1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Disprove uniform continuety

  1. Sep 11, 2011 #1
    i need to prove that [TEX]\frac{1}{\sqrt{x}}[/TEX] is not uniformly continues in (0,1)







    for epsilon=0.5



    [TEX]|\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}|=|[/TEX][TEX]]\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}\frac{\sqrt{y}+\sqrt{x}}{\sqrt {y}+\sqrt{x}}|[/TEX][TEX]=|\frac{y-x}{(\sqrt{y}-\sqrt{x})\sqrt{xy}}|[/TEX]







    i need to prove that the above exprseesion bigger then 0.5







    but i dont know what x and y to choose



    ?
     
  2. jcsd
  3. Sep 11, 2011 #2
    So, for each delta, you need to find an x and y such that |x-y| is smaller then delta but f(x)-f(y) is less than 1/2, as you have stated. But, realize two things:

    1) So long as |x-y| is smaller than delta, you can let |x-y| be anything you want (think about why this is true.) For example, if delta is bigger than zero, you can choose x and y such that (I'm going to write "d" for delta here) d/4 < |x-y| < d. In other words, you aren't proving that it is uniformly continuous, you are proving that it isn't (so think of the negation of the definition of uniform continuity.)

    2) multiply [TEX]\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{x}}[/TEX] by [TEX]\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}[/TEX] and then use the fact that |x-y| is bigger than something (see the above.)

    Does this make sense?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Disprove uniform continuety
Loading...