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Uniform Continuity

  1. Feb 12, 2008 #1
    [SOLVED] Uniform Continuity

    1. The problem statement, all variables and given/known data

    Let [tex] A \subset \mathbb{R}^n [/tex] and let [tex] f: A \mapsto \mathbb{R}^m [/tex] be uniformly continuous. Show that there exists a unique continuous function [tex] g: \bar{A} \mapsto \mathbb{R}^m [/tex] such that [tex] g(x)=f(x) \ \forall \ x \in A [/tex].

    2. Relevant equations

    The definition for uniform continuity I am using is as follows:

    Let [tex] f: A \subset \mathbb{R}^n \mapsto \mathbb{R}^m [/tex]. Then [tex] f [/tex] is uniformly continuous on [tex] A [/tex] if [tex] \forall \ \ \varepsilon > 0 \ \ \exists \ \ \delta > 0 \ \ s.t. \ \ \Vert f(x) \ - \ f(y) \Vert < \varepsilon \ \ \forall \ \ x, \ y \ \ \in A \ \ s.t. \ \ \Vert x - y \Vert < \delta [/tex].

    3. The attempt at a solution

    I've got to admit, I think I'm pretty clueless about this one. I was thinking about some function

    [tex] $ g(x)=\left\{\begin{array}{cc}x,&\mbox{ if }
    x\in \bar{A} \backslash A\\f(x), & \mbox{ if } x\in A\end{array}\right $ [/tex]. ​

    But I'm not even sure if I'm allowed to do that or if it meets the criteria. Any help leading to a correct proof would be superb, thanks!
     
  2. jcsd
  3. Feb 13, 2008 #2

    EnumaElish

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    g(x) = x (on the boundary) does not satisfy the criteria. Suppose A = [0,1), x = 1, f(1) = 10 vs. g(1) = 1.
     
  4. Feb 13, 2008 #3

    HallsofIvy

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    If x is in the closure of A, then there exist a sequence of points in A converging to x. What should f(x) be?
     
  5. Feb 13, 2008 #4
    I'm not exactly sure where you got the value for f(1) since I never gave f(x) a definition, but I'll go with you on it.

    Hmm... here's what I'm thinking now:

    If [tex] x \in \bar{A}, \ \ \exists [/tex] some sequence [tex] (x_k) \in A [/tex] s.t. [tex] \lim_{\substack{k \rightarrow \infty}} (x_k) = x [/tex]. Then [tex] f(x) = \lim_{\substack{k \rightarrow \infty}} f(x_k) [/tex] by continuity. So I'm thinking I can let [tex] g(x) = \lim_{\substack{k \rightarrow \infty}} f(x_k), \ [/tex] where [tex] (x_k) \rightarrow x [/tex]. Then that is continuous since:
    [tex]
    \begin{equation*}
    \begin{split}
    \Vert g(x) - g(y) \Vert &= \Vert \left( \lim_{\substack{k \rightarrow \infty}} f(x_k) \right) - \left( \lim_{\substack{k \rightarrow \infty}} f(y_k) \right) \Vert \\
    &= \Vert f(x) - f(y) \Vert \\ &\leq \varepsilon \ \ \forall \ \Vert x - y \Vert \leq \delta
    \end{split}
    \end{equation*}
    [/tex]
    (by uniform continuity of [tex] f [/tex]). Thus [tex] g(x) [/tex] is a continuous function satisfying the criteria.
     
  6. Feb 13, 2008 #5

    EnumaElish

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    I think you should use the triangle inequality.
     
  7. Feb 13, 2008 #6
    Use it where? Are you responding to my response to you or my response to Ivy? (Or both?)
     
  8. Feb 13, 2008 #7

    EnumaElish

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    It's even simpler than that:

    g is continuous at x if for every [itex]\epsilon[/itex] > 0 there is [itex]\delta[/itex] > 0 such that |x-xk| < [itex]\delta[/itex] implies |g(x)-g(xk)| = |g(x)-f(xk)| < [itex]\epsilon[/itex].

    Since xk ---> x, I know that for any [itex]\delta[/itex], I can find N1 such that k > N1 implies ...

    And since by definition f(xk) ---> g(x), I know that for any [itex]\epsilon[/itex], I can find N2 such that k > N2 implies ...

    Let N = max{N1, N2}, then for k > N, ...

    Uniqueness follows from uniqueness of limit.
     
    Last edited: Feb 13, 2008
  9. Feb 13, 2008 #8
    Thanks so much EnumaElish, this problem is solved.
     
  10. Feb 13, 2008 #9

    EnumaElish

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    Do you mind posting your proof so it will be a reference to others?
     
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