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Uniform continuity proof

  1. Dec 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##f:X \to Y##. Show that
    ##f## not uniform continuous on ##X## ##\Longleftrightarrow## ##\exists \epsilon > 0## and sequences ##(p_n), (q_n)## in ##X## so that ##d_X(p_n,q_n)\to 0 ## while ##d_Y(f(p_n),f(q_n))\ge \epsilon##.

    2. Relevant equations
    Let ##f:X\to Y##. We say ##f## is uniform continuous on ##X## if ##\forall \epsilon >0 \exists \delta > 0## so that
    ##d_Y(f(x),f(y))< \epsilon## ##\forall x,y\in X## for which ##d_X(x,y)< \delta##.

    3. The attempt at a solution
    I was hoping someone could take a look at my proof and check if it's correct or not. I found the second part especially hard to formulate so I'm mostly unsure about that part.

    Starting with ##\Longleftarrow##
    ##d(p_n,q_n) \to 0## means by definition that
    ##\forall \delta > 0## ##\exists N## so that ##d(p_n,q_n) < \delta## for ##n \ge N##. But then ##f## can't be uniform continuous since the points ##p_n, q_n \in X## and ##d_Y(f(p_n),f(q_n))\ge \epsilon_0## we have a counter example to
    ##d(x,y)<\delta \Longrightarrow d(f(x),f(y)< \epsilon## if we take a ##\epsilon < \epsilon_0##.

    ##\Longrightarrow##
    That ##f## is not uniform continuous implies that ##\exists \epsilon >0## so that ##\forall \delta > 0## there exists ##x,y\in X## so that
    ##d_X(x,y)< \delta \Longrightarrow d_Y(f(x),f(y))\ge \epsilon##. Let's select some ##x,y## satisfying this and set ##p_n = x## and ##q_n =y## and we're done.
     
  2. jcsd
  3. Dec 29, 2016 #2

    stevendaryl

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    Staff Emeritus
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    You're not finished. You haven't defined the sequences [itex]p_n[/itex] and [itex]q_n[/itex].

    Maybe it helps to look at a concrete case: [itex]f(x) = x^2[/itex]. This is not uniformly continuous. Pick [itex]\epsilon = 1[/itex]. Then for any [itex]\delta > 0[/itex], we can find an [itex]x[/itex] and [itex]y[/itex] such that [itex]|x-y| < \delta[/itex], but [itex]|f(x) - f(y)| > \epsilon[/itex]. You just let [itex]x = \frac{1}{\delta}[/itex], and let [itex]y = x+\frac{\delta}{2}[/itex]. Then even though [itex]|x-y| < \delta[/itex], [itex]|x^2 - y^2| = 1 + \frac{\delta^2}{4} > \epsilon[/itex].

    So it's not uniformly continuous. But how does your argument show that there is a sequence [itex]p_n[/itex] and a sequence [itex]q_n[/itex] such that [itex]|p_n - q_n| \rightarrow 0[/itex] but [itex]|(p_n)^2 - (q_n)^2| > \epsilon[/itex]?

    (It's not hard to come up with such a sequence, but it doesn't seem to follow immediately from your proof.)
     
  4. Dec 29, 2016 #3
    Good catch there! I tried to fix the problem:
    So we have ##d_X(x,y)<\delta \Longrightarrow d_Y(f(x),f(y))\ge \epsilon##.
    Set ##N= floor(1+1/\delta)## Then ##d_X(x_n,y_n)\le 1/n < \delta## for ##n\ge N## implies ##d_Y(f(x),f(y))\ge \epsilon##.
    So for each ##n## select ##x_n,y_n## satisfying the above which will then form two sequences ##x_n## and ##p_n## for which ##d_X(x_n,y_n)< 1/n \to 0##.

    For your example we could set ##p_n = n+1/n## and ##q_n = n## then ##|p_n-q_n| = 1/n \to 0## while ##|(n+1/n)^2-n^2| = 2+1/n^2 \to 2##.
     
  5. Dec 30, 2016 #4
    In the proof above we can shift the indexes to start counting from ##1,2,\dots## by setting ##p_1 = x_N, p_2 = x_{N+1}, \dots ##
     
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