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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 M > 0 and > such that | f ( x ) f ( y ) | M | x y | 1+ , x, y R . Prove that f is constant. Proof: If we divide by | x y | , we get vextendsingle vextendsingle vextendsingle vextendsingle f ( x ) f ( y ) x y vextendsingle vextendsingle vextendsingle vextendsingle M | x y | , x negationslash = y R . (1) Taking limits are x y , we get | f ( y ) | y R , which implies that f = 0, and therefore f is constant. 2. Suppose that 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 differentiable, and that g ( f ( x )) = 1 f ( x ) x ( a, b ) . Proof: Take x, y ( a, b ), with x > y . Then c ( x, y ) such that f ( x ) f ( y ) = f ( c )( x y ). Since f ( c ) > 0 and x y > 0, clearly f ( x ) 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 ] [ f ( a ) , f ( b )] is continuous, and both [ a, b ] and [ f ( a ) , f ( b )] are compact). Since f ( x ) > 0, if negationslash = x then f ( ) negationslash = f ( x ) and we know that lim x x f ( ) f ( x ) = 1 f ( x ) . Therefore, given > 0, > 0 such that vextendsingle vextendsingle vextendsingle vextendsingle x f ( ) f ( x ) 1 f ( x ) vextendsingle vextendsingle vextendsingle vextendsingle < 1 2 if | f ( x ) | < . Given such > 0, > 0 such that | g ( t ) x | < if | t f ( x ) | < . Note also that g ( f (...
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This note was uploaded on 12/27/2011 for the course MATH 118b taught by Professor Garcia during the Fall '09 term at UCSB.
- Fall '09