ORDERED VECTOR SPACES AND RIESZ SPACES

51

(ix) Let H be a Hilbert space (over the complex numbers) with elements . . and inner product ( x , y), and denote by% the real linear space of all bounded Hermitian (i.e., self-adjoint) operators A , B, . . . on H. The space&' is an ordered vector space with respect to the partial ordering in% defined by saying that A S B means that (Ax, x ) =< (Bx, x ) holds for all x E H. It can be proved that % is not a Riesz space unless in the trivial case that H is one-dimensional. There exist, however, many linear subspaces of 3? that are Riesz spaces with respect to the ordering inherited from%'. Specifically, let 9 be a non-empty subset of 2 such that the elements of 9 are mutually commuting (i.e., A B = BA for all A , B E g ) ,let Y'be the set of all elements of % that commute with 9 (i.e., each element of v'commutes with each element of 9),and let Y" be the set of all elements of % that commute with V'. It can be proved that Y" is a Riesz space such that 9 c Y" c Y'(for further details and proofs, cf. Chapter 8). (x) This example is somewhat technical in nature; the reader who is not interested in harmonic functions is advised to proceed immediately to Definition 11.3. Let G be a region in the plane and L the ordered vector space of all functions f ( x , y ) in G such that f = u1 - u2 with u l , u2 harmonic and nonnegative in G. The ordering is the pointwise ordering. The null element of L is the function identically zero, so the subset L+ of allfe L greater than or equal to the null element consists of all non-negative harmonic functions in G. Every non-negative constant function is a member of L + . Note that any u E L+ is either strictly positive or identically zero in G. In order to prove that L is a Riesz space, it is sufficient to show that among all harmonic upper bounds of two given functions u1 and u2 in L+ there is a smallest one. Indeed, suppose this has been proved and suppose that f = u1 -u2 and g = u3-u4 are arbitrary functions from L. Then f l = f + ( u 2 + u 4 ) and g1 = g (u2 u4) are in L+,and so the least upper bound hl offl and g1 in L+ exists. It follows that h, - (u2 + u4) is the least upper bound off and g in L. Hence, let uI and u2 in L+ be given. We will prove that the pointwise infimum s = s ( ~ y, ) of the set H of all continuous superharmonic functions h in G with h 2 u l , h 2 u2 is the desired harmonic least upper bound of u1 and u 2 . By definition, the continuous function h in G is superharmonic if, for any point A in G and any circle in G we have h(A) 2 h"(A), where h" is the continuous function harmonic in the interior of the circle and having the same values as h on the boundary of the circle and outside the circle. We will prove first that h E H implies h" E H . It is sufficient to show that if x, y ,

.

+ +

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ORDERED VECTOR SPACES AND RlESZ SPACES

[CH. 2 , g 11

the first circle is called C, and C , is any circle in G intersecting C, and h# is constructed from h and C1 exactly as h" was from h and C , then (h")# 5 h". For any point A in the interior of C1 but not in the interior of C we have h"(A) = h(A), and hence it follows from h" 5 h that

(h")#(A)

5 h#(A) 5 h ( A ) = h"(A).

It remains to prove that (h")#(A) S h"(A) for all points A in C n C1. Now, d = h" - (h")# is harmonic in the interior of C n C1 and non-negative on the boundary of C n C , . Since the function d attains no extremum in the interior of C n C1, we must have d 2 0 in C n C1. This is the desired result. We return to the proof that the pointwise infimum s of all functions h E H is the harmonic least upper bound of the given functions u1 and u2 in L + . Given the point A in G and E > 0, there exists a function h E H such that h(A) < s(A) E. Then h(P) < s ( A ) E holds for all P in a certain neighborhood of A since h is continuous, and hence

+

+

s ( P ) 5 h(P) < S ( A ) + E

holds for all P in this neighborhood. In order to derive an inequality in the converse direction, let C be a circle in G with center A and radius r ; for any h E H let h" be the continuous function harmonic in the interior of C and having the same values as h on the boundary of C and outside C. For any point P in the interior of C we have, by the Poisson integral formula, that h"(P) = /"K(P, 0 Q ) h ( Q ) d q,~ where

-

1 r2-AP2 K(P,Q ) = 2n PQ2

*

Note that K(A, Q) = (2n)-', and that K(P, Q ) , as Q varies on the boundary of C, varies between

_1 . r - A P

2n r + A P

and

r+AP 2n r - A P '

---1

Hence, given E > 0, there is a neighborhood of A such that if P lies in this neighborhood, then K(Py Q) 2 (1 -E)K(A, Q )

CH.2,$11]

ORDERED VECTOR SPACES AND RIFSZ SPACES

53

holds for all Q on the boundary of C. It follows that

h(P) 3 h”(P) = S,”K(p, Q ) h ( Q ) d v ~

holds for every h E H. Hence, taking the infimum on the left, we obtain that s(P) 2 (1 -&)s(A) holds for all P in the neighborhood under consideration. Combining all results, it is evident now that s is continuous in G . Observing that s = i d (h : h E H ) , it is also evident that s is superharmonic. Hence, s is continuous and superharmonic, so s E H. This implies that s is the smallest continuous and superharmonic upper bound of u1 and u2 in the region G. In order to show that s is the smallest harmonic upper bound of u1 and uz in G , it will be sufficient to prove now that s is harmonic, i.e., to prove that s ( ~ )= (2n)-1JoZns(~)dqQ

holds for every point A in G and for Q varying on any circle C in G with center A . This is so because if

> (2n1-1 [ : S ( Q ) ~ ~ Q would hold for some circle with center A , the corresponding function s“ would satisfy s“ E H and s ” ( A ) < $(A), which is impossible. This finishes the proof. The present proof, wherein it is shown first that s is continuous and secondly that s is harmonic, is due to C . Visser. The example of the harmonic functions is due to F. Riesz ([2], 1937), and he refers to a proof by means of Poincare’s method of balayage. Having proved that L is a Riesz space, we will show that, in general, not every harmonic function in the region G is a member of L. For this purpose, assume that G is the open unit circle in the complex plane, u non-negative and harmonic in G, point P is in G, and C is a circle with the origin 0 as center and radius r satisfying OP < r < 1. For Q varying on the boundary of C, we have

54

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH.2 , s 11

and so, for r 7 1, we obtain that

#(P){1 - OP} s 2 4 0 ) .

Hence, givenf E L, the set of numbersf(P){1 - OP}is bounded for P varying in the unit circle. Now, let F ( z ) = (1 -z)-' for IzI < 1. The real part f of F is harmonic in G , but the above boundedness condition is not satisfied, and so f is no member of L. A somewhat different proof for the existence of a smallest harmonic upper bound of two given non-negative harmonic functions u1, u2 in a region G is as follows. Once more, let H be the set of all continuous superharmonic functions h in G satisfying h 2 u1 and h 2 u 2 , and let the function s be the pointwise infimum of all h E H in G . Furthermore, let S be a fixed closed circular disc in G, and for each h E H let h" be the continuous function harmonic in the interior of S and equal to h outside S and on the boundary of S. The set H" of all h" is a subset of H, and s = inf (h" :h" E H " ) . Given h ; , h i E H " , let the continuous function h i in G be defined by for every point ( x , y ) outside S and on the boundary of S, and inside S the function h i is harmonic. Then h i E H", and h i hl- as well as h i S h i (indeed, within the set H" the function h i is the greatest lower bound of hl- and h i ) . We will use the notation h i = hl- A h i . Since all numbers

s

.

are 2 0, the infimum a of this set of numbers exists. Let {A,"; n = 1, 2, . .} be a sequence in H" such that SSsh,dxdy + a

as n

-,m.

On account of the property proved just above we may assume that hl- 2 h; 2 * * -.It follows that there exists a function qs 2 0 in G such that h,- 1qps holds pointwise in G . By Harnack's well-known theorem, qs is harmonic in the interior of S. We will prove that qs(x,y ) = s(x, y ) holds for all points ( x , y ) in the interior of S; it will follow that s is harmonic in the interior of S, and SO (since S may be any circle in G ) the function s is harmonic in G . For the proof that qs = s holds in the interior of S, it is sufficient to prove that 'ps 5 h" holds in the interior of S for every h" E H " . If not, there exists a

CH.2,§11]

ORDERED VECTOR SPACES AND RIESZ SPACES

55

function h i E H" such that h,' < cps holds in a subregion of the interior of S. Let kn- = hn- ~ h , 'for n = 1,2, . . ..We have kn' E H" for all n = 1,2, ..., and kn- 1'pl in the interior of S with cpl 6 cps and cpl 6 h,' in the interior of S, so 'pl(x,y) < cps(x,y) for the points of the subregion referred to. But then

which is impossible. For further details about this example, cf. Example 23.3. (xi) Let X = {x : u < x < p} be a bounded open interval in the real axis, and L the ordered vector space of all real linear functions on X. Evidently, we may just as well assume that Xis the closed interval {x : ct 5 x 6 p). The space L is a Riesz space; givenf and g in L, the function h = sup (f,g ) is the linear function determined by h(a) = m a x {f ( u ) , g(ct)} and h(P) = max {f (p), g(p)}. The present example is the one-dimensional simplification of the preceding example. For further details, cf. Example 23.3. Definition 11.3. Given the ordered vector space L, the subset Lf = { f : f E L,f 2 O] is called the positive cone of L. Elements of L' are calledpositive elements. Theorem 11.4 (Cone properties). The positive cone L i of the ordered vector space L has the following properties. ( i ) f , g E L + impIiesf+gEL+. (ii) f E L' implies af E L'for every real number a 2 0. (iii) f, - f E L+ impliesf = 0. Conversely, if L' is a mbset of the real linear space L such that L' has the properties (i), (ii) and (iii), then L is made into an ordered vector space by defining that f 6 g holds i f and only if g -f E L', and L' is then exactly the positive cone of L with respect to this partial ordering. Proof. The routine proof is left to the reader.

Theorem 11.5. Let L be an ordered vector space with positive cone L'. The following holds now. (i) f 2 g i f a n d o n l y i f f - g E L f . (ii) f 2 g ifandonly i f f = sup (f,g ) , and also ifand only i f g = inf (f,9). (iii) f 2 g i f and only if af 2 ag for a > 0, and also if and also i f a f 6 ag for a < 0. (iv) If sup (f,g ) exists, then inf (-f, - g ) exists, and

inf(-f, - 9 )

=

-sup (f,9).

56

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH.

2, 5 11

(v) Iff, g E L ,then sup (f,g ) exists in L ifand only ifinf (f,g ) exists in L, and in that case we have SUP ( f h, 9 +h ) = SUP (f, s)+h, inf ( f + h , g+h) = inf (f, g)+h for any h E L. In particular, i f L is a Riesz space then sup (f,g ) and inf (f,g ) exist for any pair of elements f, g E L. I t follows also that i f L is an ordered vector space such that sup (f,0 ) existsfor every f E L, then L is a Riesz space. (vi) If sup (f,g ) exists, then for a 2 0, SUP (af, a s ) = a SUP (f, g) for a 5 0, sup (af,ag) = a inf (f,g ) inf (af,ag) = a inf (f,g ) for a 2 0, for a 5 0. inf (af, ag) = a sup (f,g) (vii) I f L is a Riesz space andf, g, h E L , then SUP {SUP (A 91,h} = SUP {SUP (f,h), SUP (9, h)} = SUP (f,9 7 h). Similarly for inf (f,g , h). Hence, any finite set of elements in a Riesz space has a supremum and an infimum in the space.

