hmwk9sol_a

hmwk9sol_a - Math 202A Homework 9 Roman Vaisberg Problem...

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Unformatted text preview: Math 202A Homework 9 Roman Vaisberg Problem 1. (a) Let A be an algebra of continuous real-valued function on a compact space X and assume that A separates the points of X . Prove that either A = C ( X ) or there is a point p ∈ X such that f ( p ) = 0 for all f ∈ A and A = { f ∈ C ( X ) | f ( p ) = 0 } . Proof. If 1 ∈ A then A = C ( X ): Let g ∈ C ( X ), > 0 and consider the algebra B generated by A and the constant function 1. Then B = C ( X ) by the Stone-Weierstrass theorem. Thus there exists h ∈ B such that | h- g | < 2 . But then h = f 1 + c for some f ∈ A and we can let f 2 ∈ A be such that | f 2- c | < 2 . Then f 1 + f 2 ∈ A and | f 1 + f 2- g | ≤ | f 1 + c- g | + | f 2- c | < . Now suppose 1 / ∈ A and that for each x ∈ X there is an f x ∈ A such that f x ( x ) 6 = 0. For each x consider the set O x defined by O x = f- 1 x ((-∞ , 0)) if f x ( x ) < f- 1 x ((0 , + ∞ )) if f x ( x ) > We know each O x is open since each f x is continuous. Furthermore { O x } covers X and since X is compact there exists a finite subcollection { O x 1 , ··· ,O x N } which also covers X . Let g = f 2 x 1 + ··· + f 2 x N . Then g > 0 on X : let x ∈ X , then for some j , x ∈ O x j implying f 2 x j ( x ) > 0 implying g ( x ) > 0. Furthermore it is clear that g ∈ A . Now we will show that the existence of such a g implies that 1 ∈ A . Lemma 1 If Λ : [0 , 1] → [0 , 1] satisfies Λ 00 < , Λ > , Λ(0) = 0 , and Λ(1) = 1 , then for each δ > , Λ n ( x ) → 1 as n → ∞ uniformly on [ δ, 1] . Proof of Lemma : First we observe that Λ( x ) > x for x ∈ (0 , 1). Otherwise if for some x , Λ( x ) ≤ x we would have R x Λ ( t ) dt = Λ( x ) ≤ x implying that Λ ( x ) ≤ x < 1 since Λ is a decreasing function, which would imply Λ(1) = R x Λ ( t ) dt + R 1 x Λ ( t ) dt < x + (1- x ) = 1 and thus contradicting the assumption that Λ(1) = 1. Second we observe that if x < y then Λ( x ) < Λ( y ): Λ( y ) = R y x Λ ( t ) dt + Λ( x ) > Λ( x ) since Λ > 0. Now let δ ∈ (0 , 1). To show Λ n → 1 it will suffice to show Λ n ( δ ) → 1 by the second observation. Clearly the sequence { Λ n ( δ ) } has a limit, L , since it is increasing and bounded above. And since Λ itself is a continuous function Λ( { Λ n ( δ ) } ) → Λ( L ). But applying Λ to each term of the sequence does not change the sequence except for the first term and thus the limit will remain unchanged implying Λ( L ) = L . Since L cannot be 0 as δ > 0 we must have that L = 1 thus completing the proof of the lemma. Going back to the original proof we see that since X is compact g assumes its maximum and minimum on X and thus we may assume (after possible rescaling) that g takes on values in [ δ, 1] for some δ > 0....
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hmwk9sol_a - Math 202A Homework 9 Roman Vaisberg Problem...

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