sf - Math 8601. December 18, 2010. Final Exam. Problems and...

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Unformatted text preview: Math 8601. December 18, 2010. Final Exam. Problems and Solutions. #1. Let A be an arbitrary set, and for each A , let an open ball B R n be defined. Show that there is a finite or countable subset A A , such that [ A B = [ A B . Proof. The open set := [ A B = [ j =1 K j , where K j := { x : dist( x, ) 1 /j, | x | j } . Each set K j is bounded and closed, so it is compact, and it is covered by the family of open sets B , A . By the Heine-Borel property (Theorem 0.25(c)), there is a finite subset A j A such that K j is covered by B , A j . Then the set A := S j =1 A j is finite or countable, and = [ j =1 K j [ j =1 [ A j B = [ A B , and the desired property follows. #2. Show that for an arbitrary interval I := [ a,b ] (0 , ), ln x- (ln x ) I I 1 , where f I := 1 b- a b Z a f ( x ) dx. Proof. For arbitrary 0 < a < b , we have ln x- (ln x ) I = 1 b- a b Z a (ln x- ln y ) dy, | ln x- (ln x ) I | I = 1 ( b- a ) 2 b Z a Z b a (ln x- ln y ) dy dx 1 ( b- a ) 2 ZZ { a x,y b } | ln x- ln y | dxdy. One can split the square { a x,y b } into two triangles T 1 := { a x y b } and T 2 := { a y x b } . By symmetry, the integrals over T 1 and T 2 coincide, so that | ln x- (ln x ) I | 2 ( b- a ) 2 ZZ T 1 (ln y- ln x ) dxdy = 2 ( b- a ) 2 ZZ T 1 Z y x dt t dxdy = 2 ( b- a ) 2 ZZZ { a x t y 1 } t- 1 dtdxdy = 2 ( b- a ) 2 ZZ { a t y b } t- 1 Z t a dx dtdy 2 ( b- a ) 2 ZZ { a t y b } dtdy = 2 ( b- a ) 2 ZZ T 1 dtdy = 1 . 1 #3. Let f ( x ) be a function in L 1 ([0 , 1]), such that f (0) , f (1) , and f x + y 2 f ( x ) + f ( y ) 2 for all x,y [0 , 1] ....
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sf - Math 8601. December 18, 2010. Final Exam. Problems and...

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