Rudin_5 - Differentiation 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

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Diferentiation Written by Men-Gen Tsai email: [email protected] 1. Let f be defned 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 . ±ix 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 ( ), and let g be its inverse Function. Prove that g is di²erentible, and that g ± ( f ( x )) = 1 f ± ( x ) ( a < x < b ) . Proof: ±or every pair x>y in ( ), f ( x ) - f ( y )= f ± ( c )( x - y ) where y < c < x by Mean-Value Theorem. Note that c ( ) and f ± ( x ) > 0 in ( ), hence f ± ( c ) > 0. f ( x ) - f ( y ) > 0, f ( x ) >f ( y ) iF x > y , f is strictly increasing in ( ). 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 diFerentiation, h =[ f ± ( g )+ η g )]Δ g, 1 f ± ( g η g ) = Δ g h Note that f ± ( g ( x )) > 0 for all x ( a,b ) and η g ) 0 as h 0, thus, lim h 0 Δ g/h = lim h 0 1 f ± ( g η 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 ). ±ix ± > 0, and de²ne 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: ±or 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 > 0 where x < c < y for all . 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
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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 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 diFerentiable 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 de±ned and diFerentiable 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 - x ) where x < c < x 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 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 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 . Since f ± is monotonically increasing, f ± ( x ) >f ± ( c ), that is, f ± ( x ) > f ( x ) x for all 0.
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This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.

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Rudin_5 - Differentiation 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

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