{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol8 - SOLUTIONS FOR HOMEWORK 8 5.4.2(a Note that for x y...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
SOLUTIONS FOR HOMEWORK 8 5.4.2. (a) Note that, for x, y R \{ 0 } , f ( x ) f ( y ) = 1 x 2 1 y 2 = y 2 x 2 x 2 y 2 = ( y x )( y + x ) x 2 y 2 = ( y x ) parenleftBig 1 x 2 y + 1 xy 2 parenrightBig . If x, y A , then x, y 1, hence 1 / ( x 2 y ) + 1 / ( xy 2 ) 2, and | f ( x ) f ( y ) | ≤ 2 | x y | . In particular, for a given ε > 0, | f ( x ) f ( y ) | < ε when | x y | < ε/ 2. Thus, f is uniformly continuous on A . (b) For n N , let x n = 2 - n (clearly, x n B ). Then x n x n +1 = 2 - ( n +1) . However, f ( x n +1 ) f ( x n ) = 2 2( n +1) 2 2 n = 3 · 4 n . Suppose, for the same of contradiction, that f is uniformly continuous on B . Then there exists δ > 0 s.t. | f ( x ) f ( y ) | < 1 whenever x, y B satisfy | x y | < δ . Pick n N so large that 2 - n < δ . Then we must have | f ( x n ) f ( x n +1 ) | < 1, which is false. 5.4.4. Note that, for x, y R , f ( x ) f ( y ) = 1 1 + x 2 1 1 + y 2 = ( y x )( y + x ) (1 + x 2 )(1 + y 2 ) . Furthermore, (1 + x 2 )(1 + y 2 ) = 1 + x 2 + y 2 + x 2 y 2 1 + x 2 2 + 1 + y 2 2 ≥ | x | + | y | (the last inequality follows from the Arithmetic-Geometric Mean Inequality). There- fore, | f ( x ) f ( y ) | ≤ | y x | · | y + x | | x | + | y | ≤ | y x | . Thus, f is Lipschitz, hence uniformly continuous. 5.4.7. Both f and g are Lipschitz functions. Indeed, for f it is obvious. As far as g is concerned, recall that, for any t , | sin( t ) | ≤ | t | . Furthermore, sin α sin β = 2 sin α β 2 cos α + β 2 , hence vextendsingle vextendsingle sin α sin β vextendsingle vextendsingle = 2 vextendsingle vextendsingle vextendsingle sin α β 2 vextendsingle vextendsingle vextendsingle · vextendsingle vextendsingle vextendsingle cos α + β 2 vextendsingle vextendsingle vextendsingle 2 | α β | 2 = | α β | . Therefore, both f and g are uniformly continuous. However, h = fg is not uniformly continuous. Indeed, if it were, then there would exist δ > 0 s.t. | h ( x ) h ( y ) | < 1 whenever | x y | < δ . Pick N N s.t. π/N < δ . Fix M > N . For 0 k N , let x k = 2 πM π/ 2 + kπ/N . Then x 0 = 2 πM π/ 2, and 1
Background image of page 1

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

View Full Document Right Arrow Icon
2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}