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Uniform continuity statement

  1. May 24, 2015 #1
    I would appreciate it if someone could explain the steps in the reasoning of the following statement. This is not a homework assignment or anything.

    Let ##U \subseteq R^{n}## be compact and ##f:U\to R## a continuous function on ##U##. However f is not uniformly continuous.
    Then there exists an ##\epsilon>0## and sequences ##a_n## and ##b_n## such that if ## || a_n - b_n || < \frac{1}{n} ## then ##||f(a_n) - f(b_n) \ge \epsilon##.

    This statement will lead to proof that continuity on a compact interval means uniform continuity. However this is not proven yet so don't use that fact in your reasoning. Purely based on the definitions of uniform continuity.
     
  2. jcsd
  3. May 24, 2015 #2

    micromass

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    So try to prove that any function ##f:U\rightarrow \mathbb{R}^m## (for any domain ##U##) is uniform continuous if and only if for each two sequences ##(a_n)_n## and ##(b_n)_n## in ##U## holds that if ##\| a_n - b_n\|\rightarrow 0## then ##\|f(a_n) - f(b_n)\|\rightarrow 0##.
     
  4. May 24, 2015 #3
    Pick a random ##\epsilon>0##. And let ##a_n## and ##b_n## be two sequences in ##U## so that ##||a_n - b_n||## converges.

    Uniform continuity implies there exists a ##\delta## so that as long ##||a_n - b_n|| < \delta ## it follows that ##||f(a_n)-f(b_n)||< \epsilon##

    Since ##||a_n - b_n||## converges I can always find a certain ##N## so that for ##n \ge N## holds that ##||a_n - b_n|| < \delta ##

    So to summarize, I recieve an ##\epsilon## , which in turn returns a unique ##\delta##, and I can always find an ##N## so that for ##n \ge N## : ##||f(a_n)-f(b_n)||< \epsilon## which proves it one way.
     
  5. May 24, 2015 #4

    micromass

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    That's correct.
     
  6. May 24, 2015 #5
    Suggestions for the other direction of proof? It looks like it's a harder one. By the way I totally get the statement now.
     
  7. May 24, 2015 #6

    micromass

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    Use contradiction. Assume that ##f## is not uniform continuous and use that to construct sequences ##(a_n)_n## and ##(b_n)_n##.
     
  8. May 28, 2015 #7
    Hello, I don't think this is worth posting another thread so I hope it's alright if I just ask here.

    A similar problem to the last discussed matter. I need to show that these two statements are equivalent (note it's not about uniform continuity anymore though!)

    1) ##f:U->V## is continuous on some open domain ##D##

    2) For any sequence where ##x_n## converges to ##x## on ##D## , the sequence ##f(x_n)## converges to ##f(x)## on ##V##.

    I tried for a bit with assuming that ##f## is not continuous like you mentioned for the previous problem but don't get far.
     
  9. May 28, 2015 #8

    mathwonk

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    i think you can do this alone. the easy direction is 1 implies 2, assuming you are using the epsilon delta definition of continuity.
     
  10. May 29, 2015 #9
    Yeah 1 implies 2 was no problem its the other way.
     
  11. May 29, 2015 #10

    mathwonk

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    To pass from arbitrary e>0 to a sequence, one usually uses the fact that 1/n --> 0.
     
  12. May 30, 2015 #11

    WWGD

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    I think you need 1st countability of the space to go from 2 to 1, or you need something like the space being a sequential space. I mean, this is not true of all spaces.
     
  13. May 30, 2015 #12

    micromass

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    We are working in ##\mathbb{R}^n##...
     
  14. May 30, 2015 #13

    WWGD

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    OK, sorry, I did not see that. Feel free to delete my previous post if that helps PF.
     
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