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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 )...
View Full Document

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.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online