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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 α ∈ A , let an open ball B α ⊂ R n be defined. Show that there is a finite or countable subset A ⊂ A , such that [ α ∈ A B α = [ α ∈ A 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 HeineBorel 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 := ∞ S j =1 A j is finite or countable, and Ω = ∞ [ j =1 K j ⊆ ∞ [ j =1 [ α ∈ A j B α = [ α ∈ A 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  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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This note was uploaded on 02/15/2012 for the course MATH 8601 taught by Professor Staff during the Fall '08 term at Minnesota.
 Fall '08
 Staff
 Math

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