sf - Math 8601 Final Exam Problems and Solutions#1 Let A be...

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Math 8601. December 18, 2010. Final Exam. Problems and Solutions. #1. Let A be an arbitrary set, and for each α A , let an open ball B α R n be defined. Show that there is a finite or countable subset A 0 A , such that [ α A B α = [ α A 0 B α . Proof. The open set Ω := [ α A B α = [ j =1 K j , where K j := { x Ω : dist ( x, ∂ Ω) 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 α , α 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 α , α A j . Then the set A 0 := S j =1 A j is finite or countable, and Ω = [ j =1 K j [ j =1 [ α A j B α = [ α A 0 B α Ω , 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 | dx dy. 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 ) dx dy = 2 ( b - a ) 2 ZZ T 1 Z y x dt t dx dy = 2 ( b - a ) 2 ZZZ { a x t y 1 } t - 1 dt dx dy = 2 ( b - a ) 2 ZZ { a t y b } t - 1 Z t a dx dt dy 2 ( b - a ) 2 ZZ { a t y b } dt dy = 2 ( b - a ) 2 ZZ T 1 dt dy = 1 . 1
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#3. Let f ( x ) be a function in L 1 ([0 , 1]), such that f (0) 0 , f (1) 0 , and f x + y 2 f ( x ) + f ( y ) 2 for all x, y [0 , 1] . Show that f ( x ) 0 for all x [0 , 1].
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