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# Define φ e r according to the formula φ u z 1 u x 2

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Define φ : E R according to the formula φ ( u ) = Z 1 0 ( u ( x ) 2 - u ( x ) ) dx. Prove there exists a function u E at which φ ( u ) attains an absolute maximum. Solution. It is easy to see that E is closed and bounded in the norm metric. It is also easy (immediate) that it is equicontinuous. Thus it is compact. It is also quite easy to prove that φ : E R is continuous. Thus it attains a maximum value. This was, I assume, the proof that Pugh intended. However, a faster and quite elementary proof was discovered by somebody and used by several people; though in every case with a bit of something not quite right suggesting that none of the presenters was the real author. Or was she/he?

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Here is the full proof; anything added or subtracted cost points. Let u E . Then | u ( x ) | = | u ( x ) - u (0) | ≤ | x | = x for 0 x 1. Thus φ ( u ) Z 1 0 fl fl u ( x ) 2 - u ( x ) fl fl dx Z 1 0 ( u ( x ) 2 + | u ( x ) | ) dx Z 1 0 ( x 2 + x ) dx = 5 6 . Thus φ ( u ) 5 / 6 for all u E . Clearly, u 0 ( x ) = - x is in E and since φ ( u 0 ) = Z 1 0 ( x 2 + x ) dx = 5 6 , φ attains its maximum at u 0 E . 3. Suppose f : [ a, b ] R is continuous and satisfies Z b a x n f ( x ) dx = 0 for n = 0 , 1 , 2 , . . . . Prove: f ( x ) = 0 for all x [ a, b ]. Solution. Let ² > 0 be given. By Weierstrass’ Theorem, there exists a polynomial p ( x ) = n k =0 a k x k such that | p ( x ) - f ( x ) | < ² ( b - a ) M for all x [ a, b ], where M = k f k + 1, so M > 0 and | f ( x ) | ≤ M for all x [ a, b ]. Now Z b a p ( x ) f ( x ) dx = n X k =0 a k Z b a x k f ( x ) dx = 0 , thus Z b a f ( x ) 2 dx = Z b a f ( x ) 2 dx - Z b a p ( x ) f ( x ) dx = Z b a ( f ( x ) - p ( x )) f ( x ) dx Z b a | f ( x ) - p ( x ) | | f ( x ) | dx ( b - a ) · ² ( b - a ) M M = ². Thus R b a f ( x ) 2 dx < ² ; since ² > 0 is arbitrary this implies R b a f ( x ) 2 dx = 0. Since f , hence also f 2 is continuous, and f 2 0, we conclude that f 2 , hence also f , is 0. 4. Let X be a compact metric space and let A be a subalgebra of C ( X ) that separates points but does not satisfy the non-vanishing property. That is, assume A ⊂ C ( X ) satisfies: (a) f, g ∈ A implies f + g, fg ∈ A . (b) f ∈ A , c R implies cf ∈ A . (c) The set Z = { p X : f ( p ) = 0 f ∈ A} is not empty. (d) If p, q X , p / Z , there exists f ∈ A such that f ( p ) 6 = f ( q ). Prove: If f C ( K ) is such that f ( p ) = 0 for all p Z , then for every ² > 0 there is g ∈ A such that | f ( p ) - g ( p ) | < ² for all p X . 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.

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