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Unformatted text preview: Solution. I made a mistake in the statement of the exercise. The statement and the properties Z is supposed to have are contradictory except in one case: Z is a singleton set (consists of a single point). In fact, if Z has more points, then A cannot possibly separate points. If one assumes that Z is a singleton set the exercise has one small trick but is otherwise not terribly hard. Just a little bit hard, perhaps. On the other hand, if one ignores the statement and assumes that A satisfies (a)(d), things get a bit more involved. I will assume that properties (a)(d) hold, so that A does not separate points of Z if Z has more than one point, but separates points otherwise. However, I will give full 2 credit for proving the result assuming Z is a singleton set. Though it isn’t quite necessary, I will prove the singleton case first. As just said, we will prove the result assuming first that Z = { p } is a singleton set. In this case consider the set ˜ A = { c + f : c ∈ R ,f ∈ A} . It is immediate that ˜ A is an algebra and, since (for example) the constant function 1 is in it, it satisfies the nonvanishing property. It also separates points. By the Stone Weierstrass Theorem it is dense in C ( X ); in other words, given ² > 0, f ∈ C ( X ), there exists c ∈ R , g ∈ A such that  f ( p ) ( c + g ( p ))  < ²/ 2 for all p ∈ X . If f ( p ) = 0, then letting p = p in  f ( p ) ( c + g ( p ))  < ²/ 2 we get  c  < ²/ 2. Thus  f ( p ) g ( p )  ≤  f ( p ) ( c + g ( p ))  +  c  < ² 2 + ² 2 = ² for all p ∈ X . The case Z = { p } is done. For the general case, we have to be a bit more devious. We can start with a result which can either be seen as a consequence of what we just did, or proved using the same idea we used here from Weierstrass’ Approximation Theorem. Theorem 1 (Weierstrass2) Let f ∈ C ([ a,b ]) , where∞ < a < < b < ∞ and assume f (0) = 0 . For every ² > there exists a polynomial p with 0 constant term such that  f ( x ) p ( x )  < ² for all x ∈ [ a,b ] . Proof. Proof #1. Let ² > 0 be given. By the Theorem of Weierstrass there exists a polynomial q , q ( x ) = a + a 1 x + ··· + a n x n , such that  f ( x ) q ( x )  < ²/ 2 for all x ∈ [ a,b ]. Evaluating at 0 we get  a  < ²/ 2. Letting p ( x ) = a 1 x + ··· + a n x n , we have that  f ( x ) p ( x )  =  f ( x ) p ( x ) a + a  =  f ( x ) q ( x ) + a  ≤  f ( x ) q ( x )  +  a  < ² 2 + ² 2 = ². Proof #2. The set P = { p : p is a polynomial and p (0) = 0 } satisfies properties (a)(d) above with Z = { } . The result is now immediate from the Z = singleton set case. Assume now A is a subalgebra of C ( X ) (Properties (a), (b) above hold), the set Z = ∩{ f 1 ( { } ) : f ∈ A} 6 = ∅ , and p ∈ X \ Z , q ∈ X , p 6 = q implies there exists f ∈ A such that f ( p ) 6 = f ( q ). (Properties (c), (d)). Incidentally, if p / ∈ Z , there is f ∈ sA such that f ( p ) 6 = 0; this f automatically separates p from all points of...
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 Spring '11
 Speinklo
 Metric space, dx, Uniform convergence, Hpq, Stone Weierstrass theorem

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