+

Proof. (i) and (ii) are evident. (iii) We will only prove that f 2 g and -f 5 -g are equivalent. This follows easily by observing that both inequalities are equivalent tof - g 2 0. (iv) It follows from part (iii) that if h is an upper bound offand g, then -h is a lower bound of -f and -g. Hence, if h = sup (f,g ) exists, then -h = inf (-A -g). (v) If sup (f,g ) exists, then sup (f,g ) +h is evidently an upper bound of f+ h and g +h. If k is also an upper bound off +h and g +h, the element k - h is an upper bound off and g , so k - h >= sup (f,g). It follows that k 2 sup (f,g)+ h. This shows that sup (f,g ) +h is the supremum off +h andg+h, i.e., SUP ( f + h , g+h) = SUP (f,g)+h. In particular, substituting h = -f-g,

we obtain

SUP (-A -9) = SUP (f,s)-f-s. But then, in view of the preceding part (iv), the element inf (f,g ) exists and satisfies inf (f,g) = -sup (-A -g). It follows now exactly as above that g)+h inf ( f + h , g+h) = inf (f, holds for every h EL. (vi) and (vii) are now evident.

CH.2,8 111

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ORDERED VECTOR SPACES AND RIESZ SPACES

Let f be an element of the ordered vector space L with the property that sup (f,0) exists. Then, as shown in the last theorem, inf (f,0) exists, and so sup (-A 0) exists. Furthermore, the element sup (2f, 0) exists, so that by subtraction off we obtain the result that sup (f, -f) exists. Definition 11.6. I f L is an ordered vector space, and the eIement f of L has the property that sup (f,0 ) exists, we will write

Note already that inf (f,0 ) = -sup

(-I; 0) = -f -.

Theorem 11.7. If L is an ordered vector space, and the elementf of L has the property that sup (f,0 ) exists, then the following holds. (i) f +, f - EL+; f' = (-f)- and f - = (-f)+;

If1 = I-fl.

(ii) f = f'-f-

with inf(f+,f-) = 0 ; If1 = f + + f - , and so IflEL'. (iii) 0 6 If1 and 0 5 f - S Ifl. (iv) -f- 5 f 5 f + . (v) (af)' = af' and (af)- = af- for a 2 0 ; (af)' = -af- and (af)- = -af+ for a 5 0; lafl = la1 If1 for real a. (vi) Iff, g E Land both f + = sup (f,0) and g+ = sup (9,0) exist, then g+ and f - 5 g - hold, f 5 g holds if and only iff

f's

-

+

Proof. (i) Evident from the definitions. (ii) We have f -f = sup (f,0)-f = sup (0,-f) = f -, and so f = f '-f -. In addition, 0 = -f-+f- = inf(f,O)+f- = inf(f+-f-,o) +f-= inf(f+,f-). Furthermore, we have If1 = sup(f, -f) = sup(2f,O)-f = 2f+-(f+-f-) = f + + f - . (iii) It follows from f +, f - EL+ and If 1 = f + +f - that 0 S f S If 1 a n d 0 S f - 5 Ifl. (iv) It follows from f = f'-f- that -f- = f-f' S f as well as +

+

fSf+.

(v) Follows from Theorem 11.5 (vi).

58

[CH. 2, 5 11

ORDERED VECTOR SPACES AND RIESZ SPACES

(vi) Assuming that f 6 g holds, we have f + = SUP (f,0) 5 SUP ( 9 , O ) = 9+, f - =sup(-f,o)~sup(-g,o)=g-. Conversely, iff' S g + and f - 2 g-, then f = f + - f - S g + - g - = g. Theorem 11.8. If L is an ordered vector space, and the elementsf and g of L have the property that sup (f, g ) exists, then ( f - g ) ' , (g-f)' and If - g l exist, and 0) SUP (f,9)= ( f - s ) + + g = (9-f)+ +f, (4 i n f ( L 9 ) = f - ( f - g ) + = s - ( g - f ) + , (iii) SUP (f,g)+inf(f, 9) = f + g , (i.1 SUP (f, 9)- inf (f, 9) = If (v) 2 S U P ( f , d =f+s+If-sl, (vi) 2inf(f,g) =f+g-If-gl. 9 1 9

Proof. (i) We have SUP (f, 9) = SUP ( f - 9 , O)+s = ( f - d + +9. (ii) We have inf(f,g) = f+inf(O,g-f) = f - ( g - f ) - = f - V - g ) ' . (iii) Follows from (i) and (ii) by addition. (iv) Subtraction of the formulas in (i) and (ii) yields SUP (f,d - i n f ( f , 9) = ( f - d + +(s-f)+= (f-s)++(f-s)- = If-91. (v) and (vi) are obtained from (iii) and (iv) by addition and subtraction respectively. From the formulas proved in the last theorem we will derive some interesting identities, holding in any Riesz space (holding in particular, therefore, for real numbers). Theorem 11.9. If L is a Riesz space and f and g are elements of L, then

2 SUP (Ifl,191) = If+sl+lf-sl9 2inf(lfl, 191) = llf+9l-lf-9ll~ Proof. Repeated application offormula (v) in the preceding Theorem 11.8 shows that 2 SUP 191) = 2 SUP (f,-A -9) = SUP (2 SUP (f, -91,2 SUP (9,-f )> = SUP {f-s+If+sl, s-f+If+sll = If+9l+lf-sl* In order to obtain the second formula we will writef = p q and g = p -q, ( I f 1 9

9 3

+

CH.2, 5 111

ORDERED VECTOR SPACES AND RIESZ SPACES

59

so f + g = 2p and f - g = 24. Using the just obtained first formula twice, we thus obtain 2 inf (IfL Igl) = 2lfl+2Ig1-2 SUP (IfL Id) = 4 f l + 2lsl- lf+gl - If-sl = 21P +41+2lP - 41 -2lPl-2141 = 4 SUP (IPIY l4l)-21Pl-214l = 2(lPl+ 141+ llPl - 1411)-21PI -2141 = 211Pl-lqll = 112Pl-124ll =

IIf+sl-If-glI.

An alternative proof of the first formula, due to G. J. 0. Jameson, is as follows. It has to be proved that SUP (IP+4L IP-41) = lP1+141 holds for all p , 4 E L.We have SUP (IP+4L IP-41) = SUP (P+47 - P - 4 , P - 4 , -P+4)7 and by Theorem 11.5 (v) we have SUP ( P + 4 , - P + 4 ) = SUP (P7 - P ) + 4 = lPl+47 suP(-P-qyP-4) = SUP ( - P , P ) - 4 = IPI-4, so SUP (IP+41, IP-41) = SUP M + 4 Y IPI-4) = IPI+141Y once more by Theorem 11.5 (v). Observe that it follows immediately from the second formula that inf (Ifl, 191) = 0 holds if and only if If+gl = lf-91. It is not difficult to prove by more elementary means (i.e., without using the second formula) that inf (Ifl, Is]) = 0 implies If+gl = If-gl (cf. Theorem 14.4 (i)), but until rather recently the converse did not seem to be mentioned in the literature. In 1964 it was set as a problem by the present authors in “Wiskundige Opgaven”, a collection of problems edited by the Dutch Mathematical Society, to prove that If+gl = If-gl implies inf (Ifl, Igl) = 0 (Wiskundige Opgaven 21 (1964), problem 192). The original rather inelegant solution was much improved by basing the proof on the second formula in Theorem 11.9. This was done, independently, by A. de Leeuw van Weenen (Mrs. A. van Arkel), M. A. Kaashoek and N. G. de Bruyn. The elements of the positive cone L+ of the ordered vector space L will occasionally be denoted by u, v , w,. . .. The positive cone L+ is called generating if everyf E L can be written in the form f = u - v for appropriate u and v in L+.The positive cone is not always generating. By way of example, let L be real two-dimensional number space RZ with L+ = {f:f=(fi,fZ),fZ = OJ-1 2 01, i.e., Lf is the positive horizontal axis. The positive cone Lf of a Riesz space I, is always generating. Indeed, given the elementf of the Riesz space L,we

60

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH. 2, $ 11

have f = f '-f - with f +, f - EL'. In the converse direction, if the positive cone ' L of the ordered vector space L is generating, and if sup (f,g) exists for allf, g EL', then L is a Riesz space. Indeed, given arbitrary elements f and g in L,we have f = ui-u2 and g = u3-u4 with ui EL' (i = 1,2,3, 4), and so fl = f + ( u 2 + u 4 ) and g1 = g+(u2+u4) are in L + . Then h, = sup (fl,gl)exists by hypothesis. It follows that SUP (f,9) = h1-

(.2

+ U4)Y

so sup (f,g) exists for all f and g in L . This shows that L is a Riesz space. It will be proved in the next theorem that the decompositionf = f '-f has a certain minimal property. Theorem 11.10. Let L be an ordered vector space. (i) Ifu, u E L+ and sup (u, u ) exists, then 0

5 inf (u, u ) 5 sup (u, u ) 5

u+u.

(ii) Zff = u - u with u, UEL' and iff' = sup (LO)exists, then f ' S u and f - 5 v. Hence, the decomposition f = f -f- as a diference of two positive elements is a "minimal" decomposition. It was proved in Theorem 11.7 (ii) that inf (f+, f -) = 0. Conversely, iff = u-v with u, u E L + and if inf (u, u ) exists and satisfies inf (u, u ) = 0, then f = sup (f,0) exists and we have u = f + and u = f In other words, iff = u- u with inf (u, u ) = 0, thenf = u - u is the minimal decomposition.

'

-.

+

Proof. (i) Follows immediately by observing that inf (u, u ) and sup (u, u ) are in L+ and inf (u, u ) + sup (u, u ) = u + v. (ii) Let f = u - u with u, u E L + ,and let f = sup (f,0) exist. Then we have f = sup (f,0) 6 u because f 5 u as well as 0 S u. Hence also f '-f 5 u - f , i.e., f - 5 u. For the second part, assume that f = u-v with inf (u, u ) = 0. Then it is obvious that u and u are in L+,and from Theorem 11.8 (ii) it follows that +

0 = inf (u, u ) = u-(u-u)+

and so

f '

= u. But then u =

u-f = f'-f

=

u - f +,

=f - .

Exercise 11.11. Let fl,. . .,f, be elements of the Riesz space L.Prove that ISUP (fl

Y

Iinf(f19

- - .,f,)l * *

5 SUP (IfilY * * lf,l>Y .,f,)l 5 SUP (Ifll, * - KI). .Y

.Y

CH.2,511]

ORDERED VECTOR SPACES AND RIESZ SPACES

61

Exercise 11.12. Let f, g be elements of the Riesz space L. Prove that 2 SUP {inf(f, 9)s inf(-f, -911 = If+sl-lf-91, 2SUP {inf(f, -d,inf(--fgN = If-sl-lf+sl. Compare this result with the result in Theorem 11.9.

Exercise 11.13. Let L be an ordered vector space with positive cone L+. Prove that L is a Riesz space if and only if for every pairf, g E Lthere exists an element h E L such that

(f+L+) n (g+L+)= h+L+, where the addition signs denote algebraic sums (i.e., f+L+ is the set of all

f+ u for u varying in L'). Prove that L is not necessarily a Riesz space if the last condition is only satisfied for allf, g EL+,but L is a Riesz space if, in addition, ' L is generating.

Exercise 11.14. Letf, g, p , q be elements of the Riesz space L. Prove that iff +g = p q, then SUP (AP)+ inf (9,4 ) = f + 9 .

