hw8-sol

# hw8-sol - Modern Analysis Homework 8 Possible solutions All...

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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 0 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 0 ( 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 0 ( 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 0 ( c )( y - x ). But f 0 ( 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 one-to-one 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 0 [ x - δ, x + δ ]. Notice that we must have x 0 = x since f ( x 0 ) = 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 0 = 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 0 ( x ) . Hence g 0 ( f ( x )) = 1 f 0 ( x ) . 3. [11] Rudin, Ch 5, problem 3. Solution: Suppose g is a real function on R and that there exists M > 0 such that for all x R , we have | g ( x ) | ≤ M . Choose ε > 1 /M . Then f 0 ( x ) = 1 + εg 0 ( x ) > 1 + g 0 ( x ) /M > 0 for all x R . It follows that f is strictly increasing and hence one-to-one.

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