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Unformatted text preview: MATH 425b ASSIGNMENT 3 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 7 (20) Let > 0. There exists a polynomial with  P f  < (sup norm), say P ( x ) = ∑ N n =0 c n x n . f is bounded since it is continuous on the compact set [0 , 1], so there exists M such that  f ( x )  ≤ M for all x . Therefore Z 1 f ( x ) P ( x ) dx = N X n =0 c n Z 1 f ( x ) x n dx = 0 and ≤ Z 1 f ( x ) 2 dx = Z 1 f ( x ) 2 dx Z 1 f ( x ) P ( x ) dx = Z 1 f ( x )( f ( x ) P ( x )) dx ≤ Z 1  f ( x )  ( f ( x ) P ( x )  dx ≤ Z 1 M dx = M . Since is arbitrary, this shows R 1 f ( x ) 2 dx = 0. By Exercise 2 of chapter 6, this means f ( x ) 2 = 0 for all x , so f ( x ) = 0 for all x . (21) The constant function f ( e iθ ) ≡ 1 for all θ is in A , and vanishes nowhere, so A vanishes at no point of K . The identity function f ( e iθ ) = e iθ is in A , and is onetoone, so A separates points. To prove Rudin’s hint, for any function f ( e iθ ) = ∑ N n =0 c n e inθ in A we have Z 2 π f ( e iθ ) e iθ dθ = N X n =0 c n Z 2 π e i ( n +1) θ dθ = 0 . (1) For f ∈ A there exists a sequence { f n } ⊂ A with f n → f uniformly. Hence applying ( ?? ) 1 to f n , Z 2 π f ( e iθ ) e iθ dθ = Z 2 π ( f ( e iθ ) f n ( e iθ )) e iθ dθ ≤ Z 2 π  f ( e iθ ) f n ( e iθ )   e iθ  dθ ≤ 2 π  f n  → 0 as n → ∞ , (2) so we must have R 2 π f ( e iθ ) e iθ dθ = 0, for all f ∈ A . But for the particular choice....
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This note was uploaded on 06/23/2008 for the course MATH 425B taught by Professor Alexander during the Fall '07 term at USC.
 Fall '07
 Alexander
 Math

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