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Unformatted text preview: Modern Analysis, Homework 8 Possible solutions All problems are worth 5 points. 1. [12] Rudin, Ch 5, problem 1. Solution: Since  f ( x ) f ( y )  ≤ ( x y ) 2 for all x,y ∈ R , if x 6 = y , we have ≤ f ( x ) f ( y ) x y ≤  x y  Letting y → x , of course  x y  → 0 and 0 → 0, so by the squeeze theorem, we get that lim x → y f ( x ) f ( y ) x y = 0, ie f ( x ) = 0 for all x ∈ R . By the mean value theorem, this implies that f itself is constant. 2. [18] Rudin, Ch 5, problem 2. Solution: Suppose f ( x ) > 0 in ( a,b ). Notice that f is strictly increasing: given a < x < y < b , by the mean value theorem there exists c ∈ ( x,y ) such that f ( x ) f ( y ) = f ( c )( y x ). But f ( c ) > 0, so we get that f ( x ) > f ( y ). Let x ∈ ( a,b ). Choose δ > 0 such that [ x δ,x + δ ] ⊂ ( a,b ). Since [ x δ,x + δ ] is compact and connected, and since f is continuous, it follows that f ([ x δ,x + δ ]) = [ c,d ] for some f ( a ) < c,d < f ( b ). Notice actually that c < d since f is strictly increasing. Since f is continuous and onetoone and [ x δ,x + δ ] is compact, it follows from Rudin, Theorem 4.17 that g  [ a,b ] 1 is continuous. Now let { y n } be a sequence in ( a,b ) approaching f ( x ) such that y n 6 = f ( x ) for all n ∈ N . Notice that { y n } is eventually in [ c,d ] so we can assume, after possibly taking a subsequence that y n ∈ [ c,d ] \ { f ( x ) } for all n ∈ N . Since g  [ c,d ] is continuous, g ( y n ) converges to some x ∈ [ x δ,x + δ ]. Notice that we must have x = x since f ( x ) = lim n →∞ f ( g ( y n )) (by continuity of f ), which is just lim n →∞ y n = f ( x ). Since f is strictly increasing, it follows that x = x . Write g ( f ( x )) g ( y n ) f ( x ) y n = x g ( y n ) f ( x ) f ( g ( y n )) . Since g ( y n ) → x and f is differentiable at x , this limit exists and equals 1 f ( x ) . Hence g ( f ( x )) = 1 f ( x ) ....
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
 Spring '11
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