Representations of positive projections 1 Introduction - Mathematics ...

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$MATH 5b; Solutions to Homework Set 1 January 2009 1. As K is a ...$

(i). For all a b 2 L we havea + b 2(a ^ b) and so (a + b) + which implies that a + dd + dd \ b (a + b) + dd . Applying this to a = w ; f and b = w ; g we nd that ( w ; f) + dd \ ( w ; g) + dd from which (i) follows. (ii). It follows from that (( + ) w ; (f + g)) + (( + ) w ; (f + g)) + dd and this implies that (( + ) w ; (f + g)) + dd \ ( w ; f) + d 2(a + ^ b + ), (( + ) w ; (f + g)) + dd , ( w ; f) + +( w ; g) + ( w ; f) + dd + ( w ; g) + dd f +g Since p + ^ ; w ; pf is the component ofw in the band (( + ) w ; (f + g)) + dd \ ( w ; f) + d , f +g it is now clear that p + ^ ; w ; pf p g . ( w ; g) + dd . Proposition 2.3 (H. Nakano, [14]) If fg 2 L and 2 R, then p f +g =sup p f ^ p g : 2 Q + . Proof. Since the set Q of rationals is countable, the supremum on the right hand side, which we denote by p, exists in L. From (ii) of Lemma 2.2 it is clear that p p f +g . To prove the reverse inequality, take 2 Q such that and take k 2 N. De ne n =2 ;kn for all n 2 Z. Since pf # 0as n n !;1and pf " w as n !1,we have n w = _ ; _ ; f f p ; pf = w ; p n n n2Z n+1 6 n2Z ^ pf n+1
in C (w) and so it follows that f +g p ;2 ;k f +g = p ;2 ;k ^ w = _ n2Z f +g p ;2 ;k ^ ; w ; p f n ^ pf n+1 . Since ; ; 2 ;k ; n = ; n+1, Lemma 2.2 (ii) implies that for all n 2 Z. Hence, f +g p ;2 ;k ^ ; w ; p f n f +g p ;2 ;k _ n2Z p g ; n+1 p f ^ pg n+1 ; n+1 as n+1 ; n+1 2 Q and n+1 +( ; n+1) = for all n 2 Z. Since f +g p ;2 ;k " pf +g as k ! 1, it follows that pf +g p. Finally, take a sequence f ng 1 n=1 in Q such that f +g f +g n " . Then p p for all n and p n n " pf +g , so f +g p p. We may conclude that p f +g = p, by which the proof is complete. Proposition 2.4 Let L be a Dedekind -complete Riesz space with weak order unit 0 w 2 L and suppose that E is a Boolean -subalgebra ofC (w). Then L (E), as introduced in De nition 2.1, is a -complete Riesz subspace of L and CL(E) (w) = E. In particular, L (E) is Dedekind -complete and -regularly embedded in L. Proof. First we observe that L (E) isalso given by L (E) = f 2 L : q f 2E for all 2 R . Indeed, this follows immediately from the fact that if n < , n " , then q f n " pf and if n > , n # , then p f n # qf . Since p ;f = w ; q f ; for all 2 R, it is now clear that f 2 L (E) if and only if ;f 2 L (E). Moreover, if 0 < 2 R then p f = p f = and so f 2 L (E) implies that f 2 L (E). This shows that f 2 L (E) for all f 2 L (E) and all 2 R. Furthermore, it follows from Proposition 2.3 that f + g 2 L (E) for all fg 2 L (E), so L (E) is a linear subspace of L. Since p f_g = p f ^ p g for all 2 R, we see that fg 2 L (E) implies that f _ g, hence L (E) is a Riesz subspace of L. Now assume that fn 2 L (E) for n =1 2::: and that f 2 L such that fn # f in L. Then p fn " p f for all 2 R, which implies that f 2 L (E). Therefore, L (E) is a -complete Riesz subspace of L. Finally, it follows immediately from the de nition of L (E) that a component p 2 C (w) belongs to L (E) if and only if p 2E. The proof is complete. We will need some further properties of the space L (E). In particular, we will give an alternative description of the space L (E). For this purpose we give the following de nition. 7 p,
Page 1 and 2: Representations of positive project
Page 3 and 4: for all f 2 bL. We stress the fact
Page 5: that K itself is Dedekind ( -) comp
Page 9 and 10: Pe (e) =e and Pe (f) =0whenever f 2
Page 11 and 12: (i). L is locally order separable (
Page 13 and 14: this example also shows that the an
Page 15 and 16: 3 Representations Many of the famil
Page 17 and 18: Lemma 3.3 Given a Dedekind -complet
Page 19 and 20: observations that Ker ( ) = bL \ M(
Page 21 and 22: It is readily veri ed that the exis
Page 23 and 24: of 1 by CA = CA (1). Note that CA i
Page 25 and 26: for all c 2 C, bytheaveraging prope
Page 27 and 28: and, using that Qf is 1-periodic, Z
Page 29 and 30: As we have seen in Example 2.12, a
Page 31 and 32: Proposition 4.12 Let A beaDedekind
Page 33 and 34: Then there exists a representation
Page 35 and 36: can adjust the values of the functi
Page 37 and 38: so which implies that u P (p) ; 1 1
Page 39 and 40: Proof. First observe that'jK 2 K s
Page 41 and 42: Proof. Let P : Z(L) ! Z(L) be the c
Page 43 and 44: (ii). If e 1e 2 2F such that e 1 ^
Page 45 and 46: Boolean subalgebra such that CAi =
Page 47 and 48: theorem that B 1 K 1 and 1 (g) = (g
Page 49 and 50: A is a Dedekind complete f-algebra
Page 51 and 52: and (34) now follows by induction o
Page 53 and 54: and since P is strictly positive it
Page 55 and 56: Lemma 7.7 If 2 D and 0 < 1, then [q
Page 57 and 58:
Since q( ) 2 CB for all 0 1, it is
Page 59 and 60:
(ii). there exists a strictly posit
Page 61 and 62:
y Proposition 4.2 there exists a st
Page 63 and 64:
S (CB [E ). Moreover, there exists
Page 65 and 66:
Proposition 8.6 Assume that A, B an
Page 67 and 68:
Since B is Dedekind complete, it is
Page 69 and 70:
E is a Dedekind complete f-algebra
Page 71 and 72:
Then there exists a - nite measure
Page 73:
[10] Dorothy Maharam, The represent
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Representations of positive projections 1 Introduction - Mathematics ...

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