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Homework Help: Uniform Convergence

  1. Apr 3, 2007 #1
    Here is the question from the book:
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    Let [itex]f: \mathbb{R} \to \mathbb{R}[/itex] be a function. For any [itex]a\in \mathbb{R}[/itex], let [itex]f_a :\mathbb{R}\to \mathbb{R}[/itex] be the shifted function [itex]f_a(x):=f(x-a)[/itex].


    (a) Show that [itex]f[/itex] is continuous if and only if, whenever [itex](a_n)_{n=0}^{\infty}[/itex] is a sequence of real numbers which converges to zero, the shifted functions [itex]f_{a_n}[/itex] converge pointwise to [itex]f[/itex].

    (b) Show that [itex]f[/itex] is uniformly continuous if and only if, whenever [itex](a_n)_{n=0}^{\infty}[/itex] is a sequence of real numbers which converges to zero, the shifted functions [itex]f_{a_n}[/itex] converge uniformly to [itex]f[/itex].


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    [itex](\Rightarrow)[/itex] Let [itex]x_0 \in \mathbb{R}[/itex]. Suppose [itex]f[/itex] is continuous. That is, given [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that if [itex]|x-x_0| < \delta[/itex] then [itex]|f(x)-f(x_0)| < \epsilon[/itex].


    Let [itex]x_n = x_0 - a_n[/itex]. So given [itex]\delta > 0[/itex] there exists [itex]N' > 0[/itex] such that [itex]|x_n - x_0|< \delta[/itex] for all [itex]n > N'[/itex].

    Given [itex]\epsilon' > 0[/itex] take [itex]\epsilon = \epsilon'[/itex], and take [itex]N = N'[/itex].

    So by the continuity of [itex]f[/itex] we get that [itex]|x_n - x_0| < \delta[/itex] for all [itex]n>N' = N[/itex].

    Thus, [itex]|f(x_n) - f(x_0)| < \epsilon = \epsilon'[/itex].

    But, [itex]f(x_n) = f(x_0-a_n) = f_{a_n}(x_0)[/itex] for all [itex]n>N[/itex].

    So we have [itex]|f_{a_n}(x_0) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].


    Thus [itex]f_{a_n}[/itex] converges pointwise to [itex]f[/itex].



    [itex](\Leftarrow)[/itex] Suppose given [itex]a_n \to 0[/itex] that [itex]f_{a_n}[/itex] converges pointwise to [itex]f[/itex]. Given [itex]x_0[/itex] take a sequence [itex]x_n[/itex] which converges to [itex]x_0[/itex]. That is, [itex]a_n = x_0 - x_n \to 0[/itex].

    Since [itex]f_n[/itex] converges pointwise to [itex]f[/itex], given [itex]\epsilon > 0[/itex] there is some [itex]N>0[/itex] such that [itex]|f_{a_n}(x_0) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].

    But, [itex]|f_{a_n}(x_0) - f(x_0)| = |f(x_0 - a_n) - f(x_0)| = |f(x_n) - f(x_0)|[/itex].

    Hence, [itex]|f(x_n) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].

    That is, [itex]f(x_n)[/itex] converges to [itex]f(x_0)[/itex]. Thus, [itex]f[/itex] is continuous.


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    I don't think there is anything wrong with any of that (if there is, or you have any comments please say something).

    However, I am not too sure about part (b).

    The first direction is the same as in part (a), but the other direction I am not sure about.

    I have been trying just the basic epsilon-delta definition way (similar to what I did in part (a) with this direction) and not really getting anywhere (which I guess could be expected since we don't have a uniform continuity definition with sequences [at least in our book]).

    Any ideas? Thanks!


    edit... I am converting the $ signs to [itex]'s. Is there an easy (fast and painless) way to do this?
     
    Last edited: Apr 3, 2007
  2. jcsd
  3. Apr 3, 2007 #2

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    Copy into notepad, search and replace, using pre-spaces to distinguish opening from closing ones. If it's not too late.
     
  4. Apr 3, 2007 #3

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    What is your definition of uniform continuity? (Please always post relevant definitions)
     
  5. Apr 3, 2007 #4
    Cool, that worked nicely! Any ideas about the exercise now :smile:

    Sorry, I thought that was a standard definition. Here it is:

    Uniform Continuity: Let [itex]f:X\to Y[/itex] be a map from one metric space [itex](X,d_X)[/itex] to another [itex](Y,d_Y).[/itex] Then [itex]f[/itex] is uniformly continuous if, for every [itex]\epsilon > 0[/itex], there exists a [itex]\delta > 0[/itex] such that [itex]d_Y(f(x),f(x')) < \epsilon[/itex] whenever [itex]x,x'\in X[/itex] are such that [itex]d_X(x,x')<\delta[/itex].

    Here is our book's definition of uniform convergence:

    Let [itex](f^{(n)})_{n=1}^{\infty}[/itex] be a sequence of functions from one metric space [itex](X,d_X)[/itex] to another [itex](Y,d_Y)[/itex], and let [itex]f: X\to Y[/itex] be another function. We say that [itex](f^{(n)})_{n=1}^{\infty}[/itex] converges uniformly to [itex]f[/itex] on [itex]X[/itex] if for every [itex]\epsilon > 0[/itex] there exists [itex]N>0[/itex] such that [itex]d_Y(f^{(n)}(x),f(x))<\epsilon[/itex] for every [itex]n>N[/itex] and [itex]x\in X[/itex].
     
    Last edited: Apr 3, 2007
  6. Apr 3, 2007 #5

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    It is standard, it's just a lot easier to refer to a definition that's here. Remeber that the definition of continuity in terms of sequences was just a result of the more general definition of limits in terms of sequences, and you should still be able to apply that.
     
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