# solHMK11 - Section 5.4 #2) | 1 /x 2- 1 /y 2 | = | y- x || x...

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Unformatted text preview: Section 5.4 #2) | 1 /x 2- 1 /y 2 | = | y- x || x + y | / ( x 2 y 2 ) | y- x | 1 | x | y 2 + 1 x 2 | y | . If x,y [1 , then both 1 xy 2 and 1 x 2 y are less than or equal to 1. Hence, | 1 /x 2- 1 /y 2 | 2 | x- y | . So f is Lipschitz on [1 , ), and hence uniformly continuous. Suppose f were uniformly continuous on (0 , ). Then in particular, it would be uniformly continuous on (0 , 1). By the Continuous Extension Property, it could be extended to a continuous function on [0 , 1]. But continuous functions on closed bounded intervals are bounded, which f is not. Contradiction. #4) | f ( x )- f ( y ) | = | x- y | | x + y | (1+ x 2 )(1+ y 2 ) | x- y | | y | (1+ x 2 )(1+ y 2 ) + | x | (1+ x 2 )(1+ y 2 ) . If | x | 1, then | x | / ((1 + x 2 )(1 + y 2 )) 1. If | x | > 1, then | x | / ((1 + x 2 )(1 + y 2 )) < | x | / ( x 2 (1 + y 2 )) < 1 / | x | < 1. Hence, | y | (1+ x 2 )(1+ y 2 ) + | x | (1+ x 2 )(1+ y 2 ) 1 + 1 = 2. So | f ( x )- f ( y ) | 2 | x- y | . So f is Lipschitz on R , and hence uniformly continuous. #7) It is obvious that f ( x ) = x is uniformly continuous. For all x,y R , | sin( x )- sin( y ) | = 2 | sin( x- y 2 ) || cos( x + y 2 )...
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## This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.

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solHMK11 - Section 5.4 #2) | 1 /x 2- 1 /y 2 | = | y- x || x...

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