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Rudin_5

# Rudin_5 - Differentiation Written by Men-Gen Tsai email...

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Di ff erentiation Written by Men-Gen Tsai email: [email protected] 1. Let f be defined for all real x , and suppose that | f ( x ) - f ( y ) | ( x - y ) 2 for all real x and y . Prove that f is constant. Proof: | f ( x ) - f ( y ) | ( x - y ) 2 for all real x and y . Fix y , | f ( x ) - f ( y ) x - y | | x - y | . Let x y , therefore, 0 lim x y f ( x ) - f ( y ) x - y lim x y | x - y | = 0 It implies that ( f ( x ) - f ( y )) / ( x - y ) 0 as x y . Hence f ( y ) = 0, f = const . 2. Suppose f ( x ) > 0 in ( a, b ). Prove that f is strictly increasing in ( a, b ), and let g be its inverse function. Prove that g is di ff erentible, and that g ( f ( x )) = 1 f ( x ) ( a < x < b ) . Proof: For every pair x > y in ( a, b ), f ( x ) - f ( y ) = f ( c )( x - y ) where y < c < x by Mean-Value Theorem. Note that c ( a, b ) and f ( x ) > 0 in ( a, b ), hence f ( c ) > 0. f ( x ) - f ( y ) > 0, f ( x ) > f ( y ) if x > y , f is strictly increasing in ( a, b ). Let Δ g = g ( x 0 + h ) - g ( x 0 ). Note that x 0 = f ( g ( x 0 )), and thus, ( x 0 + h ) - x 0 = f ( g ( x 0 + h )) - f ( g ( x 0 )) , 1

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h = f ( g ( x 0 ) + Δ g ) - f ( g ( x 0 )) = f ( g + Δ g ) - f ( g ) . Thus we apply the fundamental lemma of di ff erentiation, h = [ f ( g ) + η ( Δ g )] Δ g, 1 f ( g ) + η ( Δ g ) = Δ g h Note that f ( g ( x )) > 0 for all x ( a, b ) and η ( Δ g ) 0 as h 0, thus, lim h 0 Δ g/h = lim h 0 1 f ( g ) + η ( Δ g ) = 1 f ( g ( x )) . Thus g ( x ) = 1 f ( g ( x )) , g ( f ( x )) = 1 f ( x ) . 3. Suppose g is a real function on R 1 , with bounded derivative (say | g | M ). Fix > 0, and define f ( x ) = x + g ( x ). Prove that f is one-to-one if is small enough. (A set of admissible values of can be determined which depends only on M .) Proof: For every x < y , and x, y R , we will show that f ( x ) = f ( y ). By using Mean-Value Theorem: g ( x ) - g ( y ) = g ( c )( x - y ) where x < c < y, ( x - y ) + (( x ) - g ( y )) = ( g ( c ) + 1)( x - y ) , that is, f ( x ) - f ( y ) = ( g ( c ) + 1)( x - y ) . ( * ) Since | g ( x ) | M , - M g ( x ) M for all x R . Thus 1 - M g ( c ) + 1 1 + M , where x < c < y . Take c = 1 2 M , and g ( c ) + 1 > 0 where x < c < y for all x, y . Take into equation (*), and f ( x ) - f ( y ) < 0 since x - y < 0, that is, f ( x ) = f ( y ), that is, f is one-to-one (injective). 2
4. If C 0 + C 1 2 + ... + C n - 1 n + C n n + 1 = 0 , where C 0 , ..., C n are real constants, prove that the equation C 0 + C 1 x + ... + C n - 1 x n - 1 + C n x n = 0 has at least one real root between 0 and 1. Proof: Let f ( x ) = C 0 x + ... + C n n +1 x n +1 . f is di ff erentiable in R 1 and f (0) = f (1) = 0. Thus, f (1) - f (0) = f ( c ) where c (0 , 1) by Mean-Value Theorem. Note that f ( x ) = C 0 + C 1 x + ... + C n - 1 x n - 1 + C n x n . Thus, c (0 , 1) is one real root between 0 and 1 of that equation. 5. Suppose f is defined and di ff erentiable for every x > 0, and f ( x ) 0 as x + . Put g ( x ) = f ( x + 1) - f ( x ). Prove that g ( x ) 0 as x + . Proof: f ( x + 1) - f ( x ) = f ( c )( x + 1 - x ) where x < c < x + 1 by Mean-Value Theorem. Thus, g ( x ) = f ( c ) where x < c < x + 1, that is, lim x + g ( x ) = lim x + f ( c ) = lim c + f ( c ) = 0 . 6. Suppose (a) f is continuous for x 0, (b) f ( x ) exists for x > 0, (c) f (0) = 0, 3

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(d) f is monotonically increasing. Put g ( x ) = f ( x ) x ( x > 0) and prove that g is monotonically increasing. Proof: Our goal is to show g ( x ) > 0 for all x > 0 g ( x ) = xf ( x ) - f ( x ) x 2 > 0 f ( x ) > f ( x ) x . Since f ( x ) exists, f ( x ) - f (0) = f ( c )( x - 0) where 0 < c < x by Mean-Value Theorem. f ( c ) = f ( x ) x where 0 < c < x . Since f is monotonically increasing, f ( x ) > f ( c ), that is, f ( x ) > f ( x ) x for all x > 0.
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