+

Exercise 11.15. Let L be the set of all real functions of finite variation on the bounded interval [a,b] in the real line with the extra property that f(a) = 0 holds for every f E L. The set L is a real linear space under pointwise addition and multiplication by real constants. Show that the set L+ of all f E L such that f is non-decreasing in [a, b ] has the cone properties. Hence, L is an ordered vector space with positive cone L + . Show that L is a Riesz space. More precisely, show that if

P = {u = t, < t , < ... < t"

= xo

5 b}

is a partition of [a, xO],and we divide the set of subintervalswhich form P into disjoint subsets P',P'' and make the sum

c'

2'

where and denote summation over the subintervals in P' and P" respectively, then

for P running through all partitions of [a, xo] and P',P" still variable for fixed P.

62

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH. 2,s 12

The example is due to L. V. Kantorovitch [l]. Note the analogy with Example 11.2 (viii). Hint: Assume first that g ( x ) = 0 for all x. It is well-known that in this particular case the number sup,.,,,, s(P', P") is exactly the positive variation P f ( x o )off at the point xo. It is also well-known that f ( x ) = P f ( x )- N,-(x), where N,- is the negative variation off, is a representation off as a difference of two functions ofL+, and also that if f ( x ) = p l ( x ) - p 2 ( x ) is another representation of this kind, then p 1 2 Pf and p 2 1 Nf. Hence f = Pr - N,is the minimal representation, so Pr = f v 0 and Nf = (-f)v 0 by Theorem 11.10 (ii). The general result about f v g follows by observing that

f v s = {(f-g)vOI+g. 12. Inequalities and distributive laws in a Riesz space

In the present section L denotes a Riesz space. Theorem 12.1 (Triangle inequality). For all f, g E L , we have

sf

(f+9)+

+

+9+,

( f + s ) -S f - + 9 - , Ilfl-lsll s If+sl s Ifl+lsl. 2 f+g and f + + g + Z 0, so f + +9+ 2 SUP (f+s,0) = ( f + 9 ) + .

Proof. We have f

+

+g+

It follows that and so

f - + g - = (-f>++(-s)+

1 (-f-9)+

= (f+9)-,

If+sl = ( f + s ) + + ( f + ss) f- f + 9 + + f - + = 9 Ifl+lsl. This implies that If1 - Igl 5 1f - g l and 191- 1 f I S I f - g l , so IIfl-lsll = suP(Ifl-191,191-lfl) S If-91This holds as well if - g is substituted for g, and so

IIfl-lsll 5 If+sl. Theorem 12.2 (Distributive laws). If D is a subfet of L such that f o = sup ( f :f E 0)exists in L , then

inf ( f o , 9) =

holds for every g in L.

SUP

{inf (f, s) : f E D>

CH.2,s 121

INEQUALITIES AND DISTRIBUTIVE LAWS IN A RIESZ SPACE

63

Similarly, i f f l = inf ( f :f E D ) exists in L, then

SUP ( f l , 9) = inf {SUP (f,9): f E Dl holdsfor every g in L.

Proof. Assume that f o = sup ( f :f E 0)exists and let g be an arbitrary element of L. Sincef o 2f holds for allf E D, we have inf (f o ,g ) 2 inf (f,g ) for all f E D, and so inf ( f o , g ) is an upper bound of the set of elements {inf (f,g) :f E D } . Let m be another upper bound of this set. Then m 2 inf (f,9) = f +9- SUP (f,91, so m-gfsup (f,g ) 2 f holds for all f E D. It follows that m-g+sup

(fo, 9) 2f

holds for all f E D, and so i.e., This shows that

m-g+sup ( f o , 9) B SUP ( f : f E 0) = f o ,

m 2 f o + g - s u P ( f o , 9 ) = inf(f0,g). inf (So,g ) = sup {inf (f,g ) :f E D } .

The second formula can be proved similarly or can be derived from the first formula by observing that f l = inf ( f :f E 0)is equivalent to -f l = sup (-f:f E 0). Corollary 12.3 (Finite distributive laws). For brevity, denote sup (f,g ) and inf (f,g ) in the present corollary by f v g and f ~g respectively. Then the formulas (fV9)Ah= (fAh)V(9Ah), hold for allf, g , h E L.

(fA9)Vh = ( f V h ) A ( V h )

It has been shown thus that every Riesz space L is a distributive lattice with respect to the partial ordering. There is in L no smallest and no largest element (unless in the trivial case that L consists only of the null element). The positive cone L+ by itself is a distributive lattice with the null element of L as smallest element; it follows that all results proved in Chapter 1 for a distributive lattice with smallest element hold in L'. In particular all results about prime ideals and the hull-kernel topology hold. We will say more about this in the next chapters.

64

[CH.2, 8 12

ORDERED VECTOR SPACES AND RIESZ SPACES

Theorem 12.4. (i) (Birkhofidentity.) For a l l f , g, h in L we have ISUP

(9, h)l

(f, +sup

+ linf (5h)-inf

(9, h)l = lf-91.

(ii) (Birkhofinequalities.) For allf, g, h in L we have ISUP (f, h)-suP (9, h)l 6 If-gl, Iinf(f, h)-inf(g, h)l 6

If-sl.

(iii) For aNf, g in L we have

If+-s+l6 lf-91

and r - g - 1 6 If-91.

Proof. (i) Applying the formula k - 4 1 = SUP @, q)-inf(p, 4 )

to the case that p = sup (f,h) and q = sup (9, h) as well as to the case that p = inf(f, h) and q = inf (9, h), we obtain

ISUP

h)l = h) A (9 V h)+ ( f A h ) V (g A h ) -fA h A g A h.

(f, h)-sup (9, h)l +linf(f, h)-inf(g, =f V h V g V h

- (fV

By means of the distributive laws, applied to the terms in the middle, this is seen to be equal to f V g V h- (fA

g) V h+ (fV g) A h - f A g A h = = { ( f v s)v h =

+ ( f v d A h ) - {(f.

{ ( f v s > + h )- {(ff4+h)

9 ) v h + (f. 9) A h )

= f v g - f A s = If-sl.

(ii) and (iii) are now evident. The theorem is due to G. Birkhoff cf. Theorem 7.8 on p. 109 of the first edition of his book on lattice theory [l](1940); the formula in (i) is not mentioned in the second edition (1948) and appears again in the third edition (1967);the formulas in (ii) and (iii) are mentioned in all the editions. Many of the formulas in this chapter, not involving scalar multiplication, hold also in ”lattice groups” even when these are not commutative, but the proofs are then sometimes more difficult. Until rather recently (1956, J. A. Kalman [l]) a proof for the Birkhoff identity in this more general case was unknown. We observe, incidentally, that the identities in Theorem 11.9 hold in a lattice group if and only if the group is commutative.

CH.2,s 121

65

INEQUALITIES AND DISTRIBUTIVE LAWS IN A RIESZ SPACE

Theorem 12.5. For all u, v, w B' L we have inf(u, w)+inf (u, w)-w 6 inf (u+v, w) 6 inf (u, w)+inf and

sup (u, u )

(0,

w)

- inf (u, v)+w _<

s sup (u, w) + sup

Ifinf ( v , w)

=

0, then

(0,

(1)

w) - w 6 sup (u + 0, w) 5 sup (24, w)+ sup ( 0 , w). (2)

inf (u+ v, w) = i d (u, w).

If inf (u, u ) = 0,then

inf (u+u, w) = inf (u, w)+inf (v, w). Proof. The last inequality in (1) is trivial. Indeed, setting 1 = inf (u, w)+ inf (u, w), we have trivially that 15 u + w and 1 U + W , so

s

1I inf(u+w, u+w) = inf (u, v)+w.

The first inequality in (1) follows then immediately. It remains only to prove that inf (u + u, w) 5 inf (u, w) + inf (u, w). (3) In virtue of one of the Birkhoff inequalities we have

o 6 inf(u+v, w)-inf

(u, w) = linf (u+v, w)-inf(u,

and it is obvious that Hence

O

w)I

6 I(u+u)-ul = 101 = v,

s inf (u+u,

w)-inf (u, w)

o 6 inf (u+v,

5 inf (u+v,

W)

5 W.

w)-inf (u, w) 6 inf (0, w),

which is the same as (3). Similarly, the only non-trivial part in the proof of (2) is the proof that sup (u, w) + sup ( 0 , w) -w I sup (u

+

0,

w).

(4)

Subtracting (3) from the identity u + u + w = (u+w)+(u+w)-w, we obtain (4). If inf (v, w) = 0, we obtain inf (u+Y, w) 6 inf (u, w) from (1). On the other hand, it follows from u + u 2 u that inf (u+u, w) 1 inf (u, w). Hence inf (u+u, w) = inf (u, w).

66

[CH.2,o 13

ORDERED VECTOR SPACES AND RIESZ SPACES

If inf (u, u ) = 0, it follows from the general formula u+v = sup (u, u)+ inf (u, u ) that now u+u = sup (u, u). Similarly u1 + u , = sup (ul, v l ) for u, = inf (u, w ) and u1 = inf (u, w ) . Hence inf(u+u,w)

= ( u + v ) ~ w= ( U V U ) A W = ( U A W ) V ( U A W )

= u1 v uI = u1

+ u1 = inf (u, w)+inf (u, w).

Exercise 12.6. Let u , , . . .,u,,, u E L+ and wi = inf ( u i , v ) for i = 1, Show that inf (ul . . . +u,, u ) = inf (wl + . . . +w,,, u).

. . .,n.

+

Exercise 12.7. Show that for allf, g , p , q E L we have inf(f, g)+inf(p, 4 ) = i n f ( f + p , f + q , 9+P, S + d , 9)+SUP (P, 4) = SUP ( f + P , f + % 9+P, 9 f 4 ) . SUP (f, 13. Suprema and infima of subsets of an ordered vector space

Let L be a n ordered vector space. Any subset of L can be considered as an indexed subset {f, : T E {T}}, where {T} is an appropriate index set.

Theorem 13.1. (i) I f f = supf, exists in L, then -f = inf (-A). Similarly, = inff, exists, then -f’ = sup (-A). (ii) Iff = supf , exists in L, then af = sup (af,) for a 2 0 and af = inf (ah)for a 5 0. Similarly, ifinff, exists. (iii) I f f = supf, exists in L, then f + h = sup ( f , + h ) for every h E L . Similarly, if inff, exists. (iv) Iff = supf, and g = sup gb exist in L, then

iff’

f+s

= sup(f,+g,). Ga

Similarly, if inf f , and inf go exist. (v) I f f = supf, exists in L, and if h E L has the property that sup (f,h ) and all sup ( f , , h ) exist in L, then SUP (f,h ) = SUP {SUP (f, h ) } . 9

T

’

In particular, iff = supf , holds and i f f and allf J exist, thenf = supf ?, . Similarly, iff’ = inff, exists in L, and if h E L has the property that inf (f’,h ) and all inf (A, h ) exist in L, then +

inf (f’,h) = inf {inf (f,,h)}. T

CH.2 , s 141

67

DISJOINTNESS

Proof. (i), (ii) and (iii) are evident. (iv) Let f = supf, and g = supg,. It is evident that f + g is an upper bound of the set of all elementsf,+g,,. If h is another upper bound of this set, we have h 2 f+g, for every Q, and so h 1f+g. This shows that

f + s = sup(f,+g,,). 7.

(v) Let f = supf,, and let h E L have the property that sup (f,h ) and all sup (A, h ) exist in L. It is evident that sup (f,h ) is an upper bound of the set of all elements sup (A, h). If k is another upper bound of this set, we have k 2 sup (f,,h ) 2 f, for all z, and so k 2 supf 7 = f. Since k 2 h is also satisfied, we have k 2 sup (f,h). This shows that SUP !f,h) = SUP {SUP (f, h ) } . 9

7

The next theorem goes in the converse direction.

