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Real analysis Help

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data

    Let f : (a, b) → R be a continuous function on (a, b) such that |f'(x)| <= 1 for all x that are elements of (a,b). Prove that f is uniformly continuous function on (a,b).


    2. Relevant equations



    3. The attempt at a solution

    Proof:For the sequence {xn}, where the limit(n→∞) xn = xo, then limit(n→∞) |((f(xn)-f(xo))/(xn-xo))| <= 1. This can be rewritten as limit(n→∞) |(f(xn)-f(xo)) <= limit(n→∞) |xn-xo|. For uniformly continuous, for all ε>0 and x,y that are elements of (a,b), there exists a δ such that |x-y|<δ implies |f(x)-f(y)|<ε. Let x=Xn and y=Xo. Then if ε=δ, then limit(n→∞) |(f(xn)-f(xo)) <= limit(n→∞) |xn-xo|<δ=ε and therefore f is uniformly continuous on (a,b)

    Any help on where to improve this or if there are any steps that are incorrect would be greatly appreciated.
     
  2. jcsd
  3. Nov 29, 2013 #2

    LCKurtz

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    My suggestion would be to abandon that approach and think about using the mean value theorem.
     
  4. Nov 29, 2013 #3

    pasmith

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    So [itex]f[/itex] is actually a differentiable function (which is sufficient, but not necessary, for [itex]f[/itex] to be continuous).

    This just says that [itex]0 \leq 0[/itex], which is true but not useful.

    You can say that for all [itex]\epsilon > 0[/itex], there exists [itex]N \in \mathbb{N}[/itex] such that if [itex]n \geq N[/itex] then
    [tex]
    -1 - \epsilon \leq f'(x_0) - \epsilon < \frac{f(x_n) - f(x_0)}{x_n - x_0} < f'(x_0) + \epsilon \leq 1 + \epsilon
    [/tex]
    so that
    [tex]
    |f(x_n) - f(x_0)| < (1 + \epsilon)|x_n - x_0|
    [/tex]
    which doesn't get you there; to take [itex]\epsilon \to 0[/itex] you must also take [itex]N \to \infty[/itex] which again gives you [itex]0 \leq 0[/itex].

    It is best to use the mean value theorem: Let [itex]x \in (a,b)[/itex] and [itex]y \in (a,b)[/itex] be arbitrary such that [itex]|x - y| < \epsilon[/itex].
    If [itex]x < y[/itex] then [itex]f[/itex] is continuous on [itex][x,y][/itex] and differentiable on [itex](x,y)[/itex]. Thus by the MVT there exists [itex]c \in (x,y)[/itex] such that
    [tex]
    f'(c) = \frac{f(y) - f(x)}{y - x}.
    [/tex]
     
    Last edited: Nov 29, 2013
  5. Nov 29, 2013 #4
    Ok so I see that I can get f(y)-f(x)=f'(c)(y-x) and then I know from the assumptions that |f'(c)|≤1. From a previous exercise I know that any contraction is uniformly continuous on the domain. (The exercise says: f is said to be a contraction if there exists a c such that 0<c<1 such that for all x,y that are elements of the domain, we have |f(x)-f(y)|≤c|x-y|). Basically, if I can get from f(y)-f(x)=f'(c)(y-x) to |f(x)-f(y)|≤c|x-y| I will be able to say that it is uniformly continuous. Is it enough to use triangle inequality to say these are equivalent?
     
  6. Nov 29, 2013 #5

    LCKurtz

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    You don't need that contraction mapping theorem. You have$$
    |f(y) - f(x)|= |f'(c)||y-x| \le 1\cdot |y-x|$$So given ##\epsilon > 0## pick ##\delta##...
     
  7. Nov 29, 2013 #6
    So if we pick δ=ε, then |f(y)−f(x)|=|f′(c)||y−x|≤1⋅|y−x|≤δ=ε and therefore |x-y|≤δ implies |f(y)−f(x)|≤ε, making it uniformly continuous
     
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