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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 HeineBorel 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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 Fall '08
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 Math

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