Theorem 13.2. r f f = supf, and f + g tisfies g = sup g,,.

=

sup,, ,, (f7+g,,) exist, then g sa-

Proof. For every fixed Q, we have f7+g,, S f + g for every z, and so supf,+g, 5 f+g, i.e., f+g, S f + g , so g,, 2 g. This shows that g is an upper bound of the set of all go. If h is another upper bound of this set, then g,, S h for all 6,so f , +go 5 f +h for all z and Q, which impliesf +g 5 f h, and so g S h. This shows that g = sup g,,.

+

Theorem 13.3. Let L be a Riesz space. r f f = supf,, then f = supf,' and f - = inff,-. Similarly, i f g = inf g7, then g+ = inf g: and g- = SUP g7- * +

Proof. Letf = supf,. It was proved in Theorem 13.1 (v) that f and from one of the distributive laws we derive

+

= sups,',

inf(f, 0 ) = SUP (inf(S7, O)}, t

i.e., -f - = sup (-L-). It follows that f - = inff,-.

14. Disjointness In the present section we assume again that L is a Riesz space.

Definition 14.1. The elements f and g of L are called di$joint ifinf (If I, Igl) 0. This will be denoted by f Ig.

=

68

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH.2,8 14

Note that for elements of L' the present definition is in agreement with the definition of disjointness in section 2 for elements in a lattice with null element. Theorem 14.2. (i) I f f l g and lhl 5 Ifl, then h I g. (ii) Ig and a is a real number, then af I g. (iii) Iffi I g andfz 19,then (A +fz)I 9. (iv) We have flg if and only iff '1g and f - I g.

vf

Proof. (i) We have

0 6 inf(Ihl, Igl) 6 inf(Ifl, Igl) = 0. (ii) Note first that for elements u, u in L' and any number b > 0 we have inf (bu, bu) = b inf (u, u ) by Theorem 11.5 (vi), so u lv implies bu Ibv. Now, flg is equivalent to If1 I 191, and so

(la1+ 1)lfl I (la1 + 1)lSl holds for every real number a. Hence, since

- If1 5 (la1 + l)lfl, it follows from part (i) that a f l (la1 + 1)lgl. Similarly, since Igl 6 (la1 + 1)1g1, lafl = la1

it follows that a f l g . (iii) Letf, Ig andf2 Ig. An application of Theorem 12.5, formula (l), shows that

inf(If1+f2l, Isl) 6 inf(If1I+lf2l, 191) inf(If1L Isl)+inf(If21,191) = o+o = 0, so (f1 + f 2 1I 9. (iv) Iff I g, thenf' Ig on account of I f ' ! = f' 5 Ifl. Similarly we find that f - I g. Conversely, iff' Ig and f - Ig, then (f' +f-)lg, i.e., If1 Ig, and s o f l g.

s

The elementfof L is said to be disjoint from the non-empty subset D of L iff Ig holds for every g E D. Furthermore, the subset D of L is said to be solid if it follows fromfE D and lhl 5 If1 that h E D. Note now already for later purposes that in virtue of (i), (ii), (iii) in the last theorem the set of all elements disjoint from a fixed nonempty subset D of L is a solid linear subspace of L. This solid linear subspace is called the disjoint complement of D , and is denoted by Dd.This terminology is in agreement with that introduced in section 2.

CH. 2,@141

69

DISJOINTNES

Theorem 14.3. If D is a subset of L such that fo = sup (f :f E D) exists in L, and iff _L g holds for all f E D, then fo Ig. Proof. We have f 'J- g and f - J- g for all f E D by Theorem 14.2 (iv). Furthermore,wehavef,f = s u p ( f ' : f ~ D ) a n d f t = i n f ( f - : f ~ D ) b y Theorem 13.3, and so inf (f:,

Igl) = sup {inf (f', 191) :f E D} = 0

by one of the distributive laws, as well as 0

5 inf(f;, Id) 5 inf(f-, Igl) = 0

for all f E D. It follows that f: I g as well as f;

Ig, so f o Ig.

Note that it follows from this theorem that the disjoint complement Dd of the subset D of L has the property that if E is a subset of Dd and fo = sup (f :f E E) exists in L,then fo E Dd. Theorem 14.4. (i) Iff J- g, then (ii)

if+sl = If-gl Iff Ig, then

= Ifl+lgl =

Ilfl-lsll

= SUP (IfL

Isl).

(f+g)+ = f + + g + = sup (f +,g'), (f+g)- =f - +g- = sup (f-,9-).

Proof. (i) It follows from and

0 = 2inf(Ifl, Isl) =

lfl+l~l--llfl--l~ll

s If+sl 5 lfI+lgL IIfl-lsll 5 If-sl 5 Ifl+lsl Ilfl-lgll

that

If+gl = If-sl = lfl+Igl = Ilfl-lsll.

Finally, we have

If I + Is1 = SUP (Ifl, Isl)+inf (If I, Isl) = SUP (IfI, Isl). (ii) It follows from inf(If1, Igl) = 0 together with 0 6 f' 5 If1 and 0 5 g' 5 191 that inf (f ', g') = 0. Hence f 'Ig', and so f '+g' =

sup (f ', 9') by part (i) of the present theorem. Similarly we obtain that f- Ig- and f -+g- = sup (f-,g-).

In order to show that (f+g)' = f + + g ' and (f+g)- = f - + g - we recall that by Theorem 12.1 we have (f+g)' 5 f'+g' and (f+g)- 5

70

ORDERED VECTOR SPACES A N D RIESZ SPACES

[CH. 2, 6 15

f - + g a y holding for arbitrary f, g E L . It will be sufficient, therefore, to prove that f l g implies f f + g + 5 ( f + g ) + and f - + g - 5 ( f + g ) - . The first inequality follows from f++g++(f+$

5f++gf+f-+9-

= Ifl+lgl = If+sl = ( f + g ) + + ( f + s ) - ,

and similarly for the second inequality.

In the converse direction we have the following theorem.

Theorem 14.5. (i) Zflf + g l = I f - g l , then f lg. (ii) Zf If+gI = sup ( I f I, Igl), then f lg. Similarly, of course, if If-gl = SUP ( I f I, Igl), thenf 19. Proof. (i) Follows from the formula 2inf(IfL Isl) = IIf+sl-If-gIl.

(ii) If I f + g l = sup (If 1, Igl), it follows from

2sup ( I f l , Igl) =

If+sl+lf-sl

that I f + g l = I f - g l , and so f l g by part (i). It will be evident from the foregoing that the notion of disjointness has some properties similar to the properties of orthogonality in Hilbert space or Euclidean space. Here is another property of that kind.

Theorem 14.6. I f ( f j : i = 1 , . . ., n ) is a system of nonzero and mutually disjoint elements of L, then this is a linearly independent system. Proof. If the system fails to be linearly independent, one of the f i , sayf,, is a linear combination of the other elements in the system, say f l = a2f2 +anfn.By Theorem 14.2 we have

+ . ..

fi

I(a2f2 + .

* *

+anL),

i.e.,fl If l . This means that inf (If l l , If l l ) = 0, i.e., If i l = 0, and sofl = 0. But this contradicts the hypotheses.

15. Monotone sequences and directed sets We begin with a definition.

Definition 15.1. The sequence (f n : n = 1 , 2, . . .) in the ordered vector space L is called increasing i f f l 5 f 2 5 . . . and decreasing l f f l 1f i 2 . . ..

CH.2 , s 151

71

MONOTONE SEQUENCES A N D DIRECTED SETS

This will be denoted by f n t or fnJ respectively. If f n t andf = supf n exists in L, we will write f, f f, or f,f, f if necessary. I f f , J and f = inf f n exists in L, we will write f, J f. Theorem 15.2. (i) I f f , f f , then A, t f for every subsequence (S,: k = 1, 2, . . .) such that n , < n2 < . . .. This holds in particular for the subsequence (f,: n 2 no), where no is an arbitrary but jixed natural number. (ii) I f f n t f and a 2 0 is real, then af. t a$ (iii) I f f n t f andgn t 9, thenfn+gn t f + g (iv) If f n t f, grit andf, +gn t f +9, then gn t 9-

Proof. (i) Let f n tf, and let (f,,:k = 1,2, . . .) be a subsequence such that n, < n2 < . . .. Evidentlyf is an upper bound of the subsequence. If g is another upper bound of the subsequence, and f , is an arbitrary member of the sequence (f n : n = 1,2, . . .), there exists a member nk, 0 of the index set (n, :k = 1, 2, . . .) such that m S n k . 0 , and hence fm

5 f n k , 5 9' 0

The inequality f , S g holds thus for all m yso f = supf , 5 g . It follows that f = supf n k . (ii) Evident. (iii) Evidentlyf n + g n is increasing and f + g is an upper bound of the set of all elementsf,+g,. If h is another upper bound of this set, and if m and n are natural numbers, say m S n, then f m + g n4 f n + g n h, and hence (in virtue of Theorem 13.1 (iv)) we have f + g = suP(fm+gn) Ihm. n

This shows that f + g = sup ( f , + g n ) ,s o f , + g , t f + g . (iv) For any fixed n we have f , + g n f m + g mfor all m fm+gn ISUP ( f m + g m : m 2 n ) m

3 n, and so

=f+g

holds for all m 2 n. It follows that SUP (fm m

: m 2 n)+gn

S f+g,

i.e.,f + g , S f + g , so gn 5 g . This shows that g is an upper bound of the set of all g,,. If h is another upper bound of this set, thenf,+g, 5 f+h holds for all n, which implies f + g 5 f + h , and so g 5 h. This shows that g = SUP 9 . 3 i.e.9 gn t 9-

72

1CH. 2.8 15

ORDERED VECTOR SPACES AND RIESZ SPACES

Theorem 15.3. If L is a Riesz space, and iff,, t f and g,, t g, then

gn1.t SUP (f, g), inf ( f n 2 gn) t inf (f, g)*

SUP ( f n ,

In particular, f n

tf

implies

f

+

and f.- 1f - .

Proof. It is evident that sup (A,g,,) is increasing and sup (f,g ) is an upper bound of the sequence of all elements sup (f,,g,,). Let h be another upper g,,) 5 h holds for all n, so bound of the sequence. Then f, 5 sup (f,,, f S h. Similarly g 5 h, and hence sup (f,g ) 5 h. This shows that sup (f,g ) is the supremum of the sequence of all elements sup (A,g,,), i.e., sup (A,g,,) t SUP (f,s). Set p = sup (f,g ) and q = inf (f,g ) , and similarly set p,, = sup (&, g,,) and q,, = inf(A,g,,) for n = 1,2,. .. Observe that p + q = f + g and p,,+ q,, = f,+gn for every n. We have p,, p as just proved, q,, is increasing and Pn+qn =fn+gn t f + g = P + q

.

by part (iii) of the preceding theorem. It follows then from part (iv) of the preceding theorem that qn t q, i.e., inf (f,,g,,) t inf (f,g). Definition 15.4. I f L is an ordered vector space, ifu,, E L f for n = 1,2, and s, = u1 + . . . + u,, satisfies s,, s, we will write u,, = s.

1:

.. .

