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Uniform Continuity 2

  1. Oct 31, 2005 #1
    "Let [tex]f:[a,b]\rightarrow [a,b][/tex] be defined such that [tex]|f(x)-f(y)|\leq a|x-y| [/tex] where 0<a<1. Prove that f is uniformly continuous and (other stuff)."

    Let e>0 and let d=e/a. Whenever [tex]0<|x-y|<d, |f(x)-f(y)|\leq a|x-y|<ad=e[/tex]. f is therefore by definition uniformly continuous.

    Did I do this right? It seems too good to be true. It doesn't seem right because I did not use the fact that 0<a<1.
     
    Last edited by a moderator: Oct 31, 2005
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  3. Oct 31, 2005 #2

    EnumaElish

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  4. Nov 1, 2005 #3

    HallsofIvy

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    Ordinary continuity says "For any x0, for any [tex]\epsilon[/tex]>0, there exist [tex]\delta[/tex]> 0 such that if [tex]|x-x_0|< \delta[/tex], then [tex]|f(x)-f(x_0)|< \epsilon[/tex]".
    Uniform continuity says "For any [tex]\epsilon[/tex]>0, there exist [tex]\delta[/tex]> 0 such that, for any x0, if [tex]|x-a_0|< \delta[/tex], then [tex]|f(x)-f(x_0)|< \epsilon[/tex]".

    The difference is that, for uniform continuity, the same [tex]\delta[/tex] works for all x0.
    If [tex]|f(x)-f(y)|< a|x-y|[/tex], then in particular, for any x0, [tex]|f(x)-f(x_0)|< a|x-x_0|[/tex], then given any [tex]\epsilon[/tex], we can choose [tex]\delta= \frac{\epsilon}{a}[/tex] for all x0.

    That's essentially what you are saying since you seem to be using a slightly different (equivalent) definition of uniform continuity.
     
  5. Nov 1, 2005 #4
    I just find it odd that it gave me a property (0<a<1) which I did not need. Thanks for your help.
     
  6. Nov 1, 2005 #5

    HallsofIvy

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    What was the "(other stuff)"? 0< a< 1 may be need for that. It means that f is a "contraction" function and, among other things, has a unique "fixed point".
     
  7. Nov 1, 2005 #6
    The other thing was to show that there exists an e in [a,b] such that f(e) = e.
     
  8. Nov 1, 2005 #7

    EnumaElish

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    Isn't there a theorem you can use?
     
  9. Nov 1, 2005 #8
    Assuming f(a) is not a and f(b) is not b, I've basically proved that a continuous function defined on [a,b] must intersect the identity function g(x)=x at least once. I haven't been able to prove uniqueness. And yes, there are lots of theorems I can use.
     
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