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Uniform Continuity

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Let f: [0,1] -> R (R-real numbers) be a continuous non constant
    function such that f(0)=f(1)=0. Let g_n be the function: x-> f(x^n)
    for each x in [0,1]. I'm trying to show that g_n converges pointwise to
    the zero function but NOT uniformly to the zero function.


    3. The attempt at a solution

    I thought: sup |g_n(x) | = sup |f(x^n)| now if I can only show that this doesn't
    tends to zero then we are done but I can't. I also have no idea how to show
    it converges pointwise to zero. Can you please help?
     
  2. jcsd
  3. Nov 12, 2008 #2

    Dick

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    Pick a definite point, say x=1/2. What's lim x^n? What's lim f(x^n)?
     
  4. Nov 12, 2008 #3
    Ok here's what I think it's a "proof".

    To show pointwise convergence:

    If x<1 then x^n converges pointwise to 0 , then because f is continuous f(x^n) -> f(0) =0.
    But f(x^n) = g_n(x) so g_n(x) ->0 if n->infinity and x<1.
    Now if x=1 then g_n(1) = f(1^n ) = f(1) =0 so g_n(1) =0.
    Conclusion: g converges pointwise everywhere to 0.

    Is this proof correct?

    Now to show it is NOT uniformly convergent.
    Let x=1/2 then x^n = (1/2)^n = 1/2^n which goes to 0.
    Now f(x^n) = f(1/2^n) but then I don't know how to compute this limit.
     
  5. Nov 12, 2008 #4

    Dick

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    That's the idea for pointwise convergence alright. To show it's not uniform go back to sup |f(x^n)| over x in [0,1]. How does that compare with sup |f(x)| over x in [0,1]?
     
  6. Nov 12, 2008 #5

    Thank you for helping, really appreciated.
    I think you suggest to show that sup |f(x^n) | >= |sup(f(x)| but I don't know how
    to prove this inequality. Any hints you can please provide?
     
  7. Nov 12, 2008 #6

    Office_Shredder

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    For every n, find an x such that xn=1/2. Since f is non-constant, you know somewhere f is non-zero, so do the same for that point
     
  8. Nov 12, 2008 #7

    Dick

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    I, in fact, think to show sup |f(x^n)|=sup |f(x)|. x goes from 0 to 1, x^n also goes from 0 to 1 for each n. f is going to range over the same values in both cases, isn't it?
     
  9. Nov 12, 2008 #8
    Thanks guys, is this correct?
    We want to show that sup | f(x^n) | = sup |f(x)|.
    Let M = sup |f(x)| which exists because f is continuous and [0,1] is compact.

    Now observe that no matter for each x in [0,1] x^n is also in [0,1].
    For simplicity let y_n = x^n. Then each y_n is an element of [0,1].
    Then by assumption M = sup | f(x) | x in [0,1], so simply "replace" x by y_n.
    Then M = sup | f(x^n) | x in [0,1] and we are done (I think)

    Is this correct?


    One more thing:
    To show x^n converges pointwise to zero we shall prove that for all e>0 there
    exists natural n>=N such that x^n< e.
    n< log(e)/log(x).

    So what natural N can we take?
     
  10. Nov 12, 2008 #9

    Dick

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    That's coming out a little confusing. I think you just say that range(f)=f([0,1]) for both f(x) and f(x^n). Since as x goes from 0 to 1, x^n also goes from 0 to 1. But for purposes of the proof, all you really need is that sup |f(x^n)| is greater than some positive constant. You could follow Office-Shredder's suggestion and just concentrate on a single nonzero point, if that makes it easier to write down. You might be getting confused in your epsilon proof because log(x) and log(e) are negative. Multiply both sides of n*log(x)<log(e) by -1 to get n>|log(e)|/|log(x)|.
     
  11. Nov 12, 2008 #10
    Thanks very much! Got it now.
     
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