Theorem 15.5. Let L be a Riesz space. (i) Given that u, v , z E L', u+u = z and 0 5 z,, z, there exist sequences 0 S u,, t u and 0 S v,, t v such that u,,+v,, = z,, for all n. In particular, if u, v E L f and u+v = zi with zi EL' for all n, there exist elements u;, vi EL' such that u = ui, v = v; and u ~ + v=~zi for all n. (ii) This can be extended to a greater number of terms; we present the formulation for three terms. I f u, v , w, z E L f , u + v + w = z and 0 5 z,, t z, there exist sequences 0 u,, t u, 0 5 v,, T v and 0 5 w,, t w such that u,,+v, w,, = z, for all n.

1: 1;

17

+

Proof. (i) Setting u,, = inf (u, z,) for all n, we have

0 5 u,, t inf (u, z) and it is evident that v, = z,,-u,, satisfies v,, Since

0 4 u,,+,-u,,

= u,

2 0 and u,,+v,,

= inf (z,,+~,u)-inf (z,,, u )

=

Iz,,+l-z,,

z,, for all n.

CH. 2,s 151

73

MONOTONE SEQUENCES AND DIRECTED SETS

by Birkhoff’s inequality, we obtain that

5 Zn+1-un+1 = un+1, so 0 5 u,,T. It follows then from u,, t u, v,,t and u,,+u,, = z,, t z vn

vn

= Zn-un

t v-

= u+

u that

(ii) Set u+u = s, so s+ w = z and 0 5 z,, t z. By part (i) there exist sequences 0 5 s,,t s and 0 5 w,, t w such that s,,+w,, = z,, for all n. Once more by part (i), there exist sequences 0 5 u,, t u and 0 5 u,, u such that u,,+ v,, = s,, for all n. The desired result follows now.

Corollary 15.6. Every Riesz space L has the following properties. ( i ) (Dominated decomposition property.) Given that

..

0 5 u 5 zi+

... +zp

with z,, E Li for n = 1, .,p , there exist elements u l , . . ., upE L i such that u = u l + . . +up and u,, 5 z,,f o r n = 1,. .,p. (ii) (Riesz interpolation property.) If u, u, z l , z2 E L+ and u+ u = z l + z 2 , there exiAt u l , u 2 , v l , u2 E L+ such that

.

.

u , + u , = u, u , + u , = z1,

u,+u, = u, uz+u2 = z2.

Definition 15.7. The indexed subset {f, : T E {z}} of the ordered vector space L is called directed upwards if for any pair zl, t 2E { t }there exists r 3 E { T } such thatf,, 2 f,,andf,, 2 f,, hold simultaneously, and {f , : t E { t } is } called directed downwards i f f o r any pair t l ,t 2E { t }there exists t 3 E { t }such that c, 5 f,, and f,, Sf,,hold simultaneously. This is denoted by f,t or f,J respectively. I f f,t and f = supf, exists in L, we will write f , t S, or f , t, f if necessary. Iff,l and f = inff, exists in L, we will writef, Jf. The upwards directed sets {f,} and {g,},indexed by means of the same index set {z}, will be called equidirected iffor any pair z1 , t 2E {T} there exists t 3 E {z} such Ithat f,, 2 f,, andf,, 2 f,, as well as g,, 2 g,, and g,, 2 g,, hold simultaneously. The upwards directed sets {f,} and {g,} are equidirected in particular iff,, 5 f,, holds if and only if g,, S gr2holds (note, however, that this is not a necessary condition for being equidirected). A similar definition holds for downwards directed sets. Monotone sequences are special examples of directed sets. Directed sets have many properties similar to properties of monotone sequences, as shown in the next theorems. Theorem 15.8. (i)

IfA t f

and a 2 0 is real, then af, t af.

74

[CH. 2 , § 15

ORDERED VECTOR SPACES A N D RIESZ SPACES

(ii) Iff, t f and g U t Y g then ( f , + g U ) t T , U (f+g)' (iii) I f {f,} and {g,} are equidirected,f, t f and g, t g, then f, g, t f g. (iv) Iff, tf, g u t and (f,+ gu) t r , u (f+g), then g u t 9. (v) If {A} and {g,} are equidirected,f,tf andf,+g, t f + g , then g, t g.

+

+

Proof. (i) Evident. (ii) Let f,t f and g,, t g. Note first that the set of all elementsf,+g,,, indexed by the set of all pairs (z, a), is directed upwards. Indeed, given (zl,al)and (z2, 02),there exists (z3,03)such that f,, 2 f,, and f,, 2 Az as well as gu, 2 g,,, and g,,, 2 g,,, and hence f,,+g,,, 2 f r , + g u l and f,, +g,,, 2f,,+guz. It was proved already in Theorem 13.1 (iv) that f+ g = sup,,,, &+go). Hence, combining these facts, we obtain that (f, +g u )

tr, u

( f +g).

(iii) Let {f,} and {g,} be equidirected,f, t f and g, t g. The set of all elementsf,+g, is now directed upwards, and evidentlyf+g is an upper bound of the set. If h is another upper bound of the set and z1,z2 E {z} are given, there exists z3 E {z} such thatf,, 2 f,, and g,, 2 gr2,and hence f,,+gzz

Sfr,+gr, S he

By Theorem 13.1 (iv) we havef+g = (f,,+g,,), so on account of f,,+grz5 h holding for all z1 ,z2 we find now that f + g S h. This shows that

f+

=

"P

r

(fT

+

ST)'

(iv) In Theorem 13.2 it was proved that i f f = supf, and f + g = sup,,,, (L+g,,), then g = sup gu.Hence, iff, If, g,, is directed upwards and (L+gu) t r , u (f+g), then g u t 9. (v) Let {f,}and {g,} be equidirected, f, f and (f,+g,) t (f+g). We will prove first that f +g is an upper bound of the set of all elementsf,, +gTz. For this purpose, given z1and z2, let z3 be such thatf,, 2f,,and g,, 2 grZ. Then f,,+g,z S f T 3 + g T 3 r f + g , which is the desired result. Sincef + g is already the supremum of the set of all elements fr+g,, it follows that

f + g = SUP (fz, +grz). T I . 72

Hence, by part (iv) above, we have g,

t g.

CH. 2 , s 151

MONOTONE SEQUENCES AND DIRECTED SETS

75

Theorem 15.9. Let L be a Riesz space. ( 9 Iff, t f and 9, t 9,then

"P (f, SU) tT, U "P (f,g), inf (f,,S U ) t T , U inf (f,g). 9

f, t f implies that sup (f,, g) t sup (f,g) and inf (A,g) inf (f,g) for every g E L. As a still more particular case, we have that f, t f impliesf,' t f f a n d f ; 1f - . (ii) I f {A} and {g,} are equidirected,f, t f and gr t g, then

I n particular,

"P (f,9,)t "P (f,g)? inf (f,ST) 7 inf (f,9). 7

9

Proof. (i) The proof for sup (f,, g,) t,, ,sup (f,g) is very similar to the proof of the corresponding statement for sequences in Theorem 15.3. For the second formula, set p = sup (f,g) and q = inf (f,9);similarly, set p,,, = sup (A,g,) and q,, ,= inf (A,g,) for all z, CT.Observe that p+q = f + g and p,, +qr,,= h + g , for all r, CT.We have p,, ,t p as observed above; furthermore. q,, ,is directed upwards and

PT,U+qT,O = (f,+gU)

t ( f + g ) = P+q

by part (ii) of the preceding theorem. It follows then from part (iv) of the preceding theorem that g,, , q, i.e., inf (A,g,) t,, inf (f,9). (ii) Similarly. Theorem 15.10. Let L be a Riesz space. Given that u, v, z E L + ,u+u = z and 0 5 z, t z, there exist directed sets 0 5 u, T u and 0 S v, t u, equidirected with {z,} in the sense that z,, S z, implies uT15 u,, and vI1 S v,,, and such that u,+ v, = z, holds for all T. This can be extended to a greater number of terms. Proof. Set u, = inf (u, z,) for all r. Then we have 0 5 u, t inf (u, z ) = u, and z,, S zI2implies uI1 S u,,. Furthermore, it is evident that v, = z,-u, satisfies v, 2 0 and u,+v, = z, for all z. For zTl5 z,, we have

0

s u,,-u,,

=

inf (z,,, u)-inf (zTl, u ) S zT2-zTI

by Birkhoff's inequality, so =

5 zT,-uUIZ

=,,T'

and hence 0 5 v,t. It follows finally from u, t u, v,t and u,+v, = z, z = u+v that u, t v.

t

76

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH. 2,515

Theorem 15.11. (i) Given the indexed subset {f,} of the Riesz space L, there exists an upwards directed subset {f,} of L such that {fu} c {A} and {f,} has the same upper bounds as {f,}. (ii) If {f,} is an upwards directed subset of the Riesz space L andAo is an arbitrary member of {A},then the set {f, :f,2 f,,} is still upwards directed and has the same upper bounds as {f,}. In particular, the set {A}has a supremum in L if and only i f the set {f,-Lo:f, 1f,,} has a supremum in L+

.

Proof. (i) For any finite subset z = (a1,. . ., 0,) of the index set {a}, let

f, = fu,,....

U"

= SUP (fu, 7

* *

.,fun).

Evidently {fu} is a subset of the thus obtained new set {A}, and {f,} and {A} have the same upper bounds. In addition, {f,} is directed upwards. Indeed, given we set

= (01 Z"

7

*

., a,),

= (a1,

Z'

= (0; ,

. . .) a;),

. . .)a, ,a; , . . .)0;),

and then we havef,., 2f, andf,.. 2 A,. (ii) It is evident that the subset {f, :f,2 A o }is still upwards directed. It remains only to prove that any upper bound h of {f, :f,Zf,,}is also an upper bound of {f,}. Given an arbitrary f,,E {A}, there exists an index z2 E (T} such thatf,, 2 f,, andf,, 2 f,,. Hence,f,? is a member of the subset {f,:f,5 f,,}, which implies thatf,, 5 h. It follows that f,, S f,, S h, and this shows that h is indeed an upper bound of {f,}. We recall that a partially ordered set is called Dedekind complete whenever every non-empty subset which is bounded from above has a supremum and every non-empty subset which is bounded from below has an infimum. It was proved in Theorem 1.2 that it is sufficient for Dedekind completeness that every non-empty subset which is bounded from above has a supremum. The first part of the last theorem shows that in a Riesz space L it is sufficient for Dedekind completeness that every non-empty upwards directed subset which is bounded from above has a supremum, and the second part of the same theorem shows that it is even sufficient that every non-empty upwards directed subset in L+ which is bounded from above has a supremum. Exercise 15.12. Show that an ordered vector space has the dominated decomposition property if and only if the space has the Riesz interpolation property (cf. Corollary 15.6 for the definitions of these properties). For any u 2 0 in the space, denote the subset (w : 0 5 w 5 u ) by [O, u ] .

CH. 2,s 151

77

MONOTONE SEQUENCES AND DIRECTED SETS

Show that the dominated decomposition property is equivalent to the property that [O, u ] [O, u ] = [O, u u J

+

+

holds for all u, u 5 0, where the addition sign on the left denotes the algebraic sum (i.e., the set of all w 1+ w2 with w1 E [0, u] and w2 E [0, u ] ) . a

Exercise 15.13. Let f, g and h be real polynomials on the interval ( x :

5 x I; b) with a and b finite, such that h 2f and h 2 g hold pointwise

on the interval and such that h is not identically equal to eitherfor g. Show the existence of a real polynomial k such that k 2f, k 5 g and k 5 h hold pointwise on the interval, and k is not identically equal to h. It follows that the ordered vector space of all real polynomials on the interval, partially ordered pointwise, is not a Riesz space. Hint: Let k = h - c ( h - f ) ( h - g ) with the positive constant c so small that 1- c(h -f)and 1 - c(h -g) are positive on the whole interval.

