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# Setting f cg dgh we have f a f p a f q b step 2 a is

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Setting f = cg + dgh we have f ∈ A , f ( p ) = a , f ( q ) = b . Step 2. A is an algebra. Same proof as in Stone Weierstrass. Left as a new exercise. I also leave as an exercise (trivial one) that all functions of ¯ A vanish on Z ; it satisfies all the properties of A . Step 3. If f ∈ A , then | f | ∈ ¯ A . Let f ∈ A and let ² > 0 be given. Consider the function φ : [ -k f k , k f k ] defined by φ ( x ) = | x | . This function is continuous and satisfies φ (0) = 0; by Theorem 1 there exists a polynomial without constant term p such that | p ( x ) - | x || < ² for all x [ -k f k , k f k ]. Because p has no constant term, p f ∈ A and we have | p ( f ( p )) - | f ( p ) || < ² for all p X . Step 4. If f, g ∈ A , then max( f, g ), min( f, g ) ¯ A . This is again the same as before: max( f, g ) = 1 2 ( f + g + | f - g | ) , min( f, g ) = 1 2 ( f + g - | f - g | ) . Step 5. Assume now given f C ( X ), f | Z = 0, and ² > 0. To complete the proof we must show there exists g ∈ A , or in ¯ A , such that | f ( p ) - g ( p ) | < ² for all p X . Fix for a moment p X . For every q X let h pq ∈ A be such that h pq ( p ) = f ( p ), h pq ( q ) = f ( q ). This can be done by Step 1: If both p, q Z , any element of A will do. If only one of p, q Z , one only needs to match the value of f at the point not in Z , the value at the point in Z , which is 0, being matched automatically. If neither p, q is in Z , Step 1 applies directly. By continuity, there is an open set V pq of X such that q V pq and h pq ( r ) > f ( r ) - ² for all r V pq . The family { V pq } q X is an open covering of the compact space X , thus there exist q 1 , . . . , q n X such that X = S n j =1 V pq j . By Step 4 (and induction), h p = max( h pq 1 , . . . , h pq n ) is in ¯ A and clearly satisfies: h p ( p ) = f ( p ), h p ( r ) > f ( r ) - ² for all r X . Step 6. Time to move p . For each p X we have h p ¯ sA as above. By continuity, for every p X there is an open set W p such that p W p and h p ( r ) < f ( r ) + ² for all r W p . As before we use compactness to get W p 1 , . . . , W p m covering X . Then g = min( h p 1 , . . . , h p m ) is in ¯ A by Step 4 applied to the closed algebra ¯ A rather than to A ; this gives that g is in the closure of the closure of A . But the closure of a closure is still just the closure. (If you close it once, and close it well, you don’t need to close it anymore.) It is clear that f ( r ) - ² < g ( r ) < f ( r ) + ² for all r X and we are done. 5. Let f ( x ) = cos x . Provide an explanation, based on fixed point theorems seen in class (or, equivalently, the proposition on page 172 of Rosenlicht) of why starting with an arbitrary x 0 R , the sequence defined by x n +1 = f ( x n ) will converge to the unique fixed point of cos. Notice that f 0 ( x ) = sin x can equal 1. Solution. Wherever x 0 R might be, cosx 0 [ - 1 , 1] and every further application of the cosine function results in values in that interval. That is, we may start with x 1 = cos x 0 instead of x 0 and then we consider cos as a map from the complete metric space [ - 1 , 1] into itself. Now [ - 1 , 1] ( - π/ 2 , π/ 2) and sine increases from sin( - 1) = - sin 1 to sin 1 in that interval, thus for all x [ - 1 , 1] | f 0 ( x ) | = | sin x | ≤ sin 1 < 1 .

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