 By R. B. Burckel

Ebook through Burckel, R. B.

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Extra resources for Characterization of C(x) among its Subalgebras

Example text

Whence = C lR (X) • Now, as observed before (III, p. lull = inft \1fIlA : f then n 1\ E A, Re f = u} u E Re A = C lR (X) is a complete (quotient) norm in C lR(X) which dominates the uniform norm. By the Open Map Theorem there is a constant II II such that K s: K II IIx. to'( IN,C(X)). teo ( IN,C(X)). ( 4) Now if IN X X IN such that nUn + i v n nA ~ (K + l) M and {un + iVn } E A. teo ( IN,C(X)) and is naturally identified with with a subalgebra thereof and (4) translates A into Re A = CJR(lN X X).

Ql 9:C lI~cll"" For E A <-nd-- , II hx /I "" s 1. I "g" .. C lR(X • =c cp( K) and ]2 E n ) hx K): say, without loss "gil .. JI (K) = 0, 1 - -1\1. take a sufficiently high power of There is a neighborhood Nx of x in which the middle inequality above persists: sup{ Ihx(Y) I: yEN } x and set h = h x3 ·hx 4 • ••• ·hxn h(K) For the function g' (x 2 ) = = 1, g' <-nd--. "g" .. , < 1. 2 Let lying in X be a compact Hausdorff space, A C(X) which separates the points of the constants. If Re A separates the points of ~: II Let f X and contains is uniformly closed in Cm(X), then X.

F is complete and E C F E C F. F If then E M such that is bounded in F so and evidently Now suppose E is complete and show first that for some (2) is dense in E = {1,2, •••• } Then By hypothesis there is a constant Therefore any bounded sequence in E are (real or complex) IN,E), the bounded functions from IN normed as usual, and in addition E,F x E F & E is dense in F. ly E E, lIyl/E < r Indeed, if this fails for every & II x -yllF < 1/2. r > 0, we can select a = F. n) Now (x n } E F [Y n } E E with (5) But for every E and ly n } E E ~ ~ for each n = 1,2, ••• means that for some constant n = 1,2, ••• Finally we use to show that each Y (2 ) x EF to prove that with .