Exercise 15.14 (F. Riesz, [2]). Let a and b be finite real numbers, and L the ordered vector space of all functions

on ( x :a 5 x 5 b), wherep and q are real polynomials such that q(x) > 0 for all x in the interval, and where the algebraic operations and the partial ordering are pointwise. Show that the positive cone L' is generating (i.e., everyf E L is of the formf = u-u with u, u EL') and L has the Riesz interpolation property, but L is not a Riesz space. Hint: Given u l , u2, ul , u2 E L + such that u1+ u2 = u1 u2 = w, let

+

xi

if

W(X)

xi

=0

for i,j = 1,2. Then w i j ( x ) = ui(x) and w i j ( x ) = uj(x). Note that wij(x) is indeed a function in L', i.e., if a 5 xo 5 b and xo is a root of multiplicity n of the equation W ( X ) = 0, then the polynomial ui(x)ui(x)contains at least the factor ( X - X ~ ) " + ~ Actually, . ui(x)uj(x)has at least the factor ( X - X ~ ) ~ " .It follows by means of the result in the preceding exercise that L is no Riesz space. Exercise 15.15. In Exercise 11.14 it was required to prove that iff+g p + q in the Riesz space L, then

=

78

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH. 2.5 16

SUP (f, PI + inf (999) = f +9. Show that this result can also easily be derived by means of the Riesz interpolation property. 16. Order convergence and relatively uniform convergence

The properties discussed in the preceding section are very much analogous to the convergence properties of monotone sequences of real numbers. An extension to not necessarily monotone sequences follows. Theorem 16.1. It is said that the sequence (f.: n = 1,2, . . .) in the Riesz space L is order convergent to the element f E L whenever there exists a sequence p,, 10 in L such that If -f,,l 6 p,, holdsfor all n. This will be denoted by f.+f. The following properties hold. (i) I f f . +f and& + g , then f = g. (ii) ? f f n l f o r f , tf, thenf. +f. (iii) If f,,? or fJ a n d f . +f, then f.? f or f.1f respectively. (iv) I f f,, +f, g,, + g, and a and b are real numbers, then af,+bg,, + af+ bg, sup (fn,g,,) + sup (f,g ) and inf (f,,, g,,) -, inf (f,9). In particular, iff, +f, thenf: + f + + , ff -i andIf,,l + Ifl. (v) Iff, -,f, thenf,, +f for every subsequence {f,,, : k = 1,2, . . .} such that nl < n, < . . ..

Proof. For (iii), observe that iff.? andf,, +f, then ( f -f,)- is monotonely 4 p,, 10. increasing, whereas on the other hand 0 S ( f - f . ) - S If-f.1 Hence ( f - f . ) - = 0, so f - f . = If-f.1. It follows that f-f,, 10. It is not true in every Riesz space L that iff. + f i n L and a,, -, a in real r,umber space, then a,,& + af. By way of example, let L be the lexicographically ordered plane, f. = f = (1, 1) for all n and a,, = n-'. The sequence {a,,&} is now decreasing, but it is not true that a,,& 10.There is an important class of Riesz spaces, however, for which f,, +f and a,, + a implies that a,f. + af. This is the class of Archimedean Riesz spaces. The Riesz space L is called Archimedean if it is true for every u E L+ that the decreasing sequence (n-lu :n = 1,2, . . .) satisfies n - ' ~10.It follows then immediately that E,,U 4 0 for every sequence (E,, : n = 1, 2, . . .) of positive real numbxs satisfying E,, 10. Properties of Archimedean Riesz spaces will be discussed in the next chapter; here we will restrict ourselves to the observation that the Riesz space L is Archimedean if and only if, given u, v E Lf such that nv 5 u holds for n = 1,2, . . ., we have v = 0. Indeed, if L is Archimedean

CH.2 , s 161

ORDER CONVERGENCE AND RELATIVELY UNIFORM CONVERGENCE

79

and 0 S nu S u holds for n = 1,2, . . ., then

0

sv

n-'u 5.0,

so v = 0. Conversely, suppose that 0 5 nu 6 u (n = 1 , 2 , . . .) implies v = 0 and, for an arbitrary uo EL', consider the sequence (n-lu0 : n = 1,2, . . .). We have to prove that any lower bound w of the sequence satisfies w 5 0, since this will imply immediately that 0 is the infimum of the sequence. For this purpose, observe that v = sup (w, 0) is still a lower bound of the sequence, so we have 0 6 nu 5 uo for n = 1,2, . . .. This implies v = 0, i.e., sup (w, 0) = 0, and hence w 6 0. One final remark about terminology. The nonzero element f in L is called infinitely small with respect to the element g in L whenever nl f I 5 Igl holds for n = 1,2, . . .. The Riesz space L, therefore, is Archimedean if and only if L contains no infinitely small elements. Besides order convergence we introduce another kind of convergence for sequences in a Riesz space. Theorem 16.2. Given the element u 2 0 in the Riesz space L, it is said that the sequence (f,: n = 1,2, . . .) in L converges u-uniformly to the element f E L whenever, for every E > 0, there exists a natural number N, such that If - f , l 5 E M holds for all n 2 N,. It is said that the sequence (f,:n = 1,2, . . .) in L converges relatively uniformly tof wheneverf, converges u-uniformly tof for some u E Li (note that we may have different ufor direrent sequences). Relatively uniform convergence o f A to f will be denoted b y f , +f(ru). In the Soviet literature relatively uniform convergence is called convergence with respect to a regulator. The following properties hold. ( i ) I f L is Archimedean a n d f , converges relatively uniformly to f, t h e n h converges in order to$ Hence, if L is Archimedean andf, converges relatively uniformly to f as well as to g, then f = g. (ii) Iff, +f (ru), gn + g(ru), and a and b are real numbers, then a f , +bg, -,af + bg(ru), SUP (f,,gn) SUP (f, g)(ru), inf (f,, g,) inf (f,g)(ru). +

+

In particular, f, -, I

f I(4.

--f

f ( r u ) implies that f,' +f '(ru), f l -,f -(ru) and

1S.I

Proof. Routine. Relatively uniform convergence is stable, i.e., it has the property that for

80

[CH.2,O 16

ORDERED VECTOR SPACES AND RlESZ SPACES

any sequence f, + 0 (ru) there exists a sequence of real numbers (A, : n = 1,2, . . .) such that 0 6 A, 00 and 1,Z.f.+ 0 (ru). Indeed, given that f, + 0 (ru) there exists a sequence of positive real numbers (E, : n = 1,2, . . .) and an element u E L + such that 8, J. 0 and If,l 5 E,U for all n, and so A, = c,,-* satisfies the abovementioned conditions. Order convergence is not necessarily stable. By way of example, letf,(for n = 1,2, .) be the element of the space 1, with the first n coordinates zero and all other coordinates equal to 1. Thenf, LO, but for any sequence of real numbers 1, satisfying 0 5 A,, t 00 it is impossible that 1,f,, + 0, simply because 1,f,is not bounded from above. The following theorem is, therefore, of some interest.

..

Theorem 16.3. In an Archimedean Riesz space order convergence is stable if and only if order convergence and relatively uniform convergence are equivalent. Proof. In an Archimedean space relatively uniform convergence implies order convergence by the preceding theorem. Assuming stability of order convergence, it will be sufficient, therefore, to prove that order convergence implies relatively uniform convergence. To this end, letf, + 0. Since order convergence is stable by hypothesis, there exists a sequence 0 < A, t 00 such that 1,f, + 0. It follows that 1,IjJ + 0, and hence the sequence

(1,,If,l : n = 1,2,. ..) is bounded, i.e., there exists an element U E L +such that A,,lf,l 5 u holds for all n. Then If,! 5 A;' u holds for all n, and so f. + 0 (ru). Conversely, if order convergence and relatively uniform convergence are equivalent, then order convergence is stable because relatively uniform convergence is so. The subset S of the Riesz space L is called order closed if for every order convergent sequence in S the order limit of the sequence is also a member of S. It is immediately verified that the empty set and the space L itself are order closed and that arbitrary intersections and finite unions of order closed sets are order closed (note that if Si, . . .,S,, are order closed and the sequence (f,, :n = 1,2,. . .) is contained in Si, then at least one of the Si contains an infinite subsequence; then use Theorem 16.1 (v)). Hence, the order closed sets are exactly the closed sets of a certain topology in L, the order topology.

u;

Theorem 16.4. (i) The order topology in L satisfies the TI-separationaxiom, i.e., every subset of L consisting of one point is order closed. (ii) The subset V of L is open in the order topology ifand only i f , for every

CH. 2 , § 161

81

ORDER CONVERGENCE AND RELATIVELY UNIFORM CONVERGENCE

.

sequence (f,: n = 1,2,. .) in L converging in order to a point f havef, E Vfor all but afinite number of thef,.

E

V, we

Proof. (i) Evident. (ii) Assume fist that V c L is open,f, +fand f E V. If it is not true that f, E V holds for all but a finite number of thef. , there exists an infinite subsequence (&, :k = 1,2,. . .) in the complement S of V, and so it follows fromf,, +fand from the fact that Sis closed thatf E S. But this is impossible since we have f E V by hypothesis. Conversely, let V c L have the property that if any sequence ( f . :n = 1,2, ., .) converges in order to an elementf E V, thenf. E Vholds for all but a finite number of thef,. In order to show that the complement S of V is closed, assume that f,E S for n = 1,2, . . . and f , + J Iff E S does not hold, thenfE V and s o f , E V for all sufficiently large n by hypothesis. This is impossible sincef, E S holds for all n. Hence we must have f E S, and this shows that S is closed. The remaining part of this section is rather technical; the reader who is not immediately interested in further properties of the order topology and the corresponding relatively uniform topology is advised to proceed to the next chapter. It follows from the last theorem that if the sequence (fn :n = 1,2, . . .) converges in order tof, thenf, converges to f in the order topology (Le., any neighborhood o f f in the order topology contains all but a finite number of the f,). In view of the definition of the order topology in L, according to which a subset o f L is closed whenever it contains all its sequential limit points with respect to order convergence, we can easily prove now that a mapping cp of L (provided with the order topology) into an arbitrary topological space X is continuous if and only iff. +f always implies that cp(f,) converges to cp(f) in the topology of the space X . Indeed, if the mapping cp is continuous, i f f , +f and if 0 c X is an open neighborhood of cp(f), then the inverse image cp-'(O) is an open neighborhood off, sofn E ~ ~ ' holds ( 0 )for all but a finite number of thef,. But then cp(f,) E 0 holds for all but a finite number of the cp(f,), which shows that cp(f,) converges to cp(f) in the topology of X . Conversely, given thatf, +falways implies that cp(f,) converges to cp(f) in the topology of X, we have to prove that cp is continuous, i.e., we have to prove that the inverse image cp-'(O) of any open set 0 c Xis open in L. To this end, letfbe a point in cp-'(O) and let f , -.f. Then cp(f,) converges in X to cp(f), so all but a finite number of the cp(f,) are in 0. It follows that all but a finite number of thef, are in cp-'(O),

82

ORDERED VECTOR SPACES AND RIESZ SPACES

[CH. 2,8 16

so in view of the above Theorem 16.4 the set rp-'(O) is open. As a particular case we obtain the result that if rp is a mapping of L into itself such thatf, +f always implies rp(f,) + rp(f), i.e., order convergence o f f , to f implies order convergence of rp(f,) to rp(f), then cp is continuous with respect to the order topology. This remark can be used in the proof of the following theorem. Theorem 16.5. (i) The order topology in ihe Riesz space L is translation invariant, i.e., for any fo E L the mappingf +f +f o of L onto itself is a homeomorphism. (ii) For any real a # 0 the mapping f + af of L onto itself is a homeomorphism. (iii) For any$xed g E L the mappings f + sup (5 g ) and f + inf (f,g ) of L into itse,f are continuous. (iv) The mappingsf +f -+ ,f +f - andf -+ If 1 of L into itselfare continuous. Proof. For (iii) and (iv), observe that fn + f implies sup sup (f,g ) , and similarly for the infimum.

