By Kallenrode

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Pj(Rj))d. pj(Qlj)Xj is dense in 'lij. pj(Qlj)Xj is dense in 'li j , for j = 1,2, it follows that there is a unitary operator U from 'iiI onto 'li2 such that (1) (A E QlI). pl(B)XI. Since CPI(QldxI is dense in 'iiI, we have UcpI(A) = cp2(cp(A))U, and (A E Qlt). 2h)U- I , and by taking strong-operator 318 NORMAL STATES It follows from (2) and (3) that the mapping R -+

Let F be the family of finite subsets of A partially ordered by inclusion. With F in F, define Elf to be ~aEll' Ea. Then Ell' E R. We note 'that Ell' is a cyclic projection in R. Indeed, with Xa a preassigned unit vector in E a(1f), let Xll' be L:aEll' Xa. Suppose a vector Ya is given for each a in F. Let Y be L:aElrYa' Choose T~ (effectively in B(Ea(1f))) such that EaT~Ea = T~, T~xa = Ya, and T~ E B(1f). Then T~ E R'. Let T' be L:aElrT~. Then T' E R' and T'xll' = y, whence Ell' is cyclic in R.

We have n Axo =L j=l n AEjl ej =L n j=l Ejl AE~l ej n = LEjlYj = LEjlE~jY j=l j=l n = LEjjY = y. j=l Thus {Axo : A E Qt} = 1t. 39. Let G be a (discrete) group and F be the set of elements in G whose conjugacy classes are finite. Show that the center C of CG (and RG) is precisely the set of elements Lx in CG such that x is constant on the conjugacy class (go) for each go in F and x(g) = 0 if 9 ri- F. Solution. 5, if Lx is in C, x is constant on each conjugacy class. Since x E 12(G), x vanishes on each infinite conjugacy class.

### A new approach to linear filtering and prediction problems by Kallenrode

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