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Unformatted text preview: Math 202A Homework 9 Roman Vaisberg Problem 1. (a) Let A be an algebra of continuous realvalued 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 StoneWeierstrass 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 : [0 , 1] [0 , 1] satisfies 00 < , > , (0) = 0 , and (1) = 1 , then for each > , n ( x ) 1 as n uniformly on [ , 1] . Proof of Lemma : First we observe that ( x ) > x for x (0 , 1). Otherwise if for some x , ( x ) x we would have R x ( t ) dt = ( x ) x implying that ( x ) x < 1 since is a decreasing function, which would imply (1) = R x ( t ) dt + R 1 x ( t ) dt < x + (1 x ) = 1 and thus contradicting the assumption that (1) = 1. Second we observe that if x < y then ( x ) < ( y ): ( y ) = R y x ( t ) dt + ( x ) > ( x ) since > 0. Now let (0 , 1). To show n 1 it will suffice to show n ( ) 1 by the second observation. Clearly the sequence { n ( ) } has a limit, L , since it is increasing and bounded above. And since itself is a continuous function ( { n ( ) } ) ( L ). But applying to each term of the sequence does not change the sequence except for the first term and thus the limit will remain unchanged implying ( L ) = L . Since L cannot be 0 as > 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 [ , 1] for some > 0....
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 Spring '10
 tao/analysis
 Algebra

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