(A, g ) +

For any subset S of L , the pseudo order closure S' of S is the set of all f E L such that there exists a sequence (f,:n = 1,2,. . .) in S converging in order tof. It is permitted that f , =f,for m # n. The order closure of S, i.e., the closure of S in the order topology, will be denoted by cl ( S ) . Evidently we have s c S' c cl ( S ) . Taking closures, we obtain cl ( S ) c cl (S') c cl ( S ) , so cl ( S ' ) = cl ( S ) . Hence, replacing S by Sf in the first formula, we obtain also that S' c (S)'c cl ( S f ) = cl (S),

and by induction it follows that

s c S'

c (9)'c S" c

. . . c cl ( S ) .

(1)

Further information is contained in the following theorem. Theorem 16.6. (i) The subset S of L is order closed if and only i f S = S', i.e., S = S' already implies that S = cl (5'). (ii) The subset S of L satisfies S' = cl (5') ifand only i f S' = (S')'. Proof. (i) We have to show that S = S' implies that S is order closed. For this purpose, assume that S = S', f.E S for n = 1,2, . . . and f, +f. Then f E S' by the definition of S', and so f E S on account of S = S'. This shows that S is order closed.

CH.2, $161

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83

(ii) If it is given that S’ = (S’)’,then S’is order closed by part (i) of the present theorem, i.e., S‘ = cl (S’). But cl ( S ’ ) = cl ( S ) holds without extra assumptions, so under the present hypotheses we have S‘ = cl (S). Conversely, given that S‘ = cl ( S ) , it follows immediately from formula (1) above that S’ = (9)’. We will prove now that, for every subset S of L, the pseudo order closure and the order closure of S are equal if and only if the space L has another remarkable property, called the diagonal gap property. The Riesz space L is said to have the diagonalproperty whenever, given any double sequence (f,k: n, k = 1, 2, . . .) in L,any sequence (fn :n = 1 , 2 , . . .) in L and any fo E L such that fnk+ f, for all n (“as k + a”)and f, +f o , there exists for every n an appropriate k = k(n) such that fn,k(,) +f o . The space L is said to have the diagonal gap property if, under the same hypotheses that f& +f, for every n a n d f , +fo, there exists a sequence f,,,k(,,) with n , < < n2 < . ., i.e., an infinite sequence containing at most one member from each sequence (fnk :k = 1,2,. . .), such that

.

(”as a”)* The diagonal property, as well as the notion of relatively uniform convergence and of stability of order convergence, was introduced by L. V. Kantorovitch ([2], 1936). He proved that in a very special class of Archimedean Riesz spaces, called regular spaces, order convergence is stable (and so order convergence and relatively uniform convergence are equivalent) and the diagonal property holds. The diagonal property and diagonal gap property will receive further attention in later chapters; at the present we will only show that the diagonal gap property is related to the order topology as follows. fn,,k(ni) + f O

+

Theorem 16.7. We have cl ( S ) = S’for every subset S of L if and only if L has the diagonal gap property. The diagonal property is, therefore, a

sujicient condition in order that cl ( S ) = S’ shall hold for every subset S of L. Actually, it will be proved later (in Theorem 68.8) that the diagonal gap property and the diagonal property are equivalent.

Proof. Assuming first that L has the diagonal gap property, we have to prove that (S‘)’ = S‘ holds for every subset S of L. Givenf o E (S‘)’,there is a sequence (f, : n = 1,2, . . .) in S‘ such thatf, +fo,and for every n there is a sequence ( f & : k = 1, 2,. . .) in S such that fnk+f, (as k + m). Then, since L has the diagonal gap property, there is a sequence f,,,k(,,) +fo,and sofoE S’.Hence, anyf, E (S‘)‘satisfiesfoE S’. This shows that (S’)’ = S‘. Conversely, assuming that (S’)’ = S’ holds for every S c L, we have to

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ORDERED VECTOR SPACES AND RIESZ SPACES

[CH.2 , s 16

prove that L has the diagonal gap property. Assume, therefore, that f, +f,, as k -P co for n = 1,2,. . and f, +fo. There are three cases to consider, (i) f, # fo for all n = 1,2, . . ., (ii) f,,= fo for only finitely many n, (iii) f,,= fo for infinitely many n. In the first case there are, for each fixed n, only a finite number off& equal tof,. These we may omit, i.e., we assume that f , k # fo holds for all n, k. Now, let S = (fnk :n, k = 1, 2, . . .). Then fo E (S)’= S, so there is a sequence in S converging in order to fo. For every n there are at most finitely many k such thatf, is a member of this sequence (otherwise there would exist a subsequence converging in order to fo as well as tof,, which is impossible on account o f f , # fo).For each n for which there exists at least onefnk in the sequence, let k(n) be the largest k such that fd is in the sequence. There are infinitely many such n, say n , , n,, . . .; we may assume that nl < n, < . ., and we have now that fn,,f(,,,) +fo. In the second case we may omit all f, and all ,Lfor the values of n satisfyingf, = fo . In the third case we restrict ourselves to the values of n satisfying f, = fo . We may assume, therefore, that fnk -P fo as k -P co for all n. Now take aa auxiliary sequence p,, 10 such that all p,, are strictly positive (if p , -1 0 is possible only whenever p,, = 0 for all sufficiently large n, our theorem is trivially true), and set g,,k = fnk +p,, for all n, k. In view of what has been proved for the first case, there exists a sequence gnf,k ( n f ) +fo. It follows easily that f,,, k(,,,) +fo . Combining all results, it has been proved thus that L has the diagonal gap property.

.

.

In is an interesting fact that tverything we have done now for order convergence can be done exactly analogously for relatively uniform convergence. The only additional phenomenon to keep in mind is that in a non-Archimedean space the limit of a relatively uniformly convergent sequence is not necessarily uniquely determined. It can happen, therefore, that f, +f ( r u ) and f, + g(ru) without f and g being equal. We briefly collect the main definitions and theorems. The subset S of the Riesz space L is called (relatively) uniformly closed whenever, for every rdatively uniformly convergent sequence in S, all relativdy uniform limits of the sequence are also members of S. The empty set and L itself are uniformly closed, and arbitrary intersections and finite unions of uniformly closed sets are uniformly closed. Hence, the uniformly closed sets are exactly the closed sets of a certain topology in L, the relatively uniform topology. If L is Archimedean, the relatively uniform topology satisfies the TI-separation axiom, i.e., every set consisting of one point is closed. Conversely, if every set consisting of one point is relatively uniformly closed,

CH. 2.5 161

ORDER CONVERGENCEAND RELATIVELY UNIFORM CONVERGENCE

85

then L is Archimedean. Indeed, if not, there exist strictly positive elements u and v in L such that no u holds for n = 1,2,. . .. It follows that the sequence ( f . :n = 1,2, . . .), with f . = 0 for all n, satisfies If.-vl S n-lu for all n, s o f , + v(ru) as well as (trivially)& + O(ru). This contradicts the hypothesis that the set ( 0 ) is relatively uniformly closed. To summarize, we have that L is Archimedean if and only if the set ( 0 ) is relatively uniformly closed. Assuming again that L is arbitrary (i.e., not necessarily Archimedean), the subset Vof L is open in the relatively uniform topology if and only if, for every sequence (f,:n = 1,2, . . .) in L which converges relatively uniformly to a point f E V,we havef. E V for all but a finite number of thef,. It follows that iff, +f(ru), thenf, converges to f in the relatively uniform topology (i.e., any neighborhood off in the relatively uniform topology contains all but a finite number of thef.). Also, a mapping cp of L (provided with the relatively uniform topology) into a topological space Xis continuous if and only i f f , +f ( r u ) always implies that cp(f,) converges to cp ( f ) in the topology of the space X. In particular, if cp is a mapping of L into itself such that f , +f ( r u ) always implies that cp(f.) + cp(f)(ru), then cp is continuous. It follows that, for f o E L and the real number a # 0 fixed, the mappings f +f +f o and f + af are homeomorphisms of L onto itself, for g E L fixed the mappingsf + sup (f,g ) and f + inf (J, 9 ) of L into itself are continuous, and finally, the mappings f +f +,f +f - and f + If I o f L into itself are continuous. For any subset S of L, the pseudo uniform closure S:, of S is the set of all f E L such that there exists a sequence (f,: n = 1,2,. . .) in S converging relatively uniformly to$ The closure of S in the relatively uniform topology will be denoted by 3. Evidently, we have

s c s;, c (s;,);,

c

. . . c s.

The diagonal property and the diagonal gap property for relatively uniform convergence are defined analogously as for order convergence. The following theorem holds now. Theorem 16.8. (i) The subset S of L is relatively uniformly closed if and only if S = SA,i.e., S = S:, already implies that S = S. (ii) The subset S of L SatisJes S:, = 3 if and only if S:, = (S;,):,,. (iii) In an Archimedean Riesz space L we have = Si, for every subset S of L if and only if L has the diagonal gap property for relatively uniform convergence. The diagonal property for relutively uniform convergence in an Archimedean space L is, therefore, a suficient condition in order that S = S;, shall holdfor every subset S of L . Actually, it will be proved later (in Theorem

s

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ORDERED VECTOR SPACES AND RlESZ SPACES

[CH.2,s 16

72.2) that, for relatively uniform convergence, the diagonal gap property and the diagonal property are equivalent. Proof. In (iii) the hypothesis that L is Archimedean is used in the part of the proof where it is assumed that S:, = (S;,):,, for every S, and where f n k -tf,(ru) and fn +fo(ru) withf, #fo for all n. Just as in Theorem 16.7 we may assume that.f,, # fo holds for all n, k. We choose S = (fnk :n, k = 1, 2, . . .). Then foE (S:,):, = S:,, so there is a sequence in S converging relatively uniformly to f o . For any fixed n, there are at most finitely many k such thatf,, is a member of the sequence. Indeed, if this is not the case, the sequence has a subsequence converging relatively uniformly to f,. The subsequence also converges relatively uniformly tofo . But we havef, # fo, and on the other hand, since L is Archimedean, every uniformly converging sequence in L has a unique limit. Hence, the assumption that for a fixed n the sequence contains infinitely manyf,, leads to a contradiction. Just as before, for each n for which there exists at least in the sequence, let k(n) be the largest k such thatf,, is in the sequence. There are infinitely many such n, say n , , n 2 , . . .; we may assume that n, < n2 < . . ., and we have now that f,,, k ( n i ) +fo. The rest of the proof is exactly as in Theorem 16.7. Let L be Archimedean. Then relatively uniform convergence of a sequencef, tofimplies order convergence off, tof, which implies immediately that every order closed set in L is relatively uniformly closed. Hence, in an Archimedean Riesz space the relatively uniform topology is stronger than the order topology. Iff, + f ( r u ) and f,+fare equivalent, the pseudo uniform closure S:, and the pseudo order closure S' of any set S are identical, and the same holds then for the topological closures S and cl (5'). Indeed, S c S c cl (S) holds always (in an Archimedean space) and, on account of the two modes of convergence (for sequences) being equivalent, the uniformly closed set S is also order closed, so must be equal to cl (S). Thus, in an Archimedean Riesz space where order convergenceand relatively uniform convergence for sequences are equivalent, the order topology and the relatively uniform topology are identical. A necessary and sufficient condition in an Archimedean space for equivalence of the two modes of convergence is that order convergence is stable (cf. Theorem 16.3). We will prove now that the diagonal gap property for order convergence is a sufficient condition.

s

Theorem 16.9. If L is Archimedean and has the diagonal gap property for order convergence, then order convergence in L is stable, and so order convergence and relatively uniform convergence are equivalent, and the order topol-

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ORDER CONVERGENCE AND RELATIVELY UNIFORM CONVERGENCE

ogy and the relatively uniform topology are identical. Furthermore, this topology has the property that the closure of any set consists of the set itself plus all its sequential limit points. Proof. It is sufficient to prove that order convergence is stable under the given conditions. Hence, given a sequence (f n :n = 1,2, . . .) in L such that f,, + 0, we have to prove the existence of real numbers A,, such that 0 A,, t co and A,fn -,0. Sincef,,+ 0 implies the existence of a sequence (p,, : n = 1, 2, . . .) in L such that Ifnl 5 p,, 10, it will be sufficient to find 0 Ant co such that A,,p,, 0. To this end, set P,,k = npk for n, k = 1, 2 , . . .. For n fixed we havep,, J. 0 as k + co;hence, in view of L having the diagonal gap property, there exists a sequence

s

s

with n, < n, <

pnr,k(nr) -, 0

. . ..

