text: {[ promptMessage ]}
  
  Bookmark it
  
  {[ promptMessage ]}
  
  # Rudin_5 - Differentiation Written by Men-Gen Tsai email...
  
  This preview shows pages 1–4. Sign up to view the full content.
  
  This preview has intentionally blurred sections. Sign up to view the full version.
  
  View Full Document
  
  This preview has intentionally blurred sections. Sign up to view the full version.
  
  View Full Document
  This is the end of the preview. Sign up to access the rest of the document.
  
  Unformatted text preview: 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 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, ≤ 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 differentible, 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 ) > 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 + h )- g ( x ). Note that x = f ( g ( x )), and thus, ( x + h )- x = f ( g ( x + h ))- f ( g ( x )) , 1 h = f ( g ( x ) + Δ g )- f ( g ( x )) = f ( g + Δ g )- f ( g ) . Thus we apply the fundamental lemma of differentiation, 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 → Δ g/h = lim h → 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 ) 6 = 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 > where x < c < y for all x, y . Take into equation (*), and f ( x )- f ( y ) < since x- y < 0, that is, f ( x ) 6 = f ( y ), that is, f is one-to-one (injective). 2 4. If C + C 1 2 + ... + C n- 1 n + C n n + 1 = 0 , where C , ..., C n are real constants, prove that the equation C + 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 x + ... + C n n +1 x n +1 . f is differentiable 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 + C 1 x + ... + C n- 1 x n- 1 + C n x n ....
  View Full Document
  
  {[ snackBarMessage ]}
  
  ### Page1 / 27
  
  Rudin_5 - Differentiation Written by Men-Gen Tsai email...
  
  This preview shows document pages 1 - 4. Sign up to view the full document.
  
  View Full Document
  Ask a homework question - tutors are online