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hw1_sol - Homework 1 – Math 118B Winter 2010 Due on...

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Unformatted text preview: Homework 1 – Math 118B, Winter 2010 Due on Thursday, January 14, 2010 1. Let f be defined for all real x , and suppose that βˆƒ M > 0 and βˆƒ Η« > such that | f ( x ) βˆ’ f ( y ) | ≀ M | x βˆ’ y | 1+ Η« , βˆ€ x, y ∈ R . Prove that f is constant. Proof: If we divide by | x βˆ’ y | , we get vextendsingle vextendsingle vextendsingle vextendsingle f ( x ) βˆ’ f ( y ) x βˆ’ y vextendsingle vextendsingle vextendsingle vextendsingle ≀ M | x βˆ’ y | Η« , βˆ€ x negationslash = y ∈ R . (1) Taking limits are x β†’ y , we get | f β€² ( y ) | ≀ βˆ€ y ∈ R , which implies that f β€² = 0, and therefore f is constant. 2. Suppose that f β€² ( 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 differentiable, and that g β€² ( f ( x )) = 1 f β€² ( x ) βˆ€ x ∈ ( a, b ) . Proof: Take x, y ∈ ( a, b ), with x > y . Then βˆƒ c ∈ ( x, y ) such that f ( x ) βˆ’ f ( y ) = f β€² ( c )( x βˆ’ y ). Since f β€² ( c ) > 0 and x βˆ’ y > 0, clearly f ( x ) βˆ’ f ( y ) > 0, hence f ( x ) > f ( y ). Since x and y were arbitrary it follows that f is strictly increasing. As a consequence, f has an inverse, g , which is also strictly increasing, and continuous (note that f : [ a, b ] β†’ [ f ( a ) , f ( b )] is continuous, and both [ a, b ] and [ f ( a ) , f ( b )] are compact). Since f β€² ( x ) > 0, if ΞΎ negationslash = x then f ( ΞΎ ) negationslash = f ( x ) and we know that lim ΞΎ β†’ x ΞΎ βˆ’ x f ( ΞΎ ) βˆ’ f ( x ) = 1 f β€² ( x ) . Therefore, given Η« > 0, βˆƒ Ξ΄ > 0 such that vextendsingle vextendsingle vextendsingle vextendsingle ΞΎ βˆ’ x f ( ΞΎ ) βˆ’ f ( x ) βˆ’ 1 f β€² ( x ) vextendsingle vextendsingle vextendsingle vextendsingle < Η« 1 2 if | ΞΎ βˆ’ f ( x ) | < Ξ΄ . Given such Ξ΄ > 0, βˆƒ Ξ· > 0 such that | g ( t ) βˆ’ x | < Ξ΄ if | t βˆ’ f ( x ) | < Ξ· . Note also that g ( f (...
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hw1_sol - Homework 1 – Math 118B Winter 2010 Due on...

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