Sincep,, is decreasing in k for n fixed, we may assume here that k(ni) increases strictly as n, increases. To summarize, we have +0

with n , < nz < . . ., k(nl) < k(n,) <

. . ..

Now, for every natural number n satisfying k(ni) 5 n < k(ni+,), let A,, t 03. Then

n i , so 0

s

IAnfnl

5

lAnpnl

A,, =

5 niPk(nr),

and so it follows easily that A,, f,,+ 0. This shows that order convergence is stable. Finally, it follows from Theorem 16.7 that the closure of any set consists of the set itself plus all its sequential limit points. In the following list several situations which may occur in an Archimedean Riesz space L are presented. The first column indicates whether order convergence and relatively uniform convergence are equivalent or not in L. In the second column is written “good” or “bad” according as pseudo order closure and order closure are the same for every subset of L or not. Similarly equiv. equiv. non-equiv. non-equiv. non-equiv. non-equiv.

good bad good good bad bad

good bad good bad good bad

Exercises 16.15 and 16.16 Exercise 16.17 impossible by Th. 16.9 impossible by Th. 16.9 Exercise 16.18 Exercise 16.19

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in the third column for the pseudo uniform closure and the uniform closure. The final column shows where an example may be found or why the listed situation is impossible. Exercise 16.10. Show that iff. + f i n the Archimedean Riesz space L and a,, + a in real number space, then ad,, + af in L. Exercise 16.11. We use the notions and notations as introduced in this section. (i) Show that if L is Archimedean, and cl (S) = S' holds for every S c L, then order convergence and relatively uniform convergence in L are equivalent, and the order topology and relatively uniform topology are identical. (ii) Show that if in the Archimedean Riesz space L the order topology and the relatively uniform topology are identical and S:, holds for every S c L , then order convergence and relatively uniform convergence are equivalent. Hint: (i) follows by combining Theorems 16.7 and 16.9. For (ii), note that the hypotheses of (ii) imply those of (i). Indeed, under the hypotheses of (ii) we have S:, = 3 = cl ( S ) and S:, c S' c cl ( S ) for every S c L , so S' = cl (S).

s=

Exercise 16.12. With the same notations, show that if S' = cl ( S ) for every S c L,and the decreasing sequence (p,, :n = 1,2,. . .) in L+ converges to zero in the order topology (i.e., every neighborhood of zero contains all but a finite number of thep,,), thenp,, 10. Show similarly that if S:, = S holds for every S c L,and the decreasing sequence (p,, :n = 1,2,. .) in L' converges to zero in the relatively uniform topology, then p,, + 0 (ru).

.

Exercise 16.13. It was proved in this section that the Riesz spaceL is Archimedean if and only if the set (0) is uniformly closed. It was observed by T. Ito [1] that L is Archimedean if and only if the positive cone L+ is uniformly closed. Present a proof. Hint: Assume L+ uniformly closed. We have to prove that (0)is uniformly closed. If not, there exists an element f # 0 in {0}iu,i.e., there exists u E Lf and a sequence of numbers E,, 4 0 such that If1 S E,,u for all n. Sincef # 0, one at least off and f - is not zero, say f+ # 0. We have['fI 5 E,,u for all n, so f E (0) and also -f + E (0) c = Lf,which is impossible. Conversely, if L is Archimedean, u,, E Lf for n = 1,2, . . . and u,, +f ( r u ) , then u,, + A so there exists a sequencep,, 10 in Lf such that u,, -f =< p,, for all n. Then 0 j u,, j f+p, for all n, sof 2 -p,, 0. This shows thatfe L'. +

+

v)

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Exercise 16.14. Show that in an Archimedean Riesz space L the following proputies are equivalent. (i) L has the diagonal property for order convergence. (ii) Order convergence in L is stable and, for any sequence (u,, : n = 1,2, .) in L + , there exists a sequence (A,, : n = 1,2, . .) of strictly positive numbers such that (A,,u, :n = 1,2,. .) is bounded from above. As observed earlier in this section, L. V. Kantorovitch [2] has introduced a special class of Archimedean Riesz spaces, called regular spaces, and he proved that these spaces have the diagonal property. He proved also that regular spaces satisfy condition (ii) of the present exercise. Hint: If (i) holds, then order convergence is stable by Theorem 16.9. Given the sequence (u,, :n = 1, 2,. . .) in L + , set fnk = k-'u,, for n, k = 1,2,. .. Then fnk J. 0 as k -+ co for every n, and so fn,&(,,) + 0 for appropriate k(n). Hence, setting A,, = {k(rz)]-', we have Anu, 0, which implies that (A,,u, :n = 1,2,. .) is bounded. Conversely, assuming (ii) to hold, we have to prove that if f n k + f , as k -+ co and f, +fo , then f,,k(") -P fo for appropriate k(n). Since order convergence and relatively uniform convergence are equivalent in view of the stability of order convergence, there exists for each n = 1,2,. . an element u,, E Lf such that f& + f , holds u,,-uniformly. Then there exist numbers A,, > 0 such that (A,,#,,:7z = 1,2, .) is bounded, say AnunS u. It follows that fnk -P f.holds u-uniformly for every n, so there exists for every n a natural number k(n) such that

..

.

.

.

-+

.

.

..

Ifn,k(n)-fnl

Finally, it follows fromf,,,(,,)-f,

+0

5 n-'u. and f , - f o

+0

that f , , & ( , , ) - f O

-+

0.

Exercise 16.15. Let L be the Riesz space of all real functions on a finite point set X,the ordering is pointwise. Show that order convergence, relatively uniform convergence and uniform convergence are equivalent, and L has the diagonal property with respect to this mode of convergence. It will be proved later (cf. section 26) that every finite dimensional Archimedean Riesz space is "isomorphic" to a space of the kind considered in the present exercise; hence, the same statements about convergence hold in every finite dimensional Archimedean Riesz space. Exercise 16.16. Let (s) be the Riesz space of all real sequences (cf. Example 11.2 (iii)), the ordering is coordinatewise. Show that order convergence and relatively uniform convergence are equivalent, and (s) has the diagonal prop-

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ORDERED VECTOR SPACES A N D RIESZ SPACES

[CH. 2 , s 16

erty with respect to this mode of convergence. Hence, pseudo closure and closure of a set in the corresponding topology are the same. Hint: In view of Theorem 16.9 it is sufficient to prove that (s) has the diagonal property for order convergence. Hence, by Exercise 16.14, it is sufficient to prove that order convergence is stable and that for any sequence (u,,:n = 1, 2, . . .)in (s)' there exists a sequence (A, :n = 1, 2, . . .) of strictly positive numbers such that (A,,u, :n = 1,2,. . .) is bounded. For the stability of order convergence it is sufficient to prove that, given f, 10, there exists u E (s)' such that f, 4 0 holds u-uniformly. The sequence u = {u(l), u(2), . . .}, defined by u(n) = nfl(n), is easily seen to satisfy this condition. Given the sequence (u, :n = 1,2,. . .) in (s)', determine A1 > 0 such that Alul(l) 6 1, then Az > 0 such that Azuz(l) S 1 and Azuz(2) -< 1, and so on. Hence, A, > 0 satisfies A,,u,(k) S 1 for k = 1,2,. . ., n. The desired majorant v of (A,,u, :n = 1,2, . . .) is found now by defining

.. u(1) = 1, v(2) = max (1, A1u1(2)}, u(3) = max (~y~lul(3),Azuz(3)}y..

Exercise 16.17. Let L be the Riesz space of all real sequencesf = {f(l), f(2), . . .} with only finitely many nonzero coordinates; the ordering is coordinatewise. Show that order convergence and relatively uniform convergence in L are equivalent, and show also that L fails to have the diagonal gap property for order convergence as well as for relatively uniform convergence. Hint: The proof that order convergence in L is stable is similar to the corresponding proof in Exercise 16.15. For the second statement it is sufficient to prove that L does not have the diagonal gap property for relatively uniform convergence. For n = 1,2, . . ., let u,, EL' have its first n coordinates equal to 1 and all other coordinates zero, and letf,, = k-lu,, for n, k = 1, 2, . . .. Then fnk 4 0 (ru) as k co for every n. Any diagonal gap sequence taken from (f,,: n, k = 1,2,. . .), however, is not bounded from above. Exercise 16.18. Let L be the Riesz space C([O,11) of all real continuous functions on the interval [0, 11, where [0, 11 is provided with its ordinary topology. Show that order convergence in L is not stable and L does not have the diagonal gap property for order convergence. In other words, order convergence and relatively uniform convergence are not equivalent,and there exists a subset of L for which the pseudo order closure is not the same as the order closure. Show that relatively uniform convergence of a sequence is the same as the familiar uniform convergence of the sequence, the relatively uniform closure of a set is the familiar uniform closure, and the relatively

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uniform topology is the norm topology with respect to the uniform norm. It follows immediately that the pseudo closure and the closure of any set with respect to this topology are the same. Derive this also by proving that L has the diagonal property with respect to uniform convergence. Show that the same facts hold in the real sequence space I , . Hint: For the proof that order convergence is not stable, consider for example the sequence (h:n = 2 , 3 , . . .), where f,,(x) = 1 on [0, n-'I, f n ( x ) = 0 on [2n-', 11 and linear on [n-', 2n-']. Thenjf, 4 0, but there exists no sequence 0 6 A,, t co such that A& -+ 0. For the proof that L does not have the diagonal gap property for order convergence, take the same sequence (fn : n = 1,2,. . .) and set fnk = nfk. Then f n k 10 as k -+ co for every n, but there is no sequencef,,,,k(nr)-,0.

Exercise 16.19. Let L b: the Riesz space of all real continuous functions with compact carrier on the real line, where the real line is provided with its ordinary topology. Show that order convergence and relatively uniform convergence are not equivalent, and show also that L fails to have the diagonal gap property for order convergence as well as for relatively uniform convergence. Hint: The proof that order convergence is not stable is similar to the proof in the preceding exercise. The proof that L does not have the diagonal gap property is (for order convergence as well as for relatively uniform convergence) similar to the proof in Exercise 16.17.